Worked examples · Sheet 3

Fourier series of polynomials on [−π, π)

Theory used: Fourier coefficients Cosine-only and sine-only series Uniform convergence for continuous piecewise smooth functions

Go to: Exercise 1 Exercise 2 Exercise 3

Exercise 1

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

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

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

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

π - x \sim π + n = 1 \sum \infty \frac{2 ( - 1 ) ^{n}}{n} sin (n x)

The function is the sum of the constant $π$ and the odd function $- x$ , therefore it is convenient to compute the coefficients by separating the two terms using the linearity of the integral. The constant contributes only to $a_{0}$ , which equals $2 π$ , hence the constant term of the series is $π$ . The $a_{n}$ with $n \geq 1$ are zero: the integral of $cos (n x)$ over one period is zero, and $x cos (n x)$ is an odd function on a symmetric interval. Only $- x$ contributes to the $b_{n}$ : integration by parts produces the boundary term $- \frac{x cos ( n x )}{n}$ , which evaluated between $- π$ and $π$ gives $- \frac{2 π ( - 1 ) ^{n}}{n}$ , while the remaining integral of $cos (n x)$ is zero. With the minus sign in front we obtain $b_{n} = \frac{2 ( - 1 ) ^{n}}{n}$ , that is, the opposite of the coefficients of the sawtooth wave $f (x) = x$ computed in the theory.

Exercise 2

Find the Fourier series of the $2 π$ -periodic extension of $f (x) = x + x^{2}$ , $x \in [- π, π)$ .

x + x^{2} = odd x + even x^{2}

a_{0} = \frac{1}{π} \int_{- π}^{π} x^{2} d x = \frac{2 π ^{2}}{3}, a_{n} = \frac{2}{π} \int_{0}^{π} x^{2} cos (n x) d x = \frac{4 ( - 1 ) ^{n}}{n ^{2}}

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

x + x^{2} \sim \frac{π ^{2}}{3} + n = 1 \sum \infty (\frac{4 ( - 1 ) ^{n}}{n ^{2}} cos (n x) + \frac{2 ( - 1 ) ^{n + 1}}{n} sin (n x))

Periodic extension of

x + x^{2}

: at the jumps

x = \pm π

the series converges to the mean of the one-sided limits,

π^{2}

(filled dots)

The function has no symmetries, but it decomposes into the odd part $x$ and the even part $x^{2}$ . Since the product of an odd function and $cos (n x)$ is odd, only $x^{2}$ contributes to the $a_{n}$ ; for the same reason only $x$ contributes to the $b_{n}$ . The coefficients of the two pieces are computed by parts on the interval $[0, π]$ : for $x^{2}$ two integrations are needed, carried out in full in the example in the theory, and the result is $a_{n} = \frac{4 ( - 1 ) ^{n}}{n ^{2}}$ ; for $x$ one is enough, and the boundary term leaves $- \frac{π cos ( nπ )}{n}$ , from which $b_{n} = \frac{2 ( - 1 ) ^{n + 1}}{n}$ , since the remaining integral of $cos (n x)$ over $[0, π]$ is zero. The final series is the sum of the two series, therefore it contains both cosines and sines.

Exercise 3

Find the Fourier series of the $2 π$ -periodic extension of $f (x) = x^{3} - π^{2} x$ , $x \in [- π, π]$ , and determine whether the convergence is uniform on $R$ .

f odd ⟹ a_{n} = 0, b_{n} = \frac{2}{π} \int_{0}^{π} (x^{3} - π^{2} x) sin (n x) d x

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

\int_{0}^{π} x^{3} sin (n x) d x = [- \frac{x ^{3} cos ( n x )}{n}]_{0}^{π} + \frac{3}{n} \int_{0}^{π} x^{2} cos (n x) d x = \frac{π ^{3} ( - 1 ) ^{n + 1}}{n} + \frac{3}{n} \cdot \frac{2 π ( - 1 ) ^{n}}{n ^{2}} = \frac{π ^{3} ( - 1 ) ^{n + 1}}{n} + \frac{6 π ( - 1 ) ^{n}}{n ^{3}}

b_{n} = \frac{2}{π} (\frac{π ^{3} ( - 1 ) ^{n + 1}}{n} + \frac{6 π ( - 1 ) ^{n}}{n ^{3}} - π^{2} \cdot \frac{π ( - 1 ) ^{n + 1}}{n}) = \frac{12 ( - 1 ) ^{n}}{n ^{3}}

x^{3} - π^{2} x \sim n = 1 \sum \infty \frac{12 ( - 1 ) ^{n}}{n ^{3}} sin (n x), uniform convergence on R

y = -12 \sin x

Periodic extension of

x^{3} - π^{2} x

, continuous because it equals

0

\pm π

, and the partial sum

- 12 sin x

(dashed), already almost overlapping

The function is odd, being a sum of odd powers of $x$ , therefore the series contains only sines. The integral of $x sin (n x)$ is computed by parts as in the previous exercises. For $x^{3} sin (n x)$ we integrate by parts once and the remaining integral contains $x^{2} cos (n x)$ , whose value $\frac{2 π ( - 1 ) ^{n}}{n ^{2}}$ on $[0, π]$ was obtained in the example in the theory on the series of $x^{2}$ (the coefficient $a_{n} = \frac{4 ( - 1 ) ^{n}}{n ^{2}}$ is equivalent to $\int_{0}^{π} x^{2} cos (n x) d x = \frac{2 π ( - 1 ) ^{n}}{n ^{2}}$ ). In the final combination the terms in $\frac{( - 1 ) ^{n + 1}}{n}$ coming from $x^{3}$ and from $- π^{2} x$ cancel, hence only the term in $\frac{( - 1 ) ^{n}}{n ^{3}}$ survives and we obtain $b_{n} = \frac{12 ( - 1 ) ^{n}}{n ^{3}}$ . Since $f (π) = f (- π) = 0$ , the periodic extension is continuous, and the function is piecewise smooth; therefore the uniform convergence theorem applies and the series converges uniformly to $f$ on $R$ , in agreement with the convergence of the series of absolute values $\sum \frac{12}{n ^{3}}$ .