×

### Let's log you in.

or

Don't have a StudySoup account? Create one here!

×

or

by: Zachary Hill

18

1

3

# Calculus II Notes Week #11 MATH 1220

Marketplace > Tulane University > Mathematics (M) > MATH 1220 > Calculus II Notes Week 11
Zachary Hill
Tulane
GPA 3.88

Get a free preview of these Notes, just enter your email below.

×
Unlock Preview

These notes cover an example of the application of the Ratio Test and the proof for a Function which is Uniformly Converged to by a Sequence of Continuous Functions is Continuous.
COURSE
Calculus II
PROF.
Benjamin Klaff
TYPE
Class Notes
PAGES
3
WORDS
CONCEPTS
Math, MATH 1220, Calculus, calculus II, Klaff
KARMA
25 ?

## Popular in Mathematics (M)

This 3 page Class Notes was uploaded by Zachary Hill on Sunday April 3, 2016. The Class Notes belongs to MATH 1220 at Tulane University taught by Benjamin Klaff in Spring 2016. Since its upload, it has received 18 views. For similar materials see Calculus II in Mathematics (M) at Tulane University.

×

## Reviews for Calculus II Notes Week #11

×

×

### What is Karma?

#### You can buy or earn more Karma at anytime and redeem it for class notes, study guides, flashcards, and more!

Date Created: 04/03/16
MATH 1220 Notes for Week #11  28 May 2016  Final day of Spring Break  29 May 2016  Group work in discussion sections  30 May 2016  ∞ (−1) xn+1 Consider  ∑ .  n=0 (2n+1)! a) What are the first five nonzero terms of this?  f0= x  x3 f1=− 3!  5 f2= x   5! x7 f3=− 7!  f = x9   4 9! b) What are the first five partial sums of this?  S = x  0 x3 S1= x − 3!  S = x − x3+ x5  2 3! 5! x3 x5 x7 S3= x − 3!+ 5!− 7!   S = x − x3+ x5− x7 + x9   4 3! 5! 7! 9! c) Is the n  term a continuous function? Why?  Yes because it’s a monomial.  d) Is the n  partial sum a continuous function? Why?  Yes because it’s an infinite polynomial.  ∞ n ∞ n (−1) x2n+1 (−1) xn+1 S(x) = ∑ (2n+1)! = n→∞ S Nx) = sin(x). S(x) = ∑ (2n+1)!  because this is how we define S(x),  n=0 n=0 and  lim S (N) = sin(x) because this is how we define lim S (x) for the giNen series.  n→∞ n→∞ e) Pick an x like 1 . If x = 1, does this series converge?  ∞ n 2n+1 ∞ n 2n+1 At x = 1,  ∑ (−1) x = ∑ (−1) 1 .  n=0 (2n+1)! n=0 (2n+1)! Ratio Test  ∞ |an+| (−1) 1n+1 Ifn→∞m |a n| < 1​, then  ∑ (2n+1)!  converges.  n=0 0 1 (−1) 1 1 (−1) 1 −1 1 | | 1 a 0 (2(0)+1)!= 1! = 1, a =1 (2(1)+1)!= 3! , … , |a |n= (2n+1)! , |an+1| = (2(n+1)+1)!, …   Then 1 lim |an+1|= lim (2(1+1)+= lim( 1 ∙ (2n+1)) = lim( 1 ∙ (2n+1)!) = lim 1 = 0 < 1,  n→∞ | an | n→∞ (2n+1)! n→∞ (2(n+1)+1)! 1 n→∞ (2n+3)(2n+2)(2n+1)! 1 n→∞ (2n+3)(2n+2) ∞ (−1) 12n+1 so  ∑  converges.   n=0 (2n+1)! f) Pick an x .  0 ∞ (−1) xn+1 ∞ (−1) 02n+1 At x = x ,0 ∑ (2n+1)! = ∑ (2n+1)!   n=0 n=0 Ratio Test  ∞ n | n+| (−1) 02n+1 If n→∞ |an | < 1​, then  ∑ (2n+1)!  converges.  n=0 x 2(n+1) +1 an+1 (2(n+1)+1)! x02n+3 (2n+1)! x0 2 1 2 Thenlim | a | = lim x 2n+1 = lim( (2n+3)! ∙ x 2n+1) = lim (2n+3)(2n+2) = x 0im (2n+3)(2n+2) = x 00) < 1 n→∞ n n→∞ (2n+1) n→∞ 0 n→∞ n→∞ ∞ (−1) x2n+1 , so  ∑ 0  converges.  n=0 (2n+1)! ∞ n 2n+1 This shows, using the ratio test with an arbitrary input x  that  ∑ (−1) x  converges for  (2n+1)! n=0 any x. Therefore (S (x)) → N(x)  ​ pointwise. ​      1 April 2016  Let I  be a closed interval. In trying to prove that if (f ) is a sequence of continuous function  n defined on I which converges uniformly to a function f , then f  is continuous on I , one gets to  a step involving 3 expressions like this:  f(x) − f (x) ,  f (x)|− f ja) , nf (j)| |f(an . jrom thnse a| | n | their properties (looking closed at n, j , x , f , a, f(a), etc.) and the hypotheses, one can  j n deduce that for every ε > 0, there’s an index J  such that if j ≥ J , then  f(x) − f(a) < ε.  | j | Statemen​ t: Let I  be a closed interval. Let (f ) be a sequence of continuous functions defined on  n I  which converges uniformly to a function f . Then f  is continuous on I .  Proof: ​ Let ε > 0 be given. Note that  > 0.   3 Let x  be a sequence of real numbers on I  such that x → a where a is a point on I .   j j ε | | ε For  , 3here’s an n  such that 3f n ≥ n , then  f (x) − f(x)3<  for any | nn I . From th|s po3nt on,  n ≥ n .   3 | | ε | | ε Then for any j ,  f(x) − f|(x) j . Thnre jx|sts 3ome index J  such that for j ≥ J ,  f (x) − f (a) < | n j n | 3 .   Since  f (x) − f(x) < , then  f (a) − f(a) < .   | ε | n | 3 | n | 3 | | | | | | Then for j ≥ J ,  f(x) − |(a) j f(x) − f |x) | f (j) − f na) j f (an − j (a) ≤nf(x) − f nx) +    | | j n j| |fn(x) j f (an + f | | − n(a) < + + = ε|   3 3 3 Then limf (x) = f(a) , so f  is continuous on I .  j→∞   j

×

×

### BOOM! Enjoy Your Free Notes!

×

Looks like you've already subscribed to StudySoup, you won't need to purchase another subscription to get this material. To access this material simply click 'View Full Document'

## Why people love StudySoup

Jim McGreen Ohio University

#### "Knowing I can count on the Elite Notetaker in my class allows me to focus on what the professor is saying instead of just scribbling notes the whole time and falling behind."

Anthony Lee UC Santa Barbara

#### "I bought an awesome study guide, which helped me get an A in my Math 34B class this quarter!"

Bentley McCaw University of Florida

#### "I was shooting for a perfect 4.0 GPA this semester. Having StudySoup as a study aid was critical to helping me achieve my goal...and I nailed it!"

Parker Thompson 500 Startups

#### "It's a great way for students to improve their educational experience and it seemed like a product that everybody wants, so all the people participating are winning."

Become an Elite Notetaker and start selling your notes online!
×

### Refund Policy

#### STUDYSOUP CANCELLATION POLICY

All subscriptions to StudySoup are paid in full at the time of subscribing. To change your credit card information or to cancel your subscription, go to "Edit Settings". All credit card information will be available there. If you should decide to cancel your subscription, it will continue to be valid until the next payment period, as all payments for the current period were made in advance. For special circumstances, please email support@studysoup.com

#### STUDYSOUP REFUND POLICY

StudySoup has more than 1 million course-specific study resources to help students study smarter. If you’re having trouble finding what you’re looking for, our customer support team can help you find what you need! Feel free to contact them here: support@studysoup.com

Recurring Subscriptions: If you have canceled your recurring subscription on the day of renewal and have not downloaded any documents, you may request a refund by submitting an email to support@studysoup.com