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Why is the volume of a sphere $\frac{4}{3}\pi r^3$? I learned that the volume of a sphere is $\frac{4}{3}\pi r^3$, but why? The $\pi$ kind of makes sense because its round like a circle, and the $r^3$ because it's 3-D, but $\frac{4}{3}$ is so random! How could somebody guess something like this for the formula?

I am no where near as proficient in math as any of the people who answered this before me, but nonetheless I would like to add a simplified version; A cylinder's volume is: $$\pi r^2h$$ A cone's volume is $\frac{1}{3}$ that of a cylinder of equal height and radius: $$\frac{1}{3}\pi r^2h$$ A sphere's volume is two cones, each of equal height and radius to that of the sphere's: $$\frac{1}{3}\pi r^2h + \frac{1}{3}\pi r^2h$$ The height of the sphere is equal to it's diameter $(r + r)$ so the earlier equation can be rewritten as; $$\frac{1}{3}\pi r^2(r + r) + \frac{1}{3}\pi r^2(r + r)$$ If we simplify it; $$\frac{1}{3}\pi r^2(2r) + \frac{1}{3}\pi r^2(2r)$$ Following the math convention of numbers before letters it changes to: $$\frac{1}{3}2\pi r^2r + \frac{1}{3}2\pi r^2r$$ Combining like terms; $$r^2\cdot r= r^3$$ and $$\frac{1}{3}\cdot 2 = \frac{2}{3}$$ The equation now becomes $$\frac{2}{3}\pi r^3 + \frac{2}{3}\pi r^3$$ Again add the like terms, being the $\frac{2}{3}$ together; $$\frac{2}{3} + \frac{2}{3} = \frac{4}{3}$$ Finally we get to how $\frac{4}{3}$ is part of the equation; $$\frac{4}{3}\pi r^3$$

Proof that the sum of two Gaussian variables is another Gaussian The sum of two Gaussian variables is another Gaussian. It seems natural, but I could not find a proof using Google. What's a short way to prove this? Thanks! Edit: Provided the two variables are independent.

I posted the following in response to a question that got closed as a duplicate of this one: It looks from your comment as if the meaning of your question is different from what I thought at first. My first answer assumed you knew that the sum of independent normals is itself normal. You have $$ \exp\left(-\frac12 \left(\frac{x}{\alpha}\right)^2 \right) \exp\left(-\frac12 \left(\frac{z-x}{\beta}\right)^2 \right) = \exp\left(-\frac12 \left( \frac{\beta^2x^2 + \alpha^2(z-x)^2}{\alpha^2\beta^2} \right) \right). $$ Then the numerator is $$ \begin{align} & (\alpha^2+\beta^2)x^2 - 2\alpha^2 xz + \alpha^2 z^2 \\ \\ = {} & (\alpha^2+\beta^2)\left(x^2 - 2\frac{\alpha^2}{\alpha^2+\beta^2} xz\right) + \alpha^2 z^2 \\ \\ = {} & (\alpha^2+\beta^2)\left(x^2 - 2\frac{\alpha^2}{\alpha^2+\beta^2} xz + \frac{\alpha^4}{(\alpha^2+\beta^2)^2}z^2\right) + \alpha^2 z^2 - \frac{\alpha^4}{\alpha^2+\beta^2}z^2 \\ \\ = {} & (\alpha^2+\beta^2)\left(x - \frac{\alpha^2}{\alpha^2+\beta^2}z\right)^2 + \alpha^2 z^2 - \frac{\alpha^4}{\alpha^2+\beta^2}z^2, \end{align} $$ and then remember that you still have the $-1/2$ and the $\alpha^2\beta^2$ in the denominator, all inside the "exp" function. (What was done above is completing the square.) The factor of $\exp\left(\text{a function of }z\right)$ does not depend on $x$ and so is a "constant" that can be pulled out of the integral. The remaining integral does not depend on "$z$" for a reason we will see below, and thus becomes part of the normalizing constant. If $f$ is any probability density function, then $$ \int_{-\infty}^\infty f(x - \text{something}) \; dx $$ does not depend on "something", because one may write $u=x-\text{something}$ and then $du=dx$, and the bounds of integration are still $-\infty$ and $+\infty$, so the integral is equal to $1$. Now look at $$ \alpha^2z^2 - \frac{\alpha^4}{\alpha^2+\beta^2} z^2 = \frac{z^2}{\frac{1}{\beta^2} + \frac{1}{\alpha^2}}. $$ This was to be divided by $\alpha^2\beta^2$, yielding $$ \frac{z^2}{\alpha^2+\beta^2}=\left(\frac{z}{\sqrt{\alpha^2+\beta^2}}\right)^2. $$ So the density is $$ (\text{constant})\cdot \exp\left( -\frac12 \left(\frac{z}{\sqrt{\alpha^2+\beta^2}}\right)^2 \right) . $$ Where the standard deviation belongs we now have $\sqrt{\alpha^2+\beta^2}$.

Why does the series $\sum_{n=1}^\infty\frac1n$ not converge? Can someone give a simple explanation as to why the harmonic series $$\sum_{n=1}^\infty\frac1n=\frac 1 1 + \frac 12 + \frac 13 + \cdots $$ doesn't converge, on the other hand it grows very slowly? I'd prefer an easily comprehensible explanation rather than a rigorous proof regularly found in undergraduate textbooks.

Let's group the terms as follows:$$A=\frac11+\frac12+\frac13+\frac14+\cdots\\ $$ $$ A=\underbrace{(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{9})}_{\color{red} {9- terms}} +\underbrace{(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+\cdots+\frac{1}{99})}_{\color{red} {90- terms}}\\+\underbrace{(\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+\cdots+\frac{1}{999})}_{\color{red} {900- terms}}+\cdots \\ \to $$ $$\\A>9 \times(\frac{1}{10})+(99-10+1)\times \frac{1}{100}+(999-100+1)\times \frac{1}{1000}+... \\A>\frac{9}{10}+\frac{90}{100}+\frac{90}{100}+\frac{900}{1000}+...\\ \to A>\underbrace{\frac{9}{10}+\frac{9}{10}+\frac{9}{10}+\frac{9}{10}+\frac{9}{10}+\frac{9}{10}+...}_{\color{red} {\text{ m group} ,\text{ and} \space m\to \infty}} \to \infty $$ Showing that $A$ diverges by grouping numbers.

Simple numerical methods for calculating the digits of $\pi$ Are there any simple methods for calculating the digits of $\pi$? Computers are able to calculate billions of digits, so there must be an algorithm for computing them. Is there a simple algorithm that can be computed by hand in order to compute the first few digits?

The first method that I applied successfully with function calculator was approximation of circle by $2^k$-polygon with approximating sides with one point on the circle and corners outside the circle. I started with unit circle that was approximated by square and the equation $\tan(2^{-k} \pi/4) \approx 2^{-k} \pi/4$, that gives $\pi \approx \frac{8}{2} = 4$ for $k=0$. I iterated the formula of tangent of half angle, that I solved applying the formula of the solution of second order equation, that was applied to the sum formula of tangent. I obtained the sequence $\pi \approx 8 \cdot 2^k \tan(2^{-k} \pi /4)/2$. The problem is that the solution formula of the second order equation has square root, that is difficult to calculate by hand. That's why I kept on searching a simple approximation method that applies addition, substraction, multiplication and division of integers. I ended up to the following calculation. This method applies Machin-like formula and was first published by C. Hutton. \begin{eqnarray} \pi & = & 4 \frac{\pi}{4} = 4 \arctan(1) = 4 \arctan\Bigg(\frac{\frac{5}{6}}{\frac{5}{6}}\Bigg) = 4 \arctan\Bigg(\frac{\frac{1}{2}+\frac{1}{3}}{1-\frac{1}{2}\frac{1}{3}}\Bigg) \\ & = & 4 \arctan\Bigg(\frac{\tan(\arctan(\frac{1}{2}))+\tan(\arctan(\frac{1}{3}))}{1-\tan(\arctan(\frac{1}{2}))\tan(\arctan(\frac{1}{3}))}\Bigg) \\ & = & 4 \arctan\Big(\tan\Big(\arctan\Big(\frac{1}{2}\Big)+\arctan\Big(\frac{1}{3}\Big)\Big)\Big) \\ & = & 4 \Big(\arctan\Big(\frac{1}{2}\Big)+\arctan\Big(\frac{1}{3}\Big)\Big) \\ & = & 4 \Big(\Big\vert_0^\frac{1}{2} \arctan(x) + \Big\vert_0^\frac{1}{3} \arctan(x)\Big) \\ & = & 4 \bigg(\int_0^\frac{1}{2} \frac{1}{1+x^2} dx + \int_0^\frac{1}{3} \frac{1}{1+x^2} dx\bigg) \\ & = & 4 \bigg(\int_0^\frac{1}{2} \sum_{k=0}^\infty (-x^2)^k dx + \int_0^\frac{1}{3} \sum_{k=0}^\infty (-x^2)^k dx \bigg) \\ & = & 4 \bigg(\sum_{k=0}^\infty \int_0^\frac{1}{2} (-x^2)^k dx + \sum_{k=0}^\infty \int_0^\frac{1}{3} (-x^2)^k dx \bigg) \\ & = & 4 \bigg(\sum_{k=0}^\infty \int_0^\frac{1}{2} (-1)^k x^{2k} dx + \sum_{k=0}^\infty \int_0^\frac{1}{3} (-1)^k x^{2k} dx \bigg) \\ & = & 4 \bigg(\sum_{k=0}^\infty \bigg\vert_0^\frac{1}{2} \frac{(-1)^k}{2k+1} x^{2k+1} + \sum_{k=0}^\infty \bigg\vert_0^\frac{1}{3} \frac{(-1)^k}{2k+1} x^{2k+1} \bigg) \\ & = & 4 \bigg(\sum_{k=0}^\infty \frac{(-1)^k}{2k+1} \frac{1}{2^{2k+1}} + \sum_{k=0}^\infty \frac{(-1)^k}{2k+1} \frac{1}{3^{2k+1}} \bigg) \\ & = & 4 \sum_{k=0}^\infty \frac{(-1)^k}{2k+1} \bigg(\frac{1}{2^{2k+1}} + \frac{1}{3^{2k+1}}\bigg) \\ & = & \sum_{k=0}^\infty \frac{4(-1)^k}{2k+1} \bigg(\frac{1}{2^{2k+1}} + \frac{1}{3^{2k+1}}\bigg). \end{eqnarray} It is the most beautiful in practice numerically applicable method I have found so far.

Sum of the alternating harmonic series $\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{k} = \frac{1}{1} - \frac{1}{2} + \cdots $ I know that the harmonic series $$\sum_{k=1}^{\infty}\frac{1}{k} = \frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \cdots + \frac{1}{n} + \cdots \tag{I}$$ diverges, but what about the alternating harmonic series $$\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{k} = \frac{1}{1} - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \frac{1}{6} + \cdots + \frac{(-1)^{n+1}}{n} + \cdots \text{?} \tag{II}$$ Does it converge? If so, what is its sum?

it is not absolutely convergent (that is, if you are allowed to reorder terms you may end up with whatever number you fancy). If you consider the associated series formed by summing the terms from 1 to n of the original one, that is you fix the order of summation of the original series, that series (which is not the original one...) converges to $\ln(2)$ See Wikipedia.

Proof that $n^3+2n$ is divisible by $3$ I'm trying to freshen up for school in another month, and I'm struggling with the simplest of proofs! Problem: For any natural number $n , n^3 + 2n$ is divisible by $3.$ This makes sense Proof: Basis Step: If $n = 0,$ then $n^3 + 2n = 0^3 +$ $2 \times 0 = 0.$ So it is divisible by $3.$ Induction: Assume that for an arbitrary natural number $n$, $n^3+ 2n$ is divisible by $3.$ Induction Hypothesis: To prove this for $n+1,$ first try to express $( n + 1 )^3 + 2( n + 1 )$ in terms of $n^3 + 2n$ and use the induction hypothesis. Got it $$( n + 1 )^3+ 2( n + 1 ) = ( n^3 + 3n^2+ 3n + 1 ) + ( 2n + 2 ) \{\text{Just some simplifying}\}$$ $$ = ( n^3 + 2n ) + ( 3n^2+ 3n + 3 ) \{\text{simplifying and regrouping}\}$$ $$ = ( n^3 + 2n ) + 3( n^2 + n + 1 ) \{\text{factored out the 3}\}$$ which is divisible by $3$, because $(n^3 + 2n )$ is divisible by $3$ by the induction hypothesis. What? Can someone explain that last part? I don't see how you can claim $(n^3+ 2n ) + 3( n^2 + n + 1 )$ is divisible by $3.$

Given the $n$th case, you want to consider the $(n+1)$th case, which involves the number $(n+1)^3 + 2(n+1)$. If you know that $n^3+2n$ is divisible by $3$, you can prove $(n+1)^3 + 2(n+1)$ is divisible by $3$ if you can show the difference between the two is divisible by $3$. So find the difference, and then simplify it, and then consider how to prove it's divisible by $3$.

Perfect numbers, the pattern continues The well known formula for perfect numbers is $$ P_n=2^{n-1}(2^{n}-1). $$ This formula is obtained by observing some patterns on the sum of the perfect number's divisors. Take for example $496$: $$ 496=1+2+4+8+16+31+62+124+248 $$ one can see that the first pattern is a sequence of powers of $2$ that stops at $16$, the second pattern starts with a prime number, in this case $31$, the rest of them are multiples of $31$, e.g. $2\cdot 31, 4\cdot 31$ and $8\cdot 31$. But I found that the pattern proceeds after $31$, for example $31=2^5-2^0$, $62=2^6-2^1$, $124=2^7-2^2$ and finally $248=2^8-2^3$, so the perfect number can be written as $$ 496=1+2+4+8+16+(2^5-2^0)+(2^6-2^1)+(2^7-2^2)+(2^8-2^3) $$ or $$ 496=(1+2+4+8+16+32+64+128+256) -(1+2+4+8). $$ So the formula follows very naturally from this. I've searched but didn't find this formulation anywhere. Well, is this something new? Has anyone seen this somewhere?

You've observed that $P_5 = 2^4(2^5-1) = 496$ can also be written as the sum of the first 9 powers of two minus the sum of the first four powers of two. Sums of powers of two Powers of two written in binary look like $1, 10, 100, 1000, \cdots$ but you can also write them like this $1, 1+1, 11+1, 111+1, \cdots$. This explains why (now in decimal notation) $1 + 2 + 4 + 8 = 15$ and the general sums of this form. Nine and four Well obvious $4 = 5-1$, and $9 = 2\cdot 5 - 1$, but where do these come from? Well just multiply out the formula! $2^{n-1}(2^n-1) = 2^{2n-1}-2^{n-1}$. This explains why the form $(1 + 2 + 4 + 8 + \cdots) - (1 + 2 + \cdots)$ works.

The Basel problem As I have heard people did not trust Euler when he first discovered the formula (solution of the Basel problem) $$\zeta(2)=\sum_{k=1}^\infty \frac{1}{k^2}=\frac{\pi^2}{6}$$ However, Euler was Euler and he gave other proofs. I believe many of you know some nice proofs of this, can you please share it with us?

There is a simple way of proving that $\sum_{n=1}^{\infty}\frac{1}{n^2} = \frac{\pi^2}{6}$ using the following well-known series identity: $$\left(\sin^{-1}(x)\right)^{2} = \frac{1}{2}\sum_{n=1}^{\infty}\frac{(2x)^{2n}}{n^2 \binom{2n}{n}}.$$ From the above equality, we have that $$x^2 = \frac{1}{2}\sum_{n=1}^{\infty}\frac{(2 \sin(x))^{2n}}{n^2 \binom{2n}{n}},$$ and we thus have that: $$\int_{0}^{\pi} x^2 dx = \frac{\pi^3}{12} = \frac{1}{2}\sum_{n=1}^{\infty}\frac{\int_{0}^{\pi} (2 \sin(x))^{2n} dx}{n^2 \binom{2n}{n}}.$$ Since $$\int_{0}^{\pi} \left(\sin(x)\right)^{2n} dx = \frac{\sqrt{\pi} \ \Gamma\left(n + \frac{1}{2}\right)}{\Gamma(n+1)},$$ we thus have that: $$\frac{\pi^3}{12} = \frac{1}{2}\sum_{n=1}^{\infty}\frac{ 4^{n} \frac{\sqrt{\pi} \ \Gamma\left(n + \frac{1}{2}\right)}{\Gamma(n+1)} }{n^2 \binom{2n}{n}}.$$ Simplifying the summand, we have that $$\frac{\pi^3}{12} = \frac{1}{2}\sum_{n=1}^{\infty}\frac{\pi}{n^2},$$ and we thus have that $\sum_{n=1}^{\infty}\frac{1}{n^2} = \frac{\pi^2}{6}$ as desired.

Solving the equation $-2x^3 +10x^2 -17x +8=(2x^2)(5x -x^3)^{1/3}$ I wanna know how to solve this equation: $-2x^3 +10x^2 -17x +8=(2x^2)(5x -x^3)^{1/3}$ I have some trouble to do that and I'd glad with any help I may get.

The algebraic $\frac{1}{12}(17 + \sqrt{97})$ is not a root of the equation \begin{eqnarray} -2 x^3 + 10 x^2 - 17 x + 8 = (2 x^2) (5 x - x^3)^{1/3} \end{eqnarray} Plugging it in, you find that the left hand side is real and equal to \begin{eqnarray} \tfrac{1}{216}(-149 - 37 \sqrt{97}) = -2.37689 \dots \end{eqnarray} The right side is \begin{eqnarray} \tfrac{1}{432} (\tfrac{1}{2}( 595 - 61 \sqrt{97})^{1/3} (17 + \sqrt{97})^2 = 1.18844 \dots + i 2.05845 \dots \end{eqnarray} Note: $595 < 61 \sqrt{97}$. I think the ambiguity lies in the fact that we have not used the third-roots of unity. Numerical computations aside, just plot the two functions. The RHS is a positive function defined only in the I and II quadrants. The LHS is cubic. There is only one real intersection point.

Probability of cumulative dice rolls hitting a number Is there a general formula to determine the probability of unbounded, cumulative dice rolls hitting a specified number? For Example, with a D6 and 14: 5 + 2 + 3 + 4 = 14 : success 1 + 1 + 1 + 6 + 5 + 4 = 17 : failure

Assuming the order matters (i,e 1+2 is a different outcome from 2+1) The probability of getting the sum $n$ with dice numbered $1,2,\dots,6$ is the coefficient of $x^n$ in $$\sum_{j=0}^{\infty}(\frac{x+x^2+x^3+x^4+x^5+x^6}{6})^j = \frac{6}{6-x-x^2-x^3-x^4-x^5-x^6}$$ Writing it as partial fractions (using roots of $6-x-x^2-x^3-x^4-x^5-x^6=0$) or using Cauchy's integral formula to find the coefficient of $x^n$, Taylor series, etc should work.

Can't Solve an Integral According to the solution manual: $\int \frac{x}{\sqrt{1-x^{4}}}dx = \frac{1}{2}\arcsin x^{2}+C$ My solution doesn't seem to be working. I know another way of solving it (setting $u=x^{2}$) but the fact that this way of solving it doesn't work bothers me. $$\text{set }u=1-x^{4}\text{ so } dx=\frac{du}{-4x^{3}} $$ $$ \begin{align*} \int \frac{x}{\sqrt{1-x^{4}}}dx &= \int \frac{x}{\sqrt{u}}dx \\ &= \int \frac{xdu}{-4x^{3}\sqrt{u}} \\ &= -\frac{1}{4} \int \frac{du}{x^{2}\sqrt{u}} \\ \end{align*} $$ $$ \text{set } v=\sqrt{u} \text{ so }du=2\sqrt{u}\,dv $$ \begin{align*} -\frac{1}{4} \int \frac{du}{x^{2}\sqrt{u}} &= -\frac{1}{2} \int \frac{dv}{x^{2}} \\ &= -\frac{1}{2} \int \frac{dv}{\sqrt{1-v^{2}}} \\ &= -\frac{1}{2} \arcsin (v) + C \\ &= -\frac{1}{2} \arcsin (\sqrt {1-x^{4}}) + C \\ \end{align*} I'll be happy to clarify any steps I took. Thanks!

Your solution is an antiderivative of the original function. You can always check whether your solution is correct by taking its derivative. This also implies that the book solution and your solution differ by a constant. For this specific problem, imagine the right triangle with sides $x^2$ and $\sqrt{1-x^4}$ and hypotenuse $1$. Then $\arcsin\sqrt{1-x^4} = \frac{\pi}{2} - \arcsin x^2$, and it should be easy to see from there how both solutions are related.

What type of triangle satisfies: $8R^2 = a^2 + b^2 + c^2 $? In a $\displaystyle\bigtriangleup$ ABC,R is circumradius and $\displaystyle 8R^2 = a^2 + b^2 + c^2 $ , then $\displaystyle\bigtriangleup$ ABC is of which type ?

$$\sin^2A+\sin^2B+\sin^2C$$ $$=1-(\cos^2A-\sin^2B)+1-\cos^2C$$ $$=2-\cos(A+B)\cos(A-B)-\cos C\cdot\cos C$$ $$=2-\cos(\pi-C)\cos(A-B)-\cos\{\pi-(A+B)\}\cdot\cos C$$ $$=2+\cos C\cos(A-B)+\cos(A+B)\cdot\cos C\text{ as }\cos(\pi-x)=-\cos x$$ $$=2+\cos C\{\cos(A-B)+\cos(A+B)\}$$ $$=2+2\cos A\cos B\cos C$$ $(1)$ If $2+2\cos A\cos B\cos C=2, \cos A\cos B\cos C=0$ $\implies $ at least one of $\cos A,\cos B,\cos C$ is $0$ which needs the respective angles $=\frac\pi2$ But we can have at most one angle $\ge \frac\pi2$ So, here we shall have exactly one angle $=\frac\pi2$ $(2)$ If $2+2\cos A\cos B\cos C>2, \cos A\cos B\cos C>0$ Either all of $\cos A,\cos B,\cos C$ must be $>0\implies$ all the angles are acute or exactly two cosine ratios $<0$ which needs the respective angles $> \frac\pi2,$ which is impossible for a triangle $(3)$ If $2+2\cos A\cos B\cos C<2, \cos A\cos B\cos C<0$ Either all the ratios $<0$ which needs the respective angles $> \frac\pi2,$ which is impossible fro a triangle or exactly one of the cosine ratios is $<0\implies $ the respective angle $> \frac\pi2,$

Rational Numbers and Uniqueness Let $x$ be a positive rational number of the form $\displaystyle x = \sum\limits_{k=1}^{n} \frac{a_k}{k!}$ where each $a_k$ is a nonnegative integer with $a_k \leq k-1$ for $k \geq 2$ and $a_n >0$. Prove that $a_1 = [x]$, $a_k = [k!x]-k[(k-1)!x]$ for $k = 2, \dots, n$, and that $n$ is the smallest integer such that $n!x$ is an integer. Conversely, show that every positive rational number can be expressed in this form in one and only one way. Note that $[x]$ is the greatest integer function. So I think there are two parts to this: (i) an inductive proof and (ii) a proof by contradiction. Would this be the correct "high level" approach to this problem?

Since $a_k \le k-1$ for $k \ge 2$ we have $$ \lfloor x \rfloor = \left\lfloor \sum_{k=1}^n \frac{a_k}{k!} \right\rfloor \le a_1 + \left\lfloor \sum_{k=2}^n \frac{k-1}{k!} \right\rfloor = a_1 $$ as the latter term is $0$ since $ \sum_{k=2}^n \frac{k-1}{k!} = \sum_{k=2}^n \lbrace \frac{1}{(k-1)!} - \frac{1}{k!} \rbrace = 1 – 1/n! < 1.$ To show $ a_m = \lfloor m! x \rfloor – m \lfloor (m-1)! x \rfloor $ for $m=2,3,\ldots,n$ we note that $$ \lfloor m! x \rfloor = \left\lfloor m! \sum_{k=1}^n \frac{a_k}{k!} \right\rfloor = A_m + \left\lfloor m! \sum_{k=m+1}^n \frac{a_k}{k!} \right\rfloor, $$ where $A_m$ is an integer given by $$A_m = m! \frac{a_1}{1!} + m! \frac{a_2}{2!} + \cdots + m! \frac{a_m}{m!}.$$ However $$ m! \sum_{k=m+1}^n \frac{a_k}{k!} \le m! \sum_{k=m+1}^n \frac{k-1}{k!} = m! \sum_{k=m+1}^n \left\lbrace \frac{1}{(k-1)!} - \frac{1}{k!} \right\rbrace $$ $$ = m! \left( \frac{1}{m!} - \frac{1}{n!} \right) = 1 - \frac{m!}{n!} < 1 $$ and hence $$ \left\lfloor m! \sum_{k=m+1}^n \frac{a_k}{k!} \right\rfloor = 0.$$ Also for $ m > 1 $ $$ m \lfloor (m-1)! x \rfloor = m \left\lfloor (m-1)! \sum_{k=1}^n \frac{a_k}{k!} \right\rfloor $$ $$=m \left\lbrace (m-1)! \frac{a_1}{1!} + (m-1)! \frac{a_2}{2!} + \cdots + (m-1)! \frac{a_{m -1}}{(m-1)!} \right\rbrace + m \left\lfloor \sum_{k=m}^n \frac{a_k}{k!} \right\rfloor $$ $$= \left( A_m - a_m \right) + m \left\lfloor \sum_{k=m}^n \frac{a_k}{k!} \right\rfloor = A_m – a_m . $$ since $$ \sum_{k=m}^n \frac{a_k}{k!} \le \sum_{k=m}^n \frac{k-1}{k!} = \frac{1}{(m-1)!} - \frac{1}{n!} < 1 .$$ Which proves that for $ m > 1 $ $$ \lfloor m! x \rfloor - m \lfloor (m-1)! x \rfloor = A_m – (A_m – a_m) = a_m . \qquad (1)$$ To prove that $n$ is the smallest integer such that $n! x $ is an integer, suppose that $ (n-1)! x $ is an integer. Then by $(1)$ we have for $n > 1$ $$ a_{n-1} = (n-1)! x – n! x < 0 $$ contradicting the fact that $a_k \ge 0 .$ And so $ (n-1)! x $ cannot be an integer, and if $ m! x $ is an integer for any $ m < n $ we have $(n-1)(n-2) \cdots m! x = (n-1)! x $ is an integer, another contradiction. Hence $ m! x $ cannot be an integer. To show that every positive rational number $x$ can be expressed in this form, let $ x = p/q, $ where $ gcd(p,q)=1 \textrm{ and } p,q \in \mathbb{N} $ and let $ n $ be the smallest integer such that $ n! p/q $ is an integer. Define $$ \begin{align*} a_1 &= \left\lfloor \frac{p}{q} \right\rfloor \\ a_m &= \left\lfloor m! \frac{p}{q} \right\rfloor - m \left\lfloor (m-1)! \frac{p}{q} \right\rfloor \quad \textrm{ for } m > 1. \quad (2) \end{align*} $$ We note that $$ \left\lfloor (n-1)! \frac{p}{q} \right\rfloor < (n-1)! \frac{p}{q} , $$ since $ (n-1)! p/q $ is not an integer, and so $a_n > 0 .$ Also, since $ (m-1)! p/q $ is not an integer, for $ m \le n $ we can write $$ (m-1)! \frac{p}{q} = N_m + r_m \quad \textrm{ where } 0 < r_m < 1 $$ and $N_m$ is an integer. Hence for $m=2,3,\ldots,n$ from $(2)$ we have $$ a_m = \lfloor mN_m + mr_m \rfloor - mN_m = \lfloor m r_m \rfloor \le m-1.$$ Note also that the $a_m$ are non-negative. Now assume that $$ \frac{p}{q} = \sum_{k=1}^n \frac{a_k}{k!} \quad (3)$$ then as the $a_k$ satisfy the conditions that each is a non-negative integer with $a_k \le k-1 $ for $k=2,3\ldots,n$ they are uniquely determined by $(1)$ and $a_1 = \lfloor p/q \rfloor .$ It only remains to prove $(3).$ To show this we note that $$ \begin{align*} \sum_{k=1}^n \frac{a_k}{k!} &= \left\lfloor \frac{p}{q} \right\rfloor + \sum_{k=2}^n \frac{a_k}{k!} \\ &= \left\lfloor \frac{p}{q} \right\rfloor + \sum_{k=2}^n \left\lbrace \frac{ \left\lfloor k! \frac{p}{q} \right\rfloor - k \left\lfloor (k-1)! \frac{p}{q} \right\rfloor }{k!} \right\rbrace \\ &= \left\lfloor \frac{p}{q} \right\rfloor + \sum_{k=2}^n \frac{ \left\lfloor k! \frac{p}{q} \right\rfloor }{k!} - \sum_{k=2}^n \frac{ \left\lfloor (k-1)! \frac{p}{q} \right\rfloor }{(k-1)!} \\ &= \frac {n! p/q}{n!} = \frac{p}{q}, \end{align*} $$ since $n! p/q$ is an integer and all the terms cancel, except the last. This completes the proof.

Proof by induction $\frac1{1 \cdot 2} + \frac1{2 \cdot 3} + \frac1{3 \cdot 4} + \cdots + \frac1{n \cdot (n+1)} = \frac{n}{n+1}$ Need some help on following induction problem: $$\dfrac1{1 \cdot 2} + \dfrac1{2 \cdot 3} + \dfrac1{3 \cdot 4} + \cdots + \dfrac1{n \cdot (n+1)} = \dfrac{n}{n+1}$$

Every question of the form: prove by induction that $$\sum_{k=1}^n f(k)=g(n)$$ can be done by verifying two facts about the functions $f$ and $g$: * *$f(1)=g(1)$ and *$g(n+1)-g(n)=f(n+1)$.

If $4^x + 4^{-x} = 34$, then $2^x + 2^{-x}$ is equal to...? I am having trouble with this: If $4^x + 4^{-x} = 34$, then what is $2^x + 2^{-x}$ equal to? I managed to find $4^x$ and it is: $$4^x = 17 \pm 12\sqrt{2}$$ so that means that $2^x$ is: $$2^x = \pm \sqrt{17 \pm 12\sqrt{2}}.$$ Correct answer is 6 and I am not getting it :(. What am I doing wrong?

You haven't done anything wrong! To complete your answer, one way you can see the answer is $6$ is to guess that $$17 + 12 \sqrt{2} = (a + b\sqrt{2})^2$$ Giving us $$17 = a^2 + 2b^2, \ \ ab = 6$$ Giving us $$a = 3, \ \ b = 2$$ Thus $$ \sqrt{17 + 12 \sqrt{2}} = 3 + 2\sqrt{2}$$ which gives $$2^x + 2^{-x} = 6$$ (And similarly for $17 - 12\sqrt{2}$) A simpler way is to notice that $(2^x + 2^{-x})^2 = 4^x + 4^{-x} + 2$.

Integrate Form $du / (a^2 + u^2)^{3/2}$ How does one integrate $$\int \dfrac{du}{(a^2 + u^2)^{3/2}}\ ?$$ The table of integrals here: http://teachers.sduhsd.k12.ca.us/abrown/classes/CalculusC/IntegralTablesStewart.pdf Gives it as: $$\frac{u}{a^2 ( a^2 + u^2)^{1/2}}\ .$$ I'm getting back into calculus and very rusty. I'd like to be comfortable with some of the proofs behind various fundamental "Table of Integrals" integrals. Looking at it, the substitution rule seems like the method of choice. What is the strategy here for choosing a substitution? It has a form similar to many trigonometric integrals, but the final result seems to suggest that they're not necessary in this case.

A trigonometric substitution does indeed work. We want to express $(a^2 + u^2)^{3/2}$ as something without square roots. We want to use some form of the Pythagorean trigonometric identity $\sin^2 x + \cos^2 x = 1$. Multiplying each side by $\frac{a^2}{\cos^2 x}$, we get $a^2 \tan^2 x + a^2 = a^2 \sec^2 x$, which is in the desired form. of (sum of two squares) = (something squared). This suggests that we should use the substitution $u^2 = a^2 \tan^2 x$. Equivalently, we substitute $u = a \tan x$ and $du = a \sec^2 x dx$. Then $$ \int \frac{du}{(a^2 + u^2)^{3/2}} = \int \frac{a \sec^2 x \, dx}{(a^2 + a^2 \tan^2 x)^{3/2}}. $$ Applying the trigonometric identity considered above, this becomes $$ \int \frac{a \sec^2 x \, dx}{(a^2 \sec^2 x)^{3/2}} = \int \frac{dx}{a^2 \sec x} = \frac{1}{a^2} \int \cos x \, dx, $$ which can be easily integrated as $$ =\frac{1}{a^2} \sin x. $$ Since we set $u = a \tan x$, we substitute back $x = \tan^{-1} (\frac ua)$ to get that the answer is $$ =\frac{1}{a^2} \sin \tan^{-1} \frac{u}{a}. $$ Since $\sin \tan^{-1} z = \frac{z}{\sqrt{z^2 + 1}}$, this yields the desired result of $$ =\frac{u/a}{a^2 \sqrt{(u/a)^2 + 1}} = \frac{u}{a^2 (a^2 + u^2)^{1/2}}. $$

How does teacher get first step? Below are the steps the teacher took to solve: $y = \sqrt{3}\sin x + \cos x$ find min and max on $[0, 2\pi)$ Step 1: = $2\sin(x + \pi/6))$ How does the teacher get this first step? Note: No calculus please.

picakhu's answer is the simplest way to see how it works having already arrived at $y=2\sin(x+\frac{\pi}{6})$ (use the identity there to expand this form). In general, given $a\sin x+b\cos x$ (let's say for $a,b>0$), it is possible to arrive at a similar equivalent form: $$\begin{align} a\sin x+b\cos x &=a\left(\sin x+\frac{b}{a}\cos x\right) \\ &=a\left(\sin x+\tan\left(\arctan\frac{b}{a}\right)\cos x\right) \\ &=a\left(\sin x+\frac{\sin\left(\arctan\frac{b}{a}\right)}{\cos\left(\arctan\frac{b}{a}\right)}\cos x\right) \\ &=\frac{a}{\cos\left(\arctan\frac{b}{a}\right)}\left(\sin x\cos\left(\arctan\frac{b}{a}\right)+\sin\left(\arctan\frac{b}{a}\right)\cos x\right) \\ &=\sqrt{a^2+b^2}\sin\left(x+\arctan\frac{b}{a}\right). \end{align}$$

$n$th derivative of $e^{1/x}$ I am trying to find the $n$'th derivative of $f(x)=e^{1/x}$. When looking at the first few derivatives I noticed a pattern and eventually found the following formula $$\frac{\mathrm d^n}{\mathrm dx^n}f(x)=(-1)^n e^{1/x} \cdot \sum _{k=0}^{n-1} k! \binom{n}{k} \binom{n-1}{k} x^{-2 n+k}$$ I tested it for the first $20$ derivatives and it got them all. Mathematica says that it is some hypergeometric distribution but I don't want to use that. Now I am trying to verify it by induction but my algebra is not good enough to do the induction step. Here is what I tried for the induction (incomplete, maybe incorrect) $\begin{align*} \frac{\mathrm d^{n+1}}{\mathrm dx^{n+1}}f(x)&=\frac{\mathrm d}{\mathrm dx}(-1)^n e^{1/x} \cdot \sum _{k=0}^{n-1} k! \binom{n}{k} \binom{n-1}{k} x^{-2 n+k}\\ &=(-1)^n e^{1/x} \cdot \left(\sum _{k=0}^{n-1} k! \binom{n}{k} \binom{n-1}{k} (-2n+k) x^{-2 n+k-1}\right)-e^{1/x} \cdot \sum _{k=0}^{n-1} k! \binom{n}{k} \binom{n-1}{k} x^{-2 (n+1)+k}\\ &=(-1)^n e^{1/x} \cdot \sum _{k=0}^{n-1} k! \binom{n}{k} \binom{n-1}{k}((-2n+k) x^{-2 n+k-1}-x^{-2 (n+1)+k)})\\ &=(-1)^{n+1} e^{1/x} \cdot \sum _{k=0}^{n-1} k! \binom{n}{k} \binom{n-1}{k}(2n x-k x+1) x^{-2 (n+1)+k} \end{align*}$ I don't know how to get on from here.

We may obtain a recursive formula, as follows: \begin{align} f\left( t \right) &= e^{1/t} \\ f'\left( t \right) &= - \frac{1}{{t^2 }}f\left( t \right) \\ f''\left( t \right) &= - \frac{1}{{t^2 }}f'\left( t \right) + f\left( t \right)\frac{2}{{t^3 }} \\ &= - \frac{1}{{t^2 }}\left\{ {\frac{1}{{t^2 }}f\left( t \right)} \right\} + f\left( t \right)\frac{2}{{t^3 }} \\ &= \left( {\frac{1}{{t^4 }} + \frac{2}{{t^3 }}} \right)f\left( t \right) \\ \ldots \end{align} Inductively, let us assume $f^{(n-1)}(t)=P_{n-1}(\frac{1}{t})f(t)$ is true, for some polynomial $P_{n-1}$. Now, for $n$, we have \begin{align} f^{\left( n \right)} \left( t \right) &= - \frac{1}{{t^2 }}P'_{n - 1} \left( {\frac{1}{t}} \right)f\left( t \right) + P_{n - 1} \left( {\frac{1}{t}} \right)f'\left( t \right) \\ &= - \frac{1}{{t^2 }}P'_{n - 1} \left( {\frac{1}{t}} \right)f\left( t \right) + P_{n - 1} \left( {\frac{1}{t}} \right)\left\{ { - \frac{1}{{t^2 }}f\left( t \right)} \right\} \\ &= - \frac{1}{{t^2 }}\left\{ {P'_{n - 1} \left( {\frac{1}{t}} \right) + P_{n - 1} \left( {\frac{1}{t}} \right)} \right\}f\left( t \right) \\ &= P_n \left( {\frac{1}{t}} \right)f\left( t \right) \end{align} Thus, \begin{align} f^{\left( n \right)} \left( t \right) = P_n \left( {\frac{1}{t}} \right)f\left( t \right) \end{align} where, $P_n \left( x \right): = x^2 \left[ {P'_{n - 1} \left( x \right) - P_{n - 1} \left( x \right)} \right]$, $P_0=1$.

decompose some polynomials [ In first, I say "I'm sorry!", because I am not a Englishman and I don't know your language terms very well. ] OK, I have some polynomials (like $a^2 +2ab +b^2$ ). And I can't decompress these (for example $a^2 +2ab +b^2 = (a+b)^2$). Can you help me? (if you can, please write the name or formula of combination (like $(a+b)^2 = a^2 +2ab +b^2$) of each polynomial. * *$(a^2-b^2)x^2+2(ad-bc)x+d^2-c^2$ *$2x^2+y^+2x-2xy-2y+1$ *$2x^2-5xy+2y^2-x-y-1$ *$x^6-14x^4+49x^2-36$ *$(a+b)^4+(a-b)^4+(a^2-b^2)^2$ Thank you! very much ....

For 1) $(a^2-b^2)x^2+2(ad-bc)x+d^2-c^2$ think about rearranging $$(a^2-b^2)x^2+2(ad-bc)x+d^2-c^2=a^2x^2+2adx+d^2-(b^2x^2+2bcx+c^2)$$ The same idea can be applied to all your questions.

How to find the sum of the following series How can I find the sum of the following series? $$ \sum_{n=0}^{+\infty}\frac{n^2}{2^n} $$ I know that it converges, and Wolfram Alpha tells me that its sum is 6 . Which technique should I use to prove that the sum is 6?

It is equal to $f(x)=\sum_{n \geq 0} n^2 x^n$ evaluated at $x=1/2$. To compute this function of $x$, write $n^2 = (n+1)(n+2)-3(n+1)+1$, so that $f(x)=a(x)+b(x)+c(x)$ with: $a(x)= \sum_{n \geq 0} (n+1)(n+2) x^n = \frac{d^2}{dx^2} \left( \sum_{n \geq 0} x^n\right) = \frac{2}{(1-x)^3}$ $b(x)=\sum_{n \geq 0} 3 (n+1) x^n = 3\frac{d}{dx} \left( \sum_{n \geq 0} x^n \right) = \frac{3}{(1-x)^2}$ $c(x)= \sum_{n \geq 0} x^n = \frac{1}{1-x}$ So $f(1/2)=\frac{2}{(1/2)^3}-\frac{3}{(1/2)^2} + \frac{1}{1/2} = 16-12+2=6$. The "technique" is to add a parameter in the series, to make the multiplication by $n+1$ appear as differentiation.

Showing $\frac{e^{ax}(b^2\cos(bx)+a^2\cos(bx))}{a^2+b^2}=e^{ax}\cos(bx).$ I've got: $$\frac{e^{ax}(b^2\cos(bx)+a^2\cos(bx))}{a^2+b^2}.$$ Could someone show me how it simplifies to: $e^{ax} \cos(bx)$? It looks like the denominator is canceled by the terms that are being added, but then how do I get rid of one of the cosines?

You use the distributive law, which says that $(X+Y)\cdot Z=(X\cdot Z)+(Y\cdot Z)$ for any $X$, $Y$, and $Z$. In your case, we have $X=b^2$, $Y=a^2$, and $Z=\cos(bx)$, and so $$\frac{e^{ax}(b^2\cos(bx)+a^2\cos(bx))}{a^2+b^2}=\frac{e^{ax}((b^2+a^2)\cos(bx))}{(a^2+b^2)}=\frac{e^{ax}((a^2+b^2)\cos(bx))}{(a^2+b^2)}=e^{ax}\cos(bx).$$

What is the simplification of $\frac{\sin^2 x}{(1+ \sin^2 x +\sin^4 x +\sin^6 x + \cdots)}$? What is the simplification of $$\frac{\sin^2 x}{(1+ \sin^2 x +\sin^4 x +\sin^6 x + \cdots)} \space \text{?}$$

What does $1 + \sin^2 x + \sin^4 x + \sin^6 x + ....$ simplify to? Or better, what does $1 + x^2 + x^4 + x^6 + ....$ simplify to? Or better, what does $1 + x + x^2 + x^3 + ....$ simplify to?

Using congruences, show $\frac{1}{5}n^5 + \frac{1}{3}n^3 + \frac{7}{15}n$ is integer for every $n$ Using congruences, show that the following is always an integer for every integer value of $n$: $$\frac{1}{5}n^5 + \frac{1}{3}n^3 + \frac{7}{15}n.$$

Lets show that $P(n)=3n^5+5n^3+7n$ is divisible by 15 for every $n$. To do this, we will show that it is divisible by $3$ and $5$ for every $n$. Recall that for a prime $p$, $x^p\equiv x \pmod{p}$. (Fermat's Little Theorem) Then, looking modulo 5 we see that $$P(n)\equiv 3n^5+7n\equiv 3n+7n=10n\equiv 0.$$ Now looking modulo 3 we see that $$P(n)\equiv 5n^3+7n\equiv 5n+7n=12n\equiv 0.$$ Thus $P(n)$ is divisible by 15 for every $n$ as desired.

Chain rule for multi-variable functions So I have been studying the multi-variable chain rule. Most importantly, and this is what I must have overlooked, is it's not always clear to me how to see which variables are functions of other variables, so that you know when to use the chain rule. For example, if you have: $$ x^2+y^2-z^2+2xy=1 $$ $$ x^3+y^3-5y=8 $$ In general, say we want to find $\frac{dz}{dt}$ but $z$ is a function of $x$, then we get: $$ \frac{dz}{dt} = \frac{dz}{dx} \frac{dx}{dt} .$$ And if $z$ is a function of both $y$ and $t$, we get: $$ \frac{dz}{dt} = \frac{dz}{dx} \frac{dx}{dt} + \frac{dz}{dy} \frac{dy}{dt}$$ In this case, we have two equations. One involving all three variables $x,y,z$ and one involving just $x,y$. Say we want to find $\frac{dz}{dx}$. What does this mean for this case? How should we interpret this rule in general?

If we have an explicit function $z = f(x_1,x_2,\ldots,x_n)$, then $$\displaystyle \frac{dz}{dt} = \frac{\partial z}{\partial x_1} \frac{dx_1}{dt} + \frac{\partial z}{\partial x_2} \frac{dx_2}{dt} + \cdots +\frac{\partial z}{\partial x_n} \frac{dx_n}{dt}$$ If we have an implicit function $f(z,x_1,x_2,\ldots,x_n) = 0$, then $$\displaystyle \frac{\partial f}{\partial z} \frac{dz}{dt} + \frac{\partial f}{\partial x_1} \frac{dx_1}{dt} + \frac{\partial f}{\partial x_2} \frac{dx_2}{dt} + \cdots +\frac{\partial f}{\partial x_n} \frac{dx_n}{dt} = 0$$ $$\displaystyle \frac{dz}{dt} = - \frac{ \frac{\partial f}{\partial x_1} \frac{dx_1}{dt} + \frac{\partial f}{\partial x_2} \frac{dx_2}{dt} + \cdots +\frac{\partial f}{\partial x_n} \frac{dx_n}{dt}}{\frac{\partial f}{\partial z} }$$ In the first example, \begin{align*} \displaystyle x^2 + y^2 - z^2 + 2y & = 1\\ \displaystyle 2x\frac{dx}{dt} + 2y\frac{dy}{dt} - 2z\frac{dz}{dt} + 2\frac{dy}{dt} & = 0\\ \displaystyle x\frac{dx}{dt} + y\frac{dy}{dt} + \frac{dy}{dt} & = z\frac{dz}{dt}\\ \displaystyle \frac{dz}{dt} & = \frac{x\frac{dx}{dt} + y\frac{dy}{dt} + \frac{dy}{dt}}{z} \end{align*} In the second example, \begin{align*} \displaystyle x^3 + y^3 - 5y & = 8\\ \displaystyle 3x^2 \frac{dx}{dt} + 3y^2 \frac{dy}{dt} - 5 \frac{dy}{dt} = 0\\ \displaystyle \frac{dy}{dt} & = \frac{3x^2}{5-3y^2} \frac{dx}{dt} \end{align*}

Simple limit, wolframalpha doesn't agree, what's wrong? (Just the sign of the answer that's off) $\begin{align*} \lim_{x\to 0}\frac{\frac{1}{\sqrt{4+x}}-\frac{1}{2}}{x} &=\lim_{x\to 0}\frac{\frac{2}{2\sqrt{4+x}}-\frac{\sqrt{4+x}}{2\sqrt{4+x}}}{x}\\ &=\lim_{x\to 0}\frac{\frac{2-\sqrt{4+x}}{2\sqrt{4+x}}}{x}\\ &=\lim_{x\to 0}\frac{2-\sqrt{4+x}}{2x\sqrt{4+x}}\\ &=\lim_{x\to 0}\frac{(2-\sqrt{4-x})(2+\sqrt{4-x})}{(2x\sqrt{4+x})(2+\sqrt{4-x})}\\ &=\lim_{x\to 0}\frac{2 \times 2 + 2\sqrt{4-x}-2\sqrt{4-x}-((\sqrt{4-x})(\sqrt{4-x})) }{2 \times 2x\sqrt{4+x} + 2x\sqrt{4+x}\sqrt{4-x}}\\ &=\lim_{x\to 0}\frac{4-4+x}{4x\sqrt{4+x} + 2x\sqrt{4+x}\sqrt{4-x}}\\ &=\lim_{x\to 0}\frac{x}{x(4\sqrt{4+x} + 2\sqrt{4+x}\sqrt{4-x})}\\ &=\lim_{x\to 0}\frac{1}{(4\sqrt{4+x} + 2\sqrt{4+x}\sqrt{4-x})}\\ &=\frac{1}{(4\sqrt{4+0} + 2\sqrt{4+0}\sqrt{4-0})}\\ &=\frac{1}{16} \end{align*}$ wolframalpha says it's negative. What am I doing wrong?

Others have already pointed out a sign error. One way to avoid such is to first simplify the problem by changing variables. Let $\rm\ z = \sqrt{4+x}\ $ so $\rm\ x = z^2 - 4\:.\:$ Then $$\rm \frac{\frac{1}{\sqrt{4+x}}-\frac{1}{2}}{x}\ =\ \frac{\frac{1}z - \frac{1}2}{z^2-4}\ =\ \frac{-(z-2)}{2\:z\:(z^2-4)}\ =\ \frac{-1}{2\:z\:(z+2)}$$ In this form it is very easy to compute the limit as $\rm\ z\to 2\:$.

Question regarding Hensel's Lemma Hensel's Lemma Suppose that f(x) is a polynomial with integer coefficients, $k$ is an integer with $k \geq 2$, and $p$ a prime. Suppose further that $r$ is a solution of the congruence $f(x) \equiv 0 \pmod{p^{k-1}}$. Then, If $f'(r) \not\equiv 0 \pmod{p}$, then there is a unique integer t, $0 \leq t < p$, such that $$f(r + tp^{k-1}) \equiv 0 \pmod{p^k}$$ given by $$t \equiv \overline{-f'(r)}\frac{f(r)}{p^{k-1}} \pmod{p}$$ where $\overline{-f'(r)}$ is an inverse of f'(r) modulo p. If $f'(r) \equiv 0 \pmod{p}$ and $f(r) \equiv 0 \pmod{p^k}$, then $f(r+tp^{k-1}) \equiv 0 \pmod{p^k}$ $\forall$ integers t. If $f'(r) \equiv 0 \pmod{p}$ and $f(r) \not\equiv 0 \pmod{p^k}$, then $f(x) \equiv 0 \pmod{p^k}$ has no solutions with $x \equiv r \pmod{p^{k-1}}$ I'm practicing solving congruence equation using Hensel's Lemma, however, I was a little confused with the last two cases. To be clear, I use this congruence equation as an example $f(x) = x^4 + 4x^3 + 2x^2 + 2x + 12 \equiv 0 \pmod{625}$ My attempt was: By inspecting all remainders of $5$, i.e $0, 1, 2, 3, 4$ \ We can see that $ x \equiv 3 \pmod{5}$ is the solution to $f(x) \equiv 0 \pmod{5}$ \ Apply Hensel's Lemma for $5^2 = 25$, we have: $$f'(x) = 4x^3 + 12x^2 + 4x + 2$$ And, $$f'(3) = 4.3^3 + 12.3^2 + 4.3 + 2 = 230 \equiv 0 \pmod{5}$$ $$f(3) = 3^4 + 4.3^3 + 2.3^2 + 2.3 + 12 = 225 \equiv 0 \pmod{5^2}$$ Hence, $$x \equiv 3 \pmod{5^2}$$ Apply Hensel's Lemma for $5^3 = 125$, we have: $$f(3) = 225 \not\equiv 0 \pmod{5^3}$$ So $f(x) \equiv 0 \pmod{5^3}$ has no solutions with $x \equiv 3 \pmod{5^2}$. \ Therefore, there are no solutions to $f(x) = x^4 + 4x^3 + 2x^2 + 2x + 12 \equiv 0 \pmod{625}$ What I understood about Hensel's Lemma is, it let us lift up the solution from $p^k$ to $p^{k + 1}$ each time we found a solution of a current $k$. But, in the case 2, when it said for all integers t, I was confused. Does it mean I can use the previous solution with $p^k$. By that I mean, if I have $x \equiv 3 \pmod{5}$, then if case 2 satisfies, then $x \equiv 3 \pmod{25}$? If the question is vague, please let me know. I will try my best to rewrite it again. Sorry for my poor English writing. Thanks,

You go off track at the word "Hence". If $f'(3)\equiv 0\pmod 5$ and $f(3)\equiv 0 \pmod{25}$ (I assume you've done this correctly; I didn't check), that means that $x\equiv 3,8,13,18,23 \pmod{25}$ are all solutions modulo 25. You only verified that there are no solutions modulo 125 which are 3 modulo 25. There may still be solutions which are 8, 13, 18, or 23 modulo 25.

Equation of the complex locus: $|z-1|=2|z +1|$ This question requires finding the Cartesian equation for the locus: $|z-1| = 2|z+1|$ that is, where the modulus of $z -1$ is twice the modulus of $z+1$ I've solved this problem algebraically (by letting $z=x+iy$) as follows: $\sqrt{(x-1)^2 + y^2} = 2\sqrt{(x+1)^2 + y^2}$ $(x-1)^2 + y^2 = 4\big((x+1)^2 + y^2\big)$ $x^2 - 2x + 1 + y^2 = 4x^2 + 8x + 4 + 4y^2$ $3x^2 + 10x + 3y^2 = -3$ $x^2 + \frac{10}{3}x + y^2 = -1$ $(x + \frac{5}{3})^2 +y^2 = -1 + \frac{25}{9}$ therefore, $(x+\frac{5}{3})^2 + y^2 = \frac{16}{9}$, which is a circle. However, I was wondering if there is a method, simply by inspection, of immediately concluding that the locus is a circle, based on some relation between the distance from $z$ to $(1,0)$ on the plane being twice the distance from $z$ to $(-1,0)$?

Just to add on to Aryabhata's comment above. The map $f(z) = \frac{1}{z}$ for $ z \in \mathbb{C} -\{0\}$, $f(0) = \infty$ and $f(\infty) = 0$ is a circle preserving homeomorphism of $\bar{\mathbb{C}}$. To see this, one needs to prove that it is continuous on $\bar{\mathbb{C}}$, and since $f(z)$ is an involution proving this would mean that its inverse is continuous as well. It is also not hard to show that $f(z)$ is bijective. Lastly use the general equation of a circle in $\bar{\mathbb{C}}$ to see that circles in $\bar{\mathbb{C}}$ are preserved under this map.

Understanding a proof by descent [Fibonacci's Lost Theorem] I am trying to understand the proof in Carmichaels book Diophantine Analysis but I have got stuck at one point in the proof where $w_1$ and $w_2$ are introduced. The theorem it is proving is that the system of diophantine equations: * *$$x^2 + y^2 = z^2$$ *$$y^2 + z^2 = t^2$$ cannot simultaneously be satisfied. The system is algebraically seen equivalent to * *$$t^2 + x^2 = 2z^2$$ *$$t^2 - x^2 = 2y^2$$ and this is what will be worked on. We are just considering the case where the numbers are pairwise relatively prime. That implies that $t,x$ are both odd (they cannot be both even). Furthermore $t > x$ so define $t = x + 2 \alpha$. Clearly the first equation $(x + 2\alpha)^2 + x^2 = 2 z^2$ is equivalent to $(x + \alpha)^2 + \alpha^2 = z^2$ so by the characterization of primitive Pythagorean triples there exist relatively prime $m,n$ such that $$\{x+\alpha,\alpha\} = \{2mn,m^2-n^2\}.$$ Now the second equation $t^2 - x^2 = 4 \alpha (x + \alpha) = 8 m n (m^2 - n^2) = 2 y^2$ tells us that $y^2 = 2^2 m n (m^2 - n^2)$ by coprimality and unique factorization it follows that each of those terms are squares so define $u^2 = m$, $v^2 = n$ and $w^2 = m^2 - n^2 = (u^2 - v^2)(u^2 + v^2)$. It is now said that from the previous equation either * *$u^2 + v^2 = 2 {w_1}^2$, $u^2 - v^2 = 2 {w_2}^2$ or * *$u^2 + v^2 = w_1^2$, $u^2 - v^2 = w_2^2$ but $w_1$ and $w_2$ have not been defined and I cannot figure out what they are supposed to be. Any ideas what this last part could mean? For completeness, if the first case occurs we have our descent and if the second case occurs $w_1^2 + w_2^2 = 2 u^2$, $w_1^2 - w_2^2 = 2 v^2$ gives the descent. Which finishes the proof.

$u^2$ and $v^2$ are $m$ and $n$, respectively, which are coprime. Then since $(u^2+v^2)+(u^2-v^2)=2u^2$ and $(u^2+v^2)-(u^2-v^2)=2v^2$, the only factor that $u^2+v^2$ and $u^2-v^2$ can have in common is a single factor of $2$. Since their product is the square $w^2$, that leaves the two possibilities given.

exponential equation $$\sqrt{(5+2\sqrt6)^x}+\sqrt{(5-2\sqrt6)^x}=10$$ So I have squared both sides and got: $$(5-2\sqrt6)^x+(5+2\sqrt6)^x+2\sqrt{1^x}=100$$ $$(5-2\sqrt6)^x+(5+2\sqrt6)^x+2=100$$ I don't know what to do now

You don't have to square the equation in the first place. Let $y = \sqrt{(5+2\sqrt{6})^x}$, then $\frac{1}{y} = \sqrt{(5-2\sqrt{6})^x}$. Hence you have $y + \frac{1}{y} = 10$ i.e. $y^2 + 1 = 10y$ i.e. $y^2-10y+1 = 0$. Hence, $(y-5)^2 =24 \Rightarrow y = 5 \pm 2 \sqrt{6}$. Hence, $$\sqrt{(5+2\sqrt{6})^x} = 5 \pm 2\sqrt{6} \Rightarrow x = \pm 2$$ (If you plug in $x = \pm 2$, you will get $5+2\sqrt{6} + 5-2\sqrt{6} $ which is nothing but $10$)

Tips for understanding the unit circle I am having trouble grasping some of the concepts regarding the unit circle. I think I have the basics down but I do not have an intuitive sense of what is going on. Is memorizing the radian measurements and their corresponding points the only way to master this? What are some ways one can memorize the unit circle? Edit for clarification: I think my trouble arises by the use of $\pi$ instead of degrees. We started graphing today and the use of numbers on the number line were again being referred to with $\pi$. Why is this?

It is probably useful to memorize a table like this: \begin{align} \theta & & \sin\theta & & \cos \theta \\ 0 & & \frac{\sqrt{0}}{2} = 0 & & \frac{\sqrt{4}}{2} = 1 \\ \frac{\pi}{6} & & \frac{\sqrt{1}}{2} = \frac{1}{2} & & \frac{\sqrt{3}}{2} \\ \frac{\pi}{4} & & \frac{\sqrt{2}}{2} & & \frac{\sqrt{2}}{2} \\ \frac{\pi}{3} & & \frac{\sqrt{3}}{2} & & \frac{\sqrt{1}}{2} = \frac{1}{2} \\ \frac{\pi}{2} & & \frac{\sqrt{4}}{2} = 0 & & \frac{\sqrt{0}}{2} = 0 \end{align}

Representation of integers How can I prove that every integer $n>=170$ can be written as a sum of five positive squares? (i.e. none of the squares are allowed to be zero). I know that $169=13^2=12^2+5^2=12^2+4^2+3^2=10^2+8^2+2^2+1^2$, and $n-169=a^2+b^2+c^2+d^2$ for some integers $a$, $b$, $c$, $d$, but do I show it? Thank you.

Hint: let $n-169 = a^2+b^2+c^2+d^2$; if $a,b,c,d \neq 0$ then ... if $d = 0$ and $a,b,c \neq 0$ then ... if $c = d = 0$ and $a,b \neq 0$ then ... if $b = c = d = 0$ and $a \neq 0$ then ... if $a = b = c = d = 0$ then - wait, that can't happen!

What is the best way to show that no positive powers of this matrix will be the identity matrix? Show that no positive power of the matrix $\left( \begin{array}{cc} 1 & 1 \\ 0 & 1 \end{array} \right)$ equals $I_2$. I claim that given $A^{n}, a_{11} = 1$ and $a_{12} >0, \forall n \in \mathbb{N}$. This is the case for $n=1$ since $A^{1} = \left( \begin{array}{cc} 1 & 1 \\ 0 & 1 \end{array} \right)$ with $1=1$ and $1>0$. Now assuming that $a_{11} = 1$ and $a_{12}>0$ for $A^{n}$ show that $A^{n+1} = A^{n}A = \left( \begin{array}{cc} a_{11} & a_{12} \\ a_{21} & a_{22} \end{array} \right)\left( \begin{array}{cc} 1 & 1 \\ 0 & 1 \end{array} \right) = \left( \begin{array}{cc} a_{11} & a_{11} + a_{12} \\ a_{21} & a_{21} + a_{22} \end{array} \right)$. According to the assumption $a_{11} = 1$ and $a_{12}>0 \Rightarrow 1+a_{12} = a_{11}+a_{12}>0$. Taken together, this shows that $A^{n} \neq I_{2} \forall n\in \mathbb{N}$ since $a_{12}\neq0 = i_{12}$. First of all, was my approach legitimate and done correctly? I suspect that I did not solve this problem as intended by the author (if at all!), could anyone explain the expected solution please? Thank you!

Your solution seems OK to me. You can also find $A^n$ explicitly: let $E=\left(\begin{array}{cc}0&1\\0&0\end{array}\right)$. Then $A=I+E$ and $E^2=0$. So $(I+nE)(I+E)=I+(n+1)E$ and so, by induction, $A^n=I+nE\ne I$ for $n\ge1$.

Solving differential equation Below is my work for a particular problem that is mixing me up, since no matter how many times, I can't get my answer to match the book solution. Given ${f}''(x)= x^{-\frac{3}{2}}$ where $f'(4)= 2$ and $f(0)= 0$, solve the differential equation. $$f'(x)= \int x^{-\frac{3}{2}} \Rightarrow \frac{x^{-\frac{3}{2}+1}}{-\frac{3}{2}+1} \Rightarrow -2x^{-\frac{1}{2}} + C$$ $$f'(4)= -2(4)^{-\frac{1}{2}}+ C= 2 \Rightarrow -4+C= 2 \Rightarrow C= 6$$ Thus, the first differential equation is $f'(x)= -2x^{-\frac{1}{2}}+6$ $$f(x)= \int -2x^{-\frac{1}{2}}+6 \Rightarrow 2(\frac{x^{-\frac{1}{2}}}{-\frac{1}{2}+1}) \Rightarrow -4x^{\frac{1}{2}}+6x+C$$ Since $f(0)= 0, C=0$, so the final differential equation should be $f(x)= -4x^{\frac{1}{2}}+6x$, but the book answer has $3x$ in place of my $6x$. Where did I go wrong?

Nothing to worry about! There is a minor slip, $-2(4)^{-1/2}=-2/2=-1$. You got $-4$ instead.

An inequality on a convex function An exercise in my textbook asked to prove the following inequality, valid for all $a,b,c,d \in R $ $$\left(\frac{a}{2} + \frac{b}{3} + \frac{c}{12} + \frac{d}{12}\right)^4 \leq \frac{a^4}{2} + \frac{b^4}{3} + \frac{c^4}{12} + \frac{d^4}{12}$$ There is a straightforward proof using Convex Functions: * *$f(x) = x^4$ is a convex function satisfying $f(\lambda x + (1-\lambda) y) \leq \lambda f(x) + (1 - \lambda)f(y)$ for all $x,y \in R$ and $\lambda \in [0,1]$ *Since $\frac{1}{2} + \frac{1}{3} + \frac{1}{12} + \frac{1}{12} = 1$, we can use the convexity property to obtain the inequality. Since this question was on the chapter about Convex Functions, I was able to find the solution quickly. However, had I seen the problem in a "standalone" manner I would probably take longer to solve it, and at least spend a lot of muscle opening up the left hand term :) My question is: What would be other ways to obtain this same result? What if someone had shown me this problem back when I was in eight grade?

By Holder $$\frac{a^4}{2} + \frac{b^4}{3} + \frac{c^4}{12} + \frac{d^4}{12}=\left(\frac{a^4}{2} + \frac{b^4}{3} + \frac{c^4}{12} + \frac{d^4}{12}\right)\left(\frac{1}{2} + \frac{1}{3} + \frac{1}{12} + \frac{1}{12}\right)^3\geq$$ $$\geq\left(\frac{|a|}{2} + \frac{|b|}{3} + \frac{|c|}{12} + \frac{|d|}{12}\right)^4\geq\left(\frac{a}{2} + \frac{b}{3} + \frac{c}{12} + \frac{d}{12}\right)^4$$ and we are done!

The locus of the intersection point of two perpendicular tangents to a given ellipse For a given ellipse, find the locus of all points P for which the two tangents are perpendicular. I have a trigonometric proof that the locus is a circle, but I'd like a pure (synthetic) geometry proof.

If all you want is a proof that the locus is a circle, we may assume that the ellipse is given by $$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1.$$ Ignoring vertical tangents for now, if a line $y=mx+k$ is tangent to the ellipse, then plugging in this value of $y$ into the equation for the ellipse gives $$\frac{x^2}{a^2} + \frac{(m^2x^2 + 2mkx + k^2)}{b^2} = 1$$ or $$(b^2 + a^2m^2)x^2 + 2a^2mkx + (a^2k^2 - a^2b^2) = 0.$$ This equation gives the two points of intersection of the line with the ellipse. If the line is tangent, then the two points must coincide, so the quadratic must have zero discriminant. That is, we need $$(2a^2mk)^2 - 4(a^2k^2 - a^2b^2)(b^2+a^2m^2) = 0$$ or equivalently, $$\begin{align*} (a^2m^2)k^2 -a^2(b^2+a^2m^2)k^2 &= -a^2b^2(b^2+a^2m^2)\\ -a^2b^2k^2&= -a^2b^2(b^2+a^2m^2)\\ k^2 &= b^2+a^2m^2\\ k &= \pm\sqrt{a^2m^2 + b^2}. \end{align*} $$ So the lines that are tangent to the ellipse are of the form $$y = mx \pm \sqrt{a^2m^2 + b^2}.$$ Since the problem is symmetric about $x$ and $y$, consider the points on the upper half plane, so that we will take the plus sign above. The tangent perpendicular to this one will therefore have equation $$y = -\frac{1}{m}x + \sqrt{\frac{a^2}{m^2} + b^2},$$ or equivalently $$my = -x + \sqrt{a^2 + m^2b^2}.$$ (We are ignoring the vertical and horizontal tangents; I'll deal with them at the end). If a point $(r,s)$ is on both lines, then we have $$\begin{align*} s-mr &= \sqrt{a^2m^2 + b^2}\\ ms + r &= \sqrt{a^2+m^2b^2}. \end{align*}$$ Squaring both sides of both equations we get $$\begin{align*} s^2 - 2mrs + m^2r^2 &= a^2m^2 + b^2\\ m^2s^2 + 2mrs + r^2 &= a^2 + m^2b^2 \end{align*}$$ and adding both equations, we have $$\begin{align*} (1+m^2)s^2 + (1+m^2)r^2 &= (1+m^2)a^2 + (1+m^2)b^2,\\ (1+m^2)(s^2+r^2) &= (1+m^2)(a^2+b^2)\\ s^2+r^2 = a^2+b^2, \end{align*}$$ showing that $(s,r)$ lies in a circle, namely $x^2+y^2 = a^2+b^2$. Taking the negative sign for the square root leads to the same equation. Finally, for the vertical and horizontal tangents, these occur at $x=\pm a$; the horizontal tangents are $y=\pm b$. Their intersections occur at $(\pm a,\pm b)$, which lie on the circle given above. So the locus of such points is contained in the circle $x^2+y^2 = a^2+b^2$. Conversely, consider a point $(r,s)$ that lies on $x^2+y^2 = a^2+b^2$. If a tangent to the ellipse $$ y = mx + \sqrt{a^2m^2 + b^2}$$ goes through $(r,s)$, then we have $$ s-mr = \sqrt{a^2m^2+b^2}.$$ Squaring both sides, we have $$s^2 - 2msr + m^2r^2 = a^2m^2 + b^2$$ or $$(a^2-r^2)m^2 +2srm + (b^2-s^2) = 0.$$ Since $r^s+s^2 = a^2+b^2$, then $a^2 - r^2 = s^2-b^2$, we we have $$(s^2-b^2)m^2 + 2srm + (b^2-s^2) = 0,$$ and if we do not have $s=\pm b$ (the horizontal/vertical tangent intersection points), then we get $$m^2 + tm - 1 = 0,\qquad\text{where } t = \frac{2sr}{s^2-b^2}.$$ So the two solutions for $m$, $m_1$ and $m_2$, satisfy $m_1m_2 = -1$, hence the two tangents are perpendicular. That is, at every point on the (upper half of the) circle, the two lines through the point that are tangent to the ellipse are perpendicular to each other. So all such points are on the circle, and all points on the circle are such points. (The circle is called the director circle of the ellipse).

Using binomial expansion to derive Pascal's rule $\displaystyle \binom{n}{k}=\binom{n-1}{k} + \binom{n-1}{k-1}$ $\displaystyle \left(1+x\right)^{n} = \left(1+x\right)\left(1+x\right)^{n-1}$ How do I use binomial expansion on the second equations for the right hand side and convert it to the first equation? The left hand side is obvious, but I'm not sure how to do the right hand side. Please give me some hints thanks

Binomial expansion of both sides of $$\left(1+x\right)^{n} = \left(1+x\right)\left(1+x\right)^{n-1}$$ gives $$\sum_{k=0}^n \binom{n}{k} x^k = \left(1+x\right)\sum_{k=0}^{n-1} \binom{n-1}{k} x^k$$ by distributivity on the right hand side we find $$\left(\sum_{k=0}^{n-1} \binom{n-1}{k} x^k \right)+\left(\sum_{k=0}^{n-1} \binom{n-1}{k} x^{k+1} \right) = \left(\sum_{k=0}^{n} \binom{n-1}{k} x^k \right)+\left(\sum_{k=0}^{n} \binom{n-1}{k-1} x^{k}\right)$$ the limits of the summations do not change the sum because $\binom{n-1}{n} = 0$, $\binom{-1}{n} = 0$. Thus we have $$\sum_{k=0}^n \binom{n}{k} x^k = \sum_{k=0}^{n} \left(\binom{n-1}{k} + \binom{n-1}{k-1}\right) x^k$$ and extracting the $x^k$ coefficients from both sides gives the identity $$\displaystyle \binom{n}{k}=\binom{n-1}{k} + \binom{n-1}{k-1}.$$

How do get combination sequence formula? What would be a closed-form formula that would determine the ith value of the sequence 1, 3, 11, 43, 171... where each value is one minus the product of the previous value and 4? Thanks!

The sequence can be written using the recurrence formula: $y_n = a+by_{n-1}$ . Then using the first 3 terms, one gets $y_1=1, a=-1, b=4 $ Sometimes it is easy to convert a recurrence formula to a closed-form, by studding the structure of the math relations. $y_1 = 1$ $y_2 = a+by_1$ $y_3 = a+by_2 = a+b(a+by_1) = a+ab+b^2y_1$ $y_4 = a+by_3 = a+b(a+by_2) = a+ab+ab^2+b^3y_1$ $y_5 = a+ab+ab^2+ab^3+b^4y_1$ there is a geometric sequence $a+ab+ab^2.. = a(1+b+b^2...) $ $ (1+b+b^2...b^n) = (1-b^{n+1}) / (1-b) $ therefore the general formula is: $y_n = a+ab+ab^2+...+ab^{n-2}+b^{n-1}y_1$ $y_n = a(1-b^{n-1}) / (1-b) + b^{n-1}y_1 $ Using the above parameters, $y_1=1 , a=-1, b=4, $ $y_n = -1(1-4^{n-1})/(1-4) +4^{n-1}y_1$ $y_n = -\frac{4^{n-1}}{3} +\frac13 +4^{n-1}y_1$ $y_n = \frac{2}{3}4^{n-1} + \frac{1}{3} $

Primitive polynomials of finite fields there are two primitive polynomials which I can use to construct $GF(2^3)=GF(8)$: $p_1(x) = x^3+x+1$ $p_2(x) = x^3+x^2+1$ $GF(8)$ created with $p_1(x)$: 0 1 $\alpha$ $\alpha^2$ $\alpha^3 = \alpha + 1$ $\alpha^4 = \alpha^3 \cdot \alpha=(\alpha+1) \cdot \alpha=\alpha^2+\alpha$ $\alpha^5 = \alpha^4 \cdot \alpha = (\alpha^2+\alpha) \cdot \alpha=\alpha^3 + \alpha^2 = \alpha^2 + \alpha + 1$ $\alpha^6 = \alpha^5 \cdot \alpha=(\alpha^2+\alpha+1) \cdot \alpha=\alpha^3+\alpha^2+\alpha=\alpha+1+\alpha^2+\alpha=\alpha^2+1$ $GF(8)$ created with $p_2(x)$: 0 1 $\alpha$ $\alpha^2$ $\alpha^3=\alpha^2+1$ $\alpha^4=\alpha \cdot \alpha^3=\alpha \cdot (\alpha^2+1)=\alpha^3+\alpha=\alpha^2+\alpha+1$ $\alpha^5=\alpha \cdot \alpha^4=\alpha \cdot(\alpha^2+\alpha+1) \cdot \alpha=\alpha^3+\alpha^2+\alpha=\alpha^2+1+\alpha^2+\alpha=\alpha+1$ $\alpha^6=\alpha \cdot (\alpha+1)=\alpha^2+\alpha$ So now let's say I want to add $\alpha^2 + \alpha^3$ in both fields. In field 1 I get $\alpha^2 + \alpha + 1$ and in field 2 I get $1$. Multiplication is the same in both fields ($\alpha^i \cdot \alpha^j = \alpha^{i+j\bmod(q-1)}$. So does it work so, that when some $GF(q)$ is constructed with different primitive polynomials then addition tables will vary and multiplication tables will be the same? Or maybe one of presented polynomials ($p_1(x), p_2(x)$) is not valid to construct field (altough both are primitive)?

The generator $\alpha$ for your field with the first description cannot be equal to the generator $\beta$ for your field with the second description. An isomorphism between $\mathbb{F}_2(\alpha)$ and $\mathbb{F}_2(\beta)$ is given by taking $\alpha \mapsto \beta + 1$; you can check that $\beta + 1$ satisfies $p_1$ iff $\beta$ satisfies $p_2$.

How do I find roots of a single-variate polynomials whose integers coefficients are symmetric wrt their respective powers Given a polynomial such as $X^4 + 4X^3 + 6X^2 + 4X + 1,$ where the coefficients are symmetrical, I know there's a trick to quickly find the zeros. Could someone please refresh my memory?

Hint: This particular polynomial is very nice, and factors as $(X+1)^4$. Take a look at Pascal's Triangle and the Binomial Theorem for more details. Added: Overly complicated formula The particular quartic you asked about had a nice solution, but lets find all the roots of the more general $$ax^{4}+bx^{3}+cx^{2}+bx+a.$$ Since $0$ is not a root, we are equivalently finding the zeros of $$ax^{2}+bx^{1}+c+bx^{-1}+ax^{-2}.$$Let $z=x+\frac{1}{x}$ (as suggested by Aryabhatta) Then $z^{2}=x^{2}+2+x^{-2}$ so that $$ax^{2}+bx^{1}+c+bx^{-1}+ax^{-2}=az^{2}+bz+\left(c-2a\right).$$ The roots of this are given by the quadratic formula: $$\frac{-b+\sqrt{b^{2}-4a\left(c-2a\right)}}{2a},\ \frac{-b-\sqrt{b^{2}-4a\left(c-2a\right)}}{2a}.$$ Now, we then have $$x+\frac{1}{x}=\frac{-b\pm\sqrt{b^{2}-4a\left(c-2a\right)}}{2a}$$ and hence we have the two quadratics $$x^{2}+\frac{b+\sqrt{b^{2}-4a\left(c-2a\right)}}{2a}x+1=0,$$ $$x^{2}+\frac{b-\sqrt{b^{2}-4a\left(c-2a\right)}}{2a}x+1=0.$$ This then gives the four roots:$$\frac{-b+\sqrt{b^{2}-4a\left(c-2a\right)}}{4a}\pm\sqrt{\frac{1}{4}\left(\frac{b-\sqrt{b^{2}-4a\left(c-2a\right)}}{2a}\right)^2-1}$$ $$\frac{-b-\sqrt{b^{2}-4a\left(c-2a\right)}}{4a}\pm\sqrt{\frac{1}{4}\left(\frac{b+\sqrt{b^{2}-4a\left(c-2a\right)}}{2a}\right)^2-1}.$$ If we plug in $a=1$, $b=4$, $c=6$, we find that all four of these are exactly $1$, so our particular case does work out.

Evaluate $\sum\limits_{k=1}^n k^2$ and $\sum\limits_{k=1}^n k(k+1)$ combinatorially $$\text{Evaluate } \sum_{k=1}^n k^2 \text{ and } \sum_{k=1}^{n}k(k+1) \text{ combinatorially.}$$ For the first one, I was able to express $k^2$ in terms of the binomial coefficients by considering a set $X$ of cardinality $2k$ and partitioning it into two subsets $A$ and $B$, each with cardinality $k$. Then, the number of ways of choosing 2-element subsets of $X$ is $$\binom{2k}{2} = 2\binom{k}{2}+k^2$$ So sum $$\sum_{k=1}^n k^2 =\sum_{k=1}^n \binom{2k}{2} -2\sum_{k=2}^n \binom{k}{2} $$ $$ \qquad\qquad = \color{red}{\sum_{k=1}^n \binom{2k}{2}} - 2 \binom{n+1}{3} $$ I am stuck at this point to evaluate the first of the sums. How to evaluate it? I need to find a similar expression for $k(k+1)$ for the second sum highlighted above. I have been unsuccessful this far. (If the previous problem is done then so is this, but it would be nice to know if there are better approaches or identities that can be used.) Update: I got the second one. Consider $$\displaystyle \binom{n+1}{r+1} = \binom{n}{r}+\binom{n-1}{r}+\cdots + \binom{r}{r}$$ Can be shown using recursive definition. Now multiply by $r!$ and set $r=2$

For the first one, $\displaystyle \sum_{k=1}^{n} k^2$, you can probably try this way. $$k^2 = \binom{k}{1} + 2 \binom{k}{2}$$ This can be proved using combinatorial argument by looking at drawing $2$ balls from $k$ balls with replacement. The total number of ways to do this is $k^2$. The other way to count it is as follows. There are two possible options either you draw the same ball on both trials or you draw different balls on both trials. The number of ways for the first option is $\binom{k}{1}$ and the number of ways for the second option is $\binom{k}{2} \times \left( 2! \right)$ Hence, we have that $$k^2 = \binom{k}{1} + 2 \binom{k}{2}$$ $$\displaystyle\sum_{k=1}^{n} k^2 = \sum_{k=1}^{n} \binom{k}{1} + 2 \sum_{k=1}^{n} \binom{k}{2} $$ The standard combinatorial arguments for $\displaystyle\sum_{k=1}^{n} \binom{k}{1}$ and $\displaystyle\sum_{k=1}^{n} \binom{k}{2}$ gives us $\displaystyle \binom{n+1}{2}$ and $\displaystyle \binom{n+1}{3}$ respectively. Hence, $$ \sum_{k=1}^{n} k^2 = \binom{n+1}{2} + 2 \binom{n+1}{3}$$ For the second case, it is much easier than the first case and in fact this suggests another method for the first case. $k(k+1)$ is the total number of ways of drawing 2 balls from $k+1$ balls without replacement where the order is important. This is same as $\binom{k+1}{2} \times \left(2! \right)$ Hence, $$\sum_{k=1}^{n} k(k+1) = 2 \sum_{k=1}^{n} \binom{k+1}{2} = 2 \times \binom{n+2}{3}$$ This suggests a method for the previous problem since $k^2 = \binom{k+1}{2} \times \left(2! \right) - \binom{k}{1}$ (It is easy to give a combinatorial argument for this by looking at drawing two balls from $k+1$ balls without replacement but hide one of the balls during the first draw and add the ball during the second draw) and hence $$\sum_{k=1}^{n} k^2 = 2 \times \binom{n+2}{3} - \binom{n+1}{2} $$

Proving $\frac{1}{\sin^{2}\frac{\pi}{14}} + \frac{1}{\sin^{2}\frac{3\pi}{14}} + \frac{1}{\sin^{2}\frac{5\pi}{14}} = 24$ How do I show that: $$\frac{1}{\sin^{2}\frac{\pi}{14}} + \frac{1}{\sin^{2}\frac{3\pi}{14}} + \frac{1}{\sin^{2}\frac{5\pi}{14}} = 24$$ This is actually problem B $4371$ given at this link. Looks like a very interesting problem. My attempts: Well, I have been thinking about this for the whole day, and I have got some insights. I don't believe my insights will lead me to a $\text{complete}$ solution. * *First, I wrote $\sin\frac{5\pi}{14}$ as $\sin\frac{9 \pi}{14}$ so that if I put $A = \frac{\pi}{14}$ so that the given equation becomes, $$\frac{1}{\sin^{2}{A}} + \frac{1}{\sin^{2}{3A}} + \frac{1}{\sin^{2}{9A}} =24$$ Then I tried working with this by taking $\text{lcm}$ and multiplying and doing something, which appeared futile. *Next, I actually didn't work it out, but I think we have to look for a equation which has roots as $\sin$ and then use $\text{sum of roots}$ formulas to get $24$. I think I haven't explained this clearly. * *$\text{Thirdly, is there a trick proving such type of identities using Gauss sums ?}$ One post related to this is: How to prove that: $\tan(3\pi/11) + 4\sin(2\pi/11) = \sqrt{11}$ I don't know how this will help as I haven't studied anything yet regarding Gauss sums.

Use $\sin(x) = \cos(\frac{\pi}2 - x)$, we can rewrite this as: $$\frac{1}{\cos^2 \frac{3\pi}{7}} + \frac{1}{\cos^2 \frac{2\pi}{7}} + \frac{1}{\cos^2 \frac{\pi}{7}}$$ Let $a_k = \frac{1}{\cos \frac{k\pi}7}$. Let $f(x) = (x-a_1)(x-a_2)(x-a_3)(x-a_4)(x-a_5)(x-a_6)$. Now, using that $a_k = - a_{7-k}$, this can be written as: $$f(x) = (x^2-a_1^2)(x^2-a_2^2)(x^2-a_3^2)$$ Now, our problem is to find the sum $a_1^2 + a_2^2 + a_3^2$, which is just the negative of the coefficient of $x^4$ in the polynomial $f(x)$. Let $U_6(x)$ be the Chebyshev polynomial of the second kind - that is: $$U_6(\cos \theta) = \frac{\sin 7\theta }{\sin \theta}$$ It is a polynomial of degree $6$ with roots equal to $\cos(\frac{k\pi}7)$, for $k=1,...,6$. So the polynomials $f(x)$ and $x^6U_6(1/x)$ have the same roots, so: $$f(x) = C x^6 U_6(\frac{1}x)$$ for some constant $C$. But $U_6(x) = 64x^6-80x^4+24x^2-1$, so $x^6 U_6(\frac{1}x) = -x^6 + 24 x^4 - 80x^2 + 64$. Since the coefficient of $x^6$ is $-1$, and it is $1$ in $f(x)$, $C=-1.$ So: $$f(x) = x^6 - 24x^4 +80x^2 - 64$$ In particular, the sum you are looking for is $24$. In general, if $n$ is odd, then the sum: $$\sum_{k=1}^{\frac{n-1}2} \frac{1}{\cos^2 \frac{k\pi}{n}}$$ is the absolute value of the coefficient of $x^2$ in the polynomial $U_{n-1}(x)$, which turns out to have closed form $\frac{n^2-1}2$.

Sum of First $n$ Squares Equals $\frac{n(n+1)(2n+1)}{6}$ I am just starting into calculus and I have a question about the following statement I encountered while learning about definite integrals: $$\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}$$ I really have no idea why this statement is true. Can someone please explain why this is true and if possible show how to arrive at one given the other?

Notice that $(k+1)^3 - k^3 = 3k^2 + 3k + 1$ and hence $$(n+1)^3 = \sum_{k=0}^n \left[ (k+1)^3 - k^3\right] = 3\sum_{k=0}^n k^2 + 3\sum_{k=0}^n k + \sum_{k=0}^n 1$$ which gives you $$\begin{align} \sum_{k=1}^n k^2 & = \frac{1}{3}(n+1)^3 - \frac{1}{2}n(n+1) - \frac{1}{3}(n+1) \\ & = \frac{1}{6}(n+1) \left[ 2(n+1)^2 - 3n - 2\right] \\ & = \frac{1}{6}(n+1)(2n^2 +n) \\ & = \frac{1}{6}n(n+1)(2n+1) \end{align}$$

Problem finding zeros of complex polynomial I'm trying to solve this problem $$ z^2 + (\sqrt{3} + i)|z| \bar{z}^2 = 0 $$ So, I know $ |z^2| = |z|^2 = a^2 + b ^2 $ and $ \operatorname{Arg}(z^2) = 2 \operatorname{Arg} (z) - 2k \pi = 2 \arctan (\frac{b}{a} ) - 2 k\pi $ for a $ k \in \mathbb{Z} $. Regarding the other term, I know $ |(\sqrt{3} + i)|z| \bar{z}^2 | = |z|^3 |\sqrt{3} + i| = 2 |z|^3 = 2(a^2 + b^2)^{3/2} $ and because of de Moivre's theorem, I have $ \operatorname{Arg} [(\sqrt{3} + i ) |z|\bar{z}^2] = \frac{\pi}{6} + 2 \operatorname{Arg} (z) - 2Q\pi $. Using all of this I can rewrite the equation as follows $$\begin{align*} &|z|^2 \Bigl[ \cos (2 \operatorname{Arg} (z) - 2k \pi) + i \sin (2 \operatorname{Arg}(z) - 2k \pi)\Bigr]\\ &\qquad \mathop{+} 2|z|^3 \Biggl[\cos \left(\frac{\pi}{6} + 2 \operatorname{Arg} (z) -2Q\pi\right) + i \sin \left(\frac{\pi}{6} + 2 \operatorname{Arg} (z) -2Q\pi\right)\Biggr] = 0 \end{align*} $$ Which, assuming $ z \neq 0 $, can be simplified as $$\begin{align*} &\cos (2 \operatorname{Arg} (z) - 2k \pi) + i \sin (2 \operatorname{Arg} (z) - 2k \pi) \\ &\qquad\mathop{+} 2 |z|\Biggl[\cos \left(\frac{\pi}{6} + 2 \operatorname{Arg} (z) -2Q \pi \right) + i \sin \left(\frac{\pi}{6} + 2 \operatorname{Arg} (z) -2Q\pi\right)\Biggr] = 0 \end{align*} $$ Now, from this I'm not sure how to go on. I tried a few things that got me nowhere like trying to solve $$ \cos (2 \operatorname{Arg}(z) - 2k \pi) = 2 |z| \cos \left(\frac{\pi}{6} + 2 \operatorname{Arg} (z) -2Q\pi\right) $$ I'm really lost here, I don't know how to keep going and I've looked for error but can't find them. Any help would be greatly appreciated.

Here is an alternative to solving it using polar form. Let $z=a+bi$, so that $\bar{z}=a-bi$ and $|z|=\sqrt{a^2+b^2}$. Then you want to solve $$(a+bi)^2+(\sqrt{3}+i)\sqrt{a^2+b^2}(a-bi)^2=0,$$ which expands to $$(a^2-b^2)+2abi+(\sqrt{3}+i)\sqrt{a^2+b^2}\left((a^2-b^2)-2abi\right)=0$$ Thus, we need both the real part and the imaginary part of the left side to be 0, i.e. $$(a^2-b^2)+\sqrt{a^2+b^2}\left(\sqrt{3}\cdot (a^2-b^2)+2ab\right)=0$$ and $$2ab+\sqrt{a^2+b^2}\left(-2ab\sqrt{3}+(a^2-b^2)\right)=0.$$ It should be possible to solve these equations by simple manipulations, though I haven't worked it out myself yet.

Proving that $ 30 \mid ab(a^2+b^2)(a^2-b^2)$ How can I prove that $30 \mid ab(a^2+b^2)(a^2-b^2)$ without using $a,b$ congruent modulo $5$ and then $a,b$ congruent modulo $6$ (for example) to show respectively that $5 \mid ab(a^2+b^2)(a^2-b^2)$ and $6 \mid ab(a^2+b^2)(a^2-b^2)$? Indeed this method implies studying numerous congruences and is quite long.

You need to show $ab(a^2 - b^2)(a^2 + b^2)$ is a multiple of 2,3, and 5 for all $a$ and $b$. For 2: If neither $a$ nor $b$ are even, they are both odd and $a^2 \equiv b^2 \equiv 1 \pmod 2$, so that 2 divides $a^2 - b^2$. For 3: If neither $a$ nor $b$ are a multiple of 3, then $a^2 \equiv b^2 \equiv 1 \pmod 3$, so 3 divides $a^2 - b^2$ similar to above. For 5: If neither $a$ nor $b$ are a multiple of 5, then either $a^2 \equiv 1 \pmod 5$ or $a^2 \equiv -1 \pmod 5$. The same holds for $b$. If $a^2 \equiv b^2 \pmod 5$ then 5 divides $a^2 - b^2$, while if $a^2 \equiv -b^2 \pmod 5$ then 5 divides $a^2 + b^2$. This does break into cases, but as you can see it's not too bad to do it systematically like this.

Find the perimeter of any triangle given the three altitude lengths The altitude lengths are 12, 15 and 20. I would like a process rather than just a single solution.

First, $\begin{aligned}\operatorname{Area}(\triangle ABC)=\frac{ah_a}2=\frac{bh_b}2=\frac{ch_c}2\implies \frac1{h_a}&=\frac{a}{2\operatorname{Area}(\triangle ABC)}\\\frac1{h_b}&=\frac{b}{2\operatorname{Area}(\triangle ABC)}\\\frac1{h_c}&=\frac{c}{2\operatorname{Area}(\triangle ABC)} \end{aligned}$ By the already mentioned Heron's formula: $\begin{aligned}\operatorname{Area}(\triangle ABC)&=\sqrt{s(s-a)(s-b)(s-c)}=\sqrt{\frac{a+b+c}2\cdot\frac{b+c-a}2\cdot\frac{a+c-b}2\cdot\frac{a+b-c}2}\Bigg/:\operatorname{Area}^2(\triangle ABC)\\\frac1{\operatorname{Area}(\triangle ABC)}&=\sqrt{\frac{a+b+c}{2\operatorname{Area}(\triangle ABC)}\cdot\frac{b+c-a}{2\operatorname{Area}(\triangle ABC)}\cdot\frac{a+c-b}{2\operatorname{Area}(\triangle ABC)}\cdot\frac{a+b-c}{2\operatorname{Area}(\triangle ABC)}}\\\frac1{\operatorname{Area}(\triangle ABC)}&=\sqrt{\left(\frac1{h_a}+\frac1{h_b}+\frac1{h_c}\right)\left(\frac1{h_b}+\frac1{h_c}-\frac1{h_a}\right)\left(\frac1{h_a}+\frac1{h_c}-\frac1{h_b}\right)\left(\frac1{h_a}+\frac1{h_b}-\frac1{h_c}\right)}\end{aligned}$ $$\implies\operatorname{Area}(\triangle ABC)=\frac1{\sqrt{\left(\frac1{h_a}+\frac1{h_b}+\frac1{h_c}\right)\left(\frac1{h_b}+\frac1{h_c}-\frac1{h_a}\right)\left(\frac1{h_a}+\frac1{h_c}-\frac1{h_b}\right)\left(\frac1{h_a}+\frac1{h_b}-\frac1{h_c}\right)}}$$ Then, plugg the values $h_a=12,h_b=15,h_c=20$ into the formula.

Determining which values to use in place of x in functions When solving partial fractions for integrations, solving x for two terms usually isn't all that difficult, but I've been running into problems with three term integration. For example, given $$\int\frac{x^2+3x-4}{x^3-4x^2+4x}$$ The denominator factored out to $x(x-2)^2$, which resulted in the following formulas $$ \begin{align*} x^2+3x-4=\frac{A}{x}+\frac{B}{x-2}+\frac{C}{(x-2)^2}\\ x^2+3x-4= A(x-2)(x-2)^2+Bx(x-2)^2+Cx(x-2)\\ x^2+3x-4= A(x-2)^2+Bx(x-2)+Cx\\\\ \text{when x=0, }A=-1 \text{ and x=2, }C=3 \end{align*} $$ This is where I get stuck, since nothing immediately pops out at me for values that would solve A and C for zero and leave some value for B. How do I find the x-value for a constant that is not immediately apparent?

To find the constants in the rational fraction $$\frac{x^2+3x-4}{x^3-4x^2+4x}=\frac{A}{x}+\frac{B}{x-2}+\frac{C}{(x-2)^2},$$ you may use any set of 3 values of $x$, provided that the denominator $x^3-4x^2+4x\ne 0$. The "standard" method is to compare the coefficients of $$\frac{x^2+3x-4}{x^3-4x^2+4x}=\frac{A}{x}+\frac{B}{x-2}+\frac{C}{(x-2)^2},$$ after multiplying this rational fraction by the denominator $x^3-4x^2+4x=x(x-2)^2$ and solve the resulting linear system in $A,B,C$. Since $$\begin{eqnarray*} \frac{x^{2}+3x-4}{x\left( x-2\right) ^{2}} &=&\frac{A}{x}+\frac{B}{x-2}+% \frac{C}{\left( x-2\right) ^{2}} \\ &=&\frac{A\left( x-2\right) ^{2}}{x\left( x-2\right) ^{2}}+\frac{Bx\left( x-2\right) }{x\left( x-2\right) ^{2}}+\frac{Cx}{x\left( x-2\right) ^{2}} \\ &=&\frac{A\left( x-2\right) ^{2}+Bx\left( x-2\right) +Cx}{x\left( x-2\right) ^{2}} \\ &=&\frac{\left( A+B\right) x^{2}+\left( -4A-2B+C\right) x+4A}{x\left( x-2\right) ^{2}}, \end{eqnarray*}$$ if we equate the coefficients of the plynomials $$x^{2}+3x-4\equiv\left( A+B\right) x^{2}+\left( -4A-2B+C\right) x+4A,$$ we have the system $$\begin{eqnarray*} A+B &=&1 \\ -4A-2B+C &=&3 \\ 4A &=&-4, \end{eqnarray*}$$ whose solution is $$\begin{eqnarray*} B &=&2 \\ C &=&3 \\ A &=&-1. \end{eqnarray*}$$ Alternatively you could use the method indicated in parts A and B, as an example. A. We can multiply $f(x)$ by $x=x-0$ and $\left( x-2\right) ^{2}$ and let $% x\rightarrow 0$ and $x\rightarrow 2$. Since $$\begin{eqnarray*} f(x) &=&\frac{P(x)}{Q(x)}=\frac{x^{2}+3x-4}{x^{3}-4x^{2}+4x} \\ &=&\frac{x^{2}+3x-4}{x\left( x-2\right) ^{2}} \\ &=&\frac{A}{x}+\frac{B}{x-2}+\frac{C}{\left( x-2\right) ^{2}},\qquad (\ast ) \end{eqnarray*}$$ if we multiply $f(x)$ by $x$ and let $x\rightarrow 0$, we find $A$: $$A=\lim_{x\rightarrow 0}xf(x)=\lim_{x\rightarrow 0}\frac{x^{2}+3x-4}{\left( x-2\right) ^{2}}=\frac{-4}{4}=-1.$$ And we find $C$, if we multiply $f(x)$ by $\left( x-2\right) ^{2}$ and let $x\rightarrow 2$: $$C=\lim_{x\rightarrow 2}\left( x-2\right) ^{2}f(x)=\lim_{x\rightarrow 2}\frac{% x^{2}+3x-4}{x}=\frac{2^{2}+6-4}{2}=3.$$ B. Now observing that $$P(x)=x^{2}+3x-4=\left( x+4\right) \left( x-1\right)$$ we can find $B$ by making $x=1$ and evaluate $f(1)$ in both sides of $(\ast)$, with $A=-1,C=3$: $$f(1)=0=2-B.$$ So $B=2$. Or we could make $x=-4$ in $(\ast)$ $$f(-4)=0=\frac{1}{3}-\frac{1}{6}B.$$ We do obtain $B=2$. Thus $$\frac{x^{2}+3x-4}{x\left( x-2\right) ^{2}}=-\frac{1}{x}+\frac{2}{x-2}+\frac{3% }{\left( x-2\right) ^{2}}\qquad (\ast \ast )$$ Remark: If the denominator has complex roots, then an expansion as above is not possible. For instance $$\frac{x+2}{x^{3}-1}=\frac{x+2}{(x-1)(x^{2}+x+1)}=\frac{A}{x-1}+\frac{Bx+C}{% x^{2}+x+1}.$$ You should find $A=1,B=C=-1$: $$\frac{x+2}{x^{3}-1}=\frac{1}{x-1}-\frac{x+1}{x^{2}+x+1}.$$

Nested Square Roots $5^0+\sqrt{5^1+\sqrt{5^2+\sqrt{5^4+\sqrt\dots}}}$ How would one go about computing the value of $X$, where $X=5^0+ \sqrt{5^1+\sqrt{5^2+\sqrt{5^4+\sqrt{5^8+\sqrt{5^{16}+\sqrt{5^{32}+\dots}}}}}}$ I have tried the standard way of squaring then trying some trick, but nothing is working. I have also looked at some nested radical previous results, but none seem to be of the variety of this problem. Can anyone come up with the answer? Thank

The trick is to pull out a $\sqrt{5}$ factor from the second term: $$ \frac{\sqrt{5^1+ \sqrt{5^2 + \sqrt{5^4 + \sqrt{5^8 + \cdots}}}}}{\sqrt{5}} = \sqrt{1 + \sqrt{1 + \sqrt{1 + \sqrt{1 + \cdots}}}}, $$ which I call $Y$ for convenience. To see why this is true, observe that $$ \begin{align*} \frac{\sqrt{5^1+x}}{\sqrt{5}} = \sqrt{1+\frac{x}{5^1}} \\ \frac{\sqrt{5^2+x}}{5^1} = \sqrt{1+\frac{x}{5^2}} \\ \frac{\sqrt{5^4+x}}{5^2} = \sqrt{1+\frac{x}{5^4}} \end{align*} $$ and so on. Applying these repeatedly inside the nested radicals gives the claim. (The wording of this explanation has been inspired mostly by a comment of @J.M. below.) Now, it only remains to compute $Y$ and $X$. By squaring, we get $$ Y^2 = 1 + Y \ \ \ \ \implies Y = \frac{\sqrt{5}+1}{2}, $$ discarding the negative root. Plugging this value in the definition of $X$, we get: $$ X = 1 + \sqrt{5} Y = 1 + \frac{5+\sqrt{5}}{2} = \frac{7+\sqrt{5}}{2} . $$ Note on the convergence issues. As @GEdgar points out, to complete the proof, I also need to demonstrate that both sides of the first equation do converge to some finite limits. For our expressions, convergence follows from @Bill Dubuque's answer to my question on the defining the convergence of such an expression. I believe that with some work, one can also give a direct proof by showing that this sequence is bounded from above (which I hope will also end up showing the theorem Bill quotes), but I will not pursue this further. Added: See @Aryabhata's answer to a related question for a hands-on proof.

Canonical to Parametric, Ellipse Equation I've done some algebra tricks in this derivation and I'm not sure if it's okay to do those things. $$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$ $$\frac{x^2}{a^2} + \frac{y^2}{b^2} = \cos^2\theta + \sin^2\theta$$ Can I really do this next step? $$\frac{x^2}{a^2} = \cos^2\theta\quad\text{and}\quad\frac{y^2}{b^2} = \sin^2\theta$$ $$x^2 = a^2\cos^2\theta\quad\text{and}\quad y^2 = b^2\sin^2\theta$$ Ignoring the negative numbers: $$x = a\cos\theta\quad\text{and}\quad y = b\sin\theta$$

The idea behind your argument is absolutely fine. Any two non-negative numbers $u$ and $v$ such that $u+v=1$ can be expressed as $u=\cos^2\theta$, $v=\sin^2\theta$ for some $\theta$. This is so obvious that it probably does not require proof. Set $u=\cos^2\theta$. Then $v=1-\cos^2\theta=\sin^2\theta$. The second displayed formula muddies things somewhat. You intended to say that if $x^2/a^2+y^2/b^2=1$, then there exists a $\theta$ such that $x^2/a^2=\cos^2\theta$ and $y^2/b^2=\sin^2\theta$. You did not mean that for any $\theta$, if $x^2/a^2+y^2/b^2=1$ then $x^2/a^2=\cos^2\theta$! But the transition from the second displayed equation to the third could be interpreted as asserting what you clearly did not intend to say. It would be better to do exactly what you did, but to use more geometric language, as follows. $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \quad\text{iff}\quad \left(\frac{x}{a}\right)^2 + \left(\frac{y}{b}\right)^2=1.$$ But the equation on the right holds iff the point $(x/a, y/b)$ lies on the unit circle. The points on the unit circle are parametrized by $(\cos \theta,\sin\theta)$, with $\theta$ ranging over $[0,2\pi)$, so the points on our ellipse are given by $x=a\cos\theta$, $y=a\sin\theta$.

Does the Schur complement preserve the partial order? Let $$\begin{bmatrix} A_{1} &B_1 \\ B_1' &C_1 \end{bmatrix} \quad \text{and} \quad \begin{bmatrix} A_2 &B_2 \\ B_2' &C_2 \end{bmatrix}$$ be symmetric positive definite and conformably partitioned matrices. If $$\begin{bmatrix} A_{1} &B_1 \\ B_1' &C_1 \end{bmatrix}-\begin{bmatrix} A_2 &B_2 \\ B_2' &C_2 \end{bmatrix}$$ is positive semidefinite, is it true $$(A_1-B_1C^{-1}_1B_1')-(A_2-B_2C^{-1}_2B_2')$$ also positive semidefinite? Here, $X'$ means the transpose of $X$.

For a general block matrix $X=\begin{pmatrix}A&B\\C&D\end{pmatrix}$, the Schur complement $S$ to the block $D$ satisfies $$ \begin{pmatrix}A&B\\C&D\end{pmatrix} =\begin{pmatrix}I&BD^{-1}\\&I\end{pmatrix} \begin{pmatrix}S\\&D\end{pmatrix} \begin{pmatrix}I\\D^{-1}C&I\end{pmatrix}. $$ So, when $X$ is Hermitian, $$ \begin{pmatrix}A&B\\B^\ast&D\end{pmatrix} =\begin{pmatrix}I&Y^\ast\\&I\end{pmatrix} \begin{pmatrix}S\\&D\end{pmatrix} \begin{pmatrix}I\\Y&I\end{pmatrix}\ \textrm{ for some } Y. $$ Hence $$ \begin{eqnarray} &&\begin{pmatrix}A_1&B_1\\B_1^\ast&D_1\end{pmatrix} \ge\begin{pmatrix}A_2&B_2\\B_2^\ast&D_2\end{pmatrix} \\ &\Rightarrow& \begin{pmatrix}S_1\\&D_1\end{pmatrix} \ge \begin{pmatrix}I&Z^\ast\\&I\end{pmatrix} \begin{pmatrix}S_2\\&D_2\end{pmatrix} \begin{pmatrix}I\\Z&I\end{pmatrix}\ \textrm{ for some } Z\\ &\Rightarrow& (x^\ast,0)\begin{pmatrix}S_1\\&D_1\end{pmatrix}\begin{pmatrix}x\\0\end{pmatrix} \ge (x^\ast,\ x^\ast Z^\ast) \begin{pmatrix}S_2\\&D_2\end{pmatrix} \begin{pmatrix}x\\Zx\end{pmatrix},\ \forall x\\ &\Rightarrow& x^\ast S_1 x \ \ge\ x^\ast S_2 x + (Zx)^\ast D_2 (Zx) \ \ge\ x^\ast S_2 x,\ \forall x\\ &\Rightarrow& S_1\ge S_2. \end{eqnarray} $$ Edit: In hindsight, this is essentially identical to alex's proof.

Proving $1^3+ 2^3 + \cdots + n^3 = \left(\frac{n(n+1)}{2}\right)^2$ using induction How can I prove that $$1^3+ 2^3 + \cdots + n^3 = \left(\frac{n(n+1)}{2}\right)^2$$ for all $n \in \mathbb{N}$? I am looking for a proof using mathematical induction. Thanks

Let the induction hypothesis be $$ (1^3+2^3+3^3+\cdots+n^3)=(1+2+3+\cdots+n)^2$$ Now consider: $$ (1+2+3+\cdots+n + (n+1))^2 $$ $$\begin{align} & = \color{red}{(1+2+3+\cdots+n)^2} + (n+1)^2 + 2(n+1)\color{blue}{(1+2+3+\cdots+n)}\\ & = \color{red}{(1^3+2^3+3^3+\cdots+n^3)} + (n+1)^2 + 2(n+1)\color{blue}{(n(n+1)/2)}\\ & = (1^3+2^3+3^3+\cdots+n^3) + \color{teal}{(n+1)^2 + n(n+1)^2}\\ & = (1^3+2^3+3^3+\cdots+n^3) + \color{teal}{(n+1)^3} \end {align}$$ QED

Solve to find the unknown I have been doing questions from the past year and I come across this question which stumped me: The constant term in the expansion of $\left(\frac1{x^2}+ax\right)^6$ is $1215$; find the value of $a$. (The given answer is: $\pm 3$ ) Should be an easy one, but I don't know how to begin. Some help please?

Sofia,Sorry for the delayed response.I was busy with other posts. you have two choices.One is to use pascals triangle and the other one is to expand using the binimial theorem. You can compare the expression $$\left ( \frac{1}{x^2} + ax \right )^6$$ with $$(a+x)^6$$ where a = 1/x^2 and x = ax,n=6.Here'e the pascal triangle way of expanding the given expression.All you need to do is to substitute the values of a and x respectively. $$(a + x)^0 = 1$$ $$(a + x)^1 = a +x a+ x$$ $$(a + x)^2 = (a + x)(a + x) = a^2 + 2ax + x^2$$ $$(a + x)^3 = (a + x)^2(a + x) = a^3 + 3a^2x + 3ax^2 + x^3$$ $$(a + x)^4 = (a + x)^3(a + x) = a^4 + 4a^3x + 6a^2x^2 + 4ax^3 + x^4$$ $$(a + x)^5 = (a + x)^4(a + x) = a^5 + 5a^4x + 10a^3x^2 + 10a^2x^3 + 5ax^4 + x^5$$ $$(a + x)^6 = (a + x)^5(a + x) = a^6 + 6a^5x + 15a^4x^2 + 20a^3x^3 + 15a^2x^4 + 6ax^5 + x^6$$ Here'e the Binomial theorem way of expanding it out. $$(1+x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3 + ...$$ using the above theorem you should get $$a^6x^6 + 6a^5x^3 + 15a^4 + \frac{20a^3}{x^3} + \frac{15a^2}{x^6}+\frac{6a}{x^9}+\frac{1}{x^{12}}$$ You can now substitute the constant term and get the desired answer

On factorizing and solving the polynomial: $x^{101} – 3x^{100} + 2x^{99} + x^{98} – 3x^{97} + 2x^{96} + \cdots + x^2 – 3x + 2 = 0$ The actual problem is to find the product of all the real roots of this equation,I am stuck with his factorization: $$x^{101} – 3x^{100} + 2x^{99} + x^{98} – 3x^{97} + 2x^{96} + \cdots + x^2 – 3x + 2 = 0$$ By just guessing I noticed that $(x^2 – 3x + 2)$ is one factor and then dividing that whole thing we get $(x^{99}+x^{96}+x^{93} + \cdots + 1)$ as the other factor , but I really don't know how to solve in those where wild guessing won't work! Do we have any trick for factorizing this kind of big polynomial? Also I am not sure how to find the roots of $(x^{99}+x^{96}+x^{93} + \cdots + 1)=0$,so any help in this regard will be appreciated.

In regard to the first part of your question ("wild guessing"), the point was to note that the polynomial can be expressed as the sum of three polynomials, grouping same coefficients: $$ P(x)= x^{101} – 3x^{100} + 2x^{99} + x^{98} – 3x^{97} + 2x^{96} + \cdots + x^2 – 3x + 2 = A(x)+B(x)+C(x)$$ with $$\begin{eqnarray} A(x) &= x^{101} + x^{98} + \cdots + x^2 &= x^2 (x^{99} + x^{96} + \cdots + 1) \\ B(x) &= - 3 x^{100} -3 x^{97} - \cdots -3 x &= - 3 x (x^{99} + x^{96} + \cdots + 1)\\ C(x) &= 2 x^{99} + 2 x^{96} + \cdots + 2 &= 2 (x^{99} + x^{96} + \cdots + 1) \\ \end{eqnarray} $$ so $$P(x) = (x^2 - 3x +2) (x^{99} + x^{96} + \cdots + 1) $$ and applying the geometric finite sum formula: $$P(x)=(x^2 - 3x +2) ({(x^{3})}^{33} + {(x^{3})}^{32} + \cdots + 1) = (x^2 - 3x +2) \frac{x^{102}-1}{x^3-1} $$ As Andre notes in the comments, your "guessing" was dictated by the very particular structure of the polynomial, you can't hope for some general guessing recipe...

Solving for Inequality $\frac{12}{2x-3}<1+2x$ I am trying to solve for the following inequality: $$\frac{12}{2x-3}<1+2x$$ In the given answer, $$\frac{12}{2x-3}-(1+2x)<0$$ $$\frac{-(2x+3)(2x-5)}{2x-3}<0 \rightarrow \textrm{ How do I get to this step?}$$ $$\frac{(2x+3)(2x-5)}{2x-3}>0$$ $$(2x+3)(2x-5)(2x-3)>0 \textrm{ via multiply both sides by }(2x-3)^2$$

$$ \frac{12}{2x-3} - (1-2x) = \frac{12 - (1+2x)(2x-3) }{2x-3} = \frac{ 12 - (2x-3+4x^2-6x)}{2x-3} $$ $$= - \frac{4x^2-4x-15}{2x-3} = - \frac{(2x+3)(2x-5)}{2x-3} $$

How can the following be calculated? How can the following series be calculated? $$S=1+(1+2)+(1+2+3)+(1+2+3+4)+\cdots+(1+2+3+4+\cdots+2011)$$

Let $S$ be our sum. Then $$S=\binom{2}{2}+\binom{3}{2}+\binom{4}{2} + \cdots + \binom{2012}{2}=\binom{2013}{3}=\frac{2013\cdot 2012\cdot 2011}{3 \cdot 2 \cdot 1}.$$ Justification: We count, in two different ways, the number of ways to choose $3$ numbers from the set $$\{1,2,3,4,\dots, n,n+1\}.$$ (For our particular problem we use $n=2012$.) First Count: It is clear that there are $\binom{n+1}{3}$ ways to choose $3$ numbers from $n+1$ numbers. Second Count: The smallest chosen number could be $1$. Then there are $\binom{n}{2}$ ways to choose the remaining $2$ numbers. Or the smallest chosen number could be $2$, leaving $\binom{n-1}{2}$ choices for the remaining $2$ numbers. Or the smallest chosen number could be $3$, leaving $\binom{n-2}{2}$ choices for the remaining $2$ numbers. And so on, up to smallest chosen number being $n-1$, in which case there are $\binom{2}{2}$ ways to choose the remaining $2$ numbers. Thus the total count is $$\binom{n}{2}+\binom{n-1}{2}+\binom{n-2}{2}+\cdots +\binom{3}{2}+\binom{2}{2}.$$ Comparing the two counts, we find that $$\binom{2}{2}+\binom{3}{2}+\binom{4}{2}+\cdots +\binom{n-1}{2}+\binom{n}{2}=\binom{n+1}{3}.$$ Comment: Similarly, it is easy to see that in general $\sum_{k=r}^n \binom{k}{r}=\binom{n+1}{r+1}.$ These natural binomial coefficient identities give a combinatorial approach to finding general formulas for the sums of consecutive squares, consecutive cubes, and so on.

Combinatorics-number of permutations of $m$ A's and at most $n$ B's Prove that the number of permutations of $m$ A's and at most $n$ B's equals $\dbinom{m+n+1}{m+1}$. I'm not sure how to even start this problem.

By summing all possibilities of $n$, we get that the number of permutations $P_n$ satisfies $$P_n = \binom{m+n}{n} + \binom{m+(n-1)}{(n-1)} + \ldots + \binom{m+0}{0} = \sum_{i=0}^n \binom{m + i}{i}$$ Note that $$\binom{a}{b} = \binom{a-1}{b} + \binom{a-1}{b-1}$$ Repeatedly applying this to the last term, we get $$\begin{array}{rcl} \binom{a}{b} &=& \binom{a-1}{b} + \binom{a-1}{b-1} \\ &=& \binom{a-1}{b} + \binom{a-2}{b-1} + \binom{a-2}{b-2} \\ &=& \binom{a-1}{b} + \binom{a-2}{b-1} + \binom{a-3}{b-2} + \binom{a-3}{b-3} \\ &=& \binom{a-1}{b} + \binom{a-2}{b-1} + \binom{a-3}{b-2} + \binom{a-4}{b-3} + \ldots \\ &=& \binom{a-1}{b} + \binom{a-2}{b-1} + \binom{a-3}{b-2} + \binom{a-4}{b-3} + \ldots + \binom{a-b-1}{0} \\ &=& \sum_{i=0}^b \binom{a-b-1+i}{i} \end{array}$$ Substituting $b$ by $a-b$ we similarly get $$\binom{a}{a-b} = \sum_{i=0}^{a-b} \binom{b-1+i}{i}$$ Replacing $b = m + 1$ and $a = n + m + 1$ we thus get $$\binom{n + m + 1}{n} = \sum_{i=0}^{n} \binom{m+i}{i} = P_n$$

Mathematical reason for the validity of the equation: $S = 1 + x^2 \, S$ Given the geometric series: $1 + x^2 + x^4 + x^6 + x^8 + \cdots$ We can recast it as: $S = 1 + x^2 \, (1 + x^2 + x^4 + x^6 + x^8 + \cdots)$, where $S = 1 + x^2 + x^4 + x^6 + x^8 + \cdots$. This recasting is possible only because there is an infinite number of terms in $S$. Exactly how is this mathematically possible? (Related, but not identical, question: General question on relation between infinite series and complex numbers).

The $n$th partial sum of your series is $$ \begin{align*} S_n &= 1+x^2+x^4+\cdots +x^{2n}= 1+x^2(1+x^2+x^4+\cdots +x^{2n-2})\\ &= 1+x^2S_{n-1} \end{align*} $$ Assuming your series converges you get that $$ \lim_{n\to\infty}S_n=\lim_{n\to\infty}S_{n-1}=S. $$ Thus $S=1+x^2S$.

Lesser-known integration tricks I am currently studying for the GRE math subject test, which heavily tests calculus. I've reviewed most of the basic calculus techniques (integration by parts, trig substitutions, etc.) I am now looking for a list or reference for some lesser-known tricks or clever substitutions that are useful in integration. For example, I learned of this trick $$\int_a^b f(x) \, dx = \int_a^b f(a + b -x) \, dx$$ in the question Showing that $\int\limits_{-a}^a \frac{f(x)}{1+e^{x}} \mathrm dx = \int\limits_0^a f(x) \mathrm dx$, when $f$ is even I am especially interested in tricks that can be used without an excessive amount of computation, as I believe (or hope?) that these will be what is useful for the GRE.

When integrating rational functions by partial fractions decomposition, the trickiest type of antiderivative that one might need to compute is $$I_n = \int \frac{dx}{(1+x^2)^n}.$$ (Integrals involving more general quadratic factors can be reduced to such integrals, plus integrals of the much easier type $\int \frac{x \, dx}{(1+x^2)^n}$, with the help of substitutions of the form $x \mapsto x+a$ and $x \mapsto ax$.) For $n=1$, we know that $I_1 = \int \frac{dx}{1+x^2} = \arctan x + C$, and the usual suggestion for finding $I_n$ for $n \ge 2$ is to work one's way down to $I_1$ using the reduction formula $$ I_n = \frac{1}{2(n-1)} \left( \frac{x}{(1+x^2)^{n-1}} + (2n-3) \, I_{n-1} \right) . $$ However, this formula is not easy to remember, and the computations become quite tedious, so the lesser-known trick that I will describe here is (in my opinion) a much simpler way. From now on, I will use the abbreviation $$T=1+x^2.$$ First we compute $$ \frac{d}{dx} \left( x \cdot \frac{1}{T^n} \right) = 1 \cdot \frac{1}{T^n} + x \cdot \frac{-n}{T^{n+1}} \cdot 2x = \frac{1}{T^n} - \frac{2n x^2}{T^{n+1}} = \frac{1}{T^n} - \frac{2n (T-1)}{T^{n+1}} \\ = \frac{1}{T^n} - \frac{2n T}{T^{n+1}} + \frac{2n}{T^{n+1}} = \frac{2n}{T^{n+1}} - \frac{2n-1}{T^n} . $$ Let us record this result for future use, in the form of an integral: $$ \int \left( \frac{2n}{T^{n+1}} - \frac{2n-1}{T^n} \right) dx = \frac{x}{T^n} + C . $$ That is, we have $$ \begin{align} \int \left( \frac{2}{T^2} - \frac{1}{T^1} \right) dx &= \frac{x}{T} + C ,\\ \int \left( \frac{4}{T^3} - \frac{3}{T^2} \right) dx &= \frac{x}{T^2} + C ,\\ \int \left( \frac{6}{T^4} - \frac{5}{T^3} \right) dx &= \frac{x}{T^3} + C ,\\ &\vdots \end{align} $$ With the help of this, we can easily compute things like $$ \begin{align} \int \left( \frac{1}{T^3} + \frac{5}{T^2} - \frac{2}{T} \right) dx &= \int \left( \frac14 \left( \frac{4}{T^3} - \frac{3}{T^2} \right) + \frac{\frac34 + 5}{T^2} - \frac{2}{T} \right) dx \\ &= \int \left( \frac14 \left( \frac{4}{T^3} - \frac{3}{T^2} \right) + \frac{23}{4} \cdot \frac12 \left( \frac{2}{T^2} - \frac{1}{T^1} \right) + \frac{\frac{23}{8}-2}{T} \right) dx \\ &= \frac14 \frac{x}{T^2} + \frac{23}{8} \frac{x}{T} + \frac{7}{8} \arctan x + C . \end{align} $$ Of course, the relation that we are using, $2n \, I_{n+1} - (2n-1) \, I_n = \frac{x}{T^n}$, really is the reduction formula in disguise. However, the trick is: (a) to derive the formula just by differentiation (instead of starting with an integral where the exponent is one step lower than the one that we're interested in, inserting a factor of $1$, integrating by parts, and so on), and (b) to leave the formula in its "natural" form as it appears when differentiating (instead of solving for $I_{n+1}$ in terms of $I_n$), which results in a structure which is easier to remember and a more pleasant way of organizing the computations.

Proof of dividing fractions $\frac{a/b}{c/d}=\frac{ad}{bc}$ For dividing two fractional expressions, how does the division sign turns into multiplication? Is there a step by step proof which proves $$\frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \cdot \frac{d}{c}=\frac{ad}{bc}?$$

Suppose $\frac{a}{b}$ and $\frac{c}{d}$ are fractions. That is, $a$, $b$, $c$, $d$ are whole numbers and $b\ne0$, $d\ne0$. In addition we require $c\ne0$. Let $\frac{a}{b}\div\frac{c}{d}=A$. Then by definition of division of fractions , $A$ is a unique fraction so that $A\times\frac{c}{d}=\frac{a}{b}$. However, $(\frac{a}{b}\times\frac{d}{c})\times\frac{c}{d}=\frac{a}{b}\times(\frac{d}{c}\times\frac{c}{d})=\frac{a}{b}\times(\frac{dc}{cd})=\frac{a}{b}\times(\frac{dc}{dc})=\frac{a}{b}$. Then by uniqueness (of $A$), $A=\frac{a}{b}\times\frac{d}{c}$.

On the meaning of being algebraically closed The definition of algebraic number is that $\alpha$ is an algebraic number if there is a nonzero polynomial $p(x)$ in $\mathbb{Q}[x]$ such that $p(\alpha)=0$. By algebraic closure, every nonconstant polynomial with algebraic coefficients has algebraic roots; then, there will be also a nonconstant polynomial with rational coefficients that has those roots. I feel uncomfortable with the idea that the root of a polynomial with algebraic coefficients is again algebraic; why are we sure that for every polynomial in $\mathbb{\bar{Q}}[x]$ we could find a polynomial in $\mathbb{Q}[x]$ that has the same roots? I apologize if I'm asking something really trivial or my question comes from a big misunderstanding of basic concepts.

Let $p(x) = a_0+a_1x+\cdots +a_{n-1}x^{n-1} + x^n$ be a polynomial with coefficients in $\overline{\mathbb{Q}}$. For each $i$, $0\leq i\leq n-1$, let $a_i=b_{i1}, b_{i2},\ldots,b_{im_i}$ be the $m_i$ conjugates of $a_i$ (that is, the "other" roots of the monic irreducible polynomial with coefficients in $\mathbb{Q}$ that has $a_i$ as a root). Now let $F = \mathbb{Q}[b_{11},\ldots,b_{n-1,m_{n-1}}]$. This field is Galois over $\mathbb{Q}$. Let $G=\mathrm{Gal}(F/\mathbb{Q})$. Now consider $$q(x) = \prod_{\sigma\in G} \left( \sigma(a_0) + \sigma(a_1)x+\cdots + \sigma(a_{n-1})x^{n-1} + x^n\right).$$ This is a polynomial with coefficients in $F$, and any root of $p(x)$ is also a root of $q(x)$ (since one of the elements of $G$ is the identity, so one of the factors of $q(x)$ is $p(x)$). The key observation is that if you apply any element $\tau\in G$ to $q(x)$, you get back $q(x)$ again: $$\begin{align*} \tau(q(x)) &= \tau\left(\prod_{\sigma\in G} \left( \sigma(a_0) + \sigma(a_1)x+\cdots + \sigma(a_{n-1})x^{n-1} + x^n\right)\right)\\ &= \prod_{\sigma\in G} \left( \tau\sigma(a_0) +\tau\sigma(a_1)x+\cdots + \tau\sigma(a_{n-1})x^{n-1} + x^n\right)\\ &= \prod_{\sigma'\in G} \left( \sigma'(a_0) + \sigma'(a_1)x+\cdots + \sigma'(a_{n-1})x^{n-1} + x^n\right)\\ &= q(x). \end{align*}$$ That means that the coefficients of $q(x)$ must lie in the fixed field of $G$. But since $F$ is Galois over $\mathbb{Q}$, the fixed field is $\mathbb{Q}$. That is: $q(x)$ is actually a polynomial in $\mathbb{Q}[x]$. Thus, every root of $p(x)$ is the root of a polynomial with coefficients in $\mathbb{Q}$. For an example of how this works, suppose you have $p(x) = x^3 - (2\sqrt{3}+\sqrt{5})x + 3$. The conjugate of $\sqrt{3}$ is $-\sqrt{3}$; the conjugate of $\sqrt{5}$ is $-\sqrt{5}$. The field $\mathbb{Q}[\sqrt{3},\sqrt{5}]$ already contains all the conjugates, and the Galois group over $\mathbb{Q}$ has four elements: the one that maps $\sqrt{3}$ to itself and $\sqrt{5}$ to $-\sqrt{5}$; the one the maps $\sqrt{3}$ to $-\sqrt{3}$ and $\sqrt{5}$ to itself; the one that maps $\sqrt{3}$ to $-\sqrt{3}$ and $\sqrt{5}$ to $-\sqrt{5}$; and the identity. So $q(x)$ would be the product of $x^3 - (2\sqrt{3}+\sqrt{5})x + 3$, $x^3 - (-2\sqrt{3}+\sqrt{5})x+3$, $x^3 - (2\sqrt{3}-\sqrt{5})x + 3$, and $x^3 - (-2\sqrt{3}-\sqrt{5})x + 3$. If you multiply them out, you get $$\begin{align*} \Bigl( &x^3 - (2\sqrt{3}+\sqrt{5})x + 3\Bigr)\Bigl( x^3 + (2\sqrt{3}+\sqrt{5})x+3\Bigr)\\ &\times \Bigl(x^3 - (2\sqrt{3}-\sqrt{5})x + 3\Bigr)\Bigl( x^3 + (2\sqrt{3}-\sqrt{5})x + 3\Bigr)\\ &= \Bigl( (x^3+3)^2 - (2\sqrt{3}+\sqrt{5})^2x^2\Bigr)\Bigl((x^3+3)^2 - (2\sqrt{3}-\sqrt{5})^2x^2\Bigr)\\ &=\Bigl( (x^3+3)^2 - 17x^2 - 2\sqrt{15}x^2\Bigr)\Bigl( (x^3+3)^2 - 17x^2 + 2\sqrt{15}x^2\Bigr)\\ &= \Bigl( (x^3+3)^2 - 17x^2\Bigr)^2 - 60x^4, \end{align*}$$ which has coefficients in $\mathbb{Q}$.

Prove for which $n \in \mathbb{N}$: $9n^3 - 3 ≤ 8^n$ A homework assignment requires me to find out and prove using induction for which $n ≥ 0$ $9n^3 - 3 ≤ 8^n$ and I have conducted multiple approaches and consulted multiple people and other resources with limited success. I appreciate any hint you can give me. Thanks in advance.

Let $f(n)=n^3-3$, and let $g(n)=8^n$. We compute a little, to see what is going on. We have $f(0) \le g(0)$; $f(1)\le g(1)$; $f(2) > g(2)$; $f(3) \le g(3)$; $f(4) \le g(4)$. Indeed $f(4)=573$ and $g(4)=4096$, so it's not even close. The exponential function $8^x$ ultimately grows incomparably faster than the polynomial $9x^3-3$. So it is reasonable to conjecture that $9n^3-3 \le 8^n$ for every non-negative integer $n$ except $2$. We will show by induction that $9n^3-3 \le 8^n$ for all $n \ge 3$. It is natural to work with ratios. We show that $$\frac{8^n}{9n^3-3} \ge 1$$ for all $n \ge 3$. The result certainly holds at $n=3$. Suppose that for a given $n \ge 3$, we have $\frac{8^n}{9n^3-3} \ge 1$. We will show that $\frac{8^{n+1}}{9(n+1)^3-3} \ge 1$. Note that $$\frac{8^{n+1}}{9(n+1)^3-3}=8 \frac{9n^3-3}{9(n+1)^3-3}\frac{8^n}{9n^3-3}.$$ By the induction hypothesis, we have $\frac{8^n}{9n^3-3} \ge 1$. So all we need to do is to show that $$8 \frac{9n^3-3}{9(n+1)^3-3} \ge 1,$$ or equivalently that $$\frac{9(n+1)^3-3}{9n^3-3} \le 8.$$ If $n\ge 3$, the denominator is greater than $8n^3$, and the numerator is less than $9(n+1)^3$. Thus, if $n \ge 3$, then $$\frac{9(n+1)^3-3}{9n^3-3} <\frac{9}{8}\frac{(n+1)^3}{n^3}=\frac{9}{8}\left(1+\frac{1}{n}\right)^3.$$ But if $n \ge 3$, then $(1+1/n)^3\le (1+1/3)^3<2.5$, so $\frac{9}{8}(1+1/n)^3<8$, with lots of room to spare.

Prove the identity $ \sum\limits_{s=0}^{\infty}{p+s \choose s}{2p+m \choose 2p+2s} = 2^{m-1} \frac{2p+m}{m}{m+p-1 \choose p}$ $$ \sum\limits_{s=0}^{\infty}{p+s \choose s}{2p+m \choose 2p+2s} = 2^{m-1} \frac{2p+m}{m}{m+p-1 \choose p}$$ Class themes are: Generating functions and formal power series.

I will try to give an answer using basic complex variables here. This calculation is very simple in spite of some more complicated intermediate expressions that appear. Suppose we are trying to show that $$\sum_{q=0}^\infty {p+q\choose q} {2p+m\choose m-2q} = 2^{m-1} \frac{2p+m}{m} {m+p-1\choose p}.$$ Introduce the integral representation $${2p+m\choose m-2q} = \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{(1+z)^{2p+m}}{z^{m-2q+1}} \; dz.$$ This gives for the sum the integral (the second binomial coefficient enforces the range) $$\frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{(1+z)^{2p+m}}{z^{m+1}} \sum_{q=0}^\infty {p+q\choose q} z^{2q} \; dz \\ = \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{(1+z)^{2p+m}}{z^{m+1}} \frac{1}{(1-z^2)^{p+1}} \; dz \\ = \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{(1+z)^{p+m-1}}{z^{m+1}} \frac{1}{(1-z)^{p+1}} \; dz.$$ This is $$\frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{(2+z-1)^{p+m-1}}{z^{m+1}} \frac{1}{(1-z)^{p+1}} \; dz \\ = 2^{p+m-1} \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{(1+(z-1)/2)^{p+m-1}}{z^{m+1}} \frac{1}{(1-z)^{p+1}} \; dz \\ = 2^{p+m-1} \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{m+1}} \frac{1}{(1-z)^{p+1}} \sum_{q=0}^{p+m-1} {p+m-1\choose q} \frac{(z-1)^q}{2^q} \; dz \\ = 2^{p+m-1} \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{m+1}} \frac{1}{(1-z)^{p+1}} \sum_{q=0}^{p+m-1} {p+m-1\choose q} (-1)^q \frac{(1-z)^q}{2^q} \; dz \\ = 2^{p+m-1} \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{m+1}} \sum_{q=0}^{p+m-1} {p+m-1\choose q} (-1)^q \frac{(1-z)^{q-p-1}}{2^q} \; dz.$$ The only non-zero contribution is with $q$ ranging from $0$ to $p.$ This gives $$ 2^{p+m-1} \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{m+1}} \sum_{q=0}^p {p+m-1\choose q} (-1)^q \frac{1}{2^q} \frac{1}{(1-z)^{p+1-q}} \; dz$$ which on extracting coefficients yields $$2^{p+m-1} \sum_{q=0}^p {p+m-1\choose q} (-1)^q \frac{1}{2^q} {m+p-q\choose p-q}.$$ Introduce the integral representation $${m+p-q\choose p-q} = \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{(1+z)^{m+p-q}}{z^{p-q+1}} \; dz.$$ This gives for the sum the integral (the second binomial coefficient enforces the range) $$2^{p+m-1} \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{(1+z)^{m+p}}{z^{p+1}} \sum_{q=0}^\infty {p+m-1\choose q}\frac{(-1)^q}{2^q} \left(\frac{z}{1+z}\right)^q \; dz \\ = 2^{p+m-1} \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{(1+z)^{m+p}}{z^{p+1}} \left(1-\frac{1}{2}\frac{z}{1+z}\right)^{p+m-1} \; dz \\ = 2^{p+m-1} \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1+z}{z^{p+1}} \left(1+z-1/2\times z\right)^{p+m-1} \; dz \\ = \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1+z}{z^{p+1}} \left(2+z\right)^{p+m-1} \; dz.$$ Extracting coefficients now yields $${p+m-1\choose p} \times 2^{m-1} + {p+m-1\choose p-1} \times 2^m.$$ This symmetric form may be re-written in an asymmetric form as follows, $${p+m-1\choose p} \times 2^{m-1} + \frac{p}{m} {p+m-1\choose p} \times 2^m \\ = 2^{m-1} \times \left(1 + \frac{2p}{m}\right) {p+m-1\choose p}$$ as claimed. The bonus feature of this calculation is that we evaluated two binomial sums instead of one. We have not made use of the properties of complex integrals here so this computation can also be presented using just algebra of generating functions. Apparently this method is due to Egorychev although some of it is probably folklore.

How do I prove equality of x and y? If $0\leq x,y\leq\frac{\pi}{2}$ and $\cos x +\cos y -\cos(x+y)=\frac{3}{2}$, then how can I prove that $x=y=\frac{\pi}{3}$? Your help is appreciated.I tried various formulas but nothing is working.

You could also attempt an geometric proof. First, without loss of generality you can assume $0 <x,y < \frac{\pi}{2}$. Construct a triangle with angles $x,y, \pi-x-y$. Let $a,b,c$ be the edges. Then by cos law, you know that $$\frac{a^2+b^2-c^2}{2ab}+ \frac{a^2+c^2-b^2}{2ac}+ \frac{b^2+c^2-a^2}{2bc}=\frac{3}{2}$$ and you need to show that $a=b=c$. The inequality is $$c(a^2+b^2-c^2)+b(a^2+c^2-b^2)+a(b^2+c^2-a^2)=3abc$$ or $$a^2b+ab^2+ac^2+a^2c+bc^2+b^2c=a^3+b^3+c^3+3abc \,.$$ This should be easy to factor, but I fail to see it....

Counting words with parity restrictions on the letters Let $a_n$ be the number of words of length $n$ from the alphabet $\{A,B,C,D,E,F\}$ in which $A$ appears an even number of times and $B$ appears an odd number of times. Using generating functions I was able to prove that $$a_n=\frac{6^n-2^n}{4}\;.$$ I was wondering if the above answer is correct and in that case what could be a combinatorial proof of that formula?

I don’t know whether you’d call them combinatorial, but here are two completely elementary arguments of a kind that I’ve presented in a sophomore-level discrete math course. Added: Neither, of course, is as nice as Didier’s, which I’d not seen when I posted this. Let $b_n$ be the number of words of length $n$ with an odd number of $A$’s and an odd number of $B$’s, $c_n$ the number of words of length $n$ with an even number of $A$’s and an even number of $B$’s, and $d_n$ the number of words of length $n$ with an odd number of $A$’s and an even number of $B$’s. Then $$a_{n+1}=4a_n+b_n+c_n\;.\tag{1}$$ Clearly $a_n=d_n$: interchanging $A$’s and $B$’s gives a bijection between the two types of word. $(1)$ therefore reduces to $$a_{n+1}=4a_n+b_n+c_n\;.\tag{2}$$ But $6^n=a_n+b_n+c_n+d_n=b_n+c_n+2a_n$, so $b_n+c_n=6^n-2a_n$, and $(2)$ becomes $$a_{n+1}=2a_n+6^n,a_0=0\;.\tag{3}$$ $(3)$ can be solved by brute force: $$\begin{align*} a_n&=2a_{n-1}+6^{n-1}\\ &=2(2a_{n-2}+6^{n-2})+6^{n-1}\\ &=2^2a_{n-2}+2\cdot 6^{n-2}+6^{n-1}\\ &=2^2(2a_{n-3}+6^{n-3})+2\cdot 6^{n-2}+6^{n-1}\\ &=2^3a_{n-3}+2^2\cdot 6^{n-3}+2\cdot 6^{n-2}+6^{n-1}\\ &\;\vdots\\ &=2^na_0+\sum_{k=0}^{n-1}2^k6^{n-1-k}\\ &=6^{n-1}\sum_{k=0}^{n-1}\left(\frac13\right)^k\\ &=6^{n-1}\frac{1-(1/3)^n}{2/3}\\ &=\frac{2^{n-1}3^n-2^{n-1}}{2}\\ &=\frac{6^n-2^n}4\;. \end{align*}$$ Alternatively, with perhaps just a little more cleverness $(1)$ can be expanded to $$\begin{align*} a_{n+1}&=4a_n+b_n+c_n\\ b_{n+1}&=4b_n+2a_n\\ c_{n+1}&=4c_n+2a_n\;, \end{align*}\tag{4}$$ whence $$\begin{align*}a_{n+1}&=4a_n+4b_{n-1}+2a_{n-1}+4c_{n-1}+2a_{n-1}\\ &=4a_n+4a_{n-1}+4(b_{n-1}+c_{n-1})\\ &=4a_n+4a_{n-1}+4(a_n-4a_{n-1})\\ &=8a_n-12a_{n-2}\;. \end{align*}$$ This straightforward homogeneous linear recurrence (with initial conditions $a_0=0,a_1=1$) immediately yields the desired result.

Prime reciprocals sum Let $a_i$ be a sequence of $1$'s and $2$'s and $p_i$ the prime numbers. And let $r=\displaystyle\sum_{i=1}^\infty p_i^{-a_i}$ Can $r$ be rational, and can r be any rational $> 1/2$ or any real? ver.2: Let $k$ be a positive real number and let $a_i$ be $1 +$ (the $i$'th digit in the binary decimal expansion of $k$). And let $r(k)=\displaystyle\sum_{n=1}^\infty n^{-a_n}$ Does $r(k)=x$ have a solution for every $x>\pi^2/6$, and how many ?

The question with primes in the denominator: The minimum that $r$ could possibly be is $C=\sum\limits_{i=1}^\infty\frac{1}{p_i^2}$. However, a sequence of $1$s and $2$s can be chosen so that $r$ can be any real number not less than $C$. Since $\sum\limits_{i=1}^\infty\left(\frac{1}{p_i}-\frac{1}{p_i^2}\right)$ diverges, consider the sum $$ S_n=\sum_{i=1}^n b_i\left(\frac{1}{p_i}-\frac{1}{p_i^2}\right) $$ where $b_i$ is $0$ or $1$. Choose $b_n=1$ while $S_{n-1}+\frac{1}{p_n}-\frac{1}{p_n^2}\le L-C$ and $b_n=0$ while $S_{n-1}+\frac{1}{p_n}-\frac{1}{p_n^2}>L-C$. If we let $a_i=1$ when $b_i=1$ and $a_i=2$ when $b_i=0$, then $$ \sum\limits_{i=1}^\infty\frac{1}{p_i^{a_i}}=\sum\limits_{i=1}^\infty\frac{1}{p_i^2}+\sum_{i=1}^\infty b_i\left(\frac{1}{p_i}-\frac{1}{p_i^2}\right)=C+(L-C)=L $$ The question with non-negative integers in the denominator: Changing $p_n$ from the $n^{th}$ prime to $n$ simply allows us to specify $C=\frac{\pi^2}{6}$. The rest of the procedure follows through unchanged. That is, choose any $L\ge C$ and let $$ S_n=\sum_{i=1}^n b_i\left(\frac{1}{i}-\frac{1}{i^2}\right) $$ where $b_i$ is $0$ or $1$. Choose $b_n=1$ while $S_{n-1}+\frac{1}{n}-\frac{1}{n^2}\le L-C$ and $b_n=0$ while $S_{n-1}+\frac{1}{n}-\frac{1}{n^2}>L-C$. If we let $a_i=1$ when $b_i=1$ and $a_i=2$ when $b_i=0$, then $$ \sum\limits_{n=1}^\infty\frac{1}{n^{a_i}}=\sum\limits_{n=1}^\infty\frac{1}{n^2}+\sum_{n=1}^\infty b_n\left(\frac{1}{n}-\frac{1}{n^2}\right)=C+(L-C)=L $$ We don't need to worry about an infinite final sequence of $1$s in the binary number since that would map to a divergent series.

How can one prove that $\sqrt[3]{\left ( \frac{a^4+b^4}{a+b} \right )^{a+b}} \geq a^ab^b$, $a,b\in\mathbb{N^{*}}$? How can one prove that $\sqrt[3]{\left ( \frac{a^4+b^4}{a+b} \right )^{a+b}} \geq a^ab^b$, $a,b\in\mathbb{N^{*}}$?

Since $\log(x)$ is concave, $$ \log\left(\frac{ax+by}{a+b}\right)\ge\frac{a\log(x)+b\log(y)}{a+b}\tag{1} $$ Rearranging $(1)$ and exponentiating yields $$ \left(\frac{ax+by}{a+b}\right)^{a+b}\ge x^ay^b\tag{2} $$ Plugging $x=a^3$ and $y=b^3$ into $(2)$ gives $$ \left(\frac{a^4+b^4}{a+b}\right)^{a+b}\ge a^{3a}b^{3b}\tag{3} $$ and $(3)$ is the cube of the posited inequality. From my comment (not using concavity): For $0<t<1$, the minimum of $t+(1-t)u-u^{1-t}$ occurs when $(1-t)-(1-t)u^{-t}=0$; that is, when $u=1$. Therefore, $t+(1-t)u-u^{1-t}\ge0$. If we set $u=\frac{y}{x}$ and $t=\frac{a}{a+b}$, we get $$ \frac{ax+by}{a+b}\ge x^{a/(a+b)}y^{b/(a+b)}\tag{4} $$ Inequality $(2)$ is simply $(4)$ raised to the $a+b$ power.

Finding the Laurent expansion of $\frac{1}{\sin^3(z)}$ on $0<|z|<\pi$? How do you find the Laurent expansion of $\frac{1}{\sin^3(z)}$ on $0<|z|<\pi$? I would really appreciate someone carefully explaining this, as I'm very confused by this general concept! Thanks

use this formula $$\sum _{k=1}^{\infty } (-1)^{3 k} \left(-\frac{x^3}{\pi ^3 k^3 (\pi k-x)^3}-\frac{x^3}{\pi ^3 k^3 (\pi k+x)^3}+\frac{3 x^2}{\pi ^2 k^2 (\pi k-x)^3}-\frac{3 x^2}{\pi ^2 k^2 (\pi k+x)^3}-\frac{x^3}{2 \pi k (\pi k-x)^3}-\frac{x^3}{2 \pi k (\pi k+x)^3}+\frac{x^2}{(\pi k-x)^3}-\frac{x^2}{(\pi k+x)^3}-\frac{\pi k x}{2 (\pi k-x)^3}-\frac{3 x}{\pi k (\pi k-x)^3}-\frac{\pi k x}{2 (\pi k+x)^3}-\frac{3 x}{\pi k (\pi k+x)^3}\right)+\frac{1}{x^3}+\frac{1}{2 x}=\csc ^3(x)$$

Is there an easy way to determine when this fractional expression is an integer? For $x,y\in \mathbb{Z}^+,$ when is the following expression an integer? $$z=\frac{(1-x)-(1+x)y}{(1+x)+(1-x)y}$$ The associated Diophantine equation is symmetric in $x, y, z$, but I couldn't do anything more with that. I tried several factoring tricks without luck. The best I could do was find three solutions such that $0<x\le y\le z$. They are: $(2,5,8)$, $(2,4,13)$ and $(3,3,7)$. The expression seems to converge pretty quickly to some non-integer between 1 and 2.

Since $$ \frac{(1-x)-(1+x)y}{(1+x)+(1-x)y} = \frac{ xy+x+y-1}{xy-x-y-1} = 1 + \frac{2(x+y) }{xy-x-y-1} $$ and $ 2x+2y < xy - x -y - 1 $ if $ 3(x+y) < xy - 1 .$ Suppose $ x\leq y$, then $ 3(x+y) \leq 6y \leq xy-1 $ if $ x\geq 7. $ So all solutions must have $0\leq x< 7 $ so it is reduced to solving $7$ simpler Diophantine equations. If $x=0 $ then $ \displaystyle z= 1 - \frac{2y}{y+1}$ so the only solutions are $ (0,0,1)$ and $ (0,1,0).$ If $x=1$ then $ \displaystyle z= -y$ so $(1,m,-m)$ is a solution for $ m\geq 1.$ If $x=2$ then $ \displaystyle z = 1 + \frac{4+2y}{y-3}$ which is an integer for $y=1,2,4,5,8,13.$ I will leave you to find the others. Each of the cases are now simple Diophantine equations.

Sketch the graph of $y = \frac{4x^2 + 1}{x^2 - 1}$ I need help sketching the graph of $y = \frac{4x^2 + 1}{x^2 - 1}$. I see that the domain is all real numbers except $1$ and $-1$ as $x^2 - 1 = (x + 1)(x - 1)$. I can also determine that between $-1$ and $1$, the graph lies below the x-axis. What is the next step? In previous examples I have determined the behavior near x-intercepts.

You can simplify right away with $$ y = \frac{4x^2 + 1}{x^2 - 1} = 4+ \frac{5}{x^2 - 1} =4+ \frac{5}{(x - 1)(x+1)} $$ Now when $x\to\infty$ or $x\to -\infty$, adding or subtracting 1 doesn't really matter hence that term goes to zero. When $x$ is quite large, say 1000, the second term is very small but positive hence it should approach to 4 from above (same holds for negative large values). The remaining part to be done is when $x$ approaches to $-1$ and $1$ from both sides. For the values $x<-1$ and $x>1$ you can show that the second term is positive and negative for $-1<x<1$. Therefore the limit jumps from $-\infty$ to $\infty$ at each vertical asymptote. Here is the whole thing.

Lambert series expansion identity I have a question which goes like this: How can I show that $$\sum_{n=1}^{\infty} \frac{z^n}{\left(1-z^n\right)^2} =\sum_{n=1}^\infty \frac{nz^n}{1-z^n}$$ for $|z|<1$?

Hint: Try using the expansions $$ \frac{1}{1-x}=1+x+x^2+x^3+x^4+x^5+\dots $$ and $$ \frac{1}{(1-x)^2}=1+2x+3x^2+4x^3+5x^4+\dots $$ Expansion: $$ \begin{align} \sum_{n=1}^\infty\frac{z^n}{(1-z^n)^2} &=\sum_{n=1}^\infty\sum_{k=0}^\infty(k+1)z^{kn+n}\\ &=\sum_{n=1}^\infty\sum_{k=1}^\infty kz^{kn}\\ &=\sum_{k=1}^\infty\sum_{n=1}^\infty kz^{kn}\\ &=\sum_{k=1}^\infty\sum_{n=0}^\infty kz^{kn+k}\\ &=\sum_{k=1}^\infty\frac{kz^k}{1-z^k} \end{align} $$

Convergence of $\lim_{n \to \infty} \frac{5 n^2 +\sin n}{3 (n+2)^2 \cos(\frac{n \pi}{5})},$ I'm in trouble with this limit. The numerator diverges positively, but I do not understand how to operate on the denominator. $$\lim_{n \to \infty} \frac{5 n^2 +\sin n}{3 (n+2)^2 \cos(\frac{n \pi}{5})},$$ $$\lim_{n \to \infty} \frac{5 n^2 +\sin n}{3 (n+2)^2 \cos(\frac{n \pi}{5})}= \lim_{x\to\infty}\frac {n^2(5 +\frac{\sin n}{n^2})}{3 (n+2)^2 \cos(\frac{n \pi}{5})} \cdots$$

Let's make a few comments. * *Note that the terms of the sequence are always defined: for $n\geq 0$, $3(n+2)^2$ is greater than $0$; and $\cos(n\pi/5)$ can never be equal to zero (you would need $n\pi/5$ to be an odd multiple of $\pi/2$, and this is impossible). *If $a_n$ and $b_n$ both have limits as $n\to\infty$, then so does $a_nb_n$, and the limit of $a_nb_n$ is the product of the limits of $a_n$ and of $b_n$, $$\lim_{n\to\infty}a_nb_n = \left(\lim_{n\to\infty}a_n\right)\left(\lim_{n\to\infty}b_n\right).$$ *If $b_n$ has a limit as $n\to\infty$, and the limit is not zero, then $\frac{1}{b_n}$ has a limit as $n\to\infty$, and the limit is the reciprocal of the limit of $b_n$: $$\lim_{n\to\infty}\frac{1}{b_n} = \frac{1}{\lim\limits_{n\to\infty}b_n},\qquad \text{if }\lim_{n\to\infty}b_n\neq 0.$$ As a consequence of $2$ and $3$, we have: * *If $\lim\limits_{n\to\infty}a_nb_n$ and $\lim\limits_{n\to\infty}a_n$ exists and is not equal to $0$, then $\lim\limits_{n\to\infty}b_n$ exists: Just write $\displaystyle b_n = \left(a_nb_n\right)\frac{1}{a_n}$ *Equivalently, if $\lim\limits_{n\to\infty}a_n$ exists and is not zero, and $\lim\limits_{n\to\infty}b_n$ does not exist, then $\lim\limits_{n\to\infty}a_nb_n$ does not exist either. So, consider $$a_n = \frac{5n^2 + \sin n}{3(n+2)^2},\qquad b_n =\frac{1}{\cos(n\pi/5)}.$$ We have, as you did: $$\begin{align*} \lim_{n\to\infty}a_n &= \lim_{n\to\infty}\frac{5n^2 + \sin n}{3(n+2)^2}\\ &= \lim_{n\to\infty}\frac{n^2\left(5 + \frac{\sin n}{n^2}\right)}{3n^2(1 + \frac{2}{n})^2}\\ &=\lim_{n\to\infty}\frac{5 + \frac{\sin n}{n^2}}{3(1+\frac{2}{n})^2}\\ &= \frac{5 + 0}{3(1+0)^2} = \frac{5}{3}\neq 0. \end{align*}$$ What about the sequence $(b_n)$? If $n=(2k+1)5$ is an odd multiple of $5$, then $$b_n = b_{(2k+1)5}\frac{1}{\cos\frac{n\pi}{5}} = \frac{1}{\cos((2k+1)\pi)} = -1;$$ so the subsequence $b_{(2k+1)5}$ is constant, and converges to $-1$. On the other hand, if $n=10k$ is an even multiple of $5$, then $$b_n = \frac{1}{\cos\frac{n\pi}{5}} = \frac{1}{\cos(2k\pi)} = 1.$$ so the subsequence $b_{10k}$ is constant and converges to $1$. Since a sequence converges if and only if every subsequence converges and converges to the same thing, but $(b_n)$ has two subsequences that converge to different things, it follows that $(b_n)$ does not converge. (It also does not diverge to $\infty$ or to $-\infty$, since there are subsequences that are constant). And so, what can we conclude, given our observations above about products of sequences?

Need help finding limit $\lim \limits_{x\to \infty}\left(\frac{x}{x-1}\right)^{2x+1}$ Facing difficulty finding limit $$\lim \limits_{x\to \infty}\left(\frac{x}{x-1}\right)^{2x+1}$$ For starters I have trouble simplifying it Which method would help in finding this limit?

$$ \begin{eqnarray} \lim \limits_{x\to \infty}\left(\frac{x}{x-1}\right)^{2x+1}=\lim \limits_{x\to \infty}\left(\frac{x-1+1}{x-1}\right)^{2x+1} =\lim \limits_{x\to \infty}\left(1+\frac{1}{x-1}\right)^{2x+1}\\= \lim \limits_{x\to \infty}\left(1+\frac{1}{x-1}\right)^{(x-1)\cdot\frac{2x+1}{x-1}} =\lim \limits_{x\to \infty}\left(1+\frac{1}{x-1}\right)^{(x-1)\cdot\frac{2x+1}{x-1}}=e^{\lim \limits_{x\to \infty}\frac{2x+1}{x-1}}=e^2 \end{eqnarray} $$

Maxima of bivariate function [1] Is there an easy way to formally prove that, $$ 2xy^{2} +2x^{2} y-2x^{2} y^{2} -4xy+x+y\ge -x^{4} -y^{4} +2x^{3} +2y^{3} -2x^{2} -2y^{2} +x+y$$ $${0<x,y<1}$$ without resorting to checking partial derivatives of the quotient formed by the two sides, and finding local maxima? [2] Similarly, is there an easy way for finding $$\max_{0<x,y<1} [f(x,y)]$$ where, $$f(x,y)=2x(1+x)+2y(1+y)-8xy-4(2xy^{2} +2x^{2} y-2x^{2} y^{2} -4xy+x+y)^{2}$$

Your first question: With a little manipulation you get that it is equivalent to $$x^2((1-x)^2+1)+y^2((1-y)^2+1) \ge 2xy[(1-x)(1-y)+1].$$ This can be obtained from addition of two inequalities $$x^2(1-x)^2+y^2(1-y)^2 \ge 2xy(1-x)(1-y)$$ $$x^2+y^2\ge 2xy.$$ Both of them are special cases of $a^2+b^2\ge 2ab$, which follows from $(a-b)^2\ge 0$. (Or, if you prefer, you can consider it as a special case of AM-GM inequality.) Note: To check the algebraic manipulations, you can simply compare the results for 2xy^2 +2x^2 y-2x^2 y^2 -4xy+x+y - ( -x^4 -y^4 +2x^3 +2y^3 -2x^2 -2y^2 +x+y) expand x^2((1-x)^2+1)+y^2((1-y)^2+1) -2xy[(1-x)(1-y)+1] Or simply subtract the two expressions: 2xy^2 +2x^2 y-2x^2 y^2 -4xy+x+y - ( -x^4 -y^4 +2x^3 +2y^3 -2x^2 -2y^2 +x+y) - [x^2((1-x)^2+1)+y^2((1-y)^2+1) -2xy[(1-x)(1-y)+1]] I did not succeed in finding similar type of solution for your second problem.

Adding a different constant to numerator and denominator Suppose that $a$ is less than $b$ , $c$ is less than $d$. What is the relation between $\dfrac{a}{b}$ and $\dfrac{a+c}{b+d}$? Is $\dfrac{a}{b}$ less than, greater than or equal to $\dfrac{a+c}{b+d}$?

One nice thing to notice is that $$ \frac{a}{b}=\frac{c}{d} \Leftrightarrow \frac{a}{b}=\frac{a+c}{b+d} $$ no matter the values of $a$, $b$, $c$ and $d$. The $(\Rightarrow)$ is because $c=xa, d=xb$ for some $x$, so $\frac{a+c}{b+d}=\frac{a+xa}{b+xb}=\frac{a(1+x)}{b(1+x)}=\frac{a}{b}$. The other direction is similar. The above is pretty easy to remember, and with that intuition in mind it is not hard to imagine that $$ \frac{a}{b}<\frac{c}{d} \Leftrightarrow \frac{a}{b}<\frac{a+c}{b+d}<\frac{c}{d} $$ and similar results.

Quadratic forms and prime numbers in the sieve of Atkin I'm studying the theorems used in the paper which explains how the sieve of Atkin works, but I cannot understand a point. For example, in the paper linked above, theorem 6.2 on page 1028 says that if $n$ is prime then the cardinality of the set which contains all the norm-$n$ ideals in $\mathbf Z[(-1+\sqrt{-3})/2]$ is 2. I don't understand why, and I am not able to relate this result to the quadratic form $3x^2+y^2=n$ used in the proof.

The main thing is that the norm of $s + t \omega$ is $s^2 + s t + t^2,$ which is a binary form that represents exactly the same numbers as $3x^2 + y^2.$ It is always true that, for an integer $k,$ the form $s^2 + s t + k t^2$ represents a superset of the numbers represented by $x^2 + (4k-1)y^2.$ For instance, with $k=2,$ the form $x^2 + 7 y^2$ does not represent any numbers $2\pmod 4,$ otherwise it and $s^2 + s t + 2 t^2$ agree. With $k=-1,$ it turns out that $x^2 - 5 y^2$ and $s^2 + s t - t^2$ represent exactly the same integers. Take $s^2 + s t + k t^2$ with $s = x - y, \; t = 2 y.$ You get $$ (x-y)^2 + (x-y)(2y) + k (2y)^2 = x^2 - 2 x y + y^2 + 2 x y - 2 y^2 + 4 k y^2 = x^2 + (4k-1) y^2.$$

Finding the limit of a sequence $\lim _{n\to \infty} \sqrt [3]{n^2} \left( \sqrt [3]{n+1}- \sqrt [3]{n} \right)$ If there were a regular square root I would multiply the top by its adjacent and divide, but I've tried that with this problem and it doesn't work. Not sure what else to do have been stuck on it. $$ \lim _{n\to \infty } \sqrt [3]{n^2} \left( \sqrt [3]{n+1}- \sqrt [3]{n} \right) .$$

$$ \begin{align*} \lim _{n\to \infty } \sqrt [3]{n^2} \left( \sqrt [3]{n+1}- \sqrt [3]{n} \right) &= \lim _{n\to \infty } \sqrt [3]{n^2} \cdot \sqrt[3]{n} \left( \sqrt [3]{1+ \frac{1}{n}}- 1 \right) \\ &= \lim _{n\to \infty } n \left( \sqrt [3]{1+ \frac{1}{n}}- 1 \right) \\ &= \lim _{n\to \infty } \frac{\sqrt [3]{1+ \frac{1}{n}}- 1 }{\frac{1}{n}} \\ &= \lim _{h \to 0} \frac{\sqrt [3]{1+ h}- 1 }{h} \\ &= \left. \frac{d}{du} \sqrt[3]{u} \ \right|_{u=1} \\ &= \cdots \end{align*} $$

Is every Mersenne prime of the form : $x^2+3 \cdot y^2$? How to prove or disprove following statement : Conjecture : Every Mersenne prime number can be uniquely written in the form : $x^2+3 \cdot y^2$ , where $\gcd(x,y)=1$ and $x,y \geq 0$ Since $M_p$ is an odd number it follows that : $M_p \equiv 1 \pmod 2$ According to Fermat little theorem we can write : $2^p \equiv 2 \pmod p \Rightarrow 2^p-1 \equiv 1\pmod p \Rightarrow M_p \equiv 1 \pmod p$ We also know that : $2 \equiv -1 \pmod 3 \Rightarrow 2^p \equiv (-1)^p \pmod 3 \Rightarrow 2^p-1 \equiv -1-1 \pmod 3 \Rightarrow$ $\Rightarrow M_p \equiv -2 \pmod 3 \Rightarrow M_p \equiv 1 \pmod 3$ So , we have following equivalences : $M_p \equiv 1 \pmod 2$ , $M_p \equiv 1 \pmod 3$ and $M_p \equiv 1 \pmod p$ , therefore for $p>3$ we can conclude that : $ M_p \equiv 1 \pmod {6 \cdot p}$ On the other hand : If $x^2+3\cdot y^2$ is a prime number greater than $5$ then : $x^2+3\cdot y^2 \equiv 1 \pmod 6$ Proof : Since $x^2+3\cdot y^2$ is a prime number greater than $3$ it must be of the form $6k+1$ or $6k-1$ . Let's suppose that $x^2+3\cdot y^2$ is of the form $6k-1$: $x^2+3\cdot y^2=6k-1 \Rightarrow x^2+3 \cdot y^2+1 =6k \Rightarrow 6 | x^2+3 \cdot y^2+1 \Rightarrow$ $\Rightarrow 6 | x^2+1$ , and $ 6 | 3 \cdot y^2$ If $6 | x^2+1 $ then : $2 | x^2+1$ , and $3 | x^2+1$ , but : $x^2 \not\equiv -1 \pmod 3 \Rightarrow 3 \nmid x^2+1 \Rightarrow 6 \nmid x^2+1 \Rightarrow 6 \nmid x^2+3 \cdot y^2+1$ , therefore : $x^2+3\cdot y^2$ is of the form $6k+1$ , so : $x^2+3\cdot y^2 \equiv 1 \pmod 6$ We have shown that : $M_p \equiv 1 \pmod {6 \cdot p}$, for $p>3$ and $x^2+3\cdot y^2 \equiv 1 \pmod 6$ if $x^2+3\cdot y^2$ is a prime number greater than $5$ . This result is a necessary condition but it seems that I am not much closer to the solution of the conjecture than in the begining of my reasoning ...

(Outline of proof that, for prime $p\equiv 1\pmod 6$, there is one positive solution to $x^2+3y^2=p$.) It helps to recall the Gaussian integer proof that, for a prime $p\equiv 1\pmod 4$, $x^2+y^2=p$ has an integer solution. It starts with the fact that there is an $a$ such that $a^2+1$ is divisible by $p$, then uses unique factorization in the Gaussian integers to show that there must a common (Gaussian) prime factor of $p$ and $a+i$, and then that $p$ must be the Gaussian norm of that prime factor. By quadratic reciprocity, we know that if $p\equiv 1\pmod 6$ is prime, then $a^2=-3\pmod p$ has a solution. This means that $a^2+3$ is divisible by $p$. If we had unique factorization on $\mathbb Z[\sqrt{-3}]$ we'd have our result, since there must be a common prime factor of $p$ and $a+\sqrt{-3}$, and it would have to have norm $p$, and we'd be done. But we don't have unique factorization in $\mathbb Z[\sqrt{-3}]$, only in the ring, R, of algebraic integers in $\mathbb Q[\sqrt{-3}]$, which are all of the form: $\frac{a+b\sqrt{-3}}2$ where $a=b\pmod 2$. However, this isn't really a big problem, because for any element $r\in R$, there is a unit $u\in R$ and an element $z\in\mathbb Z[\sqrt{-3}]$ such that $r=uz$. In particular, then, for any $r\in R$, the norm $N(r)=z_1^2+3z_2^2$ for some integers $z_1,z_2\in \mathbb Z$. As with the Gaussian proof for $x^2+y^2$, we can then use this to show that there is a solution to $x^2+3y^2=p$. To show uniqueness, you need to use properties of the units in $R$.

Is $\sin^3 x=\frac{3}{4}\sin x - \frac{1}{4}\sin 3x$? $$\sin^3 x=\frac{3}{4}\sin x - \frac{1}{4}\sin 3x$$ Is there any formula that tells this or why is it like that?

\begin{equation} \text{You can use De Moivre's identity:} \end{equation} \begin{equation} \text{Let's Call:}\\\\ \end{equation} \begin{equation} \mathrm{z=\cos x+i \sin x}\\ \mathrm{\frac{1}{z}=\cos x-i \sin x}\\ \end{equation} \begin{equation} \text{Now subtracting both equations together, we get:}\\ \end{equation} \begin{equation} \mathrm{2i\sin x=z-\frac{1}{z}}\\ \text{And we know that:}\\ \end{equation} \begin{equation} \mathrm{z^n=(cis x)^n=cis~nx}\\ \end{equation} \begin{equation} \text{So:}\\ \mathrm{2i\sin x=z-\frac{1}{z}}\Rightarrow\\ \mathrm{-8i\sin^3 x=\left (z-\frac{1}{z} \right )^{3}}\\ \end{equation} \begin{equation} \text{Expanding the RHS:}\\ \end{equation} \begin{equation} \mathrm{-8i\sin^3 x =z^3-\frac{1}{z^3}-3\left (z-\frac{1}{z}\right)}\\ \end{equation} \begin{equation} \mathrm{-8i\sin^3 x=2i\sin 3x-6i\sin x}\\ \end{equation} \begin{equation} \boxed{\boxed{\mathrm{\therefore\sin^{3} x=\frac{ 3}{4}}\sin\mathrm{x}-\frac{1}{4}\sin 3x}} \end{equation}

The square of an integer is congruent to 0 or 1 mod 4 This is a question from the free Harvard online abstract algebra lectures. I'm posting my solutions here to get some feedback on them. For a fuller explanation, see this post. This problem is from assignment 6. The notes from this lecture can be found here. a) Prove that the square $a^2$ of an integer $a$ is congruent to 0 or 1 modulo 4. b) What are the possible values of $a^2$ modulo 8? a) Let $a$ be an integer. Then $a=4q+r, 0\leq r<4$ with $\bar{a}=\bar{r}$. Then we have $a^2=a\cdot a=(4q+r)^2=16q^2+8qr+r^2=4(4q^2+2qr)+r^2, 0\leq r^2<4$ with $\bar{a^2}=\bar{r^2}$. So then the possible values for $r$ with $r^2<4$ are 0,1. Then $\bar{a^2}=\bar{0}$ or $\bar{1}$. b) Let $a$ be an integer. Then $a=8q+r, 0\leq r<8$ with $\bar{a}=\bar{r}$. Then we have $a^2=a\cdot a=(8q+r)^2=64q^2+16qr+r^2=8(8q^2+2qr)+r^2, 0\leq r^2<8$ with $\bar{a^2}=\bar{r^2}$. So then the possible values for $r$ with $r^2<8$ are 0,1,and 2. Then $\bar{a^2}=\bar{0}$, $\bar{1}$ or $\bar{4}$. Again, I welcome any critique of my reasoning and/or my style as well as alternative solutions to the problem. Thanks.

$$\begin{align} x^2 \mod 4 &\equiv (x \mod 4)(x \mod 4) \pmod 4 \\ &\equiv \begin{cases}0^2 \mod 4 \\ 1^2 \mod 4 \\ 2^2 \mod 4 \\ 3^2 \mod 4 \end{cases} \\ &\equiv \begin{cases}0 \mod 4 \\ 1 \mod 4 \\ 4 \mod 4 \\ 9 \mod 4 \end{cases} \\ &\equiv \begin{cases}0 \mod 4 \\ 1 \mod 4 \\ 0 \mod 4 \\ 1 \mod 4 \end{cases} \end{align} $$

If $n\ge 3$, $4^n \nmid 9^n-1$ Could anyone give me a hint to prove the following? If $n\ge 3$, $4^n \nmid 9^n-1$

Hint : Try to prove using induction : $1.$ $9^3 \not \equiv 1 \pmod {4^3}$ $2.$ suppose : $9^k \not \equiv 1 \pmod {4^k}$ $3.$ $9^k \not \equiv 1 \pmod {4^k} \Rightarrow 9^{k+1} \not \equiv 9 \pmod {4^k}$ So you have to prove : $ 9^{k+1} \not \equiv 9 \pmod {4^k} \Rightarrow 9^{k+1} \not \equiv 1 \pmod {4^{k+1}}$

Simple expressions for $\sum_{k=0}^n\cos(k\theta)$ and $\sum_{k=1}^n\sin(k\theta)$? Possible Duplicate: How can we sum up $\sin$ and $\cos$ series when the angles are in A.P? I'm curious if there is a simple expression for $$ 1+\cos\theta+\cos 2\theta+\cdots+\cos n\theta $$ and $$ \sin\theta+\sin 2\theta+\cdots+\sin n\theta. $$ Using Euler's formula, I write $z=e^{i\theta}$, hence $z^k=e^{ik\theta}=\cos(k\theta)+i\sin(k\theta)$. So it should be that $$ \begin{align*} 1+\cos\theta+\cos 2\theta+\cdots+\cos n\theta &= \Re(1+z+\cdots+z^n)\\ &= \Re\left(\frac{1-z^{n+1}}{1-z}\right). \end{align*} $$ Similarly, $$ \begin{align*} \sin\theta+\sin 2\theta+\cdots+\sin n\theta &= \Im(z+\cdots+z^n)\\ &= \Im\left(\frac{z-z^{n+1}}{1-z}\right). \end{align*} $$ Can you pull out a simple expression from these, and if not, is there a better approach? Thanks!

Take the expression you have and multiply the numerator and denominator by $1-\bar{z}$, and using $z\bar z=1$: $$\frac{1-z^{n+1}}{1-z} = \frac{1-z^{n+1}-\bar{z}+z^n}{2-(z+\bar z)}$$ But $z+\bar{z}=2\cos \theta$, so the real part of this expression is the real part of the numerator divided by $2-2\cos \theta$. But the real part of the numerator is $1-\cos {(n+1)\theta} - \cos \theta + \cos{n\theta}$, so the entire expression is: $$\frac{1-\cos {(n+1)\theta} - \cos \theta + \cos{n\theta}}{2-2\cos\theta}=\frac{1}{2} + \frac{\cos {n\theta} - \cos{(n+1)\theta}}{2-2\cos \theta}$$ for the cosine case. You can do much the same for the case of the sine function.

Conditional Probability Question Bowl A contains 6 red chips and 4 blue chips. Five chips are randomly chosen and transferred without replacement to Bowl B. One chip is drawn at random from Bowl B. Given that this chip is blue, find the conditional probability that 2 red chips and 3 blue chips are transferred from bowl A to bowl B. Attempt: $$P(A|B) = \frac{P(B|A)\cdot P(A)}{P(B)}$$ Let $B$ = chip is blue and $A$ = 2 red and 3 blue are chosen. $$\begin{align} &P(A) = \frac {\binom 6 2 \cdot \binom 4 3}{\binom {10} 5}\\ &P(B|A) = \frac 3 5 \end{align}$$ By Bayes Rule, $P(A|B) = \left(\dfrac 3 5\right)\cdot \dfrac{ \binom 6 2 \binom 4 3}{\binom {10} 5\cdot \dfrac{4}{10}}$. Is this correct?

There are $\frac{10!}{6!4!}$ (= 210) possible arrangements for the chips, and $\frac{5!}{2!3!}$ arrangements for the chips desired in bowl B. Any given arrangement of bowl B can occur for every corresponding arrangement in bowl A (also $\frac{5!}{2!3!}$ combinations) The total number of possiblilities with the correct bowl B is therefore $\frac{5!}{2!3!}\dot{}\frac{5!}{2!3!}=100$ Substitute P(A) = 100/210 to get P(A|B) = (3/5)(100/210)/(4/10) = 5/7, or about 71%

prove for all $n\geq 0$ that $3 \mid n^3+6n^2+11n+6$ I'm having some trouble with this question and can't really get how to prove this.. I have to prove $n^3+6n^2+11n+6$ is divisible by $3$ for all $n \geq 0$. I have tried doing $\dfrac{m}{3}=n$ and then did $m=3n$ then I said $3n=n^3+6n^2+11n+6$ but now I am stuck.

Here is a solution using induction: Let $f(x)=x^3+6x^2+11x+6$ Since we want to see if it is divisible by 3 let us assume that $f(x)=3m$. For the case where $x=0$, $f(0)=6$ which is divisible by 3. Now that we have proved for one case let us prove for the case of $f(x+1)$ $$f(x+1)=(x+1)^3+6(x+1)^2+11(x+1)+6$$ $$= x^3+3x^2+3x+1+6x^2+12x+6+11x+11+6$$ $$=(x^3+6x^2+11x+6)+3x^2+15x+18$$ And since $x^3+6x^2+11x+6=3m$ $$f(x+1)=3m+3x^2+15x+18=3(m+x^2+5x+6)$$ Which is divisible by 3.

roots of complex polynomial - tricks What tricks are there for calculating the roots of complex polynomials like $$p(t) = (t+1)^6 - (t-1)^6$$ $t = 1$ is not a root. Therefore we can divide by $(t-1)^6$. We then get $$\left( \frac{t+1}{t-1} \right)^6 = 1$$ Let $\omega = \frac{t+1}{t-1}$ then we get $\omega^6=1$ which brings us to $$\omega_k = e^{i \cdot k \cdot \frac{2 \pi}{6}}$$ So now we need to get the values from t for $k = 0,...5$. How to get the values of t from the following identity then? $$ \begin{align} \frac{t+1}{t-1} &= e^{i \cdot 2 \cdot \frac{2 \pi}{6}} \\ (t+1) &= t\cdot e^{i \cdot 2 \cdot \frac{2 \pi}{6}} - e^{i \cdot 2 \cdot \frac{2 \pi}{6}} \\ 1+e^{i \cdot 2 \cdot \frac{2 \pi}{6}} &= t\cdot e^{i \cdot 2 \cdot \frac{2 \pi}{6}} - t \\ 1+e^{i \cdot 2 \cdot \frac{2 \pi}{6}} &= t \cdot (e^{i \cdot 2 \cdot \frac{2 \pi}{6}}-1) \\ \end{align} $$ And now? $$ t = \frac{1+e^{i \cdot 2 \cdot \frac{2 \pi}{6}}}{e^{i \cdot 2 \cdot \frac{2 \pi}{6}}-1} $$ So I've got six roots for $k = 0,...5$ as follows $$ t = \frac{1+e^{i \cdot k \cdot \frac{2 \pi}{6}}}{e^{i \cdot k \cdot \frac{2 \pi}{6}}-1} $$ Is this right? But how can it be that the bottom equals $0$ for $k=0$? I don't exactly know how to simplify this: $$\frac{ \frac{1}{ e^{i \cdot k \cdot \frac{2 \pi}{6}} } + 1 }{ 1 - \frac{1}{ e^{i \cdot k \cdot \frac{2 \pi}{6}} }}$$

Notice that $t=1$ is not a root. Divide by $(t-1)^6$. If $\omega$ is a root of $z^6 - 1$, then a root of the original equation is given by $\frac{t+1}{t-1} = \omega$.

Generalization of Pythagorean triples Is it known whether for any natural number $n$, I can find (infinitely many?) nontrivial integer tuples $$(x_0,\ldots,x_n)$$ such that $$x_0^n + \cdots + x_{n-1}^n = x_n^n?$$ Obviously this is true for $n = 2$. Thanks.

These Pythagorean triples can appear in the most unexpected place. If: $a^2+b^2=c^2$ Then alignment: $N_1^3+N_2^3+N_3^3+N_4^3+N_5^3=N_6^3$ $N_1=cp^2-3(a+b)ps+3cs^2$ $N_2=bp^2+3bps-3bs^2$ $N_3=ap^2+3aps-3as^2$ $N_4=-bp^2+3(2c-b)ps+3(3c-3a-2b)s^2$ $N_5=-ap^2+3(2c-a)ps+3(3c-2a-3b)s^2$ $N_6=cp^2+3(2c-a-b)ps+3(4c-3a-3b)s^2$ And more: $N_1=cp^2-3(a+b)ps+3cs^2$ $N_2=bp^2+3bps-3bs^2$ $N_3=ap^2+3aps-3as^2$ $N_4=(3c+3a+2b)p^2-3(2c+b)ps+3bs^2$ $N_5=(3c+2a+3b)p^2-3(2c+a)ps+3as^2$ $N_6=(4c+3a+3b)p^2-3(2c+a+b)ps+3cs^2$ $a,b,c$ - can be any sign what we want. And I would like to tell you about this equation: $X^5+Y^5+Z^5=R^5$ It turns out the solution of integral complex numbers there. where: $j=\sqrt{-1}$ We make the change: $a=p^2-2ps-s^2$ $b=p^2+2ps-s^2$ $c=p^2+s^2$ Then the solutions are of the form: $X=b+jc$ $Y=-b+jc$ $Z=a-jc$ $R=a+jc$ $p,s$ - what some integers.

Showing $\gcd(n^3 + 1, n^2 + 2) = 1$, $3$, or $9$ Given that n is a positive integer show that $\gcd(n^3 + 1, n^2 + 2) = 1$, $3$, or $9$. I'm thinking that I should be using the property of gcd that says if a and b are integers then gcd(a,b) = gcd(a+cb,b). So I can do things like decide that $\gcd(n^3 + 1, n^2 + 2) = \gcd((n^3+1) - n(n^2+2),n^2+2) = \gcd(1-2n,n^2+2)$ and then using Bezout's theorem I can get $\gcd(1-2n,n^2+2)= r(1-2n) + s(n^2 +2)$ and I can expand this to $r(1-2n) + s(n^2 +2) = r - 2rn + sn^2 + 2s$ However after some time of chasing this path using various substitutions and factorings I've gotten nowhere. Can anybody provide a hint as to how I should be looking at this problem?

Let $\:\rm d = (n^3+1,\:n^2+2).\:$ Observe that $\rm \ d \in \{1,\:3,\:9\} \iff\ d\:|\:9\iff 9\equiv 0\pmod d\:.$ mod $\rm (n^3\!-a,n^2\!-b)\!:\ a^2 \equiv n^6 \equiv b^3\:$ so $\rm\:a=-1,\:b = -2\:\Rightarrow 1\equiv -8\:\Rightarrow\: 9\equiv 0\:. \ \ $ QED Or, if you don't know congruence arithmetic, since $\rm\: x-y\:$ divides $\rm\: x^2-y^2$ and $\rm\: x^3-y^3$ $\rm n^3-a\ |\ n^6-a^2,\:\ n^2-b\ |\ n^6-b^3\ \Rightarrow\ (n^3-a,n^2-b)\ |\ n^6-b^3-(n^6-a^2) = a^2-b^3 $ Note how much simpler the proof is using congruences vs. divisibility relations on binomials. Similar congruential proofs arise when computing modulo ideals generated by binomials.

How to get from $a\sqrt{1 + \frac{b^2}{a^2}}$ to $\sqrt{a^2 + b^2}$ I have the following expression: $a\sqrt{1 + \frac{b^2}{a^2}}$. If I plug this into Wolfram Alpha, it tells me that, if $a, b$ are positive, this equals $\sqrt{a^2 + b^2}$. How do I get that result? I can't see how that could be done. Thanks

$$a\sqrt{1 + \frac{b^2}{a^2}}$$ $$=a\sqrt{\frac{a^2 + b^2}{a^2}}$$ $$=a\frac{\sqrt{a^2 + b^2}}{|a|}$$ So when $a$ and $b$ are positive, $|a|=a$. Hence: $$=\sqrt{a^2 + b^2}$$ Without the assumption: $$\sqrt{a^2} =|a|=\begin{cases} a && a \geq 0\\ -a &&a < 0\\ \end{cases}$$

Is $3^x \lt 1 + 2^x + 3^x \lt 3 \cdot 3^x$ right? Is $3^x \lt 1 + 2^x + 3^x \lt 3 \cdot 3^x$ right? This is from my lecture notes which is used to solve: But when $x = 0$, $(1 + 2^x + 3^x = 3) \gt (3^0 = 1)$? The thing is how do I choose which what expression should go on the left & right side?

When $x=0$, the left side $3^0=1$, the center is $3$ as you say, and the right side is $3\cdot 3^0=3 \cdot 1=3$ so the center and right sides are equal. But you want this for large $x$, so could restrict the range to $x \gt 1$, say.

What are the possible values for $\gcd(a^2, b)$ if $\gcd(a, b) = 3$? I was looking back at my notes on number theory and I came across this question. Let $a$, $b$ be positive integers such that $\gcd(a, b) = 3$. What are the possible values for $\gcd(a^2, b)$? I know it has to do with their prime factorization decomposition, but where do I go from here?

If $p$ is a prime, and $p|a^2$, then $p|a$; thus, if $p|a^2$ and $p|b$, then $p|a$ and $p|b$, hence $p|\gcd(a,b) = 3$. So $\gcd(a^2,b)$ must be a power of $3$. Also, $3|a^2$ and $3|b$, so $3|\gcd(a^2,b)$; so $\gcd(a^2,b)$ is a multiple of $3$. If $3^{2k}|a^2$, then $3^k|a$ (you can use prime factorization here); so if $3^{2k}|\gcd(a^2,b)$, then $3^k|\gcd(a,b) = 3$. Thus, $k\leq 1$. That is, no power of $3$ greater than $3^2$ can divide $\gcd(a^2,b)$. In summary: $\gcd(a^2,b)$ must be a power of $3$, must be a multiple of $3$, and cannot be divisible by $3^3=27$. What's left? Now give examples to show all of those possibilities can occur.

Finding a simple expression for this series expansion without a piecewise definition I am doing some practice Calculus questions and I ran into the following problem which ended up having a reduction formula with a neat expansion that I was wondering how to express in terms of a series. Here it is: consider $$ I_{n} = \int_{0}^{\pi /2} x^n \sin(x) dx $$ I obtained the reduction formula $$ I_{n} = n\left(\frac{\pi}{2}\right)^{n-1} - n I_{n-1}. $$ I started incorrectly computing up to $I_{6}$ with the reduction formula $$ I_{n} = n\left(\frac{\pi}{2}\right)^{n-1} - I_{n-1} $$ by accident which ended up having a way more interesting pattern than the correct reduction formula. So, after computing $I_{0} = 1$, the incorrect reduction expansion was, $$ I_{1} = 0 \\ I_{2} = \pi \\ I_{3} = \frac{3\pi^2}{2^2} - \pi \\ I_{4} = \frac{4\pi^3}{2^3} - \frac{3\pi^2}{2^2} + \pi \\ I_{5} = \frac{5\pi^4}{2^4} - \frac{4\pi^3}{2^3} + \frac{3\pi^2}{2^2} - \pi \\ I_{6} = \frac{6\pi^5}{2^5} - \frac{5\pi^4}{2^4} + \frac{4\pi^3}{2^3} - \frac{3\pi^2}{2^2} + \pi \\ $$ Note that $\pi = \frac{2\pi}{2^1}$, of course, which stays in the spirit of the pattern. How could I give a general expression for this series without defining a piecewise function for the odd and even cases? I was thinking of having a term in the summand with $(-1)^{2i+1}$ or $(-1)^{2i}$ depending on it was a term with an even or odd power for $n$, but that led to a piecewise defined function. I think that it will look something like the following, where $f(x)$ is some function that handles which term gets a negative or positive sign depending on whether $n$ is an even or odd power in that term: $$\sum\limits_{i=1}^{n} n \left(\frac{\pi}{2} \right)^{n-1} f(x)$$ Any ideas on how to come up with a general expression for this series?

$$ \color{green}{I_n=\sum\limits_{i=2}^{n} (-1)^{n-i}\cdot i\cdot\left(\frac{\pi}{2} \right)^{i-1}} $$

Proving an algebraic identity using the axioms of field I am trying to prove (based on the axioms of field) that $$a^3-b^3=(a-b)(a^2+ab+b^2)$$ So, my first thought was to use the distributive law to show that $$(a-b)(a^2+ab+b^2)=(a-b)\cdot a^2+(a-b)\cdot ab+(a-b)\cdot b^2$$ And then continuing from this point. My problem is that I'm not sure if the distributive law is enough to prove this identity. Any ideas? Thanks!

Indeed you need distributive, associative and commutative laws to prove your statement. In fact: $$\begin{split} (a-b)(a^2+ab+b^2) &= (a+(-b))a^2 +(a+(-b))ab+(a+(-b))b^2\\ &= a^3 +(- b)a^2+a^2b+(-b)(ab)+ab^2+(-b)b^2\\ &= a^3 - ba^2+a^2b - b(ab) +ab^2-b^3\\ &= a^3 - a^2b+a^2b - (ba)b +ab^2-b^3\\ &= a^3 -(ab)b +ab^2-b^3\\ &= a^3 -ab^2 +ab^2-b^3\\ &= a^3-b^3\; . \end{split}$$

Numbers are too large to show $65^{64}+64^{65}$ is not a prime I tried to find cycles of powers, but they are too big. Also $65^{n} \equiv 1(\text{mod}64)$, so I dont know how to use that.

Hint $\rm\ \ x^4 +\: 64\: y^4\ =\ (x^2+ 8\:y^2)^2 - (4xy)^2\ =\ (x^2-4xy + 8y^2)\:(x^2+4xy+8y^2)$ Thus $\rm\ x^{64} + 64\: y^{64} =\ (x^{32} - 4 x^{16} y^{16} + 8 y^{32})\:(x^{32} - 4 x^{16} y^{16} + 8 y^{32})$ Below are some other factorizations which frequently prove useful for integer factorization. Aurifeuille, Le Lasseur and Lucas discovered so-called Aurifeuillian factorizations of cyclotomic polynomials $\rm\;\Phi_n(x) = C_n(x)^2 - n\ x\ D_n(x)^2\,$ (aka Aurifeuillean). These play a role in factoring numbers of the form $\rm\; b^n \pm 1\:$, cf. the Cunningham Project. Below are some simple examples of such factorizations: $$\begin{array}{rl} x^4 + 2^2 \quad=& (x^2 + 2x + 2)\;(x^2 - 2x + 2) \\\\ \frac{x^6 + 3^2}{x^2 + 3} \quad=& (x^2 + 3x + 3)\;(x^2 - 3x + 3) \\\\ \frac{x^{10} - 5^5}{x^2 - 5} \quad=& (x^4 + 5x^3 + 15x^2 + 25x + 25)\;(x^4 - 5x^3 + 15x^2 - 25x + 25) \\\\ \frac{x^{12} + 6^6}{x^4 + 36} \quad=& (x^4 + 6x^3 + 18x^2 + 36x + 36)\;(x^4 - 6x^3 + 18x^2 - 36x + 36) \\\\ \end{array}$$

Is $ f(x) = \left\{ \begin{array}{lr} 0 & : x = 0 \\ e^{-1/x^{2}} & : x \neq 0 \end{array} \right. $ infinitely differentiable on all of $\mathbb{R}$? Can anyone explicitly verify that the function $ f(x) = \left\{ \begin{array}{lr} 0 & : x = 0 \\ e^{-1/x^{2}} & : x \neq 0 \end{array} \right. $ is infinitely differentiable on all of $\mathbb{R}$ and that $f^{(k)}(0) = 0$ for every $k$?

For $x\neq 0$ you get: $$\begin{split} f^\prime (x) &= \frac{2}{x^3}\ f(x)\\ f^{\prime \prime} (x) &= 2\left( \frac{2}{x^6} - \frac{3}{x^4}\right)\ f(x)\\ f^{\prime \prime \prime} (x) &= 4\left( \frac{2}{x^9} - \frac{9}{x^7} +\frac{6}{x^5} \right)\ f(x) \end{split}$$ In the above equalities you can see a path, i.e.: $$\tag{1} f^{(n)} (x) = P_{3n}\left( \frac{1}{x}\right)\ f(x)$$ where $P_{3n}(t)$ is a polynomial of degree $3n$ in $t$. Formula (1) can be proved by induction. You have three base case, hence you have only to prove the inductive step. So, assume (1) holds for $n$ and evaluate: $$\begin{split} f^{(n+1)} (x) &= \left( P_{3n}\left( \frac{1}{x}\right)\ f(x) \right)^\prime\\ &= -\frac{1}{x^2}\ \dot{P}_{3n} \left( \frac{1}{x}\right)\ f(x) + P_{3n} \left( \frac{1}{x}\right)\ f^\prime (x)\\ &= \left[ -\frac{1}{x^2}\ \dot{P}_{3n} \left( \frac{1}{x}\right) +\frac{2}{x^3}\ P_{3n} \left( \frac{1}{x}\right)\right]\ f(x)\\ &= \left[ -t^2\ \dot{P}_{3n}( t) +2t^3\ P_{3n}( t)\right]_{t=1/x}\ f(x) \end{split}$$ where the dot means derivative w.r.t. the dummy variable $t$; now the function $-t^2\ \dot{P}_{3n}( t) +2t^3\ P_{3n}( t)$ is a polynomial in $t$ of degree $3n+3=3(n+1)$, therefore: $$f^{(n+1)}(x) = P_{3(n+1)} \left( \frac{1}{x}\right)\ f(x)$$ as you wanted. Formula (1) proves that $f\in C^\infty (\mathbb{R}\setminus \{0\})$. Now, for each fixed $n$, you have: $$\lim_{x\to 0} f^{(n)}(x) = \lim_{x\to 0} P_{3n}\left( \frac{1}{x}\right)\ f(x) =0$$ for $f\in \text{o}(1/x^{3n})$ as $x\to 0$. Therefore, by an elementary consequence of Lagrange's Mean Value Theorem, any derivative of your functions is differentiable also in $0$. Thus $f\in C^\infty (\mathbb{R})$.

Show that $\displaystyle{\frac{1}{9}(10^n+3 \cdot 4^n + 5)}$ is an integer for all $n \geq 1$ Show that $\displaystyle{\frac{1}{9}(10^n+3 \cdot 4^n + 5)}$ is an integer for all $n \geq 1$ Use proof by induction. I tried for $n=1$ and got $\frac{27}{9}=3$, but if I assume for $n$ and show it for $n+1$, I don't know what method to use.

${\displaystyle{\frac{1}{9}}(10^n+3 \cdot 4^n + 5)}$ is an integer for all $n \geq 1$ Proof by induction: For $n=1, {\displaystyle{\frac{1}{9}}(10^1+3 \cdot 4^1 + 5) = \frac{27}{9} = 3}$, so the result holds for $n=1$ Assume the result to be true for $n=m$, i.e. $\displaystyle{\frac{1}{9}(10^m+3 \cdot 4^m + 5)}$ is an integer To show ${\displaystyle{\frac{1}{9}}(10^{m+1}+3 \cdot 4^{m+1} + 5)}$ is an integer. $$ \begin{align*} \displaystyle{\frac{1}{9}(10^{m+1}+3 \cdot 4^{m+1} + 5) -(10^m+3 \cdot 4^m +5 )} &= \displaystyle{\frac{1}{9}\left((10^{m+1}-10^m) +3\cdot (4^{m+1}-4^m) \right)} \\ &=\displaystyle{\frac{1}{9}\left[\left(10^m(10-1)+3 \cdot 4^m (4-1) \right)\right]}\\ &= \left(10^m+4^m\right) \end{align*} $$ which is an integer, and therefore ${\displaystyle{\frac{1}{9}}(10^{m+1}+3 \cdot 4^{m+1} + 5)}$ is an integer.

How to prove a trigonometric identity $\tan(A)=\frac{\sin2A}{1+\cos 2A}$ Show that $$ \tan(A)=\frac{\sin2A}{1+\cos 2A} $$ I've tried a few methods, and it stumped my teacher.

First, lets develop a couple of identities. Given that $\sin 2A = 2\sin A\cos A$, and $\cos 2A = \cos^2A - \sin^2 A$ we have $$\begin{array}{lll} \tan 2A &=& \frac{\sin 2A}{\cos 2A}\\ &=&\frac{2\sin A\cos A}{\cos^2 A-\sin^2A}\\ &=&\frac{2\sin A\cos A}{\cos^2 A-\sin^2A}\cdot\frac{\frac{1}{\cos^2 A}}{\frac{1}{\cos^2 A}}\\ &=&\frac{2\tan A}{1-\tan^2A} \end{array}$$ Similarly, we have $$\begin{array}{lll} \sec 2A &=& \frac{1}{\cos 2A}\\ &=&\frac{1}{\cos^2 A-\sin^2A}\\ &=&\frac{1}{\cos^2 A-\sin^2A}\cdot\frac{\frac{1}{\cos^2 A}}{\frac{1}{\cos^2 A}}\\ &=&\frac{\sec^2 A}{1-\tan^2A} \end{array}$$ But sometimes it is just as easy to represent these identities as $$\begin{array}{lll} (1-\tan^2 A)\sec 2A &=& \sec^2 A\\ (1-\tan^2 A)\tan 2A &=& 2\tan A \end{array}$$ Applying these identities to the problem at hand we have $$\begin{array}{lll} \frac{\sin 2A}{1+\cos 2A}&=& \frac{\sin 2A}{1+\cos 2A}\cdot\frac{\frac{1}{\cos 2A}}{\frac{1}{\cos 2A}}\\ &=& \frac{\tan 2A}{\sec 2A +1}\\ &=& \frac{(1-\tan^2 A)\tan 2A}{(1-\tan^2 A)(\sec 2A +1)}\\ &=& \frac{(1-\tan^2 A)\tan 2A}{(1-\tan^2 A)\sec 2A +(1-\tan^2 A)}\\ &=& \frac{2\tan A}{\sec^2 A +(1-\tan^2 A)}\\ &=& \frac{2\tan A}{(\tan^2 A+1) +(1-\tan^2 A)}\\ &=& \frac{2\tan A}{2}\\ &=& \tan A\\ \end{array}$$ Lessons learned: As just a quick scan of some of the other answers will indicate, a clever substitution can shorten your workload considerably.

Residue of a pole of order 6 I am in the process of computing an integral using the Cauchy residue theorem, and I am having a hard time computing the residue of a pole of high order. Concretely, how would one compute the residue of the function $$f(z)=\frac{(z^6+1)^2}{az^6(z-a)(z-\frac{1}{a})}$$ at $z=0$? Although it is not needed here, $a$ is a complex number with $|a|<1$. Thanks in advance for any insight.

$$g(z)=\frac{1}{(z-a)(z-\frac{1}{a})}=\frac{\frac{1}{a-\frac{1}{a}}}{z-a}+\frac{\frac{-1}{a-\frac{1}{a}}}{z-\frac{1}{a}}$$ we know: $$(a+b)^n =a^n+\frac{n}{1!}a^{n-1}b+\frac{n(n-1)}{2!}a^{n-2}b^2+...+b^n$$ $$ \text{As regards }: |a|<1 $$ Taylor series of f(z) is: $$g(z)=\frac{\frac{1}{a-\frac{1}{a}}}{z-a}+\frac{\frac{1}{a-\frac{1}{a}}}{\frac{1}{a}-z}=(\frac{1}{a-\frac{1}{a}}) \left[ \frac{-\frac{1}{a}}{1-\frac{z}{a}}+\frac{a}{1-az} \right]$$ $$g(z)=(\frac{1}{a-\frac{1}{a}}) \left[ \frac{-1}{a} \sum_{n=0}^{\infty}(\frac{z}{a})^n+a \sum_{n=0}^{\infty} (az)^n \right]$$ $$f(z)=\frac{(z^6+1)^2}{az^6}g(z)=\frac{z^{12}+2z^2+1}{az^6}g(z)=\left( \frac{z^6}{a} + \frac{2}{az^4} + \frac{1}{az^6} \right)g(z)$$ $$ f(z)= \left( \frac{z^6}{a} + \frac{2}{az^4} + \frac{1}{az^6} \right) \left(\frac{1}{a-\frac{1}{a}}\right) \left[ \frac{-1}{a} \sum_{n=0}^{\infty}(\frac{z}{a})^n+a \sum_{n=0}^{\infty} (az)^n \right]$$ $$ \text{ so residue is coefficient of term }z^{-1} $$ $$ f(z)=\frac{1}{a(a-\frac{1}{a})} \left[ \frac{-1}{a}\left( \sum_{n=0}^{\infty}\frac{z^{n+6}}{a^n} +2\sum_{n=0}^{\infty}\frac{z^{n-4}}{a^n} +\sum_{n=0}^{\infty} \frac{z^{n-6}}{a^n}\right) +a \left( \sum_{n=0}^{\infty} a^nz^{n+6}+2\sum_{n=0}^{\infty} a^nz^{n-4} +\sum_{n=0}^{\infty} a^nz^{n-6} \right) \right]$$ $$ \text{residue of function at z=0 is :} $$ $$ \frac{1}{a(a-\frac{1}{a})} \left[ \frac{-1}{a}\left( 0 +2\frac{1}{a^3} +\frac{1}{a^5}\right) +a \left( 0+2a^3 +a^5 \right) \right] $$

Evaluate $\int\limits_0^{\frac{\pi}{2}} \frac{\sin(2nx)\sin(x)}{\cos(x)}\, dx$ How to evaluate $$ \int\limits_0^{\frac{\pi}{2}} \frac{\sin(2nx)\sin(x)}{\cos(x)}\, dx $$ I don't know how to deal with it.

Method 1. Let $I(n)$ denote the integral. Then by addition formula for sine and cosine, $$\begin{align*} I(n+1) + I(n) &= \int_{0}^{\frac{\pi}{2}} \frac{[\sin((2n+2)x) + \sin(2nx)]\sin x}{\cos x} \; dx \\ &= \int_{0}^{\frac{\pi}{2}} 2\sin((2n+1)x) \sin x \; dx \\ &= \int_{0}^{\frac{\pi}{2}} [\cos(2nx) - \cos((2n+2)x)] \; dx \\ &= 0, \end{align*}$$ if $n \geq 1$. Thus we have $I(n+1) = -I(n)$ and by double angle formula for sine, $$I(1) = \int_{0}^{\frac{\pi}{2}} \frac{\sin (2x) \sin x}{\cos x} \; dx = \int_{0}^{\frac{\pi}{2}} 2 \sin^2 x \; dx = \frac{\pi}{2}.$$ Therefore we have $$I(n) = (-1)^{n-1} \frac{\pi}{2}.$$ Method 2. By the substitution $x \mapsto \pi - x$ and $x \mapsto -x$, we find that $$\int_{0}^{\frac{\pi}{2}} \frac{\sin (2nx) \sin x}{\cos x} \; dx = \int_{\frac{\pi}{2}}^{\pi} \frac{\sin (2nx) \sin x}{\cos x} \; dx = \int_{-\frac{\pi}{2}}^{0} \frac{\sin (2nx) \sin x}{\cos x} \; dx.$$ Thus we have $$\begin{align*} I(n) & = \frac{1}{4} \int_{-\pi}^{\pi} \frac{\sin (2nx) \sin x}{\cos x} \; dx \\ & = \frac{1}{4} \int_{|z|=1} \frac{\left( \frac{z^{2n} - z^{-2n}}{2i} \right) \left( \frac{z - z^{-1}}{2i} \right)}{\left( \frac{z + z^{-1}}{2} \right)} \; \frac{dz}{iz} \\ & = \frac{i}{8} \int_{|z|=1} \frac{(z^{4n} - 1) (z^2 - 1)}{z^{2n+1}(z^2 + 1)} \; dz. \end{align*}$$ The last integrad has poles only at $z = 0$. (Note that singularities at $z = \pm i$ is cancelled since numerator also contains those factors.) Expanding partially, $$ \begin{align*} \frac{(z^{4n} - 1) (z^2 - 1)}{z^{2n+1}(z^2 + 1)} & = \frac{z^{2n-1} (z^2 - 1)}{z^2 + 1} - \frac{z^2 - 1}{z^{2n+1}(z^2 + 1)} \\ & = \frac{z^{2n-1} (z^2 - 1)}{z^2 + 1} - \frac{1}{z^{2n+1}} + \frac{2}{z^{2n+1}(z^2 + 1)} \\ & = \frac{z^{2n-1} (z^2 - 1)}{z^2 + 1} - \frac{1}{z^{2n+1}} + 2 \sum_{k=0}^{\infty} (-1)^{k} z^{2k-2n-1}. \end{align*}$$ Thus the residue of the integrand at $z = 0$ is $2 (-1)^n$, and therefore $$I(n) = \frac{i}{8} \cdot 2\pi i \cdot 2 (-1)^{n} = (-1)^{n-1}\frac{\pi}{2}.$$ Method 3. (Advanced Calculus) This method is just a sledgehammer method, but it reveals an interesting fact that even a nice integral with nice value at each integer point can yield a very bizarre answer for non-integral argument. By the substitution $x \mapsto \frac{\pi}{2} - x$, we have $$I(n) = (-1)^{n-1} \int_{0}^{\frac{\pi}{2}} \frac{\sin 2nx}{\sin x} \cos x \; dx.$$ Now, from a lengthy calculation, we find that for all $w > 0$, $$ \int_{0}^{\frac{\pi}{2}} \frac{\sin 2wx}{\sin x} \cos x \; dx = \frac{\pi}{2} + \left[ \log 2 - \psi_0 (1 + w) + \psi_0 \left( 1 + \frac{w}{2}\right) - \frac{1}{2w} \right] \sin \pi w.$$ Thus plugging a positive integer $n$, we obtain $$ \int_{0}^{\frac{\pi}{2}} \frac{\sin 2nx}{\sin x} \cos x \; dx = \frac{\pi}{2},$$ which immediately yields the formula for $I(n)$.

Primes modulo which a given quadratic equation has roots Given a quadratic polynomial $ax^2 + bx + c$, with $a$, $b$ and $c$ being integers, is there a characterization of all primes $p$ for which the equation $$ax^2 + bx + c \equiv 0 \pmod p$$ has solutions? I have seen it mentioned that it follows from quadratic reciprocity that the set is precisely the primes in some arithmetic progression, but the statement may require some tweaking. The set of primes modulo which $1 + \lambda = \lambda^2$ has solutions seems to be $$5, 11, 19, 29, 31, 41, 59, 61, 71, 79, 89, 101, 109, 131, 139, 149, 151, 179, 181, 191, 199, \dots$$ which are ($5$ and) the primes that are $1$ or $9$ modulo $10$. (Can the question also be answered for equations of higher degree?)

I never noticed this one before. $$ x^3 - x - 1 \equiv 0 \pmod p $$ has one root for odd primes $p$ with $(-23|p) = -1.$ $$ x^3 - x - 1 \equiv 0 \pmod p $$ has three distinct roots for odd $p$ with $(-23|p) = 1$ and $p = u^2 + 23 v^2 $ in integers. $$ x^3 - x - 1 \equiv 0 \pmod p $$ has no roots for odd $p$ with $(-23|p) = 1$ and $p = 3u^2 + 2 u v + 8 v^2 $ in integers (not necessarily positive integers). Here we go, no roots $\pmod 2,$ but a doubled root and a single $\pmod {23},$ as $$ x^3 - x - 1 \equiv (x - 3)(x-10)^2 \pmod {23}. $$ Strange but true. Easy to confirm by computer for primes up to 1000, say. The example you can see completely proved in books, Ireland and Rosen for example, is $x^3 - 2,$ often with the phrase "the cubic character of 2" and the topic "cubic reciprocity." $2$ is a cube for primes $p=2,3$ and any prime $p \equiv 2 \pmod 3.$ Also, $2$ is a cube for primes $p \equiv 1 \pmod 3$ and $p = x^2 + 27 y^2$ in integers. However, $2$ is not a cube for primes $p \equiv 1 \pmod 3$ and $p = 4x^2 +2 x y + 7 y^2$ in integers. (Gauss) $3$ is a cube for primes $p=2,3$ and any prime $p \equiv 2 \pmod 3.$ Also, $3$ is a cube for primes $p \equiv 1 \pmod 3$ and $p = x^2 + x y + 61 y^2$ in integers. However, $3$ is not a cube for primes $p \equiv 1 \pmod 3$ and $p = 7x^2 +3 x y + 9 y^2$ in integers. (Jacobi)

How to solve $\int_0^\pi{\frac{\cos{nx}}{5 + 4\cos{x}}}dx$? How can I solve the following integral? $$\int_0^\pi{\frac{\cos{nx}}{5 + 4\cos{x}}}dx, n \in \mathbb{N}$$

To elaborate on Pantelis Damianou's answer $$ \newcommand{\cis}{\operatorname{cis}} \begin{align} \int_0^\pi\frac{\cos(nx)}{5+4\cos(x)}\mathrm{d}x &=\frac12\int_{-\pi}^\pi\frac{\cos(nx)}{5+4\cos(x)}\mathrm{d}x\\ &=\frac12\int_{-\pi}^\pi\frac{\cis(nx)}{5+2(\cis(x)+\cis(-x))}\mathrm{d}x\\ &=\frac12\int_{-\pi}^\pi\frac{\cis(x)\cis(nx)}{2\cis^2(x)+5\cis(x)+2}\mathrm{d}x\\ &=\frac{1}{2i}\int_{-\pi}^\pi\frac{\cis(nx)}{2\cis^2(x)+5\cis(x)+2}\mathrm{d}\cis(x)\\ &=\frac{1}{2i}\oint\frac{z^n}{2z^2+5z+2}\mathrm{d}z \end{align} $$ where the integral is counterclockwise around the unit circle and $\cis(x)=e^{ix}$. Factor $2z^2+5z+2$ and use partial fractions. However, I only get a singularity at $z=-\frac12$ (and one at $z=-2$, but that is outside the unit circle, so of no consequence). Now that a complete solution has been posted, I will finish this using residues: $$ \begin{align} \frac{1}{2i}\oint\frac{z^n}{2z^2+5z+2}\mathrm{d}z &=\frac{1}{6i}\oint\left(\frac{2}{2z+1}-\frac{1}{z+2}\right)\,z^n\,\mathrm{d}z\\ &=\frac{1}{6i}\oint\frac{z^n}{z+1/2}\mathrm{d}z\\ &=\frac{\pi}{3}\left(-\frac12\right)^n \end{align} $$

How to get rid of the integral in this equation $\int\limits_{x_0}^{x}{\sqrt{1+\left(\dfrac{d}{dx}f(x)\right)^2}dx}$? How to get rid of the integral $\int\limits_{x_0}^{x}{\sqrt{1+\left(\dfrac{d}{dx}f(x)\right)^2}dx}$ when $f(x)=x^2$?

Summarising the comments, you'll get $$ \int\limits_{x_0}^{x}{\sqrt{1+\left(\dfrac{d}{dt}f(t)\right)^2}dt} =\int\limits_{x_0}^{x}{\sqrt{1+\left(\dfrac{d}{dt}t^2\right)^2}dt} =\int\limits_{x_0}^{x}{\sqrt{1+4t^2}dt} $$ To solve the last one substitute $t=\tan(u)/2$ and $dt=\sec^2(u)/2du$. Then $\sqrt{1+4t^2}= \sqrt{\tan^2(u)+1}=\sec(u)$, so we get as antiderviative: $$ \begin{eqnarray} \frac{1}{2}\int \sec^3(u) du&=&\frac{1}{4}\tan(u)\sec(u)+\frac{1}{4}\int \sec(u)du+\text{const.}\\ &=&\frac{1}{4}\tan(u)\sec(u)+\frac{1}{4}\log(\tan(u)+\sec(u))+\text{const.} \\ &=& \frac{t}{2}\sqrt{1+4t^2}+\frac{1}{4}\log(2t+\sqrt{1+4t^2})+\text{const.}\\ &=& \frac{1}{4}\left(2t\sqrt{1+4t^2}+\sinh^{-1}(2t) \right)+\text{const.}. \end{eqnarray} $$ Put in your limits and your done: $$ \int\limits_{x_0}^{x}{\sqrt{1+4t^2}dx}=\left[\frac{1}{4}\left(2t\sqrt{1+4t^2}+\sinh^{-1}(2t) \right) \right]_{x_0}^x $$

Root Calculation by Hand Is it possible to calculate and find the solution of $ \; \large{105^{1/5}} \; $ without using a calculator? Could someone show me how to do that, please? Well, when I use a Casio scientific calculator, I get this answer: $105^{1/5}\approx " 2.536517482 "$. With WolframAlpha, I can an even more accurate result.

Another way of doing this would be to use logarithm, just like Euler did: $$ 105^{1/5} = \mathrm{e}^{\tfrac{1}{5} \log (105)} = \mathrm{e}^{\tfrac{1}{5} \log (3)} \cdot \mathrm{e}^{\tfrac{1}{5} \log (5)} \cdot \mathrm{e}^{\tfrac{1}{5} \log (7)} $$ Use $$\log(3) = \log\left(\frac{2+1}{2-1}\right) = \log\left(1+\frac{1}{2}\right)-\log\left(1-\frac{1}{2}\right) = \sum_{k=0}^\infty \frac{2}{2k+1} \cdot \frac{1}{2^{2k+1}} = 1 + \frac{1}{12} + \frac{1}{80} + \frac{1}{448} = 1.0.83333+0.0125 + 0.0022 = 1.09803$$ $$ \log(5) = \log\frac{4+1}{4-1} + \log(3) = \log(3) + \sum_{k=0}^\infty \frac{2}{2k+1} \cdot \frac{1}{4^{2k+1}} = \log(3) + \frac{1}{2} + \frac{1}{96} +\frac{1}{2560} $$ $$ \log(7) = \log\frac{8-1}{8+1} + 2 \log(3) = 2 \log(3) - \sum_{k=0}^\infty \frac{2}{2k+1} \cdot \frac{1}{8^{2k+1}} = 2 \cdot \log(3) - \frac{1}{4} - \frac{1}{768} $$ Thus $$ \frac{1}{5} \left( \log(3) + \log(5) + \log(7)\right) = \frac{4}{5} \log(3) + \frac{1}{5} \left( \frac{1}{2} - \frac{1}{4} + \frac{1}{96} - \frac{1}{768} + \frac{1}{2560} \right) = \frac{4}{5} \log(3) + \frac{1993}{38400}= 0.9303 = 1-0.0697 $$ Now $$ \exp(0.9303) = \mathrm{e} \cdot \left( 1 - 0.0697 \right) = 2.71828 \cdot 0.9303 = 2.5288 $$