L Hôpital s Rules and Taylor s Theorem for Product Calculus

Transcription

1 Hôpital s Rules and Taylor s Theorem for Product Calculus Introduction Alex B Twist Michael Z Spivey University of Puget Sound Tacoma, Washington 9846 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 theoretical 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 df = f x) = lim 0 fx + ) fx) ) We denote the nth product derivative of f by f [n] The product integral of f > 0 over [a, b] is as follows: et P = x 0, x,, x n be a partition of [a, b] and let c k be any point on the interval [x k, x k ] et k = x k x k and let P be the maximum value of k Then the Riemann product integral of f over [a, b] is n Pafx) b = lim fc k ) k, P 0 k= provided the limit exists and is independent of the choice of P and the c k s Finally, we have the following relating the product derivative and the usual Newtonian derivative Theorem Spivey 2006) If fx) 0, then f x) exists and is nonzero if and only if f x) exists, in which case ) d f x) = exp ln fx) = e f x)/fx)

2 In Section 2 we prove some additional product differentiation rules, which lead to additional product integration rules Section 3 contains our results on l Hôpital 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 fx) > 0 and f x) > 0 and g is a differentiable function of x then Proof via Theorem d fx) gx)) d fx) gx)) = f x) gx) fx) g x) = lim 0 = lim 0 exp = exp fx + ) gx+) lim 0 = e d gx) ln fx)) fx) gx) ) ) ) fx + ) gx+) ln fx) gx) gx + ) lnfx + )) gx) ln fx) = e gx) f x) fx) +g x) ln fx) ) gx) = e ) f x)/fx) e ln fx) g x) = f x) gx) fx) g x), 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 Product Integration by Parts) et u be a product differentiable and product integrable function of x with ux) > 0 and u x) > 0 et v be a differentiable function of x Then Pdu ) v = P u v u dv ) ) 2

3 Proof This follows directly from Theorem 2 We can retain a commutative-like property in Corollary by letting vx) = ln yx) Then, using the facts that y x) = ln y x) Theorem ) and x ln y = y ln x, Corollary yx) becomes Pdu ) v = Pdu ) ln y = We also have a product chain rule u ln y Pu lny x)) = y ln u Py x) ln u Theorem 3 Product Chain Rule) et f be a product differentiable function of u with f u) > 0, and let u be a differentiable function of x Then df = f u) u x) Proof First consider the composition of our functions, fux)) Define the function fux)+t) t fux)) ) gt) =, t 0; df du, t = 0 Note that as t 0, the numerator of gt), t 0, is a product derivative Thus, gt) is continuous about t = 0 Now, letting t = u 0, we have g u) = fux)+ u) fux)) g u)df du ) u = df du ) u fux) + u) fux)) Now, considering the product derivative of fx), we can write df = lim 0 = lim 0 fux + )) fux)) fux) + u) fux)) ) ) where u = ux + ) ux) Via substitution, we can express df df = lim g u)df du ) u ) 0 = lim g u)df u du ) 0 as 3

4 ) = lim g u) lim df u lim 0 du 0 0 = df du du ) = f u) u 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) et g be a product integrable function of u, and let u be a differentiable function of x et fx) = gux)) u x) Then Pfx) = Pgux)) u x) ) = Pgu) du = C 0 Gu), where Gu) is a product antiderivative of gu) Proof If G is a product antiderivative of g, then G u) = gu), and by the product chain rule Theorem 3) we have Product integrating, we have fx) = gux)) u x) = d[gux))] du ) du = d[gux))] Pfx) = Pgux)) u x) ) = Pd[Gux))] ) = C 0 Gu) 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 Hôpital s rules and Taylor s theorem Theorem 4 Generalized Mean Value Theorem for Product Derivatives) et f be a function continuous on [a, b] and product differentiable on a, b) et g be continuous on [a, b] and differentiable on a, b) Then, if g x) 0 x a, b), fa) 0, and fb) 0, ) fb) gb) ga) = f c) g c) fa) for some c such that a < c < b 4

5 Proof et F x) = fx) fa) ) fa) fb) ) gx) ga) gb) ga) We have F a) = F b) = ; thus we may appeal to Rolle s theorem for product calculus from Spivey [4] to conclude that F c) = for some c a, b) So we have = F c) = f c) ) g c) fa) gb) ga) fb) ) g c) = f c) g c) fa) gb) ga) fb) ) fb) gb) ga) = f c) g c) fa) 3 Hôpital s Rules fx) Hôpital s rules are useful for evaluating indeterminate limits in the form lim x a gx) where, as x a, either both fx) 0 and gx) 0 or fx) and gx) Hôpital s rules can also be used to evaluate limits that result in the indeterminate forms 0 and ; however, doing so requires that the limiting expression first be put in the form 0 or by appropriate algebraic manipulations Using the product derivative, 0 though, yields explicit l Hôpital-type rules for these indeterminate forms First we need an alternate characterization of a finite limit emma We have lim x c fx) = > 0 if and only if for every ɛ > there exists δ > 0 such that if 0 < x c < δ then ɛ < fx) < ɛ Proof ) Given ɛ > 0, δ > 0 such that if 0 < x c < δ, then e ɛ < fx) < eɛ Thus we have ɛ < ln fx) ln < ɛ lim x c ln fx) = ln lim x c e ln fx) = e ln lim x c fx) = Showing the converse entails a similar argument We now present our two l Hôpital-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 Hôpital s rules See, for example, Bartle and Sherbert [, 78]) 5

6 Theorem 5 Hôpital s Rules with Product Derivatives, Part I) et a < b et f be a continuous and product differentiable function on a, b) et g be a differentiable function on a, b) such that g x) 0 x a, b) Suppose that lim x a + fx) = and lim x a + gx) = 0 a)if lim x a + f x) g x) = R +, then lim x a + fx) gx) = b)if lim x a + f x) g x) = {0, }, then limx a + fx) gx) = Proof If a < α < β < b, then by Rolle s theorem, gβ) Further, by the generalized mean value theorem for product derivatives, u α, β) such that f u) g u) ) fβ) gβ) = Case a): If R +, given some ɛ >, c a, b) such that from which we have ɛ < f u) g u) < ɛ u a, c), ɛ < Now taking the limit as α a +, we have ) fβ) gβ) < ɛ α, β a, c], α < β ɛ < fβ) gβ) < ɛ β a, c] Since ɛ > is arbitrary, we have our desired result Case b): If = + and given M > 0, then c a, b) such that from which we have If we take the limit as α a +, we have f u) g u) > M u a, c), ) fβ) gβ) > M α, β a, c], α < β fβ) gβ) > M β a, c] Since M > 0 is arbitrary, the assertion follows If = 0, the proof is similar 6

7 Theorem 6 Hôpital s Rules with Product Derivatives, Part II) et a < b et f be a continuous and product differentiable function on a, b) et g be a differentiable function on a, b) such that g x) 0 x a, b) Suppose that lim x a + gx) = ± a)if lim x a + f x) g x) = R +, then lim x a + fx) gx) = b)if lim x a + f x) g x) = {0, }, then limx a + fx) gx) = Proof Suppose that lim x a + gx) = the case for is similar) If a < α < β < b, by Rolle s theorem, gβ) Further, by the generalized mean value theorem for product derivatives Theorem 4), u α, β) such that ) fβ) gβ) = f u) g u) Case a): If >, given some ɛ >, then c a, b) such that Thus we have ɛ < ɛ < f u) g u) < ɛ u a, c) ) fβ) gβ) < ɛ α, β a, c], α < β Since gx) + as x a +, we may assume that c is such that gc) > 0 etting β = c, we have ) fc) ɛ < gc) < ɛ α a, c) ) Since gc) 0 as α a+, we may assume 0 < gc) < α a, c), from which it follows gc) = gc) > 0 α a, c) If we exponentiate ) by gc), we have Now, since gc) ) gc) ɛ d a, c) such that 0 < gc) < ) fc) < ɛ) gc) 0 and fc) as α a +, then for any δ such that 0 < δ, < δ and fc) < e δ α a, d), giving ɛ ) δ) < < ɛ) e δ α a, d) 7

8 Now take δ = min{, ln ɛ, ln ɛ ln } Then we have ɛ)e δ ɛ 2 We also have ) δ) > ɛ ɛ = δ ɛ ln ɛ ln ɛɛ ln ) = ln ɛ 2 Thus we have < ɛ 2 < ɛ 2, and because ɛ is arbitrary we have our desired result The cases for = and 0 < < are similar Case b): If =, let M > be given and c a, b) such that f u) g u) > M u a, c) Then it follows that ) fβ) gβ) > M α, β a, c], α < β 2) Since gx) as x a +, we may select c, d, d < c such that gc) > 0, fc) > / e, and 0 < gc) < α a, d) If we take β = c in 2) and exponentiate by 2 gc), we get Thus, we have ) fc) > M gc) > M > fc) M > M e Since M > is arbitrary, it follows that lim α a + for = 0 = The argument is similar 4 Product Taylor s Theorem In the final section we prove a product version of Taylor s theorem Instead of representing 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 + times continuous, product differentiable function on an open interval I containing the point c such that f [i] c) 0 for all i, 0 i n, then x I such that x > c, where for some z such that c < z < x fx) = fc)f c) x c f c) x c)2 2! f [n] c) x c)n n! R n x), R n x) = f [n+] z) x c) n+ n+)! 8

9 Proof Define the functions F t) and Gt) such that F t) = Gt) = fx) n k=0 f [k] t) x t)k k! x t)n+ n + )! Note here that, taking 0 0 =, we have F x) = and Gx) = 0 Thus, using the generalized mean value theorem for product calculus, we have for some z c, x) ) F z) G z) F x) Gx) Gc) = = F c) Gc) F c) It follows that F c) Gc) = F z) G z) dfx)) dt = n k=0 df [k] t) x t)k k! ) dt = = ) x t) n+ d n+)! dt t = z f z)f z) f z) x z) )f z) x z) f [n] z) x z)n ) n )! f [n+] z) x z)n n! = f [n+] z) Thus, we deduce that ) x z)n n! ) F c) = f [n+] z) Gc) fx) n k=0 f [k] c) x c)k k! = f n+ z) x c) n+ n+)! fx) = fc)f c) x c f c) x c)2 2! f [n] c) x c)n n! f n+ z) x c)n+ n+)! From this, it follows that R n x) = f n+ z) x c) n+ n+)! f [n+] z) x z)n n! Corollary 3 If f [n] x) exists n and is continuous on the interval I containing the point c, and if lim n R n x) =, then x > c such that x I, n fx) = lim f [k] c) x c)k k! n k=0 Proof This result follows directly from the previous theorem x z)n n! ) 9

10 References [] 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, 979 [3] Michael Grossman The First Nonlinear System of Differential and Integral Calculus MATHCO, Rockport, MA, 979 [4] Michael Z Spivey A product calculus Submitted 0