Xn+1 = Xn(2 - NXn) can be used to find the reciprocal

[ryan750]

can any1 explain why this iteration:

Xn+1 = Xn(2 - NXn)

can be used to find the reciprocal of N. I don't ned proof or to show that it does but i would like to know if sum1 can break it down and explain how it does it.

 

[Hurkyl]

What is the fixed point of the iteration?

 

[jdavel]

ryan750,

To elaborate on Hurkyl's question, not all initial guesses lead to the correct answer. What condition do you have to put on X0, for the iteration to work?

If you know how to use Excel, try writing a spread sheet that does this calculation, and then play around with different values for N and X0. You might see for yourself what makes this formula work. It's not too hard.

 

[ryan750]

ryan750,

To elaborate on Hurkyl's question, not all initial guesses lead to the correct answer. What condition do you have to put on X0, for the iteration to work?

If you know how to use Excel, try writing a spread sheet that does this calculation, and then play around with different values for N and X0. You might see for yourself what makes this formula work. It's not too hard.

it works for all values of N and u can use any value of Xn - but u would preferably choose a number that is royughly 1/n. So if N was 7 u would use 0.1. If n was 53 u would use 0.02.

 

[ryan750]

Xn+1 = Xn(2 - NXn)

=> (Xn+1)/Xn = 2 - NXn
=> NXn = 2 - (Xn+1)/Xn
=> NXn = (2Xn -Xn+1)/Xn
=> N = (2Xn - Xn+1)/(Xn)^2
=> N ~= Xn/(Xn)^2
=> N ~= 1/Xn

Which is a good estimate for the recipricol.

I found sum1 that could do it.

i didn't think about just rearranging the formula.