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 Incoming links: Fourier cosine series of a triangular wave

Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2001 / Paper 1 / 2 / Solution Created 2026-10-18 Updated 2026-10-20

On the basic interval the graph is the pair of straight segments joining

(- π, 0)

,

(0, 1)

and

(π, 0)

. Its periodic extension is an even triangular wave.

Figure 1.
The even triangular wave and its periodic extension
. Original plot marking the basic interval and the repeated linear pieces. Details.

The Fourier coefficients satisfy

b_{n} = 0

by evenness, and

a_{0} = \frac{2}{π} \int_{0}^{π} (1 - x / π) d x = 1, a_{n} = \frac{2}{π} \int_{0}^{π} (1 - x / π) cos (n x) d x = \frac{2 [ 1 - ( - 1 ) ^{n} ]}{π ^{2} n ^{2}} .

For the second formula, integration by parts gives

\int_{0}^{π} x cos (n x) d x = [(- 1)^{n} - 1] / n^{2}

. Only odd indices survive, giving this rescaled Fourier cosine series of a triangular wave:

f (x) = \frac{1}{2} + \frac{4}{π ^{2}} m = 0 \sum \infty \frac{cos [( 2 m + 1 ) x ]}{( 2 m + 1 ) ^{2}} .

The pointwise convergence of a piecewise smooth Fourier series applies throughout the interval, including the continuous joined endpoints. The coefficients also imply uniform convergence by the Weierstrass M-test. At

x = 0

, the value is one, so

m = 0 \sum \infty \frac{1}{( 2 m + 1 ) ^{2}} = \frac{π ^{2}}{8} .

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Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2005 / Paper 1 / 12 / b / Solution Created 2026-10-18 Updated 2026-10-20

Evenness removes every sine coefficient. In the period-two Fourier cosine series convention

f = a_{0} /2 + \sum_{n \geq 1} a_{n} cos (nπ x)

, the constant coefficient is

a_{0} = 2 \int_{0}^{1} (1 - x) d x = 1

. Integration by parts gives

a_{n} = 2 \int_{0}^{1} (1 - x) cos (nπ x) d x = \frac{2 [ 1 - ( - 1 ) ^{n} ]}{n ^{2} π ^{2}} .

Thus the Fourier cosine series of a triangular wave is

f (x) = \frac{1}{2} + \frac{4}{π ^{2}} m = 0 \sum \infty \frac{cos (( 2 m + 1 ) π x )}{( 2 m + 1 ) ^{2}} .

The periodic function is continuous and piecewise smooth, so the Dirichlet-Jordan convergence theorem gives its value at every point. The summable absolute coefficient bound also gives uniform convergence by the Weierstrass M-test.

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Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2005 / Paper 1 / 12 / c / Solution Created 2026-10-18 Updated 2026-10-20

Evaluate the Fourier cosine series of a triangular wave at its continuous peak

x = 0

, where

f (0) = 1

. Every cosine is one, so

1 = \frac{1}{2} + \frac{4}{π ^{2}} \sum_{m = 0}^{\infty} \frac{1}{( 2 m + 1 ) ^{2}}, m = 0 \sum \infty \frac{1}{( 2 m + 1 ) ^{2}} = \frac{π ^{2}}{8} .

The convergence established above justifies substituting this point into the series.

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Triangular wave Created 2026-10-06 Updated 2026-11-05

A triangular wave is a continuous piecewise-linear periodic function with successive rising and falling segments. For

b > 0

, the even

2 π

-periodic extension of

b x

from

[0, π]

is

P (s) = b ∣ s - 2 kπ ∣, (2 k - 1) π \leq s \leq (2 k + 1) π, k \in Z .

It has troughs of height zero at

2 kπ

, peaks of height

bπ

at

(2 k + 1) π

- b

and

b

between the corners. The function is continuous at the corners but lacks a derivative there.

Its Fourier cosine series contains a constant term and odd modes with coefficients proportional to the inverse square of the mode number. The Fourier cosine series of a triangular wave computes this for another normalization. Even periodic extensions also encode reflection in problems with Neumann boundary conditions.

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