Convergence of Fourier Series

MATH 467 Partial Differential Equations

J. Robert Buchanan

Department of Mathematics

Spring 2016

Objectives

In this lesson we will explore the questions:

▶ What are the conditions that guarantee the Fourier series

of a given function f (x) converges? ▶ If the Fourier series of a given function f (x) converges,

does it converge to the value of f (x) at a given x?

Short Answer to First Question

A Fourier series will converge for a large class of functions, though we will prove convergence only for the class of piecewise smooth functions.

Definition

A function f (x) is piecewise continuous on [a, b] (or (a, b)) if there are finitely many points a = x_{0} < x_{1} < x_{2} < x_{3} < ⋯ < x_{n} = b, such that

▶ f (x) is continuous on (x_{i−1}, x_{i}) for all i = 1, 2, …, n, ▶ the one-sided limits lim_{x→x_{i}^{−}}f (x) and lim_{x→x_{i}^{+}}f (x) exist for all

i = 1, 2, …, n − 1, and ▶ the one-sided limits lim_{x→a^{+}} f (x) and lim_{x→b^{−}}f (x) both exist.

Short Answer to First Question

A Fourier series will converge for a large class of functions, though we will prove convergence only for the class of piecewise smooth functions.

Definition

A function f (x) is piecewise continuous on [a, b] (or (a, b)) if there are finitely many points a = x_{0} < x_{1} < x_{2} < x_{3} < ⋯ < x_{n} = b, such that

▶ f (x) is continuous on (x_{i−1}, x_{i}) for all i = 1, 2, …, n, ▶ the one-sided limits lim_{x→x_{i}^{−}}f (x) and lim_{x→x_{i}^{+}}f (x) exist for all

i = 1, 2, …, n − 1, and ▶ the one-sided limits lim_{x→a^{+}} f (x) and lim_{x→b^{−}}f (x) both exist.

Remarks

▶ Function f (x) is piecewise continuous on (−∞, ∞) if it is

piecewise continuous on every finite interval [a, b]. ▶ Function f (x) is piecewise smooth on [a, b] if f′(x) is

piecewise continuous on [a, b]. ▶ Since any piecewise continuous function is integrable, the

Fourier coefficients of any piecewise continuous function are well-defined (but this is not enough to show the Fourier series converges).

Notation

For any c ∈ [a, b], define

lim_{x→c^{+}} f (x) = f (c+) and lim_{x→c^{−}} f (x) = f (c−).

Function f (x) is continuous at c if and only if f (c+) = f (c−) = f (c).

Examples

Determine which of the following functions is piecewise smooth, piecewise continuous, or neither on [−L, L] for some L > 0.

f_{1}(x) = \{ \begin{matrix}L & if −L ≤ x < 0 \\ x & if 0 ≤ x ≤ L \end{matrix}
f_{2}(x) = \{ \begin{matrix}x^{2} & if −L ≤ x < 0 \\ x − 1 & if 0 ≤ x ≤ L \end{matrix}
f_{3}(x) = cos \frac{1}{x}
f_{4}(x) = x cos \frac{1}{x}
f_{5}(x) = x^{3/5}
f_{6}(x) = x^{6/5}
f_{7}(x) = \frac{1}{x}

Dirichlet Convergence Theorem

Theorem

Assume that f (x) is a piecewise smooth function on the interval [−L, L] extended to (−∞, ∞) periodically with period 2L. Then the Fourier series of f (x) converges for all x to the value

\frac{1}{2} (f (x+) + f (x−)).

Remarks:

▶ If f is continuous at x = x_{0} then the Fourier series

converges to f (x_{0}) when x = x_{0}. ▶ If f has a jump or removeable discontinuity at x = x_{0}, the

Fourier series converges to the average of the limits of f from the left and right at x = x_{0}.

Dirichlet Convergence Theorem

Theorem

Assume that f (x) is a piecewise smooth function on the interval [−L, L] extended to (−∞, ∞) periodically with period 2L. Then the Fourier series of f (x) converges for all x to the value

\frac{1}{2} (f (x+) + f (x−)).

Remarks:

▶ If f is continuous at x = x_{0} then the Fourier series

converges to f (x_{0}) when x = x_{0}. ▶ If f has a jump or removeable discontinuity at x = x_{0}, the

Fourier series converges to the average of the limits of f from the left and right at x = x_{0}.

Example

Consider the piecewise-defined function

f (x) = \{ \begin{matrix}x & if −1 ≤ x < 0, \\ 0 & if 0 ≤ x < 1. \end{matrix}

Its Fourier series representation is

f (x) ∼ −\frac{1}{4} + ∑_{n=1}^{∞} \frac{(−1)^{n+1}}{nπ} sin(nπx)

+ ∑_{n=1}^{∞} \frac{2}{(2n − 1)^{2}π^{2}} cos ((2n − 1)πx).

Sketch the graph of the Fourier series.

Graph Application: Finding the Sum of a Series

If an infinite series is made up of Fourier coefficients for some function, the function can be used to sum the series.

Example

1. Find the Fourier series for f (x) = |x| on [−π, π].

2. Use the Fourier series and f (x) to find

∑_{n=1}^{∞} \frac{1}{(2n − 1)^{2}} .

Application: Finding the Sum of a Series

If an infinite series is made up of Fourier coefficients for some function, the function can be used to sum the series.

Example

1. Find the Fourier series for f (x) = |x| on [−π, π].

2. Use the Fourier series and f (x) to find

∑_{n=1}^{∞} \frac{1}{(2n − 1)^{2}} .

Solution

Function f (x) = |x| is an even function.

a_{0} = \frac{2}{π} ∫_{0}^{π} x dx = π
a_{n} = \frac{2}{π} ∫_{0}^{π} x cos (nx) dx = \{ \begin{matrix}−4/(n^{2}π) & if n is odd, \\ 0 & if n is even. \end{matrix}

Fourier series:

|x| = \frac{π}{2} − \frac{4}{π} ∑_{n=1}^{∞} \frac{1}{(2n − 1)^{2}} cos ((2n − 1)x)
0 = \frac{π}{2} − \frac{4}{π} ∑_{n=1}^{∞} \frac{1}{(2n − 1)^{2}} cos ((2n − 1)(0))
\frac{π^{2}}{8} = ∑_{n=1}^{∞} \frac{1}{(2n − 1)^{2}}

Application: Finding the Sum of a Series

Consider the function f (x) = \{ \begin{matrix}−1 & if −π ≤ x < 0 \\ 0 & if x = 0 \\ 1 & if 0 < x ≤ π. \end{matrix}

1. Find the Fourier series for f (x) on [−π, π].

2. Sketch the graph of the Fourier series for f (x).

3. Use the Fourier series and f (x) to find

∑_{k=1}^{∞} \frac{(−1)^{k}}{2k − 1}.

Solution

Function f (x) is an odd function.

b_{n} = \frac{1}{π} ∫_{−π}^{π} f (x) sin(nx) dx
= \frac{−1}{π} ∫_{−π}^{0} sin(nx) dx + \frac{1}{π} ∫_{0}^{π} sin(nx) dx
= \frac{2}{π} ∫_{0}^{π} sin(nx) dx
= \frac{2}{nπ}(1 − cos (nπ))
= \{ \begin{matrix}4/(nπ) & if n is odd, \\ 0 & if n is even. \end{matrix}

Fourier series:

f (x) = \frac{4}{π} ∑_{n=1}^{∞} \frac{1}{2n − 1} sin((2n − 1)x).

Graph

Question: why does the Fourier series converge to f (x) on (−π, π)?

Summing the Series

Let x = π/2.

f (x) = \frac{4}{π} ∑_{n=1}^{∞} \frac{1}{2n − 1} sin((2n − 1)x)

1 = \frac{4}{π} ∑_{n=1}^{∞} \frac{1}{2n − 1} sin ((2n − 1)\frac{π}{2})

\frac{π}{4} = ∑_{n=1}^{∞} \frac{(−1)^{n−1}}{2n − 1}

−\frac{π}{4} = ∑_{n=1}^{∞} \frac{(−1)^{n}}{2n − 1}

Let x = π/2.

f (x) = \frac{4}{π} ∑_{n=1}^{∞} \frac{1}{2n − 1} sin((2n − 1)x)

1 = \frac{4}{π} ∑_{n=1}^{∞} \frac{1}{2n − 1} sin ((2n − 1)\frac{π}{2})

\frac{π}{4} = ∑_{n=1}^{∞} \frac{(−1)^{n−1}}{2n − 1}

−\frac{π}{4} = ∑_{n=1}^{∞} \frac{(−1)^{n}}{2n − 1}

Let x = π/2.

f (x) = \frac{4}{π} ∑_{n=1}^{∞} \frac{1}{2n − 1} sin((2n − 1)x)

1 = \frac{4}{π} ∑_{n=1}^{∞} \frac{1}{2n − 1} sin ((2n − 1)\frac{π}{2})

\frac{π}{4} = ∑_{n=1}^{∞} \frac{(−1)^{n−1}}{2n − 1}

−\frac{π}{4} = ∑_{n=1}^{∞} \frac{(−1)^{n}}{2n − 1}

Partial Sums Detailed View Gibbs Phenomenon

▶ Oscillation of the partial sums of a Fourier series near a

jump discontinuity is called the Gibbs phenomenon. ▶ It was first mathematically explained by Josiah Willard

Gibbs, though others had considered it (including Albert A. Michelson of the Michelson-Morley experiment).

▶ We will outline an explanation of the Gibbs phenomenon.

Partial Sum

Denote the Nth partial sum of the Fourier series as

s_{N}(x) = \frac{4}{π} ∑_{n=1}^{N} \frac{1}{2n − 1} sin((2n − 1)x).

A calculus argument can locate the maxima in the graph of s_{N}(x) near x = 0.

s′_{N}(x) = \frac{4}{π} ∑_{n=1}^{N}cos ((2n − 1)x)
(sin x)s′_{N}(x) = \frac{4}{π} ∑_{n=1}^{N}sin x cos ((2n − 1)x)

Use a product-to-sum formula on the right-hand side.

Partial Sum

Denote the Nth partial sum of the Fourier series as

s_{N}(x) = \frac{4}{π} ∑_{n=1}^{N} \frac{1}{2n − 1} sin((2n − 1)x).

A calculus argument can locate the maxima in the graph of s_{N}(x) near x = 0.

s′_{N}(x) = \frac{4}{π} ∑_{n=1}^{N}cos ((2n − 1)x)
(sin x)s′_{N}(x) = \frac{4}{π} ∑_{n=1}^{N}sin x cos ((2n − 1)x)

Use a product-to-sum formula on the right-hand side.

Partial Sum

Denote the Nth partial sum of the Fourier series as

s_{N}(x) = \frac{4}{π} ∑_{n=1}^{N} \frac{1}{2n − 1} sin((2n − 1)x).

A calculus argument can locate the maxima in the graph of s_{N}(x) near x = 0.

s′_{N}(x) = \frac{4}{π} ∑_{n=1}^{N}cos ((2n − 1)x)
(sin x)s′_{N}(x) = \frac{4}{π} ∑_{n=1}^{N}sin x cos ((2n − 1)x)

Use a product-to-sum formula on the right-hand side.

Partial Sum

Denote the Nth partial sum of the Fourier series as

s_{N}(x) = \frac{4}{π} ∑_{n=1}^{N} \frac{1}{2n − 1} sin((2n − 1)x).

A calculus argument can locate the maxima in the graph of s_{N}(x) near x = 0.

s′_{N}(x) = \frac{4}{π} ∑_{n=1}^{N}cos ((2n − 1)x)
(sin x)s′_{N}(x) = \frac{4}{π} ∑_{n=1}^{N}sin x cos ((2n − 1)x)

Use a product-to-sum formula on the right-hand side.

Critical Numbers

(sin x)s′_{N}(x) = \frac{4}{π} ∑_{n=1}^{N}cos ((2n − 1)x) sin x

= \frac{2}{π} ∑_{n=1}^{N}(sin(2nx) − sin(2(n − 1)x))

= \frac{2}{π} sin(2Nx)

If 0 < x < π then s′_{N}(x) = 0 if and only if sin(2Nx) = 0. Thus the critical numbers in (0, π) are

(sin x)s′_{N}(x) = \frac{4}{π} ∑_{n=1}^{N}cos ((2n − 1)x) sin x

= \frac{2}{π} ∑_{n=1}^{N}(sin(2nx) − sin(2(n − 1)x))

= \frac{2}{π} sin(2Nx)

If 0 < x < π then s′_{N}(x) = 0 if and only if sin(2Nx) = 0. Thus the critical numbers in (0, π) are

(sin x)s′_{N}(x) = \frac{4}{π} ∑_{n=1}^{N}cos ((2n − 1)x) sin x

= \frac{2}{π} ∑_{n=1}^{N}(sin(2nx) − sin(2(n − 1)x))

= \frac{2}{π} sin(2Nx)

If 0 < x < π then s′_{N}(x) = 0 if and only if sin(2Nx) = 0. Thus the critical numbers in (0, π) are

Maxima or Minima?

Apply the Second Derivative Test to determine whether the critical numbers are maxima or minima.

(sin x)s′_{N}(x) = \frac{2}{π} sin(2Nx)
(cos x)s′_{N}(x) + (sin x)s′′_{N}(x) = \frac{4N}{π} cos (2Nx)

Let x = π/(2N), the critical number closest to x = 0.

(cos \frac{π}{2N} ) s′_{N} (\frac{π}{2N} ) + (sin \frac{π}{2N} ) s′′_{N} (\frac{π}{2N} ) = \frac{4N}{π} cos π
(sin \frac{π}{2N} ) s′′_{N} (\frac{π}{2N} ) = −\frac{4N}{π}

This implies s_{N}(π/2N) is a local maximum.

Value of Local Maximum

s_{N} (\frac{π}{2N} ) = \frac{4}{π} ∑_{n=1}^{N} \frac{1}{2n − 1} sin (\frac{(2n − 1)π}{2N} )
= \frac{2}{π} ∑_{n=1}^{N} \frac{2Nπ}{(2n − 1)Nπ} sin (\frac{(2n − 1)π}{2N} )
= \frac{2}{π} ∑_{n=1}^{N} \frac{2N}{(2n − 1)π} sin (\frac{(2n − 1)π}{2N} ) \frac{π}{N}

s_{N} (\frac{π}{2N} ) = \frac{4}{π} ∑_{n=1}^{N} \frac{1}{2n − 1} sin (\frac{(2n − 1)π}{2N} )
= \frac{2}{π} ∑_{n=1}^{N} \frac{2Nπ}{(2n − 1)Nπ} sin (\frac{(2n − 1)π}{2N} )
= \frac{2}{π} ∑_{n=1}^{N} \frac{2N}{(2n − 1)π} sin (\frac{(2n − 1)π}{2N} ) \frac{π}{N}

s_{N} (\frac{π}{2N} ) = \frac{4}{π} ∑_{n=1}^{N} \frac{1}{2n − 1} sin (\frac{(2n − 1)π}{2N} )
= \frac{2}{π} ∑_{n=1}^{N} \frac{2Nπ}{(2n − 1)Nπ} sin (\frac{(2n − 1)π}{2N} )
= \frac{2}{π} ∑_{n=1}^{N} \frac{2N}{(2n − 1)π} sin (\frac{(2n − 1)π}{2N} ) \frac{π}{N}

Limit of Riemann Sum

lim_{N→∞}s_{N} (\frac{π}{2N} ) = lim_{N→∞} \frac{2}{π} ∑_{n=1}^{N} \frac{2N}{(2n − 1)π} sin (\frac{(2n − 1)π}{2N} ) \frac{π}{N}

= \frac{2}{π} ∫_{0}^{π} \frac{sin x}{x} dx
≈ 1.179

Remark: thus no matter how many terms are included in the Fourier series there is an x → 0^{+} for which f (x) ≈ 1.179 > 1.

Limit of Riemann Sum

lim_{N→∞}s_{N} (\frac{π}{2N} ) = lim_{N→∞} \frac{2}{π} ∑_{n=1}^{N} \frac{2N}{(2n − 1)π} sin (\frac{(2n − 1)π}{2N} ) \frac{π}{N}

= \frac{2}{π} ∫_{0}^{π} \frac{sin x}{x} dx
≈ 1.179

Remark: thus no matter how many terms are included in the Fourier series there is an x → 0^{+} for which f (x) ≈ 1.179 > 1.

Pointwise vs. Uniform Convergence

Definition

Suppose {f_{n}}_{n=1}^{∞} is a sequence of functions defined on domain D and the sequence of values {f_{n}(x)}_{n=1}^{∞} converges for each x ∈ S ⊂ D. Then {f_{n}}_{n=1}^{∞} is said to converge pointwise on S to f defined by

f (x) = lim_{n→∞}f_{n}(x) for x ∈ S.

While the Fourier series converges pointwise to f (x) it does not converge uniformly to f (x).

Uniform Convergence

Definition

A sequence of functions {f_{n}}_{n=1}^{∞} defined on domain D converges uniformly to f on D provided there exists a sequence of positive real numbers {ϵ_{n}}_{n=1}^{∞} for which lim_{n→∞}ϵ_{n} = 0 and

|f_{n}(x) − f (x)|< ϵ_{n}

for all n ∈ ℕ and for all x ∈ D.

Remark: an infinite series converges uniformly if its sequence of partial sums converges uniformly.

Illustration

Sequence {x^{n}e^{−nx}}_{n=1}^{∞} converges uniformly to f (x) = 0 on D = [0, ∞).

Examples

▶ {x^{n}}_{n=1}^{∞} converges pointwise on [0, 1] to

f (x) = \{ \begin{matrix}if 0 ≤ x < 1 \\ 1 \\ if x = 1 \end{matrix} but does not converge uniformly. ▶ {x^{n}}_{n=1}^{∞} converges uniformly to f (x) = 0 on [0, b] for any

0 < b < 1. ▶ The infinite series ∑_{n=1}^{∞}x^{n} converges pointwise on (−1, 1) to

f (x) = \frac{x}{1 − x} but not uniformly.

Justification

▶ The Nth partial sum of the series is

f_{N}(x) = x + x^{2} + ⋯ + x^{N} = \frac{1 − x^{N+1}}{1 − x} − 1.

▶ For any −1 < x < 1,

lim_{N→∞}f_{N}(x) = lim_{N→∞}[\frac{1 − x^{N+1}}{1 − x} − 1] = \frac{1}{1 − x} −1 = \frac{x}{1 − x} = f (x),

(this proves pointwise convergence). ▶ Consider

lim_{x→1^{−}} |f_{N}(x) − f (x)| = lim_{x→1^{−}} |\frac{1 − x^{N+1}}{1 − x} − 1 − \frac{x}{1 − x} |
= lim_{x→1^{−}} \frac{x^{N+1}}{1 − x} = ∞,

so the convergence is not uniform.

Justification

▶ The Nth partial sum of the series is

f_{N}(x) = x + x^{2} + ⋯ + x^{N} = \frac{1 − x^{N+1}}{1 − x} − 1.

▶ For any −1 < x < 1,

lim_{N→∞}f_{N}(x) = lim_{N→∞}[\frac{1 − x^{N+1}}{1 − x} − 1] = \frac{1}{1 − x} −1 = \frac{x}{1 − x} = f (x),

(this proves pointwise convergence). ▶ Consider

lim_{x→1^{−}} |f_{N}(x) − f (x)| = lim_{x→1^{−}} |\frac{1 − x^{N+1}}{1 − x} − 1 − \frac{x}{1 − x} |
= lim_{x→1^{−}} \frac{x^{N+1}}{1 − x} = ∞,

so the convergence is not uniform. me

Homework

▶ Read Section x.y ▶ Exercises: