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

[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:

https://drive.google.com/open?id=0B-mOEooW03iLeGt2ZlViMzNyYTg

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.

gils...@gmail.com (MIT)
huiz...@psu.edu (HARVARD)
and...@mit.edu (MIT)
david....@math.okstate.edu (David Ullrich)
djo...@clarku.edu
mar...@gmail.com

[konyberg]

lørdag 30. september 2017 07.30.40 UTC+2 skrev John Gabriel følgende:
> 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+...

Why are you expanding the fractions?
It's like you are preparing to add them.

>
> 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

You know that you can not in general do this with series ?
KON

[konyberg]

And your arithmetic should give you 1/3.
KON

[John Gabriel]

On Saturday, 30 September 2017 03:14:20 UTC-5, konyberg wrote:
> lørdag 30. september 2017 07.30.40 UTC+2 skrev John Gabriel følgende:
> > 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+...
> Why are you expanding the fractions?
> It's like you are preparing to add them.

Because I know both series are geometric and that rearranging the terms results in a new geometric series.

> >
> > 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
> You know that you can not in general do this with series ?

Yes. Of course. You can only do it if both the series **converge**. Newton knew this. He first tried to obtain the sum of inverse sine as a general formula but realising that he was not able to do this, he knew that if he inverted the series that he would obtain a means of approximating the same by using a new series's partial sums.

[John Gabriel]

Huh? 1/2 + 1/4 + ... has a limit of 1 = 1/2 / (1 - 1/2)
1/3 + 1/9 + ... has a limit of 1/2 = 1/3 / ( 1 - 1/3 )

1/2 = 1 x 1/2

The arithmetic is correct.

[Zelos Malum]

And it isn't you dumbarse.

[John Gabriel]

What isn't you monkey? I can't interpret your nonsense question.

[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,
real numbers admiting multiplication and addition.

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

Lets first start with addition, this is simpler,
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

For the multiplication proof see also here:

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

[burs...@gmail.com]

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

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.

[burs...@gmail.com]

This can be used for a little variant of proving
the Cauchy product, we also have:

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

(s_n - s_(n-1)) * (t_n + t_(n_1))/2 +

(s_n + s_(n-1))/2 * (t_n - t_(n-1)).

Or graphically:

A A B B _
A A B B _
C C X B _
C C C X _

_ _ _ _ _

[John Gabriel]

On Saturday, 30 September 2017 13:26:37 UTC-5, burs...@gmail.com wrote:
> We can apply the theorem blindly to the two
> examples of John Gabriel, since we proved it
> already in general:

The piece of shit Cauchy wasn't even born when Newton pondered these things.
You consistently misunderstand the crux of my comments. It would help if you tried to at least improve your reading comprehension skills.

The conclusion that infino-sero (infinite series) arithmetic works on convergent series was understood by Newton. It did not require a French fool to state it "formally".

Again, you are stating very obvious facts: it's obvious that what is true for limits will be true also of their partial sums.

The irony is that this clearly nixes your wrong ideas about S = Lim S and that you still insisting a series is a limit is quite absurd. What applies to the partial sums is true ONLY because it applies to the LIMIT. Transfer this same idea to the shit you keep repeating about Cauchy sequences. Perhaps you'll learn not to be such moron all the time?

>
> 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.

Irrelevant and boring crap.

[burs...@gmail.com]

Formally you can multiply two series, even if they
are not coverging. For example by the box method,

you can multiply the following two series:

s = 1 - 1 + 1 - 1 ... = diverges

t = 0 + 1 - 1 + 1 ... = diverges

By the box method you get:

s*t = 0 + 0 + 0 + 0 + ... = 0

So Newton and all these guys that don't use convergence
criteria, might get converging series as some end-product,
maybe claiming wrong poperties about these series.

For example if Newton would believe that s had a value
and that t had a value, he would have believed that
he had found a factoring of zero:

s*t = 0

Or there are bird brains such as John Gabriel, who
believe they have found a new calculus, by not working
rigorously.

[burs...@gmail.com]

Note: product Cauchy property, uses the box method.
Cauchy product, is usually defined as triangle method.

https://en.wikipedia.org/wiki/Cauchy_product#Cauchy_product_of_two_infinite_series

[burs...@gmail.com]

Practically, in real algebra, the box method
is used and not the triangle method, see also here:

4.2 Algebraic operations on sequences
http://www.math.cornell.edu/~kahn/reals07.pdf

Proposition 5 (Page 13). Suppose that {an} and {bn} are Cauchy
sequences. Then {an}+{bn}, {an} - {bn}, and {an} * {bn}
are also Cauchy sequences.

He does also the reciprocal:

Corollary 2 (Page 24). Suppose that {an} is a very positive Cauchy
sequence such that every an is non-zero. Then, the sequence
{1/an} is a very positive Cauchy sequence.

For the later we see that |1/an - 1/am|=|(am-an)/(an*am)| =
|an-am|/|an*am|. Now we use again that the Cauchy sequence is
bounded by some A. This time we need that A is bound from below,
which gives us already:

|an-am|/|an*am| =< |an-am|/A^2

We can use M=N(d*A^2), we then have:

forall n,m>=M |an-am| =< d*A^2

Hence:

forall n,m>=M |an-am|/A^2 =< d

And therefore:

forall n,m>=M |an-am|/|an*am| =< d

And finally:

forall n,m>=M |1/an-1/am| =< d

[John Gabriel]

On Saturday, 30 September 2017 13:42:44 UTC-5, burs...@gmail.com wrote:
> Formally you can multiply two series, even if they
> are not coverging.

Tsk, tsk. No. You cannot do infino-sero arithmetic with series that do not converge. Period.

[Zelos Malum]

Actually dipshit, you can cause the question of convergence depends on a lot of factors and what converges in one instance, can diverge in another.

[John Gabriel]

I piss and shit on you retard. Chuckle.

[konyberg]

Yes. I am sorry. I looked at the series starting from n =1 , but it is n = 0.
Therefor you get 1/6 + 1/3 = 1/2
KON

[burs...@gmail.com]

Can you make an example, where two series cannot
be multiplied formally, independent of their
convergence?

For the case of Cauchy product:

s_n = sum_i=1^n a_i
t_n = sum_i=1^n b_i
(s*t)_n = sum_i=1^n c_i
where c_i=sum_k=1^i a_k*b_(i+1-k)

For the case of algebra product:

s_n = sum_i=1^n a_i
t_n = sum_i=1^n b_i
(s*t)_n = s_n*t_n

[John Gabriel]

On Tuesday, 3 October 2017 17:15:59 UTC-4, burs...@gmail.com wrote:
> Can you make an example, where two series cannot
> be multiplied formally, independent of their
> convergence?

Birdbrains. It makes no sense to multiply any two series unless they both converge. The idea is that you are multiplying the LIMITs by multiplying the partial sums.

YOU BIG MOOOOOOROOOOOOON!!!!

[burs...@gmail.com]

Can you make an example, where I cannot form
the partial sums of s*t, despite that s or t,
is not convergent?

Do you understand what we mean by formally
multiplying?

P.S.: I am refering to your claim:
So whats your example?

> On Saturday, 30 September 2017 13:42:44 UTC-5, burs...@gmail.com wrote:
> > Formally you can multiply two series, even if they
> > are not coverging.
> Tsk, tsk. No. You cannot do infino-sero arithmetic with
> series that do not converge. Period.

[John Gabriel]

On Tuesday, 3 October 2017 17:35:54 UTC-4, burs...@gmail.com wrote:
> Can you make an example, where I cannot form
> the partial sums of s*t, despite that s or t,
> is not convergent?

You may form the product Birdbrains, but it has no meaning. What part of that is still overwhelming your birdbrain circuits?

>
> Do you understand what we mean by formally
> multiplying?

Chuckle. It means you having a vasectomy as soon as possible. The world cannot afford to have another BIG MORON like you! Hurry and do it soon!

[burs...@gmail.com]

I didn't say the product has a limit, where
did I write this? Can you show me? What
are you bragging about?

Is this your parrot touret syndrom, posting
shit over other peoples posts?

[John Gabriel]

On Tuesday, 3 October 2017 17:48:47 UTC-4, burs...@gmail.com wrote:
> I didn't say the product has a limit, where
> did I write this?

Of course you didn't clueless birdbrains! I said the product must be meaningful and therefore it must be a limit. If either series is not convergent, the product is meaningless. Put down the bottle!!!

Never mind, go back to your wanking.

Did you make an appointment to get your snip-snip? Do it now!!!!

[Zelos Malum]

Whats with you and your scat and watersport fetish?

[Markus Klyver]

Den lördag 30 september 2017 kl. 07:30:40 UTC+2 skrev John Gabriel:

When we talk about infinity, it's usually understood as a limit. Particularly, this applies to infinite sums which are limits of finite sums.

[Markus Klyver]

You can, but it's no guarantee the product will converge or to what it will converge to. You can multiply two divergent series and get an convergent one, for example.

[genm...@gmail.com]

Nonsense. Provide an example you ignoramus!

[genm...@gmail.com]

I do not allow you to use concepts that are ill-formed or which I have not approved. Do you understand idiot?

You don't know what you are talking about. I do.

[Markus Klyver]

Sure, consider two infinite sums with the general terms 1/n and 1/(n+1). Now, when multiplied, all terms will be at most 1/n^2. They will also be all positive, so by the squeeze theorem the product will converge.

Den onsdag 4 oktober 2017 kl. 22:40:03 UTC+2 skrev genm...@gmail.com:
> On Wednesday, 4 October 2017 15:17:21 UTC-4, Markus Klyver wrote:

> > Den lördag 30 september 2017 kl. 07:30:40 UTC+2 skrev 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:
> > >
> > > https://drive.google.com/open?id=0B-mOEooW03iLeGt2ZlViMzNyYTg
> > >
> > > 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.
> > >
> > > gils...@gmail.com (MIT)
> > > huiz...@psu.edu (HARVARD)
> > > and...@mit.edu (MIT)
> > > david....@math.okstate.edu (David Ullrich)
> > > djo...@clarku.edu
> > > mar...@gmail.com
> >

> > When we talk about infinity, it's usually understood as a limit. Particularly, this applies to infinite sums which are limits of finite sums.
>
> I do not allow you to use concepts that are ill-formed or which I have not approved. Do you understand idiot?
>
> You don't know what you are talking about. I do.

Then please show how these concepts are ill-formed. Just stating they are does not prove it.

[Zelos Malum]

Den onsdag 4 oktober 2017 kl. 22:40:03 UTC+2 skrev genm...@gmail.com:

Gabriel, you don't get more well-formed than first order logic.

[genm...@gmail.com]

On Thursday, 5 October 2017 19:55:38 UTC-4, Markus Klyver wrote:
> Den onsdag 4 oktober 2017 kl. 22:38:55 UTC+2 skrev genm...@gmail.com:
> > On Wednesday, 4 October 2017 15:19:41 UTC-4, Markus Klyver wrote:
> > > Den tisdag 3 oktober 2017 kl. 23:30:02 UTC+2 skrev John Gabriel:
> > > > On Tuesday, 3 October 2017 17:15:59 UTC-4, burs...@gmail.com wrote:
> > > > > Can you make an example, where two series cannot
> > > > > be multiplied formally, independent of their
> > > > > convergence?
> > > >
> > > > Birdbrains. It makes no sense to multiply any two series unless they both converge. The idea is that you are multiplying the LIMITs by multiplying the partial sums.
> > > >
> > > > YOU BIG MOOOOOOROOOOOOON!!!!
> > >
> > > You can, but it's no guarantee the product will converge or to what it will converge to. You can multiply two divergent series and get an convergent one, for example.
> >
> > Nonsense. Provide an example you ignoramus!
>
> Sure, consider two infinite sums with the general terms 1/n and 1/(n+1). Now, when multiplied, all terms will be at most 1/n^2.

Prove it moron. That's just your assertion.

> They will also be all positive, so by the squeeze theorem the product will converge.

Bullshit.

[genm...@gmail.com]

The product of 1/n and 1/(n+1) does not converge to ONE unique value.

You are too stupid to understand.

[Markus Klyver]

Do you agree that 1/n is positive? Do you agree that 1/(n+1) is positive as well? Then you agree with 1/(n(n+1)) being positive as well. From n^2 ≤ (n(n+1)) it follows that 1/(n(n+1)) ≤ 1/n^2. So we have a sequence bounded from below by 0 and above by 1/n^2. We know that Σ 1/n^2 converges, hence the product of two divergent series can be convergent.

Den fredag 6 oktober 2017 kl. 16:20:19 UTC+2 skrev genm...@gmail.com:
> On Friday, 6 October 2017 10:16:51 UTC-4, genm...@gmail.com wrote:

> The product of 1/n and 1/(n+1) does not converge to ONE unique value.
>
> You are too stupid to understand.

If your statement is that the new series will not converge absolutely, I'm inclined to agree. It's fully possible it will not converge absolutely. But since every term is bounded by 1/n^2 the series cannot be greater than Σ 1/n^2 = (π^2)/6.

[John Gabriel]

Except that the product of the two series is NOT 1/(n(n+1)), you CRANK!

[Markus Klyver]

Well, it will be bounded by that if you define multiplication in the naive manner; that is you try to multiply (a_1 + a_2 + a_3 + ...)(b_1 + b_2 + b_3 + ...) as usual. For absolutely convergent series, this will work and give you the limits of respective series multiplied. For example, multiplying the Taylor series of sin(x) in this fashion will give you sin^2(x).