Full Fourier series — coefficients derivation: Question bank

Before we start, three quick reminders so the questions land.

Recall The three formulas (so you can argue against them)

On $[- L, L]$ (period $2 L$ ), using a single summation index — we call it $n$ everywhere below to avoid confusion: $a_{0} = \frac{1}{L} \int_{- L}^{L} f d x$ , and for $n \geq 1$ , $a_{n} = \frac{1}{L} \int_{- L}^{L} f cos \frac{nπ x}{L} d x$ , $b_{n} = \frac{1}{L} \int_{- L}^{L} f sin \frac{nπ x}{L} d x$ . The series is $\frac{a _{0}}{2} + n \geq 1 \sum [a_{n} cos \frac{nπ x}{L} + b_{n} sin \frac{nπ x}{L}]$ . (When two different indices appear together inside one orthogonality integral we name them $m$ and $n$ — but the coefficient we solve for is always $a_{n}$ or $b_{n}$ .)

True or false — justify

Every prompt below is a claim. Decide true/false, then give the reason.

The constant term of the series equals the average value of

f

over one period.

True — the average is

\frac{1}{2 L} \int_{- L}^{L} f d x

, and the constant term is

\frac{a _{0}}{2} = \frac{1}{2 L} \int_{- L}^{L} f d x

; they are literally the same number.

f

is odd, then every

a_{n}

(including

a_{0}

) is zero.

True —

f

odd makes

f cos \frac{nπ x}{L}

odd (even

\times

odd), so the graph on the right of

0

is the negated mirror of the left, and the two halves cancel over

[- L, L]

f

is even, then every

b_{n}

is zero.

True —

f

even makes

f sin \frac{nπ x}{L}

odd (even

\times

odd), so left and right halves cancel and each

b_{n}

integral over

[- L, L]

vanishes.

For integers

m, n \geq 1

\int_{- L}^{L} cos \frac{mπ x}{L} cos \frac{nπ x}{L} d x = 0

holds for all such pairs.

False — it is

0

only when

m \neq = n

; when

m = n (\geq 1)

the integral equals

L

, and the special case

m = n = 0

gives

\int_{- L}^{L} 1 \cdot 1 d x = 2 L

. That surviving value is exactly what lets us solve for

a_{n}

The sine–cosine integral

\int_{- L}^{L} sin \frac{mπ x}{L} cos \frac{nπ x}{L} d x

is zero only when

m \neq = n

False — it is zero for every

m, n

including

m = n

, because

sin (\cdot) cos (\cdot)

is an odd function whose two halves cancel; a sine and a cosine never overlap.

A full Fourier series must contain both sine and cosine terms.

False — "full" means we allow both; if

f

has a symmetry, one whole family drops out (an odd

f

becomes a pure sine series), yet it is still the full-interval series.

The formula

a_{n} = \frac{1}{L} \int_{- L}^{L} f cos \frac{nπ x}{L} d x

also gives the correct

a_{0}

when you plug in

n = 0

True — that is precisely why the constant is written

\frac{a _{0}}{2}

: at

n = 0

cos 0 = 1

and the formula returns

\frac{1}{L} \int_{- L}^{L} f d x = a_{0}

Two functions being orthogonal means their graphs never cross.

False — orthogonal means

\int_{- L}^{L} f g d x = 0

, i.e. the signed area under the product is zero; the graphs can and usually do cross many times while the area above the axis cancels the area below.

Multiplying

f

cos \frac{nπ x}{L}

and integrating isolates

a_{n}

because it "projects"

f

onto that one wave.

True — thinking of

\int f g

as a dot product, this is projection onto one perpendicular axis; every other basis wave contributes zero signed area, leaving only the

a_{n} L

term.

The Fourier series of the basis function

f (x) = cos \frac{3 π x}{L}

has infinitely many nonzero coefficients.

False — orthogonality kills all but

a_{3} = 1

; a basis function's own series is just itself.

Spot the error

Each line contains a mathematical or factual mistake. Say what is wrong and what it should be.

"On

[- L, L]

b_{n} = \frac{2}{L} \int_{- L}^{L} f sin \frac{nπ x}{L} d x

Wrong prefactor: the full-interval coefficient uses

\frac{1}{L}

, not

\frac{2}{L}

. The

\frac{2}{L}

only appears in half-range series on

[0, L]

, where you integrate over half the period.

"Since the average of

f

\frac{1}{2 L} \int f

, we get

a_{0} = \frac{1}{2 L} \int_{- L}^{L} f d x

Double-counted the

\frac{1}{2}

: the average equals

\frac{a _{0}}{2}

, so

a_{0} = \frac{1}{L} \int_{- L}^{L} f d x

. The

\frac{1}{2}

is already inside the constant term.

\int_{- L}^{L} cos \frac{nπ x}{L} d x = 2 L

for every

n

True only for

n = 0

; for

n \geq 1

the integral is

0

because a full period of a cosine has equal area above and below the axis. The nonzero case is the constant, not the oscillating one.

"To derive

a_{n}

, integrate the series term by term — this is always allowed."

The step assumes term-by-term integration is valid, which needs

f

to be nice enough (piecewise smooth); it is an assumption, not a free lunch, and can fail for pathological functions.

cos mπ = 1

for all integers

m

Wrong:

cos mπ = (- 1)^{m}

, so it alternates

- 1, 1, - 1, \dots

. This sign is exactly what makes even-indexed

b_{n}

vanish for the square wave.

"For the square wave,

a_{0} = \frac{1}{L} [(- 1) L + (1) L]

gives a nonzero constant."

The bracket is

- L + L = 0

, so

a_{0} = 0

; the arithmetic (and the odd symmetry) both force it to zero.

"Since

sin \frac{mπ x}{L} cos \frac{nπ x}{L}

can be positive, its integral over

[- L, L]

is positive."

No — pointwise positivity of a bit of the graph says nothing about the signed area; the product is odd, its two halves cancel exactly, so the integral is

0

Why questions

Answer the "why", not just "what".

Why do we integrate over a full period

[- L, L]

rather than part of it?

A full period is where the oscillating waves have equal area above and below the axis (so they average to zero) and where the orthogonality cancellations hold; over a partial interval those cancellations fail and the projection trick breaks.

Why is the product-to-sum identity used to prove orthogonality?

It turns a product of two cosines into a sum of single cosines, whose signed area over a full period is trivially

0

(or

L

in the matched case) — the raw product is hard to integrate directly.

Why does exactly one term survive when we multiply by

cos \frac{nπ x}{L}

and integrate?

Because

\int cos \frac{mπ x}{L} cos \frac{nπ x}{L} d x

0

for

m \neq = n

and

L

for

m = n

; the whole sum collapses to the single

a_{n} L

term.

Why write the constant as

\frac{a _{0}}{2}

instead of just

a_{0}

Pure bookkeeping: it makes the single formula

a_{n} = \frac{1}{L} \int f cos \frac{nπ x}{L} d x

valid for all

n \geq 0

, unifying the constant with the cosine coefficients.

Why can two different functions have the same set of Fourier coefficients on

[- L, L]

If they differ only at isolated points (e.g. at a jump) the integrals defining the coefficients are unchanged, so the series cannot tell them apart — it matches values in an averaged (mean-square) sense, the idea behind Parseval's theorem.

Why does symmetry (

f

even or odd) cut the work in half?

An odd

f

kills every

a_{n}

and an even

f

kills every

b_{n}

before any integration, because the relevant product is odd and its two halves cancel — so you compute only one family of coefficients.

Why are

cos \frac{nπ x}{L}

and

sin \frac{nπ x}{L}

the right building blocks for period

2 L

Each completes a whole number of cycles across a length

2 L

, so every term — and any sum of them — repeats every

2 L

, matching the periodicity of

f

Why is the complex form

\sum c_{n} e^{inπ x / L}

equivalent to the real sine–cosine form?

Euler's formula splits each exponential into a cosine plus

i

times a sine, so the complex coefficients just repackage

a_{n}, b_{n}

; both describe the same function (see Complex (exponential) Fourier series).

Edge cases

Boundary and degenerate situations the formulas quietly handle.

At a jump discontinuity, what value does the series converge to?

The midpoint of the left and right limits,

\frac{f ( x ^{-} ) + f ( x ^{+} )}{2}

, splitting the difference regardless of how

f

is defined at the jump itself.

Near a jump, the partial sums overshoot by a fixed amount that never shrinks — what is this called?

The Gibbs phenomenon: adding more terms narrows the overshoot's width but its height stays about

9%

of the jump, so the "ringing" spike never disappears — a classic trap when students expect perfect convergence everywhere.

What is the Fourier series of the zero function

f (x) = 0

Every coefficient is zero, so the series is just

0

; there is nothing to project onto and no wave is present.

For a constant function

f (x) = c

, which coefficients survive?

Only

a_{0} = \frac{1}{L} \int_{- L}^{L} c d x = 2 c

, giving the constant term

\frac{a _{0}}{2} = c

; all

a_{n}

(

n \geq 1

) and

b_{n}

vanish since

c \times

(oscillation) has equal area above and below the axis.

What happens to the

b_{n}

formula when

n = 0

sin 0 = 0

identically, so

b_{0} = 0

trivially — there is no sine constant term, which is why the sine sum starts at

n = 1

Does the series value at the endpoints

x = \pm L

equal

f (\pm L)

Generally no — the periodic extension makes

x = L

and

x = - L

the same point on the tiled line, so the series converges to the midpoint of

f (- L^{+})

and

f (L^{-})

, the "wrap-around" jump.

f

is neither even nor odd, are both sine and cosine families guaranteed nonzero?

Not guaranteed nonzero, but neither family is forced to vanish; you must actually compute both since no symmetry does the work for you.

What if

f

is continuous and

2 L

-periodic with a continuous derivative?

Then the series converges to

f

everywhere (uniformly), the cleanest case — no jump-midpoint and no Gibbs overshoot, since there are no jumps.