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Simplify[ ] affects InverseFunction[ ]?

Posted 3 years ago

Hi, I'm new here and new to Mathematica also.

So, I was defining two functions, s1 and s2. Both are the same functions but one was simplified after the integration and the other one wasn't.

So I calculate the inverse of these functions (and once they are the integral of the same function, I expect that the inverses would return the same values) and that is not happening.

I would like to understand why the Simplify cause the inverse equations to have different solutions.

I'm adding a notebook to demonstrate my question:

If anyone can help, I really would appreciate it Thank you very much in advance .

POSTED BY: Andre Andrade
3 Replies

Wow, thank you, I hadn't noticed that I have been using Mathematica for 30 years now!

POSTED BY: Gianluca Gorni
Posted 3 years ago

Thank you for your quick and precise response!

I really will be more careful about the use of the InverseFunction and certainly need to study more about complex numbers and functions.

Once again, thank you very much.

BTW: your website is great. I'm discovering a lot of interesting content there. And I only can imagine the experience you have acumulated using this amazing software for so long (since 1991, right?).

POSTED BY: Andre Andrade

The culprit is PowerExpand, not Simplify. The two forms s1 and s2 are not totally equivalent. Also, you should expect trouble when inverting a function that is not invertible over the complexes:

Solve[s1[a, b, tt] == x, tt]

You will see that there are two branches. It seems that InverseFunction chooses the first one, which involves the logarithm of a negative number (if a>0). With s2 the formulas are different:

Solve[s2[a, b, tt] == x, tt]

and InverseFunction makes its choice.

The output of InverseFunction should be carefully sanity-checked each time.

POSTED BY: Gianluca Gorni
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