Calculating the truncation error for exponential function in complex plane

Question

Having seen this question on Taylor approximation of complex exponential function, I am looking for a solution this problem and would be great if I also knew the name of the paper. It is about calculating the truncation error in the following form (below) for an entire function exponential for a given $z$ in complex plane.

For $z∈C$ and $d>0$, $||e^z−\sum_{j=0}^{d−1}z^j/j!||≤O(1)|z|^d/d!\cdot \max\{1,e^{R(z)}\}$.

It mentions that it follows from the Taylor series of the exponential function but I don't see how to derive this bound. It is trivial to derive it in real case by using remainder term from taylor theorem and using the increasing property of exponential. But complex case is confusing me.

It would be helpful if someone could show me how to obtain this bound or provide a reference. Thanks. Please do not refer me to the already provided solution on that webpage because that is something which I am not looking for.

https://math.stackexchange.com/a/2719893/527701 I have found this and I think this will be useful. So if someone can elaborate on this then that would be great

Answer 1

I have provided solution to a similar at https://math.stackexchange.com/a/3568554/527701 which might be helpful in this example

Solution: So I think I finally cracked and it goes like this. I used the result from https://math.stackexchange.com/a/2719893/527701 which was very useful. Let $$P_{d-1}(z) = \sum_{k=0}^{d-1} \frac{z^k}{k!}$$ be the truncation taylor approximation of $exp(z)$ and we are interested in calculating the truncation error i.e. $|e^z-P_{d-1}(z)|$. We cannot make use of taylor theorem for reals i.e. the remainder term of the form $R_{d-1}(z) = \frac{f^{(d)}(\xi)z^{d}}{(d)!} $ where $\xi$ is between 0 and z because this theorem doesn't hold for complex (as far as I know after my research). This is because in complex analysis, Analytical function are equivalent to Holomorphic function. I.e. if a function is analytic, then in-fact it is infinite time complex differentiable. So I wasn't sure if I can say that there is a radius $r$ such that for $\xi\in B(0,r)$, the above bound holds where $B(0,r)$ is an open ball centred at zero with radius $r$. Moving onto proving this statement.

Proof by Induction:

1) Take $d=0$ and the convention that $P_{-1}(z) = 0$, then using the result that $$e^z - P_d(z) = z\int_0^1 e^{tz} - P_{d-1}(tz) \ dt$$ we can show that $|e^z-1| \leq |z|\int_0^1|e^{tz}|dt = |z|\int_0^1|e^{tRe(z)}|dt \leq |z|e^{Re(z)}\int_0^1dt = |z|e^{Re(z)} \leq |z|.max\{1,e^{Re(z)}\} $ using the fact that exponential is an increasing function and $Re(z) <0$.

2) Assume this holds for all $d$ i.e. $|e^z - P_{d-1}(z)| \leq \frac{|z|^{d-1}.max\{1,e^{Re(z)}\}}{(d-1)!}$ and let us now consider $|e^z - P_{d}(z)|$. $$|e^z - P_d(z)| \leq |z|.max\{1,e^{Re(z)}\}\int_0^1 \frac{t^{d-1}|z|^{d-1}}{(d-1)!}dt = \frac{|z|^{d}.max\{1,e^{Re(z)}\}}{d!} $$ since $d\in\mathbb{N}$ and $O(1)$ constant is just $1$. $\quad \square$

I hope this proof is useful for others. Also please let me know if you spot any mistakes or inconsistency.

Answer 2

By the Taylor formula with integral remainder, the remainder term is $$ \frac{1}{{(d - 1)!}}\int_0^z {\mathrm{e}^t (z - t)^{d - 1} \mathrm{d}t} . $$ The path of integration is a straight segment connecting $0$ to $z$. By a simple change of variable, this is $$ \frac{{z^d }}{{(d - 1)!}}\int_0^1 {\mathrm{e}^{zs} (1 - s)^{d - 1} \mathrm{d}s} , $$ which is bounded from above in absolute value by \begin{align*} \frac{{\left| z \right|^d }}{{(d - 1)!}}\int_0^1 {\mathrm{e}^{\operatorname{Re}(z)s} (1 - s)^{d - 1} \mathrm{d}s} & \le \frac{{\left| z \right|^d }}{{(d - 1)!}}\max (1,\mathrm{e}^{\operatorname{Re}(z)} )\int_0^1 {(1 - s)^{d - 1} \mathrm{d}s} \\ & = \frac{{\left| z \right|^d }}{{d!}}\max (1,\mathrm{e}^{\operatorname{Re}(z)} ). \end{align*}