Newton-Raphson question

1. Jan 4, 2005

trap

Let f(x)=x^2 - a, where a>0. The roots of the equation f(x)=0 are +root(a), -root(a).

Show that if x1 > 0 is any initial estimate for root(a), then the Newton-Raphson method gives the iteration formula

x n-1 = 1/2( xn + a/xn ),

n>=1

2. Jan 4, 2005

trap

Sorry, the last post wasn't clear, the iteration formula is..

Xn-1 = (1/2)(Xn + a/Xn) , n >= 1

3. Jan 4, 2005

learningphysics

Shouldn't this be:
Xn+1 = (1/2)(Xn + a/Xn) , n >= 1

Did you apply the newton-raphson method Xn+1=Xn-f(xn)/f'(xn) ?

4. Jan 4, 2005

trap

yes, it's Xn-1 instead, my fault
I tried the newton's method, but still have trouble solving the question

5. Jan 4, 2005

learningphysics

Can you show what you did? The given formula arrives almost immediately by applying the method.

6. Jan 4, 2005

Integral

Staff Emeritus
Newtons Method is;

$$x_{n+1} = x_n - \frac {f(x_n)} {f'(x_n)}$$

You are given

$$f(x)= x^2 - a$$

Compute $$f'(x) = 2x$$

Now apply Newtons method:
$$x_{n+1}= x_{n}- \frac {f(x_n)} {f'(x_n)} = x_n - \frac {x_n^2 - a } {2x_n} = \frac { 2 x^2_n - x^2_n +a} {2x_n}= \frac {x^2_n + a} {2x_n} = \frac 1 2 ( x_n + \frac a {x_n})$$

Last edited: Jan 4, 2005
7. Jan 4, 2005

KataKoniK

trap, who is your TA?

8. Jan 4, 2005

trap

thx for the help guys!!

9. Jan 4, 2005

KataKoniK

??? Answer me

10. Jan 4, 2005

trap

kataKonik, hi..you must be taking the same course as i am...

11. Jan 4, 2005

KataKoniK

Yes, that is why I asked you that question.

12. Jan 4, 2005

trap

well...I'm not even sure if i can use the web to help me with these problem sets..so i just wanna keep my identity as a secret

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