The discussion centers on finding the inverse of a function that includes a floor function. Participants explore the challenges posed by the floor function's properties and consider various approaches to express the inverse, including the use of periodicity and intervals.
Participants express differing views on the existence of an inverse for functions involving the floor function. While some argue that certain forms can be inverted on specific intervals, others maintain that the floor function's properties prevent a global inverse from existing.
The discussion highlights limitations related to the definitions of functions and inverses, particularly in the context of periodicity and the properties of the floor function. There are unresolved mathematical steps regarding the generalization of inverses for functions containing the floor function.
This discussion may be useful for those interested in mathematical functions, particularly in the context of inverses, periodic functions, and the implications of non-one-to-one mappings in mathematical analysis.
Simplify it as g(x) = u - floor(u). This function is then periodic with period 1, which makes your function periodic with period T, and thus it has no global inverse. rho just moves the function around horizontally and A scales it vertically. So your function is invertible on any interval of length T, and you can add z*T to the inverse to get another solution, where z is some integer. A simple inverse for g is g-1(v) = v, where the interval is [0, 1). For your function, you would choose an interval like [0, T) and the inverse on this interval would just be the inverse of f(x) = A(x/T + rho) which is f-1(y) = T(y/A - rho).lewis198 said:This is what the equation looks like:
f(x)=A.[(x/T+rho)-floor(x/T+rho)]
I need to make x the subject of the equation, A,T and rho are constants.