The Laurent-Series is a generalisation of the power series which allows negative indices and is essential for investigating the behaviour of functions near poles.

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3
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24 views

Limit of $\frac{1}{2^n}\sum_{n=0}^{2^n-1}f(e^\frac{2k\pi}{2^n}i)$ for a complex-analytic function

Let consider a Laurent series $\displaystyle{ \sum_{k\in\mathbb Z}a_kz^k }$ with complex coefficients and converging inside the annulus $A=\{\ z\in\mathbb C\ |\ r<|z|<R\ \}$, with ...
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1answer
32 views

Is order of poles of functions determined by Laurent series?

Suppose $$f(z) = \frac{1}{(z-2)^5z}$$ is given. By looking function, i will tell there is a $5$th-order pole at $z=2$ which is in fact true. But on the other hand suppose $$f(z) = \frac{\sin ...
1
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1answer
39 views

Is there any standard method for finding the function defined by a Taylor/Laurent series?

Say you have a Taylor series defined by $$\sum_{n=0}^{\infty}a_nx^n$$ Is there any standard way to figure out what function is defined by the series? One option I see is just looking at the ...
0
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1answer
35 views

Laurent Series of $f(z) = \frac{z}{\sinh(z)}$ in the region $ 4 < |z| <5 $

Determine all coefficients, belonging to $ z^n $ with $ n<5 $, of the Laurent series of the function $f(z)=\frac{z}{\sinh(z)}$ in the region $4 < |z| <5 $. Could someone help me to find the ...
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2answers
61 views

Laurent Series at Infinity

I thought that finding the Laurent series was something that was straightforward, however, I am having some difficulty of finding the Laurent series of $$f(z) = \frac{1}{z(1-z)}$$ for $z= \infty$. ...
0
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0answers
41 views

What is Laurent series expansion of $\frac{1}{e^z-1}$ around $z=0$? [duplicate]

Consider $$f(z)=\frac{1}{e^z-1}$$ I want to expand it over $z_0=0$ in a Laurent series. We know that $$e^z=1+z+\frac{z^2}{2!}+\frac{z^3}{3!}+\cdots$$ And I know that ...
3
votes
1answer
45 views

Singularity type of $\frac{1}{z} e^{-\frac{1}{z^2}} $

I've been asked to compute the singularity type of $f(z) := \frac{1}{z}e^{-\frac{1}{z^2}} $. Here's my reasoning: $$ \frac{e^{-\frac{1}{z^2}}}{z} = z^{-1} \sum_{n=0}^\infty \big( -z^{-2} \big)^n ...
0
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2answers
73 views

The Laurent Series of $\dfrac{e^z}{z^2-1}$

The Laurent Series of $\dfrac{e^z}{z^2-1}$ At $z=1$ As we seek for powers of $z-1$, note that: $$e^z=e\cdot e^{z-1}=e(1+(z-1)+\dfrac{(z-1)^2}{2!}+\dfrac{(z-1)^3}{3!}+...)$$ So: ...
2
votes
2answers
53 views

Asymptotic Expansion of $\ f(x)=(1-\beta \frac{ log(log(x))}{log(x)})^{\beta}$

So I got this function and I'm looking for an asymptotic expansion for different values of$\ \beta > 1 $ $\ f(x)=\left(1- \beta \frac{\log \left( \log(x) \right)}{\log(x)} \right)^{\beta}$ as $\ x ...
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votes
1answer
19 views

Find the Laurent's series [closed]

Find the Laurent's series of $$f (z) = \frac1{z(1-z)^2}$$ $0<|z|<1$ $|z-1|<1$ Please help me.
0
votes
1answer
30 views

How do I write a terminating series representation of $_2 F_1(p, n+1, n+2, x)$

How do I find a terminating series representation of the hypergeometric function $_2 F_1(p, n+1, n+2, x)$, for real $p \in \mathbb{R}$ but $n \in \mathbb{Z}$, $n\geq0$? Mathematica gives (...
3
votes
2answers
52 views

Is there a simpler way to compute the residue of a function at a pole of order 3?

The function $$\frac {1}{z^2(e^{i2\pi z}-1)}$$ has a triple pole at z = 0. To compute the residue of f at z = 0, I can compute the Laurent expansion of f about z = 0, and then read off the ...
0
votes
1answer
39 views

Expanding a function into a series

I am trying to follow a proof in QFT notes, however I am unable to follow this step - it's basically Laurent/Taylor expansion but I have very little experience with it. It's claimed that: ...
2
votes
3answers
133 views

Help with this limit?

I am trying to focus on the limits of functions with similar series expansions and I stumbled on this. ...
0
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2answers
16 views

Lower bound on Indexing set of Formal Laurent Series

For a formal Laurent series defined over a ring R, we require that the indexing set is finitely bounded in the negative direction, or equivalently that the sequence of coefficients of R terminates at ...
0
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1answer
41 views

Finding the Laurent series of $f(z)=\frac{1}{(z−1)(z−2)}$ for $R =\{z∣0<|z|<1\}$ [closed]

Let $f(z)=\frac{1}{(z−1)(z−2)}$ and let $R =\{z∣0<|z|<1\}$.
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1answer
29 views

The Laurent series around $z=0$ of the function $f(z) = \frac{z}{(z-i)(z-2)}$ in the annulus $A(0,1,2)$

What I got so far: $$ \frac{z}{(z-i)(z-2)} = \frac{z}{(2-i)(z-i)} + \frac{z}{(2-i)(z-2)} $$ which is equal to $$ \frac{z}{(2-i)(z-i)} + \frac{z}{(2-i)(z-2)} = \frac{z}{(2-i)z + 1-2i} + ...
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2answers
91 views

Find the Laurent series of $f(z)=\frac{1}{z(1-z)}$

I am having difficulties finding Laurent series of the above function, around these two domains: $$0<|z-1|<1$$ and $$|z-1|>1$$ The function $f(z)$ takes the form ...
0
votes
1answer
19 views

Determine a meromorphic function satisfying certain conditions

Suppose $f$ is meromorphic on the Riemann sphere, and suppose also that $f(0) = 0$, $f(-1) = 2$, $f(3) = 3$, $f$ has a simple pole at $1$ with residue $1$, and $f$ has a triple pole at $2$ with ...
3
votes
1answer
59 views

Find the Laurent series about $z=0$

Let $f(z)=\cfrac{e^{-3z}}{z^2(z-2)^2}$, find the Laurent series about $z=0$. On the region $0<|z|<2$, I get $\cfrac{1}{(z-2)^2}=\displaystyle\sum_{n=1}^{\infty}\cfrac{nz^{n-1}}{2^{n+1}}$, ...
6
votes
1answer
72 views

When proving that f(z) is a polynomial, is it enough to consider just one point instead of keeping z arbitrary?

I think so - but I'd rather ask the MSE community too. Say I am given the bound |f(z)| < $|z|^3$, and that f is entire. Show f must be a polynomial. I used Cauchy's Integral Formula for ...
2
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0answers
23 views

How come the definition of analytic continuation doesn't require the smaller and the bigger open subsets to be connected?

The reason that is making me think that these subsets should be connected / simpled connected is because I think that the Taylor disks of convergence of f and F, which is the continuation of f to the ...
3
votes
1answer
62 views

How to solve this limit using laurent series?

$$\lim_{x\to\infty}\left(\left(\frac{x^2+5}{x+5}\right)^{3.7}+\left(\frac{x^3+5}{x+5}\right)^{1.6}\right)^{20/37}-\left(\left(x-5\right)^{3.7}+(x^2-5x+25)^{1.6}\right)^{20/37}=60$$ It is possible to ...
0
votes
0answers
17 views

Laurent series of a logarithmic derivative

Actually, the problem I have to solve is to find the number of zeroes of $f(z)=z^2-2+\frac{z-8}{z^2+2}$ within $D=\{z\in\mathbb C : 2<\left| z \right| <3 \}$. As $f(z)$ is analytic in $D$, I ...
1
vote
1answer
46 views

How to find Laurent Series for $z/(z-1)(z+4)$

How do I find the Laurent series for $$\frac{z}{(z-1)(z-4)}?$$ on: i) $0<|z-1|<5$; ii) $5<|z-1|$. I broke it up into $$\frac{1}{5}\left(\frac{4}{z+4}+\frac{1}{z-1}\right)$$ but now I am ...
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0answers
14 views

Laurent polynomial regression?

Polynomial regression is a common way of doing curvilinear regression. It is common to also use the inverse transform x^-1 (http://pareonline.net/getvn.asp?v=8&n=6). One can extend the concept ...
13
votes
3answers
588 views

Limit approach to finding $1+2+3+4+\ldots$

When exploring the divergent series consisting of the sum of all natural numbers $$\sum_{k=1}^\infty k=1+2+3+4+\ldots$$ I came across the following identity involving a one-sided limit: ...
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4answers
126 views

How does one calculate: $\left(\frac{z}{2!}-\frac{z^3}{4!}+\frac{z^5}{6!}-\cdots\right)^2$

How does one calculate: $$\left(\frac{z}{2!}-\frac{z^3}{4!}+\frac{z^5}{6!}-\cdots\right)^2$$ Is the best way to just take the first term times the following two, and the second two times the next two ...
2
votes
2answers
48 views

What is the power series expansion for Riemann-Zeta at $0$?

What are the first few terms of the Laurent series expansion of $\zeta(0)$? It gets mentioned here but they only show the first term and I am kind of confused on how they got $-1/2$.
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1answer
49 views

How to determine the Laurent expansion of $\tan{z}$ around $z=0$ that is convergent in $z=\pi$

I want to determine the Laurent expansion of $\tan{z}$ around $z=0$ that is convergent in $z=\pi$ (only the first couple of terms). Now I know that if $\sum_{n=-\infty}^{\infty}c_nz^n$ then ...
1
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1answer
32 views

Laurent series calculation(they seem to calculate it without Laurent series?)

Laurent sreies expansion of the function $f(z)=z^{-1}\sinh(z^{-1})$ about the point $0$. I thought I was meant to use this: $$f(z)=\sum_{n=0}^\infty a_n(z-z_0)^n + \sum_{n=1}^\infty ...
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1answer
36 views

How to practically classify singularities in complex analysis?

I am having trouble developing an intuition around the different types of singularity in complex analysis. The types of singularity that I am aware of are: Poles - These arise at $a_{0}$ when ...
0
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1answer
46 views

Laurent Serie of the function $f(z)=e^{z+1/z}$ around zero and prove: $\dfrac{1}{2\pi}\int_0^{2\pi}{e^{2\cos\theta}\cos{n\theta}}$

How to find the Laurent Serie of the function $f(z)=e^{z+1/z}$ around zero. Then, show that: $$\dfrac{1}{2\pi}\int_0^{2\pi}{e^{2\cos\theta}\cos{n\theta}}=\sum_{j=0}^{\infty}{\dfrac{1}{(n+j)!j!}}$$ ...
0
votes
1answer
29 views

Laurent Series of $\frac{z+1}{z(z-4)^3}$ in $0<|z-4|<4$

Find the Laurent Series of $\displaystyle \frac{z+1}{z(z-4)^3}$ in $0<|z-4|<4$ I thought about doing partial fraction decomposition first, so I'd have $\displaystyle ...
1
vote
1answer
48 views

Taylor expansion of fraction

I am trying to Taylor expand the function $f(x) = \frac{x}{x+3}\frac{1}{x-2}$ aound the point $x_0 = 2$. Clearly, the last factor explodes around this point, so I will try and expand that term. ...
2
votes
1answer
45 views

Find principal part of Laurent series

Find principal part of Laurent series: $$f(z)= \frac{1}{\sin z + \sinh z - 2z}$$ I calculate it and I have something like this: $$\frac{1}{\dfrac{2z^5}{5!}+\dfrac{2z^9}{9!}+\cdots}$$ and don't know ...
2
votes
2answers
63 views

Laurent series of cotangent

Compute the principal part of the Laurent series of $\cot(\pi z)$ on $1<|z|<2$. EDIT: After using either of the approaches below, we get that the principal part is equal to $$ \frac 1 \pi ...
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1answer
51 views

Finding the Laurent representation of a complex function

How can i find the Laurent representation fot the function: $$f(z)=\dfrac{1}{1-z^2}+\dfrac{1}{3-z}$$ In the region of: a) $\{z\in\mathbb C:1<|z|<3\}$ b) $\{z\in\mathbb C:1<|z-2|<3\}$
0
votes
1answer
40 views

Laurent series: how to join the 2 sums for $f(z)= \frac{1}{(z-1)(z+1)}$ about z = 1 for $0 < |z − 1| < 2$

We are to find the Laurent series for f(z) about $z = 1$ for $0 < |z − 1| < 2$: $f(z)= \frac{1}{(z-1)(z+1)}$ Assumptions: $|\frac{z−2}{1}| < 1 ⇔ |z − 1| < 2$ For $\frac{1}{(z-1)}$ we ...
1
vote
1answer
28 views

Laurent series to converge in $0<|z-1|<R$

Question: Determine the largest number $R$ so that the Laurent series of $$f(z)=\frac2{z^2-1} + \frac3{2z-i}$$ about $z=1$ converges for $0<|z-1|<R$. Attempt: I really don't understand this ...
1
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1answer
44 views

Classify singularities of $\frac{e^z \sin(3z)}{(z-\sqrt2)(z+\sqrt2)z^2}$

They are $0, \pm \sqrt2$. With the zero, $f(0)$ makes the numerator vanish and I have no idea how you would expand the whole function at $0$ because of the denominator. So what do you do to classify ...
0
votes
1answer
36 views

Largest $R$ value in domian $0<|z-1|<R$

Determine the largest real number $R>0$ such that the Laurent series of $$f(z)=\frac1{z-1} +\frac2{z-i}$$ about $z=1$ converges for $0<|z-1|<R$. The singularities are $1$ and $i$. But in the ...
0
votes
1answer
32 views

Laurent series in domain $|z|>0$

Find Laurent series, in powers of $z$, of $$f(z)=\frac{\sin(2z)}{z}$$ valid in the region $|z|>0$. The singularity is $0$ but $0$ isn't inside the region of the domain so what do you exactly ...
0
votes
2answers
39 views

Finding Laurent series with imaginary numbers

$$f(z)=\frac{2z}{z^2+1}=\frac1{z-i} +\frac1{z+i}$$ Find Laurent series in powers of $z$ in the domain $|z|<1$. So I got to find two Taylor series of the two terms in the function but how do you do ...
0
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0answers
41 views

Laurent series confusion

I've split it up into partial fractions and got $1/z$ - $2/(z-1)$ + $1/(z-2)$ but I'm unsure sure what to do now. I think I have done part $(i)$. I get $$z^{-1} + \sum_{n=0}^\infty ...
2
votes
1answer
46 views

Laurent series of $1/({z^3-z})$

Question: Find the Laurent series of the function $$f(z) = \frac{1}{z^3 - z}$$ at the domain $|z-1|>2$. Attempt: So we have $$\frac{1}{z(z-1)(z+1)}$$ and we only have to find a Laurent ...
0
votes
0answers
45 views

Laurent series for $\frac{2}{(z)(z-1)(z-2)}$

! So I think I am getting the hang of Laurent Series, but having a bit of trouble with one of the fractions for part a). So I split this up in to partial fractions: $\frac{1}{z} - ...
1
vote
0answers
71 views

Laurent series $\frac{-1}{z}+\frac{1}{2(z-1)} +\frac{1}{2(z+1)}$ at |z-1| > 2

This is what I tried to do. Is this correct?
0
votes
1answer
42 views

Largest number for which a laurent series converges

For part $(a)$ I got summation from $\sum^{\infty}_{n=0}(-1)^n\frac{z^{2n}}{(2n+1)!}$ Is this correct? Could someone explain how to do part (b) because I have no idea where to start Thanks
0
votes
1answer
45 views

Laurent Series expansion of $f(z)=(z-1)sin{1\over z}$

I need to find the Laurent series expansion of the function: $$f(z)=(z-1)sin{1\over z}$$ about $$A= z ∈ \Bbb C : 0<|z|<∞ $$ Any help would be appreciated!