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

  • Thread starter ryan750
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In summary, the iteration Xn+1 = Xn(2 - NXn) can be used to find the reciprocal of N. To make it work, the initial guess X0 should be roughly 1/n. This can be demonstrated through an Excel spreadsheet by playing around with different values of N and X0. The formula works by rearranging to N ~= 1/Xn, which is a good estimate for the reciprocal.
  • #1
ryan750
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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.
 
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  • #2
What is the fixed point of the iteration?
 
  • #3
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.
 
  • #4
jdavel said:
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.
 
  • #5
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.
 

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

1. How does the equation Xn+1 = Xn(2 - NXn) find the reciprocal?

The equation Xn+1 = Xn(2 - NXn) is a recursive formula that uses the previous value of Xn to calculate the next value, which is then used to calculate the next value, and so on. This process continues infinitely, approaching the reciprocal of N as the value of Xn approaches infinity.

2. What is a recursive formula?

A recursive formula is a mathematical expression that uses previous values to calculate the next value. In the case of Xn+1 = Xn(2 - NXn), the value of Xn is used to calculate the value of Xn+1, which is then used to calculate the value of Xn+2, and so on.

3. How do you use Xn+1 = Xn(2 - NXn) to find the reciprocal of a number?

To use Xn+1 = Xn(2 - NXn) to find the reciprocal of a number, simply plug in the value of N for Xn and then repeat the process, using the new value of Xn as the input for the next calculation. As the number of iterations increases, the value of Xn approaches the reciprocal of N.

4. What is the significance of the 2 in the equation Xn+1 = Xn(2 - NXn)?

The 2 in the equation Xn+1 = Xn(2 - NXn) represents the denominator of the reciprocal. As the value of Xn approaches infinity, the 2 becomes less and less significant, and the value of Xn+1 approaches the reciprocal of N.

5. Are there any limitations to using Xn+1 = Xn(2 - NXn) to find the reciprocal?

Yes, there are limitations to using Xn+1 = Xn(2 - NXn) to find the reciprocal. This equation only works for positive numbers, and the value of Xn must be greater than 1. Additionally, the number of iterations required to accurately calculate the reciprocal may be large and increase as the value of N becomes larger.

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