s01 - Ma 449: Numerical Applied Mathematics Model Solutions...

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Ma 449: Numerical Applied Mathematics Model Solutions to Homework Assignment 1 Prof. Wickerhauser Exercise 9(a*) of section 1.1, p.12 Let g(x) be the integral from 0 to x of t^2 cos(t). Then by the Second Fundamental Theorem of Calculus (2FT), we have g'(x)= x^2 cos(x). Exercise 9(b) of section 1.1, p.12 Let g(x) be the integral from 1 to x of exp(t^2). Then by the Second Fundamental Theorem of Calculus (2FT), we have g'(x)=exp(x^2). But the function we need to differentiate is f(x)=g(x^3), so we use the chain rule: (d/dx)f(x) = (d/dx)g(x^3) = 3 x^2 g'(x^3) = 3 x^2 exp(x^6). Exercise 11(b*) of section 1.1, p.12 Factor out 2/3 and apply the geometric sum formula to get 1/(1-1/3)=3/2 for the sum of 1/3^n from n=0 to infinity. The result is 1. Exercise 11(c) of section 1.1, p.12 Write 3/(n(n+1)) as 3*[ 1/n - 1/(n+1) ] to get a telescoping series: each term -1/(n+1) is cancelled by the 1/n term for the next n. Since 1/n tends to 0 as n tends to infinity, the only surviving term is 1/n for the first value of n, namely n=1. Thus the sum is 3*(1/1) = 3. Exercise 12(a) of section 1.1, p.13 Evaluate the first 4 derivatives of f(x) = x^(1/2) at x_0 = 1:: f'(x) = (1/2)x^(-1/2) ==> f'(1) = 1/2 f''(x) = (-1/4)x^(-3/2) ==> f''(1) = -1/4 f'''(x) = (3/8)x^(-5/2) ==> f'''(1) = 3/8 f''''(x) = (-15/16)x^(-7/2) ==> f''''(1) = -15/16 Also, f(1) = 1, so the 4th-degree Taylor polynomial of f is
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This note was uploaded on 05/16/2010 for the course MATH 449 taught by Professor Wickerhauser during the Fall '09 term at Washington University in St. Louis.

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s01 - Ma 449: Numerical Applied Mathematics Model Solutions...

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