White matter connectivity differences converge with gene expression in a neurodevelopmental disorder of known genetic origin

Photometric Science Alerts From Gaia

Gaia is a European Space Agency (ESA) astrometry space mission, and a successor to the ESA Hipparcos mission. The main goal of the Gaia mission is to collect high-precision astrometric data (i.e. positions, parallaxes, and proper motions) for the brightest one billion objects in the sky. This data, complemented with G band, multi-epoch photometric and low resolution (lowers) spectroscopic data collected from the same observing platform, will allow astronomers to reconstruct the formation history, structure, and evolution of the Galaxy.

In addition, the Gaia satellite is an excellent transient discovery instrument, covering the whole sky (including the Galactic plane) for the next 5 years, at high spatial resolution (50 to 100 mas, similar to the Hubble space telescope (HST)) with precise photometry (1% at G=19) and milliarcsecond astrometry (down to ∼20mag). Thus, Gaia provides a unique opportunity for the discovery of large numbers of transient and anomalous events, e.g. supernovae, black hole binaries and tidal disruption events. We discuss the validation of the alerts stream for the first six months of the Gaia observations, in particular noting how a significant ground based campaign involving photometric and spectroscopic followup of early Gaia alerts is now in place. We discuss the validation approach, and highlight in more detail the specific case of Type Ia supernova (SNe Ia) to be discovered by Gaia. The intense initial ground based validation campaign will ensure that the Gaia alerts stream for the remainder of the Gaia mission, are well classified.

Exploring a Zeta Function Extension and its Relation to Primes

# Abstract

An adapted version of the zeta function is defined as *θ*_{m}(*s*), this is used to explore the sequence of prime numbers.

Sum Over \(n\) Split into Subsets (\(n^{2}\),\(T_{n}\),\(\cdots\))

Laguerre Series: Functions in a Basis of Laguerre Polynomials

# Abstract

I show how to express functions with a known series expansion in terms of Laguerre polynomials.

# Main

We have the integral representation of the Laguerre polynomials, \begin{equation}
L_n(z)= \frac{e^z}{n!}\int_0^\infty e^{-t}t^n J_0(2\sqrt{ t z})\;dt,
\end{equation} where *J*_{0}(*x*) is a modified Bessel function, as given on the Wolfram Functions website. If we have a function *f*(*x*) that admits a series representation \begin{equation}
f(x)=\sum_{k=0}^\infty a_k x^k,
\end{equation} with some coefficients *a*_{k}, then we can define a transform on *L*[*f*] \begin{equation}
F(s)=L[f(x)]=\int_0^\infty K(x,s)f(x)\;dx.
\end{equation} If we pick the kernel function to be \begin{equation}
K(x,s)= e^{-x}J_0(2\sqrt{xs}),
\end{equation} then we have \begin{equation}
F(s) = \sum_{k=0}^\infty a_k\int_0^\infty e^{-x}J_0(2\sqrt{xs})x^k \; dx = \sum_{k=0}^\infty \frac{a_k k!}{e^s}L_k(s),
\end{equation} which is an expansion of the transformed function *F*(*s*) in terms of Laguerre polynomials. As an example, if *f*(*x*)=1, then *a*_{k} = [*k* = 0], this gives \begin{equation}
\int_0^\infty e^{-x}J_0(2\sqrt{xs})\;dx = \sum_{k=0}^\infty \frac{[k=0]k!}{e^s}L_k(s) = \frac{L_0(s)}{e^s} = e^{-s}
\end{equation} this leads to an infinitely recursive integral \begin{equation}
e^{-s} = \int_0^\infty \int_0^\infty \int_0^\infty \cdots \int_0^\infty e^{-x_n}J_0(2\sqrt{x_n x_{n-1}})\;dx_n \cdots J_0(2\sqrt{x_3x_2})\;dx_3J_0(2\sqrt{x_2x_1})\;dx_2J_0(2\sqrt{x_1s})\;dx_1
\end{equation} this should also lead to the set of results \begin{equation}
e^{-s} = \int_0^\infty e^{x_1}J_0(2\sqrt{x_1 s})\;dx_1
\end{equation} \begin{equation}
e^{-s} = \int_0^\infty\int_0^\infty e^{x_2}J_0(2\sqrt{x_2x_1})J_0(2\sqrt{x_1 s})\;dx_1dx_2
\end{equation} \begin{equation}
e^{-s} = \int_0^\infty\int_0^\infty\int_0^\infty e^{x_3}J_0(2\sqrt{x_3x_2})J_0(2\sqrt{x_2x_1})J_0(2\sqrt{x_1 s})\;dx_1dx_2dx_3
\end{equation} and so on.

If we formally define a function \begin{equation}
f(x) = \sum_{k=0}^\infty \frac{x^k}{e k!k!} = \frac{1}{e} I_0(2\sqrt{x})
\end{equation} we end up with a series expansion for *F*(*s*), starting \begin{equation}
F(s) = 1 - 2s + \frac{7}{4}s^2 - \frac{17}{18}s^3 + \frac{209}{576}s^4 - \frac{773}{7200}s^5 + \cdots
\end{equation}

It seems the inverse kernel is given by \begin{equation} K^{-1}(x,s) = \sum_{i=0}^\infty \sum_{j=0}^i \frac{(-1)^j s^j x^i}{(i-j)!j!j!} = \sum_{i=0}^\infty \frac{x^i}{i!}L_i(s) \end{equation} such that \begin{equation} f(x) = \int_0^\infty K^{-1}(x,s)F(s)\;ds \end{equation}

Hermite Series: Common Functions in a Basis of Hermite Polynomials

# Abstract

This document notes down expressions for a few common functions in terms of Hermite polynomials. There are clear symmetries between certain special functions, i.e. cos and sin, but interestingly a similar symmetry is formed between the Gaussian and the product of a Gaussian and the imaginary error function. The method of finding these expansions comes primarily from the Hermite transform. An approximate form for the log function is given, but this may have complex branching problems.

# Main

Hermite Series Couplings: We can write a few common functions in a basis of Hermite polynomials \begin{align}
e^{ax} = \sum_{n=0}^\infty \frac{a^n e^{a^2/4}}{2^n n!}H_n(x)\\
e^{x-\frac{1}{4}} = \sum_{n=0}^\infty \frac{H_n(x)}{2^n n!}\\
e^{-x^2} = \sum_{n=0}^\infty \frac{(-1)^n(2n-1)!!}{\sqrt{2}2^{2n}(2n)!}H_{2n}(x) \\
e^{-x^2} = \text{Re}\left[\sum_{n=0}^\infty \frac{i^n(n-1)!!}{\sqrt{2}2^{n}(n)!}H_{n}(x) \right] \\
?(x) = \text{Im}\left[\sum_{n=0}^\infty \frac{i^n(n-1)!!}{\sqrt{2}2^{n}(n)!}H_{n}(x) \right] \\
\cos(ax) = \text{Re}\left[\sum_{n=0}^\infty \frac{a^ni^n}{e^{a^2/4}2^{n}n!}H_{n}(x)\right]\\
\sin(ax) = \text{Im}\left[\sum_{n=0}^\infty \frac{a^ni^n}{e^{a^2/4}2^{n}n!}H_{n}(x)\right]\\
x^n = \frac{n!}{2^n} \sum_{m=0}^{\lfloor \tfrac{n}{2} \rfloor} \frac{1}{m!(n-2m)! } ~H_{n-2m}(x)
\end{align} where *H*_{n}(*x*) is a Hermite polynomial of order *n*. I want to find standard forms such that a function can be written \begin{equation}
f(x) = \sum_{n=0}^\infty c_n H_n(x)
\end{equation} for some constants *c*_{n}. It is interesting to see that there is a function marked as ?(*x*) above, which is made of the imaginary parts of the expansion of *e*^{−x2}. I would like to know what it is. We can see that \begin{align}
?(0)=0\\
?(\infty)=0
\end{align} and it has a single maximum just greater than *x* = 1. It seems we can write the Taylor expansion of this function as \begin{equation}
?(x) = \sqrt{2} x - \frac{2\sqrt{2}}{3}x^3 + \frac{4 \sqrt{2}}{15} x^5 - \frac{16 \sqrt{2}}{21} x^7 + \cdots \\
\end{equation} we can see then it looks like \begin{align}
\frac{?(x)}{\sqrt{2}} = \sum_{n=1}^\infty \frac{(-1)^{n+1} 2^{n-1}}{(2n-1)!!} x^{2n-1} \\
?(x) = \sqrt{\frac{\pi}{2}}\text{Erfi}(x)e^{-x^2}
\end{align}

For the logarithm, by experimental techniques we can formally write something like \begin{align} \log(x) = \frac{i}{2}\sqrt{\pi}(i\gamma + \pi + i\log(4)) + \sum_{n>0} \frac{(-1)^n (2n-2)! i \sqrt{\pi}}{2^{2n-1}(2n-1)!}H_{2n-1}(x) + 2\frac{(-1)^{n-1}4^{n-1}(n-1)!}{2^{2n}(2n)!}H_{2n}(x) \end{align} but it’s not clear how it converges, perhaps something slightly wrong with this. Other functions use hypergeometric expressions as coefficients, for example

\begin{equation}
\cosh(\sqrt{x}) = \sum_{n=0}^\infty \frac{_0F_2(;\frac{2n+1}{4},\frac{2n+3}{4};\frac{1}{256})}{2^n (2n)!}H_n(x)
\end{equation} by setting *x* = 0, this gives us some sums for unity\begin{equation}
\sum_{n=0}^\infty \frac{_0F_2(;\frac{2n+1}{4},\frac{2n+3}{4};\frac{1}{256})}{2^n (2n)!}\frac{2^n \sqrt{\pi}}{\Gamma(\frac{1-n}{2})} = 1
\end{equation} for the exponential expression *e*^{x − 1/4} given above we get \begin{equation}
\sum_{n=0}^\infty \frac{e^{1/4}\sqrt{\pi}}{n!\Gamma(\frac{1-n}{2})} = 1
\end{equation} also \begin{equation}
\sum_{n=0}^\infty \frac{4^n \pi}{(2n)!\Gamma((1-n)/2)\Gamma(n+1/2)} \,_0F_4(;1/4 + n/2, 1/4+n/2,3/4+n/2,3/4+n/2;1/64) = 1
\end{equation} and \begin{equation}
\frac{1}{\pi^{5/2}}=\sum_{n=0}^\infty \frac{2^{1-4n}}{(2n)!\Gamma(\frac{1}{2}-n)} \,_0\tilde{F}_4\left(;n+\frac{1}{4},n+\frac{1}{4},n+\frac{3}{4},n+\frac{3}{4};1/64\right)
\end{equation} It seems that in general we have the result \begin{equation}
D_{2n-1} e^{-x^2} = \sum_{k=0}^\infty \frac{(-1)^{n+k}(2n+2k)!}{(2k+1)!(n+k)!}x^{2k+1} = \frac{(-1)^n x (2n)!}{n!} \;_1F_1\left(n+\frac{1}{2},\frac{3}{2},-x^2\right)
\end{equation}

# Nice Relationship/Transform

Using related functions above seems to make in interesting transform kernel, however, there does not seem to be a one to one mapping between transformed functions and input functions. Some results are given below: \begin{equation} \int_{-\infty}^ \infty \frac{e^{a-\frac{x^2}{8a}}J_0(x)}{\sqrt{8\pi a}} = I_0(a) \end{equation} \begin{equation} \int_{-\infty}^ \infty \frac{e^{-a-\frac{x^2}{8a}}I_0(x)}{\sqrt{8\pi a}} = J_0(a) \end{equation} \begin{equation} \int_{-\infty}^ \infty \frac{e^{a-\frac{x^2}{8a}}e^x}{\sqrt{8\pi a}} = e^{3a} \end{equation} \begin{equation} \int_{-\infty}^ \infty \frac{e^{-a-\frac{x^2}{8a}}e^x}{\sqrt{8\pi a}} = e^a \end{equation} \begin{equation} \int_{-\infty}^ \infty \frac{e^{a-\frac{x^2}{8a}}\cos(x)}{\sqrt{8\pi a}} = e^{-a} \end{equation} \begin{equation} \int_{-\infty}^ \infty \frac{e^{-a-\frac{x^2}{8a}}\cos(x)}{\sqrt{8\pi a}} = e^{-3a} \end{equation} \begin{equation} \int_{-\infty}^ \infty \frac{e^{a-\frac{x^2}{8a}}\sin(x)}{\sqrt{8\pi a}} = 0 \end{equation} \begin{equation} \int_{-\infty}^ \infty \frac{e^{-a-\frac{x^2}{8a}}\sin(x)}{\sqrt{8\pi a}} = 0 \end{equation} \begin{equation} \int_{-\infty}^ \infty \frac{e^{a-\frac{x^2}{8a}}J_2(x)}{\sqrt{8\pi a}} = I_1(a) \end{equation} \begin{equation} \int_{-\infty}^ \infty \frac{e^{a-\frac{x^2}{8a}}J_4(x)}{\sqrt{8\pi a}} = I_2(a) \end{equation}

Inverse of General Quintic by Determinants

# Main

I will work out how to potentially express the series reversion for a general quintic (and possibly the conditions that this fails) using the determinants of matrices. Let us define our quintic polynomial \begin{equation}
P(x) = a_0 + a_1x + a_2x^2+a_3x^3 + a_4x^4 + a_5x^5
\end{equation} we can treat this is a series and do a series reversion about *a*_{0} giving \begin{equation}
P^{-1}(x) = \frac{x-a_0}{a_1} - \frac{a_2(x-a_0)^2}{a_1^3} + \frac{(2a_2^2-a_1a_3)(x-a_0)^3}{a_1^5}+ \frac{(-5a_2^3+5a_1a_2a_3-a_1^2a_4)(x-a_0)^4}{a_1^7} + \cdots
\end{equation} there is a clear pattern for the denominators and the (*x* − *a*_{0}) term. Can we express this sequence in the form \begin{equation}
P^{-1}(x) = \sum_{n=0}^\infty |M_n|\frac{(x-a_0)^{n+1}}{a_1^{2n+1}}
\end{equation} where *M*_{n} is an *n* × *n* matrix, and | ⋅ | is the determinant. Then we have *M*_{0} as the empty matrix, define the determinant |*M*_{0}|=1, then \begin{equation}
M_1=-[a_2]\\
M_2=-\begin{bmatrix} a_1 & a_2\\ 2a_2 & a_3 \end{bmatrix}\\
M_3= -\begin{bmatrix} a_1 & a_2 & 0 \\ 0 & a_1 & 5a_2 \\ a_2 & a_3 & a_4 \end{bmatrix} \\
M_4 = -\begin{bmatrix} a_1 & a_2 & 2a_3/3 & 0 \\ 0 & 6a_1 & 14a_2 & 18a_3 \\ 0 & 0 & a_1/6 & a_2 \\ a_2 & a_3 & a_4 & a_5 \end{bmatrix}\\
M_5 = -\begin{bmatrix} a_1 & a_2 & a_3 & a_4 & 0 \\ 0 & 7a_1 & 14a_2 & 7a_3 & 0 \\ 0 & 0 & a_1 & 3a_2 & a_3 \\ 0& 0 & 0 & a_1 & a_2 \\ a_2 & a_3 & a_4 & a_5 & 0 \end{bmatrix}
\end{equation} we could also swap rows and columns being careful about sign, there are likely other (potentially prettier) solutions. If we could work out a general pattern, then we would have a form for the inverse of a general quintic. However, it can be seen that if *a*_{1} → 0, the inverse function becomes ill defined.

# Inverse of Lesser Polynomials

We might quickly want to apply this method to lesser polynomials, for example the linear inverse \begin{equation}
P_1(x) = a_0 + a_1 x
\end{equation} then we have the inverse by series reversion \begin{equation}
P_^{-1}(x) = \frac{(x-a_0)}{a_1}
\end{equation} we can note this is also the first term of the quintic expression above. For the quadratic we have \begin{equation}
P_2(x) = a_0 + a_1 x + a_2 x^2
\end{equation} and by reversion the sequence \begin{equation}
P_2^{-1}(x) = \frac{(x-a_0)}{a_1} - \frac{a_2(x-a_0)^2}{a_1^3} + \frac{2a_2^2(x-a_0)^3}{a_1^5} - \frac{5a_2^3(x-a_0)^4}{a_1^7} + \cdots
\end{equation} where the coefficients 1, 1, 2, 5, ⋯ are the Catalan numbers, A000108, which gives a form for this expression \begin{equation}
P_2^{-1}(x) = \sum_{n=0}^\infty \frac{(2n)!}{n!(n+1)!}\frac{(-1)^na_2^n(x-a_0)^{n+1}}{a_1^{2n+1}} = \frac{-a_1+\sqrt{a_1^2-4a_0a_2+4a_2x}}{2a_2}
\end{equation} which clearly resembles the quadratic formula, this however gives little opportunity to yield a determinant structure, it is clear that the *a*_{2}^{n} coefficients are arriving from the determinant. Next we may try the cubic equation \begin{equation}
P_3(x) = a_0 + a_1x + a_2 x^2 + a_3 x^3
\end{equation} with reversion \begin{equation}
P_3^{-1}(x) = \frac{(x-a_0)}{a_1} - \frac{a_2(x-a_0)^2}{a_1^3} + \frac{(2a_2^2-a_1a_3)(x-a_0)^3}{a_1^5} + \frac{(-5a_2^3+5a_1a_2a_3)(x-a_0)^4}{a_1^7} + \cdots
\end{equation} this is sharing three terms with the quintic equation reversion, it also gives rise to the use of the determinants. The coefficient of *a*_{2}^{n} in each term appears to be the Catalan numbers as in *P*_{2}^{−1}(*x*). We can try to derive the series of matricies _{3}*M*_{n} which describe the cubic pattern, \begin{equation}
_3M_1=-[a_2]\\
_3M_2=-\begin{bmatrix} a_1 & 2a_2\\ a_2 & a_3 \end{bmatrix}\\
_3M_3= -\begin{bmatrix} a_1 & a_2 & 0 \\ 0 & a_1 & 5a_2 \\ a_2 & a_3 & 0 \end{bmatrix}
\end{equation} which are the same, if not similar. We now note that there seems to always be a coeffient that is the current Catalan number in the matrix, and it is on *a*_{2}. We might be better off assiging these coefficients to the lower left corner which is consistently *a*_{2},\begin{equation}
_3M_1=-[a_2]\\
_3M_2=-\begin{bmatrix} a_1 & a_2\\ 2a_2 & a_3 \end{bmatrix}\\
_3M_3= -\begin{bmatrix} a_1 & a_2 & 0 \\ 0 & a_1 & a_2 \\ 5a_2 & a_3 & 0 \end{bmatrix}\\
_3M_4= -\begin{bmatrix} 3a_1 & a_2 & 2a_3/7 & 0 \\ 0 & a_1 & a_2 & a_3 \\ 0 & 0 & a_1 & a_2\\14a_2 & a_3 & 0 & 0 \end{bmatrix}\\
_3M_5= -\begin{bmatrix} 7a_1 & -a_2 & -37a_3/126 & 0 & 0\\0 & 3a_1 & a_2 & 2a_3/7 & 0 \\ 0& 0 & a_1 & a_2 & a_3 \\0 & 0 & 0 & a_1 & a_2\\42a_2 & a_3 & 0 & 0 & 0 \end{bmatrix}
\end{equation}

# General Series

We could also attempt this from the infinite limit, and then set the corresponding coefficients to 0 afterwards. \begin{equation}
P(x) = a_0 + a_1 x + a_2 x^2 + \cdots = \sum_{k=0}^\infty a_kx^k
\end{equation} then we can describe the inverse term by \begin{equation}
P^{-1}(x) = \sum_{n=1}^\infty \frac{1}{na_1^n}\left(\sum_{s,t,u,\cdots}(-1)^{s+t+u+\cdots}\frac{n(n+1)\cdots(n-1+s+t+u\cdots)}{s!t!u!\cdots}\left(\frac{a_2}{a_1}\right)^s\left(\frac{a_3}{a_1}\right)^t\cdots\right)(x-a_0)^n
\end{equation} where the sum is over all combinations such that, *s* + 2*t* + 3*u* + ⋯ = *n* − 1 (all positive integers). [Morse Feschbach 1953] For example, when *n* = 1, the only combination is for *s* = *t* = *u* = ⋯ = 0. When *n* = 2, then then only combination is *s* = 1, for *n* = 3 we may have either, *s* = 2, *t* = 0 or *s* = 0, *t* = 1, giving two terms etc. The goal at this stage would be to develop a regular family of *n* × *n* matrices that can express these rules through their determinants. We may find that the matrices are made of the sum or product of matrices expressing each term.

A Lambert like Transformation with a Logarithmic Kernel