3 ms·
I couldn't find any information on this, but is it possible that given how nicely exponentiation and logarithms differentiate and integrate, is it possible that
by mmastrac 6mo ago
I couldn't find any information on this, but is it possible that given how nicely exponentiation and logarithms differentiate and integrate, is it possible that this operator may be useful to simplify the process of finding symbolic solutions to integrals and derivatives?
- gus_massa 6mo agoIt transform a simple expression like x+y into a long chain of "eml" applications, so: Derivatives: No. Exercise: Write the derivative of f(x)=eml(x,x) Integrals: No. No. No. Integrals of composition are a nightmare, and here they use long composition chain like g(x)=eml(1,eml(eml(1,x),1)).
- hanneshdc 6mo agoAgreed on integrals, but the derivative is relatively simple? If f(x) = exp(x) - ln(x) then f’(x) = exp(x) - 1/x, which is representable in eml form as well. To the overall point though, I don’t think it helps make derivatives easier though. To refactor a function to eml’s is far more work than refactoring into something that’s trivially differentiable with the product rule and chain rule.
- gus_massa 6mo agoYou mean f'(x) = eml(x,x) + eml(1,eml(eml(1,x)),1) + eml(eml(1, exp(eml(1, 1))),-eml(1, eml(eml(1, x))),1) and I still have to macroexpand a few x-y = eml(eml(1, exp(eml(1, x))), eml(y,1)) but I got really bored