Suppose that $f(z)= (e^z)/(1-z)$

How can I find out Power series expansion of f about $z=0$?? Is the use of cauchy product must here ? Can it be done without using cauchy product?Please help..

Many many thanks in advance.

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    $\begingroup$ Why do you want to avoid Cauchy product? $\endgroup$ – lhf Sep 3 '15 at 11:11
  • $\begingroup$ Actually I wanted to know more methods to solve the above question,which I came to know through this site.Thank you very much to all who helped me know newer ways to solve the above problem. $\endgroup$ – Suraj_Singh Sep 3 '15 at 16:06
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    $\begingroup$ @kilimanjaro I'm glad you find the answers useful! As you perhaps still don't know, you can mark one of them as accepted by clicking the check mark next to it. This will remove your question from the list of unanswered ones (and give you and the answerer reputation). For more information: meta.stackexchange.com/questions/5234/… Regarding which answer should be accepted, I would point out that lhf's answer is the only one avoiding the Cauchy product, as you seem to require in your question. $\endgroup$ – Giovanni De Gaetano Sep 4 '15 at 7:44
  • $\begingroup$ Ok,I got it! Thanks.. $\endgroup$ – Suraj_Singh Sep 4 '15 at 15:46

Your function satisfies the differential equation $$ f'(z)=f(z) \ g(z) $$ where $$ g(z)= \frac{2-z}{1-z} = 1+\frac{1}{1-z} = 2+z+z^2+z^3+z^4+z^5+\cdots $$

We can apply Leibniz rule to $f'(z)=f(z) \ g(z)$ and compute all derivatives of $f$ at $z=0$ recursively, since the derivatives of $g$ at $z=0$ are known: $$ f^{(n+1)}(0)=\sum_{k=0}^n {n \choose k} f^{(k)}(0) g^{(n-k)}(0) $$ The coefficients of the expansion of $f$ around $z=0$ are of course $\dfrac{f^{(n)}(0)}{n!}$.

This solution does not use Cauchy product but is probably more work.

  • $\begingroup$ Very nice trick! $\endgroup$ – Giovanni De Gaetano Sep 3 '15 at 15:42
  • $\begingroup$ @GiovanniDeGaetano, I had the idea to use the Leibniz rule and so defined $g$ so that it worked. Luckliy, $g$ turned out to be easy. $\endgroup$ – lhf Sep 3 '15 at 15:58
  • $\begingroup$ I'd rather use Cauchy product ,this method is tedious. $\endgroup$ – Suraj_Singh Sep 8 '15 at 19:27

Your $f(z)$ is the product of two functions with explicit power series expansion at $z=0$. Namely $$e^z = \sum_{n=0}^\infty \frac{z^n}{n!},$$ and $$\frac{1}{1-z} = \sum_{m=0}^\infty z^m.$$

Thus $$f(z) = e^z \cdot \frac{1}{1-z} = \sum_{n=0}^\infty \frac{z^n}{n!} \cdot \sum_{m=0}^\infty z^m = \sum_{j=0}^\infty z^j \left(\sum_{l=0}^j \frac{1}{l!} \right).$$

I actually don't think that the sum $\sum_{l=0}^j \frac{1}{l!}$, as you say obtained by Cauchy product, can be further simplified (except that it converges to $e$ for $j\rightarrow \infty$), but I could be mistaken.


Cauchy product is the quickest way to do this.

$$\frac{1}{1-z}=\sum_n z^n$$ $$e^z=\sum_n \frac{z^n}{n!}$$ $$\frac{e^z}{1-z}=\sum_{n=0}^\infty z^n\sum_{k=0}^n \frac{1}{k!}=\sum_{n=0}^\infty\frac{z^n}{n!}\sum_{k=0}^n \frac{n!}{k!}$$

The coefficients are $\sum_{k=0}^n \frac{n!}{k!}$ are integers that satisfy the recursion formula $F_n=n F_{n-1}+1$, $F_0=1$.

Wolfram Alpha suggests this sum equals $e \Gamma(n+1,1)$ (incomplete Gamma function).


Hint: It's useful to know that multiplication with $\frac{1}{1-z}$ means to sum up the coefficients.

Let $A(z)=\sum_{n=0}^{\infty}a_nz^n$ be a power series. We obtain when multiplying with $\frac{1}{1-z}$ \begin{align*} A(z)\frac{1}{1-z}&=\left(\sum_{k=0}^{\infty}a_zz^k\right)\left(\sum_{l=0}^{\infty}z^l\right)\\ &=\sum_{n=0}^{\infty}\left(\sum_{{k+l=n}\atop{k,l\geq 0}}a_k\right)z^n\\ &=\sum_{n=0}^{\infty}\left(\sum_{k=0}^na_k\right)z^n\\ \end{align*}

So, whenever we need the sum $\sum_{k=0}^{n}a_k$ of coefficients $a_k$ we simply have to multiply the corr. power series with $\frac{1}{1-z}$.

Keeping this in mind, you may conclude immediately \begin{align*} \frac{e^z}{1-z}=\frac{1}{1-z}\sum_{n=0}^\infty\frac{z^n}{n!} =\sum_{n=0}^{\infty}\left(\sum_{k=0}^n\frac{1}{k!}\right)z^n \end{align*}


Using the cauchy product is a good idea:

We know $$e^z = \sum_{k=0}^\infty \frac{z^k}{k!}$$ and $$\frac{1}{1-z} = \sum_{k=0}^\infty z^k \text{ for } |z| < 1.$$

So, by the cauchy product, we have around $z = 0$: $$f(z) = \left( \sum_{k=0}^\infty \frac{z^k}{k!} \right) \left(\sum_{k=0}^\infty z^k \right) = \sum_{k=0}^\infty \sum_{l=0}^k \left( \frac{1}{l!} \cdot 1 \right) z^k = \sum_{k=0}^\infty \sum_{l=0}^k \frac{1}{l!} z^k$$

I'm not aware of a simpler expression for $\sum_{l=0}^k \frac{1}{l!}$, although there might be.

You can also try to find a general expression for the $k$-th derivative of $f$, and then $f^{(k)}(0)/k!$ gives the $k$-th coefficient of the power series of $f$ around zero. Let's take a look at the first derivatives: $$f^{(1)}(z) = - \frac{e^z (z-2)}{(1-z)^2}$$ $$f^{(2)}(z) = \frac{e^z (z^2-4z+5)}{(1-z)^3}$$ $$f^{(3)}(z) = - \frac{e^z (z^3-6z^2+15z-16)}{(1-z)^4}$$

Therefore, the general form should look something like $$f^{(n)}(z) = (-1)^n \frac{p_n(z) e^z }{(1-z)^{n+1}}$$ where $p_n(z)$ is some polynomial of degree $n$. But we need the explicit form of $p_n(z)$, since the constant term changes the value of $f^{(n)}(0)$. This seems like a significantly harder problem than using the cauchy product, so I would advise against it.


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