Difference between revisions of "Euler's identity"

Revision as of 11:31, 7 July 2006

Euler's formula is $\displaystyle e^{i\theta}=\cos(\theta)+i\sin(\theta)$ . This can be shown using Taylor series for $e^x, \sin(x)$ , and $\cos(x)$ .

Proof

Note that

$e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+...=\sum_{k=0}^{\infty}\frac{x^n}{n!}$

$\sin(x)=x-\frac{x^3}{3!}+\frac{x^5}{5!}-...=\sum_{j=0}^{\infty}(-1)^{j}\frac{x^{2j+1}}{(2j+1)!}$

$\cos(x)=1-\frac{x^2}{2!}+\frac{x^4}{4!}-...=\sum_{i=0}^{\infty}(-1)^{i}\frac{x^{2n}}{(2n)!}$

(where i, j, k are just dummy variables).

The key step now is to let $x=i\theta$ and plug it into the series for $e^x$ . The result is Euler's formula above. (anyone who's willing, feel free to type up the steps).

A special, and quite fascinating, consequence of Euler's formula is the identity $e^{i\pi}+1=0$ , which relates five of the most fundamental numbers in all of mathematics: $e,i,\pi, 0$ and 1.

@@ Line 10: / Line 10: @@
 <math>\cos(x)=1-\frac{x^2}{2!}+\frac{x^4}{4!}-...=\sum_{i=0}^{\infty}(-1)^{i}\frac{x^{2n}}{(2n)!}</math>
-(where i, j, k are just [[dummy variables]]).
+(where i, j, k are just [[dummy variable]]s).
 The key step now is to let <math>x=i\theta</math> and plug it into the series for <math>e^x</math>.  The result is Euler's formula above. (anyone who's willing, feel free to type up the steps).

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Difference between revisions of "Euler's identity"

Revision as of 11:31, 7 July 2006

Proof

See Also