65

"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 ...

answered Jan 16 '14 at 6:31

Noam D. Elkies

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54

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 $\...

answered Sep 13 '19 at 12:11

Carlo Beenakker

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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 $\lambda(s)=\sum_{n=0}^\infty(2n+1)^{-s}$,

answered Dec 1 '19 at 20:58

Carlo Beenakker

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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(...

answered Sep 4 '14 at 18:23

Robert Bryant

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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 ...

answered Oct 14 '15 at 21:13

Gjergji Zaimi

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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 ...

answered Nov 20 '16 at 21:59

Christian Remling

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15

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)...

15

[a bit too long for a comment]
I understand from the question that the aim is to find a research project based on the search for a counterexample. By construction, this will mean showing that some existing paper in the literature is mistaken. That is typically not a productive way to start a project in a new field, simply because (a) if the author of that ...

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. ...

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

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 ...

answered Oct 6 '14 at 21:43

Fedor Petrov

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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

Here is a solution that I have found while working on other lattice sums.
It utilizes a very simple result:
Define $f$ by
$$f(x)=\sum_{n=0}^{\infty} (-1)^n (2n+1) e^{-\pi x (n+\frac12)^2}.$$
Then
$$\int_0^{\infty} e^{-y x} f(x)dx=\operatorname{sech}\sqrt{\pi y}.\tag{$\star$}$$
The proof is simple - just integrate term by term, and the result ...

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 ...

ca.classical-analysis-and-odes cv.complex-variables mp.mathematical-physics special-functions specific-calculation

answered May 2 '16 at 12:16

Fedor Petrov

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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 $x_j+x_j^{-1}$ ...

answered Jan 17 '19 at 13:31

Fedor Petrov

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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/...

answered Jan 17 '19 at 16:55

François Brunault

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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 $\mathbf{Q}(\zeta_p)$ ...

13

As observed in comments, we have $f(n) = \lfloor g(n) \rfloor$ where $g(n) = \frac{\alpha^n + \alpha^{-n}}{4}$ and $\alpha = 2 + \sqrt{3}$. From the recurrence $g(n+1) = 4 g(n) - g(n-1)$ we see that $g(n)$ is a half-integer when $n$ is even and an integer when $n$ is odd. In fact we see from induction that for even $n$ we have $g(n) = \frac{1}{2} \hbox{ ...

13

Here is an explicit bound. The sum $\sum_{n > N} n^{-s}$ for real $s > 1$ is bounded by the integral
$$\int_N^\infty x^{-s} = N^{1-s} / (s-1).$$
Therefore for any $N$ you have
$$0 < \zeta(s) - (1 + 2^{-s} + \cdots + N^{-s}) < N^{1-s} / (s-1).$$
E.g., with $N = 3$ you get
$$0 < \zeta(s) - 1 - 2^{-s} - 3^{-s} < 3^{1-s}/(s-1).$$

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

In [Ritt 1948], the method of J. Liouville is given for the Kepler equation.
[Ritt 1948] Ritt, J. F.: Integration in finite terms. Liouville's theory of elementary methods. 1948, page 56
A further method is the method of Rosenlicht, M.: On the explicit solvability of certain transcendental equations. Publications mathématiques de l'IHÉS 36 (1969) 15-22. It ...

12

For natural values of a, take $\displaystyle\frac{e^t-1}t=\sum_1^\infty\dfrac{t^{n-1}}{n!}$, then apply the operator $\bigg(\displaystyle\frac1t\cdot\int\bigg)$ a times to it. For $a\not\in\mathbb N$, such as $a=\dfrac14$, welcome to the “marvelous” world of fractional calculus and Riemann-Liouville integrals.
For natural values of a, the series can also be ...

12

Gradshteyn & Ryzhik equation 3.773.1 gives (for $q>0$)
$$\frac{1}{q}B(q)=\int_0^\infty \frac{\sin x}{x(q^2/4+x^2)^{1/2}}\,dx=\frac{1}{q}G^{21}_{13}\left(\frac{q^2}{16}\biggl|^1_{1/2,1/2,0}\right)$$
$$\qquad\qquad=1-\gamma+\log 4-\log q+{\cal O}(q^2)=1.80908-\log q+{\cal O}(q^2).$$
The asymptotic result desired by the OP is $B(q)/q=\pi\gamma-\log q=1....

answered Feb 17 '16 at 15:09

Carlo Beenakker

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12

Sorry, my answer is wrong. As it was pointed out by "Simply Beautiful Art" $x\ne W(z)$ but $x=e^{W(z)}.$
It is known that $$W'(z)=\frac{W(z)}{z(1+W(z))}.$$
Let $z=\log t$. Then $x=W(z)$ is a root of the equation $x^x=t$, in particular $x\log x=\log t=z.$ It means that $$W'(z)=\frac{x}{(x+1)\log t}=\frac{1}{(x+1)\log x}=\frac{1}{\log x^{x+1}},\quad x^{x+1}=e^...

answered Mar 21 '20 at 12:37

Alexey Ustinov

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12

The answer is yes indeed. It is a special case of Fox-H function, a variation of the confluent Fox-Wright $_{1}\Psi_{1}$ function (a generalization of the confluent hypergeometric function $_{1}F_{1}$) providing the inverse function. See a previous answer here for details and references. For this particular case solution is
(Setting $\alpha = a$), for $\...

11

There is a paper of G. H. Hardy, where this function is studied in great detail:
G. H. Hardy, On the integral function $ \Phi_{ a,\alpha,\beta}(z)=\sum x^n/(n+a)^{\alpha n+\beta}$, Quarterly J. Math., 5 (1906) 37, 369-378. (Collected papers of G. H.Hardy, vol. IV, p. 128).

answered Jan 7 '19 at 17:25

Alexandre Eremenko

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11

This is an entire function of order $1/\alpha$ when $\alpha>1$. So for irrational $\alpha$
it cannot satisfy any linear differential equation with polynomial coefficients.
If $0<\alpha<1$, the order is $1$ but the type is minimal, so again it cannot
satisfy any such equation. This excludes most special functions.
(But does not exclude their ...

answered Apr 10 '14 at 3:43

Alexandre Eremenko

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11

All these questions are basically answered by Floquet theory. The envelope is an exponential, and there is nothing special going on at 300, what you are seeing is just
exponential growth, and on the scale on which you are plotting everything less that $10^{49}$ just looks like zero.

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