User wadim zudilin - MathOverflow

Another Chicken or Egg: Sequence or Series

2012-08-26T10:16:24Z

This is a side question which is more motivated by teaching than research.

First, I am trying to convince myself that sequences appear before series (as numerical approximations to "interesting" quantities; on the other hand, decimal expansions -- especially infinite -- are more likely to be series).

Secondly, is it natural for sequences to be placed prior to series in a calculus course?

So, which one is more original, a sequence or a series?

After-dinner edit. We define a sequence to be... a function mapping the positive numbers to a set(?). We define a series to be... a formal infinite sum $\sum_{n=1}^\infty a_n$(?). Tell me what is your way to "define" these two guys, I do not believe they are very related.

There are no doubts that it is easier to define convergence of series via convergence of sequences, but it does not imply their "primogeniture". The notion of Cauchy sequence is an elegant way to build the apparatus of not only sequences but also of real numbers; as such it can serve as a definition of series: a series is a formal infinite sum $\sum_{n=1}^\infty a_n$, and it is called a convergent series if for any $\epsilon>0$ there exists an $N=N(\epsilon)$ such that for any $m>n>N$ the sum $|a_n+\dots+a_m|<\epsilon$. The real numbers then are nothing but representatives of equivalence classes of convergent series. (I have no desire here to expand all the details.) A sequence $b_n$ is convergent when the corresponding series $\sum_{n=1}^\infty a_n$ where $a_1=b_1$ and $a_n=b_n-b_{n-1}$ for $n\ge2$ converges. It would be honest to say that, besides the trivialities like "algebra of limits", the techniques for investigating convergence of series are quite independent from that of sequences. And it does not sound impossible to do series prior to sequences.

Historically, all these convergence/divergence issues were purely intuitive for both sequences and series, and they both were on the market for many centuries. I ask whether their exists an overwhelming historical support to the notion of sequence to lead.

Demystifying complex numbers

2010-07-01T09:17:03Z

At the end of this month I start teaching complex analysis to 2nd year undergraduates, mostly from engineering but some from science and maths. The main applications for them in future studies are contour integrals and Laplace transform, but of course this should be a "real" complex analysis course which I could later refer to in honours courses. I am now confident (after this discussion, especially after Gauss complaints given in Keith's comment) that the name "complex" is quite discouraging to average students.

Why do we need to study numbers which do not belong to the real world?

Of course, we all know that the thesis is wrong and I have in mind some examples where the use of complex variable functions simplify solving considerably (I give two below). The drawback of all them is assuming already some knowledge from students.

So I would be really happy to learn elementary examples which may convince students in usefulness of complex numbers and functions in complex variable. As this question runs in the community wiki mode, I would be glad to see one example per answer.

Thank you in advance!

Here comes the two promised example. The 2nd one was reminded by several answers and comments about relations with trigonometric functions (but also by notification "The bounty on your question Trigonometry related to Rogers--Ramanujan identities expires within three days"; it seems to be harder than I expect).

Example 1. Find the Fourier expansion of the (unbounded) periodic function $$ f(x)=\ln\Bigl|\sin\frac x2\Bigr|. $$

Solution. The function $f(x)$ is periodic with period $2\pi$ and has poles at the points $2\pi k$, $k\in\mathbb Z$.

Consider the function on the interval $x\in[\varepsilon,2\pi-\varepsilon]$. The series $$ \sum_{n=1}^\infty\frac{z^n}n, \qquad z=e^{ix}, $$ converges for all values $x$ from the interval. Since $$ \Bigl|\sin\frac x2\Bigr|=\sqrt{\frac{1-\cos x}2} $$ and $\operatorname{Re}\ln w=\ln|w|$, where we choose $w=\frac12(1-z)$, we deduce that $$ \operatorname{Re}\Bigl(\ln\frac{1-z}2\Bigr)=\ln\sqrt{\frac{1-\cos x}2} =\ln\Bigl|\sin\frac x2\Bigr|. $$ Thus, $$ \ln\Bigl|\sin\frac x2\Bigr| =-\ln2-\operatorname{Re}\sum_{n=1}^\infty\frac{z^n}n =-\ln2-\sum_{n=1}^\infty\frac{\cos nx}n. $$ As $\varepsilon>0$ can be taken arbitrarily small, the result remains valid for all $x\ne2\pi k$.

Example 2. Let $p$ be an odd prime number. For an integer $a$ relatively prime to $p$, the Legendre symbol $\bigl(\frac ap\bigr)$ is $+1$ or $-1$ depending on whether the congruence $x^2\equiv a\pmod{p}$ is solvable or not. One of elementary consequences of (elementary) Fermat's little theorem is $$ \biggl(\frac ap\biggr)\equiv a^{(p-1)/2}\pmod p. \qquad\qquad\qquad {(*)} $$ Show that $$ \biggl(\frac2p\biggr)=(-1)^{(p^2-1)/8}. $$

Solution. In the ring $\mathbb Z+\mathbb Zi=\Bbb Z[i]$, the binomial formula implies $$ (1+i)^p\equiv1+i^p\pmod p. $$ On the other hand, $$ (1+i)^p =\bigl(\sqrt2e^{\pi i/4}\bigr)^p =2^{p/2}\biggl(\cos\frac{\pi p}4+i\sin\frac{\pi p}4\biggr) $$ and $$ 1+i^p =1+(e^{\pi i/2})^p =1+\cos\frac{\pi p}2+i\sin\frac{\pi p}2 =1+i\sin\frac{\pi p}2. $$ Comparing the real parts implies that $$ 2^{p/2}\cos\frac{\pi p}4\equiv1\pmod p, $$ hence from $\sqrt2\cos(\pi p/4)\in{\pm1}$ we conclude that $$ 2^{(p-1)/2}\equiv\sqrt2\cos\frac{\pi p}4\pmod p. $$ It remains to apply ($*$): $$ \biggl(\frac2p\biggr) \equiv2^{(p-1)/2} \equiv\sqrt2\cos\frac{\pi p}4 =\begin{cases} 1 & \text{if } p\equiv\pm1\pmod8, \cr -1 & \text{if } p\equiv\pm3\pmod8, \end{cases} $$ which is exactly the required formula.

When is $n/\ln(n)$ close to an integer?

2010-06-14T03:37:37Z

As usual I expect to be critisised for "duplicating" this question. But I do not! As Gjergji immediately notified, that question was from numerology. The one I ask you here (after putting it in my response) is a mathematics question motivated by Kevin's (O'Bryant) comment to the earlier post.

Problem. For any $\epsilon>0$, there exists an $n$ such that $\|n/\log(n)\|<\epsilon$ where $\|\ \cdot\ \|$ denotes the distance to the nearest integer.

In spite of the simple formulation, it is likely that the diophantine problem is open. I wonder whether it follows from some known conjectures (for example, Schanuel's conjecture).

3-7 primes in base 10

2012-12-31T06:35:51Z

After a quick look at the sequence (of primes) A020463, $$ 3, 7, 37, 73, 337, 373, 733, 773, 3373, \dots, $$ the following question is straighforward:

Are there infinitely many primes compiled from digits 3 and 7 in base 10?

As any question about primes, this should be tough. Nevertheless I believe there exists some reasonable heuristics to answer it.

Answer by Wadim Zudilin for What is the theoretical interest of finding closed-form sols. of infinite series?

2012-08-25T11:00:09Z

I can recommend reading "Closed forms: what they are and why we care" by Jon Borwein and Richard Crandall, the article is to appear in Notices Amer. Math. Soc. 60 (2013).

Edit (Dec 2012). The paper has just appeared in the Notices: pdf. Together with the sad news about Richard Crandall: he passed away.

Complex evaluation of a classical (real) integral

2012-08-25T11:33:19Z

There are several ways to compute the classical integral $$ \int_{\mathbb R}e^{-x^2}dx=\sqrt{\pi}. $$ Probably, best known are

(1) squaring the integral with subsequent change of (now two) variables to the polar form, and

(2) the reducing to the Gamma-function at $1/2$.

I am interested though in a "complex" analysis method (namely, a use of the residue theorem) to do the job. The reason is that several integrals like $$ \int_0^\infty e^{-x^2}\cos ax\ dx \qquad\text{or}\qquad \int_0^\infty\sin x^2\ dx $$ can be computed via the residue theorem and the above integral, so I would like to avoid any reference to real analysis. Is there such a complex evaluation though?!

Answer by Wadim Zudilin for Proof of the transcendence of the Champernowne Constant with Thue-Siegel-Roth

2012-08-25T11:50:08Z

Here is an extract from van der Poorten's "Obituary. Kurt Mahler (1903--1988)" (see p.353 in J. Austral. Math. Soc. Ser. A 51 (1991)):

In a more unexpected way, Mahler's arguments led to the following amusing result: Suppose $f$ is a non-constant polynomial taking integer values at the nonnegative integers. Then the concatenated decimal $$ \phi=0.f(1)f(2)f(3)\dots $$ is transcendental. In particular Champernowne's normal number $$ 0.123\dots910111213\dots $$ is transcendental. Mahler's argument relies on the observation that one readily obtains rational approximations to $\phi$ with denominators high powers of the base 10, thus composed of the primes 2 and 5 alone. Perhaps disappointingly, Roth's definitive form of the Thue--Siegel inequalities permits a more immediate argument obviating the need for an appeal to the $p$-adic results.

This is to say that Roth's argument is more superior than Mahler's but it appeared some 20 years later...

Answer by Wadim Zudilin for Wonderful applications of the Vandermonde determinant

2011-01-06T04:21:44Z

There is an elegant (and short!) application of the (generalized) Vandermonde determinant to the famous problem of D. H. Lehmer in the article [D.C. Cantor and E.G. Straus, Acta Arith. 42 (1982/83), no. 1, 97-–100]. I put here a scan of the article together with corrections given by the authors later.

Answer by Wadim Zudilin for Titles composed entirely of math symbols

2011-07-11T13:22:46Z

"Pi" (I keep "A source book" in parentheses to hide the non-mathematical part), L. B. Berggren, J. M. Borwein, P. B. Borwein (Eds.).

"Z=60", Conference in Honor of Doron Zeilberger's 60th Birthday (this, of course, is influenced by one of ma favorite titles "$A=B$").

Removed (following the healthy criticism): "2012", a 2009 American science fiction disaster movie.

Generalizations of the Rayleigh(-Beatty) theorem

2012-01-24T08:36:36Z

For a given irrational number $\alpha>0$ and a real number $\beta$, the inhomogeneous Beatty sequence sequence $S_{\alpha,\beta}$ is the set $\lbrace\lfloor n\alpha+\beta\rfloor:n=1,2,\dots\rbrace$ (the case $\beta=0$ corresponds to a homogeneous Beatty sequence).

If $\beta=0$, the two homogeneous Beatty sequences $S_{\alpha_1,0}$ and $S_{\alpha_2,0}$ partition the set of positive integers iff $1/\alpha_1+1/\alpha_2=1$. There is also a similar result for inhomogeneous $S_{\alpha_1,\beta_1}$ and $S_{\alpha_2,\beta_2}$: assuming that neither $n\alpha_1+\beta_1$ nor $n\alpha_2+\beta_2$ is an integer for $n=1,2,\dots$, the sequences partition $\mathbb Z_{>0}$ iff $1/\alpha_1+1/\alpha_2=1$ and $\beta_1/\alpha_1+\beta_2/\alpha_2=0$.

Question. For a given $k\ge3$, what are the conditions on $\alpha_1,\dots,\alpha_k$ (and on $\beta_1,\dots,\beta_k$ in the inhomogeneous case) to ensure that the sets $S_{\alpha_i,\beta_i}$, $i=1,\dots,k$, partition the positive integers.

It looks like the book Old and new problems and results in combinatorial number theory by P. Erdős and R.L. Graham (which I do not have) mentions a version of the problem, but I am interested in some (possibly very recent) progress in the direction. My interest is motivated by the study of functional equations of the Mahler-type generating functions of the Beatty sequences.

Answer by Wadim Zudilin for Is there a set of criteria to determine whether a number is transcendental for a subset of the reals with positive Lebesgue measure?

2012-01-24T12:15:19Z

The question is indeed not well posed. Here is a nice example from [J.M. Borwein and P.B. Borwein, On the generating function of the integer part: $[n\alpha +\gamma]$, J. Number Theory 43 (1993), no. 3, 293--318], namely, part (b) of Theorem 0.4 there.

Consider the function $F(\alpha):=\sum_{n=1}^\infty\lfloor n\alpha\rfloor/2^n$ and the set of irrational numbers $A\subset\mathbb R$ that have unbounded partial quotients in their partial fraction expansions. (For example, $$ e=[2;1,2,1,1,4,1,1,6,1,1,8,1,\dots]=2+\frac1{1+\dfrac1{2+\dfrac1{\ddots}}} $$ belongs to $A$.) The set $A$ has full measure in $\mathbb R$ as its complement is of measure 0. The values $F(\alpha)$ at $\alpha\in A$ are all transcendental Liouville numbers. (Note the image set has... measure zero.)

A related variant of test for a given $\beta\in\mathbb R$ would be: if there exists $\alpha\in A$ such $\beta=F(\alpha)$, then $\beta$ is transcendental.

Is this a transcendence method for the full-measure set or a zero-measure set? $\ddot\smile$

Answer by Wadim Zudilin for Product of one minus geometric progression

2012-01-20T23:18:51Z

In a standard $q$-notation, the product of your interest is $$ (z;q) _ n:=\prod_{j=0}^{n-1}(1-zq^j). $$ The formula known as the $q$-binomial theorem expresses this product as sum: $$ (z;q) _ n=\sum _{k=0}^n {\genfrac{[}{]}{0pt}{}{m}{n}} _q (-z)^kq^{k(k-1)/2}, $$ where the $q$-binomial coefficients are defined, e.g., in this question.

Answer by Wadim Zudilin for a dilogarithm identity: known or new?

2012-01-18T13:12:45Z

All functional (i.e., depending on a parameter) relations are known to be a consequence of the 5-term relation of the dilogarithmic function; see [D. Zagier, The dilogarithm function, in: Frontiers in Number Theory, Physics and Geometry II, P. Cartier, B. Julia, P. Moussa, P. Vanhove (eds.), Springer-Verlag, Berlin-Heidelberg-New York (2006), 3-65]. In your case, I would suggest to compare first the derivatives w.r.t. $m$ and then use the fact that your identity is true for a particular value of $m$ (for example, $m=0$).

Answer by Wadim Zudilin for Closed form or/and asymptotics of a hypergeometric sum

2012-01-17T02:22:18Z

For your sequence

s:=[0, 1, 7, 39, 198, 955, 4458, 20342, 91276, 404307, 1772610, 7707106, 33278292, 142853854, 610170148, 2594956620, 10994256152, 46425048451, 195456931506, 820725032042, 3438011713540]

I use with(gfun) in Maple and then guessgf(s, x) to find out that the generating function for your sequence is $$ \frac{x(1+\sqrt{1-4x})}{2(1-4x)^2}. $$ This isn't now hard to establish rigorously by verifying that your double sum satisfies a polynomial recurrence (a multivariate version of the Gosper-Zeilberger creative telescoping).

Answer by Wadim Zudilin for Is there a closed formula for the generating function of some trinomial coefficients?

2012-01-17T01:53:35Z

What is "a way"? Of course, your question (in even more general form) was asked centuries ago and gave rise to hypergeometric series, series of the form $\sum_n c_n$ with ratio $c_{n+1}/c_n$ being a rational function of index $n$. The most convenient form is therefore the hypergeometric $_4F_3$ series expression given in Robert's answer. Note that the hypergeometric function is not algebraic (i.e., transcendental) in this case, so that no formula like $(1-4z)^{-1/2}$ can be given; all algebraic hypergeometric instances are now tabulated thanks to a fantastic result of Beukers and Heckman.

Dyson's famous 1962 paper Statistical theory of the energy levels of complex systems originated the study of constant term identities. In particular, Dyson's ex-conjecture states that for $a_1,\dots,a_n$ nonnegative integers $$ \text{constant term} \prod_{1\leq i\neq j\leq n} \biggl(1-\frac{x_i}{x_j}\biggr)^{a_i} =\frac{(a_1+a_2+\cdots+a_n)!}{a_1!a_2!\cdots a_n!}\,. $$ This could serve a different basis for other type generating functions.

Polynomials with presumably positive coefficients

2012-01-17T00:09:53Z

The $q$-Pochhammer symbol $(q) _ 0:=1$ and $(q) _ n:=\prod _ {j=1}^n (1-q^j)$ for $n > 0$ is clearly a polynomial in $q$ which has both positive and negative coefficients when $n>0$. The $q$-binomial coefficients (also known as the Gaussian polynomials) $$ \genfrac{[}{]}{0pt}{}{m}{n} ={\genfrac{[}{]}{0pt}{}{m}{n}} _q :=\frac{(q) _ m}{(q) _ n(q) _ {m-n}} $$ (zero when $n<0$ or $n>m$) are polynomials as well, of degree $n(m-n)$, in fact with all coefficients positive. The latter circumstance is a source of many other positivity (at least nonnegativity) claims like for the 2-parameter polynomial family $$ V_{n,k}(q) :=\sum_{j=0}^nq^{j^2+kj}\genfrac{[}{]}{0pt}{}{n+k-j}{k} \sum_{m=0}^{\min\lbrace j,n+k-j\rbrace}q^{m^2+km} \genfrac{[}{]}{0pt}{}{n+k-j}{m}\genfrac{[}{]}{0pt}{}{n+k}{j-m}. $$ My problem is the expected nonnegativity of another family $$ U_{n,k,t}(q) :=\sum_{j=0}^nq^{j^2+(k+t)j}\genfrac{[}{]}{0pt}{}{n+k-j}{k} \sum_mq^{m^2+(k+t)m}\genfrac{[}{]}{0pt}{}{2n+2k-j-m}{j-m}\genfrac{[}{]}{0pt}{}{n+k-j}{m}\genfrac{[}{]}{0pt}{}{n-j}{m}(q)_m $$ (note the unpleasant appearance of the $q$-Pochhammer symbol at the end) when $t=k$. The parameter $t$ is introduced by purpose, as it gives more flexibility to the polynomial family, namely, $U_{n,k,0}(q)=V_{m,k}(q)$ (a corollary of a known hypergeometric identity) is positive and $$ U_{n,k,t}(q) =q^nU_{n,k,t-1}(q)+(1-q^{k+1})U_{n-1,k+1,t-1}(q), $$ which clearly lacks of positivity because of the factor $1-q^{k+1}$. Experimental verification shows that the polynomials $U_{n,k,t}(q)$ have only nonnegative coefficients for the range $t=0,1,\dots,t_0$ where $t_0\ge2k+1$. Again, even the particular case $t=k>0$ is out of my reach.

The problem is related to the positivity issues from my earlier problem, although it will require a definite text to make the relation clear.

Answer by Wadim Zudilin for Can we always find such an irreducible polynomial of degree n where degree(p(x)-x^n)<= n/2?

2011-11-23T23:41:14Z

I leave my earlier post below where I mistakenly understood the original problem because of a bad illustrative example given there. Here I only indicate the explicit formula for the number $N_ m(p)$ of monic polynomials of exact degree $m$ irreducible over $GF(p)$: $$ N_m(p)=\frac1m\sum_{d\mid m}\mu(d)p^{m/d}. $$ This formula is again from Prasolov's Polynomials, and it seems to be absent in other posts and comments.

I assume that the irreducibility is discussed over $\mathbb Z$, otherwise $x^{400}+x^5+x^2+1$ has zero $x=1$ in $GF(2)$.

Victor Prasolov in Section 8 of his book Polynomials discusses the irreducibility of trinomials and quadrinomials, mostly based on the work [W. Ljunggren, On the irreducibility of certain trinomials and quadrinomials, Math. Scand. 8 (1960), 65--70]. One of the results from there is as follows.

Theorem. Let $n\ge2m$, $d=\text{gcd}(n,m)$, $n_1=n/d$ and $m_1=m/d$. Then the polynomial $$ g(x)=x^n+\epsilon x^m+\epsilon', \quad\text{where}\ \epsilon,\epsilon'\in\lbrace\pm1\rbrace, $$ is irreducible except for the following cases in which $n_1+m_1\equiv0\pmod3$:

(a) $n_1$ and $m_1$ are odd and $\epsilon=1$;

(b) $n_1$ is even and $\epsilon'=1$;

In all three cases (a)--(c), $g(x)$ is a product of a certain irreducible polynomial and $x^{2d}+\epsilon^m{\epsilon'}^nx^d+1$.

Corollary. If $n\not\equiv2\pmod3$, then $x^n+x+1$ is irreducible.

If $n\equiv2\pmod3$, then $x^n+x^2+1$ is irreducible.

In other words, there is an irreducible degree $n$ polynomial $g(x)$ of the wanted form such that $\deg(g(x)-x^n)\le2$, and this bound cannot be further improved.

Answer by Wadim Zudilin for area of triangle from coefficients of its cubic?

2011-11-23T10:41:23Z

There is a clear experimental maths strategy to attack a problem like this. A "generic" polynomial $(z-z_1)(z-z_2)(z-z_3)=z^3-az^2+bz-c$ can be replaced with a very concrete one: Choose $a$, $b$ and $c$ to be (at least presumably) $\mathbb Q$-algebraically independent complex numbers. Compute numerically the zeroes $z_1$, $z_2$, $z_3$ of the polynomial $z^3-az^2+bz-c$ and the corresponding value $V(z_1,z_2,z_3)$. Then we expect $V$ to satisfy an algebraic equation with coefficients from $\mathbb Q[a,b,c]$, and this can be guessed efficiently with either LLL or PSLQ.

I did not try hard after seeing Igor's response and comments, but it seems that there is no algebraic equation of degree $\le6$ for $V$.

Answer by Wadim Zudilin for Question related to the abelianization of simplectic groups

2011-11-23T02:24:47Z

Your question is related to an open problem about finiteness of the index of an "arithmetic" group in its Zariski closure (see, for example, Peter Sarnak's lecture Thin Groups and the Affine Sieve as well as arXiv:math/0605675 and arXiv:0803.3322).

Answer by Wadim Zudilin for How to show modularity of an elliptic curve?

2011-11-23T02:10:09Z

The answer is outlined in Don Zagier's 1985 paper Modular points, modular curves, modular surfaces and modular forms.

Sums of four squares and the modular invariant

2011-11-22T23:40:44Z

The question accounts my curiosity only, and may not be as deep as I think.

One of recent talks at our local seminar was devoted to the proof of the classical formula $$ F(q)=\sum_{n=0}^\infty r_4(n)q^n =\biggl(\sum_{n\in\mathbb Z}q^{n^2}\biggr)^4 =1+8\sum_{n=1}^\infty\frac{q^n}{(1+(-1)^nq^n)^2} $$ for the generating function of the number $r_4(n)$ of representations of $n$ as a sum of four squares of integers. The key ingredient is the fact that the function $F(e^{2\pi i\tau})$ is a modular form of weight 2 and level 4. As an illustrative example, the speaker gave $r_4(100)=744$ which, of course, can be also given in a more "general" form $r_4(2^\ell\cdot25)=744$ for $\ell=1,2,\dots$. This number 744 is not quite random: it appears as the constant term in the modular invariant $$ J(q)=\frac1q+\sum_{m=0}^\infty c_mq^m =\frac1q+744+196884q+21493760q^2+\cdots, $$ a (weak) modular form of weight 0 viewed as a function of $\tau$ where $q=e^{2\pi i\tau}$. The next coefficient $c_1=196884$ is not divisible by 8, hence cannot be given as $r_4(n)$ for some $n$, but then the problem of finding solutions to the equation $c_m=r_4(n)$ becomes messy. (In fact, I cannot find any other with $m\ge2$.) My particular problem is as follows.

Question. Are there (in)finitely many $m$ such that $c_m=r_4(n)$ for some $n$?

Well, it seems to be quite natural to ask even the following.

General question. Suppose $A(q)=\sum_{n\gg-\infty}a_nq^n$ and $B(q)=\sum_{m\gg-\infty}b_mq^m$ are two modular forms of different weight with, say, nonnegative coefficients and $b_{m+1}>b_m$ holding for $m\ge m_0$. Is there a way to decide whether the equation $a_n=b_m$ holds for infinitely many $m\ge m_0$?

Answer by Wadim Zudilin for An example of a beautiful proof that would be accessible at the high school level?

2011-09-08T10:12:55Z

Seeing the struggle of many students with standard trigonometry, I especially like the rational parametrization of $x^2+y^2=1$ (which is equivalent to listing all Pythagorean triples) by starting from $\sin^2\phi+\cos^2\phi=1$ and then using $$ \sin\phi=\frac{2t}{1+t^2}, \quad \cos\phi=\frac{1-t^2}{1+t^2}, \qquad t=\tan\frac{\phi}2. $$ Note that the formulas are usually used in the context of integration of rational expressions in sine and cosine.

At the same time, a more general "geometric" argument (applicable to general quadratics), due to Bachet (1620), is still at school level. Namely, fix a single rational point on the curve, $(x _ 0,y _ 0)$ say, and consider the intersection points of the curve and straight lines $y-y_0=t(x-x_0)$ with rational slope $t$ passing through the point.

A beauty here is because of variety of different geometric and analytic methods for solving a classical arithmetic problem.

Infinite games: are they well defined?

2010-07-16T14:07:39Z

It is just my curiosity about this question where we have an infinite game and (according to the answers) winning strategies for both players. I am familiar with terminating games only, and I am pretty sure that nonterminating ones can cause many paradoxes. As this is a community wiki mode, I would be happy to see formal definitions, existence of wining strategies as well as possible collisions.

Thanks in advance for keeping my mind finite!

Answer by Wadim Zudilin for Most memorable titles

2011-07-11T13:43:00Z

Why Knot? by Colin Adams.

Answer by Wadim Zudilin for Who is L. A. Balashov?

2011-07-04T20:12:55Z

It is fantastic to learn how easy people are forgotten. I made several calls to the Faculty of Mechanics and Mathematics of the Moscow Lomonosov State University (including the Human Resources and the Chair of Function Theory and Functional Analysis where Balashov worked till his sudden death) without success. I have just received a letter from Boris Sergeevich Kashin (I indicate his full name because of the question) who says that Balashov was Leonid Alekseevich (but he is not sure about the patronymic name, so he will write me back later). I will add more if some further information arrives.

Asymptotics of the $q$-harmonic series as $q\to1$

2011-07-04T09:54:26Z

The following (very simply looking!) problem occurs in regularization of the harmonic series which can be formally thought of as the limit as $q\to1$, $|q|<1$, of $$ h(q):=(1-q)\sum_{n=1}^\infty\frac{q^n}{1-q^n}. $$ I can show (with some effort) that $$ h(q)=-\log(1-q)+f(q) \qquad\text{as}\quad q\to1, \ |q|<1, $$ where $f(q)$ is a bounded function (hint: consider both $h(q)$ and $h(q^2)$ as $q\to1$). The question is whether the function $f(q)$ has a limit as $q\to1$ or not; in other words, whether $$ h(q)=-\log(1-q)+c+o(1) \qquad\text{as}\quad q\to1, \ |q|<1. $$ Then, of course, I am very much interested in the constant $c$. A straightforward computer experiment is not helpful.

Answer by Wadim Zudilin for No simple duplication formula for factorials?

2011-07-03T16:19:43Z

Armin,

Let me try to solve your original problem differently. First write the wanted polynomial in the form $f(x,y)=\sum_kx^kA_k(y)$ where the leading polynomial $A_0(y)$ is not identically zero (otherwise we can always replace $f(x,y)$ by $f(x,y)/x^\ell$ for a suitable $\ell$). Denote by $N$ the degree of the polynomial $A_0(y)$. For any prime $p>N!$ the numbers $0$ and $(-1)^kk!$, where $k=0,\dots,N-1$, are distinct residues modulo $p$, so that $p!\equiv 0\pmod p$ and $(p-k)!=(p-1)!/\prod_{j=1}^{k-1}(p-j)\equiv(-1)^k(k-1)!^{-1}\pmod p$ are pairwise noncongruent modulo $p$ as well. Substituting $x=(2p-2k)!\equiv0\pmod p$ and $y=(p-k)!$ for each $k=0,1,\dots,N$ into $f(x,y)=0$ and reducing modulo $p$, we obtain $N+1$ different solutions of the polynomial equation $A_0(x)\equiv0\pmod p$, so that all coefficients of $A_0(x)$ are divisible by $p$. Since this is true for any prime $p>N!$, the polynomial $A_0(x)$ is identically zero, which contradicts our assumption.

Is it elementary enough?

Answer by Wadim Zudilin for Bounding a hypergeometric sum

2011-07-03T17:04:19Z

Your question sounds very similar to this one, and my answer is similar as well: de Bruijn's book Asymptotic Methods in Analysis discusses this type of problems in great detail.

Write your sum as $S_k=\sum_{j=1}^ka_j$ where, as you mention, $$ \frac{a_{j+1}}{a_j} =\frac14\biggl(\frac{k-1}{j-1}-1\biggr)\biggl(\frac{n-k-1}{j-1}-1\biggr) $$ monotonically decreases (as the function of $j$) from $\infty$ to $0$. This means that $a_j$ first increases for $j\le m$ and then decreases for $j\ge m$, where $m=m(k,n)$ is determined by $a_{m+1}/a_m\approx 1$. In the your example, $m-1$ is a solution of the quadratic equation. Then the main term of the asymptotics is fully determined by the growth of $a_m$ in the sense $$ \lim_{k\to\infty}S_k^{1/k} =\lim_{k\to\infty}a_{m(k,n)}^{1/k}, $$ and then you can use Stirling's formula to determine the latter limit explicitly. (de Bruijn's book also contains details on how to compute more accurate asymptotics.)

Alternatively, you can use Zeilberger's algorithm of creative telescoping to find a recurrence equation for $S_k$; then the same asymptotics can be read from its characteristic equation.

Answer by Wadim Zudilin for $p$-adic continuity for exponents in product decomposition of the $j$-invariant

2011-07-03T08:57:16Z

I do not have Ken Ono's book "The web of modularity: arithmetic of the coefficients of modular forms and $q$-series" around, but it contains the product expansion for the Eisenstein series $$ E_4=1+240\sum_{n=1}^\infty q^n\sum_{d\mid n}d^3 $$ of the form $\prod_{n=1}^\infty(1-q^n)^{a_n}$ (the Borcherds product). The exponents $a_n$ correspond to the expansion of a certain weak modular form. Furthermore, the weight 12 cusp form $\Delta$ is defined as $q\prod_{n=1}^\infty(1-q^n)^{24}$ and, finally, $$ j=\frac{E_4^3}{\Delta}. $$ The data will hopefully explicify your $p$-adic considerations.

Addition. Example 4.8 of Ono's book discusses the product expansion of $E_4(z)$, but already Example 4.7 gives the product for $j(z)$, so I just copy the details.

Let $\theta(z)=\sum_{n\in\mathbb Z}q^{n^2}$ denote the Jacobi theta function and notation $E_k(z)\in1+q\mathbb Q[[q]]$ stand for the Eisenstein series. Define $$ \begin{aligned} f(z) &=\frac{3E_{10}(4z)\delta\theta(z)}{2\Delta(4z)} -\frac{3\theta(z)V_4(\delta E_{10}(z))}{10\Delta(4z)}-\frac{456}5\theta(z) \cr &=\frac3{q^3}-744q+80256q^4-257985q^5+5121792q^8-12288744q^9+\cdots \cr &=\frac3{q^3}+\sum_{n=1}^\infty A(n)q^n, \end{aligned} $$ where $$ \delta:\sum_{n=0}^\infty a(n)q^n\mapsto\sum_{n=0}^\infty na(n)q^n \quad\text{and}\quad V_4:\sum_{n=0}^\infty a(n)q^n\mapsto\sum_{n=0}^\infty a(n)q^{4n}. $$ Then $f(z)$ is a weight 1/2 meromorphic modular form and $$ j(z)=\frac1q\prod_{n=1}^\infty(1-q^n)^{A(n^2)}. $$

Answer by Wadim Zudilin for Are there three variable generalizations of Ramanujans theta function?

2011-07-03T13:02:03Z

I do not know what it is for, as any basic hypergeometric summation can be interpretted as a certain multi-variate "Jacobi-like" identity. Probably, the easiest 4-variate example, known in the theory of basic hypergeometric functions as Ramanujan's summation (see, for example, Section 5.2 in the $q$-Bible of Gasper and Rahman), is $$ \sum_{n\in\mathbb Z}\frac{(a;q) _ n}{(b;q) _ n}\biggl(\frac ca\biggr)^n =\frac{(q,b/a,c,q/c;q) _ \infty}{(b,q/a,c/a,b/c;q) _ \infty}, $$ where the standard notation $(a;q) _ n=\prod_{k=1}^n(1-aq^{k-1})$ is used (and extending this symbol to the negative $n$ as well). The formula is valid for $|q|<1$ and $|b|<|c|<|a|$. The Jacobi triple product identity can be realised as a limiting case of this summation.

Comment by Wadim Zudilin

2013-02-12T04:53:05Z

aglearner and Chris, I have to apologise for what is there from me: I should have put a smile at the end... Handling my comments is indeed graceful... I upvote your both comments above. I would say that it really makes some difference to me whether a question like this is asked by a pseudonym. Being "aglearner" also means justifying it is taken appropriately.

Comment by Wadim Zudilin

2013-02-11T12:44:00Z

aglearner, why ain't you agteacher (which would be a more appropriate motivation)?

Comment by Wadim Zudilin

2013-02-11T12:39:26Z

Aaron, thanks for providing the heuristics. The related calculation is unpleasantly messy...

Comment by Wadim Zudilin

2013-02-11T12:35:49Z

Wolfgang, you didn't follow Gerry's hint in your prequel which would have led you to papers of Tasoev (see also my post mathoverflow.net/questions/24958/…) and Komatsu. All your CFs were in Tasoev's PhD thesis, I am not sure though of how much of it was published.

Comment by Wadim Zudilin

2012-12-25T03:28:57Z

Noam, Thanks for providing this new life to the old question.

Comment by Wadim Zudilin

2012-08-27T09:36:36Z

I definitely like your point, Frank. Your example with $e$ could be acomplished with the limit of $(1+1/n)^n$ as $n\to\infty$, which is extremely useful in showing many other limits but at the same time impractical for actual computation of $e$. The series, on other hand, can be successfully used not only to compute the number but also to demonstrate its irrationality to a first year undergrad.

Comment by Wadim Zudilin

2012-08-27T09:28:53Z

+1 for your CW comment: I really forgot about this natural option.

Comment by Wadim Zudilin

2012-08-26T11:38:27Z

C(ompleteness) implies A(rchimedean property) but A does not imply C. It happens... What is your question?

Comment by Wadim Zudilin

2012-08-25T13:29:16Z

Américo, muito obrigado! This is essentially Mirsky's proof I mention in my remark to Gerard's answer. I guess it is the one reproduced in Jameson's book as well.

Comment by Wadim Zudilin

2012-08-25T13:10:10Z

But Mirsky's 1949 short and elegant proof (jstor.org/stable/3611303) does the job even better. Note that this reference is due to Desbrow's paper...

Comment by Wadim Zudilin

2012-08-25T13:08:20Z

GH, I meant these as two different methods: the newer version should cause no confusion.

Comment by Wadim Zudilin

2012-08-25T12:54:31Z

Fantastic! That's why I like MO (occassionally). The jstor stable link is jstor.org/stable/2588989 .

Comment by Wadim Zudilin

2012-08-25T12:45:53Z

Gerard, I believe I can upvote just for this optimistic start. The details are very welcome as well!

Comment by Wadim Zudilin

2012-08-25T12:41:23Z

Geoff, do you mean: Jameson, G. J. O. A first course on complex functions. Chapman and Hall, Ltd., London (Distributed in the U.S.A. by Barnes & Noble, Inc.) 1970 xii+148 pp.? I wonder whether I can easily find it...

Comment by Wadim Zudilin

2012-08-25T11:13:07Z

Tim, there is also an example, from December 2011, for $1/\pi^4$ due to Jim Cullen (members.bex.net/jtcullen515), another mathematics amateur; I cannot easily fine it online though.