I'm reading a paper which states the following error estimate for the Taylor approximation of the complex-valued exponential function.
For $z \in \mathbb{C}$ and $d > 0$,
$$\left| e^z – \sum_{j=0}^{d-1} \frac{z^j}{j!} \right| \le O(1)\frac{|z|^d}{d!} \cdot \max\{1, e^{\Re(z)}\}. $$
It mentions that it follows from the Taylor series of the exponential function but I don't see how to derive this bound.
It would be helpful if someone could show me how to obtain this bound or provide a reference. Thanks.
Best Answer
I don't know how this follows from the Taylor expansion but I can give different proof. Let us first prove this when $d=1$. Consider $\frac {e^{z}-1} {ze^{x}}$ where $x=Re z$. Suppose $|z|>1$ and $x>0$. This ratio is then bounded because $|e^{z}-1|\leq e^{x}+1$ and $\frac {1+e^{x}} {e^{x}}\leq 2$. It is easier to prove boundedness of $\frac {e^{z}-1} {z\max \{1, e^{x}\}}$ when $x<0$ and I will leave that part to you. We have proved that $|e^{z}-1| \leq C |z| \max \{1, e^{x}\}$ for some $C$ whenever $|z|>1$ and the case $|z| \leq 1$ is simpler. Now observe that $e^{z}-\sum\limits_{k=0}^{d-1} \frac {z^{k}} {k!}$ is obtained by repeatedly integrating $e^{z}-1$ along the line segment from $0$ to $z$. For example, $\int_0^{z} (e^{\zeta}-1)d\zeta=e^{z}-1-\frac z {1!}$, etc. So the required inequality can be derived by induction on $d$.