Possible proof of Gabriel's Theorem?
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From: Jason on 1 Mar 2005 18:22 Proof of Gabriel's Theorem: What if we simplify gabriel's proof by keeping the ens fixed and using s in the definition: i.e. we let n be fixed for each of f'(x), f'(x+w/n), f'(x+2w/n), etc. In this case, the following are true: f(x + w/s) - f(x) f'(x) = Lim ------------------ s->Infinity w/s f'(x+ w/n) = f(x + w/n + w/s) - f(x + w/n) Lim ------------------------------ s->Infinity w/s f'(x+ 2w/n) = f(x + 2w/n + w/s) - f(x + 2w/n) Lim -------------------------------- s->Infinity w/s f'(x+ 3w/n) = f(x + 3w/n + w/s) - f(x + 3w/n) Lim -------------------------------- s->Infinity w/s and so on: However, since both n and s tend to infinity, the limit is taken once over the entire sum. With n and s coinciding, we get gabriel's proof. This seems to work but is this last step legal? If not, why not, because ultimately we are trying to find the limit of the sum as n approaches infinity? Jason Wells
From: denis feldmann on 2 Mar 2005 01:29 Jason a ýcrit : > Proof of Gabriel's Theorem: > > What if we simplify gabriel's proof by keeping the ens fixed and using > s in the definition: > > i.e. we let n be fixed for each of f'(x), f'(x+w/n), f'(x+2w/n), etc. > In this case, the following are true: > > > f(x + w/s) - f(x) > f'(x) = Lim ------------------ > s->Infinity w/s > > > f'(x+ w/n) = > > > f(x + w/n + w/s) - f(x + w/n) > Lim ------------------------------ > s->Infinity w/s > > > f'(x+ 2w/n) = > > > f(x + 2w/n + w/s) - f(x + 2w/n) > Lim -------------------------------- > s->Infinity w/s > > > f'(x+ 3w/n) = > > > f(x + 3w/n + w/s) - f(x + 3w/n) > Lim -------------------------------- > s->Infinity w/s > > > and so on: > > As long as you are willing to copy this at length (and twice)... > However, since both n and s tend to infinity, the limit is taken > once over the entire sum. With n and s coinciding, we get > gabriel's proof. Why dont you take the trouble to write in full the argument following, so that those of us who didn't follow the discussion have the slightest idea of what you are talking about? And it could give you the redaction of a correct proof (or perhaps, an hint of what is still missing?) OK, I'll try to guess : you sum the equalities above, getting something like sum((f'(x+kw/n), k=0..m)= lim(s->+oo) s(f(x+mw/n+w/s)-f(x))/w (except this is wrong, of course : there is no cancelation of anything, and this last limit is infinite...) But let's pray for a miraculous cancelation and go on Now what? put m=n? => sum((f'(x+kw/n), k=0..n)= lim(s->+oo) s(f(x+w(1+1/s))-f(x))/w = lim (t->0) (1/w)* Or something deeper? Oh, I see, next you take the limit in n->oo.Why? Anyway, we get lim (n->oo) (sum((f'(x+kw/n), k=0..n))= lim(s->+oo) s(f(x+w(1+1/s))-f(x))/w At this stage, obviously, something went badly wrong... > This seems to work but is this last step legal? If not, why not, > because ultimately we are trying to find the limit of the sum as > n approaches infinity? > > Jason Wells >
From: William Elliot on 2 Mar 2005 05:18 On Tue, 1 Mar 2005, Jason wrote: > Proof of Gabriel's Theorem: > > What if we simplify gabriel's proof by keeping the ens fixed and using > s in the definition: > > i.e. we let n be fixed for each of f'(x), f'(x+w/n), f'(x+2w/n), etc. > In this case, the following are true: > > > f(x + w/s) - f(x) > f'(x) = Lim ------------------ > s->Infinity w/s > > How is this so, for when 0 < w it's lim(h->0+) (f(x+h) - f(x))/h which is not lim(h->0) (f(x+h) - f(x))/h As for example f(x) = |x| right hand f'(0) = 1 = lim(s->oo) |1/s|/(1/s) left hand f'(0) = -1 = lim(s->oo) |-1/s|/(-1/s) f'(0) undefined
From: David C. Ullrich on 2 Mar 2005 07:34 On 1 Mar 2005 15:22:38 -0800, "Jason" <logamath(a)yahoo.com> wrote: >Proof of Gabriel's Theorem: > >What if we simplify gabriel's proof by keeping the ens fixed and using >s in the definition: > >i.e. we let n be fixed for each of f'(x), f'(x+w/n), f'(x+2w/n), etc. >In this case, the following are true: > > > f(x + w/s) - f(x) > f'(x) = Lim ------------------ > s->Infinity w/s > > > f'(x+ w/n) = > > > f(x + w/n + w/s) - f(x + w/n) > Lim ------------------------------ > s->Infinity w/s > > > f'(x+ 2w/n) = > > > f(x + 2w/n + w/s) - f(x + 2w/n) > Lim -------------------------------- > s->Infinity w/s > > > f'(x+ 3w/n) = > > > f(x + 3w/n + w/s) - f(x + 3w/n) > Lim -------------------------------- > s->Infinity w/s > > > and so on: > > > However, since both n and s tend to infinity, the limit is taken > once over the entire sum. With n and s coinciding, we get > gabriel's proof. > > This seems to work but is this last step legal? _What_ last step? >If not, why not, > because ultimately we are trying to find the limit of the sum as > n approaches infinity? > > Jason Wells ************************ David C. Ullrich
From: denis feldmann on 2 Mar 2005 07:57
denis feldmann a ýcrit : > Jason a ýcrit : > >> Proof of Gabriel's Theorem: >> >> What if we simplify gabriel's proof by keeping the ens fixed and using >> s in the definition: >> >> i.e. we let n be fixed for each of f'(x), f'(x+w/n), f'(x+2w/n), etc. >> In this case, the following are true: >> >> >> f(x + w/s) - f(x) >> f'(x) = Lim ------------------ >> s->Infinity w/s >> >> >> f'(x+ w/n) = >> >> >> f(x + w/n + w/s) - f(x + w/n) >> Lim ------------------------------ >> s->Infinity w/s >> >> >> f'(x+ 2w/n) = >> >> >> f(x + 2w/n + w/s) - f(x + 2w/n) >> Lim -------------------------------- >> s->Infinity w/s >> >> >> f'(x+ 3w/n) = >> >> >> f(x + 3w/n + w/s) - f(x + 3w/n) >> Lim -------------------------------- >> s->Infinity w/s >> >> >> and so on: >> >> > > As long as you are willing to copy this at length (and twice)... > > > > >> However, since both n and s tend to infinity, the limit is taken >> once over the entire sum. With n and s coinciding, we get >> gabriel's proof. > > > Why dont you take the trouble to write in full the argument following, > so that those of us who didn't follow the discussion have the slightest > idea of what you are talking about? And it could give you the redaction > of a correct proof (or perhaps, an hint of what is still missing?) > > OK, I'll try to guess : you sum the equalities above, getting something > like > > sum((f'(x+kw/n), k=0..m)= lim(s->+oo) s(f(x+mw/n+w/s)-f(x))/w > > (except this is wrong, of course : there is no cancelation of anything, > and this last limit is infinite...) > > But let's pray for a miraculous cancelation and go on > > > Now what? put m=n? => > > sum((f'(x+kw/n), k=0..n)= lim(s->+oo) s(f(x+w(1+1/s))-f(x))/w = lim > (t->0) (1/w)* > > Or something deeper? > > Oh, I see, next you take the limit in n->oo.Why? > > Anyway, we get > > lim (n->oo) (sum((f'(x+kw/n), k=0..n))= lim(s->+oo) s(f(x+w(1+1/s))-f(x))/w > > At this stage, obviously, something went badly wrong... > > Answering my own post, I took the trouble of looking for Gabriel's theorem. Mostly, the main result looks like lim (n->oo) w/n (sum((f'(x+kw/n), k=0..n)))= f(x+w)-f(x), which is not hard to prove using Riemann sums. What I believe is the main point this absurdly long thread (or threads) try to make is to validate the (JSH-like) incredible pretention of Gabriel, trying to prove this result by, let's say, somewhat illegal means, *and* arguing at the same time that his approach is the only right way to analysis, everybody else (including Newton, Cauchy and Rieman) being dumb fools not able to see the stupid mistakes they made when "proving" elementary Calculus results. As for "Jason", I don't know. I would suspect he is Gabriel in disguise (or perhaps his smarter brother, Mycroft)... > > > > > > > > > > >> This seems to work but is this last step legal? If not, why not, >> because ultimately we are trying to find the limit of the sum as >> n approaches infinity? >> Jason Wells >> |