L’Hˆopital’s Rules and Taylor’s Theorem for Product Calculus Alex B. Twist Michael Z. Spivey University of Puget Sound Tacoma, Washington 98416 1 Introduction This paper is a continuation of the second author’s undergraduate-level survey [4] of product calculus, a variation of the standard calculus. In [4] the second author defines the product derivative and integral, describes rules for finding product derivatives and integrals, discusses applications, and proves product versions of some important theo- retical results in calculus, including a fundamental theorem. Other works on product calculus include Dollard and Friedman [2] and Grossman [3]; more can be found in the references in [2] and [4]. The major contributions of this article are product calculus versions of l’Hˆopital’s rules and Taylor’s theorem. In the process we prove an exponentiation rule, a chain rule, and a generalized mean value theorem for the product derivative. The former two imply rules for product integration by parts and product integration by substitution, respectively. We need the definitions of the product derivative and integral and a theorem from [4]. The product derivative of a function f at x is defined as 1 ∆x 1 ∗ f(x + ∆x) df dx = f (x) = lim . ∆x→0 f(x) We denote the nth product derivative of f by f [n]. The product integral of f > 0 over [a, b] is as follows: Let P = x0, x1, . , xn be a partition of [a, b] and let ck be any point on the interval [xk−1, xk]. Let ∆xk = xk − xk−1 and let kP k be the maximum value of ∆xk. Then the Riemann product integral of f over [a, b] is n b dx Y ∆xk Paf(x) = lim f(ck) , kP k→0 k=1 provided the limit exists and is independent of the choice of P and the ck’s. Finally, we have the following relating the product derivative and the usual Newto- nian derivative. Theorem 1. (Spivey 2006) If f(x) 6= 0, then f ∗(x) exists and is nonzero if and only if f 0(x) exists, in which case d 0 f ∗(x) = exp ln |f(x)| = ef (x)/f(x). dx 1 In Section 2 we prove some additional product differentiation rules, which lead to additional product integration rules. Section 3 contains our results on l’Hˆopital’s rules using the product derivative. Taylor’s theorem with the product derivative is given in Section 4. 2 More on Product Calculus Our results on l’Hˆopital’s rules and Taylor’s theorem require some results concerning the product derivative. Here we have the exponential rule, a parallel to the product rule of Newtonian calculus. Theorem 2. (Exponential Rule) If f is a product differentiable function of x with f(x) > 0 and f ∗(x) > 0 and g is a differentiable function of x then 1 0 d f(x)g(x) dx = f ∗(x)g(x)f(x)g (x). Proof. 1 1 f(x + ∆x)g(x+∆x) ∆x d f(x)g(x) dx = lim ∆x→0 f(x)g(x) 1 ! f(x + ∆x)g(x+∆x) ∆x = lim exp ln ∆x→0 f(x)g(x) g(x + ∆x) ln(f(x + ∆x)) − g(x) ln f(x) = exp lim ∆x→0 ∆x d (g(x) ln f(x)) = e dx 0 g(x) f (x) +g0(x) ln f(x) = e f(x) 0 g(x) g0(x) = ef (x)/f(x) eln f(x) = f ∗(x)g(x)f(x)g0(x), via Theorem 1. Just as the product rule for Newtonian calculus yields the technique of integration by parts, the exponential rule for product calculus produces a product integration by parts. Corollary 1. (Product Integration by Parts) Let u be a product differentiable and prod- uct integrable function of x with u(x) > 0 and u∗(x) > 0. Let v be a differentiable function of x. Then v 1 vdx u P(du dx ) = . dv dx P u dx 2 Proof. This follows directly from Theorem 2. We can retain a commutative-like property in Corollary 1 by letting v(x) = ln y(x). y0(x) ∗ ln y ln x Then, using the facts that y(x) = ln y (x) (Theorem 1) and x = y , Corollary 1 becomes ln y ln u 1 vdx 1 ln y dx u y P(du dx ) = P(du dx ) = = . Puln(y∗(x))dx Py∗(x)ln u dx We also have a product chain rule. Theorem 3. (Product Chain Rule) Let f be a product differentiable function of u with f ∗(u) > 0, and let u be a differentiable function of x. Then 1 ∗ u0(x) df dx = f (u) . Proof. First consider the composition of our functions, f(u(x)). Define the function 1 f(u(x)+t) t ( f(u(x)) ) 1 , t 6= 0; g(t) = df du 1, t = 0. Note that as t → 0, the numerator of g(t), t 6= 0, is a product derivative. Thus, g(t) is continuous about t = 0. Now, letting t = ∆u 6= 0, we have 1 f(u(x)+∆u) ∆u f(u(x)) g(∆u) = 1 df du 1 ∆u f(u(x) + ∆u) ⇔ (g(∆u)df du ) = . f(u(x)) Now, considering the product derivative of f(x), we can write 1 ∆x 1 f(u(x + ∆x)) df dx = lim ∆x→0 f(u(x)) 1 f(u(x) + ∆u) ∆x = lim ∆x→0 f(u(x)) 1 where ∆u = u(x + ∆x) − u(x). Via substitution, we can express df dx as 1 1 ∆u 1 df dx = lim ((g(∆u)df du ) ) ∆x ∆x→0 1 ∆u = lim (g(∆u)df du ) ∆x ∆x→0 3 ∆u 1 lim∆x→0 ∆x = lim g(∆u) lim df du ∆x→0 ∆x→0 1 du = (df du ) dx = f ∗(u)u0(x). The product chain rule leads directly to a product integration by substitution tech- nique, just as with the usual Newtonian calculus. We present the indefinite version; the proof of the definite version is similar. Corollary 2. (Indefinite Integration by Substitution) Let g be a product integrable func- tion of u, and let u be a differentiable function of x. Let f(x) = g(u(x))u0(x). Then dx u0(x) dx du Pf(x) = P(g(u(x)) ) = Pg(u) = C0G(u), where G(u) is a product antiderivative of g(u). Proof. If G is a product antiderivative of g, then G∗(u) = g(u), and by the product chain rule (Theorem 3) we have u0(x) 1 du 1 f(x) = g(u(x)) = (d[G(u(x))] du ) dx = d[G(u(x))] dx . Product integrating, we have dx u0(x) dx 1 dx Pf(x) = P(g(u(x)) ) = P(d[G(u(x))] dx ) = C0G(u). We now provide a generalized version of the mean value theorem for product deriva- tives presented by Spivey [4]. This theorem is pivotal to proving the product calculus versions of l’Hˆopital’s rules and Taylor’s theorem. Theorem 4. (Generalized Mean Value Theorem for Product Derivatives) Let f be a function continuous on [a, b] and product differentiable on (a, b). Let g be continuous on [a, b] and differentiable on (a, b). Then, if g0(x) 6= 0 ∀x ∈ (a, b), f(a) 6= 0, and f(b) 6= 0, 1 f(b) g(b)−g(a) 1 = f ∗(c) g0(c) f(a) for some c such that a < c < b. 4 g(x)−g(a) f(x) f(a) g(b)−g(a) Proof. Let F (x) = f(a) f(b) . We have F (a) = F (b) = 1; thus we may appeal to Rolle’s theorem for product calculus from Spivey [4] to conclude that F ∗(c) = 1 for some c ∈ (a, b). So we have g0(c) f(a) g(b)−g(a) 1 = F ∗(c) = f ∗(c) f(b) 1 1 1 f(a) g(b)−g(a) ⇒ 1 g0(c) = f ∗(c) g0(c) f(b) 1 f(b) g(b)−g(a) 1 ⇒ = f ∗(c) g0(c) . f(a) 3 L’Hˆopital’s Rules f(x) L’Hˆopital’s rules are useful for evaluating indeterminate limits in the form limx→a g(x) where, as x → a, either both f(x) → 0 and g(x) → 0 or f(x) → ∞ and g(x) → ∞. L’Hˆopital’s rules can also be used to evaluate limits that result in the indeterminate forms ∞0 and 1∞; however, doing so requires that the limiting expression first be put in 0 ∞ the form 0 or ∞ by appropriate algebraic manipulations. Using the product derivative, though, yields explicit l’Hˆopital-type rules for these indeterminate forms. First we need an alternate characterization of a finite limit. Lemma 1. We have limx→c f(x) = L > 0 if and only if for every > 1 there exists 1 f(x) δ > 0 such that if 0 < |x − c| < δ then < L < . − f(x) Proof. (⇐) Given > 0, ∃ δ > 0 such that if 0 < |x − c| < δ, then e < L < e . Thus we have − < ln f(x) − ln L < ⇒ lim ln f(x) = ln L x→c ⇒ lim eln f(x) = eln L x→c ⇒ lim f(x) = L. x→c Showing the converse entails a similar argument. We now present our two l’Hˆopital-type rules using the product derivative. As is the case with many results in [4] the proofs are not all that dissimilar from existing proofs for the usual versions of l’Hˆopital’s rules.