Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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How to evaluate the following integral involving hyperbolic functions? I've been thinking about using contour integration, but I can't seem to find the right combination of function and contour. Thanks for your attention. :)
$$
\int_{0}^{\infty} \frac {\sinh ax \sinh bx}{\cosh cx} \ dx
$$
| Consider the contour integral
$$\oint_C dz \frac{\sinh{a z} \sinh{b z}}{\sinh{c z}} $$
where $C$ is the rectangle $-R-i \pi/(2 c), R-i \pi/(2 c), R+i \pi/(2 c), -R+i \pi/(2 c)$. Note that
$$\sinh{c (x \pm i \pi/2 c)} = \pm i \cosh{c x} $$
Therefore the contour integral is
$$\int_{-R}^R dx \, \frac{\sinh{a(x-i \pi/(2 c))} \sinh{b(x-i \pi/(2 c))}}{-i \cosh{c x}}\\ +i \int_{-\pi/(2 c)}^{\pi/(2 c)} dy \, \frac{\sinh{a (R+i y)} \sinh{b (R+i y)}}{\sinh{c (R+i y)}}\\ + \int_{R}^{-R} dx \, \frac{\sinh{a(x+i \pi/(2 c))} \sinh{b(x+i \pi/(2 c))}}{i \cosh{c x}} \\-i \int_{\pi/(2 c)}^{-\pi/(2 c)} dy \, \frac{\sinh{a (R-i y)} \sinh{b (R-i y)}}{\sinh{c (R-i y)}}$$
Note that, for the second and fourth integrals to vanish as $R \to \infty$, we must have $|a|+|b| \lt |c|$; assume this is the case. The contour integral then becomes, after simplification (the algebra of which I will spare the reader) (*):
$$i 2 \cos \left(\frac{\pi a}{2 c}\right) \cos \left(\frac{\pi b}{2
c}\right) \int_{-\infty}^{\infty} dx \frac{ \sinh (a x) \sinh (b x)}{\cosh{c x}}\\-i 2 \sin \left(\frac{\pi a}{2 c}\right) \sin
\left(\frac{\pi b}{2 c}\right) \int_{-\infty}^{\infty} dx \frac{\cosh (a x) \cosh (b x)}{\cosh{c x}}$$
Note that we don't quite have the integral we want. So now consider
$$\oint_C dz \frac{\cosh{a z} \cosh{b z}}{\sinh{c z}} $$
and as a result of similar steps as above, the contour integral becomes (**)
$$i 2 \cos \left(\frac{\pi a}{2 c}\right) \cos \left(\frac{\pi b}{2
c}\right) \int_{-\infty}^{\infty} dx \frac{ \cosh (a x) \cosh (b x)}{\cosh{c x}}\\-i 2 \sin \left(\frac{\pi a}{2 c}\right) \sin
\left(\frac{\pi b}{2 c}\right) \int_{-\infty}^{\infty} dx \frac{\sinh (a x) \sinh (b x)}{\cosh{c x}}$$
We can eliminate the integral we are not seeking by multiplying $\text{(*)}$ by $\cos \left(\frac{\pi a}{2 c}\right) \cos \left(\frac{\pi b}{2
c}\right)$ and $\text{(**)}$ by $\sin \left(\frac{\pi a}{2 c}\right) \sin \left(\frac{\pi b}{2
c}\right)$. Thus, really, we are then seeking
$$\oint_C dz \, \frac{f(z)}{\sinh{c z}}$$
where
$$f(z) = \cos \left(\frac{\pi a}{2 c}\right) \cos \left(\frac{\pi b}{2 c}\right) \sinh{a z} \sinh{b z} + \sin \left(\frac{\pi a}{2 c}\right) \sin \left(\frac{\pi b}{2 c}\right) \cosh{a z} \cosh{b z} $$
The contour integral now takes the form
$$i 2 \left [\cos^2 \left(\frac{\pi a}{2 c}\right) \cos^2 \left(\frac{\pi b}{2 c}\right) - \sin^2 \left(\frac{\pi a}{2 c}\right) \sin^2 \left(\frac{\pi b}{2 c}\right) \right ] \int_{-\infty}^{\infty} dx \frac{\sinh (a x) \sinh (b x)}{\cosh{c x}}$$
This is equal to $i 2 \pi$ times the residue of $f(z) \operatorname{csch}{c z}$ at the pole $z=0$, which is simply $f(0)/c$. Using the symmetry of the integral, we finally have
$$\begin{align}\int_{0}^{\infty} dx \frac{\sinh (a x) \sinh (b x)}{\cosh{c x}} &= \frac{\pi}{2 c} \frac{\sin \left(\frac{\pi a}{2 c}\right) \sin \left(\frac{\pi b}{2 c}\right )}{\cos^2 \left(\frac{\pi a}{2 c}\right) \cos^2 \left(\frac{\pi b}{2 c}\right) - \sin^2 \left(\frac{\pi a}{2 c}\right) \sin^2 \left(\frac{\pi b}{2 c}\right) } \\ &= \frac{\pi}{4 c} \frac{\cos{\left(\frac{\pi (a-b)}{2 c}\right)}-\cos{\left(\frac{\pi (a+b)}{2 c}\right)}}{\cos{\left(\frac{\pi (a-b)}{2 c}\right)} \cos{\left(\frac{\pi (a+b)}{2 c}\right)}}\\ &= \frac{\pi}{4 c} \left [\sec{\left(\frac{\pi (a+b)}{2 c}\right)}-\sec{\left(\frac{\pi (a-b)}{2 c}\right)} \right ] \end{align}$$
where, again, $|a|+|b|\lt|c|$.
| {
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"url": "https://math.stackexchange.com/questions/781522",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Three Congruent Incircles of a divided Equilateral triangle Take an equilateral triangle with sides of unit length and choose a vertex from which to draw two cevians to the opposite side.These cevians divide the equilateral triangle into three subtriangles.
If these three subtriangles all have congruent incircles, can anyone confirm that the length of these cevians is $1/4(3^{1/3}+3^{2/3})$ and the inradius of the congruent incircles is $1/2(3^{1/2}/(3+3^{1/3}+3^{2/3}))$?
Using the Paul Yiu article at http://www.math-cs.ucmo.edu/~mjms/2003.1/pyiu.pdf, I had to solve a quartic equation to get the length of the cevians h where - $3 - 18 h - 27 h^2 + 16 h^3 + 48 h^4 = 0$. Is there a simpler method of calculating the cevian length? This question was first asked in Mathoverflow.
| Label elements of equilateral $\triangle ABC$ as shown:
Using the fact that
$$\text{area of } \triangle = \frac{1}{2} \cdot \text{inradius} \cdot \text{perimeter}$$
we have
$$\frac{2|\triangle APB|}{1+x+y} = r = \frac{2|\triangle APQ|}{1-2x+2y}$$
$$\frac{x \cdot \frac{\sqrt{3}}{2}}{1+x+y} = r = \frac{(1-2x) \cdot \frac{\sqrt{3}}{2}}{1-2x+2y}$$
so that
$$x ( 1 - 2 x + 2 y ) = ( 1 - 2 x )( 1 + x + y ) \quad\to\quad y ( 4 x - 1 ) = 1 - 2 x$$
By the Law of Cosines,
$$y^2 = 1^2 + x^2 - 2\cdot 1\cdot x\cos 60^\circ = x^2 - x + 1$$
which implies
$$( x^2 - x + 1 )( 4 x - 1 )^2 = ( 1 - 2 x )^2 \quad\to\quad x (16 x^3 - 24 x^2 + 21 x - 5 ) = 0$$
The case $x=0$ is extraneous (and evidently corresponds to the alternate problem of congruent incircles in $\triangle APQ$, $\triangle ABQ$, $\triangle ACP$), and Mathematica tells me that the sole real solution to the cubic factor is
$$x = \frac{1}{4} (2 + s - s^2) = \frac{1}{4}(2-s)(1+s)$$
where $s := 3^{1/3}$.
Then,
$$y^2 = \frac{1}{16} (6 + 3\cdot s + s^2) = \frac{1}{16}(2 s^3 + s^4 + s^2 ) = \frac{s^2}{16}(1+s)^2 \quad\to\quad y = \frac{s}{4} ( 1 + s )$$
$$r = \frac{\frac{\sqrt{3}}{2}(1-2x)}{1 -2x+2y} = \frac{\frac{\sqrt{3}}{4}s(s-1)}{s^2} = \frac{(s-1)\sqrt{3}}{4s} = \frac{(s^3-1)\sqrt{3}}{4s(1+s+s^2)} = \frac{\sqrt{3}}{2s(1+s+s^2)}$$
which match the values given in the problem.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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$\int_{0}^{\infty}\frac{x}{x^3+1}dx$ =? So guys I have this improper integral $\int_{0}^{\infty}\frac{x}{x^3+1}dx$. I checked that it converges by $ \int_{0}^{1}\frac{x}{x^3+1}dx + \int_{1}^{\infty}\frac{x}{x^3+1}dx $ and using the $\frac{c}{(x-a)^\lambda} $ and $\frac{1}{x^\lambda} $ criteria. But for finding the value I have trouble.I tried $ \int_{0}^{\infty}\frac{x}{(x+1)(x^2-x+1)}dx$ and $\int_{0}^{\frac{\pi}{2}}\frac{tg\alpha}{tg\alpha^3+1}dx$ but it gets me nowhere. Any ideas on solving it?
| Continuing Daniel Fischer's answer, we have:
$$\int_{0}^{1}\frac{dx}{x^2-x+1}=\int_{-1/2}^{1/2}\frac{dx}{x^2+3/4}=\frac{\sqrt{3}}{2}\cdot\frac{4}{3}\int_{-1/\sqrt{3}}^{1/\sqrt{3}}\frac{dx}{x^2+1}=\frac{4\sqrt{3}}{3}\arctan\frac{1}{\sqrt{3}}=\frac{2\pi}{3\sqrt{3}}.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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If $y =\sqrt{5+\sqrt{5-\sqrt{5+ \cdots}}}$, what is the value of $y^2-y$? If $$y =\sqrt{5+\sqrt{5-\sqrt{5 + \cdots}}},$$
what is the value of $y^2-y$ ?
I am unable to get the clue due to these alternative signs of plus and minus please help on this thanks...
| Indeed, there is another way:
Let $x =\sqrt{5-\sqrt{5+\sqrt{5 - \cdots}}}$,
then it's easy to see
$y^2 = 5 + x$, $x^2 = 5 - y$
then
$y^2 - x^2 = x+y$, therefore $y-x = 1$
Thus,
$$y^2 - y = y^2 - 1 - x = 5 + x - (1+x) =4$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Inverse Identity + Constant Matrix I need to invert a square symmetric matrix $$ C = c\, I+cs\, B $$
Where: (1) $B$ is a constant matrix of 1 for each entry. (2) $c$ and cs are just positive real numbers. (3) $I$ is the identity.
However, the $\det(B) = 0$ and $B$ by itself does not have an inverse, but I am sure that C must have one.
Problem: What is $C^{-1}$ in terms of $cs, c$?
| You could use the Neumann series which is the inverse, if it converges:
$$A^{-1} = \sum_{k=0}^\infty(I-A)^k\qquad\text{(if convergent)}$$
Now we apply this to $\frac1c C = I + \frac{c_s}c B$ and get using $B^k = n^{k-1} B$ for $k>0$ (since $B^2=nB$) the following
\begin{align*}
\sum_{k=0}^\infty\left(I - \frac1c C\right)^k
&= \sum_{k=0}^\infty \left(-\frac{c_s}c\right)^k B^k \\
&= \frac1n\left(\sum_{k=1}^\infty \left(-\frac{c_s n}c\right)^k\right) B + I\\
&= \frac1n\left(\frac1{1+\frac{c_s n}c} - 1\right) B + I\\
&= \frac{-c_s}{c + c_s n}B + I = \left(\frac1c C\right)^{-1},
\end{align*}
at least if $\left|\frac{c_s n}c\right| < 1$. From this it follows
$$ C^{-1} = \frac{-c_s}{c(c+c_sn)}B + \frac1c I. \tag{*}$$
Now we can see that
\begin{align}
\left(\frac{-c_s}{c(c+c_sn)}B + \frac1c I\right)\left(cI + c_s B\right)
&= \frac{-c_s}{c+c_sn}B + \frac{-c_s^2}{c(c+c_sn)}B^2 + I + \frac{c_s}{c}B\\
&= \left(\frac{-c_s}{c+c_sn} + \frac{-n c_s^2}{c(c+c_sn)} + \frac{c_s}{c}\right)B + I \\
&= \frac{-c c_s + -n c_s^2 + c_s(c+c_sn)}{c(c+c_sn)}B + I\\
&= I \\
\end{align}
independent of $\frac{c_s n}c$, as long as $c(c+c_sn)\neq 0$. So (*) holds if $c\neq0$ and $c_s\neq -\frac cn$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Points on $(x^2 + y^2)^2 = 2x^2 - 2y^2$ with slope of $1$ Let the curve in the plane defined by the equation: $(x^2 + y^2)^2 = 2x^2 - 2y^2$
How can i graph the curve in the plane and determine the points of the curve where $\frac{dy}{dx} = 1$.
My work:
First i found the roots of this equation with a change of variable $z = y^2$ and get:
and then i tried to graph the point $ x - y$ and $x + y $ but i stuck i can't graph this and find the point where the derivative is 1. Some help please.
| I won't do the graph, just the algebra.
Implicit differentiation gives
$$2(x^2+y^2)(2x+2yy')=4x-4yy'$$
so setting $y'=1$ and simplifying leads to
$$(x^2+y^2)(x+y)=x-y\qquad(*)$$
If we now rewrite the equation for the curve as
$$(x^2+y^2)^2=2(x-y)(x+y)$$
we see that multiplying both sides of equation $(*)$ by $2(x+y)$ gives
$$2(x^2+y^2)(x+y)^2=2(x-y)(x+y)=(x^2+y^2)^2$$
Ignoring the point on the curve at $(0,0)$ (where the derivative is not defined), we can cancel a $(x^2+y^2)$ from the above, leaving $2(x+y)^2=x^2+y^2$, which simplifies to $x^2+y^2=-4xy$ (which will mean that one of $x$ and $y$ will have to be positive and the other one negative). Plugging this into the original equation for the curve gives $(-4xy)^2=2x^2-2y^2$, or
$$y^2={x^2\over1+8x^2}$$
Plugging this into the original equation for the curve gives
$$\left(x^2+{x^2\over1+8x^2}\right)^2=2x^2-2{x^2\over1+8x^2}$$
Simplifying this (and keeping in mind we're already ignoring $x=0$) produces
$$16x^4-24x^2-3=0$$
The real roots for this are
$$x=\pm\sqrt{3+2\sqrt3\over4}\approx\pm1.27$$
The corresponding values for $y$ (recalling that $-4xy$ must be positive) are
$$y=\mp\sqrt{-3+2\sqrt3\over4}\approx\mp.34$$
| {
"language": "en",
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Verifying that the determinant is equal to $1!2!3!...(n-1)!$ Verifying that the determinant is equal to $1!2!3!...(n-1)!$
$$|A|=
\begin{vmatrix}
1 &1 & \dots &1\\
1 &2 & \dots &2^{n-1}\\
1 &3 & \dots &3^{n-1}\\
& & \dots\\
1 &n & \dots &n^{n-1}\\
\end{vmatrix}=1!2!3!...(n-1)!
$$
I used the definition of a determinan with Minors and Cofactors, ie.
$$|A|=\sum_{j=1}^{n-1}i^{j-1}\cdot A_{ij}=i^{1-1=0}\cdot A_{i 1}+i^{2-1=1}\cdot A_{i 2}+i^{3-1=2}\cdot A_{i 3}+\dots+i^{n-1}\cdot A_{i n-1}$$
So, what we want to prove is equal to:
$$|A|=\sum_{j=1}^{n-1}i^{j-1}\cdot A_{ij}=1!2!3! \dots (n-1)!$$
Proving for $n=1$
$$|A|=\sum_{j=1}^{n-1}i^{j-1}\cdot A_{ij}=|1|=0!=1$$
INDUCTIVE H:
Asume
$$|A|=\sum_{j=1}^{n-1}i^{j-1}\cdot A_{ij}=i^{0}\cdot A_{i 1}+i^{1}\cdot A_{i 2}+i^{2}\cdot A_{i 3}+\dots+i^{n-1}\cdot A_{i n-1}=1!2!3!...(n-1)$$
So, for $n+1$ we have:
$$|A|=\sum_{j=1}^{n}i^{j-1}\cdot A_{ij}=i^{0}\cdot A_{i 1}+i^{1}\cdot A_{i 2}+i^{2}\cdot A_{i 3}+\dots+i^{n-1}\cdot A_{i n-1}+i^{n}\cdot A_{i n}$$
INDUCTIVE STEP:
$$|A|=\sum_{j=1}^{n}i^{j-1}\cdot A_{ij}=(1!2!3! \dots (n-1)!)+i^{n}\cdot A_{i n}$$
AND AT THIS POINT, I DONT KNOW WHAT TO DO.
| We can evaluate this determinant as follows.
Let $A_n$ be the $n \times n$ matrix in question.
First, for each $j = n-1, n-2, \ldots, 1$, subtract $n$ times column $j$ from column $j+1$.
This results in the following:
\begin{align*}
|A_n| &=
\begin{vmatrix}
1 & 1-n & 1-n & \ldots & 1-n & \ldots & 1-n \\
1 & 2 - n & 2^2 - 2n & \ldots & 2^{j-1} - 2^{j-2}n & \ldots & 2^{n-1} - 2^{n-2}n \\
1 & 3 - n & 3^2 - 3n & \ldots & 3^{j-1} - 3^{j-2}n & \ldots & 3^{n-1} - 3^{n-2}n \\
\vdots & \vdots & \vdots & \ddots & \vdots & \ddots & \vdots \\
1 & i - n & i^2 - in & \ldots & i^{j-1} - i^{j-2}n & \ldots & i^{n-1} - i^{n-2}n \\
\vdots & \vdots & \vdots & \ddots & \vdots & \ddots & \vdots \\
1 & n - n & n^2 - n^2 & \ldots & n^{j-1} - n^{j-1} & \ldots & n^{n-1} - n^{n-1} \end{vmatrix} \\
&\; \\ &\; \\&\; \\ &\; \\ %vertical space
&=
\begin{vmatrix}
1 & 1-n & 1-n & \ldots & 1-n & \ldots & 1-n \\
1 & 2 - n & 2(2-n) & \ldots & 2^{j-2}(2-n) & \ldots & 2^{n-2}(2-n) \\
1 & 2 - n & 3(3-n) & \ldots & 3^{j-2}(3-n) & \ldots & 3^{n-2}(3-n) \\
\vdots & \vdots & \vdots & \ddots & \vdots & \ddots & \vdots \\
1 & i - n & i(i-n) & \ldots & i^{j-2}(i-n) & \ldots & i^{n-2}(i-n) \\
\vdots & \vdots & \vdots & \ddots & \vdots & \ddots & \vdots \\
1 & (-1) & (n-1)(-1) & \ldots & (n-1)^{j-2}(-1) & \ldots & (n-1)^{n-2}(-1) \\
1 & 0 & 0 & \ldots & 0 & \ldots & 0 \\
\end{vmatrix}
\end{align*}
By cofactor expansion along the bottom row,
\begin{align*}
&= (-1)^{n+1}
\begin{vmatrix}
1-n & 1-n & \ldots & 1-n & \ldots & 1-n \\
2 - n & 2(2-n) & \ldots & 2^{j-1}(2-n) & \ldots & 2^{n-2}(2-n) \\
2 - n & 3(3-n) & \ldots & 3^{j-1}(3-n) & \ldots & 3^{n-2}(3-n) \\
\vdots & \vdots & \ddots & \vdots & \ddots & \vdots \\
i - n & i(i-n) & \ldots & i^{j-1}(i-n) & \ldots & i^{n-2}(i-n) \\
\vdots & \vdots & \ddots & \vdots & \ddots & \vdots \\
(-1) & (n-1)(-1) & \ldots & (n-1)^{j-1}(-1) & \ldots & (n-1)^{n-2}(-1) \\
\end{vmatrix} \\
&\; \\ &\; \\&\; \\ &\; \\ %vertical space
&= (-1)^{n+1} (1-n)(2-n)(3-n)\cdots (-2)(-1) \\
&\; \\ %vertical space
& \quad \quad \quad \quad \quad \cdot \quad
\begin{vmatrix}
1 & 1 & \ldots & 1 & \ldots & 1 \\
1 & 2 & \ldots & 2^{j-1} & \ldots & 2^{n-2} \\
1 & 3 & \ldots & 3^{j-1} & \ldots & 3^{n-2} \\
\vdots & \vdots & \ddots & \vdots & \ddots & \vdots \\
1 & i & \ldots & i^{j-1} & \ldots & i^{n-2} \\
\vdots & \vdots & \ddots & \vdots & \ddots & \vdots \\
1 & (n-1) & \ldots & (n-1)^{j-1} & \ldots & (n-1)^{n-2} \\
\end{vmatrix} \\
&\; \\ %vertical space
&= (n-1)! |A_{n-1}|
\end{align*}
By induction on $n$ since $|A_1| = 1$, it follows that
$$
|A_n| = (n-1)!(n-2)!(n-3)!\cdots 2! 1!
$$
as required.
| {
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"source": "stackexchange",
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In how many ways can ww choose $10$ cards so there are $3$ exact matches? Suppose there are $20$ cards- $10$ reds numbered $1,2,\cdots,10$, and $10$ blues numbered $1,2, \cdots, 10$. In how many ways can 10 cards be picked to so that there are EXACTLY 3 MATCHES- where a match means a red card and blue card have the same number.
What I did:
Number of ways to get 3 or more matches - Number of ways to get 4 matches or more, that is:
$\dbinom{10}{3} \cdot \dbinom{14}{4} - \dbinom{10}{4}\cdot \dbinom{12}{2}$
Can someone point out what I did wrong?
| There are $\binom{10}{3}$ ways to pick the cards that will make up the three matching pairs. That is a component of your answer, so I will assume that part is clear to you. Now we count the number of ways to pick the remaining $4$ cards.
There remain $7$ "couples." We pick $4$ of these, and for each couple we choose the colour that will be used. That can be done in $\binom{7}{4}2^4$ ways, for a total of $\binom{10}{3}\binom{7}{4}2^4$.
Remark: About your answer, the issue is about the count of "bad" choices, where there are $4$ or more matches. This number is not $\binom{10}{4}\binom{12}{2}$. That expression "double-counts" the $5$ matches situations. For the $\binom{10}{4}$ part counts, among others, the situations where we picked each of the couples $1,2,3,4$, while the $\binom{12}{2}$ counts among others the situations where we picked couple $5$. But $\binom{10}{4}$ also includes picking couples $1,2,3,5$, and $\binom{12}{2}$ counts among others picking couple $4$. Thus picking the $5$ couples has been double-counted, indeed multiple-counted.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proving a metric with absolute value I need to prove that function $\mathbb R × \mathbb R → \mathbb R $ : $f(x,y) = \frac{|x-y|}{1 + |x-y|}$ is a metric on $\mathbb R$. First two axioms are trivial; it's the triangle inequality which is pain. $\frac{|x-y|}{1 + |x-y|}$ + $\frac{|y-z|}{1 + |y-z|} ≥ \frac{|x-z|}{1 + |x-z|} ⇒ \frac{|x-y| + |y-z| + 2|(x-y)(y-z)|}{1 + |x-y| + |y-z| + |(x-y)(y-z)|}≥ \frac{|x-z|}{1 + |x-z|}$, but then I am stuck. Can somebody show me way out of this?
| Since $1 + |x-y| + |y -z| \ge 1 + |x-z|$,
$$1 - \frac{1}{1 + |x-z|} \le 1 - \frac{1}{1 + |x-y| + |y -z|} \\
\implies \frac{|x-z|}{1 +|x-z|} \le \frac{|x-y| + |y-z|}{1 + |x-y| + |y-z|} \le \frac{|x-y|}{1 + |x-y|} + \frac{|y-z|}{1 + |y-z|}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Last 3 digits of $3^{999}$ I know that it's $3^{999} \mod 1000$ and since $\varphi(1000) = 400$ and $3^{400}\equiv1 \mod1000$ it will be equivalent to $3^{199} \mod 1000$ but what should I do from then? Or am I wrong about this from the start?
| Simple computation, and the fact that $\phi(8)=4$, gives
$$
\begin{align}
3^{999}
&\equiv3^3\\
&\equiv3\pmod8\tag1
\end{align}
$$
The square and multiply algorithm, and the fact that $\phi(125)=100$, gives
$$
\begin{align}
3^{999}
&\equiv3^{99}\\
&\equiv42\pmod{125}\tag2
\end{align}
$$
The Extended Euclidean Algorithm, as implemented in this answer, gives
$$
\begin{array}{r}
&&15&1&1&1&2\\\hline
1&0&1&-1&2&-3&8\\
0&1&-15&16&-31&47&-125\\
125&8&5&3&2&1&0
\end{array}\tag3
$$
which says that
$$
47\cdot8-3\cdot125=1\tag4
$$
Equation $(4)$ implies both
$$
\begin{align}
-375&\equiv1\pmod8\\
-375&\equiv0\pmod{125}
\end{align}\tag5
$$
and
$$
\begin{align}
376&\equiv0\pmod8\\
376&\equiv1\pmod{125}
\end{align}\tag6
$$
Thus, to satisfy $(1)$ and $(2)$, we compute
$$
\begin{align}
3^{999}
&\equiv42\cdot376-3\cdot375\\
&\equiv667&\pmod{1000}\tag7
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/789050",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 6
} |
Solve $\sin(2\theta) -\tan(\theta) = 0 \ $ for $ 0\leq \theta \leq 2\pi $ I want to use the fact that $$\sin(2\theta) = \frac{2\tan(\theta)}{1 + \tan^2(\theta)}$$
to solve $\sin(2\theta) -tan(\theta) = 0 \ $ for $ 0\leq \theta \leq 2\pi$
My solution:
$\frac{2tan(\theta)}{1 + tan^2(\theta)} - tan(\theta) = 0 $
so $\frac{2tan(\theta) - (1 + tan^2(\theta)) tan(\theta)}{1 + \tan^2(\theta)} = 0$
so $ 2tan(\theta) - (1 + tan^2(\theta)) tan(\theta)= 0$
$ \implies \tan^3(\theta) + \tan(\theta) = 0$ $\implies \tan(\theta) [\tan^2(\theta) + 1] = 0$ $\implies \tan(\theta) = 0 \;\textrm{or}\; \tan^2(\theta) + 1 =0 $ since $\theta$ must be real.
Then we solve $\tan(\theta) = 0$ $\implies$ $\theta = n\pi, \ \ $ $ n \in Z \ \ $ so $\theta = \pi,2\pi$
| Try this
\begin{align}
\sin2\theta&=\tan\theta\\
\frac{2\tan\theta}{1 + \tan^2\theta}&=\tan\theta\\
\frac{2}{1 + \tan^2\theta}&=1\\
\tan^2\theta-1&=0\\
(\tan\theta-1)(\tan\theta+1)&=0.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/789707",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
How prove this $\frac{r}{R}\ge\frac{AM}{AX}$ Let acute $\triangle ABC$ have largest angle $\angle A$, and let $R,r$ be the circumradius and inradius respectively. Let $M$ be the midpoint of $BC$, and $X$ be the intersection of the tangents at $B,C$ to the circumcircle of $\triangle ABC$
Show that:
$$\frac{r}{R}\ge\frac{|AM|}{|AX|}$$
Here is the original chinese version:
My method:
So now I have the following problem:
let $x,y,z>0$,and let
$$R=\dfrac{\sqrt{(x^2+y^2+z^2+2xz)(x^2+y^2+z^2-2xz)}}{2y},a=\sqrt{x^2+y^2}$$
$$r=\dfrac{2yz}{2z+\sqrt{x^2+y^2+z^2-2xz}+\sqrt{x^2+y^2+z^2+2xz}}$$
$$b=\sqrt{\dfrac{(x^2+y^2)(x^2+y^2+z^2+2xz)(x^2+y^2+z^2-2xz)}{(x^2+y^2-z^2)^2}}$$
Show that
$$\dfrac{r}{R}\ge\dfrac{a}{b}$$
I think this problem has other methods. If you find other methods, can you explain?
Thank you very much
| Let $(a,b,c) = (\angle A,\angle B,\angle C)$ in $\triangle ABC$
Let $Γ$ be the circumcircle of $\triangle ABC$ with centre $O$
Then $A,O$ are on the same side of $BC$ because $\triangle ABC$ is acute
Let $D,E$ be the intersections of the perpendicular bisector of $BC$ with $Γ$ such that $A,D$ are on the same side of $BC$
Then $(M,X;D,E) = (B,C;D,E) = -1$ by the projection through $C$
Let $F,G$ be the second intersections of $AM,AX$ with $Γ$ respectively
Then $(F,G;D,E) = (M,X;D,E) = -1$ by the projection through $A$
Thus $F,G$ are symmetric about $DE$ and hence $\angle MAE = \angle EAX$
Thus $\frac{|AM|}{|AX|} = \frac{|ME|}{|EX|}$
Also for fixed $B,C,Γ$, $r$ is minimum when $|IO|$ is maximized by Euler's Triangle Formula
Now $\angle BIE = \angle BAI + \angle IBA = \angle IAC + \angle CBI = \angle EBC + \angle CBI = \angle EBI$
Thus $|BE| = |IE|$ and by symmetry $|CE| = |IE|$
Thus $|IO|$ is maximized when $I$ is furthest from $DE$ because $I$ is on the circle through $B,C$ centred at $E$ and $O$ is between $D,E$
Thus $|IO|$ is maximized when WLOG $a = c$
If $a = c$:
$\frac{r}{R} = \sin(\frac{1}{2}a) \ge \sin(30^\circ) = \frac{1}{2}$
$|MX| = |MC| \tan(\angle BCX) = |MC| \tan(a)$
$|MC| = |ME| \tan(\angle CED) = |ME| \cot(\angle BCE) = |ME| \cot(\frac{1}{2}a)$
$\frac{|EX|}{|ME|} = \frac{|MX|}{|ME|} - 1 = \tan(a) \cot(\frac{1}{2}a) - 1 = \frac{2\sin(\frac{1}{2}a)\cos(\frac{1}{2}a)}{\cos(a)} \cdot \frac{\cos(\frac{1}{2}a)}{\sin(\frac{1}{2}a)} - 1 = \frac{1}{\cos(a)} \ge \frac{1}{\cos(60^\circ)}$
$ = 2$
Thus $\frac{r}{R} \ge \frac{|ME|}{|EX|}$
Therefore $\frac{r}{R} \ge \frac{|ME|}{|EX|}$ in all cases
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/791263",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
fractional part of Riemann zeta $\sum_{s=2}^\infty \{\zeta (s)\}=1$
$$\sum_{s=2}^\infty \{\zeta (s)\}=1$$
where $\zeta (s)$ is Riemann zeta, $\{x\}$ denotes the fractional part of the real number $x$
The problem was proposed by Michael Th. Rassias
$\{\zeta(2)\}=\frac{\pi^2}6-1,$ How to go on?
Thanks a lot!
| Since $ 1 < \zeta(s) <2$ for $s \ge 2$, it's equivalent to showing that $ \displaystyle\sum_{s=2}^{\infty} \left( \zeta(s)-1 \right)= 1$.
In which case,
$$ \sum_{s=2}^{\infty} \left( \zeta(s)-1 \right)= \sum_{s=2}^{\infty}\sum_{n=2}^{\infty} \frac{1}{n^{s}} = \sum_{n=2}^{\infty} \sum_{s=2}^{\infty} \frac{1}{n^{s}}$$
$$ = \sum_{n=2}^{\infty} \frac{\frac{1}{n^{2}}}{1-\frac{1}{n}} = \sum_{n=2}^{\infty} \frac{1}{n(n-1)}$$
$$=\sum_{n=2}^{\infty} \left( \frac{1}{n-1} - \frac{1}{n}\right) $$
$$ = \lim_{N \to \infty} \left(1- \frac{1}{2} + \frac{1}{2} -\frac{1}{3} + \ldots + \frac{1}{N-1} - \frac{1}{N} \right)$$
$$ = 1- \lim_{N \to \infty}\frac{1}{N} = 1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/792624",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
} |
proving $\tan^{-1}(x)+\tan^{-1}(y) = \pi+\tan^{-1}\left(\frac{x+y}{1-xy}\right)\;,$ when $x>0,y>0,xy>1$ (1) How ca we prove $\displaystyle \tan^{-1}(x)+\tan^{-1}(y) = \pi+\tan^{-1}\left(\frac{x+y}{1-xy}\right)\;,$ when $x>0,y>0,xy>1$
(2) How ca we prove $\displaystyle \tan^{-1}(x)+\tan^{-1}(y) = -\pi+\tan^{-1}\left(\frac{x+y}{1-xy}\right)\;,$when $x<0,y<0,xy>1$
$\bf{My\; Try::}$ My Try for $(1)\;,$ Let $\tan^{-1}(x)=A\Rightarrow x = \tan (A)$ and $\tan^{-1}(y) = B\Rightarrow y = \tan(B)$
Now $\displaystyle \tan(A+B) = \frac{\tan A+\tan B}{1-\tan A\cdot \tan B} = \frac{x+y}{1-xy}$
and Given $x>0\Rightarrow \tan A>0$ and $y>0\Rightarrow \tan B>0$ and given $xy>1$
So $\displaystyle \frac{x+y}{1-xy}<0\Rightarrow \tan (A+B)<0$
Means $(A+B)$ lies in $\bf(II)$ quadrant or in $\bf{(IV)}$ quadrant.
But above $\displaystyle \tan (A)\;\;,\tan (B)>0\Rightarrow 0<A,B<\frac{\pi}{2}$.
So $\displaystyle 0<A+B<\pi$(Means $(A+B)$ lies on $\bf{I}$ quadrant.)
Now I did not understand how can i solve it.
Help me
Thanks
| It is quite simple ;
lets consider tan(θ) = x and let it be so that tan(ϕ) = y , then we know , θ = tan⁻¹(x) and ϕ = tan⁻¹(y)
$\displaystyle\Rightarrow$ tan(θ+ϕ)= $\displaystyle(\frac{tanθ+tanϕ}{1-tanθ*tanϕ})\;$
$\displaystyle\Rightarrow$ tan(θ+ϕ) = $\displaystyle(\frac{x+y}{1-x*y})\;$ as [Note : tan(θ) = x and tan(ϕ) = y]
$\displaystyle\Rightarrow$ $\displaystyle \pi $ is an interval of tan , which means tan(a-n*$\pi$) is same as tan(a) where $n∈ℤ$
using that :
$\displaystyle\Rightarrow$ tan(θ+ϕ-$\displaystyle \pi)$ = $\displaystyle(\frac{x+y}{1-x*y})\;$
$\displaystyle\Rightarrow$ θ+ϕ-$\displaystyle \pi$ = tan⁻¹$\displaystyle(\frac{x+y}{1-x*y})\;$
and alas simply :
∴$\displaystyle \tan^{-1}(x)+\tan^{-1}(y) = \pi+\tan^{-1}\left(\frac{x+y}{1-xy}\right)\;,$ when $x>0,y>0,xy>1$
conversly using n=-1 , we can observe ,
$\displaystyle\Rightarrow$ tan(θ+ϕ+$\displaystyle \pi)$ = $\displaystyle(\frac{x+y}{1-x*y})\;$
$\displaystyle\Rightarrow$ θ+ϕ+$\displaystyle \pi$ = tan⁻¹$\displaystyle(\frac{x+y}{1-x*y})\;$
hence,
∴$\displaystyle \tan^{-1}(x)+\tan^{-1}(y) = -\pi+\tan^{-1}\left(\frac{x+y}{1-xy}\right)\;,$when $x<0,y<0,xy>1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/795394",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Integral $\int_0^1 \log \frac{1+ax}{1-ax}\frac{dx}{x\sqrt{1-x^2}}=\pi\arcsin a$ Hi I am trying to solve this integral $$
I:=\int_0^1 \log\left(\frac{1+ax}{1-ax}\right)\,\frac{{\rm d}x}{x\sqrt{1-x^2}}=\pi\arcsin\left(a\right),\qquad
\left\vert a\right\vert \leq 1.
$$
It gives beautiful result for $a = 1$
$$
\int_0^1 \log\left(\frac{1+ x}{1-x}\right)\,\frac{{\rm d}x}{x\sqrt{1-x^2}}
=\frac{\pi^2}{2}.
$$
I tried to write
$$
I=\int_0^1 \frac{\log(1+ax)}{x\sqrt{1-x^2}}dx-\int_0^1 \frac{\log(1-ax)}{x\sqrt{1-x^2}}dx
$$
If we work with one of these integrals we can write
$$
\sum_{n=1}^\infty \frac{(-1)^{n+1} a^n}{n}\int_0^1 \frac{x^{n-1}}{\sqrt{1-x^2}}dx-\sum_{n=1}^\infty \frac{a^n}{n}\int_0^1 \frac{x^{n-1}}{\sqrt{1-x^2}}dx,
$$
simplifying this I get an infinite sum of Gamma functions. which i'm not sure how to relate to the $\arcsin$ Thanks.
| The integral in question,
\begin{align}
I = \int_{0}^{1} \ln \left( \frac{1+ax}{1-ax} \right) \ \frac{dx}{x \sqrt{1-x^{2}}}
\end{align}
can be separated into the two integrals
\begin{align}
I = \int_{0}^{1} \frac{ \ln(1+ax)}{x \sqrt{1-x^{2}}} \ dx - \int_{0}^{1} \frac{ \ln(1-ax)}{x \sqrt{1-x^{2}}} \ dx
\end{align}
which will be labled $I_{1}$ and $I_{2}$. Now
\begin{align}
I_{1} &= \int_{0}^{1} \frac{ \ln(1+ax)}{x \sqrt{1-x^{2}}} \ dx \\
&= \sum_{n=1}^{\infty} \frac{(-1)^{n-1} a^{n}}{n} \ \int_{0}^{1} \frac{x^{n-1} \ dx}{\sqrt{1-x^{2}}} \\
&= - \frac{1}{4} \sum_{n=1}^{\infty} \frac{(-a)^{n}}{n} \ B(n/2, 1/2) \\
&= - \frac{1}{4} \sum_{n=1}^{\infty} \frac{(-2a)^{n} \ \Gamma^{2}(n/2)}{n!} \\
&= - \frac{1}{4} \left[ \sum_{k=0}^{\infty} \Gamma^{2}(k+1/2) \frac{(-2a)^{2k+1}}{(2k+1)!} + \sum_{k=0}^{\infty} \frac{(k!)^{2} (-2a)^{2k+2}}{(2k+2)!} \right] \\
&= - \frac{1}{4} \left[ -2\pi a \sum_{k=0}^{\infty} \frac{(1/2)_{k} (1/2)_{k} a^{2k}}{ k! (3/2)_{k}} + 4 a^{2} \sum_{k=0}^{\infty} \binom{2k+2}{k+1}^{-1} \frac{(2a)^{2k}}{(k+1)^{2}} \right] \\
&= \frac{\pi}{2} \ \sin^{-1}(a) - a^{2} \sum_{k=0}^{\infty} \binom{2k+2}{k+1}^{-1} \frac{(2a)^{2k}}{(k+1)^{2}}.
\end{align}
In a similar manor,
\begin{align}
I_{2} &= - \frac{\pi}{2} \ \sin^{-1}(a) - a^{2} \sum_{k=0}^{\infty} \binom{2k+2}{k+1}^{-1} \frac{(2a)^{2k}}{(k+1)^{2}}.
\end{align}
Since $I = I_{1} - I_{2}$ then
\begin{align}
\int_{0}^{1} \ln \left( \frac{1+ax}{1-ax} \right) \ \frac{dx}{x \sqrt{1-x^{2}}} = \pi \ \sin^{-1}(a)
\end{align}
which is the desired value.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/795493",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 5,
"answer_id": 3
} |
Gradient formula of volume of tetrahedron involving the scalar triple product Let $a,b,c,d \in \mathbb{R}^3$ be the vertices of a tetrahedron (I’m unsure whether or not the order of the vertices is important for what follows). The volume of the tetrahedron is
$$
\begin{align}
\operatorname{vol}(a,b,c,d)
&= \frac{1}{6}\big((b-a) \times (c - a)\big) \cdot (d - a) \\
&= \frac{1}{6}\det
\left|\begin{matrix}
(b-a) & (c-a) & (d-a)
\end{matrix}\right|
\end{align}
$$
The gradient can be represented nicely by
$$
\nabla\operatorname{vol}(a,b,c,d) = \frac{1}{6}\left(\begin{matrix}
(d - b) \times (c - b) \\
(c - a) \times (d - a) \\
(d - a) \times (b - a) \\
(b - a) \times (c - a) \\
\end{matrix}\right).
$$
I have verified this formula by hand, i.e. I have manually calculated $\nabla\operatorname{vol}$ componentwise and compared the result to the componentwise expansion of the given formula — but I’m not satisfied. I prefer derivation to “left-hand side equals right-hand side” verification.
Can anyone give an insightful explanation of the gradient formula?
| This comes down to a straightforward calculation. We have for $a,b,x \in \mathbb{R}^3$
$$\begin{align}
\nabla_x(a \times b)\cdot x
&= \nabla_x \big( (a_2b_3 - a_3b_2)x_1 + (a_3b_1 - a_1b_3)x_2 + (a_1b_2 - a_2b_1)x_3 \big) \\
&= a \times b
\end{align}$$
and
$$\begin{align}
\nabla_x(a \times x) \cdot b
&= \nabla_x \big( (a_2x_3 - a_3x_2)b_1 + (a_3x_1 - a_1x_3)b_2 + (a_1x_2 - a_2x_1)b_3 \big) \\
&= b \times a
\end{align}$$
and
$$\begin{align}
\nabla_x (x \times a) \cdot b
&= \nabla_x \big( (x_2a_3 - x_3a_2)b_1 + (x_3a_1 - x_1a_3)b_2 + (x_1a_2 - x_2a_1)b_3 \big) \\
&= a \times b.
\end{align}$$
From this, the three gradients $\nabla_b\operatorname*{vol}$, $\nabla_c\operatorname*{vol}$ and $\nabla_d\operatorname*{vol}$ immediately follow. For $\nabla_a\operatorname*{vol}$ we expand $\operatorname*{vol}(a,b,c,d)$ and get
$$\begin{align*}
6\cdot\operatorname*{vol}(a,b,c,d)
&= \big( (b-a) \times (c-a) \big) \cdot (d-a) \\
&= \big( b \times c - b \times a - a \times c + a \times a \big) \cdot (d - a) \\
&= (b \times c) \cdot d - (b \times a) \cdot d - (a \times c) \cdot d - (b \times c) \cdot a.
\end{align*}$$
Then use the formulae given above and obtain
$$\begin{align*}
6\cdot\nabla_a\operatorname*{vol}
&= b \times d - c \times d - b \times c && \\
&= b \times d - c \times d - b \times c - b \times b && \text{(add $0$)} \\
&= b \times d - c \times d + c \times b - b \times b && \text{(antisymmetry)} \\
&= (b - c) \times (d - b) && \text{(factor out)} \\
&= - (c - b) \times (d - b) && \\
&= (d - b) \times (c - b) && \text{(antisymmetry)}.
\end{align*}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/797845",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Finding a real root in a cubic equation? I am trying to find a real root of the following cubic.
$x^3-6x^2+14x-15=0$
I did $x=y-\frac{b}{3a}$
which is $x=y+2$
I plugged my substitution to get the depressing cubic.
$(y+2)^3-6(y+2)^2+14(y+2)-15=0$
which is
$y^3+2y=3$
then I set up the method for solving.
$y=s-t$ , first
$(3st=2)$, second
$s^3-t^3=3$
So then for the second equation I got $s=\frac{2}{3t}$
Plugged it into my third and simplified
$t^6+3t^3-\frac{8}{27}$
then Use the qudratic set $u=t^3$
so then I get
$u^2+3u-\frac{8}{27}$
Using the quadratic I got $-\frac{3}{2}+\frac{\sqrt{59/27}}{2}$
as my u which is t^3
take the cube root and I got $t=\sqrt[3]{\frac{-3}{2}+\frac{\sqrt{59/27}}{2}}$
so the s is $s=\sqrt[3]{\frac{3}{2}+\frac{\sqrt{59/27}}{2}}$
and since y=s-t you can plug than back into my $x=y+2$ equation to find x.
But I am not sure if I did this right
| Note that $x=3$ is a solution. We used the Rational Roots Theorem, since cubics in exercises often have a rational root. Exercises and real life tend to differ.
When the depressed cubic was reached (correctly) the root $y=1$ was available by inspection.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Gram-Schmidt process on complex space
Let $\mathbb{C}^3$ be equipped with the standard complex inner product. Apply the
Gram-Schmidt process to the basis: $v_1=(1,0,i)^t$, $v_2=(-1,i,1)^t$, $v_3=(0,-1,i+1)^t$ to find an orthonormal basis $\{u_1,u_2,u_3\}$.
I have found $u_1 = \dfrac{1}{\sqrt{2}} (1,0,i)^t$ and $u_2 = \dfrac{1}{\sqrt{2}}\left(\dfrac{i-1}{2},0,\dfrac{i+1}{2}\right)^t$.
I then try to use the following formulae to work out $u_3$ but I keep going wrong and can't figure out why: $w_3 = v_3 - \langle v_3, u_1\rangle u_1 - \langle v_2, u_2\rangle u_2$ and $u_3 = \dfrac{w_3}{\|w_3\|}$.
| Your $u_1$ is correct, but your $u_2$ is incorrect; as noted in the comments, $u_1\cdot u_2 \neq 0$.
Recall that the standard inner product on $\mathbb{C}^n$ is given by $\langle u, v\rangle = u\cdot\bar{v}$. With this in mind, let's calculate $u_2$. First we have
$$\langle v_2, u_1\rangle = (-1, i, 1)\cdot\overline{\frac{1}{\sqrt{2}}(1, 0, i)} = \frac{1}{\sqrt{2}}(-1, i, 1)\cdot(1, 0, -i) = \frac{-1-i}{\sqrt{2}},$$
so
$$w_2 = v_2 - \langle v_2, u_1\rangle u_1 = \left[\begin{array}{c} -1\\ i\\ 1\end{array}\right] + \frac{1+i}{\sqrt{2}}\frac{1}{\sqrt{2}}\left[\begin{array}{c} 1\\ 0\\ i\end{array}\right] = \left[\begin{array}{c} \frac{-1+i}{2}\\ i\\ \frac{1+i}{2}\end{array}\right].$$
As
\begin{align*}
\|w_2\|^2 &= \sqrt{\left|\frac{-1+i}{2}\right|^2 + |i|^2 + \left|\frac{1+i}{2}\right|^2}\\
&= \sqrt{\left(\frac{-1}{2}\right)^2 + \left(\frac{1}{2}\right)^2 + 0^2 + 1^2 + \left(\frac{1}{2}\right)^2 + \left(\frac{1}{2}\right)^2}\\
&= \sqrt{2}
\end{align*}
we have
$$u_2 = \frac{1}{\|w_2\|}w_2 = \frac{1}{\sqrt{2}}\left[\begin{array}{c} \frac{-1+i}{2}\\ i\\ \frac{1+i}{2}\end{array}\right].$$
Let's check to see if $u_2$ is orthogonal to $u_1$:
\begin{align*}
\langle u_1, u_2\rangle &= \frac{1}{\sqrt{2}}(1, 0, i)\cdot\overline{\frac{1}{\sqrt{2}}\left(\frac{-1+i}{2}, i, \frac{1+i}{2}\right)}\\
&= \frac{1}{2}(1, 0, i)\cdot\left(\frac{-1-i}{2}, -i, \frac{1-i}{2}\right)\\
&= \frac{1}{2}\left(\frac{-1-i}{2} + 0 + \frac{1+i}{2}\right)\\
&= 0.
\end{align*}
Note, we didn't have to normalise before we checked orthogonality; i.e. we could have checked $\langle u_1, w_2\rangle = 0$ instead.
I won't do the calculation of $u_3$ now. It is similar, but there are more computations. Note however that you have a typo in your formula for $w_3$; it should be
$$w_3 = v_3 - \langle v_3, u_1\rangle u_1 - \langle v_3, u_2\rangle u_2.$$
Now that you have the correct $u_2$ and the correct formula for $w_3$, the computation for $u_3$ should work out and produce $u_3 = \frac{1}{2}(i, -1-i, 1)$.
| {
"language": "en",
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"source": "stackexchange",
"question_score": "8",
"answer_count": 1,
"answer_id": 0
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Using Taylor expansion to find $\lim_{x \rightarrow 0} \frac{\exp(2x)-\ln(1-x)-\sin(x)}{\cos(x)-1}$
Use Taylor expansion to find $\displaystyle \lim_{x \rightarrow 0} \frac{\exp(2x)-\ln(1-x)-\sin(x)}{\cos(x)-1}$
I know that:
$\exp(2x) = 1 + (2x) + \frac{(2x)^2}{2!}+ O(x^3)$
$\ln(1-x) = 1-(-x) + (-x)^2 + O(x^3)$
$\sin(x) = x + O(x^3)$
$\cos(x) = 1 - \frac{x^2}{2!} + O(x^3)$
Substituting this in yields:
$\displaystyle \lim_{x \rightarrow 0} \frac{1+2x+2x^2+O(x^3) -1 - x - x^2 + O(x^3) -x + O(x^3)}{-\frac{x^2}{2!} + O(x^3)} = \lim_{x \rightarrow 0} \frac{x^2 + O(x^3)}{-\frac{1}{2}x^2 + O(x^3)} = \lim_{x \rightarrow 0} \frac{x^2[1+O(x)]}{x^2[-1/2+O(x)]} = -2$
EDIT: Progressing from $\displaystyle \lim_{x \rightarrow 0} \frac{1+2x+\frac{5}{2}x^2+O(x^3)}{-\frac{x^2}{2}+O(x^3)}$ we have,
$\displaystyle \lim_{x \rightarrow 0} \frac{1 + 5/2x^2 +O(x^3)}{-1/2x^2 + O(x^3)} = \lim_{x \rightarrow 0} \frac{x^2(1/x^2 + 5/2 + O(x)}{x^2(-1/2 + O(x))} = \lim_{x \rightarrow 0} \frac{-1/x^2 - 5/2 + O(x)}{1/2+O(x)} = \frac{\lim_{x \rightarrow 0} -1/x^2 - 5/2 + O(x) }{\lim_{x \rightarrow 0} 1/2 + O(x)} = \frac{-\infty}{1/2} = -\infty$
Is this method rigorously correct?
| Your Taylor expansion of $\ln(1-x)$ is wrong
It should be $\displaystyle \ln(1-x)=-x-\frac{x^2}{2}+O(x^3)$
Therefore, $$\frac{\exp(2x)-\ln(1-x)-\sin(x)}{\cos(x)-1}=\frac{1+2x+\frac{5}{2}x^2+O(x^3)}{\frac{-x^2}{2}+O(x^3)}$$
The last expression goes to $-\infty$
Further explanation of this last fact:
$$\frac{\exp(2x)-\ln(1-x)-\sin(x)}{\cos(x)-1}=\frac{1+2x+\frac{5}{2}x^2+O(x^3)}{\frac{-x^2}{2}+O(x^{\color{red}{4}})}=\frac{1+2x+O(x^2)}{\frac{-x^2}{2}(1+O(x^2))}$$
Since $\displaystyle\frac{1}{1+\color{red}{O(x^2)}}=1-\color{red}{O(x^2)}+O(\color{red}{O(x^2)})=1+O(x^2)$,
$$\frac{\exp(2x)-\ln(1-x)-\sin(x)}{\cos(x)-1}=\frac{\left(1+2x+O(x^2)\right)\left(1+O(x^2)\right)}{\frac{-x^2}{2}}$$
Expanding the numerator, you get $\left(1+2x+O(x^2)\right)\left(1+O(x^2)\right)=1+2x+O(x^2)$
So : $\displaystyle \frac{\exp(2x)-\ln(1-x)-\sin(x)}{\cos(x)-1}=\frac{1+2x+O(x^2)}{\frac{-x^2}{2}}$
Hence,
$$\frac{\exp(2x)-\ln(1-x)-\sin(x)}{\cos(x)-1}= \frac{-2}{x^2}-\frac{4}{x}+O(1)$$
But, at $0$, $\displaystyle \frac{1}{x}=o(\frac{1}{x^2})$
Finally $$\frac{\exp(2x)-\ln(1-x)-\sin(x)}{\cos(x)-1}= \frac{-2}{x^2}+o(\frac{1}{x^2})+O(1)$$
The last quantity goes to $-\infty$
| {
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"url": "https://math.stackexchange.com/questions/800231",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find all real solutions of $6^x+1=8^x-27^{x-1}$
Find all real solutions of $6^x+1=8^x-27^{x-1}$.
Things I tried:
We want solutions of
$$2^x3^x+1 = (2^x)^3-\frac{(3^x)^3}{27}.$$ Write $a=2^x$ and $b=3^x$. This gives
$$ab+1 = a^3-\frac{b^3}{27}$$
or
$$27ab+27=27a^3-b^3$$
How to continue?
| The equation $27ab+27=27a^3-b^3$ you have found may be written as
$$27a^3-27ab-27-b^3=0,$$
which may be viewed as a cubic in $b$. Its discriminant is
$$\Delta_b(a)=-4\cdot(-1)\cdot(-27a)^3-27\cdot(-1)^2=-(a^3+1)^2,$$
which is negative for all $a$ except for $a=-1$, when it is zero. This does not correspond to a real value of $x$, hence for all possible values of $a$ there is a unique real $b$ satisfying the polynomial, which is
$$b=3(a-1).$$
By definition we have $b=3^{\log_2a}$. So it remains to solve $3^{\log_2a}=3(a-1)$. By inspection we find the solutions $a=2$ and $a=4$ corresponding to $x=1$ and $x=2$. These are the only solutions, because such an exponential function intersects any line in at most two points.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 2
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Help me calculating chromatic polynomial of this subgraph
This is the subgraph. I just wanted to do a couple of exercises with chromatic polynomials, but around the web are pretty standard ones, so i came up with this one myself. I thought it's pretty straightforward.
It is: i color $A$ with $x$ colors, automatically $B$ and $C$ are both different, so $B$ is colored with $x-1$ colors and $C$ with $x-2$ colors and $E$ with $x-1$ colors.
This gives us that $F$ must be colored with $x-3$ colors and D with $x-2$ colors.
And it gives me polynomial equal to $$x(x-1)^2(x-2)^2(x-3)$$
But according to wikipedia, coefficient next to $[x^{n-1}]$ is equal to minus number of edges. Wolfram says that $$[x^{n-1}]x(x-1)^2(x-2)^2(x-3) = -9$$
but we have clearly 8 edges here. Seems like i counted something wrong (too much?), so i'd love to hear some hints or solutions to my problem. Thank you in advance!
| The problem here is that you haven't accounted for the fact that $E$ and $B$ might have the same color, and $E$ and $C$ might have the same color.
There are cases here that need considering:
Case 1: $E$ has the same color as $B$.
In this case, there are $x$ choices for $A$; $x-1$ for $B$ and $E$; $x-2$ for $C$; $x-1$ for $D$; and $x-2$ for $F$. So, the contribution by this case is $x(x-1)^2(x-2)^2$.
Case 2: $E$ has the same color as $C$.
In this case, there are $x$ choices for $A$; $x-1$ choices for $C$ and $E$; $x-2$ for $B$; $x-2$ for $d$; and $x-1$ for $F$. So, the contribution by this case is $x(x-1)^2(x-2)^2$ as well.
Case 3: $E$, $B$, and $C$ all have different colors.
In this case, there are $x$ choices for $A$, $x-1$ for $B$, $x-2$ for $C$, and $x-3$ for $E$; given these choices, there are $x-2$ choices for $D$ and $x-2$ choices for $F$. So, the contribution by this case is $x(x-1)(x-2)^3(x-3)$.
So, all told, the chromatic polynomial for $G$ is
$$
\begin{align*}
P_G(x)&=2x(x-1)^2(x-2)^2+x(x-1)(x-2)^3(x-3)\\
&=x(x-1)(x-2)^2(x^2-3x+4)\\
&=x^6-8x^5+27x^4-48x^3+44x^2-16x.
\end{align*}
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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The $n$'th derivative of $x^x$ I want to know the $n$'th derivative of $f(x)=x^x$.
Then, I'll calculate $f(0)$ with Taylor expansion of $f(x)$ on $a=1$.
Here is my answer, but it is unfinished.
The derivative of $f(x)=x^x$
$$\begin{align}
f'(x)&=x^x(\log x+1)\\
f''(x)&=x^x(\log x+1)^2+x^{x-1}\\
f'''(x)&=x^x(\log x+1)^3+3x^{x-1}(\log x+1)-x^{x-2}\\[5pt]
f(x)^{(4)}&=x^x(\log x+1)^4+4x^{x-1}(\log x+1)^2-4x^{x-2}(\log x+1)+3x^{x-2}+2x^{x-3}\\
f(x)^{(5)}&=x^x(\log x+1)^5+10x^{x-1}(\log x+1)^3-10x^{x-2}(\log x+1)^2+15x^{x-2}(\log x+1)\\&\quad+10x^{x-3}(\log x+1)-10x^{x-3}-6x^{x-4}\\
f(x)^{(6)}&=x^x(\log x+1)^6+15x^x(\log x+1)^4-20x^{x-2}(\log x+1)^3+45x^{x-2}(\log x+1)^2\\&\quad+30x^{x-3}(\log x+1)^2-50x^{x-3}(\log x+1)+15x^{x-3}-46x^{x-4}(\log x+1)\\&\quad+40x^{x-4}+24x^{x-5}
\end{align}$$
Taylor expansion of $f(x)=x^x$ in $a=1$
$$\begin{align}
f(x)&=\sum_{i=0}^{n-1}\frac{f^{(i)}(1)}{i!}\\[5pt]
&\qquad=\frac1{0!}+\frac1{1!}(x-1)+\frac2{2!}(x-1)^2+\frac3{3!}(x-1)^3+\frac8{4!}(x-1)^4+\frac{12}{5!}(x-1)^5\\&\qquad+\frac{54}{6!}(x-1)^6+\cdots
\end{align}$$
| $$\frac{d^n x^x}{dx^n}=\frac{d^ne^{x\ln(x)}}{dx^n}$$
Expanding $e^y$ as a series and then $\ln(x)$ as a limit:
$$\frac{d^n}{dx^n}\sum_{k=0}^\infty \frac{x^k \ln^k(x)}{k!}=\lim_{c\to0}\sum_{k=0}^\infty \frac1{k!}\frac{d^n}{dx^n}\frac{x^k(x^c-1)^k}{c^k}$$
Now use the general Leibniz rule:
$$\frac{d^n}{dx^n}x^k(x^c-1)^k=\sum_{m=0}^n\binom nm \frac{d^{n-m}}{dx^{n-m}}x^k\frac{d^m}{dx^m}(x^c-1)^k$$
and the binomial theorem:
$$\frac{d^m}{dx^m}(x^c-1)^k=\sum_{j=0}^k\binom kj (-1)^{k-j}\frac{d^m}{dx^m}x^{cj}$$
We now take the derivatives with factorial power $u^{(v)}$
$$\frac{d^v}{dx^v}x^r=x^{r-v}r^{(v)}$$
Therefore we only have a triple series:
$$\frac{d^n x^x}{dx^n}= \lim_{c\to0}\sum_{k=0}^\infty \sum_{m=0}^n\sum_{j=0}^k\frac{k!n!(-1)^{k-j}(c j)!x^{cj+k-n}}{j!m!(k-j)!(n-m)!(cj-m)!(k+m-n)!c^k}$$
Shown here when clicking “Approximate form” in the substitution section. The inner series likely has a hypergeometric answer. Additionally, $0\le j\le k\le \infty$ meaning that the $j,m$ series are interchangeable if both are infinite series. However, how would one take the limit?
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Evaluate $\int_0^1\frac{x\ln x}{(1+x^2)^2}\ dx$ $$\int_0^1\frac{x\ln x}{(1+x^2)^2}\ dx
$$
Help me please. I don't know any ways of solution. Thank you.
| Integrating by Parts,
$$I=\int \ln x\cdot\frac x{(1+x^2)^2}dx$$
$$=\ln x\int\frac x{(1+x^2)^2}dx-\int\left(\frac{d(\ln x)}{dx}\cdot \frac x{(1+x^2)^2}dx\right)dx $$
For $\displaystyle J=\int\frac x{(1+x^2)^2}dx$ set $1+x^2=u\implies2x\ dx=du$
$\displaystyle J=\int\frac{du}{2u^2}=-\frac1{2u}=-\frac1{2(1+x^2)}$
$\displaystyle\implies I=\ln x\left(-\frac1{2(1+x^2)}\right)+\int\frac1{2x(1+x^2)}\ dx$
Setting $x^2=v,$
$\displaystyle K=\int\frac1{2x(1+x^2)}\ dx=\int\frac{2x}{4x^2(1+x^2)}\ dx$
$\displaystyle4K=\int\frac{dv}{v(1+v)}=\int\frac{v+1-v}{v(1+v)}dv=\int\frac{dv}v -\int\frac{dv}{v+1}=\cdots$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 1
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How prove $\left(\frac{b+c}{a}+2\right)^2+\left(\frac{c}{b}+2\right)^2+\left(\frac{c}{a+b}-1\right)^2\ge 5$ Let $a,b,c\in R$ and $ab\neq 0,a+b\neq 0$. Find the minimum of:
$$\left(\dfrac{b+c}{a}+2\right)^2+\left(\dfrac{c}{b}+2\right)^2+\left(\dfrac{c}{a+b}-1\right)^2\ge 5$$
if and only if $$a=b=1,c=-2$$
My idea: Since
$$\left(\dfrac{b+c+2a}{a}\right)^2+\left(\dfrac{c+2b}{b}\right)^2+\left(\dfrac{c-a-b}{a+b}\right)^2$$
let
$$x=\dfrac{b+c+2a}{a},y=\dfrac{c+2b}{b},z=\dfrac{c-a-b}{a+b}$$
then I can't work.
Thank you.
| Note that "if and only if" is clearly not true, since any multiples will work as your expression is homogenous. In fact, the equality case is more general than what you listed.
Differentiating with respect to $c$ (which only appears in the numerator, hence is not too ugly), you can show that (Thanks Wolfram) the minimum occurs when
$$ c = - \frac{ b^3+3ab^2+2a^2b } {a^2+ab+b^2 }. $$
Substituting this back into the expression (and tediously expanding), we get the value of 5. Hence, the minimum is 5.
Now, verify that this is indeed the global minimum by doing all the proper checks (I'm too lazy to complete this calculus approach completely.)
| {
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"timestamp": "2023-03-29T00:00:00",
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$\int_{0}^{\infty}\int_{1}^{\infty}\frac{x^2-y^2}{(x^2+y^2)^2}dxdy$ diverges? I want someone to review my proof that $$\int_{1}^{\infty}\int_{0}^{\infty}\frac{x^2-y^2}{(x^2+y^2)^2}dxdy$$ does not converge.
To make things easier, I said let's look at the entire first quadrant and then subtract the integral over the small rectangle that we added.
move to polar coordinates:
$$\int_{0}^{\frac{\pi}{2}}\int_{o}^{\infty}r^2\frac{(\cos^2\theta-\sin^2\theta)}{r^3}drd\theta=\int_{0}^{\frac{\pi}{2}}\int_{o}^{\infty}\frac{\cos^2\theta-\sin^2\theta}{r}drd\theta=\int_{0}^{\frac{\pi}{2}}\int_{o}^{\infty}\frac{1-2\sin^2\theta}{r}drd\theta\leq\int_{0}^{\frac{\pi}{2}}\int_{o}^{\infty}\frac{1}{r}drd\theta$$
The inequality part is true since $-2\sin^2 x$ is always negative.
So:
$$\int_{0}^{\frac{\pi}{2}}\int_{o}^{\infty}\frac{1}{r}drd\theta=\frac{\pi}{2}\ln(r)|_0^{\infty}=\frac{\pi}{2}\ln(\frac{\infty}{0})=\frac{\pi}{2}\ln({\infty})=\infty$$
So over the entire first quadrant it diverges.
Is there a point to checking the small rectangle that we added? I mean, even if it diverges, the answer would still diverge.
| Since we are considering the region in the first quadrant above $y=1$, in polar coordinates we may write it as:
\begin{align*}
\int_{1}^{\infty}\int_{0}^{\infty}\frac{x^2-y^2}{(x^2+y^2)^2}\, dx\, dy &= \lim_{N\to\infty} \int_{\arcsin(1/N)}^{\pi/2} \, \int_{1/\sin\theta}^{N} \, \frac{\cos\left(2\, \theta\right)}{r}\, dr\, d\theta\\
&= \lim_{N\to\infty}\, \frac{1}{2N}\sqrt{1-\frac{1}{N^2}}+\frac{1}{2}\arcsin\frac{1}{N} - \frac{\pi}{4}\\
&= -\frac{\pi}{4}
\end{align*}
If we calculate without transformation:
\begin{align*}
\lim_{N\to\infty} \int_{1}^{N} \, \int_{0}^{N} \, \frac{x^2-y^2}{(x^2+y^2)^2}\, dx\, dy &= \lim_{N\to\infty} \arctan\left(\frac{1}{N}\right)- \frac{\pi}{4}\\
&= -\frac{\pi}{4}
\end{align*}
Using mpmath's gauss-legendre quadrature with $\mathrm{maxdegree} > 6$ indicates $\displaystyle -\frac{\pi}{2}$ as answer, and when $\mathrm{maxdegree} \le 6$ it stays close to $\displaystyle -\frac{\pi}{4}$. 'tanh-sinh' is close to neither.
Also note that if we choose
\begin{align*}
\lim_{N_1\to\infty} \lim_{N_2\to\infty} \int_{1}^{N_2} \, \int_{0}^{N_1} \, \frac{x^2-y^2}{(x^2+y^2)^2}\, dx\, dy &= \lim_{N_1\to\infty}\lim_{N_2\to\infty} \arctan\left(\frac{1}{N_1}\right) - \arctan\left(\frac{N_2}{N_1}\right)
\end{align*}
So, the actual answer depends on the ratio $\displaystyle \frac{N_2}{N_1}$. If we consider "$\infty$" in both directions to be the same, the answer is $\displaystyle -\frac{\pi}{4}$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Probability of picking three marbles in order A bag contains 9 red marbles, 5 blue marbles and 7 green marbles. Need to select one marble at a time without replacement. What is the probability that the first marble is blue, the second is blue or green an the third is red.
My Solution:
$= P(\text{blue}) * P(\text{blue} \cup \text{green}) * P(\text{red})$
$\displaystyle = {5\over 21} * ({4\over 20} + {7\over 20}) * ({9\over 19})$
Does that sound that right?
| That looks like the right answer. These are conditional probabilities, so
\begin{align*}
& \mathrm{Pr}(\text{marble 1 is blue} \cap \text{marble 2 is blue or green} \cap \text{marble 3 is red}) \\
&= \mathrm{Pr}(\text{marble 1 is blue}) \times \mathrm{Pr}(\text{marble 2 is blue or green} \mid \text{marble 1 is blue}) \times \mathrm{Pr}(\text{marble 3 is red} \mid \text{marble 1 is blue} \cap \text{marble 2 is blue or green}) \\
&= \frac{5}{9+5+7} \times \frac{(5+7)-1}{(9+5+7)-1} \times \frac{9}{(9+5+7)-2}\\
&= \frac{33}{532}.
\end{align*}
Alternatively, we can use a counting argument. There are $3! \binom{9+5+7}{3}=7980$ ordered $3$-tuples of $\{r_1,\ldots,r_9,b_1,\ldots,b_5,g_1,\ldots,g_7\}$. Of these, precisely $5 \times (5+7-1) \times 9=495$ satisfies the blue-blue/green-red pattern. We have $$\frac{5 \times (5+7-1) \times 9}{3! \binom{9+5+7}{3}}=\frac{33}{532}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/808345",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solve the following equation in positive integers $x$ and $y$ What are the solutions in positive integers of the equation:
$${1+2^x+2^{2x+1}=y^2}$$
I tried to factorize the equation but it didn't help much.
Clearly $y $ is an odd integer. Substituting $y =2n+1$, we get
$2^x+2^{2x+1}=(2n)\cdot{(2n+2)}$
$\Rightarrow (2^{x-2})\cdot(1+2^{x+1})=(n)\cdot(n+1)$
Which is the product of 2 consecutive integers. Does it help? I don't know.
| Let $2^x = z$, then: $2z^2 + z + y^2 - 1 = 0$.
Calculate $\triangle = 1^2 - 4\cdot 2\cdot (y^2 - 1) = 9 - 8y^2$.
$\triangle > 0 \iff 9 - 8y^2 > 0 \iff y^2 < \dfrac{9}{8} = 1.11$, so $y = 1$, and $\triangle = 1$. So $z = \dfrac{-1 \pm 1}{2\cdot 2} = 0$ or $-0.5$, but $z =2^x > 2$. So there is no solution.
In fact $\triangle = 1^2 - 4\cdot 2\cdot (1-y^2) = 8y^2 - 7$. Now we solve the Pell equation:
$8y^2 - 7 = k^2$ or $8y^2 - k^2 = 7$. This has an initial solution $(y_0,k_0) = (1,1)$. Using this we can generate the general solution, then we can find $z$, and then $x$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Evaluation of $\int_{0}^{1}4x^3\cdot \left\{\frac{d^2}{dx^2}\left(1-x^2\right)^5\right\}dx$ The value of $\displaystyle \int_{0}^{1}4x^3\cdot \left\{\frac{d^2}{dx^2}\left(1-x^2\right)^5\right\}dx = $
$\bf{My\; Try::}$ Let $\displaystyle I = \int 4x^3\cdot \left\{\frac{d^2}{dx^2}\left(1-x^2\right)^5\right\}dx$
Using Integration by parts, we get
$\displaystyle I = 4x^3\cdot \int\left\{\frac{d^2}{dx^2}\left(1-x^2\right)^5\right\}dx - 12\int\left\{x^2\cdot \int\left\{\frac{d^2}{dx^2}\left(1-x^2\right)^5\right\}dx\right\}dx$
$\displaystyle I = 4x^3\cdot \frac{d}{dx}(1-x^2)^5-12\int \left\{x^2\cdot \frac{d}{dx}(1-x^2)^5\right\}dx$
$\displaystyle I = 4x^3\cdot 5\cdot (1-x^2)^4\cdot -2x+120\int x^2\cdot (1-x^2)^4\cdot xdx$
Now Let $x^2=t\;,$ Then $2xdx = dt$
$\displaystyle I = -10x^4\cdot (1-x^2)^4+60\int t\cdot (1-t)^4dt=-10x^4\cdot (1-x^2)^4+60\int (1-t)t^4dt$
So $\displaystyle I = -10x^4\cdot (1-x^2)^4+60\left\{\frac{t^5}{5}-\frac{t^6}{6}\right\}+\mathbb{C}$
Is my process is right?,If not then how can we solve it.
Help Required.
Thanks
| I wonder if it would not be simpler to establish that $$\frac{d^2}{dx^2}\left(1-x^2\right)^5=80 x^2 \left(1-x^2\right)^3-10 \left(1-x^2\right)^4=-10 \left(x^2-1\right)^3 \left(9 x^2-1\right)$$ and to use the fact that $4x^3=2x^2\frac{d(x^2)}{dx}$ which makes the change of variable $x^2=t$ quite clear. Then expansion of the integrand is quite simple in terms of $t$ and integration quite easy.
| {
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Number of non-decreasing sequences How do I find the number of non-decreasing sequences of length $N$, such that all number in the sequences lie in the range $[a, b]$.
Also, the frequency of the most frequently occurring element should be unique.
For example, 1 1 2 3 is a valid sequence, but 1 1 2 2 3 is not a valid sequence, because the frequency of 1 and 2 is the same in the $2$nd sequence.
| The maximum frequency, say $k$, for a sequence can range from $1$ to
$N$. We will count the number of valid sequences for a fixed $k$
then sum over $k = 1, ... ,N$.
Let $d = b - a$.
The number, say $c$, having (uniquely) the maximum frequency $k$ can
be any number from $a$ to $b$. So we will fix $c$, count the number
of sequences for it, then subsequently sum over $c = a,...,b$, which
in fact is a simple multiplication by $d+1$ since the number of
sequences is the same for each $c$.
If number $c$ occurs exactly $k$ times then there are $N-k$ numbers
from $a$ to $b$, excluding $c$, in the sequence such that none of
them occurs $k$ or more times.
It is the same as putting $N-k$ balls into $d$ cells, with no cell
containing $k$ or more balls.
Let $S_k$ be the set of all arrangements, without the
$k$-restriction on the cells, of $N-k$ balls in $d$ cells.
$\vert S_k \vert = \binom{N-k+d-1}{d-1}\qquad\mbox{using the common
"stars and bars" argument.}$
For $i = 1,...,d$ let set $S_{ki} = \left\{s \in S_k :\mbox{
arrangement $s$ has at least $k$ balls in the $i^{th}$
cell}\right\}$
We seek the cardinality of set:
$V_k =
\left(\bigcup_{i=1}^{d}{S_{ki}}\right)^c\qquad\left(\mbox{where the
complement is wrt $S_k$}\right)$ this set $V_k$ being that of valid
sequences for fixed $k$ and $c$.
By the Inclusion-Exclusion Principle:
\begin{eqnarray*}
\vert V_k \vert &=& \vert S_k \vert - \sum_{i=1}^{d}{\vert S_{ki}
\vert} + \sum_{i<j}{\vert S_{ki} \cap S_{kj} \vert} - ... -
\left(-1\right)^{i+1} \vert \bigcap_{i=1}^{d}{S_{ki}} \vert
\end{eqnarray*}
\begin{eqnarray*}
\vert S_{ki} \vert &=& \mbox{No. of ways to put $N-2k$ balls into $b-a$ cells} \\
&&\qquad\left(\mbox{since we have at lease $k$ balls in the $i^{th}$ cell}\right) \\
&=& \binom{N-2k+d-1}{d-1}
\end{eqnarray*}
and generally, for an intersection of $n$ of the $S_{ki}$ sets,
\begin{eqnarray*}
\vert \bigcap_{j=1}^{n}{S_{ki_j}} \vert &=&
\binom{N-\left(n+1\right)k+d-1}{d-1}.
\end{eqnarray*}
Thus,
\begin{eqnarray*}
\vert V_k \vert &=& \vert S_k \vert -
\sum_{n=1}^{d}{\left(-1\right)^{n+1}\binom{d}{n}
\binom{N-\left(n+1\right)k+d-1}{d-1}} \\
\vert V_k \vert &=& \sum_{n=0}^{d}{\left(-1\right)^n\binom{d}{n}
\binom{N-\left(n+1\right)k+d-1}{d-1}}
\end{eqnarray*}
Summing over $c = a,...,b$ and over $k = 1,...,N$ gives the final
result:
\begin{eqnarray*}
Ans. &=& \left(d+1\right) \sum_{k=1}^{N}{\vert V_k \vert} \\
&=& \left(d+1\right) \sum_{k=1}^{N}{\left[ \sum_{n=0}^{d}{\left(-1\right)^n \binom{d}{n} \binom{N-\left(n+1\right)k+d-1}{d-1}}\right]}. \\
\end{eqnarray*}
As an example, let $\left[a,b\right] = \left[1,4\right]$ and $N =
3$. So $d = 3$.
There are $16$ valid sequences:
\begin{eqnarray*}
111, 112, 113, 114, 122, 133, 144, \\
222, 223, 224, 233, 244, \\
333, 334, 344, \\
444.
\end{eqnarray*}
\begin{eqnarray*}
Ans. &=& 4 \sum_{k=1}^{3}{\left[ \sum_{n=0}^{3}{\left(-1\right)^n
\binom{3}{n} \binom{5-\left(n+1\right)k}{2}}\right]}
\\ &=& 4 \sum_{k=1}^{3}{\left[ \binom{5-k}{2} -
3 \binom{5-2k}{2} + 3 \binom{5-3k}{2} - \binom{5-4k}{2}\right]}
\\ &=& 4 \left[ \binom{4}{2} -
3 \binom{3}{2} + 3 \binom{2}{2}\right] + 4 \left[
\binom{3}{2}\right] + 4 \left[ \binom{2}{2}\right] \\
&=& 4\left(6 - 9 + 3\right) + 4\left(3\right) + 4\left(1\right) \\
&=& 16
\end{eqnarray*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/808690",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
A pair of dice rolls and outcome A pair of fair dice is rolled three times and each time the two digits are added. What is the probability that a sum that is greater than or equal to 9 occurs exactly once?
My solution:
The ways to get the various totals is:
9 = 5 4, 4 5
9 = 3 6, 6 3
10 = 5 5
10 = 6 4, 4 6
11 = 5 6, 6 5
12 = 6 6
There are 10 ways to roll to get a sum of >= 9.
There are 36 different outcomes for each roll.
The probability of getting a sum of >= 9 is 10/36.
The probability of not getting a sum of < 9 is 26/36.
So for 3 rolls for a pair of dice, the probability of getting sum of >= 9 is
$3.(10/36).(26/36).(26/36)$
Correct?
| In words:
*
*Choose $1$ out of $3$ events for which the sum is greater than or equal to $9$
*Choose $2$ out of the remaining $2$ events for which the sum is smaller than $9$
Mathematically, it can be expressed as $\binom{3}{1} \cdot P(sum\geq9) \cdot \binom{2}{2} \cdot P(sum<9)$
*
*$\binom{3}{1} = \frac{3!}{1!2!} = 3$
*$\binom{2}{2} = \frac{2!}{2!0!} = 1$
*$P(sum\geq9) = P([3,6],[4,5],[4,6],[5,4],[5,5],[5,6],[6,3],[6,4],[6,5],[6,6]) = \frac{10}{36}$
*$P(sum<9) = 1-P(sum\geq9) = \frac{26}{36}$
So the answer is $3 \cdot \frac{10}{36} \cdot 1 \cdot \frac{26}{36} = \frac{65}{108}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/808803",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How to solve this weird inequality? $\frac{x-1}{x+1} < x$
Thanks!
I did the following.
$\frac{x-1}{x+1} - x< 0 /-x$
$\frac{x-1 - x(x+1)}{x+1} < 0$
$\frac{-x^2-1}{x+1} < 0$
What to do next?
| Note that the inequality can be written as
$$-(x^2 + 1) \cdot \frac{1}{x + 1} < 0$$
Now $x^2 + 1$ is always positive, so this is equivalent to studying
$$\frac{1}{x + 1} > 0$$
Can you solve this?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/809205",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Finding an exact solution to a difference equation How would I solve an equation of the form:
$u(n+1)=1/2u(n)+(1/3)^n$ when $u(0)=1$?
I got an answer of the form $u(n)= c + \sum(1/3)^j*2^{j-1}$ but believe this is incorrect?
| Consider $u_{n+1} = \dfrac{1}{2}\cdot u_n$. This gives: $u_n = 2^{-n}$, $n\geq 0$. So the general solution would be: $u_n = a\cdot 2^{-n} + b\cdot 3^{-n}$.
$u_0 = 1 \to 1 = a + b$, and $u_1 = 0.5u_0 + 1 = \dfrac{3}{2} \to \dfrac{a}{2} + \dfrac{b}{3} = \dfrac{3}{2} \to 3a + 2b = 9 \to 3a + 2(1 - a) = 9 \to a = 7$, and $b = 1 - a = 1 - 7 = -6$. Thus:
$u_n = 7\cdot 2^{-n} - 6\cdot 3^{-n}$, $n \geq 0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/809726",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Show that $f\in \mathcal{R}[0,1]$ Let $f$ be defined by
$$f(x) =\left\{\frac{1}{n}, \frac{1}{n+1} \lt x \le \frac{1}{n} \right\}$$ and $f(x) = 0$ when $x=0$ and $n \in N$. Show that $f$ is integrable and $$\int_{0}^{1}f(x)dx=\frac{\pi^2}{6}-1.$$
| Hint :
Suppose it is integrable can you relate that integral (Draw a picture) to following sum
$$\frac{1}{2}+(\frac{1}{2}-\frac{1}{3})\cdot \frac{1}{2}+(\frac{1}{3}-\frac{1}{4})\cdot \frac{1}{3}+(\frac{1}{4}-\frac{1}{5})\cdot \frac{1}{4}+\cdots$$
This should immediately give you required result.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Evaluating the integral with trigonometric integrand While solving another problem I have come across this integral which I am unable to evaluate. Can someone please evaluate the following integral? Thank you.
$$\int_0^{2\pi}\frac{1}{(2+\cos\theta)^2}\,d\theta.$$
| Here is an ugly way of doing it. Pranav's solution is much more elegant.
First substitute $u = \tan(\tfrac{\theta}{2})$ and $\mathrm{d}u = \tfrac{1}{2} \sec^2(\tfrac{\theta}{2})\mathrm{d}\theta$. Then, transform the integrand using the substitution $\sin(\theta) = \frac{2u}{u^2+1}$, $\cos(\theta) = \frac{1-u^2}{u^2+1}$ and $\mathrm{d}\theta = \frac{2\mathrm{d}u}{u^2+1}$ to get $$I = \int \frac{1}{(2+\cos(\theta))^2}\mathrm{d}\theta = \int \frac{2}{(u^2 + 1)\left(\tfrac{1-u^2}{u^2+1} +2\right)^2} \mathrm{d}u = 2\int \frac{u^2+1}{u^4 + 6u^2 +9} \mathrm{d}u.$$
Now, using partial fractions, one can rewrite this integral as
$$I = \frac{2}{3}\int\frac{1}{\tfrac{u^2}{3}+1}\mathrm{d}u - 4\int\frac{1}{(u^2+3)^2}\mathrm{d}u.$$
For the first integrand, substitute $s = \frac{u}{\sqrt{3}}$ and $\mathrm{d}s = \frac{1}{\sqrt{3}}\mathrm{d}u$ to get
$$I = \frac{2}{\sqrt{3}} \int \frac{1}{s^2+1} \mathrm{d}s - 4\int\frac{1}{(u^2+3)^2}\mathrm{d}u.$$
We know how to integrate the left side. This is just
$$I = \frac{2\arctan(s)}{\sqrt{3}} - 4\int\frac{1}{(u^2+3)^2}\mathrm{d}u.$$
For the right-hand side, we substitute $u = \sqrt{3}\tan(p)$ and $\mathrm{d}u=\sqrt{3}\sec^2(p)\mathrm{d}p$. Then, we have that $(u^2+3)^2 = (3\tan^2(p) +2)^2 = 9\sec^4(p)$ and $p = \arctan(\tfrac{u}{\sqrt{3}})$. This yields the integral
$$I = \frac{2\arctan(s)}{\sqrt{3}} - \frac{4}{3\sqrt{3}} \int \cos^2(p) \mathrm{d}p.$$
Integrating $\cos^2(p)$ is easy by using the identity $\cos^2(p) = \frac{\cos(2p) + 1}{2}$. All that remains now is to integrate this and substitute your way back to express $I$ in terms of $\theta$. Evaluate from $0$ to $2\pi$ to obtain
$$I_0^{2\pi} = \frac{4\pi}{3\sqrt{3}}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/810087",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 6,
"answer_id": 1
} |
Simplifying $(1/x-1/5 )/( 1/x^2-1/25)$ How do I get $\frac{5x}{x+5}$ from simplifying the following?
$$\frac{(\frac{1}{x}-\frac{1}{5} )}{( \frac{1}{x^2}-\frac{1}{25})}$$
My work:
I multiplied the top and bottom by the LCD: $25x^2$ (to get the same denominators).
Then I got: $\frac{25x-5x}{25-x^2}$
Then I got this for my answer: $\frac{20}{-(x-5)(x+5)}$
But the real answer is: $\frac{5x}{x+5}$
So my questions is how to I get from $\frac{20}{-(x-5)(x+5)}$ to $\frac{5x}{x+5}$?
| $$\frac{\frac1x-\frac15}{\frac1{x^2}-\frac1{25}}=\frac{\frac{5-x}{5x}}{\frac{25-x^2}{25x^2}}$$
$$=\frac{25x^2(5-x)}{5x(25-x^2)}=\frac{5x}{5+x}$$ assuming $x(5-x)\ne0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/810274",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Eigenvalues and Eigenvectors for matrix. Complex Eigenvalues How can I find out the eigenvectors for this matrix:
$$A=
\begin{pmatrix}
-3 &0&0\\
0&3&-2\\
0&1&1
\end{pmatrix}
$$
I found the eigenvalues: $\lambda_{1}=-3$, $\lambda_{2}=2-i$, $\lambda_{3}=2+i$.
The eigenvector for $\lambda_{1}=-3$ is $$\begin{pmatrix}
1\\
0\\
0
\end{pmatrix}.$$
For $\lambda_{3}=i+2$ I write $$A(u+iv)=(i+2)(u+iw),$$ but I don't know how to do$\ldots$ Thanks!
| I do not recommend separating the eigenvector into real and imaginary parts. In my opinion it only makes it more troublesome, complex numbers are perfectly good scalars to work with.
Note the following
$$\begin{align}
Av=\lambda _2v&\iff (A-\lambda _2I)v=0_{3\times 1}\\
&\iff \begin{pmatrix} -5+i & 0 & 0\\ 0 & 1+i & -2\\ 0 & 1 & -1+i\end{pmatrix}\begin{pmatrix} v_1\\ v_2\\ v_3\end{pmatrix}=\begin{pmatrix}0\\0\\0 \end{pmatrix}\\
&\iff \begin{cases}(-5+i)v_1&=0\\ (1+i)v_2-2v_3&=0\\ v_2+(-1+i)v_3&=0 \end{cases}\\
&\iff \begin{cases}v_1&=0\\ 2v_2-2(1-i)v_3&=0\\ v_2+(-1+i)v_3&=0 \end{cases}\\
&\iff \begin{cases}v_1&=0\\ 2v_2+2(-1+i)v_3&=0\\ v_2+(-1+i)v_3&=0 \end{cases}\\
&\iff \begin{cases}v_1&=0\\ 2v_2-2(1-i)v_3&=0\\ v_2+(-1+i)v_3&=0 \end{cases}\\
&\iff \begin{cases}v_1&=0\\ v_2+(-1+i)v_3&=0\end{cases}\\
&\iff (v_1, v_2, v_3)=(0,(1-i)v_3, v_3).
\end{align}$$
You can thus take the eigenpair $\left(2-i, \begin{pmatrix} 0\\ 1-i\\ 1\end{pmatrix}\right)$.
It is very important to note that given a square real matrix $M$ and $(\lambda, u)$ one of its eigenpairs, the following holds:
$$Mu=\lambda u\iff \overline{Mu}=\overline{\lambda u}\iff M\overline u=\overline \lambda \overline u.$$
What does this tell you?
| {
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"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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How to prove: prove $\frac{1+\tan^2\theta}{1+\cot^2\theta} = \tan^2\theta$ I need to prove that $\frac{1+\tan^2\theta}{1+\cot^2\theta}= \tan^2\theta.$
I know that $1+\tan^2\theta=\sec^2\theta$ and that $1+\cot^2\theta=\csc^2\theta$, making it now $$\frac{\sec^2\theta}{\csc^2\theta,}$$ but I don't know how to get it down to $ \tan^2\theta.$
HELP!!!!
I also need help proving that $\tan\theta + \cot\theta = \sec\theta\cdot\csc\theta.$
| Know that: $\tan\theta=\dfrac{1}{\cot\theta}$. Therefore:
$$\dfrac{(1+\tan^2\theta)}{(1+\cot^2\theta)}=\dfrac{(1+\tan^2\theta)}{\bigg(1+\dfrac{1}{\tan^2\theta}\bigg)}=\dfrac{(1+\tan^2\theta)}{\bigg(\dfrac{1+\tan^2\theta}{\tan^2\theta}\bigg)}=\tan^2\theta.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/810453",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 9,
"answer_id": 0
} |
Series Question: $\sum_{n=1}^{\infty}\frac{n+1}{2^nn^2}$ How to compute the following series:
$$\sum_{n=1}^{\infty}\frac{n+1}{2^nn^2}$$
I tried
$$\frac{n+1}{2^nn^2}=\frac{1}{2^nn}+\frac{1}{2^nn^2}$$
The idea is using Riemann zeta function
$$\zeta(s)=\sum_{n=1}^{\infty}\frac{1}{n^s}$$
but the term $2^n$ makes complicated. I know that
$$\sum_{n=1}^{\infty}\frac{1}{2^n}=1$$
using geometric series but I don't know how to use those series to answer the question. Any help would be appreciated. Thanks in advance.
| The series
\begin{align}
S = \sum_{n=1}^{\infty} \frac{n+1}{2^{n} \ n^{2}}
\end{align}
can be expressed as
\begin{align}
S = \sum_{n=1}^{\infty} \frac{1}{2^{n} \ n} + \sum_{n=1}^{\infty} \frac{1}{2^{n} \ n^{2}}
\end{align}
and is seen to be
\begin{align}
S = - \ln\left( 1 - \frac{1}{2} \right) + Li_{2}\left(\frac{1}{2}\right),
\end{align}
where $Li_{2}(x)$ is the dilogarithm function. Since
\begin{align}
Li_{2}\left(\frac{1}{2}\right) = \frac{\pi^{2}}{12} - \frac{1}{2} \ \ln^{2}(2)
\end{align}
then the resulting series has the value
\begin{align}
\sum_{n=1}^{\infty} \frac{n+1}{2^{n} \ n^{2}} = \frac{\pi^{2}}{12} + \ln(2) - \frac{1}{2} \ \ln^{2}(2).
\end{align}
This may also be seen in the form
\begin{align}
\sum_{n=1}^{\infty} \frac{n+1}{2^{n} \ n^{2}} = \frac{\pi^{2}}{12} + \frac{1}{2} \ \ln(2) \ \ln\left(\frac{e^{2}}{2}\right).
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/810514",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 3,
"answer_id": 0
} |
How to solve this equation with three variables with an unknown parameter using Gaussian elimination? If I've got three equations:
$$\begin{array}{ccccccc} x & + & y & + & z & = 3 \\ 2x & + & ay & - & 2z & = 4 \\ x & + & 2y & - & az & = 1 \end{array}$$
How do I solve them using Gaussian elimination. I get to this point:
$$\begin{array}{ccccccc} x & + & y & + & z & = 3 \\ & & y & - & (a+1)z & = -2 \\ & & (a-2)y & - & 4z & = -2 \end{array}$$
but I don't know where to go from here. I'd try to multiply the second by $-(a-2)$ and add it to the third but it doesn't seem to work out well. My textbook says that the next step should look like:
$$\begin{array}{ccccccc} x & + & y & + & z & = 3 \\ & & y & - & (a+1)z & = -2 \\ & & & & (a+2)(a-3)z & = 2(a-3) \end{array}$$
How do I get to that?
| This will get you to the step you asked about:
$\left[\begin{array}{ccc|c}
1 & 1 & 1 & 3\\
2 & a & -2 & 4\\
1 & 2 & -a & 1\end{array}\right]\Longrightarrow
\left[\begin{array}{ccc|c}
1 & 1 & 1 & 3\\
0 & a-2 & -4 & -2\\
0 & 1 & -a-1 & -2\end{array}\right]\Longrightarrow
\left[\begin{array}{ccc|c}
1 & 1 & 1 & 3\\
0 & 0 & -4+(a+1)(a-2) & -2+2(a-2)\\
0 & 1 & -a-1 & -2\end{array}\right]\Longrightarrow
\left[\begin{array}{ccc|c}
1 & 1 & 1 & 3\\
0 & 0 & a^2-a-6 & 2a-6\\
0 & 1 & -a-1 & -2\end{array}\right]\Longrightarrow
\left[\begin{array}{ccc|c}
1 & 1 & 1 & 3\\
0 & 1 & -a-1 & -2\\
0 & 0 & a^2-a-6 & 2a-6\end{array}\right]\Longrightarrow
\left[\begin{array}{ccc|c}
1 & 1 & 1 & 3\\
0 & 1 & -(a+1) & -2\\
0 & 0 & (a-3)(a+2) & 2(a-3)\end{array}\right]$
Then to finish the problem:
$\left[\begin{array}{ccc|c}
1 & 0 & a+2 & 5\\
0 & 1 & -(a+1) & -2\\
0 & 0 & (a+2) & 2\end{array}\right]\Longrightarrow \left[\begin{array}{ccc|c}
1 & 0 & 0 & 3\\
0 & 1 & 1 & 0\\
0 & 0 & (a+2) & 2\end{array}\right] \Longrightarrow \left[\begin{array}{ccc|c}
1 & 0 & 0 & 3\\
0 & 1 & 1 & 0\\
0 & 0 & 1 & \frac{2}{a+2}\end{array}\right] $
$\Longrightarrow \left[\begin{array}{ccc|c}
1 & 0 & 0 & 3\\
0 & 1 & 0 & \frac{-2}{a+2}\\
0 & 0 & 1 & \frac{2}{a+2}\end{array}\right] \Longrightarrow x=3 ~\&~ y=-z$
and if $a=-2$ there is no solution
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/810628",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Diophantine equation in four variables I would like to find a parametric solution for the following diophantine equation:
$-4 (1-a_1^2)(1-a_2^2) + (1+a_3^2 -a_1^2 -a_2^2)^2 = a_4^2$
Does such a solution exist?
How does one go about solving such questions systematically?
| Equation:
$q^2=(z^2+d^2-x^2-y^2)^2-4(d^2-x^2)(d^2-y^2)$
Has the solution:
$x=(p-s)sa^2+(p^2-ps+s^2)t^2-psk^2$
$y=-(p-s)sa^2+(p^2-ps+s^2)t^2-psk^2+2psak-2(p^2-ps+s^2)at$
$d=(p-s)sa^2-(p^2-ps+s^2)t^2-psk^2-2(p-s)sat+2pskt$
$z=(p-s)sa^2+(p^2-ps+s^2)t^2+psk^2-2(p^2-ps+s^2)kt-2(p-s)sak$
$q=4((p^2-ps+s^2)((p^2-s^2)k-(p^2-2ps)a)t^3+ps((p^2-s^2)t+as^2)k^3+$
$+s((p^3-3sp^2+2ps^2)t+k(p-s)s^2)a^3-(p^4-s^4)t^2k^2+p(p^3-4sp^2+6ps^2-4s^3)t^2a^2-$
$-(2p-s)s^3a^2k^2+(2p^3-3sp^2-ps^2+s^3)sakt^2-(p^3-2ps^2+2s^3)satk^2-$
$-(p^3-3sp^2+ps^2-s^3)skta^2)$
$p,s,a,t,k$ - Any integers.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/815072",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
$4 \sin 72^\circ \sin 36^\circ = \sqrt 5$ How do I establish this and similar values of trigonometric functions?
$$
4 \sin 72^\circ \sin 36^\circ = \sqrt 5
$$
| Hint: Notice that $72 = 2\cdot 36$ and $180 = 36\cdot 5$; use formulaes for halfed or doubled argument.
$$
\sin^2\frac{x}{2} = \frac{1-\cos x}{2}
$$
or
$$
\sin2x = 2\sin x\cos x
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/816195",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Link between a cubic polynomial and a trig identity Alright, so I am told to prove that:
$$\tan (3A) = \frac{3\tan(A)-\tan^3(A)}{1-3\tan^2(A)}$$
This can be pretty easily done by applying the $\tan$ addition formula, taking the angles $2A$ and $A$, upon which one then applies the $\tan$ double angle formula. Simplifying the resultant mess indeed yields the above identity. To show that I have attempted the question:
$$\tan(2A + A) = \frac{\tan(2A) + \tan(A)}{1-\tan(2A)\tan(A)}$$
Note:
$$\tan(2A)= \frac{2\tan(A)}{1-\tan^2(A)}$$
Substituting:
$$\tan(3A)=\frac{2\tan(A)/(1-\tan^2(A))+\tan(A)}{1-(2\tan(A)/(1-\tan^2(A)))\tan(A)}$$
Which simplifies to:
$$\tan(3A)=\frac{(3\tan(A)-\tan^3(A))/(1-\tan^2(A))}{(1-\tan^2(A))/(1-3\tan^2(A))}$$
The two $(1-\tan^2(A))$ cancel out and we attain our desired identity..
However, I am told to make use of the fact that the cubic polynomial
$$t^3 - 3t^2 - 3t +1 $$
factorizes into:
$$(t+1)(t^2-4t+1)$$
Clearly, the polynomial and the trig identity look similar: i.e. $3\tan = 3t$ or $1-3t^2 = 1 - 3\tan^2$, etc.
However, how can the factorization aid the proof? I am currently at a brain block, so a clue would be very much appreciated! :)
Additionally, this should not really require any use of complex numbers / De Moivre as this question comes from a chapter in which complex numbers have not been introduced.
Thank you!
| It's a lot of work to proof the identity so ;) :
Verify the following identity:
$$\tan(3a)=\frac{3\tan(a)-\tan^3(a)}{1-3\tan^2(a)}=>$$
$$\tan(3a)\left(1-3\tan^2(a)\right)=3\tan(a)-\tan^3(a)=>$$
$$\left(1-3\left(\frac{\sin(a)}{\cos(a)}\right)^2\right)\frac{\sin(3a)}{\cos(3a)}=3\frac{\sin(3a)}{\cos(3a)}-\left(\frac{\sin(a)}{\cos(a)}\right)^3=>$$
$$\frac{\sin(3a)\left(1-\frac{3\sin^2(a)}{\cos^2(a)}\right)}{\cos(3a)}=3\left(\frac{\sin(a)}{\cos(a)}\right)-\left(\frac{\sin(a)}{\cos(a)}\right)^3=>$$
$$\frac{\sin(3a)\left(1-\frac{3\sin^2(a)}{\cos^2(a)}\right)}{\cos(3a)}=\frac{3\sin(a)}{\cos(a)}-\frac{\sin^3(a)}{\cos^3(a)}=>$$
$$\frac{\sin(3a)}{\cos(3a)}\frac{\cos^2(a)-3\sin^2(a)}{\cos^2(a)}=\frac{3\sin(a)}{\cos(a)}-\frac{\sin^3(a)}{\cos^3(a)}=>$$
$$\frac{\sin(3a)\left(\cos^2(a)-3\sin^2(a)\right)}{\cos^2(a)\cos(3a)}=\frac{3\cos^2(a)\sin(a)-\sin^3(a)}{\cos^3(a)}=>$$
$$\cos^3(a)\sin(3a)\left(\cos^2(a)-3\sin^2(a)\right)=\cos^2(a)\cos(3a)\left(3\cos^2(a)\sin(a)-\sin^3(a)\right)=>$$
$$\cos(a)sin(3a)\left(\cos^2(a)-3\sin^2(a)\right)=\cos(3a)\left(3\cos^2(a)\sin(a)-\sin^3(a)\right)=>$$
$$\cos(a)\sin(3a)\left(1-\sin^2(a)-3\sin^2(a)\right)=\cos(3a)\left(3\cos^2(a)\sin(a)-\sin^3(a)\right)=>$$
$$\cos(a)\left(1-4\sin^2(a)\right)\sin(3a)=\cos(3a)\left(3\cos^2(a)\sin(a)-\sin^3(a)\right)=>$$
$$\cos(a)\left(3\cos^2(a)\sin(a)-\sin^3(a)\right)\left(1-4\sin^2(a)\right)=\cos(3a)\left(3\cos^2(a)\sin(a)-\sin^3(a)\right)=>$$
$$\cos(a)\left(1-4\sin^2(a)\right)\left(3(1-\sin^2(a))\sin(a)-\sin^3(a)\right)=\cos(3a)\left(3\cos^2(a)\sin(a)-\sin^3(a)\right)=>$$
$$\cos(a)\left(3\sin(a)-4\sin^3(a)\right)\left(1-4\sin^2(a)\right)=\cos(3a)\left(3\cos^2(a)\sin(a)-\sin^3(a)\right)=>$$
$$3\cos(a)\sin(a)-16\cos(a)\sin^3(a)+16\cos(a)\sin^5(a)=\cos(3a)\left(3\cos^2(a)\sin(a)-\sin^3(a)\right)=>$$
$$3\cos(a)\sin(a)-16\cos(a)\sin^3(a)+16\cos(a)\sin^5(a)=\cos(3a)\left(3(1-\sin^2(a))\sin(a)-\sin^3(a)\right)=>$$
$$3\cos(a)\sin(a)-16\cos(a)\sin^3(a)+16\cos(a)\sin^5(a)=\cos(3a)\left(3\sin(a)-3\sin^3(a)-\sin^3(a)\right)=>$$
$$3\cos(a)\sin(a)-16\cos(a)\sin^3(a)+16\cos(a)\sin^5(a)=\left(\cos^3(a)-3\cos(a)\sin^2(a)\right)\left(3\sin(a)-3\sin^3(a)-\sin^3(a)\right)=>$$
$$3\cos(a)\sin(a)-16\cos(a)\sin^3(a)+16\cos(a)\sin^5(a)=\left(3\sin(a)-4\sin^3(a)\right)\left(\cos(a)-4\cos(a)\sin^2(a)\right)=>$$
$$3\cos(a)\sin(a)-16\cos(a)\sin^3(a)+16\cos(a)\sin^5(a)=3\cos(a)\sin(a)-16\cos(a)\sin^3(a)+16\cos(a)\sin^5(a)$$
The left hand side and riht hand side are identical, identity has been verified
| {
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"timestamp": "2023-03-29T00:00:00",
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Applications of the identity $ab + \left(\frac{a+b}{2} - b\right)^2 = \left(\frac{a+b}{2}\right)^2$ I am reading Euclid's elements
I found the algebraic identity
$ab + \left(\frac{a+b}{2} - b\right)^2 = \left(\frac{a+b}{2}\right)^2$
I ponder on usage of this identity for $2$ hours.
but I can't click anything.
$a^2 + b^2 = c^2$ can be used when you want to know the direction between $2$ coordinates.
Any example involving this identity?
| You can use it to find all ways of writing $c=a^2-b^2$ with $a$ and $b$ positive natural numbers (or any two real numbers) for fixed $c$. To see this, your identity says:
$$ab=\left ( \frac{a+b}{2} \right )^{2} - \left ( \frac{a-b}{2} \right )^{2}$$
Now pick a number, $c=16$. Then we can factorize $c=2*8$
, so that $a=2$, $b=8$, giving $16=5^{2}-3^{2}$. This should be familiar as $3^{2}+4^{2}=5^{2}$.
We can find all solutions for fixed $c$. This trick relies on unique factorization and the fact that $(1,1)$ and $(1,-1)$ form a basis of $\mathbb{R}^{2}$. With a simple counting argument we give a formula for the number of solutions. If the prime factorization of $c$ is $2^{n_1}p_2^{n_2}...p_k^{n_k}$
$$N=\frac{1}{2}\left ( f(n_1)(n_2+1)(n_3+1)...(n_k+1)-g(c) \right )$$
Where $f(n_1)=n_1-1$ if $n_1>1$, and $1$ if $n_1=0$ and $g(c)=1$ if $c$ is a square, $0$ otherwise.
| {
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} |
$ 0< a,b,c <1\implies a+b+c-abc<2$ If $a,b,c$ are positive real numbers , all being less than $1$ , then how to prove that
$a+b+c-abc<2$ ?
| $0 \leq a,b,c \leq 1 \Rightarrow 0 \leq abc \leq 1$ and $0 \leq a+b+c \leq 3$.
It follows that $\sqrt[3]{abc} \geq abc$ and by the AM-GM inequality we know
$\dfrac{a+b+c}{3} \geq \sqrt[3]{abc} \geq abc$.
Finally
$a+b+c -abc \leq \dfrac{2}{3}(a+b+c) \leq 2$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Another inequality to prove involving arbitrary reals Let $a,b$ be non-zero reals such that $ab\ge \frac{1}{a}+\frac{1}{b}+3$ then prove the following inequality :
$$ \sqrt[3]{ab}\ge \frac{1}{\sqrt[3]{a}}+\frac{1}{\sqrt[3]{b}}$$
This one stumped me completely as usual. A solution would be welcome.
| I found another proof that establishes the equivalence between the two inequalities:
[As in my other solution, I will substitute all the variables with their cubes.]
$a^3 b^3 - \frac{1}{a^3}-\frac{1}{b^3} - 3 = ( a b - \frac{1}{a} - \frac{1}{b} ) ( a^2 b^2 + a + b + \frac{1}{a^2} + \frac{1}{b^2} - \frac{1}{ab} )$
$ = ( a b - \frac{1}{a} - \frac{1}{b} ) \left( \frac{1}{2}(ab+\frac{1}{a})^2 + \frac{1}{2}(ab+\frac{1}{b})^2+\frac{1}{2}(\frac{1}{a}-\frac{1}{b})^2 \right)$
Also $\frac{1}{2}(ab+\frac{1}{a})^2 + \frac{1}{2}(ab+\frac{1}{b})^2+\frac{1}{2}(\frac{1}{a}-\frac{1}{b})^2 \ge 0$ with equality case only when $a = b = -1$
And $a b - \frac{1}{a} - \frac{1}{b} = 3 > 0$ when $a = b = -1$
Therefore $a^3 b^3 - \frac{1}{a^3}-\frac{1}{b^3} - 3 \ge 0$ iff $a b - \frac{1}{a} - \frac{1}{b} \ge 0$
| {
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"source": "stackexchange",
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Evaluate a limit (probably involving L'Hôpital rule) Evaluate the limit:
$$\mathop {\lim }\limits_{x \to \infty } x\left( {{{\left( {1 + {1 \over x}} \right)}^x} - e} \right)$$
My attempts didn't yield a result. I'd be glad for a guidance.
Thanks!
| You just have to persist in the L'Hopital rule.
Start with
$$\frac{(1+\frac{1}{x})^x-e}{\frac{1}{x}}$$ top and bottom go to zero,
by L'Hopital we get
$$\frac{(\ln(1+\frac{1}{x})-\frac{1}{x+1})(1+\frac{1}{x})^x}{-\frac{1}{x^2}}
=(\frac{x^2}{x+1}-x^2\ln(1+\frac{1}{x}))(1+\frac{1}{x})^x$$
Now $(1+\frac{1}{x})^x\to e$ so lets set that term aside.
What remains can be written
$$\frac{\frac{1}{x+1}-\ln(1+\frac{1}{x})}{\frac{1}{x^2}}$$
top and bottom go to zero so again with L'Hopital,
$$\frac{-\frac{1}{(x+1)^2}+\frac{x}{x+1}\frac{1}{x^2}}{-\frac{2}{x^3}}=-\frac{1}{2}(\frac{x^2}{x+1}-\frac{x^3}{(x+1)^2})=-\frac{1}{2}\frac{x^2}{(x+1)^2}\rightarrow -\frac{1}{2}$$
So we get the limit $-\frac{e}{2}$
| {
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"source": "stackexchange",
"question_score": "8",
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Trying to evaluate integral$\int_0^\infty x \sqrt{1-e^{-x}}\,e^{-x}dx$ I am trying to integrating
$$
\int_0^\infty x \sqrt{1-e^{-x}}\,e^{-x}dx\equiv I
$$
but cannot get the answer, I would like a proof not a numerical answer. My attempt at proof:
$$
y=\sqrt{1-e^{-x}}\\
y(0)=0, \ y(\infty)=1\\
y^2=1-e^{-x}\\
2ydy=e^{-x}dx\\
e^{-x}=1-y^2\\
x=\ln\frac{1}{1-y^2} \rightarrow\\
I=2\int_0^1 y^2\ln\frac{1}{1-y^2} dy=\\
-2\int_0^1y^2\ln(1-y^2)\,dy=\\2\int_0^1y^2\sum_{k=1}^\infty \frac{y^{2k}}{k}=\\
2\sum_{k=1}^\infty \frac{1}{k}\int_0^1 y^{2(k+1)}dy=\\
2\sum_{k=1}^\infty \frac{1}{k(3+2k)}=\\
2\sum_{k=1}^\infty \left(\frac{1}{3k}-\frac{2}{3(2k+3)} \right)
$$
but this diverges because $\sum_k\frac{1}{k}\to\infty$? Mistakes I made...
Please help if can on doing the sum or integral. Thank you, Grazie
| You can continue with your method too. Use partial fraction decomposition this way:
$$\frac{4}{3}\sum_{k=1}^{\infty} \left(\frac{1}{2k}-\frac{1}{2k+3}\right)=\frac{4}{3}\int_0^1 \sum_{k=1}^{\infty} \left(x^{2k-1}-x^{2k+2}\right)\,dx=\frac{4}{3}\int_0^1 \,dx\left(\frac{1}{x}-x^2\right)\sum_{k=1}^{\infty} x^{2k}$$
where I used the following two:
$$\int_0^1 x^{2k-1}\,dx=\frac{1}{2k}$$
$$\int_0^1 x^{2k+2}\,dx=\frac{1}{2k+3}$$
Since $\displaystyle \sum_{k=1}^{\infty} x^{2k}=\frac{x^2}{1-x^2}$, you get:
$$\frac{4}{3}\int_0^1 \frac{1-x^3}{x}\frac{x^2}{1-x^2}\,dx=\frac{4}{3}\int_0^1 \frac{(x^2+x+1)x}{1+x}\,dx=\frac{4}{3}\int_0^1\left(x+\frac{x^3}{1+x}\right)\,dx$$
$$=\frac{4}{3}\int_0^1 \left(x+\frac{x^3+1}{1+x}-\frac{1}{1+x}\right)\,dx=\frac{4}{3}\int_0^1 \left( x^2+1-\frac{1}{x+1}\right)\,dx=\frac{4}{3}\left(\frac{4}{3}-\ln 2\right)$$
$\blacksquare$
| {
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Find the limit of $ \lim_{x \to 7} \frac{\sqrt{x+2}-\sqrt[3]{x+20}}{\sqrt[4]{x+9}-2} $ I need to evaluate the limit without using L'Hopital's rule.
$$\Large
\lim_{x \to 7} \frac{\sqrt{x+2}-\sqrt[3]{x+20}}{\sqrt[4]{x+9}-2}
$$
| Letting $x = 7 + h$, then we find: $$\displaystyle \lim_{h \to 0} \dfrac{\sqrt{h+9} - \sqrt[3]{h+27}}{\sqrt[4]{h+16} - 2} = \displaystyle \lim_{h \to 0} \dfrac{3\left(1+\dfrac{h}{18}\right) - 3\left(1+\dfrac{h}{81}\right)}{2\left(1+\dfrac{h}{64}\right) - 2} = \dfrac{112}{27},$$
where $(1+h)^{\alpha} \sim_0 1 + \alpha h$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Series Question: $\sum_{n=0}^\infty\frac{2n!}{(2n+1)!!}=\pi$ How to prove:
$$\sum_{n=0}^\infty\frac{2n!}{(2n+1)!!}=\pi$$
I tried to use property of double factorial
$$(2n+1)!!=\frac{(2n+1)!}{2^nn!}$$
then
$$\frac{2n!}{(2n+1)!!}=\frac{2^{n+1}(n!)^2}{(2n+1)!}$$
The next step I am stuck. Any help would be appreciated. Thanks in advance.
| Consider the series
\begin{align}
S = \sum_{n=0}^{\infty} \frac{ 2 \ n!}{(2n+1)!!} = 2 \sum_{n=0}^{\infty} \frac{ 2^{n} (n!)^{2} }{(2n+1)!}
\end{align}
which can quickly be cast into the form
\begin{align}
S = 2 \sum_{n=0}^{\infty} \frac{ (1)_{n} (1)_{n} }{ n! (3/2)_{n} \ 2^{n} }
\end{align}
which yields the hypergeometric form
\begin{align}
S = 2 {}_{2}F_{1} ( 1, 1; 3/2; 1/2).
\end{align}
Using the formulas
\begin{align}
{}_{2}F_{1}(1,1; c; 1/2) &= (c-1) \left[ \psi\left(\frac{c}{2}\right) - \psi\left( \frac{c-1}{2}\right) \right] \\
\psi\left(\frac{3}{4} \right) &= \frac{\pi}{2} - 3 \ln 2 - \gamma \\
\psi\left(\frac{1}{4} \right) &= - \frac{\pi}{2} - 3 \ln 2 - \gamma
\end{align}
then the series becomes
\begin{align}
S &= \psi\left(\frac{3}{4}\right) - \psi\left( \frac{1}{4}\right) = \pi.
\end{align}
Hence
\begin{align}
\sum_{n=0}^{\infty} \frac{ n!}{(2n+1)!!} = \frac{\pi}{2}.
\end{align}
| {
"language": "en",
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"source": "stackexchange",
"question_score": "6",
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Limit $\lim_{x\rightarrow\infty}1-x+\sqrt{2+2x+x^2}$ $$\lim_{x\rightarrow\infty}1-x+\sqrt{2+2x+x^2}=\lim_{x\rightarrow\infty}1-x+x\sqrt{\frac{2}{x^2}+\frac 2x+1}=\lim_{x\rightarrow\infty}1=1\neq2$$ as Wolfram Alpha state. Where I miss something?
| You have
\begin{align}
\lim_{x\rightarrow\infty}\left(1-x+\sqrt{2+2x+x^2}\right)
&=\lim_{x\rightarrow\infty}\left(1-x+x\sqrt{\frac{2}{x^2}+\frac 2x+1}\right) \\
&= \lim_{x\to\infty} \left(1+x\left(\sqrt{\frac{2}{x^2}+\frac{2}{x}+1}-1\right)\right).
\end{align}
The second term is of the form $\infty\cdot 0$; you cannot conclude that it goes to zero (and, in fact, it goes to $1$) as $x\to\infty$.
| {
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"source": "stackexchange",
"question_score": "2",
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"answer_id": 3
} |
Evaluating e using limits What algebraic operations can I use on the $RHS$ to show $RHS = LHS$
$$e=\lim_{k\to\infty}\left(\frac{2+\sqrt{3+9k^2}}{3k-1}\right)^k$$
| Hint
For large values of $k$, the following Taylor expansion can be built $$\frac{2+\sqrt{3+9k^2}}{3k-1}=1+\frac{1}{k}+\frac{1}{2 k^2}+\frac{1}{6 k^3}+\frac{1}{24 k^4}+\frac{1}{72
k^5}+O\left(\left(\frac{1}{k}\right)^6\right)$$ So $$\log\Big(\frac{2+\sqrt{3+9k^2}}{3k-1}\Big)=\frac{1}{k}+\frac{1}{180 k^5}+O\left(\left(\frac{1}{k}\right)^6\right)$$
| {
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"source": "stackexchange",
"question_score": "3",
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Proving by induction: $ \frac{1\cdot3\cdot5\cdot \ldots \cdot (2n-1)}{1\cdot2\cdot3\cdot\ldots\cdot n} \leq 2^n $ WTS $ \frac{1\cdot3\cdot5\cdot \ldots \cdot (2n-1)}{1\cdot2\cdot3\cdot\ldots\cdot n} \leq 2^n $ for all natural $n$.
Have checked $P_1$, and assumed $P_k$.
Trying the following argument:
$P_{k+1} = \frac{1\cdot3\cdot5\cdot \ldots \cdot (2k-1)\cdot(2k+1)}{1\cdot2\cdot3\cdot\ldots\cdot k \cdot (k+1)} = \frac{1\cdot3\cdot5\cdot \ldots \cdot (2k-1)}{1\cdot2\cdot3\cdot\ldots\cdot k}\cdot \frac{2k+1}{k+1} \leq 2^k \cdot\frac{2k+1}{k+1} \leq 2^{k+1}\cdot \frac{2k+1}{k+1}$ using $P_k$ and that these are all positive numbers.
Now dividing through by $\frac{2k+1}{k+1}$ would give the desired result.
I just feel like I've done something wrong here. Can anyone tell me if I've overlooked anything?
| You've basically got the right idea, except your last inequality is too weak (and not at all needed - you've basically just added an extra factor of $2$ without need). You've shown that
$$P_{k + 1} = P_k \cdot \frac{2k + 1}{k + 1}$$
Now we have that $P_k \le 2^k$ by hypothesis, and it's easy to see that
$$\frac{2k + 1}{k + 1} \le 2$$
for every $k \ge 0$. This implies $P_{k + 1} \le 2^k \cdot 2$, which is what you want.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Integration by parts - $\int \ln (2x+1) \text{dx}$ Use integration by parts to find
$\int \ln (2x+1) \text{dx}$.
So far I have:
$$x\ln(2x+1)-\int\dfrac{2x}{2x+1}dx+c$$
Using integration by substitution to find the integral
$$u=2x+1\Rightarrow\text{du}=2\text{dx}$$
$$\int\dfrac{2x}{2x+1}\cdot\dfrac{1}{2}\text{du}=\int xu^{-1}$$
$$=\int \left(\dfrac{u}{2}-\dfrac{1}{2}\right)u^{-1}\text{du}=\int\left[\dfrac{1}{2}-\dfrac{1}{2}u^{-1}\right]\text{du}$$
$$=\dfrac{1}{2}x-\dfrac{1}{2} \ln \left|2x+1\right|$$
Looking at the answer in the back, this is wrong.
The answer is $x \ln(2x+1)-x+\dfrac{1}{2}\ln(2x+1)+c$.
What have I done wrong?
| Using these steps
$u=2x+1$, $(1/2)du=dx$ and $2x=u-1$ so:
$$
\eqalign{
-\int\dfrac{u-1}u \frac12 du &= -\int\dfrac12du+\int\dfrac12u\,du \\
&= -\dfrac12(2x+1)+(1/2)\ln(2x+1).
}$$
| {
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If $\alpha = \frac{2\pi}{7}$ then the find the value of $\tan\alpha .\tan2\alpha +\tan2\alpha \tan4\alpha +\tan4\alpha \tan\alpha.$ If $\alpha = \frac{2\pi}{7}$ then the find the value of $\tan\alpha .\tan2\alpha +\tan2\alpha \tan4\alpha +\tan4\alpha \tan\alpha$
My 1st approach :
$\tan(\alpha +2\alpha +4\alpha) = \frac{\tan\alpha +\tan2\alpha +\tan4\alpha -\tan\alpha \tan2\alpha -\tan2\alpha \tan4\alpha -\tan4\alpha \tan\alpha}{1-(\tan\alpha \tan2\alpha +\tan2\alpha \tan4\alpha +\tan\alpha \tan4\alpha)} $
$\Rightarrow 0 = \frac{\tan\alpha +\tan2\alpha +\tan4\alpha -\tan\alpha \tan2\alpha -\tan2\alpha \tan4\alpha -\tan4\alpha \tan\alpha}{1-(\tan\alpha \tan2\alpha +\tan2\alpha \tan4\alpha +\tan\alpha \tan4\alpha)} $ which doesn't give me any solution.
My IInd approach :
U\sing Euler substitution :
\since $\cos\theta +i\sin\theta = e^{i\theta} $.....(i) and $\cos\theta -i\sin\theta =e^{-i\sin\theta}$....(ii)
Adding (i) and (ii) we get $\cos\theta =\frac{e^{i\theta} +e^{-i\theta}}{2}$ and subtracting (i) and (ii) we get $\sin\theta =\frac{e^{i\theta} -e^{-i\theta}}{2}$
By u\sing this we can write : $$\tan\alpha .\tan2\alpha +\tan2\alpha \tan4\alpha +\tan4\alpha \tan\alpha$$ as $$\frac{1}{4}\left[ (e^{\frac{i2\pi}{7}} -e^{\frac{-i2\pi}{7}}) (e^{\frac{i4\pi}{7}} -e^{\frac{-i4\pi}{7}}) + (e^{\frac{i4\pi}{7}} -e^{\frac{-i4\pi}{7}})(e^{\frac{i8\pi}{7}} -e^{\frac{-i8\pi}{7}}) + (e^{\frac{i8\pi}{7}} -e^{\frac{-i8\pi}{7}}) (e^{\frac{i\pi}{7}} -e^{\frac{-i\pi}{7}})\right]$$
$$\large= e^{i\frac{6\pi}{7}}-e^{\frac{i2\pi}{7}}-e^{\frac{-i2\pi}{7}} +e^{\frac{-i6\pi}{7}} +e^{\frac{i3\pi}{7}}-e^{\frac{-i5\pi}{7}}-e^{\frac{i5\pi}{7}} +e^{\frac{-3\pi}{7}} +e^0 -e^{\frac{i2\pi}{7}} -e^{\frac{-i2\pi}{7}}+e^0$$
Can anybody please suggest whether this is my correct approach or not. please guide further... Thanks.
| Not using Euler's formula which I don't think the best way for this
Let $\displaystyle a=\tan A,b=\tan2A,c=\tan4A$ where $A+2A+4A=n\pi$ where $7\nmid n$
As $\displaystyle\tan(n\pi-rA)=-\tan rA,\tan6A=-\tan A=-a$ etc.
Using Prove that $\tan A + \tan B + \tan C = \tan A\tan B\tan C,$ $A+B+C = 180^\circ$,
$\displaystyle a+b+c=abc$
Now using Sum of tangent functions where arguments are in specific arithmetic series,
$\displaystyle\tan7x=\frac{\binom71\tan x-\binom73\tan^3x+\binom75\tan^5x-\tan^7x}{1-\binom72\tan^2x+\binom74\tan^4x-\binom76\tan^6x}$
If $\displaystyle\tan7A=0,7A=m\pi$ where $m$ is any integer
$\displaystyle\implies A=\frac{m\pi}7$ where $0\le m\le6$
So, $\pm a,\pm b,\pm c,\tan0=0$ are the roots of
$\displaystyle \binom71\tan x-\binom73\tan^3x+\binom75\tan^5x-\tan^7x=0$
$\displaystyle\iff\tan^7x-21\tan^5x+35\tan^3x-7\tan x=0$
So, $\pm a,\pm b,\pm c$ are the roots of
$\displaystyle \tan^6x-21\tan^4x+35\tan^2x-7=0\ \ \ \ (1)$
Now the equation whose roots are $\pm a,\pm b,\pm c$ is
$\displaystyle(y-a)(y-b)(y-c)(y+a)(y+b)(y+c)=0$
$\displaystyle\iff(y^2-a^2)(y^2-b^2)(y^2-c^2)=0$
$\displaystyle\iff y^6-(a^2+b^2+c^2)y^4+(a^2b^2+b^2c^2+c^2a^2)y^2-a^2b^2c^2=0\ \ \ \ (2)$
If we write $\displaystyle a+b+c=abc=S$ and $\displaystyle ab+bc+ca=T,$
$\displaystyle a^2+b^2+c^2=(a+b+c)^2-2(ab+bc+ca)=S^2-2T$
and $\displaystyle a^2b^2+b^2c^2+c^2a^2=(ab+bc+ca)^2-2abc(a+b+c)=T^2-2S^2$
So, $(2)$ becomes $\displaystyle y^6-(S^2-2T)y^4+(T^2-2S^2)y^2-S^2=0$
Comparing with $\displaystyle(1), S^2=7,S^2-2T=21, T^2-2S^2=35$
Can you find the required $T$ from here?
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 0
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Expansion of $\sin^5 \theta$ using the Complex Exponential How do I expand $\sin^5\theta$ using the complex exponential, in order to obtain:
$$\frac{1}{16}\sin 5\theta - \frac{5}{16}\sin 3\theta + \frac{5}{8}\sin\theta$$
Thank you.
| Using this formula.
$$\exp({i\theta})^n = \exp(ni
\theta)$$
$$(\cos(\theta) + i\sin(\theta))^n = \cos(n\theta) + i\sin(n\theta)$$
$$(\cos(\theta) + i\sin(\theta))^5 = \cos(5\theta) + i\sin(5\theta)$$
$$\cos^5(\theta) + 5i\cos^4(\theta)\sin(\theta) -10\cos^3(\theta)\sin^2(\theta) -10i \cos^2(\theta)\sin^3(\theta) + \cdots $$ $$5\cos(\theta)\sin^4(\theta) +i\sin^5(\theta) = \cos(5\theta) + i\sin(5\theta)$$
Equate imaginary parts:
$$5\cos^4(\theta)\sin(\theta)-10 \cos^2(\theta)\sin^3(\theta) + \sin^5(\theta) = \sin(5\theta)$$
Solve for $\sin^5(\theta)$:
$$\sin^5(\theta) = \sin(5\theta)-5\cos^4(\theta)\sin(\theta)+10 \cos^2(\theta)\sin^3(\theta)$$
Use trigonometric identities:
$$\sin^5(\theta) = \dfrac{1}{16} (\sin(5 \theta)-5 \sin(3 \theta) + 10 \sin(\theta))$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/825411",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Calculate the product of these $98$ rational numbers $$\left( 1 + \frac{1}{2} \right) \left( 1 + \frac{1}{3} \right) \left( 1 + \frac{1}{4} \right) ... \left( 1 + \frac{1}{99} \right) $$
Of course I could do it factor for factor but there has got to be a more efficient way. Does anyone know it?
| We have
$$\left( 1 + \frac{1}{2} \right) \left( 1 + \frac{1}{3} \right) \left( 1 + \frac{1}{4} \right) ... \left( 1 + \frac{1}{99} \right) =\frac32\times\frac43\cdots\frac{100}{99}=\frac{100}2=50$$
by telescoping.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 4,
"answer_id": 0
} |
Prove the sum $\sum_{n=1}^\infty \frac{\arctan{n}}{n}$ diverges. I must prove, that sum diverges, but...
$$\sum_{n=1}^\infty \frac{\arctan{n}}{n}$$
$$\lim_{n \to \infty} \frac{\arctan{n}}{n} = \frac{\pi/2}{\infty} = 0$$
$$\lim_{n \to \infty} \frac{ \sqrt[n]{\arctan{n}} }{ \sqrt[n]{n} } = \frac{1}{1} = 1$$
Cauchy's convergence test undefined.
There is a $E_0\gt0$:
$$\left|\frac{\arctan{n+1}}{n+1} + \frac{\arctan{n+2}}{n+2} + ... + \frac{\arctan{n+p}}{n+p}\right| \ge \frac{\pi}{4}\left|\frac{1}{n+1} + \frac{1}{n+2} + ... + \frac{1}{n+p}\right| \ge \frac{\pi}{4} \frac{p}{n+p} (Let\, p = n) \ge \frac{\pi}{4} \frac{\bcancel{n}}{2\bcancel{n}} = \frac{\pi}{8} (\sim0.4) \ge E_0 = \frac{1}{8} \gt 0;$$
Now am I correct?
| We have
$$\frac{\arctan n }n\sim_\infty \frac{\pi}{2n}$$
and the harmonic series $\sum\limits\frac1n$ is divergent so the given series is also divergent by the asymptotic comparison.
| {
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"source": "stackexchange",
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Show that $\sum_{r-k=0}^m\left(\sum_{k=0}^n\binom nk\binom{m}{r-k}x^r\right)=\sum_{r=0}^{m+n}\left(\sum_{k=0}^r\binom nk\binom{m}{r-k}\right)x^r$
How does this hold?
$$\sum_{r-k=0}^m\left(\sum_{k=0}^n\binom nk\binom{m}{r-k}x^r\right)=\sum_{r=0}^{m+n}\left(\sum_{k=0}^r\binom nk\binom{m}{r-k}\right)x^r$$
| Consider the relation to be proved, namely,
\begin{align}
\sum_{r=0}^{m+n} \left( \sum_{k=0}^{r} \binom{n}{k} \binom{m}{r-k} \right) \ x^{r} = \sum_{k=0}^{n} \left( \sum_{r=k}^{m} \binom{n}{k} \binom{m}{r-k} \right) \ x^{r}.
\end{align}
Now consider just the left hand side, labeled $S_{L}$,
\begin{align}
S_{L} &= \sum_{r=0}^{m+n} \left( \sum_{k=0}^{r} \binom{n}{k} \binom{m}{r-k} \right) \ x^{r} \\
&= \sum_{r=0}^{m+n} \sum_{k=0}^{m+n} \binom{n}{k} \binom{m}{r} \ x^{r+k} \\
&= \left(\sum_{r=0}^{m+n} \binom{m}{r} \ x^{r} \right) \left(\sum_{r=0}^{m+n} \binom{m}{r} \ x^{k} \right) \\
&= (1+x)^{m} \ (1+x)^{n} = (1+x)^{m+n}.
\end{align}
Now consider the right-hand side, labeled $S_{R}$,
\begin{align}
S_{R} &= \sum_{k=0}^{n} \left( \sum_{r=k}^{m} \binom{n}{k} \binom{m}{r-k} \right) \ x^{r} \\
&= \sum_{k=0}^{n} \sum_{r=0}^{m-k} \binom{n}{k} \binom{m}{r} \ x^{r+k} \\
&= \sum_{k=0}^{n} \binom{n}{k} \ x^{k} \left( \sum_{r=0}^{m-k} \binom{m}{r} x^{r} + \sum_{r=m-k+1}^{m+n} \binom{m}{r} x^{r} - \sum_{r=m-k+1}^{m+n} \binom{m}{r} x^{r} \right) \\
&= \left( \sum_{k=0}^{n} \binom{n}{k} \ x^{k} \right) \left( \sum_{r=0}^{m} \binom{m}{r} \ x^{r} \right) \\
&= (1+x)^{m+n}.
\end{align}
Since $S_{L} = S_{R}$ then the identity is shown to be true.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$a,b,c>0$ , $a^a b^b c^c =1$ then $a+b+c \le 3$? Let $a,b,c$ be positive real numbers such that $a^a b^b c^c =1$ , then is it true that $a+b+c \le 3$ ?
| Let $f(x)=x \log x$ , for $x>0$. Clearly $f''(x)=\frac{1}{x}>0~$, so by Jensen's inequality we have
$$\frac{a+b+c}{3}\log\left(\frac{a+b+c}{3}\right)\leq\frac{1}{3}(a\log a+b\log b+c\log c)=
\frac{\log(a^ab^bc^c)}{3}\leq 0$$
it follows that $\log\left(\frac{a+b+c}{3}\right)\leq0$ that is
$a+b+c\leq 3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/831357",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Partial fraction (doubt) I have this partial fraction
$$\displaystyle\frac{1}{(2+x)^2(4+x)^2}$$
I tried to resolve using this method:
$$\displaystyle\frac{A}{2+x}+\displaystyle\frac{B}{(2+x)^2}+\displaystyle\frac{C}{4+x}+\displaystyle\frac{D}{(4+x)^2}$$
$$1=A(2+x)(4+x)^2+B(4+x)^2+C(4+x)(2+x)^2+D(2+x)^2$$
When x=-2
$$1=B(4-2)^2$$
$$B=\displaystyle\frac{1}{4}$$
When x=-4
$$1=D(2-4)^2$$
$$D=\displaystyle\frac{1}{4}$$
When x=0
$$1=A(2)(16)+B(16)+C(4)(4)+D(4)$$
$$1=A(32)+B(16)+C(16)+D(4)$$
Replacing the values for B y D
$$1=A(32)+4+C(16)+1$$
$$1-4-1=A(32)+c(16)$$
$$-4=A(32)+C(16)$$
How I can get the values of $A$ and $D$?
| Heaviside cover-up is an alternative technique worth exploring.
It directly gives $\begin{align}B = \frac{1}{(4+x)^2}\Bigg|_{x=-2} = \frac{1}{4} \text{ and } D = \frac{1}{(2+x)^2}\Bigg|_{x=-4} = \frac{1}{4}\end{align}\tag{1}$
Next, it requires that the $B$ and $D$ terms be moved to the other side.
The partial fraction identity then becomes
$\begin{eqnarray}\frac{A}{2+x} + \frac{C}{4+x} &= &\frac{1}{(2+x)^2(4+x)^2} - \frac{1/4}{(2+x)^2} - \frac{1/4}{(4+x)^2}\\&=&\frac{4-(4+x)^2-(2+x)^2}{4(2+x)^2(4+x)^2}\\&=&\frac{-2(8 + 6x + x^2)}{4(2+x)^2(4+x)^2}\\&=&\frac{-1/2}{(2+x)(4+x)}\end{eqnarray}$
which gives $\begin{align}A = \frac{-1/2}{(4+x)}\Bigg|_{x=-2} = -\frac{1}{4} \text{ and } C = \frac{-1/2}{(2+x)}\Bigg|_{x=-4} = \frac{1}{4}\end{align}\tag{2}$
Remark: Note how this technique eliminates the need to solve a system of linear equations for finding out the partial fraction decomposition coefficients (which can often be tedious).
See here for some more examples.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Showing $(a^{2}+2)(b^{2}+2)(c^{2}+2)\geq 9(ab+bc+ca)$
Let $a$, $b$, $c$ be nonnegative real numbers.
Prove $(a^{2}+2)(b^{2}+2)(c^{2}+2)\geq 9(ab+bc+ca)$.
| I will prove the stronger inequality:
$$(a^2+2)(b^2+2)(c^2+2)\ge 3(a+b+c)^2$$
because
$$(a^2+2)(b^2+2)=(a^2+1)(b^2+1)+a^2+b^2+3\ge (a+b)^2+\dfrac{1}{2}(a+b)^2+3
=\dfrac{3}{2}[(a+b)^2+2]$$
so
$$(a^2+2)(b^2+2)(c^2+2)\ge \dfrac{3}{2}[(a+b)^2+2](c^2+2)\ge\dfrac{3}{2}
[\sqrt{2}(a+b)+\sqrt{2}c]^2=3(a+b+c)^2$$
so
$$(a^2+2)(b^2+2)(c^2+2)\ge 3(a+b+c)^2\ge 9(ab+bc+ac)$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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Prove that $3^n > 2n^2 + 3n$ for $n \in [4,\infty) \cap\mathbb{N}$ If $n$ is a natural number $n\ge 4$, prove that $3^n > 2n^2 + 3n$
I assume I am supposed to use induction. The base $n=4$ step is clear, but how do I prove the inductive step. I tried several things including comparing $f(n+1)/f(n)$ and $f(n+1)-f(n)$ but they didn't seem to help. A hint would be great!
| Write
$$ f(n) = 3^n - 2 n^2 - 3 n$$
Then
$$ f(n+1) = 3^{n+1} - 2 \big(n+1\big)^2 - 3 \big(n+1) $$
which can be written as
$$ f(n+1) = 3 \big( 3^n - 2 n^2 - 3 n \big) + 6 n^2 + 9 n - 2 \big(n+1\big)^2 - 3\big(n+1)$$
So
$$ f(n+1) = 3 f(n) + 4 n^2 + 2 n - 5$$
Note that
$$ f(3) = 0 $$
And note that
$$ 4 n^2 + 2 n - 5 > 0 $$
for $n \ge 1$, whence
$$ f(n) > 0 $$
for $n \ge 4$,
therefore
$$ 3^n - 2 n^2 - 3 n > 0$$
for $n \ge 4$, or
$$ 3^n > 2 n^2 + 3 n$$
for $n \ge 4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/834107",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Calculate $\int_{0}^{1}\frac{\arctan(x)}{x\sqrt{1-x^2}}dx$ I am preparing for a calculus exam and I was asked to calculate $$\int_{0}^{1}\frac{\arctan(x)}{x\sqrt{1-x^2}}dx$$
Using the hint that $$\frac{\arctan(x)}{x}=\int_{0}^{1}\frac{dy}{1+x^2y^2}$$
I ran into some trouble and would appreciate help.
What I did:
I used the hint, $$\int_{0}^{1}\frac{\arctan(x)}{x\sqrt{1-x^2}}dx=\int_{0}^{1}\int_{0}^{1}\frac{1}{(1+x^2y^2)\sqrt{1-x^2}}dydx$$
Since $x,y$ move between $0,1$ I thought maybe it is best to used the transform $x=r\cos\theta$ , $y=r\sin\theta$. $r \in [0,1]$, $\theta \in [0, \frac{\pi}{2}]$
But I didn't get anything meaningful, I didn't end up an something that is easy / possible to integrate. And I honestly can't think of a way to integrate this as it is.
| The following is a different approach that doesn't use that hint.
$$ \begin{align} \int_{0}^{1} \frac{\arctan (x)}{x\sqrt{1-x^{2}}} \ dx &= \int_{0}^{1} \frac{1}{x \sqrt{1-x^{2}}} \sum_{n=0}^{\infty} \frac{(-1)^{n}}{2n+1} x^{2n+1} \ dx \tag{1}\\ &= \sum_{n=0}^{\infty} \frac{(-1)^{n}}{2n+1} \int_{0}^{1} \frac{x^{2n}}{\sqrt{1-x^{2}}} \ dx \\ &= \frac{1}{2} \sum_{n=0}^{\infty} \frac{(-1)^{n}}{2n+1} \int_{0}^{1} u^{(n+1/2)-1} (1-u)^{1/2-1} \ du \tag{2} \\ &= \frac{1}{2} \sum_{n=0}^{\infty} \frac{(-1)^{n}}{2n+1} B\left(n+\frac{1}{2}, \frac{1}{2} \right) \tag{3}\\ &=\frac{1}{2} \sum_{n=0}^{\infty} \frac{(-1)^{n}}{2n+1} \frac{\Gamma (n+\frac{1}{2}) \Gamma(\frac{1}{2})}{\Gamma (n+1)} \tag{4} \\ &= \frac{1}{2}\sum_{n=0}^{\infty} \frac{(-1)^{n}}{2n+1} \frac{\Gamma (\frac{1}{2})}{\Gamma (n+1)} \frac{\Gamma(2n) \Gamma (\frac{1}{2})}{\Gamma (n) 2^{2n-1}} \frac{2n}{2n} \tag{5} \\ &= \frac{\pi}{2} \sum_{n=0}^{\infty} \frac{(-1)^{n} \Gamma(2n+1)}{2^{2n} \ \Gamma^{2}(n+1)} \frac{1}{2n+1} \tag{6} \\ &= \frac{\pi}{2} \text{arcsinh}(1) \tag{7}\end{align}$$
$ $
(1) Maclaurin expansion of arctan(x)
(2) let $u = x^{2}$
(3) integral representation of the beta function
(4) defintion of the beta function in terms of the gamma function
(5) gamma function duplication formula
(6) $\Gamma(z+1) = z \Gamma(z)$ and $\Gamma(\frac{1}{2}) = \sqrt{\pi}$
(7) Maclaurin expansion of arcsinh(x)
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 4,
"answer_id": 3
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Let $a_1=1$, $a_2=3$ , and for $n \ge 2$ let $a_n=a_{n-1}+a_{n-2}$. Show that $a_n < \left(\frac{7}{4}\right)^n$ for all natural numbers. Let $a_1=1$, $a_2=3$ , and for $n \ge 2$ let $a_n=a_{n-1}+a_{n-2}$. Show that $a_n <
\left(\frac{7}{4}\right)^n$ for all natural numbers.
I assume I'm supposed to use induction. base step is easy. I'm stuck on how to form the inductive step. Any tips are greatly appreciated.
| Here is the inductive step for anyone reading this...
Assuming $P(k-2)$: $a_{k-2} < \left(\frac{7}{4}\right)^{k-2}$
Assuming $P(k-1)$: $a_{k-1} < \left(\frac{7}{4}\right)^{k-1}$
Definition of $a_k$: $a_k=a_{k-1}+a_{k-2}$
Combining with inductive assumptions: $a_k < \left(\frac{7}{4}\right)^{k-2} + \left(\frac{7}{4}\right)^{k-1}$
Algebraically factor out $\left(\frac{7}{4}\right)^{k-2}$: $a_k < \left(\frac{7}{4}\right)^{k-2} \cdot \left(1 + \frac{7}{4}\right)$
$a_k < \left(\frac{7}{4}\right)^{k-2} \cdot \frac{11}{4}
= \left(\frac{7}{4}\right)^k \cdot \left(\frac{4}{7}\right)^2 \cdot \frac{11}{4}
= \left(\frac{7}{4}\right)^k \cdot \frac{44}{49}
< \left(\frac{7}{4}\right)^k$
$a_k < \left(\frac{7}{4}\right)^k$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/835321",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Easier Proof of $\sin{3\theta} + \sin\theta = 2\sin{2\theta}\cos\theta$ I am curious to see whether anybody can give me a proof that takes less steps.
Here is how I did it:
$$\sin{3\theta} + \sin\theta = 2\sin{2\theta}\cos\theta$$
LHS $$\eqalign{\sin(2\theta + \theta) + \sin\theta &= \sin2\theta\cos\theta + \cos2\theta\sin\theta + \sin\theta\\
&= \sin2\theta\cos\theta + (\cos^2\theta - \sin^2\theta)\sin\theta + \sin\theta\\
&= \sin2\theta\cos\theta + \sin\theta(\cos^2\theta - \sin^2\theta + 1)\\
&= \sin2\theta\cos\theta + \sin\theta(2\cos^2\theta)\\
&= \sin2\theta\cos\theta + 2\sin\theta\cos^2\theta\\
&= \sin2\theta\cos\theta + \cos\theta(\sin\theta\cos\theta + \sin\theta\cos\theta)\\
&= \sin2\theta\cos\theta + \cos\theta(\sin2\theta)\\
&= \sin2\theta\cos\theta + \cos\theta(\sin2\theta)\\
&= 2\sin2\theta\cos\theta.}$$
| $$\sin(3\theta)+\sin(\theta)=\Im[e^{3i\theta}]+\Im[e^{i\theta}]=\Im[e^{3i\theta}+e^{i\theta}]=$$
$$=\Im[e^{2i\theta}(e^{i\theta}+e^{-i\theta})]=2\cos(\theta)\Im[e^{2i\theta}]=2\cos(\theta)\sin(2\theta)\ .$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/837430",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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$\iint_{\mathbb R^2} \frac{dx \, dy}{1+x^{10}y^{10}}$ diverges or converges? Question I'm trying to solve to prepare for an exam.
I need to find out if $\displaystyle\iint_{\mathbb R^2} \frac{dx\,dy}{1+x^{10}y^{10}}$ diverges or converges.
What I did:
I switched to polar coordinates, $x=r\cos \theta$, $y=r\sin\theta$, $J=r$
and so
$$
\begin{align}
\iint_{\mathbb R^2} \frac{dx\,dy}{1+x^{10}y^{10}} & =\int_0^{2\pi} \int_0^\infty \frac{r}{1+r^{20} \sin \theta \cos \theta} \, dr \, d\theta \\[8pt]
& < \int_0^{2\pi} \int_0^\infty \frac{1}{r^{19}\sin \theta \cos \theta} \, dr \, d\theta \\[8pt]
& = \frac{1}{2} \int_0^{2\pi} \int_0^\infty \frac{1}{r^{19}\sin 2\theta} \, dr \, d\theta \\[8pt]
& =\frac{1}{2} \int_0^{2\pi} \frac{1}{\sin 2\theta} \, d\theta \int_0^\infty \frac{1}{r^{19}}\,dr
\end{align}
$$
But note that both $\int_0^\infty \frac{1}{r^{19}}dr$ and $\int_0^{2\pi} \frac{1}{\sin 2\theta} \, d\theta$ do not converge, so we proved nothing.
So that is not the way.
Does this integral converge/ diverge? why?
And also, I'd like to know why $\int_0^{2\pi} \frac{1}{\sin 2\theta} \, d\theta$ does not converge. I know it doesn't because of wolfram, not because i showed it on paper.
| Using a tip a friend gave me: It is enough to work on the first quadrant due to symmetry.
We will instead use the transform $v=xy, u=x$, the jacobian of said transform is $\frac{1}{u}$ which can be demonstrated.
$$\iint_{\mathbb R^2} \frac{1}{1+x^{10}y^{10}}dx\,dy = \int_0^\infty \int_{0}^{\infty} \frac{1}{1+v^{10}} \frac{1}{u} \, dv \, du =\int_0^\infty \frac{1}{1+v^{10}} \, dv \int_0^\infty \frac{1}{u} \, du $$
Notice that $\frac{1}{1+v^{10}}>0$ and so $\int_0^\infty \frac{1}{1+v^{10}} \, dv>0$, and that $\int_0^\infty \frac{1}{u} \, du$ diverges.
We can infer that the original integral diverges as well.
| {
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"timestamp": "2023-03-29T00:00:00",
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$a^{12} \equiv 1 \pmod{35}$,knowing that $(a,35)=1$ Prove that $\forall a \text{ with } (a,35)=1:$
$$a^{12} \equiv 1 \pmod{35}$$
$$35 \mid a^{12}-1 \Leftrightarrow 5 \cdot 7 \mid a^{12}-1 \overset{(5,7=1)}{ \Leftrightarrow} 5 \mid a^{12}-1 \text{ and } 7 \mid a^{12}-1$$
Therefore, $\displaystyle{ a^{12} \equiv 1 \pmod{35} \Leftrightarrow a^{12} \equiv 1 \pmod 5, a^{12} \equiv 1 \pmod 7}$
$$(a,35=1) \Rightarrow (a, 5 \cdot 7)=1 \overset{(5,7)=1}{\Rightarrow } (a,5)=1 \text{ and } (a,7)=1$$
According to Fermat's theorem:
$$a^4 \equiv 1 \pmod 5$$
$$a^{12} \equiv (a^4)^3 \equiv 1 \pmod 5$$
Also:
$$a^6 \equiv 1 \pmod 7$$
$$a^{12} \equiv (a^6)^2 \equiv 1 \pmod 7$$
So,we conclude that:
$$a^{12} \equiv 1 \pmod{35}$$
Could you tell me if it is right?
| Yes we can tell. This is because we have $a^{12} \equiv 1 \pmod 5 \iff 5|(a^{12}-1).$ Similarly $a^{12} \equiv 1 \pmod 7 \iff 7|(a^{12}-1)$. Since ${\rm gcd}(5,7)=1$, we have $5\cdot 7 | (a^{12}-1)$, i.e., $a^{12} \equiv 1 \pmod {35}$.
| {
"language": "en",
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When is $(a^2+b)(b^2+a)$ a power of $2$? When is $(a^2+b)(b^2+a)$ a power of $2$?
($a$, $b$ positive integers)
I tried some values on the computer and it seems the only solutions are $a=1$, $b=1$. But I am not sure how to prove it?
Thanks for any help.
| If $a$ has lower power of $2$ than $b$:
$a+b^2$ will be an odd multiple of a power of $2$ and hence won't be a power of $2$
Similarly if $b$ has lower power of $2$ than $a$
Therefore $a,b$ must have the same power $p$ of $2$
Let $a = 2^p x$ where $x$ is odd
Let $b = 2^p y$ where $y$ is odd
Then $(a^2+b)(b^2+a) = 2^{2p} ( 2^p x^2 + y ) ( 2^p y^2 + x )$
Thus $2^p x^2$ is odd, and hence $p = 0$, and so $a,b$ are both odd
If $a = b$:
$(a^2+b)(b^2+a) = a^2 (a+1)^2$
Thus $a = 1$ otherwise $a^2$ is not a power of $2$
If $a \ne b$:
$(a-b)(a+b-1) = 2^q z$ where $q$ is positive and $z$ is odd
$a = 2^q + b$
$(a^2+b)(b^2+a) = ( 2^{2q} + 2^{q+1} b + b^2 ) ( 2^q + b^2 + b )$
Thus $2 | b$ otherwise $2^{2q} + 2^{q+1} b + b^2$ will not be a power of $2$
Contradiction
Therefore $(a,b) = (1,1)$
| {
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Inequality for harmonic means Prove that for real numbers $a_1 ,a_2 ,...,a_n >0$ the following inequality holds
$$\frac{1}{a_1 } +\frac{2}{a_1 +a_2 } +...+\frac{n}{a_1 +a_2 +...+a_n }\leq 4\cdot \left(\frac{1}{a_1} +\frac{1}{a_2 } +...+\frac{1}{a_n} \right).$$
| The linked proof of Carleman's Inequality (in the comment) indicates the method of balancing coefficients in weighted mean inequalities. In the same spirit we can show, $$\frac{1}{a_1 } +\frac{2}{a_1 +a_2 } +...+\frac{n}{a_1 +a_2 +...+a_n } < 2 \left(\frac{1}{a_1} +\frac{1}{a_2 } +...+\frac{1}{a_n} \right)$$ for positive $a_i$'s.
We choose a set of positive real numbers $x_1,x_2,\ldots,x_n$ such that, by Cauchy-Schwarz Inequality: $$(a_1+a_2+\cdots+a_k)\left(\dfrac{x_1^2}{a_1}+\dfrac{x_2^2}{a_2}++\cdots\dfrac{x_k^2}{a_k}\right) \ge (x_1+x_2+\cdots+x_k)^2$$
$$\implies \dfrac{k}{a_1+a_2+\cdots+a_k} \le \dfrac{k}{(x_1+x_2+\cdots+x_k)^2}\left(\dfrac{x_1^2}{a_1}+\dfrac{x_2^2}{a_2}++\cdots\dfrac{x_k^2}{a_k}\right)$$
for each $k=1,2,\cdots,n$.
Adding them up from $k=1$ to $n$, we get:
$$\frac{1}{a_1 } +\frac{2}{a_1 +a_2 } +...+\frac{n}{a_1 +a_2 +...+a_n } \le \dfrac{c_1}{a_1} + \frac{c_2}{a_2}+\ldots+\frac{c_n}{a_n}$$
Where, $c_k = \dfrac{kx_k^2}{(x_1+x_2+\cdots+x_k)^2} + \dfrac{(k+1)x_k^2}{(x_1+x_2+\cdots+x_{k+1})^2}+\ldots+\dfrac{nx_k^2}{(x_1+x_2+\cdots+x_n)^2}$ for each $k=1,2,\ldots,n$.
It remains to choose a set of $\{x_k\}_{k=1}^n$ such that $\max\limits_{k \in \{1,2,\cdots,n\}}\{c_k\}$ is minimized.
For example if we plug in $x_k = k$, we have,
$c_k = k^2\left(\sum\limits_{j=k}^n \dfrac{j}{(1+2+\cdots+j)^2}\right) = 4k^2\left(\sum\limits_{j=k}^n \dfrac{1}{j(j+1)^2}\right) $
$$\le 2k^2\left(\sum\limits_{j=k}^n \dfrac{2j+1}{j^2(j+1)^2}\right) = 2k^2\left(\sum\limits_{j=k}^n \dfrac{1}{j^2} - \sum\limits_{j=k}^n \dfrac{1}{(j+1)^2}\right)$$
$$ = 2k^2\left(\dfrac{1}{k^2} - \dfrac{1}{(n+1)^2}\right) < 2$$
| {
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Sketch the set $\{ z \in \mathbb{C} | \left|\frac{z-i}{z+i}\right|<1 \}$ My question is to sketch the set $\{ z \in \mathbb{C} | \left|\frac{z-i}{z+i}\right|<1\}$ in the complex plane.
I substituted $z$ for $a+bi$, but did not get anywhere:
$\left|\frac{a+(b-1)i}{a+(b+1)i}\right|<1\\
\left|\frac{(a+(b-1)i)(a-(b+1)i)}{a^2+(b+1)^2}\right|<1\\
\left|\frac{a^2-(ab+a)i+(ab-a)i+(b^2-1)}{a^2+b^2+2b+1}\right|<1\\
\left|\frac{a^2+b^2-1-2ai}{a^2+b^2+2b+1}\right|<1\\
\left|\frac{a^2+b^2-1}{a^2+b^2+2b+1}-\frac{2ai}{a^2+b^2+2b+1}\right|<1\\
\left(\frac{a^2+b^2-1}{a^2+b^2+2b+1}\right)^2+\left(\frac{2ai}{a^2+b^2+2b+1}\right)^2<1\\
\frac{a^4+2a^2+2a^2b^2-2b^2+b^4+1}{a^4+2a^2+4a^2b+2a^2b^2+4b+6b^2+4b^3+b^4+1}<1$
| Rewrite the inequality as $|z-i|<|z+i|$. Then think geometrically in terms of distances from the point $z$ to the points $\pm i$.
| {
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To find the right most non zero digit When expanded 30! ends in 7 zeroes. Find the first non zero digit from right?
| First, consider the product of elements after dividing everything by all powers of 2 and 5.
$1\cdot 2 \cdot 3 \cdot \ldots \cdot 30$
becomes:
$1 \cdot 1 \cdot 3 \cdot 1 \cdot 1 \cdot 3 \cdot 7 \cdot 1 \cdot 9 \cdot 1 \cdot
11 \cdot 3 \cdot 13 \cdot 7 \cdot 3 \cdot 1 \cdot 17 \cdot 9 \cdot 19 \cdot 1 \cdot
21 \cdot 11 \cdot 23 \cdot 3 \cdot 1 \cdot 13 \cdot 27 \cdot 7 \cdot 29 \cdot 3$
Modulo 10, and ignoring ones, this becomes $3 \cdot 3 \cdot 7 \cdot 9 \cdot 3 \cdot 3 \cdot 7 \cdot 3 \cdot 7 \cdot 9 \cdot 9 \cdot 3 \cdot 3 \cdot 3 \cdot 7 \cdot 7 \cdot 9 \cdot 3$.
Writing 7 as $-3$ and 9 as $-1$, this becomes $3^9 \cdot (-3)^5 \cdot (-1)^4 = -(3^{14}) = -(9^7) = 1$.
Now, all powers of 2 and 5 are not missing from the product. 5 is clearly the bottleneck, so notice that power of 2 in this equals $15 + 7 + 3 + 1 = 26$. Out of these 7 are taken up in the zeroes at the end. So we are left with $2^{19}$. $2^{10} = 4 \bmod 10$, and $2^9 = 2 \bmod 10$. So $2^{19} = 8 \bmod 10$. As the previous product we found was 1, the digit you want is also 8.
| {
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Show that all real roots of the polynomial $P (x) = x^5 − 10x + 35$ are negative. I got this problem out of Andreescu's Putnam and Beyond. I solved it differently from the given solution and could not understand the solution. Can you explain what is happening in the last step of the solution?
Because P (x) has odd degree, it has a real zero r. If r > 0, then by the AM–GM
inequality
$P (r)$
$ = r^5 + 1 + 1 + 1 + 2^5 − 5 · 2 · r $
$≥ 0.$ (why?)
And the inequality is strict since $1 \neq 2$. Hence r < 0, as desired.
Also, here is my own (edit: incorrect) solution:
Let p be a positive root. Then $p^5-10p+35=0$
$10p > p^5 + 35 \implies 10 > p^4 + 35/p > 2 \sqrt{35p^3} > 10p \implies p < 1$
$10p > 35 \implies p > 3.5$
Contradiction!
| That should say $P(r) = r^5+1+1+1+2^5-5\cdot 2 \cdot r$.
The AM-GM inequality applied to $\{r^5,1,1,1,2^5\}$ gives us:
$\dfrac{r^5+1+1+1+2^5}{5} \ge \sqrt[5]{r^5 \cdot 1 \cdot 1 \cdot 1 \cdot 2^5}$
$\dfrac{r^5+35}{5} \ge 2r$
$r^5+35 \ge 10r$
$r^5-10r+35 \ge 0$
Equality only holds if $r^5 = 1 = 1 = 1 = 2^5$, which gives $r = 1 = 2$, which is impossible.
Thus, $r^5-10r+35 > 0$ for any number $r > 0$. Hence $x^5-10x+35$ cannot have a positive root.
One error in your solution is that if $p$ is a root, then $10p = p^5+35$, so you can't say $10p > p^5+35$. Also, $2\sqrt{35p^3} > 10p$ is only true for $p > \dfrac{5}{7}$
| {
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How to calculate $\int\frac{1}{x + 1 + \sqrt{x^2 + 4x + 5}}\ dx$? How to calculate $$\int\frac{1}{x + 1 + \sqrt{x^2 + 4x + 5}}dx?$$ I really don't know how to attack this integral. I tried $u=x^2 + 4x + 5$ but failed miserably. Help please.
| $$\frac1{x+1+\sqrt{x^2+4x+5}}=-\frac{x+1-\sqrt{x^2+4x+5}}{2(x+2)}$$
$$=-\frac{x+2-1-\sqrt{x^2+4x+5}}{2(x+2)}$$
$$=-\frac12+\frac1{2(x+2)}+\frac{\sqrt{(x+2)^2+1}}{2(x+2)}$$
Setting $x+2=\tan y,$
$$\int\frac{\sqrt{(x+2)^2+1}}{(x+2)}\ dx=\int\frac{\sec y}{\tan y}\sec^2y\ dy$$
$$=\int\frac{dy}{\cos^2y\sin y}=\int\frac{\sin y\ dy}{\cos^2y(1-\cos^2y)}$$
Set $\displaystyle\cos y=u$
| {
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"source": "stackexchange",
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Interesting dilemma, answer not matching with stewart, My work is Included Question :
Compute flux through the upper hemisphere of $x^2+y^2+z^2 = 1$ .
Where $$\textbf{F} = \left( z^2x\right)\textbf{ i }+\left[\dfrac{1}{3}y^3+ \tan z\right]\textbf{ j } + \left(x^2z+y^2 \right)\textbf{ k }$$
ANSWER GIVEN AT BACK OF STEWART : $\dfrac{13\pi}{20}$
MY WORK
$\textbf{Divergence theorem }$
$$\textbf{Flux} = \int\int_S \textbf{F}\cdot d\textbf{S} = \int\int\int_E \text{div }\textbf{F} \hspace{2mm} dV$$
Where $S$ is a closed surface.
And $E$ is the region inside that surface.
In this problem, instead of computing the surface is not closed
But we want to use divergence theorem, because divergence of the given vector field is cute.
We will over come this problem by attaching a disk at the bottom of the hemisphere, we call this closed this closed surface $S_2$ and the disk as $S_1$
We can use divergence theorem for $S_2$
We will then find flux through $S_1$.
Then the flux through $S$ = Flux through $S_2$ $-S_1$
$$\begin{align} & \text{div }\textbf{F} = \dfrac{\partial \left( z^2x\right)}{\partial x}+\dfrac{\partial }{\partial y}\left[\dfrac{1}{3}y^3+ \tan z\right] + \dfrac{\partial \left(x^2z+y^2 \right)}{\partial z} \\
& \text{div }\textbf{F} = z^2+y^2 +x^2 \end{align}$$
$$\textbf{Compute Flux through $S_2$ using divergence theorem }$$
$$\int\int\int_E \text{div }\textbf{F} \hspace{2mm} dV= \int\int\int_E x^2+y^2+z^2 \hspace{2mm} dV$$
$ $
We can define $E$ in spherical as follows :
$$\begin{align} & \left(\rho, \theta, \phi \right)\in E \hspace{1mm} | \hspace{2mm} 0< \rho < 1, \hspace{2mm} 0< \phi < \dfrac{\pi}{2}, \hspace{2mm} 0< \theta < 2\pi \\
& \text{Therefore,} \quad \int_0^{\pi/2}\int_0^{2\pi}\int_0^{1} \rho^2 \quad (\rho) d\rho d\theta d \phi \\
& =\int_0^{\pi/2}\int_0^{2\pi}\int_0^{1} \rho^3 \hspace{2mm} d\rho d\theta d\phi \\
& =\int_0^{\pi/2}\int_0^{2\pi}\left[ \dfrac{\rho^4}{4} \right]_0^{1} \hspace{2mm} d\theta d\phi \\
& =\dfrac{1}{4}\int_0^{\pi/2}\int_0^{2\pi} d\theta d\phi = \dfrac{1}{4}\times \dfrac{\pi}{2}\times 2\pi = \dfrac{\pi^2}{4} \end{align} $$
$$\textbf{Compute Flux through $S_1$ }$$
$ $
Note that $$\int\int_{S_1} \textbf{F}\cdot d\textbf{S} = \int\int_{S_1} \textbf{F}\cdot \textbf{n}\hspace{1mm}dS$$
Note that $S_1$ is part of the plane $z=0$
$$ = \int\int_{D} \textbf{F}\cdot \textbf{n}\sqrt{\left( \dfrac{\partial z}{\partial x}\right)^2+\left( \dfrac{\partial z}{\partial y}\right)^2+1}\hspace{1mm}dA $$
Where $\textbf{n} = \textbf{k}$ [because $\textbf{n}$ is normal unit vector to $S_1$ ]
And $D$ is the region inside the circle $x^2+y^2=1$ [ Because $D$ is projection of $S_1$ on $xy$ plane ]
$$ \begin{align} & = \int\int_{D} \textbf{F}\cdot \textbf{k}\sqrt{\left( 0\right)^2+\left( 0\right)^2+1}\hspace{1mm}dA \\
& = \int\int_{D} x^2z+y^2 \hspace{1mm}dA \end{align} $$
Substitute $z=0$, since $S_1$ is part of the plane $z=0$
$$ = \int\int_{D} y^2 \hspace{1mm}dA $$
In polar coordinates the Integral becomes
$$ \begin{align} & = \int_0^{2\pi}\int_{0}^1 r^2\sin^2\theta \hspace{1mm}(r)drd\theta \\
& =\left( \int_0^{2\pi}\sin^2\theta \hspace{1mm}d\theta \right) \left(\int_{0}^1 r^3 \hspace{1mm}dr \right) \\
& =\dfrac{1}{2}\left[ \theta-\dfrac{\sin2\theta}{2}\right]_0^{2\pi} \left[ \dfrac{r^4}{4} \right]_{0}^1 \\
& =\dfrac{1}{2}\times 2\pi \times \dfrac{1}{4} = \dfrac{\pi}{4} \\ \end{align}$$
Therefore, answer should be $$\dfrac{\pi^2-\pi}{4}$$
| Your volume element for spherical coordinates is not correct. It should be $$dV=\rho^2 \sin(\varphi) \, d \rho \, d \varphi \, d \theta.$$
| {
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Find the minimum of $\displaystyle \frac{1}{\sin^2(\angle A)} + \frac{1}{\sin^2(\angle B)} + \frac{1}{\sin^2(\angle C)}$ Is it possible to find the minimum value of $E$ where
$$E = \frac{1}{\sin^2(\angle A)} + \frac{1}{\sin^2(\angle B)} + \frac{1}{\sin^2(\angle C)}$$for any $\triangle ABC$.
I've got the feeling that $\min(E) = 4$ and that the critical value occurs when $ABC$ is equilateral.
| $$E=\csc^2A+\csc^2B+\csc^2C$$
Since $f(x)=\csc^2x$ is a convex function in $(0,\pi)$, we have from Jensen's inequality:
$$\csc^2\left(\frac{A+B+C}{3}\right) \leq \frac{1}{3}\csc^2A+\frac{1}{3}\csc^2B+\frac{1}{3}\csc^2C$$
Since $A+B+C=\pi$
$$\Rightarrow \frac{\csc^2A+\csc^2B+\csc^2C}{3}\geq \frac{4}{3}$$
$$\Rightarrow \csc^2A+\csc^2B+\csc^2C \geq 4$$
And equality occurs when $A=B=C=\dfrac{\pi}{3}$.
$\blacksquare$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Verification of Stokes Theorem I want to verify Stokes Theorem for the surface
$$
\Phi = \{ (x,y,z) \in \mathbb R^3 : z = x^2 - y^2, x^2 + y^2 \le 1 \}
$$
and the vector field $F(x,y,z) := (y,z,x)$. For this I use the parametrisation
$$
\varphi(r, \theta) = \begin{pmatrix} r\cos \theta \\ r \sin \theta \\ r^2(\cos^2 \theta - \sin^2\theta) \end{pmatrix}
= \begin{pmatrix} r\cos \theta \\ r \sin \theta \\ r^2(2\cos^2\theta - 1) \end{pmatrix}
$$
with $0 \le r \le 1, 0 \le \theta < 2\pi$.
Then
$$
\varphi_r(r, \theta) = \begin{pmatrix} \cos\theta \\ \sin\theta \\ 2r(2\cos^2\theta - 1) \end{pmatrix} \qquad
\varphi_{\theta}(r,\theta) = \begin{pmatrix} -r\sin\theta \\ r\cos\theta \\ -4r^2 \cos\theta \sin\theta \end{pmatrix}
$$
and
$$
\varphi_r \times \varphi_{\theta} = \begin{pmatrix}
-2r^2\cos\theta \\
2r^2\sin\theta \\
r
\end{pmatrix}.
$$
and also
$$
\mbox{rot}(F) = \begin{pmatrix} -1\\ -1\\ -1 \end{pmatrix}
$$
So now I have everything at hand to compute the two sides of
$$
\int_{\Phi} \mbox{rot}(F) \cdot \vec{n} ~ dS = \int_{\partial \Phi} F(r) \cdot dr.
$$
For the RHS I have
$$
\begin{align*}
\int_0^{2\pi} F(\varphi(1, \theta)) \cdot \varphi_{\theta}(1, \theta)
& = r^3 \int_0^{2\pi} \cos\theta(2\cos^2\theta - 1) d\theta - r^2 \int_0^{2\pi} \sin\theta d\theta - 4r^3 \int_0^{2\pi} \cos^2\theta \sin\theta d\theta \\
& = 0
\end{align*}
$$
and for the LHS:
\begin{align*}
\int_0^1 \int_0^{2\pi} \begin{pmatrix} -1\\-1\\-1\end{pmatrix} \cdot \begin{pmatrix}
-2r^2\cos\theta \\
2r^2\sin\theta \\
r
\end{pmatrix} d\theta dr
& = - \int_0^1 \int_0^{2\pi} 2r^2(\sin(\theta)-\cos(\theta)) + r d\theta dr \\
& = - \int_0^1 \int_0^{2\pi} r d\theta dr = \pi
\end{align*}
so these are not equal. But I looked at the calculations several times and I am sure they are right, so do you see what went wrong?
| The main thing is that for your line integral you should have (remember that $r=1$)
$$\int_0^{2\pi} \big((\sin\theta)(-\sin\theta) + (2\cos^2\theta-1)(\cos\theta) + (\cos\theta)(-4\cos\theta\sin\theta)\big)\,d\theta = -\pi,$$
and you lost a minus sign in the surface integral at the end. I think you just made a copying error in part of it.
| {
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Find the exact value of $\sin (\theta)$ and $\cos (\theta)$ when $\tan (\theta)=\frac{12}{5}$ So I've been asked to find $\sin(\theta)$ and $\cos(\theta)$ when $\tan(\theta)=\cfrac{12}{5}$; my question is if $\tan (\theta)=\cfrac{\sin (\theta) }{\cos (\theta)}$ does this mean that because $\tan (\theta)=\cfrac{12}{5}$ then $\sin (\theta) =12$ and $\cos(\theta)=5$?
It doesn't seem to be the case in this example, because $\sin (\theta)\ne 12 $ and $\cos (\theta)\ne 12 $.
Can someone tell me where the error is in my thinking?
| Most of the above solutions assume, wrongly, that $\theta$ is acute, i.e. $0^{\circ} < \theta < 90^{\circ}.$ Here I give all the cases for $0^{\circ}<\theta <360^{\circ}.$ I like to use simple identities to answer these questions. We know that $1+\tan^2\theta \equiv \sec^2\theta$. If $\tan\theta = \frac{12}{5}$ then we see that:
$$\begin{eqnarray*}
1+\tan^2\theta &\equiv& \sec^2\theta \\ \\
1+\left(\frac{12}{5}\right)^{\! 2} &=& \sec^2\theta \\ \\
\frac{169}{25} &=& \sec^2\theta
\end{eqnarray*}$$
It follows that $\sec\theta = \pm \frac{13}{5}$, and hence $\cos\theta = \pm \frac{5}{13}$. The question is: which is it, plus or minus? Sadly, there is not enough information in the question.
Looking at the graph $y=\tan\theta$ we see that $\tan \theta > 0$ if $0^{\circ} < \theta < 90^{\circ}$ or $180^{\circ} < \theta < 270^{\circ}$. That means $\tan \theta = \frac{12}{5} > 0$ has solutions in the range $0^{\circ} < \theta < 90^{\circ}$ or $180^{\circ} < \theta < 270^{\circ}$. However, $\cos \theta > 0$ when $0^{\circ} < \theta < 90^{\circ}$ and $\cos \theta < 0$ when $180^{\circ} <\theta < 270^{\circ}$.
To conclude: $\cos\theta = \frac{5}{13}$ if $0^{\circ} < \theta < 90^{\circ}$ and $\cos\theta = -\frac{5}{13}$ if $180^{\circ} < \theta < 270^{\circ}$.
There are many ways to find $\sin \theta$. You could use the identity $1+\cot^2\theta \equiv \operatorname{cosec}^2\theta$, noting that $\cot$ is the reciprocal of $\tan$ and $\tan\theta = \frac{12}{5} \iff \cot\theta = \frac{5}{12}$. Alternatively, you could use your value of $\cos\theta$ along with the identity $\sin^2\theta + \cos^2\theta \equiv 1$. It's up to you; I'd use the second one.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/852337",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 9,
"answer_id": 3
} |
Solution verification: Find the orthogonal trajectories of the family of curves for $x^2 + 2y^2 = k^2$ I need help with the following question:
Find the orthogonal trajectories of the family of curves for $x^2 + 2y^2 = k^2$
I have taken the following steps, are they correct? From what I understand, I have to take the following steps
*
*First differentiate to find the differential equation
*Then write the differential equation in this
$$\frac{dy}{dx} = -\frac{1}{f(x,y)}$$
*Then solve the new equation.
So, this is what I have:
The first step is to differentiate with respect to $x$ so find $\frac{dy}{dx}$.
\begin{align}
\frac{dy}{dx} x^2 + \frac{dy}{dx} 2y^2 &= \frac{dy}{dx}k^2 \\
2x + \frac{dy}{dx}\cdot 4y &= 0 \\
\frac{dy}{dx} &= -\frac{x}{2y}
\end{align}
Now, the second step is to do the negative recipricol.
$$\frac{dy}{dx} = \frac{2y}{x}$$
The third step is to solve the newly formed differential equation.
\begin{align}
dy \frac{1}{2y} &= \frac{1}{x} dx \\
\int \frac{1}{2y} dy &= \int \frac{1}{x} dx \\
\frac{\ln{y}}{2} &= \ln{x} + C \\
\ln y &= 2 \ln x + C\\
\ln y &= \ln x^2 + C \\
y &= Ax^2 \quad
\end{align}
In this case, $A = e^C$. Are my steps correct?
Thanks a bunch for your help!
EDIT:
I have a found a minor error in the last few steps when multiplying the two. It should be:
\begin{align}
\ln y &= 2(\ln x + C) \\
\ln y &= 2\ln x + 2C \\
e^{\ln y} &= e^{\ln x^2 + 2C} \\
y &= x^2 \cdot e^{2C} \\
\end{align}
Here, we let $A = e^{2C}$ and say the final answer is $$y = Ax^2$$
| From the family of ellipses
$$x^2+2y^2=k^2 ......(1)$$
you obtained the family of parabolas
$$y=Ax^2......(2)$$
Setting $x=Y\sqrt{2}$ and $y=X/\sqrt{2}$ in (1), we obtain:
$$2Y^2+X^2=k^2 ......(3)$$
So we obtain another family of parabolas
$$Y=B_1X^2......(4)$$
Which is equivalent to
$$x=By^2......(5)$$
This is because the original family of curves are symmetric in swapping $x$ and $y\sqrt{2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/852492",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Max. and Min. value of $|z|$ in $\left|z+\frac{2}{z}\right| = 2\,$ If $z$ is a complex no. such that $\displaystyle \left|z+\frac{2}{z}\right| = 2\,$ Then find max. and min. value of $\left|z\right|$.
$\bf{My\; Try:}$ Given $\displaystyle \left|z+\frac{2}{z}\right| = 2\Rightarrow \left|z+\frac{2}{z}\right|^2 = 2^2=4$.
So $\displaystyle \left(z+\frac{2}{z}\right)\cdot \left(\bar{z}+\frac{2}{\bar{z}}\right) = 4\Rightarrow \left|z\right|^2+2\left(\frac{z}{\bar{z}}+\frac{\bar{z}}{z}\right)+\frac{1}{|z|^2} = 4$.
Now how can I find the max. and min. values of $|z|$?
Help me please.
Thanks
| Let $z=re^{i\theta}$, then your last step can be written as
$$r^2+2\cos (2\theta)+\frac{1}{r^2}=4.$$
From this we can get
\begin{align*}
\left(r+\frac{1}{r}\right)^2 & = 6-2\cos (2\theta)
\end{align*}
Thus $2 \leq \left(r+\frac{1}{r}\right) \leq 2\sqrt{2}.$ Now try to find the range of $r$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/853992",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Finding $\int_0^{2\pi}\frac{\sin t + 4}{\cos t + \frac{5}3} dt$ I'd like to ask something about the following integral:
$$
\int_0^{2\pi}\frac{\sin t + 4}{\cos t + \frac{5}3} dt
$$
I rewrote and took another variable.
$$
-i\int_0^{2\pi}\frac{e^{is}-e^{-is}+8i}{e^{is}+e^{-is} + \frac{10}3} dt
\quad = \quad
-i\int_{C(0,1)^+}\frac{z-\frac{1}{z}+8i}{z+ \frac1z+ \frac{10}3} \cdot \frac{-i}{z}dt
\quad = \quad
- \int_{C(0,1)^+} \frac{z^2-1+8iz}{z(z^2+1+\frac{10}{3} \cdot z)} dz
$$
The only root of $z(z^2+1+\frac{10}{3} \cdot z)$ inside the unit disk is $0$, so I thought that the integral would equal:
$$
Res_{z=i} \ = \ 2 \pi i \cdot \frac{0-1+8i\cdot 0}{0^2+1+\frac{10}{3}\cdot 0} \quad = \quad 2\pi i
$$
I don't understand why $i$ appears in this value. Can you please tell me what I did wrong?
| We first separate the integration into 4 intervals:
$$I=\int_0^{2\pi}\frac{\sin t + 4}{\cos t + \frac{5}3} dt=\int_0^{\pi/2}\frac{\sin t + 4}{\cos t + \frac{5}3} dt+\int_0^{\pi/2}\frac{\sin (t+(\pi/2)) + 4}{\cos (t+(\pi/2)) + \frac{5}3} dt+\int_0^{\pi/2}\frac{\sin (t+\pi) + 4}{\cos (t+\pi) + \frac{5}3} dt+\int_0^{\pi/2}\frac{\sin (t+(3\pi/2)) + 4}{\cos (t+(3\pi/2)) + \frac{5}3} dt$$
$$I=\int_0^{\pi/2}\frac{\sin t + 4}{\cos t + \frac{5}3} dt+\int_0^{\pi/2}\frac{\cos t + 4}{-\sin t + \frac{5}3} dt+\int_0^{\pi/2}\frac{-\sin t + 4}{-\cos t + \frac{5}3} dt+\int_0^{\pi/2}\frac{-\cos t + 4}{\sin t + \frac{5}3} dt$$
$$I=\int_0^{\pi/2}\frac{240}{41-9\cos(2t)}2dt=\int_0^{\pi}\frac{240}{41-9\cos s}ds$$
For this kind of problem it is usually convenient to set $\cos s=\frac{1-x^2}{1+x^2}$ where $x=\tan (s/2)$. Then $ds =\frac{2dx}{1+x^2}$.
The integral then becomes:
$$I=\int_0^{\infty}\frac{240}{9+25x^2}dx=6\pi$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/860003",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Find the Value of Integral Find the Value of $$\begin{align}I=\int_{0}^{1}\frac{\ln(x)\,dx}{1-x^2}\end{align}$$
I have tried like this: We have $$\begin{align}2I=\int_{0}^{1}\frac{\ln(x^2)\,dx}{1-x^2}=\int_{0}^{1}\frac{\ln(1-(1-x^2))\,dx}{1-x^2}\end{align}$$ So
$$2I=\begin{align}\int_{0}^{1}\frac{-(1-x^2)-\frac{(1-x^2)^2}{2}-\frac{(1-x^2)^3}{3}-\cdots }{1-x^2}\end{align}=$$
I need help from here..
| Use the series expansion for $\dfrac{1}{1-x^2}$ i.e
$\displaystyle \frac{1}{1-x^2}=\sum_{k=0}^{\infty} x^{2k}\,dx$
Hence,
$$I=\int_0^1 \ln(x)\left(\sum_{k=0}^{\infty} x^{2k}\right)\,dx=\sum_{k=0}^{\infty} \int_0^1 x^{2k}\ln x\,dx=-\sum_{k=0}^{\infty}\frac{1}{(2k+1)^2}$$
where the final integral can be evaluated using the substitution $\ln x=-t$.
Since,
$$\sum_{k=1}^{\infty} \frac{1}{k^2}=\sum_{k=1}^{\infty} \frac{1}{(2k)^2}+\sum_{k=0}^{\infty} \frac{1}{(2k+1)^2} \Rightarrow \sum_{k=0}^{\infty} \frac{1}{(2k+1)^2}=\frac{\pi^2}{6}-\frac{1}{4}\frac{\pi^2}{6}$$
$$\Rightarrow \sum_{k=0}^{\infty} \frac{1}{(2k+1)^2}=\frac{3\pi^2}{24}=\frac{\pi^2}{8}$$
Hence,
$$\boxed{I=-\dfrac{\pi^2}{8}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/860091",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Solve $\dfrac{x}{x-2}>2$ by first rewriting it in the form $\dfrac{P(x)}{Q(x)}>0$ Edit: So then is this the correct final solution? $x<4,(\infty,4), x\ne2$
I am asked to do this:
Solve $\dfrac{x}{x-2}>2$ by first rewriting it in the form $\dfrac{P(x)}{Q(x)}>0$
$$\dfrac{x}{x-2}>2$$
$$\dfrac{x}{x-2}-\dfrac{2(x-2)}{1(x-2)}>0$$
$$\dfrac{x-2x+4}{x-2}>0$$
$$\dfrac{-1(-x+4)}{x-2}>0$$
$$\dfrac{x-4}{x-2}<0(x-2)$$
$$x-4<0$$
$$x<4$$
| Your method not is correct because: if
$$\frac{4-x}{x-2}>0\equiv-1\cdot\frac{(4-x)}{x-2}<(-1)\cdot0\equiv\frac{x-4}{x-2}<0$$
then $(x-4)>0$ and $(x-2)<0$, or $(x-4)<0$ and $(x-2)>0$
solutions $2<x<4$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/861486",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Find expression for : $ S_n =\sum_{i=1}^{n} \frac{i}{i^4+i^2+1} $ I want to find a formula for the sum of this series using its general term.
How to do it?
Series
$$
S_n = \underbrace{1/3 + 2/21 + 3/91 + 4/273 + \cdots}_{n \text{ terms}}
$$
General Term
$$
S_n = \sum_{i=1}^{n} \frac{i}{i^4+i^2+1}
$$
| Set $$\color{blue}{a_k= \frac{1}{k^2 -k +1} \implies a_{k+1}= \frac{1}{(k+1)^2 -k } =\frac{1}{k^2+ k +1}} $$
But since $ (k^2 -k +1)(k^2 +k +1) =k^4 +k^2 +1$ we have,
$$ a_{k+1} -a_k = \frac{1}{k^2 +k +1}- \frac{1}{k^2 -k +1} = \frac{2k}{(k^2 -k +1)(k^2 +k +1)} =\frac{2k}{k^4 +k^2 +1}$$
Then
$$a_n-a_0 = \sum_{k=0}^{n-1}a_{k+1} -a_k=2\sum_{k=0}^{n-1}\frac{k}{k^4 +k^2 +1} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/865442",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Is this function $y=-\ln\left(1+\frac{\sin x -\cos x}{2}\right)$ convex? Is this function convex for $x\in[0,\frac{\pi}{4}]$?
$$y=-\ln\left(1+\frac{\sin x -\cos x}{2}\right)$$
without use the derivative.
| Disclaimer: it is quite a long exercise in trigonometry to avoid the use of derivatives to prove the convexity.
Since the function is continuous, you only need the midpoint-convexity in order to prove the convexity. If we take $[x,y]\subset[0,\pi/4]$, we need to prove:
$$-2\log\left(1+\frac{1}{\sqrt{2}}\sin\left(\frac{x+y}{2}-\frac{\pi}{4}\right)\right)\leq -\log\left(1+\frac{1}{\sqrt{2}}\sin\left(x-\frac{\pi}{4}\right)\right)-\log\left(1+\frac{1}{\sqrt{2}}\sin\left(y-\frac{\pi}{4}\right)\right)$$
that equivalent to:
$$\left(1+\frac{1}{\sqrt{2}}\sin\left(\frac{x+y}{2}-\frac{\pi}{4}\right)\right)^2 \geq \left(1+\frac{1}{\sqrt{2}}\sin\left(x-\frac{\pi}{4}\right)\right)\cdot\left(1+\frac{1}{\sqrt{2}}\sin\left(y-\frac{\pi}{4}\right)\right)$$
or to:
$$\left(1-\frac{1}{\sqrt{2}}\sin\left(\frac{z+w}{2}\right)\right)^2 \geq \left(1-\frac{1}{\sqrt{2}}\sin z\right)\cdot\left(1-\frac{1}{\sqrt{2}}\sin w\right)$$
for $[z=\pi/4-y,w=\pi/4-x]\subset[0,\pi/4]$. If we continue expanding we get:
$$-\frac{2}{\sqrt{2}}\sin\left(\frac{z+w}{2}\right)+\frac{1}{2}\sin^2\left(\frac{z+w}{2}\right) \geq -\frac{2}{\sqrt{2}}\frac{\sin z+\sin w}{2}+\frac{1}{2}\sin w \sin z$$
or:
$$\sqrt{2}\sin\left(\frac{z+w}{2}\right)\left(1-\cos\left(\frac{z-w}{2}\right)\right) \leq \frac{1}{2}\left(\frac{1-\cos(z+w)}{2}-\sin w \sin z\right),$$
$$2\sqrt{2}\sin\left(\frac{z+w}{2}\right)\left(1-\cos\left(\frac{z-w}{2}\right)\right) \leq \sin^2\frac{z-w}{2},$$
$$\sqrt{2}\sin\left(\frac{w+z}{2}\right) \leq \cos^2\frac{w-z}{4},$$
but if $w-z=u\in[0,\pi/4]$, then $\frac{w+z}{2}\leq \frac{\pi}{4}-\frac{u}{2}$, hence the LHS is $\leq \sqrt{2}\sin(\pi/4-u/2)$, so we just need to prove that for any $u\in[0,\pi/4]$ we have:
$$ \sqrt{2}\sin\left(\frac{\pi}{4}-\frac{u}{2}\right)\leq \cos^2\left(\frac{u}{2}\right),$$
or:
$$ \cos\frac{u}{2}-\sin\frac{u}{2}\leq \cos^2\left(\frac{u}{2}\right),$$
$$ \sin(u/2) \geq \cos(u/2)-\cos^2(u/2),$$
$$ \cot(u/4) \geq \cos(u/2), $$
That is trivial since over $[0,\pi/4]$ the LHS is greater than $\cot\frac{\pi}{16}>1$ while the RHS is obviously $\leq 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/865933",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Simplify expression $(x\sqrt{y}- y\sqrt{x})/(x\sqrt{y} + y\sqrt{x})$ I'm stuck at the expression: $\displaystyle \frac{x\sqrt{y} -y\sqrt{x}}{x\sqrt{y} + y\sqrt{x}}$.
I need to simplify the expression (by making the denominators rational) and this is what I did:
$$(x\sqrt{y} - y\sqrt{x}) \times (x\sqrt{y} - y\sqrt{x}) = (\sqrt{y} - \sqrt{x})^2$$
Divided by
$$(x\sqrt{y} + y\sqrt{x}) \times (x\sqrt{y} - y\sqrt{x} ) = (x\sqrt{y})^2$$
So I'm left with $\displaystyle \frac{(\sqrt{y} - \sqrt{x})^2}{(x\sqrt{y})^2}$.
This answer is incorrect. Can anyone help me understand what I did wrong? If there is a different approach to solve this it will also be much appreciated. Please explain in steps.
| Given the question, we can write
\begin{eqnarray}
\frac{ x\sqrt{y} - y\sqrt{x} }{ x\sqrt{y} + y\sqrt{x} }
&=& \frac{ 1 - \sqrt{y/x} }{ 1 + \sqrt{y/x} }\\
&=& \frac{ \Big( 1 - \sqrt{y/x} \Big)^2 }
{ \Big( 1 + \sqrt{y/x} \Big) \Big( 1 - \sqrt{y/x} \Big)}\\
&=& \frac{ 1 + y/x - 2\sqrt{y/x} }{ 1 - y/x }\\
&=& \frac{ x + y - 2\sqrt{xy} }{ x - y }.
\end{eqnarray}
Or
\begin{eqnarray}
\frac{ x\sqrt{y} - y\sqrt{x} }{ x\sqrt{y} + y\sqrt{x} }
&=& \frac{ \sqrt{x} - \sqrt{y} }{ \sqrt{x} + \sqrt{y} }\\
&=& \frac{ \Big( \sqrt{x} - \sqrt{y} \Big)^2 }
{ \Big( \sqrt{x} + \sqrt{y} \Big) \Big( \sqrt{x} - \sqrt{y} \Big)}\\
&=& \frac{ x + y - 2\sqrt{xy} }{ x - y }.
\end{eqnarray}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/869037",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
find a polynomial whose roots are inverse of squares of roots of $x^3+px+q$ Question is :
Given a polynomial $f(x)=x^3+px+q\in \mathbb{Q}[x]$ find a polynomial whose roots are inverse of sqares of roots of $f(x)$
Supposing $a,b,c$ as roots of $f(x)$ we have :
*
*$a+b+c=0$
*$ab+bc+ca=p$
*$abc=-q$
Now i need to know what
*
*$\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}$
*$\dfrac{1}{a^2}\cdot\dfrac{1}{b^2}+\dfrac{1}{b^2}\cdot\dfrac{1}{c^2}+\dfrac{1}{c^2}\cdot\dfrac{1}{a^2}$
*$\dfrac{1}{(abc)^2}$
All i have to do is use $(a+b+c)^2$ formula and others and conclude what those sums,products are.. I am fairly comfortable with that...
But then, this question was from a Galois theory course.. So, i some how guess there is a better way to do this...
Can some one suggest something..
| Hint: start with: $\dfrac{1}{a^2} + \dfrac{1}{b^2} + \dfrac{1}{c^2} = \dfrac{a^2b^2+b^2c^2+c^2a^2}{a^2b^2c^2} = \dfrac{(ab+bc+ca)^2 - 2abc(a+b+c)}{(abc)^2} = \dfrac{p^2 - 0}{(-q)^2} = \dfrac{p^2}{q^2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/872264",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
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Find the $n$th term of $1, 2, 5, 10, 13, 26, 29, ...$ How would you find the $n$th term of a sequence like this?
$1, 2, 5, 10, 13, 26, 29, ...$
I see the sequence has a group of three terms it repeats: Double first term to get second term, add three to get third term, repeat.
What about: $2, 3, 6, 7, 14, 15, 30,... $?
Again the sequence has a group of three terms it repeats: Add one to first term to get second term, then double second term to get third term.
How do you compute the $n$th term of such sequences directly, without iterating through all preceding terms?
| For the first sequence:
$$S_n = {2}^{(2+\lfloor\frac{n}{2}\rfloor)}+6(\frac{n}{2}-\lfloor\frac{n}{2}\rfloor-1)$$
For example:
*
*$S_1 = 2^2-3 = 1$
*$S_2 = 2^3-6 = 2$
*$S_3 = 2^3-3 = 5$
*$S_4 = 2^4-6 = 10$
*$S_5 = 2^4-3 = 13$
*$S_6 = 2^5-6 = 26$
*$S_7 = 2^5-3 = 29$
For the second sequence:
$$S_n = {2}^{\lceil\frac{n}{2}\rceil}+2(\frac{n}{2}-\lfloor\frac{n}{2}\rfloor-1)$$
For example:
*
*$S_1 = 2^2-2 = 2$
*$S_2 = 2^2-1 = 3$
*$S_3 = 2^3-2 = 6$
*$S_4 = 2^3-1 = 7$
*$S_5 = 2^4-2 = 14$
*$S_6 = 2^4-1 = 15$
*$S_7 = 2^5-2 = 30$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/872485",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 7,
"answer_id": 4
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Symmetric matrix multiplication Let $A$ and $B$ be symmetric matrices.
Prove:
*
*$AB=BA$
*$AB$ is a symmetric matrix
As for 1. due to the axiom $(AB)^T=B^T A^T$ so $AB=BA$
As for 2. I did not find any axiom that can support the claim, but from test I found that it is true for symmetric matrices when the entries on the diagonal are equal.
| Both claims are false and almost any $A$ and $B$ are counterexamples. For a specific example, you can see $$\begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} \cdot \begin{pmatrix} 1 & 2 \\ 2 & 3 \end{pmatrix} = \begin{pmatrix} 3 & 5 \\ 3 & 5 \end{pmatrix}$$ while $$\begin{pmatrix} 1 & 2 \\ 2 & 3 \end{pmatrix} \cdot \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} = \begin{pmatrix} 3 & 3 \\ 5 & 5 \end{pmatrix}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/874469",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 2,
"answer_id": 0
} |
Evaluation of $\int\frac{\sqrt{\cos 2x}}{\sin x}\,dx$
Compute the indefinite integral
$$
\int\frac{\sqrt{\cos 2x}}{\sin x}\,dx
$$
My Attempt:
$$
\begin{align}
\int\frac{\sqrt{\cos 2x}}{\sin x}\,dx &= \int\frac{\cos 2x}{\sin^2 x\sqrt{\cos 2x}}\sin xdx\\
&= \int\frac{2\cos^2 x-1}{(1-\cos^2 x)\sqrt{2\cos^2 x-1} }\sin x \,dx
\end{align}
$$
Let $\cos x = t$, so that $\sin x\,dx = -dt$. This changes the integral to
$$
\begin{align}
\int\frac{(2t^2-1)}{(t^2-1)\sqrt{2t^2-1}}\,dt &= \int\frac{(2t^2-2)+1}{(t^2-1)\sqrt{2t^2-1}}\,dt\\
&= 2\int\frac{dt}{\sqrt{2t^2-1}}+\int \frac{dt}{(t^2-1)\sqrt{2t^2-1}}
\end{align}
$$
How can I solve the integral from here?
| $\displaystyle I=\int\frac{\sqrt{2t^2-1}}{t^2-1}dt$, so $\displaystyle t=\frac{1}{\sqrt{2}}\sec\theta, dt=\frac{1}{\sqrt{2}}\sec\theta\tan\theta d\theta$ gives
$\displaystyle I=\frac{1}{\sqrt{2}}\int\frac{\sec\theta\tan^2\theta}{\frac{1}{2}\sec^2\theta-1}d\theta=\sqrt{2}\int\frac{\sec\theta\tan^2\theta}{\sec^2\theta-2}d\theta=\sqrt{2}\int\frac{\sin^2\theta\cos\theta}{\cos^2\theta(1-2\cos^2\theta)}d\theta$.
Next letting $u=\sin\theta, du=\cos\theta d\theta$ gives
$\displaystyle I=\sqrt{2}\int\frac{u^2}{(1-u^2)(2u^2-1)}du=\sqrt{2}\cdot\frac{1}{2}\int\left(\frac{1}{1-u}+\frac{1}{1+u}-\frac{1}{\sqrt{2}u+1}+\frac{1}{\sqrt{2}u-1}\right)du$
$\displaystyle=\frac{1}{\sqrt{2}}\left[\ln\left(\frac{1+u}{1-u}\right)+\frac{1}{\sqrt{2}}\ln\left(\frac{\sqrt{2}u-1}{\sqrt{2}u+1}\right)\right]+C$
$\displaystyle=\frac{1}{\sqrt{2}}\left[\ln\left(\frac{1+\sin\theta}{1-\sin\theta}\right)+\frac{1}{\sqrt{2}}\ln\left(\frac{\sqrt{2}\sin\theta-1}{\sqrt{2}\sin\theta+1}\right)\right]+C=\frac{1}{\sqrt{2}}\left[2\ln\lvert\sec\theta+\tan\theta\rvert+\frac{1}{\sqrt{2}}\ln\left(\frac{\sqrt{2}\sin\theta-1}{\sqrt{2}\sin\theta+1}\right)\right]+C$
$\displaystyle=\sqrt{2}\ln\big|\sqrt{2}t+\sqrt{2t^2-1}\big|+\frac{1}{2}\ln\bigg|\frac{\sqrt{2t^2-1}-t}{\sqrt{2t^2-1}+t}\bigg|+C$
$=\displaystyle\sqrt{2}\ln\bigg|\sqrt{2}\cos x+\sqrt{\cos 2x}\bigg|+\ln\bigg|\frac{\cos x-\sqrt{\cos 2x}}{\sin x}\bigg|+C$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/876175",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 5,
"answer_id": 1
} |
How to factorize $x^4+2x^2+4$ to a product of polynomials with real coefficients? How do you factor
$$x^4+2x^2+4 $$
so it can be written as
$$ (x^2+2x+2)(x^2-2x+2) $$
| The key idea is to complete $\,x^4\!+4\,$ to a square $\,(x^2\!+2)^2-4x^2,\,$ yielding a difference of squares
$$ x^4\!+4\, =\, (x^2\!+2)^2 - (\color{#c00}{2x})^2 =\, (x^2\!+2-\color{#c00}{2x})(x^2\!+2 + \color{#c00}{2 x})$$
The same idea works for $\,x^4+4 + 2x^2\,$ if you meant that instead of $\,x^4+4.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/876244",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Solve for $x$, $\tan x +\sec x = 2\cos x$ ; $−∞ < x < ∞$ Solve for $x$, $\tan x +\sec x = 2\cos x$ ; $−∞ < x < ∞$
$$\tan x + \sec x = 2\cos x$$
I tried changing it all to sin and cos
$$\frac{\sin x}{\cos x} + \frac{1}{\cos x} = 2\cos x$$
then I made it to one fraction
$$\frac{\sin x + 1}{\cos x} = 2 \cos x$$
Then I don't know where to go from there. Please help!
| Multiply across by $\cos x$ to get $\sin x + 1 = 2 \cos^2 x= 2 - 2 \sin^2 x$,
or $2 \sin^2 x +\sin x -1 = 0$.
Since the solutions of $2y^2+y-1 = 0$ are $\{-1, {1 \over 2} \}$, the possible solutions to the original equations are $\sin x = -1$ and $\sin x = {1 \over 2} $. However, if $\sin x = -1$, then $\cos x = 0$ and the equation is undefined, hence we ignore this possibility.
The remaining solutions are $\arcsin ( { 1\over 2}) + 2\pi n= { \pi \over 6} + 2 \pi n$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/877539",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Prove $\frac{\sin A}{\sec A+\tan A-1}+ \frac{\cos A}{\csc A+\cot A-1}=1$ $$\frac{\sin A}{\sec A+\tan A-1}+ \frac{\cos A}{\csc A+\cot A-1}=1$$
Prove that L.H.S.$=$R.H.S.
This type of questions always creates problem when in right hand side some trigonometry function is given then it is bit easy to think how to proceed further.
Can some one help me not only to solve the problem but also how to tackle this type of other problem (when R.H.S. is $1$)?
| $$\frac{\sin A}{\sec A+\tan A-1}+ \frac{\cos A}{\csc A+\cot A-1}= \frac{\sin A}{\frac{1}{\cos A}+\frac{\sin A}{\cos A}-\frac{\cos A}{\cos A}}+ \frac{\cos A}{\frac{1}{\sin A}+\frac{\cos A}{\sin A}-\frac{\sin A}{\sin A}} =$$ $$= \frac{\sin A \cos A}{1 + \sin A - \cos A} + \frac{\sin A \cos A}{1 + \cos A - \sin A} =$$ $$= \sin A\cos A\frac{1 + \sin A - \cos A + 1 + \cos A - \sin A}{(1+\sin A - \cos A)(1 + \cos A - \sin A)} = $$ $$=\frac{2\sin A \cos A}{1 + \cos A - \sin A + \sin A + \sin A\cos A -\sin^2 A -\cos A -\cos^2 A +\sin A\cos A} = $$ $$=\frac{2\sin A \cos A}{2\sin A\cos A} = 1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/877936",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 4
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In a triangle ABC, prove that cot(A/2)+cot(B/2)+cot(C/2) =cot(A/2)cot(B/2)cot(C/2) In a triangle ABC, prove that $\cot \left ( \frac{A}{2} \right )+\cot \left ( \frac{B}{2} \right )+\cot \left ( \frac{C}{2} \right )=\cot \left ( \frac{A}{2} \right )\times \cot \left ( \frac{B}{2} \right )\times \cot \left ( \frac{C}{2} \right )$. I tried all identities I know but I have no idea how to proceed.
| Geometric Proof
Let $a,b,c$ be the length of the sides $BC,CA,AB$ respectively, $p$ be the semi-perimeter, $r$ be the radius of the incircle, and $S$ be the area.
Notice that $$\cot\frac{A}{2}=\frac{p-a}{r},~~~\cot\frac{B}{2}=\frac{p-b}{r},~~~\cot\frac{C}{2}=\frac{p-c}{r}.$$
Hence, $$\cot\frac{A}{2}+\cot\frac{B}{2}+\cot\frac{C}{2}=\frac{p}{r}.$$
and $$\cot\frac{A}{2}\cot\frac{B}{2}\cot\frac{C}{2}=\frac{(p-a)(p-b)(p-c)}{r^3}.$$
Recall Heron's formula $$S=\sqrt{p(p-a)(p-b)(p-c)},$$ and another area formula$$S=pr.$$
We obtain $$\frac{(p-a)(p-b)(p-c)}{r^3}=\frac{S^2}{pr^3}=\frac{(pr)^2}{pr^3}=\frac{p}{r}.$$
As a result, $$\cot\frac{A}{2}+\cot\frac{B}{2}+\cot\frac{C}{2}=\cot\frac{A}{2}\cot\frac{B}{2}\cot\frac{C}{2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/878577",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove ${\large\int}_{-1}^1\frac{dx}{\sqrt[3]{9+4\sqrt5\,x}\ \left(1-x^2\right)^{2/3}}=\frac{3^{3/2}}{2^{4/3}5^{5/6}\pi }\Gamma^3\left(\frac13\right)$ Here is one more conjecture I discovered numerically:
$${\large\int}_{-1}^1\frac{dx}{\sqrt[3]{9+4\sqrt5\,x}\ \left(1-x^2\right)^{\small2/3}}\stackrel{\color{#808080}?}=\frac{3^{\small3/2}}{2^{\small4/3}\,5^{\small5/6}\,\pi }\Gamma^3\!\!\left(\tfrac13\right)$$
How can we prove it?
Note that $\sqrt[3]{9+4\sqrt5}=\phi^2$.
Mathematica can evaluate this integral, but gives a large expression in terms of Gauss and Appel hypergeometric functions of irrational arguments.
| As suggested by Chen Wang, this integral is related to an integral of the form
$$
\int_0^1 \frac{dx}{ \sqrt{1-x} \; x^{2/3} (1-zx)^{1/3} }
$$
which appears in this forum's
Question 879089
by the same author.
Here's a more direct route than given by Kirill, without
conversions between definite integrals and hypergeometric functions.
A differential $dx \left/ \left(\sqrt[3]{A+Bx} \, (1-x^2)^{2/3}\right)\right.$
has four poles of fractional order, two at $x = \pm 1$ of order $2/3$,
and two at $x = \infty$ and $x = -B/A$ of order $1/3$.
Hence the differential is invariant under an involution
in the form of a fractional linear transformation that switches
$-1 \leftrightarrow +1$ and $\infty \leftrightarrow -B/A$.
To exploit this symmetry it is convenient to use a coordinate
$u = (cx+1)/(x+c)$: for any $c$ we have $u=\pm 1$ at $x = \pm 1$
and $u = c$ at $x = \infty$, and we choose $c>1$ so that $u=-c$ at $x=-B/A$.
Explicitly, this change of variable gives
$$
{\large\int}_{-1}^1
\frac{dx}{\sqrt[3]{c^2+1+2cx}\ \left(1-x^2\right)^{\small2/3}}
=
{\large\int}_{-1}^1
\frac{du}{\sqrt[3]{c^2-u^2}\ \left(1-u^2\right)^{\small2/3}} .
$$
By symmetry the second $\int_{-1}^1$ is $2\int_0^1$ of the same integrand,
and then the change of variable $u^2 = t$ gives
$$
{\large\int}_0^1 \frac{dt}{\sqrt{t} \, (c^2-t)^{\small1/3} (1-t)^{\small2/3}}.
$$
Now integrating with respect to $X=1-t$ instead of $t$ gives
$$
{\large\int}_0^1 \frac{dX}{\sqrt{1-X} \; X^{2/3}
\left(c^2-1+X\right)^{\small1/3}},
$$
and we have reached an integral of the desired form.
In the present case we take $c = \frac12 \! \sqrt 5$.
Then $c^2+1 + 2cx = (9 + 4\sqrt{5}x)/4$, so our integral is
$4^{1/3}$ times the one with $\sqrt[3]{9 + 4\sqrt{5}x}$ in the denominator.
Hence $c^2-1 = 1/4$, so we gain another factor of $4^{1/3}$ and we've reached
the integral in the first display with $z=-4$. Vladimir Reshetnikov
already noted in a comment to his question that the integral seemed to
have an exact value at $z = -4$, and the analysis I gave in my answer there
(or Kirill's answer here) leads to the period of a CM elliptic curve
with $j$-invariant $146329141248 \sqrt{5} - 327201914880$; this curve is
related by $5$-isogeny with a curve of $j$-invariant zero, and thus
has a period proportional to a Beta integral.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/879854",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "29",
"answer_count": 2,
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Find modulus of $z$ given modulus of $(z-3w)/(3-z\overline{w})$ Question:
(23) If $z_1$, $z_2$ are complex numbers such that $\left|\dfrac{z_1-3z_2}{3-z_1\overline{z}_2}\right|=1$ and $|z_2|\neq 1$, then find $|z_1|$.
How would I attempt this question? I tried using values for $z_1$ and $\overline{z}_2$ but it is coming out to be extremely lengthy. There is probably a quicker solution to this. Can someone provide me with a hint to start?
| $$\frac{z_1 - 3z_2}{3 - z_1 \overline{z_2}} = \frac{\frac{z1}{3} - z_2}{1 - \frac{z1}{3} \overline{z_2}}$$
$b = \overline{z_1} / 3, a = z_2 \Rightarrow |a - \overline{b}| = |1 - ab|$ and we should find $b$.
Let's look this problem geometrically and consider complex numbers $a, b$ as two-dimensional vectors (common approach).
Main idea is that ($Q \angle W$ denotes angle between $Q$ and $W$)
$$a \angle \overline{b} = 1 \angle ab$$
Explanation: Let $\arg(a) = \alpha, \arg(b) = \beta$. Then $\arg(ab) = \alpha + \beta, \arg(1) = 0, \arg(\bar{b}) = -\beta$. Hence $a \angle \overline{b} = \alpha + \beta = 1 \angle ab$.
Let $A = |a|, B = |b|$. Having angles equation, we use law of cosines for corresponding triangles and get (|ab| = AB, |1| = 1)
$$(AB) ^ 2 + 1^2 - 2 AB \cos(\alpha + \beta) = A^2 + B^2 - 2 AB \cos (\alpha + \beta)$$
$$A^2 B^2 + 1 = A^2 + B^2$$
$$B^2 (A^2 - 1) = A^2 - 1$$
Given $1 \neq |z_2| = |a| = |A|$,
$$B^2 = 1$$
Therefore, $B = 1$ because it is be positive.
But $1 = B = |\overline{z_1} / 3| = |z_1| / 3$, so answer is 3.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/880453",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
A strange trigonometric equation Today,in our class, we received a trigonometric equation
$$\sin^{10}{x}+\cos^{10}{x}=\frac{29}{16}\cos^4{2x}$$
and the question was to find the general solution of this equation. My approach was, at First, trying to show that there were no solutions using inequalities, but I failed. So, my last method was, expanding RHS by binomial theorem, and canceling some terms out, which at last gives a quadratic in $\sin{x}\cos{x}$. But this way was too long.
Can anyone suggest or give a simpler method? I firmly believe there's one trick in ques to make it easier, which I cannot solve.
| We know that $X^5+Y^5=(X+Y)(X^4-X^3Y+X^2Y^2-XY^3+Y^4)$; if we set $X=\sin^2x$ and $Y=\cos^2x$, the left hand side becomes
$$
(\sin^2x+\cos^2x)(\sin^8x-\sin^6x\cos^2x+\sin^4x\cos^4x-\sin^2x\cos^6x+\cos^8x)
$$
or
$$
\sin^8x+\cos^8x+\sin^4x\cos^4x-\sin^2x\cos^2x(\sin^4x+\cos^4x)
$$
The first three terms can be written
$$
(\sin^4x+\cos^4x)^2-\sin^4x\cos^4x
$$
and
$$
\sin^4x+\cos^4x=(\sin^2x+\cos^2x)^2-2\sin^2x\cos^2x=1-2\sin^2x\cos^2x
$$
so we get
$$
(1-2\sin^2x\cos^2x)^2-\sin^4x\cos^4x-\sin^2x\cos^2x(1-2\sin^2x\cos^2x)
$$
which further simplifies into
$$
1-4\sin^2x\cos^2x+4\sin^4x\cos^4x-\sin^4x\cos^4x
-\sin^2x\cos^2x+2\sin^4x\cos^4x
$$
that is
$$
1-5\sin^2x\cos^2x+5\sin^4x\cos^4x
$$
or
$$
1-\frac{5}{4}\sin^2(2x)+\frac{5}{16}\sin^4(2x)
$$
Can you go on?
| {
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"url": "https://math.stackexchange.com/questions/880723",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 2
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How do you prove this integral involving the Glaisher–Kinkelin constant According to wikipedia on the page Glaisher–Kinkelin constant
$$\int_0^{1/2} \ln\Gamma(x) dx=\frac32\ln \text{A}+\frac5{24}\ln 2+\frac14\ln\pi$$
I got interested in how you possibly could prove something like that, but couldn't find any citations about it on the wiki page.
| Recall Kummer´s fourier expansion for LogGamma $0<x<1$
$$\ln\left(\Gamma(x)\right)=\frac{\ln\left(2 \pi\right)}{2}+ \sum_{k=1}^{\infty}\frac{1}{2k}\cos\left(2 \pi k x\right) + \sum_{k=1}^{\infty}\frac{\gamma+\ln\left(2 \pi k\right)}{ \pi k}\sin\left(2 \pi k x \right) \tag{1}$$
Integrating $(1)$ we obtain
$$\int_0^x\ln\left(\Gamma(u)\right)du=\frac{x \ln\left(2 \pi\right)}{2}+\frac{1}{4 \pi}\sum_{k=1}^{\infty}\frac{\sin\left(2 \pi k x \right)}{k^2} +\frac{\gamma}{2 \pi^2}\sum_{k=1}^{\infty}\frac{1}{k^2}-\frac{\gamma}{2 \pi^2}\sum_{k=1}^{\infty}\frac{\cos \left(2 \pi k x \right)}{k^2}+\frac{1}{2 \pi^2}\sum_{k=1}^{\infty}\frac{\ln \left( 2 \pi k\right)}{k^2}-\frac{1}{2 \pi^2}\sum_{k=1}^{\infty}\frac{\ln \left( 2 \pi k\right) \cos\left(2 \pi k x \right)}{k^2} \tag{2}$$
Letting $x=\frac12$ in $(2)$ we obtain
$$
\begin{aligned}
\int_0^{1/2}\ln\left(\Gamma(x)\right)dx&=\frac{ \ln\left(2 \pi\right)}{4} +\frac{\gamma}{2 \pi^2}\zeta(2)+\frac{\gamma}{2 \pi^2}\eta(2)+\frac{\ln(2\pi)}{2 \pi^2}\zeta(2)+\frac{1}{2 \pi^2}\sum_{k=1}^{\infty}\frac{\ln \left( k\right)}{k^2}+\frac{\ln \left( 2 \pi \right)}{2 \pi^2}\eta(2)-\frac{1}{2 \pi^2}\sum_{k=1}^{\infty}\frac{\ln \left( k \right) \left(-1 \right)^{k}}{k^2}\\
&=\frac{ \ln\left(2 \pi\right)}{4} +\frac{\gamma}{12 }+\frac{\gamma}{24}+\frac{\ln(2\pi)}{12 }-\frac{1}{2 \pi^2}\zeta^{\prime}(2)+\frac{\ln \left( 2 \pi \right)}{24}-\frac{1}{2 \pi^2}\eta^{\prime}(2)\\
&=\frac{\gamma}{8}+\frac{3 \ln(2 \pi)}{8}-\frac{1}{2 \pi^2} \cdot \frac{\pi^2}{6}\left(\ln(2 \pi)+\gamma-12 \ln A \right)-\frac{1}{2 \pi^2} \left(\frac{\pi^2 \ln(2 )}{12} +\frac{\pi^2}{12}\left(\ln(2 \pi)+\gamma-12 \ln A\right)\right)\\
&=\frac{\gamma}{8}+\frac{3 \ln(2 \pi)}{8}-\frac{\ln(2 \pi)}{12}-\frac{\gamma}{12}+\ln A-\frac{\ln (2)}{24}-\frac{\ln(2 \pi)}{24}-\frac{\gamma}{24}+\frac{\ln A}{2}\\
&=\left(\frac{1}{8}-\frac{1}{12}-\frac{1}{24} \right)\gamma+\left(\frac38-\frac{1}{12}-\frac{1}{24} \right)\ln(2 \pi)-\frac{\ln(2)}{24}+\ln A+\frac{\ln A}{2}\\
&=\frac{3}{2}\ln A+ \frac{6}{24}\ln (2 \pi)-\frac{\ln (2)}{24}\\
&=\frac{3}{2}\ln A+\frac{5}{24}\ln 2+\frac{1}{4}\ln \pi \qquad \blacksquare
\end{aligned}
$$
We used that $\zeta^{\prime}(2)=\frac{\pi^2}{6}\left(\ln 2 \pi + \gamma -12 \ln A \right)$ and that $\eta^{\prime}(s)=2^{1-s}\ln (2) \zeta(s)+(1-2^{1-s})\zeta^{\prime}(s)$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Series for logarithms This is more of a challenge than a question, but I thought I'd share anyway. Prove the following identities, and prove that the pattern continues.
\begin{equation*}
\sum_{n=0}^\infty\left(\frac{1}{2n+1}-\frac{1}{2n+2}\right)=\ln2
\end{equation*}\begin{equation*}
\sum_{n=0}^\infty\left(\frac{1}{3n+1}+\frac{1}{3n+2}-\frac{2}{3n+3}\right)=\ln3
\end{equation*}\begin{equation*}
\sum_{n=0}^\infty\left(\frac{1}{4n+1}+\frac{1}{4n+2}+\frac{1}{4n+3}-\frac{3}{4n+4}\right)=\ln4
\end{equation*}\begin{equation*}
\mathrm{etc.}
\end{equation*}
By the way, a good notation for discussing this problem can be found here. The problem is to prove that:
$[\overline{1,-1}]=\ln2,\;\;$ $[\overline{1,1,-2}]=\ln3,\;\;$ $[\overline{1,1,1,-3}]=\ln4,\;\;$ $[\overline{1,1,1,1,-4}]=\ln5,\;\;$ etc.
| For any $m \in \mathbb{N},$
$$\sum_{k=0}^{\infty} \Big(\frac{1}{mk+1} + \frac{1}{mk+2} + ... + \frac{1}{mk+m-1}- \frac{m-1}{mk+m}\Big)$$ $$= \lim_{n \rightarrow \infty} \Big(\sum_{k=0}^{mn} \frac{1}{mk+1} + ... + \frac{1}{mk+m-1} - \frac{m-1}{mk+m}\Big)$$$$= \lim_{n \rightarrow \infty} \Big( \sum_{k=1}^{mn} \frac{1}{k} - m \sum_{k=1}^n \frac{1}{mk} \Big)$$ $$= \lim_{n \rightarrow \infty} \Big(\sum_{k=1}^{mn} \frac{1}{k} - \sum_{k=1}^n \frac{1}{k}\Big)$$ $$= \lim_{n \rightarrow \infty} \Big( \sum_{k=1}^{mn} \frac{1}{k} - \ln(mn) + \ln(mn) - \sum_{k=1}^n \frac{1}{k} \Big)$$$$= \lim_{n \rightarrow \infty} \Big( \sum_{k=1}^{mn} \frac{1}{k} - \ln(mn) + \ln(n) - \sum_{k=1}^n \frac{1}{k} \Big) + \ln(m)$$ $$= \gamma - \gamma + \ln(m) = \ln(m),$$ where $\gamma$ is the Euler-Mascheroni constant.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/883348",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
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Prove that the expression is Independent of theta Prove that $2\sin^2 \theta +4 \cos(\theta + \alpha ) \sin \alpha \sin \theta +\cos(2(\theta + \alpha))$ is independent of $\theta$.
How do we solve such problems ?
| $$2\sin^2\theta+4\sin\alpha\sin\theta\cos(\theta+\alpha)+\cos(2\theta+2\alpha)$$
$$=2\sin^2\theta+4\sin\alpha\sin\theta(\cos\theta\cos\alpha-\sin\theta\sin\alpha)+\cos2\theta\cos2\alpha-\sin2\theta\sin2\alpha$$
$$=2\sin^2\theta+4\sin\alpha\cos\alpha\sin\theta\cos\theta-4\sin^2\alpha\sin^2\theta+(1-2\sin^2\theta)(1-2\sin^2\alpha)-4\sin\alpha\cos\alpha\sin\theta\cos\theta$$
$$=2\sin^2\theta-4\sin^2\alpha\sin^2\theta+1-2\sin^2\alpha-2\sin^2\theta+4\sin^2\alpha\sin^2\theta$$
$$=1-2\sin^2\alpha=\cos2\alpha.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/883538",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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A double sum with combinatorial factors Let $n$, $p$ and $j$ be integers. As a byproduct of some other calculations I have discovered the following identity:
\begin{equation}
\sum\limits_{p=0}^{j} \sum\limits_{p_1=0}^j \binom{p+p_1}{p_1} \binom{j}{p_1} \binom{2n-j}{j-p} \binom{n-j+p}{1+p+p_1} (-1)^p = \frac{1}{2} \binom{2n}{2j+1}
\end{equation}
I used Mathematica to check that it holds true. The problem is how do I go about proving it?
| This is not going to be a complete answer but since this approach seems to me intuitive i will show it. This is a sort of ``by induction proof''. Let us look at the lhs for the biggest possible $j$ and then go down in $j$. We hope to see a pattern as a function of $j$ and thus guess the result.
So we take $j=n-1$.We have:
\begin{equation}
S(j=n-1)= \sum\limits_{p=0}^{n-1} \sum\limits_{p_1=0}^{n-1} \binom{p+p_1}{p_1} \binom{n-1}{p_1} \binom{n+1}{n-1-p} \binom{1+p}{1+p+p_1}
\end{equation}
Since the last binomial factor on the right equals $\delta_{p_1,0}$, we have:
\begin{equation}
S(j=n-1)= \sum\limits_{p=2}^{n+1} \binom{n+1}{p} (-1)^p = \frac{1}{2} \binom{2 n}{1} = rhs
\end{equation}
Now we take $j=n-2$.We have:
\begin{equation}
S(j=n-2)= \sum\limits_{p=0}^{n-2} \sum\limits_{p_1=0}^{n-2} \binom{p+p_1}{p_1} \binom{n-2}{p_1} \binom{n+2}{n-2-p} \binom{2+p}{1+p+p_1}
\end{equation}
Since the last binomial factor on the rhs equals $\delta_{p_1,0} (2+p) + \delta_{p_1,1}$ we have:
\begin{equation}
S(j=n-2)= \sum\limits_{p=4}^{n+2} \binom{n+2}{p}\left[(p-2) + (p-3)(n-2)\right](-1)^p = \frac{1}{2} \binom{2n}{3} = rhs
\end{equation}
Now we take $j=n-3$. We have:
\begin{equation}
S(j=n-3)= \sum\limits_{p=0}^{n-3} \sum\limits_{p_1=0}^{n-3} \binom{p+p_1}{p_1} \binom{n-3}{p_1} \binom{n+3}{n-3-p} \binom{3+p}{1+p+p_1}
\end{equation}
Since the last binomial factor on the rhs equals $\delta_{p_1,0} \binom{p+3}{2} + \delta_{p_1,1} \binom{p+3}{1} + \delta_{p_1,2}$ we have:
\begin{align}
S(j=n-3)
&= \sum\limits_{p=6}^{n+3} \binom{n+3}{p}\Big[\binom{p-3}{2}
+ \binom{p-3}{1}(p-5)(n-3) \\& \hspace{3cm} + \binom{p-4}{2} \binom{n-3}{2}\Big](-1)^p = \frac{1}{2} \binom{2n}{5} = rhs
\end{align}
As we go down in $j$ we always do the sum over $p_1$ using properties of the delat function and we end up with a couple of sums over $p$ each of which are elementary. The only thing that remains is to write down all those sums and compute them. Yet, since the pattern has already emerged I quit at this stage..
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/883802",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
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