Exercise 8.4.2

Question

Verify that the series converges absolutely for all $x \in \mathbf{R}$, that $E(x)$ is differentiable on $\mathbf{R}$, and $E'(x) = E(x)$.

Ulisse Mini · Accepted Answer

Note that
$$\sum^\infty_{n=0} \left|\frac{x^n}{n!}\right| = E(\left|x\right|)$$
so we only need to show the series converges for $x \geq 0$.

Fix $x \geq 0$ and let $N > x$ for $N \in \mathbf{N}$. We have
$$E(x) = \sum^{2N}_{n=0} \frac{x^n}{n!} + \sum_{n=2N+1}^\infty \frac{x^n}{n!} \leq K + \sum_{n=2N+1}^\infty \frac{N^n}{N!(N+1)^n} = K + \frac{1}{N!}\sum_{n=2N+1}^\infty \left(\frac{N}{N+1}\right)^n$$
for some finite constant $K$. The infinite series left over is a geometric series which converges.

Term-by-term differentiation is safe to apply on power series which converge (Theorem 6.5.6), and it's clear that $E'(x) = E(x)$ when applying termwise differentiation.

Exercise 8.4.2

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Exercise 8.4.2

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