每个月都来的东西竟然来晚了……
Puzzle for July 2006.
Let f be a function and g an approximation to f. We define the relative error in the approximation g to f at a point x to be abs((g(x)-f(x))/f(x)) where abs means absolute value.
This month's problem involves finding approximations to the functions 1/x and sqrt(x). Recall Newton's method for refining an approximate root r of an equation h(r)=0 replaces r with the new approximation r-h(r)/h'(r) where h' denotes the derivative of h.
1. Let r = p*x + q be an approximation to the function 1/x on the interval (a,b). Suppose we refine r by performing one iteration of Newton's method (applied to the equation 1/r-x=0). Figure out how to choose p and q so that the maximum relative error on the interval (a,b) after refinement is minimized. What is this minimum maximum relative error on the interval (1,2)? Give your answer as a decimal value with 4 significant figures.
2. Let r = p*x + q be an approximation to the function sqrt(x) on the interval (a,b). Suppose we refine r by performing one iteration of Newton's method (applied to the equation r*r-x=0). Figure out how to choose p and q so that the maximum relative error on the interval (a,b) after refinement is minimized. What is this minimum maximum relative error on the interval (1,2)? Give your answer as a decimal value with 4 significant figures.
It is only necessary to submit the two decimal values requested above for credit. Remember leading zeros do not count as significant figures.
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