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What Is Q-Calculus?

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What Is Q-Calculus? What is q-Calculus? Anthony Ciavarella July 1, 2016 Abstract In this talk, I will present a q-analog of the classical derivative from calculus. From there, I will prove q-analogs of the binomial theorem and Taylor's theorem. If time permits, I will show some applications of the q-calculus in number theory and physics. 1 q-Derivative In classical calculus we define the derivative of a function f to be df f(x + h) − f(x) = lim dx h!0 h Equivalently, we could define the derivative to be df f(qx) − f(x) = lim dx q!1 qx − x Now one might wonder, what happens when we don't take the limit, and just use the expression inside the limit as the definition of our derivative. This leads us into the exciting world of quantum calculus, also known as q-calculus. Definition 1.1. q −differential :A q −differential of a function f is defined to be dqf = f(qx) − f(x). From here, we can define our q − derivative to be d f f(qx) − f(x) q = dqx qx − x Now for an example, d qnxn − xn qn − 1 q (xn) = = xn−1 dqx qx − x q − 1 d n n−1 Comparing this to the classical result, dx (x ) = nx , we can define a q- qn−1 dn n dk n analog of a number to be [n] = q−1 . Comparing dxn (x ) = n! and dxk (x ) = n n−k k k!x to their q-analogs motivates the following two definitions. 1 Definition 1.2. q −factorial :A q −factorial of a natural number n is defined to be [n]! = [n] × [n − 1] × ::: × [1] if n > 0 and 1 if n = 0. Definition 1.3. q − factorial : For natural numbers n and k, the q-factorial is n [n]! defined to be k = [k]![n−k]! . 2 Taylor's Theorem Now that we have a q-derivative, it is reasonable to wonder if there are the- orems analogous to those from classical calculus. In fact, there is a q-analog of Taylor's theorem, but before proving it we are going to restate the classical Taylor theorem in a slightly weaker form. Theorem 2.1 (Taylor's theorem). Let P0;P1; :::; PN be polynomials and a be a number such that: (1) P0(a) = 1 and Pn(a) = 0 for n ≥ 1. (2) deg(Pn) = n. d (3) dx Pn(x) = Pn−1(x) for n ≥ 1 Then any polynomial f of degree n can be written in the form N X dnf f(x) = Pn(x) dxn x=a n=0 Proof. Let V be the vector space of polynomials of degree ≤ N. The polyno- mials P0; ::; PN are linearly independent by (2) since their degrees are strictly PN increasing. So they form a basis of V, and we have f = n=1 cnPn. Now to solve for ck, we will differentiate k times and evaluate at a. N N dkf X dk X = cn( Pn) = cnPn−k(a) = ck dxk x=a dxk x=a n=1 n=k Now that we have Taylor's theorem in a different form, we can state the q-analog of Taylor's theorem. Theorem 2.2 (q-Taylor's theorem). Let P0;P1; :::; PN be polynomials and a be a number such that: (1) P0(a) = 1 and Pn(a) = 0 for n ≥ 1. (2) deg(Pn) = n. dq (3) Pn(x) = Pn−1(x) for n ≥ 1 dq x Then any polynomial f of degree n can be written in the form N n X dq f f(x) = Pn(x) d xn x=a n=0 q 2 Proof. The proof is identical to the proof of Theorem 2.1. Although we now have the q-analog of Taylor's theorem, we still need to (x−a)n find the polynomials Pn. Naively one would expect Pn(x) = [n]! by analogy with the classical case, but these polynomials fail to satisfy condition (3). From conditions (1) and (2), we must have P0 = 1. In order to satisfy all of the 2 dq 2 x conditions we must then have P1 = x − a. Since x = [2]x, P2 = − ax − dq x [2] a2 2 (x−a)(x−qa) [2] + a = [2] satisfies the necessary conditions. Assuming this trend Qn−1 k k=0 (x−q a) continues, a reasonable guess would be Pn = [n]! . To simplify notation, we will introduce the q-analog of (x − a)n Definition 2.1. q-analog of (x − a)n: The q-analog of (x − a)n is defined to be n Qn−1 k (x − a)q = k=0 (x − q a). Before proving that these Pn satisfies the conditions in the q-Taylor's theo- rem, we will need the following lemma. Lemma 2.3 (q-Product Rule). d d g d f q (f(x)g(x)) = f(qx) q + g(x) q dqx dqx dqx Proof. dq(f(x)g(x)) = f(qx)g(qx) − f(x)g(x) = f(qx)g(qx) − f(qx)g(x) + f(qx)g(x) − f(x)g(x) = f(qx)dqg(x) + g(x)dqf(x) (1) d (f(x)g(x)) d g d f q = f(qx) q + g(x) q dqx dqx dqx I would like to point out that this expression is symmetric in f and g, however the symmetry is hidden in the definition of the q-derivative. Theorem 2.4. dq (x − a)n = [n](x − a)n−1 dq x q q Proof. For n = 1 this is clearly true. Suppose this theorem holds for n − 1, then we have n n−1 n−1 (x − a)q = (x − a)q (x − q a) Using the product rule, dq n n−1 n−1 dq n−1 (x − a)q = (x − a)q + (qx − q a) (x − a)q dqx dqx dq n n−1 n−2 n−2 (x − a)q = (x − a)q + q(x − q a)[n − 1](x − a)q dqx 3 dq n n−1 n−1 (x − a)q = (1 + q[n − 1])(x − a)q = [n](x − a)q dqx So the theorem, holds 8n 2 N S 0 by induction. n At this point, I would like to note that the definition of (x + a)q can be ex- −n 1 tended to negative integers with the following definition, (x−a)q = −n n . (x−q a)q 3 Binomial Theorem and Euler's Identities Now that we have the q-Taylor's theorem, we can use it to prove two q-analogs of the binomial theorem. Theorem 3.1 (Gauss' Binomial Theorem). n n X k(k−1) n k n−k (x + a) = q 2 a x q k k=0 Note: This expression is not symmetric in x and a, because the second value will be multiplied by some powers of q. Proof. k dq n n−k k (1 + x)q = [n][n − 1]:::[n − k + 1](1 + x)q dqx (n−k)(n−k−1) n−k n−k−1 2 n−k−1 n−k 2 n−k (0+a)q = (a+0)(qa+0):::(q a) = 1×q×q ×:::×q ×a = q a Using the q-Taylor's theorem we have n n X (n−k)(n−k−1) n n−k k (x + a) = q 2 a x q n − k k=0 When we replace k with n − k as the variable of summation,, this simplifies to n n X k(k−1) n k n−k (x + a) = q 2 a x q k k=0 n n Since n−k = k . Theorem 3.2 (Heine's Binomial Theorem). 1 1 X [n][n − 1]:::[n − k + 1] = xk (1 − x)n [k]! q k=0 Note: this equality holds as an equivalence of formal power series so issues of convergence are ignored. 4 Proof. The proof of this theorem is completely analogous to the proof of Gauss's binomial theorem and can be found in full in Kac's textbook Quantum Calculus. Now that we have these two formulas, one might wonder what happens in the limit where n goes to infinity. In the classical case, both sides go to infinity and we extract no new information, however this is not so in the quantum case. Theorem 3.3 (Euler's Identities). 1 1 k Y k 1 X k(k−1) x (1 + q x) = (1 + x) = q 2 q (1 − q)(1 − q2):::(1 − qk) k=0 k=0 1 1 Y 1 1 X xk = = (1 − qkx) (1 − x)1 (1 − q)(1 − q2):::(1 − qk) k=0 q k=0 for jqj < 1. Proof. We have qn − 1 1 lim [n] = lim ( ) = n!1 n!1 q − 1 1 − q when jqj < 1. And n (qn − 1)(qn−1 − 1):::(qn−k+1 − 1) 1 lim = lim = n!1 k n!1 (q − 1)(q2 − 1):::(qk − 1) (1 − q)(1 − q2):::(1 − qk) Plugging these two results into Gauss' and Heine's binomial theorems gives us the desired identities. 4 Applications The two Euler identites derived using the q-calculus have many applications in number theory. A few of them will be listed here and the proofs can be found in Victor Kac's textbook Quantum Calculus. Definition 4.1. The classical partition function p(n) is defined to be the num- ber of ways that a positive integer n can be written as a sum of positive integers, 1 if n = 0 and 0 if n < 0.
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