Proof of Euler's formula

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:




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

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




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

Q.E.D.

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




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

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,


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

Q.E.D.