# Find the first four non-zero terms of the Maclaurin series of $f(z)=e^\frac{1}{1-z}$

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!

• Your lecturer is probably right. Dec 28 '20 at 13:56
• @Tavish She is, my point is that I'm confident there's a better way to solve it, rather then differentiating a function multiply times. Can't see any added value in this method. Dec 30 '20 at 13:32
• That way is better and more straightforward, in my opinion. Dec 30 '20 at 13:36

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 $$z^3$$: $$f(z)= e \cdot \left( 1 + (z + z^2 + z^3 + \ldots) + \frac 12(z^2 + 2 z^3 + \ldots) + \frac 16(z^3 + \ldots) + \ldots \right) \\ = e \cdot \left(1 + z + \frac 32 z^2 + \frac{13}{6} z^3 + \ldots\right)$$

• Yep. Wolfy says 1 + x + (3 x^2)/2 + (13 x^3)/6 + (73 x^4)/24 + O(x^5). Dec 28 '20 at 14:40
• Wolfe with an $e$?
– mjw
Dec 30 '20 at 0:19
• Thanks, that's indeed the correct answer! Although, I didn't quite catch why we're using that "trick". Why can't we write directly $f(z)=\sum_{k=0}^\infty \frac{1}{k!} (1 + z + z^2 + z^3 + \ldots )^k$ ? Dec 30 '20 at 13:36
• @Burekas: You are right, that would work here as well. Generally, when computing the Taylor series of a composition $F(G(z))$ it is simpler with $G(0) = 0$. Dec 30 '20 at 14:05

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:

$$\begin{eqnarray*} \exp \left( \frac{1}{1-z}\right) &=&\exp \left( 1+z+z^{2}+z^{3}+O\left( z^{4}\right) \right)\\&=& \exp \left( 1 \right)\exp \left( z \right)\exp \left( z^2 \right)\exp \left( z \right)\exp \left( z^3 \right) \exp \left( O(z^4) \right) \\ &=&\underset{\exp \left( 1\right) }{\underbrace{e}}\,\underset{\exp \left( z\right) }{\underbrace{\left( 1+z+\frac{1}{2}z^{2}+\frac{1}{6}z^{3}+O( z^{4})\right) }}\times \\ &&\qquad\underset{\exp \left( z^{2}\right) }{\times \underbrace{\left( 1+z^{2}+O( z^{4}) \right) }}\, \underset{\exp \left( z^{3}\right) }{ \underbrace{\left( 1+z^{3}+O( z^{4}) \right ) }} \\ &=&e\left( 1+z+\frac{1}{2}z^{2}+\frac{1}{6}z^{3}+O( z^{4}) \right) \underset{\exp \left( z^{2}\right) \exp \left( z^{3}\right) }{ \underbrace{\left( 1+z^{2}+z^{3}+O( z^{4}) \right ) }} \\ &=&e\left( 1+z+( 1+\frac{1}{2}) z^{2}+( 1+1+\frac{1}{6}) z^{3}+O( z^{4}) \right) \\ &=&e+ez+\frac{3}{2}ez^{2}+\frac{13}{6}ez^{3}+O( z^{4}) . \end{eqnarray*}$$

• You lost the multiplier $e$ at some point.
– Gary
Dec 29 '20 at 16:26
• @Gary Thanks! Fixed. Dec 29 '20 at 16:28