# Newton Raphson Method?????? HELP!

• Mar 5th 2007, 07:12 AM
bobchiba
Newton Raphson Method?????? HELP!
This is the last question on some tutorial sheets im going through and Im completely lost on this last question....

1) "Show that the problem of finding the cube root of 70 is equivalent to finding a root of (x^3 - 70 = 0). Approximate this root using the Newton Raphson Method. Choose 4 as the initial estimate. Continue the approximation until two successive approximations, rounded to 3 decimal places, are the same."

I will be 1) seriously impressed and 2) very gratefull if anyone can explain this to me.

Thaaaaaanks
• Mar 5th 2007, 08:20 AM
Aryth
Quote:

Originally Posted by bobchiba
This is the last question on some tutorial sheets im going through and Im completely lost on this last question....

1) "Show that the problem of finding the cube root of 70 is equivalent to finding a root of (x^3 - 70 = 0). Approximate this root using the Newton Raphson Method. Choose 4 as the initial estimate. Continue the approximation until two successive approximations, rounded to 3 decimal places, are the same."

I will be 1) seriously impressed and 2) very gratefull if anyone can explain this to me.

Thaaaaaanks

Well, we know that the root of a function f(x) is the zero of it's function, so we set f(x) = x^3 - 70 equal to zero, as you already have:

x^3 - 70 = 0

Now we add 70 to both sides to single out x.

x^3 = 70

Now we take the cube root of both sides:

x = cubert(70)

Thus x = the cube root of 70.

Now the Newton Raphson Method states this:

X_n+1 = X_n - f(X_n)/f'(X_n)

Where X_n+1 = first approximation and X_n is the initial estimate and f'(X_n) is the derivative of the function at the initial estimate. Since the initial estimate is 4 we just plug in and solve.

X_n+1 = 4 - (4^3 - 70)/(3*4^2)

The derivative of x^3 -70 is 3x^2 so 3(4)^2 checks out.

X_n+1 = 4.125

The cube root of 70 = 4.1213

So they are close.

If we keep using the formula, the approximations are as follows:

4.12128
4.12129

And there you go. Two consecutive approximations that are exact to 3 decimal places. If you would like to calculate and verify you may. This also seems to converge right onto the square root of 70, which is the root of the original function f(x) = x^3 - 70

I hope this helps.
• Mar 5th 2007, 08:26 AM
Jhevon
Quote:

Originally Posted by bobchiba
This is the last question on some tutorial sheets im going through and Im completely lost on this last question....

1) "Show that the problem of finding the cube root of 70 is equivalent to finding a root of (x^3 - 70 = 0). Approximate this root using the Newton Raphson Method. Choose 4 as the initial estimate. Continue the approximation until two successive approximations, rounded to 3 decimal places, are the same."

I will be 1) seriously impressed and 2) very gratefull if anyone can explain this to me.

Thaaaaaanks

First we show that finding the cube root of 70 is equivalent to finding a root of (x^3 - 70 = 0).

Since we don't know what the cube root of 70 is, we will use a variable, say x, to represent it.

Let x = cube root of 70. That is x = 70^(1/3). Raising both sides to the third power we obtain:

x^3 = 70
=> x^3 - 70 = 0 .............subtratced 70 from both sides. So finding x (which is the cube root of 70) is the same as finding the root for this equation

Now, The Newton Raphson Method says: given any estimate, called x_n, of the root of some function f(x), we can find a better approximation, called x_n+1 by using the formula:

http://www.shodor.org/UNChem/math/newton/newton.gif

here f(x) = x^3 - 70
=> f '(x) = 3x^2

We are given the first estimate x_n = 4. By the Newton Raphson Method, a better estimate, x_n+1 is given by:

x_n+1 = 4 - f(4)/f '(4) = 4 - (-6)/48 = 4 + 6/48 = 4.125

Now using a 4.125 as x_n, we can find a better estimate

x_n+1 = 4.125 - f(4.125)/f '(4.125) = 4.125 - 0.1895/50.0469 = 4.12129

Now using 4.12129 as x_n, we can find a better estimate

x_n+1 = 4.12129 - f(4.12129)/f '(4.12129) = 4.12129 - (-0.0145)/50.948 = 4.12128

Now using 4.12128 as x_n we can find a better estimate

x_n+1 = 4.12128 - f(4.12128)/f '(4.12128) = 4.121

We notice for the last 3 approximations, the answer rounded off to 3 decimal places is 4.121, so we take this as the cube root of 70.

Note, this is just to get you used to the steps, i rounded off decimal points during some internediate calculations, so the values aren't exact
• Mar 5th 2007, 08:37 AM
Soroban
Hello, bobchiba!

Quote:

1) Show that the problem of finding the cube root of 70
is equivalent to finding a root of: .x³ - 70 .= .0

Approximate this root using the Newton-Raphson Method.
Choose 4 as the initial estimate.
Continue the approximation until two successive approximations,
rounded to 3 decimal places, are the same.

Are you familiar at all with the Newton-Raphson Method?

. . . . . . . . . . . .f(x1)
. . x2 . = . x1 - -------
. . . . . . . . . . . f '(x1)

Select x1 as the first estimate of the root
. . Substitute it into the formula.
Then x2 is a better approximation of the root.

We have: .f(x) .= .x³ - 70 .and .f'(x) .= .3x²

. . . . . . . . . . . . . . . . . . . . . x³ - 70
Our formula is: . x2 . = . x1 - ---------
. . . . . . . . . . . . . . . . . . . . . . 3x²

. . . . . . . . . .4³ - 70
. . x2 .= .4 - ---------- .= .4.125
. . . . . . . . . . .3·4²

Now try: .x2 = 4.125

. . . . . . . . . . . . .4.125³ - 70
. . x3 .= .4.125 - -------------- .= .4.121288644
. . . . . . . . . . . . . .3·4.125²

Now try:. x3 = 4.12129

. . . . . . . . . . . . . . 4.12129³ - 70
. . x4 .= .4.12129 - ----------------- .= .4.1212853
. . . . . . . . . . . . . . . 3·4.12129²

The last two approximations are 4.121 to three decimal places.
. . We can stop.

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Check: .4.121³ .= .69.985... . Pretty close!