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## How is $(1+hM)^n \leq e^M$ ? [closed]

This is used in a proof in our textbook. I know that $\lim_{x \rightarrow \infty} (1+\frac{x}{n})^n = e^x$. I do not see how this would mean that $(1+hm)^n \leq e^M$. $h \leq 1$ is the grid step size. $M \geq 0$ is a constant.

The proof verfies that the forward Euler scheme for an ODE is stable. Below is the part of the proof that I don't understand 100%.

$2(1+Mh)^n\|\epsilon^{(h)}\| \leq 2e^M\|\epsilon^{(h)}\|$

Since this hold. $(1+ \frac{h}{2}M)^n \leq e^{M/2}$ and $(1- \frac{h}{2}M)^n \leq e^{-M/2}$ correct.

Thanks.

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math.stackexchange.com is the appropriate venue for this question. I've voted to close. – Ryan Budney Sep 12 2011 at 20:57