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Calculating $ x= \sqrt{4+\sqrt{4-\sqrt{ 4+\dots}}} $ If $ x= \sqrt{4+\sqrt{4-\sqrt{ 4+\sqrt{4-\sqrt{ 4+\sqrt{4-\dots}}}}}} $ then find value of 2x-1 I tried the usual strategy of squaring and substituting the rest of series by x again but could not solve.

I assume you mean $$ x=\sqrt{4+\sqrt{4-\sqrt{4+\sqrt{4-\sqrt{4\pm\ldots}}}}}$$ so that $$ \begin{align}x&=\sqrt{4+\sqrt{4-x}}\\ x^2&=4+\sqrt{4-x}\\ (x^2-4)^2&=4-x\\ 0 &= x^4-8x^2+x+12= (x^2-x-3)(x^2+x-4)\end{align}$$ Since clearly $x\ge \sqrt 4=2$, the second factor is $x^2+x-4\ge2>0$, which leaves us with the positive solution $x=\frac{1+\sqrt{13}}{2}\approx 2.3027756 $ from the first factor.

$\sin^2 \alpha + \sin^2 \beta - \cos \gamma < M$ given that the sum of the angles is $\pi$ Question: Find the least real value of $M$ such that the following inequality holds: $$\sin^2 \alpha + \sin^2 \beta - \cos \gamma < M$$ Given that $\alpha, \beta, \gamma \in \mathbb{R}^+$, $\alpha + \beta + \gamma = \pi$ My attempt: Step 1: Replace $\sin^2 t$ with $1 - \cos^2 t$ $2 - \cos^2 \alpha - \cos^2 \beta - \cos \gamma < M$ Furthermore, note that $- \cos \gamma = \cos (180 - \gamma) = \cos(\alpha + \beta)$ In addition, use this identity: $-\frac{1}{2}(\cos(2x) + \cos(2y)) = -\cos^2 x - \cos^2 y + 1$ to arrive at the following: $$1 - \frac{1}{2}(\cos(2 \alpha) + \cos(2 \beta)) + \cos(\alpha + \beta) < M$$ And, conveniently, $\frac{1}{2} (\cos(2\alpha) + \cos(2\beta)) = \cos(\alpha + \beta) \cos(\alpha - \beta)$ $$1 - (\cos(\alpha + \beta))(\cos(\alpha - \beta) - 1) < M$$ $$ (\cos (\alpha + \beta))(1 - \cos(\alpha - \beta)) < M - 1$$ From the inequality $ab \leq \frac{(a + b)^2}{4}$, we have that $$(\cos (\alpha + \beta))(1 - \cos(\alpha - \beta)) \leq \frac{(1 + 2 \sin \alpha \sin \beta)^2}{4} < \frac{1}{4}$$ Since $0 < 2 \sin \alpha \sin \beta < 2$ That's all I have so far. Is it logical to then say that $M - 1 = \frac{1}{4}$? I don't think it is because that doesn't make sense to me (it's asking for the least M, and how do i know that $\frac{1}{4}$ is minimized?), but I am not very experienced by any means in dealing with inequalities. Though I do see that $\frac{5}{4}$ is approachable with $\alpha = \frac{\pi}{3} - h, \beta = h, \gamma = \frac{2 \pi}{3}$ where $h$ is an infitesimally small number. Can anyone give me some guidance to finish up this question?

It should be a lot easier to look at the function: $$\sin^2(x)+ \sin^2(y)-\cos(\pi - x - y)$$ And note it is symmetric when interchanging $x$ and $y$, and noting that comparing it's derivatives to zero leads to $\sin(2x)=\sin(2y)$. Thus $x=y+n\pi$. Now find the maximum value of the function: $$\sin^2(x)+\sin^2(x+n \pi)-\cos(\pi-2 x-n \pi)$$ And show that $M=3$. Edit: As the OP's question constrains $x,y>0$, and since we have shown that there are no local maxima in the region, the maximum must lie on the boundary, i.e. either $x$ or $y$ must be either $0$ or $\pi$. Examine all four options and find for instance that when $y=\pi$: $$M=\max_x\ \sin^2(x)-\cos(x) = 5/4$$

Solving $\arcsin(1-x)-2\arcsin(x)=\pi/2$ \begin{eqnarray*} \arcsin(1-x)-2\arcsin(x) & = & \frac{\pi}{2}\\ 1-x & = & \sin\left(\frac{\pi}{2}+2\arcsin(x)\right)\\ & = & \cos\left(2\arcsin(x)\right)\\ & = & 1-2\left(\sin\left(\arcsin(x)\right)\right)^{2}\\ & = & 1-2x^{2}\\ x & = & 2x^{2}\\ x\left(x-\frac{1}{2}\right) & = & 0 \end{eqnarray*} So $x=0$ or $x=\frac{1}{2}$ But puttig $x=\frac{1}{2}$ in the original expression gives $-\frac {\pi} 4 \ne \frac \pi 2$ So, why do we get $x=-1/2$ as an answer?

In your first step you added an extra solution. Since $\arcsin x$ must be smaller than $\pi/2$, the first line reads: $$\arcsin(1-x)= \frac{\pi}{2}+2\arcsin(x) \le \frac{\pi}{2}$$ Thus, $x\le 0$ as well. Now, by taking the $\sin$ of both sides, you took a function that was only defined up to $x=1$ (e.g. $\arcsin(x-1)$ ) and extended it to all reals (e.g $x-1$). Here is where you added the extra solution.

Showing convergence or divergence of a sequence I need to determine if the series with $n$th term $\ln(n)e^{-\sqrt n}$ converges or diverges. I've tried numerous identities for $ln(x)$ and $e^{x}$ and various convergence tests but I'm still very stuck.

To prove the given series convergent, we use the following inequalities: * *For $ x > 1$ , $Ln(x) < x$. *$Exp(x) > 1 + x + \dfrac{x^2}{2!} + \dfrac{x^3}{3!} + \dfrac{x^4}{4!} + \dfrac{x^5}{5!}$ for $x > 0.$ Let $a(n) = \dfrac{ln(n)}{e^{\sqrt{n}}}$, then $a(n) < \dfrac{2ln(\sqrt{n})}{1 + \sqrt{n}+ \dfrac{\sqrt{n}^2}{2!} + \dfrac{\sqrt{n}^3}{3!} + \dfrac{\sqrt{n}^4}{4!} + \dfrac{\sqrt{n}^5}{5!}} < \dfrac{2 \sqrt{n}}{\dfrac{\sqrt{n}^5}{5!}}= \dfrac{1}{60n^2} = b(n)$. for n large enough. But the series whose nth term $ b(n) = \dfrac{1}{60n^2}$ converges, and by comparison test, the original series converges.

partial Fibonacci summation Let $F_{n}$ be the n-th Fibonacci number. How to calculate the summation like following: $\sum_{n \geq 0} F_{3n} \cdot 2^{-3n}$

Here's an approach via generating functions. As the Fibonacci recurrence is defined by $F_{n+2} = F_{n+1} + F_n$, we have $$\sum_{n \ge 0} F_{n+2}z^{n+2} = \sum_{n \ge 0} F_{n+1}z^{n+1}z + \sum_{n \ge 0}F_nz^nz^2$$ which with the generating function $G(z) = \sum_{n\ge0} F_n z^n$ gives $$G(z) - F_0 - F_1z = zG(z) - zF_0 + z^2G(z)$$ and therefore (using $F_0 = 0$ and $F_1 = 1$), $$G(z) - z = zG(z) + z^2G(z) \implies G(z) = \frac{z}{1 - z - z^2}.$$ This much is well-known. Now let $\omega$ be a third root of unity, so that $\omega^3 = 1$. Then $$G(z) + G(z\omega) + G(z\omega^2) = \sum_{n\ge0} F_nz^n(1 + \omega^n + \omega^{2n}) = \sum_{n\ge0} 3F_{3n}z^{3n},$$ as we have $$1 + \omega^n + \omega^{2n} = \begin{cases} 3 \text{ if $3$ divides $n$}\\0 \text{ otherwise.}\end{cases}$$ This means that the number $\sum_{n\ge0} F_{3n}2^{-3n}$ we want is $$\frac{G(z) + G(z\omega) + G(z\omega^2)}{3}$$ with $z = \frac12$. The sum turns out to be $$\frac13\left(\frac{1/2}{1-1/2-(1/2)^2} + \frac{\omega(1/2)}{1-\omega(1/2)-\omega^2(1/2)^2} + \frac{\omega^2(1/2)}{1-\omega^2(1/2)-\omega(1/2)^2}\right)$$ $$=\frac13\left(2 - \frac{14}{31}\right) = \frac{16}{31}.$$

Solving simultaneous equations in terms of variables If $x+y = m$ and $x-y=n$ then $(x^2-y^2) -2x$ is equal to in terms of $m$ and $n$ only! How do you solve?

Notice that $$(x^2 - y^2) - 2x = (x + y)(x - y) - [(x + y) + (x - y)]$$ But you know what $x + y$ and $x - y$ are ($m$ and $n$ respectively). I believe it is very, very simple to continue on from here. This is the simplest method to solve this particular problem. Of course, you could also choose to express $x$ and $y$ individually in terms of $m$ and $n$ and then substitute them in. If we do a bit of manipulation, we get $x = \frac{m + n}{2}$ and $y = \frac{m - n}{2}$. Substituting into $(x^2 - y^2) - 2x = (x + y)(x - y) - 2x$: $$\left(\frac{m + n + m - n}{2}\right)\left(\frac{m + n - m + n}{2}\right) - 2\left(\frac{m + n}{2}\right)\\ =\left(\frac{2m}{2}\right)\left(\frac{2n}{2}\right) - (m + n)\\ = mn - m - n$$ The key idea is to split $x^2 - y^2$ into $(x + y)(x - y)$ instead of actually going all the way to evaluate two squares.

Find the sum of $\binom{100}1 + 2\binom{100}2 + 4\binom{100}3 +8\binom{100}4+\dots+2^{99}\binom{100}{100}$ Find the sum of $\binom{100}1 + 2\binom{100}2 + 4\binom{100}3 +8\binom{100}4+\dots+2^{99}\binom{100}{100}$ How you guys work on with this question? With the geometric progression? Combination? Or anyother way to calculate?

$$\sum_{r=1}^{100}2^{r-1}\binom{100}r=\frac12\sum_{r=1}^{100}2^r\binom{100}r=\frac12\left[(1+2)^{100}-1\right]$$

A hint on why if $c$ is not a square in $\mathbf{F}_p$, then $c^{(p - 1)/2} \equiv -1 \mod p$ Let $\mathbf{F}_p$ be a finite field and let $c \in (\mathbf{Z}/p)^\times$. If $x^2 = c$ does not have a solution in $\mathbf{F}_p$, then $c^\frac{p - 1}{2} \equiv -1 \mod p$. I will try to prove the contrapositive: Suppose that $c^\frac{p - 1}{2} \not\equiv -1 \mod p$. We show that $x^2 = c$ has a solution in $\mathbf{F}_p$. By Fermat's Theorem, $c^{p - 1} \equiv 1 \mod p$. Then $c^{p - 1} - 1 \equiv 0 \mod p$. Then $(c^\frac{p - 1}{2} + 1)(c^\frac{p - 1}{2} - 1) \equiv 0 \mod p$. This implies that either $c^\frac{p - 1}{2} \equiv -1 \mod p$ or $c^\frac{p - 1}{2} \equiv 1 \mod p$. Hence it must be that $c^\frac{p - 1}{2} \equiv 1 \mod p$. I'm not sure how to derive an $a \in \mathbf{F}_p$ such that $a^2 = c$.

We assume $p$ is odd, and use an argument that yields additional information. There are two possibilities, $p$ is of the form $4k-1$, and $p$ is of the form $4k+1$. Let $p$ be of the form $4k-1$. If $c^{(p-1)/2}\equiv 1\pmod{p}$, then $c^{(p+1)/2}\equiv c\pmod{p}$. But $\frac{p+1}{2}=2k$, and therefore $$(c^k)^2\equiv c\pmod{p}.$$ To complete things, we show that if $p$ is of the form $4k+1$, then the congruence $x^2\equiv -1\pmod{p}$ has a solution. The argument goes back at least to Dirichlet. Suppose that $x^2\equiv -1\pmod{p}$ has no solution. Consider the numbers $1,2,\dots,p-1$. For any $a$ in this collection, there is a $b$ such that $ab\equiv -1\pmod{p}$. Pair numbers $a$ and $b$ if $ab\equiv -1\pmod{p}$. Since the congruence $x^2\equiv -1\pmod{p}$ has no solution, no number is paired with itself. The product of all the pairs is $(p-1)!$, and it is also congruent to $(-1)^{(p-1)/2}$ modulo $p$. Since $\frac{p-1}{2}$ is even, it follows that $(p-1)!\equiv 1\pmod{p}$, which contradicts Wilson's Theorem.

Help with a simple derivative I am trying to solve $\dfrac {6} {\sqrt {x^3+6}}$ and so far I made it to $6(x^3+6)^{-\frac 1 2}$ then I continued and now I have $(x^3+6)^{- \frac 3 2} * 3x^2$ and I cannot figure out what how to find the constant that should be before the parenthesis.

$$\frac{d}{dx}(\frac{6}{\sqrt{x^3+6}})= 6\frac{d}{dx}(\frac{1}{\sqrt{x^3+6}})\\ \implies 6\frac{d}{dx}(\frac{1}{\sqrt{x^3+6}})=6\frac{d}{d(x^2+6)}(\frac{1}{\sqrt{x^3+6}})\frac{d}{dx}(x^3+6)=\frac{-3}{(x^3+6)^{3/2}}\cdot 3x^2=-9\frac{x^2}{(x^3+6)^{3/2}}$$

Solve $\sin x - \cos x = -1$ for the interval $(0, 2\pi)$ We have an exam in $3$ hours and I need help how to solve such trigonometric equations for intervals. How to solve $$\sin x - \cos x = -1$$ for the interval $(0, 2\pi)$.

Method $\#1$ Avoid squaring which immediately introduces extraneous roots which demand exclusion We have $\displaystyle\sin x-\cos x=-1$ $$\iff\sin x=-(1-\cos x)\iff2\sin\frac x2\cos\frac x2=-2\sin^2\frac x2$$ $$\iff2\sin\frac x2\left(\cos\frac x2+\sin\frac x2\right)=0$$ If $\displaystyle \sin\frac x2=0,\frac x2=n\pi\iff x=2n\pi$ where $n$ is any integer If $\displaystyle\cos\frac x2+\sin\frac x2=0\iff\sin\frac x2=-\cos\frac x2$ $\displaystyle\iff\tan\frac x2=-1=-\tan\frac\pi4=\tan\left(-\frac\pi4\right)$ $\displaystyle\iff\frac x2=m\pi-\frac\pi4\iff x=2m\pi-\frac\pi2$ where $m$ is any integer Method $\#2$ Let $\displaystyle1=r\cos\phi,-1=r\sin\phi\ \ \ \ (1)$ where $r>0$ $\displaystyle\cos\phi=\frac1r>0$ and $\displaystyle\sin\phi=-\frac1r<0$ $\displaystyle\implies\phi$ lies in the fourth Quadrant On division, $\displaystyle\frac{r\sin\phi}{r\cos\phi}=-1\iff\tan\phi=-1$ $\displaystyle\implies\phi=-\frac\pi4$ $\displaystyle\sin x-\cos x=-1\implies r\cos\phi\sin x+r\sin\phi\cos x=r\sin\phi$ $\displaystyle\implies\sin(x+\phi)=\sin\phi$ $\displaystyle\implies x+\phi=r\pi+(-1)^r\phi$ where $r$ is any integer If $r$ is even $=2a$(say) $\displaystyle\implies x=2a\pi$ If $r$ is odd $=2a+1$(say) $\displaystyle\implies x+\phi=(2a+1)\pi-\phi\iff x=2a\pi+\pi-2\phi$ Method $\#3$ Using Weierstrass Substitution, $\displaystyle\frac{2u}{1+u^2}-\frac{1-u^2}{1+u^2}=-1\ \ \ \ (2)$ where $u=\tan\frac x2$ $\displaystyle\iff2u^2+2u=0\iff u(u+1)=0$ If $\displaystyle u=0,\tan\frac x2=0\iff\frac x2=b\pi$ where $b$ is any integer If $\displaystyle u=-1,\tan\frac x2=-1$ which has been addressed in Method $\#1$

Find basis so Transformation Matrix will be diagonal $e_1,e_2$ will be basis for $V$. $W$ has a basis $\{e_1+ ae_2,2e_1+be_2\}$. Choose an $a,b$ s.t. that the basis for $W$ will have a transformation matrix $T$ will be in diagonal form. $T(e_1) = 1e_1+5e_2$ $T(e_2) = 2e_1+4e_2$ $V$ and $W$ are linear spaces of dimension $2$.

In the basis for $V$, $$T_{[V]} = \left(\begin{matrix} 1 & 2 \\ 5 & 4 \end{matrix} \right)$$ If you want the transformation $T$ written in $W$'s basis to be diagonal, then you want each basis vector of $W$ to be mapped to some multiple of itself: $$T_{[W]} = \left(\begin{matrix} \lambda_1 & 0 \\ 0 & \lambda_2 \end{matrix} \right)$$ You know that $T : e_1 \mapsto e_1 + 5 e_2$ and $T : e_2 \mapsto 2e_1 + 4e_2$. Using this, you can solve for $a$ and $b$ by stipulating that: $$e_1 + a e_2 \mapsto \lambda_1 e_1 + \lambda_1 a e_2,$$ $$2 e_1 + b e_2 \mapsto \lambda_2 2 e_1 + \lambda_2 b e_2$$ Appling the map $T$: $$T_{[V]} \left( \begin{matrix} 1 \\ a \end{matrix} \right) = \lambda_1 \left( \begin{matrix} 1 \\ a \end{matrix} \right) = \left( \begin{matrix} 1 \\ 5 \end{matrix} \right) + a\left( \begin{matrix} 2 \\ 4 \end{matrix} \right)$$ $$T_{[V]} \left( \begin{matrix} 2 \\ b \end{matrix} \right) = \lambda_2 \left( \begin{matrix} 2 \\ b \end{matrix} \right) = 2\left( \begin{matrix} 1 \\ 5 \end{matrix} \right) + b\left( \begin{matrix} 2 \\ 4 \end{matrix} \right)$$ We've ended up with simultaneous equations for $\lambda_1$ and $\lambda_2$: $$\begin{cases} \lambda_1 = 1 + 2a \\ a \lambda_1 = 5 + 4a \end{cases},\quad \quad \begin{cases} 2\lambda_2 = 2 + 2b \\ b \lambda_2 = 10 + 4b \end{cases}$$ Rearranging these equations gives two quadratics: $$2a^2 - 3a - 5 = 0$$ $$b^2 - 3b - 10 = 0$$ Our solutions are $a = 5/2$ or $-1$ and $b = 5$ or $-2$. Any of these four possible configurations will make $T$ written in $W$'s basis be a diagonal matrix. As an example, set $a = -1, b = 5$. Then: $$\underbrace{\left( \begin{matrix} 1 & 2 \\ 5 & 4 \end{matrix} \right)}_{T_{[V]}} \left( \begin{matrix} 1 & 2 \\ -1 & 5 \end{matrix} \right) = \left( \begin{matrix} -1 & 12 \\ 1 & 30 \end{matrix} \right)$$ It can be seen that the first basis vector of $W$ is an eigenvector of $T$ with eigenvalue of $-1$, and the second is an eigenvector of $T$ with eigenvalue $6$. Thus: $$T_{[W]} = \left( \begin{matrix} -1 & 0 \\ 0 & 6 \end{matrix} \right)$$ With this choice for $a$ and $b$.

Find the limit of $\lim_{x\to 0}\frac{\sqrt{x^2+a^2}-a}{\sqrt{x^2+b^2}-b}$ Can someone help me solve this limit? $$\lim_{x\to0}\frac{\sqrt{x^2+a^2}-a}{\sqrt{x^2+b^2}-b}$$ with $a>0$ and $b>0$.

No need for L'Hopital - we simply multiply and divide by the conjugate radical expression: \begin{align} \frac{\sqrt{x^2+a^2}-a}{\sqrt{x^2+b^2}-b}&=\left(\frac{\sqrt{x^2+a^2}-a}{\sqrt{x^2+b^2}-b}\cdot\frac{\sqrt{x^2+a^2}+a}{\sqrt{x^2+b^2}+b}\right)\cdot\frac{\sqrt{x^2+b^2}+b}{\sqrt{x^2+a^2}+a} \\ &=\frac{x^2+a^2-a^2}{x^2+b^2-b^2}\cdot\frac{\sqrt{x^2+b^2}+b}{\sqrt{x^2+a^2}+a}= \frac{\sqrt{x^2+b^2}+b}{\sqrt{x^2+a^2}+a}\to\frac{2b}{2a}=\frac{b}{a}. \end{align}

Proof: $a^2 - b^2 = (a-b)(a+b)$ holds $\forall a,b \in R$ iff R is commutative We want to show that for some ring $R$, the equality $a^2 - b^2 = (a-b)(a+b)$ holds $\forall a,b \in R$ if and only if $R$ is commutative. Here's my proof --- I'm not sure if the first part stands up to examination. I'd be grateful if someone could take a look. Forward: $a^2 -b^2 = (a-b)(a+b) \forall a,b \in R$ implies $R$ is commutative Let $x = (a-b)$. Then \begin{align} x(a+b) &= xa+xb\\ &= (a-b)a + (a-b)b\\ &= a^2 -ba + ab - b^2\end{align} Then we note that $a^2 - ba + ab - b^2 = a^2 - b^2$ iff $-ba + ab = 0$ if and only if $ab=ba$ iff $R$ is commutative. Backwards: $R$ is commutative implies $a^2 - b^2 = (a-b)(a+b) \forall a,b \in R$. Let $x = (a+b)$. Then $(a-b)x = ax - bx = a(a+b) - b(a+b) = a^2 + ab - ba - b^2$. $R$ is commutative, so $ab-ba = 0$, so $a^2 + ab - ba - b^2 = a^2 - b^2$.

What you did is correct, as far as I can tell. Perhaps a shorter solution is that $(a^2 - b^2) - (a-b)(a+b) = ba - ab$. This is $0$ iff the $a$ and $b$ commute. So, the expression $(a^2 - b^2) - (a-b)(a+b)$ is identically $0$ iff the ring is commutative.

Solve inequality: $\frac{2x}{x^2-2x+5} + \frac{3x}{x^2+2x+5} \leq \frac{7}{8}$ Rational method to solve $\frac{2x}{x^2-2x+5} + \frac{3x}{x^2+2x+5} \leq \frac{7}{8}$ inequality? I tried to lead fractions to a common denominator, but I think that this way is wrong, because I had fourth-degree polynomial in the numerator.

HINT: As $\displaystyle x^2\pm2x+5=(x\pm1)^2+4\ge4>0$ for real $x$ we can safely multiply out either sides by $(x^2+2x+5)(x^2-2x+5)$ Then find the roots of the Quartic Equation

What is the exact coefficient of $x^{12}$ in $(2+x^3+x^6)^{10}$? What is the coefficient of $x^{12}$ in $(2+x^3+x^6)^{10}$? I figure you need to pick $x^3$ 4 times so $C(10,4)$...but what happens with the other numbers/variables??? Can someone explain to me how this is done properly? Thanks. EDIT: $(x + y)^n = C(n,k) \cdot x^{n-k} \cdot y^k$ EX: Find the term for $x^5$ in $(5-2x)^8$ Answer: $C(8,5) \cdot (-2)^5 \cdot 5^3$ How can I use this info to solve a polynomial based question such as the featured? Answer: To sum up all information provided by everyone (Thanks!!!): $(C(10,4) * 2^6) + (C(10,2) * 2^8) + (C(10,1) * C(9,2) + 2^7) = 71040 $

Hint 1: $x^{12} = x^6 x^6 = x^3 x^3 x^3 x^3 = x^3 x^3 x^6$ How many ways to pick $x^6 x^6$? Everything that's not an $x$ term is a multiplier of $2$. This would be $2^8 {10 \choose 2}$ for a total of $10$ elements. For four $x^3$ there would be $10 \choose 4$ ways to pick. Hint 2: For $x^3 x^3 x^6$ there are $10$ ways to choose the $x^6$, $10-1 \choose 2$ ways to pick $x^3$. Hopefully you can figure out the final sum of the three cases from here.

Calculate ratio $\frac{p^{3n}}{(\frac{p}{2})^{3n}}$ How do I calculate this ratio? I do not know even where to begin. $$\frac{p^{3n}}{(\frac{p}{2})^{3n}}$$ Thanks

Regarding the original question: $$ \frac{p^{3n}}{\frac{p^{3n}}{2}} = \frac{p^{3n}}{\frac{p^{3n}}{2}} \cdot 1 = \frac{p^{3n}}{\frac{p^{3n}}{2}} \cdot \frac{2}{2} = \frac{p^{3n}\cdot 2}{\frac{p^{3n}}{2} \cdot 2} = \frac{p^{3n}\cdot 2}{p^{3n}\cdot 1} = \frac{p^{3n}}{p^{3n}} \cdot \frac 2 1 = 1\cdot \frac 2 1 = 2. $$ In general we have $$ \frac{A}{\frac p q} = \frac{A\cdot q}{\frac p q \cdot q} = A\cdot\frac{q}{p}, $$ so dividing by $\frac p q$ is the same as multiplying by $\frac q p$. Using this you also get $$ \frac{p^{3n}}{\frac{p^{3n}}{2}} = p^{3n}\cdot\frac{2}{p^{3n}} = 2. $$ For your updated question we use $\left(\frac{p}{q}\right)^k = \frac{p^k}{q^k}$ to obtain $$ \frac{p^{3n}}{\left(\frac{p}{2}\right)^{3n}} = \frac{p^{3n}}{\left(\frac{p^{3n}}{2^{3n}}\right)} = p^{3n} \cdot \frac{2^{3n}}{p^{3n}} = 2^{3n}. $$

Order of operations when using evaluation bar Suppose we have the function \begin{align*} f(x) = \sin(x) \end{align*} with first derivative \begin{align*} \frac{d}{dx}f(x) = \cos(x). \end{align*} If we evaluate $f'(x)$ at $x=0$, the result depends on whether you evaluate $f(0)$ or differentiate $f(x)$ first. \begin{align*} \displaystyle \frac{d}{dx}f(x)\mid_{x = 0} = \cos(x)\mid_{x = 0} = 1\\ \displaystyle \frac{d}{dx}f(x)\mid_{x = 0} = \frac{d}{dx}f(0) = \frac{d}{dx}0 = 0 \end{align*} First question: Does this mean the following two statements are not equivalent? \begin{align*} \displaystyle \left(\frac{d}{dx}f(x)\right)\mid_{x = 0}\\ \displaystyle \frac{d}{dx}\left(f(x)\mid_{x = 0}\right) \end{align*} Second question: if so, which of the following is true, and why? \begin{align*} \displaystyle \frac{d}{dx}f(x)\mid_{x = 0} = \left(\frac{d}{dx}f(x)\right)\mid_{x = 0} \end{align*} or \begin{align*} \displaystyle \frac{d}{dx}f(x)\mid_{x = 0} = \frac{d}{dx}\left(f(x)\mid_{x = 0}\right) \end{align*}

If the following were true: $$\frac{d}{dx}f(x)\mid_{x = n} = \frac{d}{dx}\left(f(x)\mid_{x = n}\right)$$ Then it would always be 0. Why? Because once you evaluate a function of x at a particular value of x, it is no longer a function of x; it's a value (or, a constant function). The example (sin, cos) you gave is a little misleading. Consider instead the function $f(x)=x^2$ , and instead the point $x=3$ : $$\frac{d}{dx}f(x)\mid_{x = 0} = \frac{d}{dx}\left(f(x)\mid_{x = 0}\right)\\ f(x)\mid_{x = 0}=9\\ \frac{d}{dx}9=0$$

Combining results with Chinese Remainder Theorem? $9x^2 + 27x + 27 \equiv 0 \pmod{21}$ What is the "correct" way to solve this using the Chinese Remainder Theorem? How do I correctly solve this modulo $3$ and modulo $7$ without brute force?

First, modulo $3$, your congruence reduces to $0\equiv 0$, because all coefficients are multiples of $3$. Therefore there are three solutions: $x\equiv 0,1,2 (\mod 3)$. Working modulo $7$ the congruence becomes $2x^2+6x+6\equiv 0$, or $x^2+3x+3\equiv 0$, since we can multiply both sides by the inverse of $2$. To solve this, we use the quadratic formula, noting that in our case $\frac{1}{2a}=4$. Thus $x\equiv 4(-3\pm\sqrt{9-12})\equiv 4(4\pm\sqrt{4})\equiv 4(4\pm2)\equiv 3$ or $1$. Now, you have three solutions modulo $3$ and two solutions modulo $7$, so that's six combinations to feed into the Chinese Remainder Theorem. $x\equiv_3 0, x\equiv_7 1 \implies x\equiv_{21} 15 \\ x\equiv_3 1, x\equiv_7 1 \implies x\equiv_{21} 1 \\ x\equiv_3 2, x\equiv_7 1 \implies x\equiv_{21} 8 \\ x\equiv_3 0, x\equiv_7 3 \implies x\equiv_{21} 3 \\ x\equiv_3 1, x\equiv_7 3 \implies x\equiv_{21} 10 \\ x\equiv_3 2, x\equiv_7 3 \implies x\equiv_{21} 17 \\$ As noted in the comments above, since the solution is everything modulo $3$, it makes more sense in this case to just solve the problem modulo $7$. However, you wanted to see the gory details, and the above method generalizes just fine.

Proof that certain number is an integer Let $k$ be an integer and let $$ n=\sqrt[3]{k+\sqrt{k^2-1}}+\sqrt[3]{k-\sqrt{k^2-1}}+1 $$ Prove that $n^3-3n^2$ is an integer. (I have started posting any problem I get stuck on and then subsequently find a good solution to here on math.se, primarily to get new solutions which might be even better, and to get checked if my solutions are ok)

Since $$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$$ $$a+b+c=0\implies a^3+b^3+c^3=3abc$$ Let $a=\sqrt[3]{k+\sqrt{k^2-1}}$, $b=\sqrt[3]{k-\sqrt{k^2-1}}$, $c=1-n$ Clearly we have $a+b+c=0$ $$a^3+b^3+c^3=3abc\tag{1}$$ $$ab=\sqrt[3]{k+\sqrt{k^2-1}}\cdot\sqrt[3]{k-\sqrt{k^2-1}}$$ $$ab=\sqrt[3]{k^2-(\sqrt{k^2-1})^2}$$ $$ab=\sqrt[3]1 = 1$$ Substituting back in $(1)$ $$k+\sqrt{k^2-1}+k-\sqrt{k^2-1}+(1-n)^3=3(1-n)$$ $$2k+1-3n+3n^2-n^3 = 3-3n$$ $$\require{cancel}{2k+1-\cancel{3n}+3n^2-n^3 = 3-\cancel{3n}}$$ $$n^3-3n^2=2k-2$$ which is obviously an integer.

Proving that $\sum_{k=0}^n\frac{1}{n\choose k}=\frac{n+1}{2^{n+1}}\sum_{k=1}^{n+1}\frac{2^k}{k}$ I want to prove for any positive integer $n$, the following equation holds: $$\sum_{k=0}^n\frac{1}{n\choose k}=\frac{n+1}{2^{n+1}}\sum_{k=1}^{n+1}\frac{2^k}{k}$$ I tried to expand $2^k$ as $\sum_{i=0}^k{k\choose i}$ and interchange summation, also tried let $f(x)=\sum_{k=1}^{n+1}\frac{x^k}{k}$ and compute $f'(x)$. But it seems I'm not on the right path.

Note that successive terms in the sum on the left turn out have a simple expression for their sum: when $k \ne 0$, we have $$\frac{1}{\binom{n}{k-1}} + \frac1{\binom{n}{k}} = \frac{\binom{n}{k}+\binom{n}{k-1}}{\binom{n}{k-1}\binom{n}{k}} = \frac{\binom{n+1}{k}}{\binom{n}{k-1}\binom{n}{k}} = \frac{\frac{n+1}{k}\binom{n}{k-1}}{\binom{n}{k-1}\binom{n}{k}} = \frac{\frac{n+1}{k}}{\binom{n}{k}}$$ where we've used Pascal's rule and the "absorption" identity that $\binom{n}{k} = \frac{n}{k} \binom{n-1}{k-1}$ for $k \neq 0$. Applying absorption again further gives the above expression to be equal to $$\frac{\frac{n+1}{k}}{\binom{n}{k}} = \frac{\frac{n+1}{k}}{\frac{n}{k}\binom{n-1}{k-1}} = \frac{n+1}{n} \frac{1}{\binom{n-1}{k-1}}$$ This gives a strategy for evaluating the sum on the left: $$\begin{align} 2\sum_{k=0}^n \frac{1}{\binom{n}{k}} &= \sum_{k=1}^{n} \left( \frac{1}{\binom{n}{k-1}} + \frac1{\binom{n}{k}}\right) + 2 \\ &= 2 + \frac{n+1}{n} \sum_{k=1}^{n} \frac{1}{\binom{n-1}{k-1}} \end{align}$$ or, calling the left-hand-side sum as $L_n = \sum_{k=0}^n \frac{1}{\binom{n}{k}}$, we have $$2 L_n = 2 + \frac{n+1}{n} L_{n-1}$$ $$L_n = \frac{n+1}{2n}L_{n-1} + 1$$ Calling the right-hand-side term $R_n$, we have $$\begin{align} \frac{2^{n+1}}{n+1} R_n &= \sum_{k=1}^{n+1}\frac{2^k}{k} \\ &= \frac{2^n}{n} R_{n-1} + \frac{2^{n+1}}{n+1} \end{align}$$ thus $$R_n = \frac{n+1}{2n}R_{n-1} + 1$$ and both the LHS $L_n$ and RHS $R_n$ satisfy the same recurrence and have the same initial values (check for $n=1$, say), so they are equal.

Is $f$ differentiable at $(x,y)$? I am working on a practice question, and I am not sure if what I have done would be considered, 'complete justification'. I would greatly appreciate some feedback or helpful advice on how it could be better etc. The question is here: Let $f: \mathbb {R}^2 \to \mathbb{R} $ be a function defined by: $$ \ f(x,y) = \begin{cases} \frac{\sin(x^2 + y^2)}{x^2 + y^2} & \text{if } (x,y) \ne (0,0) \\ 1 & \text{if } (x,y) = (0,0) \end{cases} $$ Is $f$ diff'ble at $(x,y) = (0,0)$? Here is what I have: By definition, $f(x,y)$ is diff'ble at $(0,0)$ if $$ \lim_{(x,y) \to (0,0)} f(x,y)= \frac{f(x,y) - [f(0,0) + f_{x}(0,0)(x-0) + f_{y}(0,0)(y-0)]}{\sqrt{x^2 + y^2}} = 0\tag{*}$$ Since $f(x,y)$ is piecewise, $f_{x}(0,0)$ and $f_{y}(0,0)$ is derived from 1st principles: So, $$\begin{align}f_{x}(0,0) &= \lim_{h \to 0} \frac{\frac{\sin((0+h)^2 + (0)^2)}{(0+h)^2 + (0)^2} - f_{x}(0,0)}{h} \\ &= \lim_{h \to 0} \frac{\frac{\sin(h^2)}{h^2} - 1}{h}\\ &= \lim_{h \to 0} \frac{\sin(h^2) - h^2}{h^3} \\ &= \lim_{h \to 0} \frac{2h \cos(h^2) - 2h}{3h^2}\end{align} $$ by L'Hopital's rule, apply this twice more I can see that the limit is $0$. Similarly, $f_y(0,0)$ is derived the same way. Then from $(*)$, I have: $$f_{x}(0,0) = \lim_{h \to 0} \frac{\frac{\sin(x^2 + y^2)}{x^2 + y^2} - 1}{\sqrt{x^2 + y^2}} $$ From here I haven't had luck trying to get this to $0$. So instead, I try to make it easier. Since $f_x(0,0)$ and $f_y(0,0)$ exist we must evaluate whether $f_x(x,y)$ is continuous at $(x,y)=(0,0)$ i.e. if $$\lim_{(x,y) \to (0,0)} \frac{\sin(x^2 + y^2)}{x^2 + y^2} = 1 $$ then $f(x,y)$ is differentiable at $(0,0)$. So letting $u = x^2 + y^2$, $$\lim_{(x,y) \to (0,0)} \frac{\sin(x^2 + y^2)}{x^2 + y^2} = \lim_{ (x,y) \to (0,0)} \frac{\sin(u)}{u} = 1 $$ by L'Hopital once more. Therefore, $f$ is differentiable at (0,0). I feel like I am perhaps not completely justifying this, because I take a short cut - but is it valid? Should I perhaps use a epsilon-delta proof? Or is there a way to work with that tricky limit I have in $(*)$? Many thanks for the help in advance!

You've already found $$\lim \limits_{(x,y)\to (0,0)}\left(\dfrac{f(x,y) - [f(0,0) + f_{x}(0,0)(x-0) + f_{y}(0,0)(y-0)]}{\sqrt{x^2 + y^2}}\right)=\lim \limits_{(x,y)\to (0,0)}\left( \dfrac{\frac{\sin(x^2 + y^2)}{x^2 + y^2} - 1}{\sqrt{x^2 + y^2}}\right).$$ To evaluate $\lim \limits_{(x,y)\to (0,0)}\left( \dfrac{\frac{\sin(x^2 + y^2)}{x^2 + y^2} - 1}{\sqrt{x^2 + y^2}}\right)$ simplify the denominator and change the variables with $x=\rho \cos(\theta)$ and $y=\rho\sin(\theta)$ to get $$\begin{align}\lim \limits_{(x,y)\to (0,0)}\left( \dfrac{\frac{\sin(x^2 + y^2)}{x^2 + y^2} - 1}{\sqrt{x^2 + y^2}}\right)&=\lim \limits_{(x,y) \to (0,0)}\left(\dfrac{\sin(x^2+y^2)-(x^2+y^2)}{(x^2+y^2)^{3/2}}\right)\\ &=\lim \limits_{\rho \to 0}\left(\dfrac{\sin\left(\rho^2\right)-\rho ^2}{|\rho| ^3}\right)\\ &=\lim \limits_{\rho \to 0}\left(\dfrac{\rho ^2+o\left(\rho^2\right)-\rho^2}{|\rho|^3}\right)\\ &=0.\end{align}$$

Rationalization of $\frac{2\sqrt{6}}{\sqrt{2}+\sqrt{3}+\sqrt{5}}$ Question: $$\frac{2\sqrt{6}}{\sqrt{2}+\sqrt{3}+\sqrt{5}}$$ equals: My approach: I tried to rationalize the denominator by multiplying it by $\frac{\sqrt{2}-\sqrt{3}-\sqrt{5}}{\sqrt{2}-\sqrt{3}-\sqrt{5}}$. And got the result to be (after a long calculation): $$\frac{\sqrt{24}+\sqrt{40}-\sqrt{16}}{\sqrt{12}+\sqrt{5}}$$ which is totally not in accordance with the answer, $\sqrt{2}+\sqrt{3}-\sqrt{5}$. Can someone please explain this/give hints to me.

What I would do is multiply by the first term plus the conjugate of the last two terms. I have coloured the important parts of the following expression to make it easier to understand. $$\frac{2\sqrt{6}}{\color{green}{\sqrt{2}+\sqrt{3}+}\color{red}{\sqrt{5}}}\cdot\frac{\color{green}{\sqrt{2}+\sqrt{3}-}\color{red}{\sqrt{5}}}{\color{green}{\sqrt{2}+\sqrt{3}-}\color{red}{\sqrt{5}}}$$ $$\frac{2\sqrt{6}(\sqrt{2}+\sqrt{3}-\sqrt{5})}{(\sqrt{2}+\sqrt{3}+\sqrt{5})(\sqrt{2}+\sqrt{3}-\sqrt{5})}$$ Why do I do this, you ask? Remember the difference of squares formula: $$a^2-b^2=(a+b)(a-b)$$ I am actually letting $a=\color{green}{\sqrt{2}+\sqrt{3}}$ and $b=\color{red}{\sqrt{5}}$. Therefore, our fraction can be rewritten as: $$\frac{2\sqrt{6}\sqrt{2}+2\sqrt{6}\sqrt{3}-2\sqrt{6}\sqrt{5}}{(\color{green}{\sqrt{2}+\sqrt{3}})^2-(\color{red}{\sqrt{5}})^2}$$ $$=\frac{4\sqrt{3}+6\sqrt{2}-2\sqrt{30}}{2+2\sqrt{6}+3-5}$$ Oh. How nice. The integers in the denominator cancel out! $$\frac{4\sqrt{3}+6\sqrt{2}-2\sqrt{30}}{2\sqrt{6}}$$ Multiply by $\dfrac{2\sqrt{6}}{2\sqrt{6}}$ $$\frac{4\sqrt{3}+6\sqrt{2}-2\sqrt{30}}{2\sqrt{6}}\cdot\frac{2\sqrt{6}}{2\sqrt{6}}$$ $$=\frac{8\sqrt{3}\sqrt{6}+12\sqrt{2}\sqrt{6}-4\sqrt{30}\sqrt{6}}{(2\sqrt{6})^2}$$ $$=\frac{\color{red}{24}\sqrt{2}+\color{red}{24}\sqrt{3}-\color{red}{24}\sqrt{5}}{\color{red}{24}}$$ $$=\frac{\color{red}{24}(\sqrt{2}+\sqrt{3}-\sqrt{5})}{\color{red}{24}}$$ Cancel $24$ out in the numerator and denominator and you get: $$\sqrt{2}+\sqrt{3}-\sqrt{5}$$ $$\displaystyle \color{green}{\boxed{\therefore \dfrac{2\sqrt{6}}{\sqrt{2}+\sqrt{3}+\sqrt{5}}=\sqrt{2}+\sqrt{3}-\sqrt{5}}}$$ There is actually a much shorter way. Let's go back to the fraction $$\frac{2\sqrt{6}\sqrt{2}+2\sqrt{6}\sqrt{3}-2\sqrt{6}\sqrt{5}}{(\sqrt{2}+\sqrt{3})^2-(\sqrt{5})^2}$$ $$=\frac{2\sqrt{6}\sqrt{2}+2\sqrt{6}\sqrt{3}-2\sqrt{6}\sqrt{5}}{2+2\sqrt{2}\sqrt{3}+3-5}$$ $$=\frac{\color{red}{2\sqrt{6}}\sqrt{2}+\color{red}{2\sqrt{6}}\sqrt{3}-\color{red}{2\sqrt{6}}\sqrt{5}}{\color{red}{2\sqrt{6}}}$$ Do you see that we can factor out $2\sqrt{6}$ in the numerator? $$\frac{\color{red}{2\sqrt{6}}(\sqrt{2}+\sqrt{3}-\sqrt{5})}{\color{red}{2\sqrt{6}}}$$ Cancel $2\sqrt{6}$ in the numerator and the denominator out, and you get: $$\sqrt{2}+\sqrt{3}-\sqrt{5}$$ $$\displaystyle \color{green}{\boxed{\therefore \dfrac{2\sqrt{6}}{\sqrt{2}+\sqrt3+\sqrt5}=\sqrt{2}+\sqrt{3}-\sqrt{5}}}$$ Hope I helped!

Find the point on a plane $3x + 4y + z = 1$ that is closest to $(1,0,1)$ Is anyone able to help me with regards to this question? Find the point on a plane $3x + 4y + z = 1$ that is closest to $(1,0,1)$ http://i.imgur.com/ywdsJi7.png

Let $(x, y, z)$ be the point in question. The distance is given by $\sqrt{(x - 1)^2 + y^2 + (z - 1)^2}$. By Cauchy Schwarz, $\left((x-1)^2 + y^2 + (z-1)^2\right)(3^2 + 4^2 + 1^2) \geq (3x + 4y + z - 4)^2$, so $\left((x-1)^2 + y^2 + (z-1)^2 \right) \geq \frac{9}{26}$ Equality is reached when $\frac{x-1}{3} = \frac{y}{4} = \frac{z-1}{1}$. Solving using $3x + 4y + z = 1$ gives $\left(\frac{17}{26}, -\frac{6}{13}, \frac{23}{26} \right)$

Is this differential form closed / exact? Could you check if I calculated the exterior derivative of this differential form $\omega$ correctly? $\omega \in \Omega_2 ^{\infty} (\mathbb{R}^3 \setminus \{0\})$ $\omega = (x^2 + y^2 + z^2)^{\frac{-3}{2}}(x \mbox{d}y \wedge \mbox{d}z + y \mbox{d}z \wedge \mbox{d}x + z \mbox{d}x \wedge \mbox{d}y)$ $\mbox{d} \omega = \mbox{d}( (x^2 + y^2 + z^2)^{\frac{-3}{2}}x (\mbox{d}y \wedge \mbox{d}z) + (x^2 + y^2 + z^2)^{\frac{-3}{2}}(y \mbox{d}z \wedge \mbox{d}x ) + (x^2 + y^2 + z^2)^{\frac{-3}{2}}( z \mbox{d}x \wedge \mbox{d}y))$ We differentiate the fist summand only by $x$, second only by $y$ and the third only by $z$, because in the other cases we get forms like $\mbox{d}x \wedge \mbox{d}x$, and due to the fact that exterior derivative is antisymmentric such forms are zero. So: $\mbox{d} \omega = \left( - \frac{3}{2} (x^2 + y^2 + z^2)^{\frac{-5}{2}} 2x \cdot x + (x^2 + y^2 + z^2)^{\frac{-3}{2}}\right) \wedge \mbox{d}x \wedge \mbox{d}y \wedge \mbox{d}z + \left( - \frac{3}{2} (x^2 + y^2 + z^2)^{\frac{-5}{2}} 2y \cdot y + (x^2 + y^2 + z^2)^{\frac{-3}{2}}\right) \wedge \mbox{d}y \wedge \mbox{d}z \wedge \mbox{d}x + \left( - \frac{3}{2} (x^2 + y^2 + z^2)^{\frac{-5}{2}} 2z \cdot z + (x^2 + y^2 + z^2)^{\frac{-3}{2}}\right) \wedge \mbox{d}z \wedge \mbox{d}x \wedge \mbox{d}y$ Because $\mbox{d}x \wedge \mbox{d}y \wedge \mbox{d}z = \mbox{d}y \wedge dz \wedge dx = \mbox{d}z \wedge \mbox{d}x \wedge \mbox{d}y$ (we need two transpositions), we have: $\mbox{d} \omega = \left( -3(x^2 + y^2 + z^2) ^{\frac{-5}{2}} (x^2 + y^2 + z^2) + (x^2 + y^2 + z^2) ^{\frac{-3}{2}} \right) \mbox{d}x \wedge \mbox{d}y \wedge \mbox{d}z$ But this isn't equal to zero. Are my calculations correct? I have one more question - how do determine if this form is exact? I know a differential form $\omega \in \Omega_n (U)$ is exact if there exists $\beta \in \Omega_{n+1} (U)$ s. t. $\omega = \mbox{d} \beta$. But I don't know how to guess such $\beta$. Could you help me? $\mbox{d} \omega = \left( -3(x^2 + y^2 + z^2) ^{\frac{-5}{2}} (x^2 + y^2 + z^2) + (x^2 + y^2 + z^2) ^{\frac{-3}{2}} \right) \mbox{d}x \wedge \mbox{d}y \wedge \mbox{d}z$ Thank you. EDIT: There is a mistake in my calculations: There should be $\mbox{d} \omega = \left( -3(x^2 + y^2 + z^2) ^{\frac{-5}{2}} (x^2 + y^2 + z^2) + 3(x^2 + y^2 + z^2) ^{\frac{-3}{2}} \right) \mbox{d}x \wedge \mbox{d}y \wedge \mbox{d}z$ and this is zero, so the form is closed.

I did not find any errors in your calculations. So it looks like $\omega$ is not closed. And how to figure out if you have exact form, given it is closed? In this case you know that $H^2_{dR}(\mathbb{R}^3 \setminus \{0\}) \simeq \mathbb{R}$ There is just one closed but not exact form(up to scalar multiple). And it is(I'think) $$\alpha = \star d \frac1{\sqrt{x^2+y^2+z^2}}.$$ If you suspect that your form $\omega$ is exact, you know that $\omega + \gamma \alpha$ has to be exact for some $\gamma \in \mathbb{R}$. You can calculate integral $$\int_{S^2} \omega + \gamma \alpha$$ and find $\gamma_0$ for which is the integral zero, than $\omega + \gamma_0 \alpha$ is exact. If $\gamma_0$ happens to be zero than $\omega$ is exact. edit: Finding $\gamma_0$ is simple thanks to linearity of integral. $$ \gamma_0 = - \frac{\int_{S^2} \omega}{\int_{S^2} \alpha} $$ In most cases you just want to show that $\int_{S^2} \omega = 0$. Thanks to that you do not need to know exactly what $\alpha$ is. edit2: How to compute $\int_{S^2} \omega$ You can think of integrating 2-forms(in 3d) as integrating vector field $\vec F$. $$\int \omega = \int \vec F \cdot \vec n dA$$ $\vec n$ is outer normal. I found this question which discuss correspondence of 1,2-forms and vector fields. So we can apply this to $\int_{S^2} \omega$ $$ \int_{S^2} \omega = \int_{S^2} (x^2 + y^2 + z^2)^{\frac{-3}{2}}(x \mbox{d}y \wedge \mbox{d}z + y \mbox{d}z \wedge \mbox{d}x + z \mbox{d}x \wedge \mbox{d}y)= $$ $$ = \int_{S^2} 1 n\cdot n dA = 4 \pi $$ because $\vec n = (x,y,z)$ on unit sphere.

Prove that $\frac{x \log(x)}{x^2-1} \leq \frac{1}{2}$ for positive $x$, $x \neq 1$. I'd like to prove $$\frac{x \,\log(x)}{x^2-1} \leq \frac{1}{2} $$ for positive $x$, $x \neq 1$. I showed that the limit of the function $f(x) = \frac{x \,\text{log}(x)}{x^2-1}$ is zero as $x$ tends to infinity. But not sure what to do next.

If $x > 1$ we prove equivalent inequality: $2x \ln x \leq x^2 - 1 \iff 2x\ln x - x^2 + 1 \leq 0$. Look at $f(x) = 2x \ln x - x^2 + 1$ for $x > 1$. We have $f'(x) = 2\ln x + 2 - 2x$, and $f''(x) = \dfrac{2}{x} - 2 < 0$. So $f'(x) < f'(1) = 0$. So $f(x) < f(1) = 0$, and this means $2x\ln x \leq x^2 - 1$. If $0 < x < 1$ we prove equivalent inequality: $x^2 - 1 \leq 2x\ln x \iff x^2 - 1 - 2x\ln x \leq 0$. Look at $f(x) = x^2 - 1 - 2x\ln x$ on $0 < x < 1$. $f'(x) = 2x - 2\ln x - 2$, and $f''(x) = 2 - \dfrac{2}{x} < 0$. So $f'(x) > f'(1) = 0 \implies f(x) < f(1) = 0 \implies x^2 - 1 \leq 2x\ln x$. Done.

Prove $\int\limits_{0}^{\pi/2}\frac{dx}{1+\sin^2{(\tan{x})}}=\frac{\pi}{2\sqrt{2}}\bigl(\frac{e^2+3-2\sqrt{2}}{e^2-3+2\sqrt{2}}\bigr)$ Prove the following integral $$I=\int\limits_{0}^{\frac{\pi}{2}}\dfrac{dx}{1+\sin^2{(\tan{x})}}=\dfrac{\pi}{2\sqrt{2}}\left(\dfrac{e^2+3-2\sqrt{2}}{e^2-3+2\sqrt{2}}\right)$$ This integral result was calculated using Mathematica and I like this integral. But I can't solve it. My idea: Let $$\tan{x}=t\Longrightarrow dx=\dfrac{1}{1+t^2}dt$$ so $$I=\int\limits_{0}^{\infty}\dfrac{dt}{1+\sin^2{t}}\cdot \dfrac{1}{1+t^2}$$ then I can't proceed. Can you help me? Thank you.

Let $f : [-1,1] \to \mathbb{R}$ be any continuous even function on $[-1,1]$. Consider following integral $$I_f \stackrel{def}{=} \int_0^{\pi/2} f(\sin(\tan x)) dx = \frac12 \int_{-\pi/2}^{\pi/2} f(\sin(\tan x)) dx = \frac12 \int_{-\infty}^{\infty} f(\sin y)\frac{dy}{1+y^2} $$ where $y = \tan x$. Since $f(\cdot)$ is even, $f(\sin y)$ is periodic in $y$ with period $\pi$. We can rewrite $I_f$ as $$ I_f = \frac12 \int_{-\pi/2}^{\pi/2} f(\sin y)g(y) dy $$ where $$g(y) = \sum_{n=-\infty}^\infty \frac{1}{1+(y+n\pi)^2} = \frac{1}{2i}\sum_{n=-\infty}^\infty \left(\frac{1}{n\pi + y - i} - \frac{1}{n\pi+y + i}\right)$$ Let $z = \tan y$ and $T = -i\tan i = \tanh 1 = \frac{e^2-1}{e^2+1}$. Recall following expansion of $\cot y$, $$\cot y = \frac{1}{y} + \sum_{n \in \mathbb{Z} \setminus \{0\}}\left( \frac{1}{y+n\pi} - \frac{1}{n\pi}\right)$$ We find $$g(y) = \frac{1}{2i}(\cot(y-i) - \cot(y+i)) = \frac{1}{2i} \left(\frac{1 + izT}{z - iT} - \frac{1 - izT}{z + iT}\right) = \frac{(z^2+1)T}{z^2+T^2} $$ As a result, we can get rid of the explicit trigonometric dependence in $I_f$: $$ I_f = \frac12 \int_{-\infty}^\infty f\left(\frac{z}{\sqrt{1+z^2}}\right) \frac{(z^2+1)T}{z^2+T^2} \frac{dz}{1+z^2} = \frac{T}{2}\int_{-\infty}^\infty f\left(\frac{z}{\sqrt{1+z^2}}\right) \frac{dz}{z^2+T^2} $$ For any $a > 0$, let $b = \sqrt{a^2+1}$ and apply above formula to $f_{a}(z) \stackrel{def}{=} \frac{1}{a^2+z^2}$, we have $$\begin{align} I_{f_a} &= \frac{T}{2}\int_{-\infty}^\infty \frac{z^2+1}{a^2+b^2z^2}\frac{dz}{z^2+T^2}\\ &= \frac{T}{2(a^2-b^2T^2)}\int_{-\infty}^\infty \left(\frac{1-T^2}{z^2+T^2} - \frac{1}{a^2+b^2z^2}\right) dz\\ &= \frac{T}{2(a^2-b^2T^2)}\left((1-T^2)\frac{\pi}{T} - \frac{\pi}{ab}\right) = \frac{\pi}{2ab}\frac{b + aT}{a + bT} \end{align} $$ When $a = 1$, $b$ becomes $\sqrt{2}$ and $I_{f_1}$ reduces to the integral $I$ we want to compute, i.e. $$I = I_{f_1} = \frac{\pi}{2\sqrt{2}}\frac{\sqrt{2}+\frac{e^2-1}{e^2+1}}{1+\sqrt{2}\frac{e^2-1}{e^2+1}} = \frac{\pi}{2\sqrt{2}}\frac{(\sqrt{2}+1)e^2 + (\sqrt{2}-1)}{(\sqrt{2}+1)e^2 - (\sqrt{2}-1)} = \frac{\pi}{2\sqrt{2}}\frac{e^2 + (\sqrt{2}-1)^2}{e^2 - (\sqrt{2}-1)^2} $$

cyclic subgroup elements I'm having hard time finding elements of the cyclic subgroup $\langle a\rangle$ in $S_{10}$, where $a = (1\ 3\ 8\ 2\ 5\ 10)(4\ 7\ 6\ 9)$ This is my attempt: \begin{align} a^2 &= (1\ 8\ 5\ 10)(4\ 6\ 9) \\ a^3 &= (1\ 3\ 5\ 10)(4\ 7\ 9\ 6) \\ a^4 &= (1\ 5\ 10)(4\ 9\ 7) \\ a^5 &= (1\ 3\ 8\ 2\ 10)(7\ 6) \\ a^6 &= (1\ 8\ 10)(4\ 6\ 9) \\ a^7 &= (1\ 3\ 10)(4\ 7\ 9\ 6) \\ a^8 &= (1\ 10)(4\ 9\ 7) \\ a^9 &= (1\ 3\ 8\ 2\ 5\ 10)(7\ 6) \\ a^{10} &= (1\ 8\ 5)(4\ 6\ 9) \\ a^{11} &= (1\ 3\ 5\ 10)(4\ 7\ 9) \\ a^{12} &= (1\ 5)(4\ 9\ 7\ 6) \end{align} I suspect I already went wrong somewhere. I understand I need to get to $e = (1)$ at some point. Is there a way to check and make sure there are no mistakes when you calculate this?

If you have $$a = (1\,3\,8\,2\,5\,10)(4\,7\,6\,9)$$ that means that $a$ is the permutation that takes 1 to 3, 3 to 8, 8 to 2, and so on. The permutation $a^2$ is obtained by applying $a$ twice. Since $a$ takes 1 to 3, and then 3 to 8, $a^2$ takes 1 to 8. $$\begin{array}{ccc} a^0 & a^1 & a^2 \\ \hline 1 & 3 & 8 \\ 2 & 5 & 10 \\ 3 & 8 & 2 \\ 4 & 7 & 6 \\ 5 & 10 & 1 \\ 6 & 9 & 4 \\ 7 & 6 & 9 \\ 8 & 2 & 5 \\ 9 & 4 & 7 \\ 10 & 1 & 3 \end{array}$$ Reading off the first and last column of the first row, we have that $a^2$ takes 1 to 8, so it begins $a^2 = (1\,8\ldots)\ldots$. Reading the first and last column of the 8th row, we see that $a^2$ takes 8 to 5, so $a^2 = (1\,8\,5\ldots)\ldots$. Reading off the rest of the rows similarly, we get: $$a^2 = (1\,8\,5)(2\,10\,3)(4\,6)(7\,9).$$ Perhaps you can take it from here.

Taylor series of $\sqrt{1+x}$ using sigma notation I want help in writing Taylor series of $\sqrt{1+x}$ using sigma notation I got till $1+\frac{x}{2}-\frac{x^2}{8}+\frac{x^3}{16}-\frac{5x^4}{128}+\ldots$ and so on. But I don't know what will come in sigma notation.

The generalized binomial theorem says that $$ (1+x)^{1/2}=\sum_{k=0}^\infty\binom{1/2}{k}x^k $$ where $\binom{1/2}{0}=1$ and for $k\ge1$, $$ \begin{align} \binom{1/2}{k} &=\frac{\frac12(\frac12-1)(\frac12-2)\cdots(\frac12-k+1)}{k!}\\ &=\frac{(-1)^{k-1}}{2^kk!}1\cdot3\cdot5\cdots(2k-3)\\ &=\frac{(-1)^{k-1}}{2^kk!}\frac{(2k-2)!}{2^{k-1}(k-1)!}\\ &=\frac{(-1)^{k-1}}{k2^{2k-1}}\binom{2k-2}{k-1} \end{align} $$ Thus, $$ \begin{align} (1+x)^{1/2} &=1-\sum_{k=1}^\infty\frac2k\binom{2k-2}{k-1}\left(-\frac x4\right)^k\\ &=1-\sum_{k=0}^\infty\frac2{k+1}\binom{2k}{k}\left(-\frac x4\right)^{k+1} \end{align} $$

How prove this inequality $\frac{1}{(a+1)^2+\sqrt{2(b^4+1)}}+\frac{1}{(b+1)^2+\sqrt{2(c^4+1)}}+\frac{1}{(c+1)^2+\sqrt{2(a^4+1)}}\le\frac{1}{2}$ let $a,b,c>0$,and such $abc=1$, show that $$\dfrac{1}{(a+1)^2+\sqrt{2(b^4+1)}}+\dfrac{1}{(b+1)^2+\sqrt{2(c^4+1)}}+\dfrac{1}{(c+1)^2+\sqrt{2(a^4+1)}}\le\dfrac{1}{2}$$ My idea : Use Cauchy-Schwarz inequality,we have $$2(b^4+1)=(1+1)(b^4+1)\ge (b^2+1)^2$$ so $$\dfrac{1}{(a+1)^2+\sqrt{2(b^4+1)}}\le\dfrac{1}{(a+1)^2+(b^2+1)}\le\dfrac{1}{(a+1)^2+\dfrac{(b+1)^2}{2}}$$ then we only prove $$\Longleftrightarrow\sum_{cyc}\dfrac{2}{2(a+1)^2+(b+1)^2}\le\dfrac{1}{2}$$ $$\Longleftrightarrow\sum_{cyc}\dfrac{1}{2(a+1)^2+(b+1)^2}\le\dfrac{1}{4}$$ Then I can't.Thank you

since Use Cauchy-Schwarz inequality and AM-GM inequality,we have $$\dfrac{1}{(a+1)^2+\sqrt{2(b^4+1)}}\le\dfrac{1}{(a+1)^2+b^2+1}\le\dfrac{1}{2ab+2a+2}=\dfrac{1}{2}\cdot\dfrac{1}{ab+a+1}$$ Use follow well know reslut,if $abc=1$,then we have $$\sum_{cyc}\dfrac{1}{ab+a+1}=1$$ poof: since \begin{align*}\dfrac{1}{ab+a+1}+\dfrac{1}{bc+b+1}+\dfrac{1}{ac+c+1}&=\dfrac{1}{ab+a+1}+\dfrac{a}{abc+ab+a}+\dfrac{ab}{abc\cdot a+abc+1}\\ &=\dfrac{ab+a+1}{ab+a+1}\\ &=1 \end{align*} so the inequality we have By Done

What type of triangle satisfies the equation $\cos(A)-\cos(B)+\sin(C)=0$? A triangle with angle $A,B,C$ satisfies the equation $\cos(A)-\cos(B)+\sin(C)=0$. What type of triangle is this? Regular, acute, right, obtuse etc. I tried using sine and cosine rule, but no result.

Using Prosthaphaeresis Formulas, $$\sin C=\cos B-\cos A=2\sin\dfrac{A+B}2\sin\dfrac{A-B}2$$ Now, $\displaystyle \sin\dfrac{A+B}2=\sin\dfrac{\pi-C}2=\cos\dfrac C2$ Using double angle formula the given relation becomes, $$2\sin\dfrac C2\cos\dfrac C2=2\cos\dfrac C2\sin\dfrac{A-B}2$$ which implies $(1)$ either $\displaystyle\cos\dfrac C2=0\iff\frac C2=(2n+1)\frac\pi2\iff C=(2n+1)\pi$ where $n$ is any integer But it is impossible as $0<C<\pi$ $(2)$ or $\displaystyle\sin\dfrac C2 =\sin\dfrac{A-B}2$ $\displaystyle\implies \dfrac C2=m\pi+(-1)^m\dfrac{A-B}2\iff C=2m\pi+(-1)^m(A-B)$ where $m$ is any integer If $m$ is even $=2r$(say), $\displaystyle C=4r\pi+(A-B)\iff B+C=4r\pi+A$ $\displaystyle\implies \pi-A=4r\pi+A\iff A=\dfrac{(1-4r)\pi}2$ As $\displaystyle0<A<\pi, 0<\dfrac{(1-4r)\pi}2<\pi\iff 0<1-4r<2\iff0>4r-1>-2$ $\displaystyle\implies r=0\implies A=\dfrac{(1-4\cdot0)\pi}2$ Similarly check for odd $m=2r+1$(say)

Integrating partial fractions I have $\int{\frac{2x+1}{x^2+4x+4}}dx$ Factorising the denominator I have $\int{\frac{2x+1}{(x+2)(x+2)}}dx$ From there I split the top term into two parts to make it easier to integrate $\int{\frac{2x+1}{(x+2)(x+2)}}dx$ = $\int{\frac{A}{(x+2)}+\frac{B}{(x+2)}}dx$ =$\int{\frac{A(x+2)}{(x+2)^2}+\frac{B(x+2)}{(x+2)^2}}dx$ Therefore $2x+1 = A(x+2) +B(x+2)$ This is where I would normally use a substitution method to eliminate either the A term or B term by letting x = something like -2, which would get rid of the A and usually leave me with the B term to solve. However since they are the same I'm not sure what to do. I've been told to try evaluate the co-efficients, but am not sure how.

All you need to do is to solve this with respect to polynomials: $2x+1=Ax+2A+Bx+2B$ $2x+1=x(A+B)+(2A+2B)$ $A+B=2 \rightarrow B=2-A$ $2A+2B=1$ $2A+4-2A=1\rightarrow 4=1$ This is contradiction! You have made an mistake in step where you split the term into two fractions, you should have done it like this: $\frac{2x+1}{(x+2)^2}=\frac{A}{x+2}+\frac{B}{(x+2)^2}$ and then proceed like usual. $A(x+2)+B=2x+1$ $A=2$ $2A+B=1$ $B=-3$ $\int \frac{2}{x+2}+\frac{-3}{(x+2)^2}dx=2ln(x+2)+\frac{3}{x+2}+C$

How to find integers $x,y$ such that $1+5^x=2\cdot 3^y$ Find this equation integer solution $$1+5^x=2\cdot 3^y$$ I know $$x=1,y=1$$ is such it and $$x=0,y=0$$ I think this equation have no other solution. But I can't prove it. This problem is from Shanghai mathematics olympiad question in 2014.

If $x=0$, then $y=0$. If $y=0$, then $x=0$. Let $x,y\ge 1$. Then $1+(-1)^x\equiv 0\pmod{3}$, so $x$ is odd, so $5^x\equiv 5\pmod{8}$, so $3^y\equiv 3\pmod{8}$, so $y$ is odd. Three cases: * *$y=3m$. Then $y\equiv 3\pmod{6}$, so $3^y\equiv -1\pmod{7}$, so $5^x\equiv 4\pmod{7}$, so $x=6t+2$, contradiction (because $x$ is odd). *$y=3m+1$. Then $y\equiv 1\pmod{6}$, so $3^y\equiv 3\pmod{7}$, so $5^x\equiv 5\pmod{7}$, so $x=6t+1$, so $5^x\equiv 5\pmod{9}$, so $3^y\equiv 3\pmod{9}$, so $y=1$, so $x=1$. *$y=3m+2$. Then $3^y\equiv 9\pmod{13}$, so $5^x\equiv 4\pmod{13}$, impossible.

Geometry Right triangles in a rectangle, find the area. Please help, I've been struggling to figure out this problem for too long... Given the area of rectangle $ABCD = 1200 \text{ unit}^2$, find the area of right triangle $ABE$

We have $[ABCD]=1200$, therefore the area of $\Delta{ABD}=\dfrac{1}{2}[ABCD]=600$. Now, calculate length $AD$ and $BD$. $$ \begin{align} [ABD]&=600\\ \dfrac{1}{2}AB\cdot AD&=600\\ \dfrac{1}{2}\cdot40\cdot AD&=600\\ 20\cdot AD&=600\\ AD&=30 \end{align} $$ Using Phytagoras' formula, we get $$ BD^2=AB^2+AD^2\quad\Rightarrow\quad BD=\sqrt{40^2+30^2}=50. $$ Now, calculate length $AE$. $$ \begin{align} [ABD]&=600\\ \dfrac{1}{2}BD\cdot AE&=600\\ \dfrac{1}{2}\cdot50\cdot AE&=600\\ 25\cdot AE&=600\\ AE&=24. \end{align} $$ Again we use Phytagoras' formula to obtain $BE$. $$ AB^2=AE^2+BE^2\quad\Rightarrow\quad BE=\sqrt{AB^2-BE^2}=32. $$ Thus, the area of $\Delta{ABE}$ is $$ \begin{align} [ABE]&=\dfrac{1}{2}AE\cdot BE\\ &=\dfrac{1}{2}\cdot24\cdot 32\\ &=384. \end{align} $$

Inequality with two binomial coefficients I am having trouble seeing why $$ \binom{k}{2} + \binom{n - k}{2} \le \binom{1}{2} + \binom{n - 1}{2} = \binom{n - 1}{2} $$

Assuming $1\le n$ and $0\le k\le n$ it's equivalent to \begin{align*}k(k-1)+(n-k)(n-k-1)&\le(n-1)(n-2)\\ (n-k)(n-k-1)&\le(n-k-1+k)(n-2)-k(k-1)\\ (n-k)(n-k-1)&\le(n-k-1)(n-2)+k(n-2)-k(k-1)\\ 0&\le(n-k-1)(k-2)+k(n-k-1)\\ 0&\le(n-k-1)(k-1)\end{align*} So it holds for all $0<k<n$.

Problem with trigonometric equation I am having trouble solving this equation $$4\cdot \sin \theta + 2 \cdot \sin 2\theta =5$$ Thank you for your help.

If you put $t=\tan \frac {\theta}2$ you obtain $$4\cdot\sin \theta+4\cdot \sin \theta\cdot\cos\theta=4\left(\frac {2t}{1+t^2}\right)\left(1+\frac{1-t^2}{1+t^2}\right)=5$$ Multiply though by $(1+t^2)^2$ to obtain $$16t=5(1+t^2)^2$$From which it is clear that any solution has $t$ positive (rhs is positive), and a quick sketch graph shows there will be two solutions. $t=0, 16t=0, 5(1+t^2)^2=5\gt 0$ $t=.5, 16t=8, 5(1+t^2)^2=\frac {125}{16}\lt 8$ $t=1, 16t=16, 5(1+t^2)^2=20\gt 16$ So there is one solution for $t$ in $(0,0.5)$ and another in $(0.5,1)$ The equation can be rewritten as a quartic $$5t^4+10t^2-16t+5=0$$

How do I solve this definite integral: $\int_0^{2\pi} \frac{dx}{\sin^{4}x + \cos^{4}x}$? $$\int_0^{2\pi} \frac{dx}{\sin^{4}x + \cos^{4}x}$$ I have already solved the indefinite integral by transforming $\sin^{4}x + \cos^{4}x$ as follows: $\sin^{4}x + \cos^{4}x = (\sin^{2}x + \cos^{2}x)^{2} - 2\cdot\sin^{2}x\cdot\cos^{2}x = 1 - \frac{1}{2}\cdot\sin^{2}(2x) = \frac{1 + \cos^{2}(2x)}{2}$, and then using the $\tan(2x) = t$ substitution. But if I do the same with the definite integral, both bounds of the integral become $0$.

\begin{aligned} & \int_{0}^{2 \pi} \frac{d x}{\sin ^{4} x+\cos ^{4} x} \\ =& \int_{0}^{2 \pi} \frac{d x}{\left(\sin ^{2} x+\cos ^{2} x\right)^{2}-2 \sin ^{2} x \cos ^{2} x} \\ =& \int_{0}^{2 \pi} \frac{d x}{1-\frac{\sin ^{2} 2 x}{2}} \\ =& 16 \int_{0}^{\frac{\pi}{4}} \frac{d x}{1+\cos ^{2} 2 x} \\ =& 16 \int_{0}^{\frac{\pi}{4}} \frac{\sec ^{2} 2 x}{\sec ^{2} 2 x+1}d x \\ =& 8 \int_{0}^{\frac{\pi}{4}} \frac{d(\tan 2 x)}{\tan ^{2}(2 x)+2} \\=&4 \sqrt{2}\left[\tan ^{-1}\left(\frac{\tan 2 x}{\sqrt{2}}\right)\right]_{0}^{\frac{\pi}{4}} \\ =& 2 \sqrt{2} \pi \end{aligned}

Using Polar Integrals to find Volume of surface Here's the Question and the work that I've done so far to solve it: Use polar coordinates to find the volume of the given solid. Enclosed by the hyperboloid $ −x^2 − y^2 + z^2 = 61$ and the plane $z = 8$ Ah MathJaX is confusing by the way. Not sure how to close the text without having to add something like theta. $\int_0^{\sqrt{3}} 8r-\sqrt{61+r^2}r dr$ (with respect to $r$) I evaluated this to be $12 - 512/2 + \dfrac{61^3}{3}$ What do I put here I then took the integral with respect to $\theta$ from $0$ to $2\pi$ because it is a circle covering the whole $xy$ plane. This gave the above evaluation multiplied by $2\pi$. However this answer is wrong? Can anyone see something wrong with my reasoning?

Notice that $-x^2-y^2+z^2=61$ can be rewritten as $x^2+y^2=z^2-61$ or $$x^2+y^2=\sqrt{z^2-61}^2$$ Given a horizontal slice of the upper sheet of the hyperboloid (i.e. given $z$), it should be plain to see from the graph (and the equation that we have a circle of radius $\sqrt{z^2-61}$. The vertex of the upper sheet occurs at $(0,0,\sqrt{61})$ because $0^2+0^2-\sqrt{61}^2=61$, so $z$ will range from $\sqrt{61}$ to 8. So what we would like to evaluate would be this integral, where $r(\theta)=\sqrt{z^2-61}$, which is a constant with respect to $\theta$ $$\int_{\sqrt{61}}^8\int_0^{2\pi}\frac{1}{2}r(\theta)^2d\theta dz$$ $$=\int_{\sqrt{61}}^8\int_0^{2\pi}\frac{1}{2}\sqrt{z^2-61}^2d\theta dz$$ $$=\int_{\sqrt{61}}^8\int_0^{2\pi}\frac{1}{2}(z^2-61)d\theta dz$$ $$=\int_{\sqrt{61}}^8\frac{1}{2}(z^2-61)dz\int_0^{2\pi}1d\theta$$ $$=\pi\int_{\sqrt{61}}^8(z^2-61)dz$$ Alternatively, I think what you're trying to do is compute a volume by cylindrical shells, in which case the integral should look like this (where $h(r)$ is the height of a cylinder). $$\int_0^{\sqrt{3}}2\pi r h(r)dr$$ $$=\int_0^{\sqrt{3}}2\pi r (8-\sqrt{r^2+61})dr$$ $$=2\pi\int_0^{\sqrt{3}} (8r-r\sqrt{r^2+61})dr$$ $$=2\pi\bigg( 12 -\int_0^{\sqrt{3}} (r\sqrt{r^2+61})dr\bigg)$$ It seem that where you went wrong was in evaluating $$\int_0^{\sqrt{3}} r\sqrt{r^2+61}dr$$ $$=\int_0^{\sqrt{3}} r\sqrt{r^2+61}dr\cdot\frac{\frac{d(r^2+61)}{dr}}{\frac{d(r^2+61)}{dr}}$$ $$=\int_0^{\sqrt{3}} \frac{r\sqrt{r^2+61}}{2r}d(r^2+61)$$ $$=\frac{1}{2}\int_0^{\sqrt{3}} \sqrt{r^2+61}d(r^2+61)$$ $$=\frac{1}{2}\int_{61}^{64} \sqrt{u}du,\:u=r^2+61$$ $$=\frac{\sqrt{u}^3}{3}\bigg|_{61}^{64}$$ $$=\frac{8^3}{3} - \frac{\sqrt{61}^3}{3}$$ so the final answer would be $$\int_0^{2\pi}\int_0^{\sqrt{3}}(8r-r\sqrt{r^2+61})drd\theta=2\pi\bigg(12-\frac{8^3}{3} + \frac{\sqrt{61}^3}{3}\bigg)$$

Interesting association between tangent lines of slope one and ellipses Why is it that a tangent line with slope $1$ to an ellipse centered at the origin will have a transformation of $\pm \sqrt{a^2 +b^2}$ where $a$ and $b$ are the major and minor axis of the ellipse? For example: The tangent line of slope one to the ellipse $\frac{x^2}{9} + \frac{y^2}{4} = 1$ is $y=x+\sqrt{13}$.

The relation you are looking for can be derived algebraically. Start with some general line $y=mx+c$ for the tangent. Substitute it into the equation of your ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. This gives: $$ \frac{x^2}{a^2} + \frac{(mx+c)^2}{b^2} = 1 \\ b^2 x^2 + a^2 (m^2 x^2 + 2mxc + c^2) = a^2 b^2 \\ (b^2 + m^2 a^2)x^2 + 2a^2 mc x + a^2 (c^2 - b^2) = 0 \\ $$ Now since the line is a tangent the discriminant of this quadratic equation will be zero: $$ \Delta = (2a^2 mc)^2 - 4(b^2 + a^2 m^2)[a^2(c^2 - b^2)] = 0 \\ $$ Rearrange it to obtain the value of $c$: $$ 4a^4 m^2 c^2 - 4a^2 (b^2 c^2 - b^4 + a^2 m^2 c^2 - a^2 m^2 b^2 ) = 0 \\ b^2 c^2 - b^4 - a^2 m^2 b^2 = 0 \\ b^2 c^2 = b^4 + a^2 m^2 b^2 \\ c^2 = b^2 + a^2 m^2 \\ \therefore c = \pm \sqrt{a^2 m^2 + b^2} $$ So the tangent line equation is $y = mx \pm \sqrt{a^2 m^2 + b^2}$. In your case when the gradient $m=1$ then we have the required transformation $y=x \pm \sqrt{a^2 + b^2}$. NB: Geometrically speaking the tangents to the ellipse pass through the foci of a hyperbola with the same $a$ and $b$ values e.g. for the case of $a=3, b=2$.

How find this $5xy\sqrt{(x^2+y^2)^3}$ can write the sum of Four 5-th powers of positive integers. Find all positive integer $x,y$ such $$5xy\sqrt{(x^2+y^2)^3}$$ can write the sum of Four 5-th powers of positive integers.In other words: there exst $a,b,c,d\in N^{+}$ such $$5xy\sqrt{(x^2+y^2)^3}=a^5+b^5+c^5+d^5$$ This problem is from Math competition simulation test.I seach this problem and found this problem background is Euler's sum of powers conjecture.can see link maybe this problem is not hard.because is from competition. since $$5xy\sqrt{(x^2+y^2)^3}=5xy(x^2+y^2)\sqrt{x^2+y^2}$$ so $$x^2+y^2=m^2$$ $$x=3,y=4,m=5$$ and $$x=(a'^2-b'^2),y=2a'b',m=a'^2+b'^2$$ then I can't it Thank you for you help .

Something that might help. Using what you said, namely: $x^2+y^2=k^2$, with $x=m^2-n^2$, $y=2mn$ then the left hand side of the equation becomes: $LHS=10(m^9n-mn^9+2m^7n^3-2m^3n^7)$ Since $RHS\equiv 0$ modulo 2 and modulo 5 then we have $a+b+c+d\equiv 0 \mod{2}$ and $a+b+c+d\equiv 0 \mod{5}$

How $\tan{\frac{A}{2}}\tan{\frac{B}{2}}=\frac{1}{2}$,then find $\angle C$ In $\Delta ABC$, if $$\tan{\dfrac{A}{2}}\tan{\dfrac{B}{2}}=\dfrac{1}{2}\\\sin{\dfrac{A}{2}}\sin{\dfrac{B}{2}}\sin{\dfrac{C}{2}}=\dfrac{1}{10}$$ Find the $\angle C$ My try: since $$2\sin{\dfrac{A}{2}}\sin{\dfrac{B}{2}}=\cos{\dfrac{A}{2}}\cos{\dfrac{B}{2}}$$ $$\left(\cos{(\dfrac{A-B}{2})}-\cos{(\dfrac{A+B}{2})}\right)\sin{\dfrac{C}{2}}=\dfrac{1}{5}$$ since $$\cos{(\dfrac{A+B}{2})}=\sin{\dfrac{C}{2}}$$ so $$\left(\cos{(\dfrac{A-B}{2})}-\sin{\dfrac{C}{2}}\right)\sin{\dfrac{C}{2}}=\dfrac{1}{5}$$ Then I can't go on.

Denote $~~a=\tan\dfrac{A}{2}$, $~~b=\tan\dfrac{B}{2}$ $($let $a\le b$$)$. $\sin\dfrac{A}{2}\sin\dfrac{B}{2}\sin\dfrac{C}{2}=\dfrac{1}{10}$; $\sqrt{\dfrac{1-\cos A}{2}} \cdot \sqrt{\dfrac{1-\cos B}{2}} \cdot \sqrt{\dfrac{1-\cos C}{2}} = \dfrac{1}{10}$; $(1-\cos A) (1-\cos B)(1-\cos C) = \dfrac{2}{25}$; $(1-\cos A) ~ (1-\cos B) ~ (1+\cos(A+B)) = \dfrac{2}{25}$; $\dfrac{2a^2}{1+a^2} \cdot \dfrac{2b^2}{1+b^2} \cdot \dfrac{2(1-ab)^2}{(1+a^2)(1+b^2)} = \dfrac{2}{25}$; $\dfrac{ab|1-ab|}{(1+a^2)(1+b^2)}=\dfrac{1}{10}$. Now we get system: $$ \left\{ \begin{array}{l} ab=\dfrac{1}{2};\\ \dfrac{ab|1-ab|}{(1+a^2)(1+b^2)}=\dfrac{1}{10}. \end{array} \right. $$ $b=\dfrac{1}{2a}$ $\implies$ $\dfrac{a^2}{(1+a^2)(4a^2+1)}=\dfrac{1}{10}$ ; $4a^4-5a^2+1=0$; $(2a-1)(2a+1)(a-1)(a+1)=0$. there are $2$ positive roots for this eq.: $a=\dfrac{1}{2}$ and $a=1$. If $a=\dfrac{1}{2}$, then $b=1$, then $A = \arctan\dfrac{4}{3}=\arcsin\dfrac{4}{5}=\arccos\dfrac{3}{5}$, $~B=\pi/2$, $~C=...$; If $a=1$, then $b=\dfrac{1}{2}$ (simply permutation of $a,b$). (don't know if this is the simplest way, but at least this is $\approx$ clear :) Finally, $C=\arctan\dfrac{3}{4}=\arcsin\dfrac{3}{5}=\arccos\dfrac{4}{5} = 0.643501108793...$.

How do I solve this square root problem? I need to solve the following problem: $$\frac{\sqrt{7+\sqrt{5}}}{\sqrt{7-\sqrt{5}}}=\,?$$

\begin{align} \frac{\sqrt{7+\sqrt{5}}}{\sqrt{7-\sqrt{5}}}&=\frac{\sqrt{7+\sqrt{5}}}{\sqrt{7-\sqrt{5}}}\cdot \frac{\sqrt{7+\sqrt{5}}}{\sqrt{7+\sqrt{5}}}\\ &=\frac{(\sqrt{7+\sqrt{5}})^2}{\sqrt{(7-\sqrt{5})(7+\sqrt{5})}}\\ &=\frac{7+\sqrt{5}}{\sqrt{7^2-(\sqrt{5})^2}}\\ &=\frac{7+\sqrt{5}}{\sqrt{49-5}}\\ &=\frac{7+\sqrt{5}}{\sqrt{44}}\\ &=\frac{7+\sqrt{5}}{2\sqrt{11}}\cdot\frac{\sqrt{11}}{\sqrt{11}}\\ &=\frac{7\sqrt{11}+\sqrt{5\cdot11}}{2(\sqrt{11})^2}\\ &=\frac{7\sqrt{11}+\sqrt{55}}{22} \end{align}

Integral: $\int \frac{dx}{\sqrt{x^{2}-x+1}}$ How do I integrate this? $$\int \frac{dx}{\sqrt{x^{2}-x+1}}$$ I tried solving it, and I came up with $\ln\left | \frac{2\sqrt{x^{2}-x+1}+2x-1}{\sqrt{3}} \right |+C$. But the answer key says that the answer should be $\sinh^{-1}\left ( \frac{2x-1}{\sqrt{3}} \right )+C$. Any answer is very appreciated.

Notice $$ x^2 - x + 1 = \left(x-\frac{1}{2}\right)^2 + 1 - \frac{1}{4} = \left(x-\frac{1}{2}\right)^2 + \frac{3}{4}$$

Calculating Eigenvalues is only Assume that the following is used: $$ A = \begin{pmatrix} 0& 1&\\ 2& 3&\\ 4& 5&\\ 6& 7&\\ 8& 9& \end{pmatrix} $$ Then calculating the Coveriance matrix, which, gives me: $$ C = \begin{pmatrix} 40& 40&\\ 40& 40&\\ \end{pmatrix} $$ Then using the following: $$ det = (a+b) \cdot (a+b)-4 \cdot(a \cdot b - c \cdot c), $$ where in this case, $a = 40, b = 40, c = 40$ gives the answer: $$ \lambda_{1} = 80, \\ \lambda_{2} = 0, $$ These are therefore the correct Eigen values. However, using this formula, if I have the following: $$ A = \begin{pmatrix} -4& -2&\\ -1& -3&\\ 4& 5&\\ 6& 7&\\ 8& 9& \end{pmatrix}, $$ where the Covariance matrix is given: $$ C = \begin{pmatrix} 99.2& 103.4&\\ 103.4& 116.8&\\ \end{pmatrix}, $$ gives the Eigenvalues as: $$ \lambda_{1} = 218.119 \\ \lambda_{2} = -15.5189 $$ When the actual values are: $$ \lambda_{1} =211.774 \\ \lambda_{2} = 4.226 $$ Could anyone tell me where I am calculating this wrong please? EDIT: For $\lambda_{1} = (a + b + det)/2 \\ \lambda_{2} = (a + b - det)/2 $

Have you double-checked your calculations? The eigenvalues of any $2\times2$ real symmetric matrix $\pmatrix{a&c\\ c&b}$ are given by $$\frac{a+b\pm\sqrt{(a+b)^2-4(ab-c^2)}}2.$$ Plug in the entries of $C$, I don't find any discrepancies.

Tricky Triangle Area Problem This was from a recent math competition that I was in. So, a triangle has sides $2$ , $5$, and $\sqrt{33}$. How can I derive the area? I can't use a calculator, and (the form of) Heron's formula (that I had memorized) is impossible with the$\sqrt{33}$ in it. How could I have done this? The answer was $2\sqrt{6}$ if it helps. Edited to add that it was a multiple choice question, with possible answers: a. $2\sqrt{6}$ b. $5$ c. $3\sqrt{6}$ d. $5\sqrt{6}$

From the law of cosines ($C^2 = A^2 + B^2 - 2AB\cos \theta$), we get that $(\sqrt{33})^2 = 2^2 + 5^2 - 2 \cdot 2 \cdot 5 \cos \theta$. Simplifying this, we get $33 = 29 - 20 \cos \theta$, which means that $\displaystyle \cos \theta = -\frac{1}{5}$ Because $\cos^2 \theta + \sin^2 \theta = 1$, we get that $\displaystyle \sin^2 \theta = \frac{24}{25}$. This means that $\displaystyle \sin \theta = \frac{2\sqrt{6}}{5}$ (note that, because $0 \le \theta \le \pi$, $\sin \theta \ge 0$). The area of the triangle is $\displaystyle \frac{1}{2} A B \sin \theta = \frac{1}{2} \cdot 2 \cdot 5 \cdot \frac{2\sqrt{6}}{5} = 2\sqrt{6}$.

How to find the values of a and b? If the polynomial 6x4 + 8x3 - 5x2 + ax + b is exactly divisible by the polynomial 2x2 - 5, then find the values of a and b.

The roots of $2x^2-5$ must be roots of our polynomial: $$\left\lbrace \begin{array}{c} 6\cdot\frac{25}4+15\sqrt{\frac52}-\frac{25}2+\sqrt{\frac52}a+b=0\\ \phantom{1}\\ 6\cdot\frac{25}4-15\sqrt{\frac52}-\frac{25}2-\sqrt{\frac52}a+b=0 \end{array} \right.$$ Adding up the equations you get $$50+2b=0$$ And substracting them: $$30\sqrt{\frac52}+2\sqrt{\frac52}a=0$$ So $a=-15$, $b=-25$.

How come if $\ i\ $ not of the following form, then $12i + 5$ must be prime? I know if $\ i\ $ of the following form $\ 3x^2 + (6y-3)x - y\ $ or $\ 3x^2 + (6y-3)x + y - 1, \ \ x,y \in \mathbb{Z}^{+},i \in \mathbb Z_{\ge 0}$, then $\ 12i + 5\ $ must be composite number, because: $12(3x^2 + (6y-3)x + y - 1) + 5 = 36x^2 + (72y - 36)x + (12y - 7) = (6x + 12y - 7)(6x + 1)$ How come if $\ i\ $ not of the following form $\ 3x^2 + (6y-3)x - y\ $ and $\ 3x^2 + (6y-3)x + y - 1\ $, then $12i + 5$ must be prime? $\ x,y \in \mathbb{Z}^{+},i \in \mathbb Z_{\ge 0}$. For example: $\ 5 = \ 3*1^2 + (6*1-3)*1 - 1\ $ ,when $\ x = y = 1\ $, as proved before,$\ 5*12+5\ $ must be composite number; $\ 0, \ 1. \ 2.\ 3.\ 4\ $ can't of the following form $\ 3x^2 + (6y-3)x - y\ $ and $\ 3x^2 + (6y-3)x + y - 1\ $, $\ 12*0 + 5 = 5\ $, $\ 5 $ is prime,$\ 12*1 + 5 = 17\ $, $\ 17 $ is prime, and so on,

This is rather elementary. Suppose that $12i+5$ is composite, say $12i+5 = ab$. Then looking modulo $6$, we get that one of the elements $a,b$ is $1 \pmod{6}$ and the other is $-1 \pmod{6}$; assume w.l.o.g. that $a$ is $1 \pmod{6}$, and write $a = 6r+1$ and $b = 6s-1$. Observe that $r$ and $s$ must have different parity. Indeed, if they are both even or both odd, then computing modulo $12$, we would get that $ab \equiv -1 \pmod{12}$, which is false. If $s>r$, then we set $x = r$, and $s = r + (2y-1)$ for some positive number $y$. Then compute $ab$, to get $i = 3x^2 + (6y-3)x + y - 1$. If $s<r$, then we set $x = s$, and $r = s + (2y-1)$ for some positive number $y$. Then compute $ab$, to get $i = 3x^2 + (6y-3)x - y$.

Triple Integral Troubles I'm having trouble calculating this integral. I can do the first one just fine, but it's in simplifying and calculating the third integral where I get stuck. $16\int_0^\frac{\pi}{4}\int_0^1\int_0^{\sqrt{1-r^2cos^2(\theta)}}rdzdrd\theta$ $16\int_0^\frac{\pi}{4}\int_0^1r\sqrt{1-r^2cos^2(\theta)}drd\theta$ with u-substitution: $16\int_0^\frac{\pi}{4}\frac{-1}{3}(\frac{(1-cos^2(\theta))^(\frac{3}{2})}{cos^2(\theta)})d\theta$ I know the answer should be $16-8\sqrt{2}$

You're forgetting a "$-1$" term. Let $u = 1 - r^2\cos^2\theta$ so that $du = -2r\cos^2\theta \, dr$. Then observe that: \begin{align*} \int_0^1 r \sqrt{1 - r^2\cos^2\theta} \, dr &= \frac{1}{-2\cos^2\theta}\int_1^{1-\cos^2\theta} \sqrt{u} \, du \\ &= \frac{1}{-2\cos^2\theta}\left[\frac{u^{3/2}}{3/2}\right]_1^{1-\cos^2\theta} \\ &= \frac{1}{-3\cos^2\theta}\left[(1-\cos^2\theta)^{3/2} - 1\right] \\ \end{align*} Hence, we obtain: $$ \frac{-16}{3}\int_0^{\pi/4} \frac{(1-\cos^2\theta)^{3/2} - 1}{\cos^2\theta} \, d\theta $$ Using Wolfram|Alpha, this integral does indeed equal $16 - 8\sqrt2$, as desired. If you want to actually compute the antiderivative by hand, then here are some trig identities that will magically simplify things: \begin{align*} \frac{(1-\cos^2\theta)^{3/2} - 1}{\cos^2\theta} &= \frac{(\sin^2\theta)^{3/2} - 1}{\cos^2\theta} \\ &= \frac{\sin^3\theta - 1}{\cos^2\theta} \\ &= \frac{\sin\theta(1 - \cos^2\theta) - 1}{\cos^2\theta} \\ &= \frac{\sin\theta - \sin\theta\cos^2\theta - 1}{\cos^2\theta} \\ &= \frac{1}{\cos\theta} \cdot \frac{\sin\theta}{\cos\theta} - \sin\theta - \frac{1}{\cos^2\theta} \\ &= \sec\theta\tan\theta - \sin\theta - \sec^2\theta \end{align*} Thus, the antiderivative is: $$ \sec\theta + \cos\theta - \tan\theta $$

Using properties of determinants, show that Using Properties of determinants, show that: $$ \begin{vmatrix} a & a+b & a+2b\\ a+2b & a & a+b\\ a+b & a+2b & a \end{vmatrix} = 9b^2 (a+b) $$ I've tried it but not getting $9b^2$

Add all the three columns to get $$\left \vert \begin{bmatrix} 3a+3b & a+b & a+2b\\ 3a+3b & a & a+b\\ 3a+3b & a+2b & a\end{bmatrix}\right \vert = 3(a+b)\left \vert \begin{bmatrix} 1 & a+b & a+2b\\ 1 & a & a+b\\ 1 & a+2b & a\end{bmatrix}\right \vert$$ Subtract first row from row and first from third to get $$3(a+b)\left \vert \begin{bmatrix} 1 & a+b & a+2b\\ 1 & a & a+b\\ 1 & a+2b & a\end{bmatrix}\right \vert = 3(a+b)\left \vert \begin{bmatrix} 1 & a+b & a+2b\\ 0 & -b & -b\\ 0 & b & -2b\end{bmatrix}\right \vert = 3(a+b)(2b^2+b^2) = 9b^2(a+b)$$

Prove the identity $\frac{\cos B}{1-\tan B} + \frac{\sin B}{1-\cot B}=\sin B+\cos B$ I have worked on this identity from both sides of the equation and can't seem to get it to equal the other side no matter what I try. $\displaystyle\frac{\cos B}{1-\tan B} + \frac{\sin B}{1-\cot B} =\sin B+\cos B$

HINT: $\displaystyle\frac{\cos B}{1-\tan B}=\frac{\cos B}{1-\dfrac{\sin B}{\cos B}}=\frac{\cos^2B}{\cos B-\sin B}$ $\displaystyle\frac{\sin B}{1-\cot B}=\frac{\sin B}{1-\dfrac{\cos B}{\sin B}}=\frac{\sin^2B}{\sin B-\cos B}=-\frac{\sin^2B}{\cos B-\sin B}$

Prove or disprove inequality $\frac{a^2}{b+c}+\frac{b^2}{a+c}+\frac{c^2}{a+b}\le\frac{a^4+b^4+c^4}{2abc}$. Let $a$, $b$ and $c$ be real numbers greater than $0$. Prove inequality $$\displaystyle{\frac{a^2}{b+c}+\frac{b^2}{a+c}+\frac{c^2}{a+b}\le\frac{a^4+b^4+c^4}{2abc}}.$$

By AM-GM and C-S we obtain: $$\frac{a^4+b^4+c^4}{2abc}=\sum_{cyc}\frac{a^3}{2bc}\geq\sum_{cyc}\frac{2a^3}{(b+c)^2}=$$ $$=\frac{2}{a+b+c}\sum_{cyc}\frac{a^3}{(b+c)^2}\sum_{cyc}a\geq\frac{2}{a+b+c}\left(\sum_{cyc}\frac{a^2}{b+c}\right)^2=$$ $$=\frac{2}{a+b+c}\cdot\sum_{cyc}\frac{a^2}{b+c}\cdot\sum_{cyc}\frac{a^2}{b+c}\geq$$ $$\geq\frac{2}{a+b+c}\cdot\frac{(a+b+c)^2}{2(a+b+c)}\cdot\sum_{cyc}\frac{a^2}{b+c}=\sum_{cyc}\frac{a^2}{b+c}.$$ Done!

$l^2+m^2=n^2$ $\implies$ $lm$ is always a multiple of 3 when $l,m,n,$ are positive integers. Let $l,m,n$ be any three positive integers such that $l^2+m^2=n^2$ Then prove that $lm$ is always a multiple of 3.

Here is another approach. The general solution of this equation is: $l = x^2 - y^2$, $m = 2xy$, $n = x^2 + y^2$. So: $l\cdot m = 2\cdot (x^2 - y^2)\cdot x\cdot y = 2\cdot (x - y)\cdot (x + y) \cdot x\cdot y$. From this, we have some cases to consdier: * *$3|x$ or $3|y$ then $3|l\cdot m$. *$3 \not|x$ and $3\not|y$ then if $x \equiv y \pmod 3$, then $3|(x - y)$ and $(x - y)|l\cdot m$, so $3|l\cdot m$, but if $x \neq y \pmod 3$, then $x + y \equiv 0 \pmod 3$, and this implies that $3|l\cdot m$

How did Euler realize $x^4-4x^3+2x^2+4x+4=(x^2-(2+\alpha)x+1+\sqrt{7}+\alpha)(x^2-(2-\alpha)x+1+\sqrt{7}-\alpha)$? How did Euler find this factorization? $$\small x^4 − 4x^3 + 2x^2 + 4x + 4=(x^2-(2+\alpha)x+1+\sqrt{7}+\alpha)(x^2-(2-\alpha)x+1+\sqrt{7}-\alpha)$$ where $\alpha = \sqrt{4+2\sqrt{7}}$ I know that he had some super powers, like he was sent to us from a super intelligent alien universe just to humiliate our intelligence, but how the hell did he do that three centuries ago? :|

I like this rather old question. Here is a yet another possible way Euler could have taken: Note that $\displaystyle x^4+ax^2+b$ can be factorized easily if $\displaystyle a^2-4b\geq 0$. If, however, $\displaystyle a^2-4b\leq 0$, then \begin{align} x^4+ax^2+b&=(x^2+\sqrt{b})^2-(x\sqrt{2\sqrt{b}-a})^2\\ &=(x^2+\sqrt{b}-x\sqrt{2\sqrt{b}-a})(x^2+\sqrt{b}+x\sqrt{2\sqrt{b}-a}). \end{align} Now in $P(x)=x^4-4x^3+2x^2+4x+4$, use $x=y+1$, and proceed: \begin{align} P(y+1)&=y^4-4y^2+7\\ &=(y^2+\sqrt{7}-y\sqrt{2\sqrt{7}+4})(y^2+\sqrt{7}+y\sqrt{2\sqrt{7}+4}). \end{align} substituting $y=x-1$ we arrive at Euler's result.

Basic Combinatorics I have a basic combinatorics question I am unsure how to complete, the question is as follows: A company has 9 people in Office A, 6 in Office B and 3 in Office C. A new team of 6 people is to be formed. How many ways can the new team be formed if: a) The team includes two members from each office b) Office A is to have at least two representatives If anyone can help me with how to answer this I would be most grateful

a) There are $\binom{9}{2}$, $\binom{6}{2}$, and $\binom{3}{2}$ ways to choose $2$ persons from office $A, B, C$ respectively. So there are $\binom{9}{2}\cdot \binom{6}{2}\cdot \binom{3}{2}$ ways of choosing $6$-person teams with $2$ members from each office. b) If $A$ has $2$ members, then the other $4$ members are chosen from $B$ and $C$, and there are $9$ from $B$ and $C$ combined. So we can have $\binom{9}{4}$ choices for the $4$ persons to form $6$-person teams. So we have $\binom{9}{2}\cdot \binom{9}{4}$ choices for for this case. If $A$ has $3$ members, then similarly we have: $\binom{9}{3}\cdot \binom{9}{3}$ choices to make $6$-person teams. And continue this way until $A$ has $6$ members, then we have the total choices is: $\binom{9}{2}\cdot \binom{9}{4} + \binom{9}{3}\cdot \binom{9}{3} + \binom{9}{4}\cdot \binom{9}{2} + \binom{9}{5}\cdot \binom{9}{1} + \binom{9}{6}$.

Trying to prove that there are no p and q such that $|\sqrt5 - p/q| < 1/(7q^2)$. Like the title says, I'm having trouble proving that there are no integers p and q such that $|\sqrt5 - p/q| < 1/(7q^2)$. I was given the hint that $|(q\sqrt5 - p)(q\sqrt5 + p)| \geq 1$, but I don't quite know how that helps... Thanks!

I think you are my classmate, since we had submitted our homework a few hours ago, I would like to share my solution. Because we are studying continued fractions now, so my solution is based on it. First, we know if we can find some $(p, q)$ pair, and $q \not= 0$, by Theorem 12.18 or Corollary 12.18.1 in the textbook(Elementary Number Theory by Kenneth H. Rosen 6ed), then the most possible solutions are the convergents $C_k = \frac{p_k}{q_k}, k = 0, 1, 2, ...$ of $\sqrt{5}$. Because they are the closest rational approximation if $q$ is given. But we can show no one in those convergents satisfy this inequality. We can easily know $\sqrt{5} = [2;\bar{4}]$ and $2 + \sqrt{5} = [\bar{4}]$. When $k = 0$ and $k = 1$ we can just do some calculations: $k = 0$, $|\sqrt{5} - \frac{2}{1}| = \sqrt{5} - 2 > \frac{1}{7}$, since $5 > \frac{225}{49}$. $k = 1$, $|\sqrt{5} - \frac{9}{4}| = \frac{9}{4} - \sqrt{5} > \frac{1}{112}$, since $\frac{251}{112} > \sqrt{5}$. For $k > 1$, We know $\sqrt{5} = \frac{\alpha_{k+1}p_k + p_{k-1}}{\alpha_{k+1}q_k + q_{k-1}}$, so $$\begin{aligned} |\sqrt{5} - \frac{p_k}{q_k}| &= |\frac{\alpha_{k+1}p_k + p_{k-1}}{\alpha_{k+1}q_k + q_{k-1}} - \frac{p_k}{q_k}| \\ &= |\frac{\alpha_{k+1}p_kq_k + p_{k-1}q_k - \alpha_{k+1}p_kq_k - p_kq_{k-1}}{\alpha_{k+1}q_k^2 + q_{k-1}q_k}| \\ &= |\frac{-(p_kq_{k-1} - p_{k-1}q_k)}{\alpha_{k+1}q_k^2 + q_{k-1}q_k}| \\ &= \frac{1}{\alpha_{k+1}q_k^2 + q_{k-1}q_k} \\ &> \frac{1}{\alpha_{k+1}q_k^2 + q_k^2}\end{aligned}$$ We know $\alpha_{k+1} = [\bar{4}] = 2 + \sqrt{5}$ so $$|\sqrt{5} - \frac{p_k}{q_k}| > \frac{1}{(3 + \sqrt{5})q_k^2} > \frac{1}{7q_k^2}$$ Which completes the proof. $\blacksquare$

Find minimum value of the expression x^2 +y^2 subject to conditions Find the values of $x,y$ for which $x^2 + y^2$ takes the minimum value where $(x+5)^2 +(y-12)^2 =14$. Tried Cauchy-Schwarz and AM - GM , unable to do.

Hint: take a look at the picture below, and all the problem will vanish... In fact the picture shows the circle of equation $(x+5)^2 +(y-12)^2 =14$, and the line passing trough its centre and the origin. The question asks the minimum length of the segment whose extremities are the origin and a point on the circumference...Thus you should minimize the distance from the point on the circumference from the origin, and this is done by drawing a line passing through the centre of the square and the centre (trivial proof). So you get $$A\equiv \left(5\left(\frac{\sqrt{14}}{13}-1\right);12\left(1-\frac{\sqrt{14}}{13}\right)\right)\Rightarrow \\ \text{length of} \overline{AB} = \sqrt{\left(5\left(\frac{\sqrt{14}}{13}-1\right)\right)^2+\left(12\left(1-\frac{\sqrt{14}}{13}\right)\right)^2}=\\ \sqrt{(\sqrt{14}-13)^2}=\sqrt{14}-13$$ and finally $$\text{minimum of } x^2+y^2=\overline{AB}^2=(\sqrt{14}-13)^2=183-26\sqrt{14}$$

What is the radius of the circle? Please help with this grade nine math problem. How does one calculate the radius if the two sides of the right angle triangle are 85cm. The sides of the triangle are tangent to the circle.

It's useful to realize that the "left" and "right" radia, as drawn in the above picture, will be parallel to the respective cathetae. Then you get: $$C=\sqrt{A^2+A^2}=\sqrt{2}A$$ The height of the triangle is then: $$h=\sqrt{A^2-\left(\frac{C}{2}\right)^2}=\sqrt{A^2-\frac{A^2}{2}}=\frac{1}{\sqrt{2}}A$$ Define x-axis along the base of the triangle and y-axis along the height. Unit vectors at a 45° angle to the x-axis are given by: $$\vec{u}_1=\frac{1}{\sqrt{2}}\left({1}\atop{1}\right)~~~~~,~~~~~\vec{u}_2=\frac{1}{\sqrt{2}}\left({1}\atop{-1}\right)$$ You can check $\vec{u}\cdot\vec{u}=1$. Now use that the distance from any of the two 45° angles to the two nearest spots where the circle touches the triangle is the same, namely $C/2=A/\sqrt{2}$. With this you can establish a vectorial relation between the following vectors: $$h\vec{e}_y+\left(A-\frac{C}{2}\right)\vec{u}_2=R\vec{e}_y+R\vec{u}_1$$ Where $\vec{e}_y=(0,1)$ is the unit vector along the y-axis. This gives you two equations. The y-axis equation is: $$h-\frac{1}{\sqrt{2}}\left(A-\frac{C}{2}\right)=R+\frac{1}{\sqrt{2}}R\\\frac{A}{\sqrt{2}}-\frac{1}{\sqrt{2}}\left(A-\frac{A}{\sqrt{2}}\right)=\left(1+\frac{1}{\sqrt{2}}\right)R\\\frac{A}{2}=\left(1+\frac{1}{\sqrt{2}}\right)R\\R=\frac{A}{2+\sqrt{2}}$$ The x-axis equation is: $$\frac{1}{\sqrt{2}}\left(A-\frac{C}{2}\right)=\frac{1}{\sqrt{2}}R\\\left(1-\frac{1}{\sqrt{2}}\right)A=R\\R=\frac{\left(1-\frac{1}{\sqrt{2}}\right)\left(2+\sqrt{2}\right)}{2+\sqrt{2}}A\\R=\frac{2-\sqrt{2}+\sqrt{2}-1}{2+\sqrt{2}}A\\R=\frac{A}{2+\sqrt{2}}$$ Both answers properly agree, so that the world is a happy and sunny place.

How to finish this integration? I'm working with the integral below, but not sure how to finish it... $$\int \frac{3x^3}{\sqrt[3]{x^4+1}}\,dx = \int \frac{3x^3}{\sqrt[3]{A}}\cdot \frac{dA}{4x^3} = \frac{3}{4} \int \frac{dA}{\sqrt[3]{A}} = \frac{3}{4}\cdot\quad???$$ where $A=x^4+1$ and so $dA=4x^3\,dx$

$$\dfrac{1}{\sqrt[\large 3]{A}} = \dfrac 1{A^{1/3}} = A^{-1/3}$$ Now use the power rule. $$ \frac{3}{4} \int A^{-1/3}\,dA = \frac 34 \dfrac {A^{2/3}}{\frac 23} + C = \dfrac 98 A^{2/3} + C$$

Finding determinant for matrix using upper triangle method Here is an example of a matrix, and I'm trying to evaluate its determinant: $$ \begin{pmatrix} 1 & 3 & 2 & 1 \\ 0 & 1 & 4 & -4 \\ 2 & 5 & -2 & 9 \\ 3 & 7 & 0 & 1 \\ \end{pmatrix} $$ When applying first row operation i get: $$ \begin{pmatrix} 1 & 3 & 2 & 1 \\ 0 & 1 & 4 & -4 \\ 0 & -1 & -6 & 7 \\ 0 & -2 & -6 & -2 \\ \end{pmatrix} $$ Now, if I continue doing row operations until i get the upper triangle, the determinant will be 14 (which is said to be the correct one). $$ \begin{pmatrix} 1 & 3 & 2 & 1 \\ 0 & 1 & 4 & -4 \\ 0 & 0 & 2 & 3 \\ 0 & 0 & 0 & -7 \\ \end{pmatrix} $$ However, if I instead apply this certain operation, R4 --> (1/-2)R4... $$ \begin{pmatrix} 1 & 3 & 2 & 1 \\ 0 & 1 & 4 & -4 \\ 0 & -1 & -6 & 7 \\ 0 & 1 & 3 & 1 \\ \end{pmatrix} $$ ...and then carry on with operations, I get a different final answer: The determinant will be 7 in this case! $$ \begin{pmatrix} 1 & 3 & 2 & 1 \\ 0 & 1 & 4 & -4 \\ 0 & 0 & -1 & 5 \\ 0 & 0 & 0 & -7 \\ \end{pmatrix} $$ Could someone explain that to me - is this operation illegal? R4 --> (1/-2)R4 How so? Because i always tend to use it, just to simply things a little.

You just multiplied a row with $\frac {1}{-2}$! This will change the value of determinant. What you can do is take $-2$ common from a row and write it outside. Consider a $1\times 1$ matrix $A=[1]$. $det(A)=1$ Apply $R_!\to2R_1$ $A=[2]$ $det(A)=2$ Can you see why you cannot do it?

Find all the singularities of $f(z)= \frac{1}{z^4+1}$ and the associated residue for each singularity I know that there are poles at $$\Large{z=e^{\frac{i\pi}{4}}},$$ $$\Large{z=-e^{\frac{i\pi}{4}}},$$ $$\Large{z=e^{\frac{i3\pi}{4}}},\text{ and}$$ $$\Large{z=-e^{\frac{i3\pi}{4}}}$$ I am having trouble with the residues for each one. Are the answers just the poles but all divided by $4$? can someone help? Thanks!

Like N3buchadnezzar just said the residues are given by $$\mathrm{Res}\left(\frac{f(z)}{g(z)},z_k\right) = \frac{f(z_k)}{g'(z_k)}$$ In your case the algebra involved in the calculation may lead to many errors if you consider the residues as you listed them. I suggest you to write the singularities of $\frac{1}{z^4+1}$ as \begin{align*}z_1 &= \frac{1+i}{\sqrt{2}} & z_2&=\frac{-1+i}{\sqrt{2}}\\ z_3&=\frac{1-i}{\sqrt{2}} & z_4&=\frac{-1-i}{\sqrt{2}} \end{align*} You can find these singularities just by a simple geometric reasoning. Now i think it's easier to evaluate the residues. For intance you get \begin{align*} \mathrm{Res}\left(\frac{1}{z^4+1},z_1\right) &= \mathrm{Res}\left(\frac{1}{z^4+1},\frac{1+i}{\sqrt{2}}\right)\\ &=\frac{1}{4z^3}\mid_{\frac{1+i}{\sqrt{2}}}\\ &=\frac{1}{\frac{4(1+i)^3}{\sqrt{2}}}\\ &=\frac{1}{\frac{4(-2+2i)}{\sqrt{2}}\cdot\frac{\sqrt{2}}{\sqrt{2}}}\\ &=\frac{1}{4\sqrt{2}(-1+i)}\cdot \frac{(-1-i)}{(-1-i)}\\ &=\frac{-1-i}{8\sqrt{2}} \end{align*} You can evaluate the other residues very similar.

Evaluating the following integral: $\int\frac1{x^3+1}\,\mathrm{d}x$ How to integrate $$\int\frac1{x^3+1}~\mathrm{d}x$$ Is it possible to use Taylor expansion?

If $x^3 + 1=0$ then $x^3=-1$ so $x=-1$, at least if $x$ is real. If you plug $-1$ in for $x$ in a polynomial and get $0$, then $x-(-1)$ is a factor of that polynomial. So you have $x^3+1=(x+1)(\cdots\cdots\cdots\cdots)$. The second factor can be found by long division or other means. It is $x^2-x+1$. Can that be factored? Solving the quadratic equation $x^2-x+1=0$ yields two non-real solutions, complex conjugates of each other. Doing arithmetic or algebra with complex numbers is in many ways just like doing the same things with real numbers. But doing calculus with complex numbers opens some cans of worms that get dealt with in more advanced courses. With real numbers, the quadratic polynomial $x^2-x+1$ is irreducible, i.e. cannot be factored. The quickest way to see that is by observing that the discriminant $b^2-4ac$ is negative. A way that's not as quick but that may give some insight is completing the square: $$ x^2-x+1 = \left( x-\frac12\right)^2 + \frac 3 4. $$ Obviously this can never be $0$ when $x$ is real, so this can't be factored with real numbers. So $$ \frac{1}{x^3+1} = \frac{A}{x+1} + \frac{Bx+C}{x^2-x+1} $$ and then you have to find $B$ and $C$. Now another difficulty comes along: How to find $$ \int \frac{Bx+C}{x^2-x+1} \,dx\text{ ?} $$ Let $u=x^2-x+1$ so that $du = (2x-1)\,dx$ and you get part of it: $$ (Bx+C)\,dx = \underbrace{\frac B 2 (2x-1)\,dx} + \left(C + \frac B2\right)\,dx. $$ The substitution handles the part over the $\underbrace{\text{underbrace}}$. What about the other part? You have $$ \text{constant}\cdot\int \frac{dx}{x^2-x+1}. $$ Complete the square: $$ \int \frac{dx}{\left(x - \frac 12\right)^2 + \frac 3 4}. $$ Starts to remind you of an arctangent, but you have $3/4$ where you need $1$. $$ \int \frac{dx}{\left(x - \frac 12\right)^2 + \frac 3 4} = \frac 4 3 \int \frac{dx}{\frac43\left(x-\frac12\right)^2+1} $$ Now $$ \frac43\left(x-\frac12\right)^2+1 = \left(\frac{2x-1}{\sqrt{3}}\right)^2+1 $$ So let $w=\dfrac{2x-1}{\sqrt{3}}$ and $\sqrt{3}\,dw=dx$, and then you're almost done.

Direct proof that $n!$ divides $(n+1)(n+2)\cdots(2n)$ I've recently run across a false direct proof that $n!$ divides $(n+1)(n+2)\cdots (2n)$ here on math.stackexchange. The proof is here prove that $\frac{(2n)!}{(n!)^2}$ is even if $n$ is a positive integer (it is the one by user pedja, which got 11 upvotes). The proof is wrong because it claims that one can rewrite $(n+1)\cdots (2n)$ as $$ (n+1)(n+2)\cdots 2(n-2)(2n-1)(2n) = 2\cdots 2\cdot n!\cdot (2n-1)(2n-3)\cdots (n+1).$$ In other words, it claims that the product of the factors $2n$, $2(n-1)$, $2(n-2)$, $\ldots$, all of which are in $(n+1)\cdots(2n)$, amounts to $2^kn!$, but this is not true since the factors $2m$ under scrutiny do not start from $m=1$ but from values greater than $n$. For instance, for $n=4$, we have $(8)(7)(6)(5)=2\cdot 2\cdot 4\cdot 3\cdot 5\cdot 7$, not $(8)(7)(6)(5)=2\cdot 2\cdot 4!\cdot 5\cdot 7$. This makes me wonder two things: (1) What is a valid direct proof? (2) How many wrong proofs do go undetected here? (How many false proofs receive 10+ upvotes?) NB Not interested in any proof that uses binomial coefficients and/or the relationship $\binom{2n}{n}=\frac{(2n)!}{n!n!}$.

Here is a more direct number theoretical type proof that if $a \ge 0$ and $a_1+a_2+\cdots+a_r = n$ that $\frac{n!}{a_1!a_2! \cdots a_r!}$ is an integer. This reduces to proving $\sum \left \lfloor \frac{n}{p_i} \right \rfloor \ge \sum \left \lfloor \frac{a_1}{p_i} \right \rfloor + \sum \left \lfloor \frac{a_1}{p_i} \right \rfloor + \cdots + \sum \left \lfloor \frac{a_r}{p_i} \right\rfloor $. Lemma: $\lfloor x \rfloor + \lfloor y \rfloor \le \lfloor x+y \rfloor$ if $x,y$ are real numbers. This can be easily proved by writing $x,y$ in terms of their integer and fractional parts. Applying this to the $RHS$ gives $RHS \le \left \lfloor \frac{a_1+a_2+ \cdots + a_r}{p_i} \right \rfloor = \lfloor \frac{n}{p_i} \rfloor$ and doing this over all $i$ and all $p$ gives the desired result.

Second derivative of $\arctan(x^2)$ Given that $y=\arctan(x^2)$ find $\ \dfrac{d^2y}{dx^2}$. I got $$\frac{dy}{dx}=\frac{2x}{1+x^4}.$$ Using low d high minus high d low over low squared, I got $$\frac{d^2y}{dx^2}=\frac{(1+x)^4 \cdot 2 - 2x \cdot 4(1+x)^3}{(1+x^4)^2}.$$ I tried to simplify this but didn't get the answer which is $$\frac{d^2y}{dx^2}=\frac{2(1-3x^4)}{(1+x^4)^2}.$$ Where am I going wrong?

Alternatively, $ \large \tan (y) = x^2 \Rightarrow \sec^2 (y) \cdot \frac { \mathrm{d}y}{\mathrm{d}x} = 2x $ $ \large \Rightarrow \frac { \mathrm{d}y}{\mathrm{d}x} = \frac {2x}{\sec^2 (y)} = \frac {2x}{\tan^2 (y) + 1} = \frac {2x}{x^4 + 1} $ $ \large \Rightarrow \frac { \mathrm{d^2}y}{\mathrm{d}x^2} = \frac {2(x^4+ 1) - 2x(4x^3)}{(x^4+1)^2} = \frac {-6x^4 + 2}{(x^4 + 1)^2} $

Algebraic solution for the intersection point(s) of two parabolas I recently ran through an algebraic solution for the intersection point(s) of two parabolas $ax^2 + bx + c$ and $dx^2 + ex + f$ so that I could write a program that solved for them. The math goes like this: $$ ax^2 - dx^2 + bx - ex + c - f = 0 \\ x^2(a - d) + x(b - e) = f - c \\ x^2(a - d) + x(b - e) + \frac{(b - e)^2}{4(a - d)} = f - c + \frac{(b - e)^2}{4(a - d)} \\ (x\sqrt{a - d} + \frac{b - e}{2\sqrt{a - d}})^2 = f - c + \frac{(b - e)^2}{4(a - d)} \\ (a - d)(x + \frac{b - e}{2(a - d)})^2 = f - c + \frac{(b - e)^2}{4(a - d)} \\ x + \frac{b - e}{2(a - d)} = \sqrt{\frac{f - c + \frac{(b - e)^2}{a - d}}{a - d}} \\ x = \pm\sqrt{\frac{f - c + \frac{(b - e)^2}{a - d}}{a - d}} - \frac{b - e}{2(a - d)} \\ $$ Then solving for $y$ is as simple as plugging $x$ into one of the equations. $$ y = ax^2 + bx + c $$ Is my solution for $x$ and $y$ correct? Is there a better way to solve for the intersection points?

You lost a factor $4$ somewhere. You can simply rewrite your problem as $$(a-d)x^2+(b-e)x+(c-f)=0$$ and use the standard formula for a quadratic equation, i.e. $$x=-\frac{b-e}{2(a-d)}\pm\sqrt{\frac{(b-e)^2}{4(a-d)^2}-\frac{c-f}{a-d}}$$ Before evaluating this equation, you need to check if $a-d=0$, in which case $$x=\frac{f-c}{b-e}$$ In this case you of course need to check if $b-e=0$.

Surely You're Joking, Mr. Feynman! $\int_0^\infty\frac{\sin^2x}{x^2(1+x^2)}\,dx$ Prove the following \begin{equation}\int_0^\infty\frac{\sin^2x}{x^2(1+x^2)}\,dx=\frac{\pi}{4}+\frac{\pi}{4e^2}\end{equation} I would love to see how Mathematics SE users prove the integral preferably with the Feynman way (other methods are welcome). Thank you. (>‿◠)✌ Original question: And of course, for the sadist with a background in differential equations, I invite you to try your luck with the last integral of the group. \begin{equation}\int_0^\infty\frac{\sin^2x}{x^2(1+x^2)}\,dx\end{equation} Source: Integration: The Feynman Way

This integral is readily evaluated using Parseval's theorem for Fourier transforms. (I am certain that Feynman had this theorem in his tool belt.) Recall that, for transform pairs $f(x)$ and $F(k)$, and $g(x)$ and $G(k)$, the theorem states that $$\int_{-\infty}^{\infty} dx \, f(x) g^*(x) = \frac1{2 \pi} \int_{-\infty}^{\infty} dk \, F(k) G^*(k) $$ In this case, $f(x) = \frac{\sin^2{x}}{x^2}$ and $g(x) = 1/(1+x^2)$. Then $F(k) = \pi (1-|k|/2) \theta(2-|k|)$ and $G(k) = \pi \, e^{-|k|}$. ($\theta$ is the Heaviside function, $1$ when its argument is positive, $0$ when negative.) Using the symmetry of the integrand, we may conclude that $$\begin{align}\int_0^{\infty} dx \frac{\sin^2{x}}{x^2 (1+x^2)} &= \frac{\pi}{2} \int_0^{2} dk \, \left ( 1-\frac{k}{2} \right ) e^{-k} \\ &= \frac{\pi}{2} \left (1-\frac1{e^2} \right ) - \frac{\pi}{4} \int_0^{2} dk \, k \, e^{-k} \\ &= \frac{\pi}{2} \left (1-\frac1{e^2} \right ) + \frac{\pi}{2 e^2} - \frac{\pi}{4} \left (1-\frac1{e^2} \right )\\ &= \frac{\pi}{4} \left (1+\frac1{e^2} \right )\end{align} $$

How to find determinant of this matrix? Is there a manual method to find $\det\left(XY^{-1}\right)$ ? Let $$X=\left[ {\begin{array}{cc} 1 & 2 & 2^2 & \cdots & 2^{2012} \\ 1 & 3 & 3^2 & \cdots & 3^{2012} \\ 1 & 4 & 4^2 & \cdots & 4^{2012} \\ \vdots & \vdots & \vdots & \cdots & \vdots \\ 1 & 2014 & 2014^2 & \cdots & 2014^{2012} \\ \end{array} } \right], $$ $$Y=\left[ {\begin{array}{cc}\frac{2^2}{4} & \frac{3^2}{5} & \dfrac{4^2}{6} & \cdots & \dfrac{2014^2}{2016} \\ 2 & 3 & 4 & \cdots & 2014 \\ 2^2 & 3^2 & 4^2 & \cdots & 2014^{2} \\ \vdots & \vdots & \vdots & \cdots & \vdots \\ 2^{2012} & 3^{2012} & 4^{2012} & \cdots & 2014^{2012} \\ \end{array} } \right] $$. Thanks in advance.

Consider something a bit more general. Let $$X=\left[ {\begin{array}{cc} 1 & x_1 & x_1^2 & \cdots & x_1^{n-1} \\ 1 & x_2 & x_2^2 & \cdots & x_2^{n-1} \\ 1 & x_3 & x_3^2 & \cdots & x_3^{n-1} \\ \vdots & \vdots & \vdots & \cdots & \vdots \\ 1 & x_n & x_n^2 & \cdots & x_n^{n-1} \\ \end{array} } \right], $$ and $$Y=\left[ {\begin{array}{cc}\frac{x_1^2}{x_1+r} & \frac{x_2^2}{x_2+r} & \dfrac{x_3^2}{x_3+r} & \cdots & \dfrac{x_n^2}{x_n+r} \\ x_1 & x_2 & x_3 & \cdots & x_n \\ x_1^2 & x_2^2 & x_3^2 & \cdots & x_n^2 \\ \vdots & \vdots & \vdots & \cdots & \vdots \\ x_1^{n-1} & x_2^{n-1} & x_3^{n-1} & \cdots & x_n^{n-1} \\ \end{array} } \right]. $$ Your problem corresponds to $x_i=i+1,$ $n=2013,$ $r=2.$ Let $Y'$ be the matrix obtained by rescaling the columns of $Y$ by multiplying column $i$ by $(x_i+r)/x_i.$ So $$\det Y'=\det Y\prod_{i=1}^n\frac{x_i+r}{x_i} $$ and $$Y'=\left[ {\begin{array}{cc}x_1 & x_2 & x_3 & \cdots & x_n \\ x_1+r & x_2+r & x_3+r & \cdots & x_n+r \\ x_1(x_1+r) & x_2(x_2+r) & x_3(x_3+r) & \cdots & x_n(x_n+r) \\ \vdots & \vdots & \vdots & \cdots & \vdots \\ x_1^{n-2}(x_1+r) & x_2^{n-2}(x_2+r) & x_3^{n-2}(x_3+r) & \cdots & x_n^{n-2}(x_n+r) \\ \end{array} } \right]. $$ Now show that $$\det Y'=-r\det X,$$ from which your determinant can easily be evaluated. This is done by a series of row operations on $Y'.$ First subtract row $1$ from row $2.$ Then swap the first two rows. Then divide row $1$ by $r$. (These steps account for the factor $-r.$) We now have the matrix $$\left[ {\begin{array}{cc}1 & 1 & 1 & \cdots & 1 \\ x_1 & x_2 & x_3 & \cdots & x_n \\ x_1(x_1+r) & x_2(x_2+r) & x_3(x_3+r) & \cdots & x_n(x_n+r) \\ \vdots & \vdots & \vdots & \cdots & \vdots \\ x_1^{n-2}(x_1+r) & x_2^{n-2}(x_2+r) & x_3^{n-2}(x_3+r) & \cdots & x_n^{n-2}(x_n+r) \\ \end{array} } \right]. $$ Now subtract $r$ times row $2$ from row $3$. Then subtract $r$ times row $3$ from row $4.$ Continue in this way, finally subtracting $r$ times row $n-1$ from row $n.$ The resulting matrix will be $X^T.$

Domain of the Function Square Root of 12th Degree Polynomial Find the Domain of $$f(x)=\frac{1}{\sqrt{x^{12}-x^9+x^4-x+1}}$$ My Try: The Domain is given by $$x^{12}-x^9+x^4-x+1 \gt 0$$ $\implies$ $$x(x-1)(x^2+x+1)(x^8+1)+1 \gt 0$$ Please help me how to proceed further..

If $x$ is out of the interval $[0,1]$ all the facotrs in $$x(x-1)(x^2+x+1)(x^8+1)+1$$ are positive so we have $$x(x-1)(x^2+x+1)(x^8+1)+1\gt 0$$ But for $x\in [0,1],$ $$x(x-1)\geq -\frac{1}{4}$$ So $$x(x-1)(x^2+x+1)(x^8+1)+1\geq -\frac{1}{4}\cdot (x^2+x+1)(x^8+1)+1\geq -\frac{1}{4}\cdot\frac{3}{4}.1+1=\frac{13}{16}$$ i.e., for all $x\in \mathbb R$ $$x(x-1)(x^2+x+1)(x^8+1)+1\gt 0$$

Find a formula for $1 + 3 + 5 + ... +(2n - 1)$, for $n \ge 1$, and prove that your formula is correct. I think the formula is $n^2$. Define $p(n): 1 + 3 + 5 + \ldots +(2n − 1) = n^2$ Then $p(n + 1): 1 + 3 + 5 + \ldots +(2n − 1) + 2n = (n + 1)^2$ So $p(n + 1): n^2 + 2n = (n + 1)^2$ The equality above is incorrect, so either my formula is wrong or my proof of the implication is wrong or both. Can you elaborate? Thanks.

The issue here is that $p(n+1)$ is note the statement $$ 1+3+5+\cdots+(2n-1)+2n=(n+1)^2; $$ it is the statement $$ 1+3+5+\cdots+(2n-1)+(2n+1)=(n+1)^2. $$ Why? The left side of your formula is the sum of all odd numbers between $1$ and $2n-1$. So, when you replace $n$ by $n+1$, you get the sum of all odd numbers between $1$ and $2(n+1)-1=2n+1$.

Algebra Manipulation Contest Math Problem The question was as follows: The equations $x^3+Ax+10=0$ and $x^3+Bx^2+50=0$ have two roots in common. Compute the product of these common roots. Because $x^3+Ax+10=0$ and $x^3+Bx^2+50=0$ it means that $x^3+Ax+10=x^3+Bx^2+50$ Take $x^3+Ax+10=x^3+Bx^2+50$ and remove $x^3$ from both sides, you get $Ax+10=Bx^2+50$ or $Bx^2-Ax+40=0$ By the quadratic equation, we get $\frac {A \pm \sqrt {(-A)^2 - 4*40B}}{2B}=\frac {A \pm \sqrt {A^2 - 160B}}{2B}$ This gives us two answers: $\frac {A + \sqrt {A^2 - 160B}}{2B}$ and $\frac {A - \sqrt {A^2 - 160B}}{2B}$ $\frac {A + \sqrt {A^2 - 160B}}{2B} * \frac {A - \sqrt {A^2 - 160B}}{2B}=\frac {A^2 - {A^2 - 160B}}{4B^2}$ This simplifies as $\frac {160B}{4B^2}=\frac{40}{B}$ $\frac{40}{B}$ is an answer, but in the solutions, they expected an integer answer. Where did I go wrong?

Hint: The common roots must be both roots of $$- (x^3 + Ax +10 ) + (x^3 + Bx^2 + 50) = Bx^2 - Ax + 40 $$ Let this quadratic polynomial be denoted by $f(x)$. Hint: We have $$ f(x) ( \frac{1}{B} x + \frac{5}{4} ) = x^3 + Bx^2 + 50. $$ This gives $B^2 = 4A$ and $160=5AB$, so $5B^3 = 640 $. This gives $B = 4 \sqrt[3]{2} $, $ A = 4\sqrt[3]{4}$. This does not give me an integer answer for $ \frac{40}{B} = 5 \sqrt[3]{4}$, so perhaps they had an error?

Procedure for evaluating $\int_{x=\ -1}^1\int_{y=\ -\sqrt{1-x^2}}^{\sqrt{1-x^2}}\frac{x^2+y^2}{\sqrt{{1-x^2-y^2}}}\,dy\,dx$ While solving another problem I have come across this integral which I am unable to evaluate. Can someone please evaluate the following integral? Thank you. $$\int_{x=\ -1}^1\int_{\large y=\ -\sqrt{1-x^2}}^{\large\sqrt{1-x^2}}\frac{x^2+y^2}{\sqrt{{1-x^2-y^2}}}\,dy\,dx.$$ I know the answer is $\dfrac{4\pi}3$, but I am more interested in the procedure followed to get to this answer.

Based on the limit of integral $-\sqrt{1-x^2} < y < \sqrt{1-x^2}$ and $-1 < x < 1$, the region of integration is a unit circle in the Cartesian coordinate. See this plot to visualize the region of integration. Using polar coordinate, we have $x^2+y^2=r^2$ and the region of integration will be $0<r<1$ and $0<\theta<2\pi$. Therefore \begin{align} \int_{x=\ -1}^1\int_{\large y=\ -\sqrt{1-x^2}}^{\large\sqrt{1-x^2}}\frac{x^2+y^2}{\sqrt{{1-x^2-y^2}}}\,dy\,dx&=\int_{\theta=0}^{2\pi}\int_{r=0}^{1}\frac{r^2}{\sqrt{{1-r^2}}}\,r\ dr\,d\theta\\ &=\int_{\theta=0}^{2\pi}\,d\theta\ \int_{r=0}^{1}\frac{r^3}{\sqrt{{1-r^2}}}\ dr. \end{align} Let $r=\sin t\;\Rightarrow\;dr=\cos t\ dt$ and the corresponding region is $0<t<\dfrac\pi2$, then \begin{align} \int_{x=\ -1}^1\int_{\large y=\ -\sqrt{1-x^2}}^{\large\sqrt{1-x^2}}\frac{x^2+y^2}{\sqrt{{1-x^2-y^2}}}\,dy\,dx&=\int_{\theta=0}^{2\pi}\,d\theta\ \int_{t=0}^{\Large\frac\pi2}\frac{\sin^3t}{\sqrt{{1-\sin^2t}}}\cdot \cos t\ dt\\ &=2\pi\int_{t=0}^{\Large\frac\pi2} \sin^3t\ dt\\ &=2\pi\int_{t=0}^{\Large\frac\pi2} \sin^2t\ \sin t\ dt\\ &=2\pi\int_{t=0}^{\Large\frac\pi2} (\cos^2t-1)\ d(\cos t). \end{align} Set $\theta=\cos t$ and you can take it from here.

Integer solutions to $x^x=122+231y$ How can I find the integer solutions to the following equation (without a script or trial and error)? $$x^x=122+231y$$

The function $f_a:\Bbb N\to\Bbb Z_p$ defined as $f_a(x)=a^x$ where $a\in\Bbb Z_p^*$ is periodic and its period is a divisor of $p-1$. Then the modular equation $x^x\equiv 2 \pmod 3$ has only to be checked for $x\in\{1,\ldots,6\}$. And $$1^1\equiv 1\pmod 3$$ $$2^2\equiv 1\pmod 3$$ $$3^3\equiv 0\pmod 3$$ $$4^4\equiv 1\pmod 3$$ $$5^5\equiv 2\pmod 3$$ $$6^6\equiv 0\pmod 3$$ Therefore, if $x$ is a solution, then $x\equiv 5\pmod 6$. Now, write $x=5+6k$ and try to find solutions for $$(5+6k)^{5+6k}\equiv 3\pmod 7$$ but, by LFT, this is equivalent to $$(5+6k)^{-1}\equiv 3\pmod 7$$ or $$5+6k\equiv 5\pmod 7$$ so $k$ is a multiple of $7$, that is, $x=5+42j$. Now we have to deal with the last prime factor of $231$, that is, $11$. $$(5+42j)^{5+42j}\equiv(5-2j)^{5+2j}\equiv 1\pmod {11}$$ There are some possibilities now: * *$5-2j\equiv1\pmod{11}$ which gives $x=462m+89$. *$5-2j\equiv -1\pmod{11}$ and $5+2j$ is even. But this is impossible. *$5-2j\equiv 4,5,9\text{ or }3\pmod{11}$ and $5+2j$ is a multiple of $5$, which gives $x=2310m+5$, $845$, $1895$ or $2105$. *$5+2j$ is multiple of $10$, which is impossible. To sum up, the solutions are the positive integers of the form $$x=\left\{ \begin{array}{l} 89+462m\\ 5+2310m\\ 845+2310m\\ -405+2310m\\ -25+2310m \end{array} \right.$$ where $m$ is an integer. Some solutions for $x$: $5$, $89$, $551$, $845$, $1013$, $1475$, ...

Determine which Fibonacci numbers are even (a) Determine which Fibonacci numbers are even. Use a form of mathematical induction to prove your conjecture. (b) Determine which Fibonacci numbers are divisible by 3. Use a form of mathematical induction to prove your conjecture I understand that for part a that all multiples of 3 of n are even. So F(0),F(3),F(6)... I just don't understand how to prove it. For part B it is the same thing except multiples of 4 Please help, thank you!

Part A: Base case: $F(0) = 0$, 0 is even. $F(3) = 2$, 2 is even. Inductive Hypothesis: Assume $F(k)$ is even for some arbitrary positive integer $k$ that is divisible by 3. Want to prove: That $F(k+3)$ is even given the inductive hypothesis. \begin{align*} F(k+3) &= F(k+2) + F(k+1)\\ &= F(k+1) + F(k) + F(k+1)\\ &= F(k) + 2 F(k+1) \end{align*} We know by our hypothesis that $F(k)$ is even. We also know that $F(k+1)$ is an integer, so $2 F(k+1)$ is necessarily even by properties of integers. And because the sum of two even integers is also even, it follows that $F(k+3)$ is even. This holds for any arbitrary positive integer $k$ that is divisible by 3, hence we've proved that $F(m)$ is even for all positive integers $m$ divisible by 3. Part B: Base case: $F(0) = 0$, 0 is a multiple of 3. $F(4) = 3$, 3 is a multiple of 3. Inductive Hypothesis: Assume $F(k)$ is a multiple of 3 for some arbitrary positive integer $k$ that is divisible by 4. Want to prove: That $F(k+4)$ is a multiple of 3 given the inductive hypothesis. \begin{align*} F(k+4) &= F(k+3) + F(k+2)\\ &= F(k+2) + F(k+1) + F(k+1) + F(k)\\ &= F(k+1) + F(k) + F(k+1) + F(k+1) + F(k)\\ &= 2 F(k) + 3 F(k + 1) \end{align*} By our hypothesis, $F(k)$ is a multiple of 3, so $2 F(k)$ must also be a multiple of 3. Furthermore, we know that $F(k+1)$ is some integer, so $3 F(k+1)$ must also be multiple of 3. Hence their sum, $F(k+4)$ must also be a multiple of 3.

middle school question on geometry As you can see from the picture, the angle $A$ is $90^\circ$, and the segments $BD$ and $CE$ (which intersect at $F$) are angle bisectors of the angles $B$ and $C$, respectively. When the length of $CF$ is $\frac72$ and and the quadrilateral $BCDE$ has an area of $14$, what is the length of $BC$? This is supposedly a middle-school question, appreciate any help.

This is not a middle school level answer. But perhaps a complicated answer is better than no answer at all. Start by choosing coordinates as follows: $$ A=\begin{pmatrix}0\\0\end{pmatrix}\quad B=\begin{pmatrix}b\\0\end{pmatrix}\quad C=\begin{pmatrix}0\\c\end{pmatrix}\quad D=\begin{pmatrix}0\\d\end{pmatrix}\quad E=\begin{pmatrix}e\\0\end{pmatrix} $$ Then the area condition becomes $$\tfrac12bc-\tfrac12de=14\tag{1}$$ For the distance condition, you have to compute $$F=\frac{1}{bc-de}\begin{pmatrix}be(c-d)\\cd(b-e)\end{pmatrix}$$ by intersecting $BD$ with $CE$. As an alternative, you might intersect one of these lines with the bisector $x=y$, but I'll leave the above for now. From that you get \begin{align*} \lVert F-C\rVert &= \tfrac72 \\ \lVert F-C\rVert^2 &= \left(\tfrac72\right)^2 \\ \left(\frac{be(c-d)}{bc-de}\right)^2+\left(\frac{cd(b-e)}{bc-de}-c\right)^2 &= \left(\tfrac72\right)^2 \\ %\left(\frac{1}{bc-de}\right)^2\left((be(c-d))^2 + (cd(b-e)-c(bc-de))^2\right) %&= \left(\tfrac72\right)^2 \\ \bigl(be(c-d)\bigr)^2 + \bigl(cd(b-e)-c(bc-de)\bigr)^2 &= \left(\tfrac72(bc-de)\right)^2 \tag{2} \end{align*} The most difficult part to formulate is probably the angle bisector conditions. Start at the double angle formula for the tangens: \begin{align*} \tan2\theta &= \frac{2\tan\theta}{1-\tan^2\theta} \\ \frac cb &= \frac{2\frac db}{1-\left(\frac db\right)^2} \\ \frac cb &= \frac{2bd}{b^2-d^2} \\ c(b^2-d^2) &= 2b^2d \tag{3} \\ b(c^2-e^2) &= 2c^2e \tag{4} \end{align*} Now you have four (non-linear) equations $\text{(1)}$ through $\text{(4)}$ in four variables $b$ through $e$. Eliminate variables (e.g. using resultants) to obtain the solution: $$ b=\frac{120}{17} \quad c=\frac{161}{34} \quad d=\frac{840}{391} \quad e=\frac{644}{255} $$ Then you can compute $$ \lVert B-C\rVert=\sqrt{b^2+c^2}=\frac{17}2 $$

How to solve $\int \frac{\,dx}{(x^3 + x + 1)^3}$? How to solve $$\int \frac{\,dx}{(x^3 + x + 1)^3}$$ ? Wolfram Alpha gives me something I am not familiar with. I thought that the idea was using partial fractions because $x^3$ and $x$ are bijections, there must be a real root but it seems that Wolfram Alpha is using numerical methods to approximate the root so it's not a "nice" number. I can't thing of any substitution which could help me nor any formula to transform this denominator. The formula $\int u\,dv = uv - \int v\,du$ also yields a more complicated integral: $$ \,dv = \,dx \implies v = x \\ u = \frac{1}{(x^3 + x + 1)^3} \implies du = -3\frac{3x^2 + 1}{(x^3 + x + 1)^4} \,dx \\ \int \frac{\,dx}{(x^3 + x + 1)^3} = \frac{x}{(x^3 + x + 1)^3} + 3\int \frac{3x^3 + x}{(x^3 + x + 1)^4} \,dx \\ = \frac{x}{(x^3 + x + 1)^3} + 3\int \frac{3x^3 + x \pm 2x \pm 3}{(x^3 + x + 1)^4}\,dx \\ = \frac{x}{(x^3 + x + 1)^3} + 3\int \frac{3x^3 + 3x + 3}{(x^3 + x + 1)^4}\,dx + 3\int \frac{- 2x - 3}{(x^3 + x + 1)^4}\,dx \\ = \frac{x}{(x^3 + x + 1)^3} + 9\int \frac{\,dx}{(x^3 + x + 1)^3} - 3\int \frac{2x + 3}{(x^3 + x + 1)^4} \,dx \\ I = \int \frac{\,dx}{(x^3 + x + 1)^3} \implies \\ -8I = \frac{x}{(x^3 + x + 1)^3} - 3\int \frac{2x + 3}{(x^3 + x + 1)^4} \,dx$$ Is this one really as complicated as Wolfram Alpha "tells" me or is there some sort of "trick" which can be applied?

Hint: the polynomial $x^3+x+1$ has one real root, say $\alpha$. Then $x^3+x+1=(x-\alpha)(x^2+\alpha x+\alpha^2+1)$ and then apply integration techniques of rational expressions of polynomials with repeated factors, see for example https://math.la.asu.edu/~surgent/mat271/parfrac.pdf

Why does $\frac{(x^2 + x-6)}{x-2} = \frac{(x+3)(x-2)}{x-2}$? I'm not the best at algebra and would be grateful if someone could explain how you can get from, $$\frac{x^2 + x-6}{x-2}$$ to, $$\frac{(x+3)(x-2)}{x-2}$$

Key fact: Knowing the roots of a polynomial (where the polynomial equals zero), let us factor it. So if $n$ and $m$ are two roots of the quadratic $ax^2+bx+c$, then we can factor it as $$ax^2+bx+c=a(x-n)(x-m).$$ The roots of a quadratic can be determined using the quadratic formula: $$x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}.$$ To find out the roots of the quadratic $x^2+x-6$ use the above formula and you'll find out that they are: $-3$ and $2$. Therefore we can write our polynomial as: $$x^2+x-6=(x-(-3))(x-2)=(x+3)(x-2).$$ Hence, it follows that: $$\require{cancel}\frac{x^2 + x-6}{x-2} = \frac{(x+3)\color{red}{\cancel{\color{black}{(x-2)}}}}{\color{red}{\cancel{\color{black}{x-2}}}}=x+3.\tag{assuming $x\neq2$}$$

How to find the following sum? $\sum\limits_{n = 0}^\infty {\left( {\frac{1}{{4n + 9}} - \frac{1}{{4n + 7}}} \right)} $ I want to calculate the sum with complex analysis (residue) $$ 1 - \frac{1}{7} + \frac{1}{9} - \frac{1}{{15}} + \frac{1}{{17}} - ... $$ $$ 1 + \sum\limits_{n = 0}^\infty {\left( {\frac{1}{{4n + 9}} - \frac{1}{{4n + 7}}} \right)} = 1 - \frac{1}{7} + \frac{1}{9} - \frac{1}{{15}} + \frac{1}{{17}} - ... $$ I ask $$ f\left( z \right) = - \frac{2}{{\left( {4z + 9} \right)\left( {4z + 7}\right)}} $$ is to : $$\sum\limits_{n = - \infty }^\infty {\frac{2}{{\left( {4n + 9} \right)\left( {4n + 7}\right)}}} = \left( {\mathop {\lim }\limits_{z \to - \frac{9}{4}} \left[ {\left( {z + \frac{9}{4}} \right)\frac{{\pi \cot \left( {\pi z} \right)}}{{\left( {4z + 9} \right)\left( {4z + 7} \right)}}} \right] + \mathop {\lim }\limits_{z \to - \frac{7}{4}} \left[ {\left( {z + \frac{7}{4}} \right)\frac{{\pi \cot \left( {\pi z} \right)}}{{\left( {4z + 9} \right)\left( {4z + 7} \right)}}}\right] } \right)$$ I found: \begin{array}{l} \mathop {\lim }\limits_{z \to - \frac{9}{4}} \left[ {\left( {z + \frac{9}{4}} \right)\frac{{\pi \cot \left( {\pi z} \right)}}{{\left( {4z + 9} \right)\left( {4z + 7} \right)}}} \right] = \frac{1}{4}\mathop {\lim }\limits_{z \to - \frac{9}{4}} \left[ {\left( {z + \frac{9}{4}} \right)\frac{{\pi \cot \left( {\pi z} \right)}}{{\left( {z + \frac{9}{4}} \right)\left( {4z + 7} \right)}}} \right] \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad = \frac{1}{4}\mathop {\lim }\limits_{z \to - \frac{9}{4}} \left[ {\frac{{\pi \cot \left( {\pi z} \right)}}{{4z + 7}}} \right] = \frac{1}{4}\left[ {\frac{{ - \pi }}{{ - 2}}} \right] = \frac{\pi }{8} \\ \mathop {\lim }\limits_{z \to - \frac{7}{4}} \left[ {\left( {z + \frac{7}{4}} \right)\frac{{\pi \cot \left( {\pi z} \right)}}{{\left( {4z + 9} \right)\left( {4z + 7} \right)}}} \right] = \frac{1}{4}\mathop {\lim }\limits_{z \to - \frac{9}{4}} \left[ {\left( {z + \frac{7}{4}} \right)\frac{{\pi \cot \left( {\pi z} \right)}}{{\left( {z + \frac{7}{4}} \right)\left( {4z + 9} \right)}}} \right] \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad = \frac{1}{4}\mathop {\lim }\limits_{z \to - \frac{9}{4}} \left[ {\frac{{\pi \cot \left( {\pi z} \right)}}{{\left( {4z + 9} \right)}}} \right] = \frac{\pi }{8} \\ \end{array} \begin{array}{l} \sum\limits_{n = - \infty }^\infty {\frac{2}{{\left( {4n + 9} \right)4n + 7}}} = - \frac{\pi }{4} = - \left( {\frac{\pi }{8} + \frac{\pi }{8}} \right) \\ \Rightarrow s = 1 + \sum\limits_{n = 0}^\infty {\left( {\frac{1}{{4n + 9}} - \frac{1}{{4n + 7}}} \right)} = 1 - \frac{\pi }{8} = \frac{{7 - \pi }}{8} \\ \end{array} I have a question for the result $$\sum\limits_{n = 0}^\infty {\left( {\frac{1}{{4n + 9}} - \frac{1}{{4n + 7}}} \right)} = - \frac{1}{5} \Rightarrow s = 1 + \sum\limits_{n = 0}^\infty {\left( {\frac{1}{{4n + 9}} - \frac{1}{{4n + 7}}} \right)} = \frac{4}{5} \ne \frac{{7 - \pi }}{8}$$ thank you in advance

Here is a way to evaluate your series with the method of residues. $$\sum\limits_{n = 0}^\infty {\left( {\frac{1}{{4n + 9}} - \frac{1}{{4n + 7}}} \right)} = \sum_{n=2}^{\infty} \left(\frac{1}{4n+1} - \frac{1}{4n-1}\right) =\sum_{n=2}^{\infty} \frac{-2}{(4n)^2 - 1} = f(n)$$ Consider a function $$ f(z)= \frac{-2}{(4z)^2 - 1} $$ Now, $$\sum_{n=-\infty}^{\infty} f(n) = 2\sum_{n=2}^{\infty} \frac{-2}{(4n)^2 - 1} + f(1)+f(0)+f(-1)$$ Using residue theorem we calculate the sum of residue as , $$\sum_{n=-\infty}^{\infty} f(n) = \frac \pi 2$$ and $$f(-1) + f(0) + f(1) = \frac{26}{15}$$ Putting it together you get $$\boxed{\mathrm{Required\,Sum} = \sum_{n=2}^{\infty} \frac{-2}{(4n)^2 - 1} = \frac 1 2 \left( \frac \pi 2 - \frac {26}{15}\right)}$$

limits of function without using L'Hopital's Rule $\mathop {\lim }\limits_{x \to 1} \frac{{x - 1 - \ln x}}{{x\ln x+ 1 - x}} = 1$ Good morning. I want to show that without L'Hopital's rule : $\mathop {\lim }\limits_{x \to 1} \frac{{x - 1 - \ln x}}{{x\ln x + 1 - x}} = 1$ I did the steps $ \begin{array}{l} \mathop {\lim }\limits_{x \to 1} \frac{{x - 1 + \ln \left( x \right)}}{{x\ln \left( x \right) - x + 1}} = \mathop {\lim }\limits_{y \to 0} \frac{{y + \ln \left( {y + 1} \right)}}{{\left( {y + 1} \right)\ln \left( {y + 1} \right) - y}} \\ \ln \left( {y + 1} \right) = 1 - \frac{{y^2 }}{2} + o\left( {y^2 } \right);and\quad \mathop {\lim }\limits_{y \to 0} o\left( {y^2 } \right) = 0 \\ \Rightarrow \left( {y + 1} \right)\ln \left( {y + 1} \right) = 1 + y - \frac{{y^2 }}{2} + o\left( {y^2 } \right) \\ \mathop {\lim }\limits_{y \to 0} \frac{{y + \ln \left( {y + 1} \right)}}{{\left( {y + 1} \right)\ln \left( {y + 1} \right) - y}} = \frac{{1 + y - \frac{{y^2 }}{2}}}{{1 + y - \frac{{y^2 }}{2}}} = 1 \\ \end{array} $ help me what you please

$ \displaylines{ \left\{ \begin{array}{l} t = 1 + u \\ u \cong \ln t \\ \end{array} \right. \cr \Rightarrow \cr \mathop {\lim }\limits_{t \to 1} \left[ {\frac{{\left( {1 + t} \right)\ln t}}{{t\ln t - t + 1}}} \right] = \mathop {\lim }\limits_{u \to 2} \left[ {\frac{{\left( {2 + u} \right)u}}{{\left( {1 + u} \right)u - u}}} \right] \cr = \mathop {\lim }\limits_{u \to 2} \left[ {\frac{{2u + u^2 }}{{u^2 }}} \right] = 2 \cr \mathop {\lim }\limits_{t \to 1} \frac{{\ln t + t - 1}}{{t\ln t - t + 1}} = - \mathop {\lim }\limits_{t \to 1} \left[ {\frac{{ - \ln t - t\ln t + t\ln t - t + 1}}{{t\ln t - t + 1}}} \right] \cr = - 1 + \mathop {\lim }\limits_{t \to 1} \left[ {\frac{{\left( {1 + t} \right)\ln t}}{{t\ln t - t + 1}}} \right] \cr = - 1 + 2 = 1 \cr \mathop {\lim }\limits_{t \to 1} \frac{{\ln t + t - 1}}{{t\ln t - t + 1}} = 1 \cr} $ othere way \begin{array}{l} \mathop {\lim }\limits_{x \to 1} \frac{{x - 1 + \ln \left( x \right)}}{{x\ln \left( x \right) - x + 1}} = \mathop {\lim }\limits_{y \to 0} \frac{{y + \ln \left( {y + 1} \right)}}{{\left( {y + 1} \right)\ln \left( {y + 1} \right) - y}} \\ \ln \left( {y + 1} \right) = 1 - \frac{{y^2 }}{2} + o\left( {y^2 } \right);and\quad \mathop {\lim }\limits_{y \to 0} o\left( {y^2 } \right) = 0 \\ \Rightarrow \left( {y + 1} \right)\ln \left( {y + 1} \right) = 1 + y - \frac{{y^2 }}{2} + o\left( {y^2 } \right) \\ \mathop {\lim }\limits_{y \to 0} \frac{{y + \ln \left( {y + 1} \right)}}{{\left( {y + 1} \right)\ln \left( {y + 1} \right) - y}} = \frac{{1 + y - \frac{{y^2 }}{2}}}{{1 + y - \frac{{y^2 }}{2}}} = 1 \\ \end{array}

How to find integral $\underbrace{\int\sqrt{2+\sqrt{2+\sqrt{2+\cdots+\sqrt{2+x}}}}}_{n}dx,x>-2$ Find the integral $$\int\underbrace{\sqrt{2+\sqrt{2+\sqrt{2+\cdots+\sqrt{2+x}}}}}_{n}dx,x>-2$$ where $n$ define the number of the square I know this if $0 \le x\le 2$, then let $$x=2\cos{t},0\le t\le\dfrac{\pi}{2}$$ so $$\sqrt{2+x}=\sqrt{2+2\cos{t}}=2\cos{\dfrac{t}{2}}$$ so $$\sqrt{2+\sqrt{2+x}}=2\cos{\dfrac{t}{2^2}}$$ so $$\int\sqrt{2+\sqrt{2+\sqrt{2+\cdots+\sqrt{2+x}}}}dx=\int2\cos{\dfrac{t}{2^n}}(-2\sin{t})dt$$ and for $x\ge 2$ case, I let $x=\cosh{t}$, but for $-2\le x\le 0$ case, I can't do it.

You were too timid: For $-2\leq x\leq2$ use the substitution $$x=2\cos t\qquad(-\pi\leq t\leq 0)\ .$$ Then everything goes through as before: $$\sqrt{2+x}=\sqrt{2+2\cos t}=2\cos{t\over2},\quad \sqrt{2+\sqrt{2+x}}=\sqrt{2+\cos{t\over2}}=2\cos{t\over4}\ ,$$ etcetera.

Volume between cylinder and plane Problem: Find the volume bounded by $z = y^2, x =0, y =0, z =9-x$. My working: $z$ goes from $y^2$ to $9-x$ so these are the limits of integration. Work out the points of intersection of $9-x$ and $y^2$. When $y=0$, $9-x=0$ and $x=9$. So $x$ goes from 0 to 9. When $x=0$, $y^2 = 9$ so $y=3$ (take the positive one). So $y$ goes from 0 to 9. Then evaluate \begin{align} \int_{x=0}^{x=9} \int_{y=0}^{y=9} \int_{z=y^2}^{z=9-x} dz dy dz &= \int_{x=0}^{x=9} \int_{y=0}^{y=9} y^2 - 9 + x dy dx \\ &= \int_{x=0}^{x=9} 18+3x dx \\ &= \frac{567}{2} \end{align} My textbook says the answer is $\frac{324}{5}$. What have I done wrong?

Your limits of integration don't make sense. The region of integration is given by the set $$R = \{(x,y,z) \in \mathbb R^3 \mid (y^2 \le z \le 9-x) \cap (0 \le x \le 9) \cap (0 \le y \le 3)\}.$$ The projection of $R$ onto the $xz$-plane is simply the triangle $x \ge 0$, $z \ge 0$, $x + z \le 9$. Over this triangle, the curve $y = \sqrt{z}$ is the boundary, so the integral is $$\int_{x=0}^9 \int_{z=0}^{9-x} \int_{y=0}^\sqrt{z} 1 \, dy \, dz \, dx = \frac{324}{5}.$$ You may also project $R$ onto the $yz$-plane, in which case we would have $$\int_{y=0}^3 \int_{z=y^2}^9 \int_{x=0}^{9-z} 1 \, dx \, dz \, dy = \frac{324}{5}.$$ Projecting onto the $xy$-plane is trickier, but yields the integral $$\int_{y=0}^3 \int_{x=0}^{9-y^2} \int_{z=y^2}^{9-x} 1 \, dz \, dx \, dy = \frac{324}{5}.$$

Finding an equation of circle which passes through three points How to find the equation of a circle which passes through these points $(5,10), (-5,0),(9,-6)$ using the formula $(x-q)^2 + (y-p)^2 = r^2$. I know i need to use that formula but have no idea how to start, I have tried to start but don't think my answer is right.

I know i need to use that formula but have no idea how to start \begin{equation*} \left( x-q\right) ^{2}+\left( y-p\right) ^{2}=r^{2}\tag{0} \end{equation*} A possible very elementary way is to use this formula thrice, one for each point. Since the circle passes through the point $(5,10)$, it satisfies $(0)$, i.e. $$\left( 5-q\right) ^{2}+\left( 10-p\right) ^{2}=r^{2}\tag{1}$$ Similarly for the second point $(-5,0)$: $$\left( -5-q\right) ^{2}+\left( 0-p\right) ^{2}=r^{2},\tag{2}$$ and for $(9,-6)$: $$\left( 9-q\right) ^{2}+\left( -6-p\right) ^{2}=r^{2}.\tag{3}$$ We thus have the following system of three simultaneous equations and in the three unknowns $p,q,r$: $$\begin{cases} \left( 5-q\right) ^{2}+\left( 10-p\right) ^{2}=r^{2} \\ \left( -5-q\right) ^{2}+p^{2}=r^{2} \\ \left( 9-q\right) ^{2}+\left( 6+p\right) ^{2}=r^{2} \end{cases}\tag{4} $$ To solve it, we can start by subtracting the second equation from the first $$\begin{cases} \left( 5-q\right) ^{2}+\left( 10-p\right) ^{2}-\left( 5+q\right) ^{2}-p^{2}=0 \\ \left( 5+q\right) ^{2}+p^{2}=r^{2} \\ \left( 9-q\right) ^{2}+\left( 6+p\right) ^{2}=r^{2} \end{cases} $$ Expanding now the left hand side of the first equation we get a linear equation $$\begin{cases} 100-20q-20p=0 \\ \left( 5+q\right) ^{2}+p^{2}=r^{2} \\ \left( 9-q\right) ^{2}+\left( 6+p\right) ^{2}=r^{2} \end{cases} $$ Solving the first equation for $q$ and substituting in the other equations, we get $$\begin{cases} q=5-p \\ \left( 10-p\right) ^{2}+p^{2}-\left( 4+p\right) ^{2}-\left( 6+p\right) ^{2}=0 \\ \left( 4+p\right) ^{2}+\left( 6+p\right) ^{2}=r^{2} \end{cases} $$ If we simplify the second equation, it becomes a linear equation in $p$ only $$\begin{cases} q=5-p \\ 48-40p=0 \\ \left( 4+p\right) ^{2}+\left( 6+p\right) ^{2}=r^{2} \end{cases} $$ We have reduced our quadratic system $(4)$ to two linear equations plus the equation for $r^2$. From the second equation we find $p=6/5$, which we substitute in the first and in the third equations to find $q=19/5$ and $r^2=1972/25$, i.e $$\begin{cases} q=5-\frac{6}{5}=\frac{19}{5} \\ p=\frac{6}{5} \\ r^{2}=\left( 4+\frac{6}{5}\right) ^{2}+\left( 6+\frac{6}{5}\right) ^{2}= \frac{1972}{25}. \end{cases}\tag{5} $$ So the equation of the circle is \begin{equation*} \left( x-\frac{19}{5}\right) ^{2}+\left( y-\frac{6}{5}\right) ^{2}=\frac{1972}{25}. \end{equation*}

Help to compute sum of products I need to compute the following sum: $$(1\times2\times3)+(2\times3\times4)+(3\times4\times5)+ ...+(20\times21\times22)$$ All that I have deduced is: * *Each term is divisible by $6$. So sum is is divisible by $6$. *Sum is divisible by $5$ as 1st term is $1$ less than multiple of $5$ and second term is $1$ more than multiple of $5$. Next three terms are divisible by $5$. This cycle continues for every $5$ terms. So sum will obviously be divisible by $30$.

Here's an interesting solution: $(1\cdot2\cdot3)+(2\cdot3\cdot4)+(3\cdot4\cdot5)+\dots+(20\cdot21\cdot22)$ $\dfrac{3!}{0!}+\dfrac{4!}{1!}+\dfrac{5!}{2!}+\dots+\dfrac{22!}{19!}$ $3!\left(\dfrac{3!}{0!3!}+\dfrac{4!}{1!3!}+\dfrac{5!}{2!3!}+\dots+\dfrac{22!}{19!3!}\right)$ $3!\left(\dbinom{3}{3}+\dbinom{4}{3}+\dbinom{5}{3}+\dots+\dbinom{22}{3}\right)$ then using the hockey-stick identity, we see that this is equal to $3!\dbinom{23}{4} = 3!\dfrac{23\cdot22\cdot21\cdot20}{4\cdot3!} = \dfrac{23\cdot22\cdot21\cdot20}{4} = 53130$

Prove infinite series $$ \frac{1}{x}+\frac{2}{x^2} + \frac{3}{x^3} + \frac{4}{x^4} + \cdots =\frac{x}{(x-1)^2} $$ I can feel it. I can't prove it. I have tested it, and it seems to work. Domain-wise, I think it might be $x>1$, the question doesn't specify. Putting the LHS into Wolfram Alpha doesn't generate the RHS (it times out).

I think a less formal solution could be more understandable. consider $$ S_n= \frac{1}{x} + \frac{2}{x^2} + \frac{3}{x^3} + \frac{4}{x^4} + \dots + \frac{n}{x^n}$$ $$ xS_n = 1 + \frac{2}{x} + \frac{3}{x^2} + \frac{4}{x^3} + \dots + \frac{n}{x^{n-1}}$$ then $$xS_n - S_n = 1+ (\frac{2}{x}-\frac{1}{x})+(\frac{3}{x^2}-\frac{2}{x^2})+\dots+(\frac{n}{x^{n-1}}-\frac{n-1}{x^{n-1}}) - \frac{n}{x^n}$$ $$S_n(x-1) = 1 + \frac{1}{x} + \frac{1}{x^2}+\dots+\frac{1}{x^{n-1}} - \frac{n}{x^n}$$ Now we have a simplified the problem to one of a basic geometric series, so $$S_n(x-1) = T_{n-1} - \frac{n}{x^n}$$ where $$T_{n-1} = 1 + \frac{1}{x} + \frac{1}{x^2}+\dots+\frac{1}{x^{n-1}}$$ $$\frac{T_{n-1}}{x} = \frac{1}{x} + \frac{1}{x^2} + \frac{1}{x^3}+\dots+\frac{1}{x^{n}}$$ $$T_{n-1} - \frac{T_{n-1}}{x} = 1 + (\frac{1}{x}-\frac{1}{x})+ (\frac{1}{x^2}-\frac{1}{x^2})+\dots - \frac{1}{x^n}$$ $$T_{n-1}(1-\frac{1}{x}) = 1 - \frac{1}{x^n}$$ $$T_{n-1}(\frac{x-1}{x}) = \frac{x^n-1}{x^n}$$ $$T_{n-1} = \frac{x^n-1}{x^n}\cdot(\frac{x}{x-1})$$ $$T_{n-1} = \frac{x^n-1}{x-1}\cdot(\frac{1}{x^{n-1}})$$ $$T_{n-1} = \frac{x-\frac{1}{x^{n-1}}}{x-1}$$ Thus $S_n(x-1)$ becomes $$S_n(x-1) = \frac{x-\frac{1}{x^{n-1}}}{x-1} - \frac{n}{x^n}$$ for $|x|\gt 0$ this becomes $$\lim_{n\to\infty}S_n(x-1) = \lim_{n\to\infty}\frac{x-\frac{1}{x^{n-1}}}{x-1} - \frac{n}{x^n}$$ $$S(x-1) = \frac{x-\displaystyle\lim_{n\to\infty}\frac{1}{x^{n-1}}}{x-1} - \lim_{n\to\infty}\frac{n}{x^n}$$ $$S(x-1) = \frac{x-0}{x-1} - 0 = \frac{x}{x-1}$$ $$S = \frac{x}{(x-1)^2}$$ I used l'Hopital's rule to evaluate $\displaystyle\lim_{n\to\infty}\frac{n}{x^n}$, being an $\frac{\infty}{\infty}$ indeterminate form This helps me to understand the problem. Afterwards, I would go on to compose a more formal proof.

Prove the following Trig Identity with reciprocals Prove that: $$\frac{\tan x}{\sec x-1}+\frac{1-\cos x}{\sin x}=2\csc x$$ Help please! I tried so many things but couldn't get the LHS = RHS. A hint please?

Hint: $$ (1 - \cos x)(1 + \cos x) = 1 - \cos^2 x = \sin^2 x\\ (\sec x - 1)(\sec x + 1) = \sec^2 x - 1 = \tan^2 x $$ Further $$ \frac{\sec x + 1}{\tan x} = \frac{1+\cos x}{\sin x} = \frac{(1 + \cos x)^2}{(1 + \cos x) \sin x}\\ \frac{\sin x}{1 + \cos x} = \frac{\sin^2 x}{(1 + \cos x) \sin x} $$

Integrate $\int_0^1 \ln(x)\ln(b-x)\,\mathrm{d}x$, for $b>1$? Let $b>1$. What's the analytical expression for the following integral? $$\int_0^1 \ln(x)\ln(b-x)\,\mathrm{d}x$$ Mathematica returns the following answer: $$2-\frac{\pi^{2}}{3}b+\left(b-1\right)\ln\left(b-1\right)-b\ln b+\mathrm{i}b\pi\ln b+\frac{1}{2}b\ln^{2}b+b\mathrm{Li}_{2}\left(b\right)$$ which contains the imaginary term $\mathrm{i}b\pi\ln b$. But the actual answer is real, so this term should cancel somehow with the dilogarithm function. But I don't know how to do this.

The following is an evaluation in terms of $ \displaystyle \text{Li}_{2} \left(\frac{1}{b} \right)$, which is real-valued for $b > 1$. $$\begin{align} \int_{0}^{1} \log(x) \log(b-x) \ dx &= \log(b) \int_{0}^{1} \log(x) + \int_{0}^{1}\log(x) \log \left(1- \frac{x}{b} \right) \ dx \\ &= - \log(b) - \int_{0}^{1} \log(x) \sum_{n=1}^{\infty} \frac{1}{n} \left( \frac{x}{b}\right)^{n} \ dx \\ &= - \log(b) - \sum_{n=1}^{\infty} \frac{1}{nb^{n}} \int_{0}^{1} \log(x) x^{n} \ dx \\ &= - \log(b) + \sum_{n=1}^{\infty} \frac{1}{nb^{n}} \frac{1}{(n+1)^{2}} \\ &= - \log(b) - \sum_{n=1}^{\infty} \frac{1}{n+1} \frac{1}{b^{n}} - \sum_{n=1}^{\infty} \frac{1}{(n+1)^{2}} \frac{1}{b^{n}} + \sum_{n=1}^{\infty} \frac{1}{n} \frac{1}{b^{n}} \\ &= - \log(b) - \left(-\frac{\log(1-\frac{1}{b})}{\frac{1}{b}}-1\right) - \left(\frac{\text{Li}_{2}(\frac{1}{b})}{\frac{1}{b}} -1\right) - \log \left(1- \frac{1}{b} \right) \\ &= - \log(b) +2 + (b-1) \log \left(1-\frac{1}{b} \right) - b \ \text{Li}_{2} \left( \frac{1}{b}\right) \end{align}$$ EDIT: The answer can be written in the form $$-b \ \text{Li}_{2} \left( \frac{1}{b}\right) +2 + (b-1) \log(b-1) - b \log(b) $$ which is what Wolfram Alpha returns for specific integer values of $b$ greater than $1$. For non-integer values of $b$ greater than $1$, it manipulates the answer a bit differently.

$ 7\mid x \text{ and } 7\mid y \Longleftrightarrow 7\mid x^2+y^2 $ Show that $$ 7\mid x \text{ and } 7\mid y \Longleftrightarrow 7\mid x^2+y^2 $$ Indeed, First let's show $7\mid x \text{ and } 7\mid y \Longrightarrow 7\mid x^2+y^2 $ we've $7\mid x \implies 7\mid x^2$ the same for $7\mid y \implies 7\mid y^2$ then $ 7\mid x^2+y^2 $ * *Am i right and can we write $a\mid x \implies a\mid x^P ,\ \forall p\in \mathbb{N}^*$ Now let's show $7\mid x^2+y^2 \Longrightarrow 7\mid x \text{ and } 7\mid y$ $7\mid x^2+y^2 \Longleftrightarrow x^2+y^2=0 \pmod 7 $ for \begin{array}{|c|c|c|c|c|} \hline x& 0 & 1 & 2& 3 & 4 & 5 & 6 \\ \hline x^2& 0 & 1 & 4& 2 & 2 & 4 & 1 &\pmod 7\\ \hline y& 0 & 1 & 2& 3 & 4 & 5 & 6 \\ \hline y^2& 0 & 1 & 4& 2 & 2 & 4 & 1 & \pmod 7 \\ \hline \end{array} which means we have one possibility that $x=y= 0 \pmod 7 $ * *Am I right and are there other ways?

$►$ If $x=7x_1$ and $y=7y_1$ then $x^2+y^2=7(x_1^2+y_1^2$). $►$ If $x^2+y^2\equiv0\pmod7\iff x^2\equiv -y^2\pmod7$ then because of $\mathbb F_7^2=\{1,4,2,0\}$ and $(-1)\mathbb F_7^2=\{6,3,5,0\}$ the only possibility for $x^2\equiv -y^2\pmod7$ is that both $x^2$ and $y^2$ are equal to $0$ modulo $7$ so $x$ and $y$ are equal to $0$ modulo $7$ (since $7$ is prime). Then $x\equiv y\equiv0\pmod7$.

Show that $\dfrac{\sqrt{8-4\sqrt3}}{\sqrt[3]{12\sqrt3-20}} =2^\frac{1}{6}$ This was the result of evaluating an integral by two different methods. The RHS was obtained by making a substitution, the LHS was obtained using trigonometric identity's and partial fractions. Now I know that these two are equal, but I just can't prove it. I tried writing the LHS in terms of powers of $2$, but can't get any further to the desired result.

We have $$8 - 4\sqrt{3} = 2(4-2\sqrt{3}) = 2(3-2\sqrt{3}+1) = 2(\sqrt{3}-1)^2,$$ so $\sqrt{8-4\sqrt{3}} = \sqrt{2}(\sqrt{3}-1)$. Then we see that looking at $(\sqrt{3}-1)^3$ is a good idea: $$(\sqrt{3}-1)^3 = (4-2\sqrt{3})(\sqrt{3}-1) = 6\sqrt{3} - 10,$$ so $12\sqrt{3}-20 = 2(\sqrt{3}-1)^3$ and $\sqrt[3]{12\sqrt{3}-20} = \sqrt[3]{2}(\sqrt{3}-1)$. $\frac{\sqrt{2}}{\sqrt[3]{2}} = 2^{1/6}$ is then easy to see.

how to solve this question of polynomials Given the polynmial is exactly divided by $x+1$, when it is divided by $3x-1$, the remainder is $4$. The polynomial leaves remainder $hx+k$ when divided by $3x^2+2x-1$. Find $h$ and $k$. This is the question which is confusing me.. i have done this question like this: $p(x) = g1(x)(x+1) +0 $ $p(x) = g2(x)(3x-1) +4 $ $p(x) = g3(x)(3x+2x-1)+hx+k $ $=> p(x) = g3(x)(x+1)(3x-1) + hx+k $ now putting the value of $x$ in each $p(x) $ $p(-1) = 0 \tag{1}$ $p(1/3) = 4 \tag{2}$ $p(-1,1/3) = hx+k \tag{3}$ from equation $(2)$ and $(3)$ $4=hx+k\tag{4}$ now putting the value of $x$ in equation $(4)$. $h(-1)+k=4 => -h+k=4 \tag{5}$ $h(1/3)+k=4 => h+3k=12 \tag{6}$ now adding equations $(5)$ and $(6)$ $ h+3k-h+k = 4+12 =>4k=16 => k=4$ now puting the value of of $k$ in equation $(5)$ $h=0$ therefore my answer is $h=0,k=4$ but the answers are$ h=3, k=3$. please help me to sort out my problem

The error is that $\,p(-1) = 0\,\Rightarrow\, h(-1) + k = \color{#c00}0,\,$ not $\,\color{#c00}4.\,$ Fixing that yields the given answer. Remark $\ $ This is a special case of the Chinese Remainder Theorem (CRT) or, equivalently, Lagrange interpolation. Either of these methods can be applied to solve the general case.

Solve the integeral equation (C.S.I.R) Let $\lambda_1, \lambda_2$ be the eigen value and $f_1 , f_2$ be the coressponding eigen functions for the homogeneous integeral equation $$ \phi(x) - \lambda \int_0^1 (xt +2x^2) \phi(t) dt = 0 $$ Then * *$\lambda_1 = -18 - 6 \sqrt{10} , \lambda_2 = -18 + 6 \sqrt{10}$ *$\lambda_1 = -36 - 12 \sqrt{10} , \lambda_2 = -36 + 12 \sqrt{10}$ *$\int_0^1f_1(x)f_2(x) dx = 1$ *$\int_0^1f_1(x)f_2(x) dx = 0$ I have solved simply and get the values of $\lambda_1 \ \ and \ \ \lambda_2$ belong to (1). Please tell me about (3) and (4) option. Thank you.

Let's rewrite your equation in the following form: \begin{align} \frac{1}{\lambda}\phi(x) = x\int_{0}^{1}t\phi(t) \mathrm{d}t + 2x^{2}\int_{0}^{1} \phi(t) \mathrm{d}t = c_{1}x + c_{2}x^{2} \end{align} for $c_{1}: = \int_{0}^{1}t\phi(t) \, \mathrm{d}t$ and $c_{2}:= 2\int_{0}\phi(t)\, \mathrm{d}t$. From this we can see that $\phi$ is a quadratic polynomial with no constant term. Substituting $\phi(x) = c_{1}x + c_{2}x^{2}$, we find that \begin{align} \frac{c_{1}}{\lambda} x + \frac{c_{2}}{\lambda}x^{2} &=x \int_{0}^{1} t(c_{1}t+ c_{2}t^{2}) \, \mathrm{d}t + 2x^{2} \int_{0}^{1} c_{1}t + c_{2}t^{2} \, \mathrm{d}t \\ &=x(c_{1}/3 + c_{2}/4) + 2x^{2}(c_{1}/2 + c_{2}/3) \end{align} so that \begin{align} \frac{c_{1}}{\lambda} &= \frac{c_{1}}{3} + \frac{c_{2}}{4} \\ \frac{c_{2}}{\lambda} &= c_{1} + \frac{2c_{2}}{3} \\ \end{align} It can be verified that if both $c_{1}$ and $c_{2}$ are nonzero, then $\lambda = -18 \pm 6\sqrt{10}$. Since $c_{1} = \left(\frac{1}{\lambda} - \frac{2}{3} \right)c_{2}$ \begin{align} f_{1}(x) &= -\left(\frac{1}{18+6\sqrt{10}} + \frac{2}{3}\right)c_{2}x + c_{2}x^{2} = -\frac{1}{6}(1+\sqrt{10})c_{2}x + c_{2}x^{2} \\ f_{2}(x) &= -\left(\frac{1}{18-6\sqrt{10}} + \frac{2}{3}\right)c_{2}'x + c_{2}'x^{2} = -\frac{1}{6}(1-\sqrt{10})c_{2}'x + c_{2}'x^{2} \end{align} so that \begin{align} \int_{0}^{1} f_{1}(x)f_{2}(x) \, \mathrm{d}x &= c_{2}c_{2}' \int_{0}^{1} x^{4} -\frac{1}{3}x^{3} - \frac{1}{4}x^{4} \, \mathrm{d}x = \frac{c_{2}c_{2}'}{30} \end{align} So we can't claim that these eigenfunctions are orthogonal, or that their inner product is 1, unless we choose a scaling such that $c_{2}c_{2}' = 30$.

Meaningful lower-bound of $\sqrt{a^2+b}-a$ when $a \gg b > 0$. I know that, for $|x|\leq 1$, $e^x$ can be bounded as follows: \begin{equation*} 1+x \leq e^x \leq 1+x+x^2 \end{equation*} Likewise, I want some meaningful lower-bound of $\sqrt{a^2+b}-a$ when $a \gg b > 0$. The first thing that comes to my mind is $\sqrt{a^2}-\sqrt{b} < \sqrt{a^2+b}$, but plugging this in ends up with a non-sense lower-bound of $-\sqrt{b}$ even though the target number is positive. \begin{equation*} \big(\sqrt{a^2}-\sqrt{b} \big) - a < \sqrt{a^2+b}-a \end{equation*} How can I obtain some positive lower-bound?

The first three terms of $(1+x)^{\frac12}$ are $1 + \frac12 x - \frac18 x^2$. And you can check for yourself that $$\left(1 + \frac12 x - \frac18 x^2\right)^2 = 1 + x - \frac18 x^3 + \frac{1}{64}x^4$$ which is $\le 1+x$ whenever $\frac18 x^3 \ge \frac{1}{64}x^4$, i.e. for $0 \le x \le 8$. Now just put $x = \frac{b}{a^2}$ to get $$\sqrt{a^2+b} - a \ge \frac{b}{2a} - \frac{b^2}{8a^3}$$ whenever $0 \le b \le 8a^2$.

Please check my solution of $\int \sin^6(x)\cos^3(x) dx$ $$\int \sin^6(x)\cos^3(x) dx = \int \sin^6(x)(1-\sin^2(x))\cos(x)dx$$ $$\int \sin^6(x)\cos(x)dx - \int\sin^8x\cos xdx$$ Now, $\cos xdx = d(\sin x)$ $$\int u^6du - \int u^8du = \frac{1}{7}u^7 - \frac{1}{9}u^9 + C$$ $$\frac{1}{7}\sin^7(x) - \frac{1}{9}\sin^9(x) + C$$ However, WolframAlpha says it's: Can anyone tell if those expressions are equal?

$11 + 7 \cos 2x = 11 + 7 - 14\sin^2 x = 2(9 - 7\cos^2 x) \Rightarrow\\ \dfrac {1}{126} (11 + 7 \cos 2x) = \dfrac{1}{63}(9 - 7\sin^2x) = \boxed{\dfrac{1}{7} - \dfrac{\sin^2x}{9}}\Rightarrow\\ \\ \text{The expressions are equal.}$

How to prove $\sum_{m=1}^\infty \sum_{n=1}^\infty \frac{1}{m^2+n^2}=+\infty$ How to prove $$\sum_{m=1}^\infty \sum_{n=1}^\infty \frac{1}{m^2+n^2}=+\infty.$$ I try to do like $$\sum_{m=1}^\infty \sum_{n=1}^\infty \frac{1}{m^2+n^2}=\sum_{N=1}^\infty \sum_{n+m=N}^\infty \frac{1}{m^2+n^2}=\sum_{N=1}^\infty \sum_{m=1}^{N-1} \frac{1}{m^2+(N-m)^2}$$ $$\frac{1}{m^2+(N-m)^2}\leq \frac{2}{N^2}$$ but it doesn't work.

$$ \begin{align} \sum_{m=1}^\infty\sum_{n=1}^\infty\frac1{m^2+n^2} &\ge\sum_{m=1}^\infty\sum_{n=1}^\infty\frac1{m^2+n^2+2mn+m+n}\\ &=\sum_{m=1}^\infty\sum_{n=1}^\infty\left(\frac1{m+n}-\frac1{m+n+1}\right)\\ &=\sum_{m=1}^\infty\frac1{m+1}\\[6pt] &=\infty \end{align} $$

Power iteration If $A$ is a matrix you can calculate its largest eigenvalue $\lambda_1$. What are the exact conditions under which the power iteration converges? Power iteration Especially, I often see that we demand that the matrix is symmetric? Is this necessary? What seems to be indespensable is that there is a largest eigenvalue (absolute value is large). But what about the structure of the eigenspace of this eigenvalue? Apparently, many times it is not considered that the eigenspace to this largest eigenvalue does not need to be one-dimensional. So what happens if the eigenspace is two-dimensional. Can we still use this algorithm?

Let $$M:=\begin{pmatrix}0 & 1 \\ 1 & 0\end{pmatrix},\ \text{then}\quad M \begin{pmatrix}1\\ 1 \end{pmatrix}=1\cdot \begin{pmatrix}1\\ 1 \end{pmatrix},\ \text{and} \quad M \begin{pmatrix}1\\ -1 \end{pmatrix}=-1\cdot \begin{pmatrix}1\\ -1 \end{pmatrix} $$ Now let $x^0 =(x^0_1,x^0_2) \neq (0,0)$, then the sequence $x^{k+1} = Mx^k$ will never converge since $M$ just switch the coordinates of $x^k$. If you want a matrix with maximal eigenvalue (without absolute value) is not simple and the algorithm don't converge, consider $$A = \begin{pmatrix}0 & 1 & 0 \\ 1 &0 & 0 \\ 0& 0 &1\end{pmatrix}, \quad A\begin{pmatrix}1 \\ 1 \\ 0 \end{pmatrix}=1\cdot \begin{pmatrix}1 \\ 1 \\ 0 \end{pmatrix},\quad A\begin{pmatrix}1 \\ -1 \\ 0 \end{pmatrix}=-1\cdot \begin{pmatrix}1 \\ -1 \\ 0 \end{pmatrix},\quad A\begin{pmatrix}0 \\ 0 \\ 1 \end{pmatrix}=1\cdot \begin{pmatrix}0 \\ 0\\ 1 \end{pmatrix}, $$ and the sequence $x^{k+1} = Ax^k$ will never converge except if you start directly with a eigenvector.

How to get $(\frac{x^2}{2}+\frac{1}{2x^2})^2$ from $1+(\frac{x^2}{2}-\frac{1}{2x^2})^2$? How can I get $(\frac{x^2}{2}+\frac{1}{2x^2})^2$ from $1+(\frac{x^2}{2}-\frac{1}{2x^2})^2$? The book lists the former as the solution to that step. This is part of an arc length problem, and I think I'm just hitting a mental roadblock on solving this.

$$\begin{eqnarray*} 1 + \left(\frac{x^2}{2} - \frac{1}{2x^2}\right)^2 &=& 1 + \left(\frac{x^2}{2}\right)^2 - 2\left(\frac{x^2}{2} \cdot \frac{1}{2x^2}\right) + \left(\frac{1}{2x^2}\right)^2 \\ &=& 1 + \left(\frac{x^2}{2}\right)^2 - \frac{1}{2} + \left(\frac{1}{2x^2}\right)^2 \\ &=& \left(\frac{x^2}{2}\right)^2 + \frac{1}{2} + \left(\frac{1}{2x^2}\right)^2 \\ &=& \left(\frac{x^2}{2}\right)^2 + 2\left(\frac{x^2}{2} \cdot \frac{1}{2x^2}\right) + \left(\frac{1}{2x^2}\right)^2 \\ &=& \left(\frac{x^2}{2} + \frac{1}{2x^2}\right)^2 \end{eqnarray*}$$

Evaluation of $ \lim_{x\rightarrow \infty}\left\{2x-\left(\sqrt[3]{x^3+x^2+1}+\sqrt[3]{x^3-x^2+1}\right)\right\}$ Evaluate the limit $$ \lim_{x\rightarrow \infty}\left(2x-\left(\sqrt[3]{x^3+x^2+1}+\sqrt[3]{x^3-x^2+1}\right)\right) $$ My Attempt: To simplify notation, let $A = \left(\sqrt[3]{x^3+x^2+1}\right)$ and $B = \left(\sqrt[3]{x^3-x^2+1}\right)$. Now $$ \begin{align} 2x^2 &= A^3-B^3\\ x &= \sqrt{\frac{A^3-B^3}{2}} \end{align} $$ So the limit becomes $$\lim_{x\rightarrow \infty}\left(\sqrt{\frac{A^3-B^3}{2}}-A-B\right)$$ How can I complete the solution from this point?

$a+b=\dfrac{a^3+b^3}{a^2-ab+b^2}$ I think this identity can be used to simplify your expression. Let $a=\sqrt[3]{x^3+x^2+1}$ and $b=\sqrt[3]{x^3-x^2+1}.$ Then $a+b=\dfrac{(x^3+x^2+1)+(x^3-x^2+1)}{(x^3+x^2+1)^{2/3}-(x^3+x^2+1)^{1/3}(x^3+x^2+1)^{1/3}+(x^3+x^2+1)^{2/3}}\\=\dfrac{2(x+1/x^2)}{(1+1/x+1/x^3)^{2/3}-(1+1/x+1/x^3)^{1/3}(1+1/x+1/x^3)^{1/3}+(1+1/x+1/x^3)^{2/3}}$ As $x\rightarrow \infty$ we have $a+b\rightarrow 2x.$ Hence $$\lim_{x\rightarrow \infty}\left\{2x-\left(\sqrt[3]{x^3+x^2+1}+\sqrt[3]{x^3-x^2+1}\right)\right\}\\= \lim_{x\rightarrow \infty} {2x-(a+b)}=0$$

Basic induction proof methods so we're looking to prove $P(n)$ that $$1^2+2^3+\cdots+n^3 = (n(n+1)/2)^2$$ I know the basis step for $p(1)$ holds. We're going to assume $P(k)$ $$1^3+2^3+\cdots+k^3=(k(k+1)/2)^2$$ And we're looking to prove $P(k+1)$ What I've discerned from the internet is that I should be looking to add the next term, $k+1$, to both sides so... $$1^3+2^3+\cdots+k^3 + (k+1)^3=(k(k+1)/2)^2 + (k+1)^3$$ now I saw some nonsense since that we assumed $p(k)$ we can use it as a definition in our proof, specifically on the left hand side so since $$1^3+2^3+\cdots+k^3=(k(k+1)/2)^2$$ then $$(k(k+1)/2)^2 + (k+1)^3 = (k(k+1)/2)^2 + (k+1)^3$$ and we have our proof OK so far thats wrong so far ive figured this. $$1^3+2^3+\cdots+k^3 + (k+1)^3=((k+1)((k+1)+1)/2)^2$$ Then $$1^3+2^3+\cdots+k^3 + (k+1)^3=((k+1)((k+2)/2)^2$$ using the definition $$(k(k+1)/2)^2 + (k+1)^3 = ((k+1)((k+2)/2)^2$$ $$(k^2+k/2)^2 + (k^2+2k+1)(k+1) = (k^2+3k+2/2)^2$$ $$(k^4+k^2/4)+(k^2+2k^2+k+k^2+2k+1)= (k^4+9k^2+4/4)$$ Where should I go from here? It doesn't possibly look like these could equate, I'll keep going though

What you need to show is that $S(k-1)+k^3=S(k)$, i.e. $$\frac{(k-1)^2k^2}4+k^3=\frac{k^2(k+1)^2}4.$$ Simplifying by $\frac{k^2}4$, you get $$(k-1)^2+4k=(k+1)^2.$$ QED.

Quick floor function This isn't true, right? $$k\left\lfloor\frac n {2k}\right\rfloor\leq \left\lfloor\frac n k\right\rfloor$$ $2<k\leq \left\lfloor\dfrac {n-1} 2\right\rfloor$, $n>4$, $k,n$ are coprime.

Let $n=15$ and $k=4$. Not that $15>4$, $2<4<\left\lfloor\dfrac{15-1}{2}\right\rfloor=7$, and that $gcd(4,15)=1$ Now $$k\left\lfloor\dfrac {2n} k\right\rfloor=4\left\lfloor\dfrac {30} 4\right\rfloor=28$$ and $$\left\lfloor\dfrac n k\right\rfloor=\left\lfloor\dfrac {15}4\right\rfloor=3$$ Clearly $28>3$, so this provides a counterexample.

If $K = \frac{2}{1}\times \frac{4}{3}\times \cdots \times \frac{100}{99}.$ Then value of $\lfloor K \rfloor$ Let $K = \frac{2}{1}\times \frac{4}{3}\times \frac{6}{5}\times \frac{8}{7}\times \cdots \times \frac{100}{99}.$ Then what is the value of $\lfloor K \rfloor$, where $\lfloor x \rfloor$ is the floor function? My Attempt: By factoring out powers of $2$, we can write $$ \begin{align} K &= 2^{50}\times \left(\frac{1} {1}\times \frac{2}{3}\times \frac{3}{5}\times \frac{4}{7}\times \frac{5}{9}\times\cdots\times \frac{49}{97}\times \frac{50}{99}\right)\\ &= 2^{50}\cdot 2^{25}\times \left(\frac{1\cdot 3 \cdot 5\cdots49}{1\cdot 3 \cdot 5\cdots 49}\right)\times \left(\frac{1}{51\cdot 53\cdot 55\cdots99}\right)\\ &= \frac{2^{75}}{51\cdot 53\cdot 55\cdots99} \end{align} $$ How can I solve for $K$ from here?

If we make the problem more general and write $$\displaystyle K_n = \frac{2}{1}\times \frac{4}{3}\times \frac{6}{5}\times \frac{8}{7}\times \cdots \times \frac{2n}{2n-1}$$ the numerator is $2^n \Gamma (n+1)$ and the denominator is $\frac{2^n \Gamma \left(n+\frac{1}{2}\right)}{\sqrt{\pi }}$. So, $$K_n=\frac{\sqrt{\pi } \Gamma (n+1)}{\Gamma \left(n+\frac{1}{2}\right)}$$ Considering this expression for large values of $n$, we then have $$K_n=\sqrt{\pi } \sqrt{n}+\frac{1}{8} \sqrt{\pi } \sqrt{\frac{1}{n}}+\frac{1}{128} \sqrt{\pi } \left(\frac{1}{n}\right)^{3/2}-\frac{5 \sqrt{\pi } \left(\frac{1}{n}\right)^{5/2}}{1024}+O\left(\left(\frac{1}{n}\right)^3\right)$$ which implies $$\lfloor K_n \rfloor =\lfloor \sqrt{\pi n} \rfloor$$ which is verified for any value of $n \gt 5$.

Limit of $a(k)$ in $ \sum_{k=1}^n \frac{a_k}{(n+1-k)!} = 1 $ For n = 1, 2, 3 ... (natural number) $ \sum_{k=1}^n \frac{a_k}{(n+1-k)!} = 1 $ $ a_1 = 1, \ a_2 = \frac{1}{2}, \ a_3 = \frac{7}{12} \cdots $ What is the limit of {$ a_k $} $ \lim_{k \to \infty} a_k $ = ? I have no idea where to start.

Note that $$ \begin{align} \frac{x}{1-x} &=\sum_{n=1}^\infty x^n\\ &=\sum_{n=1}^\infty\sum_{k=1}^n\frac{a_k}{(n-k+1)!}x^n\\ &=\sum_{k=1}^\infty\sum_{n=k}^\infty\frac{a_k}{(n-k+1)!}x^n\\ &=\sum_{k=1}^\infty\sum_{n=0}^\infty\frac{a_k}{(n+1)!}x^{n+k}\\ &=\frac{e^x-1}{x}\sum_{k=1}^\infty a_kx^k\tag{1} \end{align} $$ Therefore, $$ \sum_{k=1}^\infty a_kx^k=\frac{x^2}{(e^x-1)(1-x)}\tag{2} $$ If $a_k$ limit to some $b$, then as $x\to1$ we would have $$ \begin{align} b &=\lim_{x\to1^-}\frac{\displaystyle\sum_{k=1}^\infty a_kx^k}{\displaystyle\sum_{n=0}^\infty x^k}\\ &=\lim_{x\to1^-}\frac{x^2}{e^x-1}\\[9pt] &=\frac1{e-1}\tag{3} \end{align} $$ Now that we have an idea of what the limit would be, let's try to prove it. From the defining formula, $$ a_n=1-\sum_{k=1}^\infty\frac{a_{n-k}}{(k+1)!}\tag{4} $$ where we define $a_k=0$ for $k\le0$. Thus, if we let $c_n=a_n-\frac1{e-1}$, then for $n\gt1$, we have $$ c_n=-\sum_{k=1}^\infty\frac{c_{n-k}}{(k+1)!}\tag{5} $$ Note that if $|c_k|\lt c\,(4/5)^k$ for $k\lt n$ then $$ \begin{align} |c_n| &\le\sum_{k=1}^\infty\frac{c\,(4/5)^{n-k}}{(k+1)!}\\ &=c\,(4/5)^n\sum_{k=1}^\infty\frac{(4/5)^{-k}}{(k+1)!}\\ &=c\,(4/5)^n\frac45\left(e^{5/4}-1-\frac54\right)\\[9pt] &\le c\,(4/5)^n\tag{6} \end{align} $$ Since $c_1=\frac{e-2}{e-1}$ and $c_n=-\frac1{e-1}$ for $n\le0$, we can use $c=\frac1{e-1}$ in $(6)$. Therefore, $$ \begin{align} \left|\,a_n-\frac1{e-1}\,\right| &=|c_n|\\ &\le\frac{(4/5)^n}{e-1}\tag{7} \end{align} $$ Therefore, $$ \lim_{n\to\infty}a_n=\frac1{e-1}\tag{8} $$

Definite integral $\int_0^{2\pi}(\cos^2(x)+a^2)^{-1}dx$ How do I prove the following? $$ I(a)=\int_0^{2\pi} \frac{\mathrm{d}x}{\cos^2(x)+a^2}=\frac{2\pi}{a\sqrt{a^2+1}}$$

If anyone wants to see a complex analysis solution. Let $\gamma$ be the unit$\require{autoload-all}$ circle. This proof holds for all complex $a$ such that the integral exists. $$ I(a)=\int_0^{2\pi}\!\!\! \frac{\mathrm{d}t}{\cos^2(t)+a^2}$$ $$ \toggle{ \text{Set} \; x = e^{it}\quad\enclose{roundedbox}{\text{ Click for Information }} }{ \begin{align} x &= e^{it}\\ \sin(t) &= \frac{1}{2i}\left(x-\frac{1}{x}\right)\\ \cos(t) &=\frac{1}{2}\left(x+\frac{1}{x}\right)\\ \mathrm{d}t &= \frac{-i \, \mathrm{d}x}{x} \end{align} }\endtoggle $$ $$\begin{align} I(a)&=\int_\gamma \frac{-i\,\mathrm{d}x}{x\frac{x^4+2x^2+4x^2a^2+2}{4x^2}} \\[.2cm] &=\int_\gamma \frac{-4ix\,\mathrm{d}x}{x^4+2x^2+4x^2a^2+2} \end{align}$$ The zeros of the denominator are $$x= \pm\sqrt{-2 a^2-2\sqrt{a^2} \sqrt{a^2+1}-1},\quad x = \pm\sqrt{-2 a^2+2\sqrt{a^2}\sqrt{a^2+1}-1}$$ Only the second two roots will be inside the unit circle. Call these roots $x_+$ and $x_-$ and set $\alpha = x_+^2$. $$I(a) = 2\pi i \left(\lim_{x\to x_+} \frac{-4ix(x-x_+)}{x^4+2x^2+4x^2a^2+2} + \lim_{x\to x_-} \frac{-4ix(x-x_-)}{x^4+2x^2+4x^2a^2+2}\right)$$ Factor out constants and apply L'Hopitals rule for both of the limits $$\begin{align}I(a) &= 8\pi \left(\lim_{x\to x_+} \frac{2x-x_+}{4x^3+4x+8xa^2} + \lim_{x\to x_-} \frac{2x-x_-}{4x^3+4x+8xa^2}\right)\\ &= 8\pi \left(\frac{x_+}{4x_+^3+4x_++8x_+a^2} + \frac{x_-}{4x_-^3+4x_-+8x_-a^2}\right) \\ &=8\pi \left(\frac{1}{4x_+^2+4+8a^2} + \frac{1}{4x_-^2+4+8a^2}\right)\end{align}$$ Using the fact that $x_+^2 = x_-^2=\alpha$ $$\begin{align} I(a) &= 4\pi \left(\frac{1}{\alpha+1+2a^2}\right) \\ &= \frac{4\pi}{(-2 a^2+2\sqrt{a^2}\sqrt{a^2+1}-1)+1+2a^2} \\ &= \frac{2\pi}{\sqrt{a^2}\sqrt{a^2+1}}\end{align}$$ This is the result for $a \in \mathbb{C}$. For $a \in \mathbb{R}$ we can say that $$I(a)=\int_0^{2\pi}\!\!\! \frac{\mathrm{d}t}{\cos^2(t)+a^2} = \frac{2\pi}{a\sqrt{a^2+1}}$$

Use a proof by cases to show that $\lfloor n/2 \rfloor$ * $\lceil n/2 \rceil$ = $\lfloor \frac{n^2}{4} \rfloor$ for all integers $n$. Question Use a proof by cases to show that $\lfloor n/2 \rfloor$ * $\lceil n/2 \rceil$ = $\lfloor \frac{n^2}{4} \rfloor$ for all integers $n$. My Attempt: I can only think of two cases, * *$n/2 \in \mathbb{Z}$ *$n/2 \notin \mathbb{Z}$ First case is straightforward: $\lfloor n/2 \rfloor = \lceil n/2 \rceil = n/2$, $\frac{n}{2}*\frac{n}{2} = \frac{n^2}{4}$ Second case troubled me, $\lceil n/2 \rceil = \lfloor n/2 \rfloor + 1\\ \lceil n/2 \rceil = \lfloor n/2 + 1\rfloor$ $n/2 - 1 \leq \lfloor n/2 \rfloor < n/2\\ n/2 \leq \lfloor n/2 + 1 \rfloor < n/2 + 1$ I multiply both inequalities, $\frac {n^2 - 2n}{4} \leq \lfloor n/2 \rfloor * \lfloor n/2 + 1 \rfloor < \frac{n^2 + 2n}{4}$ I need to prove that $\lfloor n/2 \rfloor * \lfloor n/2 + 1 \rfloor$ should be at least $n^2 /4$ and less than $n^2 /4 + 1$, this ensures that if I floor that, it will be $n^2/4$, but I'm lost. My second attempt, (I didn't think the top have anywhere to go). This time I used some epsilon $\epsilon \in (0, 1)$, $\lfloor n/2 \rfloor = n/2 - \epsilon\\ \lceil n/2 \rceil = n/2 + 1 - \epsilon$ $\lfloor n/2 \rfloor * \lfloor n/2 + 1 \rfloor = (n/2 - \epsilon)*(n/2 + 1 - \epsilon)\\ = n^2/4 + n/2 - n*\epsilon/2 - n*\epsilon/2 - \epsilon + \epsilon ^ 2\\ = n^2/4 + n/2 - 2n\epsilon/2 + 2\epsilon^2/2\\ = n^2/4 + \frac{n-2n\epsilon - 2\epsilon + 2\epsilon^2}{2}$ The problem now is I need to prove that $\frac{n-2n\epsilon - 2\epsilon + 2\epsilon^2}{2}$ is between 0 and 1. I don't really think this one is the solution is either so I gave up.

So you've got the case when $n$ is even. When $n$ is odd, then $\lfloor n/2 \rfloor * \lceil n/2 \rceil = \frac{n - 1}{2} * \frac{n + 1}{2} = \frac{n^2 -1}{4}$. We want to show that this equals the right hand side. Since $n^2/4 = (n/2)^2$, we can rewrite $n^2/4$ as $(m + 1/2)^2$ where $m$ is an integer, and $m = \frac{n-1}{2}$. Furthermore, $(m + 1/2)^2 = m^2 + m + 1/4$. Observe that $\lfloor m^2 + m + 1/4 \rfloor = m^2 + m$ since $m$ is an integer. Let's go write $m^2 + m$ in terms of $n$: $m^2 + m = \frac{(n-1)^2}{2^2} + \frac{n-1}{2} = \frac{n^2 - 2n + 1}{4} + \frac{2n - 2}{4} = \frac{n^2 -1}{4}$. I hope that's not too roundabout. The important thing was to show that flooring $n^2/4$ is the same as subtracting $1/4$.

Simplify [1/(x-1) + 1/(x²-1)] / [x-2/(x+1)] Simplify: $$\frac{\frac{1}{x-1} + \frac{1}{x^2-1}}{x-\frac 2{x + 1}}$$ This is what I did. Step 1: I expanded $x^2-1$ into: $(x-1)(x+1)$. And got: $\frac{x+1}{(x-1)(x+1)} + \frac{1}{(x-1)(x+1)}$ Step 2: I calculated it into: $\frac{x+2}{(x-1)(x+1)}$ Step 3: I multiplied $x-\frac{2}{x+1}$ by $(x-1)$ as following and I think this part might be wrong: * *$x(x-1) = x^2-x$. Times $x+1$ cause that's the denominator = *$x^3+x^2-x^2-x = x^3-x$. *After this I added the $+ 2$ *$\frac{x^3-x+2}{(x-1)(x+1)}$ Step 4: I canceled out the denominator $(x-1)(x+1)$ on both sides. Step 5: And I'm left with: $\frac{x+2}{x^3-x+2}$ Step 6: Removed $(x+2)$ from both sides I got my UN-correct answer: $\frac{1}{x^3}$ Please help me. What am I doing wrong?

It simplifies things a lot if you just multiply the numerator and denominator by $(x+1)(x-1)$ $$\frac{\frac{1}{x-1} + \frac{1}{x^2-1}}{x-\frac 2{x + 1}}\cdot\frac{\frac{(x+1)(x-1)}{1}}{\frac{(x+1)(x-1)}{1}} = \frac{(x+1)+1}{x(x+1)(x-1)-2(x-1)}=\frac{x+2}{(x-1)(x(x+1)-2)}$$ $$=\frac{x+2}{(x-1)(x^2+x-2)}=\frac{x+2}{(x-1)(x+2)(x-1)}=\frac{1}{(x-1)^2}$$