- #1

Anixx

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Do you have any interesting ideas about this?

https://mathoverflow.net/questions/...ithm-of-zero-properties-and-reference-request

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In summary, the conversation revolves around the idea of extending the real numbers with the logarithm of zero, as well as the validity and rigor of such an extension. One person mentions that it may be possible in a restricted sense, but questions the necessity of doing so. Another person shares their opinion that this idea is nonsensical and lacks rigor. The concept of regularization and the Cauchy principal value are also brought up as potential methods to define the logarithm of zero. The conversation ends with a reference to the Euler-Mascheroni constant and a link to a related discussion on another website.

- #1

Anixx

- 81

- 12

Do you have any interesting ideas about this?

https://mathoverflow.net/questions/...ithm-of-zero-properties-and-reference-request

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- #2

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

Do you have any interesting ideas about this?

https://mathoverflow.net/questions/...ithm-of-zero-properties-and-reference-request

-Dan

- #3

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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.Anixx said:What do you guys have to say about this Mathoverflow post?

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

Sorry, that is nonsense in my opinion, i.e. it does not make any senseThe 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##.

$$

\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 .##

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- #4

Anixx

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That the Harmonic series has the regularized value ##\gamma## is a known mathematical fact. Take the Cauchy principal value for instance:fresh_42 said: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.

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

- #5

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Yes, and this was the first time you actually mentioned whatAnixx said: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]

$$

\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

Here is a nice list of what we already know about ##\gamma ##topsquark said:But why would you need to do this?

https://en.wikipedia.org/wiki/Euler's_constant

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- #6

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$$

\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.

- #7

Anixx

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The Cauchy principal value of Zeta function at ##x=1## is ##\gamma##.fresh_42 said:If so, can you prove that this is a complex continuation of the zeta-function at ##x=1## No.

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

- #8

Anixx

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Obviously I have never defined ##\lambda=-\gamma##. If you read the post more carefully, you will see that ##-\gamma## is thefresh_42 said:This makes it obvious that we do not need a new constant which is simply the negative Euler-Mascheroni constant.

- #9

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So you want to discuss the meaning of the Cauchy principal value?Anixx said:The Cauchy principal value of Zeta function at ##x=1## is ##\gamma##.

... and the comments there. It makes no sense to discuss something with a constant reference to another website.Anixx said:Also, take a look here: https://math.stackexchange.com/a/4083388/2513

- #10

Anixx

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Cauchy principal value is a method of regularization.fresh_42 said:So you want to discuss the meaning of the Cauchy principal value?

- #11

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

- #12

Anixx

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

- #13

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Emphasis by me.Anixx said:Cauchy principal value isamethod of regularization.

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.

- #14

Anixx

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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##.fresh_42 said: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.

- #15

Anixx

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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.fresh_42 said:What for? The Cauchy principal value isn't even substitution invariant! It is not of much use.

- #16

Anixx

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Including, for instance, representation of some divergent integrals:

##\int_0^\infty \frac{\ln t}{t} \, dt=\gamma\lambda+\gamma^2/2##

- #17

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

Including, for instance, representation of some divergent integrals:

##\int_0^\infty \frac{\ln t}{t} \, dt=\gamma\lambda+\gamma^2/2##

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

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- #18

Anixx

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1. Investigation of extending realsfresh_42 said:If you want to seriously consider ##\lambda ## then you have to answer a few questions, regardless of what is written on MO.

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

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

- #19

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This is a slogan, not a goal.Anixx said:1. Investigation of extending reals

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

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.Anixx said:2. ##-\int_0^1 \frac1x dx##

It is.Anixx said:3. Hmmm. This is not a pole.

... of what?Anixx said:4. ##-\gamma ## is its finite part.

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.Anixx said:##\lambda## is negatively infinite.

Have you looked at the many formulas that contain ##\gamma ##? Look again at my Wikipedia link.Anixx said: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).

Especially, as it isn't defined!Anixx said:On the other hand ##\lambda## appears in nice relations.

- #20

Anixx

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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

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?

- #21

PeterDonis

Mentor

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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.Anixx said:What do you guys have to say about this Mathoverflow post?

- #22

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No, it is not. But anyway, why don't you repeat it here?Anixx said:1. It is described in the linked post what can be done with this extension.

You can cheat with infinite large values quite well. That's why I demand a rigorous definition.Anixx said:2. It is a divergent integral, and it is well described what is meant. You can work with infinitely large values quite well.

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

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.Anixx said:3. Logarithmic singularityis nota pole.

You can just pick a divergent series or integral and play with it. It simply does not make any sense to me.Anixx said:4. Of divergent integral ##-\int_0^1 \frac1x dx## and of divergent series ##\sum_{k=1}^\infty -\frac1k##

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

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