i need to prove that sin(x)-x=0 has a unique solution in the open interval (0,pi/2). indicate a computational procedure for finding that solution approximately. on the previous parts of the problem i proved that every contraction mapping is continuous and that every contraction on a complete metric space has a unique fixed pt, so we can use these facts. my intuition is that we'll need to set f(x) = sin x and use the contraction mapping idea, but it's hard to prove that sin x is a contraction mapping. help!!!
Here is why
so the derivative is always negative on the above interval.
Now by the Mean Value Theorem we have
let a=0 and b=x
since the derivative is always negative and x is positive we know that f(x) is negative on the interval i.e f(x) < 0
so finally we get
by the last they they are never equal, so no solution exits.
I hope this helps.