@Azure http://smbc-comics.com/comic/derivative do you know what the answer to this is?
@bj @Patashu @Azure i think you'd have to rigorously define "symbol" first? and make sure you can reduce anything to a standard form, because you can trivially add arbitrarily many symbols by just putting "+ 0 + 0 + 0 + 0 +..." at the end
some cases might be trickier though. like if you end up with 256x⁹-576x⁷+432x⁵-120x³+9x, do you then have to represent that as T₉(x)? What if the context it comes up in is entirely unrelated to Chebyshev polynomials?
-F
D(e^x)=e^x
D(e^(e^x))=e^(e^x) e^x, thanks to the chain rule
D(e^(e^(e^x)))=e^(e^(e^x)) e^(e^x) e^x, and so on
The derivative of an exponent-tower of e^e^e^...^x is a product of these towers starting from the original and going down to e^x. The number of symbols on the left side increases linearly, but the number of symbols on the right increases quadratically, so the higher you make your tower, the higher your ratio is, and you can make the ratio arbitrarily large.
Oh, cool! I like this, and it's pleasingly trivial, too.
And now that we can get an arbitrarily large ratio, the question now shifts to 'what's the fastest growing function', LMAO
https://tasvideos.org/Forum/Topics/2820?CurrentPage=129&Highlight=513930#513930 I got a response on tasvideos that aims to prove that quadratic is the best you can do.
@Felthry @Patashu @Azure Yeah, my solution falls apart if you introduce alternate notation for things. You have to define a dictionary of valid symbols for it to make sense. Of course, you can destroy any solution by just defining some new symbol that means “that mess your derivative evaluates to, so there”.