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Compute integral $\int_0^1\int_0^1\int_0^1 \sqrt{x^2+y^2+z^2} \,\mathrm{d}x\mathrm{d}y\mathrm{d}z$ How to Compute
$$\int_0^1\int_0^1\int_0^1 \sqrt{x^2+y^2+z^2} \,\mathrm{d}x\mathrm{d}y\mathrm{d}z $$
The Mathematica 9.01 give a result is $$-\frac{\pi}{24}+\frac{1}{4}(\sqrt{3}+\log(7+4\sqrt{3})).$$ I want to know how to get it.
| You are computing the average distance from a vertex in a unit hypercube. Assuming that $X,Y,Z$ are three independent random variables, uniformly distributed over $(0,1)$, and computing the probability density function of $W=X^2+Y^2+Z^2$ (that is a piecewise-simple function), it follows that
$$\mathbb{E}[\sqrt{W}] = \color{red}{\frac{\sqrt{3}}{4}+\frac{\log(2+\sqrt{3})}{2}-\frac{\pi}{24}}=0.960591956455\ldots $$
As an alternative, you may just integrate $\sqrt{x^2+y^2+z^2}$ with respect to $x$ first, getting
$\frac{1}{2} x \sqrt{x^2+y^2+z^2}+\frac{1}{2} \left(y^2+z^2\right) \log\left(x+\sqrt{x^2+y^2+z^2}\right)$, then integrate the resulting expression with respect to $y$ and $z$. Lengthy but perfectly doable.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solve $x^{3}-3x=\sqrt{x+2}$ Solve for real $x$
$$x^{3}-3x=\sqrt{x+2}$$
By inspection, $x=2$ is a root of this equation. So, I squared both sides and divided the six degree polynomial obtained by $x-2$. Then I got a quintic which I couldn't solve despite applying rational root theorem and substitutions. I believe that there must be some nice method to solve this which I can't think about. Please help. Thanks!
| You shouldn't dismiss Karo's graph. Drawing a graph to get a feel for the equation (when you don't know how to proceed) is most helpful. And this isn't nonsense, in fact the graph is the key for the solution, as it makes it clear that we can consider only $x$ for which $-2 \leq x \leq 2$. This can be conveniently rewritten as $-1\leq\frac{x}{2}\leq1$.
This inequality reminds us immediately of the one that $\sin a$ and $\cos a$ also satisfy, and indeed we can assume WLOG $\frac{x}{2}=\sin a$, for some angle $a$ in radians of course. This makes $x=2 \sin a$, so let's substitute this in the equation.
$$\begin{align}
8\sin^3 a-6\sin a&=\sqrt{2(\sin a+1)}\\
-2(-4\sin^3a+3\sin a)&=\sqrt{2(\sin a+1)}\\
-2\sin3a&=\sqrt{2(\sin a+1)}
\end{align}$$
It helped that we could factor it into $\sin 3a$, but then again we can't get rid of the square root in a nice way. This is because $\sin^2a$ is given in terms of $\cos 2a$, and not $\sin 2a$, so we can't turn the radicand into a nice square of a sine. On the other hand, $\cos^2 a$ can be expressed in terms of a cosine (indeed, we have the trigonometric identity $\cos^2a=\frac{1+\cos2a}{2}$). So we are lead to believe that the substituition $x=2\cos a$ is much more fortunate. It's worth a shot:
$$\begin{align}
8\cos^3 a-6\cos a&=\sqrt{2(\cos a+1)}\\
2(4\cos^3a-3\cos a)&=\sqrt{4\frac{(\cos a+1)}{2}}\\
2\cos3a&=\sqrt{2^2\cos^2\frac{a}{2}}\\
\cos3a &=\cos\frac{a}{2}
\end{align}$$
The equation has been successfully trivialized. We should note that imposing $0\leq a\leq\pi$, $\cos a$ will still assume all of the values it possibly could, so we impose this restriction on $a$.
Therefore, we can have $3a=\frac{a}{2}\Rightarrow a=0$, which gives us the solution $x=2\cos0=2$, or we can have the other, non-trivial solutions:
$$3a=\frac{a}{2}+2\pi\Rightarrow a=\frac{4\pi}{5}$$
which gives us $x=2\cos\frac{4\pi}{5}=-2\cos\frac{\pi}{5}$. If $0\leq a\leq\pi$, then $0\leq 3a \leq 3\pi\lt4\pi$, so we don't need to look the case $3a=\frac{a}{2}+4\pi$ or higher, and we just need to consider the last case:
$$3a=2\pi-\frac{a}{2}\Rightarrow a=\frac{4\pi}{7}$$
Which gives us the last solution, $x=2\cos\frac{4\pi}{7}=-2\cos\frac{3\pi}{7}$.
Note: This solution was inspired by the one brazilian mathematician Nicolau C. Saldanha presented in a lesson. Here's the link: http://y2u.be/jFFdOSsVGgg (someone who's fluent in Spanish shouldn't have trouble understanding the explanations). It is also shown that with some simple manipulations on the regular pentagon, one can arrive at $\cos\frac{\pi}{5}=\frac{1+\sqrt{5}}{4}$, so the second solution can be written as $x=-\frac{1+\sqrt{5}}{2}$ (which justifies the quadratic factor $x^2-x-1$ of the 6th degree polynomial you get when you square both sides, which obviously contains some extraneous roots).
I'm not sure there's an easy way to turn $\cos\frac{3\pi}{7}$ into one algebraic term (if there is one at all), so we just leave it that way. The solution set is $$x \in \left. \left \{ 2,-\frac{1+\sqrt{5}}{2},-2\cos\frac{3\pi}{7} \right. \right \}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Get variables with Matrix I try to get the variables for this equation:
$$\begin{cases}
6x_1 + 4x_2 + 8x_3 + 17x_4 &= -20\\
3x_1 + 2x_2 + 5x_3 + 8x_4 &= -8\\
3x_1 + 2x_2 + 7x_3 + 7x_4 &= -4\\
0x_1 + 0x_2 + 2x_3 -1x_4 &= 4
\end{cases}$$
So i started with:
$$ \begin{pmatrix}
6 & 4 & 8 & 17 & -20 \\
3 & 2 & 5 & 8 & -8 \\
3 & 2 & 7 & 7 & -4\\
0 & 0 & 2 & -1 &4\\
\end{pmatrix}$$
Then I continued: $$
\begin{pmatrix}
6 & 4 & 8 & 17 & -20 \\
-6 & -4 & -10 & -16 & 16 \\
-6 & -4 & -14 & -14 & 8\\
0 & 0 & 2 & -1 &4\\
\end{pmatrix}
$$
And began to count:
$$
\begin{pmatrix}
6 & 4 & 8 & 17 & -20 \\
0 & 0 & -2 & -1 & -4 \\
0 & 0
& -6 & 3 & -12\\
0 & 0 & 2 & -1 &4\\
\end{pmatrix}
$$
And finally I transformed it to:$$
\begin{pmatrix}
6 & 4 & 8 & 17 & -20 \\
0 & 0 & -2 & -1 & -4 \\
0 & 0
& -6 & 3 & -12\\
0 & 0 & 0 & 0 &0\\
\end{pmatrix}
$$
What did i wrong? I think the last row cannot be correct because:
$$(0, 0 ,0 , 0) \neq 0x_1 + 0x_2 + 2x_3 -1x_4 = 4$$
How should i solve it? Thanks
| As Git Gud points out, it should be 1, not -1. Now if you correct that and proceed further, the third row also will become zero, and you'll get
$\left(\begin{matrix}
6 & 4 & 8 & 17 & -20\\
0 & 0 & -2 & 1 & -4\\
0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0
\end{matrix}\right)$
so that
$6x_1 + 4x_2 + 8x_3 + 17x_4 = -20\\
-2x_3 + x_4 = -4$
which can be solved by back-substitution.
From the second equation, you'll get $x_4 = 2x_3 - 4$. Substitute this in the first to get $x_1$ in terms of $x_2$ and $x_3$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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If $ab=3$ and $\frac1{a^2}+\frac1{b^2}=4,$ then $(a-b)^2=\;$? If $ab=3$ and $\frac1{a^2}+\frac1{b^2}=4$, what is the value of $(a-b)^2$? I think $a^2+b^2=36$, please confirm and is it possible to to figure out one of the variables?
| $$\frac1{a^2}+\frac1{b^2}=4$$
$$a^2b^2\left(\frac1{a^2}+\frac1{b^2} \right)=4a^2b^2$$
$$b^2+a^2=(2ab)^2$$
$$a^2-2ab+b^2=(2ab)^2-2ab$$
$$(a-b)^2=(2ab)^2-2ab=(2\cdot3)^2-2\cdot3=30$$
| {
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"timestamp": "2023-03-29T00:00:00",
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prove that $ a^2 b^2 \left( {a^2 + b^2 - 2} \right) \ge \left( {a + b} \right)\left( {ab - 1} \right) $ good evening I want to show that
$(1)a;b\in\mathbb {R^*_+}:a^2 b^2 \left( {a^2 + b^2 - 2} \right) \ge \left( {a + b} \right)\left( {ab - 1} \right)
$
$
\begin{array}{l}
\frac{a}{b} + \frac{b}{a} \ge 2 \\
\frac{a}{b} + \frac{b}{a} = \frac{{a^2 + b^2 }}{{ab}} \Rightarrow a^2 + b^2 = ab\left( {\frac{a}{b} + \frac{b}{a}} \right) \ge 2ab \\
a^2 b^2 \left( {a^2 + b^2 - 2} \right) \ge a^2 b^2 \left( {2ab - 2} \right) = 2a^2 b^2 \left( {ab - 1} \right) \\
\end{array}
$
$ x;y;z\in\mathbb {R^*_+}:$
$(2)\sum_{cyc}^{ } xy(x+y-z)\ge \sqrt {3(x^3y^3+y^3z^3+z^3x^3)}$
thank you in advance
| Do you keep the first true in the case of
$a;b\in\mathbb {R^*_+} and : a^2 + b^2 \ge 2$
$ (2)x;y;z\in\mathbb {R^*_+}:$
$\sum_{cyc}^{ } xy(x+y-z)\ge \sqrt {3(x^3y^3+y^3z^3+z^3x^3)}$
Thank you for all helping and guidance. ...
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Why does $u$-substitution not work here? $$ \int{\frac{1}{2y}dy} $$
Method 1:
$$\int{\frac{1}{2y}dy} = \frac{1}{2}\int{\frac{1}{y}dy} = \frac{1}{2}\ln|y|+C$$
Method 2 ($u$-substitution):
$$\int{\frac{1}{2y}dy} = \int{\frac{1}{u}dy} = \frac{1}{2}\int{\frac{1}{u}(2)dy}= \frac{1}{2}\int{\frac{1}{u}du}= \frac{1}{2}\ln|u|+C = \frac{1}{2}\ln|2y|+C$$
$$u=2y$$
$$du = 2 dy$$
Why is method 2 wrong?
| $$\ln |2y| = \ln 2 + \ln |y|$$
$$ \frac{1}{2}\ln|2y|+C = \frac 12 \ln |y| + \underbrace{\frac 12 \ln 2 + C}_{\large =\, C'}$$
So your results are equivalent up to a constant.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/820747",
"timestamp": "2023-03-29T00:00:00",
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How to find integral $\underbrace{\int\sqrt{2+\sqrt{2+\sqrt{2+\cdots+\sqrt{2+x}}}}}_{n}dx,x>-2$ Find the integral
$$\int\underbrace{\sqrt{2+\sqrt{2+\sqrt{2+\cdots+\sqrt{2+x}}}}}_{n}dx,x>-2$$
where $n$ define the number of the square
I know this if
$0 \le x\le 2$, then let $$x=2\cos{t},0\le t\le\dfrac{\pi}{2}$$
so
$$\sqrt{2+x}=\sqrt{2+2\cos{t}}=2\cos{\dfrac{t}{2}}$$
so
$$\sqrt{2+\sqrt{2+x}}=2\cos{\dfrac{t}{2^2}}$$
so
$$\int\sqrt{2+\sqrt{2+\sqrt{2+\cdots+\sqrt{2+x}}}}dx=\int2\cos{\dfrac{t}{2^n}}(-2\sin{t})dt$$
and for $x\ge 2$ case, I let
$x=\cosh{t}$, but for $-2\le x\le 0$ case, I can't do it.
| I think you have the wrong approach, take a look at this
$$f(x) = \sqrt{2+\sqrt{2+...+\sqrt{2+x}}} $$
Square both sides
$$(f(x))^2 = 2+\sqrt{2+...+\sqrt{2+x}} $$
$$(f(x))^2 = 2+f(x) $$
$$(f(x))^2 -f(x)-2= 0 $$
This has two solutions $f(x)=2$ and $f(x)=-1$
Note that $f(x)$ cannot be -1, since square root is always positive for real numbers, and it is given that $x > -2$
Hence you have $f(x)=2$, which is independent of $x$
The answer is therefore finally 2x
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Evaluate $ \int ^2 _1 \frac{\ln x}{x^2} \text{dx}$ $\displaystyle \int ^2 _1 \dfrac{\ln x}{x^2} \text{dx}$
My working so far:
$u=\ln x, v'=x^2$
$u'=x^{-1}, v=\dfrac{1}{3}x^3$
Integrating by parts:
$\displaystyle \dfrac{1}{3}x^3\ln x-\int\dfrac{1}{3}x^3\cdot x^{-1}$
$\displaystyle \left[\dfrac{1}{3}x^3\ln x-\int\dfrac{1}{3}x^2\right]^2 _1$
$\left[\dfrac{1}{3}x^3\ln x-\dfrac{1}{9}x^3\right]^2_1$
Is this right so far? The answer is $\frac{1}{2}(1-\ln2)$ but from the integral I got I don't think I can get this answer.
| You should have $v'=x^{-2}$, so $v=-x^{-1}$, now go from there.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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What is $f '(x)$ for $f(x)=(x-3)^3$?
What is $f '(x)$ for $f(x)=(x-3)^3$?
I'm thinking it is $3x^2 - 18x + 27$
but my textbook says it is $3x^2 - 18x - 27$
| Note that
$$
f(x)=U^n\\
f^{\prime}(x)=nU^{\prime}U^{n-1}
$$
Then
$$\begin{align}
f^{\prime}(x)&=3(x-3)^{2}\\
&=3(x-3)(x-3)\\
&=3(x(x-3)-3(x-3))\\
&=3(x^2-3x-3x+9)\\
&=3(x^2-6x+9)\\
&=3x^2-18x+27\\
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/828305",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Show that $\sum_{k=1}^{n}\frac{x_k^2}{x_k^2-2x_k\cos(\frac{2\pi}{n})+1}\geq 1$
Let $n>2$ an integer and $x_k>0$ with $x_1\cdot x_2\cdots x_n=1$
Show that $$\sum_{k=1}^{n}\frac{x_k^2}{x_k^2-2x_k\cos(\frac{2\pi}{n})+1}\geq 1$$
I tried an induction without succeed, I do not really have idea to approach this inequality.
Any suggestion is appreciated.
| it is true that $n\ge 6$ the inequality is always true because
$ \frac{x_k^2}{x_k^2-2x_k\cos(\frac{2\pi}{n})+1}\geq 1$ is always true when $x_k \ge 1$, proof:
$\frac{x_k^2}{x_k^2-2x_k\cos(\frac{2\pi}{n})+1}\geq 1 \iff x_k \ge \dfrac{1}{2\cos(\frac{2\pi}{n})} \iff 1 \ge \dfrac{1}{2\cos{(\frac{2\pi}{n})}} \iff \cos{(\frac{2\pi}{n})} \ge \dfrac{1}{2} \iff \dfrac{2\pi}{n} \le \dfrac{\pi}{3} \iff n \ge 6$
so it is remain $n=3,4,5$
$n=3$ is already proved.
$f_n(x)=\frac{x^2}{x^2-2x\cos(\frac{2\pi}{n})+1}$ is mono increasing function when $n=3,4,5$.then the result is easy to induct.
| {
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"timestamp": "2023-03-29T00:00:00",
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linear recursion $y_n=A \cdot y_{n-1}$ Let $a,b, \in \mathbb{R}$. Let $x_0=a, x_1=b$ and $x_n=\frac{x_{n-1}+x_{n-2}}{2}$ for $n \geq 2$
(i) Write the recursion in the form $y_n=A \cdot y_{n-1}$ where $A$ is a $2 \times 2$ matrix and $y_i=\begin {pmatrix} x_i \\ x_{i-1} \end{pmatrix}$
(ii) Find a diagonal matrix $D$ and an invertible matrix $S$, so that $A=SDS^{-1}$
(iii) Calculate $\lim_{n \rightarrow \infty}S^{-1}A^nS$
(iv) Deduce from that $\lim_{n \rightarrow \infty}A^n$ and $\lim_{n \rightarrow \infty}x_n$
My solution so far:
(i) $\begin {pmatrix} x_i \\ x_{i-1} \end{pmatrix}=\begin {pmatrix} \frac{1}{2} & \frac{1}{2} \\ 1 & 0 \end{pmatrix} \begin {pmatrix} x_{i-1} \\ x_{i-2} \end{pmatrix}$
(ii) Eigenvalues are $\lambda_1=1$ and $\lambda_1=-\frac{1}{2}$.
Basis of $V_{\lambda_1}$ is $B_{\lambda_1} =\begin {pmatrix} 1 \\ 1 \end{pmatrix}$
Basis of $V_{\lambda_2}$ is $B_{\lambda_2} =\begin {pmatrix} -1 \\ 2 \end{pmatrix}$
$\Rightarrow D=\begin {pmatrix} 1 & 0 \\ 0 & -\frac{1}{2} \end{pmatrix}, S\begin {pmatrix} 1 & -1 \\ 1 & 2 \end{pmatrix}, S^{-1}= \begin {pmatrix} \frac{2}{3} & \frac{1}{3} \\ -\frac{1}{3} & \frac{1}{3} \end{pmatrix}$
(iii)
$\lim_{n \rightarrow \infty}S^{-1}A^nS = \lim_{n \rightarrow \infty}S^{-1}(SDS^{-1})^nS=\lim_{n \rightarrow \infty}S^{-1}(SD^nS^{-1})S=\lim_{n \rightarrow \infty}D^n= \lim_{n \rightarrow \infty} \begin {pmatrix} 1 & 0 \\ 0 & (-\frac{1}{2})^n \end{pmatrix}$
Is this correct so far? How to go on now?
| Because $\lim_{n \rightarrow \infty}(-\frac{1}{2})^n=0$, we have
$$\lim_{n \rightarrow \infty}D^n= \lim_{n \rightarrow \infty} \begin {pmatrix} 1 & 0 \\ 0 & (-\frac{1}{2})^n \end{pmatrix}=\begin {pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}$$
To simplify the notation, we Define $X^\infty:=\lim_{n \rightarrow \infty}X^n$ and $Y_\infty:=\lim_{n \rightarrow \infty}Y_n$
Then
$$A^\infty=SD^\infty S^{-1}=\begin {pmatrix} \frac{2}{3} & \frac{1}{3} \\ \frac{2}{3} & \frac{1}{3} \end{pmatrix}$$
Therefore (with $x_1=\{a,b\}$)
$$y_\infty=A^\infty.x_1$$
From any component of vector $y_\infty$, we have
$$x_\infty=\frac{2}{3}a+\frac{1}{3}b$$
This is consistent because $\lim_{n \rightarrow \infty}x_n=\lim_{n \rightarrow \infty}x_{n-1}$.
| {
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Finding $\lim_{x\rightarrow 0}\frac{10^x-2^x-5^x+1}{x\tan(x)}= $? How to calculate the following limit : $$\lim_{x\rightarrow 0}\frac{10^x-2^x-5^x+1}{x\tan(x)}$$
Thanks in advance.
| Notice that by the Taylor series we have
$$a^x=\exp(x\ln a)=_0 1+x\ln a+\frac{\ln^2 a}2x^2+o(x^2)$$
and
$$x\tan x= x^2+o(x^2)$$
so we find easily that
$$\lim_{x\to0}\frac{10^x-2^x-5^x+1}{x\tan x}=\frac{\ln^2(10)-\ln^22-\ln^25}{2}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Area of the intersection of two discs: Integral solution? Here is the problem :
We consider two circles that intersect in exactly two points. There $O_1$ the center of the first and $r_1$ its radius. There $O_2$ the center of the second and $r_2$ its radius. We note $d=O_1O_2$. Question: Express the area of the intersection of the two disks with the distances $d$, $r_1$ and $r_2$.
My proof :
I call $ A $ and $ B $ the points of intersection of the two disks and $C$ is the intersection point of $d$ and $AB$ ($O_1C=x$).
I selected a landmark $\mathcal R (O, \vec e_x , \vec e_y )$ such that $O_1$ is centered at $(0,0)$ and $(d, 0)$ for $O_2$
The equation of the two circles are $x^2+y^2=r_1^2$ and $(x-d)^2=y^2=r_2^2$.
We have $x=\frac{d^2+r_1^2-r_2^2}{2d}$, then $$y^2=r_1^2-x^2=r_1^2-\left(\frac{d^2+r_1^2-r_2^2}{2d}\right)^2$$
Writing that $AB=2y$ we have $$AB=\frac{\sqrt{4d^2r_1^2-(d^2-r_2^2+r_1^2)^2}}{d}$$
I am going to calculate the half of the requested area.
Denote $\theta$ the angle of $AO_1B$. The half of the requested area is :
$A_1$=Area (sector)$-$Area (isosceles triangle $O_1$) =$$\frac{\pi r_1^2\theta}{2\pi}-\frac{1}{2}r_1^2\sin(\theta)$$
Furthermore $$\sin\left(\frac{\theta}{2}\right)=\frac{AB}{2r_1}=\frac{\sqrt{4d^2r_1^2-(d^2-r_2^2+r_1^2)^2}}{2dr_1}\qquad \cos\left(\frac{\theta}{2}\right)=\frac{x}{r_1}$$
Therefore $$A_1=r_1^2\arccos\left(\frac{d^2-r_1^2+r_2^2}{2dr_1}\right)-\frac{(d^2-r_1^2+r_2^2)AB}{4d}$$
Using $sin(x)=2\sin(\frac{x}2)\times \cos(\frac{x}2)$.
After some algebra to compute the request area we have
$$A=r_1^2\arccos\left(\frac{d^2-r_1^2+r_2^2}{2dr_1}\right)+r_2^2\arccos\left(\frac{d^2-r_2^2+r_1^2}{2dr_2}\right)- \frac{\sqrt{Y}}{2} $$
where $$Y=(-d+r_1+r_2)(d+r_2-r_1)(d-r_2+r_1)(d+r_1+r_2)$$
My Question:
Can we proved this result with integrals for exemple ?
| I derived my own proof with simple geometry and trigonometry. I ended up with a slightly different formula:
$$
A=r_1^2 \text{arccos}\left(\frac{d^2-(r_2^2-r_1^2)}{2dr_1}\right) + r_2^2 \text{arccos}\left(\frac{d^2+(r_2^2-r_1^2)}{2dr_2}\right)-d\sqrt{r_1^2-\left(\frac{d^2-(r_2^2-r_1^2)}{2d}\right)^2}.
$$
The last term can be rewritten:
$$
-d\sqrt{r_1^2-\left(\frac{d^2-(r_2^2-r_1^2)}{2d}\right)^2} = -\frac{\sqrt{Y}}{2},
$$
with
$$
Y=(-d+r_1+r_2)(d+r_2-r_1)(d-r_2+r_1)(d+r_1+r_2).
$$
So this is very close to the first submitted equation, apart from a swap between r1 and r2 in the argument of the arccos(). It seems that this corrects the typo pointed out in the comment. For d=3, r1=1, r2=2 it gives
$$
A=r_1^2 \text{arccos}\left(1\right) + r_2^2 \text{arccos}\left(1\right)+0=0
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/833960",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Equations of lines tangent to an ellipse Determine the equations of the lines that are tangent to the ellipse $\displaystyle{\frac{1}{16}x^2 + \frac{1}{4}y^2 = 1}$
and pass through $(4,6)$.
I know one tangent should be $x = 4$ because it goes through $(4,6)$ and is tangent to the ellipse but I don't know how to find the other tangents. Any help is appreciated.
| You need a line that passes through the point $(4,6)$ and that touches the ellipse at just one point. The vertical line does that and you've already found it. Obviously there is exactly one other tangent line (and if that's not obvious to you, then draw the picture and look at it!).
Nonvertical lines passing through the point $(4,6)$ have equation $y-6=m(x-4)$.
That implies $y=mx-4m+6$, so we can put $mx-4m+6$ in place of $y$:
$$
\frac{x^2}{16} + \frac{(mx-4m+6)^2}{4} = 1.
$$
This is equivalent to
$$
\underbrace{(1+4m)}x^2 + \underbrace{-8m(4m-6)}\ x + \underbrace{4(4m-6)^2 -16} = 0.
$$
This equation is quadratic in $x$. We therefore want a quadratic equation with exactly one solution. A quadratic equation $ax^2+bx+c=0$ has exactly one solution precisely if its discriminant $b^2-4ac$ is $0$. So we have
$$
b^2-4ac = \underbrace{64m^2(4m-6)^2 - 4(1+4m)(4(4m-6)^2-16) = 0}.
$$
Now we only need to solve this last equation for $m$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/834392",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
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} |
Find the value of a + b + c + d Let $a$ and $b$ be the roots of the equation: $x^2 - 10cx - 11d = 0$ where $c$ and $d$ be the roots of $x^2 - 10ax - 11b = 0$. Find the value of $a+b+c+d$, assuming that they all are distinct.
I first tried making an equation with roots $(a+b)$ and $(c+d)$ to get the sum of the roots, however I wasn't able to solve this question using that method as the answer which I got was in terms of the variables itself.
I also tried placing $a$ into the first equation and $c$ into the second to cancel out a common term ($-10ac$), but after cancelling, I got: $(a^2 - c^2 - 11d + 11b = 0)$. Now I don't know how to move ahead.
| We have
$a+b=10c$
and
$c+d=10a$, hence
$$b=10c-a$$
and
$$d=10a-c$$
We also have $ab=-11d$ and $cd=-11b$, hence
$$10ac-a^2=-11(10a-c)$$
and
$$10ac-c^2=-11(10c-a)$$
Subtracting these gives
$$a^2-c^2=-11(10c-a)+11(10a-c)=121(a-c)$$
Assuming $a\not=c$, we get
$$a+c=121$$
Going back to the formulas for $b$ and $d$, we have
$$b+d=(10c-a)+(10a-c)=9(a+c)$$
Thus
$$a+b+c+d=(a+c)+(b+d)=(a+c)+9(a+c)=10(a+c)=1210$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/834745",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 1
} |
Prove this identity... $$\frac{\sin 2x}{1+\cos 2x} \times \frac{\cos x}{1+\cos x}=\tan\frac{x}{2}$$
This is what I've done:
$$\frac{2\sin x \cos x}{1+\cos^2 x-\sin^2 x} \times \frac{\cos x}{1+\cos x}=$$
$$\frac{2\sin x \cos x}{2\cos^2 x} \times \frac{\cos x}{1+\cos x}=$$
$$\frac{\sin x}{1+\cos x}$$
I have no idea what to do next.
edit:
Solution:
$$\frac{\sin2\frac{x}{2}}{1+\cos2\frac{x}{2}}=$$
$$\frac{2\sin\frac{x}{2}cos\frac{x}{2}}{\sin^2\frac{x}{2}+\cos^2\frac{x}{2}+\cos^2\frac{x}{2}-\sin^2\frac{x}{2}}=$$
$$\frac{2\sin\frac{x}{2}\cos\frac{x}{2}}{2\cos^2\frac{x}{2}}=$$
$$\frac{\sin\frac{x}{2}}{\cos\frac{x}{2}}=\tan\frac{x}{2}$$
| $\cos 2\theta = 2\cos^2 \theta - 1 \Rightarrow 1 + \cos 2\theta = 2\cos^2 \theta$. Therefore:
$$\dfrac{\sin 2x}{1 + \cos 2x} \times \dfrac{\cos x}{1 + \cos x}\\
= \dfrac{\not 2\sin x \not\cos x}{\not 2 \not\cos^2 x} \times \dfrac{\not\cos x}{2 \cos^2 (x/2)}\\
= \dfrac{\not 2\sin(x/2)\not\cos(x/2)}{\not 2\cos^{\not 2} (x/2)}\\
= \boxed{\tan \dfrac{x}{2}}
$$
Note: In the above, the $\not\square$s denote cancellations.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/837282",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Asymptotic value of Fibonacci numbers It is well known that $F_n\sim\frac{\phi^n}{\sqrt{5}}$, where $\phi=\frac{1+\sqrt{5}}{2}$. Does someone know a better estimate? With proof please.
I'm trying to calculate the following limit:
Let $u_1=1, u_2=C, u_3=C$ and
$$
u_n=\frac{C^{F_{n-1}}}{2^{F_{n-3}}3^{F_{n-4}}\cdots(n-2)^{F_1}}
$$ for $n\geq 4$, where
$$
C=\exp\left(\sum_{k=1}^{\infty}\frac{\log k}{\phi^k}\right).
$$
Find $\lim_{n\to\infty}u_n$.
| If we let
$$
a_n = 2^{\large F_{n-3}} 3^{\large F_{n-4}} \cdots (n-2)^{\large F_{1}} = \prod_{k=2}^{n-2} k^{\large F_{n-1-k}}
$$
then
$$
\log a_n = \sum_{k=2}^{n-2} F_{n-1-k} \log k. \tag{1}
$$
Using Binet's formula
$$
F_m = \frac{\varphi^m - (-\varphi)^{-m}}{\sqrt{5}},
$$
$(1)$ becomes
$$
\begin{align}
\sqrt{5} \log a_n &= \varphi^{n-1} \sum_{k=2}^{n-2} \varphi^{-k} \log k - (-\varphi)^{1-n} \sum_{k=2}^{n-2} (-\varphi)^{k} \log k \\
&= b_n + c_n,
\end{align}
$$
where
$$
b_n = \varphi^{n-1} \sum_{k=2}^{n-2} \varphi^{-k} \log k \qquad \text{and} \qquad c_n = -(-\varphi)^{1-n} \sum_{k=2}^{n-2} (-\varphi)^{k} \log k. \tag{2}
$$
The sum in $b_n$ converges as $n \to \infty$, so we'll rewrite it as
$$
b_n = \varphi^{n-1} \sum_{k=2}^{\infty} \varphi^{-k} \log k - \varphi^{n-1} \sum_{k=n-1}^{\infty} \varphi^{-k}\log k. \tag{3}
$$
Lemma 1:
$$
\sum_{k=n-1}^{\infty} \varphi^{-k}\log k = \frac{\varphi^{2-n}\log n}{\varphi - 1} + O\left(\frac{\varphi^{-n}}{n}\right). \tag{4}
$$
Idea of the proof: Split the sum at $k=\frac{3}{2}n$. For the sum over $k \geq \frac{3}{2} n$ use the rough estimate $\log k < k$, and for the sum over $n-1 \leq k < \frac{3}{2} n$ expand $\log k$ in series about the point $k = n-1$.
Lemma 2:
$$
\sum_{k=2}^{n-2} (-\varphi)^{k}\log k = \frac{-(-\varphi)^{n-1}\log n}{\varphi + 1} + O\left(\frac{\varphi^{n}}{n}\right). \tag{5}
$$
Idea of the proof: Split the sum at $k=\frac{1}{2}n$ and proceed as in Lemma 1.
Combining $(2)$, $(3)$, $(4)$, and $(5)$ we obtain
$$
\sqrt{5} \log a_n = \varphi^{n-1} \sum_{k=2}^{\infty} \varphi^{-k} \log k + \left(\frac{1}{\varphi + 1} - \frac{\varphi}{\varphi - 1}\right) \log n + O\left(\frac{1}{n}\right).
$$
Since $\frac{1}{\varphi + 1} - \frac{\varphi}{\varphi - 1} = -\sqrt{5}$ we can rewrite this as
$$
\log a_n = \frac{\varphi^{n-1}}{\sqrt{5}} \sum_{k=2}^{\infty} \varphi^{-k} \log k - \log n + O\left(\frac{1}{n}\right).
$$
Exponentiating this yields
$$
\begin{align}
a_n &= \frac{1}{n} \exp\left(\frac{\varphi^{n-1}}{\sqrt{5}} \sum_{k=2}^{\infty} \varphi^{-k} \log k\right)\left[1 + O\left(\frac{1}{n}\right)\right] \\
&= \frac{1}{n} \exp\left(F_{n-1} \sum_{k=2}^{\infty} \varphi^{-k} \log k\right)\left[1 + O\left(\frac{1}{n}\right)\right] \\
&= \frac{C^{\large F_{n-1}}}{n} \left[1 + O\left(\frac{1}{n}\right)\right],
\end{align}
$$
where in the second line we used the estimate $F_n = \frac{\varphi^n}{\sqrt{5}} + O(\varphi^{-n})$. Finally we have
$$
u_n = \frac{C^{\large F_{n-1}}}{a_n} = n \left[1 + O\left(\frac{1}{n}\right)\right] = n + O(1).
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/837889",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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} |
Solving $x-\sqrt{(x^2-36)} = {(x-6)^2\over 2x+12}$. I have problem with this equation:
$$x-\sqrt{(x^2-36)} = {(x-6)^2\over 2x+12}$$
Any ideas on beatiful solving?
| $$
\begin{array}{l l l}
x-\sqrt{(x^2-36)}& =& {(x-6)^2\over 2x+12}\\
(2x+12)\left(x-\sqrt{(x^2-36)}\right)&=&(x-6)^2\\
12x+2x^2+(-12-2x)\sqrt{(x^2-36)}&=&(x-6)^2\\
(-12-2x)\sqrt{(x^2-36)}&=&(x-6)^2-12x-2x^2\\
(-12-2x)^2(x^2-36)&=&\left((x-6)^2-12x-2x^2\right)^2\\
4x^4+48x^3-1728x-5184&=&x^4+48x^3+504x^2-1728x+1296\\
3x^4-504x^2-6480&=&0\\
\text{Substitute }y=x^2\\
3y^2-504y-6480&=&0\\
3(y-180)(y+12)&=&0\\
(y-180)(y+12)&=&0\\
\end{array}
$$
We now split into 2 equations
$$y-180=0\text{ or }y+12=0$$
$$y=180\text{ or }y=-12$$
$$x^2=180\text{ or }x^2=-12$$
$\sqrt{-12}$ has no solutions
$$x=\pm\sqrt{180}=\pm6\sqrt{5}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/838699",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Product in terms of $n$ of $\frac{1}{2} \cdot \frac{3}{4} \cdot \frac{5}{6} \cdot \frac{7}{8} \cdot \cdots \cdot \frac{2n-1}{2n}$ What is the following product in terms of $n$?
$$\frac{1}{2} \cdot \frac{3}{4} \cdot \frac{5}{6} \cdot \frac{7}{8} \cdot \cdots \cdot \frac{2n-1}{2n}$$
Thank you.
| As an alternative answer that has its uses, this product of fractions can be written as a definite integral:
$$\begin{align}
\frac12\frac34\frac56\frac78\dots\frac{2n-1}{2n}&=\frac{(2n-1)!!}{(2n)!!}\\
&=\frac{2}{\pi}\int_{0}^{\pi/2}\sin^{2n}{x}\,\mathrm{d}x
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/842911",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Square roots modulo powers of $2$ Experimentally, it seems like every $a\equiv1 \pmod 8$ has $4$ square roots mod $2^n$ for all $n \ge 3$ (i.e. solutions to $x^2\equiv a \pmod {2^n}$)
Is this true? If so, how can I prove it? If not, is it at least true that the maximum (over $a$) number of square roots of $a$ mod $2^n$ is bounded by some polynomial in $n$?
| When $n\ge 3$, the number of solutions of $x^2\equiv 1\pmod{2^n}$ is $4$. The solutions are $x=\pm 1\pmod{2^n}$ and $x\equiv \pm 1+2^{n-1}\pmod{2^n}$.
Proof: We want $x^2-1\equiv 0\pmod{2^n}$, that is, $(x-1)(x+1)\equiv 0\pmod{2^n}$. Since $x$ must be odd, the gcd of $x-1$ and $x+1$ is $2$. Either all of the $2$'s come from $x-1$, or all the $2$'s come from $x+1$, or $n-1$ of them come from one of $x-1$ or $x+1$, and $1$ of them comes from the other.
To connect this with square roots, note that $u$ and $v$ are square roots of $a$, then $uv^{-1}$ is a square root of $1$, and conversely. So if $a$ has a square root, it has $4$ of them.
To show all $a$ congruent to $1$ mod $8$ have a square root, we use a counting argument. Note that there are $2^{n-3}$ numbers between $1$ and $2^n-1$ that are congruent to $1$ modulo $8$, and $2^{n-1}$ odd numbers. Since the squaring function is $4$ to $1$, it must be the case that every number congruent to $1$ modulo $8$ is the square of something.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/845486",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 0
} |
Problem of quadratic equation If $\alpha$ be a root of $ 4x^2 +2x -1 = 0 $ , prove that the other root is $4\alpha^3 - 3\alpha$ . I have tried to do it but of no success.[$4\alpha^3 -2\alpha$ = $\dfrac {-1}{2}$ and $4\alpha^4 - 3\alpha^2$ = $\dfrac {-1}{4}$ ] .How to prove it?
| Someone will probably show you a smart way to do this, but ...I guess you could just use that $\alpha^2 = 1 - 2\alpha$ and then
$$\begin{align}
4(4\alpha^3 - 3\alpha)^2 + &2(4\alpha^3 - 3\alpha) - 1 \\ &= 4(4\alpha^6 + 9\alpha^2 - 24\alpha^4) + 8\alpha^3 - 6\alpha - 1 \\
&= 16\alpha^6 + 36\alpha^2 - 96\alpha^4 + 8\alpha^3 - 6\alpha -1\\
&= 16(1-2\alpha)^3 + 36(1 - 2\alpha) - 96(1-2\alpha)^2 + 8\alpha(1-2\alpha) - 6\alpha -1 \\
&= \dots
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/846203",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
Prove that $\exists\,! \,\lambda \in (1/5,1/4)$ such that $\frac{1}{2\pi}\int_0^{2\pi}e^{\sin x}\,\mathrm{d}x=e^{\lambda}$ The following question came up in chat
Prove that $\exists\,! \,\lambda \in (1/5,1/4)$ such that
$\displaystyle\frac{1}{2\pi}\int_0^{2\pi}e^{\sin x}\,\mathrm{d}x=e^{\lambda}$
Now the integral can be written as
$$
\frac{1}{2\pi}\int_0^{2\pi}e^{\sin x}\,\mathrm{d}x
= \int_0^{1}e^{\sin 2\pi x}\,\mathrm{d}x
= \sum_{k\geq 0} \int_0^{1} \frac{\sin^k(2\pi x)}{k!} \,\mathrm{d}x
= \sum_{k\geq 0} \frac{1}{4^k(k!)^2}
$$
Which is the same as $J_0(1)$, where $J$ is the bessel function of the first kind. Now one can also rewrite the problem by using $e^x = \sum_{k=0} x^k/k!$. So
\begin{array}{ccc}
e^{1/5} & < & \int_0^{1}e^{\sin 2\pi x}\,\mathrm{d}x & < & e^{1/4}\\
\sum_{k\geq0} \frac{1}{5^k k!} & < & \sum_{k\geq 0} \frac{1}{4^k(k!)^2}
& < & \sum_{k\geq0} \frac{1}{4^k k!}
\end{array}
Alas the latter sums seems just as hard to prove than the integral form.
Any help or suggestions would be appreciated.
| Anticlimax ahead:
$$\sum_{k=0}^\infty \frac{1}{4^k(k!)^2} < \sum_{k=0}^\infty \frac{1}{4^k\,k!}$$
follows since $4^k(k!)^2 \geqslant 4^k\, k!$ for all $k$ and the inequality is strict for $k\geqslant 2$.
For the inequality
$$\sum_{k=0}^\infty \frac{1}{5^k\,k!} < \sum_{k=0}^\infty \frac{1}{4^k(k!)^2},$$
we note that the terms for $k = 0$ are equal in both series, so it suffices to see
$$\sum_{k=2}^\infty \frac{1}{k!}\left(\frac{1}{5^k} - \frac{1}{4^k\,k!}\right) \leqslant \sum_{k=2}^\infty \frac{1}{5^k\,k!} \leqslant \frac{1}{50}\sum_{\nu=0}^\infty \frac{1}{15^\nu} = \frac{15}{14\cdot 50} = \frac{3}{140} < \frac{1}{20} = \frac{1}{4^1(1!)^2} - \frac{1}{5^1\,1!}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/848623",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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Matrices as linear transformations I am reading a proof which claims:
A matrix of $m\times n$ is a linear transformation from $m$
vector-space to $n$ vector-space, And therefore, by the dimension theorem: $m = \dim\ker A + \dim\Im A$
Isn't it the opposite?
For example: $A\in M_{4\times 2}(\mathbb{R})$ is a linear transformation from $\mathbb{R}^2$ to $\mathbb{R}^4$.
| Let us take an example:
$$ \begin{pmatrix} \color{blue}1& \color{red}2 & \color{green}3 \\ 5 & 6 & 7 \end{pmatrix} \begin{pmatrix} \color{blue}{10} \\ \color{red}{20} \\ \color{green}{30}\end{pmatrix} = \begin{pmatrix}\color{blue}1 \cdot \color{blue}{10} + \color{red}2\cdot \color{red}{20} +\color{green}3 \cdot \color{green}{30} \\ 380\end{pmatrix} $$
but
$$ \begin{pmatrix} \color{blue}1& \color{red}2 & \color{green}3 \\ 5 & 6 & 7 \end{pmatrix} \begin{pmatrix} \color{blue}{10} \\ \color{red}{20} \end{pmatrix} = \begin{pmatrix}\color{blue}1 \cdot \color{blue}{10} + \color{red}2\cdot \color{red}{20} +\color{green}3 \cdot \color{green}{ ? } \\ ?\end{pmatrix} $$
seems to be harder to compute. So, if $\mathbb{R}^{m\times n}$ is set the set of matrices with $m$ rows and $n$ columns then for $A \in \mathbb{R}^{m\times n}$ and $x \in \mathbb{R}^n$ we have $Ax \in \mathbb{R}^m$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/849222",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Simple Differentiation Problem Involving Area Radius and Circumference A stone is dropped into a pool of water, and the area covered by the spreading ripple increases at a rate of $4 m^2 s^{-1} $.
Calculate the rate at which the circumference of the circle formed is increasing 3 seconds after the stone is dropped.
My method:
$$ \begin{align}
& \cfrac{dA}{dt}=4\\
& A = \pi \times r \times r \ \text{so} \ \cfrac{dA}{dr} = 2 \times \pi \times r \\
& C = 2 \times \pi \times r \ \text{so} \ \cfrac{dC}{dr} =2 \times \pi \\
& \cfrac{dA}{dt} = \cfrac{dA}{dr} \times \cfrac{dr}{dt} \\
& 4 = 2 \times \pi \times (3) \cfrac{ dr}{dt} \\
& \text{so} \ \cfrac{dr}{dt} = \cfrac{2}{(3 \times \pi)} \\
& \text{now} \ \cfrac{dC}{dt} = \cfrac{dC}{dr} \times \cfrac{dr}{dt} \\
& \text{so} \ \cfrac{dC}{dt} = 2 \times \pi \times \cfrac{2}{(3 \times \pi)}
\end{align} $$
I'm getting $1.3333...$
But the answer is $2.05m/s$ , I don't understand how to get it...
| $$ \frac{dC}{dt}=\frac{dA}{dt}\frac{dC}{dA} $$
$$C=2\pi r= 2\frac{\pi r^2}{r}=2\frac{A}{r} $$ (since $A=\pi r^2 $) and
$$A=\pi r^2 \implies r=\sqrt{\frac{A}{\pi}}$$
Therefore using the previous 2 results $$C=2 \frac{A}{\sqrt{\frac{A}{\pi}}}=2\frac{A}{\sqrt{\frac{A}{\pi}}}\times \frac{\sqrt{\pi}}{\sqrt{\pi}}=2\frac{A\sqrt{\pi}}{\sqrt{A}}=2\sqrt{\pi A}$$
$$\therefore \frac{dC}{dA}=\sqrt{\frac{\pi}{A}}$$
$$\implies \frac{dC}{dt}=\frac{dA}{dt}\sqrt{\frac{\pi}{A}} $$
and since A is changing at a constant rate of $4m^2/s$, $A=4t$.
$$\frac{dC}{dt}=4 \sqrt{\frac{\pi}{4t}}$$
So at 3 seconds
$$\frac{dC}{dt}=4\sqrt{\frac{\pi}{12}}\approx2.0466 \ m/s$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/849920",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Find all values that make the expression a perfect cube Find all the positive integers $n$ such that $n^3-n$ is a perfect cube.
| If $n > 1$, then $3n^2-4n+1 = (3n-1)(n-1) > 0$.
Hence, $(n-1)^3 = n^3-3n^2+3n-1 < n^3-n < n^3$.
Since $n^3-n$ is strictly between two consecutive perfect cubes, $n^3-n$ is not a perfect cube.
The only remaining positive integer is $n = 1$, which yields $n^3-n = 0 = 0^3$.
Therefore, the only positive integer $n$ such that $n^3-n$ is a perfect cube is $n = 1$.
| {
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"timestamp": "2023-03-29T00:00:00",
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How to prove $\sum_{m=1}^\infty \sum_{n=1}^\infty \frac{1}{m^2+n^2}=+\infty$ How to prove $$\sum_{m=1}^\infty \sum_{n=1}^\infty \frac{1}{m^2+n^2}=+\infty.$$
I try to do like $$\sum_{m=1}^\infty \sum_{n=1}^\infty \frac{1}{m^2+n^2}=\sum_{N=1}^\infty \sum_{n+m=N}^\infty \frac{1}{m^2+n^2}=\sum_{N=1}^\infty \sum_{m=1}^{N-1} \frac{1}{m^2+(N-m)^2}$$
$$\frac{1}{m^2+(N-m)^2}\leq \frac{2}{N^2}$$
but it doesn't work.
| You're almost there.
$$\frac{1}{m^2 + (N-m)^2} = \frac{1}{2m^2 +N^2 -2mN} = \frac{1}{2m(m-N) +N^2}\ge \frac{1}{N^2}$$
Now $$\sum_{N=1}^\infty \sum_{m=1}^{N-1}\frac{1}{m^2 + (N-m)^2} \ge \sum_{N=1}^\infty \sum_{m=1}^{N-1}\frac{1}{N^2} = \sum_{N=1}^{\infty}\frac{N-1}{N^2}$$
Can you finish from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/851302",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
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} |
Determinant involving recurrence Evaluate
$$\left| A \right| = \left| {\matrix{
{x + y} & {xy} & 0 & \cdots & \cdots & 0 \cr
1 & {x + y} & {xy} & \cdots & \cdots & 0 \cr
0 & 1 & {x + y} & \cdots & \cdots & 0 \cr
\cdots & \cdots & \cdots & \cdots & \cdots & \vdots \cr
0 & \cdots & 0 & 1 & {x + y} & {xy} \cr
0 & \cdots & 0 & 0 & 1 & {x + y} \cr
} } \right|$$
And show that $\det(A) = \frac{x^{n+1}-y^{n+1}}{x-y}$ if $x\ne y$ and $\det(A) = (n+1)x^n$ if $x=y$.
I actually was able to get this recurrence formula:
$$D_n = (x+y)\cdot D_{n-1} + xy\cdot D_{n-2}$$
I tried to prove it by induciton, but the algebric calculation didn't bring me to the desired result.
| The problem is that the recurrence formula should be
$$D_n=(x+y)D_{n-1}-xyD_{n-2}$$ then
$$D_n=\frac{x^n-y^n}{x-y}(x+y)-xy\frac{x^{n-1}-y^{n-1}}{x-y}=
\frac{(x^n-y^n)(x+y)-xy(x^{n-1}-y^{n-1})}{x-y}=
\frac{x^{n+1}-y^{n+1}+x^ny -xy^n-x^{n}y+xy^{n}}{x-y}=
\frac{x^{n+1}-y^{n+1}}{x-y}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/852177",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Show that a specific $w$ cannot be the root of an quadratic with integer coefficients. Let $w$ be the only real root of $x^3-x-1=0$. Show that
$w$ cannot satisfy the quadratic $ax^2 + bx + c$ ,where $a,b,c\in \Bbb Z$.
I have written
$$w^3=w+1$$
but I can't go any further than this. Thank you.
| Note that, by Euclidien division
$$x^3-x-1=Q(x)(ax^2+bx+c)+x \left(\frac{b^2}{a^2}-\frac{c}{a}-1\right)+\frac{b c}{a^2}-1$$
So, if $w$ is a root of both $x^3-x-1=0$ and $ax^2+bx+c=0~$, then we would have
$$
w \left(\frac{b^2}{a^2}-\frac{c}{a}-1\right)+\frac{b c}{a^2}-1=0
$$
So if $a^2+a c-b^2\ne0~$ then
$$
w=\frac{b c-a^2}{a^2+a c-b^2}\in\mathbb{Q}
$$
which is absurd since $x^3-x-1=0~$ has no rational solutions.
Now what happens if $a^2+a c-b^2 =0~$?
In this cas we must also have
$a^2=bc$. Without loss of generality we may suppose that $\gcd(a,b,c)=1$. From
$a^2=bc$ we conclude that $\gcd(b,c)=1$, because otherwise any common prime divisor would also divide $a$.
From $\gcd(b,c)=1$ and $a^2=bc$ we conclude that
$$b=\epsilon\beta^2, c=\epsilon\gamma^2, a=\epsilon'\beta\gamma,\qquad\hbox{with $\gcd(\beta,\gamma)=1,~(\epsilon,\epsilon)'\in\{+1,-1\}^2$}$$
Replacing in $a^2+a c-b^2 =0$ we get $\gamma^2(\beta+\epsilon\epsilon'\gamma)=\beta^3$, so $\gamma|\beta^3$, but $\gcd(\beta,\gamma)=1$ so,
$\gamma=1$ and consequently $ \beta+\epsilon\epsilon'=\beta^3$. This implies that $\beta $ is an integer solution of $x^3-x-1=0$ or $x^3-x+1=0$ which is clearly absurd. So, this case cannot happen.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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} |
$13\mid4^{2n+1}+3^{n+2}$ How can I prove that $4^{2n+1}+3^{n+2}$ is always divisible by 13?
| You can use congruence:
$$
4^{2n+1}+3^{n+2}\equiv 4(16^n)+9(3^n) \equiv 4(16^n)+(13-4)(3^n) \equiv 4(16^n-3^n) \equiv 4(3^n-3^n) \equiv 0\ (\text{mod }13).
$$
Also you can prove it directly:
$$
4^{2n+1}+3^{n+2} = 4(16^n-3^n)+(13)(3^n) = 13[4(16^{n-1}+16^{n-2}3+16^{n-3}3^2+\ldots+3^{n-1})+3^n]
$$
which proves that $4^{2n+1}+3^{n+2}$ is dividable by $13$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/859523",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Calculate cosh(x) given sinh(x) Given the value of sinh(x)
for example sinh(x) = 3/2
How can I calculate the value of cosh(x) ?
| Use the identity $\cosh^2x-\sinh^2x \equiv 1$. If $\sinh x = \frac{3}{2}$ then
$$\cosh^2x - \left(\frac{3}{2}\right)^{\! 2} = 1$$
$$\cosh^2x - \frac{9}{4} = 1$$
$$\cosh^2x = \frac{13}{4}$$
It follows that $\cosh x = \pm\frac{1}{2}\sqrt{13}$. Since $\cosh x \ge 1$ for all $x \in \mathbb{R}$ we have $\cosh x = \frac{1}{2}\sqrt{13}$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 1
} |
Finding domain of $\sqrt{4-2\sqrt{x^2 - 1}}$ Find the (maximum) domain of: $\sqrt{4-2\sqrt{x^2 - 1}}$
Well, I guess it is $2 - \sqrt2 \sqrt{x^2 - 1}$ = $2- \sqrt{2(x^2 - 1)}$
from here, $x^2 - 1 \ge 0$ and then $ x \ge 1$ and $ x \ge -1$
What do you guys think?
| You are right on the first restriction, we need $|x|\ge 1$. We also need $4-2\sqrt{x^2-1}\ge 0$. This inequality is equivalent to $2\sqrt{x^2-1}\le 4$, or equivalently $x^2-1\le 4$, that is, $|x|\le \sqrt{5}$.
Thus a way of describing the (maximum) domain is that it is $\{x: 1\le |x|\le \sqrt{5}\}$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Find $c$ if $a,b, \; c$ satisfy $c = (a+bi)^3 - 107i$
Find $c$ if $a,b, \; c$ are positive integers which satisfy $c = (a+bi)^3 - 107i$
I can try expanding the cube, but that seems too direct. What other ways are there to go about this?
| Since $c \in \mathbb Z$, the right-hand side must have an imaginary component of $0$. Expanding, we get
$$c = (a+bi)^3 - 107i = (a^3-3ab^2) + (3a^2b-b^3-107)i$$
Therefore $3a^2b-b^3-107 = 0$, which implies that $(3a^2-b^2)b=107$. Since $107$ is prime, we conclude that $b$ is either $1$ or $107$.
Case 1: $b=1$. Then $3a^2 -1 = 107$, so $a=6$.
Case 2: $b=107$. Then $3a^2 - 107^2 = 1$. But there is no integer solution to this equation.
Therefore the unique solution is $(a,b,c) = (6,1,198)$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Let $p$ be a prime of the form $3k+2$ that divides $a^2+ab+b^2$ for some integers $a,b$. Prove that $a,b$ are both divisible by $p$. Let $p$ be a prime of the form $3k+2$ that divides $a^2+ab+b^2$ for some integers $a,b$. Prove that $a,b$ are both divisible by $p$.
My attempt:
$p\mid a^2+ab+b^2 \implies p\mid (a-b)(a^2+ab+b^2)\implies p\mid a^3-b^3$
So, we have, $a^{3k}\equiv b^{3k}\mod p$ and by Fermat's Theorem we have, $a^{3k+1}\equiv b^{3k+1}\mod p$ as $p$ is of the form $p=3k+2$.
I do not know what to do next. Please help. Thank you.
| You have shown that $a^3 \equiv b^3 \pmod p$. Remark that $(3, p-1)=1$ because $p=3k+2$. Thus we can write $3m + (p-1)n = 1$ for some integers $m, n$. Use this to show that $a\equiv b \pmod p$, so that $a^2+ab+b^2 \equiv 3a^2 \equiv 3b^2 \equiv 0 \pmod p$, and conclude from there.
(P.S. I have to commend you for posting your work!)
| {
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"timestamp": "2023-03-29T00:00:00",
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} |
Find a sequence Find the function for the sequence $a_0 = 0, a_1 = 1$ and $a_{n}=a_{n+10}+a_n$ for all $n>0$.
| Let's build such function:
$$
f(x) = \sum_{n=1}^\infty F_nx^n,
$$
where $F_n$ are Fibonacci numbers.
Then
$$
f(x) = x+\sum_{n=2}^\infty (F_{n-1}+F_{n-2})x^n = x + \sum_{k=1}^\infty F_kx^{k+1}+\sum_{m=0}^\infty F_mx^{m+2}
\\= x+xf(x)+x^2f(x),
$$
or much wide:
$$
f(x) = x+x^2+2x^3+3x^4+5x^5+8x^6+13x^7+\cdots \\
= x \;\;+\;\; (x^2+x^3+2x^4+3x^5+5x^6+8x^7+\cdots )\\
\qquad\qquad\quad\;\; +( x^3+x^4+2x^5+3x^6+5x^7+\cdots) \\
=x+xf(x)+x^2f(x),
$$
hence
$$
(1-x-x^2)f(x)=x,
$$
$$
f(x)=\dfrac{x}{1-x-x^2}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/867660",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
N! ends with exactly 30 zeros? How many values of N exist, such that N! ends with exactly 30 zeros?
| The number of zeros at the and of $n!$ is just the number of factors of $5$ in $n!$ (since the number of factors of $2$ in $n!$ is always larger than that). So, we need all $n$ such that the numbers $1$ up to $n$ have a total of $30$ factors of $5$. A first guess would be $30\cdot 5=150$, but then we forget that the multiples of $5^2=25$ have (at least) two factors of $5$. Since there are $5$ multiples of $25$ below $150$, we should try $n=125$. The numbers below $125$ have a total of
$$\frac{120}{5}+\left\lfloor\frac{120}{25}\right\rfloor=24+4=28<30$$
factors of $5$. If we include $125=5^3$, the number of factors increases with $3$ so it becomes $31>30$. Thus, we find that the are no $n$ such that $n!$ has $30$ factors of $5$ or s.t. $n!$ ends with $30$ zeros.
| {
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Proving that one of $a(1-b), b(1-c), c(1-a) \le \frac{1}{4}$ how can a prove that at least one of those is less than or equal to 1/4.
$$\forall a,b,c\in \mathbb R^+, \ a(1-b)\leq 1/4 \lor b(1-c) \leq 1/4 \lor c(1-a) \leq 1/4.$$
help please!
| We can assume $1-a, 1-b, 1-c \geq 0$, since otherwise we are done.
By the AM-GM inequality (see http://en.wikipedia.org/wiki/AM-GM_inequality), we have $abc(1-a)(1-b)(1-c) \leq (\frac{a+b+c+(1-a)+(1-b)+(1-c)}{6})^6= (\frac{1}{2})^6 = \frac{1}{64}$.
Then, if $a(1-b)> 1/4, b(1-c) > 1/4$ and $ c(1-a) > 1/4$, multiplying together we get $abc(1-a)(1-b)(1-c)> (\frac{1}{4})^3 = \frac{1}{64}$, which is a contradiction, and thus the result follows.
EDIT:
If you do not want to use AM-GM: Let $x \in \mathbb{R}^+ $. We have $0\leq (\sqrt x - \sqrt{(1-x)})^2 = x +(1-x) -2\sqrt{x(1-x)}$, and thus $ 2\sqrt{x(1-x)} \leq 1$, which implies $x(1-x) \leq \frac{1}{4}$. Apply this for $a,b$ and $c$, multiply together, and you get the inequality of the first paragraph.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that if $n$ is not divisible by $5$, then $n^4 \equiv 1 \pmod{5}$ Suppose $n$ is an integer which is not divisible by $5$. Prove that $n^4 \equiv 1 \pmod{5}$.
| Like Timbuc,
$$n^4-1=(n-1)(n+1)(n^2+1)=(n-1)(n+1)(n^2-4+5)$$
$$=\underbrace{(n-2)(n-1)(n+1)(n+2)}+5(n-1)(n+1)$$
As $n$ must divide exactly one of any five consecutive integers and $5\nmid n,(5,n)=1,$
$n$ must divide one of the multiplicand underbrace
| {
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"url": "https://math.stackexchange.com/questions/871353",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Evaluate $\int \frac{1}{(2x+1)\sqrt {x^2+7}}\,\text{d}x$. How to do this indefinite integral (anti-derivative)?
$$I=\displaystyle\int \dfrac{1}{(2x+1)\sqrt {x^2+7}}\,\text{d}x.$$
I tried doing some substitutions ($x^2+7=t^2$, $2x+1=t$, etc.) but it didn't work out.
| Let $x=\sqrt{7}\tan{u}\implies dx=\sqrt{7}\sec^2{u}du$
\begin{align}
I
=\int \frac{1}{(2x+1)\sqrt{x^2+7}}dx=\int \frac{\sec{u}}{(1+2\sqrt{7}\tan{u})}du\\
\end{align}
Let $t=\tan{\frac{u}{2}}\implies du=\frac{2}{1+t^2}dt$
\begin{align}
I
=\int\frac{\frac{1+t^2}{1-t^2}}{1+2\sqrt{7}\frac{2t}{1-t^2}}\frac{2}{1+t^2}dt=-2\int \frac{1}{t^2-4\sqrt{7}t-1}dt
\end{align}
Decompose the integrand into partial fractions and integrate.
\begin{align}
I=-2A\ln{(t-2\sqrt{7}-\sqrt{29})}-2B\ln{(t-2\sqrt{7}+\sqrt{29})}+c
\end{align}
where
\begin{align}
A={\rm Res}(f,2\sqrt{7}+\sqrt{29})=\frac{1}{2\sqrt{29}}, B={\rm Res}(f,2\sqrt{7}-\sqrt{29})=-\frac{1}{2\sqrt{29}}
\end{align}
Now express everything in terms of x.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the triangular matrix and determinant. I have a 4x4 matrix and I want to find the triangular matrix (lower half entries are zero).
$$A=
\begin{bmatrix}
2 & -8 & 6 & 8\\
3 & -9 & 5 & 10\\
-3 & 0 & 1 & -2\\
1 & -4 & 0 & 6
\end{bmatrix}
$$
Here are the elementary row operations I performed to get it into triangular form.
row swap rows 1 and row 4
$r_2 - 3\cdot r_1$ replacing $r_2$
$r_3 + 3\cdot r_1$ replacing $r_3$
$r_4 - 2\cdot r_1$ replacing $r_4$
I get this matrix
$$A= -
\begin{bmatrix}
1 & -4 & 0 & 6\\
0 & 3 & 5 & -8\\
0 & -12 & 1 & 16\\
0 & 0 & 6 & -4
\end{bmatrix}
$$
I then did $4\cdot r_2 + r_3$ to replace $r_3$ and got
$$A= -
\begin{bmatrix}
1 & -4 & 0 & 6\\
0 & 3 & 5 & -8\\
0 & 0 & 21 & -16\\
0 & 0 & 6 & -4
\end{bmatrix}
$$
I then did $-21\cdot r_4 + 6\cdot r_3$ to replace $r_4$ and got
$$A= -
\begin{bmatrix}
1 & -4 & 0 & 6\\
0 & 3 & 5 & -8\\
0 & 0 & 21 & -16\\
0 & 0 & 0 & -12
\end{bmatrix}
$$
I am not sure if I did this correctly but the determinant of the matrix should be -36. When I multiply the diagonal entries it isn't -36. I can't figure out what I am doing wrong.
|
"I then did -21*row 4 + 6*row 3 to replace row 4 and got"
This is a determinant altering operation and not an elementary operation.
Don't write that $A$ equals something which isn't $A$.
Picking up where you errored and using the same idea you had one gets:
$$\begin{align} \begin{bmatrix}
1 & -4 & 0 & 6\\
0 & 3 & 5 & -8\\
0 & 0 & 21 & -16\\
0 & 0 & 6 & -4
\end{bmatrix}&\leadsto \begin{bmatrix}
1 & -4 & 0 & 6\\
0 & 3 & 5 & -8\\
0 & 0 & 6\cdot 21 & -6\cdot 16\\
0 & 0 & -21\cdot 6 & (-21)\cdot (-4)
\end{bmatrix}\\
&\leadsto \begin{bmatrix}
1 & -4 & 0 & 6\\
0 & 3 & 5 & -8\\
0 & 0 & 6\cdot 21 & -16\\
0 & 0 & 0 & -12
\end{bmatrix}_.\end{align}$$
Making the proper compensation yields
$$\det(A)=-\dfrac{1\cdot 3\cdot (6\cdot 21)\cdot (-12)}{-21\cdot 6}=-36.$$
| {
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How to show that $(a+b)^p\le 2^p (a^p+b^p)$ If I may ask, how can we derive that $$(a+b)^p\le 2^p (a^p+b^p)$$ where $a,b,p\ge 0$ is an integer?
| Assume that both $A, B > 0$ and $p \geq 0$ to begin with, then we can rewrite the inequality as:
$\left(\dfrac{A}{A+B}\right)^p + \left(\dfrac{B}{A+B}\right)^p \geq 2^{-p}$. So there are $3$ cases to consider:
*
*$p = 0$, then $LHS = 2 > 1 = RHS$, and the inequality holds.
*$p = 1$, then $LHS = 1 > \dfrac{1}{2} = RHS$, and the inequality also holds.
*$p \geq 2$, then $f(x) = x^p$ is convex, so $f(x) + f(y) \geq f\left(\dfrac{x+y}{2}\right)$
with $x = \dfrac{A}{A+B}$, and $y = \dfrac{B}{A+B}$ yields the inequality.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Trigonometric Identities help How do you solve this? I can't figure out what I should do.
$$\sin ^4\left(A\right)+\cos ^2\left(A\right)=\cos ^4\left(A\right)+\sin ^2\left(A\right)$$
Also, why is this equal zero? Can someone explain how that simplifies to be zero?
$$\frac{\left(\frac{1}{\cos \left(x\right)}\right)-1}{\left(\frac{1}{\cos \left(x\right)}\right)+1}+\frac{\cos \left(x\right)-1}{\cos \left(x\right)+1}$$
| You just have to know that:
$\sin^4(a)=\sin^2(a)\sin^2(a)$
$\cos^4(a)=\cos^2(a)\cos^2(a)$
$\sin^2(a)+\cos^2(a)=1$
$$\begin{align}
\sin^4(a)+\cos^2(a)&=\cos^4(a)+\sin^2(a) \\
\sin^2(a)\sin^2(a)+\cos^2(a)&=\cos^2(a)\cos^2(a)+\sin^2(a) \\
\sin^2(a)\left(1-\cos^2(a)\right)+\cos^2(a)&=\cos^2(a)\left(1-\sin^2(a)\right)+\sin^2(a) \\
\sin^2(a)-\sin^2(a)\cos^2(a)+\cos^2(a)&=\cos^2(a)-\sin^2(a)\cos^2(a)+\sin^2(a) \\
1-\sin^2(a)\cos^2(a)&=1-\sin^2(a)\cos^2(a) \\
0&=0 \\
\end{align}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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if $\frac{1}{(1-x^4)(1-x^3)(1-x^2)}=\sum_{n=0}^{\infty}a_{n}x^n$,find $a_{n}$ Let
$$\dfrac{1}{(1-x^4)(1-x^3)(1-x^2)}=\sum_{n=0}^{\infty}a_{n}x^n$$
Find the closed form $$a_{n}$$
since
$$(1-x^4)(1-x^3)(1-x^2)=(1-x)^3(1+x+x^2+x^3)(1+x+x^2)(1+x)$$
so
$$\dfrac{1}{(1-x^4)(1-x^3)(1-x^2)}=\dfrac{1}{(1-x)^3(1+x+x^2+x^3)(1+x+x^2)(1+x)}$$
then I feel very ugly,can you someone have good partial fractions methods by hand?
because I take an hour to solve this problem.
ago I have solve $$x+2y+3z=n$$ the number of the positive integer solution $a_{n}$ I found
$$a_{n}=\left[\dfrac{(n+3)^2}{12}\right]$$
Thank you
| Here is an approach to compute $a_n$ using the observation that it is the number of ways to make change for $n$ using coins of denomination $2, 3, 4$.
Case: $n$ even
It may be noted that $3$ denomination coin can be used, but only as a multiple of $6$, so you have $n=2k=2p+4q+6r \implies k = p+2q+3r$. Thus this is the number of ways you can make change for $k=n/2$ with coins of denominations $1, 2, 3$. So
$$\sum_{k=0}^\infty a_{2k}x^k = \frac1{(1-x)(1-x^2)(1-x^3)} = \frac{(1+x+x^2+x^3+x^4+x^5)(1+x^2+x^4) (1+x^3)}{(1-x^6)^3} = \left(1+x+2 x^2+3 x^3+4 x^4+5 x^5+4 x^6+5 x^7+4 x^8+3 x^9+2 x^{10}+x^{11}+x^{12} \right)\left(\sum_{k=0}^\infty (-1)^k \binom{-3}{k} x^{6k} \right)$$
will help get $a_{2k}$.
Case: n odd
Now we need to have one of the $3$ denominations used, and the rest is the even case, so we have $a_{2k+3} = a_{2k}$ in this case.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/875792",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Area of a Curved Surface Find the area of the part o the surface $z=xy$ that lies within the cylinder $x^2+y^2=1$.
I'm not sure how to set up the surface integral to compute this.
| The formula of the are of the surface given as a graph of the function $z=f(x,y)$ over the region $(x,y) \in D$ is
$$A(S)=\iint_D \sqrt{1+f_x^2+f_y^2}dA$$
In this case $D$ is the disk of radius $1$ with center at $(0,0)$:
$D=\{(x,y): x^2+y^2 \leq 1\}$
$$z=f(x,y)=xy$$
$$f_x=y, f_y=x$$
So, we have the following:
$$A(S)=\iint_D \sqrt{1+f_x^2+f_y^2}dA=\iint_D \sqrt{1+y^2+x^2}dA$$
Now we use polar coordinates:
$$x=r \cos{\theta}, y=r \sin{\theta}, 0 \leq r \leq 1, 0 \leq \theta \leq 2 \pi$$
$$dA=dxdy=r dr d \theta$$
So we have the following:
$$A(S)=\iint_D \sqrt{1+y^2+x^2}dA=\int_0^{2 \pi} \int_0^1 \sqrt{1+r^2}rdr d \theta=\int_0^{2 \pi} \int_0^1 (1+r^2)^{1/2}rdr d \theta=\int_0^{2 \pi} \int_0^1 \frac{d}{dr} \left (\frac{1}{3}(1+r^2)^{3/2} \right ) dr d \theta =\int_0^{2 \pi} \left [ \frac{1}{3}(1+r^2)^{3/2} \right ] _0^1 d \theta=\int_0^{2 \pi} \left (\frac{1}{3} 2^{3/2}-\frac{1}{3} \right ) d \theta=\frac{1}{3} ( 2^{3/2}-1 ) 2 \pi =\frac{2 \pi }{3} ( 2 \sqrt{2}-1 )$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Trigonometric functions of the acute angle Find the other five trigonometric functions of the acute angle A, given that:
\begin{align}
&\text{a)}\ \ \sec A = 2 \\[15pt]
&\text{b)}\ \ \cos A = \frac{m^2 - n^2}{m^2 + n^2}
\end{align}
Help me. I don't know how to solve this one. Thanks.
| Honestly, I suspect that equation b is a red herring.
\begin{align}
\cos{A}
&=\frac{1}{\sec{A}}\\
&=\frac{1}{2}
\end{align}
\begin{align}
\sin{A}
&=\sqrt{1-\cos^2{A}}\\
&=\frac{\sqrt{3}}{2}
\end{align}
\begin{align}
\tan{A}
&=\frac{\sin{A}}{\cos{A}}\\
&=\sqrt{3}
\end{align}
\begin{align}
\csc{A}
&=\frac{1}{\sin{A}}\\
&=\frac{2}{\sqrt{3}}\\
\end{align}
\begin{align}
\cot{A}
&=\frac{1}{\tan{A}}\\
&=\frac{1}{\sqrt{3}}\\
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/876732",
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"source": "stackexchange",
"question_score": "1",
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When the numerator of a fraction is increased by $4$, the fraction increases by $2/3$... When the numerator of a fraction is increased by $4$, the fraction increases by $2/3$. What is the denominator of the fraction?
I tried,
Let the numerator of the fraction be $x$ and the denominator be $y$.
Accordingly, $$\frac{x+4}y=\frac xy+\frac 23$$
I am not able to find the second equation.
| $$x=\frac{a}{b}$$
$$x+\frac{2}{3}=\frac{a+4}{b} \Rightarrow \frac{a}{b}+\frac{2}{3}=\frac{a+4}{b} \Rightarrow a+\frac{2}{3}b=a+4 \Rightarrow b=\frac{12}{2}=6$$
| {
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Can a Mersenne number ever be a Carmichael number? Can a Mersenne number ever be a Carmichael number?
More specifically, can a composite number $m$ of the form $2^n-1$ ever pass the test: $a^{m-1} \equiv 1 \mod m$ for all intergers $a >1$ (Fermat's Test)?
Cases potentially proved so far: (That are never Carmichael numbers)
*
*where $n$ is odd
*where $n$ is prime
Work using "main" definition:
First off take the definition of a Carmichael number:
A positive composite integer $m$ is a Carmichael number if and only if $m$
is square-free, and for all prime divisors $p$ of $m$, it is true that $p -
1 \mid m - 1$.
Let's assume $m=2^n-1$ is squarefree. (Best case, and I believe it always is for $2^p-1$)
Take the case where $n$ (in $2^n-1$) is a prime $p$. All factors of $2^p-1$ must of the form: $2kp+1$ for some constant $k$. So will $2kp$ ever divide $2^p-2$? Factoring a $2$ out gives us $kp \mid 2^{p-1}-1$, or split into two: $k \mid 2^{p-1}-1$ and $p \mid 2^{p-1}-1$ must both be true. By Fermat's little theorem, $2^{p-1} \equiv 1 \mod p$, so $p \mid 2^{p-1}-1$ is always true.
So if $k \mid 2^{p-1}-1$ for $k = {q-1 \over p}$, is false for at least one factor $q$ of $2^p-1$, no Carmichael numbers can exist of form $2^p-1$.
Now for other cases where $n$ is composite, lets say $n=cp$, for some prime $p$, and some number $c$:
$\begin{align}2^{cp}-1&=(2^p-1)\cdot \left(1+2^p+2^{2p}+2^{3p}+\cdots+2^{(c-1)p}\right)\end{align}$
Thus: $2^{n-1} \mid 2^p-1$
Because of that, we must look at the factors of $2^p-1$ when considering if $2^{cp}-1$ is a Carmichael number. So we know those factors are already of form $2kp+1$, and then $kp \mid 2^{cp-1}-1$.
This is where I'm left. on an incomplete proof.
Using Bernoulli definition:
An odd composite squarefree number $m$ is a Carmichael number iff $m$
divides the denominator of the Bernoulli number $B_{n-1}$.
Using the Von Staudt–Clausen theorem, there may be a way to proof that that factors of the Bernoulli number denominators may never divide a mersenne number.
| Partial Proof
I have found a proof for numbers with an even number of factors, and a prime exponent. ($m = 2^p-1$ and $p > 2$)
First note that $2^p-1 \equiv 3 \mod 4$ for $p > 1$. Next note the following table for $a*b \mod 4$:
$$
\begin{array}{c|lcr}
b & a = 0 & a = 1 & a = 2 & a = 3 \\
\hline
0 & 0 & 0 & 0 & 0 \\
1 & 0 & 1 & 2 & 3 \\
2 & 0 & 2 & 0 & 2 \\
3 & 0 & 3 & 2 & 1 \\
\end{array}
$$
Because the only results $3 \mod 4$ in the table involve $3 \mod 4$ as a factor. At least one factor of $2^n-1$ must be congruent to $3 \mod 4$ because $2^n-1 \equiv 3 \mod 4$. And because $3*3 \equiv 1 \mod 4$ The amount of such factors must be odd.
Note that factors of $2^p-1$ must be of the form $2kp+1$ for some value $k$. Assuming $p$ to be odd, $p = 2a+1$, and then $2kp+1 = 2k(2a+1)+1 = 4ka+2k+1$. Next $4ka+2k+1 \equiv 2k+1 \mod 4$
Based of the fact that the amount of factors congruent to $3 \mod 4$ have to be odd, if the number of factors of $m = 2^p-1$ is even, then for at least one factor $q$, $q \equiv 1 \mod 4$. Thus $2kp+1 \equiv 1 \mod 4$, so $2k+1 \equiv 1 \mod 4$ and $2k \equiv 0 \mod 4$. Which means $k$ is even.
From a portion proved above, if there is a value $k$ in which $q = 2kp + 1$ for which $q$ is a factor of $2^p-1$, in which $k$ does not divide $2^{p-1}-1$, then $2^p-1$ is never a Carmichael number. Because an even number never divides an odd number, an even $k$ does not divide $2^{p-1}$. This even $k$ value is the one proven above.
So if the number of factors of $2^p-1$ is even, $2^p-1$ is never a Carmichael number.
| {
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Why these two series are convergent or divergent? I do not understand why $$\sum^{\infty}_{k=1} z_k = \sum^{\infty}_{k=1} \frac1k$$ is divergent but the other series $$\sum^{\infty}_{k=1} z_k = \sum^{\infty}_{k=1} \frac{(-1)^{k+1}}k$$ is convergent. For both cases the $\displaystyle\lim_{n \to +\infty} z_{n} = 0$. Could you explain please?
I prefer to come up with an all-inclusive test of convergence that accepts the second but rejects the first. Thanks.
| We do not answer the specific question, but instead look at two related series,
$$1+\frac{1}{2}+\frac{1}{2}+\frac{1}{4}+\frac{1}{4}+\frac{1}{4}+\frac{1}{4}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{16}+\cdots\tag{1}$$
and
$$1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+\frac{1}{8}-\frac{1}{8}+\frac{1}{8}-\frac{1}{8}+\frac{1}{8}-\frac{1}{8}+\frac{1}{8}-\frac{1}{16}+\cdots\tag{2}$$
Series (1) diverges. For the sum of the first $3$ terms is $2$, the sum of the first $7$ is $3$, the sum of the first $15$ is $4$, the sum of the first $31$ is $5$, and so on. The partial sums "blow up," albeit with a great deal of reluctance. It takes an awful lot of terms to get a sum of $100$.
Series (2) converges. The partial sums are $1, \frac{1}{2}, 1, \frac{3}{4}, 1, \frac{3}{4}, 1,\frac{7}{8}, 1,\frac{7}{8}$, and so on. It is clear that the partial sums approach $1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/880980",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Algebraic proof of $\tan x>x$ I'm looking for a non-calculus proof of the statement that $\tan x>x$ on $(0,\pi/2)$, meaning "not using derivatives or integrals." (The calculus proof: if $f(x)=\tan x-x$ then $f'(x)=\sec^2 x-1>0$ so $f$ is increasing, and $f(0)=0$.) $\tan x$ is defined to be $\frac{\sin x}{\cos x}$ where these are defined by their infinite series. What I have so far:
$$|z|\le1\implies\left|\sum_{n=4}^\infty\frac{z^n}{n!}\right|<\sum_{n=0}^\infty\frac{|z|^4}{4!\,5^n}=\frac{5|z|^4}{4\cdot 4!}$$
$$\left|\sin x-\Big(x-\frac{x^3}6\Big)\right|=\Im\left[\sum_{n=4}^\infty\frac{(ix)^n}{n!}\right]<\frac{5x^4}{4\cdot 4!}<\frac{x^3}6$$
$$\left|\cos x-\Big(1-\frac{x^2}2\Big)\right|=\Re\left[\sum_{n=4}^\infty\frac{(ix)^n}{n!}\right]<\frac{5x^4}{4\cdot 4!}<\frac{x^2}6$$
Thus $\sin x>x-\frac{x^3}3$ and $\cos x<1-\frac{x^2}3$, so $\tan x>x$. However, this only covers the region $x\le1$, and I still need to bound $\tan x$ on $(1,\pi/2)$. My best approximation to $\pi$ is the very crude $2<\pi<4$, derived by combining the above bounds with the double angle formulas (note that $\pi$ is defined as the smallest positive root of $\sin x$), so I can't quite finish the proof with a bound like $\sin x>1/\sqrt 2$, $\cos x\le\pi/2-x$ (assuming now $x\ge1\ge\pi/4$) because the bound is too tight. Any ideas?
| I'm adding a second answer because the method is very different.
This proof uses the double angle formulas for sine and cosine. From
$$\sin 2x=2\sin x\cos x\qquad\cos2x=2\cos^2x-1$$
we get
$$\tan2x=\frac{\sin2x}{\cos2x}=\frac{2\sin x\cos x}{2\cos^2x-1}>\frac{2x(1-x^2/3)(1-2x^2/3)}{2(1-x^2/3)^2-1},$$
using the bounds $\sin x>x(1-x^2/3)$, $1-2x^2/3<\cos x<1-x^2/3$ derived in the original post. Letting $y=x^2/3$, we have:
\begin{align}\frac{2x(1-y)(1-2y)}{2(1-y)^2-1}>2x&\iff(1-y)(1-2y)>2(1-y)^2-1\\
&\iff1-3y+2y^2>1-4y+2y^2\\
&\iff y>0.
\end{align}
Now $y=x^2/3>0$ for $x\ne0$, and the first step is justified when $2(1-y)^2-1>0$, but $0<\cos 2x<2(1-x^2/3)^2-1$ ensures that this is the case, so we can conclude $\tan2x>2x$ for all $x\in(0,1]$ such that $\cos 2x>0$, which is to say, when $2x\in(0,\pi/2)$ (since $\pi/4<1$).
PS: This answer has been successfully turned into a formal proof.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Count the number of ways n different-sided dice can add up to a given number I am trying to find a way to count the number of ways n different-sided dice can add up to a given number.
For example, 2 dice, 4- and 6-sided, can add up to 8 in 3 different ways: ($(2,6),(3,5),(4,4)$).
I've found a similar question but I cannot find a way to generalize it to my problem.
| I can't provide a nice formula as in your referenced question, but I can show you at least half of the way.
Let's start with your
Example: There are two dice with 4 and 6 sides. How many ways are there, so that rolling both dice once result in 8 pips?
We encode the dies with polynomials and use the exponents to label the pips
\begin{align*}
\text{4-sided die}&\qquad x^1+x^2+x^3+x^4\\
\text{6-sided die}&\qquad x^1+x^2+x^3+x^4+x^5+x^6
\end{align*}
If we denote with $[x^k]$ the coefficient $a_k$ of $x^k$ of a polynomial $\sum_{j=0}^{n}a_j x^j$, we observe
\begin{align*}
[x^8]&(x^1+x^2+x^3+x^4)(x^1+x^2+x^3+x^4+x^5+x^6)\\
&=[x^8]x\frac{1-x^4}{1-x}x\frac{1-x^6}{1-x}\\
&=[x^6](1-x^4)(1-x^6)\frac{1}{(1-x)^2}\\
&=[x^6](1-x^4-x^6+x^{10})\sum_{j\geq 0}\binom{-2}{j}(-x)^j\\
&=[x^6](1-x^4-x^6+x^{10})\sum_{j\geq 0}\binom{j+1}{j}x^j\tag{1}\\
&=([x^6]-[x^2]-[x^0])\sum_{j\geq 0}(j+1)x^j\\
&=7-3-1\\
&=3
\end{align*}
So, the result is $3$ in accordance with your example. Now the same generalized:
Generalisation: Let's assume there are $N$ dice with different sides. How many ways are there rolling $N$ dies once result in $K$ pips? Since we do not require pairwise different sides, we my assume $j_1$ dice with $k_1$ sides, $j_2$ dice with $k_2$ and finally $j_l$ dice with $k_l$ sides together with $j_1+\dots+j_l=N$.
We can now calculate similarly to the special case above:
\begin{align*}
[x^K]&(x^1+\cdots+x^{k_1})^{j_1}\cdot...\cdot(x^1+\cdots+x^{k_l})^{j_l}\\
&=[x^K]x^{j_1}\left(\frac{1-x^{k_1}}{1-x}\right)^{j_1}\cdot...\cdot x^{j_l}\left(\frac{1-x^{k_l}}{1-x}\right)^{j_l}\\
&=[x^{K-j_1-...-j_l}](1-x^{k_1})^{j_1}\cdot...\cdot(1-x^{k_l})^{j_l}\frac{1}{(1-x)^N}\\
&=[x^{K-j_1-...-j_l}]\prod_{m=1}^{l}\left(1-x^{k_m}\right)^{j_m}\sum_{n\geq 0}\binom{-N}{n}(-x)^n\\
&=[x^{K-j_1-...-j_l}]\prod_{m=1}^{l}\left(1-x^{k_m}\right)^{j_m}\sum_{n\geq 0}\binom{N+n-1}{n}x^n\tag{2}\\
\end{align*}
Observe, that formula $(2)$ corresponds to formula $(1)$ in the special case above. So, for specific values we can continue at least in principle as we did from $(1)$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluate $\int_{1}^{\infty} \frac{\ln{(2x-1)}}{x^2} $ $$\int_{1}^{\infty} \frac{\ln{(2x-1)}}{x^2} dx$$
My approach is to calc
$$\int_{1}^{X} \frac{\ln{(2x-1)}}{x^2} dx$$ and then take the limit for the answer when $X \rightarrow \infty$
However, I must do something wrong. The correct answer should be $2\ln(2)$.
$$\int_{1}^{X} \frac{\ln{(2x-1)}}{x^2} dx = \left[-\frac{1}{x} \ln (2x-1) \right]_{1}^{X} + \int_{1}^{X} \frac{1}{x} \times \frac{2}{2x-1} dx = -\frac{1}{X}\ln(2X-1) + 2\int_{1}^{X} -\frac{1}{x} + \frac{2}{2x-1} dx = -1\frac{1}{X}\ln(2X-1)-2\ln X+2\ln(2X-1) $$
Am I wrong? If I'm not, how to proceed?
=== EDIT ===
After the edit I wonder if this is the correct way to proceed:
$$ - \frac{1}{X}\ln(2X-1)-2\ln X+2\ln(2X-1) $$ The first part will do to zero because of $\frac{1}{X} $ so we ignore that one, the second and third part:
$$ -2\ln X+2\ln(2X-1) = 2\ln \left( \frac{2X-1}{X}\right) = 2\ln \left( 2-\frac{1}{X} \right) = 2\ln (2)$$
| Another approach :
Setting $x\mapsto\frac1x$, we will obtain
$$
\int_{1}^{\infty} \frac{\ln{(2x-1)}}{x^2}\ dx=\int_{0}^{1} \ln\left(\frac{2-x}{x}\right) dx=\int_{0}^{1} \bigg[\ln(2-x)-\ln x\bigg]\ dx.
$$
Note that
$$
\int\ln y\ dy=y\ \ln y-y+C,
$$
hence
$$
\int_{1}^{\infty} \frac{\ln{(2x-1)}}{x^2}\ dx=2\ln 2.
$$
| {
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Prove $\frac{a^3}{b^2}+\frac{b^3}{c^2}+\frac{c^3}{a^2} \geq \frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}$
If $a$, $b$ and $c$ are positive real numbers, prove that:
$$\frac{a^3}{b^2}+\frac{b^3}{c^2}+\frac{c^3}{a^2} \geq \frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}$$
Additional info:We can use AM-GM and Cauchy inequalities mostly.We are not allowed to use induction.
Things I have tried so far:
Using Cauchy inequality I can write:$$\left(\frac{a^3}{b^2}+\frac{b^3}{c^2}+\frac{c^3}{a^2}\right)(a+b+c) \geq \left(\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\right)^2$$
but I can't continue this.I tried expanded form:$$\sum \limits_{cyc} \frac{a^5c^2}{a^2b^2c^2} \geq \sum \limits_{cyc} \frac{a^3c}{abc}$$
Which proceeds me to this Cauchy:$$\sum \limits_{cyc} \frac{a^5c^2}{abc}\sum \limits_{cyc}a(abc)\geq \left(\sum \limits_{cyc}a^3c\right)^2$$
I can't continue this one too.
The main Challenge is $3$ fraction on both sides which all of them have different denominator.and it seems like using Cauchy from first step won't leads to anything good.
| By AM–GM, we have
$$14\frac{a^3}{b^2}+3\frac{b^3}{c^2}+2\frac{c^3}{a^2}\geq 19\frac{a^2}{b}\quad (1)$$
$$2\frac{a^3}{b^2}+14\frac{b^3}{c^2}+3\frac{c^3}{a^2}\geq 19\frac{b^2}{c}\quad (2)$$
$$3\frac{a^3}{b^2}+2\frac{b^3}{c^2}+14\frac{c^3}{a^2}\geq 19\frac{c^2}{a}\quad (3)$$
Add all three equations together and conclude.
| {
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Evaluate $\sum_{k=0}^{511}\frac{\sin\frac\pi{2^{11}}}{\sin\frac{(4k+1)\pi}{2^{12}}\sin\frac{(4k+3)\pi}{2^{12}}}$ I need to evaluate
$$\sum_{n=0}^{511}\frac{\sin\frac\pi{2^{11}}}{\sin\frac{(4n+1)\pi}{2^{12}}\sin\frac{(4n+3)\pi}{2^{12}}}$$
Please give me some hint!
The final answer is $2^{10}$.
By CuriousGuest's answer, we need to prove
$$\cot\frac{\pi}{2^{12}}-\cot\frac{2046\pi}{2^{12}}+\cot\frac{2\pi}{2^{12}}-\cot\frac{2047\pi}{2^{12}}=2^{10}$$
any idea?
| Hint: note that $$\frac{\pi}{2^{11}}=\frac{(4n+3)\pi}{2^{12}}-\frac{(4n+1)\pi}{2^{12}}.$$
Then use formula for $\sin(\alpha-\beta)$ and you'll get a telescopic sum.
Edit: It turns out to be not exactly telescopic, but something like
$$\cot\frac{\pi}{2^{12}}-\cot\frac{3\pi}{2^{12}}+\cot\frac{5\pi}{2^{12}}-\cot\frac{7\pi}{2^{12}}+\ldots+\cot\frac{2045\pi}{2^{12}}-\cot\frac{2047\pi}{2^{12}}.$$
This one can be handled in the following way: using that $$\cot\frac{(2^{11}-2k-1)\pi}{2^{12}}=\tan\frac{(2k+1)\pi}{2^{12}}$$
and the equality $\cot \alpha-\tan\alpha=2\cot2\alpha$, we can make this sum twice as short, then do the same thing with resulting sum, and so on 10 times, till we get the final result.
| {
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How do I go from this $\frac{x^2-3}{x^2+1}$ to $1-\frac{4}{x^2+1}$? So I am doing $\int\frac{x^2-3}{x^2+1}dx$ and on wolfram alpha it says the first step is to do "long division" and goes from $\frac{x^2-3}{x^2+1}$ to $1-\frac{4}{x^2+1}$. That made the integral much easier, so how would I go about doing that in a clear manner? Thanks in advance for the help!
| In this particular case
$$\frac{x^2-3}{x^2+1}=\frac{x^2+1-1-3}{x^2+1}=\frac{x^2+1}{x^2+1}-\frac{4}{x^2+1}=1-\frac{4}{x^2+1}$$
| {
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How to find the limit $x=\lim_{a\to{b}}{\frac{a^b-b^a}{a^a-b^b}}$ without using L'Hopital's rule? Let
$$x=\lim_{a\to{b}}{\frac{a^b-b^a}{a^a-b^b}}$$
It is very simple to solve it using L'Hopital's rule, but problem is to solve this limit without L'Hopital's rule. Is there any way to do this?
| Well, this is essentally the long form of L'Hopital, but:
$$\begin{align}\frac{a^b-b^a}{a^a-b^b} &= \frac{a^b-b^b+b^b-b^a}{a-b} \cdot\frac{a-b}{a^a-b^b}\\
&=\frac{\dfrac{a^b-b^b}{a-b} - \dfrac{b^a-b^b}{a-b}}{\dfrac{a^a-b^b}{a-b}}
\end{align}$$
The three terms here converges to the derivative of the functions $x^b$, $b^x$, and $x^x$ at $x=b$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluating a limit using L'Hôpital's rule I know that it can be also evaluate using Taylor expansion, but I am intentionally want to solve it using L'Hôpital's rule:
$$ \lim\limits_{x\to 0} \frac{\sin x}{x}^{\frac{1}{1-\cos x}} =
\lim\limits_{x\to 0}\exp\left( \frac{\ln(\frac{\sin x}{x})}{1-\cos x} \right)$$
Now, from continuity and L'hopital Rule:
$$\lim\limits_{x\to 0} \frac{\ln(\frac{\sin x}{x})}{1-\cos x} =
\lim\limits_{x\to 0} \frac{\frac{x}{\sin x}\cdot\frac{x\cos x - \sin x}{x^2}}{\sin x} =
\lim\limits_{x\to 0}\frac{\frac{x\cos x - \sin x}{x\sin x}}{\sin x}$$
This is where I got stuck.
If I'm not mistaken the limit is $-\frac{1}{3}$ so the orginial one is $e^{-\frac{1}{3}}$
What should I do different (Or what's is wrong with my calculation?)
Thanks
| Alternative approach using Taylor series:
Using $\frac{\sin x}{x} = 1 - \frac{x^2}{6} + o(x^2)$ we get $\ln(\frac{\sin x}{x}) \sim_0 \ln(1-\frac{x^2}{6})$. This expression has the Taylor expansion $-\frac{x^2}{6}+o(x^2)$.
We can approximate the denominator by $1-\cos x=\frac{x^2}{2} + o(x^2)$.
Then we have:
$\frac{\ln(\frac{\sin x}{x})}{1-\cos x} \sim_0 -\frac{x^2}{6} \frac{2}{x^2} = -\frac{1}{3}$.
This doesn't exactly answer your question, but hopefully it provides a little bit of insight. :)
| {
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Dealing with absolute values after trigonometric substitution in $\int \frac{\sqrt{1+x^2}}{x} \text{ d}x$. I was doing this integral and wondered if the signum function would be a viable method for approaching such an integral. I can't seem to find any other way to help integrate the $|\sec \theta|$ term in the numerator of the integrand.
$$ \begin{aligned} \int \dfrac{\sqrt{x^2 + 1}}{x} \text{ d}x \ \ & \overset{x=\tan \theta}= \int \dfrac{\sqrt{\sec^2 \theta} \sec^2 \theta}{\tan \theta} \text{ d}\theta \\ & \ \ = \int \dfrac{|\sec \theta| \sec^2 \theta}{\tan \theta} \text{ d}\theta \\ & \ \ = \int \dfrac{\text{sgn} (\sec \theta) \sec^3 \theta}{\tan \theta} \text{ d}\theta \\ & \ \ = \text{sgn} (\sqrt{1+x^2}) \left( - \log \left| \dfrac{\sqrt{1+x^2} + 1}{x} \right| + \sqrt{1+x^2} \right) + \mathcal{C} \end{aligned} $$
It's clear that $\text{sgn} (\sqrt{1+x^2}) = 1$ since the sign of the argument of that function is always positive and the signum function extracts the sign. So I'd leave the integral as: $$ \begin{aligned} \int \dfrac{\sqrt{x^2 + 1}}{x} \text{ d}x = \log \left| x \right| - \log \left( \sqrt{1+x^2} + 1 \right) + \sqrt{1+x^2} + \mathcal{C} \end{aligned} $$
Would that be ok?
Also, apparently Wolfram has suggested that my final result should have $x$ as opposed to $|x|$ in the argument of the first logarithm. Why is that?
Any help would be greatly appreciated!
| I want to share with you my method using rationalization. $$
\begin{aligned}
& \int \frac{\sqrt{x^{2}+1}}{x} d x \\
=& \int \frac{x^{2}+1}{x \sqrt{x^{2}+1}} d x \\
=& \int \frac{x^{2}+1}{x^{2}} d\left(\sqrt{x^{2}+1}\right) \\
=& \int\left(1+\frac{1}{x^{2}}\right) d\left(\sqrt{x^{2}+1}\right) \\
=& \int 1 d\left(\sqrt{x^{2}+1}\right)+\int \frac{1}{\left(\sqrt{x^{2}+1}\right)^{2}-1} d\left(\sqrt{x^{2}+1}\right) \\
=& \sqrt{x^{2}+1}+\frac{1}{2} \ln \left|\frac{\sqrt{x^{2}+1}-1}{\sqrt{x^{2}+1}+1}\right|+C
\end{aligned}
$$
Wish you enjoy the solution!
| {
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Prove that $\frac{a^3}{x} + \frac{b^3}{y} + \frac{c^3}{z} \ge \frac{(a+b+c)^3}{3(x+y+z)}$ a,b,c,x,y,z are positive real numbers. I stumbled upon it on some olympiad papers. Tried to AM>GM but didn't get any idea to move forward.
| Macavity's way is certainly the most elegant. Here is a way to do it just using Cauchy-Schwarz and AM-GM. Use Cauchy-Schwarz inequality to get:
$$(x+y+z)\left(\frac{a^3}{x}+\frac{b^3}{y}+\frac{c^3}{z}\right)\geq (a^{3/2}+b^{3/2}+c^{3/2})^2$$
So to prove your inequality, it suffices to show that
$$ (a^{3/2}+b^{3/2}+c^{3/2})^2 \geq \frac{(a+b+c)^3}{3}$$
which can be proved as follows. It is equivalent to (after expanding):
$$3(a^3+b^3+c^3)+6(a^{3/2}b^{3/2}+b^{3/2}c^{3/2}+c^{3/2}a^{3/2})\geq a^3+b^3+c^3 +3ab(a+b)+3bc(b+c)+3ca(c+a)+6abc$$
In other words, we want to prove:
$$ 2(a^3+b^3+c^3)+6(a^{3/2}b^{3/2}+b^{3/2}c^{3/2}+c^{3/2}a^{3/2})\geq 3ab(a+b)+3bc(b+c)+3ca(c+a)+6abc$$
Now apply AM-GM:
$$a^3+a^{3/2}b^{3/2}+a^{3/2}b^{3/2}\geq 3a^{2}b$$
$$b^3+a^{3/2}b^{3/2}+a^{3/2}b^{3/2}\geq 3ab^2$$
$$b^3+b^{3/2}c^{3/2}+b^{3/2}c^{3/2}\geq 3b^{2}c$$
$$c^3+b^{3/2}c^{3/2}+b^{3/2}c^{3/2}\geq 3bc^2$$
$$c^3+c^{3/2}a^{3/2}+c^{3/2}a^{3/2}\geq 3c^{2}a$$
$$a^3+c^{3/2}a^{3/2}+c^{3/2}a^{3/2}\geq 3ca^2$$
$$2(a^{3/2}b^{3/2}+b^{3/2}c^{3/2}+c^{3/2}a^{3/2})\geq 6abc$$
Now add all of these seven inequalities to get the desired inequality.
| {
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Explaining why $\sqrt {x^2+a} = x\sqrt{1+ \frac{a}{x^2}}$ For $x>0$. I understand the technical operation of extracting $x^2$ out of the root, but is there a way proving it?
$$\sqrt {x^2+a} = x\sqrt{1+ \frac{a}{x^2}}$$
| Assuming that $x>0$ we have that
\begin{align}
\sqrt{x^2+a} &=\\
\sqrt{x^2\left(1+\frac{a}{x^2}\right)} &=\\
\sqrt{x^2}\sqrt{1+\frac{a}{x^2}} &=\\
|x| \sqrt{1+\frac{a}{x^2}}&=\\
x\sqrt{1+\frac{a}{x^2}}.
\end{align}
In the first equality we must require that $x\neq 0$, since otherwise the division by $x^2$ is not defined. For the next equality we have used that $\sqrt{m\cdot n} = \sqrt{m}\sqrt{n}$, which holds whenever $m,n\geq 0$. For the last equality we have, in the general case, that $\sqrt{x^2} = |x|$. But since we assumed that $x>0$ we have that $|x|=x$.
| {
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Simplify the following compound fraction: $$\frac{2x+1}{\frac{3}{x^2}+\frac{2x+1}{x}}$$
My calculator says the final answer is $$\frac{x^2(2x+1)}{2x^2+x+3}$$
Please show the work. Thanks.
| First let's start with your denominator, $$\frac{3}{x^2}+\frac{2x+1}{x}$$
We can turn this into one term by multiplying the,
$$\frac{2x+1}{x} by \frac{x}{x}$$
and then combine terms to get,
$$\frac{2x^2+x+3}{x^2}$$
Now if we put this back into the first expression we now have,
$$\frac{2x+1}{\frac{2x^2+x+3}{x^2}}$$
From here we simply need to multiply the numerator and denominator by x^2 and we get,
$$\frac{x^2(2x+1)}{2x^2+x+3}$$
Hope this helps!
| {
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What is the non-trivial, general solution of these equal ratios? Provide non-trivial solution of the following:
$$\frac{a}{b+c}=\frac{b}{c+a}=\frac{c}{a+b}$$
$a=?, b=?, c=?$
The solution should be general.
| HINT:
$\displaystyle\frac{a}{b+c}=\frac{b}{c+a}=\frac{c}{a+b}=\frac{a+b+c}{b+c+(c+a)+(a+b)}$
Similarly,
$\displaystyle F=\frac{a}{b+c}=\frac{b}{c+a}=\frac{a-b}{b+c-(c+a)}$
Either $a=b$ or $\displaystyle F=-1$
Hope these should help
| {
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How to solve $\displaystyle x=\sqrt{4+\sqrt{4-\sqrt{4+\sqrt{4-x}}}}$ for $x$? How to solve $\displaystyle x=\sqrt{4+\sqrt{4-\sqrt{4+\sqrt{4-x}}}}$ for $x$?
I tried this way:
Let
$$f(x)=\sqrt{4+\sqrt{4-x}}$$
So, $x=f^2(x)=f^{2n}(x)$ where $n\in\mathbb{N}$. Then, I tried to prove that $f^k(x)=f^{k+1}(x)$ for any $k\in\mathbb{N}$, but I cannot find any easy way to prove this. If I succeed to prove this, I can write
$$x=\sqrt{4+\sqrt{4-x}}$$
because for $k=0$ I have $f^0(x)=f^1(x)$, but then it will be hard to solve $x=\sqrt{4+\sqrt{4-x}}$.
My question is: how to prove that $f^k(x)=f^{k+1}(x)$ and how to solve $x=\sqrt{4+\sqrt{4-x}}$? If you have any easier method, post your solution.
| $\displaystyle x=\sqrt{4+\sqrt{4-\sqrt{4+\sqrt{4-x}}}}$, the replace $x$ at RHS by the same equaility, we get
$$x=\sqrt{4+\sqrt{4-\sqrt{4+\sqrt{4-\sqrt{4+\sqrt{4-\sqrt{4+\sqrt{4-x}}}}}}}}$$.
Continue in this way then we know that $x$ is equal to the limit(provided it exists) of $$\sqrt{4+\sqrt{4-\sqrt{4+\sqrt{4-\sqrt{4+\sqrt{4-\sqrt{4+\sqrt{4-\cdots}}}}}}}}$$.
To compute the limit, say $l$, we have $l = \sqrt{4+\sqrt{4-l}}$, which is not hard to solve.
As for existence of the limit, remark that if $a_1(x) = \sqrt{4+\sqrt{4-x}}$ and $a_{n+1}(x) = a_n(a_1(x))$, then $|a_n(x)-a_n(y)| \leq |a_n(4) - a_n(0)|$(since $a_n(x)$ is either increasing or decreasing), which tends to $0$ by repeating using $\sqrt{x}-\sqrt{y} =\frac{x-y}{\sqrt{x}+\sqrt{y}}$. So $a_n(x)$ is Cauchy(since $a_{n+k}(x)= a_n(y)$ for some $y$) and the limit exists
| {
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Mathematical way to solve integer numbers $217 = (20x+3)r+x$ Is there any mathematical way to find the integer numbers that solve the following equation:
$$217 = (20x+3)r+x$$
| $$217 = (20x+3)r+x\\
217=20xr+3r+x$$
Now
$$(5x+\frac{3}{4})(4r+\frac{1}{5})=20xr+3r+x+\frac{3}{20}$$
Therefore
$$217+\frac{3}{20}=(5x+\frac{3}{4})(4r+\frac{1}{5})$$
Multiplying both sides by $20$ we get
$$4343=(20x+3)(20r+1)$$
There are only few ways of writing $4343$ as a product of two integers, in each case solve. Note that the product has to be of terms of the form $1 \pmod{20}$ and $3 \pmod{20}$.
P.S. Here is a faster way to get to the product:
$$217 = (20x+3)r+x$$
Note that $x=\frac{(20x+3)-3}{20}$. Therefore
$$217 = (20x+3)r+\frac{(20x+3)-3}{20}=(20x+3)(r+\frac{1}{20})-\frac{3}{20}$$
Move $\frac{3}{20}$ on the other size and multiply by $20$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluate $\int\frac{\sqrt{9-x^2}}{x^2}\mathrm dx$ I am trying to solve $$\int\frac{\sqrt{9-x^2}}{x^2}\mathrm dx$$
My answer is slightly different to the memo:
$x=3\sin\theta\quad\iff\quad\theta=\arcsin\left(\frac x 3\right)\\
\text dx=3\cos\theta\ \text d\theta\\$
$\begin{align}I&=\int\frac{3\sqrt{1-\sin^2\theta}}{3\sin^2\theta}\cdot3\cos\theta\ \text d\theta=3\int\frac{\cos^2\theta}{\sin^2\theta}\ \text d\theta=3\int\cot^2\theta\ \text d \theta\\
&=3\int\csc^2\theta\ \text d\theta - 3\int\text d \theta\\
&=-3\cot\theta-3\theta+C\\
&=-\frac{\sqrt{1-\left(\frac x 3\right)^2}}{\frac x 3}-3\arcsin\left(\frac x 3\right)+C\\
&=-\frac{3\sqrt{9-x^2}}{3x}-3\arcsin\left(\frac x 3\right)+C\\
&=-\frac{\sqrt{9-x^2}}x-3\arcsin\left(\frac x 3\right)+C \end{align}$
and the memo has $$-\frac{\sqrt{9-x^2}}x-\arcsin\left(\frac x 3\right)+C$$
Your help is appreciated!
| $$\int\frac{\sqrt{9-x^2}}{x^2}dx$$
$x=3\sin t,dx=3\cos tdt$
$$\int\frac{\sqrt{9-9\sin^2t}}{9\sin^2t}3\cos tdt=\int\frac{\cos^2t}{\sin^2t} dt\neq3\int\frac{\cos^2t}{\sin^2t} dt$$
| {
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Why is $ A_1 x + ... + A_n x^n $ a solution of $ \sum_0^{n} (-1)^n \frac{x^n}{n!} \frac{d^n y}{d x^n} = 0 $? I was playing(/fiddling) around with some maths and I saw this pattern(
where $ A_n $ is a constant.):
$ A_1 x $ is a soultion of:
$$ \frac{y}{x} - \frac{dy}{dx} = 0 $$
$ A_1 x + A_2 x^2 $ is a solution of:
$$ \frac{y}{x} - \frac{dy}{dx} + \frac{x}{2!} \frac{d^2y}{dx^2} =0 $$
$ A_1 x + A_2 x^2 + A_3 x^3 $ is a solution of:
$$ \frac{y}{x} - \frac{dy}{dx} + \frac{x}{2!} \frac{d^2y}{dx^2} - \frac{x^2}{3!} \frac{d^3y}{dx^3} =0 $$
It continues so on. Can someone prove the solution of
$ A_1 x + A_2 x^2 + A_3 x^3 + ... + A_n x^n $ is:
$$ \sum_{k=0}^{n} (-1)^k \frac{x^{k-1}}{k!} \frac{d^k y}{d x^k} = 0 $$
| For any polynomial $P(x)$ of degree at most $n$, we have $\frac{d^k}{dx^k}P(x) = 0$ for any $k > n$. As a result, for any constant $y$, we have
$$\sum_{k=0}^{n} \frac{(y-x)^k}{k!} \frac{d^k}{d x^k}P(x)
= \sum_{k=0}^{\infty} \frac{(y-x)^k}{k!} \frac{d^k}{d x^k}P(x) = P(y)$$
The last equality is true because the expression in the middle is nothing but
the Taylor expansion of $P(y)$ with respect to the point $x$.
Substitute $y = 0$, we find any polynomial $P(x)$ with $P(0) = 0$ satisfy the ODE
$$\sum_{k=0}^n \frac{(-x)^k}{k!} \frac{d^k}{dx^k} P(x) = P(0) = 0$$
| {
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Number of solutions of $x_1 + x_2 + x_3 + x_4 = 14$ such that $x_i \le 6$
Let $x_1, x_2, x_3, x_4$ be nonnegative integers.
(a) Find the number of solutions to the following equation:
$$ x_1 + x_2 + x_3 + x_4 = 14 $$
I got $17 \choose 3$ for this. Is that correct?
(b) Find the number of solutions if we add the restriction that $x_i \le 6$ for
$1 \le i \le 4$
| The answer you got for the first question is right.
For the second, call a distribution bad if one or more of the $x_i$ is $\ge 7$. Our strategy is to count the number of bads, and subtract from the answer of a).
One can have $2$ of the $x_i$ equal to $7$. This can be done in $\binom{4}{2}$ ways.
Now we count the number of bads in which only one of the $x_i$ is $\ge 7$. Which one it is can be chosen in $\binom{4}{1}$ ways. Suppose that $x_1\ge 7$. Give $7$ candies to kid 1. The remaining $7$ have to be split between the four people, with none of 2, 3, or 4 getting $7$. There are $\binom{10}{3}-3$ ways to do this.
We get a total of $\binom{4}{1}\left[\binom{10}{3}-3\right]+\binom{4}{2}$ bads.
Alternately, we use Inclusion/Exclusion more explicitly. Choose one of the $4$ to give at least $7$ to, and give her $7$. We can distribute the remaining $7$ among the $4$ people in $\binom{10}{3}$ ways. But this double-counts the $\binom{4}{2}$ ways to give $7$ to two of the variables. So the number of bads is $\binom{4}{1}\binom{10}{3}-\binom{4}{2}$.
| {
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Find the sum of the multiples of $3$ and $5$ below $709$? I just cant figure this question out:
Find the sum of the multiples of $3$ or $5$ under $709$
For example, if we list all the natural numbers below $10$ that are multiples of $3$ or $5$, we get $3$, $5$, $6$ and $9$. The sum of these multiples is $23$.
| Let $n_3 = \lfloor \frac{708}{3}\rfloor, \; n_5 = \lfloor \frac{708}{5}\rfloor, \; n_15 = \lfloor \frac{708}{15}\rfloor.\;$ Then using the hints in the comments your sum $S$ is
$$S=3\sum_{k=1}^{n_3}k + 5\sum_{k=1}^{n_5}k-15\sum_{k=1}^{n_{15}}k
=3\frac{n_3(n_3+1)}{2}+5\frac{n_5(n_5+1)}{2}-15\frac{n_{15}(n_{15}+1)}{2}$$
$$=3\frac{236\times 237}{2}+5\frac{141\times 142}{2}-15\frac{47\times 48}{2} = 117033
$$
Note added in proof: After Darth Geek's first differing answer, I verified my $S=117033$ with a small program.
| {
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How many ways are there to divide $100$ different balls into $5$ different boxes so the last $2$ boxes contains even number of balls? How many ways are there to divide $100$ different balls into $5$ different boxes so the last $2$ boxes contains even number of balls?
I tried to think about tylor function but got stuck.
Thanks.
| Let $g_{e}(x)=\displaystyle\left(1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots\right)^3\left(1+\frac{x^2}{2!}+\frac{x^4}{4!}+\cdots\right)^2=(e^x)^3\left(\frac{e^x+e^{-x}}{2}\right)^2$
$\;\;\;\;\;\;\;\;\;\;\;\;=\displaystyle e^{3x}\left(\frac{e^{2x}+2+e^{-2x}}{4}\right)=\frac{1}{4}(e^{5x}+2e^{3x}+e^x)$.
The coefficient of $x^{100}$ is given by $\displaystyle\frac{a_{100}}{100!}=\frac{1}{4}\left(\frac{5^{100}}{100!}+2\cdot\frac{3^{100}}{100!}+\frac{1}{100!}\right)$, so
$\hspace{1.5 in}$ $\displaystyle a_{100}=\frac{1}{4}(5^{100}+2\cdot3^{100}+1)$.
| {
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Find the values of $f(0)$, $f(4)$, $f(6)$ and $f(18)$
A function $f:\mathbb{R} \to \mathbb{R}$ is such that $f(2)=2$ and
$$f(x+1)+f(x-1)=\sqrt{3}f(x) \tag{1}.$$
Find the values of $f(0)$, $f(4)$, $f(6)$ and $f(18)$.
My approach: replace $x$ with $x+1$ in $(1)$ we get
$$f(x+2)+f(x)=\sqrt{3}f(x+1) \tag{2}.$$
Replace $x$ with $x+2$ in $(1)$ and using $f(x+2)$ from $(2)$ we get
$$f(x+3)+f(x+1)=\sqrt{3}f(x+2)=\sqrt{3}\left(\sqrt{3}f(x+1)-f(x)\right) $$ $\implies$
$$f(x+3)=2f(x+1)-\sqrt{3}f(x)=2f(x+1)-f(x+1)-f(x-1)=f(x+1)-f(x-1).$$
So $$f(x+3)=f(x+1)-f(x-1) \tag{3}.$$
Replace $x$ with $x+2$ in $(3)$ we get
$$f(x+5)=f(x+3)-f(x+1)=-f(x-1).$$
So
$$f(x+5)=-f(x-1)$$ $\implies$
$f(x+6)=-f(x)$ and $f(x+12)=f(x).$
So $f(x)$ is periodic with Fundamental Period $12$.
Using $x=1$ $(3)$ we get $f(0)+f(4)=2$ but I could not find the value of $f(0)$.
| Without any additional conditions on $f$, it is possible to pick $x_0, f(x_0), f(x_0+1)$ arbitrary and then use $(1)$ (i.e. $f(x) = \sqrt 3f(x-1)-f(x-2)$ to go forward $f(x) = \sqrt 3f(x+1)-f(x+2)$ to go backwards) to extend this uniquely to $x_0+\mathbb Z$. Likewise, we can prescribe arbitrary values for $f|_{(0,2]}$ and extend this uniquely to all of $\mathbb R$. Even if we want $f$ to be smooth, we can prescribe $f$ on $(\epsilon,2]$ arbitrarily. This gives us $f|_{(\epsilon-1,0]}$. If we "glue" these parts smoothly together, the resulting $f\colon\mathbb R\to\mathbb R$ is smooth.
With $a:=f(1)$ we find by the recursion that $f(0)=a\sqrt 3-2$, $f(18)=f(6)=-f(0)=2-a\sqrt 3$ and $f(4)=4-a\sqrt 3$. However, as just seen $a$ can be picked arbitrarily, even if we assume $f$ is smooth.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/905892",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Analogue of $\zeta(2) = \frac{\pi^2}{6}$ for Dirichlet L-series of $\mathbb{Z}/3\mathbb{Z}$? Consider the two Dirichlet characters of $\mathbb{Z}/3\mathbb{Z}$.
$$
\begin{array}{c|ccr}
& 0 & 1 & 2 \\ \hline
\chi_1 & 0 & 1 & 1 \\
\chi_2 & 0 & 1 & -1
\end{array}
$$
I read the L-functions for these series have special values
*
*$ L(2,\chi_1) \in \pi^2 \sqrt{3}\;\mathbb{Q} $
*$ L(1,\chi_2) \in \pi \sqrt{3}\;\mathbb{Q} $
In other words, these numbers are $\pi^k \times \sqrt{3} \times \text{(rational number)}$. Is there a way to derive this similar to to the famous $\zeta(2) = \tfrac{\pi^2}{6}$ formula?
Here are 14 proofs of $\zeta(2) = \tfrac{\pi^2}{6}$ for reference
| We see that
\begin{align}
\frac{1}{2\pi i}\oint_{C_N}\frac{\pi\cot(\pi z)}{(3z+1)^2}{\rm d}z
&=\operatorname*{Res}_{z=-1/3}\frac{\pi\cot(\pi z)}{(3z+1)^2}+\sum^\infty_{n=-\infty}\frac{1}{(3n+1)^2}\\
&=-\frac{4\pi^2}{27}+\sum^\infty_{n=-\infty}\frac{1}{(3n+1)^2}\\
&=0
\end{align}
and
\begin{align}
\frac{1}{2\pi i}\oint_{C_N}\frac{\pi\cot(\pi z)}{(3z+2)^2}{\rm d}z
&=\operatorname*{Res}_{z=-2/3}\frac{\pi\cot(\pi z)}{(3z+2)^2}+\sum^\infty_{n=-\infty}\frac{1}{(3n+2)^2}\\
&=-\frac{4\pi^2}{27}+\sum^\infty_{n=-\infty}\frac{1}{(3n+2)^2}\\
&=0
\end{align}
This implies
$$L(2,\chi_1)=\frac{1}{2}\left(\frac{4\pi^2}{27}+\frac{4\pi^2}{27}\right)=\frac{4\pi^2}{27}$$
Similarly,
\begin{align}
\frac{1}{2\pi i}\oint_{C_N}\frac{\pi\cot(\pi z)}{3z+1}{\rm d}z
&=\operatorname*{Res}_{z=-1/3}\frac{\pi\cot(\pi z)}{3z+1}+\sum^\infty_{n=-\infty}\frac{1}{3n+1}\\
&=-\frac{\pi}{3\sqrt{3}}+\sum^\infty_{n=-\infty}\frac{1}{3n+1}\\
&=0
\end{align}
and
\begin{align}
\frac{1}{2\pi i}\oint_{C_N}\frac{\pi\cot(\pi z)}{3z+2}{\rm d}z
&=\operatorname*{Res}_{z=-2/3}\frac{\pi\cot(\pi z)}{3z+2}+\sum^\infty_{n=-\infty}\frac{1}{3n+2}\\
&=\frac{\pi}{3\sqrt{3}}+\sum^\infty_{n=-\infty}\frac{1}{3n+2}\\
&=0
\end{align}
Hence
$$L(1,\chi_2)=\frac{1}{2}\left(\frac{\pi}{3\sqrt{3}}-\left(-\frac{\pi}{3\sqrt{3}}\right)\right)=\frac{\pi}{3\sqrt{3}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/906424",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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The number $(3+\sqrt{5})^n+(3-\sqrt{5})^n$ is an integer Prove by induction that this number is an integer:
$$u_n=(3+\sqrt{5})^n+(3-\sqrt{5})^n$$
Progress
I assumed that it holds for $n$ and I tried to do it for $n+1$ but the algebra gets quite messy and I'm unable to prove that the following term is an integer: $\sqrt{5}((3+\sqrt{5})^n-(3-\sqrt{5})^n)$
| Let $v_n = (3+\sqrt{5})^n - (3-\sqrt{5})^n$, then
$$
u_1 = 6, v_1 = 2\sqrt{5}
$$
And
$$
u_{n+1} = (3+\sqrt{5})(3+\sqrt{5})^n + (3-\sqrt{5})(3-\sqrt{5})^n = 6u_n + \sqrt{5}v_n
$$
$$
v_{n+1} = (3+\sqrt{5})(3+\sqrt{5})^n - (3-\sqrt{5})(3-\sqrt{5})^n = 6v_n +\sqrt{5}u_n
$$
If $v_n$ is an integer multiple of $\sqrt{5}$ and $u_n$ is an integer, then $v_{n+1}$ is an integer multiple of $\sqrt{5}$ and $u_{n+1}$ is an integer.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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Sum the series $\sum_{n = 1}^{\infty}\{\coth (n\pi x) + x^{2}\coth(n\pi/x)\}/n^{3}$ This sum is from Ramanujan's letters to G. H. Hardy and Ramanujan gives the summation formula as
\begin{align} &\frac{1}{1^{3}}\left(\coth \pi x + x^{2}\coth\frac{\pi}{x}\right) + \frac{1}{2^{3}}\left(\coth 2\pi x + x^{2}\coth\frac{2\pi}{x}\right) \notag\\
&\, \, \, \, \, \, \, \, + \frac{1}{3^{3}}\left(\coth 3\pi x + x^{2}\coth\frac{3\pi}{x}\right) + \cdots\notag\\
&\, \, \, \, \, \, \, \, = \frac{\pi^{3}}{90x}(x^{4} + 5x^{2} + 1)\notag
\end{align}
Since $$\coth x = \frac{e^{x} + e^{-x}}{e^{x} - e^{-x}} = \frac{1 + e^{-2x}}{1 - e^{-2x}} = 1 + 2\frac{e^{-2x}}{1 - e^{-2x}}$$the above sum is transformed into $$(1 + x^{2})\sum_{n = 1}^{\infty}\frac{1}{n^{3}} + 2\sum_{n = 1}^{\infty}\frac{e^{-2n\pi x}}{n^{3}(1 - e^{-2n\pi x})} + 2x^{2}\sum_{n = 1}^{\infty}\frac{e^{-2n\pi/x}}{n^{3}(1 - e^{-2n\pi/x})}$$ If we put $q = e^{-\pi x}$ we get sums like $\sum q^{2n}/\{n^{3}(1 - q^{2n})\}$ which I don't know how to sum.
It seems I am going on a wrong track. Please provide some alternative approach.
Update: All the answers given below so far use complex analyis (transforms and residues) to evaluate the sum. I am almost certain that Ramanujan did not evaluate the sum using complex analysis. Perhaps the method by Ramanujan is more like the one explained in this question. Do we have any approach based on real-analysis only?
| Yet another approach using contour integration is to integrate the function $$f(z) = \frac{\pi \cot (\pi z) \coth (\pi x z)}{z^{3}} $$ around a circle centered at the origin that avoids the poles on the real and imaginary axes.
If we let the radius of the circle go to infinity discretely, the integral will vanish.
So summing the residues, we get $$2 \sum_{n=1}^{\infty} \frac{\coth (n \pi x)}{n^{3}} + \sum_{n=1}^{\infty} \frac{\cot (\frac{in \pi}{x})}{x(\frac{in}{x})^{3}} + \sum_{n=1}^{\infty} \frac{\cot (-\frac{i n \pi}{x})}{x (-\frac{in}{x} )^{3}} + \text{Res}[f(z),0] = 0,$$
which implies
$$\sum_{n=1}^{\infty} \frac{\coth (n \pi x)}{n^{3}} + x^{2} \sum_{n=1}^{\infty} \frac{\coth(\frac{n \pi}{x})}{n^{3}} = - \frac{1}{2} \ \text{Res} [f(z),0]. $$
Expanding at the origin, we get
$$ \begin{align} \small f(z) &= \frac{\pi}{z^{3}}\left(\frac{1}{\pi z} - \frac{2 \zeta(2)}{\pi} z-\frac{2 \zeta(4)}{\pi} z^{3} + \mathcal{O}(z^{5})\right) \left(\frac{1}{\pi (xz)} + \frac{2 \zeta(2)}{\pi} (xz) - \frac{2 \zeta(4)}{\pi} (xz)^{3} + \mathcal{O}(z^{5}) \right) \\ &= \frac{1}{\pi x} \frac{1}{z^{5}} + \frac{2 \zeta(2) x^{2}-2 \zeta(2)}{\pi x} \frac{1}{z^{3}} {\color{red}{-\frac{2 \zeta(4) x^{4}+4 \zeta(2)^{2} x^{2} + 2 \zeta(4)}{\pi x}}} \frac{1}{z} + \mathcal{O}(z) .\end{align} $$
Therefore, $$ \sum_{n=1}^{\infty} \frac{\coth (n \pi x)}{n^{3}} + x^{2} \sum_{n=1}^{\infty} \frac{\coth(\frac{n \pi}{x})}{n^{3}} = \frac{\zeta(4) x^{4}+2 \zeta(2)^{2} x^{2} + \zeta(4)}{\pi x} =\frac{\pi^{3}}{90x} \left( x^{4}+5x^{2}+1 \right). $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/907480",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "25",
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Show the origin is the only critical point of $f$ $f(x,y,z)=\frac1 2(x^2+y^2+z^2)+a(xy+yz+zx)$
For $a\neq-\frac1 2$ and $a\neq1$, show that the origin is the only critical point of f.
I have found $\nabla f =\begin{pmatrix} x+a(y+z) \\ y+a(x+z) \\ z+a(x+y) \end{pmatrix} $
Equating this to zero to find critical points, I get $\begin{pmatrix} x \\ y \\ z \end{pmatrix}$ = $\begin{pmatrix} -a(y+z) \\ -a(x+z) \\ -a(x+y) \end{pmatrix}$ (1)
Subtracting each row in (1), I get: $x-y = a(x-y), x-z=a(x-z), y-z=a(y-z)$
, therefore $a=1$ gives non-zero critical points.
Then subbing $z$ in (1) into the other two equations, I get:
$x(1-a^2)=y(a^2-a)$ and $y(1-a^2)=x(a^2-a)$, then subbing y into first equation and canceling, I get:
$x(1+a)=\frac{a^2} {1+a}x$, this means that $(1+a)^2=a^2$, solving we get: $a=-\frac12$
Since, $a\neq1$ and $a\neq -\frac12 $, we can have only critical points for when $x,y,z=0$
Does this prove that this is the only set of critical points? Surely it shows that $a=1$ and $a= -\frac12 $ give critical points for $x,y,z$, but not that there are no other critical points. How would I show that there are no other critical points?
| Suppose $a\neq-\frac{1}{2}$ and $a\neq 1$. Sum the components of $\nabla f$ gives
$$
(x+y+z)+2a(x+y+z)=0
$$
If $x+y+z\neq 0$, then we would have $a=-\frac{1}{2}$. So $x+y+z=0$. Use this in $x=-a(y+z)$:
$$
x=-a(y+z)=-a(-x)=ax.
$$
If $x\neq 0$, then $a$ would be $1$. So $x=0$. Similarly, you can apply $x+y+z=0$ to $y=-a(x+z)$ to get $y=0$. And with $x=y=0$, $z$ must also be $0$.
So $x=y=z=0$ is the only solution to $\nabla f=0$ when $a\neq-\frac{1}{2}$ and $a\neq 1$.
P.s. Thanks smith for \nabla.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
If $\sin( 2 \theta) = \cos( 3)$ and $\theta \leq 90°$, find $\theta$ Find $\theta\leq90°$ if
$$\sin( 2 \theta) = \cos( 3)$$
I know that $\sin 2\theta = 2\sin\theta\cos \theta$, or alternatively, $\theta = \dfrac{\sin^{-1}(\cos 3)}{2}$.
Can somebody help me?
| $$\sin2\theta=\cos3^\circ=\sin87^\circ$$
$$\implies2\theta=180^\circ m+(-1)^m87^\circ\text{ where } m \text{ is any integer}$$
$$\implies\theta=90^\circ m+(-1)^m43.5^\circ$$
Check for even $m=2r$(say) and for odd $=2r+1$(say)
Find $m$ such that $\displaystyle\theta\le90^\circ$
Alternatively,
$$\cos3^\circ=\sin2\theta=\cos\left(90^\circ-2\theta\right)=\cos\left(2\theta-90^\circ\right)$$
$$\iff2\theta-90^\circ= 360^\circ n\pm3^\circ\text{ where } n \text{ is any integer}$$
$\displaystyle'+'\implies 2\theta= 360^\circ n+93^\circ\iff\theta=180^\circ n+46.5^\circ$
$\displaystyle'-'\implies 2\theta= 360^\circ n+87^\circ\iff\theta=180^\circ n+43.5^\circ$
Find $n$ such that $\displaystyle\theta\le90^\circ$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/908703",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 4
} |
$\ln (2 x-5)>\ln (7-2 x)$ Solve
$$\ln (2 x-5)>\ln (7-2 x)$$
The answer is given as $$3<x<7/2$$
This is what I have done $$\ln (2 x-5)-\ln (7-2 x)>0$$
$$\ln \left(\frac{2 x-5}{7-2 x}\right)>0$$
However I am not able to understand how to get to the answer provided.
| Exponentiation is a one to one, increasing function so:
$\ln (2x-5) > \ln (7-2x)\\
e^{\ln (2x-5)} > e^{\ln (7-2x)}\\
2x-5>7x-2\\
...\\
x>3$
However there are the domains to consider as well:
$2x-5>0$ gives $x>\frac{5}{2}$ and $7-2x>0$ gives $x<7/2$. We have to put all of the inequalities $x>3$ and $x>\frac{5}{2}$ and $x<\frac{7}{2}$ Together. This gives the answer $3<x<\frac{7}{2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/908819",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How prove this there many infinite numbers postive ineteger $n$ such $n^3+1|n!$ Question:
show that: there exsit infinite postive integer $n$ such
$$n^3+1|n!$$
Buy the way,I know prove this this simple problem:
prove there are many infinte postive integer numbers $n$,such
$n^2+1|n!$
This problem we only consider this pell equation
$$n^2+1=5y^2$$ have a solution $(2,1)$, so the pell equation have many infite solution.
because $$2y=2\sqrt{\dfrac{n^2+1}{5}}<n\Longrightarrow n>5, 5,y,2y\in\{1,2,3,\cdots,n\}$$
so there exsit infinte $n$ such
$$n^2+1|n!$$
But this is $n^3+1$ so I can't .Thank you
| I will prove there are infinitely many solutions of the form $n=3a^2$. For such $n$ we have $3 \nmid n+1$, hence $\gcd(n+1,n^2-n+1)=1$ and it is sufficient to show that both $n+1 \mid n!$ and $n^2-n+1 \mid n!$. We have $n+1 \mid n!$ whenever $n+1 = 3a^2+1$ is not prime, whereas $n^2-n+1 = 9a^4-3a^2+1$ factors as $(3a^2+3a+1)(3a^2-3a+1)$. Since both factors are relatively prime, we should have $3a^2+3a+1 \mid (3a^2)!$ and $3a^2-3a+1 \mid (3a^2)!$. The latter is always true, whereas $3a^2+3a+1 \mid (3a^2)!$ is true when $3a^2+3a+1$ is not prime. Therefore, it is sufficient to show that there are infinitely many $a$ for which neither $3a^2+1$ nor $3a^2+3a+1$ is prime, which should not be that hard (e.g. take $a \equiv 1 \mod 7$ and $a \equiv 2 \mod 13$).
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find $\sec \theta + \tan \theta$. If $\tan \theta=x-\frac{1}{x}$, find $\sec \theta + \tan \theta$.
This was the question ask in my unit test.
My Efforts:
$\tan^2 \theta=(x-\frac{1}{x})^2$
$\tan^2 \theta= (\frac {x^2-1}{x})^2$
Now we can use identity $\sec^2 \theta= 1 + \tan^2 \theta$.
But i am not able to get the answer using this.
I don't know the correct answer but I had got $2x\ or\ -\frac{2}{x} $, which was given wrong.
Also please tell me if there is better way to do this.
| As you have written $\sec^2 \theta = 1 + \tan^2 \theta$. From this you could get
$$ \sec \theta + \tan \theta = \pm \sqrt{1+\tan^2 \theta} + \tan \theta = \pm \sqrt{1 + \left( x - \frac{1}{x} \right)^2} + x - \frac{1}{x}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/911188",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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} |
Laplace Transform assistance Find the inverse laplace transform of: $\frac{25}{(s-1)^2(s^2+4)}$
$\frac{25}{(s-1)^2(s^2+4)}=\frac{A}{s-1}+\frac{B}{(s-1)^2}+\frac{C}{s^2 + 4}$
$$25=A(s^2+4)(s-1)+B(s^2+4)+C(s-1)^2$$
$\frac{25}{(s-1)^2(s^2+4)}=\frac{0}{s-1}+\frac{5}{(s-1)^2}+\frac{-15}{s^2 + 4}$
Maybe I don't know how to transform, but I got $5e^{\alpha t} -7.5 \sin(\alpha t)$ and this is apparently wrong. What is wrong?
| $$25=A(s^2+4)(s-1)+B(s^2+4)+(Cs+D)(s-1)^2$$ is correct. Therefore we have $$25=s^3(A+C)+s^2(-A+B-2C+D)+s(4A+C-2D)+(-4A+4B+D)$$ Now, $$A=0, -A+B+C=0,4A-2C=0, -4A+4B+C=25$$ which gives $$A=-2, C=2, D=-3, B=5$$
| {
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"timestamp": "2023-03-29T00:00:00",
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When $n$ is divided by $14$, the remainder is $10$. What is the remainder when $n$ is divided by $7$? I need to explain this to someone who hasn't taken a math course for 5 years. She is good with her algebra.
This was my attempt:
Here's how this question works. To motivate what I'll be doing,
consider \begin{equation*} \dfrac{5}{3} = 1 + \dfrac{2}{3}\text{.}
\end{equation*} This is because when 5 is divided by 3, 3 goes into 5
once (hence the $1$ term) and there is a remainder of $2$ (hence the
$\dfrac{2}{3}$ term). Note the following: every division problem can
be decomposed into an integer (the $1$ in this case) plus a fraction,
with the denominator being what you divide by (the $3$ in this
case).
So, when $n$ is divided by 14, the remainder is 10. This can be
written as \begin{equation*} \dfrac{n}{14} = a + \dfrac{10}{14}
\end{equation*} where $a$ is an integer.
We want to find the remainder when $n$ is divided by 7, which I'll
call $r$. So \begin{equation*} \dfrac{n}{7} = b + \dfrac{r}{7}\text{,}
\end{equation*} where $b$ is an integer.
Here's the key point to notice: notice that \begin{equation*}
\dfrac{n}{7} = \dfrac{2n}{14} = 2\left(\dfrac{n}{14}\right)\text{.}
\end{equation*} This is because $\dfrac{1}{7} = \dfrac{2}{14}$.
Thus, \begin{equation*} \dfrac{n}{7} = 2\left(\dfrac{n}{14}\right) =
2\left(a + \dfrac{10}{14}\right) = 2a + 2\left(\dfrac{10}{14}\right) =
2a + \dfrac{10}{7} = 2a + \dfrac{7}{7} + \dfrac{3}{7} = (2a+1) +
\dfrac{3}{7}\text{.} \end{equation*} So, since $a$ is an integer, $2a
+ 1$ is an integer, which is our $b$ from the original equation. Thus, $r = 3$.
To her, this method was not very intuitive. She did understand the explanation. Are there any suggestions for how I can explain this in another way?
| To motivate this to students who are just beginning to learn about remainders, it often helps to use real-world examples, e.g. if changing $\,n\,$ pennies (1 cent coins) into dimes (10 cent coins) leaves 8 cents, then changing these $\,n\,$ pennies into nickels (5 cent coins) leaves 3 cents, because we can first change them to dimes, then change each dime to 2 nickels, then change the remaining 8 pennies to 1 nickel and 3 pennies. Algebraically
$$\ n = 10q + 8\, =\, 5(2q) + 5 + 3\, =\, 5(2q+1) + 3$$
In modular language, $\ {\rm mod}\,\ \color{}5\!:\ \begin{array}{r}\color{#c00}{10\equiv 0}\\ \color{#0a0}{8\equiv 3}\end{array}\,\ \Rightarrow\ \begin{array}{r}\color{#c00}{10}q+\color{#0a0}8\\ \,\equiv\ \color{#c00}0\,+\,\color{#0a0}3\end{array}\ $ by $ $ Congruence Sum, Product Rules.
Most students can easily visualize the money-changing. The goal is to help them translate that intuition into rigorous algebra, or, more powerfully, congruence algebra - as above.
| {
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"timestamp": "2023-03-29T00:00:00",
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How to solve the ODE $y(y'(x)+a)=bx$ I've been very frustrated attempting to separate this guy. Since there are three separate items when you multiply through by the y, it is very difficult to make all sides work out to forms $f(y)dy$ and $f(x)dx$. I've attempted the trick of getting it in the form $y'dx/y$ to get the integral to be $\log(y)$, but haven't successfully separated it yet. Any help would be appreciated, I want to understand your attempt, not just apply it if you know what I mean.
Also, initial condition is: $ y(0)=0$
EDIT: we can try $y=Cx$ and find that it is a valid solution for specific C values.
| We have $$0 = y(y'+a) - bx = \frac{1}{2}\frac{d}{dx}(y^2 + ay - bx^2) - \frac{a}{2}x^2\frac{d}{dx}\left(\frac{y}{x}\right)$$
Let us therefore consider the function
$$h(x) = \frac{y(y'+a) - bx}{y^2 + axy - bx^2}$$
Then $h \equiv 0$ and by integrating the last expression we get
$$\int \frac{y(y'+a)-bx}{y^2 + axy - bx^2} dx = C$$
Lets rewrite this as
$$\frac{1}{2}\int \frac{\frac{d}{dx}(y^2 + axy - bx^2) - ax^2\frac{d}{dx}(y/x)}{y^2 + axy - bx^2} dx = C$$
This first part integrates to $$\frac{1}{2}\log(y^2 + axy - bx^2)$$ and for the second part we can write it as
$$-\frac{a}{2}\int \frac{\frac{d}{dx}(y/x + a/2)}{(y/x + a/2)^2 - (a^2/4 + b)} dx$$
which by making the substitution $z = \frac{y/x + a/2}{\sqrt{-a^2/4-b}}$ becomes
$$-\frac{a}{2\sqrt{a^2/4+b}}\int \frac{dz}{z^2 + 1} = -\frac{a}{2\sqrt{-a^2/4-b}}\arctan(z)$$
The solution can be written as the implicit equation
$$\frac{1}{2}\log(y^2 + axy - bx^2) -\frac{a}{2\sqrt{-a^2/4-b}}\arctan\left(\frac{y/x +a/2}{\sqrt{-a^2/4-b}}\right) = C$$
There might be some ways to simplify this further for particular choices of $a,b$, but in general it seems this is the closest you are going to get to a closed form solution.
${\bf Edit:}$
The solution above is for $4b + a^2 < 0$. In the opposite case the integral has a simpler representation. Making the substitution $z = \frac{y/x + a/2}{\sqrt{a^2/4+b}}$ instead we have
$$-\frac{a}{2\sqrt{a^2/4+b}}\int \frac{dz}{z^2 - 1} = -\frac{a}{4\sqrt{a^2/4+b}}\log \left(\frac{1-\frac{y/x + a/2}{\sqrt{a^2/4+b}}}{1+\frac{y/x + a/2}{\sqrt{a^2/4+b}}}\right)$$
so that we get
$$\frac{1}{2}\log(y^2 + axy - bx^2)-\frac{a}{4\sqrt{a^2/4+b}}\log \left(\frac{1-\frac{y/x + a/2}{\sqrt{a^2/4+b}}}{1+\frac{y/x + a/2}{\sqrt{a^2/4+b}}}\right) = C$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Prove $\frac{ab}{\sqrt{2c+a+b}}+\frac{bc}{\sqrt{2a+b+c}}+\frac{ca}{\sqrt{2b+c+a}}\le\sqrt\frac{2}{3}$ if $a+b+c=2$
Let $a$, $b$ and $c$ be positive numbers such that $a+b+c=2$. Prove: $$\frac{ab}{\sqrt{2c+a+b}}+\frac{bc}{\sqrt{2a+b+c}}+\frac{ca}{\sqrt{2b+c+a}}\le\sqrt\frac{2}{3}$$
Additional info:I'm looking for solutions and hint that using Cauchy-Schwarz, Hölder and AM-GM because I have background in them.
Things I have tried: I was thinking to making the denominator smaller using AM-GM.but I was not successful. My other idea was to re write LHS into this form.something like my idea on this question $$A-\frac{ab}{\sqrt{2c+a+b}}+B-\frac{bc}{\sqrt{2a+b+c}}+C-\frac{ca}{\sqrt{2b+c+a}}$$
But I was not able to observe something good.
I don't know this will lead to something useful but,here is my other idea. let $x=2c+a+b,y=2a+b+c,z=2b+c+a$. rewriting LHS:$$\sum\limits_{cyc}\frac{(3y-(x+z))(3z-(y+x))}{16\sqrt x} \le \sqrt\frac{2}{3}$$ $\sum\limits_{cyc}$denotes sums over cyclic permutations of the symbols $x,y,z$ . another thing I observed that $(x-y-z)^2-4(y-z)^2 = (3y-(x+z))(3z-(y+x))$
I looked at related problems and I think this and (Prove $\frac{a}{ab+2c}+\frac{b}{bc+2a}+\frac{c}{ca+2b} \ge \frac 98$) may have some common idea in proof with my question inequality.
Well,it seems like someone posted it a little after on AoPS.right now there is a solution there by $uvw$ and Cauchy-Schwarz. I post the starting part of solution that is with Cauchy (credits to arqady of AoPS). By Cauchy-Schwarz:$$\left(\sum_{cyc}\frac{ab}{\sqrt{2c+a+b}}\right)^2\leq(ab+ac+bc)\sum_{cyc}\frac{ab}{2c+a+b}$$
Hence, it remains to prove that: $$(ab+ac+bc)\sum_{cyc}\frac{ab}{2c+a+b}\leq\frac{(a+b+c)^3}{12}$$
I stuck here.
| By AM-GM and C-S we obtain:
$$\sum_{cyc}\frac{ab}{\sqrt{2c+a+b}}=\sum_{cyc}\frac{ab\sqrt{\frac{3}{8}}\cdot2\sqrt{\frac{8}{3}(2c+a+b)}}{2(2c+a+b)}\leq\sqrt{\frac{3}{32}}\sum_{cyc}\frac{ab\left(\frac{8}{3}+2c+a+b\right)}{2c+a+b}=$$
$$=\sqrt{\frac{1}{96}}\sum_{cyc}\frac{ab(4(a+b+c)+6c+3a+3b)}{2c+a+b}=\sqrt{\frac{1}{96}}\sum_{cyc}\frac{ab(10c+7a+7b)}{2c+a+b}=$$
$$=\sqrt{\frac{1}{96}}\sum_{cyc}\left(\frac{-4abc}{2c+a+b}+7ab\right)\leq\sqrt{\frac{1}{96}}\left(\frac{-36abc}{\sum\limits_{cyc}(2c+a+b)}+7(ab+ac+bc)\right)=$$
$$=\sqrt{\frac{1}{96}}\left(\frac{-9abc}{a+b+c}+7(ab+ac+bc)\right).$$
Thus, it remains to prove that
$$\sqrt{\frac{1}{96}}\left(\frac{-9abc}{a+b+c}+7(ab+ac+bc)\right)\leq\sqrt{\frac{2}{3}}$$ or
$$\frac{-9abc}{a+b+c}+7(ab+ac+bc)\leq8$$ or
$$\frac{-9abc}{a+b+c}+7(ab+ac+bc)\leq2(a+b+c)^2$$ or
$$\sum_{cyc}(2a^3-a^2b-a^2c)\geq0$$ or
$$\sum_{cyc}(a-b)^2(a+b)\geq0.$$
Done!
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $\int_0^{\pi/2}\ln^2(\cos x)\,dx=\frac{\pi}{2}\ln^2 2+\frac{\pi^3}{24}$
Prove that
\begin{equation}
\int_0^{\pi/2}\ln^2(\cos x)\,dx=\frac{\pi}{2}\ln^2 2+\frac{\pi^3}{24}
\end{equation}
I tried to use by parts method and ended with
\begin{equation}
\int \ln^2(\cos x)\,dx=x\ln^2(\cos x)+2\int x\ln(\cos x)\tan x\,dx
\end{equation}
The latter integral seems hard to evaluate. Could anyone here please help me to prove it preferably with elementary ways (high school methods)? Any help would be greatly appreciated. Thank you.
Addendum:
I also found this nice closed-form
\begin{equation}
-\int_0^{\pi/2}\ln^3(\cos x)\,dx=\frac{\pi}{2}\ln^3 2+\frac{\pi^3}{8}\ln 2 +\frac{3\pi}{4}\zeta(3)
\end{equation}
I hope someone here also help me to prove it. (>‿◠)✌
| In my post, I had found a reduction formula for the integral
$$
J_n=\int_{0}^{\frac{\pi}{2}} \ln ^{n}(\cos x) dx
$$
that
$$\begin{aligned}
J_n&= -\ln 2 J_{n-1} + (n-1) !\sum_{k=0}^{n-2} \frac{(-1)^{n-k}}{k !} \left(1-\frac{1}{2^{n-k-1}} \right)\zeta(n-k) J_k.
\end{aligned}
$$
Using it yields
$$
\begin{aligned}
\boxed{\int_{0}^{\frac{\pi}{2}} \ln ^{2}(\cos x)d x=-\ln 2 \cdot J_{1}+\frac{1}{2} \zeta(2) J_{0} =\frac{\pi \ln ^{2} 2}{2}+\frac{\pi^{3}}{24}}
\end{aligned}
$$
and
$$\begin{aligned}
&\int_{0}^{\frac{\pi}{2}} \ln ^{3}(\cos x )d x\\=&-\ln 2\left(\frac{\pi}{2} \ln^{2} 2+\frac{\pi^{3}}{24}\right)+2!\left[-\left(1-\frac{1}{2^2}\right)\zeta(3)\cdot \frac{\pi}{2}+\left(1-\frac{1}{2}\right)\zeta(2)\left(-\frac{\pi}{2}\ln 2\right)\right]\end{aligned}
$$
Simplifying yields
\begin{equation}
\boxed{-\int_0^{\frac{\pi}{2}}\ln^3(\cos x)\,dx=\frac{\pi}{2}\ln^3 2+\frac{\pi^3}{8}\ln 2 +\frac{3\pi}{4}\zeta(3)}
\end{equation}
| {
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"timestamp": "2023-03-29T00:00:00",
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How can I accurately compute $\sqrt{x + 2} −\sqrt{x}$ when $x$ is large? How can the values of the function $f(x) = \sqrt{x + 2} −\sqrt{x}$ be computed accurately when $x$ is large?
I have tried using Matllab. I am not able to understand when $x$ will be large.
| In fact, an asymptotic expansion of $f(x) = \sqrt{x+a} - \sqrt{x}$ about $x = \infty$ is $$f(x) \approx \frac{a}{2} x^{-1/2} - \frac{a^2}{8} x^{-3/2} + \frac{a^3}{16} x^{-5/2} - \frac{5a^4}{128} x^{-7/2} + \frac{7a^5}{256} x^{-9/2} - \frac{21a^6}{1024} x^{-11/2} + \cdots.$$ We can calculate this via the generalized binomial theorem $$(a+x)^r = \sum_{k=0}^\infty \binom{r}{k} a^k x^{r-k},$$ for $r = 1/2$, so that in particular, $$\begin{align*} \binom{1/2}{k} &= \frac{(1/2)(1/2-1)\cdot \ldots \cdot(1/2-k+1)}{k!} \\ &= \frac{1(-1)(-3)\cdot \ldots \cdot(3-2k)}{2^k k!} \\ &= (-1)^{k-1} \frac{(2k-2)!}{2^{2k-1} k! (k-1)!} \\ &= \frac{(-1)^{k-1}}{2^{2k-1}k } \binom{2k-2}{k-1}, \quad k = 1, 2, \ldots,\end{align*}$$ and $\binom{1/2}{0} = 1$.
| {
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"timestamp": "2023-03-29T00:00:00",
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A $(3k+3)$-node graph with degrees $k+1,\dots,k+3$ has $k+3$ degree-$(k+1)$ nodes or $k+1$ degree-$(k+2)$ nodes or $k+2$ degree-$(k+3)$ nodes?
Graph $G$ has order $n=3k+3$ for some positive integer $k$ . Every vertex of $G$ has degree $k+1$, $k+2$, or $k+3$. Prove that $G$ has at least $k+3$ vertices of degree $k+1$ or at least $k+1$ vertices of degree $k+2$ or at least $k+2$ vertices of degree $k+3$
I used the proof by cases.
In case 1: $k$ is odd I can show that $G$ has at least $k+1$ vertices of degree $k+2$. Otherwise it will contradict the theorem that say every graph have even number of odd vertices.
Same reason, in case 2: $k$ is odd I can show that at least $k+2$ vertices of degree $k+3$
I just can't figure out how to show $G$ has at least $k+3$ vertices of degree $k+1$
| Let $a,b,c$ denote the number of vertices of degree $k+1,k+2,k+3$ respectively.
Then we have $a+b+c=3k+3$
If $a\geq k+3$ or $c\geq k+2$ then we are done.
Now if $a< k+3$ and $c< k+2$ then we have to show that $b\geq k+1$.
Thus we have $a\leq k+2$ and $c\leq k+1$. Now we consider two cases.
Case $1$: k is odd.
Putting the values of $a$ and $c$ in $a+b+c=3k+3$ we have $b\geq k$. Now if $b=k$, then the number of vertices containing odd($k+2$) degree is odd. So, $b\neq k$. Thus $b\geq k+1$.
Case $2$: k is even.
Now $a+c\leq 2k+3$. Now if $a+c=2k+3$, then the number of vertices containing odd($k+1$ and $k+3$) degree is odd($2k+3$). So, $a+c\neq 2k+3$. Therefore $a+c\leq 2k+2$. Putting this value in the equation $a+b+c=3k+3$ we have $b\geq k+1$. Hence proved.
| {
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Solve $\sqrt x = x/2$. If $f(x) = \sqrt x$ and $g(x) = x/2$, find the area of this limited area between $f(x)$ and $g(x)$.
I am having trouble solving this equation $\sqrt x = x/2$ that should give me the x values.
I know that the next step after this one is to solve $\int_a^b f(x)-g(x)~dx$ where $a= x_1$ and $b=x_2$
How do I go by to solve this $\sqrt x = x/2$ equation?
Thank you!
| $$\sqrt{x}=\frac{x}{2}\\x\geq 0\\(\sqrt{x}=\frac{x}{2})^2\\x=\frac{x^2}{4}\\4x=x^2\\x(4-x)=0\\x=0\\x=4\\ \int_{0}^{4}(\sqrt{x}-\frac{x}{2})dx=\frac{x^{\frac{3}{2}}}{\frac{3}{2}}-\frac{x^2}{4}=\\(\frac{4^{\frac{3}{2}}}{\frac{3}{2}}-\frac{4^2}{4})-(\frac{0^{\frac{3}{2}}}{\frac{3}{2}}-\frac{0^2}{4})=\frac{4}{3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/921312",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Integrate $\int\frac{\cos x dx}{\cos^{3/2}2x}=\frac{\sin x}{\sqrt{\cos 2x}}+C$ Wolfram gives this nice result:
$$\int\frac{\cos x dx}{\cos^{3/2}2x}=\frac{\sin x}{\sqrt{\cos 2x}}+\text{constant}$$
I have tried writing $\cos 2x = \cos^2x - \sin^2x $ and doing Weierstrass substitution $\tan (x/2) = t$ but its getting very complicated. Any help/hints ?
| $\cos2x=1-2\sin^2 x$, let $t=\sin x$ so $dt=\cos xdx$ in numerator.
SO:
$$I=\int\frac{dt}{(1-2t^2)^{3/2}}$$
Then use the substitution $u^2=(1-2t^2)$ so $udu=-2tdt$, so $\displaystyle dt=-\frac{udu}{2t}=\frac{-udu}{\sqrt2\sqrt{1-u^2}}$:
$$I=-\int\frac{udu}{\sqrt2\sqrt{1-u^2}u^3}=\frac{-1}{\sqrt2}\int\frac{du}{u^2\sqrt{1-u^2}}$$
Then use $v=1/u$ so $\displaystyle du=-\frac1{v^2}dv$ So:
$$I=\frac1{\sqrt2}\int\frac{vdv}{\sqrt{v^2-1}}=\frac1{\sqrt2}\sqrt{v^2-1}+C$$
Now:
$$\sqrt{v^2-1}=\frac{\sqrt{1-u^2}}u=\frac{\sqrt2t}{\sqrt{1-2t^2}}=\frac{\sqrt2\sin\theta}{\sqrt{1-2\sin^2\theta}}=\frac{\sqrt2\sin\theta}{\sqrt{\cos2\theta}}$$
So:
$$\large I=\frac{\sin\theta}{\sqrt{\cos2\theta}}+C$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solving $ \frac{dy}{dx} = y^2 - 9$ $ \frac{dy}{dx} = y^2 - 9$
This is separable so I rewrite it as $ \frac{1}{(y^2 - 9)}dy = dx$ then I get
$$\int \frac{1}{(y^2 - 9)}dy = \int dx = \int 1dx = x + c,$$ for some $c \in \mathbb{R} $
The left hand side:
$$\int \frac{1}{(y^2 - 9)}dy = \int \frac{1}{(y+3)(y-3)}dy = \int \left( \frac{A}{y+3} + \frac{B}{ y-3}\right) \,dy $$
We have $A(y-3) + B(y+3) = 1 \implies A =-\frac{1}{6}, B = \frac{1}{6}$
So we get
$$ \int \left(\frac{A}{y+3} + \frac{B}{ y-3}\right) \, dy = \int \left (\frac{-\frac{1}{6}}{(y+3)} + \frac{\frac{1}{6}}{ (y-3)}\right)dy = \frac{1}{6}\int \left (\frac{1}{ (y-3)} - \frac{1}{ (y+3)}\right) dy = \frac{1}{6} \left(\int \frac{1}{ (y-3)}dy - \int \frac{1}{ (y+3)} dy\right ) = \frac{1}{6}\left ( \ln \left \vert y-3 \right\vert - \ln \left\vert y+3 \right\vert \right ) = \frac{1}{6}\ln \left \vert\frac{y-3}{y+3} \right \vert $$
So I have $\ln \left\vert\frac{y-3}{y+3}\right\vert = 6x + 6c \implies e^{\ln\left\vert\frac{y-3}{y+3}\right\vert} = \left\vert\frac{y-3}{y+3}\right\vert =e^{6x + 6c} $
Is this correct to far? If yes how do I solve this for $y$?
| $\frac{y-3}{y+3}=\pm e^{6x}e^{6c}=Ce^{6x}$ so $y-3=(y+3)Ce^{6x}$ Group the $y$ terms together to get $y(1-Ce^{6x})=3+3Ce^{6x}$. Now divide by $1-Ce^{6x}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Finding asymptotes for $f(x)=\frac{x^2+3x-10}{3x^2+13x-10}$ $$f(x)=\frac{x^2+3x-10}{3x^2+13x-10}$$
I know that the horizontal asymptote is $1/3$. To find the vertical asymptotes, I set the denominator equal to zero and used the quadratic formula, and I got $-5$ and $2/3$, and this is wrong.
How do you find the vertical asymptotes for this problem?
thanks.
| Note that
$$f(x)=\frac{(x+5)(x-2)}{(x+5)(3x-2)}.$$
Therefore
$$\lim_{x\to-5} f(x) = \lim_{x\to-5}\frac{(x+5)(x-2)}{(x+5)(3x-2)} =\lim_{x\to-5}\frac{x-2}{3x-2} = \frac{7}{17}.$$
Since this limit is actually not infinite, there is no vertical asymptote when $x=-5$.
In general, vertical asymptotes of a rational function occur at $x=a$ provided that the largest power of $(x-a)$ dividing the denominator is greater than that of the numerator. Here, $(x+5)$ divided the numerator and the denominator exactly once, so there was no asymptote when $x=-5$.
| {
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The minimum of $x^2+y^2$ under the constraints $x+y=a$ and $xy=a+3$ I solved the following problem: If $x,y,a \in \mathbb{R}$ such that $x+y=a$ and $xy=a+3$, find the minimum of $x^2+y^2$
Here is my solution. $x^2+y^2=(x+y)^2 -2xy= a^2-2a-6$. The minimum value is obtained when $a=1$ and it is $-7$. Where did I go wrong?
The answer says that the minimum is $2$ when $x=y=-1$
| Since $x,y$ are the real roots $X$ of $$(X-x)(X-y)=0\iff X^2-(x+y)X+xy=0\iff X^2-aX+(a+3)=0,$$ you need to have $$D=(-a)^2-4(a+3)\ge 0\iff (a-6)(a+2)\ge 0\iff a\le -2\ \text{or}\ a\ge 6.$$
So, considering $x^2+y^2=a^2-2a-6=(a-1)^2-7$ for $a\le -2\ \text{or}\ a\ge 6$ gives you that the minimum is $2$ at $a=-2$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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The limit of $\int_{0}^{1}\frac{\sqrt{n}}{1+n\ln(1+x^2)}dx$ as $n\to\infty$ The task is to calculate $$\lim_{n\to\infty}\int_{0}^{1}\frac{\sqrt{n}}{1+n\ln(1+x^2)}dx$$ I tried various estimates I know to find the dominating integrable function and nothing worked. Does anyone have any idea? Is this even an application of DCT or something else?
| DCT as Discrete Cosine Transform? I cannot understand what you are meaning.
Anyway, your integral converges towards $\frac{\pi}{2}$, since:
$$\int_{0}^{1}\frac{\sqrt{n}}{1+n\log(1+x^2)}\,dx \geq \int_{0}^{1}\frac{\sqrt{n}}{1+nx^2}\,dx = \arctan(\sqrt{n})=\frac{\pi}{2}+O\left(\frac{1}{\sqrt{n}}\right),$$
while the difference between the first and the second integral is bounded by:
$$\begin{eqnarray*}&&n^{3/2}(1-\log 2)\int_{0}^{1}\frac{x^4\,dx}{(1+nx^2)(1+n\log(1+x^2))}\\&\leq& n^{3/2}(1-\log 2)\int_{0}^{1}\frac{x^4 \,dx}{(1+n\log(1+x^2))^2}\\&\leq&n^{3/2}(1-\log 2)\int_{0}^{1}\frac{x^4 \,dx}{(1+nx^2-\frac{n}{2}x^4)^2}\end{eqnarray*}$$
where:
$$\frac{x^4}{(1+nx^2-\frac{n}{2}x^4)^2}\leq \frac{1}{n^2}\text{ for }x\in[0,1/\sqrt{n}],$$
$$\frac{x^4}{(1+nx^2-\frac{n}{2}x^4)^2}\leq \frac{1}{n^2(1-\frac{1}{2}x^2)^2}\leq\frac{4}{n^2}\text{ for }x\in[1/\sqrt{n},1],$$
hence:
$$\int_{0}^{1}\frac{\sqrt{n}}{1+n\log(1+x^2)}\,dx =\frac{\pi}{2}+O\left(\frac{1}{\sqrt{n}}\right).$$
| {
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Show that $\frac{(b+c)^2} {3bc}+\frac{(c+a)^2}{3ac}+\frac{(a+b)^2}{3ab}=1$ If $a^3+b^3+c^3=3abc$ and $a+b+c=0$ show that $\frac{(b+c)^2} {3bc}+\frac{(c+a)^2}{3ac}+\frac{(a+b)^2}{3ab}=1$
| since
$$LHS=\dfrac{a^2}{3bc}+\dfrac{b^2}{3ca}+\dfrac{c^2}{3ab}=\dfrac{a^3+b^3+c^3}{3abc}=\dfrac{3abc}{3abc}=1$$
I think this condition $a^3+b^3+c^3=3abc$ is redundant.because
$$a+b+c=0\Longrightarrow a^3+b^3+c^3=3abc$$
because
$$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ac)$$
| {
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How to show that $a+b> \sqrt{a^2+b^2-ab}, \qquad a, b >0$ How do you show that
$$a+b> \sqrt{a^2+b^2-ab}, \qquad a, b >0$$
I could write $\sqrt{a^2+b^2-ab}=\sqrt{(a+b)^2-3ab}$, but this seems to lead nowhere.
| $3ab>0$
Hence, $2ab>-ab$
Hence, $a^2+2ab+b^2>a^2+b^2-ab$
Hence $a+b>\sqrt{a^2+b^2-ab}$ because $a^2+b^2-ab>a^2+b^2-2ab=(a-b)^2>0$
| {
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How to solve the integral $\int \frac {(x^2 +1)}{x^4- x^2 +1} dx$ I have started this problem but I'm not completely sure I'm going down the right path with it.
So far I have completed the square in the denominator.
$x^4-x^2+1= (x^2-1/2)^2+\frac{3}{4}$
Then, let $u=x^2-\frac{1}{2}$ so $x=\sqrt(u+\frac{1}{2})$
$\int\frac{x^2+1}{x^4-x^2+1}dx = \int\frac{u+\frac{1}{2}+1}{u^2+\frac{3}{4}}du
=\int\frac{u}{u^2+\frac{3}{4}} +\frac{\frac{3}{2}}{u^2+\frac{3}{4}}du$
| As Hassan Muhammad commented, you have a mistake. Since you set $x=\sqrt{u+\frac{1}{2}}$, you then have $dx=\frac{1}{2 \sqrt{u+\frac{1}{2}}}$ and then $$I=\int \frac {(x^2 +1)}{x^4- x^2 +1} dx=\int \frac{2 u+3}{\sqrt{u+\frac{1}{2}} \left(4 u^2+3\right)}du$$ which is not very nice.
If you use Hippalectryon's idea, you end with $$I=\int \frac {(x^2 +1)}{x^4- x^2 +1} dx=\int\left(\frac{3 x^{2}}{x^{6} + 1} + \frac{1}{x^{2} + 1}\right)dx=\tan ^{-1}\left(x^3\right)+\tan ^{-1}\left(x\right)$$ and if you apply the formula $$\tan ^{-1}\left(a\right)+\tan ^{-1}\left(b\right)=\tan ^{-1}\left(\frac{a+b}{1-a b}\right)$$ you have, after simplifications, $$I=\tan ^{-1}\left(\frac{x}{1-x^2}\right)$$ which is result to which lab bhattacharjee's answer is taking you to.
| {
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"url": "https://math.stackexchange.com/questions/928040",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Closed form of $\int_0^{\frac{\pi}{3}} \log^2(\sin x) \mathrm{d}x$ Inspired by the popularity of these kind of integrals appearing on MSE lately, I actually learned new methods to attack weird integrals by studying the beautiful answers on the similar past questions, so I conjecture $$\int_0^{\frac{\pi}{3}} \log^2(\sin x) \mathrm{d}x$$ has a closed form. How would someone approach this?
PS; To be honest, I'm more interested in the methodology, as on the previous questions I lacked the mathematical background to understand them completely anyway. Which implies I don't want Cleo-like answers.
| $\def\B{{\text{B}}}\def\F{{\text{$_2$F$_1$}}}$I find the closed form is really nasty since it's involving the hypergeometric functions. Here is my approach. Let $I$ be the given integral and let $\sin x=\sqrt{t}$, then
\begin{align}
I&=\frac{1}{8}\int_0^{3/4}\frac{\ln^2t}{\sqrt{t}\cdot\sqrt{1-t}}\,dt\\
&=\frac{1}{8}\lim_{u\to\frac{1}{2}}\partial_u^2\,\B\left(\frac{3}{4};u,\frac{1}{2}\right)
\end{align}
where $\text{B}\left(z;u,v\right)$ is the incomplete beta function defined by
\begin{align}
\B\left(z;u,v\right)=\int_0^z t^{u-1}(1-t)^{v-1}\,dt
\end{align}
According to Wolfram MathWorld, incomplete beta function can represented as hypergeometric function
\begin{align}
\B\left(z;u,v\right)=\frac{z^u}{u}\F\left(u,1-v;1+u;z\right)
\end{align}
Therefore, with help of Wolfram Alpha, we arrive to the following closed-form
\begin{align}
I
=&\,\frac{1}{8}\lim_{u\to\frac{1}{2}}\partial_u^2\,\left(\frac{\left(\frac{3}{4}\right)^u}{u}\F\left(u,\frac{1}{2};1+u;\frac{3}{4}\right)\right)\\
=&\,\frac{\sqrt{3}}{16}\left[\F^{(0,1,0,0)}\left(\frac{1}{2},\frac{1}{2},\frac{3}{2},\frac{3}{4}\right)-8\,\F^{(0,0,1,0)}\left(\frac{1}{2},\frac{1}{2},\frac{3}{2},\frac{3}{4}\right)\right]+\\
&\,2\left[\F^{(0,0,2,0)}\left(\frac{1}{2},\frac{1}{2},\frac{3}{2},\frac{3}{4}\right)+2\,\F^{(0,1,1,0)}\left(\frac{1}{2},\frac{1}{2},\frac{3}{2},\frac{3}{4}\right)+\F^{(0,2,0,0)}\left(\frac{1}{2},\frac{1}{2},\frac{3}{2},\frac{3}{4}\right)\right]+\\
&\,-\frac{\sqrt{3}}{8}\ln\left(\frac{4}{3}\right)\left[2\,\F^{(0,0,1,0)}\left(\frac{1}{2},\frac{1}{2},\frac{3}{2},\frac{3}{4}\right)+\F^{(0,1,0,0)}\left(\frac{1}{2},\frac{1}{2},\frac{3}{2},\frac{3}{4}\right)-\frac{8\pi}{3\sqrt{3}}\right]\\
&\,+\frac{\pi}{12}\ln^2\left(\frac{4}{3}\right)+\frac{64\pi}{3\sqrt{3}}
\end{align}
It seems the above closed-form could be simplified into Mr. Mhenny Benghorbal's or Gahawar's answer but unfortunately I am unable to prove it. I wish I could, sorry... (╥﹏╥)
| {
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"url": "https://math.stackexchange.com/questions/928117",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Solving $y^3=x^3+8x^2-6x+8$ Solve for the equation $y^3=x^3+8x^2-6x+8$ for positive integers x and y.
My attempt- $$y^3=x^3+8x^2-6x+8$$
$$\implies y^3-x^3=8x^2-6x+8$$
$$\implies (y-x)(y^2+x^2+xy)=8x^2-6x+8$$
Now if we are able to factorise $8x^2-6x+8$ then we can compare LHS with RHS.Am I on the right track?Please help.
| Since we have$$y^3-x^3=2(4x^2-3x+4)$$
there exists an integer $k$ such that
$$y-x=2k\iff y=x+2k.$$
So, we have
$$(x+2k)^3-x^3=2(4x^2-3x+4)\iff (3 k-4) x^2+(6 k^2+3) x+4k^3-4=0\tag1$$
Now we have
$$D=(6k^2+3)^2-4(3k-4)(4k^3-4)\ge 0\iff -12 k^4+64 k^3+36 k^2+48 k-55\ge 0$$$$\iff 12 k^4-64 k^3-36 k^2-48 k+55\le 0$$
Here, let $f(k)=12 k^4-64 k^3-36 k^2-48 k+55=4k[k\{k(3k-16)-9\}-12]$.
Now, for $k\ge 6$, we have
$$3k-16\ge 2\Rightarrow k(3k-16)\ge 2k\ge 12$$$$\Rightarrow k\{k(3k-16)-9\}-12\ge 3k-12\gt 0\Rightarrow f(k)\gt 0.$$
Also, for $k\le -1$, we have
$$f(k)=12 k^4-36k^2(k+1) -48 k+55-28k^3\gt 0.$$
Hence, we have $k=0,1,2,3,4,5.$
Then, from $(1)$, you can find integer roots $x$ for each $k$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/928674",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 0
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If $x^2 + y^2 + Ax + By + C = 0 $. Find the condition on $A, B$ and $C$ such that this represents the equation of a circle.
If $x^2 + y^2 + Ax + By + C = 0 $. Find the condition on $A, B$ and $C$ such that this represents the equation of a circle.
Also find the center and radius of the circle.
Here's my solution, I'm not sure if it's correct or not (specifically the conditions on $A$, $B$ and $C$. I feel that my conditioning is invalid and that this may be some sort of triangle inequality case.
I grouped the $x^2$ and $Ax$ terms, and the $y^2$ and By terms together and equated them to -$C$.
After completing the square I am left with the following expression:
$(x + A/2)^2 + (y + B/2)^2 = -C + A^2/4 - B^2/4$
Conditions on A,B and C:
****CORRECTION MADE (Redundant inequalities need not be included)****
$C < (A^2 + B^2)/4$
Radius of the circle is $\sqrt{-C + A^2/4 + B^2/4}$
The center of the circle is $(-A/2, -B/2)$.
Correct?
| Lets' solve for:
X^2 + y^2 + ax + by + c = 0
Step 1: Add -x^2 to both sides
ax + by + x^2 + y^2 + c + - x^2 = 0 + - x^2
ax + by + y^2 + c = - x^2
Step 2: Add - y^2 to both sides
ax + by + y^2 + c + - y^2 = - x^2 + y^2
ax + by + c = - x^2 - y^2
Step 3: Add - by to both sides
ax + by + c + by = - x^2 - y^2 + - by
ax + c = - by - x^2 - y^2
Step 4: Add - c to both sides
ax + c + - c = by - x^2 - y^2 + - c
ax = - by - x^2 - y^2 - c
Step 5: Divide both sides by x.
ax/x = - by - x^2 - y^2 - c/ x
a = - by - x^2 - y^2 - c/x
Answer is:
a = - by - x^2 - y^2 - c/x
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/931215",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Meaning of congruence notation for Bernoulli Numbers I am studying Theorem 4(von Staudt's Theorem) in Borevich-Shafarevich's Number Theory(1966)(page 384) which states:
Let $p$ be a prime and $m$ an even integer. If $(p-1)\nmid m$, then
$B_m$ is $p$-integral(that is, $p$ does not appear in the denominator
of $B_m$). If $(p-1) | m$, then $pB_m$ is $p$-integral, and $$pB_m
\equiv -1 (mod\hspace{0.05in}p).$$
What's the meaning of this congruence?
For instance for $p=3$ and $m=4,$ $3B_4=3(\frac{-1}{30})=\frac{-1}{10}.$ How is $$\frac{-1}{10} \equiv -1 (mod\hspace{0.05in}3) ?$$
Thank you!
| The fraction $\frac{1}{n}$ is, by definition, the number which you can multiply $n$ by to get the answer $1$. Now the same definition works in congruence arithmetic. For example modulo $p$, as long as $n$ is not divisible by $p$ there is some number $m$ such that $nm \equiv 1$ mod $p$. In that case $\frac{1}{n}\equiv m$ (mod $p$).
There's a sign error in the statement of the theorem in that book. The formula for when $m|p-1$ (as you've now corrected) should be $pBm \equiv -1$ modulo $p$.
So for $m=4$ and $p=3$ we get $pB_4 = \frac{-1}{10} \equiv -1$ modulo 3, which makes sense since $10\equiv 1$ (mod $3$).
For another example let's look at $m=16$ and $p=5$, and $B_{16} = \frac{-3617}{510}$. We have
$$pB_m = 5 \frac{-3617}{510} = \frac{-3617}{102},$$
Now $102 \equiv 2$ (mod $5$), and $3*2 = 6 \equiv 1$, so $\frac{1}{102} \equiv 3$. We also have $-3617 \equiv -7 \equiv 3$ (mod $5$) and so indeed
$$pB_m = \frac{-3617}{102} \equiv \frac{3}{102} \equiv 3 * 3 = 9 \equiv -1 \text{(mod }5)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/931707",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Maximization of $x+y$ when each is greater than $1$ and $xy = 16$. The product of two numbers $x$ and $y$ is $16$. We know $x\ge 1$ and $y\ge 1$. What is the greatest possible sum of the two numbers?
| Hint:
Since $x \cdot y = 16$ and $x,y \geq 1 > 0$ we have $y = \frac{16}{x}$. Now can you find the maximum of the function
$$f(x) = x + \frac{16}{x}$$
for $x \geq 1$ and $\frac{16}{x} \geq 1$?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/932338",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Evaluation of $\int\frac{1}{\sin^2 x\cdot \left(5+4\cos x\right)}dx$ Evaluation of $\displaystyle \int\frac{1}{\sin^2 x\cdot \left(5+4\cos x\right)}dx$
$\bf{My\; Solution::}$ Given $\displaystyle \int\frac{1}{\sin^2 x\cdot (5+4\cos x)}dx = \int \frac{1}{(1-\cos x)\cdot (1+\cos x)\cdot (5+4\cos x)}dx$
Now Using Partial fraction for $\displaystyle \frac{1}{(1-\cos x)\cdot (1+\cos x)\cdot (5+4\cos x)}$
Now Let $\cos x= y\;,$ and Let $\displaystyle \frac{1}{(1-y)(1+y)(5+4y)} = \frac{A}{1+y}+\frac{B}{1-y}+\frac{C}{5+4y}$
after solving We Get $\displaystyle A = \frac{1}{2}$ and $\displaystyle B = -\frac{1}{18}$ and $\displaystyle C = -\frac{16}{9}$
So $\displaystyle \int \frac{1}{(1-\cos x)\cdot (1+\cos x)\cdot (5+4\cos x)}dx = \frac{1}{2}\int \frac{1}{1+\cos x}dx - \frac{1}{18}\int\frac{1}{1-\cos x}dx - \frac{16}{9}\int \frac{1}{5+4\cos x}dx$
And after that we can solve easily Like for
$\displaystyle \int\frac{1}{1+\cos x}dx = \int\frac{1-\cos x}{\sin^2 x}dx = \int \left(\csc^2 x-\csc x\cdot \cot x\right)dx = -\cot x +\csc x+\mathcal{C}$
My Question is , Is there is any other method by which we can solbe the above question.
OR without using partial fraction,
Thanks
| Doing the same as Aditya (Weierstrass substitution), you arrive to $$\int\frac{1}{\sin^2 x\cdot \left(5+4\cos x\right)}dx=\int \frac{\left(t^2+1\right)^2}{2 t^2 \left(t^2+9\right)}dt$$ Now, using partial fraction decomposition $$\frac{\left(t^2+1\right)^2}{2 t^2 \left(t^2+9\right)}=\frac{1}{18 t^2}-\frac{32}{9 \left(t^2+9\right)}+\frac{1}{2}$$ and integration is not so difficult, leading to $$\int \frac{\left(t^2+1\right)^2}{2 t^2 \left(t^2+9\right)}dt=\frac{1}{54} \left(27 t-\frac{3}{t}-64 \tan ^{-1}\left(\frac{t}{3}\right)\right)$$ or, going back to $x$ $$\int\frac{1}{\sin^2 x\cdot \left(5+4\cos x\right)}dx=\frac{1}{2} \tan \left(\frac{x}{2}\right)-\frac{1}{18} \cot
\left(\frac{x}{2}\right)+\frac{32}{27} \tan ^{-1}\left(3 \cot
\left(\frac{x}{2}\right)\right)$$
Just as Aditya said : OOps PF!
But, again, this is almost what you did well.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/932596",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Find the value of $f(x)$ for $x = 2 + 2^{2/3} + 2^{1/3}$ If $x = 2 + 2^{2/3} + 2^{1/3}$, then find the value of $f(x)=x^3 - 6x^2 + 6x$.
I am unable to get to the answer - end up with more than one term. Please help me solve this!
| We have $$x-2=2^{\frac23}+2^{\frac13}$$
Cubing we get, $$x^3-3x^2(2)+3x(2^2)+2^3=(2^{\frac23})^3+(2^{\frac13})^3+3\cdot2^{\frac23}\cdot2^{\frac13}(2^{\frac23}+2^{\frac13})$$
$$\iff x^3-6x^2+12x+8=2+2^2+6(x-2)$$
Can you take it home from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/933434",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Find the minimum value of $x+y+z.$ Let $x,y,z$ are nonegative such that $(x - y)(y - z)(z - x) \geq 1.$
Find the minimum value of $x+y+z.$
| Clearly, two among $x-y, y-z, z-x$ should be negative. WLOG let $a=x-y, b=z-y$ be positive numbers. Then $a(-b)(b-a) \ge 1 \iff a^2b \ge ab^2+1$ and we need the minimum of $a+b+3y$. Obviously it is a good idea to set $y =0$, then we are left with minimising $a+b$.
$$a^2b-b^2a-1 \ge 0 \iff a \ge \frac{b^2+\sqrt{b^4+4b}}{2b}$$
So we minimise $a+b =\frac32b+ \dfrac{\sqrt{b^4+4b}}{2b}$, which by calculus and some algebra gives $b = \sqrt[3]{\frac{3\sqrt3-5}2}$.
Therefore we have the minimum (after some more algebra and help from @Semiclassical) as $\sqrt[3]2 \sqrt3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/933830",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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