Leibniz Formula
Apr 18 2016
Verify Leibniz formula using Fourier series expansion
Let's consider a sawtooth wave function as follow.
`f(x) = x` `(-pi < x < pi)``f(x + 2 n pi) = f(x)`
We calculate the Fourier coefficients of this function.
`f(x) = a_0 / 2 + sum_(k=0) a_k cos(k x) + sum_(k=0) b_k sin(k x)`In the program, we calculate the following expressions to get the Fourier coefficients.
`a_0 = 1 / pi int_(-pi)^pi f(x) dx = 0``a_k = 1 / pi int_(-pi)^pi f(x) cos(k x) dx = 0`
`b_k = 1 / pi int_(-pi)^pi f(x) sin(k x) dx = (2 (-1)^k) / k`
Then, we get
`f(x) = sum_(k=0) ((2 (-1)^k sin(k x)) / k)`.If we substitute `x` with `pi / 2` in the above equation, we get Leibniz formula as follow.
`pi / 4 = 1 - 1 / 3 + 1 / 5 - 1 / 7 + 1 / 9 - ...`