Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Find convergence interval of $f(x) =\sum_{n=1}^∞ \left(\sin\frac{2}{n^2}\right)x^n $ $$f(x) =\sum_{n=1}^∞ \left(\sin\frac{2}{n^2}\right)x^n $$
so looks like what i have to do is use the formal
$$\lim_{n\to∞}\left|\frac{a_{n+1}}{a_n}\right|=a$$ with $a_n=\sin\frac{2}{n^2}$. And then $R=1/a$.
But I having trouble with the lim. Thank you for your help
| $f(x) =\sum_{n=1}^∞ \left(\sin\frac{2}{n^2}\right)x^n
$
Since
$\sin(x) \approx x$
for small $x$,
$\sin\frac{2}{n^2}
\approx \frac{2}{n^2}
$
so the sum behaves like
$\sum_{n=1}^∞ \frac{2x^n}{n^2}
$
and this converges for
$-1 \lt x \lt 1$
by comparison with
$\sum x^n$
and converges for
$x = \pm 1$
because it becomes
$\sum \frac{2}{n^2}
$
at $x=1$
and
$\sum \frac{2(-1)^n}{n^2}
$
at $x=-1$,
both of which converge.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3086712",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Finding the argument of a complex function I've the following transfer function:
$$H(s)=\frac{1}{as^3+bs^2+cs+1}$$
Where $a,b,c$ are all real and positive.
How can I find $\arg(H(i\omega))$? And I know that $\omega\ge0$
What I did:
$$H(i\omega)=\frac{1}{a(i\omega)^3+b(i\omega)^2+c(i\omega)+1}=\frac{1}{-a\omega^3i-b\omega^2+c\omega i+1}=$$
$$\frac{1}{1-b\omega^2+(c\omega-a\omega^3)i}$$
Now finding the argument I can write:
$$\arg(H(i\omega))=\arg(1)-\arg(1-b\omega^2+(c\omega-a\omega^3)i)=$$
$$0-\arg(1-b\omega^2+(c\omega-a\omega^3)i)=-\arg(1-b\omega^2+(c\omega-a\omega^3)i)$$
Now, how can I setup a function that depends on the value of $a,b,c,\omega$?
| Denote $\theta=\arg(H(i\omega))$ and $$r=|1-b\omega^2+(c\omega-a\omega^3)i|=\sqrt{(1-b\omega^2)^2+(c\omega-a\omega^3)^2}.$$
From $$\theta=-\arg\big(1-b\omega^2+(c\omega-a\omega^3)i\big)$$ we get $$r\cos \theta =1-b\omega^2 \quad \text{and} \quad r\sin \theta =-(c\omega-a\omega^3)$$
and then $$\tan \theta = \frac{a\omega^3-c\omega}{1-b\omega^2}.$$ Finally $$\theta = \arctan\frac{a\omega^3-c\omega}{1-b\omega^2}.$$
I assumed silently that $1-b\omega^2\neq 0.$ If it were $0,$ the argument would be clear since the beginning.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Showing that $x^2+5=y^3$ has no integer solutions. I'm trying to show that the Diophantine equation $x^2+5=y^3$ has no integer solutions using the fact that $\mathbb Z[ \sqrt{-5}]$ has class number two. I think I have the general idea, but I'm having a tough time establishing the statement in bold below. Any ideas?
The equation gives a factorization into ideals
$$
(x+\sqrt{-5})(x-\sqrt{-5})=(y)^3
$$
in $\mathbb Z[ \sqrt{-5}]$. Assuming that $(x+\sqrt{-5})$ and $(x-\sqrt{-5})$ are relatively prime, we know that $(x+\sqrt{-5})=\mathfrak a^3$ and $(x-\sqrt{-5})=\mathfrak b^3$ for some ideals $\mathfrak a, \mathfrak b$ in $\mathbb Z[ \sqrt{-5}]$. Then the classes of $\mathfrak a$ and $\mathfrak b$ have order dividing 3 in the class group, hence they must both be principal, generated by some $\alpha,\beta\in \mathbb Z[\sqrt{-5}]$, respectively, as the class number is 2. Hence $\alpha^3=x+\sqrt{-5}$ up to units ($\pm 1$), and matching up real and imaginary parts then gives a contradiction.
How can I show that the ideals $(x+\sqrt{-5})$ and $(x-\sqrt{-5})$ are relatively prime? I was trying something along the lines of letting $\mathfrak p$ be a prime of $\mathbb Z[\sqrt{-5}]$ diving both ideals, in which case it also divides their sum (gcd). Then $$2\sqrt{-5}=x+\sqrt{-5}-(x-\sqrt{-5})\in (x+\sqrt{-5},x-\sqrt{-5})\subseteq \mathfrak p,$$ so $\mathfrak p\mid (2\sqrt{-5})$... But I'm not totally sure where I'm going with this.
| If ideals $I,J$ are relatively prime then their sum generates the ring $R$, so we can show this. In particular we can show also that
$$
(a + b\sqrt {- 5})(x+ \sqrt{- 5}) + (c + d\sqrt {- 5})(x - \sqrt {- 5}) = 1
$$
for some $a,b,c,d$.
This gives the equations
$$
\begin{align}
(a+c)x - 5(b-d) &= 1\\
(a-c) + (b+d)x &= 0
\end{align}
$$
So a necessary condition is $5$ does not divide $x$. This can be checked to be true from $x^2 + 5 = y^3$, but otherwise the ideals are not necessarily relatively prime in general.
Setting
$$
c = a + (b+d)x
$$
gives us
$$
2ax = (1+5(b-d)) - (b+d)x^2
$$
Taking $\pmod 2$ we see that $x$ must be even, so to get an integer solution for $a$ we must have $1+5(b-d)$ divisible by $2x$.
The obvious way is to look at the linear diophantine solutions
$$
xu - 5v = 1
$$
We can always choose $u$ to be even, so now we can set $d=0, v=b$ to get
$$
\begin{align}
a &= \frac{1+5(b-d)}{2x} - (b+d)\frac{x}{2}\\
&= \frac{1+5v}{2x} - v\frac{x}{2}\\
&= \frac{u}{2} - v\frac{x}{2}
\end{align}
$$
Since $u/2$ is integral we just need $vx$ to be even, which is true based on an earlier assumption: $x$ must be even.
It remains to show that $x$ is indeed even.
If $x=2m+1$ is odd, then $y$ is even and
$$
x^2 + 5 = 4m^2+4m+ 6 \equiv 2\pmod 4
$$
which contradicts $y^3 \equiv 0 \pmod 4$. Hence $x$ is even and we are done, finding a solution to the initial equation as
$$
(u/2 -vx/2 + v\sqrt {-5})(x + \sqrt {-5}) + (u/2 + vx/2)(x - \sqrt {-5}) = 1
$$
(Recall that $xu-5v=1$ and $u$ chosen to be even.)
Since the two ideals generate $1$, they are relatively prime.
| {
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"timestamp": "2023-03-29T00:00:00",
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If x,y and z are positive integers and $\frac 1x + \frac 1y = \frac 1z$ then $\sqrt{x^2+y^2+z^2}$ is rational. To solve this problem I first started off by factoring to get $z^2=(x-z)(y-z)$ only to realise that this does nothing so I then tried squaring both sides to get the reciprocals of $x,y$ and $z$ squared minus $2$ over $xy$ but after this I have no idea what to do any form of help would be appreciated.(sorry for not using mathjax as my browser does not support it)
| Because $x\neq-y$, $z=\frac{xy}{x+y}$, $x^2+xy+y^2>0$ and
$$\sqrt{x^2+y^2+z^2}=\sqrt{x^2+y^2+\frac{x^2y^2}{(x+y)^2}}=$$
$$=\frac{\sqrt{(x^2+y^2+xy-xy)(x^2+y^2+xy+xy)+x^2y^2}}{x+y}=\frac{x^2+xy+y^2}{x+y}.$$
| {
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Find the smallest natural number using Chinese Remainder Theorem Question is find the smallest natural number $x$ such that,
\begin{align}
x &\equiv 1 \pmod 2 \\
x &\equiv 2 \pmod 3 \\
x &\equiv 3 \pmod 4\\
x &\equiv 4 \pmod 5\\
x &\equiv 5 \pmod 6\\
x &\equiv 6 \pmod 7\\
x &\equiv 7 \pmod 8\\
x &\equiv 8 \pmod 9\\
x &\equiv 9 \pmod {10}\\
x &\equiv 10 \pmod {11}\\
x &\equiv 11 \pmod {12}\\
x &\equiv 0 \pmod {13}\\
\end{align}
However when I use the proof of Chinese remainder theorem, I cannot even get over the first step where I must find inverse modulo: $$b_1 \frac{6227020800}{2}\equiv 1 \pmod 2$$ which I presume is $$0\cdot \frac{6227020800}{2} + 2\cdot 1 \equiv 1$$ hence 0?
I am somewhat confused. A friend did it using another method, but I would like to learn how to use the Chinese remainder theorem.
$$x\equiv a_1 b_1 \frac{M}{m_1} + \cdots + a_k b_k \frac{M}{m_k} \pmod M$$
| The Chinese remainder theorem is about relatively prime modulos. Having all the modulos from $2$ to $13$ is redundant.
$x\equiv 1 \mod 2; x \equiv 3\mod 4; x\equiv 7\mod 8$ are redundant and can be replaced with just $x \equiv 7 \pmod 8$.
Assuming the question is legitimate, we need not consider any $x \equiv j \pmod {2^km}$ and have to consider only $x\equiv l\pmod m$
i.e., this question can be reduced to.
$x\equiv 7\pmod 8$
$x \equiv 8\pmod 9$
$x \equiv 4 \pmod 5$
$x \equiv 6\pmod 7$
$x \equiv 10 \pmod {11}$
$x \equiv 0 \pmod {13}$.
All the rest are redundant, as the solution to the above is unique.
Note the solution to all but $x \equiv 0 \pmod {13}$ is $x \equiv -1 \pmod n$ for $n = 8,9,5,7,11$ so
$x\equiv -1 \pmod{8*9*5*7*11=27720}$ and $x \equiv 0 \pmod {13}$.
So you need to solve. $x = 27720k -1 = 13m$
$27720 \equiv 4 \pmod {13}$
So $x \equiv 27720k - 1\equiv 0 \pmod {13}$
$\equiv 4k -1 \equiv 0 \pmod {13}$
So $4k \equiv 1 \pmod {13}$ so we just have to find the inverse of $4 \mod {13}$
And $4 \times 10 = 40 \equiv 1 \pmod{13}$ and so $k = 10$
and $x \equiv 277199 \pmod {27720*13}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3088578",
"timestamp": "2023-03-29T00:00:00",
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Finding the derivative of a messy function I need to find the derivative of this function. I think I managed to do this, but the simplification is messy. Is there a good way to keep things organized while I go through all of the needed differentiation rules? It was hard to stay focused in even writing this in latex.
$y = \frac{\sqrt[3]{2x}\sqrt{x+5}}{(4x-1)^9}$
numerator(u)/denominator(v)
$u = \sqrt[3]{2x}\sqrt{x+5}$
$v = (4x-1)^9$
derivative of numerator(u')/denominator(v')
$u' =\frac{(2x)^{\frac{1}{3}}(x+5)^{-\frac{1}{2}}}{2} + \frac{(x+5)^{\frac{1}{2}}(2x)^{-\frac{2}{3}}}{3}$ product rule, chain rule
$v' = 36(4x-1)^8$ chain rule
quotient rule $\frac{vu' - uv'}{v^2}$
$= \frac{\bigg(\frac{(2x)^{\frac{1}{3}}(x+5)^{-\frac{1}{2}}}{2} + \frac{(x+5)^{\frac{1}{2}}(2x)^{-\frac{2}{3}}}{3} \bigg) - 36\sqrt[3]{2x}\sqrt{x+5}\ (4x-1)^8}{(4x-1)^9}$
$= \frac{\bigg(\frac{\sqrt[3]{2x}}{2\sqrt{x+5}} + \frac{\sqrt{x+5}}{6\sqrt[3]{2x}} \bigg) - 36\sqrt[3]{2x}\sqrt{x+5}\ (4x-1)^8}{(4x-1)^9}$
Suffice to say this is a mess, does a less-tedious way exist to find the derivative?
| The classical trick is to consider
\begin{align}
\ln y = \frac{1}{3}\ln(2x)+\frac{1}{2}\ln(x+5)-9\ln(4x-1)
\end{align}
then differentiate to get
\begin{align}
\frac{y'}{y} =&\ \frac{1}{3x}+\frac{1}{2(x+5)}-\frac{36}{4x-1}\\
=&\ -\frac{196x^2+1045x+10}{6x(x+5)(4x-1)}.
\end{align}
Hence it follows
\begin{align}
y' =&\ -y\frac{196x^2+1045x+10}{6x(x+5)(4x-1)} = - \frac{\sqrt[3]{2x}\sqrt{x+5}}{(4x-1)^9}\frac{196x^2+1045x+10}{6x(x+5)(4x-1)} \\
=&\ - \frac{196x^2+1045x+10}{3\cdot 2^{2/3} x^{2/3}\sqrt{x+5}(4x-1)^{10}}
\end{align}
Remark: In the grand scheme of things, it doesn't save much time in terms of getting a simplified expression. However, it does have some organizational benefits.
Remark $2$: As mentioned by egreg, we can also write
\begin{align}
\ln |y| = \ln\left|\frac{\sqrt[3]{2x}\sqrt{x+5}}{(4x-1)^9}\right| = \frac{1}{3}\ln|2x|+\frac{1}{2}\ln|x+5|-9\ln|4x-1|
\end{align}
and use the fact that
\begin{align}
\frac{d}{dx}\ln|f(x)| = \frac{f'(x)}{f(x)}.
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
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Two coins, one experiment: what is the probability that the coin lands on heads on exactly $7$ of the $10$ flips? When coin $1$ is flipped, it lands on heads with probability $0.4$; when coin $2$ is flipped, it lands on heads with probability $0.7$. One of these coins is randomly chosen and flipped $10$ times.
(a) What is the probability that the coin lands on heads on
exactly $7$ of the $10$ flips?
(b) Given that the first of these $10$ flips lands heads, what is the conditional probability that exactly $7$ of the $10$ flips land on heads?
MY ATTEMPT
(a) Let us denote by $X$ the random variable which counts the numbers of heads in 10 flips. Thus $X\sim\text{Binomial}(10,p)$, where the parameter $p$ depends on the coin that is chosen: $C = 1$ stands for the event "the first coin is chosen" and $C = 2$ stands for the event "the second coin is chosen". Based on these considerations, we have the answer to the first question, which is
\begin{align*}
\textbf{P}(X = 7) = \textbf{P}(X = 7\mid C = 1)\textbf{P}(C = 1) + \textbf{P}(X = 7\mid C = 2)\textbf{P}(C = 2)
\end{align*}
where
\begin{cases}
\textbf{P}(X = 7\mid C = 1) = \displaystyle{10\choose 7}(0.4)^{7}(0,6)^{3}\\\\
\textbf{P}(X = 7\mid C = 2) = \displaystyle{10\choose 7}(0.7)^{7}(0,3)^{3}\\\\
\textbf{P}(C = 1) = \textbf{P}(C = 2) = 0.5
\end{cases}
(b) Since $X = \displaystyle\sum_{k=1}^{10} X_{k}$, where each $X_{k}\sim\text{Bernoulli}(p)$, we are interested in the event $\textbf{P}(X = 7\mid X_{1} = 1)$, which is equivalent to
\begin{align*}
\textbf{P}(X = 7\mid X_{1} = 1) & = \frac{\textbf{P}(X = 7, X_{1} = 1)}{\textbf{P}(X_{1} = 1)} = \frac{\textbf{P}\left(\displaystyle\sum_{k=2}^{10}X_{k} = 6\right)}{\textbf{P}(X_{1} = 1)}
\end{align*}
Since $\textbf{P}(X_{1} = 1) = \textbf{P}(X_{1} = 1 | C = 1)\textbf{P}(C = 1) + \textbf{P}(X_{1} = 1 | C = 2)\textbf{P}(C = 2)$, it suffices to calculate $\textbf{P}(Y = 6)$, where $Y := \displaystyle\sum_{k=2}^{10}X_{k}$ and, consequentely, $Y\sim\text{Binomial}(9,p)$:
\begin{align*}
\textbf{P}(Y = 6) = \textbf{P}(Y = 6\mid C = 1)\textbf{P}(C = 1) + \textbf{P}(Y = 6\mid C = 2)\textbf{P}(C = 2)
\end{align*}
where
\begin{cases}
\textbf{P}(Y = 6\mid C = 1) = \displaystyle{9\choose 6}(0.4)^{6}(0,6)^{3}\\\\
\textbf{P}(Y = 6\mid C = 2) = \displaystyle{9\choose 6}(0.7)^{6}(0,3)^{3}\\\\
\textbf{P}(C = 1) = \textbf{P}(C = 2) = 0.5
\end{cases}
My question is: am I working it right or is there any conceptual misapplication?
Any contribution is appreciated. Thanks in advance!
| For part $a)$,
Let $K_1$ denotes when coin $1$ is chosen and $K_2$ denotes when coin $2$ is chosen.
The probability that the coin lands on heads on exactly $7$ of the $10$ flips is $P(H)$
$$P(H)=P(H|K_1)P(K_1)+P(H|K_2)P(K_2)$$
$$P(H)=\dbinom{10}{7}(0.4)^7(0.6)^3(0.5)+\dbinom{10}{7}(0.7)^7(0.3)^3()0.5$$
For part $b)$,
When the first flip is head, then the probability is computed from choosing $6$ head flips out of $9$ flips in both coins case. So, we get $\dbinom{9}{6}(0.4)^6(0.6)^3(0.5)+\dbinom96(0.7)^6(0.3)^3(0.5)$
Let $T$ denotes the event that the first flip is head.
$$P(H|T)=\dfrac{P(H\cap T)}{P(T)}$$
$P(T)=P(T|K_1)P(K_1)+P(T|K_2)P(K_2)=(0.4)(0.5)+(0.7)(0.5)=0.55$
$P(H\cap T)=P(H\cap T|K_1)P(K_1)+P(H\cap T|K_2)P(K_2)$
$$=\dbinom96(0.4)^6(0.6)^3(0.5)+\dbinom96(0.7)^6(0.3)^3(0.5)$$
| {
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How to solve this counting problem easily? Letters of the word $BRILLIANT$ are rearranged. What is the probability that the extreme positions are always occupied by consonants? Assume that all permutations are equally likely to occur.
My attempt:
There are 6 consonants (B, R, L, L, N and T). Select two of these in $\binom{6}{2}$ ways. These 2 letters can be permuted in 2! ways. Put these 2 selected consonants at the extremes and then the remaining 7 characters can be permuted in 7! ways. But we are over-counting since there are 2 Is and 2 Ls. I am unable to proceed further.
| In the comments, you requested a solution in which the probability is calculated by dividing the number of arrangements in which both ends are filled by consonants by the total number of arrangements.
The word BRILLIANT contains nine letters, including one B, one R, two Is, two Ls, one A, one N, and one T.
First, we count the number of distinguishable arrangements of the word BRILLIANT. Choose two of the nine positions for the Is, two of the remaining seven positions for the Ls, and then arrange the remaining five distinct letters in the remaining five positions.
There are $$\binom{9}{2}\binom{7}{2}5! = \frac{9!}{2!7!} \cdot \frac{7!}{2!5!} \cdot 5! = \frac{9!}{2!2!}$$ distinguishable arrrangements of the letters of the word BRILLIANT. The factors of $2!$ in the denominator represent the number of ways the two Is and two Ls can be permuted within an arrangement without producing an arrangement that is distinguishable from the given arrangement.
Next, we count the number of arrangements of the word BRILLIANT in which both ends are occupied by consonants. Note that this means that the three vowels must be located in the seven interior positions. Choose two of those seven positions for the Is and one of the remaining five interior positions for the A. That leaves six positions to fill, including both ends. Choose two of those six positions for the two Ls. Arrange the remaining four distinct consonants in the remaining four positions.
There are $$\binom{7}{2}\binom{5}{1}\binom{6}{2}4! = \frac{7!}{2!5!} \cdot \frac{5!}{1!4!} \cdot \frac{6!}{2!4!} \cdot 4! = \frac{7! \cdot 6 \cdot 5}{2!2!}$$
Thus, the desired probability is
$$\frac{\dbinom{7}{2}\dbinom{5}{1}\dbinom{6}{2}4!}{\dbinom{9}{2}\dbinom{7}{2}5!} = \frac{\dfrac{7! \cdot 6 \cdot 5}{2!2!}}{\dfrac{9!}{2!2!}} = \frac{7! \cdot 6 \cdot 5}{9!} = \frac{6 \cdot 5}{9 \cdot 8}$$ which agrees with the simpler calculation given by Prasad Kaunagoda.
| {
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Solve recursion $a_n=a_{n-1}-6\cdot3^{n-1}$ for $n>0, a_0=0$ $a_n=a_{n-1}-6\cdot3^{n-1}$ for $n>0, a_0=0$
So I calculate first terms
$a_0=0$
$a_1=-6$
$a_2=-24$
$a_3=-78$
I don't see any relation so
$a_n=a_{n-1}-6\cdot3^{n-1}$
$a_{n-1}=a_{n-2}-6\cdot 3^{n-2}$
. . .
$a_2=a_1-6\cdot3^{1}$
$a_1=a_0-6\cdot 3^{0}$
Not sure what to do next, Wolfram solves it in this way:
$a_n=-3\cdot(3^{n}-1)$
How do I get to this point?
| You have
$$a_n=a_n=a_{n-1}-6\cdot 3^{n-1}=a_{n-2}-6\cdot 3^{n-2}-6\cdot 3^{n-1}=\dotsm,$$
so you can prove with an easy induction that
$$a_n = a_0-6\sum_{k=0}^{n-1} 3^k=-6\frac{3^n-1}{3-1}.$$
| {
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Series area of convergence sadly I don't have any solutions. Is this correct?
$$\sum_{n=1}^\infty\frac{ 2^nx^n}{\sqrt{n^4+1}}$$
We calculate the radius of convergence:
$R=\lim_{x \to\infty}\big|\frac{a_n}{a_{ń+1}}\big|=\lim_{n\to\infty}\frac{1}{2}\sqrt{\frac{(n+1)^4+1}{n^4+1}}=\lim_{n\to\infty}\frac{1}{2}\sqrt{\frac{(1+1/n)^4+1/n^4}{1+1/n^4}}=1/2$
We check the boundary:
$x=\frac{1}{2}$
$$\sum_{n=1}^\infty\frac{2^n(\frac{1}{2})^n}{\sqrt{n^4+1}}=\sum_{n=1}^\infty\frac{1}{\sqrt{n^4+1}}$$
By comparing this to the harmonic series, we conclude that it does not converge.
$x=\frac{-1}{2}$
We get
$$\sum_{n=1}^\infty\frac{(-1)^n}{\sqrt{n^4+1}}$$
Since $\frac{1}{\sqrt{n^4+1}}\geq 0 \quad \forall n$ and $\lim_{n\to\infty} \frac{1}{\sqrt{n^4+1}} = 0$ and because $\frac{1}{\sqrt{n^4+1}}$ is monotonic decreasing (since the square root is monotonic increasing) we can conclude, using Leibniz-Test, that this series does converge.
So we get: $-\frac{1}{2}\leq x < \frac{1}{2}$
| The implication is not true for $x={1\over 2}$. By comparison test we have $$\sum_{n=1}^\infty\frac{1}{\sqrt{n^4+1}}<\sum_{n=1}^\infty\frac{1}{\sqrt{n^4}}=\sum_{n=1}^\infty\frac{1}{{n^2}}={\pi^2\over 6}$$
| {
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Solve the recurrence relation $a_n=2a_{n-1}+15a_{n-2}+8$ for $n\geq2$, $a_0=0$, $a_1=1$ My task:
$a_n=2a_{n-1}+15a_{n-2}+8$ for $n\geq2$, $a_0=0$, $a_1=1$
My solution
$x^{2}-2x-15$
$\Delta=64$
$x1=-3 $
$x2=5$
So I am gonna use following formula:
$a_n=ar^{n}+br^{n}$
$a_n=a*(-3)^{n}+b*5^{n}$
$+8$ is the problem, so I am looking for $c$ that $b_n:=a_n+c\implies b_n=2b_{n-1}+15b_{n-2}$
$$b_n=2(b_{n-1}-c)+15(b_{n-2}-c)+8+c=2b_{n-1}+15b_{n-2}+8-16c$$ I am setting $c=\frac{1}{2}$ so
$$b_n=2b_{n-1}+15b_{n-2}\implies\exists a,\,b:\,b_n=a*(-3)^{n}+b*5^{n}.$$From $b_0=\frac{1}{2},\,b_1=-\frac{1}{2}$, after finding $a,\,b$. Then $a_n=b_n-\frac{1}{2}$.
$$a=\frac{3}{8}$$
$$b=\frac{1}{8}$$
$b_2=\frac{52}{8}$
$a_2=\frac{52}{8}-\frac{1}{2}=6$
Actual $a_2=10, a_3$=43
So $a_2$ from $b_n$ method is not equal to actual $a_n$.
It means I am doing something wrong here, could anyone point out the mistake?
| Assuming $a(n) = \gamma^n$ and substituting into the homogeneous
$$
\gamma^n-2\gamma^{n-1}-15\gamma^{n-2}=0\to \gamma^n\left(1-\frac{2}{\gamma}-\frac{15}{\gamma^2}\right)=0
$$
and solving for $\gamma$ we have
$$
a_h(n) = C_1(-3)^n + C_2 5^n
$$
and the particular dictates
$$
a_p(n)-2a_p(n-1)-15a_p(n-2) = 8
$$
so making $a_p(n) = C_0$ and substituting into the particular we have
$$
C_0-2C_0-15C_0 = 8\to C_0 = -\frac 12
$$
and finally
$$
a(n) = a_h(n)+a_p(n) = C_1(-3)^n + C_2 5^n-\frac 12
$$
NOTE
$$
a(0) = C_1+C_2-\frac 12 = 0\\
a(1) = C_1(-3)+C_25-\frac 12 =1
$$
and solving for $C_1, C_2$ gives
$$
a(n) = \frac 18\left(-4+(-3)^n+3 \cdot 5^n\right)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3090006",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Proving $\sqrt3 + \sqrt[3]{2}$ to be irrational In a test I tried to solve recently I came across the following question:
Prove $$\sqrt3 + \sqrt[3]{2}$$ is irrational
I tried proving it by saying it is equal to some rational number $$\sqrt3 + \sqrt[3]{2} = \frac{m}{n}$$ and reaching a contradiction.
I tried squaring / cubing both sides and that didnt work. So how do I go about solving this problem?
| It's possible to find a polynomial that has $\sqrt{3}+\sqrt[3]{2}$ as a root, along with several other variations from choosing different square and cube roots.
$$\left((x-\sqrt{3})^3-2\right)\left((x+\sqrt{3})^3-2\right)=0$$
$$\left(x^3-3\sqrt{3}x^2+9x-3\sqrt{3}-2\right)\left(x^3+3\sqrt{3}x^2+9x+3\sqrt{3}-2\right)=0$$
$$x^6-9x^4-4x^3+27x^2-36x-23 = 0$$
By the rational root theorem, the only possible rational roots of that polynomial are $-23,-1,1,$ and $23$. $\sqrt{3}+\sqrt[3]{2}$ is between $1+1=2$ and $2+2=4$, so it isn't any of them. Also, as is easily verified, none of those possible roots actually are roots. The values at $\pm 1$ are $1-9-4+27-36-23=-44$ and $1-9+4+27+36-23=36$, while at $\pm 23$ the $x^6$ term is much larger than everything else combined.
This is not the easy way, of course.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
Find limit (type 0/0) I'm struggling to find the limit
$$I = \lim_{x\to 0}\frac{2\sqrt{1-x} - \sqrt[3]{8-x}}{x}$$
What I was trying:
$$ I = \lim_{x\to 0}\frac{1-x + 2\sqrt{1-x} + 1 - (1-x) - 1 - \sqrt[3]{8-x}}{x} $$
$$ = \lim_{x\to 0}\frac{(\sqrt{1-x}+1)^2 - (2-x)- \sqrt[3]{8-x}}{x} \qquad \quad $$
$$ = \lim_{x\to 0}\frac{(\sqrt{1-x}+1)^2 - (2-x+ \sqrt[3]{8-x})}{x} \qquad \qquad $$
$$ = \lim_{x\to 0}\frac{(\sqrt{1-x}+1)^2}{x} - \lim_{x\to 0}\frac{(2-x+ \sqrt[3]{8-x})}{x} \qquad $$
Thank all of you for your answers.
| $f(x)=2(1-x)^{1/2}$;
$g(x)=\sqrt[3]{8-x}$.
$F(x)=\dfrac{f(x)-f(0)}{x}=$
$\dfrac{2\sqrt{1-x}-2}{x}$
$G(x)=\dfrac{g(x)-g(0)}{x}=\dfrac{\sqrt[3]{8-x}-2}{x}$
$F'(0)= [-(1-x)^{-1/2}]_{x=0}=-1$.
$G'(0)=[-(1/3)(8-x)^{-2/3}]_{x=0}=-(1/3)(1/4)=-1/(12).$
Hence : $F'(0)-G'(0)= -11/(12)$.
Recall:
$\lim_{ x \rightarrow 0}\dfrac{F(x)-F(0)}{x}=F'(0)$,
similarly for $G$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3094942",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Alphabet with 6 vowels and 12 consonants, find the amount of words without two consonants in a row. I just took an exam and as usual with exams, the answers come to you when you're done with the exam and you are sitting in your favourite chair at home. I want to verify my solution as part of my learning process to learn from my mistakes in case I might want to schedule a resit
Consider an alphabet $A$ consisting of $6$ vowels and of $12$ consonants. Valid words consist of no two consonants in a row, so AART is not valid, nor is JUDITH, but JUDIT is fine and so is AAR, as is AIAIAIAIAIAIAIAIAI. $a_n$ denotes the amount of valid words.
a) find $a_0$, $a_1$, $a_2$, $a_3$
$a_0=1$, the empty word
$a_1=12+6=18$ (just one letter)
For $a_2$ we considers words like $AT$, $TA$, $IA$(different vowels) and $AA$ (same vowels)
$a_2= 2 \times 6 \cdot 12 + 5 \cdot 6 + 6=144 +30 +6=180$
We expand to three symbols by either adding a vowel to the end of a 2-letter word or by adding a vowel and consonant to a 1-letter word
$a_3=180 \cdot 6 + 6 \cdot 12 \cdot 18 =1080+1296=2376$
(b) Find a recurrence relation
(c) solve it
We make a case distinction for a valid word of length $n$, it either ends in a consonant or in a vowel. If it ends in a consonant, we must have obtained it from a valid word of length $n-2$ by placing a vowel followed by a consonant behind it. In all other situations we simply place a vowel behind a word of length $n-1$.
We get for $n\geq 2$:
$$ a_n = 6 \cdot a_{n-1} + 6 \cdot 12 \cdot a_{n-2}$$
One can verify that this indeed gives $180$ for $a_2$.
We can solve this recursion via an auxiliary equation of the form:
$$ r^2 = 6r + 6 \cdot 16 $$
$$ r^2 - 6r - 6 \cdot 16 =0$$
Which factorises as:
$$ (r-12)(r+6)=0$$
So we get solutions $a_n = A r_1^n + B r_2^n$:
$$ a_n = A \cdot 12^n + B \cdot (-6) ^n$$
We can now plug in our initial conditions $a_0=1$ and $a_1=18$
$$1=A+B$$
$$ 18= 12A - 6B=18A -6 \implies 18A=24 \implies A=\frac{4}{3}, B=-\frac{1}{3}. $$
We get:
$$ a_n = \frac{4}{3}\cdot 12^n -\frac{1}{3} (- 6)^n$$
I feel that this is probably correct, but I am unsure. Can someone please verify?
| if there are $a_n$ n-letter words, $c_n$ end in a consonant and $v_n$ end in a vowel
$a_n = c_n + v_n$
To make a n+1 letter word, we take the a_n letter words and tack a letter onto the end.
$a_n = 6c_n + 12 v_n\\
c_{n+1} = 12v_n\\
v_{n+1} = 6 c_n+ 6v_n = 6 a_n$
We probably could do away with $a_n$ and represent just with $c_n, v_n$
$\begin{bmatrix} c_{n+1}\\v_{n+1} \end {bmatrix} = \begin{bmatrix}0&12\\6 & 6\end{bmatrix}\begin{bmatrix} c_{n}\\v_{n} \end {bmatrix}$
$\begin{bmatrix} c_{n+1}\\v_{n+1} \end {bmatrix} = \begin{bmatrix}0&12\\6 & 6\end{bmatrix}^n\begin{bmatrix} c_{1}\\v_{1} \end {bmatrix}$
$\begin{bmatrix} c_{n+1}\\v_{n+1} \end {bmatrix} = \frac 13 \begin{bmatrix}1&-2\\1 & 1\end{bmatrix}\begin{bmatrix}12^n&0\\ 0& -6^n\end{bmatrix}\begin{bmatrix}1&2\\-1 & 1\end{bmatrix}\begin{bmatrix} 12\\6 \end {bmatrix}$
$a_n = \frac {4(12)^n - (-6)^n}{3}$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Solve $(1+x^2) f''(x) + 4x f'(x) + 2 f(x)=0$ via power series method. I have trouble finding a closed expression for the following problem:
$$(1+x^2) f''(x) + 4x f'(x) + 2 f(x)=0$$ where also $f'(0)=0$ and $f(0)=1$
Solve this via a power series method, so suppose $f(x)= \sum_{n=0} ^{\infty}a_n x^n$. Find its radius of convergence and a closed expression.
I tried solving this, but I get a really nasty expression. We first plug in our Ansatz into the differential eqution, with the usual derivatives:
$$ (1+x^2)\sum_{n=0} ^{\infty}n (n-1)a_n x^{n-2}+ 4x\sum_{n=0} ^{\infty}n a_n x^{n-1} + 2 \sum_{n=0} ^{\infty}a_n x^{n}=0$$
We now multiply out the factors:
$$ \sum_{n=0} ^{\infty}n (n-1)a_n x^{n-2} +\sum_{n=0} ^{\infty}n (n-1)a_n x^{n} + \sum_{n=0} ^{\infty}4n a_n x^{n} + \sum_{n=0} ^{\infty}2a_n x^{n}=0$$
We apply a shift to the first sum:
$$ \sum_{n=0} ^{\infty}(n+2) (n+1)a_{n+2} x^{n} +\sum_{n=0} ^{\infty}n (n-1)a_n x^{n} + \sum_{n=0} ^{\infty}4n a_n x^{n} + \sum_{n=0} ^{\infty}2a_n x^{n}=0$$
We now use the identity theorem for power series to compare coefficients and get for $n \geq 2$:
$$ (n+2)(n+1)a_{n+2} + n(n-1)a_n + 4n a_n +2a_n=0$$
After rewriting, we get:
$$ (n+2)(n+1)a_{n+2} + (n+2)(n+1)a_{n}=0$$
$$ a_{n+2} = - a_n$$
Which is not a nice expression, since $a_0=1$ and $a_1=0$ this tells me that $a_2=-1$, $a_4 =1$, $a_6=-1$ and all odd terms are $0$.
What am I missing here?
| We recognise this to be:
$$ 1 - x^2 + x^4 - x^6 +x^8\dots$$
Notice that:
$$ \frac{1}{1 - (-x^2)}= 1- x^2 + x^4 - x^6 +x^8 \dots $$
So our closed expression is the geometric series variation $$\frac{1}{1+x^2}$$
Its radius of convergence follows from the geometric series to be $|x^2|<1$ so $|x|<1$. This means the series has radius of convergence $1$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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The definite integrals of $e^{-x} \sin^n{(x)}$ and $e^{-x} \cos^n{(x)}$ between $0$ and $\infty$ For the integral
$$\begin{align}S_n =\int_{0}^{\infty} e^{-x} \sin^n{(x)}dx\end{align}$$
I have found a formula for $S_n$:
$$\begin{align} S_n=2^{\frac{(-1)^n-1}{2}} n!! (n-1)!! \prod_{2\le k\le n}^{{k\equiv n \mod{2}}} \frac{1}{k^2+1} \end{align}$$
By solving the following recurrence relation:
$$\begin{align} S_n=\frac{n(n-1)}{n^2+1}S_{n-2} \end{align}$$
But I cannot find a similar general formula for the integral
$$C_n=\int_{0}^{\infty} e^{-x} \cos^n{(x)}dx$$
Given that I have found the following recurrence relation for $C_n$:
$$\begin{align} C_n = \frac{1}{n^2+1}+ \frac{n(n-1)}{n^2+1}C_{n-2}\end{align}$$
Is it possible to find a general formula for $C_n$ in a similar format to that of $S_n$?
| Please, fasten your seatbelt before reading !
Using a CAS and defining
$$A_n=\frac{2^{n+2}}{\sqrt{\pi }\, n!}C_n$$ we have
$$A_n=\frac{\sqrt{\pi } \left(2 e^{\pi } (-1)^n+1+e^{2 \pi }\right) (\coth (\pi
)-1)}{\Gamma \left(\frac{n+2-i}{2} \right) \Gamma \left(\frac{n+2+i}{2}
\right)}-\, _3\tilde{F}_2\left(1,\frac{n+2-i}{2} ,\frac{n+2+i}{2}
;\frac{n+3}{2},\frac{n+4}{2};1\right) $$
where appears the regularized generalized hypergeometric function.
By magic, this produces for the $C_n$'s the sequence
$$\left\{\frac{1}{2},\frac{3}{5},\frac{2}{5},\frac{41}{85},\frac{9}{26},\frac{263}{62
9},\frac{101}{325},\frac{15357}{40885},\frac{7597}{26650},\frac{284603}{825877},
\frac{43116}{162565},\frac{38393473}{119752165}\right\}$$
As I wrote in a comment after @Song's answer, I have the vague feeling that this could write in terms of incomplete beta functions but I have not been able to do it.
Edit
Concerning the $S_n$, the formula is effectively simpler
$$S_n=\frac{e^{\pi /2} \pi \left((-1)^n+e^{\pi }\right) (\coth (\pi )-1) \,n!}{2^{n+1}\Gamma \left(\frac{n+2-i}{2} \right) \Gamma \left(\frac{n+2+i}{2}
\right)}$$
Concerning the $C_n$, starting from @Song's answer, I finally obtained the surprising
$$C_n=\frac{i^{n+1-i} } {2^{n+2}}\left(B_{-1}\left(-\frac{n-i}{2},n+1\right)-(-1)^i
B_{-1}\left(-\frac{n+i}{2},n+1\right)\right)$$
| {
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"source": "stackexchange",
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Why these $2$ methods give the exact same answer for sum of squares? Consider $n = 8$, the sum of squares from $1$ through $8$ is:
$1 \times 1 + 2 \times 2 + 3
\times 3 + 4 \times 4 + 5 \times 5 + 6 \times 6 + 7 \times 7 + 8 \times 8 = 204$.
Also, equal to
$1 \times 8 + 3 \times 7 + 5 \times 6 + 7 \times 5 + 9 \times 4 + 11 \times 3 + 13 \times 2 + 15 \times 1 = 204$.
The second one logic is that I start with $1$, then I increment by $2$ each time and subtract $1$ from the second one, until I reach $1$.
For $n = 2$.
$1 \times 1 + 2 \times 2 = 4 = 1 \times 2 + 3 \times 1$.
For $n = 3$.
$1 \times 1 + 2 \times 2 + 3 \times 3 = 1 \times 3 + 3 \times 2 + 5 \times 1$
The question is, why is it supposed to be the equal the one above? I tried it with a lot of values for $n$?
| The first sum is $\sum _{n=1} ^k n^2 $ , while the second sum is $\sum _{n=1} ^k (2n-1)(k+1-n)$ .
Let us try to relate the two. We have :-
$$\sum _{n=1} ^k(2n-1)(k+1-n) =k\sum _{n=1} ^k(2n-1) -2\cdot\sum_{n=1} ^kn^2+\sum_{n=1}^kn+\sum_{n=1}^k2n-1$$
$$=2k\cdot\frac{k\cdot(k+1)}{2} - k^2 -2\sum_{n=1}^kn^2+\frac{k\cdot(k+1)}{2} +2\cdot\frac{k(k+1)}{2}-k= \frac{k\cdot(k+1)\cdot(2k+1)}{2}-2\sum _{n=0}^kn^2= 3\cdot \sum _{n=0}^kn^2-2\sum _{n=0}^kn^2=\boxed {\sum _ {n=1} ^k n^2} $$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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How to evaluate $\int^{2\pi}_0\frac{d\theta}{b^2\cos^2\theta + a^2 \sin^2\theta}$ for $a>0$ and $b>0$
How to evaluate $\int^{2\pi}_{0}\frac{d\theta}{b^2\cos^2\theta+a^2\sin^2\theta}$ for $a,b>0$
I came across this integral while solving a line integral problem . but it doesn't looks easier to integrate . I tried substitution of $\tan x$ .but i'm not sure tan can be used as the tan is discontinuous at $\pi/2$ i tried other substitution but none of them work.
| Your method works. We have
$$\begin{align*}
\int^{2\pi}_{0}\frac{\mathrm d\theta}{b^2\cos^2\theta+a^2\sin^2\theta}&=4\int^{\frac\pi 2}_{0}\frac{\mathrm d\theta}{b^2\cos^2\theta+a^2\sin^2\theta}\\&=4\int^{\frac\pi 2}_{0}\frac{\sec^2 \theta\mathrm d\theta}{b^2+a^2\tan^2\theta}\\
&=4\int^{\infty}_{0}\frac{\mathrm du}{b^2+a^2u^2}\\&=\frac4{ab}\arctan\left(\frac{au}{b}\right)\Big|^\infty_0\\&=\frac{2\pi} {ab}.
\end{align*}$$ Here's an another approach. We have
$$\begin{align*}
\int^{2\pi}_{0}\frac{\mathrm d\theta}{b^2\cos^2\theta+a^2\sin^2\theta}&=\int^{2\pi}_{0}\frac{\mathrm d\theta}{\frac{a^2+b^2}2+\frac{b^2-a^2}2\cos(2\theta)}\\&=\int^{2\pi}_{0}\frac{\mathrm d\theta}{\frac{a^2+b^2}2+\frac{b^2-a^2}2\cos\theta}\\
&=\frac{2\pi}{\sqrt{\left(\frac{a^2+b^2}2\right)^2-\left(\frac{b^2-a^2}2\right)^2}}\\
&=\frac{2 \pi}{ab}.
\end{align*}$$ Here, $$
\int^{2\pi}_{0}\frac{\mathrm d\theta}{\alpha+\beta\cos\theta}=\frac{2\pi}{\sqrt{\alpha^2-\beta^2}}, \quad \alpha>|\beta|\tag{*}
$$ is used. (See, for example, this earlier post. The first approach can be seen as a way of showing the integral $(*)$.)
| {
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if $x \equiv y$ (mod $m$), then $ax^2 + bx + c \equiv ay^2 + by + c$ (mod $m$) By assumption $y = x + km$ ; some $k \in \mathbb{Z}$.
w.t.s $ay^2 + by + c = (ax^2 + bx + c$) + $k'm$ ; some $k' \in \mathbb{Z}$
If we substitute our assumption in for $y$ we have
$a(x + km)^2 + b(x + km) + c$
= $a(x^2 + 2xkm + k^2m^2) + bx + c + (bk)m$
= $ax^2 + 2xkma + ak^2m^2 + bx + c + (bk)m$
= $ax^2 + bx + c + (2xka + k^2ma + bk)m$
Lastly, set $k' = 2xka + k^2ma + bk$ and we have
$ay^2 + by + c = (ax^2 + bx + c$) + $k'm$
does this work? heavily realying just on closure of $\mathbb{Z}$ under + and $\cdot$
(I'm fairly new to number theory)
Also, if this has been asked please just refer me to the Paige, I couldn't finds it. I apologize in advance if it has been.
| Prove things once. Then assume they are written in stone.
Prove once that if $a\equiv a'\pmod n$ and $b\equiv b'\pmod n$ then $a \pm b\equiv a' \pm b'\pmod n$.
Then prove once that if $a \equiv a' \pmod n$ and $b\equiv b'\pmod n$ then $ab \equiv a'b'\pmod n$.
Then observe once by induction that if $a \equiv a'$ then $a^m \equiv a'^m \pmod n$ for all positive powers $m$.
And observe once for any integer $d$ that $d \equiv d \pmod n$ for any integer $n$.
Then this is ... just observation.
$x \equiv y \pmod m$ so $x^2 \equiv y^2$ and as $a \equiv a \pmod m$ then $ax^2 \equiv ay^2$ and as $b \equiv b$ and $c \equiv c$ then $bx \equiv by$ and $ax^2 + bx + c \equiv ax^2 + bx + c\pmod n$.
And that IS a complete proof.
=====
To prove these things ONCE:
$c - c = 0$ and $0 = 0*n$ for all integers $n$ so $n|c-c$ for all $n$ so
$c \equiv c \pmod n$ for any integers $c, n$.
If $n|a$ then there is an integer $k$ so that $a = kn$ and if $n|b$ then there is a $j$ so that $b = jn$ so $a\pm b = kn \pm jn = (k\pm j)n$. $k\pm j$ are both integers so
If $n|a$ and $n|b$ then $n| a\pm b$.
If $a\equiv a' \pmod n$ and $b\equiv b' \pmod n$ then $n|a-a'$ and $n|b-b'$ so $n|(a-a') \pm (b - b')=(a\pm b)-(a' \pm b')$ so
If $a \equiv a'\pmod n$ and $b \equiv b'\pmod n$ then $a\pm b \equiv a' \pm b'\pmod n$.
If $a\equiv a' \pmod n$ then there is a $k$ so that $a = a' + kn$ and if $b\equiv b' \pmod n$ then there is a $j$ so that $b = b' + jn$ so $ab = a'b' + ajn + bkn + jkn^2 = a'b' + n(aj + bk + jkn)$ and as $aj + bk + jkn$ is an integer then
If $a \equiv a'\pmod n$ and $b \equiv b'\pmod n$ then $ab \equiv a'b' \pmod n$.
If we have $a \equiv a' \pmod n$ and we have $a^k \equiv a'^k \pmod n$ for $k = 1$. If we assume that for some $k$ that $a^k \equiv a'^k$ then we have $a^{k+1} = a^ka \equiv a'^ka'= a'^{k+1} \pmod n$
So by induction $a^m \equiv a'^m\pmod n$ for all positive integer powers $m$.
....
And that's it. We have proven everything we need once. We will NEVER need to prove them again.
| {
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Depending on the parameter p in C, find all solutions (in complex numbers) In the title it's $p \in \mathbb{C}$
$\begin{cases}
x+py-z=3 \\
px+y-z=9 \\
x+y-pz=1 \\
\end{cases}$
I decided to use Cramer's rule
$det_m=\begin {bmatrix}
1&p&-1 \\
p&1&-1 \\
1&1&-p \\
\end {bmatrix}=p^3-3p+2=(p-1)^2(p+2)\neq0 \Rightarrow (p\neq1 \land p\neq-2)$
1 solution for $p \in \mathbb{C} \setminus \{1,-2\}$
$det_x=9p^2-4p-5=0 \ $ it's equal $0$ only for $p=1$
$det_y=3p^2-10p+7=0 \ $ equal $0$ only for $p=1$
$det_z=-p^2+12p-11=0 \ $ equal $0$ only for $p=1$
infinity of solutions for $p=1$
no solution for $p=-2$
So if I understand this correctly, all I have to do is to write answer:
$x=\frac{9p^2-4p-5}{p^3-3p+2}$
$y=\frac{3p^2-10p+7}{p^3-3p+2}$
$z=\frac{-p^2+12p-11}{p^3-3p+2}$
where $(p \neq 1 \land p\neq-2)\land p\in \mathbb{C}$
Did I solve this assignment correctly?
| Cramer's rule is indeed a good idea here.
The relevant data is
\begin{align*}
\det\left[\begin{array}{rrr}
1 & p & -1 \\
p & 1 & -1 \\
1 & 1 & -p
\end{array}\right] &= {\left(p + 2\right)} {\left(p - 1\right)}^{2} \\
\det\left[\begin{array}{rrr}
3 & p & -1 \\
9 & 1 & -1 \\
1 & 1 & -p
\end{array}\right] &= {\left(9 \, p + 5\right)} {\left(p - 1\right)} \\
\det\left[\begin{array}{rrr}
1 & 3 & -1 \\
p & 9 & -1 \\
1 & 1 & -p
\end{array}\right] &= {\left(3 \, p - 7\right)} {\left(p - 1\right)} \\
\det\left[\begin{array}{rrr}
1 & p & 3 \\
p & 1 & 9 \\
1 & 1 & 1
\end{array}\right] &= -{\left(p - 1\right)} {\left(p - 11\right)}
\end{align*}
So, for $p\neq -2$ and $p\neq 1$, the solution to $A\vec{x}=\vec{b}$ has coordinates
\begin{align*}
x_1 &= \frac{9 \, p + 5}{{\left(p + 2\right)} {\left(p - 1\right)}} & x_2 &= \frac{3 \, p - 7}{{\left(p + 2\right)} {\left(p - 1\right)}} & x_3 &= -\frac{p - 11}{{\left(p + 2\right)} {\left(p - 1\right)}}
\end{align*}
Now, we solve the system for $p=-2$ and $p=1$. Here, the relevant information is
\begin{align*}
\operatorname{rref}\left[\begin{array}{rrr|r}
1 & -2 & -1 & 3 \\
-2 & 1 & -1 & 9 \\
1 & 1 & 2 & 1
\end{array}\right] &= \left[\begin{array}{rrr|r}
1 & 0 & 1 & 0 \\
0 & 1 & 1 & 0 \\
0 & 0 & 0 & 1
\end{array}\right] \\
\operatorname{rref}\left[\begin{array}{rrr|r}
1 & 1 & -1 & 3 \\
1 & 1 & -1 & 9 \\
1 & 1 & -1 & 1
\end{array}\right] &= \left[\begin{array}{rrr|r}
1 & 1 & -1 & 0 \\
0 & 0 & 0 & 1 \\
0 & 0 & 0 & 0
\end{array}\right]
\end{align*}
Here, we find that the system is actually unsolvable for $p=-2$ and $p=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3104101",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How to solve this algebraic manipulation problem? Express the following expression $$E=(x^3-y^3)(y^3-z^3)(z^3-x^3)$$ in terms of $a, b$ where $a,b \in \mathbb R$ and
$$a=x^2y+y^2z+z^2x$$ $$b=xy^2+yz^2+zx^2$$
| Start: $$ a-b = xy(x-y)+yz (y-z)+zx(z-x) $$
$$= xy(x-y)+yz (y-z)+zx(z-\color{red}y)+zx(\color{red}y-x) $$
$$= (xy-zx)(x-y)+(yz -zx)(y-z)$$
$$ =x(y-z)(x-y)+z(y-x)(y-z) $$
$$ = (x-y)(y-z)(x-z)$$
So $$ E = (a-b)\underbrace{(x^2+xy+y^2)(y^2+yz+z^2)(z^2+zx+x^2)}_{A}$$
Now you have to figer out $A$. (I bet it is $3ab$.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3109226",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Three numbers a, b, c, over 1 have a sum of $\frac{6}{7}$. What is a+b+c? Given that there are three integers $a, b,$ and $c$ such that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{6}{7}$, what is the value of a+b+c?
Immediately, I see that I should combine the left hand side. Doing such results in the equation $$\frac{ab+ac+bc}{abc}=\frac{6x}{7x}.$$ This branches into two equation $$ab+ac+bc=6x$$$$abc=7x.$$ From this, I can tell that one of a, b, or c must be a multiple of 7, and the other two are a factor of $x$. Now, I do trial and error, but I find this very tiring and time consuming. Is there a better method?
Also, if you are nice, could you also help me on this($N$'s base-5 and base-6 representations, treated as base-10, yield sum $S$. For which $N$ are $S$'s rightmost two digits the same as $2N$'s?) question?
Thanks!
Max0815
| Your argument isn't correct as stated due to the issue pointed out by Dan Uznanski in the comments. But following your idea of combining things, we have
$$6abc = 7(ab + ac + bc)$$
so that $7 | abc$ and $6 | (ab + ac + bc)$. Let's assume that $a$ is divisible by $7$, so clearly $a \ge 7$ (assuming that all the numbers are positive). It then follows that
$$\frac 1 b + \frac 1 c = \frac 6 7 - \frac 1 a \ge \frac 5 7 > \frac 2 3$$
Now if $b$ and $c$ were both strictly greater than $2$, we would have a contradiction (why?). We can therefore assume $c = 2$ and reduce to
$$\frac 1 a + \frac 1 b = \frac 5 {14}.$$
Now repeat the ideas: We have
$$\frac 1 b = \frac 5 {14} - \frac 1 a \ge \frac 5 {14} - \frac 1 7 = \frac{3}{14} \implies b \le \frac{14}{3} \implies b \le 4.$$
This is now a very finite set to check.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3112263",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Find if terms are terms of the same arithmetic progression
Is it possible that numbers $\frac{1}{2}, \frac{1}{3}, \frac{1}{5}$ are (not necessarily adjacent) terms of the same arithmetic progression?
Hint: Yes. Try $\frac{1}{30}$ as a difference.
I was going back and forth how they found out that difference.
My idea was since we have an arithmetic sequence defined as $a, a+d, a+2d,...$ I thought I could solve for the difference $d=\frac{1}{30}$. Since:
$$\frac{1}{3} = \frac{1}{2}+nd$$
And
$$\frac{1}{5} = \frac{1}{3}+md$$
Then $nd = \frac{1}{3} - \frac{1}{2} = -\frac{1}{6}$ and $md = \frac{1}{5} - \frac{1}{3} = -\frac{2}{15}$
Since it is also part of the same sequence we can find:
$$nd + md = -\frac{1}{6} -\frac{2}{15} = -\frac{3}{10}$$
Now I'm stuck since I can't see how this brings me any closer to find $m, n, d$.
| Thinking about my previous answer,
it becomes obvious that
any monotone sequence
of rational numbers
can be embedded
in a linear sequence.
Let the sequence be
$(a_k)|_{k=1}^n$
with
$a_k < a_{k+1}$
and
$a_k = \dfrac{u_k}{v_k}$
with $u_k, v_k$ integers
and $v_k > 0$.
To make the
$a_k$ in the sequence
$a+nd$
with $n$ a positive integer
and $d > 0$,
let $V$ be the
least common multiple
of the $v_k$,
so that
$\dfrac{V}{v_k}
=w_k
$
is an integer.
Set $a = a_1$.
We want there to be
a $d$ and,
for each $k$,
an integer $n_k$
such that
$a+dn_k
=a_k
=\dfrac{u_k}{v_k}
$.
Multiplying by $V$,
$u_kw_k
=V(a+dn_k)
=Va_1+Vdn_k
$.
If we set
$d = \dfrac1{V}$,
then
$n_k
= u_kw_k-Va_1
$.
Done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3114457",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
} |
Evaluate $\int_{y=-\infty}^1\int_{x=0}^\infty 4xy\sqrt{x^2+y^2}\mathrm dx\mathrm dy$ Evaluate $$\int_{y=-\infty}^1\int_{x=0}^\infty 4xy\sqrt{x^2+y^2}\mathrm dx\mathrm dy$$
I tried to solve this for hours without any success. I tried substitution method of $u = x^2+y^2$ but doesn't seem to work.
Here is my attempt:
$$\int_{-\infty}^1\int_0^\infty (16x^2y^2)^{\frac{1}{2}}(x^2 + y^2)^\frac{1}{2} \; \mathrm{d}x \mathrm{d}{y} = \int_{-\infty}^1 \int_0^\infty (16x^3y^2 + 16x^2 y^3)^{\frac{1}{2}} \; \mathrm{d}x\mathrm{d}{y}.$$
Let
\begin{align}
u &= 16x^3 y^2 + 16x^2 y^3 \\
\mathrm{d}u &= (48x^2y^2 + 32xy^3) \; \mathrm{d}{x} \\
\mathrm{d}u &= 16xy^2(3x + 2y) \; \mathrm{d}{x} \\
\mathrm{d}u &= 4xy(12xy + 8y^2) \; \mathrm{d}{x} \\
4xy \; \mathrm{d}{x} &= \frac{\mathrm{d}{u}}{12xy + 8y^2}.
\end{align}
$$\int_{-\infty}^1 \int_0^\infty 4xy \sqrt{x^2 + y^2} \; \mathrm{d}x \mathrm{d}y = \int_{-\infty}^1 \int_0^\infty \frac{\mathrm{d}u}{(12xy + 8y^2)}.$$
Attaching image for reference
EDITS
Original Post asked integrate: 4xy sqrt(x^2+y^2) where 0<x, y<1, and added this image. .
Later edits misread 0<x,y<1 as if 0<x<1and 0<y<1
| Hint:
Don't switch to polar coordinates. Notice that
$$\frac d{dx}(x^2+y^2)^a=2ax(x^2+y^2)^{a-1}$$
and
$$\frac d{dy}\frac d{dx}2ax(x^2+y^2)^{a-1}=4a(a-1)xy(x^2+y^2)^{a-2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3114531",
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"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
Prove that inequality $\sqrt{2\sqrt{4\sqrt{8....\sqrt{2^n}}}} \leqslant n+1$
Let $n$ be the integer. Prove that $$\sqrt{2\sqrt{4\sqrt{8....\sqrt{2^n}}}} \leqslant n+1$$
SOURCE: BANGLADESH MATH OLYMPIAD
I am a new beginner at the infinite radical and sequence. I don't know about its basic conception.
My Attempt:
After seeing that problem, out of curiosity, I thought that I should calculate the multiplication of left side term.
Than I got something like this
$2^{\frac{1}{2}}$ × $2^{\frac{1}{2}}$ × $2^{\frac{3}{8}}$... × $2^{\frac{k}{2n}} = 2^{\left(\frac{1}{2}+\frac{1}{2}+\frac{3}{8}+...+\frac{k}{2n}\right)}$, where I denoted $2n$ = $2^\text{k}$.
But here I got stuck. The above sequence doesn't follow the pattern. Than how would I get the summation of the sequence?
Moreover, how should I approach to prove the above condition for the integer value of $n$? Any kind of help or clue will be greatly appreciated by this novice and new beginner. Thanks in advance.
| It's
$$2^{\frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^3}+...+\frac{n}{2^n}}=2^{x(x+x^2+...+x^{n})'}_{x=\frac{1}{2}}=2^{x\left(\frac{x(x^n-1)}{x-1}\right)'}_{x=\frac{1}{2}}=2^{x\left(\frac{x^{n+1}-x}{x-1}\right)'}_{x=\frac{1}{2}}=$$
$$=2^{x\cdot\frac{((n+1)x^n-1)(x-1)-x^{n+1}+x}{(x-1)^2}}_{x=\frac{1}{2}}=2^{x\cdot\frac{nx^{n+1}-(n+1)x^n+1}{(x-1)^2}}_{x=\frac{1}{2}}=2^{2\left(n\left(\frac{1}{2}\right)^{n+1}-(n+1)\left(\frac{1}{2}\right)^n+1\right)}=$$
$$=2^{\frac{n}{2^n}-\frac{n+1}{2^{n-1}}+2}=2^{2-\frac{n+2}{2^n}}.$$
Thus, it's enough to prove that
$$2^{2-\frac{n+2}{2^n}}\leq n+1,$$ which is obviously true for $n\geq3.$
But easy to check that for $n=1$ and for $n=2$ our inequality is also true.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3115879",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
find the maximum and minimum of $\sum_{i=1}^{n} (10x^3_{i}-9x^5_{i})$
Let $x_{i}\ge 0$ such that $$x_{1}+x_{2}+\cdots+x_{n}=1.$$
Find the maximum and minimum of
$$f=10\sum_{i=1}^{n}x^3_{i}-9\sum_{i=1}^{n}x^5_{i}.$$
I have proved $n=2$
$$1\le f\le\dfrac{9}{4}$$ see: wolfarma
When $n=3$, How prove that $10(a^3+b^3+c^3)-9(a^5+b^5+c^5)\le\dfrac{9}{4}$
So I suspect that the general positive integer $n\ge 4$, also has the following conclusion
$$1\le f\le \dfrac{9}{4}$$
when $$(x_{1},x_{2},\cdots,x_{n})=(1,0,0,\cdots,0),f=1$$
and
$$(x_{1},x_{2},\cdots,x_{n})=(0,\dfrac{1}{2}+\dfrac{1}{2\sqrt{3}},\dfrac{1}{2}-\dfrac{1}{2\sqrt{3}},0,\cdots,0),f=\dfrac{9}{4}$$But how to prove or reverse this conclusion?
| Assumption
If $f(x)$ is convex strictly increasing for $0 \le x \le 1$ then
$$
\min\sum_{k=1}^n f(x_k) \ \ \mbox{s. t. }\ \ \sum_{k=1}^n x_k = 1, \ \ x_k > 0
$$
has it's minimum at $ x_1 = \cdots = x_n = \frac 1n$
Using Lagrange Multipliers the problem can be stated as
$$
L(x,\lambda) = \sum_{k=1}^n f(x_k)-\lambda\left(\sum_{k=1}^n x_k -1\right)
$$
so the stationary points are the solutions for
$$
\frac{d}{dx_k}f(x_k) -\lambda = 0\\
\sum_{k=1}^n x_k - 1 = 0
$$
or
$$
x_k = \frac{\sqrt{5\pm\sqrt{5} \sqrt{5-\lambda }}}{\sqrt{15}}
$$
now assuming all $x_k = \frac{\sqrt{5+\sqrt{5} \sqrt{5-\lambda }}}{\sqrt{15}}$
so
$$
\lambda = \frac{15(2n^2-3)}{n^4}
$$
then
$$
x_k =\left\{\frac 1n,\frac{\sqrt{\sqrt{\frac{\left(n^2-3\right)^2}{n^4}}+1}}{\sqrt{3}}\right\}
$$
The last value is discarded because does not observe the restriction then we follow with $x_k = \frac 1n$ so
$$
\min \sum_{k=1}^n f(x_k) = n f\left(\frac 1n\right)
$$
Attached a plot showing the $F_n(x_n^*) = \sum_{k=1}^n f(x_n^*)$ evolution assuming $n$ continuous
NOTE
$$
f(x) = 10x^3-9 x^5
$$
is convex strictly increasing for $0\le x \le 0.5 $ so for $n \gt 2$ we have at $x^* = \frac 1n$ a local minimum.
For $n = 2$ making $x_2=\lambda x_1, \ x_3 = \mu x_1$ we have
$$
\min_{\lambda,\mu}\frac{10(1+\lambda^3+\mu^3)}{(1+\lambda+\mu)^3}-\frac{9(1+\lambda^5+\mu^5)}{(1+\lambda+\mu)^5}
$$
which gives the feasible solution
$$
x_1 = x_2 = \frac 12
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3117624",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Find all $n$ $\in$ $\Bbb Z^+$ such that: $\lfloor\frac{n}{2}\rfloor \cdot \lfloor \frac{n}{3} \rfloor \cdot \lfloor \frac{n}{4} \rfloor = n^2$ Find all the numbers $n$ $\in$ $\Bbb Z^+$ such that:
$$\left\lfloor\frac{n}{2}\right\rfloor \cdot \left\lfloor \frac{n}{3} \right\rfloor \cdot \left\lfloor \frac{n}{4} \right\rfloor = n^2$$
I never worked before with floor function so i'm not completely sure how to solve this. I think (only because $n$ $\in$ $\Bbb Z^+$), i can just multiply (skipping the floor function) and get the answer of $n=24$, but this is floor function so i don't know if there are more solutions.
Any hints?
| A lengthy approach
As $[2,3,4]=12,$ let us try with all $12$ in-congruent residues $\pmod{12}$
If $n=12k$
$6k(4k)(3k)=(12k)^2\iff k=2$
If $n=12k+1,$
$6k(4k)(3k)=(12k+1)^2$ which is untenable as LHS is divisible by $6$
If $n=12k+2,(6k+1)(4k)(2k)=(12k+2)^2\iff6k+1=2k^2$ which is even
If $n=12k+3,(6k+1)(4k+1)2k=(12k+3)^2 which is odd unlike the LHS
If $n=12k+4,(12k+4)^2=(6k+2)(4k+1)2k\iff4(3k+1)=4k+1$ which is odd unlike LHS
and so on
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3119036",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
How to proof $\frac{I_{m+1} I_{m} + I_m I_{m-1}}{2} \geq I_{m+1} I_{m-1}$ Let
$$ I_m=\int_0^\infty (1+s)^m \, {\rm e}^{-ns} \, {\rm d}s \, .$$
I've been thinking some time on the following inequality
$$I_{m} \left( I_{m+1} + I_{m-1} \right) \geq 2 \, I_{m+1} I_{m-1}$$
or in somewhat interesting parallel circuit fashion
$$\frac{1}{I_{m+1}} + \frac{1}{I_{m-1}} \geq \frac{2}{I_m} \, .$$
Is there any clever way to tackle it?
It also looks somewhat like AMGM
$$I_{m+1}^2 + I_{m-1}^2 \geq 2 \, I_{m+1}I_{m-1}$$
in a sense, where one $I_{m+1}$ and $I_{m-1}$ on the LHS is replaced by $I_m$.
$I_m$ is monotonically increasing in $m$ and log-convex (i.e. $I_{m+1} I_{m-1} \geq I_m^2$, without proof though).
It seems to be a rather tight inequality, since e.g.
$$I_{m+1} I_m + I_m I_{m-1} \geq I_{m+\frac{1}{2}}^2 + I_{m-\frac{1}{2}}^2 \geq 2 \, I_{m+\frac{1}{2}} I_{m-\frac{1}{2}} \leq 2 \, I_{m+1} I_{m-1}$$
or
$$I_m \, \frac{I_{m+1} + I_{m-1}}{2} \geq I_m^2 \leq I_{m+1} I_{m-1} \, .$$
| Expanding on marty cohen's answer, notice the recurrence
\begin{align*}
I_{m+1} = \frac{1}{n} + \frac{m+1}{n}I_m
\end{align*}
and
\begin{align*}
I_{m-1} = - \frac{1}{m} +\frac{n}{m}I_m
\end{align*}
Substituting this into the desired inequality,
\begin{align*}
I_m\left(\frac{1}{n} + \frac{m+1}{n}I_m - \frac{1}{m} + \frac{n}{m}I_m \right) \ge 2\left(\frac{1}{n} + \frac{m+1}{n}I_m\right)\left(- \frac{1}{m} + \frac{n}{m}I_m\right)
\end{align*}
or
\begin{align*}
I_m \ge \frac{\sqrt{(m - n - 2)^2 + 8m} + 3(n - m) - 2}{2[(m+1-n)^2 - (m+1)]}
\end{align*}
From here, you might try induction on $m, n$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3119610",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Integral check. Is partial fractions the only way? I'm getting a different answer from wolfram and I have no idea where. I have to integrate:
$$\int_0^1 \frac{xdx}{(2x+1)^3}$$
Is partial fractions the only way?
So evaluating the fraction first:
$$\frac{x}{(2x+1)^3} = \frac{A}{2x+1} + \frac{B}{(2x+1)^2} + \frac{C}{(2x+1)^3}$$
$$x = A(2x+1)^2 + B(2x+1) + C$$
$$x = A(4x^2 + 4x + 1) + 2Bx + B + C$$
$$x = 4Ax^2 + 4Ax + A + 2Bx + B + C$$
$$x = x^2(4A) + x(4A+2B) + A + B + C$$
$4A = 0$ and $4A+2B = 1$ and $A + B + C = 0$ so $A = 0$ and $B=\frac{1}{2}$ and $C = \frac{-1}{2}$
Is the partial fraction part right?
So then I get:
$$\int_0^1 \frac{xdx}{(2x+1)^3} = \int_0^1 \frac{dx}{(2(2x+1)^2)} - \int_0^1 \frac{dx}{2(2x+1)^3}$$
for both, using $u = 2x+1$ and $\frac{du}{dx} = 2$ and $du = 2dx$ and $\frac{du}{2} = dx$,
$$\frac{1}{4} \int \frac{du}{u^2} - \frac{1}{4} \int \frac{du}{u^3}$$
$$ = [\frac{1}{4} - u^{-1} - \frac{1}{4} \cdot \frac{1}{-2} u^{-2} ]_1^3$$
I'm going the route of changing the limits to the new u and since $u = 2x+1$, when $x = 0, u = 1$ and when $x = 1, u = 3$. Is this the right path?
finally I get
$$[\frac{-1}{4}u^{-1} + \frac{1}{8}u^{-2} ]_1^3$$
I plug in numbers but I get a different answer than wolfram...
| $$\int_0^1 \frac{xdx}{(2x+1)^3} =
\int_0^1 \frac{dx}{2(2x+1)^2} - \int_0^1 \frac{dx}{2(2x+1)^3}$$
$\displaystyle \int \dfrac{dx}{2(2x+1)^2}:$
$\qquad u=2x+1,\ x=\dfrac{u-1}{2}, \ dx = \dfrac 12 du$
$\qquad \displaystyle \int \dfrac{dx}{2(2x+1)^2} =
\int \dfrac{du}{4u^2} = -\dfrac{1}{4u}$
$\qquad x=0 \mapsto u=1, \ x=1 \mapsto u=3$
$\qquad \left[-\dfrac{1}{4u} \right]_1^3 = -\dfrac{1}{12} + \dfrac 14 = \dfrac 16$
$\displaystyle \int \dfrac{dx}{2(2x+1)^3}:$
$\qquad u=2x+1,\ x=\dfrac{u-1}{2}, \ dx = \dfrac 12 du$
$\qquad \displaystyle \int \dfrac{dx}{2(2x+1)^3} =
\int \dfrac{du}{4u^3} = -\dfrac{1}{8u^2}$
$\qquad x=0 \mapsto u=1, \ x=1 \mapsto u=3$
$\qquad \left[-\dfrac{1}{8u^2} \right]_1^3 = -\dfrac{1}{72} + \dfrac 18 = \dfrac 19$
$$\int_0^1 \frac{dx}{2(2x+1)^2} - \int_0^1 \frac{dx}{2(2x+1)^3}
= \dfrac 16 - \dfrac 19 = \dfrac{1}{18}$$
There is also another way to solve for $A, B$, and $C$.
$$\frac{A}{2x+1} + \frac{B}{(2x+1)^2} + \frac{C}{(2x+1)^3} =\frac{x}{(2x+1)^3}$$
$$A(2x+1)^2 + B(2x+1) + C = x$$
Let $x = -\dfrac 12$ and you get
\begin{align}
C &= -\dfrac 12 \\
A(2x+1)^2 + B(2x+1) - \dfrac 12 &= x \\
2A(2x+1)^2 + 2B(2x+1) &= 2x+1 \\
2A(2x+1) + 2B &= 1
\end{align}
Again, let $x = -\dfrac 12$ and you get
\begin{align}
2B &= 1 \\
B &= \dfrac 12 \\
2A(2x+1) + 1 &= 1 \\
A &= 0
\end{align}
| {
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"url": "https://math.stackexchange.com/questions/3120889",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 4
} |
Evaluate $\int_2^6 \frac{\ln(x-1)}{x^2+2x+2}dx$
Evaluate $$I=\int_2^6 \frac{\ln(x-1)}{x^2+2x+2}dx$$
I tried the substitution $t=x-1$ which gave $$I=\int_1^5 \frac{\ln(t)}{t^2+4t+5}dt$$
Then I let $u=\ln t$ which gave $$I=\int_0^{\ln 5} \frac{ue^u}{e^{2u}+4e^u+5}du$$
I don't see other useful substitutions and I don't know how to finish
| Symmetry works nice in this case.
$$I=\int_2^6 \frac{\ln(x-1)}{x^2+2x+2}dx\,\overset{x-2\rightarrow x}=\,\int_0^4 \frac{\color{red}{\ln(1+x)}}{x^2+6x+10}dx$$
With a substitution of
$$x=\frac{4-y}{1+y}\Rightarrow dx= - \frac{5}{(1+y)^2}dy$$
$$\require{cancel} \Rightarrow I= \int_0^4 \ln\left(1+\frac{4-y}{1+y}\right)\frac{\cancel{(1+y)^2}}{\cancel 5(y^2+6y+10)}\frac{\cancel 5}{\cancel{(1+y)^2}}dy\overset{y=x}=\int_0^4 \frac{\color{blue}{\ln\left(\frac{5}{1+x}\right)}}{x^2+6x +10}dx $$Adding this result with the previous integral gives:
$$\require{cancel} 2I=\int_0^4 \frac{\cancel{\color{red}{\ln(1+x)}}+\color{blue}{\ln 5-\cancel{\ln(1+x)}}}{x^2+6x+10}dx$$
$$\Rightarrow I=\frac{\ln 5}{2} \int_0^4 \frac{1}{x^2+6x+10}dx=\frac{\ln 5}{2} \arctan\left(\frac2{11}\right)$$
One can of course do directly $\displaystyle{x=\frac{t+4}{t-1}}$ when the bounds are $2$ and $6$ to arrive at the same result by the same method.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Greatest volume of sphere inscribed in a cone with given Lateral area The Lateral area of a right circular cone is $S$. Fine the Greatest volume of Sphere that can be inscribed in it.
My try:
Given that $$\pi rl=S$$
Let the radius of the sphere inscribed in it be $R$
Let the semi vertex angle of cone be $\theta$
We have height of the cone as:
$$h=R+R\csc \theta$$
Now
$$\pi^2 r^2 l^2 =S^2$$ $\implies$
$$\pi^2 r^2 (r^2+h^2)=S^2$$
$$r^2(r^2+R^2(1+\csc \theta)^2)=\frac{S^2}{\pi^2}$$
So:
$$R^2=\frac{\frac{S^2}{\pi^2r^2}-r^2}{(1+\csc \theta)^2}\tag{1}$$
Also:
$$\sin \theta=\frac{r}{l}$$
So
$$\pi r \frac{r}{\sin \theta}=S$$
$\implies$
$$\csc \theta=\frac{S}{\pi r^2}$$
Hence $(1)$ becomes:
$$R^2=\frac{r^2(S^2-r^4\pi^2)}{(\pi r^2+S)^2}$$
$$R^2=r^2\frac{(S-\pi r^2)}{(S+\pi r^2)}$$
$$\pi R^2=\pi r^2\frac{(S-\pi r^2)}{(S+\pi r^2)}$$
Letting $\pi r^2=t$ we get
$$\pi R^2=f(t)=\frac{t(S-t)}{S+t}$$
Differentiating with respect to $t$ we get:
$$f'(t)=-1+\frac{2S^2}{(S+t)^2}=0$$ $\implies$
$$t=\left(\sqrt{2}-1\right)S$$
Hence
$$\pi r^2=\left(\sqrt{2}-1\right)S$$
$$r^2=\frac{\left(\sqrt{2}-1\right)S}{\pi}$$
So we get:
$$R^2=\frac{(3-2\sqrt{2})S}{\pi}$$
Hence Maximum volume of sphere is:
$$V=\frac{4\pi R^3}{3}=0.0534 (S)^{\frac{3}{2}}$$
But i feel this approach is tedious.Any better method?
| The relation that you you desire for the radius of the incircle can found by examining the figure below. If $\theta$ is the half angle of the vertex, then clearly
$$
\sin\theta=\frac{r}{l}=\frac{R}{h-R}\\
h=\sqrt{l^2-r^2}
$$
From here you can find
$$R=\frac{rh}{l+r}=\frac{r\sqrt{l^2-r^2}}{l+r}=r\sqrt{\frac{l-r}{l+r}}$$
I believe this is what you are looking for.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Minimum value of expression having $2$ variables
Minimum value of $$\bigg(x-4-\sqrt{4-y^2}\bigg)^2+\bigg(4\sqrt{x}-y\bigg)^2$$ for real $x\geq 0,y\in[-2,2]$
Try: Using Partial derivative
$$f(x,y) = \bigg(x-4-\sqrt{4-y^2}\bigg)^2+\bigg(4\sqrt{x}-y\bigg)^2$$
$$\frac{df(x,y)}{dx}=2\bigg(x-4-\sqrt{4-y^2}\bigg)+\frac{4}{\sqrt{x}}\bigg(4\sqrt{x}-y\bigg)$$
$$\frac{df(x,y)}{dy}=-2y\frac{\bigg(x-4-\sqrt{4-y^2}\bigg)}{\sqrt{4-y^2}}-2\bigg(4\sqrt{x}-y\bigg)$$
Put $\displaystyle \frac{df}{dx} =0$ and $\displaystyle \frac{df}{dy}=0$
But these $2$ equation is tedious work
Could some help me how to solve it , Thanks in advance
| For $0\leq x\leq 4$ we have
$$f(x,y)-f(x,2)=2\left(4\sqrt{x}(2-y)+(4-x)\sqrt{4-y^2}\right)\geq 0 $$
$$f(x,y)\geq f(x,2)\geq \underset{0\leq x\leq 4}{\min }f(x,2)= \underset{0\leq x\leq 4}{\min }\left(20-16 \sqrt{x}+8 x+x^2\right)$$
For $x\geq 4$, use the Cauchy–Schwarz inequality, to show $f(x,y) \geq 36$ with $f(4,2)=36$ as follows
$$\left(x-4,4\sqrt{x}\right).\left(\sqrt{4-y^2},y\right)\leq 2(x+4)$$
$$f(x,y)-36=-16+8 x+x^2-2\left((x-4) \sqrt{4-y^2}+4 \sqrt{x} y\right)$$
$$\geq -16+8 x+x^2-4(x+4)=(x-4)(x+8)\geq 0$$
EDIT:
$\underset{0\leq x\leq 4}{\min }\left(20-16 \sqrt{x}+8 x+x^2\right)$ occurs at the zero of the cubic polynomial $x^3+8x^2+16x-16$ with $x \approx 0.7186$ and the minimum value $\approx 12.7019$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How do I find the distance from a point to a plane? I am trying to find the distance from point $(8, 0, -6)$ and plane $x+y+z = 6$. I tried solving it but I am still getting it wrong. Can anyone help me on this? Any help I would very much appreciate.
The following is my work:
$$d = \sqrt{(x-8)^2 + (y-0)^2 + (z+6)^2}$$
since $x+y+z = 6$, $z = 6-x-y$, so
\begin{align*}
d &= \sqrt{(x-8)^2 + (y-0)^2 + (-x-y+12)^2} \\
d^2 &= (x-8)^2 + (y-0)^2 + (-x-y)^2
\end{align*}
Find partial derivative $f_x$ and $f_y$ and critical points
\begin{align*}
f_x &= 2(x-8) + 2(-x-y+12) \\
&= 24-2y \quad (\text{set }= 0) \\
&= \text{critical point }y = 4 \\
f_y &= 2y + 2(-x-y+12) \\
&= 24 - 2x \quad (\text{set }= 0) \\
&= \text{critical point }x = 12 \\
\end{align*}
Plug in $x = 12$ and $y = 4$ to original equation
$$d = \sqrt{(x - 8)^2 + (4)^2 + (-12-4+12)^2} = \sqrt{48}$$
| The distance formula is $$D=\frac{|ax_0+by_0 +cz_0-d|}{\sqrt{a^2+b^2+c^2}}=\frac{4}{\sqrt3}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3124438",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find the integral $\int \frac{(\ln(x))^2}{x^3} \, dx$ $$\int \frac{(\ln(x))^2}{x^3} \, dx $$
Starting off with Integration by Parts
$$
\begin{align}
u = \ln(x)^2 &~~~ dv = x^{-3} \\\\
du = 2\ln(x)dx &~~~ v = \frac{x^{-2}}{-2}
\end{align}
$$
$$
\begin{align}
\int \frac{(\ln(x))^2}{x^3} &= (\ln(x))^2 \left( \frac{-x^{-2}}{2} \right) -\frac{2}{2} \int \frac{\ln(x)}{x^2}
\end{align}
$$
Integration by parts again...
$$
\begin{align}
u = \ln(x) &\hspace{10mm} dv = \frac{1}{x^2} \\\\
du = \frac{1}{x}dx &\hspace{10mm} v = -\frac{1}{x} \\\\
-\frac{2}{2} \int \frac{\ln(x)}{x^2} &= -\ln(x) \left(\frac{1}{x} \right) + \int \frac{1}{x^2}dx \\\\
\end{align}
$$
Integration by parts several times $\int \frac{1}{x^2}dx \rightarrow \int \frac{1}{x} \rightarrow \ln(x) + C$
Combining everything together, I get
$$
\int \frac{(\ln(x))^2}{x^3} = (\ln(x))^2 \left( \frac{-x^{-2}}{2} \right) -\ln(x) \left(\frac{1}{x} \right) - \ln(x) + C\\\\
= -\frac{(\ln(x))^2}{2x^2} - \frac{\ln(x)}{x} - \ln(x) + C
$$
Is there a shorter way I could have done this? I think I got the right answer, but I am not really sure either. Checking my answer via differentiate doesn't seem like a feasible test strategy and even without time constraints still seems too complex for my level right now (but then maybe this is why I need the practice)
| When doing integration by parts, I recommend using this little, trusty formula (the order whether $f(x)$ or $g(x)$ comes first does not matter):
$$\int f(x)g'(x)\,dx=f(x)g(x)-\int f'(x)g(x)\,dx.$$
$$
\begin{align}
\int\frac{\ln^2{x}}{x^3}\,dx
&=-\frac{1}{2}\int\ln^2{x}\left(\frac{1}{x^2}\right)'\,dx\\
&=-\frac{1}{2}\left(\frac{\ln^2{x}}{x^2}-\int\frac{1}{x^2}(\ln^2{x})'\,dx\right)\\
&=-\frac{1}{2}\left(\frac{\ln^2{x}}{x^2}-2\int\frac{\ln{x}}{x^3}\,dx\right)\\
&=-\frac{1}{2}\left(\frac{\ln^2{x}}{x^2}-\frac{2}{-2}\int\ln{x}\left(\frac{1}{x^2}\right)'\,dx\right)\\
&=-\frac{1}{2}\left(\frac{\ln^2{x}}{x^2}+\frac{\ln{x}}{x^2}-\int\frac{1}{x^2}(\ln{x})'\,dx\right)\\
&=-\frac{1}{2}\left(\frac{\ln^2{x}+\ln{x}}{x^2}-\int\frac{1}{x^3}\,dx\right)\\
&=-\frac{1}{2}\left(\frac{\ln^2{x}+\ln{x}}{x^2}-\int x^{-3}\,dx\right)\\
&=-\frac{1}{2}\left(\frac{\ln^2{x}+\ln{x}}{x^2}-\frac{1}{-3+1}x^{-3+1}\right)\\
&=-\frac{1}{2}\left(\frac{\ln^2{x}+\ln{x}}{x^2}+\frac{1}{2x^2}\right)\\
&=-\frac{1}{2}\left(\frac{2\ln^2{x}+2\ln{x}}{2x^2}+\frac{1}{2x^2}\right)\\
&=-\frac{2\ln^2{x}+2\ln{x}+1}{4x^2}+C.
\end{align}
$$
Wolfram Alpha check
| {
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"question_score": "4",
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Integral $\int_{a}^{b}\frac{x\ln(x)}{x^4+a^{2}b^{2}}dx$
How to evaluate the following integral $$\int_{a}^{b}\frac{x\ln(x)}{x^4+a^{2}b^{2}}dx?$$ where $0<a<b$
I observed that we can change the variable in a way that simplifies the expression but the coefficients of the integral becomes extremely ugly.
With the substitution $t=x^2$ it ends being $$\frac{1}{4}\int_{a^2}^{b^2} \frac{\ln t}{t^{2}+a^{2}b^{2}}dt=\frac14\int_{a^2}^{b^2}\frac{\ln t}{(t-ab)^2-2abt}dt$$
| $$I=\int_ {a}^{b}\frac{x\ln(x)}{x^4+a^{2}b^{2}}dx\overset{x^2=t}=\frac14 \int_{a^2}^{b^2} \frac{\ln t}{t^2+a^2b^2}dt $$
Let's get rid of that ugly denominator by doing a $t=y(ab)$:
$$I=\frac{ab}4\int_\frac{a}{b}^\frac{b}{a}\frac{\ln(yab)}{a^2b^2(y^2+1) }dy=\frac{1}{4ab}\int_\frac{a}{b}^\frac{b}{a}\frac{\color{blue}{\ln y+\ln(ab)}}{y^2+1 }dy$$
And now one might notice the simmetry directly. Substituting $y=\frac{1}{x}$ gives:
$$\frac{1}{4ab}\int_\frac{a}{b}^\frac{b}{a}\frac{\color{red}{\ln\left(\frac{1}{x}\right)+\ln(ab)}}{x^2+1 }dx$$
Adding the red and blue integral simiplifies quite nice the logarithm:
$$\require{cancel} 2I=\frac{1}{4ab}\int_\frac{a}{b}^\frac{b}{a}\frac{\color{red}{\cancel{\ln\left(\frac{1}{x}\right)}+\ln(ab)}+\color{blue}{\cancel{\ln x+}\ln(ab)}}{x^2+1}dx$$
$$\Rightarrow I=\frac{\ln (ab)}{4ab}\int_\frac{a}{b}^\frac{b}{a}\frac{1}{x^2+1}dx=\frac{\ln (ab)}{4ab}\left(\arctan\frac{b}{a} -\arctan \frac{a}{b}\right)$$
| {
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"source": "stackexchange",
"question_score": "2",
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Showing that there exists a positive integer $t$ such that $5^t\equiv -3\pmod {2^{n+4}}$ When I deal with number theory, I encounter a problem that seems to be easy but I can't prove.
let $n$ be a positive integer, there exists a positive integer $t$ such that
$$5^t\equiv -3\pmod {2^{n+4}}$$
| Claim 1. For any $n\ge0$, $5^{2^n}-1=2^{n+2}.odd$.
By induction, $$5^{2^{n+1}}-1=(5^{2^n}-1)(5^{2^n}+1)=2^{n+2}.odd.2.odd=2^{n+3}.odd$$ since in general $5^m+1=1^m+1=2\pmod{4}$. Of course, $5^1-1=2^2.1$.
Claim 2. For every $k$, there is an $r$ such that for any $n$, $2^k|(5^{2^{k-2}n+r}+3)$.
For $k=2$, $5^n+3=1+3=0\pmod{4}$.
Suppose $2^k|(5^{2^{k-2}n+r}+3)$; then either $2^{k+1}|(5^{2^{k-1}n+r}+3)$ satisfies the claim, or else $5^{2^{k-1}n+r}+3=2^k.odd$. In this case, \begin{align*} 5^{2^{k-2}(2n+1)+r}+3&=5^{2^{k-1}n+r}.5^{2^{k-2}}+3\\
&=(5^{2^{k-1}n+r}+3)5^{2^{k-2}}-3(5^{2^{k-2}}-1)\\
&=2^k.odd.5^{2^{k-2}}-3.2^k.odd\\
&=2^k(odd-odd)\\
&=2^{k+1}a\end{align*}
Hence for every $k\ge2$ there are an infinite number of $t$ such that $5^t=-3\pmod{2^k}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3128771",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Derivation Verification for an Algebraic Expression This problem comes from a MathOverFlow thread. Inside the thread the user Per Alexandersson mentions how they learned a technique to "simplify" $\sqrt{7+\sqrt{3}}$ from the book Algebra for Beginners, by Todhunter.
Greg Martin replied with the following formula:
$\sqrt{a+\sqrt b} = \sqrt{\frac{a-\sqrt{a^2-b}}{{2}}}+\sqrt{\frac{a+\sqrt{a^2-b}}{{2}}}$
I have three questions around this,
Question 1: Is the below derivation correct? If not, could someone show me how to patch it up?
Derivation:
\begin{align}
\sqrt{a+\sqrt b} &= \sqrt{\frac{a-\sqrt{a^2-b}}{{2}}}+\sqrt{\frac{a+\sqrt{a^2-b}}{{2}}} \\
\left(\sqrt{a+\sqrt b}\right)^{2} &= \bigg(\sqrt{\frac{a-\sqrt{a^2-b}}{{2}}}+\sqrt{\frac{a+\sqrt{a^2-b}}{{2}}}\bigg)^{2} \\
{a+\sqrt b} &= \frac{a-\sqrt{a^2-b}}{{2}} + 2\bigg(\sqrt{\frac{a-\sqrt{a^2-b}}{{2}}}\sqrt{\frac{a+\sqrt{a^2-b}}{{2}}}\bigg) + \frac{a+\sqrt{a^2-b}}{{2}} \\
{a+\sqrt b} &= \frac{a-\sqrt{a^2-b}}{{2}} + \frac{a+\sqrt{a^2-b}}{{2}} + 2\bigg( \sqrt{\frac{a-\sqrt{a^2-b}}{2}\cdot \frac{a+\sqrt{a^2-b}}{2}}\bigg) \\
{a+\sqrt b} &= \frac{a-\sqrt{a^2-b}+a+\sqrt{a^2-b}}{2}+ 2\bigg(\frac{\sqrt{b}}{\sqrt{4}}\bigg) \\
{a+\sqrt b} &= \frac{2a}{2} + 2\bigg(\frac{\sqrt{b}}{2}\bigg) \\
{a+\sqrt b} &= a + \sqrt{b}
\end{align}
Question 2: I assume that the above is if each step is reversible. But my question is (assuming the derivation is correct), how do we know each step is reversible when there are square roots involved?
Question 3: Regardless if the above derivation is correct or not, could someone show me another derivation and include the steps/explanations of how these are equal? I figure another approach might enlighten me as to how the longer expression was derived from $\sqrt{a+\sqrt b}$.
| The equation
$$\sqrt{a \pm \sqrt b} = \sqrt{\frac{a + \sqrt{a^2-b}}{{2}}} \pm \sqrt{\frac{a - \sqrt{a^2-b}}{{2}}} \tag{1}$$
is really two equations with proper choice of signs.
What we mean by that is that each square root of a positive real has two values, one positive and one negative.
There is one square root on the left side and on the right side is the sum/difference of two square roots. But given $\,0<b<a^2\,$ positive and the positive square root on the left, then the first square root on the right is greater than the second square root. Their sum must be the same as the left square root with the $+$ sign and the difference must be the same as with the $-$ sign.
When we replace $\,b\,$ with $\,-b,\,$ equation $(1)$ becomes
$$\sqrt{a \pm \sqrt -b} = \sqrt{\frac{a + \sqrt{a^2 +b}}{{2}}} \pm \sqrt{\frac{a - \sqrt{a^2+b}}{{2}}} \tag{2}$$
which is the well known formula for the square root of a complex number. In both cases, the equations can be verified by squaring both sides and simplifying and accounting for negative square roots as needed.
| {
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"timestamp": "2023-03-29T00:00:00",
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Suspended weight: Representing a vector differently leads to different answers I'm getting two different answers for the below problem depending on how I represent the relevant vector. My guess is the inverse sine function is affecting things, but I don't understand how I need to compensate.
Problem: A weight W is suspended from two lines. The force on each line are A and B at angles a, b. Solve for angle a.
Let $-W = -50\ \text{N},\ A = 35\ \text{N},\ b = 60^\circ.$ (The answer is $a = 74.42^\circ$)
\begin{align}
\pmb{A}
&= A<\cos c, \sin c> \text{OR}\\
&= A<-\cos a, \sin a>\\\\
\pmb{B}
&= B<\cos b, \sin b>\\\\
\pmb{W}
&= W<0, -1>\\
\end{align}
Here's my work for getting $\pmb{A}$:
I verified that both of these forms make the same vector:
\begin{align}
\pmb{A}
&= 35<-\cos 74, \sin 74>\\
&= 35<\cos(180-74), \sin(170-74)>\\
&= 35<-0.28, 33.6>
\end{align}
Below is the approach I used to find the correct value of $a$. It uses $\pmb{A} = A<\cos c, \sin c>\text{, where }c = 180 - a$:
x direction:
\begin{align}
0 &= A\cos c + B\cos b\\
B &= -A\frac{\cos c}{\cos b}
\end{align}
y direction:
\begin{align}
0 &= -W + A\sin c + B\sin b.\ \text{(Substitute for B.)}\\\\
W &= A\sin c -A\frac{\cos c}{\cos b}\sin b\\\\
\frac{W}{A} &= \sin c - \frac{\cos c}{\cos b}\sin b\\\\
\frac{W}{A}\cos b &= \sin c\cos b - \cos c\sin b\\
&= \sin(c - b)\\\\
c - b &= \arcsin(\frac{W}{A}\cos b)\\
180 - a - b &= \arcsin(\frac{W}{A}\cos b)\\\\
a &= 180 - b - \arcsin(\frac{W}{A}\cos b)\\
&= 180 - 60 - \arcsin(\frac{50}{35}\cos 60)\\
&= 74.42^\circ
\end{align}
Now, here is the approach yielding the incorrect value of $a$ where I began with $a$ directly. $\pmb{A} = A<-\cos a, \sin a>$:
x direction:
\begin{align}
0 &= -A\cos a + B\cos b\\
B &= A\frac{\cos a}{\cos b}
\end{align}
y direction:
\begin{align}
0 &= -W + A\sin a + B\sin b.\ \text{(Substitute for B.)}\\\\
W &= A\sin c + A\frac{\cos a}{\cos b}\sin b\\\\
\frac{W}{A} &= \sin a + \frac{\cos a}{\cos b}\sin b\\\\
\frac{W}{A}\cos b &= \sin a\cos b + \cos a\sin b\\
&= \sin(a + b)\\\\
a + b &= \arcsin(\frac{W}{A}\cos b)\\
a &= - b + \arcsin(\frac{W}{A}\cos b)\\
&= -60 + \arcsin(\frac{50}{35}\cos 60)\\
&= -14.42^\circ
\end{align}
These two answers, $a_1 = 74$ and $a_2 = -14$, make $60^\circ$ when added together. Again, my guess is the $\arcsin$ is causing issues, but I don't see what the logical error is. In an effort to see what's happening or if a compensating term would mean something to me, I wrote the equation for adding these two answers:
\begin{align}
\text{Let}\ X &= \arcsin(\frac{W}{A}\cos b)\\
a_1 + a_2 &= (180 - b - X) + (-b + X)\\
&= 180 - 2b\\
a_1 &= 180 - 2b - a_2\\
\end{align}
I'm left with the odd sensation where, looking at the algebra, it seems clear that the angles I'm getting would be different, but at the same time I would also expect that starting with an equivalent vector $\pmb{A}$, I ought to get the same final answer for the angle $a$. I've heard of compensating by 180, but $180 - 2b$ was unexpected.
Lastly, does the $2b$ suggest the discrepancy is related to the double-angle property? I looked into it and didn't feel it was leading anywhere.
| Yes, arc sine is a tricky tool. It takes a function that produces the same value infinitely many times, and supposes that you can invert that function.
The way we get this to work usually is we only count on it when the result should be between $-90$ and $90$ degrees ($-\frac\pi2$ to $\frac\pi2$ radians).
Given $b = 60$ degrees, the only way we could possibly have
$a + b = \arcsin(\text{anything})$
is if $a$ is between $-150$ and $30$ degrees. It isn't, so this doesn't work.
The "correct" method is fraught with danger too. If
the actual correct result had put $a < 30$ degrees,
that would imply $c > 150$ degrees, which would imply $c - b > 90$ degrees,
so $c - b = \arcsin(\text{whatever})$ would be impossible.
I would not use either method. I would construct the force diagram as shown below,
where the vector sum comes out to $\vec A + \vec B + \vec W = 0$
and where the vectors $-\vec B$ and $\vec W$ make an angle of $30$ degrees.
We then want to find the angle between $\vec A$ and $\vec W$ so that $A = 35.$
Based only on that information, there would be two possible solutions for $\vec A$
as shown in the figure below, but in this case only one of those solutions would have a positive vertical component (which is required by the setup of the problem),
so that is the solution to choose.
To work this out, let $\theta$ be the angle between $\vec A$ and $\vec B.$
Then by working out the components of $\vec W$ and $\vec A$ perpendicular to $\vec B,$ we get
$$ 35 \sin\theta = A\sin\theta = W \sin 30^\circ = 50 \sin 30^\circ = 25 $$
and therefore
$$ \sin\theta = \frac{25}{35} = \frac 57.$$
One solution is $\theta_1 = \arcsin \frac57 \approx 45.58^\circ.$
Another solution is $\theta_2 = 180 - \arcsin \frac57 \approx 134.42^\circ.$
Now, realizing that $a,$ $b,$ and $\theta$ are the three interior angles of the triangle in the original diagram, we have
$$ a = 180^\circ - b - \theta = 120^\circ - \theta, $$
which leads to the two answers
\begin{align}
a_1 &= 120^\circ - \theta_1 = \phantom{-}74.42^\circ, \\
a_2 &= 120^\circ - \theta_2 = -14.42^\circ. \\
\end{align}
So the two answers you found correspond to the two solutions of the triangular force diagram. Which method found which solution was simply a matter of which solution would put the quantity in the sin function (either $a + b$ or $c - b$) into the range $[90^\circ,90^\circ].$
Also notice that because $\theta_1$ and $\theta_2$ are the two solutions of
$\arcsin\frac57$ that fall in the range $[0,180^\circ],$
we have $\theta_1+\theta_2=180^\circ.$
So if if we add $a_1$ and $a_2,$ calculated according to the equations above,
We get
\begin{align}
a_1 + a_2 &= (120^\circ - \theta_1) + (120^\circ - \theta_1) \\
&= 240^\circ - (\theta_1 + \theta_2) \\
&= 240^\circ - 180^\circ \\
&= 60^\circ.
\end{align}
The angles that actually get "compensated" by $180^\circ$ are
$\theta_1$ and $\theta_2$, because those are the angles whose sine is actually taken.
By the way, if we try to generalize the problem to any given values of $b,$
$A,$ and $W,$ we can end up with no solution, or we can end up with two valid solutions to the same problem.
For example, if we merely change $A$ to $27,$ while keeping $b=60^\circ$
and $W=50,$ then $a_1$ and $a_2$ will both be positive and will represent two equally possible configurations of the lines holding up the weight.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3129686",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Triangle problem, $AC=3$ , $AB=5$ ,..., then $PH=?$ In $\triangle ABC$ :
$AC=3$ , $AB=5$ , $\angle ACB= 90 ^ \circ$,
$P$ is a point inside $\triangle ABC$ such that $PA+BC=PB+AC=PC+AB$,
$H$ is a point on the line $AB$ such that $\angle PHB=90^\circ$
then, $PH=?$
My Approach:
In Right-angled $\triangle ABC$
$$AB^2=AC^2 + BC^2 \implies BC=4$$
Let $PC=x$
then,
$$PA+4=PB+3=PC+5 $$
$$[\because PA+BC=PB+AC=PC+AB]$$
$$\implies PA=x+1 \, , \, PB= x+2$$
so,
My thoughts: First we will find the value of $x$ by equating area of triangle $ABC$,
i.e,
$$Ar[\triangle ABC]=Ar[\triangle APB]+Ar[\triangle BPC]+Ar[\triangle APC] \dots (i)$$
$Ar[\triangle ABC]$ can be simply found as: $\frac{1}{2}*3*4=6$ sq. units
For $Ar[\triangle APB]$ , we will use Heron's formula as follows:
$$s=\frac{1}{2} * [(x+1)+(x+2)+5] = x+4 $$
$$\therefore Ar[\triangle APB]=\sqrt{(x+4)(x-1)(2)(3)}=\sqrt{6(x+4)(x-1)}$$
Similarly, $Ar[\triangle BPC]=\sqrt{3(x+3)(x-1)}$
Similarly, $Ar[\triangle APC]=\sqrt{2(x+2)(x-1)}$
So, from eq.(i) , we get:
$$6=\sqrt{6(x+4)(x-1)} \, + \, \sqrt{3(x+3)(x-1)} \, + \, \sqrt{2(x+2)(x-1)} \dots (ii)$$
and then we will find $PH$ by equating area of triangle $APB$,
i.e,
$$\frac{1}{2}*5*PH=\sqrt{6(x+4)(x-1)} \dots (iii)$$
So, how to solve eq.(ii) to find the value of $x$ ? Please help...
| Continuing from your labels!
Drop perpendiculars $y$ and $z$ from the point $P$ to the sides $AC$ and $BC$, respectively, as shown:
$\hspace{1cm}$
Make up the system of equations:
$$\begin{cases}y^2+z^2=x^2\\ y^2+(3-z)^2=(x+1)^2\\ z^2+(4-y)^2=(x+2)^2\\ \frac12\cdot (3y+4z+5h)=\frac12\cdot 3\cdot 4 \ \text{(area of triangle $ABC$)}\end{cases}$$
Can you solve it?
WolframAlpha answer:
$h=\frac{132}{115}, x=\frac{29}{23},y=\frac{20}{23},z=\frac{21}{23}.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3130158",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
What is the integral of $\frac{2x+1}{x^2+1}$ I know the integral can be found by writing it as $$\frac{1}{x^2+1} +\frac{2x}{x^2+1}$$ This gives us an answer of $$\ln(x^2+4) +\frac{1}{2} \tan^{-1} \frac{x}{2}$$ However, when I let $x=2\tan u$, I get
$$\frac{1}{2} \tan^{-1} \frac{x}{2} + 2\ln\left|\text{sec(tan}^{-1} \frac{x}{2})\right|$$ This can be simplified and will give the same results for large values of $x$ but not smaller values. What am I doing wrong. Sorry for the format, I am still trying to figure out Latex.
| You were right to write the expression like
$$\frac{1}{x^2+1} +\frac{2x}{x^2+1}$$
The easiest way to find the integral is to notice that
$$\frac{2x}{x^2+1} = \frac{(x^2+1)'}{x^2+1}$$
and this is a logarithmic derivative.
Also,
$$\int \frac{dx}{x^2+1} = \arctan x + C$$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Determine all positive integer solutions for $\frac 1x + \frac 1y + \frac 1z + \frac 1{xy} + \frac1{yz} + \frac 1{xz} + \frac 1{xyz} = 1$ I need to determine all positive integer solutions for the equation:
$$\frac 1x + \frac 1y + \frac 1z + \frac 1{xy} + \frac1{yz} + \frac 1{xz} + \frac 1{xyz} = 1.$$
This is how I have tried to do it:
Mulitiplied both sides by $xyz$ to get
$$yz+xz+xy+z+x+y+1=xyz.$$
Factor it:
\begin{align}
x(y+z+1-yz)+yz+y+z &= -1 \\
x(y(1-z)+z+1)+y(1+z)+z &= -1
\end{align}
If $z=0$, we get
\begin{align}
x(y+1)+y=-1 &\iff xy+x+y=-1 \\
&\iff (x+1)(y+1)=0,
\end{align}
which gives us $x=y=-1$.
Is this all positive integer solutions? Or have I missed something?
EDIT: I am stupid.
New attempt.
If $z=1$, I get $2x+2y=-2 \iff x+y=-1$.
Then there is no solutions of positive integers for both $x$ and $y$ at the same time.
If I try for $z=2$, I get
\begin{align}
x(y(1-2)+2+1)+y(1+2)+2=-1 &\iff x(3-y)+3y+2=-1 \\
&\iff 3x+3y+3-xy=0
\end{align}
and I won't get a solution where all the variables are positive integers.
| This thing factors nicely after adding 1 to both sides. Not that it helps us much, though.
$$\left(1+{1\over x}\right)\cdot\left(1+{1\over y}\right)\cdot\left(1+{1\over z}\right)=2$$
Basically, it all boils down to brute force search which can easily be done by hand.
Let's assume WLOG $x\leqslant y\leqslant z$. Then just run through the possible values:
*
*$x=1$: LHS is too big regardless of $y$ and $z$.
*$x=2$:
*
*$y=2$: LHS is too big regardless of $z$.
*$y=3$: same as above.
*$y=4$: need to check.
*$y=5$: need to check.
*$y=6$: need to check.
*$y\geqslant7$: LHS is too small regardless of $z$.
*$x=3$:
*
*$y=3$: need to check.
*$y=4$: need to check.
*$y\geqslant5$: LHS is too small regardless of $z$.
*$x\geqslant4$: LHS is too small regardless of $y$ and $z$.
After checking each of the cases marked "need to check", you'll end up with quite a bunch of nice positive solutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3131265",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
Is $f(x)=\frac{1}{x+1} \cos x^2$ uniformly continuous?
Let $f:[0,\infty)\to \Bbb{R}$, $f(x)=\frac{1}{x+1} \cos x^2$, Is $f$ uniformly continuous?
My attempt:
Let $x,y\in [0,\infty),$ then
\begin{align*}
\Bigg|\frac{\cos x^2}{x+1}-\frac{\cos y^2}{y+1}\Bigg|
& =\Bigg|\frac{(y+1)\cos x^2-(x+1)\cos y^2}{(x+1)(y+1)}\Bigg| \\
& =\Bigg|\frac{(y \cos x^2 - x \cos y^2)+(\cos x^2 - \cos y^2)}{(x+1)(y+1)}\Bigg| \\
& \le \Bigg|\frac{y \cos x^2 - x \cos y^2}{(x+1)(y+1)}|+|\frac{\cos x^2 - \cos y^2}{(x+1)(y+1)}\Bigg| \\
& \le \Big|y \cos x^2 - x \cos y^2| + |\cos x^2 - \cos y^2\Big| \\
\end{align*}
How can I complete, please? Thanks.
| Hint
One can prove that if $x\mapsto f(x)$ is continuous on $[0,\infty )$ and $\lim\limits_{x\to \infty }f(x)$ exist, then $f$ is uniformly continuous on $[0, \infty )$. The proof goes as follow :
Let $\varepsilon >0$. Since $\lim\limits_{x\to \infty }f(x)=\ell$, there is $M$ s.t. $|f(x)-\ell|<\frac{\varepsilon}{2} $ if $x,y\geq M$. In particular, if $x,y\geq M$, then $$|f(x)-f(y)|\leq |f(x)-\ell|+|f(y)-\ell|\leq \varepsilon .$$
The fact that $f$ uniformly continuous on $[0,M]$ is a famous theorem.
Look what happen when $x\leq M\leq y$, and conclude.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3131600",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Number of possible products If I have set (2,3,4,5,6,7,8,9), how many possible products are there of any three numbers from the set. Repeats are allowed e.g. 2*2*2, 2*3*3, etc.
I don't know how to solve this other than brute force. Thx a lot.
| *This contsins some mistake. Be careful
I try writing about my trying, but you may think that is brute force. And it may have some mistake.
we should think the form of $2^a3^b5^c7^d(0\le a\le9,0\le b\le6,0\le c\le 3,0\le d\le3)$.
when $c=1,d=1$, then we can choose a number from$\{2,3,4,6,8,9\}$, so we get 6 numbers.
And think the case that $c=1,d\neq1$. If $4\le a\le6$,we can get only $(a,b)=(4,0),(5,0),(6,0),(4,1)$(for example,$2\times8\times5,4\times8\times5,8\times8\times5,6\times8\times5$) because 8 or double $4$ is necessasry to make $a$ not less than $4$.If $3\le b\le4$, also we get $(a,b)=(0,3),(0,4),(1,3)$.
And in the case that $0\le a \le 3$ and $0\le b \le 2$, other than $(a,b)=(0,0),(1,0),(0,1)$, we can get such a product(plese think a bit). So, we can know the number in the case equles to $4\times3-3=9$.
So, we now know the number in the case that $c=1,d\neq1$ equals to $9+4+3=16$, and the number in the case that $c\neq1,d=1$. It is equal. So, we can know the number in the case that $c=1$ or $d=1$. That is $16\times 2+6=38$
Let's think about the case that $c=0$ and $d=1$. The form is $2^a3^b(a\le a\le 9,0\le b\le6)$.
if $7\le a\le9$, we get only $(a,b)=(7,0),(7,1),(8,0),(9,0)$, and if $5\le b \le6$, we get only $(a,b)=(0,5),(1,5),(0,6)$.
When $0\le a\le6$ and $0\le b\le4$, then that is a little difficult.
we can choose three number,so we know $a+b\ge3$ And, except $8$, each number has only two or one prime factor. So, we can think that
\begin{equation}3\le a+b\le8(a=6)\\3\le a+b\le7(3\le a\le5)\\3\le a+b\le6(0\le a\le2) \end{equation}
and if we think in the case classification on the value of $a$, we know the full pairs of $a,b$ which meet the former equation can be gotten. The number is $3+3+4+5+4+4+4=27$, so we know the number in case that $c=0$ and $d=1$ is $27+4+3=34$. Then, we know the answer is $34+38=72$
former answer:"
Because of counting multisets, the number is less than $\binom {10}{3}=120$. But I can't get an accurate number. Sorry"
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3132834",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
calcuating residue of a complex function I need to calculate the residue of the function $\frac{(z^6+1)^2}{(z^5)(z^2-2)(z^2-\frac{1}{2})}$ at $z$=0.
z=0 is a pole of order 5 so I tried using the general formula to calculate the residue but the calculation becomes very tedious since it involves finding the fourth derivative of the complex function. I even tried writing the laurent series but that too got me nowhere. Could anyone please tell how to proceed with this problem?
| In order to get the residue of the function at $z=0$ we calculate the coefficient of $z^{-1}$ of the Laurent series expansion. We use the coefficient of operator $[z^n]$ to denote the coefficient of $z^n$.
We obtain
\begin{align*}
\color{blue}{[z^{-1}]}&\color{blue}{\frac{\left(z^6+1\right)^2}{z^5\left(z^2-2\right)\left(z^2-\frac{1}{2}\right)}}\\
&=[z^4]\frac{\left(z^6+1\right)^2}{(-2)\left(1-\frac{z^2}{2}\right)\left(-\frac{1}{2}\right)\left(1-2z^2\right)}\tag{1}\\
&=[z^4]\left(1+\frac{z^2}{2}+\frac{z^4}{4}\right)\left(1+2z^2+4z^4\right)\tag{2}\\
&=4+1+\frac{1}{4}\tag{3}\\
&\,\,\color{blue}{=\frac{21}{4}}
\end{align*}
Comment:
*
*In (1) we apply the rule $[z^{p-q}]A(z)=[z^p]z^qA(z)$. We also factor out constants to normalise terms in the denominator.
*In (2) we replace the term $(z^6+1)^2$ in the numerator with $1$ since other terms do not contribute to $[z^4]$. We also expand $\frac{1}{1-az^2}=1+az^2+a^2z^4+O(z^6)$ up to terms of $z^4$ since other terms do not contribute.
*In (3) we extract the coefficients of $z^4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3133218",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
$\lim\limits_{n \rightarrow +\infty} \frac{\ln(1+n+n^3)-3\ln(n)}{n(1-\cos(1/n^2))}$ I want to solve this limit:
$$\lim_{n \rightarrow +\infty} \frac{\ln(1+n+n^3)-3\ln(n)}{n(1-\cos(1/n^2))}$$
I have proved that $\lim\limits_{n \rightarrow +\infty} \frac{\ln(1+n+n^3)-3\ln(n)}{n} = 0$ and $\lim\limits_{n \rightarrow +\infty} \frac{1}{(1-\cos(1/n^2))}= \infty$ but I have indeterminate form. How can I solve that?
| You may proceed as follows:
*
*$\frac{\ln(1+n+n^3)-3\ln(n)}{n(1-\cos(1/n^2))} = \frac{\ln \left(1+\frac{1}{n^2}+\frac{1}{n^3} \right)}{n(1-\cos(1/n^2))}$
Now, set
*
*$n = \frac{1}{x}$ and consider the limit for $x \to 0^+$ and use
*$\cos t > 1- \frac{t^2}{2}$ for $t \in (0,\frac{\pi}{2})$
So, you get
\begin{eqnarray*} \frac{\ln \left(1+\frac{1}{n^2}+\frac{1}{n^3} \right)}{n(1-\cos(1/n^2))}
& \stackrel{n = \frac{1}{x}}{=} & \frac{x\cdot \ln (1+x^2+x^3)}{1-\cos x^2}\\
& \stackrel{\cos x^2 > 1 - \frac{x^4}{2}}{>} & \frac{2\ln (1+x^2+x^3)}{x^3} \\
& \stackrel{L'Hosp.}{\sim} & \frac{2}{1+x^2+x^3}\left(\frac{2}{x^2} + \frac{3}{x} \right) \\
& \stackrel{x \to 0^+}{\longrightarrow} & +\infty
\end{eqnarray*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3133657",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
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How to find a closed form for $\sum_{i=0}^n \binom{a+i}{b+i}i$ Wolframalpha tells me it's $$\frac{b (b + 1) \binom{a + 1}{ b + 1} - (b + n + 1) (b (n + 1) - (a + 1) n) \binom{a + n + 1}{ b + n + 1}}{(a - b + 1) (a - b + 2)}$$ but how to come up with or at least prove that?
| By way of enrichment and presenting various techniques we start by
introducting
$$S_{a,b}(n) = \sum_{q=0}^n {a+q\choose b+q} q$$
where $a\ge b$ and write with an Iverson bracket
$$\sum_{q\ge 0} {a+q\choose b+q} q
[[0\le q\le n]]
= \sum_{q\ge 0} {a+q\choose a-b} q
[z^n] \frac{z^q}{1-z}
\\ = [z^n] \frac{1}{1-z}
\sum_{q\ge 0} {a+q\choose a-b} q z^q
= [z^n] \frac{1}{1-z} [w^{a-b}] (1+w)^a
\sum_{q\ge 0} q z^q (1+w)^q
\\ = [z^n] \frac{1}{1-z} [w^{a-b}] (1+w)^a
\frac{z(1+w)}{(1-z(1+w))^2}
= [z^{n-1}] \frac{1}{1-z} [w^{a-b}] (1+w)^{a+1}
\frac{1}{(1-z-zw)^2}
= [z^{n-1}] \frac{1}{(1-z)^3} [w^{a-b}] (1+w)^{a+1}
\frac{1}{(1-zw/(1-z))^2}
\\ = [z^{n-1}] \frac{1}{(1-z)^3}
\sum_{q=0}^{a-b} {a+1\choose a-b-q}
(q+1) \frac{z^q}{(1-z)^q}
\\ = \sum_{q=0}^{a-b} {a+1\choose a-b-q}
(q+1) [z^{n-1-q}] \frac{1}{(1-z)^{q+3}}
\\ = \sum_{q=0}^{a-b} {a+1\choose a-b-q}
(q+1) {n+1\choose q+2}
\\ = (n+1) \sum_{q=0}^{a-b} {a+1\choose a-b-q}
{n\choose q+1}
- \sum_{q=0}^{a-b} {a+1\choose a-b-q}
{n+1\choose q+2}.$$
The first sum is
$$\sum_{q=0}^{a-b} {a+1\choose a-b-q} {n\choose q+1}
= [z^{a-b}] (1+z)^{a+1}
\sum_{q=0}^{a-b} {n\choose q+1} z^q.$$
We may extend $q$ to infinity because of the coefficient extractor in
front:
$$[z^{a-b}] (1+z)^{a+1}
\sum_{q\ge 0} {n\choose q+1} z^q
= [z^{a-b+1}] (1+z)^{a+1}
\sum_{q\ge 0} {n\choose q+1} z^{q+1}
\\ = [z^{a-b+1}] (1+z)^{a+1} ((1+z)^n - 1)
= {n+a+1\choose a-b+1} - {a+1\choose a-b+1}.$$
The second is
$$\sum_{q=0}^{a-b} {a+1\choose a-b-q} {n+1\choose q+2}
= [z^{a-b}] (1+z)^{a+1}
\sum_{q=0}^{a-b} {n+1\choose q+2} z^q.$$
We may once more extend $q$ to infinity because of the coefficient
extractor in front:
$$[z^{a-b}] (1+z)^{a+1}
\sum_{q\ge 0} {n+1\choose q+2} z^q
= [z^{a-b+2}] (1+z)^{a+1}
\sum_{q\ge 0} {n+1\choose q+2} z^{q+2}
\\ = [z^{a-b+2}] (1+z)^{a+1} ((1+z)^{n+1} - 1 - (n+1)z)
\\ = {n+a+2\choose a-b+2} - {a+1\choose a-b+2}
- (n+1) {a+1\choose a-b+1}.$$
Collecting and canceling we obtain at last
$$\bbox[5px,border:2px solid #00A000]{
S_{a,b}(n) = (n+1) {n+a+1\choose a-b+1}
- {n+a+2\choose a-b+2}
+ {a+1\choose a-b+2}}$$
or alternatively
$$\bbox[5px,border:2px solid #00A000]{
S_{a,b}(n) = (n+1) {n+a+1\choose n+b}
- {n+a+2\choose n+b}
+ {a+1\choose b-1}.}$$
This simplifies to the accepted answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3133878",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
If nine coins are tossed, what is the probability that the number of heads is even? If nine coins are tossed, what is the probability that the number of heads is even?
So there can either be 0 heads, 2 heads, 4 heads, 6 heads, or 8 heads.
We have $n = 9$ trials, find the probability of each $k$ for $k = 0, 2, 4, 6, 8$
$n = 9, k = 0$
$$\binom{9}{0}\bigg(\frac{1}{2}\bigg)^0\bigg(\frac{1}{2}\bigg)^{9}$$
$n = 9, k = 2$
$$\binom{9}{2}\bigg(\frac{1}{2}\bigg)^2\bigg(\frac{1}{2}\bigg)^{7}$$
$n = 9, k = 4$
$$\binom{9}{4}\bigg(\frac{1}{2}\bigg)^4\bigg(\frac{1}{2}\bigg)^{5}$$
$n = 9, k = 6$
$$\binom{9}{6}\bigg(\frac{1}{2}\bigg)^6\bigg(\frac{1}{2}\bigg)^{3}$$
$n = 9, k = 8$
$$\binom{9}{8}\bigg(\frac{1}{2}\bigg)^8\bigg(\frac{1}{2}\bigg)^{1}$$
Add all of these up:
$$=.64$$ so there's a 64% chance of probability?
| There's a way to do it with barely any maths:
It's clear that if there's an odd number of heads, there's an even number of tails and vice versa, so P(even number of heads) + P(even number of tails) = 1.
Formally rename "heads" to "tails". The problem remains unchanged.
So P(even number of heads) = P(even number of tails) = 1/2.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3134991",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "54",
"answer_count": 13,
"answer_id": 11
} |
$\sum_{k=0}^{n-1}\frac{1}{1-2x\cos\frac{2k\pi}{n}+x^2}$
Given it is to prove this $\sum_{k=0}^{n-1}\frac{1}{1-2x\cos\frac{2k\pi}{n}+x^2}=\frac{n(1+x^n)}{(1-x^n)(1-x^2)}$
For this i first tried substituting and using complex numbers like this but what after this if someone have any other method to proceed his welcome.
$A_k=1-2x\cos\frac{2k\pi}{n}+x^2=(x-\cos\frac{2k\pi}{n})^2-(i\sin\frac{2k\pi}{n})^2$ and then seperating this.$=(x-\cos\frac{2k\pi}{n}+i\sin\frac{2k\pi}{n})(x-\cos\frac{2k\pi}{n}-i\sin\frac{2k\pi}{n})$
| Let $\omega=e^{\frac{2\pi i}n}$ be the primitive $n$-th root of unity. We have for all $|x|<1$,
$$\begin{align*}
\frac{1-x^2}{1-2x\cos(\frac{2k\pi}{n})+x^2}&=\frac{\bar\omega^k(\omega^k-x)+x(\bar\omega^k-x)}{(\omega^k-x)(\bar\omega^k-x)}\\&=\frac{\bar\omega^k}{\bar\omega^k-{x}}+\frac{x}{\omega^k-x}\\&=\sum_{i=0}^\infty \left(\frac x {\bar\omega^k}\right)^i-1+\sum_{i=0}^\infty \left(\frac x {\omega^k}\right)^i\\&=-1+\sum_{i=0}^\infty (\omega^{ki}+\bar\omega^{ki})x^i.
\end{align*}$$ Using the identity $\sum_{k=0}^{n-1}\omega^{jk}=\sum_{k=0}^{n-1}e^{\frac{2\pi jk i}{n}}=n\mathbf 1_{j\in n\Bbb Z}$, we find for all $|x|<1$,
$$\begin{align*}
(1-x^2)\sum_{k=0}^{n-1}\frac{1}{1-2x\cos(\frac{2k\pi}{n})+x^2}&=-n+\sum_{i=0}^\infty x_i\left(\sum_{k=0}^{n-1}\omega^{ki}+\bar\omega^{ki}\right)\\&=-n+2n\sum_{i=0}^\infty x^i\mathbf 1_{i\in n\Bbb Z}\\&=n\left(-1+2\sum_{i=0}^\infty x^{ni}\right)\\&=n\left(-1+\frac{2}{1-x^n}\right)\\&=\frac{n(1+x^n)}{1-x^n}.
\end{align*}$$ This gives
$$
\sum_{k=0}^{n-1}\frac{1}{1-2x\cos(\frac{2k\pi}{n})+x^2}=\frac{n(1+x^n)}{(1-x^n)(1-x^2)}
$$ for all $|x|<1$. However, since both are meromorphic functions defined on an open connected domain, identity theorem implies they are equal where they are defined.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3135930",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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} |
Prove that the roots $\in \Bbb R$ of $x^3+x+1=0$ aren't rational without RRT I need to prove that the roots $\in \Bbb R$ of $x^3+x+1=0$ aren't rational. Obviously, it's easy to use the rational root theorem to prove that there are not rational solutions to this equation, but i want a different approach.
I saw a similar question here, but it didn't provide my "solution".
My try
$x^3+x+1=0$
$x^3+x=-1$
$x(x^2+1)=-1$
Here we have two options (product of two numbers to obtain a negative):
$1.$ $x \gt 0$ $∧$ $x^2+1 \lt 0$ (not possible in $\Bbb R)$
$2.$ $x \lt 0$ $∧$ $x^2+1 \gt 0$
So, we use $2.$ to prove that $x \ne 0$, then:
$x(x^2+1)=-1$
$x^2+1=\frac{-1}{x}$
$x^2=\frac{-1}{x}-1$
$x^2=-(\frac{x+1}{x})$
$x=\pm \sqrt {-(\frac{x+1}{x})}$
Then, $\frac{x+1}{x} \ge 0$ because $x \lt 0$ (the equality occurs when $x=-1$, but this doesn't satisfy the original polynomial equation) but here i'm missing the cases $-1 \lt x \lt 0$ (i don't know how to use this to prove that the root is irrational)
This implies that $2$ of the roots are complex, but there are $3$ roots to a third degree polynomial equation.
And here i'm stuck, because i don't know how to prove that the last solution is irrational.
Any hints?
Is there anyway to prove that the last root is irrational?
Is my proof good so far?
| If $x$ is real then
$-1 = x(x^2+1)
$
so
$x = \frac{-1}{x^2+1}
$
so
$-1 < x < 0$.
If $x = -c/d$
with $(c, d) = 1$,
then
$\frac{c}{d}
= \frac{1}{(c/d)^2+1}
=\frac{d^2}{c^2+d^2}
$
or
$c(c^2+d^2) = d^3$.
If a prime $p$
divides $c$,
then $p | d^3$
so $p | d$,
which contradicts
$(c, d) = 1$.
Therefore
$c = 1$,
so
$1+d^2 = d^3$
or
$1
=d^3-d^2
=d^2(d-1)
$
which can not hold
since it is false for $d = 1$
and
$d^2(d-1) > 1$
for
$d \ge 2$.
Therefore,
there is no rational root.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3139610",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
} |
Number of $\pm 1$ matrices whose determinant is negative or zero
Find the number of $3 \times 3$ matrices whose determinant is $(a)$ positive $(b)$ negative $(c)$ zero and whose elements are taken from $\{-1,1\}$.
This is what I tried:
The total number of $3\times 3$ matrices with entries in $\{ 1 , -1\}$ is $2^9$. Now counting each rows or column of matrices as a vectors with entries in $\{-1,1\}$, there are $8$ such vectors and if any two vectors are same then its determinant is $0$. There are such $24+24=48$ ways.
| Matrices with $\det <0$ are exactly those resulting from swapping the first two rows of matrices with $\det >0$ because this multiplies the determinant by $-1$, and so the number with $\det >0$ is the same as the number with $\det <0$. So you just need to count the matrices with $\det =0$.
First we count how many of the $2^4 =16$ matrices of the form $$A = \begin{pmatrix} 1 & 1 & 1 \\ 1 & a & b \\ 1 & c & d \end{pmatrix}$$ have $\det A =0$.
The determinant of $A$ is $$\det A = \left| \begin{smallmatrix} a & b \\ c & d \end{smallmatrix} \right| - \left| \begin{smallmatrix} 1 & b \\ 1 & d \end{smallmatrix} \right| + \left| \begin{smallmatrix} 1 & a \\ 1 & c \end{smallmatrix} \right| = (ad -a -d) - (bc -b -c).$$
We have the first term equal to $3$ when $(a,d)=(-1,-1)$, and equal to $-1$ in the other three cases- and the same for the second term. We therefore have one in the case $a=b=c=d=-1$ and $3 \times 3 = 9$ for the other cases, totaling $10$ such matrices $A$.
Each matrix with determinant $0$ can be uniquely made into the above form by multiplying a choice of the rows and columns by $-1$ (which doesn't change the determinant). There are $10$ matrices of determinant $0$ for each of the $2^5 = 32$ ways to choose those entries in the first row and column.
Hence, out of the $2^9 = 512$ matrices with entries in $\{ 1, -1 \}$, $320$ have $\det = 0$, $96$ have $\det > 0$ and $96$ have $\det <0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3140096",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
} |
Derivative of the definite integral I have to find the derivative for
$$\int_{-1}^x \frac{t^2}{t^2+4}dt-\int_{3}^x \frac{t^2}{t^2+4}dt$$
When I calculate their derivative separately
$$\frac {d}{dt} \left(\int_{-1}^x \frac{t^2}{t^2+4}dt\right)-\frac {d}{dt}\left(\int_{3}^x \frac{t^2}{t^2+4}dt\right)$$
it gives me the right result, that is $0$.
But according to the law of additive
\begin{aligned}
&\int_{-1}^x \frac{t^2}{t^2+4}dt-\int_{3}^x \frac{t^2}{t^2+4}dt \\
= & \int_{-1}^x \frac{t^2}{t^2+4}dt+\int_{x}^3 \frac{t^2}{t^2+4}dt \\= &\int_{-1}^3 \frac{t^2}{t^2+4}dt \\[2em]
\end{aligned}
$$d/dt\int_{-1}^3 \frac{t^2}{t^2+4}dt = \frac{3^2}{3^2+4} = \frac 9{13}$$
Thanks.
| You have to obtain the derivative $\frac d{dx}$ of $$\int_{-1}^3 \frac{t^2}{t^2+4}dt$$
so the result will be
$$\frac d{dx}\int_{-1}^3 \frac{t^2}{t^2+4}dt =\frac d{dx}(const.) = 0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3141147",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Breaking the barriers at $q=5$ and $q=13$ for $q^k n^2$ an odd perfect number with special prime $q$ Let $\sigma(x)$ be the sum of divisors of the positive integer $x$. If $\sigma(N)=2N$ and $N$ is odd, then $N$ is called an odd perfect number. The question of existence of odd perfect numbers is the longest unsolved problem of mathematics.
Euler proved that an odd perfect number, if one exists, must have the form $N = q^k n^2$ where $q$ is the special prime satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.
Broughan, Delbourgo, and Zhou prove in IMPROVING THE CHEN AND CHEN RESULT FOR ODD PERFECT NUMBERS (Lemma 8, page 7) that if $\sigma(n^2)/q^k$ is a square, then the Descartes-Frenicle-Sorli conjecture that $k=1$ holds.
So now suppose that $\sigma(n^2)/q^k$ is a square. This implies that $k=1$, and also that $\sigma(n^2) \equiv 1 \pmod 4$, since $\sigma(n^2)/q^k$ is odd and $q \equiv k \equiv 1 \pmod 4$.
The congruence $\sigma(n^2) \equiv 1 \pmod 4$ then implies that $q \equiv k \pmod 8$. (See this MO post for the details.) Substituting $k=1$, we obtain
$$q \equiv 1 \pmod 8.$$
This implies that the lowest possible value for the special prime $q$ is $17$. (That is, this argument breaks the barriers at $q=5$ and $q=13$, under the assumption that $\sigma(n^2)/q^k$ is a square.) Note that, if $q=17$, then $(q+1)/2 = 3^2 \mid n^2$.
Here is my question:
Can we push the lowest possible value from $q \geq 17$, to say, $q \geq 41$ or even $q \geq 97$, using the ideas in this post, and possibly more?
Reference:
The Abundancy index of divisors of odd perfect numbers – Part III
| Let $N=q^k n^2$ be an odd perfect number with special prime $q$.
Note that if
$$\frac{\sigma(n^2)}{q^k}=\frac{n^2}{\sigma(q^k)/2}$$
is a square, then $k=1$ and $\sigma(q^k)/2 = (q+1)/2$ is also a square.
The possible values for the special prime satisfying $q < 100$ and $q \equiv 1 \pmod 8$ are $17$, $41$, $73$, $89$, and $97$.
For each of these values:
$$\frac{q_1 + 1}{2} = \frac{17 + 1}{2} = 9 = 3^2$$
$$\frac{q_2 + 1}{2} = \frac{41 + 1}{2} = 21 \text{ which is not a square.}$$
$$\frac{q_3 + 1}{2} = \frac{73 + 1}{2} = 37 \text{ which is not a square.}$$
$$\frac{q_4 + 1}{2} = \frac{89 + 1}{2} = 45 \text{ which is not a square.}$$
$$\frac{q_5 + 1}{2} = \frac{97 + 1}{2} = 49 = 7^2$$
Thus, if $\sigma(n^2)/q^k$ is a square and we could rule out $q=17$, it would follow that $q \geq 97$.
| {
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"url": "https://math.stackexchange.com/questions/3142243",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Order of a sixth root in group G This question originates from Chapter 10, H3 of the 2nd edition of A Book of Abstract Algebra by Charles C. Pinter.
Let $a$ denote an element of a group $G$. Let $a$ have order 10. If $a$ has a sixth root in $G$, say $a=b^6$, what is the order of $b$?
Here is what I think:
Given $\operatorname{ord}(a) = 10$ and $a=b^6$,
$\qquad a^{10} = e = (b^{6})^{10} = b^{60}$
Let $\operatorname{ord}(b) = x, x$ must divide $60 \implies x\in \{1,2,3,4,5,6,10,12,15,20,30,60\}.$
*
*$x=1: b = e = b^6 = a$ but $\operatorname{ord}(a) \ne 1.$
*$x=2: b^2 = e = b^6 = a$ but $\operatorname{ord}(a) \ne 1.$
*$x=3: b^3 = e = b^6 = a$ but $\operatorname{ord}(a) \ne 1.$
*$x=4: b^4 = e = b^{12} = a^2$ but $\operatorname{ord}(a) \ne 2.$
*$x=5: b^5 = e = b^{30} = a^5$ but $\operatorname{ord}(a) \ne 5.$
*$x=6: b^6 = e = a$ but $\operatorname{ord}(a) \ne 1.$
*$x=10: b^{10} = e = b^{30} = a^5$ but $\operatorname{ord}(a) \ne 5.$
*$x=12: b^{12} = e = a^2$ but $\operatorname{ord}(a) \ne 2.$
*$x=15: b^{15} = e = b^{30} = a^5$ but $\operatorname{ord}(a) \ne 5.$
*$x=20: b^{20} = e = b^{60} = a^{10}$.
*$x=30: b^{30} = e = a^5$ but $\operatorname{ord}(a) \ne 5.$
*$x=60: b^{60} = e = a^{10}.$
Hence $x \in \{20,60\}$.
So it seems $x=20$ as $20 < 60$.
Or is there a way to rule out $20$?
| Just realize in general
If $a$ has order $n$ and $a$ has a $k$th root $b$, then $b$ has order
$\displaystyle\frac{nk}{l}$, where $n$ and $l$ are relatively prime.
Applying this to my original question where $n=10, k=6$. So
$\qquad\displaystyle\operatorname{ord}(b) = \frac{10\cdot 6}{l}\qquad$ where $10$ and $l$ are relatively prime.
$\implies l \in \{1,3,7,9\}$
*
*$\displaystyle l=1: \operatorname{ord}(b) = \frac{60}{1} = 60$
*$\displaystyle l=3: \operatorname{ord}(b) = \frac{60}{3} = 20$
*$\displaystyle l=7: \operatorname{ord}(b) = \frac{60}{7}$ (not possible since it's not an integer)
*$\displaystyle l=9: \operatorname{ord}(b) = \frac{60}{9}$ (not possible since it's not an integer)
Hence $b$ has order of either $20$ or $60$.
$\Box$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3142621",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Proving $\int_0^\infty \log\left (1-2\frac{\cos 2\theta}{x^2}+\frac{1}{x^4} \right)dx =2\pi \sin \theta$
Prove $$\int_0^\infty \log \left(1-2\frac{\cos 2\theta}{x^2}+\frac{1}{x^4} \right)dx =2\pi \sin \theta$$where $\theta\in[0,\pi]$.
I've met another similar problem,
$$ \int_0^{2\pi} \log(1-2r\cos \theta +r^2) d\theta=2\pi \log^+(r^2) $$
I am curious whether there is any relationship between them.
And I got stuck on the proposition in the title. I found that
$$1-2\frac{\cos 2\theta}{x^2}+\frac{1}{x^4} =\left(\frac{1}{x}-e^{i\theta}\right)\left(\frac{1}{x}+e^{i\theta}\right)\left(\frac{1}{x}-e^{-i\theta}\right)\left(\frac{1}{x}+e^{-i\theta}\right)$$
But I couldn't move on.
Any hints? Thanks in advance.
| To complement Zacky's answer I'll add a proof that $I(0)=0$. Applying $x\mapsto\frac{1}{x}$ for $x\ge 1$ gives $$I(x)=2\int_0^\infty\ln|1-x^{-2}|dx=2\int_0^1\left[(1+\frac{1}{x^2})\ln(1-x^2)-2\ln x\right]\\=-2\int_0^1\left[(1+x^2)\sum_{n\ge 0}\frac{x^{2n}}{n+1}+2\ln x\right]dx=-2\int_0^1\left[\sum_{n\ge 0}\left(\frac{x^{2n}}{n+1}+\frac{x^{2n+2}}{n+1}\right)+2\ln x\right]dx\\=-2\left[\sum_{n\ge 0}\left(\frac{x^{2n+1}}{(n+1)(2n+1)}+\frac{x^{2n+3}}{(n+1)(2n+3)}\right)+2x\ln x-2x\right]_0^1\\=-2\left[\sum_{n\ge 0}\left(\frac{4}{(2n+1)(2n+3)}\right)-2\right],$$which vanishes by partial fractions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3143523",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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"answer_id": 1
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Find all integers of the form $\frac{(x-1)^2(x+2)}{2x+1}$ I made a python program and I have that
\begin{array}{|c|c|}
\hline x & \frac{(x-1)^2(x+2)}{2x+1} \\\hline
-14 & 100 \\\hline
-5& 12 \\\hline
-2&0 \\\hline
-1&4 \\\hline
0&2 \\\hline
1&0 \\\hline
4&6 \\\hline
13&80 \\\hline
\end{array}
And taking very large intervals of integers this seems like the only integers of this form.
However I don't know how can i proved.
Any ideas?
| $\bmod 2x\!+\!1\!:\ \, x\equiv -\dfrac{1}2\ $ so $\ 0\equiv(x\!+\!2)(x\!-\!1)^2\equiv\dfrac{3}2 \left[\dfrac{-3}2\right]^2\equiv \dfrac{27}8\iff 27\equiv 0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3150077",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proving $\left|\begin{smallmatrix}1&1&1\\a&b&c\\a^3&b^3&c^3\end{smallmatrix}\right|=(b-a)(c-b)(c-a)(a+b+c)$
Prove that$$\begin{vmatrix}1&1&1\\a&b&c\\a^3&b^3&c^3\end{vmatrix}=(b-a)(c-b)(c-a)(a+b+c)$$
My attempt:
$$\begin{align}\begin{vmatrix}1&1&1\\a&b&c\\a^3&b^3&c^3\end{vmatrix}&=\begin{vmatrix}0&1&0\\a-b&b&c-b\\a^3-b^3&b^3&c^3-b^3\end{vmatrix}\\&=\begin{vmatrix}c-b&a-b\\c^3-b^3&a^3-b^3\end{vmatrix}\\&=(c-b)(a-b)\begin{vmatrix}1&1\\c^2+cb+b^2&a^2+ab+b^2\end{vmatrix}\\&=(c-b)(a-b)[(a^2+ab)-(c^2+cb)]\\\end{align}$$
Where did I go wrong?
| You can get it immediately:
$$\sum_{cyc}(a^3c-a^3b)=(a+b+c)(a-b)(b-c)(c-a)$$
because for $a=b$ or $a=c$ or $b=c$ our determinant is equal to zero, which gives
$$K(a+b+c)(a-b)(b-c)(c-a)$$ and it's enough to check the coefficient before $a^3c$, which is $1$, which gives $K=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3150198",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 3
} |
Evaluate $\int^{2}_{0}\frac{\tan^{-1}(x)}{1+4x}\mathrm dx$
Evaluate $\displaystyle \int^{2}_{0}\frac{\tan^{-1}(x)}{1+4x}\mathrm dx$
My effort:
\begin{align*}
I(a)&=\int^{2}_{0}\frac{\tan^{-1}(ax)}{1+4x}\mathrm dx\\
I'(a) &= \int^{2}_{0}\frac{x}{(1+4x)(1+a^2x^2)}\mathrm dx\\
I'(a) &= \frac{1}{4}\int^{2}_{0}\frac{(1+4x)-1}{(1+4x)(1+a^2x^2)}dx\\
I'(a) &= \frac{1}{4a}\tan^{-1}(2)-\frac{1}{4}\int^{2}_{0}\frac{1}{(1+4x)(1+a^2x^2)}dx
\end{align*}
Then how to proceed? Thank you.
| Hint
Try$${x\over (1+4x)(1+a^2x^2)}={{-{4\over a^2+16}\over 1+4x}}+{{a^2x+4\over a^2+16}\over 1+a^2x^2}={1\over a^2+16}\left({{-{4}\over 1+4x}}+{{a^2x+4}\over 1+a^2x^2}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3159242",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
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"answer_id": 0
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Find $\lim\limits_{x\to 0}\left (\frac{1^x+2^x+3^x+\dots+n^x}{n} \right)^{\frac1x}$ Consider the following expression.
$$\lim\limits_{x\to 0} \left (\frac{1^x+2^x+3^x+\dots+n^x}{n} \right )^{\frac 1 x}$$
How to solve this?
Let $y= \left (\frac {1^x+2^x+\cdots +n^x} {n} \right)^{1/x}$
I tried taking $\ln$ on both sides. We get that $$\ln(y)=\frac{1}{x}\ln \left (\frac {1^x+2^x+\cdots +n^x} {n} \right ).$$
Taking $\lim$ on both sides we get $$\ln(y)=\lim_{x\to 0}\frac{1}{x}\ln \left (\frac {1^x+2^x+\cdots +n^x} {n} \right ).$$
Now applying the LH rule, we get $$\ln(y)=\lim_{x\to 0}\frac{n}{1^x+2^x+\cdots +n^x}({1^x\ln(1)+\cdots +n^x\ln(n)})$$
Is this a right way to go?
| I write this answer only because the wrong one is accepted.
There is an error in your final expression. It should be:
$$\lim_{x\to 0}\frac{n}{1^x+2^x+\cdots +n^x}\frac{1^x\ln(1)+\cdots +n^x\ln(n)}{\color{red}n}=\frac{\ln n!}{n}.$$
Correspondingly:
$$
\lim\limits_{x\to 0} \left (\frac{1^x+2^x+3^x+\dots+n^x}{n} \right )^{\frac 1 x}=\sqrt[n]{n!}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3159300",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 3
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Tricky Integral -- $\int_0^1 \sqrt{x^2-4x+3} \arcsin(x)~dx$ TL;DR: I can't get a closed form for the integral below.
$$ \int_0^1 \sqrt{x^2-4x+3} \arcsin(x)~dx $$
I got an interesting question from a coworker a while ago:
Question:
The quantities $a$, $b$, and $c$ are chosen uniformly and independently from $[0, 1]$.
a) What is the probability a triangle can be constructed with $a$, $b$, and $c$ as side lengths?
b) Given we can form such a triangle, what is its expected area?
I can do a) pretty easily -- each constraint like $a < b + c$ cuts off a corner of the cube with area $1/6$, and the cut-off bits are disjoint, so the remaining area is $1/2$.
Part b) is where things get hairy. I can reduce the problem down to a single integral. I feel like it should be expressible in terms of known constants, though I admit I have no good reason to believe this.
$$
\frac{3}{40} \int_0^1 x \sqrt{3-4x+x^2} \left( \sqrt{1 - x^2} + \frac{\arcsin{x}}{x} \right)~dx
$$
That can be split into two parts:
$$
\frac{3}{40} \int_0^1 x \sqrt{(3-4x+x^2)(1 - x^2)}~dx + \frac{3}{40} \int_0^1 \sqrt{3-4x+x^2} \arcsin(x)~dx
$$
The first part can be solved exactly.
$$
\begin{align*}
\int_0^1 x \sqrt{(3-4x+x^2)(1 - x^2)}~dx &= \int_0^1 x \sqrt{(3-x)(1-x)(1-x)(1+x)}~dx \\
&= \int_0^1 x(1-x) \sqrt{(3-x)(1+x)}~dx \\
&= \frac{1}{12} (32 - 9 \sqrt{3} - 4\pi) \textrm{ by Mathematica}
\end{align*}
$$
The second part is still pretty stubborn.
Mathematica tells me the integral (without the 3/40 constant) is approximately 0.452854, but doesn't given an exact form. Does anyone have any ideas how to evaluate this further?
| This is not an answer.
We could use
$$\sqrt{x^2-4x+3}=\sum_{n=0}^\infty a_n\, x^n$$ with
$$a_n=\frac{2(2 n-3)\, a_{n-1}-(n-3)\, a_{n-2}}{3 n} \qquad \text{where}\qquad a_0=\sqrt{3}\qquad a_1=-\frac{2}{\sqrt{3}}$$ and
$$\int_0^1 x^n\arcsin(x)\,dx=\frac{\pi }{2( n+1)}-\frac{\sqrt{\pi }\,\,\Gamma \left(\frac{n}{2}+1\right)}{(n+1)^2
\,\, \Gamma \left(\frac{n+1}{2}\right)}$$ but the convergence is very slow.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
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If $u(r, \theta)$ is a solution of Laplace’s equation show that $u(\frac{1}{r}, \theta)$ is also a solution.
Suppose that $u(r, θ)$ is a solution of Laplace’s equation. Show that
$u(\frac{1}{r}, θ)$ is also a solution.
So far, I know that if $u$ satisfies Laplace's equation, then
$$\Delta u = u_{rr} + \frac{1}{r}u_r + \frac{1}{r^2}u_{\theta \theta}$$ since $u$ is given in polar coordinates. I let $w = u(\frac{1}{r}, \theta)$. I want $w$ to also satisfy Laplace's Equation. Thus $\Delta w = w_{rr} + \frac{1}{r}w_r + \frac{1}{r^2}w_{\theta \theta}$
I computed the partial derivatives of $w$ and plugged them into the above equation but can't seem to get the right result.
Thanks for the help!
| Consider a function $v(r, \theta) = u\left(q, \theta\right),$ with $q = \frac{1}{r}.$
Then, using the chain rule:
*
*$v_{r} = q_r u_q = -\displaystyle\frac{1}{r^2} u_q$
*$v_{rr} = \displaystyle\frac{2}{r^3} u_q + \frac{1}{r^4} u_{qq}$
*$v_{\theta\theta} = u_{\theta\theta}$
Then:
$$v_{rr} + \frac{1}{r}v_r + \frac{1}{r^2}v_{\theta \theta}
= \frac{2}{r^3} u_q + \frac{1}{r^4} u_{qq} -\displaystyle\frac{1}{r^3} u_q + \frac{1}{r^2}u_{\theta\theta} = \frac{1}{r^4}\left(u_{qq} + ru_q + r^2 u_{\theta\theta}\right).$$
Therefore, since $r = \frac{1}{q}$, then:
$$v_{rr} + \frac{1}{r}v_r + \frac{1}{r^2}v_{\theta \theta}
= q^4\left(u_{qq} + \frac{1}{q}u_q + \frac{1}{q^2} u_{\theta\theta}\right) .$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3163694",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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$\int_0^1\frac{\ln{x}\ln{(1+x)}}{1+x}dx$ I want to solve for the following Integral:
$$\int_0^1\frac{\ln{x}\ln{(1+x)}}{1+x}dx$$
I have tried to use:
$$\ln{(1+x)}=-\sum_{k=1}^\infty\frac{(-1)^kx^k}{k}$$
and so
$$\int_0^1\frac{\ln{x}\ln{(1+x)}}{1+x}dx=-\sum_{k=1}^\infty\frac{(-1)^k}{k}\int_0^1\frac{x^k\ln{x}}{1+x}dx$$
| Let,
\begin{align}
U&=\int_0^1\frac{\ln(1+x)\ln x}{1+x}\,dx\\
W&=\int_0^1\frac{\ln^2\left(\frac{x}{1+x}\right)}{1+x}\,dx\\
\end{align} Perform the change of variable $y=\dfrac{x}{1+x}$
\begin{align}
W&=\int_0^{\frac{1}{2}}\frac{\ln^2 x}{1-x}\,dx\\
&=\int_0^1\frac{\ln^2 x}{1-x}\,dx-\int_{\frac{1}{2}}^1\frac{\ln^2 x}{1-x}\,dx\\
\end{align} In the latter integral perform the change of variable $y=\dfrac{1-x}{x}$
\begin{align}
W&=\int_0^1\frac{\ln^2 x}{1-x}\,dx-\int_0^1\frac{\ln^2(1+x)}{x(1+x)}\,dx\\
&=\int_0^1\frac{\ln^2 x}{1-x}\,dx+\int_0^1\frac{\ln^2(1+x)}{1+x}\,dx-\int_0^1\frac{\ln^2(1+x)}{x}\,dx\\
&=\int_0^1\frac{\ln^2 x}{1-x}\,dx+\int_0^1\frac{\ln^2(1+x)}{1+x}\,dx-\Big[\ln x\ln^2(1+x)\Big]_0^1+2\int_0^1 \frac{\ln x\ln(1+x)}{1+x}\,dx\\
&=\int_0^1\frac{\ln^2 x}{1-x}\,dx+\int_0^1\frac{\ln^2(1+x)}{1+x}\,dx+2U\\
\end{align} On the other hand,
\begin{align}
W&=\int_0^1\frac{\left(\ln x-\ln(1+x)\right)^2}{1+x}\,dx\\
&=\int_0^1\frac{\ln^2 x}{1+x}\,dx+\int_0^1\frac{\ln^2(1+x)}{1+x}\,dx-2U\\
\end{align} Therefore,
\begin{align}
U&=\frac{1}{4}\left(\int_0^1\frac{\ln^2 x}{1+x}\,dx-\int_0^1\frac{\ln^2 x}{1-x}\,dx\right)\\
&=-\frac{1}{4}\int_0^1\frac{2x\ln^2 x}{1-x^2}\,dx
\end{align} Perform the change of variable $y=x^2$,
\begin{align}
U&=-\frac{1}{16}\int_0^1\frac{\ln^2 x}{1-x}\,dx\\
&=-\frac{1}{16}\times 2\zeta(3)\\
&=\boxed{-\frac{1}{8}\zeta(3)}\\
\end{align}
NB:
I assume only that,
\begin{align}\int_0^1\frac{\ln^2 x}{1-x}\,dx=2\zeta(3)\end{align}
$\displaystyle \left(\dfrac{\ln^2 x}{1-x}=\sum_{k=0}^\infty x^k\ln^2 x,x\in ]0;1]\right)$
| {
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"timestamp": "2023-03-29T00:00:00",
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maximum value of $c$
If the circles $x^2+y^2+(3+\sin \beta)x+(2\cos \alpha)y=0$ and $x^2+y^2+(2\cos \beta)x+2cy=0$ touches each other. Then maximum value of $c$ is
Center $\displaystyle C_{1}:\bigg(\frac{-3-\sin \beta}{2},\frac{-2\cos \alpha}{2}\bigg)$ and $\displaystyle r_{1}=\sqrt{\frac{(3+\sin \beta)^2+4\cos^2 \alpha}{4}}$
Center $\displaystyle C_{2}:\bigg(\frac{-2\cos \beta}{2},-\frac{2c}{2}\bigg)$ and $\displaystyle r_{2}=\sqrt{\frac{4\cos^2 \beta+4c^2}{4}}$
if circle touches each other, then distance between center of $2$ circle is sum of radius of circle
$$C_{1}C_{2}=r_{1}+r_{2}$$
$$\sqrt{(3+\sin \beta-2\cos \beta)^2+(2\cos \alpha -2c)^2}=\sqrt{(3+\sin \beta)^2+4\cos^2 \alpha}+\sqrt{4\cos^2 \beta+4c^2}$$
How do i solve it Help me please
| Observe that both circles pass through $(0,0)$, no matter what $\alpha, \beta, c$ are.
So their centers and $(0, 0)$ are on the same line, this implies
$$
\frac{2c}{2\cos\alpha}=\frac{2\cos\beta}{3+\sin\beta},
$$
that is, $c=\frac{2\cos\beta\cos\alpha}{3+\sin\beta}$; at maximum it must be $\cos\alpha=1$ (actually at $\cos\alpha=-1$ there is another maximum but it gives the same value and we can ignore it), and, by a quick calculation using derivative in $\beta$ we see it must be $\sin\beta=-\frac 13$ and $\cos\beta=\frac{\sqrt 8}{3}$ at maximum. So maximum of $c$ is $\frac {\sqrt 2}{2}$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Confusion regarding value of $f(x) = \frac{1}{\tan\left(x\right)}$ at $x=\pi/2$ So I was trying to find what $f(x) = \dfrac{1}{\tan\left(x\right)}$ evaluated at $\dfrac{\pi}{2}$ is and first I thought that it is undefined because $\tan(x)$ is undefined at $\dfrac{\pi}{2}$ but then thinking it through a bit more I tried to evaluate $\lim_{x\to\tfrac{\pi}{2}}\dfrac{1}{\tan(x)}=0$ so I was like aha there is a hole in the graph at $x=\dfrac{\pi}{2}$. Finally I realized that $\dfrac{1}{\tan(x)}$ can be written as $\dfrac{\cos(x)}{\sin(x)}$ and this is continuous at $x=\dfrac{\pi}{2}$. So what is going on? Does this mean that $\dfrac{1}{\tan(x)}\neq\dfrac{\cos(x)}{\sin(x)}$ at least for $x=\dfrac{\pi}{2}$? This mode of thinking would also imply that $\dfrac{1}{\dfrac{1}{x}}\neq x$ at least for $x = 0$.
| The problem is that $$f(x)=\tan x=\frac{\sin x}{\cos x}\;\text{ is indeed undefined at $x=\frac{\pi}{2}$ since $\cos\frac{\pi}{2}=0$}$$
However $$f(x)=\frac{1}{\tan x}=\color{blue}{\frac{\cos x}{\sin x}}$$ is defined at $x=\frac{\pi}{2}$ since the denominator equals $1$. In fact
$$\color{brown}{\frac{1}{\tan \frac{x}{2}}=\frac{\cos \frac{x}{2}}{\sin \frac{x}{2}}=\frac{0}{1}=0}$$
Your conclusion in the last sentence is correct.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3171067",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Coin flips with two fair coins and one biased coin I encountered this problem:
We have three coins, two fair coins and one coin with heads on each
face. One coin is picked randomly among those three coins and is
flipped two times. We see the sequence: H, H. What is the probability
of obtaining heads if we flip this coin one more time?
To tackle it, here's my reasoning:
*
*Find the probabilities of the coin being fair and unfair conditional on the sequence: H, H.
*Find the probability of getting a heads in the additional flip conditional on the sequence: H, H.
1)
Using Bayes formula we have:
\begin{align}
P(fair|HH) &= \frac{P(HH|fair)P(fair)}{P(HH|fair)P(fair)+P(HH|unfair)P(unfair)}
\\ &= \frac{\frac{1}{4}\frac{2}{3}}{\frac{1}{4}\frac{2}{3}+1\cdot\frac{1}{3}}
\\ &= \frac{1/6}{1/6+1/3}
\\ &= \frac{1}{3}
\end{align}
We get at the same time:
\begin{align}
P(unfair|HH) = 1 - 1/3 = 2/3
\end{align}
2) Let's denote $A$ the event of obtaining of heads at the additional flip. We have:
\begin{align}
P(A) = P(A|fair)P(fair) + P(A|unfair)P(unfair)
\end{align}
and conditioning on the sequence $H,H$ we get:
\begin{align}
P(A|HH) &= P(A|fair,HH)P(fair|HH) + P(A|unfair,HH)P(unfair|HH)
\\&= P(A|fair)P(fair|HH) + P(A|unfair) P(unfair|HH)
\end{align}
We then get:
\begin{align}
P(A|HH) &= \frac{1}{2}\frac{1}{3} + 1\cdot\frac{2}{3}
\\ P(A|HH) &= \frac{5}{6}
\end{align}
Is my reasoning correct? Is there maybe a quicker way to get the result?
| Your answer is correct. A shorter way of writing it:
Let $H_n$ denote the event that we get $H$ at the $n$-th coin toss. Hence, we want the probability $$P(H_3 \vert (H_1 \cap H_2))=\frac{P(H_1 \cap H_2 \cap H_3)}{P(H_1 \cap H_2)}=\frac{P(H_1 \cap H_2\cap H_3 \vert fair)\cdot P(fair)+ P(H_1 \cap H_2\cap H_3 \vert unfair)\cdot P(unfair)}{P(H_1 \cap H_2 \vert fair)\cdot P(fair)+ P(H_1 \cap H_2 \vert unfair)\cdot P(unfair)}=\frac{(\frac{1}{2})^3 \cdot \frac{2}{3}+1^3\cdot \frac{1}{3}}{(\frac{1}{2})^2 \cdot \frac{2}{3}+1^2\cdot \frac{1}{3}}=\frac{5}{6}.$$
| {
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"url": "https://math.stackexchange.com/questions/3173658",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Convergent value of $\sum\limits^\infty_{n=1}n\left(\frac{5}{6}\right)^{n-1}$ The sum in question is:
$$\sum^\infty_{n=1}n\left(\frac{5}{6}\right)^{n-1}$$
It passes the ratio test:
\begin{align}
&\lim_{n\rightarrow \infty}\frac{(n+1)\left(\frac{5}{6}\right)^{n}}{n\left(\frac{5}{6}\right)^{n-1}}\\
=\frac{5}{6}&\lim_{n\rightarrow \infty}\frac{(n+1)}{n}\frac{\left(\frac{5}{6}\right)^{n}}{\left(\frac{5}{6}\right)^{n}}\\
=\frac{5}{6}&\lim_{n\rightarrow \infty}(1+ \frac{1}{n})\\
=\frac{5}{6} &< 1\Rightarrow \text{convergent}
\end{align}
But now I do not know how to find the convergent value.
| Recall the geometric sum:
$$\sum_{k=0}^\infty x^n = \frac{1}{1-x}$$
Take the derivative of both sides:
$$\sum_{k=1}^\infty nx^{n-1} = \frac{1}{(1-x)^2}$$
Your sum appears when $x=5/6$. The above sums only hold if $|x|<1$ which does hold here.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3174070",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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How to solve $y'' = y^3 - y$?
Let $y$ be a function of $t$. Solve the following nonlinear ODE
$$y'' = y^3 - y$$ with the following initial conditions $y(0) = 0$ and $y'(0) = 0$.
So far, what I have is solving by integrating factor
\begin{align*}
y'y'' &= y'y^3 - y'y \\
\frac{1}{2}\left[(y')^2\right]' &= \frac{1}{4}(y^4)' - \frac{1}{2}(y^2)' \\
(y')^2 &= \frac{1}{2}y^4 - \frac{1}{2}y^2 + C_1 \\
y' &= \sqrt{\frac{1}{2}y^4 - \frac{1}{2}y^2 + C_1}
\end{align*}
Then do separation of variables and will have
$$\int\left(\frac{1}{2}y^4 - \frac{1}{2}y^2 + C_1\right)^{-1/2}dy = t$$
This is the furthest I can get. I don't see any analytical solution, other than $y(t) = 0$, to this equation.
Did I do anything wrong? Does the closed-form solution exist?
| With these initial conditions there is only the solution $y=0$ allowed. To see this, note (as you have done) that
$$ C= \frac{y'^2}{2} + \frac{y^2}{2} -\frac{y^4}{4}$$
is conserved. Indeed, we have that
$$\frac{d C}{dt} = y' (y'' + y - y^3) = 0\,.$$
From the initial conditions, we obtain that $C=0$. The function $y^2/2-y^4/4$ is monotonously increasing for $|y|\leq 1$. So, if we assume that $1\geq |y|> 0$ at some time $t>0$, we find that
$$ 0 = C = \frac{y'^2}{2} + \frac{y^2}{2} -\frac{y^4}{4} > \frac{y'^2}{2} $$
which is a contradiction as $y'^2\geq 0$. So $y=0$ is the only solution.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Are there any other methods to apply to solving simultaneous equations? We are asked to solve for $x$ and $y$ in the following pair of simultaneous equations:
$$\begin{align}3x+2y&=36 \tag1\\ 5x+4y&=64\tag2\end{align}$$
I can multiply $(1)$ by $2$, yielding $6x + 4y = 72$, and subtracting $(2)$ from this new equation eliminates $4y$ to solve strictly for $x$; i.e. $6x - 5x = 72 - 64 \Rightarrow x = 8$. Substituting $x=8$ into $(2)$ reveals that $y=6$.
I could also subtract $(1)$ from $(2)$ and divide by $2$, yielding $x+y=14$. Let $$\begin{align}3x+3y - y &= 36 \tag{1a}\\ 5x + 5y - y &= 64\tag{1b}\end{align}$$ then expand brackets, and it follows that $42 - y = 36$ and $70 - y = 64$, thus revealing $y=6$ and so $x = 14 - 6 = 8$.
I can even use matrices!
$(1)$ and $(2)$ could be written in matrix form:
$$\begin{align}\begin{bmatrix} 3 &2 \\ 5 &4\end{bmatrix}\begin{bmatrix} x \\ y\end{bmatrix}&=\begin{bmatrix}36 \\ 64\end{bmatrix}\tag3 \\ \begin{bmatrix} x \\ y\end{bmatrix} &= {\begin{bmatrix} 3 &2 \\ 5 &4\end{bmatrix}}^{-1}\begin{bmatrix}36 \\ 64\end{bmatrix} \\ &= \frac{1}{2}\begin{bmatrix}4 &-2 \\ -5 &3\end{bmatrix}\begin{bmatrix}36 \\ 64\end{bmatrix} \\ &=\frac12\begin{bmatrix} 16 \\ 12\end{bmatrix} \\ &= \begin{bmatrix} 8 \\ 6\end{bmatrix} \\ \\ \therefore x&=8 \\ \therefore y&= 6\end{align}$$
Question
Are there any other methods to solve for both $x$ and $y$?
| It is clear that:
*
*$x=10$, $y=3$ is an integer solution of $(1)$.
*$x=12$, $y=1$ is an integer solution of $(2)$.
Then, from the theory of Linear Diophantine equations:
*
*Any integer solution of $(1)$ has the form $x_1=10+2t$, $y_1=3-3t$ with $t$ integer.
*Any integer solution of $(2)$ has the form $x_2=12+4t$, $y_2=1-5t$ with $t$ integer.
Then, the system has an integer solution $(x_0,y_0)$ if and only if there exists an integer $t$ such that
$$10+2t=x_0=12+4t\qquad\text{and}\qquad 3-3t=y_0=1-5t.$$
Solving for $t$ we see that there exists an integer $t$ satisfying both equations, which is $t=-1$. Thus the system has the integer solution
$$x_0=12+4(-1)=8,\; y_0=1-5(-1)=6.$$
Note that we can pick any pair of integer solutions to start with. And the method will give the solution provided that the solution is integer, which is often not the case.
| {
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"timestamp": "2023-03-29T00:00:00",
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How to prove that $\sum\limits_{j = 1}^{n} j(j + 1) = \frac{1}{3}n(n + 1)(n + 2)\ \text{where }\ n \geq 1.$ I would like to know how it is to prove this problem:
$$\sum_{j = 1}^{n} j(j + 1) = \frac{1}{3}n(n + 1)(n + 2) \text{where } n \geq 1 \\
$$
| Observe that $$\sum\limits_{j=1}^{n} j^2 = \frac {1} {6} n(n+1) (2n+1)$$ and $$\sum\limits_{j=1}^{n} j = \frac {n(n+1)} {2}.$$
EDIT $:$
To prove the first sum consider the equality $(t+1)^3 - t^3 = 3t^2 + 3t +1.$ Take $t=1,2, \cdots, n.$ Then we get by summing the equality for $t=1,2, \cdots,n$ $$3\sum\limits_{t=1}^{n} t^2 = \sum\limits_{t=1}^{n} ((t+1)^3 - t^3) - 3 \left ( \sum\limits_{t=1}^{n} t \right ) - n.$$
Can you proceed now?
What is $\sum\limits_{t=1}^{n} ((t+1)^3 - t^3)$?
Observe that $\sum\limits_{t=1}^{n} ((t+1)^3 - t^3) = (n+1)^3 -1.$ Also you know that $\sum\limits_{t=1}^{n} t = \frac {n(n+1)} {2}.$ Now you should plug all these things to get the required answer.
So $$\sum\limits_{t=1}^{n}t^2 = \frac 1 3 \left ((n+1)^3 - 1 - \frac {3n(n+1)} {2} - n \right ).$$
I leave the further simplifications to you.
| {
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Find the remainder when $p(x)$ is divided by $x^2-a^2$ if $p(x)$ leaves remainders $a, -a$ when divided by $x+a, x-a$
Let $a\neq0$ and $p(x)$ be a polynomial of degree greater than $2$. If $p(x)$ leaves remainders $a$ and $-a$ when divided respectively by $x+a$ and $x-a$. Find the remainder when $p(x)$ is divided by $x^2-a^2$
$$
p(x)=q(x).(x+a)+r_1=q(x).(x+a)+a\quad\big[r_1=p(-a)=a\big]\\
p(x)=s(x).(x-a)+r_2=s(x).(x-a)-a\quad\big[r_2=p(a)=-a\big]\\
p(x)=t(x).(x^2-a^2)+r=t(x).(x^2-a^2)+Ax+B\\
p(a)=aA+B=-a\\
p(-a)=-aA+B=a\\
B=0,\;A=-1\implies r=Ax+B=-x
$$
I was wondering Is there another way to solve this problem ?
| Write $$p(x) = k(x)(x^2-a^2)+bx+c$$
for some $b$ and $c$. Since $$p(-a)=a \Longrightarrow a = -ab+c$$ and since $$p(a)=-a \Longrightarrow -a = ab+c$$
Solving thhis system we get $\boxed{r(x) = -x}$.
| {
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Sum of reciprocals of odd numbers that add up to 1 I am trying to find an Egyptian factions expansion of 1: all the odd denominators are distinct ones; $o_k<o_{k+1}$; the last or maximum odd number in the denominator is the product of all other odd denominators.
$$1=\dfrac{1}{o_1}+\dfrac{1}{o_2}+\cdots+\dfrac{1}{o_{n}}+\dfrac{1}{\prod\limits_{{\rm{k}} = 1}^n {{o_k}} }$$
Does this expansion exist? if not, can it be proved? if it exists, can there be any example of such an expansion?
If it allows that $n\to\infty$, then can there exist such an infinite series version expansion?
| Further to the example in the comments, where $o_1=3$ and $o_2=5$, leading to the sum being $\frac 35$, we can generalise the counter example:
Lets work firstly with $2$ numbers, $o_1=2k+1$ and $o_2=2m+1$, where $k<m$. We therefore have the sum
\begin{align}S&=\frac1{2k+1}+\frac1{2m+1}+\frac1{(2k+1)(2m+1)}\\\\
&= \frac{2m+1}{(2k+1)(2m+1)}+\frac{2k+1}{(2m+1)(2k+1)}+\frac1{(2k+1)(2m+1)}\\\\
&=\frac{2m+1+2k+1+1}{(2k+1)(2m+1)}\\\\
&=\frac{2m+2k+3}{4km+2k+2m+1}\end{align}
For this to be equal to $1$ then we must have \begin{align}2m+2k+3 &= 4km+2k+2m+1\\
3&=4km+1\\
4km &= 2\\
km &= \frac 12\end{align}
Give you have specified that $o_1$ and $o_2$ have to be odd numbers, then $k$ and $m$ must be whole numbers and we cannot satisfy the above equation, $km=\frac 12$ and the sum can never be equal to $1$.
In the case that we have $o_i$ for $i=1$ to $n$, we can say that each $o_i$ is equal to $2k_i+1$ where $k_i < k_i+1$
We can write the sum as \begin{align}S&=\frac1{2k_1+1} + \frac1{2k_2+1} + \cdots + \frac 1{2k_n+1} + \frac 1{\Pi_{i=1}^n(2k_i+1)}\\\\
&=\frac{\Pi_{i\neq 1}(2k_i+1)}{\Pi_{i=1}^n(2k_i+1)} + \frac{\Pi_{i\neq 2}(2k_i+1)}{\Pi_{i=1}^n(2k_i+1)}+\cdots+\frac{\Pi_{i\neq n}(2k_i+1)}{\Pi_{i=1}^n(2k_i+1)} + \frac 1{\Pi_{i=1}^n(2k_i+1)}\\\\
&= \frac{\Pi_{i\neq 1}(2k_i+1)+\Pi_{i\neq 2}(2k_i+1)+\cdots +\Pi_{i\neq n}(2k_i+1)+1}{\Pi_{i=1}^n(2k_i+1)}\end{align}
The proof here that the numerator and denominator cannot be equal is quite tricky so I will leave it to you for now and attempt to add a proof when I have the time later
| {
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"url": "https://math.stackexchange.com/questions/3184781",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Very indeterminate form: $\lim_{x \to \infty} \left(\sqrt{x^2+2x+3} -\sqrt{x^2+3}\right)^x \longrightarrow (\infty-\infty)^{\infty}$ Here is problem:
$$\lim_{x \to \infty} \left(\sqrt{x^2+2x+3} -\sqrt{x^2+3}\right)^x$$
The solution I presented in the picture below was made by a Mathematics Teacher
I tried to solve this Limit without using derivative (L'hospital) and Big O notation. Although I get the answer, I don't know if the technique I'm using definitely correct.
And here is my method:
$$\begin{align*}\lim_{x \to \infty} \left(\sqrt{x^2+2x+3} -\sqrt{x^2+3}\right)^x&=\lim_{x \to \infty} \left(\frac {2x}{\sqrt{x^2+2x+3} +\sqrt{x^2+3}}\right)^x\\&=\lim_{x \to \infty}\frac{1}{ \left(\frac {\sqrt{x^2+2x+3} +\sqrt{x^2+3}}{2x}\right)^x}\end{align*}$$
Then, I define a new function here
$$y(x)=\sqrt{x^2+2x+3} +\sqrt{x^2+3}-2x-1$$
We have
$$\begin{align*}
\lim _{x\to\infty} y(x)&=\lim_{x \to \infty}\sqrt{x^2+2x+3} +\sqrt{x^2+3}-2x-1\\
&=\lim_{x \to \infty}(\sqrt{x^2+2x+3}-(x+1))+(\sqrt{x^2+3}-x)\\
&=\lim_{x \to \infty}\frac{2}{\sqrt{x^2+2x+3}+x+1}+ \lim_{x \to \infty}\frac{3}{\sqrt{x^2+3}+x}\\
&=0.
\end{align*}$$
This implies that
$$\lim_{x \to \infty}\frac{2x}{y(x)+1}=\infty $$
Therefore,
$$\begin{align*}
\lim_{x \to \infty}\frac{1}{ \left(\frac {\sqrt{x^2+2x+3} +\sqrt{x^2+3}}{2x}\right)^x}&=\lim_{x \to\infty} \frac{1}{ \left(\frac{y(x)+2x+1}{2x} \right)^x}\\
&=\lim_{x \to\infty} \frac{1}{ \left(1+\frac{y(x)+1}{2x} \right)^x}\\
&=\lim_{x \to \infty}\frac{1}{\left( \left( 1+\frac{1}{\frac{2x}{y(x)+1}}\right)^{\frac{2x}{y(x)+1}}\right)^{\frac{y(x)+1}{2}}}\\
&
\end{align*}$$
Here, we define two functions:
$$f(x)=\left( 1+\frac{1}{\frac{2x}{y(x)+1}}\right)^{\frac{2x}{y(x)+1}},\quad
g(x)=\frac{y(x)+1}{2}.
$$
We deduce that,
$$
\lim_{x\to\infty} f(x)=e>0,\quad \lim_{x\to\infty} g(x)=\frac 12>0.
$$
Thus, the limit $\lim_{x\to\infty} f(x)^{g(x)} $ exists and is finite.
Finally we get,
$$\begin{align*}
\lim_{x \to \infty}\frac{1}{\left( \left( 1+\frac{1}{\frac{2x}{y(x)+1}}\right)^{\frac{2x}{y(x)+1}}\right)^{\frac{y(x)+1}{2}}}
&=\frac{1}{\lim_{x \to \infty}\left( \left( \left( 1+\frac{1}{\frac{2x}{y(x)+1}}\right)^{\frac{2x}{y(x)+1}}\right)^{\frac{y(x)+1}{2}}\right)}\\
&=\frac{1}{\left(\lim_{x\to\infty} \left( 1+\frac{1}{\frac{2x}{y(x)+1}} \right)^{\frac{2x}{y(x)+1}}\right)^{ \lim_{x\to\infty} \frac{y(x)+1}{2}}}\\
&=\frac {1}{e^{\frac12}}=\frac{\sqrt e}{e}.\\&&
\end{align*}$$
Is the method I use correct?
I have received criticisms against my work. What can I do to make the method I use, rigorous? What are the points I missed in the method?
Thank you!
| Your math looks good! I'd maybe just an extra step here and there to make it clear what your doing. Things like showing that you're multiplying by conjugates and maybe a change of variables, say $$z = \frac{2x}{y(x)+1},$$ near the end so it's a bit clearer where the $e$ comes from. Otherwise everything looks good! This is a tricky limit, I really like your solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3185610",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 6,
"answer_id": 0
} |
$\lim\limits_{n \to \infty}\sqrt{1+\sqrt{\frac{1}{2^2}+\sqrt{\frac{1}{3^2}+\cdots+\sqrt{\frac{1}{n^2}}}}}.$ Problem
Evaluate $\lim\limits_{n \to \infty}T_n$ where
$$T_n=\sqrt{1+\sqrt{\frac{1}{2^2}+\sqrt{\frac{1}{3^2}+\cdots+\sqrt{\frac{1}{n^2}}}}}.$$
Analysis
It's obvious that $T_n$ is increasing with a greater $n$, since
\begin{align*}
T_{n+1}&=\sqrt{1+\sqrt{\frac{1}{2^2}+\sqrt{\frac{1}{3^2}+\cdots+\sqrt{\frac{1}{n^2}+\sqrt{\frac{1}{(n+1)^2}}}}}}\\
&>\sqrt{1+\sqrt{\frac{1}{2^2}+\sqrt{\frac{1}{3^2}+\cdots+\sqrt{\frac{1}{n^2}+0}}}}\\
&=T_n.
\end{align*}
Moreover, we can prove that $T_n$ is bounded upward, since
\begin{align*}
T_n&=\sqrt{1+\sqrt{\frac{1}{2^2}+\sqrt{\frac{1}{3^2}+\cdots+\sqrt{\frac{1}{n^2}}}}}\\
&\leq \sqrt{1+\sqrt {1+\sqrt{1+\cdots+\sqrt{1}}}} \\
&\to \frac{\sqrt{5}+1}{2}.
\end{align*}
Therefore, $T_n$ is convergent as $n \to \infty$, by the monotonicity convergence theorem.
But where does it converge to on earth? Does the limit have a excact value? I have already computed the value using the former $20$ terms by Mathematica, it output:
| You can get an upper bound by "freezing" the denominators in the radicands at some value. If you freeze at $n=1$, you get the $(1+\sqrt{5})/2$ bound with which you proved convergence. Now consider a later upper bound in the sequence, say you "freeze" at $n=4$. Thereby
$L<\sqrt{1+\sqrt{\dfrac{1}{4}+\sqrt{\dfrac{1}{9}+\sqrt{\dfrac{1}{16}+\sqrt{\dfrac{1}{16}+\sqrt{\dfrac{1}{16}+...}}}}}}$
We render
$\sqrt{\dfrac{1}{16}+\sqrt{\dfrac{1}{16}+\sqrt{\dfrac{1}{16}+...}}}=\color{blue}{\dfrac{2+\sqrt{5}}{4}}$
by the usual fixed point method. Thereby
$L<(UB)_4=\sqrt{1+\sqrt{\dfrac{1}{4}+\sqrt{\dfrac{1}{9}+\color{blue}{\dfrac{2+\sqrt{5}}{4}}}}}$
Rounding to the next higher multiple of $0.001$:
$L<1.468$
When you "fix" the radicands at increasing values of $n$, the sequence of upper bounds you get will converge to the true limit from above, allowing you to certify arbitrary accuracy with a finite input.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3186189",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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I am getting undefined as the answer of this integral problem $\int\limits_{2}^{3}\frac{\mathrm dn}{(n-2)(3-n)}$. Am I doing something wrong?
Find $$\int\limits_{2}^{3}\frac{\mathrm dn}{(n-2)(3-n)}$$
My Attempt:
Let $$\begin{align}\frac{1}{(n-2)(3-n)}&=\frac{A}{n-2}+\frac{B}{3-n} \\ &= \frac{A(3-n)+B(n-2)}{(n-2)(3-n)}\\ \Rightarrow 1 &= A(3-n)+B(n-2) \\ &= 3A - An+Bn-2B \\ &= n(B-A)+3A-2B.\end{align}$$ Equating coefficients on both sides, we obtain $$B-A=0 \qquad and \qquad 3A-2B=1.$$ $$\therefore A=B=1.$$ $$$$ $$\begin{align}\therefore \int\limits_{2}^{3}\frac{\mathrm dn}{(n-2)(3-n)} &= \int\limits_{2}^{3}\bigg(\frac{1}{n-2}+\frac{1}{3-n}\bigg)\,\mathrm dn \\ &= \int\limits_{2}^{3}\frac{1}{n-2}\,\mathrm dn+\int\limits_{2}^{3}\frac{1}{3-n}\,\mathrm dn \\ &=\left[\log(n-2)\right]_{\small 2}^{\small 3}+\left[\log(3-n)\right]_{\small 2}^{\small 3} \\ &= \left[\log(3-2)-\log(2-2)\right]+\left[\log(3-3)+\log(3-2)\right] \\ &= \left[\log 1-\log 0\right]+\left[\log 0 - \log 1\right]\end{align}$$ but $\log 0$ is undefined, thus my answer is coming undefined. Am I doing something wrong in the solution?
Thank you in advance.
| No, there is nothing wrong. You have\begin{align}\int_2^3\frac1{(x-2)(x-3)}\,\mathrm dx&=\int_2^{5/2}\frac1{(x-2)(x-3)}\,\mathrm dx+\int_{5/2}^3\frac1{(x-2)(x-3)}\,\mathrm dx\\&=\lim_{t\to2^+}\int_t^{5/2}\frac1{(x-2)(x-3)}\,\mathrm dx+\lim_{t\to3^-}\int_{5/2}^t\frac1{(x-2)(x-3)}\,\mathrm dx.\end{align}None of these limits exist, since the function that's being integrated behaves as $\frac1{x-3}$ near $3$ and as $\frac1{x-2}$ near $2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3187072",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Evaluating line integral in first octant. I need to calculate the integral $$\int_{(0,1,0)}^{(1,0,2)} \frac{z}{y}\mathrm{d}x + (x^2+y^2+z^2)\mathrm{d}z$$ along the curve in the first octant given by $x^2+y^2=1, z = 2x$. I used the following parametrization:
$$
\left\{\begin{aligned}
&x=\sin t\\
&y=\cos t\\
& z=2\sin t\end{aligned}\right.,\quad 0 \leq t \leq \frac{\pi}{2}
$$
However, this does not give me the right answer, as this gives the answer $\frac{20}{3}$ and the correct answer is $2+\frac{3\pi}{2}$. I think I am making a mistake in my parametrization.
| Your answer is correct.
\begin{align}&\int_{(0,1,0)}^{(1,0,2)} \frac{z}{y}\mathrm{d}x + (x^2+y^2+z^2)\mathrm{d}z\\
&=\int_0^\frac{\pi}2 2\tan t \cos t+(1+4\sin^2t)2 \cos t \, dt\\
&=\int_0^\frac{\pi}2 2\sin t + 2\cos t(1+2(2\sin^2 t)) \, dt\\
&=\int_0^\frac{\pi}2 2\sin t + 2\cos t(1+2(1-\cos 2t))\, dt\\
&=\int_0^\frac{\pi}2 2\sin t + 6 \cos t -4\cos t\cos 2t \, dt \\
&= \int_0^\frac{\pi}2 2\sin t + 6 \cos t -2\left( \cos 3t+\cos t\right) \, dt \\
&= 8 -2\left( -\frac13+1\right)\\
&=8-\frac{4}3\\
&=\frac{20}3\end{align}
Now, let me try to guess where is the error made in the book, suppose the author forget to compute $\frac{dz}{dt}$, then his working looks as follows:
\begin{align}&\int_{(0,1,0)}^{(1,0,2)} \frac{z}{y}\mathrm{d}x + (x^2+y^2+z^2)\mathrm{d}z\\
&\color{red}=\int_0^\frac{\pi}2 2\tan t \cos t+(1+4\sin^2t) \, dt\\
&=\int_0^\frac{\pi}2 2\sin t + (1+2(2\sin^2 t)) \, dt\\
&=\int_0^\frac{\pi}2 2\sin t + (1+2(1-\cos 2t))\, dt\\
&=\int_0^\frac{\pi}2 2\sin t + 3 -2\cos 2t \, dt \\
&=2+\frac{3\pi}2\end{align}
Note the I purposely made the mistake at the highlighted equation for the second block of computations.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3189648",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Simplifying $\prod_{k=3}^{n-1}\cos\left(\frac{\pi}{k}\right)$ I am looking to simplify the following, without the use of capital Pi notation:
$$\prod_{k=3}^{n-1}\cos\left(\frac{\pi}{k}\right)$$
Which is meant to produce the sequence: $\left[1,\ \frac{1}{2},\ \frac{1}{2}\frac{\sqrt{2}}{2},\frac{1}{2}\frac{\sqrt{2}}{2}\frac{1+\sqrt{5}}{4},\frac{1}{2}\frac{\sqrt{2}}{2}\frac{1+\sqrt{5}}{4}\frac{\sqrt{3}}{2}...\right]$.
I have seen identities of a similar structure, such as: $$\prod_{k=1}^{n-1}\sin\left(\frac{k\pi}{n}\right) = \frac{n}{2^{n-1}},\qquad or\qquad \prod_{k=1}^{n-1}\cos\left(\frac{k\pi}{n}\right) = \frac{\sin(\frac{\pi n}{2})}{2^{n-1}}$$
But, I am versed neither in the proofs of these identities, nor the properties of the $\Pi$ notation, so I've had a lot of difficulty trying to simplify this on my own. Dealing with $k$ in the denominator instead of the numerator (like in the two aforementioned identities) is something that I am evidently unequipped to deal with on my own. Thank you to anyone willing to help me out!
| Well, you could write
$$ \prod_{k=3}^n \cos(\pi/k) = 2^{2-n} \sum_{signs} \cos\left(\pm \frac{\pi}{3} \pm \frac{\pi}{4} \pm \ldots \pm \frac{\pi}{n}\right) $$
where the sum is over all $2^{n-2}$ possible choices of the $\pm$ signs.
If $n$ is moderately large, those $\pm \pi/3 \pm \ldots \pm \pi/n$ will be rather nasty
rational multiples of $\pi$. So it's not exactly a "simplification".
For example, if $n=7$ I get
$$ 16^{-1} \left(\cos \left( {\frac {11\,\pi}{420}} \right) +\cos \left( {\frac {13\,
\pi}{140}} \right) +\cos \left( {\frac {27\,\pi}{140}} \right) +\cos
\left( {\frac {109\,\pi}{420}} \right) \\+\cos \left( {\frac {43\,\pi}{
140}} \right) +\cos \left( {\frac {179\,\pi}{420}} \right) +\cos
\left( {\frac {59\,\pi}{420}} \right) +\cos \left( {\frac {83\,\pi}{
140}} \right) \\+\cos \left( {\frac {57\,\pi}{140}} \right) +\cos
\left( {\frac {199\,\pi}{420}} \right) +\cos \left( {\frac {97\,\pi}{
140}} \right) +\cos \left( {\frac {319\,\pi}{420}} \right)\\ +\cos
\left( {\frac {31\,\pi}{420}} \right) +\cos \left( {\frac {113\,\pi}{
140}} \right) +\cos \left( {\frac {151\,\pi}{420}} \right) +\cos
\left( {\frac {153\,\pi}{140}} \right)
\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3190366",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
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What is the value of $\frac11+\frac13-\frac15-\frac17+\frac19+\frac1{11}-\dots$? The series $\sum_{k=1}^{\infty }\frac{(-1)^{k+1}}{2k-1}=\frac{1}{1}-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\dots$ converges to $\frac{\pi}{4}$. Here, the sign alternates every term.
The series $\displaystyle\sum_{k=1}^{\infty }{(-1)^{\left(k^{2} + k + 2\right)/2} \over 2k-1}=\frac{1}{1}+\frac{1}{3}-\frac{1}{5}-\frac{1}{7}+\dots$ also converges. Here, the sign alternates every two terms.
What is the convergence value, explicitly, of the second series?
The first summation is noted above, because it might be a useful information to evaluate the second summation.
| Hint:
$$\frac{1}{1}+\frac{1}{3}-\frac{1}{5}-\frac{1}{7}+\dots=\sum_{k=1}^{\infty }\frac{1}{8k-7}+\frac{1}{8k-5}-\frac{1}{8k-3}-\frac{1}{8k-1}$$
$$=\sum_{k=1}^{\infty }(\frac{1}{8k-7}-\frac{1}{8k-1})+(\frac{1}{8k-5}-\frac{1}{8k-3})$$
$$=\sum_{k=1}^{\infty }(1+\frac{1}{8k+1}-\frac{1}{8k-1})+(\frac{1}{3}+\frac{1}{8k+3}-\frac{1}{8k-3})$$
$$=\frac{4}{3}+\sum_{k=1}^{\infty }(\frac{-2}{64k^2-1}+\frac{-6}{64k^2-9})$$
$$=\frac{4}{3}-\frac{1}{32}\sum_{k=1}^{\infty }\frac{1}{k^2-\frac{1}{64}}-\frac{3}{32}\sum_{k=1}^{\infty }\frac{1}{k^2-\frac{9}{64}}$$
then use
$$\frac{1-\pi x \cot(\pi x)}{2x^2}=\sum_{k=1}^{\infty }\frac{1}{k^2-x^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3190908",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 0
} |
find the maximum value of $xy + yz +zx$ find the maximum value of $$xy + yz +zx$$given that $x+2y+z=4$
my attempt :
$(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+zx) $
or $2S=2(xy+zx+zy)=(x+y+z)^2 -x^2-y^2-z^2=(4-y)^2-x^2-y^2-z^2$
$2S=-x^2-z^2-8y+16=-x^2-z^2+4x+4z$
from the the above we can say due to symmetry maximum value occurs at $x=z$
hence $S=-x^2+4x$ whose maximum is 4
is this right or/and is there a better way ??
| Using $$\color{red}{ab\leq \frac{1}{4}(a+b)^2}\;\forall\; a,b \in \mathbb{R}$$
Equality hold when $\color{red}{a=b}$
So $$(x+y)(y+z)\leq \frac{1}{4}\bigg[(x+y)+(y+z)\bigg]^2$$
$$xy+yz+zx+y^2\leq 4\Rightarrow \color{Red}{xy+yz+zx\leq 4-y^2\leq 4}$$
equality hold when $y=0$ and $x=z=2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3192014",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Finding real matrices of order $2\times 2$ in matrix equation
Finding all real matrices $X$ of order $2\times 2$ which satisfy the equation $X^2=\begin{pmatrix}
1 & 2 \\
3 & 7
\end{pmatrix}$
My Try: Let $\displaystyle X=\begin{pmatrix} a & b\\
c & d \end{pmatrix}$. Then $\displaystyle X^2=\begin{pmatrix}a^2+bc & b(a+d)\\
c(a+d) & bc+d^2\end{pmatrix}=\begin{pmatrix}1 & 2\\ 3 & 7\end{pmatrix}$
So $a^2+bc=1\cdots (1)$ and $b(a+d)=2\cdots (2)$
And $c(a+d)=3\cdots (3)$ and $bc+d^2=7\cdots (4)$
But this is very complex method
Could some help me to solve it some easy way .Thanks
| The characteristic equation of $X$ is, by definition, $\det (X - t I_2) = 0$. The unknown is $t$, and $I_2 = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$. Since $X$ is square of dimension $2$, this equation will be of degree $2$; more specifically, it will be
$$t^2 - (\operatorname{tr} X) t + \det X = 0 \ .$$
By the Cayley-Hamilton theorem, $X$ too will verify the above equation, therefore
$$X^2 - (\operatorname{tr} X) X + (\det X) I_2 = 0 \ .$$
In this equation (in $X$, this time), we already know $X^2$. On the other hand, we have
$$(\det X)^2 = \det (X^2) = \det \begin{pmatrix} 1 & 2 \\ 3 & 7 \end{pmatrix} = 1 \ ,$$
whence we deduce that $\det X = \pm 1$.
This means that our equation in $X$ becomes
$$\begin{pmatrix} 1 & 2 \\ 3 & 7 \end{pmatrix} \pm \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = (\operatorname{tr} X) X \ .$$
This has two possible cases:
*
*$\det X = -1$, therefore $(\operatorname{tr} X) X = \begin{pmatrix} 0 & 2 \\ 3 & 6 \end{pmatrix}$. Taking the trace, we get $(\operatorname{tr} X) ^2 = 6$, whence $\operatorname{tr} X = \pm \sqrt 6$, so $X = \pm \frac 1 {\sqrt 6} \begin{pmatrix} 0 & 2 \\ 3 & 6 \end{pmatrix}$;
*$\det X = 1$, therefore $(\operatorname{tr} X) X = \begin{pmatrix} 2 & 2 \\ 3 & 8 \end{pmatrix}$. Taking the trace, we get $(\operatorname{tr} X) ^2 = 10$, whence $\operatorname{tr} X = \pm \sqrt {10}$, so $X = \pm \frac 1 {\sqrt {10}} \begin{pmatrix} 2 & 2 \\ 3 & 8 \end{pmatrix}$.
It is now easy to check that all the above 4 matrices are indeed solutions of the given equation. They are also the only ones, by the above analysis.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3192130",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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how many ways can you write a number n as a sum of 1s, 2s and 3s Given $n \in \mathbb{N}$ how many ways can one write $n=a+2b+3c$ for $a,b,c \in \mathbb{N}$.
I have an idea as if I use a a 3-tuple to represent $(a,b,c)$, I can list all of them using two functions $f((a,b,c)) = (a-2,b+1,c)$ and $g((a,b,c)) = (a-1,b-1,c+1)$. Which makes a nice pattern that should be easy to compute for any n however I'm struggling finding an actual equation that works everytime. Right now I have one that works sometimes being; Let $x = \left \lfloor{n/2}\right \rfloor $, $y = \left \lfloor{n/6}\right \rfloor $ and $z = x-y$. Then the function $f(n)$ is the count.
$f(n) = 1+x+(c/2)(c+1)+bc-b(b+1)$. However I know it's wrong.
EDIT:
Sorry forgot to add that f(n) should be a function of just n.
| Let $f(n)$ be the number of ways of making $n$ as a sum of only $1$s, $g(n)$ be the number of ways of making $n$ as a sum of only $1$s and $2$s, and $h(n)$ be the number of ways of making $n$ as a sum of only $1$s, $2$s, and $3$s.
It should be obvious that $f(n) = 1$.
$g(0) = f(0) = 1$ and $g(1) = f(1) = 1$, since there will not be any $2$s for these small numbers. For $n \ge 2$, $g(n) = f(n) + g(n - 2) = 1 + g(n-2)$.
For $n \lt 3$, $h(n) = g(n)$. For $n \ge 3$, $h(n) = g(n) + h(n-3)$.
$$
h(n) - h(n-3) = g(n)\\
g(n) - g(n-2) = 1\\
h(n) - h(n-3) - (h(n-2) - h(n-5)) = 1\\
h(n) - h(n-2) - h(n-3) + h(n-5) = 1\\
h(n-1) - h(n-3) - h(n-4) + h(n-6) = 1\\
h(n) - h(n-1) - h(n-2) - h(n-3) + h(n-3) + h(n-4) + h(n-5) - h(n-6) = 1 - 1 = 0\\
h(n) = h(n-1) + h(n-2) - h(n-4) - h(n-5) + h(n-6)\\
$$
So now we've got a linear recurrence. For the Fibonacci linear recurrence, $F(n) = F(n-1) + F(n-2)$, which has a related polynomial of $x^2 - x - 1$, and that polynomial has roots $\frac{1 \pm \sqrt{5}}{2}$, and the closed form of the Fibonacci series is based on a linear combination of exponentials of those roots, $Fib(n) = k_1 \left(\frac{1 + \sqrt{5}}{2}\right)^n + k_2\left(\frac{1 - \sqrt{5}}{2}\right)^n$.
So the polynomial associated with the $h(n)$ recurrence is $x^6 - x^5 - x^4 + x^2 + x - 1$. That factors to $ (x + \frac12 + \frac{\sqrt{3}}{2}i)(x + \frac12 - \frac{\sqrt{3}}{2}i) (x - 1)^3 (x + 1)$ That translates into a closed form expression of the form $(k_1 + k_2 n + k_3 n^2)(1)^n + k_4(-1)^n + k_5(-\frac12 - \frac{\sqrt{3}}{2}i)^n + k_6(-\frac12 + \frac{\sqrt{3}}{2}i)^n$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3196297",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 1
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Three fractions with a numerator of 1 and denominators of $a, b$, and $c$ added together equals $\frac{6}{7}$. What is $a+b+c?$
If $a, b$ and $c$ are positive integers such that $\frac{1}{a} +
\frac{1}{b} + \frac{1}{c} = \frac{6}{7}$ , then what is $a + b + c?$
I start by adding the LHS together, which results in $\frac{ab+bc+ca}{abc}=\frac{6x}{7x}$. I proceed with trial and error. I know the bottom is a multiple of 7, so thus one of the numbers must be 7. The top likewise has to be a multiple of 6. Taking me ~30 minutes to get to the 40th multiple of 7, doing this results in absolutely no progress. How would I solve this?
Thanks!
Max0815
| As you deduced, one of $a,b,c$ at least must be a multiple of $7$. Similarly, at least one of them must be a multiple of $6$, and the other two multiplied together must produce a multiple of $6$ (we don't end up having to use this information). Suppose $a=7k$. Then $\frac1a\le\frac17\implies \frac1b+\frac1c\ge\frac57\implies$ one of $b$ or $c$ must be $\le2$ (otherwise the sum of the reciprocals has no chance of reaching $\frac57$. But since the overall sum is $<1\implies a,b,c>1$. So wlog let $b=2$. Then, $$\frac1a+\frac1c=\frac5{14}$$
Then by the same argument, $\frac1c\ge\frac3{14}\implies c\le4\implies (c=3)\cup(c=4)$. Try these two, and we see $c=4$ does not work. $c=3$ gives $a=42$ as a solution. $$\frac12+\frac13+\frac1{42}=\frac67\\2+3+42=47$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3196536",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Interesting limit with Poisson and Chi-squared Distribution I am stuck with computing the following limit:
\begin{align}
\lim_{ n \to \infty} E \left[ \left(E \left[ \sqrt{ \frac{X}{n} + \frac{1}{2}} \, \Big | \, U \right] \right)^2 \right].
\end{align}
In the above expression, $X$ given $U$ follows Poisson with parameter $U$ and where $U$ is a Chi-square of degree $n$.
Here is what I tried:
Suppose we let
\begin{align}
V_n =\left(E \left[ \sqrt{ \frac{X}{n} + \frac{1}{2}} \, \Big | \, U \right] \right)^2
\end{align}
Then, using Jensen's inequality
\begin{align}
V_n \le E \left[ \frac{X}{n} + \frac{1}{2} \, \Big | U \, \right] = \frac{U}{n}+\frac{1}{2}
\end{align}
Moreover, we have that $E \left[ \frac{U}{n}+\frac{1}{2} \right]=1+\frac{1}{2}$. Therefore, by the dominated convergence theorem we have that
\begin{align}
\lim_{n \to \infty} E[ V_n ]= E[ \lim_{n \to \infty} V_n ]
\end{align}
Therefore, assuming everything up to here is correct, to compute the limit we have to find
\begin{align}
\lim_{n \to \infty} V_n&= \lim_{n \to \infty} \left(E \left[ \sqrt{ \frac{X}{n} + \frac{1}{2}} \, \Big | \, U \right] \right)^2\\
&= \left( \lim_{n \to \infty} E \left[ \sqrt{ \frac{X}{n} + \frac{1}{2}} \, \Big | \, U \right] \right)^2
\end{align}
This is the place where I am stuck. Is it simply another applications dominated convergence theorem? If so, then I think the answer is
\begin{align}
\lim_{ n \to \infty} E \left[ \left(E \left[ \sqrt{ \frac{X}{n} + \frac{1}{2}} \, \Big | \, U \right] \right)^2 \right]=\frac{1}{2}.
\end{align}
What I mean by another application of dominating convergence theorem is that for every $u>0$
\begin{align}
E \left[ \sqrt{ \frac{X}{n} + \frac{1}{2}} \, \Big | \, U=u \right] &\le \sqrt{ E \left[ \frac{X}{n} + \frac{1}{2} \, \Big | \, U=u \right]}\\
&= \sqrt{ \frac{u}{n} + \frac{1}{2} }\\
&= \sqrt{ u + \frac{1}{2}}
\end{align}
Therefore,
\begin{align}
E \left[ \sqrt{ \frac{X}{n} + \frac{1}{2}} \, \Big | \, U=u \right]= E \left[ \lim_{n \to \infty} \sqrt{ \frac{X}{n} + \frac{1}{2}} \, \Big | \, U=u \right]= \frac{1}{2}.
\end{align}
Is this a correct sequence of steps? I feel a bit uneasy about the second application of the dominated convergence theorem.
| proof: Instead of the heuristic argument I made below you can simply prove it by finding a lower bound, since you already found an upper bound of $3/2$ by Jensen's inequality. Using
$$\sqrt{1+2x}\geq \sqrt{3} \left(1+\frac{x-1}{3}\right) - (x-1)^2$$
$\forall x \geq 0$, which then becomes with $x=\frac{X}{n}$
$$\mathbb{E}\left[ \sqrt{1+\frac{2X}{n}} \right] \geq \mathbb{E}\left[ \sqrt{3} \left(1+\frac{X/n-1}{3}\right) - (X/n-1)^2 \right] \\
=-{\frac {{U}^{2}}{{n}^{2}}}+ \left( -\frac{1}{n^2} + \frac{1}{\sqrt{3}\,n} + \frac{2}{n} \right) U + \frac{2}{\sqrt{3}}-1$$
and so since
$$\mathbb{E}\left(U\right) = n \\ \mathbb{E}\left(U^2\right) = n(n+2)$$
then
$$\lim_{n\rightarrow \infty} \mathbb{E} \left[\mathbb{E}\left[ \sqrt{1+\frac{2X}{n}} \right]\right] \geq \sqrt{3}$$
which means
$$\lim_{n\rightarrow \infty} \mathbb{E} \left[\mathbb{E}\left[ \sqrt{1+\frac{2X}{n}} \right]^2\right] \geq \lim_{n\rightarrow \infty} \mathbb{E} \left[\mathbb{E}\left[ \sqrt{1+\frac{2X}{n}} \right]\right]^2 \geq \sqrt{3}^2 = 3 \tag{FKG}$$
because for any $u$ from the domain of the chi-squared distribution and since $\sqrt{1+2x}$ is increasing we have
$$\frac{{\rm d}}{{\rm d}u} \sum_{k=0}^\infty \frac{u^k {\rm e}^{-u}}{k!}\sqrt{1+2k/n}=\sum_{k=0}^\infty \frac{u^k {\rm e}^{-u}}{k!}\left(\sqrt{1+2(k+1)/n}-\sqrt{1+2k/n}\right)\geq 0$$ and so $\mathbb{E}\left[\sqrt{1+\frac{2X}{n}}\right]$ is an increasing function in $U=u$ and the reqiurements for the FKG inequality are fulfilled. $\square$
Originally I made a too long of a comment for a heuristic argument. Though I know I'm completely ignoring convergence here, when you simply manipulate the corresponding series expressions you obtain the Jensen upper bound result of $3/2$. Starting with
$$
\sqrt{1+\frac{2X}{n}}=\sum_{m=0}^{\infty} \binom{1/2}{m} \left(\frac{2X}{n}\right)^m
$$
and then calculating ${\mathbb{E}(\cdot)}$ with respect to Poisson one obtains
$$
\mathbb{E}\left(\sqrt{1+\frac{2X}{n}}\right)=\sum_{m=0}^{\infty} \binom{1/2}{m} \left(\frac{2}{n}\right)^m \mathbb{E}\left(X^m\right) \\
=\sum_{m=0}^{\infty} \binom{1/2}{m} \left(\frac{2}{n}\right)^m \sum_{i=0}^m {m \brace i} \, U^i \\
=\sum_{i=0}^\infty U^i \sum_{m=i}^{\infty} \binom{1/2}{m} \left(\frac{2}{n}\right)^m {m \brace i} \\
\stackrel{m=i+k}{=}\sum_{i=0}^\infty \left(\frac{2U}{n}\right)^i \sum_{k=0}^{\infty} \binom{1/2}{i+k} \left(\frac{2}{n}\right)^k {i+k \brace i} \, .
$$
The $i$-th moment with respect to the chi-squared distribution of degree $n$ is $$\mathbb{E}\left(U^i\right) = \frac{2^i \Gamma(i+n/2)}{\Gamma(n/2)} \sim n^i \left(1+\frac{i(i-1)}{n} + {\cal O}(1/n^2) \right)$$
which is merely an asymptotic series, but taking only the first term, squaring the previous equation and taking $\mathbb{E}(\cdot)$ we obtain
$$
\left(\sum_{i=0}^\infty 2^i \sum_{k=0}^{\infty} \binom{1/2}{i+k} \left(\frac{2}{n}\right)^k {i+k \brace i}\right)^2 \, .
$$
The terms $k>0$ vanish for $n\rightarrow \infty$, so we are left with
$$
\left( \sum_{i=0}^\infty \binom{1/2}{i} \, 2^i \right)^2 = \sqrt{1+2}^2 = 3
$$
where the series again doesn't converge, but only in the sense of analyticity. Gathering the left out overall factor $1/2$ the final result yields $3/2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3201068",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
} |
prove the equivalence of integral asymptotic $\int_0^{\frac{1}{a}}e^x\arctan(a^3x)dx = \frac{\pi}{2a}+O(\frac{1}{a^2})$, where $a\rightarrow +\infty$
I've tried to integrate by parts and make a change $e^x = 1+x+O(x^2)$ and $arctg(x) = \frac{\pi}{2}-\frac{1}{x}+O(\frac{1}{x^3})$, but didn't succeed. Can you give me a hint, please?
| First, let us replace $a = \frac{1}{b}$ (and $b \to 0$). Then we can split integral in two:
$\int_0^{b^2} e^x \arctan(\frac{x}{b^3})\, dx + \int_{b^2}^b e^x \arctan(\frac{x}{b^3})\, dx$. The function in the first integral is bounded by say $\frac{e \pi}{2}$, so the first integral value is $O(b^2)$.
For the second integral, we can write$\int_{b^2}^b e^x \arctan(\frac{x}{b^3})\, dx = \int_{b^2}^b e^x \frac{\pi}{2}\, dx + \int_{b^2}^b e^x \left(\arctan(\frac{x}{b^3}) - \frac{\pi}{2}\right)\, dx$. The former is $\frac{\pi b}{2}$.
Now we can use your asymptotic (better with explicit constant): $\left|\arctan(\frac{x}{b^3}) - \frac{\pi}{2}\right| < \frac{c b^3}{x}$ for some constant $c$.
$\int_{b^2}^b e^x \left(\arctan(\frac{x}{b^3}) - \frac{\pi}{2}\right)\, dx \leqslant
\int_{b^2}^b e^x \left|\arctan(\frac{x}{b^3}) - \frac{\pi}{2}\right|\, dx \leqslant
\int_{b^2}^b \frac{cb^3 e^x}{x}\, dx$. This can be computed explicitly, or can be upper bounded by $c b^3 \ln b = O(b^2)$ by noting $e^x$ is bounded by $1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3201546",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Combinatorics problem, right solution? We have $6$ lawyers, $7$ engineers and $4$ doctors. We plan on making a committee of $5$ people, and we want at least one person of each profession on board. So for the first place I choose an engineer, for the second a doctor and for the third a lawyer, leaving only $5$ laywers, $6$ engineers and $3$ doctors left.
For the remaining two places, I could have $2$ more people of a single profession. This is $\binom{5}{2}+\binom{6}{2}+\binom{3}{2}$ possibilities.
I could also have two people of different professions; a doctor and a laywer, $\binom{5}{1}\binom{3}{1}$; a doctor and an engineer, $\binom{6}{1}\binom{3}{1}$; or an engineer and a laywer $\binom{6}{1}\binom{5}{1}$.
This adds up to $\binom{5}{2}+\binom{6}{2}+\binom{3}{2}+\binom{5}{1}\binom{3}{1}+\binom{6}{1}\binom{3}{1}+\binom{6}{1}\binom{5}{1}=91$ possible committees.
I have two questions regarding my approach to the problem. Question $a)$ is the reasoning right, am I not overcounting? $b)$ Even if it is right, is there a simpler way to do this? You can see that the sum I end up with, though relatively simple, is quite long and tedious.
Thanks in advance!
| For groups of $6, 7$, and $4$ distinct types of members, would the counting be easier by counting all possible committees while ignoring the constraint, then subtracting the non-valid committees?
$$\binom{6 + 7 + 4}{5} - \left[ \binom{6+7}{5}+\binom{7+4}{5}+ \binom{6+4}{5} \right]$$
Non-valid committees are those which entirely omit one of the types of members.
(Ah: This doesn't work: The subtracted count is too high, as it duplicates non-valid committees. Trying again ...
$$\binom{6+7+4}{5}-
\left[
\left[ \binom{6+7}{5} - \left[\binom{6}{5}+\binom{7}{5}\right] \right] +
\left[ \binom{7+4}{5} - \left[\binom{7}{5}+\binom{4}{5}\right] \right] +
\left[ \binom{6+4}{5} - \left[\binom{6}{5}+\binom{4}{5}\right] \right]
\right] -
\left[ \binom{6}{5} + \binom{7}{5} + \binom{4}{5} \right]$$
The number of committees of five members, minus the number of committees with exactly two types of members, minus the number of committees with exactly one type of member.
For two given types of members, committees with exactly two types of members are all committees of those two types of members, minus committees of only one of the types of members, minus committees of only the other type of members.
To show the complete expression, I've left the terms which evaluate to 0, and haven't simplified.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3202071",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 4
} |
Simplifying $\cos(2\arcsin(x))$ using only pythagorean trigonometric identity I know that one can simplify $\cos(2\arcsin(x))$ using $\cos(a+b)=\cos(a)\cdot\cos(b)-\sin(a)\cdot\sin(b)$:
\begin{alignat}{1}
\cos(2\arcsin(x))&=\cos^2(\arcsin(x))-\sin^2(\arcsin(x))
\\&=1-2\sin^2(\arcsin(x))
\\&=1-2x^2
\end{alignat}
I tried to make this simplification using only $\sin^2(x)+\cos^2(x)=1$:
\begin{alignat}{1}
\cos^2(2\arcsin(x))&=1-\sin^2(2\arcsin(x))
\\ \left|\cos(2\arcsin(x))\right|&=\sqrt{1-\sin^2(2\arcsin(x))}
\\ &=\sqrt{1-\left(2x\sqrt{1-x^2}\right)^2}
\\ &=\sqrt{1-4x^2 |1-x^2|}
\\ &=\sqrt{1-4x^2(1-x^2)}
\\ &=\sqrt{1-4x^2+4x^4}
\\ &=\sqrt{\left(2x^2-1\right)^2}
\\ &=|2x^2-1|
\end{alignat}
And then I could not figure out how to proceed. So, how to get rid of $|\cdot|$ in:
$$\left|\cos(2\arcsin(x))\right|=|2x^2-1|$$
and get $1-2x^2$ ?
Note: this is a part of my attempt to solve the integral $\int x^2\cdot\sqrt{1-x^2}\,\,\mathrm dx$ by trigonometric substitution.
| At this point, we know
$$
\cos(2\sin^{-1}x) = \pm(2x^2-1).
$$
Since both these functions are continuous, we can just check one value in each region where $2x^2-1 > 0$ and $2x^2-1 < 0$ to verify the sign to use in that region. For $|x|<1/\sqrt{2}$, check $x = 0$, and for $|x|>1/\sqrt{2}$, check $x = \pm 1$. You'll find they all work out to $\cos(2\sin^{-1}x) = 1-2x^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3202812",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Completion for this proof that $\sum_{k=0}^n {n \choose k} \cos\left((k+1)x\right) = 2^n \cos^n\frac x 2 \cos\frac{(n+2)x}{2}$ There's this exercise in a textbook:
$$\sum_{k=0}^n {n \choose k} \cos\left((k+1)x\right) = 2^n \cos^n\frac x 2 \cos\frac{(n+2)x}{2}$$
I've been missing from some classes at school and I copied the notes from one of my classmates, but this proof given by my teacher seems incomplete to me, because it only considers certain cases for x. However, I ploted both sides of the equality in the title and it works for every x. So how can this proof be completed?
Proof:
$$ S_1 = \sum_{k=0}^n {n \choose k} \cos\left((k+1)x\right) \\
S_2 = \sum_{k=0}^n {n \choose k} \sin\left((k+1)x\right) \\
S_1 + iS_2 = \sum_{k=0}^n {n \choose k}\cos\left((k+1)x\right) + i\sin\left((k+1)x\right) \\
z = \cos x + i\sin x \\
z^k = \cos kx + i\sin kx \hspace{1in}\text{(Moivre)} \\
\begin{align*} S_1 + iS_2 & = \sum_{k=0}^n {n \choose k}z^{k+1} \\ &= z\sum_{k=0}^n{n \choose k}z^k \\ &= z(1+z)^n \hspace{1in} \text{(Newton)}\end{align*}$$
$$|z| = 1, \varphi_1 = arg(z) = x \\
\begin{align*}
|1+z| &= \sqrt{(1+\cos x)^2 + \sin^2 x} \\ &= \sqrt{1 + 2\cos x + \cos^2 x + \sin^2 x} \\ &= \sqrt{2 + 2\cos x} \\ &= \sqrt{2(1+\cos x)} \\ &= \sqrt{2\cdot 2\cos^2 \frac x 2} \\ &= 2\left|\cos\frac x 2\right| \\ &= \color{red}{2\cos\frac x 2} \end{align*}$$
$$ 1+z = |1+z|\left(\cos \varphi_2 + i\sin\varphi_2\right) \\
\begin{align*}
\color{red}{\varphi_2 } &\color{red}{= \arctan\frac{\sin x}{1 + \cos x}} \\
&=\arctan\frac{2\sin\frac x 2\cos\frac x 2}{2\cos^2\frac x 2} \\ &= \arctan\left(\tan\frac x 2\right) \\
&= \color{red}{\frac x 2}\end{align*}$$
$$ z(1+z)^n = \left(\cos x + i\sin x\right)\left[2\cos\frac x 2\left(\cos\frac x 2 + i\sin\frac x 2\right)\right]^n = 2^n\cos^n\frac x 2 \left[\cos\left(x + \frac{nx}{2}\right) + i\sin\left(x + \frac{nx}{2}\right)\right] = S_1 + iS_2 \\
S_1 = {\rm Re}(S_1 + iS_2) = 2^n\cos^n\frac x 2 \cos \frac {(n+2)x}{2}$$
In red, I put the parts where the teacher only considered certain cases for x.
| Hint
$\binom nk\cos(k+1)x$ is the real part of
$e^{ix}\binom nk(e^{ix})^k$
which is
$=e^{ix}(1+e^{ix})^n$
$=e^{ix(1+n/2)}\left(2\cos \dfrac x2\right)^n$
Using How to prove Euler's formula: $e^{i\varphi}=\cos(\varphi) +i\sin(\varphi)$?
$2\cos y=e^{iy}+e^{-iy}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3205309",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Solve in $C$ : $P(z)=z^4+2z^3+5z^2+4z+1=0$ where $P(i-\frac{1}{2}+i\frac{\sqrt 3}{2})=0$ Question solve in $C$ :
$P(z)=z^4+2z^3+5z^2+4z+1=0$
where $P(i+-\frac{1}{2}+i\frac{\sqrt 3}{2})=0$
My attempt :
Let $\lambda=i-\frac{1}{2}+i\frac{\sqrt 3}{2}$
$(2\lambda+1)^2=(i(2+\sqrt 3))^2$
$(2\lambda+1)^2+7)^2=(-4\sqrt 3)^2$
$({\lambda}^2+\lambda+2)^2=3$
So we find :
$\lambda^2++2\lambda^3+5\lambda^2+4\lambda+1=0$
But which step !? can be find all root of P(z) !!
| Since $P$ has real coefficients, then $z=-\frac12-i-\frac{\sqrt{3}}2i$ is also a solution to $P(z)=0.$ This means that both $z+\frac12-i-\frac{\sqrt{3}}2i$ and $z+\frac12+i+\frac{\sqrt{3}}2i$ are factors of $P(z),$ and so \begin{eqnarray}\left(z+\frac12-i-\frac{\sqrt{3}}2i\right)\left(z+\frac12+i+\frac{\sqrt{3}}2i\right) &=& z^2+z+\left(\frac12-i-\frac{\sqrt{3}}2i\right)\left(\frac12+i+\frac{\sqrt{3}}2i\right)\\ &=& z^2+z+\left(\frac12\right)^2+\left(1+\frac{\sqrt3}2\right)^2\\ &=& z^2+z+\frac14+1+\sqrt3+\frac34\\ &=& z^2+z+2+\sqrt3\end{eqnarray} is also a factor of $P(z).$ Thus, $$z^4+2z^3+5z^2+4z+1=(z^2+z+2+\sqrt3)(az^2+bz+c)$$ for some $a,b,c.$ Expand the right-hand side, and see if you can find $a,b,c.$ From there, you can use the quadratic formula on $az^2+bz+c=0$ to fully solve $P(z)=0.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3205914",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Show $f(x)=\frac{x}{1+x^{2}}$ is lipschitz continuous. I have to show $\frac{x}{1+x^{2}}$ is lipschitz continuous. Hence I have to show
$$|f(x)-f(y)|= \left| \frac{x}{1+x^{2}}- \frac{y}{1+y^{2}} \right| < M|x-y|,$$ for some $M \in \mathbb{R}$. I know I have to use some form of algebraic factorization but I don't see how. Obviously
$$\left| \frac{x}{1+x^{2}}- \frac{y}{1+y^{2}} \right|=\left| \frac{x+xy^{2}-y-yx^{2}}{(1+x^{2})(1+y^{2})} \right|
$$
but how to proceed?
Any tips or hints would be much appreciated.
| We have that
$$\begin{align}\left| \frac{x}{1+x^{2}}- \frac{y}{1+y^{2}} \right|&=\left| \frac{(1-xy)(x-y)}{(1+x^{2})(1+y^{2})} \right|\\&\leq \frac{1+|xy|}{(1+x^{2})(1+y^{2})} \cdot |x-y|\leq |x-y|
\end{align}$$
because
$$1+|xy|\leq 1+\frac{x^2+y^2}{2}\leq 1+x^2+y^2+x^2y^2= (1+x^{2})(1+y^{2}).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3206902",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Maximum area of triangle inscribed in an ellipse If a triangle is inscribed in an ellipse $\frac{x^2}{25}+\frac{y^2}{16}=1$, Find the Maximum area of Triangle
My try:
Let $A(5\cos p, 4\sin p)$, $B(5\cos q, 4\sin q)$ and $C(5\cos r, 4\sin r)$ be vertices of the triangle
Its area is:
$$\Delta=0.5 \times\begin{vmatrix}
5\cos p &4 \sin p &1 \\
5 \cos q& 4 \sin q &1 \\
5 \cos r &4 \sin r & 1
\end{vmatrix}=10\begin{vmatrix}
\cos p & \sin p &1 \\
\cos q & \sin q &1 \\
\cos r&\sin r & 1
\end{vmatrix}$$
$$\Delta=40\sin\left(\frac{p-q}{2}\right)\sin\left(\frac{q-r}{2}\right)\sin\left(\frac{r-p}{2}\right)$$
EDIT:
According to the mind blowing hint given by Mohammad Zuhair khan:
$p \gt q \gt r$
$$p=q+m$$
$$q=r+n$$
where $m,n \gt 0$
So we have
$$p-q=m$$
$$q-r=n$$
$$r-p=-(m+n)$$
Then
$$\Delta=40 \sin\left(\frac{m}{2}\right)\sin\left(\frac{n}{2}\right)
\sin\left(\frac{m+n}{2}\right)$$
Ignoring the negative sign, since its is Area
Now
$$\Delta(m,n)=10(\sin m+\sin n-\sin(m+n))$$
Using Partial differentiation for optimization we have:
$$\frac{\partial \Delta}{\partial m}=0$$
$$\cos m=\cos(m+n)$$
$$m=2\pi-m-n$$
$$2m+n=2\pi$$
Like wise by symmetry:
$$2n+m=2\pi$$
So
$$m=n=\frac{2\pi}{3}$$
Hence
$$\Delta_{max}=10(\frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{2}
+\frac{\sqrt{3}}{2})=15\sqrt{3}$$
But how to know using partial differentiation Maximum occurs?
Can any one give me a link when does partial differentiation applicable for optimization problems?
| If we shrink the ellipse together with an inscribed triangle by a factor of $1/5$ along the $x$-axis and by a factor of $1/4$ along the $y$-axis, the ratio of the area of the triangle to the area of the ellipse remains unchanged. The ellipse is now transformed into the unit circle. It suffices to find the largest triangle inscribed in the unit circle.
Take an arbitrary chord of the unit circle as a side of the triangle. The triangle has the largest possible area if it is the point on the circle which is farthest away from the chord, which is the farther (from the chord) point of intersection of the perpendicular bisector of the chord and the circle. This implies that the triangle must be isosceles.
Let $x$ be the distance of the chord from the centre, then the area $A$ of the triangle is $(1+x)\sqrt{1-x^2}=\sqrt{(1-x)(1+x)^3}$.
$\dfrac{dA^2}{dx}=3(1-x)(1+x)^2-(1+x)^3=(1+x)^2(2-4x)$.
It is easy to check that $A$ is the maximum when $x=1/2$. The maximum area is $\dfrac{3\sqrt{3}}{4}$.
The maximum area of the inscribed triangle of the ellipse is $\displaystyle \dfrac{3\sqrt{3}}{4}\times 5\times 4=15\sqrt{3}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3207133",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
primitives of $f(x)=\frac{1}{2\sqrt{x-x^{2}}}$ I have this function $f:(0,1)\rightarrow \mathbb{R},f(x)=\frac{1}{2\sqrt{x-x^{2}}}$ and I need to find the primitives of $f(x)$
So I calculated $\int \frac{1}{2\sqrt{x-x^{2}}}\,dx$ and I got $\frac{1}{2}\arcsin(2x-1)+C$ but the right answer is $\arcsin\sqrt{x}+C$ .Where's my mistake?How to start?
| The derivative of
$$
f(x)=\frac{1}{2}\arcsin(2x-1)
$$
is
$$
f'(x)=\frac{1}{2}\frac{1}{\sqrt{1-(2x-1)^2}}\cdot2
$$
The square root is
$$
\sqrt{1-4x^2+4x-1}=2\sqrt{x-x^2}
$$
so your result is correct.
Quite likely, the book used the substitution $t=\sqrt{x}$, so the integral becomes
$$
\int\frac{1}{2\sqrt{t^2-t^4}}\cdot2t\,dt=\int\frac{1}{1-t^2}\,dt=\arcsin t+C
$$
There is no problem if you instead complete the square observing that
$$
2\sqrt{x-x^2}=\sqrt{4x-4x^2}=\sqrt{1-(2x-1)^2}
$$
as you probably did.
Just by completeness, we can conclude that, over $(0,1)$,
$$
\frac{1}{2}\arcsin(2x-1)=\arcsin\sqrt{x}+c
$$
for a constant $c$. Evaluating at $x=0$ (it is possible, because the two functions can be extended by continuity to $[0,1]$) we find
$$
-\frac{\pi}{4}=c
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3211213",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
} |
$\sqrt {-i}$ by De Moivre's Theorem $\sqrt{-i} = \left(\cos(\frac{-\pi}{2})+i\sin(\frac{-\pi}{2})\right)^{\frac{1}{2}}$
according to De Moivre's Theorem
$\sqrt{-i} = \left(\cos(\frac{-\pi}{2})+i\sin(\frac{-\pi}{2})\right)^{\frac{1}{2}}$
$\sqrt{-i} = \left(\cos(\frac{-\pi}{4})+i\sin(\frac{-\pi}{4})\right)$
$\sqrt{-i} = \frac{1-i}{\sqrt{2}}$
similarly ,
$\sqrt{-i} = \left(\cos(\frac{3\pi}{2})+i\sin(\frac{3\pi}
{2})\right)^{\frac{1}{2}}$
$\sqrt{-i} = \left(\cos(\frac{3\pi}{4})+i\sin(\frac{3\pi}{4})\right)$
$\sqrt{-i} = \frac{i-1}{\sqrt{2}}$
why there is two different answer ?
| You get two answers because each complex number (except $0$) has two complex square roots. Note that one is the negative of the other. Thus squaring either of them gives the same result. Which one of them deserves to go under the name $\sqrt{-i}$, as which is forced to make do with $-\sqrt{-i}$ is more or less up to you.
That being said, I advise you to get away from using $\sqrt{\phantom 5}$ when dealing with complex numbers as soon as possible. The utility of the root symbol is not worth the confusions it can lead to.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3212615",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
} |
Find the value of $x +y$ If $a=\frac{x}{x^2+y^2}$ and $b=\frac{y}{x^2+y^2}$ then find $x+y$
I find that $x+y/y=\frac{a+b}{b}$ but the ans in the form of a and B only.
| Note that:
$$ax+by=1 \qquad \qquad (1)\\
\frac ax=\frac by=\frac{a+b}{x+y} \qquad \qquad (2)$$
From $(2)$ we also get:
$$x=\frac{a(x+y)}{a+b}; y=\frac{b(x+y)}{a+b} \qquad \qquad (3) $$
Plug $(3)$ into $(1)$:
$$\frac{a^2(x+y)}{a+b}+\frac{b^2(x+y)}{a+b}=1 \Rightarrow x+y=\frac{a+b}{a^2+b^2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3215584",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Using the fact that $\sqrt{n}$ is an irrational number whenever $n$ is not a perfect square, show $\sqrt{3} + \sqrt{7} + \sqrt{21}$ is irrational.
Question:
Using the fact that $\sqrt{n}$ is an irrational number whenever $n$ is not a perfect square, show $\sqrt{3} + \sqrt{7} + \sqrt{21}$ is irrational.
Following from the question, I tried:
Let $N = \sqrt{3} + \sqrt{7} + \sqrt{21}$. Then,
$$
\begin{align}
N+1 &= 1+\sqrt{3} + \sqrt{7} + \sqrt{21}\\
&= 1+\sqrt{3} + \sqrt{7} + \sqrt{3}\sqrt{7}\\
&= (1+\sqrt{3})(1+\sqrt{7}).
\end{align}
$$
Using the above stated fact, $\sqrt{3}$ and $\sqrt{7}$ are irrational. Also, sum of a rational and irrational number is always irrational, so $1+\sqrt{3}$ and $1+\sqrt{7}$ are irrational. Similarly, if we prove that $N+1$ is irrational, $N$ will also be proved to be irrational.
But, how do I prove that product of $1+\sqrt{3}$ and $1+\sqrt{7}$ are irrational.
| Hint $\,\sqrt{21}+\sqrt{7}+\sqrt{3}=q\,\Rightarrow\ \sqrt7(\sqrt 3+1) = q-\sqrt3 \,\Rightarrow\, \sqrt 7\in \Bbb Q(\sqrt3)$, contradiction as below.
Lemma $\rm\ \ [K(\sqrt{a},\sqrt{b}) : K] = 4\ $ if $\rm\ \sqrt{a},\ \sqrt{b},\ \sqrt{a\,b}\ $ all are not in $\rm\,K\,$ and $\rm\, 2 \ne 0\,$ in the field $\rm\,K.$
Proof $\ $ Let $\rm\ L = K(\sqrt{b}).\,$ Then $\rm\, [L:K] = 2\,$ via $\rm\,\sqrt{b} \not\in K,\,$ so it suffices to prove $\rm\, [L(\sqrt{a}):L] = 2.\,$ It fails only if $\rm\,\sqrt{a} \in L = K(\sqrt{b})\, $ and then $\rm\, \sqrt{a}\ =\ r + s\, \sqrt{b}\ $ for $\rm\ r,s\in K.\,$ But that's impossible,
since squaring $\Rightarrow \rm(1)\!:\ \ a\ =\ r^2 + b\ s^2 + 2\,r\,s\ \sqrt{b},\, $ which contradicts hypotheses as follows:
$\rm\qquad\qquad rs \ne 0\ \ \Rightarrow\ \ \sqrt{b}\ \in\ K\ \ $ by solving $(1)$ for $\rm\sqrt{b}\,,\,$ using $\rm\,2 \ne 0$
$\rm\qquad\qquad\ s = 0\ \ \Rightarrow\ \ \ \sqrt{a}\ \in\ K\ \ $ via $\rm\ \sqrt{a}\ =\ r+s\,\sqrt b = r \in K$
$\rm\qquad\qquad\ r = 0\ \ \Rightarrow\ \ \sqrt{a\,b}\in K\ \ $ via $\rm\ \sqrt{a}\ =\ s\, \sqrt{b},\ \ $times $\rm\,\sqrt{b}\quad$
Remark $ $ The Lemma generalizes to any number of sqrts). See the citations there for generalizations to $n$'th roots.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3217250",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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"answer_id": 4
} |
Polynomial equation for $z_{k}=\cos\frac{2k\pi}{5}+i\sin\frac{2k\pi}{5}$ I have $z_{k}=\cos\frac{2k\pi}{5}+i\sin\frac{2k\pi}{5}, k=1,2,3,4$.
I need to find the polynomial equation for the roots $z_k(k=1,2,3,4)$
The right answer is $x^4+x^3+x^2+x+1=0$.I tried to replace k with 1,2,3,4 to find the roots and then to use $a(x-x_1)(x-x_2)(x-x_3)(x-x_4)$ but I didn't get too far.
| Well, with
$z_k = e^{2ki\pi / 5}, \; 1 \le k \le 4, \tag 1$
we have
$z_k^5 = (e^{2ki\pi / 5})^5 = e^{2ki\pi} = 1, \; 1 \le k \le 4; \tag 2$
that is, each $z_k$ satisfies
$x^5 - 1 = (x - 1)(x^4 + x^3 + x^2 + x + 1); \tag 3$
but with
$z_k \ne 1, \; 1 \le k \le 4, \tag 4$
(3) yields
$0 = z_k^5 - 1 = (z_k - 1)(z_k^4 + z_k^3 + z_k^2 + z_k + 1) \Longrightarrow z_k^4 + z_k^3 + z_k^2 + z_k + 1 = 0, \; 1 \le k \le 4; \tag 5$
that is, each $z_k$ satisfies
$x^4 + x^3 + x^2 + x + 1 = 0. \tag 6$
Nota Bene: We observe that this little result generalizes to the case
$z_k = e^{2k i\pi / p}, \; p \in \Bbb P, \; 1 \le k \le p - 1, \tag 7$
for then we have
$z_k^p = (e^{2ki \pi / p})^p = e^{2ki \pi} = 1, \; 1 \le k \le p - 1; \tag 8$
that is, the $z_k$ satisfy the polynomial
$x^p - 1 = (x - 1)(x^{p - 1} + x^{p - 2} + \ldots + x + 1) = (x - 1) \displaystyle \sum_0^{p - 1} x^j; \tag 9$
thus,
$(z_k - 1)\displaystyle \sum_0^{p - 1} z_k^j = 0, \; 1 \le k \le p - 1; \tag{10}$
from (7) we see that
$z_k - 1 \ne 0, \; 1 \le k \le p - 1, \tag{11}$
and hence (10) yields
$\displaystyle \sum_0^{p - 1} z_k^j = 0, \; 1 \le k \le p - 1; \tag{12}$
that is, the $z_k$ all satisfy the polynomial
$x^{p - 1} + x^{p - 2} + \ldots + x + 1 = \displaystyle \sum_0^{p - 1} x^j. \tag{13}$
End of Note.
| {
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"url": "https://math.stackexchange.com/questions/3218644",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
How to simplify $\frac{1+\frac{1}{2^p} + \frac{1}{3^p} + \frac{1}{4^p} + \cdots}{1 - \frac{1}{2^p} + \frac{1}{3^p} - \frac{1}{4^p} + \cdots}$ Let $p$ is real number which satisfies$\quad p > 1$
How can I simplify the fraction
$$\frac{1+\frac{1}{2^p} + \frac{1}{3^p} + \frac{1}{4^p} + \cdots}{1 - \frac{1}{2^p} + \frac{1}{3^p} - \frac{1}{4^p} + \cdots}$$
Numerator is $\zeta(p)$ but I don't know the closed form of denominator.
Is there any idea to simplify this fraction?
| Hint. We have $p>1$. Using the absolute convergence of the series, one may consider
$$
\left(1+\frac{1}{2^p} + \frac{1}{3^p} + \frac{1}{4^p} + \cdots\right)-2\left(\frac{1}{2^p} + \frac{1}{4^p} + \frac{1}{6^p} + \cdots\right)=1 - \frac{1}{2^p} + \frac{1}{3^p} - \frac{1}{4^p} + \cdots
$$Then one may write
$$
\frac{1}{2^p} + \frac{1}{4^p} + \frac{1}{6^p} + \cdots=\frac1{2^p}\left(1+ \frac{1}{2^p} + \frac{1}{3^p} + \cdots\right)=\frac1{2^p}\cdot \zeta(p).
$$
Hope you can take it from here.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3219584",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
Find the sum of the series $\frac{3}{1!+2!+3!}+\frac{4}{2!+3!+4!}+...$upto n terms Find the sum of the series $\frac{3}{1!+2!+3!}+\frac{4}{2!+3!+4!}+...$upto n terms.
Based on my experience I presumed it is a telescopic function but not able to convert it into telescopic function.
| Let $S_n=\sum_{k=1}^n\frac{k+2}{k!+(k+1)!+(k+2)!}$.
$$
\begin{align*}
S_n & =\sum_{k=1}^n\frac{k+2}{k!+(k+1)!+(k+2)!} \\
& =\sum_{k=1}^n\frac{k+2}{k![1+(k+1)+(k+1)(k+2)]} \\
& =\sum_{k=1}^n\frac{k+2}{k![(k+2)+(k+1)(k+2)]} \\
& =\sum_{k=1}^n\frac{k+2}{k!(k+2)^2} \\
& =\sum_{k=1}^n\frac{k+1}{(k+2)!} \\
& =\sum_{k=1}^n\frac{(k+2)-1}{(k+2)!} \\
& =\sum_{k=1}^n\left[\frac{1}{(k+1)!}-\frac{1}{(k+2)!}\right] \\
& =\frac{1}{2!}-\frac{1}{(n+2)!} \\
\end{align*}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3220628",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Suppose that $a=\binom70+\binom73+\binom76,b=\binom71+\binom74+\binom77,c=\binom72+\binom75$. How to algebraically compute $a^3+b^3+c^3-3abc$?
$$
\begin{align}
a = {7 \choose 0}+{7 \choose 3}+{7 \choose 6}\\
b = {7 \choose 1}+{7 \choose 4}+{7 \choose 7}\\
c = {7 \choose 2}+{7 \choose 5}
\end{align}
$$
then $a^3+b^3+c^3-3abc$ is equal to _____.
I tried to write $a^3+b^3+c^3-3abc$ in terms of $a+b+c$ and failed.
$$
\begin{align}
a^3+b^3+c^3-3abc & = (a+b+c)(a^2+b^2+c^2-ab-bc-ca)\\
& = (2^7)((a+b+c)^2-3(ab+bc+ca))\\
& = (2^7)((2^7)^2-3(ab+bc+ca))
\end{align}
$$
I think the expression should be written in terms of another binomial series which I can not think of.
| There is no need to rewrite the expression - you can simply plug in the values of $a,b,c$ directly. Since
$$\binom{n}{k}=\frac{n!}{k!(n-k)!},$$
we get (I'll leave the algebra to you)
$$a=1 + 35 + 7 = 43$$
$$b=7 + 35 + 1 = 43$$
$$c=21 + 21 =42$$
so that $a^{3}+b^{3}+c^{3}-3abc= 128$. Note that the useful identity
$$\binom{n}{k}=\binom{n}{n-k}$$
can be used to reduce the number of computations above e.g. $\binom{7}{3}=\binom{7}{4}$ and $\binom{7}{2}=\binom{7}{5}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3221092",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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Recursion-like sequences which are hard to relate recursively Consider the sequence
\begin{align*}
a_1&=1\\
a_2&=2+\sqrt1\\
a_3&=3+\sqrt{2+\sqrt1}\\
&\kern5.5pt\vdots\\
a_n&=n+\sqrt{n-1+\sqrt{\cdots+\sqrt{1}}}.
\end{align*}
Something like this is easy to work with inductively, since we can simply relate $a_n = n+\sqrt{a_{n-1}}$, and prove things that way. But now consider something which instead unfolds "on the inside", such as
\begin{align*}
b_1&=1\\
b_2&=1+\sqrt2\\
b_3&=1+\sqrt{2+\sqrt{3}}\\
&\kern5.5pt\vdots\\
b_n&=1+\sqrt{2+\sqrt{\cdots+\sqrt{n}}},
\end{align*}
where it is not easy to relate $b_n$ to $b_{n-1}$. How does one work with such sequences, where we are typically interested in the same sorts of questions? Or another example, consider the sequence
\begin{align*}
c_1&=1\\
c_2&=1(1+2)\\
c_3&=1(1+2(1+3))\\
c_4&=1(1+2(1+3(1+4)))\\
&\kern5.5pt\vdots\\
c_n&=1(1+2(1+3(1+\cdots+(n-1)(1+n)\cdots))).
\end{align*}
Is it possible to prove inductively that $c_n = 1!+2!+\cdots+n!$, even though there is no obvious way to relate $c_n$ to $c_{n-1}$?
| $c_n$ is th Hörner factorization of :
$$1! + 2! + 3! + \cdots + n!$$
In fact :
*
* $1! + 2! + 3! = 3! + 2! + 1! = 3 \times 2 \times 1 + 2 \times 1 + 1 = ((3 + 1) \times 2 + 1) \times 1 = c_3$.
* $\begin{array}[t]{lcl}
1! + 2! + 3! + 4! & = & 4! + 3! + 2! + 1! \\[2mm]
& = & 4 \times 3 \times 2 \times 1 + 3 \times 2 \times 1 + 2 \times 1 + 1 \\[2mm]
& = & (((4 + 1) \times 3 + 1) \times 2 + 1) \times 1 = c_4
\end{array}$.
* $\begin{array}[t]{lcl}
1! + 2! + \cdots + n! & = & n! + \cdots + 2! + 1! \\[2mm]
& = & n \times (n - 1) \times \cdots \times 1 + \cdots + 2 \times 1 + 1 \\[2mm]
& = & (\dots(((n + 1) \times (n - 1) + 1) \times \cdots + 1) 2 + 1) \times 1 = c_n
\end{array}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3222089",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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Finding a substitution to an equation $ \frac{(C x)^2}{2} y''(x) + D x (1 - E x) y'(x) - F y(x) = 0 $ to transform it to Kummer's equation It seems that the ordinary differential equation
$$
\frac{(C x)^2}{2} y''(x) + D x (1 - E x) y'(x) - F y(x) = 0
$$
can be presented (suggested by Wolfram alpha) with a certain choice of $a$ and $b$ (and a change of variables) in a form
$$
z y''(x) + (b - x) y'(x) - a y(x) = 0.
$$
The latter equation is known as Kummer's equation and is well studied. However, I have not been able to verify this. I already tried a couple of elementary change of variables ($x^2=z$ for example) but they didn't get me very far.
| $$
\frac{(C x)^2}{2} y''(x) + D x (1 - E x) y'(x) - F y(x) = 0
$$
Change of function :
$$y(x)=x^a u(x)\quad;\quad y'=x^a u'+a x^{a-1}u\quad;\quad y''=x^a u''+2a x^{a-1}u'+a(a-1)x^{a-2}u$$
$\frac{(C x)^2}{2}(x^a u''+2a x^{a-1}u'+a(a-1)x^{a-2}u) + D x (1 - E x)(x^a u'+a x^{a-1}u) - Fx^a u = 0 $
$ \frac{(C x)^2}{2} u''+\left(2a \frac{(C x)^2}{2}x^{-1} + D x (1 - E x) \right)u' +\left(a(a-1)\frac{(C x)^2}{2}x^{-2} +aD x (1 - E x) x^{-1} - F \right)u = 0 $
$ \frac{C^2}{2} x^2u''+\left(a C^2 x + D x (1 - E x) \right)u' +\left(a(a-1)\frac{C^2}{2} +aD (1 - E x) - F \right)u = 0 $
$ \frac{C^2}{2} xu''+\left(a C^2 + D (1 - E x) \right)u' +\left(\frac{a(a-1)\frac{C^2}{2} +aD - F}{x} -aDE\right)u = 0 $
With $a(a-1)\frac{C^2}{2} +aD - F=0$
$$a=\frac{C^2-2D\pm \sqrt{(C^2-2D)^2+8C^2F}}{2C^2}$$
$$ \frac{C^2}{2} xu''+\left(a C^2 + D - DE x \right)u' -aDEu = 0 $$
Change of variable :
$$x=\frac{C^2}{2DE}X$$
$\frac{du}{dx}=\frac{2DE}{C^2}\frac{du}{dX} \quad;\quad \frac{d^2u}{dx^2}=\frac{4D^2E^2}{C^4}\frac{du^2}{dX^2}$
$ \frac{C^2}{2} \frac{C^2}{2DE}X\frac{4D^2E^2}{C^4}\frac{d^2u}{dX^2}+\left(a C^2 + D - DE \frac{C^2}{2DE}X \right)\frac{2DE}{C^2}\frac{du}{dX} -aDEu = 0 $
$X\frac{d^2u}{dX^2}+\left(\frac{2(aC^2+D)}{C^2} - X \right)\frac{du}{dX} -au = 0 $
$$b=\frac{2(aC^2+D)}{C^2}$$
$$X\frac{d^2u}{dX^2}+\left(b - X \right)\frac{du}{dX} -au = 0 $$
This is the Kummer's equation.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Can a certain series of integrals over $[0,\frac{1}{16}]$ be solved using integration-by-parts? I have a series of six integrals. The first, say, is
\begin{equation}
\int_ 0^{\frac {1} {16}} -
48 q^{3/2}\sqrt {18 q - 2\sqrt {17 q + 4}\sqrt {q} + 4} \mbox {d} q.
\end{equation}
Mathematica returns this as unperformed, as well as the subproblem
\begin{equation}
\int_ 0^{\frac {1} {16}} \sqrt {18 q - 2\sqrt {17 q + 4}\sqrt {q} + 4} \mbox {d} q.
\end{equation}
However, high-precision numerical integration shows that the latter integral equals $\frac{25}{204}$.
So, is this last result indicative that the original integral can be solved, in one way of another, by integration-by-parts? (I don't immediately see how.)
The other five integrands in my series--all exhibiting similar numerical phenomenon (upon division by $\frac{25}{204}$, after integration over $[0,\frac{1}{16}]$)--are
\begin{equation}
\left\{-\frac{2352}{5} q^{5/2} \sqrt{18 q-2 \sqrt{17 q+4} \sqrt{q}+4},-\frac{96}{5}
\sqrt{2} \sqrt{q} \sqrt{9 q-\sqrt{17 q+4} \sqrt{q}+2},-144 \sqrt{2} q^{3/2} \sqrt{9
q-\sqrt{17 q+4} \sqrt{q}+2},\frac{96}{5} \sqrt{2} q \sqrt{(-17 q-4) \left(-9 q+\sqrt{q
(17 q+4)}-2\right)},\frac{432}{5} \sqrt{2} q^2 \sqrt{(-17 q-4) \left(-9 q+\sqrt{q (17
q+4)}-2\right)}\right\}.
\end{equation}
Needless to say, I am interested in the evaluation of the six integrals through any appropriate methodology. (Actually, I am specifically interested in the sum of the six integrals, more so than each one itself.)
| Since $q>0$, let's rewrite the integral expression as:
$$
\sqrt{17q+4 - 2\sqrt{17q+4}\sqrt{q} + q} = \sqrt{\left(\sqrt{17q+4}-\sqrt{q}\right)^2}=\sqrt{17q+4}-\sqrt{q}.
$$
This expression is way easier to integrate.
Another problem part in your integrals is actually the same:
$$
9q - \sqrt{17q+4}\sqrt{q} + 2=\frac12\left(\sqrt{17q+4}-\sqrt{q}\right)^2.
$$
I hope you will figure out everything else.
| {
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In each cell of $n$x$n$ is a number so that in every $3$x$3$ subtable the sum of numbers is negative and the sum of all numbers is positive.
In each unit cell of a $n\times n$ table we have a number so that in every $3\times 3$ subtable the sum of numbers is negative and the sum of all numbers is positive.
For which $n\geq 4$ we can have such an arangement?
Clearly we don't have such an arangement if $3\mid n$. Now suppose $3\nmid n$. Then if $n=7$ (or $3$ or $10$ or $13$) we see that a configuration
\begin{array} {|r|r|r|r|r|r|r|}
\hline
7 & -1 & -1& 7&-1&-1&7 \\
\hline
-1 & -1 & -1& -1&-1&-1&-1 \\
\hline
-1 & -1 & -1& -1&-1&-1&-1 \\
\hline
7 & -1 & -1& 7&-1&-1&7 \\
\hline
-1 & -1 & -1& -1&-1&-1&-1 \\
\hline
-1 & -1 & -1& -1&-1&-1&-1 \\
\hline
7 & -1 & -1& 7&-1&-1&7 \\
\hline
\end{array}
works. So I tried to generalise this for an arbitrary $3n+1$, instead of $7$ we put a positive $a$ and instead of $-1$ we put a negative $b$. So I have to prove that there are such $a,b$ that satisfies $a+8b<0$ and $$(n+1)^2a+(8n^2+4n)b>0$$
for an arbitrary $n$, but I fail to do that. Any idea how to solve this.
| For every $n$ the quotient
$$
Q = \frac{8n^2 + 4n}{(n+1)^2} < 8.
$$
That means you can always find $a$ and $b$ such that
$$
Q < \frac{a}{b} < 8.
$$
$Q$ has limit $8$ so if you want integers for $a$ and $b$ they will have to be large when $n$ is large.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3228651",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.