Re: Bisection method: your recommendation

On 5 Feb, 15:04, Randy Poe <poespam-t...@xxxxxxxxx> wrote:
On Feb 5, 9:40 am, shalini.sing...@xxxxxxxxx wrote:





I need to solve this equaation:

(x1cos(p)+y1sin(p))^2 * (x2sin(p)-y2cos(p)) = (x1sin(p)-y1cos(p)) *
(x2cos(p)+y2sin(p))^2 ;

x1,x2,y1,y2 are constants

The equation would have 3 roots

1 real and two imaginary

I don't need imaginary roots.

I am planning to use bisection method-----

since i know the real root will lie between -2*pi and 2*pi

What is your comment on this?

Sure, why not? You might consider plotting it first
to estimate the region where the root lies.

          - Randy- Hide quoted text -

- Show quoted text -

And that would show starting at p=-2*pi and 2*pi will not work as they
give the same value (and so does p=0). Indeed, there are up to twelve
roots on this interval.

Instead start with p=0 and pi which will have opposite signs and from
one to three real roots between them.



.

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