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How to find the sum of the series $\sum_{k=2}^\infty \frac{1}{k^2-1}$? I have this problem :
$$S_n=\sum_{k=2}^\infty \frac{1}{k^2-1}$$
My solution
$$S_n=\sum_{k=2}^\infty \frac{1}{k^2-1} = -\frac{1}{2}\sum_{k=1}^\infty \frac{1}{k+1} -\frac{1}{k-1} = -\frac{1}{2}[(\frac{1}{3}-1)+(\frac{1}{4}-\frac{1}{2})+(\frac{1}{5}-\frac{1}{3})+(\frac{1}{6}-\frac{1}{4})+...]$$
I think that the sum should be $\frac{1}{2}$ since the limit of :
$$-\frac{1}{2}[-1+\frac{1}{k}+...+\frac{1}{n}] = \frac{1}{2}$$
But that wrong. Any ideas?
| your choice to use the "telescoping sum" technique is fine. what is wrong is just an arithmetic prob.
since the series is absolutely convergent try first rewriting as two sums:
$$
\sum_{k=1}^\infty \frac{1}{(2k)^2-1} + \sum_{k=1}^\infty \frac{1}{(2k+1)^2-1}
$$
and evaluate these separately.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/993890",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
How to prove the inequality $2\sqrt{n + 1} − 2 \le 1 +\frac 1 {\sqrt 2}+\frac 1 {\sqrt 3}+ \dots +\frac 1 {\sqrt n} \le 2\sqrt n − 1$? Prove that for any positive integer $n$,
$$2\sqrt{n + 1} − 2 \le 1 +\frac 1 {\sqrt 2}+\frac 1 {\sqrt 3}+ \dots +\frac 1 {\sqrt n} \le 2\sqrt n − 1$$
Progress
I think Riemann sum should be used for the middle term. I got the limit as $n\to \infty$ of the function inside using Riemann sum but what I cannot get are the lower and upper bound.
| There's a tricky solution using integral estimates:
$$2\sqrt{n+1}-2=\int_1^{n+1}\frac1{\sqrt n}\le\sum_{k=1}^n\frac1{\sqrt n}=1+\sum_{k=2}^n\frac1{\sqrt n}\le 1+\int_1^n\frac1{\sqrt {n}}=1+2\sqrt n-2=2\sqrt n-1$$
But you can easily proceed by induction, the base case is trivial and
$$(2\sqrt{n+1}-1)-(2\sqrt n-1)=2(\sqrt{n+1}-\sqrt n)=2\frac{(n+1)-n}{\sqrt{n+1}+\sqrt n}=\frac2{\sqrt{n+1}+\sqrt n}\ge\frac1{\sqrt{n+1}}$$
So the difference of successive right hand sides is bigger than what you add to the middle.
And similarly the other inequality:
$$(2\sqrt{n+2}-2)-(2\sqrt{n+1}-2)=2(\sqrt{n+2}-\sqrt{n+1})=\frac2{\sqrt{n+2}+\sqrt{n+1}}\le\frac1{\sqrt{n+1}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/995110",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Prove statement about determinants. $A$ is a $3\times 3$ matrix over $\mathbb{R}$, I want to show that if $$\det(A + I_3)=\det(A+2I_3),$$ then $$2\det(A+I_3) + \det(A-I_3) + 6 = 3\det A.$$
Can you help me?
| If $a,b,c$ are the eigenvalues of $A$, then $\det(A)=abc$, and $\det(A+dI)=(a+d)(b+d)(c+d)$.
So we are given that
$$
(a+1)(b+1)(c+1)=\det (A+I)=\det (A+2I)=(a+2)(b+2)(c+2),
$$
and hence
$$
(ab+bc+ca)+(a+b+c)+1=2(ab+bc+ca)+4(a+b+c)+8,
$$
or simpler
$$
(ab+bc+ca)+3(a+b+c)+7=0.\tag{1}
$$
We need to show that
$$
2\det (A+I)+\det (A-I)+6=3\det(A)
$$
which is equivalent to
$$
2(abc+(ab+bc+ca)+(a+b+c)+1)+(abc-(ab+bc+ca)+(a+b+c)-1)+6=3abc,
$$
or simpler
$$
(ab+bc+ca)+3(a+b+c)+7=0\tag{2}
$$
Clearly $(1)$ and $(2)$ are identical and hence we are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/995191",
"timestamp": "2023-03-29T00:00:00",
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Remainders on division of polynomials I am told that the remainder on division of a polynomial $p(z)$ by $z^3+z^2+z+1$ is $z^2-z+1$. I am also given that $p(1)=2$, and then asked to determine the remainder when $p(z)$ is divided by $z^4-1$.
I have expressed $p(z)$ as $p(z) = (z^3+z^2+z+1)(q(z)) + z^2-z+1$, for some polynomial $q$. And I found that $p(1)=2$ implies $q(1)=1/4$.
I then expressed $p(z)$ also as $p(z)=(z^4-1)(h(z)) + r(z)$, for some polynomial $h(z)$, where $r(z)$ is the remainder from the division of $p(z)$ by $z^4-1$.
I then divided $z^4-1$ by $z^3+z^2+z+1$ and found that $z^4-1 = (z-1)(z^3+z^2+z+1)$, so that I could express $p(z)$ as $p(z)= (z-1)(z^3+z^2+z+1)(h(z)) + r(z)$.
From then on I couldn't see how to proceed. Any hints on how to better approach this problem would be appreciated!
| We write the Euclidean division of $q(z)$ by $z-1$
$$q(z)=(z-1)u(z)+v$$
and since $q(1)=\frac14$ so $v=\frac14$ hence
\begin{align}p(z) &= (z^3+z^2+z+1)q(z) + z^2-z+1 \\&= (z^3+z^2+z+1)((z-1)u(z)+\frac14) + z^2-z+1\\&=(z^4-1)u(z)+\underbrace{z^3+\frac54z^2-\frac34z+\frac54}_{\text{the remainder}}\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/995765",
"timestamp": "2023-03-29T00:00:00",
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Implicit Diff. check I have the expression ${\dfrac{x^2+y^2}{x+y}}=3$ and I wanna find $dy/dx$. Here's my approach:
$x^2+y^2=3x+3y$
$\implies (x^2-3x)+(y^2-3y)=0$
$\implies (2x-3)+\dfrac{dy}{dx}(2y-3)=0$
${\implies \dfrac{dy}{dx}=\dfrac{3-2x}{2y-3}}$
Wolfram Alpha gives me a very different answer. This seems to work (i.e, the tangent I found at $x=3$ from this looks right), but I'm not sure.
| implicit differentiating of $\frac{x^2+y^2}{x+y}=3$ gives
$\frac{2x+2yy')(x+y)-(x^2+y^2)(1+y')}{(x^2+y^2)^2}=0$
plugging $x^2+y^2=3(x+y)$ in this term we get
$(2x+2yy')(x+y)-3(x+y)(1+y')=0$
with $x+y\ne 0$ we get $2x+2yy'-3-y'=0$ and this is your result
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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linear equations in a matrix form Considering
$$x_1 − x_2 + x_3 − x_4 = 2$$
$$x_1 − x_2 + x_3 + x_4 = 0$$
$$4x_1 − 4x_2 + 4x_3 = 4$$
$$−2x_1 + 2x_2 − 2x_3 + x_4 = −3$$
We have the following matrix
$$
\begin{pmatrix}
1 & -1 & 1 & -1 & 2 \\
1 & -1 & 1 & 1 & 0 \\
4 & -4 & 4 & 0 & 4\\
-2 & 2 & -2 & 1 & -3 \\
\end{pmatrix}
$$
Skiping some steps...
Row echelon form
$$
\begin{pmatrix}
1 & -1 & 1 & -1 & 2 \\
0 & 0 & 0 & 2 & -2 \\
0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & -1 & 1 \\
\end{pmatrix}
$$
Showing the row 2 is a multiple of row 4, and thus our system reduces to
$$
\begin{pmatrix}
1 & -1 & 1 & -1 & 2 \\
0 & 0 & 0 & -1 & 1 \\
\end{pmatrix}
$$
where we replace row 1 by (row 1)-(row 2):
$$
\begin{pmatrix}
1 & -1 & 1 & 0 & 1 \\
0 & 0 & 0 & -1 & 1 \\
\end{pmatrix}
$$
Solution :
$$x_4 = −1$$
$$x_1 = 1 + x_2 − x_3.$$
I am puzzled why did they decided to omit out row 2 & 3, instead of reducing it to Lowest Row echelon form. That way, we can solve for $x_2$ and $x_3$
| Applying Gauss-Jordan to reach an upper triangular matrix:
$$\left(\begin{array}{ccccc} \boxed{1}&-1&1&-1&2\\0&0&0&\boxed2&-2\\0&0&0&0&0\\0&0&0&0&0 \end{array}\right).$$
I have boxed two positions in the matrix. These are called pivots. Recall that when you want set all the positions under a pivot to $0$, you use that pivot to do that. The number of pivots is called the Rank of a matrix which tells you how many fixed variables there are in your equation system.
Before continuing I want to say that we are looking for values that satisfy:
$$x_1\left(\begin{array}{c}1\\0\\0\\0\end{array}\right)+x_2\left(\begin{array}{c}-1\\0\\0\\0\end{array}\right)+x_3\left(\begin{array}{c}1\\0\\0\\0\end{array}\right)+x_4\left(\begin{array}{c}-1\\2\\0\\0\end{array}\right) = \left(\begin{array}{c}2\\-2\\0\\0\end{array}\right).$$
It is obvious that we need to take $x_4$ to be $-1$. Now we have that $x_1-x_2+x_3 = 1$. It is standard to express the solution in terms of the free variables. You identify the free variables by the columns that do not have a pivot. So we will always choose to say that $x_1=1+x_2-x_3$ which says that $x_1$ and $x_4$ (that correspond to the pivot columns) are a linear combination of the other variables. In this case it is easy to solve for $x_2$ as well as for $x_3$. But that is not always the case.
If you were to have a matrix:
$$\left(\begin{array}{ccccc} \boxed{1}&-1&1&-1&2\\0&\boxed1&-1&2&-2\\0&0&\boxed1&2&3\\0&0&0&0&0 \end{array}\right),$$
then it would be more obvious to solve for $x_1$, $x_2$ and $x_3$ letting $x_4$ free. If you want to solve for $x_4$ you are using the wrong matrix.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/996544",
"timestamp": "2023-03-29T00:00:00",
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Line tangent to the circle Find the equation of the lines which are tangent to the circle with equation $x^2+y^2=9$ and parallel to the line of equation $x-y+1=0$
| Lines parallel to $x-y+1$ have equation $y=x+a$ for $a \in \mathbb{R}$, so common point of circle and $y=x+a$ are solutions of equation:
$$x^2+y^2=x^2+(x+a)^2=9$$
If $y=x+a$ is tangent line, then has only one common point with circle, so equation should has only one solution, so you must have $\Delta=0$.So:
$$x^2+(x+a)^2=x^2+x^2+2ax+a^2=9$$
$$2x^2+2ax+a^2-9=0$$
And $\Delta=(2a)^2-4 \cdot 2(a^2-9)=-4a^2+72=0$
So $a=\pm\sqrt{\frac{72}{4}}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/997100",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
find common ratio of $\sum_{k=1}^\infty \frac{1}{k(k+1)}$ I have this problem, I need to find the sum.
$$\sum_{k=1}^\infty \frac{1}{k(k+1)} = \frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\cdots+\frac{1}{k(k+1)}$$
The problem is that the ratio is not conclusive, Any idea how to find the ratio?
Thanks!
| Since you speak of the ratio test, maybe you're only concerned with whether the series converges rather than with what the sum is. That is the most that the ratio test can give you. If that is what you're concerned with, then you can say
$$
\sum_{k=1}^\infty \frac 1 {k(k+1)} \le \sum_{k=1}^\infty \frac 1 {k^2}
$$
and that converges, by a simple integral test.
However, we can also say
$$
\frac 1 {k(k+1)} = \frac{1}{k} - \frac{1}{k+1}
$$
and then we have
\begin{align}
& \overbrace{\left(\frac 1 1 - \frac 1 2\right)}^{k=1} + \overbrace{\left(\frac 1 2 - \frac 1 3\right)}^{k=2} + \overbrace{\left(\frac 1 3 - \frac 1 4\right)}^{k=3} + \cdots + \overbrace{\left( \frac 1 n - \frac 1 {n+1} \right)}^{k=n} \\[10pt]
= {} & \frac 1 1 \underbrace{{} - \frac 1 2 + \frac 1 2 }\ \ \underbrace{{}-\frac 1 3 + \frac 1 3 }\ \ \underbrace{{} - \frac 1 4 + \frac 1 4} -\cdots\cdots\underbrace{{} - \frac 1 n + \frac 1 n} - \frac 1 {n+1} \\[10pt]
= {} & 1 - \frac 1 {n+1} \to 1 \text{ as }n\to\infty.
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
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Show $1/(1+ x^2)$ is uniformly continuous on $\Bbb R$.
Prove that the function $x \mapsto \dfrac 1{1+ x^2}$ is uniformly continuous on $\mathbb{R}$.
Attempt: By definition a function $f: E →\Bbb R$ is uniformly continuous iff for every $ε > 0$, there is a $δ > 0$ such that $|x-a| < δ$ and $x,a$ are elements of $E$ implies $|f(x) - f(a)| < ε.$
Then suppose $x, a$ are elements of $\Bbb R. $
Now
\begin{align}
|f(x) - f(a)|
&= \left|\frac1{1 + x^2} - \frac1{1 + a^2}\right|
\\&= \left| \frac{a^2 - x^2}{(1 + x^2)(1 + a^2)}\right|
\\&= |x - a| \frac{|x + a|}{(1 + x^2)(1 + a^2)}
\\&≤ |x - a| \frac{|x| + |a|}{(1 + x^2)(1 + a^2)}
\\&= |x - a| \left[\frac{|x|}{(1 + x^2)(1 + a^2)} + \frac{|a|}{(1 + x^2)(1 + a^2)}\right]
\end{align}
I don't know how to simplify more. Can someone please help me finish? Thank very much.
| $$x^2 \geq 0 \implies 1+x^2 > 1 \implies \frac{1}{1+x^2} < 1$$
Using the above inequality,
$$
\begin{align}
|f(x)-f(a)| &\leq |x - a| \frac{|x + a|}{(1 + x^2)(1 + a^2)}\\
&\leq |x - a||x+a|\\
&\leq |x - a|(|x|+|a|)
\end{align}
$$
Choose $$|x-a| \leq 1 \implies |x|\leq |a|+1$$
Now,
$$
\begin{align}
|f(x)-f(a)| &\leq |x - a|(|x|+|a|)\\
&\leq |x-a|(2|a|+1)
\end{align}
$$
Choose $$|x-a| < \frac{\epsilon}{2|a|+1} $$
Now,
$$
\begin{align}
|f(x)-f(a)| &\leq \epsilon
\end{align}
$$
where $$\delta = Inf\{1, \frac{\epsilon}{2|a|+1}\}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/997602",
"timestamp": "2023-03-29T00:00:00",
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derivative of $y=\frac{x^2\sqrt{x+1}}{(x+2)(x-3)^5}$ $y=\dfrac{x^2\sqrt{x+1}}{(x+2)(x-3)^5}$
The answer is $\dfrac{x^2\sqrt{x+1}}{(x+2)(x-3)^5} \left(\dfrac{2}{x}+\dfrac{1}{2(x+1)}-\dfrac{1}{x+2}-\dfrac{5}{x-3}\right)$
I know that the quotient rule is used but I don't know how to do this problem. Would you multiply together all the terms and then differentiate?
| $$y'=\dfrac{(x^2\sqrt{x+1})'(x+2)(x-3)^5-x^2\sqrt{x+1}((x+2)(x-3)^5)'}{(x+2)^2(x-3)^{10}}=
\dfrac{((2x\sqrt{x+1}+\frac{x^2}{2\sqrt{x+1}})(x+2)(x-3)^5-x^2\sqrt{x+1}((x-3)^5)+5(x+2)(x-3)^4)}{(x+2)^2(x-3)^{10}}=\dfrac{x^2\sqrt{x+1}}{(x+2)(x-3)^5}(\dfrac{2}{x}+\dfrac{1}{\sqrt{x+1}\sqrt{x+1}}-(\dfrac{1}{x+2}+\dfrac{5}{x-3})$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1000136",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Proving/disproving that $1=\frac {1}{2}(x)^{-\frac {1}{2}} \Longleftrightarrow 2\sqrt{x}=0$ I was looking at the solution to this problem:
Find $\frac {dy}{dx}$ for $x=10+\sqrt {x^2+y^2}$.
The solution given was as follows:
$$
\frac {d}{dx}(x)=\frac {d}{dx}(10)+\frac {d}{dx}(\sqrt {x^2+y^2})
$$
$$
1=\frac {1}{2}(x^2+y^2)^{-\frac {1}{2}}(2x+2y\frac {dy}{dx})
$$
$$
2\sqrt{x^2+y^2}=2x+2y\frac {dy}{dx}
$$
$$
2\sqrt{x^2+y^2}-2x=+2y\frac {dy}{dx}
$$
$$
\frac {dy}{dx}=\frac {\sqrt{x^2+y^2}-x}{y}
$$
I understand it except for what happens between step two and three. I tried but was unable to prove to myself that
$$
1=\frac {1}{2}(x)^{-\frac {1}{2}} \Longleftrightarrow 2\sqrt{x}=0
$$
or that (I think this is equivalent)
$$
-(\frac {1}{2}(x)^{-\frac {1}{2}})+1=2\sqrt{x}
$$
Can someone show me how to prove it, or am I looking at the original problem wrong? Thank you.
| Rewriting from step 2 might help:$$1=\frac {1}{2}(x^2+y^2)^{-\frac {1}{2}}(2x+2y\frac {dy}{dx})$$$$1=\frac {1}{2\sqrt{x^2+y^2}}(2x+2y\frac {dy}{dx})$$$$1=\frac{(2x+2y\frac {dy}{dx})}{2\sqrt{x^2+y^2}}$$Now just multiply both sides by $2\sqrt{x^2+y^2}$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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I want help to Prove $\pi \tanh(\frac{\pi }{2})=\sum_{n=0}^{\infty }\frac{1}{n^2+n+0.5}$ It is difficult, so I need how to start to prove it
$$\pi \tanh(\frac{\pi }{2})=\sum_{n=0}^{\infty }\frac{1}{n^2+n+0.5}$$
| I am just mimicking the answer of @robjohn for this question, all credits go to him
let $x^2 + x + 0.5 = (x-a)(x-b)$, then we have $a+b = -1$, i.e. $b = -1-a$
\begin{align}
&\sum_{n=0}^{+\infty}\dfrac{1}{n^2 + n + 0.5} \\
= &\sum_{n=0}^{+\infty}\dfrac{1}{(n-a)(n-b)}\\
= & \dfrac{1}{a-b}\sum_{n=0}^{+\infty}\left(\dfrac{1}{n-a} - \dfrac{1}{n-b}\right)\\
= & \dfrac{1}{a-b}\sum_{n=0}^{+\infty}\left(\dfrac{1}{n-a} - \dfrac{1}{n+ 1 + a}\right)\\
= & \dfrac{1}{a-b}\sum_{n=0}^{+\infty}\left(\dfrac{1}{n-a} + \dfrac{1}{-n- 1 - a}\right)\\
= & \dfrac{1}{a-b}\sum_{n=-\infty}^{+\infty}\dfrac{1}{n-a}\\
\end{align}
And we use the fact $\sum_{n=-\infty}^\infty\frac{1}{n-a}=-\pi\cot(\pi a)$,
so we have $$\sum_{n=0}^{+\infty}\dfrac{1}{n^2 + n + 0.5} = \dfrac{1}{b-a}\pi\cot(\pi a)$$
Now plug in the value of $a$ and $b$ to conclude.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Solving equations with mod So, I'm trying to solve the following equation using regular algebra, and I don't think I'm doing it right: $3x+5 = 1\pmod {11}$
I know the result is $x = 6$, but when I do regular algebra like the following, I do not get 6:
$3x=1 - 5\pmod{11}$
$x = \dfrac{(1 \pmod{11} - 5)} 3$
So, I figured that since $1 \pmod{11} = 1$ the equation becomes
$x = \dfrac{-4} 3$
Which is not 6! I am totally lost here and would appreciate any help......
| When we are working with congruence, $\text{mod } n$, for some $n$ (here, 11), we are interested in integral relations, i.e., relationships between integer values, and instead of division, we use the fact, in this case, that $3x +5= 11k + 1$, where $k$ is any integer (the quotient), and $1$ is the remainder when $3x +5$ is divided by $11$
$$3x + 5 \equiv 1\pmod {11} \implies 3x+5 -1\equiv 1-1 = 0 \pmod{11}\iff 3x+4 \equiv 0 \pmod{11}$$
So we are looking for $x$ such that $3x+4 = 11k+ 0 = 11k$ for some integer $k$. That means we need for $\dfrac{3x+4}{11}$ to equal the integer $k$. Note that the given solution is one such solution: When $x=6$, then $3x+4 = 18+4 = 22 = 2(11)$. So when $x = 6$, $11$ divides $3x+4$ evenly, with no remainder.
There are infinitely many solutions $x$ that will meet this requirement, (test out $x = 6-11, x = 6+ 11= 17, x=6+ 2(11)$, etc.) So sometimes people use a representative congruence class $\overline x$, where $x$ is the least non-negative integer solution, and $$\overline x = \{6\pm 11k\mid k\in \mathbb Z\}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1004219",
"timestamp": "2023-03-29T00:00:00",
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Inequality proof by induction, what to do next in the step I have to prove that for $n = 1, 2...$ it holds: $2\sqrt{n+1} - 2 < 1 + \frac{1}{\sqrt2} + \frac{1}{\sqrt3} + ... + \frac{1}{\sqrt{n}}$
Base: For $n = 1$ holds, because $2\sqrt{2}-2 < 1$
Step: assume holds for $n_0$.
$2\sqrt{n+2} - 2 < 1 + \frac{1}{\sqrt2} + \frac{1}{\sqrt3} + ... + \frac{1}{\sqrt{n + 1}}$. But I do not know what to do next? How this can be proved?
| \begin{align}
\frac1{\sqrt{n+1}}+2\sqrt{n+1}-2&=\frac{1+2(n+1)}{\sqrt{n+1}}-2\\
&=\frac{\sqrt{4n^2+12n+9}}{\sqrt{n+1}}-2\\
&>\frac{\sqrt{4n^2+12n+8}}{\sqrt{n+1}}-2\\
&=\frac{2\sqrt{n^2+3n+2}}{\sqrt{n+1}}-2\\
&=\frac{2\sqrt{(n+2)(n+1)}}{\sqrt{n+1}}-2\\
&=2\sqrt{n+2}-2
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
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"answer_id": 2
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Finding the positive integer numbers to get $\frac{\pi ^2}{9}$ As we know, there are many formulas of $\pi$ , one of them $$\frac{\pi ^2}{6}=\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}......
$$
and this $$\frac{\pi ^2}{8}=\frac{1}{1^2}+\frac{1}{3^2}+\frac{1}{5^2}......$$
Now,find the positive integer numbers $(a_{0}, a_{1}, a_{2}....)$ to get $$\frac{\pi^2 }{9}=\frac{1}{a_{0}^2}+\frac{1}{a_{1}^2}+\frac{1}{a_{2}^2}....$$
| $$\begin{align}\frac{\pi^2}{9}&=1+\frac{1}{3}\sum_{k=0}^\infty\frac{1}{(k+1)^2(k+2)^2}\\&=1+\frac{1}{3(1·2)^2}+\frac{1}{3(2·3)^2}+\frac{1}{3(3·4)^2}+\frac{1}{3(4·5)^2}+\ldots\\&=1+\frac{1}{12}+\frac{1}{108}+\frac{1}{432}+\frac{1}{1200}\ldots...\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1007424",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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Trigonometric equation, missing some solutions I'm missing part of the answer, and I'm not quite sure why. The given answer doesn't even seem to hold...
Solve for x: $$\tan 2x = 3 \tan x $$
First some simplifications:
$$\tan 2x = 3 \tan x $$
$$\tan 2x - 3 \tan x = 0$$
$$\frac{\sin 2x}{\cos 2x} - \frac{3 \sin x}{\cos x} = 0$$
$$\frac{2 \sin x \cos^2x - 3 \sin x \cos 2x}{\cos 2x \cos x} = 0$$
$$\frac{\sin x(2 \cos^2x - 3 \cos 2x)}{\cos 2x \cos x} = 0$$
$$\frac{\sin x(2 \cos^2x - 3 (\cos^2 x - \sin^2 x))}{\cos 2x \cos x} = 0$$
$$\frac{\sin x(2 \cos^2x - 3\cos^2 x + \sin^2 x)}{\cos 2x \cos x} = 0$$
$$\frac{\sin x(\sin^2 x - \cos^2 x)}{\cos 2x \cos x} = 0$$
$$\frac{\sin x(\sin^2 x - \cos^2 x)}{(\sin^2 x - \cos^2 x) \cos x} = 0$$
$$\frac{\sin x}{\cos x} = 0$$
Looks much simpler. Now solving for x, since $\frac{\sin x}{\cos x} = 0 $ when $\sin x = 0$ and $\sin x = 0$ for every half rotation, the answer must be $k\pi$.
Alas, according to my answer sheet, I'm missing two values: $\frac{\pi}{6} + k\pi$ and $\frac{5\pi}{6} + k\pi$. But since $\frac{\sin(\frac{\pi}{6})}{\cos(\frac{\pi}{6})} = \frac{\sqrt3}{3}$, I'm not sure where these answers come from.
Furthermore, this is the kind of mistake I'm making all over these exercises, I'd like to avoid that, but how can I be sure I have ALL the answers needed?
| Hint:
You should use $$\tan 2x=\frac{2\tan x}{1-\tan^2x}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Volume of rotated region (integration) Let $T$ be a right-angled triangular region with vertices $(0,−b)$,$(1,0)$ and $(0,a)$ where $a$ and $b$ are positive numbers. When $T$ is rotated about the line $x=2$, it generates a solid with volume $V=\dfrac{410\pi}{27}$
Find $a$ and $b$.
Really having trouble with this one. Is $r=(2-x)$ and $h=(a+b)(1-x)$, and how do I integrate that?
Then, how do I display the result as a result of '$a$' and '$b$'?
| Your rotated volume is made of two parts ($y>0$ and $y<0$): both are made of a cylinder minus a truncated cone. Thus the volume is easy to compute, if you know $a$ and $b$: this gives you an equation between $a$ and $b$.
For the second equation, you know your triangle is right angled: the right angle must be at vertex $(1,0)$, so $(a+b)^2=1+a^2+1+b^2$ and $ab=1$.
Then solve for $a$ and $b$.
More precisely, the "upper" volume is, with $R=2$ and $r=1$ (see here),
$$V_{upper}=\pi R^2a-\frac a3(\pi R^2+\pi r^2+\pi R r)=\frac{a\pi}3(3R^2-R^2-r^2-R r)$$
Likewise, the lower volume is:
$$V_{lower}=\pi R^2b-\frac b3(\pi R^2+\pi r^2+\pi R r)=\frac{b\pi}3(3R^2-R^2-r^2-R r)$$
So the total volume is
$$V=\frac{(a+b)\pi}3(2R^2-r^2-R r)=\frac{5(a+b)\pi}3=\frac{410\pi}{27}$$
So
$$a+b=\frac{82}{9}=9+\frac{1}{9}$$
$$ab=1$$
Hence $a=9$ and $b=\frac{1}{9}$, or the other way around.
If you insist on computing an integral, you can write
$$V=\int_{-b}^a S(h)\mathrm{d}h$$
Where $S(h)$ is the area of the intersection of your volume and the plane $y=h$, hence, for $h>0$
$$S(h)=4\pi-\pi\left(1+\frac ha\right)^2$$
The first term is the area of a disc of radius $2$. The second is the area of a disc of radius $1+\frac ha$ (use for example Thales' theorem to see why).
For $h<0$, you have $S(h)=4\pi-\pi\left(1-\frac hb\right)^2$
Hence
$$V=4a\pi-\pi\int_0^a \left(1+\frac ha\right)^2\mathrm{d}h+4b\pi-\pi\int_{-b}^0 \left(1-\frac hb\right)^2\mathrm{d}h$$
$$V=4a\pi-\pi\int_0^a \left(1+\frac ha\right)^2\mathrm{d}h+4b\pi-\pi\int_0^b \left(1+\frac hb\right)^2\mathrm{d}h$$
$$V=4a\pi-\pi\left(a+\frac{a^2}{a}+\frac{a^3}{3a^2}\right)+4b\pi-\pi\left(b+\frac{b^2}{b}+\frac{b^3}{3b^2}\right)=\frac{5(a+b)\pi}{3}$$
Of course you find the same volume as in the formula above.
| {
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"timestamp": "2023-03-29T00:00:00",
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Indefinite integral of $\frac{\arctan x}{x^2+1}$ EDIT: I was studying from a site that uses really ambiguous notation so I misread $\arctan\ (x)^2$ as $\arctan\ (x^2)$. Now I can see why the integral is actually $\frac{1}{2} \arctan^2\ x + c $. Thanks to everyone who answered and corrected me!
Why is $$\int\frac{\arctan\ x}{x^2+1} dx=\frac{1}{2}\arctan(x^2)+c$$
instead of $\int u\ du$ where $u=\arctan\ x$ and $du=\frac{1}{x^2+1} dx$, thus $$\begin{align} &\int\frac{\arctan\ x}{x^2+1} dx=\int \arctan\ x\ dx \\
&x\ \arctan\ x-\int \frac{x}{x^2+1}\ dx \\
&x\ \arctan\ x -\frac{1}{2}\ln(x^2+1)+c\end{align}$$
Is there something I'm not getting? I'd like to see the steps for the correct calculation so I can better understand it. Thanks!
| instead of $\int u\ du$ where $u=\arctan\ x$ and $du=\frac{1}{x^2+1} dx$, thus $$\int\frac{\arctan\ x}{x^2+1} dx=\int u du= \frac12u^2+C=\frac12(\arctan x)^2+C$$
| {
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"timestamp": "2023-03-29T00:00:00",
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How find this maximum of this $(1-x)(1-y)(10-8x)(10-8y)$
let $x,y\in (0,1)$, and such
$$(1+x)(1+y)=81(1-x)(1-y)$$
Prove
$$(1-x)(1-y)(10-8x)(10-8y)\le\dfrac{9}{16}$$
I ask $\dfrac{9}{16}$ is best constant?
PS:I don't like Lagrange Multipliers,becasue this is Hight students problem.
My idea:
$$(1-x)(1-y)(10-8x)(10-8y)=\dfrac{1}{64\cdot 81}(8+8x)(8+8y)(10-8x)(10-8y)$$
since
$$(8+8x)(10-8x)\le\dfrac{18^2}{4}=81,(8+8y)(10-8y)\le 81$$
if and only if $$8+8x=10-8x,8+8y=10-8y\Longrightarrow x=y=\dfrac{1}{8}$$
but this not such
$$(1+x)(1+y)=81(1-x)(1-y)$$
so How find this maximum?
| Hint. Let $X=\frac{1+x}{1-x},\ Y=\frac{1+y}{1-y}$. We have $XY=81$ and $1\le X,\ 1\le Y$. Since $$(1-x)(10-8x)=\left(1-\frac{X-1}{X+1}\right)\left(10-8\frac{X-1}{X+1}\right)=\frac{4(X+9)}{(X+1)^2},$$ what we need to show is
$$\frac{(X+9)}{(X+1)^2}\cdot\frac{(Y+9)}{(Y+1)^2}\le \frac9{256}$$
and we would like to know when the equality holds.
Continuing from Hint. Let $k=X+Y$.
$$\frac{(X+9)}{(X+1)^2}\cdot\frac{(Y+9)}{(Y+1)^2}=\frac{162+9k}{(82+k)^2}.$$
Noting $18\le k\le 82$ and by differentiating the above by $k$, we find that $\frac{162+9k}{(82+k)^2}$ attains its maximum at $k=46$. The maximum value is
$$\frac{162+9\cdot 46}{(82+46)^2}=\frac{576}{128^2}=\frac{9}{256},$$
as required.
(If you don't want to use differentiation, you could instead use the AM-GM inequality:
\begin{align*}
\frac{162+9k}{(82+k)^2}&=\frac{9(k+18)}{(k+18)^2+128(k+18)+64^2}\\
&=\frac{9}{(k+18)+\frac{64^2}{k+18}+128}\\
&\le \frac{9}{2\sqrt{64^2}+128},
\end{align*}
where the last equality holds when $k+18=64$.)
| {
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"timestamp": "2023-03-29T00:00:00",
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Alternative ways to evaluate $\int^1_0 \frac{\text{Li}_2(x)^3}{x}\,dx$ In the following link here I found the integral & the evaluation of
$$\displaystyle \int^1_0 \frac{\text{Li}_2(x)^3}{x}\,dx$$
I'll also include a simpler version together with the question: is it possible to find some easy
ways of computing both integrals without using complicated sums that require multiple zeta
formulae and "never-ending long" generating functions?
$$i). \displaystyle \int^1_0 \frac{\text{Li}_2(x)^2}{x}\,dx$$
$$ii). \displaystyle \int^1_0 \frac{\text{Li}_2(x)^3}{x}\,dx$$
| Here is a slightly different approach to evaluating
$$\int^1_0 \frac{\text{Li}^2_2 (x)}{x} \, dx,$$
which, like the answer given by @Zaid Alyafeai, makes use of the result
$$\sum^\infty_{n = 1} \frac{H_n}{n^4} = 3 \zeta (5) - \zeta (2) \zeta (3).$$
We start by writing
$$I = \int^1_0 \frac{\text{Li}^2_2 (x)}{x} \, dx = \int^1_0 \frac{\text{Li}_2 (x)}{x} \cdot \text{Li}_2 (x) \, dx.$$
Integrating by parts we have
$$I = \text{Li}_2 (1) \text{Li}_3 (1) + \int^1_0 \frac{\text{Li}_3 (x) \ln (1 - x)}{x} \, dx = \zeta (2) \zeta (3) + \int^1_0 \frac{\text{Li}_3 (x) \ln (1 - x)}{x} \, dx,$$
where we have made use of the well-known results of $\text{Li}_s (0) = 0$ and $\text{Li}_s (1) = \zeta (s)$ respectively.
Integrating by parts again
$$I = \zeta (2) \zeta (3) - \int^1_0 \frac{\zeta (4) - \text{Li}_4 (x)}{1 - x} \, dx.$$
Recalling
$$\text{Li}_s (x) = \sum^\infty_{n = 1} \frac{x^n}{n^s} \quad \text{and} \quad \zeta (4) = \sum^\infty_{n = 1} \frac{1}{n^4},$$
we can write
\begin{align*}
I &= \zeta (2) \zeta (3) - \int^1_0 \frac{1}{1 - x} \left [\sum^\infty_{n = 1} \frac{1}{n^4} - \sum^\infty_{n = 1} \frac{x^n}{n^4} \right ] \, dx\\
&= \zeta (2) \zeta (3) - \sum^{\infty}_{n = 1} \frac{1}{n^4} \int^1_0 \frac{1 - x^n}{1 - x} \, dx.
\end{align*}
From the integral representation for the harmonic number $H_n$, namely
$$H_n = \int^1_0 \frac{1 - x^n}{1 - x} \, dx,$$
we can rewrite our integral as
$$I = \zeta (2) \zeta (3) - \sum^\infty_{n = 1} \frac{H_n}{n^4} = \zeta (2) \zeta (3) - \left \{3 \zeta (5) - \zeta (2) \zeta (3) \right \},$$
or
$$\int^1_0 \frac{\text{Li}_2^2 (x)}{x} \, dx = 2 \zeta (2) \zeta (3) - 3 \zeta (5).$$
| {
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"url": "https://math.stackexchange.com/questions/1016935",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
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Solve this triple integral I am trying to solve this triple integral:
$$\int_0^{ 2\pi} \int_0^4 \int_3^\sqrt{25-r^2} r \: dz \: dr \: d \theta$$.
I get stuck at
$$\int_0^{ 2\pi} \int_0^4 \ r(\sqrt{25-r^2}-3) \: dr \: d \theta$$
Should I use u-sub to find the rest? And if so, how?
| Firstly, note that
$$\int_0^{2 \pi} \int_0^4 r(\sqrt{25-r^2} -3)dr d \theta = 2 \pi \int_0^4 r(\sqrt{25-r^2} -3)dr$$
Now, call $5 \sin u = r$. So
$$\int_0^4 r(\sqrt{25-r^2} -3)dr = \int_0^{\arcsin \frac{4}{5}} 5 \sin u(5 \sqrt{1- \sin^2 u} -3) 5 \cos u \ du =$$
$$= 25 \int_0^{\arcsin \frac{4}{5}} \sin u \cos u (5 \cos u - 3) du = $$
Call $\cos u = t$ (so that $\sin u \ du = -dt$) and get
$$=25 \int_0^{\frac{3}{5}} t (3 - 5t) \ dt = \frac{9}{2}$$
So the final answer is
$$\int_0^{2 \pi} \int_0^4 r(\sqrt{25-r^2} -3)dr d \theta = 9 \pi$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1018437",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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On sums and identities I am given the following problem set:
(a) the Riemann $\zeta$-function for $s > 1$ is defined through the convergent sum: $$\zeta(s) := \sum_{n = 1}^{\infty} \frac{1}{n^s}$$
show the identity $$\sum_{n=1}^{\infty} \frac{1}{(2n-1)^2} = \frac{3}{4}\zeta(2)$$
(b) show that $$\sum_{n=1}^{\infty} \frac{1}{n(n+1)(n+2)}= \frac{1}{4}$$
(c) we denote $f_n$ as the $n^{th}$ fibonacci term. Show that $$\sum_{n=1}^{\infty} \frac{1}{f_n f_{n+2}} = 1$$
I basically need help on every of those identities since my knowledge about sums is pretty basic. thank you for your hints and help
| In the first identity you are interested in the infinite sum of just the odd integers, where $s = 2$. One way you can get this is to observe that $$\sum_{n=1}^{\infty} \frac{1}{(2n-1)^2} = \sum_{n=1}^{\infty} \frac{1}{n^2}-\sum_{n=1}^{\infty} \frac{1}{(2n)^2} \\ =\zeta(2)-\sum_{n=1}^{\infty} \frac{1}{4n^2} \\ = \zeta(2)-\frac{1}{4}\sum_{n=1}^{\infty} \frac{1}{n^2} \\ = \zeta(2)-\frac{1}{4}\zeta(2)$$ As for part $(b)$, manipulate the series $$\sum_{n=1}^{\infty} \frac{1}{n(n+1)(n+2)} = \sum_{n=1}^{\infty} \frac{1}{2n}-\frac{1}{n+1}+\frac{1}{2(n+2)} \\ = \sum_{n=1}^{\infty} \frac{1}{2n}- \sum_{n=1}^{\infty}\frac{1}{n+1}+ \sum_{n=1}^{\infty}\frac{1}{2(n+2)} \\ =\left(\frac{1}{2}+\frac{1}{4}+\sum_{n=3}^{\infty} \frac{1}{2n}\right)- \left(\frac{1}{2}+\sum_{n=3}^{\infty}\frac{1}{n}\right)+ \left(\sum_{n=3}^{\infty}\frac{1}{2n}\right) $$ For part $(c)$ remember that $f_{n+2} = f_{n+1}+f_n$
| {
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"timestamp": "2023-03-29T00:00:00",
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Tetrahedron- Inscribed Sphere In tetrahedron $ABCD, AB=BC=CA$ and $DA=DB=DC$. Given that the altitude of $ABCD$ from point $D$ is $24$ and that the radius of the inscribed sphere of $ABCD$ is $11$, determine $AB$.
| If we call $A1$, $A2$, $A3$, and $A4$ the areas of the faces of a tetrahedron, the radius $R$ of the inscribed sphere can be calculated using the formula
$$V = \frac{1}{3} R (A1 + A2 + A3 + A4)$$
where V is the volume of the tetrahedron.
In this case, calling $s$ the length of $AB=BC=CA$, we have that:
*
*the area of the base is $s^2 \frac{\sqrt{3}}{4}$;
*the volume is given by $V=\frac{1}{3} s^2 \frac{\sqrt{3}}{4} \cdot 24= 2 \sqrt{3} s^2$;
*the areas of the lateral faces are given by $\frac{s}{2} \sqrt{24^2+(s \frac{\sqrt{3}}{6})^2}=\frac{s}{2} \sqrt{576+s^2/12}$.
We then can write
$$2 \sqrt{3} s^2= \frac{1}{3} \cdot 11 (s^2 \frac{\sqrt{3}}{4}+ 3 \cdot \frac{s}{2} \sqrt{576+s^2/12})$$
which solved for $s$ gives $s=132$.
| {
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"timestamp": "2023-03-29T00:00:00",
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How prove this inequality $\left[\frac{n}{\sqrt{3}}\right]+1>\frac{n^2}{\sqrt{3n^2-5}}$
let $k$ is postive integer,and for any postive integer $n\ge 2$,
show that:
$$\left[\dfrac{n}{\sqrt{3}}\right]+1>\dfrac{n^2}{\sqrt{3n^2-5}}>\dfrac{n}{\sqrt{3}}$$
where $[x]$ is the largest integer not greater than $x$
| Let $q_n = \left\lfloor \frac{n}{\sqrt{3}}\right\rfloor + 1$. When $n \ge 11\sqrt{3}$, we have
$$\frac{n}{\sqrt{3}q_n} \ge \frac{\frac{n}{\sqrt{3}}}{\frac{n}{\sqrt{3}}+1} = 1 - \frac{\sqrt{3}}{n+\sqrt{3}} \ge \frac{11}{12}
\quad\implies\quad \frac{n^2}{3q_n^2} \ge \left(\frac{11}{12}\right)^2 > \frac{5}{6}
$$
Be definition, $\left\lfloor \frac{n}{\sqrt{3}}\right\rfloor$ is the largest integer
less than or equal to $\frac{n}{\sqrt{3}}$. This implies
$$\frac{n}{\sqrt{3}} < q_n \quad\implies\quad 3q_n^2 - n^2 > 0 \quad\implies\quad 3 q_n^2 - n^2 \ge 2$$
The last inequality is true because $3q_n^2 - n^2$ is an integer and there is no integer solution for the equation $3 q^2 - n^2 = 1$.
Combine these, we find for any $n \ge 20 > 11\sqrt{3}$, we have
$$3 n^2 - \frac{n^4}{q_n^2} = 3\left(\frac{n^2}{3q_n^2}\right)(3q_n^2 - n^2) > 3 \left( \frac{5}{6}\right) 2 = 5$$
This leads to
$$3n^2 - 5 > \frac{n^4}{q_n^2} \quad\iff\quad q_n > \frac{n^2}{\sqrt{3n^2 -5}} \quad\text{ for } n \ge 20 \tag{*1}$$
By brute force, one can verify RHS$(*1)$ also work for $2 \le n \le 19$.
As pointed out by Macavity in comment, the largest admissible $k$ for $n = 5$ is $5$.
This means the maximum value of $k$ which works for all $n$ is indeed $5$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Evaluate the surface integral from the paraboloid
Evaluate the surface integral $$\iint\limits_S xy \sqrt{x^2+y^2+1}\,\mathrm d\sigma,$$ where $S$ is the surface cut from the paraboloid $2z=x^2+y^2$ by the plane $z=1$.
Is it possible for the answer to be $0$ ? I am not so sure. Would anyone mind telling me the answer?
| Notice that your bounds are arranged as such:
\begin{equation}
2z=x^2+y^2\bigg|_{z=1}\implies y=\pm\sqrt{2-x^2},
\end{equation}
therefore the bounds on your integrals are
\begin{equation}
\int_{-\sqrt{2}}^{\sqrt{2}}\int_{-\sqrt{2-x^2}}^{\sqrt{2-x^2}}xy\sqrt{x^2+y^2+1}\:\:dy\:dx,
\end{equation}
because it appears as though you are integrating over a circular region and the bounds on the outer integral thus make this a definite integral. We may rewrite the bounds on the interior integral in terms of $y$ as well, which would thus change the order of integration. Now try integrating the interior using a u-substitution to get
\begin{align*}
\int_{-\sqrt{2}}^{\sqrt{2}}x\left(\int_{-\sqrt{2-x^2}}^{\sqrt{2-x^2}}\frac{u^{\frac{1}{2}}}{2}\:du\right)\:dx,\:\:\:\:u & =y^2+\left(x^2+1\right) \\
du & = 2y\:dy\implies\frac{du}{2y}=dy
\end{align*}
which gives us
\begin{align*}
& \int_{-\sqrt{2}}^{\sqrt{2}}x\cdot\left[\frac{\left(y^2+\left(x^2+1\right)\right)^{\frac{3}{2}}}{3}\right]_{-\sqrt{2-x^2}}^{\sqrt{2-x^2}}\:\:dx \\[4ex]
& =\int_{-\sqrt{2}}^{\sqrt{2}}\left\{x\left[\frac{\left(\left(2-x^2\right)+\left(x^2+1\right)\right)^{\frac{3}{2}}}{3}-\frac{\left(\left(2-x^2\right)+\left(x^2+1\right)\right)^{\frac{3}{2}}}{3}\right]\right\}\:dx=\boxed{0,}
\end{align*}
since what's inside the integrand cancels.
Note the symmetry of the graph:
| {
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How does $\sum_{n = 1}^{\infty} \frac{1}{n(n+1)}$ simplify? $\sum_{n = 1}^{\infty} \frac{1}{n(n+1)} = \frac{1}{1\cdot2} + \frac{1}{2\cdot3} + ...$
$= 1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + ... = 1$
My professor wrote this the other day. But I'm wondering...how does the series become $= 1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + ...$?
| Write
$$\frac{1}{n(n+1)} = \frac{1}{n}-\frac{1}{n+1},$$
split the sum into two sums, shift the index of one, recombine, simplify, and you are done.
N.B. You need to do this for partial sums and only at the end take the limit to be rigorous.
| {
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Find the volume of the region bounded by the planes $ z=8-y^2, y = 8-x^2, x=0, y=0, z=0$ I figured out the bounds for z:
$z=0$ to $z=8-y^2$
The bounds for y:
$y=0$ to $y=8-x^2$
The bounds for x:
$x=0$ to $x=\sqrt{8}$ (Since $8-x^2 = 0$)
So, the volume by using triple integral:
$\int_{0}^{2\sqrt{2}}\int_{0}^{8-x^2}\int_{0}^{8-y^2}dzdydx $
Am I right?
| For $x : 0 < x < \sqrt{8-2\sqrt{2}}$, the upper boundary in $y$ is $2\sqrt{2}$ not $8-x^2$ (which is bigger). The $z = 8 - y^2$ cylinder has "chopped off" part of the $y = 8 - x^2$ cylinder.
So the integral should be:
$I = \displaystyle\int\limits_{0}^{\sqrt{8-2\sqrt{2}}}\int\limits_{0}^{2\sqrt{2}}\int\limits_{0}^{8-y^2}dzdydx\ \ +\ \displaystyle\int\limits_{\sqrt{8-2\sqrt{2}}}^{2\sqrt{2}}\int\limits_{0}^{8-x^2}\int\limits_{0}^{8-y^2}dzdydx$
| {
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"source": "stackexchange",
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How to evaluate $\lim_{n\to\infty}\sqrt[n]{\frac{1\cdot3\cdot5\cdot\ldots\cdot(2n-1)}{2\cdot4\cdot6\cdot\ldots\cdot2n}}$ Im tempted to say that the limit of this sequence is 1 because infinite root of infinite number is close to 1 but maybe Im mising here something? What will be inside the root?
This is the sequence:
$$\lim_{n\to\infty}\sqrt[n]{\frac{1\cdot3\cdot5\cdot\ldots\cdot(2n-1)}{2\cdot4\cdot6\cdot\ldots\cdot2n}}$$
| You may write
$$
\begin{align}
\frac{1\cdot 3\cdot 5 \cdots(2n-1)}{2\cdot 4\cdot 6\cdots2n} &=\frac{1\cdot 2\cdot 3\cdot 4 \cdot 5\cdot6\cdots(2n-1)\cdot 2n}{(2\cdot 4\cdot 6\cdots (2n))^2}\\
&=\frac{(2n)!}{(2^{n} \cdot 1\cdot 2\cdot 3 \cdot 4 \cdots n)^2}\\
& =\frac{(2n)!}{2^{2n} (n!)^2 }\\
& =\frac{1}{\sqrt{\pi n}}+\mathcal{O}\left(\frac{1}{n^{3/2}}\right), \quad \text{for} \, n \, \text{great}
\end{align}
$$
where we have use Stirling's approximation, then you easily conclude, since
$$
\sqrt[n]{\frac{1\cdot 3\cdot 5 \cdots(2n-1)}{2\cdot 4\cdot 6\cdots2n}}=e^{\Large \frac{1}{n}\log{\frac{1\cdot 3\cdot 5 \cdots(2n-1)}{2\cdot 4\cdot 6\cdots2n}}}
$$
giving the value $1$ for your limit.
| {
"language": "en",
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"source": "stackexchange",
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How can I evaluate $\lim_{n \to\infty}\left(1\cdot2\cdot3+2\cdot3\cdot4+\dots+n(n+1)(n+2)\right)/\left(1^2+2^2+3^2+\dots+n^2\right)^2$? How can I evaluate this limit? Give me a hint, please.
$$\lim_{n \to\infty}\frac{1\cdot2\cdot3+2\cdot3\cdot4+\dots+n(n+1)(n+2)}{\left(1^2+2^2+3^2+\dots+n^2\right)^2}$$
| $$ \lim_{n\to \infty} \frac{\sum\limits_{k=1}^n k(k+1)(k+2)}{\left[\sum\limits_{k=1}^n k^2\right]^2}= \lim_{n\to \infty} \frac{\sum\limits_{k=1}^n k^3+3\sum\limits_{k=1}^n k^2+2\sum\limits_{k=1}^n k}{\left[\sum\limits_{k=1}^n k^2\right]^2} = \lim_{n\to \infty} \frac{\frac{n^2(n+1)^2}{4}+ \frac{n(n+1)(2n+1)}{2}+n(n+1)}{\left[\frac{n(n+1)(2n+1)}{6}\right]^2} = \lim_{n\to \infty} \frac{\frac{1}{4}+ \frac{2n+1}{2n(n+1)}+\frac{1}{n(n+1)}}{\frac{(2n+1)^2}{6^2}}= \lim_{n\to \infty} \frac{36\left(\frac{1}{4}+ \frac{1}{n+1}+\frac{1}{2n(n+1)}+\frac{1}{n(n+1)}\right)}{(2n+1)^2} =\frac{36\left(\frac{1}{4}+\frac{1}{\infty}+\frac{1}{\infty}+\frac{1}{\infty}\right)}{\infty}= 0 $$
| {
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Proving the inequality $4\ge a^2b+b^2c+c^2a+abc$ So, a,b,c are non-negative real numbers for which holds that $a+b+c=3$.
Prove the following inequality: $$4\ge a^2b+b^2c+c^2a+abc$$
For now I have only tried to write the inequality as $$4\left(\frac{a+b+c}3\right)^3\ge a^2b+b^2c+c^2a+abc$$ but I don't know what to do after that...
| So we need to show
$$4(a+b+c)^3 \ge 27(a^2b+b^2c+c^2a+abc)$$
One way is to use the cyclic symmetry and WLOG assume $a$ to be the min of $a, b, c$. Then we can write $b = a+x, c = a+y$, where $x, y \ge 0$. Now the inequality reduces to
$$9a(x^2-xy+y^2)+(x-2y)^2(4x+y) \ge 0$$
which is obvious. Also from the above, we get that equality is possible iff $x=y=0$ or when $a=0, x=2y$, i.e. when $(a, b, c)=(1, 1, 1)$ or is a permutation of $(0, 2, 1)$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Proving binomial coefficients identity: $\binom{r}{r} + \binom{r+1}{r} + \cdots + \binom{n}{r} = \binom{n+1}{r+1}$
Let $n$ and $r$ be positive integers with $n \ge r$. Prove that:
$$\binom{r}{r} + \binom{r+1}{r} + \cdots + \binom{n}{r} = \binom{n+1}{r+1}.$$
Tried proving it by induction but got stuck. Any help with proving it by induction or any other proof technique is appreciated.
| Proving by induction
inductive step
$$\begin{pmatrix}r\\r\end{pmatrix} + \begin{pmatrix}r+1\\r\end{pmatrix} + \dots + \begin{pmatrix}n\\r\end{pmatrix} =\begin{pmatrix}n\\r+1\end{pmatrix} + \begin{pmatrix}n\\r\end{pmatrix}$$ [as the identity holds for natural numbers less than n.]
We Know that, $$\begin{pmatrix}n\\r+1\end{pmatrix}+ \begin{pmatrix}n\\r\end{pmatrix} = \begin{pmatrix}n+1\\r+1\end{pmatrix}$$ see here
Hence proved.
| {
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how many positive integer solutions to the following equation? $a^2 + b^2 + 25 = ab + 5a + 5b$
I have tried looking for a factorisation that could solve this question but couldn't find anything useful - found $(a+b+5)^2$ - don't know if this is useful
The equation does look similar to an equation of a circle - can you use this idea?
| Let $x = a+b \to x^2 - 2ab + 25 = ab + 5x \to x^2 -5x + 25 = 3ab \leq 3\cdot \dfrac{(a+b)^2}{4} = \dfrac{3x^2}{4} \to 4x^2 - 20x + 100 \leq 3x^2 \to x^2 - 20x + 100 \leq 0 \to (x-10)^2 \leq 0 \to x = 10, a = b = \dfrac{10}{2} = 5$.
| {
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Use induction to prove $\sum_{k=1}^n\frac{k}{2^k}=2-\frac{n+2}{2^n}$
Use induction on $n\in\Bbb N$ to prove that $$\sum_{k=1}^n\frac{k}{2^k}=2-\frac{n+2}{2^n}\;.$$
I have got as far as to the induction step where I have:
$$S(n+1)= 2-\frac{n+3}{2^{n+1}}$$ and this should be equal to
$$S(n) +\frac{n+1}{2^{n+1}} = 2-\frac{n+2}{2^n}+\frac{n+1}{2^{n+1}}$$
Now I am kinda stuck.. all I end up with is ;
$$ ... = 2-\frac{3n+5}{2^{n+1}}\;.$$
Would someone help me out?
| When making a common denominator, don't forget to distribute the negative sign!
$$
\frac{-(n + 2)}{2^n} + \frac{n+1}{2^{n+1}}
= \frac{-2(n + 2) + (n + 1)}{2^{n+1}}
= \frac{(-2n - 4) + (n + 1)}{2^{n+1}}
= \frac{-n - 3}{2^{n+1}}
= -\frac{n + 3}{2^{n+1}}
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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How to evaluate these indefinite integrals with $\sqrt{1+x^4}$? These integrals are supposed to have an elementary closed form, but Mathematica only returns something in terms of elliptic integrals. I got them from the book Treatise on Integral Calculus by Edwards. How can we evaluate them?
$$
I = \int{\frac{\sqrt{1+x^4}}{1-x^4}dx}\\
J = \int{\frac{x^2}{(1-x^4)\sqrt{1+x^4}}} dx
$$
| The problem is on p. 319 in the 1921 edition of Volume I. Write
$$
I = \int \frac{1+x^4}{(1-x^4)} \frac{dx}{\sqrt{1+x^4}}
,\qquad
J = \int \frac{x^2}{(1-x^4)} \frac{dx}{\sqrt{1+x^4}}
,
$$
and note that $I=\frac{1}{2}(A+B)$ and $J=\frac{1}{4}(A-B)$, where
$$
A = \int\frac{1+x^2}{1-x^2} \frac{dx}{\sqrt{1+x^4}}
,\qquad
B = \int\frac{1-x^2}{1+x^2} \frac{dx}{\sqrt{1+x^4}}
.
$$
These integrals $A$ and $B$ appear in an exercise on p. 103, and can be solved by setting $z=\frac{\sqrt{1+x^4}}{x}$ (for $x > 1$, say, so that the change of variables is invertible; the final result doesn't depend on this assumption, as can be checked by differentiating it). This gives
$$
\frac{dz}{dx} = \frac{(x^2+1)(x^2-1)}{x^2 \sqrt{1+x^4}}
,\quad
z^2 = x^2 + \frac{1}{x^2}
,\quad
z^2 \pm 4 = \left( x \pm \frac{1}{x} \right)^2 = \left( \frac{x^2 \pm 1}{x} \right)^2
,
$$
so that
$$
A = -\int \frac{dz}{z^2-2}
,\qquad
B = -\int \frac{dz}{z^2+2}
,
$$
and I think you can take it from there!
(Another option would be to do as Edwards suggests on p. 311, and evaluate $A$ and $B$ by letting $z=1/(x-\frac{1}{x})$ and $z=1/(x+\frac{1}{x})$, respectively.)
| {
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Help me, a doubt $f(x)=\cot^{-1} \frac{1-x}{1+x}$ I have a doubt
$$f(x)=\cot^{-1} \frac{1-x}{1+x}$$
$$f´(x)=\frac{1}{(\frac{1-x}{1+x})^2}\cdot\frac{(-1)(1+x)-(1-x)}{1+\frac{(1-x)^2}{(1+x)^2}}$$
mm this could to be really easy but I do not understand in the first denominator gives one, someone who can explain,
| The derivative is incorrect.
The usual way is to use the chain rule on $\cot^{-1}$ then the quotient rule on the fraction. You would then get
$$f'(x)=(\cot^{-1})'\left( \frac{1-x}{1+x} \right) \cdot \left( \frac{1-x}{1+x} \right)'$$
$$=-\frac{1}{1+\left(\frac{1-x}{1+x} \right)^2} \cdot \frac{(1-x)'(1+x)-(1-x)(1+x)'}{(1+x)^2}$$
$$=-\frac{1}{1+\left(\frac{1-x}{1+x} \right)^2} \cdot \frac{-1(1+x)-(1-x)1}{(1+x)^2}$$
You can finish it from here.
| {
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How to get $\sqrt {k} + \frac{1}{\sqrt{k+1}}$ in the form $\frac{\sqrt{k^2} + 1}{\sqrt{k+1}}$? I was wondering if it is possible to get $\sqrt {k} + \dfrac{1}{\sqrt{k+1}}$ in the form $\dfrac{\sqrt{k^2} + 1}{\sqrt{k+1}}$, and if so, how?
I ask this, because I'm following this answer, and I get lost at how they arrive at $\dfrac{\sqrt{k^2} + 1}{\sqrt{k+1}}$.
Thanks.
EDIT: To further clarify what I'm not understanding, please read below:
I don't understand how the author goes from the line:
$\frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} +...+ \frac{1}{\sqrt{k}} + \frac{1}{\sqrt{k+1}} > \sqrt{k+1}$
to the line:
$\frac{\sqrt{k^2 + k} + 1}{\sqrt{k + 1}} > \frac{\sqrt{k^2} + 1}{\sqrt{k+1}}$
I would really appreciate it if someone could lead me through to that line and explain exactly what is being done on each line.
Thanks again.
| That answer says
$$\sqrt {k} + \frac{1}{\sqrt{k+1}}= \frac{(\sqrt {k})(\sqrt {k+1}) +1}{\sqrt{k+1}}=\frac{\sqrt {k(k+1}) +1}{\sqrt{k+1}}=\frac{\sqrt {k^2+\color{red}{k}} +1}{\sqrt{k+1}}\color{blue}{\ge}\frac{\sqrt {k^2} +1}{\sqrt{k+1}}$$
It's just like saying $\sqrt{a+b}\ge\sqrt{a}$ for $b\ge0$
Answer to Edit:
Author want to show that
$$\frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} +...+ \frac{1}{\sqrt{k}} + \frac{1}{\sqrt{k+1}} > \sqrt{k+1}$$
So you first assume That it is true for $n=k$ that is
$$\frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} +...+ \frac{1}{\sqrt{k}} > \sqrt{k}$$ And then You add $\dfrac{1}{\sqrt{k+1}}$ to both sides and then show that it is also true for $n=k+1$
| {
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Let f(x) be a non negative continuous function on R such that $f(x) +f(x+\frac{1}{3})=5$ then calculate ........ Problem :
Let f(x) be a non negative continuous function on R such that $f(x) +f(x+\frac{1}{3})=5$
then calculate the value of the integral $\int^{1200}_0 f(x) dx$
My approach :
Given that : $f(x) +f(x+\frac{1}{3})=5.....(1) $
We replace x with $x +\frac{1}{3}$ so we get the given equation as :
$f(x+\frac{1}{3})+f(x+\frac{2}{3}).....(2)$
Now subtracting (1) from (2) we get :
$f(x+\frac{2}{3}) = f(x) $
$\Rightarrow f(x) $ is function with period $\frac{2}{3}$
Now how to move further please help on how the period of this function will impact the limit of integration. Thanks.
| Check this
$\int_0 ^{1200}f=\int_0 ^{\frac{2}{3}}f+\int_{\frac{2}{3}}^{\frac{4}{3}}f+...$
$\int _{\frac{2}{3}}^{\frac{4}{3}}f(x)=\int_0 ^{\frac{2}{3}}f(x)$
$\int _0^ {1200} f=800\times \int _0^{\frac{2}{3}}f$
$\int _0^{\frac{2}{3}}f=\int _0^{\frac{1}{3}}f+\int _{\frac{2}{3}}^{\frac{4}{3}}f$
Let $y=x-{\frac{1}{3}} \implies dy=dx$
As $x={\frac{2}{3}} y={\frac{1}{3}};x={\frac{1}{3}}y=0$
$=\int _0^{\frac{1}{3}}f(y)dy+\int _0^{\frac{1}{3}}f(y+\frac{1}{3})dy$
$=\int _0^\frac{1}{3}5dy$
| {
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Factor $55 - 88 \sqrt{-2}$ as a product of primes in $\mathbb{Z}[\sqrt{-2}]$ To solve this problem, I let $K = \mathbb{Q}(\sqrt{-2})$, and I thought to take the norm $$N(55 - 88 \sqrt{-2}) = 55^2 + 2 \cdot 88^2 = 18513 = 3^2\cdot11^2 \cdot 17$$ If $a \in \mathbb{Z}[\sqrt{-2}]$ is irreducible, then $N(a) = p^f$, where $a$ lies over the prime $p$ and $f$ is the inertia degree of $K$ over $p$. Since the congruences $x^2 \equiv -2 \pmod 3$ and $x^2 \equiv -2 \pmod{11}$ are easily seen to be solvable (since $x^2 \equiv m \pmod p$ is solvable if and only if $m^{\frac{p-1}{2}} \equiv 1 \pmod p$), the primes $3$ and $11$ should split in $\mathbb{Z}[\sqrt{-2}]$. It was pretty easy to figure out that $$3 = (1 + \sqrt{-2})(1 - \sqrt{-2})$$ and $$11 = (3 + \sqrt{-2})(3 - \sqrt{-2})$$ and $17$ also splits as $(3 + 2 \sqrt{-2})(3 - 2 \sqrt{-2})$. So none of the primes $3, 11,$ or $17$ have inertia. Let $\sigma: K \rightarrow K$ be the unique nonidentity automorphism which is determined by $\sigma(\sqrt{-2}) = - \sqrt{-2}$.
If $\alpha_1 = 1 + \sqrt{-2}, \alpha_2 = 3 + \sqrt{-2}, \alpha_3 = 3 + 2\sqrt{-2}$, what I'm pretty sure should happen is that $55 - 88 \sqrt{-2}$ should be equal to some unit in $\mathbb{Z}[\sqrt{-2}]$, times either $\alpha_1^2$ or $\sigma \alpha_1^2$, times either $\alpha_2^2$ or $\sigma \alpha_2^2$, times either $\alpha_3$ or $\sigma \alpha_3$.
But how do I figure out which combination is right?
| $55 - 88\sqrt{-2} = 11(5 - 8\sqrt{-2})$ so $11=\alpha_2\sigma(\alpha_2)$ is a factor, rather than $\alpha_2^2$ or $\sigma(\alpha_2)^2$, and we only need to factor $5 - 8\sqrt{-2}$.
I think the best thing to do now is to try dividing by the possible factors and check if the result is in $\Bbb{Z}[\sqrt{-2}].$
$$\frac{5 - 8\sqrt{-2}}{1 + \sqrt{-2}} = \frac{(5-8\sqrt{-2})(1 - \sqrt{-2})}{3} = \frac{-11 - 13\sqrt{-2}}{3} \notin \Bbb{Z}[\sqrt{2}]$$
$$\frac{5 - 8\sqrt{-2}}{1 - \sqrt{-2}} = \frac{(5-8\sqrt{-2})(1 + \sqrt{-2})}{3} = \frac{21 - 3\sqrt{-2}}{3} = 7 - \sqrt{-2} \in \Bbb{Z}[\sqrt{2}]$$
Hence $\sigma(\alpha_1)$ is a factor and $\alpha_1$ is not so $\sigma(\alpha_1)^2$ must be a factor. Dividing by $\sigma(\alpha_1)^2$ we obtain
$$\frac{5 - 8\sqrt{-2}}{(1 - \sqrt{-2})^2} = 3 + 2\sqrt{-2} = \alpha_3$$
so $55 - 88\sqrt{-2} = \sigma(\alpha_1)^2\alpha_2\sigma(\alpha_2)\alpha_3$.
| {
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What am I doing wrong? (Trigonometric Identity) $$\frac { \cos\theta }{ 1-\sin\theta } =\frac { \sin\theta -\csc\theta }{ \cos\theta -\cot\theta } $$
Steps I took:
$$\frac { \sin\theta -\frac { 1 }{ \sin\theta } }{ \cos\theta -\frac { \cos\theta }{ \sin\theta } } $$
$$\frac { \frac { \sin^{ 2 }\theta -1 }{ \sin\theta } }{ \frac { \sin\theta \cos\theta -\cos\theta }{\sin\theta } } $$
$$\frac { \sin^{ 2 }\theta -1 }{ \sin\theta } \cdot \frac { \sin\theta }{ \sin\theta \cos\theta -\cos\theta } $$
$$\frac { (\sin^{ 2 }\theta -1) }{ (\cos\theta )(\sin\theta -1) } $$
Now where do I go from here? As far as I can tell, $\sin^{ 2 }\theta -1=-\cos^2\theta $
so this makes no sense to me.
| You've already solved the problem:
$$\begin{align}
\dfrac { (\sin^{ 2 }\theta -1) }{ (\cos\theta )(\sin\theta -1) } &= \dfrac{-\cos^2\theta}{(\cos\theta)(\sin\theta - 1)} \\
&= \dfrac{-\cos\theta}{\sin\theta-1} \\
&= \dfrac{\cos \theta}{-(\sin\theta-1)} \\
&= L.H.S.
\end{align}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Did I correctly verify the convergence of this series? I want to find if the following series is convergent.
$$\sum_{n=1}^\infty \frac{(1+\frac{1}{n})^nn^2-7n}{n^3+3n^2+1} $$
I use the asymptotic criterion for series convergence.
$$ a_n=\frac{(1+\frac{1}{n})^nn^2-7n}{n^3+3n^2+1} $$
I take such $b_n$ that $a_n$ and $b_n$ are asymptotically similar and that the convergence of $\sum_{n=1}^\infty b_n$ is known.
$$b_n=\frac{1}{n}$$
$$\lim_{n\to \infty}\frac{a_n}{b_n}=\lim_{n\to \infty}\frac{(1+\frac{1}{n})^nn^2-7n}{n^3+3n^2+1} \frac {n}{1}=\lim_{n\to \infty}\frac{(1+\frac{1}{n})^nn^3-7n^2}{n^3+3n^2+1}$$
The limit is $e$ which proves that $a_n \sim b_n$.
Then since $\sum_{n=1}^\infty \frac{1}{n} $ is divergent, so is the original series.
I'd be thankful if someone could review this and tell me if this solution is correct.
| Your conclusion is correct. You could get there using the comparison test too if you are interested. Namely, that $$1 < \left(1+\frac{1}{n} \right)^n \\ \frac{1}{n^3+3n^3+n^3}\leq \frac{1}{n^3+3n^2+1}$$ for all $n \geq 1$. Hence, $$\sum_{n=1}^\infty \frac{(1+\frac{1}{n})^nn^2-7n}{n^3+3n^2+1} \geq \sum_{n=1}^\infty \frac{n^2-7n}{n^3+3n^2+1} \\ \geq \sum_{n=1}^\infty \frac{n^2-7n}{n^3+3n^3+n^3} \\ = \sum_{n=1}^\infty \frac{n^2-7n}{5n^3} \\ = \sum_{n=1}^\infty \frac{1}{5n}-\sum_{n=1}^\infty\frac{7}{5n^2} \\ = \frac{1}{5}\sum_{n=1}^\infty \frac{1}{n}-\frac{7\pi^2}{30}$$
| {
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Proving $\lim_{x \to -2}$ $\frac{(x-1)}{(x+1)}$ = 3 by definition Here we go. $\lim_{x \to -2}$ $\frac{(x-1)}{(x+1)}$ = 3
Proof: $\vert \frac{x-1}{x+1} -3 \vert$ =...= $\frac{\vert-2\vert \vert x-(-2) \vert}{(x+1)}$$\lt \frac{1}{2} \vert x -(-2) \vert$...now let $\epsilon$ $\gt$ 0 arbitrary and $\delta$ = min{1,2$\epsilon$} then we have
$\frac{\vert-2\vert \vert x-(-2) \vert}{(x+1)}$$\lt \frac{1}{2} \vert x -(-2) \vert$$\lt$$\frac{1}{2} * 2\epsilon$ = $\epsilon$ so $0<\vert x-(-2) \vert< \delta$
yes or no or I need more
| If your $\delta$ is $\min\{1, 2\epsilon\}$ then it is not always true that $$\frac{2|x+2|}{|x+1|} < \frac{1}{2}|x+2|.$$
When $\epsilon$ is large, then $0 < |x + 2| < \delta = 1$ implies only that $-1 < x + 2 < 1$, or $-2 < x + 1 < 0$, so that $2 > |x+2| > 0$ and $1/2 < 1/|x+2|$ is not bounded above. The problem is $|x + 1|$ can get arbitrarily large because the bound, $\delta$, is not tight enough; for example, we can pick $x = -1.001$ and see that $0 < |x + 2| = 0.999 < 1$ is satisfied, but $2|x+2|/|x+1| = 1998$ is astronomical.
To remedy this, pick some $0 < c < 1$ and ask for $\delta = \min\{c, k\epsilon\}$ where $k$ is to be determined.
Now if $0 < |x + 2| < \delta$, then $x + 2 < c$, or $x + 1 < c - 1 < 0$. Taking absolute values reverses the inequality: $|x+1| > |1-c| = 1-c$. Now invert: $$\frac{1}{|x+1|} < \frac{1}{1-c}.$$
Therefore,
$$\Bigg|\frac{x-1}{x+1} - 3\Bigg| = 2\frac{|x+2|}{|x+1|} < \frac{2\delta}{1-c} \le \frac{2k\epsilon}{1-c}$$
and this is $\le \epsilon$ iff $k \le (1-c)/2$.
Everything will be neat if you pick $c = 1/2$ and $k = 1/4$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Multinomial Expansion Question: when variables appear more than once Find the coefficient of $x^{12}y^{24}$ in $(x^3 + 2xy^2 +y + 3)^{18}$.
I have been working on this problem for a while now and I cannot figure out how to use the multinomial theorem to solve it. I tried using auxiliary variables but then I did not know what the total exponent should be. I am really stuck on this.
| Break it into cases based on how the $x^{12}$ was formed from some combination of $x^3$ and $2xy^2$. In each case, you must also check that it is possible that $y$ could be raised to the 24th power using the remaining available options.
Case 1: $(x^3)^4$: You might have had an $x^{12}y^{24}$ term because of $(x^3)^4\cdot (2xy^2)^0\cdot(y)^{24}\cdot 3^n$, but we're only raising to the 18th power, so that can't happen.
Case 2: $(x^3)^3$: It could have been $(x^3)^3\cdot(2xy^2)^3\cdot (y)^{18}\cdot 3^n$. Also not possible, not large enough power available.
Case 3: $(x^3)^2$: In this case it might have been $(x^3)^2\cdot (2xy^2)^6\cdot y^{12}\cdot3^n$, again not possible. It would again require too large of a power of the original function (at least being raised to 20th power)
Case 4: $(x^3)^1$: This would imply it would be $(x^3)^1\cdot (2xy^2)^9\cdot (y)^6\cdot(3)^2$ This is infact possible.
Case 5: $(x^3)^0$: This would imply that it would be $(x^3)^0\cdot (2xy^2)^{12}\cdot (y)^0\cdot (3)^6$ This also is possible.
Now you know that it could have been formed one of two ways. Reimagining the problem instead as $(a+b+c+d)^{18}$, the question is then what are the coefficients of the $ab^9c^6d^2$ and $b^{12}d^6$ terms.
The coefficient of the $ab^9c^6d^2$ term is $\dfrac{18!}{1!9!6!2!}$ and the coefficient of the $b^{12}d^6$ term is $\dfrac{18!}{12!6!}$
So, we have $\dfrac{18!}{1!9!6!2!}(x^3)^1\cdot (2xy^2)^9\cdot (y)^6\cdot(3)^2 + \dfrac{18!}{12!6!}(2xy^2)^{12}\cdot (3)^6 = Cx^{12}y^{24}$
So, $C = \dfrac{18!}{9!6!2!}\cdot 2^9\cdot3^2 + \dfrac{18!}{12!6!}\cdot 2^{12}\cdot 3^6$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Solving $2\cos^2 x-2\sin^2 x-2\cos x=0$ $$f(x) = 2\cos^2 x-2\sin^2 x-2\cos x$$
Need values of x that which make $f(x) = 0$
Tried $a^2-b^2 = (a+b)(a-b)$ with no luck
Really just need a hint that could bring me in the right direction
Thanks
EDIT: Solution thanks to everyones help! :D
$$f(x) = 2\cos^2 x-2\sin^2 x-2\cos x$$
$$0 = 2\cos^2 x-2 + 2\cos^2 x-2\cos x$$
$$0 = 4\cos^2 x-2\cos x - 2$$
$$\cos x = 2\pm \sqrt {-2^2-4(4)(-2)\over8}$$
$$\cos x = {2 \pm 6\over 8}$$
$$\cos x = {-1\over2}, 1$$
$$x = 2n\pi \pm{2\pi\over3}$$
Thank you!
| We have $$\cos x=\cos^2x-\sin^2x=\cos2x$$
$$\implies2x=2m\pi\pm x$$ where $m$ is any integer
Taking the '-' sign, $3x=2m\pi\iff x=\dfrac{2m\pi}3\ \ \ \ (1)$
Taking the '+' sign, $x=2m\pi $ which is a proper subset of $(1)$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Laurent series of $f(z)=\frac{1}{z(z-1)}$ given four different conditions Expand $f(z)=\frac{1}{z(z-1)}$ in a Laurent series valid for the follwing annular domains.
$a)0\lt \vert z \rvert \lt 1 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,b)1\le\lvert z \rvert\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,c)0\le \lvert z-1 \rvert \lt1\,\,\,\,\,\,\,\,\,\,\,\,\,d) 1\le \lvert z-1 \rvert$
Ok, so here are some things I know, There are singularities at $z_0=0$and $z_0=1$
a) is the unit circle, shaded inside.$f(z)=\frac{1}{z(z-1)}=\frac{1}{z}*\frac{1}{z-1}=\frac{-1}{z}*\frac{1}{1-z}=\frac{-1}{z}*\sum_{n=0}^\infty Z^n=\frac{-1}{z}[z^0+z^1+z^2...]=\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,f(z)=\frac{-1}{z}-1-z-z^2-...\,$ which converges for $0\lt \vert z\rvert \lt 1$
b)is the unit circle shaded outside the circle. $f(z)=\frac{1}{z(z-1)}$$=\frac{1}{z}*\frac{1}{z-1}$$=\frac{1}{z}*\frac{1}{z(1-1/z)}=\frac{1}{z^2}*\frac{1}{1-1/z}=\frac{1}{z^2}*\sum_{n=0}^\infty(\frac{1}{z})^n=\frac{1}{z^2}[\frac{1}{z}^0+\frac{1}{z}^1+\frac{1}{z}^2+...]=\frac{1}{z^2}+\frac{1}{z^3}+\frac{1}{z^4}+...$
c)is a circle with r=1, shaded inside. Here, Im not so sure how to manipulate the function and have gotten stuck
d)is a cirlce with r=1, shaded outside
| For (c)
$$\frac{1}{z(z-1)}= \frac1{z-1}\frac{1}{1+(z-1)}= \frac1{z-1}\sum_{k=0}^{\infty}(-1)^k(z-1)^{k}$$
For (d)
$$\frac{1}{z(z-1)}= \frac{\frac1{(z-1)^2}}{1+\frac1{z-1}}=\frac1{(z-1)^2}\sum_{k=0}^{\infty}\frac{(-1)^k}{(z-1)^k}$$
| {
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"url": "https://math.stackexchange.com/questions/1056479",
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"source": "stackexchange",
"question_score": "3",
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Find the value of the sum $$\arctan\left(\dfrac{1}{2}\right)+\arctan\left(\dfrac{1}{3}\right)$$
We were also given a hint of using the trigonometric identity of $\tan(x + y)$
Hint
$$\tan\left(x+y\right)\:=\:\dfrac{\tan x\:+\tan y}{1-\left(\tan x\right)\left(\tan y\right)}$$
| Yep, sum of tangents identity is the way to go. Write $x=arctan\frac{1}{2}$ and $y=arctan\frac{1}{3}$.
Then $tan(arctan\frac{1}{2}+arctan\frac{1}{3})=\frac{tan(arctan\frac{1}{2})+tan(arctan\frac{1}{3})}{1-(tan(arctan\frac{1}{2}))(tan(arctan\frac{1}{3}))}$
This is equal to $\frac{\frac{1}{2}+\frac{1}{3}}{1-(\frac{1}{2})(\frac{1}{3})}=\frac{\frac{5}{6}}{\frac{5}{6}}=1$
Thus $tan(x+y)=1$ and $x+y=\boxed{\frac{\pi}{4}}$
| {
"language": "en",
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Find the minimum possible value of $x(1-z)+y(1-x)+z(1-y)$ It is given that $$xyz=(1-x)(1-y)(1-z)$$ and $$x, y, z \in (0,1)$$
Find the minimum possible value of the expression: $$x(1-z)+y(1-x)+z(1-y)$$
Using the AM-GM inequality concepts, I can write that the value is minimum when
$$x(1-z)=y(1-x)=z(1-y)$$
What else can I conclude from the given information?
| From $xyz=(1-x)(1-y)(1-z)$ we get
$$1-2xyz=\sum_{cyc}x(1-y)\geq 3\sqrt[3]{xyz(1-x)(1-y)(1-z)}=3\sqrt[3]{x^2y^2z^2},$$
Which, on setting $xyz=\dfrac {t^3}{8},$ gives us $4\ge t^3+3t^2$. This factors into
$$0\ge (t-1)(t+2)^2,$$
Hence $t\le 1$ and $xyz\le \dfrac 18$. So,
$$\sum_{cyc}x(1-y)=1-2xyz\ge\frac34.$$
Equality occurs at $x=y=z=\dfrac 12.$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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asymptotics of sum I wanna find asymptotic of sum below
$$\sum\limits_{k=1}^{[\sqrt{n}]}\frac{1}{k}(1 - \frac{1}{n})^k$$
assume I know asymptotic of this sum (I can be wrong):
$$\sum\limits_{k=1}^{n}\frac{1}{k}(1 - \frac{1}{n^2})^k \sim c\ln{n}$$
So I use Stolz–Cesàro theorem and wanna show that
$$\sum\limits_{k=1}^{[\sqrt{n}]}\frac{1}{k}(1 - \frac{1}{n})^k \sim c\ln{n}$$
where
$$x_n = \sum\limits_{k=1}^{[\sqrt{n}]}\frac{1}{k}(1 - \frac{1}{n})^k$$
$$x_n - x_{n-1} = \frac{1}{\sqrt{n}}(1 - \frac{1}{n})^{\sqrt{n}}$$
$$y_n - y_{n-1} = \ln(n) - \ln(n-1)$$
but
$$
\lim_{n \to \infty}
\frac{\frac{1}{\sqrt{n}}(1 - \frac{1}{n})^{\sqrt{n}}}
{\ln(n) - \ln(n-1)} = \infty
$$
What I'm doing wrong?
| One error that I see
is your computation of
$x_n-x_{n-1}$.
You have
$x_n = \sum\limits_{k=1}^{[\sqrt{n}]}\frac{1}{k}(1 - \frac{1}{n})^k$.
Note that the individual terms
depend on both $k$ and $n$.
Therefore,
$\begin{array}\\
x_n-x_{n-1}
&=\sum\limits_{k=1}^{[\sqrt{n}]}\frac{1}{k}(1 - \frac{1}{n})^k
-\sum\limits_{k=1}^{[\sqrt{n-1}]}\frac{1}{k}(1 - \frac{1}{n-1})^k\\
&=\sum\limits_{k=1}^{[\sqrt{n-1}]}\left(\frac{1}{k}(1 - \frac{1}{n})^k
-\frac{1}{k}(1 - \frac{1}{n-1})^k\right)
+\frac{[\sqrt{n}]-[\sqrt{n-1}]}{[\sqrt{n}]}(1 - \frac{1}{n})^{[\sqrt{n}]}\\
&=\sum\limits_{k=1}^{[\sqrt{n-1}]}\frac{1}{k}\left((1 - \frac{1}{n})^k
-(1 - \frac{1}{n-1})^k\right)
+\frac{[\sqrt{n}]-[\sqrt{n-1}]}{[\sqrt{n}]}(1 - \frac{1}{n})^{[\sqrt{n}]}\\
\end{array}
$
which is
(unfortunately)
a lot more complicated.
You can work with
$(1 - \frac{1}{n})^k
-(1 - \frac{1}{n-1})^k
$,
but that sum will still remain.
(added later)
Since $k < \sqrt{n}$,
$\frac{k}{n}
=O(\frac{1}{\sqrt{n}})
$
and
$\frac{k^2}{n^2}
=O(\frac{1}{n})
$,
so
$(1 - \frac{1}{n})^k
=1-\frac{k}{n}+\frac{k(k-1)}{2n^2}+O(\frac{k^3}{n^3})
=1-\frac{k}{n}+\frac{k(k-1)}{2n^2}+O(\frac{1}{n^{3/2}})
$.
Therefore
$\begin{array}\\
(1 - \frac{1}{n-1})^k-(1 - \frac{1}{n-1})^k
&=(1-\frac{k}{n-1}+\frac{k(k-1)}{2(n-1)^2}+O(\frac1{n^{3/2}}))
-(1-\frac{k}{n}+\frac{k(k-1)}{2n^2}+O(\frac1{n^{3/2}}))\\
&=(\frac{k}{n}-\frac{k}{n-1})
-(\frac{k(k-1)}{2(n-1)^2}-\frac{k(k-1)}{2n^2})+O(\frac1{n^{3/2}}))\\
&=-\frac{k}{n(n-1)}
-\frac{k(k-1)}{2}(\frac1{(n-1)^2}-\frac1{n^2})
+O(\frac1{n^{3/2}}))\\
&=-\frac{k}{n(n-1)}
-\frac{k(k-1)}{2}(\frac{2n-1}{n^2(n-1)^2})
+O(\frac1{n^{3/2}}))\\
\end{array}
$
If we sum this
we get
$\begin{array}\\
\sum\limits_{k=1}^{[\sqrt{n-1}]}\frac{1}{k}\left((1 - \frac{1}{n})^k
-(1 - \frac{1}{n-1})^k\right)
&=\sum\limits_{k=1}^{[\sqrt{n-1}]}\frac{1}{k}
\left(-\frac{k}{n(n-1)}
-\frac{k(k-1)}{2}(\frac{2n-1}{n^2(n-1)^2})
+O(\frac1{n^{3/2}}))\right)\\
&=-\sum\limits_{k=1}^{[\sqrt{n-1}]}\frac{1}{n(n-1)}
-\sum\limits_{k=1}^{[\sqrt{n-1}]} \frac{k-1}{2}(\frac{2n-1}{n^2(n-1)^2})
+O(\frac1{n})\\
&=-\frac{[\sqrt{n-1}]}{n(n-1)}
-\frac{[\sqrt{n-1}]([\sqrt{n-1}]-1}{2}(\frac{2n-1}{n^2(n-1)^2})
+O(\frac1{n})\\
&=O(\frac1{n^{3/2}})
+O(\frac1{n^{2}})
+O(\frac1{n})\\
\end{array}
$
At this point,
that $O(1/n)$ dominates,
so another term
should be taken
in the expansion of
$(1 - \frac{1}{n})^k$
to get a more accurate estimate for
$x_n-x_{n-1}$.
But this was such a pain
that I am going to leave it
at this.
| {
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"question_score": "3",
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Finding exact value of trigonometric functions I was wondering, how do I get the exact fraction (the value) of this trigonometric function:
$$\cos\left(\sin^{-1}(12/13)+\sin^{-1}(4/5)\right)$$
Usually, I would evaluate the inverse sin in degree mode and multiply (by hand) by $\pi/180$.
But in this case, I don't get exact values of angles
For example :
$$\begin{eqnarray}
\sin^{-1}(12/13)=67,38...^\circ\\
\sin^{-1}(4/5)=53,13...^\circ
\end{eqnarray}
$$
Is there any way of doing it ?
Thank you!
| We have:
$$\arcsin\frac{4}{5} = \arg(3+4i),\qquad \arcsin\frac{12}{13}=\arg(5+12i), $$
hence:
$$\arcsin\frac{4}{5}+\arcsin\frac{12}{13}=\arg((3+4i)(5+12i))=\arg(-33+56i)$$
and:
$$ \cos\left(\arcsin\frac{4}{5}+\arcsin\frac{12}{13}\right)=\frac{-33}{\sqrt{33^2+56^2}}=-\frac{33}{65}.$$
Avoiding complex numbers.
$$\arcsin\frac{4}{5} = \arctan\frac{4}{3},\qquad \arcsin\frac{12}{13}=\arctan\frac{12}{5}, $$
so:
$$\arcsin\frac{4}{5}+\arcsin\frac{12}{13}=\arctan\frac{\frac{4}{3}+\frac{12}{5}}{1-\frac{4}{3}\cdot\frac{12}{5}}=\arctan\left(-\frac{56}{33}\right) $$
and:
$$\cos\arctan\left(-\frac{56}{33}\right)=\frac{-33}{\sqrt{33^2+56^2}}=-\frac{33}{56}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Question on the sum $\sum_{n=1}^{\infty}\frac{x^n}{n} = -\ln(1-x)$ $f(x) = \displaystyle\sum_{n=1}^{\infty}\frac{x^n}{n} = x + \frac{x^2}{2} + \frac{x^3}{3} + ... = -\ln(1-x)$ for $|x| < 1$.
$f'(x) = \displaystyle\sum_{n=1}^{\infty}x^{n-1} = 1 + x + x^2 + x^3 +... = \frac{1}{1-x}$ for $|x| < 1$
If $f'(x) = \displaystyle\frac{1}{1-x}$, then $f(x) = \displaystyle\int\frac{1}{1-x}dx$
But $\displaystyle\int \frac{1}{1-x}dx = -\ln(1-x) + C$, and so $f(x) = -\ln(1-x) + C$
Do we need a definite integral here so there will be no $C$ and will agree with how $f(x)$ was originally defined? If so what would the definite integral be? Or maybe there is no definite integral at all?
| You're right: $f'(x)=\frac1{1-x}$, so $f(x)$ must be $-\ln(1-x)+C$ for some $C$. Now, to find what $C$ is, try letting $x$ be $0$ and solving for $C$.
Or, if you want to do a definite integral, start with:
$$1+t+t^2+\dotsb=\frac1{1-t}$$
Take the definite integral from $0$ to $x$:
\begin{align}
\int_0^x(1+t+t^2+\dotsb)\operatorname d\!t&=\int_0^x\frac1{1-t}\operatorname d\!t\\
x+\frac{x^2}2+\frac{x^3}3+\dotsb&=-\ln(1-x)
\end{align}
Note that if we take $x=-1$, we get the famous series for $\ln2$:
$$\ln2=1-\frac12+\frac13-\dotsb$$
which can be proven without calculus if you try hard enough.
| {
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How find $a,b$ if $\int_{0}^{1}\frac{x^{n-1}}{1+x}dx=\frac{a}{n}+\frac{b}{n^2}+o(\frac{1}{n^2}),n\to \infty$ let
$$\int_{0}^{1}\dfrac{x^{n-1}}{1+x}dx=\dfrac{a}{n}+\dfrac{b}{n^2}+o(\dfrac{1}{n^2}),n\to \infty$$
Find the $a,b$
$$\dfrac{x^{n-1}}{1+x}=x^{n-1}(1-x+x^2-x^3+\cdots)=x^{n-1}-x^n+\cdots$$
so
$$\int_{0}^{1}\dfrac{x^{n-1}}{1+x}=\dfrac{1}{n}-\dfrac{1}{n+1}+\dfrac{1}{n+2}-\cdots$$
and note
$$\dfrac{1}{n+1}=\dfrac{1}{n}\left(\frac{1}{1+\dfrac{1}{n}}\right)=\dfrac{1}{n}-\dfrac{1}{n^2}+\dfrac{1}{n^3}+o(1/n^3)$$
and simaler
$$\dfrac{1}{n+2}=\dfrac{1}{n}-\dfrac{2}{n^2}+o(1/n^2)$$
$$\dfrac{1}{n+3}=\dfrac{1}{n}-\dfrac{3}{n^2}+o(1/n^2)$$
then which term end?
| Thank you,I have use two parts integral following
$$\int_{0}^{1}x^nf(x)dx=\dfrac{f(1)}{n}-\dfrac{f(1)+f'(1)}{n^2}+o(1/n^2)$$
where $f\in C^{2}[0,1]$
$$I=\int_{0}^{1}f(x)d\dfrac{x^{n+1}}{n+1}=\dfrac{f(1)}{n+1}-\dfrac{1}{n+1}\int_{0}^{1}f'(x)d\dfrac{x^{n+2}}{n+2}$$
so
$$I=\dfrac{f(1)}{n+1}-\dfrac{f'(1)}{(n+1)(n+2)}+\dfrac{f''(\xi)}{(n+1)(n+2)(n+3)},0<\xi<1$$
since
$$\dfrac{f''(\xi)}{(n+1)(n+2)(n+3)}=c/n^3+o(1/n^3)=o(1/n^2),n\to \infty$$
and
$$\dfrac{1}{n+1}=\dfrac{1}{n}-\dfrac{1}{n^2}+o(1/n^2)$$
$$\dfrac{1}{(n+1)(n+2)}=\left(\dfrac{1}{n}-\dfrac{1}{n^2}+o(1/n^2)\right)\left(\dfrac{1}{n}-\dfrac{2}{n^2}+o(1/n^2)\right)=\dfrac{1}{n^2}+o(1/n^2)$$
so
$$\int_{0}^{1}x^nf(x)dx=\dfrac{f(1)}{n}-\dfrac{f(1)+f'(1)}{n^2}+o(1/n^2)$$
| {
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Prove by induction that an expression is divisible by 11
Prove, by induction that $2^{3n-1}+5\cdot3^n$ is divisible by $11$ for any even number $n\in\Bbb N$.
I am rather confused by this question. This is my attempt so far:
For $n = 2$
$2^5 + 5\cdot 9 = 77$
$77/11 = 7$
We assume that there is a value $n = k$ such that $2^{3k-1} + 5\cdot 3^k$ is divisible by $11$.
We show that it is also divisible by $11$ when $n = k + 2$
$2^{3k+5} + 5\cdot 3^{k+2}$
$32\cdot 2^3k + 5\cdot 9 \cdot3^k$
$32\cdot 2^3k + 45\cdot 3^k$
$64\cdot 2^{3k-1} + 45\cdot 3^k$ (Making both polynomials the same as when $n = k$)
$(2^{3k-1} + 5\cdot 3^k) + (63\cdot 2^{3k-1} + 40\cdot 3^k)$
The first group of terms $(2^{3k-1} + 5\cdot 3^k)$ is divisible by $11$ because we have made an assumption that the term is divisible by $11$ when $n=k$. However, the second group is not divisible by $11$. Where did I go wrong?
| Hint note that: if $k$ is a even number then also the next number $k+2$ is even
$$2^{3(k+2)-1}+5\cdot3^{k+2}=2^{3k-1+6}+5\cdot3^{k+2}=64\cdot2^{3k-1}+9\cdot5\cdot3^{k}$$$$=55\cdot2^{3k-1}+9\cdot2^{3k-1}+9\cdot5\cdot3^{k}=55\cdot2^{3k-1}+9\cdot(2^{3k-1}+5\cdot3^{k})$$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "11",
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ordered pairs $(A,B)$ of subsets of $X$ >such that $A\neq \phi\;,B\neq \phi$ and $A\cap B = \phi\;,$ is
Let $X$ be a set of $5$ elements. Then the number of ordered pairs $(A,B)$ of subsets of $X$
such that $A\neq \phi\;,B\neq \phi$ and $A\cap B = \phi\;,$ is
$\bf{My\; Try::}$ Let $X = \left\{1,2,3,4,5\right\}\;,$
then total no. of Subest of $X = 2^5 = 32$
(Which also contain $\phi$)
$\bullet\; $ If $A = \{1\}\;,$ Then $B=\left (\{2\}\;,\{3\}\;,\{4\}\;,\{5\}\;,\{2,3\}\;,\{2,4\}\;,\{2,5\}\;,\{3,4\}\;,\{3,5\}\;,\{4,5\}\;,\{2,3,4\}\;,\{2,3,5\}\;,\{3,4,5\}\;,\{2,4,5\}\;,\{2,3,4,5\}\right )$
Similarly for $A=\{2\}\;,A=\{3\}\;,A=\{4\}$ and $A=\{5\}\;,$ we get $15$ ordered pair for each single
element ed set $A$
$\bullet\; $ If $A = \{1,2\}\;,$ Then $B=\left(\{3\}\;,\{4\}\;,\{5\},\{3,4\}\;,\{3,5\}\;,\{4,5\}\;,\{3,4,5\}\right)$
Similarly for $A=\{1,3\}\;,A=\{1,4\}\;,A=\{1,5\}\;,A=\{2,3\}\;,A=\{2,4\}\;,A=\{2,5\}\;,A=\{3,4\}\;,A=\{3,5\}\;,A=\{4,5\}$ we get $7$ ordered pair for each double elemented set $A$
$\bullet\; $ If $A=\{1,2,3\}\;,$ Then $B=\left(\{4\}\;,\{5\}\;,\{4,5\}\right)$
So for $10$ ordered pair of $A\;,$ we get $3$ ordered pair of $B$
$\bullet \;$ If $A = \{1,2,3,4\}\;,$ Then $B = \{5\}$
So for $5$ ordered pair of $A\;,$ we get $1$ ordered pair of $B$
So Total ordered pair of $\left(A,B\right)$ is $ = \left(5\times 15\right)+\left(10 \times 7\right)+\left(10 \times 3\right)+\left(5 \times 1\right) = 75+70+30+5 = 180$
If Question is How can we solve using Combination (selection) way.
plz explain me, Thanks
| Phicar’s answer gives you a nice, short calculation if you know about Stirling numbers of the second kind. If not, you can still organize your argument a bit more efficiently.
Suppose that the set $A\cup B$ has $n$ elements; clearly $n$ must be $2,3,4$, or $5$. For each of these four possible values of $n$ we can argue as follows.
There are $\binom5n$ ways to choose the set $A\cup B$. $A$ can be any non-empty proper subset of $A\cup B$. $A\cup B$ has $2^n$ subsets, but one is empty and one is all of $A\cup B$, so there are only $2^n-2$ choices available for $A$. Thus, there are $\binom5n(2^n-2)$ ordered pairs $\langle A,B\rangle$ with $|A\cup B|=n$.
The answer, therefore, is
$$\sum_{n=2}^5\binom5n(2^n-2)=\sum_{n=2}^5\binom5n2^n-2\sum_{n=2}^5\binom5n\;.$$
Now notice that
$$\sum_{n=2}^5\binom5n=\sum_{n=0}^5\binom5n-\binom51-\binom50=2^5-5-1=26\;,$$
and
$$\begin{align*}
\sum_{n=2}^5\binom5n2^n&=\sum_{n=0}^5\binom5n2^n-\binom512-\binom50\\\\
&=\sum_{n=0}^5\binom5n2^n1^{5-n}-10-1\\\\
&=(2+1)^5-11\\\\
&=232\;,
\end{align*}$$
so the final answer is $232-52=180$.
This calculation suggests another elementary way to perform the calculation. If we temporarily allow $A$ and $B$ to be empty, we are in effect counting the ways to split $X$ into $3$ pieces, any of which may be empty. For each of the $5$ elements of $X$ we can put that element into $A$, into $B$, or into $X\setminus(A\cup B)$. This is a $3$-way choice made $5$ times, so there are $3^5=243$ ways to make it. However, some of these splits leave $A$ or $B$ or both empty. We can use an inclusion-exclusion argument to take care of them.
How many of these splits leave $A$ empty? They are the splits that put every element into $B$ or $X\setminus(A\cup B)$, so there are $2^5$ of them. There are also $2^5$ splits leaving $B$ empty, so we have to subtract $2\cdot2^5=64$. However, that subtracts the one split with $A=B=\varnothing$ twice, so we have to add it back in. The final result is
$$243-64+1=180\;.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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I am working on proving or disproving $\cos^5(x)-\sin^5(x)=\cos(5x)$ True or false? $$\cos^5(x)-\sin^5(x)=\cos(5x)$$ for all real x.
I have no idea how to prove or disprove this. I tried to expand $\cos(5x)$ using double angle formula but I wasn't sure how to go from that to $$\cos^5(x)-\sin^5(x)$$
| It is false. Quick counterexample: Note that, at $x=\frac{\pi}{5}$, we have $ \cos (5x) = -1 $, but clearly${}^\dagger$, $ \cos^5 \left( \frac {\pi}{5} \right) - \sin^5 \left( \frac {\pi}{5} \right) \ne 1 $. $\Box$
${}^\dagger$ It is clear that, since $ \frac {\pi}{5} $ is acute, $ 0 < \cos^5 \left( \frac {\pi}{5} \right), \sin^5 \left( \frac {\pi}{5} \right) < 1 $. Hence, $$ \cos^5 \left( \frac {\pi}{5} \right) - \sin^5 \left( \frac {\pi}{5} \right) < 1. $$This is how we get that it is not equal to $1$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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$\lim\limits_{n\to\infty}a^{n}\left(\frac{n+1}{n}\right)^{n^{2}}$ converges for $a$ in what range?
$\displaystyle\lim_{n\to\infty}a^{n}\left(\frac{n+1}{n}\right)^{n^{2}}$
converges for $a$ in what range?
I tried $\displaystyle\lim_{n\to\infty}\ln \left[a^{n}\left(\frac{n+1}{n}\right)^{n^{2}}\right]=\lim_{n\to\infty}\left[n\ln a+n\ln\left(1+\frac{1}{n}\right)^{n}\right]=\lim_{n\to\infty}\left[n\ln(a)+n\right]$
$\displaystyle\Rightarrow a=\frac{1}{e}$, a particular one. But how can I get the range?
| Let $c_n=a^n\left(1+\frac{1}{n}\right)^{n^2}$ and let $t=\frac{1}{a}$, so $\ln c_n=n\ln a+n^2\ln(1+\frac{1}{n})=n^2\ln(1+\frac{1}{n})-n\ln t$
where $\frac{1}{n}-\frac{1}{2n^2}<\ln(1+\frac{1}{n})<\frac{1}{n}-\frac{1}{2n^2}+\frac{1}{3n^3}\implies n-\frac{1}{2}<n^2\ln(1+\frac{1}{n})<n-\frac{1}{2}+\frac{1}{3n}$
$\implies (1-\ln t)n-\frac{1}{2}<n^2\ln(1+\frac{1}{n})-n\ln t<(1-\ln t)n-\frac{1}{2}+\frac{1}{3n}$,
$\hspace{.54 in}$ so $(1-\ln t)n-\frac{1}{2}<\ln c_n<(1-\ln t)n-\frac{1}{2}+\frac{1}{3n}$.
1) If $a>\frac{1}{e}$, then $t<e\implies(1-\ln t)n-\frac{1}{2}\to\infty\implies\ln c_n\to\infty\implies c_n\to\infty$.
2) If $a=\frac{1}{e}$, then $\ln c_n\to-\frac{1}{2}\implies c_n\to e^{-\frac{1}{2}}$.
3) If $0<a<\frac{1}{e}$, then $t>e\implies 1-\ln t<0\implies\ln c_n\to-\infty\implies c_n\to 0$.
Taking absolute values, we get that the sequence converges iff $a\in(-\frac{1}{e},\frac{1}{e}]$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Why this happen only with $4,8,12,.......$ We will take some examples to illustrate my question:
If we take a set of numbers, for example $1,2$, and $3$
$$1+2^4+3^4=98$$
$$1+2^8+3^4=338$$
$$1+2^8+3^8=6818$$
$$1+2^8+3^{12}=531698$$
We note that all numbers resulting end 8.This will not change for any selected power from the specified values(4,8,12....) in question.
another example
$$2^4+4^4+7^4+10^4=12673$$
$$2^4+4^8+7^8+10^4=5840353$$
$$2^{16}+4^8+7^8+10^4=5905873$$
$$2^{16}+4^8+7^8+10^8=105895873$$
We note that all numbers resulting end $3$
My question "Why this happen only with these numbers and how can I prove this?"
| Because for any integer $x$ whatsoever, and any $r,s\geq 1$,
$$x^{4r}\equiv x^{4s}\bmod 10$$
As an example, with $r=3$, and $s=5$ this says that
$$1^{12}\equiv 1^{20}\bmod 10,\qquad 2^{12}\equiv 2^{20}\bmod 10,\qquad 3^{12}\equiv 3^{20}\bmod 10$$
and therefore
$$\begin{align*}
(\text{last digit of $1^{12}+2^{12}+3^{12}$})&\equiv1^{12}+2^{12}+3^{12}\\
&\equiv 1^{20}+2^{20}+3^{20} \equiv (\text{last digit of $1^{20}+2^{20}+3^{20}$})\bmod 10
\end{align*} $$
To see that my claim is true, note that
$$\begin{array}{|c|c|c|c|c|c|c|c|c|c|c|}
x & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9\\\hline
x^4\bmod 10 & 0 & 1 & 6 & 1 & 6 & 5 & 6 & 1 & 6 & 1
\end{array}$$
and modulo $10$, the numbers $0,1,5,6$ are all fixed by $x\mapsto x^k$ for any $k$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Question about converting a polar equation to a rectangular equation $$\sec\theta =2$$
So I went through all the steps and got:
$$\cos\theta =\frac { 1 }{ 2 } $$
$$\sin\theta =\pm \sqrt { 1-\frac { 1 }{ 4 } } $$
$$\sin\theta =\pm \frac { \sqrt { 3 } }{ 2 } $$
$$y=\pm {\sqrt { 3 } }$$
Now why is it that the correct answer has an $x$ in it and why can't I give the answer $x^2+y^2=4$?
| $$ \sec \theta = 2 $$
$$ 1 = 2 \cos \theta $$
$$ r = 2 r\cos \theta $$
$$ \sqrt{x^2 + y^2} = 2x $$
$$ x^2 + y^2 = 4x^2 $$
$$ 3x^2 = y^2 $$
$$ y = \pm \sqrt{3} \,x$$
Another method:
$$ \cos \theta = \frac{1}{2} \Rightarrow \tan \theta = \pm \sqrt{3} = \frac{y}{x} \Rightarrow y = \pm \sqrt{3} \,x $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1068139",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Evaluating $\int_{0}^{\pi/2}\frac{x\sin x\cos x\;dx}{(a^{2}\cos^{2}x+b^{2}\sin^{2}x)^{2}}$ How to evaluate the following integral
$$\int_{0}^{\pi/2}\frac{x\sin x\cos x}{(a^{2}\cos^{2}x+b^{2}\sin^{2}x)^{2}}dx$$
For integrating I took $\cos^{2}x$ outside and applied integration by parts.
Given answer is $\dfrac{\pi}{4ab^{2}(a+b)}$.
But I am not getting the answer.
| Integrating by parts,
$$\int\frac{x\sin x\cos x}{(a^2\cos^2x+b^2\sin^2x)^2}dx$$
$$=x\int\frac{\sin x\cos x}{(a^2\cos^2x+b^2\sin^2x)^2}dx-\int\left[\frac{dx}{dx}\int\frac{\sin x\cos x}{(a^2\cos^2x+b^2\sin^2x)^2}dx\right]dx$$
Now $a^2\cos^2x+b^2\sin^2x=u\implies2(b^2-a^2)\sin x\cos x\ dx=du$
For $\displaystyle\int\frac{dx}{a^2\cos^2x+b^2\sin^2x}=\int\frac{\sec^2x\ dx}{a^2+b^2\tan^2x},$
set $b\tan x=a\tan y$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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Find an $\epsilon$ such that the $\epsilon$ neighborhood of $\frac{1}{3}$ contains $\frac{1}{4}$ and $\frac{1}{2}$ but not $\frac{17}{30}$ I am self studying analysis and wrote a proof that is not confirmed by the text I am using to guide my study. I am hoping someone might help me comfirm/fix/improve this.
The problem asks:
Find an $\epsilon$ such that $J_{\epsilon}(\frac{1}{3})$ contains $\frac{1}{4}$ and $\frac{1}{2}$ but not $\frac{17}{30}$
Here $J_\epsilon(x)$ means the $\epsilon$-neighborhood of $x$.
I know that $d\left(\frac{1}{4},\frac{1}{3}\right)<d\left(\frac{1}{2},\frac{1}{3}\right)$
$$\left|\frac{1}{2} - \frac{1}{3}\right| = \frac{1}{6}$$
I know that:
$$J_{\frac{1}{6}}(\frac{1}{3}) = \left(\frac{1}{6},\frac{1}{2}\right)$$
and so $$J_{\frac{1}{6}+\epsilon}\left(\frac{1}{3}\right) =\left(\frac{1}{6}-\epsilon,\frac{1}{2}+\epsilon\right)$$ where $\epsilon<\frac{1}{15}$ is a satisfactory solution.
Am I allowed to generalize this way with epsilon? Would the answer be better If provide some concrete value of epsilon?
| Compute the distances from $1/3$,
$$\Big|\frac14-\frac13\Big|=\frac1{12}=\frac5{60},$$
$$\Big|\frac12-\frac13\Big|=\frac1{6}=\frac{10}{60},$$
$$\Big|\frac{17}{30}-\frac13\Big|=\frac{7}{30}=\frac{14}{60}.$$
So any $$\frac{10}{60}<\epsilon\le\frac{14}{60}$$will do, say $1/5$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Closed form of $\int_0^\infty \ln \left( \frac{x^2+2kx\cos b+k^2}{x^2+2kx\cos a+k^2}\right) \;\frac{\mathrm dx}{x}$ Today I discussed the following integral in the chat room
$$\int_0^\infty \ln \left( \frac{x^2+2kx\cos b+k^2}{x^2+2kx\cos a+k^2}\right) \;\frac{\mathrm dx}{x}$$
where $0\leq a, b\leq \pi$ and $k>0$.
Some users suggested me that I can use Frullani's theorem:
$$\int_0^\infty \frac{f(ax) - f(bx)}{x} = \big[f(0) - f(\infty)\big]\ln \left(\frac ab\right)$$
So I tried to work with that way.
\begin{align} I&=\int_0^\infty \ln \left( \frac{x^2+2kx\cos b+k^2}{x^2+2kx\cos a+k^2}\right) \;\frac{\mathrm dx}{x}\\ &=\int_0^\infty \frac{\ln \left( x^2+2kx\cos b+k^2\right)-\ln \left( x^2+2kx\cos a+k^2\right)}{x}\mathrm dx\tag{1}\\ &=\int_0^\infty \frac{\ln \left( 1+\dfrac{2k\cos b}{x}+\dfrac{k^2}{x^2}\right)-\ln \left( 1+\dfrac{2k\cos a}{x}+\dfrac{k^2}{x^2}\right)}{x}\mathrm dx\tag{2}\\ \end{align}
The issue arose from $(1)$ because $f(\infty)$ diverges and the same issue arose from $(2)$ because $f(0)$ diverges. I then tried to use D.U.I.S. by differentiating w.r.t. $k$, but it seemed no hope because WolframAlpha gave me this horrible form. Any idea? Any help would be appreciated. Thanks in advance.
| First note that by substituting $x\mapsto kx$, we get
$$
\int_0^\infty\log\left(\frac{x^2+2kx\cos(a)+k^2}{x^2+2kx\cos(b)+k^2}\right)\frac{\mathrm{d}x}{x}
=\int_0^\infty\log\left(\frac{x^2+2x\cos(a)+1}{x^2+2x\cos(b)+1}\right)\frac{\mathrm{d}x}{x}
$$
Let $u=\frac{x+\cos(a)}{\sin(a)}$. Then
$$
\begin{align}
\frac{\mathrm{d}}{\mathrm{d}a}\int_0^\infty\log\left(\frac{x^2+2x\cos(a)+1}{x^2+2x+1}\right)\frac{\mathrm{d}x}{x}
&=-2\int_0^\infty\frac{\sin(a)}{x^2+2x\cos(a)+1}\,\mathrm{d}x\\
&=-2\int_0^\infty\frac{\sin(a)}{(x+\cos(a))^2+\sin^2(a)}\,\mathrm{d}x\\
&=-\frac2{\sin(a)}\int_0^\infty\frac1{\frac{(x+\cos(a))^2}{\sin^2(a)}+1}\,\mathrm{d}x\\
&=-2\int_{\cot(a)}^\infty\frac1{u^2+1}\,\mathrm{d}u\\[9pt]
&=-2a
\end{align}
$$
Integrating in $a$ gives
$$
\int_0^\infty\log\left(\frac{x^2+2x\cos(a)+1}{x^2+2x+1}\right)\frac{\mathrm{d}x}{x}
=-a^2
$$
Therefore, by subtraction,
$$
\int_0^\infty\log\left(\frac{x^2+2kx\cos(a)+k^2}{x^2+2kx\cos(b)+k^2}\right)\frac{\mathrm{d}x}{x}
=b^2-a^2
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1069376",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "29",
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Limits using Maclaurins expansion for $\lim_{x\rightarrow 0}\frac{e^{x^2}-\ln(1+x^2)-1}{\cos2x+2x\sin x-1}$ $$\lim_{x\rightarrow 0}\frac{e^{x^2}-\ln(1+x^2)-1}{\cos2x+2x\sin x-1}$$
Using Maclaurin's expansion for the numerator gives:
$$\left(1+x^2\cdots\right)-\left(x^2-\frac{x^4}{2}\cdots\right)-1$$
And the denominator:
$$\left(1-2x^2\cdots\right) + \left(2x^2-\frac{x^4}{3}\cdots\right)-1$$
$$\therefore \lim_{x\rightarrow 0} f(x) = \frac{-\dfrac{x^4}{2}}{-\dfrac{x^4}{3}} = \frac{3}{2}$$
But Wolfram gives that the limit is $3$.
I thought, maybe I used too few terms. What is a thumb rule for how many terms in expansion to use to calculate limits?
Using three terms yielded the answer $\lim_{x\rightarrow 0}f(x) = -4$.
What did I do wrong?
| Doing almost the same as mixedmath, I have very slightly different results since
$$e^{x^2} = 1+x^2+\frac{x^4}{2}+\frac{x^6}{6}+\frac{x^8}{24}+O\left(x^9\right) $$
$$\ln(1 + x^2) = x^2-\frac{x^4}{2}+\frac{x^6}{3}-\frac{x^8}{4}+O\left(x^9\right)$$
$$\cos(2x)=1-2 x^2+\frac{2 x^4}{3}-\frac{4 x^6}{45}+\frac{2 x^8}{315}+O\left(x^9\right)$$
$$2x\sin x =2 x^2-\frac{x^4}{3}+\frac{x^6}{60}-\frac{x^8}{2520}+O\left(x^9\right)$$ So, for the numerator, $$x^4-\frac{x^6}{6}+\frac{7 x^8}{24}+O\left(x^9\right)$$ and for the denominator $$\frac{x^4}{3}-\frac{13 x^6}{180}+\frac{x^8}{168}+O\left(x^9\right)$$ Performing the long division $$\frac{e^{x^2}-\log \left(x^2+1\right)-1}{2 x \sin (x)+\cos (2 x)-1}\approx 3+\frac{3 x^2}{20}+\cdots$$ For sure, the limit is the same, but pushing the expansion a little bit allows to see how is this limit approached.
If you push a little more the expansions, you should find that $$\frac{e^{x^2}-\log \left(x^2+1\right)-1}{2 x \sin (x)+\cos (2 x)-1}\approx 3+\frac{3 x^2}{20}+\frac{2391 x^4}{2800}+\cdots$$ and you should be amazed to see how this approximation is close to the expression (just for curiosity, graph on the same plot the function and this last approximation for $-1 \leq x \leq 1$).
| {
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"timestamp": "2023-03-29T00:00:00",
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Volume of solid of revolution by shell method consider the region bounded by $ \displaystyle y=4{{x}^{2}}$ and $ \displaystyle 2x+y=6$. What is the volume of solid of revolution about $\displaystyle x$-axis.
What is thought about setting the integral:
I split the region into two parts
$\displaystyle V=4\pi \int\limits_{0}^{4}{y\left( 1-\frac{\sqrt{y}}{2} \right)\,dy}+2\pi \int\limits_{4}^{9}{y\left( \frac{6-y}{2}-\frac{\sqrt{y}}{2} \right)\,dy}$
Will it work?
| I don't see why you need to split the region. Notice that $y = 4x^2$ and $y = 6 - 2x$ both intercept at: $$4x^2 = 6-2x \implies2x^2 + x - 3 =0 \implies x = \frac{-1\pm\sqrt{1+24}}{4} \implies x = \frac{-1\pm 5}{4},$$ so $x = -3/2$ and $x = 1$. Since the area of revolution of $y = f(x)$ around the $x$ axis is $$A = \pi \int_{a}^b f(x)^2 \ {\rm d}x$$ and $6 - 2x > 4x^2$ in $(-3/2, 1)$, the area you're looking for is: $$A = \pi\int_{-3/2}^1 (6-2x)^2 \ {\rm d}x - \pi \int_{-3/2}^1 (4x^2)^2 \ {\rm d}x.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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In triangle ABC, Find $\tan(A)$.
In triangle ABC, if $(b+c)^2=a^2+16\triangle$, then find $\tan(A)$ . Where $\triangle$ is the area and a, b , c are the sides of the triangle.
$\implies b^2+c^2-a^2=16\triangle-2bc$
In triangle ABC, $\sin(A)=\frac{2\triangle}{bc}$, and $\cos(A)=\frac{b^2+c^2-a^2}{2bc}$,
$\implies \tan(A)=\frac{4\triangle}{b^2+c^2-a^2}$
$\implies \tan(A)=\frac{4\triangle}{16\triangle-2bc}$. But the answer is in the form, $\frac{x}{y}$ where $x$ and $y$ are integers.
Any help is appreciated. Thanks in advance.
| You have: $\dfrac{\sin A}{\cos A} =\tan A = \dfrac{4\triangle}{16\triangle - 2bc} = \dfrac{2bc\sin A}{8bc\sin A-2bc} = \dfrac{\sin A}{4\sin A-1} \Rightarrow \cos A = 4\sin A -1 \Rightarrow \cos^2A = \left(1-4\sin A\right)^2 = 1 - 8\sin A + 16\sin^2A \Rightarrow 1 - \sin^2A = 1 - 8\sin A + 16\sin^2A \Rightarrow \sin A\left(8 - 17\sin A\right) = 0 \Rightarrow \sin A = 0 \text{ or} \text{ }\sin A = \dfrac{8}{17}$. But $\sin A > 0$, so $\sin A = \dfrac{8}{17} \Rightarrow \cos A = 4\sin A -1 = 4\cdot \dfrac{8}{17} - 1 = \dfrac{15}{17} \Rightarrow \tan A = \dfrac{\dfrac{8}{17}}{\dfrac{15}{17}} = \dfrac{8}{15}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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How does $\log(x^2 + 1)$ become $\log(2x^2)$? My textbook attempts to take the big O of $\log(x^2 +1)$. It proceeds by saying $x^2 + 1 \le 2x^2$ when $x \ge 1$. But I don't know how it came up with this idea.
Question:
Why set $x^2+1$ to a random value to be $2x^2$? Why $2$ of all numbers? Why not $x^2$ or $x^3$?
| We’re not setting $x^2+1$ to anything. We want to find an upper bound on $\log(x^2+1)$ that can be expressed as a fairly simple function of $x$. Provided that $x\ge 1$, we know that $x^2+1\le 2x^2$, so $\log(x^2+1)\le\log(2x^2)$, and $\log(2x^2)$ can be expressed as a simple function of $\log x$:
$$\log(2x^2)=\log 2+2\log x\;.$$
If $x\ge 2$, we know that $\log x\ge\log 2$, and therefore
$$\log(x^2+1)\le\log 2+2\log x\le 3\log x\;.$$
This is exactly the kind of thing that’s needed in order for us to conclude that $\log(x^2+1)$ is $O(\log x)$: we have constants $M$ and $c>0$ such that
$$\log(x^2+1)\le c\log x$$
for all $x\ge M$. Specifically, we can use $M=2$ and $c=3$.
We couldn’t have used $x^2$, because $x^2+1$ isn’t less than or equal to $x^2$. We could have used $x^3$: $x^2+1\le x^3$ for $x\ge 2$, say, so
$$\log(x^2+1)\le\log x^3=3\log x$$
for $x\ge 2$, and again we’ve shown that $\log(x^2+1)$ is $O(\log x)$.
We could have used $5x^2$: $x^2+1\le 5x^2$ for $x\ge 1$, so
$$\log(x^2+1)\le\log(5x^2)=\log 5+2\log x\;.$$
And $\log 5+2\log x\le3\log x$ for $x\ge 5$, so $\log(x^2+1)\le3\log x$ for $x\ge 5$, and again we’ve shown that $\log(x^2+1)$ is in $O(\log x)$.
| {
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Find the fraction where the decimal expansion is infinite? Find the fraction with integers for the numerator and denominator, where the decimal expansion is $0.11235.....$
The numerator and denominator must be less than $100$.
Find the fraction.
I believe I can use generating functions here to get $1+x+2x^2+3x^3+5x^4+.....$, but I do not know how to apply it.
| Like you hint, we can observe that the value
$$0.11235955056\ldots = \sum_{k = 1}^{\infty} 10^{-k} F_k,$$
where $F_k$ is the $k$th Fibonacci number (with the convention that $F_0 = 0$, $F_1 = 1$),
is simply the value of the series
$$F(x) := \sum_{k = 1}^{\infty} F_k x^k = x + x^2 + 2x^3 + 3x^4 + 5x^5 + 8x^6 + 13x^7 + \cdots$$
at $x = \frac{1}{10}$.
Hint Using the defining recurrence relation
$$F_{k + 2} = F_{k + 1} + F_k$$
(and the above convention) gives that $F(x)$ satisfies
$$F(x) = x + x F(x) + x^2 F(x).$$
Rearranging gives that on the open interval of convergence of the series, which turns out to be $(-1/\phi, 1/\phi)$---where $\phi$ is the Golden Ratio, and which in particular contains the value $\frac{1}{10}$ of interest)---we have $$F(x) = \frac{x}{1 - x - x^2}.$$ Thus, the series has value $$F\left(\tfrac{1}{10}\right) = \frac{\left(\frac{1}{10}\right)}{1 - \left(\frac{1}{10}\right) - \left(\frac{1}{10}\right)^2} = \color{#bf0000}{\boxed{\frac{10}{89}}} .$$ Since this is a rational number, its decimal expansion repeats: $$0.\overline{1123595505617977528089887640449438202247191}.$$
| {
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Transforming linear combination of the cosine and sine function In the proof of Transforming $a\cos\left(\, x\,\right)+b\sin\left(\, x\right)$ to $r\cos\left(\,\phi - x\,\right)$
\begin{align}
a\cos\left(\, x\,\right) + b\sin\left(\, x\,\right)
&=\,\sqrt{\,a^{2} + b^{2}\,}\,
\left[\,\frac{a}{\,\sqrt{\, a^{2} + b^{2}\,}\,}\,\cos\left(\, x\,\right)
+\frac{b}{\,\sqrt{\, a^{2} + b^{2}\,}\,}\,\sin\left(\, x\,\right)\,\right]\,
\\[2mm]&=\,\sqrt{\,a^{2} + b^{2}\,}\,\left[\,
\cos\left(\,\phi\,\right)\cos\left(\, x\,\right) + \sin\left(\,\phi\,\right)\sin\left(\,x\,\right)
\,\right]
\\[2mm]&=\,\sqrt{\, a^{2} + b^{2}\,}\,\cos\left(\,\phi - x\,\right)
\end{align}
why did we factor out $\,\sqrt{\, a^{2} + b^{2}\,}\,$ where did this idea came from ?. It seems out of the blue.
| Start at "the other end" of the problem. If we want
$$a\cos x+b\sin x=r\cos(\phi-x)$$
and we expand the right hand side, then what we are looking for is
$$a\cos x+b\sin x=r\cos\phi\cos x+r\sin\phi\sin x\ .$$
If this is to be true for all values of $x$ then we need
$$a=r\cos\phi\quad\hbox{and}\quad b=r\sin\phi\ .$$
Squaring and adding these equations,
$$a^2+b^2=r^2\cos^2\phi+r^2\sin^2\phi=r^2$$
and so
$$r=\sqrt{a^2+b^2}\ .$$
That's where the idea comes from to factor out $\sqrt{a^2+b^2}$.
| {
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Prove $4^k - 1$ is divisible by $3$ for $k = 1, 2, 3, \dots$ For example:
$$\begin{align}
4^{1} - 1 \mod 3 &=
\\
4 -1 \mod 3 &=
\\
3 \mod 3 &=
\\3*1 \mod 3 &=0
\\
\\
4^{2} - 1 \mod 3 &=
\\
16 -1 \mod 3 &=
\\
15 \mod 3 &=
\\3*5 \mod 3 &= 0
\\
\\
4^{3} - 1 \mod 3 &=
\\
64 -1 \mod 3 &=
\\
21 \mod 3 &=
\\3*7 \mod 3 &=
0\end{align}
$$
Define $x = \frac{4^k - 1}{3}$. So far I have:
$$k_1 \to 1 \Longrightarrow x_1 \to 1
\\
k_2 \to 2 \Longrightarrow x_2 \to 5
\\
k_3 \to 3 \Longrightarrow x_3 \to 21
\\
k_4 \to 4 \Longrightarrow x_4 \to 85$$
But then it's evident that
$$4^{k_n} = x_{n+1} - x_n$$
I don't know if this helps, these are ideas floating in my head.
| $4^k-1=(4-1)(4^{k-1}+\cdots+4+1)$.
| {
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$b_{n}$ is increasing I think there is misunderstanding in my last post because its contain three questions so i will post question by question step by step
An inequality for the product $\prod_{k=2}^{n}\cos\frac{\pi }{2^{k}}$
Let $n\geq 2\quad a_{n}=\prod\limits_{k=2}^{n}\cos\left(\dfrac{\pi }{2^{k}}\right)$ and $b_{n}=a_{n}\cos\left(\dfrac{\pi }{2^{n}}\right)$
Show that $b_{n}$ is increasing
my attempts :
*
*$b_{n+1}-b_{n}=a_{n+1}\cos\left(\dfrac{\pi }{2^{n+1}}\right)-a_{n}\cos\left(\dfrac{\pi }{2^{n}}\right)=a_{n}\cos\left(\dfrac{\pi }{2^{n}}\right)\cos\left(\dfrac{\pi }{2^{n+1}}\right)-a_{n}\cos\left(\dfrac{\pi }{2^{n}}\right)$
*$\dfrac{b_{n+1}}{b_n}=\cos(\pi/2^{n+1})$
im stuck here in these two ways
any help would be appreciated
| A variant: $$\begin{align*} b_n <b_{n+1} &\iff \cos\Bigl(\frac{\pi}{2^n}\Bigr) < \cos^2\Bigl(\frac{\pi}{2^{n+1}}\Bigr)\\ &\iff 2\cos^2\Bigl(\frac{\pi}{2^{n+1}}\Bigr)-1 < \cos^2\Bigl(\frac{\pi}{2^{n+1}}\Bigr)\\ &\iff \cos^2\Bigl(\frac{\pi}{2^{n+1}}\Bigr) <1 .\end{align*}$$
| {
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Number of $ 6 $ Digit Numbers with Alphabet $ \left\{ 1, 2, 3, 4 \right\} $ with Each Digit of the Alphabet Appearing at Least Once Find the number of 6 digit numbers that can be made with the digits 1,2,3,4 if all the digits are to appear in the number at least once.
This is what I did -
I fixed four of the digits to be 1,2,3,4 .
Now remaining 2 places can be filled with 4 digits each. Number of 6 digit numbers if two places are filled with same digit are 4 * 6!/3! and if filled by different digits are 12 * 6!/(2!*2!). Therefore, total such numbers are 2880.
But the correct answer is 1560.
Any hint would be appreciated.
| Really easy using generating functions. The generating function is:
$$G(x)=\left(x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}\right)^4$$
We need the coefficient of x^6 in that and we make use of the fact that $x^4 = x^2 \cdot x^2$
$\left(x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}\right) \left(x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}\right)=x^2+x^3+\frac{7 x^4}{12}...$
We only need to calculate the first 3 terms of that because other terms do not contribute to $x^6$
$\left(x^2+x^3+\frac{7 x^4}{12}\right) \left(x^2+x^3+\frac{7 x^4}{12}\right)=\frac{7 x^6}{12}+x^6+\frac{7 x^6}{12}=\frac{13 x^6}{6}$
Now we have the required coefficient, we just multiply by $6!$
$$6! \cdot \frac{13}{6}= 1560 $$
That is the answer.
| {
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$mn | m^2+n^2+m \implies$ $(n-1)$ is a square Let $m;n \in \mathbb{Z^+}$ such that $mn | m^2+n^2+m$
Prove that $(n-1)$ is a square number.
P/s : I don't have any ideas about this problems :(
Thanks :)
| Consider the equation $kmn = m^2+n^2+m$ for fixed $k$. Suppose this equation has solutions in $(m,n)$ in positive integers, and take the solution with $m+n$ minimized. We distinguish two cases.
*
*Suppose $m \geq n$. Note that $m$ satisfies the quadratic equation
$$ X^2 - (kn-1)X+n^2 = 0$$
which also has $X=\frac{n^2}{m}$ as a solution. It follows that $(\frac{n^2}{m},n)$ is also a solution to the original equation, whence $m+n \leq \frac{n^2}{m} + n$ implying $m=n$.
*Suppose $n \geq m$. Note that $n$ satisfies the quadratic equation
$$ Y^2 - kmY + m^2+m = 0$$
which also has $Y=\frac{m^2+m}{n}$ as a solution. It follows that $(\frac{m^2+m}{n},m)$ is also a solution to the original equation, whence $m+n \leq \frac{m^2+m}{n} + m$ or $n^2 \leq m^2+m$. Since $n \geq m$ this entails $n=m$ (for $m \leq n-1$ we have $m^2 +m \leq (n-1)^2 + (n-1) = n^2 - n < n^2$).
In both cases we find $m=n$. Since the only solution with $m=n$ is $m=n=1$ for $k=3$, we conclude that $k=3$ is the only possibility.
Now we have $3mn = m^2+n^2+m$ or $m(n-1)=(m-n)^2$. Since $m \mid n^2$, $m$ is coprime to $n-1$ so both $m$ and $n-1$ are squares.
This technique is known as Vieta jumping.
| {
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Equation with three variables I am confused as how to solve an equation with three squared variables to get its integer solutions? As:
$$x^2+y^2+z^2=200$$
Thanks!
| I would do it this way: (it takes just few minutes)
Let's assume $x<=y<=z$, without loss of generality.
$z$ must be larger than $8$, since $3*8^2 = 192 < 200$.
$z$ must be smaller than $15$, since $15^2 = 225 > 200$.
Now, there are six possibilities:
$$x^2+y^2=200-9^2=119$$
$$x^2+y^2=200-10^2=100$$
$$x^2+y^2=200-11^2=79$$
$$x^2+y^2=200-12^2=56$$
$$x^2+y^2=200-13^2=31$$
$$x^2+y^2=200-14^2=4$$
Then, according to the list of Pythagorean triples, only this solution can exist:
$$x=6, y=8, z=10$$
The list above assumes x<>0, and for x=0, we easily get two more solutions:
$$x=0, y=2, z=14$$
and
$$x=0, y=10, z=10$$
Of course, if variables can be negative, there is also a number of solutions directly derived from last three ones.
| {
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The locus of points $z$ which satisfy $|z - k^2c| = k|z - c|$, for $k \neq 1$, is a circle Use algebra to prove that the locus of points z which satisfy $|z - k^2c| = k|z - c|$, for $k \neq 1$ and $c = a + bi$ any fixed complex number, is a circle centre $O$.
Give the radius of the circle in terms of $k$ and $|c|$.
I squared both sides and got this:
$$(k^2−1)x^2+(k^2−1)y^2+(a^2+b^2-k^2a^2-k^2b^2)k^2=0$$
I might have gone wrong somewhere though.
Edit. Never mind, I didn't go wrong.
$$(k^2-1)x^2+(k^2-1)y^2-(k^2-1)k^2a^2-(k^2-1)k^2b^2=0$$
$$x^2+y^2=k^2(a^2+b^2)$$
$$r^2=k^2(a^2+b^2)$$
$$r=k|c|$$
| instead of squaring let me see if conjugates help.
$\begin{eqnarray}
0 = |z - k^2c|^2 -k^2 |z -c|^2 &=& (z - k^2 c)(\bar z - k^2 \bar c) -k^2(z-c)(\bar z - \bar c) \\
&=& z \bar z - k^2 c \bar z - k^2\bar c z + k^4 c \bar c -k^2(z \bar z - c \bar z -z \bar c + c \bar c) )\\
&=&(1-k^2)z \bar z - k^2(1-k^2)c\bar c \\
&=&(1-k^2)(|z|^2 - k^2|c|^2)
\end{eqnarray}$
$ \mbox{ if } k^2 \neq 1, \mbox{ then } |z|^2 = k^2|c|^2$
| {
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Two questions about divisible I have two questions.
1) Why the relation $a^n\equiv a^{n+4k}\pmod{10}$ is true?
2) Let $(a,90)=1$. Which number can be $ x$ in $ x\mid a^4-1$? Answer is $240$. Why?
| \begin{align}
3^0 & \equiv 1 \pmod{10} \\
3^1 & \equiv 3 \pmod{10} \\
3^2 & \equiv 9 \pmod{10} \\
3^3 & = 27 \equiv 7 \pmod{10} \\
3^4 & = 81 \equiv 1 \pmod{10}
\end{align}
We've returned to $1$ when the exponent is $4$.
Try this with other numbers is $\{1,3,7,9\}$ in place of $3$. In each case we get the $4$th power congruent to $1$. The four numbers $1,3,7,9$ are those that are coprime to $10$.
| {
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Show that f is periodic if $f(x+a)+f(x+b)=\frac{f(2x)}{2}$? Suppose $a$ and $b$ are distinct real numbers and $f$ is a continuous real function such that $\frac{f(x)}{x^2}$ goes to 0 when $x$ goes to infinity or minus infinity. Suppose that$ f(x+a)+f(x+b)=\frac{f(2x)}{2}$. How show that $f$ is periodic?
| There are not so many possibilities for equation transformations, so we go straight forward. For the determinancy assume that $a<b$.
Let $x$ be an arbitrary real number. Then
$$f(2x+a)+f(2x+b)=\frac{f(4x)}{2}.$$
But
$$f(2x+a)=2\left(f\left(x+a+\frac a2\right)+ f\left(x+b+\frac a2\right)\right),$$
$$f(2x+b)=2\left(f\left(x+a+\frac b2\right)+ f\left(x+b+\frac b2\right)\right).$$
So
$$f\left(x+a+\frac a2\right)+ f\left(x+b+\frac a2\right)+f\left(x+a+\frac b2\right)+ f\left(x+b+\frac b2\right)=\frac{f(4x)}{4}.$$
Similarly, by induction we can show that for each non-negative integer $n$
$$S_n(x)= \sum_{i=0}^{2^n-1} f\left(x+2a+\frac{ib-(i+1)a}{2^{n-1}}\right)=\frac{f(2^nx)}{2^n}.$$
Indeed, above this equality is already proved for $n\le 2$. Assume that the equality is proved for each real $x$ and a non-negative integer $n$. Then
$$\frac{f(2^{n+1}x)}{2^{n+1}}=\frac{1}{2^n}\left(f(2^nx+a)+ f(2^nx+b)\right)=$$ $$S_n\left(x+\frac{a}{2^n}\right)+ S_n\left(x+\frac{b}{2^n}\right)=$$
$$\sum_{i=0}^{2^n-1} f\left(x+\frac{a}{2^n}+2a+\frac{ib-(i+1)a}{2^{n-1}}\right)+ f\left(x+\frac{b}{2^n}+2b+\frac{ib-(i+1)a}{2^{n-1}}\right)=$$
$$\sum_{i=0}^{2^n-1} f\left(x+2a+\frac{2ib-(2i+1)a}{2^n}\right)+ f\left(x+2a+\frac{(2i+1)b-(2i+2)a}{2^n}\right)=$$ $$ \sum_{i=0}^{2^{n+1}-1} f\left(x+2a+\frac{ib-(i+1)a}{2^n}\right)=S_{n+1}(x).$$
Let $$\sigma_n(x)=\frac{b-a}{2^n} \sum_{i=0}^{2^n-1} f\left(x+2a+\frac{i(b-a)}{2^{n-1}}\right)$$ be an integral sum for the function $f$ at the segment $[x+2a, x+2b]$. Since the function $f$ is continuous on this segment, it is integrable on it, so for each real $x$ a sequence $\{\sigma_n(x)\}$ converges to an integral $\int^{x+2b}_{x+2a} f(t)dt$. Let $\tau_n(x)=\frac{b-a}{2^n}S_n(x)$. Since the function $f$ is continuous on the segment $I=[x+\min\{0,2a\}, x+\max\{0,2a\}+2(b-a)]$ , it is uniformly continuous on it. Then for each $\varepsilon>0$ there exists natural $N$ such that if $n>N$, $t,t’\in I$, and $|t-t’|<\frac{|a|}{2^{n-1}} $ then $|f(t)-f(t’)|<\varepsilon$. Then
$$|\tau_n(x)-\sigma_n(x)|=$$ $$\left|\frac{b-a}{2^n}\sum_{i=0}^{2^n-1} f\left(x+2a+\frac{ib-(i+1)a}{2^{n-1}}\right) - \frac{b-a}{2^n} \sum_{i=0}^{2^n-1} f\left(x+2a+\frac{i(b-a)}{2^{n-1}}\right)\right|=$$
$$\frac{b-a}{2^n}\left|\sum_{i=0}^{2^n-1} f\left(x+2a+\frac{ib-(i+1)a}{2^{n-1}}\right) - f\left(x+2a+\frac{i(b-a)}{2^{n-1}}\right)\right|< $$
$$\frac{b-a}{2^n}\sum_{i=0}^{2^n-1}\varepsilon=\frac{(b-a)}{2^n}\cdot 2^n\varepsilon=(b-a)\varepsilon.$$
Thus the sequence $\{\tau_n(x)\}$ converges to the integral $\int^{x+2b}_{x+2a} f(t)dt$ for each real $x$.
One the other hand
$$\tau_n(x)=\frac{b-a}{2^n}S_n(x)=\frac{b-a}{2^n}\cdot\frac{f(2^nx)}{2^n}$$
tends to zero when $n$ tends to infinity (the case $x\ne 0$ directly follows from the question condition, in the case $x=0$
it suffices to remark that $f(2^nx)=f(0)$ for all $n$).
Thus $\int^{x+2b}_{x+2a} f(t)dt=0$ for each real $x$. Differentiating both sides of this equality with respect to $x$ (this is allowed, by instance, by [Fich, 305, 12$^\circ$, p. 116]), we obtain $f(x+2b)-f(x+2a)=0$. Since $a\ne b$, the function $f$ is periodic. Since it is continuous, it is constant or has a minimal period. By Hans Engler’s comment the latter case is impossible and this constant is zero.
What’s a pity, so much efforts for zero!
References
[Fich] Grigorii Fichtenholz, Differential and Integral Calculus, vol. II, 7-th edition, M.: Nauka, 1970 (in Russian).
| {
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Proving that a number is an integer. Prove that the following number is an integer:
$$\left( \dfrac{76}{\dfrac{1}{\sqrt[\large{3}]{77}-\sqrt[\large{3}]{75}}-\sqrt[\large{3}]{5775}}+\dfrac{1}{\dfrac{76}{\sqrt[\large{3}]{77}+\sqrt[\large{3}]{75}}+\sqrt[\large{3}]{5775}}\right)^{\large{3}}$$
How can I prove it?
| Following the simplifications suggested by MathBot, we have:
$$\left( \dfrac{76}{\dfrac{1}{b-a}-ab}+\dfrac{1}{\dfrac{76}{b+a}+ab}\right)^{\large{3}}$$
Let's just take the part inside the parentheses, and put it over a common denominator.
$$\dfrac{\dfrac{76^2}{b+a} + 76 ab + \dfrac{1}{b-a} - ab}{\left(\dfrac{1}{b-a}-ab\right)\left(\dfrac{76}{b+a}+ab\right)}$$
Expand the denominator:
$$\dfrac{\dfrac{76^2}{b+a} + 75 ab + \dfrac{1}{b-a}}{\dfrac{76}{b^2-a^2}+\dfrac{ab}{b-a}- \dfrac{76ab}{b+a}-a^2b^2}$$
Put the numerator and denominator on a common denominator:
$$\dfrac{\dfrac{76^2b - 76^2 a + 75 a b^3 - 75 a^3 b + b + a}{b^2-a^2}}{\dfrac{76 + ab^2 + a^2 b - 76 ab^2 + 76a^2 b - a^2 b^4 + a^4 b^2}{b^2-a^2}}$$
Simplify:
$$\dfrac{76^2b - 76^2 a + 75 a b^3 - 75 a^3 b + b + a}{76 - 75 ab^2 + 77a^2 b - a^2 b^4 + a^4 b^2}$$
Remember that $a^3 = 75$ and $b^3 = 77$:
$$
\dfrac{(76^2 - 75^2 + 1)b + (75 \cdot 77 - 76^2 + 1) a}{76 - 75 ab^2 + 77a^2 b - 77 a^2 b + 75 a b^2}
\ = \ \dfrac{152b + 0 a}{76}
\ = \ 2b.$$
Remember we need to cube the whole thing, so the answer is
$(2b)^3 = 8 b^3 = 8\cdot 77$, or $616$.
| {
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Units in a ring of fractions
Let $R$ be a UFD and $D \subseteq R$ multiplicative set. What are the units in $D^{-1}R$?
I assume the answer should be $D^{-1}R^{\times}$, but I get stuck:
If $a/b$ is a unit, then there exists $c/d$ so that
$$\frac{a}{b} \cdot \frac{c}{d} = \frac{1}{1} \Longleftrightarrow ac = bd,$$
but I don't see what this tells me. For example, just because $ac \in D$ doesn't seem to imply anything about $a, c$.
| It is clear that $\dfrac{1}{1} = \dfrac{d}{d}, d \in D$ is the identity element of $D^{-1}R.$ We want to find the units. Let $u \in R$ is a unit. So there exists $v \in R$ such that $uv = vu = 1.$ Then obviously, $\dfrac{u}{1} \cdot \dfrac{v}{1} = \dfrac{v}{1} \cdot \dfrac{u}{1} = \dfrac{1}{1},$ showing that $\dfrac{u}{1}$ is a unit in $D^{-1}R.$ Also for each $d \in D, \dfrac{1}{d} \cdot \dfrac{d}{1} = \dfrac{d}{1} \cdot \dfrac{1}{d} = \dfrac{1}{1}.$ So, every element of the form $\dfrac{1}{d}$ and $\dfrac{d}{1}$ where $d \in D$ are units. Now suppose $\dfrac{a}{b}$ is a unit in $D^{-1}R.$ Then there exists $\dfrac{c}{d}$ such that $\dfrac{a}{b} \cdot \dfrac{c}{d} = \dfrac{c}{d} \cdot \dfrac{a}{b} = \dfrac{1}{1}.$ So we have, $ac = bd.$ This shows that both $a$ and $c$ must divide an element of $D.$ On the other hand, suppose $0 \neq a \in R$ be such that $a|d,$ for some $d \in D.$ Let $ab = d.$ Then for any $c \in D, \dfrac{a}{c} \cdot \dfrac{bc^2}{cd} = \dfrac{1}{1},$ proving that $\dfrac{a}{c}$ is a unit in $D^{-1}R.$
If $R$ is a UFD, then we can say more. Every element $d \in D$ is either a unit in $R$ of has a unique prime factorization in $R$ (up to units). Let $S$ be the multiplicative subset of $R$ generated be all the primes that occurs in the prime factorization of elements of $D.$ Then it is easy to show that $D^{-1}R \cong S^{-1}R.$ Form above we see that any element of the form $\dfrac{a}{b}, b \in S$ and $a$ divides some element of $S$ is a unit. This shows that $a \in S.$ So the units of $D^{-1}R$ are precisely of the form $\frac{u}{1}$ where $u$ is a unit in $R$ or of the form $\dfrac{a}{b}$ where both $a, b \in S.$
The last conclusion is not true in general, if we don't assume $R$ is a UFD. Let $R := k[x^2, x^3],$ where $k$ is field, i.e. $R$ is the subring of the polynomial ring $k[x]$ consisting of polynomials whose co-efficient of the degree $1$ term is zero. Then it is not a UFD ($x^6 = (x^2)^3 = (x^3)^2$). Let $D = \{1, x^2, x^4, x^6, \cdots \}.$ Then the element $\dfrac{x^3}{1}$ is a unit in $D^{-1}R: \dfrac{x^3}{1} \cdot \dfrac{x^3}{x^6} = \dfrac{1}{1},$ but $x^3 \notin D.$
| {
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"url": "https://math.stackexchange.com/questions/1089057",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Prove that $ \sum_{n=1}^\infty \ln\big(n\sin \frac{1}{n}\big)$ converges.
Prove that $\displaystyle \sum_{n=1}^\infty \;\ln\left(n\sin\frac{1}{n}\right)$ converges.
My Work:
$$\left|\ln \left(n \sin \frac{1}{n}\right)\right| \leq\left|\ln \left(n \sin \frac{1}{n^{2}}\right)\right| \leq\left|\ln \left(\sin \frac{1}{n^{2}}\right)\right|$$
I was going to use comparison test. But now stuck. Please give me a hint.
| In the same spirit as mixedmath's answer, let us massively use Taylor expansions for large values of $n$. Then $$\sin(\frac 1n)=\frac{1}{n}-\frac{1}{6 n^3}+\frac{1}{120 n^5}-\frac{1}{5040
n^7}+O\left(\left(\frac{1}{n}\right)^8\right)$$ $$n\sin(\frac 1n)=1-\frac{1}{6 n^2}+\frac{1}{120 n^4}-\frac{1}{5040
n^6}+O\left(\left(\frac{1}{n}\right)^8\right)$$ $$\log \left(n \sin \left(\frac{1}{n}\right)\right)=-\frac{1}{6 n^2}-\frac{1}{180 n^4}-\frac{1}{2835
n^6}+O\left(\left(\frac{1}{n}\right)^8\right)$$ Summing from $n=1$ to $\infty$, then $$S=\sum_{n=1}^\infty \ln\big(n\sin \frac{1}{n}\big)\approx -\frac 16 \sum_{n=1}^\infty\frac{1}{n^2}-\frac 1{180} \sum_{n=1}^\infty\frac{1}{n^4} -\frac 1{2835} \sum_{n=1}^\infty\frac{1}{n^6}$$ and the value of the successive sums are $\frac{\pi ^2}{6}$, $\frac{\pi ^4}{90}$, $\frac{\pi ^6}{945}$ which make $$S \approx -\frac{\pi ^2 \left(595350+1323 \pi ^2+8 \pi ^4\right)}{21432600}\approx -0.280527$$ while the numerical evaluation of the original sum is $-0.280556$.
Adding an extra term to the expansion would lead to $S \approx -0.280554$.
| {
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"url": "https://math.stackexchange.com/questions/1089232",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove: If $a^2+b^2=1$ and $c^2+d^2=1$, then $ac+bd\le1$ Prove: If $a^2+b^2=1$ and $c^2+d^2=1$, then $ac+bd\le1$
I seem to struggle with this simple proof. All I managed to find is that ac+bd=-4 (which might not even be correct).
| Here's another cool geometric approach if $a,b,c,d$ are positive. Consider the following diagram where $BD=1$ and is the diameter of the circle.
By Pythagoras, we have $a^2+b^2=1$ and $c^2+d^2=1$. However, by Ptolemy's Theorem, we also have $ac+bd=BD\cdot AC=AC$. Since $AC$ is any chord in a circle with diameter $1$, we have $AC\leq 1$. Thus, $ac+bd=AC\leq 1$ as desired. We can also find that equality occurs when $AC$ is also a diameter, which means $a=c$ and $b=d$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1089301",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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How do I integrate: $\int\sqrt{\frac{x-3}{2-x}} dx$? I need to solve:
$$\int\sqrt{\frac{x-3}{2-x}}~{\rm d}x$$
What I did is:
Substitute: $x=2\cos^2 \theta + 3\sin^2 \theta$. Now:
$$\begin{align}
x &= 2 - 2\sin^2 \theta + 3 \sin^2 \theta \\
x &= 2+ \sin^2 \theta \\
\sin \theta &= \sqrt{x-2} \\
\theta &=\sin^{-1}\sqrt{x-2}
\end{align}$$
and, $ cos \theta = \sqrt{(3-x)} $
$ \theta=\cos^{-1}\sqrt{(3-x)}$
The integral becomes:
$$\begin{align}
&= \int{\sqrt[]{\frac{2\cos^2 \theta + 3\sin^2 \theta-3}{2-2\cos^2 \theta - 3\sin^2 \theta}} ~~(2 \cos \theta\sin\theta)}~{\rm d}{\theta}\\
%
&= \int{\sqrt[]{\frac{2\cos^2 \theta + 3(\sin^2 \theta-1)}{2(1-\cos^2 \theta) - 3\sin^2 \theta}}~~(2 \cos \theta\sin\theta)}~{\rm d}{\theta} \\
%
&= \int\sqrt[]{\frac{2\cos^2 \theta - 3\cos^2 \theta}{2\sin^2 \theta - 3\sin^2 \theta}}~~(2 \cos \theta\sin\theta) ~{\rm d}\theta \\
%
&= \int\sqrt[]{\frac{-\cos^2 \theta }{- \sin^2 \theta}}~~(2 \cos \theta\sin\theta) ~{\rm d}\theta \\
%
&= \int \frac{\cos \theta}{\sin\theta}~~(2 \cos \theta\sin\theta)~{\rm d}\theta \\
%
&= \int 2\cos^2 \theta~{\rm d}\theta \\
%
&= \int (1- \sin 2\theta)~{\rm d}\theta \\
%
&= \theta - \frac {\cos 2\theta}{2} + c \\
%
&= \sin^{-1}\sqrt{x-2} - \frac {\cos 2(\sin^{-1}\sqrt{x-2})}{2} + c
\end{align}$$
But, The right answer is :
$$\sqrt{\frac{3-x}{x-2}} - \sin^{-1}\sqrt{3-x} + c $$
Where am I doing it wrong?
How do I get it to the correct answer??
UPDATE:
I am so sorry I wrote:
= $\int 2\cos^2 \theta .d\theta$
= $\int (1- \sin 2\theta) .d\theta$
It should be:
= $\int 2\cos^2 \theta .d\theta$
= $\int (1+ \cos2\theta) .d\theta$
= $ \theta + \frac{\sin 2\theta}{2} +c$
What do I do next??
UPDATE 2:
= $ \theta + \sin \theta \cos\theta +c$
= $ \theta + \sin \sin^{-1}\sqrt{(x-2)}. \cos\cos^{-1}\sqrt{(3-x)}+c$
= $ \sin^{-1}\sqrt{(x-2)}+ \sqrt{(x-2)}.\sqrt{(3-x)}+c$
Is this the right answer or I have done something wrong?
| $2(\cos x)^2=1+\cos(2x)$ but $2(\cos x)^2\neq1-\sin(2x)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1089410",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find the angle if the area of the two triangles are equal?
Let $I$ be the incenter of $\triangle ABC$, and $D$, $E$ be the midpoints of $AB$, $AC$ respectively. If $DI$ meets $AC$ at $H$ and $EI$ meets $AB$ at $G$, then find $\measuredangle A$ if the areas of $\triangle ABC$ and $\triangle AGH$ are equal.
I played around with GeoGebra and found out that it should be $60^{\circ}$, but am unable to prove it. I saw that $\cot \measuredangle HDA= \frac{a-b}{2r}$ and some stuff like that, but its not really helping me. Can anyone solve it, preferably by pure geometry? :)
| Here's my trigonometric endeavors:
Take $\alpha=\measuredangle AID$. Using the fact that $ID$ is the median of $AIB$, we have that $\cot \frac{A}2+\cot \alpha= \cot \frac{B}2+\cot (180^{\circ}-\alpha)$, or thus: $$\cot \alpha=\frac{\cot \frac{B}2 - \cot \frac{A}2}2=\frac{(s-b)-(s-a)}{2r}=\frac{a-b}{2r}=\frac{s(a-b)}{2[ABC]}$$
So we have:
$$8[ADH]=AH\cdot 2c\sin A=\frac{c^2}{\frac{s(a-b)}{[ABC]}+2\cot A}$$
Thus, using that $2c\sin A=\frac{4[ABC]}{b}$, we have:
$$AH=\frac{c^2}{2c\sin A\cdot(\frac{s(a-b)}{[ABC]}+2\cot A)}=\frac{c^2}{2[\frac{2s(a-b)}{b} +2c\cos A]}=\frac{c^2}{\frac{2[2s(a-b)+b^2+c^2-a^2]}{b}}$$
So: $$2[2s(a-b)+b^2+c^2-a^2]=2[(c+a+b)(a-b)+b^2+c^2-a^2]=2c[a-b+c]$$
Hence, $$\frac{AH}{AC}=\frac{c}{2[a-b+c]}$$
Wherein we can continue with Jack's solution .. but I have got a extra 2? :/
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Limit involving square roots, more than two "rooted" terms The limit is
$$\lim_{x\to\infty} \left(\sqrt{x^2+5x-2}-\sqrt{4x^2-3x+7}+\sqrt{x^2+7x+5}\right)$$
which has a value of $\dfrac{27}{4}$.
Normally, I would know how to approach a limit of the form
$$\lim_{x\to\infty}\left( \sqrt{a_1x^2+b_1x+c_1}\pm\sqrt{a_2x^2+b_2x+c_2}\right)$$
(provided it exists) by using the expression's conjugate, but this problem has me stumped.
I've considered using the conjugate
$$\sqrt{x^2+5x-2}-\sqrt{4x^2-3x+7}-\sqrt{x^2+7x+5}$$
and a term like this one,
$$\sqrt{x^2+5x-2}+\sqrt{4x^2-3x+7}-\sqrt{x^2+7x+5}$$
but that didn't seem to help simplify anything.
Edit: I stumbled across something at the last second that lets me use the conjugate approach. The expression can be rewritten as
$$\sqrt{x^2+5x-2}-\sqrt{4x^2-3x+7}+\sqrt{x^2+7x+5}\\
\sqrt{x^2+5x-2}-2\sqrt{x^2-\frac{3}{4}x+\frac{7}{4}}+\sqrt{x^2+7x+5}\\
\left(\sqrt{x^2+5x-2}-\sqrt{x^2-\frac{3}{4}x+\frac{7}{4}}\right)+\left(\sqrt{x^2+7x+5}-\sqrt{x^2-\frac{3}{4}x+\frac{7}{4}}\right)$$
which approaches
$$\frac{23}{8}+\frac{31}{8}=\frac{27}{4}$$
| $\lim_{x\to\infty} \left(\sqrt{x^2+5x-2}-\sqrt{4x^2-3x+7}+\sqrt{x^2+7x+5}\right) = 27/4 $
i will use the binomial theorem in the form
$(BIG + small)^{1/2} = BIG^{1/2} + {1 \over 2}BIG^{-1/2}small+\cdots $ to each.
$$\sqrt{x^2+5x-2} = x + {5x - 2 \over 2x} + \cdots \\
-\sqrt{4x^2-3x+7} = -\left( 2x + {-3x + 7 \over 4x} + \cdots \right) \\
\sqrt{x^2+7x+5} = x + {7x + 5 \over 2x} + \cdots \\
$$
adding the dominant terms the limit is $${5 \over 2} + {3 \over 4} + {7 \over 2} = {10 + 3 + 14\over 4} = {27 \over 4} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1090307",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Factorise a matrix using the factor theorem Can someone check this please?
$$
\begin{vmatrix}
x&y&z\\
x^2&y^2&z^2\\
x^3&y^3&z^3\\
\end{vmatrix}$$
$$C_2=C_2-C_1\implies\quad
\begin{vmatrix}
x&y-x&z\\
x^2&y^2-x^2&z^2\\
x^3&y^3-x^3&z^3\\
\end{vmatrix}$$
$$(y-x)
\begin{vmatrix}
x&1&z\\
x^2&y+x&z^2\\
x^3&y^2+xy+x^2&z^3\\
\end{vmatrix}$$
$$(y-x)(z-x)
\begin{vmatrix}
x&1&1\\
x^2&y+x&z+x\\
x^3&y^2+xy+x^2&z^2+xz+x^2\\
\end{vmatrix}$$
$$R_2=R_2-xR_1\implies\quad
(y-x)(z-x)
\begin{vmatrix}
x&1&1\\
0&y&z\\
x^3&y^2+xy+x^2&z^2+xz+x^2\\
\end{vmatrix}$$
$$R_3=R_3-x^2R_1\implies\quad
(y-x)(z-x)
\begin{vmatrix}
x&1&1\\
0&y&z\\
0&y^2+xy&z^2+xz\\
\end{vmatrix}$$
factor $x$$$\implies\quad
x(y-x)(z-x)
\begin{vmatrix}
1&1&1\\
0&y&z\\
0&y^2+xy&z^2+xz\\
\end{vmatrix}$$
$$\implies\quad x(y-x)(z-x)(yz^2-zy^2)$$
$$\implies\quad xyz(y-x)(z-x)(z-y)$$
Also I'd like practical tips on using the factor theorem for these types of questions. My understanding is that the determinant is $f(x,y,z)$ so if we hold $y$ and $z$ constant we could apply it somehow to $f(x)$ alone. I'm not that great spotting difference of squares etc and want a more fail safe alternative. Thanks in advance.
| you are not factoring the matrix. you are using the properties of the determinants to simplify. for example, you could write your first step as
$$AE = \pmatrix{x & y & z \\ x^2 & y^2 & z^2 \\ x^3 & y^3 & z^3}
\pmatrix{1 & -1 & 0\\0&1&0\\0&0&1} =
\pmatrix{x & y-x & z \\ x^2 & y^2-x^2 & z^2 \\ x^3 & y^3 -x^3& z^3} = B
$$
the $E$ matrices are called elementary column matrices and their determinants usually is the product of the diagonals. now use the product rule of the determinants $det(AE) = det(A) det(E)$ to conclude that $det(A) = det(B)$ now you start with $B$ and do further reductions.
| {
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"timestamp": "2023-03-29T00:00:00",
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Given $a+b+c$, Can I calculate $a^2+b^2+c^2$? I want to calculate $a^2 + b^2 + c^2$ when I am given $a+b+c$.
It is known that a,b,c are positive integers.
Is there any way to find that.
| No. Suppose that $a+b+c = 6$.
If $a = b = c = 2$, then $a^2 + b^2 + c^2 = 12$.
If $a = 4$ and $b = c = 1$, then $a^2 + b^2 + c^2 = 18$
The sum of the squares varies for a given sum of $a$, $b$, and $c$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1091013",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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a way to integrate: $\int (\sqrt{x} +3)/(2+ x^ \frac{1}{3}) dx$ Im looking for a way to integrate:
$$
\int \frac{ \sqrt{x} +3}{2+ x^ \frac{1}{3}} dx
$$
that would make it efficient and not too difficult...
Any suggestions?
| The substitution $x = t^6$ works. The integral becomes
$$ \int \frac{t^3 + 3}{t^2 + 2} \,6t^5\,dt = 6 \int \frac{t^8 + 3t^5}{t^2 + 2} dt = 6\int \left( t^6 - 2t^4 + 3t^3 + 4t^2 - 6t - 8 + \frac{12t+16}{t^2+2} \right) \, dt $$
You can take it from here
| {
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"timestamp": "2023-03-29T00:00:00",
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Sum: $\sum_{n=1}^\infty\prod_{k=1}^n\frac{k}{k+a}=\frac{1}{a-1}$ For the past week, I've been mulling over this Math.SE question. The question was just to prove convergence of
$$\sum\limits_{n=1}^\infty\frac{n!}{\left(1+\sqrt{2}\right)\left(2+\sqrt{2}\right)\cdots\left(n+\sqrt{2}\right)}$$
but amazingly Mathematica told me it had a remarkably simple closed form: just $1+\sqrt{2}$. After some fiddling, I conjectured for $a>1$:
$$\sum\limits_{n=1}^\infty\prod\limits_{k=1}^n\frac{k}{k+a}=\sum\limits_{n=1}^\infty\frac{n!}{(1+a)(2+a)\cdots(n+a)}=\frac{1}{a-1}$$
I had been quite stuck until today when I saw David H's helpful answer to a similar problem. I have included a solution using the same idea, but I would be interested to know if anyone has another method.
| Here is a completely elementary proof, which only needs introductory calculus concepts:
$$a_{n+1} = \frac{(n+1)!}{(a+1)(a+2)\dots(a+n+1)} = \frac{1}{a-1}\left(\frac{(n+1)!}{(a+1)(a+2) \dots (a+n)} -\frac{(n+2)!}{(a+1)(a+2) \dots (a+n+1)}\right) = b_{n+1} - b_{n+2}$$
where $a_n$ is the term of our series and $$b_n = \frac{1}{a-1}\left(\frac{n!}{(a+1)(a+2) \dots (a+n-1)}\right)$$ with $b_1 = \frac{1}{a-1}$
Thus the sum we seek, telescopes!
Giving us
$$\sum_{k=1}^{n} a_k = \sum_{k=1}^{n} (b_{k} - b_{k+1}) = b_1 - b_{n+1}$$
Thus we just need to compute $\lim b_n$
(we basically need to show that $b_n \to 0$ to match the limit being $\frac{1}{a-1}$)
Now we have that $a \gt 1$, so let $a = 1 + x$ for $x \gt 0$.
$$ (a-1)b_n =\frac{n!}{(a+1)(a+2) \dots (a+n-1)} = \frac{1}{(1 + x/2)(1+x/3)\dots(1+x/n)} $$
Let $ M = \lceil x \rceil$ and consider the product
$$p_n = \prod_{k=M}^{n} \left(1 + \frac{x}{k}\right)$$
It is enough to show that $\log p_n \to \infty$ (that proves that $b_n \to 0$).
Now we have that $\dfrac{1}{1+t} \gt 1-t$ for $0 \lt t \le 1$ and thus integrating between $0$ and $y$ (where $y \le 1$) we get that
$$ \log(1+y) \ge y - \dfrac{y^2}{2}$$
Now $$\log p_n = \sum_{k=M}^{n} \log (1 + \frac{x}{k})$$
$$ \ge \sum_{k=M}^{n} (\frac{x}{k} - \frac{x^2}{2k^2})$$
This goes to $\infty$ as the harmonic series diverges, and the sum of reciprocals of the squares converges.
Thus the sum of your sequence is $$b_1 = \frac{1}{a-1}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Area of the overlap between a triangle and a square $ABC$ is an equilateral triangle, each side has length 4.
$M$ is the midpoint of $\overline{BC}$, and $\overline{AM}$ is a diagonal of square $ALMN$. Find the area of the region common to both $ABC$ and $ALMN$.
I'm not sure how to solve this problem, or even where to start. Any hints?
| Use this diagram.
The equilateral triangle $ABC$ has sides of length $4$, so $CM=2$ and $AM=2\sqrt 3$.
The area you are trying to find is shaded and is clearly twice the area of triangle $AEM$. We know the base, $AM$, so we want the height $EF$. If we let $x=EF$ then $MF=x$ due to $45-45-90$ triangle $EFM$. We also have $AF=\sqrt 3x$ due to $30-60-90$ triangle $AEF$. Therefore we get
$$AF+MF=AM$$
$$\sqrt 3x+x=2\sqrt 3$$
$$x=\frac{2\sqrt 3}{\sqrt 3+1}$$
$$=\frac{2\sqrt 3(\sqrt 3-1)}{(\sqrt 3+1)(\sqrt 3-1)}$$
$$=\frac{6-2\sqrt 3}{2}$$
$$=3-\sqrt 3$$
Therefore the area we want is twice the area of triangle $AEM$:
$$Area=2\cdot \frac 12bh$$
$$=2\cdot \frac 12 \cdot 2\sqrt 3 \cdot (3-\sqrt 3)$$
$$=6\sqrt 3-6$$
| {
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"timestamp": "2023-03-29T00:00:00",
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$\lim_{n \rightarrow \infty} \frac{1-(1-1/n)^4}{1-(1-1/n)^3}$ Find
$$\lim_{n \rightarrow \infty} \dfrac{1-\left(1-\dfrac{1}{n}\right)^4}{1-\left(1-\dfrac{1}{n}\right)^3}$$
I can't figure out why the limit is equal to $\dfrac{4}{3}$ because I take the limit of a quotient to be the quotient of their limits.
I'm taking that $\lim_{n \rightarrow \infty}1-\left(1-\frac{1}{n}\right)^4 = 0$ and likewise that $\lim_{n \rightarrow \infty}1-\left(1-\frac{1}{n}\right)^3 = 0$, which still gives me that the limit should be 0.
| Hint: $\dfrac{1-x^4}{1-x^3}= \dfrac{(1-x)(1 + x + x^2 + x^3))}{(1-x)(1+x+x^2)} = \dfrac{1 + x + x^2 + x^3}{1 + x+ x^2}$ (provided $x \neq 1$.)
Now put $x = 1 - \frac{1}{n}$ and take $n \to \infty.$
| {
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"timestamp": "2023-03-29T00:00:00",
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How many divisors of N ended by 5 I must know how to find how many divisor of N ended by 5 ? In my exercise, I have $\ N=63'000 = 2^3*3^2*5^3*7 $ and I can found the number of divisors of N using $\ (3+1)*(2+1)*(3+1)*(1+1)=96$
Among these 96 divisors, how many ended by 5 ? How can calculate this ?
Thank you so much
| *
*Take out the $2$s, because when multiplied by $5$, the result will end with $0$.
*Find the number of divisors of $3^2\cdot7^1$, which is $(2+1)\cdot(1+1)=6$:
*
*$3^0\cdot7^0$
*$3^1\cdot7^0$
*$3^2\cdot7^0$
*$3^0\cdot7^1$
*$3^1\cdot7^1$
*$3^2\cdot7^1$
*Multiply each divisor by each one of the following $3$ powers of $5$:
*
*$5^1$
*$5^2$
*$5^3$
*Hence you have $6\cdot3=18$ divisors which end with $5$:
*
*$3^0\cdot7^0\cdot5^1$
*$3^1\cdot7^0\cdot5^1$
*$3^2\cdot7^0\cdot5^1$
*$3^0\cdot7^1\cdot5^1$
*$3^1\cdot7^1\cdot5^1$
*$3^2\cdot7^1\cdot5^1$
*$3^0\cdot7^0\cdot5^2$
*$3^1\cdot7^0\cdot5^2$
*$3^2\cdot7^0\cdot5^2$
*$3^0\cdot7^1\cdot5^2$
*$3^1\cdot7^1\cdot5^2$
*$3^2\cdot7^1\cdot5^2$
*$3^0\cdot7^0\cdot5^3$
*$3^1\cdot7^0\cdot5^3$
*$3^2\cdot7^0\cdot5^3$
*$3^0\cdot7^1\cdot5^3$
*$3^1\cdot7^1\cdot5^3$
*$3^2\cdot7^1\cdot5^3$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1102497",
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} |
Linear Combinations of Fibonacci Numbers (integer coefficients) While working on problem #2 on Project Euler, I came across the need to express $F_n$ as a linear combination of $F_{n-3}$ and $F_{n-6}$. This is relatively simple to do:
$$\begin{align} F_n &= F_{n-1}+F_{n-2}\\ &= F_{n-1}+F_{n-3}+F_{n-4}\\ &= F_{n-1}+F_{n-3}+F_{n-5}+F_{n-6}\\&= F_{n-2}+2F_{n-3}+F_{n-5}+F_{n-6}\\&= 3F_{n-3}+F_{n-4}+F_{n-5}+F_{n-6}\\&=4F_{n-3}+F_{n-6}\end{align}$$
This argument is ad hoc to an extreme, and it made me wonder about a more general conjecture:
Conjecture. Let $a,b<n$ and $a\neq b$. Then $F_n = \lambda F_{n-a} + \kappa F_{n-b}$ for some $\lambda,\kappa\in\mathbb Z$.
Is this true? If so, how can it be proven? If not, can we include some hypotheses on $a$ and $b$ that make it true?
| Note that we can't always do this with integer coefficients. For example,
$$
F_{n}=\frac52F_{n-2}+\frac12F_{n-5}\tag{1}
$$
and
$$
F_{n}=\frac{13}3F_{n-3}-\frac23F_{n-7}\tag{2}
$$
We can use the fact that
$$
\left(\frac{1\pm\sqrt5}2\right)^n=\frac1{2^n}\sum_{k=0}^{\lfloor n/2\rfloor}\binom{n}{2k}5^k\pm\frac{\sqrt5}{2^n}\sum_{k=0}^{\lfloor(n-1)/2\rfloor}\binom{n}{2k+1}5^k\tag{3}
$$
to get
$$
\begin{align}
F_n=
&\frac{\sum\limits_{k=0}^{\lfloor(b-1)/2\rfloor}\binom{b}{2k+1}5^k}
{\sum\limits_{k=0}^{\lfloor(b-a-1)/2\rfloor}\binom{b-a}{2k+1}5^k}\frac{F_{n-a}}{2^a}\\
&+\left[\sum_{k=0}^{\lfloor b/2\rfloor}\binom{b}{2k}5^k
-\frac{\sum\limits_{k=0}^{\lfloor(b-1)/2\rfloor}\binom{b}{2k+1}5^k}
{\sum\limits_{k=0}^{\lfloor(b-a-1)/2\rfloor}\binom{b-a}{2k+1}5^k}
\sum_{k=0}^{\lfloor(b-a)/2\rfloor}\binom{b-a}{2k}5^k\right]\frac{F_{n-b}}{2^b}\tag{4}
\end{align}
$$
Thus, there is always a recurrence with rational coefficients for any $0\lt a\lt b$.
Note that if we let $\psi=-1/\phi$, then both $\phi$ and $\psi$ satisfy
$$
\begin{align}
0
&=(x^n-\phi^n)(x^n-\psi^n)\\
&=x^{2n}-(\phi^n+\psi^n)x^n+(\phi\psi)^n\\
&=x^{2n}-L_nx^n+(-1)^n\tag{5}
\end{align}
$$
where $L_n$ is a Lucas Number. Therefore, the Fibonacci numbers satisfy
$$
F_n=L_kF_{n-k}-(-1)^kF_{n-2k}\tag{6}
$$
Fix $k$ and let $a_j=jk$ and $b_j=(j+1)k$. Equation $(6)$ has integer coefficients for $a_1,b_1$.
Equation $(6)$ says that if we have coefficients $\lambda_j,\kappa_j\in\mathbb{Z}$ for $a_j,b_j$, then
$$
\begin{align}
F_n
&=\lambda_jF_{n-jk}+\kappa_jF_{n-(j+1)k}\\
&=(\lambda_jL_k+\kappa_j)F_{n-(j+1)k}-(-1)^k\lambda_jF_{n-(j+2)k}\\
&=\lambda_{j+1}F_{n-(j+1)k}+\kappa_{j+1}F_{n-(j+2)k}\tag{7}
\end{align}
$$
where $\lambda_{j+1}=\lambda_jL_k+\kappa_j$ and $\kappa_{j+1}=(-1)^{k+1}\lambda_j$ are both integers for $a_{j+1},b_{j+1}$.
Note that $b_j=(j+1)k=(j+1)(b_j-a_j)$.
Using $(6)$ and $(7)$, we get a recurrence with integer coefficients if $b-a\mid b$.
In particular, given $k=b-a$ and $j=\frac{b}{b-a}-1$, we have
$$
\begin{bmatrix}\lambda\\\kappa\end{bmatrix}
=\begin{bmatrix}L_k&1\\(-1)^{k+1}&0\end{bmatrix}^j
\begin{bmatrix}1\\0\end{bmatrix}\tag{8}
$$
Since $\small\begin{bmatrix}2&1\\-1&-1\end{bmatrix}^2=\begin{bmatrix}3&1\\-1&0\end{bmatrix}$, we can apply $(8)$ even if $b-a=2$ when $b$ is odd. We deal with this in the next section.
As noted by achille hui, $b-a=2$ also allows $\lambda,\kappa\in\mathbb{Z}$. This follows from the case $b-a=1$.
If we apply $(8)$ to the case $a=b-1$, we get
$$
\begin{align}
F_n
&=F_b F_{n-b+1}+F_{b-1}F_{n-b}\\
&=F_b(F_{n-b+2}-F_{n-b})+F_{b-1}F_{n-b}\\
&=F_b F_{n-b+2}+(F_{b-1}-F_b)F_{n-b}\\
&=F_b F_{n-b+2}-F_{b-2}F_{n-b}\tag{9}
\end{align}
$$
Thus, for $a=b-2$,
$$
\begin{bmatrix}\lambda\\\kappa\end{bmatrix}
=\begin{bmatrix}F_b\\-F_{b-2}\end{bmatrix}\tag{10}
$$
Conclusion: The Conjecture, as stated, is false. However, if $b-a\mid b$ or $b-a=2$, then there are $\lambda,\kappa\in\mathbb{Z}$, given in $(8)$ or $(10)$, so that
$$
F_n=\lambda F_{n-a}+\kappa F_{n-b}\tag{11}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1103264",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 4,
"answer_id": 0
} |
Basic exercise with exponents and radicals I'm trying to solve a simple high school algebra problem, I would like to know if my result is correct.
Convert the radicals into exponents, solve and then express the result as a radical
$2\sqrt[3]{2}:\sqrt{(\dfrac{1}{2}\sqrt[5]{4})^{\frac{-1}{3}})}$
If I call $A$ to the numerator and $B$ to the denominator, then we have $$B=((\dfrac{1}{2}\sqrt[5]{4})^{\frac{-1}{3}})^{\frac{1}{2}}$$$$=(\dfrac{1}{2}\sqrt[5]{4})^{\frac{-1}{6}}$$$$=(\dfrac{1}{2})^{\frac{-1}{6}}(2^{\frac{2}{5}})^{\frac{-1}{6}}$$$$=2^{\frac{1}{6}}2^{\frac{-1}{15}}.$$
Now, using this result,we have $$\dfrac{A}{B}=\dfrac{2.2^{\frac{1}{3}}.2^{\frac{1}{15}}}{2^{\frac{1}{6}}}$$$$=2^{1-\frac{1}{6}}.2^{\frac{1}{3}}.2^{\frac{1}{15}}$$$$=2^{\frac{37}{30}}$$$$=\sqrt[30]{2^{37}}.$$
Is my solution correct? Thanks in advance.
| You could simplify the result to $2\sqrt[30]{2^7}$.
| {
"language": "en",
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"answer_count": 1,
"answer_id": 0
} |
Proving $\sin^2A \equiv \cos^2B + \cos^2C + 2 \cos A\cos B\cos C$ As the title,
By considering $\bigtriangleup$ABC, Prove
$$\sin^2A \equiv \cos^2B + \cos^2C + 2 \cos A\cos B\cos C$$
Thanks
| Assume that $A$ is the largest angle (the modifications when this is not the case should be minor). Drop a perpendicular from $A$ to the side $BC$. Let $A'$ and $A''$ be the angles formed by dividing $A$ with this perpendicular, where $A'$ is adjacent to $B$ and $A''$ is adjacent to $C$. The trigonometry of right triangles states that $$\sin A' = \cos B,\ \sin A'' = \cos C,\ \cos A' = \sin B,\ \cos A'' = \sin C.$$ Since $A = A + A''$ (in terms of angle measure) you have
$$ \sin A = \sin(A' + A'') = \sin A' \cos A'' + \cos A' \sin A'' = \cos B \sin C + \sin B \cos C,$$and thus $$\sin^2 A = \cos^2 B \sin^2 C + 2 \sin B \cos B \sin C \cos C + \sin^2 B \cos^2 C.$$
Next write $\sin^2 B = 1 - \cos^2 B$ and $\sin^2 C = 1 - \cos^2 C$ to arrive at
\begin{align*}
\sin^2 A &= \cos^2 B + \cos^2 C - 2 \cos^2B \cos^2 C + 2 \sin B \cos B \sin C \cos C \\
&= \cos^2 B + \cos^2 C + 2 (\sin B \cos B \sin C \cos C - \cos^2B \cos^2 C) \\
&= \cos^2 B + \cos^2 C + 2 \cos B \cos C (\sin B \sin C - \cos B \cos C)
\end{align*}
Finally use the fact that
$$\cos A = \cos A' \cos A'' - \sin A \sin A'' = \sin B \sin C - \cos B \cos C$$
to get
$$\sin^2 A = \cos^2 B + \cos^2 C + 2 \cos A\cos B \cos C.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Considering $ (1+i)^n - (1 - i)^n $, Complex Analysis I have been working on problems from Complex Analysis by Ahlfors, and I got stuck in the following problem:
Evaluate:
$$
(1 + i)^n - (1-i)^n
$$
I have just "reduced" to:
$$
(1 + i)^n - (1-i)^n = \sum_{k=0} ^n i^k(1 - (-1)^k)
$$
by using expansion of each term.
Thanks.
| ok. we can use the binomial theorem. we will check out for small values of $n.$
case $n = 1, \ z_1 = 1+ i -(1-i) = 2i$
case $n = 2, \ z_2 = (1+i)^2 - (1-i)^2 =(1 + 2i + i^2) -(1 - 2i + i^2) = 2(2i) = 4i $
case $n = 3, \ z_2 = (1+i)^3 - (1-i)^3 =(1 + 3i + 3i^2 + i^3) -(1 - 3i + 3i^2 - i^3) = 2(3i +i^3) = 4i$
case $n = 4, \ z_4 = (1 + i)^4 - (1 - i)^4 =2(4i + 4i^3) = 0$
case $n = 5, \
z_5 = (1 + i)^5 - (1 - i)^5 =2(5i + 10i^3 + i^5) = -8i $
case $n = 6, \
z_6 = (1 + i)^6 - (1 - i)^6 =2(6i + 20i^3 + 6i^5) = -16i$
case $n = 7, \
z_7 = (1 + i)^7 - (1 - i)^7 =2(7i + 35i^3 + 21i^5 + i^7) = -16i$
case $n = 8, \
z_8 = (1 + i)^8 - (1 - i)^8 =2(8i + 56i^3 + 56i^5 + 8i^7) = 0$
so it looks like the formula can be put in four classes:
if $n$ is a multiple of $4,$ then $z_n = 0$
if $n = 4k+1,$ then $z_n = (2i)^{(n+1)/2}$
if $n = 4k+2, 4k+3,$ then $z_n = 2(2i)^{(n+1)/2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1108585",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
} |
Solving Diophantine equation $1/x^2+1/y^2=1/z^2$ How can we find positive integers solutions $(x,y,z)$, where $\gcd(x,y,z)=1$ for the equation:
$$1/x^2+1/y^2=1/z^2$$
Can we conclude that $x$ and $y$ are not coprimes for it to have solution?
| Multiply both sides by $x^2y^2z^2$.
Then you get
$$y^2z^2+x^2z^2=x^2y^2$$
Now use that each of $x^2, y^2,z^2$ divide two of the terms hence the third.
Added Here is the rest of the solution. Let $a=gcd(x,y), b=gcd(x,z), c=gcd(y,z)$.
Then, $gcd(a,b)=1$ and hence $ab|x$. We claim $ab=x$.
Indeed write $x=abd$. Assume by contradiction that $d \neq 1$ and let $p|d$, $p$ prime.
As $x | yz$ we have $abd | yz \Rightarrow d | \frac{y}{a}\frac{z}{b}$.
Then $p$ divides either $\frac{y}{a}$ or $\frac{z}{b}$.
But then, in the first case $pa |x,y$ while in the second $pb | x,z$ contradicting the $gcd$.
Therefore $x=ab$. The same way you can prove that $y=ac, z=bc$.
Replacing in the above equation you get
$$a^2b^2c^4+a^2b^4c^2=a^4b^2c^2$$
or
$$c^2+b^2=a^2$$
this shows that $(c,b,a)$ is a primitive Pytagoreal triple and
$$x=ab \\
y=ac \\
z=bc$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1109505",
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"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
prove the given inequality (for series ) For any given $n \in \Bbb N,$ prove that, $$1+{1\over 2^3}+\cdots+{1\over n^3} <{3\over 2}.$$
| For $n\ge 3$ the sum is less than
$$1+\frac{1}{8}+\frac{1}{3}\left(\frac{1}{(2)(3)}+\frac{1}{(3)(4)}+\frac{1}{(4)(5)}+\cdots+\frac{1}{(n-1)(n)}\right).$$
Note that $\frac{1}{(2)(3)}=\frac{1}{2}-\frac{1}{3}$ and $\frac{1}{(3)(4)}=\frac{1}{3}-\frac{1}{4}$ and $\frac{1}{(4)(5)}=\frac{1}{4}-\frac{1}{5}$ and so on. It follows by cancellation (telescoping) that
$$\frac{1}{(2)(3)}+\frac{1}{(3)(4)}+\frac{1}{(4)(5)}+\cdots+\frac{1}{(n-1)(n)}\lt \frac{1}{2}.$$
Thus our full sum is less than $1+\frac{1}{8}+\frac{1}{6}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1110138",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
False proof of 0=1 using Laurent series I found the following proof that 0 = 1:
\begin{align*}
\sum_{n=-\infty}^{\infty} 0\cdot z^n = 0 = \frac{1}{z-1} + \frac{1}{1-z} = \frac{1}{z}\frac{1}{1-\frac{1}{z}} + \frac{1}{1-z} \\
= \frac{1}{z} \sum_{n=0}^{\infty}\frac{1}{z^n} + \sum_{n=0}^{\infty}z^n = \sum_{n=-\infty}^{\infty} 1\cdot z^n
\end{align*}
from which it follows that 0 = 1. In every step, I can't find an error. If we take $f(z) = \frac{1}{z-1} + \frac{1}{1-z}$, then it is holomorfic on the disk around 0 with radius 1 (B(0,1)).The principal part $h(z)$ would be $\frac{1}{z-1}$, which is also holomorfic on $\mathbb{C}\backslash B[0,1]$ and also $h(z)$ tends to 0 if $|z|$ tends to $\infty$. So I can't see what is going wrong.
| The equality $\dfrac{1}{z}\dfrac{1}{1-\frac{1}{z}} + \dfrac{1}{1-z}
= \dfrac{1}{z} \sum \limits_{n=0}^{\infty}\dfrac{1}{z^n} + \sum \limits_{n=0}^{\infty}z^n$ only holds if $z\not \in \{0,1\}$, $\left|\frac 1 z\right|<1$ and $|z|<1$, i.e., never.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1110326",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Matrix gauss continuation Suppose we have the matrix $4\times4$, $R_1 =(0,0,1,1)$, $R_2=(0,1,1,0)$, $R_3=(1,1,0,0)$, $R_4(1,0,0,0)$ and suppose we want to see the rank.The difficult part is that you cannot order them somehow to use gauss.Any ideas what should be done in such situations?
| You can subtract any row from any other without changing the rank:
$$\begin{pmatrix}0&0&1&1\\0&1&1&0\\1&1&0&0\\1&0&0&0\end{pmatrix}\sim\begin{pmatrix}0&0&1&1\\0&1&1&0\\0&1&0&0\\1&0&0&0\end{pmatrix}\sim\begin{pmatrix}0&0&1&1\\0&0&1&0\\0&1&0&0\\1&0&0&0\end{pmatrix}\sim\begin{pmatrix}0&0&0&1\\0&0&1&0\\0&1&0&0\\1&0&0&0\end{pmatrix}$$
So the rank is full.
You can also reorder the rows at the beginning as this doesn't change the rank:
$$\begin{pmatrix}1&0&0&0\\1&1&0&0\\0&1&1&0\\0&0&1&1\end{pmatrix}$$
and continue as before.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1112627",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Is this divisibility problem correct? Let $n$ be a natural number and let $1 \le a_1<a_2<...<(a_k=n)$ be all of its divisors.
Find all $n$ such that $a_2^3+a_3^2-15=n$ .
It seems impossible to find all such numbers.
| We must have $a_2 = p$, where $p$ is the smallest prime dividing $n$. Then $a_3$ must be either the next smallest prime $q$ dividing $n$, or $p^2$. But if we had $a_3 = p^2$, then from $p^2 | n$ we would obtain $p^2 | 15$, which is absurd. Therefore $a_3 = q$, and moreover $p < q$.
Thus we obtain $p^3 + q^2 - 15 = n = cpq$ for some integer $c$. The number $c$ is a divisor of $n$, hence either $c = 1$, $c = p$, or $c \geq q$.
If $c = 1$, then from $pq < q^2$, we deduce $p^3 < 15$, hence $p = 2$. In this case, $q$ must satisfy $q^2 - 2q - 7 = 0$, which is absurd.
Now assume $c \geq q$. Then we must also have $q \geq p + 2$ (since the only alternative is $p=2$, $q = 3$, and this is trivially impossible). Then
$$p^3 - 15 = cpq - q^2 \geq q^2(p-1) \geq (p + 2)^2(p-1),$$
and this inequality in $p$ has no solutions. Thus we conclude that $c = p$, hence
$$f(q) = q^2 - p^2q + p^3 - 15 = 0.$$
It is straightforward to check that the cases $p = 2,3$ are impossible, so we have $p > 3$.
The quadratic function $f(q)$ has its minimum at $q = p^2/2$, a half-integer. Moreover $f(p) = p^2 - 15 > 0$ and $f(p + 2) = -p^2 + 4p - 11 < 0$. Therefore $f(q)$ has a root between $p$ and $p + 2$ and another (symmetric) one strictly between $p^2 - p - 2$ and $p^2 - p$. Thus the only possibilities for $q$ are $q = p + 1$ and $q = p^2 - p - 1$, and we must have $2p - 14 = f(p + 1) = 0$, hence $p = 7$. Then either $q = 8$, which is absurd, or $q = 41$. Thus the only possibility is $n = p^2 q = 7^2 \cdot 41 = 2009$.
Conversely, it is clear that $n = 2009$ works.
| {
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"url": "https://math.stackexchange.com/questions/1113260",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Fredholm integral equation How can I solve the following fredholm integral equation
$$ψ(x)=x+λ\int_{0}^{2π}|x-t|ψ(t)dt$$
The kernel contains absolute value
| First we express it in the form
\begin{align}
(I-\lambda K)\psi = x
\end{align}
where
\begin{align}
K\psi(x):= \int_{0}^{2\pi} |x-t|\psi(t)\, \mathrm{d}t
\end{align}
Existence and uniqueness is guaranteed if $|\lambda| < \frac{1}{\left\|K\right\|}$, so assuming $K\colon C[0,2\pi] \to C[0,2\pi]$ we can estimate
\begin{align}
\left\|K\psi\right\|_{\infty}
&\le \max_{x\in [0,2\pi]} \int_{0}^{2\pi} |x-t| \, \mathrm{d}t \left\|\psi\right\|_{\infty} \\
&= \max_{x\in [0,2\pi]} (2\pi^2 - 2 \pi x + x^2) \left\|\psi\right\|_{\infty} \\
&= 2\pi^{2}\left\|\psi\right\|_{\infty}
\end{align}
So assuming $\lambda \in \left(-\frac{1}{2\pi^{2}}, \frac{1}{2\pi^{2}}\right)$, we have convergence of the Neumann series
\begin{align}
(I-\lambda K)^{-1} = \sum_{i=0}^{\infty} \lambda^{i}K^{i}.
\end{align}
If $f(x):=x$, then
\begin{align}
\psi
&= \sum_{i=0}^{\infty} \lambda^{i}K^{i}f \\
&= f + \lambda K[f] + \lambda^{2}K^{2}[f] + \cdots
\end{align}
Let's work out a couple of these integrals just for fun:
\begin{align}
K[f](x) &= \int_{0}^{2\pi} |x-t|t \, \mathrm{d}t \\
&= \frac{1}{3} (8\pi^{3} -6\pi^2x + x^{3})
\end{align}
And:
\begin{align}
K^{2}[f](x) &= \int_{0}^{2\pi} |x-t|\left( \frac{1}{3} (8\pi^{3} -6\pi^2t + t^{3}) \right) \, \mathrm{d}t \\
&=
\frac{1}{3} \left(x^5-2 \pi x^4-4 \pi ^2 x^3+20 \pi ^3 x^2-28 \pi ^4 x+16 \pi ^5\right)
\end{align}
To avoid pain in the ass, I calculated these terms via the following Mathematica code:
Assuming[x > 0 && x < 2 \[Pi] , Integrate[Abs[x - t]*t, {t, 0, 2*\[Pi]}]]
For $\lambda = \frac{1}{3\pi^{2}}$, the partial sums up to $\mathcal{O}(\lambda^{6})$ are shown below:
| {
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"source": "stackexchange",
"question_score": "2",
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} |
Limit of logarithms exponential $$
\lim_{x\to\infty}\biggl(\frac{\ln(x-2)}{\ln(x-1)}\biggr)^{x\ln x}.
$$
L'Hopital seems like a very hardcore solutions given the situation.Are the any other options?
| Let $$f(x) = \biggl(\frac{\ln(x-2)}{\ln(x-1)}\biggr)^{x\ln x}.$$
Then
$$
\begin{align*}
\ln f(x) &= x\ln x[\ln \ln (x-2) - \ln \ln (x-1)] \\
&= x \ln x\left[\ln \left(1 + \frac{\ln(1 - 2/x)}{\ln x}\right) - \ln \left(1 + \frac{\ln(1 - 1/x)}{\ln x}\right) \right]\\
&= x \ln x\left[\ln \left(1 - \frac{2}{x \ln x} + o \left(\frac{1}{x\ln x} \right)\right) - \ln \left(1 - \frac{1}{x \ln x} + o \left(\frac{1}{x\ln x} \right)\right)\right] \\
&= x\ln x \left[- \frac{2}{x \ln x} + \frac{1}{x \ln x} + o \left(\frac{1}{x\ln x} \right) \right] \\
&= -1 + o(1).
\end{align*}
$$
Therefore $\lim_{x \to +\infty} f(x) = e^{-1}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1116106",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
How prove $\sin \left( \alpha+\frac{\pi }{n} \right) \cdots \sin \left( \alpha+\frac{n\pi }{n} \right) =-\frac{\sin n\alpha}{2^{n-1}}$? How prove
$$\prod_{k=1}^{n}\sin \left( \alpha+\frac{\pi k }{n}\right) =-\frac{\sin n\alpha}{2^{n-1}}$$
for $n \in N$?
| Since, $\sin \theta = \dfrac{e^{i\theta}-e^{-i\theta}}{2i}$,
Hence, $$\begin{align}\prod_{j=1}^{n-1}\sin \left(\alpha+\frac{j\pi}{n}\right)&=\prod_{j=1}^{n-1}\left(\frac{e^{i(\alpha+j\pi/n)}-e^{-i(\alpha+j\pi/n)}}{2i}\right) \\&= \left(\frac{-1}{2i}\right)^{n-1}e^{-((n-1)\alpha + i\frac{(n-1)\pi}{2})}\prod_{j=1}^{n-1} \left(1-e^{2i\alpha+2ij\pi/n}\right) \tag{1} \\&= \left(\frac{-1}{2i}\right)^{n-1}e^{-((n-1)\alpha + i\frac{(n-1)\pi}{2})}\sum_{j=0}^{n-1} e^{2ij\alpha} \tag{2} \\ &= \left(\frac{-1}{2i}\right)^{n-1}e^{-((n-1)\alpha + i\frac{(n-1)\pi}{2})}.\frac{e^{2in\alpha}-1}{e^{2i\alpha}-1} \tag{3}\\&= \left(\frac{-1}{2i}\right)^{n-1}e^{-i\frac{(n-1)\pi}{2}} \frac{\sin n\alpha}{\sin \alpha} \tag{4} \\ &= \frac{\sin n\alpha}{2^{n-1}\sin \alpha}\end{align}$$
Explanations:
$(1)$ Factored out $\displaystyle \prod\limits_{j=1}^{n-1}e^{-i(\alpha+j\pi/n)}$ from the product. $$\displaystyle \prod\limits_{j=1}^{n-1}e^{-i(\alpha+j\pi/n)} = \exp\left(\frac{i\pi}{n}\sum\limits_{j=1}^{n-1} j\right) = e^{\frac{i(n-1)\pi}{2}}$$
$(2)$ Used the fact that $\displaystyle 1+z+\cdots + z^{n-1} = \prod\limits_{j=1}^{n-1} (1-z\omega^j)$, where, $\omega = e^{2i\pi/n}$ is the $n^{th}$ root of unity. Here, $z = e^{2i\alpha}$
$(3)$ $\displaystyle 1+z+\cdots + z^{n-1} = \frac{z^n-1}{z-1}$
$(4)$ $\displaystyle \frac{e^{2in\alpha}-1}{e^{2i\alpha}-1} = e^{i(n-1)\alpha}\frac{e^{in\alpha} - e^{-in\alpha}}{e^{i\alpha} - e^{i\alpha}} = e^{i(n-1)\alpha}\frac{\sin n\alpha}{\sin \alpha}$ and $i^{n-1} = e^{\frac{i(n-1)\pi}{2}}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1118062",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 0
} |
Markov chain doesn't sum up to 1
Let $\{X_n\}$ be a Markov chain on $S=\{1,2,3,4,5,6\}$ with the matrix
suppose we define a new sequence $\{Y_n\}$ by $$Y_n=\cases{1\quad X_n=1\vee X_n=2\\2\quad X_n=3\vee X_n=4\\3\quad X_n=5\vee X_n=6}$$does for $a=1$ this is a Markov chain?
I thought summing up and computing for example the first row of the stochastic matrix to better understand $\{Y_n\}$ so I did the following $$P(Y_n=1\mid Y_n=1)=P(X_n=1\vee X_n=2\mid X_n=1\vee x_n=2)=\frac{a+7}{10}\\P(Y_n=2\mid Y_n=1)=P(X_n=3\vee X_n=4\mid X_n=1\vee x_n=2)=\frac{3-a}{10}\\P(Y_n=3\mid Y_n=1)=P(X_n=5\vee X_n=6\mid X_n=1\vee x_n=2)=1$$but doesn't matter the value of a, they sum up to $2$ and not to $1$ means $$P_{Y_n}=\left(\begin{array}{ccc}
\frac{a+7}{10} & \frac{3-a}{10} & 1\\
\dots & \dots & \dots\\
\dots & \dots & \dots
\end{array}\right)
$$ Do I need to normalize the results? what am I doing wrong? Is checking that it's Markov chain is as simple as summing up each row or I need to show a specific path which doesn't meet the property of Markov chain?
| Let's write out the transition matrix of $X$ when $a = 1$: $$\mathcal P = \frac{1}{10} \begin{bmatrix} 1 & 3 & 0 & 1 & 5 & 0 \\ 2 & 2 & 1 & 0 & 1 & 4 \\ 6 & 0 & 1 & 1 & 2 & 0 \\ 6 & 0 & 1 & 1 & 0 & 2 \\ 3 & 0 & 4 & 1 & 0 & 2 \\ 3 & 0 & 1 & 4 & 2 & 0 \end{bmatrix}$$ Now split it up into $2 \times 2$ blocks, and observe for example that $$\begin{align*} \Pr[Y_2 = 1 \mid X_1 = 1] &= 0.4, \\ \Pr[Y_2 = 2 \mid X_1 = 1] &= 0.1, \\ \Pr[Y_2 = 3 \mid X_1 = 1] &= 0.5. \end{align*}$$ But in fact, the same thing is true if $X_1 = 2$; thus $Y_2 \mid Y_1 = 1$ follows the above distribution. And we can also see that the same is true for the other two pairs of rows. Therefore, $Y$ is also a Markov chain, with transition matrix $$\mathcal P^* = \frac{1}{10} \begin{bmatrix} 4 & 1 & 5 \\ 6 & 2 & 2 \\ 3 & 5 & 2 \end{bmatrix}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1118302",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
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