Prove Euler's formula for complex numbers.

Euler's formula states that e^(ix) = cos(x) + i sin(x) for any real number x.

To prove Euler's formula, we start with the power series expansions of e^(ix), cos(x), and sin(x):

e^(ix) = 1 + ix + (ix)^2/2! + (ix)^3/3! + ...
cos(x) = 1 - x^2/2! + x^4/4! - x^6/6! + ...
sin(x) = x - x^3/3! + x^5/5! - x^7/7! + ...

We can see that the power series for e^(ix) contains the power series for both cos(x) and sin(x). Therefore, we can write:

e^(ix) = cos(x) + i sin(x) + (terms involving higher powers of x)

To show that the higher powers of x are equal to zero, we use the fact that e^(ix) is periodic with period 2π. That is, e^(i(x+2π)) = e^(ix). Therefore, we can write:

e^(i(x+2π)) = cos(x+2π) + i sin(x+2π)
e^(ix) = cos(x) + i sin(x)

Subtracting these two equations, we get:

e^(i(x+2π)) - e^(ix) = cos(x+2π) - cos(x) + i(sin(x+2π) - sin(x))

Using the trigonometric identity cos(x+2π) = cos(x) and sin(x+2π) = sin(x), we simplify this to:

e^(i(x+2π)) - e^(ix) = 0

Therefore, the higher powers of x in the power series for e^(ix) must be equal to zero. This means that:

e^(ix) = cos(x) + i sin(x)

as required.