L'hôpital's Rules and Taylor's Theorem for Product Calculus 1

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’Hôpital’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’Hôpital’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 product 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 technique, 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 function 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 derivatives 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’Hôpital’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’Hôpital’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’Hôpital-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. (See, for example, Bartle and Sherbert [1, 178].)

5 Theorem 5. (L’Hôpital’sRules with Product Derivatives, Part I) Let −∞ ≤ a < b ≤ ∞. Let f be a continuous and product differentiable function on (a, b). Let g be a differentiable function on (a, b) such that g0(x) 6= 0 ∀x ∈ (a, b). Suppose that limx→a+ f(x) = 1 and limx→a+ g(x) = 0.

1 1 ∗ g0(x) + g(x) a)If limx→a+ f (x) = L ∈ R , then limx→a+ f(x) = L. 1 1 ∗ g0(x) g(x) b)If limx→a+ f (x) = L ∈ {0, ∞}, then limx→a+ f(x) = L.

Proof. If a < α < β < b, then by Rolle’s theorem, g(α) 6= g(β). Further, by the generalized mean value theorem for product derivatives, ∃u ∈ (α, β) such that

1 1 f(β) g(β)−g(α) f ∗(u) g0(u) = . f(α)

Case (a): If L ∈ R+, given some > 1, ∃ c ∈ (a, b) such that

L 1 < f ∗(u) g0(u) < L ∀ u ∈ (a, c), from which we have

1 L f(β) g(β)−g(α) < < L ∀ α, β ∈ (a, c], α < β. f(α)

Now taking the limit as α → a+, we have

L 1 < f(β) g(β) < L ∀ β ∈ (a, c]. Since > 1 is arbitrary, we have our desired result. Case (b): If L = +∞ and given M > 0, then ∃ c ∈ (a, b) such that

1 f ∗(u) g0(u) > M ∀ u ∈ (a, c), from which we have

1 f(β) g(β)−g(α) > M ∀ α, β ∈ (a, c], α < β. f(α)

If we take the limit as α → a+, we have

1 f(β) g(β) > M ∀ β ∈ (a, c].

Since M > 0 is arbitrary, the assertion follows. If L = 0, the proof is similar.

6 Theorem 6. (L’Hôpital’sRules with Product Derivatives, Part II) Let −∞ ≤ a < b ≤ ∞. Let f be a continuous and product differentiable function on (a, b). Let g be a differentiable function on (a, b) such that g0(x) 6= 0 ∀x ∈ (a, b). Suppose that limx→a+ g(x) = ±∞.

1 1 ∗ g0(x) + g(x) a)If limx→a+ f (x) = L ∈ R , then limx→a+ f(x) = L. 1 1 ∗ g0(x) g(x) b)If limx→a+ f (x) = L ∈ {0, ∞}, then limx→a+ f(x) = L.

Proof. Suppose that limx→a+ g(x) = ∞ (the case for −∞ is similar). If a < α < β < b, by Rolle’s theorem, g(α) 6= g(β). Further, by the generalized mean value theorem for product derivatives (Theorem 4), ∃ u ∈ (α, β) such that

1 f(β) g(β)−g(α) 1 = f ∗(u) g0(u) . f(α) Case (a): If L > 1, given some > 1, then ∃ c ∈ (a, b) such that

L 1 < f ∗(u) g0(u) < L ∀ u ∈ (a, c). Thus we have 1 L f(β) g(β)−g(α) < < L ∀α, β ∈ (a, c], α < β. f(α) Since g(x) → +∞ as x → a+, we may assume that c is such that g(c) > 0. Letting β = c, we have 1 L f(c) g(c)−g(α) < < L ∀ α ∈ (a, c). (1) f(α) g(c) + g(c) Since g(α) → 0 as α → a , we may assume 0 < g(α) < 1 ∀ α ∈ (a, c), from which it follows g(α) − g(c) g(c) = 1 − > 0 ∀ α ∈ (a, c). g(α) g(α) g(α)−g(c) If we exponentiate (1) by g(α) , we have

g(α)−g(c) 1 L g(α) f(α) g(α) g(α)−g(c) < < (L) g(α) . f(c)

1 g(c) g(α) + Now, since g(α) → 0 and f(c) → 1 as α → a , then for any δ such that 0 < δ ≤ 1, 1 g(c) g(α) δ ∃ d ∈ (a, c) such that 0 < g(α) < δ and f(c) < e ∀α ∈ (a, d), giving

(1−δ) L 1 < f(α) g(α) < (L) eδ ∀α ∈ (a, d).

7 1 δ 2 Now take δ = min{1, ln , ln ln L }. Then we have (L)e ≤ L . We also have L(1−δ) L L L L > δ ≥ 1 = 1 = 2 . L Lln ln L ( ln L )ln L 1 L g(α) 2 Thus we have 2 < f(α) < L , and because is arbitrary we have our desired result. The cases for L = 1 and 0 < L < 1 are similar. 1 Case (b): If L = ∞, let M > 1 be given and c ∈ (a, b) such that f ∗(u) g0(u) > M ∀ u ∈ (a, c). Then it follows that

1 f(β) g(β)−g(α) > M ∀ α, β ∈ (a, c], α < β. (2) f(α)

1 Since g(x) → ∞ as x → a+, we may select c, d, d < c such that g(c) > 0, f(c) g(α) > √ g(c) 1 1/ e, and 0 < g(α) < 2 ∀ α ∈ (a, d). If we take β = c in (2) and exponentiate by g(c) 1 − g(α) , we get 1 g(α) g(c) √ f(α) 1− > M g(α) > M. f(c) Thus, we have r 1 1 √ M f(α) g(α) > f(c) g(α) M > . e 1 g(α) Since M > 1 is arbitrary, it follows that limα→a+ f(α) = ∞. The argument is similar for L = 0.

4 Product Taylor’s Theorem

In the ﬁnal section we prove a product version of Taylor’s theorem. Instead of repre- senting a function as the sum of a number of terms with an additive remainder, the product version shows how to represent a function as the product of a number of factors with a multiplicative remainder. Theorem 7. (Product Taylor Theorem) If f is an n + 1 times continuous, product diﬀerentiable function on an open interval I containing the point c such that f [i](c) 6= 0 for all i, 0 ≤ i ≤ n, then ∀x ∈ I such that x > c,

2 n ∗ x−c ∗∗ (x−c) [n] (x−c) f(x) = f(c)f (c) f (c) 2! . . . f (c) n! Rn(x), where (x−c)n+1 [n+1] Rn(x) = f (z) (n+1)! for some z such that c < z < x.

8 Proof. Deﬁne the functions F (t) and G(t) such that f(x) F (t) = n (x−t)k Q [k] k! k=0 f (t) (x − t)n+1 G(t) = . (n + 1)! Note here that, taking 00 = 1, we have F (x) = 1 and G(x) = 0. Thus, using the generalized mean value theorem for product calculus, we have for some z ∈ (c, x)

1 1 F (x) G(x)−G(c) 1 F ∗(z) G0(z) = = F (c) G(c) . F (c) It follows that 1 1 F (c) G(c) = F ∗(z) G0(z) 1 1 (x−t)n+1 ! d d(f(x)) dt (n+1)! dt = k t = z n (x−t) 1 Q [k] k! dt k=0 d(f (t) ) 1   −(x−z)n 1 ( n! ) =  (n−1)  −(x−z) (x−z)n f ∗(z)(f ∗(z)−1f ∗∗(z)(x−z))f ∗∗(z)−(x−z) . . . f [n](z) (n−1)! f [n+1](z) n! 1 ! −(x−z)n 1 ( n! ) = (x−z)n f [n+1](z) n! = f [n+1](z). Thus, we deduce that F (c) = f [n+1](z)G(c) f(x) (x−c)n+1 ⇔ = f n+1(z) (n+1)! n (x−c)k Q [k] k! k=0 f (c) 2 n (x−c)n+1 ∗ x−c ∗∗ (x−c) [n] (x−c) n+1 ⇔ f(x) = f(c)f (c) f (c) 2! . . . f (c) n! f (z) (n+1)! .

(x−c)n+1 n+1 From this, it follows that Rn(x) = f (z) (n+1)! . Corollary 3. If f [n](x) exists ∀ n and is continuous on the interval I containing the point c, and if limn→∞ Rn(x) = 1, then ∀ x > c such that x ∈ I, n k Y [k] (x−c) f(x) = lim f (c) k! . n→∞ k=0 Proof. This result follows directly from the previous theorem.

9 References

[1] Robert G. Bartle and Donald R. Sherbert. Introduction to Real Analysis. Wiley, New York, 3rd edition, 2000.

[2] John D. Dollard and Charles N. Friedman. Product Integration with Applications to Diﬀerential Equations. Addison-Wesley, Reading, MA, 1979.

[3] Michael Grossman. The First Nonlinear System of Diﬀerential and Integral Calcu- lus. MATHCO, Rockport, MA, 1979.

[4] Michael Z. Spivey. A product calculus. Submitted.