tommy's "linear" summability method

[tommy1729]

tommy's summability method
**************************

I might not be the first to consider it so the name is perhaps disputable or temporarily anyway ,


This method is for divergeant strictly increasing positive integer sums

consider the geometric sum : 1 + 2 + 4 + 8 + ... + 2^n + ... = x

now we see 2 x = 2 + 4 + 8 + ...

notice x - 1 = 2 + 4 + 8 + ...

so

2 x = x - 1.

thus

x = -1.


lets generalize for real a > 1 :

1 + a + a^2 + ... = x

then

a x = x - 1

so

x = -1/(a-1) = 1/(1-a).

This is all old stuff.

So whats next ?

consider an entire function f(x) given by a taylor series : f_1 x + f_2 x^2 + f_3 x^3 + ... + f_i x^i + ...

such that f(i+1) > f(i) for every integer i > 0.

Notice f(0) = 0.


Now we can sum f(1) + f(a) + f(a^2) + f(a^3) + ... + f(a^n) + ... = x

Here is how

sum f(1) + f(a) + f(a^2) + f(a^3) + ... + f(a^n) + ... = x

sum_n sum_i  f_i (a^n)^i = x

so

x = sum_n sum_i  f_i (a^n)^i = sum_i sum_n  f_i (a^i)^n

= sum_i f_i * - 1/(a^i - 1) = sum_i f_i / (1 - a^i)



As example consider

dexp(x) := exp(x) - 1.

dexp(1) + dexp(2) + dexp(2^2) + dexp(2^3) + dexp(2^4) + ... = x

Then

x  = sum_i f_i /( 1 - 2^i ) = sum_i  1/( i! * (1 - 2^i) )


this clearly converges.

(I got an estimate of x = -1.19355 btw)

I call this a linear summability method because it has some linear properties.



regards


tommy1729

[tommy1729]

I want to point out some problems :

1) SUMMABILITY METHODS WORK GREAT TO GET A VALUE OF AN ANALYTIC CONTINUATION BUT OFTEN FAIL TO DETECT ( TRUE ) DIVERGEANCE !!

we still want to know about divergeance , also beyond the analytic continuation !

2) the entire function is an interpolation 

so f(2^n) is defined as an f that grows like say f(x) = exp(x).

but almost any entire f that agrees with exp(x) at integer x can be used ... but might give different results !!

So a very disturbing thing.

3)

the sum 1 + 2 + 4 + ...

is associated to the equation 2 x = x - 1

or 4 x = x - 3
8 x = x - 7 

etc

ALL of them give x = -1.

HOWEVER

the sum 1 + 3 + 9 + ...

has equations

3 x = x - 1

9 x = x - 1 - 3 = x - 4

etc

all with DIFFERENT solutions.

---

these " problems " are quite universal with summability methods.

Which is why " the master forbids it " ( being Weierstrass )

I do not forbid it , but be very careful.


regards

tommy1729

[jacob]

1 + 3 + 9.... = -0.5

[Caleb]

tommy's summability method
**************************

I might not be the first to consider it so the name is perhaps disputable or temporarily anyway ,


This method is for divergeant strictly increasing positive integer sums

consider the geometric sum : 1 + 2 + 4 + 8 + ... + 2^n + ... = x

now we see 2 x = 2 + 4 + 8 + ...

notice x - 1 = 2 + 4 + 8 + ...

so

2 x = x - 1.

thus

x = -1.


lets generalize for real a > 1 :

1 + a + a^2 + ... = x

then

a x = x - 1

so

x = -1/(a-1) = 1/(1-a).

This is all old stuff.

So whats next ?

consider an entire function f(x) given by a taylor series : f_1 x + f_2 x^2 + f_3 x^3 + ... + f_i x^i + ...

such that f(i+1) > f(i) for every integer i > 0.

Notice f(0) = 0.


Now we can sum f(1) + f(a) + f(a^2) + f(a^3) + ... + f(a^n) + ... = x

Here is how

sum f(1) + f(a) + f(a^2) + f(a^3) + ... + f(a^n) + ... = x

sum_n sum_i  f_i (a^n)^i = x

so

x = sum_n sum_i  f_i (a^n)^i = sum_i sum_n  f_i (a^i)^n

= sum_i f_i * - 1/(a^i - 1) = sum_i f_i / (1 - a^i)



As example consider

dexp(x) := exp(x) - 1.

dexp(1) + dexp(2) + dexp(2^2) + dexp(2^3) + dexp(2^4) + ... = x

Then

x  = sum_i f_i /( 1 - 2^i ) = sum_i  1/( i! * (1 - 2^i) )


this clearly converges.

(I got an estimate of x = -1.19355 btw)

I call this a linear summability method because it has some linear properties.



regards


tommy1729

This derivation is almost correct, though it possesses a problem that only emerges when considering the right functions. You noted that
\[\sum_{n=0}^\infty  f(a^n) = \sum_k f_k \sum_n a^{nk} = \sum_k f_k \frac{1}{1-a^k}\]
However, note that the last step involves evaluating an analytical continuation outside the domain of the convergence of the series. When this happens, things can do wrong. Consider \(f(z,x) = z^x\) and \[\sum (-1)^n f(z,a^n) = \sum (-1)^n z^{a^n} \text{ } {}^{"}=^{"} \text{ } \sum \frac{\ln(z)^k}{k!(1+a^k)}\] Actually, for \(a>1\) these things are not equal, Gottfried talks about the difference for \(a=2\) here: https://mathoverflow.net/questions/19866...871#198871. However, for \(a<1\), the inner series converges and so in many cases the reasoning is valid and the series converges properly. 

Now, one could stop there, and say perhaps the summation method simply does not work for \(a>1\) since we are attempting to sum divergent series. However, we don't need to be so closed minded-- maybe there is a way to recover the correct function from these evaluations. I consider this view, which you may find interesting, here: https://mathoverflow.net/questions/43869...tical-cont

If you believe the arguemnt I give in the above link is valid, then you evaulation of the dexp is not quite correct, and part of the correction factor can be obtained by evaulating the residues that emerge when \(1-2^z=0\). However,

[tommy1729]

tommy's summability method
**************************

I might not be the first to consider it so the name is perhaps disputable or temporarily anyway ,


This method is for divergeant strictly increasing positive integer sums

consider the geometric sum : 1 + 2 + 4 + 8 + ... + 2^n + ... = x

now we see 2 x = 2 + 4 + 8 + ...

notice x - 1 = 2 + 4 + 8 + ...

so

2 x = x - 1.

thus

x = -1.


lets generalize for real a > 1 :

1 + a + a^2 + ... = x

then

a x = x - 1

so

x = -1/(a-1) = 1/(1-a).

This is all old stuff.

So whats next ?

consider an entire function f(x) given by a taylor series : f_1 x + f_2 x^2 + f_3 x^3 + ... + f_i x^i + ...

such that f(i+1) > f(i) for every integer i > 0.

Notice f(0) = 0.


Now we can sum f(1) + f(a) + f(a^2) + f(a^3) + ... + f(a^n) + ... = x

Here is how

sum f(1) + f(a) + f(a^2) + f(a^3) + ... + f(a^n) + ... = x

sum_n sum_i  f_i (a^n)^i = x

so

x = sum_n sum_i  f_i (a^n)^i = sum_i sum_n  f_i (a^i)^n

= sum_i f_i * - 1/(a^i - 1) = sum_i f_i / (1 - a^i)



As example consider

dexp(x) := exp(x) - 1.

dexp(1) + dexp(2) + dexp(2^2) + dexp(2^3) + dexp(2^4) + ... = x

Then

x  = sum_i f_i /( 1 - 2^i ) = sum_i  1/( i! * (1 - 2^i) )


this clearly converges.

(I got an estimate of x = -1.19355 btw)

I call this a linear summability method because it has some linear properties.



regards


tommy1729

This derivation is almost correct, though it possesses a problem that only emerges when considering the right functions. You noted that
\[\sum_{n=0}^\infty  f(a^n) = \sum_k f_k \sum_n a^{nk} = \sum_k f_k \frac{1}{1-a^k}\]
However, note that the last step involves evaluating an analytical continuation outside the domain of the convergence of the series. When this happens, things can do wrong. Consider \(f(z,x) = z^x\) and \[\sum (-1)^n f(z,a^n) = \sum (-1)^n z^{a^n} \text{ } {}^{"}=^{"} \text{ } \sum \frac{\ln(z)^k}{k!(1+a^k)}\] Actually, for \(a>1\) these things are not equal, Gottfried talks about the difference for \(a=2\) here: https://mathoverflow.net/questions/19866...871#198871. However, for \(a<1\), the inner series converges and so in many cases the reasoning is valid and the series converges properly. 

Now, one could stop there, and say perhaps the summation method simply does not work for \(a>1\) since we are attempting to sum divergent series. However, we don't need to be so closed minded-- maybe there is a way to recover the correct function from these evaluations. I consider this view, which you may find interesting, here: https://mathoverflow.net/questions/43869...tical-cont

If you believe the arguemnt I give in the above link is valid, then you evaulation of the dexp is not quite correct, and part of the correction factor can be obtained by evaulating the residues that emerge when \(1-2^z=0\). However,

I will think about it.

But some comments first.

1) This is probably a Fubini type problem.

2) I required the numbers to be positive

3) I also want the function to be entire.

These considerations are exactly to avoid such issues , together with Eulers series accelerations ( not giving the same when they should ! ) and Riemann series theorem ( alternating series can be summed to anything by changing the order of summing )

Thanks for the comment.

And thanks to Gottfriend for these nice ideas.


regards

tommy1729

[Caleb]

tommy's summability method
**************************

I might not be the first to consider it so the name is perhaps disputable or temporarily anyway ,


This method is for divergeant strictly increasing positive integer sums

consider the geometric sum : 1 + 2 + 4 + 8 + ... + 2^n + ... = x

now we see 2 x = 2 + 4 + 8 + ...

notice x - 1 = 2 + 4 + 8 + ...

so

2 x = x - 1.

thus

x = -1.


lets generalize for real a > 1 :

1 + a + a^2 + ... = x

then

a x = x - 1

so

x = -1/(a-1) = 1/(1-a).

This is all old stuff.

So whats next ?

consider an entire function f(x) given by a taylor series : f_1 x + f_2 x^2 + f_3 x^3 + ... + f_i x^i + ...

such that f(i+1) > f(i) for every integer i > 0.

Notice f(0) = 0.


Now we can sum f(1) + f(a) + f(a^2) + f(a^3) + ... + f(a^n) + ... = x

Here is how

sum f(1) + f(a) + f(a^2) + f(a^3) + ... + f(a^n) + ... = x

sum_n sum_i  f_i (a^n)^i = x

so

x = sum_n sum_i  f_i (a^n)^i = sum_i sum_n  f_i (a^i)^n

= sum_i f_i * - 1/(a^i - 1) = sum_i f_i / (1 - a^i)



As example consider

dexp(x) := exp(x) - 1.

dexp(1) + dexp(2) + dexp(2^2) + dexp(2^3) + dexp(2^4) + ... = x

Then

x  = sum_i f_i /( 1 - 2^i ) = sum_i  1/( i! * (1 - 2^i) )


this clearly converges.

(I got an estimate of x = -1.19355 btw)

I call this a linear summability method because it has some linear properties.



regards


tommy1729

This derivation is almost correct, though it possesses a problem that only emerges when considering the right functions. You noted that
\[\sum_{n=0}^\infty  f(a^n) = \sum_k f_k \sum_n a^{nk} = \sum_k f_k \frac{1}{1-a^k}\]
However, note that the last step involves evaluating an analytical continuation outside the domain of the convergence of the series. When this happens, things can do wrong. Consider \(f(z,x) = z^x\) and \[\sum (-1)^n f(z,a^n) = \sum (-1)^n z^{a^n} \text{ } {}^{"}=^{"} \text{ } \sum \frac{\ln(z)^k}{k!(1+a^k)}\] Actually, for \(a>1\) these things are not equal, Gottfried talks about the difference for \(a=2\) here: https://mathoverflow.net/questions/19866...871#198871. However, for \(a<1\), the inner series converges and so in many cases the reasoning is valid and the series converges properly. 

Now, one could stop there, and say perhaps the summation method simply does not work for \(a>1\) since we are attempting to sum divergent series. However, we don't need to be so closed minded-- maybe there is a way to recover the correct function from these evaluations. I consider this view, which you may find interesting, here: https://mathoverflow.net/questions/43869...tical-cont

If you believe the arguemnt I give in the above link is valid, then you evaulation of the dexp is not quite correct, and part of the correction factor can be obtained by evaulating the residues that emerge when \(1-2^z=0\). However,

I will think about it.

But some comments first.

1) This is probably a Fubini type problem.

2) I required the numbers to be positive

3) I also want the function to be entire.

These considerations are exactly to avoid such issues , together with Eulers series accelerations ( not giving the same when they should ! ) and Riemann series theorem ( alternating series can be summed to anything by changing the order of summing )

Thanks for the comment.

And thanks to Gottfriend for these nice ideas.


regards

tommy1729

Is it \( f_{i+1} > f_i \) or \( f(i+1) > f(i) \) that you wanted for the positive condition? I can probably construct a counter example for either depending on your preference.
Edit: Or actually, on second thought, in either case the series won't converge in the usual sense, so there is no intersection between convergent series and instance where your summability applies. So I guess in either case I'd have to come up with an example where the series can be summed by analytical continuation and compare it there, though I somewhat doubt there are examples where functions can be analytically continued when they are summed over f(a^n) since it probably has enough symmetry to generate a natural boundary for a>1. However, I will probably still try and see if I can come up with something.

[tommy1729]

The link Caleb gave ( https://mathoverflow.net/questions/43869...tical-cont ) me reminded me of this

https://math.stackexchange.com/questions...0-ns-n-s-1

which I posted here too.

The other link he gave :

https://mathoverflow.net/questions/19866...871#198871

There is alot to say about such functions , and I and others have so in the past somewhere.

But the main thing is my method is designed for analytic continuation.

The problem he gave was at a point that was not even continu.

So I see that as a somewhat violation of conditions although I did not clearly stated the conditions.
Without the alternating part that problem probably goes away , even if we hit a natural boundary ...

***

By no means is my method considered perfect and I even doubt " perfect summability " exists since we always get undesired properties , divergeance anyway or contradiction results it seems.

It just felt like a nice idea.

***

About this 

The other link he gave :

https://mathoverflow.net/questions/19866...871#198871





Giving lim sup or lim inf can be a very tricky thing.

I have at least 2 such questions/conjectures illustrating how tricky it is , posted by my friend mick at mathstackexchange and/or mathoverflow

https://mathoverflow.net/questions/34906...n-13-a-n-2

https://mathoverflow.net/questions/32224...f-m-frac54

https://math.stackexchange.com/questions...-g-frac916

https://math.stackexchange.com/questions...f-m-frac54

and many more and I did not even mention the complicated functions.

Btw the continuum sum has been discussed here.


regards

tommy1729

[tommy1729]

The link Caleb gave ( https://mathoverflow.net/questions/43869...tical-cont ) me reminded me of this

https://math.stackexchange.com/questions...0-ns-n-s-1

which I posted here too.

The other link he gave :

https://mathoverflow.net/questions/19866...871#198871

There is alot to say about such functions , and I and others have so in the past somewhere.

But the main thing is my method is designed for analytic continuation.

The problem he gave was at a point that was not even continu.

So I see that as a somewhat violation of conditions although I did not clearly stated the conditions.
Without the alternating part that problem probably goes away , even if we hit a natural boundary ...

Well analytic at the boundary is usually hard to show, unless we get some *good* recursion and induction.
But still many mysteries about them.

regards

tommy1729

[tommy1729]

intuitively , it makes no sense to do a summability if you can easily see it does not convergeance because it is not continu.
It is defined in the neighbourhood and can be seen so that is not the issue.

It feels like asking the limit of sin(x) at +oo.
Although sin(x) is defined everywhere and analytic everywhere it makes no sense.

regards

tommy1729

[Caleb]

The link Caleb gave ( https://mathoverflow.net/questions/43869...tical-cont ) me reminded me of this

https://math.stackexchange.com/questions...0-ns-n-s-1

which I posted here too.

The other link he gave :

https://mathoverflow.net/questions/19866...871#198871

There is alot to say about such functions , and I and others have so in the past somewhere.

But the main thing is my method is designed for analytic continuation.

The problem he gave was at a point that was not even continu.

So I see that as a somewhat violation of conditions although I did not clearly stated the conditions.
Without the alternating part that problem probably goes away , even if we hit a natural boundary ...

***

By no means is my method considered perfect and I even doubt " perfect summability " exists since we always get undesired properties , divergeance anyway or contradiction results it seems.

It just felt like a nice idea.

***

About this 

The other link he gave :

https://mathoverflow.net/questions/19866...871#198871





Giving lim sup or lim inf can be a very tricky thing.

I have at least 2 such questions/conjectures illustrating how tricky it is , posted by my friend mick at mathstackexchange and/or mathoverflow

https://mathoverflow.net/questions/34906...n-13-a-n-2

https://mathoverflow.net/questions/32224...f-m-frac54

https://math.stackexchange.com/questions...-g-frac916

https://math.stackexchange.com/questions...f-m-frac54

and many more and I did not even mention the complicated functions.

Btw the continuum sum has been discussed here.


regards

tommy1729

Actually, that first link (https://math.stackexchange.com/questions...0-ns-n-s-1) is perhaps instructive because I apply a trick similar to your summability method but consideration of the poles actually becomes important. In my answer the pole actually introduces a branch cut. 

But the main thing is my method is designed for analytic continuation.

The problem he gave was at a point that was not even continu.

I'm sorry -- I don't quite understand what you mean here, are you refering to the example I gave of 
\[\sum (-1)^n x^{2^n}\] 
definitely describes a continuous and complex analytic function for \( |x|<1 \). But I do agree that it doesn't satisfy the condition of only having positive terms. I think instead 
\[\sum \frac{(x^{2^n}-1)}{\sqrt{2}^n}\]
should be slightly closer to the type of example you are looking for. We would obtain
\[\sum \frac{(x^{2^n}-1)}{\sqrt{2}^n} = \sum_{k=1}^\infty \frac{\ln(x)^k}{k!2^k(1-2^{k-1/2})}\]
But this doesn't actually match the sum. Instead, we need to pick up the residues. In fact, there is an even a significant residue on the real line that you have to pick up to make the sums even close. For instance, by considering this residue, we get a very good approximation of 
\[\sum \frac{(x^{2^n}-1)}{\sqrt{2}^n} = - \frac{\sqrt{\ln(\frac{1}{x})}}{\Gamma(3/2)}\frac{\pi}{\ln(2)}+\sum_{k=1}^\infty \frac{\ln(x)^k}{k!2^k(1-2^{k-1/2})} \]
The contribution of the extra residues is about of the order \(10^{-7}\), so its barely even perceptable.