TITLE: A followup question on surreal numbers and Collatz iteration game
QUESTION [1 upvotes]: This question is not a duplicate of the previous, because here we use a different function and get different values for the games:
Let us define a game based on the modified Collatz function $T(n) = n/2$ if $n$ is even, otherwise $=(3n+1)/2$.
Each number $n$ represents a game played by left $L$ and right $R$:
$$g(n) = \{L_n | R_n \}$$
The rules of the game are:
If $n$ is even, the left player can play the game:
$$g(n) = \{g(n/2) |  \}$$
If $n \neq 1$ is odd, the right player can play the game:
$$g(n) = \{| g((3n+1)/2) \}$$
If $n=1$, the game is over as no player can make a move:
$$g(n)=g(1) = \{| \}$$
Assuming the Collatz conjecture is true, this game starting with an arbitrary number $n$, has always an end.
So these games define surreal numbers and we could add and multiply, negate those games, which I think would be fun.
Q: How do I find the numeric value of these surreal numbers and what interpretation can be given to these values?
Here are some examples, but I do not see any pattern, where $g(n)$ denotes the numerical value of $n$:
$n,g(n)$
1 0
2 1
3 -1
4 2
5 0
6 0
7 -2
8 3
9 -1
10 1
11 -1
12 1
13 0
14 0
15 -3
16 4
17 0
18 0
19 -1
20 2
21 0
22 0
23 -2
24 2
25 -1
26 1

Some more examples in SAGEMATH
Related question:  Conway's surreal numbers and the Collatz iteration as a game?

REPLY [1 votes]: Setup
I have a characterization, but the answer is pretty complicated. These
definitions depend on the Collatz conjecture being true.
I will define a function related to the Collatz conjecture that associates
strings of "$\mathrm{E}$"s and "$\mathrm{O}$"s to the positive
integers. I will use multiplicative notation for concatenating strings,
so that $\mathrm{E}\left(\mathrm{OE}\right)=\mathrm{EOE}$ and $\mathrm{E}^{3}=\mathrm{EEE}$.
Define $C\left(1\right)$ to be the empty string. Then, for $n>1$,
define $C(n)=\begin{cases}
\mathrm{E}C\left(\frac{n}{2}\right) & \text{if }n\text{ is even}\\
\mathrm{O}C\left(\frac{3n+1}{2}\right) & \text{if }n\text{ is odd}
\end{cases}$. For example, $C\left(7\right)=\mathrm{OOOEOEEOEEE}$, with the tail
$\mathrm{OEEE}$ coming from $5\overset{\mathrm{O}}{\to}8\overset{\mathrm{E}}{\to}4\overset{\mathrm{E}}{\to}2\overset{\mathrm{E}}{\to}1)$.
We can abbreviate this as $C\left(7\right)=\mathrm{O}^{3}\mathrm{E}^{1}\mathrm{O}^{1}\mathrm{E}^{2}\mathrm{O}^{1}\mathrm{E}^{3}$.
Answer
We can express the values of $g$ in terms of $C$ in a relatively
simple way. Note that $g\left(1\right)=0$ by definition. For $n>1$,
we have the following piecewise formula, where $m\ge1$ is the length
of the initial run and $k\ge0$:
$$
g\left(n\right)=\begin{cases}
\begin{cases}
m-1 & \text{if }C\left(n\right)=\mathrm{E}^{m}\left(\mathrm{OE}\right)^{k}\mathrm{O}^{2}\cdots\\
m & \text{if }C\left(n\right)=\mathrm{E}^{m}\cdots\text{ otherwise}
\end{cases} & \text{if }n\text{ is even}\\
\begin{cases}
1-m & \text{if }C\left(n\right)=\mathrm{O}^{m}\left(\mathrm{EO}\right)^{k}\mathrm{E}^{2}\cdots\\
-m & \text{if }C\left(n\right)=\mathrm{O}^{m}\cdots\text{ otherwise}
\end{cases} & \text{if }n\text{ is odd}
\end{cases}
$$
In other words, using $\mathrm{A}$ and $\mathrm{B}$ as stand-ins
for $\mathrm{E}$ and $\mathrm{O}$ in some order, and $\left[\mathrm{A}=\mathrm{O}\right]=\begin{cases}
0 & \text{if }\mathrm{A}=\mathrm{E}\\
1 & \text{if }\mathrm{A}=\mathrm{O}
\end{cases}$, we have:
$$
g\left(n\right)=\left(-1\right)^{\left[\mathrm{A}=\mathrm{O}\right]}*\begin{cases}
m-1 & \text{if }C\left(n\right)=\mathrm{A}^{m}\left(\mathrm{BA}\right)^{k}\mathrm{B}^{2}\cdots\\
m & \text{otherwise}
\end{cases}
$$
Proof
Since $C\left(n\right)$ determines $n$, I will define $G:\text{strings}\to\mathbb{N}$
so that $g\left(n\right)=G\left(c\right)$ if $c=C\left(n\right)$.
In order to show that the above formula for $g$/$G$ is correct for
$n>1$, we can induct on the number of alternations between $\mathrm{O}$
and $\mathrm{E}$.
Base cases
Firstly, $G\left(\mathrm{E}^{m}\right)=m$ (including if $m=0$) by
a straightforward induction on $m$. And secondly, for $j\ge1$, $$G\left(\mathrm{O}^{j}\mathrm{E}^{m}\right)=\left\{ \mid G\left(\mathrm{O}^{j-1}\mathrm{E}^{m}\right)\right\} =\begin{cases}
G\left(\mathrm{O}^{j-1}\mathrm{E}^{m}\right)-1 & \text{if }G\left(\mathrm{O}^{j-1}\mathrm{E}^{m}\right)\le0\\
0 & \text{if }G\left(\mathrm{O}^{j-1}\mathrm{E}^{m}\right)>0
\end{cases}\text{.}$$ Note that we must have $m\ge3$ (as $1,2,4$ can't be reached from
an odd number $n>1$). Therefore, $G\left(\mathrm{E}^{m}\right)>0$
and $G\left(\mathrm{O}\mathrm{E}^{m}\right)=0$. By induction on $j$,
$G\left(\mathrm{O}^{j}\mathrm{E}^{m}\right)=1-j$. This agrees with
the formula since $m\ge3$, so this has the form $G\left(\mathrm{O}^{j}\left(\mathrm{EO}\right)^{0}\mathrm{E}^{2}\cdots\right)$.
Induction step
E
Assume, for induction purposes, that the formula for $G$ holds for
all strings with a given number of alternations that begin with $\mathrm{E}$,
so that $G\left(\mathrm{E}^{\ell}\cdots\right)=\begin{cases}
\ell-1 & \text{if }\mathrm{E}^{\ell}\left(\mathrm{OE}\right)^{k}\mathrm{O}^{2}\cdots\\
\ell & \text{otherwise}
\end{cases}$.
Then we have $$G\left(\mathrm{O}\mathrm{E}^{\ell}\cdots\right)=\begin{cases}
\begin{cases}
-1 & \text{if }\mathrm{O}\mathrm{E}^{\ell}\left(\mathrm{OE}\right)^{k}\mathrm{O}^{2}\cdots\\
0 & \text{otherwise}
\end{cases} & \text{if }\ell=1\\
0 & \text{if }\ell>1
\end{cases}=\begin{cases}
-1 & \text{if }\left(\mathrm{OE}\right)^{k+1}\mathrm{O}^{2}\cdots\\
0 & \text{otherwise}
\end{cases}\text{.}$$ And $G\left(\mathrm{E}\mathrm{O}\mathrm{E}^{\ell}\cdots\right)=\begin{cases}
0 & \text{if }\mathrm{E}\left(\mathrm{OE}\right)^{k+1}\mathrm{O}^{2}\cdots\\
1 & \text{otherwise}
\end{cases}$. Since those values are nonnegative, we can conclude that $G\left(\mathrm{E}^{m}\mathrm{O}\mathrm{E}^{\ell}\cdots\right)=\begin{cases}
m-1 & \text{if }\mathrm{E}^{m}\left(\mathrm{OE}\right)^{k+1}\mathrm{O}^{2}\cdots\\
m & \text{otherwise}
\end{cases}$.
For higher powers of $\mathrm{O}$, now consider $G\left(\mathrm{O}^{2}\mathrm{E}^{\ell}\cdots\right)=\begin{cases}
-2 & \text{if }\mathrm{O}\left(\mathrm{OE}\right)^{k+1}\mathrm{O}^{2}\cdots\\
-1 & \text{otherwise}
\end{cases}$. And so, for $j\ge2$, $G\left(\mathrm{O}^{j}\mathrm{E}^{\ell}\cdots\right)=\begin{cases}
-j & \text{if }\mathrm{O}^{j-1}\left(\mathrm{OE}\right)^{k+1}\mathrm{O}^{2}\cdots\\
1-j & \text{otherwise}
\end{cases}$. Since these are all negative, $G\left(\mathrm{E}\mathrm{O}^{j}\mathrm{E}^{\ell}\cdots\right)=0$,
and $G\left(\mathrm{E}^{m}\mathrm{O}^{j}\mathrm{E}^{\ell}\cdots\right)=m-1$.
Note that $\mathrm{E}^{m}\mathrm{O}^{j}\mathrm{E}^{\ell}\cdots$ has
the form $\mathrm{E}^{m}\left(\mathrm{OE}\right)^{0}\mathrm{O}^{2}\cdots$,
so this agrees with our formula.
O
The induction steps for adding two alternations that begin with $\mathrm{O}$
are completely analogous. Just flip signs and swap $\mathrm E$ and $\mathrm O$.
Commentary
The formula for $g$ depends on $C$, which basically requires us
to run through an unbounded amount of steps of the Collatz sequence
for a number, which is not ideal. It also doesn't have an obvious
tidy characterization that's not recursive, like "write the $\mathrm{EO}$
string with $1$s and $0$s in a certain way, and then do something
with the number to get the value of $G$".
Simple cases
The only thing I could think to do is collect together some simple
cases that have a fixed form; but nothing can be done for the general
case as that essentially requires you to prove Collatz.
For example, the numbers with $C\left(n\right)=\mathrm{E}^{m}$ are
the powers of $2$, with $g\left(2^{m}\right)=m$. These are A000079 in the OEIS.
The numbers of the form $\mathrm{O}\mathrm{E}^{n}$ are those of the
form $(4^{k}-1)/3$ for $k\ge1$ (A002450), and they
have $g\left(n\right)=0$.
The numbers of the form $\mathrm{O}^{2}\mathrm{E}^{m}$ are those
of the form $\left(64^{k}-10\right)/18$ for $k\ge1$ (A228871),
and they have $g\left(n\right)=-1$.
The numbers of the form $E^{m}\mathrm{O}\mathrm{E}^{\ell}$ are those
of the form $2^{m}(4^{k}-1)/3$ for $k\ge1$ (A181666),
and have $g\left(n\right)=m$.
The numbers of the form $E^{m}\mathrm{O^{2}}\mathrm{E}^{\ell}$ are
those of the form $2^{m}\left(64^{k}-10\right)/18$ for $k\ge1$ (no
OEIS entry), and have $g\left(n\right)=m-1$.
The numbers of the form $\mathrm{O}\mathrm{E}^{m}\mathrm{O}\mathrm{E}^{\ell}$
have one of the two forms $\dfrac{4^{3k+r}-4^{r+1}-6}{18}$ or $\dfrac{4^{3k+r}-4^{r+2}-48}{144}$,
etc.
Simpler Code
Your Sage code appears to be doing all of the work of the Conway notation. But since these games only have one move for one of the two players, you can calculate values of $g$ in a much more straightforward way. I imagine you could port the following code to Sage if desired:
Wolfram Language (Mathematica)
g[x_] := g[x] = If[x == 1, 0,
  If[Mod[x, 2] == 0, If[g[x/2] >= 0, g[x/2] + 1, 0],
   If[g[(3 x + 1)/2] <= 0, g[(3 x + 1)/2] - 1, 0]]];
Table[g[x],{x,1,10000}]//Print

Try it online!


TITLE: question regarding the geometric meaning of eigenvalues and eigenvectors
QUESTION [0 upvotes]: Ok so I've known how to get eigenvalues and eigenvectors for a while, but am becoming more interested in a 'simple' explanation of what actually is going on. I've looked up things on Google etc. What's confusing me is that it says x is an eigenvector of A if the image of x which is $\lambda x$ is parallel to x. Fine, but then even on wikipedia it shows a small graph where $Ax$ is lying along the vector x but it is just longer or stretched by $\lambda$. This obviously isn't parallel to x ? But x is an eigenvector?

REPLY [2 votes]: Parallel in terms of linear algebra means that the two vectors have the same direction.  This can be characterised in $\mathbb{R}^3$ or $\mathbb{R}^2$ as the cross product of the two vectors being zero.  This is equivalent to the two vectors having the same direction, as
$$||a \times b|| = ||a|| ||b||\sin \vartheta \text{,}$$
where $\vartheta$ is the angle between the two vectors (note that this definition of parallel also implies that the zero vector is parallel to all vectors).
Also note that vectors can be picked up and moved around, so two vectors lying on top of eachother, as in your example, can be made parallel in the traditional sense by simply picking one up and moving it a bit to the side.
Another way to think geometrically of eigenvectors, which makes sense in all dimensions, not just 2 or 3, is that they are scaled by the linear operator (the matrix $A$) by the value $\lambda$ (the eigenvalue).  So eigenvectors are those that have only their magnitude changed by $A$, and not their direction.


TITLE: Ricci form is closed?
QUESTION [2 upvotes]: Let $(M,g,J)$ be an almost Kähler manifold and let $\rho$ denote its Ricci form
$$
\rho(X,Y) = \operatorname{ric}^{\mbox{c}}(JX,Y)
$$
where $\operatorname{ric}^{\mbox{c}}$ is the $J$-invariant part of the Ricci tensor of the Riemannian manifold $(M,g)$.
If $\rho(X,JY)=\operatorname{ric}(X,Y)$ (i.e. $(M,g,J)$ has $J$-invariant Ricci tensor), is it true that $\rho$ is a closed $2$-form?
A second question is:
what are well known results from the theory of Kähler manifolds valid in almost Kahler manifolds with $J$-invariant Ricci tensor?

REPLY [3 votes]: For a Kähler manifold $(M^{2n}, g, J)$, the Ricci 2-form $\rho(X, Y):={\rm Ric}(JX, Y)$ is closed, of type $(1, 1)$, and represents (up to a scaling factor $\frac{1}{2\pi}$) the de Rham cohomology of the first Chern class of $(M, J)$.
For generic almost-Kähler manifolds (i.e. the almost complex structure $J$ is not integrable), none of these properties remains true for the tensor $\rho(X, Y)={\rm Ric}'(JX, Y)$, where ${\rm Ric}'$ denotes the $J$-invariant part of the Ricci tensor.
For details on almost Kähler manifolds with $J$-invariant Ricci tensor, see the paper of T. Draghici, Almost Kähler 4-manifolds with $J$-invariant Ricci tensor, Houston J. Math. 25
(1999), 133–145, or  papers of V. Apostolov, or K. Sekigawa.


TITLE: Solving the differential equation $x x''(t)=\frac{1}{t^3-t}$
QUESTION [8 upvotes]: Solve the following differential equation
$$x x''(t)=\frac{1}{t^3-t}$$
I tried to integrate both members:
$$x(t) x'(t)-\int [x'(t)]^2 dt=\frac{1}{t}+\log\left|\frac{1}{t}-1\right|$$
but the situation does not improve. At this point, I think there is a mistake in the text of exercise.
I would appreciate some help with this problem.
Thank you very much.

REPLY [0 votes]: Hint:
$xx''(t)=\dfrac{1}{t^3-t}$
$x\dfrac{d^2x}{dt^2}=\dfrac{1}{t^3-t}$
$x\dfrac{d}{dt}\left(\dfrac{dx}{dt}\right)=\dfrac{1}{t^3-t}$
$x\dfrac{d}{dx}\left(\dfrac{1}{\dfrac{dt}{dx}}\right)\dfrac{dx}{dt}=\dfrac{1}{t^3-t}$
$-x\dfrac{\dfrac{d^2t}{dx^2}}{\left(\dfrac{dt}{dx}\right)^3}=\dfrac{1}{t^3-t}$
$t(1-t^2)x\dfrac{d^2t}{dx^2}=\left(\dfrac{dt}{dx}\right)^3$


text:
\begin{document}

\maketitle

\begin{abstract}
In the present paper, we are concerned with subordination problems related to $\lambda$-Robertson function. The radii of $\lambda$-spirallikeness and starlikeness of $\lambda$-Robertson function are also determined.

\end{abstract}

\footnote[0]{Primary 30C45, Secondary 30C62, 30C75}
\footnote[0]{Key words and Phrases: Robertson functions, spirallike functions, subordination theory, radii of spirallikeness and starlikeness. }

\section{Introduction and Main Results}
Let $\mathbb{D}_{r}=\{z\in \mathbb{C}:\, |z|<r\}$ for $0<r\leq 1$ and $\mathbb{D}=\mathbb{D}_{1}$ be the unit disc. Let $\mathcal{A}$ be the family of functions $f$ analytic in $\mathbb{D}$, and $\mathcal{A}_{1}$ be the subset of $\mathcal{A}$ consisting of functions $f$ which are normalized by $f(0)=f'(0)-1=0$.
A function $f\in\mathcal{A}$ is said to be subordinate to a function $F\in\mathcal{A}$ in $\mathbb{D}$ (in symbols $f\prec F$ or $f(z)\prec F(z)$) if there exists an analytic function $\omega(z)$ on $\mathbb{D}$ with $|\omega(z)|<1$ and $\omega(0)=0$, such that
\[
f(z)=F(\omega(z))
\]
in $\mathbb{D}$. When $F$ is a univalent function, the condition $f\prec F$ is equivalent to $f(\mathbb{D})\subseteq F(\mathbb{D})$ and $f(0)=F(0)$.
 Let
\[
\mathcal{P}_{\lambda}=\left\{p\in\mathcal{A} :\,  p(0)=1,\,\, \Re e^{-i \lambda}p(z)>0 \right\}.
\]
Here and hereafter we always suppose $-\pi/2<\lambda<\pi/2$. Note that $\mathcal{P}_{\lambda}$ is a convex and compact subset of $\mathcal{A}$ which is equipped with the topology of uniform convergence on compact subsets of $\mathbb{D}$.  Since $\mathcal{P}_{0}$ is the well-known Carath{\'e}odory class, we call $\mathcal{P}_{\lambda}$ \textit{the tilted Carath{\'e}odory class by angel $\lambda$}. Some characterizations and estimates of elements in $\mathcal{P}_{\lambda}$ are known (for a short survey, see \cite{Wang}).

For a function $f\in\mathcal{A}$, let
\begin{equation}\label{Q(z)}
Q_{f}(z)=\frac{zf'(z)}{f(z)}
\end{equation}
and
\begin{equation*}
P_{f}(z)=1+\frac{zf''(z)}{f'(z)}.
\end{equation*}
It is worthwhile to note that
\[
Q_{f}(z)+\frac{zQ_{f}'(z)}{Q_{f}(z)}=P_{f}(z).
\]
These quantities are important for investigation of geometric properties of analytic functions. Next we will define two subclasses of analytic functions related to these two quantities.

A function $f\in\mathcal{A}_{1}$ is said to be a \textit{$\lambda$-spirallike function} (denoted by $f\in\mathcal{SP}_{\lambda}$) if
\[
Q_{f}\in \mathcal{P}_{\lambda}.
\]
Note that $\mathcal{SP}_{0}$ is precisely the set of starlike functions normally denoted by $\mathcal{S}^{*}$. Sprirallike functions were introduced and proved to be univalent by \v Spa\v cek \cite{Spacek} in 1932.
For general references about spirallike functions, see e.g. \cite{Duren} or \cite{AhujaSilverman}.

A function $f\in\mathcal{A}_{1}$ is said to be a \textit{$\lambda$-Robertson function} if $zf'(z)\in\mathcal{SP}_{\lambda}$, i.e.
\[
P_{f}\in \mathcal{P}_{\lambda}.
\] Let $\mathcal{R}_{\lambda}$ denote the set of these functions. Note that $\mathcal{R}_{0}$ is precisely the set of convex functions sometimes denoted by $\mathcal{K}$. Convex functions have been the subject of numerous investigations, among which the following result was proved by MacGregor \cite{MacGregor} in 1975.\\

\noindent{\bf Theorem A.}
\textit{\,\,Let $f\in \mathcal{K}$, then the subordination relation
\[
\frac{zf'(z)}{f(z)}\prec \frac{zf'_{0}(z)}{f_{0}(z)}
\]
holds, where $f_{0}(z)=z/(1-z)$}.\\

We are interested in more general subordination problems related to $\lambda$-Robertson functions. For this purpose, we first introduce some specific functions for convenience.

A distinguished member of $\mathcal{R}_{\lambda}$ is
\begin{equation*}
f_{\lambda}(z)=\frac{(1-z)^{1-2e^{i\lambda}\cos\lambda}-1}{2e^{i\lambda}\cos\lambda -1}.
\end{equation*}
A simple calculation yields
\begin{equation}\label{q(lambda)}
Q_{\lambda}(z)=\frac{zf'_{\lambda}(z)}{f_{\lambda}(z)}=\frac{e^{2i\lambda}z}{1-z-(1-z)^{1+e^{2i\lambda}}},
\end{equation}

\begin{equation*}
P_{\lambda}(z)=1+\frac{zf''_{\lambda}(z)}{f'_{\lambda}(z)}=\frac{1+e^{2i\lambda}z}{1-z}
\end{equation*}
and
\begin{equation}\label{differential2}
Q_{\lambda}(z)+\frac{zQ_{\lambda}'(z)}{Q_{\lambda}(z)}=P_{\lambda}(z).
\end{equation}

In \cite{KimSrivastava}, Kim and Srivastava posed the open problem which is an extension of Theorem A whether
\[
\frac{zf'(z)}{f(z)}\prec \frac{zf_{\lambda}'(z)}{f_{\lambda}(z)}
\]
holds for $f\in\mathcal{R}_{\lambda}$ with general $\lambda$. In other words, if
\begin{equation}\label{broit-bouquet}
q(z)+\frac{zq'(z)}{q(z)}\prec P_{\lambda}(z)
\end{equation}
in $\mathbb{D}$, then whether
\begin{equation*}
q\prec Q_{\lambda}
\end{equation*}
holds in $\mathbb{D}$ for general $\lambda$? Relation ($\ref{broit-bouquet}$) is a kind of Briot-Bouquet differential subordinations which have a surprising number of important applications in the theory of univalent functions. Many sources and references are given in \cite{MillerMocanu}.
\par In the present paper, we solve the above problem in a restricted disc and obtain the radii of spirallikeness and starlikeness for Robertson funtions as well.

\begin{thm}\label{theorem 1}
Let $q\in \mathcal{A}$ with $q(0)=1$ satisfy the differential subordination
\begin{equation}\label{p(z)}
q(z)+\frac{zq'(z)}{q(z)}\prec P_{\lambda}(z)
\end{equation}
and $R_{1}(\lambda)$ be defined by
\begin{equation}\label{R_{1}}
R_{1}(\lambda)=\sup\{r<1\,: Q_{\lambda}(rz)\prec P_{\lambda}(rz)\,\, in \, \mathbb{D}\}.
\end{equation}
Then
\[
q(z)\prec Q_{\lambda}(z)
\]
in $|z|<R_{1}(\lambda)$.
\end{thm}

By the discussion in Section 1, we can deduce the following corollary immediately from Theorem 1.
\begin{cor}\label{corollary 1}
Let $f\in \mathcal{R}_{\lambda}$, then
\[
\frac{zf'(z)}{f(z)}\prec \frac{zf_{\lambda}'(z)}{f_{\lambda}(z)}
\]
in $|z|<R_{1}(\lambda)$, where $R_{1}(\lambda)$ is given in \eqref{R_{1}}.
\end{cor}

\begin{remark}
The radius of $\lambda$-spirallikeness of $\lambda$-Robertson functions is at least $R_{1}(\lambda)$, since
\[
\frac{zf'(z)}{f(z)}\prec \frac{zf'_{\lambda}(z)}{f_{\lambda}(z)}\prec P_{\lambda}(z)
\]
in $\mathbb{D}_{R_{1}(\lambda)}$ for any $f\in\mathcal{R}_{\lambda}$.
\end{remark}

\begin{remark}

For $0\leq r<1$, let
\begin{equation*}
\begin{split}
\psi_{\lambda}(r)
&=\max_{|z|=1}|P_{\lambda}^{-1}(Q_{\lambda}(rz))/r|\\
&=\max_{|z|=1}\left|\frac{1}{r} \frac{mrz-1+(1-rz)^m}{1-(1-rz)^m}\right|\\
\end{split}
\end{equation*}
where $m=1+e^{2i\lambda}$. $\psi_{\lambda}(r)$ is an increasing function defined on $[0,1)$ with $\psi_{\lambda}(0)=0$. By the definition of subordination, $R_{1}(\lambda)$ defined in ($\ref{R_{1}}$) could be expressed in terms of $\psi_{\lambda}$:
\[
R_{1}(\lambda)=\sup\{r<1\,: \psi_{\lambda}(r)<1\}=\psi_{\lambda}^{-1}(1).
\]
\end{remark}

\begin{thm}\label{theorem 2}
Let $q\in \mathcal{A}$ with $q(0)=1$ satisfy the differential subordination
\begin{equation}
q(z)+\frac{zq'(z)}{q(z)}\prec P_{\lambda}(z),
\end{equation}
then
\[
q(R_{2}z)\prec P_{0}(z)
\]
in $\mathbb{D}$, where
\begin{equation}\label{R_{2}}
R_{2}:=R_{2}(\lambda)=\frac{2}{\sqrt{4+2\sqrt{3}|\sin(2\lambda)|}}.
\end{equation}
\end{thm}

\begin{cor}\label{corollary 2}
The radius of starlikeness of $\lambda$-Robertson functions is at least $R_{2}(\lambda)$ given in \eqref{R_{2}}.
\end{cor}

Note that $R_{1}(0)=R_{2}(0)=1$, thus both Corollary 1 and Corollary 2 imply Theorem A. Note also that in \cite{AhujaSilverman}, Ahuja and Silverman posed the problem to find the radius of starlikeness for all $\lambda$-Robertson functions. Corollary \ref{corollary 2} implies that this radius is at least
\[
R_{2}=\min\{R_{2}(\lambda):\,-\pi/2<\lambda<\pi/2\}=R_{2}(\pi/4)=\sqrt{3}-1\approx 0.732.
\]
Libera and Ziegler in \cite{Libera-Ziegler} have shown that the radius of close-to-convexity for all $\lambda$-Robertson functions is approximately 0.99097524 and the radius of convexity is $\sqrt{2}/2$.

\section{Proofs of Resluts}

In order to obtain our main results, the following lemmas are required.

\begin{lem}[\cite{Wang}]\label{lemma1}
Let $p\in \mathcal{P}_{\lambda}$, then we have
\begin{equation*}
\left|p(z)-A(r)\right|\leq B(r),
\end{equation*}
where
\[
A(r)=\frac{1+r^2e^{2i\lambda}}{1-r^2},\,\,
B(r)=\frac{2r\cos \lambda}{1-r^2}
\]
and $r=|z|<1$. Equality holds if and only if $p(z)=P_{\lambda}(xz)$ with $|x|=1$.
\end{lem}

\begin{lem}[\cite{MillerMocanu}, Lemma 2.2d]\label{lemma2}
Let $g(z)$ and $h(z)$ be in $\mathcal{A}$ with $g(0)=h(0)$. If $g\not\prec h$ in $\mathbb{D}$, then there exist two points $z_{0}$ with $|z_{0}|<1$ and $\eta_{0}$ with $|\eta_{0}|=1$ and $s\geq 1$ such that
\[
g(\mathbb{D}_{|z_{0}|})\subset h(\mathbb{D}),
\]
\[
g(z_{0})=h(\eta_{0})
\]
and
\[
z_{0}g'(z_{0})=s \eta_{0}h'(\eta_{0}).
\]
\end{lem}

The next lemma is due to Nunokawa \cite{Nunokawa}. We only quote the relevant part.
\begin{lem}[\cite{Nunokawa}]\label{lemma3}
Let $p(z)\in\mathcal{A}$ satisfy $p(0)=1$ and $p(z)\not=0$ in $\mathbb{D}$. If there exsits a point $z_{0}\in\mathbb{D}$ such that $\Re p(z)>0$ in $|z|<|z_{0}|$ and $p(z_{0})=i a$ where $a\in\mathbb{R}\setminus \{0\}$, then
\[
\frac{z_{0}p'(z_{0})}{p(z_{0})}=ik
\]
where $k\geq (a+1/a)/2$ if $a>0$ and $k\leq -(a+1/a)/2$ if $a<0$.
\end{lem}

\noindent\textit{Proof of Theorem 1.} For simplicity, we let $R=R_{1}(\lambda)$ and $p(z)=q(z)+zq'(z)/q(z)$, thus $p\in\mathcal{P}_{\lambda}$. If $q(z)\not\prec Q_{\lambda}(z)$ in $|z|<R$, then Lemma \ref{lemma2} implies the existences of $z_{0}$ with $|z_{0}|<R$, $\eta_{0}$ with $|\eta_{0}|=R$ and $s\geq 1$ such that
\begin{equation}{\label{M-M condition}}
\begin{split}
q(\mathbb{D}_{|z_{0}|})\subset Q_{\lambda}(\mathbb{D}),\\
q(z_{0})=Q_{\lambda}(\eta_{0}),\\
z_{0}q'(z_{0})=s \eta_{0}Q_{\lambda}'(\eta_{0}).\\
\end{split}
\end{equation}

Thus in view of  \eqref{differential2}, $(\ref{p(z)})$ and $(\ref{M-M condition})$, we have
\begin{equation}{\label{equality}}
\begin{split}
p(z_{0})
&=q(z_{0})+\frac{z_{0}q'(z_{0})}{q(z_{0})}\\
&=Q_{\lambda}(\eta_{0})+s\frac{\eta_{0}Q_{\lambda}'(\eta_{0})}{Q_{\lambda}(\eta_{0})}\\
&=sP_{\lambda}(\eta_{0})+(1-s)Q_{\lambda}(\eta_{0}).
\end{split}
\end{equation}

Therefore ($\ref{equality}$) and Lemma $\ref{lemma1}$ show that
\begin{eqnarray*}
|p(z_{0})-A(R)|
&=&
|s(P_{\lambda}(\eta_{0})-A(R))-(1-s)(Q_{\lambda}(\eta_{0})-A(R))|\\
&\geq &
|s(P_{\lambda}(\eta_{0})-A(R))|-(s-1)|(Q_{\lambda}(\eta_{0})-A(R))|\\
&\geq&
sB(R)-(s-1)B(R)\\
&=&
B(R)
\end{eqnarray*}
which contradicts with $p(z)\in\mathcal{P}_{\lambda}$. Therefore we get the assertion.
\qed

\noindent\textit{Proof of Theorem 2.} For simplicity, we let $R=R_{2}(\lambda)$ and $p(z)=q(z)+zq'(z)/q(z)$, thus $p\in\mathcal{P}_{\lambda}$. If $q(Rz)\not\prec P_{0}(z)$ in $\mathbb{D}$, it follows from $q(0)=1$ that there esists a point $z_{0}\in\mathbb{D}$ such that $\Re q(Rz)>0$ for $|z|<|z_{0}|$ and $q(Rz_{0})=ia$ where $a\in\mathbb{R}\setminus \{0\}$, then by Lemma \ref{lemma3}, we have
\[
\frac{Rz_{0}q'(Rz_{0})}{q(Rz_{0})}=ik,
\]
where $k\geq (a+1/a)/2$ if $a>0$ and $k\leq -(a+1/a)/2$ if $a<0$. Therefore
\[
p(Rz_{0})=q(Rz_{0})+\frac{Rz_{0}q'(Rz_{0})}{q(Rz_{0})}=ia+ik,
\]
which implies $p(Rz_{0})\in\Omega$ since $|a+k|\geq \sqrt{3}$, where $\Omega=\{it,|t|\geq\sqrt{3}\}$. Next we will show that
\[
p(\mathbb{D}_{R})\cap \Omega=\varnothing,
\]
which contradicts the above assertion. Since $p\in\mathcal{P}_{\lambda}$, it is sufficient to prove for functions $P_{\lambda}(z)$. Suppose that there is a point $z_{1}\in\mathbb{D}$ such that $P_{\lambda}(z_{1})=it_{0}$ with $|t_{0}|\geq\sqrt{3}$, then a simple calculation gives that
\[
z_{1}=\frac{it_{0}-1}{it_{0}+e^{2i\lambda}}.
\]
Hence
\[
|z_{1}|^2=\frac{t_{0}^2+1}{t_{0}^2+1-2t_{0}\sin(2\lambda)}\geq\frac{4}{4+2\sqrt{3}|\sin(2\lambda)|}=R
\]
since $|t_{0}|\geq\sqrt{3}$. Therefore $P_{\lambda}(\mathbb{D}_{R})\cap\Omega=\varnothing$. The proof is completed.
 \qed

\[
\text{Acknowledgements}
\]
The author is grateful to Professor Toshiyuki Sugawa for his guidance and useful discussions during the preparation of this paper.

config: arxiv


TITLE: Modulo multiplication of cyclic groups that are divisors of 24
QUESTION [2 upvotes]: I recently came across this reference in Wolfram Mathword regarding the modulo multiplication of cyclic groups ($M_n$)

The only ordered n for which the elements of $M_n$ are all
  self-conjugate are the divisors of $24: 1, 2, 3, 4, 6, 8, 12, 24 (OEIS
> A018253; Eggar 2000)$. These correspond to the groups$ <e>, C_2,
> C_2×C_2,$ and  $C_2×C_2×C_2.$ This also means that no modulo
  multiplication group is isomorphic to a direct product of more than
  three copies of $C_2$.

The Eggar (2000) result is this:
"The divisors of 24 greater than 1 are the only positive integers n with the property $m^2 = 1 (mod n)$ for all integer m coprime to n."
Q1: What does self-congancy $m^2=1$ actually mean in this context?  I get the math, for example in
$$Z_{12}^* = {1,5,7,11}$$
the square of all these numbers modulo-12 is $1$, but what implications or deductions follows from this?
Q2: More specifically, how does this self-conjugancy apply to models or theories built from cyclic groups? Naively, I'm thinking that any theory that needs inverses would be restricted by this since if the generators don't invert, then not all elements in the group invert.
Q3: More specific yet (or I may just be repeating myself), how should we view the limit $C_2×C_2×C_2$ and what deductions or implications can we make from this?
Any insights welcome and thank you in advance.

REPLY [1 votes]: If the bounty description means you'd like to know why the "self-inverse" condition (or what I will call the "involutory" condition) forces $m$ to be a divisor of $24$, then that question is easy to answer.
The Chinese remainder theorem implies that if $m = p_1^{e_1} p_2^{e_2} \ldots p_k^{e_k}$ is the prime factorization of $m$, then we have a canonical ring isomorphism
$$\mathbb{Z}/(m) \cong \mathbb{Z}/(p_1^{e_1}) \times \mathbb{Z}/(p_2^{e_2}) \times \ldots \times \mathbb{Z}/(p_k^{e_k})$$
which immediately implies an isomorphism of unit groups (groups of invertible elements)
$$(\mathbb{Z}/(m))^\ast \cong (\mathbb{Z}/(p_1^{e_1}))^\ast \times (\mathbb{Z}/(p_2^{e_2}))^\ast \times \ldots \times (\mathbb{Z}/(p_k^{e_k}))^\ast.$$
So the question boils down to: when is the group $(\mathbb{Z}/(p^e))^\ast$ ($p$ a prime) isomorphic to a product of copies of $C_2$?
There is a unique ring homomorphism $\phi: \mathbb{Z}/(p^e) \to \mathbb{Z}/(p)$; this map is surjective. It is well-known that the unit group $(\mathbb{Z}/(p))^\ast$ is cyclic of order $p-1$. (In fact any finite multiplicative group in a field is cyclic.) If $x \in \mathbb{Z}/(p^e)$ is any element such that $\phi(x)$ is a generator of $(\mathbb{Z}/(p))^\ast$, then $x$ is invertible and $p-1$ divides the order of $x$ -- and therefore the order of $x$ can't be $2$ if the prime $p$ is greater than $3$. Hence the only primes that can divide $m$ under the involutory condition are $2$ and $3$.
Let's now consider the case $p = 3$; what can the exponent $e$ be? Well, $4$ is invertible modulo $3^e$, and we can't have $4^2 \equiv 1 \; \text{mod}\; 3^e$ unless $3^e$ divides $15$. So the involutory condition would force $e = 0$ or $e = 1$.
Now consider the case $p = 2$. If $e \geq 4$, then we again have a surjective map $(\mathbb{Z}/2^e)^\ast \to (\mathbb{Z}/(16))^\ast$. The element $3 + (16) \in \mathbb{Z}/(16)$ has order $4$. If $x \in \mathbb{Z}/(2^e)$ is any element mapping to $3 + (16)$, then $x$ is invertible and $4$ must divide its order. So the involutory condition rules out $e \geq 4$ in the case $p = 2$.
Thus the involutory condition forces $m = 2^{e_1} \cdot 3^{e_2}$ where $e_1 \leq 3$ and $e_2 \leq 1$, so that $m$ divides $2^3 \cdot 3^1 = 24$.
(Incidentally: you can't actually mean $m^2 = 1$ in the OP and in the bounty description; you must have meant what I wrote in the opening paragraph. And you don't mean "groups that are divisors of 24"; you must have meant groups whose orders are divisors of 24. There are other abuses of language as well.)


TITLE: Commutativity relationship between the Malliavin derivative and the Skorkohod divergence operators.
QUESTION [3 upvotes]: On proposition $1.3.1$ of Nualart's book "The Malliavin Calculus and
  Related Topics" it's stated the following. Let $H$ be a real separable
  Hilbert space, and let $W=\{W(h),h\in H\}$ be a Gaussian isonormal
  process.
Let $u$ be a process of the form $$u=\sum_{j=1}^n F_j h_j,$$ where the
  $F_j$ are smooth random variables (in the Malliavin calculus sense)
  and the $h_j$ are elements on the Hilbert space $H$.
We have the following commutativity relationship between the Malliavin
  derivative and the divergence operators.
$$D^h(\delta(u)):=\langle D\delta(u),h\rangle_H=\langle u,
 h\rangle_H+\delta (D^h u) (\star)$$

My attempt:
Using the integration by parts formula I can see that
$$\delta(u)= \sum_{i=1}^n F_i W(h_i)-\sum_{i=1}^n \langle DF_i,h_i\rangle_H,$$
hence
$$D^h(\delta(u))=D^h\left(\sum_{i=1}^n F_i W(h_i)-\sum_{i=1}^n \langle DF_i,h_i\rangle_H\right)$$
The linearity of the Malliavin derivative allows us to write:
$$=\sum_i D^h(F_i W(h_i))-\sum_i D^h\langle DF_i,h_i\rangle_H$$
For the first term we can use the product rule and we obtain
$$=\langle h,u\rangle_H+ \sum_i (D^hF_i)W(h_i)-\color{red}{D^h\langle DF_i,h_i\rangle_H}.$$
At this point the author writes the red term as:
$$D^h\langle DF_i,h_i\rangle_H=\langle D(D^hF_i),h_i\rangle_H$$
(1) Why does this hold?
The definitions tells me:
$$D^h\langle DF_i,h_i\rangle= \langle D \langle DF_i,h_i\rangle, h\rangle=\langle D(D^{h_i}F_i),h\rangle.$$

(2) The Malliavin derivative is an operator that takes a real-valued random variable and returns us an $H$-valued random variable. Why  on $\star$ we have $D^h u:= \langle Du, h\rangle$? $u$ is not a real-valued random variable, how to take the Malliavin derivative of this "stochastic process"?

REPLY [2 votes]: Let me start by answering the second point. Nate Eldredge points to the right place in Nualart's book in a comment. Here I will just recreate the detail from there that if $G$ is a smooth $H$-valued random variable i.e.
$$G = \sum_{j=1}^n G_j v_j$$
where the $G_j$ are smooth random variables and $v_j \in H$ then e.g.
$$DG = \sum_{j=1}^n DG_j \otimes v_j$$
is valued in $H \otimes H$. One then checks in much the same way as the real-valued case that this defines a closable operator etc.
Now I turn to your first question.
Since $F_j$ is smooth, we can write $F_j = f_j(W(e_1), \dots, W(e_n))$ for $\{e_i: i = 1, \dots, n\}$ orthonormal in $H$ and $f_j$ smooth. Then
\begin{align}
D^h \langle DF_j, h_j \rangle =& D^h \left (\sum_{i=1}^n \partial_if_j(W(e_1),\dots,W(e_n)) \langle e_i, h_j \rangle \right)
\\
=& \sum_{k=1}^n \sum_{i=1}^n \partial_k \partial_i f_j(W(e_1), \dots, W(e_n)) \langle e_i, h_j \rangle \langle e_k, h \rangle
\\
=& \sum_{i=1}^n \sum_{k=1}^n \partial_i\partial_k  f_j(W(e_1), \dots, W(e_n))\langle e_k, h \rangle\langle e_i, h_j \rangle
\\
=& \sum_{i=1}^n \partial_i D^h F_j \langle e_i, h_j \rangle
\\
=& \langle DD^h F_j, h_j \rangle
\end{align}
as desired.


TITLE: Simplification of $\frac{\left(\frac{6\ln^22x}{2x}\right)}{\left(\frac{3}{2\sqrt{x}}\right)}$
QUESTION [0 upvotes]: My textbook goes from
$$\frac{\left( \frac{6\ln^22x}{2x} \right)}{\left(\frac{3}{2\sqrt{x}}\right)}$$
to:
$$\frac{6\ln^22x}{3\sqrt{x}}$$
I don't see how this is right. Could anyone explain?

REPLY [1 votes]: The textbook has rationalised the denominator for $\frac{3}{2\sqrt{x}}$ before simplifying further
i.e.
$$\frac{3}{2\sqrt{x}} = \frac{3\times \sqrt{x}}{2\sqrt{x}\times \sqrt{x}}= \frac{3\sqrt{x}}{2x}$$
Now Simplify:
$$ \frac{\left(\frac{6\ln^2 2x}{2x}\right)}{\left(\frac{3}{2\sqrt{x}}\right)}= \frac{\left(\frac{6\ln^2 2x}{2x}\right)}{\left(\frac{3\sqrt{x}}{2x}\right)} = \frac{6 \ln^2 2x}{3\sqrt{x}}$$


TITLE: Find the following limits. $\lim_{x\to e^+} (\ln x)^{\frac{1}{x-e}}$
QUESTION [1 upvotes]: Find the following limits.
$$\lim_{x\to e^+} (\ln x)^{\frac{1}{x-e}}$$
I need some clues or hints to get me started. I can't even make a step. Thanks stack!
I think I can bring the limit up to the power, since ln x is continuous near e.

REPLY [4 votes]: Write $y = (\ln(x))^{1\over x - e}$. Then $\ln(y) = {\ln(\ln(x))\over {x - e}}$
You are now in an $0\cdot \infty$ situation. What can you do?


TITLE: Show that $(P \rightarrow \lnot R) \lor (P \rightarrow Q)$
QUESTION [0 upvotes]: I'm having a lot of trouble proving that $(P \rightarrow \lnot R) \lor (P \rightarrow Q)$ is logically equivalent to $(P\land R) \rightarrow Q.$
I'm very new to this course, and unsure of what to do. Can anyone correct/guide/explain to me, please. This is what I've come up with. Thanks for your help.

$(p \rightarrow \lnot r) \lor (p \rightarrow q)$
$(\lnot p \lor \lnot r) \lor (\lnot p \lor q)$  [Implication]
$\lnot(p \land r) \lor (\lnot p \lor q)$   [De Morgan's]

I don't know what next. I'm thinking Distribution?

REPLY [0 votes]: How can $(P \to \lnot R) \lor (P \to Q)$ be made false?
Only if both terms of the "or" are, so iff $P \to \lnot R$ and $ P \to Q$ are false, and this happens only if $P$ is true and $R$ is true (so $\lnot R$ is false) and $Q$ is false.
How can $(P \land R) \to Q$ be false? Iff if $Q$ is false and $P \land R$ is true, so $P$ and $R$ are true and $Q$ is false.
So both statements are false exactly under the same truth assignments. Hence they are equivalent.


TITLE: Finding the angles a line makes with the coordinate planes
QUESTION [1 upvotes]: The line $\vec r=2\hat i-\hat j+3  \hat k+\lambda (\hat i+\hat j + \sqrt{2}\hat k)$ is defined to make angles $\alpha, \beta$ and $\gamma$ with the xy, yz, and the zx planes respectively.
I simplified the directional vector of the line to $2(\frac {\hat i}{2}+\frac{\hat j}{2} + \frac{\hat k}{\sqrt2})$, and thought of the angle the line makes with each coordinate plane as the complement of the angle it makes with the coordinate axis normal to it. That implies that the directional cosines give you the sines of the required angles. With this, I concluded that $$\alpha=45°, \beta= 30° , \gamma=30°$$
But apparently I've got $\alpha$ and $\gamma$ mixed up. Where did I mess up here?

REPLY [2 votes]: In general, the angle between a line with direction $\mathbf v$ and a plane with normal $\mathbf n$ is $$\theta=\arcsin\left|\frac{\mathbf v \cdot\mathbf n}{\left|\mathbf v\right| \,\left|\mathbf n\right|} \right|.$$
Since the $xy-,$ $yz-, zx-$ planes have respective normals $\begin{pmatrix}0\\0\\1\end{pmatrix},\begin{pmatrix}1\\0\\0\end{pmatrix},\begin{pmatrix}0\\1\\0\end{pmatrix},$ as pointed out by Ted, your own answers are indeed correct.
P.S. The above formula has a similar derivation as yesterday's formula for the angle between two planes.


TITLE: Find the value of $\lim_{n\to \infty}\sum_{i=0}^n\frac{n}{n^2+i^2}$
QUESTION [0 upvotes]: Find the value of $\lim_{n\to \infty}\sum_{i=0}^n\frac{n}{n^2+i^2}$ I can't solve this problem and I've seen that some books like Spivak or Apostol use integrals to solve it but my teacher said that we can't integrate, I would appreciate your help :)

REPLY [1 votes]: $$\sum_{i=0}^n\frac n{n^2+i^2}=\frac1n\sum_{i=0}^n\frac1{1+\left(\frac in\right)^2}\xrightarrow[n\to\infty]{}\int\limits_0^1\frac1{1+x^2}dx=\ldots$$


TITLE: Why is the subscheme fixed if I fix the Hilbert Polynomial?
QUESTION [0 upvotes]: Let $Y$ be a closed subscheme of $X=\mathbb P{^n}$ with
Hilbert Polynomial given by $h(m) =\binom{m+r}{r}$

Prove that $Y$ is isomorphic to $\mathbb P^r$ embedded linearly inside $X$.

Can someone help me solve this? I know for $\mathbb P^r$ that is indeed the Hilbert Polynomial.
Edit : The context of this problem is a step towards the construction of Hilbert Schemes as done in FGA explained (chapter 2).
Looking at the Hilbert polynomial I can say that the closed subscheme has degree 1 and dimension r. Let $I$ be the ideal of $Y$, I would like to show that it is generated by a single linear polynomial(?). I tried to prove that it can't be generated by polynomials of degree greater than 1. To do this I was trying to use $h_X+h_Y= h_{X{\cup}Y}+h_{X{\cap}Y}$ but I wasn't able to show it.

REPLY [1 votes]: If you knew that $Y\cong \Bbb P^r$, then you'd be much closer to a solution: a closed immersion $\Bbb P^r\to\Bbb P^n$ comes from a line bundle $\mathcal{L}$ on $\Bbb P^r$ with a choice of $n+1$ global sections which generate the line bundle at every point. Such a line bundle must be $\mathcal{O}(d)$ for some $d>0$, and I claim that the degree of the image is exactly $d$: pick a hyperplane $H$ in $\Bbb P^n$ not containing $Y$ so that $\deg H\cap Y=\deg Y$ by the generalized Bezout theorem, while the degree is also $(r-1)!$ times the leading coefficient of Hilbert polynomial of $H\cap Y$. The Hilbert polynomial of $H\cap Y$ is $\binom{m+r}{r}-\binom{m+d+r}{r}=\frac{d}{(r-1)!}m^{r-1}+\cdots$ because $H$ restricts to a hypersurface on $Y$ cut out by a polynomial of degree $d$, so we must have $\deg Y=d$. As $\deg Y=1$, this shows that $Y$ must be embedded by $\mathcal{O}(1)$, which is a linear embedding.
To show $Y\cong \Bbb P^r$, we'll first show that we have a finite map $Y\to \Bbb P^r$ and then calculate that the degree of this map is 1, showing it's an isomorphism. Let $\Bbb P^{n-r-1}\subset\Bbb P^n$ be a linear $(n-r-1)$-plane which is disjoint from $Y$, and let $\Bbb P^r\subset \Bbb P^n$ be a linear $r$-plane disjoint from $\Bbb P^{n-r-1}$. Let $\pi: \Bbb P^n\setminus \Bbb P^{n-r-1}\to\Bbb P^r$ denote the projection from $\Bbb P^{n-r-1}$ to $\Bbb P^r$. I claim that $\pi$ restricts to a finite map on $Y$. First, the fiber over $p\in\Bbb P^r$ is the intersection of $Y$ with the linear $(n-r)$-plane spanned by $p$ and $\Bbb P^{n-r-1}$: this must be a dimension-zero closed subscheme of $\Bbb P^{n-r}$ for dimension reasons, as $Y$ does not intersect $\Bbb P^{n-r-1}$ while any two closed subschemes of $\Bbb P^a$ with dimensions adding to $a$ must intersect. Thus the fiber is a finite set of points, so $\pi|_Y$ is quasi-finite. But $\pi|_Y$ is also projective, and quasi-finite + projective implies finite, so $\pi|_Y$ is actually finite.
This implies that $(\pi|_Y)_*\mathcal{O}_Y$ is a coherent sheaf of $\mathcal{O}_{\Bbb P^r}$-algebras and $Y\cong \mathop{\mathbf{Spec}} (\pi|_Y)_*\mathcal{O}_Y$. If we can show that $(\pi|_Y)_*\mathcal{O}_Y$ is of rank 1, then $(\pi|_Y)_*\mathcal{O}_Y$ is a line bundle which will imply that $Y\cong \Bbb P^r$. As the rank of $(\pi|_Y)_*\mathcal{O}_Y$ at a point $p\in\Bbb P^r$ is the degree of the intersection of $Y$ with the $(n-r)$-plane spanned by $\Bbb P^{n-r-1}$ and $p$, we need to do a little intersection theory to show that this is 1. The following argument will do the trick:
Claim. If $Y\subset \Bbb P^n$ is a closed subscheme with Hilbert polynomial $h(m)=\binom{m+r}{r}$ and $L$ is a linear subspace of dimension $n-r$ so that $Y$ and $L$ intersect properly, then $Y\cap L$ is a single point (ie a point with Hilbert polynomial $1$).
Proof. $L$ has an ideal which is generated by $r$ linear forms $h_1,\cdots,h_r$, and $Y\cap L=Y\cap V(h_1)\cap\cdots\cap V(h_r)$, so we can prove this by induction. $Y\cap V(h_1)$ has Hilbert polynomial $\binom{r+m}{r}-\binom{r+m-1}{r}=\binom{r-1+m}{r-1}$, and by induction we see that the Hilbert polynomial of $Y\cap V(h_1)\cap\cdots\cap V(h_i)$ is $\binom{r-i+m}{r-i}$. Therefore when $i=r$ this Hilbert polynomial is just $1$ and the claim is proven.


text:
\begin{document}
\newtheorem{teo}{Theorem}
\newtheorem{lem}{Lemma}
\newtheorem{prop}{Proposition}
\newtheorem{cor}{Corollary}
\title{Fractional differential equations: $\alpha$-entire
solutions, regular and irregular singularities}
\author{Anatoly N. Kochubei
\footnote{Partially supported by the Scientific Program of
the National Academy of Sciences of Ukraine, Project No. 0107U002029, by the DFG, Grant 436 UKR 113/87/01, and
by the Ukrainian Foundation
for Fundamental Research, Grant 14.1/003.}
\\ \footnotesize Institute of Mathematics,\\
\footnotesize National Academy of Sciences of Ukraine,\\
\footnotesize Tereshchenkivska 3, Kiev, 01601 Ukraine.\\
\\ \footnotesize E-mail: kochubei@i.com.ua \quad Phone: (380-44)-243-4515 \quad Fax: (380-44)-235-2010}
\date{}
\maketitle
\vspace*{2cm}
Running head:\quad  ``Fractional differential equations''

\vspace*{2cm}
\begin{abstract}
We consider fractional differential equations of order $\alpha \in (0,1)$ for functions of one independent variable $t\in (0,\infty )$ with the Riemann-Liouville and Caputo-Dzhrbashyan fractional derivatives. A precise estimate for the order of growth of $\alpha$-entire solutions is given. An analog of the Frobenius method for systems with regular singularity is developed. For a model example of an equation with a kind of an irregular singularity, a series for a formal solution is shown to be convergent for $t>0$ (if $\alpha$ is an irrational number poorly approximated by rational ones) but divergent in the distribution sense.
\end{abstract}

\vspace{2cm}
{\bf Key words: }\ fractional differential equation; Riemann-Liouville derivative; Caputo-\linebreak Dzhrbashyan derivative; regular singularity; irregular singularity

\bigskip
{\bf AMS subject classifications:} 26A33, 34M99.
\newpage
\section{INTRODUCTION}

Fractional differential equations are widely used for modeling anomalous relaxation and diffusion phenomena; see \cite{EIK,KST} for further references. Meanwhile the mathematical theory of such equations is still in its initial stage. In particular, a systematic development of the analytic theory of fractional differential equations with variable coefficients was initiated only recently, in the paper by
Kilbas, Rivero, Rodr\'\i guez-Germ\'a, and Trujillo \cite{KRRT} (see also Section 7.5 in \cite{KST}). For equations of order $\alpha \in (0,1)$, of the form
\begin{equation}
\left( \DRL u\right) (t)=a(t)u(t),\quad t>0,
\end{equation}
where $\DRL$ is the Riemann-Liouville fractional derivative, or
\begin{equation}
\left( \DC u\right) (t)=a(t)u(t),\quad t>0,
\end{equation}
where $\DC$ is the Caputo-Dzhrbashyan fractional derivative, their main results are as follows. If $a(t)=A(t^\alpha )$, and $A(z)$ is a real function possessing an absolutely convergent Taylor expansion on an interval $|z|<\theta$, then the equation (1) possesses a solution of the form
$$
u(t)=t^{\alpha -1}\sum\limits_{n=0}^\infty a_nt^{\alpha n},\quad 0<t<\theta ,
$$
while the equation (2) has a solution
$$
u(t)=\sum\limits_{n=0}^\infty b_nt^{\alpha n},\quad
0\le t<\theta .
$$
In both cases the solutions are unique, if appropriate initial conditions are prescribed.

Thus, for example, the property of $\alpha$-analyticity of the coefficient $a(t)$ (defined above) implies a similar property of a solution of the equation (2). In fact, we have $u(t)=U(t^\alpha )$ where $U$ is holomorphic in a disk $\{ z\in \mathbb C,\ |z|<\theta \}$. The coefficient $a$ may be complex-valued as well.

The above results open the way for developing a theory of $\alpha$-analytic solutions of fractional differential equations in the spirit of classical analytic theory of ordinary differential equations. Here we give some results in this direction.

If in (2) $a(t)=A(t^\alpha )$ where $A$ is an entire function, then the above results from \cite{KRRT} with $\theta =\infty$ (stated there in a weaker form, only for real arguments of analytic functions, than actually proved) show that the solution of the Cauchy problem for the equation (2) is of the form $u(t)=U(t^\alpha )$ where $U$ is an entire function. Following $\cite{KRRT,KST}$ we call such solutions $\alpha$-entire. In particular, that is true, if $A$ is a polynomial. A natural question is about the order of $U$ (here we investigate this subject just for the equation (2) since its properties are closer to those of ordinary differential equations).
It is known (see, for example, \cite{CGH} or \cite{HKR}) that every nontrivial solution of the equation $u^{(k)}(z)=A(z)u(z)$, $k\in \mathbb N$, with a polynomial coefficient $A$, is an entire function of order $1+\deg (A)/k$. In this paper we prove that the orders of the entire functions $U$ corresponding to solutions of (2) do not exceed $(1+\deg (A))/\alpha$. As $\alpha \to 1$, this agrees with the above differential equation result. On the other hand, if $\deg (A)=0$, that is $A(z)=\lambda$, $\lambda \in \mathbb C$, then $U(z)=E_\alpha (\lambda z)$, where $E_\alpha$ is the Mittag-Leffler function whose order is $1/\alpha$ \cite{EIK,KST}, which shows the exactness of our general estimate.

Next, we investigate systems of fractional equations with regular singularity, that is the equations
\begin{equation}
t^\alpha \left( \DRL u\right) (t)=A(t^\alpha )u(t)
\end{equation}
and
\begin{equation}
t^\alpha \left( \DC u\right) (t)=A(t^\alpha )u(t)
\end{equation}
where $A(z)$ is a holomorphic matrix-function. Under some assumptions, we prove that formal power series solutions of (3) and (4) converge near the origin and develop an analog of the classical Frobenius method of finding a solution. For scalar equations, the latter problem was considered in \cite{KST,RRT}.

Finally, in order to clarify characteristic features of fractional equations with irregular singularity, we study a model example, the equation
\begin{equation}
t^{2\alpha }\left( \DC u\right) (t)=\lambda u(t),\quad \lambda \in \mathbb C,
\end{equation}
where, as before, $0<\alpha <1$. Assuming that $\alpha$ is irrational and satisfies a Diophantine condition (that is $\alpha$ is poorly approximated by rational numbers), we construct a kind of a formal solution of (5) convergent for $t>0$. We prove that the series for the formal solution does not converge in the distribution sense, within a theory of distributions associated with the fractional calculus (see \cite{R,SKM}). Thus, the formal solution $u(t)$ cannot be interpreted as a distribution solution. It is interesting that $u(t)$ is closely connected with a class of analytic functions with irregular behavior introduced by Hardy \cite{H}.

\section{Preliminaries}

{\bf 2.1. Fractional derivatives and integrals \cite{EIK,KST,SKM}.} Let $\alpha \in (0,1)$ be a fixed number. {\it The Riemann-Liouville fractional integral} of order $\alpha$ of a function $\varphi \in L_1(0,T)$ is defined as
$$
\left( I_{0+}^\alpha \varphi \right) (t)=\frac{1}{\Gamma (\alpha )}\int\limits_0^t(t-\tau )^{\alpha -1}\varphi (\tau )\,d\tau ,\quad 0<t\le T.
$$
{\it The Riemann-Liouville fractional derivative} of order $\alpha$ is given by the expression
$$
\left( D_{0+}^\alpha \varphi \right) (t)=\frac{1}{\Gamma (1-\alpha )}\frac{d}{dt}\int\limits_0^t(t-\tau )^{-\alpha }\varphi (\tau )\,d\tau ,\quad 0<t\le T,
$$
that is $\left( \DRL \varphi \right)(t)=\frac{d}{dt}\left( I_{0+}^{1-\alpha }\varphi \right) (t)$, provided the fractional integral $I_{0+}^{1-\alpha }\varphi$ is an absolutely continuous function. If $\varphi$ is defined on the whole half-axis $(0,\infty )$, then $I_{0+}^\alpha \varphi$ and $\DRL \varphi$ are also defined on $(0,\infty )$. Below we will consider just this case.

Note that the Riemann-Liouville derivative is defined for some functions with a singularity at the origin. For example, if $\varphi (t)=t^d$, $d>-1$, then
\begin{equation}
\left( \DRL \varphi \right) (t)=\frac{\Gamma (d+1)}{\Gamma (d+1-\alpha )}t^{d-\alpha },
\end{equation}
so that $\DRL \varphi =0$, if $\varphi (t)=t^{\alpha -1}$. For $\varphi (t)=t^d$, $d>-1$, we have also
\begin{equation}
\left( I_{0+}^\alpha  \varphi \right) (t)=\frac{\Gamma (d+1)}{\Gamma (d+1+\alpha )}t^{d+\alpha }.
\end{equation}

The Riemann-Liouville fractional differentiation and integration are inverse to each other in the following sense. If $\varphi \in L_1(0,T)$, then $\DRL I_{0+}^\alpha  \varphi =\varphi$. The equality $I_{0+}^\alpha  \DRL \varphi =\varphi$ holds under the stronger assumption that $\varphi =I_{0+}^\alpha \psi$ with some $\psi \in L_1(0,T)$. The latter is equivalent to the conditions of absolute continuity of $I_{0+}^{1- \alpha }\varphi$ on $[0,T]$ and the equality $\left( I_{0+}^{1-\alpha }\varphi \right) (0)=0$.

Let a function $\varphi$ be continuous on $[0,T]$ and possess the Riemann-Liouville fractional derivative of order $\alpha$. The function
\begin{equation}
\left( \DC \varphi \right) (t)=\frac{1}{\Gamma (1-\alpha )}\left[ \frac{d}{dt}\int\limits_0^t(t-\tau )^{-\alpha }\varphi (\tau )\,d\tau -t^{-\alpha }\varphi (0)\right] =\left( \DRL \varphi \right) (t)-\frac{1}{t^\alpha \Gamma (1-\alpha )}\varphi (0)
\end{equation}
is called {\it the Caputo-Dzhrbashyan, or regularized, fractional derivative}. If $\varphi$ is absolutely continuous on $[0,T]$, then
\begin{equation}
\left( \DC \varphi \right) (t)=\frac{1}{\Gamma (1-\alpha )}
\int\limits_0^t(t-\tau )^{-\alpha }\varphi' (\tau )\,d\tau .
\end{equation}
In contrast to $\DRL$, $\DC$ is defined only on continuous functions and vanishes on constant functions. In most of physical applications, equations with $\DC$ are used, because a solution of an equation with the Riemann-Liouville derivative typically has a singularity at the origin $t=0$, so that the initial state of a system to be described by the equation is not defined.

Let $v(z)=\sum\limits_{n=0}^\infty c_nz^n$ be an entire function. Consider the Caputo-Dzhrbashyan derivative of the function
$$
\varphi (t)=v(t^\alpha )=\sum\limits_{n=0}^\infty c_nt^{\alpha n}.
$$
It follows from (6) and (8) that
$$
\left( \DC \varphi \right) (t)=\sum\limits_{n=1}^\infty c_n\beta (n)t^{\alpha (n-1)}=\left( \mathfrak D_\alpha v\right) (t^\alpha )
$$
where
\begin{equation}
\beta (n)=\frac{\Gamma (n\alpha +1)}{\Gamma (n\alpha +1-\alpha )},
\end{equation}
and the operator
$$
\left( \mathfrak D_\alpha v\right) (z)=\sum\limits_{n=1}^\infty c_n\beta (n)z^{n-1}
$$
is known as the Gelfond-Leontiev generalized differentiation operator (see \cite{K,SKM}; in fact, $\mathfrak D_\alpha$ is defined for wider classes of functions).

The operator, right inverse to $\mathfrak D_\alpha$, has the form
$$
\left( \mathfrak I_\alpha f\right) (z)=\frac{z}{\Gamma (\alpha )}\int\limits_0^1(1-t)^{\alpha -1}f(zt^\alpha )\,dt
$$
(see Sect. 22.3 in \cite{SKM}). It will be convenient to make a change of variables setting $z=Re^{i\theta }$, $r=Rt^\alpha$. Then
\begin{equation}
\left( \mathfrak I_\alpha f\right) (Re^{i\theta })=\frac{e^{i\theta }}{\alpha \Gamma (\alpha )}\int\limits_0^R
\left[ \left( \frac{R}r\right)^{1/\alpha}-1\right]^{\alpha -1}f(re^{i\theta })\,dr.
\end{equation}

If $f(z)=\sum\limits_{k=0}^\infty f_kz^k$, then
$$
\left( \mathfrak I_\alpha f\right) (z)=\sum\limits_{k=0}^\infty \frac{\Gamma (\alpha k+1)}{\Gamma (\alpha k+1+\alpha )}f_kz^{k+1},
$$
and it is easy to check that
\begin{equation}
\left( \mathfrak I_\alpha \mathfrak D_\alpha v\right) (z)=v(z)-v(0).
\end{equation}

\medskip
{\bf 2.2. A class of distributions.} Spaces of test functions and distributions behaving reasonably under the action of fractional integration operators were introduced by Rubin \cite{R} (for a brief exposition see also \cite{SKM}; both in \cite{SKM} and \cite{R} there are references regarding other approaches and earlier publications in this field). Proceeding from \cite{R}, it is easy to come to a class of distributions where the Caputo-Dzhrbashyan derivative $\DC$ is defined in a natural way.

Let $\mathcal S (0,\infty )$ be the Schwartz space of smooth functions on $[0,\infty )$ with rapid decay at infinity. Denote
$$
\mathcal S_+=\{ \varphi \in \mathcal S (0,\infty ):\ \varphi^{(l)}(0)=0,\ l=0,1,2,\ldots \},
$$
$$
\Phi_+^{1-\alpha }=\left\{ \varphi \in \mathcal S_+:\ \int\limits_0^\infty \varphi (x)x^{1-\alpha -k}\,dx=0,\ k=1,2,\ldots \right\} ,
$$
$$
\Phi_+^{\alpha -1}=\left\{ \varphi \in \mathcal S_+:\ \int\limits_0^\infty \varphi (x)x^{\alpha -1-k}\,dx=0,\ k=0,1,2,\ldots \right\} .
$$
These spaces are interpreted as topological vector spaces with the topologies induced from $\mathcal S (0,\infty )$; see \cite{R} for various descriptions of these topologies including the description by seminorms.

Together with the Riemann-Liouville fractional integration operator $I_{0+}^\alpha$, it is convenient to use the operator
$$
\left( I_-^\alpha \varphi \right) (t)=\frac{1}{\Gamma (\alpha )}\int\limits_t^\infty (\tau -t)^{\alpha -1}\varphi (\tau )\,d\tau ,\quad t>0.
$$
If $\varphi ,\psi$ are sufficiently good functions, for example, if $\varphi \in L_p(0,\infty )$, $\psi \in L_q(0,\infty )$, $p,q>1$, $\dfrac{1}p+\dfrac{1}q=2-\alpha$, then
$$
\int\limits_0^\infty \varphi (x)\left( I_{0+}^{1-\alpha }\psi \right) (x)\,dx=\int\limits_0^\infty \psi (x)\left( I_-^{1-\alpha }\varphi \right) (x)\,dx
$$
(see Section 2.5.1 in \cite{SKM}). We will write this in the notation
$$
\left\langle \varphi ,I_{0+}^{1-\alpha }\psi \right\rangle =\left\langle \psi ,I_-^{1-\alpha }\varphi \right\rangle .
$$

In particular, if $\psi =u'$, where $u\in \mathcal S (0,\infty )$, then
\begin{equation}
\left\langle \varphi ,\DC u\right\rangle =\left\langle u' ,I_-^{1-\alpha }\varphi \right\rangle .
\end{equation}
It is known \cite{R} that $I_-^{1-\alpha }$ acts continuously from $\Phi_+^{1-\alpha }$ onto $\Phi_+^{\alpha -1}$. Therefore the identity (13) can be used to define $\DC u$ as a distribution from $\left( \Phi_+^{1-\alpha }\right)'$, if $u\in C^1(0,\infty )$, and $u'$ has no more than a power-like growth near zero and infinity. This definition agrees with the classical one: if $u$ is continuously differentiable at the origin too, then
\begin{multline*}
\left\langle u' ,I_-^{1-\alpha }\varphi \right\rangle =\frac{1}{\Gamma (1-\alpha )}\int\limits_0^\infty u'(x)\,dx \int\limits_x^\infty (t-x)^{-\alpha }\varphi (t)\,dt \\
=\frac{1}{\Gamma (1-\alpha )}\int\limits_0^\infty \varphi (t)\,dt \int\limits_0^t (t-x)^{-\alpha }u'(x)\,dx=\left\langle \varphi ,\DC u\right\rangle
\end{multline*}
where $\DC u$ is understood in the sense of (9).

A typical example of a function from $\Phi_+^{1-\alpha }$ is the function
$$
\varkappa_\alpha (x)=x^{\alpha -2}\exp \left( -\frac{\log^2x}4\right) \sin \left( \frac{\pi}2\log x\right) .
$$
It is clear that $\varkappa_\alpha \in \mathcal S_+$. Next, if
$$
\varkappa (x)=\exp \left( -\frac{\log^2x}4\right) \sin \left( \frac{\pi}2\log x\right) ,
$$
then we can write explicitly the Mellin transform
$$
\widetilde{\varkappa}(z)=\int\limits_0^\infty x^{z-1}\varkappa (x)\,dx.
$$
Namely, by the formula (4.133.1) from \cite{GR},
\begin{equation}
\widetilde{\varkappa}(z)=2\int\limits_0^\infty e^{-t^2/4}\sinh (zt)\sin \frac{\pi t}2\,dt=2\sqrt{\pi }e^{z^2-\frac{\pi^2}4}\sin \pi z.
\end{equation}
In particular,
$$
\int\limits_0^\infty \varkappa_\alpha (x)x^{1-\alpha -k}\,dx=\widetilde{\varkappa}(-k)=0,\quad k=0,1,2,\ldots ,
$$
so that indeed $\varkappa_\alpha \in \Phi_+^{1-\alpha }$.

In order to have a full concept of a class of distributions, one needs a result regarding density of a space of test functions in some space of integrable functions. This gives a one-to-one correspondence between ordinary functions and distributions they generate. Here we present such a result though it will not be used directly in this paper. For similar properties in other situations see \cite{S}.

\medskip
\begin{prop}
The space $\Phi_+^{1-\alpha }$ is dense in $L_p((0,\infty ),t^{-1}dt)$, $1\le p<\infty$.
\end{prop}

\medskip
{\it Proof}. Let $R=\int\limits_0^\infty \varkappa_\alpha (t)t^{-1}dt$. We have
$$
R=\widetilde{\varkappa }(\alpha -2)=2\sqrt{\pi }\exp \left( (\alpha -2)^2-\frac{\pi^2}4\right) \sin \pi \alpha >0.
$$
Denote
$$
z_N(x)=\frac{N}R\varkappa_\alpha (x^N),\quad N=1,2,\ldots .
$$
Then
\begin{equation}
\int\limits_0^\infty z_N(x)x^{-1}dx=\frac{1}R\int\limits_0^\infty \varkappa_\alpha (t)t^{-1}dt=1.
\end{equation}

Suppose that $f\in L_p((0,\infty ),t^{-1}dt)$. Consider the so-called Mellin convolution
$$
\left( z_N*_Mf\right) (t)=\int\limits_0^\infty z_N(\tau )f\left( \frac{t}\tau \right) \tau^{-1}d\tau ,
$$
that is actually the convolution on the multiplicative group $(0,\infty )$ (note that $t^{-1}dt$ is a Haar measure on that group). Obviously, $z_N*_Mf\in \Phi_+^{1-\alpha }$. Denote by $\| \cdot \|_p$ the norm in $L_p((0,\infty ),\linebreak t^{-1}dt)$.

Using (15) we can write
$$
\left( z_N*_Mf\right) (t)-f(t)=\int\limits_0^\infty z_N(\tau )\left[ f\left( \frac{t}\tau \right) -f(t)\right] \tau^{-1}d\tau .
$$
By the generalized Minkowski inequality,
$$
\left\| z_N*_Mf-f\right\|_p\le \int\limits_0^\infty |z_N(\tau )|\left\{ \int\limits_0^\infty \left| f\left( \frac{t}\tau \right) -f(t)\right|^pt^{-1}dt\right\}^{1/p}d\tau .
$$

Next we use the $L_p$-continuity of shifts on the multiplicative group $(0,\infty )$ (see \cite{HR}, Theorem 20.4). For any $\varepsilon >0$, there exists $\delta >0$, such that
$$
\left\{ \int\limits_0^\infty \left| f\left( \frac{t}\tau \right) -f(t)\right|^pt^{-1}dt\right\}^{1/p}<\varepsilon ,
$$
if $|\tau -1|<\delta$. Thus,
\begin{equation}
\left\| z_N*_Mf-f\right\|_p\le 2\| f\|_p\int\limits_{|\tau -1|\ge \delta }|z_N(\tau )|\tau^{-1}d\tau +\varepsilon \int\limits_{|\tau -1|<\delta }|z_N(\tau )|\tau^{-1}d\tau .
\end{equation}

Note that
$$
\int\limits_0^\infty |z_N(\tau )|\tau^{-1}d\tau =\frac{N}R
\int\limits_0^\infty \left| \varkappa_\alpha (x^N)\right| x^{-1}dx=\frac{1}R\int\limits_0^\infty \left| \varkappa_\alpha (t)\right| t^{-1}dt=C_1
$$
where the constant $C_1$ does not depend on $N$. On the other hand,
$$
|z_N(x)|\le C_2Nx^{N(\alpha -2)}\exp \left( -\frac{N^2\log^2x}4\right) .
$$
If $|x-1|\ge \delta$, then $\log^2x\ge b>0$, so that
$$
\int\limits_{1+\delta }^\infty |z_N(x)|x^{-1}dx\le C_3e^{-\frac{N^2b}4}(1+\delta )^{N(\alpha -2)}\to 0,
$$
as $N\to \infty$, and (for $\delta <1$)
\begin{multline*}
\int\limits_0^{1-\delta }|z_N(x)|x^{-1}dx\le C_2N\int\limits_{-\infty }^{\log (1-\delta )}\exp \left\{ Nt(\alpha -2)-\frac{N^2t^2}4\right\} \,dt \\
=C_2\int\limits_{-\infty }^{N\log (1-\delta )}\exp \left\{ s(\alpha -2)-\frac{s^2}4\right\} \,ds\to 0,
\end{multline*}
as $N\to \infty$.

As a result, we see that, if $N$ is large enough, the first summand in (16) does not exceed $2\|f\|_p\varepsilon$, while the second $\le C_1\varepsilon$. Thus,
$$
\left\| z_N*_Mf-f\right\|_p\longrightarrow 0,\quad \text{as $N\to \infty$.} \qquad \blacksquare
$$

\bigskip
{\bf 2.3. On ratios of the Gamma functions.} We will often use the function
\begin{equation}
\rho (t)=\frac{\Gamma (t+1)}{\Gamma (t+1-\alpha )},\quad -1<t<\infty .
\end{equation}
Here we collect some of its properties.

If $t>\alpha -1$, the integral representation
\begin{equation}
\frac{1}{\rho (t)}=\frac{1}{\Gamma (\alpha )}\int\limits_0^\infty e^{-st}e^{(\alpha -1)s}\left( 1-e^{-s}\right)^{\alpha -1}ds
\end{equation}
holds (see Chapter 4 in \cite{Ol}). It follows from (18) that the function $t\mapsto \dfrac{1}{\rho (t)}$ is strictly monotone decreasing and $\dfrac{1}{\rho (t)}\to 0$, as $t\to \infty$. Since $\Gamma (t+1-\alpha )$ has a pole at $t=\alpha -1$, it is seen from (17) that $\rho (t)\to 0$, as $t\to \alpha -1+0$.

If $-1<t<\alpha -1$, then (by a well-known identity for the Gamma function)
$$
\rho (t)=\frac{\Gamma (-t+\alpha )}{\Gamma (-t)}\cdot \frac{\sin \pi (t+1)}{\sin \pi (t+1-\alpha )}.
$$
The integral representation for the ratio of the Gamma functions \cite{Ol} leads, after an elementary investigation, to the conclusion that $\rho (t)$ is strictly monotone increasing from $-\infty$ to 0.

Thus, we conclude that on the interval $(-1,\infty )$ the function $\rho (t)$ is strictly monotone increasing from $-\infty$ to $\infty$. The inverse function $\gamma (\lambda )$ solving the equation
$$
\rho (t)=\lambda ,\quad \lambda \in \mathbb R,
$$
is a well-defined continuous function. Note that $\rho (0)=\dfrac{1}{\Gamma (1-\alpha )}$, so that $\gamma (\lambda ) \ge 0$, if $\lambda \ge \dfrac{1}{\Gamma (1-\alpha )}$.

It is known (\cite{Ol}, Chapter 4) that
\begin{equation}
\frac{\Gamma (t+a)}{\Gamma (t+b)}\sim t^{a-b}(1+O(t^{-1})),\quad t\to \infty ,
\end{equation}
if $b>a$. In particular,
$$
\rho (t)\sim t^\alpha , \quad \text{as $t\to \infty$}.
$$

For the sequence $\beta (n)$ defined in (10) and appearing in the definition of the Gelfond-Leontiev generalized differentiation operator $\mathfrak D_\alpha$, we have $\beta (n)=\rho (\alpha n)$, so that the above asymptotics implies the relation
\begin{equation}
\beta (n)\sim Cn^\alpha ,\quad n=0,1,2,\ldots .
\end{equation}

\section{$\alpha$-Entire Solutions}

Let us consider $\alpha$-entire solutions of the equation (2) with $a(t)=A(t^\alpha )$ where $A$ is a polynomial of degree $m\ge 0$. We assume the initial condition $u(0)=u_0$.

\medskip
\begin{teo}
Under the above assumptions, the solution $u(t)$ of the equation (2) has the form $u(t)=v(t^\alpha )$ where $v$ is an entire function whose order does not exceed $(1+m)/\alpha$.
\end{teo}

\medskip
{\it Proof}. Seeking the function $v$,
we have $\left( \mathfrak D_\alpha v\right) (z)=a(z)v(z)$. Let us apply the operator $\mathfrak I_\alpha$ (see (11)) to both sides of this equality. We get from (11) and (12) that
$$
v(Re^{i\theta })-v(0)=\frac{e^{i\theta }}{\alpha \Gamma (\alpha )}\int\limits_0^R\left[ \left( \frac{R}r\right)^{1/\alpha}-1\right]^{\alpha -1}a(re^{i\theta })v(re^{i\theta })\,dr,
$$
which implies the inequality
$$
|v(Re^{i\theta })|\le |v(0)|+C\int\limits_0^R\left[ \left( \frac{R}r\right)^{1/\alpha}-1\right]^{\alpha -1}|a(re^{i\theta })||v(re^{i\theta })|\,dr
$$
(here and below we denote by the same letter $C$ various positive constants).

We have the asymptotic relations
$$
t^{\frac{1}\alpha}-1\sim \frac{1}\alpha (t-1),\quad \text{as $t\to 1+0$};
$$
$$
\left( t^{\frac{1}\alpha}-1\right)^{\alpha -1}\sim t^{\frac{\alpha -1}\alpha},\quad \text{as $t\to \infty$}.
$$
Therefore
$$
\left( t^{\frac{1}\alpha}-1\right)^{\alpha -1}\le C(t-1)^{\alpha -1}t^{2-\frac{1}\alpha -\alpha },\quad t\ge 1,
$$
so that
$$
|v(Re^{i\theta })|\le |v(0)|+CR^{2-\frac{1}\alpha -\alpha }\int\limits_0^R(R-r)^{\alpha -1}r^{\frac{1}\alpha -1}|a(re^{i\theta })||v(re^{i\theta })|\,dr.
$$

Since $a(z)$ is a polynomial of degree $m$, we get
$$
|v(Re^{i\theta })|\le |v(0)|+CR^{2-\frac{1}\alpha -\alpha }\int\limits_0^R(R-r)^{\alpha -1}r^{\frac{1}\alpha -1+m}|v(re^{i\theta })|\,dr
$$
where $C$ does not depend on $R,\theta$. Fixing $\theta$ and denoting
$$
w(r)=\frac{|v(re^{i\theta })|}{r^{2-\frac{1}\alpha -\alpha }}
$$
we come to the inequality
\begin{equation}
w(R)\le \frac{|v(0)|}{R^{2-\frac{1}\alpha -\alpha }}+C\int\limits_0^R(R-r)^{\alpha -1}r^{1-\alpha +m}w(r)\,dr.
\end{equation}
Now we are in a position to apply Henry's theorem (see Lemma 7.1.2 from \cite{He}), which states that the inequality (21) implies the inequality
$$
w(R)\le \frac{|v(0)|}{R^{2-\frac{1}\alpha -\alpha }}\mathcal E_{\alpha ,2-\alpha +m}(CR)
$$
where $\mathcal E_{\alpha ,\sigma}(s)$ is a certain function admitting the estimate
$$
\mathcal E_{\alpha ,\sigma}(s)\le Cs^{\frac{1}2\left( \frac{\alpha +\sigma -1}\alpha -\sigma \right)} \exp \left( \frac{\alpha}{\alpha +\sigma -1}s^{\frac{\alpha +\sigma -1}\alpha }\right) .
$$
Thus,
$$
|v(Re^{i\theta })|\le C\exp \left( \mu R^{\frac{1+m}\alpha }\right)
$$
for some $\mu \ge 0$, as desired. $\qquad \blacksquare$

\section{Regular Singularity}

{\bf 4.1. Formal and $\alpha$-analytic solutions.} Let us consider systems of equations of the form
\begin{equation}
t^\alpha \left( \DRL u\right) (t)=A(t^\alpha )u(t)
\end{equation}
where
$$
A(z)=A_0+\sum\limits_{m=1}^\infty A_mz^m,
$$
$A_m$ are $n\times n$ complex matrices and
$$
\| A_m\|\le M\mu^m\ (\mu >0),\quad m=0,1,2,\ldots .
$$

Suppose we have a formal series
\begin{equation}
u(t)=\sum\limits_{k=0}^\infty u_kt^{\alpha k},\quad u_k\in \mathbb C^n.
\end{equation}
Let us substitute the series (23) formally into (22). We get, in accordance with (6), that
$$
\sum\limits_{k=0}^\infty \beta (k)u_kt^{\alpha k}=\sum\limits_{m,k=0}^\infty A_mu_kt^{\alpha (m+k)}
$$
where $\beta (k)$ is the sequence (10). Collecting and comparing the terms we find that
$$
\beta (l)u_l=\sum\limits_{k=0}^l A_ku_{l-k},\quad l=0,1,2,\ldots ,
$$
or, equivalently,
\begin{equation}
A_0u_0=\frac{1}{\Gamma(1-\alpha )}u_0;
\end{equation}
\begin{equation}
\left[ A_0-\beta (l)\right] u_l=-\sum\limits_{k=1}^lA_ku_{l-k},\quad l\ge 1.
\end{equation}

It is natural to call the formal series (23) {\it a formal solution} of the system (22) if the relations (24), (25) hold.

\medskip
\begin{prop}
If a formal series (23) is a formal solution of the system (22), then the series (23) is absolutely convergent on some neighbourhood of the origin.
\end{prop}

\medskip
{\it Proof}. It follows from (19) that
$$
\| \left[ A_0-\beta (l)\right]^{-1}\|\le Cl^{-\alpha },\quad l\ge l_0.
$$
In particular, we may assume that
$$
\| \left[ A_0-\beta (l)\right]^{-1}\|\le 1,\quad l\ge l_0.
$$
Considering, if necessary, $\lambda u(t)$ instead of $u(t)$, with $|\lambda |$ small enough, we may assume that $\|u_0\|\le 1$.

Let us choose so big $r>0$ that $\|u_l\|\le r^l$ for $l\le l_0$ and
$$
M\sum\limits_{k=1}^\infty \left( \frac{\mu }r\right)^k\le 1.
$$
Then
$$
\|u_l\|\le r^l\quad \text{for all $l$.}
$$

Indeed, if this inequality is proved up to some value of $l\ge l_0$, then
$$
\|u_{l+1}\|\le \left\| \sum\limits_{k=1}^{l+1}A_ku_{l+1-k}\right\|\le M\sum\limits_{k=1}^{l+1}\mu^kr^{l+1-k}=Mr^{l+1}\sum\limits_{k=1}^{l+1} \left( \frac{\mu }r\right)^k\le r^{l+1},
$$
and the above inequality implying local convergence in (23) has been proved. $\qquad \blacksquare$

\medskip
The above arguments remain valid for systems of the form
\begin{equation}
t^\alpha \left( \DC u\right) (t)=A(t^\alpha )u(t).
\end{equation}
The only difference is that, instead of (24), we get the relation $A_0u_0=0$, just as in the classical case (see \cite{Hart}).

\bigskip
{\bf 4.2. Model scalar equations.} Consider the equation
\begin{equation}
t^\alpha \left( \DRL \varphi \right) (t)=\lambda \varphi (t),\quad \lambda \in \mathbb R.
\end{equation}
By the relation (6), a solution of the equation (27) is $\const \cdot t^{\gamma (\lambda )}$ where $\gamma$ (the inverse function to $\rho$) was defined in Section 2.3. For example, if $\lambda =0$, then we have $\gamma (0)=\alpha -1$.

If we consider an equation similar to (27), but with the Caputo-Dzhrbashyan derivative, that is
\begin{equation}
t^\alpha \left( \DC \varphi \right) (t)=\lambda \varphi (t),
\end{equation}
then the constant function is a solution of (28) for $\lambda =0$. Suppose that $\lambda \ne 0$, and $\varphi$ is a solution of (28), that is $\varphi \in C[0,T]$, the function
$$
\left( I_{0+}^{1-\alpha }\varphi \right) (t)=\frac{1}{\Gamma (1-\alpha )}\int\limits_0^t(t-\tau )^{-\alpha }\varphi (\tau )\,d\tau
$$
is absolutely continuous, and (28) is satisfied with
$$
\left( \DC \varphi \right) (t)=\frac{d}{dt}\left( I_{0+}^{1-\alpha }\varphi \right) (t)-\dfrac{1}{t^\alpha \Gamma (1-\alpha )}\varphi (0).
$$
We have
$$
\left( I_{0+}^{1-\alpha }\varphi \right) (t)=\frac{t^{1-\alpha }}{\Gamma (1-\alpha )}\int\limits_0^1(1-s)^{-\alpha }\varphi (st)\,ds,
$$
and since $\varphi$ is continuous,
$$
\left( I_{0+}^{1-\alpha }\varphi \right) (t)\longrightarrow 0,\quad \text{as $t\to +0$.}
$$

It is known \cite{SKM} that in these circumstances $I_{0+}^\alpha \DRL \varphi =\varphi$. Note also that $I_{0+}^\alpha$ transforms the function $t^{-\alpha }$ into the constant $\Gamma (1-\alpha )$. Dividing the equation (28) by $t^\alpha$ and applying $I_{0+}^\alpha$ to both sides, we find that
$$
\varphi (t)-\varphi (0)=\frac{\lambda }{\Gamma (\alpha )}\int\limits_0^t(t-\tau )^{-1+\alpha }\tau^{-\alpha }\varphi (\tau )\,d\tau =\frac{\lambda }{\Gamma (\alpha )}\int\limits_0^1(1-s)^{-1+\alpha }s^{-\alpha }\varphi (ts)\,ds.
$$
Passing to the limit, as $t\to 0$, and taking into account the continuity of $\varphi$, we obtain the identity
$$
\frac{\lambda \varphi (0)}{\Gamma (\alpha )}\int\limits_0^1(1-s)^{-1+\alpha }s^{-\alpha }ds=0,
$$
whence $\varphi (0)=0$.

Thus, for $\lambda \ne 0$, the equation (28) is equivalent to (27), if (27) is considered for continuous functions vanishing at the origin. The power solution $Ct^{\gamma (\lambda )}$ belongs to this class, if $\lambda >\frac{1}{\Gamma (1-\alpha )}$. It may be instructive to see these solutions, satisfying the equation
$$
\varphi (t)=\frac{\lambda }{\Gamma (\alpha )}\int\limits_0^t(t-\tau )^{-1+\alpha }\tau^{-\alpha }\varphi (\tau )\,d\tau ,
$$
as examples of non-uniqueness of solutions of linear Volterra integral equations occurring due to the singularity of a kernel.

\bigskip
{\bf 4.3. Systems with good spectrum.} Let us consider the equation (3) with $A(z)=\sum\limits_{m=0}^\infty A_mz^m$ where $A_m$ are complex $n\times n$ matrices, the matrix $A_0$ is Hermitian, and the series converges on a neighbourhood of the origin. Without restricting generality, we may assume that
$$
A_0=\diag (\lambda_1,\ldots ,\lambda_n),\quad \lambda_1,\ldots ,\lambda_n\in \mathbb R.
$$

Following the classical method (see, for example, \cite{B}) we look for a matrix-valued solution ({\it a fundamental solution}) of the equation (3), in the form
\begin{equation}
u(t)=S(t^\alpha )\psi (t)
\end{equation}
where $\psi (t)=\diag \left( t^{\gamma (\lambda_1)},\ldots , t^{\gamma (\lambda_n)}\right)$, $S(z)=\sum\limits_{\nu =0}^\infty \sigma_\nu z^\nu$, $\sigma_\nu$ ($\nu \ge 1$) are some unknown matrices, $\sigma_0=I$.

We have
$$
u(t)=\sum\limits_{\nu =0}^\infty \sigma_\nu \diag \left( t^{\gamma (\lambda_1)+\alpha \nu},\ldots ,t^{\gamma (\lambda_n)+\alpha \nu}\right),
$$
whence
\begin{equation}
t^\alpha \left( \DRL u\right) (t)=\sum\limits_{\nu =0}^\infty \sigma_\nu R_\nu \diag \left( t^{\gamma (\lambda_1)+\alpha \nu},\ldots ,t^{\gamma (\lambda_n)+\alpha \nu}\right)
\end{equation}
where
$$
R_\nu u =\diag \left( \frac{\Gamma (\gamma (\lambda_1)+\alpha \nu +1)}{\Gamma (\gamma (\lambda_1)+\alpha \nu +1-\alpha )},\ldots ,\frac{\Gamma (\gamma (\lambda_n)+\alpha \nu +1)}{\Gamma (\gamma (\lambda_n)+\alpha \nu +1-\alpha )}\right) .
$$
On the other hand,
\begin{equation}
A(t^\alpha )u(t)=\sum\limits_{\nu =0}^\infty \left( \sum\limits_{m=0}^\nu A_m\sigma_{\nu -m}\right)\diag \left( t^{\gamma (\lambda_1)+\alpha \nu},\ldots ,t^{\gamma (\lambda_n)+\alpha \nu}\right).
\end{equation}

Note that $R_0=\diag (\lambda_1,\ldots ,\lambda_n)=A_0$, and since $\sigma_0=I$, the coefficients corresponding to $\nu =0$ in (30) and (31) coincide. Comparing the rest of the coefficients, we obtain the following system of equations for the matrices $\sigma_k$:
\begin{equation}
\sigma_kR_k-A_0\sigma_k=\sum\limits_{l=0}^{k-1}A_{k-l}\sigma_l,\quad k\ge 1.
\end{equation}

For each $k$, the matrix equation (32) for $\sigma_k$ has a unique solution if the spectra of the matrices $R_k$ and $A_0$ are disjoint (see Appendix A.1 in \cite{B}), that is
\begin{equation}
\frac{\Gamma (\gamma (\lambda_i)+\alpha k+1)}{\Gamma (\gamma (\lambda_i)+\alpha k+1-\alpha )}\ne \lambda_j
\end{equation}
for all $i,j\in \{ 1,\ldots ,n\}$, or, equivalently, since the left-hand side of (33) equals $\rho (\gamma (\lambda_i)+\alpha k)$,
$$
\gamma (\lambda_j)-\gamma (\lambda_i)\ne \alpha k,\quad \text{for all $i,j\in \{ 1,\ldots ,n\}$}.
$$

We call our system (3) {\it a system with good spectrum}, if
\begin{equation}
\gamma (\lambda_j)-\gamma (\lambda_i)\notin \alpha \mathbb N, \quad \text{for all $i,j\in \{ 1,\ldots ,n\}$}.
\end{equation}
This definition extends the classical one \cite{B}, since for $\alpha =1$ we would have $\rho (t)=\gamma (t)=t$, and the condition (34) would mean that the eigenvalues of $A_0$ must not differ by a natural number.

\medskip
\begin{teo}
If a system (3) has a good spectrum, then it possesses a fundamental solution (29) where the series for $S(z)$ has a positive radius of convergence.
\end{teo}

\medskip
{\it Proof}. By the asymptotic relation (19),
$$
\frac{\Gamma (\gamma (\lambda_j)+\alpha k+1)}{\Gamma (\gamma (\lambda_j)+\alpha k+1-\alpha )}\sim (\alpha k)^\alpha \left( 1+O(k^{-1})\right),\quad k\to \infty ,
$$
for all $j=1,\ldots ,n$. Therefore
\begin{equation}
(\alpha k)^{-\alpha }R_k=I+O(k^{-1}), \quad k\to \infty .
\end{equation}

Let us divide both sides of the equation (32) by $(\alpha k)^\alpha$. The resulting equation, considered as a system of scalar equations for $n^2$ elements of the matrix $\sigma_k$, has the coefficients bounded in $k$ and the determinant, which is different from zero for each $k$ and tends to 1, as $k\to \infty$. This implies the estimate
\begin{equation}
\|\sigma_k\| \le ak^{-\alpha }\left\| \sum\limits_{l=0}^{k-1}A_{k-l}\sigma_l\right\| ,\quad k\ge 1,
\end{equation}
where the constant $a>0$ does not depend on $k$. It follows from (35), (36), and the convergence near the origin of the power series for $A(z)$ that
$$
\|\sigma_k\| \le a_1k^{-\alpha }\sum\limits_{l=0}^{k-1}b^{k-l}\|\sigma_l\| ,\quad k\ge 1,
$$
where $a_1$ and $b$ are positive constants independent of $k$.

Define a sequence $\{ s_k\}_0^\infty$ of positive numbers, setting $s_0=1$,
$$
s_k=a_1k^{-\alpha }\sum\limits_{l=0}^{k-1}b^{k-l}s_l, \quad k\ge 1.
$$
The induction on $k$ yields the inequality $\| \sigma_k\| \le s_k$ for all $k\ge 0$. On the other hand,
\begin{multline*}
s_{k+1}=a_1(k+1)^{-\alpha }\sum\limits_{l=0}^kb^{k+1-l}s_l
=\frac{(k+1)^{-\alpha}}{k^{-\alpha }}\left[ a_1k^{-\alpha }\left( b\sum\limits_{l=0}^{k-1}b^{k-l}s_l+bs_k\right) \right] \\
=\frac{(k+1)^{-\alpha}}{k^{-\alpha }}\left( bs_k+a_1k^{-\alpha }bs_k\right) =\frac{(k+1)^{-\alpha}b}{k^{-\alpha }}\left( 1+a_1k^{-\alpha}\right) s_k.
\end{multline*}
Therefore
\begin{equation}
\frac{s_k}{s_{k+1}}\longrightarrow b^{-1},\quad \text{as $k\to \infty$.}
\end{equation}

It follows from (37) that the series $\sum\limits_{k=0}^\infty s_kz^k$ has the convergence radius $b^{-1}$ (see Section 2.6 in \cite{WW}). Moreover, the series $\sum\limits_{\nu =0}^\infty \sigma_\nu z^\nu$ converges for $|z|<b^{-1}$. $\qquad \blacksquare$

\medskip
For the equation (26), a similar construction is valid, if we assume that $\lambda_1,\ldots ,\lambda_n\ge \dfrac{1}{\Gamma (1-\alpha )}$.

\section{Irregular Singularity: an Example}

{\bf 5.1. A formal solution.} In this section we construct a solution, in a sense to be specified, of the equation (5). Looking at classical first order equations, corresponding formally to $\alpha =1$, we have to consider the equation $t^2y'(t)=\lambda y(t)$ whose solution is $y(t)=\exp (-\lambda t^{-1})$. Therefore it is natural to seek a solution of the equation (5) in the form
\begin{equation}
u(t)=\sum\limits_{n=0}^\infty c_nt^{-n\alpha},\quad c_n\in \mathbb C.
\end{equation}

A fractional derivative of any term in (38) with $n>\alpha^{-1}$ does not make sense classically. However we may apply the distribution theory from Sect. 2.2. Below we understand the fractional derivative $\DC$ in the sense of (13).

\medskip
\begin{prop}
$\mathrm{(i)}$ If $\mu <0$, $\mu \ne -1,-2,\ldots$, then
\begin{equation}
\DC t^\mu =\frac{\Gamma (\mu +1)}{\Gamma (\mu +1-\alpha )}t^{\mu -\alpha }.
\end{equation}

$\mathrm{(ii)}$ If $k$ is a natural number, then
\begin{equation}
\DC t^{-k}=\frac{(-1)^{k-1}}{(k-1)!\Gamma (-k+1-\alpha )}t^{-k -\alpha }\log t.
\end{equation}
\end{prop}

\medskip
{\it Proof}. Let $\varphi \in \Phi_+^{1-\alpha }$. By (13),
\begin{equation}
\left\langle \DC t^\mu ,\varphi (t)\right\rangle =\mu \left\langle t^{\mu -1},\left( I_-^{1-\alpha} \varphi \right) (t)\right\rangle .
\end{equation}
It is clear that the right-hand side of (41) is an entire function of $\mu$. For $\mu >0$, by virtue of (7),
$$
\left\langle t^{\mu -1},\left( I_-^{1-\alpha} \varphi \right) (t)\right\rangle =\left\langle I_{0+}^{1-\alpha}t^{\mu -1}, \varphi (t)\right\rangle =\frac{\Gamma (\mu )}{\Gamma (\mu +1-\alpha )}\left\langle t^{\mu -\alpha },\varphi (t)\right\rangle .
$$
For $\mu <0$, $\mu \ne -1,-2,\ldots$, the analytic continuation gives the equality (39).

Next, consider the entire function
$$
F(\mu )=\langle t^{\mu -\alpha },\varphi (t)\rangle ,\quad \mu \in \mathbb C.
$$
Note that $F(-k)=0$, $k\in \mathbb N$, by the definition of the space $\Phi_+^{1-\alpha }$. We have
$$
F'(\mu )=\left\langle \log t\cdot e^{(\mu -\alpha )\log t},\varphi \right\rangle .
$$
In particular,
$$
F'(-k)=\int\limits_0^\infty t^{-\alpha -k}\log t\cdot \varphi (t)\,dt.
$$

As $\mu$ belongs to a small neighbourhood of the point $-k$, $F(\mu )=F'(-k)(\mu +k)+o(\mu +k)$. Since the residue of $\Gamma (\mu +1)$ at $\mu =-k$ equals $\dfrac{(-1)^{k-1}}{(k-1)!}$ (see Section 4.4.1 of \cite{T}), we see that the function $\Gamma (\mu +1)F(\mu )$ is holomorphic at $\mu =-k$ (in fact, it is entire) and
$$
\Gamma (\mu +1)F(\mu )|_{\mu =-k}=\frac{(-1)^{k-1}}{(k-1)!}
\int\limits_0^\infty t^{-\alpha -k}\log t\cdot \varphi (t)\,dt.
$$
Now the equality (39) implies (40). $\qquad \blacksquare$

\medskip
Returning to (38), we will formally apply $\DC$ termwise and find the coefficients $c_n$ comparing resulting terms in (5). It is clear from the equality (40) that such a procedure would fail if some of the numbers $n\alpha$ are integers. Thus, we have to assume that $\alpha$ {\it is irrational}. Using (39) we find that (formally)
$$
t^{2\alpha }\left( \DC u\right) (t)=\sum\limits_{n=1}^\infty c_n\frac{\Gamma (-n\alpha +1)}{\Gamma (-n\alpha +1-\alpha)}t^{-n\alpha +\alpha}.
$$
Substituting this into (5) we come to the recurrence relation
$$
c_{n+1}=\lambda \frac{\Gamma (1-(n+2)\alpha )}{\Gamma (1-(n+1)\alpha )}c_n,\quad n\ge 0,
$$
and it is easy to find by induction that
$$
c_n=\lambda^n\frac{\Gamma (1-(n+1)\alpha )}{\Gamma (1-\alpha )}c_0.
$$

Thus, we have found a formal solution
\begin{equation}
u(t)=\frac{c_0}{\Gamma (1-\alpha )}\sum\limits_{n=0}^\infty \Gamma (1-(n+1)\alpha )t^{-n\alpha }
\end{equation}
of the equation (5). Using the identity
$$
\Gamma (z)\Gamma (1-z)=\frac{\pi }{\sin \pi z}
$$
we can rewrite (42) in the form
\begin{equation}
u(t)=\frac{c_0\pi }{\Gamma (1-\alpha )}\sum\limits_{n=0}^\infty \frac{1}{\sin (\pi (n+1)\alpha )}\frac{t^{-n\alpha }}{\Gamma ((n+1)\alpha )}.
\end{equation}

\bigskip
{\bf 5.2. The convergence problem.} The convergence of the series (43) depends on the arithmetic properties of the irrational number $\alpha$. It was shown by Hardy \cite{H} that $\alpha$ can be chosen in such a way (to be well approximated by rational numbers) that the series (43) would diverge for small values of $t$.

An irrational number $\alpha \in (0,1)$ is said to be {\it poorly approximated by rational numbers} if there exist such $\varepsilon >0$, $c>0$ that for any rational number $\dfrac{p}q$, $p,q\in \mathbb N$,
\begin{equation}
\left| \alpha -\frac{p}q\right| \ge cq^{-2-\varepsilon }.
\end{equation}
By the Thue-Siegel-Roth theorem (see \cite{Sch}) such are all algebraic numbers.

The first statement of the next theorem is actually contained already in the paper \cite{H}.

\medskip
\begin{teo}
If $\alpha$ is poorly approximated by rational numbers, then the series (43) converges for any $t>0$. However this series diverges in the space of distributions $\left( \Phi_+^{1-\alpha }\right)'$.
\end{teo}

\medskip
{\it Proof}. It follows from (44) that
$$
|q\alpha -p|\ge cq^{-1-\varepsilon }\quad \text{for any $p\in \mathbb N$,}
$$
so that
$$
\dist ((n+1)\alpha ,\mathbb Z_+)\ge c(n+1)^{-1-\varepsilon },
$$
and taking $l\in \mathbb Z_+$, such that $|(n+1)\alpha -l|\le \dfrac12$, we find that
$$
|\sin (\pi (n+1)\alpha )|=|\sin (\pi ((n+1)\alpha -l)|\ge 2|(n+1)\alpha -l|\ge 2\dist ((n+1)\alpha ,\mathbb Z_+)\ge 2c(n+1)^{-1-\varepsilon }.
$$
Using the Stirling formula we obtain that the series (43) converges for each $t>0$.

To prove the second assertion, consider $\left\langle t^{-\alpha n},\varkappa_\alpha (t)\right\rangle$ where the function $\varkappa_\alpha$ was defined in Sect. 2.2. We have, by (14), that
\begin{multline*}
\left\langle t^{-\alpha n},\varkappa_\alpha (t)\right\rangle =\int\limits_0^\infty t^{-\alpha n}\varkappa_\alpha (t)\,dt=\widetilde{\varkappa_\alpha}(1-\alpha n)=\widetilde{\varkappa}(\alpha -1-\alpha n)\\
=2\sqrt{\pi }\exp \left( (\alpha -1-\alpha n)^2-\frac{\pi^2}4\right) \sin (\pi (\alpha -1-\alpha n)).
\end{multline*}

Now we can give a lower estimate of the coefficients in the series (43) understood in the distribution sense: there are such $a,a_1,b>0$ that
\begin{multline*}
\left| \frac{1}{\sin (\pi (n+1)\alpha )}\cdot \frac{\left\langle t^{-n\alpha },\varkappa_\alpha (t)\right\rangle }{\Gamma ((n+1)\alpha )}\right| \ge a(n+1)^{-b(n+1)}|\sin (\pi \alpha (n-1))|\exp (\alpha ^2n^2-2\alpha (\alpha -1)n) \\
\ge a_1(n+1)^{-b(n+1)-1-\varepsilon }\exp (\alpha ^2n^2-2\alpha (\alpha -1)n)\\
=a_1\exp \left\{ \alpha ^2n^2-2\alpha (\alpha -1)n-(bn+b+1+\varepsilon )\log (n+1)\right\} \to \infty ,
\end{multline*}
as $n\to \infty$. Therefore the series (43) does not converge in $\left( \Phi_+^{1-\alpha }\right)'$. $\qquad \blacksquare$

\medskip


TITLE: Could someone help me to understand how overall acceleration vary for a pendulum taking a cycloidal path?
QUESTION [0 upvotes]: For a normal pendulum experiencing simple harmonic motion (in a circular path), the acceleration of the pendulum is as follows:
$$a = \omega^2 x$$
where $\omega$ is angular frequency and $x$ is displacement from equilibrium position.
As for a cycloidal path, my gut feeling is that that acceleration of the pendulum should increase with increasing displacement from the equilibrium position. This allows the pendulum to have the same period no matter how far it was displaced from the equilibrium position. However, I am unable to find any understandable explanations or formulas that can support this, hence the question.
Thanks everyone!

REPLY [0 votes]: You've got it backwards.  If you write out the net forces on a pendulum bob oscillating in a circular path, you'll find that the acceleration of the pendulum is not $a = \omega^2 x$, but rather
$$
\alpha = - \omega^2 \sin \theta,
$$
where $\alpha$ is the (angular) acceleration and $\theta$ is the (angular) distance from equilibrium.  The minus sign indicates that the acceleration is opposite the direction of the displacement, i.e., the acceleration is always towards the equilibrium position.
Now, for small angles, with $\theta \ll 1$ (in radians), we can make the approximation that $\sin \theta \approx \theta$, and so the above equation becomes
$$
\alpha \approx - \omega^2 \theta
$$
In other words, the acceleration is approximately proportional to the displacement.  This condition is what leads to simple harmonic motion;  it follows from the fact that the only functions for which the second derivative is proportional to minus the function (i.e. $f'' \propto - f$) are sines and cosines.  However, if we actually use larger angles, the approximation $\sin \theta \approx \theta$ is no longer a good one;  and in fact, for large angles, the period of a circular pendulum is not simple harmonic motion.  For example, the period changes at larger amplitudes:

The cycloidal path, on the other hand, relies on the fact that the acceleration of the pendulum is always directly proportional to the displacement, even for large displacements from the bottom of the arc.  It's not too hard to show using free-body diagrams that acceleration due to gravity of a cart moving along a "roller coaster" path is equal to $g$ times the rate of change of the height $y$ with arc length $s$ (i.e., $a = g (dy/ds)$.)  If we want the cart to exhibit simple harmonic motion, it must be the case that $a$ is proportional to $s$ as well, i.e., $dy/ds \propto -s$.  Actually seeing that a cycloid has this property requires a fair amount of mathematics (though if anyone has a nice derivation of that, please post it as an answer alongside mine—I'd be interested to see it.)


TITLE: Proof that $\sum_{k=0}^{n}\frac{(-1)^k}{2k+1}{n\choose k}=\frac{4^n}{(2n+1){2n\choose n}}$
QUESTION [5 upvotes]: I saw in this paper the following identity: $$\sum_{k=0}^{n}\frac{(-1)^k}{2k+1}{n\choose k}=\frac{4^n}{(2n+1){2n\choose n}}$$
I have a pervious post on an integral quite closely related to this identity, but I still do not know how to derive/prove the identity. I'm really not that good at combinatorics or evaluating series, so I don't know how to start, which is why I don't have any attempts to show you. Please explain your steps thoroughly.
Edit: I know there other posts on this series, but I did not get from them the proof I was satisfied by.

REPLY [3 votes]: The Binomial Theorem says
$$
\sum_{k=0}^n(-1)^k\binom{n}{k}x^{2k}=\left(1-x^2\right)^n
$$
Integrating both sides over $[0,1]$ gives
$$
\begin{align}
\sum_{k=0}^n\frac{(-1)^k}{2k+1}\binom{n}{k}
&=\int_0^1\left(1-x^2\right)^n\,\mathrm{d}x\\
&=\frac12\int_0^1\left(1-x\right)^nx^{-1/2}\,\mathrm{d}x\\
&=\frac12\frac{\Gamma(n+1)\,\Gamma(1/2)}{\Gamma(n+3/2)}\\
&=\frac12\frac{n!}{\frac12\frac32\cdots\left(n+\frac12\right)}\\
&=\frac{2^nn!}{1\cdot3\cdots(2n+1)}\\
&=\frac{2^nn!2^nn!}{(2n+1)!}\\
&=\frac{4^n}{(2n+1)\binom{2n}{n}}
\end{align}
$$
Using the Beta Function.


TITLE: Evaluate $\int_{\pi f}^{3\pi f} dx \sqrt{\cos(x/f)}.$
QUESTION [0 upvotes]: I am trying to do the following integral:
$$\int_{\pi f}^{3\pi f} dx \sqrt{\cos(x/f)}.$$
Wolfram alpha gives me
$$\int_{\pi f}^{3\pi f} dx \sqrt{\cos(x/f)} = 4f E(2) = 2.39628f + 2.39628if,$$
where E is the elliptic function.
Mathematica also gives me the same answer. How can the integral of a real integrand with real limits be complex?

REPLY [0 votes]: Consider the problem of the antiderivative $$I=\int \sqrt{\cos \left(\frac{x}{f}\right)}\,dx=f\int \sqrt{\cos \left(y\right)}\,dy=2 f \,E\left(\left.\frac{y}{2}\right|2\right)$$ The problem is that $E(z|2)$ has real values only $-\frac \pi 4 \leq z \leq \frac \pi 4$.


text:
/-
Copyright (c) 2021 Yaël Dillies. All rights reserved.
Released under Apache 2.0 license as described in the file LICENSE.
Authors: Yaël Dillies
-/
import data.nat.interval
import data.pnat.basic

/-!
# Finite intervals of positive naturals

This file proves that `ℕ+` is a `locally_finite_order` and calculates the cardinality of its
intervals as finsets and fintypes.
-/

open finset pnat

instance : locally_finite_order ℕ+ := subtype.locally_finite_order _

namespace pnat
variables (a b : ℕ+)

lemma Icc_eq_finset_subtype : Icc a b = (Icc (a : ℕ) b).subtype (λ (n : ℕ), 0 < n) := rfl
lemma Ico_eq_finset_subtype : Ico a b = (Ico (a : ℕ) b).subtype (λ (n : ℕ), 0 < n) := rfl
lemma Ioc_eq_finset_subtype : Ioc a b = (Ioc (a : ℕ) b).subtype (λ (n : ℕ), 0 < n) := rfl
lemma Ioo_eq_finset_subtype : Ioo a b = (Ioo (a : ℕ) b).subtype (λ (n : ℕ), 0 < n) := rfl

lemma map_subtype_embedding_Icc : (Icc a b).map (function.embedding.subtype _) = Icc (a : ℕ) b :=
map_subtype_embedding_Icc _ _ _ (λ c _ x hx _ hc _, hc.trans_le hx)

lemma map_subtype_embedding_Ico : (Ico a b).map (function.embedding.subtype _) = Ico (a : ℕ) b :=
map_subtype_embedding_Ico _ _ _ (λ c _ x hx _ hc _, hc.trans_le hx)

lemma map_subtype_embedding_Ioc : (Ioc a b).map (function.embedding.subtype _) = Ioc (a : ℕ) b :=
map_subtype_embedding_Ioc _ _ _ (λ c _ x hx _ hc _, hc.trans_le hx)

lemma map_subtype_embedding_Ioo : (Ioo a b).map (function.embedding.subtype _) = Ioo (a : ℕ) b :=
map_subtype_embedding_Ioo _ _ _ (λ c _ x hx _ hc _, hc.trans_le hx)

@[simp] lemma card_Icc : (Icc a b).card = b + 1 - a :=
by rw [←nat.card_Icc, ←map_subtype_embedding_Icc, card_map]

@[simp] lemma card_Ico : (Ico a b).card = b - a :=
by rw [←nat.card_Ico, ←map_subtype_embedding_Ico, card_map]

@[simp] lemma card_Ioc : (Ioc a b).card = b - a :=
by rw [←nat.card_Ioc, ←map_subtype_embedding_Ioc, card_map]

@[simp] lemma card_Ioo : (Ioo a b).card = b - a - 1 :=
by rw [←nat.card_Ioo, ←map_subtype_embedding_Ioo, card_map]

@[simp] lemma card_fintype_Icc : fintype.card (set.Icc a b) = b + 1 - a :=
by rw [←card_Icc, fintype.card_of_finset]

@[simp] lemma card_fintype_Ico : fintype.card (set.Ico a b) = b - a :=
by rw [←card_Ico, fintype.card_of_finset]

@[simp] lemma card_fintype_Ioc : fintype.card (set.Ioc a b) = b - a :=
by rw [←card_Ioc, fintype.card_of_finset]

@[simp] lemma card_fintype_Ioo : fintype.card (set.Ioo a b) = b - a - 1 :=
by rw [←card_Ioo, fintype.card_of_finset]

end pnat

subset_name: curated


TITLE: Non-commutative property of rotation
QUESTION [1 upvotes]: Addition of angles are non-commutative in three dimensions. Hence some other angular vector quantities like angular velocity, momentum become non-commutative. What is the physical significance of this non-commutative property?

REPLY [2 votes]: I'll compare the equations of motion of a point particle and a rigid body.
To understand a free point particle, once I solve its motion in one dimension ($x(t) = v_0 t + x_0$) I can immediately solve the three dimensional problem. This is because the equations are separable and commutative: since translations along $x$ (label this symmetry $T_x$) implies a conserved quantity $p_x$, and translations along different axes are commutative ($T_x T_y = T_y T_x$), then (1) I can write any translation as $T_x T_y T_z$, and (2) I have the conserved quantities $p_x$, $p_y$, and $p_z$, which don't have anything to do with each other.
Even if I lose translational symmetry, the equations I need to solve generally look like
$$ \ddot x = f(x, y, z) $$
$$ \ddot y = g(x, y, z) $$
$$ ... $$
Now, I know that rotation about one axis (say $R_x$ is a rotation about the $x$ axis) has a conserved quantity $L_x$. That's true for all three axes. But it doesn't mean I can immediately write down the equations of motion; since $R_x R_y \neq R_y R_x$, I can't just write down three angular momenta and be done with it. The actual equations of motion are confusing because the coordinates are all mixed up. Even for a rigid body in free space, the equations we need to solve use Euler angles (this might be a better reference), which are the correct representation to find constants of motion. As you can see (when you look up Euler angles), they're coordinates called spin, nutation, precession that are defined in some funny way that seems to mix up the usual angles. However, that's the way to write down coordinates that will simplify the equations of motion and gives you conserved quantities.
In quantum mechanics, things get even more interesting. For translations, I have a quantum mechanical operator $\hat x$ that describes position. I know that $\hat p_x$ is the conjugate operator and that they don't commute ($[\hat x, \hat p_x] = i \hbar$). However, commutativity of the translational group means that most everything else you write down does commute, e.g. $[x, y] = [p_x, p_y] = [x, p_y] = 0$. This greatly simplifies the problem of solving the evolution of a point particle.
Now, the lack of commutativity of rotations gives rise to the Lie group in quantum mechanics, where the momentum operators $\hat L_x$, $\hat L_y$, and $\hat L_z$ don't commute. This leads to interesting symmetry groups in quantum mechanics, e.g. SU(N). SU(2), the symmetry of a spin-1/2 particle, turns out to be very related to SO(3), the rotational symmetries of a rigid body.


text:
\section{Conclusion}
We have demonstrated that the optimal interpolation between SAGA and gradient descent is not trivially SAGA. We thus presented formulas for the optimal probabilistic interpolation factor $q$ between SAGA and the optimal minibatch SAGA of size $\tau$. We demonstrated experimentally that the proposed SAGD is consistently better than SAGA on both synthetic and real dataset. We have also shown that the theory predicts what is observed in practice for optimal batch size $\tau^*$ of minibatch SAGA. Interestingly, experiments strongly suggest that a linear speedup is achieved with a parallel implementation.


TITLE: Powers of a unitary matrix
QUESTION [0 upvotes]: I'm trying to find the minimum exponent $M\in \mathbb{N}$, such that for a certain unitary matrix $F\in \mathbb{C}^{N}$,
$$F^M = 1_{\mathbb{C}^{N}}.$$
I don't think it matters, but it's the DFT matrix.
Now, I noticed that if $F^M = 1$, then $F^{M/2} = ({F^{M/2}})^\dagger$. However, I can't find a justification as to why M must be even. I'm not even sure if it has to be. (for the DFT matrix, M=4, though I'm still trying to figure out why it isn't 2)
Any ideas?

REPLY [4 votes]: It seems to me that this comes from the eigenvalues of a DFT matrix. Raising a matrix to the power $k$ raises its eigenvalues to the power $k$.
So, if the order of your matrix is $2$, you have $1$ eigenvalue equal to $-1$; when you have order greater than $2$, you have at least one eigenvalue equal to $-i$ (and, for order at least $4$, you have $i$ as well, but this doesn't change anything). See the table on the above link.
Notice that this does not hold for a general unitary matrix. For a counterexample, just take a 2D rotation by an angle $\varphi$ which is not of a form $\varphi = q\pi$ for $q \in \mathbb{Q}$ and no power of your matrix will ever become identity. Also, rotation for $\varphi = 2\pi/3$ will become identity when raised to the third (so, odd) power.