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When $0\le h \le 0.01$, show that $e^h$ may be replaced by $1+h$ with an error of magnitude no greater than $0.6$% of h.

use $e^{0.001} = 1.01$

What I did was :-

Hope you will be able to understand my hand writing

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    $\begingroup$ The error is not bounded by the next term; this is an error. This technique is only valid for alternating series. Instead use Taylor's theorem. $\endgroup$ – vadim123 Dec 7 '15 at 18:41
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    $\begingroup$ Use Taylor expansion upper bounded by geometric series to get: $\frac {e^h-1-h}{h}\le \frac h {2(1-h)}$. $\endgroup$ – A.S. Dec 7 '15 at 19:04
  • $\begingroup$ How to do this? $\endgroup$ – dknight Dec 8 '15 at 4:59

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