Firstly, I use this: $e^{\frac{1}{1-z}} = \sum_{n=0}^{\infty} \frac{(-1)^n \frac{1}{z}}{n!\big(1 - \frac{1}{z}\big)^n}$ Next I use the binomial expansion for $\big(1 - \frac{1}{z}\big)^n = \sum_{k=0}^{n} \frac{n! (-1)^k z^{-k}}{k!(n-k)!}$ Combining this, I have: $e^{\frac{1}{1-z}} = \sum_{n=0}^{\infty} \frac{(-1)^n \frac{1}{z}}{n! \sum_{k=0}^{n} \frac{n! (-1)^k z^{-k}}{k!(n-k)!}}$ How I can simplify it? Or maybe there is another way to get a Laurent series? Thank you in advance! Nothing on stack help me.
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Possible duplicate of Laurent Series $e^{\frac1{1-z}}$, $|z|>1$. – Nosrati Sep 27 '18 at 09:26
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I saw this question. There's not what I need. – mathmaniac Sep 27 '18 at 09:27
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You should mention that! – Nosrati Sep 27 '18 at 09:31
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I apologize for not mentioning. – mathmaniac Sep 27 '18 at 09:39
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Why apologizing? I removed duplication. – Nosrati Sep 27 '18 at 09:51
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Write $\exp{\big(\frac{1}{1-z}\big)}=\exp{\big(\frac{1/z}{1/z-1}\big)}.$ Why? Because we're going to use the generating function for the Laguerre polynomials, $$(1-y)^{-a-1}\exp{\big(\frac{x\,y}{y-1}\big)} = \sum_{n=0}^\infty L_n^{a}(x)\,y^n \quad \text{where} \quad L_n^{a}(x) = \sum_{m=0}^n\frac{(-x)^m}{m!}\binom{n+a}{n-m} $$ Set $y=1/z, \, a=-1, \, \text{and } x=1$ and the problem is finished.
user321120
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Where did we get this $$(1-y)^{-a-1}\exp{\big(\frac{x,y}{y-1}\big)}$$ And how can I understand, in general, that I need to use Laguerre polynomials? – mathmaniac Oct 09 '18 at 11:21
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1Anytime I see $1-x$ in the denominator of an argument involving an exponential, I think Laguerre polynomials. If I see $x^2,$ I think Hermite polynomials. These are some of the more common generating functions used in Mathematical Physics. Familiarity with the appendix of Gradshteyn and Rhyzik can get you part of the way, but a text on special functions which gives the derivations is also useful to read through once. – user321120 Oct 09 '18 at 19:33