TITLE: new $\arctan$ series working for any $x$?
QUESTION [2 upvotes]: Using Log series, we can write:
$$
\frac{1}{2 i} \left( \text{Log} \left( 1 + e^{\text{i2t}} \right) - \text{Log} \left( 1 + e^{\text{-i2t}} \right) \right)
=
\frac {1} {2 i}\left ( \sum _ {k = 1}^{\infty}\frac {(-1)^{k + 1}} {k} e^{\text{i2t}^k} - \sum _ {k = 1}^{\infty}\frac {(-1)^{k + 1}} {k} e^{-\text{i2t}^k} \right)
$$
$$
\frac {1}{2 i} \text {Log} \left( \frac{1 + e^{\text {i2t}}}{1 + e^{-\text {i2t}}} \right)
=
\frac{1}{2 i} \sum_{k = 1}^{\infty}\frac{(-1)^{k + 1}}{k} \left( e^{\text{i2kt}} - e^{-\text{i2kt}} \right)
$$
$$
\frac {1}{2 i} \text {Log} \left( \frac{e^{\text{i2t}} \left( e^{-\text{i2t}}+1 \right)}{1+e^{-\text{i2t}}} \right)
=
\sum_{k = 1}^{\infty} \frac{(-1)^{k + 1}}{k} \left( \frac{e^{\text {i2kt}} - e^{-\text {i2kt}}}{2 i} \right)
$$
$$
\frac{1}{2i} \text{Log} \left( e^{\text{i2t}} \right)
=
\sum_{k = 1}^{\infty}\frac {(-1)^{k + 1}} {k}\text {Sin} (2kt)
$$
$$
t
=
\sum_{k = 1}^{\infty}\frac {(-1)^{k + 1}} {k}\text {Sin} (2kt)
$$
Here is the plot of the sum in red with sucessive aproximations and y=t in blue:

Choosing $t = \text{ArcTan} (x)$, we have:
$$
\text{ArcTan} (x)
=
\sum_{k = 1}^{\infty} \frac {(-1)^{k + 1}} {k} \text{Sin} ( 2k \text{ ArcTan} x )
$$
Using $z=1 + i x=|z| e^{i \text{arcTan}x }$ and its complex conjuguate $z^*=1 - i x=|z| e^{-i \text{arcTan}x }$, with $|z|=\sqrt{1 + x^2}$, we can write:
$$
\text{Sin} ( 2k \text{ Arctan} x)
=
\frac {z^{2 k} - z^{*2 k}} {2i |z| ^{2 k}}
=
\frac {(1 + ix)^{2k} - (1 - ix)^{2k}} {2i (1 + x^2)^k}
=
\frac{\sum_{q=0}^k (-1)^q \left( \begin{array}{c} 2k \\ 2q+1 \end{array} \right) x^{2 q+1}}{\left(1+x^2\right)^k}
$$
Thus:
$$
\text{ArcTan} (x)
=
\sum_{k = 1}^{\infty} \frac {(-1)^{k + 1}} {k}
\frac{\sum_{q=0}^k (-1)^q \left( \begin{array}{c} 2k \\ 2q+1 \end{array} \right) x^{2 q+1}}{\left(1+x^2\right)^k}
$$
Here are the plots of $\text{ArcTan}(x)$ in blue, and successive aproximations to the classic ArcTan series $\text{ArcTan}(x)=\sum_{k = 0}^{\infty} \frac {(-1)^k} {2k + 1} x^{2k + 1}$ in green, working for $x<1$, and the one I gave in red, working for any $x$ as the sum goes to infinity:

is that result known?

REPLY [2 votes]: I do not know whether your series for $\arctan$ is new (it may be), but when evaluating its advantages, you should consider alternatives other than the classical McLaurin series. For example, the identity $\tan 2\theta=\frac{2\tan \theta}{1-\tan^2\theta}$ can be turned into $$\arctan x = 2\arctan \frac{x}{1+\sqrt{1+x^2}}$$
Note that the argument of $\arctan$ on the right is always less than $1$ in absolute value. Hence,
$$\arctan x = 2 \, \sum_{k=0}^\infty \frac{(-1)^k}{2k+1}\left(\frac{x}{1+\sqrt{1+x^2}}\right)^{2k+1},\quad x\in\mathbb R$$
The content of parentheses is an irrational function of $x$ but it only needs to be computed once. Partial sums up to $k=0,1,2,3$ are shown below; the purple curve is the arctangent itself.

(The horizontal  window is $[0,4]$, as in your post.)


TITLE: Can I say Axioms of Quantum Mechanics instead of Postulates?
QUESTION [1 upvotes]: I have been attempting to clarify the difference between an axiom and a postulate, but I keep running into almost contradictory answers. Some will say they're equivalent, some will  say a postulate may not known to be unprovable (cannot be derived?) while axioms are? Some will say it depends on the field of science and one is used over the other due to historical context. But I have noticed that most often, it is said to be the 'Postulates of Quantum Mechanics' and not the 'Axioms of Quantum Mechanics'. Can someone clarify for me why that is?

REPLY [1 votes]: To my understanding, an axiom is a statement that something is the case. A postulate is more like an assumption that something is the case.
When you are talking about a physical theory meant to be predictive of reality built on experimental and theoretical work (i.e. Quantum Mechanics), I don't think it makes much sense to use axioms since then you'd be claiming that reality is such and such way.


TITLE: Some confusion regarding RLC circuits
QUESTION [0 upvotes]: I'm a mathematics student, and it has been years since I last did current/voltage/resistor related subjects in high school. However, for one problem that I've encountered, I actually need to calculate a few of these. The problem is as follows:
The problem is the following. I want to find the voltage over the capacitor, and the current going through the capacitor. I found one source, telling me that I should treat the problem as if there is only one voltage source. First for $V_{1}$, and then for $V_{2}$, and combine the results. I set off to do this, starting with only considering $V_1$. I want to find the total resistance of the circuit, $R_{{\rm tot}}$, and from what I saw,
$$R_{{\rm tot}}=R_1+\frac{1}{\frac{1}{R_2}+\frac{1}{R_3}} \, ,$$
where $R_3$ would be the resistance in the "branch" (not sure about terminology) with the capacitor. However, there is no resistor in this branch. What does this mean for $R_{{\rm tot}}$, and the circuit as a whole? I feel like I might be taking the wrong approach, and I'm getting incredibly confused trying to figure out how this actually works. I would appreciate any advice on how to find the current through the capacitor, and how the voltage across the circuit actually works.

REPLY [0 votes]: This is a "homework and exercise" problem that we generally don't provide solutions to, but only guidance. So here is some guidance for you to do additional research.
If I remember my filter concepts correctly (its been a while), this appears to be what is called a low pass filter, i.e., a filter that permits low frequency signals at the input (left side) through to the output (right side) while blocking high frequencies. The capacitor is low impedance at high frequencies and inductor is low impedance at low frequencies. So the capacitor shunts out  high frequencies and the inductor blocks high frequencies. You should research the internet to learn more.
Since there is a capacitor and inductor in the circuit, you shouldn't be looking at the equivalent input resistance, but the equivalent impedance, which is the more general term that includes resistance.
The complex impedances of the capacitor and inductor, which depend on frequency, as well as resistance which does not depend on frequency need to be combined. You can look up the impedances for capacitors and inductors for sinusoidal ac voltages, and how to combine them with resistance.
If this is filter problem, V1 and V2 are not voltage sources, but represent the input (V1,I1) and output (V2,I2) of the filter. Otherwise the process of shorting each voltage source one at a time and adding the results is referred to as superposition.

If you wouldn't mind, could you verify for me whether the following is
  correct, so that I know if I am on the right track? From what I
  gathered, the current going through C1 is going to be V1/R1+V2/R2. I am suspicious of this because it looks remarkably simple.
  They, V1 and V2, are the voltages at the input and output.... In what way does that
  make it incorrect? –

It does look remarkably simple if it were based on your assumption that $V_1$ ad $V_2$ in the equation are the input and output voltages of the circuit. Unfortunately that assumption is incorrect.
It is incorrect because the voltages $V_1$ and $V_2$ in your equations are not the input and output voltages shown in the diagram, but are the voltage drops across $R_1$ and $R_2$, respectively. The following will explain.
Based on the circuit diagram, $I_1$ and $I_2$ are in the same direction through $C_1$, therefore the current in $C_1$ should be
$$I_{C_1}=I_{1}+I_{2}$$
Now the voltage drop across $R_1$ is
$$V_{R1}=I_{1}R_{1}$$
And the voltage drop across $R_2$ is
$$V_{R2}=I_{2}R_{2}$$
Putting these three equations together we get
$$I_{C1}= \frac{V_{R1}}{R_1}+\frac{V_{R2}}{R_2}$$
Which shows that the voltages in your equation are not the input and output voltages, but the voltage drops across $R_1$ and $R_2$.
Hope this helps.


TITLE: Sequential Characteristic of Riemann Integral
QUESTION [2 upvotes]: Wanted to check the validity of a proof:
Claim:
Let $f:[a,b]\to\mathbb R$ be bounded. Then f is integrable if and only if there exists a sequence of partitions $P_n$ such that
$\lim_{n\to\infty}U(f,P_n)-L(f,P_n) = 0$
My attempt at a proof:
Suppose f is integrable $\implies \forall \epsilon > 0\exists$ a partition P such that $U(f,P) - L(f,P) < \epsilon$
Let $\epsilon = \frac{1}{n} , n \in\mathbb N$
Thus for each $n \in \mathbb N$ choose $P_n$ such that
$U(f,P_n) - L(f,P_n) < \frac{1}{n} =\epsilon$
Thus $\lim_{n\to\infty}U(f,P_n)-L(f,P_n) = 0$
Then proving the contrapositive:
Suppose $f$ is not integrable
$\implies U(f) $ does not equal $L(f)$
Thus $\forall n\in \mathbb N \exists \epsilon > 0$ such that
$U(f,P_n)-L(f,P_n) > \epsilon$
Thus the claim follows.
Any help and criticism is appreciated, thanks.

REPLY [1 votes]: I have come up with an idea for the implication in the other direction:
Suppose $\lim_{n\to \infty} U(f,P_n)-L(f,P_n) = 0$
$\implies \forall \epsilon > 0 \exists N \in \mathbb N$ such that
$U(f,P_n)-L(f,P_n) < \epsilon$
$\implies \forall \epsilon > 0 \exists$ a partition $Q=P_n$ such that when $n>N$
$U(f,Q)-L(f,Q) < \epsilon$
and thus f is integrable.


TITLE: Difficulty proving a statement about polynomials using the argument principle.
QUESTION [2 upvotes]: Hello I am having some difficulty with the following problem, I believe it is meant to be a rather simple exercise.
Suppose that $P_{t}(z)$ is a continuous polynomial in $z$ for any fixed $0 \le t \le 1$ and that it is continuous in $t$, in the sense that the degree of $P_t(z)$ is bounded independent of $t$ and the coefficients are continuous functions of $t$.
Let $Z=\{(z,t):P_{t}(z)=0\}$ (note that it is closed in $\mathbb{C} \times [0,1]$ ) if $P_{t_{0}}(z_{0})=0$ and its partial derivative with respect to $z$ is not zero at $z_{0}$ then use the argument principle to show that there exists an $\epsilon \gt 0$ such that for t sufficiently close to $t_{0}$ there is a unique $z$ in the open disk with centre $z_{0}$ and radius $\epsilon$ with $P_{t}(z)=0$ , moreover what can be said about $P_{t_{0}}$ if it vanishes to order k at $z_{0}$.
So I am not sure how to proceed.
Here is what I have thought:
Well the polynomial are certainly holomorphic, so I have a result that says that $(1/2\pi)$ times the contour around the boundary integral of $\frac{p'}{p}$ will be equal to the sum of the multiplicities of the zeros minus the sum of the multiplicities of the poles. But I dont understand how I can use this to show what is being asked. Ie, I get what the general theorem states, but I am not sure how it relates to this question

REPLY [1 votes]: $P_{t_0}(z)$ has a (simple or multiple) zero at $z_0$. Thus there is an $\epsilon$ so that the disk of radius $\epsilon$ has no zero of $P_{t_0}(z)$ in its interior or its boundary (other than $z_0$). Simply because zeros are isolated.
Now the claim is that there is a $\delta$ such that for $|t-t_0|<\delta$ the polynomial $P_t(z)$ has no zero on the boundary of the circle. This is just compactness, one could say that if there were $t_n\to t_0$ with zeros $z_n$ on the boundary then some subsequence of those zeros would converge to a zero of $P_{t_0}$.
Therefore for $|t-t_0|<\delta$ the function
$$n(t)=\frac{1}{2\pi i}\int_C\frac{P^{\prime}_t(z)}{P_t(z)} dz$$ is well defined and is a continuous integer valued function of $t$.
Thus this function is constant and the polynomials $P_t(z)$ all have the same number of zeros inside the circle.
Note that in the case of a zero of multiplicity greater than $1$, this zero may split into many zeros, but the sum of their multiplicities will be constant.


TITLE: Solving the Dirichilet problem in a strip relating to conformal mappings
QUESTION [0 upvotes]: Observe that if
$$e^{i\phi}=\frac{i-e^{\pi t}}{i+e^{\pi t}}$$
then take the imaginary part and differentiate both sides to establsh the two following identities:
$$\sin{\phi}=\frac{1}{\cosh{\pi t}},\ \frac{d\phi}{dt}=\frac{\pi}{\cosh{\pi t}}.$$
I had successfully constructed the second identity by differentiating, but I don't know how to derive the second one.

REPLY [0 votes]: Rearrange the quantity on the right by multiplying the top and bottom by the complex conjugate of the bottom:
$$\frac{i-e^{\pi t}}{i + e^{\pi t}} \cdot \frac{-i+e^{\pi t}}{-i + e^{\pi t}} = \frac{(1-e^{2\pi t}) + i(2e^{\pi t})}{1 + e^{2\pi t}}$$
$$\implies \sin\phi = \text{Im}\left(\frac{i-e^{\pi t}}{i + e^{\pi t}}\right) = \frac{2e^{\pi t}}{1 + e^{2\pi t}} = \frac{2}{e^{-\pi t} + e^{\pi t}} \equiv \text{sech}(\pi t)$$
As a bonus, using
$$\cos\phi = \text{Re}\left(\frac{i-e^{\pi t}}{i + e^{\pi t}}\right) = \frac{1-e^{2\pi t}}{1 + e^{2\pi t}} = \frac{e^{-\pi t}-e^{\pi t}}{e^{\pi t} + e^{\pi t}} \equiv -\tanh(\pi t)$$
we get that
$$\frac{d}{dt}(\cos \phi) = -\pi \hspace{2 pt} \text{sech}^2(\pi t) \implies -\sin\phi \frac{d\phi}{dt} = -\text{sech}(\pi t) \frac{d\phi}{dt} = -\pi \hspace{2 pt} \text{sech}^2(\pi t)$$
$$\therefore \frac{d\phi}{dt} = \pi \hspace{2 pt}\text{sech}(\pi t)$$


TITLE: Replacing the flat part of ads metric with another flat metric.
QUESTION [0 upvotes]: In the poincare co-ordinate representation metric of the ads space has a flat part that corresponds to the minkowski metric. Can one show that this flat part can be replaced by other metric that satisfies the condition of zero curvature and still the whole ads metric satisfies vaccum Einstein's equation with negative cosmological constant? Please help me with a proof.

REPLY [0 votes]: You can easily convince yourself that if you replace the Minkowski metric with another flat, maximally symmetric 4D metric, it has no impact on the Einstein equations.
Writing this up in a formal manner is left for the OP as an excercise :P


TITLE: question about the conditional error term in linear regression
QUESTION [0 upvotes]: Suppose we are given $n$ i.i.d. random vectors $\{ y_i,X_i \}$ where $y_i$ is a random scalar and $X_i$ is a random vector. Further suppose that $\epsilon_i$ is a linear function of $\{ y_i,X_i \}$. Now, how do I show
$$
\mathbb{E}[\epsilon_i\mid {\bf X}]=\mathbb{E}[\epsilon_i\mid { X_i}]?
$$
where $\bf X$ is the random vector $(X_1,X_2,\dots,X_n)$
My attempt:
Since $\epsilon_i$ is a linear function of $\{ y_i,{X_i} \}$, we note that $\epsilon_i$ is independent of each random vector $\{ y_j,{X_j} \}$ where $j\neq i$. Now, I feel like I need to show $\epsilon_i$ is independent of the random vector $$\left( \{ y_1,{X_1} \},\{ y_2,{X_2}\},\dots,\{ y_{i-1},{X_{i-1}}\},\{ y_{i+1},{X_{i+1}}\},\dots ,\{ y_{n},{X_{n}}\}\right).$$ However, I know that merely know each component random vector of this big random vector is each independent of $\epsilon_i$ is not a sufficient condition for $\epsilon_i$ to be independent of this huge vector. Hints are appreciated.
Actually, this problem is in the context of classical linear regression model.

REPLY [1 votes]: The property mentioned in the comment is that $E[A|B,C]=E[A|B]$ when $C$ is independent of $A,B$. By definition, $E[A|B,C]$ is the a.s. unique random variable $Y$ s.t. for any $g\in\sigma(B,C)$, $E[Y1_g]-E[A1_g]=0$. So we want to show that $E[E[A|B]1_g]-E[A1_g]=0$ for any such $g$. In fact we don't need to check any $g$ in $\sigma(B,C)$ but can take a suitable subset, and something like $\pi-\lambda$ theorem or monotone class theorem extends to all of $\sigma(B,C)$. So I will just look at $g$ of the form $b\cap c$ for $b\in\sigma(B),c\in\sigma(C)$. Then $$
E[E[A|B]1_{b\cap c}]=E[E[A|B]1_b1_c]=E[E[A1_b|B]1_c]=E[E[A1_b|B]E[1_c]=E[A1_b]E[1_c]=E[A1_b1_c]=E[A1_{b\cap c}].
$$


text:
{\bf Problem.} Six cards numbered $1$ through $6$ are to be lined up in a row. Find the number of arrangements of these six cards where one of the cards can be removed leaving the remaining five cards in either ascending or descending order.

{\bf Level.} Level 5
{\bf Type.} Counting & Probability
{\bf Solution.} Realize that any sequence that works (ascending) can be reversed for descending, so we can just take the amount of sequences that satisfy the ascending condition and multiply by two.
If we choose any of the numbers $1$ through $6$, there are five other spots to put them, so we get $6 \cdot 5 = 30$. However, we overcount some cases. Take the example of $132456$. We overcount this case because we can remove the $3$ or the $2$. Therefore, any cases with two adjacent numbers swapped is overcounted, so we subtract $5$ cases (namely, $213456, 132456, 124356, 123546, 123465$,) to get $30-5=25$, but we have to add back one more for the original case, $123456$. Therefore, there are $26$ cases. Multiplying by $2$ gives the desired answer, $\boxed{52}$.

set_name: MATH


TITLE: How to prove that the principal minor criterion of Hurwitz does not always imply positive semi definiteness if principal minors are $\geq 0$?
QUESTION [0 upvotes]: Let $ n\ge2$ and $A:=((a_{ij})_{1\leq i,j\leq k})  \in \mathrm{Mat}(n\times n,\mathbb{R} )$ a symmetric matrix with the property that all principal minors are $\geq 0$ or in other words:
$\det((a_{ij})_{1\leq i,j\leq k}) \geq 0$ for $k=1,...,n$.
Does it always follow that $A$ is positive semi definite or is there for $n\geq 2$ a matrix $A$ with $ x^{t}Ax<0$ for a $x \in \mathbb{R}^n$?

REPLY [4 votes]: Consider $A=\left[ {\begin{array}{cc} 0 & 0 \\ 0 & -1 \\ \end{array} } \right]$
which meets the criteria. Now put $x=\left[ {\begin{array}{cc} 0 \\ 1 \\ \end{array} } \right]$ and $x^T Ax=-1$.


TITLE: Proof of linearity of abstract integration using simple functions
QUESTION [2 upvotes]: Proof of linearity of abstract integrals for simple functions by Prof. David Gamarnik and Prof. John Tsitsiklis - Lecture notes of Fundamentals of Probability
6-436j OCW MIT
In the proof provided above, it seems very confusing to me that the sum of two simple functions
 $g = \sum_{i=1}^n a_i \mathbf{1}_{A_i}$
and $h=\sum_{i=1}^m b_i \mathbf{1}_{B_i}$,
where $\mathbf{1}_{S}$ is the indicator function,
is given by
$f = \sum_{i=1}^n \sum_{j=1}^m (a_i+b_j) \mathbf{1}_{A_i\cap B_j}$.
For example, let $g = 3\times\mathbf{1}_{\omega\geq 0}$ and $h = 2\times\mathbf{1}_{\omega< 0}$.
If $\omega = 2$, then $g(2) = 3, h(2) = 0$, while $f(2) = 0 $ since the intersection of ${\omega\geq 0}$ and ${\omega< 0}$ is empty.
Moreover, in the third equality in the proof where $\sum_{j=1}^m\mu(A_i\cap B_j)$ is said to be $ = \mu(A_i)$ by finite additivity, how do we know that the union of all intersections of $B_j$'s with $A_i$ will cover all of $A_i$ ?
Thanks in advance!
Link to lecture notes

REPLY [0 votes]: When writing this way you should use partitions into measurable sets ${\cal A} = (A_1,...,A_n)$ and ${\cal B} = (B_1,...,B_m)$. (so that $\cup_i A_i=X$ and $\cup_j B_j=X$). Then the refinement
$$ {\cal A} \wedge {\cal B} = \{ A\cap B : A\in {\cal A}, B\in {\cal B}\}$$
is again a partition and the function $g+h$ is a simple function with respect to this refined partition. Using this writing systematically, the set of simple functions become an algebra which you then may use in monotone/dominated/etc... convergence.
Another point of view is that each measurable simple function $g$ takes finitely many values (possibly including zero) and ${\cal A}$ is the collection of (measurable) pre-images of all these values and ${\cal A}$ forms a partition.


text:
\begin{document}

\begin{abstract}

We consider the $k$-higher Mahler measure $m_k(P)$ of a
Laurent polynomial $P$ as the integral of $\log ^k \left| P \right|$ over
the complex unit circle. In this paper we derive an explicit formula
for the value of $\left| m_k(P) \right|/k!$ as $k \to \infty.$

\end{abstract}

\subjclass[2010]{11R06; 11M99.}
\keywords{Mahler measure, higher Mahler measure.}

\maketitle

\section{Introduction}

For a non-zero Laurent polynomial $P(z) \in \mathbb{C}[z, z^{-1}],$
the $k$-higher Mahler measure of $P$ is defined \cite{R2} as
\[
  m_k(P) = \int _{0}^{1} \log^k \left|P\left(e^{2\pi it}\right)\right|\dif{t}.
\]
For $k=1$ this coincides with the classical (log) Mahler measure defined as
\[
  m(P) = \log|a| + \sum_{j=1}^{n} \log \left(\max \{1,|r_j|\} \right),
          \,\,\,\mbox{for} \,\,\, P(z)=a \prod_{j=1}^{n} (z-r_j),
\]
since by Jensen's formula  $m(P)=m_1(P)$ \cite{book}.

Though classical Mahler measure was studied extensively, higher
Mahler measure was introduced and studied very recently by
Kurokawa, Lalín and Ochiai \cite{R2}  and Akatsuka \cite{R1}.
It is very difficult to evaluate $k$-higher Mahler measure for
polynomials except few specific examples shown in \cite{R1} and
\cite{R2} , but it is relatively easy to find their limiting values.

In \cite{sinha} Lalin and Sinha answered Lehmer's question
\cite{book} for higher Mahler measure by finding
 non-trivial lower bounds for $m_k$ on $\mathbb{Z}[z]$ for $k \geq 2.$

In \cite{B} it has been shown using Akatsuka's zeta function of \cite{R2} that for $|r|=1$,
$|m_k(z+r)|/k! \to 1/\pi$ as
$k \to \infty.$
In this paper we generalize this result by computing the same limit
for an arbitrary Laurent polynomial
$P(z) \in \mathbb{C}[z, z^{-1}]$ using a different technique.

\begin{thm} \label{thm:main}
Let $P(z) \in \mathbb{C}\left[z, z^{-1}\right]$ be a Laurent polynomial,
possibly with repeated roots.
Let $z_1,\dots, z_n$ be the distinct roots of $P$. Then
\[
  \lim_{k \to \infty} \frac{\left| m_k(P) \right|}{k!}
         = \frac{1}{\pi} \, \sum_{z_j \in S^1} \frac{1}{\left| P'(z_j) \right|},
\]
where $S^1$ is the complex unit circle $|z|=1$, and the right-hand side is
taken as $\infty$ if $P'(z_j)=0$ for some $z_j\in S^1$, i.e., if $P$ has a
repeated root on $S^1$.
\end{thm}

\section{Proof of the theorem}
  We first prove several lemmas which essentially show that the integrand
may be linearly approximated near the roots of $P$ on $S^1$.

\begin{lem} \label{lem1}
  Let $P(z) \in \mathbb{C}\left[z, z^{-1}\right] $ be a
  Laurent polynomial and $A \subseteq [0,1]$ be a closed
  set such that $P\left(e^{2 \pi i t}\right) \neq 0$ for
  all $t \in A.$ Then
  \[
    \lim_{k\to\infty} \frac{1}{k!} \int_A \log^k \left| P\left(e^{2 \pi i t}\right) \right| \d{t}=0
  \]
\end{lem}

\begin{proof}
  Since $A$ is closed, due to the periodicity of $e^{2\pi it}$
  and continuity of $P(e^{2\pi it})$ there exist constants $b$ and
  $B$ such that $0 < b \leq \left| P\left(e^{2\pi it}\right) \right| \leq B$ on $A$.
  Then for each positive integer $k$,
  $(\log^k \left| P\left(e^{2\pi it}\right) \right|)/k!$
  is bounded between $(\log ^k b)/k!$ and $(\log ^k B)/k!$, and therefore
  $(1/k!)\int _A\log^k \left| P\left(e^{2\pi it}\right) \right| \d{t}$
  is bounded between $(\mu A \, \log ^k b)/k!$ and
  $(\mu A \, \log ^k B)/k!,$ where $\mu A$ is the Lebesgue measure of
  $A$. The result follows by letting $k$ tend to infinity.
\end{proof}

\begin{lem} \label{lem:lapprx}
  Let $P(z) \in \mathbb{C}\left[z, z^{-1}\right] $ be a Laurent polynomial
  with a root of order one at $z_0=e^{2\pi it_0},$ and $P'(z)$ be its
  derivative with respect to $z$.
  Then for each $\varepsilon\in(0,1)$ there exists $\delta >0$ such that $|t-t_0| < \delta$ implies
  \[
    \left| 2 \pi (1-\varepsilon)(t-t_0)  P' \left( e^{2 \pi i t_0} \right) \right|
    \leq \left| P\left( e^{2 \pi i t} \right) \right|
    \leq \left| 2 \pi (1+\varepsilon)(t-t_0)  P' \left( e^{2 \pi i t_0} \right) \right|.
  \]
 \end{lem}

\begin{proof}
  Set $f(t)=P\left(e^{2 \pi i t}\right).$
  Then $f'(t_0)=2 \pi i P'\left(e^{2 \pi i t_0}\right)\ne 0$ and
  \[
    f'(t_0)=\lim_{t \to t_0} \frac{f(t)-f(t_0)}{t-t_0}.
  \]
  Since $f'(t_0)\ne 0$, it follows that for each $\varepsilon\in(0,1)$
  there exists $\delta > 0$ such that $0<|t-t_0|<\delta$ implies
  \[
    1-\varepsilon < \left| \frac{f(t)-f(t_0)}{(t-t_0)} \cdot \frac{1}{f'(t_0)} \right| < 1+\varepsilon,
  \]
  which proves the lemma since $f(t_0)=P(z_0)=0.$
\end{proof}

\begin{lem} \label{lem3}
  Let $c \neq 0,$ and $t_0 \in \mathbb{R}.$ Then for all $\varepsilon > 0,$
  \[
    \lim_{k \to \infty} \frac{1}{k!} \left|\,\,
        \int_{t_0 - \varepsilon}^{t_0 + \varepsilon} \log^k|c(t-t_0)| \d{t} \right|
     = \frac{2}{|c|}.
  \]
\end{lem}

\begin{proof}
  For $k \geq 1$ and $x>0,$ it follows from integration by parts and induction that
  \[
    \int_{0}^{x} \log^k u \d{u}
      = x \log ^k x +x \sum_{j=1}^{k} \frac{(-1)^j \, k! \, \log^{k-j} \, x}{(k-j)!}.
  \]
  Using the even symmetry of the integrand and substituting $u=|c(t-t_0)|$, we have
  \[
    \frac{1}{k!} \left|\,\,\int_{t_0 - \varepsilon}^{t_0 + \varepsilon} \log^k|c(t-t_0)| \d{t} \right|
     = \frac{2}{|c|\,k!} \left|\,\int_{0}^{|c\varepsilon|} \log^k u \,\d{u}  \right|,
  \]
  and it follows that
  \begin{eqnarray*}
    \lim_{k \to \infty} \frac{1}{k!} \left|\,\,\int_{t_0 - \varepsilon}^{t_0 + \varepsilon} \log^k|c(t-t_0)| \d{t} \right|
      &=& \lim_{k \to \infty} \frac{2}{|c|\,k!} \left|\,\int_{0}^{|c\varepsilon|} \log^k u \d{u} \right|\\
      &=& 2 \varepsilon \lim_{k \to \infty} \! \left| \frac{\log^k |c \varepsilon|}{k!} +
            \sum_{j=1}^{k} \frac{(-1)^j \log^{k-j} |c \varepsilon|}{(k-j)!} \right| \\
      &=& 2 \varepsilon \left| \sum_{n=0}^{\infty} \frac{(-1)^{n} \log^{n} |c \varepsilon|}{n!} \right| \\
      &=& 2 \varepsilon e^{- \log |c \varepsilon|} = 2/|c|.
  \end{eqnarray*}
  \end{proof}

\begin{lem} \label{lem4}
  Let $P(z) \in \mathbb{C}\left[z, z^{-1}\right] $ be a Laurent polynomial
  with a root of order one at $z_0=e^{2 \pi i t_0}.$ Then for all sufficiently
  small $\delta > 0,$
  \[
    \lim_{k \to \infty} \frac{1}{k!} \left|\,\,
      \int_{t_0-\delta}^{t_0+\delta} \log^k \left| P \left( e^{2 \pi i t} \right) \right| \d{t} \right|
      =\frac{1}{\pi \left| P'\left( e^{2\pi it_0}\right) \right|}.
  \]
\end{lem}

\begin{proof}
  First notice that since $z_0$ has order one, it cannot be a root of $P'(z).$
  Now let $\varepsilon \in (0,1).$ By Lemma \ref{lem:lapprx} there is a $\delta > 0$
  such that $|t-t_0| < \delta$ implies
  \[
    \left| 2\pi(1-\varepsilon)(t-t_0) P'\left( e^{2\pi it_0}\right)\right|
    \leq \left|P\left(e^{2\pi it}\right)\right|
    \leq \left|2\pi(1+\varepsilon)(t-t_0) P'\left(e^{2\pi it_0}\right)\right|
    \leq 1.
  \]
    Setting $c=2\pi(1-\varepsilon)P'\left(e^{2\pi it_0}\right)$
    and $d=2\pi(1+\varepsilon)P'\left(e^{2\pi it_0}\right)$ it follows that
    for $0< |t-t_0|<\delta$,
  \[
    \log |c(t-t_0)| \leq \log\left|P\left(e^{2\pi it}\right)\right|
    \leq \log|d(t-t_0)| \leq 0,
  \]
  and hence
  \[
    \left|\log^k|c(t-t_0)|\right| \geq \left|\log^k\left|P\left(e^{2\pi it}\right)\right|\right|
    \geq \left|\log^k|d(t-t_0)| \right| \geq 0,
  \]
 for all $k \in \mathbb{N}.$ Therefore,
  \[
    \int \limits_{t_0-\delta}^{t_0+\delta} \left|\log^k|c(t-t_0)|\right| \d{t}
    \geq \! \! \! \int \limits_{t_0-\delta}^{t_0+\delta} \left|\log^k\left|P\left(e^{2\pi it}\right)\right|\right|\d{t}
    \geq \! \! \! \int \limits_{t_0-\delta}^{t_0+\delta} \left|\log^k|d(t-t_0)| \right| \d{t} \geq 0.
  \]
  But on $(t_0-\delta,t_0+\delta),$ for each fixed $k$,
  either all three functions $\log^k |c(t-t_0)|$,
  $\log^k \left|P\left(e^{2\pi it}\right)\right|$ and
  $\log^k |d(t-t_0)|$ are negative (if $k$ is odd), or positive (if $k$ is even).
  So the integrals of their absolute values are equal to the
  absolute values of their integrals and therefore we have
  \[
    \left|\,\,\int \limits_{t_0-\delta}^{t_0+\delta}  \log^k |c(t-t_0)| \d{t} \right|
    \geq \left|\,\,\int \limits_{t_0-\delta}^{t_0+\delta} \log^k \left|P\left(e^{2\pi it}\right)\right| \d{t}\right|
    \geq \left|\,\,\int \limits_{t_0-\delta}^{t_0+\delta} \log^k |d(t-t_0)| \d{t} \right|.
  \]
  By Lemma \ref{lem3} it follows that
  \[
    \frac{2}{|c|}
    \geq \lim_{k \to \infty} \frac{1}{k!} \left|\,\,
          \int_{t_0-\delta}^{t_0+\delta} \log^k \left|P\left(e^{2\pi it}\right)\right|\right|
    \geq \frac{2}{|d|}.
  \]
  Since $c=2 \pi (1-\varepsilon)P' \left( e^{2 \pi i t_0} \right)$ and
  $d=2 \pi (1+\varepsilon)P' \left( e^{2 \pi i t_0} \right)$ and $\varepsilon > 0$
  is arbitrary, we are done.
\end{proof}

  With these lemmas, we now proceed to prove the main theorem.

\begin{proof}[Proof of Theorem \ref{thm:main}]
  First notice that
  \[
    \frac{m_k(P)}{k!} = \frac{1}{k!} \int_{0}^{1} \log^k \left|P\left(e^{2\pi it}\right)\right|\d{t}.
  \]
  If $P(z)$ does not have any roots on $S^1$ then choosing
  $A=[0,1]$ and applying Lemma \ref{lem1} we see that
  $ \left| m_k(P) \right|/k! \to 0$ as $k \to \infty$
  and the theorem holds in this case.

  Now let $t_1,\dots, t_m\in[0,1]$ such that
  $e^{2\pi it_1},\dots, e^{2\pi it_m}$ are the distinct roots of $P$ on $S^1$.
  Let $\delta>0$ be sufficiently small so that $|P(e^{2\pi it_j})|<1$ on
  each interval $(t_j-\delta, t_j+\delta)$, $j=1,\dots, m$, and these intervals are disjoint and define
  \[
    A=[0,1] \smallsetminus \bigcup _{j=1}^{m} (t_j -\delta, t_j + \delta).
  \]
  Using Lemma \ref{lem1}, and the fact that $\log|P(e^{2\pi it})|<0$ on $[0,1]\setminus A$,
  we find that
  \begin{eqnarray}
    \lim_{k\to\infty}\frac{|m_k(P)|}{k!}
    &=& \lim_{k\to\infty} \frac{1}{k!} \left| \int_A \log^k|P(e^{2\pi it})|\d{t}
                                            + \int_{[0,1]\setminus A} \log^k|P(e^{2\pi it})|\d{t} \right| \nonumber \\
    &=& \lim_{k\to\infty} \sum_{j=1}^m \frac{1}{k!}
           \left| \int_{t_j-\delta}^{t_j+\delta} \log^k|P(e^{2\pi it})|\d{t} \right| \label{eqn:decomp}
  \end{eqnarray}
  If $P$ has no repeated roots on $S^1$, by Lemma \ref{lem4}, this final sum is equal
  $\pi^{-1}\sum_{j=1}^m |P'(e^{2\pi it_j})|^{-1}$,
  and so the theorem is proven in this case.

  Finally, if $P$ has a repeated root on $S^1$, we may assume without loss of generality that
  $P(z_1)=P'(z_1)=0$ where $z_1=e^{2 \pi i t_1}$. With $f(t)=P(e^{2\pi it})$, we have that $f(t_1)=f'(t_1)=0$.
  Then for each $\varepsilon\in(0,1)$ there is a $\delta_\varepsilon \in (0,1)$ such that
  \[
    \left| \frac{f(t)}{t-t_1}\right| = \left| \frac{f(t)-f(t_1)}{t-t_1}\right|
      \le \varepsilon, \hspace{12pt}\mbox{ for all } 0<|t-t_1|<\delta_\varepsilon.
  \]
  It follows that $\log|f(t)|\le \log|\varepsilon(t-t_1)|<0$ for all $0<|t-t_1|<\delta_\varepsilon$,
  and so
  \[
    \left| \log^k|f(t)|\right| \ge \left| \log^k|\varepsilon(t-t_1)|\right|,
      \hspace{12pt}\mbox{ for all } 0<|t-t_1|<\delta_\varepsilon.
  \]
  We may assume that $\delta_\varepsilon < \delta$, and using \eqref{eqn:decomp} and Lemma \ref{lem3}
  deduce that
  \begin{eqnarray*}
    \lim_{k\to\infty}\frac{|m_k(P)|}{k!}
      &\ge & \lim_{k\to\infty} \left| \int_{t_1-\delta}^{t_1+\delta} \log^k|P(e^{2\pi it})|\d{t} \right| \\
      & =  & \lim_{k\to\infty} \int_{t_1-\delta}^{t_1+\delta} \left| \log^k|P(e^{2\pi it})| \right| \d{t} \\
      &\ge & \lim_{k\to\infty} \int_{t_1-\delta_\varepsilon}^{t_1+\delta_\varepsilon}
                     \left| \log^k|\varepsilon(t-t_0)| \right| \d{t} \\
      &=& \frac{2}{|\varepsilon|}.
  \end{eqnarray*}
  Since $\varepsilon\in(0,1)$ was arbitrary, the limit in question diverges to $\infty$
  and the theorem is proven.
\end{proof}

\bibliographystyle{elsarticle-num}

config: arxiv


TITLE: $a_1=2$ and $a_{n+1}=2+\frac{1}{a_n}\implies |a_{n+1}-a_n|\leq\frac14|a_{n-1}-a_n|$
QUESTION [0 upvotes]: Let $(a_n)_{n\in\mathbb{N}}$ satisfy that $a_1=2$ and $a_{n+1}=2+\frac{1}{a_n}$. Show that
  for all $n\in\mathbb{N}$ with $n\geq 2$,
  $$
|a_{n+1}-a_n|\leq\frac14|a_{n-1}-a_n|
$$

So I can show the base case to be true, but I can't see how to show the inductive step.

REPLY [0 votes]: $$\forall i\in \mathbb{N} \qquad a_{i}=2+\frac{1}{a_{i-1}}\geq2$$
$$a_{n+1}=2+\frac{1}{a_n}...(1)$$
$$a_{n}=2+\frac{1}{a_{n-1}}...(2)$$
Subtracting (2) fom (1),
$$ |a_n-a_{n-1}|=|\frac{a_n-a_{n-1}}{a_na_{n-1}}|=\frac{|a_n-a_{n-1}|}{|a_na_{n-1|}}\leq\frac{|a_n-a_{n-1}|}{2\times 2}=\frac{|a_n-a_{n-1}|}{4}$$


TITLE: Linear Algebra: how to extrapolate vector from endpoint?
QUESTION [0 upvotes]: I am in uni and was given this problem:

$T:\mathbb{R}->\mathbb{R}^4$ sends $3$ to $(2,5,1,6)$. Where does it send $7$?

I don't understand how a solution is possible. I assume I must extrapolate a vector from  $(3)$ to $(2, 5, 1, 6)$, but surely there are many ways to do it and many such vectors?

REPLY [0 votes]: From what I know, $7 \in \mathbb{R}$.
However, if your function is $\mathbb{R} \to \mathbb{R^4}$, then by the rule that $T(cx)=cT(x)$, $7$ is sent to the vector $\left(\frac{14}{3},\frac{35}{3},\frac{7}{3},14\right)$, which is $\frac{7}{3}T(3)$


TITLE: Find $f$ such that the divergence of $(f(\underline{a}\cdot \underline{x}))\underline{x}=1$
QUESTION [0 upvotes]: Let $\underline{a}\in\mathbb R^n$ be a fixed vector. Find all functions $f:\mathbb R \to \mathbb R $ such that the divergence of the vector field $(f(\underline{a}\cdot \underline{x}))\underline{x}$, $x \in \mathbb R^n$ is everywhere equal to $1$.

Attempt:
I've begun by trying to work out what the vector field I've been given 'is'. So I wrote $$G=(f(\underline{a}\cdot \underline{x}))\underline{x}=(f(\underline{a}\cdot \underline{x}))(x_1,...,x_n)=(x_1f(\underline{a}\cdot\underline{x}),...,x_nf(\underline{a}\cdot\underline{x}))=(\alpha x_1,...,\alpha x_n)$$
and I know that given a vector field $\underline{v}(\underline{x})=(v_1(x_1,...,x_n),...,v_n(x_1,...,x_n))$ the divergence is $$\frac{\partial v_1}{\partial x_1}+...+\frac{\partial v_n}{\partial x_n}$$
The problem I have is that my vector field isn't of the correct form to work out the divergence so I guess this means I've done something wrong? Can anyone help?

REPLY [0 votes]: There is nothing wrong. You must work a little bit more. In this case
$$
v_i(x)=f(a\cdot x)\,x_i=f(a_1\,x_1+\dots+a_n\,x_n)\,x_i.
$$
Now you have to calculate $\partial v_i/\partial x_i$, sum from $i=1$ to $i=n$ and equal the result to $1$.


TITLE: Baumslag-Solitar Group $G=\langle a,t \mid tat^{-1}=a^k\rangle\cong\mathbb{Z}[1/k]\rtimes\langle t\rangle$?
QUESTION [3 upvotes]: Let $G$ be the Baumslag-Solitar group $\langle a,t \mid tat^{-1}=a^k\rangle$ and
$$\mathbb{Z}[1/k]:=\left\{\frac{x}{k^n}\mid x\in\mathbb{Z},n\in\mathbb{N}\cup\{0\}\right\}.$$
I'm searching for an isomorphism $G=\langle a,t \mid tat^{-1}=a^k\rangle\cong\mathbb{Z}[1/k]\rtimes\langle t\rangle$. I only need to know the map $\Psi:\langle a\rangle^G\to\mathbb{Z}[1/k]$.
I defined $$\Psi(t^na^xt^{-n}):=\frac{x}{k^n}.$$
But for one hour I'm calculating to prove that this map induces an homomorphism, but I always get for $n>m$:
$$\Psi(t^na^xt^{-n})\cdot \Psi(t^ma^yt^{-m})=\frac{x}{k^n}+\frac{y}{k^m}=\frac{x+y\cdot k^{n-m}}{k^n}$$
and
$$\Psi(t^na^xt^{-n}\cdot t^ma^yt^{-m})=\Psi(t^ma^{(x\cdot k^{n-m})}t^{-m}t^myt^{-m})=\Psi(t^ma^{x\cdot k^{n-m}+y}t^{-m})=\frac{x\cdot k^{n-m}+y}{k^m}.$$
And these images are obvisually unequal. So what is my fault?
Is this map the wrong one?
Thanks for your help.

REPLY [6 votes]: You need to modify the definition of $\Psi$: put $$\Psi(t^na^xt^{-n})=xk^n$$ (instead of $\Psi(t^na^xt^{-n})=xk^{-n}$). The homomorphic property should follow easily along the lines of the computations given in the original question.
Alternatively, you could replace $t$ by $t^{-1}$ in the definition of the Baumslag-Solitar group, in which case the $\Psi$ as originally given is a homomorphism.


TITLE: not able to get the divisor
QUESTION [0 upvotes]: I faced one issue. The issue is as follows.
I want to divide a particular number with 7,9,11 but in every case i want to get the
remainder as 1,2 3 accordingly. Could you please help me get the number and please explain how
it works.

REPLY [2 votes]: Hint: Let the number be $A$. Consider $2A+5$. This must be a multiple of 7, 9 and 11 from the conditions.
Hence $2A+5$ is a multiple of the LCM, which is $7\times 9 \times 11 = 693$.
So for example, since $\frac{693-5}{2} = 344$, 344 satisfies the conditions in the question. Check this. Other solutions will be $344 + 693 n$, which follows from the above.

Of course, Chinese Remainder Theorem is the way to go for the general case.


TITLE: Chordless cycles and planarity in graphs
QUESTION [1 upvotes]: Let $\{C(G)\}$ be the set of chordless cycles of a graph $G$. Compare the cycles pairwise. Let $\{V\}$ represent the pairs which have exactly one vertex in common; and, let $\{P\}$ represent those pairs which have a single continuous sequence of edges, ie, a path, in common. For a pair in $\{V\}$, let $X$ denote their common vertex. Obviously, for planarity, the edges at vertex $X$ must be ordered so that the cycles do not overlap, ie., crossover. Similar restrictions obtain on the orderings at the two terminal vertices for the common path shared by a pair of cycles in $\{P\}$.
In either type of conjunction, certain edge orderings on common vertices are required for planarity of $G$. Have the Kuratowski criteria for planarity been shown to be equivalent to the two types of restrictions described above?

REPLY [1 votes]: I believe there is a characterization like the one you mention.
The de Fraysseix–Rosenstiehl planarity criterion traverses a given graph with a depth-first search and characterizes the way edges can exist in this ordering with respect to being on certain sides of other visited paths.
It is not immediately clear whether your conditions would be exactly like these ordered-edge conditions, but the fact that there is a characterization that uses edge orderings suggests that you might be able to formalize yours into something similar or equivalent.


text:
\section{From Potential Outcomes to Other Causal Frameworks}\label{sec:extension}
We proposed a method based on weighted conformal inference which
produces interval estimates of counterfactuals and individual
treatment effects under the potential outcome framework. For
randomized experiments with perfect compliance, our method has
guaranteed coverage in finite samples without any modeling assumptions
on the data generating process. For randomized experiments with
ignorable compliance or general observational studies under the strong
ignorability assumption, our method is doubly robust in the sense that
the coverage is asymptotically guaranteed if either the conditional
quantiles of potential outcomes or the propensity scores are
consistently estimated. In contrast, existing methods may suffer from
a significant coverage deficit even in simple models. Furthermore, our
framework naturally extends to other populations by modifying the
weight function according to Table \ref{tab:weight_func}.

The key observation is the covariate shift together with the
invariance of the conditional distribution: the observed distribution
of $(X, Y(1))$ is $P_{X\mid T = 1}\times P_{Y(1)\mid X}$ under
ignorability while the target distribution is
$P_{X}\times P_{Y(1)\mid X}$. Now the invariance of the conditional distribution is
also the enabling property in other frameworks of causal inference. As
a consequence, our method can be naturally extended to those settings,
and we close this paper by discussing two of them.

\subsection{Causal diagram framework}
Judea Pearl in his pioneering work \citep{pearl1995causal} introduced
a general framework of causal inference based on graphical models,
which has been widely applied, see
  e.g.~\cite{greenland1999causal, spirtes2000causation,
    richardson2013single, glymour2014discovering,
    tennant2019use}. This framework does not rely on the notion of
counterfactuals but instead defines causal effects through the
\texttt{do} operator, which modifies the observed distribution by
removing the causal paths that directly point to the intervention
variable. We refer the readers to \cite{pearl2018book} for the
philosophy and basics of causal diagrams and to \cite{pearl2016causal}
for the mathematical foundation of this framework.

We here only discuss the case where $T$ is the intervention variable,
$Y$ is the outcome variable and $X$ is a set of variables that
satisfies the back-door criterion. Roughly speaking, this means that $X$
includes all confounders and excludes all post-treatment variables. In
this case, the foundational result in \cite{pearl1995causal} shows
that
\[P_{(X, Y) \mid \mathrm{do}(T = t)} = P_{X}\times P_{Y\mid X, T = t}.\]
This is the target distribution to be inferred. In contrast, the observed distribution of $(X, Y)$ given $T = t$ is
\[P_{(X, Y)\mid T = t} = P_{X\mid T = t}\times P_{Y\mid X, T = t}.\]
Clearly, this has exactly the same structure as in the potential
outcome framework. As a consequence, weighted split-CQR can be applied
without any modification to produce doubly robust interval estimates
of $Y$ under the \texttt{do} intervention.

\subsection{Invariant prediction framework}

Invariant prediction is another framework proposed by
\cite{peters2016causal}. It is particularly powerful when there are
multiple data sources under different interventions, such as in gene
knockout experiments. Consider an outcome variable $Y$, a set of
interventions or covariates $X$ and an environment variable $E$ that
indicates the source of data. Then one basic setting under this
framework assumes that $Y\indep E \mid X$ while $X$ may depend on $E$ with
$X \mid E\sim P_{X}^{E}$. The goal is to predict the outcome under a
new environment. For simplicity, we assume that a dataset
$(X_{ij}, Y_{ij})_{i=1}^{n_{j}}$ is available for each environment
$e_{1}, \ldots, e_{J}$ and a testing dataset $(X_{i0})_{i=1}^{n_{0}}$
for the target environment $e_{0}$. Due to the invariance assumption,
on the $j$-th dataset, the observed distribution of $(X, Y)$ is
$P_{X}^{e_{j}}\times P_{Y\mid X}$ while the target distribution to be
inferred on is $P_{X}^{e_{0}}\times P_{Y\mid X}$.

Again, this is reduced to a problem of constructing valid prediction
intervals under covariate shifts. When $J = 1$, this has exactly the
same structure as in the potential outcome framework and thus weighted
split-CQR with weight function $dP_{X}^{e_{0}}(x) / dP_{X}^{e_{1}}(x)$
produces doubly robust intervals of $Y$ under environment $e_{0}$. When $J > 1$, we can in principle apply the general
weighted conformal inference techniques from
\cite{barber2019conformal}. However the weight function becomes much
more complicated than that in Algorithm \ref{algo:split-CQR}. An
alternative approach is to create a weighted population from
environments $e_{1}, \ldots, e_{J}$ with observed distribution
\[\lb\sum_{j=1}^{J}q_{j}P_{X}^{e_{j}}\rb\times P_{Y\mid X},\]
and apply Algorithm \ref{algo:split-CQR} on this pseudo dataset with weight $w(x) = 1 / \sum_{j=1}^{J}q_{j}(dP_{X}^{e_{j}} / dP_{X}^{e_{0}})(x)$. The weights $q_{j}$ can be chosen through certain balancing procedures that forces the covariate distribution to approximate $P_{X}^{e_{0}}$. We leave the formal development of this idea to future work.


TITLE: Hyphothesis test for a coin tossed 10000 times
QUESTION [0 upvotes]: For a coin, there is no information whether it is fair or not. The following two hypothesis
are supposed for getting tail : $H_0 : p = 0,5$ and $H_1 : p = 0,7$. This coin is tossed $10^4$
times and if the number of tails that appeared at the top is more than $4940$, $H_0$ is accepted. Find  $\alpha $(type 1 error).
I do not understand why $H_0 $ is accepted,since we are in the non-critical region,then we can easily find $\beta$ but not $\alpha$
$\alpha=P(H_1:\text{accepted}|H_0 : true)=\frac{P(H_1\cap H_0)}{P(H_0:\text{true})}$
$P(H_0:\text{true})=\frac{4940}{10000}=0.494$
i dont know how to proceed from here...

REPLY [0 votes]: The test statistic that you wrote is not consistent with the hypotheses you stated.  If $H_0 : p = 0.5$ and $H_1 : p = 0.7$, and $p$ is the single trial probability of flipping a tail, then the test should reject $H_0$ if more than a certain number of tails are observed.  What you have written is that $H_0$ is accepted if more than $4940$ tails are observed.
Okay, so say I take your question exactly as it is stated.  Then your type I error is the conditional probability of rejecting $H_0$ when $H_0$ is true; i.e., $p = 0.5$ but you observed more than $4940$ tails out of $10^4$ trials.  This is a binomial probability with $n = 10^4$, $p = 0.5$:  $$\alpha = \sum_{x=4941}^{10000} \binom{10000}{x} (0.5)^x (1 - 0.5)^{10000-x}.$$  Of course, we can approximate this sum using a normal approximation:  the binomial distribution can be approximated with a normal distribution with mean $np = 5000$ and variance $np(1-p) = 2500$, so $$\alpha \approx \Pr[4940.5 \le X] = \Pr[-1.19 \le Z] = 0.882977.$$  This is your probability of Type I error given the EXACT statement of your question.  But as I said earlier, this makes no sense because your test statistic is not consistent with your hypotheses.  You are basically saying that the more tails you observe, the more likely you will conclude the coin is fair.  That is why the Type I error we calculated above is so large.


text:
\begin{document}
\title{On the invariant theory of the B{\'e}zoutiant}
\author{Jaydeep V.~Chipalkatti}
\maketitle

\parbox{12cm}{ \small
{\sc Abstract.}
We study the classical invariant theory of the B{\'e}zoutiant $\R(A,B)$ for a
pair of binary forms $A,B$. It is shown that $\R(A,B)$ is determined
by the first two odd transvectants $M = (A,B)_1$, $N = (A,B)_3$, and
one can characterize the forms $M,N$ which can arise from some $A,B$.
We give formulae which express the higher odd transvectants
$(A,B)_5,(A,B)_7$ in terms of $M$ and $N$. We also describe
a `reduction formula' which recovers $B$ from $\R(A,B)$ and $A$.}

\bigskip

\parbox{12cm}{\small
Mathematics Subject Classification (2000):\, 13A50. \\
Keywords: binary forms, B{\'e}zoutiant, transvectant, covariant,
Grassmannian.}

\bigskip

\section{Introduction}
We begin by recalling the construction of the B{\'e}zoutiant of
two binary forms. Let
$\ux = (x_0,x_1), \uy=(y_0,y_1)$ be pairs of variables, and
write $\omega = x_0 y_1 - x_1y_0$. If $A,B$ are (homogeneous) forms
of order $d$ in $\ux$, their B{\'e}zoutiant is defined to be
\[ \R(A,B) = \frac{A(x_0,x_1)B(y_0,y_1) - B(x_0,x_1)A(y_0,y_1)}{\omega}. \]
Since $\R$ is symmetric in $\ux$ and $\uy$ of order $d-1$ in each,
it can be seen as a quadratic form over the vector space of
order $d-1$ binary forms.

If $V = \text{Span} \{x_0,x_1\}$, then the construction of $\R$ corresponds
to the isomorphism of $SL(V)$-representations
\[ \wedge^2 \, \Sym^d \, V \lra
 \Sym^2(\Sym^{d-1} \, V), \quad
A \wedge B \lra \R(A,B). \]
It is easy to see that
\[ \R(\alpha \, A + \beta \, B,\gamma \, A + \delta \, B) =
(\alpha \, \delta - \beta \, \gamma) \, \R(A,B), \]
i.e., up to a scalar, $\R$ depends only on the
pencil spanned by $A,B$ (denoted $\Pi_{A,B}$). Conversely,
$\R$ determines the pair $(A,B)$ up to a unimodular transformation.

B{\'e}zoutiants have been principally studied
for their use in elimination theory
(e.g., see \cite{Householder1} or \cite[vol.~I, \S 136 ff]{Netto}).
In contradistinction, our interest lies in their invariant theoretic
properties (understood in the sense of Grace and Young \cite{GrYo}).

\subsection{A summary of results}
In section \ref{section.prelim} we recall some fundamental
facts about transvectants.
We will show that $\R(A,B)$ admits a `Taylor series' in
$\omega$ as follows:
\[ \R(A,B) = c_0 \, T_1^\sympol + c_1 \, \omega^2 \, T_3^\sympol +
   c_2 \, \omega^4 \, T_5^\sympol + \dots, \]
where
\begin{itemize}
\item $T_{2r+1}$ denotes the $(2r+1)$-th transvectant of
$A,B$,
\item
$\sympol$ denotes the operation of symmetric polarization, and
\item
$c_r$ are rational constants dependent on $d$ and $r$.
\end{itemize}

{\sl Hence, from our viewpoint, a study of $\R(A,B)$ will be
tantamount to a study of the odd
transvectants $\{T_{2r+1}:r \ge 0\}$ of $A$ and $B$.}

\smallskip

In section \ref{section.wr.ode}, we formulate a second order
differential equation derived from $T_1,T_3$ whose solution
space is $\Pi_{A,B}$. This shows that the terms of degree $\le 2$ in
the Taylor series implicitly determine those of higher degree. The former
cannot be chosen arbitrarily, and we give an algebraic
characterization of terms which can so appear. Specifically, we construct a
set of joint covariants $\Phi_0,\dots,\Phi_d$ with arguments $M,N$,
with the following property:

{\sl There exist $A,B$ such that
$M = (A,B)_1, N = (A,B)_3$, if and only if
$\Phi_0(M,N) = \dots = \Phi_d(M,N) = 0$.}

\smallskip

We have remarked earlier that $\R$ determines $\Pi_{A,B}$.
Hence, given $A$ and $\R$, the form $B$ is determined up to an
additive multiple of $A$. In section \ref{section.gen.red}, we
give an equivariant formula for $B$ in terms of $A$ and $\R$.
This is called a `generic reduction formula',
in analogy with a device introduced by D'Alembert in the theory of
differential equations.

In section \ref{section.t5t7}, we use the classical Pl{\"u}cker relations
to describe formulae which calculate $T_5,T_7$ from a
knowledge of $T_1$ and $T_3$. The question of a formula in
the general case of $T_{2r+1},r \ge 4$ is left open.
Three more open problems (with some supporting examples) are
given in section~\ref{section.open}.

\medskip

{\sc Acknowledgements.} I thank my colleage
A.~Abdesselam for several helpful conversations, and the University
of British Columbia for financial support.

\section{Preliminaries} \label{section.prelim}
We will heavily use \cite{GrYo} as a standard reference for classical
invariant theory. Glenn's treatise \cite{Glenn} covers substantially
the same ground. In particular, we assume some familiarity with
transvectants, covariants, and the symbolic calculus.
A more recent exposition of this material is given in \cite{Olver}.
Basic facts about the representation theory of $SL_2$ can be found
in \cite{FH,Springer1,Sturmfels}.

The base field is throughout $\complex$.
A form will always mean a homogeneous polynomial in $\ux$.
By contrast, an $\ux\uy$-form will involve both sets of variables,
and will be homogeneous in each set. The $\ux$-degree of a form
will be called its order (to avoid conflict with \cite{GrYo}).
The order of an $\ux\uy$-form is a pair of integers.

The letter $\cons$ will stand for an
unspecified nonzero constant.

\subsection{$SL_2$-modules}
Let $V$ be a $\complex$-vector space of dimension two with the
natural action of $SL(V)$. We write $S_e$ for the
symmetric power representation $\text{Sym}^e \, V$, and
$S_e(S_f)$ for $\text{Sym}^e(\text{Sym}^f \, V)$ etc.

The $\{S_e: e \ge 0\}$ are a complete set of finite dimensional
irreducible $SL(V)$-modules. By complete reducibility,
each finite dimensional $SL(V)$-module is isomorphic to a
direct sum of the $S_e$.
If $\{x_0,x_1\}$ is a basis of $V$, then an element of $S_e$
is a form of order $e$ in $\ux$. We identify
the projective space $\P^e$ with $\P S_e$, and write
$A \in \P^e$ for the point represented by a (nonzero) form $A$.
By convention, $S_e =0$ if $e < 0$.

\subsection{Transvectants}
For integers $e,f \ge 0$, we have a decomposition of $SL(V)$-modules
\begin{equation}
S_e \otimes S_f \simeq \bigoplus\limits_{r=0}^{\min\{e,f\}}
S_{e+f-2r}.
\label{clebschgordan} \end{equation}
If $E,F$ are forms of orders $e,f$, the image of the projection
of $E \otimes F$ in the $r$-th summand is called their $r$-th transvectant,
denoted $(E,F)_r$. It is a form of order $e+f-2r$, whose coefficients are
linear in the coefficients of $E$ and $F$.
In coordinates, it is given by the formula
\begin{equation}
(E,F)_r = \frac{(e-r)!(f-r)!}{e!f!}
\sum\limits_{i=0}^r \, (-1)^i \binom{r}{i} \,
\frac{\partial^{\,r} E}{\partial x_0^{r-i} \, \partial x_1^i } \;
\frac{\partial^{\,r} F}{\partial x_0^i \, \partial x_1^{r-i}}
\label{trans.defn} \end{equation}
(The initial scaling factor is conventional, some authors choose it
differently.) In particular $(E,F)_0 = E \, F$, and
$(E,F)_1 = \cons \times$ Jacobian$(E,F)$. Note that
\begin{eqnarray}
(E,F)_r & = & (-1)^r(F,E)_r, \label{trans.1} \\
(E,F)_r & = & 0 \qquad \text{for $r > \min\{e,f\}$}. \label{trans.2}
\end{eqnarray}
If $E,F$ have the same order, then
\begin{equation} (\alpha \, E + \beta \, F,
\gamma \, E + \delta \, F)_{2r+1} =
(\alpha \, \delta - \beta \, \gamma) \, (E,F)_{2r+1},
\label{comb} \end{equation}
for arbitrary constants $\alpha,\beta,\gamma,\delta$.
This shows that the odd transvectants $(E,F)_{2r+1}$ are
{\sl combinants} of $E,F$, i.e., up to a scalar, they
depend only on the pencil spanned by $E,F$.

If $E,F$ are given {\sl symbolically}, then \cite[\S 49]{GrYo} gives
an algorithm for calculating their transvectants. See
Proposition \ref{prop.w} for a typical instance of its use.

The following lemma is elementary (see \cite[Lemma 2.2]{Goldberg1}).
\begin{Lemma} \sl Let $E,F$ be nonzero forms of order $e$ such that
$(E,F)_1=0$. Then $E = \cons F$. \qed
\label{lemma.jac} \end{Lemma}

\subsection{} \label{self-dual}
Each representation $S_e$ is self-dual, i.e.,
we have an isomorphism
\[ S_e \stackrel{\sim}{\lra} S_e^* = \Hom_{SL(V)}(S_e , \complex). \]
This map sends an order $e$ form $E$ to the functional
\[\delta_E: S_e \lra \complex, \quad F \lra (E,F)_e. \]

\subsection{The Gordan series}
Given three forms $E,F,G$, this very useful series describes certain
linear dependency relations between transvectants of the
type $((E,F)_\star,G)_\star$ and $((E,G)_\star,F)_\star$.

Let $E,F,G$ be of orders $e,f,g$ respectively, and
$a_1,a_2,a_3$ three integers satisfying the following
conditions:
\begin{itemize}
\item
$\, a_2 + a_3 \le e, \, a_1 + a_3 \le f, \, a_1 + a_2 \le g$, and
\item either
$a_1 = 0$ or $a_2 + a_3 = e$ (or both).
\end{itemize}
Then we have an identity
\begin{equation} \begin{aligned}
& \sum\limits_{i \ge 0} \, \frac{\binom{f-a_1-a_3}{i}\binom{a_2}{i}}
{\binom{e+f-2a_3-i+1}{i}} \, ((E,F)_{a_3+i},G)_{a_1+a_2-i} \\
= (-1)^{a_1} & \sum\limits_{i \ge 0} \,
\frac{\binom{g-a_1-a_2}{i}\binom{a_3}{i}}
{\binom{e+g-2a_2-i+1}{i}} \, ((E,G)_{a_2+i},F)_{a_1+a_3-i} \, .
\label{Gordan} \end{aligned}\end{equation}
Usually (\ref{Gordan}) is denoted by
$\Gordan{E}{F}{G}{e}{f}{g}{a_1}{a_2}{a_3}$.

\subsection{The Clebsch-Gordan series}
Let $\polar$ denote the polarization operator
\[ y_0 \frac{\partial}{\partial x_0} + y_1 \frac{\partial}{\partial x_1}. \]
If $E$ is a form of order $e$, then define its $m$-th polar
to be
\[ \po{E}{m} = \frac{(e-m)!}{e!} \, (\polar)^m \, E, \]
which is an $\ux\uy$-form of order $(e-m,m)$. By Euler's theorem,
we can recover $E$ from $\po{E}{m}$ by the substitution $\uy:=\ux$.
If $e$ is even, we will denote $\po{E}{e/2}$ by $E^\sympol$.
It is symmetric in $\ux, \uy$, and naturally thought of as an
element of $S_2(S_{e/2})$.

The Clebsch-Gordan series is a more precise version of
the decomposition (\ref{clebschgordan}). For
forms $E,F$ of orders $e,f$, it gives an identity
\begin{equation} E(\ux) \, F(\uy) = \sum\limits_{r \ge 0} \,
\frac{\binom{e}{r} \binom{f}{r}}{\binom{e+f-r+1}{r}} \,
\omega^r \, \po{(E,F)_r}{f-r}.
\label{CG} \end{equation}

\begin{Remark} \rm
The notional distinction between the Gordan series and
Clebsch-Gordan series is merely for convenience of reference, and
has no historical basis. In fact (\ref{CG}) directly leads
to (\ref{Gordan}) (see \cite[\S 52]{GrYo}).
\end{Remark}

Now let $\U \in S_2(S_{d-1})$. We identify $\U$ with
an $\ux\uy$-form of order $(d-1,d-1)$ which is symmetric in
both sets of variables. It can then be expressed
as a `Taylor series' in $\omega$. Define constants
\begin{equation}
c_r = \frac{2 \, \binom{d}{2r+1}^2}{\binom{2d-2r}{2r+1}} \quad
\text{for $0 \le r \le \lfloor \frac{d-1}{2} \rfloor$}.
\label{cr} \end{equation}
\begin{Proposition} \sl
There exists a unique sequence of forms
\[
U_\bullet = (U_1,U_3,\dots,U_{2r+1}, \dots), \]
where $\ord \, U_{2r+1} = 2(d-2r-1)$, such that
\[
\U = \sum\limits_{r \ge 0} c_r \, \omega^{2r} \, (U_{2r+1})^\sympol. \]
\end{Proposition}
\demo First we prove the existence.
Since $\U$ is symmetric in $\ux$ and $\uy$, it is a linear
combination of expressions of the form
\[ \langle i \, j \rangle  =
x_0^{d-1-i} x_1^i \, y_0^{d-1-j}y_1^j +
x_0^{d-1-j} x_1^j \, y_0^{d-1-i}y_1^i. \]
Let $A = x_0^{d-1-i}x_1^i, B = x_0^{d-1-j}x_1^j$, so
\[ \langle i \, j \rangle  = A(\ux)B(\uy) + B(\ux)A(\uy). \]
Rewrite the right-hand side as a sum of two Clebsch-Gordan series.
By property (\ref{trans.1}), only the
even powers of $\omega$ will survive. This shows the existence claim
for $\langle i j \rangle$, and hence in general by linearity.

Conversely, let $U_\bullet, U_\bullet'$ be two such sequences for $\U$. By
the substitution $\uy:=\ux$, we deduce $U_1 = U_1'$. Now
divide $\U - U_1$ by $\omega^2$ and again let $\uy:=\ux$ etc.,
then we successively see that $U_{2r+1} = U_{2r+1}'$ for all $r$. \qed

\smallskip

Henceforth, $A,B$ will always denote linearly independent
forms of order $d$. We will write
\begin{equation} T_i:= (A,B)_i, \quad
\Pi_{A,B}:= \text{Span}\{A,B\} \subseteq S_d.
\label{T_and_Pi} \end{equation}

\begin{Proposition} \sl
With notation as above,
\begin{equation}
\R(A,B) = \sum\limits_{r \ge 0} \, c_r \,\omega^{2r} \,
(T_{2r+1})^\sympol. \label{series.taylor} \end{equation}
\end{Proposition}
\demo Express $A(\ux)B(\uy)$ and $B(\ux)A(\uy)$ as Gordan
series and subtract. By property (\ref{trans.1}), only the odd powers
of $\omega$ will survive. Now divide by $\omega$, then they all
become even powers. \qed

\smallskip

It follows that the collection $\{ T_{2r+1}: r \ge 0\}$ determines
$\R(A,B)$. It will be shown below that the terms $r=0,1$ are already
sufficient.
\section{The Wronskian o.d.e.} \label{section.wr.ode}
\subsection{Generalities on Wronskians}
Given integers $p,q$ with $q \le p+1$, there is an
isomorphism of $SL(V)$-modules (see \cite[\S 11]{FH})
\begin{equation} \wedge^q S_p \stackrel{\sim}{\lra} S_q(S_{p-q+1}).
\label{extsym} \end{equation}
Composing it with the natural surjection
\begin{equation} S_q(S_{p-q+1}) \lra S_{q(p-q+1)},
\label{surj}\end{equation}
we get the Wronskian map
\[ \Theta: \wedge^q S_p \lra S_{q(p-q+1)}. \]
If $F_1, \dots, F_q$ are order $p$ forms, then
their Wronskian
$\Theta(F_1 \wedge \dots \wedge F_q)$ is given
by the $q \times q$ determinant
\[ (i,j) \lra
\frac{\partial^{\,q-1} F_i}{\partial x_0^{q-j} \, \partial x_1^{j-1}}
\quad 1 \le i,j \le q. \]
The crucial property of the construction is that
$\Theta$ is nonzero on decomposable tensors, i.e.,
$\Theta(F_1 \wedge \dots \wedge F_q)=0 \iff
F_1 \wedge \dots \wedge F_q=0 \iff $
the $F_i$ are linearly dependent.

\subsection{}
Now let $A,B,F$ be of order $d$, with Wronskian
\[ \W = \Theta(A \wedge B \wedge F) =
 \left| \begin{array}{ccc}
A_{x_0^2} & A_{x_0x_1} & A_{x_1^2} \\
B_{x_0^2} & B_{x_0x_1} & B_{x_1^2} \\
F_{x_0^2} & F_{x_0x_1} & F_{x_1^2}
\end{array} \right|. \]
We will evaluate $\W$ symbolically. Let us write
\begin{equation} A = \alpha_x^d, \; B = \beta_x^d, \; F = f_x^d.
\label{sym.exp} \end{equation}
As usual, $\alpha_x$ stands for the symbolic linear form
$\alpha_0 x_0 + \alpha_1 x_1$, and
$\br{\alpha}{\beta}$ for $\alpha_0 \beta_1 - \alpha_1 \beta_0$ etc.

\begin{Lemma} \sl With notation as above,
\begin{equation}
\frac{1}{(d^2-d)^3} \,
\W = \br{\alpha}{\beta}\br{\alpha}{f}\br{\beta}{f} \,
\alpha_x^{d-2} \, \beta_x^{\, d-2} \, f_x^{\, d-2}.
\label{sym.wronskian} \end{equation}
\end{Lemma}
\demo Differentiating (\ref{sym.exp}), we get expressions such as
\[ A_{x_0x_1} = d(d-1) \, \alpha_x^{d-2} \, \alpha_0 \, \alpha_1. \]
Substitute these into $\W$ and factor out
$\alpha_x^{d-2} \, \beta_x^{d-2} \, f_x^{d-2}$. We are left
with a Vandermonde determinant which evaluates to
$\br{\alpha}{\beta}\br{\alpha}{f}\br{\beta}{f}$. \qed

Now we will rewrite $\W$ in terms of transvectants.
\begin{Proposition} \sl
With notation as in (\ref{T_and_Pi}), we have an identity
\[ \frac{1}{(d^2-d)^3} \, \W = (T_1,F)_2 - \frac{d-2}{4d-6} \, T_3 \, F. \]
\label{prop.w} \end{Proposition}
\demo
Symbolically, the transvectants can be written as
\[
T_1 = \br{\alpha}{\beta} \, \alpha_x^{\,d-1} \beta_x^{\,d-1}, \quad
T_3 = \br{\alpha}{\beta}^3 \, \alpha_x^{\,d-3} \beta_x^{\,d-3}. \]
First we calculate the transvectant $(T_1,F)_2$ using the
algorithm given in (see \cite[\S 49]{GrYo}):
\begin{itemize}
\item
Calculate the second polar $T_1$. It is equal to
\[ \begin{aligned}
{} & \frac{(2d-4)!}{(2d-2)!} \, (\polar)^2 \, T_1  =
\frac{1}{(2d-2)(2d-3)} \br{\alpha}{\beta} \, \alpha_x^{\,d-3} \beta_x^{\,d-3}
\times \\
& \{(d-1)(d-2) \, \alpha_x^2 \, \beta_y^2 +
 2 (d-1)^2 \, \alpha_x \, \beta_x \, \alpha_y \, \beta_y +
 (d-1)(d-2) \, \beta_x^2 \, \alpha_y^2 \}.
\end{aligned} \]
\item
Make substitutions $\alpha_y:=\br{\alpha}{f},\beta_y:=\br{\beta}{f}$,
and multiply by $f_x^{\,d-2}$. The result is
\begin{equation} \begin{aligned}
(T_1,F)_2 = &
\frac{d-2}{4d-6} \br{\alpha}{\beta} \,
\alpha_x^{\,d-3} \beta_x^{\,d-3} f_x^{\,d-2} \times  \\
& \{ \br{\beta}{f}^2 \, \alpha_x^2 + \frac{2d-2}{d-2}
\br{\alpha}{f} \br{\beta}{f} \,
\alpha_x \beta_x + \br{\alpha}{f}^2 \, \beta_x^2 \}.
\end{aligned} \label{t1f2} \end{equation}
\end{itemize}

We would like to compare (\ref{sym.wronskian}) and (\ref{t1f2}),
so we will rewrite both of them in terms of
{\sl standard monomials} (see \cite[Ch.~3]{Sturmfels}). Order the variables as
$\alpha < \beta < f < x$.
The monomial $\br{\beta}{f} \, \alpha_x$ is nonstandard, so
use the Pl{\"u}cker syzygy to rewrite it as
\[ \br{\beta}{f} \, \alpha_x =
\br{\alpha}{f} \, \beta_x - \br{\alpha}{\beta} f_x. \]
Substitute this into the right hand sides of (\ref{sym.wronskian}) and
(\ref{t1f2}). Subtracting the two expressions, we get
\[ (T_1,F)_2 - \frac{1}{(d^2-d)^3} \W =
\frac{d-2}{4d-6} \, \br{\alpha}{\beta}^3 \, \alpha_x^{d-3}\beta_x^{d-3}f_x^d
= \frac{d-2}{4d-6} \, T_3 \, F. \]
This completes the proof. \qed

\subsection{}
If $M,N$ are forms of orders $2d-2,2d-6$ respectively, then
we define
\begin{equation}
\psi_{M,N}(F) := (M,F)_2 - \frac{d-2}{4d-6} \, N \, F.
\label{psi} \end{equation}
For fixed $M,N$, we are interested in the differential equation
\begin{equation} \psi_{M,N}(F) = 0. \label{diffeq} \end{equation}
We may call this the Wronskian (second order) ordinary differential
equation with parameters $M,N$. (It is always assumed that $M \neq 0$,
otherwise the equation is of no interest.)
The following corollary is immediate.
\begin{Corollary} \sl If $F$ is of order $d$, then
$F \in \Pi_{A,B}$ iff $\psi_{T_1,T_3}(F) = 0$.
\label{cor.T1T3} \end{Corollary}
\demo Indeed, $\psi_{T_1,T_3}(F)=0$ iff $A,B,F$ are linearly
dependent. \qed

\smallskip

Hence, given $T_1,T_3$, the pair $\{A,B\}$ is determined up to a
unimodular transformation (cf.~(\ref{comb})). It follows that $T_1,T_3$
together determine all the $T_{2r+1}$.

\begin{Proposition} \sl
Let $M,N$ be of orders $2d-2,2d-6$. Assume that (\ref{diffeq}) has
two linearly independent solutions $A,B$ of order $d$. Then there
exists a nonzero constant $\lambda$ such that
$M = \lambda \, T_1, \; N = \lambda \, T_3$.
\label{prop.pencil} \end{Proposition}
\demo
Multiply the identities $\psi_{M,N}(A)=0,\psi_{M,N}(B)=0$
by $B,A$ respectively and subtract, this gives
$ B\, (M,A)_2 =  A \, (M,B)_2$. Now the Gordan series
\[ \Gordan{A}{M}{B}{d}{2d-2}{d}{0}{0}{2}, \quad
   \Gordan{B}{M}{A}{d}{2d-2}{d}{0}{0}{2} \]
respectively give identities
\[ \begin{aligned}
(A,M)_2 \, B & = (AB,M)_2 + ((A,B)_1,M)_1 + \frac{d}{4d-2} (A,B)_2 \,M \\
(B,M)_2 \, A & = (BA,M)_2 + ((B,A)_1,M)_1 + \frac{d}{4d-2} (B,A)_2 \,M.
\end{aligned} \]
Subtracting and using property (\ref{trans.1}) for $A,B$, we get
$((A,B)_1,M)_1 =0$. Now $(A,B)_1 \neq 0$ since $A,B$ are
independent, but then Lemma \ref{lemma.jac} implies
that $M = \lambda \, (A,B)_1$ for some $\lambda$. Finally
\[ \frac{d-2}{4d-6} \, N A = (M,A)_2 = \lambda \, (T_1,A)_2 =
\lambda \, \frac{d-2}{4d-6} \, T_3 \, A, \]
hence $N = \lambda \, T_3$. \qed

\subsection{}
We have shown that the following conditions are equivalent for
the pair $(M,N)$.
\begin{itemize}
\item[(i)]
There exist $A,B$ such that
$M = (A,B)_1, N = (A,B)_3$.
\item[(ii)]
There exist $A,B$ such that
\[ \R(A,B) = c_0 \, M^\sympol + c_1 \, \omega^2 \, N^\sympol + O(\omega^4). \]
\item[(iii)]
The dimension of the kernel of the map
\[ \psi_{M,N}: S_d \lra S_{3d-6} \]
is at least $2$ (and then it is exactly $2$).
\end{itemize}
We can now construct the covariants $\Phi_r$ as in the
introduction. Clearly (iii) is equivalent to the condition that the map
\[ \wedge^d \psi_{M,N}:
\wedge^d S_d \lra \wedge^d (S_{3d-6}) \]
be zero. Identify $\wedge^d S_d$ with $S_d$ via (\ref{extsym}).
Let $f_1$ denote the image of $\wedge^d \psi_{M,N}$ via the
isomorphism
\[ \Hom_{SL(V)}(S_d, \wedge^d S_{3d-6}) \simeq
\Hom_{SL(V)}(\complex, \wedge^d S_{3d-6} \otimes S_d).\]
Consider the composite morphism
\[ \complex \stackrel{f_1}{\lra}
\wedge^d S_{3d-6} \otimes S_d \stackrel{f_2}{\lra}
S_d (S_{2d-5}) \otimes S_d \stackrel{f_3}{\lra}
S_{d(2d-5)} \otimes S_d, \]
where $f_2$ comes from the isomorphism (\ref{extsym}), and
$f_3$ from the natural surjection (\ref{surj}). For each
$0 \le r \le d$, we have projection maps
\[ \pi_r: S_{d(2d-5)} \otimes S_d \lra S_{d(2d-4)-2r} \]
induced by the decomposition (\ref{clebschgordan}).

Define $\Phi_r(M,N)$ to be the image of $1 \in \complex$
via the map $\pi_r \circ f_3 \circ f_2 \circ f_1$. This is
a joint covariant of $M,N$ of order $d(2d-4)-2r$.
We will describe it in coordinates.
For $0 \le i \le d$, define
\[ w_i = (-1)^i \frac{1}{i!(d-i)!} \,
\Theta( \, \bigwedge\limits_{\stackrel{s=0}{s \neq i}}^d
\psi_{M,N}(x_0^s \, x_1^{d-s})), \]
which is an element of $S_{d(2d-5)}$. Then
\[ (f_3 \circ f_2 \circ f_1) (1) =
\sum\limits_{i=0}^d w_i \otimes x_0^i \, x_1^{d-i}, \quad
\text{and} \quad
\Phi_r = \sum\limits_{i=0}^d \, (w_i, x_0^i \, x_1^{d-i})_r. \]
All of this is straightforward and follows by chasing through the
$f_i$. Each $\Phi_r$ has total degree $d$ in the coefficients
of $M,N$ (because $w_i$ does).

\begin{Theorem} \sl
Let $M,N$ be orders $2d-2,2d-6$ respectively.
Then the pair $(M,N)$ satisfies the (equivalent) conditions
(i)--(iii) if and only if
\[ \Phi_r(M,N) =0 \quad \text{for $0 \le r \le d$.} \]
\end{Theorem}
\demo If (iii) holds, then $f_1=0$, which  shows the
`only if' part. Conversely, assume that all
the $\Phi_r$ vanish. Then $(f_3 \circ f_2 \circ f_1)(1) =0$, which
implies that all the $w_i$ vanish. By the fundamental property
of Wronskians, the forms
\[ \psi_{M,N}(x_0^s \, x_1^{d-s}), \quad 0 \le s \le d, \, s \neq i \]
are linearly dependent for any $i$. But then the map
$\wedge^d \psi_{M,N}$ is zero on every basis element of $\wedge^d S_d$,
hence it is zero. This implies (iii).
\qed

\subsection{The incomplete Pl{\"u}cker imbedding} \label{incomplete.plucker}
The fact that $\R$ is determined by $T_1,T_3$ has the following geometric
interpretation. Assume $d \ge 3$, and let
$\G = G(2,S_d)$ denote the Grassmannian of two-dimensional
subspaces in $S_d$. (See \cite[Lecture 6]{JoeH} for generalities on
Grassmannians.) The line bundle $\O_{\G}(1)$ has
global sections
\[ H^0(\G,\O_\G(1)) \simeq \wedge^2 S_d \simeq S_2(S_{d-1})
\simeq \bigoplus\limits_{r \ge 0} S_{2d-4r-2}. \]
The usual Pl{\"u}cker imbedding is given by the complete
linear system $|\O_\G(1)|$.
Consider the subspace $W = S_{2d-2} \oplus S_{2d-6}
\subseteq H^0(\O_\G(1))$.

\begin{Proposition} \sl
The map
\[ \mu : \G \lra \P \, W, \quad
\P \, \Pi_{A,B} \lra [ \, T_1 \oplus T_3 \, ] \]
is an isomorphic imbedding.
\end{Proposition}
The usual conventions (\cite[Ch.~II, \S 7]{Ha}) dictate that the
imbedding is in $\P W^*$, but note the self-duality in \S \ref{self-dual}.

\smallskip

\demo We have already shown that $\mu$ is a set-theoretic injection.
To complete the proof, it suffices to show that it is an injection
on tangent spaces at every point (cf.~\cite[Ch.~II, Prop.~7.3]{Ha}).
The Zariski tangent space to $\G$ at $\Pi = \Pi_{A,B}$ is
canonically isomorphic to $\Hom(\Pi,S_d/\Pi)$
(see \cite[Lecture 16]{JoeH}).
Let $\alpha: \Pi \lra S_d/\Pi$ be a tangent vector, and say
\[ \alpha(A) = Q + \Pi, \quad \alpha(B) = P + \Pi, \]
for some forms $P,Q$ of order $d$.

The tangent space to $\P W $ at $[T_1 \oplus T_3]$ is
isomorphic to $W/\langle T_1 \oplus T_3 \rangle$.
Let $d \mu: T_{\G,\Pi} \lra T_{\P,\mu(\Pi)}$ denote the induced map
on tangent spaces. Then $d\mu(\alpha)$ is the element
\[ ((A,P)_1+(Q,B)_1) \oplus ((A,P)_3+(Q,B)_3) \in W \]
considered modulo $T_1 \oplus T_3$. (To see this,
let $\epsilon$ be an `infinitesimal'. Now
expand $(A + \epsilon \, Q, B + \epsilon \, P)_i, i=1,3$, and
set $\epsilon^2 = 0$.)

We would like to show that $d\mu$ is injective, hence
suppose that $d\mu(\alpha)=0$. Then there exits a constant $c$ such that
\[ \begin{aligned}
(A,P)_1 + (Q,B)_1 & = c \, (A,B)_1, \\
(A,P)_3 + (Q,B)_3 & = c \, (A,B)_3. \end{aligned} \]
Substitute $P + c \, B$ for $P$ (which does not change $\alpha$), then
\[ (A,P)_1 = (B,Q)_1, \quad (A,P)_3 = (B,Q)_3. \]
If the first pair is zero, then $P,Q$ are respectively
constant multiples of $A,B$, hence $\alpha=0$. If not, then
$\Pi_{A,P} = \Pi_{Q,B}$ by Corollary \ref{cor.T1T3}. But this implies
$\Pi_{A,B} = \Pi_{P,Q}$, again forcing $\alpha=0$. \qed

\begin{Remark} \rm
Let ${\mathfrak a} \subseteq \text{Sym}^\bullet \, W$ denote the ideal
generated by the coefficients of $\Phi_0,\dots,\Phi_d$, and
$J$ the homogeneous ideal of the image $\mu(\G) \subseteq \P W$.
Since ${\mathfrak a}$ defines the image set-theoretically,
$(\sqrt{\mathfrak a})_{\text{sat}} = J$. Already for $d=3$
these ideals differ (since ${\mathfrak a}$ is generated in
degree $3$ and $J$ in degree $2$), but I do not know if one can state
a more precise relation in general.
\end{Remark}

\section{Generic reduction formulae} \label{section.gen.red}
\subsection{}
We begin with the example which eventually led to
the main result of this section. Let $A,B$ be of order $2$.
The series $\Gordan{A}{B}{A}{2}{2}{2}{0}{1}{1}$ implies the relation
\[ ((A,B)_1,A)_1 + \frac{1}{2} \, (A,B)_2 \, A =
\frac{1}{2} \, (A,A)_2 \, B; \]
which can be rewritten as
\[ -\frac{2}{(A,A)_2} (A,T_1)_1 = B - \frac{(A,B)_2}{(A,A)_2} \, A.
\]
Hence, given $\R$ (which involves only $T_1$ in this case) and $A$,
the function
\begin{equation} (A,T_1) \lra  -\frac{2}{(A,A)_2} (A,T_1)_1
\label{red1} \end{equation}
recovers $B$ up to an additive multiple of $A$.
(Since $\R(A,B + \cons A) = \R(A,B)$, the last proviso is
indispensable.)
We will show that there exist such formulae for every $d$.
\begin{Remark} \rm
We may call (\ref{red1}) a reduction formula in the following
sense. If we are given a linear second order
o.d.e., together with one of its solutions, then a second solution
can be found by the method of `reduction of order'
(see \cite[\S 44]{Rainville}). In our case, we are to find $B$,
given the equation $\psi_{T_1,T_3}(F)=0$ with one solution $A$.
However, this analogy is inexact in two respects:
\begin{itemize}
\item
our formula will involve all the $\{T_{2r+1}\}$, and
not merely $T_1,T_3$,
\item
the process is algebraic and involves no integration.
\end{itemize}
Moreover, the formula is generic in the sense that it is only defined over
an open subset, e.g., the set $\{A \in \P^2: (A,A)_2 \neq 0\}$ above.
\end{Remark}

\subsection{}
Throughout this section we assume that $A,B$ are order $d$ forms
whose coefficients are algebraically independent indeterminates.
Write
\begin{equation}
A = \sum\limits_{p=0}^d \binom{d}{p} \, a_p \, x_0^{d-p} \, x_1^p.
\label{exp.A} \end{equation}
Let $\J$ be an invariant of $A$ of degree (say) $n$. We define its
first evectant (cf.~\cite{Sylvester3}) to be
\begin{equation}
\Ev_\J = \frac{(-1)^d}{n} \sum\limits_{q=0}^d \,
(-1)^q \, \frac{\partial \J}{\partial a_q} \, x_0^q \, x_1^{d-q},
\label{Ev} \end{equation}
it is a covariant of degree-order $(n-1,d)$. The initial scaling
factor is chosen so as to make the following lemma true:
\begin{Lemma} \sl
We have an identity $(\Ev_\J,A)_d  = \J$.
\label{EV} \end{Lemma}
\demo
Substitute (\ref{exp.A}) and (\ref{Ev}) in formula (\ref{trans.defn}).
We get a nonzero term whenever $p=q$ and $i=d-p$, hence
\[ \begin{aligned} (\Ev_\J,A)_d  & =
\frac{(-1)^{2d}}{n \, (d \, !)^2} \, \sum\limits_{p=0}^d \,
(p!(d-p)!)^2 \binom{d}{d-p}^2 \, a_p \, \frac{\partial \J}{\partial a_p} \\
& = \frac{1}{n} \sum\limits_p  \, a_p \, \frac{\partial \J}{\partial a_p}
= \J,  \end{aligned} \]
the last equality is by Euler's theorem. \qed

\smallskip

Now our generic reduction formula is as follows. Let
\begin{equation}
\beta(A,\R) = - \frac{1}{\J}
\sum\limits_{r \ge 0} \, c_r \, (\Ev_\J,T_{2r+1})_{d-2r-1},
\end{equation}
with the $c_r$ as in (\ref{cr}).
\begin{Theorem} \sl
With notation as above,
\[ \beta(A,\R) = B - \frac{(\Ev_\J,B)_d}{\J} \, A. \]
\label{Theorem.red.formula} \end{Theorem}
Hence, as long as $A$ stays away from the hypersurface $\{ \J =0\}$,
we can recover $B$ from $A$ and $\R(A,B)$.

\begin{Remark} \rm
If $d$ is even, then we can take $\J$ to be the unique
degree two invariant $(A,A)_2$.
There is no invariant in degrees $\le 3$ if $d$ is odd, but then
there exists a degree four invariant $\J = ((A,A)_{d-1},(A,A)_{d-1})_2$.
\end{Remark}

\subsection{}
The proof of the theorem will emerge from the discussion below.
The element $A\wedge B \in \wedge^2 S_d$ defines a map
\[ \sigma_{A \wedge B}: S_d \lra S_d, \quad
F \lra (F,B)_d \, A - (F,A)_d \, B. \]
We identify the codomain of $\sigma=\sigma_{A \wedge B}$ with
$S_d^*$ as in \S \ref{self-dual}.

\begin{Lemma} \sl With the convention above, $\sigma$
is skew-symmetric, i.e.,
\[
\delta_{\sigma(F)}(G) = - \delta_{\sigma(G)}(F),
\quad \text{for $F,G \in S_d$.} \]
\end{Lemma}
\demo
Unwinding the definitions, this becomes
\[ (F,B)_d (A,G)_d - (F,A)_d (B,G)_d =
- \{(G,B)_d (A,F)_d - (G,A)_d(B,F)_d \}, \]
which is obvious. \qed

\begin{Lemma} \sl
With notation as above,
\begin{equation}
\sigma(F) = [ \, (F,\R)_{d-1}^\ux \, ]_{\uy:=\ux}.
\label{sigma.F.R} \end{equation}
\end{Lemma}
The right hand side of this equation is interpreted as follows:
calculate the $(d-1)$-th transvectant of $F$ and $\R$ as
$\ux$-forms (treating the $\uy$ in $\R$ as constants). This produces
an $\ux\uy$-form of order $(1,d-1)$; finally replacing $\uy$ by
$\ux$ gives a form of order $d$.

\smallskip

\demo
We will calculate both sides symbolically. Let
$A = \alpha_x^d, \; B = \beta_x^d, \; F = f_x^d$, then
\[ \begin{aligned}
\R & = \frac{A(\ux)B(\uy) - A(\uy)B(\ux)}{\omega}
= \frac{\alpha_x^d \, \beta_y^d - \alpha_y^d \, \beta_x^d}{\omega} \\
& = \frac{
(\alpha_x \, \beta_y - \alpha_y \, \beta_x)
\sum\limits_{i=0}^{d-1} \,
(\alpha_x \, \beta_y)^{d-1-i}(\alpha_y \, \beta_x)^i}
{\omega} \\
& = \br{\alpha}{\beta} \,
\sum\limits_i (\alpha_x \, \beta_y)^{d-1-i}(\alpha_y \, \beta_x)^i.
\end{aligned} \]

Now calculate the $(d-1)$-th transvectant of $F$
with each summand in the last expression. (We have agreed to
treat $\alpha_y,\beta_y$ as constants). Using the
algorithm of \cite[\S 49]{GrYo},
\[ (f_x^{\,d},\alpha_x^{d-1-i} \, \beta_x^i)_{d-1} =
(-1)^{d-1} \br{\alpha}{f}^{d-1-i} \, \br{\beta}{f}^i \, f_x. \]
Hence,
\begin{equation}
[\, (F,\R)_{d-1}^\ux \, ]_{\uy:=\ux} =
(-1)^{d-1} \br{\alpha}{\beta} \, f_x \sum\limits_i \,
\br{\alpha}{f}^{d-1-i} \, \br{\beta}{f}^i \,
\alpha_x^i \, \beta_x^{d-1-i}.
\label{FR} \end{equation}
Now directly from the definition,
\[ \begin{aligned} \sigma(F) & =
\{ \br{f}{\beta}^d \, \alpha_x^d - \br{f}{\alpha}^d \, \beta_x^d \} \\
& = (-1)^d
\{ \br{\beta}{f}^d \, \alpha_x^d - \br{\alpha}{f}^d \, \beta_x^d \} \\
& = (-1)^d \{\br{\beta}{f} \, \alpha_x - \br{\alpha}{f} \, \beta_x \}
\sum\limits_i \, \br{\alpha}{f}^{d-1-i} \, \br{\beta}{f}^i \,
\alpha_x^i \, \beta_x^{d-1-i}.
\end{aligned} \]
Since $\br{\beta}{f}\alpha_x - \br{\alpha}{f}\beta_x
= -\br{\alpha}{\beta}f_x$, the last
expression is identical to (\ref{FR}). \qed

\smallskip

\begin{Lemma} \sl
Let ${\mathbf T}$ be an arbitrary form of order $2d-4r-2$. Then
\[ [\, (F,\omega^{2r} \, T^\sympol)_{d-1}^\ux \,]_{\uy:=\ux}
= (F,{\mathbf T})_{d-2r-1}. \]
\end{Lemma}
\demo Let ${\mathbf T} = t_x^{\, 2d-4r-2}$, so that
$T^\sympol = t_x^{\, d-2r-1} \, t_y^{\, d-2r-1}$. Then make a
calculation as in the previous lemma. \qed

\smallskip

Now substitute the Taylor series (\ref{series.taylor}) into the
right hand side of (\ref{sigma.F.R}), and use the previous lemma.
This gives the formula
\begin{equation}
\sigma(F) = \sum\limits_{r \ge 0} \, c_r \, (F,T_{2r+1})_{d-2r-1}.
\label{sigma_and_T} \end{equation}
Now specialize to $F = \Ev_\J$. Then
\[
\sigma(\Ev_\J) = (\Ev_\J,B)_d \, A - (\Ev_\J,A)_d \, B =
(\Ev_\J,B)_d \, A - \J \, B, \]
hence
\[ \beta(A,\R) =
-\frac{1}{\J} \, \sigma(\Ev_\J) = B - \frac{(\Ev_\J,B)_d}{\J} \, A. \]
This completes the proof of Theorem \ref{Theorem.red.formula}. \qed

\section{Formulae for $T_5$ and $T_7$} \label{section.t5t7}
We have observed that $T_1,T_3$ determine
the higher odd transvectants $T_{2r+1}$. However this dependence is
rather indirect, and it is unclear if one can give a formula for the latter
in terms of the former. In this section we give such
explicit formulae for $T_5$ and $T_7$.
\subsection{The Pl{\"u}cker relations}
Let
\[ \G \subseteq \P (\wedge^2 S_d) =
\P ( \, \bigoplus\limits_{r \ge 0} \, S_{2d-4r-2})
\]
be the usual Pl{\"u}cker imbedding, and let $\I$ denote the
homogeneous ideal of the image. It is well-known that
$\I$ is generated by its quadratic part $\I_2$,
usually called the module of Pl{\"u}cker relations.
\begin{Lemma} \sl
As $SL(V)$-modules, $\I_2 \simeq \wedge^4 S_d$.
\end{Lemma}
\demo
Consider the short exact sequence
\[
0 \ra \I_2 \ra H^0(\O_{\P(\wedge^2 S_d)}(2)) \ra
H^0(\O_{\G}(2)) \ra 0. \]
(The exactness on the right comes from the projective normality of
the imbedding.) Using the plethysm formula of
\cite[\S I.8, Example 9]{MacDonald}, the middle term is
isomorphic to
\[ S_2(\wedge^2 S_d) \simeq S_{(2,2)} (S_d) \oplus \wedge^4  S_d. \]
By the Borel-Weil theorem (see \cite[p.~687]{Porras}),
$H^0(\O_\G(2)) \simeq S_{(2,2)}(S_d)$. This completes the
proof. \qed

\smallskip

Each Pl{\"u}cker relation corresponds to an algebraic identity
between the $\{T_{2r+1}\}$. To make this more
precise, let $\{M_{2r+1}:r \ge 0\}$ be generic forms of orders $2d-4r-2$, and
$S_e \stackrel{\xi}{\hookrightarrow} \I_2$ an inclusion of
$SL(V)$-modules. Then $\xi$ corresponds to a joint covariant
$\Xi \, (M_1,M_3,\dots)$ of order $e$ and total degree two in the
$\{M_{2r+1} \}$, such that
\[ \Xi \, (T_1,T_3,\dots) = 0, \qquad \text{for any $A,B$ of order $d$.}
\]

\begin{Example} \rm
Assume $d=4$. In this case $\I_2 \simeq S_4$, so we look for
an order $4$ covariant in $M_1,M_3$. There are
three `monomials' of total degree $2$ and order $4$, namely
$(M_1,M_1)_4,(M_1,M_3)_2,M_3^2$. Our covariant must be a linear
combination of these, i.e.,
\[ \Xi(M_1,M_3) = \alpha_1 \, (M_1,M_1)_4 +
\alpha_2 \, (M_1,M_3)_2 + \alpha_3 \, M_3^2, \]
for some constants $\alpha_i$.

Now specialize to $A = x_0^4,B = x_1^4$, and use formula
(\ref{trans.defn}) to calculate $T_1,T_3$ and $\Xi$ explicitly.
Since $\Xi(T_1,T_3)$ must vanish identically, its coefficients
give $5$ linear equations for the $\alpha_i$.
Solving these (they must admit a nontrivial solution), we deduce that
\[ [\alpha_1:\alpha_2:\alpha_3] = [25:-10:-4], \]
which determines $\Xi$ (of course, up to a scalar). This
`method of undetermined coefficients' (specializing the
forms followed by solving linear equations) will be liberally
used in the sequel.
\end{Example}
\begin{Example} \rm
For $d=3$, the Grassmannian is a quadric hypersurface
defined by
\[ \Xi(M_1,M_3) = (M_1,M_1)_4 - \frac{1}{6} \, M_3^2. \]
\end{Example}

\subsection{} We begin with a technical
lemma about the irreducible submodules of $\I_2$.
\begin{Lemma} \sl If $d \ge 4$, then there exists exactly one copy
each of the modules $S_{4d-12},S_{4d-16}$ inside $\I_2$.
\end{Lemma}
\demo There are isomorphisms
\[ \I_2 \simeq \wedge^4 S_d \simeq S_4(S_{d-3}) \simeq S_{d-3}(S_4), \]
where the second isomorphism is from (\ref{extsym}), and the
third is Hermite reciprocity. Hence we may as well work with
$S_{d-3}(S_4)$. Now the following are in bijective correspondence
(see \cite{Littlewood1} for details):
\begin{itemize}
\item inclusions $S_e \subseteq S_{d-3}(S_4)$ of $SL(V)$-modules,
\item
covariants of degree-order $(d-3,e)$ (distinguished up to scalars)
for binary quartics,
\end{itemize}
Fortunately, a complete set of generators for the covariants of
binary quartics is known (see \cite[\S 89]{GrYo}).
It contains five elements, conventionally called $f,H,t,i,j$, having
degree-orders
\[ (1,4), \; (2,4), \; (3,6), \; (2,0), \; (4,0). \]
(It is unnecessary for us to know how they are defined.)
Each covariant of quartics is a polynomial in the elements of this set.

Now it is elementary to see that only one expression of
degree-order $(d-3,4d-12)$ is possible, namely $f^{d-3}$.
Similarly, the only possible expression for degree-order
$(d-3,4d-16)$ is $f^{d-5} \, H$. Hence there is exactly
one copy each of $S_{4d-12}$ and $S_{4d-16}$. \qed

\subsection{}
We will find the joint covariant $\Xi$ corresponding to
$S_{4d-12} \subseteq \I_2$. We look for degree two monomials
of order $4d-12$ in the $\{T_{2r+1}\}$; any such monomial must be
of the form
\[(T_{2a+1},T_{2b+1})_s, \]
where
\begin{itemize}
\item
$(2d-4a-2)+(2d-4b-2) -2s = 4d-12$,
\item
$a,b \le \lfloor \frac{d-1}{2} \rfloor$,
\item
$s \le \min \{2d-4a-2,2d-4b-2 \}$, and
\item
if $a=b$, then $s$ is even.
\end{itemize}
The first condition comes from the order, the rest are forced by
properties (\ref{trans.1}),(\ref{trans.2}) of transvectants.
Sifting through these conditions gives only four possibilities, namely
\[ (T_1,T_1)_4, \; (T_1,T_3)_2, \;
   T_3^2, \; T_1T_5. \]
Hence we have an identical relation of the form
\[ \alpha_1 \, (T_1,T_1)_4 + \alpha_2 (T_1,T_3)_2 +
   \alpha_3 \, T_3^2 - \alpha_4 \, T_1 \, T_5 =0. \]
Specialize $A,B$ successively to the pairs
\[ (x_0^d,x_1^d), \; (x_0^{d-1}x_1,x_1^d), \;
(x_0^{d-2}x_1^2,x_1^d), \; (x_0^{d-1}x_1,x_0 \, x_1^{d-1}), \; \]
and use the method of undetermined coefficients.
Up to a scalar, the solution is
\[ \begin{array}{ll}
\alpha_1 = -\frac{2(2d-3)^2}{d(d-2)} &
\alpha_2 = \frac{4(2d-3)(d-3)}{d(d-2)} \\
\alpha_3 = 1 &
\alpha_4 = \frac{(d-3)(d-4)(2d-3)^2}{d(2d-5)(2d-7)(d-2)}.
\end{array} \]
This gives a formula for $T_5$.
\begin{Theorem} \sl Assume $d \ge 5$, then
\[
T_5 = \frac{1}{T_1} \,
( \, \frac{\alpha_1}{\alpha_4} \, (T_1,T_1)_4
 +\frac{\alpha_2}{\alpha_4} \, (T_1,T_3)_2
 +\frac{\alpha_3}{\alpha_4} \, T_3^2).
\]
\end{Theorem}
We can make a similar argument with $S_{4d-16}$, this leads to
a formula for $T_7$. Define
\[ \begin{array}{ll}
\beta_1 = - \frac{8(2d-5)(2d-7)(2d-3)}{(d-1)(4d-13)} &
\beta_2 = - \frac{60(2d-7)(2d-5)}{d(d-1)(4d-13)} \\
\beta_3 = \frac{12(2d-3)(d-5)}{d(4d-13)} &
\beta_4 = \frac{20(2d-5)(2d-7)(d-3)}{(d-1)(4d-13)(2d-3)} \\
\beta_5 = 1  &
\beta_6 = \frac{(d-5)(d-6)(2d-3)(2d-5)}{d(d-1)(2d-9)(2d-11)}
\end{array} \]

\begin{Theorem} \sl Assume $d \ge 7$, then
\[
\begin{aligned}
T_7 = \frac{1}{T_1}
(& \, \frac{\beta_1}{\beta_6} \, (T_1,T_1)_6 +
 \frac{\beta_2}{\beta_6} \, (T_1,T_3)_4 +
 \frac{\beta_3}{\beta_6} \, (T_1,T_5)_2 +  \\
 & \frac{\beta_4}{\beta_6} \, (T_3,T_3)_2 +
 \frac{\beta_5}{\beta_6} \, T_3 \, T_5).
\end{aligned} \]
\end{Theorem}
(Of course we can substitute for $T_5$ using the previous
result, but this would make the formula very untidy.)

This method breaks down for higher transvectants, so a new idea
will be needed for the general case.
My colleage A.~Abdesselam, when shown the formulae above,
remarked that the coefficients look very similar to those appearing in the
classical hypergeometric series. Perhaps there is something
to this suggestion.

\section{Open problems} \label{section.open}
This section contains a series of miscellaneous calculations
and examples, all of them for small specific values of $d$.
They should serve simultaneously as a source of
open questions and further lines of inquiry.
\subsection{The Jacobian predicate}
Let $A,M$ be forms of orders $d,2d-2$. Consider the
following predicate
\[ J(A,M): \text{there exists an order $d$ form $B$ such that
$(A,B)_1 = M$.} \]
If $J(A,M)$ holds, then
$(A,M)_2 = \cons \, T_3 \, A$, hence $A$ must divide $(A,M)_2$.
We will see below that this condition is sufficient for $d=2,3$,
but not for $d=4$.

\begin{Proposition} \sl Assume $d=2$. Then
\[ J(A,M) \iff (A,M)_2=0. \]
\end{Proposition}
\demo
The forward implication is clear.
For the converse, assume $(A,M)_2=0$. Then
$\Gordan{A}{M}{A}{2}{2}{2}{0}{1}{1}$ implies that
$((A,M)_1,A)_1 = -\frac{1}{2} \, (A,A)_2 \, M$.
If $(A,A)_2 \neq 0$, then let
\[ B = \frac{2}{(A,A)_2} \, (A,M)_1. \]

If $(A,A)_2 =0$, then by a change of variable,
we may assume $A = x_0^2$. Then $(A,M)_2 = 0$ implies
that $M = c_1 \, x_0^2 + c_2 \, x_0 x_1$. Now
let $B = c_1 \, x_0 x_1 + \frac{c_2}{2} \, x_1^2$.
In either case, $(A,B)_1=M$. \qed

\begin{Proposition} \sl
Assume $d=3$, then
\[ J(A,M) \iff ((A,M)_2,A)_1=0. \]
\end{Proposition}
\demo By Lemma \ref{lemma.jac}, $((A,M)_2,A)_1=0$ iff $(A,M)_2 = \cons \, A$.
This shows the forward implication.

Conversely, assume that $(A,M)_2 = c \, A$ for some
constant $c$. I claim that the map
\[ \psi_{M,6 \, c}: S_3 \lra S_3, \quad
F \lra (M,F)_2 - c \, F \]
is skew-symmetric. Indeed,
\[ \delta_{\psi_{M,6 \, c}(F)}(G)
= ((M,F)_2,G)_3 - c \, (F,G)_3. \]
Using $\Gordan{M}{F}{G}{4}{3}{3}{1}{2}{2}$, this can be
transformed into
\[ -((M,G)_2,F)_3 + c \, (G,F)_3 = - \delta_{\psi_{M,6 \, c}(G)}(F).
\]
This proves the claim, and implies that the rank of $\psi_{M,6 \, c}$
must be even. Suppose that $A$ and another form $B$ span its
kernel. Then by Proposition \ref{prop.pencil}, $(A,B)_1 = M$
(after multiplying $B$ by a constant if necessary). \qed

\begin{Example} \rm
Assume $d=4$, and let $A = (x_0x_1)^2, M = (x_0x_1)^3$. Then
$A$ divides $(A,M)_2 = \cons \,(x_0x_1)^3$. However there exists no
$B$ such that $(A,B)_1 = M$. Indeed,
\[ (A,B)_1 = \cons  x_0 \, x_1(x_1 B_{x_1} - x_0 B_{x_0}) =
(x_0 \, x_1)^3 \]
would imply $x_1 B_{x_1} - x_0 B_{x_0} = \cons  (x_0x_1)^2$.
But then $B = \cons (x_0x_1)^2$, which is absurd.
\end{Example}

The two propositions above suggest the following natural problem:
\begin{Problem} \rm
Find a (finite) number of joint covariants of $A,M$
which simultaneously vanish iff $J(A,M)$ holds.
\end{Problem}

\subsection{The resultant}
Let $\Res(A,B)$ denote the resultant of $A,B$. Up to a scalar, it is
equal to the discriminant of $\R(A,B)$ (regarded as a quadratic form).
Since the latter implicitly depends only on $T_1,T_3$, the
following problem is natural:
\begin{Problem} \rm
Give an explicit formula (in a reasonable sense)
for $\Res(A,B)$ as a joint invariant of $T_1$ and $T_3$.
\end{Problem}
For instance, if $d=2$ then $\cons \Res(A,B) = (T_1,T_1)_2$.
\begin{Proposition} \sl If $d=3$, then
\[ \cons \Res(A,B) = T_3 \, (T_1,T_1)_4 - 6 \, (T_1,(T_1,T_1)_2)_4. \]
\end{Proposition}
\demo
By construction, $\Res = \Res(A,B)$ is joint invariant of total degree $3$
in $T_1,T_3$. Every joint invariant is a linear combination of
compound transvectants (see \cite[p.~92]{Glenn}), hence $\Res$ is a
linear combination of terms of the form
\[ (X_1,(X_2,X_3)_a)_b, \]
where $a,b$ are integers, and each $X_i$ stands for
either $T_1$ or $T_3$. Since the total order must be zero,
$\sum\limits_i \ord X_i = 2(a+b)$. Using
properties (\ref{trans.1}),(\ref{trans.2}), we are left with only
two possibilities, namely
\[ (T_3,(T_1,T_1)_4)_0, \quad (T_1,(T_1,T_1)_2)_4. \]
Now specialize to
$A = x_0 x_1(x_0-x_1), B = x_0(x_0 + x_1)(x_0+2x_1)$ and
use the method of undetermined coefficients. \qed

\subsection{The `minimal equation' for $T_3$}
Consider the following equivalence relation on pairs $(A,B)$ of
independent order $d$ forms:
\[ (A,B) \sim (\alpha A + \beta B, \gamma A + \delta B)
\quad \text{if $\alpha \delta - \beta \gamma =1$.}
\]
An equivalence class determines and is determined by $T_1,T_3$.
Let $\F$ denote the set of equivalence classes, and
consider the map
\[ \pi: \F \lra {\mathbb A}^{2d-1}, \quad
(A,B) \lra T_1. \]
It is known that $\pi$ has finite fibres, and the cardinality of
the general fibre is equal to the Catalan number
$\rho(d) = \frac{1}{d}\binom{2d-2}{d-1}$
(see \cite[Theorem 1.3]{Goldberg1}).

Now assume $d=4$, then $\rho(4) =5$. Let $A,B$ be forms of order $4$
with {\sl indeterminate} coefficients, and write
\[ T_1 = \sum_i \binom{6}{i} u_i \, x_0^{6-i}x_1^i, \quad
   T_3 = \sum_j \binom{2}{j} v_j \, x_0^{2-j}x_1^j,
\]
where $u_i,v_j$ are functions of the coefficients of $A,B$.
The map $\pi$ corresponds to a degree $5$ field extension
$K \subseteq L$, where
\[ K = \complex(u_0,\dots,u_6),
\quad L = K(v_0,v_1,v_2).\]

We recall the concept of a seminvariant of a form: it is an expression
in the coefficients of the form which remains unchanged by a substitution
\begin{equation}
x_0 \lra x_0 + c \, x_1, \; x_1 \lra x_1; \quad c \in \complex.
\label{subsU} \end{equation}
An alternative is to define it as the leading coefficient
of a covariant (see \cite[\S 32]{GrYo}). Let
\begin{equation}
v_0^5 + \sum\limits_{i=1}^5 \, l_i \,
v_0^{5-i} = 0, \qquad l_i \in K,
\label{mineq} \end{equation}
denote the unique minimal equation of $v_0$ over $K$.
Firstly, since $v_0$ is a seminvariant of $T_3$
and substitutions in (\ref{subsU}) must leave (\ref{mineq}) unchanged,
all the $l_i$ are seminvariants of $T_1$.
Secondly, by the main theorem of \cite[\S 33]{GrYo}, any algebraic relation
between the seminvariants translates into a relation between the corresponding
covariants. That is to say, we must have an identity
\begin{equation}
 T_3^5 + \sum\limits_{i=1}^5 \, \Lambda_i \, T_3^{\, 5-i} = 0,
\label{mineq.cov} \end{equation}
where $\Lambda_i$ are covariants of $T_1$, and (\ref{mineq.cov}) reduces
to (\ref{mineq}) by the substitution $x_0:=1,x_1:=0$.
By homogeneity, $\Lambda_i$ must have degree-order $(i,2i)$.
\subsection{}
A complete set of generators for the ring of covariants of
order $6$ forms is given in \cite[\S 134]{GrYo}. It is then
a routine matter to identify the $\Lambda_i$ by the method
of undetermined coefficients. I omit all calculations and
merely state the result. Define the following covariants of $T_1$.
\[ \begin{array}{lll}
q_{20} = (T_1,T_1)_6, & q_{24} = (T_1,T_1)_4, & q_{28} = (T_1,T_1)_2, \\
q_{32} = (T_1,q_{24})_4, & q_{36} = (T_1,q_{24})_2, &
q_{38} = (T_1,q_{24})_1, \\
q_{44} = (T_1,q_{32})_2.
\end{array} \]
These are all taken from the table in \cite[p.~156]{GrYo},
but the notation is modified so that $q_{ab}$ is of
degree-order $(a,b)$. There can be no covariant of degree-order $(1,2)$,
hence $\Lambda_1 = 0$. The others are
\[ \begin{aligned}
\Lambda_2 & = -\frac{125}{8} \, q_{24} \\
\Lambda_3 & = \frac{625}{24} \, q_{36} + \frac{125}{36} \, T_1 \, q_{20} \\
\Lambda_4 & = \frac{3125}{48} \, q_{24}^2 -\frac{625}{96} \, q_{20} \, q_{28}
         -\frac{3125}{96} \, T_1 \, q_{32} \\
\Lambda_5 & = \frac{3125}{64} \, T_1 \, q_{44} +
         \frac{3125}{64} \, q_{32} \, q_{28}
        -\frac{3125}{16} \, q_{36} \, q_{24}
        -\frac{3125}{192} \, T_1 \, q_{20} \, q_{24}.
\end{aligned} \]

\smallskip

\begin{Problem} \rm
Find the equation analogous to (\ref{mineq.cov}) for arbitrary $d$.
\end{Problem}

\bibliographystyle{plain}