Possible proof of Gabriel's Theorem?
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From: David C. Ullrich on 7 Mar 2005 08:33 On 6 Mar 2005 20:00:10 -0800, "Jason" <logamath(a)yahoo.com> wrote: >Actually, they all use a mesh value. However, the riemann integral is >an *approximation*. Gabriel's average tangent/derivative theorem is not >an approximation - it is *natural integration*. You can take any >definite integral and compute it's value using the average derivative. > >So you do not have the time to *decipher* gabriel's math yet you >question its validity? Strange, I did not know that math needs >deciphering for it is either true or false. If you have not looked at >something, why do you by default cast doubt on its truth? You are >simply following in the footsteps of most fools on this forum. The >majority of posts including those by math professors demonstrate they >(the real trolls) did not even bother to so much as read gabriel's >stuff. All I have seen so far is sheer arrogance, stupidity and the >utmost ignorance. > >Now allow me to critisize your post: > >> I don't claim to be an expert on analysis, nor do I have the time to >> decipher Gabriel's (pseudo?) math, but the Riemann integral does not >> necessarily use a mesh value. Most analysis texts these days seem to >> define the upper integral F as inf U(g,P) and the lower integral f as >> sup L(g,P). If F=f, then the function g is integrable. > >Most analysis texts are a load of rubbish and are taught by insecure >individuals the likes of whom can be found on this forum. What are you >trying to say? What does *integrable* mean? It's not an English word. >Let's suppose you mean that g is a function and that by *integrable* >you mean it can be integrated. So what is P then? Do you think that >most people who read this forum know what you mean? Yes, most people reading this thread know what "integrable" means. The fact that you obviously don't is one of the many things that makes it all amusing enough to persuade some people to keep reading your posts. Definition: Suppose that f:[a,b] -> R is bounded. We say f is (Riemann) integrable if for every epsilon > 0 there exists delta > 0 such that if a = t_0 < ... < t_n = b, I_j = [t_{j-1}, t_j], M_j is the sup of f on I_j and m_j is the inf of f on I_j then the sum of (M_j - m_j)*(length(I_j)) is less than epsilon. You really didn't know that? If you don't know what the word "integrable" means then the fact that you've been replying to posts where people have been discussing the concept seems very curious. ************************ David C. Ullrich
From: Tim Peters on 7 Mar 2005 10:34 [David C. Ullrich] >> Definition: Suppose that f:[a,b] -> R is bounded. We say f is >> (Riemann) integrable if for every epsilon > 0 there exists >> delta > 0 such that if a = t_0 < ... < t_n = b, >> I_j = [t_{j-1}, t_j], M_j is the sup of f on I_j and m_j is >> the inf of f on I_j then the sum of (M_j - m_j)*(length(I_j)) >> is less than epsilon. >> You really didn't know that? If you don't know what the word >> "integrable" means then the fact that you've been replying >> to posts where people have been discussing the concept seems >> very curious. .... [Jason] > The epsilon-delta definition is a load of hogwash in my opinion. In > this case of a constant functioin it is stating that f(x) = c is *not* > integrable since no epsilon > 0 exists. Of course every epsilon > 0 exists, so "no epsilon > 0 exists" is nonsensical. Maybe you meant to say that, given an epsilon > 0, no delta exists satisfying the condition? Nope, a constant function f(x)=c is especially _easy_ this way. Then M_j - m_j = c-c = 0 regardless of the partition, so the sum of (M_j - m_j)*length(I_j) is 0 regardless of the partition, which is indeed less than epsilon. It doesn't matter which delta you pick in this case; e.g., always picking delta = b-a, regardless of epsilon, is enough to show the Riemann integrability of a constant function. > I hinted at this earlier but seeing you were all so stupid, I simply > avoided it. Thank you for your attempted kindness.
From: Angus Rodgers on 7 Mar 2005 10:41 On Mon, 07 Mar 2005 07:25:32 -0600, David C. Ullrich <ullrich(a)math.okstate.edu> wrote: >On 6 Mar 2005 16:43:32 -0800, "Jason" <logamath(a)yahoo.com> wrote: > >>Truth is you may end up convincing me >>that gabriel is indeed wrong because I still have some questions. What >>really keeps me going is the fact neither you nor anyone else has been >>able to disprove or prove *anything* aside from rhetoric, ignorant >>posts and irrelevant troll-like comments. > >I don't believe that anyone has _said_ that his theorem is false >as stated. But nobody has given a proof of it, and people >_suspect_ it's false, for example if there does indeed exist >a non-constant differentiable f such that f'(r) = 0 for all >rational r, as I suspect, then the theorem's certainly false. Yikes! I wasn't expecting that. I killfiled Jason in exasperation some time ago, and I haven't been following these threads regularly, but this made me sit up and take notice. You're saying that, in spite of the strange presentation on Gabriel's website, and in spite of Jason's many hilarious misconceptions (about constant functions not being differentiable, and so on), there is a coherent conjecture here that hasn't been settled? I don't suppose you could save me the effort of deciphering Gabriel's handwriting by pointing me to an article where this conjecture has been competently formulated, could you? Sorry I haven't been paying attention. I wasn't expecting actually to have to look at the maths, and I only hope my rusty analysis is up to the job. I'd better get that hat and humble pie ready! -- Angus Rodgers (angus_prune@ eats spam; reply to angusrod@) Contains mild peril
From: Jesse F. Hughes on 7 Mar 2005 11:57 "Jason" <logamath(a)yahoo.com> writes: >> "Jason" <logamath(a)yahoo.com> writes: >>>What does *integrable* mean? It's not an English word. [...] > Actually I know exactly what it is. I am questioning the relevance of > his post. I am not sure he knows exactly what he is talking about. You got a curious way of expressing yourself. -- 17:49 3/4/05: "The proof is actually not hard, and it is perfect." 07:25 3/5/05: "Nope. I made a mistake." 11:06 3/5/05: "Maybe I screwed up[...] Otherwise, um, it's very easy to factor." 11:48 3/5/05: "The answer is just that simple." -- JSH: A day in the life.
From: Jesse F. Hughes on 7 Mar 2005 13:17
"Jason" <logamath(a)yahoo.com> writes: > To what do I owe this pleasure? :-) Are you the famous JSH? I'm merely his sycophant. I'm the unfamous JFH, not the famous JSH, but sometimes my .sig conveys the wrong message. -- Jesse F. Hughes "Well, I guess that's what a teacher from Oklahoma State University considers proper as Ullrich has said it, and he is, in fact, a teacher at Oklahoma State University." -- James S. Harris presents a syllogism |