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

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David C. Ullrich

Posts: 3,085
Registered: 12/13/04
Re: 1.37 - Quiz on Cauchy's Kludge.
Posted: Jun 19, 2014 5:44 PM
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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<|x-c|<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. x-c = 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.

>>
>>





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