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.

  • $\begingroup$ You do not need asymptotics. Only the Taylor expansion of $e^z$ with Lagrange remainder. $\endgroup$ – gammatester Jul 23 '15 at 13:18
  • $\begingroup$ What is your definition for $e^z$? $\endgroup$ – Siméon Jul 23 '15 at 13:34
  • $\begingroup$ What is asymptotics but sexed-up Taylor expansions? $\endgroup$ – Neal Jul 23 '15 at 13:39

We have

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


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

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 $K$.

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  • $\begingroup$ Thank you, great solution. $\endgroup$ – Aleksandar Jul 23 '15 at 20:53

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