Full Fourier series — coefficients derivation

What we are trying to do

WHAT we want: formulas for $a_0, a_n, b_n$. WHY these basis functions: $\cos\frac{n\pi x}{L}$ and $\sin\frac{n\pi x}{L}$ all have period dividing $2L$, so any sum of them is $2L$-periodic — matching $f$.

The key tool: Orthogonality (derive it first)

Everything rests on these integrals over a full period $[-L, L]$. Let $m, n$ be positive integers.

HOW to derive one (the cosine–cosine case). Use the product-to-sum identity $\cos A\cos B = \tfrac{1}{2}\big[\cos(A-B)+\cos(A+B)\big].$ So with $A=\frac{m\pi x}{L},\,B=\frac{n\pi x}{L}$: $\int_{-L}^{L}\cos\frac{m\pi x}{L}\cos\frac{n\pi x}{L}\,dx = \frac12\int_{-L}^{L}\cos\frac{(m-n)\pi x}{L}\,dx + \frac12\int_{-L}^{L}\cos\frac{(m+n)\pi x}{L}\,dx.$

Why this step? Splitting the product into two simple cosines makes each integral trivial.

• If $m\neq n$: both $m-n$ and $m+n$ are nonzero integers $k$, and $\int_{-L}^{L}\cos\frac{k\pi x}{L}\,dx = \frac{L}{k\pi}\sin\frac{k\pi x}{L}\Big|_{-L}^{L} = \frac{L}{k\pi}\big[\sin k\pi - \sin(-k\pi)\big]=0.$ So the whole thing is $0$.
• If $m=n$: the first integrand is $\cos 0 = 1$ giving $\frac12\cdot 2L = L$, and the second ($k=2n\neq0$) gives $0$. Total $= L$. ✔

The sine–sine and sine–cosine cases follow identically using $\sin A\sin B=\frac12[\cos(A-B)-\cos(A+B)]$ and $\sin A\cos B=\frac12[\sin(A+B)+\sin(A-B)]$.

Deriving the coefficient formulas (from scratch)

Start by assuming the series converges to $f$ and that we may integrate term by term: $f(x) = \frac{a_0}{2} + \sum_{n=1}^{\infty}\left[ a_n \cos\frac{n\pi x}{L} + b_n \sin\frac{n\pi x}{L}\right].$

Step 1 — find $a_0$

Integrate both sides over $[-L,L]$: $\int_{-L}^{L} f\,dx = \frac{a_0}{2}\underbrace{\int_{-L}^{L}1\,dx}_{2L} + \sum_n a_n\underbrace{\int_{-L}^{L}\cos\frac{n\pi x}{L}dx}_{0} + \sum_n b_n\underbrace{\int_{-L}^{L}\sin\frac{n\pi x}{L}dx}_{0}.$ Why this step? Integrating kills every oscillating term (they have zero average over a full period), leaving only the constant. $\Rightarrow \int_{-L}^{L} f\,dx = a_0 L \quad\Longrightarrow\quad \boxed{a_0 = \frac1L\int_{-L}^{L} f(x)\,dx.}$

Step 2 — find $a_m$ (for $m\ge 1$)

Multiply both sides by $\cos\frac{m\pi x}{L}$, then integrate: $\int_{-L}^{L} f\cos\frac{m\pi x}{L}\,dx = \frac{a_0}{2}\underbrace{\int\cos\frac{m\pi x}{L}dx}_{0} + \sum_n a_n\int\cos\frac{n\pi x}{L}\cos\frac{m\pi x}{L}dx + \sum_n b_n\underbrace{\int\sin\frac{n\pi x}{L}\cos\frac{m\pi x}{L}dx}_{0}.$ Why this step? By orthogonality, in the $a_n$ sum only the $n=m$ term survives (equals $L$); all others vanish, and every $b_n$ term vanishes too. $\Rightarrow \int_{-L}^{L} f\cos\frac{m\pi x}{L}\,dx = a_m L \quad\Longrightarrow\quad \boxed{a_m = \frac1L\int_{-L}^{L} f(x)\cos\frac{m\pi x}{L}\,dx.}$

Step 3 — find $b_m$ (for $m\ge 1$)

Multiply by $\sin\frac{m\pi x}{L}$ and integrate. Same reasoning: only the $n=m$ sine–sine term survives ($=L$): $\boxed{b_m = \frac1L\int_{-L}^{L} f(x)\sin\frac{m\pi x}{L}\,dx.}$

Worked examples

Common mistakes

Active recall

Connections

• Orthogonality of functions — the inner-product viewpoint that makes projection work.
• Half-range Fourier series (sine and cosine) — special cases on $[0,L]$ with factor $\frac2L$.
• Even and odd functions — symmetry shortcuts for coefficients.
• Convergence of Fourier series (Dirichlet conditions) — when the series actually equals $f$.
• Complex (exponential) Fourier series — repackaging $a_n,b_n$ as $c_n=\frac1{2L}\int f e^{-in\pi x/L}dx$.
• Separation of variables for the heat equation — where Fourier coefficients fix the initial condition.
• Parseval's theorem — energy relation $\frac1L\int f^2 = \frac{a_0^2}{2}+\sum(a_n^2+b_n^2)$.

Concept Map

Hinglish (regional understanding)