Question

What do I need to do to the equation
sqrt(x^2-x+1)=2sin(pix)
to be able to apply Newton's Method?

Answer 1

write it as
\[f(x) = 2 \sin(\pi x) - (x^2-x+1)^{\frac{1}{2}} \]

and solve using
\[ x_{n+1} = x_n - \frac{f(x)}{f'(x)} \]
you will need a rough idea of where the root(s) are. i.e a guess for x0

Answer 2

x^2-x+1 is a "cup shaped" parabola with a minimum value at -b/(2a) = 1/2
and so sqrt( ) will have a min value at 0.707

so if it intersects the sin, it will be when the sin is positive, between 0 and pi
I would use x0=0 and x0= pi as the starting values, and expect to find 2 roots.

Answer 3

Okay. Thank you very much for the help!