# Numerical Analysis

• Oct 10th 2007, 04:45 AM
taypez
Numerical Analysis
I am trying to find the positive root to:

1-4xcos(x)+2x^2 + cos(2x)=0.

I've tried graphing the function and separating into 2 separate functions to find an intersection, but I'm not finding that it has a positive root.
Any NA root finding method can be used to find the root.

Any ideas or clues to what I may be doing wrong?

Thanks.
• Oct 10th 2007, 05:02 AM
CaptainBlack
Quote:

Originally Posted by taypez
I am trying to find the positive root to:

1-4xcos(x)+2x^2 + cos(2x)=0.

I've tried graphing the function and separating into 2 separate functions to find an intersection, but I'm not finding that it has a positive root.
Any NA root finding method can be used to find the root.

Any ideas or clues to what I may be doing wrong?

Thanks.

If it has roots they are all near 0.7391, and it could well be a multiple root.

RonL
• Oct 10th 2007, 05:06 AM
taypez
Can you give me a hint as to how you came up with this approximation?
• Oct 10th 2007, 05:09 AM
topsquark
Quote:

Originally Posted by taypez
Can you give me a hint as to how you came up with this approximation?

Are you sure you graphed it?

You could try an interval search (estimate where the root is and whittle this interval down by successive approximations) or something like the Newton-Raphson method.

My calculator is coming up with two very close roots near where CaptainBlack estimated one at.

-Dan
• Oct 10th 2007, 10:56 AM
CaptainBlack
Quote:

Originally Posted by topsquark
Are you sure you graphed it?

You could try an interval search (estimate where the root is and whittle this interval down by successive approximations) or something like the Newton-Raphson method.

My calculator is coming up with two very close roots near where CaptainBlack estimated one at.

-Dan

Please tell us how close, I get x=0.73908517, is a root to as high precission as I can get, and at the same time a minima.

RonL
• Oct 11th 2007, 05:26 AM
topsquark
Quote:

Originally Posted by CaptainBlack
Please tell us how close, I get x=0.73908517, is a root to as high precission as I can get, and at the same time a minima.

RonL

Specifically I get
x = 0.7390851382956
x = 0.7390851224003

I didn't solve it in the graphing mode, I made the calculator figure it out numerically. (It's a TI-92, I don't know what algorithm it uses to solve the equations.)

-Dan