23. The Finite Fourier Transform and the Fast Fourier Transform Algorithm 23.1 Introduction: Fourier Series Early in the Nineteenth Century, Fourier studied sound and oscillatory motion and conceived of the idea of representing periodic functions by their coefficients in an expansion as a sum of sines and cosines rather than their values. He noticed that for example, you can represent the shape of a vibrating string of length L, fixed at its ends as y(x) = ΣΣΣΣκ = 1,κκκ= 1,= 1,= 1,∞ ak sin (2ππππkx/L) The coefficients, ak, contain important and useful information about the quality of the sound that the string produces that is not easily accessible from the ordinary y = f(x) representation of the shape of the string. This kind of representation of a function is called a Fourier Series, and there is a tremendous amount of mathematical lore about properties of such series and for what classes of functions they can be shown to exist. One particularly useful fact about them is how we can obtain the coefficients ak from the function following from the orthogonality property of sines: ∫∫∫∫x= 0,L sin (2πkx/L) sin (2πjx/L) dx = To see this notice that the product of these sines can be written as a constant multiple of the difference between cos(2π(k+j)x/L) and cos(2π(k-j)x/L), and each of these cosines integrates to 0 over this range. By multiplying the expression for y(x) above by 2πjx/L and integrating the result from 0 to L we get then the expression aj = (1/π)∫∫∫∫x= 0,000,,,L f(x) sin (2πjx/L) dx.Fourier series represent only one of many alternate ways we can represent a function. We can derive formulae for coefficients in a series by introducing an appropriate weight function in the integral and obtaining an orthogonality relation among the functions. 23.2 The Fourier Transform Given a function f(x) defined for all real x, we can give an alternative represent to it as an integral rather than as an infinite series, as follows. f(x) = ∫∫∫∫ eikx g(k) dk where the integral is over all real values of k. The representation of f by the function g is called a Fourier transform of f, and it is very important tool used in physics. One reason for this is that exponential functions eikx, which f is written as an integral sum of, are eigenfunctions of the derivative. That is, the derivative, acting on an exponential, merely multiplies the exponential by ik. This makes the Fourier transform a useful tool in investigating differential equations. Another example of the Fourier transform’s applications can be found in quantum mechanics. We can represent the state of a particle in a physical system as a wave function ϕ(x), and the probability that the particle in this state has a position lying between x and x+dx is |ϕ(x)|2 dx. The same state can also be represented by its wave function in momentum space, and that wave function of the variable p, is a constant multiple of the Fourier transform of ϕ(x): ψψψψ(p) = c ∫∫∫∫ eikx ϕϕϕϕ(x)dx. We can invert the Fourier Transform in much the same way that we can invert Fourier Series. The resulting formula is g(k) = (1/2π) ∫∫∫∫ e-ikx f(x) dx The integration is over all real values of x. 23.3 The Finite Fourier Transform Given a finite sequence consisting of n numbers, for example the coefficients of a polynomial of degree n-1, we can define a Finite Fourier Transform that produces adifferent set of n numbers, in a way that has a close relationship to the Fourier Transform just mentioned. I like to look at it backwards. Suppose we have a polynomial p of degree n-1. It can be described by its coefficients, {aj} with p(x) = ΣΣΣΣj= 0 to n-1 ajxj. We can also represent p by giving its values at any n arguments {p(xk)}. This can be done as follows. Observe first that the polynomial of degree n-1 f(xj)((x-x1) /(xj-x1))*. . .(exclude (x-xj) /(xj-xj)) . . . *((x-xn)/ (xj-xn)) takes the value f(xj) at x=xj and is 0 at all other of our arguments xk. We can recover p from its values by summing similar terms over all j. To evaluate a polynomial of degree n-1 at n values requires an order of n2 actions: n evaluations each of n terms. Similarly the procedure just described for recovering p from its values requires at least n2 operations to obtain all the coefficients of p. When you evaluate a polynomial at an argument x whose magnitude is not close to 1, the powers of x that are big dominate those that are small by overwhelming them so much that you have to worry about losing the smaller terms entirely from round off errors. The finite Fourier transform can be defined as the act of evaluating a polynomial of degree n-1 at n roots of unity, that is, at n solutions to the equation xn=1. This transform can be performed upon polynomials with coefficients in any field in which this equation has n solutions, which will happen when there is a primitive n-th root of unity in the field. (This means a number such that xn = 1 but xk is not 1 for any k between 1 and n-1.) The n roots of unity are then the various powers of the primitive root. When does this happen? It does for complex numbers, in which case we have e2ππππi/n which is cos(2ππππ/n)+isin(2ππππ/n) as primitive n-th root of unity. But it also happens for remainders on dividing by a prime number of the form kn+1. In such fields there is a primitive kn-th root of unity and hence a primitive n-th root of unity (such as the k-th power of the former.)The analogy between this finite transform and the Fourier transform is most apparent when we use complex numbers. Then, if the coefficients of the polynomial are {aj}, the evaluations become p e2ππππik/n = Σj aj (cos(2πjk/n)) + iΣj aj (sin(2πjk/n)). (This is why we say that we are doing things backwards. It is the aj which are analogous to the Fourier coefficients for the function p.) Let z be our primitive n-th root of unity. Transforms of this kind can be defined for any value of n. And there is a symmetric form for the inverse transformation which takes the values, {p(zk)}, which we shall abbreviate as {pk}, and produces the aj, so there is no significant difference between defining this transform forward or backward. We can obtain the inverse transformation by multiplying each pk by z-sk, and summing over the n values of k. We get… ΣΣΣΣj,k aj zjk*z-sk or ΣΣΣΣj aj (ΣΣΣΣk z(j-s)k) or ΣΣΣΣj aj ts-j, where tr is our old friend the sum of the r-th powers of the n