CONVOLUTION DEMYSTIFIED which is past my pain threshold. You've all noticed long ago that the way a degree-n monomial is created is when TERRY A. LORING a degree-r and a degree-s monomial comeX together and n r + s = n: The coe±cient of x is thus arbs: Thus r+s=n the quadratic term in fg is (a0b2 + a1b1 + a2b0) : Indeed 1. You've convolved before [fg](x) = (a0b0) = (a b + a b ) x Convolution is made mysterious, but it comes up in a 0 1 1 0 2 familiar place: polynomial multiplication. = (a0b2 + a1b1 + a2b0) x 2 3 3 Consider f(x) = a0 + a1x + a2x + a3x and g(x) = = (a1b2 + a2b1 + a3b0) x 2 b0 + b1x + b2x : The annoying way to compute and fg; 4 = (a2b2 + a3b1) x which is a polynomial, and to ¯nd all its coe±cients, is 5 like this: = (a3b2) x [fg](x) = f(x)g(x) X ¡ ¢ ¡ ¢ 2 3 2 The meaning of a sum like arbs is to create all = a0 + a1x + a2x + a3x b0 + b1x + b2x r+s=n ¡ 2¢ = a0 b0 + b1x + b2x possible pairs (r; s) add these all together. There are ¡ ¢ 2 only the order in a ¯nite sum of reals in irrelevant, and + a1 b0 + b1x + b2x ¡ 2¢ yet most people feel better if an order is forced. Thus + a2 b0 + b1x + b2x such a sum is usually rewritten to imply an order: ¡ 2¢ + a3 b0 + b1x + b2x X Xn 2 5 = a0b0 + a0b1x + a0b2x + ::: + a3b2x arbs = arbn¡r r+s=n r=0 Date: August 28, 2003. We're almost to convolution. Next we pad the polynomials f and g with zeros. For De¯nition 1. A doubly-in¯nite sequence x = hxni (in- j ·¡1 and for j ¸ 4; set aj = 0: For k ·¡1 and k ¸ 3 dex set the integers), has ¯nite support is there are inte- set bk = 0: Then (for all real values of x; except perhaps gers C1 and C2 such that zero) we have (n < C1 or n > C2) ) xn = 0: X1 j f(x) = ajx (Equivalently, all but ¯nitely many of the terms in the j=¡1 sequence are zero.) and A good convention with a doubly in¯nite sequence is X1 k to mark the term of index zero with an underline. Thus g(x) = bkx k=¡1 z = h::: 0; 0; ¡1; 2; 1; 0; 0;:::i (You can ignore ² and ± for these sums. Just toss the is the sequence that has all terms zero except terms that equal zero and sum the ¯nitely many terms that remain.) What's more, we see that z¡1 = ¡1; z0 = 2; z1 = 1: X1 n To give x a playmate, we let [fg](x) = cnx n=¡1 w = h::: 0; 0; 1; 0; 1; 0; 0;:::i: where, for all n; De¯nition 2. Suppose x and y are double-in¯nite se- quences of ¯nite support. Then the convolution x ¤ y is X1 again a double-in¯nite sequence, where cn = arbn¡r r=¡1 X1 [x ¤ y]n = xryn¡r r=¡1 2. Convolving sequences So, for example, We'll consider all types of bi-in¯nite sequences later. Here's a good place to start: z ¤ w = h::: 0; ¡1; 2; 0; 2; 1; 0;:::i: 2 It is no coincidence that the following if true for all then real x 6= 0 : ¡ ¢ ¡ ¢ © (f) = ha¡2; a¡1; a0; a1; a2; i: ¡x¡1 + 2 + x x¡1 + 0 + x (It is simple once you look past the notation. It just = ¡x¡2 + 2x¡1 + 0 + 2x + x2 strips o® the coe±cients.) Here's the type of formula that creates all the fuss: 3. A homorphism (y) ©(fg) = ©(f) ¤ ©(g): Mathematicians are always excited when one opera- tions is turned into another by a function. For example, We would rather have the reverse formula, that ex- the log function is beloved precisely because of what it plains the convolution of sequences in terms of the point- does to products: wise product of functions. That is not really possible if ln(ab) = ln(a) + ln(b) for positive a and b: we stick with real variables. But for a pure mathemati- cian, (y) is inspiration enough to pursue this subject. A pure mathematician is excited because the structure of the product operation can be explained via the structure of addition. An applied mathematician gets excited an 4. Differentiation made \easy" invents the sliderule. We will see that the Fourier Transform for R takes as All the variations on the Fourier transform perform input certain functions F : R ! R and returns a new a similar trick, converting regular multiplication (point- function ©(f): R ! R: Of the many formulas that hold, wise multiplication) of functions into convolution, and here's one: vice-versa. We found something very close to the Fourier transform above. ©(f 0)(x) = x©(f)(x) Consider the set P of all functions from (¡1; 0] [ In words, the Fourier transform converts di®erentiation [0; 1) to(¡1; 1) that are polynomials in x and x¡1: Let into multiplication by x: S denote the set of all doubly in¯nite real sequences of (Multiplying by x is easier than taking derivatives, ¯nite support. We make the following simple de¯nition: hence \di®erentiation made easy." Of course, the Fourier If Transform is harder than di®erentiation, so this is a bit ¡2 ¡1 1 2 f(x) = ::: + a¡2x + a¡1x + a0 + a1x + a2x ::: of a mirage.) 3 This formula opens lots of doors. Using the inverse of the Fourier Transform, ©¡1; we can get this: Dnf(x) = ©¡1(xn©(f)(x)) for n = 1; 2;::: Now, just because we can, we put in a non-integer for n and call this a de¯nition. In particular, for n = 1=2; we de¯ne D1=2(f) = ©¡1(x1=2©(f)(x)): We now have the order \1/2" derivative of (some) functions! Studying this is outside the scope of this class. I just present this to you to give you more of a sense of why the Fourier Transform is so essential in pure math. 4.