All Questions
12
questions
1
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0
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44
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Multiplication of multiple summations of complex functions
I have a series that looks like $\sum_{l,m,n}\frac{A^{l}B^{m}C^{n}}{l!m!n!}$ where $A$ is a complex function and $B$ and $C$ are real functions. The summation is finite up to some cutoff $p$. $A$, $B$,...
0
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2
answers
52
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What is the Radius of Convergene of $f(z) = \sum_{n=0}^\infty \frac1{4^n}z^{2n+1}$
Here's what I have:
$f(z) = z + \frac14z^3+\frac1{4^2}z^5+...$
So, my coefficients are either $0$ or $\frac1{4^n}$ with $\frac1{4^n}$ being the supremum.
So, $\limsup \limits_{n \to \infty} |c_n|^{\...
0
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1
answer
57
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Find the radius of convergence of $\sum_{i=0}^\infty a_n$ where $\sum_{i=0}^\infty 2^n a_n$ converges, but $\sum_{i=0}^\infty (-1)^n2_na_n$ diverges [closed]
From this example: $\sum_{n=1}^\infty a_n$ converges and $\sum_{n=1}^\infty|a_n|$ diverges. Then Radius of convergence?
I believe I'm supposed to leverage these two statements to show that $R \leq |z|$...
0
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1
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45
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Show $\liminf_n |\sum^n a_kb_{n-k}|^{-\frac{1}{n}} = \liminf_n |a_n|^{-\frac{1}{n}}$?
Suppose for some complex $a_n$ and $b_n$ such that $\liminf_n |a_n|^{-\frac{1}{n}} = \liminf_n |b_n|^{-\frac{1}{n}}= R$. Show that $\liminf_n |\sum^n a_kb_{n-k}|^{-\frac{1}{n}} = R$?
I'm somehow ...
3
votes
0
answers
85
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Replacing $n!$ with Stirling's approximation in $e^x = \sum_n \frac{x^n}{n!}$
I was wondering if there is a closed-form expression for
$$\sum_{n=0}^{\infty} \frac{x^n}{e^{-n}n^n},$$
although I expect there is none because Mathematica cannot compute it. However, from Stirling'...
4
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0
answers
824
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Laurent expansion for $\sqrt{z(z-1)}$
Let $f(z) = \sqrt{z(z-1)}$. The branch cut is the real interval $[0,1]$, and $f(z)>0$ for real $z$ that are greater than 1. I need to find the first few terms of the Laurent expansion of $f(z)$ for ...
1
vote
0
answers
229
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Difficult De Moivre's theorem question involving series
Use De Moivre's theorem to show that
$$\sin (2m+1) \theta = (\sin^{2m+1} \theta) \cdot P_m (\cot^2 \theta)$$ for $0 < \theta < \pi/2$, where
$$P_m(x) = \sum_{k=0}^{m} (-1)^k C^{2m+1}_{2k+1} ...
0
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0
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226
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Proving an identity with geometric series.
I've been at this for MANY hours and I think it's time I sought help.
Question: Given $k = \frac{2 \pi}{Na}\left ( p-\frac{N}{2} \right )$, prove that $\sum_{k=1}^{N}e^{ika\left ( n-m \right )}=N\...
1
vote
0
answers
201
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Find the Laurent Series expansion.
Find the Laurent series expansion of $$f=\frac {2z}{(z-1)(z-3)}$$ at the $1<|z|<3$.
Sol.: My centre is around zero.Ill use the known geometric series around zero.
$$f(z)=\frac {2z}{(z-1)(z-3)}=...
0
votes
1
answer
171
views
Close form of a power series starting at $n=2$
This is the power series I am looking at $\sum_{n=2}^{\infty}{n(n-1)z^n}$. I want to find the closed form of this power series.
This is my approach, if I divide the power series by $z^2$, then I ...
0
votes
1
answer
70
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What is the series expansion of $f(z)\cdot\exp\left({s\,\log(z)}\right)$?
For analytic $f$, how can I represent the expression $f(z)\cdot\exp\left({s\,\log(z)}\right)$, i.e. $f(z)\cdot z^s$ in the form
$$\sum_{n}^\infty\left(\sum_{k}^\infty a_k s^k\right)z^n,$$
at least ...
0
votes
2
answers
126
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Prove that $f(z) = \sum\limits_{k = 1}^\infty \frac{z^{2^k}}{2^k}$ is continuous in the closed unit disc and holomorphic inside it.
I have started off by assuming that there is a disc of radius $r$ for which $|z|<r$ for $r \in (0,1)$ and $z \in D_r$.
This implies that $|z|^{2^k} < r^{2^k}$
And after that, I don't know ...