lecture 7: Trigonometric Interpolation

Transcription

1 lecture : Trigonometric Interpolation 9 Trigonometric interpolation for periodic functions Thus far all our interpolation schemes have been based on polynomials However, if the function f is periodic, one might naturally prefer to interpolate f with some set of periodic functions To be concrete, suppose we have a continuous p-periodic function f that we wish to interpolate at the uniformly spaced points k = pk/n for k =,, n with n = We shall build an interpolant as a linear combination of the p-periodic functions p-periodic means that f is continuous throughout IR and f () = f ( + p) for all IR The choice of period p makes the notation a bit simpler, but the idea can be easily adapted for any period b () =, b () =sin(), b () =cos(), b () =sin(), b () =cos() Note that we have si interpolation conditions at k for k =,,, but only five basis functions This is not a problem: since f is periodic, f ( )= f ( ), and the same will be true of our p-periodic interpolant: the last interpolation condition is automatically satisfied We shall construct an interpolant of the form such that t () = Â c k b k () k= t ( j )= f ( j ), j =,, To compute the unknown coefficients c,,c, set up a linear system as usual, b ( ) b ( ) b ( ) b ( ) b ( ) c f ( ) b ( ) b ( ) b ( ) b ( ) b ( ) c f ( ) b ( ) b ( ) b ( ) b ( ) b ( ) c = f ( ), b ( ) b ( ) b ( ) b ( ) b ( ) c f ( ) b ( ) b ( ) b ( ) b ( ) b ( ) f ( ) c which can be readily generalized to accommodate more interpolation points We could solve this system for c,,c n, but we prefer to epress the problem in a more convenient basis for the trigonometric functions Recall Euler s formula, e iq = cos(q)+i sin(q), You would b () = sin(), b () =cos(), etc: one function for each additional interpolation point Generally you would use an odd value of n, to include pairs of sines and cosines which also implies that e iq = cos(q) From these formulas it follows that i sin(q) To prove this, write the Taylor epansion of e iq, then separate the real and imaginary components to give Taylor epansions for cos(q) and sin(q) span{e iq,e iq } = {cos(q), sin(q)}

2 Note that we can also write b () = e i Putting these pieces together, we arrive at an alternative basis for the trigonometric interpolation space: span{, sin(), cos(), sin(), cos()} = span{e i,e i,e i,e i,e i } The interpolant t n can thus be epressed in the form t () = Â g k e ik = Â g k (e i ) k k= k= This last sum is written in a manner that emphasizes that t is a polynomial in the variable e i, and hence t n is a trigonometric polynomial In this basis, the interpolation conditions give the linear system e i e i e i e i e i e i e i e i e i e i e i e i e i e i e i e i e i e i e i e i e i e i e i e i e i g g g g g = f ( ) f ( ) f ( ) f ( ) f ( ) again with the natural generalization to larger odd integers n At first blush this matri looks no simpler than the one we first encountered, but a fascinating structure lurks Notice that a generic entry of this matri has the form e`i k for ` = (n )/,, (n )/ and k =,, n Since k = pk/n, rewrite this entry as, e`i k = e i k ` = e pik/n ` = e pi/n k` = w k`, where w = e pi/n is an nth root of unity In the n = case, the linear system can thus be written as () w w w w w w w w w w w w w w w w w w w w w 8 w w w w 8 g g g g g = f ( ) f ( ) f ( ) f ( ) f ( ) This name comes from the fact that w n = Denote this system by Fg = f Notice that each column of F equals some (entrywise) power of the vector w w w w w In other words, the matri F has Vandermonde structure From our past eperience with polynomial fitting addressed in Section, we

3 might fear that this formulation is ill-suited to numerical computations, ie, solutions g to the system Fg = f could be polluted by large numerical errors In the language of numerical linear algebra, we might fear that the matri F is ill-conditioned, ie, the condition number kfkkf k is large Before jumping to this conclusion, eamine F F To form F note F is the conjugate-transpose of F: that w ` = w`, so F = F T, w w w w w 8 w w w w w so (F ) j,k = F k,j w w w w w w w w w w F F = w w w w w w w w w w w w w w w w w w w w w w w w w 8 w 8 w w w w 8 The (`, k) entry for F F thus takes the form (F F)`,k = w + w (k `) + w (k `) + w (k `) + w (k `) On the diagonal, when ` = k, we simply have (F F) k,k = w + w + w + w + w = n On the off-diagonal, use w n = to see that all the off diagonal entries simplify to (F F)`,k = w + w + w + w + w, ` = k You can think of this last entry as n times the average of w, w, w, w, and w, which are uniformly spaced points on the unit circle, shown in the plot to the right As these points are uniformly positioned about the unit circle, their mean must be zero, and hence We thus must conclude that (F F)`,k =, ` = k F F = ni, thus giving a formula for the inverse: - - w w w w - - w = w F = n F The system Fg = f can be immediately solved without the need for any factorization of F: g = n F f The ready formula for F is reminiscent of a unitary matri In fact, the matrices p F and p F n n Q C n n is unitary if and only if Q = Q, or, equivalently, Q Q = I

4 are indeed unitary, and hence kn / Fk = kn / F k = From this we can compute the condition number of F: kfk kf k = n kfk kf k = kn / Fk kn / F k = This special Vandermonde matri is perfectly conditioned! One can easily solve the system Fg = f to high precision The key distinction between this case and standard polynomial interpolation is that now we have a Vandermonde matri based on points e i k that are equally spaced about the unit circle in the comple plane, whereas before our points were distributed over an interval on the real line This distinction makes all the difference between an unstable matri equation and one that is not only perfectly stable, but also forms the cornerstone of modern signal processing In fact, we have just computed the Discrete Fourier Transform (DFT) of the data vector f ( ) f ( ) f ( n ) The matri -norm is defined as kfk = ma = kfk kk, where the vector norm on the right hand side is the Euclidean norm kyk =  y k / =(y y) / k The -norm of a unitary matri is one: If Q Q = I, then kqk = Q Q = = kk, so kqk = The coefficients g (n )/,,g (n )/ that make up the vector g = n F f are the discrete Fourier coefficients of the data in f From where does this name derive? 9 Connection to Fourier series In a real/functional analysis course, one learns that a p-periodic function f can be written as the Fourier series f () =  c k e ik, k= where the Fourier coefficients c` are defined via c k := Z p f ()e ik p Notice that g k =((/n)f f) k is an approimation to this c k : g k = n = n n  f (`)w `k `= n  f ( k )e (pk/n)i` = n n  f ( k )e i` k k= k= To ensure pointwise convergence of this series for all [, p], f must be a continuous p-periodic function with a continuous first derivative The functions e k () = e ik / p p form an orthonormal basis for the space L [, p] endowed with the inner product Z p ( f, g) = f ()g() d The Fourier series is simply an epansion of f in this basis: f = Â( f, e k ) e k k

5 Now use the fact that f ( )e i` = f ( n )e i` n to view the last sum as a composite trapezoid rule approimation of an integral: pg` = p n Z p = pc` f ( )e i` n + Â f ( k )e i` k + f ( n)e i` n k= f ()e i` d The composite trapezoid rule will be discussed in Chapter The coefficient g` that premultiplies e i` in the trigonometric interpolating polynomial is actually an approimation of the Fourier coefficient c` Let us go one step further Notice that the trigonometric interpolant t n () = (n )/ Â g k e ik k= (n )/ is an approimation to the Fourier series f () = Â c k e ik k= obtained by () truncating the series, and () replacing c k with g k To assess the quality of the approimation, we need to understand the magnitude of the terms dropped from the sum, as well as the accuracy of the composite trapezoid rule approimation g k to c k We will thus postpone discussion of f () t n () until we develop a few more analytical tools in the net two chapters 9 Computing the discrete Fourier coefficients Normally we would require O(n ) operations to compute these coefficients using matri-vector multiplication with F, but Cooley and Tukey discovered in 9 that given the amazing structure in F, one can arrange operations so as to compute g = n F f in only O(n log n) operations: a procedure that we now famously call the Fast Fourier Transform (FFT) We can summarize this section as follows The FFT of a vector of uniform samples of a p-periodic function f gives the coefficients for the trigonometric interpolant to f at those sample points These coefficients approimate the function s Fourier coefficients Apparently the FFT was discovered earlier by Gauss, but it was forgotten, given its limited utility before the advent of automatic computation Jack Good (Bletchley Park codebreaker and, later, a Virginia Tech statistician) published a similar idea in 98 Good recalls: John Tukey (December 9) and Richard L Garwin (September 9) visited Cheltenham and I had them round to steaks and fries on separate occasions I told Tukey briefly about my FFT (with little detail) and, in Cooley and Tukey s well known paper of 9, my 98 paper is the only citation See D L Banks, A conversation with I J Good, Stat Sci (99) 9

6 Eample 8 (Trig interpolation of a smooth periodic function) Figure shows the degree n =,, 9 and trigonometric interpolants to the p-periodic function f () =e cos()+sin() Notice that although all the interpolation points are all drawn from the interval [, p) (indicated by the gray region on the plot), the interpolants are just as accurate outside this region In contrast, a standard polynomial fit through the same points will behave very differently: (nonconstant) polynomials must satisfy p n ()! as! Figure shows this behavior for n = : for [, p], the polynomial fit to f is about as accurate as the n = trigonometric polynomial in Figure Outside of [, p],the polynomial is much worse Eample 9 (Trig interpolation of non-smooth function) Figure shows that a standard (non-perioidic) polynomial fit to a periodic function can yield a good approimation, at least over the interval from which the interpolation points are drawn Now turn the tables: how well does a (periodic) trigonometric polynomial fit a smooth but non-periodic function? Simply take f () = on [, p], and construct the trigonometric interpolant as described above for n = The top plot in Figure shows that t gives a very poor approimation to f, constrained by design to be periodic even though f is not The fact that f (p) = f () acts like a discontinuity, vastly impairing the quality of the approimation The bottom plot in Figure repeats this eercise for f () =( p) Since in this case f () = f (p) we might epect better results; indeed, the approimation looks quite reasonable Note, however, that f () = f (p), and this lack of smoothness severely slows convergence of t n to f as n! Figure contrasts this slow rate of convergence with the much faster convergence observed for f () =e cos()+sin() used in Figure Clearly the periodic interpolant is much better suited to smooth f 9 Fast MATLAB implementation MATLAB organizes its Fast Fourier Transform is a slightly different fashion than we have described above To fit with MATLAB, reorder the unknowns in the system () to obtain () w w w w w w w w w w w w w w w w w w w w w w w 8 w 8 w g g g g g = f ( ) f ( ) f ( ) f ( ) f ( ),

7 f () t () Figure : Trigonometric interpolant to p-periodic function f () =e cos()+sin(), using n =,, 9 and points uniformly spaced over [, p) ({ k } n k= for k = pk/n) Since both f and the interpolant are periodic, the function fits well throughout IR, not just on the interval for which the interpolant was designed - -: : : : f () t () - -: : : : - -: : : : t 9 () f () f () t () - -: : : :

8 8 p () f () Figure : Polynomial fit of degree n = through uniformly spaced grid points,, n for j = pj/n, for the same function f () =e cos()+sin() used in Figure In contrast to the trigonometric fits in the earlier figure, the polynomial grows very rapidly outside the interval [, p] Moral: if your function is periodic, fit it with a trigonometric polynomial - -: : : : f () = t () - -: : : : f () = ( p) Figure : Trigonometric polynomial fit of degree n = through uniformly spaced grid points,, n for j = pj/n, for the non-periodic function f () = (top) and for f () =( p) (bottom) By restricting the latter function to the domain [, p], one can view it as a continuous periodic function with a jump discontinuity in the first derivative The interpolant t seems to give a good approimation to f, but the discontinuity in the derivative slows the convergence of t n to f as n! 8 t () - -: : : : which amounts to reordering the columns of the matri in () You can obtain this matri by the command ifft(eye(n)) For n =, ifft(eye()) = w w w w w w w w w w w w w w w w w w w w w w w 8 w 8 w

9 9 ma f () t n() [,p] f () =( p) f () =e cos()+sin() Figure : Convergence of the trigonometric polynomial interpolants to f () =e cos()+sin() and f () =( p) For the first function, convergence is etremely rapid as n! The second function, restricted to [, p], can be viewed as a continuous but not continuously differentiable function Though the approimation in Figure looks good over [, p], the convergence of t n to f is slow as n! - - n Similarly, the inverse of this matri can be computed from fft(eye(n)) command: w w w w w fft(eye())/ = w w w w w w w w w w 8 w w w w w 8 w w w w w We could construct this matri and multiply it against f to obtain g, but that would require O(n ) operations Instead, we can compute g directly using the fft command: g = fft(f)/n Recall that this reordered vector gives g = g g g g g, which must be taken into account when constructing t n Eample (MATLAB code for trigonometric interpolation) We close with a sample of MATLAB code one could use to construct the interpolant t n for the function f () = e cos()+sin() First we present a generic code that will work for any (real- or complevalued) p-periodic f Take special note of the simple one line command to find the coefficients g

10 f ep(cos()+sin(*)); n = ; k = [:n-] **pi/n; = linspace(,*pi,) ; tn = zeros(size()); % define the function % # terms in trig polynomial (must be odd) % interpolation points % fine grid on which to plot f, t_n % initialize t_n gamma = fft(f(k))/n; % solve for coefficients, gamma for k=:(n+)/ tn = tn + gamma(k)*ep(i*(k-)*); end for k=(n+)/+:n tn = tn + gamma(k)*ep(i*(-n+k-)*); end plot(,f(), b- ), hold on plot(, tn, r- ) % add in gamma_, gamma_, gamma_{(n-)/} % add in gamma_{-(n-)/},, gamma_{-} % plot f % plot t_n In the case that f is real-valued (as with all the eamples shown in this section), one can further show that g k = g k, indicating that the imaginary terms will not make any contribution to t n Since for k =,, (n )/, g k e ik + g k e ik = Re(g k ) cos(k) Im(g k ) sin(k), the code can be simplified slightly to construct t n as follows gamma = fft(f(k))/n; tn = gamma()*ep(i**); for k=:(n+)/ end tn = tn + *real(gamma(k))*cos((k-)*) % solve for coefficients, gamma % initialize t_n() = gamma_ - *imag(gamma(k))*sin((k-)*); % eploit gamma_{-k} = conj(gamma_k)