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.