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• CommentAuthorAnixx
• CommentTimeNov 3rd 2010

I wonder why my answer to this question was deleted: http://mathoverflow.net/questions/17605/how-to-solve-ffx-cosx

Although I did not provide the solution to the exact question by the asker, to find the solution f(f(x))=cos(x), he asked also "Is there a general solution strategy to equations of this kind?" and I gave the exact formula for the solution of f(f(x)=sin(x), and made a plot. I wonder why my question was so downvoted and deleted after all?
1.
I don't know why your question was deleted but you didn't really supply any details regarding when your procedure works and when it does not, so in that regard I imagine people deemed it not really an answer to the question. Formulas without any reasoning behind them aren't particularly useful.
• CommentAuthorAnixx
• CommentTimeNov 3rd 2010

The formula works well and converges for most functions which superfunction growth rate is below factorial. In fact it works excellent for the function sin(x).
• CommentAuthorAnixx
• CommentTimeNov 3rd 2010

• CommentAuthorAnixx
• CommentTimeNov 3rd 2010

• CommentAuthorRyan Budney
• CommentTimeNov 3rd 2010 edited

"Works well" isn't an actual criterion, but a vague assertion. Could you give precise conditions where there's a proof that it does the job, and a reference for the proof, or supply the proof? If you had originally answered with this level of precision I doubt you'd have these kinds of problems.
2.
To respond to your last comment "exact question, exact answer..." I think some people have grown tired of commenting on your answers and questions, as they tend to result in long arguments that if resolved, get resolved by 3rd parties. Many people (including me) don't want to see MO work this way.
3.

First of all, here is the answer, since right now only 10K+ users can see it. (The original had a plot in it, which I haven't copied.)

You can use this formula to solve your equation:

$$f^{[1/2]}(x) = \sum_{m=0}^{\infty} \binom{1/2}{m} \sum_{k=0}^m \binom{m}{n} (-1)^{m-k} f^{[k]}(x)$$

where $f^{[k]}(x)$ is a $k$-th iterate of $x$. Here is a plot for $\sin x$.

But for cosine this method apparently doesn't work.

I would not have deleted this answer, were I a moderator. But it falls into the frustrating category, which I know very well from grading exams, of the answer which mentions some useful knowledge but does not show how to actually apply it to the problem, or an awareness of when it can be applied.

It is interesting to know that this formula exists, and it is new to me. So I think there is some valuable information in this answer.

The obvious problem on looking at this formula is that it is not clear when it converges, either in the sense of analysis or in the ring of formal power series. Even if $f(x)=0$, computing any term of $f^{[1/2]}$ appears to require evaluating an infinite sum. If $f(0)=0$, and $|f'(0)|<1$, then I'm pretty sure I could show the sum converges (warning, I have not checked all the details here), but when $f'(1)=1$ even that isn't clear to me, so I don't know whether your formula is even meaningful for the case of sine where you compute it.

Ironically, if you (Anixx) had thought about these issues, you would see that the case of cosine is less subtle. Since cosine has a fixed point at approximately 0.739, where the derivative is clearly small, you could just shift your power series to be centered around this point and obtain a nice sum which would probably converge quickly. This would build nicely on Gerald Edgar's answer, which uses the same idea of shifting to a fixed point, but does not obtain an explicit solution.

As I said, I would not have deleted this answer, because it contributes valuable information. But I think it generally suffers from a willingness to thinking about mathematics as formulas, rather than thinking about when those formulas are meaningful.

4.
My guess is that your answer got flagged as spam. If enough people flag an answer as spam, then it gets deleted.
• CommentAuthorAnixx
• CommentTimeNov 3rd 2010

It is difficult to predict what can result in the long arguments, but if people have question such as "when does it converge", the questions can be answered individually, if the answer exists. I saw no comment with such questions, my answer was simply deleted without a comment.
5.

As an example of the subtleties which can arise, I was recently reminded of http://www.ams.org/mathscinet-getitem?mr=97532 , which shows that there is no analytic function f, convergent in a neighborhood of 0, such that f(f(z))=e^z-1. This despite the fact that there is a perfectly good formal power series solution.

6.
Anixx, it's your track-record that's getting people's attention more than your individual answers at this point. By and large issues such as convergence are things people expect you would address when you offer an answer. By not supplying this kind of information people presume it will be like in your previous threads -- if people ask, you will argue first before doing anything else. I suggest you avoid supplying formulaic answers without giving detailed descriptions of when you can prove convergence.
• CommentAuthorAnixx
• CommentTimeNov 3rd 2010

@ David Speyer

This is not power series, so I fail to see how it can be centered around fixed point. By which variable?
7.

Most mathematicians take the view that, when preparing a communication, one should anticipate the arguments that will arise, and address them proactively. This is why math papers are hard to read, but also why one mathematician can write a paper which is useful to thousands.

To me, there is a big difference between what I will criticize in a comment, and what I would vote to delete. I wouldn't delete this, but I would have left comments asking about the points I raised above (and, also, asking for a citation for this result). MO software allows for collaborative editing and incremental improvement, and I think we should take advantage of that. But I also try to write answers which are as thorough and clear as I can make them before revision. When someone seems to be paying no attention to these concerns, I am not surprised that it annoys other mathematicians.

• CommentAuthorAnixx
• CommentTimeNov 3rd 2010

> I suggest you avoid supplying formulaic answers without giving detailed descriptions of when you can prove convergence.

And I suggest you let the asker to decide if the answer was useful.

Besides this, this is a very general formula, and proving convergence for all cases may be impossible. Anyway this proof does not belong here, it is better targeted in a general discussion of convergence of Newton's series. Only one thing is worth mention here, that the formula converges when Newton series of superfunction converges.
8.
Let a be the fixed point of cos. Then

$$\cos(a+z) -a = (\sin a) * z - (\cos a) z^2/2 - (\sin a) z^3/6 + (\cos a) z^4/24 + (\sin a) z^5/120 - \cdots.$$

Take $f(z) = \cos(a+z)-a$ and use your formula to compute $g$ such that $g(g(z))= f(z)$. I am pretty sure that (1) On some neighborhood of a, your formula converges, and gives a g such that $g(g(z))=f(z)$. (2) If you expand your formula out term by term, it gives a convergent sum for each each coefficient of the power series of $g$, and the resulting formal power series converges to $g$.

Then $h(z) := a+g(z-a)$ is a solution of $h(h(z)) = \cos z$, near $z=a$.
9.
I would allow the questioner the ability to vote on this, which is part of why I don't think the answer should have been deleted. However, I also want to build a repository of knowledge which would be useful for the future, which is why I think the answer should be improved.
• CommentAuthorAnixx
• CommentTimeNov 3rd 2010

@ David Speyer

No, David, this seems does not work. The result I got is the same as applying the formula without shifting.
10.
Interesting. Well, as I've been saying, I haven't checked everything, so maybe there is a subtlety I'm not getting. I don't have time to track it down, but I'd certainly be interested if someone did.
11.

@Anixx: You asked why your question got deleted, and Ryan has given you what to me sounds like a very plausible explanation -- there's clearly something undesirable about the answer (hence the 4 downvotes), and discussing with the matter with you has proven in the past to be too burdensome to be worth the effort. Your discussion with David is only furthering Ryan's point in this regard. If you take the time to include some sort of theoretical framework (proof of convergence, references, even a heuristic argument, etc.), rather than simply an isolated claim, and proactively discuss potential issues that arise with your answer, I think you will find that you are met with much less resistance.

• CommentAuthorAnixx
• CommentTimeNov 3rd 2010

I am not against discussion ever. But how thw anser can be discussed if it is deleted?
12.
Oh, wait I do see it, I think. $|f'(0)|<1$ is NOT the right condition.

Suppose that $f(r)=r$ and $f'(r)=a$. Then $f^n(a+x) = a + b r^n x + O(r^{2n})$, for $x$ near zero and for some constant $b$. So, near $0$, the inner sum is about
$$\sum_{k=0}^m (-1)^k \binom{m}{k} a + bx \sum_{k=0}^m (-1)^k \binom{m}{k} r^k = 0 + b (1-r)^m.$$ (for m>0)

So this term is small, not when $r$ is small (as I had been expecting) but when $(1-r)$ is small. In particular, since $\binom{1/2}{m} = o(s^m)$ for any $s>1$, we expect convergence when $|1-r|<1$. (This is still heuristic, since I haven't been careful about how I cut off the $O(r^{2k})$ term, but it's a better heuristic than I had before.) In particular, since the derivative at cosine near its fixed point is about -0.7, we get that $1-r$ is about $1.7$ and the sum blows up.

So it looks like this formula is optimized for dealing with fixed points where the derivative is near 1, not near 0. Interesting. And making it especially interesting to know why it fails for $e^z-1$.
• CommentAuthorAnixx
• CommentTimeNov 3rd 2010

How better this would be if this discussion was on the page of the question, not here...
• CommentAuthorAnixx
• CommentTimeNov 3rd 2010

@ David Speyer

For e^z-1 this formula works, at least for negative x, see my answer there: http://mathoverflow.net/questions/4347/ffxexpx-1-and-other-functions-just-in-the-middle-between-linear-and-expo/42592#42592
• CommentAuthorAnixx
• CommentTimeNov 3rd 2010

The series diverge for Cos(x) die to Runge pohenomenon http://en.wikipedia.org/wiki/Runge's_phenomenon
13.

I'm going to regret this, but I've reposted your answer, because I don't think it should have been deleted, merely criticized.

Now, can I make some points?

(1) Note that my rewrite actually addresses the relevant context, and states which things are theorems and which ones only sometimes work. It provides a useful link to Wikipedia, where the reader can start finding references to the literature. This wasn't that hard for me, all I had to do was to google Newton Series and skim a few screens of text. It would have been much easier for you.

You definitely seem to know a lot of mathematics. But you also need to learn how to present it so it is useful to other mathematicians.

(2) It is rude to rely on others to do this sort of work. Now, it is not as rude on the web as it is in print. Here, once one person does this effort, everyone can benefit from it. But it still should be the obligation of the original author to clean up their work as much as they can, rather than relying on their readers to do so.

I have a stack of exams to grade today. When I do that, I will go through every problem as carefully as I can, try to work out what is meant by ambiguous sentences, and do my best to suggest more precise phrasings. This is important work, and I try to do it well. When you're not paying me, I shouldn't have to do that. After I grade that stack of exams, I have a paper to edit. I will be going through that paper trying to make every point as precise as possible, so that no one else has to guess what I meant.

(3) In light of the stack of the exams, and the paper, I'm bowing out of this thread.

14.
Quote from Anixx, several messages back:

"And I suggest you let the asker to decide if the answer was useful.

Besides this, this is a very general formula, and proving convergence for all cases may be impossible. Anyway this proof does not belong here, it is better targeted in a general discussion of convergence of Newton's series. Only one thing is worth mention here, that the formula converges when Newton series of superfunction converges. "

Anixx, MO is a site where the design is so that the community decides how valuable a contribution is. The primary role the question asker has is whether or not to accept. If you don't like that you have a fundamental disagreement with the design of the forum. Flagging, up and down votes are available to the entire community and we encourage people to vote their minds within certain acceptable bounds of course.

Getting to your 2nd quoted comment, if you read my original response (above) I did not ask for you to give a statement of the form "this series converges if and only if..." but to at least begin to address the standard issues people have with your posts.

The impression you're giving many people is you tend to supply formulas without any justification, and when people ask you to supply details, such as here or in previous threads, you argue rather than cooperate. This is why people are frustrated with your posts. I think I speak for many people when I say I'd really appreciate your responses if you began to address the typical issues people have with your posts. But formulas accompanied with computer-generated plots where it's not clear if the associated formulas make sense, large numbers of people on this forum will always have trouble with this.
• CommentAuthorWill Jagy
• CommentTimeNov 3rd 2010

My comment at David's re-post:

Well, Anixx, you have a **public relations** opportunity here. Find a proof of convergence for your formula and put it here. You have Gerald Edgar's example. Do not wait for somebody else to do it. Find the proof yourself and *learn something.* If you cannot prove it, say so here. If you can prove it, post an answer with *proof of convergence.* Do not wait for somebody else to do the work. Stop posting wrong formulas or wrong graphs or formulas without proof of convergence. Stop **guessing.** Do it perfectly and prove it. You have our attention already.
• CommentAuthorAnixx
• CommentTimeNov 3rd 2010

• CommentAuthorWill Jagy
• CommentTimeNov 3rd 2010

But no proof. There is a good, very short book about the habit of speaking without really caring whether one speaks the truth,

http://press.princeton.edu/titles/7929.html

I had hoped you were capable of struggling towards a proof and waiting to post anything when that was completely done, a few days with either a proof or a thoughtful discussion of difficulties involved. I weary of this. I have no idea why you keep posting this stuff, but I can't stop you. And I suppose I cannot stop you from complaining when people point out the standards of rigor adhered to by mathematicians.
• CommentAuthorAnixx
• CommentTimeNov 3rd 2010

It is pity that you perceive this exact and general answer as a bullshit. No other answer in that question gives a single general formula for half-iterate.
15.

I was the person who deleted the answer. It was problematic for a few reasons:

1. The first claim was blatantly false: the formula that Anixx wrote does not work for the equation f(f(x)) = cos x, because it doesn't converge.
2. That same formula was already given a couple weeks ago in this answer, where it still fails to be applicable. Anixx had basically established a pattern of writing the same formula in places where it was unhelpful, so I decided it was spam.
3. Anixx made no mention of convergence, when analytic issues like convergence become central whenever you have an infinite sum.

I see that Anixx is still claiming to have provided an "exact and general answer" and a "single general formula" when the formula completely fails to answer the question that was given. In modern mathematics, our norms of discussion dictate that when we provide a formula as a solution, we must present a convincing argument that the formula is well-defined and solves the problem. If no such argument is given, then the formula is not a complete solution. The answers that Anixx gave may have been acceptable 250 years ago, but now we know that an explicit formula can fail badly when we leave its domain of applicability.

I think I was wrong to delete Anixx's roughly identical answer to Kevin Buzzard's question, since that is where the formula belongs (together with a suitable discussion of convergence). I see that David Speyer has reposted a cleaner version.

• CommentAuthorAnixx
• CommentTimeNov 3rd 2010 edited

> The first claim was blatantly false: the formula that Anixx wrote does not work for the equation f(f(x)) = cos x, because it doesn't converge.

It was not claimed that the formula does work for cos(x). The opposite: it was noted that it does not work for this function. But the questioner asked for a method of approaching similar equations.

> That same formula was already given a couple weeks ago in this answer, where it still fails to be applicable.

It is fully applicable there also as it provides the solution to the question.

> Anixx had basically established a pattern of writing the same formula in places where it was unhelpful,

The fact that this formula can be applied for answering different questions does not make it spam.

> formula completely fails to answer the question that was given.

The formula answers the question. Not completely, but for a large class of equations.

> we must present a convincing argument that the formula is well-defined and solves the problem.

I think the convergence analysis cannot be made for a general case. It should be performed case-by-case. I even do not know if there a strict criterion when Runge phenomenon emerges. This does not make the Newton Series "unacceptable". If there is no general criterion of convergence, just accept that the formula works when converges.

> I think I was wrong to delete Anixx's roughly identical answer to Kevin Buzzard's question, since that is where the formula belongs

In that question there are concrete sub-questions, all of which already answered. The questioner seems does not ask for a general solution.
16.
Anixx, you're making assertions but not backing them up with anything. You need to supply a proof of an appropriate theorem if you want to convince people "The formula answers the question. Not completely, but for a large class of equations."
• CommentAuthorEmerton
• CommentTimeNov 3rd 2010

I agree with David Speyer's decision to repost Annix's answer, and I appreciate his additions to the answer, which help to explain it. I don't think that these answers should be deleted simply because they don't demonstrate convergence. Annix's answers have certainly taught me about a method and some phenomena of which I was previously unaware.

If people don't want to engage with Annix, they are welcome not to. But as far as I am aware, he is not spamming the site; rather, he seems to be posting his answers to questions where they are relevant. (In particular, his repost in the cosine question addresses the part of the original question asking about general methods.)

• CommentAuthorRon Maimon
• CommentTimeJul 29th 2011

This whole episode is somewhat shameful for this site. Anixx's formula is obviously formally correct, but it is very difficult to make precise, because it is using an operator on maps. To make it rigorously correct, you can work inside a Witt algebra of infinitesimal conformal maps from C to C, but this obviously just a germ of the intended domain. Simeon Hellerman did something similar years ago, in private. These types of formal relations are miracles of nature, and to deny their existence because they are hard to justify in full generality is the worst sort of mathematical censorship.
17.

What's shameful about having an minor over-reaction by a moderator corrected?

A common and difficult moderating situation is when a user does a series of several problematic things and the moderator reacts to the whole situation, while people coming along only see part of it. Annix has a history of making genuinely false statements (this is a separate issue than using and clearly stating when certain steps are difficult to justify rigorously), which is why his posts attracted reasonable moderator attention.

• CommentAuthorRon Maimon
• CommentTimeJul 30th 2011

What is shameful is that the discussion is not about the mathematics, it is about the person who came up with it. If Ghenghis Khan signed up to this webside and came up with an interesting original idea, that idea should be evaluated based on it's mathematical content, even if he gave 20 false proofs of the twin prime conjecture first.

Anixx sometimes makes mistakes. So does every other human being. What makes even his answers interesting is that he is bold and original. This means that his ideas do not have the support of a big mainstream orthodoxy, and this makes them fruitful, even if they are not always 100% precise on the first go.

The orthodoxy protection system is very strong here, and this is shameful. The whole topic of fractional iteration is filled with many ridiculously up-voted no-go theorems which are totally idiotic, but they reinforce the orthodoxy that this is something that one should not study. For example, someone proved that there is no entire function f such that f(f(x))=exp(x). No duh, since f(x)=log(f(exp(x))), and log has a branch cut (this is the essence of his proof). But this answer is up-voted 28 times (and accepted), when it is completely off the mark. The same argument would give you that exp doesn't have an analytic inverse.

The functional square root of exp(x) is perfectly well defined (but poorly studied, and without an orthodoxy to defend it), and if I had to guess about its analytic properties, it requires the same branch cutting as log does and no more.

Further, minor problems, like the difficulty with decreasing fixed points and with cuts hitting quadratic fixed points are blown out of proportion. Please try to have a little more respect for original ideas, especially less studied ones.
18.
@Ron, relax. This forum isn't perfect and no amount of analogies to Genghis Khan is going to correct for that. I'd like to suggest turning down the hyperbole and trying to be more productive. This thread has been dead for a long time and it's not clear what purpose it serves in dredging it back up.
19.

It is crucial to having a functioning website that we keep our signal to noise ratio high. If we needed to wade through 20 false proofs of the twin prime conjecture every day MO would be useless.