John Gabriel <email@example.com> wrote in news:firstname.lastname@example.org:
> On Sunday, 23 February 2014 08:59:57 UTC+2, Wizard-Of-Oz wrote: >> John Gabriel <email@example.com> wrote in >> news:firstname.lastname@example.org: > >> http://www.spacetimeandtheuniverse.com/math/4507-0-999-equal-one- 560.h >> tml#post27297 Let f(x) = x^3 >> Using Cauchy's definition: >> f'(x) = 3x^2 + 3xh + h^2 >> We can write that as >> f'(x) = 3x^2 + Q(x,h) where Q(x,h) = 3xh + h^2 >> Set h = 0 and we get >> f'(x) = 3x^2 >> Using New Calculus: >> f'(x) = 3x^2 + 3x(n-m) + m^2 -mn + n^2 >> We can write that as >> f'(x) = 3x^2 + Q(x,m,n) where Q(x,m,n) = 3x(n-m) + m^2 -mn + n^2 >> Set m = 0 and we get >> f'(x) = 3x^2 + Q(x,0,n) where Q(x,0,n) = 3xn + n^2 > >> Which is the same as Cauchy > > That is not correct.
Yes it is. I note that you did not point out any errors in the above
> 1. The New Calculus does not use infinity, limits, infinitesimals or > any other ill-formed concept.
Neither did I above. Yet you claim it is not correct.
> 2a) m, n and x have a special relationship in the new calculus and can > be used to find the gradient of the tangent line.
As can old calculus. Nothing new there.
> 2b) h and x have no relationship and no bearing to the tangent line.
As with the new calculus, set what you call the Q function to zero and you get the tangent. There is no difference there between old and new, except you add an extra variable that gets set to zero.
> 3) The New Calculus does not contain any kludges such as division by 0 > in order to find the general derivative. Cauchy's definition is a > kludge.
There is no difference in the 'kludges'. In both cases we rewrite the function and then substitute h=0 or m=n=0 accordingly when there is no division by h or m, n
>>(just the change in pronumeral name from h to n, which is >>insignificant) > > Nonsense, nothing to do with pronumeral name/s.
That's what I said. So you claim what I said is nonsense and then agree with it. Typical of a troll.
> h is in no way related > to n.
It is identical when m = 0
> In fact Cauchy's definition cannot be translated to the New > Calculus.
As I already said, the new calculus with m = 0 is just the old calculus. And as you set m = 0 to get yout answer, there is effectively no difference between them
> If you set h=m+n,
Noone said that you do. That's your own stupid idea.
> you get f'(x) = lim (m+n->0) [ f(x+m+n) - f(x)] / > (m+n) which is hogwash and completely unrelated to the New Calculus.
I already showed setting m = 0 in yours give the identical results to 'old' calculus. There is nothing new of note in the new calculus
> One can however arrive at the New Calculus definition by using the > Mean Value Theorem: > > If f'(x) =[ f(b)-f(a) ] / (b - a) then let b-a=m+n and b=x+n: > > So, f'(x)= [ f(x+n)-f(x-m) ] / (m+n)
So you just plagiarised the mean value theorum and claimed it as your own. Nice of you to admit it.
>> Set n = 0 and we get >> f'(x) = 3x^2 >> There is nothing new about this so-called 'New Calculus'. It adds >> complexity, is defined in fewer placed and gives us nothing new or >> interesting. > > That's just your ignorant opinion.
No, it is fact.
You do not allow a differential at a point of inflection, so there are fewer places your method can supposedly work. And having both m and n is more complex than just having h. You get the same answers that the 'old' calculus gives, so there is nothing new.
> From your past comments, it is easy > to see that for the most part, you keep getting things wrong.