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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 coefficient 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 find all its coefficients, 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 finite 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 Definition 1. A doubly-infinite x = hxni (in- j ≤ −1 and for j ≥ 4, set aj = 0. For k ≤ −1 and k ≥ 3 dex set the ), has finite 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. X∞ j f(x) = ajx (Equivalently, all but finitely many of the terms in the j=−∞ sequence are zero.) and A good convention with a doubly infinite sequence is X∞ k to mark the term of index zero with an underline. Thus g(x) = bkx k=−∞ z = h... 0, 0, −1, 2, 1, 0, 0,...i (You can ignore ² and δ for these sums. Just toss the is the sequence that has all terms zero except terms that equal zero and sum the finitely many terms that remain.) What’s more, we see that z−1 = −1, z0 = 2, z1 = 1. X∞ n To give x a playmate, we let [fg](x) = cnx n=−∞ w = h... 0, 0, 1, 0, 1, 0, 0,...i. where, for all n, Definition 2. Suppose x and y are double-infinite se- quences of finite support. Then the convolution x ∗ y is X∞ again a double-infinite sequence, where cn = arbn−r r=−∞ X∞ [x ∗ y]n = xryn−r r=−∞ 2. Convolving So, for example, We’ll consider all types of bi-infinite 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 : ¡ ¢ ¡ ¢ Φ(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 off the coefficients.) Here’s the type of formula that creates all the fuss: 3. A homorphism (†) Φ(fg) = Φ(f) ∗ Φ(g). Mathematicians are always excited when one opera- tions is turned into another by a . 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, (†) 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 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 Φ(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. Φ(f 0)(x) = xΦ(f)(x) Consider the set P of all functions from (−∞, 0] ∪ In words, the Fourier transform converts differentiation [0, ∞) to(−∞, ∞) that are polynomials in x and x−1. Let into multiplication by x. S denote the set of all doubly infinite real sequences of (Multiplying by x is easier than taking derivatives, finite support. We make the following simple definition: hence “differentiation made easy.” Of course, the Fourier If Transform is harder than differentiation, 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, Φ−1, we can get this: Dnf(x) = Φ−1(xnΦ(f)(x)) for n = 1, 2,... Now, just because we can, we put in a non- for n and call this a definition. In particular, for n = 1/2, we define D1/2(f) = Φ−1(x1/2Φ(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.

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