Jackson 2.8, Why is the first term in Fourier Series separate? In 3ed of Jackson E&M, page 68 details orthogonal & orthonormal functions. The particular example of sines and cosines as orthonormal functions over a period interval are brought up:
$$ \sqrt\frac{2}{a}\sin(\frac{2\pi mx}{a}), \sqrt\frac{2}{a}\cos(\frac{2\pi mx}{a}). $$
He then states, "...and for m = 0, the  cosine function is $\frac{1}{\sqrt a}$." I don't understand why this is the case.
I thought that maybe this has something to do with the first term of the fourier series, but performing the fourier trick and manipulating the equations in the section haven't helped me understand. If we accept that for m = 0, the cosine function is in fact $\frac{1}{\sqrt a}$, then we could write the first term of the fourier series as $A_0$ instead of $\frac{1}{2}A_0$, I think. But I'm not sure if this is the point.
Any help would be greatly appreciated.
 A: I haven't read the book in a while, but I will attempt to answer, as I believe I have understood what the confusion is. I might be wrong and if this is the case (and hence my reply does not answer your question), please do not hesitate to comment.
Jackson wants to choose a set of functions that are orthogonal to one another and a complete set (i.e. to be able to write any function as a linear combination of any function existing in that set). The collection of the cosine functions $\cos(2\pi mx/a),\ m=1,2,...$ are indeed orthogonal to one another, but they are not a complete set! In order to become a complete set, we also insert the constant function (which can be thought of as the function corresponding to $m=0$).
Remember, we choose the set of the functions to be orthonormal. This means that we require the inner product of each function with itself to be equal to one. Hence the $\sqrt{\frac{2}{a}}$ factor for the $m\ne0$ cosine functions. The normalization factor corresponding to the $m=0$ case is simply $\frac{1}{\sqrt{a}}$, since it is the only normalization factor for which $\int_0^adx\psi_{m=0}^*\psi_{m=0}=1$. Including a factor of $\sqrt{2}$ would not result to the inner product above being equal to one!!
So, in short, the reason of choosing the normalization in that way is simply because Jackson wants to work with a normalized set of orthogonal functions.
