Thursday

July 31, 2014

July 31, 2014

Posted by **Julie** on Sunday, September 28, 2008 at 4:28pm.

- newton's method -
**Damon**, Sunday, September 28, 2008 at 6:56pmg(x) = f(x)-R

look for zero of g(x)

Xn+1 = Xn -g(Xn)/g'(Xn)

here we have

Xn+1 = Xn -(f(Xn)-R)/f'(Xn)

Xn+1 =Xn -f(Xn)/f'(Xn) + R/f'(Xn)

so here

f(x)/f'(x) = cos x sin x

R/f'(x) = - R x cos^2 x

so

f' = 1/(x cos^2x)

f = cos x sin x * 1/(x cos^2 x) = sin x/x cos x

I am not sure I have all your parentheses right and it does not quite check out but that is the idea.

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