S sheva2291 May 2010 8 0 May 15, 2010 #1 Use Newton's method to approximate the given number correct to eight decimal places 130Radical130

M mathaddict Sep 2008 1,261 539 West Malaysia May 15, 2010 #2 sheva2291 said: Use Newton's method to approximate the given number correct to eight decimal places 130Radical130 Click to expand... Use the Newton method to approximate \(\displaystyle \sqrt{130}\) , then multiply this approximation with 130 . then let \(\displaystyle x=\sqrt{130} \) square them , and let \(\displaystyle f(x)=x^2-130\) then find f'(x) , and let the initial approximation be 10 , make use of \(\displaystyle X_{n+1}=X_n-\frac{f(x)}{f'(x)} \) Reactions: HallsofIvy

sheva2291 said: Use Newton's method to approximate the given number correct to eight decimal places 130Radical130 Click to expand... Use the Newton method to approximate \(\displaystyle \sqrt{130}\) , then multiply this approximation with 130 . then let \(\displaystyle x=\sqrt{130} \) square them , and let \(\displaystyle f(x)=x^2-130\) then find f'(x) , and let the initial approximation be 10 , make use of \(\displaystyle X_{n+1}=X_n-\frac{f(x)}{f'(x)} \)

H HallsofIvy MHF Helper Apr 2005 20,249 7,909 May 16, 2010 #3 Unless, of course, you mean \(\displaystyle \sqrt[130]{130}\). In that case, use Newton's method with \(\displaystyle f(x)= x^{130}- 130\). Reactions: mathaddict

Unless, of course, you mean \(\displaystyle \sqrt[130]{130}\). In that case, use Newton's method with \(\displaystyle f(x)= x^{130}- 130\).