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Euler’s Formula by Taylor Series

April 5th, 2009 |

Author: Merrin

Euler’s Formula is a very useful formula for calculation as well as understanding the links between algebra, trigonometry, calculus, and complex variables. Euler’s Formula is

$e^{ix} = \sin x + i \cos x$

We can derive this formula with our knowledge of the Taylor series of sine, cosine, and exp(x).

Example 1. Write the Maclaurin series for sine, cosine, and exp(x).
Solution 1. Recall that the Maclaurin series of exp(x) is just

$e^{x} =1+x+\cfrac{1}{2} x^{2} +\cfrac{1}{3!} x^{3} +...+\cfrac{x^{n} }{n!} +...$

The first derive of exp(x) is itself. The second derivative is itself and so on.

$f^{(n)} (0)=1\qquad a_{n} =\cfrac{f^{(n)} (0)}{n!}$

The derivatives of sine evaluated at zero are as follows. The pattern repeats after four derivatives.

$\sin 0=0\quad \sin'0=\cos 0=1\quad \sin''0=-\sin 0=0$

$\sin'''0=-\cos 0=-1$

The derivatives of cosine evaluated at zero are as follows. This pattern repeats every four derivatives.

$\cos 0=1\quad \cos'0=-\sin 0=0\quad \cos''0=-\cos 0=-1$

$\cos'''0=\sin 0=0$

From the pattern of the derivatives we can see that

Later we will learn a general formula for the nth derivative of sine or cosine see if you can figure one out.

${\sin x=\displaystyle \sum _{n=0}^{\infty }\cfrac{\sin ^{(n)} (0)x^{n} }{n!} =x-\cfrac{1}{3!} x^{3} +\cfrac{1}{5!} x^{5} -...}$
${\cos x=\displaystyle \sum _{n=0}^{\infty }\cfrac{\cos ^{(n)} (0)x^{n} }{n!} =1-\cfrac{1}{2!} x^{2} +\cfrac{1}{4!} x^{4} -...}$

Now we have all the pieces to build Euler’s formula. Plugging ix into the Maclaurin series for exp(x) we get

${e^{ix} }={1+ix+\cfrac{1}{2} (ix)^{2} +\cfrac{1}{3!} (ix)^{3} +...+\cfrac{(ix)^{n} }{n!} +...}$
$= {1+ix-\cfrac{1}{2} x^{2} +\cfrac{1}{3!} (-1)ix^{3} +\cfrac{1}{4!} x^{4} +...}$
$={1-\cfrac{1}{2!} x^{2} +\cfrac{1}{4!} x^{4} -...+i\left(x-\cfrac{1}{3!} x^{3} +\cfrac{1}{5!} x^{5} -...\right)}$
${e^{ix} }={\cos x+i\sin x}$