Possible proof of Gabriel's Theorem?
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From: denis feldmann on 4 Mar 2005 12:53 Jason a ýcrit : > Denis, > > >>Sure. Just try it for f(x)=|x| (Hint : w/n ->0 *and stays >0*) > > > Okay, let x = 0 and w = 1: > > > |0 + 1/n| - |0| 1/n > ----------------- = ---- = 1 > 1/n 1/n > > Now let w = -1; > > |0 - 1/n| - |0| 1/n > ----------------- = ---- = -1 > - 1/n -1/n > > So you have exactly the same result as you would for the classic > definition: > > > |0 + 1| - |0| > -------------- = 1 > 1 > > |0 - 1| - |0| > -------------- = -1 > -1 > > See, in this particular function it does not even matter if w goes to > zero or not. > > Jason Wells > Oh, I see. Where on earth was w->0 included in your "definition"? So you get f'(x)= lim (w->0) lim (n->+oo) f(x+w/n)-f(x))/(w/n) , right? But if you include limits towards 0, what's wrong with the much simpler (and usual) f'(x)=lim (w->0) lf(x+w)-f(x))/w ; may I ask?
From: Jason on 4 Mar 2005 12:57 > Does it? What does it says for f(x)=|x| (at 0)? Or for f(x)=exp(-1/x) ? Of course it does. Check it out! > You did. So what? Did you bother to read David's answer? (Hint : f' > continuous is stronger than f differentiable) Absolute rubbish! This is not about one being stronger than the other. It is not required that f be differentiable at x+w. Yes, it is different to David's statement. (Hint: Read gabriel's theorem) > I did. I note you never answered that. For your information, here it is > again : by the fundamental theorem, f(x+w)-f(x)= integral(f'(t)dt , > t=x..x+w) Then we use the definition of integral, by riemann sum with > fixed step w/n, and that's it. You did nothing of the sort! This is the original result which says nothing about the *average tangent* or *average derivative* - did you get this? Furthermore, it is not a riemann sum because a riemann sum is finite. Neither is it a riemann integral! So you have not shown me anything at all. > How condescending of you. The point is if you write >> f(x+2w/n)-f(x+w/n) >> Lim ------------------ = f'(x+2w/n), >> n->oo w/n > you make an absurd mistake. If you wait until the whole double limit is > taken, you make an interversion of limits which is absolutely illegal I'll ignore the condescending comment. I do not write that. This is what David wrote and *misunderstood* just as he has been misunderstanding almost everything else from the beginning. So gabriel makes an absurd mistake here - why? > If you wait until the whole double limit is taken, you make an interversion of limits which is absolutely illegal Oh really, this makes no sense to me whatsoever. What are you saying exactly? What on earth is an *interversion*? Is this a French word because it does not exist in the English language? Jason Wells
From: Jason on 4 Mar 2005 13:02 > Oh, I see. Where on earth was w->0 included in your "definition"? So you > get f'(x)= lim (w->0) lim (n->+oo) f(x+w/n)-f(x))/(w/n) , right? > But if you include limits towards 0, what's wrong with the much simpler > (and usual) f'(x)=lim (w->0) lf(x+w)-f(x))/w ; may I ask? w->0 is not included in my definition. It's part of the classical definition. And no, I don't get what you stated: f'(x)= lim (w->0) lim (n->+oo) f(x+w/n)-f(x))/(w/n) This looks like absolute nonsense. The fact of the matter is that you are making posts without carefully thinking about what is written. What do you think if you just slow down a bit and try to see what I am saying instead of impulsively posting new comments?
From: David C. Ullrich on 5 Mar 2005 07:26 On 4 Mar 2005 06:55:33 -0800, "Jason" <logamath(a)yahoo.com> wrote: >No. By derivative, I mean: > > f(x + w/n) - f(x) > f'(x) = Lim ------------------ > n->oo w/n > >w is the width of the interval >n is the number of partitions > >The above definition is *correct* and it produces the same end results >as the clasic definition. No, it defines a function of two variables x and w. >> > 1 n-1 ws >> > AFD(f) = Lim - SIGMA f'(x+ --- ) --- R.H.S >> > n->oo n s=0 n >> >> As has been pointed out many times, _if_ you assume in >> addition that f' is continuous (and assuming that the >> correction above is what you really meant) then the >> result is a trivial consequence of the fundamental >> theorem of calculus. If you assume only that f is >> differentiable then I actually doubt that the theorem >> is true, although I don't have a counterexample handy. > >Did you even bother reading my proof David? I think I stated it is >continuous everywhere and differentiable everywhere except possibly >at x+w. 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. >And once again David, please show me how this is the same as >the fundametal theorem of calculus. 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'. >Thus far you have only been able >to haw-hem. Show me where you have seen this result before. Can you >do what gabriel has done with the classic definition? If yes, please >show me! > >> Let's see about the proof: >> This step is simply wrong. For example, >> >> f(x+2w/n)-f(x+w/n) >> Lim ------------------ + >> n->oo w/n >> >> is not f'(x + w/n), it is actually f'(x) (at least >> if f is continuously differentiable.) > >Now you are beginning to get it David, but you are not quite there I >think. I am not saying that the above result is equal to f'(x + w/n). >It is not actually f'(x) either after we equate both limits. >This is part of the result of the Average Sum Theorem. The above result >is obtained *only* once we let n and t run through to infinity. > >> You can't just drop the inner limit that way - as I said >> yesterday, interchanging limits is the hard part of most >> theorems in analysis, just assuming it works with no >> justification is a way to prove anything, including >> things that are false. > >This part seems rather straight forward: if we consider that both >limits >are taken as n and t approaches infinity, we can drop the inner limit. You have no idea what you're talking about. ************************ David C. Ullrich
From: Jason on 5 Mar 2005 11:12
> 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. > 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. 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. 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! Jason Wells. |