Can step functions have inverse relationships?

[AJKing]

If a function s(t) exists, does a function t(s) always exist?

Are there functions with no inverse relationships?

Suppose

$s = \int^t_a e^{u^2} du$

Can there be a t(s)?

 

[jedishrfu]

Look at the trig functions for sin and cos over 0 to 2pi. They are clearly functions if inverted will map to two angles for a given sin or cos value. So the inverse is not a function.

For the sin if you restrict it 0 to pi/2 then its invertible and similarly for cos if you restrict it to 0 to pi.

http://en.wikipedia.org/wiki/Inverse_function

Try drawing a graph of e^u^2 and estimate the area under the curve and see if its invertible.

 

[Mark44]

If a function s(t) exists, does a function t(s) always exist?

Let's revise your notation a bit to make things more understandable.
Suppose y = s(t) is a function. "s" is just the name of the function that maps values of t to values of y. Many functions do not have inverses that are themselves functions. A very simple example is y = f(x) = x². Because f is not one-to-one, f does not have an inverse.

Are there functions with no inverse relationships?

Suppose

$s = \int^t_a e^{u^2} du$

Can there be a t(s)?

 

[AJKing]

A very simple example is y = f(x) = x². Because f is not one-to-one, f does not have an inverse.

Hmm, what about a solution involving step functions?

x = u₀ y^1/2-(1-u₀)y^1/2

Wolfram visual.