The statement is the following: Let $P_n(z) = 1 + z + \frac{z^2}{2!} + … + \frac{z^n}{n!}$ be the $n$ first terms of the Taylor series for $e^z$ centered at the origin. Then, prove that $\forall n \in \mathbb{N}$, and $\forall z \in \mathbb{C} : Re(z)< 0$, this inequality holds:
$|e^z – P_n(z)|<|z|^{n+1}$
I am trying to use the integral formula for the remainder:
https://en.wikipedia.org/wiki/Taylor%27s_theorem#Taylor's_theorem_in_complex_analysis
And then the fact that in the half plane with $Re(z) < 0$, the exponential map has modulus less than one. But this bound is maybe too coarse, because I cannot reach the conclusion. Any help will be much appreciated.
Best Answer
Use $z\int_0^{1} \{e^{tz} -P_{n-1}(tz)\}dt =e^{z} -P_{n}(z)$. A simple induction argument gives the desired inequality. [To begin with $|\int_0^{1}e^{tz}dt| \leq 1$ for $\Re (z) <0$ so $|e^{z} -1| \leq |z|$