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Prove that $\tan(75^\circ) = 2 + \sqrt{3}$ My (very simple) question to a friend was how do I prove the following using basic trig principles:
$\tan75^\circ = 2 + \sqrt{3}$
He gave this proof (via a text message!)
$1. \tan75^\circ$
$2. = \tan(60^\circ + (30/2)^\circ)$
$3. = (\tan60^\circ + \tan(30/2)^\circ) / (1 - \tan60^\circ \tan(30/2)^\circ) $
$4. \tan (30/2)^\circ = \dfrac{(1 - \cos30^\circ)}{ \sin30^\circ}$
Can this be explained more succinctly as I'm new to trigonometry and a little lost after (2.) ?
EDIT
Using the answers given I'm almost there:
*
*$\tan75^\circ$
*$\tan(45^\circ + 30^\circ)$
*$\sin(45^\circ + 30^\circ) / \cos(45^\circ + 30^\circ)$
*$(\sin30^\circ.\cos45^\circ + \sin45^\circ.\cos30^\circ) / (\cos30^\circ.\cos45^\circ - \sin45^\circ.\sin30^\circ)$
*$\dfrac{(1/2\sqrt{2}) + (3/2\sqrt{2})}{(3/2\sqrt{2}) - (1/2\sqrt{2})}$
*$\dfrac{(1 + \sqrt{3})}{(\sqrt{3}) - 1}$
*multiply throughout by $(\sqrt{3}) + 1)$
Another alternative approach:
*
*$\tan75^\circ$
*$\tan(45^\circ + 30^\circ)$
*$\dfrac{\tan45^\circ + \tan30^\circ}{1-\tan45^\circ.\tan30^\circ}$
*$\dfrac{1 + 1/\sqrt{3}}{1-1/\sqrt{3}}$
*at point 6 in above alternative
| $$
\begin{array}{l}
\tan (75^{\circ})=\frac{\sin (75^{\circ})}{\cos (75^{\circ})}=\frac{\cos (15^{\circ})}{\sin (15^{\circ})}=\frac{\cos (30^{\circ} / 2)}{\sin (30^{\circ} / 2)}=\sqrt{\frac{1+\cos (30^{\circ})}{1-\cos (30^{\circ})}} \\
=\sqrt{\frac{1+\frac{\sqrt{3}}{2}}{1-\frac{\sqrt{3}}{2}}}=\sqrt{\frac{2+\sqrt{3}}{2-\sqrt{3}}}=\sqrt{\frac{(2+\sqrt{3})^{2}}{(2-\sqrt{3})(2+\sqrt{3})}}=2+\sqrt{3}
\end{array}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/360747",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 4,
"answer_id": 3
} |
Power Series Solution for $e^xy''+xy=0$ $$e^xy''+xy=0$$
How do I find the power series solution to this equation, or rather, how should I go about dealing with the $e^x$? Thanks!
| $e^xy''+xy=0$
$y''+xe^{-x}y=0$
Let $y=\sum\limits_{n=0}^\infty\dfrac{a_nx^n}{n!}$ ,
Then $y'=\sum\limits_{n=0}^\infty\dfrac{na_nx^{n-1}}{n!}=\sum\limits_{n=1}^\infty\dfrac{na_nx^{n-1}}{n!}=\sum\limits_{n=1}^\infty\dfrac{a_nx^{n-1}}{(n-1)!}$
$y''=\sum\limits_{n=1}^\infty\dfrac{(n-1)a_nx^{n-2}}{(n-1)!}=\sum\limits_{n=2}^\infty\dfrac{(n-1)a_nx^{n-2}}{(n-1)!}=\sum\limits_{n=2}^\infty\dfrac{a_nx^{n-2}}{(n-2)!}$
$\therefore\sum\limits_{n=2}^\infty\dfrac{a_nx^{n-2}}{(n-2)!}+xe^{-x}\sum\limits_{n=0}^\infty\dfrac{a_nx^n}{n!}=0$
$\sum\limits_{n=2}^\infty\dfrac{a_nx^{n-2}}{(n-2)!}+x\left(\sum\limits_{n=0}^\infty\dfrac{(-1)^nx^n}{n!}\right)\left(\sum\limits_{n=0}^\infty\dfrac{a_nx^n}{n!}\right)=0$
By the formula in http://en.wikipedia.org/wiki/Cauchy_product#Series,
$\sum\limits_{n=2}^\infty\dfrac{a_nx^{n-2}}{(n-2)!}+x\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{a_kx^k(-1)^{n-k}x^{n-k}}{k!(n-k)!}=0$
$\sum\limits_{n=2}^\infty\dfrac{a_nx^{n-2}}{(n-2)!}+x\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^{n-k}a_kx^n}{k!(n-k)!}=0$
$\sum\limits_{n=2}^\infty\dfrac{a_nx^{n-2}}{(n-2)!}+x\sum\limits_{n=0}^\infty\left(\sum\limits_{k=0}^n\dfrac{(-1)^{n-k}n!a_k}{k!(n-k)!}\right)\dfrac{x^n}{n!}=0$
$\sum\limits_{n=2}^\infty\dfrac{a_nx^{n-2}}{(n-2)!}+\sum\limits_{n=0}^\infty\left(\sum\limits_{k=0}^n(-1)^{n-k}C_k^na_k\right)\dfrac{x^{n+1}}{n!}=0$
$\sum\limits_{n=2}^\infty\dfrac{a_nx^{n-2}}{(n-2)!}+\sum\limits_{n=3}^\infty\left(\sum\limits_{k=0}^{n-3}(-1)^{n-k-3}C_k^{n-3}a_k\right)\dfrac{x^{n-2}}{(n-3)!}=0$
$a_2+\sum\limits_{n=3}^\infty\left(\dfrac{a_n}{n-2}+\sum\limits_{k=0}^{n-3}(-1)^{n-k-3}C_k^{n-3}a_k\right)\dfrac{x^{n-2}}{(n-3)!}=0$
$\therefore\begin{cases}a_2=0\\\dfrac{a_n}{n-2}+\sum\limits_{k=0}^{n-3}(-1)^{n-k-3}C_k^{n-3}a_k=0\end{cases}$
But the recurrence relation is still very difficult to solve.
| {
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"url": "https://math.stackexchange.com/questions/362089",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
$f: \mathbb{R} \to \mathbb{R}$ satisfies $(x-2)f(x)-(x+1)f(x-1) = 3$. Evaluate $f(2013)$, given that $f(2)=5$ The function $f : \mathbb{R} \to \mathbb{R}$ satisfies $(x-2)f(x)-(x+1)f(x-1) = 3$. Evaluate $f(2013)$, given that $f(2) = 5$.
| Hint 1: $3=(x+1)-(x-2)$
$(x-2)f(x)-(x+1)f(x-1)=3\Leftrightarrow (x-2)f(x)-(x+1)f(x-1)=(x+1)-(x-2)\Leftrightarrow (x-2)(f(x)+1)-(x+1)(f(x-1)+1)=0$
Hint 2: Is there a function closely related to $f$ that would verify a simpler equation?
$g(x)=f(x)+1$
$ $
$(x-2)g(x) - (x+1)g(x-1)=0 \Leftrightarrow (x-2)g(x)=(x+1)g(x-1) \Leftrightarrow g(x)=\cfrac{x+1}{x-2}g(x-1)$
$g(x)=\cfrac{x+1}{x-2}g(x-1)=\cfrac{x+1}{x-2}\cfrac{x+1-1}{x-2-1}g(x-1-1) = \left(\prod\limits_{k=0}^{n-1} \cfrac{x+1-k}{x-2-k}\right) g(x-n)$
Now give the good value to $n$, simplify the product and you'll have the expression of $g$ you need.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/367473",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Difficulties performing Laurent Series expansions to determine Residues The following problems are from Brown and Churchill's Complex Variables, 8ed.
From §71 concerning Residues and Poles, problem #1d:
Determine the residue at $z = 0$ of the function $$\frac{\cot(z)}{z^4} $$
I really don't know where to start with this. I had previously tried expanding the series using the composition of the series expansions of $\cos(z)$ and $\sin(z)$ but didn't really achieve any favorable outcomes. If anyone has an idea on how I might go about solving this please let me know. For the sake of completion, the book lists the solution as $-1/45$.
From the same section, problem #1e
Determine the residue at $z = 0$ of the function $$\frac{\sinh(z)}{z^4(1-z^2)} $$
Recognizing the following expressions:
$$\sinh(z) = \sum_{n=0}^{\infty} \frac{z^{(2n+1)}}{(2n +1)!}$$
$$\frac{1}{1-z^2} = \sum_{n=0}^{\infty} (z^2)^n $$
I have expanded the series thusly:
$$\begin{aligned} \frac{\sinh(z)}{z^4(1-z^2)} &= \frac{1}{z^4} \bigg(\sum_{n=0}^{\infty} \frac{z^{(2n+1)}}{(2n +1)!}\bigg) \bigg(\sum_{n=0}^{\infty} (z^2)^n\bigg) \\ &= \bigg(\sum_{n=0}^{\infty} \frac{z^{2n - 3}}{(2n +1)!}\bigg) \bigg(\sum_{n=0}^{\infty} z^{2n-4} \bigg) \\\end{aligned} $$
I don't really know where to go from here. Any help would be great.
Thanks.
| $$\cos z=1-\frac{z^2}{2}+\frac{z^4}{24}-\ldots\;,\;\;\;\sin z=z-\frac{z^3}{6}+\frac{z^5}{120}-\ldots\implies$$
$$\cot z=\frac{\left(1-\frac{z^2}{2}+\frac{z^4}{24}-\ldots\right)}{z\left(1-\left(\frac{z^2}{6}-\frac{z^4}{120}\right)+\mathcal O(z^6)\right)}=$$
$$=\frac{1}{z}\left(1-\frac{z^2}{2}+\frac{z^4}{24}-\ldots\right)\left(1+\left(\frac{z^2}{6}-\frac{z^4}{120}\right)+\left(\frac{z^2}{6}-\frac{z^4}{120}\right)^2+\ldots\right)=$$
$$=\frac{1}{z}\left(1-\frac{z^2}{2}+\frac{z^4}{24}-\ldots\right)\left(1+\frac{z^2}{6}-\frac{z^4}{120}+\frac{z^4}{36}-\frac{z^6}{360}+\frac{z^8}{120^2}+\ldots\right)$$
$$=\frac{1}{z}\left(1-\frac{z^2}{3}-\frac{z^4}{45}+\ldots\right)=\frac{1}{z}-\frac{z}{3}-\frac{z^3}{45}+\mathcal O(z^5)$$
Multipying the above by $\,\displaystyle{\frac{1}{z^4}}\;$ renders the residue $\,\displaystyle{-\frac{1}{45}}\,$ .
Notice we only use the powers necessary to calculate the coefficient of $\,z^{-1}\,$ . All the rest is unimportant to us.
| {
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"url": "https://math.stackexchange.com/questions/367940",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Finding the limit of function - exponential one Find the value of $\displaystyle \lim_{x \rightarrow 0}\left(\frac{1+5x^2}{1+3x^2}\right)^{\frac{1}{\large {x^2}}}$
We can write this limit function as :
$$\lim_{x \rightarrow 0}\left(1+ \frac{2x^2}{1+3x^2}\right)^{\frac{1}{\large{x^2}}}$$
Please guide further how to proceed in such limit..
| $$\lim_{x \rightarrow 0}\left(1+ \frac{2x^2}{1+3x^2}\right)^{\frac{1}{x^2}}$$
$$=\lim_{x \rightarrow 0}\left(\left(1+ \frac{2x^2}{1+3x^2}\right)^{\huge{\frac{1+3x^2}{2x^2}}}\right)^{\huge{\frac2{1+3x^2}}}$$
$$=\left(\lim_{\frac{2x^2}{1+3x^2} \rightarrow 0}\left(1+ \frac{2x^2}{1+3x^2}\right)^{\huge{\frac{1+3x^2}{2x^2}}}\right)^{\huge\lim_{x \rightarrow 0}{\frac2{1+3x^2}}}=e^2$$
as $x\to0\implies \frac{2x^2}{1+3x^2}\to0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/368382",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Hyperbolic Functions
Hey everyone, I need help with questions on hyperbolic functions.
I was able to do part (a).
I proved for $\sinh(3y)$ by doing this:
\begin{align*}
\sinh(3y) &= \sinh(2y +y)\\
&= \sinh(2y)\cosh(y) + \cosh(2y)\sinh(y)\\
&= 2\sinh(y)\cosh(y)\cosh(y) + (\cosh^2(y)+\sinh^2(y))\sinh(y)\\
&= 2\sinh(y)(1+\sinh^2(y)) + (1+\sinh^2(y) + \sinh^2(y))\sinh(y)\\
&= 2\sinh(y) + 2\sinh^3(y) + \sinh(y) +2\sinh^3(y)\\
&= 4\sinh^3(y) + 3\sinh(y).
\end{align*}
Therefore, $0 = 4\sinh^3(y) + 3\sinh(y) - \sinh(3y)$.
I have no clue what to do for part (b) and part (c) but I do see similarities between part (a) and part(b) as you can subtitute $x = \sinh(y)$.
But yeah, I'm stuck and help would be very much appreciated.
| In the identity that you proved, put $\sinh 3y=2$. Then if $x=\sinh y$, the identity says that $4x^3+3x-2=0$. So we are almost finished, we have shown this $x$ is a solution of the equation.
Note that $\sinh t$ is a strictly increasing function, and that $\sinh t$ is large negative when $t$ is large negative, and large positive when $t$ is large positive. So it has an inverse function $\sinh^{-1}$, defined everywhere. We have $\sinh 3y=2$ if and only if $3y=\sinh^{-1} 2$ if and only if $y=\frac{1}{3}\sinh^{1}2$. Finally, to get $x$, take the $\sinh$ of this.
Now start from $ax^3+bx+c=0$. We want to manipulate this equation so that it will look like $4t^3+3t-d=0$, so that we can use the same trick.
We will make the substitution $x=kt$ for some constant $k$. This yields the equation $ak^3t^3+bkt +c=0$. We want the lead coefficient to be $4$. So multiply through by $\frac{4}{ak^3}$. We get the equation
$$4t^3+\frac{4b}{ak^2}t+\frac{4b}{ak^3}c=0.\tag{$1$}$$
We want the coefficient of $t$ to be $3$. So we need $\frac{4b}{ak^2}=3$. That gives
$$k=2\sqrt{\frac{b}{3a}}.$$
To solve the cubic $(1)$ using the $\sinh$ method, let $\sinh 3y=-\frac{4b}{ak^3}c$, where $k$ is as just calculated.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/369339",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Finding coefficient of generating functions I have the equation
$$(1+x+x^2+\ldots+x^k+\ldots)(1+x^2+x^4+\ldots+x^{2k}+\ldots)(x^2+x^3)$$
how of I find the coefficent of $x^{24}$. I know to condense this down to
$$\frac1{1-x}\cdot\frac1{1-x^2}\cdot x(1+x)$$
but I don't know what to do after that
| Hint: There is cancellation, $(1-x^2)=(1-x)(1+x)$. And $x^2+x^3=x^2(1+x)$.
So we end up with $\dfrac{x^2}{(1-x)^2}$.
Now everything is straightforward. The coefficients for $\dfrac{1}{(1-x)^2}$ can be found by a direct computation of $(1+x+x^2+\cdots)(1+x+x^2+\cdots)$, or by noticing that $\dfrac{1}{(1-x)^2}$ is the derivative of $\frac{1}{1-x}$, and differentiating $1+x+x^2+x^3+\cdots$ term by term.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/372031",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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What will be the units digit of $7777^{8888}$? What will be the units digit of $7777$ raised to the power of $8888$ ?
Can someone do the math with explaining the fact "units digit of $7777$ raised to the power of $8888$"?
| $$7777^{1} \equiv 7 \pmod{10}$$
$$7777^{2} \equiv 9 \pmod{10}$$
$$7777^{3} \equiv 3 \pmod{10}$$
$$7777^{4} \equiv 1 \pmod{10}$$
$$7777^{5} \equiv 7 \pmod{10}$$
And the relation continues, in general you see that:
$$7777^{4n + 1} \equiv 7 \pmod{10}$$
And
$$7777^{4n} \equiv 1 \pmod{10}$$
$$\frac{8888}{4} = 2222$$
Hence, $7777^{8888} \equiv 1 \pmod{10}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/372518",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
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Prove this inequality $a \sqrt{1-b^2}+b\sqrt{1-a^2}\le1 $ Prove that for $a,b\in [-1,1]$:
$$a\sqrt{1-b^2}+b\sqrt{1-a^2}\leq 1$$
| Method 1:
HINT:
As $1-b^2\ge 0\implies -1\le b\le 1,$ let $ b=\sin B$
Similarly, $a=\sin A$
$\implies a\sqrt{1-b^2}+b\sqrt{1-a^2}=\sin A\cos B+\cos A\sin B=\sin(A+B)$
Method 2:
Let $a\sqrt{1-b^2}+b\sqrt{1-a^2}=y$
Squaring we get, $y^2=a^2(1-b^2)+b^2(1-a^2)+2ab\sqrt{(1-a^2)(1-b^2)}$
So, $1-y^2$
$=(1-a^2)(1-b^2)+(ab)^2-2ab\sqrt{(1-a^2)(1-b^2)}$
$=\left(\sqrt{(1-a^2)(1-b^2)}-ab\right)^2\ge0$ for real $a,b$
$\implies y^2\le 1\implies y\le1$
Method 3:
$$\left(a\sqrt{1-b^2}\pm b\sqrt{1-a^2}\right)^2+\left(a\cdot b\mp \sqrt{(1-a^2)(1-b^2)}\right)^2$$
$$=a^2(1-b^2)+b^2(1-a^2)\pm2ab\sqrt{(1-a^2)(1-b^2)}+a^2b^2+(1-a^2)(1-b^2)\mp2ab\sqrt{(1-a^2)(1-b^2)}=1$$
Now, if $p^2+q^2=1$ where $p,q$ are real, $q^2\ge0\implies p^2=1-q^2\le1$
So, each of $a\sqrt{1-b^2}\pm b\sqrt{1-a^2},a\cdot b\mp \sqrt{(1-a^2)(1-b^2)}$ is $\le1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/375260",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
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Fractional Trigonometric Integrands
*
*$$∫\frac{a\sin x+b\cos x+c}{d\sin x+e\cos x+f}dx$$
*$$∫\frac{a\sin x+b\cos x}{c\sin x+d\cos x}dx$$
*$$∫\frac{dx}{a\sin x+\cos x}$$
What are the relations between the numerator in the denominator, and what is the general pattern to solve these type of questions?
| I did not treat this question like a homework question, since it is not tagged homework.
What are the relations between the numerator in [and] the denominator
Both are linear functions of $\sin x,\cos x$
what is the general pattern to solve these type of questions?
The universal standard substitution to evaluate an integral of a rational fraction in $\sin x,\cos x$, i.e. a rational fraction of the form
$$R(\sin x,\cos x)=\frac{P(\sin x,\cos x)}{Q(\sin x,\cos x)},$$
where $P,Q$ are polynomials in $\sin x,\cos x$ is a trigonometric substitution known as the Weierstrass substitution
$$\tan \frac{x}{2}=t,\qquad x=2\arctan t.\tag{*}$$
Differentiating both sides of $(^*)$ w.r.t. $t$, we get
$$\frac{dx}{dt}=\frac{2}{1+t^2},\qquad dx=\frac{2}{1+t^2}dt.$$
Since we know$^1$ from trigonometry that $$\cos x =\frac{1-\tan ^{2}\frac{x }{2}}{1+\tan ^{2}\frac{
x}{2}}=\frac{1-t^2}{1+t^2},\qquad \sin x =\frac{2\tan \frac{x }{2}}{1+\tan ^{2}
\frac{x }{2}}=\frac{2t}{1+t^2},$$
we see that in general, the integrand becomes a rational fraction in $t$, whose standard integration technique is the partial fractions decomposition:
$$\int R(\sin x,\cos x)\, dx=\int R\Big(\frac{2t}{1+t^2},\frac{1-t^2}{1+t^2} \Big)\frac{2}{1+t^2}\, dt.$$
For instance this substitution in the 2nd. integral yields
$$
\begin{eqnarray*}
\int \frac{a\sin x+b\cos x}{c\sin x+d\cos x}dx &=&\int \dfrac{a\dfrac{2t}{
1+t^{2}}+b\dfrac{1-t^{2}}{1+t^{2}}}{c\dfrac{2t}{1+t^{2}}+d\dfrac{1-t^{2}}{
1+t^{2}}}\dfrac{2}{1+t^{2}}dt \\
&=&2\int \frac{bt^{2}-2at-b}{\left( dt^{2}-2ct-d\right) \left(
1+t^{2}\right) }dt.
\end{eqnarray*}
$$
--
$^1$ A possible proof is the following one, which uses the double-angle formulas and the identity $\cos ^{2}\frac{x}{2}+\sin ^{2}\frac{x}{2}=1$:
$$
\begin{eqnarray*}
\cos x &=&\cos ^{2}\frac{x}{2}-\sin ^{2}\frac{x}{2}=\frac{\frac{\cos ^{2}
\frac{x}{2}-\sin ^{2}\frac{x}{2}}{\cos ^{2}\frac{x}{2}}}{\frac{\cos ^{2}
\frac{x}{2}+\sin ^{2}\frac{x}{2}}{\cos ^{2}\frac{x}{2}}}=\frac{1-\tan ^{2}
\frac{x}{2}}{1+\tan ^{2}\frac{x}{2}}, \\
&& \\
\sin x &=&2\sin \frac{x}{2}\cos \frac{x}{2}=\frac{\frac{2\sin \frac{x}{2}
\cos \frac{x}{2}}{\cos ^{2}\frac{x}{2}}}{\frac{\cos ^{2}\frac{x}{2}+\sin ^{2}%
\frac{x}{2}}{\cos ^{2}\frac{x}{2}}}=\frac{2\tan \frac{x}{2}}{1+\tan ^{2}
\frac{x}{2}}.
\end{eqnarray*}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Finding the remainder when $2^{100}+3^{100}+4^{100}+5^{100}$ is divided by $7$
Find the remainder when $2^{100}+3^{100}+4^{100}+5^{100}$ is divided by $7$.
Please brief about the concept behind this to solve such problems. Thanks.
| Fermat's little theorem says that $a^6\equiv 1 \pmod 7$ whenever $7$ does not divide $a$. So $2^{100}+3^{100}+4^{100}+5^{100}\equiv 2^4+3^4+4^4+5^4 \pmod 7$
| {
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"url": "https://math.stackexchange.com/questions/377378",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find all matrices $A$ of order $2 \times 2$ that satisfy the equation $A^2-5A+6I = O$
Find all matrices $A$ of order $2 \times 2$ that satisfy the equation
$$
A^2-5A+6I = O
$$
My Attempt:
We can separate the $A$ term of the given equality:
$$
\begin{align}
A^2-5A+6I &= O\\
A^2-3A-2A+6I^2 &= O
\end{align}
$$
This implies that $A\in\{3I,2I\} = \left\{\begin{pmatrix}
3 & 0\\
0 & 3
\end{pmatrix},
\begin{pmatrix}
2 & 0\\
0 & 2
\end{pmatrix}\right\}$.
Are these the only two possible values for $A$, or are there other solutions?If there are other solutions, how can I find them?
| $A^2 - 5A + 6 = 0$ is equivalent to $(A-2)(A-3) = 0$, which is equivalent to $Sp(A) \subset \{2, 3\}$.
Three cases are possible :
*
*$Sp(A) = \{2\}$, i.e. $A = 2I$
*$Sp(A) = \{3\}$, i.e. $A = 3I$
*$Sp(A) = \{2, 3\}$, i.e. $A$ is similar to $\begin{pmatrix}
2 & 0\\
0 & 3
\end{pmatrix}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/379076",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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How do evaluate an inequality that involves a fractional part? I am stuck on how to evaluate whether the following condition is true:
Let $\{k\}$ be the fractional part of a real number such that $\{k\} = k - \lfloor{k}\rfloor$.
if $\{\frac{x}{2}\} < \frac{1}{2} + \frac{\{x\}}{2}$, does it then follow that:
$$\{\frac{x+1}{2}\} \ge \frac{1}{2} + \frac{\{x+1\}}{2}$$
I can see that:
$\{x+1\} = \{x\}$
if $\{\frac{x}{2}\} < \frac{1}{2}$, $\{\frac{x+1}{2}\} = \{\frac{x}{2}\} + \frac{1}{2}$
$\frac{\{x\}}{2} < \frac{1}{2}$
So, it follows that $\frac{1}{2} + \frac{\{x\}}{2} < 1$
I can that it is true in the case where $\{x\} = \frac{1}{2}$ since:
$\{\frac{1}{4}\} < \frac{1}{2} + \frac{\{\frac{1}{2}\}}{2}$ and $\{\frac{3}{4}\} \ge \frac{1}{2} + \frac{\{\frac{1}{2} + 1\}}{2}$
Can anyone help me to evaluate this condition is always true?
Thanks,
-Larry
| Let $x = n + r,$ where $n = \lfloor x \rfloor$ and $r = \{x\}.$ If $n$ is odd, then $\left\{\frac{n+r}{2}\right\} = \frac{1}{2} + \frac{r}{2},$ so the inequality fails. Hence $n$ is even. Then $n + 1$ is odd, whence $\left\{\frac{n+1+r}{2}\right\} = \frac{1}{2} + \frac{r}{2} = \frac{1}{2} + \frac{\{n+1+r\}}{2},$ so in fact, it is simply an equality.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/380817",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Limit $\frac{\tan^{-1}x - \tan^{-1}\sqrt{3}}{x-\sqrt{3}}$ without L'Hopital's rule. Please solve this without L'Hopital's rule? $$\lim_{x\rightarrow\sqrt{3}} \frac{\tan^{-1} x - \frac{\pi}{3}}{x-\sqrt{3}}$$
All I figured out how to do is to rewrite this as $$\frac{\tan^{-1} x - \tan^{-1}\sqrt{3}}{x-\sqrt{3}}$$
Any help is appreciated!
| Putting $$\frac\pi3-\tan^{-1}x=\theta\implies \tan^{-1}x=\frac\pi3-\theta$$
So, $$x=\tan\left(\frac\pi3-\theta\right)\text{ and }x-\sqrt3=\tan\left(\frac\pi3-\theta\right)-\tan\frac\pi3=\frac{\sin(\frac\pi3-\theta-\frac\pi3)}{\cos\frac\pi3\cos \left(\frac\pi3-\theta\right)}=-\frac{2\sin\theta}{\cos\left(\frac\pi3-\theta\right)}$$
$$\text{So,}\lim_{x\rightarrow\sqrt{3}} \frac{\tan^{-1} x - \frac{\pi}{3}}{x-\sqrt{3}}=\lim_{\theta\to0}\frac{-\theta}{-\frac{2\sin\theta}{\cos\left(\frac\pi3-\theta\right)}}=\frac12 \cdot \frac{\lim_{\theta\to0}\cos\left(\frac\pi3-\theta\right)}{\lim_{\theta\to0}\frac{\sin\theta}\theta}=\frac{\cos\frac\pi3}{2\cdot1}=\frac1{2\cdot2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/382684",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove by mathematical induction for any prime number$ p > 3, p^2 - 1$ is divisible by $3$? Prove by mathematical induction for any prime number $p > 3, p^2 - 1$ is divisible by $3$?
Actually the above expression is divisible by $3,4,6,8,12$ and $24$.
I have proved the divisibility by $4$ like:
$$
\begin{align}
p^2 -1 &= (p+1)(p-1)\\
&=(2n +1 +1)(2n + 1 - 1)\;\;\;\text{as $p$ is prime, it can be written as $(2n + 1)$}\\
&= (2n + 2)(2n)\\
&= 4(n)(n + 1)
\end{align}
$$
Hence $p^2 - 1$ is divisible by 4.
But I cannot prove the divisibility by $3$.
| If $p>3$ is a prime then $3$ does not divide $p$ i.e. $3$ and $p$ are relatively prime , now $p-1$ , $p , p+1$ are three consecutive integers , so one of them must be divisible by $3$ , hence $3$ must divide their product i.e. , $p(p-1)(p+1)=p(p^2-1)$ , but $3$ and $p$ are relatively prime , so $3$ must divide $p^2-1$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Evaluating Complex Integral. I am trying to evaluate the following integrals:
$$\int\limits_{-\infty}^\infty \frac{x^2}{1+x^2+x^4}dx $$
$$\int\limits_{0}^\pi \frac{d\theta}{a\cos\theta+ b} \text{ where }0<a<b$$
My very limited text has the following substitution:
$$\int\limits_0^\infty \frac{\sin x}{x}dx = \frac{1}{2i}\int\limits_{\delta}^R \frac{e^{ix}-e^{-ix}}{x}dx \cdots $$
Is the same of substitution available for the polynomial? Thanks for any help. I apologize in advance for slow responses, I have a disability that limits me to an on-screen keyboard.
| $f(z) := \frac{1}{a\cos z+b}$
Let $I$ be the integral in question. Let's double the integral, to get
$$2I =\int_0^{2\pi} f(z)\, dz$$
Let $C$ be the contour of the unit circle $|z|=1$ in the positive direction. Then using this, we have
$$2I = \oint_C \frac{1}{a\left(\frac{z+z^{-1}}{2}\right)+b}\frac{dz}{iz} = -2 i \oint_C \frac{dz}{az^2+2bz+a}$$
The last integral can be evaluated by residues. The poles of the function are at
$$b_\pm = \frac{-b\pm \sqrt{b^2-a^2}}{a}$$
Based on $b>a>0$, we find the only pole in the contour $C$ is $b_+$. The residue there is:
$$z_+ = \operatorname*{Res}_{z=b_+}\frac{1}{az^2+2bz+a} = \frac{1}{2\sqrt{b^2-a^2}}$$
Then
$$2I = -2 i \left(\frac{2 \pi i}{2\sqrt{b^2-a^2}}\right) = \frac{2\pi}{\sqrt{b^2-a^2}}$$
Divide by $2$ and done.
$$f(z) := \frac{z^2}{1+z^2+z^4} = \frac{z^2}{(z^2-z+1)(z^2+z+1)}$$
Poles of $f$ occur at, using the quadratic formula:
$$b_{1} = \frac{1+\sqrt 3i}{2}$$
$$b_{2} = \frac{-1+\sqrt 3i}{2}$$
$$b_{3} = \frac{1-\sqrt 3i}{2}$$
$$b_{4} = \frac{-1-\sqrt 3i}{2}$$
Using the canonical semicircle contour ($Re^{i\theta}$ for $\theta \in [0,\pi]$) over the upper half plane, it is easily seen that the integral of $f(z)$ over the arc disappears as $R \to \infty$. Then we only need to find the residues of $b_1$ and $b_2$ (using the first or second formula here):
$$z_1 = \operatorname*{Res}_{z=b_1}f(z)= -\frac{1}{4}-\frac{i}{4\sqrt 3}$$
$$z_2 = \operatorname*{Res}_{z=b_2}f(z)= \frac{1}{4}-\frac{i}{4\sqrt 3}$$
$$\oint_C f(z)\, dz = \int_{-\infty}^\infty f(x)\, dx = 2 \pi i (z_1+z_2) =-2 \pi i \frac{i}{2\sqrt 3} = \frac{\pi}{\sqrt 3}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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} |
Maclaurin expansion of Sin(Sin(x)) I want to calculate the limit,
$$ \lim_{x\to 0} \frac{sin(sin x) - sin x}{x^3}$$
and doing so using Maclaurin expansion.
Now $sin x$ expands to $x -\frac{x^3}{3!}x^3 + O(x^5)$
Which would give $sin(sinx)= (x -\frac{1}{3!}x^3 + O(x^5)) -\frac{1}{3!}(x -\frac{x^3}{3!}x^3 + O(x^5))^3 + O(x^5)$
This expression should simplify to (I have the answer) $x - \frac{1}{3}x^3 + O(x^5)$ But i cannot see how this step comes about. Particulary, how do I handle the term $(x -\frac{x^3}{3!}x^3 + O(x^5))^3$?
Thanks a lot!
Alexander
| You are on the right track.
Since $\sin x=x-\frac{x^3}{6}+O(x^5)$, we have
$$
\begin{align}
\sin(\sin x)&=x-\frac{x^3}{6}+O(x^5)-\frac{1}{6}(x-\frac{x^3}{6}+O(x^5))^3+O((x-\frac{x^3}
{6}+O(x^5))^5)\\
&\text{Expanding $(x-\frac{x^3}{6}+O(x^5))^3=x^3+O(x^5)$, you get}\\
&=x-\frac{x^3}{6}-\frac{1}{6}x^3+O(x^5).
\end{align}
$$
Now I believe you can continue.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Integrating a school homework question. Show that $$\int_0^1\frac{4x-5}{\sqrt{3+2x-x^2}}dx = \frac{a\sqrt{3}+b-\pi}{6},$$ where $a$ and $b$ are constants to be found.
Answer is: $$\frac{24\sqrt3-48-\pi}{6}$$
Thank you in advance!
| Note first that the derivative of the quadratic in the denominator is $-2x+2$. It would be great if the numerator were $4x-4$, because then we could set $u=3+2x-x^2$, $du=(-2x+2)dx$, and write the indefinite integral as
$$\int\frac{-2}{\sqrt u}du=-2\int u^{-1/2}du\;.$$
Unfortunately, the numerator is actually $4x-5=-2(-2x+2)-1$. The trick is to split the integral in two:
$$\int_0^1\frac{4x-5}{\sqrt{3+2x-x^2}}dx=\int_0^1\frac{4x-4}{\sqrt{3+2x-x^2}}dx-\int_0^1\frac{dx}{\sqrt{3+2x-x^2}}\;.$$
We’ve already sorted out how to deal with the first of these integrals, so that leaves only the second. Complete the square:
$$3+2x-x^2=-(x^2-2x-3)=-\big((x-1)^2-4\big)=2^2-(x-1)^2\;,$$
so
$$\int_0^1\frac{dx}{\sqrt{3+2x-x^2}}=\int_0^1\frac{dx}{\sqrt{2^2-(x-1)^2}}\;,$$
which can be handled with a standard trig substitution.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Closed form for $\sum_{n=1}^\infty\frac{(-1)^n n^4 H_n}{2^n}$ Please help me to find a closed form for the sum $$\sum_{n=1}^\infty\frac{(-1)^n n^4 H_n}{2^n},$$ where $H_n$ are harmonic numbers: $$H_n=\sum_{k=1}^n\frac{1}{k}=\frac{\Gamma'(n+1)}{n!}+\gamma.$$
| A related problem. Here is another approach. Recalling the generating function of the harmonic numbers
$$ \sum_{n=1}^{\infty} H_n x^n = \frac{\ln(1-x)}{x-1} \implies (xD)^4\sum_{n=1}^{\infty} H_n x^n=(xD)^4 \frac{\ln(1-x)}{x-1}, $$
$$ \implies \sum_{n=1}^{\infty} H_n n^4 x^n = {\frac {x \left( 1+11\,x+11\,{x}^{2}+{x}^{3} \right) \ln \left( 1-x
\right) }{ \left( -1+x \right) ^{5}}}+{\frac {x \left( -1-27\,{x}^{2}
-18\,x-4\,{x}^{3} \right) }{ \left( -1+x \right) ^{5}}}.$$
Substituting $x=-\frac{1}{2}$ in the above identity gives the desired result
$$ \sum_{n=1}^\infty\frac{(-1)^n n^4 H_n}{2^n}={\frac {28}{243}}-{\frac {10}{81}}\,\ln \left( \frac{3}{2} \right). $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/385067",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "26",
"answer_count": 3,
"answer_id": 1
} |
Please correct my work, finding Eigenvector Determine whether matrix A is a Diagonalizable. if it is , determine matrix P that Diagnolizes it and compute $P^{-1}AP$.
$$A=
\begin{bmatrix}
+3 & +2\\
-2 & -3\\
\end{bmatrix}
$$
$$A-\ell I =
\begin{bmatrix}
+3-\ell & +2\\
-2 & -3-\ell\\
\end{bmatrix}
$$
Then Determinant should be zero :
$$
\begin{vmatrix}
+3-\ell & +2\\
-2 & -3-\ell\\
\end{vmatrix}=(3-\ell)(-3-\ell)+4=0 \to \ell^2-5=0 \to \ell_{1}=\sqrt 5,\ell_{2}=-\sqrt5 ,
$$
$$
\begin{bmatrix}
3-\ell_{1}& +2\\
-2 & -3-\ell{1}\\
\end{bmatrix}=
\begin{bmatrix}
3-\sqrt 5=0.76& +2\\
-2 & -3-\sqrt 5=-5.24\\
\end{bmatrix}\to{R_1\Leftarrow\Rightarrow R2 \mapsto}
\begin{bmatrix}
-2 & -5.24\\
0.76& +2\\
\end{bmatrix}\to{R_1=R1/{-2}\mapsto}
\begin{bmatrix}
1 & 2.62\\
0.76& +2\\
\end{bmatrix}\to{R_2=R2-0.76R1\mapsto}
\begin{bmatrix}
1 & 2.62\\
0 & 0.009\\
\end{bmatrix}\to{R_2=R2/0.009\mapsto}
\begin{bmatrix}
1 & 2.62\\
0 & 1\\
\end{bmatrix}\to{R_1=R1-2.62R1\mapsto}
\begin{bmatrix}
1 & 0 | 0\\
0 & 1 | 0\\
\end{bmatrix}
$$
$then$
$$ V_1= \begin{bmatrix}
0 \\
0 \\
\end{bmatrix}
$$
I am stuck here , matrix of $0,0$ is the right answer Eigenvector 1 ?
| The eigenvalues are correct. However you go wrong in the computation of the eigenvectors:
$$
\begin{bmatrix}
3-\sqrt{5} & 2\\
-2 & -3-\sqrt{5}
\end{bmatrix}
$$
Divide the first row by $3-\sqrt{5}$, which is the same as multiplying it by $(3+\sqrt{5})/2$, getting
$$
\begin{bmatrix}
1 & \frac{3+\sqrt{5}}{2}\\
-2 & -3-\sqrt{5}
\end{bmatrix}
$$
Now adding to the second row the first one multiplied by $2$ brings the matrix in the form
$$
\begin{bmatrix}
1 & \frac{3+\sqrt{5}}{2}\\
0 & 0
\end{bmatrix}
$$
so you know that one eigenvector is
$$
\begin{bmatrix}
-\frac{3+\sqrt{5}}{2}\\
1
\end{bmatrix}
$$
The computations for the other eigenvalue are similar
$$
\begin{bmatrix}
3+\sqrt{5} & 2\\
-2 & -3+\sqrt{5}
\end{bmatrix}
$$
$$
\begin{bmatrix}
1 & \frac{3-\sqrt{5}}{2}\\
-2 & -3+\sqrt{5}
\end{bmatrix}
$$
$$
\begin{bmatrix}
1 & \frac{3-\sqrt{5}}{2}\\
0 & 0
\end{bmatrix}
$$
So the other eigenvector you're looking for is
$$
\begin{bmatrix}
-\frac{3-\sqrt{5}}{2}\\
1
\end{bmatrix}
$$
Of course a matrix that diagonalizes $A$ is
$$
P=
\begin{bmatrix}
-\frac{3+\sqrt{5}}{2} & -\frac{3-\sqrt{5}}{2}\\
1&1
\end{bmatrix}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/385972",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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How to simplify $\frac{(\sec\theta -\tan\theta)^2+1}{\sec\theta \csc\theta -\tan\theta \csc \theta} $ How to simplify the following expression :
$$\frac{(\sec\theta -\tan\theta)^2+1}{\sec\theta \csc\theta -\tan\theta \csc \theta} $$
| Here is a detailed solution.Ready, set, go!
$$\require{cancel}\begin{align}\frac{\left(\sec\theta-\tan\theta\right)^2+1}{\sec\theta\csc\theta-\tan\theta\csc\theta}\\&=\frac{\sec^2\theta-2\sec\theta\tan\theta+\color{blue}{\tan^2\theta+1}}{\sec\theta\csc\theta-\tan\theta\csc\theta}\\&=\frac{\sec^2\theta-2\sec\theta\tan\theta+\color{blue}{\sec^2\theta}}{\sec\theta\csc\theta-\tan\theta\csc\theta}\tag{1}\label{ko-eq1}\\&=\frac{2\sec^2\theta-2\sec\theta\tan\theta}{\sec\theta\csc\theta-\tan\theta\csc\theta}\\&=\frac{2\sec\theta\cancel{\left(\sec\theta-\tan\theta\right)}}{\csc\theta\cancel{\left(\sec\theta-\tan\theta\right)}}\\&=\frac{2\sec\theta}{\csc\theta}\\&=\frac{\frac{2}{\cos\theta}}{\frac{1}{\sin\theta}}\\&=\frac{2}{\cos\theta}\sin\theta\\&=\frac{2\sin\theta}{\cos\theta}\\&=2\tan\theta\end{align}$$
In equation $\eqref{ko-eq1}$, $\tan^2\theta+1=\sec^2\theta$.I hope this helps.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove $\int_0^{\infty } \frac{1}{\sqrt{6 x^3+6 x+9}} \, dx=\int_0^{\infty } \frac{1}{\sqrt{9 x^3+4 x+4}} \, dx$ Prove that:
$(1)$$$\int_0^{\infty } \frac{1}{\sqrt{6 x^3+6 x+9}} \ dx=\int_0^{\infty } \frac{1}{\sqrt{9 x^3+4 x+4}} \ dx$$
$(2)$$$\int_0^{\infty } \frac{1}{\sqrt{8 x^3+x+7}} \ dx>1$$
What I do for $(1)$ is (something trival):
$$\int_0^{\infty } \frac{1}{\sqrt{6 x^3+6 x+9}} \, dx=\int_0^{\infty } \frac{1}{\sqrt{x^3+36 x+324}} \, dx$$
$$\int_0^{\infty } \frac{1}{\sqrt{9 x^3+4 x+4}} \, dx=\frac{\sqrt3}3\int_0^{\infty } \frac{1}{\sqrt{x^3+4 x+12}} \, dx$$
so it remains to prove that
$$\int_0^{\infty } \frac{1}{\sqrt{x^3+36 x+324}} \, dx=\frac{\sqrt3}3\int_0^{\infty } \frac{1}{\sqrt{x^3+4 x+12}} \, dx$$
Thanks in advance!
| I modified the lower bound for the integral (2) though it is still not exactly 1. Still I'm giving my bounds.
\begin{equation*}
\int_{0}^\infty \frac{1}{\sqrt{(8x^3+x+7)}}dx= \int_{0}^1 \frac{1}{\sqrt{(8x^3+x+7)}}dx+\int_{1}^\infty \frac{1}{\sqrt{(8x^3+x+7)}}dx=I_1+I_2
\end{equation*}
Now, for $x \in [0,1],\ x^2\geq x^3$, Hence it follows, \begin{equation*}
I_1=\int_{0}^1 \frac{1}{\sqrt{(8x^3+x+7)}}dx > \int_{0}^1 \frac{1}{\sqrt{(8x^2+x+7)}}dx
\end{equation*}
which can be simplified, by standard methods, to $$\int_{0}^1 \frac{1}{\sqrt{(8x^2+x+7)}}dx=\frac{1}{2\sqrt{2}}\ln\left(\frac{17+16\sqrt{2}}{1+4\sqrt{14}}\right)\approx 0.321387 $$
Using the transformation $y=1/x$ $$I_2=\int_{0}^1 \frac{1}{\sqrt{x(7x^3+x^2+8)}}dx$$
and using the fact that for all $x \in [0,1],\ x^2\geq x^3$, we get, $$I_2>\int_{0}^1 \frac{1}{\sqrt{8x(x^2+1)}}dx$$ Using the transformation $x=\tan(\theta/2)$, we get $$I_2> 1/4\int_{0}^{\pi/2}\frac{d\theta}{\sqrt{\sin{\theta}}}=\frac{1}{8}\beta\left(\frac{1}{4},\frac{1}{2}\right)=\frac{\left(\Gamma\left(\frac{1}{4}\right)\right)^2}{8\sqrt{2\pi}}\approx 0.655514$$ Hence, my bound is
$$\int_{0}^\infty \frac{1}{\sqrt{(8x^3+x+7)}}dx>0.321387+0.655514=0.976901$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Calculate the distance between the points $(1, 2, \dots, n)$ and $(2, 3, \dots, n, 1)$ I know that the operation to find the distance between two vectors is:
$$\sqrt{(b_1-a_1)^2+(b_2-a_2)^2+...+(b_n-a_n)^2}.$$
So the distance between $(7, 5, 3, 1)$ and $(1, 3, 5, 7)$ is:
$$\sqrt{(1-7)^2+(3-5)^2+(5-3)^2+(7-1)^2} = \sqrt{80}.$$
The distance between $(1, 2, \dots, n)$ and $(0, 0, \dots, 0, 0)$ is:
$$\sqrt{(0-1)^2+(0-2)^2+\dots+(0-n-1)^2+(0-n)^2} = \|(1, 2,\dots, n)\|.$$
But what about the distance between $(1, 2, \dots,n)$ and $(2,3, \dots,n, 1)$?
$$\sqrt{(2-1)^2+(3-2)^2+\dots+(n-n-1)^2+(1-n)^2} = \sqrt{1 + 1 +\dots+ 1 + (1-n)^2} = ?$$
I think the result is $n-1$, because it adds one to one up to $\sqrt{(1-n)^2} = n-1$, but I'm not sure.
Someone can help me solve this?
| Note that
$$\underbrace{1 + 1 + \dots + 1}_{n-1} + (1 - n)^2 = (n - 1) + (n - 1)^2 = (n-1)[1 + (n - 1)] = (n-1)n$$
where the first equality used the fact that
$$(1 - n)^2 = [(-1)(n-1)]^2 = (-1)^2(n-1)^2 = (n-1)^2.$$
Therefore your calculation shows that
$$\|(2, 3, \dots, n, 1) - (1, 2, \dots, n-1, n)\| = \sqrt{n(n-1)}.$$
| {
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Computing $\int_0^{\pi\over2} \frac{dx}{1+\sin^2(x)}$? How would you compute$$\int_0^{\pi\over2} \frac{dx}{1+\sin^2(x)}\, \, ?$$
| $$\int_0^{\frac \pi 2} \frac{1}{1 + \sin^2x}dx = \int_0^{\frac \pi 2} \frac{2}{3 - \cos 2x} dx = \int_0^\pi \frac{d\theta}{3 - \cos \theta } = \frac 12\int_0^{2\pi}\frac{d\theta}{3 - \cos \theta}$$
To evaluate $\displaystyle \int_0^{2\pi}\frac{1}{3 - \cos \theta}d\theta$ let $\displaystyle z = e^{i\theta} \implies \cos \theta = \frac{z^2 + 1}{2z}, d\theta = \frac{1}{iz}dz$
$$\int_0^{2\pi}\frac{1}{3 - \cos \theta}d\theta = \frac 2{i}\oint_{|z| = 1} \frac{1}{-z^2 + 6z - 1}dz $$
The poles are $3 - 2 \sqrt 2$ and $3 + 2 \sqrt 2$, and since $3 + 2 \sqrt 2 > 1$,
$$\frac 2{i}\oint_{|z| = 1} \frac{1 }{-z^2 + 6z - 1}dz = 4\pi \text{Res}\left[ \frac{1}{-z^2 + 6z - 1} , 3 - 2 \sqrt 2\right] = \frac{\pi}{\sqrt 2} $$
So $\displaystyle \int_0^{\frac \pi 2} \frac{1}{1 + \sin^2x}dx = \frac{\pi }{2 \sqrt 2}$
| {
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$x^4+y^4 \geq \frac{(x^2+y^2)^2}{2}$ I'm doing some exercise to prepare for my multivariable analysis exam. I didn't understand the second part of this question.
Given the function
$$f(x,y)=(x^2+y^2+1)^2 - 2(x^2+y^2) +4\cos(xy)$$
Prove that the taylor polynomial of degree $4$ of $f$ is equal to
$5+x^4+y^4$.
First, $4\cos(xy) = 4 - 2(xy)^2 + 4R_3 $
$(x^2+y^2+1)^2=x^4+2 x^2 y^2+2 x^2+y^4+2 y^2+1$
Therefore: $(x^2+y^2+1)^2 - 2(x^2+y^2)=x^4+2 x^2 y^2+y^4+1$
Therefore: $f(x,y)=x^4+y^4+5+4R_3$
I don't know exactly why I can now conclude that Taylor Polynomial of degree 4 must be $5+x^4+y^4$, but I don't know exactly why.
Now the second question is: $x^4+y^4 \geq \frac{(x^2+y^2)^2}{2}$
New edit
I understand this now thanks to hint of Hagen von Eitzen, thanks !
The third question is:
Determine what kind of stationary point you have in $(0,0)$.
| Hint: The difference left minus right is $\frac12(x^4-2x^2y^2+y^4)$
| {
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"timestamp": "2023-03-29T00:00:00",
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How to integrate $\int \sqrt{x^2+a^2}dx$ $a$ is a parameter. I have no idea where to start
| Integrating by parts
$$I=\int \sqrt{x^2+a^2}\cdot1dx$$
$$=\sqrt{x^2+a^2}\int dx-\int\left( \frac{d \sqrt{x^2+a^2}}{dx}\int dx\right)dx$$
$$=x\sqrt{x^2+a^2}-\int\frac{x^2 dx}{\sqrt{x^2+a^2}}$$
Now, $$\int\frac{x^2 dx}{\sqrt{x^2+a^2}}=\frac{x^2+a^2-a^2}{\sqrt{x^2+a^2}}dx=I-a^2\int\frac{dx}{\sqrt{x^2+a^2}}$$
Put $x=a\tan \theta $ in
$$\int\frac{dx}{\sqrt{x^2+a^2}}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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If $2n+1$ and $4n+3$ are prime, then $2n-1$ and $4n+1$ are not when $n>2$ How do you prove that, for $n>2$, if $2n+1$ and $4n+3$ are prime numbers, then $2n-1$ and $4n+1$ are composite numbers?
| HINT: $3$ divides $2n(2n+1)(2n-1)$ and $3$ divides $(4n+1)(4n+2)(4n+3)$
As $n>2, 4n+1>2n+1>3$
Hence, $4n+1, 2n+1$ are not divisible by $3$
If $4n+3$ is prime, $3$ does not divide $n\implies 3$ divide $2n-1$
If $2n+1$ is prime, $3$ does not divide $2(2n+1)\implies 3$ divide $4n+1$
which can also be derived from $4n+1=2(2n-1)+3$
| {
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General Solution of Diophantine equation Having the equation:
$$35x+91y = 21$$
I need to find its general solution.
I know gcf $(35,91) = 7$, so I can solve $35x+917 = 7$ to find $x = -5, y = 2$. Hence a solution to $35x+91y = 21$ is $x = -15, y = 2$.
From here, however, how do I move on to finding the set of general solutions? Any help would be very much appreciated!
Cheers
| On division by $7$ we get $5x+13y=3$
Now, by observation $13-5\cdot2=3$
So, $5x+13y=3=13-5\cdot2$
$\implies 5(x+2)=13(1-y)\implies \frac{5(x+2)}{13}=1-y$ which is an integer
$\implies 13$ divides $5(x+2)$
$\implies 13$ divides $(x+2)$ as $(5,13)=1$
So, $x+2=13a$ where $a$ is any integer, $x=13a-2$
So, $13y=3-5x=3-5(13a-2)=13(1-5a)\implies y=1-5a$
We can use the convergent property to a continued fraction as follows:
$$\frac{13}5=2+\frac35=2+\frac1{\frac53}=2+\frac1{1+\frac23}=2+\frac1{1+\frac1{\frac32}}=2+\frac1{1+\frac1{1+\frac12}}$$
So, the previous convergent of $\frac{13}5$ is $2+\frac1{1+\frac11}=\frac52$
$\implies 13\cdot2-5\cdot5=1$
So, we can write $5x+13y=3=3(13\cdot2-5\cdot5)$
Can you take it from here?
| {
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Comparing $\sqrt{1001}+\sqrt{999}\ , \ 2\sqrt{1000}$ Without the use of a calculator, how can we tell which of these are larger (higher in numerical value)?
$$\sqrt{1001}+\sqrt{999}\ , \ 2\sqrt{1000}$$
Using the calculator I can see that the first one is 63.2455453 and the second one is 63.2455532, but can we tell without touching our calculators?
| This is connected to some of the other answers, but I thought it might be worth mentioning to show the utility of approximation formulas. The first two derivatives of $ \ f(x) = x^{1/2} \ $ at $ \ a = 1000 \ $ are $ \ f'(1000) = \frac{1}{2} \cdot 1000^{-1/2} \ $ and $ \ f''(1000) = -\frac{1}{4} \cdot 1000^{-3/2} \ $ (we won't need these to be evaluated). The Taylor series for $ \ x^{1/2} \ $ about $ \ a = 1000 \ $ is then
$$f(x) \ = \ 1000^{1/2} \ + \ \frac{1}{2\cdot 1000^{1/2}} (x - 1000) \ - \ \frac{1}{4\cdot 1000^{3/2}} (x-1000)^2 \ + \ \ldots $$
Hence,
$$999^{1/2} \ + \ 1001^{1/2} \ \approx$$
$$[ \ 1000^{1/2} \ + \ \frac{1}{2\cdot 1000^{1/2}} (-1) \ - \ \frac{1}{4\cdot 1000^{3/2}} (-1)^2 \ + \ \ldots $$
$$ + \ 1000^{1/2} \ + \ \frac{1}{2\cdot 1000^{1/2}} (1) \ - \ \frac{1}{4\cdot 1000^{3/2}} (1)^2 \ + \ \ldots \ ] $$
$$= \ 2 \cdot 1000^{1/2} \ - \ \frac{2}{4\cdot 1000^{3/2}} \ - \ \ldots \ , $$
with all the terms in odd powers of $ \ (x-1000) \ $ cancelling and all of the higher-order terms in even powers also being negative. Because the two square-roots are being added, it is not sufficient to only carry this calculation to linear terms ("first-order") ; this is hinted at by the fact that OP found by calculator how close the two values are.
ADDED: The linear terms are sufficient, though, to show that $ \ \sqrt{1001} \cdot \sqrt{999} < 1000 \ . $
| {
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Help me prove this inequality : How would I go about proving this?
$$ \displaystyle\sum_{r=1}^{n} \left( 1 + \dfrac{1}{2r} \right)^{2r} \leq n \displaystyle\sum_{r=0}^{n+1} \displaystyle\binom{n+1}{r} \left( \dfrac{1}{n+1} \right)^{r}$$
Thank you! I've tried so many things. I've tried finding a series I could compare one of the series to but nada, I tried to change the LHS to a geometric series but that didn't work out, please could someone give me a little hint? Thank you!
| And here's the proof why $(1+1/x)^x$ is concave for $x\le 1$ (the case $x\ge 1$ is already proven by @TCL).
We study the second derivative with respect to $x$, it is equal to (thanks to TCL) to
$\displaystyle \left(1+ \frac 1x\right)^x\left(\left(\ln (1+1/x)-\frac{1}{1+x}\right)^2-\frac{1}{x(1+x)^2}\right)$;
First of all, $\ln (1+1/x)\ge\frac{1}{1+x}$; we say that $y=1/x\ge 1$ and then study $(\ln (1+y)-\frac{y}{1+y})$ - this function is positive in $y=1$ and it's derivative with respect to $y$ is positive for $y\ge 1$, hence the function itself is always positive; therefore $\sqrt{\left( \ln (1+1/x)-\frac{1}{1+x}\right)^2}= \ln (1+1/x)-\frac{1}{1+x}$.
Now we study the sign of $\displaystyle \left(\left(\ln (1+1/x)-\frac{1}{1+x}\right)^2-\frac{1}{x(1+x)^2}\right)$; it's equivalent to study the sign of
$\displaystyle \ln (1+1/x)-\frac{1}{1+x} -\frac{1}{\sqrt x(1+x) } $. Once again, we change $y=1/x\ge 1$:
$\displaystyle \ln (1+y)-\frac{y}{1+y} -\frac{y\sqrt y}{ (1+y) } $.
In $y=1$ the last expression is negative (we have concavity for $x\ge 1$, after all); its derivative for $y\ge 1$ is (after several manipulations)
$\displaystyle 0.5(1+y)^{-2}(2y -3\sqrt y - y^{3/2})\le 0$,
which ensures that $ \ln (1+1/x)-\frac{1}{1+x} -\frac{1}{\sqrt x(1+x) } $ remains negative for $y\ge 1$; therefore, $(1+1/x)^x$ is concave for $x\le 1$. QED
| {
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Calculating $\sqrt{28\cdot 29 \cdot 30\cdot 31+1}$ Is it possible to calculate $\sqrt{28 \cdot 29 \cdot 30 \cdot 31 +1}$ without any kind of electronic aid?
I tried to factor it using equations like $(x+y)^2=x^2+2xy+y^2$ but it didn't work.
| \begin{align}
&\text{Let }x=30
\\ \\
\therefore&\ \ \ \ \ \sqrt{(x-2)(x-1)x(x+1)+1}
\\ \\
&=\sqrt{[(x-2)(x+1)[(x-1)x]+1}
\\ \\
&=\sqrt{(x^2-x-2)((x^2-x)+1}
\\ \\
&=\sqrt{(x^2-x)^2-2(x^2-x)+1}
\\ \\
&=\sqrt{(x^2-x-1)^2}
\\
&=x^2-x-1
\\
&=30^2-30-1
\\
&=\boxed{869}
\end{align}
| {
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How to solve this integral easily: $\int \frac{x\cdot \sqrt[3]{x+2}}{x+\sqrt[3]{x+2}} dx$ I am trying to solve this integral $$\int\frac{x\cdot \sqrt[3]{x+2}}{x+\sqrt[3]{x+2}} dx$$
I can do it by brute force (means using a substitution then long division and then substitutions again) but it's too long (suspiciously long solution). Is there any better way to solve it? I guess it should be. If you could provide me with the process that leads to the answer that would really help.
| Substitute $\sqrt[3]{x+2}=u$ therefore $\frac{du}{dx} =\dfrac{1}{3\sqrt[3]{(x+2)^2}}$
Your integral will become : 3$\int\dfrac{u^3(u^3-2)}{u^3+u-2}du $
= $3\int [u^3+ \dfrac{5u-2}{4(u^2+u+2)}-u-\dfrac{1}{4(u-1)}]du$
=$3\int u^3du + \dfrac{3}{4}\int(\dfrac{5(2u+1)}{2(u^2+u+2)}-\dfrac{9}{2(u^2+u+2)})du - \dfrac{3}{4}\int\dfrac{1}{u-1}du -3\int u.du$
Now you can easily see that differentiation of $u^2+u+2 = 2u+1$ so you can substitute $u^2+u+2 = t$ then proceed you will get your result.
In case of further clarification do let me know...
| {
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Evaluating $\int_0^\infty \frac{\log (1+x)}{1+x^2}dx$ Can this integral be solved with contour integral or by some application of residue theorem?
$$\int_0^\infty \frac{\log (1+x)}{1+x^2}dx = \frac{\pi}{4}\log 2 + \text{Catalan constant}$$
It has two poles at $\pm i$ and branch point of $-1$ while the integral is to be evaluated from $0\to \infty$. How to get $\text{Catalan Constant}$? Please give some hints.
| Letting $x=\tan \theta$ yields
$$\begin{aligned}\int_{0}^{\infty} \frac{\ln (1+x)}{1+x^{2}}dx&=\int_{0}^{\frac{\pi}{2}}[\ln (\cos \theta+\sin \theta) d \theta-\ln (\cos \theta)] d \theta \\&= \underbrace{\int_{0}^{\frac{\pi}{4}}2\ln (\cos \theta+\sin \theta)}_{L} d \theta-\underbrace{\int_{0}^{\frac{\pi}{2}} \ln (\cos \theta) d \theta}_{-\frac{\pi}{2} \ln 2}\end{aligned}$$
For the integral $L$, $$
\begin{aligned}
L &=\int_{0}^{\frac{\pi}{4}} \ln (1+\sin 2 \theta) d \theta \\
&=\frac{1}{2} \int_{0}^{\frac{\pi}{2}} \ln (1+\sin \theta) d \theta\\
&=\frac{1}{2} \int_{0}^{\frac{\pi}{2}} \ln (1+\cos \theta) d \theta \\
&=\frac{1}{2} \int_{0}^{\frac{\pi}{2}} \ln \left(2 \cos ^{2} \frac{\theta}{2}\right) d \theta\\&= \frac{\pi}{4} \ln 2+2 \int_{0}^{\frac{\pi}{4}} \ln (\cos \theta) d \theta\\&= \frac{\pi}{4} \ln 2+2\left(\frac{G}{2}-\frac{\pi}{4} \ln 2\right)\\&= -\frac{\pi}{4} \ln 2+G
\end{aligned}
$$
where the last integral comes from my post and $G$ is the Catalan’s constant.
Now we can conclude that $$
\boxed{\int_{0}^{\infty} \frac{\ln (1+x)}{1+x^{2}}dx= \frac{\pi}{4} \ln 2+G}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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How to evaluate $\sqrt[3]{a + ib} + \sqrt[3]{a - ib}$? The answer to a question I asked earlier today hinged on the fact that
$$\sqrt[3]{35 + 18i\sqrt{3}} + \sqrt[3]{35 - 18i\sqrt{3}} = 7$$
How does one evaluate such expressions? And, is there a way to evaluate the general expression
$$\sqrt[3]{a + ib} + \sqrt[3]{a - ib}$$
| Consider $a^3 =35+ 18i\sqrt{3}$ and $b^3=35- 18 i\sqrt{3}$
You are supposed to find $a+b$
$(a+b)^3=a^3+b^3+3ab(a+b) \implies(a+b)^3-3ab(a+b)=a^3+b^3$
Let $a+b=x \implies x^3-3ab(x)=70$ and $ab= [(35-18i\sqrt3)(35+18i\sqrt3)]^{1/3}=13$
You get $x^3-39x-70 =0$, product of the roots is $70$ and sum of the roots is $0$. You can do clever factorization or guess the roots(It has only 8 divisors). SO, number of possible should be just $8 \choose 3$. I kinda flicked your trick. ;)
| {
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"timestamp": "2023-03-29T00:00:00",
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How prove this $\int_{0}^{\infty}\sin{x}\sin{\sqrt{x}}\,dx=\frac{\sqrt{\pi}}{2}\sin{\left(\frac{3\pi-1}{4}\right)}$ Prove that
$$\int_{0}^{\infty}\sin{x}\sin{\sqrt{x}}\,dx=\frac{\sqrt{\pi}}{2}\sin{\left(\frac{3\pi-1}{4}\right)}$$
I have some question. Using this, find this integral is not converge, I'm wrong?
Thank you everyone
| Numerical calculation shows that the graph of
$$ y = \int_{0}^{x} \sin t \sin \sqrt{t} \, dt $$
is given by
Though a graph cannot constitute a proof, it strongly suggests that $y$ cannot converge as $x \to \infty$. Indeed, we can show that
$$ y(x) = -\cos x \sin \sqrt{x} + \frac{\sqrt{\pi}}{2} \sin \left( \frac{3\pi - 1}{4} \right) + o(1) $$
as $x \to \infty$. Thus in ordinary sense, the integral
$$ \int_{0}^{\infty} \sin x \sin \sqrt{x} \, dx $$
diverges. But if we understand the integral in Abel summation senas as follows
$$ \int_{0}^{\infty} \sin x \sin \sqrt{x} \, dx := \lim_{s \to 0^{+}} \int_{(0, \infty)} \sin x \sin \sqrt{x} \; e^{-sx} \, dx, $$
then the oscillating part vanishes and we obtain the identity. In fact, it converges to the proposed limit as weak as in Cesaro summation sense.
Now let us show that the integral converges in Able summation sense. By some calculation, we have
\begin{align*}
\int_{0}^{\infty} \sin x \sin \sqrt{x} \; e^{-sx} \, dx
&= \int_{0}^{\infty} 2x \sin x \sin (x^2) \; e^{-sx^{2}} \, dx \\
&= \int_{0}^{\infty} x \Re \left( e^{-i(x^2-x)} - e^{-i(x^2+x)} \right) e^{-sx^{2}} \, dx \\
&= \Re \int_{-\infty}^{\infty} x e^{-(s+i)x^2 + ix} \, dx.
\end{align*}
By noting that
$$ z e^{-(s+i)z^2 + iz} = z \exp \left\{ -(s+i) \left( z - \tfrac{i}{2(s+i)} \right)^2 + \tfrac{i}{4(1-is)} \right\} $$
is an entire function with a nice vanishing speed as $ \left| \Re z \right| \to \infty$, we find that we can shift the contour of integration so that
\begin{align*}
\int_{0}^{\infty} \sin x \sin \sqrt{x} \; e^{-sx} \, dx
&= \Re \left[ e^{\frac{i}{4(1-is)}} \int_{-\infty}^{\infty} \left(x + \frac{i}{2(s+i)} \right) e^{-(s+i)x^2} \, dx \right] \\
&= \Re \left[ e^{\frac{i}{4(1-is)}} \frac{i}{2(s+i)} \int_{-\infty}^{\infty} e^{-(s+i)x^2} \, dx \right] \\
&= \Re \left[ e^{\frac{i}{4(1-is)}} \frac{i}{2(s+i)} \frac{\sqrt{\pi}}{\sqrt{s + i}} \right].
\end{align*}
Taking $s \to 0^{+}$, we find that in Abel summation sense,
\begin{align*}
\int_{0}^{\infty} \sin x \sin \sqrt{x} \, dx
&= \frac{\sqrt{\pi}}{2} \Re \left( e^{\frac{i-i\pi}{4}} \right)
= \frac{\sqrt{\pi}}{2} \cos \left( \frac{1-\pi}{4} \right)
= \frac{\sqrt{\pi}}{2} \sin \left( \frac{3\pi - 1}{4} \right)
\end{align*}
as desired.
| {
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Solve the equation $\sqrt{3x-2} +2-x=0$
Solve the equation: $$\sqrt{3x-2} +2-x=0$$
I squared both equations $$(\sqrt{3x-2})^2 (+2-x)^2= 0$$
I got $$3x-2 + 4 -4x + x^2$$
I then combined like terms $x^2 -1x +2$
However, that can not be right since I get a negative radicand when I use the quadratic equation.
$x = 1/2 \pm \sqrt{((-1)/2)^2 -2}$
The answer is 6
| If you meant
$$\sqrt{3x-2}+2-x=0\implies \sqrt{3x-2}=x-2\implies3x-2=(x-2)^2=x^2-4x+4\implies$$
$$\implies x^2-7x+6=0$$
Now just check that $x^2-7x+6=(x-6)(x-1)$ ...and remember to check at the end whether both solutions of this quadratic are actually solutions of your original equation, since when squaring some mess can happen there. For example, one of the soltuions of the quadratic is not a solution of your equation.
| {
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How to show $x^4 - 1296 = (x^3-6x^2+36x-216)(x+6)$ How to get this result: $x^4-1296 = (x^3-6x^2+36x-216)(x+6)$?
It is part of a question about finding limits at mooculus.
| Here's a harder and similar way, but it still makes sense if you don't have what Julien said memorized.
$$x^4 - 1296 = x^4 - 6^4 \Rightarrow x^4 = 6^4$$Now two roots that seem obvious are $6$ and $-6$. But $x^4 = x^4 \cdot 1 = x^4 \cdot i^4 = (xi)^4$. So two more roots are $6i$ and $-6i$.
Hence, the factorization is $$\begin{aligned}&(x - 6)(x + 6)(x+6i)(x - 6i) \\ =& (x-6)(x+6)(x^2+36) \\ =& (x+6)(x^3 - 6x^2 +36x-216)\end{aligned}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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How can I prove this closed form for $\sum_{n=1}^\infty\frac{(4n)!}{\Gamma\left(\frac23+n\right)\,\Gamma\left(\frac43+n\right)\,n!^2\,(-256)^n}$ How can I prove the following conjectured identity?
$$\mathcal{S}=\sum_{n=1}^\infty\frac{(4\,n)!}{\Gamma\left(\frac23+n\right)\,\Gamma\left(\frac43+n\right)\,n!^2\,(-256)^n}\stackrel?=\frac{\sqrt3}{2\,\pi}\left(2\sqrt{\frac8{\sqrt\alpha}-\alpha}-2\sqrt\alpha-3\right),$$
where
$$\alpha=2\sqrt[3]{1+\sqrt2}-\frac2{\sqrt[3]{1+\sqrt2}}.$$
The conjecture is equivalent to saying that $\pi\,\mathcal{S}$ is the root of the polynomial
$$256 x^8-6912 x^6-814752 x^4-13364784 x^2+531441,$$
belonging to the interval $-1<x<0$.
The summand came as a solution to the recurrence relation
$$\begin{cases}a(1)=-\frac{81\sqrt3}{512\,\pi}\\\\a(n+1)=-\frac{9\,(2n+1)(4n+1)(4 n+3)}{32\,(n+1)(3n+2)(3n+4)}a(n)\end{cases}.$$
The conjectured closed form was found using computer based on results of numerical summation. The approximate numeric result is $\mathcal{S}=-0.06339748327393640606333225108136874...$ (click to see 1000 digits).
| This is a very interesting question. Since it is different from similar question of this kind in that a new technique is involved I will provide the answer. We have:
\begin{eqnarray}
&&\sum\limits_{n=1}^\infty \frac{(4n)!}{\Gamma(n+\frac{2}{3}) \Gamma(n+\frac{4}{3})(n!)^2 (-256)^n}=\\
&&
\frac{1}{(-\frac{1}{3})!}\frac{1}{(-\frac{2}{3})!} \sum\limits_{n=1}^\infty \binom{4 n}{n} \cdot \underbrace{\frac{1}{n+\frac{1}{3}}}_{\int\limits_0^1\theta^{n-\frac{2}{3}} d\theta}\cdot \left( -\frac{3^3}{4^4}\right)^n=\\
&&\frac{1}{(-\frac{1}{3})!}\frac{1}{(-\frac{2}{3})!}\int\limits_0^1 \theta^{-\frac{2}{3}} \cdot \left\{ \frac{x[\theta](1+\theta \frac{27}{256}x[\theta]^3)}{1+4\theta \frac{27}{256}x[\theta]^3}-1\right\} d\theta=\\
&&\frac{1}{(-\frac{1}{3})!}\frac{1}{(-\frac{2}{3})!}\left\{-\frac{4 \cdot 2^{\frac{2}{3}}}{3}\cdot \int\limits_1^{x[1]} \frac{1}{x^{\frac{4}{3}}} \cdot \frac{1}{(1-x)^{\frac{2}{3}}}dx-3\right\}\\
&&\frac{1}{(-\frac{1}{3})!}\frac{1}{(-\frac{2}{3})!}\left\{4 \cdot 2^{\frac{2}{3}}(\frac{1-x[1]}{x[1]})^{\frac{1}{3}}-3\right\}\\
&&\frac{\sqrt{3}}{2 \pi}\left\{4 \cdot 2^{\frac{2}{3}}(\frac{1-x[1]}{x[1]})^{\frac{1}{3}}-3\right\}\\
\end{eqnarray}
In the first line we used elementary properties of factorials. In the second line we used my answer to Closed form solutions for a family of hypergeometric sums. .
Here the function$x[\theta]$ is defined as a solution to the following polynomial equation:
\begin{equation}
1-x[\theta] - \theta \frac{27}{256} x[\theta]^4=0
\end{equation}
where out of the four different solutions we take the solution that is the closest to unity. In the third line we substituted for $x=x[\theta]$, in the fourth line we evaluated the integral and finally in the fifthe line we simplified the result.
Here :
\begin{equation}
x[1]=\frac{2 \sqrt{2+\left(1+\sqrt{2}\right)^{2/3} \left(4 \sqrt{\frac{2}{1+\sqrt{2}-\sqrt[3]{1+\sqrt{2}}}}-2\right)}-2 \sqrt{2 \left(\left(1+\sqrt{2}\right)^{2/3}-1\right)}}{3 \sqrt[6]{1+\sqrt{2}}}
\end{equation}
| {
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Solving $\sqrt{3x^2-2}+\sqrt[3]{x^2-1}= 3x-2 $ How can I solve the equation $$\sqrt{3x^2-2}+\sqrt[3]{x^2-1}= 3x-2$$
I know that it has two roots: $x=1$ and $x=3$.
| Consider the functions: $f_{1},f_{2}:D\rightarrow\textbf{R}$, where $D = (-\infty,-\sqrt{\frac{2}{3}}]\cup[\sqrt{\frac{2}{3}}, \infty) $,
$f_{1}(x)={\sqrt{3X^{2}-2}}+\sqrt[3]{x^{2}-1}$ and $f_{2}(x)=3x-2$.
For $x\leq-\sqrt{\frac{2}{3}}$ $f_{1}$ is decreasing, $f_{2}$ is increasing and $f_{1}(-\sqrt{\frac{2}{3}})>f_{2}(-\sqrt{\frac{2}{3}})$, the equation has no negative solutions.
For $x\geq\sqrt{\frac{2}{3}}$, $f_{1}$ is concave ( $f''_{1}(x) < 0$ ) and $f_{2}$ is linear function, the equation has at most two roots, which are $x_{1}=2$ and $x_{2}=3$.
| {
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Show that $n \ge \sqrt{n+1}+\sqrt{n}$ (how) Can I show that:
$n \ge \sqrt{n+1}+\sqrt{n}$ ?
It should be true for all $n \ge 5$.
Tried it via induction:
*
*$n=5$: $5 \ge \sqrt{5} + \sqrt{6} $ is true.
*$n\implies n+1$:
I need to show that $n+1 \ge \sqrt{n+1} + \sqrt{n+2}$
Starting with $n+1 \ge \sqrt{n} + \sqrt{n+1} + 1 $ .. (now??)
Is this the right way?
| Here is another way:
Define $f(x)=x-\sqrt{x}-\sqrt{x+1}$. We need to show that $f(x)\ge 0$ for all $x\ge 5$.
Since
$$f'(x)=1-\frac{1}{2\sqrt{x}}-\frac{1}{2\sqrt{x+1}} \ge 0, \quad \forall x\ge 1$$
the function is increasing on $[1,\infty)$. As $f(5)\ge 0$, the result follows.
| {
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If $a,b,c \in R$ are distinct, then $-a^3-b^3-c^3+3abc \neq 0$. If $a,b,c \in R$ are distinct, then $-a^3-b^3-c^3+3abc \neq 0$.
I think it is trivial because they are distinct.
So I wonder just saying "Since they are distinct" is enough to prove it?
Of course there could be several more detailed versions but I just want to know that reasoning is true or not.
| HINT:
$$a^3+b^3+c^3-3abc$$
$$=(a+b)^3-3ab(a+b)+c^3-3abc=(a+b)^3+c^3-3ab(a+b+c)$$
$$=(a+b+c)\{(a+b)^2-(a+b)c+c^2\}-3ab(a+b+c)$$
$$=(a+b+c)\{(a+b)^2-(a+b)c+c^2-3ab\}$$
$$=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$$
$$=(a+b+c)\frac{\{(a-b)^2+(b-c)^2+(c-a)^2\}}2$$
which will be $>=<0 $ according as $a+b+c>=<0$ as for real distinct $a,b,c,$ each of $(a-b)^2,(b-c)^2,(c-a)^2>0$
| {
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Does $\int_{0}^{\infty} \frac{\sin (\tan x)}{x} \ dx $ converge? Does $ \displaystyle \int_{0}^{\infty} \ \frac{\sin (\tan x)}{x} dx $ converge?
$ \displaystyle \int_{0}^{\infty} \frac{\sin (\tan x)}{x} \ dx = \int_{0}^{\frac{\pi}{2}} \frac{\sin (\tan x)}{x} \ dx + \sum_{n=1}^{\infty} \int_{\pi(n-\frac{1}{2})}^{\pi(n+\frac{1}{2})} \frac{\sin (\tan x)}{x} \ dx $
The first integral converges since $\displaystyle \frac{\sin (\tan x)}{x}$ has a removable singularity at $x=0$ and is bounded near $ \displaystyle \frac{\pi}{2}$.
And $ \displaystyle \int_{\pi(n-\frac{1}{2})}^{\pi(n+\frac{1}{2})} \frac{\sin (\tan x)}{x} \ dx$ converges since $\displaystyle \frac{\sin (\tan x)}{x}$ is bounded near $\pi(n-\frac{1}{2})$ and $\pi(n+\frac{1}{2})$.
But does $ \displaystyle \sum_{n=1}^{\infty} \int_{\pi(n-\frac{1}{2})}^{\pi(n+\frac{1}{2})} \frac{\sin (\tan x)}{x} \ dx$ converge?
| Yes.
Note that
$$\begin{align} I_n:=\int_{\pi(n-\frac12)}^{\pi(n+\frac12)}\frac{\sin(\tan(x))}{x}\,\mathrm dx&=\int_0^{\frac\pi2}\left(\frac1{n\pi+x}-\frac1{n\pi-x}\right)\sin(\tan(x))\,\mathrm dx\\&=\int_0^{\frac\pi2}\frac{-2x}{n^2\pi^2-x^2}\sin(\tan(x))\,\mathrm dx\\
\end{align}$$
With $A:=\int_0^{\frac\pi2}\max\{-2x\sin(\tan (x)),0\}\,\mathrm dx$, $B:=\int_0^{\frac\pi2}\min\{-2x\sin(\tan (x)),0\}\,\mathrm dx$ (which both converge), we can thus estimate
$$ \frac{1}{n^2\pi^2-\pi^2/4}B+\frac{1}{n^2\pi^2-0}A\le I_n\le \frac{1}{n^2\pi^2-\pi^2/4}A+\frac{1}{n^2\pi^2-0}B,$$
The difference between these bounds and $\frac1{n^2\pi^2}(A+B)$ is governed by
$$\frac{1}{n^2\pi^2-\frac{\pi^2}{4}}-\frac1{n^2\pi^2}=\frac1{4\pi^2 n^4-\pi^2 n^2},$$
hence
$$ I_n=\frac1{n^2\pi^2}(A+B)+O(n^{-4}).$$
We conclude that $\sum I_n$ converges at least as good as $\sum\frac1{n^2}$.
| {
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Is it possible to simplify $\frac{\Gamma\left(\frac{1}{10}\right)}{\Gamma\left(\frac{2}{15}\right)\ \Gamma\left(\frac{7}{15}\right)}$? Is it possible to simplify this expression?
$$\frac{\displaystyle\Gamma\left(\frac{1}{10}\right)}{\displaystyle\Gamma\left(\frac{2}{15}\right)\ \Gamma\left(\frac{7}{15}\right)}$$
Is there a systematic way to check ratios of Gamma-functions like this for simplification possibility?
| Amazingly, this can be greatly simplified. I'll state the result first:
$$\frac{\displaystyle\Gamma\left(\frac{1}{10}\right)}{\displaystyle\Gamma\left(\frac{2}{15}\right)\Gamma\left(\frac{7}{15}\right)} = \frac{\sqrt{5}+1}{3^{1/10} 2^{6/5} \sqrt{\pi}}$$
The result follows first from a version of Gauss's multiplication formula:
$$\displaystyle\Gamma(3 z) = \frac{1}{2 \pi} 3^{3 z-1/2} \Gamma(z) \Gamma\left(z+\frac13\right) \Gamma\left(z+\frac{2}{3}\right)$$
or, with $z=2/15$:
$$\Gamma\left(\frac{2}{15}\right)\Gamma\left(\frac{7}{15}\right) = 2 \pi \,3^{1/10} \frac{\displaystyle\Gamma\left(\frac{2}{5}\right)}{\displaystyle\Gamma\left(\frac{4}{5}\right)}$$
Now use the duplication formula
$$\Gamma(2 z) = \frac{1}{\sqrt{\pi}}\, 2^{2 z-1} \Gamma(z) \Gamma\left(z+\frac12\right)$$
or, with $z=2/5$:
$$\frac{\displaystyle\Gamma\left(\frac{2}{5}\right)}{\displaystyle\Gamma\left(\frac{4}{5}\right)} = \frac{\sqrt{\pi} \, 2^{1/5}}{\displaystyle\Gamma\left(\frac{9}{10}\right)}$$
Putting this all together, we get
$$\frac{\displaystyle\Gamma\left(\frac{1}{10}\right)}{\displaystyle\Gamma\left(\frac{2}{15}\right)\Gamma\left(\frac{7}{15}\right)} = \frac{\displaystyle\Gamma\left(\frac{1}{10}\right) \Gamma\left(\frac{9}{10}\right)}{\sqrt{\pi^3} \, 2^{6/5} \, 3^{1/10}}$$
And now, we may use the reflection formula:
$$\Gamma(z) \Gamma(1-z) = \frac{\pi}{\sin{\pi z}}$$
With $z=1/10$, and noting that
$$\sin{\left(\frac{\pi}{10}\right)} = \frac{\sqrt{5}-1}{4} = \frac{1}{\sqrt{5}+1}$$
the stated result follows. This has been verified numerically in Wolfram|Alpha.
| {
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Last digits of a power of 2
Prove that there exists a power of 2 such that the last 1000 digits in its decimal representation are all 1 and 2.
One fact that I think can be used in this problem: if $2^{n}=\cdots dn$ where $d$ is the digit to the left of $n$, then $2^{dn}=\cdots dn$ (A concept that was used in MMO 1978).
Furthermore I have a feeling that reducing $n$ to its binary or ternary base may be of some help.
If one feels that the question is wrong then please mention why this is never possible.
Thanks!
| Let $n = 1000$. Since $\Bbb Z/10^n \Bbb Z = (\Bbb Z/2^n \Bbb Z)\times(\Bbb Z/5^n \Bbb Z)$, a first step is to inquire about the possible values of powers of $2$ modulo $2^n$ and $5^n$.
The first one is easy : If $k\ge n$, then $2^k \equiv 0 \pmod {2^n}$. And with enough luck, we may find a solution with $k \ge 1000$ so let's not think about $k<1000$ for now.
As for the second one, one can show that the order of $2$ is $4.5^{n-1}$ in the multiplicative group $(\Bbb Z/5^n \Bbb Z)^*$, and so $2$ generates that group :
First, we check that $2$ is of order $4$ modulo $5$, and that $2^4 = 3.5 + 1 \neq 1 \pmod {5^2}$.
Since $(2^4)^5 \equiv 3.5^2 + 1 \pmod {5^3} \equiv 1 \pmod {5^2}$, $2$ is of order $20$ modulo $5^2$. And so on, $(2^4)^{5^k} \equiv 3.5^{k+1} + 1 \pmod {5^{k+2}} \equiv 1 \pmod 5^{k+1}$, hence $2$ is of order $4.5^k$ modulo $5^{k+1}$.
This shows that the powers of $2$ modulo $5^n$ are exactly the invertible classes modulo $5^n$, i.e. those not divisible by $5$.
Finally we have to show that there is a class modulo $10^n$ containing only the digits $1$ and $2$ such that it is not divisible by $5$ (easy !), and it is divisible by $2^n$
In fact, we can show that forall $x \in \Bbb Z/2^n \Bbb Z$, there is a unique $y \in \Bbb Z/10^n \Bbb Z$ such that $y$ only uses the digits $1$ and $2$ and $y \equiv x \pmod {2^n}$:
This is true for $n=1$, if $x=1 \pmod {2}$ we pick $y=1 \pmod {10}$ and if $x=0 \pmod {2}$ we pick $y=2 \pmod {10}$.
Suppose $n>1$ and let $x' \in \Bbb Z/2^{n-1}\Bbb Z$. Using the induction hypothesis we have a unique $y' \in \Bbb Z/10^{n-1}\Bbb Z$ such that $y' \equiv x' \pmod {2^{n-1}}$. Above $y'$ there are two allowed classes mod $10^n$, $1y$ and $2y$. Since their difference is $10^{n-1} \equiv 2^{n-1} \pmod 2^n$, they are distinct and they give modulo $2^n$ the two classes above $x'$
Applying this proof we get a number $...1212122112$ with $1000$ digits which is divisible by $2^{1000}$ modulo $10^{1000}$, not divisible by $5$, and hence is a power of $2$ modulo $10^{1000}$.
| {
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Prove inequality: $74 - 37\sqrt 2 \le a+b+6(c+d) \le 74 +37\sqrt 2$ without calculus
Let $a,b,c,d \in \mathbb R$ such that $a^2 + b^2 + 1 = 2(a+b),
c^2 + d^2 + 6^2 = 12(c+d)$, prove inequality without calculus (or
langrange multiplier): $$74 - 37\sqrt 2 \le a+b+6(c+d) \le 74
+37\sqrt 2$$
The original problem is find max and min of $a+b+6(c+d)$ where ...
Using some calculus, I found it, but could you solve it without calculus.
| HINT:
So, $(a-1)^2+(b-1)^2=1^2$ and we can set $a=1+\cos A,b=1+\sin A$
So, $a+b=2+(\cos A+\sin A)=2+\sqrt2\cos\left(A-\frac\pi4\right)$
As $-1\le \cos\left(A-\frac\pi4\right)\le 1, 2-\sqrt2\le a+b\le 2+\sqrt2 $
Similarly, $(c-6)^2+(d-6)^2=6^2$ and we can set $c=6+6\cos B,d=6+6\sin B$
| {
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Evaluate : $\int^{\frac{\pi}{2}}_0 \frac{\cos^2x\,dx}{\cos^2x+4\sin^2x}$ Evaluate: $$\int^{\frac{\pi}{2}}_0 \frac{\cos^2x\,dx}{\cos^2x+4\sin^2x}$$
First approach :
$$\int^{\frac{\pi}{2}}_0 \frac{\cos^2x\,dx}{\cos^2x+4(1-\cos^2x)}$$
$$=\int^{\frac{\pi}{2}}_0 \frac{\cos^2xdx}{4 - 3\cos^2x}$$
$$=\int^{\frac{\pi}{2}}_0 \frac{1}{3}\{\frac{4-3\cos^2x-4}{4 - 3\cos^2x}\}\,dx$$
$$=\int^{\frac{\pi}{2}}_0 \frac{1}{3}\{ 1- \frac{4}{4 - 3\cos^2x}\}\,dx$$
$$=\int^{\frac{\pi}{2}}_0 \frac{1}{3} 1\,dx- \int^{\frac{\pi}{2}}_0 \frac{1}{3} \frac{4\sec^2x}{4\sec^2x - 3}\,dx$$
$$=\int^{\frac{\pi}{2}}_0 \frac{1}{3} 1\,dx- \int^{\frac{\pi}{2}}_0 \frac{1}{3} \frac{4
\sec^2x}{4(1+\tan^2x) - 3}\,dx$$
Now I can easily put $\tan x = t$ and I get $\sec^2x \,dx =dt$
Second approach :
$$\int^{\frac{\pi}{2}}_0 \frac{\cos^2x\,dx}{\cos^2x+4\sin^2x}$$
Dividing numerator and denominator by $\cos^2x$ we get :
$$=\int^{\frac{\pi}{2}}_0 \frac{dx}{1 +4\tan^2x}$$
$$=\int^{\frac{\pi}{2}}_0 \frac{dx}{(4)\{\frac{1}{4} +\tan^2x\}}$$
$$=\int^{\frac{\pi}{2}}_0 \frac{dx}{(4)\{\{\frac{1}{2}\}^2 +(\tan x)^2\}}$$
Can we apply this formula of integral here :
$$\int \frac{1}{a^2+x^2}dx = \frac{1}{a}\tan^{-1}\frac{x}{a}$$
I tried but its not working here, I think doing some manipulation we can implement this here.. Please suggest thanks...
| Use the fact that
$$3 \int_0^{\pi/2} dx \frac{\cos^2{x}}{\cos^2{x}+4 \sin^2{x}} = 3 \int_0^{\pi/2} dx \frac{\cos^2{x}}{4-3 \cos^2{x}} = 4 \int_0^{\pi/2} \frac{dx}{4-3 \cos^2{x}} - \int_0^{\pi/2} dx$$
So consider the first integral on the RHS:
$$4 \int_0^{\pi/2} \frac{dx}{4-3 \cos^2{x}} = 2 \int_0^{2 \pi} \frac{dy}{5-3 \cos{y}}$$
We may evaluate this latter integral using residue theory. Let $z=e^{i y}$, $dy = -i dz/z$ so that
$$2 \int_0^{2 \pi} \frac{dy}{5-3 \cos{y}} = -i 4 \oint_{|z|=1} \frac{dz}{10z-3(z^2+1)}$$
The denominator has zeroes at $z=1/3$ and $z=3$; of these, only $z=1/3$ lies within the unit circle. Thus the integral is
$$(i 2 \pi) \frac{-i 4}{10-6 (1/3)} = \pi$$
Then
$$\int_0^{\pi/2} dx \frac{\cos^2{x}}{\cos^2{x}+4 \sin^2{x}} = \frac{\pi-(\pi/2)}{3} = \frac{\pi}{6}$$
ADDENDUM
Alternatively, you may substitute $t=\tan{(y/2)}$, $dy = 2 dt/(1+t^2)$, $\cos{y} = (1-t^2)/(1+t^2)$, and then
$$2 \int_0^{2 \pi} \frac{dy}{5-3 \cos{y}} = \int_0^{\infty} \frac{dt}{(1/4)+t^2} = \frac{\pi}{2} 2 = \pi$$
| {
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Prove $x^2+y^2+z^2 \ge 14$ with constraints Let $0<x\le y \le z,\ z\ge 3,\ y+z \ge 5,\ x+y+z = 6.$ Prove the inequalities:
$I)\ x^2 + y^2 + z^2 \ge 14$
$II)\ \sqrt x + \sqrt y + \sqrt z \le 1 + \sqrt 2 + \sqrt 3$
My teacher said the method that can solve problem I can be use to solve problem II. But I don't know what method that my teacher talking about, so the hint is useless, please help me.
Thanks
| I) We have $2x+y=12-(y+z)-z \leq 12-5-3=4$. Thus by AM-GM inequality $2xy \leq (\frac{2x+y}{2})^2 \leq 4$. Finally by QM-AM inequality $$x^2+y^2+z^2=(x+y)^2+z^2-2xy \geq 2(\frac{(x+y)+z}{2})^2-2xy=18-2xy \geq 14$$
II) Let $s=\sqrt{x}+\sqrt{y}+\sqrt{z}$. Then $$s^2=x+y+z+2(\sqrt{xy}+\sqrt{xz}+\sqrt{yz})=6+2(\sqrt{xy}+\sqrt{xz}+\sqrt{yz})$$
$$(\frac{s^2-6}{2})^2=xy+xz+yz+2s\sqrt{xyz}$$
We have by I) $xy+xz+yz=\frac{(x+y+z)^2-(x^2+y^2+z^2)}{2}=\frac{36-(x^2+y^2+z^2)}{2} \leq 11$. Also, $6x+3y+2z=36-3(y+z)-z \leq 36-3(5)-3=18$ so by AM-GM inequality $36xyz \leq (\frac{6x+3y+2z}{3})^3 \leq 216$, so $xyz \leq 6$. Thus
$$(\frac{s^2-6}{2})^2 \leq 11+2\sqrt{6}s$$
$$s^4 \leq 12s^2+8+8\sqrt{6}s=12(s+\frac{\sqrt{6}}{3})^2$$
$$s^2 \leq 2\sqrt{3}(s+\frac{\sqrt{6}}{3})=2\sqrt{3}s+2\sqrt{2}$$
$$(s-\sqrt{3})^2 \leq 2\sqrt{2}+3=(1+\sqrt{2})^2$$
$$s \leq 1+\sqrt{2}+\sqrt{3}$$
| {
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Finding the root of a degree $5$ polynomial $\textbf{Question}$: which of the following $\textbf{cannot}$ be a root of a polynomial in $x$ of the form $9x^5+ax^3+b$, where $a$ and $b$ are integers?
A) $-9$
B) $-5$
C) $\dfrac{1}{4}$
D) $\dfrac{1}{3}$
E) $9$
I thought about this question for a bit now and can anyone provide any hints because I have no clue how to begin to eliminate the choices?
Thank you very much in advance.
| If you want to try process of elimination, an easy place to start is by assuming that either $a = 0$ or $b = 0$. (Hey, zero is an integer!)
If you set $a = 0$, then you get $b = -9x^5$. Because $\mathbb{Z}$ is closed under multiplication, if $x$ is an integer, then so is $b$, and so you have a valid $(a, b)$ pair with $x$ as a root. Thus, (A), (B), and (E) cannot be the right answer. However, for $x = \frac{1}{4}$ or $x = \frac{1}{3}$, you'd get $b = \frac{-9}{1024}$ or $b = \frac{-1}{27}$, respectively, so $b \notin \mathbb{Z}$, and these aren't valid solutions. So far, (C) and (D) are still possible answer choices.
If you set $b = 0$, then you get $9x^5 + ax^3 = 0$, which factors to $x^3(9x^2+a) = 0$, so either $x = 0$ (which isn't one of the answer choices) or $a = -9x^2$. If $x = \frac{1}{3}$, then $a = -1$, which is an integer, so that rules out answer choice (D). But if $x = \frac{1}{4}$, then $a = \frac{-9}{16}$, a non-integer, so (C) is still in the running.
To double-check that (C) is the correct answer, plug in $x = \frac{1}{4}$ into the original equation, to get $\frac{9}{1024} + \frac{1}{64}a + b = 0$. Moving the constant to the right and multiplying by 64 gives $a + 64b = \frac{-9}{16}$. Regardless of the specific values of $a$ and $b$, the left-hand side is an integer but the right-hand side is not. This is a contradiction, so $\frac{1}{4}$ cannot be a root of the polynomial if $a, b \in \mathbb{Z}$.
| {
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If $a, b$ are positive integers, does $\;b\mid(a^2 + 1)\implies b\mid (a^4 + 1)\quad?$ If $a, b$ are positive integers, does $\;b\mid(a^2 + 1)\implies b\mid (a^4 + 1)\;$?
Explain if this is true or not. If no, give a counterexample.
| Suppose $b = 5, a = 3$:
$$5\mid (3^2 + 1)\;\;\text{but}\;\;5 \not\mid (3^4 + 1)$$
Or, simply choose $a = 2, b= 5$ and again, $$5\mid (2^2 + 1)\;\;\text{but}\;\;5 \not\mid (2^4 + 1)$$
What is true is that $b\mid (a^2 + 1) \implies b\mid(a^2 + 1)^2$,
but note that $$\begin{align}\;(a^2 + 1)^2 &= (a^4 + 2a^2 + 1) \\ \\ &= \left[(a^4 + 1) + 2a^2\right]\\ \\ &\neq (a^4 + 1)\end{align}$$
| {
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Evaluate $\int_0^\infty \frac{\log(1+x^3)}{(1+x^2)^2}dx$ and $\int_0^\infty \frac{\log(1+x^4)}{(1+x^2)^2}dx$ Background: Evaluation of $\int_0^\infty \frac{\log(1+x^2)}{(1+x^2)^2}dx$
We can prove using the Beta-Function identity that
$$\int_0^\infty \frac{1}{(1+x^2)^\lambda}dx=\sqrt{\pi}\frac{\Gamma \left(\lambda-\frac{1}{2} \right)}{\Gamma(\lambda)} \quad \lambda>\frac{1}{2}$$
Differentiating the above equation with respect to $\lambda$, we obtain an expression involving the Digamma Function $\psi_0(z)$.
$$\int_0^\infty \frac{\log(1+x^2)}{(1+x^2)^\lambda}dx = \sqrt{\pi}\frac{\Gamma \left(\lambda-\frac{1}{2} \right)}{\Gamma(\lambda)} \left(\psi_0(\lambda)-\psi_0 \left( \lambda-\frac{1}{2}\right) \right)$$
Putting $\lambda=2$, we get
$$\int_0^\infty \frac{\log(1+x^2)}{(1+x^2)^2}dx = -\frac{\pi}{4}+\frac{\pi}{2}\log(2)$$
Question:
But, does anybody know how to evaluate $\displaystyle \int_0^\infty \frac{\log(1+x^3)}{(1+x^2)^2}dx$ and $\displaystyle \int_0^\infty \frac{\log(1+x^4)}{(1+x^2)^2}dx$?
Mathematica gives the values
*
*$\displaystyle \int_0^\infty \frac{\log(1+x^3)}{(1+x^2)^2}dx = -\frac{G}{6}+\pi \left(-\frac{3}{8}+\frac{1}{8}\log(2)+\frac{1}{3}\log \left(2+\sqrt{3} \right) \right)$
*$\displaystyle \int_0^\infty \frac{\log(1+x^4)}{(1+x^2)^2}dx = -\frac{\pi}{2}+\frac{\pi \log \left( 6+4\sqrt{2}\right)}{4}$
Here, $G$ denotes the Catalan's Constant.
Initially, my approach was to find closed forms for
$$\int_0^\infty \frac{1}{(1+x^2)^2(1+x^3)^\lambda}dx \ \ , \int_0^\infty \frac{1}{(1+x^2)^2(1+x^4)^\lambda}dx$$
and then differentiate them with respect to $\lambda$ but it didn't prove to be of any help.
Please help me prove these two results.
| Another approach for evaluating the second integral using contour integration that avoids having to deform the contour around branch cuts is to consider $$ \displaystyle f(z) = \frac{\log(z+ e^{i \pi /4})}{(1+z^{2})^{2}}$$ and integrate around a contour that consists of the line segment $[-R,R]$ and the upper half of $|z|=R$.
Then letting $R \to \infty$,
$$ \begin{align} &\int_{-\infty}^{0} \frac{\log(x+e^{i \pi /4})}{(1+x^{2})^{2}} \ dx + \int_{0}^{\infty} \frac{\log(x+e^{i \pi /4})}{(1+x^{2})^{2}} \ dx \\ &= 2 \pi i \ \text{Res}[f(z),i] \\ &= 2 \pi i \lim_{z \to i} \frac{d}{dz} \frac{\log(z+e^{i \pi /4})}{(z+i)^{2}} \\ &=2 \pi i \lim_{z \to i} \left(\frac{1}{(z+e^{i \pi /4})(z+i)^{2}} - \frac{2 \log(z+e^{i \pi /4})}{(z+i)^{3}} \right) \\ &= 2 \pi i \left(- \frac{1}{4} \frac{\sqrt{2}}{1+i(1+\sqrt{2})} + \frac{\log|i+e^{i \pi /4}| + i \arg (i +e^{i \pi/4}) }{4i}\right)\\ &= 2 \pi i \left(\frac{1-\sqrt{2}+i}{8} + \frac{\frac{1}{2} \log (2+\sqrt{2})+ i \frac{3 \pi}{8}}{4i} \right) \\ &= \frac{\pi}{4} \Big(\log(2+\sqrt{2})-1 \Big) + \frac{i\pi}{4} \left(1-\sqrt{2}+\frac{3 \pi}{4} \right) . \end{align}$$
But notice that $$ \begin{align} &\text{Re} \left( \int_{-\infty}^{0} \frac{\log(z+e^{i \pi /4})}{(1+x^{2})^{2}} \ dx + \int_{0}^{\infty} \frac{\log(z+e^{i \pi /4})}{(1+x^{2})^{2}} \ dx \right) \\ &= \text{Re} \left(\int_{0}^{\infty} \frac{\log(-u + e^{i \pi /4})}{(1+u^{2})^{2}} \ du + \int_{0}^{\infty} \frac{\log(z+e^{i \pi /4})}{(1+x^{2})^{2}} \ dx \right) \\ &= \int_{0}^{\infty} \frac{\log|-u + e^{i \pi /4}|}{(1+u^{2})^{2}} \ du + \int_{0}^{\infty} \frac{\log |x+e^{i \pi /4}|}{(1+x^{2})^{2}} \ dx\\ &= \int_{0}^{\infty} \frac{\frac{1}{2} \log(x^{2}-\sqrt{2}x+1) + \frac{1}{2} \log(x^{2}+\sqrt{2}x+1)}{(1+x^{2})^{2}} \ dx \\ &= \frac{1}{2} \int_{0}^{\infty} \frac{\log(1+x^{4})}{(1+x^{2})^{2}} \ dx. \end{align}$$
Therefore,
$$ \begin{align} \int_{0}^{\infty} \frac{\log(1+x^{4})}{(1+x^{2})^{2}} \ dx &= \frac{\pi}{2} \Big(\log(2+\sqrt{2}) -1 \Big) \\ &= \frac{\pi}{2} \Big(\frac{1}{2} \log \big((2+\sqrt{2})^{2} \big) -1 \Big) \\ &= \frac{\pi}{2} \Big(\frac{\log(6+4\sqrt{2})}{2} -1\Big) \\ &= - \frac{\pi}{2} + \frac{\pi \log(6+4\sqrt{2})}{4}. \end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/414642",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "31",
"answer_count": 6,
"answer_id": 1
} |
Show that $\frac{x}{1-x}\cdot\frac{y}{1-y}\cdot\frac{z}{1-z} \ge 8$.
If $x,y,z$ are positive proper fractions satisfying $x+y+z=2$, prove that $$\dfrac{x}{1-x}\cdot\dfrac{y}{1-y}\cdot\dfrac{z}{1-z}\ge 8$$
Applying $GM \ge HM$, I get $$\left[\dfrac{x}{1-x}\cdot\dfrac{y}{1-y}\cdot\dfrac{z}{1-z}\right]^{1/3}\ge \dfrac{3}{\frac 1x-1+\frac 1y-1+\frac 1z-1}\\=\dfrac{3}{\frac 1x+\frac 1y+\frac 1z-3}$$
Then how to proceed. Please help.
| Two proofs of the inequality have been posted. The following is simply a comment. In the attempted solution of the OP, the inequality
$$\left(\frac{x}{1-x}\cdot\frac{y}{1-y}\cdot \frac{z}{1-z}\right)^{1/3}\ge \frac{3}{\frac{1}{x}+\frac{1}{y}+\frac{1}{z}-3}\tag{1}$$
has been proved.
One can complete things by showing, under the hypothesis $x+y+z=2$, that the right-hand side of (1) is $\ge 2$.
However, that is not true. For instance, take $x=\frac{5}{6}$, $y=\frac{5}{6}$, and $z=\frac{2}{6}$. Then
$$ \frac{3}{\frac{1}{x}+\frac{1}{y}+\frac{1}{z}-3}=\frac{18}{10}\lt 2.$$
So too much has been given away in producing (1): there is probably no direct path from it to the desired result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/424529",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 4,
"answer_id": 1
} |
Show that $(1+x)(1+y)(1+z)\ge 8(1-x)(1-y)(1-z)$.
If $x>0,y>0,z>0$ and $x+y+z=1$, prove that $$(1+x)(1+y)(1+z)\ge 8(1-x)(1-y)(1-z).$$
Trial: Here $$(1+x)(1+y)(1+z)\ge 8(1-x)(1-y)(1-z) \\ \implies (1+x)(1+y)(1+z)\ge 8(y+z)(x+z)(x+y)$$ I am unable to solve the problem. Please help.
| since
$$t=(x+y)(y+z)(x+z)\le\left(\dfrac{2(x+y+z)}{3}\right)^3=\dfrac{8}{27}$$
and we
have
$$(1-x)(1-y)\ge 0$$
then
$$(1+x)(1+y)\ge2(x+y)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/425134",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Evaluating $\int_0^\infty \frac{dx}{1+x^4}$. Can anyone give me a hint to evaluate this integral?
$$\int_0^\infty \frac{dx}{1+x^4}$$
I know it will involve the gamma function, but how?
| HINT:
Putting $x=\frac1y,dx=-\frac{dy}{y^2}$
$$I=\int_0^\infty\frac{dx}{1+x^4}=\int_\infty^0\frac{-dy}{y^2\left(1+\frac1{y^4}\right)}$$
$$=-\int_\infty^0\frac{y^2dy}{1+y^4}=\int_0^\infty\frac{y^2dy}{1+y^4} \text{ as } \int_a^bf(x)dx=-\int_b^af(x)dx$$
$$I=\int_0^\infty\frac{y^2dy}{1+y^4}=\int_0^\infty\frac{x^2dx}{1+x^4}$$
$$\implies 2I=\int_0^\infty\frac{dx}{1+x^4}+\int_0^\infty\frac{x^2dx}{1+x^4}=\int_0^\infty\frac{1+x^2}{1+x^4}dx=\int_0^\infty\frac{\frac1{x^2}+1}{\frac1{x^2}+x^2} dx$$
Now the idea is to express the denominator as a polynomial of $\int \left(\frac1{x^2}+1 \right)dx =x-\frac1x=u\text{(say)}$
$$\text{The denominator =}\frac1{x^2}+x^2=\left(x-\frac1x\right)^2+2=u^2+2$$
Now, complete the definite integral with $u$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/426152",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 7,
"answer_id": 6
} |
Need help in proving that $\frac{\sin\theta - \cos\theta + 1}{\sin\theta + \cos\theta - 1} = \frac 1{\sec\theta - \tan\theta}$ We need to prove that $$\dfrac{\sin\theta - \cos\theta + 1}{\sin\theta + \cos\theta - 1} = \frac 1{\sec\theta - \tan\theta}$$
I have tried and it gets confusing.
| multiply numerator and denominator by $(1-\sin\theta)$.
$$\dfrac {\sin\theta-\cos\theta+1}{\sin\theta+\cos\theta-1}\times \dfrac{1-\sin\theta}{1-\sin\theta}$$
$$\dfrac {\sin\theta-\sin^2\theta-\cos\theta+\sin\theta\cdot\cos\theta+1-\sin\theta}{(\sin\theta+\cos\theta-1)(1-\sin\theta)}$$
$$\dfrac {1-\sin^2\theta-\cos\theta+\sin\theta\cdot\cos\theta}{(\sin\theta+\cos\theta-1)(1-\sin\theta)}$$
$$\dfrac {\cos^2\theta-\cos\theta+\sin\theta\cdot\cos\theta}{(\sin\theta+\cos\theta-1)(1-\sin\theta)}$$
$$\dfrac {\cos\theta(\cos\theta+\sin\theta-1)}{(\sin\theta+\cos\theta-1)(1-\sin\theta)}$$
$$\dfrac {\cos\theta}{1-\sin\theta}$$
$$\dfrac {1}{\dfrac{1-\sin\theta}{\cos\theta}}$$
$$\dfrac {1}{\dfrac{1}{\cos\theta}-\dfrac{\sin\theta}{{\cos\theta}}}$$
$$\dfrac {1}{\sec\theta-\tan\theta}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/426981",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 4
} |
Lagrange multiplier method, find maximum of $e^{-x}\cdot (x^2-3)\cdot (y^2-3)$ on a circle I attempted to design an exercise for my engineer students and couldn't solve it myself. Maybe here are some experts in calculus who have some better tricks than I do:
The exercise would be to find the maxima of $e^{-x}(x^2-3)(y^2-3)$ on the circle $x^2+(y-1)^2=4$.
Now using the Lagrange multiplier method this amounts to solving the following system of equations:
$$\begin{align*}
e^{-x}(y^2-3)(-x^2+2x+3)+\lambda\cdot 2x&=0\\
e^{-x}(x^2-3)(2y)+\lambda\cdot 2(y-1)&=0\\
x^2+(y-1)^2&=4
\end{align*}$$
I did not succeed to find the solutions and also my standard online calculor didn't.
Now I thought, this is partly because of the $e^{-x}$-term, so it would be good if one could eliminate it. Noting that $x^2+(y-1)^2-4=0$ iff $e^{-x}\cdot (x^2+(y-1)^2-4)=0$ we can instead use Lagrange multipliers on this condition. This amounts to solving the easier system:
$$\begin{align*}
(y^2-3)(-x^2+2x+3)+\lambda(-x^2+2x-(y-1)^2+4)&=0\\
(x^2-3)y+\lambda(y-1)&=0\\
x^2+(y-1)^2-4&=0
\end{align*}$$
In fact I could still not solve it, but the computer could (but the form is not very nice).
So the question is now two-fold:
*
*Do you have any ideas how to solve either of the systems?
*If not, do you have any ideas how to tweak it a little bit (preferrable on the circle condition side) so that it becomes easier to solve?
| Let us parametrize the circumfererence:$x=2\cos(\phi),y=1+2\sin(\phi)$. Then the function
$$f(\phi):=e^{-2\cos(\phi)}(4\cos^2(\phi)-3)(4\sin^2(\phi)+4\sin(\phi)-2)$$ should be maximized on $[0,2\pi]$.
The equation $f'(\phi)=0$ is equivalent to
$$
32 \sin^3 ( \phi ) \cos ^2(
\phi ) -32\cos ( \phi ) \sin^3
\left( \phi \right) +
$$
$$
32 \cos^2 ( \phi ) \sin ^2( \phi) +
32 \cos ^3( \phi ) \sin ( \phi
) -24 \sin^3 ( \phi ) -
$$
$$
32\,\cos
\left( \phi \right) \sin^2 \left( \phi \right) -16
\, \cos^2 \left( \phi \right) \sin \left( \phi
\right) +
$$
$$
16\, \cos ^3\left( \phi \right) -24\,
\sin^2 \left( \phi \right) -
$$
$$
8\,\cos \left( \phi
\right) \sin \left( \phi \right) +12\,\sin \left( \phi \right) -12\,
\cos \left( \phi \right) =0.
$$
Its roots are expressed in terms of polynomials of higher degrees (One of these equals 10.).
The numerical solution with Maple produces $\max_{\phi \in [0,2\pi]}f(\phi)=13.75012514$ at $\phi=4.105026133$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/429939",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 0
} |
Is there an inverse to Stirling's approximation? The factorial function cannot have an inverse, $0!$ and $1!$ having the same value. However, Stirling's approximation of the factorial $x! \sim x^xe^{-x}\sqrt{2\pi x}$ does not have this problem, and could provide a ballpark inverse to the factorial function. But can this actually be derived, and if so how? Here is my work:
$$
\begin{align}
y &= x^xe^{-x}\sqrt{2\pi x}\\
y^2 &= 2\pi x^{2x + 1}e^{-2x}\\
\frac{y^2}{2\pi} &= x^{2x + 1}e^{-2x}\\
\ln \frac{y^2}{2\pi} &= (2x + 1)\ln x - 2x\\
\ln \frac{y^2}{2\pi} &= 2x\ln x + \ln x - 2x\\
\ln \frac{y^2}{2\pi} &= 2x(\ln x - 1) + \ln x
\end{align}
$$
That is as far as I can go. I suspect the solution may require the Lambert W function.
Edit: I have just realized that after step 3 above, one can divide both sides by e to get
$$\left(\frac{x}{e}\right)^{2x + 1} = \frac{y^2}{2e\pi}$$
Can this be solved?
| It is better to start with
$$
x! \approx \left( {\frac{{x + \frac{1}{2}}}{\mathrm{e}}} \right)^{x + \frac{1}{2}} \sqrt {2\pi } .
$$
Setting
$$
y = \left( {\frac{{x + \frac{1}{2}}}{\mathrm{e}}} \right)^{x + \frac{1}{2}} \sqrt {2\pi } ,
$$
we find
$$
\frac{1}{\mathrm{e}}\log \left( {\frac{y}{{\sqrt {2\pi } }}} \right) = \frac{{x + \frac{1}{2}}}{\mathrm{e}}\log \left( {\frac{{x + \frac{1}{2}}}{\mathrm{e}}} \right) = \exp \left( {\log \left( {\frac{{x + \frac{1}{2}}}{\mathrm{e}}} \right)} \right)\log \left( {\frac{{x + \frac{1}{2}}}{\mathrm{e}}} \right).
$$
We use the Lambert $W$-function defined by $W(z)\mathrm{e}^{W(z)} = z$ for $z>0$. From the above we see that
$$
\log \left( {\frac{{x + \frac{1}{2}}}{\mathrm{e}}} \right) = W\!\left( {\frac{1}{\mathrm{e}}\log \left( {\frac{y}{{\sqrt {2\pi } }}} \right)} \right).
$$
Whence
$$
\frac{{x + \frac{1}{2}}}{\mathrm{e}} = \exp \left( {W\!\left( {\frac{1}{\mathrm{e}}\log \left( {\frac{y}{{\sqrt {2\pi } }}} \right)} \right)} \right) = \frac{{\frac{1}{\mathrm{e}}\log \left( {\frac{y}{{\sqrt {2\pi } }}} \right)}}{{W\!\left( {\frac{1}{\mathrm{e}}\log \left( {\frac{y}{{\sqrt {2\pi } }}} \right)} \right)}},
$$
i.e.,
$$
x = \frac{{\log \left( {\frac{y}{{\sqrt {2\pi } }}} \right)}}{{W\!\left( {\frac{1}{\mathrm{e}}\log \left( {\frac{y}{{\sqrt {2\pi } }}} \right)} \right)}} - \frac{1}{2}.
$$
For example, if $y=720$ then
$$
x = \frac{{\log \left( {\frac{{720}}{{\sqrt {2\pi } }}} \right)}}{{W\!\left( {\frac{1}{\mathrm{e}}\log \left( {\frac{{720}}{{\sqrt {2\pi } }}} \right)} \right)}} - \frac{1}{2} \approx 5.99658,
$$
which is a very good approximation.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/430167",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 3,
"answer_id": 1
} |
Calculating 7^7^7^7^7^7^7 mod 100 What is
$$\large 7^{7^{7^{7^{7^{7^7}}}}} \pmod{100}$$
I'm not much of a number theorist and I saw this mentioned on the internet somewhere. Should be doable by hand.
| $7^4 = 2401 \equiv 1 \pmod{100}$, so you only need to calculate $7^{7^{7^{7^{7^7}}}} \pmod{4}$. We know that $7 \equiv -1 \pmod 4$ and $7^{7^{7^{7^7}}}$ is odd, so $7^{7^{7^{7^{7^7}}}} \equiv -1 \equiv 3 \pmod 4$, and then $$7^{7^{7^{7^{7^{7^7}}}}} \equiv 7^3 \equiv 43 \pmod {100}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/430633",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 3,
"answer_id": 1
} |
Induction and convergence of an inequality: $\frac{1\cdot3\cdot5\cdots(2n-1)}{2\cdot4\cdot6\cdots(2n)}\leq \frac{1}{\sqrt{2n+1}}$ Problem statement:
Prove that $\frac{1*3*5*...*(2n-1)}{2*4*6*...(2n)}\leq \frac{1}{\sqrt{2n+1}}$ and that there exists a limit when $n \to \infty $.
, $n\in \mathbb{N}$
My progress
LHS is equivalent to $\frac{(2n-1)!}{(2n)!}=\frac{(2n-1)(2n-2)(2n-3)\cdot ....}{(2n)(2n-1)(2n-2)\cdot ....}=\frac{1}{2n}$ So we can rewrite our inequality as:
$\frac{1}{2n}\leq \frac{1}{\sqrt{2n+1}}$ Let's use induction:
For $n=1$ it is obviously true. Assume $n=k$ is correct and show that $n=k+1$ holds.
$\frac{1}{2k+2}\leq \frac{1}{\sqrt{2k+3}}\Leftrightarrow 2k+2\geq\sqrt{2k+3}\Leftrightarrow 4(k+\frac{3}{4})^2-\frac{5}{4}$ after squaring and completing the square. And this does not hold for all $n$
About convergence: Is it not enough to check that $\lim_{n \to \infty}\frac{1}{2n}=\infty$ and conclude that it does not converge?
| About the original attempt:
Your first "rewrite" (immediately above (1))is incorrect:
$$\frac{1}{\sqrt{3n+1}} \times \frac{2(n+1)-1}{2(n+1)} \leq \frac{1}{\sqrt{3(n+1)+1}}$$
is NOT equivalent to, nor does it entail the rewrite I referenced:
$$\underbrace{\frac{\sqrt{3n+1}}{3n+1}}_{\large =\,\frac 1{\sqrt{3n+1}}} \times \frac{2(n+1)-1}{2(n+1)} \leq \frac{\sqrt{3n+1}}{\sqrt{3(n+1)+1}}$$
I presume you meant to multiply both sides of the inequality with $\sqrt{3n+1}$, which is valid, but note that in doing so, $\sqrt{3n+1}\cdot \dfrac 1{\sqrt{3n+1}} = 1 \neq \dfrac {\sqrt{3n+1}}{3n+1}$.
Suggestion on your revised work:
You've done fine so far. So we can pick up from $(3)$:
Rewrite $(3)$ as follows: $$\sqrt{3n+4}(2n+1) \leq (2n+2)\sqrt{3n+1} \iff \sqrt{(3n+4)(2n+1)^2} \leq \sqrt{(2n+2)^2(3n+1)}$$ Expand the factors under each square root sign, and unless I made a mistake, you should end up with the true statement:
$$\sqrt{12n^3 + 28n^2 + \color{blue}{\bf 19n} +4}\leq \sqrt{12n^3 + 28n^2 + \color{blue}{\bf 20n} + 4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/431234",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Probability Question about Tennis Games! $2^{n}$ players enter a single elimination tennis tournament. You can assume that the players are of equal ability.
Find the probability that two particular players meet each other in the tournament.
I could't make a serious attempt on the question, hope you can excuse me this time.
| First off, an even number of players is a necessary but not sufficient condition for a full balanced tournament. Fir this you need $P=2^R$ where $R$ is the number of rounds (including finals, semis etc). I will assume this is what you are after.
Without loss of generality, let Player 1 occupy the first slot in the draw.
For the players to meet in the first round Player 2 must be in Slot 2, for round 2 - slot 3 or 4, round 3 - slot 5 to 8, etc. In general, for the players to meet in round $r\in 1,\dots,R$, there are $2^{r-1}$ slots available. Given that Player 2 is randomly allocated to one of $2^R-1$ slots, the probability of a potential meeting in round $r$ is
$$S(r)=\frac{2^{r-1}}{2^R-1}$$
The probability that each player makes round $r$ is $\left(\frac{1}{2}\right)^{r-1}$.
The probability that both do is therefore $\left(\frac{1}{2}\right)^{2(r-1)}$
So, the probability that they meet in round $r$ is
$$\begin{align}
M(r)&=\left(\frac{1}{2}\right)^{2(r-1)}\frac{2^{r-1}}{(2^R-1)}\\
&=\left(\frac{1}{2}\right)^{(r-1)}\frac{1}{(2^R-1)}
\end{align}$$
And that they meet in any round is
$$\begin{align}
M&=\sum_{r=1}^R \left(\frac{1}{2}\right)^{r-1} \frac{1}{(2^R-1)}\\
&=\sum_{r=0}^{R-1} \left(\frac{1}{2}\right)^{r} \frac{1}{(2^R-1)}\\
&=\frac{1-\left(\frac{1}{2}\right)^R}{1-\frac{1}{2}}\frac{1}{(2^R-1)}\\
&=2\frac{2^R-1}{2^R}\frac{1}{(2^R-1)}\\
&=2^{1-R}\\
\end{align}$$
Sanity check!
For 2 players $R=1, p=1$; 4 players $R=2, p=\frac{1}{2}=\frac{1}{3}\text{Rd1}+\frac{1}{6}\text{Rd2}$; 8 players $R=2, p=\frac{1}{4}=\frac{1}{7}\text{Rd1}+\frac{1}{14}\text{Rd2}+\frac{1}{28}\text{Rd3}$ - Sanity check passed.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/433430",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 2
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Find the determinant by using elementary row operations I'm having a problem finding the determinant of the following matrix using elementary row operations. I know the determinant is -15 but confused on how to do it using the elementary row operations.
Here is the matrix
$$\begin{bmatrix} 2 & 3 & 10 \\ 1 & 2 & -2 \\ 1 & 1 & -3 \end{bmatrix}$$
Thank you
| Note that the determinant of a lower (or upper) triangular matrix is the product of its diagonal elements. Using this fact, we want to create a triangular matrix out of your matrix
\begin{bmatrix} 2 & 3 & 10 \\ 1 & 2 & -2 \\ 1 & 1 & -3 \end{bmatrix}
So, I will start with the last row and subtract it from the second row to get
\begin{bmatrix} 2 & 3 & 10 \\ 0 & 1 & 1 \\ 1 & 1 & -3 \end{bmatrix}
Now, I want to get rid of the $2$ in the first row. I thus multiply the last row by $2$ and subtract it from the first row to obtain:
\begin{bmatrix} 0 & 1 & 16 \\ 0 & 1 & 1 \\ 1 & 1 & -3 \end{bmatrix}
Finally, I subtract the second row from the first one to obtain
\begin{bmatrix} 0 & 0 & 15 \\ 0 & 1 & 1 \\ 1 & 1 & -3 \end{bmatrix}
We now have
$$\det \begin{bmatrix} 2 & 3 & 10 \\ 1 & 2 & -2 \\ 1 & 1 & -3 \end{bmatrix} = \det \begin{bmatrix} 0 & 0 & 15 \\ 0 & 1 & 1 \\ 1 & 1 & -3 \end{bmatrix}$$
Now, I will transform the RHS matrix to an upper diagonal matrix. I can exchange the first and the last rows. Exchanging any two rows changes the sign of the determinant, and therefore
$$\det \begin{bmatrix} 2 & 3 & 10 \\ 1 & 2 & -2 \\ 1 & 1 & -3 \end{bmatrix} = -\det \begin{bmatrix} 1 & 1 & -3 \\ 0 & 1 & 1 \\ 0 & 0 & 15 \end{bmatrix}$$
The matrix on the RHS is now an upper triangular matrix and its determinant is the product of its diagonal elements, which is $15$. With the minus sign, the $\det$ of our initial matrix is thus $-15$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/433870",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 0
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Finding Potential with d dimensions terms Lagrangian for a spherically-symmetric, real scalar field in
d spatial dimensions,
$$L=c_d \int r^{d-1}dr\left[ \frac{1}{2} \dot\phi^2 - \frac{1}{2} \left(\frac{\partial \phi}{\partial r} \right)^2 -V(\phi)\right] \tag{1}$$
where $$v= m^2\phi^2$$, $$c_d = 2π^{d/2} /Γ(d/2)$$ is the unit-
sphere volume in d dimensions.
The solution of $\phi$ is,
$$\phi(r,t) = A(t)P(r,R)= A(t)e^{\frac{−r^2}
{R2}}\tag{2}$$
A solution for a scalar field in d dimensions
$$\phi(r,t)= A(t) e ^\frac{-r^2}{R^2}$$
If we define a potential,
$$V = \phi^2- \phi^3+ \frac{\phi^4}{4}$$
then how do we get by integrating the above equation on the space d dimensions? $$V(A)= (1+\frac{d}{2R^2})A^2-\left(\frac{2}{3}\right)^\frac{d}{2} A^3+ \frac{A^4}{2^\frac{d+4}{2}}$$
| I think I have understood what you want to achieve. You want to obtain a Lagrangian for $A(t)$ with a suitable potential. Using $\int\limits_{0}^{\infty}r^{d-1}e^{-ar^{2}} dr=\frac{1}{2} a^{-\frac{d}{2}}\Gamma\left( \frac{d}{2} \right)$ we find by inserting $\phi(r,t)=A(t)e^{-\frac{r^{2}}{R^{2}}}$ that
$$ \begin{aligned} L(A) &= c_{d} \int\limits_{0}^{\infty} r^{d-1} \left( \frac{1}{2} \dot{A}^{2} \exp\left(-\frac{2 r^{2}}{R^{2}}\right) - \frac{1}{2}A^{2} \left(\frac{4r^{2}}{R^{4}}\right)\exp\left(-\frac{2 r^{2}}{R^{2}}\right) - A^{2} \exp\left(-\frac{2 r^{2}}{R^{2}}\right) + A^{3} \exp\left(-\frac{3 r^{2}}{R^{2}}\right) - \frac{1}{4} A^{4} \exp\left(-\frac{4 r^{2}}{R^{2}}\right) \right) dr \\
&= \frac{2\pi^{d/2}}{\Gamma\left( \frac{d}{2} \right)} \left( \dot{A}^{2} \frac{1}{4} \left( \frac{R^{2}}{2} \right)^{\frac{d}{2}}\Gamma\left( \frac{d}{2} \right) -A^{2} \frac{1}{R^{4}} \left( \frac{R^{2}}{2} \right)^{\frac{d+2}{2}} \underbrace{\Gamma\left( \frac{d+2}{2} \right)}_{=\frac{d}{2}\Gamma\left( \frac{d}{2} \right)} - A^{2} \frac{1}{2} \left( \frac{R^{2}}{2} \right)^{\frac{d}{2}}\Gamma\left( \frac{d}{2} \right) + A^{3} \frac{1}{2} \left( \frac{R^{2}}{3} \right)^{\frac{d}{2}}\Gamma\left( \frac{d}{2} \right) -A^{4} \frac{1}{8} \left( \frac{R^{2}}{4} \right)^{\frac{d}{2}}\Gamma\left( \frac{d}{2} \right) \right) \\
&= \pi^{d/2} R^{d} \left( \dot{A}^{2} \frac{1}{2} \left( \frac{1}{2} \right)^{\frac{d}{2}} -A^{2} \frac{1}{ R^{2}} \left( \frac{1}{2} \right)^{\frac{d+2}{2}} d - A^{2} \left( \frac{1}{2} \right)^{\frac{d}{2}}\ + A^{3} \left( \frac{1}{3} \right)^{\frac{d}{2}} -A^{4} \frac{1}{4} \left( \frac{1}{4} \right)^{\frac{d}{2}} \right) \\ &= \left( \frac{\pi R^{2}}{2} \right)^{\frac{d}{2}} \left( \frac{1}{2} \dot{A}^{2} - \left(1+ \frac{d}{2 R^{2}} \right) A^{2} + A^{3} \left( \frac{2}{3} \right)^{\frac{d}{2}} -A^{4} \left( \frac{1}{2} \right)^{\frac{d+4}{2}} \right). \end{aligned}$$
Dropping the overall constant $\left( \frac{\pi R^{2}}{2} \right)^{\frac{d}{2}}$ one thus has the potential $V(A)=\left(1+ \frac{d}{2 R^{2}} \right) A^{2} - A^{3} \left( \frac{2}{3} \right)^{\frac{d}{2}} +A^{4} \left( \frac{1}{2} \right)^{\frac{d+4}{2}}$ which coincides with the expression you were looking for.
Remark: You mentioned a mass term $v=m^{2}\phi^{2}$ which somehowdoes not appear in your Lagrangian. I also think you might want to add certain coupling constants to adjust the units in $V(\phi)$. However, those manipulations are easily achieved.
| {
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"url": "https://math.stackexchange.com/questions/434298",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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simple limit but I forget how to prove it I have to calculate the following limit
$$\lim_{x\rightarrow -\infty} \sqrt{x^2+2x+2} - x$$
it is in un undeterminated form.
I tried to rewrite it as follows:
$$\lim_{x\rightarrow -\infty} \sqrt{x^2+2x+2} - \sqrt{|x|^2}$$
but seems a dead road.
Can anyone suggest a solution?
thanks for your help
| Assuming you meant $\sqrt{x^2+2x+2} + x$ (as $\sqrt{x^2+2x+2} - x \to +\infty$ when $x\to-\infty$):
Another option would be to use asymptotics and known Taylor expansions (at $0$): for $x\to-\infty$,
$$
\begin{align*}
\sqrt{x^2+2x+2} + x &= |x|\sqrt{1+\frac{2}{x}+\frac{2}{x^2}} - |x| \\
&= |x|\left( 1+\frac{1}{2}\cdot\frac{2}{x} + o\left(\frac{1}{x}\right) - 1 \right) \\
&= \frac{|x|}{x} + o(1) \sim -1
\end{align*}
$$
so the limit is $-1$.
| {
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"timestamp": "2023-03-29T00:00:00",
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The golden ratio and a right triangle Assume the square of the hypotenuse of a right triangle is equal to its perimeter and one of its legs is $1$ plus its inradius(the radius inside the circle inscribed inside the triangle.) Find an expression for the hypotenuse $c$ in terms of the golden ratio.
| Draw a picture. Let $r$ be the radius of the incircle, and let the legs be $r+x$ and $r+y$. We are told that one of the legs, say $r+x$, is equal to $r+1$, so $x=1$. Thus the hypotenuse is $1+y$.
The condition that the square of the hypotenuse is equal to the perimeter says that
$(1+y)^2=2r+2y+2$, which simplifes to $y^2=2r+1$.
The Pythagorean Theorem says that
$$(1+y)^2=(1+r)^2+(r+y)^2,$$
which simplifies to
$$y(1-r)=r^2+r.$$
Substitute $\sqrt{2r+1}$ for $y$, and square both sides. We get
$$(1-r)^2(2r+1)=(r^2+r)^2,$$
which miraculously simplifies to
$$r^4+4r^2-1=0.$$
Solve. We get $r^2=\sqrt{5}-2$, and it's over.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Partial fraction integration $\int \frac{dx}{(x-1)^2 (x-2)^2}$ $$\int \frac{dx}{(x-1)^2 (x-2)^2} = \int \frac{A}{x-1}+\frac{B}{(x-1)^2}+\frac{C}{x-2}+\frac{D}{(x-2)^2}\,dx$$
I use the cover up method to find that B = 1 and so is C. From here I know that the cover up method won't really work and I have to plug in values for x but that won't really work either because I have two unknowns. How do I use the coverup method?
| We can use the following method
$$\frac1{(x-1)^2(x-2)^2}=\frac{\{(x-1)-(x-2)\}^2}{(x-1)^2(x-2)^2}=\frac1{(x-2)^2}+\frac1{(x-1)^2}-2\frac1{(x-1)(x-2)}$$
$$\frac1{(x-1)(x-2)}=\frac{(x-1)-(x-2)}{(x-1)(x-2)}=\frac1{x-2}-\frac1{x-1}$$
Alternatively,
Put $x-2=y$ to ease of calculation
$$1=A(y+1)y^2+By^2+C(y+1)^2y+D(y+1)^2$$
$$\implies 1=D+y(C+2D)+y^2(A+B+2C+2D)+y^3(A+C)$$
As this is an identity, we can compare the coefficients of the different powers of $y$
Comparing the coefficients of $y^0,D=1$
Comparing the coefficients of $y,C+2D=0\implies C=-2D=-2$
Comparing the coefficients of $y^3,A+C=0\implies A=-C=2$
Comparing the coefficients of $y^2,A+B+2C+2D=0$
May I leave this for you to find $B?$
| {
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"timestamp": "2023-03-29T00:00:00",
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Show that $4x^2+3$ has at least a prime divisor of the form $12n+7$ If $x$ is not divisible by $3$, how to prove that $4x^2+3$ has at least a prime divisor of the form $12n+7$?
Thanks.
| *
*Fact $0$: A number of the form $4x^2 + 3$ is odd, hence all its prime divisors must necessarily be odd.
*Fact $1$: If $x$ is not a multiple of $3$, then $x^2 \equiv 1 \pmod{3}$, and therefore $4x^2 + 3$ is not a multiple of $3$.
*Fact $2$: If $p$ is a prime dividing a number of the form $4x^2 + 3$ where $x$ is not a multiple of $3$, then $-3$ is a quadratic residue modulo $p$, that is
$$1 = \left(\frac{-3}{p}\right) = \left(\frac{p}{3}\right),$$
the latter by quadratic reciprocity. That means that $p \equiv 1 \pmod{3}$.
Hence a prime dividing a number $4x^2 + 3$ where $x \not\equiv 0 \pmod{3}$ must be either $\equiv 1 \pmod{12}$ or $\equiv 7 \pmod{12}$.
But the number $4x^2 + 3 \equiv 7 \pmod{12}$, so not all its prime divisors can be $\equiv 1 \pmod{12}$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Trigonometry Equations. Solve for $0 \leq X \leq 360$, giving solutions correct to the nearest minute where necessary,
a) $\cos^2 A -8\sin A \cos A +3=0$
Can someone please explain how to solve this, ive tried myself and no luck. Thanks!
| $$\frac{1+\cos2A}2-4(\sin2A)+3=0$$
$$\implies \cos2A-8\sin2A+7=0$$
Putting $1=r\cos B,8=r\sin B $ where $r>0$
Squaring we get $r^2=8^2+1^2=65\implies r=\sqrt{65}$ and $\cos B=\frac1{\sqrt{65}}$
So, $\cos2A-8\sin2A=r(\cos 2A\cos B-\sin2A\sin B)=\sqrt{65}\cos(2A+\arccos \frac1{\sqrt{65}})$
$\implies \cos(2A+\arccos \frac1{\sqrt{65}})=-\frac7{\sqrt{65}} $
$\implies 2A+\arccos \frac1{\sqrt{65}}=2n\pi\pm \arccos(\frac{-7}{\sqrt{65}})$ where $n$ is any integer
Taking '+' sign, $2A=2n\pi+(\arccos \frac{-7}{\sqrt{65}}-\arccos \frac1{\sqrt{65}})$
$=2n\pi+\arccos\left(\frac{1(-7)+8\cdot4}{65}\right)$ (Using $\arccos x-\arccos y=\arccos\left(xy+\sqrt{(1-x^2)(1-y^2)}\right) $)
$\implies 2A=2n\pi+\arccos\frac{25}{65}=2n\pi+\arccos\frac5{13}$
So, $A=n\pi+\frac12\arccos\frac5{13}$
If $\frac12\arccos\frac5{13}=C, \cos 2C=\frac5{13}\implies 2\cos^2C-1=\frac5{13}\implies \cos C=\pm\frac3{\sqrt{13}}$
$\implies C=\arccos(\pm \frac3{\sqrt{13}})$
$\implies A=n\pi+\arccos(\pm \frac3{\sqrt{13}})$
$\implies \cos A=\pm \frac3{\sqrt{13}}$ and $\sin A=\pm\sqrt{1-\cos^2A}=\pm\frac2{\sqrt{13}}$
Observe that $(\cos A,\sin A)=\pm(\frac3{\sqrt{13}},\frac2{\sqrt{13}})$ satisfies the given eqaution
Similarly, for the '-' sign
| {
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Find volume of the body $V = \{ z = \sqrt{6-x^2-y^2}, z = x^2 + y^2 \}$
Find volume of the body $V = \{ z = \sqrt{6-x^2-y^2}, z = x^2 + y^2 \}$
Now what I said is:
$$V = \iint_{D} {\sqrt{6-x^2-y^2} - x^2 - y^2 dxdy}$$.
But when I wanted to get what $D$ is, I intersected the two $z$ functions to get $\sqrt{6-x^2-y^2} = x^2+y^2$.
But what is this $D$? It is certainly not a circle.
| An idea: the intersection of
$$z=x^2+y^2\;,\;\;z=\sqrt{6-x^2-y^2}\implies z=\sqrt{6-z}\implies 0=z^2+z-6=(z+3)(z-2)$$
Since $\,z=-3\,$ is absurd (why?), we then have $\;2=z=x^2+y^2\;$ and you can now get easily your limits in the $\,xy-plane\;$ by means of this canonical circle.
| {
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Why does this polynomial related to Hamming Codes have integer coefficients? Question: Why does
$$
\frac{(1 + x)^{2^k - 1} - (1 - x^2)^{2^{k-1} - 1}(1-x)}{2^k}
$$
have integer coefficients?
Details: For a question I'm thinking about, I needed to know all the real numbers $c$ such that the generating function
$$
p(x) = \frac{(1 + x)^{2^k - 1}}{2^k} - c \cdot(1 - x^2)^{2^{k-1} - 1}(1-x)
$$
is a polynomial with integer coefficients and constant term $0$ or $1$.
Plugging in $x = 0$ reveals that $c = -\frac{1}{2^k}$ or $c = 1 - \frac{1}{2^k}$.
Both of these $c$ values seem to work empirically for all $k$, but it is unclear to me why this is the case. Binomial formula gives the coefficient of $x^i$ (for, say, $i$ even) as
$$
\frac{{{2^k - 1} \choose {i}} - (-1)^{i/2}{2^{k-1} - 1 \choose i/2} }{2^k}
$$
But I don't think this expression is nice.
Also, a search on oeis reveals that $p(x)$ is related to Hamming Codes.
| Note that
$$
\frac{(1+x)^{2^{k+1}-1}-(1-x^2)^{2^k-1}(1-x)}{2^{k+1}}
=(1+x)^{2^k-1}\frac{(1+x)^{2^k}-(1-x)^{2^k}}{2^{k+1}}
$$
$\frac{(1+x)^{2^k}-(1-x)^{2^k}}{2}$ is the odd part of $(1+x)^{2^k}$. Therefore, we just need to show that $\left.2^k\,\middle|\,\binom{2^k}{j}\right.$ when $j$ is odd. The number of factors of $2$ in $\binom{2^k}{j}$ is the sum of the bits in the binary representation of $j$ plus the sum of the bits in the binary representation of $2^k-j$ minus the sum of the bits in the binary representation of $2^k$. When $j$ is odd, this is $k$.
100000000000 $\leftarrow2^k$ has $k$ zeros on the right
001100100101 $\leftarrow j$ is odd, so it has a one on the right
010011011011 $\leftarrow2^k-j$ is the complement of $j$ plus one (has a one on the right)
Except for the rightmost bit, all bits of $j$ and $2^k-j$ are complements. Since the rightmost bit is one in both, the sum of the bits in $j$ and $2^k-j$ is $k+1$. The sum of the bits in $2^k$ is $1$. Thus, there are $k$ factors of $2$ in $\binom{2^k}{j}$. That is, $\left.2^k\,\middle|\,\binom{2^k}{j}\right.$ when $j$ is odd.
| {
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Prove that $\sin 10^\circ \sin 20^\circ \sin 30^\circ=\sin 10^\circ \sin 10^\circ \sin 100^\circ$? $\sin 10^\circ \sin 20^\circ \sin 30^\circ=\sin 10^\circ \sin 10^\circ \sin 100^\circ$
This is a competition problem which I got from the book "Art of Problem Solving Volume 2". I'm not sure how to solve it because there's just so many possibilities of using most of the trig identities that I don't know which path to take. One way I tried is canceling out the $\sin 10$ on both sides:
$\sin 20^\circ \sin 30^\circ= \sin 10^\circ \sin 100^\circ$
$2 \sin 10^\circ \cos 10^\circ \sin 30^\circ=\sin 10^\circ \sin 100^\circ$
$2 \sin 80^\circ \sin 30^\circ= \sin 100^\circ$
$2 \sin 80^\circ \sin 30^\circ= \sin 30^\circ \cos 70^\circ+\sin 70^\circ \cos 30^\circ$
$2 \sin 80^\circ \sin 30^\circ=\sin 20^\circ \sin 30^\circ+\sin 70^\circ \sin 60^\circ$
$2 \sin 80^\circ \sin 30^\circ=\sin 20^\circ \sin 30^\circ + 2\sin 70^\circ \sin 30^\circ \cos 30^\circ$
$2 \sin 80^\circ \sin 30^\circ=\sin 20^\circ \sin 30^\circ + 2\sin 70^\circ \sin 30^\circ \sin 60^\circ$
$1=\large \frac {\sin 20^\circ}{2 \sin 80^\circ}+ \frac {\sin 70^\circ \sin 60^\circ}{\sin 80^\circ}$
Now I set $\sin^2 \theta=\large \frac {\sin 20^\circ}{2 \sin 80^\circ}$ and $\cos^2 \theta =\large \frac {\sin 70^\circ \sin 60^\circ}{\sin 80^\circ}$
I solves for $\sin \theta$ and $\cos \theta$ and set up a right triangle, which then led me to the equation
$\sin 70^\circ \sin 60^\circ=\sin 80^\circ-\frac 12 \sin 20^\circ$
Another approach I tries is from the beginning having $\sin 10^\circ \sin 20^\circ \sin 30^\circ=\sin^2 10^\circ \sin 100^\circ$ and the using the power-reducing formula, but that also got me nowhere. Any help is appreciated. Thanks.
| Big hint of the day:
*
*Always try to simplify the expression first, before doing anything else
*What's the value of $\sin 30^\circ$?
*You can simplify $\sin 100^\circ$ by noticing that $\sin(90^\circ + \alpha) = \cos \alpha$.
Everything should be easy from here. :)
| {
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Proof that $ \lim_{x \to \infty} x \cdot \log(\frac{x+1}{x+10})$ is $-9$ Given this limit:
$$ \lim_{x \to \infty} x \cdot \log\left(\frac{x+1}{x+10}\right) $$
I may use this trick:
$$ \frac{x+1}{x+1} = \frac{x+1}{x} \cdot \frac{x}{x+10} $$
So I will have:
$$ \lim_{x \to \infty} x \cdot \left(\log\left(\frac{x+1}{x}\right) + \log\left(\frac{x}{x+10}\right)\right) = $$
$$ = 1 + \lim_{x \to \infty} x \cdot \log\left(\frac{x}{x+10}\right) $$
But from here I am lost, I still can't make it look like a fondamental limit. How to solve it?
| By the Lagrange mean value theorem, for $0<a\le b$, we have
$$
\frac{b-a}{b}\le\log(b)-\log(a)\le\frac{b-a}{a}
$$
So,
$$
L(x)=x\log\left(\frac{x+1}{x+10}\right)=-x\log\left(\frac{x+10}{x+1}\right)=
-x(\log(x+10)-\log(x+1))\\
\implies -9\frac{x}{x+10}\le L(x)\le -9\frac{x}{x+1}
$$
which shows $L(x)\stackrel{x \to \infty}{\to} -9$, by the squeeze principle (theorem).
| {
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$x \sin x=2$ why is my proof that there no solutions wrong? $\frac 12 x \sin x=1$ . Let's look at a right triangle with base $x$ and altitude $\sin x$ . Then our equation is for the area of this triangle. Let the sides of the triangle be $a=x$ , $b=\sqrt {x^2+sin^2 x}$ , and $c= \sin x$ . According to wikipedia, Heron's formula can be written as $$A=\large \frac { \sqrt {4a^2c^2-(a^2+b^2-c^2)^2}}{4}$$
Plugging in:
$4=\large \sqrt{4x^2 \sin^2 x-(x^2+x^2+\sin^2 x-\sin ^2 x)^2}$
$4=x^2 \sin^2 x -x^4$
$x^2(x^2- \sin^2 x)=-4$
$x^2$ will always be positive, and $\sin^2 x$ is never greater than $x^2$ , so this equation can have no real solutions. The original has solutions, so why is this wrong?
| This particular wikipedia formula is wrong.
It should be correctly either
$$A=\large \frac { \sqrt {4a^2b^2-(a^2+b^2-c^2)^2}}{4}$$
or
$$A=\large \frac { \sqrt {4a^2c^2-(a^2-b^2+c^2)^2}}{4}\,.$$
Mind the symmetry..
| {
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Find the coefficient of $x^{20}$ in $(x^{1}+⋯+x^{6} )^{10}$ I'm trying to find the coefficient of $x^{20}$ in
$$(x^{1}+⋯+x^{6} )^{10}$$
So I did this :
$$\frac {1-x^{m+1}} {1-x} = 1+x+x^2+⋯+x^{m}$$
$$(x^1+⋯+x^6 )=x(1+x+⋯+x^5 ) = \frac {x(1-x^6 )} {1-x} = \frac {x-x^7} {1-x}$$
$$(x^1+⋯+x^6 )^{10} =\left(\dfrac {x-x^7} {1-x}\right)^{10}$$
But what do I do from here ? any hints ?
Thanks
| The coefficient of $x^{10}$ in $(1+ x + \ldots + x^5)^{10}$ is equal to the number of integers $0 \leq x_i \leq 5 $ such that $\sum_{i=1}^{10} x_i= 10$.
We apply the Principle of Inclusion and exclusion, to deal with the restriction of $x_i \leq 5$.
If the only restriction is $0 \leq x_i$ then there are ${10 + 9 \choose 9 } $ solutions by the bars and stars method (sum of 10 non-negative integers is 10).
If $x_1 \geq 6$, then we substitute $x_1 = 6 + x_1 ^*$, and there are ${4 + 9 \choose 9}$ solutions by the stars and bars method (sum of 10 non-negative integers is 4).
Observe that we can't have 2 terms which are more than $6$.
Hence, by PIE, the coefficient is ${ 19 \choose 9} - 10 { 13 \choose 9}$, which is 85228.
| {
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Rationalizing a numerator I'm having trouble rationalizing a numerator with radicals. After multiplying the conjugate I get 0. Does anyone know where I went wrong?
\begin{align}
\frac{\sqrt{2+y} + \sqrt{2 - y}}{y} & = \left(\frac{\sqrt{2+y} + \sqrt{2 - y}}{y}\right) \left(\frac{\sqrt{2+y} - \sqrt{2 - y}}{\sqrt{2+y} - \sqrt{2 - y}}\right)\\
& = \frac{\sqrt{2+y}\sqrt{2+y} - \sqrt{2+y}\sqrt{2-y} + \sqrt{2-y}\sqrt{2+y} - \sqrt{2-y}\sqrt{2-y}}{y\sqrt{2+y} - y\sqrt{2-y}}\\
& =\frac{2 + y - 2 - y}{y\sqrt{2+y} - y\sqrt{2-y}} = 0
\end{align}
| Careful:
$$(a-b)(a+b)=a^2-b^2\implies $$
$$\left(\sqrt{2+y}+\sqrt{2-y}\right)\left(\sqrt{2+y}-\sqrt{2-y}\right)=(2+y)-(2-y)=2+y-2+y=2y$$
you forgot to change the second$\,-y\,$ into $\,+y\,$ ....
| {
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Show $\cos(x+y)\cos(x-y) - \sin(x+y)\sin(x-y) = \cos^2x - \sin^2x$
Show $\cos(x+y)\cos(x-y) - \sin(x+y)\sin(x-y) = \cos^2x - \sin^2x$
I have got as far as showing that:
$\cos(x+y)\cos(x-y) = \cos^2x\cos^2y -\sin^2x\sin^2y$
and
$\sin(x+y)\sin(x-y) = \sin^2x\cos^2y - \cos^2x\sin^2y$
I get stuck at showing:
$\cos^2x\cos^2y -\sin^2x\sin^2y - \sin^2x\cos^2y - \cos^2x\sin^2y = \cos^2x - \sin^2x$
I know that $\sin^2x + \cos^2x = 1$ and I have tried rearranging this identity in various ways, but this has not helped me so far.
| Looking at
$\cos^2x\cos^2y -\sin^2x\sin^2y - \sin^2x\cos^2y + \cos^2x\sin^2y$ (The last is a negative times a negative that forms a positive I believe.
You could pull $\sin^2x$ from the middle two terms to get this expression:
$ -\sin^2x\sin^2y - \sin^2x\cos^2y
= -\sin^2x(\sin^2y+\cos^2y) = -\sin^2x $
There is a similar reduction with the first and last terms around $\cos^2x$ that should make this appear easier.
| {
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"question_score": "3",
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Evaluating $\int\cos\theta~e^{−ia\cos\theta}~\mathrm{d}\theta$ Is anybody able to solve this indefinite integral :
$$
\int\cos\theta~e^{\large −ia\cos\theta}~\mathrm{d}\theta
$$
The letter $i$ denotes the Imaginary unit;
$a$ is a constant;
Mathematica doesn't give any result.
Thanks for any help you would like to provide me.
| $\int\cos\theta~e^{−ia\cos\theta}~d\theta$
$=\int\cos\theta\cos(a\cos\theta)~d\theta-i\int\cos\theta\sin(a\cos\theta)~d\theta$
$=\int\cos\theta\sum\limits_{n=0}^\infty\dfrac{(-1)^na^{2n}\cos^{2n}\theta}{(2n)!}d\theta-i\int\cos\theta\sum\limits_{n=0}^\infty\dfrac{(-1)^na^{2n+1}\cos^{2n+1}\theta}{(2n+1)!}~d\theta$
$=\int\sum\limits_{n=0}^\infty\dfrac{(-1)^na^{2n}\cos^{2n+1}\theta}{(2n)!}d\theta-i\int\sum\limits_{n=0}^\infty\dfrac{(-1)^na^{2n+1}\cos^{2n+2}\theta}{(2n+1)!}~d\theta$
For $\int\cos^{2n+1}\theta~d\theta$ , where $n$ is any non-negative integer,
$\int\cos^{2n+1}\theta~d\theta$
$=\int\cos^{2n}\theta~d(\sin\theta)$
$=\int(1-\sin^2\theta)^n~d(\sin\theta)$
$=\int\sum\limits_{k=0}^nC_k^n(-1)^k\sin^{2k}\theta~d(\sin\theta)$
$=\sum\limits_{k=0}^n\dfrac{(-1)^kn!\sin^{2k+1}\theta}{k!(n-k)!(2k+1)}+C$
For $\int\cos^{2n+2}\theta~d\theta$ , where $n$ is any non-negative integer,
$\int\cos^{2n+2}\theta~d\theta=\dfrac{(2n+2)!\theta}{4^{n+1}((n+1)!)^2}+\sum\limits_{k=0}^n\dfrac{(2n+2)!(k!)^2\sin\theta\cos^{2k+1}\theta}{4^{n-k+1}((n+1)!)^2(2k+1)!}+C$
This result can be done by successive integration by parts, e.g. as shown as similar as http://hk.knowledge.yahoo.com/question/question?qid=7012022000808
$\therefore\int\sum\limits_{n=0}^\infty\dfrac{(-1)^na^{2n}\cos^{2n+1}\theta}{(2n)!}d\theta-i\int\sum\limits_{n=0}^\infty\dfrac{(-1)^na^{2n+1}\cos^{2n+2}\theta}{(2n+1)!}~d\theta$
$=\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^{n+k}n!a^{2n}\sin^{2k+1}\theta}{(2n)!k!(n-k)!(2k+1)}-\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{i(-1)^n(2n+2)(k!)^2a^{2n+1}\sin\theta\cos^{2k+1}\theta}{4^{n-k+1}((n+1)!)^2(2k+1)!}-\sum\limits_{n=0}^\infty\dfrac{i(-1)^n(2n+2)a^{2n+1}\theta}{4^{n+1}((n+1)!)^2}+C$
$=\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^{n+k}n!a^{2n}\sin^{2k+1}\theta}{(2n)!k!(n-k)!(2k+1)}-\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{i(-1)^n(n+1)(k!)^2a^{2n+1}\sin\theta\cos^{2k+1}\theta}{2^{2n-2k+1}((n+1)!)^2(2k+1)!}-\sum\limits_{n=0}^\infty\dfrac{i(-1)^n(n+1)a^{2n+1}\theta}{2^{2n+1}((n+1)!)^2}+C$
$=\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^{n+k}n!a^{2n}\sin^{2k+1}\theta}{(2n)!k!(n-k)!(2k+1)}-\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{i(-1)^n(k!)^2a^{2n+1}\sin\theta\cos^{2k+1}\theta}{2^{2n-2k+1}n!(n+1)!(2k+1)!}-\sum\limits_{n=0}^\infty\dfrac{i(-1)^na^{2n+1}\theta}{2^{2n+1}n!(n+1)!}+C$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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convert ceil to floor Mathematically, why is this true?
$$\left\lceil\frac{a}{b}\right\rceil= \left\lfloor\frac{a+b-1}{b}\right\rfloor$$
Assume $a$ and $b$ are positive integers.
Is this also true if $a$ and $b$ are real numbers?
| $$\left\lceil \frac { a }{ b } \right\rceil =\frac { a }{ b } +1-\left( \frac { a }{ b } \mod\ 1 \right) ,\quad \quad (1)$$
$$\left\lfloor \frac { a+b-1 }{ b } \right\rfloor =\frac { a+b-1 }{ b } -\left( \frac { a+b-1 }{ b } \mod\ 1 \right) =\frac { a }{ b } +1-\frac { 1 }{ b } -\left( \frac { a }{ b } -\frac { 1 }{ b } \mod\ 1 \right) ,\quad \quad (2)$$
For the sake of simplicity write $$\left( \frac { a }{ b } \mod\ 1 \right) =k$$
The equations $(1)$ and $(2)$ are equal: $$\frac { a }{ b } +1-k=\frac { a }{ b } +1-\frac { 1 }{ b } -k+\left( \frac { 1 }{ b } \mod\ 1 \right) $$
Finally, $$\frac { 1 }{ b } =\left( \frac { 1 }{ b } \mod 1 \right) .$$ For $b>1$ your claim holds.
I think that something is missing. Until the next edit...
| {
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Proving : $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge4(a+b+c)$ For $a,b,c > 0$ and $ab+bc+ca+2abc=1$, how to prove that:
$$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge4(a+b+c) \, ?$$
| The condition is equivalent to $$\frac{a}{a + 1} + \frac{b}{b + 1} + \frac{c}{c + 1} = 1$$ then take $x = \frac{a}{a + 1}$, $y = \frac{b}{b + 1}$, $z = \frac{c}{c + 1}$ and the condition becomes $$x + y + z = 1$$ So we have $a = \frac{x}{1 - x} = \frac{x}{y + z}$ and similarly (by cyclic permutations) for other variables, on the other hand it is easily verified that $a = \frac{x}{y + z}$, $b = \frac{y}{z + x}$, $c = \frac{z}{x + y}$ satisfy the given condition, thus the condition is equivalent to saying that there exist positive real numbers $x,y,z$ such that $a = \frac{x}{y + z}$, $b = \frac{y}{z + x}$, $c = \frac{z}{x + y}$. Substituting this into the inequality we get that it is equivalent to $$\frac{y + z}{x} + \frac{z + x}{y} + \frac{x + y}{z} \geq 4 \cdot \left(\frac{x}{y + z} + \frac{y}{z + x} + \frac{z}{x + y}\right)$$ But this is easily proved because by Cauchy-Schwarz inequality we have $$(y + z)\left(\frac{x}{y} + \frac{x}{z}\right) \geq \left(\sqrt{x} + \sqrt{x}\right)^2 = 4x \Leftrightarrow \frac{x}{y} + \frac{x}{z} \geq \frac{4x}{y + z}$$ Analogously we have $$\frac{y}{x} + \frac{y}{z} \geq \frac{4y}{z + x} \text{ and } \frac{z}{x} + \frac{z}{y} \geq \frac{4z}{x + y}$$ Adding these last three inequalities yields the desired.
| {
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$5^m = 2 + 3^n$ help what to do how to solve this for natural numbers $5^m = 2 + 3^n$ i did this
$5^m = 2 + 3^n \Rightarrow 5^m \equiv 2 \pmod 3 \Rightarrow m \equiv 1 \pmod 2$
now if i put it like this
$ 5^{2k+1} = 2 + 3^n $
what to do ??
now is this right another try :
$ m = n \Rightarrow 5^n - 3^n = 2 = 5 - 3 \Rightarrow (m_1,n_1)= (1,1) $
we can prove by induction :
$5^n - 3^n > 2 \quad \forall n > 1 \Rightarrow m = n = 1 $
another case :
$m > n \Rightarrow 5^m > 5^n \geq 3^n+2 \quad \forall n \geq 1 \Rightarrow 5^m - 3^n > 2 \Rightarrow \emptyset$
$m < n \Rightarrow no \ sol. $ by putting some values
| Assume there is a solution with $n>1$. Then $ 5^m \equiv 2 \pmod 9 \Rightarrow m \equiv 5 \pmod 6$.
Then $ 5^m \equiv 3 \pmod 7 \Rightarrow 3^n \equiv 1 \pmod 7 \Rightarrow n \equiv 0 \pmod 3.$
Also, $ 5^m \equiv 5 \text { or } 8 \pmod {13} \Rightarrow 3^n \equiv 3\text { or } 6 \pmod {13} \Rightarrow n \equiv 1 \pmod 3.$
This contradiction means that $n$ cannot be greater than $1$ and so $m=n=1$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Power series for $(1+x^3)^{-4}$ I am trying to find the power series for the sum $(1+x^3)^{-4}$ but I am not sure how to find it. Here is some work:
$$(1+x^3)^{-4} = \frac{1}{(1+x^3)^{4}} = \left(\frac{1}{1+x^3}\right)^4 = \left(\left(\frac{1}{1+x}\right)\left(\frac{1}{x^2-x+1}\right)\right)^4$$
I can now use
$$\frac{1}{(1-ax)^{k+1}} = \left(\begin{array}{c} k \\ 0 \end{array}\right)+\left(\begin{array}{c} k+1 \\ 1 \end{array}\right)ax+\left(\begin{array}{c} k+2 \\ 2 \end{array}\right)a^2x^2+\dots$$
on the $\frac{1}{1+x}$ part but I am not sure how to cope with the rest of formula.
| \begin{gather}(1+x^3)^{-4}=(1+t)^\alpha=1+\alpha t+\frac{\alpha(\alpha-1)}{2!}t^2+\dots=\\ =1+\frac{(-4)}{1!}x^3+\frac{(-4)(-4-1)}{2!}x^6+\dots
=\sum_{k=0}^\infty(-1)^{k}\frac{(3+k)!}{3!k!}x^{3k}\end{gather}
| {
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for every integer $n \ge 1$ one has the equality Could any one help me?
For every integer $n \ge 1$ one has the equality:
$$ 1-{1\over 2}+ {1\over 3}-{1\over 4}+\dots+{1\over 2n-1}-{1\over 2n}={1\over n+1}+{1\over n+2}+\dots +{1\over 2n}.$$
Should I proceed by Induction?
| Here's another way to do the problem,$$1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots-\frac{1}{2n}\\=1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{2n}-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\cdots+\frac{1}{2n}\right)\\=1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{2n}-\left(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}\right)\\=\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{2n}$$
If I am not mistaken,this is the Catalan's Identity.
| {
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Double Integral Over Region Common to Two Circles Evaluate $\iint\frac{(x^2+y^2)^2}{x^2y^2} dx dy$ over the region common to the circles $x^2+y^2=7x$ and $x^2+y^2=11y$.
| You may use the Cylindrical coordinates to find the value of the integrals. This, as @Ron's Caesarian's approach, will get a bit messy but it works as well. We have two circles on $z=0$ intersected each other, so $$7r\cos\theta=7x=x^2+y^2=11y=11r\sin\theta\longrightarrow\theta_{0}=\tan^{-1}\left(\frac{7}{11}\right)\cong0.56$$ Now we have $$I=\int_{\theta=0}^{\theta_0}\int_{r=0}^{11\sin\theta}\frac{r^4}{r^4\cos^2\theta\sin^2\theta}rdrd\theta+\int_{\theta=\theta_0}^{\pi/2}\int_{r=7\cos\theta}^{0}\frac{r^4}{r^4\cos^2\theta\sin^2\theta}rdrd\theta$$
| {
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show that $\int_{-\infty}^{\infty} \frac {(\sin x) (x^2+a^2)}{x(x^2+b^2)}dx=\frac{\pi(a^2+e^{-b}(b^2-a^2))}{b^2}$ show that $$\int_{-\infty}^{\infty} \frac {(\sin x) (x^2+a^2)}{x(x^2+b^2)}dx=\frac{\pi(a^2+e^{-b}(b^2-a^2))}{b^2}$$
for every $a,b>0$
thanks for all
| Consider the contour integral
$$\oint_C dz \frac{(z^2+a^2) e^{i z}}{z (z^2+b^2)}$$
where $C$ is the semicircular contour of radius $R$ in the upper half-plane, with an additional semicircular contour of radius $\epsilon$ centered at the origin, jutting into the upper half-plane.
The contour integral is equal to
$$\int_{-R}^{-\epsilon} dx \, \frac{(x^2+a^2) e^{i x}}{x (x^2+b^2)} + i \epsilon \int_{-\pi}^0 d\phi e^{i \phi} \frac{(a^2+\epsilon^2 e^{i 2 \phi}) e^{i \epsilon e^{i \phi}}}{\epsilon e^{i \phi} (b^2+\epsilon^2 e^{i 2 \phi})}+\\\int_{\epsilon}^R dx \, \frac{(x^2+a^2) e^{i x}}{x (x^2+b^2)}+i R \int_0^{\pi} d\theta e^{i \theta}\frac{(a^2+R^2 e^{i 2 \theta}) e^{i R e^{i \theta}}}{Re^{i \theta} (b^2+R^2 e^{i 2 \theta})} $$
We take the limit as $R \to \infty$ and $\epsilon \to 0$ and get
$$PV \int_{-\infty}^{\infty} dx \, \frac{(x^2+a^2) e^{i x}}{x (x^2+b^2)} - i \pi \frac{a^2}{b^2}$$
Note that the magnitude of fourth integral vanishes as
$$2 \int_0^{\pi/2} e^{-R \sin{\theta}} \le 2 \int_0^{\pi/2} e^{-2 R \theta/\pi} \le \frac{\pi}{R}$$
as $R \to \infty$. By the residue theorem, the contour integral is equal to $i 2 \pi$ times the residue of the pole at $z=i b$. Therefore
$$PV \int_{-\infty}^{\infty} dx \, \frac{(x^2+a^2) e^{i x}}{x (x^2+b^2)} = i \pi \frac{a^2}{b^2} - i 2 \pi \frac{(a^2-b^2) e^{-b}}{2 b^2}$$
Similarly,
$$PV \int_{-\infty}^{\infty} dx \, \frac{(x^2+a^2) e^{-i x}}{x (x^2+b^2)} = -i \pi \frac{a^2}{b^2} + i 2 \pi \frac{(a^2-b^2) e^{-b}}{2 b^2}$$
Therefore,
$$\int_{-\infty}^{\infty} dx \, \frac{(x^2+a^2) \sin{x}}{x (x^2+b^2)} = \pi \frac{a^2}{b^2} - \pi \frac{(a^2-b^2) e^{-b}}{ b^2}$$
which is equivalent to the stated result.
| {
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$24\mid n(n^{2}-1)(3n+2)$ for all $n$ natural problems in the statement. "Prove that for every $ n $ natural, $24\mid n(n^2-1)(3n+2)$"
Resolution:
$$24\mid n(n^2-1)(3n+2)$$if$$3\cdot8\mid n(n^2-1)(3n+2)$$since$$n(n^2-1)(3n+2)=(n-1)n(n+1)(3n+2)\Rightarrow3\mid n(n^{2}-1)(3n+2)$$and$$8\mid n(n^{2}-1)(3n+2)?$$$$$$
Would not, ever succeeded without the help of everyone that this will post tips, ideas, etc., etc.. Thank you.
| To prove $8|n(n^2-1)(3n+2)$ you can simply break the problem in three cases:
Case 1 $n$ odd.
Then $n^2-1=(n-1)(n+1)$ is the product of two consecutive even numbers. Thus one must be divisible by $4$ and the other by $2$.
Case 2 $n$ multiple of 4.
Then $4|n$ and $2|3n+2$.
Case 3 $n$ even but not divisible by 4.
Then $2|n$ and $n-2$ is divisible by $4$. Hence
$$4|3n+2=4n-(n-2) \,.$$
| {
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Show that $\sin3\alpha \sin^3\alpha + \cos3\alpha \cos^3\alpha = \cos^32\alpha$
Show that $\sin3\alpha \sin^3\alpha + \cos3\alpha \cos^3\alpha = \cos^32\alpha$
I have tried $\sin^3\alpha(3\sin\alpha - 4 \sin^3\alpha) = 3\sin^4\alpha - 4\sin^6\alpha$ and $\cos^3\alpha(4\cos^3\alpha - 3\cos\alpha) = 4\cos^6\alpha - 3\cos^4\alpha$ to give
$$\sin3\alpha \sin^3\alpha + \cos3\alpha \cos^3\alpha = 3\sin^4\alpha - 4\sin^6\alpha + 4\cos^6\alpha - 3\cos^4\alpha$$
I can't work out how to simplify this to $\cos^32\alpha$.
I also noticed that the LHS of the question resembles $\cos(A-B)$, but I can't figure a way of making that useful.
| Let $\cos^2\alpha=a,\sin^2\alpha=b\implies a+b=1, a-b=\cos2\alpha$
$ \sin3\alpha\sin^3\alpha + \cos3\alpha\cos^3\alpha = $
$ = 3\sin^4\alpha - 4\sin^6\alpha + 4\cos^6\alpha - 3\cos^4\alpha = $
$ = 3b^2-4b^3+4a^3-3a^2=4(a^3-b^3)-3(a^2-b^2)=(a-b)\{4(a^2+ab+b^2)-3(a+b)\} $
Now, $4(a^2+ab+b^2)-3(a+b)=4\{(a+b)^2-ab\}-3=4(1-ab)-3=1-4ab$
$=1-4\cos^2\alpha\sin^2\alpha=1-(2\sin\alpha\cos\alpha)^2=1-\sin^22\alpha=\cos^22\alpha$
| {
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Recurrence relation for $n$ numbers in which no 3 consecutive digits are the same. I am stuck on trying to find (and solve) a recurrence relation to find all n-digit numbers in which no 3 consecutive digits are the same. These numbers are in decimal expansion.
Now I first imagine trying this for finding the number of n-digit numbers that don't have two consecutive digits. If I am correct I have 10 choices for the first (we are in decimal expansion so 0 counts), then 9 choices for the second, 9 for the third and so on. So we would have $10*9^{n-1}$ possible numbers. This looks like it works for n = 2 as we would have 10 numbers {00,11,..,99} that are repeated. I however wonder about 00, if that should be counted here, but for n > 2 I could have .001 etc. I however figure that if .00 is not a valid number in decimal expansion then it also won't have to be removed. Hence I have 90 numbers which work. I can't quite see in to the n = 3 case though - it is this next digit (and so on) that boggle me.
Now I rigged up a recurrence relation that seems to work for the n = 1 and n = 2 case in the no consecutive 2 digit case. It is:
$T(n) = 10^n - T(n-1) -1$ where T(0) = 0.
$T(1) = 10^1 - T(0) - 1 = 9$ Note that 0 is not counted for n = 1 as it is .0 (hence the -1 included)
$T(2) = 10^2 - T(1) - 1 = 10^2 - (10^1 - T(0) - 1) - 1 = 100 - (10 - 0 - 1) - 1 = 90$
Since I don't know what T(3) is supposed to be I didn't bother putting it in. If this Recurrence relation works, perhaps someone would have a combinatorial reason why it works?
I however then am still stuck on the non 3 consecutive digit case and finding a recurrence relation for it.
Thanks for any thoughts,
Brian
| We can capture the essence of this problem by introducing two sequences, namely the sequence $a_n$ that counts the number of $n$-digit numbers that do not end in a repeated digit, starting with $a_2 = 9\times 10 -9 = 81$ and the number $b_n$ of $n$-digit numbers that end in two repeated digits, starting with $b_2 = 9.$
The problem definition translates straightforwardly into a pair of recurrences, namely
$$a_{n+1} = 9 a_n + 9 b_n
\quad \text{and} \quad
b_{n+1} = a_n.$$
We are interested in the quantity $$a_n+b_n.$$
By substitution we obtain
$$a_{n+1} = 9 a_n + 9 a_{n-1}$$
with characteristic equation
$$ x^2 = 9 x + 9$$
whose roots are
$$\rho_{1,2} = \frac{9}{2} \pm \frac{3}{2} \sqrt{13}.$$
Solving for $c_{1,2}$ in the system
$$ c_1\rho_1^2 + c_2\rho_2^2 = 81
\quad \text{and} \quad
c_1\rho_1^3 + c_2\rho_2^3 = 810$$
we obtain
$$ c_{1,2} = \pm \frac{3}{\sqrt{13}}$$
and hence
$$ a_n =
\frac{3}{\sqrt{13}} \left( \frac{9}{2} + \frac{3}{2} \sqrt{13}\right)^n
- \frac{3}{\sqrt{13}} \left( \frac{9}{2} - \frac{3}{2} \sqrt{13}\right)^n.$$
Since $$a_n+b_n = a_n + a_{n-1} = \frac{1}{9} a_{n+1}$$
the final answer is
$$a_n + b_n =
\frac{1}{3\sqrt{13}} \left( \frac{9}{2} + \frac{3}{2} \sqrt{13}\right)^{n+1}
-
\frac{1}{3\sqrt{13}} \left( \frac{9}{2} - \frac{3}{2} \sqrt{13}\right)^{n+1}.$$
The first few values are
$$ 90, 891, 8829, 87480, 866781, 8588349, 85096170, 843160671, 8354311569.$$
This is sequence A057092 from the OEIS.
| {
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find the inverse of $x^2 + x + 1$ In $\mathbb{F}_2[x]$ modulo $x^4 + x + 1$
find the inverse of $x^2 + x + 1$
not 100% sure but here what i have:
user euclid algorithm:
$x^4 + x + 1 = (x^3 + 1)(x + 1) + x$
$(x^3 + 1) = x * x * x + 1$
$1 = (x^3 + 1) - x * x * x $
| Note that:
$(x^{2}+x+1)(x^{2}+x+1)=x^{2}+x$
$(x^{2}+x)(x^{2}+x+1)=x^{4}+x=1$. To get this I used that $x^{4}+x+1=0$ and that the coefficients are in $\mathbb{F}_{2}$.
So the inverse is $(x^{2}+x+1)^{2}=x^{2}+x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/457594",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Find $\int \sin(2x)\sqrt{1+3\cos^{2}(x)}\ dx$ Is it humanly possible to integrate the following equation?
$$\int \sin(2x)\sqrt{1+3\cos^{2}(x)}\ dx$$
| Let $u=\cos^2(x)$, therefore $du=-2\cos(x)\sin(x)\ dx$:
$$\int\sin(2x)\sqrt{1+3\cos^2(x)}\ dx=-\int\sqrt{1+3u}\ du$$
Now, let $v=1+3u$ and $dv=3\ du$:
\begin{align*}
-\int\sqrt{1+3u}\ du&=-\frac13\int\sqrt v\ dv\\
&=-\frac{2v^{3/2}}9+C,\text{ where $C$ is a constant}\\
&=-\frac29(1+3u)^{\frac32}+C\\
&=\boxed{-\frac29(1+3\cos^2(x))^{\frac32}+C}
\end{align*}
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Approximation to the Lambert W function If:
$$x = y + \log(y) -a$$
Then the solution for $y$ using the Lambert W function is:
$$y(x) = W(e^{a+x})$$
In a paper I'm reading, I saw an approximation to this solution, due to "Borsch and Supan"(?):
$$y(x) = W(e^{a+x}) \approx x\left(1 - \frac{\log x - a}{1+x}\right)$$
Any idea how this approximation was derived?
| We can use a procedure known as "bootstrapping" to determine an approximation for the Lambert $W$ function. Let's go back to its definition.
For $x > 0$ the equation
$$
we^w = x
$$
has exactly one positive solution $w = W(x)$ which increases with $x$. Note that $(w,x) = (1,e)$ is one such solution, so if $x > e$ then $w > 1$. By taking logarithms of both sides of the equation we get
$$
\log w + w = \log x
$$
or
$$
w = \log x - \log w. \tag{1}
$$
When $x > e$ we therefore have
$$
w = \log x - \log w < \log x.
$$
In other words, our first approximation is that
$$
1 < w < \log x \tag{2}
$$
when $x > e$. We then have
$$
0 < \log w < \log\log x,
$$
and plugging this into $(1)$ yields
$$
\log x - \log \log x < w < \log x, \tag{3}
$$
where the left side is positive for $x > 1$. Taking logarithms as before yields
$$
\log\log x + \log\left(1 - \frac{\log\log x}{\log x}\right) < \log w < \log\log x,
$$
and upon substituting this back into $(1)$ we get
$$
\log x - \log\log x < w < \log x - \log\log x - \log\left(1 - \frac{\log\log x}{\log x}\right).
$$
Since $w = W(x)$ we have shown that
$$
\log x - \log\log x < W(x) < \log x - \log\log x - \log\left(1 - \frac{\log\log x}{\log x}\right) \tag{4}
$$
for $x > e$.
In your particular case we're interested in $W(e^{x+a})$, for which we have
$$
x+a - \log(x+a) < W(e^{x+a}) < x+a - \log(x+a) - \log\left(1 - \frac{\log(x+a)}{x+a}\right)
$$
for $x+a > 1$. In this sense we have
$$
W(e^{x+a}) \approx x+a - \log(x+a) = x\left(1 - \frac{\log(x+a) - a}{x}\right) \tag{5}
$$
when $x+a$ is large. Now by applying Taylor series a couple times we see that, for $x$ large and $a \ll x$,
$$
\begin{align}
\frac{\log x - a}{x+1} &= \frac{\log x - a}{x} \cdot \frac{1}{1+\frac{1}{x}} \\
&\approx \frac{\log x - a}{x} \left(1-\frac{1}{x}\right) \\
&= \frac{\log x - a}{x} - \frac{\log x - a}{x^2} \\
&= \frac{\log(x+a-a) - a}{x} - \frac{\log x - a}{x^2} \\
&= \frac{\log(x+a) + \log\left(1-\frac{a}{x+a}\right) - a}{x} - \frac{\log x - a}{x^2} \\
&= \frac{\log(x+a) - a}{x} + \frac{\log\left(1-\frac{a}{x+a}\right)}{x} - \frac{\log x - a}{x^2} \\
&\approx \frac{\log(x+a) - a}{x} - \frac{a}{x(x+a)} - \frac{\log x - a}{x^2} \\
&\approx \frac{\log(x+a) - a}{x}.
\end{align}
$$
We may then conclude from $(5)$ that
$$
W(e^{x+a}) \approx x \left(1 - \frac{\log x - a}{x+1}\right)
$$
for $x$ large and $a \ll x$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "27",
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Find the sum : $\frac{1}{\cos0^\circ\cos1^\circ}+\frac{1}{\cos1^\circ \cos2^\circ} +\frac{1}{\cos2^\circ \cos3^\circ}+......+$ Find the sum of the following :
(i) $$\frac{1}{\cos0^\circ \cos1^\circ}+\frac{1}{\cos1^\circ\cos2^\circ} +\frac{1}{\cos2^\circ \cos3^\circ}+......+\frac{1}{\cos88^\circ \cos89^\circ}$$
I tried :
$$\frac{1}{\cos1^\circ}\left[\frac{\cos(1^\circ-0^\circ)}{\cos0^\circ\cos1^\circ} + \frac{\cos(2^\circ-1^\circ)}{\cos1^\circ\cos2^\circ}+...\right]$$
= $$\frac{1}{\cos1^\circ}\left[\frac{\cos1^\circ\cos0^\circ}{\cos0^\circ\cos1^\circ} - \frac{\sin1^\circ \sin0^\circ}{\sin0^\circ\cos1^\circ} + \frac{\cos2^\circ \cos1^\circ}{\cos1^\circ\cos2^\circ} -\frac{\sin2^\circ \sin1^\circ}{\cos1^\circ\cos2^\circ}...\right]$$
For this, as well, I am not getting any pattern to solve further. Please suggest, thanks.
| HINT:
$$\frac{\sin(A-B)}{\cos A\cos B}=\frac{\sin A\cos B-\cos A\sin B}{\cos A\cos B}=\tan A-\tan B$$
If $A= (n+1)^\circ,B=n^\circ$
$$\frac{\sin 1^\circ}{\cos (n+1)^\circ\cos n^\circ}=\tan(n+1)^\circ-\tan n^\circ $$
Put $n=0,1,2,\cdots,87,88$ and add to find the series to be Telescopic
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
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Please help on this Probability problem
A bag contains 5 red marbles and 7 green marbles. Two marbles are drawn randomly one at a time, and without replacement. Find the probability of picking a red and a green, without order.
This is how I attempted the question: I first go $P(\text{Red})= 5/12$ and $P(\text{Green})= 7/11$ and multiplied the two: $$\frac{7}{11}\times \frac{5}{12}= \frac{35}{132}$$
Then I got $P(\text{Green})= 7/12$ and $P(\text{Red})= 5/11$ $\implies$
$$\frac{5}{11} × \frac{7}{12}= \frac{35}{132}$$
So I decided that $$P(\text{G and R}) \;\text{ or }\; P(\text{R and G}) =\frac{35}{132} + \frac{35}{132} =\frac{35}{66}$$ Is this correct?
| Your method is correct. For more complicated problems of the same general kind, one might take a slightly different approach.
Imagine that the $12$ marbles are distinct, they have different driver license numbers. There are $\binom{12}{2}$ equally likely ways to choose $2$ marbles from the $12$.
There are $\binom{5}{1}$ ways to choose a red marble, and $\binom{7}{1}$ ways to choose a green marble. Thus there are $\binom{5}{1}\binom{7}{1}$ ways to choose a red and a green. It follows that our probability is
$$\frac{\binom{5}{1}\binom{7}{1}}{\binom{12}{2}}.$$
Harder! However, suppose we have $25$ marbles, $10$ red and $15$ green. We choose (without replacement) $8$ marbles. What is the probability that we get $3$ red and $5$ green? The same analysis shows that the probability is
$$\frac{\binom{10}{3}\binom{15}{5}}{\binom{25}{8}}.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Expected number of people sitting in the right seats. There was a popular interview question from a while back: there are $n$ people getting seated an airplane, and the first person comes in and sits at a random seat. Everyone else who comes in either sits in his seat, or if his seat has been taken, sits in a random unoccupied seat. What is the probability that the last person sits in his correct seat?
The answer to this question is $1/2$ because everyone looking to sit on a random seat has an equal probability of sitting in the first person's seat as the last person's.
My question is: what is the expected number of people sitting in their correct seat?
My take: this would be $\sum_{i=1}^n p_i$ where $p_i$ is the probability that person $i$ sits in the right seat..
$X_1 = 1/n$
$X_2 = 1 - 1/n$
$X_3 = 1 - (1/n + 1/n(n-1))$
$X_4 = 1 - (1/n + 2/n(n-1) + 1/n(n-1)(n-2))$
Is this correct? And does it generalize to $X_i$ having an $\max(0, i-1)$ term of $1/n(n-1)$, a $\max(0, i-2)$ term of $1/n(n-1)(n-2)$ etc?
Thanks.
| Correct answer to incorrect question: please see second answer
The answer is $\frac{1}{2}$ as was said.
The general pattern is that for $n$ people, there is a $\frac{1}{n}$ probability of success, $\frac{1}{n}$ probability of failure, and an $\frac{n-2}{n}$ probability that the problem repeats itself on the $n-1$ scale.
Case $n=2$: The probability that the first picks the correct seat is $\frac{1}{2}$, and then the last person sits in proper seat with probability 1. The probability that the first picks the wrong seat is $\frac{1}{2}$, and then the last person sits in proper seat with probability 0. So the probability in total is:
$$
\frac{1}{2}\cdot1+\frac{1}{2}\cdot0=\frac{1}{2}
$$
Case $n=3$: The probability that the first picks the correct seat is $\frac{1}{3}$, and then the last person sits in proper seat with probability 1. The probability that the first picks the last person's seat is $\frac{1}{3}$, and then the last person sits in proper seat with probability 0. The probability that the first picks the middle person's seat is $\frac{1}{3}$, and now there is a $\frac{1}{2}$ that the second person picks the last seat or the first seat (Case 2) since two non-last people switching seats means everyone else takes their own seat. So:
$$
\frac{1}{3}\cdot1+\frac{1}{3}\cdot0+\frac{1}{3}\cdot\frac{1}{2} = \frac{2}{6} + \frac{1}{6} = \frac{3}{6} = \frac{1}{2}
$$
Proof by induction.
Assume it holds true for $n$ that the probability of the last person sitting in the proper seat is $\frac{1}{2}$. Now if there are $n+1$ people, we have a $\frac{1}{n+1}$ chance that the last-seat probability is 1 (correct seat), a $\frac{1}{n+1}$ chance that the last-seat probability is 0 (last seat), and an $\frac{n-1}{n+1}$ chance that the probability is $\frac{1}{2}$, since we know the $n$ case.
$$
\frac{1}{n+1} + \frac{0}{n+1} + \frac{n-1}{2(n+1)}\\
=\frac{2}{2(n+1)}+\frac{n-1}{2(n+1)}\\
=\frac{n+1}{2(n+1)}\\
=\frac{1}{2}
$$
QED
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
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Simplify : $( \sqrt 5 + \sqrt6 + \sqrt7)(− \sqrt5 + \sqrt6 + \sqrt7)(\sqrt5 − \sqrt6 + \sqrt7)(\sqrt5 + \sqrt6 − \sqrt7) $ The question is to simplify $( \sqrt 5 + \sqrt6 + \sqrt7)(− \sqrt5 + \sqrt6 + \sqrt7)(\sqrt5 − \sqrt6 + \sqrt7)(\sqrt5 + \sqrt6 − \sqrt7)$ without using a calculator .
My friend has given me this challenge . I solved it by expanding $$(a+b+c)(-a+b+c)(a-b+c)(a+b-c) = -a^4+2a^2b^2+2a^2c^2-b^4+2b^2c^2-c^4$$ and then substituting $a,b,c=\sqrt 5 , \sqrt6 , \sqrt7$ respectively to get the answer $104$ .
But I suppose there is a more elegant and easy way to solve this problem .
Can anyone find it ?
| Given that your original numbers are square roots and your expanded equation involves only even powers of your original numbers, you could say:
$$A=a^2=5, B=b^2=6, C=c^2=7$$
and simplify your equation to:
$$2(AB+AC+BC) - (A^2+B^2+C^2) = 2(30+35+42) - (25+36+49) = 214-110 = 104$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
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How prove this $|\{n\sqrt{3}\}-\{n\sqrt{2}\}|>\frac{1}{20n^3}$ Prove that
$$|\{n\sqrt{3}\}-\{n\sqrt{2}\}|>\dfrac{1}{20n^3}$$
let $t=\{n\sqrt{2}\}-\{n\sqrt{3}\}$ and $k=[n\sqrt{3}]-[n\sqrt{2}]$
then we have $$t=k-(\sqrt{3}-\sqrt{2})n=k-\sqrt{5-2\sqrt{6}}n\neq 0$$
so
\begin{align*}
t&=\dfrac{(k-(\sqrt{3}-\sqrt{2})n)(k-(\sqrt{3}+\sqrt{2})n)(k+(\sqrt{3}-\sqrt{2})n)(k+(\sqrt{3}+\sqrt{2})n)}{(k-(\sqrt{3}+\sqrt{2})n)(k+(\sqrt{3}-\sqrt{2})n)(k+(\sqrt{3}+\sqrt{2})n)}\\
&=\dfrac{k^4-10k^2n^2+n^4}{(t-2\sqrt{2}n)(t+2(\sqrt{3}-\sqrt{2})n)(t+2\sqrt{3}n)}
\end{align*}
notice that
$$|t-2\sqrt{2}n|\le 2\sqrt{2}n+\dfrac{1}{20}\le(2\sqrt{2}+\dfrac{1}{20})n$$
$$|t+2(\sqrt{3}-\sqrt{2})n|\le2(\sqrt{3}-\sqrt{2})n+\dfrac{1}{20}\le(2\sqrt{3}-2\sqrt{2}+\dfrac{1}{20})n$$
$$|t+2\sqrt{3}n|\le2\sqrt{3}n+\dfrac{1}{20}\le(2\sqrt{3}+\dfrac{1}{20})n$$
so
$$|t|\ge\dfrac{1}{(2\sqrt{2}+\dfrac{1}{20})(2\sqrt{3}+2\sqrt{2}-\dfrac{1}{20})(2\sqrt{3}+\dfrac{1}{20})n^3}>\dfrac{1}{7n^3}$$
so
$$|t|\ge\min{\left(\dfrac{1}{20},\dfrac{1}{7n^3}\right)}\ge\dfrac{1}{20n^3}$$
\
This post https://math.stackexchange.com/questions/465419/how-prove-this-t-2-sqrt2n-le2-sqrt2n-frac120 is not true? so This methods is wrong, so How prove it? Thank you
| If the statement is false, then there exists a $n \in \mathbb{Z}_{+}$ such that
$$\left|\{n\sqrt{3}\} - \{n\sqrt{2}\}\right| \le \frac{1}{20n^3}\tag{*1}$$
This in turn implies existences of $m \in \mathbb{Z}$ and
$\delta \in [ -\frac{1}{20n^3}, \frac{1}{20n^3} ]$ such that:
$$n (\sqrt{3} - \sqrt{2}) = m + \delta$$
Consider the polynomial $f(x) = x^4 - 10x^2 + 1$, it has the factorization:
$$f(x) = (x - \sqrt{3}-\sqrt{2})(x - \sqrt{3} + \sqrt{2})(x + \sqrt{3} - \sqrt{2})( x + \sqrt{3} + \sqrt{2})$$
We have:
*
*$f(\frac{m+\delta}{n}) = f(\sqrt{3} - \sqrt{2}) = 0$.
*$f(\frac{m}{n}) \ne 0$ because the roots of $f$ are all irrational.
*$n^4 f(\frac{m}{n}) = m^4 - 10 m^2 n^2 + n^4 \in \mathbb{Z}$.
(2) and (3) together implies $$n^4 \left|f(\frac{m}{n})\right| \ge 1\quad\iff\quad \left|f(\frac{m}{n})\right| \ge \frac{1}{n^4}$$
On the other hand, (1) and Mean value theorem implies existence of $\xi \in [0,1]$ such that
$$f(\frac{m}{n}) = f'(\frac{m+\xi\delta}{n}) \frac{\delta}{n}$$
Notice
$$|\delta| \le \frac{1}{20n^3} \quad\implies\quad |\frac{\xi\delta}{n}| \le \frac{1}{20}$$
A plot of $f'(x)$ tells us $|f'(x)| \le 7.2$ whenever $\left|x - (\sqrt{3}-\sqrt{2})\right| \le \frac{1}{20}$. From this, we obtain a contradiction:
$$\frac{1}{n^4} \le \left|f(\frac{m}{n})\right| = \left| f'(\frac{m+\xi\delta}{n}) \frac{\delta}{n}\right| \le 7.2\left|\frac{\delta}{n}\right| \le \frac{7.2}{20n^4}$$
As a result, the original assumption $(*1)$ is false and we have
$$\left|\{n\sqrt{3}\} - \{n\sqrt{2}\}\right| > \frac{1}{20n^3}$$
In fact, we can improve the constant in above inequality form $20$ to something around $8.4033154$.
| {
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How find this $3\sqrt{x^2+y^2}+5\sqrt{(x-1)^2+(y-1)^2}+\sqrt{5}(\sqrt{(x-1)^2+y^2}+\sqrt{x^2+(y-1)^2})$ find this follow minimum
$$3\sqrt{x^2+y^2}+5\sqrt{(x-1)^2+(y-1)^2}+\sqrt{5}\left(\sqrt{(x-1)^2+y^2}+\sqrt{x^2+(y-1)^2}\right)$$
I guess This minimum is $6\sqrt{2}$
But I can't prove,Thank you
| Too long for a comment, I wonder if the problem can be solved geometrically.
Let $A,B, C, D$ be the points of coordinates $(0,0), (1,1), (1,0)$ and $(0,1)$ and let $P$ be the point in plane of coordinates $(x,y)$.
The problem asks you to find the minimum of
$$3PA+5PB+\sqrt{5}(PC+PD) (*)$$
The issue now is that we cannot split the problem in two halves: $\sqrt{5}(PC+PD) \geq \sqrt{5} CD$ with equality if and only if $P \in (C,D)$, but $3PA+5PB \geq 3AB$ with equality if and only if $P=B$.
But one might be able to find the min of $(*)$ geometrically...Or maybe physically (put four objects of weights $3,5,\sqrt{5}, \sqrt{5}$ at $A,B,C,D$ then $P$ must have some physical meaning).
| {
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Problem 2-7 in Spivak One is asked to show that $ \sum\limits_{i=1}^{n} k^{p}$ (typo on $i$?) can always be written in the form
$$\frac{n^{p+1}}{p+1}+An^{p}+Bn^{p-1}+Cn^{p-2}+\cdots.$$
The solution states:
The proof is by complete induction on $p$. The statement is true for $p=1$, since
$$\sum\limits_{k=1}^{n} k= \frac{n(n+1)}{2}= \frac{n^{2}}{2}+n.$$
Suppose that the statement is true for all natural numbers $\leq p$. The binomial theorem yield the equations
$$(k+1)^{(p+1)}-k^{p+1}=(p+1)k^{p}+ \textrm{terms involving lower powers of }k.$$
Adding for $k=1,\dots, n,$ we obtain
$$\frac{(n+1)^{p+1}}{p+1}=\sum\limits_{k=1}^{n} k^{p} + \textrm{terms involving} \sum\limits_{k=1}^{n} k^{r} \textrm{ for } r<p.$$
By assumption we can write each $\sum\limits_{k=1}^{n} k^{r}$ as an expression involving powers $n^{s}$ with $s\leq p$. It follows that
$$\sum\limits_{k=1}^{n} k^{p}=\frac{(n+1)^{p+1}}{p+1} + \textrm{ terms involving powers of }n \textrm{ less than } p+1.$$
What I tryed based on it and more explicitly:
$$\begin{align}(k+1)^{p+1} &={p+1 \choose p+1}k^{p+1}+{p+1 \choose p}k^{p}+ \cdots +{p+1 \choose 1}k + {p+1 \choose 0}k^{0}\\
(k+1)^{p+1}&= 1\cdot k^{p+1}+{p+1 \choose p}k^{p}+ \cdots +{p+1 \choose 1}k + {p+1 \choose 0}k^{0}\\
(k+1)^{p+1} -k^{p+1} &= (p+1)k^{p}+ \cdots +(p+1)k + 1\cdot k^{0}\\
\end{align}$$
Adding for $k=1,\dots, n,$ is something like
$$\begin{align}
{2^{p+1}} -1^{p+1} &= (p+1)1^{p}+ \cdots +(p+1)1+ 1\cdot 1^{0}\\
{3^{p+1}} - {2^{p+1}} &= (p+1)2^{p}+ \cdots +(p+1)2+ 1\cdot 2^{0}\\
& \vdots\\
(n+1)^{p+1} -{n^{p+1}} &= (p+1)n^{p}+ \cdots +(p+1)n+ 1\cdot n^{0}\\
\hline & \hline\\
(n+1)^{p+1} &= (p+1)\sum\limits_{k=1}^{n} k^{p}+ \cdots +(p+1)\sum\limits_{k=1}^{n} k^{1}+ \sum\limits_{k=1}^{n} k^{0} + k^{0}\\
\frac{(n+1)^{p+1}}{(p+1)} &= \sum\limits_{k=1}^{n} k^{p}+ \frac{{p+1 \choose p-1}}{(p+1)}\sum\limits_{k=1}^{n}k^{p-1} + \cdots +\frac{(p+1)}{(p+1)}\sum\limits_{k=1}^{n} k^{1}+ \frac{1}{(p+1)}(\sum\limits_{k=1}^{n} k^{0} + k^{0})\\
\end{align}$$
And assuming the proposition as true for $p-1$ we could write
$$\begin{align}
\frac{(n+1)^{p+1}}{(p+1)} &= \sum\limits_{k=1}^{n} k^{p}+ \textrm{terms involving powers of } n\leq p.\\
\sum\limits_{k=1}^{n} k^{p} &= \frac{(n+1)^{p+1}}{(p+1)}+ \textrm{terms involving powers of } n\leq p.\\
\end{align}$$
The question is:
How do $(n+1)^{p+1}$ instead of $n^{p+1}$ in the result still makes the proof valid?
| If I understand correctly you are concerned that the last part of the solution gives:
$\sum_{k=1}^{n}k^{p}=\frac{(n+1)^{p+1}}{p+1}+$ terms involving powers of $n\le p$
but the original question asked for:
$\sum_{k=1}^{n}k^{p}=\frac{n^{p+1}}{p+1}+$ terms involving powers of $n\le p$.
But this is not a problem since $(n+1)^{p+1}=n^{p+1}+$ terms involving powers of $n\le p$ by the binomial theorem. You can combine the constants of these new terms with the ones given in the solution. More explicitly, we will have:
$\sum_{k=1}^{n}k^{p}=\frac{(n+1)^{p+1}}{p+1}+\sum_{q=0}^{p}A_{q}n^{q}=\frac{n^{p+1}}{p+1}+\frac{1}{p+1}\sum_{q=0}^{p}\binom{p+1}{q}n^{q}+\sum_{q=0}^{p}A_{q}n^{q}=\frac{n^{p+1}}{p+1}+\sum_{q=0}^{p}\bigg(\frac{\binom{p+1}{q}}{p+1}+A_{q}\bigg)n^{q}$
| {
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System of quadratic equations $x^2 + y = 4$ and $x + y^2 = 10$ How would you solve the following system of equations:
$$
x^2 + y = 4 \\
x + y^2 = 10
$$
Thanks very much!
I tried defining y in terms of x and then inserting in to the second equation:
$$
y = 4 - x^2 \\
x + (4 - x^2)^2 = 10
$$
Expand the second equation:
$$
x + 16 - 8x^2 + x^4 = 10
$$
Rearrange the terms:
$$
x^4 - 8x^2 +x + 6 = 0
$$
I tried factoring out this polynomial to simplify it for solving, but didn't succeed :(
| $$y=4-x^2$$
put it in 2nd equation
$$x+(4-x^2)^2=10$$
$$x+16+x^4-8x^2=10$$
$$x^4-8x^2+x+6=0$$
$$x^4-x^3+x^3-7x^2-x^2+7x-6x+6=0$$
$$x^4-x^3+x^3-x^2-7x^2+7x-6x+6=0$$
$$x^3(x-1)+x^2(x-1)-7x(x-1)-6(x-1)=0$$
$$(x-1)(x^3+x^2-7x-6)=0\implies x-1=0\implies x=1$$
solve the above equation
one of the value of x=1 and y=3
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/467229",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Between Mertens' theorems It is well-known that
$$
\sum_{p\le x}\frac{\log p}{p}=\log x+O(1)
$$
and
$$
\sum_{p\le x}\frac1p=\log\log x+M+o(1).
$$
What is the order of
$$
\sum_{p\le x}\frac{\sqrt{\log p}}{p}
$$
?
| Let's calculate:
$$\begin{align}
\sum_{p\leqslant x} \frac{\sqrt{\log p}}{p} &= \sum_{n \leqslant x} \frac{\sqrt{\log n}}{n}\left(\pi(n) - \pi(n-1)\right)\\
&= \sum_{n \leqslant x}\frac{\sqrt{\log n}}{n}\pi(n) - \sum_{n \leqslant x-1}\frac{\sqrt{\log (n+1)}}{n+1}\pi(n)\\
&= \sum_{n \leqslant x} \left(\frac{\sqrt{\log n}}{n} - \frac{\sqrt{\log (n+1)}}{n+1}\right)\pi(n) - \frac{\sqrt{\log x}}{x}\pi(x) + O(1)\\
&= \sum_{n\leqslant x} \frac{\sqrt{\log n} - n\left(\sqrt{\log (n+1)} - \sqrt{\log n}\right)}{n(n+1)}\pi(n) + O(1).
\end{align}$$
The first $O(1)$ comes from replacing $\lfloor x\rfloor$ with $x$ for the term after the sum. Since $\frac{\sqrt{\log x}}{x}\pi(x) \sim \frac{1}{\sqrt{\log x}}$, we can absorb it in the $O(1)$ in the next line.
Now we split the sum, the second part is bounded, we have
$$n\left(\sqrt{\log (n+1)} - \sqrt{\log n}\right) = \frac{n(\log (n+1) - \log n)}{\sqrt{\log (n+1)} + \sqrt{\log n}} = \frac{n\log (1 + \frac1n)}{\sqrt{\log (n+1)} + \sqrt{\log n}} \approx \frac{1}{2\sqrt{\log n}},$$
and thus
$$\begin{align}
\sum_{n\leqslant x} \frac{n\left(\sqrt{\log (n+1)} - \sqrt{\log n}\right)}{n(n+1)}\pi(n) &\leqslant \sum_{n\leqslant x} \frac{\pi(n)}{2n(n+1)\sqrt{\log n}}\\
&\leqslant \sum_{n\leqslant x} \frac{1}{(n+1)(\log n)^{3/2}}\\
&= O(1).
\end{align}$$
So it remains to look at
$$\sum_{n \leqslant x} \frac{\sqrt{\log n}}{n(n+1)}\pi(n).$$
Using $\displaystyle\pi(x) = \frac{x}{\log x} + O\left(\frac{x}{(\log x)^2}\right)$, we observe that the second term again yields a sum of $1/\left(n(\log n)^{3/2}\right)$ and hence remains bounded. For the remaining part, we find
$$
\sum_{2 \leqslant n \leqslant x} \frac{\sqrt{\log n}}{n(n+1)}\frac{n}{\log n}
= \sum_{2 \leqslant n \leqslant x} \frac{1}{(n+1)\sqrt{\log n}}
$$
sandwiched:
$$\int_3^{x+1} \frac{dt}{(t+1)\sqrt{\log t}} \leqslant \sum_{2 \leqslant n \leqslant x} \frac{1}{(n+1)\sqrt{\log n}} \leqslant \int_2^{x} \frac{dt}{(t+1)\sqrt{\log t}}$$
Approximating,
$$\int_a^b \frac{dt}{t\sqrt{\log t}} = \int_{\log a}^{\log b} \frac{du}{\sqrt{u}} = 2\left(\sqrt{\log b} - \sqrt{\log a}\right),$$
with the difference to the original integrand being of the order $t^{-2}(\log t)^{-1}$ with a finite integral.
Altogether,
$$\sum_{p\leqslant x} \frac{\sqrt{\log p}}{p} = 2\sqrt{\log x} + O(1).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/467486",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Derivative of $\frac {x\cdot\left(1 - 3x\right)}{\sqrt{x-1}}$ Problem. Find the first derivative of $$ \dfrac {x \left( 1 - 3x \right)}{\sqrt{x-1}} $$
Work. Let $u = x-1$ and $y = \dfrac {(u+1)(-3u-2)}{\sqrt{u}} $
Using the chain rule, I got$$\dfrac{(-9x^2-5x+2)}{(2(x-1)^\frac{3}{2})}$$
But the answer is $$\dfrac{(-9x^2+13x+2)}{\left(2(x-1)^\frac{3}{2}\right)}$$
I'm not sure what I did wrong, maybe something related to the $(-3u-2)$?
| You don't have to use the Chain Rule. You can, instead, use the Product Rule.
$ f(x) = \dfrac {x \cdot (1 - 3x)}{\sqrt{x-1}} = \dfrac {x-3x^2}{\sqrt{x-1}} $
$ f'(x) = \dfrac {\sqrt {x-1} \cdot (1-6x) - (x-3x^2) \cdot \left( \sqrt{x-1} \right)'}{x-1} $
$ f'(x) = \dfrac {\sqrt{x-1} \cdot (1-6x) - (x-3x^2) \cdot \dfrac {1}{2\sqrt{x-1}}}{x-1} $
I used the Chain Rule here only to find the derivative of $ \sqrt {x-1} $, which, it seems, you know how to do. I'll leave the rest as an exercise for you -- just simplify. You should get the right answer. Good luck!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/467592",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Can someone explain this trigonometric limit without L'Hopital? I can not solve this limit:
$$\lim \limits_{x\to 0}\frac{x^2}{1-\sec(x)}$$
$$\lim \limits_{x\to 0} \frac{x^2}{1-\sec(x)}=\lim \limits_{x\to 0}\frac {x^2}{1-\sec(x)}\cdot{\frac{1+\sec(x)}{1+\sec(x)}}=\lim \limits_{x\to 0}\frac{x^2(1+\sec(x))}{1-\sec^2(x)}=\lim \limits_{x\to 0}\frac{x^2(1+\sec(x))}{-\tan^2(x)}$$
| $\lim \limits_{x\to 0} \frac{x^2}{1-\sec x}= \lim \limits_{x\to 0} \frac {x^2}{1-\sec x}\cdot{\frac{1+\sec(x)}{1+\sec(x)}}=\lim \limits_{x\to 0} \frac{x^2(1+\sec(x))\cos^2 x}{\cos^2 x-1}=\lim \limits_{x\to 0} \frac{x^2(1+\sec(x))\cos^2 x}{-\sin^2(x)}$
Now if you know the limit of $\frac {\sin x}x$ the other terms are well behaved.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/468674",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Calculate $\int \frac{dx}{x\sqrt{x^2-1}}$ I am trying to solve the following integral
$$\int \frac{dx}{x\sqrt{x^2-1}}$$
I did the following steps by letting $u = \sqrt{x^2-1}$ so $\text{d}u = \dfrac{x}{\sqrt{{x}^{2}-1}}$ then
\begin{align}
&\int \frac{\sqrt{x^2-1} \, \text{d}u}{x \sqrt{x^2-1}} \\
&\int \frac{1}{x} \text{d}u \\
&\int \frac{1}{\sqrt{u^2+1}} \text{d}u\\
\end{align}
Now, this is where I am having trouble. How can I evaluate that? Please provide only hints
Thanks!
EDIT:
The problem specifically states that one must use substitution with $u = \sqrt{x^2-1}$. This problem is from the coursera course for Single Variable Calculus.
| $$u^{2}=x^{2}-1\Rightarrow xdx=udu\\
\int \frac{dx}{x\sqrt{x^2-1}}=\int \frac{xdx}{(x^{2}-1+1)\sqrt{x^2-1}}\\
=\int\frac{udu}{(u^{2}+1)u} =\int\frac{du}{u^{2}+1}=\arctan(u)+c=\arctan\sqrt{x^{2}-1}+c$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/468727",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 7,
"answer_id": 5
} |
Subsets and Splits