Possible proof of Gabriel's Theorem?
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From: Mike Oliver on 8 Mar 2005 17:16 Randy Poe wrote: > Jason wrote: > >>Q: If f' exists everywhere need f' be continuous? >>A: No. >> >>I cannot think of *one* example where this is true. Again you are >>stating something irrelevant and skirting the main issue at hand. > > > Surely you're not saying you can't think of a function > which exists everywhere but is not continuous. > > If the left and right limits of f(x) as x->x0 are not > the same, f is not continuous. > > Consider: > 1. Does a step function exist everywhere? > 2. Is the step function continuous? > 3. Can you think of a function whose derivative > is a step function? That's not going to get the desired example. A step function has no antiderivative (though it has an "almost-everywhere antiderivative" or whatever the correct term is). One standard example of an everywhere-differentiable function whose derivative is not continuous is: | x^2 sin(1/x) , x != 0 f(x) = | | 0 , x = 0
From: Mike Oliver on 8 Mar 2005 17:25 Jesse F. Hughes wrote: > I'm the unfamous JFH, Are you sure you're not *in*famous? That's when you're *more* than famous, as in the *in*famous El Guapo.
From: Jason on 8 Mar 2005 18:21 Ullrich: > Curious that you didn't notice the _error_ in the version I posted then. Jason: >I do not fully agree with it. As I stated earlier, I don't like to talk about constant >functions such as f(x) = c for c some constant as being differentiable. >They are not differentiable in my opinion. I don't pay much attention when you write incomprehensible junk. So you made a typo - big deal. I overlooked it because it was not important. See, this is what I mean about you: You are not interested in answering any questions, only in deviating from the main issue and knit-picking at stuff which really has nothing to do with the subject. You picked up where some twit had left off. Don't you have an opinion of your own? Show me step by step where gabriel's theorem is wrong or shut up. I meant that a delta > 0 does not exist, therefore we don't even need to look for an epsilon. since length(I_j) < delta and delta is eventually zero, no delta > 0 exists. This is a common problem with this sort of epsilon-delta analysis. Delta has to eventually be zero or else the integral is incomplete and thus not *integrable*.
From: John Gabbriel on 8 Mar 2005 22:10 Jason wrote: > > Silly boy. How hard is it to comprehend the fact that there could be > > more than one John Gabriel in this world? Why, there are at least 10 > in > > the state of Texas alone. > > There are a lot of fools in this world. How hard is it for you to > comprehend that you are one? Taking your maturity into account, a reply to this would be: "It takes one to know one". Giggle Giggle.. tee hee.. Seriously though, Bubba, ever thought of seeing a doctor? Maybe if you take JSH along you will get a discount. Heck, the doctor might even say that you are right. Good luck.
From: David C. Ullrich on 9 Mar 2005 08:38
On 8 Mar 2005 13:24:44 -0800, "Jason" <logamath(a)yahoo.com> wrote: >Q: If f' exists everywhere need f' be continuous? >A: No. > >I cannot think of *one* example where this is true. Fasinating. A few days ago you said you _had_ thought of such an example. (I pointed out that that must be a lie since you couldn't possibly have thought of such a thing given how confused you are about everything. I guess I was right, it was a lie.) An example has been given _many_ times in this thread: f(x) = x^2 sin(1/x) for x <> 0, f(0) = 0. >Again you are >stating something irrelevant and skirting the main issue at hand. ************************ David C. Ullrich |