Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Radius of a largest circle inscribed under $y=\frac{1}{(1+x^2)^n}$, closed form The curve $y=\frac{1}{1+x^2}$ has an obvious connection to circles, because it's the derivative of the arctangent function.
Besides, if we inscribe a circle under it, its radius is exactly $R=\frac{1}{2}$, so it takes exactly one fourth of the full area under the curve.
Actually, I don't know an easy way to find $R$ even in this simple case. So, let's consider how I solve the more general problem.
Find the largest circle fitting between the curve $y=y (x)$ and the line $y=0$
$$y(x)=\frac{1}{(1+x^2)^n}~~~~~~ n=1,2,3,\dots$$
The distance from the circle origin $(0,R)$ to the curve is the minimum of a function:
$$s(x)=\sqrt{x^2+\left(R-\frac{1}{(1+x^2)^n} \right)^2}$$
$$s(x)'=0$$
The condition for the inscribed circle is $s(x)=R$.
Let's denote:
$$t=1+x^2$$
Then from the above we obtain the system of equations:
$$t^{2n+1}+2n~R~t^n-2n=0$$
$$t^{2n+1}-t^{2n}-2~R~t^n+1=0$$
This is how I got the solution for $n=1$ (using Mathematica). Other solutions do not have obvious closed forms. Here are the numerical roots I've got with Mathematica:
$$R_2=0.4735710971151933$$
$$R_3=0.4401444298014721$$
$$R_4=0.41216506385826285$$
And so on.
The questions I ask:
Can there be closed forms for $R$ for $n \neq 1$? How to find them?
Is there an easier way to find $R$? At least for some $n$?
Edit
The most simple equation for $t$, as far as I can see, is:
$$(n+1)~t^{2n+1}-n~t^{2n}-n=0$$
I think I might ask a separate question about this equation. It's easy to solve by Newton-Raphson, but can it have closed form solutions for any $n$?
| The radius of curvature of a function $f(x)$ is
$$ \rho =- \frac{ \left( 1 + \left(\frac{{\rm d}}{{\rm d}x}f(x)\right)^2 \right)^\frac{3}{2} } { \frac{{\rm d}^2}{{\rm d}x^2} f(x) } $$
NOTE: If the curve is given implicitly then $$\rho = \frac{ (x'^2+y'^2)^\frac{3}{2}}{y' x'' - y'' x'}$$
Case 1
Consider $f(x) = \frac{1}{1+x^2}$
$$ \rho = -\frac{ \left(1+\frac{4 x^2}{(1+x^2)^4} \right)^\frac{3}{2}} { \frac{2(3 x^2-1)}{(1+x^2)^3} } $$
and for $x=0$ the circle is $\rho =\frac{1}{2}$ as you expect.
Case 2
Consider $f(x) = \frac{1}{(1+x^2)^n}$
The radius of curvature is
$$ \rho = \frac{ \left( 4 n^2 x^2 + (1+x^2)^{2(n+1)}\right)^\frac{3}{2} }{ \tfrac{2 n (1-x^2 (2n+1))}{(1+x^2)^{-(2n+1)}}} $$
and for $x=0$ the curvature is simply ${\rho = \frac{1}{2n}}$
Now the circle is going to be tangent to the curve (one of the curvature circles) and its center is going to be located at
$$ x_c = x - \frac{ y' (1+y'^2)}{y''} \\ y_c = y + \frac{1+y'^2}{y''} $$
here $y' = \frac{{\rm d}}{{\rm d}x} f(x)$ and $y'' = \frac{{\rm d}^2}{{\rm d}x^2} f(x)$
For the circle to be tangent to $y=0$, its center has to be at $y_c = \rho$ or
$$ y + \frac{1+y'^2}{y''} = - \frac{(1+y'^2)^\frac{3}{2}}{y''} $$
For $y=\tfrac{1}{1+x^2}$ my CAS says the solution (besides $x=0$) is the roots of
$$ 2 x^{16} + 15 x^{14} + 48 x^{12} + 97 x^{10} + 126 x^8+81 x^6-20 x^4 -33 x^2 = 12$$
$ x = 0.759119999241623 $ or $\rho = 4.32460278011942$, but that ends up being a circle over the curve.
For the general case of $y = \frac{1}{(1+x^2)^n}$ the equation to be solved is
$$ \left( (1+x^2)^{2(n+1)} + 4 n^2 x^2\right)^\frac{3}{2} + (1+x^2)^{3(n+1)} + 2 n ( 1+x^2)^{n+1} \left( 4 n x^2 + x^2 -1 \right) = 0 $$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "5",
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Prove $\cos(2\theta) + \cos\left(2 \left(\frac{\pi}{3} + \theta\right)\right) +\cos\left(2 \left(\frac{2\pi}{3} + \theta\right)\right) = 0$ It's been a while since I've done trig proofs. I know that
$$\cos(2\theta) + \cos\left(2 \left(\frac{\pi}{3} + \theta\right)\right) + \cos\left(2 \left(\frac{2\pi}{3} + \theta\right)\right) = 0$$
is true. This can be easily seen by plugging in values, for example where $\theta=0$ we get
$$1 -\frac{1}{2} -\frac{1}{2} = 0$$
and the same can be seen for any $\theta$. Is there a straight forward way to prove this statement is true.
| $$\cos(2\theta) + \cos\left(2 \left(\frac{\pi}{3} + \theta\right)\right) +\cos\left(2 \left(\frac{2\pi}{3} + \theta\right)\right)=\cos{2\theta}+\cos({\frac{2\pi}{3}+2\theta})+\cos({\frac{4\pi}{3}}+2\theta)$$
Expanding this we get $\cos{2\theta}+\cos({\frac{2\pi}{3}+2\theta})+\cos({\frac{4\pi}{3}+2\theta})=\cos{2\theta}+-\frac{1}{2}\cos{2\theta}-\frac{\sqrt{3}}{2}\sin{2\theta}-\frac{1}{2}\cos{2\theta}+\frac{\sqrt{3}}{2}\sin{2\theta}$
which equals $0$
| {
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Probability function on $\mathbb N$ - no convergence to $1$?
Consider a box containing one red ball and one black ball. If we draw
a black ball, we put it back and add another black ball. If we draw
the red ball, the experiment is over.
What is the probability $p_n$ that the red ball is drawn in the $n$-th
drawing? Show that it's a probability function.
My thoughts:
In the first drawing, there are only two balls (red, black). So the probability is
$$p_1 = \frac{1}{2}$$
In the second drawing, if we didn't draw the red ball yet, the probability would be
$$p_2 = \frac{1}{3}\left( 1 - \frac{1}{2} \right)$$
because there are three balls now, and we multiply the probability of drawing the red ball with the counter probability $p_1^c = 1 - p_1$ from the first step.
This procedure leads to:
\begin{align}
p_3 & = \frac{1}{4} - \frac{1}{24} \\[4pt]
p_4 & = \frac{1}{5} - \frac{5}{120} \\
&\vdots \\
p_n & = \frac{1}{n+1} - \frac{1}{(n+1)!}
\end{align}
But this doesn't seem to be a probability function on $\mathbb N$:
$$\sum_{n \in \mathbb N} p_n = \sum_{n \in \mathbb N} \frac{1}{n+1} - \frac{1}{(n+1)!} = \sum_{n \in \mathbb N} \frac{n! - 1}{(n+1)!} = \infty$$
Can you help me find the mistake?
| \begin{align}
p_n & = \left( 1 - \frac 1 2 \right)\left( 1 - \frac 1 3 \right)\left( 1 - \frac 1 4 \right)\left( 1 - \frac 1 5 \right) \cdots\left( 1 - \frac 1 n \right) \frac 1 {n+1} \\[10pt]
& = \left(\frac 1 2 \cdot\frac 2 3 \cdot \frac 3 4 \cdot\frac 4 5 \cdots \frac{n-1} n\right)\cdot \frac 1 {n+1} \\[10pt]
& =\frac 1 {n(n+1)} = \frac 1 n - \frac 1 {n+1}
\end{align}
These add up to $1$ because the sum telescopes:
\begin{align}
\left(1 - \frac 1 2 \right) + \left(\frac 1 2 - \frac 1 3 \right) + \left( \frac 1 3 - \frac 1 4 \right) + \cdots + \left( \frac 1 n - \frac 1 {n+1} \right) = 1 - \frac 1 {n+1} \to 1\text{ as } n\to\infty.
\end{align}
| {
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Trigonometric inequality I'm trying to solve the following inequality, but I can't seem to be able to factor it:
$$5\sin^2{}x>\cos{x}(3\sin{x}+2\cos{x})$$
$$5\sin^2x-3\cos{x}\sin{x}-2\cos^2{x}>0$$
$$5(1-\cos^2{x})-3\cos{x}\sin{x}-2\cos^2{x}>0$$
Any hints on how can I solve it?
| Just to see a different approach. we can write the inequality as:
$$
5\sin^2 x-5\sin x \cos x+2 \sin x \cos x-2 \cos^2 x >0
$$
$$
(\sin x- \cos x)(5 \sin x+2 \cos x)>0
$$
now let $\delta$ such that:
$$
\cos \delta= \frac {5}{\sqrt{5^2+2^2}}\qquad \sin \delta= \frac {2}{\sqrt{5^2+2^2}}
$$
the inequality becomes:
$$
\sqrt{29}(\sin x - \cos x)(\cos \delta \sin x+\sin \delta \cos x)>0
$$
$$
(\sin x - \cos x)\sin (\delta+x)>0
$$
that you can solve with simple trigonometry.
| {
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Transformation of positive semi-definite matrices Let $a,b,c,d,e$ be positive reals such that the following matrix is positive semi-definite:
$$
\begin{pmatrix} a+4b+6c+4d+e & a+3b+3c+d & a+2b+c \\
a+3b+3c+d & a+2b+c & a+b \\
a+2b+c & a+b & a \\ \end{pmatrix}
$$
Does it follow that also the following matrix is positive semi-definite?
$$
\begin{pmatrix} e & d & c \\
d & c & b \\
c & b & a \\ \end{pmatrix}
$$
[By iterated subtraction of rows and columns, the two matrices have the same determinant; but do these operations preserve also this stronger property?]
| Let
$$
A = \begin{pmatrix} a+4b+6c+4d+e & a+3b+3c+d & a + 2b+c \\a+3b+3c+d & a+2b+c & a+b \\ a+2b+c & a+b & a\end{pmatrix}, \quad
B = \begin{pmatrix} e & d & c \\ d & c & b \\ c & b & a \end{pmatrix}.$$
Then $A=UBU^T$ and $B=VAV^T$, where
$$U = \begin{pmatrix} 1 & 2 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{pmatrix}, \quad
V = U^{-1} = \begin{pmatrix} 1 & -2 & 1 \\ 0 & 1 & -1 \\ 0 & 0 & 1\end{pmatrix} $$
Therefore $A$ and $B$ represent the same quadratic form, but in different bases, hence $A$ is positive semi-definite if and only if $B$ is positive semi-definite.
Added to answer: $x^TBx = (V^Tx)^TA(V^Tx) \geq 0$, hence $B$ is positive semi-definite.
| {
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Proving that $\cos\frac{2\pi}{13}+\cos\frac{6\pi}{13}+\cos\frac{8\pi}{13}=\frac{\sqrt{13}-1}{4}$ How can I prove that:
$\cos\frac{2\pi}{13}+\cos\frac{6\pi}{13}+\cos\frac{8\pi}{13}=\frac{\sqrt{13}-1}{4}$
Without using complex numbers?
I tried to raise by 2 and to multipy by 2, and got:
$2y^2=3+3\cos\frac{4\pi}{13}+2\cos\frac{10\pi}{13}+\cos\frac{12\pi}{13}+ 2\cos\frac{14\pi}{13}+2y$
But I'm stuck from here.
Thanks.
| Let $\cos\frac{2\pi}{13}+\cos\frac{6\pi}{13}+\cos\frac{8\pi}{13}=x$ and
$\cos\frac{4\pi}{13}+\cos\frac{10\pi}{13}+\cos\frac{12\pi}{13}=y$.
Hence, $x=2\cos\frac{5\pi}{13}\cos\frac{3\pi}{13}+\cos\frac{6\pi}{13}>0$.
Now, $$x+y=\cos\frac{2\pi}{13}+\cos\frac{4\pi}{13}+\cos\frac{6\pi}{13}+\cos\frac{8\pi}{13}+\cos\frac{10\pi}{13}+\cos\frac{12\pi}{13}=$$
$$=\tfrac{2\sin\frac{\pi}{13}\cos\frac{2\pi}{13}+2\sin\frac{\pi}{13}\cos\frac{4\pi}{13}+2\sin\frac{\pi}{13}\cos\frac{6\pi}{13}+2\sin\frac{\pi}{13}\cos\frac{8\pi}{13}+2\sin\frac{\pi}{13}\cos\frac{10\pi}{13}+2\sin\frac{\pi}{13}\cos\frac{12\pi}{13}}{2\sin\frac{\pi}{13}}=$$
$$=\tfrac{\sin\frac{3\pi}{13}-\sin\frac{\pi}{13}+\sin\frac{5\pi}{13}-\sin\frac{3\pi}{13}+\sin\frac{7\pi}{13}-\sin\frac{5\pi}{13}+\sin\frac{9\pi}{13}-\sin\frac{7\pi}{13}+\sin\frac{11\pi}{13}-\sin\frac{9\pi}{13}+\sin\frac{13\pi}{13}-\sin\frac{11\pi}{13}}{\sin\frac{\pi}{13}}=-\frac{1}{2}$$
and $$xy=\left(\cos\frac{2\pi}{13}+\cos\frac{6\pi}{13}+\cos\frac{8\pi}{13}\right)\left(\cos\frac{4\pi}{13}+\cos\frac{10\pi}{13}+\cos\frac{12\pi}{13}\right)=$$
$$=\frac{3}{2}\left(\cos\frac{2\pi}{13}+\cos\frac{4\pi}{13}+\cos\frac{6\pi}{13}+\cos\frac{8\pi}{13}+\cos\frac{10\pi}{13}+\cos\frac{12\pi}{13}\right)=-\frac{3}{4}.$$
Id est, $x$ and $y$ are roots of the following equation
$$z^2+\frac{1}{2}z-\frac{3}{4}=0$$
and since $x>0$, we obtain $x=\frac{\sqrt{13}-1}{4}$.
Done!
| {
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Proving that $8^n - 3^n$ is divisible by $5$ I really should be able to do this but I don't know why I can't figure it out.
My problem is that I have to prove $8^n - 3^n$ is divisible by $5$.
So what I did is I tried it for $n=1, n=2, n=3$ and so on and all the numbers I tried, the expression was divisible by 5.
So then what I did is I said "assume it's true for $n=g$"
So the expression turns into $8^g - 3^g$ is divisible by 5
Then I wanted to try it for $g+1$ so:
$8^{g+1} - 3^{g+1} $
$(8^g * 8) - (3^g * 3)$
Where should I go from here?
| Recall the difference of $n$th powers factorization.
\begin{equation}
a^n - b^n = (a-b)(a^{n-1} + a^{n-2}b + \cdots + ab^{n-2} + b^{n - 1})
\end{equation}
Therefore
\begin{equation}
8^n - 3^n = (8 - 3)(8^{n - 1} + 3 \cdot 8^{n - 2} + \cdots + 8 \cdot 3^{n - 2} + 3^{n - 1}) = 5M
\end{equation}
where $M = 8^{n - 1} + 3 \cdot 8^{n - 2} + \cdots + 8 \cdot 3^{n - 2} + 3^{n - 1}$. Since $M$ is an integer it is clear that $8^n - 3^n$ is divisible by 5.
| {
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For what value of $x$ the following series will converge $\sin x+2 \sin \frac x 3+4\sin \frac x 9+\ldots $? For what value of $x$ the following series will converge $\sin x+2 \sin \frac x 3+4\sin \frac x 9+8\sin \frac x {27}+\ldots $?
Work:
\begin{align}
\sin x+2 \sin \frac x 3+4\sin \frac x 9+8\sin \frac x {27}+\ldots & =\sum_{n=0}^\infty 2^n \sin \frac {x}{3^n}
\end{align}
Now, applying root test, we will get,
$$\limsup_{n\to \infty}\left| 2^n \sin \frac {x}{3^n} \right|^{\frac 1 n}=2\limsup_{n\to \infty}\left|\sin \frac {x}{3^n} \right|^{\frac 1 n} $$
After that, how to proceed to get the value of $x$ for which the series converges.
| $$\lim_{n\to \infty} \left| \sin {x \over 3^n} \right|^{1/n} =
\lim_{n\to \infty}
\left| \sin {x \over 3^n} \over {x \over 3^n}\right|^{1/n} \cdot \lim_{n\to \infty} \left| {x \over 3^n}\right|^{1/n}
= 1^0 \cdot \lim_{n\to \infty} \left| {x \over 3^n}\right|^{1/n}.
$$
You should be able to do the rest.
| {
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Prove $\int_0^a (a^2 - x^2)^n dx = a^{2n+1}\cdot\frac{(2n)!!}{(2n+1)!!}$ Prove
$$\int_0^a (a^2 - x^2)^n dx = a^{2n+1}\cdot\frac{(2n)!!}{(2n+1)!!}$$ for $ n \in N $
My idea is to use binomial theorem and integrate partly, but I have not succeeded yet. Are there some hints?
| Note: @Heropup's approach seems to be the most adequate. But you can also succeed by applying the binomial theorem.
We obtain
\begin{align*}
\int_{0}^a\left(a^2-x^2\right)^n\,dx&=\int_0^a\sum_{k=0}^n\binom{n}{k}\left(-x^2\right)^k\left(a^2\right)^{n-k}\,dx\\
&=\sum_{k=0}^n\binom{n}{k}(-1)^ka^{2n-2k}\int_0^ax^{2k}\,dx\\
&=\sum_{k=0}^n\binom{n}{k}(-1)^ka^{2n-2k}\frac{1}{2k+1}a^{2k+1}\\
&=a^{2n+1}\sum_{k=0}^n(-1)^k\binom{n}{k}\frac{1}{2k+1}\tag{1}
\end{align*}
There is a nice identity called Melzak's formula which can be used to solve (1).
Melzak's formula: Let $f(x)$ be a polynomial in $x$ of degree $n$, i.e.
\begin{align*}
f(x)=\sum_{k=0}^na_kx^k
\end{align*}
Let $y$ be an arbitrary complex number. Melzak's formula states that for $y \ne 0,-1,-2,\ldots,-n$
\begin{align*}
f(x+y)&=y\binom{y+n}{n}\sum_{k=0}^n(-1)^k\binom{n}{k}\frac{f(x-k)}{y+k}\tag{2}\\
\end{align*}
The whole chapter 7 in H.W.Goulds Combinatorial Identities for Stirling Numbers is devoted to this formula. An easily accessible proof can be found in New proofs of Melzak's identity by U. Abel, H.W. Gould and J. Quaintance.
Setting $f(x)=1$ and $y=\frac{1}{2}$ we obtain from (2)
\begin{align*}
\sum_{k=0}^n(-1)^k\binom{n}{k}\frac{1}{2k+1}=\frac{1}{\binom{\frac{1}{2}+n}{n}}
\end{align*}
Since
\begin{align*}
\binom{\frac{1}{2}+n}{n}&=\frac{\left(n+\frac{1}{2}\right)\left(n-\frac{1}{2}\right)\cdots\left(n+\frac{1}{2}-(n-1)\right)}{n!}\\
&=\frac{1}{2^n}\cdot\frac{\left(2n+1\right)\left(2n-1\right)\cdots\left(3\right)}{n!}\\
&=\frac{(2n+1)!!}{2^nn!}\\
&=\frac{(2n+1)!!}{(2n)!!}
\end{align*}
the claim follows.
| {
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Sums of the series $1 + (x^2) / 3! +( x^4) / 5! +\cdots$ How can I compute sum of the series ; $$1 + \frac{x^2}{3!}+\frac{x^4}{5!}+\frac{x^6}{7!}+\frac{x^8}{9!}+\cdots$$
I tried to divide it to two pieces such that $$f(x) = 1+\frac{x}{2!}+\frac{x^2}{3!}+\frac{x^3}{4!}+\cdots = \frac{e^x-x}{x}$$ and that $$g(x) = \frac{x}{2!}+\frac{x^3}{4!}+\frac{x^5}{6!}+\frac{x^7}{8!}+\cdots $$
so that answer is equal to $f(x)-g(x)$ but I couldn't proceed since I cannot find any expression for $g(x)$
| Note that $xf(x)=\frac{1}{2}(e^x-e^{-x})=\sinh x$.
| {
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Definite integral of $\int_0^\pi \frac{d\theta}{a+\sin^2 \theta} \quad (a>0)$ Definite integral of $$\int_0^\pi \frac{d\theta}{a+\sin^2 \theta} \quad (a>0)$$
I was attempting to do a trig substitution but that did not work out for me I believe because of the $a$. Looking at this to begin with I thought it was going to go smoothly but I keep getting stuck.
| Let's find
$$
\int_0^{2\pi} \frac{1}{a+\sin^2\theta}d\theta
$$
instead. Let $z=e^{i\theta}$, then $dz=ie^{i\theta}d\theta=izd\theta$, so
\begin{align}
\int_0^{2\pi} \frac{1}{a+\sin^2\theta}d\theta &=\int_0^{2\pi} \frac{1}{a+\left(\frac{e^{i\theta}-e^{-i\theta}}{2i}\right)^2}d\theta\\
&=\int_C \frac{1}{a+\left(\frac{z-\frac{1}{z}}{2i}\right)^2}\frac{dz}{iz}\\
&=\int_C \frac{-4z}{i(z^4-(4a+2)z^2+1)}dz,
\end{align}
where $C$ is a unit circle.
$f(z)=\dfrac{-4z}{i(z^4-(4a+2)z^2+1)}$ has four simple poles, at
$$
z=\pm\sqrt{2a+1+\sqrt{4a^2+4a}},\pm\sqrt{2a+1-\sqrt{4a^2+4a}}
$$
Since $a>0$, $z=\pm\sqrt{2a+1+\sqrt{4a^2+4a}}$ is outside $C$, and others are inside $C$. Find $\operatorname{Res}f(\pm\sqrt{2a+1-\sqrt{4a^2+4a}})$: With routine calculations, we get
\begin{align}
\operatorname{Res}(f;\sqrt{2a+1-\sqrt{4a^2+4a}})&=\lim_{z\to\sqrt{2a+1-\sqrt{4a^2+4a}}}(z-\sqrt{2a+1-\sqrt{4a^2+4a}})f(z)\\
&=\frac{1}{i\sqrt{4a^2+4a}}
\end{align}
and
$$
\operatorname{Res}(f;-\sqrt{2a+1-\sqrt{4a^2+4a}})=\frac{1}{i\sqrt{4a^2+4a}}.
$$
Therefore,
\begin{align}
\int_C f(z)dz&=2\pi i( \operatorname{Res}(f;\sqrt{2a+1+\sqrt{4a^2+4a}})+\operatorname{Res}(f;-\sqrt{2a+1+\sqrt{4a^2+4a}}))\\
&=2\pi i\cdot \frac{2}{i\sqrt{4a^2+4a}}\\
&=\frac{2\pi}{\sqrt{a^2+a}}.
\end{align}
Since $\sin^2\theta=\sin^2(\pi+\theta)$,
$$
\int_0^{\pi} \frac{1}{a+\sin^2\theta}d\theta=\frac{1}{2}\int_0^{2\pi} \frac{1}{a+\sin^2\theta}d\theta=\frac{\pi}{\sqrt{a^2+a}}.
$$
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How do I show that $\frac a{1 - a^2} + \frac b{1 - b^2} + \frac c{1 - c^2} \ge \frac {3 \sqrt 3}2$ For $0 \lt a, b, c \lt 1$, if $ab + bc + ca = 1$, show that $$\frac a{1 - a^2} + \frac b{1 - b^2} + \frac c{1 - c^2} \ge \frac {3 \sqrt 3}2.$$
I want to use trigonometric substitution:
For the angles $A, B, C$ of any acute triangle, $$\tan A + \tan B + \tan C = \tan A \tan B \tan C,$$ $$\frac 1{\tan A \tan B} + \frac 1{\tan B \tan C} + \frac 1{\tan C \tan A} = 1.$$
Also, $\tan A, \tan B, \tan C \gt 0$. So I substitute $a, b, c$ for $\frac 1{\tan A}, \frac 1{\tan B}, \frac 1{\tan C}$ respectively. Then the inequality in question becomes $$\frac {\tan A}{1 - \tan^2 A} + \frac {\tan B}{1 - \tan^2 B} + \frac {\tan C}{1 - \tan^2 C} \le -\frac {3 \sqrt 3}2.$$
Here $A, B, C \not = \frac {\pi}4$ since $a, b, c \not = 1$.
By the trigonometric identity $\tan 2A = \frac {2 \tan A}{1 - \tan^2 A}$, we have
$$\tan 2A + \tan 2B + \tan 2C \le -3 \sqrt 3,$$
where $0 \lt A, B, C \lt \frac {\pi}2$, $A, B, C \not = \frac {\pi}4$, and $A + B + C = \pi$.
How do I proceed?
Edit: The restriction $a, b, c \lt 1$ was added after the question had received some answers, thanks to Michael Rozenberg, who pointed out this mistake.
| If you only want to prove it with AM-GM, I will give one proof using AM-GM and Cauchy-Schwarz.
Note that the given condition implies that $a+b+c\geq\sqrt 3$. In addition, I will use the following two inequalities. $$(a+b+c)(ab+bc+ca)\leq a^3+b^3+c^3+6abc$$ and $$abc\leq\big(\dfrac{ab+bc+ca}{3}\big)^{\frac{3}{2}}.$$ Now,
$$\sum\limits_{a,b,c} \dfrac{a}{1-a^2}\sum\limits_{a,b,c} a(1-a^2)\geq (a+b+c)^2.$$ But $\sum\limits_{a,b,c} a(1-a^2) = (a+b+c)(ab+bc+ca)-a^3-b^3-c^3\leq abc\leq \dfrac{6}{3\sqrt 3}=\dfrac{2}{\sqrt 3}$. These two combined together implies that $$\sum\limits_{a,b,c} \dfrac{a}{1-a^2}\geq\dfrac{(a+b+c)^2}{\dfrac{2}{\sqrt 3}}\geq\dfrac{3 \sqrt 3}{2}.$$
The first inequality is called Schur's inequality.
EDIT.
Actually an even easier solution is obtained by using $$\dfrac{a}{1-a^2}+\dfrac{9}{4}a(1-a^2)\geq 3a,$$ by AM-GM. Thus, $\sum\limits_{a,b,c} \dfrac{a}{1-a^2}\geq\dfrac{3}{4}\sum\limits_{a,b,c} a+\dfrac{9}{4}\sum\limits_{a,b,c} a^3\geq\dfrac{3}{4}\cdot\sqrt 3 + \dfrac{9}{4}\cdot\dfrac{1}{\sqrt 3} = \dfrac{3 \sqrt 3}{2}$. That $a+b+c\geq\sqrt 3$ and $a^3+b^3+c^3\geq\dfrac{1}{\sqrt 3}$ are trivial from the given condition.
| {
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"timestamp": "2023-03-29T00:00:00",
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If $f(x)=x^3+ax^2+bx+5\sin^2x$ is a strictly increasing function on the set of real numbers then $a$ and $b$ must satisfy the relation If $f(x)=x^3+ax^2+bx+5\sin^2x$ is a strictly increasing function on the set of real numbers then $a$ and $b$ must satisfy the relation
$(A)a^2-3b+15\leq0$
$(B)a^2-3b+20\leq0$
$(C)a^2-3b+25\leq0$
$(D)a^2-3b+30\leq0$
Since $f(x)=x^3+ax^2+bx+5\sin^2x$ is a strictly increasing function.
So $f'(x)=3x^2+2ax+b+5\sin2x>0$
I am stuck here.
| Since
\begin{align*}
3x^2+2ax+b+5\sin2x \geq 3x^2+2ax+b-5.
\end{align*}
Consider the quadratic expression $3x^2+2ax+b-5$. For this to be strictly positive, the discriminant should be $\leq 0$. Thus
\begin{align*}
4a^2-12(b-5) & \leq 0\\
a^2-3b+15 & \leq 0.
\end{align*}
| {
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"timestamp": "2023-03-29T00:00:00",
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How many non-negative integer solutions of $x_1 + x_2 + x_3 + x_4 = 28$ are there with $x_{1} \leq 6, x_{2} \leq 10, x_{3} \leq 15, x_{4} \leq21$? How many non-negative integer solutions of $x_1 + x_2 + x_3 + x_4 = 28$ are there with $x_{1} \leq 6, x_{2} \leq 10, x_{3} \leq 15, x_{4} \leq21$?
Attempt:
$x_{1} \geq 7$
$x_{2} \geq 11$
$x_{3} \geq 16$
$x_{4} \geq 22$
$N(U)$ = ${28 + 4 - 1 \choose 4-1}$
$N(x_1)$ = ${28-7 + 4 - 1 \choose 4-1}$
$N(x_2)$ = ${28-11 + 4 - 1 \choose 4-1}$
$N(x_3)$ = ${28-16 + 4 - 1 \choose 4-1}$
$N(x_4)$ = ${28-22 + 4 - 1 \choose 4-1}$
$N(x_1\cap x_2)$ = ${28-7-11 + 4 - 1 \choose 4-1}$
$N(x_1\cap x_3)$ = ${28-7-16 + 4 - 1 \choose 4-1}$
$N(x_1\cap x_4)$ = ${28-7-22 + 4 - 1 \choose 4-1}$
$N(x_2\cap x_3)$ = ${28-11-16 + 4 - 1 \choose 4-1}$
$N(x_2\cap x_4)$ = ${28-11-22 + 4 - 1 \choose 4-1}$
$N(x_3\cap x_4)$ = ${28-11-22 + 4 - 1 \choose 4-1}$
Answer = $N(U) - N(x_1) - N(x_2) -N(x_3) -N(x_4)+ N(x_1\cap x_2)+ N(x_1\cap x_3)+ N(x_1\cap x_4)+ N(x_2\cap x_3)+ N(x_2\cap x_4)+ N(x_3\cap x_4)$
| HINT
Use generating functions. For $x_1$ you get
$$
1 +x + \ldots x^6 = \frac{x^7-1}{x-1}
$$
and for the rest of the variables similarly, and then you need to find the coefficient of $x^{28}$ in the product of these 4 generating functions...
| {
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Maclaurin series of $x^3/(e^x-1)$ how would i taylor expand $f(x)=\frac{x^3}{e^x-1}$ around $x=0$?
I was thinking of writing
$\frac{x^3}{e^x-1}\approx\frac{x^3}{1+x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}+\dots}$
$~~~~~~~~= \frac{1}{\frac{1}{x^3}+\frac{1}{x^2}+\frac{1}{2x}+\frac{1}{6}+\frac{x}{24}+\frac{x^2}{120}+\dots}$
$~~~~~~~~=6\bigg(\frac{1}{1+(\frac{6}{x^3}+\frac{6}{x^2}+\frac{3}{x}+\frac{x}{4}+\frac{x^2}{20}+\dots)}\bigg)$
And then use the geometric series $\frac{1}{1+x}=1-x+x^2-x^3+\dots$ but that didn't get ge the right answer..
| Do you probably know that $$\frac{x}{e^x-1} = \sum_{n=0}^{+\infty} B_n \frac{x^n}{n!},$$ where $B_n$ are the Bernoulli's numbers. Hence $$\frac{x^3}{e^x-1} = \sum_{n=0}^{+\infty} B_n \frac{x^{n+2}}{n!}=x^2-\frac{x^3}{2}+\frac{x^4}{12}+ \dots.$$
| {
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conjectured general continued fraction for the quotient of gamma functions Given complex numbers $a=x+iy$, $b=m+in$ and a gamma function $\Gamma(z)$ with $x\gt0$ and $m\gt0$, it is conjectured that the following general continued fraction which is symmetric on $a$ and $b$ is true
$$\frac{\displaystyle\Gamma\left(\frac{a+3b}{4(a+b)}\right)\Gamma\left(\frac{3a+b}{4(a+b)}\right)}{\displaystyle\Gamma\left(\frac{3a+5b}{4(a+b)}\right)\Gamma\left(\frac{5a+3b}{4(a+b)}\right)}=\cfrac{4(a+b)}{a+b+\cfrac{(2a)(2b)} {3(a+b)+\cfrac{(3a+b)(a+3b)}{5(a+b)+\cfrac{(4a+2b)(2a+4b)}{7(a+b)+\ddots}}}}\tag{1}$$
And can be further generalised to
$$\frac{1}{\sqrt{c}}\tan\left(\frac{b-a}{b+a}\tan^{-1}\left(\frac{\sqrt{c}}{d}\right)\right)=\cfrac{(b-a)}{d(a+b)+\cfrac{c(2a)(2b)} {3d(a+b)+\cfrac{c(3a+b)(a+3b)}{5d(a+b)+\cfrac{c(4a+2b)(2a+4b)}{7d(a+b)+\ddots}}}}\tag{2}$$
Corollaries:
(i) Specializing to $a=1/2$ and $b=2z/n+1/2$,we obtain
the continued fraction for $\tan\left(\frac{z\pi}{4z+2n}\right)$ found in this post after applying the functional equation of the gamma function
(ii) and specializing further to $a=-1/2$ and $b=2z/n+3/2$,we obtain immediately the continued fraction for $\cot\left(\frac{z\pi}{4z+2n}\right)$ found here after applying the functional equation of the gamma function.This continued fraction was proved by @GEdgar.
(iii) letting $2a=m-n$ and $2b=m+n$,we find $$\displaystyle\tan\left(\frac{\pi n}{4m}\right)=\cfrac{n}{m+\cfrac{m^2-n^2} {3m+\cfrac{4m^2-n^2}{5m+\cfrac{9m^2-n^2}{7m+\cfrac{16m^2-n^2}{9m+\ddots}}}}}$$
From which we obtain its hyperbolic companion $$\displaystyle\tanh\left(\frac{\pi n}{4m}\right)=\cfrac{n}{m+\cfrac{m^2+n^2} {3m+\cfrac{4m^2+n^2}{5m+\cfrac{9m^2+n^2}{7m+\cfrac{16m^2+n^2}{9m+\ddots}}}}}$$
(iv)and if $2a=-n$ and $2b=2m+n$,then it follows that
$$\displaystyle\tan\left(\frac{\pi(m+n)}{4m}\right)=\frac{1+\tan\Big(\frac{\pi n}{4m}\Big)}{1-\tan\Big(\frac{\pi n}{4m}\Big)}=\cfrac{m+n}{m+\cfrac{(-n)(2m+n)} {3m+\cfrac{(m-n)(3m+n)}{5m+\cfrac{(2m-n)(4m+n)}{7m+\cfrac{(3m-n)(5m+n)}{9m+\ddots}}}}}$$
Q: Is the conjectured general continued fraction true (for all complex numbers $a$,$b$ with $x\gt0$ and $m\gt0$)?
| Continued fraction (2) can be simplified as
$$
\tan\left(\alpha\tan^{-1}z\right)=\cfrac{\alpha z}{1+\cfrac{\frac{(1^2-\alpha^2)z^2}{1\cdot 3}} {1+\cfrac{\frac{(2^2-\alpha^2)z^2}{3\cdot 5}}{1+\cfrac{\frac{(3^2-\alpha^2)z^2}{5\cdot 7}}{1+\ddots}}}}\tag{2a}
$$
This is a special case of the following continued fraction due to N$\ddot{\text{o}}$rlund (B.Berndt, Ramanujan's notebooks, vol.2)
To obtain (2a) from (21.6) set $\beta=-\alpha,\gamma=\frac{1}{2},\frac{x(1-x)}{(1/2-x)^2}=z^2$. Now one can apply the following formulas
$$
\, _2F_1\left(a,-a;\frac{1}{2};x\right)=\cos \left(2 a \sin ^{-1}\left(\sqrt{x}\right)\right)
$$
$$
-2a^2\, _2F_1\left(1+a,1-a;\frac{3}{2};x\right)=\frac{d}{dx}\left[\, _2F_1\left(a,-a;\frac{1}{2};x\right)\right]
$$
$$
2 \sin ^{-1}\left(\sqrt{x}\right)=\tan^{-1}\frac{\sqrt{x(1-x)}}{1/2-x}
$$
to complete the proof.
| {
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"timestamp": "2023-03-29T00:00:00",
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Triangular Numbers and Sum of Two Squares "If n is a triangular number, show that each of the three consecutive integers, $8n^2, 8n^2+1, 8n^2+2$ can be written as a sum of two squares."
I have spend hours working on this problem and cannot seem to get anywhere with it. I was advised to start with $8n^2+1$ and work through it just using algebra to express this as a sum of two squares, but I am really struggling. Can anyone help?
| $8n^2$ is trivial, since $8n^2=(2n)^2+(2n)^2$.
For the first triangular numbers, we have
*
*$8\cdot1^2+1=9=0^2+3^2$
*$8\cdot3^2+1=73=3^2+8^2$
*$8\cdot6^2+1=289=8^2+15^2$
*$8\cdot10^2+1=801=15^2+24^2$
*...
Note that the numbers $0,3,8,15,24,\ldots$ form the sequence $\{n^2-1\}$.
Do something similar for $8n^2+2$.
EDIT (answering an OP's comment):
*
*$10=1^2+3^2$
*$74=5^2+7^2$
*$290=11^2+13^2$
*$802=19^2+21^2$
*...
The "small" squares sequence is $1,5,11,19,\ldots$, which is $2(n-0.5)^2+0.5$.
| {
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How to find x such that $\sqrt{x}= \sqrt{a-x} + \sqrt{b-x} + \sqrt{c-x}$? Let $a,b,c \in R$ are given. How to find x such that $\sqrt{x}= \sqrt{a-x} + \sqrt{b-x} + \sqrt{c-x}$? Is there a simple way?
| As I wrote in a comment, to get a polynomial in $x$, you must square multiple times. For starting $$(\sqrt{x}- \sqrt{a-x})^2=(\sqrt{b-x} + \sqrt{c-x})^2$$ $$a-2 \sqrt{x} \sqrt{a-x}=2 \sqrt{b-x} \sqrt{c-x}+b+c-2 x$$ $$a-b-c+2 x=2 \sqrt{x} \sqrt{a-x}+2 \sqrt{b-x} \sqrt{c-x}$$ Squaring again $$(a-b-c+2 x)^2=4(\sqrt{x} \sqrt{a-x}+ \sqrt{b-x} \sqrt{c-x})^2$$ Continue the same way, pushing everytime the radicals to the lhs.
Hoping no mistakes (not sure !), you should end with something like $$80 x^4-64 (a+b+c)x^3 +8 \left(a^2+6 a (b+c)+b^2+14 b c+c^2\right)x^2-64 a b c x+\left(a^2-2 a (b+c)+b^2+6 b
c+c^2\right)^2=0$$ Now, you have a quartic polynomial ... which can be solced using radicals.
I wish you a very good time !!
By the way, do not forget that squaring introduces extra roots and some ot them need to be discarded later after checks (in your case, if solution exists it must be between $0$ and the minimum of $a,b,c$).
| {
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Find the area inside of a equilateral triangle Can anyone help me with this problem? I assume that similar triangles are used to solve the problem, but I can't find the solution.
Problem: ABC is an equilateral triangle with sides equal to 2 cm. BC is extended its own length to D, and E is the midpoint of AB. ED meets AC at F. Find the area of the quadrilateral BEFC in square centimeters in simplest radical form.
| Note that $\angle A = \pi/3$ because $\triangle ABC$ is equilateral. Hence we can compute:
$$ S_{\triangle ABC} = \frac{2\times 2 \sin(\pi/3)}{2} = \sqrt{3} $$
and
$$ S_{\triangle AEF} = \frac{1\times \overline{AF}\sin(\pi/3)}{2}. $$
So we have to calculate $\overline{AF}$, but it is a simple application of Menelao's Theorem:
$$1=\frac{\overline{BE}\cdot \overline{AF}\cdot \overline{CD}}{\overline{AE}\cdot\overline{FC}\cdot\overline{BD}} = \frac{1\cdot x\cdot 2}{1\cdot (2-x)\cdot 4}, $$
where $x=\overline{AF}$. Solvin this equation for $x$ gives $x=4/3$
Finally, the area $S$ we are looking for is the diference $S_{\triangle ABC}-S_{\triangle AEF}$. Then
$$ S = \sqrt{3}-\frac{4\sqrt{3}}{4\cdot 3} = \frac{2\sqrt{3}}{3}. $$
| {
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"timestamp": "2023-03-29T00:00:00",
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Transversals of Latin Squares According to this thesis, page $28$, the following Latin Square has $3$ $0$-s transversals: $$\begin{bmatrix}1 & 2 & 3 & 4 & 5\\ 2 & 4 & 1 & 5 & 3\\ 3 & 5 & 4 & 2 & 1\\ 4 & 1 & 5 & 3 & 2\\5 & 3 & 2 & 1 & 4\end{bmatrix} \implies \begin{bmatrix}1 & & & &\\ & & & 5 & \\ & & 4 & & \\ & & & & 2\\ & 3\end{bmatrix}, \begin{bmatrix}& & & 4 &\\2 & & & &\\ & & & & 1 \\ & & 5 & &\\ & 3\end{bmatrix}, \begin{bmatrix}& & & &5\\& & 1 & &\\ & & & 2 &\\4 & & & &\\& 3\end{bmatrix}.$$ The definition of a $0$-s transversal for a Latin square of order $n$ is
a set of $n$ ordered triples such that the first and second entries are the rows and columns respectively in which the values $1,\ldots,n$ occur exactly once and the third entry of the triple is the value, of which there are $n$ distinct values.
Basically, we need to visit each row and column only once and we must have $5$ distinct symbols at the end. I can represent each transversal as $$t_1 = \{(1,1,1),(2,4,5),(3,3,4),(4,5,2),(5,2,3)\}$$ $$t_2 = \{(1,4,4),(2,1,2),(3,5,1),(4,3,5),(5,2,3)\}$$ $$t_3 = \{(1,5,5),(2,3,1),(3,4,2),(4,1,4),(5,2,3)\}$$ So why are these the only three? How do I know there are only three of them?
| A latin square is represented as $t_i=(r,c,s)$ which means that r represents row
of the latin square, c represents column of the latin square and s is the number
whose location is $row=r\;\;and\;\;column=c$.
| {
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What is the quickest method of finding the inverse laplace transform of $ 7e^{-6}/(s^2+6)^4 $ Solving for a differential equation, I found this which I need to obtain the inverse laplace of. What method should I use to solve this as quick as possible? Partial fractions and convolution seem to both take quite a long time.
| You could start with a few derivatives:
$$\frac d{ds}\frac1{s^2+6}=-\frac{2s}{(s^2+6)^2}$$
$$\frac{d^2}{ds^2}\frac1{s^2+6}=\frac6{(s^2+6)^2}-\frac{48}{(s^2+6)^3}$$
$$\frac d{ds}\frac s{s^2+6}=-\frac1{s^2+6}+\frac{12}{(s^2+6)^2}$$
$$\frac{d^2}{ds^2}\frac s{s^2+6}=\frac{2s}{(s^2+6)^2}-\frac{48s}{(s^2+6)^3}$$
$$\begin{align}\frac{d^3}{ds^3}\frac s{s^2+6}&=-\frac{1728}{(s^2+6)^4}+\frac{288}{(s^2+6)^3}-\frac6{(s^2+6)^2}\\
&=-\frac{1728}{(s^2+6)^4}-6\frac{d^2}{ds^2}\frac1{s^2+6}+\frac{30}{(s^2+6)^2}\\
&=-\frac{1728}{(s^2+6)^4}-6\frac{d^2}{ds^2}\frac1{s^2+6}+\frac52\frac d{ds}\frac s{s^2+6}+\frac52\frac1{s^2+6}\end{align}$$
Or
$$\frac{7e^{-6}}{(s^2+6)^4}=-\frac{7e^{-6}}{1728}\left\{\frac{d^3}{ds^3}\frac s{s^2+6}+6\frac{d^2}{ds^2}\frac1{s^2+6}-\frac52\frac d{ds}\frac s{s^2+6}-\frac52\frac1{s^2+6}\right\}$$
Now we can find
$$f(t)=-\frac{7e^{-6}}{1728}\left\{-t^3\cos(\sqrt6t)+\sqrt6t^2\sin(\sqrt6t)+\frac52\cos(\sqrt6t)-\frac{5\sqrt6}{12}\sin(\sqrt6t)\right\}$$
Just look at the method; don't trust the arithmetic.
EDIT: As one might have suspected, there were arithmetic mistakes. After correction, my results now agree with Wolfram|Alpha.
| {
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How do I show that $\frac {a^2}b + \frac {b^2}c + \frac {c^2}d + \frac {d^2}a \ge 4$ for $a^2 + b^2 + c^2 + d^2 = 4$? Let $a, b, c, d$ be positive real numbers such that $a^2 + b^2 + c^2 + d^2 = 4$, show that $$\frac {a^2}b + \frac {b^2}c + \frac {c^2}d + \frac {d^2}a \ge 4.$$
My try:
$$\frac {a^2}b + \frac {b^2}c + \frac {c^2}d + \frac {d^2}a \ge a + b + c + d,$$
yet $$a + b + c + d \le \sqrt{4(a^2 + b^2 + c^2 + d^2)} = 4.$$
Thus, direct application of Cauchy-Schwarz inequality is too weak. I tried other methods but with no significant progress:
$$(\frac {a^2}b + \frac {b^2}c + \frac {c^2}d + \frac {d^2}a)^2 \ge \frac {(a^{4/3} + b^{4/3} + c^{4/3})^3}{a^2 + b^2 + c^2 + d^2} = \frac {(a^{4/3} + b^{4/3} + c^{4/3})^3}4.$$
I also observed that
$$(\frac {a^2}b + \frac {b^2}c + \frac {c^2}d + \frac {d^2}a) + (\frac {a^2}c + \frac {b^2}d + \frac {c^2}a + \frac {d^2}b) + (\frac {a^2}d + \frac {b^2}a + \frac {c^2}b + \frac {d^2}c) + (\frac {a^2}a + \frac {b^2}b + \frac {c^2}c + \frac {d^2}d) = 4 (\frac 1a + \frac 1b + \frac 1c + \frac 1d) \ge 16,$$
since $$\frac 1a + \frac 1b + \frac 1c + \frac 1d \ge \sqrt{\frac {(1 + 1 + 1 + 1)^3}{a^2 + b^2 + c^2 + d^2}} = 4.$$
Now my work might seem stupid or off-topic here, but I provide it here because I wish any of these attempts will lead to a solution. Any hints will be appreciated.
| by the AM-GM inequality:
Given that: 4= aˆ2 + bˆ2 + cˆ2 + dˆ2 ===> 1 = 1/4 (aˆ2 + bˆ2 + cˆ2 + dˆ2) ≥ 4^√((aˆ2)(bˆ2)(cˆ2)(dˆ2))= √(abcd)
since: 1≥ √(abcd)
(aˆ2+bˆ2+cˆ2+dˆ2)/4 (aˆ2/b + bˆ2/c + cˆ2/d + dˆ2/a)≥ 4ˆ√(abcd)(aˆ2 + bˆ2 + cˆ2 + dˆ2)ˆ4 = 4√(abcd)≥ 4
Hence (aˆ2/b + bˆ2/c + cˆ2/d + dˆ2/a) ≥ 4
| {
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} |
Average Value of a Line Integral I'm having quite a hard time calculating the average value of a line integral.
Given the surface $f(x,y) = \sqrt{16 + 36y^{2/3}}$ and the curve $y = x^{3/2}$, I need to calculate the average value of the integral of the surface for $0 \leq x \leq 13$
I start by parameterizing the curve and the surface for $0 \leq t \leq 13:
$\begin{align}
r(t) &= \langle t, t^{3/2} \rangle \\
f(t) &= \sqrt{16+36t} \\
\end{align}$
And calculate a few things I'll need later:
$\begin{align}
r'(t) &= \langle 1, \frac{3}{2}t^{1/2} \rangle \\
\left|r'(t)\right| &= \sqrt{ 1 + \frac{9}{4}t } \\
\end{align}$
Next, I calculate the line integral:
$\begin{align}
&\int_0^{13} f(t) \left|r'(t)\right| dt \\
&\int_0^{13} \sqrt{16+36t} \sqrt{ 1 + \frac{9}{4}t }\ dt \\
2 &\int_0^{13} \sqrt{4+9t} \sqrt{ 1 + \frac{9}{4}t }\ dt \\
2 &\int_0^{13} \sqrt{ \frac{4}{4} (4+9t) } \sqrt{ 1 + \frac{9}{4}t }\ dt \\
4 &\int_0^{13} \sqrt{ 1 + \frac{9}{4}t } \sqrt{ 1 + \frac{9}{4}t }\ dt \\
4 &\int_0^{13} 1 + \frac{9}{4}t\ dt \\
4 &\left( t + \frac{9}{8}t^2\right|_0^{13} \\
4 &\left( 13 + \frac{9}{8}13^2\right) \\
&\frac{1625}{2} \\
\end{align}$
Then, I calculate the length of the curve:
$\begin{align}
L &= \int_0^{13} \sqrt{1 + [r'(t)]^2}\ dx \\
L &= \int_0^{13} \sqrt{1 + 1 + \frac{9}{4}t}\ dx \\
L &= \int_0^{13} \sqrt{2 + \frac{9}{4}t}\ dx \\
\end{align}$
Letting $u = 2 + 9/4 t$
$\begin{align}
L &= \frac{4}{9} \int_2^{125\ /\ 4} u^{1/2}\ du \\
L &= \frac{8}{27} \left( u^{3/2} \right|_{\ 2}^{\ 125\ /\ 4} \\
L &= \frac{8}{27} \left[ \left(\frac{125}{4}\right)^{3/2} - 2^{3/2} \right] \\
L &= \frac{625 \sqrt{5} - 16 \sqrt{2} }{27} \\
\end{align}$
Then, dividing the integral by the length of the curve gives a gnarly, incorrect mess.
What am I not understanding here? Are there algebra errors? Errors in the calculus?
| you mess with the integral of the curve, have mix dx with t.
$$r'(t)=\frac {dr}{dt}=\hat T \frac {ds}{dt}$$
$$ds=\sqrt{1+\frac 94 t^2}dt$$
That is easy solve with trig substitution.
$$\int_0^{13} \sqrt{1+\frac 94 t^2}dt=128.1379$$
$$ ave(f(x,y))=\frac {1625}{2\cdot 128.1379}= 6.3408 $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1744155",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Sum of cube roots of a quadratic If $a$ and $b$ are the roots of $x^2 -5x + 8 = 0$. How do I find $\sqrt[3]{a} + \sqrt[3]{b}$ without finding the roots?
I know how to evaluate $\sqrt[2]{a} + \sqrt[2]{b}$ by squaring and subbing for $a+b$ and $ab$ via sum and product of roots. But for this question, if I cube $\sqrt[3]{a} + \sqrt[3]{b}$ I'm left with radicals which are difficult to resolve, e.g. $\sqrt[3]{a^2b}$
How should I go about approaching this problem?
Edit: I've also tried letting $\sqrt[3]a+\sqrt[3]b=m$, which makes $a+b+3m\sqrt[3]{ab}=m^3$ (by rising everything to the power of $3$ and then substituting $\sqrt[3]a+\sqrt[3]b=m$ again), if that is of any help.
| The solutions of the quadratic equation are not real numbers, so $\sqrt[3]{a}$ and $\sqrt[3]{b}$ are not canonically determined; therefore, the sum can take up to 9 different values.
Since $ab = 8$, once you choose one of the three possible values for $\sqrt[3]{a}$, you may want to consider only the value $\sqrt[3]{b} = 2/\sqrt[3]{a}$ for the second root. In that case, let $S = \sqrt[3]{a} + \sqrt[3]{b}$. Then
$S^3 = a+b + 3\sqrt[3]{ab} S$, so $S$ is a solution of the cubic equation $S^3 = 6S+5 \Longleftrightarrow (S+1)(S^2-S-5) = 0$. The solutions are $S_1=-1$, $S_2=\frac{1}{2}(1-\sqrt{21})$, and $S_3 = \frac{1}{2}(1+\sqrt{21})$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1744538",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Proving a trigonometric identity:
Prove that $\sin \frac{{2\pi }}{7} + \sin \frac{{4\pi }}{7} + \sin
\frac{{8\pi }}{7} = \frac{{\sqrt 7 }}{2}$.
I have tried to square both side and got ${\sin ^2}\frac{{2\pi }}{7} + {\sin ^2}\frac{{4\pi }}{7} + {\sin ^2}\frac{{8\pi }}{7} = \frac{7}{4}$. But I cannot proceed further. Any help would be appreciated.
| If $7y=(2n+1)\pi$ where $7\nmid(2n+1)$
Using Prosthaphaeresis & Double angle Formulas
$$F=\sin2y+\sin4y+\sin8y$$
$$=2\sin\dfrac{2y+8y}2\cos\dfrac{8y-2y}2+2\sin2y\cos2y$$
$$=2\sin5y\cos3y+2\sin2y\cos2y$$
As $\cos3y=\cdots=-\cos4y,\sin5y=\sin2y,$
$$F=2\sin2y(\cos2y-\cos4y)=2\sin2y(2\sin3y\sin y)$$
Using Prove that $\prod_{k=1}^{n-1}\sin\frac{k \pi}{n} = \frac{n}{2^{n-1}}$,
can you prove $$\prod_{k=1}^m\sin\dfrac{k\pi}{2m+1}=+\dfrac{\sqrt{2m+1}}{2^m}$$
Here $m=3$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1745060",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Prove that the identity is true for all natural numbers for the identity:
$$\frac{n!}{x(x+1)(x+2)...(x+n)} = \frac{A_0}{x+0} + \frac{A_1}{x+1}+...+ \frac{A_n}{x+n}$$
prove
$$A_k= (-1)^kC(n,k)$$
I think this might work by induction, but i am not able to arrive at a final answer. please help!!!
| As what you tagged, we prove it by induction. For $n=0$, then the result is
trivial. Suppose that the result holds for some $n\ge 0$, then for $n+1$,
\begin{align}
&\frac{(n+1)!}{x(x+1)(x+2)\cdots[x+(n+1)]}\\
&\quad\quad=\frac{n!}{x(x+1)(x+2)\cdots(x+n)}\cdot\frac{n+1}{x+n+1}\\
&\quad\quad=\left(\frac{A_0}{x+0}+\frac{A_1}{x+1}+\cdots+\frac{A_n}{x+n}\right)\cdot\frac{n+1}{x+n+1}\\
&\quad\quad=\frac{(n+1)A_0}{(x+0)(x+n+1)}+\frac{(n+1)A_1}{(x+1)(x+n+1)}+\cdots+\frac{(n+1)A_n}{(x+n)(x+n+1)}\\
&\quad\quad=\left[\frac{(n+1)A_0}{x+0}-\frac{(n+1)A_0}{x+n+1}\right]\cdot\frac{1}{n+1}
+\left[\frac{(n+1)A_1}{x+1}-\frac{(n+1)A_1}{x+n+1}\right]\cdot\frac{1}{n}
\\&\quad\quad\,\,\,\,\,\,\,+\cdots+\left[\frac{(n+1)A_n}{x+n}-\frac{(n+1)A_n}{x+n+1}\right]\\
&\quad\quad=\frac{A_0}{x+0}+\frac{\frac{n+1}{n}A_1}{x+1}+\cdots+\frac{(n+1)A_n}{x+n}
-\frac{A_0+\frac{n+1}{n}A_1+\cdots+(n+1)A_n}{x+n+1}\\
&\quad\quad=\frac{A_0'}{x+0}+\frac{A_1'}{x+1}+\cdots+\frac{A_n'}{x+n}
+\frac{A_{n+1}'}{x+n+1},
\end{align}
where $A_0',A_1',\ldots,A_{n+1}'$ denotes the new coefficient in terms of the case $n+1$. Now, by the assumption, we check that
\begin{align*}
A_k'=\frac{n+1}{n-k+1}A_k=\frac{n+1}{n-k+1}(-1)^k{n\choose k}=(-1)^k{n+1\choose k}\quad\mbox{for }0\le k\le n,
\end{align*}
and
\begin{align*}
A_{n+1}'
&=-\left[A_0+\frac{n+1}{n}A_1+\cdots+(n+1)A_n\right]\\
&=-\sum_{k=0}^n(-1)^k\frac{n+1}{n-k+1}{n\choose k}\\
&=-\sum_{k=0}^n(-1)^{k}{n+1\choose k}\\
&=-\sum_{k=0}^{n+1}(-1)^{k}{n+1\choose k}+(-1)^{n+1}\\
&=(-1)^{n+1}.
\end{align*}
This completes the proof.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1746533",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Can someone explain how the cartesian equation is formed? Can someone explain how the cartesian equation of $r = 1 - \cos (\theta)$ is $x^4 + y^4 + 2x^2y^2 + 2x^3 + 2xy^2 - y^ 2 = 0$ ?
| Since my comment was pseudo-copied to an answer (i.e. an answer identical to my comment was posted) I thought I'd just post a full solution completing that comment.
*
*Remember:
$$\left\{\begin{array}{c}
r\cos\theta=x \\
r\sin\theta=y \\
r=\sqrt{x^2+y^2} \\
\theta=\arctan\frac{y}{x}\text{ well, in some cases}
\end{array}\right.;$$
*Multiply by $r$ to get $r^2=r-r\cos\theta$;
*Substitute the above equations to get $x^2+y^2=\sqrt{x^2+y^2}-x$;
*Carry $x$ over to the LHS and square to get $x^2+y^2=(x^2+y^2+x)^2$;
*Expand the square on the right and get $x^4+y^4+x^2+2x^2y^2+2xy^2+2x^3=x^2+y^2$;
*$x^2$'s cancel out, $y^2$ goes to the left side, and we get:
$$x^4+y^4+2x^3-y^2+2x^2y^2+2xy^2=0.$$
Which is precisely what we wanted.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1746797",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
What is the solution of $\sin z=\cosh 4$? What is the solution of $\sin z=\cosh 4$?
By putting $z=x+iy$ I managed to find that the real part of $z$ is $x= \frac \pi 2+2n\pi $, but the imaginary part is contradictory giving negative value of $\cosh x$.
The answer is given to be $\frac \pi 2+2n\pi \pm \cosh 4$.
| Well, $\cosh 4$ is just a number, so we apply inverse functions... \begin{align*}
\sin z &= \cosh 4 \\
z = \sin^{-1} \cosh 4 + 2\pi k &\text{ or } z = \pi - \sin^{-1} \cosh 4 + 2 \pi k, \text{ for any integer $k$}.
\end{align*} Suppose we already know $\sin^{-1}\cosh 4 = \frac{\pi}{2} - 4\mathrm{i}$. Then $$
z = \frac{\pi}{2} - 4\mathrm{i} + 2\pi k \text{ or } z = \frac{\pi}{2} + 4\mathrm{i} + 2\pi k, \text{ for any integer $k$}.
$$
We could perhaps simplify this to $z = \frac{\pi}{2} \pm 4\mathrm{i} + 2\pi k, \text{ for any integer $k$}$. This is as close as I can get any answer to the form you gave (since $\cosh 4 = 27.308\dots \neq 4\mathrm{i}$).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1751184",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find the possible values of $x$ if $2^{2x+1} = 3(2^x) -1$
Find the possible values of $x$ if $2^{2x+1} = 3(2^x) -1$
I know that $x=0$ and $x=-1$ are possible values of $x$ by looking at the equation. I need help understanding how to use logarithms to solve questions of this type. Here is what I'm doing, where am I going wrong?:
$$2^{2x+1} = 3(2^x) -1$$
Can be written as $$ 3(2^x) - 2^{2x+1} =1$$
Taking logarithms of each side (and here is where I think I go wrong):
$$[x \ln(2) + \ln(3)] - [2x \ln(2) + \ln(2)] = \ln(1)$$
$$[x \ln(2) + \ln(3)] - [2x \ln(2) + \ln(2)] = 0$$
$$-x \ln(2) + \ln(3) - \ln(2) = 0$$
| $$2^{2x+1} = 3(2^x) -1$$
$$2\cdot2^{2x}-3\cdot2^x+1=0$$
$2^x=t>0$
$$2t^2-3t+1=0$$
$t=1$ or $t=\frac12$
$2^x=2^{-1}$ or $2^x=1$
$x=-1$ or $x=0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1752706",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Eigenvalues and Eigenvectors relating to orthogonal basis and diagonal matrices Find the eigenvalues and eigenvectors of the matrix.
$$A =
\begin{bmatrix}
1 & 1 & 0 \\
1 & 0 & -1\\
0 & -1 & 1
\end{bmatrix}$$
As we have seen in the lectures, these eigenvectors form an orthogonal basis with respect to the standard inner product $\mathbb{C}^3$ . By considering a basis transformation to an orthonormal basis of eigenvectors find a diagonalizing matrix $P$, and hence $B = P^{-1}AP$ where $B$ is diagonal. (Hint: $P^{-1} = P^{T}$ for an orthonormal basis to another.)
I've only got 1 eigenvalue to be $\lambda = 1, -1, 2$ with their corresponding eigenvectors. I am not sure where to go from here. Any help would be appreciated!! Thank you.
| To find the eigenvalues you can use the characteristic polynom :
$$det \left( \begin{bmatrix}
1-X & 1 & 0 \\
1 & 0-X & -1\\
0 & -1 & 1-X
\end{bmatrix}\right)=det \left( \begin{bmatrix}
1-X & 0 & 1-X \\
1 & -X & -1\\
0 & -1 & 1-X
\end{bmatrix}\right)=(1-X)det \left( \begin{bmatrix}
1 & 0 & 1 \\
1 & -X & -1\\
0 & -1 & 1-X
\end{bmatrix}\right)=(1-X)det \left( \begin{bmatrix}
1 & 0 & 0 \\
1 & -X & -2\\
0 & -1 & 1-X
\end{bmatrix}\right)=(1-X)(-X(1-X)-2)=(1-X)(X^2-X+2)=-(X-1)(X+1)(X-2)$$
So the eigen values are $-1,1,2$.
Can you find the eigen vectors from there ?
For $-1$. Let $x \in \mathbb{C}^3$ such as :
$$Ax=-x\iff A\begin{bmatrix}
x_1 \\
x_2 \\
x_3
\end{bmatrix}=\begin{bmatrix}
-x_1 \\
-x_2 \\
-x_3
\end{bmatrix} \iff \begin{bmatrix}
x_1+x_2 \\
x_1-x_3 \\
-x_2+x_3
\end{bmatrix}=\begin{bmatrix}
-x_1 \\
-x_2 \\
-x_3
\end{bmatrix} $$
So $x_2=-2x_1$ and $x_3=-x_1$. So the eigenvector associated to $-1$ is $\begin{bmatrix}
1 \\
-2 \\
-1
\end{bmatrix}$.
Do the same for the two other eigenvalues($1$ then $2$), concatenate the vectore you obtain in the order of obtention (first column is the vector associated with $-1$) and you obtain $P$ such as $A=PDP^{-1}$, where $D=\begin{bmatrix}
-1 & 0 & 0 \\
0 & 1 & 0\\
0 & 0 & 2
\end{bmatrix}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1753239",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Solving generalized determinant related How to solve following determinant by applying suitable elementary row/column transformations to obtain characteristic polynomial?
\begin{align*}
\left\vert
\begin{matrix}
-\lambda & 0 & 1 & 0 & 0 & \cdots & 0\\
0 & -\lambda & 0 & 1 & 1 &\cdots & 1\\
1 & 0 & -\lambda & 1 & 1 &\cdots & 1\\
0 & 1 & 1 & -\lambda & 0 &\cdots & 0\\
0 & 1 & 1 & 0 & -\lambda &\cdots & 0\\
\vdots & \vdots & \vdots & \vdots& \vdots &\ddots & \vdots \\
0 & 1 & 1 & 0 & 0 &\cdots & -\lambda
\end{matrix}
\right\vert_{n+3} & =0 \\
\end{align*}
| @joriki's proof considered the case $\lambda\ne 0$.
Now, for the case that $\lambda=0$, we have
\begin{align}
\left\vert
\begin{matrix}
0 & 0 & 1 \\
0 & 0 & 0 \\
1 & 0 & 0
\end{matrix}
\right\vert=0\quad\mbox{and}\quad
\left\vert
\begin{matrix}
0 & 0 & 1 &0\\
0 & 0 & 0 &1\\
1 & 0 & 0 &1\\
0 & 1& 1 & 0
\end{matrix}
\right\vert=
\left\vert
\begin{matrix}
0 & 0 & 1 \\
1 & 0 & 1 \\
0 & 1& 0
\end{matrix}
\right\vert=1,
\end{align}
and for $n\geq 2$,
\begin{align}
\left\vert
\begin{matrix}
0 & 0 & 1 & 0 & 0 & \cdots & 0\\
0 & 0 & 0 & 1 & 1 &\cdots & 1\\
1 & 0 & 0 & 1 & 1 &\cdots & 1\\
0 & 1 & 1 & 0 & 0 &\cdots & 0\\
0 & 1 & 1 & 0 & 0 &\cdots & 0\\
\vdots & \vdots & \vdots & \vdots& \vdots &\ddots & \vdots \\
0 & 1 & 1 & 0 & 0 &\cdots & 0
\end{matrix}
\right\vert_{n+3}=0
\end{align}
because the last $n$ columns are identical.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1753665",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Rational Expression equivalent form EDIT: I know how to find the answer, but does anyone know why plugging in numbers for x does not work?
The Question: If the rational expression $\frac {3x^2}{3x-1}$ is rewritten in the equivalent form $\frac {\frac 13}{3x-1}+A$, what must expression A be in terms of x?
The four answer choices: A) $x+ \frac 13$ B) $x+1$ C) $x-1$ D) $x-3$
The Answer:
A) $x+ \frac 13$
I don't really know the way this question is "supposed" to be solved, so I just tried plugging in some numbers.
*
*Plugging in 0 for x, I would get $0=\frac {-1}3+ A$
*Plugging in 1 for x, I would get $\frac 32=\frac 23 + A$.
*Plugging in 2 for x, I would get $\frac {12}{5}=\frac 53 +A$.
The equation for $x=0$ makes it seem like $A$ really is the correct answer, but plugging in 1 and 2, I wasn't sure which one, if any were correct. Pretty sure I'm just missing something really obvious here, but asking anyway. Answer Key contains the answer listed above.
| $$\frac{3x^2}{3x-1} = \frac{0+3x^2}{3x-1}=\frac{(\frac{1}{3}-\frac{1}{3})+3x^2}{3x-1}=\frac{\frac{1}{3}+(-\frac{1}{3}+3x^2)}{3x-1}=\dots$$
Noting that
$$x = \frac{3x}{3}=\frac{3x+0}{3}=\frac{3x+(-1+1)}{3}=\frac{(3x-1)+1}{3}$$
Then we have
$$A = \frac{3x^2-\frac{1}{3}}{3x-1} = \frac{3\bigg(\frac{(3x-1)+1}{3}\bigg)^2-\frac{1}{3}}{3x-1}=\frac{\frac{(3x-1)^2}{3}+\frac{2(3x-1)}{3}+\frac{1}{3}-\frac{1}{3}}{3x-1}=x+\frac{1}{3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1753771",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 7,
"answer_id": 6
} |
Solve given equation $4^{(x-2)(x+3)} - 64^{(x-3)} = 0?$
Solve given equation $4^{(x-2)(x+3)} - 64^{(x-3)} = 0?$
My attempt:
I've attempted to solve this question, but isn't it impossible to solve, i.e has already been simplified completely?
$4^{(x-2)(x+3)} - 64^{(x-3)} = 0$
Because $(x-2)(x+3) = x^2 +x - 6?$
So adding those two powers together would mean that they are unable to be solved for $x$?
Thanks
| From
$$
4^{x^2+x-6}-64^{x-3}=0
$$ you get
$$
2^{2(x^2+x-6)}-2^{6(x-3)}=0
$$ or
$$
2^{2(x^2+x-6)}=2^{6(x-3)}
$$ equivalently
$$
2(x^2+x-6)-6(x-3)=0
$$
Can you take it from here?
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Contour Integration of $\sin^2(x)/(1+x^2)$ How should I calculate this integral
$$\int\limits_{-\infty}^\infty\frac{\sin^2x}{(1+x^2)}\,dx\quad?$$
I have tried forming an indented semicircle in the upper half complex plane using the residue theorem and I tried to integrate along a curve that went around the complex plane and circled the positive real axis (since the integrand is even). Nothing has worked out for me.
| First we simplify the integral the following way using the simple trigonometric identity $\sin^2 x = \frac 12 - \frac 12 \cos{2x}$: $$\begin{align*}\int_{-\infty}^{\infty} \frac{\sin^2 x}{1+x^2}\,dx &= \frac 12 \int_{-\infty}^{\infty} \frac{dx}{1+x^2} - \frac 12 \int_{-\infty}^{\infty} \frac{\cos {2x}}{1+x^2}\,dx\\&= \frac{\pi}{2} - \frac 12 \int_{-\infty}^{\infty} \frac{\cos {2x}}{1+x^2}\,dx\end{align*}$$
Now to solve this second integral we note that it is equivalent to $\int_{-\infty}^{\infty} \frac{e^{2ix}}{1+x^2}\,dx$ as the imaginary part is odd so it's integral evaluates to 0. We now consider the semi circle contour integral of radius $R$ as $R \to \infty$ in the top half plane for this function traveling in an anti-clockwise direction, encompassing the pole at $z = i$.
$$\oint \frac{e^{2iz}}{1+z^2}\,dz = \int_{-R}^{R}\frac{e^{2iz}}{1+z^2}\,dz + \int_{\gamma} \frac{e^{2iz}}{1+z^2}\,dz$$ where $\gamma$ is the semi-circle arc contour. By Cauchy's integral formula we have $$\oint \frac{e^{2iz}}{1+z^2}\,dz = \oint \frac{\frac{e^{2iz}}{z+i}}{z-i}\,dz = 2\pi i \frac{e^{-2}}{2i} = {\pi \over e^2}$$
Now we look at the semi-circle arc contour. $$\lvert z^2 + 1 - 1 \rvert \leq \lvert z^2 + 1 \rvert + 1 \implies \lvert z^2 + 1 \rvert \geq R^2 - 1\\\left\lvert e^{2iz}\right\rvert \leq 1$$
Hence by the estimation lemma:
$$\left\lvert \int_{\gamma} \frac{e^{2iz}}{1+z^2}\,dz \right\rvert \leq \frac{\pi R}{R^2 - 1}$$
Finally, taking the limit as $R \to \infty$ gives us
$$\int_{-\infty}^{\infty} \frac{e^{2ix}}{1+x^2}\,dx = {\pi \over e^2} \\ \implies \int_{-\infty}^{\infty} \frac{\sin^2 x}{1+x^2}\,dx = \frac{\pi}{2} - \frac{\pi}{2e^2}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Prove that $\sin^2\frac{A}{2}\csc2A$, $\sin^2\frac{B}{2}\csc2B$, $\sin^2\frac{C}{2}\csc2C$ are in harmonic progression
If sides $a,b,c$ of $\triangle ABC$ are in arithmetic progression (AP), then prove that
$$\sin^2\frac{A}{2}\csc2A, \quad\sin^2\frac{B}{2}\csc2B, \quad \sin^2\frac{C}{2}\csc2C$$ are in harmonic progression (HP).
In order for
$$\sin^2\frac{A}{2}\csc2A, \quad \sin^2\frac{B}{2}\csc2B, \quad \sin^2\frac{C}{2}\csc2C$$
to be in HP, their reciprocals
$$\frac{\sin 2A}{\sin^2\frac{A}{2}}, \quad \frac{\sin 2B}{\sin^2\frac{B}{2}}, \quad \frac{\sin 2C}{\sin^2\frac{C}{2}}$$
need to be in AP. That is,
$$2\frac{\sin 2B}{\sin^2\frac{B}{2}}=\frac{\sin 2A}{\sin^2\frac{A}{2}}+\frac{\sin 2C}{\sin^2\frac{C}{2}} \tag{1}$$
We are given that $2b=a+c$; thus, by the Sine Rule,
$$2\sin B=\sin A+\sin C \tag{2}$$
I do not know how to prove equation $(1)$ from equation $(2)$.
| $$\dfrac1{\sin^2\frac{A}{2}\csc2A}=\dfrac{2\sin A\left(1-2\sin^2\frac{A}{2}\right)}{\sin^2\frac{A}{2}}=4\cot\dfrac A2-4\sin A$$
So, we need to prove $\cot\dfrac A2$ etc. are in A.P.
Now use $\cot\dfrac A2=\dfrac{s(s-a)}{\triangle}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1755129",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Systems of equation
Find non-negative solutions of systems of equations:
$$\begin{cases}
x^2y^2+1=x^2+xy
\\
y^2z^2+1=y^2+yz
\\
z^2x^2+1=z^2+zx
\end{cases}
$$
My work so far:
1) $(1;1;1) - $ solution.
2) $(y^2-1)x^2-yx+1=0$
$D=y^2-4(y^2-1)=-3y^2+4\ge0 \Rightarrow 0\le y \le \frac2{\sqrt3}$
3) $x^2y^2+1-(y^2z^2+1)=x^2+xy-(y^2+yz)$
$y^2(x-z)(x+z)=(x-y)(x+y)+y(x-z)$
| Put $y=hx$. Then the first equation gives $h^2x^4-(h+1)x^2+1=0$. Hence $(h+1)^2\ge (2h)^2$, so $h\le 1$. Similarly, putting $z=ky$ we get $k\le 1$ from the second equation. But then $x=\frac{1}{hk}z$ and the third equation gives $hk\ge1$, so we must have $x=y=z$. Now the first equation gives $x^4-2x^2+1=0$ so $x^2=1$ and hence $x=1$. So the unique solution is $x=y=z=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1755276",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Volume from equation $(x ^2+ y ^2 + z ^2 ) ^2 = xyz$ How can you calculate the volume of the shape represented by the following equation: $$(x ^2+ y ^2 + z ^2 ) ^2 = xyz$$
I tried converting it to polar form (so $r = \sin^2\left(\theta\right)\cdot\cos\left(\theta\right)\cdot\sin\left(\phi\right)\cdot\cos\left(\phi\right)$) and integrate over $\phi$ and $\theta$ with range $2\pi$, however, that didn't work out. How can I approach this?
| $$ (x^2+y^2+z^2)^2 = xyz$$
$$\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} r \cos \varphi \sin \psi \\ r \sin \varphi \sin \psi \\ r \cos \psi \end{pmatrix} $$
$$ r^4 =| r^3 \sin( \varphi) \cos( \varphi) \cos( \psi) \sin^2 (\psi) |$$
$$ r = |\sin( \varphi) \cos( \varphi) \cos( \psi) \sin^2 (\psi)| $$
$${\rm d}V = r^2 \sin(\psi) {\rm d}r {\rm d}\psi {\rm d}\varphi $$
$$ V = \iiint {\rm d}V = \int \limits_0^{2\pi} \int \limits_0^\pi \int_0^{|\sin( \varphi) \cos( \varphi) \cos( \psi) \sin^2 (\psi)|} r^2 \sin\psi\; {\rm d}r\,{\rm d}\psi\,{\rm d}\varphi $$
$$ V = \int \limits_0^{2\pi} \int \limits_0^\pi \frac{\sin\psi}{3} | \sin\varphi \cos\varphi \sin^2 \psi \cos \psi|^3\,{\rm d}\psi\,{\rm d}\varphi \\ = \int \limits_0^{2\pi} \frac{\sin^2\varphi \cos^2\varphi |\sin \varphi \cos \varphi|}{60}\;{\rm d}\varphi$$
According to Wolfram
$$ V = \frac{1}{180}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1755640",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Stirling on ${2n-1 \choose n}$ I'm trying to find an expression for
$${2n-1 \choose n}$$ using Stirling's approximation
$$k!\sim \sqrt{2\pi k}(\frac{k}{e})^k.$$
I see
$${2n-1 \choose n}\approx \frac{1}{\sqrt{2\pi}}\sqrt{\frac{2n-1}{n(n-1)}}\frac{(2n-1)^{2n-1}}{n^n(n-1)^{n-1}}$$
but cannot get much further. I could do
\begin{align*}
{2n-1 \choose n}&\approx \frac{1}{(2n-1)}\frac{1}{\sqrt{2\pi}}\sqrt{\frac{2n-1}{n(n-1)}} \frac{(2n-1)^{n-1}}{n^n}\frac{(2n-1)^{n-1}}{(n-1)^{n-1}} \\
&= \frac{1}{(2n-1)}\frac{1}{n\sqrt{2\pi}}\sqrt{\frac{2n-1}{n(n-1)}}(\frac{2n-1}{n})^{n-1}(\frac{2n-1}{n-1})^{n-1}
\end{align*}
but I don't see how that should work. I have done the calculation before and used $(\frac{n-1}{n})^n$=e for large n but $2n-1$ does not really help for this.
| It looks like it's asymptotic to $\dfrac1{\sqrt{\pi n}} 2^{2n-1}$. For example, write
$$\left(\frac{2n-1}n\right)^n\left(\frac{2n-1}{n-1}\right)^{n-1} = \left(\frac{2n-1}{2n}\right)^n\left(\frac{2(n-1)+1}{2(n-1)}\right)^{n-1}\cdot 2^n\cdot 2^{n-1}.$$
Then note
$$\left(\frac{2n-1}{2n}\right)^n = \left(\big(1-\frac1{2n}\big)^{2n}\right)^{1/2} \sim e^{-1/2},$$
and so on.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1756038",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
How to write the following polynomial in $(1-\frac{x}{a}) (1-\frac{x}{b}) (1-\frac{x}{c}) (1-\frac{x}{d})$? I was given the following problem:
Write the polynomial $f(x) = \frac{1}{24} \displaystyle \prod_{i \mathop = 1}^4 (x-i)$ in the form $(1-\frac{x}{a}) (1-\frac{x}{b}) (1-\frac{x}{c}) (1-\frac{x}{d})$
So far I have done this:
$f(x) = \frac{1}{24} \displaystyle \prod_{i \mathop = 1}^4 (x-i) = \frac{1}{24}(x-1)(x-2)(x-3)(x-4)
= \frac{1}{24}(1-x)(2-x)(3-x)(4-x)
= (\frac{1-x}{1}) (\frac{2-x}{2}) (\frac{3-x}{3}) (\frac{4-x}{4})
= ({1-x}) (1-\frac{x}{2}) (1-\frac{x}{3}) (1-\frac{x}{4})$
EDIT: Seems like I got it, thanks for all the help!
| hint:roots of the polynomial have to be same , what are the roots of l.h.s and r.h.s and then compare
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1757075",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Evaluation of $ \int_{0}^{\pi}\ln(5-4\cos x)\,dx$
Evaluation of $\displaystyle \int_{0}^{\pi}\ln(5-4\cos x)dx = \int_{0}^{\pi}\ln(5+4\cos x)dx$
$\bf{My\; Try::}$ Let $\displaystyle I(a,b) = \int_{0}^{\pi}\ln(a+b\cos x)dx$
Then $$\frac{d}{db}(a,b) = \frac{d}{db}\left[\int_{0}^{\pi}\ln(a+b\cos x)dx\right]db$$
So $$I'(a,b) = \int_{0}^{\pi}\frac{\cos x}{a+b\cos x}dx = \frac{1}{b}\int_{0}^{\pi}\frac{(a+b\cos x)-a}{a+b\cos x}dx$$
So we get $$I'(a,b) = \frac{\pi}{b}-\frac{a}{b}\int_{0}^{\pi}\frac{1}{a+\cos x}dx$$
Using half angle formula $\displaystyle \tan x = \frac{1-\tan^2 x/2}{1+\tan^2 x/2}$
so we get $$I'(a,b) = \frac{\pi}{b}-\frac{a}{b}\int_{0}^{\pi}\frac{\sec^2 x/2}{(a+b)+(a-b)\tan^2 x/2}dx$$
Now Put $\tan x/2= t\; $ Then $\displaystyle \sec^2 \frac{x}{2}dx = 2dt$
So we get $$I'(a,b) = \frac{\pi}{b}-\int_{0}^{\infty}\frac{1}{(a+b)+(a-b)t^2}=\frac{\pi}{b}-\frac{a\pi}{2b\sqrt{a^2-b^2}}$$
So we get $$I(a,b) = \pi\ln|b|-\frac{\pi a}{2}\left[-\frac{\ln|b^2-a^2|+2\ln |b|}{2a^2}\right]$$
So we put $a = 5$ and $b=4$
We get $$I(5,4) = \pi\cdot \ln (5)-\frac{\pi}{2}\left[\frac{-\ln(9)+2\ln(5)}{2\cdot 5}\right]$$
So We get $$I(5,4) = \left[\frac{18\ln(5)+\ln(9)}{20}\right]\cdot \pi$$
I did not understand where i have done mistake, Help me
Thanks
| This approach seems really good, but the first mistake was in
$$\int_0^{\pi}\frac1{a+b\cos x}dx=\frac1a\int_0^{\pi}\frac1{1+e\cos x}dx$$
Where $e=\frac ba$. If you used the eccentric anomaly at this point,
$$\sin E=\frac{\sqrt{1-e^2}\sin x}{1+e\cos x}$$
So that $$dE=\frac{\sqrt{1-e^2}}{1+e\cos x}dx$$
You would get
$$\int_0^{\pi}\frac1{a+b\cos x}dx=\frac1a\int_0^{\pi}\frac{dE}{\sqrt{1-e^2}}=\frac{\pi}{\sqrt{a^2-b^2}}$$
So you were off by a factor of $2$ at this point. Next, if we let $b=a\sin\theta$, then
$$\begin{align}\int\frac{db}{b\sqrt{a^2-b^2}}&=\frac1a\int\csc\theta\,d\theta=-\frac1a\ln\left|\csc\theta+\cot\theta\right|+C_1(a)\\
&=-\frac1a\ln\left(\frac ab+\frac{\sqrt{a^2-b^2}}b\right)+C_1(a)\\
&=-\frac1a\ln\left(a+\sqrt{a^2-b^2}\right)+\frac1a\ln b+C_1(a)\end{align}$$
So you would arrive at
$$I(a,b)=\pi\ln\left(a+\sqrt{a^2-b^2}\right)+C(a)$$
At $b=0$ we get $\pi\ln a=\pi\ln2a+C(a)$ so $C(a)=-\pi\ln2$. Then
$$I(5,4)=\pi\ln8-\pi\ln2=\pi\ln4$$
Numerical quadrature confirms this result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1757448",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
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} |
Evaluate $\int \frac{1-x}{(1+x)\sqrt{x+x^2+x^3}}dx$ Evaluate $$\int \frac{1-x}{(1+x)\sqrt{x+x^2+x^3}}dx$$ i used substitution $x=\tan^2 y$ so $dx=2\tan y \sec^2 y dy$ so the integral becomes
$$I=\int\frac{2\cos 2y\: \tan y\: \sec^2 y \:dy}{\sqrt{\tan^2 y+\tan^4 y+\tan^6 y}}=\int\frac{2\cos 2y \:\sec^2 y\: dy}{\sqrt{1+\tan^2 y+\tan^4 y}}$$ I was stuck here
| Let $$I = \int\frac{1-x}{(1+x)\sqrt{x+x^2+x^3}}dx =\int\frac{(1-x^2)}{(x^2+2x+1)\sqrt{x+x^2+x^3}}dx$$
Now again Reaaranging we get $$I = -\int\frac{\left(1-\frac{1}{x^2}\right)}{\left(x+\frac{1}{x}+2\right)\sqrt{x+\frac{1}{x}+1}}dt$$
Now put $\displaystyle x+\frac{1}{x}+1=u^2\;,$ Then $\displaystyle \left(1-\frac{1}{x^2}\right)dt = 2udu$
So we get $$I = -2\int\frac{1}{u^2+1}du = -2\tan^{-1}(u)+\mathcal{C} = -2\tan^{-1}\left(\sqrt{x+\frac{1}{x}+1}\right)+\mathcal{C}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1757843",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
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What's the solution to this puzzle? I saw this on Instagram with no solution and was wondering what the answer is. I got $33$.
$$1+4=5$$
$$2+5=12$$
$$3+6=21$$
$$8+11=?$$
| $a_n + b_n = c_n$
should in fact be interpreted as
$a_n \cdot b_n + n = c_n$
with
$b_n = a_n + 3$
and the following recursive relation for $a_n$
$a_n = a_{n-1} \cdot a_{n-2} + (n-2),\:\:$ $n \geq 3$
$a_0 = 0, \:\:\:\:a_1 = 1, \:\:\:\:a_2 =2$
$a_1 \cdot b_1 + 1 = c_1$
$a_{2} \cdot b_{2} + 2 = c_{2}$
$a_{3} \cdot b_{3} + 3 = c_{3}$
$a_{4} \cdot b_{4} + 4 = c_{4}$
give us
$1+4=5$
$2+5=12$
$3+6=21$
$8+11=c_4$
and
$c_{4}$
$= a_{4} \cdot b_{4} + 4 = 92$
or to demonstrate the recursion
$c_{4}$
$= (a_{3} \cdot a_{2} + 2) \cdot b_{4} + 4 $
$= (3 \cdot 2 + 2) \cdot 11 + 4$
$= 8 \cdot 11 + 4 = 92$
This results in the following sequence
$1+4=5$
$2+5=12$
$3+6=21$
$8+11=92$
$200 + 203 = 40605$
$4824 + 4827 = 23285454$
$964808 + 964811 = 930857371296$
$\vdots$ $\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\vdots$
Im pretty sure this is what they had in mind :)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1760086",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Probability - die - The number of throws until a $5$ and a $6$ have been obtained. An unbiased die is thrown repeatedly until a 5 and a 6 have been obtained. the random variable M denotes the number of throws required. For example, for the sequence of results 6,3,2,3,6,6,5, the value of M is 7. Calculate P(M=r).
| Let's say the $r$-th one is $6$. Then in the previous $r-1$ throw, you must have at least one $5$. That means at the same time you cannot have $6$. So you can have one $5$ and $(r-2)$ $1-4$, or two $5$ and $(r-3)$ $1-4$, ... etc.
So the probability is
$$P=\frac{1}{6}\sum_{i=1}^{r-1}C^{r-1}_i \left(\frac{1}{6}\right)^i\left(\frac{2}{3}\right)^{r-2-i}$$
$$=\frac{1}{6}\bigg[\left(\frac{1}{6}+\frac{2}{3}\right)^{r-1}-\left(\frac{2}{3}\right)^{r-1}\bigg]$$
$$=\frac{1}{6}\frac{5^{r-1}-4^{r-1}}{6^{r-1}}$$
Finally, switching the role of $5$ and $6$ means doubling the probability, so
$$P=\frac{1}{3}\frac{5^{r-1}-4^{r-1}}{6^{r-1}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1761410",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
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} |
Area enclosed by the curve $\lfloor |x''| \rfloor +\lfloor |y''| \rfloor = 2$
The area enclosed by the curve $$\bigg\lfloor \frac{|x-1|}{|y-1|}\bigg\rfloor +\bigg\lfloor \frac{|y-1|}{|x-1|}\bigg\rfloor = 2\;,$$ Where $-2 \leq x,y\leq 0$
$\bf{My\; Try::}$ Let $x-1=x'$ and $y-1 = y'\;,$ Then $$\bigg\lfloor \frac{|x'|}{|y'|}\bigg\rfloor+\bigg\lfloor \frac{|y'|}{|x'|}\bigg\rfloor = 2\Rightarrow \bigg\lfloor \left|\frac{x'}{y'}\right|\bigg\rfloor+\bigg\lfloor \left|\frac{y'}{x'}\right|\bigg\rfloor=2$$
So here $-3\leq x',y'\leq -1.$ Now Put $\displaystyle \frac{x'}{y'}=x''$ and $\displaystyle \frac{y'}{x'} = y''$ and Here $\displaystyle \frac{1}{3}\leq x'',y''\leq 3$
So we get $$\lfloor |x''| \rfloor +\lfloor |y''| \rfloor = 2$$
Is my Process is Right , If not then how can I calculate it, Help me
Thanks.
|
Is my Process is Right
I think it is right.
We can separate it into cases as the following :
$$\left\lfloor\left|\frac{x-1}{y-1}\right|\right\rfloor+\left\lfloor\left|\frac{y-1}{x-1}\right|\right\rfloor=2$$
$$\iff \left(\left\lfloor\left|\frac{x-1}{y-1}\right|\right\rfloor,\left\lfloor\left|\frac{y-1}{x-1}\right|\right\rfloor\right)=(0,2),(1,1),(2,0)$$
since both $|(x-1)/(y-1)|$ and $|(y-1)/(x-1)|$ are positive.
Case 1 :
$$\begin{align}&\left(\left\lfloor\left|\frac{x-1}{y-1}\right|\right\rfloor,\left\lfloor\left|\frac{y-1}{x-1}\right|\right\rfloor\right)=(0,2)\\&\iff 0\le\left|\frac{x-1}{y-1}\right|\lt 1\quad\text{and}\quad 2\le\left|\frac{y-1}{x-1}\right|\lt 3\\&\iff 0\le\left|\frac{x-1}{y-1}\right|\lt 1\quad\text{and}\quad \frac 13\lt\left|\frac{x-1}{y-1}\right|\le \frac 12\\&\iff \frac 13\lt\left|\frac{x-1}{y-1}\right|\le \frac 12\\&\iff \frac 13\lt\frac{x-1}{y-1}\le \frac 12\\&\iff \frac 13(y-1)\gt x-1\ge \frac 12(y-1)\\&\iff 3x-2\lt y\le 2x-1\end{align}$$
Case 2 :
$$\begin{align}&\left(\left\lfloor\left|\frac{x-1}{y-1}\right|\right\rfloor,\left\lfloor\left|\frac{y-1}{x-1}\right|\right\rfloor\right)=(1,1)\\&\iff 1\le\left|\frac{x-1}{y-1}\right|\lt 2\quad\text{and}\quad 1\le\left|\frac{y-1}{x-1}\right|\lt 2\\&\iff 1\le\left|\frac{x-1}{y-1}\right|\lt 2\quad\text{and}\quad \frac 12\lt\left|\frac{x-1}{y-1}\right|\le 1\\&\iff \left|\frac{x-1}{y-1}\right|=1\\&\iff \frac{x-1}{y-1}=1\\&\iff y=x\end{align}$$
Case 3 : By symmetry about $y=x$,
$$\begin{align}&\left(\left\lfloor\left|\frac{x-1}{y-1}\right|\right\rfloor,\left\lfloor\left|\frac{y-1}{x-1}\right|\right\rfloor\right)=(2,0)\\&\iff 3y-2\lt x\le 2y-1\\&\iff \frac 12x+\frac 12\le y\lt \frac 13x+\frac 23\end{align}$$
Therefore, we want to find the area of the following two triangles in red.
$\qquad\qquad\qquad$
Hence, from $A(-2,0),B(-2,-1/2),C(-1,0)$, the answer is
$$2\times [\triangle{ABC}]=2\times\frac 12\times (-1-(-2))\times (0-(-1/2))=\color{red}{\frac 12}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1763707",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
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} |
f(x) is a real valued function,$ f(0) = 78$ and $f(x+2) -f(x) \leq 3 * 2^x$ and $f(x+6) -f(x) \geq 63 * 2^x$ $f(78) = k-1+ 2^k$. find k Let $f(x)$ be a real valued function such that $ f(0) = 78$ and $f(x+2) -f(x) \leq 3 * 2^x$ and $f(x+6) -f(x) \geq 63 \cdot 2^x$ $f(78) = k-1+ 2^k$. Find $k$.
I calculated $f(2)$, $f(4)$ and $f(6)$. I found the value of the function for $x=6$. What should be done further to solve the problem.
| Using the first inequality three times we get that
\begin{align}
f(x+2) - f(x) &\leq 3\cdot 2^x\\
f(x+4) - f(x+2) &\leq 3\cdot 2^{x+2}\\
f(x+6) - f(x+4) &\leq 3\cdot 2^{x+4}
\end{align}
Adding all these inequalities yields
$$f(x+6)-f(x) \leq 3\cdot(1+2^2+2^4)\cdot 2^x = 63\cdot 2^x. $$
Together with the other given inequality this gives equality, i.e.
$$f(x+6)-f(x) = 63\cdot 2^x,$$
or equivalently
$$f(x+6) = f(x) + 63\cdot 2^x. $$
You can now apply this multiple times starting with the known value when $x=0$ until you reach the value for $f(78) = f(13\cdot 6)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1770131",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to diagonalize a matrix $A$ and then compute $A^{10}$?
Diagonalize the following matrix $$A = \begin{bmatrix} -2 & -8 & -12\\ 1 & 4 & 4\\ 0 & 0 & 1\end{bmatrix}$$ and compute $A^{10}$.
I have found that the eigenvalues of $A$ are $\lambda _1 = 0$, $\lambda _2 = 1$, $\lambda _1 = 2$. Then, I found the following eigenvectors who form my matrix $P$.
$$v_1=\left (\begin{matrix}1\\ 1\\ 1\end{matrix} \right ), \qquad v_2=\left (\begin{matrix}-1/4\\ 0\\ -1/2\end{matrix} \right ), \qquad v_3=\left (\begin{matrix}0\\ -1/4\\ 0\end{matrix} \right )$$
At the end my matrix
$$P=\left ( \begin{matrix}
1 & 1 &1 \\
-\frac{1}{4}& 0 &-\frac{1}{2} \\
0 &-\frac{1}{4} & 0
\end{matrix} \right )$$
In linear algebra, two $n$-by-$n$ matrices $A$ and $B$ are called similar if $B=P^{-1}AP$. I get that
$$B=\left ( \begin{matrix}
0 & 0 & 0\\
0 &1 &0 \\
0& 0 & 2
\end{matrix} \right )$$
So is this all good? How can I use that to get value of $A^{10}$?
| Notice the following: $B = P^{-1}AP \iff A = PBP^{-1}$.
Now consider $A^2$. We see that $A^2 = (PBP^{-1})(PBP^{-1})$. But since $PP^{-1} = I$, this simplifies to $A^2 = PB^2P^{-1}$. Inductively it should be clear that $A^n = PB^nP^{-1}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1771988",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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} |
Eigenvalues and Eigenvectors to find $A\textbf{v}$ If $A$ is a $3\times 3$ matrix with eigenvector $\begin{bmatrix} 3\\ 0\\ -2\end{bmatrix}$ corresponding to an eigenvalue of $5$ and $\begin{bmatrix} -1\\ 2\\ 7\end{bmatrix}$ corresponding to an eigenvalue of $2$, and $\textbf{v}=\begin{bmatrix}10 \\ 4 \\10\end{bmatrix}$ find $A\textbf{v}$.
My attempt at solving this problem was to begin by stating that $Ax=\lambda x$, Where $\lambda$ is the eigenvalue and $x$ is the eigenvector and then saying that $$A=5*(3,0,-2)+2*(-1,2,7)=(13,4,4)=A.$$ Then I followed by saying that $Av=(13,4,4)*(10,4,10)=186$ but I'm sure If I'm moving in the right direction..
| $Ax = λx \Leftrightarrow (Α-λ)x = 0$
Let :
$ A = \left( \begin{array}{ccc}
a_1 & a_2 & a_3 \\
a_4 & a_5 & a_6 \\
a_7 & a_8 & a_9 \end{array} \right) $
Then, $ \left( \begin{array}{ccc}
a_1 & a_2 & a_3 \\
a_4 & a_5 & a_6 \\
a_7 & a_8 & a_9 \end{array} \right)
\left( \begin{array}{ccc}
3 \\
0 \\
-2 \end{array} \right) = 5 \left( \begin{array}{ccc}
3 \\
0 \\
-2 \end{array} \right) $
and
$ \left( \begin{array}{ccc}
a_1 & a_2 & a_3 \\
a_4 & a_5 & a_6 \\
a_7 & a_8 & a_9 \end{array} \right)
\left( \begin{array}{ccc}
-1 \\
2 \\
7 \end{array} \right) = 2 \left( \begin{array}{ccc}
-1 \\
2 \\
7 \end{array} \right) $
Proceed first of all by solving the system that is created by the matrix equations above and you should be able to find the matrix $A$. After finding $A$, do the multiplication $Av$, where $v =[10,4,10]^T$ and find the solution to your exercise.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1774869",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Methods of Characteristics I have a problem solving the ODE associated with the question, any help will be greatly appreciated.
Use method of characteristics to solve the problem
$(x-y)\dfrac{\partial u}{\partial x}+(x+y)\dfrac{\partial u}{\partial y}=\alpha u$
where $\alpha$ is a constant, with initial condition $u(x,0)=x^2,x\geq0$.
By considering your solution explain :
(i) Why initial conditions cannot be specified on the whole $x$-axis
(ii) why a single valued solution is not possible if $\alpha\neq u$
From what I have learnt, I substituted
$\dfrac{dx}{ds}=x-y$ and $\dfrac{dy}{ds}=x+y$ , but I am not sure how to proceed from here.
| If I well understand, they are two questions :
First : solving the PDE . Second : "By considering your solution explain" the two items (i) and (ii).
HINT : Solving the PDE with method of characteristics.
$(x-y)\dfrac{\partial u}{\partial x}+(x+y)\dfrac{\partial u}{\partial y}=\alpha u \quad$ where $\alpha$ is a constant.
$$\frac{dx}{x-y}=\frac{dy}{x+y}=\frac{du}{\alpha u}$$
First characteristic equation from
$\frac{dx}{x-y}=\frac{dy}{x+y}=\frac{xdx+ydy}{x^2+y^2}=\frac{du}{\alpha u}$
$\frac{1}{2}\ln|x^2+y^2|=\frac{1}{\alpha}\ln|u|+$constant.
$$(x^2+y^2)^{\alpha /2}u=c_1$$
Second characteristic equation from $\frac{dx}{x-y}=\frac{dy}{x+y}$
In order to solve this homogeneous ODE, change of function : $y=xf(x)$
$\frac{dx}{x-xf}=\frac{fdx+xdf}{x+xf}\quad\to\quad \frac{1-f}{1+f^2}f'=x$
$x^2+\ln(f^2+1)-2\tan^{-1}(f)=c_2$
$$x^2+\ln\left(\frac{y^2}{x^2}+1\right)-2\tan^{-1}\left(\frac{y}{x}\right)=c_2$$
The general solution of the PDE, expressed on implicit form is :
$$\Phi\left( \left((x^2+y^2)^{\alpha /2}u\right) \: , \: \left(x^2+\ln\left(\frac{y^2}{x^2}+1\right)-2\tan^{-1}\left(\frac{y}{x}\right)\right) \right)=0$$
where $\Phi(X,Y)$ is any differentiable function of two variables.
Solving the implicit equation for $X$ leads to the explicit equation :
$(x^2+y^2)^{\alpha /2}u=F\left(x^2+\ln\left(\frac{y^2}{x^2}+1\right)-2\tan^{-1}\left(\frac{y}{x}\right) \right)$
where $F$ is any differentiable function.
$$u(x,y)=(x^2+y^2)^{-\alpha /2}F\left(x^2+\ln\left(\frac{y^2}{x^2}+1\right)-2\tan^{-1}\left(\frac{y}{x}\right) \right)$$
According to the boundary condition : $u(x,0)=x^2=(x^2)^{-\alpha /2}F\left(x^2\right) \quad\to\quad F(x^2)=(x^2)^{1+\frac{\alpha}{2}}$
$$u(x,y)=(x^2+y^2)^{-\alpha /2} \left(x^2+\ln\left(\frac{y^2}{x^2}+1\right)-2\tan^{-1}\left(\frac{y}{x}\right) \right)^{1+\frac{\alpha}{2}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1775321",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Evaluating an indefinite cosine integral Evaluate $$\int\frac{1+2\cos x}{(\cos x+2)^2} dx$$ The problem looks easy but I don't think it is. Please just don't give hints for possible substitutions as I myself have tried quite many and all of them are just taking too long to solve. Thanks.
| $$\int \frac { 1+2\cos x }{ (\cos x+2)^{ 2 } } dx=2\int { \frac { \cos ^{ 2 }{ x } +\sin ^{ 2 }{ x } +2\cos { x } }{ { \left( \cos x+2 \right) }^{ 2 } } } dx=\int { \frac { \cos { x } \left( \cos { x } +2 \right) +\sin ^{ 2 }{ x } }{ { \left( \cos x+2 \right) }^{ 2 } } } dx=\\ =\int { \frac { { { \left( \sin { x } \right) }^{ \prime }\left( \cos { x } +2 \right) -\sin { x{ \left( \cos { x } +2 \right) }^{ \prime } } } }{ { \left( \cos x+2 \right) }^{ 2 } } } dx=\int { d\left( \frac { \sin { x } }{ { \cos x+2 } } \right) } =\frac { \sin { x } }{ { \cos x+2 } } +C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1775673",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Show that $\pi=\frac{22}{7}-\frac{\pi}{4}\int_0^{\infty}\frac{e^{-2x\pi}\left( 1-e^{-\frac{x\pi}{2} } \right)^4}{\cosh\left(\frac{x\pi}{2}\right)}dx$ An integral from maths world; pi formulas (50),
$$\pi=\frac{22}{7}-\int_0^1\frac{x^4(1-x)^4}{1+x^2}dx$$
We found another similar integral to it, via experimental with wolfram integrator,
$$\pi=\frac{22}{7}-\frac{\pi}{4}\int_0^{\infty}\frac{e^{-2x\pi}\left( 1-e^{-\frac{x\pi}{2} } \right)^4}{\cosh\left(\frac{x\pi}{2}\right)}dx$$
Can anyone help us to prove it?
| Note that $\displaystyle\cosh x=\frac{e^x+e^{-x}}{2}$. Let
$\displaystyle t=e^{-\frac{x\pi}{2}}$, then $\displaystyle dt=-\frac{\pi}{2}e^{-\frac{x\pi}{2}}dx$. Also, we have $t=1$ as $x=0$, and $t\to0$ as $x\to\infty$. Therefore
\begin{align}
\frac{\pi}{4}\int_0^\infty\frac{e^{-2x\pi}\left( 1-e^{-\frac{x\pi}{2} } \right)^4}{\cosh\left(\frac{x\pi}{2}\right)}dx
&=\frac{\pi}{4}\int_0^\infty\frac{e^{-2x\pi}\left( 1-e^{-\frac{x\pi}{2} } \right)^4}{\frac{e^{\frac{x\pi}{2}}+e^{-\frac{x\pi}{2}}}{2}}dx\\
&=\frac{\pi}{2}\int_0^\infty\frac{e^{-2x\pi}\left( 1-e^{-\frac{x\pi}{2} } \right)^4}{e^{\frac{x\pi}{2}}+e^{-\frac{x\pi}{2}}}\left(-\frac{2}{\pi}e^{\frac{x\pi}{2}}\right)\left(-\frac{\pi}{2}e^{-\frac{x\pi}{2}}dx\right)\\
&=-\int_0^\infty\frac{e^{-2x\pi}\left( 1-e^{-\frac{x\pi}{2} } \right)^4}{1+e^{-x\pi}}\left(-\frac{\pi}{2}e^{-\frac{x\pi}{2}}dx\right)\\
&=-\int_1^0\frac{t^4\left( 1-t\right)^4}{1+t^2}dt\\
&=\int_0^1\frac{t^4\left( 1-t\right)^4}{1+t^2}dt.
\end{align}
| {
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"url": "https://math.stackexchange.com/questions/1775865",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
For $a,b,c$ positive real numbers can it be true that: $(ab+bc+ca)^3 \ge (a^2+2b^2)(b^2+2c^2)(c^2+2a^2)$ For $a,b,c$ positive real numbers can it be true that: $$(ab+bc+ca)^3 \ge (a^2+2b^2)(b^2+2c^2)(c^2+2a^2)$$
It really seems unlikely because it reminds me the rearrangement inequality when the direction is flipped.
| $\prod\limits_{cyc}(a^2+2b^2)\geq(ab+ac+bc)^3\Leftrightarrow\sum\limits_{cyc}(2a^4b^2+4a^4c^2-a^3b^3-3a^3b^2c-3a^3c^2b+a^2b^2c^2)\geq0$,
which is true by Schur and AM-GM:
$\sum\limits_{cyc}(2a^4b^2+4a^4c^2-a^3b^3-3a^3b^2c-3a^3c^2b+a^2b^2c^2)\geq$
$\geq\sum\limits_{cyc}(2a^4c^2+3a^3b^3-3a^3b^2c-3a^3c^2b+a^2b^2c^2)=$
$=3\sum\limits_{cyc}(a^3b^3-a^3b^2c-a^3c^2b+a^2b^2c^2)+2\sum\limits_{cyc}(a^4c^2-a^2b^2c^2)\geq0$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluating the rational integral $\int \frac{x^2+3}{x^6(x^2+1)}dx $
Evaluate $$\int \frac{x^2+3}{x^6(x^2+1)}dx .$$
I am unable to break into partial fractions so I don't think it is the way to go. Neither is $x=\tan \theta$ substitution. Please give some hints. Thanks.
| That rational expression, like all rational expressions, can be broken into partial fractions. You may be making the common mistake of trying to change the expression to
$$\frac{x^2+3}{x^6(x^2+1)}=\frac{A}{x^6}+\frac{Bx+C}{x^2+1}$$
However, you need to include all the lesser powers of the irreducible factors. In other words, you want
$$\frac{x^2+3}{x^6(x^2+1)}=\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x^3}+\frac{D}{x^4}+\frac{E}{x^5}+\frac{F}{x^6}+\frac{Gx+H}{x^2+1}$$
Try that, then come back if you cannot finish.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1776507",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Use the sine Fourier series for $x$ and $x^2$ to show $1-\frac{1}{3^3} + \frac{1}{5^3} - \frac{1}{7^3} + ... = \frac{\pi ^3}{32}$ We were given that $x=2(\sin x - \frac{\sin 2x}{2}+ \frac{\sin 3x}{3}-\dots)$ and I computed that $x^2=\sum_{n=1}^{\infty}\frac{(-1)^n(4-2\pi ^2 n^2)-4}{\pi n^3}\sin(nx)$.
How can I use these two to show that $1-\frac{1}{3^3} + \frac{1}{5^3} - \frac{1}{7^3} + \dots = \frac{\pi ^3}{32}$
| HINT:
First show that the function $f(x)=x^3-\pi^2 x$ has Fourier Series representation for $x\in [-\pi,\pi]$
$$x^3-\pi^2 x=12 \sum_{n=1}\frac{(-1)^n}{n^3}\sin(nx) \tag 1$$
Then, let $x=\pi/2$.
SPOILER ALERT: Scroll over the highlighted area to reveal the solution
We can find the Fourier series for $x^3$ by integrating the given Fourier Series for $x$. To that end, we proceed. First, note that the Fourier Series representation of $x$ on $[-\pi,\pi]$ is $$x=2\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n}\,\sin(nx) \tag 1$$Integrating $(1)$ we find that $$\frac12 x^2 =2\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^2}\, (1-\cos(nx)) \tag 2$$Integrating $(2)$, we find that $$\frac16 x^3=2x\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^2}\, -2\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^3}\, \sin(nx) $$Recall that $\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^2}=\frac{\pi^2}{12}$ (this can easily be shown by looking at the even and odd components of the Basel Series $\sum_{n=1}^\infty\frac1{n^2}$). Then, rearranging terms yields $(1)$. Letting $x=\pi/2$, we obtain the coveted identity $$\sum_{n=1}^\infty \frac{(-1)^{n-1}}{(2n-1)^3}=\frac{\pi^3}{32}$$
| {
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"url": "https://math.stackexchange.com/questions/1780126",
"timestamp": "2023-03-29T00:00:00",
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Solution to $\sqrt{x-2} = 3- 2\sqrt{ x}$ The above question is from Serge Lang's basic mathematics. The question asks if there are any values of x which satisfy the above equation.
Serge Lang's answer key states that there is no solution.
From the equation, I squared both sides, then used the quadratic formula to find $\sqrt{ x} = 2-\sqrt{1/3}$ or $2+\sqrt{1/3}$.
I substituted the former into the equation, and found that it would equate $\sqrt{x-2}$ to a negative answer, and hence is wrong since $\sqrt{x-2}$ refers to the positive root of $x-2$.
For the second answer, I substituted and found that both sides of the equation equaled out. I drew $y=\sqrt{x-2}$ and $y=\sqrt{x}$ in a graphing calculator, and found they intersected correctly at a point.
Yet the book states there are no solutions. What went wrong?
| Set $\sqrt{x}=t$, so the equation becomes $\sqrt{t^2-2}=3-2t$. We have some conditions, though: we need $t\ge0$, $t^2\ge2$ and $3-2t\ge0$, that can be put together as $\sqrt{2}\le t\le 3/2$.
Now we can square and get $t^2-2=9-12t+4t^2$, so the quadratic
$$
3t^2-12t+11=0
$$
whose roots are
$$
\frac{6-\sqrt{3}}{3},\qquad\frac{6+\sqrt{3}}{3}
$$
Note that
$$
\sqrt{2}<\frac{6-\sqrt{3}}{3}<\frac{3}{2}<\frac{6+\sqrt{3}}{3}
$$
so $t=(6-\sqrt{3})/3$ is a solution, that translates to
$$
x=\left(\frac{6-\sqrt{3}}{3}\right)^2=\frac{13-4\sqrt{3}}{3}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1784486",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Show that $\frac{d^n}{dx^n}\left[ \frac{1}{1+x^2} \right] = \frac{(-1)^nn!}{(x^2+1)^{\frac{n+1}{2}}}\sin[(n+1)\arctan x]$ We have, $$\frac{1}{1+x^2}=\frac{A}{x-i}+\frac{B}{x+i} $$
From this we get $A=\frac{1}{2i}, B=-\frac{1}{2i}$.
So,
$$\frac{1}{1+x^2}=\frac{1}{2i}\left[\frac{1}{x-i} + \frac{1}{x+i} \right] $$
Now,
\begin{align*}\left(\frac{1}{1+x^2}\right)^{(n)}&=\frac{1}{2i}\left[\left(\frac{1}{x-i} \right)^{(n)} + \left( \frac{1}{x+i}\right)^{(n)} \right] \\
&= \frac{1}{2i}\left[ \frac{(-1)^{n}n!}{(x-i)^{n+1}} - \frac{(-1)^{n}n!}{(x+i)^{n+1}}\right] \\
&= \frac{(-1)^{n}n!}{2i} \left( \frac{(x+i)^{n+1}-(x-i)^{n+1}}{(x^2+1)^{n+1}}\right)
\end{align*}
Then we have, $$ (x+i)^{n+1}=(x^2+1)^{\frac{n+1}{2}}\left(\cos((n+1)\cdot \mathrm{arccot}x) + i \sin((n+1)\mathrm{arccot}x) ) \right)$$
$$ (x-i)^{n+1}=(x^2+1)^{\frac{n+1}{2}}\left(\cos((n+1)\cdot \mathrm{arccot}(-x)) + i \sin((n+1)\mathrm{arccot}(-x)) ) \right)$$
But this is where I'm stuck, because we are not winning $\arctan$, maybe it is some elementary trygonometry identity that I'm missing.
| Because of $x+i=\sqrt{x^2+1}e^{i \arctan(x)}$ we get $\frac{1}{x+i}=\frac{1}{\sqrt{x^2+1}}e^{-i \arctan(x)}$. Then continue as you have done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1786053",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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$q$-digamma function evaluation What is the value of $\psi_2^{(0)}(1)$, where $\psi_q^{(0)}(z)$ is the $q$-digamma function?
My attempt:
\begin{align*}
\psi_2^{(0)}(z) &=\frac{1}{\Gamma_2(z)}\frac{d\Gamma_2(z)}{dz}
\\&=\frac{(2^z\ ;2)_\infty}{(2\ ;2)_\infty(1-2)^{1-z}}\cdot\frac{d}{dz}\left((1-2)^{1-z}\frac{(2\ ;2)_\infty}{(2^z\ ;2)_\infty}\right)
\\
&=\frac{1}{(-1)^{1-z}}\prod_{n=0}^\infty\frac{1-2^{n+z}}{1-2^{n+1}}\cdot\frac{d}{dz}(-1)^{1-z}\prod_{n=1}^\infty\frac{1-2^{n+1}}{1-2^{n+z}},
\end{align*}
where $\Gamma_q(z)$ is the q-gamma function, and $(a\ ;q)_k$ is the q-Pochhammer symbol. Evaluating at $1$, we see
$$\begin{align*}
\psi_2^{(0)}(1)&=\frac{1}{(-1)^0}\prod_{n=0}^\infty\frac{1-2^{n+1}}{1-2^{n+1}}\cdot\frac{d}{dz}(-1)^{1-z}\prod_{n=1}^\infty\frac{1-2^{n+1}}{1-2^{n+z}}\Bigr|_{z=1}
\\&=\frac{d}{dz}(-1)^{1-z}\prod_{n=1}^\infty\frac{1-2^{n+1}}{1-2^{n+z}}\Bigr|_{z=1}
\\&=\frac{d}{dz}\exp\left((\ln(-1)^{1-z}+\sum_{n=0}^\infty\ln\left(\frac{1-2^{n+1}}{1-2^{n+z}}\right)\right)\Bigr|_{z=1}
\\&=\exp\left(\ln(-1)^{1-z}+\sum_{n=0}^\infty\ln\left(\frac{1-2^{n+1}}{1-2^{n+z}}\right)\right)\left(i\pi+\sum_{n=0}^\infty\frac{2^{n+z}\ln2}{1-2^{n+z}}\right)\Bigr|_{z=1}
\\&=i\pi+\ln2\sum_{n=0}^\infty\frac{2^{n+1}}{1-2^{n+1}}
\end{align*}$$
which clearly diverges.... and even if it converged, I would have an imaginary part. Wolfram Alpha gives only an approximation of about $-0.7671026$, so what am I missing here?
| Consider $D_{x} a^{x}$:
\begin{align}
D_{x} a^{x} &= D_{x} e^{x \, ln(a)} \\
&= \ln(a) \, e^{x \, ln(a)} \\
&= a^{x} \, \ln(a).
\end{align}
Now,
\begin{align}
\frac{d}{dz}\left((1-q)^{1-z}\frac{(q\ ;q)_\infty}{(q^z\ ;q)_\infty}\right) &= (1-q)^{1-x} \, \frac{d}{dx} \left( \frac{(q\ ;q)_\infty}{(q^z\ ;q)_\infty}\right) - (1-q)^{1-x} \, \ln(1-q) \, \frac{(q\ ;q)_\infty}{(q^z\ ;q)_\infty}
\end{align}
This is the process presented by the proposer. It is correct. What must be considered now is $0 < q <1$, $q > 1$. For these cases see formulas (1.4) and (1.5) of Some inequalities of the q-Digamma function
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Compute the minimum value of $a^n + b^n + c^n$ subject to $a^2 + b^2 + c^2 = 1 $
Assume that $a,b,c$ are non-negative real numbers and $n$ is a natural number $n \ge 3$. What is $f(n)=$ the minimum value of $a^n + b^n + c^n$ ?
I find ;
$$f(3) = \frac{1}{\sqrt{3}}\qquad f(4) = \frac{1}{3}$$
then I guess
$$f(n) = \left(\frac{1}{\sqrt{3}} \right)^n \times 3 $$
Is that true?
| Set $(x_1, x_2, x_3) = (a, b, c)$. Applying Holder's inequality to $(x_1^2, x_2^2, x_3^2)$ and $ (1, 1, 1)$
with
$$
p = \frac n2 \, , \quad q = \frac{n}{n-2}
$$
gives
$$
1 = \sum_{k=1}^3 x_k^2 \cdot 1 \le \left( \sum_{k=1}^3 x_k^n \right)^{\frac 2n} \cdot \left( \sum_{k=1}^3 1 \right)^{\frac {n-2}{n}} \\
= \left( \sum_{k=1}^3 x_k^n \right)^{\frac 2n} \cdot 3 ^{\frac {n-2}{n}}
$$
which implies
$$
x_1^n + x_2^n + x_3^n\ge \left(\frac 13 \right)^{\frac {n-2}{2}} =
\frac{3}{(\sqrt 3)^n}
$$
and equality holds for $x_1 = x_2 = x_3 = \frac{1}{\sqrt 3}$,
so this is the minimum value, as you conjectured.
| {
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"timestamp": "2023-03-29T00:00:00",
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express the value of an expression as a common factor In the following problem, by adding $0.141414$..., $0.414141..., 0.151515...$, and $0.515151...$, I get $1.111....$ Then the expression becomes square root ($11 \times 1.1111$....). My answer is $11 \times \sqrt{0.1010101...}$, which is different from the answer sheet. Any help is highly appreciated.
What is the value of the following expression? Express your answer as a common fraction.
| I like @ncmathsadist's suggestion. If you can reason that $0.\overline{14} + 0.\overline{41} = 0.\overline{5}$ and $0.\overline{15} + 0.\overline{51} = 0.\overline{6}$ Then think about it:
\begin{align}
&0.555 \\
+&0.666 \\\hline
&1.221
\end{align}
Clearly if it repeats forever, $0.\overline{5} + 0.\overline{6} = 1.\overline{2}$. Then we can solve by doing the following:
$$
x = 1.\overline{2} \\
10x = 12.\overline{2}\\
10x - x = 9x \rightarrow 9x = 12 - 1 = 11
$$
Therefore $x = 1.\overline{2} = \frac{11}{9}$. Then we have:
$$
\sqrt{11\cdot\frac{11}{9}} = \sqrt{\frac{11^2}{3^2}} = \frac{11}{3} = 3\frac{2}{3}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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Two real numbers, $x$ and $y$, satisfy the condition $x + y = 2 $. Show $xy(x^2+y^2) \leq 2$
Question: Two real numbers, $x$ and $y$, satisfy the condition $x + y = 2 $.
Show $xy(x^2+y^2) \leq 2$
What I have attempted:
Consider $$x+y=2$$
$$ \Leftrightarrow (x+y)^2 = 2^2 $$
$$ \Leftrightarrow (x+y)^2 = 4 $$
Notice that $$ x^2 + y^2 = (x+y)^2 - 2xy $$
But $(x+y)^2 = 4$
$$ \Longrightarrow x^2 + y^2 = 4-2xy $$
Now we have this inequality; $$xy(x^2+y^2) \leq 2$$
$$\Leftrightarrow xy(4-2xy) \leq 2 $$
$$ \Leftrightarrow 2xy(2-xy) \leq 2 $$
$$ \Leftrightarrow 2(2xy-x^2y^2) \leq 2 $$
Now I am stuck, how should I continue also is there another way to approach this question?
| Hint: Consider $f(x)= x(2-x)(x^2+(2-x)^2)$. Then $f'(x)=-8 (x-1)^3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1789868",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 2
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Finding taylor expansion of $\cos^2x$ and $\sin^2x$ My task is this:
Find the taylor-series of $\cos^2x$ and $\sin^2x$.
My work so far:
We know that $\cos^2x \backslash \sin^2x = \frac{1\pm \cos 2x}{2}$, and the series for $\cos x = \sum_{n=0}^\infty\frac{(-1)^nx^{2n}}{(2n)!} \implies \cos 2x = \sum_{n=0}^\infty\frac{(-1)^n(2x)^{2n}}{(2n)!}$. Which leads us to $$\cos^2x \backslash\sin^2x = \frac{1\pm\sum_{n=0}^\infty\frac{(-1)^n(2x)^{2n}}{(2n)!}}{2} = \frac{1}{2}\left(1\pm\sum_{n=0}^\infty\frac{(-1)^n(2x)^{2n}}{(2n)!}\right).$$
Is it correct?
| Since you know the Taylor expansion of $\sin x$ and $\sin^2x = (\sin x)^2$ you can use the Cauchy product:
\begin{align*}
\sin^2x &= \left(\sum_{k=0}^\infty \frac{(-1)^k}{(2k+1)!}x^{2k+1}\right)^2 \\
&= \sum_{k=0}^\infty c_k,
\end{align*}
where
\begin{align*}
c_k &= \sum_{i=0}^k \frac{(-1)^i}{(2i+1)!}\frac{(-1)^{k-i}}{(2k - 2i + 1)!} x^{2i+1}x^{2k-2i + 1} \\
& = (-1)^kx^{2k+2}\sum_{i=0}^k \frac{1}{(2i+1)!(2k+2-(2i+1))!} \\
& = \frac{(-1)^kx^{2k+2}}{(2k+2)!}\sum_{i=0}^k \frac{(2k+2)!}{(2i+1)!(2k+2-(2i+1))!} \\
& = \frac{(-1)^kx^{2k+2}}{(2k+2)!}\sum_{i=0}^k \binom{2k+2}{2i+1} \\
& = \frac{(-1)^kx^{2k+2}}{(2k+2)!} 2^{2k+1} =\frac12 \frac{(-1)^k(2x)^{2k+2}}{(2k+2)!}
\end{align*}
Putting it together yields
\begin{align*}
\sin^2x &= \frac12\sum_{k=0}^\infty \frac{(-1)^k (2x)^{2k+2}}{(2k+2)!}
\end{align*}
By shifting the summation index down one, we arrive at the very same thing that you derived (maybe a bit cleverer than me).
Analog $\cos^2x$ can be done.
| {
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"url": "https://math.stackexchange.com/questions/1792681",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Show geometrically or algebraically $(\sqrt2-1)a+(\sqrt3-1)bPythagoras theorem
$$a^2+b^2=c^2$$
Show geometrically (Addressing to @blue a Trigonographer)
(1)$$(\sqrt2-1)a+(\sqrt3-1)b<c$$
Or algebraically (general users)
(@BLue the Trigonographer)
Expand (1)
$a\sqrt2+b\sqrt3<a+b+c$
Letting P=a+b+c be the perimeter of triangle ABC and where $A_1=a\sqrt2$, $A_2=b\sqrt3$ are two areas.
We can say $A_1+A_2<P$
I just wonderly can it be construct geometrically to show this inequality.
I have seem on his site, an amazing diagrams and beautiful proof via diagrams
| Using Schwarz inequality we get $$\left(\sqrt 2 - 1\right)a + \left(\sqrt 3 - 1\right)b \le \sqrt{\left(\left(\sqrt 2 - 1\right)^2 + \left(\sqrt 3 - 1\right)^2\right)\left(a^2+b^2\right)} = \sqrt{7 - 2\sqrt 2 - 2 \sqrt 3}c.$$
Therefore it is enough to check that $7 - 2\sqrt 2 - 2\sqrt 3 < 1$ which I leave as an exercise.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1792775",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
General Principles of Solving Radical Equations What are the general ways to solve radical equations similar to questions like
$\sqrt{x+1}+\sqrt{x-1}-\sqrt{x^2 -1}=x$ $\sqrt{3x-1}+\sqrt{5x-3}+\sqrt{x-1}=2\sqrt2$$\sqrt{\frac{4x+1}{x+3}}-\sqrt{\frac{x-2}{x+3}}=1$
Are there just a few known ways to solve them? How do you know the best way to solve such questions? I have trouble with a lot of square root equations, and when I ask them on this site, I get good answers, but for one question. I was wondering if there were any general principles of solving such questions.
| I would write the equation in the form $$\sqrt{x+1}+\sqrt{x-1}=x+\sqrt{x^2-1}$$
with $$x\geq 1$$ after squaring one times and isolating the square root we get
$$2\sqrt{1-x^2}(1-x)=2x^2-2x-1$$ squaring again we obtain
$$(x^2-1)(2-2x)^2=(2x^2-2x-1)^2$$ this gives the equation $$4x-5=0$$ and we get $$x=\frac{5}{4}$$ fulfills our equation.
in the second equation we get $$x=1$$
write your third equation in the form
$$\sqrt{\frac{4x+1}{x+3}}=1+\sqrt{\frac{x-2}{x+3}}$$ we get after squaring two times
$$\left(\frac{2x}{x+3}\right)^2-4\left(\frac{x-2}{x+3}\right)=0$$
after simplifying we obtain$$-4\,{\frac {x-6}{ \left( x+3 \right) ^{2}}}=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1793591",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
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Conic Sections: Hyperbola (Finding the Locus) This is a multipart question so bear with me until I get to the part where I am stuck on.
$H$: $xy=c^2$ is a hyperbola
(i) Show that $H$ can be represented by the parametric equations $x=ct$ , $y= \frac{c}{t}$.
If we take $y= \frac{c}{t}$ and rearrange it to $t= \frac{c}{y}$ and subbing this into $x=ct$
$$ x = c(\frac{c}{y}) $$
$$ \therefore xy = c^2 $$
(ii) Find the gradient of the normal to $H$ at the point $T$ with the coordinates $(ct, \frac{c}{t})$
As $xy = c^2$
$$ \Leftrightarrow y = c^2 x^{-1} $$
$$ \Leftrightarrow y' = -\frac{c^2}{x^2} $$
$$ \Leftrightarrow y'_{x=ct} = \frac{-1}{t^2} $$
Hence gradient of the normal is $m = t^2$
The normal to H at the point $T$ meets $H$ again at $P(cp , \frac{c}{p})$
(iiI) By finding the gradient of the line TP and comparing it with the gradient from (ii), find a relationship between $t$ and $p$
So repeating steps of (ii) again
$xy = c^2$
$$ \Leftrightarrow y = c^2 x^{-1} $$
$$ \Leftrightarrow y' = -\frac{c^2}{x^2} $$
$$ \Leftrightarrow y'_{x=cp} = \frac{-1}{p^2} $$
Hence gradient of the normal is $$m = p^2$
Comparing gradients
$$ p^2 = t^2 $$
$$ \Leftrightarrow |p| = |t| $$ Now do we assume $t>0$ and $p>0$? which gives us
$$ \Leftrightarrow p = t $$ as our relationship.
Now this is where I am mostly stuck on
(iv) Hence or otherwise by finding the parametric equations of $M$, the midpoint of $TP$ in terms of $t$, show that the coordinates of $M$ lie on the curve $4x^3y^3 + c^2(x^2-y^2)^2 = 0$
Note: This can also be stated as "Find the Locus of M"
So basically what I have done is to find the normal line of TP
Which is given as
$$ (y-y_1)=m(x-x_1) $$
Because they want everything in terms of $t$ , let $p=t$
$$ (y-\frac{c}{t}) = t^2(x-ct) $$
$$ \Leftrightarrow (y-\frac{c}{t}) = t^2x - ct^3 $$
$$ \Leftrightarrow y = t^2x - ct^3 + \frac{c}{t} $$
Now this is stuck how should I continue?
| Part (iii):
As pointed out by @mathlove, $$p=-1/t^3$$.
Part (iv): Coordinates of $M$ are given by
$$x=\frac 12 \left(ct-\frac c{t^3}\right)\qquad\Rightarrow \frac {2x}c=t-\frac 1{t^3}=X\\
y=\frac 12 \left(\frac ct-ct^3\right)\qquad\Rightarrow \frac {2y}c=\frac 1t-t^3=Y\\\\
\frac XY=-\frac 1{t^2}\\
XY=2-t^4-\frac 1{t^4}=-\left(t^2-\frac 1{t^2}\right)^2=-\left(-\frac YX+\frac XY\right)^2\\
(XY)^3=-(X^2-Y^2)^2\\
\left(\frac {2x}c\cdot \frac {2y}c\right)^3=-\left(\left(\frac{2x}c\right)^2-\left(\frac{2y}c\right)^2\right)^2\\
4x^3y^3=-c^2(x^2-y^2)^2\qquad\blacksquare$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find all $x,y,z \in \mathbb{Z^{+}}$ such that $20^x+15^y=2015^z$ Find all $x,y,z \in \mathbb{Z^{+}}$ such that $20^x+15^y=2015^z$
I was checking modulo $4$ to see that $y,z$ must have the same parity.
Then took two cases when both $y,z$ are even and when both are odd. But it is difficult to find solutions
| $20^x+15^y=2015^z$ equiv. $4^x5^x+ 3^y 5^y=403^z 5^z$.
We get $x=y$ or $x=z$ or $y=z$ because of divisibility by 5 to a certain power.
(Suppose, let's say, $x < y < z$ then $4^x =403^z 5^{z-x} - 3^y 5^{y-x}$ impossible because the left side is multiple of 5)
Case 1) $x=y$
This case $z \ge x$ (otherwise $(3^x + 4^x)5^{x-z}=403^z$, impossible because $403$ is not divisible by $5$)
and $3^x + 4^x=403^z5^{z-x} \tag1$
But $3^x + 4^x < 2\cdot4^x< 403^z$ so (1) cannot hold
Case 2) $y=z$
This case $x \ge y$ and $3^x5^{x-y} + 4^y=403^y \tag2$
From (2) $ 3^x5^{x-y}=403^y - 4^y=399 \cdot A = 3\cdot7\cdot19\cdot A \tag3$
So the equality (3) cannot hold.
Case 3) $x=z$
This case $y \ge x$ and $3^x + 4^y5^{y-x}=403^x \tag4$
Applying modulo $3$ to (4) we get $2^{y-x}=1 \mod 3$ so $y-x=2k, k \in \mathbb{N}$.
Also from (4) $ 4^y5^{y-x}=403^x - 3^x=400 \cdot A \tag5=4^25^2\cdot A$
So far, this case I couldn't get further.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1797074",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Fake proofs using matrices Having gone through the 16-page-list of questions tagged fake-proofs, and going though both the relevant MSE Question and Wikipedia page, I didn't find a single fake proof that involved matrices.
So the question (or challange) here is: what are some fake proof using matrices?
In particular, the fake proof should use a property, an operation, ..., specific to matrices (or at least not present in $\mathbb{R}$ or $\mathbb{C}$), e.g.
*
*Noncommutativity
*(Non-)existence of an inverse
*Matrix sizes
*Operations as $\det$, $\text{trace}$, ...
*Eigenvalues and diagonalization
*Matrix decompositions and normal forms
*...
Note: It does not matter if the result being "proven" is correct or not. The fallacy in the proof itself is what matters.
Examples:
*
*Proof that 1 = 0
Proof: it is a well-known fact that $(x+y)^2 = x^2 + 2xy + y^2$.
Now let
$$x = \begin{pmatrix}0 & 1\\ 0 & 0 \end{pmatrix},\;\;y = \begin{pmatrix}1 & 0\\ 0 & 0 \end{pmatrix}.$$
On the one hand, we have that
$$ (x+y)^2 = \begin{pmatrix}1 & 1\\ 0 & 0 \end{pmatrix}^2 = \begin{pmatrix}1 & 1\\ 0 & 0 \end{pmatrix},$$
on the other hand we have
$$x^2 + 2xy + y^2 = \begin{pmatrix}0 & 0\\ 0 & 0 \end{pmatrix} + 2\begin{pmatrix}0 & 0\\ 0 & 0 \end{pmatrix} + \begin{pmatrix}1 & 0\\ 0 & 0 \end{pmatrix} = \begin{pmatrix}1 & 0\\ 0 & 0 \end{pmatrix}.$$
Since two matrices are equal if and only if all their entries are equal, we conclude that $1 = 0$.
The mistake here is that $x$ and $y$ do not commute. Thus $(x+y)^2 = x^2 + xy + yx + y^2 \neq x^2 + 2xy + y^2$.
*Proof that 2 = 0
Proof: We know that $\det (AB) = \det (BA)$, since $$\det (AB) = (\det A) (\det B) = (\det B) (\det A) = \det (BA).$$
Now consider the matrices
$$ A = \begin{pmatrix}1 & 0 & 1\\ 0 & 1 & 0 \end{pmatrix}, \; \; B = \begin{pmatrix}1 & 0\\ 0 & 1\\ 1 & 0 \end{pmatrix}.$$
We have that
$$AB = \begin{pmatrix}2 & 0\\ 0 & 1 \end{pmatrix}, \;\; BA = \begin{pmatrix}1 & 0 & 1\\ 0 & 1 & 0 \\1 & 0 & 1\end{pmatrix}.$$
Hence $\det (AB) = 2$ and $\det (BA) = 0$, therefore $2 = 0$.
The mistake here is that $\det$ is defined for square matrices only, and thus $\det AB = \det BA$ only holds in general if $A$ and $B$ are square.
| Here is a fake proof of a true theorem, Cayley-Hamilton.
So the characteristic polynomial of a matrix $A$ is $f_A(t) = \textrm{det}(tI-A)$. Then $f_A(A) = \textrm{det}(AI-A) = \textrm{det}(A-A) = \textrm{det}(0) = 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1801950",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "28",
"answer_count": 2,
"answer_id": 1
} |
Chain rule to differentiate $\sin ^2\frac{x}{2}$ I have this equation $$\sin ^2(\frac{x}{2})$$
Using the chain rule $
M'(N(x)).N'(x)$:
$$\begin{align*}
&M= (\sin \frac{x}{2})^2 \\
&N= \frac{x}{2}\end{align*}$$
That makes
$$2\sin \frac{x}{2}*\frac{1}{2}$$ or $$\sin \frac{x}{2}$$
Going again, we should have
$$\frac{1}{2} \cos \frac{x}{2}$$ or $$\frac{\cos \frac{x}{2}}{2}$$
Yet solution is giving me
$$(\sin \frac{x}{2})(\cos \frac{x}{2})$$
Why...?
| Let $u=sin(\frac{x}2)$
Apply the chain rule we get $\frac{d}{dx}sin^2(\frac{x}2)=\frac{d}{du}(u^2)\frac{d}{dx}(sin(\frac{x}2))$
$\frac{d}{du}(u^2)=2u$
$\frac{d}{dx}(sin(\frac{x}2))=\frac12cos(\frac{x}2)$
So $$\frac{d}{du}(u^2)\frac{d}{dx}(sin(\frac{x}2))=2u(\frac12)cos(\frac{x}2)=ucos(\frac{x}2)=sin(\frac{x}2)cos(\frac{x}2)$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How can you prove that $1+ 5+ 9 + \cdots +(4n-3) = 2n^{2} - n$ without using induction? Using mathematical induction, I have proved that
$$1+ 5+ 9 + \cdots +(4n-3) = 2n^{2} - n$$
for every integer $n > 0$.
I would like to know if there is another way of proving this result without using PMI. Is there any geometric solution to prove this problem? Also, are there examples of problems where only PMI is our only option?
Here is the way I have solved this using PMI.
Base Case: since $1 = 2 · 1^2 − 1$, the formula holds for $n = 1$.
Assuming that the
formula holds for some integer $k ≥ 1$, that is,
$$1 + 5 + 9 + \dots + (4k − 3) = 2k^2 − k$$
I show that
$$1 + 5 + 9 + \dots + [4(k + 1) − 3] = 2(k + 1)^2 − (k + 1).$$
Now if I use hypothesis I observe.
$$
\begin{align}
1 + 5 + 9 + \dots + [4(k + 1) − 3]
& = [1 + 5 + 9 + \dots + (4k − 3)] + 4(k + 1) −3 \\
& = (2k^2 − k) + (4k + 1) \\
& = 2k^2 + 3k + 1 \\
& = 2(k + 1)^2 − (k + 1)
\end{align}
$$
$\diamond$
| You can use the “trick” attributed to Gauss for finding the sum of the first $n$ integers. Call your sum $S_n$, and consider this.
$$\begin{align}
&\quad S_n=+1&+ 5 +& \quad\cdots&+(4n-7)&+(4n-3)\\
+&\quad S_n=+(4n-3)&+(4n-7)+&\quad\cdots&+5&+1
\\ &\hline\\
\quad&2S_n\quad =+\underbrace{(4n-2)}_{1^{st}} &+\underbrace{(4n-2)}_{2^{nd} }+&\quad\underbrace{\cdots}_{3^{rd}\mbox{ to }{(n-2)}^{nd}}&+\underbrace{(4n-2)}_{(n-1)^{st}}&+\underbrace{(4n-2)}_{n^{th}}\\\end{align}$$
Thus $2S_n$ equals $n(4n-2)$, and $S_n=2n^2-n$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1802846",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "53",
"answer_count": 14,
"answer_id": 9
} |
Integrate $\int e^x\frac{1+\sin x}{1+\cos x}dx$ Integrate
$$\int e^x\cdot\frac{1+\sin x}{1+\cos x}\,dx$$
My try;
First step:
I let
$$\frac{1+\sin x}{1+\cos x} = u$$
$$e^x = v$$
and then I applied integration by parts:
$$\frac{1+\sin x}{1+\cos x}=u \implies du=\frac{\sin x+\cos x+1}{(1+\cos x)^2}dx$$
$$v = e^x \implies dv = e^x$$
and
$$\int e^x\cdot\frac{1+\sin x}{1+\cos x}\,dx=e^x\cdot\frac{1+\sin x}{1+\cos x}-\int e^x\cdot\frac{\sin x+\cos x+1}{(1+\cos x)^2}dx$$
| $$\int e^x\cdot\dfrac{1+\sin x}{1+\cos x}\,dx = \int e^x\cdot\dfrac{1+2\sin\left(\frac{x}{2}\right)\cos\left(\frac{x}{2}\right)}{2\cos^2\left(\frac{x}{2}\right)}\,dx\\=\int e^x\cdot\left(\frac{1}{2}\sec^2\frac{x}{2} + \tan\frac{x}{2}\right)dx$$
Expand,
$$\int e^x\cdot\frac{1}{2}\sec^2\frac{x}{2}\,dx + \int e^x\cdot\tan\frac{x}{2}\,dx$$
Integrate by parts, the integral: $$\int e^x\cdot\tan\frac{x}{2} dx$$
You should get the final answer as
$$ e^x\cdot\tan\frac{x}{2} + c$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1803339",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Minimal polynomial of $\sqrt[3]{2} + \sqrt{3}$
Suppose I want to find the minimal polynomial of the number $\sqrt[3]{2} + \sqrt{3}$.
Now that means I want to find a unique polynomial that is irreducible over $\Bbb Q$ such that $f(x)=0$. Now I know that because $\sqrt[3]{2} + \sqrt{3}$ belongs to $\Bbb Q( \sqrt[3]{2} , \sqrt{3})$ it might be degree $2$, $3$ or $6$ and does not belong to $\Bbb Q( \sqrt[3]{2})$ so it cannot be of degree $3$ or $\Bbb Q( \sqrt{3})$ so it cannot be of degree $2$.
So it is of degree $6$. I think my sayings are a bit intuitive and not formal and lack rigorous.
Couldn't it belong to another extension of degree $2$? or $3$? I can't answer that. Why checking only those $2$ is enough? Or is it wrong at all to say that?
| Let $a=\sqrt[3]{2} + \sqrt{3}$. Notice that
$$(a-\sqrt{3})^3=2=a^3-3\sqrt 3 a^2+9a-3\sqrt 3
= a^3+9a-\sqrt 3 (3a^2+3) \tag 1$$
therefore
$$\sqrt 3 = \frac{a^3+9a-2}{3a^2+3} \tag 2$$
In particular, $\Bbb Q(a)$ contains $\Bbb Q(\sqrt 3)$ and also contains $\Bbb Q(a-\sqrt 3) = \Bbb Q(\sqrt[3]{2})$.
Therefore your intuition is correct: the degree of $\Bbb Q(a)$ is a multiple of $3$ and a multiple of $2$ (over $\Bbb Q$).
The degree of the minimal polynomial of $a$ over $\Bbb Q$ is then at
least $6$.
From $$(a^3+9a-2)^2 = [\sqrt 3 (3a^2+3)]^2 \tag 3$$
you get a monic polynomial $P \in \Bbb Q[X]$ of degree $6$, such that $P(a)=0$. Thus $P$ is the minimal polynomial of $a$ over $\Bbb Q$.
Here is the minimal polynomial $P(X)$ of $a$ over $\Bbb Q$ :
$P(x) = x^6-9 x^4-4 x^3+27 x^2-36 x-23$.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "5",
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Prove that $U$ is a vector-subspace
If $U$ is the set of all matrices that are commutative with the matrix
$A$, show that $U$ is a vector subspace of the space
$M^\mathbb{R}_{3\times 3}$
$$A=\begin{pmatrix}2&0&1\\ 0&1&1\\ 3&0&4\end{pmatrix}$$
Also show if it contains the $$\operatorname{span}(I,A,A^2,...)$$ and then find the dimension of $U$ and the base.
Can someone explain how would I prove that $U$ is a subspace and maybe point out a good example/text book that has lots of similar examples for me to practice upon.
P.S. I know that for $U$ to be a subspace it has to be "closed" for the operations of addition and multiplication. But I have no clue on how to prove it.
| Suppose $B,C \in U$. Then $$(B+C)A = BA + CA = AB + AC = A(B+C).$$ So $B+C\in U$. Similarly if $c \in \mathbb{R}$, $$(cB)A = c(BA) = c(AB) = A(cB).$$
So $U$ is a subspace.
Now note that $IA = AI$ so $I \in U$. Also $A^kA = A^{k+1} = AA^k$. So $A^k \in U$ for all $k > 1$. Since U is a subspace, and $I, A, A^2,\dots \in U$, $$span(I,A,A^2,\dots) \subseteq U.$$
Now let's tackle the commuting. There is some general theory that can help us here, I will only use the Jordan form (though we could be even more slick about our approach).
Let J denote the Jordan Form of A. $$J = \begin{pmatrix}
1 & 1 & 0 \\
0 & 1 & 0 \\
0 & 0 & 5
\end{pmatrix}_.$$
$A$ is similar to $J$, that is there is an invertible matrix $S$ such that $A = SJS^{-1}$. Here $$S = \begin{pmatrix}
0 & -1 & 4 \\
1 & 0 & 3 \\
0 & 1 & 12\\
\end{pmatrix}_.$$
Suppose $B \in U$. Let $C = S^{-1}BS$. Then $JC = S^-1ASS^{-1}BS = S^{-1}ABS = S^{-1}BAS = S^{-1}BSS^{-1}AS = CJ$. So $C$ commutes with $J$. Similarly if you have a matrix that commutes with $J$, it corresponds to an element of $U$. So instead of trying to find matrices that commute with $A$, we can work with $J$ and then pass through the change of coordinates to get a matrix that commutes with $A$.
Now here is where I will work harder rather than smarter. We could do less work because of the block form of $J$, but I will take a less elegant approach.
Take an arbitrary matrix:
$$\begin{pmatrix}
a & b & c \\
d & e & f \\
g & h & i \\
\end{pmatrix}_.$$
Let's see how it interacts with J:
$$\begin{pmatrix}
a & b & c \\
d & e & f \\
g & h & i \\
\end{pmatrix} \begin{pmatrix}
1 & 1 & 0 \\
0 & 1 & 0 \\
0 & 0 & 5
\end{pmatrix} = \begin{pmatrix}
a & a+b & 5c \\
d & d+e & 5f \\
g & g+f & 5i \\
\end{pmatrix}_.$$
$$\begin{pmatrix}
1 & 1 & 0 \\
0 & 1 & 0 \\
0 & 0 & 5
\end{pmatrix} \begin{pmatrix}
a & b & c \\
d & e & f \\
g & h & i \\
\end{pmatrix} = \begin{pmatrix}
a + d & b + e & c + f\\
d & e & f\\
5g & 5h & 5i\\
\end{pmatrix}$$
Setting these matrices equal to each other (hence forcing commutativity) yields a bunch of equations. Solving these equations gives: $d = g = f = h = c = 0$, $a = e$, $b$ and $i$ are free.
So our arbitrary matrix looks like
$$\begin{pmatrix}
a & b & 0\\
0 & a & 0\\
0 & 0 & i\\
\end{pmatrix}$$
So a basis of the space of matrices commuting with $J$ is $$\begin{pmatrix}
1 & 0 & 0\\
0 & 1 & 0\\
0 & 0 & 0\\
\end{pmatrix}_, \begin{pmatrix}
0 & 1 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0 \\
\end{pmatrix}_, \begin{pmatrix}
0 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 1 \\
\end{pmatrix}_.$$
Now a basis of $U$ is $$S\begin{pmatrix}
1 & 0 & 0\\
0 & 1 & 0\\
0 & 0 & 0\\
\end{pmatrix}S^{-1}_, \ S\begin{pmatrix}
0 & 1 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0 \\
\end{pmatrix}S^{-1}_, \ S\begin{pmatrix}
0 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 1 \\
\end{pmatrix}S^{-1}_.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1804756",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Integration with function as boundary I have to find the function F defined trough the integral:
A) $\displaystyle F(x)= \int_0^{1-x} (1-2t+3t^2) \, dt $
B) $\displaystyle F(x)= \int_x^{x^2} (\sqrt t + \sqrt{t}^3) \, dt $
C) $\displaystyle F(x)= \int_0^{x^2} (1/2-\sin t) \, dt $
I tried solving the integrals:
A) $\displaystyle F(y)= \int_0^{1-x} (1-2t+3t^2) \, dt = -x^3+2x^2-2x+1$
B) I do not know how to integrate this...
C) $ \displaystyle F(x)= \int_0^{x^2} (1/2-\sin t) \, dt = \frac{x^2}{2} +\cos(x^2)-1$
Is this correct ? Please if you can explain me how to solve it if I am wrong and why... Thanks for your time and help.
| (A) looks correct. (B) is very straightforward once you realize $ \sqrt t $ = $t^ \frac{1}{2}$ and similarly, $ \sqrt (t^3) $ = $t^ \frac{3}{2}$.So now it's simply a matter of using the ordinary integration rules:
$\displaystyle F(x)= \int_x^{x^2} (\sqrt t + \sqrt{t}^3) \, dt $ =
$\displaystyle F(x)= \int_x^{x^2} (t^ \frac{1}{2} + t^ \frac{3}{2})\,dt = \frac{2}{3} t^ \frac{3}{2} + \frac{2}{5} t^ \frac{5}{2}|_{x}^{x^2}=\frac{2}{3}(x^ 3) + \frac{2}{5} (x^ 5))- (\frac{2}{3} x^ \frac{3}{2} + \frac{2}{5} x^ \frac{5}{2})$
You could rearrange (B) to collect like terms,but in this case I don't think it matters unless your prof is anal about layout. (C) also looks correct.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find all oddly even numbers $m$ such that $m^{2^n - 1} - 1 = (m - 1)(m^n + p^k)$ Find all natural numbers $m \equiv 2 \pmod{4}$ such that there exist exist two natural numbers $n, k$ and a prime number $p$ with $\dfrac{m^{2^{n}-1}-1}{m-1}=m^n+p^k$.
My solution:
The equation lead to $n \geq 2 $,hence $LHS \equiv3$ mod4.So,$p^k \equiv 3$ mod 4. Thus, $p \equiv 3$ mod 4. Then,
$m^{2^{n}-1}+(1-m)(m^{n+1}+1)= m(m-1)p^k$
Set $n+1=2^l(2h+1)$ hence $l<n$ and $m^{2^l}+1 \mid m^{2^n}-1$.We also have $$m^{2^l}+1 \mid m^{n+1}+1$$.Thus, $m^{2^l}+1 \mid m(m-1)p^k$. Because $\gcd \left( m(m-1), m^{2^l}+1 \right)=1$ so $m^{2^l}+1 \mid p^k$.
If $l \ge 1$ then always exist a prime number $q \equiv 1 \pmod{4}$ and $ q|m^{2^l}+1$( a well-known problem), contradiction.
Thus, $l=0$ and $m+1 \mid p^k$.So, $m=p^l-1$ , $2 \nmid l$
We'll prove that for each number $m=p^l-1$ there exist $n,k,p$.Really, we choose $n=2,k=l$
| If $n=1$, $m+p^k=1$. Contradiction.
Let $n\ge2$.
$$\frac{m^{2^n-1}-1}{m-1}=m^{2^n-2}+\ldots+m^2+m+1=m^n+p^k$$
$4$ divides every power of $m$ except $m$ thus $LHS\equiv m+1\equiv 3\pmod4$.
LHS is odd, and $m$ is even. So,$p$ is odd. If $2|k$, $p^k\equiv 1\pmod4$ and $4|m^n$. So,$RHS\equiv 1\pmod4$. Contradiction. So, $k$ is odd.
If $n=2$, we need $m+1=p^k$. There are infinitely many solutions with $m=p-1$, where $p$ is an arbitrary prime of the form $4k+3$, and $k$ is an arbitrary odd integer.
Suppose, $n\ge3$. Then,
we have $p^k\equiv 3\pmod4$ and $p^k\equiv 1\pmod m$. If $d=\gcd(n,2^n-2)>1$, then, for any $q|\frac{m^{d}-1}{m-1}$,
$$m^n\equiv 1\equiv \frac{m-1}{m-1}\equiv \frac{m^{2^n-2}m-1}{m-1}\pmod q$$
So, $q|p^k$, thus, $q=p$. So, $q<m^n$ and $q^k>m^{2^n-2}$, implies, $k>\frac{2^n-2}n$. In particular, $k\ne1$.
Moreover, $q=p$ is true for any divisor $q$ of ${m^d-1\over m-1}$, ${m^d-1\over m-1}$ should be a power of $q$. Which implies $d$ is a prime number by Zsigmondy's Theorem.(Exception for Zsigmondy: $m=2$,$d=6$ does not work here.)
Note: I'm aware I'm assuming something weird. But, this actually works for $n$ prime power or $\gcd(6,n)>1$. I tried to factorize $p$ and this looked like the pattern and unfortunately, the smallest $n$ that does not satisfy this property is $35$, which is out of my computer's ability to calculate.
| {
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"url": "https://math.stackexchange.com/questions/1806542",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
$\lim_{n\to\infty} \dfrac{1^3+2^3+3^3+...+n^3}{(n^2+1)^2}$ using the Squeeze Theorem Evaluate
$$\lim_{n\to\infty} \dfrac{1^3+2^3+3^3+...+n^3}{(n^2+1)^2}$$
I know that the standard way would be to use the Closed Form of$$1^3+2^3+3^3+...+n^3$$ However I've been trying to practise using the Squeeze Theorem and thus tried to solve it using the Squeeze Theorem. However, I seem to have made a mistake as I don't get the expected answer.
$$$$My approach:$$$$
Taking $$f(n)=\dfrac{1^3+2^3+3^3+...+n^3}{(n^2+1)^2}$$ I identified the general term of $f(n)$ as $\dfrac{k^3}{(n^2+1)^2}$ where $1\le k\le n$
$$$$
Clearly $$\dfrac{1^3}{(n^2+1)^2}\le \dfrac{k^3}{(n^2+1)^2}\le \dfrac{n^3}{(n^2+1)^2}$$
Thus $$\dfrac{1^3}{(n^2+1)^2}+\dfrac{1^3}{(n^2+1)^2}+... \text{n times}\le f(n)\le \dfrac{n^3}{(n^2+1)^2}+\dfrac{n^3}{(n^2+1)^2}+... \text{n times}$$
$$\dfrac{n}{(n^2+1)^2}\le f(n)\le \dfrac{n\times n^3}{(n^2+1)^2}$$
$$\dfrac{n}{(n^2+1)^2}\le f(n)\le \dfrac{n^4}{(n^2+1)^2}$$
$$$$However on taking the limits as $n\to \infty$, I'm not getting the correct answer.
$$$$Could somebody please show me where I've gone wrong? Thanks!
| We have $$(r + 1)^{4} - r^{4} = 4r^{3} + 6r^{2} + 6r + 1 > 4r^{3}\tag{1}$$ and $$r^{4} - (r - 1)^{4} = 4r^{3} - 6r^{2} + 6r - 1 < 4r^{3}\tag{2}$$ and putting $r = 1, 2, \ldots, n$ and adding these equations we get $$\frac{n^{4}}{4} < S = \sum_{r = 1}^{n}r^{3} < \frac{(n + 1)^{4} - 1}{4}\tag{3}$$ and then we can see that squeeze theorem can be easily applied to get desired limit as $1/4$.
Update: OP is aware of the exact formula for sum $$S = \sum_{r = 1}^{n}r^{3}$$ and hence he knows that the limit is $1/4$. Since we are supposed to use the Squeeze theorem we must be able to bound the sum $S$ with two functions $f(n), g(n)$ such that $$f(n) < S < g(n)$$ and $$\frac{f(n)}{(n^{2} + 1)^{2}} \to \frac{1}{4}$$ and $$\frac{g(n)}{(n^{2} + 1)^{2}} \to \frac{1}{4}$$ Thus we want the bounds $f(n), g(n)$ to be of type $n^{4}/4$. In your approach the bounds are either too high or too low. Thus the upper bound $g(n)$ for $S$ in your answer looks like $n^{4}$ and lower bound $f(n)$ for $S$ looks like $n$. So from this you can only conclude that desired limit lies between $0$ and $1$.
Now the procedure to get the bounds comes from the technique used to obtain the closed form for $S$. This is done by calculating the differences $$(r + 1)^{4} - r^{4}$$ and $$r^{4} - (r - 1)^{4}$$ Any of the relations $(1)$ or $(2)$ of my answer can be used to derive an exact closed form for $S$, but we only need an approximation so the identities in $(1), (2)$ are given the shape of inequalities by neglecting terms containing $r^{2}, r$ and constant term. By summing these inequalities for $r = 1, 2, \ldots, n$ I don't get an exact formula for $S$ but rather very close bounds for $S$ and that's what we need here.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1808381",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Induced metric on a one-sheet hyperboloid I am trying to find the induced metric on a one-sheet hyperboloid. Suppose we use cylindrical coordinates $(r, \theta, z)$ for the ambient space in which the hyperboloid is embedded. The hyperboloid itself then consists of all points such that
$$ r^2 = R^2 + z^2 $$
where $R$ is the minimum radius, at the throat. On the hyperboloid itself, we define the following coordinates $(\rho, \varphi)$:
\begin{align}
r &= \sqrt{R^2 + \rho^2} \\
\theta &= \varphi \\
z &= \rho
\end{align}
We can now find the induced metric using the formula
$$g_{ab} = g_{\mu\nu} \partial_a X^\mu \partial_b X^\nu$$
where $a$ and $b$ describe the indices of the coordinates of the submanifold (in this case, the hyperboloid), while the $X^\mu$ and $X^\nu$ encode the embedding into the ambient space.
First, the nonzero components of the ambient metric tensor in cylindrical coordinates are
\begin{align}
g_{rr} &= 1 \\
g_{\theta \theta} &= r^2 \\
g_{zz} &= 1 \\
\end{align}
Thus we have
\begin{align}
g_{ab} &= g_{rr} \partial_a X^r \partial_b X^r + g_{\theta \theta} \partial_a X^\theta \partial_b X^\theta + g_{zz} \partial_a X^z \partial_b X^z \\
&= \partial_a X^r \partial_b X^r + r^2 \partial_a X^\theta \partial_b X^\theta + \partial_a X^z \partial_b X^z
\end{align}
Hence the nonzero components of the metric tensor on the hyperboloid are
\begin{align}
g_{\rho \rho} &= \partial_\rho X^r \partial_\rho X^r + r^2 \partial_\rho X^\theta \partial_\rho X^\theta + \partial_\rho X^z \partial_\rho X^z \\
&= \left(\partial_\rho \sqrt{R^2 + \rho^2}\right)^2 + (\partial_\rho \rho)^2 \\
&= \left(\frac{\rho}{\sqrt{R^2 + \rho^2}}\right)^2 + 1^2 \\
&= \frac{\rho^2}{R^2 + \rho^2} + 1
\end{align}
and
\begin{align}
g_{\varphi \varphi} &= \partial_\varphi X^r \partial_\varphi X^r + r^2 \partial_\varphi X^\theta \partial_\varphi X^\theta + \partial_\varphi X^z \partial_\varphi X^z \\
&= 0 + r^2 \cdot 1^2 + 0 \\
&= r^2
\end{align}
The cross-terms $g_{\rho \varphi} = g_{\varphi \rho}$ vanish. Is this derivation correct?
| Yes, your derivation is correct.
Incidentally, you can avoid expressing the Euclidean metric in cylindrical coordinates: Parametrize the hyperboloid by
$$
\Phi(\rho, \varphi)
= \left(\sqrt{R^{2} + \rho^{2}} \cos\varphi, \sqrt{R^{2} + \rho^{2}} \sin\varphi, \rho\right).
$$
The partials are
\begin{align*}
\Phi_{\rho}(\rho, \varphi)
&= \left(\frac{\rho}{\sqrt{R^{2} + \rho^{2}}} \cos\varphi, \frac{\rho}{\sqrt{R^{2} + \rho^{2}}} \sin\varphi, 1\right), \\
\Phi_{\varphi}(\rho, \varphi)
&= \left(-\sqrt{R^{2} + \rho^{2}} \sin\varphi, \sqrt{R^{2} + \rho^{2}} \cos\varphi, 0\right),
\end{align*}
so pulling back the Euclidean metric gives the components
\begin{align*}
g_{\rho\rho} &= \Phi_{\rho} \cdot\Phi_{\rho} = \frac{\rho^{2}}{R^{2} + \rho^{2}} + 1, \\
g_{\rho\varphi} &= \Phi_{\rho} \cdot\Phi_{\varphi} = 0, \\
g_{\varphi\varphi} &= \Phi_{\varphi} \cdot\Phi_{\varphi} = \rho^{2}.
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1808588",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Minimal generation of $S_n$ For a given $n$, the symmetric group $S_n$ is generated by permutations $p_1$, $p_2$, and $p_3$. Two of them generate subgroups $P_{23}$, $P_{13}$, $P_{12}$, which have group orders $a$, $b$, and $c$.
What is the minimal $max(a,b,c)$ for various $n$?
| The answer is $8$ for all $n$ with $8 \le n \le 26$ (as verret observed, it is $10$ for $n=7$), and I conjecture that it is $8$ for all larger $n$, with $S_n$ being a quotient of $$\langle x,y,z \mid x^2=y^2=z^2=(xy)^4=(yz)^4=(xz)^4=1 \rangle.$$
(It is possible that this is a known result.)
There are lots of solutions. Here are some for $n=6,7,8,9$ as requested.
$$n=6: (1, 2)(3, 4)(5, 6),\
(1, 2)(3, 6),\
(2, 3),$$
$$n=7: (1, 7)(2, 4)(5, 6),\
(1, 2),\
(1, 7)(2, 3)(4, 6),$$
$$n=8: (1, 8)(2, 3)(6, 7),\
(1, 7)(2, 4)(3, 5),\
(1, 2),$$
$$n=9: (1, 2)(3, 4),\
(2, 4)(5, 6)(7, 8),\
(3, 5)(4, 7)(6, 9).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1808657",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How $2 \sqrt[3]{\frac{x^2y^2z^2}{(x+y)(y+z)(z+x)}}$ and $\frac{3xyz}{x\sqrt{yz}+y\sqrt{zx}+z\sqrt{xy}}$ are related for $x,y,z \geq 0$? The first expression is a heometric mean of pairwise harmonic means and obeys the following inequality:
$$\sqrt[3]{xyz} \geq \color{blue}{ 2 \sqrt[3]{\frac{x^2y^2z^2}{(x+y)(y+z)(z+x)}}} \geq \frac{3xyz}{xy+yz+zx}$$
The second expression is a harmonic mean of pairwise heometric means and obeys the same inequality:
$$\sqrt[3]{xyz} \geq \color{blue}{ \frac{3xyz}{x\sqrt{yz}+y\sqrt{zx}+z\sqrt{xy}}} \geq \frac{3xyz}{xy+yz+zx}$$
I want to know how these two expressions are related ($\leq$ or $\geq$) for $x,y,z \geq 0$ .
Edit
Since the realtion is homogenous, we can take $xyz=1$ without loss of generality, then we just need to compare:
$$\frac{\sqrt[3]{(x+y)(y+z)(z+x)}}{2} \text{ ? } \frac{\sqrt{x}+\sqrt{y}+\sqrt{z}}{3}$$
Or rather:
$$\frac{\sqrt[3]{(x+y)(y+1/(xy))(1/(xy)+x)}}{2} \text{ ? } \frac{\sqrt{x}+\sqrt{y}+\sqrt{1/(xy)}}{3}$$
| I will prove
$$\frac{3xyz}{x\sqrt{yz}+y\sqrt{zx}+z\sqrt{xy}}\ge 2 \sqrt[3]{\frac{x^2y^2z^2}{(x+y)(y+z)(z+x)}}$$
Let $x=a^2,y=b^2,z=c^2$. Then, after taking third powers, our inequality becomes
$$\frac{27a^6b^6c^6}{(a+b+c)^3a^3b^3c^3}\ge 8{\frac{a^4b^4c^4}{(a^2+b^2)(b^2+c^2)(c^2+a^2)}}$$
After expanding, one needs to show
$$27(a^2+b^2)(b^2+c^2)(c^2+a^2)\ge 8(a+b+c)^3abc\qquad (1)$$
or
$$27\sum_{sym}u^4v^2+54a^2b^2c^2\ge 4\sum_{sym}u^4vw+24\sum_{sym}u^3v^2w+48a^2b^2c^2$$
After canceling both sides,
$$27\sum_{sym}u^4v^2+\sum_{sym}u^2v^2w^2\ge 4\sum_{sym}u^4vw+24\sum_{sym}u^3v^2w$$
By AM-GM,
$$\sum_{sym}u^4v^2+\sum_{sym}u^2v^2w^2\ge \sum_{sym}u^3v^2w$$
And by Muirhead's inequality
$$\sum_{sym}u^4v^2\ge \sum_{sym}u^3v^2w$$
$$\sum_{sym}u^4v^2\ge \sum_{sym}u^4vw$$
Summing these all up gives us the desired result. Observe that $(1)$ can be proved in million different ways, I just like Muirhead's Inequality.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Show that matrix of $SO_3(R)$ has the form $\left(\begin{smallmatrix}1&0&0\\0&\cos\theta&-\sin\theta\\0&\sin\theta&\cos\theta\end{smallmatrix}\right)$ I would like to show that matrix of $SO_3(\mathbb R)$ are of the form $$\begin{pmatrix}1&0&0\\0&\cos\theta&-\sin\theta\\0&\sin\theta&\cos\theta\end{pmatrix}$$
in a certain basis. So first, I'm trying to show that a matrix of $SO_3(\mathbb R)$ is of the form $$\begin{pmatrix}1&0&0\\0&a&b\\0&c&d\end{pmatrix},$$
and thus since $ad-bc=1$ and that $$1=\begin{pmatrix}1&0&0\\0&a&b\\0&c&d\end{pmatrix}\begin{pmatrix}1&0&0\\0&a&b\\0&c&d\end{pmatrix}^T=\begin{pmatrix}a&b\\c&d\end{pmatrix}\begin{pmatrix}a&b\\c&d\end{pmatrix}^T,$$
we will get that $a^2+c^2=1$, $a=d$ and $c=-b$.
So, if $\lambda=1$ is a eigen value, we will have that $Au=u$ and $A^Tu=u$ for a certain $u$. Then, in the basis $\{u,a,b\}$, we will have that $$A=\begin{pmatrix}1&a&b\\0&c&d\\0&e&f\end{pmatrix}$$
and $$A^T=\begin{pmatrix}1&A&B\\0&C&D\\0&E&F\\\end{pmatrix},$$
we finally get $A=a=B=b=0$, $C=c$, $D=d$, $E=e$ and $F=f$ and thus the claim follow. Now, how can I show that $1$ is really an eigevalue ?
| Since $A$ is orthogonal, the eigenvalues of $A$ must be $\pm 1$. Consider the characteristic polynomial $\chi_A(x)$ of $A \in \operatorname{SO}_3(\mathbb{R})$. This is a real polynomial of degree three and so must have a real root. If this root is $\lambda_1 = 1$, we are done. If this root is $\lambda_1 = -1$, write
$$\chi_A(x) = (x + 1)(x^2 + bx + c) = (x - \lambda_1)g(x). $$
The polynomial $g$ is a real polynomial of degree two. Now we have two options
*
*If $g$ factors as $(x - \lambda_2)(x - \lambda_3)$ then $\lambda_i \in \{ \pm 1 \}$ and we can't have $\lambda_2 = \lambda_3 = -1$ because then $\det(A) = \lambda_1 \lambda_2 \lambda_3 = (-1)^3 = -1$.
*If $g$ is irreducible, we will have $g(x) = (x - \lambda_2)(x - \overline{\lambda_2})$ and $\det(A) = \lambda_1 \lambda_2 \overline{\lambda_2} = - |\lambda_2|^2 \neq 1$ - a contradiction.
Thus, in any case, we have shown that $A$ must have $1$ as an eigenvalue.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove $\frac{\sqrt[3]{(x+y)(y+z)(z+x)}}{2} \geq \sqrt{\frac{xy+yz+zx}{3}}$ for $x,y,z \geq 0$ We have geometric mean of pairwise arithmetic means on the left, which obeys the following inequality:
$$\frac{x+y+z}{3} \geq \color{blue}{ \frac{\sqrt[3]{(x+y)(y+z)(z+x)}}{2} } \geq \sqrt[3]{xyz}$$
And on the right we have root-mean-square of geometric means, obeying the same inequality:
$$\frac{x+y+z}{3} \geq \color{blue}{\sqrt{\frac{xy+yz+zx}{3}} } \geq \sqrt[3]{xyz}$$
This time I checked with Wolfram Alpha first, and apparently, the inequality in the title is true for all $x,y,z \geq 0$:
$$\frac{\sqrt[3]{(x+y)(y+z)(z+x)}}{2} \geq \sqrt{\frac{xy+yz+zx}{3}}$$
How do we prove this inequality for for all $x,y,z \geq 0$?
We need to prove:
$$27(x+y)^2(y+z)^2(z+x)^2 \geq 64(xy+yz+zx)^3$$
Expanding directly leads to a very complicated expression, and overall this inequality is a little tight. I have not been able to prove it by AM-GM or Cauchy.
| First we prove a very close inequality:
$$27(x+y)^2(y+z)^2(z+x)^2 \geq 63(xy+yz+zx)^3+27x^2y^2z^2$$
The proof is as follows. By direct expansion we see:
$$(x+y)(y+z)(z+x)=(x+y+z)(xy+yz+zx)-xyz$$
$$(x+y)^2(y+z)^2(z+x)^2= \\ =(x+y+z)^2(xy+yz+zx)^2-2xyz(x+y+z)(xy+yz+zx)+x^2y^2z^2$$
Using Cauchy-Schwartz we can transform the second term:
$$(x+y+z)(xy+yz+zx) \geq 9xyz$$
$$(x+y)^2(y+z)^2(z+x)^2 \geq \\ \geq (x+y+z)^2(xy+yz+zx)^2-\frac{2}{9}(x+y+z)^2(xy+yz+zx)^2+x^2y^2z^2$$
$$(x+y)^2(y+z)^2(z+x)^2 \geq \frac{7}{9}(x+y+z)^2(xy+yz+zx)^2+x^2y^2z^2= \\ = \frac{7}{9}(x^2+y^2+z^2+2(xy+yz+xz))(xy+yz+zx)^2+x^2y^2z^2 \geq \\ \geq \frac{7}{3}(xy+yz+zx)^3+x^2y^2z^2$$
Finally we get:
$$3(x+y)^2(y+z)^2(z+x)^2 \geq 7(xy+yz+zx)^3+3x^2y^2z^2$$
Now we prove the original inequality.
First, let's get back and rearrange the original equality we used:
$$(x+y)(y+z)(z+x)+xyz=(x+y+z)(xy+yz+zx)$$
Now let's square this and denote:
$$A=(x+y)(y+z)(z+x)$$
$$B=(xy+yz+zx)$$
$$A^2+2xyzA+x^2y^2z^2=(x+y+z)^2B^2$$
$$A^2=(x+y+z)^2B^2-2xyzA-x^2y^2z^2 \geq 3B^3-\frac{2}{\sqrt{27}} \sqrt{B^3} A-A^2+\frac{7}{3}B^3$$
$$A^2+\frac{1}{\sqrt{27}} \sqrt{B^3} A-\frac{8}{3}B^3 \geq 0 \tag{1}$$
Here we used the inequality from the previous section and two well known inequalities:
$$x^2y^2z^2 \leq A^2-\frac{7}{3}B^3$$
$$27x^2y^2z^2 \leq B^3$$
$$(x+y+z)^2 \geq 3B$$
$(1)$ is a quadratic inequality, and is easily solved:
$$A \geq \frac{8}{\sqrt{27}} \sqrt{B^3}$$
The original inequality obviously follows.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1812595",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
How to divide the following polynomial and factor it? The question is
$$ (2x^3+3x^2-39x-20) / (x-4) $$
I divided the following and got this as the answer
$$ 2x^2+9x+3-8/(x-4))$$
I thought that this was the answer, but when i looked at the answer sheet the answer was this $$(x-4)(x+5)(2x+1)$$
I think the answer should've been factored but i don't know how to factor this
$$ (2x^3+3x^2-39x-20) / (x-4) $$
and also i have a general question about factoring these types of problems
$$ax^3+bx^2+cx+d$$
which method would i choose, because i should be getting 3 answers and I don't know how to do that.
| $4$ is a root of $3x^3+3x^2-39x-20$, hence it is divisible by $x-4$. To obtain the quotient, you can use Horner's scheme:
$$\begin{array}{rrcrcrr}
&2&&3&&-39&&-20\\
\hline
&&&8&&44&&20\\
&&\!\!\nearrow\!\!&&\!\!\nearrow\!\!\!\!&&\!\!\nearrow\!\!\!\!\\
{}\times 4&\color{red}{2}&&\color{red}{11}&&\color{red}{5}&&0
\end{array}$$
Hence $\;2x^3+3x^2-39x-20=(2x^2+11x+5)(x-4)$.
The quadratic polynomial has discriminant equal to $81$, hence its roots are $-5$ and $-\frac12$, so it factors as $(2x+1)(x+5)$ and
$$\frac{2x^3+3x^2-39x-20}{x-4}=\frac{(2x+1)(x+5)(x-4)}{x-4}=(2x+1)(x+5)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1812671",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Find $f'(x)$ at given value of x Find $f'(x)$ at the given value of x
$f(x)=\sqrt{x+2}$
Find $f'(7)$
My question for this one is do I approach this question by trying to find the derivative of the initial equation and then once I have found the derivative do I simply plug in 7 for x and solve?
So once I start solving I would get $\frac{\sqrt{x+h+2}-\sqrt{x+2}}{h}$
$\lim_{h\to0} \frac{\sqrt{x+h+2}-\sqrt{x+2}}{h} = \lim_{h\to0} \frac{\sqrt{x+h+2}-\sqrt{x+2}}{h} \cdot \frac{\sqrt{x+h+2}+\sqrt{x+2}}{\sqrt{x+h+2}+\sqrt{x+2}}$
From that point I would get
$\frac{(x+h+2-x-2)}{h\sqrt{x+h+2}+\sqrt{x+2}}$
| Hint: Since you're using the difference quotient, you want to get rid of the square roots in the numerator by using the numerator's conjugate:
$$\lim_{h\to0} \frac{\sqrt{x+h+2}-\sqrt{x+2}}{h} = \lim_{h\to0} \frac{\sqrt{x+h+2}-\sqrt{x+2}}{h} \cdot \frac{\sqrt{x+h+2}+\sqrt{x+2}}{\sqrt{x+h+2}+\sqrt{x+2}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1812791",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
An Euler squared sum Let $\mathcal{H}_n$ denote the $n$-th harmonic number. Evaluate the following sum
$$\mathcal{S}=\sum_{n=1}^{\infty} \frac{(-1)^{n-1} \mathcal{H}_n^2}{n+1}$$
Here $\mathcal{H}_n^2$ is the square harmonic number, e.g $\left ( \sum \limits_{k=1}^{n} \frac{1}{k} \right )^2$. I don't know how to tackle this. One idea of mine was the following:
\begin{align*}
\mathcal{S} &=\sum_{n=1}^{\infty} (-1)^{n-1} \mathcal{H}_n^2 \int_{0}^{1}x^n \, {\rm d}x \\
&= \int_{0}^{1}\sum_{n=1}^{\infty} (-1)^{n-1} \mathcal{H}_n^2 x^n \, {\rm d}x
\end{align*}
and the last sum evaluates to ?
Motivation: The series comes from this question . I tried a different approach. This is what I tried:
\begin{align*}
\int_{0}^{1} \frac{\log(1+x) \log(1-x)}{1+x} \, {\rm d}x&= \int_{0}^{1}\log(1-x) \sum_{n=1}^{\infty} (-1)^{n-1} \mathcal{H}_n x^n \, {\rm d}x\\
&=\sum_{n=1}^{\infty} (-1)^{n-1} \mathcal{H}_n \int_{0}^{1}x^n \log(1-x) \, {\rm d}x \\
&= -\sum_{n=1}^{\infty} (-1)^{n-1} \mathcal{H}_n \cdot \frac{\mathcal{H}_{n+1}}{n+1} \\
&= -\sum_{n=1}^{\infty} (-1)^{n-1} \mathcal{H}_n \left ( \frac{\mathcal{H}_n + \frac{1}{n+1}}{n+1} \right )\\
&= -\sum_{n=1}^{\infty} \frac{(-1)^{n-1} \mathcal{H}_n^2}{n+1} - \sum_{n=1}^{\infty} \frac{(-1)^{n-1} \mathcal{H}_n}{\left ( n+1 \right )^2}
\end{align*}
The right hand Euler sum appears to be elementary. I have not evaluated it though. But now we known that:
$$\bbox[blue, 2pt]{\color{white}{-\sum_{n=1}^{\infty} \frac{(-1)^{n-1} \mathcal{H}_n^2}{n+1} - \sum_{n=1}^{\infty} \frac{(-1)^{n-1} \mathcal{H}_n}{\left ( n+1 \right )^2} = \frac{1}{3}\log^3 2-\frac{\pi^2}{12}\log 2+\frac{\zeta(3)}{8}}}$$
from the linked question. So, I expect both of these series to have a closed form. Any help how to proceed with the first series? If anyone wishes to give a go for the second (right hand Euler sum) be my guest.
Addendum: For the second Euler sum we have that:
\begin{align*}
-\sum_{n=1}^{\infty} \frac{(-1)^{n-1} \mathcal{H}_n}{\left ( n+1 \right )^2} &= \sum_{n=1}^{\infty} (-1)^{n-1} \mathcal{H}_n \int_{0}^{1}x^n \log x \, {\rm d}x\\
&=\int_{0}^{1}\log x \sum_{n=1}^{\infty} (-1)^{n-1} \mathcal{H}_n x^n \, {\rm d}x \\
&= \int_{0}^{1}\frac{\log (1+x) \log x}{1+x}\, {\rm d}x\\
&=-\frac{\zeta(3)}{8}
\end{align*}
The integral is elementary. Thus, it appears that the sum I seek evaluates to
$$\frac{1}{3} \log^3 2 - \frac{\pi^2}{12} \log 2 + \frac{\zeta(3)}{8} = -\sum_{n=1}^{\infty} \frac{(-1)^{n-1} \mathcal{H}_n^2}{n+1} - \frac{\zeta(3)}{8} \Rightarrow \\\\\\
\Rightarrow \sum_{n=1}^{\infty} \frac{(-1)^{n-1} \mathcal{H}_n^2}{n+1}= -\frac{1}{3} \log^3 2 +\frac{\pi^2}{12} \log 2 - \frac{\zeta(3)}{4}$$
Of course this is an indirect way of evaluating the sum. Any other way of tackling it?
| Some preliminary lemma.
Lemma 1.
$$ \sum_{n\geq 1} H_n x^n = \frac{\log(1-x)}{1-x}. $$
Lemma 2. By Lemma $1$,
$$ \sum_{n\geq 1}\frac{H_n}{n+1} x^{n+1} = \frac{1}{2}\log^2(1-x),\qquad \sum_{n\geq 1}\frac{H_n+H_{n+1}}{n+1}x^{n}=\frac{-x+\log^2(1-x)+\text{Li}_2(x)}{x}.$$
Lemma 3. Since $H_{n+1}^2-H_n^2 = \frac{H_n+H_{n+1}}{n+1}$,
$$ \sum_{n\geq 1}H_{n}^2 x^n = \frac{\log^2(1-x)+\text{Li}_2(x)}{1-x}.$$
Lemma 4. By Lemma 3,
$$ \sum_{n\geq 1}\frac{(-1)^{n+1} H_{n}^2}{n+1} = -\int_{0}^{1}\frac{\log^2(1+x)+\text{Li}_2(-x)}{1+x}\,dx=-\frac{\log^3(2)}{3}-\color{red}{\int_{0}^{1}\frac{\text{Li}_{2}(-x)}{1+x}\,dx}.$$
The problem boils down to the evaluation of the last integral. By integration by parts, it is:
$$ \color{red}{\int_{0}^{1}\frac{\text{Li}_{2}(-x)}{1+x}\,dx}=-\frac{\pi^2}{12}\log(2)+\color{blue}{\int_{0}^{1}\frac{\log^2(1+x)}{x}\,dx}\tag{1} $$
but:
$$ \color{blue}{\int_{0}^{1}\frac{\log^2(1+x)}{x}\,dx} = -2\int_{0}^{1}\frac{\log(1+x)\log(x)}{1+x}\,dx=\color{blue}{\frac{\zeta(3)}{4}}\tag{2}$$
and the proof is complete.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1814467",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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} |
Prove $\int_{0}^{\infty}e^{-nx}\sin^{2k}xdx={(2k)!\over n\prod_{j=1}^{k}(n^2+4j^2)}$
$$I=\int_{0}^{\infty}e^{-nx}\sin^{2k}xdx={(2k)!\over n\prod_{j=1}^{k}(n^2+4j^2)}\tag1$$
Recall
$$\sin^{2k}(x)={1\over 2^{2k}}{2k\choose k}+{2\over 2^{2k}}\sum_{j=0}^{k-1}(-1)^{k-j}{2k\choose j}\cos[(2k-2j)x]$$
$$I={1\over 2^{2k}}{2k\choose k}\int_{0}^{\infty}e^{-nx}dx+{2\over 2^{2k}}\sum_{j=0}^{k-1}(-1)^{k-j}{2k\choose j}\int_{0}^{\infty}e^{-nx}\cos[(2k-2j)x]dx\tag2$$
$$I={1\over 2^{2k}n}{2k\choose k}+{2\over 2^{2k}}\sum_{j=0}^{k-1}(-1)^{k-j}{2k\choose j}\int_{0}^{\infty}e^{-nx}\cos[(2k-2j)x]dx\tag3$$
$$\int_{0}^{\infty}e^{-nx}\cos[(2k-2j)]dx={n\over n^2+(2k-2j)^2}$$
$$I={1\over 2^{2k}n}{2k\choose k}+{2\over 2^{2k}}\sum_{j=0}^{k-1}(-1)^{k-j}{2k\choose j}\cdot{n\over n^2+(2k-2j)^2}\tag4$$
How can I simplify the LHS sum to the RHS product?
$${1\over 2^{2k}n}{2k\choose k}+{2\over 2^{2k}}\sum_{j=0}^{k-1}(-1)^{k-j}{2k\choose j}\cdot{n\over n^2+(2k-2j)^2}={(2k)!\over n\prod_{j=1}^{k}(n^2+4j^2)}\tag5$$
| Following Olivier's idea in the comments, let $I_k = \int_0^\infty e^{-nx} \sin^{2k}\! x \ dx$ for $k\geq1$.
$$\begin{align}
I_k &= \int_0^\infty e^{-nx} \sin^{2k}\! x \ dx \\
&=\left[-\frac{1}{n}e^{-n x}\sin^{2k}\! x\right]_0^\infty-\int_0^\infty -\frac{2k}{n}e^{-nx} \sin^{2k-1}\! x \cos x\ dx \\
&=\frac{2k}{n} \int_0^\infty e^{-n x}\sin^{2k-1}\!x\cos x\ dx \\
&=\frac{2k}{n}\left(\left[-\frac{1}{n}e^{-n x}\sin^{2k-1}\!x\cos x\right]_0^\infty-\int_0^\infty -\frac{1}{n}e^{-n x}\frac{d}{dx}(\sin^{2k-1}\!x\cos x)\ dx \right) \\
&=\frac{2k}{n^2}\int_0^\infty e^{-n x}\left((2k-1)\sin^{2k-2}\!x \cos^2 x-\sin^{2k}\!x\right)\ dx \\
&=\frac{2k}{n^2}\int_0^\infty e^{-n x}\left((2k-1)\sin^{2k-2}\!x (1-\sin^2 x)-\sin^{2k}\!x\right)\ dx \\
&=\frac{2k}{n^2}\int_0^\infty e^{-n x}\left((2k-1)\sin^{2k-2}\!x-2k\sin^{2k}\!x\right)\ dx \\
&=\frac{2k\times(2k-1)}{n^2}I_{k-1}-\frac{2k \times2k}{n^2}I_{k}\\
n^2 I_k &= 2k(2k-1)I_{k-1}-4k^2I_k \\
(n^2+4k^2)I_k&=2k(2k-1)I_{k-1} \\
I_k &= \frac{2k(2k-1)}{n^2+4k^2} I_{k-1}
\end{align}$$
We also need to check $I_0$:
$$\begin{align}I_0 &= \int_0^\infty e^{-nx}dx \\
&= \left[-\frac{1}{n}e^{-nx}\right]_0^\infty \\
&= \frac{1}{n}\end{align}$$
So, $I_1 = \frac{2\times1}{n^2+4} \frac{1}{n} = \frac{2!}{n^2+4} \frac{1}{n}$, similarly $I_2 = \frac{4\times3}{n^2+4\times2^2} \frac{2!}{n^2+4} \frac{1}{n} = \frac{4!}{(n^2+4\times2^2)(n^2+4)} \frac{1}{n}$, and so on. You can follow the pattern (or, more formally, use induction) to see that
$$I_k = \frac{(2k)!}{n \prod_{j=1}^k (n^2 + 4j^2)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1814913",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
$5$ is quadratic residue mod $p$ if and only if $ p\equiv \pm 1, \pm 9 \pmod {20}$ 5 is quadric residue mod p if and only if $ p\equiv +/- 1, +/-9 \pmod {20}$
$$(5/p)=(p/5)$$
$p\equiv 1 \pmod 4$ ⟹ $1,5,9,13,17 \pmod {20}$
$p\equiv 1 \pmod 5$ ⟹ $1,6,11,16 \pmod {20}$
then $p\equiv 1 \pmod {20}$
$p\equiv 1 \pmod 4$ ⟹ $1,5,9,13,17 \pmod {20}$
$p\equiv 4 \pmod 5$ ⟹ $4,9,14,19 \pmod {20}$
then $p\equiv 9 \pmod {20}$
And now I should show that $$(5/p)=-(p/5)$$
$p\equiv 3 \pmod 4$ ⟹ $3,7,11,15,19 \pmod {20}$
and p is congruent to 2 mod 5 it's p ≡ 7 mod 20
and p ≡3 mod 5 it's p ≡3 mod 20
And it doesn't +/-1 and +/-9 Why?
Please of help.
| Because $5\equiv 1 \pmod 4$, we have $(5/p)=(p/5)$ by quadratic reciprocity, so there is no second case. The squares mod $5$ are $\pm 1$. Modulo $20$, this gives $1,4,6,9,11,14,16,19$. However, because primes are odd, this eliminates all the even numbers in the list, leaving just $1,9,11,19$, which is the same as $\pm 1, \pm 9 \pmod{20}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1815227",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
prove if n - natural number divide number $34x^2-42xy+13y^2$ then n is sum of two square number prove if n - natural number divide number $34x^2-42xy+13y^2$ where x,y are relatively prime then n is sum of two square number.
I don't know what is going on in this exercise. I will be grateful for explanation.
| They are asking about Gauss reduction. The given form is positive and of discriminant $−4.$ That means there is an invertible (determinant 1) change of variables in $SL_2 \mathbb Z,$ call it
$$
P =
\left(
\begin{array}{rr}
\alpha & \beta \\
\gamma & \delta
\end{array}
\right)
$$
that takes one form to the other, and preserves gcd. If $u=2x+3y, v=3x+5y,$ then $$34u^2−42uv+13v^2=x^2+y^2$$
Backwards, $$(5x−3y)^2+(−3x+2y)^2=34x^2−42xy+13y^2.$$
Here we are looking at
$$
H =
\left(
\begin{array}{rr}
34 & -21 \\
-21 & 13
\end{array}
\right)
$$ and
$$
P =
\left(
\begin{array}{rr}
2 & 3 \\
3 & 5
\end{array}
\right)
$$
Taking
$$
Q = P^{-1} =
\left(
\begin{array}{rr}
5 & -3 \\
-3 & 2
\end{array}
\right),
$$
The first displayed line equation $u=2x+3y, v=3x+5y,$ $34u^2−42uv+13v^2=x^2+y^2$ is
$$ P^T H P = I, $$ the second is
$$ Q^T Q = H. $$ The second ought to make you think of Cholesky, only with integers required.
It happens that $P$ and so $Q$ are symmetric so the transpose signs are redundant, not important.
EDIT: Gauss reduction is a step by step procedure intimately linked with the Euclidean algorithm. At each step, we are changing the symmetric matrix in the middle by multiplying on the right by either some
$$
P_j =
\left(
\begin{array}{rr}
1 & \beta \\
0 & 1
\end{array}
\right)
$$
or
$$
P_j =
\left(
\begin{array}{rr}
0 & -1 \\
1 & 0
\end{array}
\right)
$$
and keeping the running product $P_1 P_2 \cdots P_r = P.$ The changes on the triple of coefficients go this way:
$$ \langle 34, -42, 13 \rangle, $$
$$ \langle 13, 42, 34 \rangle, $$
$$ \langle 13, -10, 2 \rangle, $$
$$ \langle 2, 10, 13 \rangle, $$
$$ \langle 2, 2, 1 \rangle, $$
$$ \langle 1, -2, 2 \rangle, $$
$$ \langle 1, 0, 1 \rangle. $$
The particular string $P_1 P_2 \cdots P_6 = P$ that accomplishes this is
$$
\left(
\begin{array}{rr}
0 & -1 \\
1 & 0
\end{array}
\right)
\left(
\begin{array}{rr}
1 & -2 \\
0 & 1
\end{array}
\right)
\left(
\begin{array}{rr}
0 & -1 \\
1 & 0
\end{array}
\right)
\left(
\begin{array}{rr}
1 & -2 \\
0 & 1
\end{array}
\right)
\left(
\begin{array}{rr}
0 & -1 \\
1 & 0
\end{array}
\right)
\left(
\begin{array}{rr}
1 & 1 \\
0 & 1
\end{array}
\right) =
\left(
\begin{array}{rr}
2 & 3 \\
3 & 5
\end{array}
\right)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1816083",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Prove that $ A=\sqrt{\frac{(b-c)^2}{a^2}+\frac{(c-a)^2}{b^2}+\frac{(a-b)^2}{c^2}} $ is also rational number. Let $a,b,c\in\mathbb{Q}$ distinct and none of them equal to $0$ satisfying $\frac{a^2}{(b-c)^2}+\frac{b^2}{(c-a)^2}+\frac{c^2}{(a-b)^2}\leq 2. $ Prove that $ A=\sqrt{\frac{(b-c)^2}{a^2}+\frac{(c-a)^2}{b^2}+\frac{(a-b)^2}{c^2}} $ is also rational number. Is there a simply way?
| $$\sum_{cyc}\frac{a^2}{(b-c)^2}-2=\frac{\left(a^3-a^2 b-a b^2+b^3-a^2 c+3 a b c-b^2 c-a c^2-b c^2+c^3\right)^2}{(a-b)^2 (a-c)^2 (b-c)^2}$$
It follows that the LHS can be $\leq 0$ in very few cases:
$$ abc=(a+b-c)(a-b+c)(-a+b+c).$$
Then it is not difficult to prove that such identity grants $\sum_{cyc}b^2 c^2 (b-c)^2$ to be a rational square.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1816338",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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An example of a non-abelian group of order $p^3$ Let $p$ be prime and consider the set of all matrices $$\begin{bmatrix} 1 & a & b \\ 0 & 1 & c \\ 0 & 0 & 1 \end{bmatrix}$$ where $a,b,c \in \mathbb{Z_p}$. This set forms a non-abelian group of order $p^3$.
Any ideas how to prove using group actions?
| $${{H}_{p}}=\left\{ \left. \left( \begin{matrix}
1 & a & b \\
1 & 1 & c \\
1 & 1 & 1 \\
\end{matrix} \right)\,\, \right|\,\,\,a,b,c\in {{Z}_{{{p}^{3}}}} \right\}
$$
$$\left( \begin{matrix}
1 & a & b \\
0 & 1 & c \\
0 & 0 & 1 \\
\end{matrix} \right)\left( \begin{matrix}
1 & a' & b' \\
0 & 1 & c \\
0 & 0 & 1 \\
\end{matrix}' \right)=\left( \begin{matrix}
1 & a+a' & b+b'+ac' \\
0 & 1 & c+c' \\
0 & 0 & 1 \\
\end{matrix} \right)\ne \left( \begin{matrix}
1 & a+a' & b+b'+a'c \\
0 & 1 & c+c' \\
0 & 0 & 1 \\
\end{matrix} \right)
$$
$$\left( \begin{matrix}
1 & a & b \\
0 & 1 & c \\
0 & 0 & 1 \\
\end{matrix} \right)\left( \begin{matrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1 \\
\end{matrix} \right)=\left( \begin{matrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1 \\
\end{matrix} \right)\left( \begin{matrix}
1 & a & b \\
0 & 1 & c \\
0 & 0 & 1 \\
\end{matrix} \right)=\left( \begin{matrix}
1 & a & b \\
0 & 1 & c \\
0 & 0 & 1 \\
\end{matrix} \right)
$$
$${{\left( \begin{matrix}
1 & a & b \\
0 & 1 & c \\
0 & 0 & 1 \\
\end{matrix} \right)}^{-1}}=\left( \begin{matrix}
1 & -a & ac-b \\
0 & 1 & -c \\
0 & 0 & 1 \\
\end{matrix} \right)
$$
let
$$G=\left\{ \left. \left( \begin{matrix}
x & y \\
0 & 1 \\
\end{matrix} \right)\, \right|\,\,x,y\in {{Z}_{{{p}^{2}}}}\,,\,\,x=1\,\,\bmod \,p\, \right\}
$$Indeed we can say $H_p$ and $G$ are isomorphic!
| {
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"url": "https://math.stackexchange.com/questions/1817166",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Clarification regarding integral of $\arcsin (\sqrt x)$ I need to integrate $\arcsin (\sqrt x)$. I tried using integration by parts but going wrong somewhere. Please help.
\begin{align}
&x\arcsin (\sqrt x)-\frac {1}{2}\int \frac{x}{\sqrt{x-x^2}}\,dx \\
&=x\arcsin (\sqrt x)-\frac {1}{2}\int \frac{x-1/2}{\sqrt{1/4-(x-1/2)^2}}dx+\frac {1}{2}\int \frac{1/2}{\sqrt{1/4-(x-1/2)^2}}\,dx\\
&=x\arcsin (\sqrt x)+\frac {1}{2}\sqrt {x-x^2}+ \frac {1}{4}\int \frac{1}{\sqrt{1/4-(x-1/2)^2}}\,dx\\
&=x\arcsin (\sqrt x)+\frac {1}{2}\sqrt{(x-x^2)}+ (1/4)\arcsin ((x-1/2)/(1/2))
\end{align}
But the answer given in my book is $-1/2\arcsin (\sqrt x)(1-2x)+1/2\sqrt x\sqrt{1-x}$
Btw I took 1 as the first function while applying integration by parts.
| You've made a mistake with the signs.
$$-\frac {1}{2}\int \frac{x}{\sqrt{x-x^2}}\,dx = -\frac 12 \left(\int \frac{x-\frac12}{\sqrt{\frac 14-(x-\frac 12)^2}}dx+\int \frac{\frac 12}{\sqrt{\frac 14-(x-\frac 12)^2}}\,dx\right)$$
Notice the negative factors into the 2nd integral as well, which you overlooked.
Differentiating the result with this issue resolved yields the correct answer.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Linear Relations $y=mx+b$ and graphing lines Consider the linear relation $2x - 3y = 12$.
a) Find $y$ when $x = 3$.
b) Solve the original equation in the question for $y$.
c) Use your equation from b) to find y when $x = 3$.
d) Do you prefer finding the answer using step a), or do you prefer finding the answer by combining steps b) and c)? Explain
How do I do this?!
| Part a. To find $y$ when $x=3$ we begin by substituting in the value $3$ where we see $x$ in the equation: $$2x-3y=12 \Rightarrow \\2\cdot3 - 3y = 12 \Rightarrow \\6-3y = 12.$$ Next we simplify, first by moving the $6$ over by subtracting it on both sides, then by dividing out the coefficient $-3$. That is: $$6-3y=12 \Rightarrow \\ -3y = 12-6 \Rightarrow \\-3y = 6 \Rightarrow \\y = \frac{6}{-3} = -2.$$
Part b. We do the same thing, but without substituting first: $$2x - 3y = 12 \Rightarrow \\ -3y = 12 - 2x \Rightarrow \\ y = \frac{12}{-3} - \frac{2}{-3}x = \frac{2}{3}x - 4.$$
Part c. Simply plug 3 into the equation you just got, that is: $$y = \frac{2}{3}x - 4 \Rightarrow \\ y = \left(\frac{2}{3}\cdot 3\right) - 4 = 2 - 4 = -2.$$
Part d. Using b and c is typically prefered, especially if you need to check the value at multiple points, that way you aren't doing the same work over and over!
| {
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"source": "stackexchange",
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Prove that $\lfloor\frac{n+1}{2}\rfloor+\lfloor\frac{n+2}{4}\rfloor+\lfloor\frac{n+4}{8}\rfloor+\lfloor\frac{n+8}{16}\rfloor+ \dots=n$
Prove $$\left[\dfrac{n+1}{2}\right]+\left[\dfrac{n+2}{4}\right]+\left[\dfrac{n+4}{8}\right]+\left[\dfrac{n+8}{16}\right] + \dots=n$$
where $[x]=\lfloor x\rfloor$
$$$$
It was suggested that somehow I use the identity $[x]=\left[\dfrac x2\right]+\left[\dfrac{x+1}{2}\right]$$$$$After struggling for a while, I realised I wasn't getting anywhere using his hint, probably because I couldn't really understand how I was to use it. Instead I tried to use the Squeeze Theorem by rewriting the $nth$ term of the series (referred to later as S) as$$$$ $$t_n=\left[\dfrac{n+2^k}{2^{k+1}}\right] \text{ where } 0\le k<\infty$$
$$\Rightarrow \dfrac{n+2^k}{2^{k+1}}-1<\left[\dfrac{n+2^k}{2^{k+1}}\right]\le \dfrac{n+2^k}{2^{k+1}}$$
$$$$
$$ \lim_{k\to \infty}(k+1)\left(\dfrac{n+2^k}{2^{k+1}}-1\right)<S\le \lim_{k\to \infty}(k+1)\left( \dfrac{n+2^k}{2^{k+1}}\right)$$
However these bounds are too loose as the limits diverge to $-\infty$ and $\infty$ respectively.
$$$$
Could somebody please show me how to prove the series is equal to $n$, either through the given hint, or through the selection of tighter bounds for the Squeeze Theorem? Many thanks!
| $$\begin{align}
\qquad\qquad\qquad \qquad\qquad\qquad\qquad\qquad\left[ \frac{x+1}2\right]&=\left[x\right]-\left[\frac x2\right]\end{align}$$
Put $x=\dfrac n{2^r}$:
$$\begin{align}
\left[ \frac{\frac n{2^r}+1}2\right]&=\left[\frac n{2^r}\right]-\left[\frac {\frac n{2^r}}2\right]\\
\left[\frac{n+2^r}{2^{r+1}}\right]&=\left[\frac n{2^r}\right]-\left[\frac n{2^{r+1}}\right]\\
r=0:\qquad \left[\frac{n+1}{2}\right]&=\;\left[n\right]\;-\left[\frac n{2}\right]\\
r=1:\qquad \left[\frac{n+2}{4}\right]&=\left[\frac n{2}\right]-\left[\frac n{4}\right]\\
r=2:\qquad \left[\frac{n+4}{4}\right]&=\left[\frac n{4}\right]-\left[\frac n{8}\right]\\
r=3:\qquad \left[\frac{n+8}{16}\right]&=\left[\frac n{8}\right]-\left[\frac n{16}\right]\\
&\vdots\\
&\vdots\\
\text{Summing:}
\left[\dfrac{n+1}{2}\right]+\left[\dfrac{n+2}{4}\right]+\left[\dfrac{n+4}{8}\right]+\left[\dfrac{n+8}{16}\right] + \dots&=\left[n\right]\\
&=n\qquad\blacksquare
\end{align}$$
| {
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"url": "https://math.stackexchange.com/questions/1821536",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
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"answer_id": 3
} |
Prove that the determinant is $(a-b)(b-c)(c-a)(a+b+c)$ I have the determinant :
\begin{vmatrix}
1 &1 &1 \\
a &b &c \\
a^3 &b^3 &c^3 \\
\end{vmatrix}
How do I prove that this determinant is equal to
$$ (a-b)(b-c)(c-a)(a+b+c) $$
| HINT:
Use $$C_1'=C_1-C_3$$ and $$C_2'=C_2-C_3$$
where $C_r$ is the $r$th column, $C_r'$ is the resultant $r$th column
See also: Factorise the determinant $\det\Bigl(\begin{smallmatrix} a^3+a^2 & a & 1 \\ b^3+b^2 & b & 1 \\ c^3+c^2 & c &1\end{smallmatrix}\Bigr)$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Lines tangent to two circles I'm trying to find the lines tangent to two circles. I've seen several examples but with poorlyy explained methods. Given the circle
$(x-x_{0})^2+(y-y_{0})^2=r_{1}^2$
and the the line equation
$y=ax+b$
one of the method is based on the relation
$(ax_{0}+b-y_{0})^2=(a^2+1)r_{1}^{2}$
Can you tell me what relation is this? How it was obtained? Thank you.
| Consider a circle $C$
$$
(X-x_0)^2+(Y-y_0)^2=r^2
$$
and a line $L$ given by $Y=aX+b$. Then if $C$ and $L$ intersect each other at a point $(x,y)$, then since $(x,y)\in L$ you have $y=ax+b$ and since $(x,y)\in C$ we have $(x-x_0)^2+(ax+b-y_0)^2=r^2$. Therefore
$$
\Longrightarrow (a^2+1)x^2 +2x[x_0+a(b-y_0)]+x_0^2+(b-y_0)^2-r^2=0
$$
This is a quadratic equation with solutions
$$
x=\frac{-(x_0+a(b-y_0)\pm \sqrt{(x_0+a(b-y_0))^2+(a^2+1)(r^2-x_0^2-(b-y_0)^2)}}{a^2+1}
$$
The $\Delta$ of this quadratic equation (the term under the square root) must vanish if the line $L$ is tangent to $C$ because otherwise the line intersects $C$ at two points ($\Delta>0$) or no points $(\Delta<0)$. Hence we must have
$$
\begin{aligned}
0=\Delta=2ax_0(b-y_0)&=(x_0+a(b-y_0))^2+(a^2+1)(r^2-x_0^2-(b-y_0)^2)\\
&= x_0^2+2ax_0(b-y_0)+a^2(b-y_0)^2+(a^2+1)(r^2-x_0^2-(b-y_0)^2)\\
&=(1+a^2)r^2-(b-y_0)^2-a^2x_0^2+2ax_0(b-y_0)\\
&=(1+a^2)r^2 - (ax_0+b-y_0)^2\Longrightarrow\\
(ax_0+b-y_0)^2&=(1+a^2)r^2
\end{aligned}
$$
What this means is that the line $L$ given via $y=ax+b$ is tangent to $C$ if and only if $a,b$ satisfy
$$
(ax_0+b-y_0)^2=(1+a^2)r^2
$$
Now if $C'$ is another circle with center $O'=(x'_0,y'_0)$ and radius $r'$, and if $L$ is also tangent to $C'$ then $a,b$ are the mutual solution to the equations
$$
\begin{cases}
(ax_0+b-y_0)^2=(1+a^2)r^2\\
(ax_0'+b-y_0')^2=(1+a^2)r'^2
\end{cases}
$$
Solve this to find your tangent line. This is of course assuming the tangent line is not $x=c$ because we took our line equation to be $y=ax+b$. You need to carry out similar but simpler analysis to check if $x=0$ is a double tangent.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Help to simplify $\arctan\left(\frac{\sqrt{1 + x^2} -1}{x}\right)$ Can someone help me simplify the argument of $\arctan$ in this problem ?
$$\arctan\left(\frac{\sqrt{1 + x^2} -1}{x}\right)$$
| It might be helpful to take the inverse of this function by solving for $x$.
We have the following:
$$y=\arctan\left(\frac{\sqrt{1+x^2}-1}{x}\right)$$
Take the $\tan$ of both sides:
$$\tan y=\frac{\sqrt{1+x^2}-1}{x}$$
Muultiply both sides by $x$ and add $1$:
$$x\tan y+1=\sqrt{1+x^2}$$
Square both sides:
$$x^2\tan^2 y+2x\tan y+1=1+x^2$$
Subtract both sides by $1+x^2$:
$$x^2(\tan^2 y-1)+2x\tan y=0$$
Now, clearly, the original function is undefined for $x=0$, so we can conclude that $x \neq 0$. Therefore, divide by $x$:
$$x(\tan^2 y-1)+2\tan y=0$$
Subtract both sides by $2\tan y$ and divide by $\tan^2 y-1$:
$$x=-\frac{2\tan y}{\tan^2 y-1}$$
Now, this looks very similar to the $\tan(2y)$ identity. However, the denominator is flipped and there is a $-$ sign out front. Therefore, distribute the negative across the denominator:
$$x=\frac{2\tan y}{1-\tan^2 y}=\tan 2y$$
Now, we can solve this equation for $y$ to get a much simpler form of our original equation. Take the $\arctan$ of both sides:
$$\arctan x=2y$$
Divide both sides by $2$ and switch the sides of the equation:
$$y=\frac{\arctan x}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1824815",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Evaluate the contour integral (Most likely without Green's Theorem) $\int_{c}\frac{-y}{x^2+y^2}dx + \frac{x}{x^2+y^2}dy$ where $C$ is the triangle with vertices at $(5,5), (-5,5),$ and $(0,-5)$ traversed counterclockwise. (Hint: Be careful about the hypotheses of any theorem you use).
My attempt: It appears clear that we cannot use Green's theorem here, as the functions are not defined at the origin. Hence, we can break the line integral into three distinct line integrals. For the line between $(5,5)$ and $(-5,5)$, we can substitute $y=5$ into the integral and solve, for the line between $(0,-5)$ and $(5,5)$ we can substitute $y= 2x-5$, and for the final line we can substitute $y = -2x-5$.
My trouble comes with attempting to evaluate the first integral
$\int_{-5}^5\frac{-5}{x^2+25}dx + \frac{x}{x^2+25}dy$.
Can I simplify this to
$\int_{-5}^5\frac{1}{x+5}dx$?
This would give an integral that does not converge, I believe.
Any help greatly appreciated!
| Parameterizing by using the $x$ coordinate as the parameter, we have
$$\begin{align}
\oint_{\partial \mathbb{D}} \left(\frac{-y}{x^2+y^2}\,dx+\frac{x}{x^2+y^2}\,dy\right)&=\int_0^5\left(\frac{-(2x-5)}{x^2+(2x-5)^2}\,dx+\frac{x}{x^2+(2x-5)^2}\,2\,dx\right)\\\\&+\int_{5}^{-5}\left(\frac{-5}{x^2+25}\,dx\right)\\\\&+\int_{-5}^0 \left(\frac{-(-2x-5)}{x^2+(2x+5)^2}\,dx+\frac{x}{x^2+(2x+5)^2}\,(-2)\,dx\right)\\\\&=5\int_0^5 \frac{1}{x^2+(2x-5)^2}\,dx\\\\
&+10\int_0^5 \frac{1}{x^2+25}\,dx\\\\
&+5\int_{-5}^0 \frac{1}{x^2+(2x+5)^2}\,dx\\\\
&=10\int_0^5 \left(\frac{1}{x^2+(2x-5)^2}+\frac{1}{x^2+25}\right)\,dx \\\\
&=10\left(\frac{3\pi}{20}+\frac{\pi}{20}\right)\\\\
&=2\pi
\end{align}$$
NOTE:
We could have exploited Green's Theorem by deforming the original contour with the classical "keyhole" that removes the origin with a circular contour of radius $\epsilon$. Then, transforming to polar coordinates reveals
$$\begin{align}
\oint_{\partial \mathbb{D}} \left(\frac{-y}{x^2+y^2}\,dx+\frac{x}{x^2+y^2}\,dy\right)&=\int_{\sqrt{x^2+y^2}=\epsilon} \left(\frac{-y}{x^2+y^2}\,dx+\frac{x}{x^2+y^2}\,dy\right)\\\\
&=\int_0^{2\pi }\frac{\epsilon}{\epsilon^2}\,\epsilon\,d\phi\\\\
&=2\pi
\end{align}$$
as expected!
| {
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"url": "https://math.stackexchange.com/questions/1825874",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Integral involving Legendre polynomial and $x^n$ I am trying to show that,
\begin{align}
I = \int_{-1}^1 x^nP_n(x)\,\mathrm{d}x = \frac{2^{n+1}n!n!}{(2n+1)!}
\end{align}
So far I have done the following. Rodrigues formula is as follows:
\begin{align*}
P_n(x) = \sum_{k=0}^N \frac{(-1)^k (2n-2k)!}{2^nk!(n-k)!(n-2k)!} x^{n-2k}
\end{align*}
where,
\begin{align*}
\begin{aligned}
N&=n/2, && \text{if} \quad n=\text{even} \\
N&=(n-1)/2, && \text{if} \quad n=\text{odd}
\end{aligned}
\end{align*}
Substitute Rodrigues formula,
\begin{align*}
I
&=
\int_{-1}^1 x^n \sum_{k=0}^N \frac{(-1)^k (2n-2k)!}{2^nk!(n-k)!(n-2k)!} x^{n-2k} \,\mathrm{d}x \\
&=
\sum_{k=0}^N \frac{(-1)^k (2n-2k)!}{2^nk!(n-k)!(n-2k)!} \int_{-1}^1 x^{2n-2k} \,\mathrm{d}x \\
&=
\sum_{k=0}^N \frac{(-1)^k (2n-2k)!}{2^nk!(n-k)!(n-2k)!} \left. \frac{x^{2n-2k+1}}{2n-2k+1}\right\rvert_{-1}^1 \\
&=
\sum_{k=0}^N \frac{(-1)^k (2n-2k)!}{2^nk!(n-k)!(n-2k)!} \frac{1 - (-1)^{2n-2k+1}}{2n-2k+1} \\
&=
\sum_{k=0}^N \frac{(-1)^k (2n-2k)!}{2^nk!(n-k)!(n-2k)!} \frac{1 + (-1)^{2n-2k}}{2n-2k+1}
\end{align*}
Since $2n-2k$ is even,
\begin{align*}
I
&=
\sum_{k=0}^N \frac{(-1)^k (2n-2k)!}{2^nk!(n-k)!(n-2k)!} \frac{2}{2n-2k+1} \\
&=
\sum_{k=0}^N \frac{2^{1-n}(-1)^k (2n-2k)!}{k!(n-k)!(n-2k)!(2n-2k+1)}
\end{align*}
Can someone give me a hint about how to proceed or do I need to prove with another way?
| It is faster to exploit the generating function that comes from Rodrigues' formula or Bonnet's recursion formula:
$$\frac{1}{\sqrt{1-2xt+t^2}} = \sum_{n=0}^\infty P_n(x) t^n\tag{1} $$
By replacing $t$ with $xt$ we have:
$$ \int_{-1}^{1}\frac{dx}{\sqrt{1-2x^2 t+x^2 t^2}}=\sum_{n\geq 0}\left(\int_{-1}^{-1}x^n P_n(x)\,dx\right)t^n\tag{2} $$
so our integrals can be computed through the Taylor series of the LHS, regarded as a function of $t$.
The LHS of $(2)$ is:
$$ \frac{2\arcsin\!\sqrt{t(2-t)}}{\sqrt{t(2-t)}}=\frac{4\arcsin\sqrt{\frac{t}{2}}}{\sqrt{t(2-t)}} \tag{3} $$
and your initial claim turns out to be equivalent to a well-known identity.
Truth to be told, it is also very interesting sideways: some series related with the $\Gamma^2$ function can be easily computed by exploiting the mentioned property of Legendre polynomials.
Using Rodrigues' formula we have:
$$\int_{-1}^{1}x^n P_n(x)\,dx = \frac{1}{2^n n!}\int_{-1}^{1}x^n\cdot\frac{d^n}{dx^n}(x^2-1)^n\,dx \stackrel{RIBP}{=}\frac{1}{2^n}\int_{-1}^{1}(1-x^2)^n\,dx\tag{4}$$
where $\text{RIBP}$ stands for repeated integration by parts.
But the RHS of $(4)$ is a value of the Euler's beta function:
$$\frac{1}{2^n}\int_{-1}^{1}(1-x^2)^n\,dx=\frac{1}{2^n}\int_{0}^{1}z^{-1/2}(1-z)^n\,dz = \frac{\Gamma\left(\frac{1}{2}\right)\Gamma(n+1)}{2^n\,\Gamma\left(n+\frac{3}{2}\right)}\tag{5}$$
and we are equally done, since:
$$\frac{\Gamma\left(\frac{1}{2}\right)\Gamma(n+1)}{2^n\,\Gamma\left(n+\frac{3}{2}\right)} = \frac{2\cdot n!}{(2n+1)!!} = \frac{2^{n+1} n!^2}{(2n+1)!}.\tag{6}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Finding partial fractions expansions mentally On a problem on a test, my students were asked to find $\displaystyle\int\frac{6x^4-7x^3-13x-6}{x^3-2x^2} dx$,
and one student began by writing $\displaystyle\int\frac{6x^4-7x^3-13x-6}{x^2(x-2)} dx=\int\left(\frac{3}{x^2}+6x+\frac{2}{x-2}+\frac{8}{x}+5\right)dx$.
My question is how someone can get this result without doing any division or partial fraction decomposition -- what techniques can be used to get this?
| At a schematic level, what makes this problem "ugly" is the $x-2$ in the denominator. So the problem solving strategy is to isolate the $x-2$; the rest should then be easy to do mentally. To this end, you know from the principle of partial fraction decomposition that $\frac{6x^4-7x^3-13x-6}{x^2(x-2)}$ can be written in the form $\frac{p(x)}{x^2} + \frac{c}{x-2}$ for some constant $c$. Thus, $$6x^4-7x^3-13x-6 = p(x)(x-2) + cx^2,$$ so in particular $6x^4-7x^3 - cx^2 -13x-6$ is divisible by $x-2$.
There is a standard algorithm for determining when a polynomial $a x^4 + bx^3 + cx^2 + dx+e$ is divisible by $x-2$: It is divisible if and only if $e+2(d+2(c+2(b+2(a)))) = 0$. In the case at hand, this translates to $0=-6+2(-13+2(-c+2(-7+2(6)))) = 8-4c$, which means $c=2$.
Now we can determine $p(x) = \frac{6x^4-7x^3-2x^2-13x-6}{x-2} = 6x^3 + 5x^2 + 8x + 3$. Dividing this by $x^2$ gives all the other terms in the decomposition of the original fraction.
There are really only two computations to do here: figuring out $c$, and dividing $6x^4-7x^3-2x^2-13x-6$ by $x-2$, both of which can reasonably be done in one's head.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "12",
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How do you simplify this square root of sum: $\sqrt{7+4\sqrt3}$? I came around this expression when solving a problem.
$$\sqrt{7+4\sqrt{3}}$$
WolframAlpha says it equals $2+\sqrt{3}$. We can confirm it like this
$$\left(2+\sqrt{3}\right)^2 \;=\; 4+4\sqrt{3} + 3 \;=\; 7 + 4\sqrt{3}.$$
However, the only way I can think of how to simplify that expression in hand is guessing. Is there a better way of calculating square root of a sum like that one?
| ${\sqrt {7 + 4{\sqrt3}}} = {\sqrt {7 + \sqrt{48}}}$
The latter can be solved by the formula:
$\sqrt {a + \sqrt{b} } = \sqrt{ {a + \sqrt{a^2 -b}}\over2} +\sqrt{ {a - \sqrt{a^2 -b}}\over2} $
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
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"answer_id": 6
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Solving the given differential equation. We need to solve : $$ ( \sqrt{x+y} + \sqrt{x-y}) \,dx + ( \sqrt{x-y} - \sqrt{x+y})\,dy=0$$
I tried as follows :
$$ \frac{dy}{dx} = \frac{ \sqrt{x+y} + \sqrt{x-y}}{\sqrt{x-y} - \sqrt{x+y}}$$
And hence letting $ y = vx$ yields :
$$ v + x \: \frac{dv}{dx} = \frac{\sqrt{1+v} + \sqrt{1-v}}{\sqrt{1-v} - \sqrt{1+v}}$$
Continuing from here yields a very complicated integral , is there a simpler way to proceed ?
| Here is an alternate approach:
\begin{equation}
( \sqrt{x+y} + \sqrt{x-y}) \,dx + ( \sqrt{x-y} - \sqrt{x+y})\,dy=0
\end{equation}
Before making the substitution $y=xv$ one should first multiply by the conjugate of the second term and simplify the result to obtain
\begin{equation}
( x + \sqrt{x^2-y^2}) \,dx -y\,dy=0
\end{equation}
Then the $y=vx$ substitution gives, after simplification
\begin{equation}
( x-xv^2+x\sqrt{1-v^2}) \,dx -vx^2\,dv=0
\end{equation}
dividing by $x$ and separating variables and doing a bit of rationalizing of the denominator yields an expression solved by elementary integrals.
\begin{equation}
\frac{1}{x}\,dx+\left(\frac{1}{v}-\frac{1}{v\sqrt{1-v^2}} \right)\,dv=0
\end{equation}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Using the definition of a definite integral find $ \int_2^4 (3x^2-2)dx $ Find, with the definition of a definite integral, where $\bar{x}i$ is the right sum of each subinterval.
$$
\int_2^4 (3x^2-2)dx
$$
So I start here...
$$
\Delta xi = 4-2/n = \frac{2}{n}
$$
For the right sum:
$$
\bar{x}i= 1 +i\Delta xi = 1+\frac{2i}{n}
$$
We have:
$$
f(\bar{x}i) = (3x^2-2) = f(\bar{x}i) = (3(1+\frac{2i}{n})^2-2)
$$
$$
f(\bar{x}i) = (\frac{12i^2}{n^2}+\frac{12i}{n}+1)
$$
With the sum of Riemann:
$$
SR= \sum f(\bar{x}i)\Delta xi = \sum (\frac{12i^2}{n^2}+\frac{12i}{n}+1)* \frac{2}{n}
$$
$$
SR = \sum (\frac{24i^2}{n^3}+\frac{24i}{n^2}+\frac{2}{n})
$$
$$
SR= \frac{12}{n^2}\sum i^2 + \frac{12}{n}\sum i + \frac{2}{n}\sum 1
$$
$$
SR= \frac{12}{n^2}(\frac{n(n+1)(2n+1)}{6}) + \frac{12}{n}(\frac{(n(n+1))}{2}) + \frac{2}{n} (n)
$$
$$
SR= \frac{8n^2+8n+4}{n^2} + \frac{12n+12}{n} + 2
$$
$$
SR= \frac{4}{n^2} + \frac{20}{n} + 22
$$
And finally:
$$\lim_{n\rightarrow \infty}\sum f(\bar{x})\Delta xi=\frac{4}{n^2} + \frac{20}{n} + 22 = 22$$
This is my answer. But the answer from online calculators are different (52).. Can anyone spot my mistakes? Thank you.
$$$$
Edit: Okay thank you guys! My mistake right at the beginning... where it should have started at 2 and not 1. Thanks again.
| \begin{align}
& f({{x}_{i}})=3{{\left( 2+\frac{2i}{n} \right)}^{2}}-2=12\left( 1+\frac{2i}{n}+\frac{{{i}^{2}}}{{{n}^{2}}} \right)-2=10+\frac{24i}{n}+\frac{12{{i}^{2}}}{{{n}^{2}}} \\
& \Delta x\sum\limits_{i=1}^{n}{f({{x}_{i}})=\frac{2}{n}}\left( 10n+12(n+1)+\frac{2(n+1)(2n+1)}{n} \right) \\
& \underset{n\to \infty }{\mathop{\lim }}\,\Delta x\sum\limits_{i=1}^{n}{f({{x}_{i}})=20+24+}8=52 \\
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1829019",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove that the equation of the cone is $yz(\frac{b}{c}+\frac{c}{b})+zx(\frac{c}{a}+\frac{a}{c})+xy(\frac{a}{b}+\frac{b}{a})=0$ The plane $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$ cuts the coordinate axes in $A,B,C$. Prove that the lines passing through the origin and intersecting the circle $ABC$ generate the cone with equation
$$yz(\frac{b}{c}+\frac{c}{b})+zx(\frac{c}{a}+\frac{a}{c})+xy(\frac{a}{b}+\frac{b}{a})=0$$
My Attempt:
The equation of the generator line passing through the origin and direction cosines $l,m,n$ is $\frac{x}{l}=\frac{y}{m}=\frac{z}{n}$
Cone is a surface generated by the lines called generators passes through fixed point called vertx on the fixed curve called guiding curves.
I am not able to find the equation of the guiding curve the circle $ABC$.Hence i could not find out the equation of the cone.
Please help.
| I do not know if it is really helpful, but the center $P$ of the circle has (horrible) coordinates $\dfrac{1}{2(a^2b^2+b^2c^2+c^2a^2)}\begin{pmatrix} a^3(b^2+c^2) \\ b^3(c^2+a^2) \\ c^3(a^2+b^2) \end{pmatrix}$.
Indeed, in the plane, we want to find the circle passing through the vertices of the triangle $ABC$. The center $P$ of this circle is the intersection of the perpendicular bisectors of $[AB]$ and $[AC]$. The middle $M$ of $[AB]$ has coordinates $\begin{pmatrix} \frac{a}{2} \\ \frac{b}{2} \\ 0 \end{pmatrix}$. We want the vectors $\overrightarrow{PM}$ and $\overrightarrow{AB}$ to be perpendicular, so we get the equation $ax-by = \dfrac{a^2-b^2}{2}$. Similarly for $[AC]$ we get $ax-cz = \dfrac{a^2-c^2}{2}$. Do not forget the third equation $\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}=1$. Thus we have a linear system to solve, and the unique solution is the aforementioned one.
ADDENDUM after edit:
Let $M=\begin{pmatrix} x\\ y\\ z\end{pmatrix}$ and $t=\dfrac{x}{a}+\dfrac{y}{b} + \dfrac{z}{c}$. The line $(OM)$ cuts the plane if and only if $t \neq 0$. If so, the point of intersection is $M'=\dfrac{1}{t} M_0$. Then $M'$ is on the circle if and only if $\|P-M'\|^2 = \|P-A\|^2$. Now we have (with the obvious abuse of notation about norms of vectors and scalar product) \begin{align*} &\|P-M'\|^2 = \|P-A\|^2 \iff P^2-2P\cdot\dfrac{M}{t} + \dfrac{M^2}{t^2} = P^2 - 2aP_1+a^2 \\
&\iff M^2-2tP\cdot M = (a^2-2aP_1)t^2 \\
&\iff x^2+y^2+z^2-\dfrac{\dfrac{x}{a}+\dfrac{y}{b} + \dfrac{z}{c}}{a^2b^2+b^2c^2+c^2a^2}\left(xa^3(b^2+c^2)+yb^3(c^2+a^2)+zc^3(a^2+b^2)\right) = \dfrac{a^2b^2c^2}{a^2b^2+b^2c^2+c^2a^2}\left(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2} + \dfrac{2xy}{ab}+\dfrac{2yz}{bc}+\dfrac{2zx}{ca}\right)\end{align*}
You can check that the terms in $x^2$, $y^2$ and $z^2$ vanish, and that the remaining terms give you the desired equation. For example, when putting all the terms in the RHS, the coefficient of $xy$ is $\dfrac{1}{a^2b^2+b^2c^2+c^2a^2}\left(2abc^2+\dfrac{b^3(c^2+a^2)}{a} + \dfrac{a^3(b^2+c^2)}{b}\right) = \dfrac{1}{a^2b^2+b^2c^2+c^2a^2} \dfrac{2a^2b^2c^2+b^4(c^2+a^2)+a^4(b^2+c^2)}{ab} = \dfrac{a^2+b^2}{ab} = \dfrac{a}{b}+\dfrac{b}{a}$.
I have to admit this is a bit cumbersome...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1829379",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solve $x^2+x+1\equiv0\pmod5$
Solve:
$$x^2+x+1\equiv0\pmod5$$
My attempt:
Our proffesor told us that if we have $ax^2+bx+c\equiv\pmod p$ we need to multiply by $4a$, to get form of $(\text{ something})^2\equiv D\pmod p$.
$p$ is a prime and $\gcd(4a,p)=1$, so I tried to do that here:
$$4x^2+4x+4\equiv 0 \pmod 5$$
But now
$$(2x+1)^2+3\equiv 0 \pmod 5$$
$$(2x+1)^2\equiv -3 \pmod 5$$
| More generally, there's this trick for solving quadratic congruences: if $\gcd(n,2a)=1$, then:
$$ax^2+bx+c\equiv 0\pmod{n}$$
$$\stackrel{\cdot 4a}\iff (2ax+b)^2\equiv b^2-4ac\pmod{n}$$
In this case,
$$x^2+x+1\equiv 0\pmod{5}$$
$$\stackrel{\cdot 4}\iff (2x+1)^2\equiv -3\equiv 2\pmod{5},$$
contradiction, because $2$ is not a quadratic residue mod $5$ (to see this, notice $(5k\pm 1)^2\equiv 1\pmod{5}$, $(5k\pm 2)^2\equiv 4\pmod{5}$, $(5k)^2\equiv 0\pmod{5}$).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1830591",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
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