Prove there exists a constant $K>0$ such that $|e^z-1-z-\frac{z^2}{2}|

Question

The title says it all however:

Prove that there exists a positive constant $K$ such that $|e^z-1-z-\frac{z^2}{2}|<K|z^3|$ when $|z|$ is sufficiently small.

Or in other words prove $e^x=1+x+\frac{x^2}{2}+O(x^3)$ by the way I have no background in asymptotics so complicated solutions will have no benefit for me.

You do not need asymptotics. Only the Taylor expansion of $e^z$ with Lagrange remainder. — gammatester, Jul 23 '15 at 13:18

score 5 · Accepted Answer · answered Jul 23 '15 at 13:18

5

We have

$$e^z - 1 - z - \frac{z^2}{2} = \sum_{n = 3}^\infty \frac{z^n}{n!},$$

hence

$$\biggl\lvert e^z - 1 - z - \frac{z^2}{2}\biggr\rvert \leqslant \sum_{n = 3}^\infty \frac{\lvert z\rvert^n}{n!}.$$

If we require $\lvert z\rvert < 1$, then we have $\lvert z\rvert^n < \lvert z\rvert^3$ for all $n \geqslant 3$, and that gives you an explicit $K$.

answered Jul 23 '15 at 13:18

Daniel Fischer

206,697

Thank you, great solution. – Aleksandar Jul 23 '15 at 20:53

Prove there exists a constant $K>0$ such that $|e^z-1-z-\frac{z^2}{2}|

1 Answers1