MATH 124B: HOMEWORK 2

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1 MATH 24B: HOMEWORK 2 Suggested due date: August 5th, 26 () Consider the geometric series ( ) n x 2n. (a) Does it converge pointwise in the interval < x <? (b) Does it converge uniformly in the interval < x <? (c) Does it converge in the L 2 sense in the interval < x <? (2) Let f(x) be a function on (, L) that has a continuous derivative and satisfies the periodic boundary conditions. Let a n be the Fourier cosine coefficients and b n be the Fourier sine coefficients of f(x) and let a n and b n be the corresponding Fourier coefficients of its derivative f (x). Show that a n = nπb n L and b n = nπa n L for n. Deduce from this that there is a constant k independent of n such that a n + b n k n for all n. Note, this does not mean that the differentiated series converges. (3) If f(x) is a piecewise continuous function in [, L], show that its indefinite integral F (x) = f(s)ds has a Full Fourier series that converges pointwise. (4) Write this convergent series for f(x) explicitly in terms of the Fourier coefficients a, a n and b n of f(x). Why does this imply that we can integrate the terms of the Fourier series term by term? (5) Find the sum n 6. (6) Prove the inequality L derivative f (x) is continuous. (f (x)) 2 dx (f(l) f()) 2, for any real function f(x) whose (7) Show that if f(x) is a C function in [, π] that satisfies the periodic boundary condition and if f(x) =, then π f 2 dx f 2 dx. This inequality is known as Wirtinger s inequality and is used in the proof of the isoperimetric inequality.

2 2 MATH 24B: HOMEWORK 2. a. We will show that Using the formula we have Hence, Solutions ( x 2 ) n = + x 2. t n = tn+ t ( x 2 ) n = ( x2 ) N+ + x 2. ( x 2 ) n + x 2 = x2n+2 + x. 2 Since x, the right hand side goes to as N. b. We have sup ( x 2 ) n + x 2 2 [,] (why does the above hold independent of N?) hence does not converge uniformly. c. The L 2 norm is x 4N+4 ( + x 2 ) 2 dx 2 as N. (Why does the first inequality hold?) 2. The coefficients are given by { a n = L b n = L x 4N+4 dx = f(x) cos( nπ L x)dx f(x) sin( nπ L x)dx and { a n = L f (x) cos( nπ L )dx L b n = L f (x) sin( nπx)dx. L L Integrating by parts, a n = L = L = nπ L 2 = nπ L b n f (x) cos( nπ L x)dx f(x)(cos( nπ L x)) dx f(x) sin( nπ L x)dx 2 4N + 5

3 MATH 24B: HOMEWORK 2 3 (why does the boundary term in the integration by parts vanish?) Similar for the other case. Therefore, we have a n + b n = L nπ ( a n + b n ) Now a n = L L f (x) cos( nπ L x)dx f (x) dx < hence we obtain a constant independent of n. 3. Since F (x) is differentiable, with F (x) = f(x) piecewise continuous, we can apply the pointwise convergence theorem for classical Fourier series. 4. Let A n, B n be the Fourier coefficients of F (x). A n = L = nπ = nπ F (x) cos( nπ L x)dx = L nπ b n ( ) f(s)ds d(sin( nπ L x)) f(x) sin( nπ L x)dx (why does the boundary term for integration by parts vanish?) and similarly we can show Therefore F (x) = 2 A + = 2 A + B n = L nπ a n A n cos( nπ L x) + B n sin( nπ L x) L nπ b n cos( nπ L x) + L nπ a n sin( nπ L x). now if we formally integrate f, assuming that a =. Then we have, f(s)ds = a n cos( nπ L s) + b n sin( nπ L s)ds which equals F (x) except for a constant. In fact, if a, then the indefinite integral is no longer a Fourier series, however the convergence of the infinite sum is guaranteed.

4 4 MATH 24B: HOMEWORK 2 5. This can be done in a number of ways. We will need to use the fact that n = π2 2 6 and n = π First compute the Fourier sine series for x2 on the interval (, ), which gives the coefficients { 2 m even mπ A m =. By Parseval s identity, we have Therefore, m even 2m 2 π 2 8 m 3 π 3 A m 2 sin 2 (mπx)dx = m= x 4 dx 4 m 2 π2 + ( 4 m 2 π 32 2 m 4 π + 64 ) = 2 4 m 6 π 6 5 Now the /m 2 term is known and the odd part of /m 4 can be computed from the whole series by hence m= odd m 4 = π4 96. Hence 64 m + 4 (2m) = π4 4 9 m= π 6 m 6 = 5 or m = π6. Since the whole series is the even and the odd terms and the even terms are 6 96 (2m) = 6 64 m therefore 6 m= m= m 6 = π Apply Cauchy-Schwarz with f and. 7. From the assumption, we know that for the Fourier coefficient of f, A =. By Parseval s equality, we have f 2 dx = π ( A n 2 + B n 2 ))

5 MATH 24B: HOMEWORK 2 5 Note that X2 n = π for X n = cos(nx) and X n = sin(nx). It was shown before that A n = n B n and B n = n A n hence π ( A n 2 + B n 2 )) π ( A n 2 + B n 2 )) = (f ) 2 dx Note that A = f dx = f(π) f() = by the periodic boundary condition. (Further consideration: When is the inequality an equality? Which part of the proof will give us an idea of what type of function will be an equality? The equality case gives us a hint as to why this inequality is related to the isoperimetric inequality.)