Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
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 posit... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/687173",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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$\sin^2 \alpha + \sin^2 \beta - \cos \gamma < M$ 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 attem... | 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:
$$\sin^2(x)+\sin^2(x+n \pi)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/687974",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Solving $\arcsin(1-x)-2\arcsin(x)=\pi/2$ \begin{eqnarray*}
\arcsin(1-x)-2\arcsin(x) & = & \frac{\pi}{2}\\
1-x & = & \sin\left(\frac{\pi}{2}+2\arcsin(x)\right)\\
& = & \cos\left(2\arcsin(x)\right)\\
& = & 1-2\left(\sin\left(\arcsin(x)\right)\right)^{2}\\
& = & 1-2x^{2}\\
x & = & 2x^{2}\\
x\left(x-\frac{1}{2}\right) &... | In your first step you added an extra solution.
Since $\arcsin x$ must be smaller than $\pi/2$, the first line reads:
$$\arcsin(1-x)= \frac{\pi}{2}+2\arcsin(x) \le \frac{\pi}{2}$$
Thus, $x\le 0$ as well.
Now, by taking the $\sin$ of both sides, you took a function that was only defined up to $x=1$ (e.g. $\arcsin(x-1)$... | {
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Showing convergence or divergence of a sequence I need to determine if the series with $n$th term $\ln(n)e^{-\sqrt n}$ converges or diverges. I've tried numerous identities for $ln(x)$ and $e^{x}$ and various convergence tests but I'm still very stuck.
| To prove the given series convergent, we use the following inequalities:
*
*For $ x > 1$ , $Ln(x) < x$.
*$Exp(x) > 1 + x + \dfrac{x^2}{2!} + \dfrac{x^3}{3!} + \dfrac{x^4}{4!} + \dfrac{x^5}{5!}$ for $x > 0.$
Let $a(n) = \dfrac{ln(n)}{e^{\sqrt{n}}}$, then $a(n) < \dfrac{2ln(\sqrt{n})}{1 + \sqrt{n}+ \dfrac{\sqrt{n}^2}... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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partial Fibonacci summation Let $F_{n}$ be the n-th Fibonacci number.
How to calculate the summation like following:
$\sum_{n \geq 0} F_{3n} \cdot 2^{-3n}$
| Here's an approach via generating functions. As the Fibonacci recurrence is defined by $F_{n+2} = F_{n+1} + F_n$, we have
$$\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... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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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 $y$ ... | {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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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 ... | $$\sum_{r=1}^{100}2^{r-1}\binom{100}r=\frac12\sum_{r=1}^{100}2^r\binom{100}r=\frac12\left[(1+2)^{100}-1\right]$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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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 th... | 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
$$(c^k)^2\equ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/698066",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Help with a simple derivative I am trying to solve $\dfrac {6} {\sqrt {x^3+6}}$
and so far I made it to $6(x^3+6)^{-\frac 1 2}$
then I continued and now I have $(x^3+6)^{- \frac 3 2} * 3x^2$
and I cannot figure out what how to find the constant that should be before the parenthesis.
| $$\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}}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 1
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Solve $\sin x - \cos x = -1$ for the interval $(0, 2\pi)$ We have an exam in $3$ hours and I need help how to solve such trigonometric equations for intervals.
How to solve
$$\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$$
If $\displaystyle \sin\frac x2=0,\frac x2=n\pi... | {
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"timestamp": "2023-03-29T00:00:00",
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"answer_id": 6
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Find basis so Transformation Matrix will be diagonal $e_1,e_2$ will be basis for $V$. $W$ has a basis $\{e_1+ ae_2,2e_1+be_2\}$. Choose an $a,b$ s.t. that the basis for $W$ will have a transformation matrix $T$ will be in diagonal form.
$T(e_1) = 1e_1+5e_2$
$T(e_2) = 2e_1+4e_2$
$V$ and $W$ are linear spaces of dimensio... | In the basis for $V$,
$$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{matr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/702722",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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"answer_id": 1
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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... | {
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"url": "https://math.stackexchange.com/questions/703116",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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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 someo... | 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.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/703185",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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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
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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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) \... | 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 $10$ way... | {
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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 ... | {
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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{ali... | 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 lit... | {
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"answer_id": 1
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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 ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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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 m... | 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 $a+b+c=0$
$$a^3+b^3+c^3=3abc\tag{1}$$
$$ab=\sqrt[3]{k+\sqrt{k^2-1}}\cdot\sqrt[3]{k-\sqrt{k^2-1}}$$
$$ab=\sqrt[3]{k^2-(\sqrt{k^2-1})^2}$$
$... | {
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"timestamp": "2023-03-29T00:00:00",
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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 int... | 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
$$\frac{1}{\binom{n}{k-1}} + \frac1{\binom{n}{k}} = \frac{\binom{n}{k}+\binom{n}{k-1}}{\binom{n}{k-1}\binom{n}{k}} = \frac{\binom{n+1}{k}}{\binom{n}{k-1}\binom{n}{k}} = \frac{\frac{n+1}{k}\binom{n}... | {
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"url": "https://math.stackexchange.com/questions/719121",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Is $f$ differentiable at $(x,y)$? I am working on a practice question, and I am not sure if what I have done would be considered, 'complete justification'. I would greatly appreciate some feedback or helpful advice on how it could be better etc. The question is here:
Let $f: \mathbb {R}^2 \to \mathbb{R} $ be a function... | You've already found
$$\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).$$
To evaluate $\lim \limits_{(x,y)\to (0,0)}\left( \dfrac{\frac{\si... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/719511",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Rationalization of $\frac{2\sqrt{6}}{\sqrt{2}+\sqrt{3}+\sqrt{5}}$
Question:
$$\frac{2\sqrt{6}}{\sqrt{2}+\sqrt{3}+\sqrt{5}}$$ equals:
My approach:
I tried to rationalize the denominator by multiplying it by $\frac{\sqrt{2}-\sqrt{3}-\sqrt{5}}{\sqrt{2}-\sqrt{3}-\sqrt{5}}$. And got the result to be (after a long calculat... | What I would do is multiply by the first term plus the conjugate of the last two terms. I have coloured the important parts of the following expression to make it easier to understand.
$$\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... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/720867",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 4,
"answer_id": 2
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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}... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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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{... | 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... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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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 ... | 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 $0 < x < 1$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/725680",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
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 in... | 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 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/730876",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "52",
"answer_count": 3,
"answer_id": 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... | 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. $$\begin{array}{ccc}
a^0 & a^1 & a^2 \\ \hline
1 & 3 & 8 \\
2 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/731792",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Taylor series of $\sqrt{1+x}$ using sigma notation I want help in writing Taylor series of $\sqrt{1+x}$ using sigma notation
I got till
$1+\frac{x}{2}-\frac{x^2}{8}+\frac{x^3}{16}-\frac{5x^4}{128}+\ldots$ and so on.
But I don't know what will come in sigma notation.
| The generalized binomial theorem says that
$$
(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{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/732540",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 1
} |
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 :... | 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}+\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/737557",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
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 $\display... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/738925",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
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)... | 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}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/741861",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
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 $x$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/747805",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
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
$$
BD^2=AB^2+AD^2\quad\Rightarr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/752186",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Inequality with two binomial coefficients I am having trouble seeing why
$$
\binom{k}{2} + \binom{n - k}{2} \le \binom{1}{2} + \binom{n - 1}{2} = \binom{n - 1}{2}
$$
| Assuming $1\le n$ and $0\le k\le n$ it's equivalent to
\begin{align*}k(k-1)+(n-k)(n-k-1)&\le(n-1)(n-2)\\
(n-k)(n-k-1)&\le(n-k-1+k)(n-2)-k(k-1)\\
(n-k)(n-k-1)&\le(n-k-1)(n-2)+k(n-2)-k(k-1)\\
0&\le(n-k-1)(k-2)+k(n-k-1)\\
0&\le(n-k-1)(k-1)\end{align*}
So it holds for all $0<k<n$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/753799",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Problem with trigonometric equation I am having trouble solving this equation
$$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$$From which it is clear that any solution has $t$ positive (rhs is positive), and a quick s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/754542",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
How do I solve this definite integral: $\int_0^{2\pi} \frac{dx}{\sin^{4}x + \cos^{4}x}$? $$\int_0^{2\pi} \frac{dx}{\sin^{4}x + \cos^{4}x}$$
I have already solved the indefinite integral by transforming $\sin^{4}x + \cos^{4}x$ as follows:
$\sin^{4}x + \cos^{4}x = (\sin^{2}x + \cos^{2}x)^{2} - 2\cdot\sin^{2}x\cdot\cos^{2... | \begin{aligned}
& \int_{0}^{2 \pi} \frac{d x}{\sin ^{4} x+\cos ^{4} x} \\
=& \int_{0}^{2 \pi} \frac{d x}{\left(\sin ^{2} x+\cos ^{2} x\right)^{2}-2 \sin ^{2} x \cos ^{2} x} \\
=& \int_{0}^{2 \pi} \frac{d x}{1-\frac{\sin ^{2} 2 x}{2}} \\
=& 16 \int_{0}^{\frac{\pi}{4}} \frac{d x}{1+\cos ^{2} 2 x} \\
=& 16 \int_{0}^{\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/754750",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 3
} |
Using Polar Integrals to find Volume of surface Here's the Question and the work that I've done so far to solve it:
Use polar coordinates to find the volume of the given solid.
Enclosed by the hyperboloid $ −x^2 − y^2 + z^2 = 61$ and the plane $z = 8$
Ah MathJaX is confusing by the way. Not sure how to close the text w... |
Notice that $-x^2-y^2+z^2=61$ can be rewritten as $x^2+y^2=z^2-61$ or
$$x^2+y^2=\sqrt{z^2-61}^2$$
Given a horizontal slice of the upper sheet of the hyperboloid (i.e. given $z$), it should be plain to see from the graph (and the equation that we have a circle of radius $\sqrt{z^2-61}$.
The vertex of the upper sheet oc... | {
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"url": "https://math.stackexchange.com/questions/756305",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Interesting association between tangent lines of slope one and ellipses Why is it that a tangent line with slope $1$ to an ellipse centered at the origin will have a transformation of $\pm \sqrt{a^2 +b^2}$ where $a$ and $b$ are the major and minor axis of the ellipse?
For example: The tangent line of slope one to the e... | The relation you are looking for can be derived algebraically. Start with some general line $y=mx+c$ for the tangent. Substitute it into the equation of your ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. This gives:
$$
\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... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/758638",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
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
$$5xy\sqrt{(x^2+y^2)^3}=a^5+b^5+c^5+d^5$$
Thi... | 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 $a+b+c+d\equiv 0 \mod{5}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/760071",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
} |
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
$$2\sin{\dfrac{A}{2}}\sin{\dfrac{B}{2}}=\cos{\dfrac{A}{2}}\cos{\dfra... | Denote $~~a=\tan\dfrac{A}{2}$, $~~b=\tan\dfrac{B}{2}$ $($let $a\le b$$)$.
$\sin\dfrac{A}{2}\sin\dfrac{B}{2}\sin\dfrac{C}{2}=\dfrac{1}{10}$;
$\sqrt{\dfrac{1-\cos A}{2}} \cdot \sqrt{\dfrac{1-\cos B}{2}} \cdot \sqrt{\dfrac{1-\cos C}{2}} = \dfrac{1}{10}$;
$(1-\cos A) (1-\cos B)(1-\cos C) = \dfrac{2}{25}$;
$(1-\cos A) ~ (... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/767109",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
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}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/770259",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
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 a... | Notice
$$ x^2 - x + 1 = \left(x-\frac{1}{2}\right)^2 + 1 - \frac{1}{4} = \left(x-\frac{1}{2}\right)^2 + \frac{3}{4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/771306",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 0
} |
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:
$$
det = (a+b)... | Have you double-checked your calculations? The eigenvalues of any $2\times2$ real symmetric matrix $\pmatrix{a&c\\ c&b}$ are given by
$$\frac{a+b\pm\sqrt{(a+b)^2-4(ab-c^2)}}2.$$
Plug in the entries of $C$, I don't find any discrepancies.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/772500",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Tricky Triangle Area Problem This was from a recent math competition that I was in. So, a triangle has sides $2$ , $5$, and $\sqrt{33}$. How can I derive the area? I can't use a calculator, and (the form of) Heron's formula (that I had memorized) is impossible with the$\sqrt{33}$ in it. How could I have done this? The ... | From the law of cosines ($C^2 = A^2 + B^2 - 2AB\cos \theta$), we get that $(\sqrt{33})^2 = 2^2 + 5^2 - 2 \cdot 2 \cdot 5 \cos \theta$.
Simplifying this, we get $33 = 29 - 20 \cos \theta$, which means that $\displaystyle \cos \theta = -\frac{1}{5}$
Because $\cos^2 \theta + \sin^2 \theta = 1$, we get that $\displaystyle ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/773504",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 5,
"answer_id": 1
} |
How to find the values of a and b? If the polynomial 6x4 + 8x3 - 5x2 + ax + b is exactly divisible by the polynomial 2x2 - 5, then find the values of a and b.
| The roots of $2x^2-5$ must be roots of our polynomial:
$$\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... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/775258",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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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 ... | 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 parit... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/775648",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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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)}... | 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\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/777798",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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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... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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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:
$\displaystyle\frac{\cos B}{1-\tan B}=\frac{\cos B}{1-\dfrac{\sin B}{\cos B}}=\frac{\cos^2B}{\cos B-\sin B}$
$\displaystyle\frac{\sin B}{1-\cot B}=\frac{\sin B}{1-\dfrac{\cos B}{\sin B}}=\frac{\sin^2B}{\sin B-\cos B}=-\frac{\sin^2B}{\cos B-\sin B}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/779760",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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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:
$$\frac{a^4+b^4+c^4}{2abc}=\sum_{cyc}\frac{a^3}{2bc}\geq\sum_{cyc}\frac{2a^3}{(b+c)^2}=$$
$$=\frac{2}{a+b+c}\sum_{cyc}\frac{a^3}{(b+c)^2}\sum_{cyc}a\geq\frac{2}{a+b+c}\left(\sum_{cyc}\frac{a^2}{b+c}\right)^2=$$
$$=\frac{2}{a+b+c}\cdot\sum_{cyc}\frac{a^2}{b+c}\cdot\sum_{cyc}\frac{a^2}{b+c}\ge... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/781405",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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$l^2+m^2=n^2$ $\implies$ $lm$ is always a multiple of 3 when $l,m,n,$ are positive integers. Let $l,m,n$ be any three positive integers such that $l^2+m^2=n^2$
Then prove that $lm$ is always a multiple of 3.
| Here is another approach. The general solution of this equation is:
$l = x^2 - y^2$, $m = 2xy$, $n = x^2 + y^2$. So: $l\cdot m = 2\cdot (x^2 - y^2)\cdot x\cdot y = 2\cdot (x - y)\cdot (x + y) \cdot x\cdot y$. From this, we have some cases to consdier:
*
*$3|x$ or $3|y$ then $3|l\cdot m$.
*$3 \not|x$ and $3\not|y$ t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/783106",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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How did Euler realize $x^4-4x^3+2x^2+4x+4=(x^2-(2+\alpha)x+1+\sqrt{7}+\alpha)(x^2-(2-\alpha)x+1+\sqrt{7}-\alpha)$? How did Euler find this factorization?
$$\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 s... | 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}... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 5,
"answer_id": 0
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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... | 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 ch... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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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(E... | {
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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 circumferen... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 2
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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$$
Define x-axis a... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to finish this integration? I'm working with the integral below, but not sure how to finish it...
$$\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$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/792086",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Finding determinant for matrix using upper triangle method Here is an example of a matrix, and I'm trying to evaluate its determinant:
$$
\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 ... | 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?
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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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... | 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 $\frac{1}{z^4+1}$... | {
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"timestamp": "2023-03-29T00:00:00",
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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 facto... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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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 proo... | 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}... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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"answer_id": 3
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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 answe... | 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... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/798277",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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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:
$$
ax^2 - dx^2 + bx - ex + c - f = 0 \\
x^2(a -... | You lost a factor $4$ somewhere. You can simply rewrite your problem as
$$(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=\fr... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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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 a... | 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
$$\int_{-\infty}^{\infty} dx \, f(x) g^*(x) = \frac1{2 \pi} \int_{-\inft... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "37",
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"answer_id": 2
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How to find determinant of this matrix? Is there a manual method to find $\det\left(XY^{-1}\right)$ ?
Let $$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 & \cdo... | 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
$$... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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"answer_id": 0
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Domain of the Function Square Root of 12th Degree Polynomial Find the Domain of $$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}$$
i.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/805281",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find a formula for $1 + 3 + 5 + ... +(2n - 1)$, for $n \ge 1$, and prove that your formula is correct. I think the formula is $n^2$.
Define $p(n): 1 + 3 + 5 + \ldots +(2n − 1) = n^2$
Then $p(n + 1): 1 + 3 + 5 + \ldots +(2n − 1) + 2n = (n + 1)^2$
So $p(n + 1): n^2 + 2n = (n + 1)^2$
The equality above is incorrect, so e... | The issue here is that $p(n+1)$ is note the statement
$$
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.
$$
Why? The left side of your formula is the sum of all odd numbers between $1$ and $2n-1$. So, when you replace $n$ by $n+1$, you get the sum of all odd numbers b... | {
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Algebra Manipulation Contest Math Problem The question was as follows:
The equations $x^3+Ax+10=0$ and $x^3+Bx^2+50=0$ have two roots in common. Compute the product of these common roots.
Because $x^3+Ax+10=0$ and $x^3+Bx^2+50=0$ it means that $x^3+Ax+10=x^3+Bx^2+50$
Take $x^3+Ax+10=x^3+Bx^2+50$ and remove $x^3$ from ... | Hint: The common roots must be both roots of
$$- (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. $$
This gives $B^2 = 4A$ and $160=5AB$, so $5B^3 = 640 $. This gives $B = 4 \sqrt[3]{2} $,... | {
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"timestamp": "2023-03-29T00:00:00",
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Procedure for evaluating $\int_{x=\ -1}^1\int_{y=\ -\sqrt{1-x^2}}^{\sqrt{1-x^2}}\frac{x^2+y^2}{\sqrt{{1-x^2-y^2}}}\,dy\,dx$ While solving another problem I have come across this integral which I am unable to evaluate. Can someone please evaluate the following integral? Thank you.
$$\int_{x=\ -1}^1\int_{\large y=\ -\sqr... | 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... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/811349",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
} |
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
$$1^1\equiv 1\pmod 3$$
$$2^2\equiv 1\pmod 3$$
$$3^3\equiv 0\pmod 3$$
$$4^4\equiv 1\pm... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/813736",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Determine which Fibonacci numbers are even (a) Determine which Fibonacci numbers are even. Use a form of mathematical induction to prove your conjecture.
(b) Determine which Fibonacci numbers are divisible by 3. Use a form of mathematical induction to prove your conjecture
I understand that for part a that all multiple... | Part A:
Base case:
$F(0) = 0$, 0 is even.
$F(3) = 2$, 2 is even.
Inductive Hypothesis:
Assume $F(k)$ is even for some arbitrary positive integer $k$ that is divisible by 3.
Want to prove: That $F(k+3)$ is even given the inductive hypothesis.
\begin{align*}
F(k+3) &= F(k+2) + F(k+1)\\
&= F(k+1) + F(k) + F(k+1)\\
&= F(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/815004",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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middle school question on geometry
As you can see from the picture, the angle $A$ is $90^\circ$, and the segments $BD$ and $CE$ (which intersect at $F$) are angle bisectors of the angles $B$ and $C$, respectively. When the length of $CF$ is $\frac72$ and and the quadrilateral $BCDE$ has an area of $14$, what is the le... | This is not a middle school level answer. But perhaps a complicated answer is better than no answer at all. Start by choosing coordinates as follows:
$$
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=\begi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/815487",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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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... | 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
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/816377",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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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: $$x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}... | {
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"source": "stackexchange",
"question_score": "1",
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How to find the following sum? $\sum\limits_{n = 0}^\infty {\left( {\frac{1}{{4n + 9}} - \frac{1}{{4n + 7}}} \right)} $ I want to calculate the sum with complex analysis (residue)
$$
1 - \frac{1}{7} + \frac{1}{9} - \frac{1}{{15}} + \frac{1}{{17}} - ...
$$ $$
1 + \sum\limits_{n = 0}^\infty {\left( {\frac{1}{{4n + 9}} -... | 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
$$ f(z)= \frac{-2}{(4z)^... | {
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"question_score": "5",
"answer_count": 4,
"answer_id": 3
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limits of function without using L'Hopital's Rule $\mathop {\lim }\limits_{x \to 1} \frac{{x - 1 - \ln x}}{{x\ln x+ 1 - x}} = 1$ Good morning.
I want to show that without L'Hopital's rule :
$\mathop {\lim }\limits_{x \to 1} \frac{{x - 1 - \ln x}}{{x\ln x + 1 - x}} = 1$
I did the steps
$
\begin{array}{l}
\mathop {\lim... | $
\displaylines{
\left\{ \begin{array}{l}
t = 1 + u \\
u \cong \ln t \\
\end{array} \right. \cr
\Rightarrow \cr
\mathop {\lim }\limits_{t \to 1} \left[ {\frac{{\left( {1 + t} \right)\ln t}}{{t\ln t - t + 1}}} \right] = \mathop {\lim }\limits_{u \to 2} \left[ {\frac{{\left( {2 + u} \right)u}}{{\left( {1 +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/818908",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
How to find integral $\underbrace{\int\sqrt{2+\sqrt{2+\sqrt{2+\cdots+\sqrt{2+x}}}}}_{n}dx,x>-2$ Find the integral
$$\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}$$
so... | You were too timid: For $-2\leq x\leq2$ use the substitution
$$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}\ ,$$
etcetera.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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"answer_id": 0
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Volume between cylinder and plane Problem: Find the volume bounded by $z = y^2, x =0, y =0, z =9-x$.
My working:
$z$ goes from $y^2$ to $9-x$ so these are the limits of integration.
Work out the points of intersection of $9-x$ and $y^2$. When $y=0$, $9-x=0$ and $x=9$. So $x$ goes from 0 to 9. When $x=0$, $y^2 = 9$ so $... | Your limits of integration don't make sense. The region of integration is given by the set $$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)\}.$$ The projection of $R$ onto the $xz$-plane is simply the triangle $x \ge 0$, $z \ge 0$, $x + z \le 9$. Over this triangle, ... | {
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"url": "https://math.stackexchange.com/questions/823468",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Finding an equation of circle which passes through three points How to find the equation of a circle which passes through these points $(5,10), (-5,0),(9,-6)$
using the formula
$(x-q)^2 + (y-p)^2 = r^2$.
I know i need to use that formula but have no idea how to start, I have tried to start but don't think my answer is... |
I know i need to use that formula but have no idea how to start
\begin{equation*}
\left( x-q\right) ^{2}+\left( y-p\right) ^{2}=r^{2}\tag{0}
\end{equation*}
A possible very elementary way is to use this formula thrice, one for each point. Since the circle passes through the point $(5,10)$, it satisfies $(0)$, i.e.
$$... | {
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"question_score": "7",
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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 $5$ ... | 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}... | {
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"source": "stackexchange",
"question_score": "7",
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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... | 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
$$xS_n - S_n = 1+ (\frac{2}{x}-\frac{1}{x})+(\frac{3}{x^2}-\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/829168",
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"source": "stackexchange",
"question_score": "4",
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Prove the following Trig Identity with reciprocals Prove that:
$$\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}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/829622",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Integrate $\int_0^1 \ln(x)\ln(b-x)\,\mathrm{d}x$, for $b>1$? Let $b>1$. What's the analytical expression for the following integral?
$$\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... | The following is an evaluation in terms of $ \displaystyle \text{Li}_{2} \left(\frac{1}{b} \right)$, which is real-valued for $b > 1$.
$$\begin{align} \int_{0}^{1} \log(x) \log(b-x) \ dx &= \log(b) \int_{0}^{1} \log(x) + \int_{0}^{1}\log(x) \log \left(1- \frac{x}{b} \right) \ dx \\ &= - \log(b) - \int_{0}^{1} \log(x... | {
"language": "en",
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"source": "stackexchange",
"question_score": "7",
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$ 7\mid x \text{ and } 7\mid y \Longleftrightarrow 7\mid x^2+y^2 $
Show that
$$ 7\mid x \text{ and } 7\mid y \Longleftrightarrow 7\mid x^2+y^2 $$
Indeed,
First let's show
$7\mid x \text{ and } 7\mid y \Longrightarrow 7\mid x^2+y^2 $
we've $7\mid x \implies 7\mid x^2$ the same for $7\mid y \implies 7\mid ... | $►$ If $x=7x_1$ and $y=7y_1$ then $x^2+y^2=7(x_1^2+y_1^2$).
$►$ If $x^2+y^2\equiv0\pmod7\iff x^2\equiv -y^2\pmod7$ then because of $\mathbb F_7^2=\{1,4,2,0\}$ and $(-1)\mathbb F_7^2=\{6,3,5,0\}$ the only possibility for
$x^2\equiv -y^2\pmod7$ is that both $x^2$ and $y^2$ are equal to $0$ modulo $7$ so $x$ and $y$ are e... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
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"answer_id": 2
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Show that $\dfrac{\sqrt{8-4\sqrt3}}{\sqrt[3]{12\sqrt3-20}} =2^\frac{1}{6}$ This was the result of evaluating an integral by two different methods. The RHS was obtained by making a substitution, the LHS was obtained using trigonometric identity's and partial fractions.
Now I know that these two are equal, but I just can... | We have
$$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,$$
so $12\sqrt{3}-20 = 2(\sqrt{3}-1)^3$ and $\sqrt[3]{12\sqrt{3}-20}... | {
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"source": "stackexchange",
"question_score": "6",
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"answer_id": 3
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how to solve this question of polynomials
Given the polynmial is exactly divided by $x+1$, when it is divided by
$3x-1$, the remainder is $4$. The polynomial leaves remainder $hx+k$ when divided by
$3x^2+2x-1$. Find $h$ and $k$.
This is the question which is confusing me.. i have done this question like this:
$p(... | The error is that $\,p(-1) = 0\,\Rightarrow\, h(-1) + k = \color{#c00}0,\,$ not $\,\color{#c00}4.\,$ Fixing that yields the given answer.
Remark $\ $ This is a special case of the Chinese Remainder Theorem (CRT) or, equivalently, Lagrange interpolation. Either of these methods can be applied to solve the general case.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solve the integeral equation (C.S.I.R) Let $\lambda_1, \lambda_2$ be the eigen value and $f_1 , f_2$ be the coressponding eigen functions for the homogeneous integeral equation
$$ \phi(x) - \lambda \int_0^1 (xt +2x^2) \phi(t) dt = 0 $$
Then
*
*$\lambda_1 = -18 - 6 \sqrt{10} , \lambda_2 = -18 + 6 \sqrt{10}$
*$\lamb... | Let's rewrite your equation in the following form:
\begin{align}
\frac{1}{\lambda}\phi(x) = x\int_{0}^{1}t\phi(t) \mathrm{d}t + 2x^{2}\int_{0}^{1} \phi(t) \mathrm{d}t = c_{1}x + c_{2}x^{2}
\end{align}
for $c_{1}: = \int_{0}^{1}t\phi(t) \, \mathrm{d}t$ and $c_{2}:= 2\int_{0}\phi(t)\, \mathrm{d}t$. From this we can see t... | {
"language": "en",
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"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Meaningful lower-bound of $\sqrt{a^2+b}-a$ when $a \gg b > 0$. I know that, for $|x|\leq 1$, $e^x$ can be bounded as follows:
\begin{equation*}
1+x \leq e^x \leq 1+x+x^2
\end{equation*}
Likewise, I want some meaningful lower-bound of $\sqrt{a^2+b}-a$ when $a \gg b > 0$.
The first thing that comes to my mind is $\sqrt{a... | The first three terms of $(1+x)^{\frac12}$ are $1 + \frac12 x - \frac18 x^2$. And you can check for yourself that $$\left(1 + \frac12 x - \frac18 x^2\right)^2 = 1 + x - \frac18 x^3 + \frac{1}{64}x^4$$
which is $\le 1+x$ whenever $\frac18 x^3 \ge \frac{1}{64}x^4$, i.e. for $0 \le x \le 8$.
Now just put $x = \frac{b}{a^2... | {
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"source": "stackexchange",
"question_score": "1",
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Please check my solution of $\int \sin^6(x)\cos^3(x) dx$ $$\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... | $11 + 7 \cos 2x = 11 + 7 - 14\sin^2 x = 2(9 - 7\cos^2 x) \Rightarrow\\
\dfrac {1}{126} (11 + 7 \cos 2x) = \dfrac{1}{63}(9 - 7\sin^2x) = \boxed{\dfrac{1}{7} - \dfrac{\sin^2x}{9}}\Rightarrow\\
\\
\text{The expressions are equal.}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/850618",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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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 $$\sum_{m=1}^\infty \sum_{n=1}^\infty \frac{1}{m^2+n^2}=\sum_{N=1}^\infty \sum_{n+m=N}^\infty \frac{1}{m^2+n^2}=\sum_{N=1}^\infty \sum_{m=1}^{N-1... | $$
\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}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/851302",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 0
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Power iteration If $A$ is a matrix you can calculate its largest eigenvalue $\lambda_1$. What are the exact conditions under which the power iteration converges? Power iteration
Especially, I often see that we demand that the matrix is symmetric? Is this necessary?
What seems to be indespensable is that there is a larg... | Let $$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} $$
Now let $x^0 =(x^0_1,x^0_2) \neq (0,0)$, then the sequence $x^{k+1... | {
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"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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How to get $(\frac{x^2}{2}+\frac{1}{2x^2})^2$ from $1+(\frac{x^2}{2}-\frac{1}{2x^2})^2$? How can I get $(\frac{x^2}{2}+\frac{1}{2x^2})^2$ from $1+(\frac{x^2}{2}-\frac{1}{2x^2})^2$?
The book lists the former as the solution to that step.
This is part of an arc length problem, and I think I'm just hitting a mental roadbl... | $$\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{... | {
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"source": "stackexchange",
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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 ... | $a+b=\dfrac{a^3+b^3}{a^2-ab+b^2}$
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^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/851849",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 3
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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... | 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.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/852689",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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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... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/853700",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
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?
M... | 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, $$K_n=\frac{\sqrt{\pi } ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/855990",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 5,
"answer_id": 4
} |
Limit of $a(k)$ in $ \sum_{k=1}^n \frac{a_k}{(n+1-k)!} = 1 $ For n = 1, 2, 3 ... (natural number)
$ \sum_{k=1}^n \frac{a_k}{(n+1-k)!} = 1 $
$ a_1 = 1, \ a_2 = \frac{1}{2}, \ a_3 = \frac{7}{12} \cdots $
What is the limit of {$ a_k $}
$ \lim_{k \to \infty} a_k $ = ?
I have no idea where to start.
| Note that
$$
\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}
$$
Theref... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/862914",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Definite integral $\int_0^{2\pi}(\cos^2(x)+a^2)^{-1}dx$ How do I prove the following?
$$ 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... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/863339",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 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, ... | 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{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/864140",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
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-\... | 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)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/866935",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
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Fractions in Questions and Answers
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