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Calculus II Notes Week #11

by: Zachary Hill

Calculus II Notes Week #11 MATH 1220

Marketplace > Tulane University > Mathematics (M) > MATH 1220 > Calculus II Notes Week 11
Zachary Hill
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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.
Calculus II
Benjamin Klaff
Class Notes
Math, MATH 1220, Calculus, calculus II, Klaff
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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.

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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


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