Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Functional expression for the sum of a certain power series Does anybody recognize the following power series together with a functional expression for the sum:
$$
\sum_{n = 0}^{\infty} \left( \begin{array}{c} 2n \\ n \end{array} \right) x^n
$$
| I wanted to elaborate on the alternate derivation Jack suggested. I'm sure there's a shorter way, but here it goes.
One can use induction to prove $$\frac{1}{4^n}\binom{2n}{n} = \frac{2}{\pi}\int_{0}^{\pi/2}\cos^{2n}(\theta)\,d\theta.$$
The $n=0$ case is clear. For the inductive step note that
$$\binom{2n+2}{n+1} = 4\f... | {
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"timestamp": "2023-03-29T00:00:00",
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If $a+b+c=abc$ then $\sum\limits_{cyc}\frac{1}{7a+b}\leq\frac{\sqrt3}{8}$ Let $a$, $b$ and $c$ be positive numbers such that $a+b+c=abc$. Prove that:
$$\frac{1}{7a+b}+\frac{1}{7b+c}+\frac{1}{7c+a}\leq\frac{\sqrt3}{8}$$
I tried C-S:
$$\left(\sum_{cyc}\frac{1}{7a+b}\right)^2\leq\sum_{cyc}\frac{1}{(ka+mb+c)(7a+b)^2}\sum_{... | This inequality is something hard
$ a\geq b\geq\sqrt{3}\geq c>0$ such that $abc=a+b+c$ and constraints $(C)$ and $(2)$ and $(3)$ then we have :
$$\frac{1}{7b+a}+\frac{1}{7a+c}+\frac{1}{7c+b}\leq \left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\frac{1}{1+7\left(\frac{\frac{b}{a}+\frac{a}{c}+\frac{c}{b}+\frac{a}{b}+\f... | {
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"timestamp": "2023-03-29T00:00:00",
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Explain why twice the sum $\binom{12}{0} + \binom{12}{1} + \binom{12}{2} + \binom{12}{3} + \binom{12}{4} + \binom{12}{5}$ is $2^{12}-\binom{12}6$ Can someone explain how is the RHS concluded? I did with sample numbers and it is all correct. but I can't figure out how C(12,6) comes to play.
$$
\binom{12}{0} + \binom{12}... | Suppose we toss a fair coin for $12$ times. The number of combinations is $2^{12}$.
Now notice the number of outcomes $N(H>T)$ such that the total number of Head is more than Tail is same as that of Tail more than Head $N(H<T)$, by symmetry.
We know $N(H>T)$ is the total number of choosing less than $6$ Tails, which is... | {
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Solve boundary value problem $x^2y'' + 3xy' + y = x^2$ Problem
$x^2y'' + 3xy' + y = x^2$
BC: $y(1) = 0, y(e) = 0$
I tried to first solve for the homogenous solution, but I get a really weird equation to solve for values of r. How do you solve the above equation?
My Work
$y=e^{rx} \\
x^2r^2 + 3xr + 1 = x^2 \\
x^2r^2 + ... | $$x^2\frac{d^2y}{dx^2} +3x\frac{dy}{dx} + y = x^2$$
This is known as an Euler Equation - differential equations where the degree of $x$ matches the order of the derivative. They can be solved using the change of variable $x= e^z$
So we get $$x\frac{dy}{dx} = \frac{dy}{dz}$$
and $$x^2\frac{d^2y}{dx^2}= \frac{d^2y}{dz^2... | {
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$\sqrt{x^{2}+\frac{1}{n}}$ converges uniformly to $\left|x\right|$ Im trying to prove that $f_{n}\left(x\right)=\sqrt{x^{2}+\frac{1}{n}}$ converges uniformly to $f(x) = \left|x\right|$ in $[-1,1]$.
So for evary $\varepsilon$ exists $N \in \mathbb{N}$ s.t for all $n>N$ and for all $x \in [-1,1]$ $\left|f_{n}\left(x\... | You can notice that
$$
\sqrt{x^2+\frac1n}\le\sqrt{x^2+2\frac{|x|}{\sqrt n}+\frac1n}=\sqrt{\left(|x|+\frac1{\sqrt n}\right)^2}=|x|+\frac1{\sqrt n}.
$$
So $0<f_n(x)-|x|\le\dfrac1{\sqrt n}$. Hence, the uniform convergence (on all $\mathbb R$, not only on $[-1,1]$).
| {
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I need some help finding my mistake for solving a second order recurrence relation I have the recurrence relation: $$ a_n = a_{n-1} + 2a_{n-2} ; a_0 = 2, a_1=1$$
This is what I did:
$$ let \ \ g(x) = \sum a_n x^n \ \ then; \\
g(x) -a_0 -a_1x = \sum_{n=2}^\infty a_nx^n \\
\implies g(x) -2 -x = \sum_{n=2}^\infty (a_{n-1... | @WW1 says it all. You mistook $2$ for $1$ in what is supposed to be
$g(x)−2−x=x(g(x)−2)+2x^2g(x)$
Checking:
$g(x)=(2-x)/(1-x-2x^2)=(1-2x+1+x)/(1-x-2x^2)$
$g(x)=1/(1+x)+1/(1-2x)$
which you will see, gives the right answer when we work out the Taylor series and read off the coefficients for each power of $x$.
| {
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The $n$-th derivative of the reciprocal of a function and a binomial identity While I was looking for an answer to this MSE post in order to prove
\begin{align*}
\frac{d^n}{dx^n}\left(\frac{1}{1-e^{-x}}\right)=(-1)^n\sum_{j=1}^n{n\brace j}j!\frac{e^{-jx}}{\left(1-e^{-jx}\right)^{j+1}}\tag{1}
\end{align*}
with the numb... | Extracting coefficients on $[y^m]$ where $0\le m\le n$ we see that we
have to prove that
$$\sum_{j=1}^n (-1)^j {n+1\choose j+1}
\sum_{k=1}^j {j\choose k} (-1)^k k^n {n-j\choose m-k} (-1)^{m-k}
\\ = \sum_{j=1}^n (-1)^j {n-j\choose m-j} (-1)^{m-j}
\sum_{k=1}^n (-1)^{k} {j\choose k} k^n$$
or alternatively
$$\sum_{j=1}^n... | {
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Is there $n\in\mathbb N$ for which $3n+1$ and $2n+1$ are perfect squares and $5n+3$ is a prime?
Is there $n\in\mathbb N$ for which $3n+1$ and $2n+1$ are perfect squares and $5n+3$ is prime?
My trying:
We know that $n=3n+1-(2n+1)$ is not prime.
But how it can help for $5n+3$?
$\mathbb N=\{1,2,...\}$.
Thank you!
| Lets take $2n+1=a^2$ and $3n+1=b^2$,WLOG assume $a,b>0$ then $4a^2-b^2=(2a-b)(2a+b)=5n+3$ this is prime only if $2a-b=1$ and $2a+b$ is prime.
Take $b=2a-1$ then $$3n+1=4a^2-4a+1\\3n=4(2n+1)-4\sqrt{2n+1}\\5n+4=4\sqrt{2n+1}\\25n^2+40n+16=16(2n+1)\\25n^2+8n=0$$so either $n=0$ or $n=-\frac{8}{25}$,but $0,-\frac{8}{25}\not\... | {
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Help finding integer solutions of equation. There are infinitely many integer solutions for the equation $4x + 6y = 8 $
My work:
$2x+3y=4$ and $2x=4-3y$ so $x=2-(\frac 32)y$
Similarly $y=\frac 43 - \frac 23 x$ are integer solutions of equation.
But correct answer is $x = −4 + 3t, y = 4 − 2t$ for all integers $t$. Help ... | There are infinitely many solutions to the equation $4x+6y=8$ since $(4,6) =2$ and $2\mid8$. We use Euclidean Algorithm to determine $m,n$ such that $4m+6n=2$. Here $m=-1,n=1$ and $8=2.4$.
Thus $x_0 = 4(-1)$ and $y_0=4.1=4$ is a particular solution.
The solutions are given by $x=−4+3t,y=4−2t$ for all integers $t$.
| {
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How to prove that $\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n} \le \frac{n}{n+1}$? I have a series $$a_n = \frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n} \quad n \ge 1$$
For example, $a_3 = \frac{1}{4}+\frac{1}{5}+\frac{1}{6}$.
I need to prove that for $n \ge 1$:
$$a_n = \frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n} \l... | One question you seem to have is:
How do we know that $$ \frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n} $$ means
$$
\sum_{i=1}^n\frac{1}{n+i}?$$
The answer is that we don't for sure though experience suggests it does.
So if that IS the intended meaning then
here is no ambiguity to
$$
\sum_{i=1}^n\frac{1}{n+i} \le \s... | {
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How do you derive the following two summations?
I tried using the geometric series formula, $$ \sum_{n=0}^{k-1} ar^{n}=a\left (\frac{1-r^{n}}{1-r} \right ) $$, but I got $1-2\left ( \frac{1}{2}^{n-1} \right )$ for the first one and $2-2\left ( \frac{1}{2}^{n-1} \right )$ for the second.
Edit: I derived the first answ... | Your calculations are correct. You just got which problems you were addressing muddled up.
Notice that $\sum\limits_{k=0}^{n-1}2^{-k} = \dfrac{1-2^{-n}}{1-2^{-1}} = (2-2^{n-1})$ answers the second problem, not the first.
And likewise $\sum\limits_{k=1}^{n-1}2^{-k} = \sum\limits_{k=0}^{n-1}2^{-k} -1 = (1-2^{n-1})$ an... | {
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show that inequality $(x-3)^4+(y-3)^4+(z-3)^4\ge 193$ Let $x,y,z\in R$,and such
$$xy+yz+xz=-1$$ show that
$$(x-3)^4+(y-3)^4+(z-3)^4\ge 193$$
it seem use Cauchy-Schwarz inequality to solve it?But I try sometime can't get this answer,even now I can't find this equality when $=$?
| Let $x+y+z=3u$, $xy+xz+yz=3v^2$, where $v^2$ can be negative, and $xyz=w^3$.
Hence, the expression $\sum\limits_{cyc}(x-3)^4$ is a linear expression of $w^3$,
which says that it gets a minimal value for an extremal value of $w^3$,
which happens for equality case of two variables.
Let $y=x$. Thus, $z=-\frac{1+x^2}{2x}$... | {
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"timestamp": "2023-03-29T00:00:00",
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Prove $1 + \frac{n}{2} \leq 1+ \frac{1}{2} +\frac{1}{3} +\cdots + \frac{1}{2^n}$ for all natural numbers $n$ Definitions
*
*$H_n = 1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}$ for all $n \in \mathbb{N}$
The Question
*
*Prove $1 + \frac{n}{2} \leq H_{2^n}$ for all $n \in \mathbb{N}$
My Work
*
*
Base Case:... | If $n\leq 2^k$, then $$\frac{1}{2^k+n}\geq \frac1{2^k+2^k}=\frac1{2^{k+1}}$$ Doing this substitution, we now have a sum of $2^k$ identical fractions, which may then be simplified greatly.
| {
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How have I incorrectly computed $\int \frac{x^{2}+4}{x(x-1)^{2}} dx$? My work:
$\displaystyle \int \frac{x^{2}+4}{x(x-1)^{2}}dx$$ =\int \frac{A}{x} dx+\int \frac{B}{x-1}dx +\int \frac{C}{(x-1)^{2}}dx$
$x^{2} + 4 = A(x-1)^{2} + Bx(x-1) + C(x) $
$x^2 + 4 = x^{2}(A+B) + x(-2A+C) + (A-B)$
equating coefficients:
$1 = A + B ... | $x$ has a $-B$ coefficient as well, so $0 = - 2A - B - C$. And $4 = A-B$ should just be $4=A$. It looks like you mistakenly did $Bx(x-1) = Bx^2 - B$ instead of $Bx^2 - Bx$.
Here's a handy tip for dealing with partial fraction decompositions.
When you get to $x^{2} + 4 = A(x-1)^{2} + Bx(x-1) + C(x) $, you can exploit ... | {
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Alternating sum of reciprocals Find the infinite sum: $\frac{1}{(1)(2)}+\frac{1}{(3)(4)}+\frac{1}{(5)(6)}+\cdots$.
I thought of expanding it as $\frac{1}{1}-\frac{1}{2}+\frac{1}{3}- \cdots$, but it would yield nothing useful as this runs into the non-convergence issue.
| We have,
$$\sum_{n \geq 1, \text{odd}} \frac{1}{n(n+1)}$$
$$=\sum_{n \geq 1, \text{odd}} (\frac{1}{n}-\frac{1}{n+1} )$$
$$=\sum_{n \geq 1} \frac{1-(-1)^n}{2} (\frac{1}{n}-\frac{1}{n+1})$$
$$=\frac{1}{2}\sum_{n \geq 1} (\frac{1}{n}-\frac{1}{n+1})-\frac{1}{2} \sum_{n \geq 1} \frac{(-1)^n}{n}+\frac{1}{2} \sum_{n \geq 1} \... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Integrate $e^{2x} \cdot \sin 3x dx$ using integration by parts technique I'm trying to integrate $e^{2x} \cdot \sin 3x dx$ using integration by parts technique. According to some websites my answer is incorrect.
My attempt was:
$\int e^{2x}\sin (3x) dx = [u=e^{2x}, du = 2e^{2x} , dv= \sin 3x, v= -1/3\cos 3x] = uv -\int... | In your second integration by parts you forgot the factor of $\frac{2}{3}$ from the first one, so
\begin{align*}
... &= -\frac{1}{3}e^{2x}\cos 3x + \frac{2}{3}\left(\frac{1}{3}e^{2x}\sin 3x - \frac{2}{3}\int e^{2x}\sin 3x\,dx\right) \\
&= -\frac{4}{9}\int e^{2x}\sin 3x\,dx - \frac{1}{3}e^{2x}\cos 3x + \frac{2}{9}... | {
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Solve $x^x\equiv1 \pmod {14}$ I know that the order $x$ must be relatively prime to $14$ in order to have a solution so do I just check $1,3,5,9,11,13$ and see which ones raised to themselves are congruent to $1$ mod $14$?
| This answer uses results mentioned in the comments above (specifically @RobertIsrael)
You are correct that you need only consider bases which are relatively prime to $14$. In other words, you only need to consider the classes of
$$
1,3,5,9,11,13.
$$
Next, we should compute the orders of each of these. Using $o(n)$ fo... | {
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How to evaluate this definite integral? Here is the integral we have to evaluate:
$$\int_0^4\sqrt{x^2+4}\,dx+\int_2^{2\sqrt{5}}\sqrt{x^2-4}\,dx$$
After observing, i realise that i can't evaluate these integrals from area of circle, I say that $u=\sqrt{x^2+4}$. Then i can say $dx=\frac{udu}{\sqrt{u^2-4}}$. The first t... | Hint -
1.) $\int \sqrt{x^2+a^2} = \frac 12 x \sqrt{x^2+a^2} - \frac{a^2}2 sinh^{-1}\frac xa + c$
Or
$= \frac 12 x \sqrt{x^2+a^2} + \frac{a^2}2 \ln|x + \sqrt{x^2+a^2}+c$
2.) $\int \sqrt{x^2-a^2} = \frac 12 x \sqrt{x^2-a^2} - \frac{a^2}2 cosh^{-1}\frac xa + c$
Or
$= \frac 12 x \sqrt{x^2-a^2} - \frac{a^2}2 \ln|x + \sqrt... | {
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Show $\lim_{(x,y)\to(0,0)}\frac{x^3-xy^3}{x^2+y^2}=0$ I need to show if the following limit is exists (and exists $0$),
$$
f(x,y)=\frac{x^3-xy^3}{x^2+y^2},
$$
for $\vec x\to 0$. I tried out the following:
$$
\frac{x^3-xy^3}{x^2+y^2}\leq\frac{x^3-xy^3}{x^2}=1-\frac{y^3}{x^2},
$$
but obviously I'm stuk here. A similar ap... | With polar coordinates we have $$x=r\cos { \theta ,y=r\sin { \theta } } \\ \lim _{ (x,y)\to (0,0) } \frac { x^{ 3 }-xy^{ 3 } }{ x^{ 2 }+y^{ 2 } } =\lim _{ r\rightarrow 0 } \frac { { r }^{ 3 }\cos ^{ 3 }{ \theta } -{ r }^{ 4 }\cos { \theta \sin ^{ 3 }{ \theta } } }{ { r }^{ 2 } } =0$$
| {
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Can anything interesting be said about this fake proof? The Facebook account called BestTheorems has posted the following. Can anything of interest be said about it that a casual reader might miss?
Note that \begin{align}
\small 2 & = \frac 2{3-2} = \cfrac 2 {3-\cfrac2 {3-2}} = \cfrac 2 {3 - \cfrac 2 {3 - \cfrac 2 {3-... | For any $a$ and $b$, finding $n$ and $k$ such that
$a=\dfrac{n}{k-a}$
$b=\dfrac{n}{k-b}$
"proves" that $a=b\;\;\forall a,b\in\mathbb{N}$.
| {
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"answer_id": 2
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Expand binomially to prove trigonometric identity Prompt: By expanding $\left(z+\frac{1}{z}\right)^4$ show that $\cos^4\theta = \frac{1}{8}(\cos4\theta + 4\cos2\theta + 3).$
I did the expansion using binomial equation as follows
$$\begin{align*}
\left(z+\frac{1}{z}\right)^4 &= z^4 + \binom{4}{1}z^3.\frac{1}{z} + \bino... | Following your idea, you can suppose that $|z|=1$ and then $z=\text{cis}\theta$ and $z^{-1}=\text{cis}(-\theta)$
$$z+z^{-1}=2\cos\theta \to (z+z^{-1})^4=2^4\cos^4\theta$$
on the other hand:
$$z^2+z^{-2}=(z+z^{-1})^2-2=2^2\cos^2\theta-2=2\cos2\theta$$
$$z^4+z^{-4}=(z^2+z^{-2})^2-2=(2\cos2\theta)^2-2=2\cos4\theta$$
Repla... | {
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"question_score": "3",
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Deriving formula from geometric relations If we know that
and we have this triangle
Then how can we derive this relation?
($\alpha$ and $\beta$ are assumed to be small, so that $\tan(\alpha) \approx \alpha$ and $\tan(\beta) \approx \beta$). We also know that $\theta = \alpha + \beta$, so
$$\alpha \approx \frac{4GM}{... | $$\tan \alpha=\frac{b}{d} \text{ and } \tan \beta=\frac{b}{D}$$
so,
$$\frac{D/d}{D+d}=\frac{1}{b}\frac{\tan \alpha/\tan \beta}{\frac{1}{\tan \alpha}+\frac{1}{\tan \beta}}=\frac{1}{b}\frac{\tan^2 \alpha}{\tan \alpha+\tan \beta}$$
and,
$$\frac{4GM}{c^2}\frac{D/d}{D+d}=\theta\cdot b\cdot \frac{1}{b}\frac{\tan^2 \alpha}{\t... | {
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Can someone please check if my reasoning for this proof is valid or not? I have already seen the other questions about this proof. I'm just trying a different sort of method, though I'm not sure if it's valid or not.
Context for the main question: Prove by induction $2^n\gt n^3$ for $n\ge10$
Obviously, the base case wo... | It's good but it'd be easier (both to do and follow) to go forward:
$(n+1)^3 = $$n^3 + 3n^2 + 3n + 1 $$< n^3 + 3n^2 + 3n^2 + 3n^2 = $$n^3 + 9n^2 < n^3 + n*n^2 = $$n^3 + n^3 =2n^3 $$< 2*2^n = 2^{n+1}$
| {
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"answer_id": 2
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Suppose $x^2+y^2=4$ and $z^2+t^2=9$ and $xt+yz=6$. What are $x,y,z,t$? Let $x^2+y^2=4$ and $z^2+t^2=9$ and $xt+yz=6$.
What are $x,y,z,t$?
| Putting the parametric equation for both circles
$$x=2\sin a,y=2\cos a,z=3\sin b,t=3\cos b\\$$From this you get$$xt-yz=6\sin a\cos b+6\sin b\cos a=6\sin(a+b)=6\\\sin(a+b)=1\\a+b=\frac{\pi}{2}+2k
\pi$$
Which implies $x=2\sin(\pi/2-b)=2\cos b$ and $y=2\cos(\pi/2-b)=2\sin b$
From this we see that $(x,y,z,t)=(2\cos b,2\sin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2222245",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Finding Recurrence relations for combinatorics problems After you graduate you accept a job that promises a starting salary of $40,000$ and a raise at the end of each year equal to $5\%$ of your current salary plus $1000$. For example, your raise at the end of the first year is $3000$. Let $S_n$ be your salary after $n... | The recurrence relation
\begin{align*}
S_0 & = 40,000\\
S_{n + 1} & = S_n \cdot 1.05 + 1000
\end{align*}
that you stated is correct.
Let's look at the first few terms of the sequence.
\begin{align*}
S_1 & = S_0 \cdot 1.05 + 1000\\
& = 40,000 \cdot 1.05 + 1000\\
S_2 & = S_1 \cdot 1.05 + 1000\\
& = (40,000 \cdot... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2229999",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find coefficients of $x^{2012}$ in $(x+1)(x^2+2)(x^4+4)\cdots (x^{1024}+1024)$ Find coefficients of $x^{2012}$ in $(x+1)(x^2+2)(x^4+4)\cdots (x^{1024}+1024)$
Attempt: i have break $2012$ in to sum of power of $2$
as $2012 = 2^{10}+2^{9}+2^{8}+2^7+2^6+2^4+2^3+2^2$
but wan,t be able to go further, could some help me , t... | What is the coefficient of $x^5$ in $(x+1)(x^2+2)(x^4+4)(x^8+8)(x^{16}+16)$?
$5$ in binary is $101_2$, or $5 = 2^0 + 2^2$
This means the coefficient of $x^5$ will be formed by looking at the term $x^5 = x^{2^0}x^{2^2} = x \cdot x^4$ and multiplying it by the constant term contributed from everything else in the express... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2232769",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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Proving $\sum_{i=1}^{n} {\frac{1}{4i^2-1}} = \frac{n}{2n +1}$ for all $n \in N$ using mathematical induction I'm having a horrible time completing this since this is the first time I'm learning mathematical induction.
So far I have this:
Let $P(n)$ be:
$P(n):\sum_{i=1}^{n} {\frac{1}{4i^2-1}} = \frac{n}{2n +1}$
For the ... | for the sum $$\sum_{i=1}^{k+1}\frac{1}{4i^2-1}$$ we can write:
$$\frac{n}{2n+1}+\frac{1}{4(n+1)^2-1}$$
and this must be $$\frac{n+1}{2(n+1)+1}$$
can you proceed?
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Solve$\int\frac{x^4}{1-x^4}dx$ Question: Solve $\int\frac{x^4}{1-x^4}dx.$
My attempt:
$$\int\frac{x^4}{1-x^4}dx = \int\frac{-(1-x^4)+1}{1-x^4}dx = \int 1 + \frac{1}{1-x^4}dx$$
To integrate $\int\frac{1}{1-x^4}dx,$ I apply substitution $x^2=\sin\theta.$ Then we have $2x \frac{dx}{d\theta} = \cos \theta.$ which implies... | HINT:$$\frac{1}{1-x^4}=\frac{1}{2}\left(\frac{1}{1+x^2}+\frac{1}{1-x^2}\right)$$
You handle these. I guess?
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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How to show that $\mathbb{Z}\left[\frac{1 + \sqrt{5}}{2}\right]$ is finitely generated? We can say that $\mathbb{Z}\left[\cfrac{1 + \sqrt{5}}{2}\right]$ is finitely generated if minimal polynomial of $\cfrac{1 + \sqrt{5}}{2}$ is in $\mathbb{Z}[X]$. After some calculations it can be shown that $f(X) = X^2 - X - 1 \in \... | You might want to check that
$$\left(\frac{1+\sqrt{5}}2\right)^2=\frac{3+\sqrt5}2,$$
$$\left(\frac{1+\sqrt{5}}2\right)^3=2+\sqrt5.$$
$$\left(\frac{1+\sqrt{5}}2\right)^4=\frac{7+3\sqrt5}2$$
etc.
Indeed you can prove (by induction maybe) that
$$\left(\frac{1+\sqrt{5}}2\right)^n=\frac{a_n+b_n\sqrt5}2$$
where $a_n$ and $b_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2240560",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find all $n$ for which $504$ doesn't divide $n^8-n^2$ I have seen a similar problem where someone showed that it does divide $n^9-n^3$ but here one has to show that it isn't the case for $n^8-n^2$ for all $n$.
| As I said, $n^8-n^2=n^2(n^6-1)$ and $504=2^3\times 3^2\times 7$.
If $n$ is not a multiple of $7$, then $n^6-1$ is divisible by $7$ by Fermat and on the other hand if $n$ is a multiple of $7$, $n^2$ is divisible by $7$.
As for the powers of $2$, if $n$ is even, $n^2$ is divisible by $8$ if $n \equiv 0(\textrm{mod}\hspac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2241697",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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$ \sum_{n=0}^\infty \Bigg[ \frac{(-1)^n}{(2n+1)^3} + \frac{1}{(2n+1)^2}\Bigg] $ find the sum of the series, no solution was provided.
$$ \sum_{n=0}^\infty \Bigg[ \frac{(-1)^n}{(2n+1)^3} + \frac{1}{(2n+1)^2}\Bigg] = \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)^3} + \sum_{n=0}^\infty \frac{1}{(2n+1)^2}$$
I've tried a few ... | In this mathstackexchange post, an answer shows how a Fourier expansion of $x(1-x)$ on $[0,1]$ gives $$\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)^3}=\dfrac{\pi^3}{32}.$$ On the other hand, we have $$\sum_{n=0}^\infty \frac{1}{(2n+1)^2}=\sum_{n=0}^\infty \frac{1}{n^2}-\sum_{n=0}^\infty \frac{1}{(2n)^2}=\frac{\pi^2}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2241807",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Sum of all possible angles. If $$\tan\left(\frac{\pi}{12}-x\right) , \tan\left(\frac{\pi}{12}\right) , \tan \left(\frac{\pi}{12} +x\right)$$ in order are three consecutive terms of a GP then what is sum of all possible values of $x$.
I am not getting any start, can anybody provide me a hint?
| Clearly, $x=n\pi$ (where $n$ is any integer) is a trivial solution.
Otherwise
$$\dfrac{\tan\dfrac\pi{12}}{\tan\left(\dfrac\pi{12}-x\right)}=\dfrac{\tan\left(\dfrac\pi{12}+x\right)}{\tan\dfrac\pi{12}}$$
$$\iff\dfrac{\sin\dfrac\pi{12}\cos\left(\dfrac\pi{12}-x\right)}{\cos\dfrac\pi{12}\sin\left(\dfrac\pi{12}-x\right)}=\df... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2242177",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find $\sin A$ and $\cos A$ if $\tan A+\sec A=4 $ How to find $\sin A$ and $\cos A$ if
$$\tan A+\sec A=4 ?$$
I tried to find it by $\tan A=\dfrac{\sin A}{\cos A}$ and $\sec A=\dfrac{1}{\cos A}$, therefore
$$\tan A+\sec A=\frac{\sin A+1}{\cos A}=4,$$
which implies
$$\sin A+1=4\cos A.$$
Then what to do?
| Rearranging the identity $1 + \tan^2A = \sec^2A$ yields
$$\sec^2A - \tan^2A = 1$$
Factoring yields
$$(\sec A + \tan A)(\sec A - \tan A) = 1$$
Since $\sec A + \tan A = 4$, we have
$$4(\sec A - \tan A) = 1$$
which yields the system of equations
\begin{align*}
\sec A + \tan A & = 4\\
\sec A - \tan A & = \frac{1}{4}
\en... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2243571",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Irreducibility of $x^{11}+x^{10}+x^{9}+\dots+x+1$
Prove that$$g(x)=x^{11}+x^{10}+x^{9}+\cdots+x+1$$ is irreducible over $\mathbb{Q}$.
This is my (incorrect-see comments) attempt
$$g(x)=\frac{(x-1)(x^{11}+x^{10}+x^{9}+\cdots+x+1)}{x-1}=\frac{x^{12}-1}{x-1},$$
So$$g(x+1)=\frac{(x+1)^{12}-1}{x}=x^{11}+12x^{10}+66x^{9}+\... | Hint:
It is not true because:
$$
(x^{12}-1)=(x^6+1)(x^6-1)=(x^6+1)(x^3+1)(x^3-1)=
$$
$$
=(x^6+1)(x+1)(x^2-x+1)(x-1)(x^2+x+1)
$$
and
$$
(x^{12}-1)=(x-1)(x^{11}+x^{10}+ \cdots +x+1)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2246173",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Calculating $\int_{0}^{\frac{\pi}{6}} \cos{(x)} \sqrt{2\sin (x)+1} dx$ I am trying to calculate the value of the following
$$\int_{0}^{\frac{\pi}{6}} cosx \sqrt{2sin x+1} dx$$
I used a substitution method.
$$u = 2 \sin (x) + 1$$
$$\frac{du}{dx} = 2\cos (x)$$
$$\frac{u}{2 \cos (x)}du = dx$$
hence
$$\int_{0}^{\frac{\... | You had this:
$$\frac{du}{dx}=2\cos(x)$$
You should've moved the $dx$ to the other side to get
$$du=2\cos(x)\ dx$$
but you had an extra $u$. Follow this, and the rest of your work is fine.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2246334",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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The number of solutions of the equation $\tan x +\sec x =2\cos x$ lying in the interval $[0, 2\pi]$ is: The number of solutions of the equation $\tan x +\sec x =2\cos x$ lying in the interval $[0, 2\pi]$ is:
$a$. $0$
$b$. $1$
$c$. $2$
$d$. $3$
My Attempt:
$$\tan x +\sec x=2\cos x$$
$$\dfrac {\sin x}{\cos x}+\dfrac {1}{... | Following on from $$(\sin x +1)(2\sin x-1)=0$$
Note it's useful to sketch the graph of $\sin x$ in the given interval:
We get that either:
$\sin x=-1\implies x=\frac{3\pi}{2},\quad$or
$\sin x =\frac{1}{2}\implies x=\frac{\pi}{6}\quad\text{or}\quad x=\frac{5\pi}{6}$
Now we must be careful since we had $\cos x$ in the d... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Linear transformation problem M2x2 to P2 Can you guys help me with this question?
Let $T:{M_{2\times2}} \to {P_2}$ be defined by $$T\begin{pmatrix} a&b\\c&d \end{pmatrix}=(a+b-c-d)t^2+ (c+d)t+ (a+b). $$ Find the matrix of T with respect to the standard bases for ${M_{2\times 2}}$ and ${P_2}$.
The standard bases for ... | The matrix of a linear transformation comes from expressing each of the basis elements for the domain in terms of basis elements for the range upon applying the transformation.
For example,
$$T \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} = 1\cdot 1 + 0\cdot t + 1 \cdot t^2.$$
Thus the first column of the matrix for $T... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2249106",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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What minimum number (A) can be taken so that (A)^N is larger than the product of N numbers? Given a sequence of N numbers say 2,8,4,7,6,5. How can we calculate a minimum number say A such that AN is greater than the product of 2*8*4*7*6*5 = 13440?
So the minimum number satisfying the above condition is 5.
As 56 = 15625... | The geometric mean will do it: in your case we have
$\sqrt[6]{2 \cdot 8 \cdot 4 \cdot 7 \cdot 6 \cdot 5} = (2 \cdot 8 \cdot 4 \cdot 7 \cdot 6 \cdot 5)^{\frac{1}{6}} = 4.87603...$ so, as you noted, $5^6 > 2 \cdot 8 \cdot 4 \cdot 7 \cdot 6 \cdot 5$ but $4^6 < 2 \cdot 8 \cdot 4 \cdot 7 \cdot 6 \cdot 5$.
In general, if w... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2249332",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Naive Bayes Classification Example Given the following data:
$$\begin{array}{c|c|c|c|c|}
\text{Instance} & \text{A} & \text{B} &\text{C} &\text{Class} \\ \hline
\text{1} & 1 & 2 & 1 & 1 \\ \hline
\text{2}& 0 & 0 & 1 & 1 \\ \hline
\text{3} & 2 & 1 & 2 & 2 \\ \hline
\text{4} & 1 & 2 & 1 & 2 \\ \hline
\text{5} & 0 & 1 & ... | For $C_1$, by the assumption of Naive Bayesian Classifier, we have
$$
P(A = 1, B = 2, C=2 \mid C_1) = P(A = 1 \mid C_1) \cdot P(B = 2 \mid C_1) \cdot P(C = 2 \mid C_1)
$$
Take $P(A = 1 \mid C_1)$ as an example. There are $4$ training records of $C_1$, among which there are $2$ records with $A = 1$. Therefore, $P(A = 1 ... | {
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"timestamp": "2023-03-29T00:00:00",
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$2x^2 + 3x +4$is not divisible by $5$ I tried by $x^2 \equiv 0, 1, 4 \pmod 5$ but how can I deal with $3x$?
I feel this method does not work here.
| The method does work. How did you deal with $2x^2$ anyway? You know that $x^2 \equiv 0, 1, 4 \pmod 5$. So you just double those, like this: $0, 2, 8$, rewrite as $0, 2, 3$.
Likewise with $3x$, you just have to triple $0, 1, 2, 3, 4$ to get $0, 3, 6, 9, 12$ which we rewrite as $0, 3, 1, 4, 2$.
And lastly $4 \equiv 4 \pm... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Number of possible values of $n$ of an AP The sum of the first $n$ terms of an AP whose first term is a (not necessarily positive) integer and the common difference is $2$, is known to be 153. If $n>1$, then the number of possible values of $n$?
My approach :
$153= \frac{n}{2}[ 2a+ (n-1)\cdot2]$
from this we get;
$(a-1... | Some initial observations might help.
As the AP terms sum to $153$ and the common difference is $2$, the AP must consist of sequential odd numbers. Also, the $153= 3\cdot 3\cdot 17$ hence factors are $1,3,9,17,51,153$.
Consider an AP, $AP1$, with positive first term $2r+1$ and with $m$ terms.
Consider its "mirror AP"... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2251529",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Permutations avoiding repeated elements in a row Assume a working set of: {A, B, C}
I'm creating a new set of 10 elements using the working set. The only restriction is that no single element can repeat more than 4 in a row.
For example:
Valid: AAABBCABAA
Invalid: AAAAABCABC
Valid: AAAABCABCA
Invalid: BABAACCCCC
How ma... | The following answer is long but introduces what I feel is a wonderfully elegant method which itself introduces the use of regular expressions to find the generating function that enumerates valid strings.
Let's imagine listing all the different valid strings and let's separate the strings with "+"s. We will call thi... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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If $ab$ is a square number and $\gcd(a,b) = 1$, then $a$ and $b$ are square numbers.
Let $n, a$ and $b$ be positive integers such that $ab = n^2$. If $\gcd(a, b) = 1$, prove that there exist positive integers $c$ and $d$ such that $a = c^2$ and $b = d^2$
So far I have tried this:
Since $n^2 = ab$ we have that $n = \s... | Consider the prime factorization of $ab=n^2$. Each prime factor appears an even number of times because $n^2$ is a square (each prime factor of $n$ is repeated twice in the factorization of $n^2$).
The condition $\gcd(a,b)=1$ means that $a$ and $b$ have no common prime factors.
That means each prime factor of $a$ itsel... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2253940",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $\lim_{n \rightarrow \infty} (a_{n+1}-\frac{a_n}{2})=0$ then show $a_n$ converges to $0$. I have been stuck on this question for a while now. I have tried many attempts. Here are two that I thought looked promising but lead to a dead end:
Attempt 1:
Write out the terms of $b_n$:
$$b_1=a_{2}-\frac{a_{1}}{2}$$
$$b_2=a... | Let $b_n = a_{n+1} - \frac{1}{2}a_n$. Then
$$ a_n = b_{n-1} + \frac{b_{n-2}}{2} + \cdots + \frac{b_1}{2^{n-2}} + \frac{a_1}{2^{n-1}}. $$
Since $(b_n)$ converges, there exists $M$ such that $|a_1| \leq M$ and $|b_n| \leq M$ for all $n$. Thus for any fixed $m$ and for any $n > m$, we have
$$ |a_n| \leq \Bigg| \underbrace... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2254694",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 0
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Convergence of $\int^\infty_0\frac{1}{1+x^2\sin^2(5x)}\,dx$ I need to find out whether the following improper integral converges:
$$\int^\infty_0\frac{1}{1+x^2\sin^2(5x)}\,dx$$
I tried two comparison tests that failed, any ideas?
| Note that we can write
$$\begin{align}
\int_0^{N\pi/5}\frac{1}{1+x^2\sin^2(5x)}\,dx&=\frac15\int_0^{N\pi}\frac{1}{1+(x/5)^2\sin^2(x)}\,dx\\\\
&=\frac15\int_0^{\pi/2}\frac{1}{1+(x/5)^2\sin^2(x)}\,dx\\\\
&+\frac15\sum_{n=1}^{N-1}\int_{(n-1/2)\pi}^{(n+1/2)\pi}\frac{1}{1+(x/5)^2\sin^2(x)}\,dx\\\\
&+\frac15\int_{(N-1/2)\pi}... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Exact period of simple pendulum. Edit: Here is in depth derivation.
Suppose the pendulum is composed of a string of length $L$ and has a point mass of mass $m$ at the end of the string. Say we incline it at an angle $\theta_0 \in (0,\pi)$ counterclockwise from horizontal (counterclockwise counted positive and clockwise... | Hint: Write $-g\sin \theta=L \theta''=L\dfrac{d^2\theta}{dt^2}$ and
$$-g\sin \theta\dfrac{d\theta}{dt}=L\dfrac{d^2\theta}{dt^2}\dfrac{d\theta}{dt}$$
$$-2g\sin \theta d\theta=L\times2\dfrac{d^2\theta}{dt^2}\dfrac{d\theta}{dt}dt=L\Big[\left(\dfrac{d\theta}{dt}\right)^2\Big]'dt$$
after integration
$$2g\cos\theta=L\left(\d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2257095",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Write the function $\frac{1}{(z+1)(3-z)}$ as a Laurent series. $$f(z)=\frac{1}{(z+1)(3-z)}=\frac{1}{4z+4} + \frac{1}{12-4z}$$
$$\frac{1}{4z+4}=\frac{1}{4z}\frac{1}{1-\frac{-1}{z}}=\frac{1}{4z}\sum_{k=0}^{\infty} \left(\frac{-1}{z}\right)^k$$
$$\frac{1}{12-4z}=\frac{1}{12}\frac{1}{1-\frac{z}{3}}=\frac{1}{12}\sum_{k=0}^{... | I can write the expression as:
$$\frac{1}{(z+1)(-z+3)} = \frac{-1}{(z+1)(z-3)}$$
Use partial fractions:
$$\frac{-1}{(z+1)(z-3)} = \frac{A}{z+1}+\frac{B}{z-3}$$
$$-1 = A(z-3)+B(z+1)$$
If z=3, then
$$-1 = B(3+1)$$
$$-1 = B(4)$$
$$B=\frac{-1}{4}$$
If z=1, then
$$-1 = A(1-3)$$
$$-1 = A(-2)$$
$$1 = A(2)$$
$$A = 1/2$$
So, w... | {
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"url": "https://math.stackexchange.com/questions/2257748",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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"answer_id": 1
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Proving that $3^n+7^n+2$ is divisible by 12 for all $n\in\mathbb{N}$. Can someone help me prove this? :( I have tried it multiple times but still cannot get to the answer.
Prove by mathematical induction for $n$ an element of all positive integers that $3^n+7^n+2$ is divisible by 12.
| A general rule of thumb for induction is to try to get
$3^{n+1} + 7^{n+1}+2 = somethingtodo with(3^n + 7^n+2) = something to do with(multiple of 12) = multiple of 12$
So $3^{n+1} + 7^{n+1} +2 = $
$3*3^n + 7*7^n + 2=$
$3*3^n + 3*7n + 4*7^n + 2= $
$3(3^n + 7^n) + 4*7^n + 2 = $
$3(3^n + 7^n + 2) - 6 + 4*7^n + 2 = $
$3(mul... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2258457",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
Show this matrix is invertible Consider the matrix
\begin{equation}
S_n =
\begin{bmatrix}
\frac{1}{2!} & \frac{1}{3!} & \cdots & \frac{1}{(n+1)!} \\
\frac{1}{3!} & \frac{1}{4!} & \cdots & \frac{1}{(n+2)!} \\
\vdots & \vdots & \ddots & \vdots \\
\frac{1}{(n+1)!} & \frac{1}{(n+2)!} & \cdots & \frac{1}{2n!}... | For each $k\le n$, define a $k\times k$ matrix as follows:
$$
A_k=\begin{bmatrix}
\dfrac{(n+1)!}{(n-k+2)!} &\dfrac{(n+2)!}{(n-k+3)!} &\cdots &\cdots &\dfrac{(n+k)!}{(n+1)!}\\
\dfrac{(n+1)!}{(n-k+3)!} &\dfrac{(n+2)!}{(n-k+4)!} &\cdots &\cdots &\dfrac{(n+k)!}{(n+2)!}\\
\vdots &\vdots &\ddots &\vdots &\vdots\\
\dfrac{(n+1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2260024",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Infinite series of $\lim_{n \to \infty}\sum_{r=n^2}^{(n+2)^2} \frac{1}{\sqrt{r}}$ Find Value of $$S=\lim_{n \to \infty}\sum_{r=n^2}^{(n+2)^2} \frac{1}{\sqrt{r}}$$
$$S=\frac{1}{\sqrt{n^2}}+\frac{1}{\sqrt{n^2+1}}+\frac{1}{\sqrt{n^2+2}}+\cdots +\frac{1}{\sqrt{n^2+4n+4}}$$
But here i cannot express above sum in form of $$... | As you already wrote $\lim_{n \to \infty}\sum_{r=n^2}^{(n+2)^2} \frac{1}{\sqrt{r}}$,
$$S=\frac{1}{\sqrt{n^2}}+\frac{1}{\sqrt{n^2+1}}+\frac{1}{\sqrt{n^2+2}}+\cdots +\frac{1}{\sqrt{n^2+4n+4}}$$
When you take $n^2$ common out from the square root then your eq. becomes:
$$S=\frac{1}{n}\lim {n \to \infty}[1+ \frac{1}{\sqrt(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2260158",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Three quadratic equations have positive roots The three quadratic equations $ax^2-2bx+c=0$, $bx^2-2cx+a=0$ and $cx^2-2ax+b=0$ has both roots positive. Then which of the following is/are true
A) $a^2=bc$
B) $b^2=ac$
C) $c^2=ab$
D) $a=b=c$
I assumed $a>0$ so parabola is open upwards and hence it should definitely cut pos... | Since the equations have two roots, the leading coefficients must be non-$0\,$, so $abc \ne 0$. Assume WLOG that $a \gt 0\,$ (otherwise use the same argument for $a,b,c \mapsto -a,-b,-c\,$). Since both roots of each equation are positive, it follows by Vieta's relations that $b \gt 0 , c \gt 0\,$ as well, so in the end... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2263638",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Value of integral under certain condition If f is a real valued function and satisfies $f (x)+f (1-\frac {1}{x})=\arctan (x) $ for all $x $ except $0$ then find the least integer greater than or equal to $N=\int _0 ^1 f (x)dx $ . I havent done any progress . What I tried was putting $x $ as $\frac {x-1}{x} $ to get... | Using your hints, we have, substituting $x$ with $\frac{x-1}{x}$ and then with $\frac{1}{1-x}$, $$ f(x) + f\left( 1 - \frac{1}{x}\right) = \arctan(x)$$ $$ f\left( \frac{x-1}{x}\right) + f\left(\frac{1}{1-x} \right) = \arctan\left(\frac{x-1}{x}\right)$$ and $$ f(x) + f \left( \frac{1}{1-x} \right) = \arctan\left( \frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2264333",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
$\lim_{n\rightarrow \infty } (\frac{(n+1)^{2n^2+2n+1}}{(n+2)^{(n+1)^2}\cdot n^{n^2}})$ The given question is asking me to test the series for convergence or divergence.
$$\sum_{n=1}^{\infty } (\frac{n}{n+1})^{n^2}$$
My attempt, I'm using ratio test and I'm stuck at the limits part.
I don't know how to evaluate $\lim_{n... | If you want to deal with the limit you posted, consider $$u_n= \left(\frac{n}{n+1}\right)^{n^2}\implies \log(u_n)=n^2 \log\left(\frac{n}{n+1}\right)=n^2 \log\left(1-\frac{1}{n+1}\right)$$ Now, use Taylor expansion and get $$\log(u_n)=-n+\frac{1}{2}-\frac{1}{3 n}+\frac{1}{4
n^2}+O\left(\frac{1}{n^3}\right)$$ Similarl... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2266675",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Maximum of $x^3+y^3+z^3$ with $x+y+z=3$ It is given that, $x+y+z=3\quad 0\le x, y, z \le 2$ and we are to maximise $x^3+y^3+z^3$.
My attempt : if we define $f(x, y, z) =x^3+y^3 +z^3$ with $x+y+z=3$ it can be shown that,
$f(x+z, y, 0)-f(x,y,z)=3xz(x+z)\ge 0$ and thus $f(x, y, z) \le f(x+z, y, 0)$. This implies that $f... | Here is how I would solve it, First, I would eliminate x.
\begin{eqnarray*}
x &=& 3 - y - z \\
f(x,y,z) &=& (3 - y - z)^3 + y^3 + z^3
\end{eqnarray*}
then I would apply the second derivative test. More information about the second derivative test can be found at the following URL:
http://faculty.csuci.edu/brian.sitting... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2268524",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
Area between $y^2=2ax$ and $x^2=2ay$ inside $x^2+y^2\le3a^2$ I need to find the area between $y^2=2ax$ and $x^2=2ay$ inside the circle $x^2+y^2\le3a^2$. I know it's an integral but I can't seem to find the right one.
| The entire figure scales proportionally to $a$, so we may set $a=1$ and multiply the area thus obtained by $a^2$ at the end.
The desired area when $a=1$ is twice the area of the blue region $B$ above plus the area of the circular sector $C$ bounded by the black lines $y=\sqrt2x$ and $x=\sqrt2y$. $B$ is bounded by $y=\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2268705",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
If $\sin x + \sin^2 x =1$ then find the value of $\cos^8 x + 2\cos^6 x + \cos^4 x$ If $\sin x + \sin^2 x =1$ then find the value of $\cos^8 x + 2\cos^6 x + \cos^4 x$
My Attempt:
$$\sin x + \sin^2 x=1$$
$$\sin x = 1-\sin^2 x$$
$$\sin x = \cos^2 x$$
Now,
$$\cos^8 x + 2\cos^6 x + \cos^4 x$$
$$=\sin^4 x + 2\sin^3 x +\sin^... | Even shorter:
You know $\sin x = \cos^2 x$. Then
$\cos^8 x + 2 \cos^6 x + \cos^4 x = \sin^4 x + 2 \sin^3 x + \sin^2 x = (\sin^2 x +\sin x)^2 = 1^2 = 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2269298",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 2
} |
Prove if $x > 3$ and $y < 2$, then $x^{2} - 2y > 5$ My solution is:
Multiply $x > 3$ with $x$, yielding $x^{2} > 9$
Multiply $y < 2$ with $2$, yielding $2y < 4$
Thus, based on the above $2$ yielded inequalities, we can prove that if $x > 3$ and $y < 2$, then $x^{2} - 2y > 5$.
Is this a correct proofing steps?
| For a roundabout way to prove it, which is overkill in this case, but may prove useful in other cases, note that the blue terms are positive since $x -3\gt 0$ and $2-y \gt 0\,$, therefore:
$$
x^2-2y =\left((x-3)+3\right)^2 - 2\left(-(2-y)+2\right) = \color{blue}{(x-3)^2} + 6\color{blue}{(x-3)} + \color{red}{9} +2\color... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2269402",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
Calculate $\sqrt{2i}$ I did:
$\sqrt{2i} = x+yi \Leftrightarrow i = \frac{(x+yi)^2}{2} \Leftrightarrow i = \frac{x^2+2xyi+(yi)^2}{2} \Leftrightarrow i = \frac{x^2-y^2+2xyi}{2} \Leftrightarrow \frac{x^2-y^2}{2} = 0 \land \frac{2xy}{2} = 1$
$$\begin{cases}
\frac{x^2-y^2}{2} = 0 \\
xy = 1\\ \end{cases} \\
=\... | $2i=1+2i+i^2=(1+i)^2$. The roots are $\pm(1+i)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2273345",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Find the domain of $\arcsin\left(\frac{3-x}{\sqrt{9-x^2}}\right)$
Find the domain of
$$\arcsin\left(\dfrac{3-x}{\sqrt{9-x^2}}\right)$$
This question is given in my book. The answer is as follows:
For the function to be defined, $-1\le3-x\le1$ and $9-x^2>0$. Solve these inequalities and take their common region.
... | For $x\ne3,\dfrac{3-x}{\sqrt{9-x^2}}=\sqrt{\dfrac{3-x}{3+x}}$
Now we need $$-1\le\sqrt{\dfrac{3-x}{3+x}}\le1\iff0\le\dfrac{3-x}{3+x}\le1$$
If $\dfrac{3-x}{3+x}=0\implies x=3\ \ \ \ (1)$
$\dfrac{3-x}{3+x}>0\iff(3-x)(3+x)>0\iff(x-3)(x+3)<0\iff-3<x<3\ \ \ \ (2)$
$\dfrac{3-x}{3+x}\le1\iff0\ge\dfrac{3-x}{3+x}-1=\dfrac{3-x-(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2274127",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Completing the square of $x^2 - mx = 1$ is not giving me the right answer. This is my attempt
$$
\begin{align}
x^2 - mx &= 1 \\
x^2 - mx - 1 &= 0 \\
\left(x^2 - mx + \frac{m^2}{4} - \frac{m^2}{4}\right) - 1 &= 0 \\
\left(x^2 - mx + \frac{m^2}{4}\right) - \frac{m^2}{4} - 1 &= 0 \\
\left(x^2 - mx + \f... | When you combined $$-\frac{m^2}{4} - \frac{4}{4}$$ into one fraction, you wrote it as $$-\frac{m^2-4}{4}.$$ You should have gotten $$-\frac{m^2+4}{4}$$ to make both terms appropriately negative.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2275086",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Given $a+b+c+d=4$ To find Minimum value of $\frac{a}{b^3+4}+\frac{b}{c^3+4}+\frac{c}{d^3+4}+\frac{d}{a^3+4}$ Given $a+b+c+d=4$ where $a,b,c,d \in \mathbb{R^{+}}$
find Minimum value of $$S=\frac{a}{b^3+4}+\frac{b}{c^3+4}+\frac{c}{d^3+4}+\frac{d}{a^3+4}$$
I have no clue to start...any hint?
| By AM-GM $$\sum_{cyc}\frac{a}{b^3+4}=\frac{a+b+c+d}{4}-\sum_{cyc}\left(\frac{a}{4}-\frac{a}{b^3+4}\right)=$$
$$=1-\sum_{cyc}\frac{b^3a}{4\left(\frac{b^3}{2}+\frac{b^3}{2}+4\right)}\geq1-\frac{b^3a}{4\cdot3\sqrt[3]{\left(\frac{b^3}{2}\right)^2\cdot4}}=$$
$$=1-\sum_{cyc}\frac{b^3a}{12b^2}=1-\frac{1}{12}(ab+bc+cd+da)=$$
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2275743",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Indentify the plane defined by $Re(z\overline{z})=\frac{lm(\overline{z})}{2}$ I tried:
$$Re(z\overline{z})=\frac{lm(\overline{z})}{2} \Leftrightarrow \\
Re((x+yi)(x-yi))=\frac{lm(x-yi)}{2} \Leftrightarrow \\
Re(x^2+y^2)= \frac{lm(x-yi)}{2} \Leftrightarrow \\
x^2 = \frac{-y}{2} \Leftrightarrow \\
-2x^2 = y$$
But my book... | $$Re(z\overline{z})=\frac{lm(\overline{z})}{2} \Leftrightarrow \\
Re((x+yi)(x-yi))=\frac{lm(x-yi)}{2} \Leftrightarrow \\
Re(x^2+y^2)= \frac{lm(x-yi)}{2} \Leftrightarrow \\
x^2 + y^2 = \frac{-y}{2} \Leftrightarrow \\
x^2 + (y+\frac{1}{4})^2 = \frac{1}{16}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2279988",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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How to evaluate the closed form for $\int_{0}^{\pi/2}{\mathrm dx\over \sin^2(x)}\ln\left({a+b\sin^2(x)\over a-b\sin^2(x)}\right)=\pi\cdot F(a,b)?$ Proposed:
$$\int_{0}^{\pi/2}{\mathrm dx\over \sin^2(x)}\ln\left({a+b\sin^2(x)\over a-b\sin^2(x)}\right)=\pi\cdot F(a,b)\tag1$$
Where $a\ge b$
Examples:
Where $F(1,1)= \s... | You may like this method. Let
$$I(b)=\int_{0}^{\pi/2}{\mathrm dx\over \sin^2(x)}\ln\left({a+b\sin^2(x)\over a-b\sin^2(x)}\right) $$
and then
\begin{eqnarray}
I'(b)&=&\int_{0}^{\pi/2}\bigg(\frac1{a+b\sin^2(x)}+\frac1{a-b\sin^2(x)}\bigg) \mathrm dx\\
&=&\frac{\pi}{2\sqrt a}(\frac1{\sqrt{a+b}}+\frac{1}{\sqrt{a-b}}).
\end... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2283669",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 0
} |
Solving System of Congruences Solve
$x^{} \equiv 5 \pmod {7}$
$x^{} \equiv 5 \pmod {9}$
$x^{7} \equiv 5 \pmod {32}$
This is what I have done so far:
$x=7k+5$
Substitute in x in the second congruence statement
$7k+5 \equiv 5 \pmod {9}$
Solving this results in
$k \equiv 0 \pmod {9}$
then
$k=9j$
$x=63j+5$
I'm lost fro... | By Euler's totient theorem, for all odd integer $x$:
$$x^{16} \equiv 1 \pmod{32}$$
Trivially, $x$ is odd, for if $x$ is even, then so would be $x^7$.
Then,
$$\begin{array}{rcll}
x^7 \equiv 5 & \pmod{32} \\
(x^7)^7 \equiv 13 & \pmod{32} \\
x^{49} \equiv 13 & \pmod{32} \\
x^{16\times3+1} \equiv 13 & \pmod{32} \\
(x^{16})... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2284458",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Binomial expansion of negative exponents. Let's say I have to expand $(1+x)^{-1}$ using binomial expansion.
Using the theorem, I get:
$$(1+x)^{-1} = 1-x+x^2-x^3+x^4-x^5+x^6+....+{\infty}$$
Substituting $x$ for $1$, I get:
$$\frac{1}{2}= 1-1+1-1+1-1+1+....+{\infty}$$
A similar result arises with higher power of the expo... | The equality $$\sum_{n=0}^\infty x^n = \frac{1}{x+1}$$
is only true if $|x<1|$, so you cannot substitute $x=1$ into it and expect the result to work.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2284792",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
How many solutions does the equation $x^2-y^2=5^{29}$ have, given that $x$ and $y$ are positive integers?
How many solutions does the equation $x^2-y^2=5^{29}$ have, given that $x$ and $y$ are positive integers?
Someone told me that it has $(29+1)/2=15$ solutions. How come? Any other method to solve this?
| Note $x^2-y^2=(x+y)(x-y)$ so both of these must be factors of $5^{29}$, i.e. powers of $5$.
Let $x+y=5^a$ for some $0 \leq a \leq 29$. Then by the factorisation above, $x-y=5^{29-a}$ and this always has a solution in rational numbers, namely $x=\frac{5^a+5^{29-a}}{2}$, $y=\frac{5^a-5^{29-a}}{2}$.
We do require these to... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2286212",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Inequality Question-Maximum Problem:
Let $a, b, c, d, e, f$ be nonnegative real numbers such that $a^2 + b^2 + c^2 + d^2 + e^2 + f^2 = 6$ and $ab + cd + ef = 3$. What is the maximum value of $a+b+c+d+e+f$?
How would I do this? Would we need to use Cauchy-Schwarz or any of those types of inequalities?
Edit: My questio... | Use Cauchy-Schwarz on the vectors
$$x = \begin{pmatrix}a+b \\
c+d \\
e+f\end{pmatrix}, \quad y = \begin{pmatrix}1 \\
1 \\
1\\ \end{pmatrix}$$
Then $$x\cdot x = (a+b)^2 + (c+d)^2 + (e+f)^2 \\= a^2 + b^2 + c^2 + d^2 + e^2 + f^2+ 2ab+2cd+2ef \\=6+2\times3 \\=12$$
Also, $y\cdot y = 3$, and $x \cdot y = a+b+c+d+e+f$, the q... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2289400",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Sum of all possible values of $\gcd(a-1,a^2+a+1)$
Find the sum of all possible values of
$$\gcd(a-1,a^2+a+1)$$
where $a$ is a positive integer.
| HINT:
Let $a-1=b\iff a=?$
$$a^2+a+1=(b+1)^2+b+1+1=b^2+3b+3$$
$$(a-1,a^2+a+1)=(b,b^2+3b+3)=(b,3)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2295979",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Finding the probability generating function of $P(X=n)$ You toss a fair coin repeatedly until heads appears three times.
Suppose the third head appears on the $X$-th toss. Find the
probability distribution of $X$, that is find a formula for $F_{X}(n) =
P(X = n)$. Hence find a formula for the generating function
$P_{X}(... | The problem is that the series is not geometric, but we could try to turn it into a geometric:
$$P_{X}(s) = \sum_{n=3}^\infty \frac{1}{2^n}\binom{n-1}{2}s^n =\frac{s^3}{2^3} \sum_{m = 0}^\infty \binom{m+2}{2}\frac{s^m}{2^m}$$
$$= a\sum_{m = 0}^\infty (m+2)(m+1)\frac{s^m}{2^m},\text{ where }a=\frac{s^3}{2^4}$$
Let $S_n=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2296306",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Finding Singular Value of an SVD Say you're given the following SVD:
$B=
\begin{bmatrix}
1 & -2 & -2 \\
-6 & 3 & -6 \\
\end{bmatrix}
=
\begin{bmatrix}
0 & d \\
1 & e\\
\end{bmatrix}
\begin{bmatrix}
9 & 0 & 0 \\
0 & f & g \\
\end{bmatrix}
\begin{bmatrix}
-2/3 & -1/3 & a \\
1/3 & 2/3 & ... | This is a nice problem to probe understanding of the singular value decomposition. My upvote goes to @Roland for his succinct answer. The following elaboration is to help other readers who may need help with other parts.
Problem
Input matrix:
$$
\mathbf{A} =
\left[
\begin{array}{rrr}
1 & -2 & -2 \\
-6 & 3 & -6 \\
\... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Maximum of $xy(1-\frac{z^2}{(x+y)^2})+yz(1-\frac{x^2}{(y+z)^2})+zx(1-\frac{y^2}{(z+x)^2})$ Let $x,y,z>0$ and $x^2+y^2+z^2=1$. What is the maximum value of
$$xy\left(1-\frac{z^2}{(x+y)^2}\right)+yz\left(1-\frac{x^2}{(y+z)^2}\right)+zx\left(1-\frac{y^2}{(z+x)^2}\right)?$$
We can get an upper bound of $1$ if we throw away... | For $x=y=z=\frac{1}{\sqrt3}$ we get a value $\frac{3}{4}$.
We'll prove that it's a maximal value.
Indeed, we need to prove that
$$\sum_{cyc}xy\left(1-\frac{z^2}{(x+y)^2}\right)\leq\frac{3}{4}(x^2+y^2+z^2)$$ or
$$\sum_{cyc}\left(\frac{1}{4}z^2-\frac{z^2xy}{(x+y)^2}\right)\leq\sum_{cyc}(x^2-xy)$$ or
$$\sum_{cyc}\frac{z^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2302111",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Q: Convergence of improper double integral Consider the improper double integral
$$I=\iint_{x^2+y^2>0}\frac{sin(x^2+y^2)}{(x^2+4y^2)^{\alpha}}\ dx\ dy$$
I want to determine the range of $\alpha$ , such that the integral above is convergent. Using transformation to polar coordinates I get $
I=\iint_{r>0}\frac{rsin(r^... | It is more practical to "confine the mess to the numerator" then switch to polar coordinates:
$$\begin{eqnarray*} I &=& \frac{1}{2}\iint_{\mathbb{R}^2\setminus\{(0,0)\}}\frac{\sin\left(x^2+\frac{1}{4}y^2\right)}{(x^2+y^2)^{\alpha}}\,dx\,dy \\[0.2cm]&=&\frac{1}{2}\,\text{Im}\int_{0}^{+\infty}\int_{0}^{2\pi}\frac{e^{i\rh... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2303418",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How to find $\theta$ bounds when calculating the volume enclosed between sphere $x^2+y^2+z^2=4a^2$ and cylinder $x^2+(y-a)^2=a^2$?
I need to find $\theta$ bounds when calculating the volume enclosed between sphere $x^2+y^2+z^2=4a^2$ and cylinder $x^2+(y-a)^2=a^2$. The final answer must be that volume=$\frac{48\pi-64}{... | Step 2 is incorrect because both limits vanish $E$ at $0, 2\pi$ . Steps 3 and 4 are correct evaluating between limits $ \int _{\pi/2}^0 $ and multiplying by $4.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2305889",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Convergence of the following series. My professor asked me to prove that the series $\sum_{n=1}^{\infty}\frac{1}{\sqrt{n}(\sqrt{2n}+\sqrt{2n+2})}$ is convergent and find the sum.
But I ended up proving it divergent. Here's my work :
Notice that $\sqrt{1+\frac{1}{n}}<\frac{4}{\sqrt{2}}-1$ for all positive integer $n$. T... | Yes the series diverges,
as $2n + 2 \geq n $ and $\sqrt{2n} \leq 2\sqrt{n}$,so $\sqrt{2n} + \sqrt{2n+2} \leq 2\sqrt{n} + \sqrt{n} = 3 \sqrt{n}$,
So,$\frac{1}{(\sqrt{2n}+\sqrt{2n+2})} \geq \frac{1}{3\sqrt{n}}$,
$\frac{1}{\sqrt{n}(\sqrt{2n}+\sqrt{2n+2})} \geq \frac{1}{3n}$ or $\frac{1}{3n} \leq \frac{1}{\sqrt{n}(\sqrt{2n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2306416",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find this limit of an integral Find $$\lim_{n\to\infty}n\left(\int_{0}^{\frac{\pi}{2}}(\sin{x})^{\frac{1}{n}}(\cos{x})^{\frac{1}{2n}}dx-\frac{\pi}{2}\right)$$
it is obvious
$$\int_{0}^{\frac{\pi}{2}}(\sin{x})^{\frac{1}{n}}(\cos{x})^{\frac{1}{2n}})dx=B\left(\dfrac{n+1}{2n},\dfrac{2n+1}{4n}\right)$$
| $\lim_{n\to\infty}n^2\left(\int_{0}^{\frac{\pi}{2}}(\sin{x})^{\frac{1}{n}}(\cos{x})^{\frac{1}{2n}}dx-\frac{\pi}{2}\right)
$
I'll proceed very naively.
Since,
for large $n$,
$z^{1/n}
=e^{\ln z/n}
=1+\frac{\ln z}{n}+\frac{\ln^2 z}{2n^2}+O(\frac1{n^3})
=1+\frac{\ln z}{n}+O(\frac1{n^2})
$,
$\begin{array}\\
(\sin{x})^{\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2306526",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Area between $r=4\sin(\theta)$ and $r=2$ I'm trying to find the area between $r=4\sin(\theta)$ and $r=2$.
I found the points of intersections to be $\pi/6,5\pi/6$. Which implies the area is $$A=\frac{1}{2}\int_{\pi/6}^{5\pi/6}(4\sin(\theta))^2-2^2d\theta.$$
Is this correct? Or did I find the area for the following reg... | You are just intersecting two circles with the same radius, going through the center of each other.
The area of a circle sector with radius $R=2$ and amplitude $60^\circ$ is $\frac{1}{6}\pi R^2=\frac{2\pi}{3}$, while the area of an equilateral triangle with side length $2$ is given by $\sqrt{3}$, hence the area of the ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2307272",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Modulus Function issue.
I'm asked to solve $$|x+1|>x^2-5$$
My attempt,
For my basic inequality skill, this is a very easy question.
Since $$|x|=\left\{\begin{matrix}
x, x \geq0 & \\
-x,x<0&
\end{matrix}\right.$$
Then for $x<-1$
$x^2+x-4<0$
$\frac{-1-\sqrt{17}}{2}<x<\frac{-1+\sqrt{17}}{2}$
Then for $x\geq-1$
$x+... | It's $x+1>x^2-5$ or $x+1<-x^2+5$ without any additional cases.
because if $x^2-5<0$ then the inequality is obviously true.
We get $-2<x<3$ or $\frac{-1-\sqrt{17}}{2}<x<\frac{-1+\sqrt{17}}{2}$, which gives your answer:
$$\frac{-1-\sqrt{17}}{2}<x<3$$
I think your solution is right, but your way is bad.
For example, sol... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2307887",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Find $\lim_{x\rightarrow \frac{\pi}{4}}\frac{\cos(2x)}{\sin(x-\frac{\pi}{4})}$ without L'Hopital I tried:
$$\lim_{x\rightarrow \frac{\pi}{4}}\frac{\cos(2x)}{\sin(x-\frac{\pi}{4})} =
\frac{\frac{\cos(2x)}{x-\frac{\pi}{4}}}{\frac{\sin(x-\frac{\pi}{4})}{x-\frac{\pi}{4}}}$$ and $$\begin{align}\frac{\cos(2x)}{x-\frac{\pi}... | Let $u = x-\pi/4$ then you want $$\lim_{u\to 0} \frac{\cos(2(u + \pi/4))}{\sin u}=-\lim_{u \to 0} \frac{\sin 2u}{\sin u } = -\lim_{u \to 0} 2\cos u=-2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2311265",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
Calculate the value of the series $\,\sum_{n=1}^\infty\frac{1}{2n(2n+1)(2n+2)}$ Calculate the infinite sum
$$\dfrac{1}{2\cdot 3\cdot 4}+ \dfrac{1}{4\cdot 5\cdot 6}+\dfrac{1}{6\cdot 7\cdot 8}+\cdots$$
I know this series is convergent by Comparison Test, but I can't understand how can I get the value of the sum.
Is th... | Rewrite the sum $\sum_{i=1}^{\infty }\frac {1}{(2i)(2i+1)(2i+2)}$ as
$$ \sum_{i=1}^{\infty }\frac {(2i+1)-2i}{(2i)(2i+1)(2i+2)} = \sum_{i=1}^{\infty }\frac {1}{(2i)(2i+2)} -\sum_{i=1}^{\infty }\frac {1}{(2i+1)(2i+2)} $$
Or using partial fractions
$$ \frac{1}{4} \sum_{i=1}^{\infty} \left(\frac{1}{i} - \frac{1}{i+1}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2312143",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
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Show that $f(x)=x^2+x+4$ is irreducible over $\mathbb{Z}_{11}$ I know this can be done by evaluating $f$ at the points $0,1,...10$ to check if $f$ has a linear factor.
Is there any other shorter way?
| The discriminant of the quadratic form $u^2 + uv + 4 v^2$ is $-15.$
$$ (-15|11) = (-4|11) = (-1|11) (4|11) = (-1|11)=-1, $$
the last part because $11 \equiv 3 \pmod 4.$
Therefore, if
$$ u^2 + uv + 4 v^2 \equiv 0 \pmod {11}, $$ it follows that both $u,v$ are divisible by $11.$ General theorem with proof at Prime divis... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2312350",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
$f(\frac{1}{n})=a_n$ for which sequence holomorphic? Let $D=\{z\in\mathbb{C}\colon |z|<1\}$. Is there a holomorphic function $f\colon D\rightarrow\mathbb{C}$ with $f(\frac{1}{n})=a_n$ $(n=2,3,4,...)$ where $a_n$ is one of the sequences ?
a) $a_n=0,\frac{1}{2},0,\frac{1}{4},0,\frac{1}{6},...$
b) $a_n=\frac{1}{2},\frac... | If our function is holomorphic the derivatives at zero will all exist, and and any sequence of points approaching zero should give us the same derivative. So what happens for a)?
$\displaystyle\lim_{n \to \infty}\frac{f(\frac{1}{2(n+1)})-f(\frac{1}{2n})}{\frac{1}{2(n+1)}-\frac{1}{2n}}=\displaystyle\lim_{n \to \infty}\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2317308",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Inverse of the matrix $$U(a,b)=\left(\begin{matrix}a&b&b&b\\b&a&b&b\\b&b&a&b\\b&b&b&a\end{matrix}\right)$$
Is there an easy way to find the inverse of $U(1,2)$ , a trick to solve this problem easy?(its part of an exam with answers)
| Note that
$$U(a,b)U(c,d) = U(ac+3bd, bc+ad+2bd)$$
This can be seen because, if we let $A$ be the matrix each of whose entries are $1$'s, since $U(a,b)=(a-b)I+bA$, and since $A^2=4A$
$$((a-b)I+bA)((c-d)I+dA)$$
$$ = (a-b)(c-d)I+(b(c-d)+d(a-b)+4bd)A$$
$$= ((ac+3bd)-(bc+ad+2bd))I+(bc+ad+2bd)A$$
So, if $U(a,b)^{-1}$ exists... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2318984",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 4,
"answer_id": 3
} |
equation of common tangent touching circle and the parabola I am trying to find the equation of the common tangent touching the circle $(x-3)^2+y^2=9$ and the parabola $y^2=4x$ above the x-axis is :
Equation of tangent on parabola: $y=mx+a/m$ here a = 1;
so $y = mx + 1/m$
centre of the circle (3,0) and r = 3. Now
dista... | \begin{align}
\frac{\displaystyle \left|3m+\frac{1}{m}\right|}{\sqrt{m^2+1}}&=3\\
\left(3m+\frac{1}{m}\right)^2&=9(m^2+1)\\
9m^2+6+\frac{1}{m^2}&=9m^2+9\\
\frac{1}{m^2}&=3\\
m&=\frac{\pm1}{\sqrt{3}}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2323384",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Show that $(2,0,4) , (4,1,-1) , (6,7,7)$ form a right triangle What I tried:
Let $A(2,0,4)$, $B(4,1,-1)$, $C(6,7,7)$ then
$$\vec{AB}=(2,1,-5), \vec{AC}=(4,7,3), \vec{BC}=(2,6,8)$$
Then I calculated the angle between vectors:
$$\begin{aligned}
\alpha_1 &= \cos^{-1}\left(\frac{(2,1,-5)(4,7,3)}{\sqrt{2^2+1^2+(-5)^2}\sqrt{... | It's enough to show the following:
$$(2,1,-5)\cdot(4,7,3)=0$$
and we are done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2324551",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 4,
"answer_id": 3
} |
Numbers of the form $8k^2-1$ in a sequence defined by $a_0=-1$, $a_1=1$ and $a_{n+2}=6a_{n+1}-a_n$.
Suppose a sequence $\{a_n\}_{n\in\mathbb{N}}$ is defined by $a_0=-1$, $a_1=1$ and $$a_{n+2}=6a_{n+1}-a_n$$ for all $n\in\mathbb{N}$. Find all $n$ such that $a_n$ is of the form $8k^2-1$ for some $k\in\mathbb{N}$.
The p... | Work in progress ...
Proposition 1. $a_{2k} \equiv -1 \pmod{8}$
Proof. Leaving the theoretical material aside, characteristic polynomial of the recurrent sequence is
$$x^2-6x+1=0$$
with the solutions $x_1=3-2\sqrt{2},x_2=3+2\sqrt{2}$, thus
$$a_n=A(3-2\sqrt{2})^n+B(3+2\sqrt{2})^n$$
or, given the initial conditions
$$a_n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2325887",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
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If the tangent at $P$ of the curve $y^2=x^3$ intersects the curve again at $Q$ If the tangent at $P$ of the curve $y^2=x^3$ intersects the curve again at $Q$ and the straight lines $OP,OQ$ make angles $\alpha,\beta$ with the $x$-axis where $O$ is the origin then $\frac{\tan\alpha}{\tan\beta}=$
$(A)-1$
$(B)-2$
$(C)2$
$(... | Simplify the expression a bit.
Notice that tangent of an angle is just the slope it makes.
So, we need to find the ratio of the slopes.
Let $P$ be $(1,1)$.
We have the derivative of $y=\displaystyle \sqrt{x^3} = \frac32\sqrt{x}$ .
Therefore, the tangent line is $(y-1)=\frac{3}{2}(x-1)$.
Because $y^2=x^3$ is concave upw... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2326898",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Integrating the Fourier series to find the Fourier series of $\frac{1}{2}x^2$
The function $\phi(x) = x$ on the interval $[-l,l]$ has the Fourier series
$$x = \frac{2 l}{\pi}\sum_{m=1}^{\infty}\frac{(-1)^{m+1}}{m}\sin\left(\frac{m\pi x}{l} \right) = \frac{2 l}{\pi}\left(\sin\left(\frac{\pi x}{l} \right) - \frac{1}{2... | You're trying to integrate $x$ to $x^2/2$, implying you want to use the indefinite operator $\int dx$, not a definite integration on a period. This gives you a series of cosines, viz. $$\frac{x^2}{2}=A-\frac{2l^2}{\pi^2}\sum_{m\ge 1}\frac{\left( -1\right)^{m+1}}{m^2}\cos\frac{m\pi x}{L}$$ for some constant $A$ obtainab... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2327636",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Find the limit $\lim\limits_{n\to \infty} \int_{0}^{n} {1\over{1+n^2\cos^2x}}\,dx$ $$\lim_{n\to \infty} \int_{0}^{n} {1\over{1+n^2\cos^2x}}\, dx$$
Some help please. I don't have any idea. Thank you!
| Using the formula for the tangent of a sum, we get
$$
\arctan(\alpha\tan(x))=x+\arctan\left(\frac{(\alpha-1)\tan(x)}{1+\alpha\tan^2(x)}\right)
$$
Since $\left|\frac{(\alpha-1)\tan(x)}{1+\alpha\tan^2(x)}\right|\le\frac{|\alpha-1|}{2\sqrt{\alpha}}$, arctan never has to go through a singularity, so this is a nice, smooth ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2328468",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
Complex integral Let $C$ denote the unit circle centered at the origin in Complex Plane
What is the value of
$$ \frac{1}{2\pi i}\int_C |1+z+z^2 |dz,$$ where the integral is taken anti-clockwise along $C$?
*
*0
*1
*2
*3
What I have answered is 0 because it seems like $f(z)$ is analytic at 0 hence by Cauchy's T... | We want the integral
$$\frac{1}{2\pi i}\int_C |1+z+z^2|dz=\frac{1}{2\pi}\int_0^{2\pi}|1+z+z^2|z~d\theta$$
since $z=e^{i\theta}$.
Now consider the absolute value portion,
$$
\begin{align}
|1+z+z^2|
&=\sqrt{(1+z+z^2)(1+z+z^2)^*}\\
&=\sqrt{(1+z+z^2)(1+z^{-1}+z^{-2})}\\
&=\sqrt{(1+z+z^2)(1+z^{-1}+z^{-2})\frac{z^2}{z^2}}\\
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2328658",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Finding values of $t$ I have this equation -
$3t^{\frac{1}{2}} - \frac{2}{5} t^{\frac{-3}{2}} = 0 $
I'm struggling on how to find the values of $t$
First, I power both sides by $2$ to make the power be a whole number .
I get
$3t - \frac{2}{5} t^{-3} = 0 $
From here I'm stunned and stuck . Can I get a hint ! Thank... | You can't square.
$(a + b)^2 \ne a^2 + b^2$ but instead equals $a^2 + 2ab + b^2$ so that won't make things simpler.
Also squaring both both sides of an equation will add extraneous solutions. For example, the problem $x + 1 = 2$ has one solution: $x = 1$. But if I chose to square both sides $(x+1)^2 = 4^2$ then $x^2 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2330637",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Given that $\sum^{n}_{r=-2}{r^3}$ can be written in the form $an^4+bn^3+cn^2+dn+e$, show that: $\sum^{n}_{r=0}{r^3}=\frac14n^2(n+1)^2$ Question:
Given that $\displaystyle\sum^{n}_{r=-2}{r^3}$ can be written in the form $an^4+bn^3+cn^2+dn+e$, show that: $$\sum^{n}_{r=0}{r^3}=\frac14n^2(n+1)^2$$
Attempt:
Substituting $... | $$\begin{align}
&n=0: &e&=-9\tag{1}\\\\
&n=1: &a+b+c+d&=1\tag{2}\\
&n=-1: &a-b+c-d&=0\tag{3}\\\\
\frac 12[(2)+(3)]: &&a+c&=\frac 12\tag{4}\\
\frac 12[(2)-(3)]: &&b+d&=\frac 12\tag{5}\\\\
&n=2: &16a+8b+4c+2d&=9\tag{6}\\
&n=-2: &16a-8b+4c-2d&=1\tag{7}\\\\
\frac 12 [(6)+(7)]: &&16a+4c&=5
\quad\Rightarrow a=\frac 14, c=\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2330733",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Can a solved Sudoku game have an invalid region if all rows and columns are valid? Given a $9 \times 9$ solved Sudoku game with $3 \times 3$ regions, is it possible that one (or more) of the regions are invalid if all rows and columns are valid (i.e. have a unique sequence of $1-9$)?
| Yes, it can happen that all $3 \times 3$ regions are invalid:
\begin{array}{|ccc|ccc|ccc|} \hline
1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\
2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 1 \\
3 & 4 & 5 & 6 & 7 & 8 & 9 & 1 & 2 \\ \hline
4 & 5 & 6 & 7 & 8 & 9 & 1 & 2 & 3 \\
5 & 6 & 7 & 8 & 9 & 1 & 2 & 3 & 4 \\
6 & 7 & 8 & 9 & 1 & 2 & ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2331022",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 1,
"answer_id": 0
} |
Irreducible polynomials in GF(p) Given is the polynomial $\varphi(X)=X^4+1$.
Now there are two tasks:
(1) Show, that $ \varphi(X)$ is reducible in $\mathbb F_p [X]$, where $p$ is prime number with $p \equiv1$ (mod 4).
(2) Show, that $ \varphi(X)$ is reducible in $\mathbb F_p [X]$, where $p$ is prime number with $p \e... | The two congruences give you
$$b^2 \equiv -1 \pmod{p}$$
This is the same as asking when is $-1$ a quadratic residues mod $p$? If you are familiar with Legendre symbol, then perhaps you might have seen
$$\left(\frac{-1}{p}\right)=(-1)^{\frac{p-1}{2}}=\begin{cases}1 & \text{ if } p \equiv 1 \pmod{4}\\-1 & \text{ if } p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2334085",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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The perimeter is equal to the area The measurements on the sides of a rectangle are distinct integers. The perimeter and area of the rectangle are expressed by the same number. Determine this number.
Answer: 18
It could be $4*4$ = $4+4+4+4$ but the answer is 18.
Wait... Now that I noticed, the sides are different ... | Let $a$ and $b$ be the sides of the rectangle. We know the perimeter of a rectangle is $ 2(a+b) $, and the area is$(ab)$.
Since these two expressions have to be equal, let us solve the equation:
$$2(a+b) = ab$$
$$2a = ab - 2b$$
$$2a = b(a-2)$$
$$b = \frac{2a}{a-2}$$
Since $b$ is an integer, $\frac{2a}{a-2}$ must be on... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2334207",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Find the derivative of the inverse function, $Dg(0,1)$. Let $f:\mathbb{R}^2\to \mathbb{R}^2$ be defined by the equation
\begin{equation} f(x,y)=(x^2-y^2,2xy)\end{equation}
Parts (a) and (b) of the problem asked me to show that $f$ is one-to-one on the set $A$ consisting of all $(x,y)$ with $x>0$, and to find set $B=f(A... | First, fix the differential ($f(x,y)=(u(x,y),v(x,y))\;;u=x^2-y^2\;;v=2xy$):
$$Df=
\begin{pmatrix}
\partial u/\partial x &\partial u/\partial y\\
\partial v/\partial x & \partial v/\partial y
\end{pmatrix}=
\begin{pmatrix}
2x &-2y\\
2y & 2x
\end{pmatrix}$$
Now, with the invaluable aid of the commentators, the inverse is... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2335443",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Proving $\frac{2\sin x- \sin 2x}{2\sin x+ \sin 2x} = \left(\tan\frac{x}{2}\right)^2$
How do I prove this equality?
$$\frac{2\sin x- \sin 2x}{2\sin x+ \sin 2x} = \left(\tan\frac{x}{2}\right)^2$$
I have come this far by myself:
$$\begin{array}{llll}
\dfrac{2\sin x- \sin 2x}{2\sin x+ \sin 2x} &= \dfrac{2\sin x- 2\sin x... | Please see the attached photo for a solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2335520",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 5
} |
If $z^4 + \frac1{z^4}=47$ then find the value of $z^3+\frac1{z^3}$ If $z^4 + \dfrac {1}{z^4}=47$ then find the value of $z^3+\dfrac {1}{z^3}$
My Attempt:
$$z^4 + \dfrac {1}{z^4}=47$$
$$(z^2+\dfrac {1}{z^2})^2 - 2=47$$
$$(z^2 + \dfrac {1}{z^2})^2=49$$
$$z^2 + \dfrac {1}{z^2}=7$$
How do I proceed further??
| Try this notation for less clutter:
Let $$S_n=z^n+\frac 1{z^n}$$
It can be easily shown that
$$S_n^2=S_{2n}+2$$
Hence
$$S_4=S_2^2-2=47
\qquad \Rightarrow S_2=7\\
S_2=S_1^2-2=7\qquad \Rightarrow S_1=3$$
Also, $$S_aS_b=S_{a+b}S_{a-b}$$
Putting $a=3, b=1$,
$$S_3S_1=S_4+S_2\\
S_3=\frac {S_4+S_2}{S_1}=\frac {47+7}3=\color... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2335712",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
"answer_id": 6
} |
Find the magnitude of the vertex angle of an isosceles triangle of the given area $A$ Find the magnitude of the vertex angle $\alpha$ of an isosceles triangle with the given area $A$ such that the radius $r$ of the circle inscribed into the triangle is maximal.
My attempt:
| We'll prove that for all triangle
$$r\leq\sqrt{\frac{A}{3\sqrt3}}.$$
Indeed, let $AB=c$, $AC=b$ and $BC=a$.
Hence, $A=\frac{1}{2}ra+\frac{1}{2}rb+\frac{1}{2}rc$, which gives $r=\frac{2A}{a+b+c}$.
Thus, we need to prove that
$$\frac{2A}{a+b+c}\leq\sqrt{\frac{A}{3\sqrt3}}$$ or
$$12\sqrt3A\leq(a+b+c)^2.$$
But by Heron f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2337999",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 0
} |
Proving trigonometric identity $\frac{\sin(A)}{1+ \cos(A )}+\frac{1+ \cos(A )}{\sin(A)}=2 \csc(A)$
$$
\frac{\sin(A)}{1+\cos(A)}+\frac{1+\cos(A)}{\sin(A)}=2\csc(A)
$$
\begin{align}
\mathrm{L.H.S}&= \frac{\sin^2A+(1+\cos^2(A))}{\sin(A)(1+\cos(A))} \\[6px]
&= \frac{\sin^2A+2\sin(A)\cos(A)+\cos^2(A)+1}{\sin(A)(1+\cos(A))... | $$\frac{\sin^2A+(1+\cos(A))^2}{\sin(A)(1+\cos(A))} = \frac{\sin^2A+1+2\cos(A) + \cos^2(A)}{\sin(A)(1+\cos(A))}$$
Now, usse $$\sin^2(A) + \cos^2(A) = 1.$$
Take it from here.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2339468",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 1
} |
Remainder and long division I have been thinking about this. Lets say we take 7 divided by 3, we know the remainder is 1. However, if we let x=7 and x-4=3, and we take x/(x-4), after performing long division the remainder is 4. Why is it not 1?
| Notice:
$7=4*3+(-5) $
$7=3*3+(-2) $
$7=2*3+1$
$7=1*3+4$
$7=0*3+7$
$7=-1*3+10$
Etc. So which one qualifies as "the" remainder? Why is it $1$? Whis it not $4$, $10$ or $-2$?
Well, because the remainder is defined to be at least $0$ but strictly less than the divisor.
The 1 is "the" remainder. The rest are congruences... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2340147",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.