Divergent Series and Analytical Continuation (LONG post)

[tommy1729]

Apart from the idea of reflection formula's ( which may or may not be a good idea )

I want to take for example the prime zeta function P(s).

It is well known to have formulas that converge for Re(s) > 1 or Re(s) > 0.

There also exists a formula for Re(s) > 1/2 :

Assuming the RH then \[P(s) = s \int_2^\infty \pi(x) x^{-s-1}dx = s \int_2^\infty (\pi(x)-Li(x)) x^{-s-1}dx+  L(s) \\  = s \int_2^\infty (\pi(x)-Li(x)) x^{-s-1}dx+L(2) + \log(s-1)+ \int_2^s \frac{2^{1-u}-1}{u-1}du\] where the latter integrals converge and are analytic for Re(s) > 1/2.

\[ Li(x) = \int_2^x \frac{dt}{\log t}\] \[L(s) = s\int_2^\infty Li(x) x^{-s-1}dx  = \int_2^\infty \frac{x^{-s}}{\log x}dx \] \[ = L(2)+\int_2^s L'(u)du = L(2) + \log(s-1)+ \int_2^s \frac{2^{1-u}-1}{u-1}du\]

since \[L'(s) = -\int_2^\infty x^{-s}dx = \frac{2^{1-s}}{s-1}\]

I guess that is clear to all here.



Now the natural boundary at Re(s) = 0 is made completely of log singularities getting dense.

Maybe we should make distinctions of the type of natural boundaries we are getting.

I mean for instance 

g(x) = (1 - x)(1 - x^3)(1 - x^5)(1 - x^7)... 

or h(x) = 1 + x^2 + x^2^2 + x^2^3 + ...

are having "different" natural boundaries, like accumilations of zeros.

i SAID MAYBE lol



But now,

what is the value of P(s) for Re(s) < 0 ??

OR is this type of natural boundary unsuitable because it has logs instead of poles and zero's ??
AND IF UNSUITABLE , WHAT DOES THAT MEAN ??  no continuation for some but for others we do ?

I will definitely have a talk about that with my friend mick.


I want to point out that the derivative of the prime zeta has an infinite amount of poles instead of logs.

and the inverse of the derivative of prime zeta has an infinite amount of zero's on Re(s) = 0.


Im holding back on making conjectures , im a bit confused.


regards

tommy1729

This is a good question. Let me try to answer it by sharing my motivation in studying the series I decided to study in the post.

Analytical continuation beyond natural boundaries is hard! I don't think anyone knows how to do it in general. I suspect it can't be done in general-- because I don't think it would be meaningful to analytically continue a series that has purely random coefficents for instance. Since the problem is so hard, I'm choosing to study the easiest possible example I could come up with. 

The examples I study have a really nice property. For instance, consider the following series 
\[f(x)=\sum_{n=0}^\infty \frac{x^n}{1+x^n} \frac{1}{2^n}\]
This series has a natural boundary, since 1+x^n provides a dense set of poles. So, it cannot be analytically continued to |x|>1. However, the series is still well-defined outside of |x|>1. In particular, I can compute f(2) by just plugging into the series. This gives me a very natural candidate for a definition of \( f(x)\) outside the boundary.

So, I choose to study these much easier functions, and try to analyze what sort of properties they have. My goal looks like this
\[ \text{Get a bunch of very easy examples of functions with natural boundaries } \to \]
\[\text{ Study those examples in depth, and understand the mechanism underlying how those functions behave } \to \]
\[\text{ Try to generalize that mechanism into harder functions }\]

The prime zeta function is definitely in the category of "harder functions." I don't know if logarithmic singularities will cause the continuation to behave differently than regular poles-- that's something I will only find out once I've studied the easy examples in depth!

So, its not neccesariy that the prime zeta function has properties that make it unsuitable for continuation-- its that I haven't yet figured out the right way to do continuation beyond natural boundaries.

\[f(x)=\sum_{n=0}^\infty \frac{x^n}{1+x^n} \frac{1}{2^n}\]

Your example is why I talk about reflection formula's.

Notice f(oo) = limit of 

\[f(oo)=\sum_{n=0}^\infty \frac{oo^n}{1+oo^n} \frac{1}{2^n}\]

=

\[f(oo)=\sum_{n=0}^\infty 1 *  \frac{1}{2^n}\]

Which means f(oo) = 2.

Now think about 

f(0) = 0

because

\[f(0)=\sum_{n=0}^\infty \frac{0^n}{1+0^n} \frac{1}{2^n}\]

and 

\[f(1/x)=\sum_{n=0}^\infty \frac{x^{-n}}{1+x^{-n}} \frac{1}{2^n}\]

which converges for x =/= 0.

But we know f(1/0) = f(oo) = 2.

Now 1/x takes from inside the natural function boundary to outside.

and there is only one boundary.

This makes formulas like 

f(1/x) = g( f(x) )

for some g maybe a good idea.

and the idea mainly comes from the shape of the boundary and not so much of the function itself.

Although a natural boundary being the unit circle does not prove the existance of a g function , and i guess with little effort we can find examples where g exists and where not.

Riemann mapping function is maybe key here since it can map any natural boundary to the unit circle.
and then from there we might get to find a valid g function if it exists.

However assume a subset D of the unit disk where x is univalent.

now if x is element of D and f(1/x) is not univalent we have a problem constructing a function g.

 another idea is 

f(conjugate 1/x ) = g( f(x) )

with similar logic.

I mentioned these before in pieces here and there in other threads.
I wanted to make things more formal and clear.

Now there is more to the shape of the boundary than meets the idea i think.

dirichlet series converges beyond a straight line.
So the boundary is a line.

Now this relates to a zeta like function in particular , and dirichlet generalized sums , functions that contain n^s.

This means functions like that ( prime zeta , the function mick posted on MSE etc ) are naturally related to bernouilli numbers !

because basically the bernouilli numbers ARE the zeta function.

So , what tf am I saying ?

I think it is natural to use bernouilli numbers when the functions contain n^s.

And so every type of function or more so boundary has its own special numbers.

So when does it depend rather on function and when more on boundary ? 

well that probably depends on how many times you use resummation; double sums , triple sums , etc.

; if your n^s terms occur as main or as one of the inner sums.


***

you used sum of products.

I consider the idea of sums of compositions , which might relate to the riemann mapping idea.

But more research needs to be done.

Ofcourse one can consider 

sum f(n)^2 , sum f(n)^3 etc

and have a taylor logic.

***

***

you mentioned lambert series.

i agree , but admit at the moment i do not have much to say about them.

it is a mystery for now.

***

I want to add more


f(conjugate 1/x ) = g( f(x) )

this is like having the boundary as an actual mirror.

the reflection is continu at the boundary.

continu points within natural boundaries are actually a thing.



Now lets look at boundary being the imag line.

and it has a continu point at 0.

suppose we also " should " have f( - s) = f(s) like we did in mick's post.

IF f(x) is (real) continu at x = 0 and for real x > 0 the function is analytic then

when 

f(-s) = f(s) " should " be true 

claiming

f(x) = f(-x) 

ACTUALLY works for real x.

but may or may not work for the upper plane.

f(s) = f (- conj s)

then again might work !!

which now relates to the circle idea above.

But "f(s) = f(-s)" does not always imply f(0) = 0 if f ( y i) for real y is not analytic EVEN IF F(0) IS ANALYTIC.


OK TIME FOR AN EXAMPLE

very similar to the function mick posted on MSE :


\[f(s)=\sum_{n=1}^\infty \frac{1}{n^s +n^{-s}} \frac{1}{n^2}\]

Now " clearly "

f(-s) = f(s) or it should.

But it is not analytic at Re(s) = 0 anywhere.

however it is real-continu at f(0).

But is f(0) = f(-0) = 0 ?

\[f(s)=\sum_{n=1}^\infty \frac{1}{n^s +n^{-s}} \frac{1}{n^2}\]

\[f(0)=\sum_{n=1}^\infty \frac{1}{n^0 +n^{-0}} \frac{1}{n^2}\]

\[f(0)=\sum_{n=1}^\infty \frac{1}{2} \frac{1}{n^2}\]

so f(0) = zeta(2)/2 = pi^2 / 12.

not zero.

This function converges better than the one mick posted btw.
which explains morally the real-continu and definite value at 0.

notice real-continu , not complex continu ofcourse.

but f(0) is not 0 and this leads to issues.
however adding constants makes this true, that is unlikely to be the " magic solution " ofcourse.

Lets analyse the value at f(0) and the function f(s) and in particular resummation and the f(-s) = f(s) idea ;


f(s) = zeta(3s + 2) - zeta(5s + 2) + zeta(7s + 2) - ... or something like that.
( see the MSE post by mick and caleb answer there )

while 

f(-s) = zeta(-3s + 2) - zeta(-5s + 2) + zeta(-7s + 2) - ... or something like that.

The cesaro sum of those two ( f(s) , f(-s) ) should be equal.
( since they agree on 1+1+1+... )

But 

zeta(3s +2) - zeta(-3s + 2) is not 0.

and it is unlikely that 

  zeta(3s + 2) - zeta(5s + 2) + zeta(7s + 2) - ... - (  zeta(-3s + 2) - zeta(-5s + 2) + zeta(-7s + 2) - ... ) = 0

even when rearranging terms.

the symmetry F(-s) = F(s) is not within those zeta terms.

in fact even a generalization where +2 and /n^2 is replaced by + l and /n^l would probably not make it happen.


If however 
 zeta(-3s + 2) - zeta(-5s + 2) + zeta(-7s + 2) - ...

or the other one EACH equal 0 then it must be true.

But this means f(0) = 0.

aha.

but also

zeta((2k+1)s + 2) for s = 0 = zeta(2).

then

f(0) = zeta(2) + zeta(2) + ...

which makes no sense.

Now your relection formula needs to be consistant with all of this ofcourse.

So this shows there is probably no reflection formula for this one.

YET

if we " naively ? " plug in -2 + i or we plug in +2 + i

we get similar results ( values ) for

f( -2 + i )

or f( +2 + i )

and their conjugates.


something to consider.


another example to study is 

\[f(s)=\sum_{n=1}^\infty \frac{1}{n^s  - n^{-s}} \frac{1}{n^2}\]

***

The bernouilli numbers came from summing n^s for s positive !

this is remarkable because we use them here mainly on the other side , for the standard def of the functions.

I said before the type of numbers relate to the boundary shape.

But apparantly on the divergeant side !


So

studying truncated sums such as n^s that diverge to oo are cruxial.

In combination with the riemann mapping and the analogue of truncated sums and the shape of boundary we should be able to arrive at the related special numbers for a given boundary.

However all sums are related to bernouilli numbers. see the indefinite sum formulas.

which means that bernouilli is somewhat universal !

and thus zeta method desummations are somewhat universal and should give same results as compared to other methods.


The riemann mapping and symmetry like -x , 1/x , conjugate

relate the circle to the line and likewise the 

taylor/laurent to dirichlet or fourier.

So i know i mentioned dirichlet stuff and not the taylor stuff like you did also as 50 percent of your cases , but i think they are related.

However an exp does not equal 0 , that might be a bit of a distinction.

All of math is related.


Maybe we can find 

f(-s) = f(s)

or 

f(-s) = - f(s)

with natural boundaries.

maybe we should ??


Finally higher exponential levels should probably be considered a resummation.
So the lower ones might be more important.



regards

tommy1729