Find the first four non-zero terms of the Maclaurin series of $f(z)=e^\frac{1}{1-z}$
Sophia Terry 
Consider the following complex function:
$$f(z)=e^\frac{1}{1-z}$$
Find the first four non-zero terms of the Maclaurin series for $f(z)$, so around $z_0=0$.
I've asked my lecturer what's the best way to solve this and she simply said "just differentiate the function 4 times and find its value at $z=0$". I'm refusing to accept this and I'm sure there's a smarter way to approach this.
Unfortunately I'm clueless about this one and have no idea how to even start. I tried thinking of a function $g(z)$, which I know its Maclaurin series already, such that $g(z)f(z)=1$ and then comparing coefficients but no luck of finding one.
Any help would be grealy appreciated!
$\endgroup$ 32 Answers
$\begingroup$One can compose the Taylor series for the exponential function with the geometric series. The “trick” is to write$$ \frac{1}{1-z} = 1 + \frac{z}{1-z} $$where the second term on the right has a Taylor series without constant term, so that it can be substituted into the exponential series:$$ f(z)=e^{1/(1-z)} = e \cdot e^{z/(1-z)} = e \cdot \sum_{k=0}^\infty \frac{1}{k!} \left( \frac{z}{1-z}\right)^k \\ = e \cdot \sum_{k=0}^\infty \frac{1}{k!} z^k(1 + z + z^2 + z^3 + \ldots )^k $$Now collect all terms with powers from $z^0$ up to $z^3$:$$ f(z)= e \cdot \left( 1 + (z + z^2 + z^3 + \ldots) + \frac 12(z^2 + 2 z^3 + \ldots) + \frac 16(z^3 + \ldots) + \ldots \right) \\ = e \cdot \left(1 + z + \frac 32 z^2 + \frac{13}{6} z^3 + \ldots\right) $$
$\endgroup$ 4 $\begingroup$From the Maclaurin series of $$\dfrac{1}{1-z}=1+z+z^{2}+z^{3}+O\left( z^{4}\right) $$ and $$\exp(z)= 1+z+\dfrac{1}{2!}z^{2}+\dfrac{1}{3!}z^{3}+O( z^{4}) ,$$ we can obtain the expansion of $\exp \left( \frac{1}{1-z}\right)$ as explained in the following steps:
\begin{eqnarray*} \exp \left( \frac{1}{1-z}\right) &=&\exp \left( 1+z+z^{2}+z^{3}+O\left( z^{4}\right) \right)\\&=& \exp \left( 1 \right)\exp \left( z \right)\exp \left( z^2 \right)\exp \left( z \right)\exp \left( z^3 \right) \exp \left( O(z^4) \right) \\ &=&\underset{\exp \left( 1\right) }{\underbrace{e}}\,\underset{\exp \left( z\right) }{\underbrace{\left( 1+z+\frac{1}{2}z^{2}+\frac{1}{6}z^{3}+O( z^{4})\right) }}\times \\ &&\qquad\underset{\exp \left( z^{2}\right) }{\times \underbrace{\left( 1+z^{2}+O( z^{4}) \right) }}\, \underset{\exp \left( z^{3}\right) }{ \underbrace{\left( 1+z^{3}+O( z^{4}) \right ) }} \\ &=&e\left( 1+z+\frac{1}{2}z^{2}+\frac{1}{6}z^{3}+O( z^{4}) \right) \underset{\exp \left( z^{2}\right) \exp \left( z^{3}\right) }{ \underbrace{\left( 1+z^{2}+z^{3}+O( z^{4}) \right ) }} \\ &=&e\left( 1+z+( 1+\frac{1}{2}) z^{2}+( 1+1+\frac{1}{6}) z^{3}+O( z^{4}) \right) \\ &=&e+ez+\frac{3}{2}ez^{2}+\frac{13}{6}ez^{3}+O( z^{4}) . \end{eqnarray*}
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