starting with the following function:

f(x)= -1 x < 0
= 0 x = 0
= 1 x > 0

notice that f(x) is odd and as such, it's

fourier series expansion will only have

sin()

coeficients:

+inf
f(x)= Σ an * sin(2*n*Pi*x/L)
n=1

where

a

n -

fourier series coeficients yet to be determined

L -

length of the

interval
a

n (the

fourier series coeficients) are calculated as:

a

n=2/L * integral(a, a+L, f(x)*sin(2*n*pi*x/L), x)

if we calculate the

expansion from a=-1 to a+L=1 we get (L=2):

a

n=2/2*integral(-1,1,f(x)*sin(2*n*pi*x), x)

noticing that f(x) and sin() are

odd functions, their product will be

even, and as such we can say that

integral(-1,1, ..., x)=2*integral(0,1, ..., x)

then:

a

n =2*integral(-1,1,f(x)*sin(2*n*pi*x), x)

= 2/(n*pi) * (1-(-1)^n)
= 0 n even
4/(n*pi) n odd

f(x) then becomes:

+inf
f(x) = Σ 2/(n*pi) * (1-(-1)^n) * sin(2*n*Pi*x/2)
n=1

changing

variables n=2k+1 (getting rid of the even terms (which are 0)):

+inf
f(x) = Σ 4/((2k+1)*pi) * sin((2k+1)*Pi*x)
k=0

if we get the value of f(x)at x=1/2 (which we know to be 1):

+inf
1 = Σ 4/((2k+1)*pi) * sin(k*pi+Pi/2)
k=0
+inf
= Σ 4/((2k+1)*pi) * (-1)^k
k=0

divide by 4/pi:

+inf
pi/4 = Σ (-1)^k/(2k+1)
k=0
= 1 - 1/3 + 1/5 - 1/7 + ....

just thought it'd be interesting to know how to

__actually__ get the result.

note though that this

converges very

slowly. Check the

pi metanode for faster ways of calculating pi.