Requesting constructive criticism for my paper

[mathhabibi]

TL;DR Summary

I got my first paper approved on ArXiV but I want to know how to improve.

Hello PF!

This is my very first post here. Just yesterday my paper was accepted by ArXiV, called "A Simple Continuation for Partial Sums". If you have time (it's 14 pages) you can take a look at it here. I was just interested in ways I could improve my paper or if it was completely useless in the first place. Ask me any questions about it if you need to.

Thanks for your time

 

[FactChecker]

There might be things that I don't understand.
1) Right at the beginning, I expected some context. What is f? What are k, n, and L? Real? Complex? Matrices?
2) In the statement of Theorem 1, do you prove that the infinite summation makes sense?
3) In Theorem 1, for n=1, does that say that
f(1) = L + (f(1)-f(2)) + (f(2)-f(3)) + (f(3)-f(4)) +.... =? L+f(1)
If f(1) = L+f(1) doesn't that mean that L=0?

 

[mathhabibi]

There might be things that I don't understand.
1) Right at the beginning, I expected some context. What is f? What are k, n, and L? Real? Complex? Matrices?
2) In the statement of Theorem 1, do you prove that the infinite summation makes sense?
3) In Theorem 1, for n=1, does that say that
f(1) = L + (f(1)-f(2)) + (f(2)-f(3)) + (f(3)-f(4)) +.... =? L+f(1)
If f(1) = L+f(1) doesn't that mean that L=0?

1) As stated in the paper, L is constant, k is an index. I was not specific about n (thanks for that) it should be a variable. f is a function that approaches L, which I haven’t been able to extend to complex values yet.
2) If n is an integer it converges. If f is Monotonic (sufficient not necessary) then the infinite series converges if k+n is in the domain of f (ie every term is defined).
3) Yes the reason I pointed this out was because a professor came up with this. However this counter example isn’t true because of the Riemann rearrangement theorem.

You will realize after this that the Main result is trivial but the theory behind it is quite dense.

 

[FactChecker]

1) As stated in the paper, L is constant, k is an index.

Is that later in the paper? I don't see L or k mentioned anywhere before they are used. They should be defined, before they are used, like $L \in \mathbb R$ and $k \in \mathbb N$.

I was not specific about n (thanks for that) it should be a variable. f is a function

Apparently, it is a function defined on the natural numbers? Where is that stated? All the definitions of f, L, k, n, should be included at the beginning of the statement of Theorem 1 before they are used.

that approaches L, which I haven’t been able to extend to complex values yet.
2) If n is an integer it converges. If f is Monotonic (sufficient not necessary) then the infinite series converges if k+n is in the domain of f (ie every term is defined).
3) Yes the reason I pointed this out was because a professor came up with this. However this counter example isn’t true because of the Riemann rearrangement theorem.

You will realize after this that the Main result is trivial but the theory behind it is quite dense.

I was just trying to help, but I can not understand the beginning.

 

[mathhabibi]

Is that later in the paper? I don't see L or k mentioned anywhere before they are used. They should be defined, before they are used, like $L \in \mathbb R$ and $k \in \mathbb N$.

Apparently, it is a function defined on the natural numbers? Where is that stated? All the definitions of f, L, k, n, should be included at the beginning of the statement of Theorem 1 before they are used.

I was just trying to help, but I can not understand the beginning.

So I did not define that L is real. In fact I was unclear specifying all the variables you were asking about. Thanks for that, will be noted. Also I was saying that once the formula makes sense to you it will be very obvious and you’d wonder how it hasn’t been stated before. I’ve never found it before after an Approach0 search so I assumed no one knows it.

 

[FactChecker]

Clearly, you have put a lot of work into this. I can tell you that writing a publishable theoretical math paper is a painful, tedious, thankless job. This type of writing is a skill that takes time to develop. My initial advice is to be as clear as you can about what every entity mentioned is before you use them. Do not be discouraged.

 

[mathhabibi]

Clearly, you have put a lot of work into this. I can tell you that writing a publishable theoretical math paper is a painful, tedious, thankless job. This type of writing is a skill that takes time to develop. My initial advice is to be as clear as you can about what every entity mentioned is before you use them. Do not be discouraged.

Yes it was hard, it took me the collective work of half a year in just fourteen pages. I really appreciate your last sentence though, because whenever I talked about potential research ideas math hobbyists would absolutely dismantle me lol.

 

[fresh_42]

TL;DR Summary: I got my first paper approved on ArXiV but I want to know how to improve.

Hello PF!

This is my very first post here. Just yesterday my paper was accepted by ArXiV, called "A Simple Continuation for Partial Sums". If you have time (it's 14 pages) you can take a look at it here. I was just interested in ways I could improve my paper or if it was completely useless in the first place. Ask me any questions about it if you need to.

Thanks for your time

I am not convinced that your result is even true.

It looks to me as if the neglect of dealing with any orders of summation and arbitrary summands $f(n)$ is in contradiction to Riemann's series theorem. However, you didn't say what $f$ is, a continuous function, or only a discrete summand in which case the notation $a_k$ would be better. Also, you use $x$ as a discrete value. That's confusing, too. There are name conventions in mathematics. If you break them, you need to explicitly list their new usage.

Those questions have to be addressed before I take the effort to construct a counterexample.

 

[mathhabibi]

I am not convinced that your result is even true.

It looks to me as if the neglect of dealing with any orders of summation and arbitrary summands $f(n)$ is in contradiction to Riemann's series theorem. However, you didn't say what $f$ is, a continuous function, or only a discrete summand in which case the notation $a_k$ would be better. Also, you use $x$ as a discrete value. That's confusing, too. There are name conventions in mathematics. If you break them, you need to explicitly list their new usage.

Those questions have to be addressed before I take the effort to construct a counterexample.

The value $x$ was only used for the proof, it doesn’t matter for the result (it doesn’t appear in Theorem 1). Notice that the purpose of this theorem is to provide a continuation for the LHS of theorem 1, so if I only cared about the LHS then $a_k$ would be proper notation, but that’s not the case. The proof is assuming that f is defined on the integers, but after the proof ends I prove that it is sufficient for f to be monotonic for the infinite sum to converge when $n$ is not an integer.

Also f doesn’t need to be continuous, but that affects the domain. And if it is only discrete then it is useless.

Notes to self: Be much more clear on the assumptions for Theorem 1 and it’s proof.

 

[FactChecker]

I am still concerned about my point #3 in post #2. The case of n=1 in Theorem 1 seems to imply that L=0 in all cases, which can not be universally true. So I doubt that Theorem 1 can be correct.

 

[mathhabibi]

I am still concerned about my point #3 in post #2. The case of n=1 in Theorem 1 seems to imply that L=0 in all cases, which can not be universally true. So I doubt that Theorem 1 can be correct.

Yes but the reorganization of parantheses isn’t allowed for infinite series. I.e. associative property doesn’t work anymore.

 

[FactChecker]

Yes but the reorganization of parantheses isn’t allowed for infinite series. I.e. associative property doesn’t work anymore.

I guess that is fine as long as you justify, in your proof, your choice of that grouping when other groupings might give other results. I'll take your word for it that you do that. I am reluctant to deal with summations that are not absolutely convergent and have very little experience with them. It seems that they can be arranged to get anything you want. So I will leave this to others.

 

[mathwonk]

the first theorem seems to say that if a sequence f(N)-->L, as N-->∞, (N taking on values of positive integers), then for all positive integers n, f(N+1)+....+f(N+n)-->nL, as N-->∞. (Just put the series over on the LHS and simplify the Nth partial sum.)

This seems both true and elementary. I.e. if the elements of the sequence converge to L, then the sum of n consecutive elements converges to nL. If this is the main result, you may want to reconsider your submission.

 

[mathhabibi]

If this is the main result, you may want to reconsider your submission.

By main result I meant the base of the paper. If you continue to read then you’ll find Taylor series derived from this formula, derivatives, PDEs, a new conjecture, and so on. But what makes a simple formula unpublishable? Also I will use the feedback in this post to see whether this should enter a journal or not, so if my paper isn’t worth it then I will just leave it alone.

Also the main result here isn’t that the sum of n elements that approach L is nL, although it is a help with making the formula.

 

[fresh_42]

You should definitely define your variables and letters! I'm still not sure what $f$ is or - in case it is arbitrary - use it to find a counterexample by the vanishing mass at infinity, reordering the infinite sum, or similar "tricks"; will say: does the RHS even converge? And you should avoid $x$ if you mean a natural number or integer. Otherwise, what does $\sum_{k=1}^x$ mean, $\sum_{k=1}^{\lfloor x \rfloor}$ maybe?

By "main result" we usually mean the central new theorem. It is what everybody reads who opens the pdf or the journal. It decides whether people keep reading or turn away. A simple fact as "main theorem" turns away readers. I even thought it was a high school project and you were just playing around with series. I personally do not like W.L.O.G., and I hate it when it comes in caps. Caps mean shouting. I'm always skeptical when such phrases occur, especially so early in the text. I wonder whether they hide something. Why can't you just assume it if $x\to \infty$ is the central argument anyway?

 

[mathhabibi]

You should definitely define your variables and letters! I'm still not sure what $f$ is or - in case it is arbitrary - use it to find a counterexample by the vanishing mass at infinity, reordering the infinite sum, or similar "tricks". And you should avoid $x$ if you mean a natural number or integer. Otherwise, what does $\sum_{k=1}^x$ mean, $\sum_{k=1}^{\lfloor x \rfloor}$ maybe?

By "main result" we usually mean the central new theorem. It is what everybody reads who opens the pdf or the journal. It decides whether people keep reading or turn away. A simple fact as "main theorem" turns away readers. I even thought it was a high school project and you were just playing around with series. I personally do not like W.L.O.G., and I hate it when it comes in caps. Caps mean shouting. I'm always skeptical when such phrases occur, especially so early in the text. I wonder whether they hide something. Why can't you just assume it if $x\to \infty$ is the central argument anyway?

f is arbitrary, as I’ve said it is any function that takes discrete values. Good advice about $x$, I will take note of that. Also I thought that taking caps when letters stand for words is correct grammatically, thanks about that too. The reason why I put wlog is to justify splitting the sums the way I did it, but I guess it’s enough to write that I want to take $x\to\infty$.

 

[fresh_42]

f is arbitrary, as I’ve said it is any function that takes discrete values. Good advice about $x$, I will take note of that. Also I thought that taking caps when letters stand for words is correct grammatically, thanks about that too. The reason why I put wlog is to justify splitting the sums the way I did it, but I guess it’s enough to write that I want to take $x\to\infty$.

You could write "Wlog." as it is at the beginning of a sentence, but you do not need it! Simply assume $x>n+1$ although I think $x>n$ will do. You are free to choose $x$ appropriately. Phrases like wlog or obvious are traps. I personally try to avoid them, but this could be a matter of taste. I spent too much time finding out what so obvious was, and whether wlog hides an entire branch of a proof, in this case $x<n.$ Fact is, it is not wlog, it is simply unnecessary, and that's something else. You cause people to think about an irrelevant case.

 

[mathhabibi]

You cause people to think about an irrelevant case.

Oh god, that’s terrible. I highly appreciate your feedback.

 

[fresh_42]

It is late at night here so maybe I scribbled something wrong. We have
\begin{align*}
\sum_{k=1}^{n}f(k)&\stackrel{!}{=}nL+\sum_{k=1}^{n}(f(k)-f(k+n))+\sum_{k=n+1}^{\infty }(f(k)-f(k+n))\\
\sum_{k=1}^{n}f(k)&\stackrel{!}{=}nL+\sum_{k=1}^{n}f(k) +\sum_{k=n+1}^{\infty }f(k) -\sum_{k=1}^{\infty }f(k+n)\\
nL+\sum_{k=n+1}^{\infty }f(k)&\stackrel{!}{=} \sum_{k=1}^{\infty }f(k+n)=\sum_{k=n+1}^{\infty }f(k)
\end{align*}
Why doesn't that imply $L=0$?

 

[mathhabibi]

It is late at night here so maybe I scribbled something wrong. We have
\begin{align*}
\sum_{k=1}^{n}f(k)&\stackrel{!}{=}nL+\sum_{k=1}^{n}(f(k)-f(k+n))+\sum_{k=n+1}^{\infty }(f(k)-f(k+n))\\
\sum_{k=1}^{n}f(k)&\stackrel{!}{=}nL+\sum_{k=1}^{n}f(k) +\sum_{k=n+1}^{\infty }f(k) -\sum_{k=1}^{\infty }f(k+n)\\
nL+\sum_{k=n+1}^{\infty }f(k)&\stackrel{!}{=} \sum_{k=1}^{\infty }f(k+n)=\sum_{k=n+1}^{\infty }f(k)
\end{align*}
Why doesn't that imply $L=0$?

Try this trick for $\frac1x$. The infinite series doesn’t need to absolutely converge. Also you forgot $-\sum_{k=1}^nf(k+n)$ in the second equation.

I’ve tried this formula too many times (around half a year to be precise), finding a counter example would be nearly impossible.

 

[fresh_42]

Try this trick for $\frac1x$. The infinite series doesn’t need to absolutely converge. Also you forgot $-\sum_{k=1}^nf(k+n)$ in the second equation.

I don't think so.

\begin{align*}
&\text{the statement in your paper}\\
\sum_{k=1}^{n}f(k)&\stackrel{!}{=}nL+\sum_{k=1}^{n}(f(k)-f(k+n))+\sum_{k=n+1}^{\infty }(f(k)-f(k+n))\\
&\text{the parentheses resolved and sorted by the argument of }f \\
\sum_{k=1}^{n}f(k)&\stackrel{!}{=}nL+\sum_{k=1}^{n}f(k) +\sum_{k=n+1}^{\infty }f(k) -\sum_{k=1}^{\infty }f(k+n)\\
&\text{intermediate step 1, subtraction of the finite sum on both sides}\\
0&\stackrel{!}{=}nL+\sum_{k=n+1}^{\infty }f(k) -\sum_{k=1}^{\infty }f(k+n)\\
&\text{intermediate step 2, putting the negative sum on the other side}\\
\sum_{k=1}^{\infty }f(k+n)&\stackrel{!}{=}nL+\sum_{k=n+1}^{\infty }f(k)\\
&\text{intermediate step 3, changing sides}\\
nL+\sum_{k=n+1}^{\infty }f(k)&\stackrel{!}{=}\sum_{k=1}^{\infty }f(k+n)\\
&\text{intermediate step 4, changing index } p=k+n\\
nL+\sum_{k=n+1}^{\infty }f(k)&\stackrel{!}{=}\sum_{p=n+1}^{\infty }f(p)\\
&\text{finally re-substituting }k=p\\
nL+\sum_{k=n+1}^{\infty }f(k)&\stackrel{!}{=} \sum_{k=n+1}^{\infty }f(k)\\
&\text{subtraction of the same infinite sum on both sides}\\
nL&\stackrel{!}{=}0 \\
&\text{assume }n>0\\
L&\stackrel{!}{=}0
\end{align*}

 

[mathhabibi]

I don't think so.

\begin{align*}
&\text{the statement in your paper}\\
\sum_{k=1}^{n}f(k)&\stackrel{!}{=}nL+\sum_{k=1}^{n}(f(k)-f(k+n))+\sum_{k=n+1}^{\infty }(f(k)-f(k+n))\\
&\text{the parentheses resolved and sorted by the argument of }f \\
\sum_{k=1}^{n}f(k)&\stackrel{!}{=}nL+\sum_{k=1}^{n}f(k) +\sum_{k=n+1}^{\infty }f(k) -\sum_{k=1}^{\infty }f(k+n)\\
&\text{intermediate step 1, subtraction of the finite sum on both sides}\\
0&\stackrel{!}{=}nL+\sum_{k=n+1}^{\infty }f(k) -\sum_{k=1}^{\infty }f(k+n)\\
&\text{intermediate step 2, putting the negative sum on the other side}\\
\sum_{k=1}^{\infty }f(k+n)&\stackrel{!}{=}nL+\sum_{k=n+1}^{\infty }f(k)\\
&\text{intermediate step 3, changing sides}\\
nL+\sum_{k=n+1}^{\infty }f(k)&\stackrel{!}{=}\sum_{k=1}^{\infty }f(k+n)\\
&\text{intermediate step 4, changing index } p=k+n\\
nL+\sum_{k=n+1}^{\infty }f(k)&\stackrel{!}{=}\sum_{p=n+1}^{\infty }f(p)\\
&\text{finally re-substituting }k=p\\
nL+\sum_{k=n+1}^{\infty }f(k)&\stackrel{!}{=} \sum_{k=n+1}^{\infty }f(k)\\
&\text{subtraction of the same infinite sum on both sides}\\
nL&\stackrel{!}{=}0 \\
&\text{assume }n>0\\
L&\stackrel{!}{=}0
\end{align*}

You are only allowed to split the sum when the infinite series of f(k) converges, which happens only when $L=0$. I got a little scared lol

 

[fresh_42]

You are only allowed to split the sum when the infinite series of f(k) converges, which happens only when $L=0$. I got a little scared lol

I'll try the vanishing mass at infinity tomorrow. This lecture note is in the wrong language but the formulas are independent of language: http://scratchpost.dreamhosters.com/math/HM3-D-2x2.pdf
Something like that.

I first thought that you need to impose a condition on $f$ such that Riemann's series theorem doesn't apply. But this is basically the situation with $L=0.$ I still wonder, however, whether the RHS is always convergent for arbitrary $f$ and a fixed $n.$

Another idea is to check functions $f(k)=\sum_{m=1}^k \dfrac{(-1)^{m+1}}{m}$ where $\lim_{n \to \infty}f(n)$ depends on the ordering of the sum.

 

[mathhabibi]

I still wonder, however, whether the RHS is always convergent for arbitrary $f.$

Great question. I don't know exactly what the weakest conditions are on $f$, but if $n$ is an integer, the infinite series will converge (equal to the LHS). And if $f$ is monotonic then it's sufficient to make the infinite series converge (by simple bounding). However, $f$ doesn't need to be monotonic (try $f(x)=\frac{\sin x}x$ or $f(x)=-\frac{\cos(\pi x)}x$ for example) as long as every term is defined.

 

[mathwonk]

Of course I could be wrong, but I believe, as I remarked (equivalently) in post #13, the Nth partial sum of the series in theorem #1, equals:
f(1)+....+f(n) -[ f(N+1)+....+f(N+n)], which converges to f(1)+....+f(n) -nL, as N-->∞.

Since by definition, the sum of a series is the limit of its sequence of partial sums, the claim in the theorem seems correct, but elementary.

Indeed, to me this is a nice example to illustrate why you cannot reorder such series to claim that L=0.

The series is apparently an example, slightly tweaked, of a simple telescoping series of form (a1-a2) + (a2-a3) +....+(an-an+1) +...........
whose nth partial sum is a1-an+1, and which therefore converges to a1-L, where L is the limit of the sequence {an}.

(In theorem #1, (a1+...+an) plays the role of a1 in this basic example.)

If indeed interesting consequences are derived from this, I would suggest writing something like: " we intend to show how to derive some interesting consequences from the following simple result," rather than stating that this fact is the most fundamental formula in the paper. After reading this, I myself was not motivated to read further. I apologize if this is unhelpful.

Perhaps I should mention that I did not originally read past line 3 of the paper (the stated formula). When I affirmed it after a few minutes' thought, I turned away without giving the paper a further chance.

 

[mathhabibi]

If indeed interesting consequences are derived from this, I would suggest writing something like: " we intend to show how to derive some interesting consequences from the following simple result," rather than stating that this fact is the most fundamental formula in the paper. After reading this, I myself was not motivated to read further. I apologize if this is unhelpful.

Perhaps I should mention that I did not originally read past line 3 of the paper (the stated formula). When I affirmed it after a few minutes' thought, I turned away without giving the paper a further chance.

Oh yes, this is very helpful. Before submitting to a journal I will have to completely change the format of my paper (which is so far the only problem). Thank you for your feedback.

 

[mathwonk]

note also, that it is well known that every sequence can be written as the sequence of partial sums of a series, i.e. the sequence a1, a2, a3, a4,...,an,... is the sequence of partial sums of the series a1 + (a2-a1) + (a3-a2) + (a4-a3) +.....+(an - an-1) +.... Theorem 1 is essentially just an application of this observation.

The series for Euler's number e at the bottom of page 9, is an example of this well known fact, applied to the famous expression for e, as the limit of the sequence { (1 + 1/n)^n }. Hence you may wish to be more reserved when asserting such results are new. Best wishes with your efforts.

 

[mathhabibi]

The series for Euler's number e at the bottom of page 9

So you read the properties section? What do you think about it?

 

[sbrothy]

Now, I'm just an uneducated hack and the contents of your paper flies way way over my head. What kinda jumped out at me though, was the relatively few citations. You mention Riemann, Euler, MacLauri and Ramanjuan but don't cite them. Now this may be quite normal among you physicists as these people may be so common you just per convention don't cite them (as you probably wouldn't cite Newton for using his basic work).

To wit, these authors in particular may be dead but it occurs to me that if you yourself want to be cited (as I'm quuite sure you want :P) I'm sure reciprocation will be appreciated. Maybe it's a good habit to get into. If for nothing else then to show that you're actually aware you're standing on these people's shoulders.

Just my 2 pretty worthless cents. Nevermind.

 

[mathhabibi]

You mention Riemann, Euler, MacLauri and Ramanjuan but don't cite them.

I forgot that I mentioned Ramanujan, lol. As you said, their results are very famous so we can use them without citation. Except the Ramanujan one, because not many know the Ramanujan summation. So I will cite his formula when I replace this submission.

 

[sbrothy]

Heh, I'm immensely proud I could contribute with anything at all. That will keep me going all week. :)

 

[mathhabibi]

Another idea is to check functions $f(k)=\sum_{m=1}^k \dfrac{(-1)^{m+1}}{m}$ where $\lim_{n \to \infty}f(n)$ depends on the ordering of the sum.

I didn’t see this edit, I devoted half a subsection on this specific function (the alternating harmonic numbers) and came up with a conjecture on its zeroes and a reflection formula for it

 

[Mark44]

You mention Riemann, Euler, MacLauri and Ramanjuan but don't cite them.

Maybe something from Ramanujan, but Riemann, Euler, and Maclaurin don't need specific citations, IMO. Every calculus book mentions the Riemann integral and Maclaurin series, as well as Euler's equation involving $e, \pi, -1, i$ and $0$ without citing the work in which these first appeared.

 

[mathhabibi]

Maybe something from Ramanujan, but Riemann, Euler, and Maclaurin don't need specific citations, IMO. Every calculus book mentions the Riemann integral and Maclaurin series, as well as Euler's equation involving $e, \pi, -1, i$ and $0$ without citing the work in which these first appeared.

No, there isn't a Riemann integral or Maclaurin series. It's the Riemann zeta function and Euler-Maclaurin formula. But again, these two are pretty well-known already.

 

[Mark44]

No, there isn't a Riemann integral or Maclaurin series. It's the Riemann zeta function and Euler-Maclaurin formula. But again, these two are pretty well-known already.

I gave those as examples, but my point was that these names are so well-known, that citations aren't necessary.