Methods of Approximation
Date: 7/31/96 at 14:7:48 From: Albert Nicorgua Subject: Methods of Approximation Hi, My teacher used Newton's Method of Approximation to solve "Y to the power of Y = 25." Do you know of another way to solve the above equation? --Thanks, Andy
Date: 9/1/96 at 18:5:51 From: Doctor Jerry Subject: Re: Methods of Approximation You wish to solve the equation x^x-25=0. By Newton's Method, I obtain x=2.9632.... Another method is called the Bisection Method. Suppose you want to find a zero of a continous function f(x). The idea of the Bisection Method is that if f has a xero in an interval [a,b] with midpoint m, then f must have a zero in either the left or right half of [a,b]. If we choose the first interval so that f(a)*f(b)<0 and let h=m-a, then if f(m-h)*f(m)>0, replace the interval [a,b] = [m-h,m+h] by [m,m+h]; otherwise, replace [m-h,m+h] by [m-h,m]. Repeat this algorithm until the containing interval is sufficiently small. This method is not as fast as Newton's Method, but can be applied to any continuous function, not just differentiable functions. For the function f(x) = x^x-25, here are a few results. Take the intial interval to be [2.5,3.5]. So m=3 and h=0.5. Successive function values, values of m, and h are -15.11 2.75 0.25 -8.85 2.875 0.125 -4.175 2.9375 0.0625 You can see that the midpoints are approaching 2.96... -Doctor Jerry, The Math Forum Check out our web site! http://mathforum.org/dr.math/
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