TITLE: D'Alembert criterion
QUESTION [0 upvotes]: can someone explain to me the middle steps between (1), (2), (3)?
This exercise is used as an example of the convergence of series, and I understand the criterion. I don't follow the middle steps.

Thank you in advance.

REPLY [0 votes]: $$\lim_{n\to \infty}{\left|{(n+1)^3+\sqrt{n+1}\over (2+4i)^{n+1}}\right|\over \left|{n^3+\sqrt{n}\over (2+4i)^{n}}\right|}{=\lim_{n\to \infty}{\left|{(n+1)^3+\sqrt{n+1}}\right|\over \left|{n^3+\sqrt{n}}\right|}\cdot \left|{{1\over  (2+4i)^{n+1}}\over {1\over  (2+4i)^{n}}}\right|\\=\lim_{n\to \infty}{\left|{(n+1)^3+\sqrt{n+1}}\right|\over \left|{n^3+\sqrt{n}}\right|}\cdot \left|{{(2+4i)^{n}\over  (2+4i)^{n+1}}}\right|\\=\lim_{n\to \infty}{\left|{(n+1)^3+\sqrt{n+1}}\right|\over \left|{n^3+\sqrt{n}}\right|}\cdot {1\over |2+4i|}\\=1\cdot {1\over \sqrt{20}}\\={1\over \sqrt{20}}}$$


TITLE: Let $n=pg$, then we get $p-1 \mid n-1$ because there are multiplicative groups of integers modulo n
QUESTION [1 upvotes]: I have one more question:
Let $n=pq$ with $p,q \in \mathbb{P}$, then we have $p-1 \mid n-1$ and $q-1 \mid n-1$. I don't understand the reason the author tells me: "because there are multiplicative groups of integers modulo $p$ and modulo $q$."  Is there any way to proof the existence of the prime integers?
Any help is appreciated.

REPLY [4 votes]: This just is not true.  Let $q=3$, $p=5$ and $n=3\cdot 5 =15$.  Certainly $p-1$ does not divide $n-1$ as $4$ does not divide $14$.
Another example is $q=5$, $p=7$, $n=35$.  Then neither of $p-1|n-1$ or $q-1|n-1$ hold.
When is it true?:  I claim that if $n=pq$ where $p,q$ are primes, and both $p-1|n-1$, $q-1|n-1$ hold, then we must have $q=p$.
Proof: Notice that we can write $$n-1=pq-1=p-1+p(q-1).$$  This means that if $p-1|n-1$, since $p$ and $p-1$ are relatively prime, we must have $$p-1|\left(p-1+p(q-1)\right)\Rightarrow p-1|q-1.$$  By symmetry, if $q-1|n-1$ as well then we have $q-1|p-1$. Hence they both divide each other so $q-1=p-1$, and we see they are equal.


TITLE: Prove a unique point satisfies $\theta f(p)=f(q)$ if $\theta: C(X)\rightarrow C(Y)$ denotes an isomorphism of complex algebra
QUESTION [1 upvotes]: Let $X$ and $Y$ be compact Hausdorff spaces, and $\theta: C(X)\rightarrow C(Y)$ denotes an isomorphism (just in algebraic sense) of complex algebra. Let $p\in Y$, I want to show that there is a unique point $q\in X$ s.t.
$$\theta f(p)=f(q), f\in C(X)$$

REPLY [0 votes]: This question can be found on Arveson's 'A short course on spectral theory', with the same unicity assumption. I don't see why it should be unique :
Note first that $\theta$ sends non-invertible elements to non-invertible elements and invertibles to invertibles (being an isomorphism), thus $$range(f)= \sigma(f)= \sigma(\theta f)= range(\theta f)$$
If $p\in Y$ is any point, $\theta f(p)\in range(f)$ so $\exists \ q\in X: \theta f(p)= f(q)$ .
This point need not be unique, take for example the unit element, the constant function $f:x\to 1$, you have $\theta f= 1_{C(Y)}$ and for any $p\in Y$ all of the points in $X$ satisfy $1_{C(Y)}(p)= f(q)$, so it suffice to take your $X$ to have more than one element.


TITLE: Frechet differentiable implies continuity in $\mathbb{R} ^n$
QUESTION [3 upvotes]: Need help completing a proof. The statement is as follows:

Let $E \subset \mathbb{R} ^m$ be an open set. If  $f:E \longrightarrow \mathbb{R}^n $ is Frechet differentiable at $a \in E$, then it is continuous at a.

Proof: Let $f:E \longrightarrow \mathbb{R}^n $ be Frechet differentiable. Then  for $a \in E$ ,   $\exists$  a linear map $A:\mathbb{R}^m \longrightarrow \mathbb{R}^n $ s.t. $\forall \epsilon>0$, $\exists\delta >0$, if $h \in \mathbb{R}^m$ and $0<||h||<\delta$, then $$||f(a+h)-f(a)-Ah||<\epsilon||h||$$ We then have for $$||f(a+h)-f(a)||\leq||h||\frac{||f(a+h)-f(a)-Ah||}{||h||}+||Ah||$$ by the triangle inequality. But I get stuck here. I feel that I'm very close to finishing it but just cant see it. Any hints or help will be appreciated!

REPLY [2 votes]: Suppose $f$ is Frechet differentiable at $a$, with derivative $A$.  Then we know that $$\lim_{h\to0} \frac{\|f(a+h)-f(a)-Ah\|}{\|h\|} = 0.$$  Define a function $\eta$ by $\eta(h) = {\|f(a+h)-f(a)-Ah\|}\big/{\|h\|} $ if $h\ne0$ and $\eta(0)=0$.  By the differentiability of $f$ at $a$ the function $\eta(h)$ is continuous at $h=0$.  (This is only a restatement of the differentiablilty of $f$, not a deep theorem.) So we know $$\tag{*}f(a+h) = f(a) + Ah +\|h\|\eta(h).$$ What we want is $\lim_{h\to0}f(a+h)=f(a)$.  But this can be read off of (*): the right hand side of (*) is  $f(a)$ plus terms that manifestly converge to $0$ as $h\to0$:  $Ah\to0$ as $h\to0$, and $\|h\|\to0$ and $\eta(h)\to0$ as $h\to 0$, and so does their product $\|h\|\eta(h)$, too, etc.


TITLE: Alternative to Choleski Decomposition for Correlation Matrix
QUESTION [1 upvotes]: Let $\Sigma$ be a correlation matrix, ie. symmetric. The Choleski decompositon gives upper triangular $A$ such that $A^TA = \Sigma$. Instead of upper triangularity, we are looking for $A$ that is not upper triangular, instead having each column sum to the same value and each entry $ \ge 0$. I've already managed to get this working in two dimensions with a simple solver, but three and up seem challenging. Is there a known method that would produce the desired matrix?
The goal is to apply a correlation matrix to random vectors with a "balanced" distortion effect on the points, rather than leaving the first vector the same and heavily altering the last few.

REPLY [3 votes]: While the question of positiveness was answered by Noah the other question (of equal column- (or row?) sums in A) can be dealt with the concept of Givens-rotation, where you iteratively rotate pairs of columns (or rows?) to a predefined optimization criterion. The angle for the pairwise column-rotations must then be determined by a criterion which minimizes the the difference of the sums. This should be similar to the centroid-rotation, but where we compute the maximizing criteria in the centroid-rotation we need the minimizing criteria now. Perhaps I can support this with an example later....
[update] I've got the rotation criterion for each elementary Givens-rotation. Let's -for my convenience- assume, that by $ \Sigma = A \cdot A^T$ the cholesky-factor $A$ of $ \Sigma$ is lower triangular. Then we iterate pairwise column-rotations in $A$ to pairwise and finally to completely equal column-sums. One such pairwise rotation is called "elementary Givens rotation".
The criterion-formula for the sought c as cosine of the rotation-angle and s as sine of the rotation-angle where x denotes the column with the lower index ("to the left") and y denotes the column with the higher index ("to the right") of one pairwise rotation and k goes over all row-indices is:
$$ \sum_k (x_k \cdot c - y_k \cdot s )  = \sum_k (x_k \cdot s + y_k \cdot c )$$
From this we can compute c and s as
$$ c \cdot \sum x_k  - s \sum y_k    = s \cdot \sum x_k + c \cdot \sum y_k  \\
  c \cdot \sum (x_k - y_k) =  s \sum (y_k + x_k) \\
  c = { a  \over z } \qquad s = { b  \over z } \\
\text{where } a=\sum (y_k + x_k) \qquad b=\sum (x_k - y_k) \qquad z = \sqrt{a^2+b^2}$$
Then the (elementary Givens-) rotation of the columns X and Y is
$$ X' = X\cdot c - Y \cdot s \qquad Y'=X\cdot s + Y \cdot c $$
One iteration-pass is then to apply this computation to all combinations of pairs of columns, where only always the X column means that of the smaller index.
Here is one example with a random correlation matrix, its initial cholesky-factor A and the vector SU containing the column-sums (which shall be made equal by the rotations):
        |   1,0000    0,0000      0,0000      0,0000 |
        |   0,7139    0,7003      0,0000      0,0000 |
 A=     |   0,9022    0,3439      0,2603      0,0000 |
        |   0,5865    0,6970      0,3367      0,2382 |
            =====================================
SU=     |   3,2026    1,7412      0,5970      0,2382 |

One rotation on the leftmost columns gives
        |   0,9590    0,2835      0,0000      0,0000 |
        |   0,4861    0,8739      0,0000      0,0000 |
  A=    |   0,7677    0,5856      0,2603      0,0000 |
        |   0,3649    0,8347      0,3367      0,2382 |
==========================================================
 SU=    |   2,5777    2,5777      0,5970      0,2382 |

and after 5 complete iteration passes I get
        |   0,7133    0,0320      0,4429      0,5422 |
        |   0,4761    0,7200      0,2739      0,4241 |
 A=     |   0,5372    0,3850      0,6116      0,4349 |
        |   0,1242    0,7138      0,5225      0,4495 |
 =========================================================
 SU=    |   1,8508    1,8508      1,8508      1,8508 |

where we see, that all column-sums are equal.


TITLE: What is the proper term for a function where domain and codomain coincide?
QUESTION [1 upvotes]: What is the proper term for a function where domain and codomain coincide?

E.g. in programming languages a function f : Int => Int or f : Double => Double.
Thanks.

REPLY [1 votes]: I would like to add this as a comment, but I don't have enough reputation to do so.
A function $f:S\to S$ can be called an endomorphism in Set, and one doesn't have to worry about the structure-preservation that is implied by -morphism because sets have no additional structure to preserve.
However, it depends on the context: if I want to talk about any function $f:V\to V$, $V$ being a vector space, and I were to use the term endomorphism, it would be understood that $f$ is a linear transformation, because the morphisms in $K$-Vect are linear transformations.


text:
\section{Introduction}\label{sec:intro}

\subsection{Background and motivation}

Let $K$ be an algebraically closed field of characteristic $p>0$.
A nonsingular curve over $K$ is called \emph{superspecial} (resp.\ \emph{supersingular}) if its Jacobian variety is isomorphic (resp.\ isogenous) to a product of supersingular elliptic curves.
Superspecial curves are not only theoretically interesting in algebraic geometry and number theory but also have many applications in coding theory, cryptology, and so on, because they tend to have many rational points and their Jacobian varieties have large endomorphism rings.
However, it is not always easy to find such curves,
and there are only finitely many superspecial curves for a given genus and characteristic.
One method of constructing
superspecial curves is to consider fiber products of superspecial curves of lower genera.
In this paper, we demonstrate that this method can be efficient by considering the simplest
example in which the genus is at least $4$:
the case of Howe curves.
A \emph{Howe curve}\footnote{So named by Senda and the first two authors in \cite{KHS20}.} is a curve of genus $4$ obtained as the fiber product of two genus-$1$ double covers $E_1\to\bbP^1$ and $E_2\to\bbP^1$.
In \cite{Howe}, the third author studied these curves in order to quickly construct genus-$4$ curves with many rational points.

\subsection{Related works}

The reason that we consider the case of genus $g\ge 4$ is that the enumeration of the isomorphism classes of superspecial curves with $g\le 3$ has already been done, by Deuring~\cite{Deuring} for $g=1$, by Ibukiyama, Katsura, and Oort \cite{IKO} for $g=2$, and by Brock~\cite{Brock} for $g=3$; see also Ibukiyama~\cite{Ibukiyama} and Oort~\cite{Oort91} for the existence of such curves for $g=3$.
In contrast to the case $g\le 3$, the existence or non-existence of a superspecial curve of genus $4$ in general characteristic is an open problem, although some results for specific small $p$ are known;
see \cite[Theorem~1.1]{Ekedahl} for the non-existence for $p\le 3$ and \cite[Theorem~B]{KH17} for the non-existence for $p=7$.
As for enumeration, computational approaches have been proposed recently in \cite{KH18}, \cite{KH17}, and \cite{KH19} in the case of genus~$4$. The main strategy common to
these papers is to parametrize a family of curves (canonical curves in
the first two papers, hyperelliptic curves in the third), and then to
find the superspecial curves $X$ in these families
by computing the zeros of a multivariate system derived from the condition that the Cartier--Manin matrix of $X$ is zero.
With computer algebra techniques such as Gr\"{o}bner bases, the authors
of these papers enumerated superspecial canonical
curves for $p \leq 11$ in \cite{KH18} and \cite{KH17} and superspecial hyperelliptic curves for $p \leq 23$ in~\cite{KH19}.
However, results for larger $p$ have not been obtained yet due to the cost of solving multivariate systems, and no complexity analysis is given in \cite{KH18}, \cite{KH17}, or~\cite{KH19}.

Now we  turn our attention to Howe curves.
Recently, it was proven in \cite{KHS20} that there exists a supersingular Howe curve in every positive characteristic.
In particular, the authors of \cite{KHS20} reduce the existence of such a curve
to the existence of a zero of a certain multivariate system, as follows:
They study a family of Howe curves realized as
$E_1\colon z^2=f_1(x)$ and $E_2\colon w^2=f_2(x)$ for cubic polynomials $f_1$ and $f_2$
parametrized by elements $(\lambda : \mu : \nu)$ of $\bbP^2$.
Let $C$ be the genus-$2$ curve $y^2 = f_1f_2$.
The supersingularity of $H$ is equivalent to
that of $E_1$, $E_2$ and $C$,
because there exists an isogeny of
$2$-power degree from the Jacobian $J (H)$ to $E_1 \times E_2 \times J(C)$~\cite[Theorem~2.1]{Howe}).
Thus, once supersingular isomorphism classes of $E_1$ and $E_2$ are given,
finding supersingular curves $H$ is reduced to finding values of the parameter $(\lambda : \mu : \nu)$ that satisfy a multivariate system
derived from the supersingularity of $C$.
The authors of \cite{KHS20} deduced the existence of such a zero $(\lambda : \mu : \nu)$ from various algebraic properties of the defining polynomials of the system.

The above reduction is applicable also for the superspecial case, but the method used in \cite{KHS20} to prove the existence of solutions does not carry over well.
For this reason, the superspecial case is still open, and we are left to ask:
For which primes $p>7$ are there superspecial Howe curves in characteristic~$p$?

\subsection{Our contribution}

We study the existence of superspecial Howe curves by creating efficient
algorithms to produce and enumerate them. The following theorems summarize
some of what we have found.

\begin{theorem}\label{thm:main1}
For every prime $p$ with $7 < p < 20000$ or with $p\equiv 5\bmod 6$, there exists a superspecial Howe curve in characteristic~$p$.
\end{theorem}

\begin{theorem}\label{thm:main2}
For every prime $p$ with $7 < p \le 199$, the number of isomorphism classes of superspecial Howe curves in characteristic $p$ is given in
Table~\textup{\ref{table:enumerate}}.
\end{theorem}

\begin{table}[ht]
\caption{For each prime $p$ from $11$ to $199$, we give the number $n(p)$ of superspecial Howe curves over $\overline{\bbF}_p$ and the ratio of $n(p)$
to the heuristic prediction $p^3/1152$ (see Section~\ref{sec:second}).}
\label{table:enumerate}
\vspace{-1ex}
\begin{center}
\renewcommand{\arraystretch}{0.9}
\begin{tabular}{rrrrrrrrrrr}
\toprule
  $p$ &  $n(p)$ &\small    Ratio &\hbox to 1em{}&   $p$ &  $n(p)$ &    Ratio &\hbox to 1em{}&   $p$ &  $n(p)$ &    Ratio \\
\cmidrule{1-3}\cmidrule{5-7}\cmidrule{9-11}
 $11$ &     $4$ &  $3.462$ &&  $67$ &   $260$ &  $0.996$ && $137$ &  $2430$ &  $1.089$ \\
 $13$ &     $3$ &  $1.573$ &&  $71$ &   $742$ &  $2.388$ && $139$ &  $2447$ &  $1.050$ \\
 $17$ &    $10$ &  $2.345$ &&  $73$ &   $316$ &  $0.936$ && $149$ &  $3082$ &  $1.073$ \\
 $19$ &     $4$ &  $0.672$ &&  $79$ &   $595$ &  $1.390$ && $151$ &  $3553$ &  $1.189$ \\
 $23$ &    $33$ &  $3.125$ &&  $83$ &   $655$ &  $1.320$ && $157$ &  $3427$ &  $1.020$ \\
 $29$ &    $45$ &  $2.126$ &&  $89$ &   $863$ &  $1.410$ && $163$ &  $3518$ &  $0.936$ \\
 $31$ &    $59$ &  $2.281$ &&  $97$ &   $802$ &  $1.012$ && $167$ &  $6268$ &  $1.550$ \\
 $37$ &    $41$ &  $0.932$ && $101$ &  $1207$ &  $1.350$ && $173$ &  $4780$ &  $1.064$ \\
 $41$ &   $105$ &  $1.755$ && $103$ &  $1151$ &  $1.213$ && $179$ &  $5771$ &  $1.159$ \\
 $43$ &    $79$ &  $1.145$ && $107$ &  $1237$ &  $1.163$ && $181$ &  $5419$ &  $1.053$ \\
 $47$ &   $235$ &  $2.608$ && $109$ &  $1193$ &  $1.061$ && $191$ &  $9610$ &  $1.589$ \\
 $53$ &   $167$ &  $1.292$ && $113$ &  $1323$ &  $1.056$ && $193$ &  $6298$ &  $1.009$ \\
 $59$ &   $259$ &  $1.453$ && $127$ &  $2013$ &  $1.132$ && $197$ &  $6839$ &  $1.030$ \\
 $61$ &   $243$ &  $1.233$ && $131$ &  $2606$ &  $1.335$ && $199$ &  $8351$ &  $1.221$ \\
\bottomrule
\end{tabular}
\vskip 3ex
\end{center}
\end{table}

The upper bounds on $p$ in these two theorems can easily be increased. For example, on a 2.8 GHz Quad-Core Intel Core i7 with 16GB RAM, computing the
$8351$ superspecial Howe curves in characteristic $199$ using method (B) below
took $124$ seconds in Magma. Finding examples of superspecial Howe curves for
every $p$ between $7$ and $20000$ took $680$ minutes on the same machine.

In this paper we discuss two strategies, (A) and (B) below, to find superspecial Howe curves. We also show how isomorphisms between Howe curves can be easily detected from the data that defines them (C).

\subsection*{(A) \texorpdfstring{$(E_1,E_2)$}{(E1,E2)}-first, using Cartier--Manin matrices.}
In this strategy, we use the same realization of Howe curves as in \cite{KHS}, that is,
the fiber product of
$E_1\colon z^2y = x^3 + A_1 \mu^2 xy^2 + B_1 \mu^3y^3$ and
$E_2\colon w^2y = (x-\lambda)^3 + A_2 \mu^2 (x-\lambda)y^2 + B_2 \mu^3y^3$
over $\bbP^1=\Proj K[x,y]$.
We enumerate pairs $(E_1,E_2)$ of supersingular elliptic curves
so that $C$ is superspecial.
We first discuss the field of definition of superspecial Howe curves (cf.\;Proposition \ref{prop:field_of_def}), which enables us to reduce the size of our search space drastically.
Specifically, the coordinates $A_1$, $B_1$, $A_2$, $B_2$, $\lambda$, $\mu$, $\nu$ belong to $\bbF_{p^2}$, whereas in the supersingular case~\cite{KHS20} these coordinates can generate larger
subfields of $\overline{\bbF}_{p}$.
For the test of superspeciality, we use the criterion that the Cartier--Manin matrix of $C$ must be zero \cite[Lemma~1.1(i)]{IKO}.
This reduces the enumeration problem to solving a system of algebraic equations. See Section \ref{sec:first} for the details of this strategy, including a complexity analysis.

\subsection*{(B) \texorpdfstring{$C$}{C}-first, using Richelot isogenies.}
The second strategy first enumerates
superspecial curves $C\colon y^2=f(x)$ of genus $2$ with $f(x)$ of degree $6$
and then enumerates decompositions $f(x)=f_1(x)f_2(x)$ with $f_i(x)$ of degree $3$
so that there is a $b$ that makes both curves $E_i\colon y^2 = (x-b)f_i(x)$ supersingular.
The moduli space of curves of genus $2$ is of dimension $3$.
As this dimension is bigger than the space of $(\lambda:\mu:\nu)\in \bbP^2$ considered in (A), this strategy, a priori, looks inefficient. But, surprisingly, we conclude that
strategy (B) enumerates superspecial Howe curves much more efficiently than does (A).
The advantage of (B) comes from making use of Richelot isogenies.
Specifically, we construct some superspecial curves of genus $2$ by
gluing supersingular elliptic curves together along their $2$-torsion
\cite[\S3]{HLP}, and then produce more such curves by applying Richelot
isogenies to the curves already produced. This procedure terminates because there
are only finitely many superspecial curves of genus $2$, and a recent result
of Jordan and Zaytman~\cite[Corollary~18]{JZ} shows that we obtain all isomorphism classes of superspecial curves of genus $2$ in this way.

\subsection*{(C) A new isomorphism test for Howe curves.}
Strategy (A) above produces many
not-necessarily-distinct Howe curves, so to prevent overcounting we are left with the task of
producing a unique representative for each isomorphism class. As every Howe curve is canonical (cf.\;Proposition \ref{HoweIsCanonical}), one may check whether two Howe curves are isomorphic by using the isomorphism test for canonical curves given in \cite[\S6.1]{KH17}, whose implementation is found in \cite[\S4.3]{KH18}. This turns out to be very costly, because it uses many
Gr\"obner basis computations. Our Corollary~\ref{C:isomorphism} gives a
much simpler isomorphism test, based on the observation that a Howe curve
is completely determined (up to isomorphism) by the degree-$2$ map to a genus-$2$ curve it is provided with by virtue of its definition as a fiber product. This isomorphism test is added on as a separate step in strategy (A), but is baked into the algorithm we use for strategy (B).

\subsection*{Acknowledgments}

The first and the second authors thank Everett Howe for
joining as the third author; he told them the second strategy (B), which had not been considered in the earlier version \cite{KH20}.
The third author thanks Professors Kudo and Harashita for inviting him to join them in this work.
The authors thank the referees for their careful reading and for helpful suggestions and comments.
This work was supported by JSPS Grant-in-Aid for Scientific
Research (C) 17K05196, JSPS Grant-in-Aid for Research Activity Start-up 18H05836
and 19K21026, and JSPS Grant-in-Aid for Young Scientists 20K14301.

config: arxiv


TITLE: Solve $ y'(x)+2y(x)+3y(-x)=0$
QUESTION [1 upvotes]: I just arrived to an extremely weird ODE and I was wondering how could I solve it: $$
y'(x)+2y(x)+3y(-x)=0.
$$
Actually, I am not even sure if it is possible to prove that it has a unique solution (provided an initial condition). I tried to put it on wolfram alpha but I didn't get anything. My first thought was to use "separation of variables" method but I don't think that it will help because of the term $y(-x)$. Actually I am completely clueless.
Edit: I am actually wondering if this ODE has non-symmetric solutions, that is, neither odd nor even solutions.

REPLY [2 votes]: Decompose $y(x)$ as a sum of its even and odd part:
$$y(x) = y_e(x) + y_o(x)\quad\text{with}\quad \begin{cases} y_e(x) &= \frac12(y(x) + y(-x))\\ y_o(x) &= \frac12(y(x) - y(-x))\end{cases}$$
The ODE becomes
$$(y_e+y_o)' + 2(y_e+y_o) + 3(y_e-y_o) = 0\tag{*1a}$$
Subsititute $x$ by $-x$ and notice under such a change, $\frac{d}{dx}$ and $y_o$ picks up a minus sign while $y_e$ remains the same. We get
$$-(y_e - y_o)' + 2(y_e - y_o) + 3(y_e + y_o) = 0\tag{*1b}$$
Combined $(*1a)$ and $(*1b)$, we get
$$y'_o + 5y_e = y'_e - y_o = 0\quad\implies\quad y''_e + 5y_e = 0
$$
This implies $$y_e(x) = A \cos(\sqrt{5}x) + B\sin(\sqrt{5}x)$$
for suitably chosen constants $A, B$. Since $y_e(x)$ is even, $B = 0$
and hence $$y_e(x) = A\cos(\sqrt{5}x)
\quad\implies\quad y_o(x) = y'_e(x) = -A\sqrt{5}\sin(\sqrt{5}x)$$
Since $A = y_e(0) = y(0)$, we have
$$y(x) = y(0)(\cos(\sqrt{5}x) - \sqrt{5}\sin(\sqrt{5}x))
$$
Up to an overall scaling factor $A = y(0)$, this solution is unique.


TITLE: Integral $\int _0^1\frac{\ln \left(x\right)\text{Li}_2\left(x\right)}{x\left(4\pi ^2+\ln ^2\left(x\right)\right)}\:\mathrm{d}x$
QUESTION [8 upvotes]: A comrade sent me this conjecture
$$\int _0^1\frac{\ln \left(x\right)\operatorname{Li}_2\left(x\right)}{x\left(4\pi ^2+\ln ^2\left(x\right)\right)}\:\mathrm{d}x=3\zeta (2)\left(4\ln \left(A\right)-1\right)$$
$A$ here represents the Glaisher–Kinkelin constant
, but this time I truly have no idea how to approach the problem, my mate said that it can even be generalized like
$$\int _0^1\frac{\ln \left(x\right)\operatorname{Li}_2\left(x\right)}{x\left(4\pi ^2n^2+\ln ^2\left(x\right)\right)}\:\mathrm{d}x,\:n\in\mathbb{N}$$
I tried using Landen's identity among other functional equations but neither worked, I also tried to apply differentiation under the integral sign
$$\mathcal{J}(s)=\int _0^1\frac{x^{s-1}\operatorname{Li}_2\left(x\right)}{4\pi ^2+\ln ^2\left(x\right)}\:\mathrm{d}x$$
But as you can see it's hopeless, how else can I proceed?

REPLY [12 votes]: $$\int_0^1\frac{\ln x\color{blue}{\operatorname{Li}_2(x)}}{x(4\pi^2+\ln^2 x)}dx\overset{\large x\to e^{-x}}=-\int_0^\infty \color{blue}{\sum_{n=1}^\infty \frac{1}{n^2}e^{-nx}}\frac{x}{4\pi^2+x^2}dx$$
$$\small \overset{nx\to x}=-\frac14\int_0^\infty \color{red}{\sum_{n=1}^\infty \frac{1}{n^2}}\frac{xe^{-x}}{\color{red}{(n\pi)^2+\left(\frac{x}{2}\right)^2}}dx=-\frac14\color{red}{\frac{\pi^2}{2}}\int_0^\infty \color{red}{\left(\frac{1}{\left(\frac{x}{2}\right)^4}+\frac{1}{3\left(\frac{x}{2}\right)^2}-\frac{\coth \left(\frac{x}{2}\right)}{\left(\frac{x}{2}\right)^3}\right)}xe^{-x}dx$$
$$=\frac{\pi^2}{2}\int_0^\infty \left(\frac{4}{x^2(1-e^{-x})}-\frac{1}{3x}-\frac{2}{x^2}-\frac{4}{x^3}\right) e^{-x}dx=\frac{\pi^2}{2}I(0)=\boxed{3\zeta(2)\left(4\ln(A)-1\right)}$$
The red sum can be deduced like in the first part of this answer and also above we introduced  the $I(s)$ integral similarly as seen here.
$$I(s)=\int_0^\infty \left(\frac{4}{x^2(1-e^{-x})}-\frac{1}{3x}-\frac{2}{x^2}-\frac{4}{x^3}\right)x^s e^{-x}dx$$
$$=\int_0^\infty \left(4\sum_{n=1}^\infty  x^{s-2}e^{-nx}-\frac13 x^{s-1}e^{-x}-2 x^{s-2}e^{-x}-4 x^{s-3}e^{-x}\right)dx$$
$$=4\Gamma(s-1)\zeta(s-1)-\frac13\Gamma(s)-2\Gamma(s-1)-4\Gamma(s-2)$$
$$=\Gamma(s+1)\left(\color{blue}{\frac{4\zeta(s-1)+\frac13}{(s-1)s}}\color{red}{-\frac{\frac{1}{3}}{(s-1)s}-\frac{1}{3s}-\frac{2}{(s-1)s}-\frac{4}{(s-2)(s-1)s}}\right)$$
$$\Rightarrow I(0)=\lim_{s\to 0} I(s)=\color{blue}{-4\zeta'(-1)}\color{red}{-\frac23}=4\ln(A)-1$$
Above follows from the relation of the
derivatives of the Zeta function with the Glaisher's constant.


text:
\label{sec:hyperanalytic}
We first review and examine the notion of orthogonality in the
case of complex signals before we consider how to construct a signal
orthogonal to a given complex signal.
\begin{definition}
\label{def:orthogonality}
Two complex signals $x(t)$ and $y(t)$ are orthogonal if:
$\int\limits_{-\infty}^{\infty}\!x(t)\cconjugate{y}(t)\,\dt = 0$,
where $\star$ denotes the complex conjugation.
\end{definition}
Orthogonality is invariant to multiplication of either
signal by any real or complex constant.
This is evident from the definition, since if
the integral is zero for a given pair of $x(t)$ and $y(t)$
it must also be zero for $\zeta y(t)$ or $\zeta x(t)$,
where $\zeta$ is a complex constant.
Therefore, given a pair of orthogonal signals,
one of the signals may be rotated by an arbitrary angle
in the complex plane (constant with respect to time),
and still remain orthogonal to the other.
Obviously the same applies for a real constant,
so scaling one or both of the signals does not alter their orthogonality.

Consider the integral in Definition \ref{def:orthogonality}
separated into real and imaginary parts:
\begin{align}
\label{eqn:seporthoreal}
\int\limits_{-\infty}^{\infty}\!(x_r(t)y_r(t)+x_i(t)y_i(t))\,\dt&=0\\
\label{eqn:seporthoimag}
\int\limits_{-\infty}^{\infty}\!(x_i(t)y_r(t)-x_r(t)y_i(t))\,\dt&=0
\end{align}
where $x_r(t)$ (resp. $y_r(t)$) is the real part of $x(t)$
(resp. $y(t)$) and $x_i(t)$ (resp. $y_i(t)$) is the imaginary part of $x(t)$
(resp. $y(t)$).
There are two significantly different ways in which these integrals
can vanish.
One way is for all four of the products
within them to vanish when integrated:
\begin{align*}
\int\limits_{-\infty}^{\infty}\!x_r(t)y_r(t)\,\dt=0\qquad
\int\limits_{-\infty}^{\infty}\!x_i(t)y_i(t)\,\dt=&0\\
\int\limits_{-\infty}^{\infty}\!x_i(t)y_r(t)\,\dt=0\qquad
\int\limits_{-\infty}^{\infty}\!x_r(t)y_i(t)\,\dt=&0
\end{align*}
This can only happen if both complex signals have an imaginary
part orthogonal to their real part\footnote{An example of such
a signal can be created numerically by taking four singular vectors
from a singular value decomposition of a real matrix, since the
singular vectors are mutually orthogonal.}. In this case only,
orthogonality is invariant to complex conjugation of either or
both of the complex signals.

In more general cases the integrals in \eqref{eqn:seporthoreal}
and \eqref{eqn:seporthoimag} vanish because the product terms
within them integrate to equal values and appropriate signs
(opposite signs in the first case, the same sign in the second).

Of course, a hybrid of these possibilities is also possible,
and indeed occurs in the next section: one integral vanishes
because each of the product terms within it integrates to zero,
and the other vanishes because the product terms integrate to
the same value with appropriate sign, and cancel.

\subsection{Hilbert transform}
\label{sec:hilbert}
\begin{figure}
\setlength{\unitlength}{0.01\columnwidth}
\thicklines
\begin{picture}(100,30)
\put(70,13){\circle{8}}
\put(70,13){\makebox(0,0){$\times$}}
\put(70,4){\vector(0,1){5}}
\put(70,0){\makebox(0,3){$-1$}}
\put(30,20){\framebox(20,10){$\mathcal{H}$}}
\put(30,8){\framebox(20,10){$\mathcal{H}$}}
\put(15,13){\vector(1,0){15}}
\put(15,25){\vector(1,0){15}}
\put(50,13){\vector(1,0){16}}
\put(50,25){\vector(1,0){35}}
\put(74,13){\vector(1,0){10}}
\put(8,25){\makebox(0,0){$z_r(t)$}}
\put(8,13){\makebox(0,0){$z_i(t)$}}
\put(85,25){\makebox(15,0){$o_r(t)$}}
\put(85,13){\makebox(15,0){$o_i(t)$}}
\end{picture}
\caption{\label{fig:dual_hilbert}Construction of the orthogonal signal using a pair of Hilbert transforms.}
\end{figure}

An orthogonal signal may be constructed using a pair of
classical Hilbert transforms as shown in Figure~\ref{fig:dual_hilbert}.
\begin{lemma}
\label{lemma:hilbertcross}
Given two arbitrary real-valued functions $f(t)$ and $g(t)$
the following holds, where $\hilbert{x(t)}$ is the Hilbert transform
of $x(t)$, provided the Hilbert transforms exist:
\begin{equation}
\label{eqn:hilbertcross}
\int\limits_{-\infty}^{\infty}f(t)\hilbert{g(t)}\,\dt
\,+\!
\int\limits_{-\infty}^{\infty}\hilbert{f(t)}g(t)\,\dt
=0
\end{equation}
\end{lemma}
\begin{IEEEproof}
We make use of the inner product form of Parseval's Theorem
\cite[\S\,27.9 equation (27.34)]{Jordan:2008} or
\cite[Theorem 7.5]{Champeney:1987}:
\begin{equation}
\label{eqn:innerparseval}
\int\limits_{-\infty}^{\infty}x(t)\cconjugate{y}(t)\,\dt
=
\int\limits_{-\infty}^{\infty}X(\nu)\cconjugate{Y}(\nu)\,\dnu
\end{equation}
where $X(\nu)$ is the Fourier transform of $x(t)$ and similarly for $y(t)$.
Note that, if $y(t)$ is real, as it is in what follows, $Y(\nu)$ must still
be conjugated.
We also make use of the Fourier transform of a Hilbert transform
\cite[(2.13-8, p\,68)]{Stark:1979}:
$\ffourier{\hilbert{x(t)}}=-i\sign(\nu)X(\nu)$, where $X(\nu)=\ffourier{x(t)}$.
Using \eqref{eqn:innerparseval}, we rewrite \eqref{eqn:hilbertcross} as:
\[
\int\limits_{-\infty}^{\infty}\!\cconjugate{F}(\nu)i\sign(\nu)G(\nu)\,\dnu
+
\!\!\int\limits_{-\infty}^{\infty}\!i\sign(\nu)F(\nu)\cconjugate{G}(\nu)\,\dnu=0
\]
which may be simplified to:
\[
i\!\int\limits_{-\infty}^{\infty}\!\sign(\nu)\left[\cconjugate{F}(\nu)G(\nu)
                                                  +F(\nu)\cconjugate{G}(\nu)
                                            \right]\,\dnu=0
\]
Since $\cconjugate{F}(\nu)G(\nu)$ and $F(\nu)\cconjugate{G}(\nu)$ are
conjugates, their imaginary parts cancel and the term inside the brackets
is real.
Since $F(\nu)$ and $G(\nu)$ are the Fourier transforms of real signals,
their real parts are even functions of $\nu$.
Therefore when multiplied by the signum function, they integrate to zero.
\end{IEEEproof}
\begin{proposition}
\label{prop:dual_hilbert}
A complex signal $o(t)$ constructed from an arbitrary complex signal
$z(t)$ using two classical Hilbert transforms operating independently
on the real and imaginary parts,
as represented in Figure~\ref{fig:dual_hilbert} and below:
\begin{equation}
\label{eqn:orthogonal}
o(t) = \hilbert{z_r(t)} - i\hilbert{z_i(t)}
\end{equation}
is orthogonal to $z(t)$, that is:

$\int\limits_{-\infty}^{\infty}z(t)\cconjugate{o}(t)\,\dt = 0$

\end{proposition}

\begin{IEEEproof}
First consider the real part of the integral:
\begin{equation*}
\int\limits_{-\infty}^{\infty}\left[z_r(t)o_r(t)+z_i(t)o_i(t)\right]\,\dt = 0
\end{equation*}
and write it as two integrals with the real and imaginary parts of $o(t)$
replaced by their equivalents from \eqref{eqn:orthogonal}:
\begin{equation*}
\int\limits_{-\infty}^{\infty}z_r(t)\hilbert{z_r(t)}\,\dt\,-\!
\int\limits_{-\infty}^{\infty}z_i(t)\hilbert{z_i(t)}\,\dt =0
\end{equation*}
Both integrals evaluate to zero as a consequence of orthogonality between:
$z_r(t)$ and its Hilbert transform;
and $z_i(t)$ and its Hilbert transform.
Now consider the imaginary part of the integral:
\begin{equation*}
\int\limits_{-\infty}^{\infty}\left[z_i(t)o_r(t)-z_r(t)o_i(t)\right]\,\dt =0
\end{equation*}
and replace the real and imaginary parts of $o(t)$ as before:
\begin{equation*}
\int\limits_{-\infty}^{\infty}z_i(t)\hilbert{z_r(t)}\,\dt\,+\!
\int\limits_{-\infty}^{\infty}z_r(t)\hilbert{z_i(t)}\,\dt =0
\end{equation*}
This is true by Lemma~\ref{lemma:hilbertcross}.

\end{IEEEproof}

\subsection{Complex Fourier transform pair}
An alternative method to construct an orthogonal signal is shown in
Figure~\ref{fig:complex_ft} using a complex Fourier transform pair operating
directly on the complex signal, rather than Hilbert transforms
operating on the real and imaginary parts separately.
The Fourier spectrum of $z(t)$ is multiplied by a signum function times
$i$, equivalent to multiplication of positive frequencies by $e^{i\pi/2}$
and negative frequencies by $e^{-i\pi/2}$.
This has the effect of rotating positive frequency coefficients by
$\pi/2$ and negative frequency coefficients by $-\pi/2$.
An inverse Fourier transform then yields the conjugate of the orthogonal
signal.
\begin{proposition}
\label{prop:dual_fft}
A complex signal $o(t)$ constructed from an arbitrary complex signal
$z(t)$ using a complex Fourier transform as represented in
Figure~\ref{fig:complex_ft} and below, is identical to the orthogonal
signal in Proposition~\ref{prop:dual_hilbert}, equation \eqref{eqn:orthogonal}.
\begin{equation}
\label{eqn:dual_fft_orthogonal}
\cconjugate{o}(t) = \ifourier{-i\sign(\nu)Z(\nu)}
\end{equation}
where $\fourier$ denotes a Fourier transform and $Z(\nu)=\ffourier{z(t)}$.
\end{proposition}
\begin{IEEEproof}
From equation \eqref{eqn:orthogonal}, $\cconjugate{o}(t)=\hilbert{z_r(t)} + i\hilbert{z_i(t)}$.
Taking the Fourier transforms of the Hilbert transforms (see Lemma \ref{lemma:hilbertcross}),
we have:
\begin{align*}
\ffourier{\cconjugate{o}(t)}&=-i\sign(\nu)Z_r(\nu)-i\sign(\nu)Z_i(\nu)\\
                            &=-i\sign(\nu)Z(\nu)
\end{align*}
from the linearity of the Fourier transform.

\end{IEEEproof}
\begin{figure}
\setlength{\unitlength}{0.01\columnwidth}
\thicklines
\begin{picture}(100,21)
\put(50,16){\circle{8}}
\put(50,16){\makebox(0,0){$\times$}}
\put(50,6){\vector(0,1){6}}
\put(50,0){\makebox(0,5){$-\i\sign(\nu)$}}
\put(15,11){\framebox(20,10){$\fourier$}}
\put(65,11){\framebox(20,10){$\fourier^{-1}$}}
\put(0,16){\makebox(7,0){$z(t)$}}
\put(7,16){\vector(1,0){8}}
\put(35,16){\vector(1,0){11}}
\put(35,16){\makebox(11,6){$Z(\nu)$}}
\put(54,16){\vector(1,0){11}}
\put(85,16){\vector(1,0){6}}
\put(91,16){\makebox(9,0){$\cconjugate{o}(t)$}}
\end{picture}
\caption{\label{fig:complex_ft}Construction of the orthogonal signal using a complex Fourier transform.}
\end{figure}

\subsection{Discussion}
We have shown in this section how to construct a complex signal $o(t)$
orthogonal to an arbitrary complex signal $z(t)$ using classical Hilbert
or complex Fourier transforms.
The notion of analytic signal, however, requires more than just an
orthogonal signal --- generalising from the classical case, we must
treat the original and orthogonal complex signals as a pair in order
to define the generalisations of envelope and phase.
In the next section we introduce the quaternion Fourier transform,
and show that this permits us to construct a \emph{hyperanalytic signal}
from which both the original signal and the orthogonal signal
defined in the current section may be extracted.
Using the quaternion hyperanalytic signal we then show how to
construct the envelope and phase in the complex case.


TITLE: Geodesics of this metric
QUESTION [6 upvotes]: I have to calculate the geodesics of the metric:
$$\left(\matrix {1 &0\\0& x^2 }\right)$$
I've been able to derive its equations, which are:
$$\ddot x -x\dot y ^2=0$$
$$\ddot y+\frac{2}{x}\dot x\dot y=0$$
It's easy to check that lines of constant $y$ are solutions of that equation: all $(v_0t+x_0,y_0)$ satisfy the above, so geodesics between two points of the form $(x_1,y_1)$, $(x_2,y_1)$ are straight lines, but I can't get the general solutions.
Any help? Thanks in advance.
BTW, To avoid calculations on your side, the Riemann tensor, and therefore the Ricci tensor and scalar curvature, vanish., so $R^2$ with that metric is flat

REPLY [3 votes]: Note that
$$ \ddot{y} + \frac{2}{x}\dot{x}\dot{y} = \frac{1}{x^2}\frac{d}{dt}\left(\dot{y}x^2\right) = 0$$
so we can now solve for $\dot{y}$:
$$\dot{y} = \frac{C}{x^2}$$
which gives
$$\ddot{x} - x\dot{y}^2 = \ddot{x} - \frac{C^2}{x^3} = 0 $$
This is solved by
$$x(t) = \pm\frac{\sqrt{C^2 + a^2(t+b)^2}}{\sqrt{a}}$$
for integration constants $a,b$. The last equation I leave for you to solve
$$y(t) - y(0) = \int_0^t\frac{C a}{C^2 + a^2(t+b)^2}dt $$


TITLE: Unsure how to prove that this function is homeomorphism:
QUESTION [1 upvotes]: Consider the sets: $S^{1} = \{(x,y) \in \mathbb{R^2}:x^2+y^2=1\}$ and $Q = \{(x,y)\in\mathbb{R}^2:|x|+|y|=1\}$ and the function: $f:Q\to S^1$ given by:
$f(x,y) = (\frac{x}{\sqrt{x^2+y^2}},\frac{y}{\sqrt{x^2+y^2}})$. I want to show that this function is a homeomorphism. Given the definition of a homeomorphism is a continuous, bijective function with a continuous inverse. To prove it is bijective I have said: Let $(a,b),(c,d) \in Q$ so that: $|a|+|b|=1$ and $|c|+|d|=1$. Suppose that $f(a,b) = f(c,d)$. From this I have gathered that $\frac{a}{b} = \frac{c}{d}$, but I don't know how to deduce that $a = c$ and $b = d$. Furthermore it is very difficult to find an inverse.

REPLY [0 votes]: The easy form: after proving that $f$ is bijective (for example, like in the answer of Eric Towers) use that Continuous Bijection from ($T_1$) Compact to Hausdorff is Homeomorphism. The essential idea: $f^{-1}$ is continuous because $f$ is closed because $f$(compact) $=$ compact.


TITLE: Having trouble with composite functions, inverse functions. Have an example question from upcoming test.
QUESTION [1 upvotes]: Here is a look at what I am doing. The I can evaluate the first one no problem. The second one is causing me problems and the solution given isn't helping me to get my head around it. Can someone please try and give me the steps and an explanation? Thanks

REPLY [1 votes]: If $y=f_1(x)=3x+2$, then $y-2=3x$ and $x=\frac13(y-2)$, so $f^{-1}(y)=\frac13(y-2).$
To calculate $f_1^{-1}(f_2(1.9))$, first calculate $f_2(1.9)=4\times1.9^3=27.436$.
Now calculate $f_1^{-1}(f_2(1.9))=f_1^{-1}(27.436)=\frac13(27.436-2)\approx8.479.$

REPLY [1 votes]: For the second problem, you are given
\begin{align}
f_{1}(x) &=3x+2 \\
f_{2}(x) &=4x^3
\end{align}
and you are asked to find $$(f_{1}^{-1}\circ f_{2})(1.9).$$  To do this, you will need to first find the inverse function of $f_{1}(x)$.  To do that, we can take the equation $y = 3x+2$ and solve for $y$:
\begin{align}
y &= 3x + 2\\
y - 2 &=3x\\
\frac{1}{3}(y - 2) &= x.
\end{align}
This tells you that the inverse function is given by the equation $$f_{1}^{-1}(y) = \frac{1}{3}(y-2),$$
or to keep the input notation the same with all your functions, we can write this as $$f_{1}^{-1}(x) = \frac{1}{3}(x-2)$$ as they did in the solution.  Then, to evaluate \begin{align}
(f_{1}^{-1}\circ f_{2})(1.9) &= f_{1}^{-1}(f_{2}(1.9)).
\end{align}
This means we first need to evaluate $f_{2}(1.9)$.  This gives us $$f_{2}(1.9) = 4(1.9)^{3} = 27.436,$$ which leaves us with $f_{1}^{-1}(27.436)$.  To evaluate that, we use the inverse function we found, so $$f_{1}^{-1}(27.436) = \frac{1}{3}(27.436 - 2) = 8.478$$ to three decimal places.  So, all in all, we have that $$(f_{1}^{-1}\circ f_{2})(1.9) = 8.478$$ to three decimal places.


TITLE: How to prove that $\lim_{(x,y)\to\ (0,1)} \frac{x y - y}{x^2 + y^2 - 2y + 1}$ does not exist?
QUESTION [1 upvotes]: I tried $y=mx+1$, considering $(\frac{1}{n}, 1 + \frac{1}{n})$ and etc.

REPLY [0 votes]: When $x=0$, the term reduces to
$$
\frac{-y}{(y-1)^2}.
$$
Hence, as $y\to 1$, above diverges to $-\infty$.


TITLE: Help with set theory proof
QUESTION [1 upvotes]: Here is a problem from Munkres Topology 2nd Edition. The question is,
"Here is a "proof" that every relation $C$ that is both symmetric and transitive is also reflexive: "Since $C$ is symmetric, $aCb$ implies $bCa.$ Since $C$ is transitive, $aCb$ and $bCa$ together imply $aCa$, as desired." Find the flaw in this argument. "
Here is my proof. My proff says that my proof is in correct, but I can't seem to find the error.
The reflexive property is not dependent upon transitivity or symmetry; it is a unique property. Claiming that transitivity implies the relation is reflexive is the flaw of the "proof." Consider the relation $\sim$ on a subset $C$ of the Cartesian product $AxA.$ Let $C={a,b,c}$ and $$D={(a,b),(b,a),(a,c),(b,c),(c,a),(c,b)}.$$ Clearly, $\forall a,b,c\in C$,$x\sim y$ implies $y\sim x$. Also,  $\forall a,b,c\in C, x\sim y$ and $y\sim z$ implies $x\sim z,$ but $a\not\sim a$ since $(a,a)\not\in D$, which shows the relation is not reflexive.  Therefore, a transitive, symmetric relation is not reflexive in general.
Thanks a lot for the help!

REPLY [2 votes]: The relation you defined isn't transitive: $a\sim b$ and $b\sim a$, so if this relation were transitive then we would have $a\sim a$. As you point out, $a\not\sim a$, hence $D$ is not transitive.


text:
\begin{document}

\maketitle

\begin{abstract}
\noindent
Turing's method uses explicit bounds on $|\int_{t_{1}}^{t_{2}} S(t)\, dt|$, where $\pi S(t)$ is the argument of the Riemann zeta-function. This article improves the bound on $|\int_{t_{1}}^{t_{2}} S(t)\, dt|$ given in \cite{TrudgianTuring}.
\end{abstract}

\section{Introduction}\label{intro}
Let $\zeta(s)$ be the Riemann zeta-function, and let $N(T)$ denote the number of zeroes of $\zeta(s)$ with $0<\Re(s)<1$ and $0<\Im(s)< T$. One seeks to calculate $N(T)$ as follows.

First one finds zeroes by locating sign changes of a real-valued function the zeroes of which agree with the non-trivial zeroes of the zeta-function.
\begin{comment}
This can be done using Gram's Law, the Riemann--Siegel formula and Euler--Maclaurin summation or by some sophisticated bisection techniques and the fast Fourier transform --- see, e.g., \cite{Platt}.
\end{comment}
This gives one a lower bound on the number of zeroes of $\zeta(s)$ with $0<\Im(s)<T$.

To check whether this initial analysis has omitted some zeroes one employs Turing's method. This was first annunciated by Turing \cite{Turing} in 1953 and has been used extensively since then. Recently, another method has been deployed by B\"{u}the \cite{Buethe}.

To apply Turing's method one needs good explicit bounds on
$$\bigg|\int_{t_{1}}^{t_{2}} S(t)\, dt\bigg|,$$
for $t_{2}>t_{1}>0$, where $\pi S(t)$ is defined to be the argument of $\zeta(\frac{1}{2} + it)$. For a complete definition and a brief history of the problem, see \cite[\S 1]{TrudgianTuring} and \cite[Ch.~7]{Edwards}.

This article improves \cite{TrudgianTuring} and contains frequent references to the results therein. The main result is
\begin{thm}\label{main}
\begin{equation}\label{postman}
\bigg| \int_{t_{1}}^{t_{2}} S(t)\, dt\bigg| \leq 1.698 + 0.183\log\log t_{2}+ 0.049\log t_{2},
\end{equation}
for $t_{2}>t_{1} > 10^{5}.$ If the right-side of (\ref{postman}) is replaced by $a + b \log\log t_{2} + c \log t_{2}$, one may use Table \ref{tabled} on page \pageref{tabled} for more specific values of $a, b$ and $c$.
\end{thm}
In \cite{TrudgianTuring} the main result followed from Lemma 2.8 and Lemma 2.11 which concerned respectively obtaining an upper and a lower bound for $\Re\log\zeta(s)$ for $\Re(s)\geq \frac{1}{2}$. This article refines only the upper bound. Theorem \ref{main} improves on Theorem 2.2 in \cite{TrudgianTuring} for all $t_{2} \geq 10^{5}$.

The idea in this article is to use more sophisticated estimates on $\zeta(\sigma + it)$ for $\frac{1}{2} \leq \sigma \leq 1$; these estimates have been given in \cite{PlattTrudgian} and \cite{TrudZeta}. A bound on $|\zeta(s)|$ is given in \S \ref{bound}, a proof of Theorem 1 is given in \S \ref{proof}, and some concluding remarks are provided in \S \ref{conc}.

\subsection*{Acknowledgements}
I should like to thank Jan B\"{u}the for detailed discussions on this problem.

\section{Bounding $|\zeta(\sigma + it)|$ across the strip $\frac{1}{2}\leq\sigma\leq 1+ \delta$}\label{bound}

Using the inequality $\log (1+ x) \leq x$ it is easy to see that
\begin{equation}\label{eq1}
\log|Q_{0} + \sigma + it| - \log t \leq \frac{1}{2} \left( \frac{\sigma_{1} + Q_{0}}{t_{0}}\right)^{2},
\end{equation}
for $\sigma \leq \sigma_{1}$ and $t\geq t_{0}$ and any $Q_{0}\geq 0$. With the trivial observations $\log|Q_{0} + \sigma + it| \geq \log t$ and $|\arg (Q_{0} + \sigma + it)| \leq \frac{\pi}{2}$ at hand, we may apply (\ref{eq1}) to see that
\begin{equation}\label{zero}
|\log (Q_{0} + \sigma + it)| \leq (1+ a_{0}) \log t, \quad\quad (\sigma \leq \sigma_{1})
\end{equation}
where
\begin{equation*}\label{a0def}
a_{0} = a_{0}(\sigma_{1}) = \frac{\sigma_{1} + Q_{0}}{2 t_{0}^{2} \log t_{0}} + \frac{\pi}{2\log t_{0}} + \frac{\pi(\sigma_{1} + Q_{0})^{2}}{4 t_{0} \log^{2} t_{0}}.
\end{equation*}

Suppose that
\begin{equation}\label{int}
|\zeta(\tfrac{1}{2} + it)| \leq k_{1} t^{k_{2}} (\log t)^{k_{3}}, \quad (t\geq t_{1}), \quad\quad |\zeta(1+ it)|\leq k_{4} \log t^{k_{5}}, \quad (t\geq t_{2}).
\end{equation}
Consider the function $h(s) = (s-1) \zeta(s)$, which is entire. Once we are able to exhibit bounds for $|h(s)|$ using the information in (\ref{int}) we can apply a version of the Phragm\'{e}n--Lindel\"{o}f principle to bound $|\zeta(s)|$.
Using Lemma 3 in \cite{TrudgianS2} and (\ref{zero}) and (\ref{eq1}) we may prove
\begin{lem}\label{lem1}
Let $h(s) = (s-1) \zeta(s)$, and let $\delta$ be a positive real number. Furthermore, let $Q_{0}\geq 0$ be a number for which
\begin{equation*}\label{one}
\begin{split}
|h(\tfrac{1}{2} +it)| &\leq k_{1} |Q_{0} + \tfrac{1}{2} + it|^{k_{2} + 1} (\log| Q_{0} + \tfrac{1}{2} + it|)^{k_{3}}\\
|h(1 + it)& \leq k_{4}|Q_{0} + 1+ it| (\log|Q_{0} + 1+ it|)^{k_{5}} \\
|h(1 + \delta + it)| &\leq \zeta(1+ \delta)|Q_{0} + 1+ \delta + it|,
\end{split}
\end{equation*}
for all $t$.
Then for $\sigma \in[\frac{1}{2}, 1]$ and $t\geq t_{0}$,
\begin{equation}\label{p1}
 |\zeta(s)| \leq \alpha_{1} k_{1}^{2(1-\sigma)} k_{4}^{2(\sigma- \frac{1}{2})} t^{2k_{2}(1-\sigma)} (\log t)^{2(k_{3}(1-\sigma) + k_{5}(\sigma - \frac{1}{2}))},
 \end{equation}
where
\begin{equation*}\label{ce}
\alpha_{1} = (1+ a_{1}(1+ \delta, Q_{0}, t_{0}))^{k_{2} + 1} (1+a_{0}(1+ \delta, Q_{0}, t_{0}))^{k_{3} + k_{5}},
\end{equation*}
and
\begin{equation*}
a_{0}(\sigma, Q_{0}, t) = \frac{\sigma + Q_{0}}{2 t^{2} \log t} + \frac{\pi}{2\log t} + \frac{\pi(\sigma + Q_{0})^{2}}{4 t \log^{2} t}, \quad a_{1}(\sigma, Q_{0}, t) = \frac{\sigma + Q_{0}}{t}.
\end{equation*}
Whereas for $\sigma\in[1, 1+ \delta]$ and $t\geq t_{0}$,
\begin{equation}\label{p2}
|\zeta(s)| \leq \alpha_{2} k_{4}^{\frac{1+ \delta - \sigma}{\delta}} \zeta(1+ \delta)^{\frac{\sigma -1}{\delta}} (\log t)^{k_{5}(\frac{1+ \delta -\sigma}{\delta})},
\end{equation}
where
\begin{equation*}\label{a3}
\alpha_{2} = (1+ a_{1}(1+ \delta, Q_{0}, t_{0}))(1 + a_{0}(1+ \delta, Q_{0}, t_{0}))^{k_{5}}.
\end{equation*}
Finally, for all $\sigma\in[\frac{1}{2}, 1+ \delta]$ and $t\geq t_{0}$ we have
\begin{equation*}\label{kite}
|\zeta(s)| \leq \left(1+ a_{1}(1+ \delta, Q_{0})\right)^{k_{2} + 1}\left(1+ a_{0}(1+ \delta, Q_{0})\right)^{k_{3} + k_{5}}k_{1} t^{k_{2}} (\log t)^{k_{3}},
\end{equation*}
provided that
\begin{equation}\label{conditions}
t^{k_{2}} (\log t)^{k_{3} - k_{5}} \geq \frac{k_{4}}{k_{1}}, \quad
 t \geq \exp\left\{ \left(\frac{\zeta(1+ \delta)}{k_{4}}\right)^{\frac{1}{k_{5}}}\right\}.
\end{equation}
\end{lem}
\begin{proof}
In applying Lemma 3 of \cite{TrudgianS2} to $h(s)$ we need to relate $|Q_{0} + s|$ and $|s-1|$ to $t$. We simply note that
$$ \bigg| \frac{Q_{0} + s}{s-1}\bigg| \leq \frac{|Q_{0} + s|}{t} \leq 1+ \frac{\sigma + Q_{0}}{t_{0}} = 1 + a_{1}(\sigma, Q_{0}) \leq 1+ a_{1}(1+ \delta, Q_{0}),$$
in both regions $\sigma \in [\frac{1}{2}, 1]$ and $\sigma \in[1, 1+ \delta]$.
Since $a_{0}$ and $a_{1}$ are small for any respectable value of $t_{0}$ we throw away some information in the exponents of $1+ a_{1}$ and $1+ a_{0}$. For example, in proving (\ref{p1}) we arrive at
$$(1+ a_{1})^{2k_{2}(1- \sigma) +1} (1+ a_{0})^{2k_{3}(1-\sigma) + 2k_{5}(\sigma - \frac{1}{2})}.$$
Rather than retain this dependence on $\sigma$ in the exponents, we simply bound $1-\sigma$  and $\sigma - \frac{1}{2}$ by $\frac{1}{2}$. A similar procedure is applied to prove (\ref{p2}).

To prove the bound in the region $\sigma\in[\frac{1}{2} , 1+ \delta]$ we note that the bounds in (\ref{p1}) and (\ref{p2}) are decreasing in $\sigma$ if the inequalities in (\ref{conditions}) are met. Finally, the bound in (\ref{p1}), evaluated at $\sigma = \frac{1}{2}$ exceeds the bound in (\ref{p2}), evaluated at $\sigma = 1$. This completes the lemma.
\end{proof}
It is worth recording values of $k_{1}, \ldots, k_{5}$, which we do in
\begin{cor}\label{jeep}
For $\sigma\in[\frac{1}{2}, 1+ \delta]$ and $t\geq t_{0}$ we have
\begin{equation*}\label{zephyr}
|\zeta(s)| \leq 0.732(1 + a_{1}(1+ \delta, 5, t_{0}))^{7/6} (1+ a_{0}(1+ \delta,5, t_{0}))^{2} t^{1/6} \log t,
\end{equation*}
provided that
\begin{equation*}\label{boreas}
t\geq \max\{ 1.16, \exp[4\zeta(1+ \delta)/3]\}.
\end{equation*}
\end{cor}
\begin{proof}
In \cite{TrudZeta} it was shown that $|\zeta(1+ it)| \leq \frac{3}{4} \log t$ for $t\geq 3$. In \cite{PlattTrudgian} it was shown that
$|\zeta(\frac{1}{2} + it)| \leq 0.732 |4.678+it|^{\frac{1}{6}} \log |4.678+it|$ for all $t$. One may therefore choose
\begin{equation}\label{values}
(k_{1}, k_{2}, k_{3}, k_{4}, k_{5}, Q_{0}) = (0.732, \tfrac{1}{6}, 1, \tfrac{3}{4}, 1, 5)
\end{equation}
 in Lemma \ref{lem1}, which proves the corollary.
\end{proof}
Although we shall use (\ref{values}) in our computation  we proceed with the variables $(k_{1}, \ldots, Q_{0})$ as parameters.
We remark that, although Corollary \ref{jeep} is not used in this article, it is derived at very little additional cost and should prove useful for related problems.

\section{Proof of Theorem \ref{main}}\label{proof}
We are now able to proceed to the proof of Theorem \ref{main}. We need to give an upper bound for
$$I = \Re\int_{\frac{1}{2} + it}^{\infty + it} \log \zeta(s)\, ds.$$
To that end, we shall write
\begin{equation}\label{L2}
I \leq \int_{\frac{1}{2} + it}^{1+ it} \log |\zeta(s)|\, ds  +  \int_{1 + it}^{1+ \delta+ it} \log |\zeta(s)|\, ds + \int_{1+ \delta}^{\infty} \log |\zeta(\sigma)| \, d\sigma,
\end{equation}
and apply (\ref{p1}) to the first integral in (\ref{L2}) and (\ref{p2}) to the second. This gives us
\begin{lem}\label{second}
For $t\geq t_{0}$ we have
\begin{equation*}
I\leq A_{1} + B_{1} \log\log t + C_{1} \log t,
\end{equation*}
where
\begin{equation*}
\begin{split}
A_{1} &= \int_{1+\delta}^{\infty} \log |\zeta(\sigma)| \, d\sigma
+ (\frac{k_{2}}{4} + \frac{1}{2} + \delta) \log (1+ a_{1}(1+ \delta, Q_{0}, t_{0}))
+ \frac{1}{4} \log k_{1} \\
&
+ (\frac{k_{3}}{4} + \frac{k_{5}}{4} + \frac{k_{5}\delta}{2}) \log (1+ a_{0}(1+ \delta, Q_{0}, t_{0}))
+ (\frac{1}{4} + \frac{\delta}{2})\log k_{4}
+ \frac{\delta}{2} \log \zeta(1+ \delta)
\end{split}
\end{equation*}
and
\begin{equation*}
B_{1} = \left( \frac{k_{3} + k_{5}}{4} + \frac{\delta k_{5}}{2}\right), \quad C_{1} = \frac{k_{2}}{4}.
\end{equation*}
\end{lem}
The term corresponding to $C_{1}$ in Lemma 2.8 of \cite{TrudgianTuring} is, $k_{2}/4+ \delta k_{2}/2$. Since we are not permitted to take $\delta$ too small, lest the integral in $A_{1}$ become too large, this represents a considerable qualitative saving. This is due entirely to estimating $|\zeta(s)|$, not in one go over $\sigma\in[\frac{1}{2}, 1+ \delta]$ as in \cite{TrudgianTuring} but by using Lemma \ref{lem1}.
We combine Lemma \ref{second} with Lemma 2.11 in \cite{TrudgianTuring} to obtain
\begin{thm}\label{T2}
For $t_{2}\geq t_{1} \geq 10^{5}$,
\begin{equation*}
\bigg| \int_{t_{1}}^{t_{2}} S(t)\, dt\bigg| \leq a + b\log\log t_{2}+ c\log t_{2},
\end{equation*}
where
\begin{equation}\label{fina}
\begin{split}
\pi a &= \int_{1+ \delta}^{\infty} \log |\zeta(\sigma)| \, d\sigma
+ (\frac{k_{2}}{4} + \frac{1}{2} + \delta) \log (1+ a_{1}(1+ \delta, Q_{0}, 10^{5})) + \frac{1}{4} \log k_{1}  \\
&
+ (\frac{k_{3}}{4} + \frac{k_{5}}{4} + \frac{k_{5}\delta}{2}) \log (1+ a_{0}(1+ \delta, Q_{0}, 10^{5}))
+ (\frac{1}{4} + \frac{\delta}{2})\log k_{4}
+ \frac{\delta}{2} \log \zeta(1+ \delta)\\
 &+ d^{2} \log 4\left\{ -\frac{\zeta'(\frac{1}{2} + d)}{\zeta(\frac{1}{2} + d)} - \frac{1}{2} \log 2\pi + \frac{1}{4}\right\} + \frac{d^{2}}{2} \log \pi - \frac{1}{2} \int_{1+ 2d}^{\infty} \log \zeta(\sigma)\, d\sigma \\
 & + \int_{\frac{1}{2} + d}^{\infty}\log \zeta(\sigma)\, d\sigma - \frac{1}{2} \int_{1+ 2d}^{1+ 4d} \log \zeta(\sigma)\, d\sigma
+ \int_{\frac{1}{2} + d}^{\frac{1}{2} + 2d} \log \zeta(\sigma)\, d\sigma + 3\times 10^{-4},
\end{split}
\end{equation}
and
\begin{equation}\label{finb}
\pi b = \left( \frac{k_{3} + k_{5}}{4} + \frac{\delta k_{5}}{2}\right), \quad\quad \pi c = \frac{k_{2}}{4} + \frac{d^{2}}{2}(\log 4 -1).
\end{equation}
\end{thm}
\subsection{Computation}
Before we commence an analysis of the coefficients appearing in Theorem \ref{T2} we make the following observation. One may replace the values of $(k_{1}, k_{2}, k_{3})$ in (\ref{values}) by
\begin{equation}\label{bounds}
(k_{1}, k_{2}, k_{3}) = \left(\frac{4}{(2\pi)^{\frac{1}{4}}}, \frac{1}{4}, 0\right),
\end{equation}
which appear in \cite[Lem.\ 2]{Lehman}. The values in (\ref{bounds}) are obtained using the approximate functional equation of $\zeta(s)$: the values in (\ref{values}) are obtained using exponential sums. The value $k_{2}= \frac{1}{4}$ follows from convexity theorems. We call  (\ref{bounds}) the convexity result, and (\ref{values}) the sub-convexity result.

As in Theorem 2.12 in \cite{TrudgianTuring} no term in either (\ref{fina}) or (\ref{finb}) depends on both $\delta$ and $d$. We can run two one-dimensional optimisations on each of $a$, $b$ and $c$. In Table \ref{tabled} we compare the results obtained from the convexity result (C), the sub-convexity result (SC), and the coefficients in Theorem 2.2 of \cite{TrudgianTuring}, when $t_{1} \approx T$. The values of $\delta, d, a, b$ and $c$ correspond to the sub-convexity result. We find that the sub-convexity result overtakes the convexity result when $T\geq 2.85\times 10^{10},$ which is, just barely, beneath the height to which the Riemann hypothesis has been verified --- see \cite{Plattarxiv}.

The values used in Theorem \ref{main} are taken from the row $T = 10^{10}$. It should be stressed that all of the results in this table are valid for $t_{1}\geq 10^{5}$. The stated values of $a, b$ and $c$  are those that are close to the best values obtainable by this method when $t_{1} \approx T$.

\begin{table}[ht]
\caption{Comparison of bounds for $|\int_{t_{1}}^{t_{2}} S(t)\, dt|\leq a + b \log\log t_{2} + c \log t_{2}$}
\centering
\begin{tabular}
{c c c c c c c c c}
\hline\hline
$T$ & Thm 2.2 & C & SC & $d$ & $\delta$ & $a$ & $b$ & $c$  \\[0.5ex]\hline
$10^{5}$ & 2.747 & 2.629 & 2.658 & 0.883 & 0.279 & 1.457 & 0.204 & 0.062 \\
$10^{6}$ & 2.883 & 2.800 & 2.827 & 0.845 & 0.237 & 1.520 & 0.197 & 0.058 \\
$10^{7}$ & 3.018 & 2.959 & 2.982& 0.817 & 0.206 & 1.573 & 0.192 & 0.055 \\
$10^{8}$ & 3.154 & 3.110 & 3.128& 0.795 & 0.182 & 1.620 & 0.189 & 0.053 \\
$10^{9}$ & 3.290 & 3.255 & 3.266& 0.777 & 0.163 & 1.661 & 0.186 & 0.051  \\
$10^{10}$ & 3.426 & 3.395 & 3.398& 0.762 & 0.148 & 1.698 & 0.183 & 0.049  \\
$10^{11}$ & 3.562 & 3.530 & 3.526 & 0.749 & 0.135 & 1.733 & 0.181 & 0.048 \\
$10^{12}$ & 3.698 & 3.663 & 3.649 & 0.738 & 0.124 & 1.764 & 0.179 & 0.047 \\
$10^{13}$ & 3.834 & 3.792 & 3.770 & 0.729 & 0.115 & 1.792 & 0.178 & 0.046 \\
$10^{14}$ & 3.969 & 3.919 & 3.887 & 0.720 & 0.107 & 1.820 & 0.177 & 0.046 \\
$10^{15}$ & 4.105 & 4.044 & 4.002& 0.713 & 0.100 & 1.844 & 0.176 & 0.045  \\
 \hline\hline
\end{tabular}
\label{tabled}
\end{table}

\section{Conclusion}\label{conc}
It seems difficult to improve substantially on Theorem \ref{main}.

Given that the improvements obtained in this paper are only modest, and since further improvements would require a lot of effort in estimating $\zeta(\frac{1}{2} +it)$ or $\zeta(1+ it)$, it seems hopeless to try to improve this part of the argument.
\begin{comment}
To illustrate the point further, suppose the Lindel\"{o}f hypothesis were true and one had $k_{2} = 0$. Keeping all other constants the same we would then have, for example $|\int_{t_{1}}^{t_{2}}S(t)\, dt|\leq 3.093$ where $T\approx 10^{15}$.
\end{comment}

One could try one's luck at reducing the term $\log 4$ that appears in both (\ref{fina}) and (\ref{finb}). This comes from Lemma 4.4 in \cite{Booker}.  Reducing this would have a more profound influence on bounding $|\int_{t_{1}}^{t_{2}}S(t)\, dt|$ than better bounds for $|\zeta(s)|$.

Finally, it is worth considering Theorems 3.3 and 4.3 in \cite{TrudgianTuring}, which relate to Dirichlet $L$-functions and Dedekind zeta-functions. Both of these could be improved, in line with this article, were one in possession of explicit estimates on the lines $\sigma = \frac{1}{2}$ and $\sigma = 1$.
One such estimate, bounding $|L(1+ it, \chi)|$ for $L(s)$ a Dirichlet $L$-function, appears in \cite{DudekL}. It is possible that this could be used to obtain an improvement to Theorem 3.3 in \cite{TrudgianTuring}.

\bibliographystyle{plain}
\bibliography{themastercanada}

\end{document}