Math 424 Midterm. π t if t π. ( sin(πλ/2) (λ/2) = 1 2π. ( sin π(λ + α) (λ + α) if t 2π g α (t) = 0 if t > 2π. 2 e. f (t)e iλt dt

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1 Math 424 Midterm Instructions: You may use your textbook and notes, but should not use other texts or consult anyone except me about material on this exam. In solving any question or part of a question, the results of other questions or other parts may be used, even if you did not succeed in solving those other questions or parts. (e.g. You may use the result of question 1(b) in any other question, even if you did not do 1(b).) This exam must be returned within 48 hours of collection, either by ing a scanned copy (in a standard format such as pdf, jpeg, gif, tiff) or by returning it directly to me. One final comment - the last question is harder than the others, and the final part of that question is for bonus points only (it is a bit beyond what I would expect of students in this course, but if you want to try it, go right ahead!). 1. Let (a) Show that (b) Is ˆ f L 2? If so, compute its L 2 -norm. (c) Show that if then π t if t π f (t) if t > π f ˆ(λ) 1 ( ) 2 sin(πλ/2) (λ/2) ( sin π(λ + α) ĝ α (λ) (λ + α) ( ) π 2 π t 2 e iαt if t g α (t) if t > ) 2 Solution: (a) This is just integration by parts. It is easiest if you use e iλt cos λt i sin λt. Then the symmetry of f ensures that the sin terms cancel, and f ˆ(λ) () 1/2 () 1/2 2 f (t)e iλt dt f (t) cos(λt) dt () 1/2 2 (π t) cos(λt) dt [ (π t) sin(λt) () 1/2 2 cos(λt) λ λ [ 2 () 1/2 2 cos(πλ) λ ] λ 2 () 1/2 2 2 sin2 (πλ/2) λ 2 ) 2 1 ( sin(πλ/2) (λ/2) ] π

2 where we used the identity cos(2θ) 1 2 sin 2 θ. (b) We use Plancherel s theorem, which says that the L 2 norms of f and ˆ f are equal. Clearly it is easier to calculate the L 2 norm of f, which is quick to compute using symmetry: so that ˆ f 2 π 3 3. f [ 1 (π t) 2 dt 2 (π t)3 3 (c) The easy way to do this is to use the formulas for how translations and modulations affect the Fourier transform. For example, the 6th formula on page 1 in your book tells you that the Fourier transform of e iat f (t) at λ is just f ˆ(λ + a). The 7th one tells you how to deal with dilations. Recognizing that the formula given for g α is just the constant π/2, times the modulation e αt applied to the dilation f (t/2), we can compute as follows: [ π F 2 f ( t ) ] e iαt π [( t )] (λ) 2 2 F ( ) λ + α 2 π 2 2 F [ f (t)]( 2 ( λ + α ) ) just by substitution into the formula for ˆ f. ] π f ˆ(2λ + 2α) ( ) 2 sin π(λ + α) (λ + α) π (a) Let f be a differentiable function on [π, π] which satisfies f () f (π). Suppose also that f (x) α n e inx. Show that the n-th complex Fourier coefficient of the derivative f is equal to inα n. (Warning - you cannot just differentiate the series term by term, because this relies on somewhat delicate convergence issues for the series and these issues may get messed up if the terms are multiplied by something as large as in) (b) The previous part of this problem suggests a way to define a derivative of a function f L 2 ([, π]), even if the function is not differentiable in the classical sense. We know we can write f (x) α n e inx with convergence in L 2 by taking α n () 1 f (x)e inx dx. Suppose we can also find some other mathematical object g (that need not be a function) which has the property that g(x)e inx dx inα n. Then it seems reasonable to call g a Fourier derivative of f. Show that, in this sense, the Fourier derivative of the function 1 if x π/2 f (x) 1 if π/2 < x π is a linear combination of two Dirac masses. (Recall that the Dirac mass δ a (x) is a mathematical version of the point mass. It has the property that if φ is any continuous function, then δ a (x)φ(x) φ(a). Page 2

3 Solution: (a) This is virtually identical to a calculation done in the book, but on [, π] instead of R. Simply compute () 1 f (x)e inx dx [ () 1 f (t)e int] π () 1 f (x)( in)e inx dx () 1 f (π) ( e inπ e inπ) + (in)() 1 f (x)e inx dx inα n because e inπ e inπ 2i sin(nπ) for all n and the second term is the formula for the n-th Fourier coefficient of f. (b) We first want to know the Fourier coefficients of f. α n () 1 f (x)e inx dx /2 e inx dx + 2 ( e inπ/2 e inπ/2) in /2 /2 e inx dx e inx dx π/2 Now the coefficients corresponding to the Dirac mass δ a (x) are δ a (x)e inx dx e ina. so we can get the terms occurring in inα n by taking a to be π/2 (with coefficient 2) and /2 (with coefficient -2). This suggests that we should set g 2(δ π/2 δ /2 ), and a quick calculation shows that with this choice of g we have the Fourier coefficients of g are inα n, so that g is the Fourier derivative of f. 3. Let f be a piecewise differentiable function on R with support in a finite interval. It is a fact (that you need not prove) that then F(x) f (x + k) is a piecewise differentiable function. k (a) Show that F(x) is periodic with period. (b) Let ˆ f be the Fourier transform of f on R. Show that F(x) 1 ˆ f (n)e inx Hint: Show that they have the same Fourier series on [, π]. Why is this sufficient? Page 3

4 (c) Using the formula you get from the previous part when x, and the result of Problem 1(c) above, give a formula for 1 (n + α) 2 that is valid if α R \ Z. Solution: (a) This is easy, because for any x F(x + ) f (x + + k) k j f (x + j) F(x) by making the change of variables k + 1 j. (b) Since F is -periodic and differentiable, we know from Theorem 1.3 in your book that the Fourier series for F converges uniformly to F, i.e. F(x) ˆF(n)e inx where ˆF(x) () 1 π f (x)e inx dx is the Fourier coefficient of F as a periodic function on [, π]. Comparing this to the equation we want to prove, it clearly suffices to show that () 1/2 f ˆ(n) ˆF(n). After sorting out the factors of this is just f (x)e inx dx F(x)e inx dx but having got this far we need only make an easy computation, because F(x)e inx dx k k (2k+1)π k (2k 1)π f (x + k)e inx dx f (x + k)e inx dx f (y)e iny dy f (y)e iny dy where we made the change of variables y x + k and used that e in(y k) e iny, and the fact that that intervals [(2k 1)π, (2k + 1)π) are disjoint and fill out all of R as k runs from to. (c) When x, the formula we have is that k f (k) F() 1 ˆ f (n) Page 4

5 Let us first work out what the right side gives when applied to ĝ α as in Problem 1(c). In that case ( ) 2 sin π(n + α) ĝ α (n) sin2 (πα) (n + α) (n + α) 2 because sin(πn + θ) sin θ if n is even and sin θ if n is odd. Squaring makes both equal to sin θ. We therefore see that ĝ α (n) sin 2 (πα) (n + α) 2 sin2 (πα) 1 (n + α) 2 Computing the left side of the equation is even easier, because we want k g α (k), but the only one of these where g α is non-zero is g α () 3 /2. Equating sides in the formula above we have, hence sin 2 (πα) 1 (n + α) 2 π2 1 (n + α) 2 π 2 sin 2 (πα). Since we are dividing by sin 2 (πα) this formula requires that α R \ Z. The only thing that remains to mention is that we used the result of part (a) of this problem in our argument, and for that purpose we needed to know that the function we started with was piecewise differentiable and had finite support. Since both are true for g α there is no problem, and we are done. 4. Suppose we have a closed curve in R 2, and we let (x(t), y(t)) be its unit speed parametrization. All this means is that the functions x(t) and y(t) are differentiable on [, L], where L is the length of the curve, that the point with coordinates (x(t), y(t)) traces out the curve when t runs from to L, and that (x (t)) 2 + (y (t)) 2 1 at every value of t. The last condition is the unit speed condition, because (x (t), y (t)) is the (vector) rate of change of position with respect to time, and the condition says this vector has length 1. Since the curve is closed it must have the same starting and ending point, so x() x(l) and y() y(l). (a) If we write the Fourier series Show that x(t) α n e int/l and y(t) n 2( α n 2 + β n 2) 1/2 L. β n e int/l Hint: Use Problem 2(a) and Parseval s equation. Do it first for the case L and then change variables very carefully. Page 5

6 (b) Recall that there is a formula for the area enclosed by the curve using Green s theorem from calculus. This formula is A L x(t)y (t)dt By expressing the area using the Fourier coefficients α n and β n, show that A π n 2( α n 2 + β n 2) and conclude that A L2 4π. (c) By considering what form the Fourier series must have in order to make the inequality in part (b) into an equality, show that if A L2 then the curve is a circle. 4π Solution: (a) What we know is that (x (t)) 2 + (y (t)) 2 1 for all t, so that L ( (x (t)) 2 + (y (t)) 2) L dt dt L If L then we can use Parseval s equation and the fact (from Problem 2(a) that the Fourier coefficients x (t) are inα n to write Doing the same for y (t) we get L (x (t)) 2 dt x (t), x (t) inα n 2 n 2 α n 2 (x (t)) 2 dt + (y (t)) 2 dt n 2( α n 2 + β n 2) which is what we wanted in this case, because it shows the sum is 1, so its square root is 1 which is consistent with the desired result. To get this to work for general L we have to rescale. One way is to define X(s) α n e ins and Y(s) β n e ins so that if s t/l then X(s) x(t) and Y(s) y(t). Notice that the above argument with the Parseval equation still applies to X(s) and Y(s), so we have (X (s)) 2 ds + (Y (s)) 2 ds n 2( α n 2 + β n 2) Now all we need to do is change variables, because when s runs from to, t runs from to L. We have ds (/L)dt, and using that X(s) x(t) we get X (s) d dt d X(s) ds ds dt x(t) L x (t) Page 6

7 so (x (t)) 2 dt (/L)(X (s)) 2 ds, and similarly (y (t)) 2 dt (/L)(Y (s)) 2 ds. It follows that L L ( (x (t)) 2 + (y (t)) 2) L ( (X (s)) 2 + (Y (s)) 2) ds L n 2( α n 2 + β n 2) from which the desired result follows immediately. (b) This is a similar computation, using the same change of variables and Parseval s theorem to see that A L x(t)y (t) dt X(s) Y (s) L Lds X(s)Y (s) ds α n inβ n Now we want to know that this is less than π n 2( α n 2 + β n 2), so it seems natural to compare terms. We know that always 2 α n β n α n 2 + β n 2 by completing the square. Since it is also true that in n 2, we see that for each value of n 2inα n β n n 2 2 α n β n n 2 ( α n 2 + β n 2) which establishes the inequality between the sums. The result A L 2 /4π 2 follows by the previous part. (c) In order for the equality A L 2 /4π 2 to hold, we must have equality in each step of the inequalities we used. In particular we need that 2 α n β n α n 2 + β n 2, which means that ( α n β n ) 2, so α n β n. Also we need that in n 2, so the only terms in the sum can be n, n 1 and n 1; all other values of α n and β n must be zero. This means that x(t) α 1 e it/l + α + α 1 e it/l y(t) β 1 e it/l + β + β 1 e it/l and since x and y are real valued we must have that α 1 ᾱ 1 and β 1 β 1. In particular α 1 α 1 and β 1 β 1, and we already know α 1 β 1, so all four of these coefficients have equal length. Substituting this into the sum in part (a) we find that (4 α 1 2 ) 1/2 L, so α 1 L/4π. Let us call this number r, and rewrite the coefficients in terms of r. The fact that α 1 ᾱ 1 and β 1 β 1 means that these numbers have the form α 1 re iθ β 1 re iφ α 1 re iθ β 1 re iφ for some angles θ and φ. Actually we can say more. Going back to the condition for equality we can now write it as 2r 2 ( α β 1 2)2 2iα 1 β 1 2ir 2 e iθ+φ so that θ + φ /2 (actually I am cheating here - it is also possible for it to be π/2 because the area could be signed). Therefore we finally have formulas for both x(t) and y(t) that are as Page 7

8 simple as possible. Substituting all we have found out, we get x(t) α + r ( e i θ + t/l e θ t/l) α + L 2 cos(θ + t/l) 4π α + L cos(θ + t/l) y(t) β + r ( e i θ π/2 + t/l e θ+π/2 t/l) β + L 2 cos(θ π/2 + t/l) 4π β + L sin(θ + t/l) from which it is obvious that the curve is a circle with center (α, β ) and radius L/. Page 8