Can All Functions Be Explicitly Inverted?

[eljose]

Let,s suppose we want to obtain the inverse of the functions:

y=\frac{sin(x)}{x} y=cos(x)+x or y=\int_{c}^{x}dt/logt

as you can check you can,t explicitly get g(y)=x from y=f(x)..then how would you manage to get it?..i have heard about Lagrange inverse series theorem to invert a series..but what happens if the function is not analytic on the whole real line?..for example includes terms in the form |x|, lnx, 1/x or x^{r} with r a real number.

 

[HallsofIvy]

Most of the functions you mention do not have inverses defined for all real numbers. How are you restricting them so that they do?

 

[tim_lou]

you can use some sorts of integrals to find inverse functions.
1)
y'=\frac{xcosx-sinx}{x^2}
let g be the inverse function,
\frac{dg}{dx}=\frac{1}{y'(y)}
substitute y=\frac{sinx}{x}

integrate it and get g... well its going to be messy. however, as hallsofivy said, these functions have restrictions... so be careful.

 

[matt grime]

Let,s suppose we want to obtain the inverse of the functions:

y=\frac{sin(x)}{x} y=cos(x)+x or y=\int_{c}^{x}dt/logt

as you can check you can,t explicitly get g(y)=x from y=f(x)..then how would you manage to get it?..i have heard about Lagrange inverse series theorem to invert a series..but what happens if the function is not analytic on the whole real line?..for example includes terms in the form |x|, lnx, 1/x or x^{r} with r a real number.

You are confusing existence with, well, who knows what, evaluation perhaps. *If* a function has an inverse (i.e. if it is bijective) then its inverse is what it is. Constructing the value of the inverse evaluated at any given point, or expressing it in a nice elementary function form is strictly a different issue from saying you can't get its inverse explicitly (you always can get its inverse explicitly...).