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Topic: The objects that Newton played with were called infinite series
but had ZERO to do with infinity. The name infinite series is a misnomer.

Replies: 9   Last Post: Oct 6, 2017 1:37 AM

 Messages: [ Previous | Next ]
 bursejan@gmail.com Posts: 5,511 Registered: 9/25/16
Re: The objects that Newton played with were called infinite series
but had ZERO to do with infinity. The name infinite series is a misnomer.

Posted: Sep 30, 2017 2:26 PM

We can apply the theorem blindly to the two
examples of John Gabriel, since we proved it

s = 1/2 + 1/4 + 1/8 + ... = 1

t = 1/3 + 1/9 + 1/27 + ... = 1/2

Now the partial sums are:

s_n = 1 - 2^(-n)

t_n = 1/2*(1 - 3^(-n))

The product partial sums are:

s_n*t_n = 1/2*(1 - 2^(-n) - 3^(-n) + 6^(-2n))

Lets show it as a sum again:

s_n*t_n - s_(n-1)*t_(n-1) = 1/2 (2^(-n) + 2 3^(-n) - 35 6^(-2 n))

So we get:

s*t = 7/72 + 577/2592 + 9253/93312 + 146413/3359232 + .. = 1/2

Note we dont do the sum this way (triangle):

* * * _ _
* * _ _ _
* _ _ _ _
_ _ _ _ _
_ _ _ _ _

What happens is doing the sum this way (boxes):

* * * _ _
* * * _ _
* * * _ _
_ _ _ _ _
_ _ _ _ _

To see how the box is built, consider:

A A B _ _
A A B _ _
C C D _ _
_ _ _ _ _
_ _ _ _ _

Or algebraically:

s_n*t_n - s_(n-1)*t_(n-1) =

(s_n - s_(n-1)) * t_(n_1) +

s_(n-1) * (t_n - t_(n-1)) +

(s_n - s_(n-1))*(t_n - t_(n-1)) +

It could be that Newton use the triangle and
not the boxes. The boxes are obviously in relationship
to Cauchy. How to go about showing Cauchy with the
triangles I dunnot know right now.

Am Samstag, 30. September 2017 20:00:01 UTC+2 schrieb burs...@gmail.com:
> Well there is these rules:
>
> (lim n->oo an)*(lim n->oo bn) = lim n->oo (an*bn)
>
> (lim n->oo an)+(lim n->oo bn) = lim n->oo (an+bn)
>
> Thats why Cauchy sequences are clause concerning
> multiplication and additiona. Then it is only a small
> step to see that Cauchy sequences limits are real numbers,
>
> For a proof that multiplication {an*bn} and addition
> {an+bn} are also Cauchy if the original {an} and {bn}
> where Cauchy, just observer:
>
> Proof:
>
> By definition, since we assume {an} Cauchy, for every e,
> there is an N, such:
>
> forall n,m >= N(e) |an-am| =< e
>
> And the same for {an}, since we assume it Cauchy as
> well, for every d there is an M, such that:
>
> forall n,m >= M(d) |bn-bm| =< d
>
> Now assume we have a further epsilon g, lets see whether
> we can admister an index K and J, such that:
>
> forall n,m >= K |(an*bn)-(am*bm)| =< g
>
> forall n,m >= J |(an+bn)-(am+bm)| =< g
>
> we have |(an+bn)-(am+bm)| = |an-am+bn-bm|, by
> the triangle inequality, we have:
>
> |an-am+bn-bm| =< |an-am|+|bn-bm|
>
> Now use J=max(N(g/2),M(g/2)), we clearly have:
>
> forall n,m >=J |an-am| =< g/2
>
> forall n,m >=J |bn-bm| =< g/2
>
> Amd therefore:
>
> forall n,m >=J |an-am|+|bn-bm| =< g
>
> And hence as required:
>
> forall n,m >=J |(an+bn)-(am+bm)| =< g
>
> Now lets try multipiclation: We have |(an*bn)-(am*bm)|=
> |an*(bn-bm)+bm*(an-am)|, by the triangle inequality,
> we have:
>
> |(an*bn)-(am*bm)| =< |an*(bn-bm)|+|bm*(an-am)|
>
> = |an|*|bn - bm| + |bm|*|an-am|
>
> Now we use the property that Cauchy sequence are bounded,
> they never tend to +oo or -oo, there fore we have
> A with |an|=<A and B with |bm|=<B irrespective of the
> index n or m. And hence:
>
> =< A*|bn - bm| + B*|an-am|
>
> Now use K=max(N(g/(2*B)),M(g/(2*A))), we clearly have:
>
> forall n,m >=K |an-am| =< g/(2*B)
>
> forall n,m >=K |bn-bm| =< g/(2*A)
>
> Or respectively:
>
> forall n,m >=K B*|an-am| =< g/2
>
> forall n,m >=K A*|bn-bm| =< g/2
>
> Amd therefore:
>
> forall n,m >=K A*|bn-bm|+B*|an-am| =< g
>
> And hence as required:
>
> forall n,m >=K |(an*bn)-(am*bm)| =< g
>
>
> Proving that product of two Cauchy sequences is Cauchy
> https://math.stackexchange.com/questions/376324/proving-that-product-of-two-cauchy-sequences-is-cauchy
>
> Cauchy Sequence: Multiplication Property
> https://math.stackexchange.com/questions/309867/cauchy-sequence-multiplication-property
>
> Am Samstag, 30. September 2017 07:30:40 UTC+2 schrieb John Gabriel:

> > The objects that Newton played with were called infinite series but had ZERO to do with infinity. The name infinite series is a misnomer.
> >
> > s = 1/2+1/4+1/8+... = 3/6+3/12+3/24+...
> > t = 1/3+1/9+1/27+... = 2/6+2/18+2/54+...
> >
> > s * t = 6/36 + 6/108 + 3/108 + 3/108 + 1/12 + 3/324 + 1/24 +1/72 + 6/1296+...
> >
> > If my arithmetic is correct, then you end up getting:
> >
> > s * t = 6/36 + 12/108 + 24/324 + ... = 1/2
> >
> > So all Newton did was work with the LIMITS. Nothing with infinity. By taking sufficient terms he was able to calculate the product of the limits. So strictly speaking he is not multiplying series at all, ONLY some of the partial sums and from these obtaining the limit.
> >
> > Newton used this approach in determining sine series through inversion. He knew that he might end up with a series that could no longer be summed as in the case of these example geometric series, but he also knew that if he could find a pattern, then he would be able to approximate the sine ratio.
> >
> > This is hard evidence that it's a very bad idea to define S = Lim S.
> >
> > How arc length was derived:
> >
> >
> > No doubt the majority of the morons on this site will not be able to produce sufficient inference to reach an AHA moment. The orangutans will simply dismiss all of this without any serious study or consideration. Too bad.
> >
> > Comments are unwelcome and will be ignored.
> >
> > Posted on this newsgroup in the interests of public education and to eradicate ignorance and stupidity from mainstream mythmatics.
> >
> > gilstrang@gmail.com (MIT)
> > huizenga@psu.edu (HARVARD)
> > andersk@mit.edu (MIT)
> > david.ullrich@math.okstate.edu (David Ullrich)
> > djoyce@clarku.edu
> > markcc@gmail.com