Just for kicks and giggles, I'm going to define a function f ( r ) such that f ( r ) is the area of a circle with radius r.
And, I'm going to define it in terms of an integral
The details are here
(other such sites are also available)
It turns out that f ( r ) = pi r^2
Now, can I find an inverse for this function which is defined in terms of an integral (of course it's an integral- its the area of a curve)
Well, yes I more or less can ( if we decide that either r in always >0 or that a negative radius is permitted as an alternative to a positive one)
I can calculate the radius ( r ) of a circle, given the area (a)
So, in this (trivial) case I can find an inverse fro a function that is defined in terms of an integral.
(It's entirely possible that someone better at maths than I cam could find more exciting examples but this one will do)
OK, That was easy, so I will now try something a bit more complex.
I will define a function, g, of two variables x and y such that g(x,y) = x+y,x+y
Sorry for my clumsy explanation- what I mean is the equivalent of multiplication by the matrix
That matrix has the not very rare property of having no inverse.
Now imagine that I define a further function h(x,y) in such a way that it depends on an integral of g(x,y). It doesn't have to be anything clever.
The integral of g(x,y) dy,dx over some range will do nicely.
Now I contend that since the calculation of the intermediate i.e. g(x,y) is not invertible, the integral of that function also isn't invertible.
So, in this case it is not possible to invert a function defined in terms of an integral.
So, the answer to the question "Is there a way to invert a function defined by an integral?" is
You can if it's f(x), but you can't if it's h(x,y)
Incidentally, the choice of a function in two variables makes Fiveworld's solution a lot more interesting- it would need a 2 D array, but not all 2 d arrays have an inverse.
If you had thought about what he said- rather than discounting it rudely, you would have got to at least part of the answer.
(It's possible that someone is going to point out that I'm playing fast and loose with the definition of "function" here by sticking arrays into it. OK, fair cop, but the OP started it by introducing integrals)
Edited by John Cuthber, 20 April 2017 - 07:21 PM.
What's this signature thingy then? Did you know Santa only brings presents to people who click the + sign? -->