Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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How to prove $e^x\ge \left(1+\frac xn\right)^n$ for any real numbers $x, n > 0$ Can someone provide a detailed proof? I saw a proof here
$$\begin{align}
\frac{e_{n+1}(x)}{e_n(x)}&=\frac{\left(1+\frac x{n+1}\right)^{n+1}}{\left(1+\frac xn\right)^n}\\\\
&=\left(1+\frac{-x}{(n+x)(n+1)}\right)^{n+1}\left(1+\frac xn\right) \tag 1\\\\
&\ge \left(1+\frac{-x}{n+x}\right)\left(1+\frac xn\right)\tag 2\\\\
&=1
\end{align}$$
where in going from (1) to (2) we used Bernoulli's Inequality. Note
that (2) is valid whenever $n>−x$ or $x>−n$. Since $e_n(x)$
monotonically increases and is bounded above by $e^x$, then $$e^x\ge
\left(1+\frac xn\right)^n \tag 3$$ for all $n\ge 1$.
But I don't know how do we get $(1)$.
| Note that we have
$$\begin{align}
\frac{\left(1+\frac{x}{n+1}\right)^{n+1}}{\left(1+\frac{x}n\right)^n}&=\left(\frac{1+\frac{x}{n+1}}{1+\frac xn}\right)^{n+1}\left(1+\frac{x}n\right)\\\\
&=\left(\frac{\frac{n+1+x}{n+1}}{\frac {n+x}n}\right)^{n+1}\left(1+\frac{x}n\right)\\\\
&=\left(\frac{n(n+1+x)}{(n+1)(n+x)}\right)^{n+1}\left(1+\frac{x}n\right)\\\\
&=\left(1+\frac{n(n+1+x)-(n+1)(n+x)}{(n+1)(n+x)}\right)^{n+1}\left(1+\frac{x}n\right)\\\\
&=\left(1+\frac{-x}{(n+1)(n+x)}\right)^{n+1}\left(1+\frac{x}n\right)\\\\
\end{align}$$
as was to be shown!
| {
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"url": "https://math.stackexchange.com/questions/2668555",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Given the list of six positive integers 9, 13, $a$, 12, $b$, and 8, how many ordered pairs $(a,b)$ will result in a mean of 10 for the entire list? Given the list of six positive integers $9$, $13$, $a$, $12$, $b$, and $8$, how many ordered pairs $(a,b)$ will result in a mean of $10$ for the entire list?
I took the sum of these six numbers divided by 6 and set it equal to 10. Then I got the equation $a+b=18$. I found there are $17$ pairs that satisfy these conditions, examples $(1,17),(2,16),(3,15),(4,14),\dots,(15,3),(16,2),(17,1)$. Am I missing anything or does this make sense?
| We want $\frac{8+9+12+13+a+b}{6} = 10$. We have $8+9+12+13 = 42$, so we are looking for two positive integers $a$ and $b$ that sum to $60 - 42 = 18$. Because we are looking at ordered pairs, $a = 1$, $b=17$ is a distinct solution from $a = 17$, $b= 1$. There are $17$ possible values of $a$ for which there is a corresponding value of $b$ (that is, $a$ can equal any integer from $1$ to $17$, inclusive), so our final answer is $\boxed{17}$.
| {
"language": "en",
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Sum of a trigonometric series involving sin and tan I'm trying to prove that the following expression is true:
$$\sum_{k=1}^{N-1} \frac{\sin\left(\frac{\pi k n}{N}\right)}{\tan\left(\frac{\pi k}{2N}\right)} = N-n$$
I think that substituting $\tan\left(\frac{\pi k}{2N}\right)=\frac{\sin\left(\frac{\pi k}{N}\right)}{1+\cos\left(\frac{\pi k}{N}\right)}$ might simplify the problem, but I still end up with a series with trigonometric term in the denominator, which I'm not sure how to solve.
Many thanks,
Calum
| Use $e^{i\theta}=\cos(\theta)+i\sin(\theta)$, $\theta\in\textbf{R}$ to write
$$
\cos\left(\theta\right)=\frac{e^{i\theta}+e^{-i\theta}}{2}\textbf{, }\theta\in\textbf{R}.
$$
Hence
$$
\cos\left(\frac{\pi k}{N}\right)\cos\left(\frac{n\pi k}{N}\right)=
$$
$$
=\frac{1}{4}e^{-ik\pi/N-ikn\pi/N}+\frac{1}{4}e^{ik\pi/N+ikn\pi/N}+\frac{1}{4}e^{-ik\pi/N+ikn\pi/N}
$$
$$
+\frac{1}{4}e^{ik\pi/N-ikn\pi/N}.\tag{id}
$$
If we set
$$
A(n):=\sum^{N-1}_{k=0}\cos\left(\frac{\pi k}{N}\right)\cos\left(\frac{n\pi k}{N}\right)\textrm{, }n=0,1,2,\ldots,
$$
then from (id) we have (for $n$ nonnegative integer), (we collect the conjugates to find the real parts):
$$
A(n)=\frac{N}{2}\textrm{ if }n=1;1\textrm{ if }n=even;0\textrm{ if }n=odd\neq 1.
$$
Using $\cos^2(\theta)=\frac{1+\cos(2\theta)}{2}$, we get
$$
\sum^{N-1}_{k=0}\cos^2\left(\frac{\pi k}{2N}\right)\cos\left(\frac{n\pi k}{N}\right)=
$$
$$
=\frac{N+1}{2}\textrm{ if }n=0;\frac{N+2}{4}\textrm{ if }n=1;\frac{1}{2}\textrm{ if }n=2,3,4,\ldots\tag{1}
$$
But
$$
\frac{\sin\left(\frac{\pi k n}{N}\right)}{\tan\left(\frac{\pi k}{2N}\right)}=\frac{\left(1+e^{i k\pi/N}\right)\left(e^{-nik\pi/N}-e^{nik\pi/N}\right)}{2\left(1-e^{ik\pi/N}\right)}
$$
and if $\zeta=e^{-\pi i/N}$, then
$$
e^{-nik\pi/N}-e^{nik\pi/N}=\zeta^{nk}-\zeta^{-nk}=\zeta^{-nk}\left(\zeta^{2nk}-1\right)=
$$
$$
=\zeta^{-nk}(\zeta^{2k}-1)\left(1+\zeta^{2k}+\zeta^{4k}+\ldots+\zeta^{2(n-1)k}\right).\tag{2}
$$
Hence
$$
\frac{\sin\left(\frac{\pi k n}{N}\right)}{\tan\left(\frac{\pi k}{2N}\right)}=\frac{1}{2}\left(\zeta^k+1\right)^2\zeta^{-nk}\left(1+\zeta^{2k}+\zeta^{4k}+\ldots+\zeta^{2(n-1)k}\right).\tag{3}
$$
i) If $n$ is odd, then easily we have
$$
\sum^{n-1}_{j=0}\zeta^{2kj}=\zeta^{(n-1)k}\left(1+2\sum^{\frac{n-1}{2}}_{j=1}\cos\left(\frac{2jk\pi}{N}\right)\right)\tag{4}
$$
and
$$
\frac{1}{2}(\zeta^k+1)^2 \zeta^{-nk}\zeta^{(n-1)k}=1+\cos\left(\frac{k\pi}{N}\right)=2\cos^2\left(\frac{k\pi}{2N}\right)\tag{5}
$$
Hence relation (3) becomes
$$
\frac{\sin\left(\frac{\pi k n}{N}\right)}{\tan\left(\frac{\pi k}{2N}\right)}
=2\cos^2\left(\frac{\pi k}{2N}\right)\left(1+2\sum^{\frac{n-1}{2}}_{j=1}\cos\left(\frac{2jk\pi}{N}\right)\right)\tag{6}
$$
Summing (6) and using (1), we get
$$
\sum^{N-1}_{k=0}\frac{\sin\left(\frac{\pi k n}{N}\right)}{\tan\left(\frac{\pi k}{2N}\right)}=
\sum^{N-1}_{k=0}2\cos^2\left(\frac{k\pi}{2N}\right)\left(1+2\sum^{\frac{n-1}{2}}_{j=1}\cos\left(\frac{2jk\pi}{N}\right)\right)=
$$
$$
=2\left(\frac{N+1}{2}+2\frac{n-1}{2}\cdot\frac{1}{2}\right)=N+n\tag{7}
$$
ii) If $n$ is even, then
$$
\sum^{n-1}_{j=0}\zeta^{2kj}=2\zeta^{(n-1)k}\sum^{\frac{n}{2}}_{j=1}\cos\left(\frac{(2j-1)k\pi}{N}\right)\tag{8}
$$
Hence using (3),(8) and (1) again we get
$$
\sum^{N-1}_{k=0}\frac{\sin\left(\frac{\pi k n}{N}\right)}{\tan\left(\frac{\pi k}{2N}\right)}=\sum^{N-1}_{k=0}4\cos\left(\frac{k\pi}{2N}\right)^2\left(\sum^{\frac{n}{2}}_{j=1}\cos\left(\frac{(2j-1)k\pi}{N}\right)\right)=
$$
$$
=4\left(\frac{N+2}{4}+\frac{1}{2}\left(\frac{n}{2}-1\right)\right)=N+n\tag{9}
$$
Also it is (easily)
$$
\lim_{k\rightarrow 0}\frac{\sin\left(\pi k n/N\right)}{\tan\left(\pi k/(2N)\right)}=2n.\tag{10}
$$
From (7),(9) and (10), we get the result.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Calculating $\int^1_0\frac{x \log x}{1 + x^2} dx$ Trying to reinvent the spirit of Calculus, I was trying to Understand the Leibniz Rule of Differentiation under Integration sign. To check my understanding, I was considering this integral:
$$ I(b) = \int^1_0 \frac{x \log(b + x)}{1 + x^2} dx $$
After some simplification, I came upon the Integral in question:
$$ 2I(1) = \frac{1}{4} \log^2{2} + \frac{\pi^2}{16} + 2\int^1_0\frac{b \log b}{1 + b^2} db $$
Now I am a bit confused about how to progress. Maxima shows the answer to be $-\frac{\pi^2}{48}$.
This is not a homework problem.
| This calculation may contains some convergence problem, but it'll be resolved easily.
Using integration by parts, we have
$$
I(0)=\int_{0}^{1}\frac{x\log x}{1+x^{2}}dx = \left[\frac{1}{2}\log(1+x^{2})\log x\right]_{0}^{1} - \int_{0}^{1} \frac{1}{2}\log(1+x^{2})\frac{dx}{x}
$$
Since
$$
\lim_{x\to 0} \log(1+x^{2})\log x = \lim_{x\to 0} \frac{\log(1+x^{2})}{x^{2}}\cdot x\cdot x\log x = 0,
$$
we have
$$
I(0) = -\frac{1}{2}\int_{0}^{1} \frac{\log(1+x^{2})}{x}dx.
$$
Now use substitution $x^{2} = t$, then
$$
I(0) = -\frac{1}{4} \int_{0}^{1}\frac{\log(1+t)}{t}dt
$$
Now use the Taylor series of $\log(1+t)$,
$$
I(0) = -\frac{1}{4} \int_{0}^{1} \frac{1}{t}\left(t-\frac{t^{2}}{2}+\frac{t^{3}}{3}-\cdots\right)dt = -\frac{1}{4}\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^{2}} = -\frac{1}{4}\left(\sum_{n=1}^{\infty}\frac{1}{n^{2}}-2\sum_{n=1}^{\infty} \frac{1}{(2n)^{2}}\right) = -\frac{1}{4}\cdot \frac{1}{2}\zeta(2) = -\frac{\pi^{2}}{48}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2671916",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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How many different positive integers can be obtained as a sum of some or all off the numbers: $1,3,5,10,25$? How many different positive integers can be obtained as a sum of some or all of the numbers:
$1,3,5,10,25$?
I have just started discrete mathematics and I am having problems with some tasks. This is one of them. I don't really know what should be my starting approach. Could someone explain how to solve a problem like this one?
| A sum is of the form $a*1 + b*3 + c*5 + d*10 + e*25$ where $a,b,c,d,e$ are either $0$ or $1$. There are $2$ choices for each $a,b,c,d,e$ so there are $2^5 = 32$ possible sums. But $0+0+0+0+0 = 0$ is not acceptable so there are only $2^5 - 1 = 31$ possible sums.
But we need to worry about if there are two different ways of writing sums that add to the same number.
$1 + 3 + 5 + 10 = 19 < 25$ so if the sum is more than or equal to $25$ then $e$ must be $1$ and if the sum is less then $25$ then $e$ must be $0$ so if two different sums add to the some thing then the $e$ value in each sum must be the same.
So if $a*1 + b*3 + 5*c + 10*d + 25*e = h*1 + i*3+k*5 + l*10 + m*25$ then $e=m$ and $a*1 + b*3 + c*5 + d*10 = h*1 + i*3 + k*5 +l*10 = N$.
We do the same argument again. $1 + 3 + 5 = 9 < 10$ so if $N \ge 10$ then $d=l=1$ and if $N < 10$ then $d=l=0$ and we have
$a*1 + b*3 + c*5 = h*1 +i*3 + k*5 = M$ and as $1 + 3 = 4 < 5$ we have $c = k$ and as $1 < 3$ we have $b = i$ and therefore $a=h$.
So any possible number can be reached only one way. So there are $31$ distinct sums.
....
It's worth thinking about writing them out and comparing them to the binary representations of $1$ to $31$. It should build intuition:
$1 = 1_2 \to 1$
$2 = 10_2 \to 3$
$3 = 11_2 \to 3+1 = 4$
$4 = 100_2 \to 5 = 5$
$5 = 101_2 \to 5 + 1 = 6$
....
$29 = 11101_2 \to 25 + 10 + 5 + 1 = 41$
$30 = 11110_2 \to 25 + 10 + 5 + 3 = 43$
$31 = 11111_2 \to 25 + 10 + 5 + 3 + 1 = 44$.
It's interesting that there are only $13$ numbers that aren't possible. We can't get $2$ so we can get any $25*a + 10*b + 5*c + 2$ so that rules out, $2,7,12,...,42$ and that accounts for $9$ of the numbers. The rest are that $20... 24$ are impossible.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Can we calculate $ i\sqrt { i\sqrt { i\sqrt { \cdots } } }$? It might be obvious that $2\sqrt { 2\sqrt { 2\sqrt { 2\sqrt { 2\sqrt { 2\sqrt { \cdots } } } } } } $ equals $4.$ So what about $i\sqrt { i\sqrt { i\sqrt { i\sqrt { i\sqrt { i\sqrt { \cdots } } } } } } \text{ ?} $ The answer might be $-1$, but I'm not sure as $i$ is not a real number. Can anyone help?
| \begin{eqnarray*}
x= a\sqrt { a\sqrt { a\sqrt { a\sqrt { a\sqrt { a\sqrt { \cdots } } } } } } \\
x=a^{ 1+1/2+1/4+1/8+\cdots} \\
x=a^2
\end{eqnarray*}
So it would seem that
\begin{eqnarray*}
i\sqrt { i\sqrt { i\sqrt { i\sqrt { i\sqrt { i\sqrt { \cdots } } } } } }=\color{red}{-1}.
\end{eqnarray*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2672742",
"timestamp": "2023-03-29T00:00:00",
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If $A+B+C=180^{\circ}$, then show that: $\cos B=\sin A\sin C-\cos A\cos C$ Here is the question :
If $A+B+C = 180^{\circ}$, then show that: $\cos B = \sin A \sin C - \cos A \cos C$.
EDIT : Here is my reviewed working :
$$
\cos B=-\cos (A+C)
$$
Since $$\space A+B+C = 180^\circ, \space B =180^\circ-(A+C)$$
And
$$\begin{align}
-\cos B &=\cos (180+B) \\
-\cos B &=\cos(180+(180-(A+C)) \\
-\cos B &=\cos(360-(A+C)) \\
-\cos B &=\cos(A+C) \\
-\cos B &=\cos A \cos C - \sin A \sin C \\
\cos B &= \sin A \sin C - \cos A \cos C \\
\cos B &= -\cos (A+C)
\end{align}$$
Can someone confirm that my working is correct?
Thanks!
| Your first step $\cos(B) = \cos(A+C)$ is incorrect. It should be $\cos(B) = -\cos(A+C)$.
| {
"language": "en",
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Showing that $\frac 1 {(x+1)^2}=\frac 1 {(x+1)(x+2)}+\frac {1!} {(x+1)(x+2)(x+3)} + \frac {2!} {(x+1)(x+2)(x+3)(x+4)} + \cdots$
Prove that if $x>-1$ then
$$\frac 1 {(x+1)^2}=\frac 1 {(x+1)(x+2)}+\frac {1!} {(x+1)(x+2)(x+3)} + \frac {2!} {(x+1)(x+2)(x+3)(x+4)} + \cdots$$
Could I have a hint for this? I tried writing the RHS as a power series and coming up with a differential equation but it doesn't really help as I end up with an intractable integral.
| Hint: This is equivalent to
$$\frac{1}{x} = \frac{0!}{(x+1)}+\frac{1!}{(x+1)(x+2)}+\frac{2!}{(x+1)(x+2)(x+3)}+\cdots.$$
Let $$f_n(x)=\frac{n!}{x(x+1)\cdots(x+n)}.$$
We have that
$$f_n(x)=f_{n-1}(x)-f_{n-1}(x+1),$$
and we wish to prove that
$$\frac{1}{x}=\sum_{n=0}^{\infty} f_n(x+1).$$
Can you take it from here?
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that the series $\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{\sqrt{n}}$ is convergent by Cauchy Criterion. Prove that the series
$\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{\sqrt{n}}=1-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{4}}+\frac{1}{\sqrt{5}}-\cdots$ is convergent by Cauchy Criterion
Let $\epsilon >0$. We will find $N \in \mathbb{N}$ such that when $n>m \geq N$ then $|s_n-s_m|< \epsilon$. Where $|s_n-s_m|=|a_{m+1}-a_{m+2}+a_{m+3}- \cdots \pm a_n|$. Note that $(\frac{1}{\sqrt{m+1}} - \frac{1}{\sqrt{m+2}}+\frac{1}{\sqrt{m+3}}-\frac{1}{\sqrt{m+4}} \cdots + \frac{1}{{\sqrt{n}}}or +(\frac{1}{\sqrt{n-1}}-\frac{1}{\sqrt{n}}) >0$
Also note we have that $\frac{1}{\sqrt{m+1}} -(\frac{1}{\sqrt{m+2}}-\frac{1}{\sqrt{m+3}})-(\frac{1}{\sqrt{m+4}}-\frac{1}{\sqrt{m+5}})\cdots -\frac{1}{\sqrt{n}}or-(\frac{1}{\sqrt{n-1}}-\frac{1}{\sqrt{n}})$
So $0< \frac{1}{\sqrt{m+1}} -\frac{1}{\sqrt{m+2}}+ \cdots \frac{1}{\sqrt{n}}<\frac{1}{\sqrt{m+1}}<\frac{1}{\sqrt{m}}$ Then as soon as $\frac{1}{\sqrt{m}}< \epsilon$ we will have that $\frac{1}{\sqrt{m+1}} -\frac{1}{\sqrt{m+2}}+ \cdots \frac{1}{\sqrt{n}}< \epsilon$. Thus choose $N \in \mathbb{N}$, where $N> \frac{1}{\epsilon^2}$. Therefore if $n>m\geq N$ then we know that $|a_{m+1}-a_{m+2}+a_{m+3}- \cdots \pm a_n|< \frac{1}{\sqrt{m}}\leq \frac{1}{\sqrt{N}} < \epsilon$.
| Note that the partial sums alternately overshoot and undershoot the limit.
| {
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Prove for any odd integer $n$, $n^2-1\equiv 1\pmod{8}$ Let, $\exists k \in \mathbb{Z}, n=2k+1\implies n^2-1= 4k(k+1) \pmod{8}$
Now, if $k$ is odd, $k+1$ is even, and vice-versa, so $4k(k+1) \pmod{8}=\forall k\, \exists k' \in \mathbb{Z},\,8k' \pmod{8}$
Unable to proceed further.
| The source of confusion stated the following:
$2\mid n \,\Rightarrow\, 8\mid \color{#c00}{n^3},\,$ else $\,(n,2)=1 \,\Rightarrow\, 8\mid \color{#0a0}{n^2-1}\ $ by $\,{\rm odd}^2\equiv \{\pm1,\pm3\}^2\equiv 1 \pmod{\!8}$
That is they claim that if $n$ is odd, then $8|n^2-1$.
$8|n^2-1$ is equivalent to $$n^2-1 \equiv 0 \pmod{8}$$
Their argument is if $n$ is odd, that is $n \in \{ \pm1, \pm 3\}$, then we have $n^2\equiv 1 \pmod{8}$.
Remark:
My own understanding of $n^2-1$ is $n^2-1=(n-1)(n+1)$, which is a product of two consecutive even numbers, of which one of them must be a multiple of $4$. Hence the product must be a multiple of $8$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Standard Deviation for sums of fair dice given the number of dice, and the number of sides on each die $n$ fair die are rolled, and each dice has $x$ sides, with the numbers on the sides going from $1$ to $x$, and with each side having a different number from the other sides. how do I figure out the standard deviation for the probability of the sums of the numbers that the die land on?
| I am assuming you mean the standard deviation of the sum of $n$ dice, each with $x$ sides.
Let the random variables $X_1,X_2,X_3,..,X_n$ denote the results on the first, second, third$,...,n^{th}$ dice. The $X_i$ are independent. Because of this, we can simply take
$$\begin{align*}
Var(X_1+X_2+X_3+...+X_n)
&=Var(X_1)+Var(X_2)+Var(X_3)+...+Var(X_n)\\\\
&=nVar(X_1)
\end{align*}$$
since the $X_i$ are identically distributed.
To calculate the variance of $X_1$, we calculate $E(X_1^2)-E(X_1)^2$
We have
$$\begin{align*}
E(X_1^2)
&=\frac{1}{x}\left(1^2+2^2+\cdots+x^2\right)\\\\
&=\frac{1}{x}\left(\frac{x\cdot(x+1)\cdot(2x+1)}{6}\right)\\\\
&=\frac{(x+1)\cdot(2x+1)}{6}
\end{align*}$$
and
$$\begin{align*}
E(X_1)
&=\frac{1}{x}\left(1+2+\cdots+x\right)\\\\
&=\frac{1}{x}\left(\frac{x\cdot (x+1)}{2}\right)\\\\
&=\frac{x+1}{2}
\end{align*}$$
Thus
$$\begin{align*}
Var(X_1+...+X_n)
&=nVar(X_1)\\\\
&=n\cdot\left(\frac{(x+1)\cdot(2x+1)}{6}-\left(\frac{x+1}{2}\right)^2\right)
\end{align*}$$
Finally, we take the square root to obtain the standard deviation
$$\sigma=\sqrt{n\cdot\left(\frac{(x+1)\cdot(2x+1)}{6}-\left(\frac{x+1}{2}\right)^2\right)}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Mathematical Induction prove that $n^3+5n$ is divisible by $6$ Sorry, I know this will be a duplicate on the site but the other solution I found confusing and the method look completely different to what I was taught.
Prove that $n^3 + 5n$ is divisible by $6$ by using induction
The question is
Prove by mathematical Induction $(n^3+5n)$ is divisible by $6$
Here is what I have done
Assume $n=k$ is true
$\sum_{1}^{k} k =\dfrac{(k^3+5k)}{6}$
Now assume $n=k+1$ is true
$\sum_{1}^{k+1} k+1 =\dfrac{(k+1)^3+5(k+1)}{6}$
Then now
$\dfrac{(k+1)^3+5(k+1)}{6}=\sum_{1}^{k} k + (k+1)$
$\dfrac{(k+1)^3+5(k+1)}{6}=\dfrac{(k^3+5k)}{6} + (k+1)$
$\dfrac{(k+1)^3+5(k+1)}{6}=\dfrac{(k^3+5k)}{6} + \dfrac{(6k+6)}{6}$
$\dfrac{(k+1)^3+5(k+1)}{6}=\dfrac{(k^3+11k+6)}{6}$
But the other side doesnt equate (LHS)
$\dfrac{(k^3+3k^2+3k+1)+5k+5}{6}$
$\dfrac{(k^3+3k^2+8k+6)}{6}\ne\dfrac{(k^3+11k+6)}{6}$
I hope my method was clear enough so you can see where I went wrong. It would be much more use to me if you solved it as I learn better from looking at solutions and then applying them to other questions.
| As an alternative for induction step assume $n^3+5n=6k$ is true and we need to prove that $(n+1)^3+5(n+1)=6h$.
Let observe that
$$(n+1)^3+5(n+1)=n^3+3n^2+3n+1+5n+5=3n^2+3n+6+(n^3+5n)=\\=3n^2+3n+6(k+1)$$
then we need to show that $3n^2+3n$ is divisibleby $6$ wich can be easily verified by plugging $n=2r$ and $n=2s+1$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Find Laurent series about $z_0=0$ for $\frac{1}{z-1} -\frac{1}{z+1}$ in $2<|z|<\infty$ Find Laurent series about $z_0=0$ for $\frac{1}{z-1} -\frac{1}{z+1}$ in $2<|z|<\infty$.
I know how to do this for $|z|>1$ but I don't understand how I should set it up for $|z|>2$.
Here is how I would do it for $|z|>1$.
$\frac{1}{z-1}= \frac{1}{z}\frac{1}{1-1/z}=\frac{1}{z} \sum_{n=0}^\infty (\frac{1}{z})^n$
And similarly $\frac{1}{z+1}=\frac{1}{z} \frac{1}{1-(-1/z)} = \frac{1}{z} \sum_{n=0}^\infty \frac{(-1)^n}{z^n}$
Then subtract the two sums. Is it the same answer for |z|>2?
| HINTS:
$$\frac{1}{z+b}=\frac{1}{z-c+(b+c)}=\left(\frac{1}{b+c}\right)\,\left(\frac{1}{1+\left(\frac{z-c}{b+c}\right)}\right)$$
and
$$\frac{1}{z+b}=\frac{1}{z-c+(b+c)}=\left(\frac{1}{z-c}\right)\,\left(\frac{1}{1+\left(\frac{b+c}{z-c}\right)}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2691314",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Proving ${l\choose n}+nl-{d+n\choose d}<0$ for $n\geq 6$, $d\geq 3$, $lLet $n\geq 6$ and $d\geq 3$. Prove that if $l<d+n-1$, then
$${l\choose n}+nl-{d+n\choose d}<0.$$
This is the necessary condition in my paper. Experiments show that it is true and $<0$. I tried by expanding but could not find any special for showing that it is $<0$. Thanks for any idea.
| Because$$
\binom{l}{n} < \binom{d + n - 1}{n}, \quad nl < n(d + n - 1),
$$
then it suffices to prove that$$
\binom{d + n}{d} \geqslant \binom{d + n - 1}{n} + n(d + n - 1).
$$
Since\begin{align*}
&\mathrel{\phantom{=}}{} \binom{d + n}{d} - \binom{d + n - 1}{n} = \binom{d + n}{d} - \binom{d + n - 1}{d - 1}\\
&= \binom{d + n - 1}{d} = \frac{1}{d!} (d + n - 1) \cdots (n + 1)n\\
&= \frac{1}{d(d - 1)}·\binom{d + n - 2}{d - 2}·n(d + n - 1),
\end{align*}
then it suffices to prove that$$
\binom{d + n - 2}{d - 2} \geqslant d(d - 1),
$$
which is true in that\begin{align*}
\binom{d + n - 2}{d - 2} &\geqslant \binom{d + 5 - 2}{d - 2} = \binom{d + 5 - 2}{5}\\
&= \frac{1}{5!} (d + 3)(d + 2)(d + 1)·d(d - 1)\\
&\geqslant \frac{1}{5!} (3 + 3)(3 + 2)(3 + 1)·d(d - 1) = d(d - 1).
\end{align*}
| {
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Why do I get that $0 = 1$ when trying to prove that $\forall n\in\mathbb{N}, \ 3 + 2\left(n + \sum_{i=1}^{2n-1}(i+1)\right) = (2n+1)^2\,?$ I was playing around with numbers and by looking at some similar observations, I found an interesting pattern. For a natural number $n$, it seemed like the following was always true: $$3 + 2\left(n + \sum_{i=1}^{2n-1}(i+1)\right) = (2n+1)^2.$$
How must we be able to prove this, though?
My Attempt:
Assume that the equation is true for all natural $n$, then we can similarly write the same for $n+1$.
$$\begin{align} 3 + 2\left(n+1+\sum_{i=1}^{2(n+1)-1}(i+1)\right) &= \big(2(n+1)+1\big)^2 \\ \Leftrightarrow 3 + 2\left(n+1+\sum_{i=1}^{2n+1}(i+1)\right) &= 3 + 2n + 2\left(1+\sum_{i=1}^{2n+1}(i+1)\right) \\ &= (2n+3)^2 \\ \Leftrightarrow (2n+3)^2 - (2n + 3) &= (2n+3)(2n + 3 - 1) \\ &= 2(2n+3)(n+1) \\ &= 2\left(1+\sum_{i=1}^{2n+1}(i+1)\right) \\ \Leftrightarrow (2n+3)(n+1) &= 1+\sum_{i=1}^{2n+1}(i+1) \\ &= \sum_{i=0}^{2n+1}(i+1) \\ &= \sum_{i=1}^{2n+1}i.\end{align}$$
I know that, $$\sum_{i=1}^ki = \frac{k(k+1)}{2}.\tag{The Triangular Numbers}$$ Therefore, by substituting $k = 2n+1$, we get $$\begin{align} \sum_{i=1}^{2n+1}i &= \frac{(2n+1)(2n+2)}{2} \\ &= \frac{2(n+1)(2n+1)}{2} \\ &= (n+1)(2n+1).\end{align}$$ It follows, then, that $(2n+3)(n+1) = (2n+1)(n+1)$ which means that $2n+3 = 2n+1$ and absurdly we get that $n=n+1$ or $0=1$.
I obviously did something wrong, but I don't know where. Can someone help me out please?
Thank you in advance.
| The formula you have found is just simply a way of rewriting $(2n+1)^2$.
$$3+2\left(n+\sum_{i=1}^{2n-1}I+1\right)=3+2\left(n+2n-1+\frac{(2n-1)(2n)}{2}\right)$$
$$=3+2(3n-1+2n^2-n)=4n^2+4n+1=(2n+1)^2$$
Therefore
$$3+2\left(n+\sum_{i=1}^{2n-1}I+1\right)\equiv (2n+1)^2$$
| {
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"timestamp": "2023-03-29T00:00:00",
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How to use integration by parts to solve an integral? I'm having trouble with solving this integral by parts. Here is what I have so far. Can anyone please help me out?
Solve the integral $\int x^3\sqrt{x^2-1}dx$ by parts, choosing u = $x^2$ and
$dv = x\sqrt{x^2-1} dx$
$du_1 = 2xdx$
$v_1$ = $\int x\sqrt{x^2-1}dx$ $ t = x^2-1$ $dt = 2x dx$
$v_1$ = $\frac{1}{2}\int \sqrt{t}dt$
$v_1$ = $\frac{1}{2}(\frac{2}{3}t^\frac{3}{2}) + c$
$v_1$ = $\frac{1}{3} (x^2-1)^\frac{3}{2} + c$
$uv - \int vdu$
= $\frac{1}{3} (x^2-1)^\frac{3}{2}x^2 - \int 2x\frac{1}{3} (x^2-1)^\frac{3}{2}dx$
$u_2 = 2x$ $dv_2 = \frac{1}{3} (x^2-1)^\frac{3}{2}dx$
$du_2 = 2dx$
| You don't have to use integration by parts for $\displaystyle \frac{1}{3}\int 2x (x^2-1)^\frac{3}{2}dx$.
$$\displaystyle \frac{1}{3}\int 2x (x^2-1)^\frac{3}{2}dx=\frac{1}{3}\int(x^2-1)^\frac{3}{2}d(x^2-1)$$
So the integral is
\begin{align*}
\frac{1}{3} (x^2-1)^\frac{3}{2}x^2 - \frac{1}{3}\int(x^2-1)^\frac{3}{2}d(x^2-1)&=\frac{1}{3} (x^2-1)^\frac{3}{2}x^2 - \frac{2}{15}(x^2-1)^\frac{5}{2}+C\\
&=\frac{1}{15}(x^2-1)^\frac{3}{2}[5x^2-2(x^2-1)]+C\\
&=\frac{1}{15}(x^2-1)^\frac{3}{2}(3x^2+2)+C\\
\end{align*}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove by Induction: Summation of Factorial (n! * n) Prove by induction (weak or strong) that:
$$(1! \cdot 1) + (2! \cdot 2) + \cdots + (n! \cdot n) =\sum_{k=1}^nk!\cdot k= (n + 1)! - 1$$
My base case is:
$n = 1$, which is true.
And my Inductive Hypothesis is:
$(1! \cdot 1) + (2! \cdot 2) + \cdots + (k! \cdot k) = (k + 1)! - 1$
After that, I'm trying to show the $(k + 1)$-stage where:
$(1! \cdot 1) + (2! \cdot 2) + \cdots + (k! \cdot k) + ((k + 1)! \cdot (k + 1)) = ((k + 1) + 1)! - 1$
Which simplifies to:
$(1! \cdot 1) + (2! \cdot 2) + \cdots + (k! \cdot k) + ((k + 1)! \cdot (k + 1)) = (k + 2)! - 1$
I see that I can substitute in my Inductive Hypothesis but where I'm stuck is manipulating the LHS to be equal to the RHS after that:
$(k + 1)! - 1 + ((k + 1)! \cdot (k + 1)) = (k + 2)! - 1$
| \begin{align*}
(1!\cdot 1) + (2!\cdot 2) + \cdots + (k!\cdot k) + ((k+1)!\cdot(k+1))
&= (k+1)!-1 + ((k+1)!\cdot(k+1)) \\
&= ((k+1)!\cdot(k+2))-1 \\
&= (k+1)!-1.
\end{align*}
| {
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"timestamp": "2023-03-29T00:00:00",
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How does $n^4 + 6n^3 + 11n^2 + 6n + 1 = (n^2 + 3n + 1)^2?$ C++ student here, not quite familiar with these type of expressions. Can someone explain how does this work? I'm familiar with $(a+b)^2$ etc. mathematics but this seems to be like $(a+b+c)^2$ and having searched online, the opened form for this formula doesn't look much alike. Any help will be appreciated. Thanks!
| When you expand something squared, you multiply each term by each other term, so in this case you have $$\begin{aligned}(n^2+3n+1)^2&=n^2\cdot n^2 + n^2\cdot 3n+n^2\cdot 1\\&\;+3n\cdot n^2+3n\cdot3n+3n\cdot1\\&\;+1\cdot n^2+1\cdot3n+1\cdot1\\&=n^4+3n^3+n^2+3n^3+9n^2+3n+n^2+3n+1\\&=n^4+6n^3+11n^2+6n+1\end{aligned}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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I try to find sum of $\sum_{i=n}^{2n-1} 3+4i$ I try to find sum of $\sum_{i=n}^{2n-1} 3+4i$
$\sum_{i=n}^{2n-1} 3+ \sum_{i=n}^{2n-1} 4i$ =$3 (2n-1-n+1) + 4 \sum_{i=n}^{2n-1} i $ -> am I right?
if I change the index of $4 \sum_{i=n}^{2n-1} i $ to $4 \sum_{i=0}^{2n-1-n} i+n $ am I right?
$4 \sum_{i=0}^{2n-1-n} i+n $
$4 (\sum_{i=0}^{n-1} i+ \sum_{i=0}^{n-1} n )$ = $4(\frac{n(n-1)}{2}+n(n-1))$=$2(3n^2-3n)=6n^2-6n$
total= $3 (2n-1-n+1) + 4 \sum_{i=n}^{2n-1} i $=$3n+ 6n^2-6n $, am I right???
| Note $4 (\sum_{i=0}^{n-1} i+ \sum_{i=0}^{n-1} n )=4 \left( \frac{n(n-1)}{2}+\color{red}{n^2}\right)$ since from $0$ to $n-1$ there are $n$ numbers.
Another comment is use braces
$4 \sum_{i=0}^{2n-1-n} i+n $ should be $4 \sum_{i=0}^{2n-1-n} (i+n) $
Alternative approach: Note that you are summing an arithmetic progression.
There are a total of $(2n-1-n+1)=n$ terms.
The first term is $3+4n$, the last term is $3+4(2n-1)=8n-1$.
Hence the sum is $\frac{n}2 \cdot (3+4n+8n-1)=\frac{n}2(12n+2)=n(6n+1)$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Multivariable delta-epsilon limit proofs The Question asks: Find the delta epsilon limit if it exist
$$ \lim \limits_{(x, y) \to (0, 0)} \ \frac{x^2 y^3}{2x^2 + y^2} = 1$$
What I've done so far:
$$ \mid {\frac{x^2 y^3}{2x^2 + y^2} - 1 \mid} < \epsilon \quad \quad \quad \quad0 < \sqrt{x^2 + y^2} < \delta $$
Now, since $x ^2 \leq 2x^2 + y^ \implies \frac{x^2}{2x^2 + y^2} \leq 1$
Then:
$$ \frac{x^2 \mid y^3 \mid}{2x^2 + y^2} \leq \mid y^3 \mid = \sqrt{y^2}^3 = \quad ...$$
Here is where I am stuck, I can't make $\sqrt{y^2}^3$ into $\sqrt{ x^2 + y^2} $.
How should I proceed further?
Using Stewart's Calculus (8th ed) 14.2 Limit proof method
| In fact
$$\left|\frac{x^2y^3}{2x^2+y^2}\right| \leq \frac{1}{2}|y^3|$$ which is equivalent to
$$2x^2\leq 2x^2+y^2.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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"answer_id": 1
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Solve $\left ( 1- \sqrt{2}\sin x \right )\left ( \cos 2x+ \sin 2x \right )= \frac{1}{2}$ Solve
$$\left ( 1- \sqrt{2}\sin x \right )\left ( \cos 2x+ \sin 2x \right )= \frac{1}{2}$$
Now I did not understand how can i solve that.
I have tried substituting $\cos(2x)=\cos^2(x)−\sin^2(x)$ and$\,$ $\sin(2x)=2\sin(x)\cos(x)$,
the equation is now $(1−\sqrt2\sin(x))(\cos^2(x)−\sin^2(x)+2\sin(x)\cos(x))=\frac12$
Help Required
Thanks
| Multiplying by $1+\sqrt 2 \sin x$ on both sides and using $1-2 \sin^2 x = \cos 2x$ we get
$\cos 2x (\cos 2x + \sin 2x ) = \dfrac{1+\sqrt 2 \sin x}{2}$
or $1+\cos 4x + \sin 4x = 1+\sqrt 2 \sin x$
or $\dfrac{\sin 4 x + \cos 4x}{\sqrt 2} = \sin x$
or $\sin \left(4x+\dfrac{\pi}{4} \right) = \sin x$ which can be easily solved
| {
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"url": "https://math.stackexchange.com/questions/2707293",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find the rational number of a, b, c, solving $\sqrt[3]{\sqrt[3]{2}-1}=\sqrt[3]{a}+ \sqrt[3]{b}+\sqrt[3]{c}$ I try as following
let
\begin{eqnarray}
x= \sqrt[3]{a} \\
y= \sqrt[3]{b} \\
z= \sqrt[3]{c} \\
x+y+z = \sqrt[3]{\sqrt[3]{2}-1 }\\
\end{eqnarray}
We know that
\begin{equation}
x^3+y^3+z^3 = (x+y+z)^3-3(x+y)(x+z)(y+z)
\end{equation}
that turns out to be
\begin{equation}
(x+y+z-1)(x+y+z)(x+y+z-1)=3(x+y)(x+z)(y+z)
\end{equation}
plug in $x+y+z$,
we will get
\begin{equation}
\sqrt[3]{2}-1 - \sqrt[3]{\sqrt[3]{2}-1 } = 3(x+y)(x+z)(y+z)
\end{equation}
From now I stuck to solve for the rational numbers of $a,b,c$. Could anyone show me the way how to continue to solve it?
| I try this:
$y=\sqrt[3]{2}$,$y^3=2$, $x= \sqrt[3]{\sqrt[3]{2}-1}$
Therefore, $x^3=y-1$
$y^3-1=(y-1)(y^2+y+1)=1$
Because $y^2+y+1=\frac{3y^2+3y+3}{3}=\frac{y^3+3y^2+3y+1}{3}=\frac{(y+1)^3}{3}$
and $y^3+1=3=(y+1)(y^2-y+1)$
Hence
$x^3=y-1=\frac{1}{y^2+y+1}=\frac{3}{(y+1)^3}=\frac{1}{9}(y^2-y+1)$
Therefore, based on the definition of $y$, we can obtain
$a=\frac{4}{9}$,$b=-\frac{2}{9}$, and $c=\frac{1}{9}$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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"answer_id": 2
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Visualizing $3^3+4^3+5^3=6^3$ I was wondering how would the visualization of $3^3+4^3+5^3=6^3$ look? I tried some 3D sketches but when calculating none work. Any ideas would be highly appreciated.
| $$ 5^3 + 3^3 + 4^3 = 5^3 + 3 \cdot (3^2 + 4^2) + 4^2 = 5^3 + 3 \cdot 5^2 + 3 \cdot 5 + 1 = (5 + 1)^3 $$
So a cube of side $6$ units can be made by taking a $\color{green}{\text{cube of side 5 units} }$, adding $ \color{blue}{\text{3 5-by-5 squares}}$ at the top, right and left, then adding $\color{red}{\text{3 lines of length 5}}$ along $3$ edges and finally adding $\color{black}{ \text{1 cube} }$ at the corner.
Source of picture of cube
| {
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"timestamp": "2023-03-29T00:00:00",
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$A289B$ is divisible by $90$, then what is $A+B$? It is known that $A289B$ is divisible by $90$, are with $A$ and $B$ as digits. What is $A+B$?
My approach :
If $A289B$ is divisible by $90$, then the unit-value must be a zero, $B=0$.
So now we have $A2890$, which can be written as
$$ A2890 = A\times10000 + 2890 $$
since it is divisible by $90$, the sum of the remainders of $A \times 1000$ and $2890$ being divided by 90, must be divisible by 90.
We have $mod(2890, 90) = 10$.
Now we may find $A$ such that $mod(A \times 10000, 90) = 80$.
$$mod(10000, 90) = 10 \implies 10000 = (90 \times 111 + 10) $$
so $A$ must be $8$, since
$$ 8 \times 10000 + 2890 = 8 \times (90 \times 111 + 10) + (90 \times 32 + 10) = 90 \times (888+32) + 90 $$
is divided by 90.
$$A+B = 8$$
Is this the best approach..? Thanks.
| You can simply take the sum of digits $A+2+8+9+0 \equiv A + 1 \equiv 0 \pmod 9$, so as to know that $A \equiv 8 \pmod 9$ is the desired solution because $10^k \equiv 1 \pmod 9$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Partial Derivative of arctan Given that $$f(x,y)=\tan^{-1}\left(\frac{x+y}{1-xy}\right)$$
Find $f_x(x,y)$
My attempt,
$$
\begin{aligned}
f_x(x,y)&=\frac{(1-xy)(1)-(x+y)(-y)}{(1-xy)^2}\cdot\frac{1}{1+\left(\frac{x+y}{1-xy}\right)^2}\\
&=\frac{1+y^2}{(1-xy)^2+(x+y)^2}\\
&=\frac{1+y^2}{1+y^2+x^2+x^2y^2}
\end{aligned}
$$
But the given answer is $\frac{1}{x^2+1}$.
How?
| Hint:
$$1+x^2+y^2+x^2y^2=(1+x^2)+y^2(1+x^2)=?$$
Alternatively, see Inverse trigonometric function identity doubt: $\tan^{-1}x+\tan^{-1}y =-\pi+\tan^{-1}\left(\frac{x+y}{1-xy}\right)$, when $x<0$, $y<0$, and $xy>1$
| {
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limit of a recursive sequence_1
I have this sequence
$$ \begin{cases}a_1= \alpha +2 & \alpha >0\\
a_{n+1}=\left(a_n- \frac{1}{n}\right)^n+ \frac{1}{n+1} & n \ge1\end{cases}$$
and I want to know
$\lim_\limits{n\rightarrow \infty} a_n$
I can prove by induction that the sequence is a monotone increasing.
$$
a_1= \alpha +2
$$
$$
a_2= \alpha +\frac{3}{2}
$$
$$
a_3= \alpha^2+2\alpha +\frac{4}{3}
$$
$$
a_4= (\alpha^2+2\alpha +1)^3+\frac{1}{4}
$$
if we suppose that $$a_n>a_{n-1} \rightarrow a_n=\left(a_{n-1}- \frac{1}{n-1}\right)^{n-1}+ \frac{1}{n}>a_{n-1} \rightarrow \left(a_{n-1}- \frac{1}{n-1}\right)^{n-1}>a_{n-1}- \frac{1}{n}$$
Then, $$a_{n+1}=\left(a_n- \frac{1}{n}\right)^n+ \frac{1}{n+1}=a_{n+1}=\left(\left(a_{n-1}- \frac{1}{n-1}\right)^{n-1}+ \frac{1}{n}-\frac{1}{n}\right)^n+\frac{1}{n+1}=\left(a_{n-1}- \frac{1}{n-1}\right)^{n(n-1)}+\frac{1}{n+1}>\left(a_{n-1}- \frac{1}{n}\right)^{n}+\frac{1}{n+1}$$
$$a_{n+1}>\left(a_{n-1}- \frac{1}{n}\right)^{n}+\frac{1}{n+1}=a_n \rightarrow a_{n+1}>a_n$$
We can say that $a_{n+1}>a_n>0$
$$\lim_{n\rightarrow \infty} a_n=\sigma$$ can be a $\alpha \in R$ or $\infty$.
In particular we can say that:
$$a_{n+1}=\left(a_n- \frac{1}{n}\right)^n+ \frac{1}{n+1} \rightarrow a_{n+1}- \frac{1}{n+1}=\left(a_n- \frac{1}{n}\right)^n \rightarrow \sigma= \sigma^n \rightarrow \sigma=+ \infty$$
| This series diverges towards infinity as you noted. In this case, instead of a good old "$\varepsilon - \delta$" proof, we need a $K-\delta$ proof.
Note first that:
$$
a_n >\frac{n+1}{n} \Rightarrow a_{n+1}>2+\frac{1}{n+1}
$$
Because:
$$
a_{n+1} = \left(a_n-\frac{1}{n} \right)^n +\frac{n+2}{n+1}> \left(\frac{n+1}{n}-\frac{1}{n} \right)^n +\frac{n+2}{n+1}= (1)^n +\frac{n+2}{n+1}>2+\frac{1}{n+1}
$$
So $a_n>2$ for $\alpha>0$ and $n>2$.
Therefore:
$$
a_{n+1} > \left(2+\frac{1}{n}-\frac{1}{n} \right)^n +\frac{n+2}{n+1}> 2^n>K
$$
So for all $n>\log_2(K)$, $a_n>K$, and hence $a_n$ diverges towards positive infinity.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2710604",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Finding an implicit solution of the ODE $y' (x) = \frac{y^2-xy}{2xy^3+xy+x^2}$ I am currently trying to find an implicit solution of the nonlinear ODE:$$y' (x) = \frac{y^2-xy}{2xy^3+xy+x^2}$$
But to be quite honest, I am a little bit lost. I am pretty new to this field and have not learned many methods of solving differential equations so far. What I tried was, to reformulate the problem as $$xy-y^2+y'(2xy^3+xy+x^2)=0$$
and check if this is an exact equation. However, the integrability criterion does not hold: $$\frac d{dx}(2xy^3+xy+x^2) \neq \frac{d}{dy}(xy-y^2)$$
Therefore, I tried to find an integrating factor to make this equation exact. The approach that my integrating factor $m(x,y)$ only depends on $x$ or only depends on $y$ seems to fail. Another approach I have seen is to set $m(x,y) = x^\alpha y^\beta$ as the integrating factor and then find out what $\alpha $ and $\beta$ have to be. Unfortunately, when using this approach, I end up with utterly long equations, which do not seem easily solveable ( I managed to actually find values for $\alpha$ and $\beta$ using Maple and then verify that the ODE multiplied by that factor is exact numerically, but I do not think this is the way to go here.)
Any help would be greatly appreciated!
| $\dfrac{dy}{dx}=\dfrac{y^2-xy}{2xy^3+xy+x^2}$
$(y^2-xy)\dfrac{dx}{dy}=x^2+(2y^3+y)x$
$(y-x)\dfrac{dx}{dy}=\dfrac{x^2}{y}+(2y^2+1)x$
This belongs to an Abel equation of the second kind.
Let $u=y-x$ ,
Then $x=y-u$
$\dfrac{dx}{dy}=1-\dfrac{du}{dy}$
$\therefore u\left(1-\dfrac{du}{dy}\right)=\dfrac{(y-u)^2}{y}+(2y^2+1)(y-u)$
$u-u\dfrac{du}{dy}=\dfrac{u^2}{y}-(2y^2+3)u+2y(y^2+1)$
$u\dfrac{du}{dy}=-\dfrac{u^2}{y}+2(y^2+2)u-2y(y^2+1)$
Let $u=\dfrac{v}{y}$ ,
Then $\dfrac{du}{dy}=\dfrac{1}{y}\dfrac{dv}{dy}-\dfrac{v}{y^2}$
$\therefore\dfrac{v}{y}\left(\dfrac{1}{y}\dfrac{dv}{dy}-\dfrac{v}{y^2}\right)=-\dfrac{v^2}{y^3}+\dfrac{2(y^2+2)v}{y}-2y(y^2+1)$
$\dfrac{v}{y^2}\dfrac{dv}{dy}-\dfrac{v^2}{y^3}=-\dfrac{v^2}{y^3}+\dfrac{2(y^2+2)v}{y}-2y(y^2+1)$
$\dfrac{v}{y^2}\dfrac{dv}{dy}=\dfrac{2(y^2+2)v}{y}-2y(y^2+1)$
$v\dfrac{dv}{dy}=2y(y^2+2)v-2y^3(y^2+1)$
Let $s=y^2+2$ ,
Then $\dfrac{dv}{dy}=\dfrac{dv}{ds}\dfrac{ds}{dy}=2y\dfrac{dv}{ds}$
$\therefore2yv\dfrac{dv}{ds}=2y(y^2+2)v-2y^3(y^2+1)$
$v\dfrac{dv}{ds}=(y^2+2)v-y^2(y^2+1)$
$v\dfrac{dv}{ds}=sv-(s-2)(s-1)$
Let $t=\dfrac{s^2}{2}$ ,
Then $\dfrac{dv}{ds}=\dfrac{dv}{dt}\dfrac{dt}{ds}=s\dfrac{dv}{dt}$
$\therefore sv\dfrac{dv}{dt}=sv-(s-2)(s-1)$
$v\dfrac{dv}{dt}=v-s+3-\dfrac{2}{s}$
$v\dfrac{dv}{dt}=v\pm\sqrt{2t}+3\pm\sqrt{\dfrac{2}{t}}$
This belongs to an Abel equation of the second kind in the canonical form.
Please follow the method in https://arxiv.org/ftp/arxiv/papers/1503/1503.05929.pdf
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2710702",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Sum of the first integer powers of $n$ up to k Pascal's triangle has a lot of interesting patterns in it; one of which is the triangular numbers and their extensions. Mathematically:
$$\sum_{n=1}^k1=\frac{k}{1}$$
$$\sum_{n=1}^kn=\frac{k}{1}\cdot\frac{k+1}{2}$$
$$\sum_{n=1}^kn^2=\frac{k}{1}\cdot\frac{k+1}{2}\cdot\frac{2k+1}{3}$$
At first, we could guess that the next summation is:
$$\sum_{n=1}^kn^3 ?=\frac{k}{1}\cdot\frac{k+1}{2}\cdot\frac{2k+1}{3}\cdot\frac{3k+1}{4}$$
Yet this is off. However, it is off geometrically. Notice:
$$\left(\sum_{n=1}^kn^3\right)-\frac{k}{1}\cdot\frac{k+1}{2}\cdot\frac{2k+1}{3}\cdot\frac{3k+1}{4}=error$$
$$k=1, r=0$$
$$k=2, r=0.25$$
$$k=3, r=1$$
$$k=4, r=2.5$$
$$k=5, r=5$$
$$k=6, r=8.75$$
...
Consider the ratios of the errors:
$$er(k)=\frac{r(k+1)}{r(k)}$$
$$k=1, r=udf$$
$$k=2, r=4$$
$$k=3, r=2.5$$
$$k=4, r=2$$
$$k=5, r=1.75$$
$$k=6, r=1.6$$
Then, rewriting the error as a function of n starting at k = 5:
$$1.75=2.5-\frac{1.5}{2}$$
$$1.6=2.5-\frac{1.5}{2}-\frac{1.5}{10}$$
$$1.5=2.5-\frac{1.5}{2}-\frac{1.5}{10}-\frac{1.5}{15}$$
$$1.42857=2.5-\frac{1.5}{2}-\frac{1.5}{10}-\frac{1.5}{15}-\frac{1.5}{21}$$
The denominators in the series are from pascals triangle: (3rd columns, or dependent again on the triangular numbers)
Then the total formula for equating the two is:
$$\left(\frac{k}{1}\cdot\frac{\left(k+1\right)}{2}\cdot\frac{\left(2k+1\right)}{3}\cdot\frac{\left(3k+1\right)}{4}\right)-\left(\sum_{n=1}^kn^3\right)+\frac{1}{24}\left(k-1\right)k\left(k+1\right)=0$$
Super interesting!
At least, I thought it was interesting how this the error is related back to the previous power's formula. Am I missing something obvious? Any input is greatly appreciated. (I'm not smart, so in the likely event I missed something obvious try not to be too harsh)
Update:
For the next power (4), I found the formula with trial and error:
$$\left(\frac{k}{1}\cdot\frac{\left(k+1\right)}{2}\cdot\frac{\left(2k+1\right)}{3}\cdot\frac{\left(3k+1\right)}{4}\cdot\frac{\left(4k+1\right)}{5}\right)+\frac{1}{24}\left(k-1\right)k\left(k+1\right)+\frac{1}{12}\left(k-1\right)k\left(k+1\right)k$$
Any ideas on power (5) and so on? I'll continue to try and generalize it.
| The general result is called Faulhaber's Formula.
A hint in the general direction.
Let $f_i(x)=x(x+1)(x+2)\cdots(x+(i-1)),$ and $f_0(x)=1.$ We call $f_i$ the "rising factorial," and is sometimes written $x^{(i)}.$
Note the property that $f_{i+1}(x)-f_{i+1}(x-1)=(i+1)f_i(x).$
So, we can telescope the sum:
$$\sum_{n=1}^{k}f_i(n)=\frac{1}{i+1}\sum_{n=1}^{k}\left(f_{i+1}(n)-f_{i+1}(n-1)\right)= \frac{f_{i+1}(k)}{i+1}.$$
since $f_{i+1}(0)=0.$
Now, what you can see is that since $f_i(x)$ is of degree $i$ and monic (the lead coefficient is $1$) we can write:
$$x^i = f_i(x)+O(x^{i-1})$$
where the rest is a polynomial.
So $$\sum_{n=1}^{k} n^i =\frac{f_{i+1}(k)}{i+1}+O(k^{i})=\frac{k^{i+1}}{i+1}+O(k^{i}).$$
Now, your polynomial $$\frac{x(x+1)(2x+1)(3x+1)\cdots((ix+1)}{(i+1)!}=\frac{x^{i+1}}{i+1}+O(x^i)$$ too.
So these are generally going to grow similarly.
But if we look at the second term, then we get, for $i>0:$
$$x^{i}=f_i(x)-\frac{i(i-1)}{2}f_{i-1}(x)+O(x^{i-2})$$
So:
$$\sum_{n=0}^{k} n^i = \frac{f_{i+1}(k)}{i+1}-\frac{(i-1)f_{i}(k)}{2} +O(k^{i-1})$$
A little fiddling gives you that when $i>0$:
$$\sum_{n=1}^{k} n^i = \frac{k^{i+1}}{i+1} +\frac{k^i}{2}+O(k^{i-1})$$ for $i>0.$
The coefficient $k^{i}$ in your polynomial is $\frac{1}{i+1}\left(1+\frac{1}{2}+\cdots+\frac1{i}\right)\sim \frac{\log i}{i+1}.$ So your second coefficient goes to $0$ as $i\to\infty$, so the difference between the sum and your polynomial will approach $\frac{1}{2}x^i.$
Looking for the general solution is tricky, but we can find an approach for each $i$. For example, let $i=5.$
We can work out that $$x^5=f_5(x)-10f_4(x)+25f_3(x)-15f_2(x)+f_1(x),$$ so we get:
$$\sum_{n=1}^{k} n^5 = \frac{1}{6}f_6(k)-2f_5(k)+\frac{25}{4}f_4(k)-5f_3(k)+\frac{1}{2}f_2(k)$$
Then we can expand this out to get the final form.
Note that since $n^i$ is zero a $n=0$ when $i>0$, there is never an $f_0$ term in the expression for $n^i$ in terms of $f_j,$ and hence the result of the sum will only have terms $f_j$ with $j\geq 2$, and hence the resulting polynomial will be divisible by $k(k+1)$ when $i>0.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2713657",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
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Evaluate $\int_{0}^{\frac{\pi}{2}}\frac{x^2}{ \sin x}dx$ I want to evaluate $$\int_{0}^{\frac{\pi}{2}}\frac{x^2}{ \sin x}dx$$
First,I tried to evaluate like this:
$$\int_{0}^{\frac{\pi}{2}}\frac{x^2}{ \sin x}dx=\int_{0}^{\frac{\pi}{2}}x^2\left(\frac{1+\cos x}{\sin x}\right)\frac{dx}{1+\cos x}=\int_{0}^{\frac{\pi}{2}}x^2\left(\frac{1+\cos x}{\sin x}\right)d\left(\frac{\sin x}{1+\cos x}\right)$$
$$=\int_{0}^{\frac{\pi}{2}}x^2d\log\left(\frac{\sin x}{1+\cos x}\right)=x^2\log\left(\frac{\sin x}{1+\cos x}\right)|_{0}^{\frac{\pi}{2}}-2\int_{0}^{\frac{\pi}{2}}x\log\left(\frac{\sin x}{1+\cos x}\right)dx$$
$$=0+2\int_{0}^{\frac{\pi}{2}}x\log\left(\frac{1+\cos x}{\sin x}\right)dx=2\int_{0}^{\frac{\pi}{2}}x\log\left(1+\cos x\right)dx-2\int_{0}^{\frac{\pi}{2}}x\log\left(\sin x\right)dx$$
$$=2\int_{0}^{\frac{\pi}{2}}x\log\cot \left(\frac{x}{2}\right)dx=8\int_{0}^{\frac{\pi}{4}}x\log\cot xdx$$
but I can't proceed next step,help me,thanks.
| As pointed out within the other answers we want to prove that
$$\mathfrak{I}=\int_0^{\pi/2}\frac{x^2}{\sin x}~dx=2\pi G-\frac72\zeta(3)$$
As the OP showed $\mathfrak{I}$ can be reduced to a linear combination of $x$ and the function $\log(\cot x)$
$$\mathfrak{I}=\int_0^{\pi/2}\frac{x^2}{\sin x}~dx=8\int_0^{\pi/4}x\log(\cot x)~dx$$
By applying the definition of the cotangent function followed up by the usage of the well-known Fourier series expansions of $\log(\cos x)$ and $\log(\sin x)$ this can be further simplified. Therefore we get
$$\small\begin{align}
\mathfrak{I}=8\int_0^{\pi/4}x\log(\cot x)~dx&=8\left[\int_0^{\pi/4}x\log(\cos x)~dx-\int_0^{\pi/4}x\log(\sin x)~dx\right]\\
&=8\left[\int_0^{\pi/4}x\left(-\log(2)-\sum_{n=1}^{\infty}(-1)^n\frac{\cos(2nx)}{n}\right)~dx-\int_0^{\pi/4}x\left(-\log(2)-\sum_{n=1}^{\infty}\frac{\cos(2nx)}{n}\right)~dx\right]\\
&=8\left[-\sum_{n=1}^{\infty}\frac{(-1)^n}{n}\underbrace{\int_0^{\pi/4}x\cos(2nx)~dx}_I+\sum_{n=1}^{\infty}\frac{1}{n}\underbrace{\int_0^{\pi/4}x\cos(2nx)~dx}_I\right]\\
\end{align}$$
The inner integral $I$ can be easily evaluated using IBP which leads to
$$I=\int_0^{\pi/4}x\cos(2nx)~dx=\frac{\pi}{8n}\sin\left(n\frac{\pi}2\right)-\frac1{(2n)^2}$$
Plugging this into our original formula and followed by a little bit of algebraic manipulation we get
$$\small\begin{align}
\mathfrak{I}&=8\left[-\sum_{n=1}^{\infty}\frac{(-1)^n}{n}\left(\frac{\pi}{8n}\sin\left(n\frac{\pi}2\right)-\frac1{(2n)^2}\right)+\sum_{n=1}^{\infty}\frac{1}{n}\left(\frac{\pi}{8n}\sin\left(n\frac{\pi}2\right)-\frac1{(2n)^2}\right)\right]\\
&=2\left[\sum_{n=1}^{\infty}\frac{(-1)^n}{n^3}-\sum_{n=1}^{\infty}\frac{1}{n^3}\right]+\pi\left[\sum_{n=1}^{\infty}\frac{1}{n^2}\sin\left(n\frac{\pi}2\right)-\sum_{n=1}^{\infty}\frac{(-1)^n}{n^2}\sin\left(n\frac{\pi}2\right)\right]
\end{align}$$
The first terms can be evaluated in terms of the Riemann Zeta Function $\zeta(s)$ and the Dirichlet Eta Function $\eta(s)$ whereas for the second term we have to consider some basic properties of the Sine function. For $n\in\mathbb{N}>0$ the function $\sin\left(n\frac{\pi}2\right)$ will be zero for all even $n$ and $-1$ and $1$ respectively for odd $n$ starting with $\sin\left(\frac{\pi}2\right)=1$ for $n=1$. Therefore all even terms vanish while the odd ones will remain with a oscillating negative sign. This leads to
$$\small\begin{align}
\mathfrak{I}&=2\left[\sum_{n=1}^{\infty}\frac{(-1)^n}{n^3}-\sum_{n=1}^{\infty}\frac{1}{n^3}\right]+\pi\left[\sum_{n=1}^{\infty}\frac{1}{n^2}\sin\left(n\frac{\pi}2\right)-\sum_{n=1}^{\infty}\frac{(-1)^n}{n^2}\sin\left(n\frac{\pi}2\right)\right]\\
&=2[-\eta(3)-\zeta(3)]+\pi\left[\sum_{n=0}^{\infty}\frac{1}{(2n+1)^2}(-1)^n-\sum_{n=0}^{\infty}\frac{(-1)^{2n+1}}{(2n+1)^2}(-1)^n\right]\\
&=-2[(1-2^{-2})\zeta(3)+\zeta(3)]+2\pi\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)^2}\\
\Leftrightarrow\mathfrak{I}&=-\frac72\zeta(3)+2\pi G
\end{align}$$
where within the last step the functional relation between the Riemann Zeta Function and the Dirichlet Eta Function aswell as the series defintions of Catalan's Constant $G$ where used.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2714146",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 9,
"answer_id": 3
} |
Simplistic Odd Collatz formulas I've been experimenting with the Collatz for a couple of months.
I just want to give you some formulas and show tables that give the same results for odd inputs. I have also several new functions, but ill only provide a few of them here. I tweaked the formulas. It seems very strange that all of them give the same outputs. My notes are a bit messy, so bear with me if I didnt provide all the details.
\begin{array}{|c|c|c|c|}\hline
n & 3n & 3n+1 & \frac{3n+1}{2} \\ \hline
1 & 3 & 4 & 2 \\ \hline
3 & 9 & 10 & 5 \\ \hline
5 & 15 & 16 & 8 \\ \hline
7 & 21 & 22 & 11 \\ \hline
9 & 27 & 28 & 14 \\ \hline
11 & 33 & 34 & 17 \\ \hline
13 & 39 & 40 & 20 \\ \hline
\end{array} orginal collatz formula, dividing by two.
\begin{array}{|c|c|c|c|}\hline
n & 2n & \lfloor\frac{n}{2}\rfloor & 2n-\lfloor\frac{n}{2}\rfloor \\ \hline
1 & 2 & 0 & 2 \\ \hline
3 & 6 & 1 & 5 \\ \hline
5 & 10 & 2 & 8 \\ \hline
7 & 14 & 3 & 11 \\ \hline
9 & 18 & 4 & 14 \\ \hline
11 & 22 & 5 & 17 \\ \hline
13 & 26 & 6 & 20 \\ \hline
\end{array} a new(?) formula.
\begin{array}{|c|c|c|c|}\hline
n & \lfloor\frac{n}{2}\rfloor & n+1 & \lfloor\frac{n}{2}\rfloor + n+1 \\ \hline
1 & 0 & 2 & 2 \\ \hline
3 & 1 & 4 & 5 \\ \hline
5 & 2 & 6 & 8 \\ \hline
7 & 3 & 8 & 11 \\ \hline
9 & 4 & 10 & 14 \\ \hline
11 & 5 & 12 & 17 \\ \hline
13 & 6 & 14 & 20 \\ \hline
\end{array} yet another.
Here $n$ is a positive integer. The floor function omits the fractional part if the result is real.
The functions from the tables above:
$f(n) = \frac{3n+1}{2}$
$f(n) = 2n-\lfloor\frac{n}{2}\rfloor$
$f(n) = \lfloor\frac{n}{2}\rfloor + n+1$
The latter formula is the one i've been working on lately.
We can change it again,
$\lfloor\frac{n}{2}\rfloor+n+1 = \frac{n}{2}+\frac{2n}{2}+\frac{2}{2} = \frac{3n+2}{2}=\lfloor\frac{3n}{2}\rfloor+1$.
That gets us back to the formula in the first table. They look quite similar.
The function
$f(n) = 2n-\lfloor\frac{n}{2}\rfloor$
is also quite interesting. In a "geometric" standpoint we can look at it like doubling the $n$ and then removing the half of the total, here's a simple figure showing this:
I have never seen this formula before, and wonder if I can use this notation:
$f(n)=\lfloor\frac{n}{2}\rfloor+\begin{cases} n+1 & n\equiv 1\pmod2 \\ 0 & n\equiv 0\pmod2 \end{cases}$
See, for odd and even we can halve the input in one go. And then add the result of the cases.
| There is nothing wrong with using an alternate form of the original Collatz ($3x+1$) rule, as long as you have a reason for the modification.
Without using a floor function, it is much easier to show that your method is equivalent to the $\frac{(3n+1)}{2}$ rule. The floor function truncates 0.5 after dividing an odd integer in half, therefore removing 1 from the odd integer in the first place and then dividing by 2 will provide the same effect. (ex. $\lfloor$9/2$\rfloor$ = 4.5 - 0.5 = 4, and (9-1)/2 = 4.)
With that in mind, here is a proof to validate your method:
$\frac{n-1}{2}$ + $n$ + $1$
= $\frac{n-1}{2}$ + $2n\over2$ + $2\over2$
= $\frac{n - 1 + 2n + 2}{2}$
= $\frac{n + 2n + (-1) + 2}{2}$
= $\frac{3n+1}{2}$
The expressions $\frac{3n+1}{2}$ and $\frac{n-1}{2}$ + $n$ + $1$ are equivalent.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2714923",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How to factorize $2x^2-9x+9$ by completing the square? I know that $x^2-bx+c=(x-k)^2=x^2-2kx+k^2$ if it is a complete square. If not we create one by adding and subtracting $\left(\frac{b}{2}\right)^2$
I tried $$2\left(x^2-\frac{9}{2}x+\frac{9}{2}\right)=2\left(x^2-\frac{9}{2}x+\left(\frac{9/2}{2}\right)^2-\left(\frac{9/2}{2}\right)^2+\frac{9}{2}\right)$$
What am I missing and what is the best way to factorize when we have coefficients?(Is this the right term, ($a x^2$)?)
| $$2x^2-9x+9\tag1$$
Factorise
$$2\left(x^2-{9\over 2}x+{9\over 2}\right)\tag2$$
$$2\left(x^2-{9\over 2}x+{9\over 2}\right)=2\left[\left(x-{9\over 4}\right)^2+{9\over 2}-\left(9\over 4\right)^2\right]\tag3$$
$$2\left(x^2-{9\over 2}x+{9\over 2}\right)=2\left[\left(x-{9\over 4}\right)^2-{9\over 16}\right]\tag4$$
$$2x^2-9x+9=2\left(x-{9\over 4}\right)^2-{9\over 8}\tag5$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2721535",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 2
} |
Show the equivalence of norms on polynomial space Let $\|ax+b\|^2:=2a^2+3b^2+2ab$ be a norm on the space of all polynomials of degree one, where $a,b\in\mathbb{R}$.
Hence, $\|ax+b\|_{L^2} ^2:=\int_0^1(ax+b)^2\,\mathrm{d}x$ be another norm ($L^2$-norm).
I want to show equivalence, in the sense that, for all polynomials of degree one
$\exists\, c>0\colon \|\cdot\|\le c\|\cdot \|_{L^2}$ holds and vice versa.
The first part was not too hard,
\begin{align*}
\|ax+b\|_{L^2}^2 &= \int_0^1 (ax+b)^2\,\mathrm{d}x = a^2/3+ab+b^2\le (a+b)^2-ab\le (a+b)^2+\frac{a^2+b^2}{2}\\
&\le \frac{3}{2}a^2+2ab+\frac{3}{2}b^2\le 2a^2+2ab+3b^2=\|ax+b\|^2
\end{align*}
Is there any way to do it better? I noticed, that I got $c=1$, is that in some way significant?
The other way was resistant against all my attempts.
$\|ax+b\|^2=2a^2+2ab+3b^2=\ ...$
Here, I thought that finding a real constant $\alpha$ such that $\alpha(a^3/3+ab+b^2)\ge 2a^2+2ab+3b^2$ would yield success, since then I can find a suitable antiderivative to connect the norms. I was not successful, nevertheless.
| \begin{align*}
\|ax+b\|_{L^2}^2 &= \int_0^1 (ax+b)^2\,\mathrm{d}x = a^2/3+ab+b^2\le a^2 + ab + b^2
\le
2(a^2+ab+b^2) \le 2a^2+2ab+3b^2=\|ax+b\|^2
\end{align*}
Therfore c is atleast $2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2721870",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Basis for an eigenspace Let $A=\begin{pmatrix}
6&3&-8\\
0&-2&0\\
1&0&-3\\
\end{pmatrix}$. I found that the two distinct eigenvalues are $\lambda=5,-2$. Now, I am asked to find a basis for each eigenspace corresponding to each eigenvalue. I found the basis for the eigenspace corresponding to $\lambda=5$ to be $(8,0,1)$. For the eigenspace corresponding to $\lambda=-2$, I get a basis different than the key; below is my work.
We say $\vec{x}$ is an eigenvector of a given matrix A if and only if A$\vec{x}=\lambda\vec{x}$ for some eigenvalue $\lambda$. Thus, we solve for $\vec{x}$. $$(A-\lambda I)\vec{x}=\vec{0}$$ $$\begin{pmatrix}
6&3&-8\\
0&-2&0\\
1&0&-3\end{pmatrix}\vec{x}-\begin{pmatrix}
-2&0&0\\
0&-2&0\\
0&0&-2\\
\end{pmatrix}\vec{x}=\vec{0}$$ $$\begin{pmatrix}
8&3&-8\\
0&0&0\\
1&0&-1\\
\end{pmatrix}\vec{x}=\vec{0}$$ This equates to to solving the linear system: $$\begin{pmatrix}
8&3&-8&0\\
0&0&0&0\\
1&0&-1&0\\
\end{pmatrix}$$
Interchanging rows $1$ and $3$: $$\begin{pmatrix}
1&0&-1&0\\
0&0&0&0\\
8&3&-8&0\\
\end{pmatrix}$$
Eliminating below $1$ in first column: $$\begin{pmatrix}
1&0&-1&0\\
0&0&0&0\\
0&3&0&0\\
\end{pmatrix}$$
Interchanging rows $2$ and $3$ and rescaling: $$\begin{pmatrix}
1&0&-1&0\\
0&1&0&0\\
0&0&0&0\\
\end{pmatrix}$$
Thus I would conclude my basis for the eigenspace corresponding to $\lambda=-2$ is $(1,0,0)$. However, the book insists the basis is $(1,0,1)$. Are there any errors in my work? Thanks in advance!
Edit: I now see my error. The basis is indeed $(1,0,1)$.
| From here,
$$\begin{pmatrix}
8&3&-8\\
0&0&0\\
1&0&-1\\
\end{pmatrix}\vec{x}=\vec{0}$$
we see that $\vec{x}=(1,0,1)$ is the correct solution.
Indeed from here
$$\begin{pmatrix}
8&3&-8&0\\
0&0&0&0\\
1&0&-1&0\\
\end{pmatrix}\stackrel{R_1-8R_3}\to \begin{pmatrix}
0&3&0&0\\
0&0&0&0\\
1&0&-1&0\\
\end{pmatrix}
\implies x_2=0 \quad x_1=x_3$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2723404",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Proving expression$\geq 2$, where variables are $x,y,z$
If $x,y,z$ are distinct real number. Then prove that $$\bigg(\frac{x}{y-z}\bigg)^2+\bigg(\frac{y}{z-x}\bigg)^2+\bigg(\frac{z}{x-y}\bigg)^2\geq 2$$
Try : let $\displaystyle \frac{x}{y-z}=p\Rightarrow x=py-pz\Rightarrow x-py+pz=0$
And let $\displaystyle \frac{y}{z-x}=q\Rightarrow y=qz-qx\Rightarrow qx+y-qz=0$
let $\displaystyle \frac{z}{x-y}=r\Rightarrow z=xr-yr\Rightarrow xr-yr-z=0$
So $$\begin{bmatrix}
1& -p &p \\
q& 1 & -q\\
r& -r & -1
\end{bmatrix}\begin{bmatrix}
x\\
y\\
z
\end{bmatrix}=\begin{bmatrix}
0\\
0\\
0
\end{bmatrix}$$
Which is $AX=B$
So for solution of equation , $|A|=pq+qr+rp+1=0$
Now using $(p+q+r)^2=p^2+q^2+r^2+2(pq+qr+rp)$
So $$p^2+q^2+r^2\geq 2$$
What i have used above is right or wrong. If it is right then i did not understand when will the equality hold
If it is wrong then could someone explain me a right solution, thanks
| $$\bigg(\frac{x}{y-z}\bigg)^2+\bigg(\frac{y}{z-x}\bigg)^2+\bigg(\frac{z}{x-y}\bigg)^2-2=\frac{\left(\sum\limits_{cyc}(x^3-x^2y-x^2z+xyz)\right)^2}{\prod\limits_{cyc}(x-y)^2}\geq0.$$
Also, we can use the following way.
Since $$\sum_{cyc}\left(\frac{x}{y-z}\cdot\frac{y}{z-x}\right)=\sum_{cyc}\frac{xy(x-y)}{\prod\limits_{cyc}(x-y)}=\frac{(x-y)(x-z)(y-z)}{(x-y)(y-z)(z-x)}=-1,$$ we obtain
$$\bigg(\frac{x}{y-z}\bigg)^2+\bigg(\frac{y}{z-x}\bigg)^2+\bigg(\frac{z}{x-y}\bigg)^2-2=\left(\sum_{cyc}\frac{x}{y-z}\right)^2\geq0.$$
The last solution it's just your work:
$$p^2+q^2+r^2=p^2+q^2+r^2+2(pq+pr+qr)+2=(p+q+r)^2+2\geq2.$$
The equality occurs for example, for $x=-y$ and $z=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2724332",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Get variable value from Logarithm? I have below equation.
$$ x = \left(\frac{1}{-2 \log_{10}\left(9 + \frac{46}{x}\right)}\right)^2 $$
I am not very familiar with Logarithm.
How can I grab value of $x$ from this equation?
| For $x>0$ write the equation
$$x = \left(\frac{1}{-2 \log_{10}\left(9 + \frac{46}{x}\right)}\right)^2$$
as
$$\dfrac{1}{-2\sqrt{x}}=\log_{10}\left(9 + \frac{46}{x}\right)=\dfrac{1}{\ln10}\ln\left(9 + \frac{46}{x}\right)$$
or
$$\dfrac{\ln10}{-2\sqrt{x}}=\ln\left(9 + \frac{46}{x}\right)=\ln9+\ln\left(1 + \frac{46}{9x}\right)$$
An approximation could be find with the series
$$\ln(1+z)=z-\dfrac{1}{2}z^2+\dfrac{1}{3}z^3-\dfrac{1}{4}z^4+\dfrac{1}{5}z^5+\cdots$$
by getting some terms!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2724844",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
How has this result been obtained?? We have the quadratic equation
$$ 2y^2 -(1+x)y + x = 0 $$
The author derives the following result from the above equation -
$$ y' = \frac{y-1}{4y-1-x} = \frac{(x-3)y-x+1}{x^2-6x+1} $$
Now, I understand the first part. It has been obtained by simply differentiating w.r.t to $x$ and then solving for $y'$. But, I am unable to obtain the second part.
I have tried lot of different substitutions, but haven't been able to arrive at the result. Please help!!
| There's a way to simplify the denominator without explicitly solving for $y$. Note that
\begin{align}
\big(4y-(1+x)\big)^2 &= 16y^2 - 8(1+x)y + (1+x)^2 \\
&= 8\big(2y^2 - (1+x)y\big) + (1+x)^2 \\
&= -8x + (1+x)^2 \\
&= x^2 - 6x + 1
\end{align}
Therefore
\begin{align}
y' = \frac{y-1}{4y-(1+x)} &= \frac{(y-1)\big(4y-(1+x)\big)}{\big(4y-(1+x)\big)^2} \\
&= \frac{4y^2 - 4y - (1+x)y+(1+x)}{x^2-6x+1} \\
&= \frac{2\big(2y^2-(1+x)y\big)+(1+x)y - 4y+ (1+x)}{x^2-6x+1} \\
&= \frac{-2x+(1+x)y-4y+1+x}{x^2-6x+1} \\
&= \frac{1-x+(x-3)y}{x^2-6x+1}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2728065",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Recreational math problem: $f(x)=\frac{x^2}{2x+101}$ Here is a math problem (just for fun) for the residents of MSE to enjoy:
Let $f(z)$ be defined as
$$f(z)=\frac{z^2}{2z+101}$$
Find a 4-cycle of $f$ - that is, find four distinct complex numbers $z_1,z_2,z_3,z_4$ such that
$$f(z_1)=z_2$$
$$f(z_2)=z_3$$
$$f(z_3)=z_4$$
$$f(z_4)=z_1$$
The answer (my answer, at least) is a bit messy, so watch out! After a correct answer is posted, I will post my solution method.
Enjoy!
| HINT:
Let's do some reductions of $f$:
$$f(101 x) = \frac{(101 x)^2}{2\cdot 101 x + 101}= 101\cdot \frac{x^2}{2x+1}$$
so we have
$$f= s \circ g \circ s^{-1}$$ where $s(x) = 101 x$ and $g(x)=\frac{x^2}{2x+1}$. Further
$$g(x)=\frac{x^2}{2x+1}= \frac{x^2}{(x+1)^2 - x^2}= \frac{1}{(1+\frac{1}{x})^2 -1}= t^{-1}\circ h \circ t(x)$$
where $t(x) = \frac{1}{x}+1$ and $h(x)=x^2$. Finally
$$f=s \circ t^{-1} \circ h \circ t \circ s^{-1}= u \circ h \circ u^{-1}$$
where $u(x)=s\circ t^{-1}(x)=\frac{101}{x-1}$. In other words, $f\circ u = u\circ h$,
$$f(\frac{101}{x-1})=\frac{101}{x^2-1}$$
So $f(x)$ is conjugate to the function $h(x)=x^2$. One checks easily that $x$ starts a cycle of length $4$ for $h$ if and only if $u(x)$ starts a cycle of length $4$ for $f$.
To find cycles of length $4$ for $h$, one looks for solutions of
$x^{16}=x$ that are not solutions of $x^4=x$. That means, $x$ satisfies $x^{15}=1$ but not $x^{3}=1$. There are $15-3=12$ such solutions,
$$x= \exp \frac{2 k \pi i}{15}, \ \ k = 1,2,3,4,6,7,8,9,11,12,13,14$$
They break into $3$ cycles of length $4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2730159",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
Probability - Peculiar die
A peculiar die has the following properties: on any roll the probability of rolling either a $4$, a $6$, or a $1$ is $1/2$, just as it is with an ordinary die. Moreover, the probability of rolling either a $1$, a $3$, or a $2$ is again $1/2$. However, the probability of rolling a $1$ is $5/16$, not $1/6$ as one would expect of an ordinary fair die.
From what you know about this peculiar die, compute the following:
a. The probability of rolling either a $5$, a $3$, or a $2$;
b. The probability of rolling a $5$.
My answer
Given $P(1) = \frac{5}{16}$, then
$$P(2) + P(3) + P(1) = P(2) + P(3) + \frac{5}{16} = \frac{1}{2} \iff P(2) + P(3) = \frac{3}{16}$$
Similarly, using $P(4) + P(6) + P(1) = \frac{1}{2}$, I find
$$P(6) + P(4) = \frac{3}{16}$$
b. $P(5) = 1 - \left(\frac{5}{16} + \frac{3}{16} + \frac{3}{16}\right) = \frac{5}{16}$
a. $P(5) + P(3) + P(2) = \frac{5}{16} + \frac{3}{16} = \frac{1}{2}$
Are my calculations correct?
| I agree with your solution. I wrote a very similar solution below.
Let $E_j$ be the event that we roll a $j$. Then
$$
P(E_2 \cup E_3) = P(E_1 \cup E_2 \cup E_3) - P(E_1) = \frac12 - \frac{5}{16} = \frac{3}{16}.
$$
It follows that
\begin{align}
P(E_5) &= 1 - P(E_1 \cup E_2 \cup E_3 \cup E_4 \cup E_6) \\
&= 1 - P(E_1 \cup E_4 \cup E_6) - P(E_2 \cup E_3) \\
&= 1 - \frac12 - \frac{3}{16} = \frac{5}{16}.
\end{align}
Finally, we can put these pieces together to see that
$$
P(E_2 \cup E_3 \cup E_5) = P(E_2 \cup E_3) + P(E_5) = \frac{3}{16} + \frac{5}{16} = \frac12.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2730443",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Evaluate $\lim_{x \to \infty} e^{-x^2} \int_x^{x + \ln(x)/x} e^{t^2} dt $
Evaluate
$$\lim \limits_{x \to \infty} \frac{1}{e^{x^2}}\int \limits_x^{x + \frac{\ln x}{x}} e^{t^2} dt.$$
I should probably use such inequality:
$$ \int \limits_x^{x + \frac{\ln x}{x}} e^{t^2} dt > \frac{e^{x^2} \ln x}{2x} \text { for } x \ge 1,$$
which gives me that the numerator goes to infinity. I used L'Hopital rule, but without any conclusions.
$$\frac{d}{dx}(\int \limits_x^{x + \frac{\ln x}{x}} e^{t^2} dt) = e^{x^2} (-1 + e^{\frac{\ln^2(x)}{x^2}} (1 + x^2) - e^{\frac{\ln^2(x)}{x^2}} \ln(x))$$
and
$$\frac{d}{dx}(e^{x^2}) = 2 x e^{x^2}.$$
Please help.
| Let $$I(x) =\int \limits_x^{x + \frac{\ln x}{x}} e^{t^2} dt \Rightarrow I(x) \geq \frac{\ln x}{x}e^{x^2} \stackrel{x\rightarrow\infty}{\longrightarrow}\infty$$
So, we have with $\frac{I(x)}{e^{x^2}}$ a L'Hospital case of $\frac{\infty}{\infty}$ for $x\rightarrow\infty$:
$$\frac{I(x)}{e^{x^2}} \sim \frac{e^{(x^2+2\ln x+ \frac{\ln^2x}{x^2})}(\frac{x^2-\ln x + 1}{x^2}) - e^{x^2}}{2xe^{x^2}}=\frac{e^{\frac{\ln^2x}{x^2}}(x^2-\ln x +1)}{2x}- \frac{1}{2x} = \frac{1}{2}e^{\frac{\ln^2x}{x^2}} \left(x - \frac{\ln x}{x} + \frac{1}{x} \right)- \frac{1}{2x}\stackrel{x\rightarrow\infty}{\longrightarrow}\infty$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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If a,b, and c is real, and following the equation, solve for
If $a,b,c$ are real and aren't 0, that complete the following equations
$a^2 + a = b^2$
$b^2 + b = c^2$
$c^2 + c = a^2$
Solve for $(2a+c)(2b+a)(2c+b)$,
What I've done:
*
*Adding all of the equation of it and substact with $(a^2 + b^2 + c^2)$ of both side of the equation
$a+b+c=0$
*Substituting $a = -(b+c); b=-(c+a); c=-(a+b)$ to equation $(1),(2),(3)$ respectively:
$c^2 -b -c + 2bc= 0$
$a^2 - c -a + 2ac = 0$
$b^2 -b -a +2ab = 0$
*Expanding the equation into $4a^2c+2a^2b+9acb+4ab^2+2ac^2+4c^2b+2cb^2$
then re-arranged by it's constant
$4(a^2c+ab^2+c^2b)+2(a^2b+ac^2+cb^2)+9abc$
focused on
$2(a^2b+ac^2+cb^2)$
which can be written as
$2((a)ab+(c)ac+(b)cb)$
and subsituting $a, b, c$ that inside parentheses
$2(-(b+c)(ab)-(a+b)ac-(a+b)cb)
-2(ab^2 + a^2c + cb^2 +3abc)$
removing $3abc$, and $Cb^2$ from the parentheses
$-2(ab^2 + a^2c) - 6abc -cb^2$
so I get
$(4-2)(a^2c+ab^2+cb^2)+9abc-6abc$
$2(a^2c+ab^2+cb^2)+3abc$
But that's seems I can go to, I didn't find anything that could be useful anymore
| Hah!
$a + b + c = 0$
And $a = b^2 - a^2 = (b-a)(b+a) = (b-a)(-c) = c(a-b)$
$b = c^2 - b^2 = (c-b)(c+b) = (c-b)(-c) = a(b-c)$
$c = a^2 - c^2 = (a-c)(a+c) = (a-c)(-b) = b(c-a)$.
and $(2a + c) = (a +(a+c)) = (a-b) = \frac ac$
$(2b + a) = (b + (b + a)) = (b-c) = \frac ba$
$(2c + b) = (c + (c+b)) = (c -a) = \frac cb$
So
$(2a+c)(2b+a)(2c+b) = \frac ac\frac ba \frac cb = 1$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Can this equation be solved by radicals? $x^6+x^5+2x^4-2x^3-x^2+1=0$ Can this equation be solved by radicals? If it can be solved, what is the method? At least, I will resolve it myself if you specify reference or source.
$$x^6+x^5+2x^4-2x^3-x^2+1=0$$
| Yes because$$x^6+x^5+2x^4-2x^3-x^2+1=0\iff (x^3-1)^2+x^2(x^3-1)+2x^4=0$$ It follows solving as a quadratic in $x^3-1$
$$2(x^3-1)=-x^2\pm\sqrt{-7}x\iff2x^3+x^2\mp\sqrt{-7}x^2-2=0$$ so you can apply Cardano's formula.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2734487",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove that $9^n+4^{n+1}$ is a multiple of $5$ for all $n \in \mathbb{N}$
Prove that $9^n+4^{n+1}$ is a multiple of $5$ for all $n \in \mathbb{N}$
Proof. So i'm going to prove this by induction. The first case when $n=1$ is trivial since:
$$9+16=25,$$
implying $ 5 \mid 25$.
Now we need to show is divisible when $n=k+1$. We will use this later on, but $$9^k+4^{k+1}=5k.$$
The
n,
$$\begin{align}9^{k+1}+4^{k+2}
&= 9 \cdot 9^k+4 \cdot 4^{k+1} \\
&= 9 \cdot 9^k+(9-5) \cdot 4^{k+1} \\
&=9 \cdot 9^k+9 \cdot 4^{k+1} -5 \cdot 4^{k+1} \\
&= 9(9^k + 4^{k+1})-5 \cdot 4^{k+1} \\
&=\underbrace{9(5k)-5 \cdot 4^{k+1}}_{\text{This is where the $5k$ comes in}} \\
&= 5(9k-4^{k+1}),\end{align}$$
thus, the original expression a multiple of $5$.
Is my induction correct?
Edit: I see several answers that took a different approach, all is welcome it really helps me see it in a different way. Thank You!
| Or another way:
$$9^n + 4^{n+1} = (10-1)^n +(5-1)^{n+1}=10c+(-1)^n + 5d+(-1)^{n+1}=10c+5d = 5e$$
Edit:
Another similar way:
$$9^n + 4^{n+1}= 9^n+4\cdot(9-5)^n = 9^n+4(9^n+5c)= (1+4)9^n + 20c = 5e$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2735856",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 3
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Showing that $\frac{1}{2\pi}\int^{2\pi}_{0}(2\cos\theta)^{2n}d\theta=\frac{(2n)!}{n!n!}$ As the title suggests, I would like to show that: $$\frac{1}{2\pi}\int^{2\pi}_{0}(2\cos\theta)^{2n}d\theta=\frac{(2n)!}{n!n!}$$ for every positive integer $n$.
How can I go about doing that? I understand the that there are already solutions for the problem: $\frac{1}{2\pi}\int^{2\pi}_{0}(\cos\theta)^{2n}d\theta= {2n \choose n} \frac{\pi}{2^{2n-1}}$. Is there any modifications I can do to obtain the desired solution?
| You can solve the integral by induction. First of all, you may notice that
\begin{align*}
\frac{1}{2\pi}\int_0^{2\pi}(2\cos\theta)^2 d\theta&=\frac{1}{\pi}\Big[\theta+\cos\theta\sin\theta\Big]_0^{2\pi}=2=\frac{2!}{1!1!}
\end{align*}
Thus the case $n=1$ holds. Now, integrating by parts:
\begin{align*}
\frac{1}{2\pi}\int_0^{2\pi}(2\cos\theta)^{2n} d\theta&=\frac{2^{2n}}{2\pi}\Big(\Big[\cos^{2n-1}\theta\sin\theta\Big]_0^{2\pi}+(2n-1)\int_0^{2\pi}\sin^2\theta\cos^{2n-2}\theta d\theta\Big)\\
&=\frac{2^{2n}(2n-1)}{2\pi}\int_0^{2\pi}(\cos^{2n-2}\theta-\cos^{2n}\theta)d\theta\\
\end{align*}
then
\begin{align*}
\frac{1}{2\pi}\int_0^{2\pi}(2\cos\theta)^{2n} d\theta+\frac{(2n-1)}{2\pi}\int_0^{2\pi}(2\cos\theta)^{2n}d\theta&=\\ \frac{2n}{2\pi}\int_0^{2\pi}(2\cos\theta)^{2n}d\theta&=\\
&=\frac{2^{2n}(2n-1)}{2\pi\cdot2^{2n-2}}\int_0^{2\pi}(2\cos\theta)^{2n-2}d\theta\\
\end{align*}
that is
\begin{align*} \int_0^{2\pi}(2\cos\theta)^{2n} d\theta &=\frac{2\pi}{2n}\cdot\frac{2^{2n}(2n-1)}{2\pi\cdot2^{2n-2}}\int_0^{2\pi}(2\cos\theta)^{2n-2} d\theta=\\
&=\frac{2^2(2n-1)}{2n}\cdot\frac{(2n-2)!}{(n-1)!(n-1)!}=\frac{2n(2n-1)!}{n!n!}=\frac{(2n)!}{n!n!}
\end{align*}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 2
} |
How to find value of floor function of a given number? I am considering numbers $n = 3m + 1$ and $p = 3q$, where $m$ and $q$ are some positive integers.
I want to find the value of $\lfloor \frac{3m + 1}{2} \rfloor$ and $\lfloor \frac{3q}{2} \rfloor$. I know somewhere the answer depends on the values of $m$ and $q$.
For a particular number I can write but my mind is not working to find the generalised value.
Kindly help.
| Assume that $m=2k+1$ ($k$ is a positive integer), then
$$\lfloor \frac{3m + 1}{2} \rfloor=\lfloor \frac{3(2k+1)+1}{2} \rfloor=\lfloor \frac{6k+4}{2} \rfloor=\lfloor 3k+2 \rfloor=3k+2.$$
If $m$ is an even number or $m=2k$, then
$$\lfloor \frac{3m + 1}{2} \rfloor=\lfloor \frac{3(2k)+1}{2} \rfloor=\lfloor \frac{6k+1}{2} \rfloor=\lfloor 3k+0.5 \rfloor=3k.$$
This is because when $k$ is an integer, $3k<3k+0.5<3k+1.$
For $\lfloor \frac{3q}{2} \rfloor$, you can also do similarly by splitting into cases whether $q$ is an even or odd number.
| {
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"timestamp": "2023-03-29T00:00:00",
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Integral $\int \frac {dx}{\sin^2 x + \tan^2x}$
I am trying to find the following integral:
$$\int \frac {dx}{\sin^2 x + \tan^2x}$$
I have tried the common thing to do when encountering rational functions that contains rational functions and converting everything in terms of $\tan \frac{x}{2}$ then substituting it.
$$\tan\frac{x}{2}=t\Rightarrow \sin x=\frac{2t}{1+t^2},\ \cos x=\frac{1-t^2}{1+t^2},\ dx=\frac{2}{1+t^2}dt$$
$$\Rightarrow \int \frac{dx}{\sin^2 x + \frac{\sin^2 x}{\cos^2 x}}=\int \frac{\frac{2}{1+t^2}}{\left(\frac{2t}{1+t^2}\right)^2+\left(\frac{2t}{1-t^2}\right)^2}dt$$
However I am stuck. Is there perhaps an easier way to approach it?
| $$\begin{align}\int\dfrac{dx}{\sin^2{(x)}+\tan^2{(x)}}\cdot\dfrac{\sec^2{x}}{\sec^2{x}}&=\int\dfrac{\sec^2{x}\ dx}{\tan^2{x}+\sec^2{x}\tan^2{x}}\\
&=\int\dfrac{\sec^2{x}\ dx}{\tan^2{x}+(\tan^2{x}+1)\tan^2{x}}\\
&=\int\dfrac{\sec^2{x}\ dx}{2\tan^2{x}+\tan^4{x}}\end{align}$$
Let $u=\tan{x}$ and $du=\sec^2{x}\ dx$
$$\begin{align}
&=\int\dfrac{du}{u^4+2u^2}\\
&=\int\left(\dfrac{1}{2u^2}-\dfrac{1}{2(u^2+2)}\right)du\\
&=-\dfrac{1}{2}\int\dfrac{du}{u^2+2}+\dfrac{1}{2}\int\dfrac{du}{u^2}\\
&=-\dfrac{1}{2}\int\dfrac{du}{2(\frac{u^2}{2}+1)}+\dfrac{1}{2}\int\dfrac{du}{u^2}\\
&=-\dfrac{1}{4}\int\dfrac{du}{\frac{u^2}{2}+1}+\dfrac{1}{2}\int\dfrac{du}{u^2}\end{align}$$
Let $s=\frac{u}{\sqrt{2}}$ and $ds=\frac{du}{\sqrt{2}}$.
\begin{align}
&=-\dfrac{1}{2\sqrt{2}}\int\dfrac{ds}{s^2+1}+\dfrac{1}{2}\int\dfrac{du}{u^2}\\
&=-\frac{\arctan{s}}{2\sqrt{2}}+\dfrac{1}{2}\int\dfrac{du}{u^2}\\
&=-\frac{\arctan{s}}{2\sqrt{2}}-\frac{1}{2u}+C\\
&=-\frac{\sqrt{2}u\arctan{\frac{u}{\sqrt{2}}+2}}{4u}+C\\
&=-\frac{1}{4}\left(\sqrt{2}\arctan{\frac{\tan{x}}{\sqrt{2}}\tan{x}+2} \right)\cot{x}+C
\end{align}
Which is equal to
$$-\frac{1}{4}\left(\sqrt{2}\arctan{\frac{\tan{x}}{\sqrt{2}}+2\cot{x}}\right)+C.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2745598",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
How many solution does $1/a+1/b+1/c+1/d=1$ have? From my friend, he gives me a competition question:
"How many solution $(a,b,c,d)$ does $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}=1$ have where $a,b,c,d$ are positive integers? (the size of $a,b,c,d$ doesn't matter, either one can be the biggest or smallest, and they are not necessarily distinct)"
I want to ask if there is any solution shorter than mine? I think mine is too long, and maybe yields a wrong answer.
My solution:
WLOG, let $a\leq b\leq c\leq d$
$$1=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\leq \frac{4}{a}$$
$$a\leq4$$
Because $a=1$ yields no solution, so consider $a=2,3,4$
Case 1:$a=2$, then $\frac{1}{b}+\frac{1}{c}+\frac{1}{d}=\frac{1}{2}$
Do that again: $\frac{1}{2}=\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\leq\frac{3}{b}$, so $b\leq 6$.
Let $b=6$, then $\frac{1}{c}+\frac{1}{d}=\frac{1}{3}$, going to $$(c-3)(d-3)=9$$ $$(c,d)=(4,12),(6,6)$$ so in the case have: $(a,b,c,d)=(2,6,4,12),(2,6,6,6)$ then eliminate some case not satisfy $a\leq b\leq c\leq d$
Then going through when $b=5$,$b=4$,$b=3$... yields $184$ distinct solutions.
Case 2: Following the same procedure as Case 1... yields $18$ solutions.
Case 3: As above... yields only a solution which is $(4,4,4,4)$
Conclude it, the equation has $203$ solutions.
That is my solution, I wrote it using one and a half piece of A4 paper, I have recently tried $abcd=abc+abd+acd+bcd$ but don't know how to continue, or should I use Vieta theorem?
---After first edit---
According to Robert Z, I had miscount quadruplet $(3,4,4,6)$ which add up the count to $215$ solutions.
-- After last edit --
Seems like there is no faster solution, I will close this question and marked as solved. Thanks to everyone who spend effort to my question.
| In my opinion your approach is fine and I am not aware of a faster method.
However I got a different number of solutions.
Assuming that $a\leq b\leq c\leq d$ then
1) if $a=2$ and $b=3$ then $(c,d)$ can be
$$(7,42),\;(8,24),\;(9,18),\;(10,15),\;(12,12).$$
2) if $a=2$ and $b=4$ then $(c,d)$ can be
$$(5,20),\;(6,12),\;(8,8).$$
3) if $a=2$ and $b\geq 5$ then $(b,c,d)$ can be
$$(5,5,10),\;(6,6,6).$$
4) if $a=3$ then $(b,c,d)$ can be
$$(3,4,12),\;(3,6,6),\;(4,4,6).$$
5) if $a=4$ then $(b,c,d)=(4,4,4)$.
Hence rearranging the $14$ ordered solutions we find the total number of solutions:
$$6\cdot 4!+5\cdot \frac{4!}{2}+1\cdot \frac{4!}{2!2!}+1\cdot \frac{4!}{3!}+1=215.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2746906",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Taylor series $\sqrt{1-2x+x^2+o(x^3)}$ I need to solve this asymptotic expansion for $x\to0$
$$\sqrt{1-2x+x^2+o(x^3)}$$
This is the expansion to use:
$$\sqrt{1+t} = 1+ \frac{t}{2}-\frac{t^2}{8}+\frac{t^3}{16}+o(x^3)$$
let:
$$t = -2x+x^2+o(x^3)$$
My question is, should I stop to the first order or keep going?
How can I decide when to stop?
I have tried first and second order and I get:
$$\sqrt{1+t}= 1+ \frac{-2x+x^2+o(x^3)}{2}+o(-2x+x^2+o(x^3) = 1-x+\frac{1}{2}x^3+o(x^3)$$
The second grade:
$$\sqrt{1+t}= 1+ \frac{-2x+x^2+o(x^3)}{2} - \frac{\big(-2x+x^2+o(x^3)\big)^2}{8} +o\big(-2x+x^2+o(x^3)\big)^2 = 1-x+4x^2+\frac{1}{2}x^3-4x^3-\frac{1}{8}x^4+x^4+o(x^4)$$
| Since $1-2x+x^2 = (1-x)^2$, it is more practical to check which coefficient $K$ ensures
$$\left[1-x+K x^2+o(x^3)\right]^2 = 1-2x+x^2+o(x^3) $$
but the coefficient of $x^2$ in the LHS is $2K+1$, hence $K$ has to be $0$.
Just a different way for putting Lubin's argument.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2747341",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Stuck calculating the derivative of $f(x)=\log_{10}{\frac{x}{1+\sqrt{5-x^2}}}$. I have to calculate the derivative of this:
$$f(x)=\log_{10}{\frac{x}{1+\sqrt{5-x^2}}}$$
But I'm stuck. This is the point where I have arrived:
$$f'(x) = \frac{(1+\sqrt{5-x^2})(\sqrt{5-x^2})+x^2}{x(\ln 10)(1+\sqrt{5-x^2})(\sqrt{5-x^2})}$$
How can I simplify? I didn't include all the passages.
| Since the denominator is positive (when it's defined, that is, for $-\sqrt5\le x\le \sqrt5 $), the function is only defined for $0<x\le \sqrt5 $, so you can write it as
$$
f(x)=\log_{10}x-\log_{10}(1+\sqrt{5-x^2})
$$
which should simplify the computation very much. Recall that the derivative of $\log_{10}{x}$ is
$$
\frac{1}{x\log10}
$$
(natural logarithm, write it ln if you prefer) and apply the chain rule.
You will get
$$
\frac{1}{\log10}\left(
\frac{1}{x}-\frac{1}{1+\sqrt{5-x^2}}\frac{-x}{\sqrt{5-x^2}}
\right)
$$
Then it's just simplifications, if you really want to do more than that:
$$
\frac{1}{\log10}\frac{\sqrt{5-x^2}+5-x^2+x^2}{x\sqrt{5-x^2}(1+\sqrt{5-x^2})}
=
\frac{1}{\log10}\frac{\sqrt{5-x^2}+5}{x\sqrt{5-x^2}(1+\sqrt{5-x^2})}
$$
The derivative doesn't exist at $x= \sqrt5 $.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2748629",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
If $x$, $y$, and $z$ are real numbers such that $x+y+z=8$ and $x^2+y^2+z^2=32$, what is the largest possible value of $z$? I tried swapping $z$ from the first equation to the second, and got $$x^2 + x y - 8 x + y^2 - 8 y + 16=0$$ Not sure where to go from there, and if I'm on the right track at all.
| \begin{align*}
x^2+(8-x-z)^2+z^2&=32\\
x^2-(8-z)x+z^2-8z+16&=0
\end{align*}
As $x$ is real,
\begin{align*}
[-(8-z)]^2-4(1)(z^2-8z+16)&\ge0\\
3z^2-16z&\le 0\\
0\le z&\le\frac{16}{3}
\end{align*}
The largest value of $z$ is $\displaystyle \frac{16}{3}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2750329",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 0
} |
Find all positive integers $n$ such that $n^2 + 1$ is divisible by $n+ 1$
Find all positive integers $n$ such that $n^2 + 1$ is divisible by $n+ 1$
I know a lot about number theory, for example conjectures, theories, equations and so on. But I just can't solve this type of problems. Which algorithm should i use in order to solve them?
| Another proof:
$$gcd(n+1,n^2+1)=gcd(n+1, n^2+1-n(n+1))=gcd(n+1,1-n)=gcd(n+1,1-n+(1+n))=gcd(n+1,2).$$
If $n$ is even, then $n+1$ is odd. Then $gcd(n+1,2)=1$ (is this clear?), this is:
$gcd(n+1,n^2+1)=1$ so that $n+1,n^2+1$ are relative primes.
If $n$ is odd, then $n+1$ is even, so $gcd(n+1,2)=2$, thus $gcd(n+1,n^2+1)=2$.
Remember that: If $a>0$, then $gcd(a,b)=a \iff a \vert b. $
Then $n+1 \vert n^2+1 \iff gcd(n+1,n^2+1)=n+1,$ then $n+1$ must be equal to $2$, then $n=1$ is the only positive integer that satisfies the property.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2751864",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 1
} |
To find a relation between $a,b,c$ given a quadratic and a cubic epression If $x^2+px+1$ is a factor of $ax^3+bx+c$ then which of the following is true:
a)$a^2-b^2=ac$
b)$a^2-c^2=ab$
c)$a^2+b^2=bc$
d)$a^2+c^2=ab$
I assumed the roots of $x^2+px+1$ to be $k$ and $\frac{1}{k}$ since product of roots is 1 . $ax^3+bx+c$ would have roots $k$, $\frac{1}{k}$ and $r$. So
$$(k)( \frac{1}{k})(r)=-c \Rightarrow -c=r$$
So $-c$ is a root and hence $$a(-c)^3+b(-c)+c=0 \Rightarrow -ac^2-b+1=0$$ as $c\neq0$ but there is no such solution so how should I proceed??
| $ax^3+bx+c$ is the product of $x^2+px+1$ and a linear polynomial. This polynomial must be $ax+c$.
$$ax^3+bx+c=(ax+c)(x^2+px+1)$$
So, $c+ap=0$ and $a+pc=b$.
$$0=c(c+ap)=c^2+acp=c^2+a(b-a)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2751970",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
AC^2 Algebraic simplifications Suppose I know that:
$AB^2=DF^2+CF^2$;
$AD^2=BE^2+CE^2$;
$AC^2=AE^2+CE^2$;
$AC^2=AF^2+CF^2$
And also:
$DF=AF-AD$ and $BE=AE-AB$
How can I prove that $AB * AE + AD * AF = AC^2$?
I start like this, but then I get stucked:
$AB*AE + AD*AF = AB*(BE+AB) + AD*(DF+AD) =
AB*BE + AB^2 + AD*DF + AD^2 =
AB*BE + AB^2 + AD*DF + BE^2 + CE^2 =
(AE - BE)*BE + AB^2 + AD*DF + BE^2 + CE^2 =
AE*BE - BE^2 + AB^2 + AD*DF + BE^2 + CE^2 =
AE*BE + AB^2 + AD*DF + CE^2 =
AE*(AE - AB) + AB^2 + AD*DF + CE^2 =
AE^2 - AE*AB + AB^2 + AD*DF + CE^2...$
Can anyone give a hint how to continue?
Thanks!
| Let $AB=x$, $BE=y$, $AD=p$, $DF=q$, now from the first $4$ equations we have
\begin{eqnarray*}
p^2+(x+y)^2=y^2+CE^2+(x+y)^2=y^2+AC^2 \\=y^2+(p+q)^2+CF^2=y^2+p^2+2pq+x^2
\end{eqnarray*}
and so $xy=pq$.
Now
\begin{eqnarray*}
AB \times AE+ AD \times AF = x(x+y)+p(p+q)
\end{eqnarray*}
and
\begin{eqnarray*}
AC^2 = (x+y)^2+CE^2=(x+y)^2+p^2-y^2
\end{eqnarray*}
and thus $\color{red}{AB \times AE+ AD \times AF =AC^2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2752097",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Solve a trigonometric equation: $|1-2\sin^2 x|=|\cos x|$ I have difficulty in solving this equation:
$$|1-2\sin^2 x|=\lvert\cos x\rvert.$$
What are the indications to solve the exercise correctly?
| $1 - 2\sin^2 x = 2\cos^2 x - 1\\
2\cos^2 x - 1 = \pm \cos x\\
2\cos^2 x \pm\cos x - 1 = 0$
Apply the quadratic formula:
$\cos x = \pm \frac 14 \pm \frac 34\\
\cos x = \pm \frac 12, \pm 1 $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2753786",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
Find the polynomial if remainder is given If $f$ is a quintic polynomial which leaves remainder $1$ when divided by $(x-1)^3$, and $-1$ when divided by $(x+1)^3$ , then find the value of first derivative of $f$ at $x=2$.
My approach
Let $$ f = A(x-1)^5 + B(x-1)^4 + C(x-1)^3 +1 $$
Also
$$ f = A(x+1)^5 + D(x+1)^4 + E(x+1)^3 -1$$
Is this method correct? Also I want to discuss other methods to solve these kind of problems.
| Another approach is the extended Euclidean algorithm to find the quintic $f,$ although there is now the possibility of fractions:
$$ \left( x^{3} + 3 x^{2} + 3 x + 1 \right) \left( \frac{ 3 x^{2} - 9 x + 8 }{ 16 } \right) - \left( x^{3} - 3 x^{2} + 3 x - 1 \right) \left( \frac{ 3 x^{2} + 9 x + 8 }{ 16 } \right) = \left( 1 \right) $$
=================================================
$$ \left( x^{3} + 3 x^{2} + 3 x + 1 \right) $$
$$ \left( x^{3} - 3 x^{2} + 3 x - 1 \right) $$
$$ \left( x^{3} + 3 x^{2} + 3 x + 1 \right) = \left( x^{3} - 3 x^{2} + 3 x - 1 \right) \cdot \color{magenta}{ \left( 1 \right) } + \left( 6 x^{2} + 2 \right) $$
$$ \left( x^{3} - 3 x^{2} + 3 x - 1 \right) = \left( 6 x^{2} + 2 \right) \cdot \color{magenta}{ \left( \frac{ x - 3 }{ 6 } \right) } + \left( \frac{ 8 x }{ 3 } \right) $$
$$ \left( 6 x^{2} + 2 \right) = \left( \frac{ 8 x }{ 3 } \right) \cdot \color{magenta}{ \left( \frac{ 9 x }{ 4 } \right) } + \left( 2 \right) $$
$$ \left( \frac{ 8 x }{ 3 } \right) = \left( 2 \right) \cdot \color{magenta}{ \left( \frac{ 4 x }{ 3 } \right) } + \left( 0 \right) $$
$$ \frac{ 0}{1} $$
$$ \frac{ 1}{0} $$
$$ \color{magenta}{ \left( 1 \right) } \Longrightarrow \Longrightarrow \frac{ \left( 1 \right) }{ \left( 1 \right) } $$
$$ \color{magenta}{ \left( \frac{ x - 3 }{ 6 } \right) } \Longrightarrow \Longrightarrow \frac{ \left( \frac{ x + 3 }{ 6 } \right) }{ \left( \frac{ x - 3 }{ 6 } \right) } $$
$$ \color{magenta}{ \left( \frac{ 9 x }{ 4 } \right) } \Longrightarrow \Longrightarrow \frac{ \left( \frac{ 3 x^{2} + 9 x + 8 }{ 8 } \right) }{ \left( \frac{ 3 x^{2} - 9 x + 8 }{ 8 } \right) } $$
$$ \color{magenta}{ \left( \frac{ 4 x }{ 3 } \right) } \Longrightarrow \Longrightarrow \frac{ \left( \frac{ x^{3} + 3 x^{2} + 3 x + 1 }{ 2 } \right) }{ \left( \frac{ x^{3} - 3 x^{2} + 3 x - 1 }{ 2 } \right) } $$
$$ \left( x^{3} + 3 x^{2} + 3 x + 1 \right) \left( \frac{ 3 x^{2} - 9 x + 8 }{ 16 } \right) - \left( x^{3} - 3 x^{2} + 3 x - 1 \right) \left( \frac{ 3 x^{2} + 9 x + 8 }{ 16 } \right) = \left( 1 \right) $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2755018",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Evaluating $\lim _{x\to \infty} (\sqrt[3]{(x+1)(x+2)(x+3)}-x)$
Evaluate: $\lim _{x\to \infty}( \sqrt[3]{(x+1)(x+2)(x+3)}-x)$
I set $y= x-3$ for simplification and then tried to solve $y\to \infty$.
I tried to use the tayor expansion of the cubic function f(y) in the cuberoot but that didn't help.
How do I approach this problem then?
| Using general Binomial expansion, $((x+1)(x+2)(x+3))^\frac{1}{3}-x = ((x+2)^3-(x+2))^\frac{1}{3}=(x+2)-\frac{1}{3}(x+2)^{-2}(x+2)-\frac{1}{9}(x+2)^{-5}(x+2)^3+..-x=2-P(x+2)$,
where the power of the polynomial $P(x+2)$ is negative. Letting $x \to \infty$, we get the limit 2.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2756029",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
Without using the Rule of Sarrus, prove that: Without using the Rule of Sarrus, prove that:
$$\left|
\begin{matrix}
(b+c)&(a-b)&a \\
(c+a)&(b-c)&b \\
(a+b)&(c-a)&c \\
\end{matrix}\right|=3abc-a^3-b^3-c^3$$
My Approach:
$$LHS=
\left|
\begin{matrix}
(b+c)&(a-b)&a \\
(c+a)&(b-c)&b \\
(a+b)&(c-a)&c \\
\end{matrix}\right|$$
$$C_1\to C_1+C_2$$
$$=
\left|
\begin{matrix}
(c+a)&(a-b)&a \\
(a+b)&(b-c)&b \\
(b+c)&(c-a)&c \\
\end{matrix}\right|$$
$$C_1\to C_1-C_3$$
$$=
\left|
\begin{matrix}
c&(a-b)&a \\
a&(b-c)&b \\
b&(c-a)&c \\
\end{matrix}\right|$$
How do I complete the rest?
| \begin{align}
\left|
\begin{matrix}
(b+c)&(a-b)&a \\
(c+a)&(b-c)&b \\
(a+b)&(c-a)&c \\
\end{matrix}\right|&=
\left|
\begin{matrix}
a+b+c&a-b&a \\
a+b+c&b-c&b \\
a+b+c&c-a&c \\
\end{matrix}\right|\\
&= (a+b+c)\left|
\begin{matrix}
1&a-b&a \\
1&b-c&b \\
1&c-a&c \\
\end{matrix}\right|\\
&= (a+b+c)\left|
\begin{matrix}
1&-b&a \\
1&-c&b \\
1&-a&c \\
\end{matrix}\right|\end{align}
Can you continue from here?
| {
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"url": "https://math.stackexchange.com/questions/2756209",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to find the sum $1+\frac{1}{2}-\frac{1}{4}-\frac{1}{5}+\frac{1}{7}+\frac{1}{8}-\frac{1}{10}-\frac{1}{11}+\cdots =\ ?$ Let $\phi(x)=\begin{cases}0, & 0\lt x\lt 1\\ 1, & 1\lt x\lt3 \end{cases}$
We have that the Fourier cosine series is given by $$\phi(x)=\begin{cases}0, & 0\lt x\lt1\\ \frac{4}{3}+\displaystyle\sum_{m=1}^{\infty}\frac{-2\sin\frac{m\pi}{3}}{m\pi}\cos\frac{m\pi x}{3}, & 1\lt x\lt3 \end{cases}$$
Put $x=0$ to find the sum
$\displaystyle 1+\frac{1}{2}-\frac{1}{4}-\frac{1}{5}+\frac{1}{7}+\frac{1}{8}-\frac{1}{10}-\frac{1}{11}+\cdots$
I tried the following
$$\phi(0)=\frac{4}{3}+\sum_{m=1}^{\infty}\frac{-2\sin\frac{m\pi}{3}}{m\pi}\\=\frac43-2\frac{\sin\frac{\pi}{3}}{\pi}-\frac{\sin\frac{2\pi}{3}}{\pi}-2\frac{\sin\pi}{3\pi}-\frac{\sin\frac{4\pi}{3}}{2\pi}-\cdots\\=\frac{4}{3}-\frac{\sqrt3}{\pi}-\frac{\sqrt3}{2\pi}-0+\frac{\sqrt3}{4\pi}\dots=\frac{4}{3}-\frac{\sqrt3}{\pi}(1+\frac{1}{2}-\frac{1}{4}\dots)=\ ? $$
And I'm stuck here,
What can I do here?
I greatly appreciate any assistance you may provide.
| The key observation, which will be exploited below in computing the series
$$
S=\sum_{n=1}^\infty\frac{\sigma_n}{n},
$$
where $\sigma_n$ is repeating sequence of $(1,1,0,-1,-1,0)$, is that
$$
\sigma_n=\frac{2}{\sqrt3}\sin\frac{\pi n}{3}.
$$
It follows:
$$
S=\frac{2}{\sqrt3}S',\text{ with } S'=\sum_{n=1}^\infty\frac{\sin\frac{\pi n}{3}}{n}.\tag{1}
$$
It appears to be simpler to compute directly $S'$. Two approaches to perform this are given below.
The first way, suggested in question, exploits the Fourier series of the even periodic function:
$$
\phi(x)=\begin{cases}
0& -1<x<1\\
1& -3<x<-1\text{ or } 1<x<3
\end{cases};\quad\phi(x+6)=\phi(x),
$$
which is:
$$
\phi(x)=\frac{2}{3}-\sum_{n=1}^\infty\frac{2\sin\frac{\pi n}{3}}{\pi n}\cos\frac{\pi n x}{3}.\tag{2}
$$
Substituting in (2) $x=0$ one obtains:
$$
0=\phi(0)=\frac{2}{3}-\sum_{n=1}^\infty\frac{2\sin\frac{\pi n}{3}}{\pi n}=
\frac{2}{3}-\frac{2}{\pi}S'\Rightarrow S'=\frac{\pi}{3}.
$$
The second way is more direct as it does not rely on a foreknowledge of an appropriate function for Fourier series. Observe that for $|x|<1$:
$$
\sum_{n=0}^\infty{x^n\sin\frac{\pi n}{3}}=\frac{1}{2i}\left(\frac{1}{1-xe^{\frac{\pi i}{3}}}-\frac{1}{1-xe^{-\frac{\pi i}{3}}}\right)
=\frac{x\sin\frac{\pi}{3}}{1-2x\cos\frac{\pi}{3}+x^2}=\frac{\sqrt{3}}{2}\frac{x}{1-x+x^2}.
$$
Thus,
$$
S'=\sum_{n=1}^\infty\frac{\sin\frac{\pi n}{3}}{n}=\sum_{n=1}^\infty\sin\frac{\pi n}{3}\int_0^1 x^{n-1}dx=\frac{\sqrt3}{2}\int_0^1\frac{dx}{1-x+x^2}=\frac{\pi}{3}.
$$
Finally the sum in question is computed using (1) as:
$$
S=\frac{2}{\sqrt3}S'=\frac{2\pi}{3\sqrt3}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2759901",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 2,
"answer_id": 0
} |
$\lim_{n\rightarrow \infty} \frac{1^4+2^4+...+n^4}{n^5}=\frac{1}{5}$ using integral(lower-/upper- or Riemann-sums) how can I show that $$
\lim_{n\rightarrow \infty} \frac{1^4+2^4+...+n^4}{n^5}=\frac{1}{5}$$
using integral(lower-/upper- or Riemann-sums).
I tried the following:
$\lim_{n\rightarrow \infty} \frac{1^4+2^4+...+n^4}{n^5}= \lim_{N\rightarrow\infty}\frac{1}{N^5} \lim_{N \rightarrow\infty} \sum_{n=1}^Nn^4=\frac{\frac{N^5}{5}}{N^5}=\frac{1}{5}$
But it feels like I am missing some steps here, especially why $\lim_{N \rightarrow \infty}\sum_{n=1}^N n^4=\frac{N^5}{5}$
How can I improve my try?
Thank you in advance!
| Hint
$$\frac{\sum_{k=1}^n k^4}{n^5}=\frac{1}{n} \sum_{k=1}^n \left(\frac{k}{n} \right)^4=\frac{1}{n} \sum_{k=1}^n f\left(\frac{k}{n} \right)$$
with $f(x)=x^4$, so you can recognize a Riemann sum.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2761700",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
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What is the value of $\sin 1 ^\circ \sin3^\circ\sin5^\circ \sin 7^\circ \sin 9^\circ \cdots \sin 179^\circ $?
What is the value of
$\sin 1 ^\circ \sin3^\circ\sin5^\circ \sin 7^\circ \sin 9^\circ \cdots \sin 179^\circ $ ?
The question is indeed intriguing. We could start by condensing it using $\sin \theta = \sin (180-\theta)$, This reduces the problem as the products till $89^\circ$. But that doesn't help proceed.
Thanks in advanced.
| We have to calculate $$\prod^{45}_{k=1}\sin^2((2k-1)^\circ)$$
Because $\sin(180-\theta)=\sin(\theta)$
Now Let $$P=\prod^{45}_{k=1}\sin((2k-1)^\circ)$$
Then $$\prod^{45}_{k=1}\sin(2k^\circ)\cdot P=\prod^{45}_{k=1}\sin((2k-1)^\circ)\cdot \prod^{45}_{k=1}\sin(2k^\circ)$$
So $$\prod^{45}_{k=1}\sin(2k^\circ)\cdot P=\frac{1}{2^{44}}\cdot \frac{1}{\sqrt{2}}\cdot \prod^{45}_{k=1}\sin(2k^\circ)$$
So we get $$P=\frac{1}{2^{\frac{89}{2}}}$$
So $$\prod^{45}_{k=1}\sin^2((2k-1)^\circ)=\frac{1}{2^{89}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2766630",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
} |
Find the parametrization of the curve resulting from intersection of a function and a curve I have the following function $f(x,y) = 2-x^2-4y^2$ and the surface $2x+4y+z-1 = 0.$
How do i go about finding the parametrization of the curve resulting from intersection of these surfaces? I see that $f(x,y)$ is the equation of an ellipsoid. I have tried to set $ f(x,y) = z$ and go from there but i cant seem to find any parameterization.
| You have
*
*$z = f(x,y)= 2- x^2-4y^2$
*$2x+4y+z-1 = 0 \Leftrightarrow z=1-2x-4y$
It follows
$$1-2x-4y = 2- x^2-4y^2 \Leftrightarrow x^2-2x+4y^2-4y = 1$$
Square completion gives
$$(x-1)^2-1 +(2y-1)^2-1 = 1 \Leftrightarrow (x-1)^2 +(2y-1)^2 = 3 \Leftrightarrow$$ $$ \left( \frac{x-1}{\sqrt{3}} \right)^2 + \left( \frac{2y-1}{\sqrt{3}} \right)^2 = 1$$
Now, set
$$\frac{x-1}{\sqrt{3}} = \cos \phi,\; \frac{2y-1}{\sqrt{3}} = \sin \phi$$
Solve for $x$ and $y$ and plug it into one of the given equations to get a parametrization for $z$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
$\lim_{(x,y)\rightarrow0} \frac{x\ln(1+x^3)}{y(x^2+y^2)}$ doesn't exist (?) Is what I did correct?
We have: $\lim_{(x,y)\rightarrow0} \frac{x\ln(1+x^3)}{y(x^2+y^2)}$
We know that as $x\rightarrow 0$ $\frac{\ln(1+x^3)}{x^3}=1$ , can be proven using taylor series..
So we have: $\lim_{(x,y)\rightarrow0} \frac{\ln(1+x^3)}{x^3} \frac{x^4}{yx^2+y^3}$
Pose $y=Cx^2$ we get: $$\lim_{(x,y)\rightarrow0} \frac{\ln(1+x^3)}{x^3} \frac{x^4}{Cx^4+C^3x^6}=\lim_{(x,y)\rightarrow0} \frac{\ln(1+x^3)}{x^3} \frac{x^4}{x^4(C+C^3x^2)}=\lim_{(x,y)\rightarrow0} \frac{\ln(1+x^3)}{x^3} \frac{1}{(C+C^3x^2)}=\frac{1}{(C+C^3x^2)}$$
Which is dependent on the constant $C$, Hence the limit doesn't exist.
What i'm doubtful about is the usage of $\lim_{x\rightarrow 0}\frac{\ln(1+x^3)}{x^3}=1$ in double variable limits, What i thought was since $x$ is still tending towards zero, and these terms are independent of y I can use it.
Thanks for your help!
| Yes you derivation is correct, indeed since $x\to 0$
$$\frac{\ln(1+x^3)}{x^3}\to1$$
it is true and we can conclude that for the trajectories $y=Cx^2\to 0$
$$\lim_{(x,y)\rightarrow0} \frac{x\ln(1+x^3)}{y(x^2+y^2)}=\frac1C$$
then the limit doesn't exist.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2766888",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to solve the integration How to find $$ \frac{\int_0^{\pi}x^3\log(\sin x)\,dx}{\int_0^{\pi} x^2 \log(\sqrt{2} \sin x)\,dx} $$
I couldn’t resolve it by using integration by parts.
| Using Fourier series, we see that
\begin{align}
\log(\sin x)= -\log 2-\sum^\infty_{k=1}\frac{\cos(2kx)}{k}
\end{align}
for $0\le x\le \pi$. Hence we have
\begin{align}
\int^\pi_0x^3\log(\sin x)\ dx =&\ -\log 2\int^\pi_0x^3\ dx -\sum^\infty_{k=1} \frac{1}{k}\int^\pi_0 x^3\cos(2kx)\ dx\\
=&\ -\frac{\pi^4}{4}\log 2 - \frac{3\pi^2}{4}\sum^\infty_{k=1} \frac{1}{k^3} = -\frac{\pi^4}{4}\log 2- \frac{3\pi^2}{4}\zeta(3).
\end{align}
On the other hand, we have
\begin{align}
\int^\pi_0 x^2\log(\sqrt{2}\sin x)\ dx =&\ \int^\pi_0 \frac{1}{2}x^2\log 2+ x^2\log(\sin x)\ dx\\
=&\ \frac{\pi^3}{6}\log 2 + \int^\pi_0 x^2\log(\sin x)\ dx\\
=&\ \frac{\pi^3}{6}\log 2 -\log 2\int^\pi_0x^2\ dx -\sum^\infty_{k=1} \frac{1}{k}\int^\pi_0 x^2\cos(2kx)\ dx\\
=&\ -\frac{\pi^3}{6}\log 2 -\frac{\pi}{2}\sum^\infty_{k=1} \frac{1}{k^3}= -\frac{\pi^3}{6}\log 2 - \frac{\pi}{2}\zeta(3).
\end{align}
Then we see that
\begin{align}
\frac{\int^\pi_0 x^3 \log(\sin x)\ dx}{\int^\pi_0 x^2\log(\sqrt{2}\sin x)\ dx} = \frac{-\frac{\pi^4}{4}\log 2- \frac{3\pi^2}{4}\zeta(3)}{-\frac{\pi^3}{6}\log 2 - \frac{\pi}{2}\zeta(3)} = \frac{3\pi}{2}.
\end{align}
| {
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"url": "https://math.stackexchange.com/questions/2768626",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Finding Sum of $\frac{1}{4!}+\frac{4!}{8!}+\frac{8!}{12!}+\frac{12!}{16!}+\cdots \cdots \cdots \infty$ Finding Sum of $$\frac{1}{4!}+\frac{4!}{8!}+\frac{8!}{12!}+\frac{12!}{16!}+\cdots \cdots \cdots \infty\; \bf{terms}$$
Try: Writting it as $$\sum^{\infty}_{r=0}\frac{(4r)!}{(4r+4)!}=\sum^{\infty}_{r=0}\frac{1}{(4r+1)(4r+2)(4r+3)(4r+4)}$$
$$\frac{1}{3}\sum^{\infty}_{r=0}\bigg[\frac{1}{(4r+1)(4r+2)(4r+3)}-\frac{1}{(4r+2)(4r+3)(4r+4)}\bigg]$$
$$\frac{1}{6}\bigg[\bigg(\frac{1}{(4r+1)(4r+2)}-\frac{1}{(4r+2)(4r+3)}\bigg)-\bigg(\frac{1}{(4r+2)(4r+3)}-\frac{1}{(4r+3)(4r+4)}\bigg)\bigg]$$
I did not understand how can i solve further, thanks
| As Piyush Divyanakar remarks, the sum is
$$\frac16\sum_{r=0}^\infty\left(
\frac{1}{4r+1}-\frac{3}{4r+2}+\frac{3}{4r+3}-\frac{1}{4r+4}\right).$$
This equals
$$\frac16\sum_{r=0}^\infty\int_0^1(t^{4r}-3t^{4r+1}+3t^{4r+2}-t^{4r+3})\,dt
=\frac16\int_0^1\frac{(1-t)^3}{1-t^4}\,dt.$$
You just have to do this integral...
| {
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"url": "https://math.stackexchange.com/questions/2776111",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Prove the equality: $\det\left[\begin{smallmatrix} -2a &a+b &a+c \\ b+a&-2b &b+c \\ c+a&c+b &-2c \end{smallmatrix}\right] = 4(a+b)(b+c)(c+a)$
i have to prove:
$$\begin{vmatrix}
-2a &a+b &a+c \\
b+a&-2b &b+c \\
c+a&c+b &-2c
\end{vmatrix} = 4(a+b)(b+c)(c+a)$$
I have tried many calculations between the rows and columns of the determinant to get to the answer i want to solve the exercise however none of them gave me something right. Can someone help?
| Brute force: expand using Cramer's rule:
$$
\begin{split}
\det A = &-2a\left( (-2b)(-2c) - (b+c)^2 \right)\\
&-(b+a)\left( (-2a)(-2c) - (a+c)^2\right)\\
&+(c+a)\left( (a+b)(b+c)-(-2b)(a+c)\right)
\end{split}
$$
and simplify out.
Similarly multiply out the RHS. It is painful, but maybe less painful than looking for the clever transformations.
| {
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"url": "https://math.stackexchange.com/questions/2782450",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find $a\in\mathbb{R}$ such that the image of $f(x)=\frac {x^2+ax+1}{x^2+x+1}$ is included in $[0,2]$.
Find $a\in\mathbb{R}$ such that the image of $f(x)=\frac {x^2+ax+1}{x^2+x+1}$ is included in $[0,2]$.
My attempt:
We have: $f'(x)=-\frac{(a-1)(x^2-1)}{(x^2+x+1)^2}\implies x = 1$ and $x = -1$ points of extrema.
then for $a\geq 1$:
so then $$2-a=0\implies a=2$$ and $$\frac {a+2}3=2\implies a=4.$$
and for $a\leq1:$
so then $$2-a=3\implies a=-1$$ and $$\frac {a+2}3=0\implies a=-2.$$
Now my answers are in the type of interval. How do I know which interval to choose?
| Lemma: If $x^2+mx+n\geq 0$ for all $x$ then discriminant $m^2-4n \leq 0$.
First note that $x^2+x+1>0$ for all $x$ (since discriminant =$-3$)
$$0\leq f(x)\leq 2 \implies 0\leq x^2+ax+1\leq 2x^2+2x+2$$
*
*From $0\leq x^2+ax+1$ we get $a^2-4\leq 0$ so $|a|\leq 2$ so $\boxed{-2\leq a\leq 2}$.
*From $x^2+ax+1\leq 2x^2+2x+2$ we have $0\leq x^2+(2-a)x+1$ so $(a-2)^2-4\leq 0$ so $|a-2|\leq 2$ so $\boxed{0\leq a\leq 4}$.
Thus $a\in[0,2]$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2783982",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
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Show $\frac{1}{1-\frac{x}{3-x}+\frac{x}{4-x}}$ is equivalent to $1+\frac{1}{2}\left(\frac{1}{2-x}-\frac{3}{6-x}\right)$ for $\lvert x\rvert < 1$ I have been asked to show that $$\frac{1}{1-\frac{x}{3-x}+\frac{x}{4-x}}$$ is equivalent to writing $$1+\frac{1}{2}\left(\frac{1}{2-x}-\frac{3}{6-x}\right)$$
From here I just tried to work out the bottom of the first fraction which I found to be $\frac{(3-x)(4-x)-x}{(3-x)(4-x)}$ now taking the reciprocal gives me $\frac{(3-x)(4-x)}{(3-x)(4-x)-x}$.
I did try factorising the bottom to get to $\frac{(3-x)(4-x)}{(x-6)(x-2)}$ and then using partial fractions gets me to $\frac{-15}{2(x-6)}+\frac{1}{2(x-2)}$ which is definitely not where I want to be.
I feel like I need to work from $\frac{(3-x)(4-x)}{(3-x)(4-x)-x}$ and somehow pull out a $1$ here but not entirely sure how.
Would really appreciate if anyone could help me.
Thank you.
| Hint:
\begin{align}
\frac{(3-x)(4-x)}{(3-x)(4-x)-x}&=\frac{(3-x)(4-x)}{x^2-8x+12}=\frac{(3-x)(4-x)}{(x-2)(x-6)} \\
&=1+\frac x{(x-2)(x-6)} \qquad\text{(by Euclidean division)}\\
&=1+\frac A{x-2}+\frac B{x-6}\qquad\text{(partial fractions)}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2787517",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
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Prove that $a^3+b^3=10^5$ has no positive integer solution.
Prove that there do not exist two positive integers, $a$ and $b$, such that $$a^3+b^3=100\,000$$
I tried to use congruence modulo $7$ and some other modulo, but it does not seem to work. If possible, please prove it with modular congruences.
| The possible cubic residues modulo $19$ are:
$$1,7,8,11,12,18$$
Further, $100000\equiv 3\pmod{19}$
The only two residues whose sum is equivalent to $3$ would be $11$ and $11$.
This implies that both $a$ and $b$ are one of $5\pmod{19},16\pmod{19}$ and $17\pmod{19}$
Further suppose that $a\geq b$. This implies that $a^3\geq 50000$ and $b^3\leq 50000$
Since $(16+2\cdot 19)^3>100000$ and since $(17+19)^3<50000$, this would imply that $a^3=(5+2\cdot 19)^3=79507$ as this is the only cube of a number from one of those equivalence classes in the desired range but then $b^3=100000-79507=20493$ is not a perfect cube. Hence, no solutions exist.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2789586",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
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What is the probability of the drawn ball from U2 being white? This question is already posted here ,but i want to know flaw in my approach.
Question
Let $U_1$ and $U_2$ be two urns such that $U_1$ contains $3$ white and $2$ red balls, and $U_2$ contains only $1$ white ball.
A fair coin is tossed:
*
*If head appears then $1$ ball is drawn at random from $U_1$ and put into $U_2$.
*If tail appears then $2$ balls are drawn at random from $U_1$ and put into $U_2$.
Now $1$ ball is drawn at random from $U_2$,What is the probability of the drawn ball from $U_2$ being white?
My Approach
There are $2$ cases.
*
*$\text{Case 1(Head Appears)}$
If Head Appears,then only $1$ ball will be drawn from $U_1$(white/red) to $U_2$
Current possible status of $U_2$
$WW,RW$
i.e $$\frac{1}{2}\times( \frac{2}{2}+\frac{1}{2} )$$
*
*$\text{Case 2(Tail Appears)}$
If Tail Appears,then $2$ balls will be drawn from $U_1$(WW,WR,RR) to $U_2$
Current possible status of $U_2$
$WWW,WRW,RRW$
i.e $$\frac{1}{2}\times( \frac{1}{1}+\frac{2}{3} +\frac{1}{3})$$
Total Probability =$\text{Case 1+case 2}$
=$$\frac{1}{2}\times( \frac{2}{2}+\frac{1}{2} )+\frac{1}{2}\times( \frac{1}{1}+\frac{2}{3} +\frac{1}{3})$$
But my answer is coming $>1$,So it is absolutely wrong.But i am not getting where i am doing wrong.
Please help
| Following your approach it should be:
Case 1 (Head Appears): $U_2=WW$ with probability $\frac{1}{2}\cdot\frac{3}{5}=\frac{3}{10}$ and $U_2=RW$ with probability $\frac{1}{2}\cdot\frac{2}{5}=\frac{2}{10}$.
Case 2 (Tail Appears): $U_2=WWW$ with probability $\frac{1}{2}\cdot\frac{3}{10}=\frac{3}{20}$, $U_2=RWW$ with probability $\frac{1}{2}\cdot\frac{6}{10}=\frac{6}{20}$ and $U_2=RRW$ with probability $\frac{1}{2}\cdot\frac{1}{10}=\frac{1}{20}$.
Hence the probability of the drawn ball from $U_2$ being white is:
$$\frac{3}{10}+\frac{1}{2}\cdot \frac{2}{10}+\frac{3}{20}+\frac{2}{3}\cdot \frac{6}{20}+\frac{1}{3}\cdot \frac{1}{20}=\frac{23}{30}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2789965",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Cyclotomic polynomials and products of cosines problem I've run into an inconsistency I can't figure out while trying to find products of the cosine of various roots of unity.
For example: $\cos(\frac{2\pi}{5})\cdot cos(\frac{4\pi}{5}) \cdot cos(\frac{6\pi}{5}) \cdot cos(\frac{10\pi}{5})$ Multiply everything by $\frac{\sin(\frac{2\pi}{5})}{sin(\frac{2\pi}{5})}$ and apply the trig rules and you get $\frac{1}{16}$
(which is correct)
These are all part of the 5th roots of unity which are solutions to $z^5 - 1 = 0$ and all of the form $\cos(x) + i\sin(x)$ So you can substitute that in expand and replace $\sin(x)^2 = 1 - cos(x)^2$ and split by the real and imaginary parts. For the reals you get $16\cos(x)^5 -10\cos(x))^3 + 5\cos(x) - 1 = 0$ and by the vieta formula this also implies that product of the cosine of the 5 roots is $\frac{1}{16}$ One root is 1 so this is consistent with first method.
But then I was looking at $\Phi_5 = z^4 + z^3 + z^2 + z + 1$ The four root besides one are roots of this polynomial and I should be able to do the same substitution from above.
let c = cos(x) and s =sin(x)
$(c + is)^4 + (c + is)^3 + (c + is)^2 + c + is + 1 = 0$
$(c^4 + 4ic^3s -6c^2s^2 -4ics^3 + s^4) + (c^3 + 3ic^2s - 3cs^2 - is^3) + (c^2 + 2ics - s^2) + (c + is) + 1 = 0$
For the real part:
$c^4 -6c^2s^2 + s^4 + c^3 - 3cs^2 + c^2 - s^2 + c + 1 = 0$
Then substituting $s^2 = 1 - c^2$ I get
$8\cos(x)^4 + 4\cos(x)^3 - 6\cos(x)^2 - 2\cos(x) + 1 = 0$ which implies that the product of the cosines is $\frac{1}{8}$
What's going on?
| $$z=\cos(2k\pi/5)+i\sin(2k\pi/5)$$
($k\in\{1,2,3,4\}$) are the zeros of $z^4+z^3+z^2+z+1$. Then
$w=z+1/z=2\cos(2k\pi/5)$ and
$$0=z^2+z+1+z^{-1}+z^{-2}=(z+z^{-1})^2+(z+z^{-1})-1=w^2+w-1.$$
Each $\cos(2k\pi/5)$ is a zero of
$$4x^2+2x-1$$
and so a zero of
$$(4x^2+2x-1)^2=16x^4+16x^3-4x^2-4x+1.$$
I suspect this is the equation you were after.
[content added in edit]
Your equation is
$$8y^4+4y^3-6y^2-2y+1=0.$$
This factors as
$$(4y^2+2y-1)(2y^2-1)$$
so has zeros $\cos(2\pi/5)$, $\cos(4\pi/5)$, $1/\sqrt2$ and $-1/\sqrt2$.
From this we correctly deduce
$$\left(\cos\frac{2\pi}5\right)\left(\cos\frac{4\pi}5\right)
\left(\frac1{\sqrt2}\right)\left(-\frac1{\sqrt2}\right)=\frac18.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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let $f(x) = (x-1)\ln x$, and given $0 < a < b$. If $f(a) = f(b)$, prove that $\frac{1}{\ln a}+\frac{1}{\ln b} < \frac{1}{2}$ Let $f(x) = (x-1)\ln x$, and given $0 < a < b$.
If $f(a) = f(b)$, how to prove that $\frac{1}{\ln a}+\frac{1}{\ln b} < \frac{1}{2}$?
| Proof.
Let $f(x) = (x-1)\ln x$.
Clearly, $f(1) = 0$, $\lim_{x\to 0^{+}} f(x) = \infty$, and $\lim_{x\to \infty} = \infty$.
We have $f'(x) = \ln x + 1 - \frac{1}{x}$. Clearly, $f(x)$ is strictly decreasing on $(0, 1)$, and strictly increasing on $(1, \infty)$. Thus, $0 < a < 1$ and $1 < b$.
We have
$$
\frac{1}{\ln a} + \frac{1}{\ln b} = \frac{a-1}{(a-1)\ln a} + \frac{b-1}{(b-1)\ln b} = \frac{a+b-2}{(b-1)\ln b}.
$$
It suffices to prove that
$$a < \frac{1}{2}(b-1)\ln b - b + 2. \tag{1}$$
We split into two cases:
*
*$b \ge \mathrm{e}^2$: Then $\frac{1}{2}(b-1)\ln b - b + 2 \ge 1$. Clearly, (1) is true.
*$1 < b < \mathrm{e}^2$: It is not difficult to prove that $0 < \frac{1}{2}(b-1)\ln b - b + 2 < 1$.
Since $f(x)$ is strictly decreasing on $(0, 1)$, it suffices to prove that
$$f(a) > f(\tfrac{1}{2}(b-1)\ln b - b + 2)$$
or
$$f(b) > f(\tfrac{1}{2}(b-1)\ln b - b + 2)$$
or
$$(b-1)\ln b - \left(\tfrac{1}{2}(b-1)\ln b - b + 1\right)\ln \left( \tfrac{1}{2}(b-1)\ln b - b + 2 \right) > 0$$
or
$$\frac{1}{2}(b-1)\left[ 2\ln b + (2 - \ln b)\ln \left(1 - \tfrac{1}{2}(b-1)(2-\ln b)\right)\right] > 0.$$
It suffices to prove that
$$2\ln b + (2 - \ln b)\ln \left(1 - \tfrac{1}{2}(b-1)(2-\ln b)\right) > 0.$$
Let $b = \mathrm{e}^y$. Then $0 < y < 2$. It suffices to prove that
$$2y + (2-y)\ln \left(1 - \tfrac{1}{2}(\mathrm{e}^y-1)(2-y)\right) > 0$$
or
$$\frac{2y}{2-y} + \ln \left(1 - \tfrac{1}{2}(\mathrm{e}^y-1)(2-y)\right) > 0.$$
Denote the LHS by $g(y)$. We have
$$g'(y) = \frac{(y^3-5y^2+12y-12)\mathrm{e}^y - y^2 + 12}{(2-y)^2(2 - (\mathrm{e}^y-1)(2-y))}.$$
It is not difficult to prove that $(y^3-5y^2+12y-12)\mathrm{e}^y - y^2 + 12 > 0$
for all $y$ in $(0, 2)$.
Thus, $g'(y) > 0$ for all $y$ in $(0, 2)$. Also, $g(0) = 0$.
Thus, $g(y) > 0$ for all $y$ in $(0, 2)$.
We are done.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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} |
Integrate $\int \frac{dx}{\sqrt{(x-a)(b-x)}}$
Integrate $\int \frac{dx}{\sqrt{(x-a)(b-x)}}$ where $b>a$
My Attempt
$$
\begin{align}
&\int \frac{1}{\sqrt{(x-a)(b-x)}}dx=\int\frac{dx}{\sqrt{-x^2+(a+b)x-ab}}\\
&=\int\frac{dx}{\sqrt{-(x^2-2.\frac{a+b}{2}x+ab+\frac{(a+b)^2}{4}-\frac{(a+b)^2}{4})}}\\
&=\int\frac{dx}{\sqrt{\frac{a^2+b^2+2ab}{4}-ab-\big[x-\frac{a+b}{2}\big]^2}}=\int\frac{dx}{\sqrt{\frac{(a-b)^2}{4}-\big(x-\frac{a+b}{2}\big)^2}}\\
&=\sin^{-1}\frac{x-\frac{a+b}{2}}{\frac{b-a}{2}}+C\color{red}{=\sin^{-1}\frac{2x-(a+b)}{b-a}+C}
\end{align}
$$
My reference shows the solution $I=2\sin^{-1}\sqrt{\frac{x-a}{b-a}}$, yet why am I getting a different solution or both the same ?
Thanx@lab bhattacharjee for the hint.
$$
\begin{align}
\sin^{-1}\frac{2x-(a+b)}{b-a}+C&=\frac{\pi}{2}-\cos^{-1}\frac{2x-(a+b)}{b-a}+C\\&=\frac{\pi}{2}-\pi+\cos^{-1}\frac{(a+b)-2x}{b-a}+C_1\\
&=\cos^{-1}\frac{(a+b)-2x}{b-a}+C_2
\end{align}
$$
Let,
$$
y=\cos^{-1}\frac{(a+b)-2x}{b-a}\implies\cos y=\frac{(a+b)-2x}{b-a}\\
2\sin^2\frac{y}{2}=1-\cos y=1-\frac{(a+b)-2x}{b-a}=\frac{b-a-a-b+2x}{b-a}=\frac{2x-2a}{b-a}\\
\sin^2\frac{y}{2}=\frac{x-a}{b-a}\implies \sin\frac{y}{2}=\sqrt{\frac{x-a}{b-a}}\\
\frac{y}{2}=\sin^{-1}\sqrt{\frac{x-a}{b-a}}\implies\boxed{y=2\sin^{-1}\sqrt{\frac{x-a}{b-a}}}
$$
| Use the Euler substition $$\sqrt{(x-a)(x-b)}=(x-a)t$$ Then is $$x=\frac{at^2+b}{1+t^2}$$ and then is $$dx=\frac{2(a-b)t}{(1+t^2)^2}dt$$
| {
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Solving the system $5(\sin x + \sin y) = 1$ and $5(\sin 2x + \sin 2y) = 1$ To find the general solution $(x,y)$ satisfying the system of equations
\begin{align} 5(\sin x + \sin y) &= 1 \\ 5(\sin 2x + \sin 2y) &= 1
\end{align}
I applied $\sin C + \sin D$ rule and then divided these two equations, then I am stuck at
$$\cos\frac{x-y}{2} = 2\cos \frac{x+y}{2}\cos(x-y).$$
I do not know what to do further.
| Starting from Alex Francisco's answer
$$16 (u^2)^3 - \left( \frac{4}{25} + 17 \right) (u^2)^2 + \left( \frac{4}{25} + 4 \right) u^2 - \frac{1}{25} = 0$$ let $x=u^2$ to get the cubic
$$400 x^3-429 x^2+104 x-1=0$$ and use the trigonometric method for three real roots. This would give the nasty
$$x_k=\frac{143}{400}+\frac{7}{200} \sqrt{\frac{403}{3}} \cos \left(\frac{2 \pi }{3}k-\frac{1}{3} \cos
^{-1}\left(\frac{89421
\sqrt{{3}}}{138229\sqrt{403}}\right)\right)\qquad (k=1,2,3)$$ which are all positive. Then six solutions for $u$ and $v$.
| {
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"source": "stackexchange",
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} |
find range for dependent
There are three vector $a$, $b$, $c$ in three-dimensional real vector space, and the inner product between them $a\cdot a=b\cdot b = a\cdot c= 1, a\cdot b= 0, c\cdot c= 4$. When setting $x = b\cdot c$, answer the following question: when $a, b, c$ are linearly dependent, find all possible values of $x$.
(dot here means dot product)
For dependent condition
$$\begin{align}
(a×b)·c &= 0\\
a·(b×c) &= 0\\
a(bc \sin θ)&=0
\end{align}$$
So $\theta = 0, \pi$.
Then
$$\begin{align}
x&=|b||c| \cos \theta \\
x&=2 \cos \theta \\
\implies x &= 2 \cos 0, x = 2 \cos \pi \\
x &= \mp 2
\end{align}$$
Am I right using triple cross product? or should i find $\theta$?
| Linear dependence places the vectors in the same plane. This makes the problem easier because we can focus on the angle.
From $c\cdot c=4$ we get that $|c|=2$.
So $a\cdot c=2\cos\theta=1\implies\cos\theta=\frac{1}{2}\implies\theta=60^{\circ}$.
From $b\cdot b=1$ and $a\cdot b=0$ we get that $b$ is a unit vector perpendicular to $a$.
Since the angle between $a$ and $c$ is $60^{\circ}$, the angle between $b$ and $c$ is either $30^{\circ}$ or $150^{\circ}$.
Therefore, $x=2\cos{30^{\circ}}$ or $x=2\cos{150^{\circ}}$ so $x=\pm \sqrt{3}$
| {
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"url": "https://math.stackexchange.com/questions/2798026",
"timestamp": "2023-03-29T00:00:00",
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Finding the probability that the matrix has full rank
There are only two entries, $0$ and $1$, over $\mathbb{Z}_2$. Thus,
only $16$ possible $2\times 2$ matrices over $Z_2$, and $6$ of them
have full rank:
$\begin{pmatrix}0&1\\ 1&0\end{pmatrix} \quad \begin{pmatrix}1&1\\
> 1&0\end{pmatrix} \quad \begin{pmatrix}0&1\\ 1&0\end{pmatrix} \quad
> \begin{pmatrix}0&1\\ 1&1\end{pmatrix} \quad \begin{pmatrix}1&1\\
> 0&1\end{pmatrix} \quad \begin{pmatrix}1&0\\ 0&1\end{pmatrix}$
Randomly generate a $n\times n$ matrix over $\mathbb {Z}_2$ (where n
is big, say, $1000$). What's the probability that the matrix has full
rank?
I'd want to talk something regarding to this question.
I found the probability is equal to:
$\dfrac{(2^n-1)(2^n-2)(2^n-2^2)...(2^n-2^{n-1})}{2^n} \tag{1}$
On other hand, what's difference between $2^n$ and $2^{n^2}$?
$\dfrac{(2^n-1)(2^n-2)(2^n-2^2)...(2^n-2^{n-1})}{2^{n^2}} \tag{2}$
Can you tell whether or not I'm wrong?
Regards!
| The second formula seems correct to me. The rows of the matrix must be linearly independent. The first row can be any vector but the zero vector, so there are $2^n-1$ choices. Suppose that $k$ linearly independent rows have been added. The next row can be any vector not in the linear span of the the first $k$ rows. Since these form a $k-$dimensional vector space over GF(2), there are $2^n-2^k$ choices for row $k+1.$
Since there are $2^{n^2}$ possible $n\times n$ matrices, the denominator should be $2^{n^2}$.
| {
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"question_score": "2",
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For $0 < x < \pi/2 $, prove that $x + \frac{x^{3}}{3} <\tan x $ I proved as follows:
Let $f(x) = \tan x - x - \frac{x^{3}}{3}.$
Then $f '(x) = \sec^{2}x - 1 - x^{2}$ and $f''(x) = 2\sec^{2}x\tan x-x > \tan x - x,$ since $\\tan x > x,$ $f''(x) > 0$ so $f'(x)$ is increasing.
Since $f '(0) = 0,$ $f '(x) > 0$ for all $x \in (0, \frac{\pi}{2}).$
So $f(x)$ is increasing. since $f(0) = 0,$ $f(x) > 0$ for all $x \in (0, \frac{\pi}{2}).$
But my proof seems complex. Is there any simpler way to prove this?
| Differentiate the given inequation three times
$$1+x^2\stackrel?<\tan^2x+1,$$
$$x\stackrel?<\tan x(\tan^2x+1),$$
$$1\stackrel?<(\tan^2x+1)^2+2\tan^2x(\tan^2x+1).$$
The final one is obviously true and equality holds at $x=0$, so that the second holds, by integration. That one holds at $x=0$, so that the first holds. And as the first also holds at $x=0$, so does the given one.
You can generalize the reasoning with more terms of the Taylor expansion.
| {
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"question_score": "2",
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Find $x^5+x^{-5}$ given the value of $x^2+x^{-2}$.
Find $x^5+\dfrac1{x^5}$ in its simplest form given that $x^2+\dfrac1{x^2}=a$ for $a,x>0$.
Attempt:
We write $$x^2+\frac1{x^2}=a\implies x^4-ax^2+1=0\implies x^5=ax^3-x$$ and $$\frac1{x^2}=a-x^2\implies \frac1{x^4}=a^2-2ax^2+x^4\implies \frac1{x^5}=\frac{a^2}x-2ax+x^3$$ so $$\begin{align}x^5+\frac1{x^5}&=(1+a)x^3-(1+2a)x+a^2\cdot\frac1x\\&=(1+a)x^3-(1+2a)x+a^2(ax-x^3)\\&=(1+a-a^2)x^3-(1+2a-a^3)x\\\implies x^5+\frac1{x^5}&=(a^2-a-1)x(a+1-x^2)\end{align}$$ But is this in its simplest form?
| $$
x+\frac{1}{x} = b\Rightarrow \left(x+\frac{1}{x}\right)^2=b^2
$$
then
$$
x^2+\frac{1}{x^2}= b^2-2 = a\Rightarrow b = \pm\sqrt{a+2}
$$
$$
\left(x^2+\frac{1}{x^2}\right)^2=a^2\Rightarrow x^4+\frac{1}{x^4} = a^2-2
$$
$$
\left(x^2+\frac{1}{x^2}\right)^2\left(x+\frac{1}{x}\right) = x^3+\frac{1}{x^3}+x+\frac{1}{x}
$$
$$
\left(x^4+\frac{1}{x^4}\right)^2\left(x+\frac{1}{x}\right) =x^5+\frac{1}{x^5}+x^3+\frac{1}{x^3}
$$
| {
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"source": "stackexchange",
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Question on an identity the representation of the Riemann Zeta function using Bernoulli numbers I'm reading Serre's A Course in Arithmetic.
In Section 4.1, we are showing that the Riemann Zeta function has a representation using the Bernoulli numbers.
The trick was to take the logarithmic derivative of
$$ \sin(z) = z \prod\limits_{n = 1}^{\infty} \left(1 - \frac{z^2}{n^2 \pi^2} \right). $$
I did this and got $z \cot z = 1 + 2 \sum\limits_{n = 1}^{\infty} \frac{z^2}{z^2 - n^2 \pi^2}$, as written in Serre.
But now Serre claims that
$$ z \cot z = 1 + 2 \sum_{n = 1}^{\infty} \frac{z^2}{z^2 - n^2 \pi^2} = 1 - 2 \sum_{n = 1}^{\infty} \sum_{k = 1}^{\infty} \frac{z^{2k}}{n^{2k} \pi^{2k}}. $$
How did we get the last equality?
I thought there was some geometric series tricks going on but I failed to discover the trick after a while of hard trying. Thanks for any help!
| Note that and using some of the below, you should be ale to recover the steps in the last line above.
\begin{align}
1+2z^2\sum_{n=1}^{\infty} \frac{1}{z^2-n^2}&= 1-2z^2 \sum_{n=1}^{\infty}\frac{1}{n^2}\frac{1}{1-z^2/n^2} \\
&=1-2z^2\sum_{n=1}^{\infty}\frac{1}{n^2}\sum_{k=0}^{\infty}\left(\frac{z}{n}\right)^{2k} \\
&= 1-2 \sum_{n=1}^{\infty}z^{2k+2}\sum_{n=1}^{\infty}\frac{1}{n^{2k+2}} \\
&=1-2 \sum_{\text{even}\, k \geq 2}\zeta(k)z^k
\end{align}
| {
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Is the Method of Frobenius Appropriate for this DE? If so, how to proceed? To not bore with motivation, I'll get straight to the point. I have struggled for quite some time finding analytical solutions to the differential equation
$$y'' + \frac{a}{x}y' - \frac{b}{x^2\left(1 - \frac{c}{x^{n-2}}\right)}y = 0,\tag{1}$$ where $a,b,c$ are positive constants, and $n \geq 4$ a positive integer. I am interested in the behavior of this equation around $x = x_p \equiv c^{n-2}$. My approach was to use Frobenius' method, and posit the two solutions
$$y_1(x) = \sum_{k \ge 0}a_k(x - x_p)^{k+1} \hspace{0.2cm}\text{ and } \hspace{0.2cm}y_2(x) = C\log(x-x_p)y_1(x) + \sum_{k\ge0}b_k(x - x_p)^k,$$ where $a_k,b_k$ are to be determined constants, with $a_0,b_0 \neq 0$, and $C$ a constant that may be zero. [These solutions were obtained by first computing the indices of the associated indicial equation for $x = x_p$ (they turn out to be 0 and 1).] I arrive at some trouble when trying to deduce recurrence relations for $a_k$ and $b_k$. I end up multiplying (1) through by $x^2(1 - \frac{x}{x^{n-2}})$ to simplify the form, but I then obtain the unforgiving expression (for $y_1$)
$$x^2\left(1 - \frac{c}{x^{n-2}}\right)\sum_{k\ge 0}a_k(k+1)k(x-x_p)^{k-1} + ax\left(1 - \frac{c}{x^{n-2}}\right)\sum_{k\ge 0}a_k(k+1)(x-x_p)^k - b\sum_{k\ge 0}a_k(x-x_p)^k = 0$$
It is not at all clear to me how to write this solely in terms of $(x-x_p)$ factors, with no stray factors of $x$ or $(x + x_p)$, etc. Perhaps there is a better approach to solving (1)? Or maybe this method is fine, but I'm not seeing how to proceed. Any input is appreciated.
| $y''+\dfrac{a}{x}y'-\dfrac{b}{x^2\left(1-\dfrac{c}{x^{n-2}}\right)}y=0$
$x^2\dfrac{d^2y}{dx^2}+ax\dfrac{dy}{dx}-\dfrac{bx^{n-2}}{x^{n-2}-c}y=0$
With reference to http://eqworld.ipmnet.ru/en/solutions/ode/ode0216.pdf:
Let $r=x^{n-2}$ ,
Then $\dfrac{dy}{dx}=\dfrac{dy}{dr}\dfrac{dr}{dx}=(n-2)x^{n-3}\dfrac{dy}{dr}$
$\dfrac{d^2y}{dx^2}=\dfrac{d}{dx}\left((n-2)x^{n-3}\dfrac{dy}{dr}\right)=(n-2)x^{n-3}\dfrac{d}{dx}\left(\dfrac{dy}{dr}\right)+(n-2)(n-3)x^{n-4}\dfrac{dy}{dr}=(n-2)x^{n-3}\dfrac{d}{dr}\left(\dfrac{dy}{dr}\right)\dfrac{dr}{dx}+(n-2)(n-3)x^{n-4}\dfrac{dy}{dr}=(n-2)x^{n-3}\dfrac{d^2y}{dr^2}(n-2)x^{n-3}+(n-2)(n-3)x^{n-4}\dfrac{dy}{dr}=(n-2)^2x^{2n-6}\dfrac{d^2y}{dr^2}+(n-2)(n-3)x^{n-4}\dfrac{dy}{dr}$
$\therefore x^2\left((n-2)^2x^{2n-6}\dfrac{d^2y}{dr^2}+(n-2)(n-3)x^{n-4}\dfrac{dy}{dr}\right)+a(n-2)x^{n-2}\dfrac{dy}{dr}-\dfrac{bx^{n-2}}{x^{n-2}-c}y=0$
$(n-2)^2x^{2n-4}\dfrac{d^2y}{dr^2}+(n-2)(a+n-3)x^{n-2}\dfrac{dy}{dr}-\dfrac{bx^{n-2}}{x^{n-2}-c}y=0$
$(n-2)^2r^2\dfrac{d^2y}{dr^2}+(n-2)(a+n-3)r\dfrac{dy}{dr}-\dfrac{br}{r-c}y=0$
$(n-2)^2r(r-c)\dfrac{d^2y}{dr^2}+(n-2)(a+n-3)(r-c)\dfrac{dy}{dr}-by=0$
Which reduces to Gaussian hypergeometric equation.
| {
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Find the value of $S$ in term of $k$ (telescoping sums) Let $k=\frac{1}{1\times2}+\frac{1}{3\times4}+\frac{1}{5\times6}+\cdots+\frac{1}{2549\times2550}$.
Find the value of $S=\frac{1275}{1276}+\frac{1276}{1277}+\frac{1277}{1278}+\cdots+\frac{2548}{2549}$ in term of $k$.
I've tried to write the terms of $S$ in the following way:
\begin{align}
\frac{1275}{1276} & =\frac{1}{1\times2}+\frac{1}{2\times3}+\cdots+\frac{1}{1275\times1276} \\
\frac{1276}{1277} & =\frac{1}{1\times2}+\frac{1}{2\times3}+\cdots+\frac{1}{1276\times1277} \\
& \,\,\,\vdots \\
\frac{2548}{2549} & =\frac{1}{1\times2}+\frac{1}{2\times3}+\cdots+\frac{1}{2548\times2549}
\end{align}
But it's just not enough, because in $k$ we don't have $\frac{1}{2\times3}$, $\frac{1}{4\times5}$, etc.
Can you please help me solve the problem? Thanks!
| $k=\frac{1}{1\times2}+\frac{1}{3\times4}+\frac{1}{5\times6}+\cdots+\frac{1}{2549\times2550} = \frac{1}{1}-\frac{1}{2} + \frac{1}{3}-\frac{1}{4} + ... + \frac{1}{2549}-\frac{1}{2550}$
Collecting even and odd and with some manipulation, we get
$k = H_{2550} - H_{1275}$ where $H_n$ = harmonic number
$S = 1 - \frac{1}{1276} + 1 - \frac{1}{1277} + ... + 1 - \frac{1}{2549}$
Collecting the ones and with some manipulation we get:
$S = 1274 - (H_{2550} - H_{1275}) + \frac{1}{2550} = 1274 - k + \frac{1}{2550}$
| {
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How do I calculate how many ways 14 non-attacking bishops can be placed on a chessboard? It's possible to place up to 14 non-attacking bishops on a chessboard.
How can I calculate the number of valid configurations, without writing a program to brute force it?
I've checked Non Attacking Chess Pieces, but it doesn't cover the variation I'm interested in, and a quick Google hasn't yielded useful results.
UPDATE:
Proof that only 14 bishops can be placed:
If we divide the chessboard into diagonals, and we treat diagonal 1 and 15 as the same diagonal, then the maximum number of bishops per diagonal is 1, therefore 14.
Here is an example configuration of 14 bishops to show it's possible:
NOTE: Not a duplicate of this question because I am asking about the number of arrangements of 14 bishops, not the maximum number of bishops on a chess board.
| Let's convert the chessboard into points: (rank,file). The normal chessboard refers to file A through H. We will use 1 through 8. Consider the number rank minus file:
$$\begin{matrix} & & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\ \hline 8 & | & 7 & 6 & 5 & 4 & 3 & 2 & 1 & 0 \\ 7 & | & 6 & 5 & 4 & 3 & 2 & 1 & 0 & -1 \\ 6 & | & 5 & 4 & 3 & 2 & 1 & 0 & -1 & -2 \\ 5 & | & 4 & 3 & 2 & 1 & 0 & -1 & -2 & -3 \\ 4 & | & 3 & 2 & 1 & 0 & -1 & -2 & -3 & -4 \\ 3 & | & 2 & 1 & 0 & -1 & -2 & -3 & -4 & -5 \\ 2 & | & 1 & 0 & -1 & -2 & -3 & -4 & -5 & -6 \\ 1 & | & 0 & -1 & -2 & -3 & -4 & -5 & -6 & -7\end{matrix}$$
Notice how these numbers are constant on the diagonals?
Next consider rank + file:
$$\begin{matrix} & & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\ \hline 8 & | & 9 & 10 & 11 & 12 & 13 & 14 & 15 & 16 \\ 7 & | & 8 & 9 & 10 & 11 & 12 & 13 & 14 & 15 \\ 6 & | & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14 \\ 5 & | & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 \\ 4 & | & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 \\ 3 & | & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 \\ 2 & | & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\ 1 & | & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9\end{matrix}$$
Notice how these numbers are constant on the opposite diagonals? So, you are looking for 14 pairs of numbers from 1 through 8 such that you get 14 distinct pairs (rank-file,rank+file) such that no other pair shares the same rank-file or rank+file.
I went through my old notes to see where I remember a similar problem to this. It may have been the seminar I took that discussed block designs. I did not actually take a class that focused on block designs, so I do not recall the set up. I think this approach could be interesting (and give more details about configurations), but also far more time consuming and maybe not worth it. Specifically, this is a problem for a Transversal Design.
| {
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"source": "stackexchange",
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Prove identity: $\sin \alpha= \frac{2\tan \frac{\alpha}{2}}{1+\tan^2 \frac{\alpha}{2}} $ $$\sin \alpha= \frac{2\tan \frac{\alpha}{2}}{1+\tan^2 \frac{\alpha}{2}} $$
I am having a problem proving this identity. I write tan like $\frac{\sin \frac{\alpha}{2}}{\cos \frac{\alpha}{2}}$ and the squared one in the same way.
I eventually get $$\frac {2\sin \frac{\alpha}{2} \cos \frac{\alpha}{2}}{\sin^2 \frac{\alpha}{2} + \cos^2 \frac{\alpha}{2}} $$
And I am stuck...
| $$\sin\alpha=2\sin\dfrac{\alpha}{2}\cos\dfrac{\alpha}{2}=\frac{2\tan\dfrac{\alpha}{2}}{\sec^2 \dfrac{\alpha}{2}}=\frac{2\tan\dfrac{\alpha}{2}}{1+\tan^2 \dfrac{\alpha}{2}}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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If $4|n(n-1)$ then $4|n$ or $4|(n-1)$ Since $4$ is not prime, we can't use Euclid's lemma. Assume $4$ does not divide $n$, we'd like to show that $4|(n-1)$.
Suppose $n=4a+b$ where $b\neq 0$. It suffices to prove that $b=1$, but I feel like this is a dead end. Is my setup wrong?
| Method 1: You can't use Euclid's Lemma with $4$ but you can use it with $2$. So either i) $2^2|n$ or ii) $2|n$ and $2|n-1$ or iii) $2^n|n-1$. And ii) $2|n$ and $2|n-1$ is impossible.
Method 2: $n$ and $n - 1$ are relatively prime. So If $2|n$ then $2\not \mid n-1$ so $2^2|n$. Likewise if $2|n-1$ then $2\not \mid n$ so $2^2|n-1$.
Method 3:
Corollary to Euclids lemma: If $m|ab$ then there exist $j,k$ so that $m = jk$ and $j|a$ and $k|b$. If $\gcd(a,b) = 1$ then $\gcd(j,k) = 1$.
(Can you prove that)
Thus $4|n(n-1)$ means either $4|n$ or $4|n-1$.
Method 4:
$n \equiv 0, 1,2,3 \mod 4$.
If $n\equiv 0 \mod 4$ then $n(n-1)\equiv 0*3\equiv 0 \mod 4$
If $n \equiv 1\mod 4$ then $n(n-1)\equiv 1*0 \equiv 0 \mod 4$
If $n \equiv 2 \mod 4$ then $n(n-1) \equiv 2*1 \equiv 2 \mod 4$.
If $n \equiv 3 \mod 4$ then $n(n-1) \equiv 3*2\equiv 2 \mod 4$.
So if $4|n(n-1)$ then either $n\equiv 0 \mod 4$ or $n\equiv 1 \mod 4$.
| {
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Linear equation help $$\frac{1-x}{4} + \frac{5x+1}{2} = 3 - \frac{2(x+1)}{8}$$
I got x=1 but the book says x=4/5 and I don't understand how to get to that, I tried working backwards too but I just can't figure it out. Any help would be appreciated, thanks.
Here's my work
using 8 as the lcm
$$
\begin{align}
\frac 81(\frac {(1-x)}4) + \frac 81(\frac {(5x+1)}2) &= \frac 81(3) - \frac 81\frac {2(x+1)}8 \\
2(1-x) + 4(5x+1) &= 24 - 2x + 2 \\
2 - 2x + 20x + 4 &= 24 - 2x + 2 \\
6 + 18x &= 26 -2x \\
20x &= 20 \\
x &= 1 \\
\end{align}
$$
| Get everything over as a fraction with a denominator of $8$ and equate, so:
$$\frac{2(1-x)}{8}+\frac{4(5x+1)}{8}=\frac{24}{8}-\frac{2(x+1)}{8}$$
Thus we have:
$$2-2x+20x+4=24-2x-2$$
See if you can go from there.
| {
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When is the given matrix not diagonizable?
Given the following matrix $$A = \begin{bmatrix} 0 & 1\\ bk+b & -bk\end{bmatrix}$$ where $b \neq 0$ and $k \neq 0$, find when the given matrix is not diagonizable.
My initial thought would be when the matrix doesn't have $2$ linearly independent eigenvectors.
In this case it would be when the eigenvalue has a multiplicity of 2.
I.e. $\lambda_1=\lambda_2 $.
I find the eigenvalues:
$
\lambda_1=\frac{-bk+\sqrt{(bk)^2-4\cdot(-bk-b)}}{2}\\
\lambda_2=\frac{-bk-\sqrt{(bk)^2-4\cdot(-bk-b)}}{2}
$
I then let them equal each other:
$
(1)\qquad\frac{-bk+\sqrt{(bk)^2-4\cdot(-bk-b)}}{2}=\frac{-bk-\sqrt{(bk)^2-4\cdot(-bk-b)}}{2}\\
\quad\\
\quad\\
(2)\qquad\sqrt{(bk)^2-4\cdot(-bk-b)}=-\sqrt{(bk)^2-4\cdot(-bk-b)}
$
Here my initial thought would be that the only case this is true if the term under the square root is equal zero.
$
b^2k^2+4bk+4b=0\\
b(bk^2+4k+4)=0
$
Since $b\neq0$ I need the term inside the parantheses to be equal 0.
$
bk^2+4k+4=0
$
I'm a bit lost from here.
I know the solution is
$
-bk-k=(\frac{-bk}{2})^2
$.
The solution was provided by someone else here on Stack Exchange, he used the determinant and the trace of the matrix to come to that conclusion, which I can't seem to replicate.
| $$b\neq 0, k\neq 0, A=\begin{pmatrix}
0 & 1\\
bk+b & -bk
\end{pmatrix}\Rightarrow \det (A-xI)=\begin{vmatrix}
-x & 1\\
bk+b & -bk-x
\end{vmatrix}=0\Rightarrow$$
$$\Rightarrow x^2+bkx-(bk+b)=0\Rightarrow x=\dfrac{-bk\pm \sqrt{b^2k^2+4bk+4b}}{2}$$
$$x_1=x_2\Rightarrow b^2k^2+4bk+4b=0\Rightarrow b^2k^2=-4(bk+b)\Rightarrow \left( \dfrac{bk}{2}\right)^2=-bk-b$$
Let us now look at the eigenvectors.
$$x=-\dfrac{bk}{2}\Rightarrow A-xI=\begin{pmatrix}
\frac{bk}{2} & 1\\
-\left( \frac{bk}{2}\right)^2 & -\frac{bk}{2}
\end{pmatrix}$$
$$\begin{pmatrix}
\frac{bk}{2} & 1\\
-\left( \frac{bk}{2}\right)^2 & -\frac{bk}{2}
\end{pmatrix}\begin{pmatrix}
x\\
y
\end{pmatrix}=\begin{pmatrix}
0\\
0
\end{pmatrix}\Rightarrow \left\{ \begin{array}{lcc}
\dfrac{bk}{2}x+y=0 \\
\\ -\left( \dfrac{bk}{2}\right)^2x-\dfrac{bk}{2}y=0
\end{array}
\right.\Rightarrow bkx+2y=0\Rightarrow$$
$$E_{-\frac{bk}{2}}=\langle (-2,bk)\rangle \Rightarrow \dim \left( E_{-\frac{bk}{2}}\right) =1$$
Not diagonalizable.
| {
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Congruence 10 modulo 7 I'm reading some proofs for divisibility rules and see that $10 \equiv -1 \pmod{11}$, then $10^2 \equiv 1 \pmod{11}$ and etc., it's clear for me, but I can't understand the situation with modulo 7. If we have $10 \equiv 3 \pmod{7}$ how can we get the rest of the values? They are $10^2 \equiv 2 \pmod{7}$, $10^3 \equiv -1 \pmod{7}$ and etc.
I'm sorry if it's a stupid question, but I want to figure out this very well.
| As a start, we use the rule that
$$a \equiv b \pmod{n} \Rightarrow a^k \equiv b^k \pmod{n},\forall k\in \mathbb{N} \tag{1}$$
Hint
$$10 \equiv 3 \pmod{7} \Rightarrow 10^2 \equiv 3^2 \pmod{7} \Rightarrow \\
10^2 \equiv 9\equiv 7+2\equiv 2 \pmod{7}$$
or
$$10^3 \equiv 3^3 \pmod{7} \Rightarrow \\
10^3 \equiv 27\equiv 28-1\equiv 4\cdot7-1\equiv -1 \pmod{7}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find all real values of $a$ such that $\sum_{n=0}^{\infty} \sin(\pi\sqrt{a^2+n^2})$ diverges To find all $a$ that makes the series diverge, it is sufficient to make the limit of $\sin(\pi\sqrt{a^2+n^2})$ not exist or exists but not equal to $0$ as $n\to\infty.$
I suspect only $a=0$ would converge the series, and the limit of $\sin(\pi\sqrt{a^2+n^2})$ would not even exist if $a\neq 0.$ .
| HINT:
Observe that we can write
$$\begin{align}
\sin(\pi\sqrt{a^2+n^2})&=\sin\left(n\pi\sqrt{1+\frac{a^2}{n^2}}\right)\\\\
&=\sin\left(n\pi+\frac{a^2\pi}{2n}+O\left(\frac1{n^3}\right)\right)\\\\
&=(-1)^n\sin\left(\frac{a^2\pi}{2n}+O\left(\frac1{n^3}\right)\right)
\end{align}$$
Can you finish now? Is there any real value for $a$ for which this series diverges?
| {
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"timestamp": "2023-03-29T00:00:00",
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Remainder of trinomial What will be the remainder if I divide $(a+b+c)^{333}-a^{333}-b^{333}-c^{333}$ by $(a+b+c)^3-a^3-b^3-c^3$.
I have tried trinomial expention. But its still too big for long division. Is there a shorter method for this problem?
| Since $$(a+b+c)^3-a^3-b^3-c^3=3(a+b)(a+c)(b+c)$$ and
$$(a+b+c)^{333}-a^{333}-b^{333}-c^{333}=0$$ for $a=-b$, for $a=-c$, for $b=-c$
and $(a+b+c)^{333}-a^{333}-b^{333}-c^{333}$ is divided by $3$,
we obtain that the remainder is $0$.
| {
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Sequence $(f_n)$ equibounded, equicontinuous but with not uniformly convergent subsequence
Let $f_n:[0,\infty)\rightarrow R$, given by $f_n(x) = \sin(\sqrt{x+4\pi^2n^2})$. Prove that $(f_n)$ is equicontinuous and equibounded, but has no uniformly convergent subsequence.
I'm having trouble to prove that this sequence is equicontinuous and has no uniformly convergent subsequence.
Clearly, $|f_n(x)\rvert= \lvert\sin\sqrt{x+4\pi^2n^2}|\leq1$ for every $x$ and every $n\in N$. So $(f_n)$ is equibounded.
Now in order to prove that it is equicontinuous, what I have noted so far using the mean value theorem is that:
$$|f_n(x)-f_n(y)|=$$
$$|\sin\sqrt{x+4\pi^2n^2}-\sin\sqrt{y+4\pi^2n^2}|\leq\bigg|\frac{\cos\sqrt{c_n+4\pi^2n^2}}{2\sqrt{c_n+4\pi^2n^2}}\bigg||x-y|\leq\bigg|\frac{1}{2\sqrt{c_n+4\pi^2n²}}\bigg||x-y|$$
but I can't control the term $\bigg|\frac{1}{2\sqrt{c_n+4\pi^2n²}}\bigg|$ properly, since it is dependent of $n$.
Also, any help on how to prove that there is no uniformly convergent subsequence?
Thanks in advance!
| Let's prove that this sequence of function is equicontinuous. Indeed fix $\epsilon > 0$, then consider $\delta > 0$ for a $\delta$ to be chosen shortly. Now consider $|f_n(x) - f_n(y)|$ = $|\sin(\sqrt{x+4 \pi^2n^2}) - \sin(\sqrt{y+4 \pi^2n^2})|$. Indeed, now (assume $x < y$) by MVT we see there exists a $\xi \in (x,y)$ such that $|\sin(\sqrt{x+4 \pi^2n^2}) - \sin(\sqrt{y+4 \pi^2n^2})|$ = $\frac{\cos(\sqrt{4 \pi^2 n^2 + \xi}}{2 \sqrt{4 \pi^2n^2 + \xi}}$$|x-y|$ $\leq$ $\frac{1}{2 \sqrt{4 \pi^2n^2 + \xi}}|x-y|$. Notice that $x \in [0, \infty)$, and as $\xi \geq 0$, we have $2 \sqrt{4 \pi^2n^2 + \xi}$ $\geq$ $2\sqrt{4\pi^2n^2}$. In particular, we also have $2\sqrt{4\pi^2n^2} \geq 2\sqrt{4 \pi^2}$ for any $n \in \mathbb{N}$. Hence, we have the following inequality $|\sin(\sqrt{x+4 \pi^2n^2}) - \sin(\sqrt{y+4 \pi^2n^2})|$ $\leq$ $\frac{1}{2 \sqrt{4\pi^2}}|x-y|$ < $\frac{1}{2 \sqrt{4\pi^2}}\delta$. Now choose $\delta = 2\epsilon\sqrt{4 \pi^2} $ and we have equicontinuity.
Sorry, I didn't originally see the no uniformly convergent subsequence. Let's assume for contradiction that there exists a sequence $\{f_{n_k}(x)\} \rightarrow f$ uniformly. In particular, $\sup_{x \in [0,\infty))} |f_{n_k}(x) - f(x)| \rightarrow 0$. As the space of continuous bounded functions is complete, this is possible if and only if the sequence is cauchy. E.g. $\sup_{x \in [0,\infty))}|f_{n_k}(x) - f_{m_k}(x)| \rightarrow 0$. Hence, there exists an $N \in \mathbb{N}$ such that for any $n,m \geq N$,$\sup_{x \in [0,\infty))}|f_{n_k}(x) - f_{m_k}(x)| < \frac{1}{2}$. In particular, $|\sin(\sqrt{x + 4 \pi^2n_k^2}) - \sin(\sqrt{x + 4 \pi^2m_k^2})| < \frac{1}{2} $ for every $x \in [0,\infty)$. Let $x = 12 \pi^2n_k^2 $, then $\sin(\sqrt{12 \pi^2n_k^2 + 4 \pi^2n_k^2}) = \sin(\sqrt{16 \pi^2n_k^2}) = 0 $. So we have $|\sin(\sqrt{12\pi^2n_k^2 + 4 \pi^2m_k^2})| < \frac{1}{2} \Rightarrow |\sin(2\pi\sqrt{3n_k^2 + m_k^2})| < \frac{1}{2}$ for any $m_k > N$, which is clearly a false statement due to the periodicity of our function with period 1. We can choose a $m_k$ to make this a false statement.
I hope this helps.
| {
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Show that the substitution $x=X-1$ and $y=Y+3$ turns $\frac{dy}{dx}=...$ into a homogenous equation. $$\frac{dy}{dx}=\frac{4x-y+7}{2x+y-1}$$
After substitution of the given variable I get $$\frac{dY}{dX} = \frac{4-\frac{Y}{X}}{2+\frac{Y}{X}}$$ which seems to give a homogenous equation. (There is no answer given for this part of the question)
But then the next question says find the particular solution with $x=0$ and $y=3$ giving your answer in the form $f(x,y)=c$. Then after this point I get an answer that is completely different to the answer given in the book.
Let $Y=vX$. Then $\frac{dY}{dX} = X\frac{dv}{dX}+v$
Replace, then $$X\frac{dv}{dX}+v=\frac{4-v}{2+v}$$
Making the Integrals
$$\int \frac{1}{\frac{4-v}{2+v}-v}dv=\int\frac{1}{X}dX$$
$$\frac{3\ln(1-v)}{5}-\frac{2\ln(4+v)}{5}-\ln(X)=c$$
Replacing everything back to $y$ and $x$.
$$\frac{3\ln(1-\frac{y-3}{x+1})}{5}-\frac{2\ln(4+\frac{y-3}{x+1})}{5}-\ln(x+1)=c$$
According to this $c=-\frac{2\ln(4)}{5}$
Even before the replacement stages the answer is already completely different to that of the answer given.
What did I do wrong?
Just for reference, answer given is $(x-y+4)^3(4x+y+1)^2 = 16$.
| Solving for $u,v$
$$
4x-y+7=v\\
2x+y-1=u
$$
and substituting into
$$
\frac{dy}{dx} = \frac{4x-y+7}{2x+y-1}\rightarrow \frac{dv}{du} = \frac{4u-v}{2u+v}\;\;\;\mbox{which is homogeneous}
$$
Now making $z = v = \lambda u$ and correspondingly $dz = \lambda du+ud\lambda$ we have the new DE
$$
\frac{dv}{du} = \frac{dz}{du} = \lambda + u \frac{d\lambda}{du} = \frac{4-\lambda}{2+\lambda}
$$
or
$$
u\frac{d\lambda}{du} = \frac{4-\lambda}{2+\lambda}-\lambda
$$
This is a separable DE so
$$
\frac{du}{u} = \frac{d\lambda}{\frac{4-\lambda}{2+\lambda}-\lambda}
$$
and solving
$$
\log u = -\frac{3}{5} \log (1-\lambda )-\frac{2}{5} \log (\lambda +4)+C_0
$$
or
$$
u = \frac{C_1}{(1-\lambda)^{\frac 35}(\lambda+4)^{\frac 25}}
$$
or
$$
u^5 = \frac{C_2}{(1-\lambda)^{3}(\lambda+4)^{2}}
$$
etc.
| {
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Lagrange multipliers exercise Given the function $$f(x,y,z)=xy+xz+yz$$ I have to find its extremes on the set $E=\{(x,y,z)\in\mathbb R|x^2+y^2+z^2=1\}$.
Given the fact that $E$ is the border of a sphere, I can conclude that it is a compact set and because $f$ is continuous then there will be some extremes.
I proceed with the system of equations given by $\nabla f=\lambda\nabla g$ and the constraint $g(x,y,z)=x^2+y^2+z^2-1$ obtaining
$$\cases{y+z=2\lambda x\\x+z=2\lambda y\\x+y=2\lambda z\\x^2+y^2+z^2=1}\Rightarrow\cases{x+y+z=(2\lambda+1) x\\x+y+z=(2\lambda+1) y\\x+y+z=(2\lambda+1) z\\x^2+y^2+z^2=1}$$
Then for $\lambda\ne-\frac 1 2$ I obtain $x=y=z$ and putting it in the last equation I obtain that the points of extreme are in $x=y=z=\pm\frac 1 {\sqrt 3}$ and then I can analyze these points.
If on the other hand $\lambda=-\frac 1 2$ using the first system of equations I obtain
$$\cases{y+z=-x\\x+z=-y\\x+y=-z\\x^2+y^2+z^2=1}\Rightarrow\cases{x+y+z=0\\x^2+y^2+z^2=1}$$ which leads to $x=-y-z$ and produces a conic with equation $y^2+z^2+yz=\frac 1 2$. I think that means that the points of extreme of $f$ in this case are on that conic, but then how can I analyze these points?
| On the intersection of the plane $x+y+z=0$ and the unit sphere,
$$xy+xz+yz=\frac12\left((x+y+z)^2-(x^2+y^2+z^2)\right)=\frac{0-1}2=-\frac12.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Floor function of a real number The floor of a real number $x$, denoted by $⌊x⌋$, is the largest integer that is less than or equal to $x$. The ceiling of a real number x, denoted by $⌈x⌉$, is the smallest integer that is greater than or equal to $x$. Let $x$ and $y$ be any real numbers. Let $n, m$ and $k$ be any integers.
*
*$⌊x⌋ \le x < ⌊x⌋+1$, equivalently $x-1 < ⌊x⌋ \le x$.
*$⌈x⌉-1 < x \le ⌈x⌉$, equivalently $x \le ⌈x⌉ < x+1$.
*If $n \le x < n+1$ (or $x-1 < n \le x$), then $n=⌊x⌋$; similarly, if $n-1 < x \le n$ (or $x \le n < x+1$), then $n=⌈x⌉$.
*If $x \le y$, then $⌊x⌋ \le ⌊y⌋$ and $⌈x⌉ \le ⌈y⌉$. Similarly for $x \ge y$.
*If $n \le x$, then $n \le ⌊x⌋$. If $n \ge x$, then $n \ge ⌈x⌉$.
*If $x$ has an integer value, then $⌊x⌋=x=⌈x⌉$. If x has a non-integer value, then $⌈x⌉=⌊x⌋+1$.
Furthermore, we have the well-known inequality: $ \frac {x-1}{x} \le \ln x \le \frac {x^2 -1}{2x} \le x-1$
I am trying to find all pairs of positive real numbers $(x,y)$ such that $ \ln(⌊x+y⌋ )=⌊y⌋+⌊x⌋,$ but the above identities are hard to follow. Thank you
|
The following does not use the fact that $e$ is transcendental.
First , note that $$\tag 1\lfloor x+y\rfloor \le\lfloor x\rfloor +\lfloor y\rfloor +1$$
for all real $x,y$.
Take the exponential of both sides of teh target equation to arrive at the equivalent condition
$$ \lfloor x+y\rfloor =e^{\lfloor x\rfloor}e^{\lfloor y\rfloor}$$
Using the mother of all $e$-inequalities,
$$\tag{$\star$} e^t\ge 1+t\qquad\text{with equality iff }t=0,$$
we find (as $x,y$ are positive by assumption)
$$\tag2 \begin{align}\lfloor x\rfloor+\lfloor y\rfloor +1&\ge \lfloor x+y\rfloor \\&=e^{\lfloor x\rfloor} e^{\lfloor y\rfloor}\\&\ge(1+\lfloor x\rfloor)(1+\lfloor y\rfloor)\\&=1+\lfloor x\rfloor+\lfloor y\rfloor +\lfloor x\rfloor\lfloor y\rfloor\end{align}$$
so that we must have $\lfloor x\rfloor\lfloor y\rfloor =0$ and the inequality in $(2)$ is sharp. by the sharpness condition in $(\star)$, we conclude that $$\lfloor x\rfloor =\lfloor y\rfloor=0$$
and by exploiting $(2)$, now an equality,
$$ \lfloor x+y\rfloor =\lfloor x\rfloor+\lfloor y\rfloor +1=1.$$
hence a necessary condition is that $$\fbox{$\quad0\le x<1, \quad 0\le y<1, \quad1\le x+y.\quad\vphantom\int$}$$
As this implies $x+y<2$, we see immediately that these are also sufficient as we compute $\ln\lfloor x+y\rfloor=\ln 1=0$ and $\lfloor x\rfloor+\lfloor y\rfloor = 0+0=0$.
| {
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Pedal form to polar form of an ellipse How do I convert the equation
$$\frac{a^2b^2}{p^2}=a^2+b^2-\frac{1}{u^2}$$
into the following equivalent form?
$$u^2=\frac{\sin^2 \theta}{b^2}+\frac{\cos^2 \theta}{a^2}$$
where $$\frac{1}{p^2}=u^2+\left( \frac{du}{d\theta} \right)^2$$
EDIT:
I have tried and found out
$$\frac{uab}{\sqrt{u^2(a^2+b^2)-1-u^4a^2b^2}}\, du = d\theta$$
How do I integrate and express the result in terms of $\sin \theta$ and $\cos \theta$?
$p$ is the perpendicular distance from $O$ to the tangent line to $C$ at the point in case of pedal equation of a curve. It is converted to $u$ as stated above.
| Assume $a>b>0$,
\begin{align}
t &= \sqrt{a^2\sin^2 \theta+b^2\cos^2 \theta} \\
t^2 &= (a^2-b^2)\sin^2 \theta+b^2 \\
\sin^2 \theta &= \frac{t^2-b^2}{a^2-b^2} \\
t^2 &= a^2+(b^2-a^2)\cos^2 \theta \\
\cos^2 \theta &= \frac{a^2-t^2}{a^2-b^2} \\
dt &=
\frac{(a^2-b^2)\sin \theta \cos \theta}
{\sqrt{a^2\sin^2 \theta+b^2\cos^2 \theta}} \, d\theta \\
&=\frac{\sqrt{(t^2-b^2)(a^2-t^2)}}{t} \, d\theta \\
d\theta &= \frac{t \, dt}{\sqrt{(t^2-b^2)(a^2-t^2)}} \\
\end{align}
Now
\begin{align}
d\theta &= \frac{abu}{\sqrt{(a^2 u^2-1)(1-b^2 u^2)}} \, du \\
&= \frac{abu \times (ab\, du)}{\sqrt{(a^2 b^2 u^2-b^2)(a^2-a^2b^2u^2)}} \\
&= \frac{t \, dt}{\sqrt{(t^2-b^2)(a^2-t^2)}} \\
dt &= ab \, du
\end{align}
With boundary conditions:
*
*$\theta=0$ , $u=\dfrac{1}{r}=\dfrac{1}{b}$
*$\theta=\dfrac{\pi}{2}$ , $u=\dfrac{1}{r}=\dfrac{1}{a}$
Therefore,
$$\fbox{$u=\frac{\sqrt{a^2\sin^2 \theta+b^2\cos^2 \theta}}{ab}$}$$
Alternatively,
\begin{align}
\int \frac{t\, dt}{\sqrt{(a^2-t^2)(t^2-b^2)}} &=
\tan^{-1} \sqrt{\frac{t^2-b^2}{a^2-t^2}} \\
\int \frac{u\, du}{\sqrt{(a^2 u^2-1)(1-b^2 u^2)}} &=
\frac{1}{ab} \tan^{-1}
\left( \frac{b}{a} \sqrt{\frac{a^2 u^2-1}{1-b^2 u^2}} \right)
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2834619",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove $(1-x)\ln(1-x)+(1+x)\ln(1+x)\leq 2x^2$ for $0
Find the smallest value of $c$ such that $(1-x)\ln(1-x)+(1+x)\ln(1+x)\leq c x^2$ holds for $0<x<1$.
I saw the curve and realized this is true for $c=2$. How can I prove it? What is the smallest $c$ that still makes the inequality valid?
I think we should use the Taylor expansions of the $\ln$s.
| The radius of convergence of the Maclaurin series of $(1-x)\log(1-x)+(1+x)\log(1+x)$ is one.
In explicit terms
$$\begin{eqnarray*} (1-x)\log(1-x)+(1+x)\log(1+x) &=& x^2+\frac{x^4}{6}+\frac{x^6}{15}+\frac{x^8}{28}+\ldots\\&=&x^2+\sum_{n\geq 2}\frac{x^{2n}}{n(2n-1)} \end{eqnarray*}$$
hence
$$ f(x)=\frac{(1-x)\log(1-x)+(1+x)\log(1+x)}{x^2} = 1+\sum_{n\geq 1}\frac{x^{2n}}{(n+1)(2n+1)} $$
is an increasing function on $(0,1)$, ranging from $1$ to $\color{red}{2\log 2}$, which is the optimal $c$-constant.
An improved inequality is $f(x)\leq x^2+(2\log 2-1)x^4$. An improved lower bound is $f(x)\geq \frac{x^2}{1-x^2/6}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2835098",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Use the Gauss elimination method in order to find $u$ in $\mathbb{R}^2$ such that $T(u)=\left(\begin{matrix} 3 & 1& 4 \end{matrix}\right)^T$ Consider the linear map $T:\mathbb{R}^2->\mathbb{R}^3$ with
$$T\left(\begin{matrix} x \\ y \end{matrix}\right)=\left(\begin{matrix} x-2y \\ 2x+y \\ -5x+8y \end{matrix}\right)$$ Use the Gauss elimination method in order to find $u$ in $\mathbb{R}^2$ such that $T(u)=\left(\begin{matrix} 3 \\ 1 \\ 4 \end{matrix}\right)$
My try:
$$\left(\begin{matrix} 1 & -2 & 3 \\ 2 & 1 & 1 \\ -5 & 8 & 4 \end{matrix}\right)_{R_2->R_2-2R_1}$$
$$=\left(\begin{matrix} 1 & -2 & 3 \\ 0 & 5 & -5 \\ -5 & 8 & 4 \end{matrix}\right)_{R_3->R_3+5R_1}$$
$$=\left(\begin{matrix} 1 & -2 & 3 \\ 0 & 5 & -5 \\ 0 & -2 & 19 \end{matrix}\right)_{R_2->\frac{R_2}{5}}$$
$$=\left(\begin{matrix} 1 & -2 & 3 \\ 0 & 1 & -1 \\ 0 & -2 & 19 \end{matrix}\right)_{R_3->R_3+2R_2}$$
$$=\left(\begin{matrix} 1 & -2 & 3 \\ 0 & 1 & -1 \\ 0 & 0 & 17 \end{matrix}\right)_{R_3->\frac{R_3}{17}}$$
$$=\left(\begin{matrix} 1 & -2 & 3 \\ 0 & 1 & -1 \\ 0 & 0 & 1 \end{matrix}\right)_{R_2->R_2+R_3}$$
$$=\left(\begin{matrix} 1 & -2 & 3 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}\right)_{R_1->R_1-3R_3}$$
$$=\left(\begin{matrix} 1 & -2 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}\right)_{R_1->R_1+2R_2}$$
$$=\left(\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}\right)$$
Then how to find $u$?
| What you need to solve is the system of linear equations:
$$\left\{\begin{align} x-2y&=3, \\ 2x+y&=1, \\ -5x+8y&=4. \end{align}\right.$$
The point of the Gauss elimination method (I recommend you check out this Wikipedia page!) is that we "encode" a system of equations by its augmented matrix, and then we apply the so-called elementary row operations to simplify it in a sense. Each row of the augmented matrix represents the respective equation. And each operation on the matrix corresponds to what we can do to equations in the system.
You correctly set up the augmented matrix as
$$\begin{bmatrix} 1 & -2 & 3 \\ 2 & 1 & 1 \\ -5 & 8 & 4 \end{bmatrix},$$
but why? (Disclaimer: traditionally we would separate the last column of an augmented matrix with a bar or a dashed line, but I don't remember how to do it in LaTeX.) Look at the coefficients in the first equation and at the numbers in the first row:
$$x-2y=3 \quad \text{or} \quad \color{green}{1}x+\color{blue}{(-2)}y=\color{magenta}{3} \quad \text{is represented by} \quad \begin{matrix} \color{green}{1} & \color{blue}{-2} & \color{magenta}{3} \end{matrix}.$$
Similarly for all the other rows and equations. Then, as I said, each step of the Gauss elimination method is equivalent to doing something to the equations of the original system. Each new matrix you obtain corresponds in the same way to a new system of equations, but all these systems are equivalent to the original one — in the sense that the solution set is preserved. For example, the second matrix that you obtained describes the follows system of equations:
$$\begin{bmatrix} 1 & -2 & 3 \\ 0 & -5 & 5 \\ -5 & 8 & 4 \end{bmatrix} \quad \text{represents} \quad \left\{\begin{align} x-2y&=3, \\ -5y&=5, \\ -5x+8y&=4. \end{align}\right.$$
And this new system is equivalent to the original one.
Now look at the last row of the last matrix and think what is the equation represented by that row:
$$\begin{matrix} \color{green}{0} & \color{blue}{0} & \color{magenta}{1} \end{matrix} \quad \text{represents} \quad \color{green}{0}x+\color{blue}{0}y=\color{magenta}{1}.$$
What values of $x$ and $y$ make this equation true?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2835392",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
How to express $\frac{\frac{1}{2}}{\sqrt[4]{\frac{3}4}-\sqrt[4]{\frac{1}{4}}}$ into $\frac{1}{2}\sqrt{5+3\sqrt{3}+2\sqrt{12+7\sqrt{3}}}$ I am just wondering how to express $\frac{\frac{1}{2}}{\sqrt[4]{\frac{3}4}-\sqrt[4]{\frac{1}{4}}}$ into $\frac{1}{2}\sqrt{5+3\sqrt{3}+2\sqrt{12+7\sqrt{3}}}$ (This one is taken from Wolfram BTW)
Clearly, $\frac{\frac{1}{2}}{\sqrt[4]{\frac{3}4}-\sqrt[4]{\frac{1}{4}}}$ can be rewritten as $\frac{1}{2(\sqrt[4]{\frac{3}{4}}-\sqrt[4]{\frac{1}{4}})}$, but what should I do next?
I am a little inexperienced in manipulating surds, so some helps are very much appreciated
Thank you!
| You will probably have to use the identity:
$$\dfrac{1}{x-y} = \dfrac{(x+y)(x^2+y^2)}{x^4-y^4} = \dfrac{x^3+xy^2+x^2y+y^3}{x^4-y^4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2838072",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Eliminating $\theta$ from $a\cos(\theta-\alpha)=x$ and $b\cos(\theta-\beta)=y$ Eliminate $\theta$ in following equations
$$\begin{align}
a \cos(\theta-\alpha) &= x \\
b \cos(\theta- \beta) &=y
\end{align}$$
I am trying to solve this problem but still I am unable to get the perfect answer
I added both the equations but it transformed it to
$2 \cos(\theta+(\alpha + \beta)/2)$
| \begin{align}
\frac{x}{a}+\frac{y}{b} &= \cos (\theta-\alpha)+\cos (\theta-\beta) \\
&=2\cos \frac{\alpha-\beta}{2} \cos \frac{\alpha+\beta-2\theta}{2} \\
\frac{x}{a}-\frac{y}{b} &= \cos (\theta-\alpha)-\cos (\theta-\beta) \\
&=-2\sin \frac{\alpha-\beta}{2} \sin \frac{\alpha+\beta-2\theta}{2} \\
1 &=
\left(
\frac{\frac{x}{a}+\frac{y}{b}}{2\cos \frac{\alpha-\beta}{2}}
\right)^2+
\left(
-\frac{\frac{x}{a}-\frac{y}{b}}{2\sin \frac{\alpha-\beta}{2}}
\right)^2 \\
\sin^2 (\alpha-\beta) &=
\frac{x^2}{a^2}-\frac{2xy\cos (\alpha-\beta)}{ab}+\frac{y^2}{b^2}
\end{align}
*
*The curve is known as Lissajous figure (with same frequencies).
*The area bounded by the ellipse is $A=\pi ab\sin (\beta-\alpha)$.
*$A>0$ gives anti-clockwise trace whereas $A<0$ for clockwise.
*When the curve degenerates to a line segment, $A=0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2838356",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Solutions of $x^2 -xy + y^2 \equiv 0 \pmod{43}$
$x^2 -xy + y^2 \equiv 0 \pmod{43}$
We could transform this into
$\frac{x^3 + y^3}{x+y} \equiv 0 \pmod{43}$
and thus
$x^3 + y^3 \equiv 0 \pmod{43}$
From what I could manage so far, I know that if $x = 3k$ , $y= 3k+1$, as $43 = 3k+1$.
Also there are no solutions for $43$ itself, $84$ and $127$.
How should I proceed?
| The polynomial $p(T)=T^2-T+1$ is the sixth cyclotomic polynomial. Meaning that its zeros have order six.
Because $43-1=6\cdot7$ an element $a$ of the cyclic group $\Bbb{Z}_{43}^*$ is of order six if and only if it is a seventh power but is not a 14th or a 21st power.
Let's just list seventh powers and do a bit of screening:
$2^7=128\equiv-1$, no good. $3^7\equiv37$, this works because its square is $37^2\equiv-7$ and its cube is $\equiv-1$ so it has order six. Therefore $37$ is one zero of $p(T)$. The sum of the zeros is, by Vieta, equal to $1$, so the other zero is $1-37=7$.
Therefore
$$p(T)=(T-7)(T-37).$$
This settles your question. Whenever $x=ay$ we have
$$x^2-xy+y^2=y^2(a^2-a+1)=y^2p(a).$$
The solutions are thus $x=7y$ and $x=37y$, $y$ can be anything.
These are all the solutions because if we write $a=x/y$ the above calculation shows that $p(a)=0$. If $y=0$ then the only solution is obviously $x=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2840674",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
} |
Find all $z$ for which $z^2+2z+2$ is real positive. Find all $z$ for which $z^2+2z+2$ is real positive.
My Attempt :
Let $z=x+iy, (x,y\in R)$:
$$z^2+2z+2= (x^2-y^2+2ixy) +2(x+iy)+2$$
$$=(x^2-y^2+2x+2)+i(2xy+2y)$$
| Let $p$ be a real number.
$$(z+1)^2+1=p^2\iff z=\pm\sqrt{p^2-1}-1.$$
If $|p|<1$, $z=\pm i\sqrt{1-p^2}-1$ runs from $-1-i$ to $-1+i$,
If $|p|\ge1$, $z=\pm\sqrt{p^2-1}-1$, runs from $-\infty$ to $\infty$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2842927",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 8,
"answer_id": 6
} |
Find a close formula using generating functions I want to find the sum:
$$\sum \limits_{k=0}^n {n\choose k}2^{k-n}$$
I did it by using the binomial theorem and I got
$$\sum \limits_{k=0}^n {n\choose k}2^{k-n}
= 2^{-n} \sum \limits_{k=0}^n {n \choose k}2^k
= \left(\frac{1}{2}\right)^n3^n$$
I would like to know how to do it using generating functions. I have tried some calculations but I am stuck. This is what I have so far.
$$\sum \limits_{k=0}^n {n\choose k}2^{k-n}
=\sum \limits_{n\ge0} \sum \limits_{k\ge0} {n \choose k} 2^{k-n} x^n
=\sum \limits_{k\ge0} 2^k \sum \limits_{n\ge0} {n \choose k}\left(\frac{x}{2}\right)^n$$
| Another look at the problem, using the fact that
$${\displaystyle {\binom {n}{k}}={\binom {n-1}{k-1}}+{\binom {n-1}{k}}}$$
we can construct the recurrence relation for the sequence:
$$a_n=\sum \limits_{k=0}^n {n\choose k}2^{k-n}=
{n\choose 0}2^{-n}+\sum \limits_{k=1}^{n-1} {n\choose k}2^{k-n}+{n\choose n}2^{0}=\\
{n\choose 0}2^{-n}+\sum \limits_{k=1}^{n-1} \left({n-1\choose k-1}+{n-1\choose k}\right)2^{k-n}+{n\choose n}2^{0}=\\
{n\choose 0}2^{-n}+\sum \limits_{k=1}^{n-1} {n-1\choose k-1}2^{(k-1)-(n-1)}+\sum \limits_{k=1}^{n-1}{n-1\choose k}2^{k-n}+\color{red}{{n\choose n}2^{0}}=\\
\color{blue}{{n\choose 0}2^{-n}}+\sum \limits_{k=0}^{n-2} {n-1\choose k}2^{k-(n-1)}+\color{red}{{n-1\choose n-1}2^{0}}+\sum \limits_{k=1}^{n-1}{n-1\choose k}2^{k-n}=\\
a_{n-1}+\color{blue}{\frac{1}{2}{n-1\choose 0}2^{-(n-1)}}+\frac{1}{2}\sum \limits_{k=1}^{n-1}{n-1\choose k}2^{k-(n-1)}=\\
a_{n-1}+\frac{1}{2}a_{n-1}=\frac{3}{2}a_{n-1}$$
This recurrence can be solved by induction, using generating functions or characteristic polynomials.
With generating functions, given $a_0=1$
$$f(x)=a_0+\sum\limits_{n=1}a_nx^n=1+\frac{3}{2}x\sum\limits_{n=1}a_{n-1}x^{n-1}=\\
1+\frac{3}{2}x\sum\limits_{n=0}a_{n}x^{n}=
1+\frac{3}{2}xf(x)$$
leading to
$$f(x)=\frac{1}{1-\frac{3}{2}x}=\sum\limits_{n=0}\color{red}{\left(\frac{3}{2}\right)^n}x^n$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2843760",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Find the inverse function of $f(x) = \frac{x^3+3x}{2}$
Find the inverse function of $f(x) = \frac{x^3+3x}{2}$
My thought on this are as follows:
I've solved a similar problem where I had to find the inverse of:
$$
g(x) = \sqrt[3]{x+\sqrt{x^2-1}} + \sqrt[3]{x-\sqrt{x^2-1}}
$$
So let $a = \sqrt[3]{x+\sqrt{x^2-1}}$ and $b = \sqrt[3]{x-\sqrt{x^2-1}}$, then
$y = a + b$ and hence:
$$
y^3 = (a+b)^3 = a^3 + b^3 + 3ab(a+b) = \\
= 2x+3ab(a+b)=2x+3(a+b)
$$
But $y = a + b$ and then
$$
y^3 = 2x + 3y \iff x = \frac{y^3-3y}{2}
$$
This function is very similar to the one in the title.
The answer says the inverse $f^{-1}(x) = \sqrt[3]{x+\sqrt{x^2+1}} + \sqrt[3]{x-\sqrt{x^2+1}}, \; x\in \mathbb R$. So knowing the answer it's easy to show that it indeed appears to be the inverse of $f(x)$ using the approach above. But how could I achieve the same result given the inverse function is unknown?
| Hint:
Let $$x:=\sqrt[3]u-\frac1{\sqrt[3]u}$$
Then
$$x^3+3x=u-3\sqrt[3]u+\frac3{\sqrt[3]u}-\frac1u+3\sqrt[3]u-\frac3{\sqrt[3]u}=u-\frac1u.$$
Now solve the equation
$$u-\frac1u=2y$$
for $u$ (it can be reduced to a quadratic one).
In the end,
$$x=\sqrt[3]{u(2y)}-\frac1{\sqrt[3]{u(2y)}}.$$
Notice that the product of the two $u$ roots is $-1$, so that the solution can also be expressed as the sum of the cubic roots.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2843861",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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