I am a bit confused about how to approach this question / not sure if I'm on the right track.
I know we can write $e^z = \sum_{n=0}^{\infty} \frac{z^n}{n!}$. Further:
$e^z$
$= e^{z_0}(e^{z-z_0})$
$= e^{z_0} \sum_{n=0}^{\infty} \frac{(z – z_0)^n}{n!}$
Replacing $z_0$ with $1$ we get:
$e^z= e^{1} \sum_{n=0}^{\infty} \frac{(z – 1)^n}{n!}$
Is the above equation the correct Taylor Series for $e^z$ around $z_0 = 1$?
Also: if $|z-1| < R$ for some fixed $R > 0$ then:
$e^{1} \sum_{n=0}^{\infty} \frac{|(z – 1)^n|}{n!} \leq e \sum_{n=0}^{\infty} \frac{|R^n|}{n!}$
Using the ratio test:
$L = \lim_{n \to \infty}|\frac{R^{n+1}}{(n+1)!}*\frac{n!}{R^n}| = \frac{R}{n+1}$
Since $L = 0$ for any fixed $R$, does this mean that the Taylor Series converges everywhere?
I'm a little confused on how to put everything together, so any insight is appreciated!
Best Answer
That's the correct series. You can also get it directly from the formula:
$$ f(z) = \sum_{n = 0}^\infty \frac{f^{(n)}(1)}{n!}(z - 1)^n. $$
Both methods are good.
For the second question, you have $$ e^{1} \sum_{n=0}^{\infty} \frac{|(z - 1)^n|}{n!} \leq e \sum_{n=0}^{\infty} \frac{R^n}{n!} $$
and the ratio test shows that this converges. Thus the Taylor series converges absolutely for every $|z - 1| < R$, for every $R > 0$. Thus it converges for every $z \in \mathbb C$ (since you can always find an $R$ such that $|z - 1| < R$). This is perfectly valid reasoning.
A common argument here is exactly what you did, but phrased using the Weierstrass M-test. Namely, take
$$ f_n(z) = \frac{e^1}{n!}(z - 1)^n \text{ and } M_n = \frac{e^1}{n!} R^n $$
Then on the set $\{|z - 1| \le R\}$,
$$ |f_n(z)| \le M_n $$
Since the sum $\sum_{n \ge 0} M_n$ converges, the M-test says that
$$ \sum_{n = 0}^\infty f_n(z) = \sum_{n = 0}^\infty \frac{e^1}{n!}(z - 1)^n $$
converges absolutely and uniformly on the same set, $\{|z - 1| \le R\}$. Absolute convergence you've seen, uniform convergence implies, for instance, that the resulting sum is continuous everywhere.