Step one: The Taylor Series
The Taylor Series is an
infinite series that rewrites any
function f(x) as
∞ f(n)(x0) (x-x0)n
Σ ---------------
n=0 n!
If you don't know what that means, there's no way in hell I can explain it to you here. However, I will clarify that f(n)(x) means "the nth derivative of f(x)".
If none of this makes sense to you, you might want to look at:
Important Taylor Series #1: f(x)=ex
The function ex is its own derivative--this is an extremely unique property of ex that I will not prove here.
For this Taylor series, I will use x0=0. (¬e;: The x0=0 case of the Taylor Series is called the Maclurin Series.) Since when f(x)=ex, all f(n)(x)=f(x), and f(x0)=f(0)=e0=1, the term f(n) always cancels to 1 and drops out of the equation completely. Also, (x-x0)=(x-0)=(x). Therefore:
∞
f(x)=ex=x0/0!+x1/1!+x2/2!+x3/3!...= Σ xn/n!
n=0
Important Taylor Series #2 and 3: f(x)=sin(x) and f(x)=cos(x)
Once again, x0=0. I will not derive this particular series here, but if you plug in a number for a few terms you can satisfy yourself that it works.
∞
f(x)=sin(x)=x-x3/3!+x5/5!-x7/7!...= Σ (-1)nx2n+1/(2n+1)!
n=0
And for f(x)=cos(x):
∞
f(x)=sin(x)=1-x2/2!+x4/4!-x6/6!...= Σ (-1)nx2n/(2n)!
n=0
Trust me, these work. You can derive them with only a little trouble, if you want--the key is that sin(x0)=0 and cos(x0)=1.
Step two: exi
If we
want to find
eπi, first we have to find a
general expression for
exi ∀ x.
Plug in xi for x in the ex Taylor Series.
f(xi)=exi=(xi)0/0!+(xi)1/1!+(xi)2/2!+(xi)3/3!+ (xi)4/4!...
=i0x0/0!+i1x1/1!+i2x2/2!+i3x3/3!+i4x4/4!
Now, we'd like to get rid of as many i's as possible--remember, i=√(-1). Therefore, exponents of i are on a four-step cycle:
i0 =1
i1 =i
i2=√-1)2 =-1
i3=√-1)2i=-1i =-i
( i4=√-1)22=-12 =1 )
If this doesn't make any sense to you, read up on:
Using these
identities, we can substitute into the Taylor series for
f(x)=
eix:
f(x)=exi=x0/0!+ix1/1!-x2/2!-ix3/3!+x4/4!
Separating out therealand imaginary terms:
=(x0/0!-x2/2!+x4/4!...)+i(x1/1!-x3/3!+x5/5!...)
Look
familiar? You'll notice that the first series in
parentheses
is the same as the cosine Taylor series.
The second one is the sine series. How
handy!
∞ ∞
=( Σ (-1)nx2n+1/(2n+1)!)+i( Σ (-1)nx2n/(2n)!)
n=0 n=0
exi=cos(x)+isin(x)
Step three: eπi
From here, it is a simple matter to
substitute π in for x in our newly derived
equation.
exi=cos(x)+isin(x)
eπi=cos(π)+isin(π)
As
we all know, cos(π)=-1 and sin(π)=0. If
you didn't know that, you should read:
Therefore:
eπi=cos(π)+isin(π)=-1+i0=-1
eπi=-1
Addendum: There was quite a horrible mistake in this node that was only corrected because of a helpful msg I got from
ttsen483. Thanks, and shame on the rest of us for not noticing it earlier.
(The equation is still true and will be for the forseeable future. And that, folks, is why math is better than fizziks.)