Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
$1-x+x^2-x^3+..(-1)^nx^n$ I have the following sum:
$1-x+x^2-x^3+..(-1)^nx^n, x\neq -1$
So what I thought was separating it in two cases like this:
Case 1. n is even
$$
1+x^2+x^4+...+x^n-x(1+x^2+...+x^n)
$$
Which I can turn into $\frac{1-x^{n+2}}{1-x^2}-\frac{x(1-x^{n+2})}{1-x^2}=\frac{1-x^{n+2}}{1-x^2}(1-x)$
Case 2. n... | It should be $$\frac{(1-x)(1-(-1)^{n+1}x^{n+1})}{1-x^2} = \frac{1-(-1)^{n+1}x^{n+1}}{1+x}$$
But why make it complicated? This is just a geometric series $\sum_{i=0}^n (-x)^i$.
| {
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"url": "https://math.stackexchange.com/questions/2339661",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Find the smallest number which when divided by 3, 5 and 7 leaves remainders 2, 4 and 6 respectively Find the smallest number which when divided by 3, 5 and 7 leaves remainders 2, 4 and 6 respectively
I took out the LCM and I thought 105 would be the answer but answer is 104. Can anyone please explain the approch behind... | From Euclid's Algorithm we have $\color{blue}{2 \times 5} - \color{red}{3 \times 3}=1$ showing that $3$ & $5$ are coprime.
To solve
\begin{eqnarray*}
x \equiv \color{red}{2} \pmod{3} \\
x \equiv \color{blue}{4} \pmod{5}
\end{eqnarray*}
$x$ is construct as $x= \color{red}{2} \times \color{blue}{ 2 \times 5} - \color{blu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2341101",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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What is the minimum value of $\frac{a_1}{a_2} + \frac{a_2}{a_3} +\cdots + \frac{a_n}{a_1} $ Let $a_1,a_2,\cdots,a_n$ are positive real numbers.
Question: what is the minimum value of
$$\frac{a_1}{a_2} + \frac{a_2}{a_3} +\cdots + \frac{a_n}{a_1} $$
Thought: I have no clue how to proceed. Tried some standard inequalities... | By the inequality of arithmetic and geometric means we have that $$\frac{a_{1}}{a_{2}} + \frac{a_{2}}{a_{3}}+\dots +\frac{a_{n}}{a_{1}} \geq n \, \left( \frac{a_{1}\cdot a_{2}\cdot \dots \cdot a_{n-1} \cdot a_{n}}{a_{2}\cdot a_{3} \dots \cdot a_{n} \cdot a_{1}}\right)^{1/n} = n $$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solve $\frac{x+12}{x-3} \gt |x-1| + 1$ Solve $\frac{x+12}{x-3} \gt |x-1| + 1$
I finally got $\frac{(x-6)(x+2)}{x-3} \lt 0$
That means the solutions are $3\lt x \le 6$ and $x\lt -2$ Am I right?
| We can rearrange this equation to give us
\begin{align}\frac{x+12}{x-3}&>|x-1|+1\\
\frac{x+12}{x-3}-1&>|x-1|\\
\frac{x+12}{x-3}-\frac{x-3}{x-3}&>|x-1|\\
\frac{x+12-(x-3)}{x-3}&>|x-1|\\
\frac{15}{x-3}&>|x-1|\end{align}
Clearly $x\neq 3$
We now need to consider three intervals: $(-\infty,1]$, $[1,3)$, $(3,\infty)$
When ... | {
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How many ways are there to cover a $2\times 16$ rectangle with $2\times 2,$ $2\times 3$ and $2\times 4$ rectangles?
How many ways are there to cover a $2\times 16$ rectangle with $2\times 2,$ $2\times 3$ and $2\times 4$ rectangles?
I already dealt with a similar problem, which is how many ways are there to cut a $1\t... | We are dealing with a $2 \times 16$ grid. The first column of two cells can be filled by a $2 \times 2$, a $2 \times 3$ or a $2 \times 4$ grid. Let us call $f(n)$ the number of ways to fill a $2 \times n$ grid. We then have:
$$f(n) = f(n-2) + f(n-3) + f(n-4), n \geq 4$$
We also know that:
$$f(0) = 1$$
$$f(1) = 0$$
$$f(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2346232",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Maximum value of a function. I am not able to check double derivative. Can someone explain me how $\sin^p x \cos^q x$ attains maximum at $\tan^2 x = \frac pq$.
I am not able to check whether double derivative is positive or negative.
Question
Show that $$\sin^p\theta\cos^q\theta$$ attains a maximum when $$\theta=\tan^... | Let $f(x) = \sin^{p}(x) \, \cos^{q}(x)$ then
\begin{align}
f'(x) &= f(x) \, (-q \, \tan(x) + p \, \cot(x)) \\
&= -q \, \cot(x) \, f(x) \, \left( \tan^{2}(x) - \frac{p}{q} \right)
\end{align}
Now, the maximum of a function is defined by $f'(x) = 0$ then, seeking values other than $f(x) = 0$,
$$\tan^{2}(x) = \frac{p}{q}$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2346890",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
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Let $a,b,c$ be roots of $x^3+px+r=0$. Find the cubic whose roots are $(a-b)^2$,$ (b-c)^2$ and $(c-a)^2$ Question
Let $a,b,c$ be roots of $x^3+px+r=0$. Then find the cubic whose roots are $(a-b)^2, (b-c)^2$ and $ (c-a)^2$
Attempt
I have tried using Vieta's formulas to compute coefficients of the sought cubic.
For sum of... | $\sum_{cyc}(a-b)^2$ is a symmetric polynomial of degree two, so it must be a polynomial in the coefficients of $x^3+px+r$. The only possibilities to form a polynomial of degree $2$ are $(a+b+c)^2$ and $ab+bc+ca=p.$ Since all terms containing the factor $a+b+c$ must vanish, it can be only a multiple of $p$. From the spe... | {
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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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The number $m+n^2$ is divisible by $m+n$. Prove that the number $m+n^3$ is divisible by $m+n$.
$m$ and $n$ are positive integer numbers; $m+n\mid m+n^2$.
Prove that $m+n\mid m+n^3$
So far I have done:
If $m+n\mid m+n^2$, then $m+n\mid (m+n^2)-(m+n)=n^2-n=n(n-1)$.
$m+n^3=n(n^2-n)+(m+n^2)$ and each component is divis... | Hint: If $m + n^2$ is divisible by $m + n$ then so is $(m + n^2) - (m + n) = n^2 - n$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2352289",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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$\text{If } x(x+y+z)=20,y(x+y+z)=30 \text{ and } z(x+y+z)=50 \text{ then the value of } 2(x+y+z) is:$
If $x(x+y+z)=20$, $y(x+y+z)=30$ and $z(x+y+z)=50$ then what is the value of $2(x+y+z)$?
Ans. $20$
I have tried the following:
$$ \frac{20}{x}=\frac{30}{y}=\frac{50}{z}$$
From which I get:
$$x:y:z=2:3:5$$
Now,
$$2(x+y... | Consider instead $$x(x+y+z)+y(x+y+z)+z(x+y+z)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2353505",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Rotate $xy=1$ by $\frac{\pi}{4}$ in a negative (clockwise) direction. I was studying hyporbolae for the first time and noticed that $y=\frac{1}{x}$ is a rotated hyperbola. I had seen equations like $y=\frac{1}{x}$ before but never noticed they where hyperbolae.
Anyway using geometry and the general form of a hyperbola,... | You can't get a 4th degree equation by an affine transformation of coordinates.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2354835",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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if a,b,c are roots of a cubic equation then for the following question... If $a, b, c$ are roots of $x^3 -3x^2 + 2x +4 = 0$ and
$$y= 1 + \frac{a}{x-a} + \frac{bx}{(x-a)(x-b)} + \frac{cx^2}{(x-a)(x-b)(x-c)}$$
then value of $y$ at $x=2$ is:
| $(x-a)(x-b)(x-c)+a(x-b)(x-c)+bx(x-c)+cx^2=x^3,$
which says that
$$y=\frac{x^3}{(x-a)(x-b)(x-c)}.$$
Thus, $$y(2)=\frac{8}{(2-a)(2-b)(2-c)}=\frac{8}{8-4\cdot3+2\cdot2-(-4)}=2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2355843",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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How to calculate $\sum_{r=1}^\infty\frac{8r}{4r^4+1}$?
Calculate the following sum: $$\frac{8(1)}{4(1)^4+1} + \frac{8(2)}{4(2)^4+1} +\cdots+ \frac{8(r)}{4(r)^4+1} +\cdots+ \text{up to infinity}$$
MY TRY:- I took $4$ common from the denominator. and used $a^2+b^2=(a+b)^2-2ab$. It gave me two brackets, whose subtractio... | Let the general term be $\frac {8r}{4r^4+1}=\frac {2r}{r^4+(\frac {1}{2})^2}=\frac {2r}{(r^2+\frac {1}{2})^2-r^2} =\frac {2r}{(r^2+\frac {1}{2}-r)(r^2+\frac {1}{2}+r)} $ now $2r=r^2+\frac {1}{2}+r-(r^2+\frac {1}{2}-r) $ so iur series is a telescoping one and is equal to $S=\sum _0^{\infty} \frac {1}{r^2+\frac {1}{2}-r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2357340",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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"answer_id": 3
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Find the domain of $f(x)=\log_{10}(1-\log_7(x^2-5x+13))+\arccos\left(\frac{3}{2+\sin\frac{9\pi x}{2}}\right)$ Find the domain of definition of $$f(x)=\log_{10}(1-\log_7(x^2-5x+13))+\arccos\left(\frac{3}{2+\sin\frac{9\pi x}{2}}\right)$$
I found the domain of $\log_{10}(1-\log_7(x^2-5x+13))$ to be $x\in(2,3)$ and the do... | $\left|\dfrac3{2+\sin{\frac{9\pi x}2}}\right|\leq1$
$⇔3\leq|2+\sin\frac{9\pi x}2|$
$⇒\sin\frac{9\pi x}2=1$
$⇒\frac{9\pi x}2=2k\pi+\frac{\pi}2$
$⇒x=\dfrac{1+4k}9$
$k=5,6$ satisfy that condition.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2357644",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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An expression involving the roots of a quadratic polynomial I have the equation $$ 9x^2 - 11x + 1 = 0 $$ whose two roots are $ \alpha $ and $ \beta $ .
I need to evaluate $$ \frac 1 {(9\alpha-11)^2} + \frac{11\beta - 1} 9$$
What I've tried
*
*Expanded the denominator and add them, but nothing simplifies and I get e... | Since $9\alpha^2-11\alpha+1=0=9\beta^2-11\beta+1$, by Vieta's theorem we have
$$ \frac{1}{(9\alpha-11)^2}+\frac{(11\beta-1)}{9} = \alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta=\left(\frac{11}{9}\right)^2-2\cdot\frac{1}{9}=\color{red}{\frac{103}{81}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2357760",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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let $a,b,c \in \mathbb{R^+} \ \ a+b+c =1$ Then prove that : $a^3+b^3+c^3 \geq \dfrac{1}{3}(a^2+b^2+c^2)$
Let $\{a,b,c\}\subset\mathbb{R^+}$ such that $a+b+c =1$. Prove that :
$$a^3+b^3+c^3 \geq \dfrac{1}{3}(a^2+b^2+c^2)$$
$$a^3+b^3+c^3-3abc=(x+b+c)(a^2+b^2+c^2-(ab +ac+bc))$$
$$a^2+b^2+c^2=(a+b+c)^2-2(ac+bc+ab)$$
N... | We need to prove that
$$3(a^3+b^3+c^3)\geq(a^2+b^2+c^2)(a+b+c)$$
and since $(a^2,b^2,c^2)$ and $(a,b,c)$ are the same ordered, our inequality
it's just Chebyshov's inequality:
$$3(a^2\cdot{a}+b^2\cdot{b}+c^2\cdot{c})\geq(a^2+b^2+c^2)(a+b+c).$$
Done!
| {
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"timestamp": "2023-03-29T00:00:00",
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Set of $x$ such that $\sum \frac{nx^n}{n^2 + x^{2n}}$ converges What is the radius of convergence of $$\sum_{n=1}^\infty \frac{nx^n}{n^2 + x^{2n}}?$$ Here is what I have so far. With the ratio test I want to find values of $x$ for which
$$ \lim_{n\to\infty}x \frac{n+1}{n} \frac{n^2 + x^{2n}}{(n+1)^2 + x^{2(n+1)}} < 1.$... | Let's apply the ratio test
$$
\frac{a_{n+1}}{a_{n}} = x\frac{n+1}{n}\frac{n^2 + x^{2n}}{(n+1)^2+x^{2n+2}} =\frac{1}{x}\frac{n+1}{n}\frac{1 + (n/x^n)^2}{1+([n+1]/x^{n+1})^2}
$$
If $|x| < 1$, then clearly the first expression goes to $x$ as $n\rightarrow \infty$, and if $|x|>1$, clearly the second expression goes to $x^{... | {
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"timestamp": "2023-03-29T00:00:00",
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let $a_1,a_2,. . . ,a_n \in \mathbb{R^+} $ then prove that :
Let $a_1,a_2,. . . ,a_n \in \mathbb{R^+} $. Prove that :
$$\frac{ a_{1} }{ a_{2} ^{2}+ a_{3} ^{2}+\cdots+ a_{n} ^{2}}
+\frac{ a_{2} }{ a_{1} ^{2}+ a_{3} ^{2}+\cdots+ a_{n} ^{2}}
+\cdots+\frac{ a_{n} }{ a_{1} ^{2}+ a_{2} ^{2}+\cdots+ a_{n-1} ^{2}}
\geq \fr... | Notice that $$\left( \frac{a_1}{a_2^2 + a_3^2 + ... + a_n^2} + \frac{a_2}{a_1^2 + a_3^2 + ... + a_n^2} + \ ... \ \frac{a_n}{a_1^2 + a_2^2+... + a_{n-1}^2}\right)(a_1 + a_2 + ... + a_n) \ge \left( \sqrt{\frac{a_1^2}{a_2^2 + a_3^2 + ... + a_n^2}} + \sqrt{\frac{a_2^2}{a_1^2 + a_3^2 + ... + a_n^2}} + \ ... \ \sqrt{\frac{a_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2361447",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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How to deduce the following equation from the other? How, from the equation $(1)$
$(1) a^{2x} + 2a^xb^y + b^{2y} = m - 2a^x - 2b^y$
can we deduce the equation $ (2) a^x + b^y = \sqrt{m + 1} - 1$
I know that the right hand side of $(1)$ is the perfect square $(a^x + b^y)^2 $, but this doesn't help me to deduce $(2)$.
... | Write $a^x+b^y=s$, then the rearranged form of $(1)$ you have provided is $s(s+2)=m$ or $s^2+2s-m=0$. It remains to solve the quadratic equation for $s$, yielding
$$s=\frac{-2\pm\sqrt{4+4m}}2=-1\pm\sqrt{m+1}$$
Taking the $+$ in the $\pm$ gives $(2)$: $s=a^x+b^y=\sqrt{m+1}-1$.
| {
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"url": "https://math.stackexchange.com/questions/2362297",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Number of different natural numbers which are smaller than two hundred million and using only the digits $1$ or $2$ is Number of different natural numbers which are smaller than two hundred million and using only the digits $1$ or $2$ is:
These numbers will be either 1 digit,2 digits,.......9 digits.Number of such $9$... | The number of different natural numbers which are smaller than $200,000,000$ and use only the digits $1$ or $2$ is
$$2^8 + 2^8 + 2^7 + 2^6 + 2^5 + 2^4 + 2^3 + 2^2 + 2^1$$
as you stated. However,
\begin{align*}
2^8 + 2^8 + 2^7 + 2^6 + 2^5 + 2^4 + 2^3 + 2^2 + 2^1 & = 2^8 + \sum_{k = 1}^{8} 2^k\\
& = 2^8 + 2\sum_{k = 1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2362416",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Showing $3^x+5^x$ is injective How can we show that
$$f(x)=3^x+5^x$$
is injective?
I know that $f$ is strictly increasing in $R$ so $f(x) = k, k \in R$ has only one root.
But I was looking for a proper proof, something I tried was:
Let $x,y \in \mathbb R$. Then for $f$ to be injection, $$3^{x}+5^{x} = 3^{y}+5^{y}$$ mus... | Suppose $3^x+5^x=3^y+5^y.$
Now $$3^x+5^x=3^y+5^y\Rightarrow 3^x-3^y=-(5^x-5^y)$$
Case 1: $x\geq y$, then $5^x-5^y\geq 0\Rightarrow 3^x-3^y=-(5^x-5^y)\leq 0$. But $3^x-3^y\geq 0$. Which lead to $3^x=3^y\Rightarrow x=y$
Case 2: $x\leq y$, then $5^x-5^y\leq 0\Rightarrow 3^x-3^y=-(5^x-5^y)\geq 0$. But $3^x-3^y\leq 0$. W... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Why, after squaring, is the smallest root of the quartic the solution of the real problem? Looking at this post, I considered the more general problem of the solution of equation $$x+\sqrt{x+\sqrt{a+x}}=a\tag 1$$ where $a> 0$ .
Using successive squaring processes, this leads to the quartic
$$x^4-2 (2
a+1) x^3+\left... | If the real and positive solutions from the 4th-order equations are $x_1 < x_2 < x_3 < x_4$ and we introduce
$$
y_i^{(\alpha,\beta)} = x_i + \alpha \sqrt{x_i + \beta \sqrt{x_i + a}}
$$
where $\alpha,\beta = \pm$ and only take the positive roots. For each $(\alpha,\beta)$ there is a unique $x_i$ that will result in $y_i... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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3 variable multiplication with 1 constraint lagrange multiplier Using Lagrange multipliers, I need to calculate all points $(x,y,z)$ such that $$x^4y^6z^2$$ has a maximum or a minimum subject to the constraint that $$x^2 + y^2 + z^2 = 1$$
So, $f(x,y,z) = x^4y^6z^2 $
and $g(x,y,z) = x^2 + y^2 + z^2 - 1$
then i've done t... | The minimal value is $0$ for $x=0$.
The maximal value we can find by AM-GM:
$$x^4y^6z^2=108\left(\frac{x^2}{2}\right)^2\left(\frac{y^2}{3}\right)^3z^2\leq108\left(\frac{2\cdot\frac{x^2}{2}+3\cdot\frac{y^2}{3}+z^2}{6}\right)^6=\frac{1}{432}.$$
The equality occurs for $\frac{x^2}{2}=\frac{y^2}{3}=z^2$ and $x^2+y^2+z^2=1$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2364728",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove the inequality $\sqrt{\frac{x}{y^2+z^2}}+\sqrt{\frac{y}{z^2+x^2}}+\sqrt{\frac{z}{x^2+y^2}}\ge \frac{2\sqrt{6}}3$ with $x+y+z=3,x,y,z\ge 0$
Let $x\geq0$, $y\geq0$ and $z\ge 0$ such that $x+y+z=3$. Show that
$$\sqrt{\dfrac{x}{y^2+z^2}}+\sqrt{\dfrac{y}{z^2+x^2}}+\sqrt{\dfrac{z}{x^2+y^2}}\ge \dfrac{2\sqrt{6}}{3}.$... | Holder helps!
$$\left(\sum_{cyc}\sqrt{\frac{x}{y^2+z^2}}\right)^2\sum_{cyc}x^2(y^2+z^2)\geq(x+y+z)^3.$$
Thus, it remains to prove that
$$(x+y+z)^3\geq\frac{8}{3}\sum_{cyc}x^2(y^2+z^2)$$ or
$$(x+y+z)^4\geq16\sum_{cyc}x^2y^2,$$ which is true by AM-GM:
$$(x+y+z)^4=\left(\sum_{cyc}(x^2+2xy)\right)^2\geq\left(2\sqrt{\sum_{c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2365883",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
What is the best way to solve modular arithmetic equations such as $9x \equiv 33 \pmod{43}$? What is the best way to solve equations like the following:
$9x \equiv 33 \pmod{43}$
The only way I know would be to try all multiples of $43$ and $9$ and compare until I get $33$ for the remainder.
Is there a more efficient wa... | $43$ is prime. And even if it weren't, $\gcd(9,43)= 1$.
That means we know that $9^{-1}_{43}$ exists. That is there is a number so that $9\cdot 9^{-1}_{43} \equiv 1 \pmod {43}$. And we could calculate (or guess) that $9^{-1}_{43}\equiv 24$ (because $9\cdot 24\equiv 1 \pmod {43}$) and we can solve $9x \equiv 33\pmod... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2367841",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
Contour integration getting wrong result.
By means of contour integration, evaluate
$$
\int_0^\infty \frac{(\log_e(u))^2}{u^2+1} \, du \, .
$$
When we apply Cauchy Residue theorem, we get $+i$ as a valid pole.
The residue for this pole is $\frac{(i(\pi)^2)}{8}$.
The final answer is $2(\pi)i *$ (sum of residues), ... | As far as I can see, you've accurately calculated the integral
$$\int_{-\infty}^\infty{\frac{\log^2{u}}{u^2+1}du} = -\frac{\pi^3}{4}$$
I suspect the problem comes because you're assuming that
$$\int_{-\infty}^\infty{\frac{\log^2{u}}{u^2+1} du} = 2 \int_{0}^\infty{\frac{\log^2{u}}{u^2+1}du} $$
This is not true since, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2375006",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Probability that at least $2$ people will not receive any ace.
I've a deck with 52 french cards ($13$ values for each of $4$ suits) and $4$ players. Randomly dealing out all cards, what's the probability that at least
$2$ people will not receive any ace?
My try:
$$p=\frac{\frac{4!}{2!}\binom{48}{13,13,12,10}+\binom... | There are $\binom42$ ways to choose two of the people who have 26
available "slots" between them.
Consider only the distribution of aces, where the rest of the cards go doesn't matter.
The probability that all the aces fall in one or other of the 26
slots of the chosen two is
$\binom42\cdot \frac{26}{52}\cdot\frac{25... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2375800",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Find the inverse laplace transform of complicated rational function
Find the inverse laplace $$\mathcal{L}^{-1}\left\{ \frac{2s + 1}{2s^2 + s + 2} \right\}$$
I don't see a simple way to take the inverse laplace transform, convolution theorem?
| This is a case where you "massage" the expression to fit it into a more standard form
First get rid of the leading coefficient in the denominator.
$$ \frac{2s+1}{2s^2+s+2}=\frac{s+\frac{1}{2}}{s^2+\frac{1}{2}s+1}$$
Then complete the square on the denominator.
$$ \frac{s+\frac{1}{2}}{s^2+\frac{1}{2}s+1}=\frac{s+\frac{1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2376264",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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Let $\ a,b,c$ be integers such that $ \ a^{2} + b^{2} = c^{2}$. If $c$ is divisible by $3$, prove that $a$ and $b$ are both divisible by $3$. Question:
Let $\ a,b,c$ be integers such that $ \ a^{2} + b^{2} = c^{2}$. If $c$ is divisible by $3$, prove that $a$ and $b$ are both divisible by $3$.
My attempt:
Proof by cont... | Assume that the remainders of the divisions $a:3$ and $b:3$ are $r$ and $s$, respectively. Write $a=3p+r$, $b=3q+s$
Then
$$a^2+b^2=9(p^2+q^2)+6(pr+qs)+r^2+s^2$$
Since $r$ and $s$ are $0$, $1$ or $2$, $r^2+s^2$ is $0$, $1$, $4$, $2$, $5$ or $8$. Can you finish?
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Divide triangle into two equivalent parts by line perpendicular to the base. Originally the problem wants me to divide into 3 equivalent(equal areas) parts but I thought that I should do 2 first. Tried 2 and could not after many attempts. DF below is movable so it is not really the correct solution.
| The Math
\begin{align}
\text {Area of }\triangle \text{ ABH} &= \frac{mbh}{2} \\[1.5ex]
\text {Area of }\triangle \text{ AXY} &= \frac{mx^2}{2} \\[1.5ex]
\text{For } \triangle \text { AXY to be half the area of } \triangle { ABH:} \\[1.5ex]
\frac{mx^2}{2} &= \frac{1}{2} \times \frac{mbh}{2} \\[1.5ex]
\text{giving } x ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2380604",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Find $(a^2+b^2),$ where $(a+b)=\dfrac{a}{b}+\dfrac{b}{a}$ The question is the same .Find $a^2+b^2$. I think we have to find $a$ and $b$ firstly. Given that $a$ and $b$ are integers.
| The condition gives $$(a+b)ab=a^2+b^2$$ or
$$(b-1)a^2+b^2a-b^2=0.$$
If $b=1$ we obtain $a=1$ and $a^2+b^2=2$.
If $b\neq1$ we need $b^4+4b^2(b-1)=m^2$, where $m\in\mathbb N$ or
$$b^2+4b-4=n^2,$$
where $n\in\mathbb N$ or
$$(b+2)^2=8+n^2$$ or
$$(n-b-2)(n+b+2)=-8$$ and since $n>0$, we obtain two cases only:
*
*$n-b-2=4$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2381988",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
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Probability of truth in a chain of statements
Three individuals $A$,$B$ and $C$ tell the truth with probability $1/3$.
(A) $C$ makes a statement and $A$ claims that it is true. What is the probability that the statement is true.
(B) $C$ makes a statement and $A$ tells you that $B$ claims the statement is true. What is... | Careful. We need to distinguish between "$X$ tells the truth" and "$X$ claims that $Y$ told the truth". The former always happens with probability $1/3$, while the latter depends on whether or not $Y$ is true or false. Let's revisit part (a) and redefine our events. Define:
*
*$X_C$: $C$ makes a true statement.
*$X... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2382204",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Given $\lim\limits_{x\to 0} \frac {x(1+a\cos x)-b\sin x}{x^3}=1$, what is the value of $a+b$?
Given that
$$\lim\limits_{x\to 0} \frac {x(1+a\cos x)-b\sin x}{x^3}=1$$
What is the value of $a+b$
My try
$\lim\limits_{x\to 0} (\frac {x(1+a\cos x)}{x^3}-\frac {b\sin x}{x^3})=1$
$\lim\limits_{x\to 0} (\frac {(1+a×cos(... | We need two limits below (which are easily obtained and the second one necessitates the use of Taylor series or L'Hospital's Rule) $$\lim_{x\to 0}\frac{1-\cos x} {x^{2}}=\frac{1}{2},\,\lim_{x\to 0}\frac{x-\sin x} {x^{3}}=\frac{1}{6}$$ The limit in question can be written as $$\lim_{x\to 0}(a+1)\cdot\frac{x-\sin x} {x^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2382306",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
How to solve $y''+y=\cos x$?
Solve $y''+y=\cos x$.
After first solving the homogeneous equation we know that the solution to it is $y=c_1\cos x+c_2\sin x$.
We can guess that the private solution to non-homogeneous equation will be of form: $y_p=x(A_1\cos x+A_2\sin x)$.
Then:
$$
y_p'=A_1\cos x-A_1x\sin x+A_2\sin x+A_2... | So the corresponding auxiliary equation to $y''+y=\cos x$ is $m^2+1=0$, so $$y_c=c_1 \cos x + c_2 \sin x,$$
so things are fine so far. Now since our RHS is $\cos x$, like you said, we assume that the particular solution is of the form $A \sin x+B \cos x$. But since $A\sin x$ is already accounted for in $y_c$, we take ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2382387",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
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Given $a+b+c=0$ find the value of $\big(\frac{b-c}{a}+\frac{c-a}{b}+\frac{a-b}{c}\big)\big(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}\big)$ I already have a solution, which is correct, it is $9$. I'm just wondering if there is a simpler method.
First we just expand and find
$3+\frac{b-c}{a}\big(\frac{b}{c-a}+\frac{c}{a-... | It is an olympiad question. Otherwise picking suitable numbers will do:
$$a=1,b=2,c=-3$$
$$S=\left(5-2+\frac13 \right)\left(\frac15 -\frac12+3\right)=\frac{10}{3}\cdot \frac{27}{10}=9.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2382672",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
} |
Evaluating a Discontinuous Integral How do you evaluate this integral? $$\int _0 ^4\frac{dx}{x^2 - 2x - 3}$$
My work:
The expression $x^2 - 2x - 3$ is discontinuous at $x = 3$ in the interval $x = 0$ to $x = 4$, so I got to integrate the expression like this:
$$\int _0 ^4\frac{dx}{x^2 - 2x - 3}$$ is equal to
$$\int _0 ... | In fact, you should use the expression for integral of the function $\dfrac{1}{x^2-a^2}$:
$$
\int_3^4 \frac{dx}{(x-1)^2-2^2} = \int_2^3 \frac{dx}{x^2-2^2} = \frac{1}{4}\lim_{x\to 2^-}\left (\log\frac{1}{5} -\log\frac{x-2}{x+2}\right ).
$$
The above limit does not exist.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2385253",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
A Curious binomial identity While playing around with random binomial coefficients , I observed that the following identity seems to hold for all positive integers $n$:
$$ \sum_{k=0}^{2n} (-1)^k \binom{4n}{2k}\cdot\binom{2n}{k}^{-1}=\frac{1}{1-2n}.$$
However, I am unable to furnish a proof for it ( though this result i... | When going through Jack's nice answer I did some intermediate steps to better see what's going on. Here is a somewhat more elaborated version, which might also be convenient for other readers.
We obtain
\begin{align*}
\color{blue}{\sum_{k=0}^{2n}}&\color{blue}{(-1)^k\binom{4n}{2k}\binom{2n}{k}^{-1}}\\
&=\sum_{k=0}^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2385949",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Conversion of Parametric Equation to a Rectangular Equation I was looking for the rectangular equation of the given parametric equation $$ x = \tan \theta$$
$$ y = \tan 2\theta$$
My work:
$$x = \tan \theta$$
$$\theta = \arctan x$$
Then, substituting it to $ y = \tan 2\theta$, it becomes:
$$y = \tan 2\theta$$
$$y = \tan... | From trigonometry we know that $\displaystyle \tan(\alpha+\beta) = \frac{\tan\alpha + \tan\beta}{1 - \tan\alpha\tan\beta}.$
If $\alpha$ happens to be the same as $\beta$ then this says $\displaystyle\tan(\alpha+\alpha) = \frac{\tan\alpha + \tan\alpha}{1-\tan\alpha\tan\alpha} = \frac{2\tan\alpha}{1 - \tan^2\alpha}.$
In ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2386249",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
What is the Galois group of $x^6+x^5-8x^4-5x^3+19x^2+4x-11=0$? A while ago I happened on this equation: $$f(x)=x^6+x^5-8x^4-5x^3+19x^2+4x-11=0.$$
$f$ does not have linear factors, as can be seen by drawing its graph. I am not certain about whether it has quadratic or cubic factors over $\mathbb{Q}$, but over $\mathbb{Q... | Irreducible over the rationals. Galois group as Robert said; I added two degree six polynomials where the roots are also real of modest absolute value, but come out very pretty, method originally due to Gauss.
================================================================
? f = x^6 + x^5 - 8 * x^4 - 5 * x^3 + 19 * x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2387139",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Closed-form for Floor Sum 1 Does a closed form exist for the following sum?
$$\sum_{k=0}^n \lfloor \sqrt{k} + \sqrt{k + n} \rfloor$$
If not, why is this sum so radically different than the sums below?
Closed forms do exist for the following sums*:
$$\sum_{k=0}^n \lfloor \sqrt{k + n} \rfloor$$
$$\sum_{k=0}^n \lfloor \sq... | The "difference" is actually that
$$
\eqalign{
& \left\lfloor {x + y} \right\rfloor = \cr
& = \left\lfloor x \right\rfloor + \left\lfloor y \right\rfloor + \left\lfloor {\left\{ x \right\} + \left\{ y \right\}} \right\rfloor = \cr
& = \left\lfloor x \right\rfloor + \left\lfloor y \right\rfloor + \le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2390080",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Formulation of the Wallis product for $\pi/2$ using the $\Gamma$ function I am trying to write the Wallis Product (WP) using the $\Gamma$ function, thereby hoping for either a new identity or a simple derivation of the formula, but I get a result I am not sure its correct.
This is the WP
$$ \frac{\pi}{2} = \frac{2}{1}\... | By Stirling,
$$\frac{\Gamma(n)}{\Gamma(n-\frac12)}\approx\frac{\sqrt{2\pi n}\left(\dfrac ne\right)^n}{\sqrt{2\pi (n-\frac12)}\left(\dfrac {n-\frac12}e\right)^{n-\frac12}}.$$
After simplifications,
$$\frac1{\left(1-\dfrac1{2n}\right)^n}\left(\frac {n-\frac12}e\right)^{1/2}\to e^{1/2}\sqrt {\frac ne}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2390493",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Find the curve of intersection between $x^2 + y^2 + z^2 = 1$ and $x+y+z = 0$
Find the curve of intersection between $x^2 + y^2 + z^2 = 1$ and $x+y+z = 0$
My attempt:
*
*$x^2 + y^2 + z^2 = 1$
*$x+y+z=0$
$$(2) \implies z = -(x+y)$$
$$(1) \implies x^2+y^2+(x+y)^2 = 1$$
$$2x^2 + 2y^2 + 2xy = 1$$
This is the curve i... | $2x^2+2xy+2y^2=1$ can be rewritten as $(2x+y)^2+3y^2=2$ and this can be parameterised
\begin{eqnarray*}
\frac{2x+y}{\sqrt{2}} &=& \cos \theta \\
\sqrt{\frac{3}{2}} y &=& \sin \theta . \\
\end{eqnarray*}
now substitute these into $x+y+z=0$ and we have
\begin{eqnarray*}
x&=&\frac{1}{\sqrt{2}} \cos \theta- \frac{1}{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2390584",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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Prove $1 + \cos 2C - \cos 2A - \cos 2B = 4\sin A\sin B\cos C$ for $A$, $B$, $C$ the angles of a triangle Given that $A$,$B$ and $C$ are angles of a triangle, show that
$$1 + \cos 2C - \cos 2A - \cos 2B = 4\sin A\sin B\cos C$$
| Note that
$$\cos u + \cos v = 2\cos((u + v)/2)\cos((u − v)/2)$$
and
$$\cos u - \cos v = -2\sin((u + v)/2)\sin((u − v)/2).$$
Hence
\begin{align*}
1 + \cos 2C - \cos 2A - \cos 2B&=1 + \cos (2\pi-2(A+B)) - 2\cos(A+B)\cos(A-B)\\
&=2\cos^2 (A+B) - 2\cos(A+B)\cos(A-B)\\
&=2\cos(A+B)(\cos(A+B)-\cos(A-B))\\
&=2(-\cos C)(-2\sin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2393062",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
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prove that $(x^2 + y^2 - z^2)^2 = 4x^2y^2$. We have
$$x\cos \theta+y\cos \phi = -z\cos \psi \tag 1$$
$$x\sin \theta+y\sin \phi = -z\sin \psi \tag 2$$
$$x\sec \theta+y\sec \phi = -z\sec \psi \tag 3$$
and we have to prove that
$$(x^2 + y^2 - z^2)^2 = 4x^2y^2$$
squaring (1) & (2), and adding them we have
$$x^2 + y^2 - z^2... | We denote $a,b,c$ the angles $\theta, \phi, \psi$. The condition is
$$\begin{bmatrix} \cos a &\cos b &\cos c\\\sin a &\sin b &\sin c\\\sec a &\sec b &\sec c\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}0\\0\\0\end{bmatrix}$$ The determinant of the matrix is equal to
$$\frac{\sin(b-a)}{\cos c}+\frac{\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2393564",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Correct logic of permuting 5 men and 5 women to find probability of different highest women rank The problem reads like this:
Problem
Five men and $5$ women are ranked according to their scores on an examination. Assume that no two scores are alike and all $10!$ possible rankings are equally likely. Let $X$ denote t... | I would like to suggest another way of getting the results. For me this way is "combinatorically more intuitive":
*
*If $X = i$, then there are $10-i$ ranks left for the remaining $4$ women. So. there are $\binom{10-i}{4}$ ways of choosing 4 further ranks.
*As each permutation of the $5$ men and $5$ women gives ano... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2399038",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Exponentiating Matrices - Am I doing it correctly? I recently learned how to exponentiate matrices by extending the power series of $e^x$ to matrices:
$$e^A=A^0+A^1+\frac{A^2}{2!}+\frac{A^3}{3!}+...$$
I thought that this was pretty cool, so I decided to try it with the matrix
$$A=\begin{pmatrix}1&1\\1&1\end{pmatrix}$$
... | Your method is correct, as you've been told. But here's another way of doing this that may interest you. Let$$M=\frac1{\sqrt2}\begin{pmatrix}1&-1\\1&1\end{pmatrix}.$$Then$$M^{-1}=\frac1{\sqrt2}\begin{pmatrix}1&1\\-1&1\end{pmatrix}\text{ and }M^{-1}.A.M=\begin{pmatrix}2&0\\0&0\end{pmatrix}.$$Therefore$$M^{-1}.e^A.M=e^{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2399180",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Solving an infinite product of consecutive square roots Given $a$ and $b$ calculate $ab$
$$a=\sqrt{7\sqrt{2\sqrt{7\sqrt{2\sqrt{...}}}}}$$
$$b=\sqrt{2\sqrt{7\sqrt{2\sqrt{7\sqrt{...}}}}}$$
I simplified the terms and further obtained that $ab$ is equal to:
$$ab=2^{\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}...}\cdot7... | $$a=\sqrt{7\sqrt{2\sqrt{7\sqrt{2\sqrt{...}}}}}$$
$$a^2=7\sqrt{2\sqrt{7\sqrt{2\sqrt{...}}}}$$
$$a^4=98\sqrt{7\sqrt{2\sqrt{...}}}$$
so
$$a^4=98a$$
and, assuming $a$ is nonzero,
$$a=\sqrt[3]{98}$$
$$b=\sqrt{2\sqrt{7\sqrt{2\sqrt{7\sqrt{...}}}}}$$
$$b^2=2\sqrt{7\sqrt{2\sqrt{7\sqrt{...}}}}$$
$$b^4=28\sqrt{2\sqrt{7\sqrt{...}}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2399381",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 4,
"answer_id": 3
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If $h(r, \theta) = f(r \cos \theta, r \sin \theta)$, show that $ f_{xx}+f_{yy} = h_{rr} + \frac{1}{r} h_r + \frac{1}{r^2} h_{00}$
If $h(r, \theta) = f(r \cos \theta, r \sin \theta)$, show that
$$ f_{xx}+f_{yy} = h_{rr} + \frac{1}{r} h_r + \frac{1}{r^2} h_{00}$$
Hint: Rewrite the defining equation as $f(x,y) = h(r... | Let $f(x,y)=f(r\cos\theta ,r\sin\theta )=h(r,\theta ).$
Then, $$\nabla _{(r,\theta )}h=\begin{pmatrix}\cos\theta &\sin\theta \\ -r\sin\theta &r\cos\theta \end{pmatrix}\nabla f\implies \nabla f=\frac{1}{r}\begin{pmatrix}r\cos\theta &-\sin\theta \\r\sin\theta &\cos\theta \end{pmatrix}\nabla _{(r,\theta )}h=:\begin{pmatri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2401057",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
$x^4-2x^3+mx^2-2x+1=0$, $m \in R$, $x_1=x_2 \in R-\{-1\}$ Given the equation: $x^4-2x^3+mx^2-2x+1=0$, $m \in R$
To which interval does $m$ belong, so that the equation has $x_1=x_2 \in \mathbb R\setminus\{-1\}$, provided that $x_1,x_2,x_3,x_4$ are the roots of the equation.
Since I got this from Vieta's:
$$2x_1+x_3+x_4... | We can get your result also by the following way.
$$(x^4-2x^3+mx^2-2x+1)'=0$$ or
$$4x^3-6x^2+2mx-2=0$$ or
$$2x^4-3x^3+mx^2-x=0$$ or
$$mx^2=-2x^4+3x^3+x,$$ which gives
$$x^4-2x^3+(-2x^4+3x^3+x)-2x+1=0$$ or
$$x^4-x^3+x-1=0$$ or
$$(x^3+1)(x-1)=0,$$
which gives $x=1$ and $m=2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2401598",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Is there a pair of numbers $a,b\in\Bbb{R}$ such that $\frac{1}{a+b}=\frac{1}{a}+\frac{1}{b}$? I'm asked in an exercise from an algebra textbook if there exists a pair of numbers $a,b\in\Bbb{R}$ such that $\frac{1}{a+b}=\frac{1}{a}+\frac{1}{b}$
I'm trying to prove that such pair of numbers does not exist, but I'm not su... | If $a$ and $b$ have the same sign, the magnitude of $\frac 1{a+b}$ is less than either $\frac 1a$ or $\frac 1b$ so they must have opposite signs. By symmetry we can demand $a$ be positive. If $a \gt -b, \frac 1{a+b}$ is positive while $\frac 1a +\frac 1b$ is negative. The opposite happens if $a \lt -b$ so there is no s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2402803",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Evaluate the following complex integrals using parameterisation $\int \limits_{\beta} f(z) dz$ Evaluate the following complex integrals using parameterisation
$$\int \limits_{\beta} f(z) dz$$
where $\beta$ represents the line segment from $-i$ to $2+5i$, followed by the line segment from $2+5i$ to $5i$, and $f(x+yi)=iy... | So you have an integral $\int_\beta f(z)\, dz$ where $\beta$ represents two line segments. First, let's parameterize each one:
$C_1$ goes from $-i$, which means the point is $(0,-1)$, to $2+5i$, which means the point is $(2,5)$, so:
\begin{align}
C_1(t) &= (0,1) + t[(2, 5)-(0,-1)] \\
C_1(t) &= (2t, 6t-1).
\end{align}
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2403512",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Find the value $\binom {n}{0} + \binom{n}{4} + \binom{n}{8} + \cdots $, where $n$ is a positive integer. Given
$$(1+x)^n= \binom {n}{0} + \binom{n}{1} x+ \binom{n}{2} x^2+ \cdots + \binom {n}{n} x^n.$$
Find the value $\binom {n}{0} + \binom{n}{4} + \binom{n}{8} + \cdots $, where $n$ is a positive integer.
I tried to u... | This answer repeatedly takes advantage of one fact,
$$g(k)=\frac{(-1)^k+1}{2}$$
Returns $1$ for a nonnegative even integer, and $0$ for a nonnegative odd integer.
Our sum is,
$$\sum_{k=0,\text{even}}^{n/2} {n \choose 2k}$$
$$=\sum_{k=0}^{n/2} \frac{1+(-1)^k}{2}{n \choose 2k}$$
$$=\frac{1}{2}\left(\sum_{k=0}^{n/2} {n ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2404878",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 2,
"answer_id": 0
} |
Mathematical Olympiad Treasures Problem 1.92 how does this relate to geometric sequence You are supposed to show that if
$a,b,c$ are nonzero real numbers
and
$(ab+bc+ca)^3 =abc(a+b+c)^3$
$a, b, c$ are terms of a geometric sequence.
One given solution is that:
$(ab+bc+ca)^3 −abc(a+b+c)^3 = (a^2 −bc) (b^2 −ac) (c^2 −a... | You got $(ab+bc+ca)^3 −abc(a+b+c)^3 = (a^2 −bc) (b^2 −ac) (c^2 −ab)$. Now since $(ab+bc+ca)^3 =abc(a+b+c)^3$, it implies $(ab+bc+ca)^3 −abc(a+b+c)^3=0$. Then;
$$(a^2 −bc) (b^2 −ac) (c^2 −ab)=0$$
This implies one of $(a^2 −bc),(b^2 −ac),(c^2 −ab)$ must be zero. Then we have $a^2 =bc$ or $b^2 =ac$ or $c^2 =ab$.
If ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2405384",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Demonstrate that all integers of a certain form are divisible by primes of a certain other form I wish to demonstrate that for any $n\in\mathbb{N}$ all prime divisors of a number $n^2 + n + 1$ are either equal to $3$ or of the form $3k + 1$ with positive integer $k$.
Thus far, I have tried checking cases. If $n$ is of ... | Disclaimer: The following answer is based on finite fields, where the polynomial $x^2+x+1$ should ring some bells.
We have
$$n^3-1 = (n-1)(n^2+n+1).$$
Now let $p$ be a prime of the form $p = 3k+2$ for some positive $k$ and assume that $p$ divides $n^2+n+1$ and thus also divides $n^3-1$. Then we have that
$$n^3 \equiv... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
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Proving that for $\{ a,b\} \subset \Bbb{R^{+}}$; $a+b=1 \implies a^2+b^2 \ge \frac{1}{2}$
For $\{ a,b\} \subset \Bbb{R^{+}}$; $a+b=1 \implies a^2+b^2 \ge
\frac{1}{2}$
I'm trying to prove this in the following way, but I'm not sure if it's correct. Could anyone please check it and see if it's okay?
$a+b=1 \implies (a... | It is ok. You could have also put $b=1-a$ and minimize a quadratic function
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2406268",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 2
} |
For $08$.
For $0<x<\dfrac{\pi}{4}$ prove that
$$\frac{\cos x}{\sin^2x(\cos x-\sin x)}>8$$
I am trying to use the following for x $\in$ (0,$\frac{\pi}{4}$)
Cos x $\in$ ($\frac{1}{√2}$,1)
(sinx)^2 $\in$ (0,$\frac{1}{2}$)
(cosx-sinx) $\in (0,1)$
| Let me try. We have $$\frac{\cos x}{\sin^2 x (\cos x - \sin x)} = \frac{1+\tan^2 x}{\tan^2 x(1-\tan x)}.$$
Note that $0 < x < \dfrac{\pi}{4}$, so $0 < \tan x < 1$,
$$\tan x(1-\tan x) \leq \frac{1}{4}(\tan x + 1-\tan x)^2 = \frac{1}{4}.$$
So we have $$\text{LHS} \geq 4\frac{1+\tan^2x}{\tan x} \geq 8.$$
The equality happ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Proving that $\{x\in\Bbb{R}\mid 1+x+x^2 = 0\} = \varnothing$ without the quadratic formula and without calculus I'm asked to prove that $\{x\in\Bbb{R}\mid 1+x+x^2 = 0\} = \varnothing$ in an algebra textbook.
The formula for the real roots of a second degree polynomial is not introduced yet. And the book is written wit... | If
$x^2 + x + 1 = 0, \tag 1$
then
$x^2 + x + \dfrac{1}{4} = -\dfrac{3}{4}; \tag 2$
but
$(x + \dfrac{1}{2})^2 = x^2 + x + \dfrac{1}{4}, \tag 3$
so
$(x + \dfrac{1}{2})^2 = -\dfrac{3}{4}; \tag 4$
but no real has a negaitive square, so . . .
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2410300",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 14,
"answer_id": 1
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Write $(x^2 + y^2 + z^2)^2 - 3 ( x^3 y + y^3 z + z^3 x)$ as a sum of (three) squares of quadratic forms The quartic form
$$(x^2 + y^2 + z^2)^2 - 3 ( x^3 y + y^3 z + z^3 x)$$
is non-negative for all real $x$, $y$, $z$, as one can check (with some effort). A theorem of Hilbert implies that there exist quadratic forms $... | $$(x^2 + y^2 + z^2)^2 - 3 ( x^3 y + y^3 z + z^3 x)=\frac{1}{2}\sum_{cyc}(x^2-y^2-xy-xz+2yz)^2=$$
$$=\frac{1}{6}\sum_{cyc}(x^2-2y^2+z^2-3xz+3yz)^2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2410994",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 2
} |
Monotonicity of $f(x) =\sin(\ln(x))-\cos(\ln(x))$ Find the interval in which $f(x) =\sin(\ln(x))-\cos(\ln(x))$ is increasing.
After differentiating we get $$f'(x) = \frac{\cos\left(\ln(x)\right)}{x} +\frac{\sin\left(\ln(x)\right)}{x}$$
Now how do we analyze this expression?
| $$f'(x)=\frac{\sin \log x+\cos \log x}{x}$$
$\log x$ is an increasing function. Substitute $u=\log x$. Denominator is positive because $x$ is argument of logarithm.
$\sin u + \cos u > 0\to \sin u > - \cos u$
$\sin u = -\cos u$ when $\tan u=-1$ when $u=\dfrac{3\pi}{4}+k\pi,\forall k\in\mathbb{Z}$
therefore $\sin u > - \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2411354",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 3
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Factoring a polynomial with possibly repeated root I've been trying to factor $x^3 - 8x^2 + 17x - 4$. One of factors is $x - 4$ so I got $(x - 4)(x^2 -4x + 1)$, but I don't know how to factor the second part $x^2 -4x + 1$.
I also have a similar problem with $x^3(x^3 - 3x + 2)$. I'm supposed to get $x^3(x - 1)^2 (x + 2... | $$x^2-4x+1=x^2-4x+4-3=(x-2)^2-(\sqrt3)^2=(x-2-\sqrt3)(x-2+\sqrt3)$$
$$x^3-3x+2=x^3-2x^2+x+2x^2-4x+2=(x^2-2x+1)(x+2)=(x-1)^2(x+2)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2411693",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Finding the minimum value of $a^2+b^2+c^2$
Let $a$, $b$ and $c$ be $3$ real numbers satisfying $2 \leq ab+bc+ca$. Find the minimum value of $a^2+b^2+c^2$.
I've been trying to solve this, but I don't really know how to approach this. I thought of $(a+b+c)^2 = a^2+b^2+c^2 + 2(ab+bc+ca)$, but that gives me $a+b+c$, wh... | Famke's answer is the simplest; however, we can use a variational argument, as well.
For all variations that maintain $ab+bc+ca=2$, we have
$$
(b+c)\,\delta a+(a+c)\,\delta b+(a+b)\,\delta c=0\tag{1}
$$
To minimize $a^2+b^2+c^2$, we must have
$$
2a\,\delta a+2b\,\delta b+2c\,\delta c=0\tag{2}
$$
for all variations that... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2413134",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 0
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M tosses $7$ fair coins and has $M$ heads. $A$ tosses $6$ fair coins and has $A$ heads. Find probability $P(M>A).$ $M$ tosses $7$ fair coins and has M heads. A tosses $6$ fair coins and has $A$ heads. Find probability $P(M>A)$.
I suppose that both distributions are binominal, but I don't know what to do next.
| The values $6$ and $7$ are sufficiently small that you could calculate this by hand.
The probabilty that $A$ gets zero heads is $ \frac{1}{64}$ and $M$ get one or more heads is $\frac{127}{128}$.
The probabilty that $A$ gets one head is $ \frac{6}{64}$ and $M$ get two or more heads is $\frac{120}{128}$.
...
The probabi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2413547",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Partial Fractions Decomposition $\frac{6x^2 - 29x - 29}{(x+1)(x-3)^2}$ explanation repeated factors I am trying to solve the fraction $$\frac{6x^2 - 29x - 29}{(x+1)(x-3)^2}$$
into partial fractions.
Now, I thought it could be solved into the following
$$\frac{6x^2 - 29x - 29}{(x+1)(x-3)^2} = \frac{A}{x+1} + \frac{B}{(... | When you decompose, the degree of the numerator will be less than the degree of the denominator. How much less is yet to be determined. Assume that it is one degree less, and then if you get a zero coefficient, so be it.
If you had something like...$\frac {P(x)}{(x^2 +x + 1)(x-1)}$
your first step would be $\frac {Ax... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2413640",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluate the sum $\sum_{n=1}^{b}n\binom{a}{n}\binom{b}{n}$ Let $a$ and $b$ be natural numbers such that $a \geq b \geq 1$. How can we evaluate the following sum?
$$ \sum_{n=1}^{b}n\binom{a}{n}\binom{b}{n}$$
| $$\begin{align}
\sum_{n=1}^b n\binom an\binom bn
&=\sum_{n=1}^b \binom an\cdot n\binom bn\\
&=\sum_{n=1}^b \binom an\cdot b\binom {b-1}{n-1}\\
&=b\sum_{n=1}^b \binom a{a-n}\binom {b-1}{n-1}\\
&=b\binom {a+b-1}{a-1}
&&\text{(Vandermonde)}\\
&=b\binom {a+b-1}{b}\\
&=(a+b-1)\binom {a+b-2}{b-1} \\
&=(a+b-1)\binom {a+b-2}{a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2413868",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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If $abcd=1$ then $a^4b+b^4c+c^4d+d^4a\geq a+b+c+d$
I want to prove that for all $a>0$, $b>0$, $c>0$ and $d>0$ with $abcd=1$: $$a^4b+b^4c+c^4d+d^4a\geq a+b+c+d.$$
I think that this should be provable by AM-GM inequality, but I could not manage to prove it. Can you give me a hint?
Best wishes
| The hint:
Write
$$xa^4b+yb^4c+zc^4d+td^4a\geq a^2bcd,$$
where $x$, $y$, $z$ and $t$ are non-negatives such that $x+y+z+t=1$
and after using AM-GM solve the system, which you got.
I got $x=\frac{23}{51}$, $y=\frac{7}{51}$, $z=\frac{11}{51}$ and $t=\frac{10}{51}$,
which gives the following magical solution in one line.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2413948",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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I want to find all the rational solutions of $x^2+y^3=z^3$ I want to find all the rational solutions of $x^2+y^3=z^3$,
and any help would be appreciated.
My answer is as follows：
$$x^2=z^3-y^3=(z-y)(z^2+zy+y^2)$$
Let $x=ab,a,b\in\Bbb Q$
when
$$ z-y=a^2,z^2+zy+y^2=b^2$$
$$z^2+z(z-a^2)+(z-a^2)^2=b^2$$
Sort it out, and ... | EDIT: Let's first consider the more symmetric equation
$$x^2=y^3+z^3\tag1.$$If $y=0$, we see that gives the solutions $$(x,y,z)=(t^3,0,t^2)$$ for rational $t$. If $y\neq0$, we can set $x=ry$ and $z=ty$ with rational $r,t$. That gives $r^2=y(1+t^3)$, i.e. $r=0$, giving $t=-1$ and thus the solutions $$(x,y,z)=(0,y,-y),$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2414088",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Sequence : $f(f(n))+f(n+1)=3n$ Does there exist a function $f : \mathbb{Z}^+ \to \mathbb{Z}^+$ such that $f(f(n))+f(n+1)=3n$, $\forall n \in \mathbb{Z}^+$ ?
My attempt :
Substitute $n=1$, $f(f(1))+f(2)=3$ so $f(2) \in \{1$ or $2\}$
If $f(2)=1, f(f(1))=2$
Substitute $n=2$, $f(f(2))+f(3)=6$ so $f(1)+f(3)=6$
Substitut... | We will show that there is no such function.
Since $f(2)+ff(1)=3$, we have two case:
$f(2)=2$:
$ff(2)+f(3)=6\implies f(3)=4$
$ff(3)+f(4)=9\implies 2f(4)=9$.
The last one is impossible hence $f(2)=1$.
$f(2)=1:$
$$
ff(2)+f(3)=6\implies f(3)=6-f(1)\\
ff(3)+f(4)=9\implies f(6-f(1))+f(4)=9.
$$
First of all $f(1)\neq 1,2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2414224",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
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Need help with level 3 Calculus problem
Two curves $$C_1: ([f(y)]^{2/3} + ([f(x)]^{1/3}) = 0\quad\mbox{and}\quad C_2: [f(y)]^{2/3}+ [f(x)]^{2/3} = 12, $$ satisfying the relation $$ (x-y)f(x+y)-(x+y)f(x-y) = 4xy(x^2-y^2).$$
1.) Evaluate the area bounded by $C_1$ and $C_2$.
2.) Evaluate the area bounded by $C_2$ and $|x... | According to Functional Equation : If $(x-y)f(x+y) -(x+y)f(x-y) =4xy(x^2-y^2)$ for all x,y find f(x).
the general solution of the function equation
$$(x-y)f(x+y)-(x+y)f(x-y) = 4xy(x^2-y^2)$$
is $f(x)=x^3+kx$ with $k\in\mathbb{R}$. So the problem seems to be ambiguous because $f$ (and therefore also the curves $C_1$ an... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2414433",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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Show that $x^4-x^2+1$ is irreducible over $\mathbb{Q}$ My attempts:
I cannot apply the Eisenstein's criteria here, because there is no prime number that divides the constant term i.e. $1$ Taking a translation of the form $x \rightarrow x+a$ does not solve this issue either.
Next, I tried the mod tests: $\operatornam... | There are no rational roots, so no linear factors.
If $p(x)$ is a factor of $x^4-x^2+1$ then $p(-x)$ is, too.
If $x^4-x^2+1 = (x^2+a)(x^2+b)$ then $x^2-x+1=(x+a)(x+b)$. Show that $x^2-x+1$ is irreducible.
On the other hand, you'd have to have $x^4-x^2+1=(x^2+ax+b)(x^2-ax+b)$ where $b^2=1$ and $a\neq 0$. This means that... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2414579",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 5,
"answer_id": 3
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Evaluate $\int \frac{\cos 2x \: dx}{3 \sin x+4 \cos x}$ Evaluate
$$I=\int \frac{\cos 2x \: dx}{3 \sin x+4 \cos x}$$
My Try:
$$I=\int \frac{(\cos x-\sin x)(\cos x+\sin x)dx}{3 \sin x+4 \cos x}$$
we have $$\cos x-\sin x=\frac{1}{25}(3 \sin x+4 \cos x)+\frac{7}{25}(3 \cos x-4 \sin x)$$ and
$$\cos x+\sin x=\frac{7}{25}... | HINT: set $$\sin(x)=2\,{\frac {\tan \left( x/2 \right) }{1+ \left( \tan \left( x/2
\right) \right) ^{2}}}
$$
$$\cos(x)={\frac {1- \left( \tan \left( x/2 \right) \right) ^{2}}{1+ \left(
\tan \left( x/2 \right) \right) ^{2}}}
$$
your Integrand is given by $$-1/2\,{\frac { \left( {t}^{2}+2\,t-1 \right) \left( {t}^{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2415567",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
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Prove that $\left|\begin{smallmatrix}a&-b&-c&-d\\b&a&-d&c\\c&d&a&-b\\d&-c&b&a\end{smallmatrix}\right|=(a^2+b^2+c^2+d^2)^2$
Let $a, b, c, d \in \mathbb K$ where $\mathbb K$ is a field. Prove that
$$\det \begin{bmatrix}
a & -b & -c & -d\\
b & a & -d & c\\
c & d & a & -b\\
d & -c & b & a
\end{bmatrix} = (a^2+b^2+c^2+d... | Generally， if $B$ is symmetric such that $A^TB=BA$, then
$$ \det \begin{bmatrix}
A & B \\
-B & A \\
\end{bmatrix} = \det (AA^T +BB^T). $$
In fact
$$ \begin{bmatrix}
A & B \\
-B & A \\
\end{bmatrix}\begin{bmatrix}
A & B \\
-B & A \\
\end{bmatrix}^T=\begin{bmatrix}
A & B \\
-B & A \\
\end{bmatrix}\begin{bmatrix}
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2416817",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
} |
Prove $\ln(1+x)\geq x-\frac{x^2}{2}$
When $x\geq0$ prove that:
$$\ln(1+x)\geq x-\frac{x^2}{2}.$$
My effort: From the Taylor series:
$$\ln(1+x)= x-\frac{x^2}{2}+\frac{x^3}{3}-\cdots$$
I don't know how to continue from here.
| You want to prove $\ln(1+x)\ge x-\frac{x^2}{2}$ for all $x\ge 0$.
To prove it for $0\le x<1$, we'll use Taylor series. It converges because $-1<x\le 1$. See, e.g., here for more information. Or here.
$$\ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\cdots$$
$\ln(1+x)\ge x-\frac{x^2}{2}$ is equivalent to $\frac{x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2417934",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 2
} |
Find $\frac{1}{x_1^3} + \frac{1}{x_2^3} + \frac{1}{x_3^3}$ for $ax^3 + bx^2 + cx + d$ Using Vieta's formulas, I can get $$\begin{align} \frac{1}{x_1^3} + \frac{1}{x_2^3} + \frac{1}{x_3^3} &= \frac{x_1^3x_2^3 + x_1^3x_3^3 + x_2^3x_3^3}{x_1^3x_2^3x_3^3} \\&= \frac{x_1^3x_2^3 + x_1^3x_3^3 + x_2^3x_3^3}{x_1^3x_2^3x_3^3} \\... | If $x \neq 0$ is a solution to $at^3+bt^2+ct+d=0$ then since $a+b(\frac{1}{x})+c(\frac{1}{x})^2+d(\frac{1}{x})^3=0$, $\frac{1}{x}$ is a solution to $dt^3+ct^2+bt+a=0$. Thus if we have $\frac{1}{x_i}=y_i$ for $i=1,2,3$,
\begin{align*}
\frac{1}{x_1^3}+\frac{1}{x_2^3}+\frac{1}{x_3^3}&=y_1^3+y_2^3+y_3^3\\
&=(y_1+y_2+y_3)\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2418954",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Why the minimum of this multivariate polynomial is not a critical point? I have this multivariate polynomial function
$$\begin{align*} f(L,x,y,z) &= L^2 + (1 - L x)^2 + (1 - L y)^2 + (1 - L z)^2 \\
&+ (2 - L x y)^2 + (2 - L x z)^2 + (2 - L y z)^2 + (3 - L x y z)^2
\end{align*}$$
To compute its minimum an idea is to co... | Reduce[Grad[L^2+(1-L x)^2+(1-L y)^2+(2-L x y)^2+(1-L z)^2+(2-L x z)^2+(2-L y z)^2+
(3-L x y z)^2,{L,x,y,z}]=={0,0,0,0}&&L!=0,{L,x,y,z}]
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2421516",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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The ellipse $x^2+2y^2=2.$ Question: The ellipse $x^2+2y^2=2$ is given by $x=a\cos{t}, \quad y=b\sin{t}, \quad t\in[0,2\pi],$ for what values of $a$ and $b$?
a) $(a,b)=(2,1)$
b) $(a,b)=(1,2)$
c) $(a,b)=(1,\sqrt{2})$
d) None of the above.
Attempt: Substitute the trig-values for $x$ & $y$ in the elliptic equation:
$$x^2+2... | Putting $t=0$ and $t=\frac{\pi}{2}$ into the parametric equation $(x,y) = (a\cos t, b\sin t)$, you'll see that both $(a,0)$ and $(0,b)$ have to be points on the ellipse. Since these points have to satisfy $x^2+2y^2=2$, plugging in these coordinates tells you that $a^2=2$ and $2b^2=2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2421949",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Diagonalization of a quadratic form I have to diagonalize the following quadratic form:
$$
Q(x,y,z,t)=x^2+y^2+2xy+2xz+2xt+2yz
$$
I want to complete the squares, so I observe that the first three terms are the square $(x+y)^2$. Then I sum and subtract the quantities $2x^2$, $2z^2$, $y^2$ and $t^2$. Now, how do I have to... | Consume all the terms containing $x$ first ... etc ...
\begin{eqnarray*}
(x+y+z+t)^2-(y+z+t)^2+(y+z)^2-z^2
\end{eqnarray*}
Edit:
\begin{eqnarray*}
\begin{pmatrix}1 & 1 & 1&1 \\0 & 1 & 1&1 \\0 & 1& 1&0 \\0&0&1&0 \\\end{pmatrix}^{T}
\begin{pmatrix}1 & 0 & 0&0 \\0 & -1 & 0&0 \\0 & 0& 1&0 \\0&0&0&-1 \\\end{pmatrix}
\begin{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2422711",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Proving $\frac12\cdot\frac34\cdot\dots\cdot\frac{2n-1}{2n}\leq\frac1{\sqrt{3n+1}} ,\;\forall n\in\mathbb{N}$ using induction Base case. Let $n=1$, then $\frac12\leq\frac1{\sqrt{3+1}}$.
Induction step. Let's assume the inequality is true for some $k\in\mathbb{N}$.
We need to show that it's true for $k+1$, i.e. $\frac12\... |
So now I need to show that $\frac1{\sqrt{3k+1}}\cdot\frac{2k+1}{2k+2}\leq\frac1{\sqrt{3k+4}}$. How should I do this?
Squaring and clearing denominators yields $(2k+1)^2 (3k+4) \leq (2k+2)^2 (3k+1)$. Expanding yields $12k^3+28k^2+19k+4 \leq 12k^3+28k^2+20k+4$. This simplifies to $19 k \leq 20k$, which is trivially tr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2423691",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
If the roots of the equation $p(q-r)x^2+q(r-p)x+r(p-q)=0$ are equal, show that $\dfrac {1}{p}+\dfrac {1}{r}=\dfrac {2}{q}$.
If the roots of the equation $p(q-r)x^2+q(r-p)x+r(p-q)=0$ are equal, show that $$\dfrac {1}{p}+\dfrac {1}{r}=\dfrac {2}{q}$$
My Attempt:
$$p(q-r)x^2+q(r-p)x+r(p-q)=0$$
Comapring with $ax^2+bx+c=... | Dividing by $p^2q^2r^2$
$$=\frac{1}{p^2} +\frac{2}{pr}+\frac{1}{r^2}-\frac{4}{qr}+\frac{4}{q^2}-\frac{4}{pq}$$
$$=\Bigl(\frac{1}{p}\Bigl)^2+\Bigl(\frac{-2}{q}\Bigl)^2+\Bigl(\frac{1}{r}\Bigl)^2$$
$$2\Bigl( \frac{1}{p}\Bigl)\Bigl(\frac{1}{r}\Bigl)+2\Bigl(\frac{1}{p}\Bigl)\Bigl(\frac{-2}{q}\Bigl)$$
$$+2\Bigl(\frac{1}{r}\B... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2423905",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
Find limit of the function $$\lim_{x\to 0^+} \frac { \sqrt {x+bx^2}-\sqrt x}{bx^{3/2}}$$
I tried taking x common and all but it doesn't work.Are there are any shortcuts for finding the answer?Any help is greatly appreciated
Thanks in advance
| First of all $b$ cannot be zero.
By multiplying both numerator and denominator by $\sqrt{x+bx^2}+\sqrt{x}$
the function becomes:
$$\frac{(\sqrt{x+bx^2}+\sqrt{x})(\sqrt{x+bx^2}-\sqrt{x})}{(\sqrt{x+bx^2}+\sqrt{x})bx^{3/2}}=\frac{bx^2}{(\sqrt{x+bx^2}+\sqrt{x})bx^{3/2}}$$ $$=\frac{\sqrt{x}}{\sqrt{x+bx^2}+\sqrt{x}}=\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2424139",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
How to prove that $\frac{b-a}{\sqrt{1-a^2}}<\sin^{-1}b-\sin^{-1}a<\frac{b-a}{\sqrt{1-b^2}}$ by using Mean Value Theorem? Given Question is Prove that
$$\frac{b-a}{\sqrt{1-a^2}}<\sin^{-1}b-\sin^{-1}a<\frac{b-a}{\sqrt{1-b^2}}$$
when $0<a<b<1$ using Mean Value Thoerem.
I considered $f(x)=\sin^{-1}x $ when $x\in(0,b)$ and... | According to MVT, there exists $c$, $a\leq c\leq b$ such that
$$f(b)-f(a)=f'(c)(b-a)$$
so with $\dfrac{d}{dx}\sin^{-1}x=\dfrac{1}{\sqrt{1-x^2}}$ we see
$$\sin^{-1}b-\sin^{-1}a=\dfrac{1}{\sqrt{1-c^2}}(b-a)$$
step by step
\begin{align}
a\leq &c\leq b\\
a^2\leq &c^2\leq b^2\\
1-b^2\leq &1-c^2\leq 1-a^2\\
\dfrac{1}{\sqrt{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2427411",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
How to show that the deltoid is a plane algebraic curve of degree $4$?
How can I show that a deltoid is a plane algebraic curve of degree 4?
I have searched that the parametric equation for deltoid is given by
$$\begin{align}
x&= 2 \cos t + \cos 2t \\
y&=2 \sin t - \sin 2t
\end{align}$$
So, by using some trigonometri... | Write the equations as
$
x=2c^2+2c-1\\
y^2=-4c^4+8c^3-8c+4
$
where $c=\cos t$.
Feeding
Eliminate[{x==2c+2c2-1,y2=-4c^4+8c^3-8c+4},c]
to WA
gives
$$
x^4 - 8 x^3 + x^2 (2 y^2 + 18) + 24 x y^2 = -y^4 - 18 y^2 + 27
$$
You can do this by hand by computing the resultant of $2c^2+2c-1-x$ and $-4c^4+ 8 c^3 -8c+4-y^2$, seen as... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2427627",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
express the union and intersections of the circles How to express the union and intersections of the circles $ \ x^2+(y+1)^2=9$ and $x^2+(y-1)^2=9 \ $ in terms of set comprehension notation.
Answer:
Let ,
$ A=\{(x,y): \ x^2+(y+1)^2=9 \} \ \ and \ \ B=\{(x,y): \ x^2+(y-1)^2=9 \} \ $
Thus,
$A \cup B \ =\{(x,y): x=\pm... | $x=\pm \sqrt{9-(y+1)^2} \ \cup \ x=\pm \sqrt{9-(y-1)^2}$
is meaningless, because the $\cup$ operator is an operator on sets, not equations. If anything, it should say
$x=\pm \sqrt{9-(y+1)^2} \ \lor \ x=\pm \sqrt{9-(y-1)^2}$
since $\lor$ is the symbol for "or".
Other than that, I see no easy way to describe the set, ex... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2430453",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Eliminate $t$ from the equations: $x = \frac{1}{t} - t \, , \, y = \frac{1}{t^2} - 1$ $$x = \frac{1}{t} - t \tag{1}$$
$$y = \frac{1}{t^2} - 1 \tag{2}$$
$$\frac{1}{t^2} = y + 1$$
$$t^2 = \frac{1}{y + 1}$$
$$t = \pm\sqrt{\frac{1}{y + 1}}$$
$$x = \frac{1 - t^2}{t}$$
$$y = \frac{1 - t^2}{t^2}$$
$$\frac{x}{y} = \frac{1 - t^... | Alternatively:
$$x=\frac{1}{t}-t \Rightarrow x^2=\frac{1}{t^2}-2+t^2=(y+1)-2+\frac{1}{y+1} \Rightarrow $$
$$y^2-x^2y-x^2=0 \Rightarrow y=\frac{x^2\pm \sqrt{x^4+4x^2}}{2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2431894",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Are there multiple ways to translate a permutation $\alpha$ to a permutation $\beta$? Are there multiple ways to translate a permutation $\alpha$ to a permutation $\beta$?
I came across the question below and found multiple solutions for $\gamma$. My approach to finding solutions is not very rigorous, but I did check ... | Yes, if two permutations are conjugate they are conjugate in multiple ways. The cycle structure is preserved by conjugation but beyond that there are two kinds of freedom in the choice of the conjugating permutation: first, the cycles may be "spun" arbitrarily. And equal cycles may be permuted. In your case $\alpha$ a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2431981",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove $ \ \forall n \ge 2, \ 2^{n} + 3^{n} < 4^{n}$ Question:
Prove $ \ \forall n \ge 2, \ 2^{n} + 3^{n} < 4^{n}$
My attempt:
Base case is trivial.
Suppose $ \ n \ge 2$ and $ \ 2^{n} + 3^{n} < 4^{n}$
Then,
$2^{n+1} + 3^{n+1} = 2.2^{n} + 3.3^{n} = 2.2^{n} + 2.3^{n} + 3^{n} = 2(2^{n} + 3^{n}) + 3^{n} <2(4^{n}) + 3^{n} $... | You can say $$2\cdot 2^n + 3\cdot 3^n < 4\cdot 2^n + 4\cdot 3^n = 4(2^n+3^n) < 4\cdot 4^n = 4^{n+1}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2432215",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Show that $(a+b)^7-a^7-b^7=7ab(a+b)(a^2+ab+b^2)^2$
Show that $(a+b)^7-a^7-b^7=7ab(a+b)(a^2+ab+b^2)^2$
In the list of questions proposed in the "Meeting for Training for the Brazilian Olympiad", 2013. No answer provided. Could solve some problems in that list but got stuck in this one. My developments are going into v... | As Mr. Chip suggested there’s no need to drag around $a$ and $b$ here.
So I'm gonna proceed in this fashion:
$$(1+x)^7-1^7-x^7=7x(1+x)(1+x+x^2)^2$$
Now we divide by $(1+x)$ or we can factor it out:
Note that various divisions like that may call into question the rigor here. But it should be as easy as pie for you (I'... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2432992",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
Solve identity : $\frac{1+\sin x\cos x}{\cos^3 x - \sin^3 x}+\frac{1}{\sin x+\cos x}+\frac{\sin^2x-2\cos x-1}{\cos^2x-\sin^2x}=\frac{1}{\tan^2x-1}$ $$\frac{1+\sin x\cos x}{\cos^3 x - \sin^3 x}+\frac{1}{\sin x+\cos x}+\frac{\sin^2x-2\cos x-1}{\cos^2x-\sin^2x}=\frac{1}{\tan^2x-1}$$
I am doing this identity for about an h... | Use $a^3-b^3=(a-b)(a^2+ab+b^2)$
Therefore we have,
$$\cos^3x-\sin^3x=(\cos x-\sin x)(1+\sin x \cos x)$$
Your expression in LHS
$$\frac{1+\sin x\cos x}{\cos^3 x - \sin^3 x}+\frac{1}{\sin x+\cos x}+\frac{\sin^2x-2\cos x-1}{\cos^2x-\sin^2x} $$
Simplifies to
\begin{align}
\text{LHS}
&=\frac{1}{\cos x - \sin x}+\frac{1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2435095",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Evaluate the limit $\lim_{x\to a} (a-x) \tan \frac {\pi x}{2a}$
Evaluate:
$$\lim_{x\to a} (a-x) \tan \dfrac {\pi x}{2a}.$$
My attempts:
$$=\lim_{x\to a} (a-x) \dfrac {\sin \left(\dfrac {\pi x}{2a}\right)}{\cos \left(\dfrac {\pi x}{2a}\right)}=1\cdot\lim_{x\to a} \dfrac {a-x}{\cos \left(\dfrac {\pi x}{2a}\right)}.$$... | Let $y = \frac{\pi x}{2a}$
\begin{align} \lim_{x \to a} \frac{a-x}{\cos \left( \frac{\pi x}{2a}\right)} &= \lim_{y \to \frac{\pi}{2}} \frac{\pi a-(2ay)}{\pi \cos(y)}\\
&=\frac{2a}{\pi} \lim_{y \to \frac{\pi}{2}} \frac{\frac{\pi}2-y}{\cos(y)}\\
&=\frac{2a}{\pi} \lim_{y \to \frac{\pi}{2}} \frac{\frac{\pi}2-y}{\cos\left(y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2437105",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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Finding all non-negative integers solutions to $x_1+x_2+x_3+...+x_6=20$ such that $x_{2n+1} \le x_{2n+2}$ for $0 \le n \le 2$ Solve $x_1+x_2+x_3+x_4+x_5+x_6=20$ such that $x_{2n+1}\leq x_{2n+2}, 0\leq n \leq2$
Edit : I solved it.
Let $0 \le a,b,c \le 20$ such that $0 \le a+b+c \le 20$
$x_1+x_2+x_3+x_4+x_5+x_6=20 \;and... | I verified your solution a different way:
Let $N(i)$ denote the number of ways to have $x_{1}+x_{2}=i$ with $x_{1}\leq x_{2}$. It is easy to show that $N(i)=\lfloor\frac{i}{2}\rfloor+1$, for $0\leq i$.
Your problem now reduces to computing the following sum:
$\sum_{(k_{1},k_{2},k_{3})\in K}N(k_{1})\cdot N(k_{2})\cdot... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2437307",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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$T(n)=2T(n-1)+n,T(1)=1,n\ge2$
Solve the recurrence $T(n)=2T(n-1)+n,T(1)=1,n\ge2$
My approach:
$$T(n)=2T(n-1)+n$$
$$T(n)=2(2T(n-2)+(n-1))+n$$
$$T(n)=4T(n-2)+2(n-1)+n\dots$$
$$T(n)=2^kT(n-k)+2^{k-1}(n-k+1)+\dots+2^{k-k}n$$
From initial conditions, $k=n-1$:
$$T(n)=2^{n-1}1+2^{n-2}2+\cdot\cdot\cdot\cdot+2^{n-n}n$$
Now wh... | $$f(x)=x2^{n-1}+x^22^{n-2}+...+x^n=2^n\left(\frac{x}{2}+\left(\frac{x}{2}\right)^2+...+\left(\frac{x}{2}\right)^n\right)=$$
$$=2^n\cdot\frac{\frac{x}{2}\left(\left(\frac{x}{2}\right)^n-1\right)}{\frac{x}{2}-1}=\frac{x^{n+1}-x2^n}{x-2}.$$
Now, calculate $f'(1).$
I got $$1\cdot2^{n-1}+2\cdot2^{n-2}+...+n\cdot2^0=f'(1)=2^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2438841",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Determine all real $x$ such that $\arccos{\frac{1-x^2}{1+x^2}}=-2\arctan{x}.$ I just need you guys to check this solution and tell me where I can improve. The key questions to be answered are:
*
*Are things clear?
*Is there any unnecessary redundancy?
*Any logical fallacies?
*Quality of stringency and mathematic... | Your solution seems good. Here's an alternative way.
It is easy to see that
$$
-1\le\frac{1-x^2}{1+x^2}\le1
$$
for every $x$.
The derivative of
$$
f(x)=\arccos\frac{1-x^2}{1+x^2}
$$
is
\begin{align}
f'(x)&=-\frac{1}{\sqrt{1-\dfrac{(1-x^2)^2}{(1+x^2)^2}}}
\frac{-2x(1+x^2)-2x(1-x^2)}{(1+x^2)^2}
\\[6px]
&=\frac{1+x^2}{|2x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2439554",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
If $ax+by=7$, .
If
$$ax+by=7$$
$$ax^2+by^2=49$$
$$ax^3+by^3=133$$
$$ax^4+by^4=406$$
then find the value of
$$2014(x+y-xy) - 100(a+b)$$
My attempt:
$$ax^2+by^2=49$$
$$ax^2+by^2=(ax+by)^2$$
$$ax^2+by^2=a^2x^2+2abxy+b^2y^2$$
$$ax^2-a^2x^2+by^2-b^2y^2=2abxy$$
$$ax^2(1-a)+by^2(1-b)=2abxy$$
| The structure of the equations indicate that $7$, $49$, $133$, $406$, ... is a sequence that is defined as a homogeneous linear difference equation of order $2$ with the characteristic equation $\lambda^2+c_1\lambda +c_0=0$ with the roots $x$ and $y$. Therefore we have
\begin{eqnarray*}
133 + \;\;49c_1 + \;\;7c_0 &=& ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2440302",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Infinite sum of $\cot^{-1} (n^2 + 3/4)$. I am trying to find the infinite sum
$$\sum_{n=1}^\infty \cot^{-1} (n^2 + ( \frac{3}{4})),$$
I tried to get a telescopic series but I couldn't find one.
| As $\cot(A-B)=\dfrac{\cot A\cot B+1}{\cot B-\cot A}$
$$\dfrac{4n^2+3}4=1+\dfrac{4n^2-1}4=1+\dfrac{2n+1}2\cdot\dfrac{2n-1}2$$
Again,$\dfrac{2n+1}2-\dfrac{2n-1}2=1$
So, we can write $\cot^{-1}\left(n^2+\dfrac34\right)=\cot^{-1}\left(\dfrac{1+\dfrac{2n+1}2\cdot\dfrac{2n-1}2}{\dfrac{2n+1}2-\dfrac{2n-1}2}\right)$
$=\cot^{-1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2440420",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How can I solve this equation for real numbers? How can I solve this equation for real numbers?
$$(x+2)^4+x^4=82.$$
I tried $(x+2)^4+x^4-82=
2x^4 + 8 x^3 + 24 x^2 + 32 x - 66=0$
It is very difficult to continue.
| Two solutions of $82=x^4+(x+2)^4$ are given by $x=1$ and $x=-3$. It follows that
$$ x^4+(x+2)^4-82 = (x-1)(x+3) q(x) $$
where $q(x)$ is a quadratic polynomial. Find it, solve $q(x)=0$ and you have the complete set of solutions ($q(x)=2\left[(x+1)^2+10\right]$, so the only real solutions are the trivial ones).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2441965",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Proving an identity in floor function For every positive integer $n$ ,show that
$$\lfloor{\sqrt{n}+\sqrt{n+1}}\rfloor=\lfloor{\sqrt{4n+1}}\rfloor=\lfloor{\sqrt{4n+2}}\rfloor=\lfloor{\sqrt{4n+3}}\rfloor$$
My Attempt:
$$\left(\sqrt{n}+\sqrt{n+1}\right)^2=2n+1+2\sqrt{n(n+1)}$$
Now, $\sqrt{n(n+1)}$ is geometric mean of $n... | Note that for an integer, $x$, $$x^2 \equiv 0 \pmod 4$$ or
$$x^2 \equiv 1 \pmod 4$$
Hence there is no complete square between $4n+1$ and $4n+3$.
Hence $\lfloor \sqrt{4n+1} \rfloor = \lfloor \sqrt{4n+3} \rfloor$, the overall task can be completed by your earlier result of $4n+1 < (\sqrt{n} + \sqrt{n+1})^2 < 4n+3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2442533",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the maximum natural number $m$ such that $n^3-n$ is divisible by $m$ $\forall$ $n$ $\ge$1 Prove your assertion. Find the maximum natural number $m$ such that $n^3-n$ is divisible by $m$ $\forall$ $n$$\ge$1 Prove your assertion.
I guess $m$$=$1 because if $m$ divides $n^3-n$ $\Rightarrow$ $m$ divides $n(n-1)(n+1)$ ... | We have $n^3 -n = n\cdot\left(n^2 - 1\right) = n\cdot(n-1)\cdot(n+1)$. These are three consecutive numbers so at least one of them has to be divisible by $3$ and one of them has to be divisible by $2$. Hence, $m$ has to be at least $2\cdot 3 = 6$.
However, for each prime $p \in \mathbb P$ with $p > 3$, we can find an $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2443001",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Integral of irrational function $\int \frac{dx}{x\left(\sqrt[3]{x}+\sqrt[5]{x^2}\right)}$
$$\int \frac{dx}{x\left(\sqrt[3]{x}+\sqrt[5]{x^2}\right)}$$
I have tried to solve the integral using, substitution but it doesn’t lead to the correct result.
Do you folks have any ideas about which technique I should use to solv... | Plug $x=u^{15}$ so $dx=15u^{14}$
$\sqrt[3]{x}=u^5;\;\sqrt[5]{x^2}=u^6$
Integral becomes
$$15\int \frac{du}{u^7+u^6} =15\int \frac{du}{u^6 (u+1)}=15\int\left[\frac{1}{u^6}-\frac{1}{u^5}+\frac{1}{u^4}-\frac{1}{u^3}+\frac{1}{u^2}+\frac{1}{u+1}-\frac{1}{u}\right]\,du$$
Indeed with partial fraction we have:
$$\frac{a}{u}+\f... | {
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"url": "https://math.stackexchange.com/questions/2443335",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Is this a Taylor series? $\ln(x) +1 = \sum_{n=0}^{\infty} \frac{n+1}{n!} \cdot \frac{(\ln(x))^n}{x}$ Can you provide a proof of this identity using only calculus?
$$\ln x + 1 = \sum_{n=0}^{\infty} \frac{n+1}{n!} \cdot \frac{(\ln x)^n}{x}$$
By the way, here is how I arrived at it:
There is string of length $x$ units. Se... | For, $x>0$, $\text{ln}x =z \in (-\infty, \infty)$.
So, we can write the r.h.s as ${e^{-z}}{\sum_{n=0}^{\infty} \frac{n+1}{n!}z^n}$.
Now, ${\sum_{n=0}^{\infty} \frac{n+1}{n!}z^n}$
$=\frac{d (\sum_{n=0}^{\infty} \frac{z^{n+1}}{n!})}{dz}$
$=\frac{d {e^{z}}z}{dz}= e^{z}+ze^{z}$
So, ${e^{-z}}{\sum_{n=0}^{\infty} \frac{n+1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2446094",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
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How to solve :$\operatorname{arcsin}x-\operatorname{arcsin}\left(x/2\right)=\operatorname{arcsin}\left(\frac{x\sqrt3}2\right)$ $$\operatorname{arcsin}x-\operatorname{arcsin}\left(x/2\right)=\operatorname{arcsin}\left(\frac{x\sqrt3}2\right)$$
Why can't I figure this one out??
Is it possible to cancel out the arcsins?
I ... | Use $\sin(A-B)$ identity and the fact that $f(f^{-1}(x)) = x$:
$$x\sqrt{1-\frac{x^2}{4}}-\frac{x}{2}\sqrt{1-x^2} = \frac{x\sqrt 3}{2}$$
So $x=0$ is one solution. Now
$$\sqrt{4-x^2}-\sqrt{1-x^2} = \sqrt{3}$$
Now let $1-x^2 = t$, so that our equation becomes:
$$\sqrt{3+t}-\sqrt{t} = \sqrt{3}$$
Upon squaring, we get:
$$3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2447406",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Guess explicit formula for sequence and then prove it is correct just trying to work through the following problem but have come to point where I don't know what to do.
Info: Let $a_0, a_1, a_2, ...$ be the sequence defined recursively as $a_0 = 0, a_k = k + a_{k-1}$ for each integer $k\ge 1$.
Question is: Using this i... | Induction isn't actually needed here. Notice that each term is a partial sum, in particular:
$$a_n = \sum_{k=0}^n k.$$
You guessed that $a_n = \frac{n^2 +n}{2}$, which is correct, so you can just show that
$$\sum_{k=0}^n k = \frac{n^2 +n}{2}.$$
The usual way is to add the sum here to itself but write it backwards once... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2447627",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove $\frac{n-1}{n}+\frac{n-1}{n}\frac{n-3}{n-2}+ \frac{n-1}{n}\frac{n-3}{n-2}\frac{n-5}{n-4} + ... = \frac{n-1}{3}$ without induction I found this identity using Maple. Is there a (simple) way to prove it without using induction? Using induction, the proof is quite easy.
Prove for odd $n$ that
$$\sum_{k=1}^{(n+1)/2}\... | We need to prove that
$$\frac{n-1}{n}+\frac{n-1}{n}\frac{n-3}{n-2}+ \frac{n-1}{n}\frac{n-3}{n-2}\frac{n-5}{n-4} + ... = \frac{n-1}{3}$$ or
$$1+\frac{n-3}{n-2}+ \frac{n-3}{n-2}\frac{n-5}{n-4} + ... = \frac{n}{3}$$ or
$$\frac{n-3}{n-2}+ \frac{n-3}{n-2}\frac{n-5}{n-4} + ... = \frac{n-3}{3}$$ or
$$1+\frac{n-5}{n-4} +\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2450038",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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