Euler’s Formula

Statement of Euler’s Formula
Proof of Euler’s Formula
- Using Taylor Series
- Using Derivative of Euler’s equation
Corollaries
- Cosine in terms of exponentials
- Sine in terms of exponentials

Statement of Euler’s Formula

Euler’s formula is an extremely important formula that has many uses in mathematics and all of its applications. A statement of this formula is as follows:

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

Proof of Euler’s Formula

Using Taylor series

To prove Euler’s formula, we make use of the Taylor series expansions of e^x, cos^x, sin^x and e^ix

$e^x \approx 1 + x + \frac{x^2}{2!} +\frac{x^3}{3!} + \frac{x^4}{4!} +\frac{x^5}{5!} +\frac{x^6}{6!} +\frac{x^7}{7!} + \dots$

$\cos x \approx 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \dots$

$\sin x \approx x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \dots$

$e^{ix} \approx 1 + ix + \frac{(ix)^2}{2!} +\frac{(ix)^3}{3!} + \frac{(ix)^4}{4!} +\frac{(ix)^5}{5!} +\frac{(ix)^6}{6!} +\frac{(ix)^7}{7!} + \dots$

Before we proceed, recall that $i=\sqrt{-1}$ . Therefore,

$i^2=-1$
$i^3=-i$
$i^4=-1^2=1$
$i^5=i^2\cdot i^3=(-1)(-i)=i$
$i^6=(i^3)^2=-1$
$i^7=i\cdot i^6=-i$

Utilizing this information in our expression for $e^{ix}$ :

$e^{ix} \approx 1 + ix + \frac{-1x^2}{2!} +\frac{-ix^3}{3!} + \frac{1x^4}{4!} +\frac{ix^5}{5!} +\frac{-1x^6}{6!} +\frac{-ix^7}{7!} + \dots$

Rearranging terms, we get

$e^{ix} \approx \underbrace{1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \dots}_{\cos x} + i\underbrace{\left(x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \dots\right)}_{\sin x}$

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

As discussed in the page on this site dealing with the Taylor’s Series, to prove that $e^{ix}$ equals its Taylor series, we would have to prove that the error, $R$ , between $e^{ix}$ and its Taylor series goes to 0 as the number of terms in the Taylor series goes to infinity. Since the maximum value that the $(n+1)^{\text{th}}$ derivative of $e^{ix}$ can take is 1, we have

$\begin{array}{rcl} \left| R \right| &\leq& \frac{M}{(n+1)!}\left| x^{n+1} \right|\\ \, &\,& \, \\ \left| R \right| &\leq& \frac{1}{(n+1)!}\left| x^{n+1} \right|\\ \end{array}$

$\lim_n\to\infty \left| R \right| &\leq& \lim_n\to\infty\frac{\left| x^{n+1}\right|}{(n+1)!} = 0$

This is true for any value of $x$ we take the Taylor series around. Therefore, $e^{ix}$ equals its Taylor series for all $x$ .

Using derivative of Euler’s equation

Proof taken from http://math2.org/math/oddsends/complexity/e%5Eitheta.htm

If $f(x) = \cos x + i\sin x$ then the derivative of $e^{ix}$ is

$f^\prime(x) = -\sin x + i\cos x = if(x)$

Define a function, $g(x)$ with the property that, just like $f(x)$

$\frac{dg}{dx} = ig(x)$

Now we solve this equation.

$\frac{dg}{g(x)} = i\,dx$

$\int \frac{dg}{g(x)} = \int i\,dx$

$\ln \left| g \right| = ix + C$

$e^{\ln \left| g \right|} = e^{ix} + C = e^Ce^{ix}$

Let $e^C = C_2$ . We get

$\left| g \right| = C_2e^{ix}$

$g = C_3e^{ix}$

We need to see if there is any value of the constant, C_3, that makes $f(x) = g(x)$ . Set $x=0$ . We have

$f(0) = \cos 0 + i\sin 0 = 1 + 0 = 1$

$g(0) = C_3e^{i0} = C_3e^0=C_3$

So $g(x) = f(x)$ when $C_3 = 1$ . Plug 1 into the equation for $g(x)$

$g(x) = 1\cdot e^{ix} = e^{ix} = f(x)$

Corollaries

There are 2 important corollaries that can be derived directly from Euler’s formula. To prove these corollaries, we need 2 trigonometric identities:

$\cos -x = \cos x$
$\sin -x = -\sin x$

A diagram that illustrates these identities can be found at the following site:

https://www.themathpage.com/aTrig/functions-angle.htm#theorem2

Cosine in terms of exponentials

Euler’s formula states:

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

Now substitute $-ix$ for $ix$ . We get

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

Now add the 2 equations:

$\begin{array}{cccccc} \, & e^{ix} & = & \cos x & + & i\sinx\\ + & e^{-ix} & = & \cos x & - & i\sinx\\ \hline\\ \, & e^{ix} + e^{-ix} &=& 2\cos x &+& 0\\ \, &\,& \,& \, & \, & \, \\ \, & \frac{e^{ix} + e^{-ix}}{2} & = & \cos x &\,& \, \end{array}$

Sine in terms of exponentials

To prove this corollary, we perform subtraction on the 2 equations we added above:

$\begin{array}{cccccc} \, & e^{ix} & = & \cos x & + & i\sin x\\ - & e^{-ix} & = & \cos x & - & i\sin x\\ \hline\\ \, & e^{ix} - e^{-ix} &=& 0 &+& 2i\sin x\\ \, &\,& \,& \, & \, & \, \\ \, & \frac{e^{ix} - e^{-ix}}{2i} & = & \sin x &\,& \, \end{array}$

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