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Topic:
Proof of easy case of change of variables
Replies:
5
Last Post:
Jun 8, 2013 5:31 PM




Re: Proof of easy case of change of variables
Posted:
Jun 8, 2013 1:24 PM


On Fri, 7 Jun 2013 14:51:38 0700 (PDT), pepstein5@gmail.com wrote:
>On Friday, June 7, 2013 5:10:08 PM UTC+1, dull...@sprynet.com wrote: >> On Thu, 6 Jun 2013 15:00:04 0700 (PDT), pepstein5@gmail.com wrote: >> >> >> >> >In Principles of Mathematical Analysis, Rudin defines a function f from R^n to R^n as being "primitive" if, for all x, at least n1 of the coordinates of >> >> >(x  f(x)) are equal to 0. >> >> >> >> Not quite  the quantifiers are in a different order. You realy might >> >> give page numbers, btw. >> >> >> >> > (I didn't encounter anyone else using this definition when I did a google search so I'm guessing (with barely more than 50% confidence) that this usage is nonstandard). >> >> > >> >> >I'm trying to follow Rudin's proof of the change of variable formula for integration in the context of multivariable calculus. >> >> >He proves it for the case n = 1 and then asserts that it is clear for primitive functions. However, it's not clear to me. Can anyone give any pointers about how to >> >> >progress from the n = 1 case to the case of more general primitive functions? >> >> >> >> Has he proved a "Fubini" sort of theorem yet, to the effect that the >> >> integral of a function of several variables is equal to an iterated >> >> onevariable integral? Ie for n = 2, int f = int(int f(x,y) dx)dy? >> > >Thanks, David. The page numbers are as follows in the 3rd edition. For the definition of primitive functions, see 10.5 (9), page 248. >The change of variable theorem is theorem 10.9 on page 252. Rudin's version of Fubini is theorem 10.2 on page 246. >(So that's a yes to your question of whether he's done Fubini yet.) Rudin's version of Fubini is based on the Riemann integral of a > continuous function on a multidimensional cell. He proves that the order in which you do the integrations doesn't matter. However, >unlike other authors, he doesn't (at least not at the point of the text where I'm at) go on to show that, whatever order you evaluate the > integrals, the result is the same as the ndimensional measure. Indeed, at this point, only onedimensional integrals are defined, and a kdimensional integral >simply means to perform a onedimensional integral k times. > >I don't think I did put the quantifiers in the wrong order. I said "...for all x, at least n1 of the coordinates of (x  f(x)) are equal to 0.)" for the definition of being primitive. > >For all x (i.e every single relevant x or every x in R ^ n), when you do f(x), you change at most one coordinate (i.e, at least n1 of the coordinates of (x  f(x) ) are equal to 0. > >I'd be grateful if you could say a bit more about why you think my definition is wrong.
For heaven's sake. I _said_ the problem was with the order of the quantifiers. Look at the two versions of the definition, and you can _see_ that the quantifier order is different! LOOK:
(i) There exists j such that for every x: f(x)_k = x_k if k <> j.
(ii) For every x there exists j such that: f(x)_k = x_k if k <> j.
The right definition is (i). You said (ii), and you've said so again. They're different. Different because of the order of the quantifiers.
(Yes, I know that (i) is what you meant. At least I hope so. I didn't comment on what you meant...)
>Admittedly, it may have been careless to say "for all x" instead of "for all x in R ^ n" but I don't think you're pedantic so I don't think that this is your objection. > >Ok. I think I get it now. Integrate over each interval in turn, but for the first interval of integration, choose the dimension in which the coordinate of f(x) differs from x.
With (ii) what you just said makes no sense: "Choose the dimension in which the coordinate of f(x) differs from x", for _what_ x? Which dimension you're talking about varies from this x to that x.
You also need to note that, taking n = 2 to make things easier to type and to read, if
f(x,y) = (g(x,y), y)
then J_f = the partial of g with respect to x.
>The case n = 1 has been proven earlier and induction on the dimension can be used. > >Paul Epstein > > >



