Advanced Analysis Math 121, Fall 2004 Solutions, Midterm 1

1. Answer the following questions. Give a brief justification of your answer.

(a) What is the (smallest) period of the function cos (5πx)?

(b) What is the value of

∫_{0}^{10} cos^{2} (5πx) dx?

(c) What is the value of

∫_{0}^{10} cos (5πx) cos (30πx) dx?

Solution.

• (a) The function cos(kx) has period 2π/k, so cos (5πx) has period 2/5.

• (b) The integration interval is an integer multiple of the period, since 10 = 25 · (2/5). Therefore, since the average value of cos^{2} x over a period is 1/2, we have

∫_{0}^{10} cos^{2} (5πx) dx = 10 · \frac{1}{2} = 5.

• (c) The functions cos (5πx) and cos (30πx) are orthogonal on any period, so

∫_{0}^{10} cos (5πx) cos (30πx) dx = 0.

2. Suppose that f(x) is defined by

f(x) = x^{2} for 0 < x < π.

Sketch graphs of the following extensions of f(x) for −3π < x < 3π:

(a) the periodic extension f_{p}(x) with period π;

(b) the even periodic extension f_{e}(x) with period 2π;

(c) the odd periodic extension f_{o}(x) with period 2π.

Solution.

• Graphs omitted.

3. Suppose that f(x) is the 2π-periodic function defined by

f(x) = x for −π < x < π.

(a) What are the Fourier cosine coefficients a_{n} of f(x)? Explain how you know the answer without evaluating any integrals.

(b) What does the Fourier series of f(x) converge to at x = π/2, and x = π? Why?

(c) Compute the Fourier sine coefficients b_{n} of f(x), and write out the Fourier series expansion of f(x).

Solution.

• (a) Since f is odd, the Fourier cosine coefficients are zero, so a_{n} = 0 for all n ≥ 0.

• (b) The function f is piecewise smooth. According to Dirichlet’s theorem, its Fourier series converges to f(x) at points x where f is continuous, and to the average of the left and right hand limits of f at points where f has a jump discontinuity. Since f(x) is continuous at x = π/2 and f(π/2) = π/2, the Fourier series converges to π/2 at x = π/2. At x = π, the function f has a jump discontinuity with left-hand limit equal to π and right-hand limit equal to −π. Hence the Fourier series converges to 0.

• (c) The Fourier sine coefficients of f are given by

b_{n} = \frac{1}{π} ∫_{−π}^{π} x sin nx dx
= \frac{2}{π} ∫_{0}^{π} x sin nx dx
= \frac{2}{π} [−\frac{x cos nx}{n} + \frac{sin nx}{n^{2}}]_{0}^{π}
= \frac{2}{π} [−\frac{π cos nπ}{n} ]
= \frac{2}{n}(−1)^{n+1}

Thus,

f(x) = 2 (sin x − \frac{1}{2} sin 2x + \frac{1}{3} sin 3x − \frac{1}{4} sin 4x + …)

4. (a) State the orthogonality relations for the functions sin(nπx) on the interval 0 < x < 1, where n = 1, 2, 3, . . ..

(b) Use these orthogonality relations to derive an expression for the coefficients b_{n} in the Fourier sine expansion

f(x) = ∑_{n=1}^{∞} b_{n} sin nπx

of a function defined in 0 < x < 1.

Solution.

• (a) Since sin mπx and sin nπx are orthogonal for n ≠ m, and the average value of sin^{2} x over a period is 1/2, the orthogonality relations are

∫_{0}^{1} sin mπx sin nπx dx = \{ \begin{matrix}0 & for m ≠ n, \\ 1/2 & for m = n. \end{matrix}

• (b) Replacing the summation variable n by m, we have

f(x) = ∑_{m=1}^{∞} b_{m} sin mπx.

Multiplying this equation by sin nπx, where n is a fixed but arbitrary positive integer, integrating the result over 0 < x < 1, and using the orthogonality relations, we get

∫_{0}^{1} f(x) sin nπx dx = ∫_{0}^{1} (∑_{m=1}^{∞} b_{m} sin mπx) sin nπx dx

= ∑_{m=1}^{∞} b_{m} (∫_{0}^{1} sin mπx sin nπx dx)
= \frac{1}{2}b_{n}.

Hence,

b_{n} = 2 ∫_{0}^{1} f(x) sin nπx dx.

5. The 2π-periodic function defined by

f(x) = |x| −π < x < π

has the Fourier series expansion

f(x) = \frac{π}{2} − \frac{4}{π} (cos x + \frac{1}{3^{2}} cos 3x + \frac{1}{5^{2}} cos 5x + \frac{1}{7^{2}} cos 7x + …) .

Apply Parseval’s theorem to this function, and use the result to deduce the sum of the infinite series

∑_{n odd} \frac{1}{n^{4}} = 1 + \frac{1}{3^{4}} + \frac{1}{5^{4}} + \frac{1}{7^{4}} + ….

Solution.

• According to Parseval’s theorem, if

f(x) = \frac{1}{2}a_{0} + ∑_{n=1}^{∞} a_{n} cos nx,

then

\frac{1}{2π} ∫_{−π}^{π} f(x)^{2} dx = \frac{1}{4}a_{0}^{2} + \frac{1}{2} ∑_{n=1}^{∞} a_{n}^{2}.

We have

\frac{1}{2π} ∫_{−π}^{π} f(x)^{2} dx = \frac{1}{2π} ∫_{−π}^{π} |x|^{2} dx
= \frac{1}{π} ∫_{0}^{π} x^{2} dx
= \frac{1}{π} [\frac{1}{3}x^{3}]_{0}^{π}
= \frac{π^{2}}{3} .

From the Fourier series expansion of f, we have

a_{0} = π, a_{n} = \{ \begin{matrix}−4/(πn^{2}) & for n ≥ 1 odd, \\ 0 & for n ≥ 2 even. \end{matrix}

It follows that

\frac{π^{2}}{3} = \frac{π^{2}}{4} + \frac{8}{π^{2}} ∑_{n odd} \frac{1}{n^{4}} .

Hence,

∑_{n odd} \frac{1}{n^{4}} = \frac{π^{4}}{96}.

6. Suppose that a function f(x) has the real Fourier series expansion

f(x) = \frac{1}{2} + ∑_{n=1}^{∞} e^{−n^{2}} \{cos nx + n sin nx\} .

(a) Find the complex Fourier series expansion of f, of the form

f(x) = ∑_{n=−∞}^{∞} c_{n}e^{inx}.

(b) Does this Fourier series converge slowly or rapidly? How smooth do you think f(x) is? That is, how many orders of derivatives of f(x) do you expect to be continuous? Explain your answer briefly. (No proofs required.)

Solution.

• Using Euler’s equation, we get

f(x) = \frac{1}{2} + ∑_{n=1}^{∞} e^{−n^{2}} \{cos nx + n sin nx\}

= \frac{1}{2} + ∑_{n=1}^{∞} e^{−n^{2}} \{(\frac{e^{inx} + e^{−inx}}{2} ) + n (\frac{e^{inx} − e^{−inx}}{2i} )\}

= \frac{1}{2} + \frac{1}{2} ∑_{n=1}^{∞} e^{−n^{2}} \{(1 − in) e^{inx} + (1 + in) e^{−inx}\}

= \frac{1}{2} ∑_{n=−∞}^{∞} (1 − in) e^{−n^{2}}e^{inx}.

Hence, the complex Fourier coefficients of f are given by

c_{n} = \frac{1}{2} (1 − in) e^{−n^{2}}.  (1)

• Alternative proof. We have

c_{n} = \frac{1}{2} (a_{n} − ib_{n}) for n ≥ 0,

where a_{n} and b_{n} are the Fourier cosine and sine coefficients of f, respectively.

From the Fourier series, we have

a_{n} = e^{−n^{2}}, b_{n} = ne^{−n^{2}}.

Note that a_{0} = 1, so the expression for a_{n} also holds when n = 0. Thus,

c_{n} = \frac{1}{2} (1 − in) e^{−n^{2}} for n ≥ 0.

Also,

c_{−n} = \overline{c_{n}}
= \frac{1}{2} (1 + in) e^{−n^{2}},

so the same formula holds for n < 0, and we get (1).

• The Fourier coefficients of f tend to zero exponentially quickly as n → ∞, so the the Fourier series converges very rapidly. Since n^{p}e^{−n^{2}} converges to zero rapidly as n → ∞ for every p = 1, 2, 3 …, the series obtained from the one for f by term-by-term differentiation converges uniformly, and we expect that f has continuous derivatives of all orders (as, in fact, it does).