1
$\begingroup$

How to prove the following:$|e^{-i\left}-1|\leq|\left|$ Where $x,h\in \mathbb{R^n}$, and $\left<\,\cdot\mid\cdot\,\right>$ is the standard inner product.

It looks like a beginning of Taylor series but I can't see how to get the inequality.

2 Answers 2

0

We actually have to prove that for a real number $t$, $|e^{it}-1|\leqslant |t|$. To see that, write $e^{it}-1=\int_0^tie^{is}ds,$ then use triangle inequality and the fact that $|ie^{is}|\leqslant 1$.

  • 0
    Nothing to do with Taylor series then...thanks!2012-11-09
1

The inequality $ |e^{i \varphi} - 1| \leqslant |\varphi| $ simply states that the length of an arc in a unit circle is greater or equal to the length of the chord with the same endpoints. It is a well known fact )