Performing symbolic math steps is often related to pattern recognition. In theory, ChatGPT could be doing a good job here. I wanted to find the inverse of \[f(x) = {\mathrm{sign}}(x) \log(1+|x|)\] This function is a form of a signed logarithmic scaling. So, let's see what ChatGPT is telling us:

The idea to split this into two branches is excellent. Strictly speaking, we should handle \(x=0\) also in one of the branches, but let's ignore that for now. More worisome, somehow, it looses the \({\mathrm{sign}}(x)\) function in step 2 (needed for the \(x\lt 0\) branch). So, let's inform ChatGPT about this:

It says, "Let's include the \({\mathrm{sign}}(x)\) factor," but then it does not actually do that.

Doing the inverse by hand yields: \[f^{-1}(x) = {\mathrm{sign}}(x)(\exp(|x|)-1)\]

Hmm. ChatGPT does not really understand much of my feedback. It is rather polite, but it is just ignoring what I say.

### Update 1: Wolfram Alpha

inverse of sign(x)*log(1+|x|)

- The inverse function is incorrect
- and the blue line is not the plot of \(e^x-1\).

### Update 2: Sage

### Update 3: Google Gemini

### Conclusion

### References

- https://www.wolframalpha.com/
- ChatGPT Gets Its “Wolfram Superpowers”!, https://writings.stephenwolfram.com/2023/03/chatgpt-gets-its-wolfram-superpowers/
- https://sagecell.sagemath.org/

The details... It knows the overall picture, but when it gets to the actual steps it completely messes up. Confidently. Like the student who turns in a very clean-looking work which is utter nonsense. If you just glance at it you may think that the student is good and surely just messed up a sign somewhere. But if you read the work carefully you realize he has no clue.

ReplyDeleteI think ChatGPT is useless for mathematics.

For the record, any decent symbolic math package could do that 30 years ago.

A comment on Wolfram's software: Generally better is to use the "Reduce"-function for such inverses, which should consider all cases. The current version of Wolfram Alpha yields with input "Reduce[ y == Sign[x] Log[1 + Abs[x]], x, Reals ]" indeed only the result you mention above. But Wolfram Mathematica (the non-free software; ver 12.1) yields correctly: (y < 0 && x == 1 - E^-y) || (y == 0 && x == 0) || (y > 0 && x == -1 + E^y).

ReplyDelete