0
$\begingroup$

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.

1 Answers 1

1

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.

  • 0
    OK, but if one insists that the coefficients be given in closed form....2012-12-24