Right inverses

Showcase_22 · #1 $Old$ 11-19-2008, 10:28 AM

Quote:

let $f: \Re \rightarrow [-1,1]$ be defined by $f(x)=sin x$ . How many distinct right inverses acan you find?

I started with the definition of a right inverse.

$f(g(y))=y \ \forall \ y\in [-1,1]$

$sin(g(y))=y$

$g(y)=arcsin (y)$

Isn't this also it's inverse? Not only that, there is no room to find more so my answer is just one distinct right inverse.

Is something going wrong or should it turn out like this?

HallsofIvy · #2 $Old$ 11-19-2008, 12:56 PM

Quote:

Originally Posted by Showcase_22 $View Post$

I started with the definition of a right inverse.

$f(g(y))=y \ \forall \ y\in [-1,1]$

$sin(g(y))=y$

$g(y)=arcsin (y)$

Isn't this also it's inverse? Not only that, there is no room to find more so my answer is just one distinct right inverse.

Is something going wrong or should it turn out like this?

No, that's not right. sin(x) is periodic so it does not have a true inverse. If we restrict x to between $-\pi/2$ and $\pi/2$ then we have arcsin(x), the "principal" value, is its inverse. For example if f(x)= sin(x) and g(x)= arcsin(x), then $f(g(\sqrt{2}/2))= sin(\pi/4)= \sqrt{2}/2$ . But if we define $g(x)= Arcsin(x)+ \pi/2$ we have, for example, that $f(g(\sqrt{2}/2))= sin(\pi/4+ \pi/2)= sin(3\pi/4)= \sqrt{2}/2$ also. You can continue doing that by adding or subtracting integer multiples of $\pi$ .

Showcase_22 · #3 $Old$ 11-19-2008, 01:38 PM

Quote:

let $f: \Re \rightarrow [-1,1]$ be defined by $f(x)=sin x$ . How many distinct right inverses can you find?

ah yes! I forgot that it has to be a one to one function to be an inverse. So should I have done the question like this?:

$f(g(y)+2n \pi)=y \ \forall \ y\in [-1,1]$

$sin(g(y)+2n \pi)=y$

$g(y)=arcsin(y)+2k \pi$

and this would result in an infinite number of right inverses.