Divergent Series and Analytical Continuation (LONG post)

[tommy1729]

Damn I'm sorry guys, I've done a terrible job of explaining what I'm trying to do here. Let me try to rectify that by explaining my motivation, goals and thoughts.

First, let me share my mindset. I have been curious about natural boundaries for a long time. But, when I tend to research them, I usually find the following opinion by other intelligent mathemtician: Studying functions beyond natural boundaries is not meaningful-- the answer is not unique, and there is nothing useful that can be done with such objects. Studying functions beyond natural boundaries has no applications-- it is useless. 

Many mathematicians have a similar opinion about divergent series in general. But, as I'm sure all of us here are aware, divergent series actually do have fruitful applications and can help solve meaningful problems. For instance, the study of differential equations gives us real problems whose solutions are best understood in terms of divergent series.  

However, I am embrassed to admit that even after being curious about generalized analytical continuation (GAC) (i.e. continuation beyond natural boundaries), I had never seen any real problem whose solution was best understood in terms of GAC. I had seen a few connections to problems in non-linear differential equations, but that was about it. Therefore, I had gradually become more and more convinced that the people were right-- GAC is useless. 

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My opinion changed recently, when I saw the MO discussion of the function \( f(x)=\sum (-1)^n x^{2^n} \). If we view this series as part of a more general \( F(a,x)=\sum (-1)^n x^{a^n} \), then finding the value of \( f(x) \) REQUIRES evaluating \( F(a,x) \) beyond its natural boundary. However, in the context of this problem, what we really want to do is to get a nicer form of  \(f(x)\), one thats much easier to evaluate. So, we want to find a different way to evaluate f(x) outside its natural boundary. Here's a really useful representation of \( f(x) \) when \( |a|<1 \)
\[ F(a,x) = \sum_{n=0}^\infty \frac{\ln(x)^k}{k!(1+a^k)}\]
But, as you all saw in the original post, this series doesn't converge to \( f(x) \) if we plug in \( F(2,x) \). Now, this is very perplexing, and when I saw this, I wondered what the relationship between \( f(x) \) and \(F(2,x) \) is. More precisely, I wanted to know, what is \(f(x)-F(2,x)\). As you saw in the post, the answer is related to the residues of inside \( F(a,x) \)

Why is this important? Because, this is an application of GAC on a real problem! It would be incredibly hard to understand the relationship between \(f(x)\) and \(F(2,x)\) without appealing to continuation beyond natural boundaries. 

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Now, I want to make something clear-- in the post, I'm not making any claims in the post! I realize now it seems like I'm claiming "This is the right way to continue these series beyond their natural boundaries." It was not my intent to say this. It was to say "Here are two functions that are related to to each other. Its a pretty tricky task to figure out exactly what their difference it. However, if we view these functions from the perpsective of GAC, then it becomes easy to solve the problem. Therefore, GAC is a useful thing to study, since it helps us solve at least one specific type of problem (that problem being finding the difference between two related sums)"  

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I should emphasize that this is why I'm choosing series that converge everywhere. Its so that the question can be analyzed without appealing to any GAC. For real values of x, the difference 
\[ \sum (-1)^n x^{2^n} -\sum_{n=0}^\infty \frac{\ln(x)^k}{k!(1+a^k)}\]
is just a plain, well-defined expression. No tricks, no magic, no continuation beyond boundaries--no BS. I can plug the expression into a calculator to get the answer. HOWEVER, the point of the post is to say that we can also find the value of this expression in a precise and closed form way by considering GAC and picking up residues in the right way.

Examples such as these shatter the opinion that GAC is useless or hopelessly non-unique. The difference between those two expressions is a real number, so in the context of this problem, the only valid continuation is the one that gives you the correct real number. 

Therefore, to Tommy's point, 

So we end up

f( -s ) =/= f(s).

HOWEVER

we could also just plug in values and then

f( s ) = f( -s )

and then criticize the zeta expansion as unvalid.

YOU picked the first.

But Is there proof it is better ??

Does that even make the zeta expansion valid or is it still not valid ? ( afteral the boundary still exists ! )

If the zeta expansion is not valid , your motivation is weak.
And even if it is valid , your motivation is still weak.

And what do you do here ( with your function ) ??

YOU SAY JUST PLUG IN THE VALUE !

I'm NOT claiming either continuation is better. I'm just trying to analyze what we have to do to make the continuation consistent with the approach in which we just plug in numbers and evaluating. In other words, when I have the sum  
\( \sum (-1)^n x^{2^n} \) I want to know the right way to continue 
\[ \sum_{n=0}^\infty \frac{\ln(x)^k}{k!(1+a^k)}\]
So that it agrees with \( \sum (-1)^n x^{2^n} \). This doesn't mean this is the objectively right continuation of \( \sum_{n=0}^\infty \frac{\ln(x)^k}{k!(1+a^k)} \), its just the one that agrees with \( \sum (-1)^n x^{2^n} \), since the whole point of the original problem was to find a nicer form for \( \sum (-1)^n x^{2^n} \). 

Tommy, you are completely right that this is an unfinished theory, this post is only my first exploration. However, I hope you can see that the theory is NOT un-falsifiable. The entire point was of this exploration is to find cases like 
\[ \sum (-1)^n x^{2^n} -\sum_{n=0}^\infty \frac{\ln(x)^k}{k!(1+a^k)}\]
where I can directy plug into a calculator to be able to figure out if certain continuations were correct. In fact, in my post, you'll that in case (1) I get the wrong answer, so there is still more to understand what is going on in those cases. 

Let me know if this makes sense to you guys-- I feel I still haven't done a great job explaining what happening, so let me know about any questions

yes i see.

thats partially why i mentioned continu at the 0.

But you also mentioned divergent functions like theta and prime zeta and such where plug-in is problematic.

You then used double sums and made fubini's head explode.

I will have to think about these.

notice you gave a solution to mick's MSE that included not plugging in despite being well defined locally if we did.

the issue was the resummation of zeta , which you somewhat solved. but not by plug-in. two skools of thought ?

ON THE OTHER HAND those two might be the only ways to make sense of it, unless we resummation overdose maybe.


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multivariable analogues of your theory could imo help to do number theory. IN FACT I THINK most problems can be restated as such.

However that requires more insight and might not be easier than the original statements.

Im talking density conjectures here , not collatz.

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also 

as you well know it is not certain resummation always works to get values.

Im aware you are aware of the unfinished state of the theory.

thanks for your attention.


regards

tommy1729