Let's consider set of all functions that can be expressed in quadratures (not sure if 'quadratures' is right word in english. I mean functions that are built from standard functions like + - * / log, pow). I know this set is infinite uncountable size.

As far as i know, derivative of all of these functions can also be expressed in quedratures. But not every integral can.

Is set of derivatives is 'bigger' than set of integrals for both internal and result functions can be expressed in quadratures?

For example, set of natural numbers (positive integers) is of the same size as integers and even of the same size as rational numbers since we can assign unique natural number for every integer value and rational value. But the set of irrational numbers is bigger than all previous ones and it is uncountable.


It seems to me that (leaving aside the constant of integration) differentiation gives a one-one onto map from the set of closed-form functions to the set of functions that can be integrated in closed form, so the two sets have the same cardinality. Now if we take into account the constant of integration, this map becomes continuum-to-one, but if the set of closed forms is an uncountable infinity, this does not affect the conclusion of my argument.

  • $\begingroup$ Not sure if I understood you correctly. While agree that every closed-form function has one-to-one relation with closed-form derivative functions, it is not true in reverse direction: not every function that has closed-form derivative, can be integrated in a closed form. Though that doesn't mean the cardinality is really greater... The proof isn't clear enough for me =( $\endgroup$ – Sasha Dec 23 '12 at 13:56
  • $\begingroup$ I understand the main idea and it is great. But at some time of calculating every 'chain' of derivatives we'll come to some function that already was used as integral-base in another chain. For example, function 'y=0' is derivative for infinite number of functions. The same with 'y=exp(x)'. Maybe all of them are o(N) of the count of all functions and that's why you 'leave aside the constant of integration'. But it is not obvious. $\endgroup$ – Sasha Dec 23 '12 at 14:11
  • $\begingroup$ You say there is an uncountable infinity of closed forms (I'm not sure I agree, but that's another issue). In the same way that if there is a 2-to-1 map between infinite sets then there is also a 1-to-1 map, if there is a continuum-to-1 map between uncountable sets then there is also a 1-to-1 map. $\endgroup$ – Gerry Myerson Dec 23 '12 at 16:37
  • $\begingroup$ If functions didn't have queofficients, I would agree that the count of closed form functions is countable. But quoefficients make them uncountable. $\endgroup$ – Sasha Dec 24 '12 at 16:16
  • $\begingroup$ OK, but if one insists that the coefficients be given in closed form.... $\endgroup$ – Gerry Myerson Dec 24 '12 at 16:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.