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Topic: John Gabriel's Thread on Mathematics.
Replies: 231   Last Post: Mar 22, 2014 9:23 PM

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 Wizard-Of-Oz Posts: 404 Registered: 12/28/13
Re: John Gabriel's Thread on Mathematics.
Posted: Feb 24, 2014 10:06 PM
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John Gabriel <thenewcalculus@gmail.com> wrote in
news:bca901e4-e31e-46c1-970f-b18dfdf980a3@googlegroups.com:

> On Monday, 24 February 2014 20:41:47 UTC+2, Spac...@hotmail.com
> wrote:

>> this excerpt shows that you Rock cosmoS, and John is his own worst
>> trolL
>>
>>
>>

>> > And the same division by zero that you complained about above.
>> > Using

>>
>> >
>>
>> > limits avoids the actual division by zero, but you don't use
>> > limits, so

>>
>> >
>>
>> > it is a real problem for you.
>>
>> >
>>
>> >
>>
>> >
>>
>> > > Morover, any valid (m,n) pair
>>
>> >
>>
>> >
>>
>> >
>>
>> > How do you determine a valid pair?
>>
>> >
>>
>> >
>>
>> >
>>
>> > > will still work in the New Calculus gradient, that is, f'(x) = 2x
>> > > + n

>>
>> >
>>
>> > > - m, which gives the general derivative if m=n=0, and also the
>> > > numeric

>>
>> >
>>
>> > > derivative when x=c.
>>
>> >
>>
>> >
>>
>> >
>>
>> > What c? There was no mention of a c there.
>
> Let the endpoints of a parallel secant line on a function f be [c-m,
> f(c-m)] and [c+n, f(c+n)].
>
> The gradient of any such parallel secant line is given by:
>
> f'(c)= (f(c+n)-f(c-m))/(m+n) [A]

That is also the forlmula for a non-parallel secant line

There is nothing in that formula that forces it to be parallel.

f'(c) is then not the derivative of f at c, it is the slow of some
arbitrary secant at c.

Your labelling is dishonest and deliberately misleading

> Now, m+n divides f(c+n)-f(c-m) exactly, that is,

Define "exactly". Prove your assertion

> f'(c) = gradient(c) + Q(c,m,n) [B]

So it is the gradient plus some other number. So you are saying that
the derivative at c is not the same as the gradient at c.

> where gradient(c) is the expression of the gradient in terms of c
> only. Q(c,m,n)=0 in [B].

How do you know? It will depend on the value of m and n

So your method then is to try to divide and simplify (f(c+n)-f(c-m))/
(m+n) and then hope that all the terms in m and n can be grouped
together. And that those terms in m and n will end up as zero for some
m and n. And that what is left will be the gradient.

Where is your rigorous justification for such a simplification to always
be possible?

Where is your rigorous justification for the Q function to have 0 as a
possible value.

Where is your rigorous justification for the terms other than the Q
function to be the gradient at c.

Where is your rigorous justification for there to be no divisions by
zero when you subsitute values for c, m and n into Q.

> However, any (m,n) pair of any parallel
> secant line will satisfy Q(c,m,n)=0.

Where is your rigorous justification for Q being zero for secants
parallel to the gradient (slope or derivative) of f at c?.

> Moreover, the same (m,n) will
> produce the gradient regardless of whether [A] or [B] is used.

Where is your rigious justification for [A] to be able to be expressed
in form [B] for every function f?

> Let's do an example:
>
> f(x)=x^3.
>
> Suppose we want to find the derivative at x=3.
>
> f'(x) = { 3x^2(m+n) + 3x(n-m)(m+n)+(m^2 - mn + n^2)(m+n) } / (m+n)
> [A]

Only for SOME values of m and n

> f'(x) = 3x^2 + 3x(n-m)+m^2 - mn + n^2 [B]

Only for SOME values of m and n

> Or
>
> f'(x) = gradient(x) + Q(x,m,n) [C]

Only for SOME values of m and n

Where is your rigious justifications that the terms you get after you
groups terms in m and n as Q(x,m,n) will be the gradient?

> where gradient(x)=3x^2 and Q(x,m,n)= 3x(n-m)+m^2 - mn + n^2
>
> To find any valid pair, we use x=3 with either of m or n. So let's
> choose m=1.
>
> Then,
>
> Q(3,1,n)=3(3)(n-1)+(1)^2 - (1)n + n^2 = 9n-9+1-n+n^2=n^2+8n-8=0
>
> So, n=-4+2sqrt(6) or n=-4-2sqrt(6)

So you find m and n pairs by assuming the Q(x,m,n) will be zer, and then
assuming that the resultant secant will have the same slope as the f(x)
curve at x

> Since -4+2sqrt(6) lies to the right of x=3, the (m,n) pair we require
> is (1,-4+2sqrt(6)).
>
> Substituting this into [A], we have f'(3)=27 as expected.
>
> We could also have used [B] with the pair (1,-4+2sqrt(6)):
>
> f'(3)=3(3)^2+3(3)(-4+2sqrt(6)-1)+(1)^2-(1)(-4+2sqrt(6))+(-4+2sqrt(6))^

2
> [B]
>
> f'(3)=27+9(-5+2sqrt(6))+1-(-4+2sqrt(6))+16-16sqrt(6)+24
>
> f'(3)=27-45+18sqrt(6)+5-2sqrt(6)+16-16sqrt(6)+24
>
> f'(3)=27-45+18sqrt(6)+5-2sqrt(6)+16-16sqrt(6)+24
>
> f'(3)=27 as expected.
>
> Of course you do not need to go through all this because
>
> f'(x) = gradient(x) + Q(x,m,n) = gradient(x)
>
> So, that's how you can find any valid (m,n) pair you like.

There is no point in finding m and n, if you are simply deliberately
chosing m and n that make Q(x,m,n) = 0 and then substituting those m and
n into Q(x,m,n) to get 0.

So you method is, get a formula for the slope of any secant, and express
its endpoints as offsets from c (m and n). You simplify the expression
and hope that you end up with a term not involving the offset and
another term that do involved the offsets. Then you hope that the first
term is actually the gradient and that the second term can give a value
of zero for some values or m and n. And you hope that there is no
division be zero in either term.

And all that is supposed to be rigorous? Where is your rigorous proofs
that each of the assumption or hopes in the above is valid?

Date Subject Author
2/22/14 thenewcalculus@gmail.com
2/23/14 Brian Q. Hutchings
2/23/14 Brian Q. Hutchings
2/23/14 Wizard-Of-Oz
2/23/14 Wizard-Of-Oz
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2/24/14 Wizard-Of-Oz
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2/26/14 thenewcalculus@gmail.com
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2/24/14 Brian Q. Hutchings
2/23/14 Art Hopkins
2/23/14 thenewcalculus@gmail.com
2/23/14 thenewcalculus@gmail.com
2/23/14 Dirk Van de moortel
2/23/14 thenewcalculus@gmail.com
2/23/14 thenewcalculus@gmail.com
2/23/14 thenewcalculus@gmail.com
2/24/14 thenewcalculus@gmail.com
2/24/14 Wizard-Of-Oz
2/24/14 Giddy Armstrong
2/24/14 thenewcalculus@gmail.com
2/24/14 thenewcalculus@gmail.com
2/26/14 thenewcalculus@gmail.com
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2/26/14 YBM
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2/26/14 Roland Franzius
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2/26/14 thenewcalculus@gmail.com
2/26/14 thenewcalculus@gmail.com
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2/26/14 Inverse 18 Mathematics
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2/26/14 Inverse 18 Mathematics
2/26/14 Karl-Olav Nyberg
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2/26/14 Wizard-Of-Oz
2/26/14 Brian Q. Hutchings
2/26/14 Brian Q. Hutchings
2/26/14 thenewcalculus@gmail.com
2/26/14 Wizard-Of-Oz
2/26/14 thenewcalculus@gmail.com
2/26/14 Wizard-Of-Oz
2/26/14 thenewcalculus@gmail.com
2/26/14 thenewcalculus@gmail.com
2/27/14 Wizard-Of-Oz
2/27/14 thenewcalculus@gmail.com
2/27/14 David C. Ullrich
2/27/14 thenewcalculus@gmail.com
2/28/14 David C. Ullrich
2/28/14 Brian Q. Hutchings
2/28/14 Peter Percival
2/28/14 fom
2/28/14 Brian Q. Hutchings
2/28/14 John Gabriel
3/1/14 David C. Ullrich
3/1/14 ross.finlayson@gmail.com
3/1/14 thenewcalculus@gmail.com
3/1/14 thenewcalculus@gmail.com
3/1/14 Karl-Olav Nyberg
3/1/14 thenewcalculus@gmail.com
3/2/14 Wizard-Of-Oz
3/2/14 David C. Ullrich
3/2/14 thenewcalculus@gmail.com
3/2/14 David C. Ullrich
3/2/14 John Gabriel
3/3/14 John Gabriel
3/3/14 David C. Ullrich
3/3/14 thenewcalculus@gmail.com
3/4/14 David C. Ullrich
3/4/14 thenewcalculus@gmail.com
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3/4/14 Karl-Olav Nyberg
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3/5/14 David C. Ullrich
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3/6/14 David C. Ullrich
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3/6/14 Brian Q. Hutchings
3/3/14 Roland Franzius
3/3/14 thenewcalculus@gmail.com
3/3/14 ross.finlayson@gmail.com
3/4/14 ross.finlayson@gmail.com
3/3/14 David C. Ullrich
3/3/14 thenewcalculus@gmail.com
3/3/14 Roland Franzius
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3/4/14 thenewcalculus@gmail.com
3/8/14 Peter Percival
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3/8/14 Virgil
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3/9/14 Peter Percival
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