Why is the zeroth coefficient in a Fourier series divided by 2?

Question

I just learned that $a_0$ is basically the average of a function $f(x)$ on the interval $[-\pi, \pi]$, and that a Fourier series is given by $$ f(x) = a_0 + \sum_{n=1}^\infty (a_n \sin(nx) + b_n\cos(nx))$$

where $a_0 = \dfrac{1}{2\pi} \int_{-\pi}^{\pi} f(x) dx $. However, on other sources online I usually see that the $a_0$ is divided by 2. What is the reason for this?

Answer 1

If you look carefully at these other sources you will find that they define $a_0$ differently. If you work out $a_0$ their way and divide by $2$, then work it out your way and don't divide by $2$, you will find that you get the same constant term in the Fourier series.

Answer 2

It is better to write it with $\frac{a_0}{2}$ because you can use the general formula for $a_k=\frac{2}{T}\int_c^{c+T}f(t)\cos(k\omega t)dt$

Answer 3

Consider expressing the Fourier series as $$f(x) = \sum_{n=-\infty}^{+\infty}(a_n\cos nx+b_n\sin nx)$$ where $a_n=\frac{1}{2\pi}\int_{-\pi}^{+\pi}f(t)\cos nxt\ dt$ and $b_n=\frac{1}{2\pi}\int_{-\pi}^{+\pi}f(t)\sin nxt\ dt$.

Note that $\cos\phi$ is an even function, i.e. $\cos(-\phi)=\cos\phi$, and therefore $a_{-n} = a_n$. Similarly, $\sin\phi$ is an odd function, i.e. $\sin(-\phi)=-\sin\phi$, and therefore $b_{-n} = -b_n$. Thus, we can combine the terms with the positive and negative $n$ of the same absolute value, which will result in doubled coefficients $2a_n$ and $2b_n$ for $n>0$, while the terms with $n<0$ will be dropped. Since the terms with $n=0$ do not have a pair, they will not be doubled, so $a_0$ will remain the same, and $b_0$ can be omitted since $b_0=0$.