## FANDOM

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### Using Taylor series

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

$i^0=1 \,$
$i^1=i \,$
$i^2=-1 \,$
$i^3=-i \,$
$i^4=1 \,$
$i^5=i \,$

and so on. The functions ex, cos(x) and sin(x) (assuming x is real) can be written as:

$e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots$
$\cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots$
$\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots$

and for complex z we define each of these function by the above series, replacing x with iz. This is possible because the radius of convergence of each series is infinite. We then find that

$e^{iz} = 1 + iz + \frac{(iz)^2}{2!} + \frac{(iz)^3}{3!} + \frac{(iz)^4}{4!} + \frac{(iz)^5}{5!} + \frac{(iz)^6}{6!} + \frac{(iz)^7}{7!} + \frac{(iz)^8}{8!} + \cdots$
$= 1 + iz - \frac{z^2}{2!} - \frac{iz^3}{3!} + \frac{z^4}{4!} + \frac{iz^5}{5!} - \frac{z^6}{6!} - \frac{iz^7}{7!} + \frac{z^8}{8!} + \cdots$
$= \left( 1 - \frac{z^2}{2!} + \frac{z^4}{4!} - \frac{z^6}{6!} + \frac{z^8}{8!} - \cdots \right) + i\left( z - \frac{z^3}{3!} + \frac{z^5}{5!} - \frac{z^7}{7!} + \cdots \right)$
$= \cos (z) + i\sin (z) \,$

The rearrangement of terms is justified because each series is absolutely convergent. Taking z = x to be a real number gives the original identity as Euler discovered it.

### Using calculus

Define the function $f$ by

$f(x) = \frac{\cos x+i\sin x}{e^{ix}}. \$

This is allowed since the equation

$e^{ix}\cdot e^{-ix}=e^0=1 \$

implies that $e^{ix}$ is never zero.

The derivative of $f$ is, according to the quotient rule:

 $f'(x)\,$ $= \displaystyle\frac{(-\sin x+i\cos x)\cdot e^{ix} - (\cos x+i\sin x)\cdot i\cdot e^{ix}}{(e^{ix})^2} \$ $= \displaystyle\frac{-\sin x\cdot e^{ix}-i^2\sin x\cdot e^{ix}}{(e^{ix})^2} \$ $= \displaystyle\frac{-\sin x-i^2\sin x}{e^{ix}} \$ $= \displaystyle\frac{-\sin x-(-1)\sin x}{e^{ix}} \$ $= \displaystyle\frac{-\sin x+\sin x}{e^{ix}} \$ $= 0 \$

Therefore, $f$ must be a constant function. Thus,

 $f(x)=f(0)=\frac{\cos 0 + i \sin 0}{e^0}=1$ $\frac{\cos x + i \sin x}{e^{ix}}=1$ $\displaystyle\cos x + i \sin x=e^{ix}$

### Using ordinary differential equations

Define the function $f(x)$ by

$f(x) \equiv e^{ix} .\$

Considering that $i$ is constant, the first and second derivatives of $f(x)$ are

$f'(x) = i e^{ix} \$
$f''(x) = i^2 e^{ix} = -e^{ix} \$

because $i^2 = -1$ by definition. From this the following 2nd order linear ordinary differential equation is constructed:

$f''(x) = -f(x) \$

or

$f''(x) + f(x) = 0. \$

Being a 2nd order differential equation, there are two linearly independent solutions that satisfy it:

$f_1(x) = \cos(x) \$
$f_2(x) = \sin(x). \$

Both $\cos(x)$ and $\sin(x)$ are real functions in which the 2nd derivative is identical to the negative of that function. Any linear combination of solutions to a homogeneous differential equation is also a solution. Then, in general, the solution to the differential equation is

 $f(x)\,$ $= A f_1(x) + B f_2(x) \$ $= A \cos(x) + B \sin(x) \$

for any constants $A$ and $B.$ But not all values of these two constants satisfy the known initial conditions for $f(x)$:

$f(0) = e^{i0} = 1 \$
$f'(0) = i e^{i0} = i \$.

However these same initial conditions (applied to the general solution) are

$f(0) = A \cos(0) + B \sin(0) = A \$
$f'(0) = -A \sin(0) + B \cos(0) = B \$

resulting in

$f(0) = A = 1 \$
$f'(0) = B = i \$

and, finally,

$f(x) \equiv e^{ix} = \cos(x) + i \sin(x). \$