Possible proof of Gabriel's Theorem?
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From: Larry Hammick on 3 Mar 2005 09:59 "Jason" > Proof of Gabriel's Theorem: Still putting lipstick on this pig? :)
From: Jason on 3 Mar 2005 12:14 David C. Ullrich wrote: > On 2 Mar 2005 07:15:30 -0800, "Jason" <logamath(a)yahoo.com> wrote: > > >> > However, since both n and s tend to infinity, the limit is taken > >> > once over the entire sum. With n and s coinciding, we get > >> > gabriel's proof. > >> > > >> > This seems to work but is this last step legal? > > > >> _What_ last step? > > > >David, > > > >The last step being that since n coincides with s and the limits are > >both taken over infinity, the inner limit falls away to be replaced by > >the outer limit. > > > >What do you think? > > I think two things: > > First, I think that you think I know what you're talking > about. There are no inner and outer limits in any of the > posts in this thread, nor any sums visible - if you think > people are going to try to make sense of your corrections > to the incoherent things we can find elsewhere you're wrong - > you should post the entire proof in a coherent form. > > Also, I _know_ that in general interchanging limits is > the hard part when you're proving almost anything in > analysis - if you just assume that that works you're > usually sweeping the entire proof under the rug. > > >Jason > > > ************************ > > David C. Ullrich Yes, I thought you had looked at gabriel's stuff so I assumed you knew what I was talking about. I will post an entire proof which I think might be correct in the next few days.
From: Jason on 3 Mar 2005 13:43 Here is my attempt to prove Gabriel's Theorem: Have changed some of his wording and am calling it the AFD rather than ATG as he calls it. 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 [or AFD(f)] defined as follows: w is the width of the interval n is the number of partitions f(x + w/n) - f(x) AFD(f) = ------------------ --- L.H.S w/n 1 n-1 ws AFD(f) = Lim - SIGMA f'(x+ --- ) --- R.H.S n->oo n s=0 n Proof: Start with RHS. 1 AFD(f) = Lim - [ f'(x) + f'(x+w/n) + f'(x+2w/n) + ... n->oo n f'(x+(n-2)w/n) + f'(x+(n-1)w/n)] Now we simplify Gabriel's proof by keeping the ens fixed and using t in the definition: i.e. we let n be fixed for each of f'(x), f'(x+w/n), f'(x+2w/n), etc. In this case, the following are true: f(x+w/t)-f(x) f'(x) = Lim ------------- t->oo w/t f'(x+w/n) = f(x+w/n+w/t)-f(x+w/n) Lim --------------------- t->oo w/t f'(x+2w/n) = f(x+2w/n+w/t)-f(x+2w/n) Lim ----------------------- t->oo w/t f'(x+ 3w/n) = f(x+3w/n+w/t)-f(x+3w/n) Lim ----------------------- t->oo w/t and so on: Now, 1 f(x+w/t)-f(x) AFD(f) = Lim - [ Lim ------------- + n->oo n t->oo w/t f(x+w/n+w/t)-f(x+w/n) Lim --------------------- + t->oo w/t f(x+2w/n+w/t)-f(x+2w/n) Lim ----------------------- + t->oo w/t f(x+3w/n+w/t)-f(x+3w/n) Lim ----------------------- + t->oo w/t + ... f(x+(n-2)w/n+w/t)-f(x+(n-2)w/n) Lim ------------------------------- + t->oo w/t f(x+(n-1)w/n+w/t)-f(x+(n-1)w/n) Lim ------------------------------- ] t->oo w/t Looking at the above we have two limits to infinity. i.e. n and t. However, since both n and t tend to infinity, the outer limit is taken once over the entire sum. With n and t coinciding, we have: 1 f(x+w/n)-f(x) AFD(f) = Lim - [ Lim ------------- + n->oo n n->oo w/n f(x+2w/n)-f(x+w/n) Lim ------------------ + n->oo w/n f(x+3w/n)-f(x+2w/n) Lim ------------------- + n->oo w/n f(x+4w/n)-f(x+3w/n) Lim ------------------- + n->oo w/n + ... f(x+(n-1)w/n)-f(x+(n-2)w/n) Lim --------------------------- + n->oo w/n f(x+w)-f(x+(n-1)w/n) Lim -------------------- ] n->oo w/n Which leads to (**) : 1 f(x+w/n)-f(x) AFD(f) = Lim - [ ------------- + n->oo n w/n f(x+2w/n)-f(x+w/n) ------------------ + w/n f(x+3w/n)-f(x+2w/n) ------------------- + w/n f(x+4w/n)-f(x+3w/n) ------------------- + w/n + ... f(x+(n-1)w/n)-f(x+(n-2)w/n) --------------------------- w/n f(x+w)-f(x+(n-1)w/n) -------------------- ] w/n Some more simplication: 1 f(x+w)-f(x) AFD(f) = Lim - [ ----------- ] n->oo n w/n Then, 1 n f(x+w)-f(x) AFD(f) = Lim -. [ ----------- ] n->oo n w So, f(x+w)-f(x) AFD(f) = Lim [ ----------- ] n->oo w And finally, f(x+w)-f(x) AFD(f) = ----------- w which is the desired result. (**) Dropping the inner limit is what I am not sure about. Jason Wells
From: denis feldmann on 4 Mar 2005 01:55 Jason a ýcrit : > Here is my attempt to prove Gabriel's Theorem: > > Have changed some of his wording and am calling it > the AFD rather than ATG as he calls it. > > 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 > [or AFD(f)] defined as follows: > > w is the width of the interval > n is the number of partitions > > f(x + w/n) - f(x) > AFD(f) = ------------------ --- L.H.S > w/n Wrong LHS; as said below, you meant in fact f(x + w) - f(x) AFD(f) = ------------------ w > > 1 n-1 ws > AFD(f) = Lim - SIGMA f'(x+ --- ) --- R.H.S > n->oo n s=0 n > > > Proof: > > Start with RHS. > > > 1 > AFD(f) = Lim - [ f'(x) + f'(x+w/n) + f'(x+2w/n) + ... > n->oo n > > f'(x+(n-2)w/n) + f'(x+(n-1)w/n)] > > > Now we simplify Gabriel's proof by keeping the ens fixed and using > t in the definition: > i.e. we let n be fixed for each of f'(x), f'(x+w/n), f'(x+2w/n), etc. > In this case, the following are true: > > > f(x+w/t)-f(x) > f'(x) = Lim -------------- > t->oo w/t > > This xorks only *on the right* of x > f'(x+w/n) = > > > f(x+w/n+w/t)-f(x+w/n) > Lim --------------------- > t->oo w/t > > > f'(x+2w/n) = > > > f(x+2w/n+w/t)-f(x+2w/n) > Lim ----------------------- > t->oo w/t > > > f'(x+ 3w/n) = > > > f(x+3w/n+w/t)-f(x+3w/n) > Lim ----------------------- > t->oo w/t > > > and so on: > > > Now, > 1 f(x+w/t)-f(x) > AFD(f) = Lim - [ Lim ------------- + > n->oo n t->oo w/t > > > > f(x+w/n+w/t)-f(x+w/n) > Lim --------------------- + > t->oo w/t > > > f(x+2w/n+w/t)-f(x+2w/n) > Lim ----------------------- + > t->oo w/t > > > f(x+3w/n+w/t)-f(x+3w/n) > Lim ----------------------- + > t->oo w/t > > + ... > > f(x+(n-2)w/n+w/t)-f(x+(n-2)w/n) > Lim ------------------------------- + > t->oo w/t > > > f(x+(n-1)w/n+w/t)-f(x+(n-1)w/n) > Lim ------------------------------- ] > t->oo w/t > > > Looking at the above we have two limits to infinity. > i.e. n and t. However, since both n and t tend to infinity, > the outer limit is taken once over the entire sum. > With n and t coinciding, we have: This is illegal (but works in this case for some other reason, as the final result is correct) : compare with lim x->0 lim y->0 x-y/x+y =1 and lim y->0 lim x->0 x-y/x+y = - 1 ; you cannot say let's x=y in the inner limit, can you ? > > > 1 f(x+w/n)-f(x) > AFD(f) = Lim - [ Lim ------------- + > n->oo n n->oo w/n > > > > f(x+2w/n)-f(x+w/n) > Lim ------------------ + > n->oo w/n > > > f(x+3w/n)-f(x+2w/n) > Lim ------------------- + > n->oo w/n > > > f(x+4w/n)-f(x+3w/n) > Lim ------------------- + > n->oo w/n > + ... > > f(x+(n-1)w/n)-f(x+(n-2)w/n) > Lim --------------------------- + > n->oo w/n > > > f(x+w)-f(x+(n-1)w/n) > Lim -------------------- ] > n->oo w/n > > > > Which leads to (**) : > > 1 f(x+w/n)-f(x) > AFD(f) = Lim - [ ------------- + > n->oo n w/n > > > > f(x+2w/n)-f(x+w/n) > ------------------ + > w/n > > > f(x+3w/n)-f(x+2w/n) > ------------------- + > w/n > > > f(x+4w/n)-f(x+3w/n) > ------------------- + > w/n > + ... > > f(x+(n-1)w/n)-f(x+(n-2)w/n) > --------------------------- > w/n > > > f(x+w)-f(x+(n-1)w/n) > -------------------- ] > w/n > > > Some more simplication: > > 1 f(x+w)-f(x) > AFD(f) = Lim - [ ----------- ] > n->oo n w/n > > > Then, > > 1 n f(x+w)-f(x) > AFD(f) = Lim -. [ ----------- ] > n->oo n w > > > So, > > f(x+w)-f(x) > AFD(f) = Lim [ ----------- ] > n->oo w > > And finally, > > f(x+w)-f(x) > AFD(f) = ----------- > w > > which is the desired result. > > (**) Dropping the inner limit is what I am not sure about. It's even worse : in your notation, n has disappeared when taking the inner limit, so you certainly cannot dropped it. > > Jason Wells >
From: David C. Ullrich on 4 Mar 2005 07:08
On 3 Mar 2005 10:43:57 -0800, "Jason" <logamath(a)yahoo.com> wrote: >Here is my attempt to prove Gabriel's Theorem: > >Have changed some of his wording and am calling it >the AFD rather than ATG as he calls it. > >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 Exactly what do you mean by "derivative" here? You've given several mutually inconsistent definitions at various times in these threads. For now I'll just assume that you mean the same thing as everyone else by "derivative". >[or AFD(f)] defined as follows: > >w is the width of the interval >n is the number of partitions > > f(x + w/n) - f(x) > AFD(f) = ------------------ --- L.H.S > w/n As Denis said, surely you meant (f(x+w) - f(x))/w here. > 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. Let's see about the proof: >Proof: > >Start with RHS. > > > 1 > AFD(f) = Lim - [ f'(x) + f'(x+w/n) + f'(x+2w/n) + ... > n->oo n > > f'(x+(n-2)w/n) + f'(x+(n-1)w/n)] > > >Now we simplify Gabriel's proof by keeping the ens fixed and using >t in the definition: >i.e. we let n be fixed for each of f'(x), f'(x+w/n), f'(x+2w/n), etc. >In this case, the following are true: > > > f(x+w/t)-f(x) > f'(x) = Lim ------------- > t->oo w/t > > > f'(x+w/n) = > > > f(x+w/n+w/t)-f(x+w/n) > Lim --------------------- > t->oo w/t > > > f'(x+2w/n) = > > > f(x+2w/n+w/t)-f(x+2w/n) > Lim ----------------------- > t->oo w/t > > > f'(x+ 3w/n) = > > > f(x+3w/n+w/t)-f(x+3w/n) > Lim ----------------------- > t->oo w/t > > > and so on: > > > Now, > 1 f(x+w/t)-f(x) > AFD(f) = Lim - [ Lim ------------- + > n->oo n t->oo w/t > > > > f(x+w/n+w/t)-f(x+w/n) > Lim --------------------- + > t->oo w/t > > > f(x+2w/n+w/t)-f(x+2w/n) > Lim ----------------------- + > t->oo w/t > > > f(x+3w/n+w/t)-f(x+3w/n) > Lim ----------------------- + > t->oo w/t > > + ... > > f(x+(n-2)w/n+w/t)-f(x+(n-2)w/n) > Lim ------------------------------- + > t->oo w/t > > > f(x+(n-1)w/n+w/t)-f(x+(n-1)w/n) > Lim ------------------------------- ] > t->oo w/t > > > Looking at the above we have two limits to infinity. > i.e. n and t. However, since both n and t tend to infinity, > the outer limit is taken once over the entire sum. > With n and t coinciding, we have: > > > 1 f(x+w/n)-f(x) > AFD(f) = Lim - [ Lim ------------- + > n->oo n n->oo w/n > > > > f(x+2w/n)-f(x+w/n) > Lim ------------------ + > n->oo w/n > > > f(x+3w/n)-f(x+2w/n) > Lim ------------------- + > n->oo w/n > > > f(x+4w/n)-f(x+3w/n) > Lim ------------------- + > n->oo w/n > + ... > > f(x+(n-1)w/n)-f(x+(n-2)w/n) > Lim --------------------------- + > n->oo w/n > > > f(x+w)-f(x+(n-1)w/n) > Lim -------------------- ] > n->oo w/n 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.) > > Which leads to (**) : > > 1 f(x+w/n)-f(x) > AFD(f) = Lim - [ ------------- + > n->oo n w/n > > > > f(x+2w/n)-f(x+w/n) > ------------------ + > w/n > > > f(x+3w/n)-f(x+2w/n) > ------------------- + > w/n > > > f(x+4w/n)-f(x+3w/n) > ------------------- + > w/n > + ... > > f(x+(n-1)w/n)-f(x+(n-2)w/n) > --------------------------- > w/n > > > f(x+w)-f(x+(n-1)w/n) > -------------------- ] > w/n > > > Some more simplication: > > 1 f(x+w)-f(x) > AFD(f) = Lim - [ ----------- ] > n->oo n w/n > > > Then, > > 1 n f(x+w)-f(x) > AFD(f) = Lim -. [ ----------- ] > n->oo n w > > > So, > > f(x+w)-f(x) > AFD(f) = Lim [ ----------- ] > n->oo w > > And finally, > > f(x+w)-f(x) > AFD(f) = ----------- > w > > which is the desired result. > > (**) Dropping the inner limit is what I am not sure about. 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. > Jason Wells ************************ David C. Ullrich |