Extending reals with logarithm of zero

[Anixx]

What do you guys have to say about this Mathoverflow post?
Do you have any interesting ideas about this?
https://mathoverflow.net/questions/...ithm-of-zero-properties-and-reference-request

 

[topsquark]

What do you guys have to say about this Mathoverflow post?
Do you have any interesting ideas about this?
https://mathoverflow.net/questions/...ithm-of-zero-properties-and-reference-request

It sounds like it might be possible in some restricted sense. But why would you need to do this?

-Dan

 

[fresh_42]

What do you guys have to say about this Mathoverflow post?

I think $\zeta(-1)=-1/12$ is already adventurous enough and this is an attempt trying to imitate Ramanujan's result here. However, while $\zeta(-1)=-1/12,$ can rigorously be deduced, I cannot see any rigor deduction that avoids the pole of the zeta-function at $x=1$ in what is said on MO.

To me, it is just playing around with terms and infinity disrespecting the analytical background. E.g.
$$
H_n=\sum_{k=1}^n\dfrac{1}{k}=\log n +\gamma +O\left(n^{-1}\right)
$$
reads on MO

The Maclaurin series of the function $\ln (x+1)$ at $x=-1$ is the Harmonic series with negative sign, thus, we can represent $\lambda=-\sum_{k=1}^\infty \frac1k$. Since the Harmonic series has the regularized value of $\gamma$ (Euler-Mascheroni constant), the regularized value (finite part) of $\lambda$ is $-\gamma$.

Sorry, that is nonsense in my opinion, i.e. it does not make any sense to me. How does this regularization work? What is the finite part of infinity? We can yield literally every result we like as long as we avoid any rigor. All I can see is
$$
\lambda =\log (0)=\displaystyle{\lim_{n \to \infty}}\log(n^{-1})=-\lim_{n \to \infty}\log (n)=\gamma -H_{\infty }=\gamma -\infty
$$
which is still a pole of the zeta-function at $x=1\, : \, \lambda = -\zeta(1)=\gamma -\infty =-\infty .$

 

[Anixx]

I think $\zeta(-1)=-1/12$ is already adventurous enough and this is an attempt trying to imitate Ramanujan's result here. However, while $\zeta(-1)=-1/12,$ can rigorously be deduced, I cannot see any rigor deduction that avoids the pole of the zeta-function at $x=1$ in what is said on MO.

That the Harmonic series has the regularized value $\gamma$ is a known mathematical fact. Take the Cauchy principal value for instance:
https://www.wolframalpha.com/input?i=Limit[(Zeta[1-h]+Zeta[1+h])/2,+h->0]

The Ramanujan summation gives the same result: https://math.stackexchange.com/a/650515/2513

 

[fresh_42]

That the Harmonic series has the regularized value $-\gamma$ is a known mathematical fact. Take the Cauchy principal value for instance:
https://www.wolframalpha.com/input?i=Limit[(Zeta[1-h]+Zeta[1+h])/2,+h->0]

Yes, and this was the first time you actually mentioned what regularization would mean in this context. Before this quote, it was only a handwavy term without substance. Now, you want to define
$$
\lambda =:\log(0) =:-\displaystyle{\lim_{h \to 0}}\left(\dfrac{\zeta(1-h)-\zeta(1+h)}{2}\right)=:-\gamma
$$
This makes it obvious that we do not need a new constant which is simply the negative Euler-Mascheroni constant.

Do you want to discuss in how far a definition $\log(0)=-\gamma$ would make sense? If so, can you prove that this is a complex continuation of the zeta-function at $x=1?$ No. And here is where we are back at

But why would you need to do this?

Here is a nice list of what we already know about $\gamma$
https://en.wikipedia.org/wiki/Euler's_constant

 

[fresh_42]

The problem lies in the formula
$$
\gamma =\lim_{s \to 1} \left(\zeta(1)-\dfrac{1}{s-1}\right)
$$
Just because $\gamma$ is the difference between two diverging sequences, it doesn't allow us to take it to define another infinite quantity.

 

[Anixx]

Euler-Mascero

If so, can you prove that this is a complex continuation of the zeta-function at $x=1$ No.

The Cauchy principal value of Zeta function at $x=1$ is $\gamma$.

Also, take a look here: https://math.stackexchange.com/a/4083388/2513

 

[Anixx]

This makes it obvious that we do not need a new constant which is simply the negative Euler-Mascheroni constant.

Obviously I have never defined $\lambda=-\gamma$. If you read the post more carefully, you will see that $-\gamma$ is the finite part (which is the same as regularized value) of $\lambda$.

 

[fresh_42]

The Cauchy principal value of Zeta function at $x=1$ is $\gamma$.

So you want to discuss the meaning of the Cauchy principal value?

Also, take a look here: https://math.stackexchange.com/a/4083388/2513

... and the comments there. It makes no sense to discuss something with a constant reference to another website.

 

[Anixx]

So you want to discuss the meaning of the Cauchy principal value?

Cauchy principal value is a method of regularization.

 

[fresh_42]

Obviously I have never defined $\lambda=-\gamma$. If you read the post more carefully, you will see that $-\gamma$ is the finite part (which is the same as regularized value) of $\lambda$.

See the quotation in post #3. If you now attach a different meaning to "we can represent" other than equality, then this is exactly what I meant by a lack of rigor.

 

[Anixx]

... and the comments there. It makes no sense to discuss something with a constant reference to another website.

I made a link to the post that has no comments. It describes another way to see how the regularized value of the integral $\int_0^1 \frac1x dx$ is $\gamma$.

 

[fresh_42]

Cauchy principal value is a method of regularization.

Emphasis by me.

So? I still try to figure out what you are aiming at. Giving $\log(0)$ some finite value? What for? The Cauchy principal value isn't even substitution invariant! It is not of much use IMO.

 

[Anixx]

See the quotation in post #3. If you now attach a different meaning to "we can represent" other than equality, then this is exactly what I meant by a lack of rigor.

In that quotation I wrote that we can represent $\lambda=-\sum_{k=1}^\infty \frac1k$, yes. The regularized value (the finite part) of the right hand is $-\gamma$.

 

[Anixx]

What for? The Cauchy principal value isn't even substitution invariant! It is not of much use.

As you can see from the original post, while $\gamma$ is neither a period (member of ring of periods), nor an EL-number, $\lambda$ is both. Thus it has more nice properties.

 

[Anixx]

Also, from that same post we can see that in dual numbers $\varepsilon^\varepsilon=1+\varepsilon(1+\lambda)$, and many other nice properties.
Including, for instance, representation of some divergent integrals:
$\int_0^\infty \frac{\ln t}{t} \, dt=\gamma\lambda+\gamma^2/2$

 

[fresh_42]

Also, from that same post we can see that in dual numbers $\varepsilon^\varepsilon=1+\varepsilon(1+\lambda)$, and many other nice properties.
Including, for instance, representation of some divergent integrals:
$\int_0^\infty \frac{\ln t}{t} \, dt=\gamma\lambda+\gamma^2/2$

If you want to seriously consider $\lambda$ then you have to answer a few questions, regardless of what is written on MO.

1. What is your goal?
2. How is $\lambda$ defined?
3. Why did you choose $\log(0)$ and not any other pole of another function?
4. How is it different from $-\gamma$?
5. What do we get, what we don't already know from $\gamma$?
6. What should renormalization here provide?
7. Do you want to discuss Cauchy principal values?
8. Do you want to discuss Ramanujan sums?
9. Do you want to discuss regularization in general?
10. Do you want to discuss Riemann's zeta function?

 

[Anixx]

If you want to seriously consider $\lambda$ then you have to answer a few questions, regardless of what is written on MO.

1. What is your goal?
2. How is $\lambda$ defined?
3. Why did you choose $\log(0)$ and not any other pole of another function?
4. How is it different from $-\gamma$?
5. What do we get, what we don't already know from $\gamma$?
6. What should renormalization here provide?
7. Do you want to discuss Cauchy principal values?
8. Do you want to discuss Ramanujan sums?
9. Do you want to discuss regularization in general?
10. Do you want to discuss Riemann's zeta function?

1. Investigation of extending reals
2. $-\int_0^1 \frac1x dx$
3. Hmmm. This is not a pole.
4. $-\gamma$ is its finite part. $\lambda$ is negatively infinite.
5. To be honest, $\gamma$ is a constant that does not have a lot of nice properties (except being the real(finite)part of logarithm of zero). On the other hand $\lambda$ appears in nice relations.
6. If you mean regularization, it provides the finite part (in this case, real part).
7. No
8. No
9. No
10. No

 

[fresh_42]

1. Investigation of extending reals

This is a slogan, not a goal.

What does $\mathbb{R}\cup \{\lambda\}$ do?

2. $-\int_0^1 \frac1x dx$

This is merely a sequence of symbols. If we read it as an integral, then this is not defined as the integral does not converge. You cannot work with infinity this way.

3. Hmmm. This is not a pole.

It is.

4. $-\gamma$ is its finite part.

... of what?

$\lambda$ is negatively infinite.

So you want to extend the real numbers by $\pm \infty$? This is a well-known concept, even if not generally suited to solve problems.

5. To be honest, $\gamma$ is a constant that does not have a lot of nice properties (except being the real(finite)part of logarithm of zero).

Have you looked at the many formulas that contain $\gamma$? Look again at my Wikipedia link.

On the other hand $\lambda$ appears in nice relations.

Especially, as it isn't defined!

 

[Anixx]

1. It is described in the linked post what can be done with this extension.

2. It is a divergent integral, and it is well described what is meant. You can work with infinitely large values quite well.

3. Logarithmic singularity is not a pole.

4. Of divergent integral $-\int_0^1 \frac1x dx$ and of divergent series $\sum_{k=1}^\infty -\frac1k$

4a. "So you want to extend the real numbers by $\pm\infty$ ?" Obviously, not. Do you think, the $\pm\infty$ compactification is the only one possible? Have you heard about Hardy fields or formal power series?

 

[PeterDonis]

What do you guys have to say about this Mathoverflow post?

It looks to me like either you're rediscovering something that mathematicians already know, in which case you should be able to find a mainstream reference somewhere, or you're doing personal research, which is off limits here at PF--if you think you've made some new mathematical discovery, you need to publish it in a journal, not here.

 

[fresh_42]

1. It is described in the linked post what can be done with this extension.

No, it is not. But anyway, why don't you repeat it here?

2. It is a divergent integral, and it is well described what is meant. You can work with infinitely large values quite well.

You can cheat with infinite large values quite well. That's why I demand a rigorous definition.

So far you defined $\lambda =-\int_0^1 t^{-1}dt$ which is not a well-defined expression. It is symbolism.

3. Logarithmic singularity is not a pole.

In my language, it is a pole. Maybe not in English although I doubt that. But it doesn't really matter what you call it. It is a location where the logarithm is not defined. Call it as you like.

4. Of divergent integral $-\int_0^1 \frac1x dx$ and of divergent series $\sum_{k=1}^\infty -\frac1k$

You can just pick a divergent series or integral and play with it. It simply does not make any sense to me.

4a. "So you want to extend the real numbers by $\pm\infty$ ?" Obviously, not. Do you think, the $\pm\infty$ compactification is the only one possible? Have you heard about Hardy fields or formal power series?

I haven't seen any formal power series in the entire discussion until now.