Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Roll a fair die and denote the outcome by $Y$ . Then flip $Y$ fair coins and let $X$ denote the number of tails observed.
Roll a fair die and denote the outcome by $Y$. Then flip $Y$ fair coins and let $X$ denote the number of tails observed.
*
*Find the probability mass function of $X$
*Compute the expectation of ... | If Y represents the outcome of a fair die and X represents the number of tails observed then,
$ \ 0 \leq x \leq y$ and $1 \leq y \leq 6 \ $
$ \displaystyle P(X = x|Y=y) = {y \choose x} \cdot \frac{1}{2^y}$
So the table that you have put together is correct but you missed $x = 0$, probability of which is same as for $x ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4156538",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Paradox of $-i$ seemingly equal to $1$ via the Wallis product for $\pi$ and the Euler sine product Assuming $x$ is a real variable throughout,$$\frac{\sinh(ix)}{i}=\sin(x)$$
$$\frac{\sinh(\pi ix)}{\pi ix} = \frac{\sin(\pi x)}{\pi x}$$
$$\frac{e^{\pi ix}-e^{-\pi ix}}{2\pi ix}=\prod_{n=1}^{\infty}\left(1-\frac{x^2}{n^2}\... | when you let $x=\frac{1}{2}$ in $\frac{e^{\pi ix}-e^{-\pi i x}}{2\pi i x}$ your numerator should be $2i$ not 2
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4158071",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Calculating the following integral: $\int_0^{\frac{z}{1+z}}\frac{-X+z-Xz}{z}dX$ I'm trying to calculate the following integral: $\int_0^{\frac{z}{1+z}}\frac{-X+z-Xz}{z}dX$ however, I seem to be getting the wrong answer and would like some support on where I went wrong. The answer should be: $\frac{z}{2(z+1)}$
What I ha... | Everything is correct besides your last equality:
\begin{align}
\left(\frac{-(1+z)}{z}\cdot \frac{(\frac{z}{1+z})^2}{2}\right)+ \frac{z}{1+z} &= \left(\frac{-(1+z)}{z}\cdot \frac{z^2}{2(1+z)^2}\right)+ \frac{z}{1+z} \\
&= \frac{-z}{2(1+z)}+ \frac{z}{1+z}\\
&= \frac{-z}{2(1+z)} + \frac{2z}{2(1+z)} \\
&= \frac{z}{2(z+1)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4158338",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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About the inequality $x^{x^{x^{x^{x^x}}}} \ge \frac12 x^2 + \frac12$
Problem: Let $x > 0$. Prove that
$$x^{x^{x^{x^{x^x}}}} \ge \frac12 x^2 + \frac12.$$
Remark 1: The problem was posted on MSE (now closed).
Remark 2: I have a proof (see below). My proof is not nice.
For example, we need to prove that
$\frac{3x^2 - 3}... | A partial answer
First Fact :
For $x\in(0,1)$ we have :
$$x^{x^{x^{x^{x^{x}}}}}\geq x^{x^{x}}$$
Proof: see the Reference in my other answer .
Second Fact
For $x\in(0,1)$ we have :
$$ x^{x^{x}}\geq x^{\left(1+\left(x-1\right)x\right)}$$
Hint :use Bernoulli's inequality.
Third Fact
For $x\in[0.65,1]$ we have :
$$x^{\... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Integral $\int x^2\sqrt{1-x^2} dx$ I tried to solve the integral $$\int x^2\sqrt{1-x^2} dx.$$
(I know there are already solutions here on MSE but I need some help to find my error)
Substituting $x(y):=\cos(y)$ and using trigonometirc rules yields:
$$\int \cos^2(y)\sqrt{1-\cos(y)^2}(-1)\sin(y) dy= (-1)\int \cos^2(y)\si... | $$\left(-\frac{1}{8}\right)\left(\arccos(x)+\sqrt{1-x^2}x\left(1-2x^2\right)\right)+C\\=\left(-\frac{1}{8}\right)\left(\frac{\pi}2-\arcsin(x)+\sqrt{1-x^2}x\left(1-2x^2\right)\right)+C\\=-\frac{\pi}{16}+\frac{\arcsin(x)}{8}-\frac{1}{8}\sqrt{1-x^2}x(1-2x^2)+C\\=\frac18\left(\arcsin(x)-\sqrt{1-x^2}x(1-2x^2)\right)+C_1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4168212",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the number of 5-digit number divisible by 6 which can be formed using $0,1,2,3,4,5$ if repetition of digits is not allowed.
Find the number of $5$-digit number divisible by $6$ which can be formed using $0,1,2,3,4,5$ if repetition of digits is not allowed.
I started by considering the following cases:
*
*Unit ... | The sum of the digits must be divisible by $3$, so either $0$ or $3$ must be absent (since $0+1+2+3+4+5=15$ is divisible by $3$).
If $0$ is absent, then we have a permutation of $12345$ where the last digit is $2$ or $4$, giving $(2)(4!)=(2)(24)=48$ possibilities.
If $3$ is absent, then we have a permutation of $01245$... | {
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"timestamp": "2023-03-29T00:00:00",
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Proving $\sum_{k=1}^{n}\cos\frac{2\pi k}{n}=0$ I want to prove that the below equation can be held.
$$\sum_{ k=1 }^{ n } \cos\left(\frac{ 2 \pi k }{ n } \right) =0, \qquad n>1 $$
Firstly I tried to check the equation with small values of $n$
$$ \text{As } n=2 $$
$$ \cos\left(\frac{ 2 \pi \cdot 1 }{ 2 } \ri... | Observe that
$$\sum_{k=1}^n\cos(\frac{2\pi k}{n})=\operatorname{Re}\left(\sum_{k=1}^ne^{\frac{2\pi ki}{n}}\right).$$
Consider the sum
$$\sum_{k=1}^ne^{\frac{2\pi ki}{n}}=\sum_{k=1}^n(e^{\frac{2\pi i}{n}})^k$$
and use the fact that
$$1+\sum_{k=1}^n r^k =\frac{1-r^{n+1}}{1-r}$$
to show it's zero.
| {
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"url": "https://math.stackexchange.com/questions/4170548",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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If $3\sin x +5\cos x=5$, then prove that $5\sin x-3\cos x=3$
If $3\sin x +5\cos x=5$ then prove that $5\sin x-3\cos x=3$
What my teacher did in solution was as follows
$$3\sin x +5\cos x=5 \tag1$$
$$3\sin x =5(1-\cos x) \tag2$$
$$3=\frac{5(1-\cos x)}{\sin x} \tag3$$
$$3=\frac{5\sin x}{(1+\cos x)} \tag4$$
$$5\sin x-3\... | Yes you are right,he/she should have broken the problem into two cases.
Case $1.$ When $\sin x\ne 0$, then according to your teacher $5\sin x-3\cos x=3$.
Case $2.$ If $\sin x=0$, then from the given equation we get $\cos x=1$. Plugging these in to $5\sin x-3\cos x$ we get the required value as $-3$.
We can do the same ... | {
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"url": "https://math.stackexchange.com/questions/4171907",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solve $y' = (1+\frac{y-1}{2x})^2$ Solve $y' = (1+\frac{y-1}{2x})^2$
My first thought was to expand to see if I can get a linear form:
$$y' = 1 + \frac{y-1}{x} + (\frac{y-1}{2x})^2 = 1 + \frac{y-1}{x} + \frac{y^2-2y+1}{4x^2}$$
$$y' = 1 + \frac{y}{x} -\frac{1}{x} + \frac{y^2}{4x^2} + \frac{-2y}{4x^2} + \frac{1}{4x^2} ... | $$y' = \left(1+\frac{y-1}{2x}\right)^2$$
Substitute :
$$z=1+\frac{y-1}{2x}$$
$$y'=2(z-1+xz')$$
Then the DE is separable.
$$2xz'=(z-1)^2+1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4173844",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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Looking for other approaches to evaluate $\lim _{x\to 0^+}\frac{\sin 3x}{\sqrt{1-\cos ^3x}}$ Here is my approach: $$\lim _{x\to 0^+}\frac{\sin 3x}{\sqrt{1-\cos ^3x}}=\lim _{x\to 0^+}\frac{1}{\sqrt{\cos^2x+\cos x+1}}\times\frac{\sin 3x}{\sqrt{1-\cos x}}$$
$$=\lim _{x\to 0^+}\frac1{\sqrt3}\times\frac{\sin 3x}{\sqrt{1-\c... | I would write something like that,
$$\sin(3x) \sim 3x$$
$$\cos^3(x) = \left(1-\dfrac{x^2}{2}+o(x^2)\right)^3 = 1-3\dfrac{x^2}{2}+o(x^2)$$
Hence,
$$1-\cos^3(x) \sim 3\dfrac{x^2}{2} $$
Hence,
$$\sqrt{1-\cos^3(x)} \sim \sqrt{3/2}x $$
Finally,
$$\dfrac{\sin(3x)}{\sqrt{1-\cos^3(x)}} \sim \dfrac{3}{\sqrt{3/2}} \sim \sqrt{6}... | {
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"url": "https://math.stackexchange.com/questions/4179100",
"timestamp": "2023-03-29T00:00:00",
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Let $α,β$ be the roots of $x^2-x+p=0$ and $γ,δ$ be the roots of $x^2-4x+q = 0$ where $p$ and $q$ are integers. If $α,β,γ,δ$ are in GP then $p + q$ is?
Let $α,β$ be the roots of $x^2-x+p=0$ and $γ,δ$ be the roots of
$x^2-4x+q = 0$ where $p$ and $q$ are integers. If $α,β,γ,δ$ are in GP
then $p + q$ is ?
My solution app... | In the first approach: when you solve for $r$ you would have got an equation $r^4=4$. This gives you $r^2=2$ and $r^2=-2$. You have only considered the first case (i.e. when $r^2=2$). If you consider the second case $r^2=-2$, then you get $r=\pm i\sqrt{2}$. In which case you will get $p+q=-34$. So both approaches will ... | {
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"url": "https://math.stackexchange.com/questions/4180595",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solve : $(2x - 4y + 5) \frac{dy}{dx} + x - 2y + 3 = 0 , y(2) = 2.5$ I have to find the solution of following differential equation:
$$(2x - 4y + 5) \frac{dy}{dx} + x - 2y + 3 = 0 , y(2) = 2.5$$
It can be re-written as
$$(2(x - 2y) + 5) \frac{dy}{dx} + (x - 2y) + 3 = 0$$
let $u = (x-2y)$
so, on differentiating both si... | The answer was given by Lalit Tolani. This is an additional information.
From the implicit equation
$$4x+8y+\ln(4x-8y+11)=28$$
the explicit solution $y(x)$ cannot be expressed with a finite number of elementary functions. It can be analytically expressed on closed form thanks to a special function namely the Lambert's ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4180707",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Let $K/F$ be a finite extension of finite fields and suppose $K=F[\alpha]$ with $\alpha^{15}=1$ then $\mid K:F \mid \leq 4$.
Let $K/F$ be a finite extension of finite fields and suppose $K=F[\alpha]$. If $\alpha^{15}=1$ then show that $\mid K:F \mid \leq 4$.
Let $F=\mathbb F_q$, where $q$ is power of some prime $p$, ... | Claim:$\;$If $K$ is a field with an element $\alpha$ and a subfield $F$ such that
*
*$F$ is a finite field.$\\[4pt]$
*$K=F(\alpha)$.$\\[4pt]$
*$\alpha^{15}=1$.
then $[K:F]\le 4$.
Proof:
For the subfield $F(\alpha^3,\alpha^5)$ of $K$, we have
\begin{align*}
&
\frac{\alpha^5}{\alpha^3}\in F(\alpha^3,\alpha^5)
\\[4... | {
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"url": "https://math.stackexchange.com/questions/4181785",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Mathematicals inequalities For $$x,y,z>0 $$
Prove that
$$(2xyz)^2 \ge (x^3+y^3+z^3+xyz)(x+y-z)(y+z-x)(z+x-y)$$
I have tried a famous inequalities: $$(x+y-z)(y+z-x)(z+x-y) \le xyz$$
So the problem is:
$$3xyz \ge x^3+y^3+z^3$$
But in fact, this isn’t true
Help me plz
| If $\prod\limits_{cyc}(x+y-z)<0$ so the inequality is obvious.
Thus, it's enough to assume that $\prod\limits_{cyc}(x+y-z)\geq0.$
Now, since $x+y-z<0$ and $x+z-y<0$ gives $x<0$, which is a contradiction, we can assume that
$x+y-z=c\geq0,$ $x+z-y=b\geq0$ and $y+z-x=a\geq0$ and we need to prove that
$$(a+b)^2(a+c)^2(b+c)... | {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $(n+1)a\leq a^{n+1}+n, \forall a,n\in\mathbb{N}$. We can start from the fact that:
\begin{align*}
0\leq a^n + a^{n-1} +\ldots + a^2 + a-n,\forall a,n\in\mathbb{N}.
\end{align*}
The above is true, since if $a>1$, then $a^n>n, \forall n \in\mathbb{N}$. Also if $a = 1$, then we will have $1 + 1 + \ldots + 1... | Alternatively, the result follows from the Bernoulli’s inequality:
$$[1+(a-1)]^{n+1}\geq 1+(n+1)(a-1),$$ which is true for all real numbers $a\geq 0$, and all integers $n\geq -1$. (Exclude the case when $a=0,n=-1$ if necessary to avoid the ambiguity $0^0$.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4187115",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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Cartesian product of finitely generated integral monoids I would like to understand the fact that the Cartesian product of finitely generated integral monoids is also a finitely generated monoid.
$\textbf{Definition:}$ A monoid is a set of vectors in Q and is defined under addition. When it only contains integer vecto... | Take any $\begin{pmatrix} r^1 \\ r^2\end{pmatrix} \in M_1\times M_2$. Since $M_1$ is generated by $r_1^1,\dots,r_{n_1}^1$ and $M_2$ is generated by $r_1^2,\dots,r_{n_2}^2$, we may write
\begin{align}
r^1 &= k_1^1 r_1^1 + \cdots + k_{n_1}^1 r_{n_1}^1 \\
r^2 &= k_1^2 r_1^2 + \cdots + k_{n_2}^2 r_{n_2}^2
\end{align}
for s... | {
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"timestamp": "2023-03-29T00:00:00",
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The $x^2+ax+b = 0$ has only one solution and this is $x = -1+1/2$. What are the values of $a$ and $b$? This is a two part question. I was able to solve the first part. I need help with the second part.
a) The equation $x^2 + ax + b = 0$ has solutions $x = 2$ and $x = -5$. Find $a$ and $b$.
I was able to solve this one.... | *
*
$$\begin{align}&x^2+ax+b \\
\iff &(2x+a)^2=a^2-4b\end{align}$$
If the quadratic equation has one real root, then $\Delta=a^2-4b=0$.
This implies,
$$\begin{cases}2x+a=0 \\a^2=4b \end{cases}\implies \begin{cases} a=-2x \\ b=x^2\end{cases}$$
*
If the quadratic has one real root, then, $x_1=x_2$.
By the Vieta's f... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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$\lim_{x\to 0^+}\dfrac{\tan^2(\frac1{\sqrt{1-x^2}}-1)}{\left(1-\cos(\sqrt{2x})\right)^n}=a$. What is the value of $a+n$?
Suppose $\lim_{x\to
0^+}\dfrac{\tan^2(\frac1{\sqrt{1-x^2}}-1)}{\left(1-\cos(\sqrt{2x})\right)^n}=a$.
What is the value of $a+n$ ?
$1)\frac74\qquad\qquad2)\frac94\qquad\qquad3)\frac{15}4\qquad\qquad... | We have, from Newton's series:
$$(1-x^2)^{-\frac 12}=1+\frac {x^2}{2}+O(x^4)$$
Also, Taylor series for $\tan x$ yields:
$$\tan x=x+\frac {x^3}{3}+O(x^5)$$
Hence, $$\tan\left(\frac {1}{\sqrt {1-x^2}}-1\right)=\frac {x^2}{2}+O(x^4)$$
Squaring results in:
$$\tan^2\left(\frac {1}{\sqrt{1-x^2}}-1\right)=\frac {x^4}{4}+O(x^6... | {
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"timestamp": "2023-03-29T00:00:00",
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Showing $\sum_{n=1}^\infty\frac1{n^2}\left(1+\frac1{2^2}+\cdots+\frac1{n^2}\right)=\frac{7\pi^4}{360}$ Help me calculate the amount in a simpler way:
\begin{aligned}
\mathcal{S}&=\sum\limits_{n=1}^{\infty }\frac{1}{n^2}\left ( 1+\frac{1}{2^2}+\ldots+\frac{1}{n^2} \right )=\sum\limits_{n=1}^{\infty }\frac{H^{\left ( 2 ... | You don’t need any integrals.
$$S=\sum_i \sum_{j<=i} 1/(i^2 j^2)= \sum_i \sum_j 1/(2i^2 j^2) + \sum_i 1/(2i^4) =(\zeta(2)^2 +\zeta(4))/2= (\pi^4/36+\pi^4/90)/2=7\pi^4/360$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4193006",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
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Proving $\frac{2(1+y)\sqrt{1+x}+y\sqrt{1+y}}{2(1+x)\sqrt{1+y}+x\sqrt{1+x}} = \frac{1}{(1+x)^2}$ with $x\sqrt{1+y}+y\sqrt{1+x}=0$ I want to find $\frac{dy}{dx}$ in $x\sqrt{1+y}+y\sqrt{1+x}=0$ and I proceed through 2 different ways expecting the same answer.
Method:-1
$x\sqrt{1+y}=-y\sqrt{1+x}$
$\implies x^2(1+y)=y^2(1+x... | Your expression is $$\frac{2(1+y)\sqrt{1+x}+y\sqrt{1+y}}{2(1+x)\sqrt{1+y}+x\sqrt{1+x}}$$
Taking $\sqrt{1+y}$ common from numerator and $\sqrt{1+x}$ common from denominator, we get $$\frac{\sqrt{1+y}}{\sqrt{1+x}}\cdot\frac{2\sqrt{1+y}\sqrt{1+x}+y}{2\sqrt{1+x}\sqrt{1+y}+x}$$
Given that $x\sqrt{1+y}+y\sqrt{1+x}=0\implies\... | {
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"timestamp": "2023-03-29T00:00:00",
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Simple proof that $\sum_{k = 1}^\infty \sum_{n = 1}^\infty a_k b_n / (k + n) \lesssim (\sum_k a_k^2)^{1/2} (\sum_n b_n^2)^{1/2}$ I have a complicated proof involving real interpolation + restricted bounds on operators, but I can only imagine there is a simpler proof of this statement involving some combinations of basi... | To avoid the ambiguity about the convergence, assume $a_n,b_n \ge 0$ and $\sum a_n^2 =1$.
We have:
$$
\begin{align}
\left(\sum_{k\ge 1}\sum_{n \ge 1} \frac{a_kb_n}{n+k}\right)^2 &\le \left( \sum_{k \ge 1} a_k^2 \right)\left( \sum_{k \ge 1} \left( \sum_{n \ge 1} \frac{b_n}{n+k} \right)^2 \right)
\\
& \le \sum_{k \ge 1}... | {
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"timestamp": "2023-03-29T00:00:00",
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Prove $\sum_{n=2}^{\infty}\frac{1}{n^2+e}<\frac{1}{2}$ Of course, you can use the following formula
$$\sum_{n = 1}^\infty \frac{1}{n^2 + a^2} = \frac{\pi\coth(\pi a)}{2a} - \frac{1}{2a^2},$$
but which is too "advanced". We want to find a solution only depending on inequality estimation only.
Maybe, we can obtain
\begin... | I only use $\sum_{n=1}^\infty \frac{1}{n^k}$ for $k=2,4,6$.
Using $\frac{1}{n^2}-\frac{1}{n^2+e} = \frac{e}{n^2(n^2+e)}$, it suffices to show that $\sum_{n=2}^\infty \frac{1}{n^2(n^2+e)} > \frac{\pi^2-9}{9e}$. Then using $\frac{1}{n^4}-\frac{1}{n^2(n^2+e)} = \frac{e}{n^4(n^2+e)}$, it suffices to show $\sum_{n=2}^\infty... | {
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If the roots of the equation $x^4 + ax^3 + bx^2 + cx + d = 0$ are in geometric progression then, If the roots of the equation $x^4 + ax^3 + bx^2 + cx + d = 0$ are in geometric progression then,
$a) b^2 = ac$
$b) a^2 = b$
$c) c^2 = a^2d$
Using Vieta's relations, finding values of coefficients in terms of the assumed roo... | Here is the OP's original method. Assume the roots are $k, kr, kr^2, kr^3$. Then:
$$k(1+r+r^2+r^3) = -a$$
$$k^2r+k^2r^3+k^2r^4+k^2r^3+k^2r^4+k^3r^5 = b$$
$$k^3r^3 + k^3r^4+k^3r^5+k^3r^6 = k^3 r^3(1+r+r^2+r^3) = -c$$
$$k^4r^6 = d$$
By comparing the maximum powers of $r$, a) is not correct ($r^{10}$ vs $r^9$) and so is b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4197109",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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For each value for $a$ and $b$, being real numbers, $a^2 + b^2 \ge ab$ For each value for $a$ and $b$, being real numbers, $a^2 + b^2 \ge ab$
Should I solve this by replacing all possible real number formulas in these to be able to prove this? As an example for odd numbers, by replacing $2x + 1$, etc...?
| Try to write it as $a^2+b^2-ab=\frac{2a^2+2b^2-2ab}{2}=\frac{a^2+b^2+(a-b)^2}{2}$. Now all three terms are $\geq 0$, so you get $a^2+b^2\geq ab$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4198705",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Help with $\int _0^{\infty }\frac{\sinh \left(x\right)}{x\cosh ^2\left(x\right)}\:\mathrm{d}x$ I want to know how to prove that
$$\int _0^{+\infty }\frac{\sinh \left(x\right)}{x\cosh ^2\left(x\right)}\:\mathrm{d}x=\frac{4G}{\pi }$$
Here $G$ denotes Catalan's constant, I obtained such result with the help of mathematica... | I present another approach, utilising a well-known property of the Laplace Transform i.e.
$$\int_{0}^{+\infty} f\left(x\right) g\left(x\right) \, dx = \int_{0}^{+\infty} \left(\mathcal{L} f\right)\left(y\right)\left(\mathcal{L}^{-1} g\right)\left( y\right) \, dy$$
Letting $f\left(x\right) = \tanh \left(x\right) \text{s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4199018",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "25",
"answer_count": 6,
"answer_id": 4
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Convergence of $\sum_{n=0}^{\infty} \frac{4^n}{3^n+7^n}$ I need help with this.
$\sum_{n=0}^{\infty} \frac{4^n}{3^n+7^n}$
I know that it converges but i can not proove why.
I tried to rewrite it, it seems to be a geometric serie. I tried to do a common factor between $3^n+7^n \rightarrow 3^n(1+\frac{7^n}{3^n})$
So I ... | Comparison test: Observe that
$$\frac{4^n}{3^n+7^n} \le \frac{4^n}{7^n} = \left(\frac47\right)^n$$
which is a convergent geometric series
Root test:
Observe that
$$\frac{4}{\sqrt[n]{7^n+7^n}}\le \sqrt[n]{\frac{4^n}{3^n + 7^n}} \le \frac{4}{\sqrt[n]{7^n}} \\
\frac{4}{7\sqrt[n]{2}}\le \sqrt[n]{\frac{4^n}{3^n + 7^n}} \le... | {
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"url": "https://math.stackexchange.com/questions/4200990",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Order on three real numbers given they are roots of a cubic, their sum and sum of pairwise product
Let $a,b,c$ be three real numbers which are roots of a cubic polynomial and satisfy $a+b+c=6$ and $ab+bc+ca=9.$ Suppose $a<b<c.$ Show that $$0<a<1 <b<3<c<4$$
Source: ISI Bmath Entrance 18-Jul-2021
Let the polynomial be ... | I approached the problem differently. It was a long write-up, here are the key points:
▪︎Define a cubic $f(x)=x^3-6x^2+9x+t$. It is clear that this is the cubic with roots $a,b,c$.
▪︎Try to put a bound on $t$ for which three distinct roots are obtained. The graphical argument can be used. The defined $f(x)$ is merely a... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that $2^n + 5^n + 56$ is divisible by $9$, where $n$ is an odd integer
Prove that $9 \mid2^n + 5^n + 56$ where n is odd
I have proved this by using division into cases based on the value of $n\bmod3$ but it seems a little bit clumsy to me and I wonder if there are other ways to prove it, probably by using modul... | Here's one way (not sure if it is the easiest, though).
Let us start with a lemma which you may easily prove by induction.
Lemma: For each natural number $k$, $2^{2k+1}+1$ and $2^{2k}-1$ are divisible by $3$.
Next, we have the following
Claim:
For $n=2k+1$, one has
$$2^n+5^n+56\equiv-2(2^{2k+1}+1)(2^{2k}-1)\mod 9.$$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4206716",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 8,
"answer_id": 3
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Evaluate $\iiint_{V} (x^2+y^2+z^2)\,dx\,dy\,dz$ in the common part of $2az > x^2+y^2$ and $x^2+y^2+z^2 < 3a^2$ Evaluate $$\iiint_{V} x^2+y^2+z^2 \,dx\,dy\,dz$$
Where $V$ (the integration region) is the common part of the paraboloid $x^2 + y^2 \leq 2az$ and the sphere $x^2+y^2+z^2 \leq a^2$.
I first found the intercept ... | You are correct, and the person who said the greatest lower limit is $z=0$ is wrong:
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4208625",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Extrema of $f(x,y)=(1-x^2-y^2)\cdot xy$ subject to $x^2+y^2\leq 1$ I want to determine the extrema of $f(x,y)=(1-x^2-y^2)\cdot xy=xy-x^3y-xy^3$ subject to $x^2+y^2\leq 1$.
We use Lagrange Multipliers to check the critical points on the circle $x^2+y^2=1$.
To check the critical points inside the circle, $x^2+y^2<1$, we... | The poser of this problem seems to have been having a little fun, as the behavior of this function $ \ f(x,y) \ = \ xy·(1 - x^2 - y^2) \ $ is a bit peculiar. The factor $ \ 1 - x^2 - y^2 \ $ has "four-fold" symmetry about the origin, but the factor $ \ xy \ $ "breaks" this down to just "diagonal" symmetry. So we ma... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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$\tan\frac{\pi}{9} +4\sin\frac{\pi}{9} = \sqrt 3 $
Prove that $$ \tan\frac{\pi}{9} +4\sin\frac{\pi}{9} = \sqrt 3 $$
There seem to be a lot of similar identities that are provable, for example, by using roots of unity. However, here I cannot get things to work out nicely.
If $u=e^{\frac{2\pi i}{9}} $, then $$i\left(\... | Here's a purely trigonometric solution, in case you're interested:
$$\tan 20°+4\sin 20°=\frac {\sin 20°+2(2\sin 20° \cos 20°)}{\cos 20°}=\frac {(\sin 20°+\sin 40°)+\sin 40°}{\cos 20°}=\frac {2\frac 12 \cos 10°+\cos 50°}{\cos 20°}=\frac {2\cos 30°\cos 20°}{\cos 20°}=2 \frac {\sqrt 3}{2}=\sqrt 3$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4212380",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
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$a^a\cdot{b^b}\ge \bigl(\frac{a+b}{2}\bigl)^{a+b}\ge{a^b}\cdot{b^a}$ If $a$ and $b$ are positive rational numbers, prove that
$$a^a\cdot b^b\ge \left(\frac{a+b}{2}\right)^{a+b} \ge a^b \cdot{b^a}$$
My try:
consider $\frac{a}{b}$ and $\frac{b}{a}$ be two positive numbers with associated weights $b$ and $a$.
Then $\displ... | From weighted AM-GM, we have $$\left(\frac{\frac 1a\cdot a+\frac 1 b\cdot b}{a+b}\right )^{a+b}\geq \frac{1}{a^ab^b}\\ \implies \left(\frac{2}{a+b}\right)^{a+b}\geq \frac{1}{a^ab^b}$$which is equivalent to the first inequality.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4214747",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
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Solving $2x^3-5ix^2+3x+4i=0$, need $(21x_1-1)^{-2}+(21x_2-1)^{-2}+(21x_3-1)^{-2}$ I have this polynomial from Problemas selectos (Lumbreras editors):
$$2x^3-5ix^2+3x+4i=0$$
I make $x=it$. then:
$$2(it)^3-5i(it)^2+3it+4i=0$$
$$i(-2it^3+5it^2+3it+4)=0$$
$$2t^3-5t^2-3t-4=0$$
$$8t^3-20t^2-12t-16=0$$
$$(2t)^3-3(2t)^2(5/3)+(... | Assume $x$ is pure imaginary.
$x=iy$, with $y$ real.
We have:
$$- 2i y^3 +5 i y^2 +3iy +4 i =0$$
or
$$2 y^3 -5 y^2 -3y -4 =0$$
This leads to the solution
$$y \approx 3.17174.$$
The other solutions are
$$y \approx -0.335868 \mp 0.719557 i$$
and the solutions to the original problem are $x=iy.$
There are techniques to ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4216153",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Computing $I=\int \frac{dx}{(x+1)(3x^2+3x+1)^{1/3}}$ I am trying to solve this integral and find its primitive, unfortunately I have not been successful. I made some variable changes, but I think it has become even more complicated.
My solution is as follows:
$$I=\int \frac{dx}{(x+1)(3x^2+3x+1)^{1/3}}$$
$$u=x+\frac{1}{... | \begin{gather*}
\int \frac{dx}{( x+1)\left( 3x^{2} +3x+1\right)^{\frac{1}{3}}}\\
=\int \frac{dx}{( x+1)\left( x^{3} +3x^{2} +3x+1-x^{3}\right)^{\frac{1}{3}}}\\
=\int \frac{dx}{( x+1)\left(( x+1)^{3} -x^{3}\right)^{\frac{1}{3}}}\\
=\int \frac{dx}{( x+1)^{2}\left( 1-\left(\frac{x}{x+1}\right)^{3}\right)^{\frac{1}{3}}}\\
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4217921",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
The Diophantine equation $x^5-2y^2=1$ I'm trying to solve the Diophantine equation $x^5-2y^2=1$.
Here's my progress so far. We can write the Diophantine equation as
$$\frac{x-1}{2}\cdot(x^4+x^3+x^2+x+1)=y^2.$$
If $x\not\equiv1\pmod{5}$, then $\gcd(\frac{x-1}{2},x^4+x^3+x^2+x+1)=1$, so both $\frac{x-1}{2}$ and $x^4+x^3+... | You don't really say if you just want to know the solutions, or if you want a nice elementary argument for why the solutions are only $(3, \pm 11)$, if you just want a proven answer and not an elementary argument the following works, its a bit overkill but its easier than thinking if you know these methods already:
The... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4218564",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
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Finding foot of perpendicular of centre of ellipse on its variable tangent
Find the locus of the foot of perpendicular from the centre of the ellipse $${x^2\over a^2} +{y^2\over b^2} = 1$$
on the chord joining the points whose eccentric angles differ by $π/2.$
Ref:Locus of foot of perpendicular of origin from tangent ... | You are missing a $(u^2+v^2)$ when you add $(ax)^2$ and $(by)^2$.
Just in case, you want to finish continuing from the point you got to,
$(ax)^2 + (by)^2 = (ab)^4 (u^2+v^2) \left[ \frac{u^2 +v^2}{b^2 u^2 +a^2v^2} \right]^2$
$ = \cfrac{(ab)^4}{2} \left[ \frac{(u/v)^2 +1}{b^2 (u/v)^2 + a^2} \right]^2$
Now note that $\c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4218853",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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To prove: $\cot^{-1}7+\cot^{-1}8+\cot^{-1}18 = \cot^{-1}3$ To prove: $\cot^{-1}7+\cot^{-1}8+\cot^{-1}18 = \cot^{-1}3$
My Attempt:
First Method: we know that $\cot^{-1}x = \tan^{-1}\frac{1}{x}$ for $x>0$ and $\tan^{-1}x+\tan^{-1}y = \tan^{-1}\frac{x+y}{1-xy}, xy<1$
Now $\cot^{-1}7+\cot^{-1}8+\cot^{-1}18$ = $\tan^{-1}\fr... | I can't resist, even though your request for a critique makes this response somewhat off-topic. Let the 3 LHS angles be denoted as $a,b,c$, and use the formula
$$\cot(a + b) = \frac{[\cot(a)\cot(b)] - 1}{\cot(a) + \cot(b)}.$$
This gives $\cot(a + b) = \frac{55}{15} = \frac{11}{3}.$
Therefore,
$$\cot[(a+b) + c] = \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4220297",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
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Find all primes $p$ and $r$ such that $ pr+1+r^3=p^2 $ I tried some small values, and I found that $p=7$ and $r=3$ was a solution.
I ve also find a way to factorise the equation:
$$ p(p-r)=(r+1)(r^2-r+1)$$
Since we know that $ p > r+1 $ then $$p| (r^2\color{blue}{-}r+1) $$
That's all what i've found , thank you in adv... | As you've already noted,
$$p(p - r) = (r + 1)(r^2 - r + 1) \tag{1}\label{eq1A}$$
Also, there's a positive integer $k$ such that
$$p \mid (r^2 - r + 1) \; \; \to \; \; r^2 - r + 1 = kp \tag{2}\label{eq2A}$$
Similar to how Find all pairs of prime numbers $p$ and $q$ such that $\,p^2-p-1=q^3.$ that Dietrich Burde's questi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4222384",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
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Finding the anti-derivative of $ \frac{e^{-c y^2 }}{y\sqrt{y^2-1}}$ I am trying to evaluate the integral
\begin{align}
\frac{1}{2\sqrt{2}\pi}\int_{0^{-}}^{t} ds \ \frac{e^{-x^2/2S^2(t,s) }}{\Sigma(s) S(t,s)}
\end{align}
where $S(t, s) = 2D(t-s)+\frac{\Sigma(s)}{2}$ and $\Sigma(s)= \sigma^2+2Ds$. I found a rather neat c... | As others have pointed out, no closed-form solution in terms of elementary functions can be found. However, if you can live with the error function and the Owen T function then a closed-form solution can be found in terms of these special functions.
I will assume the error function $\operatorname{erf} (x)$ is well know... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4223576",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 0
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Integrate $ \int \frac{2\tan x+1}{\sqrt{\tan^2 x+2\tan x+2}} \, \mathrm dx $ I tried writing $\tan^2 x+2\tan x+2$ as $(\tan x+1)^2+1$ and letting $\tan\theta = \tan x+1$ yields
$$ \int \frac{2\tan\theta-1}{\tan^2\theta-2\tan\theta+2}\,\sec\theta\,\mathrm d\theta $$
Then letting $ t = \tan(\frac{\theta}{2})$ yields
$$ \... | $$\int\dfrac{2\tan x+1}{\sqrt{\tan^2x+2\tan x+2}}~dx=\int\dfrac{2(\tan x+1)-1}{\sqrt{(\tan x+1)^2+1}}~dx$$
Let $u=\tan x+1$ ,
Then $x=\tan^{-1}(u-1)$
$dx=\dfrac{1}{(u-1)^2+1}~du$
$$\therefore\int\dfrac{2(\tan x+1)-1}{\sqrt{(\tan x+1)^2+1}}~dx=\int\dfrac{2u-1}{((u-1)^2+1)\sqrt{u^2+1}}~du$$
Introduce the Euler substituti... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Inductively prove $1 + \frac12 + \frac14 + \cdots + \frac{1}{2^n} = 2 - \frac{1}{2^n}$. Inductively prove that the formula holds for all $n\in\Bbb{N}$:
$$1+\frac{1}{2}+\frac{1}{4}+\cdots+\frac{1}{2^n}=2-\frac{1}{2^n}.$$
What I have so far:
base: n = 1: $$1+\frac{1}{2}=2-\frac{1}{2}=1.5$$
inductionstep: n = k: $$1+\frac... | Use the following structure:
Step 1. $n = 0$, means $1 = 2 - \frac{1}{2^0} = 2 - 1 = 1$, which is true. We can proceed to induction step.
Step 2. Use simple induction: $p(n) \to p(n + 1)$. So, if:
$$1 + \dots + \frac{1}{2^n} = 2 - \frac{1}{2^n}$$
We have:
$$1 + \dots + \frac{1}{2^n} + \frac{1}{2^{n + 1}} = 2 - \frac{1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4225701",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 1
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How to find the maximum value of $\frac{2\sin\theta\cos\theta}{(1+\cos\theta)(1+\sin\theta)}$? Let $\theta\in\left(0,\frac{\pi}{2}\right)$, then find the maximum value of $\dfrac{2\sin\theta\cos\theta}{(1+\cos\theta)(1+\sin\theta)}$
I found the derivative, which is equal to $\dfrac{(\cos\theta−\sin\theta)(\sin\theta+\c... | The first derivative is:
$\frac{2\sqrt{2}cos(\theta+\frac{\pi}{4})}{(1+\cos\theta)(1+\sin\theta)}$,
which becomes null in:
$\theta=\frac{5\pi}{4}$, $\theta=\frac{-3\pi}{4}$, $\theta=\frac{\pi}{4}$.
The maximum is for
$\theta=\frac{\pi}{4}$ and,
and is equal to $-4\sqrt{2}+6$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4230143",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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Proof by induction: induction hypothesis question In this question I found online:
*
*Show that $$ S(n):0^2 + 1^2 + 2^2 + · · · + n^2 =
\frac{n(n + 1)(2n + 1)}{6}$$
I don't understand why for S(k+1) they wrote:
$$S(k+1):1^2+2^2+3^2+⋯+k^2+(k+1)^2=\frac{(k+1)(k+2)(2(k+1)+1)}{6}$$
instead of:
$$S(k+1):1^2+2^2+3^2+⋯+(k+1... | The sum you show is simply going up to $k+1$, which means the preceding term is $k^2$, and the final term is $(k+1)^2$. In the other proof you reference there is no sum.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4231436",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Troubleshooting a trigonometry/geometry question - spot the (silly) mistake! I'm back again!
Again, another error - not sure if it's them or me this time...
Here is the question
In order to calculate the shaded region, I added some lines to the diagram. I realise that there are other ways to solve this (and indeed the... | Another way to compute the area is to add the part on the right of chord $CD$
and the part on the left of chord $CD$.
The portion of the shaded region on one side of the chord is called a circular segment.
You can compute the area of the circular segment on the right by finding the area of the circular sector between $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4233346",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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The solution set of the equation $|\lfloor x|x|-7\rfloor|<5$ is? My solution approach :-
$|\lfloor x|x|-7\rfloor|<5$
When $x\geq 0$, then;
$|\lfloor x^2-7\rfloor|<5$
$\Rightarrow -5 \lt \lfloor x^2-7\rfloor \lt5$
After this I do know that I'll have to take another case in which $x\lt0$ but I am not able to get how to s... | So far, when $x\geq 0$ you got here
$$
-5 \lt \lfloor x^2-7\rfloor \lt5
$$
This means $\lfloor x^2-7\rfloor \in \{-4,-3,-2,-1,0,1,2,3,4\}$ or in other words $x^2 - 7 \in [-4,5)$.
Therefore we have
$$
x^2 \in [3,12),
$$
or
$$
x\in[\sqrt{3},2\sqrt{3}).
$$
Note that $x\in(-2\sqrt{3}, -\sqrt{3}]$ is excluded because we onl... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4235720",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Generalization of Remainder Theorem Technique Generalization of this question
$$\frac{x^{2021}}{x^3 +x^2+x+1}$$
We wish to determine the remainder for the expression above. As in the linked question, write
$$x^{2021}=({x^3 +x^2+x+1})P(x)+ R(x)$$
To eliminate the term with P(x) let
$$x^3 +x^2+x+1=0\Rightarrow x^4 =1$$
W... | As opposed to blindly using "eliminate the term", a simple explanation of why that works is just algebraic manipulation:
$$\begin{align} & x^{2021} \\
= & (x^4 -1 ) A(x) + x \\
= & (x^3+x^2+x+1) B(x) + x.\\
\ \\
& x^{2023} \\
= & (x^4 -1 ) C(x) + x^3 \\
= & (x^3+x^2+x+1) D(x) + x^3 \\
= & (x^3+x^2+x+1)E(x) + (-x^2-x-1)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4237919",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
USAMO 1973 (Simultaneous Equations)
Determine all the roots, real or complex, of the system of simultaneous equations (USAMO 1973/4)
$$x+y+z=3$$
$$x^2+y^2+z^2=3$$
$$x^3+y^3+z^3=3$$
Multiply equation I by 2 and subtract it from Equation II:
$$x^2-2x+y^2-2y+z^2-2z=-3$$
Complete the square for all three variables:
$$(x-... | The nicer solutions have been posted already, so here is a different one, just for fun.
Moving the constant terms $\,3\,$ to the LHS, distributing and factoring, the system can be written as:
$$
\begin{cases}
(x-1) & +\; (y-1) & +\; (z-1) & = 0 \\
(x-1) \cdot (x+1) & +\; (y-1) \cdot (y+1) & +\; (z-1) \cdot (z+1) & ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4239747",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Finding the inverse of a block $2\times2$ square matrix $ \begin{bmatrix} I & A \\ A^T & 0 \end{bmatrix} $ where $A$ is a square invertible matrix. The hint I got was to rewrite the original matrix as a product of 3 matrices and use the property for inverse of product of matrices $(XYZ)^{-1} = Z^{-1} Y^{-1} X^{-1}$
$\b... | Your product formula is incorrect. Computing the product you have written yields
$$
\begin{bmatrix} I & 0 \\ A^T & 0 \end{bmatrix} \begin{bmatrix} I & 0 \\ 0 & -A^T A \end{bmatrix} \begin{bmatrix} I & A \\ 0 & I \end{bmatrix} =
\begin{bmatrix}
I & A\\A^T & A^TA
\end{bmatrix}.
$$
A correct formula is
\begin{align}
\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4240327",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
What will be the maximum value of expression: 7sin²x + 5cos²x +√2[sin(x/2) + cos(x/2)]? This is what I've managed to do:
$=5 + 2\sin²x +√2[\sin(x/2) + \sin(π/2 - x/2)]$
$=5 + 2\sin²x + √2[2.\sin((x/2 + π/2 -x/2)/2)·\cos((x/2 - π/2 + x/2)/2)]$
$=5 + 2\sin²x + √2[2\sin(π/4)\cos(x/2 - π/4)]$
$=5 + 2\sin²x + 2\cos(x/2 - π/... | HINT
I would start with noticing that
\begin{align*}
\sqrt{2}\left[\sin\left(\frac{x}{2}\right) + \cos\left(\frac{x}
{2}\right)\right] & = 2\times \left[\frac{1}{\sqrt{2}}\times\sin\left(\frac{x}{2}\right) + \frac{1}{\sqrt{2}}\times\cos\left(\frac{x}{2}\right)\right]\\\\
& = 2\sin\left(\frac{x}{2} + \frac{\pi}{4}\righ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4242603",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Prove that $\int_0^1\frac{x^{n+1}}{x+1}dx<\frac{1}{2(n+1)}$ Prove that $$\int\limits_0^1\frac{x^{n+1}}{x+1}dx<\frac{1}{2(n+1)}$$
On simplifying by parts we get:
$$\int\limits_0^1\frac{x^{n+1}}{x+1}dx=\frac{1}{2(n+2)}+\int\limits_0^1\frac{x^{n+2}}{(x+1)^2(n+2)}dx$$
Thus if we prove that$$\max\left(\displaystyle\int\limi... | Inspired by the other answers: $f(x) = \frac{x}{1+x}$ is concave on $[0, \infty)$, and therefore
$$
f(x) \le f(1) + (x - 1) f'(1) = \frac 1 4(1+x)
$$
for $x \ge 0$. It follows that
$$
\int_0^1 \frac{x^n}{1+x}\, dx \le \frac 1 4 \int_0^1 \left(x^{n+1} + x^{n+2} \right) \, dx \\
= \frac 1 4 \left( \frac 1{n+2} + \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4245686",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Stuck solving limit without L'Hospital's rule We need to solve limit:
$$\lim_{x \to \frac{\pi}{3}} \frac{\ln{(2\sqrt{3}\sin{x}-2})}{4\cos^2{x}-1}$$ without using derivation (L'Hospital's rule). With the substitution $t = 4\cos^2{x}-1$, I got it to $$\lim_{t \to 0} \frac{\ln{(\sqrt{9-3t}-2)}}{t}$$ but can't progress any... | Here's another way. First use $\cos^2{x}=1-\sin^2{x}$ and
let $y=\ln{(2\sqrt{3}\sin{x}-2})$, $\frac{e^{y}+2}{2\sqrt{3}}=\sin{x}$, $y\to0$ as $x\to\frac{\pi}{3}$
$$\begin{align}
\lim_{x \to \frac{\pi}{3}} \frac{\ln{(2\sqrt{3}\sin{x}-2})}{4\cos^2{x}-1}
&=\lim_{x \to \frac{\pi}{3}} \frac{\ln{(2\sqrt{3}\sin{x}-2})}{4(1-\si... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4251455",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Integrate $\int_{0}^{\infty}\frac{\tan^{-1}x}{x\left(1+x^{2}\right)}dx$ Q:
Integrate $I=\int_{0}^{\infty}\frac{\tan^{-1}x}{x\left(1+x^{2}\right)}dx$
My Approach:
Put $$\tan^{-1}x=t\to x=\tan t$$
Also we have, $$\frac{dx}{1+x^{2}}=dt$$
We get, $$I=\int_{0}^{\frac{\pi}{2}}\frac{t}{\tan t}dt$$
I'm stuck here, how do I p... | $$\begin{align*}
\int_0^\infty \frac{\tan^{-1}(x)}{x(1+x^2)} \, dx &= \int_0^1 \frac{\tan^{-1}(x)}{x(1+x^2)} \, dx + \int_1^\infty \frac{\tan^{-1}(x)}{x(1+x^2)} \, dx \\[1ex]
&= \int_0^1 \frac{\tan^{-1}(x)}{x(1+x^2)} \, dx - \int_1^0 \frac{\tan^{-1}\left(\frac1x\right)}{\frac1x\left(1+\frac1{x^2}\right)} \frac{dx}{x^2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4253958",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
} |
Find the sum of series $ \sum_{n=1}^{\infty}\frac{1}{n^3(n+1)^3}$ Let it be known that $$\sum_{n=1}^{\infty}\frac{1}{n^2}=\frac{\pi^2}
{6}.$$ Given such—find $$\sum_{n=1}^{\infty}\frac{1}{n^3(n+1)^3}$$
Attempt:
I have tried using the fact that $\displaystyle \frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}$ and then expandi... | HINT:
Use the partial fraction decomposition
$$\frac{1}{n^3(n+1)^3} = \left(\frac{1}{n^3} - \frac{1}{(n + 1)^3}+ \frac{6}{n} - \frac{6}{n+1}\right) - \left(\frac{3}{n^2} + \frac{3}{(n + 1)^2}\right) $$
The sum equals $(1+6) -( 3 \zeta(2) +3 \zeta(2)- 3)= 10 - \pi^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4254127",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 0
} |
Solving $\frac{(1+x)(1+2x)(1+3x)}{(4+x)(4+2x)(4+3x)}=4 $. Simply bringing it to a common denominator does not lead me to success How can I solve this equation?
$$\frac{(1+x)(1+2x)(1+3x)}{(4+x)(4+2x)(4+3x)}=4 $$
Simply bringing it to a common denominator does not lead me to success
What I tried
| If you insist on not dealing with a cubic equation, then maybe this can work. Notice:
$$\dfrac{(1+x)(1+2x)(1+3x)}{(4+x)(4+2x)(4+3x)} = \dfrac{(1+x)(-1-2x)(1+3x)}{(4+x)(-4-2x)(4+3x)} = \dfrac{a(b-c)(c-4)}{b(a-c)(c-1)} = 4$$
where $c = 3x+5$, $a = x+1$ and $b = x+4.$ If you expand this quadratic in $c:$
$$c^2(a-4b)+c(3ab... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4254604",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
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Making $6$ digits numbers by the digits $2,3,3,5,5,5,5$
How many $6$ digits number we can generate by the digits $2,3,3,5,5,5,5$ ?
To solve this problem I considered three cases because we have $7$ digits and looking for six digit number.
*
*Numbers without $2\Rightarrow\quad\dfrac{6!}{4!2!}=15$
*Numbers without ... | \begin{array}{|c|c|c|c|}
\hline\hline
2s&3s&5s&6-\text{digit numbers}\\
\hline
1&1&4&\frac{6!}{4!}=30\\
\hline
1&2&3&\frac{6!}{2!3!}=60\\
\hline
0&2&4&\frac{6!}{2!4!}=15
\end{array}
\begin{array}{|c|c|}
\hline
\hline
\text{Total 6-digit numbers}&30+60+15=105
\end{array}
Your calculations are correct, as shown in the ta... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4258048",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Convergence of $1 - \sqrt{ \frac{2}{n-1} } \cdot \frac{ \Gamma(n/2) }{ \Gamma( \frac{n-1}{2} ) } $ as $n \to \infty$ I stumbled upon this problem while reading about the bias of the sample standard deviation.
How to show that:
$$\bigg(1 - \sqrt{ \frac{2}{n-1} } \cdot \frac{ \Gamma(n/2) }{ \Gamma( \frac{n-1}{2} ) } \bi... | By Stirling's formula we have
$$1 - \sqrt{ \frac{2}{n-1} } \cdot \frac{ \Gamma(n/2) }{ \Gamma( \frac{n-1}{2} ) } \sim 1 - \sqrt{ \frac{2}{n-1} } \cdot \frac{ \sqrt{2\pi \left(\frac n 2 -1\right)}\left(\frac {\frac n 2 -1} e\right)^{\frac n 2 -1} }{\sqrt{2\pi \left(\frac n 2 -\frac 32\right)}\left(\frac {\frac n 2 -\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4258430",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
The series representation of $\sum _{i=0}^{n-1} (-1)^{i+1} \left(\frac{i}{n}\right)^{r}$ Lets consider the:
$$\left(\frac{1}{n}\right)^r-\left(\frac{2}{n}\right)^r+\left(\frac{3}{n}\right)^r-...+\left(\frac{n-1}{n}\right)^r$$
Trying to find any formal serise representation for variable $n$. Have tried the Euler–Maclaur... |
Hint: We can write the expression as
\begin{align*}
\color{blue}{\sum_{i=0}^{n-1}(-1)^{i+1}\left(\frac{i}{n}\right)^r}
&=\frac{1}{n^r}\sum_{i=0}^{n-1}(-1)^{i+1}i^r\\
&=\frac{1}{n^r}\left(\sum_{i=0}^{n-1}i^r-\sum_{i=0}^{\left\lfloor\frac{n-1}{2}\right\rfloor}(2i)^r\right)\tag{1}\\
&\,\,\color{blue}{=\frac{1}{n^r}\left(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4259826",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Struggling to solve an equation containing a floor function I am trying to solve the following equation ($x\in \mathbb R)$
$$
2\lfloor{(x+1)^2+8}\rfloor=(x+1)(2x+3) \tag{1}
$$
I have tried substituting $x$ with its split form (integer part ($n$) + decimal part ($\alpha$)), i.e., $x=n+\alpha$ with $n=\lfloor x \rfloor.$... | Let $x$ be an integer.
Set $y = x+1$, then we have $2\lfloor y^{2}+8 \rfloor=y(2y+1)=2y^{2}+y. $
Since $\lfloor a+c \rfloor = \lfloor a \rfloor +c$ for all integers $a$, $2\lfloor y^{2}+8 \rfloor=2(\lfloor y^{2} \rfloor + 8)=2\lfloor y^{2} \rfloor +16=2y^{2}+y$ from above.
If $y > 16$, then $2\lfloor y^{2} \rfloor = 2y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4260648",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Prove that $\sqrt[3]{a} + \sqrt[3]{b} + \sqrt[3]{c} \ge ab + bc + ca$ For all positive $a,b,c $ satisfying $a+b+c = 3$,Prove:
$$
\sum_{cyc} \sqrt[3]{a} \ge \sum_{cyc} ab
$$
This is a hard problem and I tried it myself, but it's really hard without using advanced techniques(e.g. EV theorem or HCF theorem).
Is there any ... | pqr method:
Using AM-GM, we have
$$\sqrt[3]{a} = \frac{a}{\sqrt[3]{a \cdot a \cdot 1}} \ge \frac{a}{(a + a + 1)/3} = \frac{3a}{2a + 1}.$$
It suffices to prove that
$$\frac{3a}{2a + 1} + \frac{3b}{2b + 1} + \frac{3c}{2c + 1} \ge ab + bc + ca.$$
Let $p = a + b + c = 3, q = ab + bc + ca, r = abc$.
It suffices to prove tha... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4261358",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Eliminating $\theta$ from $\cos^3\theta +a\cos\theta =b$ and $\sin^3\theta +a\sin\theta =c$
Eliminate $\theta$ from the equations.
$$\cos^3\theta +a\cos\theta =b$$
$$\sin^3\theta +a\sin\theta =c$$
Can anyone solve this question?
| $\def\¿{\mathcal}$
Maybe I am complicating this, yet another method.
I will use these identities:
$\cos^6x+\sin^6x=1-\frac34\sin^22x$
$\cos^6x-\sin^6x=\cos 2x(1-\frac14\sin^22x)$
$\cos^4x+\sin^4x=1-\frac12\sin^22x$
$\cos^4x-\sin^4x=\cos2x$
Let
$c=\cos x$,
$s=\sin x$,
$\¿ C=\cos 2x$,
$\¿S=\sin2x$
I use $d$ instead of ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4265926",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
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solve the equation $1+2z+2z^2+\ldots +2z^{n-1}+z^n=0$ im trying to solve the equation $$(E)\quad 1+2z+2z^2+\ldots +2z^{n-1}+z^n=0$$
attempt : because $1$ isnt a solution we have $$\begin{aligned}
1+2 z+2 z^{2}+\cdots+2 z^{n-1}+z^{n} &=2\left(z^{0}+z+z^{2}+\cdots+z^{n-1}\right)-1+z^{n} \\
&=2 \frac{1-z^{n}}{1-z}-1+z^{... | Hint: \begin{align}
z^{n+1}+z^n-z-1&=z^n(z+1)-(z+1)\\
&=(z^n-1)(z+1)
\end{align}
Also, I think you should justify why the second equality holds, i.e. explain why you can assume $1-z\neq 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4269596",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Find the largest value of $P=\frac{3x+2y+1}{x+y+6}$ I know there has been a similar question here, but my question is a little bit different
The Problem:
Given that $3(x+y)=x^2+y^2+xy+2$,
then find the maximum value of
\begin{align}
P=\frac{3x+2y+1}{x+y+6}
\end{align}
What I was thinking about is that I'm trying to tra... | $3(x+y)=x^{2}+y^{2}+xy+2$ must be a conic section because it can be written in the form $Ax^2+Bxy+Cy^2+Dx+Ey+F = 0$. It also is symmetric across the line $y=x$ as it is invariant after swapping $x$ and $y$.
Now if the maximum value is $P$, then $3x+2y+1 = P(x+y+6)$, equation $1$. This can be seen below:
If we can rota... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4271791",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Isn't my book doing this math about differentiation wrongly? Problem:
Differentiate with respect to $x$: $\ln(e^x(\frac{x-1}{x+1})^{\frac{3}{2}})$
My attempt:
Let,
$$y=\ln(e^x(\frac{x-1}{x+1})^{\frac{3}{2}})$$
Both $e^x$ and $\frac{x-1}{x+1}$ are positive: $e^x$ can never be negative, and you can take the square root o... |
My book's attempt:
$$y=x+\frac{3}{2}\ln(\frac{x-1}{x+1})$$
$$=x+\frac{3}{2}(\ln(x-1)-\ln(x+1))\tag{1}\\\ldots$$
$$\frac{dy}{dx}=\frac{x^2+2}{x^2-1}$$
is line $(1)$ valid? I think my book's assumption that $(x-1)$ & $(x+1)$ must be positive is unfounded. $\frac{x-1}{x+1}$ can be positive even when both $(x-1)$ & $(x+1)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4274506",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 2,
"answer_id": 1
} |
$3 \sin x + 4 \cos y = 5$, $4 \sin y + 3 \cos x = 2$ How to find $\sin x$, $\sin y$, $\cos x$, $\cos y$, 2020 contest question https://www.cemc.uwaterloo.ca/contests/past_contests/2020/2020CSMC.pdf
https://www.cemc.uwaterloo.ca/contests/past_contests/2020/2020CSMCSolution.pdf
Question 5 from 2020 CSMC math contest:
$$3... | We have $4\cos y=5-3\sin x$. Similarly we have $4\sin y=2-3\cos x$. Squaring and add:
$$
16=(5-3\sin x)^2+(2-3\cos x)^2=38-30\sin x-12\cos x
$$
so $15\sin x+6\cos x=11$
which gives you possible $\sin x,\cos x$ (i.e., solve the system $15\sin x+6\cos x=11, \sin^2 x+\cos^2 x=1$ to get $\sin x=\frac{55\pm 4\sqrt{35}}{87... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4282724",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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How to use the epsilon delta definition to prove that $\lim\limits_{x\to 1} \frac{x^3-1}{x-1} = 3$
Not sure if I am doing this right, however, this is what I have:
Let $\epsilon > 0$. We need to find a $\delta > 0$ such that $0<|x-1|<\delta$ leads to the conclusion $|f(x)-3|<\epsilon$.
$$|\frac{x^3-1}{x-1} - 3| < \epsi... | Following @Ben's comment...
We want to find $\delta$ such that $|x-1|<\delta \Rightarrow |(x-2)(x-1)|<\epsilon$.
As $|(x-2)(x-1)|=|x-2||x-1|\leq(|x-1|+3)|x-1|$, because of the triangle inequality..then it is enough to find $\delta$ such that
$|x-1|<\delta \Rightarrow (|x-1|+3)|x-1|<\epsilon$
Notice that $(|x-1|+3)|x-1|... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4283629",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
If $\frac{x^2-yz}{a^2-bc}=\frac{y^2-zx}{b^2-ca}=\frac{z^2-xy}{c^2-ab}$, prove that $\frac{x}{a}=\frac{y}{b}=\frac{z}{c}$.
Problem: Let $a, b, c, x, y, z$ be real numbers
such that $abc \ne 0$, $ ~ a + b + c \ne 0$, $~ (a^2 - bc)(b^2 - ca)(c^2 - ab)\ne 0$, $xyz\ne0$, $x+y+z\ne0$, $(x-y)^2+(y-z)^2+(z-x)^2\ne0$
and $$\fr... | An idea would be to show that if
$$[p\colon q\colon r] = [x^2 - y z \ \colon y^2 - x z\ \colon z^2 - x y]$$
then
$$[x\ \colon y \colon z] = [p^2 - q r\ \colon q^2 - p r\ \colon r^2 - p q]$$
In other words, the map from $\mathbb{P}^2$ to itself given by
$$[x\ \colon y \colon z]\mapsto [x^2 - y z \ \colon y^2 - x z\ \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4286648",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
What's the measure of the segment $ON$ in the circle below? For reference:On the circle with center $O$ and perpendicular diameters
perpendiculars $AC$ and $BD$ a string $AE (E~ in~ \overset{\LARGE{\frown}}{BC})$ is drawn
such that AE intersects $BD$ at $M$; if $BM = MO$. Calculate $ON$, if $OA = 12$. ($N =ED\cap AC$).... | Denote $\angle OAM =\alpha, \angle EAB =\beta.$
We have $\tan\alpha =\frac{|MO|}{|OA|} =\frac{1}{2}$ thus $\cos\alpha =2\sin\alpha$ and hence $$\sin\alpha =\frac{\sqrt{5}}{5} , \cos\alpha =\frac{2\sqrt{5}}{5}.$$
Now $$\sin\beta =\sin (\frac{\pi }{4} -\alpha ) =\frac{\sqrt{2}}{2}\cdot \frac{2\sqrt{5}}{5} -\frac{\sqrt{2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4286822",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Find $f: \mathbb R \to \mathbb R$ which satisfies $f\left(xf(y)-y^2\right)=(y+1)f(x-y)$
Find $f: \mathbb R \to \mathbb R$ which satisfies $f\left(xf(y)-y^2\right)=(y+1)f(x-y)$.
My attempt:
\begin{align}
&P(x, -1): f\bigl(xf(-1)-1\bigr)=0. \\
&\text{If } f(-1) \ne 0 \implies x \leftarrow \frac {x}{f(-1)}: f(x-1)=0 \Ri... | For any $y\ne -1$, choose $x$ such that $x f(y) -y^2=-1$. This is possible because, when $f$ is not constant, its only zero is $-1$. Then,
$$
(y+1) f\left(\frac{y^2-1}{f(y)}-y\right) = 0\Leftrightarrow \frac{y^2-1}{f(y)}-y = -1,
$$
This means that any non-constant solution satisfies $f(y)=y+1$ for any $y\ne -1$. Finall... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4287783",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Dealing more efficiently with fractional forms in system of equations As an example, suppose we have to solve the following system of two equations and two unknowns:
$$
\begin{cases}
-\frac{10}{x}-\frac{8}{y} &= \frac{8}{3} \\
-\frac{6}{x}+\frac{6}{y} &= -\frac{1}{3}
\end{cases}
$$
My approach and solution
I opted ... | As requested in a comment, here are the details of what I suggested in a comment. I've arranged the work to minimize the arithmetic details. In fact, I not only didn't have to resort to a calculator, but I did not have to resort to any "paper and pencil" multiplication. For example, $81 \cdot 2$ comes up at one point, ... | {
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"url": "https://math.stackexchange.com/questions/4288341",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Prove that $f(x)^2+f(x+1)^2 = f(2x+1)$ for the function $f$ satisfying $f(x)=f(x-1)+f(x-2), f(1)=f(2)=1.$
Prove that $f(x)^2+f(x+1)^2 = f(2x+1)$ for the function $f$ satisfying $f(x)=f(x-1)+f(x-2), f(1)=f(2)=1.$
I know how to prove it, but it is interesting so I am posting it.
The first hint is:
\begin{align} &\text... | My answer above:
\begin{align} &f(x)=f(x-1)+f(x-2) = f(2)f(x-1)+f(1)f(x-2). \ \\ \ \\ &f(x-1)=f(x-2)+f(x-3) \\ &\Rightarrow f(x)=f(2)(f(x-2)+f(x-3))+f(1)f(x-2)=(f(1)+f(2))f(x-2) \\ &+f(2)f(x-3)=f(3)f(x-2)+f(2)f(x-3). \ \\ \ \\ &f(x-2)=f(x-3)+f(x-4). \\ &\Rightarrow f(x)=f(3)(f(x-3)+f(x-4))+f(2)f(x-3)=(f(2)+f(3))f(x-3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4290814",
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"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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General form for $\sum_{n=1}^{\infty} (-1)^n \left(m n \, \text{arccoth} \, (m n) - 1\right)$ I'm wondering if there is a general form for the following sum:
$$\sum_{n=1}^{\infty} (-1)^n \left(m n \, \text{arccoth} \, (m n) - 1\right)$$ for $m \in \mathbb{N}$
I have obtained the following closed-forms for these speci... | Assume that $m >1$.
Similar to my answer for the non-alternating version of the series, we can exploit the fact that $\csc(\pi z)$ has the Laurent series expansion $$\csc(\pi z) = \frac{1}{\pi z} + \frac{2}{\pi z} \sum_{n=1}^{\infty} \eta(2n)z^{2n}, \quad 0 <|z| < 1, $$ where $\eta(z)$ is the Dirichlet eta function.
Th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4290994",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 3,
"answer_id": 1
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Proving $\sum_{n=1}^\infty \frac{1}{(n^2+a^2)^2}=\frac{\pi}{4a^3}\left[\coth(\pi a)+\frac{\pi a}{\sinh^2 \pi a}-\frac{2}{\pi a}\right]$ Use the function $$f(z)=\frac{\pi\text{cosec}(\pi z)}{z^2+a^2}$$
to show that
$$\sum_1^\infty \frac{1}{(n^2+a^2)^2}=\frac{\pi}{4a^3}\left[\coth(\pi a)+\frac{\pi a}{\sinh^2 \pi a}-\frac... | Without the hint.
$$\frac{1}{(n^2+a^2)^2}=\frac{i}{4 a^3 (n+i a)}-\frac{i}{4 a^3 (n-i a)}-\frac{1}{4 a^2 (n+i
a)^2}-\frac{1}{4 a^2 (n-i a)^2}$$ Computing the partial sums
$$\sum_{n=1}^p\frac{i}{4 a^3 (n+i a)}=-\frac{i (\psi ^{(0)}(i a+1)-\psi ^{(0)}(i a+p+1))}{4 a^3}$$
$$\sum_{n=1}^p\frac{i}{4 a^3 (n-i a)}=-\frac{i ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4291745",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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What is my mistake in this counting problem?
Question: How many 6-digit numbers, written in decimal notation, have at least one 1, one 2, and one 3 among their digits?
Define $A, B, C$ to be the set of $6$-digit numbers with at least one $1$, at least one $2$, and at least one $3$, respectively.
We may count the card... | We have to find $|A \cap B \cap C|$.
Note that the complement of $A$ is the set of $6$-digit numbers which have no digit $1$, therefore $|A^c|=8\cdot 9^5$. Hence in your attempt you should have $|A|=9\cdot 10^5-8\cdot 9^5$ which less than $8\cdot5\cdot10^4 + 10^5$ because you are overcounting! For instance the number $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4291911",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Number of solutions of $\cos^5x+\cos^5\left( x+\frac{2\pi}{3}\right) + \cos^5\left( x+\frac{4\pi}{3}\right) =0$
Solve in the interval $[0,2\pi]$ : $$\cos^5x+\cos^5\left( x+\frac{2\pi}{3}\right) + \cos^5\left( x+\frac{4\pi}{3}\right)=0 $$
I tried expanding the L.H.S by applying the formula of $\cos(A+B)$ but it result... | Clearly the roots of $$4c^3-3c-\cos3x=0$$ are $c_r\cos(x+2r\pi/3), r=0,1,2$
Let $p=c_0,c_1=q,c_2=r$
$p+q+r=0\implies p^3+q^3+r^3=3pqr$
$pq+qr+rp=-3/4\implies p^2+q^2+r^2=0^2-2(-3/4)$
$pqr=\dfrac{\cos3x}4$
As $4c^5=3c^3+c^2\cos3x,$
$4(p^5+q^5+r^5)$
$=3(p^3+q^3+r^3)+(p^2+q^2+r^2)\cos3x$
Replace the values of $p^3+q^3+r^3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4293179",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
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Find $a\in \mathbb{R}$ such that a root of $ax^3-13x^2+(15a)x-25$ is $2+i$ Find $a\in \mathbb{R}$ such that a root of the polynomial $$p(x)=ax^3-13x^2+(15a)x-25$$ is $2+i$
Solution:
$q(x)=\frac{p(x)}{a}$
$=x^3-\frac{13}{a}x^2+15x-\frac{25}{a}$
$p(x)$ and $q(x)$ have the same roots, call them $r_1, r_2, r_3$
$r_1=2+i$. ... | $(x-2)^2 = -1 $ or $x^2 - 4x + 5 = 0$ represents the pair of complex roots.
$$\begin {aligned}
p(x) &=ax^3-13x^2+(15a)x-25 \\
& = (ax^3 - 4ax^2 + 5ax) - ((13-4a) x^2 - 10ax + 25) \\
& = ax (x^2 - 4x + 5) - 5 \left(\frac{13-4a}{5} x^2 - 2ax + 5 \right)
\end{aligned}$$
That gives $a = 2$ as the solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4295130",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Inequalities when matrix is negative definite
Let $$A = \begin{pmatrix} x&1&y \\ 1&-1&0 \\ y&0&-2 \end{pmatrix} $$ $A$ is negative definite $\iff x<a \text{ and } bx^2 + cy^2 + dx + 2 < 0$. Find $a,b,c,d$
I know that for a matrix to be negative definite all of its eigenvalues must be negative (I thought this may lin... | $A$ is negative definite iff $(-A)$ is positive definite. To test the positive definiteness of $(-A)$, we have to calculate all three principal determinants of $(-A)$
$-A = \begin{bmatrix} -x && - 1 && - y \\ -1 && 1 && 0 \\ - y && 0 && 2 \end{bmatrix}$
$\Delta_1 = - x \gt 0 \Longleftrightarrow x \lt 0$
$\Delta_2 = ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4297791",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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$\int_c 3y dx +5x dy + \frac{2x+3}{z^2} dz$ for the intersection between two surfaces For the closed curve, $c = \{(x,y,z) \vert x^2 + y^2 - z^2 =0\} \cap \{(x,y,z) \vert (x-1)^2 + y^2 =4\}$
Find the $\int_c 3y dx +5x dy + \frac{2x+3}{x^2 + y^2} dz$
First I focused the surface $\{(x,y,z) \vert x^2 + y^2 - z^2 =0\}$ and... | $x^2 + y^2 - z^2 = 0$ represents a cone with axis along the $z$ axis, and $(x-1)^2 + y^2 = 4 $ is a right circular cylinder with axis parallel to the $z$ axis and passing through $(1, 0)$ and a radius of $2$.
Let $x = 1 + 2 \cos \theta , y = 2 \sin \theta $, then this point lies on the cylinder, the intersection with t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4303867",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Determining all $(a,b)$ on the unit circle such that $2x+3y+1\le a(x+2)+b(y+3)$ for all $(x,y)$ in the unit disk In the middle of another problem, I came up with the following inequality which needed to be solve for $(a, b)$ : $$2x+3y+1\le a(x+2)+b(y+3)$$ for all $(x, y)\in\mathbb{R}^2$ with $x^2+y^2\le1.$
Here the sol... | Since $(a,b)$ lies on the unit circle and $(a,b)$ satisfies the linear inequality, it must be tangent to the circle. So first find the tangent line to the unit circle at $(\cos t, \sin t)$. This is:
$$y - \sin t = (-\cot t)(x - \cos t) \implies y = -(\cot t)x + \csc t.$$
Now the equality case of the condition $2x+3y+1 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4305595",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Integrate: $\frac{x^2+x-1}{x^2-1}$ with respect to $x$ Ok so here is where I am up to:
I can find some similarity with the numerator and denominator and managed to reduce to the following:
$\int\frac{x^2+x-1}{x^2-1}dx = \int\frac{x^2-1}{x^2-1}+\frac{x}{x^2-1}dx$
$=\int1+\frac{x}{x^2-1}dx.$
I tried to reduce it further ... | Yeah!
My first mathSE integration problem answer, in I don't remember how long.
$\displaystyle \frac{1}{x^2-1} = \frac{1}{2}
\times \left[\frac{1}{x-1} - \frac{1}{x+1}\right].$
$\displaystyle \frac{x}{x^2-1} = \frac{1}{2}
\times \left[\frac{x}{x-1} - \frac{x}{x+1}\right].$
This equals $\displaystyle
\frac{1}{2} \lef... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4307299",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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convert hyperbola in rectangular form to polar form i am trying to convert the rectangular equation of a conic (hyperbola) to a polar form.
the rectangular equation is:
$$3y^2 - 16y -x^2 + 16 = 0.$$
i substituted $r\sin\theta$ for y and $r\cos\theta$ for x, and tried to simplify, but I am stuck.
i tried substituting $1... | Staring with $3 y^2 - 16 y - x^2 + 16 = 0 $
Put it first in the standard format. Complete the square in $y$
$ 3 \left(y - \dfrac{8}{3} \right)^2 - \dfrac{64}{3} - x^2 + 16 = 0 $
$ 3 \left(y - \dfrac{8}{3} \right)^2 - x^2 = \dfrac{16}{3} $
$ \dfrac{9}{16} \left( y - \dfrac{8}{3} \right)^2 - \dfrac{3}{16} x^2 = 1 $
Thus... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4308224",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the number of ways in which the number 30 can be partitioned into three unequal parts.(Please use multinomial theorem) OfLet $a,b,c$ be the parts such that $a<b<c$
Now, let $a−b=x, b−c=y,$ implies $x,y>0$
$⇒a+b+c=30\\
⇒(b+x)+b+c=30\\
⇒x+2(c+y)+c=30\\
⇒x+2y+3c=30$ ,
$c≤27$
Sum =30,Co-eff =$1,2,3$
$(x... | You are pretty much on the right track. The last part is pretty tedious, so it might be better that you just use Wolfram Alpha or possibly some other program. I think you defined your variables a little wrong. For example, if $a<b$, then $a-b<0$. However, you defined $x=a-b$, but you are saying that $x>0$. It looks lik... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4310314",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Number of real solutions of $\begin{array}{r} {\left[\frac{2 x+1}{3}\right]+\left[\frac{4 x+5}{6}\right]} =\frac{3 x-1}{2} \end{array}$ Solve for $x \in \mathbb{R}$ $$\begin{array}{r}
{\left[\frac{2 x+1}{3}\right]+\left[\frac{4 x+5}{6}\right]}
=\frac{3 x-1}{2}
\end{array}$$ where $[x]$ denotes greatest integer less th... | Hint :
$$ \lfloor a \rfloor + \lfloor a + \frac{1}{2} \rfloor = \lfloor 2a \rfloor $$
This follows from Hermite's identity.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4311152",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
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Rationalize the denominator of $\frac{4}{9-3\sqrt[3]{3} + \sqrt[3]{9}}$ Rationalize the denominator of $\frac{4}{9-3\sqrt[3]{3}+\sqrt[3]{9}}$
I keep making a mess of this. I tried vewing the denominator as
$a +\sqrt[3]{9}$, where $a=9-3\sqrt[3]{3}$ and secondly as
$b -3\sqrt[3]{3}+\sqrt[3]{9}$, where $b=9$.
Then using ... | in general we may factor the norm form
$ x^3 + d y^3 + d^2 z^3 - 3dxyz$
as
$$ x^3 + d y^3 + d^2 z^3 - 3dxyz = $$ $$ \left( x+ d^{1/3}y + d^{2/3} z \right) \left( x^2 + d^{2/3} y^2 + d^{4/3} z^2 - dyz -d^{2/3}zx - d^{1/3} x y \right) $$
so that
$$ \frac{1}{ x+ d^{1/3}y + d^{2/3} z} = $$ $$ \frac{x^2 + d^{2/3} y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4315306",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Seeking for help to find a formula for $\int_{0}^{\pi} \frac{d x}{(a-\cos x)^{n}}$, where $a>1.$ When tackling the question, I found that for any $a>1$,
$$
I_1(a)=\int_{0}^{\pi} \frac{d x}{a-\cos x}=\frac{\pi}{\sqrt{a^{2}-1}}.
$$
Then I started to think whether there is a formula for the integral
$$
I_n(a)=\int_{0}^{\p... | Express the integral as
\begin{align}
\int_{0}^{\pi} \frac{d x}{(a-\cos x)^{n}}
=\sum_{k=0}^{[\frac{n-1}2]} \frac{\binom{n-1}{2k}a^{n-2k-1}}{(a^2-1)^{n-1/2}}\int_0^{\pi}\cos^{2k}x \ dx
= \frac{\pi P(a)}{\left(a^{2}-1\right)^{n-{1}/{2}}}
\end{align}
which leads to
$$P(a) = \sum_{k=0}^{[\frac{n-1}2]}
\binom{n-1}{2k} \fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4315858",
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"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
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Choosing at least 2 women from 7 men and 4 women In how many different ways can we choose six people, including at least two women, from a group made up of seven men and four women?
Attempt:
As we have to have at least two women in the choices, then $\displaystyle\binom{4}{2}$, leaving a total of $4$ out of $9$ people ... | Consider three cases:
*
*choosing 2 women and 4 men
*choosing 3 women and 3 men
*choosing 4 women and 2 men
So the total number of arrangements is:
$\binom{4}{2} \times \binom{7}{4}+ \binom{4}{3} \times \binom{7}{3}+ \binom{4}{4} \times \binom{7}{2}=371$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4317983",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 3
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Solve indefinite integral $\int\frac{x^2}{1-x^2+\sqrt{1-x^2}}dx$ $$\int\frac{x^2}{1-x^2+\sqrt{1-x^2}}dx$$
I multiply the integral so that I can get $-x^2$ in the numerator. I then expand the fraction so I can split the integral into easier integrals.
$$-\int\frac{-x^2}{1-x^2+\sqrt{1-x^2}}dx$$
$$-\int\frac{1-x^2+\sqrt{1... | The terms $\sqrt{1-x^2}$ motivates us to use change of variable by trigonometric functions. The term $\sqrt{1-x^2}$ naturally implies $x\in[-1,1]$ and so we can assume $x=\sin(t)$, $t\in[-\frac{\pi}{2},\frac{\pi}{2}]$. Then
\begin{align*}
\int\frac{x^2}{1-x^2+\sqrt{1-x^2}}\,dx&=\int\frac{\sin^2(t)}{\cos^2(t)+\cos(t)}\,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4319573",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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How does $\tan^2(x) \sec(x) + \sec^3(x)$ turn in to $2\sec^3(x) - \sec(x)$ Can someone explain how $\tan^2 $ disappeared and $\sec^3$ turn into $2\sec^3$ ???
The derivatives of the function $2\sec(x)\tan(x)$ is apparently $2(-\sec(x) + 2\sec^3(x))$
| Need these
$$
\tan=\frac{\sin}{\cos} \tag{$\color{green}{\blacksquare}$}
$$
$$
\sec=\frac{1}{\cos} \tag{$\color{blue}{\blacksquare}$}
$$
$$
\sin^2+\cos^2=1\longrightarrow \sin^2=1-\cos^2 \tag{$\color{red}{\blacksquare}$}
$$
Just use properties and algebra
\begin{align*}
\tan^2\sec+\sec^3
&=
\left(\frac{\sin}{\cos}\righ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4320342",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Limit without hospital's rule $\lim_{x\rightarrow 0} (\frac{1}{x^2} - \frac{x}{\sin^3(x)})$
Calcule $\lim_{x\rightarrow 0} (\frac{1}{x^2} - \frac{x}{\sin^3(x)})$ without using L'Hospital
I first was trying to replace $x = \frac{\pi}{2} + u$ but I can't find a rule or eliminate anything.
After that I try using cubes:
... | $$
\begin{align}
\lim_{x\to 0}\left(\frac{1}{x^2}-\frac{x}{\sin^3x}\right) &= \lim_{x\to 0}\left(\frac{\sin^3x-x^3}{x^2\sin^2x}\right) \\
&= \lim_{x\to 0}\left(\frac{x^3}{\sin^3x}\right)\left(\frac 1{x^2}\right)\left(\frac{\sin^3x}{x^3}-1\right) \\
&=1\cdot\lim_{x\to 0}\left(\frac{1}{x^2}\right)\left(\frac{\sin x}{x}-1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4320754",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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About an inequality wich is an upper bound for Am-Gm. Hi it's a upper bound for New bound for Am-Gm of 2 variables :
Problem :
Let $0<a\leq 1$ and $x\geq 1$ such that $2\leq a+x\leq 5$ then (dis)prove :
$$\left(1+\frac{1}{x+a}\left(\frac{4\left(x+a\right)^{2}}{\left(x+a+2\right)^{2}}-1\right)\right)\left(x+a-\frac{2\le... | Partial answer :
We prove a first step :
we have for $x\in[1,2-a]$ and $0.5\leq a\leq 1$ :
$$g(x)\leq x+a-1$$
or we need to show :
$$f(x)=\left(a+1\right)\ln\left(a\right)+\left(x+1\right)\ln\left(x\right)-\left(a+x\right)\ln\left(a+x-1\right)$$
It's easy using a computer or WA to see that the function is convex on the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4323015",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Using limits properties when proving using epsilon-delta definition. I want to prove $\lim\limits_{x \to 5} \sqrt{2x+6} = 4$
using the epsilon-delta definition.
My intuition to proving this is to use the root rule and square both ends of the equation to end up with:
$\lim\limits_{x \to 5} 2x+6 = 16$
I would then contin... | Note: This is extra detailed so you can use it as a template for similar problems.
When trying to use the $\epsilon, \delta$ method to show that
$$\lim_{x\to a}f(x)=L$$
Suppose that for some punctuated interval $I=(a-t)\cup(a+t)$ we can find an upper bound $B$ such
$$ \left|\frac{f(x)-L}{x-a}\right|\le B $$
for $x\in I... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4324207",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
$10$ distinct color balls and we want to divide them into $3$ identical boxes with hard restrictions and probability
We have $10$ distinct color balls and we want to divide them into $3$ identical boxes .We decide that two of the groups can contain at most $3$ balls and the rest one contain at most $4$ balls. Moreover... | First question:
The first term is incorrect. It should be,
$ \displaystyle \binom{8}{1}\binom{7}{3}\binom{4}{4}=280$
Unlike second term, it should not be divided by $2!$ as the boxes having $3$ balls each are different groups - one of them has $A$ and $B$ and the other is without them.
So the answer should be $560$
Sec... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4325294",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Diophantine problem: Find all pairs of $(x, y)$ such that $x^3-4xy+y^3=-1$
Find all pairs of integers $x, y$ such that:
$$x^3-4xy+y^3=-1$$
My analysis:
*
*$x$ and $y$ can't be both negative because all terms would be negative and $(x,y)=(-1,-1) \implies (-1)^3-4(-1)(-1)+(-1)^3 =-6 $ would already be too small.
*R... | Hint: A messy quadratic is better than a simple cubic while dealing with integers.
$$x+y=s\implies x^3-4x(s-x)+(s-x)^3=-1 $$
Simplifying, $$ (3s+4)x^2-(3s^2+4s)x+s^3+1=0$$
The discriminant must be non-negative, $$\Delta=-(3s+4)(s^3-4s^2+4)\ge 0$$
If $s\in (-\infty,-2]$ or $s\in [4,\infty)$, $\Delta<0$. Therefore... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4326149",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Taylor expansion of $\sin \pi z$ at $z = -1$. Taylor expansion of $\sin \pi z$ at $z = -1$ is
$$\sin\pi z = -\sin(\pi(z+1)) = -\sum_{n=0}^\infty \frac{(-1)^n\pi^{2n+1}}{(2n+1)!}(z+1)^{2n+1}$$ so that
$$\sin\pi z = \sum_{n=0}^\infty \frac{(-1)^{n+1}\pi^{2n+1}}{(2n+1)!}(z+1)^{2n+1}. \tag{$\dagger$}$$
But if I try this
\b... | You can continue by changing the order of summation and then splitting the sum into two parts:
\begin{align*}
& \sum\limits_{n = 0}^\infty {\frac{{( - 1)^{n + 1} \pi ^{2n + 1} }}{{(2n + 1)!}}\left( {\sum\limits_{k = 0}^{2n + 1} {( - 1)^k \binom{2n + 1}{k}(z + 1)^k } } \right)}
\\ & = \sum\limits_{k = 0}^\infty {\lef... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4329877",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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$1^2 - 2^2 + 3^2 - 4^2 + \dots + 1999^2$ using closed form formula for sum of squares of first $n$ natural numbers The question is simple, the solution, not so much
Q.Find the sum of the given expression $1^2- 2^2 + 3^2 - 4^2 + \dots + 1999^2$
My idea is we know $1^2 + 2^2 + 3^2 + 4^2 + \dots + n^2 = \frac{n(n + 1)(2n ... | If you first add all the squares:
$$
1^2 + 2^2 + 3^2 + 4^2 + 5^2 + 6^2 + \cdots
$$
and then subtract the even squares:
$$
\phantom{1^2} - 2^2\phantom{ + 3^2} - 4^2 \phantom{+ 5^2} - 6^2 \phantom{-}\cdots
$$
then you are left with only the odd squares:
$$
1^2 \phantom{+ 2^2} + 3^2 \phantom{+ 4^2} + 5^2 \phantom{+ 6^2} +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4335944",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 0
} |
$a,b>0$ then: $\frac{1}{a^2}+b^2\ge\sqrt{2(\frac{1}{a^2}+a^2)}(b-a+1)$
Let $a,b>0$. Prove that: $$\frac{1}{a^2}+b^2\ge\sqrt{2\left(\frac{1}{a^2}+a^2\right)}(b-a+1)$$
Anyone can help me get a nice solution for this tough question?
My approach works for 2 cases:
Case 1: $b-a+1>0$ then squaring both side, we get equival... | Due to @Calvin Lin, I will post solution based on his hint later.
Now I get solution by AM-GM. Notice that: $\frac{1}{a^2}+a^2=\left(\frac{1}{a}+a\right)^2-2=\left(\frac{1}{a}+a+\sqrt{2}\right)\left(\frac{1}{a}+a-\sqrt{2}\right)$
Then using AM-GM: $$2\frac{\sqrt{2}\frac{1}{a^2}}{\sqrt{\frac{1}{a^2}+a^2}}+\frac{\frac{1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4338382",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Calculate Indefinite Integral $\int \frac{x^5(1-x^6)}{(1+x)^{18}}dx$ The following integration is given by Wolfram Alpha
$$\int \frac{x^5(1-x^6)}{(1+x)^{18}}dx=\frac{x^6(28x^4+16x^3+39x^2+16x+28)}{168(1+x)^{16}}.$$
My question is: what is the best (meaning least work), method to achieve this result by hand ? There are ... | $$\int \frac{x^5 \left(1-x^6\right)}{(x+1)^{18}}\,dx=\frac{P(x)}{(x+1)^{17}}$$
$$\frac{x^5 \left(1-x^6\right)}{(x+1)^{18}}=\frac{(x+1) P'(x)-17 P(x)}{(x+1)^{18}}$$ So, $P(x)$ is a polynomial of degree $11$.
Let
$$P(x)=\sum_{n=0}^{11} a_n\,x^n$$ Expand and group powers to get
$$0=(a_1-17 a_0)+(2 a_2-16 a_1) x+(3 a_3-15 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4339068",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
} |
If $\frac{n^2}{3} < xy \leq n^2 - 5$, then does it follow that $n > \max(x,y)$? My question here is pretty basic:
If $\frac{n^2}{3} < xy \leq n^2 - 5$, then does it follow that $n > \max(x,y)$?
Here, $n, x, y$ are positive integers.
MY ATTEMPT
Let $n, x, y$ be positive integers such that
$$\frac{n^2}{3} < xy \leq n^2... | For every $n\ge 3$, a counter example is given by $x=n-2$ and $y=n+1$
indeed in that case we have
\begin{equation}
\frac{n^2}{3}< n^2 - n - 2 = x y \le n^2 - 5
\end{equation}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4340994",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
prove that $(p,mp]$ will always contain at least $y-1$ cubes
If $n, m\in \mathbb{Z}_{>0}$ and $(n, mn]$ contains $y$ cubes, prove that $(p,mp]$ will always contain at least $y-1$ cubes (integers $a$ so that $a=k^3$ for some integer k) for all $p\in [n,\infty)$.
This works for small examples, such as for $n=7, m=4$ wh... | The statement to prove is obviously true for $y = 0$ and $y = 1$, so consider just $y \gt 1 \; \to \; y - 1 \gt 0$. Since $(n, mn]$ contains $y$ cubes, this means there's an $a \in \mathbb{Z}_{>0}$ such that
$$n \lt a^3 \lt (a + (y - 1))^3 \le mn \tag{1}\label{eq1A}$$
Since the ratio of the upper & lower limits, i.e., ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4342530",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Integration of Bounded Region I'm trying to solve this problem:
The region of integration is the triangle $D$ with vertexes $A(0,0),B(1,1),C(10,1)$.
Find the solution of $\iint_D \sqrt{x^2-y^2}\,dx\,dy$.
MY SOLUTION (as @ryang suggested):
We can write the triangle as $D=\{(x,y)\in\mathbb R^2|0\le y\le 1,y\le x \le 10y\... | MY SOLUTION (as @ryang suggested):
We can write the triangle as $D=\{(x,y)\in\mathbb R^2|0\le y\le 1,y\le x \le 10y\}$.
After that the integral becomes:
$$\int_0^1\int_y^{10y} \sqrt{x^2-y^2}\,dx\,dy$$
We can solve first this indefinite integral $$\int\sqrt{x^2-y^2}\,dx$$
where $y\in[0,1]$ is a constant.
So$$\begin{alig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4343599",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
How to solve system of linear equations that came up in looking at a probability problem, any tricks? The following system of $k$ equations came up in a probability problem I was looking at:
$$(n+1)x_1 - x_2 = 1, \quad x_1 + nx_2 - x_3 = 1 , \quad x_1 + nx_3 - x_4 = 1, \quad x_1 + nx_4 - x_5 = 1, \quad \ldots \quad x_1... | Note that we have $\forall r\in [2,k]$
$$x_r=nx_{r-1}+x_1-1$$
With the condition that $x_k=\frac{2-x_1}{n}$. Note that we can actually solve this recursion for a general formula. Shift the indices up by $1$ to get that
$$x_{r+1}=nx_r+x_1-1$$
and subtract the two equations to get
$$x_{r+1}-(n+1)x_r+nx_{r-1}=0$$
Note tha... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4344226",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
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.