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Finding all natural $n$ such that $2^n+2^{2n} +2^{3n}$ has only $2$ prime factors. Find all natural $n$ such that $2^n+2^{2n} +2^{3n}$ has only $2$ prime factors.
I've tried checking the first 6-7 $n$'s on wolframalpha, but I don't see any patterns for even nor odd $n$'s. At first I thought for all odd $n$'s it was divisible by $2,3,7$, but $n=5$ doesn't work ($n=1,3,7$ do work). How are these $n$ found?
| This is just a sketch of a partial answer...
$$2^n+2^{2n}+2^{3n}=2^n(1+2^n+2^{2n})$$
so you need to check if $1+2^n+2^{2n}$ is of the form $p^k$. First note that, if $n$ is even
$$1+2^n+2^{2n}\equiv 1 + 1 + 1\equiv 0\bmod 3$$
so, if $n$ is even, you need
$$2^n+2^{2n}=3^a-1$$
if $a$ is odd, then $3^a-1\equiv -1-1\equiv -2\bmod 4$, so $a$ is even. Then, by LTE
$$v_2(3^a-1)=v_2(3+1)+v_2(3-1)+v_2(a)-1=2+v_2(a)$$
but also $v_2(2^n+2^{2n})=n$ so $n=2+v_2(a)$, hence $2^{n-2}\vert a$, but then
$$2^n+2^{2n}\leq 3^{2n+1} < 3^{2^{n-2}}$$
for $n\geq 5$. So no solutions for $n$ even greater than $5$.
Moreover, if $3\not\vert n$, then $2^n\equiv 2,4\bmod 7$, so $1+2^n+2^{2n}\equiv 0\bmod 7$. The same reasoning as before says that $1+2^n+2^{2n}$ is not a power of $7$ if $n\geq 5$ and coprime with $3$.
We are left with odd multiples of $3$. We see that for $n=3$ we get $73$, which is prime; repeating the previous reasoning, we exclude the numbers of the form $9k+3$ and $9k+6$, but we have still to check odd multiples of $9$.
This strategy doesn't seem promising, because $9$ is a solution, $27$ is not but in the factorization of the number we obtain with $n=27$ other primes ($4$-digit primes!) show up...
| {
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"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Limit of a Sequence involving 3rd root I'm not finding any way to simplify and solve the following limit:
$$
\lim_{n \to \infty} \sqrt{n^2+n+1}-\sqrt[3]{n^3+n^2+n+1}
$$
I've tried multiplying by the conjugate, but this give a more complex limit.
| Notice the $n^{th}$ term of the sequence can be rewritten as
$$
\sqrt{n^2+n+1} - \sqrt[3]{n^3 + n^2 + n + 1}
= \sqrt{\frac{n^3-1}{n-1}}-\sqrt[3]{\frac{n^4-1}{n-1}}
= n \left[\sqrt{\frac{1-n^{-3}}{1-n^{-1}}} - \sqrt[3]{\frac{1-n^{-4}}{1-n^{-1}}}\right]
$$
When one expand what's in the square/cubic roots in the rightmost expression
as a series in $n^{-1}$, the numerator contributes nothing up to order $O(n^{-3})$.
For the denominators, we have
$$\frac{1}{\sqrt{1-n^{-1}}} \sim 1 + \frac12 n^{-1} + O(n^{-2})
\quad\text{ AND }\quad
\frac{1}{\sqrt[3]{1-n^{-1}}} \sim 1 + \frac13 n^{-1} + O(n^{-2})$$
As a result, the $n^{th}$ term has the dependence:
$$n \left[
\left(1 + \frac12 n^{-1} + O(n^{-2})\right) -
\left(1 + \frac13 n^{-1} + O(n^{-2})\right)
\right]
= \frac{1}{6} + O(n^{-1})
$$
This immediately tells us the limit is $\frac16$ as $n \to \infty$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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"answer_count": 3,
"answer_id": 1
} |
Integral $\int\limits_{\sqrt{2}}^{2}\frac1{t^3\sqrt{t^2-1}}$ This is my very first trigonometric substitution integral, so please bear with me here. I would mostly like verification that my methods are sound.
$$\int\limits_{\sqrt{2}}^{2}\dfrac{1}{t^3\sqrt{t^2-1}}\text{ d}t\text{.}$$
Because of the $\sqrt{t^2-1} = \sqrt{t^2-1^2}$ in the integrand, I set $t = 1\sec(\theta) = \sec(\theta)$, and $\theta = \sec^{-1}(t)$. Thus $\text{d}t = \sec(\theta)\tan(\theta)\text{ d}\theta$ and
$$\begin{align}\int\limits_{\sqrt{2}}^{2}\dfrac{1}{t^3\sqrt{t^2-1}} \text{ d}t&= \int\limits_{\pi/4}^{\pi/3}\dfrac{1}{\sec^{3}(\theta)\sqrt{\sec^{2}(\theta)-1}}\sec(\theta)\tan(\theta)\text{ d}\theta \\
&= \int\limits_{\pi/4}^{\pi/3}\dfrac{1}{\sec^{2}(\theta)}\text{ d}\theta \\
&= \int\limits_{\pi/4}^{\pi/3}\cos^{2}(\theta)\text{ d}\theta \\
&= \dfrac{1}{2}\int\limits_{\pi/4}^{\pi/3}[1+\cos(2\theta)]\text{ d}\theta \\
&= \dfrac{1}{2}\left\{\dfrac{\pi}{3}-\dfrac{\pi}{4}+\dfrac{1}{2}\left[\sin\left(\dfrac{2\pi}{3}\right) - \sin\left(\dfrac{\pi}{2}\right)\right]\right\}\\
&= \dfrac{1}{2}\left[\dfrac{\pi}{12}+\dfrac{1}{2}\left(\dfrac{\sqrt{3}}{2}-1\right)\right] \\
&= \dfrac{\pi}{24}+\dfrac{1}{4}\left(\dfrac{\sqrt{3}-2}{2}\right)\text{.}
\end{align}$$
A quick side question: without access to a calculator and assuming I have the values of $\sin(x)$ memorized for $x = 0, \dfrac{\pi}{6}, \dfrac{\pi}{4}, \dfrac{\pi}{3}, \dfrac{\pi}{2}$, what is the easiest way to find $\sin\left(\dfrac{2\pi}{3}\right)$?
| Your working looks alright.$$\sin \theta= \sin (\pi - \theta)$$
Is another identity.
| {
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"answer_id": 0
} |
How can I determine if the sequence $\frac{n^{2} + 3^{n}}{n^{4} + 2^{n}}$ converges or diverges? Determine whether the sequence $\frac{n^{2} + 3^{n}}{n^{4} + 2^{n}}$ converges. If it converges find the limit and if it diverges determine whether it has an infinite limit.
Proof: let $a_{n} = \frac{n^{2} + 3^{n}}{n^{4} + 2^{n}}$
Divide by $n^{2}$ so you have $\frac{1+\frac{3^n}{n^2}}{\frac{2^{n}}{n^{2}}+n^{2}}$
Therefore it converges to 0 ?
Is that right?
| your sequence is equivalent to $a_n=\frac{\frac{n^2}{2^n}+\left(\frac{3}{2}\right)^n}{1+\frac{n^4}{2^n }}$ this goes to infinity for $n$ tends to infinity
| {
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Show that $F = \{a + b\sqrt{5} | a, b ∈ \mathbb Q\}$ is a field Question:
Show that $F = \{a + b\sqrt{5} | a, b ∈ \mathbb Q\}$ is a field under the
operations - addition and multiplication where addition is given by:
$(a + b\sqrt{5})+(c + d\sqrt{5}) = (a + c) + (b + d)\sqrt{5}$ and
multiplication is given by $(a + b\sqrt{5})(c + d\sqrt{5}) = (ac+5bd)
+ (ad+bc)\sqrt{5}$
My work:
Since $\mathbb Q$ is a field, addition and multiplication defined by $(a + b\sqrt{5})+(c + d\sqrt{5}) = (a + c) + (b + d)\sqrt{5}$ and $(a + b\sqrt{5})(c + d\sqrt{5}) = (ac+5bd) + (ad+bc)\sqrt{5}$ will produce elements in $F$ therefore $F$ is closed under multiplication and addition.
Because $F$ is a subset of $\mathbb R$ the operation on $F$ correspond to the usual operations on $\mathbb R$ so the associative, commutative and distributive conditions are inherited from $\mathbb R$
The additive identity is $0+0\sqrt{5}$ because $(a + b\sqrt{5})+(0+0\sqrt{5}) = (a + b\sqrt{5})$
The multiplicative identity is $1+0\sqrt{5}$ because $(a + b\sqrt{5})(1+0\sqrt{5}) = (a + b\sqrt{5})$
The additive inverse is $((-a) + (-b)\sqrt{5})$ because $(a + b\sqrt{5})+((-a) + (-b)\sqrt{5}) = 0$
The multiplicative inverse is where I went wrong, and not quite sure what to do, I think I have to find something such that $(a + b\sqrt{5})\times? = 1 = 1+0\sqrt{5}$ I thought of just putting $$\frac{1+0\sqrt{5}}{a+b\sqrt(5)}$$ but I was told this wasn't right so not sure what else to do.
If anyone can help me with checking the multiplicative inverse condition that would be really appreciated.
| Every non-zero element of $\mathbb Q[√5]$ is a unit. And other proof you have already shown. That's why we can conclude it's a field.
| {
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"answer_count": 4,
"answer_id": 2
} |
How to evaluate $\int_0^1\frac{\tanh ^{-1}(x)\log(x)}{(1-x) x (x+1)} \operatorname d \!x$? How to evaluate the following integral
$$\int_0^1\frac{\tanh ^{-1}(x)\log(x)}{(1-x) x (x+1)} \operatorname d \!x $$
The numerical result is $= -1.38104$ and when I look at it, I have no idea how to work on it. Could you provide me hits to evaluate the integral above?
Thank you.
| Your integral can be rewritten as
\begin{align}
\int^1_0\frac{{\rm artanh}\ {x}\ln{x}}{x(1-x)(1+x)}{\rm d}x
=\int^1_0\frac{{\rm artanh}\ {x}\ln{x}}{x}{\rm d}x+\int^1_0\frac{x{\rm artanh}\ {x}\ln{x}}{1-x^2}{\rm d}x
\end{align}
The first integral is
\begin{align}
\int^1_0\frac{{\rm artanh}\ {x}\ln{x}}{x}{\rm d}x
=&\chi_2(x)\ln{x}\Bigg{|}^1_0-\int^1_0\frac{\chi_2(x)}{x}{\rm d}x\\
=&-\chi_3(1)=-\frac{7}{8}\zeta(3)
\end{align}
The second integral is
\begin{align}
\int^1_0\frac{x{\rm artanh}\ {x}\ln{x}}{1-x^2}{\rm d}x
=&\sum^\infty_{n=0}\sum^n_{k=0}\frac{1}{2k+1}\int^1_0 x^{2n+2}\ln{x}\ {\rm d}x\\
=&\sum^\infty_{n=0}\frac{\frac{1}{2}H_n-H_{2n+1}}{(2n+3)^2}\\
=&\frac{1}{2}\sum^\infty_{n=1}\frac{H_n}{(2n+1)^2}-\sum^\infty_{n=1}\frac{H_{2n}}{(2n+1)^2}
\end{align}
For the first sum, consider $\displaystyle f(z)=\frac{\left(\gamma+\psi_0(-z)\right)^2}{(2z+1)^2}$. At the positive integers,
\begin{align}
\sum^\infty_{n=1}{\rm Res}(f,n)
=&\sum^\infty_{n=1}\operatorname*{Res}_{z=n}\left[\frac{1}{(2z+1)^2(z-n)^2}+\frac{2H_n}{(2z+1)^2(z-n)}\right]\\
=&2\sum^\infty_{n=1}\frac{H_n}{(2n+1)^2}-\frac{7}{2}\zeta(3)+4
\end{align}
At $z=0$,
\begin{align}
{\rm Res}(f,0)=\operatorname*{Res}_{z=0}\frac{1}{z^2(2z+1)^2}=-4
\end{align}
At $z=-\frac{1}{2}$,
\begin{align}
{\rm Res}\left(f,-\tfrac12\right)=\frac{1}{4}\frac{{\rm d}}{{\rm d}z}(\gamma+\psi_0(-z))^2\Bigg{|}_{z=-\frac{1}{2}}=\frac{\pi^2}{2}\ln{2}
\end{align}
Since the sum of residues is zero,
$$\sum^\infty_{n=1}\frac{H_n}{(2n+1)^2}=\frac{7}{4}\zeta(3)-\frac{\pi^2}{4}\ln{2}$$
For the second sum, consider $\displaystyle f(z)=\frac{\pi\cot(\pi z)(\gamma+\psi_0(-2z))}{(2z+1)^2}$. Note that the poles at the half integers are cancelled by the zeroes of $\pi\cot(\pi z)$. At the positive integers,
\begin{align}
\sum^\infty_{n=1}{\rm Res}(f,n)
=&\sum^\infty_{n=1}\operatorname*{Res}_{z=n}\left[\frac{1}{2(2z+1)^2(z-n)^2}+\frac{H_{2n}}{(2z+1)^2(z-n)}\right]\\
=&\sum^\infty_{n=1}\frac{H_{2n}}{(2n+1)^2}-\frac{7}{4}\zeta(3)+2
\end{align}
At the negative integers,
\begin{align}
\sum^\infty_{n=1}{\rm Res}(f,-n)
=&\sum^\infty_{n=1}\frac{H_{2n+1}}{(2n+1)^2}+1\\
=&\sum^\infty_{n=1}\frac{H_{2n}}{(2n+1)^2}+\frac{7}{8}\zeta(3)
\end{align}
At $z=0$,
\begin{align}
{\rm Res}(f,0)=-2
\end{align}
Since the sum of residues is zero,
$$\sum^\infty_{n=1}\frac{H_{2n}}{(2n+1)^2}=\frac{7}{16}\zeta(3)$$
Hence, if I had not made any mistakes, the original integral is simply
$$\int^1_0\frac{\operatorname{artanh}{x}\ln{x}}{x(1-x)(1+x)}{\rm d}x=-\frac{7}{16}\zeta(3)-\frac{\pi^2}{8}\ln{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/986904",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "5",
"answer_count": 1,
"answer_id": 0
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Simplfy trigonometric functions by only considering integer inputs? I have the below function which only takes integer input,
$$ 2 \sqrt{3} \sin \left(\frac{\pi t}{3}\right)+\sqrt{3} \sin \left(\frac{2 \pi t}{3}\right)-\sqrt{3} \sin \left(\frac{4 \pi t}{3}\right)+6 \cos \left(\frac{\pi t}{3}\right)+\cos \left(\frac{2 \pi t}{3}\right)+\cos \left(\frac{4 \pi t}{3}\right) $$
The resulting output from $t=0$ to $t=10$ is $\{8, 8, -4, -4, -4, -4, 8, 8, -4, -4, -4\}$, and this pattern repeats indefinitely.
Is there a systematic way of reducing this expression using the fact we only consider integer input?
| Notice that
$$\begin{align}\sin\left(\frac{4\pi t}{3}\right) &= \sin\left(\frac{6\pi t}{3} - \frac{2\pi t}{3}\right)\\
&= \sin\left(2\pi t - \frac{2\pi t}{3}\right)\\
&= - \sin \left(\frac{2\pi t}{3}\right)\end{align}$$
For the cosine part,
$$\begin{align}\cos\left(\frac{4\pi t}{3}\right) &= \cos\left(\frac{6\pi t}{3} - \frac{2\pi t}{3}\right)\\
&= \cos\left(2\pi t - \frac{2\pi t}{3}\right)\\
&= \cos \left(\frac{2\pi t}{3}\right)\end{align}$$
If we apply these two identitiesto the given expression, then we have
$$ 2 \sqrt{3} \sin \left(\frac{\pi t}{3}\right)+2\sqrt{3} \sin \left(\frac{2 \pi t}{3}\right)+6 \cos \left(\frac{\pi t}{3}\right)+2\cos \left(\frac{2 \pi t}{3}\right)\\ $$
With the so-called 'R-Formulae', this becomes
$$\sqrt{48}\sin\left(\frac{\pi t}{3} + \arctan{\frac{3}{\sqrt{3}}}\right) + \sqrt{16}\sin\left(\frac{2\pi t}{3} + \arctan{\frac{1}{\sqrt{3}}}\right)\\
= {4\sqrt{3}\sin\left(\frac{\pi t}{3} + \frac{\pi}{3}\right) + 4\sin\left(\frac{2\pi t}{3} + \frac{\pi}{6}\right)}$$
$$= {4\sqrt{3}\sin\left(\frac{(t + 1)\pi t}{3}\right) + 4\sin\left(\frac{(4t + 1)\pi t}{6}\right)}$$
... and this is as far as I got.
I am unsure what you mean by "systematic". If you are referring to a general approach of summing up $\sin$'s and $\cos$'s that does not depend on one's observation skills, then perhaps you can make use of complex numbers.
Sum up the sines by summing up their corresponding complex numbers and taking the imaginary part of the result, and then do the same for the cosines, this time taking the real part of the result.
| {
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"answer_id": 0
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How to show that $\int_0^1 dx \frac{1+x^a}{(1+x)^{a+2}} = \frac{1}{a+1}$? From numerical evidence it appears that whenever the integral converges, $$J_a :=\int_0^1 dx \frac{1+x^a}{(1+x)^{a+2}} = \frac{1}{a+1}.$$
For $a \in \mathbb{N}$, I was able to prove this using induction (see below). How can we prove it for non-integer $a$?
Integrating by parts,
$$\begin{align}
J_n &= \left.-\frac{1}{n+1} \frac{1+x^n}{(1+x)^{n+1}}\right\lvert_0^1+ \frac{n}{n+1} \int_0^1 dx \frac{x^{n-1}}{(1+x)^{n+1}}
\\&=\frac{1}{n+1}\left(1-2^{-n} \right) + \frac{n}{n+1} \int_0^1 dx \left[ \frac{1+x^{n-1}}{(1+x)^{n+1}} - \frac{1}{(1+x)^{n+1}} \right]
\\&=\frac{n}{n+1}J_{n-1} + \frac{1}{n+1} \left[\left(1-2^{-n} \right) -\int_0^1 dx \frac{n}{(1+x)^{n+1}} \right]
\\&=\frac{n}{n+1}J_{n-1}.
\end{align}.
$$
Because $J_0 = 1$, we therefore have $$J_n = \frac{1}{n+1} \mathrm{for\,} n \in \mathbb{N}.$$
| In general, an integral representation of the beta function is $$B(x,y) = \int_{0}^{1} \frac{t^{x-1} + t^{y-1}}{(1+t)^{x+y}} \ dt \ , \ \text{Re} (x), \text{Re}(y) >0 . $$
This can be derived by expressing the more well-known integral representation $$ B(x,y) = \int_{0}^{\infty}\frac{t^{x-1}}{(1+t)^{x+y}} \ dt \tag{1}$$ as
$$\begin{align} B(x,y) &= \frac{1}{2} \int_{0}^{\infty} \frac{t^{x-1}+t^{y-1}}{(1+t)^{x+y}} \ dt \\ &= \frac{1}{2} \int_{0}^{1} \frac{t^{x-1}+t^{y-1}}{(1+t)^{x+y}} \ dt + \frac{1}{2} \int_{1}^{\infty} \frac{t^{x-1}+t^{y-1}}{(1+t)^{x+y}} \ dt \end{align}$$
and then making the substitution $t = \frac{1}{u}$ in the second integral to get that
$$ \begin{align} B(x,y) &= \frac{1}{2} \int_{0}^{1} \frac{t^{x-1}+t^{y-1}}{(1+t)^{x+y}} \ dt + \frac{1}{2} \int_{0}^{1} \frac{u^{y-1}+u^{x-1}}{(u+1)^{x+y}} \ du \\ &= \int_{0}^{1} \frac{t^{x-1}+t^{y-1}}{(1+t)^{x+y}} \ dt. \end{align}$$
Therefore,
$$ \int_{0}^{1} \frac{1+x^{a}}{(1+x)^{a+2}} \ dx = B(1,a+1) = \frac{\Gamma(a+1)}{\Gamma(a+2)} = \frac{1}{a+1}.$$
$ $
$(1)$ http://mathworld.wolfram.com/BetaFunction.html (22)
| {
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"answer_id": 1
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Find the sum of an infinite series of Fibonacci numbers divided by doubling numbers. How would I find the sum of an infinite number of fractions, where there are Fibonacci numbers as the numerators (increasing by one term each time) and numbers (starting at one) which double each time as the denominators?
I'm assuming this has something to do with limits.
| Using Moivre-Binet
$$
F_n = \frac{\phi^n - \psi^n}{\phi - \psi}
$$
with $\phi$ and $\psi$ being the positive and negative root of the golden ratio equation $x^2 - x - 1 = 0$,
we get
\begin{align}
\sum_{n=0}^\infty \frac{F_n}{2^n}
&=
\frac{1}{\phi - \psi} \sum_{n=0}^\infty \frac{\phi^n-\psi^n}{2^n} \\
&=
\frac{1}{\phi - \psi} \sum_{n=0}^\infty \left(\frac{\phi}{2}\right)^n -
\frac{1}{\phi - \psi} \sum_{n=0}^\infty \left(\frac{\psi}{2}\right)^n \\
&=
\frac{1}{\phi - \psi} \frac{1}{1 - \frac{\phi}{2} } -
\frac{1}{\phi - \psi} \frac{1}{1 - \frac{\psi}{2} } \\
&=
\frac{2}{\phi - \psi} \left(
\frac{1}{2 - \phi} - \frac{1}{2 - \psi}
\right) \\
&=
\frac{2}{\phi - \psi} \left(
\frac{(2-\psi) - (2-\phi)}{(2 - \phi)(2-\psi)}
\right) \\
&=
\frac{2}{\phi - \psi} \left(
\frac{\phi - \psi}{4 - 2 \phi - 2 \psi + \phi \psi}
\right) \\
&=
\frac{2}{4 - 2 (\phi + \psi) + \phi \psi} \\
&=
\frac{2}{4 - 2 + (-1)} \\
&= 2
\end{align}
Note: The geometric series converge because $\lvert \phi/2 \rvert < 1$ and $\lvert \psi/2 \rvert < 1$.
Appendix:
$$
x^2 - x - 1 = 0 \iff x = \frac{1 \pm \sqrt{5}}{2}
$$
Thus
$$
\phi + \psi = \frac{1 + \sqrt{5}}{2} + \frac{1 - \sqrt{5}}{2} = 1
$$
and
$$
\phi \psi = \frac{1 + \sqrt{5}}{2} \frac{1 - \sqrt{5}}{2} = \frac{1 - 5}{4} = -1
$$
| {
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"answer_id": 0
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Prove $\int_0^{2\pi}\frac{3a\sin^2\theta}{(1-a\cos \theta)^4}$ or $\int_0^{2\pi}\frac{\cos\theta}{(1-a\cos\theta)^3}=\frac{3a\pi}{(1-a^2)^{5/2}}$ While doing some mathematical modelling of planetary orbits I have come up with two definite integrals $D_1$ and $D_2$ which appear to produce the same result R when tested with various values of $a$ ( where $0<a<1$).
$$
D_1
\, =\,
\int_0^{2\pi}f_1\,\mathrm{d}\theta
\, =\,
\int_0^{2\pi}\frac{3a\sin^2\theta}{(1-a\cos \theta)^4}\mathrm{d}\theta
\,=\,
\frac{3a\pi}{(1-a^2)^{5/2}}
\, =\,R
$$
and
$$D_2\, =\,\int_0^{2\pi}f_2\,\mathrm{d}\theta
\, =\,
\int_0^{2\pi}\frac{\cos \theta}{(1-a\cos \theta)^3}\,\mathrm{d}\theta
\, =\,
\frac{3a\pi}{(1-a^2)^{5/2}}
\, =\,R$$
The hypothesis: $D_1$ = $D_2$ has been proved in a separate question Prove $\int_0^{2\pi}\frac{3a\sin^2\theta}{(1-a\cos \theta)^4}\mathrm{d}\theta = \int_0^{2\pi}\frac{\cos \theta}{(1-a\cos \theta)^3}\,\mathrm{d}\theta$ .
The remaining hypotheses $D_1$ = $R$ and $D_2$ = $R$ have not been proved. So the question is:-
Prove $D_1$ = $R$ or $D_2$ = $R$.
Only one proof is required because the other can then be obtained from $D_1$ = $D_2$.
For information
WolframAlpha computes expressions for the indefinite integrals $I_1,I_2$ as follows:-
$$I_1
\, =\,
\int\frac{3a\sin^2\theta}{(1-a\cos \theta)^4}\mathrm{d}\theta
\,=\,
$$
$$constant1 + \frac
{a\,\sqrt{a^2-1}\sin\theta\,[-(2a^3+a)\cos^2\theta+3(a^2+1)cos\theta+a(2a^2-5)]}
{2(a^2-1)^{5/2}(a\cos\theta-1)^3}
$$
$$-\frac
{6a\,(a\cos\theta-1)^3\,\tanh^-1
\left(
\frac{(a+1)\tan(\theta/2)}{\sqrt{a^2-1}}
\right)
}
{(2(a^2-1)^{5/2}\,(a\cos\theta-1)^3}
$$
and
$$I_2
\, =\,
\int\frac{\cos \theta}{(1-a\cos \theta)^3}\,\mathrm{d}\theta
\, =\,
$$
$$constant2 -
\frac
{2a^2\sin\theta-sin\theta}
{2(a^2-1)^2(a\cos\theta-1)}
-\frac
{\sin\theta}
{2(a^2-1)(a\cos\theta-1)^2}
$$
$$
-\frac
{3a\tanh^-1\left(\frac{(a+1)\tan(\theta/2)}{\sqrt{a^2-1}}\right)}
{(a^2-1)^{5/2}}
$$
Note that the final terms of each expression ( i.e. the terms involving $\tanh^{-1} $ and $\tan$ ) are equivalent to each other.
Also, note that
$$\int\frac{\cos\theta}{(1-a\cos\theta)^3}\,d\theta=
\frac{-\sin\theta}{(1-a\cos\theta)^3}
+\int \frac{3a\sin^2\theta}{(1-a\cos\theta)^4}\,d\theta.
$$
Written with StackEdit.
UPDATE 20141028
I have accepted TenaliRaman's answer. I don't yet understand all the steps but his helpful exposition gives me confidence that with time I can understand it because the methods cited (binomials, series) are ones I have learned (at high school).
The answer of M.Strochyk also appears to give a good proof. But the residue method is too advanced for me to understand at present.
UPDATE 20220713
I have now accepted Quanto's answer (because it is simple enough for me to understand). I have also added an answer based on Quanto's but with the intermediate steps written out.
| According to Brian Bóruma's suggestion, integral can be evaluated using the residue theorem
$$\int\limits_0^{2\pi}\frac{\cos \theta}{(1-a\cos \theta)^3}\,{d\theta}=\dfrac{1}{2}\int\limits_{|z|=1}{\dfrac{z+\frac{1}{z}}{\left[1-\frac{a}{2}\left(z+\frac{1}{z}\right)\right]^3} \frac{dz}{iz}}=\\
=-4i\int\limits_{|z|=1}{\dfrac{z^3+z}{\left(-az^2+2z-a\right)^3}dz}= \\
=-4i\int\limits_{|z|=1}{\dfrac{z^3+z}{-a^3\left(z-z_1\right)^3 \left(z-z_2\right)^3}dz}=-\dfrac{8\pi}{a^3}{\operatorname{Res}\limits_{z=z_1}{\dfrac{z^3+z}{\left(z-z_1\right)^3 \left(z-z_2\right)^3}}},$$
where $z_1=\dfrac{1-\sqrt{1-a^2}}{a}, \;\; z_2=\dfrac{1+\sqrt{1-a^2}}{a}.$ Only the point $z_1$ lies in the unit disc and it is a third-order pole for the integrand function.
The residue can be calculated in a standard way:
$$ \operatorname{Res}\limits_{z=z_1}{\dfrac{z^3+z}{\left(z-z_1\right)^3 \left(z-z_2\right)^3}}=\\= \dfrac{1}{2!} \lim\limits_{z\to z_1}{\dfrac{d^2}{dz^2}\left(\dfrac{\left(z-z_1\right)^3 \left(z^3+z \right)}{\left(z-z_1\right)^3 \left(z-z_2\right)^3} \right)}
=\dfrac{1}{2!} \lim\limits_{z\to z_1}{\dfrac{d^2}{dz^2}\left(\dfrac{z^3+z }{ \left(z-z_2\right)^3} \right)} = \\
=\left. \left[\frac{3 z}{{\left(z - \frac{1+\sqrt{1-a^{2}}} {a}\right)}^{3}} - \frac{3 {\left(3 z^{2} + 1\right)}}{{\left(z - \frac{1+\sqrt{1-a^{2}} }{a}\right)}^{4}} + \frac{6 {\left(z^{3} + z\right)}}{{\left(z - \frac{1+\sqrt{1-a^{2}}}{a}\right)}^{5}}\right] \right|_{z=z_1}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/990813",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 0
} |
If $f(x) = -3x^3 + 2x^2$, find $f(-1)$ and $f(1/2)$. I hope you can help me out
For number 8 and 9 find each value if $f(x) = -3x^3 + 2x^2$
8) $f(-1)$
9) $f(1/2)$
| For x=-1 we write $f(-1)=-3*(-1)^3+2*(-1)^2=3+2=5$
For x=1/2 we write $f\left(\frac{1}{2}\right)=-3\left(\frac{1}{2}\right)^3+2\left(\frac{1}{2}\right)^2=\frac{1}{8}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/991481",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
If I roll three dice at the same time, how many ways the sides can sum up to $13$? If I rolled $3$ dice how many combinations are there that result in sum of dots appeared on those dice be $13$?
| Let (x, y, z) be the numbers showing on the 3 dice.
We want x + y + z = 13.
Assuming the dice are distinguishable, the possibilities are:
(1, 6, 6)
(2, 5, 6), (2, 6, 5)
(3, 4, 6), (3, 5, 5), (3, 6, 4)
(4, 3, 6), (4, 4, 5), (4, 5, 4), (4, 6, 3)
(5, 2, 6), (5, 3, 5), (5, 4, 4), (5, 5, 3), (5, 6, 2)
(6, 1, 6), (6, 2, 5), (6, 3, 4), (6, 4, 3), (6, 5, 2), (6, 6, 1)
So, there are 21 different combinations.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/992125",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 6,
"answer_id": 2
} |
Error in solution for a separable differential equation I "solved" the differential equation $x'=x^2-1$ a couple of months ago, now I checked the solution with wolfram and it seem I was wrong...
According to Wolfram the solution should be $x(t)=\displaystyle\frac{1-e^{2(t+C)}}{1+e^{2(t+C)}}$, but mine is $x(t)=\displaystyle\frac{1+e^{2(t+C)}}{1-e^{2(t+C)}}$.
What I did...
$$\frac{dx}{dt}=x^2-1\\t=\int\frac{1}{x^2-1}dx\\=\frac{1}{2}\int\frac{1}{x-1}dx-\frac{1}{2}\int\frac{1}{x+1}dx\\=\frac{1}{2}(\log(x-1)-\log(x+1))-C\\=\log\sqrt{\frac{x-1}{x+1}}-C\\ \implies \log\sqrt{\frac{x-1}{x+1}}=t+C\\ \implies \sqrt{\frac{x-1}{x+1}}=e^{t+C}\\ \implies \frac{x-1}{x+1}=e^{2(t+C)}\\ \implies x-1=xe^{2(t+C)}+e^{2(t+C)}\\ \implies x(1-e^{2(t+C)})=1+e^{2(t+C)}\\ \implies x=\frac{1+e^{2(t+C)}}{1-e^{2(t+C)}}$$.
I can't spot the error.
| The solutions $x(t)=\displaystyle\frac{1-e^{2(t+C_1)}}{1+e^{2(t+C_1)}}$ and $x(t)=\displaystyle\frac{1+e^{2(t+C_2)}}{1-e^{2(t+C_2)}}$ are the same in fact, but with different $C$. Note that $C$ can be complex. With this form of writing we have the relationship $C_2=C_1+i\pi$
Another equivalent form is $x(t)=\displaystyle\frac{1-c_1e^{2t}}{1+c_1e^{2t}}=\displaystyle\frac{1+c_2e^{2t}}{1-c_2e^{2t}}$ where $c_2=-c_1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/992653",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to prove the lower bound of $\frac{x^2}{\sin^2x}$? How to prove $$1+\frac{x^2}{3}\leq \frac{x^2}{\sin^2x}, x\in (0,\pi/2)?$$
I do want to show it by intermediate value theorem as
$$\frac{1}{\sin^2x}-\frac{1}{x^2}
=\frac{2}{\xi^3}(x-\sin x)>\frac{2(x-\sin x)}{x^3}.$$
However, this may corrupt, since the rhs $<=\frac{1}{3}$.
| You might prefer to prove
$$ 1+\frac{x^2}{3} \le e^{x^2/3} \le \left(\frac{x}{\sin x}\right)^2 $$
For the second inequality, the slickest proof is perhaps
$$ \frac{\sin x}{x} = \prod_{n=1}^\infty \left(1-\frac{x^2}{\pi^2 n^2}\right)
\le \prod_{n=1}^\infty \exp\left(-\frac{x^2}{\pi^2 n^2}\right)
= \exp\left(-\frac{x^2}{\pi^2} \sum_{n=1}^\infty \frac1{n^2}\right)
= e^{-x^2/6}
$$
but this requires a couple big guns in the first and last steps. A more elementary, but tedious, proof can be had by comparing power series:
$$ \frac{\sin x}{x} \le 1 - \frac{x^2}{6} + \frac{x^4}{120}
\le 1 - \frac{x^2}{6} + \frac{x^4}{72} - \frac{x^6}{1296}
\le e^{-x^2/6} $$
The first and last inequalities here come from the usual bounds for alternating series when the omitted terms decrease in absolute value (which they do here if $x$ is not very big); the middle inequality is routine algebra, and is valid for $|x|\le\frac6{\sqrt5}$, which covers the interval you want.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/994322",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Limit with Arctan Here's a hard limit I've been trying to answer for a while :
$$\lim_{x\rightarrow 1} \dfrac{-2x\arctan{x} + \dfrac{\pi}{2}}{x-1}$$
I've tried all the tricks that the teacher has taught us and still nothing, I even tried to factor the top by $(x-1)$ but still nothing. Can I get some help on how to evaluate this limit please?
| \begin{align*}
\frac{\frac\pi2 - 2x\arctan x}{x-1}
&= 2 \frac{\frac\pi4 - x\arctan x}{x-1} \\
&= 2\left(\frac{\frac\pi4 - \arctan x}{x-1} - \arctan x\right) \\
&= 2\left(\frac{\arctan 1 - \arctan x}{x-1} - \arctan x\right) \\
&= 2\left(\frac{\arctan\left(\frac{1-x}{1+x}\right)}{x-1} - \arctan x\right) \\
&= 2\left(\frac{\arctan\left(\frac{1-x}{1+x}\right)}{\left(\frac{1-x}{1+x}\right)}\cdot\frac{-1}{1+x} - \arctan x\right)
\end{align*}
Recall that $\lim_{x\to 0} \frac{\sin x}{x} = 1$; from this it follows that $\lim_{x\to 0} \frac{\tan x}{x} = 1$; from that it follows that $\lim_{x\to 0} \frac{x}{\arctan x} = 1$; and note that $\frac{1-x}{1+x}\to 0$ as $x\to 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/998894",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
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How many distinct solutions are there? Suppose you put the numbers $1,2,\cdots ,10$ in each of the boxes below
such that every connected row and column sum to the same number. How many distinct solutions are there? (By distinct we mean disregarding cyclic permutations, reflections, etc.)
I tried this problem and found only one solution, namely $4,5, 1,10,8,2,3,6,9,7$ when read clockwise starting on the top leftmost square. I tried to prove it's unique but failed (miserably).
From experience, I know algebra isn't the best way to approach these types of problems, due to the number of variables and the symmetry of the equations. When trying to find $3\times3$ magic squares, for example, it'd be best to brute force all the possibilities rather than trying to solve a $9$-variable system.
So does anyone know how to solve this (preferably without guess-and-check)? Any help is welcomed and appreciated.
| There are ten solutions
\begin{align*}
&1,4,5,9,7,3,2,6,8,10 \\
&1,4,7,6,10,2,3,5,8,9 \\
&1,5,6,7,9,3,2,4,10,8 \\
&1,5,7,6,9,4,2,3,10,8 \\
&2,3,6,9,7,4,1,5,10,8 \\
&2,3,8,5,9,4,1,7,6,10 \\
&3,1,10,5,8,6,2,4,7,9 \\
&3,2,5,10,6,4,1,7,8,9 \\
&6,1,5,10,4,8,2,3,9,7 \\
&6,3,5,8,4,10,1,2,9,7
\end{align*}
which turn into $40$ distinct solutions when the numbers in the middles of the upper and/or lower rows are reversed. These are all the solutions, regarding rotations and reflections as the same.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/999217",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
} |
If $\gcd(7,abc)=1$ and $a^2+b^2=c^2$, prove that $7$ divides $a^2-b^2$ The only information I have on this problem is that for $a^2+b^2=c^2$ that
$$
a = st, b = \frac{s^2-t^2}{2}, c = \frac{s^2+t^2}{2}
$$
and that $\gcd(7,abc)=1$ gives $7x + abcy = 1$
I have no idea how to proceed, so any help welcomed
| From what is given, $a,b,c$ are not multiples of $7$. From $(\pm1)^2\equiv 1\pmod 7$, $(\pm2)^2\equiv 4\pmod 7$, $(\pm3)^2\equiv 2\pmod 7$., we see that the only few values for $a^2,b^2,c^2\pmod7$ are possible and the only combinations leading to $a^2+b^2=c^2$ are $1+1\equiv 2$ and $2+2\equiv4$ and $4+4\equiv1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/999854",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
I am having trouble with this integral from the 2012 MIT Integration Bee: $\int\frac{dx}{(1+\sqrt{x})\sqrt{x-x^2}}$ $$\int\frac{dx}{(1+\sqrt{x})\sqrt{x-x^2}} $$
Could someone explain to me how to integrate this integral.
Thank you and cheers.
| $\displaystyle{\int\frac{dx}{(1+\sqrt{x})\sqrt{x-x^2}}}$
Let's put $x=\sin^{2}t$, then
\begin{align*}
\int\frac{dx}{(1+\sqrt{x})\sqrt{x-x^2}}&=\int{\frac{2\sin t\cos t dt}{(1+\sin t)\sqrt{\sin^2 t-\sin^4 t}}} \\
&=\int{\frac{2\sin t\cos t dt}{(1+\sin t)\sqrt{\sin^2 t(1-\sin^2 t)}}}\\
&=\int{\frac{2\sin t\cos t dt}{(1+\sin t)\sin t\cos t}}\\
&=\int{\frac{2dt}{1+\sin t}}\\
&=\int{\frac{2(1-\sin t)dt}{\cos^2 t}}=2\int{\sec^2 tdt}-2\int{\sec t\tan tdt}\\
&=2\tan t - 2\sec t + C\\
&=\frac{2(\sin t - 1)}{\cos t} + C\\
&=\frac{2(\sqrt{x}-1)}{\sqrt{1-x}}+C
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1000886",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
} |
How to prove the inequality $ \frac{a}{\sqrt{1+a}}+\frac{b}{\sqrt{1+b}}+\frac{c}{\sqrt{1+c}} \ge \frac{3\sqrt{2}}{2}$ How to prove the inequality $$ \frac{a}{\sqrt{1+a}}+\frac{b}{\sqrt{1+b}}+\frac{c}{\sqrt{1+c}} \ge \frac{3\sqrt{2}}{2}$$
for $a,b,c>0$ and $abc=1$?
I have tried prove $\frac{a}{\sqrt{1+a}}\ge \frac{3a+1}{4\sqrt{2}}$
Indeed,$\frac{{{a}^{2}}}{1+a}\ge \frac{9{{a}^{2}}+6a+1}{32}$
$\Leftrightarrow 32{{a}^{2}}\ge 9{{a}^{2}}+6a+1+9{{a}^{3}}+6{{a}^{2}}+a$
$\Leftrightarrow 9{{a}^{3}}-17{{a}^{2}}+7a+1\le 0$
$\Leftrightarrow 9{{\left( a-1 \right)}^{2}}\left( a+\frac{1}{9} \right)\le 0$ (!)
It is wrong. Advice on solving this problem.
| Let$$f(a,b,c)=\frac{a}{\sqrt{1+a}}+\frac{b}{\sqrt{1+b}}+\frac{c}{\sqrt{1+c}}$$
$$g(a,b,c)=abc-1=0$$
Using Lagrange Multiplier
$$\large\frac{\frac{\partial f}{\partial a}}{\frac{\partial g}{\partial a}}=
\frac{\frac{\partial f}{\partial b}}{\frac{\partial g}{\partial b}}=
\frac{\frac{\partial f}{\partial c}}{\frac{\partial g}{\partial c}}=k
$$
We get $$\frac{a+2}{2bc(a+1)^{3/2}}=\frac{b+2}{2ac(b+1)^{3/2}}=\frac{c+2}{2ab(c+1)^{3/2}}=k$$
by solving this for $a,b,c$ we get $$a=b=c$$ and from constraint $g(a,b,c)=0$ we get $$a=b=c=1$$
$$f_{min}=\frac{1}{\sqrt{1+1}}+\frac{1}{\sqrt{1+1}}+\frac{1}{\sqrt{1+1}}=\frac{3}{\sqrt2}=\frac{3\sqrt{2}}{2}$$
$$f(a,b,c)=\frac{a}{\sqrt{1+a}}+\frac{b}{\sqrt{1+b}}+\frac{c}{\sqrt{1+c}}\ge f_{min}$$
$$f(a,b,c)=\frac{a}{\sqrt{1+a}}+\frac{b}{\sqrt{1+b}}+\frac{c}{\sqrt{1+c}}\ge \frac{3\sqrt{2}}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1000997",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 7,
"answer_id": 4
} |
solve quadratic equation I'm trying to solve the following equation $2t^2 + t - 3 = 0$
I start by dividing by 2, $t^2 + \frac {t}{2} - \frac {3}{2} = 0$
Then I solve for t $t = - \frac{ \frac {1}{2} }{2} \binom{+}{-} \sqrt{(\frac {1}{2})^2 + \frac {3}{2}}$
$t = - \frac{1}{4} \binom{+}{-} \sqrt{(\frac {1}{4}) + \frac {6}{4}}$
I calculate $t = - \frac{1}{4} \binom{+}{-} \frac {\sqrt7}{4}$
$t_1 = - \frac{1}{4} + \frac {\sqrt7}{4}$
$t_2 = - \frac{1}{4} - \frac {\sqrt7}{4}$
But according to wolframalpha it's suppose to be
$t_1 = 1$
$t_2 = - \frac {3}{2}$
Can't figure out where did I go wrong in my calculation?
| Well actually you have applied the quadratic formula wrong. The roots of the equation $$ax^2+bx+c=0$$ is given by
$$\alpha, \beta ={-b \pm \sqrt {b^2-4ac} \over 2a}$$
So for the equation $$t^2+ \frac12t - \frac32=0$$
the roots become $$\alpha, \beta={- \frac12 \pm \sqrt { (\frac12)^2+4(\frac32)} \over 2}$$
which gives the roots as $1, -1.5$
But in cases like these, I'd suggest an alternative method.
the equation can be re-written as $$2t^2-2t+3t-3=0$$
which can be written as
$$2t(t-1)+3(t-1)=0$$
Now on taking $t-1$ as the common factor, we can write $$(t-1)(2t+3)=0$$
which gives the desired roots.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1001077",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Blackboard operation $x,y,z\rightarrow x,y,1/(zx+zy)$ The three numbers $2,3,6$ are written on the blackboard. In each move, we can pick any two numbers, say $x,y$, and replace the third number $z$ by $1/(zx+zy)$. Using finitely many operations, is it possible to obtain the three numbers $2,3,4$?
In the first move, we can get $2,3,1/30$, and if we fix $2,3$ again, we get back $2,3,6$. To prove that it's impossible to reach $2,3,4$, we might need to define some potential function. But it's not clear which.
[Source: Russian competition problem]
| The invariance is motivated by $ \frac{ x+y} { z ( x + y) } = \frac{ 1}{z} $.
This suggests that we want $ xz, yz, \frac{1}{z}$ as terms in the invariance. Of course, it needs to be cyclic, so the potential invariance has the form
$$ A ( xy+yz+zx) + B ( \frac{1}{x} + \frac{1}{y} + \frac{1}{z} ) $$
Hint: The invariance is
$$ f(x,y,z) = ( xy+yz+zx) + ( \frac{1}{x} + \frac{1}{y} + \frac{1}{z} ) $$
You can verify this because
$$ f( x,y,\frac{1}{ z (x+y) }) = ( xy + \frac{1}{z} ) + ( \frac{1}{x} + \frac{1}{y} + zx+zzy) $$
Now, verify that $ f(2, 3, 4) \neq f (2,3, 6 ) $.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1001324",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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How prove this $x^3+y^3+z^3+3\ge 2(x^2+y^2+z^2)$ Question:
let $x,y,z>0$ and such $xyz=1$, show that
$$x^3+y^3+z^3+3\ge 2(x^2+y^2+z^2)$$
My idea: use AM-GM inequality
$$x^3+x^3+1\ge 3x^2$$
$$y^3+y^3+1\ge 3y^2$$
$$z^3+z^3+1\ge 3z^2$$
so
$$2(x^3+y^3+z^3)+3\ge 3(x^2+y^2+z^2)$$
But this is not my inequality,so How prove it? I know this condition is very important.but how use this condition? and this inequality is stronger
| I have been trying to find a more aesthetic answer, but until I do, I will post a variational approach.
I have simplified the argument a bit, but I am still looking for a simpler approach.
To minimize $x^3+y^3+z^3-2x^2-2y^2-2z^2$ given $xyz=1$, we need to find $x,y,z$ so that for all variations $\delta x,\delta y,\delta z$ that maintain $xyz=1$; that is,
$$
\frac{\delta x}{x}+\frac{\delta y}{y}+\frac{\delta z}{z}=0\tag{1}
$$
we have $x^3+y^3+z^3-2x^2-2y^2-2z^2$ is stationary:
$$
(3x^2-4x)\,\delta x+(3y^2-4y)\,\delta y+(3z^2-4z)\,\delta z=0\tag{2}
$$
Standard orthogonality arguments imply that there is a $\lambda$ so that $x,y,z$ satisfy
$$
3t^2-4t=\frac\lambda{t}\implies3t^3-4t^2=\lambda\tag{3}
$$
Since $3t^3-4t^2$ decreases on $\left[0,\frac89\right]$ and increases for $t\gt\frac89$, for any value of $\lambda$, there can be at most two positive values for $x,y,z$; thus, two must be the same. Say $y=x$ and $z=x^{-2}$. Both $x$ and $x^{-2}$ must satisfy $(3)$, therefore,
$$
3x^3-4x^2=3x^{-6}-4x^{-4}\tag{4}
$$
Using Sturm's Theorem, we see that $3x^9-4x^8+4x^2-3$ has only one real root; that is $x=1$. Plugging this into the expression to be minimized gives a minimum of $-3$. This means
$$
x^3+y^3+z^3+3\ge2(x^2+y^2+z^2)\tag{5}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1002529",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 4,
"answer_id": 0
} |
Solving $T(n)= 2T(n/2) + \sqrt{n}$ without master theorem (algebraically & recurrence tree) $$T(n)= 2T(n/2) + \sqrt{n}$$
This recurrence was in a stackoverflow question, and I want to solve it without relying on the master method. The solution was given, but wolframAlpha gives a slightly different equation:
$$(1) \quad T(n)= n/2 + (1+\sqrt2)(n-\sqrt{n})$$
My Attempt
This was an image that one of the replier's had:
From the recursion tree, I get that the recursion is $\sqrt{n} + 2\sqrt{n/2} + 4\sqrt{n/4} +...$
Reduced it becomes $\sqrt{n} + \sqrt{2n} + \sqrt{4n} +...$
I'm pretty sure this summation happens log(n) times, which would make the equation be:
$$(2) \quad T(n)= \sum_{k=0}^{log(n)} \sqrt{2^kn}$$
Now I have no idea. I don't know how to solve a sum with log(n) bounds or if this is even set up right. How do I get from #2 to #1?
| There is another closely related recurrence that admits an exact
solution. Suppose we have $T(0)=0$ and for $n\ge 1$ (this gives
$T(1)=1$)
$$T(n) = 2 T(\lfloor n/2 \rfloor) + \lfloor \sqrt{n} \rfloor.$$
These would seem to be reasonable assumptions since the work step of a recursive algorithm involves a discrete number of operations, i.e. an integer.
Furthermore let the base two representation of $n$ be
$$n = \sum_{k=0}^{\lfloor \log_2 n \rfloor} d_k 2^k.$$
Then we can unroll the recurrence to obtain the following exact
formula for $n\ge 1$
$$T(n) = \sum_{j=0}^{\lfloor \log_2 n \rfloor}
2^j\Bigg\lfloor
\sqrt{\sum_{k=j}^{\lfloor \log_2 n \rfloor} d_k 2^{k-j}}
\Bigg\rfloor.$$
Now to get an upper bound consider a string of digits with value one
to obtain
$$T(n) \le \sum_{j=0}^{\lfloor \log_2 n \rfloor}
2^j \sqrt{\sum_{k=j}^{\lfloor \log_2 n \rfloor} 2^{k-j}}
= \sum_{j=0}^{\lfloor \log_2 n \rfloor}
2^j \sqrt{2^{\lfloor \log_2 n \rfloor +1 - j} -1}
\\ < \sum_{j=0}^{\lfloor \log_2 n \rfloor}
\sqrt{2}^{2j} \sqrt{2^{\lfloor \log_2 n \rfloor +1 - j}}
= \sum_{j=0}^{\lfloor \log_2 n \rfloor}
\sqrt{2}^{\lfloor \log_2 n \rfloor +1+j}
\\ = \frac{\sqrt{2}^{\lfloor \log_2 n \rfloor +1}-1}{\sqrt{2}-1}
\sqrt{2}^{\lfloor \log_2 n \rfloor +1}.$$
Note that this bound suffers from a defect namely that when $j$ is
close to $\lfloor \log_2 n \rfloor$ in the second sum the error, which
is comparatively large, gets magnified by a factor of $2^j.$
Nonetheless it suffices for the asymptotics.
This bound is actually attained and cannot be improved upon, just like
the lower bound, which occurs with a one digit followed by zeroes to
give
$$T(n) \ge \sum_{j=0}^{\lfloor \log_2 n \rfloor}
2^j \left(\sqrt{2^{\lfloor \log_2 n \rfloor-j}} -1 \right)
= - \left(2^{\lfloor \log_2 n \rfloor+1} - 1\right) +
\sum_{j=0}^{\lfloor \log_2 n \rfloor} \sqrt{2}^{2j}
\sqrt{2}^{\lfloor \log_2 n \rfloor -j}
\\ = 1 - 2^{\lfloor \log_2 n \rfloor+1}
+ \sum_{j=0}^{\lfloor \log_2 n \rfloor}
\sqrt{2}^{\lfloor \log_2 n \rfloor +j}
\\ = 1 - 2^{\lfloor \log_2 n \rfloor+1}
+ \frac{\sqrt{2}^{\lfloor \log_2 n \rfloor +1}-1}{\sqrt{2}-1}
\sqrt{2}^{\lfloor \log_2 n \rfloor}.$$
To do the asmptotics observe that the dominant term in the upper bound
is in $\sqrt{2}^{2\lfloor \log_2 n \rfloor}$ and in the lower bound it
is also in $\sqrt{2}^{2\lfloor \log_2 n \rfloor}$ and additionally in
$2^{\lfloor \log_2 n \rfloor}.$
Joining the these terms from the upper and the lower bound we obtain
the asymptotics
$$\sqrt{2}^{2\lfloor \log_2 n \rfloor}
\in \Theta\left(2^{\log_2 n}\right)
= \Theta\left(n\right).$$
Observe that there is a lower order term
$$\sqrt{2}^{\lfloor \log_2 n \rfloor}
\in \Theta\left(2^{1/2 \log_2 n}\right)
= \Theta\left(\sqrt{n}\right).$$
The above is in agreement with what the Master theorem would produce.
Remark.
This is how we would produce a better upper bound.
Start from
$$\sum_{j=0}^{\lfloor \log_2 n \rfloor}
2^j \sqrt{2^{\lfloor \log_2 n \rfloor +1 - j} -1}
= \sum_{j=0}^{\lfloor \log_2 n \rfloor}
2^{\lfloor \log_2 n \rfloor -j}
\sqrt{2^{j+1} -1}
\\ = 2^{\lfloor \log_2 n \rfloor}
\sum_{j=0}^{\lfloor \log_2 n \rfloor}
\sqrt{2}^{-2j}
\sqrt{2^{j+1} -1}
= 2^{\lfloor \log_2 n \rfloor}
\sum_{j=0}^{\lfloor \log_2 n \rfloor}
\sqrt{2}^{1-j}
\sqrt{1-2^{-j-1}}
.$$
The remaining sum term is readily seen to converge to a constant,
which is about $4.210094.$
Here is another computation in the same spirit
MSE link.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Is it possible to write a sum as an integral to solve it? I was wondering, for example,
Can:
$$ \sum_{n=1}^{\infty} \frac{1}{(3n-1)(3n+2)}$$
Be written as an Integral? To solve it. I am NOT talking about a method for using tricks with integrals.
But actually writing an integral form. Like
$$\displaystyle \sum_{n=1}^{\infty} \frac{1}{(3n-1)(3n+2)} = \int_{a}^{b} g(x) \space dx$$
What are some general tricks in finding infinite sum series.
| In such cases, the partial fractions of general term (i.e. $n^{th}$ term ) of the infinite-series are very useful.
Given that $$\sum_{n=1}^{\infty}\frac{1}{(3n-1)(3n+2)}=\sum_{n=1}^{\infty} T_{n}$$
Where, $T_{n}$ is the $n^{th}$ term of the given series which can be easily expressed in the partial fractions as follows $$T_{n}=\frac{1}{(3n-1)(3n+2)}$$$$=\frac{1}{3}\left(\frac{1}{3n-1}-\frac{1}{3n+2}\right)$$ Now, we have $$\sum_{n=1}^{\infty}\frac{1}{(3n-1)(3n+2)}$$$$=\frac{1}{3}\sum_{n=1}^{\infty} \left(\frac{1}{3n-1}-\frac{1}{3n+2}\right) $$ $$=\frac{1}{3} \lim_{n\to \infty} \left[\left(\frac{1}{2}-\frac{1}{5}\right)+\left(\frac{1}{5}-\frac{1}{8}\right)+\left(\frac{1}{8}-\frac{1}{11}\right)+\! \cdot \! ........ +\left(\frac{1}{3n-4}-\frac{1}{3n-1}\right)+\left(\frac{1}{3n-1}-\frac{1}{3n+2}\right)\right]$$ $$=\frac{1}{3} \lim_{n\to \infty} \left[\frac{1}{2} -\frac{1}{3n+2}\right]$$ $$=\frac{1}{3} \left[\frac{1}{2} -0\right]$$ $$=\color{blue}{\frac{1}{6}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1004081",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "57",
"answer_count": 7,
"answer_id": 2
} |
$10\sin(x)\cos(x) = 6\cos(x)$ In order to solve
$$10\sin x\cos x = 6\cos x$$
I can suppose: $\cos x\ne0$ and then:
$$10\sin x = 6\implies \sin x = \frac{6}{10}\implies x = \arcsin \frac{3}{5}$$
And then, for the case $\cos x = 0$, we have:
$$x = \frac{\pi}{2} + 2k\pi$$
Is my solution rigth? Because I'm getting another strange things at wolfram alpha
| Clearly either $\sin x = \frac{3}{5}$ or $\cos x = 0$. Now you have to find out all the possible values for $x$.
*
*If $\sin x = \frac{3}{5}$, then $x=\arcsin \frac{3}{5} + 2k\pi$ ... or $x = \pi - \arcsin \frac{3}{5} + 2k\pi$.
*If $\cos x = 0$, then $x = \frac{\pi}{2} + 2k\pi$ ... or $x = -\frac{\pi}{2} + 2k\pi$. This is the same as $x = \frac{\pi}{2} + k\pi$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1004592",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Finding the positive integer numbers to get $\frac{\pi ^2}{9}$ As we know, there are many formulas of $\pi$ , one of them $$\frac{\pi ^2}{6}=\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}......
$$
and this $$\frac{\pi ^2}{8}=\frac{1}{1^2}+\frac{1}{3^2}+\frac{1}{5^2}......$$
Now,find the positive integer numbers $(a_{0}, a_{1}, a_{2}....)$ to get $$\frac{\pi^2 }{9}=\frac{1}{a_{0}^2}+\frac{1}{a_{1}^2}+\frac{1}{a_{2}^2}....$$
| $$\frac{\pi ^2}{9}=\frac{\pi ^2}{6}\times 6 \times \frac{1}{9}=(\frac{1}{3^2}+\frac{1}{6^2}+\frac{1}{9^2}+
\cdots)\times 6$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1007424",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 6,
"answer_id": 1
} |
Recurrence Fibonacci Sequence Proof I'm having troubles proving that in a fibonacci sequence if n is divisible by four, then Fn is divisible by three
So when Fn is 6, n is 8 and so on. I was thinking maybe I could use mod 3 or mod 4 but don't really know what to do with it.
| This follows from the matrix formulation, which is well worth knowing and easily proved:
$$
\begin{pmatrix}1&1\\1&0\end{pmatrix}^n=
\begin{pmatrix}F_{n+1}&F_n\\F_n&F_{n-1}\end{pmatrix}
$$
We will prove by induction that $3$ divides $F_{4k}$.
Let
$
A=\begin{pmatrix}1&1\\1&0\end{pmatrix}
$.
Then $A^{4(k+1)}=A^{4k}A^4$ :
$$
\begin{pmatrix}F_{4(k+1)+1}&F_{4(k+1)}\\F_{4(k+1)}&F_{4(k+1)-1}\end{pmatrix}
=
\begin{pmatrix}F_{4k+1}&F_{4k}\\F_{4k}&F_{4k-1}\end{pmatrix}
\begin{pmatrix}F_{5}&F_4\\F_4&F_{3}\end{pmatrix}
$$
and so, by looking at position $2,1$,
$$
F_{4(k+1)}=F_{4k}F_{5}+F_{4k-1}F_4
$$
By induction, $F_4=3$ divides $F_{4k}$ and so $F_4=3$ divides $F_{4(k+1)}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1008144",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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Question about $\frac {\Gamma'(z+1)}{\Gamma(z+1)}$ If $\psi (z)= \log\Gamma(z+1)$
Prove that : $$\psi(n)+\gamma=1+\frac{1}{2}+\cdots+\frac{1}{n}$$
My Proof :
$$\psi (z)= \frac {\Gamma'(z+1)}{\Gamma(z+1)}=-\frac{1}{z+1}-\gamma + \sum_{n=1}^\infty \frac{1}{n} - \frac{1}{n+z+1} $$
$$= \sum_{n=0}^{\infty}\frac{1}{n+1}\frac{1}{n+z+1} -\gamma \ $$
at $z=n$
$$ \psi(n)+\gamma = \sum_{n=0}^{\infty} \frac{1}{n+1}-\frac{1}{2n+1} $$
$$ \psi(n)+\gamma =(1-1)+\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{5}\right)+\left(\frac{1}{4}-\frac{1}{7}\right)+\cdots$$
$$=\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\cdots+\frac{1}{2n}+\cdots $$
$$\psi(n)+\gamma=\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\cdots+\frac{1}{2n}+\cdots $$
where is the error in my proof ?
| One error:
From
$\psi(z) = \sum_{n=0}^{\infty}\left(\frac{1}{n+1}-\frac{1}{n+z+1}\right) -\gamma $
you set $z = n$
to get
$\psi(n)+\gamma = \sum_{n=0}^{\infty} \left(\frac{1}{n+1}-\frac{1}{2n+1}\right)$,
You use $n$ both as an argument of $\psi$ and as an index of summation.
You can't do that.
(added later)
From this,
if you set
$z = m$
where $m$ is a positive integer,
$\psi(m) +\gamma
= \sum_{n=0}^{\infty}(\frac{1}{n+1}-\frac{1}{n+m+1} )
= \sum_{n=0}^{m-1}\frac{1}{n+1}
= \sum_{n=1}^{m}\frac{1}{n}
$,
since all terms from
$n=m+1$
get cancelled out.
(added even later)
Here is why the cancellation happens:
Suppose $N$ is a large integer.
Then
$\begin{array}\\
\sum_{n=0}^{N}(\frac{1}{n+1}-\frac{1}{n+m+1} )
&=\sum_{n=0}^{N}\frac{1}{n+1}-\sum_{n=0}^{N} \frac{1}{n+m+1}\\
&=\sum_{n=1}^{N+1}\frac{1}{n}-\sum_{n=m+1}^{N+m+1} \frac{1}{n}\\
&=\left(\sum_{n=1}^{m}\frac{1}{n}+\sum_{n=m+1}^{N+1}\frac{1}{n}\right)
-\left(\sum_{n=m+1}^{N+1} \frac{1}{n}+\sum_{n=N+2}^{N+m+1} \frac{1}{n}\right)\\
&=\sum_{n=1}^{m}\frac{1}{n}+\sum_{n=N+2}^{N+m+1} \frac{1}{n}\\
\end{array}
$
and the right hand sum
is less than
$\frac{m}{N}$
since there are $m$ terms
each of which is
less than
$\frac1{N}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1008240",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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} |
Prove that $\frac{1}{a^3(b+c)}+\frac{1}{b^3(a+c)}+\frac{1}{c^3(a+b)}\ge \frac32$
$a,b,c$ are positive reals with $abc = 1$. Prove that
$$\frac{1}{a^3(b+c)}+\frac{1}{b^3(a+c)}+\frac{1}{c^3(a+b)}\ge \frac32$$
I try to use AM $\ge$ HM.
$$\frac{\dfrac{1}{a^3(b+c)}+\dfrac{1}{b^3(a+c)}+\dfrac{1}{c^3(a+b)}}3\ge \frac{3}{a^3(b+c)+b^3(a+c)+c^3(a+b)}$$
Then how I proceed.
| $$ AM \ge GM $$
$$\frac{\frac{1}{a_{3}(b+c)} + \frac{1}{b_{3}(a+c)} + \frac{1}{c_{3}(a+b)}}{3} \ge \sqrt[3]{\frac{1}{(a+b)(b+c)(c+a)}}$$
$$ \frac{1}{a_{3}(b+c)} + \frac{1}{b_{3}(a+c)} + \frac{1}{c_{3}(a+b)} \ge \frac{3}{\sqrt[3]{2abc+ab^2 + ba^2 + ac^2 + ca^2 + bc^2 + cb^2}}$$
$$AM \ge GM $$
$$ \frac{2abc+ab^2 + ba^2 + ac^2 + ca^2 + bc^2 + cb^2}{8} \ge \sqrt[8]{a^8b^8c^8}$$
$$ \ge \frac{3}{\sqrt[3]{8\sqrt[8]{a^8b^8c^8}}}$$
$$ \ge \frac{3}{\sqrt[3]{8}}$$
$$ \ge \frac{3}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1009022",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
What is the limit for the radical $\sqrt{x^2+x+1}-x $? I'm trying to find oblique asmyptotes for the function $\sqrt{x^2+x+1}$ and I manage to caclculate that the coefficient for the asymptote when x approaches infinity is 1.
But when i try to find the m-value for my oblique asymptote by taking the limit of:
$$
\lim_{x\to\infty}\sqrt{x^2+x+1}-x=m
$$
I'm stuck.
How do i simplify this expression to find the limit?
I've tried manipulating the radical by converting it to the denominator:
$$\lim_{x\to\infty}\frac{x^2+x+1}{\sqrt{x^2+x+1}}-x.$$
Or by multiplying both terms with the conjugate:
$$\lim_{x\to\infty}\frac{x^2+x+1-x^2}{\sqrt{x^2+x+1}}$$
But in neither case do I know how to take the limit. Any help would be greatly appreciated.
| We have
$$\sqrt{x^2 + x + 1} - x = \frac{x^2 + x + 1 - x^2}{\sqrt{x^2 + x + 1} + x} = \frac{x + 1}{\sqrt{x^2 + x + 1} - x}.$$
L'Hopital's rule then gives
$$\lim_{x\to\infty} \sqrt{x^2 + x + 1} - x = \lim_{x\to\infty}\left(1 + \frac{2x+1}{2\sqrt{x^2 + x + 1}}\right)^{-1} = \frac{1}{2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1009110",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 5
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Wrong applying of simple Chinese Remainder Theorem problem What am I doing wrong?
So for the following equations
$$
\begin{align}
(*) \left\{
\begin{array}{l}
2x\equiv 3\pmod 5 \\
4x\equiv 2\pmod 6 \\
3x\equiv 2\pmod 7
\end{array} \right.
\end{align}
$$
and $N =\mathrm{lcm}\langle5,6,7\rangle = 210$, giving
$N_1 = \frac{210}{5} = 42, \enspace N_2 = \frac{210}{6} = 35, \enspace N_3 = \frac{210}{7} = 30 $.
$$
\begin{align}
42z_1&\equiv 1\pmod 5\Rightarrow\enspace\enspace\;2z_1\equiv 1\pmod 5\Rightarrow\enspace &&\overline{z_1}=\overline{3}\\
35z_2&\equiv 1\pmod 6\Rightarrow\enspace-1z_2\equiv 1\pmod 6\Rightarrow\enspace &&\overline{z_2}=\overline{-1}\\
30z_3&\equiv 1\pmod 7\Rightarrow\enspace\enspace\;2z_3\equiv 1\pmod 7\Rightarrow\enspace &&\overline{z_1}=\overline{4}
\end{align}
$$
So the solution should be
$$
\begin{align}
\overline{x} &= \overline{3\times42\times3} + \overline{2\times35\times(-1)}+\overline{2\times30\times4}\\
&= \overline{378-70+240}\\
&= \overline{548}\\
&= \overline{128}
\end{align}
$$
Which is clearly wrong, so I'm wondering which additional steps I need to take to get to the correct answer.
Thanks in advance.
| You seem to have solved $$\left\{\begin{array}{c} x\equiv3 \pmod{5}\\x\equiv 2\pmod{6}\\x\equiv 2 \pmod{7}\end{array}\right.$$
But you need to take into account the coefficients on $x$ in each congruence.
$2x\equiv 3\pmod{5}$ means $x\equiv 4\pmod{5}$
$4x\equiv 2\pmod{6}$ means $x\equiv 2 \mbox{ or } 5 \pmod{6}$
$3x\equiv 2\pmod{7}$ means $x\equiv 3\pmod{7}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1009179",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
positive fractions, denominator 4, difference equals quotient (4,2) are the only positive integers whose difference is equal to their quotient. Find the sum of two positive fractions, in their lowest terms, whose denominators are 4 that also share this same property.
| I prove your first sentence below:
If $x, y\in\mathbb Z$ satisfy $x-y=x/y,$ then we have :
$$x(y-1)=y^2.$$
Thus, if a prime $p$ divides $y-1,$ then $p$ divides $y$ as well, a contradiction as $y$ and $y-1$ are coprime. Hence no prime divides $y-1,$ i.e. $y=2.$ Therefore $x(2-1)=4.$
As to the second sentence, if $(x-y)/4=x/y,$ then we have:
$$x(y-4)=y^2.$$
Since any common divisor of $y$ and $y-4$ must divide $4,$ denoting $\gcd(y,y-4)=g,$ we consider three cases:
I. $g=1.$
We have already shown that, in this case, $y-4=1,$ i.e. $y=5$ and $x=25.$
II. $g=2.$
In this case $y-4$ cannot have any odd prime divisor, and $4$ cannot divide $y-4$ either, thus $y-4=2,$ i.e. $y=6$ and $x=18.$
III. $g=4.$
Again no common odd prime divisor of $y$ and $y-4$ can occur, while $4$ divides $y-4.$ Hence $y-4=4,$ i.e. $y=8,$ and $x=16.$
So you know all the positive fractions with the required property, and it remains to calculate the sum of two of them, which I left to you.
Hope this helps.
Edit:
The fractions are supposed to be in the lowest terms, thus the cases II. and III. do not count at all. Hence the only pair of fractions with the required property is $(\frac{x}{4},\frac{y}{4})=(\frac{25}{4},\frac{5}{4}),$ in which case the sum is $\frac{15}{2}.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1010288",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
} |
Evaluate $\lim_{x \to 0} \frac{\sqrt{1 + \tan x} - \sqrt{1 + \sin x}}{x^3}$ What I attempted thus far:
Multiplying by conjugate
$$\lim_{x \to 0} \frac{\sqrt{1 + \tan x} - \sqrt{1 + \sin x}}{x^3} \cdot \frac{\sqrt{1 + \tan x} + \sqrt{1 + \sin x}}{\sqrt{1 + \tan x} + \sqrt{1 + \sin x}} = \lim_{x \to 0} \frac{\tan x - \sin x}{x^3 \cdot (\sqrt{1 + \tan x} + \sqrt{1 + \sin x})}$$
factor out $\sin x$ in the numerator
$$\lim_{x \to 0} \frac{\sin x \cdot (\sec x - 1)}{x^3 \cdot (\sqrt{1 + \tan x} + \sqrt{1 + \sin x})}$$
simplify using $\lim_{x \to 0} \frac{\sin x}{x} = 1 $
$$\lim_{x \to 0} \frac{\sec x - 1}{x^2 \cdot (\sqrt{1 + \tan x} + \sqrt{1 + \sin x})}$$
From here I don't see any useful direction to go in, if I even went in an useful direction in the first place, I don't know.
I suspect that this could be evaluated using the definition of derivatives, if so, or not, any suggestions?
| $ \lim_{x\to 0} \frac { \sqrt{1+\tan(x)} - \sqrt{1+sin(x)}} { x^3 } = \lim_{x\to 0} \frac {tan(x) - sin(x)} {x^3(\sqrt{1+\tan(x)} + \sqrt{1+sin(x)})} = \lim_{x\to 0} \frac {sin(x)(\frac {1} {cos(x)} - 1)} {x^3(\sqrt{1+\tan(x)} + \sqrt{1+sin(x)})} = \lim_{x\to 0} \frac {1 - cos(x)} {cos(x)x^2(\sqrt{1+\tan(x)} + \sqrt{1+sin(x)})} = \lim_{x\to 0} \frac {x^2} {2cos(x)x^2(\sqrt{1+\tan(x)} + \sqrt{1+sin(x)})} = \frac {1} {4} $
Intermediate calculations:
$$ \lim_{x\to0} \frac {2(1 - cos(x))} {x^2} = 1 $$
$$ \lim_{x\to0} cos(x) = 1 $$
$$ \lim_{x\to0} \frac {sin(x)}{x} = 1 $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1011290",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
Simplfiying $x(5xy+2x-1)=y(5xy+2y-1)$ I want to simplify: $x(5xy+2x-1)=y(5xy+2y-1)$ to $(x-y)(5xy+something-1)=0$ but I can't figure out what to do with the $2x$ and $2y$ on both sides.
| \begin{eqnarray}
&&x(5xy+2x-1)=y(5xy+2y-1)\\
&\Rightarrow&(5x^2y-5xy^2)+(2x^2-2y^2)-(x-y)=0\\
&\Rightarrow&5xy(x-y)+2(x-y)(x+y)-(x-y)=0\\
&\Rightarrow&(x-y)(5xy+2x+2y-1)=0
\end{eqnarray}
Moreover, note that $5xy+2x+2y-1=5(x+\frac{2}{5})(y+\frac{2}{5})-\frac{9}{5}$, you can simplifying as following (if you want):
$(x-y)[(x+\frac{2}{5})(y+\frac{2}{5})-\frac{9}{25}]=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1013696",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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How many 10-letter words are there in which each of the letters e,n,r,s occur at most once? Solve with a generating function. My solution was
$$g(x)=\left(\frac{x^0}{0!} + \frac{x^1}{1!}\right)^4 \left(\frac{x^0}{0!} + \frac{x^1}{1!} + \frac{x^2}{2!} + ... \right)^{26-4}$$
$$g(x) = (1 + x)^4e^{22x}$$
$$g(x) =\left( \binom{4}{0}x^0 + ... +\binom{4}{r}x^r... + \binom{4}{4}x^4\right)\sum\nolimits 22^r \frac{x^r}{r!}$$
Then we are trying to find the coefficient of the tenth term I denote as $a_{10}$
$$g(x) = h(x)f(x)$$ where $h(x) = (1 +x)^4$ and $f(x) = e^{22x}$
$$g_{10} =h_0f_{10} + h_1f_{9} + h_2f_{8} + h_3f_{7} + h_4f_{6} $$
$$ = \binom{4}{0}22^{10} + \binom{4}{1}22^{9} + \binom{4}{2}22^{8} + \binom{4}{3}22^{7} + \binom{4}{4}22^{6} $$
Now this is not correct and instead is supposes to be:
$$ = \binom{4}{0}\frac{10!}{10!}22^{10} + \binom{4}{1}\frac{10!}{9!}22^{9} + \binom{4}{2}\frac{10!}{8!}22^{8} + \binom{4}{3}\frac{10!}{7!}22^{7} + \binom{4}{4}\frac{10!}{6!}22^{6} $$
1) The correct expression seems to include $\binom{4}{r}\frac{10!}{(10-r)!}$ for evaluating the coefficients of $h(x)$ but why is this?
2) If this if correct, why are we using two different values for $n$? We make $n=4$ for evaluating the coefficient of $(1+x)^4$ but then somehow incorporate $n=10$ for $P(n,r)$
| In your computation you are missing the factor $\frac{1}{i!}$ for $f_i$ coefficients.
Fixing this, what you should be looking for is: $P(n=10,r=4)$ (this should answer your second question)that is:
$$\sum_{i=0}^{r}h_i f_{n-i}x^n= \sum_{i=0}^{r} \binom{r}{i} \frac{22^{n-i}}{(n-i)!}x^n$$
substituting for $r=4$ and $n=10$ you will get the expected result you mentioned except for the $10!$ factor, which is the result of the permutation of 10 letters.
So the final answer would be:
$$ = 10!\bigg( \binom{4}{0}\frac{1}{10!}22^{10} + \binom{4}{1}\frac{1}{9!}22^{9} + \binom{4}{2}\frac{1}{8!}22^{8} + \binom{4}{3}\frac{1}{7!}22^{7} + \binom{4}{4}\frac{1}{6!}22^{6})\bigg) $$
| {
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"url": "https://math.stackexchange.com/questions/1014133",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Evaluating $\int_0^{\infty} \log(\sin^2(x))\left(1-x\operatorname{arccot}(x)\right) \ dx$ One of the ways to compute the integral
$$\int_0^{\infty} \log(\sin^2(x))\left(1-x\operatorname{arccot}(x)\right) \ dx=\frac{\pi}{4}\left(\operatorname{Li_3}(e^{-2})+2\operatorname{Li_2}(e^{-2})-2\log(2)-\zeta(3)\right)$$
is to make use of the series of $\log(\sin(x))$, but the result I got after doing that wasn't that friendly.
Is it possible to find a neat way of evaluating the integral?
| We start with: $\displaystyle \int_0^1 \frac{t^2}{x^2+t^2}\,dt = 1-x\tan^{-1}\frac{1}{x}$
Then,
$\displaystyle \begin{align} \int_0^{\infty} \log (2\sin x)\left(1-x\tan^{-1}\frac{1}{x}\right)\,dx &= \int_0^{\infty}\int_0^1 \frac{t^2\log (2\sin x)}{t^2+x^2}\,dt\,dx\\&= -\sum\limits_{n=1}^{\infty} \int_0^1 t^2\int_0^{\infty} \frac{1}{n}\frac{\cos 2nx}{t^2+x^2}\,dx\,dt\tag{1}\\&= -\frac{\pi}{2}\sum\limits_{n=1}^{\infty} \frac{1}{n}\int_0^1 t^2\frac{e^{-2nt}}{t}\,dt\\&= -\frac{\pi}{2}\sum\limits_{n=1}^{\infty} \frac{1}{n}\left(\frac{1}{4n^2} - \frac{e^{-2n}}{4n^2} - \frac{e^{-2n}}{2n}\right)\\&= -\frac{\pi}{2}\left(\frac{1}{4}\operatorname{Li}_3(1)-\frac{1}{4}\operatorname{Li}_3(e^{-2})-\frac{1}{2}\operatorname{Li}_2(e^{-2})\right)\end{align}$
where, we used: $\displaystyle \int_0^{\infty} \frac{\cos ax}{b^2+x^2}\,dx = \frac{\pi e^{-ab}}{2b}$ in $(1)$.
Combining with the fact that: $\displaystyle \int_0^{\infty} \left(1-x\tan^{-1}\frac{1}{x}\right)\,dx = \frac{\pi}{4}$ we get the desired result.
| {
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"url": "https://math.stackexchange.com/questions/1015639",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
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Mathematical Induction Proof - Exponent with n in denominator Use mathematical induction to prove the following:
$$\frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^3}+...+\frac{n}{2^n}=2-\frac{n+2}{2^n}; n ∈ N $$
I am having trouble figuring out how to solve this with an exponent in the denominator.
$$2-\frac{n+2}{2^n}+\frac{n+1}{2^{n+1}} $$
| $$\begin{align*}\underbrace{\left(\frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^3}+...+\frac{n}{2^n}\right)}_{=2-\frac{n+2}{2^n}}+\frac{n+1}{2^{n+1}}&=\left(2-\frac{n+2}{2^n}\right)+\frac{n+1}{2^{n+1}}\\&=2-\left(\frac{2(n+2)}{2^{n+1}}-\frac{n+1}{2^{n+1}}\right)\\\\&=2-\frac{(n+1)+2}{2^{n+1}}\end{align*}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1015830",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Limit of Sum to Infinity Is $ \Sigma_{n=0}^\infty (\sqrt[3]{n^3+1} - n)$ convergent or divergent?
For expressions of the form $\sqrt{n^2+1} - n$, I believe the common trick is to multiply by the "conjugate" $\frac{\sqrt{n^2+1} + n}{\sqrt{n^2+1} + n}$.
Is there a similar trick for other roots (i.e. not square roots) as well?
Thanks.
| $$\sqrt[3]{n^3+1}-n=\\\sqrt[3]{n^3+1}-n \cdot \frac{\sqrt[3]{(n^3+1)^2}+n\sqrt[3]{(n^3+1)^1}+n^2}{\sqrt[3]{(n^3+1)^2}+n\sqrt[3]{(n^3+1)^1}+n^2}\\=\\\frac{n^3+1-n^3}{\sqrt[3]{(n^3+1)^2}+n\sqrt[3]{(n^3+1)^1}+n^2}\\=\frac{1}{\sqrt[3]{(n^3+1)^2}+n\sqrt[3]{(n^3+1)^1}+n^2}\\<\frac{1}{\sqrt[3]{(n^3)^2}+n\sqrt[3]{(n^3)^1}+n^2}\\\\\frac{1}{3n^2} $$
Also
$$ \sum_{n=1}^{\infty}\frac{1}{3n^2}=\\\frac{1}{3}\sum_{n=1}^{\infty}\frac{1}{n^2}=\\\frac{1}{3}\frac{\pi^2}{6}$$
Thus the original series is convergent.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $\lim _{n\to \infty \:}\left(\sqrt{n^2+3}-\sqrt{n^2+1}\right)=0$ Prove that $\lim _{n\to \infty \:}\left(\sqrt{n^2+3}-\sqrt{n^2+1}\right)=0$. I’m new to the subject and the square roots are throwing me a bit off.
| You first have to multiply the top and bottom by the conjuage, which is $\sqrt{n^2+3} + \sqrt{n^2+1}$.
So you get $$\dfrac{(\sqrt{n^2+3}-\sqrt{n^2+1})(\sqrt{n^2+3}+\sqrt{n^2+1})}{\sqrt{n^2+3}+\sqrt{n^2+1}} = \dfrac{n^2+3 - (n^2+1)}{\sqrt{n^2+3} + \sqrt{n^2+1}} = \dfrac{2}{\sqrt{n^2+3} + \sqrt{n^2+1}}$$
Clearly, the $\lim_{n\to\infty}$ of this expression $ = 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1019907",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Solve for constants: Derivatives using first principles
*
*Question
Find the values of the constants $a$ and $b$ such that $$\lim_{x \to 0}\frac{\sqrt[3]{ax + b}-2}{x} = \frac{5}{12}$$
*
*My approach
*
*Using the definition of the derivative, $$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$$
*I view limit as a derivative of a function $f$ at some value, let's call that value $c$, as follows :$$f'(c) = \lim_{x\to 0}\frac{f(c + x) - f(c)}{x} = \frac{\sqrt[3]{ax + b} - 2}{x} = \frac{5}{12}$$
*Now I deduce the following: $$f(c + x) = \sqrt[3]{ax + b}$$ and $$f(c) = 2$$
*Use limits as follows: $$\lim_{x\to 0} f(c + x) = f(c) = 2$$ that is, $$\lim_{x\to 0} f(c + x) = \lim_{x\to 0} \sqrt[3]{ax + b} = \sqrt[3]{b} = 2$$ now solve for $b$, $$\sqrt[3]{b} = 2 \Leftrightarrow b = 8$$
*Since I know that $$f(c + x) = \sqrt[3]{ax + 8}$$, I can solve for $a$, which is $$a = \frac{[f(c+x)]^3 - 8}{x} = \frac{[f(c+x)]^3 - [f(c)]^3}{x}$$
*Let $g(x) = [f(x)]^3$, such that $$g'(x) = 3\cdot [f(x)]^2 \cdot f'(x)$$
so $$g'(c) = 3\cdot [f(c)]^2 \cdot f'(c) = 3 \cdot 4 \cdot \frac{5}{12} = 5$$
*Rephrase $g'(c)$ using first principles such that $$g'(c) = \lim_{x \to 0}\frac{g(c + x)- g(c)}{x}= \lim_{x \to 0}\frac{[f(c + x)]^3 - [f(c)]^3}{x} = \lim_{x \to 0} a = 5$$
*Since $a$ is a constant, $\lim_{x \to 0} a = a$, that is, $$a = 5$$
*My solution: $b = 8, a = 5$.
Please have a look at my approach and give me any hints/suggestions regarding the solution and/or steps taken.
| Your approach looks fine, and it is explained well. Here is a variant of your solution, that uses similar ideas:
1) $\displaystyle\lim_{x\to0}(\sqrt[3]{ax+b}-2)=\lim_{x\to0}\bigg(\frac{\sqrt[3]{ax+b}-2}{x}\cdot x\bigg)=\frac{5}{12}\cdot 0=0,\;\;\;$ so $\sqrt[3]{b}=2$ and $b=8$.
2) Let $g(x)=\sqrt[3]{ax+8}.\;\;\;$ Then $\displaystyle g^{\prime}(0)=\lim_{x\to0}\frac{g(x)-g(0)}{x-0}=\frac{\sqrt[3]{ax+8}-2}{x}=\frac{5}{12}$,
so $g^{\prime}(x)=\frac{1}{3}(ax+8)^{-2/3}\cdot a\implies\frac{1}{3}\cdot\frac{1}{4}a=\frac{5}{12}\implies a=5$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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prove equation equals $-1$ I was wondering if it was possible to prove that
$\left( \frac{( a^2 - c^2 + (\frac{d}{c}a)^2-d^2)}{ \sqrt{(a-c)^2+(\frac{d}{c}a)-d)^2} \sqrt{(a+c)^2+(\frac{d}{c}a)+d)^2}} \right) = -1$ when $a,d,c \in [-1,1]$ and $|a|<c$, and $|b|<d$?
Attempt: By Asdrul, I know that
$$\left[(a-c)^{2}+\bigg(\frac{d}{c}a-d\bigg)^{2}\right]\cdot\left[(a+c)^{2}+\bigg(\frac{d}{c}a+d\bigg)^{2}\right]=\frac{1}{c^4}( c^{2}+d^{2}) ^{2}\left( a^2-c^2\right) ^{2}$$
So applying this, I have that the above equals
$\frac{c^2(a^2-c^2+(\frac{da}{c})^2-d^2)}{(c^2+d^2)|a^2-c^2|}$
| Hint: note that
$$\left[(a-c)^{2}+\bigg(\frac{d}{c}a-d\bigg)^{2}\right]\cdot\left[(a+c)^{2}+\bigg(\frac{d}{c}a+d\bigg)^{2}\right]=\frac{1}{c^4}( c^{2}+d^{2}) ^{2}\left( a^2-c^2\right) ^{2}$$
| {
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"url": "https://math.stackexchange.com/questions/1021090",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Calculating the value of $\frac{a-d}{b-c}$ If $\frac{a-b}{c-d}=2$ and $\frac{a-c}{b-d} = 3$ then determine the value of:
$$\frac{a-d}{b-c}$$
Where $a,b,c,d$ are real numbers.
Can someone please help me with this and give me a hint? I tried substitutions and solving them simultaneously but I couldn't determine this value. Please help.
| You have $\displaystyle{\frac{a-b}{c-d}}=2$ and $\displaystyle{\frac{a-c}{b-d}}=3$, hence
$$a-b = 2c-2d\\a-c = 3b-3d$$
By subtracting,
$$c-b=2c-3b+d$$
Or, adding $2b-2c$ to both terms,
$$b-c=d-b$$
Then
$$2=\frac{a-b}{c-d}=\frac{a-d+d-b}{c-b+b-d}$$
In denominator, $c-b$ and $b-d$ are equal, so
$$2=\frac12\frac{a-d}{c-b}+\frac12\frac{d-b}{b-d}=\frac12\frac{a-d}{c-b}-\frac12$$
Therefore,
$$\frac{a-d}{c-b}=5$$
Or
$$\frac{a-d}{b-c}=-5$$
Alternate solution, write the following as a system of equations where $c$ and $d$ are the unknowns:
$$a-b = 2c-2d\\a-c = 3b-3d$$
$$2c-2d=a-b\\-c+3d=3b-a$$
By adding the first row to twice the second,
$$-2d+6d=a-b+6b-2a$$
$$d=\frac{5b-a}{4}$$
Then
$$2c=a-b+2d=\frac{2a-2b+5b-a}{2}$$
$$c=\frac{3b+a}{4}$$
You then plug these values in your fraction,
$$\frac{a-d}{b-c}=\frac{4a-5b+a}{4b-3b-a}=\frac{5a-5b}{b-a}=-5$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1021567",
"timestamp": "2023-03-29T00:00:00",
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Complex numbers, polynomials Let $a$ be complex number such that $a^5 + a + 1 = 0$. What are possible values of $a^2(a - 1)$?
I have tried to find $a$. Is there any way to find it?
| $$
\begin{align}
a^5+a+1&=(a^2+a+1)(a^3-a^2+1)\\
&=(a^2+a+1)(a^2(a-1)+1)
\end{align}$$
If $a$ vanishes the second factor then $a^2(a-1)=-1$.
If $a$ vanishes the first factor, then divide $a^2(a-1)$ by $a^2+a+1$. We get $$a^3-a^2 = (a-2) × (a^2+a+1)+a+2$$
Therefore $a^2(a-1)=a+2$. We know that the roots of $a^2+a+1$ are $\frac{-1\pm\sqrt{1-4}}{2}$. Then only need to add $2$ to them.
In conclusion, the possible values are
$$-1, \frac{3\pm\sqrt{-3}}{2}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Showing two things are equal by Fourier series Given the Fourier series for the function:
$$f(x) = x+\frac14x^2 \quad -\pi\leq x \lt \pi$$
$$f(x)=f(x+2\pi) \quad -\infty \leq x \lt \infty$$
is
$$\frac{\pi^2}{12}+\sum \limits_{n=1}^\infty (-1)^n \left(\frac{\cos(nx)}{n^2}-\frac{2\sin(nx)}{n}\right)$$
show that$$\sum \limits_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6}$$
Now I hope this doesn't mean I have to calculate the Fourier series for this, since it seems to be given, since I had written down the Fourier coefficient for $a_0$ and most of $a_n$ and realised that this will simply be too time consuming in an exam!
If I sub in $x=2\pi$ for $f(2\pi)=2\pi + \pi^2$ and the fourier gives:
$$\frac{\pi^2}{12}+\sum \limits_{n=1}^\infty (-1)^n \left(\frac{\cos(2n\pi)}{n^2}-\frac{2\sin(2n\pi)}{n}\right)=\frac{\pi^2}{12}+\sum \limits_{n=1}^\infty (-1)^n \left(\frac{1}{n^2}\right)$$
$$2\pi + \pi^2=\frac{\pi^2}{12}+\sum \limits_{n=1}^\infty (-1)^n \frac{1}{n^2}$$
$$2\pi+\frac{11\pi^2}{12}=\sum \limits_{n=1}^\infty (-1)^n \frac{1}{n^2}$$
So this isn't right. Perhaps another value for $x$, how about $x=\pi$
$$\pi + \frac{\pi^2}{4}=\frac{\pi^2}{12}+\sum \limits_{n=1}^\infty (-1)^n \left(\frac{\cos(n\pi)}{n^2}-\frac{2\sin(n\pi)}{n}\right)$$
$$\pi + \frac{\pi^2}{4}=\frac{\pi^2}{12}+\sum \limits_{n=1}^\infty (-1)^n \left(\frac{(-1)^n}{n^2}\right)$$
$$\pi+ \frac{\pi^2}{6}=\sum \limits_{n=1}^\infty\frac{1}{n^2}$$
Almost correct, but I have an extra $\pi$ term. Is there a way I could have gotten this on the first guess, and how do I show it?
| Why not substitute $\;x=0\;$ to make things easier (though perhaps slightly longer)?
$$f(x)=x+\frac{x^2}4\implies f(0)=0=\frac{\pi^2}{12}+\sum_{n=1}^\infty (-1)^n\frac1{n^2}\implies$$
$$\implies\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n^2}=\frac{\pi}{12}$$
But
$$\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n^2}=1-\frac14+\frac19-\frac1{16}+\ldots=\sum_{n=1}^\infty\frac1{(2n-1)^2}-\sum_{n=1}^\infty\frac1{(2n)^2}=$$
$$=\sum_{n=1}^\infty\frac1{(2n-1)^2}-\frac14\sum_{n=1}^\infty\frac1{n^2}$$
and
$$\sum_{n=1}^\infty\frac1{n^2}=\sum_{n=1}^\infty\frac1{(2n-1)^2}+\sum_{n=1}^\infty\frac1{(2n)^2}\implies\frac34\sum_{n=1}^\infty\frac1{n^2}\stackrel *=\sum_{n=1}^\infty\frac1{(2n-1)^2}$$
So we finally get:
$$\frac{\pi^2}{12}=\sum_{n=1}^\infty\frac1{(2n-1)^2}-\frac14\sum_{n=1}^\infty\frac1{n^2}\stackrel *=\frac12\sum_{n=1}^\infty\frac1{n^2}$$
and the result follows at once.
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluating $\lim_{(x,y)\rightarrow (0,0)} \frac{(xy)^3}{x^2+y^6}$ $$\lim_{(x,y)\rightarrow (0,0)} \frac{(xy)^3}{x^2+y^6}$$
I don't really know how to do, but I was trying to do like that:
$a=x$,
$b=y^2$
then I was trying to do this
$$\lim_{(x,y)\rightarrow (0,0)} \frac{ab}{a^2+b^2}$$
then I don't know no more how to do...
| Conversion into polar coordinates can be of help as well. Letting $x=r\cos\theta$ and $y=r\sin\theta$, we can write the limit as follows:
$$\lim_{r\to 0}\dfrac{r^3\cos^3\theta \cdot r^3\sin^3\theta}{r^2\cos^2\theta+r^6\sin^6\theta}=\lim_{r\to 0}\dfrac{r^3(\cos^3\theta\sin^3\theta)}{r^2(\cos^2\theta+r^4\sin\theta)}=\lim_{r\to 0}\dfrac{r(\cos^3\theta\sin^3\theta)}{(\cos^2\theta+r^4\sin\theta)}=0$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to find the splitting field of $X^4-10X^2+1$? How to find the splitting field of $X^4-10X^2+1$ ?
I found the roots
\begin{align*}
X^4-10X^2+1=0&\iff (X^2-5)^2-24=0\\
&\iff X^2-5=\pm 2\sqrt 6\\
&\iff X^2=5\pm 2\sqrt 6\\
&\iff X\in\left\{\sqrt{5+2\sqrt 6},\sqrt{5-2\sqrt 6},-\sqrt{5+2\sqrt 6},-\sqrt{5-2\sqrt 6}\right\}
\end{align*}
But I'm not able to continue...
| Hint: $(\sqrt 3 + \sqrt 2)^2 = 5 + 2\sqrt6$ and $(\sqrt 3 - \sqrt 2)^2 = 5 - 2\sqrt 6$, so your splitting field contains $\sqrt 2$ and $\sqrt 3$.
| {
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"timestamp": "2023-03-29T00:00:00",
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How can I evaluate $\lim_{n \to\infty}\left(1\cdot2\cdot3+2\cdot3\cdot4+\dots+n(n+1)(n+2)\right)/\left(1^2+2^2+3^2+\dots+n^2\right)^2$? How can I evaluate this limit? Give me a hint, please.
$$\lim_{n \to\infty}\frac{1\cdot2\cdot3+2\cdot3\cdot4+\dots+n(n+1)(n+2)}{\left(1^2+2^2+3^2+\dots+n^2\right)^2}$$
| Note that
$$(1^2+2^2+\ldots+n^2)^2\ge 1^4+2^4+\ldots+n^4$$
Now we can use Stolz–Cesàro theorem (kind of a discrete l'Hôpital's rule):
$$\lim_{n\to\infty}\frac{1\cdot2\cdot3+2\cdot3\cdot4+\ldots+n(n+1)(n+2)}{1^4+2^4+\ldots+n^4}=\lim_{n\to\infty}\frac{n(n+1)(n+2)}{n^4}=0$$
And conclude the original limit was also $0$.
| {
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Order of group from its presentation Say we want to determine the order of a group generated by $x$ and $y$ who satisfy $x^2y = xy^3 = 1$.
Ok so it would be nice to know the order of $x$ and $y$ respectively. We can readily conclude from the equations above that $x^2 = y^{-1}$ and $x^{-1} = y^3$.
Next we can derive the relationship $x = y^2$ and plugging this into the given equations we get $y^5 = 1$, hence $y$ is of order $5$.
Now since $x = y^2 = y \cdot y$, we have $y^{-1} x = y$ and we knew $y^{-1} = x^2$ so we get $x^3 = y$. But since $y^5 = 1$ we get $x^{15} = 1$. Hence the $x$ is of order $15$.
$x$ and $y$ generate the group so every element can be written as $x^a y^b$ where $a$ and $b$ range from 0 to 14 and 0 to 5 respectively. But the group is not necessarily abelian, so are we not missing out elements of the form $y^kx^j$?
Not sure how to conclude the argument, help much appreciated!
| $x^2 = y^{-1}, x = y^{-3} \Rightarrow x^6 = y^{-3} = x \Rightarrow x^5 = 1 \Rightarrow |x| = 5 \Rightarrow |x^2| = 5 \Rightarrow |y^{-1}| = 5 \Rightarrow |y| = 5.$ Now $y \in <x> \Rightarrow$ the group is the cyclic group of order $5$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1026969",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the limit $\lim_{(x,y,z)\to(0,0,0)}\frac{xy+xz+yz}{x^2+y^2+z^2}$ Find the limit if it exists
$$\lim_{(x,y,z)\to(0,0,0)}\frac{xy+xz+yz}{x^2+y^2+z^2}$$
| Let's see the limit in the plane $z=0$. So,
\begin{align*}
\lim_{(x, y, z)\rightarrow (0,0,0)} \frac{xy + yz + xz}{x^2 + y^2 + z^2} &= \lim_{(x, y) \rightarrow (0,0)} \frac{xy} {x^2 + y^2}
\end{align*}
Now analyzing by paths, we have to take the path $y=mx$, with $m>0$.
\begin{align*}
\require{cancel}
\lim_{(x, y)\rightarrow (0,0)} \frac{xy}{x^2 + y^2} &= \lim_{x\rightarrow 0} \frac{mx^2}{x^2 + m^2x^2}\\
& = \lim_{x\rightarrow 0} \frac{mx^2}{x^2(1 + m^2)}\\
& = \lim_{x\rightarrow 0} \frac{m\cancel{x^2}}{\cancel{x^2}(1 + m^2)}\\
& = \frac{m}{1 + m^2}.
\end{align*}
The above tells us that the limit is not unique, because it varies according to the value that $m$ takes. Therefore the limit $\displaystyle \lim_{(x, y, z) \rightarrow (0,0,0)}\frac{xy + yz + xz}{x^2 + y^2 + z^2}$ does not exist.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Comparison between lebesgue integral and riemann integral of $f(x)=x^2$ in $[0,2]$ If we have an example $f(x)=x^2$ let's say for $[0,2]$.
In lebesgue integral, I already use a sequence of function $f_n(x)$ as approximation to $f(x)$ ($f_n(x)$ converges to $f(x)$) which is stated by $f_n(x)=\sum_{k=1}^{2.2^n}\frac{k-1}{2^n}1_{\left\{\left(\sqrt{\frac{k-1}{2^n}},\sqrt{\frac{k}{2^n}}\right]\right\}}$. Then we know that the step after that is by using monotone convergence theorem which say $\lim_{n\rightarrow\infty}\int_{R}f_n=\int_{R}f$. Then I got the limit of integral of $f_n$ as $\lim_{n\rightarrow\infty}\sum_{k=1}^{2.2^n}\frac{k-1}{2^n}(\sqrt{\frac{k}{2^n}}-\sqrt{\frac{k-1}{2^n}})$.
When I try to figure out this limit of sequence of function. It just ended up not in the right answer it should be (the answer in Riemann integral is : $\frac{8}{3}$. So by using lebesgue integral, we must have the same answer as riemann integral is).
Please help for proof.
| Since $f(2) = 2^2 = 4$, the upper limit for the sum should be $4 \cdot 2^n$:
$$f_n(x) = \sum_{k=1}^{4 \cdot 2^n}\frac{k-1}{2^n}1_{E_n},$$
where
$$E_n = \left\{x : \frac{k-1}{2^n} \leqslant f(x)\leqslant \frac{k}{2^n}\right\}=\left\{x : \sqrt{\frac{k-1}{2^n}} \leqslant x\leqslant \sqrt{\frac{k}{2^n}}\right\}.$$
Then the Lebesgue integral is
$$\int f= \lim_{ n \rightarrow \infty}\sum_{k=1}^{4 \cdot 2^n}\frac{k-1}{2^n}m(1_{E_n})$$
with
$$\sum_{k=1}^{4 \cdot 2^n}\frac{k-1}{2^n}m(1_{E_n})=\sum_{k=1}^{4 \cdot 2^n}\frac{k-1}{2^n}\left(\sqrt{\frac{k}{2^n}}-\sqrt{\frac{k-1}{2^n}}\right)\\=\frac{1}{2^n\sqrt{2^n}}\sum_{k=1}^{4 \cdot 2^n}(k-1)\left(\sqrt{k}-\sqrt{k-1}\right).$$
We can simplify this using summation by parts:
$$\sum_{k=1}^{m}a_k(b_{k+1}-b_k)= a_{m+1}b_{m+1}-a_1b_1 - \sum_{k=1}^{m}b_{k+1}(a_{k+1}-a_k)$$
Summing by parts, we obtain
$$\frac{1}{2^n\sqrt{2^n}}\sum_{k=1}^{4 \cdot 2^n}(k-1)\left(\sqrt{k}-\sqrt{k-1}\right) \\=\frac{1}{2^n\sqrt{2^n}}\left[4 \cdot 2^n\sqrt{4 \cdot 2^n}-\sum_{k=1}^{4 \cdot 2^n}\sqrt{k}\right]\\=8-\frac{1}{2^n}\sum_{k=1}^{4 \cdot 2^n}\sqrt{\frac{k}{2^n}}.$$
Hence,
$$\int f= 8-\lim_{ n \rightarrow \infty}\frac{1}{2^n}\sum_{k=1}^{4 \cdot 2^n}\sqrt{\frac{k}{2^n}}.$$
The limit on the RHS can be recognized as the limit of a Riemann sum for $\sqrt{x}$ on the interval $[0,4]$.
Whence,
$$\int f= 8-\int_0^4\sqrt{x} \, dx = 8-\left.\frac{2}{3} x^{3/2}\right|_0^4=\frac{8}{3}$$
| {
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"url": "https://math.stackexchange.com/questions/1028293",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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First Order Ordinary Differential Equation by Any Method (1R-24) I just need for someone to check my work and suggest a better way to solve this if one exists. I can use any method but not numerical or any other iterative series approximation. The following documents my process of attempting to solve this problem. It does not appear to be separable, reducible to separable such as with an appropriate change of variable like the u-substitution $u=\tfrac{y}{x}$, or linear, or reducible to linear? But is it exact or can it be made to be exact?
$$y' = \frac{y}{x + (y + 1)^2}$$
Rearrange terms and expand then test for total differential (exactness):
$$Mdx + Ndy = 0$$
$$-ydx + (x + y^2 + 2y + 1)dy = 0$$
$$ \frac {\partial M}{\partial y} = -1, \frac {\partial N}{\partial x} = 1$$
It is not exact but can it be made to be exact with an integrating factor in x? Here $P$ is $M$ and $Q$ is $N$. $P$ and $Q$ are used to imply that ODE is not exact but is being worked on to find an integrating factor.
$$ \frac {1}{F} \frac {dF}{dx} = \frac {1}{Q} \left[ \frac {\partial P}{\partial y} - \frac {\partial Q}{\partial x} \right]$$
$$ \frac {1}{F} \frac {dF}{dx} = \frac {1}{x + y^2 + 2y + 1} \left[ -1 - 1 \right] = \frac {-2}{x + y^2 + 2y + 1}$$
This will not work because of the y-terms. Plus it would be far too complex to use. We need to find a simpler integrating factor. How about deriving one in y? Like before $P$ is $M$ and $Q$ is $N$.
$$ \frac {1}{F} \frac {dF}{dy} = \frac {1}{P} \left[ \frac {\partial Q}{\partial x} - \frac {\partial P}{\partial y} \right]$$
$$ \frac {1}{F} \frac {dF}{dy} = \frac {1}{-y} \left[ 1 - (-1) \right] = - \frac {2}{y}$$
It appears we may have made a breakthrough in finding an integrating factor in y.
$$ \int \frac {dF}{F} = -2 \int \frac {dy}{y}$$
$$ln|F| = -2 \ln|y|$$
$$F = y^{-2}$$
So far so good. Let us now multiply the rearranged ODE by $F$ and test once more for exactness. As before $P$ is $M$ and $Q$ is $N$.
$$FPdx + FQdy = 0$$
$$y^{-2}(-y)dx + y^{-2}(x + y^2 + 2y + 1)dy = 0$$
$$-\frac {1}{y}dx + \left(\frac {x}{y^2} + 1 + \frac {2}{y} + \frac {1}{y^2}\right)dy = 0$$
$$ \frac {\partial M}{\partial y} = \frac {1}{y^2}, \frac {\partial N}{\partial x} = \frac {1}{y^2}$$
The integrating factor worked. We now have exactness and can finally proceed in solving this ODE. We may begin with either term $M$ or $N$.
$$u(x,y) = \int Mdx = - \int \frac {1}{y}dx = - \frac {x}{y} + k(y)$$
$$ \frac {\partial u}{\partial y} = \frac {\partial \left[-\frac{x}{y} + k(y)\right]}{\partial y} = \frac {x}{y^2} + k(y)' = N = \frac {x}{y^2} + 1 + \frac {2}{y} + \frac {1}{y^2}$$
We are now at the home stretch. We just need to isolate and solve for $k(y)$ and insert it back into our premature solution $u(x,y)$.
$$ k(y)' = 1 + \frac {2}{y} + \frac {1}{y^2}$$
$$ k(y) = \int \left[1 + \frac {2}{y} + \frac {1}{y^2}\right]dy = y + 2\ln|y| - \frac {1}{y} + c$$
$$u(x,y) = - \frac {x}{y} + k(y) = - \frac {x}{y} + y + 2\ln|y| - \frac {1}{y} + c$$
$$y + 2\ln|y| - \frac {1}{y}(x + 1) = c$$
I believe I may have solved it but I just need someone to check my work.
| You could check by yourself your result :
$$y+2\ln|y|-\frac{1}{y}(x+1)=c$$
$$x=y^2-2y\ln|y|-cy-1$$
$$dx=(2y-2\ln|y|-2-c)dy$$
Bringing it back into the ODE :
$$-ydx + (x + y^2 + 2y + 1)dy = 0$$
leads to :
$$-y\big((2y-2\ln|y|-2-c)dy\big) + \big((y^2-2y\ln|y|-cy-1) + y^2 + 2y + 1\big)dy = $$
after simplification $$=0$$
So, the result is correct.
| {
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"source": "stackexchange",
"question_score": "1",
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How do I express the sum $(1+k)+(1+k)^2+\ldots+(1+k)^N$ for $|k|\ll1$ as a series? Wolfram Alpha provides the following exact solution
$$ \sum_{i=1}^N (1+k)^i = \frac{(1+k)\,((1+k)^N-1)}{k}.$$
I wish to solve for $N$ of the order of several thousand and $|k|$ very small (c. $10^{-12}).$ When I do this on a computer in excel the software cannot handle it (because of truncation of significant figures) and the results are nonsense.
I thought to approximate the result using the first few terms of a series in increasing powers of $k$. I can multiply out the first few terms and examine the patterns in the following pyramid...
$$(1+k)^1 = k +1 $$
$$(1+k)^2 = k^2 +2k +1 $$
$$(1+k)^3 = k^3 +3k^2 +3k +1 $$
$$(1+k)^4 =k^4+4 k^3+6 k^2+4 k+1$$
$$(1+k)^5 =k^5+5 k^4+10 k^3+10 k^2+5 k+1$$
$$(1+k)^6 =k^6+6 k^5+15 k^4+20 k^3+15 k^2+6 k+1$$
So for example, for $N=3$ we would obtain the sum
$$
S= k^3 +4k^2 +6k +1
$$
The results suggest a solution with a pattern of the form
$$
S = a + bk^1 +ck^2+dk^3...
$$
I can see that $a=N$. The other coefficients increase monotonously and it might be possible to determine a formula for the coefficients from the pattern. Although the general pattern is not convergent, it is possible that for certain restricted ranges of $N$ and $k$ a convergent formula could be obtained. If so then it is possible that a useful approximation of S can be obtained just by evaluating the first few terms in the series.
But is there a well-known general formula for the terms in this series or can one be derived algebraically from the original formula?
UPDATE
following on from the answer by User73985...
$$ S=\sum_{i=1}^N (1+k)^i = N + \sum_{j=2}^{N+1}\binom{N+1}{j}k^{j-1}$$
So
$$ S= N + \sum_{j=2}^{N+1}\frac{(N+1)!}{(N+1-j)!j!}k^{j-1}$$
then
$$ S= N
+ \frac{(N+1)!}{(N-1)!2!} k^{1}
+ \frac{(N+1)!}{(N-2)!3!} k^{2}
+ \frac{(N+1)!}{(N-3)!4!} k^{3} +...
$$
giving
$$ S= N
+ \frac{(N+1)(N)}{2!} k^{1}
+ \frac{(N+1)(N)(N-1)}{3!} k^{2}
+ \frac{(N+1)(N)(N-1)(N-2)}{4!} k^{3} +...
$$
thus
$$ S= N
+ \frac{N^2+N}{2} k^{1}
+ \frac{N^3-N}{6} k^{2}
+ \frac{N^4-2 N^3-N^2+2 N}{24} k^{3} +...
$$
For $N=1 to 10,000$ and $k= 2.40242 * 10^{-12}$ this formula can be truncated to
$$ S = N + \frac{N^2+N}{2} k^{1}
$$
and then gives results very close to those expected. Because $k$ is so small relative to $n$ the terms in higher powers of $k$ can be ignored. Note that the coefficient of $k^1$ is consistent with that found by examination of the coefficients in the "pyramid" presented above.
| There is nowhere a converging series in sight. From the data given it seems that $N$ is large and
$$p:=Nk$$
is very small. Therefore we may write
$$S={1+k\over k}\bigl((1+k)^N-1\bigr)={1+k\over k}\left(\bigl(1+{p\over N}\bigr)^N-1\right)\doteq{1+k\over k}(e^p-1)\ .$$
If $N$ is not in the thousands use the first few terms of the binomial series:
$$\left(1+k\right)^N-1=\sum_{j=1}^\infty{N\choose j}k^j\ .$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1029172",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluate $\int_0^\infty \frac{(\log x)^2}{1+x^2} dx$ using complex analysis How do I compute
$$\int_0^\infty \frac{(\log x)^2}{1+x^2} dx$$
What I am doing is take
$$f(z)=\frac{(\log z)^2}{1+z^2}$$
and calculating
$\text{Res}(f,z=i) = \dfrac{d}{dz} \dfrac{(\log z)^2}{1+z^2}$
which came out to be $\dfrac{\pi}{2}-\dfrac{i\pi^2}{8}+\dfrac{i\pi}{2}$
Im not too sure how to move on from here. the given answer is $\dfrac{\pi^3}{8}$
Any help will be appreciated. thank you in advanced.
| I would consider the following contour integral
$$\oint_C dz \frac{\log^3{z}}{1+z^2} $$
where $C$ is a keyhole contour of outer radius $R$ and inner radius $\epsilon$, about the positive real axis. Without explicitly showing that the integrals over the outer and inner circles vanish as $R\to\infty$ and $\epsilon \to 0$, the integral is equal to
$$\int_0^{\infty} dx \frac{\log^3{x}-(\log{x}+i 2 \pi)^3}{1+x^2} = -i 6 \pi \int_0^{\infty} dx \frac{\log^2{x}}{1+x^2} + 12 \pi^2 \underbrace{\int_0^{\infty} dx \frac{\log{x}}{1+x^2}}_{0} + \\i 8 \pi^3 \underbrace{\int_0^{\infty} dx \frac{1}{1+x^2}}_{\pi/2} $$
The contour integral is also equal to the sum of the residues of the integrand at the poles $z=\pm i$:
$$-i 6 \pi \int_0^{\infty} dx \frac{\log^2{x}}{1+x^2} + i 4 \pi^4 = i 2 \pi \left (\frac{-i \pi^3/8}{2 i} + \frac{-i 27\pi^3/8}{-2 i} \right ) = i \frac{13 \pi^4}{4}$$
Note that the keyhole contour demands that $\arg{z} \in [0,2\pi]$. Thus,
$$\int_0^{\infty} dx \frac{\log^2{x}}{1+x^2} = \frac{\pi^3}{8} $$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
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Help with Linear Transformation of a multivariate normal Given X ~ $N_2$ (μ, Σ)$
Find the Distribution of
$$
\begin{pmatrix}
X+Y \\
X-Y
\end{pmatrix}
$$
Show independence if $Var(X) = Var(Y)$
Attempt:
Given proper of Multitvariate Normal Transformations $N_m$ (Aμ, $AΣA^t$ )
Using $$ A =
\begin{pmatrix}
1&1 \\
1 &-1
\end{pmatrix}* \begin{pmatrix}
X \\
Y
\end{pmatrix} =
\begin{pmatrix}
X+Y \\
X-Y
\end{pmatrix}\begin{pmatrix}
1&1 \\
1 &-1
\end{pmatrix} =
$$
The general Variance-covariance matrix of a bivariate normal,multiplied by $AΣA^t$
$$ A =
\begin{pmatrix}
1&1 \\
1 &-1
\end{pmatrix}* \begin{pmatrix}
δ_x^2&δ_{xy} \\
δ_{xy} &δ_y^2
\end{pmatrix} *
\begin{pmatrix}
1&1 \\
1 &-1
\end{pmatrix} =
\begin{pmatrix}
4δ&0 \\
0 &0
\end{pmatrix} $$
Given $δ_{xy} = δ_x^2 = δ_y^2$
My Questions:
i. Do the zeros in the final Variance-Covariance matrix sufficiently show independence?
ii. Does $δ_y^2 = 0$ effect the pdf, ie making it degenerate?
Where the bivariate pdf is
$\frac{1}{2\piδ_yδ_x(1-\rho^2)} e^{\frac{q}{2(1-\rho^2)}}$
$q = [(\frac{x-μ_x}{δ_x})^2 -2\rho(\frac{y-μ_y}{δ_x})(\frac{y-μ_y}{δ_y}) + (\frac{y-μ_y}{δ_y})^2]$
By $AΣA^t$ $\rho = 0$ and $δ_y^2 = 0 $
$q = [(\frac{x-μ_x}{δ_x})^2 + (\frac{y-μ_y}{0})^2]$ which is problematic.
I'm not sure where I am going wrong here...
| *
*Yes, for a Gaussian random vector $(X,Y)$, it is know that $X$ and $Y$ are independent if and only if the covariance $\text{cov}(X,Y)$ equals $0$. Since the entries on the off-diagonal are exactly the covariance, this means that $X$ and $Y$ are indeed independent.
*Well, if $\delta_y^2=0$ (which is equivalent to $\text{var}(Y)=0$), then $Y$ is degenerate Gaussian, i.e. $Y=\mathbb{E}Y$ almost surely.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1032365",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Induction Proof that $\sum_{i=0}^n 3^{n-i} {n \choose i} = \sum_{i=0}^n (-1)^i 5^{n-i} {n \choose i}$ Show that for all $n\geq0$
$$\binom{n}{0}3^n+\binom{n}{1}3^{n-1}+\dotsc+ \binom{n}{n-1}3^{1}+\binom{n}{n} $$
$$= \binom{n}{0}5^n-\binom{n}{1}5^{n-1}+\binom{n}{2}5^{n-2}-\binom{n}{3}5^{n-3}+\dotsc (-1)^n\binom{n}{n}$$
I believe this needs to be proved by induction but I'm not sure how to do it.
Help is appreciated!
| We have
$$(x+a)^n=\sum_{k=0}^{n}\binom{n}{k}x^ka^{n-k}$$
Now by putting $x=3$ and $a=1$ we get
$$\begin{align}(3+1)^n&=\sum_{k=0}^{n}\binom{n}{k}(3)^k(1)^{n-k}\\
&=\binom{n}{0}3^n+\binom{n}{1}3^{n-1}+\dotsc+ \binom{n}{n-1}3^{1}+\binom{n}{n}\\
\end{align}$$
And by putting $x=5$ and $a=-1$ we get
$$\begin{align}(5+(-1))^n&=\sum_{k=0}^{n}\binom{n}{k}(5)^k(-1)^{n-k}\\
&=\binom{n}{0}5^n-\binom{n}{1}5^{n-1}+\binom{n}{2}5^{n-2}-\binom{n}{3}5^{n-3}+\dotsc (-1)^n\binom{n}{n}\\
\end{align}$$
Since $$(3+1)^n=(5+(-1))^n=(5-1)^n=4^n$$
we get
$$4^n=\sum_{k=0}^{n}\binom{n}{k}3^k1^{n-k}=\sum_{k=0}^{n}\binom{n}{k}5^k(-1)^{n-k}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Double Integral $\int\limits_0^a\int\limits_0^a\frac{dx\,dy}{(x^2+y^2+a^2)^\frac32}$ How to solve this integral?
$$\int_0^a\!\!\!\int_0^a\frac{dx\,dy}{(x^2+y^2+a^2)^\frac32}$$
my attempt
$$
\int_0^a\!\!\!\int_0^a\frac{dx \, dy}{(x^2+y^2+a^2)^\frac{3}{2}}=
\int_0^a\!\!\!\int_0^a\frac{dx}{(x^2+\rho^2)^\frac{3}{2}}dy\\
\rho^2=y^2+a^2\\
x=\rho\tan\theta\\
dx=\rho\sec^2\theta \, d\theta\\
x^2+\rho^2=\rho^2\sec^2\theta\\
\int_0^a\!\!\!\int_0^{\arctan\frac{a}{\rho}}\frac{\rho\sec\theta}{\rho^3\sec^3\theta}d\theta \, dy=
\int_0^a\!\!\!\frac{1}{\rho^2}\!\!\!\int_0^{\arctan\frac{a}{\rho}}\cos\theta \, d\theta \, dy=\\
\int_0^a\frac{1}{\rho^2}\sin\theta\bigg|_0^{\arctan\frac{a}{\rho}} d\theta \, dy=
\int_0^a\frac{1}{\rho^2}\frac{x}{\sqrt{x^2+\rho^2}}\bigg|_0^ady=\\
\int_0^a\frac{a}{(y^2+a^2)\sqrt{y^2+2a^2}}dy$$
Update:
$$\int_0^a\frac{a}{(y^2+a^2)\sqrt{y^2+2a^2}}dy=\frac{\pi}{6a}$$
| Let us consider your last integral $$I=\int_0^a\frac{a}{(y^2+a^2)\sqrt{y^2+2a^2}}dy$$ Changing variable $y=a z$, it becomes $$I=\frac 1 a \int_0^1 \frac{dz}{\left(z^2+1\right) \sqrt{z^2+2}}$$ Now, make a change of variable such that $$z=\frac{\sqrt{2} t}{\sqrt{1-t^2}}$$ (it did not come immediately to my mind, I must confess) $$dz=\frac{\sqrt{2}}{\left(1-t^2\right)^{3/2}}$$ and so $$\int \frac{dz}{\left(z^2+1\right) \sqrt{z^2+2}}=\int\frac{dt}{1+t^2}=\tan ^{-1}(t)=\tan ^{-1}\left(\frac{z}{\sqrt{z^2+2}}\right)$$ Now, use the bounds for the integral.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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"answer_id": 0
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Get the numbers from (0-30) by using the number $2$ four times How can I get the numbers from (0-30) by using the number $2$ four times.Use any common mathematical function and the (+,-,*,/,^)
I tried to solve this puzzle, but I couldn't solve it completely. Some of my results were:
$$2/2-2/2=0$$ $$(2*2)/(2*2)=1$$ $$2/2+2/2=2$$ $$2^2-2/2=3$$ $$\frac{2*2}{2}+2=4$$ $$2^2+2/2=5$$ $$2^2*2-2=6$$ $$\frac{2^{2*2}}{2}=8$$ $$(2+2/2)^2=9$$ $$2*2*2+2=10$$ $$2*2*2*2=16$$ $$22+2/2=23$$ $$(2+2)!+2/2=25$$ $$(2+2)!+2+2=28$$
| Though I suspect this may be pushing common functions..
$2^2 + 2 + \Gamma(2) = 7$
$22/{\sqrt{2}^2} = 11$
$\int_{2/2}^{22}dx =21$
$\int_{2-2}^{22}dx =22$
$(2+2)! + \sqrt{2+2} = 26$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1034122",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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} |
Solve the recurrence relation $a_n=3a_{n-1}+n^2-3$, with $a_0=1$. Solve the recurrence relation $a_n=3a_{n-1}+n^2-3$, with $a_0=1$.
My solutions: the homogeneous portion is $a_n=c3^n$, and the inhomogeneous portion is $a^*_n=-1/2n^2-3/4n+9/8$.
This results in a final recurrence relation of
$$a_n=-\frac{1}{8}3^n-\frac{1}{2}n^2-\frac{3}{4}n+\frac{9}{8}.$$
I am just wondering if someone could check my work to make sure I have the procedure correct.
| $$ a_n = 3a_{n-1} + n^2 - 3 = 3^ka_{n-k} + n^2 + 3(n - 1)^2 + \ldots + 3^{k - 1}(n - k + 1)^2 - (3 + \ldots + 3^k) $$
$$ a_n = 3^ka_{n - k} + \sum_{m=0}^{k-1}3^m(n - m)^2 - \sum_{m = 0}^{k-1}3^{m+1}$$
Setting $k = n$, we have
$$a_n = 3^n + \sum_{m=0}^{n-1}3^m(n - m)^2 - \sum_{m = 0}^{n-1}3^{m+1} = \frac{1}{2}(-n^2 - 3n + 3^{n+1} - 3) - \frac{3}{2}(3^n - 1) + 3^n.$$
The last was calculated using numerical software. Simplifying we have,
$$\frac{1}{2}(-n^2 - 3n + 3^{n+1} - 3) - \frac{3}{2}(3^n - 1) + 3^n = 3^n - \frac{1}{2}n^2 - \frac{3}{2}n.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1035625",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
How to solve this equation if we can't use Chinese remainder theorem. Let consider:
$$\begin{cases}6x \equiv 2 \mod 8\\ 5x \equiv 5\mod 6 \end{cases}$$
We can't use Chinese remainder theorem because $\gcd(8,6) = 2 > 1$
Help me.
| $$
\begin{array}{l}
\left\{ \begin{array}{l}
6x \equiv 2\left[ 8 \right] \\
5x \equiv 5\left[ 6 \right] \\
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
3x \equiv 1\left[ 4 \right] \\
x \equiv 1\left[ 6 \right] \\
\end{array} \right. \\
\Rightarrow \left\{ \begin{array}{l}
x \equiv - 1\left[ 4 \right] \\
x \equiv 1\left[ 6 \right] \\
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
x \equiv 3\left[ 4 \right] \\
x \equiv 1\left[ 6 \right] \\
\end{array} \right. \\
4k + 3 = 6m + 1 \Leftrightarrow 1 = 3m - 2k \\
\end{array}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1039430",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 4
} |
Infimum of $\left\{\frac{n}{n^2+1}\:\:;\:n\:\in \mathbb N\right\}$ with a proof Consider $A=\left\{\frac{n}{n^2+1}\:\:;\:n\:\in \:N\right\}$. I need to find and prove $\inf(A)$.
So I know that I need to prove that for every $\epsilon > 0$ exists some $a\:\in A$ such that $a\:=\:\frac{n_0}{n_0^2\:+1}$ for some $n_0$, so $\inf(A)\:+\:\epsilon \:>\:a$
What is the method? How to choose what $\epsilon$ is bigger than?
| Revised to match corrected question and the OP’s background.
Let $a_n=\dfrac{n}{n^2+1}$. Then
$$\begin{align*}
\frac{n+1}{(n+1)^2+1}-\frac{n}{n^2+1}&=\frac{n+1}{n^2+2n+2}-\frac{n}{n^2+1}\\\\
&=\frac{(n+1)(n^2+1)-n(n^2+2n+2)}{(n^2+1)(n^2+2n+2)}\\\\
&=\frac{n^3+n^2+n+1-(n^3+2n^2+2n)}{(n^2+1)(n^2+2n+2)}\\\\
&=\frac{-n^2-n+1}{(n^2+1)(n^2+2n+2)}\;.
\end{align*}$$
The denominator is always positive, so this is negative when the numerator is negative, i.e., when $-n^2-n+1<0$. This occurs when $n^2+n>1$, i.e., when $n(n+1)>1$, and this is clearly true for every positive integer $n$. Thus, we know that $a_1>a_2>a_3>\ldots\;$.
How small can these fractions get? For every positive integer $n$ we have
$$0<a_n=\frac{n}{n^2+1}<\frac{n}{n^2}=\frac1n\;.\tag{1}$$
Thus, $a_n$ is always bigger than $0$, but by taking $n$ big enough, we can get it as close to $0$ as we want. That says that $\inf A$ ought to be $0$. Now you just have to show that if $\epsilon>0$, there is an $n$ such that $a_n<\epsilon$. The inequality $(1)$ should help you do that.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1040358",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Using an Integral to Solve for a Variable a I am struggling to use the following equation:
$$
\int_0^a \sqrt{a^2-x^2}\,\,\text{sgn}(|x|-1)\, dx = 0
$$
where $a > 1$, to deduce that $a = \text{cosec}(\frac{\pi}{4} - \frac{\alpha}{2})$, where $\alpha$ satisfies $\alpha = \cos(\alpha)$.
I integrate the integrand, via
$$
\int_0^a \sqrt{a^2-x^2}\,\,\text{sgn}(|x|-1)\, dx = -\int_0^1 \sqrt{a^2-x^2}\, dx + \int_1^a \sqrt{a^2-x^2}\, dx
$$
But once I calculate those integrals I cannot seem to get any closer to the answer.
Any help would be great.
| Your question is now: If $\alpha =2\sqrt{1-\frac{1}{a^{2}}}$ how can one
prove that $\cos \alpha =\alpha ?$
Your question in the title is :'' solve for a variable $a$ ''
it means that: the problem is: solve for the variable $a$
the equation $\cos \alpha =\alpha ,$ where $a>1.$ So, lets go:
From $\frac{\alpha }{2}=\sqrt{1-\frac{1}{a^{2}}}$ then $\cos (\frac{\alpha }{%
2})=\frac{1}{a}$ (figure may help i will attach it below). But $\cos ^{2}(%
\frac{\alpha }{2})=\frac{1+\cos \alpha }{2}=\frac{1}{a^{2}}$ hence $\cos
\alpha =\frac{2}{a^{2}}-1.$ Therefore $\cos \alpha =\alpha $ if and only if $%
\frac{2}{a^{2}}-1=2\sqrt{1-\frac{1}{a^{2}}},$ or $\sqrt{1-\frac{1}{a^{2}}}=%
\frac{1}{a^{2}}-\frac{1}{2}.$ So solving this equation and taking only
solution $>1$ one obtains $a=\frac{\sqrt{2}}{\sqrt[4]{3}}\simeq
1.07>1.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1040513",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Evaluation of $\int \frac{x^4}{(x-1)(x^2+1)}dx$ Evaluation of $\displaystyle \int \frac{x^4}{(x-1)(x^2+1)}dx$
$\bf{My\; Try::}$ Let $$\displaystyle I = \int\frac{x^4}{(x-1)(x^2+1)}dx = \int \frac{(x^4-1)+1}{(x-1)(x^2+1)}dx = \int\frac{(x-1)\cdot (x+1)\cdot (x^2+1)}{(x-1)(x^2+1)}+\int\frac{1}{(x-1)(x^2+1)}dx$$
So $\displaystyle I = \int (x+1)dx+J\;\,\;,$ Where $\displaystyle J = \int\frac{1}{(x-1)(x^2+1)}dx$
Now can we solve $J$ without using Partial fraction.
If yes then plz explain me, Thanks
| You are just a baby step away from the answer...: $\dfrac{1}{(x-1)(x^2+1)} = \dfrac{1}{2}\left(\dfrac{1}{x-1} - \dfrac{x}{x^2+1} - \dfrac{1}{x^2+1}\right)$. Can you continue?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1041893",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Finding the limit of a series
I got the n'th term but clueless about what to do next
| We have
$$t_r = \dfrac{r(r+1)(2r+1)/6}{r^2(r+1)^2/4} = \dfrac23 \dfrac{2r+1}{r(r+1)} = \dfrac23 \left(\dfrac1r + \dfrac1{r+1}\right)$$
Hence,
\begin{align}
S_n & = \sum_{r=1}^n (-1)^r t_r = \sum_{r=1}^n (-1)^r \dfrac23 \left(\dfrac1r + \dfrac1{r+1}\right) = \dfrac23 \left(\sum_{r=1}^n \dfrac{(-1)^r}r + \sum_{r=1}^n \dfrac{(-1)^r}{r+1} \right)\\
& = \dfrac23 \left(-\dfrac11 + \sum_{r=2}^n \dfrac{(-1)^r}r + \sum_{r=1}^{n-1} \dfrac{(-1)^r}{r+1} + \dfrac{(-1)^n}{n+1} \right)\\
& = \dfrac23 \left(-1 + \sum_{r=1}^{n-1} \dfrac{(-1)^{r+1}}{r+1} + \sum_{r=1}^{n-1} \dfrac{(-1)^r}{r+1} + \dfrac{(-1)^n}{n+1}\right)\\
& = -\dfrac23 \left(1+ \dfrac{(-1)^{n+1}}{n+1}\right)
\end{align}
Now conclude the limit of $S_n$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1043678",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Integral of $\frac{1}{\sqrt{x^2-x}}dx$ For a differential equation I have to solve the integral $\frac{dx}{\sqrt{x^2-x}}$. I eventually have to write the solution in the form $ x = ...$ It doesn't matter if I solve the integral myself or if I use a table to find the integral. However, the only helpful integral in an integral table I could find was:
$$\frac{dx}{\sqrt{ax^2+bx+c}} = \frac{1}{\sqrt{a}} \ln \left|2ax + b +2\sqrt{a\left({ax^2+bx+c}\right)}\right|$$
Which would in my case give:
$$\frac{dx}{\sqrt{x^2-x}} = \ln \left|2x -1 + 2\sqrt{x^2-x}\right|$$
Which has me struggling with the absolute value signs as I need to extract x from the solution. All I know is that $x<0$ which does not seem to help me either (the square root will only be real if $x<-1$).
Is there some other formula for solving this integral which does not involve absolute value signs or which makes extracting $x$ from the solution somewhat easier? Thanks!
| $x^2-x\ge0\iff x\not\in(0,1)$. Thus, we need to choose whether $x\ge1$ or $x\le0$.
For $x\ge1$, let $x=u+\frac12$ and $u=\frac12\sec(\theta)$. We can assume that $\theta\in[0,\frac\pi2)$ since that gives a full range for $x\ge1$. As is shown in this answer, $\int\sec(\theta)\,\mathrm{d}\theta=\log(\sec(\theta)+\tan(\theta))+C$. Therefore,
$$
\begin{align}
\int\frac{\mathrm{d}x}{\sqrt{x^2-x}}
&=\int\frac{\mathrm{d}u}{\sqrt{u^2-\frac14}}\\
&=\int\frac{\sec(\theta)\tan(\theta)\,\mathrm{d}\theta}{\tan(\theta)}\\[9pt]
&=\log(\sec(\theta)+\tan(\theta))+C\\[9pt]
&=\log\left(2x-1+2\sqrt{x^2-x}\right)+C
\end{align}
$$
Since $2x-1+2\sqrt{x^2-x}\ge0$ for all $x\ge1$, absolute values are not needed.
For $x\le0$, we can assume $\theta\in(\frac\pi2,\pi]$. Then, we get
$$
\int\frac{\mathrm{d}x}{\sqrt{x^2-x}}=\log\left(1-2x-2\sqrt{x^2-x}\right)+C
$$
Since $1-2x-2\sqrt{x^2-x}\ge0$ for all $x\le0$, absolute values are not needed.
The absolute values are needed only if we want to try to give a solution for all $x\not\in(0,1)$. However, since the interval of integration cannot span $(0,1)$, we must have either $x\ge1$ or $x\le0$.
Since you have stated that $x\lt0$ in your question, that means we can use the solution for $x\lt0$:
$$
\int\frac{\mathrm{d}x}{\sqrt{x^2-x}}=\log\left(1-2x-2\sqrt{x^2-x}\right)+C
$$
and we can solve for $x$. If
$$
u=\log\left(1-2x-2\sqrt{x^2-x}\right)
$$
Then
$$
\begin{align}
e^u&=1-2x-2\sqrt{x^2-x}\\
\left(e^u+2x-1\right)^2&=4x^2-4x\\
e^{2u}+(4x-2)e^u+1&=0\\
x&=\frac12-\frac{e^u+e^{-u}}4\\
&=\frac12(1-\cosh(u))
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1044834",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 7,
"answer_id": 2
} |
Prove that $\sum_{i = 2^{n-1} + 1}^{2^n}\frac{1}{a + ib} \ge \frac{1}{a + 2b}$
Prove that $$\sum_{i = 2^{n-1} + 1}^{2^n}\frac{1}{a + ib} \ge \frac{1}{a + 2b}$$
I tried to to prove the above statement using the AM-HM inequality:
$$\begin{align}\frac{1}{2^n - 2^{n-1}}\sum_{i = 2^{n-1} + 1}^{2^n}\frac{1}{a + ib} &\ge \frac{2^n - 2^{n-1}}{\sum_{i = 2^{n-1} + 1}^{2^n}(a + ib)}\\
\sum_{i = 2^{n-1} + 1}^{2^n}\frac{1}{a + ib} &\ge \frac{(2^n - 2^{n-1})^2}{\frac{2^n -2^{n-1}}{2}(2a + (2^n + 2^{n-1} + 1)b)}\\
&=\frac{2^{n+1} - 2^n}{2a + (2^n + 2^{n-1} + 1)b}\end{align}$$
after which I am more or less stuck. How can I continue on from here, or is there another method?
| A generalized version of the result might be easier to prove:
For every positive integers $k$ and $m$, $$\sum_{i=k+1}^{k(m+1)}\frac1{a+ib}\geqslant\frac{m}{a+(m+1)b}.$$
The question asks about the case $k=2^{n-1}$ and $m=1$.
To prove the claim, note that $a+ib\leqslant a+k(m+1)b$ for every $i$ used in the sum $S$ of the LHS and that the sum $S$ has $km$ terms hence $$S\geqslant\frac{km}{a+k(m+1)b}=\frac{m}{(m+1)b}\,\left(1-\frac{a}{a+k(m+1)b}\right).$$ The RHS is an increasing function of $k$ hence it is at least equal to its value when $k=\color{red}{\bf1}$, that is, $$S\geqslant\frac{\color{red}{\bf1}\cdot m}{a+\color{red}{\bf1}\cdot (m+1)\,b}=\frac{m}{a+(m+1)b}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1044910",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Need help setting this up: find $I_z$ the given lamina with uniform density of 1 Find $I_z$ for the given lamina with $\rho=1$,
$z = x^2 + y^2 \;\text{ and }\; 0 ≤ z ≤ h$.
I tried to set it up the following, but I am not sure if this is correct:
I know $\int_{0}^{h}dz, z-x^2-y^2=0$ and $I_z=\int_{s}\int x^2+y^2ds$
but how do I setup the rest?
$$
I_z=\int_{0}^{h}\int
$$
| Hint:
Since
\begin{align*}
I_z&=\int\int_S\left(x^2+y^2\right)\mathrm ds
\end{align*}
and the shape $S$ of the lamina is given by $z=x^2+y^2$ with $0\leq z\leq h$, so
\begin{align*}\mathrm ds &=\left[\left(\frac{\partial z}{\partial x}\right)^2+\left(\frac{\partial z}{\partial y}\right)^2+1\right]^{1/2}\mathrm d A\\
&=\sqrt{4x^2+4y^2+1}\;\mathrm d A
\end{align*}
It follows
\begin{align*}
I_z&=\int\int_S\left(x^2+y^2\right)\sqrt{4x^2+4y^2+1}\;\mathrm d A
\end{align*}
We can work with polar coordinates setting $x=r\cos\theta$, $y=r\sin\theta$, for $0\leq r\leq \sqrt{h}$, $0\leq \theta \leq 2\pi$ and $\mathrm d A=r\mathrm dr\mathrm d \theta$.
Then
\begin{align*}
I_z&=\int_{\theta=0}^{\theta=2\pi}\int_{r=0}^{r=\sqrt{h}}{r^2\sqrt{4r^2+1}\,r\mathrm dr\mathrm d \theta}\\
&=\int_{\theta=0}^{\theta=2\pi}\int_{r=0}^{r=\sqrt{h}}{r^3\sqrt{4r^2+1}\,\mathrm dr\mathrm d \theta}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1045767",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Proving formula for sum of squares with binomial coefficient $$\sum_{k=0}^{n-1}(k^2)= \binom{n}{3} + \binom{n+1}{3}$$
How should I prove that it is the correct formula for sum of squares?
Should I use induction to prove the basis? Any help is appreciated.
| Yes you can. Base case is easy. After it, you have to prove following implication.
$$\left(\underbrace{\sum_{k=0}^{n-1}(k^2)= \binom{n}{3} + \binom{n+1}{3}}_{\mathrm{Assumption}}
\right) \Longrightarrow \left(
\underbrace{\sum_{k=0}^{n}(k^2)= \binom{n+1}{3} + \binom{n+2}{3}}_{\mathrm{Thesis}}
\right)
$$
Now, using assumption transform thesis.
$$\begin{split}
\sum_{k=0}^{n}(k^2)
= n^2 + \sum_{k=0}^{n-1}(k^2)
&= n^2 + \binom{n}{3}+\binom{n+1}{3}\\
&= n^2 + \frac{n!}{3!\cdot (n-3)!} +\binom{n+1}{3}\\
&=n^2 + \frac{n \cdot (n-1) \cdot (n-2)}{6} + \binom{n+1}{3}\\
&=\frac{6n^2 + \left(n^3-3n^2+2n\right)}{6}+ \binom{n+1}{3}\\
&=\frac{n(n+1)(n+2)}{3!}+ \binom{n+1}{3}\\
&=\frac{(n+2)!}{3! \cdot \left((n+2)-3\right)!}+ \binom{n+1}{3}\\
&= \binom{n+2}{3} + \binom{n+1}{3}
\end{split}$$
$\mathscr{Q.E.D.}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1046693",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
Gradient of $x^T B^T B x - x^T B^T b - b^T Bx$ I want to compute the gradient $\nabla_x f(x)$ of $f(x) = x^T B^T B x - x^T B^T b - b^T Bx$ with respect to the vector $x$. So far I have tried below. But when I try to add them together, I couldn't see they come together.
Edits
Now I think I got it. They does come together as the following answers. Since $C=B^TB$, $C$ is symmetric. Then $c_{ij} = c_{ji}$, so
$$\frac{\partial D}{\partial x} = 2\Big[\sum_{i=1}^n c_{1i} x_i \ \ \sum_{i=1}^n c_{2i} x_i \ \ \cdots \ \ \sum_{i=1}^n c_{ni} x_i\Big] =2 C x.$$
Hence $$\frac{\partial x^TBb}{\partial x} = \Big[
\sum_{i=1}^n b_{i1} b_i \ \
\sum_{i=1}^n b_{i2} b_i \ \
\ \ \cdots \ \ \sum_{i=1}^n b_{in} b_i \ \
\Big] = B^T b.$$
So $\frac{d}{dx} (x^T B^T B x - 2x^T B^T b) = 2B^T B - 2B^Tb$.
Thank you!
Let $C = B^T B$.
\begin{align*}
x^T C x
=&
\begin{pmatrix} x_1 & x_2 & \cdots & x_n\end{pmatrix}
\begin{pmatrix}
c_{11} & c_{12} & \cdots & c_{1m}\\
c_{21} & c_{22} & \cdots & c_{2m}\\
\vdots & \vdots& \ddots & \vdots \\
c_{n1} & c_{n2} & \cdots & c_{nm}\\
\end{pmatrix}
\begin{pmatrix} x_1 \\ x_2 \\ \vdots \\x_n\end{pmatrix}\\
=&
\begin{pmatrix} x_1 & x_2 & \cdots & x_n\end{pmatrix}
\begin{pmatrix}
c_{11} x_1 + c_{12} x_2 + \cdots c_{1n} x_n\\
c_{21} x_1 + c_{22} x_2 + \cdots c_{2n} x_n\\
\vdots\\
c_{n1} x_1 + a_{n2} x_2 + \cdots a_{nn} x_n\\
\end{pmatrix}\\
=& x_1(c_{11} x_1 + c_{12} x_2 + \cdots c_{1n} x_n)+x_2(c_{21} x_1 + c_{22} x_2 + \cdots c_{2n} x_n) + \cdots + x_n(c_{n1} x_1 + a_{n2} x_2 + \cdots a_{nn} x_n)
\end{align*}
Since the derivative of the scalar $D = x^T C x$ by a vector $x$ is
$$\frac{\partial D}{\partial x} = \bigg(\frac{\partial D}{\partial x_1} \frac{\partial D}{\partial x_2} \cdots
\frac{\partial D}{\partial x_n}\bigg),$$
I have:
\begin{align*}
\frac{\partial D}{\partial x_1}
&= c_{11} x_1 + c_{12} x_2 + \cdots c_{1n} x_n
+ c_{11} x_1
+ c_{21} x_2
+ \cdots + c_{n1} x_n = \sum_{i=1}^n (c_{1i} + c_{i1})x_i
\end{align*}
Hence $$\frac{\partial D}{\partial x} = \Big[\sum_{i=1}^n (c_{1i} + c_{i1})x_i \ \ \sum_{i=1}^n (c_{2i}+c_{i2})x_i \ \ \cdots \ \ \sum_{i=1}^n (c_{ni}+c_{in})x_i\Big].$$
\begin{align*}
x^T B b
=&
\begin{pmatrix} x_1 & x_2 & \cdots & x_n\end{pmatrix}
\begin{pmatrix}
b_{11} b_1 + b_{12} b_2 + \cdots b_{1n} b_n\\
b_{21} b_1 + b_{22} b_2 + \cdots b_{2n} b_n\\
\vdots\\
b_{n1} b_1 + b_{n2} b_2 + \cdots b_{nn} b_n\\
\end{pmatrix}\\
=&
x_1(b_{11} b_1 + b_{12} b_2 + \cdots b_{1n} b_n)+
x_2(b_{21} b_1 + b_{22} b_2 + \cdots b_{2n} b_n)+\cdots+
x_n(b_{n1} b_1 + b_{n2} b_2 + \cdots b_{nn} b_n)
\end{align*}
Hence $$\frac{\partial x^TBb}{\partial x} = \Big[
\sum_{i=1}^n b_{1i} b_i \ \
\sum_{i=1}^n b_{2i} b_i \ \
\ \ \cdots \ \ \sum_{i=1}^n b_{ni} b_i \ \
\Big].$$
| Hint: For a scalar $\epsilon$ and appropriate sized vectors $x,y$ we have:
$f(x+\epsilon y) = (x+\epsilon y)^TB^TB(x+\epsilon y) - (x+\epsilon y)^TB^Tb - b^TB(x+\epsilon y)$
$= x^TB^TBx + \epsilon y^TB^TBx + \epsilon x^TB^TBy + \epsilon^2 y^TB^TBy - x^TB^Tb - \epsilon y^TB^Tb - b^TBx - \epsilon b^TBy$
$= (x^TB^TBx - x^TB^Tb - b^TBx) + \epsilon(y^TB^TBx+x^TB^TBy-y^TB^Tb-b^TBy) + \epsilon^2y^TB^TBy$
$= f(x) + \epsilon(2y^TB^TBx-2y^TBb) + \epsilon^2\text{stuff}$
where we have used the fact that $x^TB^TBy = y^TB^TBx$ and $b^TBy = y^TB^Tb$, since the transpose of a scalar is itself.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1047905",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Compute $\int \frac{\mathrm{d}x}{49x^2+1}$ So I tried solving this by taking a substitute for the integrand, $t=49x$, so its derivative is $dx = \frac {dt} {49}$. Then you insert it into the integrand and get $$\int \frac{\mathrm dt}{49(t^2 +1)} = \frac{1}{49}\int\frac{1}{t^2 +1}dt = \frac{1}{49}\arctan t + c = \frac{1}{49}\arctan 49x + c $$
Why is this not correct? Why do I have to take substitute $t=7x$ and then do the thingy? I don't know what I'm doing incorrect.
| $$\begin{align}
F(x) &= \int \left(\frac{1}{49x^2 + 1} \right)dx \\
&= \int \left(\frac{1}{(7x)^2 + 1} \right)dx \\
t &= 7x \\
\left(7x \right)dx &= dt \\
\left(\frac{d}{dx}(7x) \right)dx &= dt \\
\left(7*\frac{d}{dx}(x) \right)dx &= dt \\
\left(7*\frac{dx}{dx} \right)dx &= dt \\
\left(7 \right)dx &= dt \\
dx &= \left(\frac{1}{7} \right)dt \\
F(t) &= \int \left( \frac{1}{t^2 + 1}*\frac{1}{7} \right)dt \\
&= \frac{1}{7}*\int \left( \frac{1}{t^2 + 1} \right)dt \\
&= \frac{1}{7}*\arctan(t) \\
t &= 7x \\
F(x) &= \frac{1}{7}*\arctan(7x) \\
F(x) &= \bf \left[ \frac{1}{7}*\arctan(7x) + C \right] \\
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1048548",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
Summation notation for divided factorial. I have the following sum
$$5\cdot4\cdot3+5\cdot4\cdot2+5\cdot4\cdot1+5\cdot3\cdot2+5\cdot3\cdot1+$$$$5\cdot2\cdot1+4\cdot3\cdot2+4\cdot3\cdot1+4\cdot2\cdot1+3\cdot2\cdot1$$
It is basically $5!$ divided by two of the numbers in the factorial. So
$$\frac{5!}{1\cdot2}+\frac{5!}{1\cdot3}+\frac{5!}{1\cdot4}+...+\frac{5!}{3\cdot5}+\frac{5!}{4\cdot5}$$
Is there a way to write this as a single summation?
| You can write it as a single sum as follows
$$\frac{5!}{1\cdot2}+\frac{5!}{1\cdot3}+\frac{5!}{1\cdot4}+...+\frac{5!}{3\cdot5}+\frac{5!}{4\cdot5}=5!\sum_{1\le i <j\le 5}\frac{1}{i\cdot j}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1049060",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Linear Algebra - Find inverse of $A$ I have this problem :
$$A = \left(\begin{array}{ccc}
3 & -1 & 1 \\
2 & 0 & 1 \\
1 & -1 & 2 \end{array}\right) $$
1) Show that $A^3-5A^2+8A-4I=0$.
2) Using (1) To find $A^{-1}$.
I did (1) show that is correct, Usually to find inverse I use the Identity Matrix.
Anyhow for this problem, I thought of something like that :
$$A^3-5A^2+8A-4I=0 \implies
\\A^3-5A^2+8A=4I \implies
\\\frac{A^3-5A^2+8A}{4}=I \implies
\\\frac{A^3}{4}-\frac{5}{4}A^2+2A=I \implies
\\ A\left(\frac{A^2}{4}-\frac{5}{4}A+2I\right)=I$$
Then I think that : $$A^{-1}=\left(\frac{A^2}{4}-\frac{5}{4}A+2I\right)$$
I checked if $A^{-1}A=I$, but this is not the case.
Any ideas? Thanks!
| You have
$A^3 - 5 A^2 + 8A = 4I $
Left multiply both sides by $A^{-1}$. Use the fact that matrix multiplication is associative, and $A^{-1}A=I$,
$$
A^2 - 5A + 8I = 4A^{-1}
$$
which gives you an $A^{-1}$ of
$$
\frac{1}{4}
\begin{bmatrix}
1 & 1 & -1 \\
-3 & 5 & -1 \\
-2 & 2 & 2
\end{bmatrix}
$$
You can now check that $A A^{-1}$ is indeed $I$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1050111",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solve the linear homogeneous recurrence relation with constant coefficients $$9a_{n} = 6a_{n-1}-a_{n-2}, a_{0}=6, a_{1}=5$$
So
$$x^n = (6x^{n-1}-x^{n-2})\div9$$
thus
$$[x^2 = (6x-1)\div9] \equiv [x^2 - \frac{2}{3}x + \frac{1}{9} = 0], x=\frac{1}{3}$$
also
$$a_{2}=\frac{8}{3}, a_{3}=\frac{31}{27}$$
How do I plug in that x/root to solve for the given recurrence relation?
I tried
$$9a_{n} = 6(\frac{1}{3})^{n-1} - (\frac{1}{3})^{n-2}$$
which for n=1 gives 5... which is correct, for a_1 though, not 9a_1? So that's not right.
| Since $a_n=c\left(\frac{1}{3}\right)^n+dn\left(\frac{1}{3}\right)^n$, $a_0=2\implies c=6$ and $a_1=5\implies d=9$, so
$\displaystyle a_n=\frac{6}{3^n}+\frac{9n}{3^n}=\frac{2+3n}{3^{n-1}}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1050440",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that $1 + 4 + 7 + · · · + 3n − 2 = \frac {n(3n − 1)}{2}$ Prove that
$$1 + 4 + 7 + · · · + 3n − 2 =
\frac{n(3n − 1)}
2$$
for all positive integers $n$.
Proof: $$1+4+7+\ldots +3(k+1)-2= \frac{(k + 1)[3(k+1)+1]}2$$
$$\frac{(k + 1)[3(k+1)+1]}2 + 3(k+1)-2$$
Along my proof I am stuck at the above section where it would be shown that:
$\dfrac{(k + 1)[3(k+1)+1]}2 + 3(k+1)-2$ is equivalent to $\dfrac{(k + 1)[3(k+1)+1]}2$
Any assistance would be appreciated.
| Call $S = 1 + 4 + \ldots + [3n-2]$.
Add the numbers in reverse direction: $S = [3n-2] + [3n-5] + \ldots + 1$.
Add the two equations term by term: $2S = (1 + [3n-2]) + (4 + [3n - 5]) + \ldots + ([3n-2]+1) = n (3n-1)$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
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Show $\sum_{n=1}^\infty\frac{1}{n^2+3n+1}=\frac{\pi\sqrt{5}}{5}\tan\frac{\pi\sqrt{5}}{2}$. How to show that
$$\sum_{n=1}^\infty\frac{1}{n^2+3n+1}=\frac{\pi\sqrt{5}}{5}\tan\frac{\pi\sqrt{5}}{2}$$
?
My try:
We have
$$n+3n+1=\left(n+\frac{3+\sqrt{5}}{2}\right)\left(n+\frac{3-\sqrt{5}}{2}\right),$$
so
$$\frac{1}{n^2+3n+1}=\frac{2}{\sqrt{5}}\left(\frac{1}{2n+3-\sqrt{5}}-\frac{1}{2n+3+\sqrt{5}}\right).$$
Then, I don't know how to proceed.
| I think we can make some use of the residue theorem. Write $n^2+3 n+1 = (n+3/2)^2-5/4$ and the sum is
$$\sum_{n=1}^{\infty} \frac1{\left (n+\frac{3}{2} \right )^2-\frac{5}{4}} = \frac12 \sum_{n=-\infty}^{\infty} \frac1{\left (n+\frac{3}{2} \right )^2-\frac{5}{4}} - 1 +1$$
(To get the doubly infinite sum, I had to add back the $n=0$ and $n=-1$ terms, which happen to sum to zero.)
The sum on the RHS may be attacked via the residue theorem, using the following:
$$\sum_{n=-\infty}^{\infty} f(n) = - \pi \sum_k \operatorname*{Res}_{z=z_k} [f(z)\cot{\pi z} ] $$
where $z_k$ is a non-integer pole of $f$. The poles are at
$$z_{\pm} = -\frac{3}{2} \pm \frac{\sqrt{5}}{2} $$
The sum is then equal to
$$\frac{\pi}{2 \sqrt{5}} \left [\cot{\left (\frac{3 \pi}{2} - \frac{\sqrt{5} \pi}{2} \right )} - \cot{\left (\frac{3 \pi}{2} + \frac{\sqrt{5} \pi}{2} \right )} \right ] = \frac{\pi}{\sqrt{5}}\tan{\frac{\sqrt{5}\pi}{2}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1053683",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Find limit of $\frac {1}{x^2}- \frac {1}{\sin^2(x)}$ as x goes to 0 I need to use a taylor expansion to find the limit.
I combine the two terms into one, but I get limit of $\dfrac{\sin^2(x)-x^2}{x^2\sin^2(x)}$ as $x$ goes to $0$. I know what the taylor polynomial of $\sin(x)$ centered around $0$ is… but now what do I do?
| We have:
$$\lim_{x\to 0} \frac{\sin^2 x -x^2}{x^2\sin^2 x}$$
Since direct substitution yields $\frac{0}{0}$, we use L'Hospital's rule:
$$\lim_{x\to 0} \frac{2\sin x \cos x - 2x}{2x\sin^2 x + 2x^2\sin x \cos x }$$
Using it again:
$$\lim_{x\to 0} \frac{2\cos2x - 2}{x^2(2\cos^2 x-2\sin^2 x) + 2\sin^2 x +8x\sin x \cos x }$$
And again:
$$\lim_{x\to 0} \frac{-4\sin2x}{6x(2\cos^2 x-2\sin^2 x) + 12x\sin x \cos x -8x^2\sin x \cos x }$$
One more to go:
$$\lim_{x\to 0} \frac{-8\cos2x}{12(2\cos^2 x-2\sin^2 x) + 64x\sin x \cos x +x^2(8\cos^2 x-8\sin^2 x)} = \frac{-8}{24} = -\frac{1}{3}$$
Keep in mind this is a much longer method. L'Hospital's Rule is a simple yet sometimes very lengthy derivation of limits such as these. Any alternatives seen could most likely be easier to do.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1053754",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Show that for a $2\times 2$ matrix $A^2=0$ Assuming that given a $2\times2$ matrix $A$ with the property $A^3=0$ show that $A^2=0$.
Okay so this question is messing with me. When it says $A^3=0$ do you think it means the determinant? How would one approach this? Any help would be much appreciated
| It does not mean the determinant. It means that if you multiply $A$ with itself three times, you get the zero matrix. The nullspace of $A^2$ is contained in the nullspace of $A^3$, i.e. if $A^2 x = 0$, then clearly $A^3 x = A(A^2 x) = 0$. Particularly, this means that the nullspace can only increase as you apply more powers of $A$. The nullspace of $A^2$ can be zero dimensional (just the zero vector), one dimensional or two dimensional since $A$ is a $2\times 2$ matrix.
$A^2$ cannot have trivial nullspace (consisting of just the zero vector) because this would imply that $A$ is invertible and so $A^3$ is invertible, particularly $AA^2$ couldn't be the zero matrix since the zero matrix is not invertible. This gives a contradiction.
If $A^2$ has a one dimensional nullspace, $A$ would also have a one dimensional nullspace and the nullspaces would be the same. Since they have the same nullspaces, it follows that $A^2A$ has only a one dimensional nullspace, which is again a contradiction since we need that $A^3$ is the zero matrix (that is, $A^3$ has a two dimensional nullspace).
Thus we must conclude that $A^2$ has a two dimensional nullspace, i.e. $A$ is the zero matrix.
From @amcalde's comment, I decided to work out the exact expression for $A^3$ to argue that $A^2$ must be the zero matrix. Suppose $A = \left(\begin{array}{cc} a & b \\ c & d\end{array}\right)$, then
$$A^2 = \left(\begin{array}{cc} a^2 + bc & ab + bd \\ ac + cd & bc + d^2\end{array}\right)$$
and additionally
$$A^3 = \left(\begin{array}{cc} a(a^2+bc)+b(ac+cd) & a(ab+bd)+b(bc+d^2) \\ c(a^2+bc)+d(ac+cd) & c(ab+bd)+d(bc+d^2)\end{array}\right)$$
Since $A^3 = 0$, we have the following equations:
$$a(a^2+bc) + b(ac+cd) = 0\tag{1}$$
$$a(ab+bd)+b(bc+d^2) = 0\tag{2}$$
$$c(a^2+bc)+d(ac+cd) = 0\tag{3}$$
$$c(ab+bd) + d(bc+d^2) = 0\tag{4}$$
Suppose now that not all of $a^2+bc$, $ac+cd$, $ab+bd$ and $bc+d^2$ are zero. Particularly, suppose $a^2+bc\neq 0$. If $ac+cd = 0$, then $a=0$ by equation $1$ and so by equation $2$, since $b\neq 0$, $bc+d^2 = 0$. Thus by equation $4$, we must have that $ab+bd = 0$ (since $c\neq 0$). However since $a=0$ and $b\neq 0$, we conclude that $d=0$. Since $bc+d^2=0$ and $d=0$, we conclude that $bc=0$ which gives us a contradiction.
Suppose again that $a^2+bc\neq 0$. If $ab+bd = 0$, then either $b=0$ or $a=-d$. In the first case, $a\neq 0$. By equation $4$, we get that $d=0$ and then by equation $1$, we get $a=0$ which is a contradiction since $a^2+bc=0$. In the second case, we get that $bc+d^2\neq 0$. By equation $4$, it follows that $d=0$. By equation $2$, it follows that $b=0$ which in turn gives us that $a=0$ by equation $1$. This is again a contradiction since $a^2+bc\neq 0$.
Repeating this rather yucky process, you can see that all of $a^2+bc$, $ac+cd$, $ab+bd$ and $bc+d^2$ are zero. That is to say that if $A^3 = 0$, $A^2$ must also be the zero matrix. This is a much more tedious approach as you see.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1053886",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solving $n^5+n^4-3=x^2\pmod p$ Prove that for every odd prime number $p$ there is a natural number $n$ such that the equation $n^5+n^4-3=x^2\pmod p$ has no solutions.
So we have to understand that for each $p$ we can find $n$ such that the Legendre symbol $\left(\dfrac{n^5+n^4-3}{p}\right) = -1$. For $p=4k+3$ we can take $n=1$. How to deal with $4k+1$ case?
| The curve $x^2=n^5+n^4-3$ has genus $2$, and so the Hasse-Weil bound implies that the number of ordered pairs $(x,n)$ modulo $p$ satisfying the equation is between $p-4\sqrt p$ and $p+4\sqrt p$. There can be at most five values of $n$ for which the right-hand side is congruent to $0\pmod p$; for all other values of $n$, if $x$ satisfies the congruence then so does $-x$. Therefore the number of distinct $n\pmod p$ that can appear is at most $5 + \frac12(p+4\sqrt p-5)$. This quantity is less than $p$ as soon as $p\ge29$, and so there exists $n\pmod p$ such that the equation has no solutions. And the result can be checked by hand for $3\le p\le 23$.
| {
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"url": "https://math.stackexchange.com/questions/1054002",
"timestamp": "2023-03-29T00:00:00",
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What is the value of the expression $2x^2 + 3xy – 4y^2$ when $x = 2$ and $y = - 4$? What is the value of the expression $2x^2 + 3xy – 4y^2$ when $x = 2$ and $y = - 4$?
I'm not good at algebra so please explain in easy to understand steps.
Thanks
| Subbing in $x=2$ and $y=-4$ into the equation gives:
$2(2^2) +3(2)(-4)-4((-4)^2)=2(4) + (-24) -4(16)=8-24-64=8-88=-80$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1055124",
"timestamp": "2023-03-29T00:00:00",
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Simplifying $\frac{x^6-1}{x-1}$ I have this:
$$\frac{x^6-1}{x-1}$$
I know it can be simplified to $1 + x + x^2 + x^3 + x^4 + x^5$
Edit : I was wondering how to do this if I didn't know that it was the same as that.
| Hint:- $$y^3-1=y^2(y-1)+y(y-1)+(y-1)=(y-1)\left(y^2+y+1\right)$$
Solution:-
$y=x^2\implies x^6-1=\left(x^2-1\right)\left(x^4+x^2+1\right)=(x-1)(x+1)\left(x^4+x^2+1\right)$ $$\boxed{\therefore\dfrac{x^6-1}{x-1}=(x+1)\left(x^4+x^2+1\right)=x^5+x^4+x^3+x^2+x+1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1055671",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Use Laplace Transform to solve the following IVP: I know that this is a somewhat simple problem but I have been having trouble coming up with the little "tricks" that help with Laplace.
The problem is:
$y''+2y' +5y = e^{-t}\sin(2t)$ where $y(0) = 2, y'(0) = -1$
Attempt at Solution
$(s^2+2s+5)Y = \frac{2}{(s+1)^2+4} + (s+2)(2) + 1(-1)$
$Y = \frac{2}{(s^2+2s+5)^2}+\frac{2s+3}{s2+2s+5}$
$Y = \frac{2}{(s^2+2s+5)^2}+2(\frac{s+1}{s^2+2s+5})+\frac{1}{s^2+2s+5}$
$Y = \frac{2}{(s^2+2s+5)^2}+\frac{1}{2}\sin(2t)+2e^{-t}\cos(2t)$
And I am stuck here.
The answer given in the book is as follows:
Y = $\frac{5}{8}e^{-t}\sin(2t)+2e^{-t}\cos(2t)-\frac{1}{4}te^{-t}\cos(2t)$
Any help would be greatly appreciated.
| We can take inverse Laplace transform by using the Bromwich integral.
That is,
\begin{align}
\mathcal{L}^{-1}\{Y(s)\} &= \frac{1}{2\pi i}\lim_{T\to\infty}\int_{\gamma -iT}^{\gamma +iT}Y(s)e^{st}ds\\
&= \sum\text{Res}
\end{align}
where
$$
Y(s) = \frac{2}{(s^2 + 2s + 5)^2} + \frac{2s + 3}{s^2 + 2s + 5}
$$
Then the poles of $s$ are at $s = -1\pm 2i$. We can then evaluate the integrals by splitting them into two. In the first one, the poles are of order two.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Complex number calculation I'm supposed to show $ z^{10} $, when z = $ \frac{1+ \sqrt{3i} }{1- \sqrt{3i} } $
I can work it out to $ \frac{(1+\sqrt{3}\sqrt{i})^{10}}{(1-\sqrt{3}\sqrt{i})^{10}} $
However this is inconclusive because I need to show $ z^{10} $ in the form x+yi, and I can't figure out the real and imaginary parts from from this answer because of the exponent.
What must I do?
| It seems to me that @MichaelAlbanese is right. It should be
$ \frac{1+\sqrt{3} i}{1-\sqrt{3} i} $. After removing any complex part from denominator we get
$$
\frac{1+\sqrt{3} i}{1-\sqrt{3} i} = \frac{1+\sqrt{3} i}{1-\sqrt{3} i} \frac{1+\sqrt{3} i}{1+\sqrt{3} i} = \frac{-2 + 2 \sqrt{3} i}{4} = -\frac{1 - \sqrt{3} i}{2}
$$
We can transfer it into polar form, which is $ {e}^{\frac{2}{3} \pi i} $
And now the answer is quite easy to find
$$
{{e}^{\frac{2}{3} \pi i}}^{10} = {e}^{10 \frac{2}{3} \pi i} = {e}^{\frac{20}{3} \pi i} = {e}^{\frac{18 + 2}{3} \pi i} = {e}^{{6 \pi i} + {\frac{2}{3} \pi i}} = {e}^{6 \pi i} {e}^{\frac{2}{3} \pi i} = {e}^{\frac{2}{3} \pi i}
$$
So the final answer is that $ z^{10} = z = -\frac{1 - \sqrt{3} i}{2}$
BTW, if really $ z = \frac{1+\sqrt{3 i}}{1-\sqrt{3 i}} $ you have big troubles because its polar form is something like that:
$$
z = \frac{\sqrt{{\left( \frac{\sqrt{3}}{\sqrt{2}}+1\right) }^{2}+\frac{3}{2}}\,{e}^{i\,\left( \mathrm{atan}\left( \frac{\sqrt{3}}{\sqrt{2}\,\left( \frac{\sqrt{3}}{\sqrt{2}}+1\right) }\right) +\mathrm{atan}\left( \frac{\sqrt{3}}{\sqrt{2}\,\left( 1-\frac{\sqrt{3}}{\sqrt{2}}\right) }\right) +\pi \right) }}{\sqrt{{\left( 1-\frac{\sqrt{3}}{\sqrt{2}}\right) }^{2}+\frac{3}{2}}}
$$
and the answer is
$$
z^{10} = \frac{{\left( {\left( \frac{\sqrt{3}}{\sqrt{2}}+1\right) }^{2}+\frac{3}{2}\right) }^{5}\,{e}^{i\,\left( \mathrm{atan}\left( \frac{\mathrm{sin}\left( 10\,\mathrm{atan}\left( \frac{\sqrt{3}}{\sqrt{2}\,\left( \frac{\sqrt{3}}{\sqrt{2}}+1\right) }\right) \right) }{\mathrm{cos}\left( 10\,\mathrm{atan}\left( \frac{\sqrt{3}}{\sqrt{2}\,\left( \frac{\sqrt{3}}{\sqrt{2}}+1\right) }\right) \right) }\right) -\mathrm{atan}\left( \frac{\mathrm{sin}\left( 10\,\left( -\mathrm{atan}\left( \frac{\sqrt{3}}{\sqrt{2}\,\left( 1-\frac{\sqrt{3}}{\sqrt{2}}\right) }\right) -\pi \right) \right) }{\mathrm{cos}\left( 10\,\left( -\mathrm{atan}\left( \frac{\sqrt{3}}{\sqrt{2}\,\left( 1-\frac{\sqrt{3}}{\sqrt{2}}\right) }\right) -\pi \right) \right) }\right) \right) }}{{\left( {\left( 1-\frac{\sqrt{3}}{\sqrt{2}}\right) }^{2}+\frac{3}{2}\right) }^{5}}
$$
(thanks Maxima…)
| {
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Prove that $\sqrt{n^2 + 2}$ is irrational
Suppose $n$ is a natural number. Prove that $\sqrt{n^2 + 2}$ is irrational.
From looking at the expression, it seems quite obvious to me that $\sqrt{n^2 + 2}$ will be irrational, since $n^2$ will be a natural number, and after adding $2$ to it, $n^2 + 2$ will no longer be a perfect square.
From that logic, I tried letting $n^2 + 2$ = ${a^2\over b^2}$ to find a contradiction, but I wasn't able to formulate a proof that worked. Any suggestions as to how I should approach this?
| $$\begin{array}{l}
\sqrt {2 + n^2 } = \frac{p}{q};\quad and\quad p \wedge q = 1 \\
\sqrt {2 + n^2 } = \frac{p}{q} \Leftrightarrow 2 + n^2 = \left( {\frac{p}{q}} \right)^2 \\
\Leftrightarrow 2 = \left( {\frac{p}{q}} \right)^2 - n^2 \\
\Leftrightarrow 2 = \left( {\frac{p}{q} - n} \right)\left( {\frac{p}{q} + n} \right) \\
\Leftrightarrow 2q^2 = \left( {p - nq} \right)\left( {p - nq} \right) \cdots \left( * \right) \\
\left( * \right) \Rightarrow \left\{ \begin{array}{l}
\exists m_0 \in\mathbb{ N} :p - nq = m_0 q \\
or \\
\exists m_1 \in \mathbb{N} :p + nq = m_1 q \\
\end{array} \right. \\
\Rightarrow \left\{ \begin{array}{l}
\exists m_0 \in \mathbb{N} :p = \left( {n + m_0 } \right)q \\
or \\
\exists m_1 \in \mathbb{N} :p = \left( { - n + m_1 } \right)q \\
\end{array} \right. \\
\left\{ \begin{array}{l}
p = \left( {n + m_0 } \right)q \Rightarrow q/p \\
or \\
p = \left( { - n + m_1 } \right)q \Rightarrow q/p \\
\end{array} \right. \\
\end{array}$$
contradiction with $\quad p \wedge q = 1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1060433",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
Laurent Series, region of convergence I want to find the laurent series for
$$
f(z) = \frac{z}{z^2 - (1+i)z +i}
$$ in powers of $z-1$ and find the region of convergence. I am not quite sure how to do this.
I know that
$$
f(z) = \frac{z}{(z-1)(z-i) }
$$
but I do not know where to go from here. Any help would be great! Thanks!
| First rewrite $f$:
$$
f(z) = \frac{z}{z^2 - (1+i)z +i} = \frac{z}{(z-1)(z-i)} = \left(\frac{A}{z-1} + \frac{B}{z-i}\right) = \frac{1}{1-i}\left(\frac{1}{z-1} - \frac{i}{z-i}\right).
$$
Since you want the expansion around $1$, let's write everything in terms of $(z-1)$:
$$
\begin{align}
f(z) &=\frac{1}{1-i}\left(\frac{1}{z-1} - \frac{i}{(z-1)+(1-i)}\right)\\
\end{align}
$$
Note that $f$ has two singularities, one at $1$ and the other at $i$, so if you want the Laurent expansion around $1$ it has to be done for two different cases: $|z-1|<\sqrt{2}$ and $|z-1|>\sqrt{2}$. Then, for $|z-1|<\sqrt{2}$ we have that $|z-1|<|1-i|$, so:
$$
\begin{align}
f(z) & = \frac{1}{1-i}\left(\frac{1}{z-1} - \frac{i}{1-i}\frac{1}{1 + \frac{z-1}{1-i}}\right)\\
&= \frac{1}{1-i}\left(\frac{1}{z-1} - \frac{i}{1-i}\sum_{k=0}^\infty (-1)^k\left(\frac{z-1}{1-i}\right)^k\right).
\end{align}
$$
Analogously if $|z-1|>|1-i|=\sqrt{2}$:
$$
\begin{align}
f(z) & = \frac{1}{1-i}\left(\frac{1}{z-1} - \frac{i}{z-1}\frac{1}{1 + \frac{1-i}{z-1}}\right)\\
&= \frac{1}{1-i}\left(\frac{1}{z-1} - \frac{i}{z-1}\sum_{k=0}^\infty (-1)^k\left(\frac{1-i}{z-1}\right)^k\right).
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1061716",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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Find cosine of acute angles in a right triangle.
If sides of a right triangle are in Geometric Progression, then find the cosines of acute angles of the triangle.
[Answer] $\frac{\sqrt{5}-1}{2}$,$\sqrt\frac{\sqrt{5}-1}{2}$
My work:
Using Pythagoras Theorem, $a^{2}+a^{2}r^{2}=a^{2}r^{4}$
$1+r^{2}=r^{4}$
$r^{4}-r^{2}-1=0$
Using quadratic equation, $r^{2}=\frac{1\pm\sqrt{1+4}}{2}$
(-) sign rejected because squre cannot be nagetive
so, $r^{2}=\frac{1+\sqrt{5}}{2}$
In right triangle ABC,
$\cos(A)=\frac{ar}{ar^{2}}$=$\frac{1}{r}$=$\frac{1}{\sqrt\frac{1+\sqrt{5}}{2}}$ after rationalization $\sqrt{\frac{2(1-\sqrt{5})}{(1+\sqrt{5})(1-\sqrt{5})}}$ negative in denominator
$\cos(C)=\frac{a}{ar^{2}}$=$\frac{1}{r^{2}}$=$\frac{1}{\frac{1+\sqrt{5}}{2}}$= $\frac{2(1-\sqrt{5})}{(1+\sqrt{5})(1-\sqrt{5})}$=$\frac{2(1-\sqrt{5})}{-4}$= $\frac{\sqrt{5}-1}{2}$
I cannot understand what is wrong with my calculations. Any help is appreciated.
Thanks in advance.
| You actually do have the right answers; they just merely look different. A quick check on wolfram alpha will tell you that they are equivalent.
First answer: http://cl.ly/image/1C3F0O3U1A16
Second answer: http://cl.ly/image/0E0X3z3q0M0B
Both of these calculations can be very quickly justified by multiplying both sides of the equation by the denominator and noticing that the conjugates cancel out all of the squareroots.
| {
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"url": "https://math.stackexchange.com/questions/1061819",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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Existence of Holomorphic function (Application of Schwarz-Lemma) Let, $D=\{z\in \mathbb C:|z|<1\}$. Which are correct?
*
*there exists a holomorphic function $f:D \to D$ with $f(0)=0$ & $f'(0)=2$.
*there exists a holomorphic function $f:D \to D$ with $f\left(\dfrac{3}{4}\right)=\dfrac{3}{4}$ & $f'\left(\dfrac{2}{3}\right)=\dfrac{3}{4}$.
*there exists a holomorphic function $f:D \to D$ with $f\left(\dfrac{3}{4}\right)=-\dfrac{3}{4}$ & $f'\left(\dfrac{3}{4}\right)=-\dfrac{3}{4}$
*there exists a holomorphic function $f:D \to D$ with $f\left(\dfrac{1}{2}\right)=-\dfrac{1}{2}$ & $f'\left(\dfrac{1}{4}\right)=1$.
With the help of Schwarz lemma & its applications, we find that $(1)$ is false & $(3)$ is true.
But, I can not think about options $(2)$ & $(4)$.
| We want to determine whether for given $a,b,c,d$, there exists a holomorphic $f\colon D \to D$ with
*
*$f(a) = b$, and
*$f'(c) = d$.
A typical way to attack such a problem is the Schwarz-Pick lemma, resp. its differential version
$$\frac{\lvert f'(z)\rvert}{1 - \lvert f(z)\rvert^2} \leqslant \frac{1}{1-\lvert z\rvert^2}\tag{1}$$
for $z\in D$ when $f\colon D\to D$ is holomorphic, and if we have equality at one point, then $f$ is an automorphism of $D$.
In our case, we must check whether
$$\lvert d\rvert \leqslant \frac{1 - \lvert f(c)\rvert^2}{1-\lvert c\rvert^2}\tag{2}$$
for some holomorphic $f\colon D\to D$ with $f(a) = b$. If $(2)$ doesn't hold for any such $f$, then $(1)$ tells us that no $f$ with the prescribed properties exists, and if there is such an $f$ that $(2)$ holds, we often get enough restrictions from $(2)$ that constructing a function with the desired properties or a proof that no such function exists are easier.
In case 4., $f\colon D\to D$ with $f\left(\frac{1}{2}\right) = - \frac{1}{2}$ and $f'\left(\frac{1}{4}\right) = 1$, the Schwarz-Pick lemma tells us that we must have $\left\lvert f\left(\frac{1}{4}\right)\right\rvert \geqslant \frac{1}{4}$ since the hyperbolic distance between $f\left(\frac{1}{4}\right)$ and $-\frac{1}{2}$ can be at most equal to the hyperbolic distance between $\frac{1}{4}$ and $\frac{1}{2}$. On the other hand, $(2)$ tells us that we must have $\left\lvert f\left(\frac{1}{4}\right)\right\rvert \leqslant \frac{1}{4}$ in order to have the right hand side $\geqslant 1$. The only point in $D$ satisfying both requirements is $-\frac{1}{4}$, so if an $f$ with the desired properties exists, we must have $f\left(-\frac{1}{4}\right) = -\frac{1}{4}$, and since equality holds in $(1)$ then, it follows that $f(z) = -z$. But then we have $f'\left(\frac{1}{4}\right) = -1$, so there is no holomorphic $f\colon D \to D$ with $f\left(\frac{1}{2}\right) = -\frac{1}{2}$ and $f'\left(\frac{1}{4}\right) = 1$.
For case 2., $f\colon D\to D$ with $f\left(\frac{3}{4}\right) = \frac{3}{4}$ and $f'\left(\frac{2}{3}\right) = \frac{3}{4}$, the Schwarz-Pick lemma is not as effective. From it, we obtain the bounds $$\frac{2}{3} \leqslant \left\lvert f\left(\frac{2}{3}\right)\right\rvert \leqslant \sqrt{\frac{7}{12}},$$
which don't narrow down the possibilities for $f$ much. However, with so much space to play, we suspect that such an $f$ exists. To find one, we move the fixed point of $f$ to $0$ and consider $g = T_{3/4}\circ f \circ T_{-3/4}$, where
$$T_w \colon z \mapsto \frac{z-w}{1-\overline{w}\cdot z}.$$
We want $f$ to "shrink the unit disk towards $\frac{3}{4}$", so we make the ansatz $g(z) = c\cdot z$ for some $c\in (0,1)$ which we want to determine so that $f'\left(\frac{2}{3}\right) = \frac{3}{4}$. So we try
$$f(z) = T_{-3/4}\left(c\cdot T_{3/4}(z)\right).$$
We have $T_{3/4}(2/3) = -\frac{1}{6}$, hence
$$f'\left(\frac{2}{3}\right) = T_{-3/4}'\left(-\frac{c}{6}\right)\cdot c \cdot T_{3/4}'\left(\frac{2}{3}\right).$$
Since
$$T_w'(z) = \frac{(1-\overline{w}z) +\overline{w}(z-w)}{(1-\overline{w}z)^2} = \frac{1-\lvert w\rvert^2}{(1-\overline{w}z)^2},$$
we compute
$$T_{3/4}'(2/3) = \frac{7/16}{(1/2)^2} = \frac{7}{4};\qquad T_{-3/4}'(-c/6) = \frac{7/16}{(1-c/8)^2} = \frac{28}{(8-c)^2}$$
and find that $c$ should satisfy
$$\frac{49 c}{(8-c)^2} = \frac{3}{4}.$$
Solving the quadratic equation gives the solution
$$c = \frac{122 - 14\sqrt{73}}{3} \approx 0.7946491885181928.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1063584",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 1,
"answer_id": 0
} |
Solve cubic equation $x^3-9 x^2-15x-6 =0$ without going Cardano
Solve the cubic equation for $x\in\mathbb{R}$ $$x^3-9 x^2-15x-6 =0$$
Note that the only real solution is $x=3+2\sqrt[3]{7}+\sqrt[3]{7^2}$. Given the regularity of this solution, can we solve for it constructively, without going full Cardano?.
Also, can we prove that there is only one real solution without using the discriminant?
| If $y=x-3$ and $y^3-42y=105$, then take $y=t+r\implies t^3+r^3+(3tr-42)y=105$, then:
$$3tr-42=0$$$$t^3+r^3=105$$
$$3tr-42=0\implies tr=14\implies t^3r^3=2744$$
Then $t^3$, and $r^3$ are roots of:
$$Z^2-105Z+2744=0$$
Then $Z_1=49$ and $Z_2=56$, $\implies t^3=49$ and $r^3=56$, $\implies t=\sqrt[3]{7^2}$ and $r=2\sqrt[3]{7}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1065891",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 2
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Formula for tangent derivatives, how to prove? How to prove?
$$(\tan x)^{(s-1)}=\pi^{-s}\Gamma(s)\left(\zeta\left(s, \frac12-\frac x\pi\right)+(-1)^s\zeta\left(s, \frac12+\frac x\pi\right)\right) $$
| We start with the following claim (I hope it has been proven on this web site, but I couldn't find a link):
$$\sum_{k=-\infty}^\infty\frac{1}{(k-\frac{x}{\pi})^2}=\frac{\pi^2}{\sin^2(x)}$$
Because
$$\frac{d}{dx}\tan(x)=\frac{1}{\cos^2(x)}=\frac{1}{\sin^2(x-\frac{\pi}{2})}$$
it follows that
$$\frac{d}{dx}\tan(x)=\frac{1}{\pi^2}\sum_{k=-\infty}^\infty\frac{1}{(k+\frac{1}{2}-\frac{x}{\pi})^2}$$
Differentiating $s-1$ times gives
$$\tan(x)^{(s-1)}=\pi^{-s}\Gamma(s)\sum_{k=-\infty}^\infty\frac{1}{(k+\frac{1}{2}-\frac{x}{\pi})^{s}}$$
The Hurwitz zeta function is the following:
$$\zeta(s,q)=\sum_{k=0}^\infty\frac{1}{(k+q)^s}$$
Rewriting the summation above:
\begin{align}
\sum_{k=-\infty}^\infty\frac{1}{(k+\frac{1}{2}-\frac{x}{\pi})^{s}}&=\sum_{k=1}^\infty\frac{1}{(-k+\frac{1}{2}-\frac{x}{\pi})^{s}}+\sum_{k=0}^\infty\frac{1}{(k+\frac{1}{2}-\frac{x}{\pi})^{s}}\\
&=\sum_{k=0}^\infty\frac{1}{(-k-\frac{1}{2}-\frac{x}{\pi})^{s}}+\sum_{k=0}^\infty\frac{1}{(k+\frac{1}{2}-\frac{x}{\pi})^{s}}\\
&=\sum_{k=0}^\infty\frac{1}{(-1)^s(k+\frac{1}{2}+\frac{x}{\pi})^{s}}+\sum_{k=0}^\infty\frac{1}{(k+\frac{1}{2}-\frac{x}{\pi})^{s}}\\
&=(-1)^s\zeta\left(s,\frac{1}{2}+\frac{x}{\pi}\right)+\zeta\left(s,\frac{1}{2}-\frac{x}{\pi}\right)
\end{align}
Therefore we conclude:
$$\tan(x)^{(s-1)}=\pi^{-s}\Gamma(s)\left[(-1)^s\zeta\left(s,\frac{1}{2}+\frac{x}{\pi}\right)+\zeta\left(s,\frac{1}{2}-\frac{x}{\pi}\right)\right]$$
Inspired by page 22 of this: http://www.staff.science.uu.nl/~ban00101/funcr2014/fr_opgaven_2014.pdf
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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(Non?)-uniqueness of sums of squares (I've had almost no exposure to number theory, so please keep answers as elementary as possible.)
Write $\mathbb{N} = \{0,1,2,3,\ldots\}$ for the natural numbers. Then every element of $\mathbb{N}$ can be expressed as a sum of squares. For example:
$$6 = 1+1+1+1+1+1$$
Usually, we can get away with fewer terms in our sum. For instance, in the preceeding example, we can get down from 6 terms to 3 by writing $6 = 4+1+1.$ So by the potency of $n \in \mathbb{N}$, let us mean the least $k \in \mathbb{N}$ such that $n$ can be expressed as a sum of $k$-many squares. Hence the potency of $6$ is $3.$
Question. Let $n \in \mathbb{N}$ denote a natural number with potency $k$. Suppose $x$ and $y$ are sequences of length $k$ in
$\mathbb{N}$ such that $$n=\sum_{j = 1}^k x_j^2 = \sum_{j = 1}^k y_j^2.$$
Are the sequences $x$ and $y$ necessarily equal up to a reordering of their terms?
I'm guesssing not, but haven't been able to find a counterexample.
Further information.
*
*If $n$ is a prime of potency $2$, then the sequences $x$ and $y$ are equal up to reordering; see here.
*Here we learn that the corresponding statement for higher powers is false, since $1^3+5^3+5^3=2^3+3^3+6^3.$
| There are many counterexamples, of which the smallest is $$5^2 + 5^2 = 7^2 + 1^2.$$
(I earlier stated that $25$ was the smallest counterexample, since $3^2+4^2 = 0^2+5^2$, but in your terminology, it has a potency of 1, not 2, so it is not a counterexample.)
Brahmagupta's identity shows that if $x$ and $y$ are each expressible as a sum of two squares, then $xy$ is (almost always) expressible as a sum of two squares in more than one way, because $$\begin{align}
(a^2 + b^2)(c^2 + d^2) & =
(ac-bd)^2 + (ad+bc)^2 \\
&= (ac+bd)^2 + (ad-bc)^2
\end{align}$$
The counterexample of $50$ I gave above illustrates this, since $50 = 5\cdot 10 = (1^2+2^2)(1^2+3^2)$; you can use the Brahmagupta identity to find many similar counterexamples such as $(1^2+2^2)\cdot (2^2+3^2) = 1^2+8^2 = 4^2+7^2, $ etc. By repeating this process, one can find numbers like $50\cdot 65 = 1^2 + 57^2 = 15^2+55^2 = 21^2 + 53^2 = 35^2 + 45^2 $ that have a potency of $2$ and are expressible as $a^2+b^2$ in arbitrarily many ways.
As André Nicolas points out, Lagrange proved that every positive integer has a potency of at most 4; the example of $7$ shows that this bound is attained. The number 28 is the smallest number with a potency of 4 that can be expressed as a sum of four positive squares in multiple ways:
$$\begin{align}
28 & = 5^2 + 1^2 + 1^2 + 1^2 \\
& = 3^2 + 3^2 + 3^2 + 1^2 \\
& = 4^2 + 2^2 + 2^2 + 2^2
\end{align}$$
Other small examples include $31, 39, 47, 55, 60, $ and $63$, which is a sum of four positive squares in four ways.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1068887",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
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} |
Number of Sets of Partitions I looked at the partitions of numbers, like let's say $n=5$. You get
$$
\begin{eqnarray}
5&=&5\\
\hline
&=&4+1\\
&=&3+2\\
\hline
&=&3+1+1\\
&=&1+2+2\\
\hline
&=&2+1+1+1\\
\hline
&=&1+1+1+1+1\\
\end{eqnarray}
$$
where I grouped the partitions according to their distribution (i.e. appearance) of summands. So you get $5$ sets.
Is it possible to get the number of sets for general $n$?
ANOTHER EDIT, thanks to Brian
If $\pi$ is a partition of $n$, let $M_\pi$ be the multiset of pieces, and let $\sigma_\pi$ be the sequence of multiplicities of $M_\pi$ listed in non-decreasing order. Then partitions $\pi$ and $\pi'$ are in the same set if $\sigma_\pi=\sigma_{\pi'}$. Thus, $\pi=1+3+3$ and $\pi'=2+2+3$ are in the same set, because $M_\pi=[1,3,3]$, so $\sigma_\pi=\langle 1,2\rangle$, and $M_{\pi'}=[2,2,3]$, so $\sigma_{\pi'}=\langle 1,2\rangle$ as well.
EDIT
Here's another example for $n=6$:
$$
\begin{eqnarray}
6 = 6\\
\hline
5 + 1 = 6\\
4 + 2 = 6\\
\hline
3 + 3 = 6\\
\hline
4 + 1 + 1 = 6\\
\hline
3 + 2 + 1 = 6\\
\hline
2 + 2 + 2 = 6\\
\hline
3 + 1 + 1 + 1 = 6\\
\hline
2 + 2 + 1 + 1 = 6\\
\hline
2 + 1 + 1 + 1 + 1 = 6\\
\hline
1 + 1 + 1 + 1 + 1 + 1 = 6
\end{eqnarray}
$$
so we have $10$ sets...
| As pointed out, there is no easy closed form for the partition function without any restrictions. However, given some restrictions, there are some nicer closed forms. For example, if we let a composition of $n$ be a partition of $n$ in which order matters, then let $P(n)$ be the number of compositions of $n$ which only use $1$ and $2$ as its parts and let $Q(n)$ be the number of compositions of $n$ which only use odd parts. Then, it can be easily proven that $P(n)=F_{n+1}$ and $Q(n)=F_n$, where $F_n$ is the $n$th Fibonacci number.
| {
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"url": "https://math.stackexchange.com/questions/1069663",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "7",
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Prove that $c_n = \frac1n \bigl(\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + \cdots + \frac{1}{\sqrt{n}} \bigr)$ converges I want to show that $c_n$ converges to a value $L$ where:
$$c_n = \frac{\large \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + \cdots + \frac{1}{\sqrt{n}}}{n}$$
First, it's obvious that $c_n > 0$.
I was able to show using the following method that $c_n$ is bounded:
$$c_n = \frac{\large \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + \cdots + \frac{1}{\sqrt{n}}}{n} < \overbrace{\frac{\large \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} + \cdots + \frac{1}{\sqrt{2}}}{n}}^{n-1 \text{ times}} = \frac{\large {n - 1}}{n\sqrt{2}} < \frac{1}{\sqrt{2}}$$
So now we know that $\large {0 < c_n < \frac{1}{\sqrt{2}}}$.
I know from testing for large values of $n$ that $c_n \to 0$.
What's left is actually finding a way to show this.
Any hints?
| We have by the Riemann sum
$$c_n=\frac1n\sum_{k=2}^n\frac1{\sqrt k}=\frac1{n^{3/2}}\sum_{k=2}^n\frac1{\sqrt{\frac kn}}\sim\frac1{\sqrt n}\int_0^1\frac{dx}{\sqrt x}=\frac2{\sqrt n}$$
so clearly $c_n$ tends to $0$ but we found also that
$$\lim_{n\to\infty}c_n\sqrt n=2$$
which we can't find it by applying the Cesàro's theorem.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1070575",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 6,
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Double integral with two parameters $\int_{x_1=1}^{x_2=2}\int_{y_1=0}^{y_2=x}\arctan\left(\frac{y}{x}\right)\,dx\,dy$ Given the following integral:
$$\int_{y_1=0}^{y_2=x}\,dy\int_{x_1=1}^{x_2=2}\arctan\left(\frac{y}{x}\right)\,dx$$
I thought of using $u$-substitution:
$$\begin{align}
u &= \frac{y}{x} \\
w &= \arctan(u) \\
\frac{\partial w}{\partial y} &= -\frac{x}{y^2 \cdot (1 + u^2)} \\
\frac{\partial w}{\partial x} &= \frac{1}{y \cdot (1 + u^2)}
\end{align}$$
But then... how do I use the partial derivatives for integration?
| $$
\int \arctan(\frac{y}{x})\;dx, \;\text{ Let } u=\arctan\frac{y}{x}\; \Rightarrow \;
du = \frac{1}{(\frac{y}{x})^2+1}\cdot \frac{-y}{x^2}dx \; \Rightarrow \; du = \frac{-y}{x^2+y^2}dx \\ \text{and let } dv=dx \; \Rightarrow \; v=x \\ \text{Therefore} \\ \int u \;dv = uv \; - \; \int v\;du \;\; \Rightarrow \; \; \int \arctan\frac{y}{x}dx = \; x\cdot \arctan\frac{y}{x} \; + \; \int \frac{xy}{x^2+y^2}dx
$$
To solve
$$
\int \frac{xy}{x^2+y^2}dx
$$
we can let $w = x^2+y^2$, where $\frac{1}{2}dw = x\;dx$
Which gives us
$$
\int \frac{xy}{x^2+y^2}dx \; = \; \frac{1}{2}y\int\frac{1}{w}dw \; = \frac{1}{2}y\cdot \ln|w| \; = \frac{1}{2}\cdot y \cdot \ln|x^2+y^2|
$$
giving us the final result of
$$
\int \arctan\frac{y}{x}dx = x\cdot \arctan\frac{y}{x} \; + \; \frac{1}{2} \cdot y \cdot \ln|x^2+y^2|+C
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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If a,b,c are in AP and $a^2,b^2,c^2$ are in HP, then prove either $a=b=c$ or $a,b,- \frac c2 $ are in GP As the title says.
Although first part of the proof is obvious, I'm still able to prove it.
And for the second part, I'm essentially trying to prove $b^2=-c/a$ (which is possible only when c<0 Xor a<0).
The relations found by me are: $a^2+c^2+2ac=4b^2$ and $b^2=\left(\frac {2a^2c^2}{a^2+c^2} \right)$.
Which provides me with:
$(a^2+c^2+2ac)/4=(2a^2c^2)/(a^2+c^2)$
I don't think this would lead me to the answer. Any help would be appreciated.
| The AP condition gives $2b = a+c$ and the HP condition gives
$$\frac2{b^2}= \frac1{a^2}+\frac1{c^2} \iff \frac8{(a+c)^2}=\frac{a^2+c^2}{a^2c^2} \iff (a-c)^2(a^2+4ac+c^2)=0$$
Now either $(a-c)^2=0 \implies a=b=c$
or $a^2+4ac+c^2 = 0 \implies (a+c)^2+2ac=0 \implies 2b^2+ac=0 \implies a, b, -\dfrac{c}2$ are in GP.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1071661",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Give the explicit form of the following parametrized surface Let $\boldsymbol{X}:\boldsymbol{R}^2\to \boldsymbol{R}^3$ be the paramtrized surface given by$$\boldsymbol{X}(s,t)=(s^2-t^2,s+t,s^2+3t)$$
I'm trying to describe the parametrized surface by an equation of the form $z=f(x,y)$
| You have three equations: $x=s^2-t^2$, $y=s+t$ and $z=s^2+3t$. You need to eliminate $s$ and $t$.
Rearranging $y=s+t$ gives $t=y-s$. Substituting this into the other two equations gives
$$\begin{eqnarray*}
x &=& 2ys-y^2 \\
z&=& s^2-3s+3y
\end{eqnarray*}$$
Now we can rearrange $x = 2ys-y^2$ to solve for $s$. We get $\displaystyle{s=\frac{x+y^2}{2y}}$.
Finally, substituting this into $z= s^2-3s+3y$ gives an equation in $x$, $y$ and $z$:
$$z = \left(\frac{x+y^2}{2y}\right)^{\!2} - 3\left(\frac{x+y^2}{2y}\right) + 3y$$
This assumes that $y \neq 0$. If $y=0$ then $s=-t$, meaning that $x=0$ and $z=t^2+3t$. Here $t$ is arbitrary. Completing the square gives
$$z=\left(t+\frac{3}{2}\right)^{\!2} - \frac{9}{4}$$
This tells is that for al $t$, $z \ge -\frac{9}{4}$. So, when $x=y=0$, we have to add the semi-axis $z \ge -\frac{9}{4}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1073552",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Calculating a limit of integral Computing the limit: $$\lim_{n\rightarrow\infty}\left(\frac{1}{3\pi}\int_\pi^{2\pi}\frac{x}{\arctan(nx)} \ dx\right)^n$$
I made the substiution $t=nx$ then, we have: $$I=\frac{1}{n^2}\int_{n\pi}^{2n\pi}\frac{t}{\arctan t}dt$$ where $I$ is the inside integral. How would you continue?
| Asymptotics of the integral:
When $x$ is large,
\begin{align}
\frac{1}{\arctan{x}}
=&\frac{2}{\pi}\frac{1}{1-\frac{2}{\pi}\arctan\left(\frac{1}{x}\right)}\tag1\\
=&\frac{2}{\pi}\frac{1}{1-\frac{2}{\pi}\left(\frac{1}{x}-\frac{1}{3x^3}+\cdots\right)}\tag2\\
=&\frac{2}{\pi}\left(1+\frac{2}{\pi}\left(\frac{1}{x}-\frac{1}{3x^3}+\cdots\right)+\frac{4}{\pi^2}\left(\frac{1}{x}-\frac{1}{3x^3}+\cdots\right)^2+\cdots\right)\tag3\\
=&\frac{2}{\pi}+\frac{4}{\pi^2 x}+\frac{8}{\pi^3 x^2}+\mathcal{O}\left(x^{-3}\right)\tag4
\end{align}
Then
\begin{align}
\frac{1}{3\pi}\int^{2\pi}_\pi\frac{x}{\arctan(nx)}{\rm d}x
=&\frac{1}{3\pi}\int^{2\pi}_\pi\left(\frac{2x}{\pi}+\frac{4}{\pi^2 n}+\frac{8}{\pi^3 n^2x}+\cdots\right)\ {\rm d}x\\
=&\ 1+\frac{4}{3\pi^2n}+\frac{8\ln{2}}{3\pi^4n^2}+\mathcal{O}\left(n^{-3}\right)
\end{align}
Computing the limit:
Your limit is thus
\begin{align}
\lim_{n\to\infty}\left(\frac{1}{3\pi}\int^{2\pi}_\pi\frac{x}{\arctan(nx)}{\rm d}x\right)^n
=&\ \exp\left\{\lim_{n\to\infty}n\ln\left(\frac{1}{3\pi}\int^{2\pi}_\pi\frac{x}{\arctan(nx)}{\rm d}x\right)\right\}\tag5\\
=&\ \exp\left\{\lim_{n\to\infty}n\left(\frac{4}{3\pi^2n}+\mathcal{O}\left(n^{-2}\right)\right)\right\}\tag6\\
=&\ \large{\color{red}{\exp\left(\frac{4}{3\pi^2}\right)}}\normalsize\approx1.144645419236050\tag7\\
\end{align}
Numerical Verification:
Using Mathematica I am getting
$$\left(\frac{1}{3\pi}\int^{2\pi}_\pi\frac{x}{\arctan(999999999x)}{\rm d}x\right)^{999999999}\approx1.144645419247325$$
which is consistent with the derived result.
Explanation:
$(1)$: Used the fact that $\displaystyle\arctan{x}=\frac{\pi}{2}-\arctan\left(\frac{1}{x}\right)$.
$(2)$: Used the series for $\arctan{x}$.
$(3)$: Used a geometric series.
$(4)$: Expanded the terms.
$(5)$: Used the fact that $\displaystyle\lim_{n\to\infty}a_n^{b_n}=\exp\lim_{n\to\infty}b_n\ln(a_n)$.
$(6)$: Used the series for $\ln(1+x)$.
$(7)$: As $n\to\infty$, only the constant term remains.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1077167",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 1
} |
Integral with contour integration I want to evaluate the integral:
$$\int_{-\infty}^{0}\frac{2x^2-1}{x^4+1}\,dx$$
using contour integration.
I re-wrote it as: $\displaystyle \int_{0}^{\infty}\frac{2x^2-1}{x^4+1}\,dx$. I am considering of integrating on a semicircle contour with center at the origin. I considered the function $\displaystyle f(z)=\frac{2z^2-1}{z^4+1}$ which has $4$ simple poles but only two of them lie on the upper half plane and included in the contour which are: $\displaystyle z_1=\frac{1+i}{\sqrt{2}}, \;\; z_2=\frac{-1+i}{\sqrt{2}}$.
The residue at $\displaystyle z_1$ equals $\displaystyle \mathfrak{Res}\left ( f; z_1 \right )=-\frac{2i-1}{2\sqrt{2}}$ while the residue at $z_2$ equals $\displaystyle -2\sqrt{2}i-2\sqrt{2}$. (if I have done the calculations right)
Now, I don't know how to continue. Should I find the residues at the other poles as well and the say $\displaystyle \oint_{C}f(z)=2\pi i \sum res$ where $C$ is the semicircle contour and then expand it? That is:
$$\oint_{C}f(z)\,dz=\int_{0}^{a} + \int_{{\rm arc}}$$
Then let $a \to +\infty$ then than arc integral would go to zero.
But I don't know how to proceed.
I had dealt with this integral with residues converting it into a minus infinity to infinity integral but with contours I am having a bit of problem.
Therefore I'd like some help.
| Let's try and hope I have not made any mistakes here.
We will calculate the integral:
$$\int_{-\infty}^{0}\frac{2x^2-1}{x^4+1}\,dx$$
Since the integrand function is even we can rewritte it as:
$$\int_{-\infty}^{\infty}\frac{2x^2-1}{x^4+1}\,dx=2\int_{-\infty}^{0}\frac{2x^2-1}{x^4+1}\,dx$$
Considering the function $\displaystyle f(z)=\frac{2z^2-1}{z^4+1}$ we see that is has $4$ simple poles which are $\displaystyle z_1=\frac{1+i}{\sqrt{2}}, \; z_2=\frac{1-i}{\sqrt{2}}, \; z_3=\frac{-1+i}{\sqrt{2}}, \; z_4=\frac{-1-i}{\sqrt{2}}$
Only two of them lie on the upper plane (meaning they have positive imaginery part)
Those are $z_1, \; z_3$.
The residues at $z_1, \; z_3$ are:
$$\begin{aligned}
\mathfrak{Res}\left ( f; z_1 \right ) &=\lim_{z\rightarrow z_1}(z-z_1)f(z) \\
&= \lim_{z\rightarrow \frac{1+i }{\sqrt{2}}}\left ( z-z_1 \right )\frac{2z^2-1}{(z-z_1)\left ( z-\frac{1-i}{\sqrt{2}} \right )\left ( z-\frac{-1+i}{\sqrt{2}} \right )\left ( z-\frac{-1-i}{\sqrt{2}} \right )}\\
&= \cdots\\
&= \frac{3\sqrt{2}}{8} -i \frac{\sqrt{2}}{8}\\
\end{aligned}$$
and
$$\begin{aligned}
\mathfrak{Res}\left ( f; z_3\right ) &=\lim_{z\rightarrow z_3}(z-z_3)f(z) \\
&= \lim_{z\rightarrow \frac{-1+i }{\sqrt{2}}}\left ( z-z_3 \right )\frac{2z^2-1}{\left ( z-\frac{1+i}{\sqrt{2}} \right )\left ( z-\frac{1-i}{\sqrt{2}} \right )\left ( z-z_3 \right )\left ( z-\frac{-1-i}{\sqrt{2}} \right )}\\
&= \cdots\\
&= -\frac{3\sqrt{2}}{8} -i \frac{\sqrt{2}}{8}\\
\end{aligned}$$
Now we are integrating on a semicircle contour with center at the origin and radius $R$. We know that $\displaystyle \oint_{\gamma}f(z)\,dz=2\pi i \left ( \frac{3\sqrt{2}}{8} -i \frac{\sqrt{2}}{8} -\frac{3\sqrt{2}}{8}-i\frac{\sqrt{2}}{8} \right )=\frac{\pi}{\sqrt{2}}$ .
Expanding the contour integral we have that:
$$\oint_{\gamma}f(z)\,dz=\int_{-a}^{a}+\int _{{\rm arc}}$$
Letting $a \to +\infty$ we have that the arc integral goes to zero, therefore we get the value of the integral is $\dfrac{\pi}{\sqrt{2}}$ thus the original integral is $\dfrac{\pi}{2\sqrt{2}}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1079750",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Why the differentiation of $e^x$ is $e^x?$ $$\frac{d}{dx} e^x=e^x$$
Please explain simply as I haven't studied the first principle of differentiation yet, but I know the basics of differentiation.
| Have a look at the series representation of $e^x$ which is
$$e^x=\sum_{n=0}^{\infty}\frac{x^n}{n!}=1+x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}+\frac{x^5}{120}+\dots$$
Taking derivative of this gives
$$\left(e^x\right)'=\left(\sum_{n=0}^{\infty}\frac{x^n}{n!}\right)'=\left(1+x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}+\frac{x^5}{120}+\dots\right)'$$
$$=1'+x'+\left(\frac{x^2}{2}\right)'+\left(\frac{x^3}{6}\right)'+\left(\frac{x^4}{24}\right)'+\left(\frac{x^5}{120}\right)'+\dots$$
$$\implies (e^x)'=\sum_{n=0}^{\infty}\left(\frac{x^n}{n!}\right)'$$Then, differentiating term by term gives us
$$(e^x)'=0+1+x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}+\dots$$
$$=1+x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}+\frac{x^5}{120}+\dots$$
$$=e^x$$
$$\implies \frac{d}{dx}e^x=e^x$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1079821",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 9,
"answer_id": 4
} |
Integrating $\int_0^\pi \frac{x\cos x}{1+\sin^2 x}dx$ I am working on $\displaystyle\int_0^\pi \frac{x\cos x}{1+\sin^2 x}\,dx$
First: I use integrating by part then get
$$ x\arctan(\sin x)\Big|_0^\pi-\int_0^\pi \arctan(\sin x)\,dx $$
then I have $\displaystyle -\int_0^\pi \arctan(\sin x)\,dx$ because $x\arctan(\sin x)\Big|_0^\pi$ is equal to $0$
However, I don't know how to integrate $\displaystyle -\int_0^\pi \arctan(\sin x)\,dx$
Can someone give me a hint?
Thanks
| Let be
$$\begin{align} I=\int_0^\pi \frac{x\cos x}{1+\sin^2 x}\operatorname{d}x&=-\int_0^\pi \arctan(\sin x)\operatorname{d}x \\
&=-2\int_0^{\pi/2} \arctan(\sin x)\operatorname{d}x\\
&=-2\int_0^1\frac{\arctan t}{\sqrt{1-t^2}}\operatorname{d}t
\end{align}$$
Let $$ \displaystyle I(a) = \int_{0}^{1} \frac{\arctan at}{\sqrt{1-t^{2}}} \ dt.$$
Differentiating under the integral,
$$ \begin{align} I'(a) &= \int_{0}^{1} \frac{t}{(1+a^{2}t^{2})\sqrt{1-t^{2}}} \ dt \\ &= \frac{1}{a \sqrt{1+a^{2}}} \text{arctanh} \left( \frac{a}{\sqrt{1+a^{2}}} \right)\\
&= \frac{1}{a \sqrt{1+a^{2}}} \text{arcsinh}(a) .
\end{align}$$
Integrating back,
$$ \begin{align} I(1)-I(0) = I(1) &= \int_{0}^{1} \frac{\text{arcsinh}(a)}{a \sqrt{1+a^{2}}} \ da \\ &= - \text{arcsinh}(a) \text{arcsinh}(\frac{1}{a}) \Big|^{1}_{0} + \int_{0}^{1} \frac{\text{arcsinh}(\frac{1}{a})}{\sqrt{1+a^{2}}} \ da \\ &= - \text{arcsinh}^{2}(1) + \int_{0}^{1} \frac{\text{arcsinh}(\frac{1}{a})}{\sqrt{1+a^{2}}} \ da \\ &= - \ln^{2}(1+\sqrt{2}) + \int_{0}^{1} \frac{\text{arcsinh}(\frac{1}{a})}{\sqrt{1+a^{2}}} \ da . \end{align}$$
Now let $ \displaystyle w = \frac{1}{a}$.
Then
$$ I(1) = - \ln^{2}(1+\sqrt{2}) + \int_{1}^{\infty} \frac{\text{arcsinh}(w)}{w \sqrt{1+w^{2}}}$$
$$ = - \ln^{2}(1+\sqrt{2}) + I(\infty) - I(1) .$$
Therefore,
$$ \begin{align} I(1) &= - \frac{\ln^{2}(1+\sqrt{2})}{2} + \frac{I(\infty)}{2} \\ &= - \frac{\ln^{2}(1+\sqrt{2})}{2} + \frac{\pi}{4} \int_{0}^{1} \frac{1}{\sqrt{1-t^{2}}} \ dt \\ &= - \frac{\ln^{2}(1+\sqrt{2})}{2} + \frac{\pi^{2}}{8} . \end{align}$$
Finally
$$\color{blue}{\int_0^\pi \frac{x\cos x}{1+\sin^2 x}\operatorname{d}x=-2I(1)=\ln^{2}(1+\sqrt{2})-\frac{\pi^{2}}{4}.}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1081039",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "22",
"answer_count": 3,
"answer_id": 2
} |
Inequality involving $\frac{\sin x}{x}$ Can anybody explain me, why the following inequality is true?
$$\sum_{k=0}^{\infty} \int_{k \pi + \frac{\pi}{4}}^{(k+1)\pi-\frac{\pi}{4}} \left| \frac{\sin \xi}{\xi} \right| \, \text{d} \xi \geq \sum_{k=0}^{\infty} \frac{\left| \sin\left( (k+1) \pi - \frac{\pi}{4} \right)\right|}{(k+1) \pi - \frac{\pi}{4}} \ \frac{\pi}{2} $$
The question is motivated from the following calculation
$$
\sum_{k=0}^{\infty} \int_{k \pi}^{(k+1) \pi} \left| \frac{\sin \xi}{\xi} \right| \, \text{d} \xi \geq \sum_{k=0}^{\infty} \int_{k \pi + \frac{\pi}{4}}^{(k+1)\pi-\frac{\pi}{4}} \left| \frac{\sin \xi}{\xi} \right| \, \text{d} \xi \geq \\
\geq \sum_{k=0}^{\infty} \frac{\left| \sin\left( (k+1) \pi - \frac{\pi}{4} \right)\right|}{(k+1) \pi - \frac{\pi}{4}} \ \frac{\pi}{2} = \sum_{k=0}^{\infty} \frac{\left| \sin\left( (k+1) \pi - \frac{\pi}{4} \right)\right|}{2(k+1) - \frac{1}{2}} = \\
= \sum_{k=0}^{\infty} \frac{\sqrt{2}}{2} \frac{1}{2(k+1) - \frac{1}{2}} = \sum_{k=0}^{\infty} \frac{\sqrt{2}}{4k+3} \geq \sum_{k=0}^{\infty} \frac{\sqrt{2}}{4k} = \\
= \frac{\sqrt{2}}{4}\sum_{k=0}^{\infty} \frac{1}{k} = \infty
$$
which shows that $\frac{\sin \xi}{\xi} \notin \mathcal{L}^1(\mathbb{R})$.
| This uses the fact that $$\int_a^b |f(x)|\ge (b-a)\inf_{x\in(a,b)} |f(x)|$$and that $\left|\frac{\sin x}x\right|$ is minimised on $(k\pi +\frac\pi4, (k+1)\pi -\frac\pi4)$ when $x=(k+1)\pi -\frac\pi4$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1081309",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Is there a "counting groups/committees" proof for the identity $\binom{\binom{n}{2}}{2}=3\binom{n+1}{4}$? This is exercise number $57$ in Hugh Gordon's Discrete Probability.
For $n \in \mathbb{N}$, show that
$$\binom{\binom{n}{2}}{2}=3\binom{n+1}{4}$$
My algebraic solution:
$$\binom{\binom{n}{2}}{2}=3\binom{n+1}{4}$$
$$\binom{\frac{n(n-1)}{2}}{2}=\frac{3n(n+1)(n-1)(n-2)}{4 \cdot 3 \cdot 2}$$
$$2\left(\frac{n(n-1)}{2}\right)\left(\frac{n(n-1)}{2}-1\right)=\frac{n(n+1)(n-1)(n-2)}{2}$$
$$2n(n-1)\frac{n^2-n-2}{2} = n(n+1)(n-1)(n-2)$$
$$n(n-1)(n-2)(n+1)=n(n+1)(n-1)(n-2)$$
This finishes the proof.
I feel like this is not what the point of the exercise was; it feels like an unclean, inelegant bashing with the factorial formula for binomial coefficients. Is there a nice counting argument to show the identity? Something involving committees perhaps?
| $\binom{\binom n2}2$ counts pairs of (distinct) 2-element subsets of $n$-element set. Union of such pair is either 4-element set (and each 4-element set is counted 3 times: there are 3 ways to divide 4-set into 2 pairs) or 3-element set (and each 3-element set is also counted 3 times). That gives $3\binom n4+3\binom n3=3\binom{n+1}4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1081417",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
} |
Find the greatest common divisor of $2003^4 + 1$ and $2003^3 + 1$ Find the greatest common divisor of $2003^4 + 1$ and $2003^3 + 1$ without the use of a calculator. It is clear that $2003^4+1$ has a $082$ at the end of its number so $2003^4+1$ only has one factor of 2, while $2003^3+1$ has a $028$ at the end of its number so $2003^3+1$ has 2 factors of 2. The answer is supposed to be 2, but at this point it could be greater than 2. Any ideas? This was on a previous qualifying exam at BU.
| Let $d=\text{ GCD }(2003^3+1,2003^4+1)$, then $d$ divides $2003(2003^3+1)-(2003^4+1)=2003-1=2002=2\cdot 7\cdot 11 \cdot 13$, and since both of $2003^3+1$ and $2003^4+1$ are even we have $d\geq 2$.
Since $2003=2002+1=2\cdot 7\cdot 11 \cdot 13 + 1$ it follows $2003^4+1$ gives residue $2$ when is divided by $7, 11$ or $13$. Then $d|2002$ but $d$ doesn't divide $7,11$ or $13$. It follows $d=2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1081675",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
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