Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
How do you evaluate limit of $\frac{\sqrt{1+x^2} - \sqrt {1+x}}{\sqrt {1+x^3} - \sqrt {{1+x}}}$ when $x$ tends to $0$? I tried rationalization method and got as $\frac{x^2-x}{\sqrt {1+x^3} - \sqrt {1+x} ({\sqrt{1+x^2} + \sqrt {1+x}})}$. But i feel the denominator having power of 3 I may be doing it wrong.The answer sho... | Use definition of derivation:
\begin{align}
\frac{\sqrt{1+x^2} - \sqrt {1+x}}{\sqrt {1+x^3} - \sqrt {{1+x}}}
&=
\frac{\frac{\sqrt{1+x^2}-1}{x} - \frac{\sqrt {1+x}-1}{x}}{\frac{\sqrt {1+x^3}-1}{x} - \frac{\sqrt {{1+x}}-1}{x}}\\
&=
\frac{(\sqrt{1+x^2})'|_{x=0} - (\sqrt {1+x})'|_{x=0} }{(\sqrt {1+x^3})'|_{x=0} - (\sqrt {... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2852294",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
} |
Show $\frac{1}{b+c+d} + \frac{1}{a+c+d} + \frac{1}{a+b+d} + \frac{1}{a+b+c} \ge \frac{16}{3(a+b+c+d)}$.
If $a,b,c,d > 0$ and distinct then show that
$$
\frac{1}{b+c+d} + \frac{1}{a+c+d} + \frac{1}{a+b+d} + \frac{1}{a+b+c}
\ge \frac{16}{3(a+b+c+d)}
$$
I tried using HM < AM inequality but am missing on $16$. Probably... | You can assume $a+b+c+d=1$, so we need to prove $$E:=\sum {1\over 1-x} \geq 16/3$$
Now, since for $x\in (0,1)$ we have $${1\over 1-x}\geq {16x\over 9}+{8\over 9}$$ we get $$E\geq {16\over 9}(a+b+c+d)+4{8\over 9} = {48\over 9}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2854810",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Impulse train: why an indeterminate result? I have an impulse train given by
$$\frac{1}{R+1}+\frac{\sum_{k=1}^R \cos(\frac{2k \pi x}{R+1})}{R+1}$$
It seems obvious to me that, for $x=0$, the function returns $1$. This is because $\cos (0)=1$, and we therefore end up with $\frac{1}{R+1}+\frac{R}{R+1}=\frac{R+1}{R+1}=1$.... | I set the problem in Wolfram Development Platform and the limit is properly given.
The only thing I noticed is that the simplification of the result is not as good as it could be since, using another CAS,
$$\sum_{k=1}^R \cos \left(\frac{2 k\pi x}{R+1}\right)=\frac{1}{2} \left(\sin \left(\frac{\pi (2 R+1) x}{R+1}\rig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2859803",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Algebraic Inequality involving AM-GM-HM If $$a,b,c \;\epsilon \;R^+$$
Then show, $$\frac{bc}{b+c} + \frac{ab}{a+b} + \frac{ac}{a+c} \leq \frac{1}{2} \Bigl(a+b+c\Bigl)$$
I have solved a couple of problems using AM-GM but I have always struggled with selecting terms for the inequality.
I tried writing $a+b+c$ as $x$ and... | Hint: Remember that $HM(a, b)$ can also be written as $\dfrac {2ab}{a+b}$, which looks a lot like the LHS...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2861098",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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the maximum value of $\frac{\sin A}{A}+\frac{\sin B}{B}+\frac{\sin C}{C}$
For any acute angled triangle ABC , find the maximum value of $\frac{\sin A}{A}+\frac{\sin B}{B}+\frac{\sin C}{C}$ .
Attempt:
As $A+B+C=\pi$
$C=\pi -(A+B)$
After differentiating it
$dA+dB+dC=0$
Now :
$\frac{\sin A}{A}+\frac{\sin B}{B}+\frac{\s... | Here's one approach to doing this:
*
*Take $f(x)=\frac{\sin x}{x}$ and show that $f''(x)<0$ for $x\in(0,\pi/2)$.
*Use Jensen's inequality to conclude that for any $A,B,C\in(0,\pi/2)$ with $A+B+C$ fixed, $f(A)+f(B)+f(C)$ is maximised when they are all equal.
(Further hint for 1: write $f''(x)$ as a fraction and di... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2862101",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Polynomial problem with unknown coefficients $a, b, c$ $p(x)=x^3 +ax^2+bx+c$, where $a,b,c$ are distinct non-zero integers. Suppose $p(a)=a^3$ and $p(b)=b^3$, find $p(13)$.
Ok so I've gotten $a^3-b^3=a^3-b^3+a^3-ab^2+ab-b^2$,
So $(a-b)[a(a+b)+b]=0$, so $b=-\frac{a^2}{a+1}$ and then getting $c=-\frac{a^4}{a+1}$ does no... | Put $n=a+1$, then: $$b={-a^2\over a+1} = -{(n-1)^2\over n} = -{n^2-2n+1\over n} = -n+2-{1\over n} \in\mathbb{Z}$$
So $$n\mid 1\implies n =\pm 1 \implies a=0 \vee a= -2 $$
Now if $a=0$ then $b=0$ and this can't be, so $a=-2$ and $b=4$ and $c= 16$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2865053",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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In how many ways a student can get $2m $ Marks An examination contains four Question papers each paper carrying maximum marks as $m$. Find number of ways a student appearing for all the four papers gets a total of $2m$ Marks.
I used generating Polynomial method that is to find coefficient of $x^{2m}$ in
$$(1+x+x^2+\cdo... | Let us consider the marks in all the $4$ papers as $a,b,c,d$; $0\le a,b,c,d\le m$
$$a+b+c+d=2m$$
Now we need to find the integer solution of the above equation in the given condition.
Number of solutions $=\dbinom{2m+4-1}{4-1}=\dbinom{2m+3}{3}$, but note that this also contains those solutions in which any variable is... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2867227",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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A problem related with limit and continuity.
What is the limit as $x$ tends to $1$ in the function $$\frac{x+x^2+x^3+\cdots+x^n-n}{x-1}\quad?$$
I tried factoring out the $x$ term then again rewriting the factored term. But nothing came out at all. Could anyone help me out?
| Define
$$f(x)=x+x^2+x^3+\cdots+x^n.$$
Then $$f'(x)=1+2x+3x^2+\cdots+nx^{n-1}.$$
It follows that
\begin{align*}
\lim_{x \to 1}\frac{x+x^2+x^3+\cdots+x^n-n}{x-1}&=\lim_{x \to 1}\frac{f(x)-f(1)}{x-1}=f'(1)\\&=1+2+3+\cdots+n\\&=\frac{n(n+1)}{2}.\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2875023",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
Prove that inequality $1+\frac{1}{2\sqrt{2}}+...+\frac{1}{n\sqrt{n}}<2\sqrt{2}$ Let $n$ is a natural number. Prove that inequality $$1+\frac{1}{2\sqrt{2}}+\frac{1}{3\sqrt{3}}+...+\frac{1}{n\sqrt{n}}<2\sqrt{2}$$
My try: $$\frac{1}{n\sqrt{n}}=\frac{\sqrt{n}}{n^2}<\frac{\sqrt{n}}{n\left(n-1\right)}=\sqrt{n}\left(\frac{1}... | I think your way gets something wrong.
By your work:
For all $n\geq2$ we obtain:
$$\frac{1}{n\sqrt{n}}<\left(\frac{1}{\sqrt{n-1}}-\frac{1}
{\sqrt{n}}\right)\frac{\sqrt{n}+\sqrt{n-1}}{\sqrt{n-1}}=$$
$$=\left(\frac{1}{\sqrt{n-1}}-\frac{1}{\sqrt{n}}\right)\left(1+\sqrt{\frac{n}{n-1}}\right)\leq\left(\frac{1}{\sqrt{n-1}}-\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2876221",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 1
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Incircle bisectors and related measures
This question was inspired by
in-triangle-abc-d-is-a-point-on-ac...,
show-that-am2-pp-a.
Cevians $|AD_a|=d_a$,
$|BD_b|=d_b$, $|CD_c|=d_c$
divide $\triangle ABC$ into three pairs of triangles,
($\triangle ABD_a$, $\triangle AD_aC$),
($\triangle BCD_b$, $\triangle BD_bA$),
and
(... | Accidentally, I've found this open-access reference:
Yiu, Paul. The Congruent-Incircle Cevians of a Triangle. Missouri J.
Math. Sci. 15 (2003), no. 1, 21--32. doi:10.35834/2003/1501021.
https://projecteuclid.org/euclid.mjms/1567216820
For such cevians
they use a term "the congruent-incircle cevians of a triangle".
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2876870",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to find $ f(x)$ if $f(1-f(x))=x$ for all $x$ $\in \mathbb{R}$ How can I determine $ f(x)$ if $f(1-f(x))=x$ for all real $x$?
I have already recognized one problem caused from this: it
follows that $ f(f(x))=1-x $, which is discontinuous. So how can I construct a function $f(x)$?
Best regards and thanks,
John
| This is a partial answer.
We know that $f(x)$ is invertible, because $f^{-1}(x)=1-f(x),$ from the original; from here we get the very interesting relationship of $f(x)+f^{-1}(x)=1.$ Suppose we try to find out what $f(0)$ is (set it equal to $a$). By repeated alternating applications of $f$ and the equation $f^{-1}(x)=1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2878508",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Which is the value of $t$ ? It is given $t\in \{0,1,2,3,4\}$, a group $G$ and an element $a$ of $G$ such that $a^8=a^t$ and $a^{88}=a^{74}$ in $G$. If $a^2\neq a^0$ which is $t$ ?
For $a\neq 0$ we have the following:
\begin{align*}a^{88}=a^{74}&\Rightarrow a^{14}=a^0\Rightarrow a^{2\cdot 8-2}=a^0\\ & \Rightarrow \lef... | We know that $a^{14} = 1$, so the order of $a$ divides $14$.
We know that $a^2 \ne 1$, so the order of $a$ does not divide $2$.
We know that $a^{8-t} = 1$, so the order of $a$ is at most $8-t$, i.e. at most $8$.
Therefore, the order of $a$ is $7$, so $a^8 = a$, so $t = 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2879407",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Real value of equation $(x-\frac{1}{x})^\frac{1}{2}+(1-\frac{1}{x})^\frac{1}{2}=x$ Find the real value of x in the equation $(x-\frac{1}{x})^\frac{1}{2}+(1-\frac{1}{x})^\frac{1}{2}=x$
I tried to square the whole term and after expansion not getting the result.
| We may assume $x>0$. From $\sqrt{x^2-1}+\sqrt{x-1}=x\sqrt{x}$, subtract $\sqrt{x-1}$ from both side and squaring gives us $$x^2-1=x^3+x-1-2x\sqrt{x^2-x}$$
Simplifying, $$(x^2-x)-2\sqrt{x^2-x}+1=0$$
Can you go on from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2883280",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to proof the reason? I have this statement:
If $\frac{a}{b} = \frac{c}{d},$ prove that
$\frac{a+b}{a-b}=\frac{c+d}{c-d}$
I tried to add 1, multiply 1 and nothing.
My development was:
$\frac{a}{b} - \frac{b}{b} = \frac{c}{d} - \frac{d}{d}$
$\frac{a-b}{b} = \frac{c-d}{d}$
$\frac{b}{a-b} = \frac{d}{c-d}$ (I raised... | Let, $\frac{a}{b}=\frac{c}{d}=k$ or, $a=bk,c=dk$. Now consider the LHS:$$\frac{a+b}{a-b}=\frac{bk+b}{bk-b}=\frac{k+1}{k-1}$$
Also, $$\frac{c+d}{c-d}=\frac{dk+d}{dk-d}=\frac{k+1}{k-1}~~~~~~~~~\boxed{}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2883782",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Convergence of $\sum_{n=1}^{\infty}\left(\frac{\sin(2\alpha^n)}{\alpha^n}-2\right)$ Determine values of $\alpha \in \left[\frac{1}{2} ; \frac{3}{2}\right]=:I$ s.t. the following series converges:
$$\sum_{n=1}^{\infty}\left(\frac{\sin(2\alpha^n)}{\alpha^n}-2\right)$$
Am I doing it wrong or this series fails the limit te... | Let $a_n = \frac{\sin(2\alpha^n)}{\alpha^n}-2$ be the general term for your series.
If $\alpha = 1$, then the $a_n$ are constant and equal to $\sin(2) - 2 \neq 0$, so the series diverges.
If $\alpha > 1$, then $\lim_{n\to\infty} a_n = -2$.
Since it is not $0$, the series does not converge.
Now, suppose $\alpha<1$, and ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2884994",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find the oblique asymptote of $\sqrt{x^2+3x}$ for $x\rightarrow-\infty$ I want to find the asymptote oblique of the following function for $x\rightarrow\pm\infty$
$$f(x)=\sqrt{x^2+3x}=\sqrt{x^2\left(1+\frac{3x}{x^2}\right)}\sim\sqrt{x^2}=|x|$$
For $x\rightarrow+\infty$ we have:
$$\frac{f(x)}{x}\sim\frac{|x|}{x}=\frac{x... | $$f(x)=\sqrt{x^2+3x}=|x|\left(1+{3\over x}\right)^{1\over2}$$ which, as $x\rightarrow -\infty$, is equal to $$-x\left(1-{3\over{2x}}\right)^{1\over2} \sim -x\left(1+{3\over{2x}}\right)= -x -{3\over2}$$ using Taylor or Bernoulli
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2886545",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Is $f_{n}\left(x\right)=\sin\left(x+\frac{x^{2}}{n}\right)$ uniformly convergent on $\left[0,\:2\pi\right]$ Is $f_{n}\left(x\right)$ uniformly convergent on $\left[0,\:2\pi\right]$?
\begin{equation}
f_{n}\left(x\right)=\sin\left(x+\frac{x^{2}}{n}\right)
\end{equation}
We can see that $f_{n}\left(x\right)$ converges to ... | Use that
$$\sin\left(x+\frac{x^{2}}{n}\right)=\sin x\cos \dfrac{x^2}{n}+\cos x\sin \dfrac{x^2}{n}.$$
Thus
\begin{align}\left|\sin\left(x+\frac{x^{2}}{n}\right)-\sin x\right|&= \left|sin x\cos \dfrac{x^2}{n}+\cos x\sin \dfrac{x^2}{n}-\sin x\right| \\ &\le |\sin x|\left|1-\cos\dfrac{x^2}{n}\right|+|\cos x|\left|\sin\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2887058",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove that $\frac{1+2\sin{\theta}\cos{\theta}}{\cos^2{\theta}-\sin^2{\theta}}=\frac{1+\tan{\theta}}{1-\tan{\theta}}$ Prove that $$\frac{1+2\sin{\theta}\cos{\theta}}{\cos^2{\theta}-\sin^2{\theta}}=\frac{1+\tan{\theta}}{1-\tan{\theta}}$$
Here's my attempt
$$\require{cancel}\text{Left - Right} = \frac{(1+2\sin{\theta}\cos... | $\dfrac{1+\tan t}{1-\tan t} =$
$\dfrac{\cos t + \sin t}{\cos t -\sin t}=$
$\dfrac{(\cos t+\sin t)^2}{\cos^2 t - \sin^2t}=$
$\dfrac{1+2\sin t \cos t}{\cos^2 t-\sin^2 t}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2887740",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 4
} |
Find $z^3+bz^2+c=0$ Find $b,c\in \mathbb{R}$ where $z^3+bz^2+c=0$
And $z_1=(-\sqrt{2}-\sqrt{2}i)^5$ and $z_2=(-\sqrt{2}+\sqrt{2}i)^5$
We have $z_1=(-\sqrt{2}-\sqrt{2}i)^5=32e^{i\frac{15\pi}{4}}$
and $z_2=(-\sqrt{2}+\sqrt{2}i)^5=32e^{-i\frac{15\pi}{4}}$
Can we conclude straight away something about $b,c$?
| Using Vieta's formulas, we get that the product $z_1 z_2 z_3$ of the roots is equal to $-c$. Hence $z_3 = -2^{-10}c$. Plugging this value into the equation you get
$$-2^{-30} c^3 + 2^{-20}bc^2 +c=0$$ As $c \neq 0$ we get the equation $-c^2+2^{10}bc + 2^{30}=0 \tag{1}.$
Also $$z_1z_2+z_1z_3+z_2z_3=2^{10}-2^{10}c(z_1+z_2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2888065",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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Find $\lim_{x \to 0 }\frac{(1+x)^{(1/2)} -1}{(1+x)^{(1/3)} -1}$ Find the limit without using L'hopital
$$\lim_{x \to 0 }\frac{(1+x)^{(1/2)} -1}{(1+x)^{(1/3)} -1}$$
I have tried multiply by the conjugate of the denominator and numerator but didn't work
Any hint would be appreciated
Thank you
| There are two conjugates you need to multiply:
\begin{align*}
\lim_{x\to0} \frac{\sqrt{1 + x} - 1}{\sqrt[3]{1 + x} - 1} &= \lim_{x\to0} \frac{\sqrt{1 + x} - 1}{\sqrt[3]{1 + x} - 1} \cdot \frac{\sqrt{1 + x} + 1}{\sqrt{1 + x} + 1} \cdot \frac{\sqrt[3]{1 + x}^2 + \sqrt[3]{1 + x} + 1}{\sqrt[3]{1 + x}^2 + \sqrt[3]{1 + x} + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2888412",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 1
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Trouble with argument in a complex number We have:
$z_1 = 2 + 2i$ and $z_2 = -1 -\sqrt{3} i$
I am asked for obtaining arg$\left(\frac{z_1}{z_2}\right)$ and arg($z_1z_2$)
As we know:
$$\theta = \tan^{-1} \left(\frac{y}{x}\right)$$
The division of complex numbers is:
$$\frac{z_1}{z_2} = z_1z_2^{-1} = \frac{z_1z_2^*}{z_... | I do not really know if you wanted this. But, try using the following steps to find the value
$$tan^{-1} \left( \dfrac{-1 - \sqrt{3}}{- 1 + \sqrt{3}} \right) = \tan^{-1} \left( \dfrac{1 + \sqrt{3}}{1 - \sqrt{3}} \right)$$
$$\therefore \tan^{-1} \left( \dfrac{-1 - \sqrt{3}}{-1 + \sqrt{3}} \right) = \tan^{-1} \left( \dfr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2889136",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Calculate correlation coefficient for discrete random variable From a population consisting of the numbers: $\lbrace 1,2 \ldots 10 \rbrace$, two samples are chosen from it without replacement. If the random variable denoting the first choice is X and the second choice is $Y$, what is the correlation coefficient ($\rho$... | Assuming $X$ is uniform on $\{1, 2, \dots, 10\}$ we have
$$
\mathbb E X = \frac{1 + 2 + \dots + 10}{10} = 5.5, \quad \mathbb{V}ar X = \frac{1^2 + \dots + 10^2}{10} - \mathbb E X^2 = 8.25.
$$
To compute the expected value of $Y$ write
$$
\mathbb E Y = \sum_{x \in [10]} \mathbb E[Y ~|~ X=x] \mathbb{P}(X = x),
$$
where ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2890628",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Solving $|z-4i|=2|z+4|$ $$\begin{align}
|z-4i|&=2|z+4|\\[4pt]
|x+yi-4i|&=2|x+yi+4|\\[4pt]
|x+i(y-4)|&=2|(x+4)+iy|\\[4pt]
\sqrt{x^2+(y-4)^2}&=2\sqrt{(x+4)^2+y^2}\\[4pt]
(\sqrt{x^2+(y-4)^2})^2&=(2\sqrt{(x+4)^2+y^2})^2\\[4pt]
x^2+y^2-8y+16&= 4(x^2+8x+16+y^2)\\[4pt]
x^2+y^2-8y+16&= 4x^2+32x+64+4y^2\\[4pt]
0&= 3x^2+3y^2+32x... | Here's another approach altogether:
$$\begin{align}
|z-4i|=2|z+4|
&\iff\left|z-4i\over z+4 \right|=2\\
&\iff{z-4i\over z+4 }=2e^{i\theta}\quad\text{for some }\theta\in\mathbb{R}\\
&\iff z={8e^{i\theta}+4i\over1-2e^{i\theta}}
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2891300",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 0
} |
Find $\lim\limits_{t\to\infty}x(t)$ if $x'= (x-y)(1-x^2-y^2)$, $y' = (x+y)(1-x^2-y^2)$ Suppose $x_0,y_0$ are reals such that $x_0^2+y_0^2>0.$ Suppose $x(t)$ and $y(t)$ satisfy the following:
*
*$\frac{dx}{dt} = (x-y)(1-x^2-y^2),$
*$\frac{dy}{dt} = (x+y)(1-x^2-y^2),$
*$x(0) = x_0,$
*$y(0) = y_0.$
I am asked to f... | We first pass to polar coordinates, viz:
$x = r\cos \theta; \; y = r \sin \theta; \tag 1$
$\dot x = \dot r \cos \theta - r\dot \theta \sin \theta; \tag 2$
$\dot y = \dot r \sin \theta + r \dot \theta \cos \theta; \tag 3$
$\begin{pmatrix} \dot x \\ \dot y \end{pmatrix} = \begin{bmatrix} \cos \theta & -r \sin \theta \\ \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2892740",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
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Find all $n$ such that $x+y$ is a power of $2$ whenever $xy=n$.
Let $n$ be a positive integer. $n$ is called a golden integer if $n$
is composite, and if we write $n$ as $n=xy$, where $x$ and $y$ are
positive integers, then $x+y$ is a power of two ($x+y=2^{r}$). Find
all golden integers.
It is obvious by defini... | If $n$ is a golden number and $n=xy$ is a decomposition with $x,y>1$, then indeed
$$x+y=2^b\qquad\text{ and }\qquad xy=2^a-1,$$
for positive integers $a$ and $b$. Note that $b<a$ because
$$2^b=x+y<xy=2^a-1.$$
Suppose $a$ is prime, say $a=p$. Then $x$ and $y$ divide $2^p-1$ and hence${}^1$ $x\equiv y\equiv1\pmod{p}$. Th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2892990",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Reversing digits of power of 2 to yield power of 7 Are there positive integers $n$ and $m$ for which reversing the base-10 digits of $2^n$ yields $7^m$?
I've answered this question for powers of 2 and powers of 3 in the negative. Permutations of the digits of a number divisible by 3 yields another number divisible by ... | Some basic observations:
$n$ must be even: one has $2^n=7^m=1 \pmod{3}$ and so $n$ is even.
$n-m$ must be divisible by 3: one has $2^n=7^m=(-2)^m \pmod{9}$ and so $2^{n-m}=(-1)^m \pmod{9}$ and so $n-m$ is divisible by 3.
$n-2m$ is divisible by 10 as the following two paragraph show.
If the number of digits is even, th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2893768",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Evaluate $\int \frac{\sqrt{1+x^8} dx}{x^{13}}$
Evaluate: $\displaystyle\int \frac{\sqrt{1+x^8}}{x^{13}}dx$
My attempt:
I have tried substituting $1+x^8=z^2$ but that did not work. I also tried writing $x^{13}$ in the denominator as $x^{16}.x^{-3}$ hoping that it would bring the integrand into some form but that too d... | The substitution $x=\tan^{1/4}t$ gives $$\int\frac{\sqrt{1+x^8}}{x^{13}}dx=\int\frac{1}{4}\sin^{-4}t\cos tdt=-\frac{1}{12}\sin^{-3}t+C=-\frac{1}{12}\bigg(\frac{x^8}{1+x^8}\bigg)^{-3/2}+C.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2893991",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 0
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Can't understand how $\frac{1}{2}{ 2k+4 \choose k+2} = \frac{1}{2}(2k+4)!(k+2)!^2$ is correct I'm following my teacher's notes and I cannot understand how $\frac{1}{2}{ 2k+4 \choose k+2} = \frac{1}{2}(2k+4)!(k+2)!^2$ is correct. Shouldn't it be $\frac{1}{2}{ 2k+4 \choose k+2} = \frac{1}{2}\frac{(2k+4)!}{(k+2)!^2}$ ?
No... | Indeed by definition of the binomial coefficient
$$\frac{1}{2}\binom{2k+4}{k+2}=\frac{1}{2}\frac{(2k+4)!}{(k+2)!^2}.$$
Plugging in any value, for example simply $k=0$, shows that
$$\frac{1}{2}\frac{(2k+4)!}{(k+2)!(k+2)!}\neq\frac{1}{2}(2k+4)!(k+2)!^2,$$
so this is indeed a misprint as suggested in the comments.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2894400",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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On Lame's Theorem I was trying to study Lame's theorem but I'm a little bit stuck on one specific step... I write here the theorem and the proof:
Lame's theorem:
using the Euclidean algorithm to find the greatest common divisor of two positive integers has number of divisions less than or equal five times the number... | You can easily expand out the polynomial
$({1+\sqrt{5}})^{5}$ and confirm it is greater than $10*2^{5}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2895532",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to calculate the limit $\lim\limits_{n\to\infty} \int\limits_0^1 \frac{n(2nx^{n-1}-(1+x))}{2(1+x)}\,dx$? How to calculate the limit $\lim\limits_{n\to\infty}\displaystyle\int_0^1\dfrac{n(2nx^{n-1}-(1+x))}{2(1+x)} dx$?
I have to calculate the limit when solving
Find $a,b$ for $\displaystyle\int_0^1 \dfrac{x^{n-1}}{... | Your evaluation of $a$ is correct. As regards $b$, note that, by integration by parts applied twice, we have that
$$\begin{align}
\int_0^1 \dfrac{x^{n-1}}{x+1} dx&=\frac{1}{2n}+\frac{1}{n}\int_0^1 \frac{x^{n}}{(x+1)^2} dx
\\
&=\frac{1}{2n}+\frac{1}{4n(n+1)}+\frac{2}{n(n+1)}\int_0^1 \frac{x^{n+1}}{(x+1)^3} dx.\end{align... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2895891",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
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How do I show that $\cos^4x=\frac{1}{8}\cos(4x)+\frac{1}{2}\cos(2x )+\frac{3}{8}$ I know how to prove that
$$\cos^2x=\frac{1}{2}+\frac{1}{2}\cos(2x)$$
by substituting $\cos(2x)$ with $2\cos^2x-1$ according to the double angle identity
$$\cos(2x)=2\cos^2x-1$$
However, how do I do that for $\cos^4x$?
Do I do it by wri... | As an alternative we can use that
$$\cos x=\Re(e^{ix})$$$$\implies \cos 4x=\Re(e^{i4x})=\Re[(\cos x+i\sin x)^4]=\cos^4 x-6\cos^2x \sin^2x+\sin^4x$$
that is
$$\cos 4x=\cos^4 x-6\cos^2x \sin^2x+\sin^4x$$
$$\cos 4x=\cos^4 x-6\cos^2x (1-\cos^2 x)+(1-\cos^2 x)^2$$
$$\cos 4x=\cos^4 x-6\cos^2x+6\cos^4x+1-2\cos^2x+\cos^4 x$$
$... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find value of $\prod(1+\alpha^2)$ if $\alpha,\beta,\gamma,\delta$ are the roots of equation $x^4+4x^3-6x^2+7x-9$ then find $\prod(1+\alpha^2)$
I know
*
*$\sum \alpha=-4$
*$\sum \alpha\beta=-6$
*$\sum \alpha\beta\gamma=-7$
*$\alpha\beta\gamma\delta=-9$
I know the sum and products and other things about the roots... | Transforming roots from $\alpha$ to $1+\alpha^2$,
$$y=1+\alpha^2$$
$$\alpha=\sqrt{y-1}$$
As $\alpha$ satisfies our given polynomial, Substitute $\alpha$ into that.
We get new equation with roots $1+\alpha^2,1+\beta^2,\ldots$
$$(x-1)^2+4(x-1)\sqrt{x-1}-6(x-1)+7\sqrt{x-1}-9=0$$
Rearranging and squaring,
$$(x-1)^2(x-7)^2+... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "5",
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The four digit number $AABB$ can be divided by $6$ without remainder. Determine the greatest and least value of $AABB$.
The four digit number $AABB$ can be divided by $6$ without remainder. Determine the greatest and least value of $AABB$.
On the condition that a number can be divided by $6$, it must be divisible by ... | You have shown that we must have $B$ even and $A+B$ divisible by $3$. To make $AABB$ as large as possible, we'd like to have $A=9$ If this is so, can we find an even number $B$ such that $3|(9+B)?$ Similarly, to make $AABB$ as small as possible, we'd like to have $A=1$. What are the choices for $B?$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How do you calculate this infinite series $\sum_{n=0}^\infty \frac{\binom{2n+1}{n-p}}{(4^n)(2n+1)(1.25)^{2n+1}}$?
Evaluate the infinite series $$\sum_{n=0}^\infty \frac{\binom{2n+1}{n-p}}{(4^n)(2n+1)(1.25)^{2n+1}}$$
On a calculator, for $p=0$ this tends to $1$. For $p=1$ it tends to $\frac{1}{12}$. For all values of... | Hint. Note that
$$\begin{align}\sum_{n=0}^\infty \frac{\binom{2n+1}{n-p}}{(4^n)(2n+1)(1.25)^{2n+1}}&=
2\sum_{n=0}^\infty \frac{\binom{2n+1}{n+p+1}}{(2n+1)(2.5)^{2n+1}}\\
&=\frac{4}{5}\sum_{n=0}^\infty \frac{\binom{2n}{n+p}x^n}{n+p+1}
\end{align}$$
with $x=4/25$.
For $p=0$, recall that, for $|x|<1/4$,
$$\sum_{n=0}^{\inf... | {
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Prove that if $a+b+c+d=4$, then $(a^2+3)(b^2+3)(c^2+3)(d^2+3)\geq256$
Given $a,b,c,d$ such that $a + b + c + d = 4$ show that $$(a^2 + 3)(b^2 + 3)(c^2 + 3)(d^2 + 3) \geq 256$$
What I have tried so far is using CBS:
$(a^2 + 3)(b^2 + 3) \geq (a\sqrt{3} + b\sqrt{3})^2 = 3(a + b)^2$
$(c^2 + 3)(d^2 + 3) \geq 3(c + d)^2$
$... | Alternatively, you could solve
$min_{(a, b, c, d) |a+b+c+d=4} f(a, b, c, d),$
$f$ being your term with the squares.
The convexity guarantees a global minimum. And the symmetry of the problem will yield the answer.
Edit
According to the comments, $f$ is not convex but it is subharmonic (thanks to Jack D'Aurizio), i.e.... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
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Solving an equality with 3 equations, and 3 variables Following is the question asked in a recent aptitude exam:
Given that :
$$ a+b+ab=10\\ b+c+bc=20 \\ c+a+ac=30$$
What is the value of $a+b+c+abc$ ?
I can solve it by finding the individual values by:
$$ a = \frac{10-b}{b+1} ,\\ c = \frac{20-b}{b+1}$$
Putting these ... | Just add one
Add one to each equation and factorize to get :
$$
(a+1)(b+1) = 11 \\
(b+1)(c+1) = 21 \\
(a+1)(c+1) = 31 \\
$$
Now, set $x ,y,z = a+1,b+1,c+1$ respectively.This gives $xy = 11, yz = 21,zx = 31$, so $x^2y^2z^2 = 11 \times 21\times 31$ by multiplying these.
Now divide this equation by the other equations su... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Given $\sin(t) + \cos(t) = a$, derive an expression in '$a$' for $(\cos(t))^4 + (\sin(t))^4$ I received this question from a student's trigonometry review assignment. After spending an embarrassing amount of time on it, I consulted others and learned that nobody has been able to solve this problem for multiple semester... | $$(\sin t + \cos t )^2= a^2\rightarrow \sin t\cos t=\dfrac{a^2-1}{2}$$
then
$$\cos^4t + \sin^4t=(\cos^2t + \sin^2t)^2-2\cos^2t \sin^2t=1-2\left(\dfrac{a^2-1}{2}\right)^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2905869",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 0
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Evaluating $\frac{1}{\sin(2x)} + \frac{1}{\sin(4x)} + \frac{1}{\sin(8x)} + \frac{1}{\sin(16x)}$
Evaluate
$$\dfrac{1}{\sin(2x)} + \dfrac{1}{\sin(4x)} + \dfrac{1}{\sin(8x)} + \dfrac{1}{\sin(16x)}$$
It would be tough for us to solve it using trigonometric identities. There should be strictly an easy trick to proceed.
Re... | $$\sin(A-B)=\sin A \cos B-\cos A \sin B$$
$$\frac{1}{\sin 2x}=\dfrac{\sin(2x-x)}{\sin2x\sin x}=\cot x-\cot2x$$
$$\frac{1}{\sin 4x}=\dfrac{\sin(4x-2x)}{\sin4x\sin 2x}=\cot 2x-\cot4x$$
$$\frac{1}{\sin 8x}=\dfrac{\sin(8x-4x)}{\sin8x\sin 4x}=\cot 4x-\cot8x$$
$$\frac{1}{\sin 16x}=\dfrac{\sin(16x-8x)}{\sin16x\sin 8x}=\cot 8x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2907344",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Geometric series: two ways to tackle same problem giving me issues I am trying to get a closed form expression for the following sum:
$$x + 2x^2 + 3x^3 + \cdots + nx^n$$
so by perturbation method:
$$S = x + 2x^2 + 3x^3 + \cdots + nx^n$$
$$xS= x^2 + 2x^3 + \cdots + (n-1)x^n + nx^{n+1}$$
$$(1-x)S = x + x^2 + x^3 + \... | $$S = x + 2x^2 + \cdots + nx^n$$ implies $$(1-x)S = (x + x^2 + \cdots + x^n) - nx^{n+1}.$$ This much is clear for you. If you do not already know the sum for a finite geometric series, this too can be obtained by the same perturbation method you just applied to $S$. For instance, let $$G = 1 + x + x^2 + \cdots + x^n... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Evaluate $ \lim_{x \to 0} \left( {\frac{1}{x^2}} - {\frac{1} {\sin^2 x} }\right) $ $$\lim_{x\to0}\left({\frac{1}{x^2}}-{\frac{1}{\sin^2x}}\right)$$
Using the L'Hospital Rule I obtained the value $-1/4$, but the answer is given to be $-1/3$. I can't find the mistake. Here's what I did; please point out the mistake.
\beg... | Hint: Write the function as $$\frac{\sin^2(x)-x^2}{x^4}\times \frac{x^2}{\sin^2(x)}$$ Otherwise use the Talor's expantion if you know it.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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If $\sin^8(x)+\cos^8(x)=48/128$, then find the value of $x$? If $$\sin^8(x)+\cos^8(x)=48/128,$$ then find the value of $x$? I tried this by De Moivre's theorem:
$$(\cos\theta+i\sin\theta)^n=\cos(n\theta)+i\sin(\theta)=e^{in\theta}$$
But could not proceed further please help.
| Assuming $x\in \Bbb R.$ For brevity let $c=\cos x$ and $s=\sin x.$ Let $p=c^2s^2.$ We have $$ c^8+s^8=\frac {3}{8}\iff$$ $$ 1=(c^2+s^2)^4=(c^8+x^8)+c^2s^2(4c^4+6c^2s^2+4s^4)=$$ $$=\frac {3}{8}+c^2s^2(4(c^2+s^2)^2-2c^2s^2)=$$ $$=\frac {3}{8}+c^2s^2(4-2c^2s^2)\iff$$ $$\iff(0\leq p\leq 1\land \frac {5}{16}=2p-p^2)$$ $$\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2910845",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 2
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Find sum of all integer solution of $\frac{2}{a} + \frac{3}{b} + \frac{4}{c} + \frac{5}{d} = \frac{1}{14}$ Given $\frac{2}{a} + \frac{3}{b} + \frac{4}{c} + \frac{5}{d} = \frac{1}{14}$ , which $a,b,c ,d$ are an positive integer , all solution of $a,b,c,d$ are members of Set "S" ,then find sum all of members in se... | For Question 2, there are infinitely many integer solutions. Take $$(a,b,c,d):=(4,-7,5k,-4k)\,,\text{ where }k\in\mathbb{Z}_{\neq 0}\,.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2911617",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How can I prove this sequence convergent? I am trying to prove the sequence $(u_n)$ so that $u_1=1$ and $u_{n+1}=1+\dfrac{1}{u_n}$ is convergent. With some first items, I guess that the subsequence $(u_{2n})$ is decreasing and the subsequence $(u_{2n-1})$ is increasing, but I can't proof this. How can I proof?
| Hint: We can write
$$u_{3} = 1 + \frac{1}{1 + \frac{1}{u_1}} = 1 + \frac{u_1}{u_1+1} =
\frac{u_1+1}{u_1+1} + \frac{u_1}{u_1+1}
= \frac{2u_1+1}{u_1+1}$$
$$u_{4} = 1+\frac{u_1+1}{2u_1+1} =
\frac{2u_1+1}{2u_1+1} + \frac{u_1+1}{2u_1+1} =
\frac{3u_1+2}{2u_1+1}$$
$$u_{5} = 1 + \frac{2u_1+1}{3u_1+2} =
\frac{3u_1+2}{3u_1+2} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2913302",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
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Find all non-negative integer $n$ that satisfy $f(x+1)+f(x-1)=\sqrt nf(x)$
Find all non-negative integer $n$ that there exists a non-periodic function $f:\Bbb R->\Bbb R\quad f(x+1)+f(x-1)=\sqrt nf(x)\forall x$.
My attempt:
$f(x+2)+f(x)=\sqrt n\cdot f(x+1)$
$\sqrt n \cdot (f(x+1)+f(x+3))=n\cdot f(x+2)$
$f(x+2)+f(x+4... | Hint.
Assuming $f(x) = \lambda^x$ we have
$$
\lambda^{x+1}+\lambda^{x-1} = \sqrt{n}\lambda^x\Rightarrow \left(\lambda+\frac{1}{\lambda}=\sqrt n\right)\lambda^x
$$
hence
$$
\lambda = \frac 12\left(\sqrt n\pm\sqrt{n-4}\right)
$$
| {
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"url": "https://math.stackexchange.com/questions/2917573",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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$\frac {x_1^3}{y_1}+\frac {x_2^3}{y_3}+\ldots+\frac {x_n^3}{y_n}\leq \frac {a^4+b^4}{ab (a^2+b^2)}(x_1^2+x_2^2+\ldots+x_n^2) $
Let $b>a>0$ and $x_1, x_2,\ldots,x_n,y_1, y_2,\ldots,y_n\in [a,b]$. If $$x_1^2+x_2^2+\ldots+x_n^2=y_1^2+y_2^2+\ldots+y_n^2\,,$$
then
$$\frac {x_1^3}{y_1}+\frac {x_2^3}{y_3}+\ldots+\frac {... | Note that an equality condition is possible. Condider $n$ even. Let half of the $x_i$ equal to $a$ and let the corresponding $y_i$ equal to $b$. For the other half of the variables, exchange $a$ and $b$. Then the condition $x_1^2+x_2^2+\ldots+x_n^2=y_1^2+y_2^2+\ldots+y_n^2 = \frac{n}2 (a^2 + b^2)$ holds. Inserting in ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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I was asked to evaluate the determinant of the I was asked to evaluate the determinant of the $n$x$n$ matrix
$$ A=
\begin{bmatrix}
x & 2 & 2 & \cdots & 2 \\
2 & x & 2 & \cdots & 2 \\
2 & 2 & x & \cdots & 2 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
2 & 2 & 2 & \cdots & x \\
\end{bmatrix}
$$
I tried starting from ... | We add all rows to the first one, then factor $x+2(n-1)$ from this first row, than use it (doubled) to eliminate in the other rows. With this strategy we are done quickly:
$$
\begin{aligned}
\det A
&=
\begin{vmatrix}
x & 2 & 2 & \cdots & 2 \\
2 & x & 2 & \cdots & 2 \\
2 & 2 & x & \cdots & 2 \\
\vdots & \vdots & \vdots ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2919579",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Find the plane $P$ passing through the origin such that the three planes $P$, $P_1=(x+y+z=1)$ and $P_2= (x-y+z=2)$ meet along a line in R3. What I did is find the equation of a line of the intersection like
$$(x+y+z=1)+t(x-y+z=2)=0$$
Since it passes through the origin, I substitute $x=0$, $y=0$, $z=0$ into the equation... | Adding x + y + z = 1 and x - y + z = 2 gives 2x + 2z = 3 or z = 3/2- x. putting that into x+ y+ z= 1, x+ y+ 3/2- x= y+ 3/2= 1 so y= -3/2. Taking t= x as parameter, the line of intersection is x= t, y= -3/2, z= 3/2- t. Any plane that contains the origin can be written as z= Ax+ By. In order that it contains that lin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2924426",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Extend $f(x)=\frac{6x^3-11x^2-16x-4}{2x^2-5x-2}$ in a way it is continuous $\forall x \in \mathbb{R}$ Problem
If it is possible to extend $$f(x)=\frac{6x^3-11x^2-16x-4}{2x^2-5x-2}$$ in a way it is continuous $\forall x \in \mathbb{R}$. Extend all $x \not\in X$ where $X$ is domain of $f$. Also define domain $X$.
Atte... | You can simply perform polynomial division to get
$$6x^3 -11x^2 -16x -4 = 3x(2x^2-5x-2) + 4x^2 - 10x-4 =\\
= 3x(2x^2-5x-2) + 2(2x^2 - 5x-2) = (3x+2)(2x^2 - 5x-2)$$
This explains the fact that the function has finite limits at points where the denominator $2x^2-5x-2$ is zero.
Thus, the desired extension is
$$f(x)=3x+2$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2927664",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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How to find the value of $r$ such that $\frac{1}{\binom{9}{r}} - \frac{1}{\binom{10}{r}} = \frac{11}{6\binom{11}{r}}$? Find the value of $r$:
$$\frac{1}{\dbinom{9}{r}} - \frac{1}{\dbinom{10}{r}} = \frac{11}{6\dbinom{11}{r}}$$
I'm not sure where I should take this problem in order to isolate $r$. I seem to get many fact... | Since
$$\binom{n}{k} =
\begin{cases}
\dfrac{n!}{k!(n - k)!} & \text{if $0 \leq k \leq n$}\\[2 mm]
0 & \text{if $k > n$}
\end{cases}$$
we require that $k \leq 9$ since otherwise we would be dividing by $0$.
\begin{align*}
\frac{1}{\binom{9}{k}} - \frac{1}{\binom{10}{k}} & = \frac{11}{6\binom{11}{k}}\\
\frac{1}{\dfrac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2929538",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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The probability of getting a 7 This problem is probably too easy for all you math geniuses out there, but I am having trouble understanding it, so here it is:
An athlete has 10 cards, numbered 1 to 10. Every day, he picks a card randomly and does that number of pushups. He doesn't replace the card, and picks until all... | First calculate the relative probabilities of getting a $2, 3$ or $4$ card sum of $12$.
These are:$$\frac{4}{\binom{10}{2}};\frac{7}{\binom{10}{3}}; \frac{2}{\binom{10}{4}} = \frac{4}{45}; \frac{7}{120}; \frac{2}{210} = \frac{448}{5040}; \frac{294}{5040}; \frac{48}{5040}$$
Given the sum is $12$, the actual probabilitie... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2930879",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Recurrence relation for Pell's equation $x^2-2y^2=1$ I am wondering how to find the recurrence relation for solutions for $x$ in the Pell's equation $x^2-2y^2=1$.
I know the formula for the general term.
It is $$\frac{(3+2\sqrt2)^n+(3-2\sqrt2)^n}{2}$$ for $x_n$, the $n^{th}$ smallest solution for $x$.
Any help would be... | The fundamental solution is $9-8 = 1,$ meaning $3^2 - 2 \cdot 2^2 = 1.$ As a result, we have the matrix
$$
A =
\left(
\begin{array}{cc}
3 & 4 \\
2 &3
\end{array}
\right)
$$
which solves the automorphism relation, $A^T H A = H,$ where
$$
H =
\left(
\begin{array}{cc}
1 & 0 \\
0 & -2
\end{array}
\right)
$$
That is
$$ (3x+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2932750",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Eigenvalues of $aI + bJ$ Just as context, this problem arose when I was looking at the adjacency matrices of moore graphs of diameter 2.
Given that $I$ is the identity matrix and $J$ is the all 1 matrix. I have constructed some matrix $aI + bJ$. I am completely stuck in terms of how to find the eigenvalues of this matr... | here is a matrix that shows a basis of the eigenvectors as columns, these being perpendicular to each other as well. Below is the 10 by 10 case. In smaller dimension, take the upper left square corner.
$$
\left( \begin{array}{rrrrrrrrrr}
1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 \\
1 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2938635",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Prove that $n$ is divisible by $6$.
If the quadratic equations $x^2-mx+n=0$ and $x^2+mx-n=0$ both have integral roots, prove that $6|n$.
I've proved that $3|n$, and that $2|n$ for odd $m$, but I can't seem to prove it for even $m$.
Please help.
| Hint: Since both eqaution have intergal solution we have $$m^2-4n=a^2\;\;\;{\rm and}\;\;\;\;m^2+4n =b^2$$
for some integers $a,b$.
So $2m^2 =a^2+b^2$ implies
a) if $3\mid m$ then $3\mid a$ and $3\mid b$ implies $3\mid a^2-b^2 =4n$ so $3\mid n$
b) if $3\nmid m$ then $a^2\equiv_3 1$ and $b^2\equiv_3 1$ implies $$4n =a^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2939037",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Is there any mistake in my approach for solving $ \int_0^{\pi/2} \frac{ \cos x}{3 \cos x + \sin x} \, dx $ ?? I had to evaluate this integral .
$$
\int_0^{\pi/2} \frac{\cos x}{3 \cos x + \sin x} \, dx
$$
Here is how I proceeded
Dividing $N^r$ And $D^r$ by $\cos^3 x$
$$
\int_0^{\pi/2} \frac{ \sec^2 x}{3 \sec^2 x + \tan... | $$\int_0^\infty \frac{ 1 }{(1+t^2)(t+3)} \, dt =\class{steps-node}{\cssId{steps-node-1}{\dfrac{1}{10}}}{\displaystyle\int_0^\infty}\dfrac{1}{t+3}\,\mathrm{d}t-\class{steps-node}{\cssId{steps-node-2}{\dfrac{1}{10}}}{\displaystyle\int_0^\infty}\dfrac{t-3}{t^2+1}\,\mathrm{d}t\\=\dfrac{\ln\left(\left|t+3\right|\right)}{10}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2941008",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
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Show that $\frac{\sqrt{2}+\sqrt{2+\sqrt{3}}}{\sqrt{2}-\sqrt{2+\sqrt{3}}}=-3-2\sqrt{3}$
Show that $$\frac{\sqrt{2}+\sqrt{2+\sqrt{3}}}{\sqrt{2}-\sqrt{2+\sqrt{3}}}=-3-2\sqrt{3}.$$
My attempt: I could find a way to develop the LHS by un-nesting the double radical, figuring out that
$$\sqrt{2+\sqrt{3}}=\frac{\sqrt{6}+\sqr... | The simple, "natural" solution is using direct radical denesting, as noted by the OP. But any algebraic solution will likely involve some form of "denesting" eventually, as noted by Misha Lavrov in a comment. For example, the following is a rather convoluted, very "artificial" solution using polynomial resultants.
Let ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2941552",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Determining the minimum value of the function $y = x + 2\sqrt{x^2 - \sqrt{2}x + 1}$ I am curious whether there is an algebraic verification for $y = x + 2\sqrt{x^2 - \sqrt{2}x + 1}$ having its minimum value of $\sqrt{2 + \sqrt{3}}$ at $\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{6}}$. I have been told the graph of it is that ... | Here is a "non-calculus" way that plays the whole show back to AMGM. It is a bit cumbersome but works.
I prefer giving all stepwise substitutions to show how to bring the whole expression back to hyperbolic functions where AMGM suddenly gives all. The basic idea behind it is that $\cosh t = \sqrt{\sinh^2 t + 1}$:
$$\co... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2942263",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 2
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Find the equation in spherical coordinates of $x^2 + y^2 – z^2 = 4$. Find the equation in spherical coordinates of $x^2 + y^2 – z^2 = 4$.
$$\begin{align}
x^2 + y^2 &= r^2\sin^2(\theta)\\
z^2 &= r^2 \cos(\theta) \\
x^2 + y^2 - z^2&=r^2(\sin^2(\theta) - \cos^2(\theta)) = 4
\end{align}$$
Thus, $$-r^2(\cos(2\theta)) = 4$$
... | The equation given is correct, but we can simplify further. Multiply by $-\sec 2\theta$ and take the square root to solve for $r$ which is $>0$. Thus $r=2\sqrt{-\sec(2\theta)}$.
Note that the result appears to be the square root of a negative number ... unless the secant function is negative. Recalling where the cos... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2943176",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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The ways of covering a $4\times 4$ square by $1\times 2$ colored dominoes I'm stuck with this question
We have eight $1\times 2$ tiles that each one of them has one $1\times 1$ blue square and one $1\times 1$ red square. We want to cover a $4\times 4$ area with these tiles in a way that every row and every column of t... | It actually should not be too bad (if a little tedious) to compute case-by-case with judicious use of symmetry.
I'll use a $4 \times 4$ matrix with entries $b$ (for blue), $r$ (for red) and $\cdot$, to denote the number of tilings of the $\cdot$ spaces given the colours of the $b$ and $r$ spaces. By symmetry,
the cas... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2943291",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 1,
"answer_id": 0
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Finding the complex square roots of a complex number without a calculator
The complex number $z$ is given by $z = -1 + (4 \sqrt{3})i$
The question asks you to find the two complex roots of this number in the form $z = a + bi$ where $a$ and $b$ are real and exact without using a calculator.
So far I have attempted to ... | That the square roots are $\pm(\sqrt 3 + 2i)$ can be seen by elementary algebra and trigonometry as follows.
\begin{align}
& \left|-1 + i4\sqrt 3\right| = \sqrt{(-1)^2 + (4\sqrt 3)^2 } = 7. \\[10pt]
\text{Therefore } & -1+i4\sqrt 3 = 7(\cos\varphi + i\sin\varphi). \\[10pt]
\text{Therefore } & \pm\sqrt{-1+i4\sqrt 3} = \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2943851",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 4
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A hotel has 6 single rooms, 6 double rooms and 4 rooms in which 3 persons can stay. In how many ways can 30 persons be accommodated in this hotel? I wanted to know if my answer and the method I used is correct or not...
First, we pick $6$ out of $30$ people for the $6$ single rooms = $^{30}C_6$
They can be in any of ro... | As stated in the comments, the flaw in your answer is that while you treated the rooms as distinguishable for the single rooms, you did not do so for the rooms that accommodate more than one person.
In how many ways can a hotel with six single rooms, six double rooms, and four rooms in which three people can stay assi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2945660",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Probability kth outcome is not a repetition
A fair die is tossed $k$ times. What is the probability that the kth
outcome is ${\bf not}$ a repetition?
attempt.
First, the size of our sample space is $6^k$. We only have six possible outcomes, so for any outcome, say $O$, that occur at the kth try, we want the previous ... | Your approach looks sensible, but would lead to $\dfrac{5^{k-1}}{6^{k-1}}$ for one die
Another way to get there would be consider the possible outcomes $1,2,3,4,5,6$ for the $k$th die and combine the probabilities of these with the conditional probabilities of the previous dies being different so $\frac16\left(\frac56\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2946763",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Triangle numbers that are squares of triangle numbers. What are the triangle numbers the are squares of other triangle numbers? I have found $1^2=1$ and $6^2=36$, but other than these examples I can't find any other triangle numbers that are squares of other triangle numbers, and I used a program to check this idea int... | Note that $n$ and $n+1$ are coprime, as are $k$ and $k+1$. We can have both sides zero if $n=0,-1, k=0,-1$. Otherwise we must have either $n=k^2,2(n+1)=(k+1)^2$ or $n+1=k^2,2n=(k+1)^2$. The first gives
$$n=k^2\\2(n+1)=(k+1)^2\\2k^2+2=k^2+2k+1\\k^2-2k+1=0\\k=1\\n=1$$ while the second gives $$n+1=k^2\\2n=(k+1)^2\\2k^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2948216",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Maximum and minimum absolute value of a complex number
Let, $z \in \mathbb C$ and $|z|=2$. What's the maximum and minimum value of $$\left| z - \frac{1}{z} \right|? $$
I only have a vague idea to attack this problem.
Here's my thinking :
Let $z=a+bi$
Exploiting the fact that, $a^2+b^2=4$
We get $z-\dfrac{1}{z}=a-\df... | Hint: $z - \frac 1z = z - \frac {\overline z}{z\overline z} = z - \frac {\overline z}{|z|^2} = z - \frac {\overline z}4 = \frac 34 Re(z) -i\frac 54 Im(z)$[1]
Let $Re(z) = a$ and $Im(z) = b$ which can be any values so that $a^2 + b^2 = 4$.
So you are being asked to find the maximum and minimum possible values of $\sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2948812",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Polar integral $\int_{r=0}^{\infty} \frac{b r dr}{(r^2+b^2)^{(3/2)}(r^2+a^2)^{(1/2)}}$ We need to solve above integral which I obtained after manipulation. Now I used $r^2 = u $ to get:
$$\int_0^\infty \frac{bdu}{2 (u+b^2)^{(3/2)}(u+a^2)^{(1/2)}}$$
which is converted to:
$$\int_0^\infty \frac{bdu}{2 (u+b^2) \sqrt{\left... | HINT:
Denote $I$ the given integral. Integration by parts leads to a function and an integral that is a constant multiple of $I.$
Solution (not accomplished, based on the above hint)
For
$$I(b,a)=\int_{r=0}^{\infty} \frac{b r\; dr}{(r^2+b^2)^{(3/2)}(r^2+a^2)^{(1/2)}}$$ apply integration by parts considering $u(r)=\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2955315",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Find the remainder when $13^{13}$ is divided by $25$. Find the remainder when $13^{13}$ is divided by $25$.
Here is my attempt, which I think is too tedious:
Since $13^{2} \equiv 19 (\text{mod} \ 25),$ we have $13^{4} \equiv 19^{2} \equiv 11 (\text{mod} \ 25)$ and $13^{8} \equiv 121 \equiv 21 (\text{mod} \ 25).$
Finall... | $\!\bmod 25\!:\ 13\equiv \overbrace{2^{\large -1}\!\equiv 3^{\large -3}}^{\Large 2\ \, \equiv\,\ 3^{\LARGE 3}}$ $\Rightarrow 13^{\large 13}\!\equiv 3^{\large -39}\!\equiv 3,\ $ by $\ 3^{\large 40}\!\equiv\!\!\! \overbrace{(3^{\large \color{#c00}{20}})^{\large 2}\!\equiv 1}^{\large\quad\ \ \color{#c00}{20}\ =\ \phi(25)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2956834",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
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solve for $x$: $(x^4-13x^2+36)^4+|x^2+x-6|+\sqrt{x^3-7x+6}=0$ There is an equation that I think it is complicated ,a little!
$$(x^4-13x^2+36)^4+|x^2+x-6|+\sqrt{x^3-7x+6}=0$$
Actually we must solve for $x$ here.
I want you to hint me how can I simplify the equation and solve it.
| Hint:
Observe that for real $x$
each of the summands must be $=0$
Now $x^2+x-6=(x+3)(x-2)$
$$x^4-13x^2+36=(x^2-4)(x^2-9)$$
$$x^3-7x+6=x^3-2^3-7(x-2)$$ or try divide $x^3-7x+6$ by $x-2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2963146",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Need a hint on how to solve this inequality I want to show that, whenever $\frac{u_1^2}{p^2}+\frac{u_2^2}{q^2} \leq 1$ and $\frac{v_1^2}{p^2}+\frac{v_2^2}{q^2} \leq 1$, then
$$
\frac{(\lambda \, u_1 + (1 - \lambda) \, v_1)^2}{p^2} + \frac{(\lambda \, u_2 + (1 - \lambda) \, v_2)^2}{q^2} \leq 1
$$
for all $0 \leq \lamb... | Using simpler variable names,
we are given
$\frac{a^2}{p^2}+\frac{c^2}{q^2}
\le 1$
and
$\frac{b^2}{p^2}+\frac{d^2}{q^2}
\le 1$.
$\begin{array}\\
\frac{(ra + (1 - r) b)^2}{p^2} + \frac{(rc + (1 - r)d)^2}{q^2}
&=\frac{r^2a^2+2r(1-r)ab+(1-r)^2b^2}{p^2} + \frac{r^2c^2+2r(1-r)cd+(1-r)^2d^2}{q^2}\\
&=r^2(\frac{a^2}{p^2}+\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2963819",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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find the answer in terms of $a$ and $b$ only ($a, b$ are roots of $\ x^4 + x^3 - 1 = 0$ If $a$ and $b$ are the two solutions of $\ x^4 + x^3 - 1 = 0$ , what is the solution of $\ x^6 + x^4 + x^3 - x^2 - 1 = 0$ ?
Well I am not able to eliminate or convert $\ x^6$. Please help.
| Easy to see that $ab<0$.
Let $ab=x$.
Thus, we have $$(a^4+a^3)(b^4+b^3)=1$$ or
$$a^3b^3(ab+a+b+1)=1$$ or
$$x^3(x+a+b+1)=1$$ or
$$a+b=\frac{1-x^3-x^4}{x^3}.$$
Also, we have
$$\frac{a^4+a^3-1-(b^4+b^3-1)}{a-b}=0$$ or
$$a^3+a^2b+ab^2+b^3+a^2+ab+b^2=0$$ or
$$(a+b)^3-2ab(a+b)+(a+b)^2-ab=0,$$ which gives $$\left(\frac{1-x^3-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2964172",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
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Solve $\sqrt{\frac{4-x}x}+\sqrt{\frac{x-4}{x+1}}=2-\sqrt{x^2-12}$ The equation is
$$\sqrt{\frac{4-x}x}+\sqrt{\frac{x-4}{x+1}}=2-\sqrt{x^2-12}$$
I tried squaring both left side and right side then bringing them to same numerator but got lost from there ... any ideas of how should this be solved?
I got to:
$$2\cdot\sqr... | Left side exist only iff $${\frac{4-x}x}\geq 0\iff x\in (0,4]$$
and $${\frac{x-4}{x+1}}\geq 0\iff x\in (-\infty,-1)\cup [4,\infty)$$
So the only legitimate value for left side is $4$ which works.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2964275",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
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Glueing two affine curves along a map Let $C = V(y^2 - x^4 - 1) \subset \Bbb A^2$ and set
$C_0 = C \times \{0\}, C_1 = C \times \{1\} \subset \Bbb A^3$.
Consider the space $X$ obtained by quotienting $C_0 \sqcup C_1$ but the relation $((x, y) ; 0) \sim (1/x, y/x^2)$ for all $y$ and all $x \neq 0$.
According Miranda II... | Let $S = \frac{k[X,Y,Z]}{(Y^2 - (X^4 + Z^4))}$, where we give $Y$ weight $2$, and $X$ and $Z$ weight $1$. Then the polynomial $F = Y^2 - (X^4 + Z^4)$ is weighted homogeneous of degree $4$, so the quotient $S$ is a graded ring. Thus the completed curve $\overline{C}$ is given by the equation $Y^2 = X^4 + Z^4$ in the wei... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2965725",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
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Tips on additional methods to solve inequality involving multiple variables So I want some feedback about my proof of the following implication:
Let $a,b,c,d\in\mathbb{Z}$ where $a,b,c,d>0$. Prove that if $\frac{a}{b}<\frac{c}{d}$, then
$$\frac{a}{b}<\frac{a+c}{b+d}<\frac{c}{d}$$
My attempt went as follows:
If $\frac... | I think your proof is right, but I like the following way.
$$\frac{a+c}{b+d}-\frac{a}{b}=\frac{bc-ad}{b(b+d)}>0$$ and
$$\frac{c}{d}-\frac{a+c}{b+d}=\frac{bc-ad}{d(b+d)}>0.$$
| {
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"url": "https://math.stackexchange.com/questions/2966689",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find the value of $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$
Suppose $(a, b, c)\in\Bbb R^3$, $a,b,c$ are all nonzero, and we have $
\sqrt{a+b}+\sqrt{b+c}=\sqrt{c+a}$. Find the value of $$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}.$$
Here is my attempt:
$$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{bc+ac+ab}{abc}.$$
I am having tr... | Square both sides to get
$$b=-\sqrt{(a+b)(b+c)}$$
square again
$$0=ab+ac+bc$$
divide by $abc$
and finish.
| {
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"source": "stackexchange",
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Show that $z_n$ is a perfect square if $z_0 = z_1 = 1$ and $z_{n+1} = 7z_n − z_{n−1} − 2$ Let $z_0 = z_1 = 1$ and $$z_{n+1} = 7z_n − z_{n−1} − 2$$ for all positive integers $n$. How is it possible to show that $z_n$ is a perfect square for all $n$?
| I answered the last one this way, let me make it easier reading
LEMMA if $$w_{n+3} - (M+1)w_{n+2} + (M+1)w_{n+1} - w_n = 0,$$ then $w_{n+2} - M w_{n+1} + w_n$ is CONSTANT
PROOF: $$ (w_{n+3} - M w_{n+2} + w_{n+1}) - w_{n+2} + M w_{n+1} - w_n = 0 \; , $$
$$w_{n+3} - M w_{n+2} + w_{n+1} = w_{n+2} - M w_{n+1} + w_{n} $$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2970014",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
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Continuous Bijection From $S^1$ to $[0,2\pi)$ I came up with the following (I believe) continuous bijection $\theta: S^1 \rightarrow [0,2\pi)$:
$\begin{align} \theta(x,y) & =
\begin{cases}
\arctan(\frac{y}{x}) & x > 0, 0 \leq y < 1 \\
\frac{\pi}{2} & x = 0, y = 1 \\
\arctan(\frac{-x}{y}) + \frac{\p... | you cannot find such a bijection since the image of a compact set by a continuous function is compact, $S^1$ is compact and $[0,2\pi)$ is not compact.
| {
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Evaluate $\lim_{n\to \infty}\left(\frac{1}{\sqrt{n(n+1)}}+\frac{1}{\sqrt{(n+1)(n+2)}}+\cdots+\frac{1}{\sqrt{(2n-1)2n}}\right)$ I want to evaluate $$\lim_{n\to \infty}\left(\frac{1}{\sqrt{n(n+1)}}+\frac{1}{\sqrt{(n+1)(n+2)}}+\cdots+\frac{1}{\sqrt{(2n-1)2n}}\right).$$
I have computed
\begin{align*}
a_{n+1}-a_n & = \frac... | HINT
The hint provided by Chinnapparaj R is the key point, to evaluate the bounding sum we can use that
$$\frac1n+\frac1{n+1}+\ldots+\frac1{2n}=\sum_{k=1}^n \frac{1}{n+k}=\frac1n \sum_{k=1}^n \frac{1}{1+\frac kn}$$
which is a Riemann's sum.
| {
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Sum of the series $(i)$ and $(ii)$ Find the sum of the series: $(i)1+\frac{2}{9}+\frac{2.5}{9.18}+\frac{2.5.8}{9.18.27}+\cdots +\infty$$(ii)1+\frac{3}{4}+\frac{7}{16}+\frac{13}{64}+\cdots +\infty$The answer providing my book is :$(i)\frac{9}{4}^{\frac{1}{3}},(ii)\frac{8}{3}$.i couldn't think of how to start.Because i d... | I still don't know what it's supposed to be .
1) I supposed that numerator always increased by next even number : $3 = 1 + \color{blue}{2}$, then $7 = 3 +\color{blue}{4}$ and so on.
And denominator is $4^{k}$.
So we have $\displaystyle \sum_{k=0}^{\infty} \frac{k(k+1)+1}{4^{k}} = \sum_{k=0}^{\infty} \frac{k(k+1)}{4^{k}... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Finding the laurent series with an added imaginary component Find the laurent series of
$$f(z)=\frac{i}{z^2-iz+2}$$ with the values of the region $a,b$ in which it is valid $$ a<|z-1|<b $$
My attempt at a solution:
I did a partial fraction expansion and found
$$ f(z)=\frac{1}{3(z-2i)} - \frac{1}{3(z+i)} $$
Expanding... | You have;
$$f(z)=\frac{1}{3(1-2i)}\frac{1}{(1-(-\frac{(z-1)}{1-2i})}-\frac{1}{3(1+i)}\frac{1}{(1-(-\frac{z-1}{1+i}))}$$
Expanding in a geometric series gives;
$$f(z)=\frac{1}{3(1-2i)}\sum_{n=0}^{\infty}\left(\frac{z-1}{2i-1}\right)^{n} -\frac{1}{3(1+i)}\sum_{n=0}^{\infty}\left(-\frac{z-1}{1+i}\right)^{n}$$
A geometric ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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Functional equation - Cyclic Substitutions Please help solve the below functional equation for a function $f: \mathbb R \rightarrow \mathbb R$:
\begin{align}
&f(-x) = -f(x) , \text{ and } f(x+1) = f(x) + 1, \text{ and } f\left(\frac 1x\right) = \frac{f(x)}{x^2} \\
&\text{ for all } x \in \mathbb R \text{ and } x \ne 0 ... | I don't know if this is what is meant with "cyclic substitutions", but it is a solution.
First, we observe that
$$
f(0)=0, f(1)=1, f(-1)=-1
$$
has to be true. Also, it suffices to determine $f(x)$ for $x>0$, because the rest follows from the condition $f(-x)=-f(x)$.
Let $x>0$.
Applying some of the conditions, we have
$... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Counting Credit Cards The credit cards (VISA CARDS & MASTER CARDS) numbers have the following properties;
Let $N$ be the card number
*
*$N$ has $16$ digits; $N = a_{1}a_{2}a_{3}...a_{16}$ where $a_{k}$ is the $k$-th digit of $N$
*$a_1$ $\neq 0$
*$2(a_1+a_3+a_5 + ... + a_{15}) + (a_{2}+a_{4}+a_{6}+...+a_{16})$ $+$... | The answer is $9\cdot10^{14}$. That is, there are $9$ choices for $a_1$ followed by $10$ choices for each of $a_2$ to $a_{15}$, and ending with the unique value
$$a_{16}=-2(a_1+a_3+\cdots+a_{15})-(a_2+a_4+\cdots+a_{14})-\text{count}(\ge5;a_1,a_3\ldots,a_{15})\mod 10$$
Remark: The OP's formula looks different from the ... | {
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find all integers $n$ such that $2^{n-1}*n+1$ is a perfect square. Clearly $n=0$
and i found that also $n=5$ gives a perfect square
And By representing the two functions , we found that there are only two solutions that are $n=0,5$
But I don't know how to prove that using elementary number theory.
| Robert Israel has already provided a nice answer, but it seems that you don't get it.
The idea in this answer is the same as the one in his answer. One can make his argument a bit simpler.
Let us prove that $2^{n-1}n+1$ is not a perfect square for $n\gt 5$.
Suppose that there exist positive integers $n,a$ such that $n\... | {
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Determine the set of all $x∈ℝ$ satifying $x^2 + |x-1| ≤ 1$ Determine the set of all $x∈ℝ$ satifying $x^2 + |x-1| ≤ 1$
My working out so far:
$|x-1|$ =
$x-1$ when $x≥1$
$1-x$ when $x<1$
Then we compute $x^2 + |x-1|$ as follows:
$x^2 + |x-1|$ =
$x^2 +x -1$ when $x≥1$
$x^2 - x + 1$ when $x<1$
So we have the following 2 ... | You may rewrite the inequality first:
$$x^2 + |x-1| \leq 1 \Leftrightarrow |1-x| \leq 1-x^2 = (1-x)(1+x)$$
*
*$\color{blue}{x= 1}$ is an obvious solution.
For $x \neq 1$ you get
$$|1-x| \leq (1-x)(1+x) \stackrel{\color{blue}{x \neq 1}}{\Longleftrightarrow} \begin{cases} 1 \leq -(1+x) & \mbox{ for } x > 1 \\ 1\leq ... | {
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Applying rules of algebra when working with multiplication and exponents I'm taking an online course and help is hard to find. This specific problem has to do with recurrence relation. I apologize for being too general but I'm just looking for help in how to go about solving this problem.
The entire problem is:
Let ... | I start from what you wrote down:
$$5d_{k-1}-6d_{k-2}=5\cdot3^{k-1}-6\cdot3^{k-2}-5\cdot2^{k-1}+6\cdot2^{k-2}$$
and I want to get to $d_k=3^k-2^k$.
I rewrite $3^{k-1}=3\cdot3^{k-2}$, and similarly for $2^{k-1}$:
$$5\cdot3\cdot3^{k-2}-6\cdot3^{k-2}-5\cdot2\cdot2^{k-2}+6\cdot2^{k-2}$$
$$=15\cdot3^{k-2}-6\cdot3^{k-2}-10\c... | {
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"timestamp": "2023-03-29T00:00:00",
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Proving $5 \mid (n^5-n)$ for all $n \in \mathbb{Z}^+$
Prove for all $n \in \mathbb{Z}^+$ that $5 \mid (n^5-n)$
My proof
Basis step: Since $5 \mid (1^5-1) \iff 5 \mid 0$ and $5 \mid 0$ is true, the statement is true for $n=1$.
Inductive step: Assume the statement is true for $n = k$; that is, assume that $5 \mid (k^... | In $\mathbb{Z}_5$ the integers mod $5$, which is a field, all elements obey $x^5=x$, or $0= x^5 -x$, from which the statement follows right away.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Prove that $\sqrt[3]{\frac{1}{9}}+\sqrt[3]{-\frac{2}{9}}+\sqrt[3]{\frac{4}{9}}$ is a root for $x^3+\sqrt[3]{6}x^2-1$
Prove that
$$c=\sqrt[3]{\frac{1}{9}}+\sqrt[3]{-\frac{2}{9}}+\sqrt[3]{\frac{4}{9}}$$
is a root for
$$P(x)=x^3+\sqrt[3]{6}x^2-1$$
Source: list of problems for math contest preparation.
I have n... | I am going to expand the hint provided in the comments: by letting $\omega=e^{2\pi i/3}$, one if free to conjecture (especially if he/she knows how the Galois group of a cubic polynomial is usually structured) that the algebraic conjugates of
$$ \alpha_1 = \frac{1}{\sqrt[3]{9}}\left[1+\sqrt[3]{-2}+\sqrt[3]{4}\right] $... | {
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Solution to differential equation system and solution to its conversion into 2nd order differential equation Following is the differential equation system with IVP
$\vec{x}'=\small\begin{pmatrix}3&-9\\4&-3\end{pmatrix}\vec{x},
\vec{x}(0)=\small\begin{pmatrix}2\\-4\end{pmatrix}$ The particular solution to this differ... | We have the general case of
$$x' = a x + b y \\ y' = c x + d y$$
Taking the derivative of the first equation yields
$$x'' = a x' + b y' = a x' + b(c x + d y)$$
However, from the first equation, we also have that $y = \dfrac{1}{b}(x' - ax)$, for $b \neq 0$, and upon substituting
$$x'' = a x' + b c x + d (x' - a x) = 3x'... | {
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Trying to prove $\log\left(n!\right)$ is greater than or equal to $\left(\frac{n}{2}\log\left(n\right)\right)$ I'm trying to prove that $\log\left(n!\right)$ is greater than $\left(\frac{n}{2}\log\left(n\right)\right)$ and I'm kinda stuck.
I could only prove $\log\left(n!\right)\ge\ \left(\frac{n}{2}\log\left(\frac{n}... | You're almost there -- just don't discard the second half of the sum (and deal with the "good" term $\log n$ separately):$$\begin{align}
\log(n!) &= \sum_{k=1}^n \log k = \log n + \sum_{k=\frac{n}{2}+1}^{n-1} \log k + \sum_{k=2}^{\frac{n}{2}}\log k \\
&\geq \log n + \left(\frac{n}{2}-1\right)\log \frac{n}{2} + \left(\f... | {
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Easy way of tackling $\sqrt{x^2}+x=\sqrt{5}$ Solve for x
$$\sqrt{x^2}+x=\sqrt{5}\tag1$$
$$\sin(\sqrt{x^2}+x)=\sin({\sqrt{5}})\tag2$$
$$\sin(\sqrt{x^2})\cos(x)+\cos(\sqrt{x^2})\sin(x)=\sin(\sqrt{5})\tag3$$
$$\sin(\sqrt{x^2})+\cos(\sqrt{x^2})\tan(x)=\sin(\sqrt{5})\sec(x)\tag4$$
$$\sin(\sqrt{x^2})+\cos(\sqrt{x^2})\sqrt{\s... | There is another way to solve this.
Using the fact that $\sqrt{x^2} = |x|$. Your equation turns out to be $$|x|+x=\sqrt{5}$$
First look for solutions where $x\geq 0$ in this case you get $x+x=\sqrt{5}$ and so $x=\frac{\sqrt{5}}{2}$.
Now if $x<0$ then $|x|=-x$ and so you get $-x+x=\sqrt{5}$ and clearly no $x$ satisfies ... | {
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$\epsilon-\delta$ proof that $f(x) = x \sin(1/x)$ on $(0,1)$ is uniformly continuous I have to prove
$$
f: (0,1) \to [-1,1], \ \ f(x)=x\sin(1/x)
$$
using $\epsilon-\delta$.
Try
Fix $\epsilon >0$. I have
$$
\left|x \sin(1/x) - y \sin (1/y)\right| \le |\sin (1/x)| |x-y| + |y| |\sin (1/x) - \sin(1/y)|
$$
I can control $|... | Using the prosthaphaereis formula
$$|y| \left|\sin \frac{1}{x} - \sin\frac{1}{y} \right| = 2|y|\left|\sin \frac{x^{-1}-y^{-1}}{2} \right| \left|\cos \frac{x^{-1}+y^{-1}}{2} \right| \leqslant 2|y|\left|\frac{x^{-1} - y^{-1}}{2} \right| \\\leqslant \frac{|x-y|}{|x|}$$
WLOG assume $x > y$ -- otherwise, work with the alter... | {
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Finding the side length of an equilateral triangle having three inscribed $120^\circ$ sectors in a certain arrangement
How do i even start this question? I thought of using length of tangent equal from exterior point but still of no use.
|
Let $G$ be the common centroid for the two triangles.
Pick an edge, say $AB$, let $H$ be the vertex on the small triangle which is the
apex for the circular sector touching edge $AB$. Let $G'$ and $H'$ be the foot of
$G$ and $H$ on edge $AB$.
It is clear the distance of $G$ to line $AB$ is
$$|GG'| = |HH'| - |GH|\cos... | {
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Proving a sequence is bounded from below Let $$a_1=3\quad , \quad a_{n+1}=\frac{3a_n}{4}+\frac{1}{a_n}$$
Show that $a_n$ converges.
I know that I need to prove that $a_n$ is monotonic and bounded.
I've assumed that the limit exists, made the calculation and get that $L=2$, hence claiming two claims:
*
*$a_n\ge 2$ (... | To show boundedness from below, you may use AM-GM:
$$\frac{3a_n}{4}+\frac{1}{a_n}\geq 2 \sqrt{\frac{3a_n}{4}\cdot \frac{1}{a_n}}= \sqrt{3}$$
For convergence you may proceed as follows:
*
*$f(x) = \frac{3x}{4} + \frac{1}{x}$ has a fixpoint for $x^{\star} = 2$
*$f'(x) = \frac{3}{4}-\frac{1}{x^2}$
*A quick calculatio... | {
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Problems with proof by induction $\frac1{1\times2} + \frac1{2\times3} + \dots + \frac1{n(n+1)} = \frac1{n+1}$?
$$\frac1{1\times2} + \frac1{2\times3} + \dots + \frac1{n(n+1)} = \frac1{n+1}$$
Prove for $n=1$: $$\frac1{1\times2}=\frac1{1+1}=\frac12$$
Hip: $$\frac1{1\times2} + \frac1{2\times3} + \dots + \frac1{n(n+1)} = ... | $\frac {1}{1\times 2}=1-1/2$
$\frac{1}{2\times 3}=1/2 - 1/3$
........
$\frac {1}{n(n+1)} =1/n- \frac {1}{n+1}$
Add them up and cancel the middle terms to get $1-\frac {1}{n+1}=\frac {n}{n+1}$
| {
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Given $3^x = 12$ find $(\sqrt{4/3})^\frac{1}{x-2}$ Given $3^x = 12$ find $$\left(\sqrt{\dfrac{4}{3}}\right)^\dfrac{1}{x-2}$$ in simple form.
I've faced this problem, in a high school book, but failed to solve.
I've tried to calculate
*
*$3^{x-2} = \dfrac{12}{9}$
*$x -2 = \log_3 \dfrac{12}{9}$
*$\dfrac{1}{(x -2)} ... | You have found $3^{x-2}=\frac{12}{9}=\frac{4}{3}$. So replace the $\frac{4}{3}$ in your expression: $$\left(\sqrt{\frac43}\right)^{1/(x-2)}=\left(\sqrt{3^{x-2}}\right)^{1/(x-2)}=\cdots$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Evaluating a Limit using Riemann integration We want to evaluate
$$ \lim_{n \to \infty} \sum_{i=1}^{3n} \sqrt{36 - \left(
\dfrac{i-\frac{1}{6}}{n} \right)^2 } \cdot \frac{1}{n} $$
I have spent almost an hour in this problem. Here is my thought
First, I tried to evaluate by comparing with an integral of the form $\in... | You may proceed as follows:
$$\sum_{i=1}^{3n} \sqrt{36 - \left(
\dfrac{i}{n} \right)^2 } \cdot \frac{1}{n} \leq \sum_{i=1}^{3n} \sqrt{36 - \left(
\dfrac{i-\frac{1}{6}}{n} \right)^2 } \cdot \frac{1}{n} \leq \sum_{i=1}^{3n} \sqrt{36 - \left(
\dfrac{i-1}{n} \right)^2 } \cdot \frac{1}{n}$$
$$3\sum_{i=1}^{3n} \sqrt{36 - ... | {
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"timestamp": "2023-03-29T00:00:00",
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Proving a set of vectors is a basis for the quotient map between two vector spaces I want to see if my work is justifiable. I am tasked with the following:
I will neglect to prove (a), as the work for this is fairly straight forward. I will center my attention on (b).
$$\text{Implication is} \ \mathscr {B}_{\mathbb R^... | It's all correct.
Note that the 4 given vectors form a basis of $\Bbb R^4$, two of them spans the kernel, so the (images of) the other two will be a basis of the quotient space.
| {
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"url": "https://math.stackexchange.com/questions/3002639",
"timestamp": "2023-03-29T00:00:00",
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Maximum value of $x$ when equality is given $$ x + y = \sqrt{x} + \sqrt{y} $$
Find maximum value of $x$. $x$ and $y$ are reals.
| Try this method of completing the square.
Let $a=\sqrt x$ and $b=\sqrt y$ so that $a^2+b^2=a+b$ and $4a^2-4a+1+4b^2-4b+1=2$ ie $(2a-1)^2+(2b-1)^2=2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3002761",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 0
} |
$\frac{1}{15}<(\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot \cdot \cdot \cdot\frac{99}{100})<\frac{1}{10}$. Show that
$$\frac{1}{15}<(\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot \cdot \cdot\cdot\frac{99}{100})<\frac{1}{10}$$
My attempt: This problem is from a text book where is introduced as: https://en.wikip... | Just a thought, that may be worth mentioning:
The expression:
$$p=(\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot \cdot \cdot\cdot\frac{99}{100})$$
We could use the following identities:
Product of $n$ odd numbers =
$$p_o = \frac{(2n!)}{(n!)2^{n}}$$
Product of $n$ even numbers =
$$p_e = (n!)2^{n}$$
The first $4$ te... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3004689",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 3
} |
$\frac{|a|}{|b-c|} + \frac{|b|}{|c-a|} + \frac{|c|}{|b-a|} \geq 2$ If $a, b, c$ are distinct real numbers then you demonstrate that:
$$ S=\frac{|a|}{|b-c|} + \frac{|b|}{|c-a|} + \frac{|c|}{|b-a|} \geq 2.$$
Using inequality $ |x-y|\leq |x|+|y|$ we showed that $ S >\frac{3}{2}.$
For $b = 2a, c = 3a, S=5,$ that is, the ... | Let $\frac{a}{b-c}=x$, $\frac{b}{c-a}=y$ and $\frac{c}{a-b}=z$.
Thus, $$xy+xz+yz=\sum_{cyc}\frac{ab}{(b-c)(c-a)}=\frac{\sum\limits_{cyc}ab(a-b)}{(a-b)(b-c)(c-a)}=\frac{(a-b)(a-c)(b-c)}{(a-b)(b-c)(c-a)}=-1.$$
Id est,
$$\sum_{cyc}\left|\frac{a}{b-c}\right|=\sqrt{\left(|x|+|y|+|z|\right)^2}=\sqrt{x^2+y^2+z^2+2\sum\limits_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3005207",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Maximize $x_1^3+x_2^3+\cdots + x_n^3$ This is from a Brazilian math contest for college students (OBMU):
Given a positive integer $n$, find the maximum value of
$$x_1^3+x_2^3+ \cdots + x_n^3$$
where $x_j$ is a real number for all $j \in \{1,2,\cdots, n\}$ such that $x_1 + x_2 + \cdots + x_n = 0$ and $x_1^2 + x_2^2 + ... | I come up with a solution using Lagrange Multipliers (one of the comments and one of answers suggests to use this). First, we need to realize that the domain
$$ S =\{ (x_1,x_2,\cdots,x_n) \in {\mathbb{R}}^n | x_1 + x_2 + \cdots + x_n = 0 \text{ and } x_1^2 + x_2^2 + \cdots + x_n^2 = 1\} $$
is closed (since the function... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3005707",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Lyapunov stability of 4x4 matrix. Consider the following continuous-time state space representation of the form:
$\frac{d}{dx}x(t) = Ax(t)+Bu(t), \quad y(t)=Cx(t), \quad t\in \mathbb{R}^{+}$
$A=\begin{bmatrix}-1&3&0&0\\-3&-1&0&0\\0&0&0&3\\0&0&-3&0 \end{bmatrix} \quad B = \begin{bmatrix}0\\1\\0\\0 \end{bmatrix} \quad C=... | Write $I_n$ for a $n \times n$ identity matrix.
Take $P = (1/2)I_4$ and $V(x) = x^T P x$, which is clearly positive definite. Now calculate the directional derivative:
$$
\begin{align}
\dot{V}(x) &= \dot{x}^T P x + x^T P \dot{x} \\
&= x^T A^T P x + x^T P A x \\
&= x^T(A^T P + P A) x \\
&= x^T Q x,
\end{align}
$$
inser... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3007700",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find the sum: $\sum_{n=2}^\infty \frac{1}{n^2-1}$
Evaluate : $$\sum_{n=2}^\infty \frac{1}{n^2-1}$$
I've tried to rewrite the questions as $$\sum _{n=2}^{\infty \:\:}\left(-\frac{1}{2\left(n+1\right)}+\frac{1}{2\left(n-1\right)}\right)$$ but still couldnt't get any answer as when I substitute numbers $(n)$ into th... | More generally,
if
$s_m(n)
=\sum_{k=m+1}^n \frac{1}{k^2-m^2}
$
then,
if $n > 3m$,
$\begin{array}\\
s_m(n)
&=\sum_{k=m+1}^n \frac{1}{k^2-m^2}\\
&=\sum_{k=m+1}^n \frac1{2m}(\frac{1}{k-m}-\frac1{k+m})\\
&=\frac1{2m}\sum_{k=m+1}^n \frac{1}{k-m}-\frac1{2m}\sum_{k=m+1}^n\frac1{k+m}\\
&=\frac1{2m}(\sum_{k=1}^{2m} \frac{1}{k}+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3009920",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"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.