I am trying to use a Taylor Expansion to expand a function about a singular point z_0 to obtain a Laurent Series about the point z_0. As an example, consider the function
f(z) = exp(2z) / (1-z)^3 about z_0=1
First, I converted the denominator from (1-z)^3 to -(z-1)^3. I've done Taylor Expansions about non-singular points many times using this process. According to this post, it is impossible to do a Taylor Expansion about a singular point. Yet, typing the operation into Wolfram Alpha produces the correct Laurent Series.
How can I manually reproduce the output returned by Wolfram Alpha? I'd appreciate even the smallest hint, as my problem is that evaluating the function and a few consequent derivatives at z=z_0 returns .../0, as is easier to see here.
This could be a too simplistic answer.
Let $z=1-x$ which makes $$f=\frac{e^{2-2 x}}{x^3}=e^2\,\frac{e^{-2 x}}{x^3}$$ Expand the exponential as a Taylor series $$e^{-2 x}=\sum_{n=0}^\infty \frac{(-2x)^n}{n!}=\sum_{n=0}^\infty \frac{(-1)^n 2^n x^n}{n!}$$ which make $$f=e^2\,\sum_{n=0}^\infty \frac{(-1)^n 2^n x^{n-3}}{n!}=e^2\,\sum_{n=0}^\infty \frac{(-1)^n 2^n(1-z)^{n-3}}{n!}=e^2\,\sum_{n=0}^\infty \frac{(-1)^{2n-1} 2^n(z-1)^{n-3}}{n!}$$ and then the result.