MA113_2010SPRING_HW4_PROFSOLN_[0]

MA113_2010SPRING_HW4_PROFSOLN_[0] - MA113 Calculus II The...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: MA113 Calculus II The Cooper Union, Spring 2010 Prof. Alex Casti casti@cooper.edu Homework 4 Solutions Chapter 11 problems (Stein and Barcellos) Section 11.1-2-1.5-1-0.5 0.5 1 1.5 2-2-1.5-1-0.5 0.5 1 1.5 2 x y Problem 7, section 11.1 Stein & Barcellos tan-1 (x) P 1 (x;0) P 3 (x;0) Figure 1: Figure for problem 7, section 11 . 1. Comparison of f ( x ) = tan- 1 x with the Taylor polynomials P 1 and P 3 . P 2 is not shown since P 2 = P 3 . 7 Compute the Taylor polynomials P 1 ( x ;0 ) , P 2 ( x ;0 ) , and P 3 ( x ;0 ) for tan- 1 x and graph them for comparison. The N th-order Taylor polynomial of f ( x ) at x = a is de- fined by P N ( x ; a ) = N n = f ( n ) ( a ) n ! ( x- a ) n . For this case we choose a = 0. In calculating the 3 rd- order Taylor polynomial we need all derivatives up to the third evaluated at x = 0: f ( ) ( x ) = tan- 1 x f ( ) ( ) = f ( 1 ) ( x ) = 1 1 + x 2 f ( 1 ) ( ) = 1 f ( 2 ) ( x ) =- 2 x ( 1 + x 2 ) 2 f ( 2 ) ( ) = f ( 3 ) ( x ) = 6 x 2- 2 ( 1 + x 2 ) 3 f ( 3 ) ( ) =- 2 . The Taylor polynomials are then P 1 ( x ;0 ) = x P 2 ( x ;0 ) = x P 3 ( x ;0 ) = x- 1 3 x 3 . A comparison of tan- 1 x with the Taylor polynomials P 1 ( x ;0 ) and P 3 ( x ;0 ) is shown in figure 1. 14 Compute the Maclaurin series of the function f ( x ) = ln ( 1- x ) . The Maclaurin series for f ( x ) is the Taylor series expansion about x = 0 and is defined by f ( x ) = n = f ( n ) ( ) n ! x n . We need to compute all the derivatives of f ( x ) = ln ( 1- x ) and evaluate them at x = 0. After noting that d f dx = 1 x- 1 , the derivatives are 1 f ( 1 ) ( x ) = 1 x- 1 f ( 2 ) ( x ) =- 1 ( x- 1 ) 2 f ( 3 ) ( x ) = 2 ( x- 1 ) 3 f ( 4 ) ( x ) =- 3 2 ( x- 1 ) 4 f ( 5 ) ( x ) = 4 3 2 ( x- 1 ) 5 . . . . . . . . . f ( n ) ( x ) = (- 1 ) n + 1 ( n- 1 ) ! ( x- 1 ) n f ( n ) ( ) =- ( n- 1 ) ! ( n 1 ) f ( n ) ( ) n ! =- 1 n . Also, note that f ( ) ( ) = ln ( 1 ) = 0. It follows that the Maclaurin series is ln ( 1- x ) = n = 1- x n n =- x- x 2 2- x 3 3- ... 27 Consider the function f ( x ) = 1 + x .-1-0.5 0.5 1 0.5 1 1.5 x y Problem 27, section 11.1 Stein & Barcellos y = 1 + x P 4 ( x ;0) Figure 2: Figure for problem 27, section 11 . 1. Comparison of f ( x ) = 1 + x with the Taylor polynomial P 4 ( x ;0 ) . (a) Find P 4 ( x ;0 ) . To find the 4 th Taylor polynomial P 4 ( x ;0 )= 4 k = f ( k ) ( ) k ! x k we need the first four derivatives of f ( x ) = 1 + x : f ( ) ( x ) = 1 + x f ( ) ( ) = 1 f ( 1 ) ( x ) = 1 2 ( 1 + x )- 1 2 f ( 1 ) ( ) = 1 2 f ( 2 ) ( x ) =- 1 4 ( 1 + x )- 3 2 f ( 2 ) ( ) =- 1 4 f ( 3 ) ( x ) = 3 8 ( 1 + x )- 5 2 f ( 3 ) ( ) = 3 8 f ( 4 ) ( x ) =- 15 16 ( 1 + x )- 7 2 f ( 4 ) ( ) =- 15 16 P 4 ( x ;0 ) = 1 + 1 2 x- 1 8 x 2 + 1 16 x 3- 5 128 x 4 ....
View Full Document

Page1 / 26

MA113_2010SPRING_HW4_PROFSOLN_[0] - MA113 Calculus II The...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online