Evaluate coefficients of complex power series

Question

Evaluate the coefficients of this power series $$\frac{e^{tz}}{1-z}= \sum_{n=0}^{\infty} c_n (t) z^n$$

I rewrote it as $e^{tz} = (1-z) \sum_{n=0}^{\infty} c_n z^n = \sum_{n=0}^{\infty} (c_n - c_{n-1}) z^n$, with some manipulation of the index of summation.

By comparison to the power series of exponential I find the recursion $$c_n - c_{n-1} = \frac{t^n}{n!}$$

How to find now the value of $c_1$ and $c_0$ (or $c_0$ and $c_{-1}$)?

Answer 1

By plugging $z=0$ into $f(z)$ we can find $c_0$ and so $$c_0=\frac{e^{t\cdot0}}{1-0}=1$$ $$c_n=c_{n-1}+\frac{t^n}{n!}$$ $$\therefore c_n=\sum_{k=0}^n \frac{t^k}{k!}$$

Answer 2

With Cauchy - product we have

$$\frac{e^{tz}}{1-z}= (\sum_{n=0}^{\infty} \frac{t^n}{n!}z^n)(\sum_{n=0}^{\infty}z^n)=\sum_{n=0}^{\infty}\phi(n,t,z)$$

with $\phi(n,t,z)=\sum_{k=0}^n\frac{t^k}{k!}z^k \cdot z^{n-k}=(\sum_{k=0}^n\frac{t^k}{k!})z^n.$

Answer 3

Your calculation $$ e^{tz} = (1-z) \sum_{n=0}^{\infty} c_n z^n = \sum_{n=0}^{\infty} (c_n - c_{n-1}) z^n $$ is only valid if you define $c_{-1} = 0$. Then $$ \begin{align} c_{-1} &= 0 \\ c_n - c_{n-1} &= \frac{t^n}{n!} \text{ for } n \ge 0 \, . \end{align} $$ determines all $c_n$ uniquely.

If you want to avoid negative indices then you have to do the index manipulations carefully: $$ e^{tz} = (1-z) \sum_{n=0}^{\infty} c_n z^n = \sum_{n=0}^{\infty} c_n - \sum_{n=0}^{\infty} c_n z^{n+1} = \sum_{n=0}^{\infty} c_n - \sum_{n=1}^{\infty} c_{n-1} z^{n} = c_0 + \sum_{n=1}^{\infty} (c_n - c_{n-1}) z^n $$ And now the comparison of the coefficients of the power series gives $$ \begin{align} c_0 &= 1 \\ c_n - c_{n-1} &= \frac{t^n}{n!} \text{ for } n \ge 1 \, . \end{align} $$ which again determines all $c_n$ uniquely.