Deriving the Divide and Average Method from Newton-Raphson Formula

  • Thread starter thedc
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In summary: This conversation is about using the Divide and Average Method to compute the square root of 2. The method is derived from the Newton-Raphson Formula, which is used to find the roots of a function. The method involves finding the zero of the function F(x), which in this case is F(x) = x^2 - 2. The iterates for the solution of x are given by the expression x_{nu+1} = x_{nu}-\frac{F(x_{nu})}{F'(x_{nu})}. The key to solving the problem is to manipulate this expression to find the square root of 2.
  • #1
thedc
6
0

Homework Statement



[Divide and average Method] Square root of 2 was computed using the formula

Xi+1 = 1/2 ( Xi + 2/Xi).------------ (1)

Derive this method from the Newton-Raphson Formula

Homework Equations





The Attempt at a Solution



Im boggled at how to derive this solution.

The equation for Newton Raphson is

F'(Xi) = (F(Xi) - 0) / Xi -(Xi+1) ------------- (2)

which can be rearranged to

Xi+1 = Xi - F(Xi) / F'(Xi)--------- (3)

does this mean that i take the derivative of the equation (1)?

(Xi+1)' =1/2(Xi+2/Xi)
= ??
 
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  • #2
thedc said:

Homework Statement



[Divide and average Method] Square root of 2 was computed using the formula

Xi+1 = 1/2 ( Xi + 2/Xi).------------ (1)

Derive this method from the Newton-Raphson Formula

Homework Equations


The Attempt at a Solution



Im boggled at how to derive this solution.

The equation for Newton Raphson is

F'(Xi) = (F(Xi) - 0) / Xi -(Xi+1) ------------- (2)

which can be rearranged to

Xi+1 = Xi - F(Xi) / F'(Xi)--------- (3)

does this mean that i take the derivative of the equation (1)?

(Xi+1)' =1/2(Xi+2/Xi)
= ??

You have
[tex]x^2 = 2[/tex]
or
[tex]x = \frac{2}{x}[/tex]
So,
[tex]f(x) = \frac{2}{x}[/tex]
Derive your equations from there.
 
Last edited:
  • #3
I still don't get it, do i take the derivative of 2/x?

that would be f(x)'=-2(1/x^2)
 
  • #4
thedc said:
I still don't get it, do i take the derivative of 2/x?

that would be f(x)'=-2(1/x^2)

What is the Newton-Raphson method?
 
  • #5
CEL said:
What is the Newton-Raphson method?

Good job guy.

/s
 
  • #6
I am having the same problem.

Newton Raphson method: Xof(i+1) = xi - f(x)/f(x)'

it is used to find roots by iteration
 
  • #7
Hidemons said:
I am having the same problem.

Newton Raphson method: Xof(i+1) = xi - f(x)/f(x)'

it is used to find roots by iteration

Write your equation in the form y = f(x).
Calculate f'(x).
Choose a starting value for x0.
If y - f(x0) < tolerance then end
else
Calculate x1 using Newton-Raphson formula.
Iterate
 
  • #8
I think there is sufficient confusion amoung these posts to warrent another (hopefully non-confusing) post

thedc: For Newton-Raphson, you are looking for the zero of a function (F), hence, you need to express the function (F) such that F(x) = 0.

In your original post, you desire to find the answer to [itex]x[/itex] for [itex]x = \sqrt{2}[/itex]. Consider the more general solution for [itex]x[/itex] with [itex]x = \sqrt{A}[/itex] for some positive [itex]A[/itex].

Question: How can we express a function, [itex]F(x)[/itex], such that it results in [itex]F(x)=0[/itex] for this problem?

Answer: Look at the [itex]x = \sqrt{A}[/itex]. This is really the same as finding [itex]x^2[/itex] such that [itex]x^2 = A[/itex]. Hence, one selection of [itex]F(x)[/itex] might be [itex]F(x) = x^2 - A=0[/itex].

This is the "F" that is needed in the N-R method. The iterates for the solution of x are as follows:

[tex]
x_{\nu+1} = x_{\nu}-\frac{F(x_{\nu})}{F'(x_{\nu})}
[/tex]

Here, [itex]F'(x)[/itex] is shorthand to mean [itex]\frac{d\,}{dx}F(x)[/itex]. Also, in your case, the vale of A is A=2. You will need an initial estimate [itex]x_{0}[/itex] to start this procedure.



The key to achieve the end goal of your exercise is to do some algebra on the resulting iterate expresion once you take the derivative of F and substitute it into the expression.

Hope this helps.
 
Last edited:

1. What is the purpose of the "Deriving the Divide and Average Method from Newton-Raphson Formula"?

The purpose of deriving the divide and average method from the Newton-Raphson formula is to provide a more efficient and accurate method for finding roots of a polynomial equation. This method involves dividing the polynomial by its derivative and averaging the resulting values to get closer and closer to the root.

2. How does the Divide and Average Method improve upon the Newton-Raphson Formula?

The Divide and Average Method improves upon the Newton-Raphson formula by reducing the number of iterations needed to reach the root. It also helps to avoid convergence issues that can occur with the original formula.

3. What is the mathematical derivation of the Divide and Average Method?

The mathematical derivation of the Divide and Average Method involves substituting the original Newton-Raphson formula into itself and simplifying the resulting expression. This leads to the division of the polynomial by its derivative and the averaging of the resulting values.

4. Can the Divide and Average Method be applied to all types of polynomial equations?

Yes, the Divide and Average Method can be applied to all types of polynomial equations as long as they have a single real root. It is especially useful for complex and high-degree polynomials.

5. Are there any limitations or drawbacks to using the Divide and Average Method?

One limitation of the Divide and Average Method is that it may not work for polynomials with multiple roots or roots that are close together. It can also be more computationally expensive compared to other root-finding methods. Additionally, the method may not converge if the initial guess is too far from the root.

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