Possible proof of Gabriel's Theorem?
in [Math]
|
Prev: Contractible metric space
Next: Notation question
From: denis feldmann on 6 Mar 2005 00:08 Jason a ýcrit : >>Given an interval [x;x+w] subdivided into n equal parts >>and a function f which is continuous on [x;x+w] and >>differentiable on [x;x+w), the secant gradient or mean >>value of f is equal to the average first derivative of f > > > Ullrich replied: > >>That does not state that f' is continuous. > > > Well, if it does not imply f' is continuous, then all your real > analysis goes down the toilet! > And you call yourself a college math professor? > > Giggle, giggle. You put your foot in it again!! > Dod you send that one before or after the one where we showed you f' can be not continuous (like x^2*sin(1/x^2) if x<>0, 0 if x=0)? If after, you are a troll ; if before, don't you think you could have been, well, a little bit less petulant in your answer? And you ask apologies from me? (remember, I have no connection with David Ullrich)
From: on 6 Mar 2005 07:28 In article <1110039173.479820.271700(a)f14g2000cwb.googlegroups.com>, "Jason" <logamath(a)yahoo.com> writes: >> No, it defines a function of two variables x and w. > >Once again: No David! Both you and Yan got this wrong. It is a function >of *one* variable, i.e. x. w which is the width is *constant*. The only >thing that's changing is w/n, not w. But if w is a constant, then you have to tell us what its value is, don't you? If I want to use this definition to compute a derivative of some function f at some point x, then how do I decide which w to use? Perhaps you will answer something like it does not matter which w I choose, provided that f is continuous in [x,x+w]. But if that is the case, then how do I know that I will not get different answers for the derivative, depending on whether I choose, say w = 1 or w = 1/2 ? Derek Holt.
From: David C. Ullrich on 6 Mar 2005 08:34 On 5 Mar 2005 08:12:53 -0800, "Jason" <logamath(a)yahoo.com> wrote: >> No, it defines a function of two variables x and w. > >Once again: No David! Both you and Yan got this wrong. It is a function >of *one* variable, i.e. x. w which is the width is *constant*. The only >thing that's changing is w/n, not w. If this is a function of two >variables, then so is the classic definition. How is it a function of >two variables in your mind? Is w a variable in the classic definition? >Most certainly not! > >You just don't seem to get this, do you? What is bothering you about >this? >You have not grasped this since the beginning. Guffaw. >> You stated this: >> >> >Given an interval [x;x+w] subdivided into n equal parts >> >and a function f which is continuous on [x;x+w] and >> >differentiable on [x;x+w), the secant gradient or mean >> >value of f is equal to the average first derivative of f >> That does not state that f' is continuous. > >Oh yes it does! What an uninformed thing of you to say. Let's see: > >If f is differntiable everywhere, this means that f' exists everywhere >in the interval! And if f' exists everywhere in the interval, then it >is continuous everywhere in the interval. Give me one example of where >this is untrue. Your ignorance of calculus is astounding. It's a very standard example: Let f(x) = x^2 sin(1/x^2) for x <> 0, f(0) = 0. Then f'(x) exists for every x, although f' is not continuous. >Gabriel states that *only* f'(x+w) need not exist. > >> I didn't say it was the same, I said it was a trivial >> consequence of ftc. It is, for exactly the reason various >> people have explained to you: that sum is a Riemann sum >> for the integral of f'. > >By saying it is a trivial consequence of the ftc, you are effectively >saying it is the same. No. >It is not a Riemann sum either for it if were, >then >it would be *finite*. It's actually infinite because it is summed over >as n tends to infinity. You're gibbering here - the fact that we're taking a limit as n -> infinity does not mean that n is infinite. The sum sum_j=1^n f'(x+jw/n)/(w/n) _is_ a Riemann sum for int_x^{x+w} f'. >Riemann sums are approximations, this is not an >approximation, it is *exact*. The Riemann sum becomes a *riemann >integral* as the size of each partition approaches 0. >It would be nice to use the riemann integral except that it is not >easily understood by students and you cannot use it to prove the >*missing link* which I believe is Gabriel's average tangent theorem. >Just try showing how > > x+w > f(x+w) - f(x) = INT f'(x) dx > x > >using the riemann sum! However, gabriel's theorem fits exactly between >the LHS and RHS in the above formula: > > x+w > f(x+w) - f(x) = w * ATG(f) = INT f'(x) dx > x > >No real analysis is required here. Furthermore, the riemann sum is not >really a riemann sum, but an archimedean sum. it was Archimedes who >discovered the integral and used the first methods of exhaustion. Sorry >to >downplay your German roots a little! > >> You have no idea what you're talking about. > >I don't think so Daivd. Maybe you can stop being so narrow minded and >hardheaded and look at this theorem without any preconceived ideas. >Perhaps if you try, you might see something you did not see before. >The average tangent theorem is a small step to the average sum theorem >which I find fascinating but do not completely understand. It is a very >interesting theorem. I believe that had anyone known gabriel's theorems >back then, real analysis may not even have been around today. Careful >David, you may end up looking the fool in a few years time! Just curious: Why are you continuing this series of posts? I mean by now it must be clear that _everyone_ disagrees. So that proves that we're all stupid and you're the only one who can see the truth, fine. Why are you wasting your breath on us? >Jason Wells. ************************ David C. Ullrich
From: David C. Ullrich on 6 Mar 2005 08:38 On 5 Mar 2005 08:38:18 -0800, "Jason" <logamath(a)yahoo.com> wrote: >Okay David, > > Have just thought of an example where f' may not be continuous. That's a lie and you know it. Your weak understanding of these things could not possibly allow you to think of an example. You mean you just _saw_ an example in some book or somewhere online, and although you didn't understand it you decided for some reason it must be right. Now I'm curious about something. In your previous post you said this, when I said that f differentiable does not imply that f' is continuous: "Oh yes it does! What an uninformed thing of you to say. Let's see: If f is differntiable everywhere, this means that f' exists everywhere in the interval! And if f' exists everywhere in the interval, then it is continuous everywhere in the interval. Give me one example of where this is untrue. Gabriel states that *only* f'(x+w) need not exist." Now you've decided that there _is_ such an example. What I'm curious about is whether this has any affect on your estimation of who's the "uninformed" one here. > So let's just go ahead and say that Gabriel's theorem should include >a statement about f' being continuous everywhere except at x+w where it >need >not even exist according to Gabriel's theorem. Fine, let's say that. Then the amazing theorem _is_ a trivial consequence of the fundamental theorem of calculus. And I mean _trivial_: it follows from nothing but ftc plus the _definition_ of the integral. >Jason Wells ************************ David C. Ullrich
From: David C. Ullrich on 6 Mar 2005 08:43
On 5 Mar 2005 15:20:09 -0800, "Jason" <logamath(a)yahoo.com> wrote: >>Given an interval [x;x+w] subdivided into n equal parts >>and a function f which is continuous on [x;x+w] and >>differentiable on [x;x+w), the secant gradient or mean >>value of f is equal to the average first derivative of f > >Ullrich replied: >> That does not state that f' is continuous. > >Well, if it does not imply f' is continuous, then all your real >analysis goes down the toilet! >And you call yourself a college math professor? > >Giggle, giggle. You put your foot in it again!! Facinating. A minute ago I saw a post from _you_ admitting that in fact f' need not be continuous: "Have just thought of an example where f' may not be continuous." So far you've called me uninformed for my opinion on this question. Above you say it makes my real analysis go down the toilet and question whether I should be a professor. Now it turns out that you were wrong about this (and you _admit_ you were wrong). But there's been no apology yet for the things you said. You complain a lot about people being insulting to _you_. This is one reason people call you a crackpot - it's precisely typical crackpot behavior: When a crackpot complains that people are whatever to him, you can be certain that in fact _he_ is being whatever to them. ************************ David C. Ullrich |