Possible proof of Gabriel's Theorem?
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From: David C. Ullrich on 9 Mar 2005 08:40 On 8 Mar 2005 15:21:44 -0800, "Jason" <logamath(a)yahoo.com> wrote: >Ullrich: >> Curious that you didn't notice the _error_ in the version I posted >then. > >Jason: >>I do not fully agree with it. As I stated earlier, I don't like to >talk about constant >>functions such as f(x) = c for c some constant as being >differentiable. >>They are not differentiable in my opinion. > >I don't pay much attention when you write incomprehensible junk. So you >made a typo - big deal. I overlooked it because it was not important. >See, this is what I mean about you: You are not interested in answering >any questions, only in deviating from the main issue and knit-picking >at stuff which really has nothing to do with the subject. You picked up >where some twit had left off. Don't you have an opinion of your own? >Show me step by step where gabriel's theorem is wrong or shut up. > >I meant that a delta > 0 does not exist, therefore we don't even need >to look for an epsilon. > >since length(I_j) < delta and delta is eventually zero, no delta > 0 >exists. > >This is a common problem with this sort of epsilon-delta analysis. > >Delta has to eventually be zero or else the integral is incomplete and >thus not *integrable*. Uh, that's nonsense. There's nothing in the definition about delta being eventually zero. ************************ David C. Ullrich
From: Jason on 9 Mar 2005 13:58 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. This is a good example of real analysis definitions shooting themselves in the foot real badly. As for Riemann Sums/Integrals: Riemann sums and riemann Integrals are not the same as Gabriel's theorem. Let's do some comparisons: Let DELTAx = (b-a)/n An Archimedean/Riemann Sum is defined as follows: n SIGMA f(x ) DELTAx A] s=1 s An Archimedean/Riemann Integral is defined as follows: Let DELTAx = (b-a)/n n Lim SIGMA f(x ) DELTAx B] DELTAx s=1 s We can immediately disregard A] because it does not contain a limit but Gabriel's theorem does. So Gabriel's theorem is not a riemann sum. What about B] ? Aside from the summation limits which are different to gabriel's theorem and these do make a *difference*, the Archimedean/riemann integral is summing products containing f and not f'. Gabriel's theorem uses f'. Furthermore, f(x+w) - f(x) *is not equal to* n Lim SIGMA f(x ) DELTAx DELTAx s=1 s So now if you say, so what, just use f' instead of f. Well, then you have the summation limits which are *incorrect*. Using different summation limits means the two results (gabriel/riemann) will get closer the more partial sums are calculated but the riemann integral will always be *wrong*! Why? Gabriel's theorem states that f must be differentiable at x=0 and that it is *required* in the average sum. I see you can write Matlab programs. Why don't you run a few small tests instead of going off on a tangent every time? Jason Wells
From: David C. Ullrich on 9 Mar 2005 18:15 On 9 Mar 2005 10:58:52 -0800, "Jason" <logamath(a)yahoo.com> 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. Huh? >This is a >good example of real analysis definitions shooting themselves in the >foot real badly. > > >As for Riemann Sums/Integrals: >Riemann sums and riemann Integrals are not the same as Gabriel's >theorem. Let's do some comparisons: > > > Let DELTAx = (b-a)/n > > An Archimedean/Riemann Sum is defined as follows: > > n > SIGMA f(x ) DELTAx A] > s=1 s > > > An Archimedean/Riemann Integral is defined as follows: > > Let DELTAx = (b-a)/n > > n > Lim SIGMA f(x ) DELTAx B] > DELTAx s=1 s > > > We can immediately disregard A] because it does not contain > a limit but Gabriel's theorem does. So Gabriel's theorem is not a >riemann sum. > What about B] ? > > Aside from the summation limits which are different to > gabriel's theorem and these do make a *difference*, the > Archimedean/riemann integral is summing products containing > f and not f'. Gabriel's theorem uses f'. 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. > Furthermore, > > > f(x+w) - f(x) *is not equal to* > > n > Lim SIGMA f(x ) DELTAx > DELTAx s=1 s > > > So now if you say, so what, just use f' instead of f. Well, then >you have the summation limits which are > *incorrect*. Using different summation limits means the two results >(gabriel/riemann) will get closer > the more partial sums are calculated but the riemann integral will >always be *wrong*! > Why? Gabriel's theorem states that f must be differentiable at x=0 >and that it is *required* in the average sum. > > I see you can write Matlab programs. Huh? >Why don't you run a few small >tests instead of going off on a tangent every time? Huh? > Jason Wells ************************ David C. Ullrich
From: Jason on 9 Mar 2005 21:57 > 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! Huh? Jason Wells
From: William Hughes on 10 Mar 2005 00:26
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 |