Basic Fourier Series question

You're close but not quite there with that answer - it should be

[latex] \displaystyle a_{0} = \frac{\pi}{2}. [/latex]

The [latex] a_{0} [/latex] term in a Fourier series is simply the average value of the function over one full period.

So [latex] f=a_{0} [/latex] is the constant valued function which gives you the same area over one full period as the function in question.

Ah OK, yes you are correct. I forgot that some people only half the value of the [latex] a_{0} [/latex] term in the final formula, instead of calculating the actual average/mean term directly. (This is so you can use the [latex] a_{n} [/latex] formula for obtaining the [latex] a_{0} [/latex] term, instead of having to remember a separate formula for [latex] a_{0} [/latex].)

Provided in your notes there's a piece of text saying "for odd k only" or something similar, it actually works out the same whichever formula you use.

If you look at the [latex] a_{n} [/latex] term in those notes, they have

[latex] \displaystyle a_{n} = \frac{\left2((-1)^{n}-1)\right)}{\pi n^2}[/latex]

for all even values of n, this is zero, so we can just rewrite this as

[latex] \displaystyle \underbrace{a_{n}}_{n{\subscriptsize \hbox{ odd}}} = \frac{2(-2)}{\pi n^2} = \frac{-4}{\pi n^2}. [/latex]

since for all the odd values, the -1-1 add to give negative -2.

So it's the same thing at the end of the day.

Great. Glad to be of help.