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modelans - Natural Sciences Tripos Part IB Numerical...

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Natural Sciences Tripos Part IB Numerical Methods Model answers to some questions 6.1 Roots of an integrated quantity Consider the integral ( 29 ( 29 F x f a x = ξ ξ d , where f ( x ) is an easily evaluated function. (a) Show how F ( x ) may be calculated from f ( x ) for some arbitrary value of x using the Trapezium Rule with the interval [ a,x ] subdivided into n– 1 subintervals. (b) Derive the error term in this approximation and show how Romberg Integration may be used to improve the accuracy of the solution given estimates of F ( x ) obtained from n– 1 and 2 n– 1 subintervals. Show that this estimate is equivalent to calculating the integral using Simpson’s Rule. (c) Suppose we wish to find the value of x such that F ( x ) = 1 . Using the Newton-Raphson method, predict the location of the root using a single iteration. You may assume F ( x ) is a monotonically increasing function of x from x = a to the neighbourhood of the root x = x* . Why is this assumption important? (d) Describe how the second and subsequent iterations may be calculated. Discuss how precisely the integral F ( x ) should be evaluated for each iteration and suggest an appropriate method for achieving this. Answer: (a) Let G n ( x ) be our approximation to F ( x ) using the Trapezium Rule with n- 1 subintervals each of a size ∆ξ = ( x-a ) / ( n- 1) . [Suggestion: draw a sketch of some function f( x ) showing how G n ( x ) is obtained using the Trapezium Rule. You may wish to state the integral from ξ to ξ + ∆ξ and then sum for the compound rule, or simply quote the compound rule as here] Then ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 [ ] G x f a f a f a f x f x n = + + + + + + - + ∆ξ ∆ξ ∆ξ ∆ξ 2 2 2 2 2 K (b) To obtain the error in G n ( x ) , consider first the error in one of the subintervals from ξ to ξ + ∆ξ , say, using a Taylor Series expansion of f we can see ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 f f f f f O f f f f O ′ = + + ′′ + ′′′ + = + + ′′ + ′′′ + + ξ ξ ξ ξ ξ ξ ξ ξ ξ ξ ξ ξ ξ d d ∆ξ ∆ξ ∆ξ ∆ξ ∆ξ ∆ξ ∆ξ ∆ξ ∆ξ ∆ξ ∆ξ 2 3 4 0 2 3 4 5 2 6 2 6 24
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