Newton's Algorithm method??? Help Please...?

Question

c(sub n+1)=c(sub n) - f(c(sub n))/f'(c(sub n)) is the formula.

The Problem
Approximate the root of y=x^3+x^2-5, using c(sub 0)=1 as the initial guess.

I know the concept of repeating this process until the approximation becomes closer and closer, but I am confused as to what values you plug in where in the formula. Thanks for your help.

and where you get the values from...

Haha. Yeah, thanks. I just found it and know where the values go now. Thanks.

Answer 1

Check out wikipedia. It's got a sketch of the method and will definitely help you:

http://en.wikipedia.org/wiki/Newton's_method

Answer 2

Newton's method is used to solve f(x) = 0.

Here, f(x) = x³ + x² - 5

So, f '(x) = 3x² + 2x

and Newton's formula is

c[n+1] = c[n] - ( c[n]³ + c[n]² - 5 ) / ( 3c[n]² + 2c[n] )

starting with c[0] = 1

So, substitute n=0 to get

c[1] = c[0] - ( c[0]³ + c[0]² - 5 ) / ( 3c[0]² + 2c[0] )

= 1 - ( 1³ + 1² - 5 ) / ( 3 × 1² + 2 × 1 )
= 1.6

Now use n=1 and put 1.6 into right side and find value to get c[2] = 1.44779... , then c[3] = 1.4335 ... . c[5] is accurate to 15 decimals.

You can check your answer using WIMS. Choose Online Calculators then Function Calculator. Enter x^3+x^2-5 and choose "curve of f" and "search for roots" between 1 and 2 and you can get the root to a large number (200) of decimals.

1.4334276638638198085767292449690246079953341717726396083341001213866745413973842083273169403061057006279297022623406124991463397597060654661611827523551440974131389484091037892927925165296381987697646