# Summing divergences and the real projective line

1. Dec 13, 2013

### japplepie

The real projective line states that there is not difference between positive and negative infinity (maybe except the path needed to be taken to "reach" either one of them) and they are actually connected.

There are a lot of ways to get a definite sum from a divergent series; one of which is the algebraic way (my personal favorite).

note:j(x)=x^0+x^1+x^2+x^3+...
Using the algebraic method I could derive the sum of j(2) as shown below:
j(2)=1+2+4+8+16+...
j(2)=1+2j(2)
j(2)= -1= 1+2+4+8+16+...

Does this mean that j(2) diverges so much that it went pass infinity from the positive side and landed on a definite negative point?
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If the statement above is true, then the idea of < and > falls apart.

Example:j(2)+j(1/2)-1=0 ( the extra -1 here is to account for the (1/2)^0 )
A sum of all positive numbers that equals 0.

A notion of < and > that would work in this scenario is by measuring how far it "moved" in the RPL.
j(2)+j(1/2)-1 "moved" and circled back to 0, but 0 never "moved".
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j(4)=1+4+16+64+256+...
j(2)=1+2+4+8+16+32+64+128+256...=(1+2)+(4+8)+(16+32)+(64+128)+...
j(2)=3+12+48+192+...=3(1+4+16+64+...)
j(2)=3j(4)

S2:={1,2,4,8,16,32,64,128,...}
S4:={1,4,16,64,256,...}
S2\S4={2,8,32,128,...} which is S4 multiplied by 2 element-wise
S2-S4=2(S4) (treat it like a sum)
S2=3(S4)

Does this also mean that in j(x), the smaller x (x>=1), the bigger* it is.
Bigger* meaning it "moved" more.

2. Dec 13, 2013

### Office_Shredder

Staff Emeritus
3. Dec 13, 2013

### gopher_p

The real projective line is the one point compactification of the real line. It is a topological space, not a sentient being with the ability to communicate. It does not state anything.

In this space, there is no "positive" infinity or "negative" infinity. There are not two points to differentiate. There is one new point. Saying that this point is different from itself in some way is ridiculous.

This method assumes a lot of things about infinite sums; namely that this one in particular is a number and that all of the algebraic properties of finite addition and multiplication can be extended to infinite addition, whatever the hell that is. Since you haven't stipulated what infinite addition is nor why we should accept the aforementioned assumptions, there is no way to tell what your result "means".

Why? There are two very obvious linear orderings of the real projective line which extend the usual ordering on the reals.

Maybe you mean that this new ordering doesn't "play with" addition on the real projective line the same way the old ordering did on the real line. But regardless of where you put ∞ in this ordering, under the usual assignment of a+∞=∞ the theorems involving order and addition are still true.

So this is really about how your infinite addition interacts with the ordering. Again, until you tell us what infinite addition is, we can't confirm any observation regarding how the order is broken.

4. Dec 13, 2013

### japplepie

It means that the idea behind it states ...

This method assumes a lot of things about infinite sums; namely that this one in particular is a number and that all of the algebraic properties of finite addition and multiplication can be extended to infinite addition, whatever the hell that is. Since you haven't stipulated what infinite addition is nor why we should accept the aforementioned assumptions, there is no way to tell what your result "means".

And yes infinite operations are still very shady.