User alexandre eremenko - MathOverflow

Answer by Alexandre Eremenko for How I can solve this functional equation

2013-06-17T05:29:37Z

When $a=0$ there are infinitely many solutions, when $a\neq 0$ there are no solutions.

Proof. Consider the function $h(z)=P(z-1)\exp g(z-1)$. Then your equation implies that $h(z)-h(-z)=a.$ And in the opposite direction: if $h$ is any entire function, with finitely many zeros, and $h(z)-h(-z)=a$, then $h(z)=P(z-1)\exp g(z-1)$ where $P$ and $g$ satisfy your equation.

Now when $a=0$ any even function $h$ with finitely many zeros gives a solution. If $a\neq 0$, there are no solutions $h$, even analytic at $0$. Because $h(z)-h(-z)$ is odd, and constant $a$ is an even function.

Answer by Alexandre Eremenko for Calculus book in the spirit of the 18th century

2013-06-14T07:01:06Z

Here is an outstanding modern example: http://www.pdmi.ras.ru/~olegviro/Shchepin/index.html On Euler's footsteps, by Evgeny Shchepin.

On a more advanced level, MR1656255 Stalker, John Complex analysis, Fundamentals of the classical theory of functions. Birkhäuser Boston, Inc., Boston, MA, 1998. It is an interesting, unusual book, though I strongly disagree with many statements in it.

Answer by Alexandre Eremenko for What are some examples of mathematicians who had an unconventional education?

2013-06-12T18:37:06Z

I read 14 previous answers and did not find the most evident example: Ramanujan:-)

Answer by Alexandre Eremenko for Where to look for corrections of papers?

2013-06-04T20:24:29Z

If the author finds a mistake (or someone finds and tells to the author) s/he usually publishes a correction in the same journal. If the journal is reviewed by Mathscinet or Zbl, they usually review the correction as well, so you can find it. Papers posted on arxiv are usually corrected, before or after publication. Be sure that you read the latest posted version. Another place to look for corrections is the author's personal web page.

Some authors have Collected Works published. This is another place to look for corrections.

Answer by Alexandre Eremenko for How to find a topic to do research with as a Post-Doc?

2013-06-04T20:10:10Z

You should read literature in the area of your previous research, in search of interesting unsolved problems. Also attend conferences and talk to specialists in your field. When I was on this stage of my career, I found surveys with lists of unsolved problems in my field, and tried to solve them. Now I make such lists myself, to help young researchers, and I suppose they exist in every field of mathematics. You have to read a lot to become an expert in your area.

Answer by Alexandre Eremenko for Boundedness of Solutions to $\Delta u = f u$ on $R^2$

2013-05-29T21:21:26Z

The answer to the first question is yes. If $u$ is positive and $f$ non-negative then the RHS is non-negative, thus $u$ is subharmonic. Subharmonic function bounded from above must be constant ("Liouville's theorem" for subharmonic functions). If it is constant then the LHS iz $0$, then RHS is $0$ and this is a contradiction.

Answer by Alexandre Eremenko for Triangle area on surfaces of constant curvature

2013-05-12T12:29:16Z

M. Berger, Geometrie, vol. V. MR0536874

Edit. Let me sketch a proof for the spherical triangle. Let the sphere have area $4\pi$. First you derive the area of diangle. It is $2\alpha$, where $\alpha$ is the angle, by completely elementary reasons. Now consider a triangle. Extend its sides to three full great circles. These three circles make several diangles and two equal triangles (the second one is centrally symmetric to the original one). Make a picture showing how these three circles partition the sphere. As the areas of all diangles are known the area of a triangle is simply derived by the exclusion-inclusion formula!

Notice: this proof is truly elementary in the sense that it only uses the existence of the area for a diangle and triangle, its invariance with respect to rotations, and finite additivity. Euclid COULD give a rigorous proof of this. As rigorous as his investigation of areas of Euclidean triangles.

Answer by Alexandre Eremenko for Jordan curve theorem: Can every point on the curve be reached from each region?

2013-05-10T18:51:07Z

Yes. One way to prove this (and the Jordan theorem too) is to use Complex Variables:-) A good reference is Milnor, MR2193309 Dynamics in one complex variable. Third edition. Annals of Mathematics Studies, 160. Princeton University Press, Princeton, NJ, 2006.

A chapter in this book contans the best exposition of these questions that I know.

Answer by Alexandre Eremenko for Closed form solution to an iterative equation.

2013-05-06T14:53:46Z

There is no closed form except for the cases $a=0,1$. But you can find the asymptotic behavior. See, for example Fatou, Sur les equations fonctionnelles, Bull Soc. Math. France, 47 (1919), section 8 and further. Available here:

http://archive.numdam.org/ARCHIVE/BSMF/ BSMF_1919_47/BSMF_1919_47_161_0/BSMF_1919_47_161_0.pdf

Answer by Alexandre Eremenko for Classic applications of Baire category theorem

2013-05-04T19:14:18Z

Yes, there are many applications besides proving the existence of objects with pathological properties. I recommend the nice book MR0584443 Oxtoby, John C. Measure and category. A survey of the analogies between topological and measure spaces.

Answer by Alexandre Eremenko for Asymptotic series

2013-05-02T12:54:31Z

There are many modern books, for example,

MR1317343 Balser, Werner From divergent power series to analytic functions. Theory and application of multisummable power series. Lecture Notes in Mathematics, 1582. Springer-Verlag, Berlin, 1994.

MR1250603 Candelpergher, B.; Nosmas, J.-C.; Pham, F. Approche de la résurgence. Actualités Mathématiques. Hermann, Paris, 1993.

The very basic idea is the following: You frequently obtain divergent series, a) as formal solutions of differential (or functional) equations, b) as perturbation series when you vary a linear operator.

The question is whether these series have any relation to actual solutions of the problem. It often turns out that they are asymptotic series, and moreover, that they are "Borel summable". Borel summation is a procedure using a form of Laplace transform that under certain conditions recovers the function from its formal asymptotic series.

Univalent functions with non-negative coefficients

2013-04-29T18:19:53Z

Is anything non-trivial known about univalent functions with non-negative coefficients?

Let $U$ be the unit disc, and $f$ a univalent (=injective) holomorphic function, $f(0)=0$, $f'(0)>0$. There is a one-to-one correspondence between such functions and simply connected regions $D$ in the plane, containing $0$. The function is real iff $D$ is symmetric with respect to the real line.

What can be said about $D$ if it is known that coefficients are non-negative? (Except the trivial fact that the boundary point of the largest modulus is on the positive ray).
What can be said about non-negative coefficients of a power series if it is known that it represents a univalent function?
Is it possible to parametrize this class?

I am especially interested in the case when the positive ray is contained in $D$, that is $f(1)=\infty$.

Answer by Alexandre Eremenko for Good book on Calculus of Variations

2013-04-29T17:57:01Z

A famous (and remarkable) text is by L C Young, lectures on the calculus of variations and optimal control theory, MR0259704.

Answer by Alexandre Eremenko for Relation of degree and genus of superelliptic curves

2013-04-24T19:57:59Z

$2g=da+ma_1-2a-d-m+2$, where $a=a_1m,$ and $m$ is the g.c.d ($d,a$).

Edit. Let $\chi(S)=2-2g$ be the Euler characteristic. Hurwitz formula gives $$\chi(S)=2a-r,$$ where $r$ is the ramification: a branch point of order $k$ contributes $k-1$ to $r$. As your polynomial has $d$ simple roots, these roots contribute $(a-1)d$ to the ramification. To find ramification at infinity we write our equation $w^a=P_d(z)$ as $w^a=z^dh(z)=u^d$, where $h$ is a holomorphic function at $\infty$, $h(\infty)\neq 0$, and $u$ a germ of a meromorphic function with a simple pole at $\infty$. The last equation factors into irreducible factors: $$w^a-u^d=\prod_c (w^{a_1}-cu^{d_1}),$$ where $c$ are the roots of unity of degree $m$, $m$ is the greatest common factor of $a$ and $d$, $d=md_1,$ and $a=ma_1$.

From this we conclude that our Riemann surface $S$ has $m$ ramification points of order $a_1$ at $\infty$. So we obtain $$2-2g=2a-(a-1)d-m(a_1-1).$$

Rational functions with a common iterate

2012-12-01T20:59:21Z

Let $f$ and $g$ be two rational functions. To avoid trivialities, we suppose that their degrees are at least $2$. We say that they have a common iterate if $f^m=g^n$ for some positive integers $m,n$, where $f^m$ stands for the $m$-th iterate.

1. Can one describe/classify all such pairs?

This is probably very hard, and perhaps there exists no simple answer. But here is a simpler question:

2. Is there an algorithm which finds out whether two rational functions have a common iterate or not ?

I mean, I give you two rational functions, say with integer coefficients, and you tell me whether they have a common iterate or not. Perhaps using a super-computer...

Motivation. J. F. Ritt, (Permutable rational functions. Trans. Amer. Math. Soc. 25 (1923), no. 3, 399-448) gave a complete classification/description of all commuting pairs of rational functions (that is $f(g)=g(f)$)... except when they have a common iterate. I gave a completely different proof of Ritt's theorem, but again it does not apply to the case when $f$ and $g$ have a common iterate (MR1027462).

Polynomial pairs (commuting, or with a common iterate) are completely described in MR1501149 Ritt, J. F. On the iteration of rational functions. Trans. Amer. Math. Soc. 21 (1920), no. 3, 348-356, in the very end of this paper.

What is the exact relation between permutable pairs and pairs with a common iterate ?

3. If two functions have a common iterate, must they commute?

Or perhaps they must, but with explicitly listed exceptions? A positive answer to this will solve problem 2 above. See also my "answer" to http://mathoverflow.net/questions/48818 for an additional motivation.

EDIT. And one more question:

4. Can one describe commuting functions that have a common iterate?

This would complete Ritt's description of commuting functions.

Answer by Alexandre Eremenko for Monodromy of "complex Schwarz-Christoffel maps

2013-04-19T18:15:35Z

Monodromy is affine in the general case. It is generated by finitely many elliptic transformations of the form $az+b$. Proof: your function satisfies the differential equation $u^{\prime\prime}=Ru',$ where $R$ is rational. The general solution of this equation is obtained from a particular solution by the formula $au+b$.

Computing this monodromy explicitly is as difficult as in the classical case: you need to compute the integrals of your $u$ from $x_j$ to $x_k$.

Answer by Alexandre Eremenko for Great mathematics books by pre-modern authors

2013-04-19T13:41:50Z

Here is an incomplete list of pre-1900 books that I read, enjoyed and strongly recommend (I apologize for some repetitions):

Collected works of Archimedes.
Ptolemy, Almagest (yes, this is a math book:-)
Kepler, Stereometry of wine barrels.
Newton's Principia,
Complete works of Abel and Riemann, Laguerre and Stieltjes.
Gauss, General investigation of curved surfaces (available in English)
Fourier, Analytic theory of heat.
Fourier, Analyse des equations determinees (this is a rare book. Available on my web page).
Complete works of Chebyshev (available in Russian and French)
Maxwell, Treatease on Electricity and Magnetism. (There is a nice paper of F. Dyson, Missed opportunities, where he explains how much Mathematics would gain if mathematicians read this book. I completely agree with Dyson).
Painleve, Lecons, sur la theorie analytique des equations differentielles, professees a Stockholm, 1897.
Picard and Poincare, of course...

BTW, I disagree with designation "pre-modern" for the period before 1900. From my point of view, "modern period" begins with Abel. There is no substantial difference between Laguerre or Stiletjes and XX mathematics.

Answer by Alexandre Eremenko for A question from complex analysis

2013-04-18T17:53:52Z

The answer is no. For example if $n=1$, there is no such number. Or take any $n$, any $\alpha_j$ and multiply your first sum on $s^\beta$ where $\beta$ is very small positive. Then the second sum will be with $\beta_j=\alpha_j+\beta$, and the sums will have common zeros.

Answer by Alexandre Eremenko for Representation of all pass transfer functions/inner functions as Blaschke product.

2013-04-18T12:17:55Z

Atkinson, Discrete and continuous boundary problems, page 8.

Answer by Alexandre Eremenko for Weierstrass factorization with $L^2$ estimates?

2013-04-17T19:54:27Z

You have to specify what you mean by $L^2$. Is this $L^2$ with respect to Lebesgue measure (area) in $D$? Whatever you mean by $L^2$, the answer is "no". The reason is Jensen's formula. It says that a function which has too many zeros must grow fast.

If you want to solve it in weighted $L^2$ space, then your weight must be related to the growth rate of the set $X$. If instead you want to fix the weight in $L^2$, the conditions on $X$ will come from the Jensen formula. If you are interested in $L^2$ without weight, look in the books about Bergman space. There you can find the exact conditions on $X$.

Answer by Alexandre Eremenko for textbooks on asymptotic expansions

2013-04-16T21:29:03Z

There is a very large literature on asymptotic expansions, including books. What is the best books depends on your needs. A comprehensive (advanced) book oriented at physicists and applied mathematicians is

MR0499926 Dingle, R. B. Asymptotic expansions: their derivation and interpretation. Academic Press [A subsidiary of Harcourt Brace Jovanovich, Publishers], London-New York, 1973.

Another good book for physicists/engineers is

MR1721985 Bender, Carl M.; Orszag, Steven A. Advanced mathematical methods for scientists and engineers. I. Asymptotic methods and perturbation theory. Springer-Verlag, New York, 1999.

Books more oriented at pure mathematicians is

MR0435697 Olver, F. W. J. Asymptotics and special functions. Computer Science and Applied Mathematics. Academic Press [A subsidiary of Harcourt Brace Jovanovich, Publishers], New York-London, 1974.

and (an older book)

MR0115035 Ford, Walter B. Studies on divergent series and summability & The asymptotic developments of functions defined by Maclaurin series. Chelsea Publishing Co., New York 1960 x+342

Answer by Alexandre Eremenko for Newton integration without integration

2013-04-16T21:04:40Z

What does this procedure have to do with Newton? Just curious.
If you approximate with "polygonal" (=piecewise linear) functions in the most natural way, that is take points $(x_k,f(x_k))$ and connect them with straight line segments, what you obtain is the "trapezoid rule" for approximate evaluation of integrals.
Of course, the trapezoid rule is "a theorem of Adam"; already Gauss knew much more sophisticated rules.
Convergence of the procedure with any continuous function and any (reasonable) choice of $x_k$ is a trivial exercise for modern students, and it was probably in 1905. I looked at the paper, and it seems to me that in it, Lebesgue only proposes a simple way to TEACH the integral. There is nothing really new in this paper.

Answer by Alexandre Eremenko for Solution of an infinite differential system

2013-04-15T05:09:08Z

This is not true. Take $r=2$, $f(z)=e^z$, $P_0=R'-R^{\prime\prime}$, $P_1=R^{\prime\prime}-R$, $P_2=R-R'$.

Answer by Alexandre Eremenko for Mean value theorem for harmonic functions on ellipsoid

2013-04-11T19:22:02Z

Let me expand Aaron's answer: there is a mean value theorem with any centrally symmetric surface. You integrate your harmonic function on the surface against the harmonic measure at the center, and you recover the value of your function at the center. You can also generalize this to non centrally symmetric surfaces, but the statement becomes a bit longer.

Harmonic measure on a surface can be defined by this property, and the fact is that it exists for all reasonable surfaces. You can generalize even further, and dispose of the surface:-) Just consider measures such that convolution with a harmonic function reproduces this harmonic function. (They are called Jensen measures if I remember correctly).

EDIT: I remembered incorrectly: Jensen's measure at $x$ is a measure such that $$u(x)\geq\int ud\mu$$ for all superharmonic functions. The measures I was writing about apparently have no name.

Answer by Alexandre Eremenko for Is there a deep reason for the fecundity of involutions?

2013-04-04T22:38:17Z

I don't know the true philosophical reason, but $Z_2$ symmetry is really omnipresent in Matematics and in the nature. For example, most animals, including practically all vertebrate animals (like ourselves) have aproximately $Z_2$ symmetric bodies, and no larger group. This suggests that $Z_2$ was the favorite group of the Creator, at least in that period of his activity when we was creating advanced animals:-)

If you prefer Evolution, this $Z_2$ symmetry must somehow be explained by the survival of the fittest. I don't know exactly how, but this suggests that this is a very important group. Notice that plants, mushrooms, and simplest animals usually do not have it.

As a result of this (2-fold symmetry of animal bodies) we tend to like this kind of symmetry. Look at all our technology: cars, ships, airplanes, etc. They all have 2-fold symmetry, at least from outside (like our bodies, they also have this symmetry only outside). Once my friend, an airspace engineer, told me that there was a project of an airplane which did not have this outside 2-fold symmetry. The project was rejected for the only reason that "no one will want to fly in such an airplane". I am serious: http://en.wikipedia.org/wiki/Oblique_wing

In mathematics, from my personal perspective, it is $z\mapsto\overline{z}$ first of all. (Once I even proposed to my co-author to call one of our papers "Some applications of representation theory of $Z_2$"); the paper was full of different representations of this group, We were working on real algebraic geometry.)

This very same symmetry $z\mapsto\overline{z}$ is also hidden in Hermitian symmetry, $C^*$ algebras, all sorts of "duality" everywhere, etc. Which suggests that the Creator of the Universe always had a strong bias in favor of this particular group.

Answer by Alexandre Eremenko for A graduate course on Sturm Liouville theory?

2013-04-11T19:08:30Z

A more advanced/comprehensive course can be based on Atkinson's book Discrete and continuous boundary problems. No functional analysis is required, neither for Hilbert-Courant nor for Atkinson. (When Courant wrote the first volume of HC, functional analysis did not exist yet:-)

Another book which studies some of these questions in depth (and without any functional analysis) is Gantmakher-Krein, Oscillation matrices and kernels. (Gantmaher, Gantmacher...)

Is it worth offering such a course? The answer depends on whether you can find enough students who will enroll. This depends on your local conditions. Of course this is a beautiful and useful subject and worth learning.

Answer by Alexandre Eremenko for complex dynamics in several variables

2013-04-11T19:00:05Z

There is a book by Erik Fornaess and Nessim Sibony, MR1363948, survey papers of the same authors, MR1810536, MR1748606, MR1285389 and on various specific questions I also recommend papers of Misha Lyubich with various co-authors, and Eric Bedford and John Smillie, especially "Polynomial diffeomorphisms" in 8 parts.

Answer by Alexandre Eremenko for Strong notions of general position

2013-04-11T12:11:21Z

If there is a family of objects parametrized by points $a$ in some topological space $X$ they usually say that "an object is in general position" (or "generic") to mean that $a$ belongs to an open dense subset of $X$. This notion of course depends on topology on $X$. When $X$ is an algebraic variety, say over $R$ or $C$, one can consider two natural topologies: the usual one and Zariski one. I think this applies to all examples given in the question and in the answer of Sandor.

Answer by Alexandre Eremenko for Is it possible to get another math PhD

2013-04-10T12:25:36Z

I suppose that if you want a second PhD in Math you have to hide the fact that you already have one. But I do not recommend you to do this. What will you do with your existing publications ? In my university at least, it is a policy not to accept to the PhD program those who already have a PhD. (I don't know whether this policy is written or unwritten. And how strict it is. Probably they do not ask explicitly. And I know people who actually had a PhD (in another country) and who were accepted to our PhD program. They did not mention in their CV that they had a PhD.
It is probably true that most people employed by the "second-tier" math departments are from "top" universities. But it is not true that the hiring committees do not consider other applications. I know many people in "top" universities who have PhD from little-known places.

Answer by Alexandre Eremenko for It this set a Riesz Basis of $L^2(0,\pi)$

2013-04-04T23:48:21Z

I suppose that you have an extra $\pi$ somewhere: {$\sin \pi nx$} is a Riesz basis on $(0,1)$, not $(0,\pi)$.
After you correct your question, this is not a Riesz basis when $a>0$ is sufficiently large. This follows from characterization of Riesz bases by Pavlov, Sov. Math. Dokl. 20:4, 1979, 655–659, or Semmler, Ann. Acad. Sci. Fenn. Math. 35 (2010), no. 1, 23–46.

Comment by Alexandre Eremenko

2013-06-12T18:32:10Z

Learning some subject from Bourbaki is an interesting idea:-) Some later editions publish the list of misprints/errors in the previous editions (at least this is the case in the Russian translations). When you accumulate a long list, I suggest that you send it to the publisher, if he is still alive. (The author died long ago, I suppose; there was even a rumor of the official funeral).

Comment by Alexandre Eremenko

2013-06-09T06:43:42Z

Alex, You are right. I will not do this anymore.

Comment by Alexandre Eremenko

2013-06-05T19:56:24Z

I also do. I also replace the arxiv version if I find any serious misprint, even after the paper is published. And every time I asked a journal to publish a correction, it did.

Comment by Alexandre Eremenko

2013-05-16T20:41:05Z

Anton, sorry I looked at 119953 and I don't understand your objection. In elementary geometry we deal with areas of polygons. The area is defined by a) finite additivity, b) monotonicity, invariance with respect to motion, c) the area of the unit square is 1. From this it is easy to derive that the area of a polygon exists and is unique. And I believe Euclid did it rigorously. Kiselev (who wrote the common Russian high school geometry text) did it rigorously, and I studied this in 8-th grade. What's wrong with all this?

Comment by Alexandre Eremenko

2013-05-14T21:21:37Z

This is by Abel's theorem.

Comment by Alexandre Eremenko

2013-05-14T21:10:34Z

Sorry, I was using Russian edition, where this is called Chapter V. Now I checked the original, and in the original it is VOLUME V. And unfortnately I did not find an English translation:-(

Comment by Alexandre Eremenko

2013-05-14T20:47:30Z

Anton: I disagree with what you say. The area of a TRIANGLE is an elementary notion. (The theory of areas of triangles in Euclid is completely rigorous, by all modern standards.) And the formula has a really elementary proof.

Comment by Alexandre Eremenko

2013-04-26T14:17:27Z

I will explain if you vote up my answer:-)

Comment by Alexandre Eremenko

2013-04-20T13:27:15Z

And in general, Stirling formula (asymptotic expansion) holds as $|z|\to\infty$ uniformly with respect to $\arg z$ in every angle of the form $|\arg z|<\pi-\epsilon,\; \epsilon>0$.

Comment by Alexandre Eremenko

2013-04-19T13:15:29Z

This is a reasonable question, especially now, when it is corrected. Please don't close it.

Comment by Alexandre Eremenko

2013-04-19T01:25:26Z

Of course one can easily modify the statement to eliminate these simple counterexamples but I leave this to the author.

Comment by Alexandre Eremenko

2013-04-18T17:43:29Z

Medvedev is an historian, not a real mathematician, so it is possible that he confused something. The book on the history of integral that I have (by I. Pesin, who is a mathematician) does not mention this paper of Lebesgue. But of course I can say nothing definite without seeing Medvedev's book.

Comment by Alexandre Eremenko

2013-04-18T17:39:07Z

Interesting. Unfortunately I do not have Medvedev's book, and on the Internet I found it for $229, and my interest 8is not sufficient to pay this amount to satisfy it:-(

Comment by Alexandre Eremenko

2013-04-18T03:59:05Z

On my opinion, this is a legitimate and important question. These discussions are common, and sometimes even happen on the pages of BAMS. I propose to reopen.

Comment by Alexandre Eremenko

2013-04-18T03:53:24Z

What about Medvedev? Does he say anything about this paper of Lebesgue?