Solution to Homework_1_S09

Solution to Homework_1_S09 - Solution to Homework #1-...

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Solution to Homework #1- EGM6341 1. (a) Assume ƒ(x) is continuous on a ≤ x ≤ b, and consider the average = = n j j x f n S 1 ) ( 1 with all points x j in the interval [a, b]. Show that ) ( ζ f S = for some ζ in [a, b]. Soln: Because ƒ(x) is continuous in the interval [a, b], there are two numbers in the interval m and M such that m ≤ ƒ(x) ≤ M for all x in the interval. Then nm ≤ = n j j x f 1 ) ( ≤ nM m = n j j x f n 1 ) ( 1 M By the intermediate value theorem (IVT), for every value of y such that m ≤ y ≤ M, there will be a value ζ in the interval [a, b] such that ƒ(ζ) = y Thus, there exists a value in the interval ζ such that ƒ(ζ) = = n j j x f n 1 ) ( 1 3. (b) For small values of δ, measure the relative error in sin(x) = x by using x x x x x x - - ) sin( ) sin( ) sin( 0 x Bound this modified relative error for δ x . Choose δ to make this error less than 0.01, corresponding to a 1 percent error. (Hint: use Taylor series expansion for f(x) = sin(x) =…; do NOT interpret f(x) as ) sin( ) sin( x x x - ) Soln: Using Taylor Series expansion, we can find the error between the function and its approximation as x x x E - = ) sin( ) ( ... ! 7 ! 5 ! 3 ... ! 7 ! 5 ! 3 7 5 3 7 5 3 + - + - = - + - + - = x x x x x x x x To bound the relative error, we need 01 . 0 | ! 3 | ) sin( | ... ! 7 ! 5 ! 3 | 3 7 5 3 < + - + - x x x x x x => 06 . 0 2 x => 2449 . 0 = x Comment: using TS expansion twice (once to replace sin(x) by x in the denominator and then replace sin(x) by ... ! 7 ! 5 ! 3 7 5 3 + - + - x x x x in the numerator) allows us to determine the value of rather easily without solving the transcendental equation.
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(b’) In case some of you may have interpreted the original function as ) sin( ) sin( ) ( x x x x f - = and the approximate function as x x x - ) sin( , then the solution is as follows: Using Taylor Series expansion, we can find the error between the function and its approximation as ) sin( ) sin( ) ( sin ] ) [sin( ) sin( ) sin( ) sin( ) ( 2 x x x x x x x x x x x x x x x E + - - = - - - = ) ! 7 ! 5 ! 3 ( ) ! 7 ! 5 ! 3 ( ) ! 7 ! 5 ! 3 ( ) ! 7 ! 5 ! 3 ( 7 5 3 7 5 3 2 7 5 3 7 5 3 x x x x x x x x x x x x x x x x x x - + - - + - + - + - - - + - = ) ! 7 ! 5 ! 3 1 ( ! 7 120 6 ) ! 5 ! 3 2 ! 7 2 120 2 6 2 36 ( ! 7 120 6 6 4 2 2 8 6 4 2 8 8 6 4 6 2 8 6 4 x x x x x x x x x x x x x x x x x - + - - + - + - - + - + - - + - ) 6 1 ( 1 ) 360 36 ( ) ! 7 ! 5 ! 3 1 ( 360 36 2 6 4 6 4 2 2 8 6 x x x x x x x x x - + - - + - + - ) 6 1 ( ) 10 1 ( ) 36 ( 2 2 4 x x x - - - = ) 15 1 )( 36 ( 2 4 x x + - = (this is the absolute error) Now, we can also expand the original function f(x) as ... 6 1 ... 20 1 6 ... ! 7 ! 5 ! 3 ... ! 7 ! 5 ! 3 ) sin( ) sin( ) ( 2 2 2 7 5 3 7 5 3 + - + - - = + - + - + - + - = - x x x x x x x x x x x x x x f = ...) 60 7 1 ( 6 2 2 + + - x x Thus the relative error is = + - - - - = = ...) 20 1 ( 6 ...) 10 1 )( 36 ( ) ( ) ( . 2
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This note was uploaded on 04/23/2009 for the course EGM 6341 taught by Professor Mei during the Spring '09 term at University of Florida.

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Solution to Homework_1_S09 - Solution to Homework #1-...

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