Fourier series coefficient justification

Question

My textbook gives the following justification when introducing Fourier series: (NOTE: I have excluded the intermediate calculations that are irrelevant to this specific question).

(1) $ f(x) = \dfrac{a_0}{2} + \sum_{n = 1}^{\infty} \left( a_n\cos(nx) + b_n\sin(nx) \right) \forall x \in [-\pi,\pi]$

(3) $a_0 = \dfrac{1}{\pi} \int^\pi_{-\pi} f(x) dx$

(7) $a_n = \dfrac{1}{\pi} \int^\pi_{-\pi} f(x) \cos(nx) dx$

By (3), formula (7) is also valid for $n = 0$; this is the reason for writing the constant term in (1) as $\dfrac{a_0}{2}$ rather than $a_0$.

---

The justification for writing the constant term in (1) as $\dfrac{a_0}{2}$ rather than $a_0$ does not provide enough information to make sense (to an introductory reader). I would greatly appreciate it if people could please take the time to elaborate on this justification and clarify it for an introductory reader such as myself.

Answer 1

We have:

$$f(x) = \frac{a_0}{2}+\sum_{n=1}^{\infty} a_n\cos(nx) + b_n\sin(nx),$$

then we can write:

$$ \int_{-\pi}^{\pi}f(x) = \int_{-\pi}^{\pi}\frac{a_0}{2} dx + \sum_{n=1}^{\infty}a_n \int_{-\pi}^{\pi}\cos(nx) dx +\sum_{n=1}^{\infty}b_n \int_{-\pi}^{\pi}\sin(nx) dx $$

where $\int_{-\pi}^{\pi}\cos(nx) = 0$ and $\int_{-\pi}^{\pi}\sin(nx) = 0$. Then, $$ \implies a_0 = \frac{1}{\pi} \int_{-\pi}^{\pi}f(x)dx.$$

Here, we also can find the formula for

$$ \int_{-\pi}^{\pi}f(x)\cos(mx) = \frac{a_0}{2}\int_{-\pi}^{\pi}\cos(mx) dx + \sum_{n=1}^{\infty}a_n \int_{-\pi}^{\pi}\cos(nx)\cos(mx) dx \\+\sum_{n=1}^{\infty}b_n \int_{-\pi}^{\pi}\sin(nx)\cos(mx) dx $$

where $$ \int_{-\pi}^{\pi}\cos(nx)\cos(mx) dx = \begin{cases} \pi & n = m \\ 0 & n \neq m \end{cases}, $$

$$ \int_{-\pi}^{\pi}\cos(mx) dx = \frac{1}{m}\sin{mx}|_{-\pi}^{\pi} = 0,$$ $$ \int_{-\pi}^{\pi}\sin(nx)\cos(mx) dx = 0$$ here, $\sin(nx)\cos(mx)$ is an odd function, then we have:

$$\implies a_n = \frac{1}{\pi} \int_{-\pi}^{\pi}f(x)\cos(nx)dx.$$

If, from the first, you use $a_0$ instead of $\frac{a_0}{2}$, then your formulas would be different from each other by a factor: $\frac{1}{2}$. However, in your textbook and also above, it seems for simplification and uniform formulation of the two $a_0$ and $a_n$, it is provided that way. Now, you need to consider (memorize) one formula for $a_n$; then, for $a_0$, just put $n=0$ to make the formula ready for it for calculation.

Answer 2

I guess you are familiar with linear algebra. Let $V$ be a $n$-dimensional space over the real numbers, and let $( , )$ be an inner product on $V$. Let $\{\varphi_i\}_{i=1}^n$ be an orthogonal basis of $V$, in the sense that $(\varphi_i,\varphi_j)=0$ for $i\neq j$. Given any vector $v\in V$, we can express $v$ in terms of this basis: $$v=\sum_{i=1}^n a_i\varphi_i.$$ Now we can express $a_j$ in terms of the inner product and $v$ and $\varphi_j$. In fact we have $$(v,\varphi_j)=(\sum_{i=1}^n a_i\varphi_i,\varphi_j)=\sum_{i=1}^na_i(\varphi_i,\varphi_j) =a_j(\varphi_j,\varphi_j)$$ which gives $$a_j=\frac{(v,\varphi_j)}{(\varphi_j,\varphi_j)}.$$

Now in the case of Fourier series, our vector space $V$ consists of functions on $[-\pi,\pi]$, and the inner product on two such functions $f,g$ is given by $$(f,g)=\int_0^1fg\ dx.$$ And an orthogonal basis of this space is given by $\{\frac{1}{2},\cos x, \cos 2x,\cdots, \sin x,\sin 2x,\cdots\}$. Thus by our reasoning above, we can express our function $f$ in terms of this basis: $$f=a_0\cdot\frac{1}{2}+\sum_{n=1}^\infty(a_n\cos nx+b_n\sin nx)$$ and by the formula $a_j=\frac{(v,\varphi_j)}{(\varphi_j,\varphi_j)}$ we have $$a_0=\frac{(f,\frac{1}{2})}{(\frac{1}{2},\frac{1}{2})} =\frac{\int_{-\pi}^{\pi}\frac{1}{2}f(x)\ dx}{\int_{-\pi}^{\pi}(\frac{1}{2})^2\ dx}=\frac{1}{\pi}\int_{-\pi}^{\pi}f(x)\ dx,$$ $$a_n=\frac{(f,\cos nx)}{(\cos nx,\cos nx)}=\frac{\int_{-\pi}^{\pi}f(x)\cos nx\ dx}{\int_{-\pi}^{\pi}(\cos nx)^2\ dx} =\frac{1}{\pi}\int_{-\pi}^{\pi}f(x)\cos nx\ dx$$ $$b_n=\frac{(f,\sin nx)}{(\sin nx,\sin nx)}=\frac{\int_{-\pi}^{\pi}f(x)\sin nx\ dx}{\int_{-\pi}^{\pi}(\sin nx)^2\ dx} =\frac{1}{\pi}\int_{-\pi}^{\pi}f(x)\sin nx\ dx.$$ Now if you pick the basis $\{1,\cos x, \cos 2x,\cdots, \sin x,\sin 2x,\cdots\}$ instead and express $$f=a_0\cdot 1+\sum_{n=1}^\infty a_n\cos nx+b_n\sin nx,$$ then a similar computation shows that $$a_0=\frac{(f,1)}{(1,1)} =\frac{\int_{-\pi}^{\pi}f(x)\ dx}{\int_{-\pi}^{\pi}\ dx}=\frac{1}{2\pi}\int_{-\pi}^{\pi}f(x)\ dx,$$ $$a_n=\frac{(f,\cos nx)}{(\cos nx,\cos nx)}=\frac{\int_{-\pi}^{\pi}f(x)\cos nx\ dx}{\int_{-\pi}^{\pi}(\cos nx)^2\ dx} =\frac{1}{\pi}\int_{-\pi}^{\pi}f(x)\cos nx\ dx$$ $$b_n=\frac{(f,\sin nx)}{(\sin nx,\sin nx)}=\frac{\int_{-\pi}^{\pi}f(x)\sin nx\ dx}{\int_{-\pi}^{\pi}(\sin nx)^2\ dx} =\frac{1}{\pi}\int_{-\pi}^{\pi}f(x)\sin nx\ dx.$$

Both of these two bases work, but the convention of using the basis $\{\frac{1}{2},\cos x, \cos 2x,\cdots, \sin x,\sin 2x,\cdots\}$ instead of $\{1,\cos x, \cos 2x,\cdots, \sin x,\sin 2x,\cdots\}$ is simply because the expression $a_0=\frac{1}{\pi}\int_{-\pi}^{\pi}f(x)\ dx$, compared to $a_0=\frac{1}{2\pi}\int_{-\pi}^{\pi}f(x)\ dx$, looks like the expression $a_n=\frac{1}{\pi}\int_{-\pi}^{\pi}f(x)\cos nx\ dx$ for $n\geq 1$, and thus can be combined together to form only one formula $$a_n=\frac{1}{\pi}\int_{-\pi}^{\pi}f(x)\cos nx\ dx \text{ for all } n\geq 0.$$