Can anybody explain me this fourier series page in simple steps

Fourier Series for Even Functions

For an even function f(t), defined over the range -L to L (i.e. period = 2L), we have the following handy short cut.
Since

http://www.intmath.com/Fourier-series/Image58.gif

and

f(t) is even,

it means the integral will have value 0. (See Properties of Sine and Cosine Graphs.)
So for the Fourier Series for an even function, the coefficient bn has zero value:

bn = 0

So we only need to calculate a0 and an when finding the Fourier Series expansion for an even function f(t):
http://www.intmath.com/Fourier-series/Image91.gif http://www.intmath.com/Fourier-series/Image92.gif
An even function has only cosine terms in its Fourier expansion:

http://www.intmath.com/Fourier-series/Image93.gif

Specally how he has written ao/2 in last equation. Thanks

The first term is a constant and you can call it whatever you want.

From the start point of

,

integrating between and produces the result



If the coefficients are calculated (by multiplying both sides by and integrating betwen and ), you find that the fraction in front of each integral is .

So, you have a choice. Either leave things as they are in which case one of the integrals has at the front and the others have . Or, call the constant , in which case the 2's cancel and you have the consistency of all of the integrals having at the front.

Some choose the first option others choose the second.