User:Holmansf/Euler

Definitions of complex exponentiation
The exponential function ex for real values of x may be defined in a few different equivalent ways (see Characterizations of the exponential function). Several of these methods may be directly extended to give definitions of ez for complex values of z simply by substituting z in place of x and using the complex algebraic operations. In particular we may use either of the two following definitions which are equivalent. From a more advanced perspective, each of these definitions may be interpreted as giving the unique analytic continuation of ex to the complex plane.

Power series definition
For complex z


 * $$e^z = 1 + \frac{z}{1!} + \frac{z^2}{2!} + \frac{z^3}{3!} + \cdots = \sum_{n=0}^{\infty} \frac{z^n}{n!} ~.$$

Using the ratio test it is possible to show that this power series has an infinite radius of convergence, and so defines ez for all complex z.

Limit definition
For complex z


 * $$e^z = \lim_{n \rightarrow \infty} \left(1+\frac{z}{n}\right)^n ~.$$

Proofs
Various proofs of the formula are possible.

Using power series
Here is a proof of Euler's formula using power series expansions as well as basic facts about the powers of i:


 * $$\begin{align}

i^0 &{}= 1, \quad & i^1 &{}= i, \quad & i^2 &{}= -1, \quad & i^3 &{}= -i, \\ i^4 &={} 1, \quad & i^5 &={} i, \quad & i^6 &{}= -1, \quad & i^7 &{}= -i, \\ \end{align}$$

and so on. Using now the power series definition from above we see that for real values of x


 * $$\begin{align}

e^{ix} &{}= 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!} + \frac{(ix)^8}{8!} + \cdots \\ &{}= 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!} + \frac{x^8}{8!} + \cdots \\ &{}= \left( 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \frac{x^8}{8!} - \cdots \right) + i\left( x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots \right) \\ &{}= \cos x + i\sin x. \end{align}$$

In the last step we have simply recognized the Taylor series for sin(x) and cos(x). The rearrangement of terms is justified because each series is absolutely convergent.

Using calculus
Several other proofs are based on the following identity obtained by differentiating the power series definition of eix. Indeed, since this series converges absolutely for all complex numbers we can differentiate it term by term to obtain



\begin{align} \frac{\mathrm{d}}{\mathrm{d}x} e^{ix} & = \frac{\mathrm{d}}{\mathrm{d}x} \sum_{n=0}^{\infty} \frac{(ix)^n}{n!} \\ & = i\sum_{n=1}^\infty i^{n-1} \frac{x^{n-1}}{(n-1)!} \\ & = i e^{ix}. \end{align} $$ Now we define the function



\begin{align} f(x) = (\cos x - i \sin x) \cdot e^{ix}. \end{align} $$

The derivative of ƒ(x) according to the product rule (note that the product rule can be proved to hold for complex valued functions of a real variable using precisely the same proof as in the real case) is:


 * $$\begin{align}

\frac{d}{dx}f(x) &{}= (\cos x - i\sin x)\cdot\frac{d}{dx}e^{ix} + \frac{d}{dx}(\cos x - i\sin x)\cdot e^{ix} \\ &{}= (\cos x - i\sin x)(i e^{ix}) + (-\sin x - i\cos x)\cdot e^{ix} \\ &{}= (i\cos x + \sin x - \sin x - i\cos x)\cdot e^{ix} \\ &{}= 0. \end{align} $$

Therefore, ƒ(x) must be a constant function in x. Because ƒ(0)=1 in fact ƒ(x) = 1 for all x, and so multiplying by cos x + i sin x, we get


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

Using differential equations
Here is another proof that follows from the differential identity above. Define a new function ƒ(x) of the real variable x as


 * $$ \begin{align} f(x) = \cos x + i \sin x. \end{align} $$

Then we may check that



\begin{align} \frac{\mathrm{d}f}{\mathrm{d}x}(x) & = -\sin x + i \cos x \\ & = i f(x). \end{align} $$

Thus ƒ(x) and eix satisfy the same system of ordinary differential equations (here the complex values are considered as points in the plane ℝ2). If we also note that both functions are equal to 1 at x = 0, then by the uniqueness of solutions to ordinary differential equations (see Picard–Lindelöf theorem and note the comments concerning global uniqueness in the proof section there) they must be equal everywhere.