It's because the inverse sine is multivalued. When you attempt to evaluate your calculator will return a negative value whereas what you want is the value that is further on (if that makes sense).
The iteration that you need is
from which
Hi! I was asked to find an approximation for a root of on the interval [1, 2] using the fixed point iteration method, with p0=1.
So, I got
However, that way I just get numbers very close to zero (which is a solution for the equation, but it's not on the interval I want).
Anyway, I checked the answer on the back of the book and they used and got
I just don't know from where does that +2 come? Could anyone explain me, please?
It's because the inverse sine is multivalued. When you attempt to evaluate your calculator will return a negative value whereas what you want is the value that is further on (if that makes sense).
The iteration that you need is
from which