Worked examples · Sheet 2

Fourier series of even functions

Theory used: Cosine-only and sine-only series Fourier coefficients Coefficients of a trigonometric polynomial

Exercise 1

Find the Fourier series of the $2 π$ -periodic extension of $f (x) = ∣ x ∣$ , $x \in [- π, π]$ .

f even ⟹ b_{n} = 0, a_{0} = \frac{2}{π} \int_{0}^{π} x d x = \frac{2}{π} \cdot \frac{π ^{2}}{2} = π

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

∣ x ∣ \sim \frac{π}{2} - k = 0 \sum \infty \frac{4}{π ( 2 k + 1 ) ^{2}} cos ((2 k + 1) x)

y = \dfrac{\pi}{2} - \dfrac{4}{\pi} \cos x

Triangle wave, the periodic extension of

∣ x ∣

, and the partial sum

\frac{π}{2} - \frac{4}{π} cos x

(dashed)

The absolute value is an even function, therefore the $b_{n}$ are zero and the series contains only cosines, with the $a_{n}$ given by twice the integral over $[0, π]$ , where $∣ x ∣ = x$ . The coefficient $a_{0}$ equals $π$ , hence the constant term is $\frac{π}{2}$ , the mean of $∣ x ∣$ over one period. For $n \geq 1$ we integrate by parts: the term with $sin (nπ)$ is zero and what remains is the integral of $sin (n x)$ , which equals $\frac{1 - ( - 1 ) ^{n}}{n ^{2}}$ after division by $n$ . Therefore $a_{n}$ is zero for even $n$ and equals $- \frac{4}{π n ^{2}}$ for odd $n$ : the series contains only the cosines of the odd harmonics, with negative sign.

Exercise 2

Find the Fourier series of $f (x) = cos^{2} x$ .

cos^{2} x = \frac{1 + cos ( 2 x )}{2} = \frac{1}{2} + \frac{1}{2} cos (2 x)

a_{0} = 1, a_{2} = \frac{1}{2}, a_{n} = 0 (n \neq = 0, 2), b_{n} = 0

cos^{2} x \sim \frac{1}{2} + \frac{1}{2} cos (2 x)

The cosine double-angle formula rewrites $cos^{2} x$ as $\frac{1}{2} + \frac{1}{2} cos (2 x)$ , which is already a trigonometric polynomial. Since the Fourier coefficients of a trigonometric polynomial coincide with its coefficients, no integral needs to be computed: the Fourier series has only the terms $\frac{a _{0}}{2} = \frac{1}{2}$ and $a_{2} cos (2 x) = \frac{1}{2} cos (2 x)$ , and it coincides with the function at every point. It is always worth checking whether the given function reduces to a trigonometric polynomial by an identity, before setting up the integrals.

Exercise 3

Find the Fourier series of $f (x) = ∣ sin x ∣$ .

f even, π -periodic, a_{0} = \frac{2}{π} \int_{0}^{π} sin x d x = \frac{2}{π} [- cos x]_{0}^{π} = \frac{4}{π}

a_{n} = \frac{2}{π} \int_{0}^{π} sin x cos (n x) d x = \frac{1}{π} \int_{0}^{π} (sin ((1 + n) x) + sin ((1 - n) x)) d x = \frac{1}{π} (\frac{1 - ( - 1 ) ^{1 + n}}{1 + n} + \frac{1 - ( - 1 ) ^{1 - n}}{1 - n}) (n \neq = 1)

n odd ⟹ a_{n} = 0, n even ⟹ a_{n} = \frac{2}{π} (\frac{1}{1 + n} + \frac{1}{1 - n}) = - \frac{4}{π ( n ^{2} - 1 )}

a_{1} = \frac{2}{π} \int_{0}^{π} sin x cos x d x = \frac{1}{π} \int_{0}^{π} sin (2 x) d x = 0

∣ sin x ∣ \sim \frac{2}{π} - k = 1 \sum \infty \frac{4}{π ( 4 k ^{2} - 1 )} cos (2 k x)

y = \left|\sin x\right|

y = \dfrac{2}{\pi} - \dfrac{4}{3 \pi} \cos \left(2 x\right)

y = ∣ sin x ∣

has period

π

; the partial sum

\frac{2}{π} - \frac{4}{3 π} cos (2 x)

(dashed) contains only a cosine of period

π

The function $∣ sin x ∣$ is even, therefore the $b_{n}$ are zero; on $[0, π]$ the sine is nonnegative and the absolute value can be removed. The coefficient $a_{0}$ equals $\frac{4}{π}$ , hence the mean of the function is $\frac{2}{π}$ . For the $a_{n}$ the product $sin x cos (n x)$ is transformed into the sum of two sines with the product-to-sum formulas, and each sine is integrated directly; the case $n = 1$ must be treated separately because the second term would have a zero denominator, and the direct computation with $sin x cos x = \frac{sin ( 2 x )}{2}$ gives $a_{1} = 0$ . Since $(- 1)^{1 + n} = - (- 1)^{n}$ , for odd $n$ the numerators vanish and for even $n$ both equal $2$ ; the sum of the two fractions gives $a_{n} = - \frac{4}{π ( n ^{2} - 1 )}$ . Only the even harmonics remain in the series, written with $n = 2 k$ : the result reflects the fact that $∣ sin x ∣$ has period $π$ , therefore only the cosines of compatible period appear in its expansion.