Possible proof of Gabriel's Theorem?
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From: Jason on 5 Mar 2005 11:38 Okay David, Have just thought of an example where f' may not be continuous. 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. Jason Wells
From: denis feldmann on 5 Mar 2005 15:12 Jason a ýcrit : >>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! Where is w in the classic definition? > > You just don't seem to get this, do you? What is bothering you about > this? > You have not grasped this since the beginning. > > Someone indeed don't grasp things >>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. Troll, or clueless; probably both. See below 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. That's the stupid clueless thing alluded above Give me one example of where > this is untrue. Try f(x)= 0 for x=0, f(x)=x^2sin(1/x) for x<>0 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. 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. Yhis ends the debate. I will probably read you a bit longer for fun; (and to see if you have any coherent answer to the classical function above), but don't expect any answer. Good luck wiith your life, as your maths are irredeemably lost. 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! Troll, hubris, and JSH-like behaviour. You have got a Hammer, too? > > > Jason Wells. >
From: Jason on 5 Mar 2005 16:51 > Where is w in the classic definition? f(x+w) - f(x) f'(x) = Lim ------------- w->0 w Are you sure you know how to read English? Do you understand what is being written here or are you just posting stuff because you have nothing better to do? > Someone indeed don't grasp things Yes, someone is not grasping anything from what I see. > Troll, or clueless; probably both. See below Okay, unless you apologize for your foul language and your tantrums, I will no longer respond to any of your posts. See if you can understan this. Perhaps I should translate it into French for you? > That's the stupid clueless thing alluded above > Try f(x)= 0 for x=0, f(x)=x^2sin(1/x) for x<>0 You obviously did not read my second post to Ullrich where I said f' is continuous but does not have to continuous at x+w and in fact does not even have to exist there. Go back and reread it and this time you must *think* a little bit. > Yhis ends the debate. I will probably read you a bit longer for fun; > (and to see if you have any coherent answer to the classical function > above), but don't expect any answer. Good luck wiith your life, as your > maths are irredeemably lost. Read as much as you like and post as much as you like. If you do not apologize for calling me a troll and clueless, you no longer exist as far as I am concerned. Arrevoir my little frog!! Jason Wells.
From: Jason on 5 Mar 2005 18:20 >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!!
From: Jason on 5 Mar 2005 18:56
So you think I don't know what I am talking about? Well, you probably need to study what a riemann integral is and then compare it with gabriel's average derivative and you will see they are vastly different. Riemann's integral is a joke next to gabriel's average derivative. Gabriel does not use a mesh value and gabriel's result if applied to numeric integration/differentiation yields a far better result than anything riemann ever dreamt of! In fact the Lebesgue integral definition is not as strong as gabriel's ATT either. See David, you need to do your homework carefully before you continue to blabber all the junk you have been blabbering out on this forum. Evidently *you* have no idea what you are talking about!! Jason |