Consider the following complex function:


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!

  • $\begingroup$ Your lecturer is probably right. $\endgroup$
    – Tavish
    Dec 28 '20 at 13:56
  • $\begingroup$ @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. $\endgroup$
    – DannyBoy
    Dec 30 '20 at 13:32
  • $\begingroup$ That way is better and more straightforward, in my opinion. $\endgroup$
    – Tavish
    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 $z^3$: $$ f(z)= e \cdot \left( 1 + (z + z^2 + z^3 + \ldots) + \frac 12(z^2 + 2 z^3 + \ldots) + \frac 16(z^3 + \ldots) + \ldots \right) \\ = e \cdot \left(1 + z + \frac 32 z^2 + \frac{13}{6} z^3 + \ldots\right) $$

  • $\begingroup$ Yep. Wolfy says 1 + x + (3 x^2)/2 + (13 x^3)/6 + (73 x^4)/24 + O(x^5). $\endgroup$ Dec 28 '20 at 14:40
  • $\begingroup$ Wolfe with an $e$? $\endgroup$
    – mjw
    Dec 30 '20 at 0:19
  • $\begingroup$ 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 $f(z)=\sum_{k=0}^\infty \frac{1}{k!} (1 + z + z^2 + z^3 + \ldots )^k$ ? $\endgroup$
    – DannyBoy
    Dec 30 '20 at 13:36
  • $\begingroup$ @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 $G(0) = 0$. $\endgroup$
    – Martin R
    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*}

  • 1
    $\begingroup$ You lost the multiplier $e$ at some point. $\endgroup$
    – Gary
    Dec 29 '20 at 16:26
  • $\begingroup$ @Gary Thanks! Fixed. $\endgroup$ Dec 29 '20 at 16:28

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