Possible proof of Gabriel's Theorem?
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From: David C. Ullrich on 10 Mar 2005 09:25 On 9 Mar 2005 18:57:37 -0800, "Jason" <logamath(a)yahoo.com> wrote: >> Uh, right. The integral of f involves riemann sums involving f. >> The integral of g would involve riemann sums that have g in them. >> I wonder what the riemann sums approximating the integral of >> f' would look like? Hint: they have f' in them. > >The mean value theorem has f and the integral has f'. >Hint: Gabriel's theorem links the two! I don't know why you're still talking about Gabriel's "theorem". It's simply _false_ - there have been at least two counterexamples posted in the last day or so. >Huh? > >Jason Wells ************************ David C. Ullrich
From: William Hughes on 10 Mar 2005 19:51 Jason wrote: > William Hughes wrote: > > Jason wrote: > > > No. That's where the definition is a load of non-sense because the > > > function is integrated > > > over the entire interval. This means that delta must eventually be > > > zero. > > > > Poppycock. There is nothing in the definition that says > > that delta must eventually become zero, however, according to the > > definition constant functions are integrable. > > > > "the function is integrated over the entire interval" > > is nonsense. What is the difference between integrating > > over an interval and integrating over an "entire interval" ? > > > > > > -William Hughes > > When you integrate over an interval, there are *no holes* in the > interval, therefor delta is eventually zero regardless of what > the definition says. In fact we are only interested in the result > once delta is zero, else we do not obtain the correct result. More drivel. What are "holes"? Why does making delta 0 get rid of them? To repeat: What is the difference between integrating over an interval and integrating over an "entire interval" ? > > The problem with *epsilon-delta* logic is that its not supposed > to use *infinitesimals* but every epsilon-delta argument does > exactly this: "Give me any delta/epsilon and I can always find a > smaller delta/epsilon greater than zero..." No. Every epsilon-delta argument has the form: Given an epsilon > 0 I can find a delta > 0 such that if foo < delta, bar < epsilon. Finished. Two real numbers greater than 0. No infinitesimals. Nothing indeed about epsilon being small. We only talk about epsilon being small when discussing the meaning/motivation of the argument. We say: because the argument works with any epsilon, it must work with any arbitrarily small epsilon. But by arbitrarily small we mean -given any real number t>0 we can find an epsilon (possibly different for each t) such that epsilon < t We do not mean -we can find an epsilon > 0 (not dependent on t) such that given any real number t>0, epsilon < t (Indeed, in the second case epsilon must be an infinitesimal. However, we don't go there.) Note again: There is nothing about epsilon being small in the original argument. The argument states that epsilon is arbitrary as long as it is a real number greater than 0. There is no mention of infinitesimal. > But this is what an > infinitesimal was initially defined to be: a number greater than zero > but less than every positive number. Correct (almost. replace "every positive number" with something like "1/N for every integer N", infinitesimals can be positive and they are numbers) but irrelevant. Epsilon-delta arguments do not talk about "a number greater than zero but less than every positive number" <more drivel snipped> -William Hughes
From: Jason on 10 Mar 2005 20:46 > No. Every epsilon-delta argument has the form: > > Given an epsilon > 0 I can find a delta > 0 > such that if foo < delta, bar < epsilon. > Finished. Two real numbers greater than 0. No infinitesimals. > Nothing indeed about epsilon being small. What if my epsilon is infinitesimal - can you find another smaller epsilon? Answer is no. > (Indeed, in the second case epsilon must be an infinitesimal. > However, we don't go there.) Exactly because you are talking rubbish and you know it. > Note again: There is nothing about epsilon being small in the > original argument. The argument states that epsilon is arbitrary > as long as it is a real number greater than 0. There > is no mention of infinitesimal. Of course there is no talk of epsilon being small because then the definition would not work, would it? > > But this is what an > > infinitesimal was initially defined to be: a number greater than zero > > but less than every positive number. > Correct (almost. replace "every positive number" with something like > "1/N for every integer N", infinitesimals can be positive and > they are numbers) but irrelevant. Epsilon-delta arguments do not talk > about "a number greater than zero but less than every positive number" Bullshit! You don't need to replace anything. It's perfectly correct and far easier to understand than to say what you have said. See, you are a typical product of a real analysis brainwashing class. You are not even able to state what you learned correctly. Just look at the above paragraph - it is filled with errors! You tried to be smart but you are showing how stupid you really are. Jason Wells
From: W. Dale Hall on 10 Mar 2005 21:49 Jason wrote: >>No. Every epsilon-delta argument has the form: >> >> Given an epsilon > 0 I can find a delta > 0 >> such that if foo < delta, bar < epsilon. >>Finished. Two real numbers greater than 0. No infinitesimals. >>Nothing indeed about epsilon being small. > > > What if my epsilon is infinitesimal - can you find another smaller > epsilon? Answer is no. > This is a non-issue: there is no infinitesimal real number. The difference of any two real numbers is real, and the difference of a larger minus a smaller number is a positive real, not an infinitesimal. If you care to dispute that fact, perhaps you'd be kind enough to display an infinitesimal real number. It'll be easy to test, since any infinitesimal A satisfies the following condition: n*A < 1. for all positive integers n. ... the rest deleted ... > > Jason Wells > Dale.
From: William Hughes on 10 Mar 2005 21:52
Jason wrote: Naughty, naughty. You snipped a whole bunch of stuff without indicating this fact. How about going back and answering the questions. Stop ducking the question: What is the difference between integrating over an interval and integrating over an "entire interval? > > No. Every epsilon-delta argument has the form: > > > > Given an epsilon > 0 I can find a delta > 0 > > such that if foo < delta, bar < epsilon. > > Finished. Two real numbers greater than 0. No infinitesimals. > > Nothing indeed about epsilon being small. > > What if my epsilon is infinitesimal - can you find another smaller > epsilon? Answer is no. One: epsilon is not and cannot be an infinitesimal unless you use a definition under which the real numbers contain infinitesimals. This is not the standard definition (and is not necessary, the whole point about epsilon-delta is to avoid infinitesimals). Two: even if epsilon is an infinitesimal it is easy to find a smaller epsilon, try epsilon/2. (You can do epsilon-delta stuff using infinitesimals, the point is that you don't have to.) Naughty again. Not only did you cut out an important part of the argument without any indication; you cut out important context. > > > (Indeed, in the second case epsilon must be an infinitesimal. > > However, we don't go there.) > > Exactly because you are talking rubbish and you know it. > > > Note again: There is nothing about epsilon being small in the > > original argument. The argument states that epsilon is arbitrary > > as long as it is a real number greater than 0. There > > is no mention of infinitesimal. > > Of course there is no talk of epsilon being small because then the > definition would not work, would it? > What definition? Epsilon-delta arguments work just fine. There is nothing about epsilon being small because there is no need. Epsilon is arbitrary, becaue there is a need. > > > But this is what an > > > infinitesimal was initially defined to be: a number greater than > zero > > > but less than every positive number. > > Correct (almost. replace "every positive number" with something like > > "1/N for every integer N", infinitesimals can be positive and > > they are numbers) but irrelevant. Epsilon-delta arguments do not > talk > > about "a number greater than zero but less than every positive > number" > > Bullshit! You don't need to replace anything. An infinitesimal cannot be "a number greater than zero but less than every positive number." because an infinitesimal is a positive number. A number, even an infinitessimal, cannot be less than itself. - "William Hughes" > It's perfectly correct > and far easier to understand than to say what you have said. See, you > are a typical product of a real analysis brainwashing class. You are > not even able to state what you learned correctly. Just look at the > above paragraph - it is filled with errors! You tried to be smart but > you are showing how stupid you really are. > > Jason Wells |