Quickies
e^(pi i) + 1 = 0

Why does eπi + 1 = 0?

This page is just a collection of a couple of answers on the LiveJournal Mathematics Community in a thread about eπi + 1 = 0. Soon, I will whip them into a more coherent form.


In collegiate calculus, you probably learned about something called Taylor series. You can use Taylor series to make polynomial approximations of infinitely-differentiable functions. If you take the Taylor series out infinitely, you actually have the function. The general form for a Taylor series is this:

∑ f(n)(x0) * (x - x0)n / n!

where the sum goes from n=0 to n=infinity and f(n)(x0) means the n-th derivative of f(x) evaluated at x0. So, as long as there is some x0 at which we know infinitely many derivatives of f(x), we're set.

Also, note that f(0)(x) is just f(x). And, also note that 0! is defined to be one.

Fortunately, we know all of the derivatives of ex, sin x, and cos x.

All of the derivatives of ex are equal to ex. And, e0 = 1. So, one convenient Taylor series expansion of ex is:

ex = ∑ xn / n! = 1 + x + x2/2 + x3/6 + x4/24 + ...

The derivatives of sin x are a bit more tricky.

f(0)(x) = sin x
f(1)(x) = cos x
f(2)(x) = -sin x
f(3)(x) = -cos x
f(4)(x) = sin x

And, from there the pattern repeats... f(k+4)(x) = f(k).

Since sin 0 = 0 and cos 0 = 1, we can then make a Taylor series for sin x:

sin x = ∑ (-1)n x2n+1/ (2n+1)! = x - x3/6 + x5/120 + ...

And, in a similar manner, we can make a Taylor series for cos x:
cos x = ∑ (-1)n x2n/ (2n)! = 1 - x2/2 + x4/24 + ...

You can try taking these sums out a few places to verify to yourself that they really work for real numbers.

Now, comes the fun part. What happens when you put in a complex number into these polynomials instead? The relevant interesting thing happens when you put ix into the series expansion for eix. The powers of i repeat every fourth power, just like the derivatives of sin and cos repeated every fourth time.

i0 = 1
i1 = i
i2 = -1
i3 = -i
ik+4 = ik

So, when we plug ix into the expansion for ex, we get:
1 + ix + (ix)2/2 + (ix)3/6 + (ix)4/24 + (ix)5/120 - ...

= 1 + ix - (x)2/2 - i(x)3/6 + (x)4/24 + i(x)5/120 - ...

And, you can collect all the terms that contain an i to get:
( 1 - x2/2 + x4/24 - ... ) + i ( x - x3/6 + x5/120 - ... )

And that is just a series representation of cos x + i sin x.

Another sort of way to go about it:

That would have to go something like this, right:

f(x) = cos x + i sin x
f'(x) = -sin x + i cos x = i f(x)
f'(x) = i f(x)
dy/dx = i y
dy/y = i dx
∫ dy/y = ∫ i dx
ln |y| = ix + c
|f(x)| = A exi

and since f(0) must be one, then A must be one. But, I'm uncomfortable with the absolute value signs there.