Math 56 Homework 4 Michael Downs

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1 Fourier series theory. As in lecture, let be the L 2 -norm on (, 2π). (a) Find a unit-norm function orthogonal to f(x) = x on x < 2π. [Hint: do what you d do in vector-land, eg pick some simple new function and project away its f component]. (b) Combine the Fourier series for the 2π-periodic function defined by f(x) = x for x < 2π that we computed on the worksheet with Parseval s relation to evaluate m= m 2 (c) Compute the Fourier series for the 2π-periodic function defined by f(x) = x 2 in π x < π. [Hint: shift the domain of integration to a convenient one.] Comment on how ˆf m decays for this continuous (but not C ) function compared to the discontinuous function in (b). (d) Take a general f written as a Fourier series, and take the derivative of both sides. What then is the mth Fourier coefficient of f? Prove that if f is bounded, ˆf m = O(/ m ), or better. (e) Use the previous idea to prove a bound on the decay of Fourier coefficients when f has k bounded derivatives. What bound follows if f C (arbitrarily smooth)? Can you give this a name?. (a) Let f(x) = x and g(x) =, x < 2π, be two vectors in the inner product space of 2π-periodic continuous functions. A function ĝ(x) orthogonal to f(x) can be obtained from f(x) and g(x) via the Gram-Schmidt process: ĝ(x) = (x, ) (x, x) x = 2π2 x 8π 2 = 4π x ĝ(x) still has to be normalized. ĝ(x)ĝ(x) dx = π so the final unit-length function 2 ( 2 orthogonal to f(x) is ĝ(x) = π x). 4π (b) Recall that the fourier coefficients for the 2π-periodic function defined by f(x) = x for x < 2π are: { π m = ˆf m = i m m and so: { ˆf 2 π 2 m = m = m m 2 By Parseval s equality we have that f 2 = 2π m= ˆf 2 m. Now, f 2 = x 2 dx =

2 8π which allows us to produce the following series of equalities: 8π = 2π 4π 2 = n= m= π 2 = 2 m 2 m= π 2 6 = m 2 m= ˆf m 2 n 2 + π2 + m= m 2 Thus m= m 2 = π2 6. (c) Consider the 2π-periodic function defined by f(x) = x 2 in π x < π. First let y = x + π. Then y < 2π and we can write f(x) = f(y π) = (y π) 2. This allows us to compute the fourier coefficients using the appealing integration bounds to 2π. First compute ˆf : ˆf = 2π = π2 (y π) 2 dy Next compute ˆf m for m : ˆf m = e im(y π) (y π) 2 dy 2π = e imy e imπ (y π) 2 dy 2π = eimπ 2π = ( )m 2π e imy (y π) 2 dy e imy (y π) 2 dy Some omitted tedious calculations reveal that Thus: ˆf m = 2π { π 2 m = ( ) m 2 m 2 m e imy (y π) 2 dy resolves to 2 m 2. 2

3 This gives us the fourier series for f(x) = x 2 = m= ˆf m e imx. It is interesting to note that the ˆf m for this continuous function decay as the inverse square of m while the ˆf m for the discontinuous function in (b) only decay as the inverse of m. This is perhaps due to the fact that a fourier series never really converges near the discontinuities (Gibbs phenomenon). I ll now confirm that this infinite series actually is the desired result. It can be shown that for real coefficients ˆf m, m= ˆf m e imx simplifies to the entirely real sum ˆf + 2 m= ˆf m cos(mx). Thus: f(x) = x 2 = π2 + 4 m= Plotting terms of this series gives the graph: ( ) m cos(mx) m 2 () Which is what we wanted. It s also interesting to note that () gives us the identity that ( ) n n= = π2. n 2 2 (d) An arbitrary function f written in its fourier series: f(x) = m= ˆf m e imx. Taking derivatives of both sides: d dx f(x) = d dx = = m= m= m= ˆf m e imx ˆf m d dx eimx ( ˆf m im)e imx From this we can write the mth fourier coefficient of f is ˆf m = ˆf m im. Suppose f is bounded. Then f 2 is bounded as well. By Parseval s equality, f 2 =

4 2π m= ˆf m 2 so that infinite sum is bounded as well. This implies that ˆf m as m. Thus ˆf m im as m. ˆf m im = ˆf m i m = ˆf m m. Then ˆf m m as m. Then, by definition, ˆf ( ) m = o. (e) Now suppose f has k bounded derivatives. By the logic in (d), the fourier coefficients (k) for the kth derivative can be expressed as ˆf m = ˆf m i k m k. Also by the logic in (d), if f (k) (k) is bounded then ˆf m as m so ˆf m m k as m. Thus if f has k bounded derivatives then ˆf ( m = o ). If f is infinitely differentiable then m k ˆf m = o( ) for all integer k >. I call this super-algebraic convergence because I m k noticed parallels with this and the definition of super exponential convergence. The positions of the rate and number of terms appear to be swapped. m Getting to know your DFT. Use numerical exploration followed by proof (each proof is very quick): (a) Produce a color image of the real part of the DFT matrix F for N = 256. Explain one of the curves seen in the image. (b) What is F 2? [careful: matrix product, also don t forget the -indexing]. What does F 2 do to a vector? (this should be very simple!) Now, for general N, prove your claim [Hint: use ω] (c) What then is F 4? Prove this. (d) What are the eigenvalues of F? Use your previous result to prove this. (e) What is the condition number of F? Prove this using a result from lecture. 2. (a) Code: % code to e x p l o r e DFT m atrices n = 256; F = dftmtx ( n ) ; % the 5 matrix % t h i s produces a 2D image o f the r e a l part o f the DFT matrix 7 Z = r e a l (F) ; g = l i n s p a c e (, n,n ) ; 9 [X,Y] = meshgrid ( g, g ) ; s u r f (X,Y, Z) ; imagesc ( g, g, Z) ; c o l o r b a r ; t i t l e ( s t r c a t ( v i s u a l i z a t i o n o f r e a l part o f d f t matrix n =, num2str ( n ) ) ) ; 4

5 produces the picture: visualization of real part of dft matrix n = Zoomed in we can pay particular attention to the curve featured in the corners: visualization of real part of dft matrix n = This is the real part of the DFT matrix where each entry F mj = ω mj with ω = e 2πi N. Each entry is cos( 2πmj ). As the indices increase, the frequencies become higher and N higher. This can be seen in the picture above. The image of the matrix suggests a sinusoidal figure with higher and higher frequencies as the indices m and j become larger. 5

6 (b) For N = 256, F 2 appears to be a matrix with 256 at entry F,, 256 at every entry starting at F 2,256 and going diagonally down to the left ending at F 256,2, and at every other entry. I note that the indices (with subtracted) for the nonzero entries sum to mod N. Left multiplication by F 2 in this particular case appears to fix the first entry, reverse the order of the remaining entries, and then scale the entire vector by N: x x F 2 x 2. = N x 256 (2). x 256 x 2 General proof (keeping in mind that the indices here are actually i+, j+, and k+): Fij 2 = N k= F ikf kj = N k= ω (i+j)k so by the sum lemma: { Fij 2 N i + j (mod N) = otherwise (c) Matlab shows that F 4 is the identity matrix scaled by For general N, it should be the N N identity matrix scaled by N 2. If we left multiply a column vector by F 2 it fixes the first entry, reverses the order of the remaining entries, and then scales each entry by N. Left multiplying by F 2 again we get the original vector except scaled by N 2. By this argument F 4 = N 2 I N N. (d) Via matlab the eigenvalues appear to be 6, 6, 6i, and 6i with various multiplicites (or possibly all the same, I didn t check). For general N let v be one of its eigenvectors. Then F 4 v = λ 4 v. But also by (c) F 4 v = N 2 v so N 2 v = λ 4 v thus N 2 = λ 4. From this we have λ = N, N, i N, i N. (e) Matlab shows that the condition number is. Proving this for the general case: we know that the norm of F is the square root of the largest eigenvalue of F T F. We also know that F is symmetric (F T F = F 2 ) and that its eigenvalues are λ = N, N, i N, i N. Squaring F squares its eigenvalues so we have that the eigenvalues of F 2 are λ = N, N. Thus F = N. Inverting F inverts its eigenvalues so we have that the largest eigenvalues of (F ) 2 are,. Therefore N N F = N. In conclusion, κ(f ) = N N =. 6

7 The power of trigonometric interpolation, i.e. using just a few samples of a periodic function to reconstruct the function everywhere. (Applications to modeling data, etc.) (a) Let s interpolate f(x) = e sin x. For N = 4, by using the /N-weighted samples at the nodes x j = 2πj/N, j =,..., N, and fft to do the DFT, find f m. Plot their magnitudes on a log vertical scale vs m =,..., N. Relate to #(e). By what m have the coefficients decayed to ε mach times the largest? (This is the effective band-limit of the function at this tolerance.) (b) Using f m as good approximations to the true Fourier coefficients in N/2 < m < N/2, plot the intepolant given by this truncated Fourier series, on the fine grid :e-:2*pi. Overlay the samples Nf j = f(x j ) as blobs. [Hint: Remember the negative m s are wrapped into the upper half of the DFT output. Debug until the interpolant passes through the samples] (c) By looping over the above for different N, make a labeled semi-log plot of the maximum error (taken over the fine grid) between the interpolant and f, vs N, for even N between 2 and 4. At what N is convergence to ε mach reached? (Pretty amazing, eh?) Relate to (a).. (a) Code: % code to e x p l o r e i n t e r p o l a t i o n with DFTs N = 4 ; % number o f sample p o i n t s g x ) exp ( s i n ( x ) ) ; % f u n c t i o n we re sampling 5 % get the N samples 7 x = :2 pi /N: 2 pi (N )/N; f = g ( x ). /N; 9 % take the f f t and observe the magnitudes o f the c o e f f i c i e n t s f t = f f t ( f ) ; mag = abs ( f t ) ; 5 m = : : ( N ) ; 7 semilogy (m, mag) ; 9 t i t l e ( magnitudes o f f f t approximated c o e f f i c i e n t s vs m ) ; x l a b e l ( m ) ; 2 y l a b e l ( magnitude o f f m ) ; outputs: 7

8 2 magnitudes of fft approximated coefficients vs m 2 4 magnitude of f m m f(x) = e sin(x) is infinitely differentiable so it should exhibit the kind of convergence in (e). Looking at the plot, it appears better than all types of algebraic convergence, too (magnitudes appear to be concave down). It appears that, by m = 5, the magnitude of the coefficients has decayed to ε mach. (b) Code (continuation of above): % s e e how w e l l the i n t e r p o l a t e d f u n c t i o n works i n t e r x ) r e a l (sum( f t ( (N/2 + ) :N). exp( i x (N/2: :) ) ) + sum( f t ( :N/2). exp ( i x ( : (N/2 ) ) ) ) ) ; 5 gr = : e :2 pi ; f i g u r e ; 7 p l o t ( gr, arrayfun ( i n t e r, gr ), Color, green ) ; hold on ; 9 p l o t ( x, f N, Marker, o, Color, red, L i n e S t y l e, none ) ; hold o f f ; x l a b e l ( x ) ; y l a b e l ( f ( x ) ) ; t i t l e ( I n t e r p o l a t e d approximation o f exp ( s i n ( x ) ) p a s s i n g through sample p o i n t s ) outputs: 8

9 Interpolated approximation of exp(sin(x)) passing through sample points f(x) x (c) Code (continuation of above): % now s e e which N i s n e c e s s a r y f o r emach e r r o r maxerrors = z e r o s (, N/2) ; n = 2 : 2 :N; 5 f o r j = 2 : 2 :N x = :2 pi / j : 2 pi ( j )/ j ; 7 f = g ( x ). / j ; f t = f f t ( f ) ; 9 i n t e r x ) r e a l (sum( f t ( ( j /2 + ) : j ). exp( i x ( j /2: :) ) ) + sum( f t ( : j /2). exp ( i x ( : ( j /2 ) ) ) ) ) ; % t h i s i s bad coding e r r o r = max( abs ( g ( gr ) arrayfun ( i n t e r, gr ) ) ) ; maxerrors ( j /2) = e r r o r ; end 5 f i g u r e ; semilogy (n, maxerrors ) ; 7 x l a b e l ( N ) ; y l a b e l ( max e r r o r ) ; 9 t i t l e ( Error in truncated N term approximation ) outputs: 9

10 2 Error in truncated N term approximation 2 4 max error N This seems to agree with part (a). In (a), the magnitude of the coefficients decayed to ε mach around m = 5. Here we have ε mach precision starting around N meaning the truncated approximation goes to about m = 5 in either direction. Let s prove that, amongst all trigonometric polynomials of degree at most N/2, the N/2- truncated Fourier series for f is the best approximation to f in the L 2 (, 2π) norm. (a) Let g be a general trig. poly. which can be written g = n N/2 ( ˆf n + c n )e inx for some coefficient deviations c n. Recall f = ˆf n= n e inx. Write the squared L 2 -norm of the error which is the difference of g and f: error 2 = f g 2 = ˆf n e inx n >N/2 n N/2 c n e inx (b) Using the definition of the norm, expand this expression for error 2 (you should obtain four terms, two of which will vanish). Whatever you do, do not multiply any of the big sums out. Can you use this expanded form to bound error 2 below by ˆf n >N/2 n e inx 2? (c) What does this say about how small the error can get? From this lower bound conclude that the error is minimized precisely when each c n =, ie when g is the truncated Fourier series of f. 2.

11 4. (a) Apparently this part was already completed in the question. (b) (Note: ᾱ denotes the complex conjugate of α). Let α = ˆf n e inx () β = n >N/2 n N/2 c n e inx (4) (5) α β 2 = (α β, α β) = = (ᾱ β)(α β) dx ᾱα dx ᾱβ dx βα dx + ββ dx ᾱβ dx and βα dx are both because the summation indices never correspond and every e inx, e imx 2π, m n are orthogonal. ᾱα dx = α 2 and, as shown on a worksheet, ββ dx = 2π n N/2 c n 2. So error 2 = ˆf n >N/2 n e inx 2 + 2π n N/2 c n 2 and error 2 ˆf n >N/2 n e inx 2. (c) This says that the smallest the error can get is ˆf n >N/2 n e inx 2 which is the twoway tail missing from the truncated fourier series. This is achieved when each c n =. In other words, when we use the truncated fourier approximation. A quickie on run-times. Consider the matvec problem computing y = Ax for A an n-by-n matrix. Write a little code using random A and x for n = 4, which measures the runtime of Matlab s native A*x, and your own naive double loop to compute the same thing. Express your answers in flops (flop per sec), and give the ratio. Marvel at how fast the built-in library is. 5. Code: % measuring speed o f matlab l i b r a r y vs naive implementation o f matrix % m u l t i p l i c a t i o n 5 n = 4; A = randn ( n ) ; 7 x = randn (n, ) ; y2 = z e r o s (n, ) ; 9

12 % l i b r a r y speed t i c ; y = A x ; l s = toc ; 5 % naive speed t i c ; 7 f o r i = : n sum = ; 9 f o r j = : n sum = sum + A( i, j ) x ( j, ) ; 2 end y2 ( i ) = sum ; 2 end ns = toc ; 25 f p r i n t f ( Naive Speed : \n ) ; 27 f p r i n t f ( %g f l o p s \n, / ns ) ; f p r i n t f ( Library Speed : \n ) ; 29 f p r i n t f ( %g f l o p s \n, / l s ) ; f p r i n t f ( The l i b r a r y speed i s %g times f a s t e r \n, ns / l s ) ; outputs: >> HW4Q5 Naive Speed: flops Library Speed:.44 flops The library speed is times faster 2

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