61

"Numerology" such as you've observed is explained in the paper Gross, B.H., and Zagier, D.: On singular moduli, J. reine angew. Math. 355 (1985), 191$-$220. MR772491 (86j:11041) which gives more generally the factorizations of the constant terms of the minimal polynomials of $j(\tau) - j(\tau')$ where $\tau,\tau'$ are quadratic imaginaries not ...


42

The intuition may be helped by considering the generalized Euler constant function $$\gamma(z)=\sum_{n=1}^\infty z^{n-1}\left(\frac{1}{n}-\ln\frac{n+1}{n}\right),\;\;|z|\leq 1.$$ Its values include the Euler constant $\gamma=\gamma(1)$ and the "alternating Euler constant" $\ln 4/\pi=\gamma(-1)$. So any general integral formula or recursion relation for $\...


37

[Edited to outline the end of the argument that $f(-M) \rightarrow 0$ (and to correct a few typos etc. while I'm at it)] Yes, $F(x) \rightarrow 0$ from below as $x \rightarrow -\infty$. The convergence is slow, and precise asymptotic analysis seems to be somewhat annoying because it involves the lower branch of the Lambert W function. The massive ...


32

This question asks in effect to show that $\eta^n$ is a $\pm p^{n/2}$ eigenfunction for the Hecke operator $T_p$. The claim holds because each of these $\eta^n$ happens to be a CM form of weight $n/2$, and $p$ is inert in the CM field ${\bf Q}(i)$ or ${\bf Q}(\sqrt{-3})$. In plainer language, the sum over $k$ takes the $q$-expandion $$ \eta(\tau)^n = \sum_{...


29

Radial Fourier transforms provide a good, consistent perspective on most of the theory. The Fourier transform $\widehat{f}(t)$ of a function $f \colon \mathbb{R}^n \to \mathbb{R}$ is given by the integral of $f(x) e^{2\pi i \langle x,t \rangle} \, dx$ over $x \in \mathbb{R}^n$. If $f$ is a radial function (i.e., $f(x)$ depends only on $|x|$), then we can ...


23

A course of Modern Analysis by Whittaker and Watson. No, I'm not kidding! They have a wonderful chapter that ends with a wonderful classification of the main special functions according to their singularity patterns. The best modern textbook really is A=B (cited already). The modern approach to all special functions is to realize that they are a very ...


21

I'm surprised no one has posted a proof by double counting yet. First, I will rewrite the sum as $$F(n,k)=\sum_{j=0}^k\binom{2n+1}{2n-2j+1}\binom{n-j}{n-k}.$$ This counts the number of ways I can pick $2n-2j+1$ squares out of a $1\times (2n+1) $ grid, color them alternately black, white, black... etc. and then place a mark on $n-k$ white squares. By ...


21

It seems that the non-linear differential equation satisfied by the Lambert $W$ function is simple enough for this question to have already been answered. This paper proves that the Lambert $W$ is non-elementary by appealing to a result of Rosenlicht (1969): Bronstein, M., Corless, R. M., Davenport, J. H. and Jeffrey, D. J. (2008) Algebraic properties ...


20

[Edited again to give a second identity relating $E$ to eta products] Continued fraction or not, an expression $q^{\frac{(r-s)^2}{8(r+s)}} f(\pm q^r, \pm q^s)$ is a modular form of weight $1/2$ for all integers $r,s$ with $r+s>0$, because it is a sum $\sum_{n=-\infty}^\infty \pm q^{(cn+d)^2}$ with rational $c,d$ and periodic signs. Therefore the quotient ...


19

There are many, many things that can be said here! I think there are two slightly different mechanisms that conjure up Bessel functions of various types, namely, Euclidean Laplacians and separation of variables, and $SL_2(\mathbb R)$ (and orthogonal groups $O(n,1)$) Casimir or Laplace-Beltrami operator and separation of variables. The fact that there cannot ...


19

The paper entitled Euler constant as a renormalized value of Riemann zeta function at its pole by Andrei Vieru contains a derivation of the first formula in the OP (Benoȋt Cloitre's formula), and a method to obtain variations thereof, such as as well as similar iterates in terms of the Dirichlet function $\lambda(s)=\sum_{n=0}^\infty(2n+1)^{-s}$,


16

I wish to add a remark to Noam D. Elkies' beautiful answer. From the integral representation for $f$, putting $e^{-t}=s$ in the integral, $$f(-x)=1-x\int_0^\infty e^{-xte^{-t}} e^{-t}dt = 1-x\int_0^1 s^{sx}ds\, ,$$ so that, for $x\to \infty$, $ f(-x)=o(1)$ is equivalent to $$\int_0^1 xu(s)^xds=1+o(1)\, ,$$ where $u\in C([0,1])$ is the function $u(s):=s^s$. ...


16

I would guess that you are just seeing the effects of a dynamical system with a hyperbolic fixed point. Consider the matrix equation $$ A'(x) = \begin{pmatrix} 0 & 1\\ \cos(x) & 0\end{pmatrix} A(x) $$ with initial condition $A(0) = I_2$. Because of the $2\pi$-periodicity of $\cos(x)$, this fundamental solution clearly satisfies $$ A(x+2\pi) = A(x)A(...


16

Let us consider the case when $n$ is odd. Let $$P(x):=\exp\left(\frac{ixl}{2}\right)\left(\exp\left(\frac{ix}{2}\right)-\exp\left(\frac{-ix}{2}\right)\right),$$ and notice that the degree of $P$ as an exponential polynomial equals $$\max\left(\left|\frac{l+1}{2}\right|,\left|\frac{l-1}{2}\right|\right)\leq\frac{n-1}{2}.$$ Therefore this is the unique ...


16

This is response to QUESTION 1. As Fedor pointed out, we're dealing with the Chebyshev polynomials $P_n(2\cos t)=\sin nt/\sin t$. So we must show that if $$ \sum_{n=1}^N \sin nt = 0 , \quad\quad\quad\quad (1) $$ then also $\sum_{n=1}^N \sin^m nt = 0$ for any odd exponent $m\ge 1$. We may take $0<t<\pi/2$. Also, the sum in (1) can of course be ...


14

It looks to me like you don't even need the five-term identities. If I denote the arguments of your dilogarithms by $a_1$, $a_2$, $a_3$ and $a_4$ I find that $(1-a_1) a_4 = 1$ and $a_2 a_3 = -1 + a3$. Then I think using the simple formulas $$ \text{Li}_2(z)=-\text{Li}_2(1-z)-\log (1-z) \log (z)+\frac{\pi ^2}{6}, $$ and $$ \text{Li}_2(z)=-\text{Li}_2\...


14

Both are equal to $(\pi^2-6\log^22)/12$. The inner sum on the right $$\sum_{n=0}^\infty\frac{(-1)^m}{(n+1)(n+m+2)}=\frac{(-1)^m}{m+1}\sum_{n=0}^\infty \Bigl(\frac{1}{n+1}-\frac{1}{n+m+2}\Bigr)= \frac{(-1)^m}{m+1}\Bigl(1+\frac12+\frac13+\cdots+\frac{1}{m+1}\Bigr)$$ Therefore the sum on the right is equal to $$R:=\sum_{m=0}^\infty \int_0^1(-1)^m x^m\Bigl(1+\...


14

The sum $$\sum_{k=0}^\infty \frac{(a+k)!\,(b+k)!}{k!\,(a+b+c+k+1)!}z^k.$$ is not only a generalized hypergeometric series; it's the original ungeneralized Gauss hypergeometric series, $$\frac{\Gamma(a+1)\,\Gamma(b+1)}{\Gamma(a+b+c+2)}{}_2F_1\left({a+1,b+1\atop a+b+c+2}\ \Big |\, z\right).$$ Chapter 15 of Abramowitz and Stegun is on hypergeometric functions. ...


14

The first formula is trivial. $$f(s)= \frac1{s-1}+\gamma +O(s-1)$$ $$g(z)=1+2^{-z}+3^{-z}+4^{-z}+O(5^{-z})=1+2^{-z}(1+(3/2)^{-z}+(4/2)^{-z}+O(5/2)^{-z})$$ $$f(g(z)) = \frac1{2^{-z}(1+(3/2)^{-z}+(4/2)^{-z}+O(5/2)^{-z})}+\gamma+O(2^{-z})$$ $$=2^z \left(1-(3/2)^{-z}-(4/2)^{-z}+O(5/2)^{-z}+(O(3/2)^{-z})^2\right)+\gamma + O(2^{-z})$$ $$=2^z- (3/4)^{-z}-1+O(9/8)...


13

Hypergeometric functions arise as matrix coefficients of representations of Lie groups. This is my formal answer, but I will also describe very informally why I believe this is an answer to your question. (By this I roughly mean: why this might have triggered interest of 19th century mathematicians who didn't have the language to be aware that this is what ...


13

(Intended as a comment to Noam Elkies' response.) My colleague Marco Streng was kind enough to point out that according to "Eta Products and Theta Series Identities", a book of Guenther Koehler MR2766155: "Serre [128] proved that the Fourier series of a modular form f is lacunary if and only if f is of CM-type, i.e., if f is a linear combination of Hecke ...


13

This inequality follows from (6.8) in Positive Jacobi polynomial sums II by Askey and Gasper, which says that $$ \sum_{k=0}^n (-1)^k L_k(x) = \sum_{j=0}^{\lfloor n/2 \rfloor} \frac{2^{-n} (2j)!}{j!^2 (n-2j)!} H_{n-2j}^2\left(\sqrt{x/2}\right). $$ They give a couple of earlier references for this identity, but I haven't looked them up. (So I'm just giving a ...


13

Let's try to understand in which sense this equality $$\sum_{k=-\infty}^{\infty}k^nz^k=\left(z\frac{d}{dz}\right)^n\sum_{k=-\infty}^{\infty}z^k=\left(z\frac{d}{dz}\right)^n 0=0$$ may be understood. Of course, there is no convergence in usual sense on the unit circle (because terms do not tend to 0). Actually, we do not use analysis at all, as it is purely ...


13

Note that $$\frac{\Gamma(z) \Gamma(1-z)}{\Gamma(2z) \Gamma(1 - 2z)} = 2 \cos(\pi z),$$ and the RHS is transcendental for any non-rational algebraic number $z$ (by the Gelfond–Schneider theorem). So $\Gamma$ certainly won't preserve any number field $K$. It's most likely true that $\Gamma(z)$ is transcendental for algebraic $z \notin \mathbf{Z}$, but I'm ...


13

Yes. Denote $p=\sinh s$, then $s\in (-\infty,\infty)$, $\sqrt{p^2+1}=\cosh s$, $dp=\cosh s\, ds$. Next, denote $r=\sqrt{\tau^2+x^2}$, $\tau=r\cos \varphi$, $x=r\sin \varphi$, where $\varphi\in (-\pi/2,\pi/2)$. So our integral rewrites as $$\int_{-\infty}^\infty ds \exp(-r\cosh(s-i\varphi)).$$ It does not depend on $\varphi$ as may be seen from the ...


13

First of all, we use the formula $$ D:=\det [x_j^k+x_j^{-k}]_{j,k=0,\dots,m-1}=\prod_{l<j}(x_j+x_j^{-1}-x_l-x_l^{-1})=\prod_{l<j} (x_j-x_l)(1-x_j^{-1}x_l^{-1}). $$ This follows from the observation that $x^k+x^{-k}=p_k(x+x^{-1})$ for a polynomial $p$ of degree $k$ with leading coefficient 1, so our matrix is the Vandermonde matrix for $x_j+x_j^{-1}$ ...


13

The conjecture is not true, as some examples show. Let $D(p)$ denote your number. For primes $p \equiv 1 \textrm{ mod } 4$, we have $D(29)=8$, $D(37)=37$, $D(41)=121$ while $h(29)=h(37)=h(41)=1$. For primes $p \equiv 3 \textrm{ mod } 4$, we have $D(23)=3$, $D(31)=9$, $D(43)=211$, $D(47)=695$ while $h(23)=h(31)=3$, $h(43)=1$, $h(47)=5$. The following Pari/...


13

If by "the class number $h(p^*)$ of the quadratic field $\mathbb{Q}(\sqrt{p^*})$" you mean "the minus class number $h^{-}$ of $\mathbf{Q}(\zeta_p)$" and if by " a possible new formula for the class number" you mean "an elementary formula for $h^{-}$ known to Kummer (obtained by considering the ratio of the zeta functions of $\mathbf{Q}(\zeta_p)$ ...


12

This discussion makes me even firmer in my opinion that introducing fancy notation for special functions and making long lists of related formulae in reference books makes more harm than good and that we would know much more about and be at more ease with them if everybody had to start from the basics and deal with bare definitions every time he needed to ...


12

The five term relation comes from the fact that the sum of the volumes of tetrahedra $ABCD$ and $ABCE$ equals the sum of the volumes of the three tetrahedra $ABDE, ACDE, BCDE.$ One can think of $ABCDE$ as a degenerate four-dimensional simplex.


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