My Math Forum power series

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 February 21st, 2012, 03:50 AM #1 Newbie   Joined: Feb 2012 Posts: 17 Thanks: 0 power series show that e^i?=cos???+i sin?? ?, using the power series for e^x, cos?x and sin?x
 February 21st, 2012, 04:07 AM #2 Global Moderator   Joined: Dec 2006 Posts: 20,942 Thanks: 2210 I think you intended Euler's formula: $e^{i\theta}\,=\,\cos\theta\,+\,i\sin\theta.$
 February 21st, 2012, 07:19 AM #3 Senior Member   Joined: Jul 2011 Posts: 227 Thanks: 0 Re: power series We have to prove: $e^{ix}=\cos(x)+i\sin(x)$ Using the Maclaurin series: $\cos(x)=\sum_{k=0}^{+\infty} \frac{(-1)^k x^{2k}}{(2k)!}=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\ldots$ $\sin(x)=\sum_{k=0}^{+\infty} \frac{(-1)^k x^{2k+1}}{(2k+1)!}=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\ldots$ $\cos(x)+i\sin(x)= \left(1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\ldots\right)+i\left(x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\ldots\right)$ $=1+ix-\frac{x^2}{2!}-\frac{ix^3}{3!}+\frac{x^4}{4!}+\frac{ix^5}{5!}-\frac{x^6}{6!}-\frac{ix^7}{7!}+\ldots$ $=\sum_{k=0}^{+\infty}\frac{(ix)^k}{k!}=e^{ix}$

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