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Topic:
1.37  Quiz on Cauchy's Kludge.
Replies:
7
Last Post:
Jun 20, 2014 4:26 PM




Re: 1.37  Quiz on Cauchy's Kludge.
Posted:
Jun 19, 2014 5:44 PM


On Thu, 19 Jun 2014 11:24:16 0700, John Gabriel wrote:
> On Thursday, June 19, 2014 7:33:28 PM UTC+2, dull...@sprynet.com wrote: >> On Wed, 18 Jun 2014 11:09:14 0700 (PDT), John Gabriel >> >> <thenewcalculus@gmail.com> wrote: >> >> >> >> >On Wednesday, 18 June 2014 18:03:10 UTC+2, dull...@sprynet.com wrote: >> >> >> On Wed, 18 Jun 2014 06:30:21 0700 (PDT), John Gabriel >> >> >> > >> >> >Mainstream mythmaticians 'define' the derivative as follows: >> >> >> > >> >> > f ' (x) = Lim (h>0) {f(x+h)f(x)} / h >> >> >> > >> >> >1. They treat the RHS as a standard limit where they must guess L, >> >> >so that >> >> >> > >> >> > 0<xc<delta <=> 0<{f(x+h)f(x)} / h  L<epsilon >> >> >> > >> >> The fact that you still haven't got such a simple thing straight >> >> looks bad. >> >> >> > >> >I've corrected you a few times. That you still don't get it means you >> >are very dense. >> >> >> > >> >> Two errors: The limit is as h > 0, not x > c. >> >> >> > >> >Huh? Exactly the same thing. xc = h >> >> >> > >> >> And it => in the definition, not <=>: >> >> >> > >> >Wrong. >> >> >> >> For heaven's sake find a calculus book. > > I have read thousands. Since I was 14 years. I know very well what > mainstream mythmatics says. > > Once ? is expressed in terms of ?, then you can use <=>. The implication > goes both ways.
In general it's impossible to express epsilon in terms of delta. Which is irrelevant  it's => in the definition.
>> >> >2. The RHS returns a value, so it's a mystery why they assign that >> >> >value to f'(x) which may not even be defined. >> >> >> > >> >> You're making no sense here. >> >> >> The RHS is the _definition_ of f'(x); if the RHS exists then yes >> >> f'(x) is defined. >> >> >> > >> >Bullshit. I can define f(x)=x^2 and f'(3)=0. In which case, the RHS >> >still produces the limit, but f'(3) =/= RHS. >> >> >> >> No, if the definition is as above, which by the way it _is_, then you >> can't define f'(3) = 0. > > Sure I can. Hey, it's a definition. :) You did it earlier with your > x^2sin(1/x) example. Same thing really.
Nope. The problem is you're giving two different definitions for f'(3). Once you define it as above you can't define it as something else.
> >> >> >They know this, so they stipulate the condition (rule) clearly. >> >> >> >3. It's know that in the definition of a limit, a function need not >> >> >be continuous or even defined at that point. >> >> >> > >> >> Yes, and once we correct your confusion over h > 0 versus >> >> >> x > c we see that this definition is a good example why that's >> >> >> so important. >> >> >> > >> >They're exactly the same. >> >> >> > >> >> Here x is fixed, and we're talking about a certain function of h, >> >> namely >> >> >> >> > >> >> g(h) = (f(x+h)f(x))/h. >> >> >> > >> >You are very confused. We are talking about g(c) = (f(c+h)f(c))/h. >> >> >> > >> >> Oh, now I get it. You're talking about two different definitions of >> >> >> the derivative simulatneously  the real one and yours. >> >> >> > >> >You don't get it and you never have. >> >> >> > >> >> You should use a different word for yours. > > Oh, quit being such a pain Ullrich! > > >> >> > >> >> >[C] Cauchy's kludge is a failed attempt at defining the gradient of >> >> >a tangent line at a given point c in an interval (a,b) because it >> >> >requires that a derivative exist at every point in the interval >> >> >containing c. >> >> >> > >> >> Erm, no, it requires no such thing. >> >> >> > >> >Yes, it does. The link to the applet shows this dynamically. >> >> >> > >> >> >[D] Mythmaticians have tried to dismiss the definition of tangent >> >> >line in order to support their flawed thinking. >> >> >> >> >> >> >[E] John Gabriel's New Calculus solves all these problems  it is >> >> >the first and only rigorous calculus in human history. >> >>



