Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Fourier transform of squared exponential integral $\operatorname{Ei}^2(-|x|)$ Let $\operatorname{Ei}(x)$ denote the exponential integral:
$$\operatorname{Ei}(x)=-\int_{-x}^\infty\frac{e^{-t}}tdt.$$
Now consider the function $\operatorname{Ei}(-|x|)$.
Its Fourier transform is
$$\frac1{\sqrt{2\pi}}\int_{-\in... | I changed my evaluation slightly, and I was able to get the result in a very simple form.
First notice that $$ \int_{-\infty}^{\infty} \text{Ei}^{2}(-|x|) e^{ikx} \, dx = 2 \int_{0}^{\infty} \text{Ei}^{2}(-x) \cos(kx) \, dx. $$
Then integrating by parts, and assuming for now that $k >0$,
$$ \begin{align}\int_{-\infty... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/957431",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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Prove inequality formula by induction my question is from Apostol's Vol. 1 One-variable calculus with introduction to linear algebra textbook.
Page 35. Exercise 1. Prove the following formula by induction: $$1^3+2^3+3^3+\cdots+(n-1)^3<\frac{n^4}{4}<1^3+2^3+3^3+\cdots+n^3$$
The attempt at a solution: First I do the left... | Assuming $\sum_{i=1}^{n-1} i^3 < \frac{n^4}{4}$, we see that
\begin{align*}
\sum_{i=i}^n i^3 &= \sum_{i=1}^{n-1} i^3 + n^3 \\
&< \frac{n^4}{4} + n^3 = \frac{n^4+4n^3}{4} \\
&< \frac{n^4 + 4n^3 + 6n^2 + 4n + 1}{4} \\
&= \frac{(n+1)^4}{4}.
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/957571",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluating $\;\int x\cos(5x-1)\,\mathrm{d}x$ Integration by parts:
1) let $u = x$, $du = dx$, $v = \frac{1}{5} \sin (5x- 1)$, $dv = \cos (5x -1) dx$
$udv = x \frac{1}{5} \sin x ( 5x- 1) - \frac{1}{5} \int \sin (5x-1) dx$.
2) let $u = \sin (5x -1), du = \frac{1}{5} \cos(5x-1) dx , v = x, dv = dx$
Is my $du$ correct in ... | The expression should be $$I = \int u\,dv = uv -\int v\,du $$
which is what I suspect you meant, since you correctly obtained
$$I = \frac 15 x \sin(5x - 1) - \frac 15\int \sin(5x - 1)\,dx $$
From there, using substitution (see note below), we get
$$ I = \frac 15 x \sin(5x - 1) - \frac 1{25}( - \cos(5x - 1))+ C$$
$$= \f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/958454",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Integration Question: Completing the Square/Trig Sub yields a different answer than integral table. After the completing the square,
$$\int \frac{dx}{x^2 + 2x - 3}$$
becomes,
$$ \int \frac{dx}{(x+1)^2 - 4}$$
The integral table in my book says the antiderivative is,
$$\frac{1}{2a} ln \, \Biggl\lvert \frac{x-a}{x+a} \Big... | $$-\frac{1}{2} ln \, \Biggl\lvert \frac{x+3}{\sqrt{x^2+2x-3}} \Biggr\rvert + \, C $$
You can take the square of the term in the absolut value signs. At the same time you take the square root. Remember the rule $log|x^a|=a\cdot log|x|$
$$-\frac{1}{2} ln \, \Biggl\lvert \left( \frac{(x+3)^2}{x^2+2x-3 } \right) ^{1/2}\Big... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Write it as an element of this ring? Since the degree of the irreducible polynomial $x^3+2x+2$ over $\mathbb{Q}[x]$ is odd, it has a real solution , let $a$. I am asked to express $\displaystyle{\frac{1}{1-a}}$ as an element of $\mathbb{Q}[a]$.
I have done the following:
Since $a$ is a real solution of $x^3+2x+2$, we h... | $$
a^3+2a+2=0\\
a^3+2a-3=-5\\
(a-1)(a^2+a+3)=-5\\
\frac{1}{5}(a^2+a+3)=\frac{1}{1-a}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/961296",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find the limit of $\sqrt{x^2+x-1} +x$ as $x\to-\infty$ Find the limit of $\sqrt{x^2+x-1} +x$ as $x\to-\infty$.
My solution:
multiplying by: $\displaystyle\frac{\sqrt{x^2+x-1}-x}{\sqrt{x^2+x-1}-x}$
Which gives us: $\displaystyle\frac{x-1}{\sqrt{x^2+x-1}-x}$
dividing by $\sqrt{x^2}$ gives:
$\displaystyle \frac{1}{\sqrt{... | $\sqrt{x^2}$ is a positive number and since we are taking the limit as $x\to -\infty$, we have $\sqrt{x^2} = -x$. Your limit will be negative since $x^2>x^2+x-1$ as $x\to-\infty$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/961741",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
} |
$ \lim\limits_{x \rightarrow \ + \infty}(\sqrt{x^2 + 2x} - \sqrt{x^2 - 7x})$ Find the limit: $$ \lim_{x \rightarrow \ + \infty}(\sqrt{x^2 + 2x} - \sqrt{x^2 - 7x})$$
I did the following:
\begin{align}
(\sqrt{x^2 + 2x} - \sqrt{x^2 - 7x}) = \frac{(\sqrt{x^2 + 2x} - \sqrt{x^2 - 7x})}{1} \cdot \frac{(\sqrt{x^2 + 2x} + \sqrt... | The mean value theorem gives $f(x+h) = f(x) + f'(x)h + {1 \over 2} f''(\xi) h^2$, where $\xi \in [x,x+h]$ (adjusted appropriately for sign of $h$).
This gives $|\sqrt{1+\theta} - (1+ {1 \over 2} \theta) | \le {1 \over 8} \theta^2$.
We have (if $x>0$),$\sqrt{x^2+2x} - \sqrt{x^2-7x} = x(\sqrt{1+{2 \over x}}-\sqrt{1-{7 \o... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/962458",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 5
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Make x the subject given the formula for y I am given the following formula:
$$y=\frac{x}{a}+\sqrt{\frac{x}{b}}.$$
I want to make x the subject.
I rearranged the equation and got to:
$$y^{4}=x(\frac{y^{2}}{2}+2y^{2})-x^{2}.$$
and I don't know where to go from here. May be this is the wrong rearrangement.
The answer acc... | Hint: Set $u=y^2$ and solve the quadratic equation for $u$.
Then your equation is rewritten as
$$
u^2=\frac{5x}{2}u-x^2,
$$
which by completing the square is written as
$$
\Bigg(u-\frac{5}{4}x\Bigg)^2 - \frac{9}{16}x^2=0,
$$
and thus you have
$$
u-\frac{5}{4}x = \pm\frac{3}{4}x,
$$
or
$$
u = \pm\frac{3}{4}x+\frac{5}{4}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/963563",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Proving Identities. I tried to solve it but I cant get the answer. How to prove this by using a hand?
*
*$$ \sec^2x + \csc^2x = \sec^2x \csc^2x $$
*$$ \frac{\sec\theta + 1}{\sec\theta - 1} = \frac{1 + \cos\theta}{1 - \cos\theta}$$
*$$ \frac{1 - \cot^2\theta}{1 + \cot^2\theta} = \sin^2\theta - \cos^2\theta $$
Any... | I think that #1 is an unfair question, because there is nothing to prove. For if we can assume that
$$\cos^2\theta+ \sin^2\theta=1$$
$$\tan^2\theta+ 1=\sec^2\theta$$
$$\text{and }\cot^2\theta+ 1=\csc^2\theta$$
Then it should be natural to assume that
$$\sec^2\theta+\csc^2\theta=\sec^2\theta\csc^2\theta$$
Consider a rig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/963661",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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study of two sequences I need to study wether thos two sequences converge or not.
1) $u_n=n\sum_{k=1}^{2n+1} \frac{1}{n^2+k}$
2) $v_n=\frac{1}{n}\sum_{k=0}^{n-1} \cos(\frac{1}{\sqrt{n+k}})$
For the first i get it converges to $0$ by just expressing the sum as a fraction (not very smart imo).
| *
*Note that $(n^{2} +k) \leq (n^{2}+2n+1)$ whenever $k \in\left\{1,2,\cdots,(2n+1)\right\}$. So we have that $\frac{1}{n^{2}+k} \geq \frac{1}{n^{2}+2n+1} =\frac{1}{(n+1)^{2}}$. Similarly $n^{2}+1 \leq n^{2}+k$ and so we have that $\frac{1}{n^{2}+1}\geq \frac{1}{n^2+k}$. So we get that $$\sum_{k=1}^{2n+1} \frac{1}{n^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/963807",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Equation of line passing through point. The straight line $3x + 4y + 5 = 0 $ and $4x - 3y - 10 = 0$ intersect at point $A$. Point $B$ on line $3x + 4y + 5 = 0 $ and point C on line $4x - 3y - 10 = 0$ are such that $d(A,B)=d(A,C)$.
Find the equation of line passing through line $\overline{BC}$ if it also passes through... | inspection shows the lines are perpendicular and meet at $(-1,2)$.
define variables $(w,z)$ by:
$$
\begin{pmatrix} w \\ z \end{pmatrix} = \begin{pmatrix} 3 & 4 \\ 4 & -3 \end{pmatrix}
\left( \begin{pmatrix} x \\ y \end{pmatrix} - \begin{pmatrix} 1 \\ -2 \end{pmatrix}\right)
$$
the equations of the lines are now $w=0$ a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/965015",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to simplify $\sqrt[3]{29\sqrt{2}-45}-\sqrt[3]{29\sqrt{2}+45}$ I in trouble simplifying this:
$$\sqrt[3]{29\sqrt{2}-45}-\sqrt[3]{29\sqrt{2}+45}$$
couldn't find a solution. Can you help?
| Those expressions usually come from Cardan's formula
$$
\sqrt[3]{\sqrt{\frac{q^2}{4}+\frac{p^3}{27}}-\frac{q}{2}}-
\sqrt[3]{\sqrt{\frac{q^2}{4}+\frac{p^3}{27}}+\frac{q}{2}}
$$
(see Wikipedia).
So we need $45=q/2$ or $q=90$ and so
$$
\frac{p^3}{27}+\frac{90^2}{4}=2\cdot29^2,
$$
that is,
$$
\frac{p^3}{27}=-343=-7^3
$$
wh... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/970165",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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What is the values of $a$ and $b$ without using the L'Hôpital's Rule Suppose that $$\frac{(2x)^x-2} {a(x-1)+b(x-1)^2}\to 1$$ as $x \to 1$. ThenWhat is the values of $a$ and $b$ without using the L'Hôpital's Rule?
Thanks for your help!
| Factor an $x-1$ out of the denominator to get $$\frac{(2x)^x-2} {a(x-1)+b(x-1)^2} = \frac{(2x)^x - 2}{x-1} \cdot \frac{1}{a + b(x-1)}$$ so that (as long as $a \not= 0$)$$\lim_{x \to 1} \frac{(2x)^x-2} {a(x-1)+b(x-1)^2} = \left.\frac{d}{dx} (2x)^x \right|_{x=1} \cdot \frac 1a.$$
But $\dfrac{d}{dx} (2x)^x = (2x)^x (1 + ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proof by induction for $ 2^{k + 1} - 1 > 2k^2 + 2k + 1$ for $k > 4$ I was given this proof for hw.
Prove that $ 2^{k + 1} - 1 > 2k^2 + 2k + 1$
So, far I've gotten this
Basis:
$k = 5$, $2^{5 + 1} - 1 > 2\cdot5^2 + 2\cdot5 + 1$ => $63 > 61$ (So, the basis holds true)
Hypothesis:
for all $k > 4$, $ 2^{k + 1} - 1 > 2k^2 +... | Inductive step
$$2^{k+2}-1=2\times 2^{k+1}-1>2(2k^2+2k+2)-1>2(k+1)^2+2(k+1)+1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/979355",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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I can't figure out this simplification in a differential equation I was watching PatrickJMT's video on first-order differential equations and while I think I should see what he's doing on the left side here from line one to line two, I just can't. I ran it past my roommate as well without any luck :(
Elaboration: I und... | The second line is just the power rule $(uv)' = u'v + uv'$, but performed in reverse. If we expand it out starting from the second line you'll be able to see what he did:
$$
\begin{align}
\frac{d}{dx}\left[(1+x^2)y\right] & = y\frac{d}{dx}(1+x^2) + (1+x^2)\frac{dy}{dx} \\
& = 2xy + (1+x^2)\frac{dy}{dx} \\
& = \left(\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/979783",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proof modular equality by induction I'm trying to prove using induction that $5^{2^{x-2}} \equiv 1 + 2^x \pmod{2^{x+1}}$
So far, I have:
*
*Base case: $x = 2, 5 \equiv 5 \pmod{8}$, It is true. $x = 3, 25 \equiv 9 \pmod{16}$, It is true.
*Inductive step: let $x = n$ Assume $5^{2^{n-2}} \equiv 1 + 2^n \pmod{2^{n+1}}$... | $$ \begin{align*}\begin{split}
5^{2^{n-1}} &= (5^{2^{n-2}})^2 \\
&= (1 + 2^{n}+2^{n+1}a)^2 \\
&= 1+ (2^n)^2+(2^{n+1}a)^2+
2\cdot2^n+2\cdot2^{n+2}a+2\cdot2^{n}\cdot2^{n+1}a\\
&=1+2^{n+1}+2^{n+2}b
\end{split}\end{align*}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Composing Piecewise Functions I was wondering how to compose piecewise functions.
On a practice exam I was reading, a question asks what F(F(x)) will look like if F(x)= 2x if x<1/2 and = 2-2x if x>=1/2.
Would I just substitute the original parts into themselves? (like 2(2x)=4x and 2-2(2-2x)=4x-2?) On the solution to th... | Our function F is given as:
$F(x)=\left\{\begin{matrix} 2x & x<1/2 \\ 2-2x & x \ge 1/2 \end{matrix}\right.$
So for inputs on both sides of the breakpoint $x=1/2$, it gives outputs in the range $\left(-\infty,1\right]$. Which means that when you compose it with itself, you have to consider four possible events:
*
*$x... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the tangen to $\cos(\pi \cdot x)$ I have the following assignment.
Find the tangent to $y=f(x)=\cos(\pi \cdot x)$ at $x=\displaystyle\frac{1}{6}$.
First step would be to take the derivative of $f(x)$
$f'(x)= -\pi \sin(\pi \cdot x)$
Then I put the $x$-value into $f'(x)$ to find the slope
$f'(\displaystyle\frac{1}{6... | You are correct, and so is your book. The two equations are equivalent.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/984565",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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A inverse Trigonometric multiple Integrals How to calculate the closed form of the integral
$$\int\limits_0^1 {\frac{{\int\limits_0^x {{{\left( {\arctan t} \right)}^2}dt} }}{{x\left( {1 + {x^2}} \right)}}} dx$$
| Use integration by parts, taking $\displaystyle \int_{0}^{x} (\arctan t)^2 dt$ as the first function and $ \dfrac{1}{x(x^2+1)} $ as the second function to get:
$ \displaystyle \int \dfrac{dx}{x(x^2+1)} = \dfrac{1}{2} \ln\bigg(\dfrac{x^2}{x^2+1}\bigg)$
Hence,
$I = \bigg|\dfrac{1}{2} \ln \bigg(\dfrac{x^2}{x^2+1}\bigg) \... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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For the series $S = 1+ \frac{1}{(1+3)}(1+2)^2+\frac{1}{(1+3+5)}(1+2+3)^2$...... Problem :
For the series $$S = 1+ \frac{1}{(1+3)}(1+2)^2+\frac{1}{(1+3+5)}(1+2+3)^2+\frac{1}{(1+3+5+7)}(1+2+3+4)^2+\cdots $$ Find the nth term of the series.
We know that nth can term of the series can be find by using $T_n = S_n -S_{n-1... | First we should note that $$1+2+3+...+n=\dfrac{n(n+1)}{2}$$ and $$1+3+5+...+(2n-1)=n^2.$$ Therefore general term in your series become to $$T_n=\dfrac{(1+2+3+..+n)^2}{1+3+5+...+(2n-1)}=\dfrac{(n+1)^2}{4}$$ $$S_n=\sum_{k=1}^nT_k\\=\sum_{k=1}^n\dfrac{(k+1)^2}{4}\\=\sum_{k=1}^{n+1}\dfrac{k^2}{4}-\dfrac{1}{4}\\=\dfrac{(n+1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/988628",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Evaluation of $\int_0^{\pi/4} \sqrt{\tan x} \sqrt{1-\tan x}\,\,dx$ How to evaluate the following integral
$$\int_0^{\pi/4} \sqrt{\tan x} \sqrt{1-\tan x}\,\,dx$$
It looks like beta function but Wolfram Alpha cannot evaluate it. So, I computed the numerical value of integral above to 70 digits using Wolfram Alpha and I u... | To get rid of both square root, I use the substitution$ \tan x = \sin^2\theta$. Then we rewrite the integral I as \begin{array}{l}
\displaystyle I=\int_{0}^{\frac{\pi}{2}} \sin \theta \sqrt{1-\sin ^{2} \theta} \cdot\frac{2 \sin \theta \cos \theta d \theta}{1+\sin ^{4} \theta} \\
=2 \displaystyle \int_{0}^{\frac{\pi}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/989021",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "26",
"answer_count": 8,
"answer_id": 6
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Show that complex numbers are vertices of equilateral triangle 1)Show if $|z_1|=|z_2|=|z_3|=1$ and $z_1+z_2+z_3=0$ then $z_1,z_2,z_3$ are vertices of equilateral triangle inscribed in a circle of radius.
I thought I can take use from roots of unity here, since $|z_1|=|z_2|=|z_3|=1$ they lie at circle at radius $1$ but ... | Let:
$z_{1} = a + ib \\ z_2 =p+iq \\ z_{3} = x + iy$
Taking the square of modulus:
$a^ 2 +b^ 2 =p^ 2 +q^ 2 =x^ 2 +y^ 2 =1$
Equating the real and the imaginary part of $z_1 + z_2 + z_3$ to 0,
$a + p + x = b + q + y = 0$
$
|z_{1} - z_{2}|^ 2 \\
=(a - p) ^ 2 + (b - q) ^ 2 \\
=2(a^ 2 + p^ 2 )-(a+p)^ 2 + 2(b ^ 2 + q ^ 2) - ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/992262",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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Row space of matrix
Hello! I am working on some differential equations homework and it is saying that my answer for this question is wrong, and i am not sure as to why. First I reduced the matrix A and then I read the first row of the reduced matrix into the given u=[], and the second row into v=[]. When that answer w... | $\begin{equation} \begin{aligned} \pmatrix{ 3 & 3 & 5 & -3 & -2 \\ 2 & 1 & -3 & -1 & 5 \\ 4 & 5 & 13 & -5 & -9} = & \pmatrix{ 3 & 3 & 5 & -3 & -2 \\ 2 & 1 & -3 & -1 & 5 \\ 0 & 3 & 19 & -3 & -19} \\ = & \pmatrix{ 3 & 3 & 5 & -3 & -2 \\ 0 & -1 & -\frac{19}{3} & 1 & \frac{19}{3} \\ 0 & 3 & 19 & -3 & -19} \\ = & \pmatrix{ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/992337",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove that $1+4+7+...+(3n-2) = \frac{n}{2}(3n-1)$ Using induction
prove that $1+4+7+...+(3n-2) = \frac{n}{2}(3n-1) \forall n \in \mathbb{N}$
Attempt:
Let $n =1$ so $3(1)-2 = 1$ and $\frac{1}{2}(3(1)-1)=1$
Assume true at $n=k$ so $3k-2 = \frac{k}{2}(3k-1)$
What do I do next? Here's where I'm stuck:
Let $n=k+1$ So $3(k+... | If you can avoid induction, you can use Gauss's trick:
$S=1+4+7+\cdots+(3n-8)+(3n-5)+(3n-2)$
$S=(3n-2)+(3n-5)+(3n-8)+\cdots+7+4+1$
$2S=(3n-1)+(3n-1)+\cdots+(3n-1)$
$2S=n(3n-1)$
$S = \frac{n}{2}(3n-1)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/992827",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Partial Fractions Decomposition I am failing to understand partial fraction decomposition in cases like the following:
Provide the partial fraction decomposition of the following:
$$\frac{x+2}{(x-4)^3(x^2 + 4x + 16)}$$
I see this and I think of
$$\frac{A}{x-4} + \frac{Bx+C}{(x-4)^2} + \frac{Dx^2 + Ex + F}{(x-4)^3} + \... | You may see the redundancy of the $Bx$ term as follows:
\begin{align}
\frac{Bx + C}{(x - 4)^2}
& = \frac{Bx}{(x - 4)^2} + \frac{C}{(x - 4)^2} \\
& = \frac{B(x - 4)}{(x - 4)^2} + \frac{4B}{(x - 4)^2} + \frac{C}{(x - 4)^2}\\
& = \frac{B}{x - 4} + \frac{4B + C}{(x - 4)^2}.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/994252",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
Why does this method of solution for this system of equations yield an incorrect answer? We are required to solve the following system of equations:
$$x^3 + \frac{1}{3x^4} = 5 \tag1$$
$$x^4 + \frac{1}{3x^3} = 10 \tag2$$
We may multiply $(1)$ by $3x^4$ throughout and $(2)$ by $3x^3$ throughout (as $0$ is not a solution ... | Solutions to (3) and (4) are equivalent to solutions of $15x^4 - 30x^3 = 0$ and one of (3) or (4). But the solution $x = 2$ to $15 x^4 - 30x^3 = 0$ does not satisfy (3) [or equivalently (4)], as $3*2^7 + 1 \neq 15*2^4$.
You shouldn't be surprised that a system of two equations in one unknown may be inconsistent, i.e. ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/994471",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
How integrate $ \iint_{D} (\frac{x^2}{x^2+y^2})dA, \ \ \ \ D: x^2+y^2=a^2 \ \ and \ \ x^2+y^2=b^2, \ \ 0I'm trying to resolve this integral
$$
\iint_{D} (\frac{x^2}{x^2+y^2})dA, \ \ \ \ D: x^2+y^2=a^2 \ \ and \ \ x^2+y^2=b^2, \ \ 0<a<b
$$
I tried with polar coordinates:
$$
x = r\cos{\theta} \\
y = r\sin{\theta} \\
Ja... | This is just annulus with inner radius $a$ and outer radius $b$, so $a \le r \le b$ and $0 \le \theta \le 2\pi$. The integral becomes
$$ \int_0^{2\pi} \int_a^b \frac{r^2 \cos^2 \theta}{r^2} r\,dr\,d\theta
\\= \int_0^{2\pi} \int_a^b r \cos^2 \theta \, dr \, d\theta
\\= \int_0^{2\pi} \frac{1}{2}(b^2 - a^2) \cos^2 \the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/997649",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
integral calculation is wrong. Why? $$\int \sqrt{1-x^2} dx = \int \sqrt{1-\sin^2t} \cdot dt= \int \sqrt {\cos^2 t} \cdot dt= \int \cos t \cdot dt = \sin t +C = x +C$$
The answer is wrong. Why?
| Here's a way to do the integral without trigonometric substitution. All you need to know is the antiderivative $$\int\frac{\mathrm{d}x}{\sqrt{1-x^2}}=\arcsin{x}+\color{grey}{constant},$$ as well as how to integrate by parts.
$$\begin{align}
\int\sqrt{1-x^2}\,\mathrm{d}x
&=\int\frac{1-x^2}{\sqrt{1-x^2}}\,\mathrm{d}x\\
&... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1000037",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Questions about the computation of a limit. By L'Hôpital's rule, it is easy to see that
$$
\lim_{x \to 0} \frac{\tan x - x}{x^2 \sin x} = 1/3.
$$
But if we use the following procedure to compute the limit, we get a wrong answer. The procedure is in the following.
$$
\lim_{x \to 0} \frac{\tan x - x}{x^2 \sin x} =
\li... | What you did:
$${\sin x} \sim x \implies
\frac{\sin x}{\cos x} - x \sim \frac{x}{\cos x} - x
$$
is wrong, because you added equivalents: you can multiply by $1/\cos x$, to get
$$
{\sin x} \sim x \implies \frac{\sin x}{\cos x} \sim \frac x{\cos x}
$$
but you can't fo the manipulation with the sum:
$$
\frac{\sin x}{\co... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1000750",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
How to prove the inequality $ \frac{a}{\sqrt{1+a}}+\frac{b}{\sqrt{1+b}}+\frac{c}{\sqrt{1+c}} \ge \frac{3\sqrt{2}}{2}$ How to prove the inequality $$ \frac{a}{\sqrt{1+a}}+\frac{b}{\sqrt{1+b}}+\frac{c}{\sqrt{1+c}} \ge \frac{3\sqrt{2}}{2}$$
for $a,b,c>0$ and $abc=1$?
I have tried prove $\frac{a}{\sqrt{1+a}}\ge \frac{3a+1}... | Consider the function for positive $x$:
$$f(x) = \frac{x}{\sqrt{1+x}}-\frac1{\sqrt 2}-\frac3{4\sqrt 2}\log x$$
Note that $f(x) \ge 0 \implies f(a)+f(b)+f(c) \ge 0 \implies $ the given inequality. Now
$$f'(x) = \frac{4x^2-3\sqrt2 (x+1)^{3/2}+8x}{8x(x+1)^{3/2}}$$
We need to check the sign of the numerator, $4(x+1)^2-3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1000997",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 7,
"answer_id": 1
} |
What is the value of $\cos\left(\frac{2\pi}{7}\right)$? What is the value of $\cos\left(\frac{2\pi}{7}\right)$ ?
I don't know how to calculate it.
| I suggest you have a look at http://mathworld.wolfram.com/TrigonometryAnglesPi7.html which clearly explains the problem.
As you will see, $\cos\left(\frac{2\pi}{7}\right)$ is the solution of $$8 x^3+4 x^2-4 x-1=0$$ Using Cardano, you will get $$\cos\left(\frac{2\pi}{7}\right)=\frac{1}{6} \left(-1+\frac{7^{2/3}}{\sqrt[3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1001267",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Ellptic\Jacobi theta function and its residue integral The Ellptic\Jacobi theta function is given by
\begin{align}
\theta_1(\tau|z)&=\theta_1(q,y)=-iq^{\frac{1}{8}}y^{\frac{1}{2}}\prod_{k=1}^{\infty}(1-q^k)(1-yq^k)(1-y^{-1}q^{k-1}) \\
&= -i\sum_{n\in \mathbb{Z}}(-1)^n e^{2\pi i z(n+\frac{1}{2})} e^{\pi i \tau(n+\fr... | From Antonio DJC's comment i explicit compute
\begin{align}
&\theta_1(q, y)= -i y^{\frac{1}{2}}q^{\frac{1}{8}} \prod_{n=1}^\infty \left(1-q^n\right)\left(1-y q^n\right) \left(1-y^{-1}q^{n-1} \right) \\
&\phantom{\Theta(q,y)}= -i y^{\frac{1}{2}}q^{\frac{1}{8}}(1-y^{-1}) \prod_{n=1}^\infty \left(1-q^n\right)\left(1-y q... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1004285",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Examples where it is easier to prove more than less Especially (but not only) in the case of induction proofs, it happens that a stronger claim $B$ is easier to prove than the intended claim $A$ (e.g. since the induction hypothesis gives you more information). I am trying to come up with exercises for beginner students... | Here are two examples which might be useful
*
*Example 1: Generalize in order to use techniques from analysis resp. formal operator techniques
Problem: Evaluate
$$\sum_{k=1}^{n}\frac{k^2}{2^k}$$
This can be done by instead evaluating
$$\sum_{k=1}^{n}k^2x^k$$
For $x\neq 1$ we know that
$$S(x)=\sum_{k=1}^{n}x^k=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1007256",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "38",
"answer_count": 17,
"answer_id": 11
} |
If $9\sin\theta+40\cos\theta=41$ then prove that $41\cos\theta=40$. I tried it this way:
$$ 40\cosθ+9\sinθ=41 $$
$$ 9\sinθ=41-40\cos\theta $$
Squaring both the sides:
$$81\sin^2\theta=1681+1600\cos^2\theta-2\cdot 40\cdot 41 \cos\theta$$
$$81-81 \cos^2\theta= 1681+1600\cos^2\theta-3280 \cos\theta$$
$$168... | This has already been mentioned multiple times, but for completeness:
Let $\sin \theta = x, \cos \theta = y$. Then we have to solve:
$$9x+40y=41 \quad \quad x^2 + y^2 = 1$$
Now from the second equation, $(9x)^2 + 81y^2 = 81$, or $(41 - 40y)^2 + 81y^2 = 81$. This gives:
$$(40^2+9^2)y^2 - 2 \cdot 41 \cdot 40y + (41^2-9^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1008487",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 5
} |
What is the limit for the radical $\sqrt{x^2+x+1}-x $? I'm trying to find oblique asmyptotes for the function $\sqrt{x^2+x+1}$ and I manage to caclculate that the coefficient for the asymptote when x approaches infinity is 1.
But when i try to find the m-value for my oblique asymptote by taking the limit of:
$$
\lim_... | $$\lim_{x->\infty}\sqrt{x^2+x+1}-x$$
$$=\lim_{x->\infty}\left(\sqrt{x^2+x+1}-x\right)\cdot\frac{\sqrt{x^2+x+1}+x}{\sqrt{x^2+x+1}+x}$$
$$=\lim_{x->\infty}\frac{1+x}{\sqrt{x^2+x+1}+x}$$
The limit of a sum is the sum of the limits
$$=\lim_{x->\infty}\frac{1}{\sqrt{x^2+x+1}+x} + \lim_{x->\infty}\frac{x}{\sqrt{x^2+x+1}+x}$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1009110",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 2
} |
Examine limit of sequence I have to examine limit of following sequence $\{a_n\}_{ n \ge 1}$ $a_n = \sqrt[n]{\sum_{k=1}^{n}{(2 -\frac{1}{k})^k}}$. We know that $\lim_{n \to \infty} \sqrt[n]{a} = 1$ for $a > 0$ and we know that $\sqrt[n]{\sum_{k=1}^{n}{(2 -\frac{1}{k})^k}} > 0$ but sequence $a_n$ is increasing (I cannot... | I think you can get your way around this one using limited expansion.
Here I use : $ln(1-u) = -u -\frac{u^2}{2} +o(u^2)$ ; and $\exp(u) = 1+ u +o(u)$ , when $ u \rightarrow 0$
Let : $u_k = (2-\frac{1}{k})^k = 2^k*(1-\frac{1}{2k})^k = 2^k*\exp{[k*ln(1-\frac{1}{2k})]} = 2^k*\exp[-\frac{1}{2} -\frac{1}{8k} + o(\frac{1}{k}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1009169",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
} |
Number of real roots of the equation $(6-x)^4+(8-x)^4 = 16$
Find number of real roots of the equation $(6-x)^4+(8-x)^4 = 16$
$\bf{My\; Try::}$ Let $f(x) = (6-x)^4+(8-x)^4\;,$ and we have to find real values of $x$ for
which $f(x) = 16$. Now we will form Different cases.
$\bf{\bullet \; }$ If $x<6$ or $x>8\;,$ Then $... | expanding and factorizing we obtain
$2\, \left( x-6 \right) \left( x-8 \right) \left( {x}^{2}-14\,x+56
\right)
=0$
from here you will get all solutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1010622",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
Evaluate $\lim_{x \to 0} \frac{\sqrt{1 + \tan x} - \sqrt{1 + \sin x}}{x^3}$ What I attempted thus far:
Multiplying by conjugate
$$\lim_{x \to 0} \frac{\sqrt{1 + \tan x} - \sqrt{1 + \sin x}}{x^3} \cdot \frac{\sqrt{1 + \tan x} + \sqrt{1 + \sin x}}{\sqrt{1 + \tan x} + \sqrt{1 + \sin x}} = \lim_{x \to 0} \frac{\tan x - \s... | Hint: You are doing well. Now multiply top and bottom by $\sec x+1$, and note that $\sec^2 x-1=\tan^2 x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1011290",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
If $\sin A+\sin B =a,\cos A+\cos B=b$, find $\cos(A+B),\cos(A-B),\sin(A+B)$ If $\sin A+\sin B =a,\cos A+\cos B=b$,
*
*find $\cos(A+B),\cos(A-B),\sin(A+B)$
*Prove that $\tan A+\tan B= 8ab/((a^2+b^2)^2-4a^2)$
| $$a^2+b^2=2+2\cos(A-B)\implies\cos(A-B)=?$$
$$ab=(\sin A+\sin B)(\cos A+\cos B)=\sin(A+B)+\frac{\sin2A+\sin2B}2$$
$$\implies ab=\sin(A+B)[1+\cos(A-B)]\implies \sin(A+B)=?$$
$$b^2-a^2=\cos2A+\cos2B+2\cos(A+B)=2\cos(A+B)[\cos(A-B)]+1$$
$$\implies\cos(A+B)=?$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1012928",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
prove that $2^n+2^{n-1}+2^{n-2}+8^n-8^{n-2}$ is a multiple of 7 Prove that a number $2^n+2^{n-1}+2^{n-2}+8^n-8^{n-2}$ is a multiple of 7 for every natural $n\ge2$. I am not sure how to start.
| This is relatively easy to solve even without noticing some clever factorization, simply by checking several possible cases.
It suffices to check what are remainder of powers of $2$ modulo $7$.
You can see that:
$2^0 \equiv 1 \pmod 7$
$2^1 \equiv 2 \pmod 7$
$2^2 \equiv 4 \pmod 7$
$2^3 \equiv 1 \pmod 7$
$2^4 \equiv 2 \p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1015460",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Need help in figuring out what I am doing wrong when solving for n.. Here is the expression that I am trying to solve for n:
$$ \frac{4}{16+n} = \frac{10}{16+n} \frac{10}{16+n}$$
I am doing the following:
\begin{align}
\frac{4}{16+n} & = \frac{100}{(16 + n)^2} \\[8pt]
\frac{4}{16+n} & = \frac{100}{16^2 + 32n + n^2} \\[... | There's simple method
$$\frac{4}{16+n} = \frac{10}{16+n} \times \frac{10}{16+n}$$
$n\neq-16$
$${4} = \frac{10\times10}{16+n} $$
$$64+4n = 100 $$
$$36=4n$$
$$n=9$$
Well let's figure out your mstake,
$$\frac{4}{16+n} = \frac{100}{n^2+32n+256} $$
$n\neq-16$
$$n^2+32n+256=25(16+n)$$
$$n^2+7n-144=0$$
$$(n-9)(n+16)$$
since $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1016234",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Calculating limit-help needed (difference of cubes) I'm stuck trying to find this limit:
$$\lim_{x\to \infty} (\sqrt[3]{(x-2)^2(x-1)}-x)$$
Thanks in advance!
| If the problem had a square root instead of a cube root, you probably know how to proceed. For example,
\begin{align*}
\lim_{x\to\infty} \big( \sqrt{(x-2)(x-1)} - x \big) &= \lim_{x\to\infty} \frac{(\sqrt{(x-2)(x-1)} - x)(\sqrt{(x-2)(x-1)} + x)}{\sqrt{(x-2)(x-1)} + x} \\
&= \lim_{x\to\infty} \frac{(x-2)(x-1) - x^2}{\sq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1017011",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Optimization problem - Trapezoid under a parabola recently I've been working on a problem from a textbook about Optimization. The result that I get is $k = 8$, even thought the answer from the textbook is $k = \frac{32}{3}$
The problem follows:
--
The x axis interepts the parabola $12-3x^2$ at the points $A$ and $B$, a... | The trapezoid area is
$$A_{T} = \frac{(B+b) \cdot h}{2} = \frac{\left(4+2 \cdot \sqrt{\frac{12-k}{3}} \right)\cdot k}{2}$$
$$A_{T}' = 0 $$
$$ 2 + \frac{\sqrt{12-k}}{\sqrt{3}} - \frac{k}{\sqrt{3} \cdot 2 \cdot \sqrt{12-k}} = 0$$
$$4 \sqrt{3 \cdot (12-k)} + 24 - 3k = 0$$
$$4 \sqrt{3 \cdot (12-k)} = 3k - 24$$
Squaring bo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1025897",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Why is Implicit Differentiation needed for Derivative of $y = \arcsin (2x+1)$? my function is: $y = \arcsin (2x+1)$ and I want to find its derivative.
My approach was to apply the chain rule:
$$y' = \frac{dg}{du} \frac{du}{dx}$$
with $g = \arcsin(u)$ and $u = 2x+1$.
$$g' = \frac{1}{\sqrt{1-u^2}}.$$
${u}' = 2$.
My so... | $$\frac{2}{\sqrt{1-(1+2x)^2}}=\frac{2}{\sqrt{1-4x^2+
-4x-1}}=\frac{2}{\sqrt{-4x^2-4x}}=\frac{2}{\sqrt{4(-x^2-x)}}=\frac{2}{2\sqrt{-x^2-x}}=\frac{1}{\sqrt{-x^2-x}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1029724",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
How many zeros does $f(x)= 3x^4 + x + 2 $ have?
How many zeros does this function have?
$$f(x)= 3x^4 + x + 2 $$
| For $x\ge -2$, we have $(x+2)\ge 0$ and $3x^4>0$, so $f(x)>0$.
For $x<-2$, we have $3x^4+x=x(3x^3+1)=(-x)(-1-3x^3)>(2)(-1+3\cdot 2^3)=46$, so $f(x)>0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1031259",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
How to evaluate $\int_0^1 (\arctan x)^2 \ln(\frac{1+x^2}{2x^2}) dx$
Evaluate
$$
\int_{0}^{1} \arctan^{2}\left(\, x\,\right)
\ln\left(\, 1 + x^{2} \over 2x^{2}\,\right)\,{\rm d}x
$$
I substituted $x \equiv \tan\left(\,\theta\,\right)$ and got
$$
-\int^{\pi/4}_{0}\theta^{2}\,{\ln\left(\, 2\sin^{2}\left(\,\theta\,... | I dont'have (yet) a complete solution for now.
Let $I=\displaystyle \int_0^1 (\arctan(x))^2 \ln\Big(\dfrac{1+x^2}{2x^2}\Big)dx$
$I=\displaystyle \int_0^1 (\arctan(x))^2\ln(1+x^2)dx-2\int_0^1 (\arctan(x))^2\ln(x)dx-\ln(2)\int_0^1 (\arctan(x))^2dx$
I know how to compute the last one ;)
Let $J=\displaystyle \int_0^{\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1032483",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 5,
"answer_id": 3
} |
Find the domain of $\frac{x}{\sqrt{6x^2+3x+3/4}+x}$. My attempt:
Let's assume that $\sqrt{6x^2+3x+\frac{3}{4}}+x$ $>$ 0$$
\rightarrow\sqrt{6x^2+3x+\frac{3}{4}} > -x
\rightarrow {6x^2+3x+\frac{3}{4}} > x^2\\
\rightarrow {5x^2+3x+\frac{3}{4}} > 0\\$$
If x $<$ 0 then $x^2$ $>$ 0, and ${5x^2+3x+\frac{3}{4}} > 0$... | We have two cases to check.
First, $$6x^2 + 3x + \frac{3}{4} \ge 0$$
It is $\ge 0$ for all $x$.
Next, $$6x^2 + 3x + \frac{3}{4} - x\ne 0$$
$$6x^2 + 2x + \frac{3}{4} \ne 0$$
Solving we get complex values for $x$. Therefore, no real values exist for $x$.
Thus, you are correct. The domain is: $(-\infty, \infty)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1033629",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Evaluation of $\lim_{x\rightarrow \infty}\sqrt{x^2+x+1} -\lfloor \sqrt{x^2+x+1 }\rfloor\;,$ Evaluation of $\displaystyle \lim_{x\rightarrow \infty}\sqrt{x^2+x+1} -\lfloor \sqrt{x^2+x+1 }\rfloor\;,$ where $\lfloor x \rfloor$ represent floor function of $x$.
$\bf{My\; Try}::$
$\bullet\; $If $x\in \mathbb{Z}\;,$ and $x\r... | $\forall a\in(0,1),n\in N^+$,let$x_n$ be the positive zero of :
$$x^2+x+1=(n+a)^2$$
then $x_{n+1}>x_n$,and $x_n>n+a-1$,so$\lim_{n\to+\infty}x_n=+\infty$,but:
$$\sqrt{x_n^2+x_n+1}-[x_n^2+x_n+1]\equiv a$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1036936",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
valid proof of series $\sum \limits_{v=1}^n v$ $$\sum \limits_{v=1}^n v=\frac{n^2+n}{2}$$
please don't downvote if this proof is stupid, it is my first proof, and i am only in grade 5, so i haven't a teacher for any of this 'big sums'
proof:
if we look at $\sum \limits_{v=1}^3 v=1+2+3,\sum \limits_{v=1}^4 v=1+2+3+4,\su... | $\displaystyle\sum_{v=1}^nv=\dfrac{1}{2}\displaystyle\sum_{v=1}^n2v=\dfrac{1}{2}\displaystyle\sum_{v=1}^n\left((v+1)^2-v^2-1\right)=\dfrac{1}{2}\left((n+1)^2-(n+1)\right)-=\dfrac{1}{2}n(n+1)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1037736",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 1
} |
$sup_{x,y\in R}{(\cos{x^2}+\cos{y^2}-\cos{xy})}-\inf{(\cos{x^2}+\cos{y^2}-\cos{(xy)})}=6$
let $x,y\in R$,prove or disprove
$$sup_{x,y\in R}{(\cos{x^2}+\cos{y^2}-\cos{xy})}-\inf_{x,y\in R}{(\cos{x^2}+\cos{y^2}-\cos{(xy)})}=6$$
I think we must show that
$$\cos{x^2}+\cos{y^2}-\cos{xy}\in (-3,3)$$
it is clear
$$|\cos{x... | Let $\displaystyle f(x,y)=\cos(x^2)+\cos(y^2)-\cos(xy)$. It is clear that $-3\leq f(x,y)\leq 3$ for all $x,y$.
a) Let $\displaystyle x_n=\sqrt{2\pi n}$, $\displaystyle y_n=\sqrt{2\pi(n+1)}$. We have $\cos(x_n^2)=\cos(y_n^2)=1$. Put $z_n=x_ny_n$. We have $\displaystyle z_n=2\pi \sqrt{n^2+n}=2\pi n\sqrt{1+\frac{1}{n}}$. ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1038187",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Calculate the probability that the sum of 3 fair dice is at least 13 How would I approach this. I'm more concerned with method than answer.
| Can be done using generating functions which reduces down to polynomial multiplication.
$G(x)=\left(\frac{x}{6}+\frac{x^2}{6}+\frac{x^3}{6}+\frac{x^4}{6}+\frac{x^5}
{6}+\frac{x^6}{6}\right)^3 $
$G(x)$ models throwing 3 fair dice.
When that is expanded.
$\frac{x^3}{216}+\frac{x^4}{72}+\frac{x^5}{36}+\frac{5 \
x^6}{108}+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1042137",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Find the value of $x,y,z$ that minimize $3x^2+2y^2+z^2+4xy+4yz$ under constraint $x^2+y^2+z^2 = 1$ Find the value of $x,y,z$ that minimize $3x^2+2y^2+z^2+4xy+4yz$ under constraint $x^2+y^2+z^2 = 1 $
i am learning algebra and think a way solve this problem by matrix
| by Lagrange Multiplier Method we get $$3x^2+2y^2+z^2+4xy+4yz\geq -1$$ and the equal sign holds if $$x=-\frac{1}{3},y=\frac{2}{3},z=-\frac{2}{3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1044832",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Is it correct? $\tan x + \cot x \ge 2$ proof The question is to prove $\tan x + \cot x \ge 2$ when $x$ is an acute angel.
This is what I did
$$\begin{align}
\tan x + \cot x &\ge 2\\
\frac{1}{\sin x \cos x} &\ge 2\\
\left(\frac{1}{\sin x \cos x}\right) - 2 &\ge 0\\
\left(\frac{1 - 2\sin x \cos x}{\sin x \cos x}\right) &... | Let $ y = \tan x + \cot x $.
Then $ y' = \sec^{2} x - \csc^{2} x $.
So when $ y' = 0 $, $ x = \pi/4 $, for $x$ acute.
You can prove this is a minimum by considering $ y' $ for $ x = \pi/6 $ and $ x = \pi/3 $.
Thus $ y_{min} = \tan \pi/4 + \cot \pi/4 = 1 + 1 = 2.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1046560",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 8,
"answer_id": 5
} |
Proving that $x=\arccos(\sqrt{\sin\theta})$ is $\sin(x+iy)=\cos\theta+i\sin\theta$ Given:
$$\sin(x+iy)=\cos\theta+i\sin\theta$$
To prove:
$$x=\arccos (\sqrt{\sin\theta})$$
How I tried:
$$\begin{align*}
\sin x \cosh y &= \cos\theta \\
\cos x \sinh y &= \sin\theta
\end{align*}$$
Then tried to use logarithm of hyperbolic ... | given:
$sin(x+iy)=e^{i\theta}$
$sin(x)*cosh(y)+icos(x)*sinh(y)=cos(\theta)+isin(\theta)$
comparing the the real and imaginary part
$\implies sin(x)*cosh(y)=cos(\theta)$
&
$\implies cos(x)*sinh(y)=sin(\theta)$
consider hyperbolic form of $sin^2y +cos^2y=1 $
$\cosh^2y-sinh^2y=1$
put values of $sinh(y) \& cosh(y)$ in a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1046792",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
showing that cos(y)-cos(x)=2*sin((x+y)/2)*sin((x-y)/2) how can one show that $\cos(y)-cos(x)=2sin(\frac{x-y}{2})sin(\frac{x+y}{2})$ by just using trigonometrical equalities like additional theorems?
I was trying to show it by rewriting $cos(y)=cos(0.5y+0.5y)$ and then using the additional theorems, but somehow I do not... | I think the easier way to do it is by working with the right hand side, instead of the left hand side, making use of the addition formula for sine:
$$\begin{eqnarray}2\sin\left(\frac{x-y}{2}\right)\sin\left(\frac{x+y}{2}\right) &=& 2\left(\sin\left(\frac{x}{2}\right)\cos\left(\frac{-y}{2}\right)+\cos\left(\frac{x}{2}\r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1047400",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Compute the resultant of the polynomials $f(x)=x^2y+3xy-1$ and $g(x)=6x^2+y^2-4$ Consider these two polynomials:
$$f=x^2y+3xy-1$$
$$g=6x^2+y^2-4$$
I need to compute their resultant, denoted in my textbook as $h=Res(f,g,x)$.
Here's where I need help: setting up the Sylvester Matrix. I could be missing something obviou... | Write your polynomials as
$$
yx^2+ (3xy)-1
$$
and
$$
-6x^2+ (y^2-4)
$$
As polynomials in $x$, these have degree two, so your matrix should be a $4 \times 4$ matrix.
$$
\begin{pmatrix}
y & 3y & -1 & 0 \\
0 & y & 3y & -1 \\
-6 & 0 & y^2-4 & 0 \\
0 & -6 & 0 & y^2-4
\end{pmatrix}
$$
Taking the determinant gives
$$
y^{6}-6... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1048049",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Principle of inclusion and exclusion How many integer solutions can we have for the equation $x_1 + x_2 + x_3 = 18$, if $0 \leq x_1 \leq 6$, $4 \leq x_2 \leq 9$ and $7 \leq x_3 \leq 14$?
Using the Principle of inclusion and Exclusion: I have found
$S_0$ = $\binom{9}{7} $
However, I am stuck as to how to go about cal... | This is the stars and bars problem. You can set up an equivalent question. Subtract out $4$ from both sides so that $0 \leq x_{2} \leq 5$. Similarly, subtract out $7$ so $0 \leq x_{3} \leq 7$. This leaves us with $x_{1} + x_{2} + x_{3} = 7$.
We can use a generating function to give us our inclusion-exclusion formula. N... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1049058",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
variation of parameters for a Cauchy-Euler problem: where am I wrong? The problem is this one
$$x^2y'' - xy' -3y = 5x^4.$$
Then the complementary solution I get is
$$c_1x + c_2x\ln x.$$
Then after that the particular solution is
$$\ln x e^{4x} \left(\frac{x}{4} - \frac{1}{16}\right) - \frac{e^{4x}}{16} - \frac{1}{16} ... | A.N.Other method is
$$
x^2y'' - xy' - 3y = 5x^4
$$
we can realise thath
$$
\dfrac{d}{dx}x^2y' = x^2y'' + 2xy'\implies x^2y'' = \dfrac{d}{dx}x^2y' - 2xy'
$$
so we can insert into the original equation to yeild
$$
\dfrac{d}{dx}x^2y' - 2xy' - xy' -3y = \dfrac{d}{dx}\left(x^2y'\right) - 3(xy' + y) = 5x^4\\
\dfrac{d}{dx}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1050424",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Evaluating $\int_0^1\frac{x^{-a}-x^{a}}{1-x}\,\mathrm dx$ How to evaluate following integral
$$\int_0^1\frac{x^{-a}-x^{a}}{1-x}\,\mathrm dx$$
I tried the feynman way
$$\begin{align}
I'(a)&=\int_0^1\frac{-x^{-a}\ln x-x^{a}\ln x}{1-x}\,\mathrm dx\\
&= \int_0^1\ln x\left(\frac{-x^{-a}-x^{a}}{1-x}\right)\,\mathrm dx\\
... | We can do it without using The Feynman Way
$$\begin{align}
\int_0^1 \frac{x^{-a}-x^a}{1-x}\,\mathrm dx
&=\int_0^1 \left[\frac{x^{-a}}{1-x}-\frac{-x^a}{1-x}\right]\,\mathrm dx\tag{1}\\
&=\int_0^1\sum_{k=1}^\infty\left(x^{k-a-1}-x^{k+a-1}\right)\,\mathrm dx\tag{2}\\
&=\sum_{k=1}^\infty\int_0^1\left(x^{k-a-1}-x^{k+a-1}\ri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1051134",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 4
} |
Another infinite series involving Gamma function It's not hard to see that Mathematica expresses this series in terms of hypergeometric function,
but how about finding a way of expressing the result in terms of elementary functions only?
Is that possible? How would you recommend me to proceed?
$$\sum_{n=1}^{\infty} (-1... | $$\begin{eqnarray*}S=\sum_{n=1}^{+\infty}(-1)^{n+1}\frac{n\,\Gamma(2n-1)}{4^n\,\Gamma(2n+1/2)}&=&\sum_{n=1}^{+\infty}\frac{(-1)^{n+1}n}{4^n\Gamma(3/2)}B(2n-1,3/2)\\&=&\frac{2}{\sqrt{\pi}}\sum_{n=1}^{+\infty}\frac{(-1)^{n+1}n}{4^n}\int_{0}^{1}x^{2n-2}(1-x)^{1/2}\,dx\\&=&\frac{1}{2\sqrt{\pi}}\int_{0}^{1}\sum_{n=0}^{+\inf... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1052829",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Understanding the limit $\lim_{t \to \infty} \int_0^\infty \frac{t}{t^2 + x} \sin(1/x)dx$ Prove :
$$\lim_{t\to +\infty}\int_{0}^{\infty}\frac{t}{t^2 +x} \sin\frac{1}{x} dx=0$$
Maybe dominated convergence theorem? Who can give a proof?
Thanks!
| We can assume $t>1$ - we just throw out the beginning...
Since for $y<1$, $\sin y<y$, thus for $x>1$
$$\sin\frac{1}{x}<\frac{1}{x}$$
By AM-GM:
$$|\frac{t}{t^2+x}|<\frac{1}{t+\frac{x}{t}}<\frac{1}{2\sqrt{x}}$$
Also note that for $x\leq1$:
$$\frac{1}{1+x}-\frac{t}{t^2+x}=\frac{t^2-t+t(1-x)}{(1+x)(t^2+x)}>0$$
So we have ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1053748",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Using the half angle formula Find $\sin\frac { x }{ 2 }, \cos\frac { x }{ 2 }$, and $\tan\frac { x }{ 2 }$ from $\sin x=\frac { 3 }{ 5 } ,\quad 0°<x<90°$
What I did:
$$\sin\frac { x }{ 2 } =\sqrt { \frac { 1-\frac { 4 }{ 5 } }{ 2 } } =\sqrt { 10 } $$
$$\cos\frac { x }{ 2 } =\sqrt { \frac { 1+\frac { 4 }{ 5 } }{ 2 } ... | As $0<\dfrac x2<45^\circ\implies$ all the trigonometric ratios $>0$
$$\sin^2\frac x2=\frac{1-\cos x}2=\cdots=\frac1{10}\implies\sin\frac x2=+\frac1{\sqrt{10}}=\frac{\sqrt{10}}{10}$$
Similarly for $\cos\dfrac x2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1054110",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Solving $\sin\theta -1=\cos\theta $
Solve$$\sin\theta -1=\cos\theta $$
Steps I took to solve this:
$$\sin^{ 2 }\theta -2\sin\theta +1=1-\sin^2\theta $$
$$2\sin^{ 2 }\theta -2\sin\theta =0$$
$$(2\sin\theta )(\sin\theta -1)=0$$
$$2\sin\theta =0, \sin\theta -1=0$$
$$\quad \sin\theta =0, \sin\theta =1$$
$$\theta =0+\pi ... | $$\begin{align}
\sin(\theta)-\cos(\theta)&=1\\
\sin(\theta)\frac{1}{\sqrt{2}}-\cos(\theta)\frac{1}{\sqrt{2}}&=\frac{1}{\sqrt{2}}\\
\sin(\theta)\sin(\pi/4)-\cos(\theta)\cos(\pi/4)&=\frac{1}{\sqrt{2}}\\
-\cos(\theta+\pi/4)&=\frac{1}{\sqrt{2}}\\
\cos(\theta+\pi/4)&=-{\frac{1}{\sqrt{2}}}\\
\theta+\pi/4&=(2k+1)\pi\pm\frac{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1057685",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 2
} |
evaluation of the integral of a certain logarithm I come across the following integral in my work
$$\int_a^\infty \log\left(\frac{x^2-1}{x^2+1}\right)\textrm{d}x,$$
with $a>1$.
Does this integral converge ? what is its value depending on $a$ ?
| We have:
$$ I = \int_{\frac{a^2-1}{a^2+1}}^{1}\frac{\log x}{(1-x)^{3/2}(1+x)^{1/2}}\,dx $$
where the integrand function is integrable over $(0,1)$. Moreover:
$$ 0 > I > \int_{0}^{1}\frac{\log x}{(1-x)^{3/2}(1+x)^{1/2}} = -\frac{\pi}{2}-\log 2.$$
Integration by parts gives:
$$ I = x\,\left.\log\frac{x^2-1}{x^2+1}\right|... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1059759",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Vector analysis - Curl of vector
How to prove it? I have tried several times to solve it, but I still get stuck everytime.
| First Method (long)
Observe that $\operatorname{curl}(\pmb{a}\times\pmb{b})=\nabla\times (\pmb{a}\times\pmb{b})$ where $\nabla=\pmb{i}\frac{\partial}{\partial x}+\pmb{j}\frac{\partial}{\partial y}+\pmb{k}\frac{\partial}{\partial z}$.
Recalling that $\pmb{a}\times\pmb{b}=\left|\matrix{\pmb i & \pmb j & \pmb k\\
a_x & a_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1060622",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Extremely difficult log integral, real methods only $$\int_{0}^{1}\frac{x^2 + x\log(1-x)- \log(1-x) - x}{(1-x)x^2} dx$$
I tried this:
$$M_1 = \int_{0}^{1} \frac{1}{1-x} \cdot \left(\frac{x^2 + x\log(1-x) - \log(1-x) - x)}{x^2}\right) dx$$
$$M_1 = \int_{0}^{1} \frac{x^2 + x\log(1-x) - \log(1-x) - x}{(1-x)x^2} dx$$
$$M_1... | You can write the numerator as :
$$
x^2 +x\log(1-x) -\log(1-x)-x = (1 - \log(1-x))(1-x) +(x-1)(x+1)
$$
so that it remains to evaluate the integral:
$$
\int_0^1 \frac{1-\log(1-x)-1-x}{x^2} \mathrm dx = - \int_0^1 \frac{\log(1-x)+x}{x^2} \mathrm dx
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1060937",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Evaluation of $\int_0^{2\pi} \frac{1}{1+8\cos^2(\theta)}d\theta$ with Cauchy's residue Theorem I have to proof
$$\int_0^{2\pi} \frac{1}{1+8\cos^2(\theta)}d\theta = \frac{2\pi}{3}$$
with Cauchy's residue Theorem. I have showed it, but in my solution, there comes $-\frac{2\pi}{3}$. I Show you my solution:
I know that $... | It is better to notice that, by replacing $\theta$ with $\arctan t$:
$$ I = 4\int_{0}^{\pi/2}\frac{d\theta}{1+8\cos^2\theta} = 4\int_{0}^{+\infty}\frac{dt}{(1+t^2)(1+\frac{8}{1+t^2})}=2\int_{-\infty}^{+\infty}\frac{dt}{9+t^2}$$
and the last integral is very easy to compute with the residue theorem, or without:
$$2\int_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1063983",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
prove that $\sqrt{2} \sin10^\circ+ \sqrt{3} \cos35^\circ= \sin55^\circ+ 2\cos65^\circ$ Question:
Prove that: $\sqrt{2} \sin10^\circ + \sqrt{3} \cos35^\circ = \sin55^\circ + 2\cos65^\circ$
My Efforts:
$$2[\frac{1}{\sqrt{2}}\sin10] + 2[\frac{\sqrt{3}}{2}\cos35]$$
$$= 2[\cos45 \sin10] + 2[\sin60 \cos35]$$
| As Deepak wrote it's easier to start with the right hand side, but you could go ahead with your approach too, as shown by Arian.
\begin{equation*}
\sqrt{2}\sin 10{{}^\circ}
+\sqrt{3}\cos 35{{}^\circ}
=\sin 55{{}^\circ}+2\cos 65{{}^\circ}\tag{1}
\end{equation*}
Here is a variant that uses the cosine addition formula f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1067644",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Find maximum without calculus Let $f:(0,1]\rightarrow\mathbb{R}$ with $f(x)=2x(1+\sqrt{1-x^2})$. Is it possible to find the maximum of this function without calculus? Possibility through some series of inequalities?
| $(f(x))^2 = 4x^2\left(1+\sqrt{1-x^2}\right)^2$. Let $y = \sqrt{1-x^2} \Rightarrow x^2 = 1 - y^2 \Rightarrow (f(x))^2 = 4(1-y^2)(1+y)^2 = 4(1-y)(1+y)^3$.
Apply AM-GM inequality we have:
$2 = (1-y) + \dfrac{1+y}{3} + \dfrac{1+y}{3} + \dfrac{1+y}{3} \geq 4\sqrt[4]{\dfrac{(1-y)(1+y)^3}{27}} \Rightarrow \dfrac{16\times 27}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1069438",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 0
} |
Prove that $c_n = \frac1n \bigl(\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + \cdots + \frac{1}{\sqrt{n}} \bigr)$ converges I want to show that $c_n$ converges to a value $L$ where:
$$c_n = \frac{\large \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + \cdots + \frac{1}{\sqrt{n}}}{n}$$
First, it's obvious that $c_n > 0$.
I was abl... | We have $c_n>0$ and
$$\begin{align}c_{n+1}-c_n&=\frac{1}{n(n+1)}\left[\frac{n}{\sqrt{n+1}}-\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}-\cdots-\frac{1}{\sqrt{n}}\right]\\
&<\frac{1}{n(n+1)}\left[\frac{n}{\sqrt{n+1}}-\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n}}-\cdots-\frac{1}{\sqrt{n}}\right]\\
&=\frac{1}{n(n+1)}\left[\frac{n}{\sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1070575",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 6,
"answer_id": 1
} |
Compute the area of a parallelogram defined by a particular construction I got stuck with this mathematical task. Can someone help me how to solve this problem?
I need to find the F(area) value. It is kind of a thinking task
Context
The problem is extracted from a German book "Paul Eigenmann, Geometrische Wiederholung... |
If you don't want to rely on any geometric knowledge, or at least almost on none, then you can attack such a problem using coordinates.
$$
A=\begin{pmatrix}0\\0\end{pmatrix}\qquad
B=\begin{pmatrix}50\\0\end{pmatrix}\qquad
C=\begin{pmatrix}57\\h\end{pmatrix}\qquad
D=\begin{pmatrix}7\\h\end{pmatrix}
$$
The circular arcs... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1070909",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How to prove $ \lim_{n\to\infty}\sum_{k=1}^{n}\frac{n+k}{n^2+k}=\frac{3}{2}$? How to prove that
$\displaystyle \lim_{n\longrightarrow\infty}\sum_{k=1}^{n}\frac{n+k}{n^2+k}=\frac{3}{2}$?
I suppose some bounds are nedded, but the ones I have found are not sharp enough (changing $k$ for $1$ or $n$ leads to the limit being... | Rewrite the expression as $\displaystyle\sum_{k=1}^{n}\dfrac{n+k}{n^2+k} = \dfrac{1}{n}\sum_{k=1}^{n}\dfrac{1+\frac{k}{n}}{1+\frac{k}{n^2}}$.
For all $1 \le k \le n$ we have $\dfrac{n}{n+1}\left(1 + \dfrac{k}{n}\right) = \dfrac{1 + \frac{k}{n}}{1+\frac{1}{n}} \le \dfrac{1+\frac{k}{n}}{1+\frac{k}{n^2}} \le \dfrac{1 + \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1075305",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
3D coordinates of circle center given three point on the circle. Given the three coordinates $(x_1, y_1, z_1)$, $(x_2, y_2, z_2)$, $(x_3, y_3, z_3)$ defining a circle in 3D space, how to find the coordinates of the center of the circle $(x_0, y_0, z_0)$?
| I think using a projection into 2D might be the easiest way to actually calculate. If the points are $A, B, C$ then find
$$\begin{align}
\mathbf{u_1} & = B-A \\
\mathbf{w_1} &= (C-A) \times \mathbf{u_1} \\
\mathbf{u} & = \mathbf{u_1} / | \mathbf{u_1} | \\
\mathbf{w} & = \mathbf{w_1} / | \mathbf{w_1} | \\
\mathbf{v} & ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1076177",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 11,
"answer_id": 1
} |
Prove without induction $2^n \mid (b+\sqrt{b^2-4c})^n + (b-\sqrt{b^2-4c})^n $
Prove $2^n \mid (b+\sqrt{b^2-4c})^n + (b-\sqrt{b^2-4c})^n $ for all $n\ge 1$ and $b,c$ are integers.
Is it possible to prove this without induction?
| The sequence given by:
$$ a_n = \left(b+\sqrt{b^2-4c}\right)^n + \left(b-\sqrt{b^2-4c}\right)^n $$
satisfies the recurrence relation:
$$ a_{n+2} = \color{red}{2}b\cdot a_{n+1} - \color{red}{4}c\cdot a_{n}.$$
Since $\nu_2(a_n)\geq n$ holds for $n=0$ and $n=1$, it holds for every $n$ by the previous relation.
Ok, this is... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1079369",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 1
} |
Finding $ \lim_{n \rightarrow \infty} \prod_{j=1}^n \frac{2j-1}{2j}$ Finding
$$\lim_{n \rightarrow \infty} \prod_{j=1}^n \frac{2j-1}{2j}$$
Suspecting that the limit is 0, but how do I show this? Was able to get an upper bound of $ \frac{1}{2\sqrt{e}} $ easily but it's not useful. Apparently this is a hard problem.
| You may write
$$
\begin{align}
\frac{1\cdot 3\cdot 5 \cdots(2n-1)}{2\cdot 4\cdot 6\cdots2n} &=\frac{1\cdot \color{blue}2\cdot 3\cdot \color{blue}4 \cdot 5\cdot\color{blue}6\cdots(2n-1)\cdot \color{blue}{2n}}{(2\cdot 4\cdot 6\cdots (2n))^\color{blue}2}\\
&=\frac{(2n)!}{(2^{n} \cdot 1\cdot 2\cdot 3 \cdot 4 \cdots n)^2}\\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1080437",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
How to express the equations as the Square root Like? $$2\sin \frac{\pi}{16}= \sqrt{2-\sqrt{2+\sqrt{2}}}$$
What law is need to be applied here? Do I have to make the $\frac{\pi}{16}$ in a form that will be give us $\sqrt{2}$ like sin 45 degree?
| Since, $\sin{\dfrac{\theta}{2}}=\sqrt{\dfrac{1-\cos\theta}{2}}$ and $\cos{\dfrac{\theta}
{2}}=\sqrt{\dfrac{1+\cos\theta}{2}}$
$\sin{\dfrac{\theta}{4}}=\sqrt{\dfrac{1-\cos{\dfrac{\theta}{2}}}{2}}$
$\sin{\dfrac{\theta}{4}}=\sqrt{\dfrac{1-\sqrt{\dfrac{1+\cos{\theta}}{2}}}{2}}$
Putting, $\theta=\dfrac{\pi}{4}$
$\sin{\dfrac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1081294",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Is there a lower bound for the following rational expression, in terms of $b^2$? In the following, $b$ and $r$ are both positive integers.
Is there a lower bound for the following rational expression, in terms of $b^2$?
$$L = \frac{b^2\left(2^{r+1} b^2(2^r-1) + 3\cdot 2^r - 2\right)}{\left(2^{r+1} - 1\right)(2^r b^2 + ... | $$\frac{b^2\left(2^{r+1} b^2(2^r-1) + 3\cdot 2^r - 2\right)}{\left(2^{r+1} - 1\right)(2^r b^2 + 1)}=\frac{2^{r+1}(2^r-1)b^4 + (3\cdot 2^r - 2)b^2}{\left(2^{r+1} - 1\right)2^r(b^2 + 2^{-r})}=\frac{b^4 + \frac{3\cdot 2^r - 2}{2^{r+1}(2^r-1)}b^2}{\left(1 +\frac{2^r}{2^r-1}\right)(\frac{1}{2}b^2 + 2^{-r-1})}\geq b^2\frac{b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1087127",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Let $C$ be the curve of intersection of the plane $x+y-z=0$ with ellipsoid $\frac{x^2}4+\frac{y^2}5+\frac{z^2}{25}=1$. Let $C$ be the curve of intersection of the plane $x+y-z=0$ and the ellipsoid $$\frac{x^2}4+\frac{y^2}5+\frac{z^2}{25}=1$$ Find the points on $C$ which are farthest and nearest from the origin
When dea... | You can simplify the problem by using $z=x+y$.
Then you minimize
$$x^2+y^2+(x+y)^2+\lambda\left(\frac{x^2}4+\frac{y^2}5+\frac{(x+y)^2}{25}-1\right).$$
by
$$\begin{cases}x+x+y+\lambda\left(\dfrac{x}4+\dfrac{x+y}{25}\right)=0,
\\y+x+y+\lambda\left(\dfrac{y}5+\dfrac{x+y}{25}\right)=0.\end{cases}$$
Eliminating $\lambda$, y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1087814",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Can one always map a given triangle into a triangle with chosen angles by means of a parallel projection? This is something that seems to be true from experience by playing with shadows from the sun:
If one cuts a paper triangle, he can turn it in a way to make its shadow be a triangle of any given angles (of course, n... | We'll take the special case of projection along the $z$-axis to a target triangle $\triangle A^\prime B^\prime C^\prime$ inscribed in the unit circle. Without loss of generality, we can take the triangle $\triangle ABC$ making the projection to be
$$A = (1,0,0) \qquad B = (\cos2C^\prime,\sin2C^\prime, p) \qquad C = (\c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1088066",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 4,
"answer_id": 3
} |
Prove that $\int_{-\infty}^{+\infty} \frac{1}{x^4+x^2+1}dx = \frac{\pi}{\sqrt{3}}$ Good evening everyone,
how can I prove that
$$\int_{-\infty}^{+\infty} \frac{1}{x^4+x^2+1}dx = \frac{\pi}{\sqrt{3}}\;?$$
Well, I know that $\displaystyle\frac{1}{x^4+x^2+1} $ is an even function and the interval $(-\infty,+\infty)$ is sy... | \begin{align}
\int_{-\infty}^{\infty} \frac{1}{x^4+x^2+1}\,\mathrm dx&=2\int_{0}^{\infty} \frac{1}{x^4+x^2+1}\,\mathrm dx\\[7pt]
&=2\int_{0}^{\infty} \frac{1}{\left(x-\frac{1}{x}\right)^2+3}\cdot\frac{\mathrm dx}{x^2}\\[7pt]
&=2\int_{0}^{\infty} \frac{1}{\left(y-\frac{1}{y}\right)^2+3}\,\mathrm dy\tag1\\[7pt]
&=\int_{-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1090056",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 2
} |
limit of recursive function I've been struggling with this for a while now and can't seem to get it right.
Let $f(x)= \frac{x^2+3}{2x}$ from $0 $ to $\infty $. Show that $0\lt x \lt \sqrt{3}$ implies $\sqrt{3} \lt f(x)$
and show if we define $x_{n+1} = f(x_n)$ then
$$ \lim_{n\to\infty}x_n =\sqrt{3}$$ when $x_0 \gt 0$... | Note that $f(x)-\sqrt{3} = {(x-\sqrt{3})^2 \over 2x} \ge 0$. So we have $x_k \ge \sqrt{3}$ for $k \ge 1$.
Since $f(x)-\sqrt{3} = {x-\sqrt{3} \over 2x} (x-\sqrt{3})$, this suggests that we look at $\phi(x) = {x-\sqrt{3} \over 2x}$. It is straightforward to show that $\phi$ is strictly increasing for $x >0$, and $\lim_{x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1090998",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Show that $(\ln a)^k \neq k \ln a $ I have a question that I am not sure how to answer:
Show that $(\ln a)^k \neq k \ln a $
| I will show that
if $k > 1$
then $a \le e$.
In other words,
if $a > e$
there is no $k \ge 1$
that satisfies this.
(This is just playing around
with algebra and
elementary calculus.)
If
$(\log a)^k=k\log a$,
then
$(\log a)^{k-1}=k$
or
$\log a
=k^{1/(k-1)}
$.
Therefore,
for any given $k$,
there is at most one $a$.
Let
$f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1092558",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Proving of $\cos (\frac{2}{3})>\frac{\pi }{4}$ How can I prove that $$\cos \left(\frac{2}{3}\right)>\frac{\pi }{4}$$
| Use Carlson inequality
$$\arccos{x}>\dfrac{6\sqrt{1-x}}{2\sqrt{2}+\sqrt{1+x}},0\le x<1$$
then let
$x=\frac{\pi}{4}$,then we have
$$\arccos{\dfrac{\pi}{4}}>\dfrac{6\sqrt{1-\dfrac{\pi}{4}}}{2\sqrt{2}+\sqrt{1+\frac{\pi}{4}}}\approx 0.66741060\cdots>\dfrac{2}{3}$$
see wolf
so
$$\cos{\dfrac{2}{3}}>\dfrac{\pi}{4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1095125",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 1
} |
How to solve for $y$: $x^2-3xy+y^2-x+4=0$ Please, I want to know the method to arrive at the solution $y=\dfrac{-3x \pm \sqrt{5x^2+4x-16}}{2}$ , step by step.
Thank you in advance.
| $$\text{Hint: }$$
$$
\\x^2 - 3xy + y^2 - x + 4 = 0 \\ \Updownarrow \\ y^2 + (-3x)y +(x^2 -x+4) = 0
$$
$$$$
$$\text{Now solve the quadratic equation with: }$$
$$\begin{align}
a &= 1 \\ b &= -3x \\ c &= (x^2 - x + 4)
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1098017",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Nasty Limit of sum at infinity $$\lim_{n \to \infty} \left (\sum_{i=1}^n \frac{a\left[\left(\frac{b}{a}\right)^{\frac{i}{n}}-\left(\frac{b}{a}\right)^{\frac{i-1}{n}}\right]}{a\left(\frac{b}{a}\right)^{\frac{i-1}{n}}}\right) $$
How would I even begin to approach this??
Edit:
So the hints helped... I ended up with $\spac... | $$\begin{align}
&\lim_{n \to \infty} \left( \sum_{i=1}^n \frac{a \left[ \left(\frac{b}{a}\right)^{\frac{i}{n}} - \left(\frac{b}{a}\right)^{\frac{i-1}{n}}\right]}{a\left(\frac{b}{a}\right)^{\frac{i-1}{n}}}\right) \\ =
&\lim_{n \to \infty} \left( \sum_{i=1}^n \frac{ \left(\frac{b}{a}\right)^{\frac{i}{n}} - \left(\frac{b}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1098995",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Need hint for $\lim_{x\to 0} \frac{(x+1)^\frac{1}{x}-e}{x}$ Please give me some hints or solution for this limit. I had it on my exam and I'm curious how to solve it.
$$
\lim_{x\to 0} \frac{(x+1)^\frac{1}{x}-e}{x}
$$
| Easiest way is by series expansion:
$$\frac{1}{x}\log(1+x) = \frac{1}{x} \left( x - \frac{x^2}{2} + O(x^3) \right) =
\left( 1 - \frac{x}{2} + O(x^2) \right)
$$
$$
(x+1)^{1/x} = e^{\frac{1}{x}\log(1+x) } = e^{\left( 1 - \frac{x}{2} + O(x^2) \right)}
= \sum_{n=0}^\infty \frac{\left( 1 - \frac{x}{2} + O(x^2) \right)^n}{n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1101831",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Integer root of a quadratic Determine the sum of all (distinct) positive integers $ n$ , such that for some integer $a$,
$$ n^2 -an + 6a = 0. $$
| The equation is equivalent to
$$
a(n-6) = n^2.
$$
So we are looking for all $n$ such that $n-6$ divides $n^2$.
Polynomial long division of $n^2$ by $n-6$ gives $n^2 = (n-6)(n+6) + 36$, so equivalently, $n-6$ divides $36$.
Enumeration of the divisors of $36$ yields
$$
n-6\in\{\pm 1,\pm 2,\pm 3,\pm 4,\pm 6,\pm 9,\pm 12,\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1102382",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Pythagorean triples So I am given that $65 = 1^2 + 8^2 = 7^2 + 4^2$ , how can I use this observation to find two Pythagorean triangles with hypotenuse of 65.
I know that I need to find integers $a$ and $b$ such that $a^2 + b^2 = 65^2$, but I don't understand how to derive them from that observation.
Here is my attempt.... | If we solve $C=m^2+n^2$ for $n$, we get $n=\sqrt{C-m^2}$. We can then try values of $m$ to see which one(s) yield a positive integer for $n$, where
$$\text{To ensure }n<m:\biggl\lceil\sqrt{\frac{C}{2}}\space\space\biggr\rceil \le m\le\bigl\lfloor\sqrt{C}\bigr\rfloor:\text{ to ensure }n\in\mathbb{N}$$
$$\text{In the exa... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1106333",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
} |
Quadric and tangents planes Let $Q$ be the quadratic $x^2 + 4xy - 2y^2 + 6z^2 + 2y +2z = 0$
*
*Prove that $Q$ is a cone and find its vertex.
*Write the tangent plane $A$ to the cone in $(0,0,0)$ and say which kind of conic is the intersection of $Q$ with $A$.
*Find an ellipse and a parabola on the cone.
Am a bit... | *
*With $A=\begin{pmatrix}1&2&0\\2&-2&0\\0&0&6\end{pmatrix}$ and $K=\begin{pmatrix}0&2&2\end{pmatrix}$, the equation is
${\bf x}^TA{\bf x}+K{\bf x}=0$. $A$ has eigenvalues $2,-3,6$ and the eigenvector matrix is $P=\begin{pmatrix}\frac{2\sqrt{5}}{5}&-\frac{\sqrt{5}}{5}&0\\\frac{\sqrt{5}}{5}&\frac{2\sqrt{5}}{5}&0\\0&0&... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1106989",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Hypergeometric 2F1 with negative c I've got this hypergeometric series
$_2F_1 \left[ \begin{array}{ll}
a &-n \\
-a-n+1 &
\end{array} ; 1\right]$
where $a,n>0$ and $a,n\in \mathbb{N}$
The problem is that $-a-n+1$ is negative in this case. So when I try to use Gauss's identity
$_2F_1 \left[ \begin{array}{ll}
a & b \\
... | Using this identity to express the hypergeometric function as a Gegenbauer function, and this identity which gives the value of the Gegenbauer function evaluated at $1$, the hypergeometric function in question may then be expressed as a ratio of gamma functions whose arguments are each positive integers. These can be r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1107317",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Mathematical induction involving inequalities and congruences I have the following two problems:
"Prove each of the following statements by induction for all positive integers $n$:"
*
*$2\cdot7^n \equiv 2^n\cdot(2+5n) \bmod 25 \quad$ <-- I have been going at this question for a couple of hours and can't seem to come... | $\underline{\text{Problem 1:}}$ For each positive integer $n$, let $M(n)$ be the statement that
$$
2\cdot 7^n \equiv 2^n\cdot(2+5n)\pmod{25}.
$$
Base step: $M(1)$ says that $14\equiv 14 \pmod{25}$, which is true.
Inductive step: Fix $k\geq 1$ and assume that
$$
M(k):\; 2\cdot 7^k \equiv 2^k\cdot(2+5k)\pmod{25}
$$
holds... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1107388",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
How to integrate $\int \frac{dx}{\sqrt{ax^2-b}}$ So my problem is to integrate
$$\int \frac{dx}{\sqrt{ax^2-b}},$$
where $a,b$ are positive constants. What rule should I use here? Should substitution be used or trigonometric integrals?
The solution should be:
$$\frac{\log\left(\sqrt{a}\sqrt{ax^2-b}+ax\right)}{\sqrt{a}}... | The question says "Should substitution be used or trigonometric integrals?". But a trigonometric substitution is a substitution, not an alternative to substitution.
$$
\int \frac{dx}{\sqrt{ax^2-b}} = \int\frac{dx/\sqrt{b}}{\sqrt{\frac a b x^2 - 1}} = \int \frac{\sec\theta\tan\theta\,d\theta/\sqrt{a}}{\sqrt{\sec^2\thet... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1108404",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
If $x^2+x+1=0 $, then find the value of $(x^3+1/x^3)^3$?
If $x^2+x+1=0 $, then find the value of $(x^3+1/x^3)^3$
This thing doesn't make sense how should I use first identity to find the second one.
| $x^3-1 = (x-1)(x^2+x+1) = 0$, so $x^3=1$. That's all you need.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1110433",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Limit $\lim_\limits{x\to0} \frac{\ln\left(x+\sqrt{1+x^2}\right)-x}{\tan^3(x)}$ Evaluate the given limit:
$$\lim_{x\to0} \frac{\ln\left(x+\sqrt{1+x^2}\right)-x}{\tan^3(x)} .$$
I've tried to evaluate it but I always get stuck... Obviously I need L'Hôpital's Rule here, but still get confused on the way. May someone show m... | i will use the maclaurin expansion for $\sqrt{1+x}, \ln(1+x), \tan(x)$
$\begin{align}
\ln[x + (1 + x^2)^{1/2}] &= \ln[x + 1 + \frac{1}{2}x^2 -\frac{1}{8}x^4 + \cdots]\\
&=\ln(1 + x + \frac{1}{2}x^2 - \frac{1}{8}x^4 + \cdots) \\
&=(x + \frac{1}{2}x^2 - \frac{1}{8}x^4 + \cdots)-\frac{1}{2} \{x + \frac{1}{2}x^2 - \frac{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1113338",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 7,
"answer_id": 1
} |
Commutative matrices Knowing that $AB=BA$, find the matrices that commute with the matrix
\begin{pmatrix}
1 & 2 \\
3 & 4 \\
\end{pmatrix}
I have assumed that multiplying matrix
$\begin{pmatrix}
a & b \\
c & d \end{pmatrix}
$ by the first one should be equal to multiplying the first one by $\begin{pmatrix}
... | You have figured it out. The equations you end up with:
2c=3b
and
a+c=d
are correct. Basically this means that you can specify any TWO values of the matrix arbitrarily and then the remaining two are fixed by the above equations. For example, suppose you want to specify a and b. The, rewrite the equations like this:
c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1113699",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Chance of winning simple dice game Tossing two fair dice, if the sum is 7 or 11, then I win; if the sum is 2, 3 or 12, then I lose; if the sum is one of rest of numbers then I toss the two dices again. What is probability of winning?
What I have tried: every tossing, p(win)=$\frac{2}{9}$, and p(lose)=$\frac{1}{9}$, p(... | With your edited solution i think you are now correct with 2/3.
Since $$\frac{\frac{2}{9}}{\frac{2}{9}+\frac{1}{9}}=\frac{2}{3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1118352",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Prove that if $({x+\sqrt{x^2+1}})({y+\sqrt{y^2+1}})=1$ then $x+y=0$ Let
$$\left({x+\sqrt{x^2+1}}\right)\left({y+\sqrt{y^2+1}}\right)=1$$
Prove that $x+y=0$.
This is my solution:
Let
$$a=x+\sqrt{x^2+1}$$
and
$$b=y+\sqrt{y^2+1}$$
Then $x=\dfrac{a^2-1}{2a}$ and $y=\dfrac{b^2-1}{2b}$. Now $ab=1\implies b=\dfrac1a$. Then ... | Given the symmetry of the problem, wlog $x\ge y$. $x=y=0$ is a solution, from now on let us assume that $x>0$.
Let $f$ be the real-valued map $w\mapsto w+\sqrt{w^2+1}$. If $w>0$ then $f(w)>1$, which is enough to prove that $x>0>y$. Even better (without deriving it), $f$ is strictly increasing on the positive reals.
Def... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1118742",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 5,
"answer_id": 2
} |
How to integrate $(x^2 - y^2) / (x^2 + y^2)^2$ How do I integrate
$$\int \int \frac{(x^2 - y^2)}{(x^2 + y^2)^2} dx dy?$$
The WolframAlpha page gives
$$
c_1 + c_2 + \tan^{-1}(x/y).
$$
And I kind of specifically need
$$
\int_{0}^{x} \frac{(x^2 - y^2)}{(x^2 + y^2)^2} dy.
$$
Note
*
*I want to know integration techniqu... | The main integral
$$I = \int\frac{x^2-y^2}{(x^2+y^2)^2}\,dx$$
can be solved using a tan substitution.
\begin{align*}
x &= y\tan\theta \\
dx &= y\sec^2\theta\,d\theta
\end{align*}
So that
\begin{align*}
I &= \int \frac{y^2(\tan^2\theta -1)}{y^4\sec^4\theta}\,y\sec^2\theta\,d\theta \\
&= \frac{1}{y}\int \frac{\tan^2\thet... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1119784",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Show that if $m^2 + n^2 $ is divisible by $4$, then $mn$ is also divisible by $4$. Show that if $m$ and $n$ are integers such that $m^2 + n^2 $ is divisible by $4$, then $mn$ is also divisible by $4$.
I am not sure where to begin.
| Note that $n^2$ is even if and only if $n$ is even.
Since $m^2+n^2$ is even, you see that either both $m,n$ are even, or both are odd.
If both $m$, $n$ are even, then $mn$ is divisible by $4$.
If both $m$, $n$ are odd, you have $m=2k+1$, $n=2l+1$ so $m^2+n^2=4k^2+4k+4l^2+4l+2$ is not divisible by $4$. This gives the re... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1120919",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Proving that if $a,b,c\in \mathbb N$ and $a^2+b^2=c^2$ then $abc$ is even.
Let $a,b,c\in \mathbb N$ and $a^2+b^2=c^2$ then $abc$ is even.
My attempt:
If one or two numbers of $a,b,c$ are even then we're done, so we'll have to show that at least one of them is even.
Suppose $a,b$ are odd, then $a^2,b^2$ are odd, so $... | Notice that a square of an integer is either $0$ or $1$ modulo $4$ therefore. If $$c^2\equiv0\mod{4}\Rightarrow c\equiv0\mod{2}\Rightarrow abc\equiv0\mod{2}$$
If $$c^2\equiv1\mod{4}\Rightarrow a^2\equiv0, b^2\equiv1\mod{4}\Rightarrow a\equiv0\mod{2}\Rightarrow abc\equiv0\mod{2}$$
or
$$a^2\equiv1, b^2\equiv0\mod{4}\Righ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1121637",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
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