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Calculus II Notes Week #7

by: Zachary Hill

Calculus II Notes Week #7 MATH 1220

Marketplace > Tulane University > Mathematics (M) > MATH 1220 > Calculus II Notes Week 7
Zachary Hill
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These notes cover the Equilibrium Theory, the definition of sequential differentiability, the Product Law Proof, and the Reciprocal Law Proof.
Calculus II
Benjamin Klaff
Class Notes
Math, calculus II, Calculus, MATH 1220




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This 4 page Class Notes was uploaded by Zachary Hill on Sunday February 28, 2016. The Class Notes belongs to MATH 1220 at Tulane University taught by Benjamin Klaff in Spring 2016. Since its upload, it has received 72 views. For similar materials see Calculus II in Mathematics (M) at Tulane University.

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Date Created: 02/28/16
MATH 1220 Notes for Week #7  22 February 2016  Warm­Up  ● Remind yourself of the of the statement of the Intermediate Value Theorem (IVT):  ○ on a closed interval [a, b] where a < b and f(a) < 0, f(b) > 0, and f  is  continuous, there exists a number z such that f(z) = 0  ○ note: z is not unique; example is f(x) = sin(x) because sin(x) = 0 on the period kπ  Equilibrium Theory  Suppose f : [0, 1] → [0,1]. f(0) / 0 and f(1) / 1 . Then there exists a real number c in [0, 1]  such that f(c) = c. This is called a fixed point of f . If you overlay the graph y = x , the graph of f   cannot satisfy the previously laid out properties without crossing y = x where all points are fixed  points.    Proof:  ● Define a new function h(x) = x − f(x) for x in the closed interval [0, 1].  ○ Prove h  is continuous:  ■ The definition of continuity is: in (x )  is a sequence such that  nim(x ) = a,  n→∞ then  n→∞ f(n ) = f(a).  ■ So if (xn)  is a sequence such that  lim(n ) = a, then  n→∞ n→∞ h(n ) n→∞im[nx ) − n(x )n→∞ lnm(xn→∞− lnm f(x ).  ■ Since we defined  lim(x ) and know f(x ) is continuous, then  n→∞ n n lim(xn) − lim fnx ) = (a) − f(a) = h(a).  n→∞ n→∞ ■ Then ln→∞h(x n = h(a), so h is continuous.  ○ Prove h(0) < 0 :  ■ h(0) = 0 − f(0).  ■ Since f(0) falls on [0, 1] and f(0)/ 0 , then f(0) > 0.  ■ So h(0) = 0 − f(0) < 0 .  ○ Prove h(1) > 0 :  ■ h(1) = 1 − f(1).  ■ Since f(1) falls on [0, 1] and f(1)/ 1 , then f(1) < 1.  ■ So h(1) = 1 − f(1) > 0 .  ● Then, by IVT, there is some point z such that h(z) = 0. Then h(z) = z − f(z) = 0 , so  f(z) = z.      24 February 2016  Let’s figure out what the definition of (sequential) differentiability should be.  f(x+h)−f(x) ● Not it: Ifn→∞im (x+h)−x exists, we call it f′(x).  ● Definition of (sequential) differentiability:​ For every sequence (x ) innthe interval I   f(x )−f(a) converging to a such that x = n / a for all n, f (a) = lim n .  n→∞ xn−a Show that a function f  defined on an interval I  and differentiable at a number a in I is  continuous at a.  ● Suppose (x ) ns the sequence of real numbers in I  such that  lim(x ) = a andnx = n / a for  n→∞ f(n )−f(a) all n. Then lim x −a  exists, and we will call it L .  n→∞ n ● Then  f(n )−f(a) n→∞[fnx )−f(a)] 0 ∙ L =n→∞m(xn− a) n→∞ xn−a = n→∞(x n a) lim(n − ) = n→∞[f(x n − f(a)] = n→∞ f(xn) − n→∞ f(a).  n→∞ ● So 0 = lim f(x n − lim f(a)   n→∞ n→∞ n→∞ f(xn) = n→∞ f(a)  n→∞ f(xn) = f(a).  ● Since this is true for arbitrary  n→∞(x n = a, provided x = n / a , we conclude f  is continuous  at a  Let f(x) = x . Show that f  is differentiable at any point  ● Let (xn)  be a sequence converging to a such that x = n / a for any n .  f(n )−f(a) n −a ● Then lim x −a = lim x −a= lim 1 = 1 , so f  is differentiable for any a.  n→∞ n n→∞ n n→∞       26 February 2016  Product Law Proof  If f(x) and g(x) are differentiable at a , then (f(x) ∙ g(x)) ′= f (x)g(x) + g ′(x)f(x) evaluated at a.  Proof:  ● Let (x )n be a sequence converging to a such that x = n / a for any n .  ● Then lim f(n )gnx )−f(a)g(= lim f(xn)g(n )−fnx )g(a)+n(x )g(a)−f(a  (a) n→∞ xn−a n→∞ xn−a f(xn)g(n )−fnx )g(a) f(n )g(a)−f(a)g(a)  = ln→∞ xn−a + ln→∞ xn−a   g(xn)−g(a) f(xn)−f(a)  = lim[f(x ) n xn−a ] + lim[g(a) xn−a ]  n→∞ n→∞ g(x )−g(a) f(x )−f(a)  = f(a) lim n + g(a) lim n   n→∞ xn−a n→∞ xn−a  = f(a)g (a) + g(a)f (a).  Reciprocal Law Proof  If g(x) is differentiable at a, then ( 1 )′= −g′(a2.  g(a) [g(a)] Proof:  ● Let (x )n be a sequence converging to a such that x = n / a for any n .  g(a)−g(x ) g(n )g(a) g(n )g(a) ● Then lim xn−a = lim xn−a   n→∞ n→∞ g(a)−g(x )   = lim n   n→∞ g(xn)g(a)(n −a)   = lim −[g(xn)−g(a)]  n→∞ g(xn)g(a)(n −a) 1 1 −[g(xn)−g(a)]   = ln→∞ g(xn)n→∞ g(a)n→∞m xn−a   1 1   = g(a) ∙g(a)∙ [− g (a)]  −g′(a)   = [g(a)] .   


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