Half-range sine and cosine series: Question bank

Before you start, refresh three foundations this page leans on: Even and Odd Functions (the symmetry that deletes half the coefficients), Dirichlet Conditions (what the series does at a jump), and the Fourier Series — full range we specialise from.

True or false — justify

An "extension" means the invented values we place on $[- L, 0]$ so that a function known only on $[0, L]$ becomes a full-period function on $(- L, L)$ . Keep that picture in mind for every item.

A half-range sine series and a half-range cosine series of the same

f

agree on

(0, L)

True — both reproduce the same

f (x)

on the open interval

0 < x < L

; they only differ in the invented values outside

[0, L]

(one odd, one even), so on the physical interval they represent the identical function.

The half-range sine series can represent

f (x) = 5

(a nonzero constant) exactly at every point of

[0, L]

False — the odd extension forces the series to

0

x = 0

(odd functions have

f (0) = 0

), and the periodic extension jumps at

x = 0, L

, so the sine series equals

5

only on the open

(0, L)

, converging to the midpoint at the ends.

Because

a_{n} = 0

for a sine series, the sine series has no constant (average) term

True — the odd extension is antisymmetric about the origin, so its average over a full period is

0

; there is simply no

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

to carry.

If two functions have the same values on

[0, L]

they must have the same half-range sine coefficients

True —

b_{n} = \frac{2}{L} \int_{0}^{L} f sin \frac{nπ x}{L} d x

depends only on

f

[0, L]

; whatever happens elsewhere is irrelevant because we never integrate there.

The period of the extended function in a half-range expansion is

L

False — the period is

2 L

; we know only half of that period (namely

[0, L]

), which is exactly why it is called "half-range."

For an even extension, the sine coefficients

b_{n}

vanish because

f (x) sin \frac{nπ x}{L}

is odd

True — (even)

\times

(odd)

=

odd, and an odd function integrated over the symmetric interval

[- L, L]

gives

0

, killing every

b_{n}

The half-range cosine series always converges at least as fast as the sine series for the same

f

False in general — it converges faster only when the even extension is smoother than the odd one; if

f

already vanishes at the ends, the odd extension can be the continuous one and win.

Spot the error

Each statement below contains a flaw a careful student can catch. Find it and say why it breaks.

f (x) = x

[0, π]

has sine coefficients

b_{n} \sim 1/ n^{2}

, so the series converges quickly."

Wrong decay rate — the odd extension of

x

is a sawtooth with a jump at

x = π

, and jump discontinuities give

b_{n} \sim 1/ n

, the slow rate. The

1/ n^{2}

rate belongs to the cosine series (whose even extension

∣ x ∣

is continuous).

"To get the sine series I use

b_{n} = \frac{1}{L} \int_{- L}^{L} f sin \frac{nπ x}{L} d x

with the standard formula."

You cannot —

f

is undefined on

[- L, 0]

, so that integral has no integrand there. The valid route is to choose the odd extension, which turns

\frac{1}{L} \int_{- L}^{L}

into

\frac{2}{L} \int_{0}^{L}

; the factor

2

is the fingerprint of that choice.

"The cosine series of

f (x) = x

[0, π]

\sum a_{n} cos n x

with no separate constant term."

Missing the

\frac{a _{0}}{2} = \frac{π}{2}

term — the even extension of

x

∣ x ∣

, whose average over a period is nonzero, so the constant term is essential; dropping it shifts the whole series down by

π /2

"At

x = π

the sine series of

x

[0, π]

equals

π

, matching

f (π) = π

No — the periodic odd extension jumps from

+ π

(just left of

π

) to

- π

(just right), so by the midpoint rule the series converges to

\frac{1}{2} (π + (- π)) = 0

, not

π

"Since sine series suit fixed ends, I'll use a sine series for an insulated rod."

Backwards — insulated (zero-slope) ends are Neumann conditions, which need functions with zero derivative at the ends.

cos \frac{nπ x}{L}

has zero slope at

x = 0, L

, so cosine is the match; sine matches zero-value (Dirichlet) ends.

"Doubling the integral

(\frac{2}{L} \int_{0}^{L})

is just a convention; the plain

\frac{1}{L} \int_{0}^{L}

would also work."

No — the factor

2

comes from folding a symmetric integrand: for an even integrand

\int_{- L}^{L} = 2 \int_{0}^{L}

. Using

\frac{1}{L} \int_{0}^{L}

halves every coefficient and the series would reconstruct

f /2

, not

f

Why questions

Why does forcing

f

to be odd delete every cosine coefficient?

Because

cos

is even, so (odd

f

)

\times

(even

cos

) is an odd integrand, and any odd function integrated over the symmetric

[- L, L]

cancels to

0

— leaving only the sine (odd

\times

odd

=

even) terms.

Why are we free to choose the extension at all?

The boundary conditions of the PDE only specify

f

on the physical interval

[0, L]

; nothing constrains

x < 0

, so the values there are ours to invent — we pick the symmetry that makes the series match the BCs.

Why does the smoothness of the extension control how fast the coefficients decay?

A jump in the periodic extension forces slow

1/ n

decay, a continuous-but-kinked one gives

1/ n^{2}

, and each extra continuous derivative buys another power of

1/ n

— Fourier coefficients "feel" the worst discontinuity of the extension.

Why does the sine series automatically satisfy

u (0) = u (L) = 0

Every basis function

sin \frac{nπ x}{L}

is zero at

x = 0

and at

x = L

(since

sin nπ = 0

), so any sum of them vanishes there — exactly what fixed-end (Dirichlet) problems demand.

Why does the cosine series carry a

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

term while the sine series does not?

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

is the average value of

f

; an even extension keeps this average, but an odd extension has average

0

by antisymmetry, so its constant term is absent.

Why does the cosine basis pair naturally with insulated (Neumann) ends?

\frac{d}{d x} cos \frac{nπ x}{L} = - \frac{nπ}{L} sin \frac{nπ x}{L}

is zero at

x = 0, L

, so each cosine has zero slope at the ends — matching a "no heat flow through the boundary" condition.

Why can two very different-looking extensions give the same answer on

[0, L]

Both are built to reproduce

f

[0, L]

by construction; the extension only decides the behaviour outside that interval, which the physics never sees.

Edge cases

A "degenerate" input is one where the general rule needs a second look — a constant, a zero, an endpoint, or a $\to \infty$ limit. These are where marks are lost.

What is the half-range sine series of

f (x) = 0

Every

b_{n} = \frac{2}{L} \int_{0}^{L} 0 d x = 0

, so the series is identically

0

— the trivial but valid consistency check.

Take

f (x) = 1

[0, L]

: does its cosine series have any

cos

terms?

No — the even extension of a constant is still the constant, so

a_{0} /2 = 1

and all

a_{n} = 0

for

n \geq 1

; the "series" is just the constant

1

, the pure average.

For the sine series of

f (x) = 1

[0, L]

, what happens at the endpoints?

The odd extension jumps from

- 1

+ 1

across

x = 0

and again at

x = L

, so the series converges to

0

(the midpoint) at both ends, even though

f = 1

on the open interval.

n \to \infty

, what must

b_{n}

(or

a_{n}

) do for any legitimate

f

satisfying Dirichlet's conditions?

They must tend to

0

(Riemann–Lebesgue) — coefficients that don't vanish would mean the series cannot converge; the rate of that decay is what smoothness controls.

f

is already odd on

[- L, L]

but you only feed it

[0, L]

and take a cosine series, what do you get?

You get the cosine series of its even reflection, not of

f

itself off

[0, L]

; on

(0, L)

it still equals

f

, but the reconstructed function outside is the mirror image, not the original odd

f

At a point where the periodic extension is continuous, what does the series converge to?

Exactly

f (x)

— the midpoint rule only bites at jumps; where the extension is continuous, Dirichlet's theorem gives pointwise convergence to the function value.

What half-range series would you choose for a rod with one fixed end (

u = 0

) and one insulated end (

u_{x} = 0

Neither pure sine nor pure cosine — mixed boundary conditions need the odd-quarter/eighth-range basis like

sin \frac{( 2 n - 1 ) π x}{2 L}

, which vanishes at one end and has zero slope at the other. (See Neumann Boundary Conditions and Heat Equation — separation of variables.)

Connections

Half-range sine and cosine series (index 4.7.6) — the parent note these traps drill.
Even and Odd Functions — the symmetry that every "why" answer above leans on.
Dirichlet Conditions — the midpoint rule behind all the endpoint edge cases.
Fourier Series — full range — where the doubled-integral factor of $2$ comes from.
Heat Equation — separation of variables — Dirichlet ends $\to$ sine.
Wave Equation on a finite string — fixed ends $\to$ sine.
Neumann Boundary Conditions — insulated ends $\to$ cosine.