Possible proof of Gabriel's Theorem?
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From: Jason on 4 Mar 2005 09:55 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. > > 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. And once again David, please show me how this is the same as the fundametal theorem of calculus. 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.
From: Jason on 4 Mar 2005 09:57 No, I meant what you see. Read my response to Ullrich please. BTW: You are incorrect in saying that it *only works from right hand side*.
From: denis feldmann on 4 Mar 2005 11:45 Jason a ýcrit : > 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. Does it? What does it says for f(x)=|x| (at 0)? Or for f(x)=exp(-1/x) ? > > >>> 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 did. So what? Did you bother to read David's answer? (Hint : f' continuous is stronger than f differentiable) And once again David, please show me how this is the same as > the fundametal theorem of calculus. 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. 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, 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 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. Is you say so. But then, why ask us? I told you yesterday that actually even what you were writing had no sense (mismatch of free and bound variables), let alone being allowed... >
From: denis feldmann on 4 Mar 2005 11:46 Jason a ýcrit : > No, I meant what you see. Read my response to Ullrich please. > > BTW: You are incorrect in saying that it *only works from right hand > side*. > Sure. Just try it for f(x)=|x| (Hint : w/n ->0 *and stays >0*)
From: Jason on 4 Mar 2005 12:33
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 |