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$e^\frac1z$ is not holomorphic at $z=0$, but it is known that it can be expanded as $$e^\frac1z=1+\frac1z+\frac1{2!z^2}+\frac1{3!z^3}+\cdots$$ The coefficients of this Laurent expansion are computed the same way as Taylor's. The question is how is that possible? If function is not holomoprhic at $z=0$, then it's not true that it is holomophic at $|z|<R$ and Taylor's coefficients can not be used. Please someone explain.

jh2279
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VladislavsByckov
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    Use the power series for $e^w$ and set $w=1/z$, knowing that it is invalid for $z=0$. Then use the uniqueness of the Laurent series. – wj32 Nov 06 '12 at 02:47
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    Taylor series is always an analytic function throughout its disk of convergence. Laurent series is a generalization of Taylor series for functions with singularities. It just turn out that Laurent series looks like Taylor series in this case. – glebovg Nov 06 '12 at 03:14
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    They are not Taylor coefficients. If $f(z)=e^{1/z}$, then $f^{(k)}(0)$ does not exist for $k\geq 0$. What is it that doesn't seem possible? There is a general integral formula for Laurent series coefficients, but they are instead often found using some other known series, like in this case, where $f(z) = a_0 + a_1z +a_2z^2+\cdots$ is valid for all $z\in\mathbb C$, and it follows that $f(1/z)= a_0 + a_1/z +a_2/z^2+\cdots$ is valid for $z\neq 0$. Your $a_0$ should be $1$. – Jonas Meyer Nov 06 '12 at 04:26
  • Can someone point me somewhere that derives the exponential series without using taylor series formulae? – zzz Nov 18 '14 at 02:25

3 Answers3

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I'm pretty sure most of former answers didn't get the key point.

The Laurent series expansion is defined on a "deleted neighborhood" around a singularity, in this case, $\{z: 0<\lvert z-0\rvert< R \}$. In this deleted neighborhood, $e^{1/z}$ is analytic. So for any point in this neighbourhood, we can expand $e^{z}$ first and then substitute $1/z$ in.

As you can see in the Laurent expansion you gave, you can plug in any $z$ arbitrarily close to zero, calculate the infinite sum, and get a finite and well defined value. The laurent series doesn't give any information about the behavior of a function "on" its singularities.

dustin
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Joey
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  • Can we use the Laurent series for $e^{-1/x}$, and if we can do we need to specify any conditions(because it is not complex)? – Priya Jan 27 '21 at 01:19
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If $f$ is an entire function then $$ f(z)=a_0 +a_1z+a_2z^2+\ldots. $$ From Cauchy's integral formula $a_k=\frac{1}{2 \pi i} \oint_\gamma\frac{f(z)}{z^{k+1}}dz, \ k\in \mathbb{N}$ where $\gamma$ is the unit circle centered at zero. This means that for all $z\in\mathbb C$ $$ f(z)=a_0 +a_1z+a_2z^2+\ldots. $$ In particular the above equality holds for all non-zero complex numbers.

In our case the function $f(z)=e^z$ is entire with $$ f(z)=\sum_{n=0}^\infty\frac{z^n}{n!}. $$ Since the above equality holds for all non-zero complex numbers it follows that $$ e^{\frac{1}{z}}=f\left(\frac{1}{z}\right)=\sum_{n=0}^\infty\frac{1}{n!z^n},\qquad \forall z\in\mathbb C. $$ The above formula for (the Laurent series for) $f(1/z)$ was derived from the Taylor series of $f(z)$ by substituting $1/z$ for $z$ since the Taylor series formula holds for all non-zero complex numbers.

P..
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You could also re-index to obtain

$$\sum_{n=-1}^\infty \frac{1}{(n+1)!z^{n+1}}.$$

That way you can readily tell what the $a_{-1}$ term will be.

MyMathYourMath
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