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Solving a recurrence with a term $T(\frac n 2 + 2)$ I'm stuck trying to solve the following recurrence:
$$\begin{align*}
T(n) &= 4T(\frac n 2 + 2) + n : n > 8\\
T(n) &= 1 : n \leq 8
\end{align*}$$
In particular, I'm not sure how to deal with the expression $T(\frac n 2 + 2)$. I'm trying to use a recursion tree. I've found that the "cost" at the $k$th level of the tree is
$$4^k\left(\frac n {2^k} + \sum^k_{i = 1} 2^{2- i}\right)$$
by looking at the cost of the first few levels of the tree, which I understand to be:
$$4(\frac n 2 + 2)\\
16(\frac n {2^2} + 2 + 1)\\
64(\frac n {2^3} + 2 + 1 + \frac 1 2)\\
256(\frac n {2^4} + 2 + 1 + \frac 1 2 + \frac 1 4)$$
etc. So the whole cost would be
$$T(n) = \sum_{k = 1}^{\log_2 n} 4^k\left(\frac n {2^k} + \sum^k_{i = 1} 2^{2- i}\right)$$
but I'm not at all clear how to go from this to a nice, closed-form expression, or how to deal with the value of $T(n)$ over the range $[0, 8]$.
What's the right way to approach this?
| For the complementary solution part,
$T_c(n)=4T_c\left(\dfrac{n}{2}+2\right)$
$T_c(2^n+4)=4T_c\left(\dfrac{2^n+4}{2}+2\right)$
$T_c(2^n+4)=4T_c(2^{n-1}+4)$
$T_c(2^n+4)=4^nC$
$T_c(n)=(n-4)^2C$
For the particular solution part,
Let $T_p(n)=An+B$ ,
Then $An+B-4\left(A\left(\dfrac{n}{2}+2\right)+B\right)\equiv n$
$-An-8A-3B\equiv n$
$\therefore\begin{cases}-A=1\\-8A-3B=0\end{cases}$
$\begin{cases}A=-1\\B=\dfrac{8}{3}\end{cases}$
$\therefore T(n)=(n-4)^2C-n+\dfrac{8}{3}$
$T(8)=1$ :
$16C-\dfrac{16}{3}=1$
$C=\dfrac{19}{48}$
$\therefore T(n)=\dfrac{19(n-4)^2}{48}-n+\dfrac{8}{3}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/974636",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Is $\frac{1}{\lfloor \frac{a}{b} \rfloor}=\lceil \frac{b}{a} \rceil$?? Is $\frac{1}{\lfloor \frac{a}{b} \rfloor}=\lceil \frac{b}{a} \rceil$? assuming that $\frac{a}{b}>1$.
| In the case $\displaystyle\frac{a}{b}\ge2$ we get $\displaystyle\lfloor{\frac{a}{b}\rfloor}\ge2\Rightarrow\frac{1}{\lfloor{\frac{a}{b}\rfloor}}\le\frac{1}{2}$ and $\displaystyle\frac{b}{a}\le\frac{1}{2}\Rightarrow\lceil{\frac{b}{a}\rceil}=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/976820",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Logic behind the multiplicative form of "Gauss trick"? I was going through $ \lim\limits_{n \to{+}\infty}{\sqrt[n]{n!}}$ is infinite
I don't follow why/how this is true $(n!)^2 = (1 \cdot n) (2 \cdot (n-1)) (3 \cdot (n-2)) \cdots ((n-2) \cdot 3) ((n-1) \cdot 2) (n \cdot 1)\ge n^n$
I read that At least half the terms in the product for n! will be at least n/2 .. so ?
Please advise.
| If $k \le \frac{n-1}2$, then $4k^2 \le n^2 - 2n + 1$, and so
\begin{align}
\left(\frac{n + 1}2 + k\right)\left(\frac{n + 1}2 - k\right)
& = \left(\frac{n+1}2\right)^2 - k^2 \\
& = \frac{n^2 + 2n + 1 - 4k^2}{4} \\
& \ge \frac{n^2 + 2n + 1 - (n^2 - 2n + 1)}{4} \\
& = n.
\end{align}
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Show that if x,y are and $ x^4y^2+x^2+2x^3y+6x^2y+8 \leq 0 $ then $x \geq -1/6 $ Show that if x,y are real and $ x^4y^2+x^2+2x^3y+6x^2y+8 \leq 0 $ then $x \geq -1/6 $
So far I've tried factoring $x^2$ and throwing the 8 on the LHS, but can't get to the needed result. Help would be appreciated
| Completing the square we have $$(x^2y)^2+2(x^2y)x+x^2+2(x^2y)3+3^2+6x\le(1+6x)$$
$$\iff1+6x\ge(x^2y+x+3)^2$$ which is $\ge0$ for real $x,y$
| {
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"timestamp": "2023-03-29T00:00:00",
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How do you cancel the common factor of $\frac{\sqrt[3]{x}+2\sqrt{x}-3}{x-1}$ when faced with this kind of limit: $\lim _{x\to 1}\left(\frac{\sqrt[3]{x}+2\sqrt{x}-3}{x-1}\right)$
i know i can use l'hopital's rule to solve it because we get $\frac{0}{0}$ replacing $x$ with $1$, still i want to learn the other method which is cancelling the common factor, in this case i know the common factor is $x-1$
with normal polynomial function i use Euclidean division, tried it with square root but didn't seem to work.
a limit calculator gave me this answer:
$\mathrm{Cancel\:the\:common\:factors\:of}\:\frac{\sqrt[3]{x}+2\sqrt{x}-3}{x-1}$ = $\frac{2\sqrt[3]{x}+3x^{\frac{1}{6}}+3}{x^{\frac{5}{6}}+x^{\frac{2}{3}}+\sqrt{x}+\sqrt[3]{x}+x^{\frac{1}{6}}+1}$
but i don't understand yet how it is done.
| It'll be much easier to visualize if you rewrite $x^{1/6}$ as $y$
$$\frac{2y^3+y^2-3}{y^6-1}=\frac{(y-1)(2y^2+3y+3)}{(y-1)(y^5+y^4+y^3+y^2+y+1)}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Boolean Simplification of $ (a+b) \cdot (a \cdot c + a \cdot \overline{c}) + a \cdot b + b $ Below is my simplification, but my truth tables don't line up, but I can't find my error.
$ (a+b) \cdot (a \cdot c + a \cdot \overline{c}) + a \cdot b + b $
$ (a+b) \cdot a \cdot (c + \overline{c}) + a \cdot b + b $
$ (a+b) \cdot a + a \cdot b + b $
$ (a+b) \cdot a \cdot b $
| I think that $a\cdot b + b = b$. So I think your fourth line should be $(a+b)\cdot a + b$.
| {
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How does one evaluate $\int \frac{\sin(x)}{\sin(5x)} \ dx$
*
*The below problem is taken from Joseph Edwards book Integral Calculus for beginners.
How does one show:
$$5 \int \frac{\sin(x)}{\sin(5x)} \ dx= \sin\left(\frac{2\pi}{5}\right) \cdot \log\left\{\frac{\sin\left(x-\frac{2\pi}{5}\right)}{\sin\left(x+\frac{2\pi}{5}\right)}\right\} -\sin\left(\frac{\pi}{5}\right) \cdot \log\left\{\frac{\sin\left(x-\frac{\pi}{5}\right)}{\sin\left(x+\frac{\pi}{5}\right)}\right\} $$
*
*Splitting $\sin{(5x)}$ as $\sin{(4x+x)}$ doesn't seem to be of much help since then we have a big term in the denominator after expansion.
| Wolfram Mathematica $9.0$ is able to evaluate this indefinite integral. Here is the output
\begin{equation}
\sqrt{\frac{5+\sqrt{5}}{2}}{\rm{artanh}}\left(\frac{\left(\sqrt{5}-3\right)\tan x}{\sqrt{10-2\sqrt{5}}}\right)+\sqrt{\frac{5-\sqrt{5}}{2}}{\rm{artanh}}\left(\frac{\left(\sqrt{5}+3\right)\tan x}{\sqrt{10+2\sqrt{5}}}\right)+C
\end{equation}
The complete proof can be downloaded here.
Using the facts
\begin{align}
\sin\left(\frac{\pi}{5}\right)&=\frac{1}{2}\sqrt{\frac{5-\sqrt{5}}{2}}\\[10pt]
\cos\left(\frac{\pi}{5}\right)&=\frac{\sqrt{5}+1}{4}\\[10pt]
\sin\left(\frac{2\pi}{5}\right)&=\frac{1}{2}\sqrt{\frac{5+\sqrt{5}}{2}}\\[10pt]
\cos\left(\frac{2\pi}{5}\right)&=\frac{\sqrt{5}-1}{4}\\[10pt]
\sin\left(\alpha\pm\beta\right)&=\sin\alpha\cos\beta\pm\cos\alpha\sin\beta\\[10pt]
{\rm{artanh}}\,\theta&=\frac{1}{2}\ln\left(\frac{1+\theta}{1-\theta}\right)\,,\quad\mbox{for}\,|\theta|<1
\end{align}
We can derive the following answer
\begin{equation}
5 \int \frac{\sin(x)}{\sin(5x)} \ dx= \sin\left(\frac{2\pi}{5}\right) \cdot \ln\left\{\frac{\sin\left(x-\frac{2\pi}{5}\right)}{\sin\left(x+\frac{2\pi}{5}\right)}\right\} -\sin\left(\frac{\pi}{5}\right) \cdot \ln\left\{\frac{\sin\left(x-\frac{\pi}{5}\right)}{\sin\left(x+\frac{\pi}{5}\right)}\right\}+C
\end{equation}
| {
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"timestamp": "2023-03-29T00:00:00",
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Assume that 495 divides the integer $\overline{273x49y5}$ where $x,y \in \{0,1,2...9\}$. Find $x$ and $y$. So, I know that $495 = 5\times 9\times 11$. So then, if that's the case, then the number $\overline{273x49y5}$ must be divisible by $495$ if and only if it is divisible by $5$ and $9$ and $11$.
Then, I know that $5$ divides it for any number $x$ and $y$ because it only has to divide the digits number which is $5$.
To see if it divides by $9$, the sum of the digits must be divisible by $9$, and as for $11$, the alternating sum of digits must be divisible by $11$.
So, to test it by $9$, then, $2+7+3+x+4+9+y+5 = 30+x+y$ has be divisble by $9$. So, the only numbers $x$ and $y$ that make $30+x+y$ divisible by $9$ are $x= \{0,1,2,3,4,5,6\}$ and $y=\{0,1,2,3,4,5,6\}$.
But $11$ is what gets me. To make it divisible by $11$, then the alternating sums must be divisible by $11$. So then, $-2+7-3+x-4+9-y+5= 12+x-y$ must be divided by $11$. Then, the only numbers $x$ and $y$ that make it divisible by $11$ must be $x=\{0,1,2,\ldots,8\}$ and $y=\{1,2,\ldots,9\}$, if I did my math right.
Next, I assume we must then find numbers in $x$ and $y$ that satisfy the division by $9$ and $11$. So, if that is the case, then I know that since the number must be divisible by $9$ and $11$, then the numbers $x$ and $y$ must fit between $0$ to $6$ because that's $6$ is the highest number $x$ or $y$ can be in the sequence to divide the number. So, how do I check to see which numbers both satisfy division of $9$ and $11$?
| $30+x+y$ doesn't have to equal 9; it has to equal a multiple of 9. So it could be $36$ or $45$ or $54\ldots$. And $x$ could be $7$, if $y$ is 8, for example. But they can't be anything at all; if $x$ is 7 then $y$ must be $8$; no other value of $y$ works. And if $x$ is 3 then $y$ must be $3$ also. So make a table that shows for each $x$, what can $y$ be, if $30+x+y$ is to be a multiple of 9.
Then do the same for the $12+x-y$ must be divisible by $11$: if $x$ is 1, then $y$ must be $2$, and so on.
Then see which pairs are in both tables.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Finding the area between two equations The question asks to find the area of the part of the first Quadrant that lies between
$x+4y-5=0$ and $xy=1$.
I began by first changing the equations to functions of $y$:
$x+4y-5=0$ to $y=\frac{5-x}{4}$ and $xy=1$ to $y=\frac{1}{x}$
I am not sure how to find the area, can you help me?
| Set them equal to one another to find the intersection:
\begin{equation}
\frac{\left(5-x\right)}{4}=\frac{1}{x}\implies 5x-x^2=4\implies 0=x^2-5x+4,
\end{equation}
\begin{equation}
0=\left(x-4\right)\left(x-1\right)\implies x=\left\{4,1\right\}.
\end{equation}
And this matches the below graph:
Now to find the area between them you must subtract the area under the lower curve from the area under the upper-curve:
\begin{equation}
\int\limits_{1}^{4}\left(\frac{5-x}{4}-\frac{1}{x}\right)dx=\int\limits_1^4\left(\frac{5}{4}-\frac{x}{4}\right)dx-\int\limits_1^4\left(\frac{1}{x}\right)dx=\left[\frac{5x}{4}-\frac{x^2}{2}-\log\left|x\right|\right]_1^4
\end{equation}
and I'll let you evaluate...
EDIT:
You may also integrate with respect to $y$ by putting the values of $x$ back into the equations above and solving for the associated $y$ values and using those for your limits of integration. Then you will notice that the order of subtraction swaps, so you are finding the integral of the yellow curve in the graph above minus the integral of the blue curve. These will yield the same answer.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Weak form of $-\frac{d}{dx} \left[(1+x) \frac{du}{dx}\right]$ I am doing some self studying on J.N. Reddy's book, An Introduction to the finite element method, 1st edition. (This was the only one available at the local library).
In exercise 2.12, one reads
Compute the coefficient matrix and the right size of the N-parameter Ritz system of the equation
\begin{aligned}
\frac{d}{dx} \left((1+x) \frac{du}{dx}\right) & = 0 \\ 0<x<1
\end{aligned}
with conditions
$u(0)$ = 1 and $u(1)=1$.
In the answer, it says that the right-size N parameter is
\begin{aligned}
l_i = \frac{1}{(i+1)(i+2)}
\end{aligned}
However, I always find $l_i = 0$.
My procedure was the following.
Using Ritz method,
\begin{aligned}
0 = -\int_0^1 v\frac{d}{dx}\left((1+x)\frac{du}{dx}\right)dx
\end{aligned}
Integrating by parts,
\begin{aligned}
0 = -\int_0^1 v\frac{d}{dx}\left((1+x)\frac{du}{dx}\right)dx = -v(1+x)\frac{du}{dx} |_0^1 + \int^1_0 (1+x)\frac{dv}{dx}\frac{du}{dx}dx
\end{aligned}
As $u$ is specified at $x=0$ and $x=1$, $v(0) = v(1) = 0$. Therefore
\begin{aligned}
-v(1+x)\frac{du}{dx} |_0^1 = 0
\end{aligned}
And we are left with
\begin{aligned}
0 = -\int_0^1 v\frac{d}{dx}\left((1+x)\frac{du}{dx}\right)dx = \int^1_0 (1+x)\frac{dv}{dx}\frac{du}{dx}dx
\end{aligned}
Thus the Bilinear form is
\begin{aligned}
B(u,v) = \int^1_0 (1+x)\frac{dv}{dx}\frac{du}{dx}dx
\end{aligned}
and
\begin{aligned}
l(v) = 0
\end{aligned}
Could any one tell me what I got wrong?
| I understood my mistake, and it was caused by misunderstanding the notation.
Given an approximate answer
\begin{align}
u_N = \phi_0 + \sum_{i=1}^N c_i\phi_i
\end{align}
We cannot say that $l_i = l(\phi_i)$, but
\begin{align}
l_i = l(\phi_i) - B(\phi_i, \phi_0)
\end{align}
That way, choosing $\phi_0 = x$ and $\phi_i = x^i(1-x)$, we find
\begin{align}
l_i =-B(\phi_i, \phi_0) = -\int_0^1 x (1-x) x^i dx = -\int_0^1 x^{i+1}-x^{i+2} dx = -\frac{1}{i+1} + \frac{1}{i+2} = \frac{1}{(i+1)(i+2)}
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
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How to determine the Jordan form and give a Jordan base for a matrix? given is
$\begin{pmatrix} 3&0&-1&0&0 \\ 1&3&0&1&0 \\ 0&0&3&0&0 \\ 0&0&0&3&0 \\ 0&0&0&0&-3 \end{pmatrix}$
I have to determine the jordan form and also give a jordanbase.
I got this so far:
*
*Eigenvalues are
$l_1 = 3, l_2 = -3$ whereas $P_A(x) = (x-3)^4 * (x+3)$.
*Eigenvectors are
for $l_1 = 3$:
$(0,1,0,0,0)$ and $(-1,0,0,1,0)$
for $l_2 = -3$:
$(0,0,0,0,1)$
*This means, that we still need two vectors. How can I get them?
*How can I get then the jordanbase?
| To get the full base, you need the generalized eigenvectors. In this case, we do so as follows:
I define
$$
A =
\pmatrix{
3&0&-1&0&0 \\
1&3&0&1&0 \\
0&0&3&0&0 \\
0&0&0&3&0 \\
0&0&0&0&-3
}
$$
We find non-zero solutions to the following:
$$
(A - 3I)x = \pmatrix{0\\1\\0\\0\\0} \implies
x = \pmatrix{1\\0\\0\\0\\0}\\
(A - 3I)x = \pmatrix{-1\\0\\0\\1\\0} \implies
\text{no solution}\\
(A - 3I)x = \pmatrix{1\\0\\0\\0\\0} \implies
x = \pmatrix{0\\0\\-1\\0\\0}\\
$$
And we now have a base. Our Jordan form $J = S^{-1}AS$ must be
$$
J = \pmatrix{3&1\\&3&1\\&&3\\&&&3\\&&&&-3}
$$
with base
$$
S = \pmatrix{0&1&0&-1&0\\1&0&0&0&0\\0&0&-1&0&0\\0&0&0&1&0\\0&0&0&0&1}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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How to find the equation of lines passing through the origin and perpendicular to the lines $xy-3y^2+y-2x+10=0$ Problem :
How to find the equation of lines passing through the origin and perpendicular to the lines $xy-3y^2+y-2x+10=0$
My working on this :
Two lines are perpendicular if the sum of coefficient of $x^2$ and $y^2$ is equal to zero.
| HINT:
$$3y^2-y(x+1)+2x-10=0$$
$$y=\frac{x+1\pm\sqrt{(x+1)^2-12(2x-10)}}{2\cdot3}=\frac{x+1\pm\sqrt{x^2-22x+121}}6=\frac{x+1\pm(x-11)}6$$
Can you determine the gradients from here and reach the destination?
| {
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Confusing algebra rule: why $\frac{7^{n+1}-1}{6} + 7^{n+1} = \frac{7^{n+2}-1}{6}$? Math rule I don't understand.
My discrete math midterm is tomorrow and I'm studying proof styles. I came across a rule (algebra maybe?) I don't quite understand and I was hoping someone could explain it step by step for me.
$$\frac{7^{n+1}-1}{6} + 7^{n+1} = \frac{7^{n+2}-1}{6}$$
I guess I can memorize it, but could someone show me how it works step by step?
Thanks
| Do you know the rule for the sum of a finite geometric series?
$$1 + a + a^2 + \cdots + a^n = \frac{a^{n+1}-1}{a-1}$$
Now take $a=7$:
$$\begin{align}
1 + 7 + 7^2 + \cdots + 7^n\hphantom{+7^{n+1}} &= \color{maroon}{\frac{7^{n+1}-1}{6}} \\
1 + 7 + 7^2 + \cdots + 7^n+7^{n+1} &= \color{darkblue}{\frac{7^{n+2}-1}{6}} \\
\end{align}
$$
The second line is the same as the first line, but with $7^{n+1}$ added:
$$\color{maroon}{\frac{7^{n+1}-1}{6}} + 7^{n+1} = \color{darkblue}{\frac{7^{n+2}-1}{6}}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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$f\left(x + \frac1x\right)= x^3+x^{-3},$ find $f(x)$ $$f\left(x + \frac1x\right)= x^3+x^{-3},$$ find $f(x)$.
What i do know at this state is that..
express x as a function of y :
$y= x + 1/x$
$x^2−xy+1=0$
Quad formula:
$x= (y ± \sqrt {y^2-4}) / 2$
When i substitute this into the original equation, i can't solve it.
| Hint: $$\left(x + \frac1x\right)^3=x^3+3x^2\frac1x+3x\frac{1}{x^2}+\frac{1}{x^3}$$
| {
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Find one set of solutions for the following system: Find one set of solutions for the following system:
\begin{cases}
1+a^2+d^2=3+b^2+e^2=3+c^2+f^2 \\
1+ab+de=0 \\
ac+df=0 \\
bc+ef=0 \\
\end{cases}
| One way is to follow the mathematics of wishful thinking. Here, I wishfully think that I can choose many things to be zero. For instance, if $d = e = c = 0$, then the last two equations hold trivially and we are instead left with the system
$$\begin{cases} 1 + a^2 = 3 + b^2 = 3 + f^2 \\
1 + ab = 0
\end{cases}$$
If $f = b$, then $f$ is covered. So really, we just have the pair $1 + ab = 0$ and $1 + a^2 = 3 + b^2$. Fortunately, the first means that $a = -1/b$. Plugging into the other gives
$$1 + \frac{1}{b^2} = 3 + b^2.$$
Rearranging and calling $B = b^2$ gives
$$B^2 + 2B - 1 = 0,$$
with solutions $B = -1 \pm \sqrt 2$. Let's choose the positive square root, $B = 1 + \sqrt 2$. Then $b = \sqrt{1 + \sqrt 2}$, and $a = -1/b$.
So in total,
$$\begin{cases}
a = \frac{-1}{\sqrt{1 + \sqrt 2}}\\
b = \sqrt{1 + \sqrt{2}} \\
c = 0 \\
d = 0 \\
e = 0 \\
f = \sqrt{1 + \sqrt{2}}
\end{cases}$$
is a solution. Of course, this is not linear-algebra, nor algebra-precalculus, but I liked the challenge of finding a single solution to an otherwise untenable problem.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/988461",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
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} |
Solving $x^5+x^4-12x^3-21x^2+x+5=0$ $$x^5+x^4-12x^3-21x^2+x+5=0$$
I think it can be solved by trigonometric ways, but how?
| Inspecting the discriminant and the small primes where the polynomial splits suggests that its splitting field is the class field of modulus $31$ corresponding to the subgroup $\{\pm 1, \pm 5, \pm 6\}$ of $(\Bbb Z/31\Bbb Z)/\{\pm 1\}$.
Which means that the splitting field is $\Bbb Q\left(2\left(\cos\frac{2\pi}{31}+\cos\frac{10\pi}{31}+\cos\frac{12\pi}{31}\right)\right)$.
Then I computed the minimal polynomial of this quantity (pure curiosity), and ended up on your polynomial exactly.
So the roots are $2\left(\cos\frac{2\pi}{31}+\cos\frac{10\pi}{31}+\cos\frac{12\pi}{31}\right)$, and its conjugates $$2\left(\cos\frac{4\pi}{31}+\cos\frac{20\pi}{31}+\cos\frac{24\pi}{31}\right) , \\
2\left(\cos\frac{8\pi}{31}+\cos\frac{22\pi}{31}+\cos\frac{14\pi}{31}\right), \\
2\left(\cos\frac{16\pi}{31}+\cos\frac{18\pi}{31}+\cos\frac{28\pi}{31}\right),\\
2\left(\cos\frac{30\pi}{31}+\cos\frac{26\pi}{31}+\cos\frac{6\pi}{31}\right).$$
I have no idea on how to guess this without the heavy machinery.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/989183",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 2,
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} |
Prove $\lim_{n \rightarrow \infty} \frac{1}{n}\cdot \frac{3 + \frac{1}{n}}{4 - \frac{1}{n}} = 0 $ Prove $\lim_{n \rightarrow \infty} \frac{1}{n}\cdot \frac{3 + \frac{1}{n}}{4 - \frac{1}{n}} = 0 $
Let $\epsilon > 0$ be arbitrary. I want to find $N$ such that $n \in \mathbb{N}$ guarantees $ \left | \frac{1}{n}\cdot \frac{3 + \frac{1}{n}}{4 - \frac{1}{n}} - 0 \right | < \epsilon$.
$\Leftrightarrow \frac{1}{n}\cdot \frac{3 + \frac{1}{n}}{4 - \frac{1}{n}} < \epsilon$
$\Leftarrow \frac{1}{n} \cdot \frac{4}{3} < \epsilon \Leftrightarrow n > \frac{4}{3} \cdot \frac{1}{\epsilon} \Rightarrow n \geq N$
Take $N = \left \lfloor \frac{4}{3} \cdot \frac{1}{\epsilon} \right \rfloor + 1$
In the above proof, I'm confused as to how $\frac{1}{n}\cdot \frac{3 + \frac{1}{n}}{4 - \frac{1}{n}} < \epsilon$
$\Leftarrow \frac{1}{n} \cdot \frac{4}{3} < \epsilon$. How did $\frac{3 + \frac{1}{n}}{4 - \frac{1}{n}}$ become $\frac{4}{3} $?
| Hint: $\frac{3 + \frac{1}{n}}{4 - \frac{1}{n}}$ decreases when $n$ increases to infinity.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/991315",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Using differentiation under integral sign to calculate $\int^{\pi/2}_0\frac{\log(1+a\sin\phi)}{\sin\phi}d\phi$ I want to calculate the integral $$\int^{\pi/2}_0\frac{\log(1+\sin\phi)}{\sin\phi}d\phi$$ using differentiation with respect to parameter in the integral $$\int^{\pi/2}_0\frac{\log(1+a\sin\phi)}{\sin\phi}d\phi$$
I know that I have to solve from differentiate under the integral and I must use a suitable substitution for integrands involving trigonometric functions but I can't complete the solution. Could you help me?
| Define
$$I(a) = \int_0^{\pi/2} d\phi \frac{\log{(1+a\sin{\phi})}}{\sin{\phi}} $$
Then
$$I'(a) = \int_0^{\pi/2} \frac{d\phi}{1+a\sin{\phi}} $$
To evaluate the latter integral, sub $t=\tan{\phi/2}$. Then $d\phi = 2/(1+t^2) dt$ (why?), and $\sin{\phi} = 2 t/(1+t^2)$ (why again?), and the integral is
$$I'(a) = 2 \int_0^1 \frac{dt}{1+t^2} \frac1{1+2 a t/(1+t^2)} = 2 \int_0^1 \frac{dt}{1+2 a t+t^2}$$
This is easily evaluated by completing the square in the denominator:
$$\begin{align}I'(a) &= 2 \int_0^1 \frac{dt}{(t+a)^2+1-a^2} \\ &= \frac{2}{\sqrt{1-a^2}} \left [\arctan{\frac{t+a}{\sqrt{1-a^2}}} \right ]_0^1 \\ &=\frac{2}{\sqrt{1-a^2}} \left [\arctan{\frac{1+a}{\sqrt{1-a^2}}}- \arctan{\frac{a}{\sqrt{1-a^2}}}\right ]\\ &= \frac{2}{\sqrt{1-a^2}} \arctan{}\frac{\sqrt{1-a^2}}{1+a+a^2} \end{align}$$
Now we must integrate with respect to $a$. To do this, we integrate over each of the separate terms in the penultimate line. To begin
$$\int da \, \frac{2}{\sqrt{1-a^2}} \arctan{\frac{a}{\sqrt{1-a^2}}}$$
Sub $a=\sin{\theta}$, and it should be straightforward to see that this antiderivative is simply
$$2 \int d\theta \, \theta = \theta^2 = \arcsin^2{a} $$
(Yes I am ignoring the constant of integration for now.)
Now the second piece:
$$\int da \, \frac{2}{\sqrt{1-a^2}} \arctan{\frac{1+a}{\sqrt{1-a^2}}}$$
Sub $a=\cos{\theta} $. Note that the argument of the arctan becomes $\cot{(\theta/2)}$, so the integral becomes
$$-2 \int d\theta \left (\frac{\pi}{2} - \frac{\theta}{2} \right ) = \frac12 \theta^2 - \pi \theta = \frac{\pi}{2} \arcsin{a} + \frac12 \arcsin^2{a}-\frac{3 \pi^2}{8}$$
Including the constant of integration, then,
$$I(a) = \frac12 \theta^2 - \pi \theta = \frac{\pi}{2} \arcsin{a} - \frac12 \arcsin^2{a}-\frac{3 \pi^2}{8} + C$$
Given $I(0)=0$, then $I(a) = \frac{\pi}{2} \arcsin{a} - \frac12 \arcsin^2{a}$. The desired integral is $I(1)$, so that
$$\int_0^{\pi/2} d\phi \frac{\log{(1+\sin{\phi})}}{\sin{\phi}} = \frac{\pi^2}{4} - \frac{\pi^2}{8} = \frac{\pi^2}{8} $$
EDIT
I should have mentioned above that $a$ is restricted to values such that $|a| \le 1$ in the above analysis.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/992100",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 3,
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} |
Calculus - Prove that $\{a_n\}$ converges I need to prove that $a_n$ converges :
$$a_n=\frac{1}{2+1}+\frac{1}{2^2+2}+\cdots+\frac{1}{2^n+n}$$
My solution
Let choose $\epsilon=1$, Exist $N>0$ so all $N \leq x$ appiles $|a_n-L|<\epsilon$
Therefore :
$$\left|\frac{1}{2^n+n}-L\right|<\epsilon=1$$
$$L-1<\frac{1}{2^n+n}<L+1$$
Therefore the function is bounded.
Also $a_n$ is increasing since $a_{n+1}-a_n>0$
$$a_n = \frac{1}{2^n+n}$$
$$a_{n+1} = \frac{1}{2^{(n+1)}+n+1}$$
$$a_{n+1}-a_n=\frac{1}{2^n+n}-\frac{1}{2^{(n+1)}+n+1}=\frac{2^{n+1}+n+1-2^n-n}{(2^n+n)(2^{n+1}+n+1)}=\frac{2^{n+1}+1-2^n}{(2^n+n)(2^{n+1}+n+1)}>0 (1 \leq n)$$
As well it clear that the function is continuous since $1,2^n,n$ are continuous.
Therefore it possible to conclude that the function is converges.
Is this proof perfect?
Thanks!
| The question is equivalent to establishing the convergence of the series
$$
\sum_{n=1}^\infty \frac{1}{2^n+n},
$$
and perhaps the easiest way to do this is using the Comparison Test, i.e., observe that
$$
\frac{1}{2^n+n}<\frac{1}{2^n},
$$
and use the fact that $\sum_{n=1}^\infty\frac{1}{2^n}<\infty$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/992167",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Posible combinations of available rooms I have $x$ people to divide over different type of rooms.
Possible rooms:
*
*$1$ person bedroom normal view;
*$1$ person bedroom sea view;
*$2$ person bedroom
For $x = 4$
Possible combinations are
$$(0,0,2),
(0 ,2, 1),
(2 ,0, 1),
(1 ,1, 1),
(4 ,0, 0),
(3 ,1, 0),
(2 ,2, 0),
(1 ,3 ,0),
(0 ,4, 0)$$
Note : Plenty of rooms are available.
How can I get all posible combinations?
| You can solve this using generating function such as this
Bedroom 1 can be 0,1,2,3,4
Bedroom 2 can be 0,1,2,3,4
Bedroom 3 can be 0,2,4
Translating this into generating function and the number of combinations would be the coefficient of $x^4$
Thus the above problem could be translated into $(1+x+x^2+x^3+x^4)^2(1+x^2+x^4)$ and finding the coefficient of $x^4$
The product is given from Wolfram Alpha as
$x^{12}+2 x^{11}+4 x^{10}+6 x^9+9 x^8+10 x^7+11 x^6+10 x^5+$***9***$x^4+6 x^3+4 x^2+2 x+1$
Thanks
Satish
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/994837",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
How do we find the $n+2$th term of the series $1 + (3+5) + (7+9+11)+\dots$? We have the series $1 + (3+5) + (7+9+11)+\dots$.
We need to find the $n+2$th term and hence summation of the series up to this term.
However hard we try we do not seem to be able to fit this series to a pattern.
| $$a_{1}=1=1^3\\a_{2}=3+5=8=2^3\\a_{3}=7+9+11=27=3^3\\a_{4}=13+15+17+19=64=4^3\\...\\a_{n}=n^3\\so\\a_{n+2}=(n+2)^3$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/996859",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Difficult denomiator rationalization questions These are two questions from a competitive exam involving irrationals where I am supposed to simplify it to match one of the given options.
QUESTION 1: The value of $$ \frac {2 (\sqrt 2+ \sqrt6)}{3(\sqrt {2 + \sqrt 3})} + \sqrt {2 + \sqrt 3}+ \sqrt {2 - \sqrt 3}$$
is
A. $\frac {3+4 \sqrt 6}{3}$
B. $\frac {4+3 \sqrt 6}{3}$
C. $\frac {3+4 \sqrt 6}{4}$
D. $\frac {4- 3\sqrt 6}{3}$
I have been able to solve this far:
Taking denominator common, I get
$$ \frac {2 (\sqrt 2+ \sqrt6) + 3(\sqrt {2 + \sqrt 3})(\sqrt {2 + \sqrt 3})+ 3(\sqrt {2 + \sqrt 3})\sqrt {2 - \sqrt 3}}{3(\sqrt {2 + \sqrt 3})}$$
After which:
$$ \frac {2 (\sqrt 2+ \sqrt6) + 3( {2 + \sqrt 3})+ 3(\sqrt {2^2 - \sqrt 3^2})}{3(\sqrt {2 + \sqrt 3})}$$
which gives
$$ \frac {2 (\sqrt 2+ \sqrt6) + 3( {2 + \sqrt 3})+ 3}{3(\sqrt {2 + \sqrt 3})}$$
I multiply by the conjugate of the irrational term in denominator.
$$ \frac {2 (\sqrt 2+ \sqrt6)(\sqrt {2 - \sqrt 3}) + 3( {2 + \sqrt 3})\sqrt {2 - \sqrt 3}+ 3(\sqrt {2 - \sqrt 3})}{3(\sqrt {2 + \sqrt 3})(\sqrt {2 - \sqrt 3})}$$
Upon simplification,
$$ \frac {2 (\sqrt 2+ \sqrt6)(\sqrt {2 - \sqrt 3}) + 3(\sqrt 3)\sqrt {2 - \sqrt 3}+ 9(\sqrt {2 - \sqrt 3})}{3}$$
Beyond this, I am not able to work out a solution. Any hints (please explain the hint slightly) are welcome.
QUESTION 2: The value of
$$ \sqrt {43-12 \sqrt 7} - \frac {2}{\sqrt {16+6 \sqrt 7}}$$
is:
A. $-3$
B. $3 $
C. $2 \sqrt 7 -3 $
D. $- (2 \sqrt 7 +3) $
I have been able to solve this far:
Taking denominator common:
$$ \frac {\sqrt {43-12 \sqrt 7}(\sqrt {16+6 \sqrt 7}) - 2}{\sqrt {16+6 \sqrt 7}}$$
Upon simplification I get:
$$ \frac {\sqrt {184 + 66\sqrt 7} - 2}{\sqrt {16+6 \sqrt 7}}$$
Multiplying by the conjugate of the denominator:
$$ \frac {\sqrt {184 + 66\sqrt 7}(\sqrt {16-6 \sqrt 7}) - 2(\sqrt {16-6 \sqrt 7)}}{\sqrt {16+6 \sqrt 7}(\sqrt {16-6 \sqrt 7)}}$$
Simplification gives:
$$ \frac {\sqrt {172 - 48\sqrt 7}- 2(\sqrt {16-6 \sqrt 7)}}{\sqrt 4}$$
When I typed this into a software I got
$\sqrt {16-6 \sqrt 7} = 3 - \sqrt 7 $ and
$ \sqrt {172 - 48\sqrt 7} = 12 - 2 \sqrt7 $
How do we get these?
| You can even express whole root of x-√y in the form of √a-√b . Eg-whole root of 3-√2 can be written as √2-√1
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1000107",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Solving a limit without l'Hopital My friend passed me a limit that (in his opinion) is resovable using only derivatives or similar method.
Here is the limit:
$$\lim_{x\to0} \frac{\ln(1+x+x^2) - (e^x-1)}{x\sin(x)}$$
I tried to solve it using basic limits (the limit form clearly shows parts of notable limits) but my solution is $1$ that is wrong. Can someone solve it avoiding derivative methods or similar methods? Or my friend is right? Sorry for my english and thank you in advance (if what I've written is not enough please advice me)
I'm supposed to solve it knowing only basic limits and algebric manipulation nothing more. Just as someone that started to do limits in his first steps
| You can answer the question using standard definitions of Taylor series expansions:
$\ln \left(1+x\right) = x - \dfrac{x^{2}}{2} + \dfrac{x^{3}}{3} - \ldots$ ,
$e^{x} = 1+x+\dfrac{x^{2}}{2!} + \ldots$
and
$\sin \left (x\right) = x - \dfrac{x^{3}}{3!} + \ldots$
Now the only expression we need to worry about expanding is the $\ln$ term
$$\ln \left(1+x+x^{2}\right) = \left(x+x^{2}\right) - \dfrac{\left(x+x^{2}\right)^{2}}{2} + \dfrac {\left(x+x^{2}\right)^{3}}{3} - \ldots$$
$$= x + x^{2} - \dfrac{x^{2}}{2}\left(1+x\right)^{2} + \dfrac{x^{3}}{3}\left(1+x\right)^{3} + \ldots$$
which after some expanding and collecting terms gives:
$$\ln \left(1+x+x^{2}\right) = x + \dfrac{1}{2}x^{2} - \dfrac {2}{3}x^{3} + \ldots$$
We substitute this series and the other series into the expression concerned:
$$\lim_{x\rightarrow 0} \dfrac {\ln \left(1+x+x^{2}\right) - e^{x} + 1}{x\sin \left (x\right)} = \lim_{x\rightarrow 0}\dfrac{x+\dfrac{1}{2}x^{2}-\dfrac{2}{3}x^{3}-\left(1+x+\dfrac{x^2}{2!}+\dfrac{x^{3}}{3!}+\ldots\right) + 1}{x\left(x-\dfrac{x^{3}}{6}+\ldots\right)}$$
$$ = \lim_{x\rightarrow 0} \dfrac{x+\dfrac{1}{2}x^{2}-\dfrac{2}{3}x^{3}- x - \dfrac{x^{2}}{2} - \dfrac{x^{3}}{6}-\ldots}{x^{2}-\dfrac{x^{4}}{6}+\ldots}$$
$$ = \lim_{x\rightarrow 0} \dfrac{- \dfrac{5}{6}x^{3} + \ldots}{x^{2}-\dfrac{x^{4}}{6}+\ldots}$$
Divide through by the $x^{2}$ in the fraction to get:
$$\lim_{x\rightarrow 0} \dfrac {\ln \left(1+x+x^{2}\right) - e^{x} + 1}{x\sin \left (x\right)} = \lim_{x\rightarrow 0} \dfrac{ - \dfrac{5}{6}x + \ldots}{1 - \dfrac{x^{2}}{6}+\ldots} = 0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1001249",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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Summation of the series$\sum_{n=1}^\infty\frac{1}{n^2+4}$ Evaluate the sum of the following series
$$\sum_{n=1}^\infty\frac{1}{n^2+4}$$
I saw a video in youtube where it is solved using complex analysis. What other method can be used to solve this?
| After asking the same question to my prof, he get's this:
$$\sum_{n=1}^\infty\frac{1}{n^2+4}=$$
$$\frac{1}{8}\left(2\pi coth(2\pi )-1\right)=$$
$$\frac{1}{8}\left(2\pi \left(1+\frac{2}{-1+e^{4\pi}}\right)-1\right)=$$
$$\frac{1}{8}\left(2\pi +\frac{4\pi}{e^{4\pi}-1}-1\right)=$$
$$\frac{1+2\pi +2\pi e^{4\pi}-e^{4\pi}}{8e^{4\pi}-8}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1003861",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Integration $\int\frac{\sqrt{x+4}}x dx$ by partial fraction Here's what I came up with:
$$\int\frac{\sqrt{x+4}}x dx$$
for $ u = \sqrt { x + 4 } $
$$=\int\frac{u}{u^2-4}\;2u\;dx$$
$$=2\int\frac{u^2}{(u-2)(u+2)}\;dx$$
$$=\frac{A}{u-2}\;+\;\frac{B}{u+2}$$
$$\implies u^2 = A(u+2)+B(u-2)$$
$$\implies u = \pm2\;\implies\;A=1\;and\;B=-1$$
$$=\int\frac{1}{u-2}\;-\;\frac{1}{u+2}\;du$$
$$=2\ln\left|\frac{u-2}{u+2}\right|$$
$$=2\ln\left|\frac{\sqrt{x+4}-2}{\sqrt{x+4}+2}\right|+C$$
However, when I check with the book, they have the answer as:
$$2\sqrt{x+4}\;+\;2\ln\left|\frac{\sqrt{x+4}-2}{\sqrt{x+4}+2}\right|+C$$
Did I do something wrong or are these two answer practically the same?
| As the highest powers of $u$ in the numerator & in the denominator are same $$\frac{u^2}{(u-2)(u+2)}=A+\frac B{u+2}+\frac C{u-2}$$
Clearly, $A=1$ as the coefficients of the highest powers of $u$ in the numerator & in the denominator are also same
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1008476",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Tricky binomial coefficients problem Let $k$ be a positive integer and let $n = 6k - 1$. Let
$$S(n)=\sum_{j=1}^{2k-1} (-1)^{j+1} {{n}\choose{3j-1}}$$
How do you prove that $S(n)$ is never zero?
| By way of enrichment here is another algebraic proof using basic
complex variables.
Suppose we first compute
$$Q(k) = \sum_{q=1}^{2k} (-1)^{q+1} {6k-1\choose 3q-1}.$$
Introduce the integral representation
$${6k-1 \choose 6k-3q}
= \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{6k-1}}{z^{6k-3q+1}} \; dz.$$
We use this to obtain an integral for the sum. Note that when $q>2k$
the integral becomes zero and for $q=0$ is zero as well. Therefore
we may extend the sum from $q$ being zero to infinity to obtain
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{6k-1}}{z^{6k+1}}
\sum_{q\ge 0} (-1)^{q+1} z^{3q} \; dz
\\ = - \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^{6k-1}}{z^{6k+1}} \frac{1}{1+z^3} \; dz.$$
Introduce the roots where $p=0,1,2$
$$\rho_p = \exp(\pi i/3 + 2\pi i p/3)$$
Using partial fractions by the Cauchy Residue Theorem on simple poles
we thus obtain
$$\frac{1}{1+z^3} =
\sum_\rho \frac{1}{z-\rho}
\times \mathrm{Res}_{z=\rho} \frac{1}{1+z^3}
\\= \sum_\rho \frac{1}{z-\rho} \frac{1}{3\rho^2}
= \sum_\rho \frac{1}{z-\rho} \frac{\rho}{3\rho^3}
= -\frac{1}{3} \sum_\rho \frac{\rho}{z-\rho}.$$
Substituting this into the integral we have
$$ \frac{1}{3} \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^{6k-1}}{z^{6k+1}}
\sum_\rho \frac{\rho}{z-\rho} \; dz
\\ = - \frac{1}{3} \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^{6k-1}}{z^{6k+1}}
\sum_\rho \frac{1}{1-z/\rho}\; dz.$$
The integral is prepared for coefficient extraction which gives
(use $\rho^6 = 1$)
$$-\frac{1}{3}
\sum_\rho \sum_{q=0}^{6k-1} {6k-1\choose q} \rho^{-(6k-q)}
\\ = -\frac{1}{3}
\sum_\rho \rho^{-6k}
\sum_{q=0}^{6k-1} {6k-1\choose q} \rho^q
= -\frac{1}{3}
\sum_\rho \rho^{-6k} (1+\rho)^{6k-1}
\\ = -\frac{1}{3}
\sum_\rho (1+\rho)^{6k-1}.$$
The term for $\rho_0 = -1$ is zero and we are left with
$$-\frac{1}{3}
\left((1+\rho_1)^{6k-1} + (1+\rho_2)^{6k-1}\right).$$
Note that
$$|1+\rho_{1,2}| = \sqrt{(3/2)^2 + (\sqrt{3}/2)^2} = \sqrt{3}
\quad\text{and}\quad
\arg(1+\rho_{1,2}) = \pm\frac{\pi}{6}.$$
This finally yields the value
$$-\frac{1}{3} \sqrt{3}^{6k-1}
\left(\exp(i\pi/6\times(6k-1))
+ \exp(-i\pi/6\times(6k-1)) \right)
\\ = -\frac{1}{3} (-1)^k \frac{27^k}{\sqrt{3}}
\times 2 \cos(\pi/6)
= -\frac{1}{3} (-1)^k \times 27^k.$$
The question asked to show that
$$Q(k) - (-1)^{2k+1}{6k-1\choose 6k-1}
= -\frac{1}{3} (-1)^k \times 27^k + 1$$
is never zero. This follows by inspection at this point with the first
term having modulus at least nine and the second one being one.
Apparently this method is due to Egorychev.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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$\sec\theta + \tan\theta=4$ find $\cos\theta$ given: $\theta\neq90$ I tried the following :
\begin{align}\sec\theta + \tan\theta&=4\\
\frac1{\cos\theta} + \frac{\sin\theta}{\cos\theta}&=4\\
\frac{1+\sin\theta}{\cos\theta}&=4\\
\frac{1+\sin\theta}4&=\cos\theta\end{align}
now don't know how to evaluate further ?
| one way is to use the half angle formule like @Mark Bennet suggested. then you have
$$\sec(\theta)=\frac{1+t^2}{1-t^2},\qquad\tan(\theta)=\frac{2t}{1-t^2}$$ where $t=\tan\left(\frac{\theta}{2}\right)$. This results in the following equationt
$$\frac{1+2t+t^2}{1-t^2}=\frac{(1+t)^2}{(1-t)(1+t)}=\frac{1+t}{1-t}=4\\t=\frac{3}{5}\rightarrow\theta=2\arctan\left(\frac{3}{5}\right)$$
and $$\cos(\theta)=\frac{1-t^2}{1+t^2}=\frac{1-\frac{9}{25}}{1+\frac{9}{25}}=\frac{8}{17}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1010820",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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How to prove $\sin(\frac{1}x) < \frac{1}x$ for all $x$ greater than $1$. How to prove $\sin(\frac{1}x) < \frac{1}x$ for all $x$ greater than $1$?
I was thinking using slopes but I get a contradiction (i.e. $\cos(\frac{1}x) > 1$) when I do some of the algebra.
| Let $x>1$ and consider $$\sin(y) = \sum_{n=0}^\infty \frac{(-1)^ny^{2n+1}}{(2n+1)!}$$ Substitute $y = \frac{1}{x}$: $$\sin\left(\frac{1}{x}\right) = \sum_{n=0}^\infty \frac{(-1)^n}{x^{2n+1}(2n+1)!} = \frac{1}{x}-\frac{1}{3!x^3}+\frac{1}{5!x^5}-\frac{1}{7!x^7}+\frac{1}{9!x^9}- \dots \\ = \frac{1}{x}-\left(\frac{1}{3!x^3}-\frac{1}{5!x^5}\right)-\left(\frac{1}{7!x^7}-\frac{1}{9!x^9}\right)- \dots$$ now observe that each quantity in parentheses is positive, so we can think of $\sin\left(\frac{1}{x}\right)$ as $\frac{1}{x}$ minus an infinite amount of positive quantities. It follows that $$\frac{1}{x}> \frac{1}{x}-\left(\frac{1}{3!x^3}-\frac{1}{5!x^5}\right)-\left(\frac{1}{7!x^7}-\frac{1}{9!x^9}\right)- \dots = \sin\left(\frac{1}{x}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1011011",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Nonlinear diophantine equations $x^2+2y=z^2$ and $y^2+2x=w^2$ I am asked to find two sets of positive numbers $x$ and $y$, such that both $x^2+2y$ and $y^2+2x$ are perfect squares.
I found a general solution to either single equation, but it seems impossible to satisfy them both. Can it be done, and why or why not?
UPDATE: What if $x$ and $y$ are positive rational numbers?
| $x^2 + 2y = z^2 \to x^2 < z^2 \to x < z \to x+1 \leq z \to (x+1)^2 \leq z^2$
$y^2 + 2x = w^2 \to y^2 < w^2 \to y < w \to y+1 \leq w \to (y+1)^2 \leq w^2$.
Thus: $(x+1)^2 + (y+1)^2 \leq z^2+w^2$ (*)
But: $x^2+2y+y^2+2x = z^2 + w^2 \to (x+1)^2+(y+1)^2 = z^2+w^2+2 > z^2 + w^2$ (**).
(*) and (**) contradict each other. So there is no natural solutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1011453",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
proving $\sum_{n=0}^{\infty }\frac{1}{(2n+1)(2n+2)}=\sum_{n=0}^{\infty }\frac{4}{(4n+1)(4n+2)(4n+3)}$ I want to prove $$\sum_{n=0}^{\infty }\frac{1}{(2n+1)(2n+2)}=\sum_{n=0}^{\infty }\frac{4}{(4n+1)(4n+2)(4n+3)}$$
| $$\frac{1}{(2n+1)(2n+2)}=\frac{1}{2n+1}-\frac{1}{2n+2},$$
hence:
$$\sum_{n=0}^{+\infty}\frac{1}{(2n+1)(2n+2)}=\sum_{n\geq 1}\frac{(-1)^{n-1}}{n}=\int_{0}^{1}\sum_{n\geq 0}(-x)^{n}\,dx=\int_{0}^{1}\frac{dx}{1+x}=\log 2,$$
while:
$$\frac{4}{(4n+1)(4n+2)(4n+3)}=\frac{2}{4n+1}-\frac{2}{2n+1}+\frac{2}{4n+3}$$
so:
$$\begin{eqnarray*}\sum_{n\geq 0}\frac{4}{(4n+1)(4n+2)(4n+3)}&=&2\int_{0}^{1}\sum_{n\geq 0}(x^{4n}-2x^{4n+1}+x^{4n+2})\,dx\\&=&2\int_{0}^{1}\frac{(1-x)^2}{1-x^4}\,dx\\&=&2\int_{0}^{1}\frac{1-x}{(1+x)(1+x^2)}\,dx\end{eqnarray*}$$
and by setting $x=\frac{1-y}{1+y}$ we get that the last integral equals:
$$ \int_{0}^{1}\frac{2y}{y^2+1}\,dy=\log 2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1013197",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solving for $f$ when $f(x + y) = f(x) + f(y) + axy$ where $a$ is a real number Let $f$ be a real valued continuous function such that $f(x + y) = f(x) + f(y) + axy$ where $a$ is a real number.
We have to show that $f(x) = \frac{a}{2}x^2 + bx$ where $b = f(1) - \frac{a}{2}$
Here is my solution so far:
Let $x = y = 0$ . This gives $f(0) = 0$
Now putting $y = -x$ gives $ax^2 = f(x) + f(-x)$. This gives $a = f(1) + f(-1)$
Also by letting $x = y$ we obtain $f(2x) = 2f(x) + ax^2$
After juggling with these equations I am not able to show that $f(x)$ is a quadratic equation.
Any help would be appreciated.
| Let $g(x) = f(x) - \frac{a}{2}x^2$.
Then, subtracting $\frac{a}{2}x^2 + \frac{a}{2}y^2$ from both sides of the given expression, we get :
$f(x+y) - \frac{a}{2}x^2 - \frac{a}{2}y^2 -axy = f(x) - \frac{a}{2}x^2 +f(y) - \frac{a}{2}y^2 $
$f(x+y) - (\frac{a}{2}(x+y))^2 = f(x) - \frac{a}{2}x^2 + f(y) - \frac{a}{2}y^2$
$g(x+y) = g(x) + g(y)$
This equation is known to have the solution $g(x) = bx$. (Cauchy's equation)
Thus, $f(x)$ has the solution of the form $\frac{a}{2}x^2 + bx$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1013295",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Find $\lim\limits_{x \to \infty} 2x(\sqrt{x-1} - \sqrt{x+5})$ $\lim\limits_{x \to \infty} 2x(\sqrt{x-1} - \sqrt{x+5})$
For what i've found the part in brackets is an indeterminate form. I've tried to multiply the bracket part by $\frac{\sqrt{x - 1} + \sqrt{x+5}}{\sqrt{x - 1} + \sqrt{x+5}}$ , then multiply the numerator by $2x$. I don't know what to do next.
| $$\begin{align}2x (\sqrt{x-1}-\sqrt{x+5}) \cdot \frac{\sqrt{x - 1} + \sqrt{x+5}}{\sqrt{x - 1} + \sqrt{x+5}}& = \dfrac{2x((x-1)-(x+5))}{{\sqrt{x - 1} + \sqrt{x+5}}}\\ \\ &= \dfrac{-12x}{{\sqrt{x - 1} + \sqrt{x+5}}}\tag{1}\end{align}$$
Now you can divide the numerator and denominator by $x$ to get $$\lim_{x\to \infty}\dfrac{-12}{{\sqrt{\frac 1x - \frac 1{x^2}} + \sqrt{\frac 1x +\frac 5{x^2}}}}$$
Alternatively, if it's not clear to you that the limit is $-\infty$, you could instead divide the numerator and denominator $(1)$ by $\sqrt x$ to get $$\begin{align} \lim_{x\to \infty} \dfrac{-12\sqrt x}{\sqrt{1-\frac 1x} + \sqrt{1 +\frac{5}{x}}} & =\lim_{x\to \infty} \dfrac{-12\sqrt x}{2} \\ & = \lim_{x\to \infty} -6\sqrt x = -\infty\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1016805",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Linear Algebra Characteristic Polynomials Let $p(t) = t^n+a_{n-1}t^{n-1}+a_{n-2}t^{n-2} + \cdots + a_1t+a_0$. Show that the characteristic polynomial of the matrix A below
\begin{bmatrix}
0 & 0 & \cdots & & & -a_0\\
1 & 0 & & & & -a_1\\
0 & 1 & \ddots & & & \vdots\\
\vdots & & \ddots &\ddots & & \vdots\\
\vdots & & & \ddots & 0 & -a_{n-2}\\
0 & 0 & \cdots & \cdots & 1 & -a_{n-1}
\end{bmatrix}
is the polynomial $p$.
Please help, currently I got that the $\det(A)=-a_0$
| Define
$$
e_{1} =\left[\begin{array}{c}1 \\ 0 \\ 0 \\ \vdots \\ 0\end{array}\right]
e_{2} =\left[\begin{array}{c}0 \\ 1 \\ 0 \\ \vdots \\ 0\end{array}\right],
\cdots,
e_{n} =\left[\begin{array}{c}0 \\ 0 \\ 0 \\ \vdots \\ 1\end{array}\right].
$$
By the definition of your matrix $A$,
$$
\begin{align}
Ae_{1} & = e_{2} \\
A^{2}e_{1} & = e_{3} \\
\vdots \\
A^{n-1}e_{1} & = e_{n} \\
A^{n}e_{1} & = -a_{0}e_{1}-a_{1}e_{2}-\cdots-a_{n-1}e_{n}
\end{align}
$$
It follows that
$$
(A^{n}+a_{n-1}A^{n-1}+\cdots+a_{1}A+a_{0}I)e_{1}=0.
$$
Define $p(\lambda)=\lambda^{n}+a_{n-1}\lambda^{n-1}+\cdots+a_{1}\lambda+a_{0}$. Then $p(A)e_{1}=0$ and no lower order monomial $q$ can have this property. Because $p(A)e_{1}=0$, then $p(A)=0$ because
$$
p(A)e_{k} = p(A)A^{k-1}e_{1}=A^{k-1}p(A)e_{1} = 0,\;\;\; k=2,3,\cdots,n.
$$
The minimal polynomial for $A$ must be of order $n$ because it cannot be of higher order, and there is no lower order monomial $q$ for which $q(A)e_{1}=0$. So $p$ is the minimal polynomial for $A$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1017542",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that $\frac {n!}{(k! 2^k)(n-2k)!} \neq {n \choose 2}$ for $\forall k \neq 1$ where $n \neq 6$. Prove that $$\frac {n!}{(k! 2^k)(n-2k)!} \neq {n \choose 2}$$ for $\forall k > 1$ where $n \neq 6$.
I have expanded it it seems to be not equal but couldn't find the proper logic.
| $$\dfrac {n!}{(k! 2^k)(n-2k)!} = \dfrac{(2k)!}{(k! 2^k)}\dfrac {n!}{(2k)!(n-2k)!} = \dfrac{(2k)!}{(k! 2^k)}{n\choose 2k}$$
$\dfrac{(2k)!}{(k! 2^k)} = 1\cdot 3\cdot5\cdots(2k-1)$ is a positive integer strictly greater than $1$ when $k>1$
${n\choose 2k}\geq{n\choose 2}$ when $ k >1 $ and $2k\neq n-1,n$
So $\dfrac{(2k)!}{(k! 2^k)}{n\choose 2k} > {n\choose 2}$ when $ k >1 $ and $2k\neq n-1,n$
If $2k = n-1$(so $n = 2k+1 >3$), we have ${n\choose 2k} = {n\choose n-1} = n$, $\dfrac{(2k)!}{(k! 2^k)} = 1\cdot 3\cdot5\cdots(2k-1)\geq (n-2)$, it's easy to verify $n(n-2) > {n\choose 2}$ when $n > 3$
If $2k = n$(so $n\geq 4$), we have ${n\choose 2k} = {n\choose n} = 1$, $\dfrac{(2k)!}{(k! 2^k)} = 1\cdot 3\cdot5\cdots(2k-1)\geq (n-3)(n-1)$, it's easy to verify $(n-3)(n-1) > {n \choose 2}$ when $n >6$
Now verify the result for $n=4, k=2$ to conclude
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1019214",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find the exact closed from expression of $1^2 + 3^2 + 5^2 + · · · + (2n + 1)^ 2$ I know the above expression equals to $\frac{n(2n−1)(2n+1)}{3}$, but how exactly can i come up with something from scratch?
| $$\sum_{k=0}^n(2k+1)^3=\sum_{k=0}^n(2k+1+2)^3-(2n+3)^3+1$$
$$\sum_{k=0}^n(2k+1)^3=\sum_{k=0}^n[(2k+1)^3+6(2k+1)^2+12(2k+1)+8]-(2n+3)^3+1$$
$$\sum_{k=0}^n[6(2k+1)^2+12(2k+1)+8]-(2n+3)^3+1=0$$
$$\begin{align}\sum_{k=0}^n(2k+1)^2&=\frac16\left((2n+3)^3-12\sum_{k=0}^n(2k+1)-8\sum_{k=0}^n1-1\right)\\
&=\frac16\left((2n+3)^3-12\sum_{k=0}^n2k-12\sum_{k=0}^n1-8\sum_{k=0}^n1-1\right)\\
&=\frac16\left((2n+3)^3-24\sum_{k=0}^nk-20\sum_{k=0}^n1-1\right)\\
&=\frac16\left((2n+3)^3-24\sum_{k=0}^nk-20\sum_{k=0}^n1-1\right)\\
&=\frac16\left((2n+3)^3-12n(n+1)-20(n+1)-1\right)\\
&=\frac{8n^3+24n^2+22n+6}{6}\\
\end{align}$$
$$\sum_{k=0}^n(2k+1)^2=\frac{8n^3+24n^2+22n+6}{6}=\frac{(n+1)(2n+1)(2n+3)}{3}\,\,\,\,$$
$$\sum_{k=1}^{n}(2k+1)^2=\frac{(n+1)(2n+1)(2n+3)}{3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1020031",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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Calculation of $ \cos\left(\frac{2\pi}{7}\right)+\cos \left(\frac{4\pi}{7}\right)+\cos \left(\frac{6\pi}{7}\right)$ Calculation of $\displaystyle \cos\left(\frac{2\pi}{7}\right)+\cos \left(\frac{4\pi}{7}\right)+\cos \left(\frac{6\pi}{7}\right)$ and $\displaystyle \cos\left(\frac{2\pi}{7}\right)\times \cos\left(\frac{4\pi}{7}\right) \times \cos\left(\frac{6\pi}{7}\right)$ is
$\bf{My\; Try::}$ Let $\displaystyle \frac{2\pi}{7}=\phi\;,$ Then $3\phi = 2\pi-4\phi\Rightarrow \cos \left(3\phi\right) = \cos \left(2\pi-4\phi\right)=\cos \left(4\phi\right)$
So $4\cos^3 \phi -3\cos \phi = 2\left[2\cos^2 \phi - 1\right]^2-1=2\left[4\cos^4 \phi+1-4\cos^2 \phi\right]-1 = 8\cos^4 \phi-8\cos^2 \phi+1$
So $8\cos^4\phi-4\cos^3 \phi-8\cos^2 \phi+3\cos \phi+1=0$
Now I did not understand how can i solve after that
Help me
Thanks
| The three corresponding nodes in the unit radius tetradecagon (14-gon) can be regarded as the tops of three isosceles with top angle $\alpha = \frac{\pi}{7}$, due to the equal chords:
Summing the bases of these isosceles yields the 'radius' of the 14-gon:
$$ \color{green}{g} \color{black}{+} \color{blue}{b} \color{black}{+} \color{red}{r} \color{black}{ = 1}$$
But we also have that:
$$\begin{align}\cos\left(\color{blue}{ \frac{2\pi}{7} } \right) &= \color{green}{g}\color{black}{ \,\,\,+ \,\,\,\frac{1}{2}}\color{blue}{b} \\ \cos\left(\color{green}{\frac{4\pi}{7} } \right) &= -\frac{1}{2}\color{green}{g} \\ \cos\left(\color{red}{\frac{6\pi}{7} } \right)&= -\color{green}{g} \color{black}{\,\,\,- \,\,\,} \color{blue}{b}\color{black}{ \,\,\,- \,\,\, \frac{1}{2}}\color{red}{r} \end{align}$$
Adding these gives
$$\cos\left(\color{blue}{ \frac{2\pi}{7} } \right) + \cos\left(\color{green}{\frac{4\pi}{7} } \right) + \cos\left(\color{red}{\frac{6\pi}{7} } \right) = -\frac{1}{2}\left(\color{green}{g} \color{black}{+} \color{blue}{b} \color{black}{+} \color{red}{r}\right)= -\frac{1}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1021947",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
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Find Min $P=\frac{a^4}{a+2b}+\frac{b^4}{b+2c}+\frac{c^4}{c+2a}$ Give $a,b,c>0$ and $ab^2+bc^2+ca^2=1$
Find Min $P=\frac{a^4}{a+2b}+\frac{b^4}{b+2c}+\frac{c^4}{c+2a}$
| $$P = \sum_{cyc}\left(a^3-2\frac{a^3b}{a+2b} \right) \ge \sum_{cyc}\left(a^3-2\frac{a^3b}{3\sqrt[3]{ab^2}} \right)=\sum_{cyc}\left(a^3-\frac23 a^{8/3}b^{1/3}\right)$$
Now from Rearrangement (or using AM-GM), we have $\sum_{cyc} a^3 \ge \sum_{cyc} a^{8/3}b^{1/3}$ and $\sum_{cyc} a^3 \ge \sum_{cyc} ab^2$, and using these we have
$$P \ge \frac13 \sum_{cyc} ab^2=\frac13$$
as equality is reached when $a=b=c=\frac1{\sqrt3}$, this is indeed the minimum.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1023887",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Evaluation of $\int_0^1 \frac{x^3}{2(2-x^2)(1+x^2) + 3\sqrt{(2-x^2)(1+x^2)}}\,\mathrm dx$ How does one evaluate the following integral?
$$\int_0^1 \frac{x^3}{2(2-x^2)(1+x^2) + 3\sqrt{(2-x^2)(1+x^2)}}\,\mathrm dx$$
This is a homework problem and I have been evaluating this integral for hours yet no success so far. I have tried to rationalize the integrand by multiplying it with
$$\frac{2(2-x^2)(1+x^2) - 3\sqrt{(2-x^2)(1+x^2)}}{2(2-x^2)(1+x^2) - 3\sqrt{(2-x^2)(1+x^2)}}$$
but the integrand is getting worse. I have tried to use trigonometric substitutions like $x=\tan\theta$ and $x=\sqrt{2}\sin\theta$, but I cannot rid off the square root form. I have also tried to use hyperbolic trigonometric substitutions but the thing does not get any easier neither also substitution $y=x^2$ nor $y=\sqrt{(2-x^2)(1+x^2)}$. Using integration by parts is almost impossible for this one. I have also tried to use the tricks from this thread, but still did not get anything. No clue is given. My professor said, we must use clever substitutions but I cannot find them. Any idea or hint? Any help would be appreciated. Thanks in advance.
Edit :
The answer I got from my Prof is $\dfrac{3-2\sqrt{2}}{6}$.
| Let $x=\sin y$, and note that $(2-x^2)(1+x^2)$ simplifies to $2+\sin^2y\cos^2y$. Then integral becomes
$$I=\int_0^{\pi/2}\frac{\sin^3y\cos y}{2(2+\sin^2y\cos^2y)+3\sqrt{2+\sin^2y\cos^2y}}dy.$$
Now set $y=z-\frac{\pi}{2}$, to observe that
$$I=I_0=\int_0^{\pi/2}\frac{\cos^3y\sin y}{2(2+\sin^2y\cos^2y)+3\sqrt{2+\sin^2y\cos^2y}}dy.$$
With $\cos^3y\sin y+\cos y\sin^3 y=\cos y\sin y$ you obtain (with $2I=I+I_0$)
$$\begin{align}I&=\frac{1}{2}\int_0^{\pi/2}\frac{\sin y\cos y}{2(2+\sin^2y\cos^2y)+3\sqrt{2+\sin^2y\cos^2y}}dy\\
&=\frac{1}{2}\int_0^{\pi/2}\frac{\sin y}{8+\sin^2y+3\sqrt{8+\sin^2y}}dy\end{align}$$
The rest should be manageable.
Addendum:
$$\begin{align}I&=\int_0^{\pi/2}\frac{2\sin y}{17-\cos 2y+3\sqrt{34-2\cos2y}}dy\\
&=\int_0^{\pi/2}\Big(\frac{3\sec y \tan y}{\sqrt{34-2\cos 2y}}-\frac{1}{2}\sec y \tan y \Big) dy\\
&=0-\lim_{y\rightarrow 0}(-\frac{1}{2}\sec y+\frac{\sqrt{17-\cos 2y}\sec y}{6\sqrt{2}})
\end{align}$$
which results in what we need.
The first part of the second integral is clear. For the the second part note that
$$\begin{align}\sec y\frac{3 \tan y}{\sqrt{34-2\cos 2y}}&=\sec y \Big( \frac{\sin 2y}{6\sqrt{34-2\cos 2y}}+\frac{\sqrt{34-2\cos 2y}\tan y}{12}\Big)\\
&=\Big(\frac{d \frac{\sqrt{17-\cos 2y}}{6\sqrt{2}}}{dy}\Big)\sec y + \Big(\frac{d \sec y}{dy}\Big) \frac{\sqrt{17-\cos 2y}}{6\sqrt{2}}
\end{align}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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prove that $a^2yz+b^2zx+c^2xy\leq 0$ for all real $x,y,x$ such that $x+y+z=0$ prove that $a^2yz+b^2zx+c^2xy\leq 0$ for all real $x,y,x$ such that $x+y+z=0$
Given that $a,b,c$ are positive real numbers such that the sum of any two is greater than third.
We have $x+y+z=0\implies x+y=-z\implies xz+yz=-z^2$
What will be the next?
| We need to prove that $$-a^2y(x+y)-b^2x(x+y)+c^2xy\leq0$$ or
$$b^2x^2+(a^2+b^2-c^2)xy+a^2y^2\geq0$$
and since $b>0$, it's enough to prove that
$$(a^2+b^2-c^2)^2-4a^2b^2\leq0$$ or
$$(a+b+c)(a+b-c)(a+c-b)(b+c-a)\geq0$$
and we are done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1026702",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Proving the inequality $\frac{9}4\ + \frac{3abc}4\ \ge ab+bc+ca$ If a,b,c are non-negative real numbers for which holds that $a+b+c=3$ then prove the following inequality: $$\frac{9}4\ + \frac{3abc}4\ \ge ab+bc+ca$$
| If we consider the homogeneous form of our inequality:
$$\frac{9}{4}\left(\frac{a+b+c}{3}\right)^3+\frac{3}{4}abc\geq (ab+ac+bc)\left(\frac{a+b+c}{3}\right)$$
we just have to prove that for any triple $(a,b,c)$ of non-negative real numbers
$$ a^3+b^3+c^3+3abc \geq ab(a+b)+bc(b+c)+ac(a+c)\tag{1} $$
holds, but $(1)$ is exactly Schur's inequality.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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How prove this inequality$4(a^2+b^2+c^2)+9a^2b^2c^2\ge 21$ let $a,b,c\in R$,and such $$ab+bc+ac=3$$
show that
$$4(a^2+b^2+c^2)+9a^2b^2c^2\ge 21$$
if this equlity condition is $a,b,c$ be postive numbers,and I can use $pqr$ to prove it.
But this is $a,b,c$ are real numbers
let $a+b+c=p,ab+bc+ac=q=3,abc=r$
so
$$a^2+b^2+c^2=p^2-6$$
then
$$4(a^2+b^2+c^2)+9a^2b^2c^2=4p^2-24+9r^2$$
since $a,b,c\in R$,so I can't use this important schur inequality：
$$9r\ge 4pq-p^3,a,b,c>0$$
| Let $a+b+c=3u$, $ab+ac+bc=3v^2$, $abc=w^3$ and $u^2=tv^2$.
Hence, our inequality is equivalent to $w^6\geq v^4(5v^2-4u^2)$.
Since, the inequality does not depend on substitution $a\rightarrow-a$, $b\rightarrow-b$ and $c\rightarrow-c$, we can assume $u\geq0$.
Easy to show that $(a-b)^2(a-c)^2(b-c)^2\geq0\Leftrightarrow3u^2v^4-4v^6-4u^3w^3+6uv^2w^3-w^6$ or
$3uv^2-2u^3-2\sqrt{(u^2-v^2)^3}\leq w^3\leq3uv^2-2u^3+2\sqrt{(u^2-v^2)^3}$.
For $t\geq\frac{5}{4}$ the inequality is obviously true.
But for $1\leq t\leq\frac{5}{4}$ it remains to prove that
$2\sqrt{(u^2-v^2)^3}+v^2\sqrt{5v^2-4u^2)}\leq3uv^2-2u^3$
or
$u^2(3v^2-2u^2)^2\geq\left(2\sqrt{(u^2-v^2)^3}+v^2\sqrt{5v^2-4u^2)}\right)^2$
or
$t(3-2t)^2\geq\left(2\sqrt{(t-1)^3}+\sqrt{5-4t}\right)^2$
or
$t-1\geq4\sqrt{(t-1)^3(5-4t)}$ or
$(t-1)^2(8t-9)^2\geq0$. Done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1028826",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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} |
find the last two digits of $2^{250}$. Suppose we want the last two digits of $3^{250}$, one can use the theorem $a^{\phi(n)}\cong 1(\mod n)$ whenever $(3,n)=1$. But instead, if i have $2^{250}$, how do i solve this problem, because here $(2,100)\neq 1$.
notation
$(a,b)=$the gcd of $a$ and $b$
| Nowhere near as neat as Exodd's solution but I thought I'd post this because I think it gives a handy insight into the behaviour of some powers of $2$ modulo $100$.
I exploited the fact that $2^{14} = 16384 \equiv -16 = -2^4 \pmod{100}$ and $2^{16} = 65536 \equiv -64 = -2^6 \pmod{100}$, giving $2^{30} = 2^{14} \cdot 2^{16} \equiv (-2^4)\cdot (-2^6) = 2^{10} \pmod{100}$.
So $2^{250} = 2^{30\cdot 8 + 10} \equiv (2^{10})^8\cdot 2^{10} = 2^{90} = (2^{30})^3 \equiv (2^{10})^3 = 2^{30} \equiv 2^{10} = 1024 \equiv 24 \pmod{100}$.
As I said, not as elegant, and perhaps not very general, but I put it down here in case it gives the original asker or others a little more insight.
| {
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"url": "https://math.stackexchange.com/questions/1033100",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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A die whose score cannot be as before (Markov chains) A die is "fixed" so that each time it is rolled the score cannot be the same as the preceding score, all other scores having probability $1/5$. Given that the first score is 6, what is the probability that the $n$th score is 6 and what is it if the $n$th score is 1?
HINT: You can simplify things by selecting an appropriate state-space; do you really need a 6-state chain to answer the question?
Define $u_n(j) := P(X_n = j \mid X_1 = 6) = P_6 (X_n = j)$ and note that
$P_6(X_n = j) = \frac{1}{5} P_6(X_{n-1} \neq 6) = \frac{1}{5} (1 - P_6(X_{n-1} = 6))$
So $u_n = \frac{1}{5} (1 - u_{n-1})$ and solving the difference equation for the cases $j=6,1$ we have
$u_n(6) = \frac{1}{6} (1+5(-\frac{1}{5})^{n-1})$
$u_n(1) = \frac{1}{6} (1-(-\frac{1}{5})^{n-1})$
However, I feel that by using difference equations to obtain the solutions I am missing the point of the question and the hint. I tried selecting several state-spaces but couldn't get anywhere. Can anyone shed some light on this?
| Hint: Define the random variable with $Y_n \in \{0,1\}$ (indicator variable with $2$ states) as follows $$Y_n=\begin{cases}1,& X_n=6\\0,&X_n\neq6\end{cases}$$ with transition matrix $$\mathbf P_{(1)}=\begin{array}{r|cc|r}&0&1&\\\hline0&\frac{4}{5}&\frac{1}{5}\\1&1&0 \end{array}$$ Initially $Y_0=1$. You want to find the probability $$P(Y_n=1|Y_0=1)$$ which can be done by calculating the matrix $\mathbf P_{(n)}=P^n$.
Indeed using diagonalization method (or preferably online calculators...) you find that $$\mathbf P_{(1)}=\begin{array}{r|cc|r}&0&1&\\\hline0&\frac{5}{6}+\frac{1}{6}\cdot\left(-\frac{1}{5}\right)^n&\frac{1}{6}-\frac{1}{6}\cdot\left(-\frac{1}{5}\right)^n\\1&\frac{5}{6}-\frac{5}{6}\cdot\left(-\frac{1}{5}\right)^n&\frac{1}{6}+\frac{5}{6}\cdot\left(-\frac{1}{5}\right)^n \end{array}$$
which gives you that $$p_{1,1}^{(n)}=\frac{1}{6}+\frac{5}{6}\cdot\left(-\frac{1}{5}\right)^n$$ and $$p_{1,0}^{(n)}=\frac{5}{6}-\frac{5}{6}\cdot\left(-\frac{1}{5}\right)^n$$ confirming your results (note that you have to multiply $p_{1,0}^{(n)}$ with $1/5$ to get the probability that $X_n=1$, since $Y_n=0$ implies only that $X_n\neq0$ and then there is equal probability $1/5$ for each result in $\{1,2,3,4,5\}$).
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Calculate $\log_{a}(ab)$, when $\log_{ab}(b)$ is known
If you know that $\log_{ab}(b) = k$, calculate $\log_{a}(ab)$.
Last time I was asked two times about this problem. $a,b$ was given, constant, such that $a,b \in \mathbb{Z} \wedge a,b > 1 \wedge \gcd(a,b) = 1$. What is strategy to solving this problem?
In fact it was a little bit harder - numbers occurred at different powers, eg. $\log_{25}50$, when $\log_{10}64 = k$, but it doesn't change a lot and this topic is about strategy to solve similar problem.
As someone asked again (example) I was a little bit confused, because problem is easy, and I thought, I should make some .pdf, but maybe it helps someone here. Alternative solutions are welcome.
| Strategy should be clear. Logarithms have to have same bases. We will use the basic facts.
$$\begin{split}
&\left(\forall x,y \in \mathbb{R}^{+}\smallsetminus\lbrace1\rbrace \right)
\left(\log_{x}{y} = \frac{1}{\log_{y}{x}} \right)\\
&\left(\forall x \in \mathbb{R}^{+}\smallsetminus\lbrace1\rbrace \right)
\left(\forall y,z \in \mathbb{R}^{+} \right)
\left(\log_{x}{yz} = \log_{x}{y} + \log_{x}{z} \right)\\
&\left(\forall x \in \mathbb{R}^{+}\smallsetminus\lbrace1\rbrace \right)
\left(\log_{x}{x} = 1 \right)
\end{split}$$
It is easer to change known logarithm.
$$\begin{align*}\begin{split}
k = \log_{ab}b = \frac{1}{\log_{b}ab} = \frac{1}{\log_{b}a + \log_{b}b} =
\frac{1}{\log_{b}a + 1} &= k &\Longleftrightarrow\\
k(\log_{b}a + 1) &= 1 &\Longleftrightarrow \\
\log_{b}a &= \frac{1-k}{k} &\Longleftrightarrow\\
\frac{1}{\log_{a}b} &= \frac{1-k}{k} & \Longleftrightarrow\\
\log_{a}{b} &= \frac{k}{1-k}
\end{split}\end{align*}$$
Now we just have to transform searched logarithm.
$$ \log_a{ab} = \log_{a}{a} + \log_{a}{b} = 1 + \log_{a}{b} =
1 + \frac{k}{1-k} = \frac{1}{1-k}$$
And for given example. Now we have to use more facts.
$$\begin{split}
&\left(\forall x \in \mathbb{R}^{+}\smallsetminus\lbrace1\rbrace \right)
\left(\forall \alpha \in \mathbb{R} \right)
\left(\log{x^{\alpha}} = \alpha \cdot \log{x} \right)\\
&\left(\forall x,y \in \mathbb{R}^{+}\smallsetminus\lbrace1\rbrace \right)
\left(\forall \alpha \in \mathbb{R} \right)
\left(\log_{x^{\alpha}}{y} = \frac{1}{\alpha} \cdot \log{y} \right)
\end{split}$$
$$\begin{align*}\begin{split}
k = \log64 = \log 2^{6} = 6 \log 2 = \frac{6}{\log_{2}(2 \cdot 5)} =
\frac{6}{\log_2{2} + \log_2{5}} =
\frac{6}{1 + \log_2{5}} &= k &\Longleftrightarrow\\
k(1 + \log_{2}5) &= 6 &\Longleftrightarrow\\
\log_{2}5 &= \frac{6-k}{k} &\Longleftrightarrow\\
\frac{1}{\log_{5}{2}} &= \frac{6-k}{k} &\Longleftrightarrow\\
\log_{5}{2} &= \frac{k}{6-k}
\end{split}\end{align*}$$
And transform searched logarithm $\log_{25}{50}$.
$$\begin{align*}\begin{split}
\log_{25}{50} = \log_{25}(25 \cdot 2) &= \log_{25}(25) + \log_{25}(2)\\
&= 1 + \log_{5^2}2 = 1 + \frac{1}{2}\log_{5}{2}\\
&= 1 + \frac{k}{2 \cdot (6-k)}\\
&= \frac{12-k}{12-2k}
\end{split}\end{align*}$$
So the answer is $\log_{25}50 = \frac{12-k}{12-2k}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1035042",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proving the convergence of this series I have the series $\displaystyle\sum_{n=1}^{\infty}{\frac{6\sqrt{n}+5}{2n^2-n}}$
I am sure that this series converges, but I need to prove this and would like to use the comparison test to do so.
I have rearranged
$$
\frac{6\sqrt{n}+5}{2n^2-n} = \frac{6}{\sqrt{n}(2n-1)}+\frac{5}{2n-1} = \frac{1}{2n-1}(\frac{6}{\sqrt{n}}+5)
$$
but I am unsure how to use the comparison test in this case?
| Instead of using the Basic Comparison Test, let us try and use another approach: The Limit Comparison test.
We are given the series $$\displaystyle \sum_{n=1}^\infty \frac{6\sqrt{n}+5}{2n^2 +n}$$
Let $\displaystyle a_n =\frac{6\sqrt{n}+5}{2n^2 +n} = \frac{1}{n^{\frac{3}{2}}}\times\frac{(6+\frac{5}{\sqrt{n}})}{(2+\frac{1}{n})} \ $
(Obtained by removing highest power in both numerator and denominator and simplifying)
Now choose $\displaystyle b_k = \frac{1}{n^{\frac{3}{2}}} \implies \sum_{k=1}^\infty b_k$ converges, since it is a Hyperharmonic Series with $p>1$.
Consider \begin{align}\frac{a_k}{b_k} &= \frac{6+\frac{5}{\sqrt n}}{2+ \frac{1}{n}} \xrightarrow{n \to \infty}\frac{6}{2}=3 >0\end{align}
Thus, by the Limit Comparison Test, we know $\displaystyle \sum_{n=1}^\infty \frac{6\sqrt{n}+5}{2n^2 +n}$ converges.
| {
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"timestamp": "2023-03-29T00:00:00",
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GCD of two polynomials Find the greatest common divisor of each of the following pairs $p(x)$ and $q(x)$ of polynomials. $p(x) = 7x^3 + 6x^2 −8x+ 4$ and $q(x) = x^3 + x− 2$, where $p(x), q(x) \in Q[x]$.
I divided $p(x)$ by $q(x)$ and I got $7$ with the remainder $r_1(x) = 6x^2 - 15x + 18$
then I got completely confused and don't know how to find the $\gcd$ of them.
Please help. Thank you!
| The basic of Euclidean algorithm more e.g. here. Having the polynomials $p,q$. We calculate
\begin{alignat*}{2}
p &= s_0\cdot q & &+ r_0\\
q &= s_1\cdot r_0 & &+ r_1\\
r_0 &= s_2 \cdot r_1 & &+ r_2\\
&\vdots & &\\
r_{n-3} &= s_{n-1}\cdot r_{n-2} & &+ r_{n-1}\\
r_{n-2} &= s_{n}\cdot r_{n-1} & &+ r_{n}.\\
\end{alignat*}
Each line means dividing one polynomial by other (you dividing $p$ by $q$ is the first row). $s_i$ and $r_j$ are also polynomials. $r_j$ are the remainders after the division. We do those steps while $r_n \neq 0$. If $r_n = 0$ then $\gcd{(p,q)} = r_{n-1}$.
Euclidean algorithm for $p(x), q(x)$.
$$
7x^3 + 6x^2 - 8x+4 = 7\cdot (x^3+x-2) + 6x^2-15x + 18
$$
So according to notation in the algorithm above $s_0(x) = 0$ and $r_0(x) = 6x^2-15x + 18$. No we divide $q(x)$ by $r_0(x)$
$$
x^3+x-2 = \left(\frac{1}{6}x + \frac{5}{12}\right)\cdot (6x^2-15x) + \frac{17}{4}x - \frac{19}{2}.
$$
So $s_1(x) = \frac{1}{6}x + \frac{5}{12}$ and $r_1(x) = \frac{17}{4}x - \frac{19}{2}$. Can you continue in this algorithm?
| {
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Limit of $\sqrt[n]{(x+1)...(x+n)} - x$ as $x \to +\infty$ Let $n \in \mathbb{N}^{\ast}$. I want to determine the following limit :
$$ \lim \limits_{x \to +\infty} \sqrt[n]{(x+1)\ldots(x+n)} - x.$$
Let $x = \frac{1}{t}$ with $t \to 0$. It is equivalent to the following limit :
$$ \lim \limits_{t \to 0} \sqrt[n]{\displaystyle \frac{(t+1)\ldots(t+n)}{t^{n}}} - \frac{1}{t}. $$
$$
\begin{align*}
\sqrt[n]{\displaystyle \frac{(t+1)\ldots(t+n)}{t^{n}}} - \frac{1}{t} &= {} \frac{1}{t} \sqrt[n]{(t+1)\ldots(t+n)} - \frac{1}{t} \\[2mm]
&= \frac{1}{t} \sqrt[n]{ n! + u(t) } - \frac{1}{t} \\[2mm]
\end{align*}
$$
where $u(0)=0$ and $\displaystyle \lim \limits_{t \to 0} u(t) = 0$. Since $\displaystyle \sqrt[n]{1+t} = 1 + \frac{t}{n} + o(t)$ as $t \to 0$, $$\displaystyle \sqrt[n]{n! + u(t)} = \sqrt[n]{n!} \; \sqrt[n]{\displaystyle 1 + \frac{u(t)}{n!} } = \sqrt[n]{n!} \bigg( 1 + \frac{u'(0) t}{n(n!)} + o(t) \bigg).$$
I do not see how to go on from there !
| We have
$$\sqrt[n]{a} - \sqrt[n]{b} = \dfrac{a-b}{\sum_{k=0}^{n-1} a^{k/n}b^{(n-1-k)/n}}$$
In our case, $a = (x+1)(x+2)\cdots(x+n)$ and $b=x^n$. We have $a-b = \frac{n(n+1)}2x^{n-1} + O(x^{n-2})$. The denominator is $nx^{n-1} + O(x^{n-2})$. Hence, we have the expression as
$$\frac{n(n+1)/2 + O(1/x)}{n + O(1/x)} \to \frac{(n+1)}2 \text{ as }x \to \infty$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Proof that $ \forall x,y \in \mathbb{R} \qquad x^2+y^2+(x-1)(y-1)>0 $ How to proof simply that
$$
\forall x,y \in \mathbb{R} \qquad x^2+y^2+(x-1)(y-1)>0
$$
| $$x^2+y^2+(x-1)(y-1) = x^2 + y^2 + xy - x -y+1 = \dfrac{(x+y)^2}2 + \dfrac{(x-1)^2}2 + \dfrac{(y-1)^2}2$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove the inequality: $\frac{a}{c+a-b}+\frac{b}{a+b-c}+\frac{c}{b+c-a}\ge{3}$ Prove the inequality:
$\frac{a}{c+a-b}+\frac{b}{a+b-c}+\frac{c}{b+c-a}\ge{3}$
Where $a,b,c$ are sides of a triangle.
It is clear that $c+a-b$ is positive but how to use it?
| $$a+b-c=x>0$$
$$c+a-b=y>0$$
$$b+c-a=z>0$$
The original inequality becomes
$$\frac{x+y}{2y}+\frac{x+z}{2x}+\frac{y+z}{2z}\geq 3$$
$$\frac{x}{2y}+\frac{z}{2x}+\frac{y}{2z}\geq \frac{3}{2}$$ which is trivial by AM-GM inequality
| {
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"timestamp": "2023-03-29T00:00:00",
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$\forall\ x,y,z\in \mathbb{R}$ Show that: $|x+y|+|y+z|+|x+z|\leq |x+y+z|+|x|+|y|+|z|$
$\forall\ x,y,z\in \mathbb{R}$ Show that: $$|x+y|+|y+z|+|x+z|\leq |x+y+z|+|x|+|y|+|z|$$
i tired,
i notice that $x,y,z$ plays a symmetrical role in the inequality
notice also that
\begin{align*}
|x+y|+|y+z|+|x+z|\leq |x+y+z|+|x|+|y|+|z| & \Longleftrightarrow \\
(|x+y|-|x|)+(|y+z|-|y|)+(|x+z|-|z|) \leq |x+y+z|
\end{align*}
note that $\forall a,b\in \mathbb{R}\quad |a|-|b|\leq |a+b| $
then $$(|x+y|-|x|)+(|y+z|-|y|)+(|x+z|-|z|) \leq |x|+|y|+|z|$$
i'm stuck here
any help would be appreciated!
| Let $a=x+y$, $b=x+z$ and $c=y+z$. Then, the inequality to be shown
can be rewritten as
$$
2(|a|+|b|+|c|) \leq |a+b+c|+|a+b-c|+|a-b+c|+|-a+b+c| \tag{1}
$$
Let us put $f(a,b,c)=|a+b+c|+|a+b-c|+|a-b+c|+|-a+b+c|$. It is easy to
see that $f$ is even in each variable, so that $f(a,b,c)=f(|a|,|b|,|c|)$.
We may therefore assume that $a,b,c$ are all nonnegative, so that it suffices
to show that
$$
2(a+b+c) \leq |a+b+c|+|a+b-c|+|a-b+c|+|-a+b+c|\tag{2}
$$
But (2) follows from the triangle inequality, since
$$
2(a+b+c)=(a+b+c)+(a+b-c)+(a-b+c)+(-a+b+c) \tag{3}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "6",
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How to evaluate the integral of $\sqrt{\sin\sqrt x}\cos \sqrt x / ( 1+x^2)$? $$
\int \frac{\sqrt{\sin\sqrt x}\cos \sqrt x}{1+x^2} dx
$$
I have tried combinations of $x=t^2$, integration by parts, $\tan\left(\dfrac u2\right)$ substitutions it got even more complicated. Is there a way to evaluate this integral with elementary techniques?
| Hint:
Let $t=\sqrt x$ ,
Then $x=t^2$
$dx=2t~dt$
$\therefore\int\dfrac{\sqrt{\sin\sqrt x}\cos\sqrt x}{1+x^2}dx$
$=\int\dfrac{\sqrt{\sin t}\cos t}{t^4+1}dt$
$=\int\dfrac{\sqrt{\sin t}}{t^4+1}d(\sin t)$
$=\int\dfrac{2}{3(t^4+1)}d\left(\sin^\frac{3}{2}t\right)$
$=\dfrac{2\sin^\frac{3}{2}t}{3(t^4+1)}-\int\sin^\frac{3}{2}t~d\left(\dfrac{2}{3(t^4+1)}\right)$
$=\dfrac{2\sin^\frac{3}{2}t}{3(t^4+1)}+\int\dfrac{8t^3\sin^\frac{3}{2}t}{3(t^4+1)^2}dt$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Finding $\lim_{x\rightarrow 0}\frac{x}{2}\sqrt{\frac{1+\cos(x)}{1-\cos(x)}}$ We know that $$\sqrt{\frac{1-\cos(x)}{1+\cos(x)}}=\tan({x}/{2})$$ so we can change the above function to another form as follow $$\frac{x}{2}\sqrt{\frac{1+\cos(x)}{1-\cos(x)}}=\frac{(x/2)}{\tan(x/2)}$$ If we use the Taylor series of $\tan(x/2)$ we will get
$$\frac{(x/2)}{\tan(x/2)}=\frac{(x/2)}{\frac{x}{2}+\frac{x^3}{24}+\frac{x^5}{240}+....}=\frac{1}{1+\frac{x^2}{12}+\frac{x^4}{120}+...}$$ Now we will find the limit for the two same functions
$$\lim_{x\rightarrow 0^{+}}\frac{1}{1+\frac{x^2}{12}+\frac{x^4}{120}+...}=1$$
$$\lim_{x\rightarrow 0^{-}}\frac{1}{1+\frac{x^2}{12}+\frac{x^4}{120}+...}=1$$
$$\lim_{x\rightarrow 0^+}\frac{x}{2}\sqrt{\frac{1+\cos(x)}{1-\cos(x)}}=1$$
$$\lim_{x\rightarrow 0^-}\frac{x}{2}\sqrt{\frac{1+\cos(x)}{1-\cos(x)}}=-1$$
My question is "Which solution is right?"
| To begin with, it's $\sqrt{\frac{1-\cos{x}}{1+\cos{x}}}=\left| \tan{\left( \frac{x}{2}\right)} \right|$.
Anyway, you could have answered without expanding into Taylor series.
$$ \frac{\frac{x}{2}}{\left| \tan{\left( \frac{x}{2}\right)} \right|} = \frac{\frac{x}{2}\left| \cos{\frac{x}{2}} \right|}{\left| \sin{\left( \frac{x}{2}\right)} \right|} = \frac{\frac{x}{2}}{\left| \sin{\frac{x}{2}} \right|}\left| \cos{\frac{x}{2}} \right|$$
You know that $\lim_{x \to 0}\left| \cos{\frac{x}{2}} \right| = 1$ and it is easy to prove that $\lim_{x \to 0}\frac{x/2}{\left| \sin{x/2} \right|}$ doesn't exists. In fact, its lateral limits exist, and are 1 from one side and -1 from the other!
| {
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"timestamp": "2023-03-29T00:00:00",
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How to determine without calculator which is bigger, $\left(\frac{1}{2}\right)^{\frac{1}{3}}$ or $\left(\frac{1}{3}\right)^{\frac{1}{2}}$ How can you determine which one of these numbers is bigger (without calculating):
$\left(\frac{1}{2}\right)^{\frac{1}{3}}$ , $\left(\frac{1}{3}\right)^{\frac{1}{2}}$
| $((\frac{1}{2})^{\frac{1}{3}})^6=(\frac{1}{2})^2=\frac{1}{4}$
$((\frac{1}{3})^{\frac{1}{2}})^6=(\frac{1}{3})^3=\frac{1}{27}$
So as it is obvious from the above relations, $((\frac{1}{2})^{\frac{1}{3}})^6>((\frac{1}{3})^{\frac{1}{2}})^6$, so we can say $(\frac{1}{2})^{\frac{1}{3}}>(\frac{1}{3})^{\frac{1}{2}}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1041684",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
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Minimal Polynomial of $a +b\sqrt{2}$ as a function of a, b ∈ $\mathbb Q$
Determine the minimal polynomial over $\mathbb Q$ of $a +b\sqrt{2}$ as a function of a, b ∈ $\mathbb Q$.
Let $x=a+b\sqrt{2}$
If $b=0$ then the minimal polynomial is $x-a$
if not, then $x-a=b\sqrt{2}\iff(x-a)^2-2b^2=x^2-2ax+a^2-2b^2=0$
Is the polynomial further reducible, can I use Eisenstein ?
$p|-2a$ and $p^2\not|a^2-2b^2$
for $p=2, a=4, b=2$ it does not work, or does it ?
| The easiest way is just high-school mathematics, I think: Take $(X-(a+b\sqrt2))(X-(a-b\sqrt2))=X^2-2aX+(a^2-2b^2)$. The discriminant is $(-2a)^2-4\cdot1\cdot(a^2-2b^2)=8b^2$, which is not a square unless $b=0$. In that reducible case, then the two factors I wrote are equal, so you take one of them, $X-a$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Integral of $(\log x)^3$ (Spivak's Calculus, Chapter 19, Problem 3v)
$$\int (\log x)^3\,dx$$
I solved the problem by applying integration by parts twice:
First, I write the integrand as $(\log x)^2\cdot\log x$, and then I set $u=(\log x)^2 \Rightarrow du=\frac{2\log x}{x}dx$ and $dv=\log x\,dx \Rightarrow v=x\log x-x$:
$$\begin{align}\int (\log x)^2\cdot\log x\,dx&=x(\log x)^3-x(\log x)^2-\left(\int \frac{2\log x}{x}(x\log x-x)\,dx\right)\\&=x(\log x)^3-x(\log x)^2-\left(\int 2(\log x)^2\,dx-\int 2\log x\,dx\right)\\&=x(\log x)^3-x(\log x)^2-2\int (\log x)^2\,dx+\underbrace{\int 2\log x\,dx}_{=\,2x\log x-2x} \end{align}$$
Now, the integral of $(\log x)^2$ needs to be solved by integration by parts. I set $u=\log x\Rightarrow du=\frac{dx}{x}$ and $dv=\log x\,dx\Rightarrow v=x\log x-x$:
$$\begin{align}\int (\log x)^2\,dx&=x(\log x)^2-x\log x-\left(\int \frac{x\log x-x}{x}\,dx\right)\\&=x(\log x)^2-x\log x-\left(\int \log x\,dx-\int dx\right)\\&=x(\log x)^2-x\log x-(x\log x-2x)\\&=x(\log x)^2-2x\log x+2x\end{align}$$
Substituting the result of $\int (\log x)^2\,dx$ into the solution:
$$\begin{align}\int (\log x)^2\cdot\log x\,dx&=x(\log x)^3-x(\log x)^2-2(\overbrace{x(\log x)^2-2x\log x+2x}^{\,=\int\,(\log x)^2dx})+2x\log x-2x\\&=x(\log x)^3-x(\log x)^2-2x(\log x)^2+4x\log x-4x+2x\log x-2x\\&=x(\log x)^3-3x(\log x)^2+6x\log x-6x\end{align}$$
Therefore $$\int (\log x)^3\,dx=x(\log x)^3-3x(\log x)^2+6x\log x-6x$$
(I omitted the annoying $+C$ in my calculations)
My question is: Is there any more elegant and quicker solution?
| Let $x = e^y$, $dx = e^y \, dy$, giving $$\int (\log x)^3 \, dx = \int y^3 e^y \, dy.$$ Now use tabular integration: $$\begin{array}{|cc|c|} \hline y^3 & e^y & + \\ 3y^2 & e^y & - \\ 6y & e^y & + \\ 6 & e^y & - \\ \hline \end{array}$$ to obtain $$\int y^3 e^y \, dy = y^3 e^y - 3y^2 e^y + 6y e^y - 6 e^y + C.$$ Substituting back for $x$ gives the desired result.
Another example of tabular integration: compute $$\int (x^4 + x^2) e^{-3x} \, dx.$$ We write $$\begin{array}{|cc|c|} \hline x^4 + x^2 & -\frac{1}{3}e^{-3x} & + \\ 4x^3 + 2x & \frac{1}{9} e^{-3x} & - \\ 12x^2 + 2 & -\frac{1}{27} e^{-3x} & + \\ 24x & \frac{1}{81} e^{-3x} & - \\ 24 & -\frac{1}{243} e^{-3x} & + \\ \hline \end{array}$$ to get $$\begin{align*} \int (x^4 + x^2) e^{-3x} \, dx &= -\frac{1}{3}(x^4 + x^2)e^{-3x} - \frac{1}{9}(4x^3 + 2x)e^{-3x} - \frac{1}{27}(12x^2 + 2)e^{-3x} \\ &\quad - \frac{1}{81}(24x)e^{-3x} - \frac{24}{243}e^{-3x} + C \\ &= -\frac{e^{-3x}}{81} (27x^4 + 36x^3 + 63x^2 + 42x + 14) + C. \end{align*}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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How to add/subtract complex rational expressions? I'm studying for my Precalculus final and have noticed I still don't fully grasp performing basic operations on complex rational expressions, or finding if any values must be restricted from the domain. For example:
$$\frac{3}{5x+2}+\frac{5x}{25x^2-4}.$$
I know if this was a division problem, I would first factor out any exponential values of x and then multiply the first expression by the reciprocal of the second complex rational expression and then simplify.
Is there a similar process for addition/subtraction?
| Find the (least) common denominator for $$\frac{3}{5x+2}+\frac{5x}{25x^2-4}.$$ Note that $$25x^2 - 4 = (5x-2)(5x+2),$$ and so, $$d = 25x^2 - 4 = (5x-2)(5x+2), \quad d\neq \pm \frac 25$$ is the common denominator, because each of $5x+2$ and $(5x+2)(5x-2)$ divide $d$.
Write each fraction using the common denominator (let's just call it $d$ for now). Once you've done that, find the sum:
$$\text{E.g.,}\;\;\frac ad + \frac cd = \frac {a+c}{d}$$
That gives us $$\begin{align} \frac{3}{5x+2}+\frac{5x}{25x^2-4} & =\frac{3(5x-2)}{(5x - 2)(5x+2)}+\frac{5x}{(5x-2)(5x+2)}\\ \\& = \dfrac{3(5x-2) + 5x}{(5x-2)(5x+2)}\end{align}$$
Now simplify.
Added note: Don't forget that we need to keep in mind that the sum is not defined when $(5x+2)(5x-2) = 0$, i.e., it is not defined when $x = \pm \frac 25$. But that should not be surprising, neither of the original summands is defined at $x = -\frac 25$, and additionally, the second summand is not defined at $x = \frac 25$. So it makes sense that the function, as a sum, is not defined at $\pm \frac{2}{5}$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Investigate convergence of the following series Investigate the corvergence of the following series?
$1+\frac{1}{3}-\frac{1}{2}+\frac{1}{5}+\frac{1}{7}-\frac{1}{4}+\frac{1}{9}+\frac{1}{11}-\frac{1}{6}+\frac{1}{13}+\frac{1}{15}-\frac{1}{8}+ \ldots$
$1+\frac{1}{2}-\frac{1}{3}-\frac{1}{4}-\frac{1}{5}+\frac{1}{6}+\frac{1}{7}-\frac{1}{8}-\frac{1}{9}-\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}-\frac{1}{15}+ \ldots$
I need a hit about which cirterion to use in each of those cases and i do not know how to find a general term of those twos sequence?
| For the second, compute
$$\frac1{5n+1}+\frac1{5n+2}-\frac1{5n+3}-\frac1{5n+4}-\frac1{5n+5}=\frac{-625n^4+\cdots}{3125n^5+\cdots}$$
and compare with harmonic series.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Prove $1 + \frac{1}{4} + \frac{1}{9} + \cdots + \frac{1}{n^2} < 2 - \frac{1}{n}$ by Induction The Question
Prove $1 + \frac{1}{4} + \frac{1}{9} + \cdots + \frac{1}{n^2} < 2 - \frac{1}{n}$ where $n\ge2$ and $n$ is an integer by Induction
My Work
Basis Step:
1 + $\frac{1}{4} = \frac{5}{4}$
$2-\frac{1}{2} = \frac{3}{2}$
$\frac{5}{4}<\frac{3}{2}$
Inductive Hypothesis:
$1+\frac{1}{4}+\frac{1}{9} + \cdots + \frac{1}{k^2} < 2 - \frac{1}{k}$
Induction Step
We must show $1+\frac{1}{4}+\frac{1}{9} + \cdots + \frac{1}{k^2} + \frac{1}{(k+1)^2} < 2 - \frac{1}{(k+1)}$
$1+\frac{1}{4}+\frac{1}{9} + \cdots + \frac{1}{k^2} < 2 - \frac{1}{k}$
$2 - \frac{1}{k} + \frac{1}{(k+1)^2} = 2 - \frac{(k^2+2k+1)+k}{k(k+1)^2} = 2-\frac{k^2 + 3k + 1}{k^3 +2k^2+k}$
My Problem
I can't seem to complete the proof. Any hints on how to change my approach for success?
| $\displaystyle 2-\frac{k^2+3k+1}{k^3+2k^2+k}=$
$\displaystyle 2-\frac{k+3+1/k}{k^2+2k+1}=$
$\displaystyle 2-\frac{k+3+1/k}{(k+1)^2}=$
$\displaystyle 2-\frac{k+1+2+1/k}{(k+1)^2}=$
$\displaystyle =2-\frac{1}{k+1}-\frac{2+1/k}{(k+1)^2}<$
$\displaystyle 2-\frac{1}{k+1}$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Why are $x$, $x^3+1$ and $x^2+x+1$ always mutually co-prime, for any natural number $x$? I just read that for a natural number $x$, the three numbers $x$, $x^3 + 1$ and $x^2 + x + 1$ are all mutually co-prime.
I couldn't find a reason why this is true. OK, any of them does not divide either of the other two, but is it enough to conclude that their GCD is 1?
Couldn't they have any common factors?
Also on what basis, are two consecutive integers co-prime? I know they cannot have a common factor but don't know why.
| We see that the gcd of $m$ and $n$ always divides any expression of the form $am+bn$ since $m=ck, n=cj\implies c|(am+bn)=c(ak+bj)$. Being coprime means the gcd is $1$, so let's show this.
Case 1: For $m=x$ and $n=x^3+1$ we see with $a=-x^2$ and $b=1$ we have
$$-x^3+x^3+1=1$$
so they are coprime.
Case 2: For $m=x$ and $n=x^2+x+1$ we see $a=-(x+1)$ and $b=1$ gives
$$-x^2-x+x^2+x+1=1$$
which verifies coprimality.
Case 3: Finally with $m=x^2+x+1$ and $n=x^3+1$ we see that $a=(x-1)$ and $b=-1$ gives us
$$-x^3+1+x^3+1=2$$
so that $\gcd(x^2+x+1,x^3+1)\big|2$. But $x^2+x+1$ is always odd, so the gcd must be $1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1048254",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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System of 4 quartic equations \begin{align*}a &=\sqrt{4+\sqrt{5+a}},\\ b &=\sqrt{4-\sqrt{5+b}},\\ c &=\sqrt{4+\sqrt{5-c}},\\ d &=\sqrt{4-\sqrt{5-d}}.\end{align*}
Compute $abcd$.
I set up each as a quartic and got $a^4-8a^2-a+11=0$ and similar equations for $b,c,d$, but am unable to figure out how to format the roots as a product of all of them. Could someone give me a bit of a hint? Thank you.
| Notice that $a, b, c, d$ are the roots of the equation
$$(x^2 - 4)^2 = 5 + x$$
$$x^2 - 4 = \pm\sqrt{5 + x}$$
$$x^2 = 4 \pm \sqrt{5 + x}$$
$$x = 4 \pm \sqrt{5 + x}, -(4 \pm \sqrt{5 + x})$$
$$x = 4 \pm \sqrt{5 + x}, 4 \pm \sqrt{5 - x}$$
Now, the quartic equation corresponding to the first equation is, as you've expanded,
$$x^4 - 8x^2 - x + 11 = 0$$
By Vieta's Formulas, the product of the roots ($a, b, c, d$) is given by
$$abcd = (-1)^4\frac{11}{1} = 11$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1049196",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Lower estimate for $(\frac{\ln(1+2x)}{\ln(1+x)}-1)(1+2x)^{1/2}$ where $x>0$ I want to prove that:
$$\left(\frac{\ln(1+2x)}{\ln(1+x)}-1\right)(1+2x)^\frac{1}{2}\geq 1$$
where $x>0$. Any help appreciated. Thanks!
| Here's a not very beautiful solution.
As $x > 0$, it is equivalent to show that
$$\log\left(1 + \frac{x}{1 + x}\right)\sqrt{1 + 2 x} - \log(1 + x) \geq 0.$$
As LHS $\rightarrow 0$ as $x\rightarrow 0$, it is enough to show that the LHS is increasing as a function of $x$. That is, that its derivative is greater than $0$ for $x > 0$. Differentiating, we obtain the inequality
$$\frac{1 - \sqrt{1 + 2 x} + (1 + x) \log\left(1 + \frac{x}{1 + x}\right)}{(1 + x) \sqrt{1 + 2 x}} \geq 0.$$
It is of course enough to show that
$$1 - \sqrt{1 + 2 x} + (1 + x) \log\left(1 + \frac{x}{1 + x}\right)\geq 0.$$
By the same argument (i.e., limit when $x\rightarrow 0$ is $0$, so it is enough to show LHS is increasing as a function of $x$),
differentiating, it is enough to show that
$$\frac{1}{1 + 2 x} - \frac{1}{\sqrt{1 + 2 x}} + \log\left(1 + \frac{x}{1 + x}\right) \geq 0.$$
By the same argument yet again (sorry!), it is enough to show that
$$\frac{1 - \sqrt{1 + 2 x} + x (3 + 2 x)}{(1 + x) (1 + 2 x)^{\frac{5}{2}}} \geq 0.$$
So it is enough to show that
$$1 - \sqrt{1 + 2 x} + x (3 + 2 x) \geq 0.$$
But this one is easy!
| {
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"timestamp": "2023-03-29T00:00:00",
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How to integrate $\int \frac{1}{\sqrt{1+29x^2+100x^4}}dx$ and $\int \frac{1}{\sqrt{1-2x^2-8x^4}}dx$ using elliptic functions? How to integrate $$\int \frac{1}{\sqrt{1+29x^2+100x^4}}dx$$ and $$\int\frac{1}{\sqrt{1-2x^2-8x^4}}dx$$ using elliptic functions?
I have tried to use them, but I got incorrect formula $$\frac{1}{\sqrt{2}}F(arctan(\sqrt{2}x)∣3)$$ for the first one (second argument should be less or equal 1).
Could anyone solve it?
Thanks
| I'm not totally sure how to do it by hand, but using the identity
$$\int\frac{1}{\sqrt{a+bx^2+cx^4}}\mathrm{d}x=-\frac{i \sqrt{\frac{\sqrt{b^2-4 a c}+b+2 c x^2}{\sqrt{b^2-4 a c}+b}} \sqrt{\frac{2 c
x^2}{b-\sqrt{b^2-4 a c}}+1} F\left(i \sinh ^{-1}\left(\sqrt{2}
\sqrt{\frac{c}{b+\sqrt{b^2-4 a c}}} x\right)|\frac{b+\sqrt{b^2-4 a c}}{b-\sqrt{b^2-4 a
c}}\right)}{\sqrt{2} \sqrt{\frac{c}{\sqrt{b^2-4 a c}+b}} \sqrt{a+b x^2+c x^4}}$$
one obtains
$$\int\frac{1}{\sqrt{1+29x^2+100x^4}}\mathrm{d}x=-\frac{i \sqrt{4 x^2+1} \sqrt{25 x^2+1} F\left(i \sinh ^{-1}(5 x)|\frac{4}{25}\right)}{5
\sqrt{100 x^4+29 x^2+1}}$$
and
$$\int\frac{1}{\sqrt{1-2x^2-8x^4}}\mathrm{d}x=\frac{1}{2} F\left(\sin ^{-1}(2 x)|-\frac{1}{2}\right)$$
which agrees with numerical experiments.
| {
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"timestamp": "2023-03-29T00:00:00",
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The value of $b$, if $\lim_{x \to 0}(\frac{ae^x-b\cos x+ce^{-x}}{x\sin x})=2$ Question :
The value of $b$, if
$$\lim_{x \to 0^+}(\frac{ae^x-b\cos x+ce^{-x}}{x\sin x})=2$$
Solution : $$= \lim_{x \to 0^+} \frac{\log_e\left[\frac{4^x-3^x}{4^x-1}\right]}{x}(4^x-1)$$
$$= \lim_{x \to 0^+} \frac{\left[1-\frac{3^x-1}{4^x-1}\right]}{\left(\frac{3^x-1}{4^x-1}\right)}\times \frac{3^x-1}{4^x-1}\times \frac{4^x-1}{x}$$
I am unable to take it further. Please help on this thanks
| We start with Taylorexpanding the numerator and denominator:
\begin{align}
&\lim_{x \to 0^+}\left(\frac{ae^x-b\cos(x)+ce^{-x}}{x\sin(x)}\right)\\
=&\lim_{x \to 0^+}\left(\frac{a+ax+\frac{1}{2}ax^2-b(1-\frac{1}{2}x^2)+c-cx+\frac{1}{2}cx^2+O(x^3)}{x^2+O(x^3)}\right)\\
=&\lim_{x \to 0^+}\left(\frac{a+c-b+(a-c)x+\frac{1}{2}(a+b+c)x^2+O(x^3)}{x^2+O(x^3)}\right)=2\\
\end{align}
From this follows $a+c-b=0$ and $a+b+c=4$. We conclude that $b=2$
| {
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"url": "https://math.stackexchange.com/questions/1055810",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Evaluate $\int\sqrt[5]{\frac{x+5}{x-5}}\,\mathrm dx$ How to compute this integral? I stuck at a point where I get $\displaystyle\int\frac{1}{t^5-1}+ \cdots
$ $$\int\sqrt[5]{\frac{x+5}{x-5}}\,\mathrm dx$$ using $\displaystyle t=\sqrt[5]{\frac{x+5}{x-5}}$
| Substituting $t=\sqrt[\Large5]{\frac{x+5}{x-5}}$,
$$
\begin{align}
\int\sqrt[\Large5]{\frac{x+5}{x-5}}\,\mathrm{d}x
&=\int t\,\mathrm{d}\frac{10}{t^5-1}\\
&=\frac{10t}{t^5-1}-10\int\frac{\mathrm{d}t}{t^5-1}\\
\end{align}
$$
Substituting $\sin(2\pi/5)u=t-\cos(2\pi/5)$ and $\sin(4\pi/5)v=t-\cos(4\pi/5)$,
$$
\begin{align}
&\frac1{t^5-1}\\
&=\frac15\sum_{k=0}^4\frac{e^{2\pi ik/5}}{t-e^{2\pi ik/5}}\\
&=\frac{2t\cos(2\pi/5)-2}{t^2-2t\cos(2\pi/5)+1}+\frac{2t\cos(4\pi/5)-2}{t^2-2t\cos(4\pi/5)+1}+\frac1{t-1}\\
&=2\frac{\cos(2\pi/5)(t-\cos(2\pi/5))-\sin^2(2\pi/5)}{(t-\cos(2\pi/5))^2+\sin^2(2\pi/5)}\\
&+2\frac{\cos(4\pi/5)(t-\cos(4\pi/5))-\sin^2(4\pi/5)}{(t-\cos(4\pi/5))^2+\sin^2(4\pi/5)}\\
&+\frac1{t-1}\\[6pt]
&=2\frac{\cot(2\pi/5)u-1}{u^2+1}+2\frac{\cot(4\pi/5)v-1}{v^2+1}+\frac1{t-1}
\end{align}
$$
so that
$$
\begin{align}
\int\frac1{t^5-1}\,\mathrm{d}t
&=\cos(2\pi/5)\log(u^2+1)-2\sin(2\pi/5)\arctan(u)\\
&+\cos(4\pi/5)\log(v^2+1)-2\sin(4\pi/5)\arctan(v)\\[6pt]
&+\log(t-1)\\[6pt]
&+C
\end{align}
$$
Therefore, using the substitutions above, which are unwieldy to write, but simple to compute,
$$
\begin{align}
\int\sqrt[\Large5]{\frac{x+5}{x-5}}\,\mathrm{d}x
&=\frac{10t}{t^5-1}\\[3pt]
&-10\cos(2\pi/5)\log(u^2+1)+20\sin(2\pi/5)\arctan(u)\\[6pt]
&-10\cos(4\pi/5)\log(v^2+1)+20\sin(4\pi/5)\arctan(v)\\[6pt]
&-10\log(t-1)\\[6pt]
&-10\,C
\end{align}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Proof that $\int_{0}^{1}\frac{dx}{1+x^6}=\frac{\pi+\sqrt3\log(2+\sqrt3)}{6}$ without residues. How do you prove that
$$\int_{0}^{1}\dfrac{dx}{1+x^6}=\frac{\pi+\sqrt3\log(2+\sqrt3)}{6}$$
My steps:
First sub $\displaystyle u=x^3, \sqrt[3]u=x, dx=\dfrac{u^{-2/3}}{3} du\implies\dfrac{1}{3}\int_{0}^{1}\dfrac{u^{-2/3}}{1+u^2}$
Then sub $\displaystyle u=\tan\theta, du=\sec^2\theta d\theta\implies \dfrac{1}{3}\int_{0}^{\pi/4}\sec^{-4/3}\theta d\theta$
After which I am stuck...
| $$\begin{align}
\int_0^1\frac{1}{1+x^6} \,\mathrm dx
&=\frac{1}{2}\left[ \int_0^1 \frac{(1-x^2+x^4)+x^2+(1-x^4)}{(1+x^2)(1-x^2+x^4)} \,\mathrm dx \right]\tag{1}\\
&=\frac{1}{2}\left[\int_0^1 \frac{1}{1+x^2} \,\mathrm dx+ \int_0^1 \frac{x^2}{1+x^6} \,\mathrm dx + \color{grey}{\int_0^1 \frac{1-x^2}{1-x^2+x^4} \,\mathrm dx}\right] \tag{2}\\
&=\frac{1}{2}\left[\frac\pi4+ \frac\pi{12} +\frac{\log(2+\sqrt{3})}{\sqrt{3}} \right] \tag{3}\\
&=\frac12\left[\frac{\pi+\sqrt3\log(2+\sqrt{3})}{3} \right] \tag{4}\\
\end{align}$$
$$\int_0^1\frac{1}{1+x^6} \,\mathrm dx
=\frac{\pi+\sqrt3\log(2+\sqrt{3})}{6}$$
$\text{Explanation : }(3)$
Substituting $\displaystyle t=x+\frac1x\iff \,\mathrm dt=\left(1-\frac1{x^2}\right)\,\mathrm dx$ in last integral
$$\begin{align}
\color{grey}{J}
&=\color{grey}{\int_0^1 \frac{1-x^2}{1-x^2+x^4} \,\mathrm dx}
=\int_0^1 \frac{\frac{1}{x^2}-1}{x^2-1+\frac{1}{x^2}}\,\mathrm dx
=\int_2^\infty \frac{1}{t^2-3}\,\mathrm dt\\
&=\frac{1}{2\sqrt{3}}\log\left(\frac{2+\sqrt{3}}{2-\sqrt{3}}\right)
=\color{grey}{\frac{\log(2+\sqrt{3})}{\sqrt{3}}}\\
\end{align}$$
| {
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"url": "https://math.stackexchange.com/questions/1058535",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Integral of $\dfrac{\cos(x)}{5+3\cos(x)}$ I was doing
$$\int\!\mathrm{d}x \dfrac{\cos(x)}{5+3\cos(x)}$$
and using the substitution $\cos(\theta) = \dfrac{1-t^2}{1+t^2},\quad t = \tan\left(\dfrac{\theta}{2}\right)$
$$\dfrac{\dfrac{1-t^2}{1+t^2}}{5+3\dfrac{1-t^2}{1+t^2}} =
\dfrac{\dfrac{1-t^2}{1+t^2}}{5\dfrac{1+t^2}{1+t^2}+3\dfrac{1-t^2}{1+t^2}} =
\dfrac{\dfrac{1-t^2}{1+t^2}}{\dfrac{5(1+t^2)+3(1-t^2)}{1+t^2}} =
\dfrac{\dfrac{1-t^2}{1+t^2}}{\dfrac{5+5t^2+3-3t^2}{1+t^2}}\\[4em]
\dfrac{\dfrac{1-t^2}{1+t^2}}{\dfrac{8+2t^2}{1+t^2}} =
\dfrac{\dfrac{1-t^2}{1+t^2}}{\dfrac{2(4+t^2)}{1+t^2}} =
\dfrac{1-t^2}{1+t^2}\times\dfrac{1+t^2}{2(4+t^2)} =
\dfrac{1-t^2}{2(4+t^2)} =
\dfrac{1}{2}\dfrac{1-t^2}{4+t^2}$$
So now I'm left with $\dfrac{1}{2}\int\!\mathrm{d}x\dfrac{1-t^2}{4+t^2}$, but I don't know how to proceed from here...
| $$\frac{\cos x}{5+3\cos x}=\frac13\left(1-5\cdot\frac1{5+3\cos x}\right)$$
$$\int\frac1{5+3\cos x}dx=\int\frac{1+t^2}{5(1+t^2)+3(1-t^2)}\frac{2t\ dt}{1+t^2}$$
$$=\int\frac{dt}{2^2-t^2}=\frac14\int\left(\frac1{2-t}+\frac1{2+t}\right)dt=\cdots$$
Hope the rest can be handled easily.
| {
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"timestamp": "2023-03-29T00:00:00",
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Alternate way to Prove or disprove $6\mid n(n+1)(n+2)$ This is my proof, I'm wondering if I'm correct, and how to do without induction.
My Work
Basis Step
$$\frac{(1)(2)(3)}{6} = 1$$
Inductive Hypothesis
Assume that $\dfrac{k(k+1)(k+2)}{6} = d$ where $d \in \mathbb Z$
Induction Step
We must show $\frac{(k+1)(k+2)(k+3)}{6}$ is an integer
$$\frac{(k+1)(k+2)(k+3)}{6} = \frac{(k+1)(k+2)}{6} \cdot (k+3)$$
We know that:
$\dfrac{(k+1)(k+2)}{6} = \dfrac{1}{k}\cdot\dfrac{k(k+1)(k+2)}{6}$ by our inductive hypothesis we know this to be an integer
Therefore, by the closure property of integers under multiplication and addition, $\dfrac{(k+1)(k+2)}{6} \cdot (k+3)$ must be integer.
| Without induction:
At least one of $n, n+1, n+2$ is even and exactly one is divisible by three. Hence $2 \ | \ n(n+1)(n+2)$ and $3 \ | \ n(n+1)(n+2)$. As $2$ and $3$ are co-prime, this means $6 \ | \ n(n+1)(n+2)$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove $x + \frac{1}{x} \geq 2$ for $x>0$. Proof that $x+\frac{1}{x}\geq2$ for $x>0$
Would this be correct?
$x*(x+\frac{1}{x}\geq2)$
$x^2+1\geq2x$
$x^2-2x+1\geq2x-2x$
$x^2-2x+1\geq0$
Plug in 1 for x:
$(1)^2-2(1)+1\geq0$
$1-2+1\geq0$
$0\geq0$ Therefore, $x+\frac{1}{2}\geq2$ is true for $x>0$
| Proof by contradiction:
Assume that the hypothesis is wrong. Therefore, $ \exists\ x \gt 0 $ such that; $$ x + \dfrac{1}{x} \lt 2 $$
Now, follow along:
$$\begin{align}
x + \dfrac{1}{x} &\lt 2 \\
x^2 + 1 &\lt 2x \tag{$ \because x \gt 0 $} \\
x^2 - 2x + 1 &\lt 0 \\
(x - 1)^2 &\lt 0
\end{align}$$
which is a contradiction as a perfect square can never be negative for real numbers.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Calculate $\sum_{j=0}^k\binom {2k+1}{2j+1}^2=?$ Knowing that:
$${2k\choose k}=\sum_{j=0}^k{k\choose j}^2.$$
calculate the sums:
$$\sum_{j=0}^k\binom {2k+1}{2j+1}^2=?$$
Any sugestions please?
Thanks in advance.
| Using the similar combinatorial arguments as for the first identity, we get
$$\sum_{j=0}^k\binom {2k+1}{2j+1}^2= \sum_{j=0}^k\binom {2k+1}{2j+1}\binom {2k+1}{2k -2j} = \dfrac{\binom {4k+2}{2k+1}}{2}$$
To pick $2k+1$ elements out of $4k+2$ elements, we can either pick an odd number of elements from the first $2k+1$ ones and an even number of elments from the last $2k+1$ ones, or pick an even number of elements from the first $2k+1$ ones and an odd number of elments from the last $2k+1$ ones. Both ways represent half of the total possibilities, i.e. $\frac{\binom {4k+2}{2k+1}}{2}$
Then to pick an odd number of elements from the first $2k+1$ ones and an even number of elments from the last ones such that in total we pick $2k+1$ elements, we have $\sum_{j=0}^k\binom {2k+1}{2j+1}\binom {2k+1}{2k -2j}$ ways, so we conclude the result.
Another way to say this is to simply remark that
\begin{align}2\sum_{j=0}^k\binom {2k+1}{2j+1}^2 &=\sum_{j=0}^k\binom {2k+1}{2j+1}\binom {2k+1}{2k -2j} + \sum_{j=0}^k\binom {2k+1}{2k -2j}\binom {2k+1}{2j+1} \\
&= \sum_{j=0}^{2k+1}\binom {2k+1}{j}\binom {2k+1}{2k+1 -j} \\
&= \sum_{j=0}^{2k+1}\binom {2k+1}{j}^2 =\binom {4k+2}{2k+1}
\end{align}
The last equality is the given identity
| {
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"timestamp": "2023-03-29T00:00:00",
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Determining the last two digits of $229^{10} +37^{10}$ Determine the last two digits of: $229^{10}+37^{10}.$
I do not want to use the Euler-totient function or the carmichael function please!
Thanks
| Applying binomial theorem, one can rewrite
$$
229^{10}+37^{10} = (230-1)^{10} + (40-3)^{10} \\
=230^{10} - \binom{10}{1}230^9+\binom{10}{2}230^8- \ldots +\binom{10}{8}230^2-\binom{10}{9}230+1\\
+40^{10} - \binom{10}{1}40^9 3+\binom{10}{2}40^8 3^2- \ldots +\binom{10}{8}40^2 3^8-\binom{10}{9}40\cdot 3^9+3^{10}\\
=(......00 -10\cdot230+1)+(.......00-10\cdot40\cdot3^9+3^{10}),
$$
so the last two digits of $$229^{10}+37^{10}$$ are the same as the last two digits of $$1^{10}+3^{10}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "4",
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Getting the multiplicative inverse of a polynomial I have a polynomial $m(x)= x^2 + x + 2$ that's irreducible over $F=\mathbb{Z}/3\mathbb{Z}$. I need to calculate the multiplicative inverse of the polynomial $2x+1$ in $F/(m(x))$.
I'd normally use division with two polynomials like this, but it doesn't work here. So how can I get the inverse?
| Working in $(\mathbb{Z}/3\mathbb{Z})[[x]]/\langle x^2 + x + 1 \rangle$,
one (awful) way to do it is to notice that $2x + 1 = 1 - x$, so
$$(2x+1)^{-1} = \frac{1}{1-x} = 1 + x + x^2 + x^3 + \dotsb. $$
But $1 + x + x^2 = -1 = 2$
and, since $x^2 = -x - 2 = 1-x$,
$$x^3 + \dotsb = \frac{x^3}{1-x} = \frac{x^3}{x^2} = x,$$
giving $(2x+1)^{-1} = 2 + x$.
| {
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"timestamp": "2023-03-29T00:00:00",
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$2x^2+ 3y^2+4z^2 =1$ find the maximum of $4x+3y+2z$ If $2x^2+ 3y^2+4z^2 =1$ find the maximum of $4x+3y+2z$.
This is a question from a regional math olympiad and thus there must exist solutions without application of calculus.
I have no idea how to begin.
| Apply Cauchy-Schwarz Inequality: $\left(4x+3y+2z\right)^2 = \left(2\sqrt{2}\cdot \sqrt{2}x+\sqrt{3}\cdot \sqrt{3}y+1\cdot 2z\right)^2 \leq \left((2\sqrt{2})^2+(\sqrt{3})^2+1^2\right)\cdot\left(2x^2+3y^2+4z^2\right) = 12 \Rightarrow 4x+3y+2z \leq 2\sqrt{3} = \text{Maximum Value}$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Proving limit through definition Prove
$$\lim_{x\to 2}\frac{x^2+4}{x+2}=2$$
through definition.
My solution:
Fix $\epsilon >0$ and find $\delta$
\begin{align}
0<|x-2|<\delta &\Rightarrow \left| \frac{x^2+4}{x+2}-2 \right| < \epsilon\\
&\Rightarrow\left|\frac{x(x-2)}{x+2}\right| < \epsilon
\end{align}
Let $\delta <1$, then $0<|x-2|<1$ then $x\in (1,3)$ and $x>0$ and $x+2>0$
$$0<|x-2|<1 \Rightarrow \frac{x}{x+2}|x-2|<\epsilon$$
In conclusion $\delta :=\min \left\{ 1,\frac{5\epsilon}3\right\}$
Unfortunately the answer is not correct (according to my book). It says $\delta :=\min \left\{ 1,\epsilon \right\}$. Where did I make a mistake?
| Given that $x \in (1,3)$, we want to find an upper bound for:
$$
\frac{x}{x+2}
$$
To increase the value of this fraction, we can maximize the numerator and minimize the denominator. Indeed, since $x < 3$ and $x + 2 > 3$, we have that:
$$
\frac{x}{x+2} < \frac{3}{3} = 1
$$
Thus, we have that:
$$
\frac{x}{x+2}|x-2| < |x -2| < \epsilon
$$
provided that we chose $\delta :=\min \left\{ 1,\epsilon \right\}$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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More rigorous method for this elementary problem? The problem is:
Find all real values of $x$ such that $$(5+2\sqrt{6})^x+(5-2\sqrt{6})^x=2\sqrt{3}$$
One solution I received was as follows:
$5+2\sqrt{6}$ can be expressed as $(\sqrt{3}+\sqrt{2})^2$, and $5-2\sqrt{6}$ as $(\sqrt{3}-\sqrt{2})^2$. This means we want to solve $$(\sqrt{3}+\sqrt{2})^{2x}+(\sqrt{3}-\sqrt{2})^{2x}=2\sqrt{3}$$
for $x$. Note that $$(\sqrt{3}+\sqrt{2})+(\sqrt{3}-\sqrt{2})=2\sqrt{3}$$
so $x=\dfrac{1}{2}$ is a solution.
The problem with the solution is that it doesn't prove that $\dfrac{1}{2}$ is the only solution, not to mention its absolute inelegancy. Is there a more mathematically rigorous way of doing this?
I don't mind if higher algebra is used. I'm currently attempting to self-teach myself group theory, so including some of that in a solution would do me good.
| Let $(5+2\sqrt6)^x=a$
As $(5+2\sqrt6)(5-2\sqrt6)=1, (5-2\sqrt6)^x=a^{-1}$
$$\implies a+\frac1a=2\sqrt3\iff a^2-2\sqrt3a+1=0$$
$$\implies a=\frac{2\sqrt3\pm\sqrt{12-4}}2=\sqrt3\pm\sqrt2$$
If $a=\sqrt3+\sqrt2, (\sqrt3+\sqrt2)^{2x}=(\sqrt3+\sqrt2)^1$
If $a=\sqrt3-\sqrt2, (\sqrt3+\sqrt2)^{2x}=(\sqrt3-\sqrt2)^1=(\sqrt3+\sqrt2)^{-1}$ as $(\sqrt3-\sqrt2)(\sqrt3+\sqrt2)=1$
What can we conclude if $a^x=a^y$ with $a\ne\pm1,0$
One observation: If $f(x)=a^x+\dfrac1{a^x}, f\left(-x\right)=f(x)$
So, if $\dfrac12$ is a solution, $-\dfrac12$ will also be a solution and of course vice versa.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to evaluate this infinite sum I want to find $\int_0^1(1-x)^{\frac{1}{3}}x^{1/3}dx$.
From binomial theorem, $(1-x)^{\frac{1}{3}}= \sum_{0}^\infty (-x)^n\binom{\frac{2}{3}}{n}$.
Then $\int_0^1(1-x)^{\frac{1}{3}}x^{1/3}dx= \sum_{0}^\infty \frac{(-1)^n\binom{\frac{2}{3}}{n}}{n+\frac{4}{3}}$.
i don't know how to evaluate $\sum_{0}^\infty \frac{(-1)^n\binom{\frac{2}{3}}{n}}{n+\frac{4}{3}}$ .
Thank you.
| Without going through the Binomial theorem for fractional values, to can use the fact that what you want to evaluate is, up to the normalizing factor, the integral of the pdf $\beta_{\frac{4}{3}, \frac{4}{3}}$ of a $\operatorname{Beta}\!\left(\frac{4}{3}, \frac{4}{3}\right)$ distribution.. Thus, the integral you have is equal to
$$
\int_0^1 \operatorname{B}\!\left(\frac{4}{3}, \frac{4}{3}\right)\beta_{\frac{4}{3}, \frac{4}{3}}(x)dx = \operatorname{B}\!\left(\frac{4}{3}, \frac{4}{3}\right)
$$
where $\operatorname{B}\!\left(\frac{4}{3}\right) = \frac{\Gamma(4/3)\Gamma(4/3)}{\Gamma(8/3)}$ (which is in turn roughly 0.53).
| {
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$ \int_{ABC} f = \int_{CDA} f $ Problem from this year's MIT-PRIMES application:
Let $f$ be a continuous function on the plane. In any rectangle $ABCD$ so that $AB$ is parallel to the $x$-axis and $B$ has a greater $y$-coordinate than $C$, we have
$$\displaystyle\int_{ABC} f = \displaystyle\int_{CDA} f$$
Prove that $f$ is constant.
The deadline has passed so I guess it is okay to post. I was looking for a nice solution. Thank you. :)
| Let the vertices of the rectangle $ABCD$ be located at $A=(a,n)$, $B=(b,n)$, $C=(b,m)$, and $D=(a,m)$, with $a\le b$ and $m\le n$. Then, note that the equation of line $\overline{AC}$ is $$ y = \frac {m-n}{b-a} x + \left( n - \frac {m-n}{b-a} a \right), $$so we get that $$ \displaystyle\int_{ABC} f = \displaystyle\int_{a}^{b} \displaystyle\int_{\frac{m-n}{b-a} t + \left( n - \frac {m-n}{b-a} \right)}^{n} f(x) \, \mathrm{d}x \, \mathrm{d}t $$and $$ \displaystyle\int_{CDA} f = \displaystyle\int_{a}^{b} \displaystyle\int_{m}^{\frac{m-n}{b-a} t + \left( n - \frac {m-n}{b-a} \right)} f(x) \, \mathrm{d}x \, \mathrm{d}t. $$From this, note that $$ \begin {align*} \displaystyle\int_{ABC} f = \displaystyle\int_{CDA} f &\iff \displaystyle\int_{a}^{b} \displaystyle\int_{\frac{m-n}{b-a} t + \left( n - \frac {m-n}{b-a} \right)}^{n} f(x) \, \mathrm{d}x \, \mathrm{d}t - \displaystyle\int_{a}^{b} \displaystyle\int_{m}^{\frac{m-n}{b-a} t + \left( n - \frac {m-n}{b-a} \right)} f(x) \, \mathrm{d}x \, \mathrm{d}t = 0 \\&\iff \displaystyle\int_{a}^{b} \left[ \displaystyle\int_{m}^{\frac{m-n}{b-a}t + \left( n - \frac {m-n}{b-a} \right)} f(x) \, \mathrm{d}x - \displaystyle\int_{\frac{m-n}{b-a} t + \left( n - \frac {m-n}{b-a} \right)}^{n} f(x) \, \mathrm{d}x \right] \, \mathrm{d}t = 0 \\&\iff \displaystyle\int_{a}^{b} \left[ \displaystyle\int_{m}^{\frac{m-n}{b-a} t + \left( n - \frac {m-n}{b-a} \right)} f(x) \, \mathrm{d}x + \displaystyle\int_{n}^{\frac {m-n}{b-a} t + \left( n - \frac {m-n}{b-a} \right)} f(x) \, \mathrm{d}x \right] \, \mathrm{d}t = 0. \end {align*} $$We claim that, if $g(t)$ is a real-valued, continuous function with the property that
$$\int_a^bg(t)\,\text{d}t=0,$$
for all real numbers $a,b$, then $g(t)=0$, for all $t$. To prove this, note that, if $G(t)$ is an antiderivative of $g(t)$, then $ G(b) - G(a) = 0 $, giving $G(b)=G(a)$, for all $a,b$, so $G(t)$ is a constant, giving $G'(t)=g(t)=0$, for all $t$.
Therefore, letting $$ g(t) = \displaystyle\int_{m}^{\frac{m-n}{b-a} t + \left( n - \frac {m-n}{b-a} \right)} f(x) \, \mathrm{d}x + \displaystyle\int_{n}^{\frac {m-n}{b-a} t + \left( n - \frac {m-n}{b-a} \right)} f(x) \, \mathrm{d}x, $$for any rectangle $ABCD$ with vertices at $A=(a,n)$, $B=(b,n)$, $C=(b,m)$, and $D=(a,m)$, $$ \displaystyle\int_{ABC} f = \displaystyle\int_{CDA} f, $$if and only if $$ g(t) = \displaystyle\int_{m}^{\frac{m-n}{b-a} t + \left( n - \frac {m-n}{b-a} \right)} f(x) \, \mathrm{d}x + \displaystyle\int_{n}^{\frac {m-n}{b-a} t + \left( n - \frac {m-n}{b-a} \right)} f(x) \, \mathrm{d}x = 0, $$for any selection of $a,b,m,n\in\mathbb{R}$ such that $a\le b$ and $m\le n$.
By the fundamental theorem of calculus, $$ g'(t) = 2 \cdot \frac {m-n}{b-a} \cdot f \left( \frac {m-n}{b-a} t + \left( n - \frac {m-n}{b-a} \right) \right) - f(m) - f(n) = 0. $$Rearranging, we get $$ f \left( \frac {m-n}{b-a} t + \left( n - \frac {m-n}{b-a} \right) \right) = \frac {f(m) + f(n)}{2} \cdot \frac {b-a}{m-n}. $$
Therefore, since the right hand side of the above equation is a constant, $g'(t)=0$ if and only if $f$ is a constant. Hence, if $g(t)=r$ is a constant function, then $f$ is a constant function. We wish to prove that $g(t)=0$, for all $t$; i.e. $r=0$. It suffices to show that if $f$ is a constant function, then $g(t)=0$. To do this, we prove that if $g(t)=r$, a constant, then $r=0$.
Let $f(x)=c$. Then, $$ \begin {align*} g(t) &= \displaystyle\int_{m}^{\frac{m-n}{b-a} t + \left( n - \frac {m-n}{b-a} \right)} c \, \mathrm{d}x + \displaystyle\int_{n}^{\frac{m-n}{b-a} t + \left( n - \frac {m-n}{b-a} \right)} c \, \mathrm{d}x \\&= c \cdot \left( 2 \cdot \frac {m-n}{b-a}t - 2 \cdot \frac {m-n}{b-a} - \left( m - n \right) \right), \end {align*} $$which is equal to $0$ for all $t$ if and only if $m-n=0$, in which case $g(t)=0$.
Hence, $g(t)=0$ if and only if $f$ is a constant, so we are done. $\blacksquare$
| {
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Show that polynomial is reducible Show that $p(x)$ = $x^3 + 3x^2 + 2x + 4$ is reducible in $\mathbb{Z}$$/$$7$$\mathbb{Z}$
Is the approach for this to factor it and then find a root? I'm a little confused on how to start. Any tips appreciated.
| Another way:
\begin{align}p(x) &= x^3 + 3x^2 + 2x + 4\\
& = x^3 - 4x^2 + 2x + 4\\
& = x^3 - 2x^2 - 2x^2 + 4x - 2x + 4\\
& = x^2(x - 2) - 2x(x - 2) - 2(x - 2)\\
& = (x^2 - 2x - 2)(x - 2).
\end{align}
| {
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How to solve these equations for x and y.. equations are
$(x-y)(x+2y)(2x+y) = 20$
and
$x^2+xy+y^2 = 7$
i want the METHOD not the solutions
| You are given:$$(x-y)(x+2y)(2x+y)=20\tag{1}$$$$x^2+xy+y^2=7\tag{2}$$Expand the last two terms in (1) to get:$$(x+2y)(2x+y)=2x^2+5xy+2y^2=2(x^2+xy+y^2)+3xy\tag{3}$$Substitute (2) into (3) to get:$$(x+2y)(2x+y)=2(7)+3xy=14+3xy\tag{4}$$Substitute (4) into (1) to get:$$(x-y)(14+3xy)=20\tag{5}$$Then use the hint given by @Leox, i.e. let:$$a=x-y\tag{6}$$$$b=xy\tag{7}$$To get:$$a(14+3b)=20$$$$\therefore 14+3b=\frac{20}{a}\tag{8}$$If it is integer solutions that you are looking for then $a$ must be evenly divisible into $20$ which means $a$ can only be from the set: $1,-1,2,-2,4,-4,5,-5,10,-10,20,-20$
Use each value of $a$ in (8) to see which ones give an integer solution for $b$.
Finally use (6) and (7) to then find $x$ and $y$.
| {
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How to compute the integral $\int^{\pi/2}_0\ln(1+\tan\theta)d\theta$? How to compute the integral $\int^{\pi/2}_0\ln(1+\tan\theta)d\theta$. If we let $t=\tan\theta$, then the integral becomes to
$$\int^{\pi/2}_0\ln(1+\tan\theta)d\theta=\int_0^\infty\frac{\ln(1+t)}{1+t^2}dt$$.
Can we calculate this integral explicitly?
| $$\begin{align}\int_0^{\pi/2} d\theta \, \log{(1+\tan{\theta})} &=\int_0^{\pi/2} d\theta \, \log{(\sin{\theta}+\cos{\theta})} - \int_0^{\pi/2} d\theta \, \log{(\cos{\theta})} \\ &= \int_0^{\pi/2} d\theta \, \log{\left [\sqrt{2}\cos{\left (\theta-\frac{\pi}{4} \right )}\right ]} - \int_0^{\pi/2} d\theta \, \log{(\cos{\theta})} \\ &= \frac{\pi}{4} \log{2} + \int_0^{\pi/2} d\theta \, \log{\left [\cos{\left (\theta-\frac{\pi}{4} \right )}\right ]} - \int_0^{\pi/2} d\theta \, \log{(\cos{\theta})}\\ &= \frac{\pi}{4} \log{2} + \int_{-\pi/4}^{\pi/4} d\theta \, \log{\left (\cos{\theta}\right )} - \int_0^{\pi/2} d\theta \, \log{(\cos{\theta})}\\ &= \frac{\pi}{4} \log{2} + \int_{0}^{\pi/4} d\theta \, \log{\left (\cos{\theta}\right )} - \int_{\pi/4}^{\pi/2} d\theta \, \log{(\cos{\theta})} \\ &= \frac{\pi}{4} \log{2} + \int_{0}^{\pi/4} d\theta \, \log{\left (\cos{\theta}\right )} - \int_{0}^{\pi/4} d\theta \, \log{\left (\sin{\theta}\right )}\end{align} $$
Now use the Fourier series representations:
$$-\log(\sin(\theta))=\sum_{k=1}^\infty\frac{\cos(2k \theta)}{k}+\log(2)$$
and
$$-\log(\cos(\theta))=\sum_{k=1}^\infty(-1)^k\frac{\cos(2k \theta)}{k}+\log(2)$$
Substituting, exchanging the respective sums and integrals, we get
$$\begin{align}\int_0^{\pi/2} d\theta \, \log{(1+\tan{\theta})} &= \frac{\pi}{4} \log{2} + \sum_{k=1}^{\infty} \frac1{2 k^2} \left [1-(-1)^k \right ] \sin{\frac{\pi}{2} k} \\ &= \frac{\pi}{4} \log{2} + \sum_{k=1}^{\infty} \frac{(-1)^k}{(2 k+1)^2} \\ &= \frac{\pi}{4} \log{2} + G\end{align} $$
where $G$ is Catalan's constant.
| {
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"url": "https://math.stackexchange.com/questions/1073332",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
For an ellipse with minor radius $b$, show that the product of distances from the foci to any tangent line is $b^2$
Consider the ellipse with equation:
$$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$
How do we prove the product of distances from the foci to any tangent line is $b^2$?
$$F_1A \cdot F_2B = b^2$$
| WLONG, we can represent any point on the given Ellipse as $x=a\cos A,y=b\sin A$
So, the equation of the tangent at $(a\cos A,b\sin A)$ will be $\dfrac{x\cdot a\cos A}{a^2}+\dfrac{y\cdot b\sin A}{b^2}=1$
$$\iff bx\cos A+ay\sin A-ab=0$$
The coordinates of the foci are $(\pm ae,0)$ where $e$ is the eccentricity
So, the distances will be $$\frac{|\pm ae\cdot b\cos A-ab|}{\sqrt{(b\cos A)^2+(a\sin A)^2}}$$ and consequently their product will be
$$=a^2b^2\cdot\frac{(1-e\cos A)(1+e\cos A)}{(b\cos A)^2+(a\sin A)^2} =a^2b^2\cdot\frac{(1-e^2\cos^2A)}{a^2(1-e^2)\cos^2A+a^2\sin^2A}=b^2$$
as $b^2=(1-e^2)a^2$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to solve the differential equation $(2x^3y)\:\text{dy}+(1-y^2)(x^2y^2+y^2-1)\:\text{dx}=0$?
Solve
$$(2x^3y)\:\text{dy}+(1-y^2)(x^2y^2+y^2-1)\:\text{dx}=0$$
I tried the substitution $y^2=t$ ; $2y\:\text{dy}=\text{dt}$ to get
$$(x^3)\:\text{dt}+(1-t)[(x^2+1)t-1]\:\text{dx}=0$$
However, I don't know how to proceed further.
| since
$$2x^3yy'+(1-y^2)(x^2y^2+y^2-1)=0$$
then we have
$$2x^3yy'=(y^2-1)^2+(y^2-1)x^2y^2=(y^2-1)^2+(y^2-1)x^2(y^2-1)+x^2(y^2-1)$$
so
$$x^3(y^2-1)'=(y^2-1)^2+(y^2-1)^2x^2+x^2(y^2-1)$$
let
$y^2-1=u$,then we have
$$\dfrac{du}{dx}=\left(\dfrac{1}{x}+\dfrac{1}{x^3}\right)u^2+\dfrac{u}{x}$$
this is Bernoulli equation
$$-\dfrac{d(1/u)}{dx}=\dfrac{1}{x}\cdot\dfrac{1}{u}+\dfrac{1}{x^3}+\dfrac{1}{x}$$
then It is easy solve it
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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If we know $x+y+z=1$, $x^2+y^2+z^2=2$, and $x^3+y^3+z^3=3$, how to find $x^4+y^4+z^4$? Let $x$, $y$, and $z$ be such that
$$\begin{align*}
x+y+z&=1\\
x^2+y^2+z^2&=2 \\
x^3+y^3+z^3&=3
\end{align*}$$
Then $x^4+y^4+z^4=?$
| Hint:
let $a_{n}=x^n+y^n+z^n$
then we have
$$a_{n+2}=(x+y+z)a_{n+1}-(xy+yz+xz)a_{n}+xyza_{n-1}$$
it is easy to find
$$xy+yz+xz=\dfrac{1}{2}[(x+y+z)^2-(x^2+y^2+z^2)]=-\dfrac{1}{2}$$
$$x^3+y^3+z^3-3xyz=(x+y+z)^3-3(xy+yz+xz)(x+y+z)$$
then we have
$$xyz=\dfrac{1}{6}$$
so
$$a_{n+2}=a_{n+1}+\dfrac{1}{2}a_{n}+\dfrac{1}{6}a_{n-1}$$
so
$$x^4+y^4+z^4=a_{4}=a_{3}+\dfrac{1}{2}a_{2}+\dfrac{1}{6}a_{1}=\dfrac{25}{6}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Finding $\lim_{x\to 0}\frac{\sin(x+x^3/6)-x}{x^5}$ I'm trying to find the limit of this expression:
$$\lim_{x\to0}\frac{\sin\left(x+x^3/6\right)-x}{x^5}$$
My solution is as follows:
$$
\begin{align}
\lim_{x\to0}\frac{\sin\left(x+x^3/6\right)-x}{x^5}&=\lim_{x\to0}\frac{1}{x^5}\cdot\!\!\left[\frac{(x+x^3/6)\sin(x+x^3/6)}{(x+x^3/6)}-x\right]\\
&=\lim_{x\to0}\frac{1}{x^5}\cdot\!\!\left[(x+x^3/6)-x\right]\\
&=\lim_{x\to0}\frac{x^3}{6x^5}\\
&=\lim_{x\to0}\frac{1}{6x^2}=+\infty
\end{align}
$$
But Wolfram Alpha finds the limit as $-\frac{3}{40}$. Where is my mistake?
| This limit can be calculated by applying (fiercely :)) L'Hospital's rule three times in succession:
$$\lim_{x\to0}\frac{\sin\left(x+x^3/6\right)-x}{x^5}=$$I.$$=\lim_{x\to0}\frac{(1+x^2/2)\cos\left(x+x^3/6\right)-1}{5x^4}=$$
II.$$=\lim_{x\to0}\frac{x\cos\left(x+x^3/6\right)-(1+x^2/2)^2\sin\left(x+x^3/6\right)}{20x^3}=$$
III.$$=\lim_{x\to0}[\frac{\cos\left(x+x^3/6\right)-x(1+x^2/2)\sin\left(x+x^3/6\right)}{60x^2}-$$$$-\frac{2x(1+x^2/2)\sin\left(x+x^3/6\right)+(1+x^2/2)^3\cos\left(x+x^3/6\right)}{60x^2}]=$$
$$=\frac{1}{60}\lim_{x\to0}[\frac{\cos\left(x+x^3/6\right)(1-(1+x^2/2)^3)}{x^2}-\frac{3x(1+x^2/2)\sin\left(x+x^3/6\right)}{x^2}]=$$$$= \frac{1}{60}(-\frac{3}{2}-3) =-\frac{3}{40}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Solutions of $z^6 + 1 = 0$ Solve:
$$z^6 + 1 = 0$$
That lie in the top region of the plane.
We know that:
$$(z^2 + 1)(z^4 - z^2 + 1) = 0$$
$$z = -i, i$$
We need to solve:
$$((z^2)^2 - (z)^2 + 1) = 0$$
$$z = \frac{1 \pm \sqrt{-3}}{2}$$
But this is incorrect. How to do this then?
| The simplest way of solving this equation is the method based on DeMoivre's Formula that Lab Bhattacharjee outlined.
That said, you can make your method work. You found the roots $z \pm i$ by setting the factor $z^2 + 1$ equal to zero. As Rasolnikov and 5xum noted, you should have obtained
$$z^2 = \frac{1 \pm \sqrt{-3}}{2}$$
when you set the factor $z^4 - z^2 + 1$ equal to zero.
Let $z = a + bi$, with $a, b \in \mathbb{R}$. Then
\begin{align*}
z^2 & = \frac{1 \pm i\sqrt{3}}{2}\\
(a + bi)^2 & = \frac{1 \pm i\sqrt{3}}{2}\\
a^2 + 2abi - b^2 & = \frac{1 \pm i\sqrt{3}}{2}
\end{align*}
Equating real and imaginary parts yields
\begin{align*}
a^2 - b^2 & = \frac{1}{2}\tag{1}\\
2ab & = \pm\frac{\sqrt{3}}{2}\tag{2}
\end{align*}
Solving equation 2 for $b$ yields
$$b = \frac{\pm\sqrt{3}}{4a}\tag{3}$$
Substituting this expression in equation 1 yields
\begin{align*}
a^2 - \frac{3}{16a^2} & = \frac{1}{2}\\
16a^4 - 3 & = 8a^2\\
16a^4 - 8a^2 & = 3\\
16a^4 - 8a^2 + 1 & = 4 && \text{complete the square}\\
(4a^2 - 1)^2 & = 4\\
4a^2 - 1 & = \pm 2\\
4a^2 & = 3 && \text{since $a \in \mathbb{R}$}\\
a^2 & = \frac{3}{4}\\
a & = \pm \frac{\sqrt{3}}{2}
\end{align*}
Substituting this expression into equation 3 yields the four roots
$$z = \pm\frac{\sqrt{3}}{2} \pm \frac{1}{2}i$$
of the equation $z^4 - z^2 + 1 = 0$. As you can check, these roots correspond to the values $n = 0, 2, 3, 5$ in the formula Lab provided.
| {
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"timestamp": "2023-03-29T00:00:00",
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use comparison test to show divergence or convergence I'm not sure if my reasoning is correct.
a) $\displaystyle \sum_{n=2}^{\infty} \frac{\ln^5{(2n^7+13)+10\sin(n)}}{n\cdot \ln^6{(n^\frac{7}{8}}+2\sqrt{n}-1)\cdot\ln{\ln{(n+(-1)^n}})} = \sum_{n=2}^{\infty} a_n $
here I think it suffice to show that for big enough $n$ we have all logs $>1$ and $\sin n \ge -1$ so we have $\displaystyle \sum_{n=2}^{\infty}\frac{-9}{n} \le a_n $ so $\sum_{n=2}^{\infty} a_n$ diverges.
b) $\displaystyle \sum_{n=2}^{\infty} (\frac{n^2+3n+10}{n^2+5n+17})^{n^2(\sqrt{n+1}-\sqrt{n-1})}= \sum_{n=2}^{\infty} b_n$
we have $\displaystyle \sum_{n=2}^{\infty} b_n= \sum_{n=2}^{\infty}(1-\frac{2n+7}{n^2+5n+17})^{\frac{2n^2}{\sqrt{n+1}+\sqrt{n-1}}}$ and here I don't know how to compare it
| Part (b):
The exponent of $b_n$ satisfies
$$\frac{2n^2}{\sqrt{n+1}+\sqrt{n-1}}< \frac{2n^2}{\sqrt{n+1}}< 2n\sqrt{n}.$$
For sufficiently large $n$
$$1-\frac{2n+7}{n^2+5n+17}= 1-\frac{2}{n}\frac{1+7/(2n)}{1+5/n+17/n^2}< 1 - \frac{1}{n},$$
using
$$\lim_{n \to \infty}\frac{1+7/(2n)}{1+5/n+17/n^2} = 1 \implies\frac{1+7/(2n)}{1+5/n+17/n^2} > \frac1{2}.$$
Using the inequality $\displaystyle (1+x/n)^{n+1}> e^x,$ we have
$$\left(1 - \frac{1}{n}\right)^n= \left(1 + \frac{1}{n-1}\right)^{-n}<e^{-1}.$$
Hence, for sufficiently large n,
$$b_n < \left(1 - \frac{1}{n}\right)^{2n\sqrt{n}}< e^{-2\sqrt{n}}.$$
Note that $e^n$ grows faster than any power of $n$. We easily can find a dominating convergent p-series.
$$e^{2\sqrt{n}} = \sum_{k=0}^{\infty}\frac{(2\sqrt{n})^k}{k!}> \frac{(2\sqrt{n})^4}{4!}=\frac{2n^2}{3}\\ \implies e^{-2\sqrt{n}} < \frac{3}{2}n^{-2}.$$
Therefore, by the comparison test
$$\sum_{n=2}^{\infty} b_n < \sum_{n=2}^{\infty} e^{-2\sqrt{n}}< \frac{3}{2}\sum_{n=2}^{\infty} n^{-2} = \frac{3}{2}\left(\frac{\pi^2}{6}-1\right).$$
Part (a):
Note that for sufficiently large $n$
$$\frac{\ln^5{(2n^7+13)+10\sin(n)}}{\ln^5{(n^\frac{7}{8}}+2\sqrt{n}-1)} > \frac{\ln^5{(2n^7+13)-10}}{\ln^5{(n^\frac{7}{8}}+2\sqrt{n}-1)} > 1.$$
Also
$$ \lim_{n \to \infty}\frac{(n^{\frac{7}{8}}+2\sqrt{n}-1)}{n} = 0, \\ n+ (-1)^n \leqslant n+1,$$
and for sufficiently large $n$ we have
$$ \ln(n^{\frac{7}{8}}+2\sqrt{n}-1) < \ln n, \\ \ln \ln (n+ (-1)^n) \leqslant \ln \ln (n+1),$$
Thus,
$$a_n = \frac{\ln^5{(2n^7+13)+10\sin(n)}}{n\cdot \ln^6{(n^\frac{7}{8}}+2\sqrt{n}-1)\cdot\ln{\ln{(n+(-1)^n}})} > \frac{1}{n \cdot \ln n \cdot \ln \ln (n+1)} \\ > \frac{1}{(n+1) \cdot \ln (n+1) \cdot \ln \ln (n+1)} .$$
The series with terms given by the RS of the above inequality diverges by the integral test, since
$$\int_2^{\infty}\frac{dx}{(x+1) \cdot \ln (x+1) \cdot \ln \ln (x+1) }= \lim_{x \to \infty}[\ln \ln \ln(x+1)- \ln \ln \ln (3)] = \infty.$$
| {
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How many solutions of the equation $x^2-y^2=1$ are there with $x,y\in\mathbb F_{p^n}$?
Let $p>2$ be an odd prime. Let $\mathbb F_{p^n}$ be the field with $p^n$ elements. How many solutions of the equation $x^2-y^2=1$ are there with $x,y\in\mathbb F_{p^n}$?
My work:
Char $F=p$.
$x^2-y^2=(x+y)(x-y)=1$. Since we are looking for $x,y\in F$, $x+y,x-y\in F$ and if $x+y=\alpha$, $x-y=\alpha^{-1}$. Hence, $\displaystyle x=\frac{\alpha+\alpha^{-1}}{2}, y=\frac{\alpha-\alpha^{-1}}{2}$. I was stuck afterwards since I cannot find the distinct sets of $x,y$ from these relations. Can anyone please give me a hint?
| If you want something even more explicit than André's mention of the fact, note that we have
$$\begin{pmatrix} 1 & -1 \\ 1 & 1\end{pmatrix}\begin{pmatrix} x \\ y\end{pmatrix}=\begin{pmatrix} \alpha \\ \alpha^{-1}\end{pmatrix}$$
since the characteristic is not $2$ and the determinant of that matrix is $2$, we see that
$$2^{-1}\begin{pmatrix} 1 & 1 \\ -1 & 1\end{pmatrix}\begin{pmatrix} \alpha \\ \alpha^{-1}\end{pmatrix}=\begin{pmatrix} x \\ y\end{pmatrix}$$
So that establishes the bijection, in particular since there are $p^n-1$ elements $\alpha\in F$ so that $\alpha^{-1}$ exists, we have there are $p^n-1$ total solutions.
Incidentally, if the characteristic is $2$, this is even easier: $a^2-b^2=(a-b)^2$ so it's easy to see that all you need to do is solve
$$(x-y)^2-1^2=(x-y-1)^2=0.$$
Of course in this case this clearly has $p^n$ total solutions, all given by $(x, x-1)$ and $x\in F$.
| {
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For any triangle with sides $a$, $b$, $c$, show that $a^2b(a-b)+b^2c(b-c)+c^2a(c-a)\ge 0$
For any triangle with sides $a$, $b$, $c$, prove the inequality $$a^2b(a-b)+b^2c(b-c)+c^2a(c-a)\ge 0 .$$
This is IMO 1983 problem 6.
I tried substituting $a=x+y$, $b=y+z$, $c=z+x$ but well it doesn't help in any sense except wasting 3 pages that lead to nothing (please don't mind the joke). Using $a=2R\sin A$, $b=2R\sin B$, $c=2R\sin C$ also didn't lead to anything for me. Could you give me a hint for finding the proper substitution?
| Substitution $$a=y+z,b=x+z,c=x+y$$ we have to prove that
$$\frac{x^2}{y}+\frac{y^2}{z}+\frac{z^2}{x}\geq x+y+z$$
Using Cauchy Schwarz in Engel form we get
$$\frac{x^2}{y}+\frac{y^2}{z}+\frac{z^2}{x}\geq \frac{(x+y+z)^2}{x+y+z}=x+y+z$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Given $f(x) = \frac{x^2}{x-1}$, I need to find $f(2x-3)$ I'm not clear how to proceed:
Given $f(x) = \dfrac{x^2}{x-1}$, I need to determine what $f(2x-3)$ is equal to.
| Given $f(x) = \dfrac{x^2}{x-1}$, we want to find $f(2x-3)$.
To do so, we substitute $2x-3$ every place we see an x in $f(x),$ and simplify:
$$f(2x-3) = \dfrac{(2x -3)^2}{(2x-3)-1}= \dfrac{4x^2 - 12x + 9}{2x-4}$$
| {
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Is there a faster way to factor $X^{12}-1$ over $\mathbb{F}_5[X]$? Problem: Factor $X^{12}-1$ into irreducibles in $\mathbb{F}_5[X]$. This problem appeared on a past qual and took me awhile to do. While I solved it, I'll need to be able to do problems like this a lot faster to have a chance of passing. At the very least, I was wondering whether there was a quick way to tell that $X^4-X^2+1$ is not irreducible in $\mathbb{F}_5[X]$.
My solution: I first factored this polynomial over $\mathbb{Z}[X]$, as $$(X-1)(X+1)(X^2+X+1)(X^2+1)(X^2-X+1)(X^4-X^2+1)$$ via the formula $X^n-1 = \prod\limits_{d \mid n} \Phi_d(X)$, where $\Phi_d$ is the $d$th cyclotomic polynomial. The first $5$ irreducibles can easily be checked to remain irreducible when going modulo $5$ (by checking if they have a root), so the hard part of the problem is checking whether the $12$th cyclotomic polynomial, $X^4-X^2+1$, remains irreducible in $\mathbb{F}_5[X]$.
It's easy to see that $\Phi_{12}$ has no roots in $\mathbb{F}_5$, so it would have to be a product of two quadratics. I supposed it did factor, as $$X^4-X^2+1 = (X^2 + aX+b)(X^2+cX+d)$$ I expanded the right hand side as $X^4 + (a+c)X^3 + (b+ac+d)X^2 + (bc+ad)X + bd$, so I had the relations
(i): $a+c = 0$
(ii) $b+ac+d =-1$
(iii) $bc+ad=0$
(iv) $bd = 1$.
I began by considering the case where $a \neq 0$ (I had already checked that $\Phi_{12}$ couldn't factor as a product $(X^2 +e)(X^2+f)$), so that also $c \neq 0$. I multiplied (iii) by $d$ to get $c+ad^2 = 0$, and subtracted (i) from this to get $ad^2-a = 0$, or $d^2 = 1$.
So $d = \pm 1$. First I supposed $d = 1$, so that also $b=1$. Then $a, c$ were nonzero elements satisfying $a+c = 0$ and $ac + 2 = -1$, or $ac = 2$. For $ac = 2$ we can only have $\{a,c\} = \{3,4\}$ or $\{a,c\} = \{1,2\}$, and in neither of those cases could we have $a+c = 0$.
Next I supposed $d = -1$, so that also $b = -1$. Then $a, c$ were nonzero elements satisfying $ac=1$ and $a+c = 0$. This relation is satisfied by $a = 2, b = 3$. So I looked at the product $$(X^2+2x-1)(X^2+3X-1)$$ which expands as $X^4+5X^3+4X^2-5X+1$, which is $X^4-X^2+1$ in $\mathbb{F}_5[X]$.
| Rewrite $\,\,X^4 -X^2+1$ as
$$
(X^2-1)^2+X^2=(X^2-1)^2-4X^2=(X^2-2X-1)(X^2+2X-1).$$
| {
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Integral question showing the primitive functions differ only by a constant? $$\int \frac{dx}{\sqrt{x^2-6x+13}}$$
$$\int \frac{dx}{\sqrt{\left(x-3\right)^2+4}}$$
It can be solved by
Method 1
Let
$$x-3=2\tan u$$
$$dx=2\sec^2 u\,du$$
Therefore, using the trigonometric identity $\tan^2 u+1=\sec^2 u$
$$\int \frac{2\sec^2 u\,du}{\sqrt{4\tan^2 u +4}}$$
$$\int \frac{2\sec^2 u\,du}{|2\sec u |}$$
$$\int |\sec u|\,du$$
To focus on the problem , I skipped the steps in evaluating the sec integral by using this
$$\int \left|\sec u\right|\,du=\begin{cases}\int -\sec u\,du, & \sec u<0\\ \int \sec u\,du, & \sec u>0 \end{cases} = \begin{cases}-\ln |\sec u+\tan u |+C, & \sec u<0\\ \ln |\sec u+\tan u |+C, & \sec u>0 \end{cases}$$
$$=|\ln |\sec u+\tan u ||+C$$
Reverting to $x$ variables
$$\left|\ln\left|\sec \left(\arctan \frac{x-3}{2} \right) +\tan \left(\arctan \frac{x-3}{2} \right) \right|\right|+C$$
Method 2
Let
$$x-3=2\sinh u$$
$$dx=2\cosh u\,du$$
Therefore, using the trigonometirc identity $\sinh^2 u+1=\cosh^2 u$
$$\int \frac{2\cosh u\,du}{\sqrt{4\cosh^2 u +4}}$$
$$\int \frac{2\cosh u\,du}{|2\cosh u|}=\left\{\begin{matrix}\int -du,\cosh u<0\\ \int du,\cosh u>0 \end{matrix}\right.=\left\{\begin{matrix}-u+C,\cosh u<0\\ u+C,\cosh u>0 \end{matrix}\right.$$
$$=|u|+C$$
Reverting to $x$ variables
$$\left|\sinh^{-1} \left(\frac{x-3}{2}\right) \right|+C=\left|\ln \left|\left(\frac{x-3}{2}\right)+\sqrt{\left(\frac{x-3}{2}\right)^2+1}\right|\right|$$
I understand that in general, by using the results here, we can prove in general that all primitive of a function f must differ only by a constant
But
$$\left|\ln\left|\sec \left(\arctan \frac{x-3}{2} \right) +\tan \left(\arctan \frac{x-3}{2} \right) \right|\right|+C$$
and
$$|\sinh^{-1} \left(\frac{x-3}{2}\right)|+C$$
look really different. how to show that they are really the same thing save for a constant?
| Hint: $\tan\left(\tan^{-1}\theta\right){}={}\theta\,\,$, $\,\,\,\sec\left(\tan^{-1}\theta\right){}={}\sqrt{1+\tan^2\left(\tan^{-1}\theta\right)}\,$ and finally, $\theta{}={}\sinh y{}={}\dfrac{e^{y}{}-{}e^{-y}}{2}$ for some $y$.
| {
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"url": "https://math.stackexchange.com/questions/1088393",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Finding a value from 5 systems of equations of 5 variables(CHMMC 2014) $$\text{For } a_1\cdots a_5\in \mathbb{R},$$
$$\frac{a_1}{k^2+1}+\cdots+\frac{a_5}{k^2+5}=\frac{1}{k^2}$$
$$\forall k=\{2,3,4,5,6\}$$
$$\text{Find }\frac{a_1}2+\cdots+\frac{a_5}6$$
The Provided explanation/official solution was
$$\text{Solution }1:\large{\frac{65}{72}}$$
Please don't mark this as off-topic, I have no clue where to start. Clearly it would not be feasible to compute $a_1\cdots a_5$.
| Starting from $\displaystyle \frac{a_1}{k^2+1}+\cdots+\frac{a_5}{k^2+5}=\frac{1}{k^2}$ we would like to plug in $k^2=1$ and get
$$ \frac{a_1}{2}+\cdots+\frac{a_5}{6}=1$$
In fact, the correct answer is slightly less. We can add any function rational $f(k)$ which vanishes on all of $k^2 = 2,3,4,5,6$. So are are going to define $f(x) = K \cdot \prod_{k=2}^6 (x - k^2)$ where $g(x)$ can be any polynomial.
$$ \frac{a_1}{x^2+1}+\cdots+\frac{a_5}{x^2+5}=\frac{1+ K \cdot \prod_{k=2}^6 (x - k^2)}{x^2} \tag{$\ast$}$$
If we set $x= 0$ the right side is undefined, but we can cross multiply:
\begin{eqnarray} x^2 \left( \frac{a_1}{x^2+1}+\cdots+\frac{a_5}{x^2+5} \right)&=&1+ K \cdot \prod_{k=2}^6 (x - k^2) \\
0&=&1 - K \cdot 6!^2
\end{eqnarray}
We have that $\boxed{K = \frac{1}{6!^2}}$ Then plug $x=1$ into $(\ast)$
$$ \frac{a_1}{2}+\cdots+\frac{a_5}{6}=\frac{1+ K \cdot \prod_{k=2}^6 (x - k^2)}{x^2} = 1 - \frac{(3\cdot 4 \cdot 5 \cdot 6 \cdot 7)(1\cdot 2 \cdot 3 \cdot 4 \cdot 5)}{6!^2 }$$
This argument is not quite seamless as the RHS is off by factor of 6.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1089289",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Interesting and fun eigenvalue problem for a matrix $B \in M_{n}(\mathbb{Q})$ I've been thinking about ways on how to tackle this particular linear algebra problem related to eigenvalues from an old algebra prelim. The problem is that suppose we are given a matrix $B \in M_{n}(\mathbb{Q})$ such that $B^5 =I$, i.e. the $n \times n$ identity matrix, and no eigenvalue of $B$ is equal to $1$. We have to show that $n$ is divisible by $4$.
My attempt: I take it that $B^5 =I \implies B^5 -I=0$ (the zero matrix). Now by definition, the eigenvalues of a matrix $B$ are the roots of the characteristic polynomial $c_B(x)$ of $B$ and we also know that $c_B(x)$ annihilates the matrix $B$, i.e. $c_B(B)=0$. Now in this problem, the key condition is that $1$ cannot be an eigenvalue of $B$, so does this mean, via the factorization $$x^5 -1= (x-1)(x^4 +x^3 +x^2 +x+1)$$ over $\mathbb{Q}$, that $x^4 + x^3 + x^2 +x+1$ divides $c_B(x)$, which has degree $n$? Is this enough to say that $n$ is divisible by $4$, or are there any other necessary things to be considered carefully before coming to that conclusion?
| Since $B^5 = I$, the minimal polynomial of $B$ over the rationals divides
$x^5 - 1$. Now $x^5 - 1 = (x - 1)(x^4 + x^3 + x^2 + x + 1)$, where
$x^4 + x^3 + x^2 + x + 1$ is irreducible (it's the fifth cyclotomic polynomial), and since $1$ is not an eigenvalue the minimal polynomial can only be $x^4 + x^3 + x^2 + x + 1$. All irreducible factors of the characteristic polynomial divide the minimal polynomial, so the characteristic polynomial is
a power of $x^4 + x^3 + x^2 + x + 1$, and this implies that $n$ is a multiple of $4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1095933",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Show that $\sqrt[3]{3\sqrt{21} + 8} - \sqrt[3]{3\sqrt{21} - 8} = 1$
Show that
$$\sqrt[3]{3\sqrt{21} + 8} - \sqrt[3]{3\sqrt{21} - 8} = 1$$
Playing around with the expression, I found a proof which I will post as an answer.
I'm asking this question because I would like to see if there are alternative solutions which are perhaps faster / more direct / elementary / elegant / methodical / insightful etc.
| let $\sqrt[3]{\sqrt{a}+b}-\sqrt[3]{\sqrt{a}-b}=1$
so that $a=189$ and $b=8$
$(\sqrt[3]{\sqrt{a}+b}-\sqrt[3]{\sqrt{a}-b})^3=1$
$\sqrt{a}+b-3(\sqrt[3]{(\sqrt{a}+b)^2}\sqrt[3]{(\sqrt{a}-b)})+3(\sqrt[3]
{(\sqrt{a}+b)}\sqrt[3]{(\sqrt{a}-b)^2})-\sqrt{a}+b=1$
$2b-3(\sqrt[3]{(\sqrt{a}+b)^2}\sqrt[3]{(\sqrt{a}-b)})+3(\sqrt[3]{(\sqrt{a}+b)}\sqrt[3]{(\sqrt{a}-b)^2})=1$
$ 2b-3(\sqrt[3]{(\sqrt{a}+b)}\sqrt[3]{(\sqrt{a}-b)})[ \sqrt[3]{\sqrt{a}+b} -\sqrt[3]{\sqrt{a}-b}]=1$
$2b-3(\sqrt[3]{\sqrt{a}+b}.\sqrt[3]{\sqrt{a}-b})(1))=1$
$2b-3\sqrt[3]{a-b^2}=1$
$2*8-3\sqrt[3]{189-8^2}=1\rightarrow 16-15=1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1097558",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 5
} |
Evaluate $\frac{1}{4}\int \frac{dx}{(kx^{2}+ax+b)^{2}}$ I would like this integral:
$$\frac{1}{4}\int \frac{dx}{(kx^{2}+ax+b)^{2}}$$
I used substitution $x+ \frac{a}{2k} = \frac{\sqrt{4kb-a^{2}}}{2k} \tan \theta$.
This will reduce the denominator to
$$
\left(\left(\frac{\sqrt{4kb-a^{2}}}{2k}\tan\theta \right)^{2} + \frac{4kb-a^{2}}{4k^{2}}\right)^{2} = \left(\frac{4kb-a^{2}}{4k^{2}}\right)^{2} \sec^{4}\theta ,$$
and integral is reduced to
$$\frac{1}{4k^{2}}\frac{16k^{4}}{(4kb-a^{2})^{\frac{3}{2}}2k} \int \cos^{2} \theta d\theta = \frac{2k}{(4kb-a^{2})^{\frac{3}{2}}}\left(\frac{\theta}{2}+\frac{1}{4}\sin(2\theta)+c\right)$$
$$= \frac{2k}{(4kb-a^{2})^{\frac{3}{2}}} \left(\frac{\arctan\left(\frac{2k(x+\frac{a}{2k})}{\sqrt{4kb-a^{2}}}\right)}{2}+\frac{\sin(2 \arctan\left(\frac{2k(x+\frac{a}{2k})}{\sqrt{4kb-a^{2}}}\right)}{4}\right).$$
Is there a simpler solution?
| Based on your work, I'm making the assumption that determinant $\Delta = a^2 - 4kb < 0$
To make this simple, I'm going to write
$$ kx^2 + ax + b = k \left(x + \frac{a}{2k} \right)^2 + \frac{4kb - a^2}{4k} = k(u^2 + m^2) $$
where $u = x + \frac{a}{2k}$ and $m = \frac{\sqrt{4kb - a^2}}{2k}$. We have
$$ \int \frac{dx}{(kx^2 + ax + b)^2} = \frac{1}{k^2} \int \frac{du}{(u^2 + m^2)^2} $$
The trick is performing integration by parts on this integral
$$ \int \frac{du}{u^2 + m^2} \tag{1}$$
We should get
$$ \begin{align} \int \frac{du}{u^2 + m^2}
&= \frac{u}{u^2 + m^2} + \int \frac{2u^2}{(u^2 + m^2)^2} \\
&= \frac{u}{u^2 + m^2} + \int \frac{2(u^2 + m^2) - 2m^2}{(u^2 + m^2)^2} \\
&= \frac{u}{u^2 + m^2} + 2 \int \frac{du}{u^2 + m^2} - 2m^2\int\frac{du}{(u^2+m^2)^2}
\end{align} $$
Rearranging the terms
$$\int \frac{du}{(u^2 + m^2)^2} = \frac{u}{2m^2(u^2 + m^2)} + \frac{1}{2m^2} \int \frac{du}{u^2 + m^2}$$
As you might know, $(1)$ is just the arctangent function:
$$ \int\frac{du}{u^2 + m^2} = \frac{1}{m}\,\arctan \left(\frac{u}{m} \right) + C $$
And so
$$\int \frac{du}{(u^2 + m^2)^2} = \frac{u}{2m^2(u^2 + m^2)} + \frac{1}{2m^3} \,\arctan \left(\frac{u}{m} \right) + C$$
After some algebra:
$$ \int\!\frac{dx}{(kx^2 + ax + b)^2} = \frac{2kx + a}{(4kb-a^2)(kx^2 + ax + b)} + \frac{4k}{(4kb - a^2)^{3/2}}\, \arctan \left(\frac{2kx + a}{\sqrt{4kb - a^2}} \right) + C $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1101714",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to prove $\int_0^1\frac{x^3\arctan x}{(3-x^2)^2}\frac{\mathrm dx}{\sqrt{1-x^2}}=\frac{\pi\sqrt{2}}{192}\left(18-\pi-6\sqrt{3}\,\right)$? How to prove the following result?
$$\int_0^1\frac{x^3\arctan x}{(3-x^2)^2}\frac{\mathrm dx}{\sqrt{1-x^2}}=\frac{\pi\sqrt{2}}{192}\left(18-\pi-6\sqrt{3}\,\right)$$
For my part no idea?
| Although integration by parts will work, a nice general way to attack the integral is by introducing
$$
J(a,b) := \int_0^1 \frac{x \arctan(ax)}{1-b^2 x^2} \frac{dx}{\sqrt{1-x^2}},
$$
so $J(0,b) = 0$ and
$$
\begin{align}
\partial_a J &= \int_0^1 \frac{x^2}{(1-b^2 x^2)(1+a^2x^2)} \frac{dx}{\sqrt{1-x^2}}
= \int_0^{\pi/2} \frac{du\sec^2 u}{(\sec^2 u -b^2)(\sec^2 u +a^2)}
\\&= \int_0^{\infty} \frac{dt}{(t^2 + 1-b^2)(t^2 + 1+a^2)} = \frac{\pi/2}{\sqrt{1-b^2}\sqrt{1+a^2}} \frac{1}{\sqrt{1-b^2}+\sqrt{1+a^2}},
\end{align}
$$
where $ x = \cos u, t = \tan u$, and the last equality follows easily using partial fractions.
Hence
$$
\begin{align}
J(1,b) &= \frac{\pi/2}{\sqrt{1-b^2}} \int_0^1 \frac{da}{\sqrt{1+a^2}} \frac{1}{\sqrt{1-b^2}+\sqrt{1+a^2}}
\\&= \left.\frac{\pi/2}{\sqrt{1-b^2}} \frac{1}{b} \left[\arctan\left(\frac{b}{\sqrt{1-b^2}}\frac{\sqrt{1+a^2}}{a} \right)- \arctan \frac b a\right] \right\lvert_{a=0}^{a=1}
\\&= \frac{\pi/2}{\sqrt{1-b^2}} \frac{1}{b} \left[\arctan\sqrt\frac{2b^2}{1-b^2}- \arctan b\right].
\end{align}
$$
This implies that
$$
\begin{align}
&2b^{-3} \int_0^1 \frac{x^3 \arctan(x)}{(b^{-2} - x^2)^2} \frac{dx}{\sqrt{1-x^2}} = \partial_b J(1,b)
\\&= \frac{\pi}{2b^2(1-b^2)^{3/2}} \left\{b\frac{ \sqrt 2\sqrt{1- b^2}+2}{1+b^2}+\left(1-2b^2\right) \left[\arctan b-\arctan \sqrt{\frac{2b^2}{1-b^2}}\right]-b\right\}.
\end{align}
$$
Plugging in $b = \dfrac{1}{\sqrt 3}$ simplifies this a lot, giving
$$
\int_0^1 \frac{x^3 \arctan(x)}{(3 - x^2)^2} \frac{dx}{\sqrt{1-x^2}} =\frac{\pi \sqrt 2}{192}\left[18-\pi - 6\sqrt 3 \right],
$$
as desired.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1101966",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
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Solving $\frac{a}{b} + \frac{b}{c} + \frac{c}{a} = m \in \mathbb{Z}$, $\frac{a}{c} + \frac{c}{b} + \frac{b}{a }= n \in \mathbb{Z}$ Whether non-zero integers $a, b, c$ with the property that $$\frac{a}{b} + \frac{b}{c} + \frac{c}{a} = m \in \mathbb{Z}$$ and $$\frac{a}{c} + \frac{c}{b} + \frac{b}{a }= n \in \mathbb{Z}$$ Calculate all possible values for $m + n$.
| let integer $a,b,c$ such $\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}$ and $\dfrac{a}{c}+\dfrac{c}{b}+\dfrac{b}{a}$ is also integer, then we have
$$|a|=|b|=|c|$$
proof: WLOG:we may assume $gcd(a,b,c)=1$,Indeed,we can consider$\dfrac{a}{d},\dfrac{b}{d},\dfrac{c}{d}$
where $\gcd{(a,b,c)}=d$
a contradiction,then $a,b,c$ least one $\neq\pm 1$,WLOG $a\neq\pm 1$,and let prime number $p|a$,then
$$p|abc\left(\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\right)=a^2c+b^2a+c^2b$$
so
$p|c^2b$
so $p|b$ or $p|c$
WLOG: $p|b,p\nmid c$,and let $v_{p}(a)=r,v_{p}(b)=s,r\le s$,then we have
$$p^{r+s}|abc\left(\dfrac{a}{c}+\dfrac{c}{b}+\dfrac{b}{a}\right)=a^2b+c^2a+b^2c$$
But $$p^{s+r}\nmid c^2a,p^{r+s}\nmid b^2c,p^{r+s}|a^2b$$
a contradiction.
so
$$|a|=|b|=|c|$$
then easy to find $m,n$
| {
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"url": "https://math.stackexchange.com/questions/1102537",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
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How to calculate $\lim \limits_{x \to 0}{\frac{\sqrt{1 + x + x^2} - 1}{x}}$? I try to calculate $\lim \limits_{x \to 0}{\frac{\sqrt{1 + x + x^2} - 1}{x}}$. I've got $\frac{\sqrt{1 + x + x^2} - 1}{x} = \sqrt{\frac{1}{x^2} + \frac{1}{x} + x} - \frac{1}{x}$ but I don't know what to do next.
| Hint:
Multiply the numerator and the denominator by $$\sqrt{1+x+x^2}+1.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1102631",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Fibonacci sequences and related series
Let $\{a_n\}$ be a sequence such that $a_1=a_2=1$ and $a_{n+1}=a_n+a_{n-1}$ for $n\geq 2$. Prove that $\displaystyle \sum_{n=1}^\infty \frac{1}{a_n}$ converges.
My work:
Let $b_n=\frac{1}{a_n}$. Then I proved that $|\frac{b_{n+1}}{b_n}|<1$. I was going to use ratio test. But how can I show that limsup $|\frac{b_{n+1}}{b_n}|\neq1$? Can anyone please help me?
| A cute answer starts with Cassini's identity
$$
F_{n+1}F_{n-1} - F_n^2 = (-1)^n
$$
which is easy to prove by induction.
Consider the ratio $\frac{F_{n+1}}{F_n}$. If we show this is always strictly greater than one for $n>1$, then limsup $|\frac{b_{n+1}}{b_n}| : n > 1$ is less than $1$.
For $n$ odd, $\frac{F_{n+1}}{F_n} > \frac{F_{n-1}}{F_{n-2}}$ because
$$
\frac{F_{n+1}}{F_n} = \frac{F_n}{F_{n-1}}-\frac{1}{F_{n-1}} = \frac{F_{n-1}}{F_{n-2}} - \frac{1}{F_{n-1}} + \frac{1}{f_{n-2}} > \frac{F_{n-1}}{F_{n-2}}-\frac{1}{F_{n-2}} + \frac{1}{f_{n-2}} = \frac{F_{n-1}}{F_{n-2}}$$
So the smallest value of $\frac{F_{n+1}}{F_n}$ for odd $n$ greater than 1 occurs at $n=3$, and is $\frac{3}{2}$.
Further, for all even $n$, $\frac{F_{n+1}}{F_n} > \frac{F_{n}}{F_{n-1}}$ again by Cassini's identity.
So liminf $\frac{F_{n+1}}{F_n} = \frac{3}{2}$ and
$|\frac{b_{n+1}}{b_n}| > \frac{2}{3}$ for all $n > 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1103402",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Polar Integration of $ r = 2\cos(\theta)$ $ r = 2\cos(\theta)$ has the graph
I want to know why the following integral to find area does not work $$\int_0^{2 \pi } \frac{1}{2} (2 \cos (\theta ))^2 \, d\theta$$
whereas this one does:
$$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2} } \frac{1}{2} (2 \cos (\theta ))^2 \, d\theta$$
Why do the limits of integration have to go from $-\frac{\pi}{2}$ to $\frac{\pi}{2}$? Doesn't going from $0$ to $2\pi$ also sweep out the full circle?
| Start with $\theta=0$ you get $r=2$ then move to $\theta=\frac{\pi}{2}$ you will get $r=0$ (then you have from here the upper half of the circle). From $\frac{\pi}{2}$ to $\pi$ you have a negative $r$ which don't have really much sense. Then you have to consider it from $\frac{3\pi}{2}$ to $2\pi$ to complete the lower half.
$$\int_{0}^{\frac{\pi}{2} } \frac{1}{2} (2 \cos (\theta ))^2 \, d\theta+\int_{\frac{3\pi}{2}}^{2\pi} \frac{1}{2} (2 \cos (\theta ))^2 \, d\theta=\int_{0}^{\frac{\pi}{2} } \frac{1}{2} (2 \cos (\theta ))^2 \, d\theta+\int_{-\frac{\pi}{2}}^{0} \frac{1}{2} (2 \cos (\theta ))^2 \, d\theta$$
$$=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{1}{2} (2 \cos (\theta ))^2 \, d\theta$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1103478",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to prove $(\frac{1}{5^3}-\frac{1}{7^3})+(\frac{1}{11^3}-\frac{1}{13^3})+(\frac{1}{17^3}-\frac{1}{19^3})+...=(1-\frac{\pi ^3}{18\sqrt{3}})$ How to prove
$$ \sum_{k=1}^\infty \left[\frac{1}{(6k-1)^3} - \frac{1}{(6k+1)^3}\right] = 1 - \frac{\pi^3}{18\sqrt{3}}$$
I think this equality likes the Dirichlet Beta function. The numerical value is checked but I don't have the proving. Any help
| Notice$\color{blue}{^{[1]}}$
$$\sum_{k=1}^\infty \left( \frac{1}{(6k-1)^3} - \frac{1}{(6k+1)^3}\right)
= \sum_{\substack{k=-\infty\\ k\ne 0}}^\infty \frac{1}{(6k-1)^3}
= 1 - \frac{1}{6^3}\sum_{k=-\infty}^\infty \frac{1}{(\frac16-k)^3}$$
Recall the infinite product expansion of $\sin x$
$$\sin x = x \prod_{k=1}^\infty \left( 1 - \frac{x^2}{k^2\pi^2}\right)$$
If one take logarithm and differentiate, one obtain an expansion of $\cot x$
$$\cot x = \sum_{k=-\infty}^\infty \frac{1}{x - k\pi}
\quad\iff\quad
\sum_{k=-\infty}^\infty \frac{1}{x-k} = \pi\cot(\pi x)\tag{*1}
$$
Differentiate the expansion on the right two more times, we get$\color{blue}{^{[2]}}$
$$\sum_{k=-\infty}^\infty \frac{1}{(x-k)^3}
= \frac12 \frac{d^2}{dx^2} \left[ \sum_{k=-\infty}^\infty \frac{1}{x-k} \right]
= \frac{\pi}{2} \left[ \frac{d^2}{dx^2}\cot(\pi x) \right]
= \frac{\pi^3 \cos(\pi x)}{\sin(\pi x)^3}$$
As a result, the sum we want is
$$\sum_{k=1}^\infty \left( \frac{1}{(6k-1)^3} - \frac{1}{(6k+1)^3}\right)
= 1 - \frac{\pi^3}{6^3}\frac{\cos\frac{\pi}{6}}{\sin(\frac{\pi}{6})^3} \
= 1 - \frac{\pi^3}{18\sqrt{3}}
$$
Notes
*
*$\color{blue}{[1]}$ - Infinite sum of the form $\sum\limits_{k=-\infty}^\infty (\cdots)$ should be interpreted as $\lim\limits_{N\to\infty}\sum\limits_{k=-N}^N (\cdots)$.
*$\color{blue}{[2]}$ - To those who are not comfortable with the use of differentiation of
an expansion. An alternate approach is start from the more well known expansion $(*1)$,
compute a contour integral of the form:
$$\frac{1}{2\pi i}\int_{|z|=R} \frac{\pi\cot(\pi z)}{(\frac16 - z)^3} dz$$
and show it vanishes as $R \to \infty$.
The sum of the residues from $z \in \mathbb{Z}$ will be equal to the sum $\sum\limits_{k=-\infty}^\infty \frac{1}{\left(\frac16 - k\right)^3}$. It will be compensate by the residue
of the pole at $z = \frac16$. Since the pole at $z = \frac16$ is a triple one, its contribution will be proportional to $\frac{d^2}{dx^2}\pi\cot(\pi x)$.
| {
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"url": "https://math.stackexchange.com/questions/1104391",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "11",
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