3
$\begingroup$

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!

$\endgroup$
3
  • $\begingroup$ 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
5
$\begingroup$

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
  • $\begingroup$ 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
1
$\begingroup$

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*}

$\endgroup$
2
  • 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.