# [SOLVED] Newton-Raphson method

• Aug 7th 2008, 06:40 PM
sinewave85
[SOLVED] Newton-Raphson method
Ok, I need to solve this:

$0 = \frac{(\sqrt {1 - x^2}) \arccos{x} - (1 + x^2) \arctan{x}}{(1 + x)(\sqrt{1 - x^2})}$

by the Newton-Raphson method.

So do I take the derivative of

$0 = \frac{(\sqrt {1 - x^2}) \arccos{x} - (1 + x^2) \arctan{x}}{(1 + x)(\sqrt{1 - x^2})}$

or of

$0 = (\sqrt {1 - x^2}) \arccos{x} - (1 + x^2) \arctan{x}$ ?

Please say it's the latter. If I have to take it on the first, I think I will go mad.
• Aug 7th 2008, 06:50 PM
Coffee Cat
Quote:

Originally Posted by sinewave85
Please say it's the latter. If I have to take it on the first, I think I will go mad.

Unfortunately, you have to take the derivative of the larger, longer term.

$f(x) = \frac{(\sqrt {1 - x^2}) \arccos{x} - (1 + x^2) \arctan{x}}{(1 + x)(\sqrt{1 - x^2})}$

And you do not assume that f(x) = 0 at the start, because it contradicts the Newton-Raphson method.
• Aug 7th 2008, 06:52 PM
sinewave85
Quote:

Originally Posted by Coffee Cat
Unfortunately, you have to take the derivative of the larger, longer term.

$f(x) = \frac{(\sqrt {1 - x^2}) \arccos{x} - (1 + x^2) \arctan{x}}{(1 + x)(\sqrt{1 - x^2})}$

And you do not assume that f(x) = 0 at the start, because it contradicts the Newton-Raphson method.

*groans* Ok, thanks anyway.
• Aug 7th 2008, 06:59 PM
ticbol
Quote:

Originally Posted by sinewave85
Ok, I need to solve this:

$0 = \frac{(\sqrt {1 - x^2}) \arccos{x} - (1 + x^2) \arctan{x}}{(1 + x)(\sqrt{1 - x^2})}$

by the Newton-Raphson method.

So do I take the derivative of

$0 = \frac{(\sqrt {1 - x^2}) \arccos{x} - (1 + x^2) \arctan{x}}{(1 + x)(\sqrt{1 - x^2})}$

or of

$0 = (\sqrt {1 - x^2}) \arccos{x} - (1 + x^2) \arctan{x}$ ?

Please say it's the latter. If I have to take it on the first, I think I will go mad.

On your two options, I want to say that you have to take it on the former...except that the the LHS should be f(x) and not zero as posted...because you need the f'(x) in the Newton-Raphson method.

The second equation is not equivalent to the first equation.

But, question, is the (1 +x^2) in the numerator really not the same as the (1 +x) in the denominator?
Because if both were (1 +x^2), or both were (1 +x), then you can decompose the RHS of the first equation into two simpler fractions that might not make you mad to differentiate.
• Aug 7th 2008, 07:17 PM
sinewave85
Thanks for the help! I think figured out how to rewrite the equation to make taking the derivative a little easier. Does anyone know of a program to put the original function and the derivative into in order to do the iterations? I have been trying to do them by a combination of pencil/paper and graphing calculator, but that is getting really complicated.
• Aug 7th 2008, 07:19 PM
sinewave85
Quote:

Originally Posted by ticbol
On your two options, I want to say that you have to take it on the former...except that the the LHS should be f(x) and not zero as posted...because you need the f'(x) in the Newton-Raphson method.

The second equation is not equivalent to the first equation.

But, question, is the (1 +x^2) in the numerator really not the same as the (1 +x) in the denominator?
Because if both were (1 +x^2), or both were (1 +x), then you can decompose the RHS of the first equation into two simpler fractions that might not make you mad to differentiate.

Yes, that was just a typo -- they are both (1 + x^2), and I put it into two simpler fractions.