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by: Zachary Hill

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# Calculus II Notes Week #13 MATH 1220

Marketplace > Tulane University > Mathematics (M) > MATH 1220 > Calculus II Notes Week 13
Zachary Hill
Tulane
GPA 3.88

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These notes cover a continuation of discussion about Power series, the introduction and definition of Taylor series and their application, and bump functions.
COURSE
Calculus II
PROF.
Benjamin Klaff
TYPE
Class Notes
PAGES
5
WORDS
CONCEPTS
Math, MATH1220, Klaff, Calculus, calculusii
KARMA
25 ?

## Popular in Mathematics (M)

This 5 page Class Notes was uploaded by Zachary Hill on Sunday April 17, 2016. The Class Notes belongs to MATH 1220 at Tulane University taught by Benjamin Klaff in Spring 2016. Since its upload, it has received 14 views. For similar materials see Calculus II in Mathematics (M) at Tulane University.

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Date Created: 04/17/16
MATH 1220 Notes for Week #13  11 April 2016  Did an extended practice problem during class. A large part of the class was spent attempting to  solve these problems in small groups and ensure other groups understood them. The remainder  of class was spent by Dr. Klaff discussing the importance of understanding this and practicing it.  ∞ xn+1 ● Given f(x) = ∑ 4 n4 :  n=1 ○ Find the first 4 nonzero terms (or the first 4 terms not including those that,  equaling zero, would not appear in the summation)  x1+1 x2+1 x3+1 x4+1 x2 x3 x4 x5 x2 x3 x4 x5 ■ 4 1 4+ 4 24+ 4 34 + 4 44 = 4∙1 + 16∙16 + 64∙81 + 256∙256 = 4 + 256 + 5184 + 65,536   x2 x3 x4 x5 ■ Note: it is generally enough to use  4 14+ 4 24 + 4 34 + 4 44, this evaluation  was just done in order to show the final value of the terms  ○ Then the 3rd term is:  4 ■ x3 4  4 3 th th ■ Note: the n  term is not always to the n  power  ∞ xn+1 ○ Use the ratio test to find for what values of x the function f(x) = ∑ 4 n4 is well  n=1 defined (converges)  |n+1| ■ Recall the ratio test states if lim n→∞ |an | = L < 1, then the infinite series  converges   x(n+1)+1 |4(n+(n+1)| | xn+2 4 n4| | x (1)n4| | xn4 | ■ Then when lim | xn+1 | = lim | 4(n+1(n+1)4 ∙xn+1| = lim | 4(n+1)4∙ 1 | = lim | 4(n+1)4| =  n→∞ | 4 n | n→∞ | | n→∞ | | n→∞ | | | n4 | | | | | lim | 3 n4 2 |= | |lim | 4 3n42 | = | |lim | 3 12 | =  |4|n→∞ |n +4n +6n +4n+1 | |4|n→∞ |n +4n n4n +4n|1 4 |n→∞ |1+4 n6 +n4+ n4 n | | | ∞ | || | = | |< 1, f(x) = ∑ xn+1 is well defined  |4||1 | |4| 4 n4 n=1 ■ Then the radius of convergence (the values for which the infinite series  converges) is |x| < 4        12 April 2016  Did an exploration into power series and discussed Taylor series.  ∞ ● A power series for f(x) = 1  is  ∑ x  on |x| < 1  1−x n=0 ∞ n ∞ n ● If you replace x with − x, then g(x) = 1  and its power series is  ∑ (− x) = ∑ (− 1) x  on  n 1+x n=0 n=0 |x| < 1  ∞ ● If we replace the top 1 with any constant a , then j(x) = a  and its power series is  ∑ ax   n 1−x n=0 on |x| < 1  ● Others:  ∞ n ∞ ○ h(x) = 1 2;  ∑ (x ) = ∑ x  on |x| < 1  1−x n=0 n=0 ∞ n ∞ 2n ○ k(x) = 1 2;  ∑ (− x ) = ∑ (− 1) x   on |x| < 1  1+x n=0 n=0 1 ∞ n 1 ○ p(x) = 1−4x ;  ∑ (4x)  on |4x| < 1 ⇒ |x| <   4 n=0 1 ∞ (−x) n ∞ 1 n 2n (−x)2| | −x 2| ○ q(x) = 1+x2;  ∑ ( 4 ) = ∑ ( ) 4− x)  on | 4 | < 1 ⇒ (|)2< 1|⇒ |x| < 2  4 n=0 n=0 | | ● If you can get a function in the form f(x) = a m  then it can be represented by the power  1−cx ∞ series  ∑ (cx )  on the radius of convergence |cx | < 1  m n=0 ● These functions are differentiable and integrable on their domains; the derivatives and  antiderivatives are computed “term­by­term” (as if they were polynomials) and the  interval of convergence doesn’t change  ● Examples:      1 1 ○ ∫1−xdx = −∫duu=− ln(u) + C =− ln(1 − x) + C       ○ The power series for     ∞   2 3 4 n+1 − ln(1 − x) = ∫ 1 dx = ∑∫x dx = 1 + ∫ + x + x + ... + x dx = x + x + x + x + ... +x  on   1−x  n=0   2 3 4 n+1 |x| < 1    1 ○ arctan(x) = ∫1+x2dx =     ∞ n   n 3 5 7 (−x )+1 ∫ ∑ (− x ) dx = 1 ∫ x + x − x + ... + (− x ) dx = x − x3 + 5 − 7 + ... + n+1  on |x| < 1   n=0   ∞ x xn ● Recall the power series (and actually Taylor series) for e = exp(x) = ∑ n!  n=0 ∞ −x (−x) ● If v(x) = e , then the power series is  ∑ n!   n=0 ∞ x xn ● If z(x) = xe , then the power series is x ∑ n!  n=0 ∞ (n) f′(0) f′′(0)2 f′′′(0) f (0) n ● Taylor series are series such that f(x) = f(0) + 1! x + 2! x + 3! + ... = ∑ n! x  or more  n=0 ∞ (n) generically f(x) = ∑ f (a(x − a)   n! n=0 ∞ ● f(x) = e = ∑ xn= 1 + x + x + x3 + ... because all derivatives of exp(x) equal exp(x) and at  n! 2! 3! n=0 x = 0 they all equal 1        13 April 2016  Worked on Taylor series approximations about x = 0  of a function defined on (− R, R) and  proved the geometric series is a Taylor series.  ● Taylor approximations:  (0) f (0) = f(0)  y0= f(0)  (1) f (0) = f (0)  y1= f(0) + f (0)x  (2) f′′(0)2 f (0) = f ′(0)  y2= f(0) + f (0)x + 2! x   f (0) = f ′(0)  y = f(0) + f (0)x + f′′(0x + f′′′(x   3 2! 3! th ● f(x) near x = 0  to be modeled by a n  degree polynomial (derivatives through order n  at x = 0  for f(x) and the modeling function are the same), then  f′′(0)2 f′′′(03 fn(0) n y = f(0) + f ′(0)x + 2! x + 3! x + ... + n! x   ● For f(x) = 1−x   (0) f (0) = f(0)= 1−0 = 1  y0= 1  (1) −2 f (0) = f (0) =− (1 − 0) (0 − 1) = 1  y1= 1 + 1x  (2) −3 2 2 f (0) = f ′(0) =− 2(1 − 0) (− 1) = 2  y2= 1 + x + x 2! (3) −4 2 6 3 f (0) = f ′(0) =− 3(2)(1 − 0) (− 1) = 6   y3= 1 + x + x + x  3! (4) −5 2 3 24 4 f (0) = f ′′(0) =− 4(6)(1 − 0) (− 1) = 24  y4= 1 + x + x + x + 4!x   ∞ ● Then f(x) = 1  has the Taylor series 1 + x + x + x + x + ... = ∑ x   n 1−x n=0       15 April 2016  We continued discussing Taylor series and learned about bump functions.  ● Suppose we’re interested in finding a good polynomial approximation to some function  f(x) near x = 0  ● Then we’re interested in Taylor polynomials of some order which approximate f as well  as can be done near x = 0  ● Suppose we want “the best” degree 5 polynomial that approximates f(x) = sin(x)near  x = 0; we use the derivative info of f(x) = sin(x)at x = 0  to tell us the coefficients  f (0) = sin(0) = 0  y = 0  0 f (0) = cos(0) = 1  y = x  1 1! f (0) =− sin(0) = 0  y2= x + x  2 2! (3) f (0) =− cos(0) =− 1  y3= x + 3!x   (4) 1 3 0 4 f (0) = sin(0) = 0  y4= x − x3! x  4! (5) 1 3 1 5 f (0) = cos(0) = 1  y5= x − x3! x  5! ○ Then the best degree 5 polynomial to approximate f(x) = sin(x) is  y (x) = x − x + 1 x   5 6 120 ● There exists functions such as (imagine a smooth line)      ○ Has derivatives of all orders  (n) 2 n ○ f (0) = 0 ⇒ y (xn = 0 + 0x + 0x + ... + 0x = 0  ○ Appears this is the zero function (f(x) = 0), but it’s not; doesn’t play nice with  Taylor series

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