Questions tagged [laurent-series]

This tag is for questions about finding a Laurent series of functions and their convergence. 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.

Laurent series: Suppose that $~f(z)~$ is analytic on the annulus $~A : r_1 <|z − z_0| < r_2~$. Then $~f(z)~$ can be expressed as a series $$f(z) = \sum_{n=1}^{\infty}\frac{b_n}{(z-z_0)^n}+\sum_{n=0}^{\infty}a_n(z-z_0)^n$$ The coefficients have the formulas $$a_n=\frac{1}{2\pi i}\int_\gamma \frac{f(w)}{(w-z_0)^{n+1}}\, dw$$and$$b_n=\frac{1}{2\pi i}\int_\gamma {f(w)}{(w-z_0)^{n-1}}\, dw$$where $~\gamma~$ is any circle $~|w − z_0| = r~$ inside the annulus, i.e. $~r_1 < r < r_2~.~$

The entire series is called the Laurent series for $~f~$ around $~z_0~$.

Notes: $~(a)~~~$ The series $$\sum_{n=0}^{\infty}a_n(z-z_0)^n$$ is called the analytic or regular part of the Laurent series.

$(b)~~~$ The series $$\sum_{n=1}^{\infty}\frac{b_n}{(z-z_0)^n}$$ is called the singular or principal part of the Laurent series.

$(c)~~~$ Since $~f(z)~$ may not be analytic (or even defined) at $~z_0~$ we don’t have any formulas for the coefficients using derivatives.

Remarks:

  • The series $$\sum_{n=0}^{\infty}a_n(z-z_0)^n$$converges to an analytic function for $~|z − z_0| < r_2~$.
  • The series $$\sum_{n=1}^{\infty}\frac{b_n}{(z-z_0)^n}$$converges to an analytic function for $~ |z − z_0| > r_1~$.
  • Together, the series both converge on the annulus $~A~$ where $~f~$ is analytic.

The Laurent series is calculated over contour integrals of counterclockwise self-avoiding rectifiable paths of the function. For holomorphic functions the Taylor series and Laurent series are identical.

The Laurent series has a principal part, which consists entirely of negative-degree terms. When the principal part vanishes (there are no negative indices) the function is holomorphic; when it is an infinite sum the function has an essential pole.

Reference:

https://en.wikipedia.org/wiki/Laurent_series

1827 questions
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Laurent series and regions

Some help to understand why this is true would be amazing; For the function $$\frac{1}{z-1}\;\;,$$ the laurent series is different depending on the region. 1) for the region $|z|<1$, you rewrite the function as $\frac{-1}{1-z}$, and the laurent…
armara
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Remainder for Laurent expansion

The error of a finite-order Taylor expansion can be estimated by various remainders, Cauchy, Lagrange, etc. Is there any similar remainder for Laurent expansion? I tried to search in google, but did not find a result
user26143
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Finding the Laurent series of a nontrivial function

Could anyone show me how to find the Laurent series of $$ f(z) = \frac{\operatorname{Log}(z)}{(z-1)(z-5)} $$ centered at $5$? I know how to find it for $\operatorname{Log}(z)$, using the geometric series and then integrating. I'm not sure how to…
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Conceptual question regarding Laurent series

So I have a physics background, not math, but i'm trying to understand at least conceptually WHY laurent series are computed in the way they are. For example, take the series expansion of $$\frac{z}{(z-1)(z-2)}.$$ In this example they break it up…
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Laurent series expansion of given function

How to find the Laurent series for |z|>1 for $$\dfrac{e^{1/z}}{z^2-1} ?$$ Firstly, how to decide if given function has laurent series expansion in specified domain or not? I just did the long division .but I am not sure if that is the expected…
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help with laurent series

The book says: The Laurent series converges and represents $F(z)$ in the open annulus obtained by continuously increasing the radius of $C_2$ and decreasing the radius of $C_1$ until each of $C_1$ and $C_2$ reaches a point where $F(z)$ is…
bruno
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Laurent series of the function $f(z)=\frac{1}{z^2(e^{z}-e^{-z})}$

Find the first three nonzero terms of the Laurent series of $f(z)=\frac{1}{z^2(e^{z}-e^{-z})}$ in the annulus $0 \leq \lvert z \rvert \leq \pi$. Now, instead of $f(z)$ I found the Taylor series of its reciprocal. Then I tried to multiply it by the…
Tim
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Find Laurent series of $\frac{x^2}{(1+x^2)^2}$

I would like to find the Laurent series of the function $$ f(x)=\frac{x^2}{(1+x^2)^2}. $$ I already know that there are two poles of order $2$, namely $x=\pm i$. And I also know that, by partial fraction…
Rhjg
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find the Laurent series expansion of $f(z) = \frac{1}{(z + 1)(z +3)}$ for the region $0 \lt |{z + 1}| \lt 2$

I have to find the Laurent series expansion of $$f(z) = \frac{1}{(z + 1)(z +3)}$$ for the region $$0 \lt |{z + 1}| \lt 2$$ Using partial fration f(z) can written as: $$f(z) = \frac{1}{2} \frac{1}{z + 1} - \frac{1}{2} \frac{1}{z + 3}$$ since we have…
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How to grasp the relationship between Laurent series depending on the region they are developed at

in trying to understand how to set up the Laurent series for a fractional expression, I have the given function \begin{equation} \frac{1}{(z-2)}-\frac{1}{(z-1)} \end{equation} The Laurent series of each of the two fractions are evaluated over the…
Luthier415Hz
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Laurent Series for different domains

I would like to determine the Laurent series for the function $$f(z) = \frac{1}{z(z-2)^3}$$ for two different domains $$\vert{z}\vert<2$$ and $$\vert{z}\vert>2$$ But, I am unsure when I am computing the Laurent series when I am to take into…
Kyle
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Finding the first four terms of the Laurent series

I want to find the first 4 terms of the Laurent series expansion of the expression $\frac{1}{e^z-1}$, around $z_0=0$. I tried expanding first $e^z-1=(1+\frac{z}{1}+\frac{z^2}{2!}+\frac{z^3}{3!}+\frac{z^4}{4!}+...)-1$ and then consider putting this…
Esteban
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Expanding $f\left ( z \right )=\frac{e^{az}}{1+e^{z}}$ about $z= i\pi$

$$f\left ( z \right )=\frac{e^{az}}{1+e^{z}} ,\left ( a\in\left ( 0,1 \right ) \right )$$ The point $z=i\pi$ is one of the nonremovable singularities of this function. In order to expand it about that point I introduced the change of variables…
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Difference between expanding $\frac{1}{1-x}$ around $x=1$ and $\frac{1}{1-e^x}$ around $x=0$

The title pretty much gives it away already, I'm trying to find the difference between Laurent expanding $\frac{1}{1-x}$ around $x=1$ and $\frac{1}{1-e^x}$ around $x=0$. The first expansion does not work out very well, while the second one can be…
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Computing the Laurent series for $1/z^2$ about $z_0=1$

I'm trying to compute the Laurent series of $f(z) = 1/z^2$ about the point $z_0=1$. Looking at my notes, it appears that I need to compute a series for $|z-1| < 1$ and one for $|z-1|>1$, due to the singularity at the point $1$. Could someone show me…
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