TITLE: Write $-\Delta u-(1-u^2-v^2)u=0$ and $-\Delta v-(1-u^2-v^2)v=0$ in terms of $w=u^2+v^2$
QUESTION [0 upvotes]: Considering the following system:
Let $\Omega=B_1(0)$
\begin{align*}
\begin{cases}
-\Delta u-(1-u^2-v^2)u=0\\
-\Delta v-(1-u^2-v^2)v=0\\
v|_\Omega=u|_\Omega=0
\end{cases}
\end{align*}
Let $w:=u^2+v^2$, I am trying to show $|w|_{L^\infty(\bar{\Omega})}\leq1$.
To do this, I try to rewrite the system in terms of $w$.
Noticing $\frac{1}{2} \Delta w=u\Delta u+v\Delta v+ (\partial_i u)^2+(\partial_i v)^2$. By multiplying the first equation by $u$ and the second equation by $v$, We have $\frac{1}{2}\Delta w+(1-w)w=(\partial_i u)^2+(\partial_i v)^2$. But I got stuck here. Any help would be appreciated.

REPLY [0 votes]: I'm going to suppose that $u,v$ are classical solutions, that is, $u,v\in C^2(B_1)\cap C(\overline{B_1})$. Then we also have that $w\in C^2(B_1)\cap C(\overline{B_1})$. As you have established, $$\frac 1 2 \Delta w+(1-w)w=\vert \nabla u \vert^2+ \vert \nabla v \vert^2\geqslant 0 \qquad \text{in }B_1 \tag{$\ast$}$$ (I assume you that you meant to sum over $i$). Moreover, $$w=0\qquad \text{on } \partial B_1.$$
Suppose that, for the sake of contradiction, $\max_{B_1}w >1$. Since $w$ is continuous, there exists $x_\star \in \overline{B_1}$ such that $w(x_\star)=\max_{B_1}w.$ In fact, since $w=0$ on $\partial B_1$, we must have $x_\star \in B_1$. Then, since $w$ attains its maximum at $x_\star$, we have that $\Delta w (x_\star )\leqslant 0$ and so  $$\frac 1 2 \Delta w(x_\star)+(1-w(x_\star))w(x_\star)\leqslant(1-w(x_\star))w(x_\star)<0  $$ since $w(x_\star)>1$. This contradicts ($\ast$). Thus, we have that $$0\leqslant w\leqslant 1 \qquad \text{in } \overline{B_1}. $$


TITLE: Stokes' Theorem - Vector Field
QUESTION [2 upvotes]: I am having problems trying to verify Stokes' theorem (below) as part of a question.
$$\iint_{S} \text{curl} \vec F \cdot d\vec S=\oint_{c} \vec F \cdot d\vec r$$
The vector field in question is $\vec F=(xy,z^3,xz^2)$ and I am looking over a triangle plane that joins $(0,0,1)$ , $(0,4,1)$ and $(4,0,1). $
Computing the surface integral (over the triangle plane) does not seem to yield the same answer as the closed line integral (taking around the edge of the triangle) so I know I am making a mistake somewhere!
My working is below:
Surface Integral:
Function of plane: $$f(z) = z-1$$
Unit normal to plane:
$$\vec n = \frac{\nabla f}{\mod{(\nabla f)}} = (0,0,1)$$
Also,
$$dS =  dA(0,0,1) \cdot \vec n = dA = dxdy$$
Curl of vector field:
$$ \text{curl} \vec F = (-3z^2,-z^2,-x)$$
Working out the surface integral:
$$ \int \int _ S  \text{curl} \vec F \cdot d\vec S = \int \int _ S  (-3z^2,-z^2,-x) \cdot (0,0,1) dxdy$$
$$ = \int \int _ S  -x dxdy $$
Limits are $x = [0,4]$,$y=[4-x,4]$
$$ = \int^4_0\int^{4-x}_4  -x dydx $$
$$ = \int^4 _0 x^2 dx=\frac{64}{3}$$
Line Integral
Traversing the edges in an anticlockwise direction, I have split the triangle into three lines. $l_1$ from $(0,0,1)$ to $(0,4,1)$ , $l_2$ from $(0,4,1)$ to $(4,0,1)$ and $l_3$ from $(4,0,1)$ to $(0,0,1)$ .
Also,
$$d\vec r = (dx,dy,dz)$$
$l_1$
$$ \int_{l_1}\vec F \cdot d\vec r = \int_{l_1} (xy,z^3,xz^2) \cdot (dx,dy,dz)$$
$$ = \int_{l_1} xydx+z^3dy+xz^2dz $$
Across $l_1$ $x=0$ $dx=0$ $z=1$ $dz=0$ ($y$ varies).
$$ = \int_{l_1} dy $$
$$ = \int_{0}^4 dy $$
$$ = 4 $$
$l_3$
$$ \int_{l_3}\vec F \cdot d\vec r = \int_{l_3} (xy,z^3,xz^2) \cdot (dx,dy,dz)$$
$$ = \int_{l_3} xydx+z^3dy+xz^2dz $$
Across $l_3$ $y=0$ $dy=0$ $z=1$ $dz=0$ ($x$ varies).
$$ = 0 $$
$l_2$
$$ \int_{l_2}\vec F \cdot d\vec r = \int_{l_1} (xy,z^3,xz^2) \cdot (dx,dy,dz)$$
$$ = \int_{l_2} xydx+z^3dy+xz^2dz $$
Across $l_2$ $x$ and $y$ vary, $y=4-x$ $dy=-dx$ $z=1$ $dz=0$ .
$$ = \int_{l_3} x(4-x)dx - \int_{l_3} dx$$
$$ = \int_{0}^4 4x-x^2-1dx $$
$$ = 32-\frac{64}{3}-4 $$
$$ = \frac{20}{3} $$
And so
$$\oint_{c} \vec F \cdot d\vec r = 4 + \frac{20}{3} = \frac{32}{3}$$
Can anyone see where I am going wrong? Hopefully, I am missing something relatively simple!
Thanks!

REPLY [0 votes]: The surface associated to your triangle is defined by: $\lambda_1\in[0,1],\ \lambda_2\in[0,1]$ such that $\lambda_1+\lambda_2\le 1$
$$
t(\lambda_1,\lambda_2)=\lambda_1(v_2-v_1)+\lambda_2(v_3-v_1)+v1=(4\lambda_2,4\lambda_1,1)
$$
the Curvilinear abscissa are:
$$
\gamma_1(\lambda_1,\lambda_2)=\frac{\partial t}{\partial_{\lambda_1}}=(0,4,0)
$$
$$
\gamma_2(\lambda_1,\lambda_2)=\frac{\partial t}{\partial_{\lambda_2}}=(4,0,0)
$$
Thus surface element $dS$ is
$$
dS=\gamma_1\wedge \gamma_2 d\lambda_1d\lambda_2=(0,0,-16) d\lambda_1d\lambda_2
$$
So
\begin{align}
\int\int curl F.dS&=\int_{\lambda_1=0}^1\int_{\lambda_2=0}^{1-\lambda_1} \underbrace{(-3z^2,-z^2,-x)\circ t(\lambda_1,\lambda_2)}_{(-3,-1,-4\lambda_2)} \cdot \underbrace{dS(\lambda_1,\lambda_2)}_{(0,0,-16) d\lambda_1d\lambda_2} \\
&=\int_{\lambda_1=0}^1\int_{\lambda_2=0}^{1-\lambda_1}64\lambda_2d\lambda_1d\lambda_2 \\ &=64\int_{\lambda_1=0}^1\frac{1}{2}(1-\lambda_1)^2d\lambda_1\\&=\frac{32}{3}
\end{align}


TITLE: A question regarding the axioms of Separation and Replacement in ZFC
QUESTION [2 upvotes]: It is well known that the axioms of Separation and Replacement are axiom schema, that is, they subsume an infinite number of first-order axioms for ZFC.  My question is, how large an infinity is this, or, perhaps better, how large an infinity should this be.  One can naively imagine this infinity to be as large as Ord, the proper class of all ordinals. If it were, could ZFC then be considered 'complete' contrary to the Godel incompleteness theorems?  Could one argue that ZFC could not have Ord many instances of Separation and Replacement?

REPLY [1 votes]: To add to Asaf's answer, although each schema has only countably many instances in the sense of containing only countably many axioms, if we think of an "instance" of separation or replacement as being determined by a logical formula and a set parameter for that formula, then there would indeed be different "instances" for each ordinal.  This is not the standard usage of the word, however.
Note also that comparing the size of schemas with the size of sets or classes is problematic because the former (as Asaf says) is described in the meta-language and the latter is described in the object language.  The distinction becomes important when considering models of set theory, where the class of ordinals of a model may be a countable set (in the universe $V$) and also the model and the universe may not agree on what it means to be an axiom of ZFC.


TITLE: Consider the second-order equation $ \displaystyle \frac{d^2y}{dt^2}+3\frac{dy}{dt}+2y=0 $
QUESTION [0 upvotes]: Consider the second-order equation
$$\frac{d^2y}{dt^2}+3\frac{dy}{dt}+2y=0 $$
Let $v$ be a new variable that represents $ \frac{dy}{dt}$ and reduce reduce this equation to a first-order system in terms of $y$ and $v$. Find the solution to this system that satisfies the initial condition $(y(0),v(0))=(1,-2)$
What will be $y(t)$?
What will be $v(t)$?
I assume that I can rewrite the expression $ \frac{dy}{dt}$ as $v$ so I get that
$ \displaystyle \frac{d^2y}{dt^2}+3v+2y=0 $ is it the good way?

REPLY [0 votes]: $$\frac{d^2y}{dt^2}+3\frac{dy}{dt}+2y-0$$
Notice that is this a second order homogeneous differential equation with constant coefficients.
$1,3,2$
We can use the Method of auxiliary equation to solve it.
$$y=e^{mt}$$
$$\frac{dy}{dt}=me^{mt}$$
$$\frac{d^2y}{dt^2}=m^2e^{mt}$$
Substitute this into the differentia equation we then have
$$m^2e^{mt}+3me^{mt}+2e^{mt}=0$$
Provided $e^{mt}\neq 0$
$$m^2+3m+2=0$$
$$(m+1)(m+2)=0$$
$m=-1,-2$
The fundamental set of solution is then
$$y=c_1e^{-t}+c_2e^{-2t}$$
where $c_1,c_2$ are constants.
Hope that you can take it from here.


TITLE: Question regarding Inequation with multivariable functions
QUESTION [1 upvotes]: In order to simplify the notation consider: $$ x=[x_1,x_2,...x_n] $$
Consider the following inequation:
$$ s(x)(u(x)+A(x))<0 $$
My goal is to choose the function u(x) such that the inequation holds.
s(x) and A(x) are known.
For instance:
if s>0:
 $$u(x)+A(x)<0 $$
if s<0:
$$ u(x)+A(x)>0 $$
This problem seems quite easy to solve.
if s>0:
 $$u(x)<-A(x) $$
and if s<0:
$$u(x)>-A(x) $$
However, the textbook solution is quite different and I would like to know why:
Here is the solution:
if s>0
$$u(x)<-\begin{vmatrix}
A(x)
\end{vmatrix} $$
if s<0
$$u(x)>\begin{vmatrix}
A(x)
\end{vmatrix} $$

REPLY [1 votes]: When $s(x)\ne 0$, there is a lot of freedom to choose the function $u$. Both choices are correct, but your is done in a most general framework, which also shows that the textbook choice is correct.


text:
\begin{document}
\title{\vspace*{1.5cm}Symmetries of vector exterior differential systems and
the inverse problem in second-order  Ostrohrads'kyj mechanics}
\author{R.~Ya.~Matsyuk\\\small email:~\texttt{romko.b.m@gmail.com}\\[-2\jot]\small \url{http://www.iapmm.lviv.ua/12/eng/files/st_files/matsyuk.htm}}
\date{\em Institute for Applied Problems in Mechanics and Mathematics}
\maketitle\thispagestyle{firstpage}
\begin{abstract}
Symmetries of variational problems are considered as symmetries of vector bundle valued exterior differential systems. This approach is then applied to third order ordinary variational equations of motion of the semi-classical spinning particle.
\end{abstract}
It is common in the similarity theories of differential equations
to consider transformations which  leave invariant some prolonged manifolds
in jet-spaces of appropriate order within which the integral manifolds
of a given system of partial differential equations lie~\cite{Ovs}.
From this invariance it follows that the solutions of the system in charge
transform but into some other solutions. However, we are not sure enough
that the prolonged manifold is ``densely'' covered by the integral manifolds.
If not, then some additional symmetries may occur. In this paper we
consider a symmetry of a differential equation be defined in general as
a generator of such a transformation, which carries every solution into
nothing more than some other solution.

Then we reformulate this  into slightly different language of the algebraic
invariance of some vector valued  exterior differential system. This has
an advantage that we can use Lagrange multiplies method to solve the problem.
In the case of ordinary differential equations the two approaches (algebraic
and the general one) coincide due to the completeness of Pfaff systems
(that is that every differential form, annulled by every solution of the
system expands by the Lagrange multiplies into the elements of the differential
ideal, generated by the system itself).

The Euler-Lagrange equations of the variational calculus (of arbitrary order)
naturally fall into the framework of vector bundle valued exterior differential systems
 when the dependent variables are globally segregated from the independent
ones~\cite{Kal}. This is the case of the field theory. In general such
segregation depends on the local chart
\,$\mu:\,M \ra\R^{p}\times\R^{q}$ \,
which constitutes a way of introducing a fibred structure in the manifold $M$.
In applying infinitesimal considerations we can use this structure every time the
manifold $M$ is not endowed by a fibred structure in an intrinsic manner.
If $Y$ denotes a fibred manifold over a base $Z$, we take
\,$Y_s\df J_sY$ \,
to mean the space of all the jets of order $s$ of cross-sections of the surmersion
\,$\pi:\,Y\ra Z$. Then
\,$\mu_{_{J}}:\,Y_s\ra J_s(\R^{p}\times\R^{q})\approx J_s(\R^p,\R^q)$ \,
will denote the standard $s$-order prolongation of the local chart $\mu$.

\paragraph*{1.~Vector bundle valued differential systems.}
Let
\,$E\ra B$,
\,$E_{\scriptscriptstyle1}\ra B_{\scriptscriptstyle1}$,
\,$F\ra X$,
\,$F_{\scriptscriptstyle1}\ra X_{\scriptscriptstyle1}$,
and
\,$E'\ra B$ \,
be some fibre bundles, let
\,$\Gamma (E)$ \,
denote the set of smooth cross-sections of the fibred manifold $E$ and
let
\,$\F (B)$ \,
stand for the ring of smooth functions defined on the manifold $B$. We shall
denote the tensor product of the inverse image of vector bundles with respect to
some manifold morphisms
\,$f:\,B\ra X$ \,
and
\,$f_{\scriptscriptstyle1}:\,B\ra X_{\scriptscriptstyle1}$ \,
by $\otimes_{_{B}} $. If the manifold $B$ is fibred over some manifold $N$,
the semi-basic differential forms over the manifold $B$ with values in the
vector bundle $F$ are defined as smooth cross-sections of the vector bundle
\,$\wedge T^{\ast}N\otimes_{_{B}}\F$.
They constitute an
\,$\F (B)$-submodule, denoted by
\,$\F_{_{B}}(N,F)$,
of the graded module
\,$\F(B,F)=\Gamma(\wedge T^{\ast}B\otimes_{_{B}}F)$.
Every bundle homomorphism
\,$h:\,F\ra F_{{\scriptscriptstyle1}}$ \,
over the base homomorphism
\,$\ul{h}:\,X\ra X_{\scriptscriptstyle1}$ \,
satisfying
\,$f_{\scriptscriptstyle1}=\ul{h}\circ f$ \,
obviously defines the moduli homomorphism
\,$h_{{\SS \#}}:\,\F(B,F)\ra\F(B,F_{\scriptscriptstyle1})$.
The canonical vector bundle pairing
\,$E\times\Hom(E,E')\ra E'$ \,
defines an exterior product
\,$\wedge:\,\F(B,E)\times\F(B,E^{\ast}\otimes E')\ra\F(B,E')$.
If we think of a cross-section
\,$\bmat{\varphi}\in\F^{\scriptscriptstyle0}(B,E^{\ast}\times E')$ \,
as a $B$-morphism
\,$\widehat{\phi}:\,E\ra E'$,
then
\,$\widehat{\phi}_{\SS \#}\bmat{\omega}=\bmat{\omega}\wedge\bmat{\varphi}$ \,
for any
\,$\bmat{\omega}\in\F(B,E)$.

The vector bundle
\,$\End(E)$ \,
is a bundle of algebras with the composition rule that of the superposition
of endomorphism. This fibrewise composition defines the structure of a
graded non-commutative algebra in the
\,$\F(B)$
module of
\,$\End (E)$-valued differential forms which we denote by
\,$\F(E,\End E)$.
The vector bundle $E$ may be viewed as the bundle of
\,$\End(E)$-moduli with respect to the pairing
\,$E\times\End(E)\ra E$.
This pairing makes the
$\F(B)$-module
\,$\F(B,E)$ \,
into a graded module over the algebra
\,$\F(B,\End E)$.

Now let
\,${\sf h}:\,E_{\scriptscriptstyle1}\ra E$ \,
be a vector bundle homomorphism over a base morphism
\,$g:\,B_{\scriptscriptstyle1}\ra B$ \,
and let
${\sf h}_{\scriptscriptstyle1}$
denote the corresponding induced $B_{\scriptscriptstyle1}$-homomorphism of vector bundles
\,${\sf h}_{\scriptscriptstyle1}:\,E_{\scriptscriptstyle1}\ra g^{-{\scriptscriptstyle1}}E$.
The dual vector bundle homomorphism
\,${\sf h}_{\scriptscriptstyle1}{}^\ast:\,(g^{-{\scriptscriptstyle1}}E)^\ast\ra E_{\scriptscriptstyle1}{}^\ast$ \,
denotes, by virtue of identification
\,$(g^{-{\scriptscriptstyle1}}E)^\ast\approx g^{-{\scriptscriptstyle1}}(E^\ast)$,
some $g$-comorphism
\,${\sf h}^\ast:\, g^{-{\scriptscriptstyle1}}(E^\ast)\ra E_{\scriptscriptstyle1}{}^\ast$.
If set
\,$E=TB$,
\,$E_{\scriptscriptstyle1}=TB_{\scriptscriptstyle1}$,
one gets the action
${\sf h}^{\SS \#}$ of ${\sf h}$
on the ordinary differential forms on~$B$. If
${\alpha}_g$ stand for the inverse image of the cross-section
\,$\alpha\in\F(B,\R)$,
then
\,${\sf h}^{\SS \#}{}{\alpha}=(\wedge {\sf h})^\ast\circ\alpha_g\in\F(B_{\scriptscriptstyle1},\R)$.
Given some cross-section
\,$\bmat{\omega}\in\F(B,E)$ \,
and recalling the notion of its inverse image
\,${\bmat{\omega}}_g\in\Gamma(\wedge T^\ast B\otimes_{B_{\scriptscriptstyle1}}E)$,
the differential form
\,$g^\star\bmat{\omega}\in\F(B_{\scriptscriptstyle1},E)$ \,
is being defined by the superposition of mappings:
\,$g^\star\bmat{\omega}=\big((\wedge Tg)^\ast\otimes {\sf id}\big)\circ\bmat{\omega}_g$.
If the morphism $g$ has a canonical prolongation to a vector bundle
$g$-comorphism
\,${\sf k}:\,g^{-{\scriptscriptstyle1}}E\ra E_{\scriptscriptstyle1}$,
then the notation
\,$g^{\SS \#}\bmat{\omega}$ \, will mean the differential form
\,${\sf k}_{\SS \#}g^\star\bmat{\omega}=\big((\wedge Tg)^\ast\otimes {\sf k}\big)\circ\bmat{\omega}$.

The functor $g^\star$ is consistent with the tensor product and contraction
operations. Let throughout this paragraph the manifold $B$ be parallelizable
and the module
\,$\Gamma (E)$ \,
be free. Then, if
\,$\bmat{\omega}=\alpha\otimes\bmat{\gamma}\in\F (B,E)\approx\F(B,\R)\otimes\Gamma (E)$ \,
and if
\,$\bmat{\Omega} =\beta\otimes\bmat{\varphi}\in\F (B,E^\ast\otimes E')\approx\F (B,\R)
\otimes \Gamma (E^\ast\otimes E')$,
then for the inverse images of the cross-sections
$\bmat{\omega}$,
$\bmat{\Omega}$, and
\,$\bmat{\omega}\wedge\bmat{\Omega} =(\alpha\wedge\beta )\otimes\langle{\bmat{\gamma}}_g,\bmat{\varphi}\rangle$ \,,
it is true that
\,$(\bmat{\omega}\wedge\bmat{\Omega})_g=({\alpha}_g\wedge {\beta}_g)\otimes
\langle {\bmat{\gamma}}_g,{\bmat{\varphi}}_g\rangle\in
\Gamma (\wedge T^\ast B{\otimes}_{_{B_1}}E')$.
Composing with
\,$(\wedge Tg)^\ast$ \,
one gets
\,$g^\star (\bmat{\omega}\wedge\bmat{\Omega} )=g^\star\bmat{\omega}\wedge g^\star\bmat{\Omega}$.

A one-form  $\bmat{\vartheta}\in\F^{\scriptscriptstyle1}(B,TB) $ may be thought of as a $B$-homomorphism of vector bundles
\,$\widehat{\theta}:\,TB\ra TB$; \,
then the dual homomorphism
\,$\widehat{\theta}^\ast:\,T^\ast B \ra TB$ \,
acts over the cross-sections from
\,$\Gamma (T^\ast B)\equiv\F^{\scriptscriptstyle1}(B,\R)$ \,
in an obvious manner and is being extended to
\,$\F(B,\R)$ \,
as a differentiation of degree $0$ acting trivially on the ring of functions.
 We denote this extended action by $\bmat{\vartheta}\bar{\wedge}$.
Again within local
considerations for a parallelizable $B$ and $\Gamma (E)$ being free,
we can identify
\,$\F(B,\End E)$ \,
with the tensor product
\,$\F (B,\R)\otimes_{_{\hbox{\sfrak F}\,{\Ss(B)}}}\Gamma (\End E)$ \,
and define
\,$\bmat{\vartheta}\bar{\wedge}(\alpha\otimes\bmat{\varphi} )=(\bmat{\vartheta}\bar{\wedge}\alpha)
\otimes{\sf id}(\bmat{\varphi})$.
Consider the ideal
\,$\I\equiv\I (B)\equiv\I (\bmat{\vartheta})=\sum\limits_{d>0}\I^{(d)}=
\bmat{\vartheta}\bar{\wedge}\F (B,\R)$ \,
in the algebra
\,$\F (B,\R)$.
Consider also the \,$\F (B,\R)$-submodule
\,$\I (B, \End E)=\bmat{\vartheta}\bar{\wedge}\F (B,\End E)$ \,
in the
\,$\F (B,\R)$-module
\,$\F (B,\End E)$.
We denote by
\,$\I (B,E)$ \,
the submodule
\,$\F (B,E)\wedge\I$ \,
in the module
\,$\F(B,E)$ \,
over the algebra
\,$\F (B,\R)$.
It is clear that
\,$\I (B,E)$ \,
is also a submodule over the algebra
\Far{\End E} \,
and that
\,$\I (B,E)=\F (B,E)\wedge\I(B,\End E)$.
In fact,
\,$\I(B,E)$ \,
is generated over
\Far{\R} \,
by the
\,$\F^{\scriptscriptstyle0}(B,\End E)$-submodule of one-forms
\begin{displaymath}
\P (B,E)=\bmat{\vartheta}\bar{\wedge}\F^{\scriptscriptstyle1}(B,E)=\F^{\scriptscriptstyle0}(B,E)\wedge\P,
\end{displaymath}
where
\,$\P\equiv\P(B)\equiv\I^{({\scriptscriptstyle1})}=\bmat{\vartheta}\bw\F^{\scriptscriptstyle1}(B,\R)$ \,
is an \,$\F(B)$-submodule, that is the ordinary Pfaff system.

\begin{definition}
An exterior differential system \,$\S\equiv\S(F)$ \,
(with values in the vector bundle $F$) is an
\Far{\End F}-submodule of the module
\Far{F}. A Pfaff system is an exterior differential system, generated by
one-forms. Let a  manifold~$Z$ be given. We call the germ of an immersion
\,$\sigma:\,Z\ra B$ \,
be a solution of the exterior differential system $\S$ if
\,$\sigma^\star\S=0$.
Two exterior differential systems (in general, over different manifolds) are
called equivalent if the sets of their solutions coincide (in general,
are isomorphic).
\end{definition}

Let over a manifold $B$ be given an exterior differential system $\S$
with values in the vector bundle $E$ and an exterior differential system
$\S'$ with values in the vector bundle $E'$ and let
\,$\sol (\S)$\,
denote the set of solutions of the system $\S$. If
\,$\S\wedge\F (B,E^\ast\otimes E')\supset\S'$,
then
\,$\sol (\S)\subset\sol (\S')$.

\begin{definition}
Exterior differential systems $\S$ and $\S'$ are algebraically equivalent,
if at the same time both
\,$\S\wedge\F(B,E^\ast\otimes E')\supset\S'$
and
\,$\S'\wedge\F (B,E^{\prime\ast}\otimes E)\supset\S$.
\end{definition}

\begin{lemma}
Two Pfaff systems which are algebraically equivalent, are equivalent.
\end{lemma}

This follows from the completeness of Pfaff systems.

Let $\widetilde{\S}$ denote the image of the exterior differential system $\S$
under the projection
\,$ j:\,\F (B,E)\ra\F (B,E)/\I (B,E)$.
In view of the inclusion
\,$\I (B,E)\wedge\F (B,E^\ast\otimes E')\subset\I (B,E')$ \,
the action $\wedge$ of the elements from
\Far{E^\ast\otimes E'} \,
over the quotient module is defined and if for this action \,$\S\wedge\F (B,E^\ast\otimes E')\supset \S'$,
then also
\,$\widetilde{\S}\wedge\F (B,E^\ast\otimes E')\supset\widetilde{\S'}$.

\begin{definition}
A solution of the system
\,$(\S,\bmat{\vartheta})$ \,
is a germ of an immersion
\,$\sigma:\,Z\ra B$,
such that
\,$\sigma^\star j^{-{\scriptscriptstyle1}}(\widetilde{\S})=0$.
\end{definition}

Let over the manifold $B_{\scriptscriptstyle1}$ be specified a  differential form
\,$\bmat{\vartheta}_{\scriptscriptstyle1}\in\F (B_{\scriptscriptstyle1},TB_{\scriptscriptstyle1})$ \,
such that
\,$g^\star\bmat{\vartheta} =\bmat{\vartheta}_{\scriptscriptstyle1}\bw\bmat{\varrho}\in\P (B_{\scriptscriptstyle1},g^{-{\scriptscriptstyle1}}TB)$ \,
for some
\,$\bmat{\varrho}\in\F (B_{\scriptscriptstyle1},TB)$.
Taking into account the definitions of the actions $g^\star$,
$\bmat{\vartheta}_{\scriptscriptstyle1}\bw$ and identifying the differential forms $\bmat{\vartheta}$,
$\bmat{\vartheta}_{\scriptscriptstyle1}$, and $\bmat{\varrho}$ with the corresponding vector bundle
homomorphisms $\widehat{\theta}$, $\widehat{\theta}_{\scriptscriptstyle1}$ and
\,$\widehat{\rho}:\, TB_{\scriptscriptstyle1}\ra g^\ast TB$ \,
we have
\,$\widehat{\theta}_g\circ (Tg)_{\scriptscriptstyle1}=\widehat{\rho}\circ\widehat{\theta}_{\scriptscriptstyle1}$,
where $\widehat{\theta}_g$ is the inverse image of the homomorphism $\widehat{\theta}$
with respect to the morphism $g$. For the conjugated homomorphisms we
also have
\,$(Tg)^\ast\circ\widehat{\theta}_g{}^\ast =\widehat{\theta}_{\scriptscriptstyle1}{}^\ast\circ\widehat{\rho}^{\,\ast}$.

In view of the definition of the action $\widehat{\rho}^{\,\SS \#}$
on the differential form
\,$\alpha\in\F^{{\scriptscriptstyle1}}(B,\R)$ \,
it holds that
\begin{displaymath}
g^\star (\bmat{\vartheta}\bw\alpha )=(Tg)^\ast\circ\widehat{\theta}_g{}^\ast\circ\alpha_g=
\widehat{\theta}_{\scriptscriptstyle1}{}^\ast\circ\widehat{\rho}^{\,\ast}\circ\alpha_g=
\bmat{\vartheta}_{\scriptscriptstyle1}\bw
\widehat{\rho}^{\,\ast}\alpha\in\P_{\scriptscriptstyle1}\equiv\P (B_{\scriptscriptstyle1}).
\end{displaymath}
It follows that
\,$g^\star\P\subset\P_{\scriptscriptstyle1}$,
thus
\,$g^\star\I\subset\I_{\scriptscriptstyle1}$ \,
and also
\begin{displaymath}
g^\star\I (B,E)=g^\star\F (B,E)\wedge g^\star\I\subset\F (B_{\scriptscriptstyle1},g^{-{\scriptscriptstyle1}}E)
\wedge\I (B_{\scriptscriptstyle1},g^{-{\scriptscriptstyle1}}E),
\end{displaymath}
so the action $g^\star$ of the homomorphism $g$ over the quotient module
\Far{E/\I (B,E)} \,
is defined.

\paragraph*{2.~The Lie derivative.}
Let
\,$w:\,B\ra W$\,
be a morphism of manifolds and let $w_{t}$ denote its deformation. Then
\,$\dot{w}_t(0):\,B\ra TW$ \,
is a lift of the morphism $w$. If the manifold $W$ is fibred over the
manifold $B$ and if every $w_{t}$ is a cross-section, then
\,$\dot{w}_{t}(0)$ \,
belongs to the vertical tangent  bundle
\,$VW\subset TW$.
Let $\xi$ and $\bar{\eta}$ be some vector fields on the manifolds $B$ and $W$
respectively. The Lie derivative~\cite{Kol} of the morphism $w$
with respect to the pair of vector fields $\xi$, $\bar{\eta}$ is defined by
 the expression
\,${\bf L}_{\xi,\be}w =(Tw)\circ\xi-\be\circ w$.
If $w$ is a cross-section and if the vector field $\be$ is projectible onto the
vector field $\xi$, then the Lie derivative
\,${\bf L}_{\xi,\be}w$ \,
belongs to $VW$. Let under this assertions
\,$\exp_{_{B}}t\be$ \,
denote the one-parameter induced family of diffeomorphism over $B$,
corresponding to
\,$\exp t\be$ \,
and let
\begin{displaymath}
w_t=\exp_{_{B}}(-t\be)\circ w_{\exp t\xi}=\exp(-t\be)\circ w\circ\exp t\xi.
\end{displaymath}
Differentiating at
\,$t=0$ \,
we obtain
\begin{displaymath}
\dot{w}_t(0)=-(\exp t\be)\circ w +(Tw)\circ(\exp t\xi)\dot{{}}=
{\bf L}_{\xi,\be}\,w.
\end{displaymath}
If the fibre bundle $W$ is a vector bundle, then
\,$VW\approx W\times_{_{B}}W$ \,,
the first component of
\,$\dot{w}_t(0)$ \,
coincides with the initial cross-section $w$, and the second one is identified with the
Lie derivative
\,${\bf L}_{\xi,\be}\,w$ \,
itself. In a special case setting
\,$W=\wedge T^\ast B\otimes E$ \,
and taking $\be$ to be build up from the standard lift to the cotangent bundle of the
vector field $\xi$ together with some vector field $\eta$ on $E$,
projectible into $\xi$, we agree with truncated notation of
\,${\bf L}_{\xi,\eta}\df{\bf L}_{\xi,\be}$.

\begin{definition}
A diffeomorphism
\,$g:\,B\ra B$ \,
is called a symmetry ({\it vs}  an algebraic symmetry) of the exterior differential
system $\S(E)$, if the exterior differential systems
$g^\star\S$ and $\S$ itself are equivalent ({\it vs} algebraically equivalent).
A vector field $\xi$ on the manifold $B$ is called an infinitesimal
symmetry ({\it vs} an algebraic infinitesimal symmetry) of the exterior differential
system $\S$ if for all $t$ the transformation
\,$\exp t\xi$ \,
 is a symmetry ({\it vs} an algebraic symmetry) of the system $\S$.
\end{definition}

That $\xi$ is an algebraic infinitesimal symmetry of the exterior differential
system $\S$ means that
\begin{displaymath}
(\exp t\xi)^\star\S\subset\S\wedge\F\big(B,E^\ast\otimes(\exp t\xi)^{\SS-1}E\big).
\end{displaymath}
In particular, whatever the vector field
\,$\eta\in\Gamma(TE)$,
projectible into the vector field $\xi$ be, the inclusions
{
\setlength{\arraycolsep}{0pt}
\begin{eqnarray*}
(\exp -t\eta)_{_{B^{\SS\#}}}(\exp t\xi)^{\SS-1}\S&\subset&
\S\wedge\F\big(B,E^\ast\otimes
(\exp t\xi)^{\SS-1}E\wedge\F^{\scriptscriptstyle0}(B,{(\exp t\xi)^{\SS-1}}^\star(E^\ast\otimes E)\big)\\
&\subset&\S\wedge\F(B,E^\ast\otimes E)=\S
\end{eqnarray*}
}
hold.
By differentiating with respect to the parameter $t$ we conclude that if
$\xi$ is the infinitesimal symmetry of the system $\S$, then
\begin{displaymath}
{\bf L}_{\xi,\eta}\S=\frac{d}{dt}(\exp t\xi)^\star\S(0)\subset\S
\end{displaymath}
for every vector field $\eta$, projectible into the field $\xi$.
In the case of a trivial vector bundle $E$, one can put
\,$\eta=(\xi,0)$ \,
and use the notation
\,${\bf L}_{\xi,0}$ \, in place of
\,${\bf L}_{\xi,(\xi,0)}$.

\begin{definition}
A diffeomorphism
\,$g:\,B\ra B$ \,
is called a symmetry ({\it vs} algebraic symmetry) of the system
\,$(\S,\bmat{\vartheta})$,
if
\,$g^\star\I\subset\I$ \,
and if the exterior differential systems
\,$g^\star\widetilde{\S}$ \,
and $\widetilde{\S}$ itself are equivalent ({\it vs} algebraically equivalent).
\end{definition}

An (algebraic symmetry of system
\,$(\S,\bmat{\vartheta})$ \,
is symmetry ({\it vs} an algebraic symmetry) of the exterior differential system
\,$\S+\I(B,E)$ \,
and {\it vice versa}. In that case
\,${\bf L}_{\xi}\I\subset\I$ \,
and
\,${\bf L}_{\xi,\eta}\widetilde{\S}\subset\widetilde{\S}$, that is
\begin{equation}\label{1}
{\bf L}_{\xi,\eta}\S\subset\S+\I(B,E)\,.
\end{equation}

\paragraph*{3.~Symmetries of the Euler-Lagrange equations.}
On the manifold $Y_{s+{\scriptscriptstyle1}}$ of the jets of order $s+1$ of the cross-sections
of the fibred manifold
\,$\pi:\,Y\ra Z$ \,
there exists a canonical contact differential form
\,$\bmat{\vartheta}_s\in\F_{s+{\scriptscriptstyle1}}(Y_s,V_s)$, semi-basic with respect to the
projection
\,$\pi_s^{s+{\scriptscriptstyle1}}:\,Y_{s+{\scriptscriptstyle1}}\ra Y_s$ \,
which takes values in the vector bundle $V_s$ of vertical tangent vectors
to the surmersion
\,$\pi^{s}:\,Y_s\ra Z$ \,
and such that Pfaff system
\,$\P_s=\bmat{\vartheta}_s\bw{\F}_{s+{\scriptscriptstyle1}}^{\scriptscriptstyle1}(Y_s,\R)$ \,
is nothing but the Cartan co-distribution.
(We use subscript $s$ instead of $Y_s$ where possible).
For an open set
\,$U_s\subset Y_s$ \,
and applying the projection
\,$\pi_v^s:\,Y_s\ra Y_v$ \,
let us put
\,$\I_v(U_s)=\I(\left.\pi_v^s\right|_{_{U_s}}{\!\!\!\!}^\star\,\bmat{\vartheta}_v)$ \,
and let
\,$\upsilon_{_{J}}:\,Y_s\ra J(\R^p,\R^q)$ \,
be the canonical lift of a local fibred chart $\upsilon$ on the manifold $Y$.
A lagrangian is an element $\widetilde{\lambda}$ of the quotient sheaf, generated by
the quotient moduli
\,$\F_{_{U_r}}^p(Y,\R)/\I_{\scriptscriptstyle0}^{(p)}(U_r)$,
which we shall briefly denote by
\,${\cF}_{\,r}^{\,p}(Y,\R)/{\cI}_{\,\scriptscriptstyle0}^{\,(p)}(Y_r)$.
(We denote  the corresponding sheaves by the calligraphic characters).
There can always be found, at least locally, a  representative $\lambda$, semi-basic with
respect to the projection
\,$\pi^r:\,Y_r\ra Z$,
that is
\,$\lambda\in{\cF}_r{}^p(Z,\R)$.
The Euler-Lagrange expressions which correspond to the Lagrange density
\,$\upsilon_{_J}{}^{-{\scriptscriptstyle1}}{}^\star\lambda\in{\cF}_r{}^p(\R^p,\R)$ \,
naturally take the shape of the components of the local expression of
some differential $p$-form
\begin{equation}\label{2}
\bmat{\varepsilon}\in{\cF}_{\,2r}^{\,p}(Z,V_{\scriptscriptstyle0}{}^\ast)
\end{equation}
corresponding to the lagrangian $\lambda$~\cite{Kaz,MatStud}.

{\em
The symmetries of the Euler-Lagrange equations are nothing else but the
symmetries of the system
\,$(\S_{_{\bmat{\varepsilon}}},\bmat{\vartheta}_{2r-{\scriptscriptstyle1}})$,
where the module $\S_{_{\bmat{\varepsilon}}}$ is generated by the vector bundle valued differential
form $\bmat{\varepsilon}$.
}

From here on we shall deal with variational calculus in one independent
variable. As an example we consider a system of third-order ordinary
differential equations
\begin{equation}\label{18}
{{\sf E}}_{a}=0\,.
\end{equation}
Let \,$Y=\R^p\times\R^q$, and let the
canonical coordinates in the manifold $J_{r}(\R^{1};\R^{q})$ be denoted by
$t, \ \hbox{\ssb x}=({\sf x}^{a}), \ \hbox{\ssb v}=({\sf v}^{a}),
\ \hbox{\ssb v}^{{\prime}}=({{\sf v}'}^{a}), \, \dots, \hbox{\ssb v}^{\scriptscriptstyle(r-1)}
=({{\sf v}^{\scriptscriptstyle(r-1)}}^{a})$.

This corresponds to \,$p=1$, $k=q$ \, in~(\ref{2}).
We introduce a vector valued differental one-form
\begin{equation}\label{19}
{\bmat{\epsilon}}={{\sf E}}_{a}\,{\bf d}{{\sf x}}^{a}\otimes{\bf d}t\,,
\end{equation}
where the expressions ${{\sf E}}_{a}$ are the Euler-Poisson
expressions. Applying the general criterion of \cite{Tul} for an arbitrary system
of differential equations to be a system of Euler-Poisson equations, it was
established in \cite{Thesis,MatMet20} that the vector expression ${\hbox{\ssb E}}=\{{{\sf E}}_{a}\}$
in (\ref{18}) must take the shape of
\begin{equation}\label{20}
{\hbox{\ssb E}}=\A{\,\bkey.\,}{\hbox{\ssb v}}^{{{\prime}}{{\prime}}}{\,+\,}({\hbox{\ssb v}}^{{\prime}}{\!\bkey.\,}{\bpartial}_{\eisf v})\,
\A{\,\bkey.\,}{\hbox{\ssb v}}^{{\prime}}{\,+\,}\B{\,\bkey.\,}{\hbox{\ssb v}}^{{\prime}}{\,+\,}{\hbox{\ssb c}}\,,
\end{equation}
where  the skew-symmetric matrix $\A$, the matrix $\B$, and the column vector
${\hbox{\ssb c}}$ depend on the variables $t$, ${\hbox{\ssb x}}$, ${\hbox{\ssb v}}={d{\hbox{\ssb x}}}/dt$,
and satisfy the following system of partial differential equations in $t$, ${{\sf x}}^a$,
and ${{\sf v}}^a$~\cite{Thesis,Dokl}
{\openup1\jot
\begin{equation}\label{21}
\begin{array}{c}
       \partial_{_{_{_{{\hbox{\sssm v}}}}}}{\!}_{[a}{}{{\sf A}}_{bc]}=0 \\
       2\,{{\sf B}}_{[ab]}-3\,{\bf D_{_{\bmat 1}}}{\kern.01667em}{{\sf A}}_{ab}=0 \\
       2\,\partial_{_{_{_{{\hbox{\sssm v}}}}}}{\!}_{[a}{}{{\sf B}}_{b]c}
-4\,\partial_{_{_{_{{\hbox{\sssm x}}}}}}{\!}_{[a}{}{{\sf A}}_{b]c}
+{\partial_{_{_{_{{\hbox{\sssm x}}}}}}{\!}_{c}}{\,}{{\sf A}}_{ab}
+2\,{\bf D_{_{\bmat 1}}}{\kern.01667em}{\partial_{_{_{_{{\hbox{\sssm v}}}}}}{\!}_{c}}{\,}{{\sf A}}_{ab}=0 \\
       {\partial_{_{_{_{{\hbox{\sssm v}}}}}}{\!}_{(a}}{}{{\sf c}}_{b)}
-{\bf D_{_{\bmat 1}}}{\kern.01667em}{{\sf B}}_{(ab)}=0\\

       2\,{\partial_{_{_{_{{\hbox{\sssm v}}}}}}{\!}_{c}}{\,}\partial_{_{_{_{{\hbox{\sssm v}}}}}}{\!}_{[a}{}{{\sf c}}_{b]}
-4\,\partial_{_{_{_{{\hbox{\sssm x}}}}}}{\!}_{[a}{}{{\sf B}}_{b]c}
+{{\bf D_{_{\bmat 1}}}}^{2}{\,}{\partial_{_{_{_{{\hbox{\sssm v}}}}}}{\!}_{c}}{\,}{{\sf A}}_{ab}
+6\,{\bf D_{_{\bmat 1}}}{\kern.0334em}\partial_{_{_{_{{\hbox{\sssm x}}}}}}{\!}_{[a}{}{{\sf A}}_{bc]}=0 \\
       4\,\partial_{_{_{_{{\hbox{\sssm x}}}}}}{\!}_{[a}{}{{\sf c}}_{b]}
-2\,{\bf D_{_{\bmat 1}}}{\kern.0334em}\partial_{_{_{_{{\hbox{\sssm v}}}}}}{\!}_{[a}{}{{\sf c}}_{b]}
-{{\bf D_{_{\bmat 1}}}}^{3}{\,}{{\sf A}}_{ab}=0\,.
\end{array}
\end{equation}
}
In~(\ref{21}) $\mathbf{D_{{s}}}$ denotes the generators of Cartan distribution,
{\setlength{\arraycolsep}{0pt}
\begin{eqnarray*}
{\bf D_{_{\bmat 2}}}={\hbox{\ssb v}}^{{\prime}}{\!\bkey.\,}{\bpartial}_{\eisf v}
{\,+\,}&{\bf D_{_{\bmat 1}}}&\,,\\
&{\bf D_{_{\bmat 1}}}&=\partial_{t}{\,+\,}{\hbox{\ssb v}}{\,\bkey.\,}{\bpartial}_{\eisf x}\,.
\end{eqnarray*}

Let $\bmat{\vartheta}_{_{\bmat 2}}$, $\bmat{\vartheta}_{_{\bmat 3}}$ denote the canonical contact
forms
\begin{eqnarray*}
\bmat{\vartheta}_{_{\bmat 3}}
=\frac\partial{\partial {{{\sf v}}'}^{a}}\otimes({\bf d}{{\sf v}'}^{a}
-{{\sf v}''}^{a}{\bf d}t){\,+\,}&\bmat{\vartheta}_{_{\bmat 2}}&\,,  \\
&\bmat{\vartheta}_{_{\bmat 2}}&
=\frac\partial{\partial {{\sf v}}^{a}}\otimes({\bf d}{\sf v}^{a}
-{{\sf v}'}^{a}{\bf d}t)
{\,+\,}\frac\partial{\partial {{\sf x}}^{a}}\otimes({\bf d}{\sf x}^{a}
-{\sf v}^{a}{\bf d}t) \,.
\end{eqnarray*}
Along with the differential form ${\bmat{\epsilon}}$,
we introduce another one,
\begin{eqnarray*}
{\underline{\bmat{\epsilon}}}
={\sf A}_{ab}\,{\bf d}{\sf x}^{a}\otimes{\bf d}{{\sf v}'}^{b}
{\,+\,}&{\sf K}&_{a}\,{\bf d}{\sf x}^{a}\otimes{\bf d}t\,,\\
&\hbox{\ssb K}&
=({\hbox{\ssb v}}^{{\prime}}{\!\bkey.\,}{\bpartial}_{\eisf v})\,
\A{\,\bkey.\,}{\hbox{\ssb v}}^{{\prime}}{\,+\,}\B{\,\bkey.\,}{\hbox{\ssb v}}^{{\prime}}{\,+\,}{\hbox{\ssb c}}\,.
\end{eqnarray*}
}
Exterior differential systems, generated by the forms ${\bmat{\epsilon}}$ and
${\underline{\bmat{\epsilon}}}$, are equivalent:
\begin{equation}\label{33}
{\underline{\bmat{\epsilon}}}-{\bmat{\epsilon}}
=
\bmat{\vartheta}_{_{\bmat 3}}\bw({\sf A}_{ab}\,{\bf d}{\sf x}^{a}\otimes{\bf d}{{\sf v}'}^{b})\,.
\end{equation}

Now it is time to put in the concept of {\it symmetry}. Let
\begin{equation}\label{22}
{\x}=\hbox{\frak t}\frac\partial{\partial t}{\,+\,}{\hbox{\frak x}^{a}}\frac\partial{\partial{\sf x}^{a}}
\end{equation}
denote a generator of some local group of transformations of the manifold \,$\R\times\R^q$,
its successive prolongations to the space $J_{s}(\R;\R^q)$
denoted by ${\x}_{_{\bmat s}}$,
\[
{\x}_{_{\bmat 2}}=\hbox{\frak v}^{a}\frac\partial{\partial{{\sf v}'}^{a}}
{\,+\,}{\x}_{_{\bmat 1}}\,.
\]

The demand that the exterior differential system, generated by the vector
valued differential form ${\underline{\bmat{\epsilon}}}$, be invariant under the
infinitesimal transformation ${\x}$ incarnates in consistency with
(\ref{1}) into the following
equation
\begin{equation}\label{23}
{\bf L}({\x}_{_{\bmat 2}})({\underline{\bmat{\epsilon}}})={\bf\Xi}{\,\bkey.\,}
{\underline{\bmat{\epsilon}}}{\,+\,}\bmat{\vartheta}_{_{\bmat 2}}\bw{\bmat{\omega}}\,,
\end{equation}
where the elements of a matrix
${\bf\Xi} \in \F^{0}\big(J_{2}(\R;\R^{q});\Hom({\R^{q}}^{\ast};{\R^{q}}^{\ast})\big)$
and the coefficients of a $\pi^{2}_{1}$-horizontal ${\R^{q}}^{\ast}$-valued one-form
${\bmat{\omega}}\in \F^{1}_{2}\big(J_{1}(\R;\R^{q});{\R^{q}}^{\ast}\big)$
depend upon the variables $t$, $\hbox{\ssb x}$, $\hbox{\ssb v}$, and ${\hbox{\ssb v}}^{{\prime}}$.
Both ${\bf\Xi}$ and ${\bmat{\omega}}$ play the role of Lagrange multipliers.
Splitting equation~(\ref{23})
with respect to independent differentials
${\bf d}t$, ${\bf d}{\sf x}^{a}$, ${\bf d}{\sf v}^{a}$, and
${\bf d}{{\sf v}'}^{a}$, gives the following system of partial differential
equations
{\openup1\jot
\begin{eqnarray}\label{24}
{\bf L}({\x}_{_{\bmat 1}}){\sf A}_{ab}&=&{\Xi_{a}}^{c}\,{\sf A}_{cb}
-{\sf A}_{ac}{\displaystyle\frac\partial{\partial{{\sf v}'}^{b}}\hbox{\frak v}^{c}}\\
{\bf L}({\x}_{_{\bmat 2}}){\sf K}_{a}&=&{\Xi_{a}}^{b}\,{\sf K}_{b}
-{\sf A}_{ab}{\bf D_{_{\bmat 2}}}\hbox{\frak v}^{b}
-{\sf K}_{a}{\bf D_{_{\bmat 1}}}\hbox{\frak t}\,.
\end{eqnarray}
}
\subparagraph*{Variational problems in parametric form.}
A variational problem in parametric form is a variational problem, posed on the
manifold \,$J_r(Z,M)$, so we have to put \,$Y=Z\times M$, the dimension of the
manifold $M$ equal to \,$p+q$ with $p=1$, and \,$k=p+q=1+q$ \, in~(\ref{2}).
Canonical coordinates in the manifold $J_{r}(\R;\R^{1+q})$ are denoted by
$\tau, \ \bmat x=(x^{\rho}), \ \bmat u=(u^{\rho}), \ \bmat{\dot{ u}}
=({\dot{u}}^{\rho}), \, \dots,{\stackrel{\scriptscriptstyle r-1}{\bmat u}}
=({\stackrel{\scriptscriptstyle r-1}{u}}^{\rho})$.
The manifold of $r^{\rm th}$-order velocities, $T^{1}_{r}M$, is defined as
$T^{1}_{r}M=J_{r}(\R;M)_{0}$. There exists an obvious isomorphism
$J_{r}(\R;M)\approx\R\times T^{1}_{r}M$.
Coordinates in the manifold $T^{1}_{r}M$ are denoted by
$\bmat x, \bmat u,
\bmat{\dot{ u}}, \dots,{\stackrel{\scriptscriptstyle r-1}{\bmat u}}$.
If $M$ is a (pseu\-do-\negthinspace\nolinebreak) \negthinspace\negthinspace Eu\-c\-li\-d\-ean $n$-dimensional space
(of an arbitrary signature), the Hodge operator ``$\ast$'' is defined as
$(\ast w)_{\rho_{k+1} \dots \rho_{n}}=\frac1{(n-k)!}
e_{\rho_{1} \dots \rho_{n}}w^{\rho_{1} \dots \rho_{k}}$.

Consider now a variational problem in parametric form, set by a Lagrangian
$$
\ell(\tau, x^{\rho}, u^{\rho}, \dots,{\stackrel{\scriptscriptstyle r-1}{u}}{}^{\rho}){\bf d}\tau
$$
on the space $J_{r}(\R;M)$. As long as we constrain ourselves only
to the case of autonomous Euler-Poisson equations,
\begin{equation}\label{25}
{\cal E}_{\rho}=0\,,
\end{equation}
the differential form
\begin{equation}\label{27}
\bmat{\varepsilon}={\cal E}_{\rho}\,{\bf d}x^{\rho}\otimes{\bf d}\tau\,,
\end{equation}
may {\em globally\/} be deprived of the factor ${\bf d}\tau$, constituting
thus a {\em globally\/} defined $T^{\ast}M$-valued density
\begin{equation}\label{28}
\bmat e={\cal E}_{\rho}{\bf d}x^{\rho}\,.
\end{equation}

Let $C^1_rM$ denote the manifold of the $r^{\rm th}$ order contact germs
of one-dimensional submanifolds of $M$.
The projection $\wp:T^{1}_{r}M\setminus0\to C^{1}_{r}M$ can be employed to
generate an autonomous variational problem set over $T^{1}_{r}M$ from every one
variational problem over $C^{1}_{r}M$.

\begin{lemma}[\cite{Kal}]
In terms of a local chart, if in (\ref{19}) the local semi-basic differential
form ${\bmat{\epsilon}}$ corresponds to the Lagrangian
$$
{\lambda}=L{\bf d}t\,,
$$
then the vector valued density
\begin{equation}\label{29}
\bmat e=-u^{a}({\sf E}_{a}\circ\wp){\bf d}x^{\scriptscriptstyle0}
{\,+\,}u^{\scriptscriptstyle0}({\sf E}_{a}\circ\wp){\bf d}x^{a}
\end{equation}
corresponds to the Lagrangian
$$
\ell(\tau, x^{\rho}, u^{\rho}, \dots,{\stackrel{\scriptscriptstyle r-1}{u}}{}^{\rho}){\bf d}\tau
={\cal L}(x^{\rho}, u^{\rho}, \dots,{\stackrel{\scriptscriptstyle r-1}{u}}{}^{\rho}){\bf d}\tau
$$
with the Lagrange function
\begin{equation}\label{30}
{\cal L}=u^{\scriptscriptstyle0}L{\,\circ\,}\wp\,.
\end{equation}
\end{lemma}

\paragraph{4.~The classical spinning particle.}
Let in (\ref{22}) generator ${\x}$ correspond to the (pseu\-do-\negthinspace\nolinebreak) \negthinspace\negthinspace or\-tho\-go\-nal
transformations of a four-dimensional (pseu\-do-\negthinspace\nolinebreak) \negthinspace\negthinspace Eu\-c\-li\-d\-ean space. In this case,
equations (\ref{24}) have no solutions of the third order, that is to say,
no invariant system of Euler-Poisson equations (\ref{18}) consists of at least one equation
of the third order.

Nevertheless, if we allow a vector parameter $\bmat s=(s^{\scriptscriptstyle0},\hbox{\ssb s})$
(that is, a constant
quantity, transforming as a four-vector under the action of the (pseu\-do-\negthinspace\nolinebreak) \negthinspace\negthinspace or\-tho\-go\-n\-al
group) enter into expressions (\ref{20}), then equations (\,\ref{21}\,\&\ref{24}\,)  turn out to
possess a family of solutions, which depend on $\bmat s$ and actually contain some
third-order derivatives. In~\cite{Thesis,Dokl,GR,NTSh2014} we succeeded to obtain the following one (in an unessential
assumption $g^{\scriptscriptstyle00}=+1$)
\begin{eqnarray}
\lefteqn{
\hbox{\ssb E}\;=\;
\frac{\hbox{\ssb v}^{{{\prime}}{{\prime}}}\times(\hbox{\ssb s}
-s_{\scriptscriptstyle0}\hbox{\ssb v})}
{\bigl[(1+\hbox{\ssb v}{{\bcdot}}\hbox{\ssb v})(s_{\scriptscriptstyle0}{}^{2}
+\hbox{\ssb s}{{\bcdot}}\hbox{\ssb s})-(s_{\scriptscriptstyle0}+\hbox{\ssb s}{{\bcdot}}\hbox{\ssb v})^{2}\bigr]^{3/2}}
}\nonumber\\
&&\hfill\;-\;3\,\frac{(s_{\scriptscriptstyle0}{}^{2}+\hbox{\ssb s}{{\bcdot}}\hbox{\ssb s})
\;\hbox{\ssb v}^{{\prime}}\!{{\bcdot}}\hbox{\ssb v}-(s_{\scriptscriptstyle0}+\hbox{\ssb s}{{\bcdot}}\hbox{\ssb v})\;\hbox{\ssb s}{{\bcdot}}\hbox{\ssb v}^{{\prime}}  }
{\bigl[(1+\hbox{\ssb v}{{\bcdot}}\hbox{\ssb v})(s_{\scriptscriptstyle0}{}^{2}
+\hbox{\ssb s}{{\bcdot}}\hbox{\ssb s})-(s_{\scriptscriptstyle0}+\hbox{\ssb s}{{\bcdot}}\hbox{\ssb v})^{2}\bigr]^{5/2} }
\;\hbox{\ssb v}^{{\prime}}\times(\hbox{\ssb s}-s_{\scriptscriptstyle0}\hbox{\ssb v})
\hfill\nonumber\\
&&\hfill\label{35}\;+\;m\,\frac{(1+\hbox{\ssb v}{{\bcdot}}\hbox{\ssb v})\,\hbox{\ssb v}^{{\prime}}
-(\hbox{\ssb v}^{{\prime}}\!{{\bcdot}}\hbox{\ssb v})\,\hbox{\ssb v}}
{(1+\hbox{\ssb v}{{\bcdot}}\hbox{\ssb v})^{3/2}(s_{\scriptscriptstyle0}{}^{2}+\hbox{\ssb s}{{\bcdot}}\hbox{\ssb s})^{3/2}}\;.
\end{eqnarray}

Applying the prescription (\ref{29}) the above expression (\ref{35}) produces the corresponding
four-dimensional expression and thus the desired Euler-Poisson equations
\begin{equation}\label{36}
\fbox{$
\displaystyle
{\bmat{\mathcal E}}=
\frac{\ast\,\bmat{\ddot{ u}}\wedge\bmat u\wedge\bmat s}
{\|\bmat s\wedge\bmat u\|^3}
\;-\;3\,\frac{\ast\,\bmat{\dot{ u}}\wedge\bmat u\wedge\bmat s}
{\|\bmat s\wedge\bmat u\|^5}\,(\bmat{\dot{ u}}\wedge\bmat s){{\bcdot}}(\bmat u\wedge\bmat s)
+\frac m{\|{\bmat s}\|^3}\left[\frac{\bmat{\dot{ u}}}{\|\bmat u\|}
\,-\,\frac{\bmat{\dot{ u}}{{\bcdot}}\bmat u}{\|\bmat u\|^3}\,\bmat u
\right]=\boldsymbol0
$}
\end{equation}

Assume ${\x}_{_{\bmat 3}}$ denote the third-order prolongation of the generator
${\x}$ (\ref{22})
of (pseu\-do-\negthinspace\nolinebreak) \negthinspace\negthinspace or\-tho\-go\-nal transformations with the group parameters
$\hbox{\ssb n}$, $\hbox{\ssb q}$ to the space $J_{3}(\R;\R^3)$,
\begin{eqnarray}\label{37}
{\x}_{_{\bmat 3}}&=&-\,(\hbox{\ssb q}{{\bcdot}}\hbox{\ssb s})\,\partial_{_{{\scriptstyle s}^{\scriptscriptstyle0}}}
\,+\,s^{\scriptscriptstyle0}\,(\hbox{\ssb q}\,\bkey.\,\bpartial_{_{\eisf s}})
\,+\,\bkey[\hbox{\ssb n},\hbox{\ssb s},\bpartial_{_{\eisf s}}\bkey]
\,-\,(\hbox{\ssb q}{{\bcdot}}\hbox{\ssb x})\,\partial_{_{\scriptstyle t}}
\,+\,t\,(\hbox{\ssb q}\,\bkey.\,\bpartial_{_{\eisf x}})
\,+\,\bkey[\hbox{\ssb n},\hbox{\ssb x},\bpartial_{_{\eisf x}}\bkey]
\nonumber\\
&+&(\hbox{\ssb q}\,\bkey.\,\bpartial_{_{\eisf v}})
\,+\,(\hbox{\ssb q}{{\bcdot}}\hbox{\ssb v})\,(\hbox{\ssb v}\,\bkey.\,\bpartial_{_{\eisf v}})
\,+\,\bkey[\hbox{\ssb n},\hbox{\ssb v},\bpartial_{_{\eisf v}}\bkey]
\nonumber\\
&+&2\,(\hbox{\ssb q}{{\bcdot}}\hbox{\ssb v})\,(\hbox{\ssb v}^{{\prime}}\!\bkey.\,\bpartial_{_{\eisf v^\prime}})
\,+\,(\hbox{\ssb q}{{\bcdot}}\hbox{\ssb v}^{{\prime}})\,(\hbox{\ssb v}\,\bkey.\,\bpartial_{_{\eisf v^\prime}})
\,+\,\bkey[\hbox{\ssb n},\hbox{\ssb v}^{{\prime}}\!,\bpartial_{_{\eisf v^\prime}}\bkey]\,.
\end{eqnarray}

The following assertions are true:
{\em
\begin{enumerate}
\item
Vector quantity ${\bmat{\mathcal E}}$ in (\ref{36}) constitutes a system of Euler-Poisson expressions;
\item
Let vector field ${\x}_{_{\bmat 3}}$  be given by (\ref{37}), and let $\hbox{\ssb E}$
be given by (\ref{35}) (recall that (\ref{35}) is nothing more but merely (\ref{36}),
parametrized with respect to the coordinate time $t=x^{\scriptscriptstyle0}$).
The relation
$$
{\bf L}({\x}_{_{\bmat 3}})(\hbox{\ssb E})\;=\;\hbox{\ssb n}\times\hbox{\ssb E}
\;+\;(\hbox{\ssb q}{{\bcdot}}\hbox{\ssb v})\;\hbox{\ssb E}
\;-\;(\hbox{\ssb v}{{\bcdot}}\hbox{\ssb E})\;\hbox{\ssb q}
$$
proves the invariance of the Euler-Poisson equations
(\ref{36}), and at the same time shows that whatever a possible
Lagrange function for (\ref{36}) might exist, it by no means will reveal invariance
under (pseu\-do-\negthinspace\nolinebreak) \negthinspace\negthinspace or\-tho\-go\-nal transformations, even in the generalized sense (that
is up to a total derivative term).
\item
Equations (\ref{36}) describe (in metric signature $2$) the motion of relativistic spinning
free particle with constant spin four-vector $\bmat s$ and with the rest mass
$$
m_{\scriptscriptstyle0}\;=\;m\left[1\,-\,\frac{(\bmat s{{\bcdot}}\bmat u)^{2}}
{(\bmat s{{\bcdot}}\bmat s)\,(\bmat u{{\bcdot}}\bmat u)}\right]^{3/2}_{^{\ \ \ {\displaystyle.}}}
$$
\end{enumerate}
}

\subparagraph{A comment on the order of the Euler-Lagrange form.}
Euler-Lagrange equations are polynomial with respect to the derivatives of orders
greater than the maximal order of the derivatives which enter in the corresponding
Lagrange function. It was this property that inspired us to try to diminish
the order of the underlying manifold $J_{s}(\R,\R^{q})$ from $s=3$ to
$s=2$ by means of introducing  the differential form
${\underline{\bmat{\epsilon}}}$ in (\ref{33})
rather than ${\bmat{\epsilon}}$.

config: arxiv


TITLE: Show that if $|ε|$ is small enough, then all solutions are bounded of the ODE
QUESTION [1 upvotes]: I am trying to show that if $|ε|$ is small enough, then all solutions are bounded of the ODE:
$$x'' +  [1 + ε \sin 3t]  x = 0$$
I can't figure out how to do this because I can't find a way to solve the equation. Do I need to just show that $x''+x=0$ solutions are bounded so then a small epsilon won't mess up the boundedness?

REPLY [3 votes]: Truly, the differential equation :
$$x'' +  [1 + ε \sin (3t)]  x = 0$$
cannot be solved in terms of standard mathematical functions.
Multiplying both sides by $x'$, we get :
$$x''\cdot x' + [1 + ε \sin (3t)]  x\cdot x'=0 \Rightarrow \frac{1}{2}[(x')^2]'+[1+ε\sin(3t)]\cdot \frac{1}{2}(x^2)'=0 $$
Integrating both sides from $0$ to $t$, we'll get :
$$[x'(t)]^2+ \int_0^t [1+ε\sin(3s)]\cdot \frac{1}{2}(x^2)'ds = [x'(0)]^2$$
For small enough $|ε|$, note that $\int_0^t [1+ε\sin(3s)]\cdot \frac{1}{2}(x^2)'ds \geq 0$, so :
$$[x'(t)]^2 \leq     [x'(0)]^2$$
Does this lead us to a conclusion about $x(t)$ ?


TITLE: A weak version of Markov-Kakutani fixed point theorem
QUESTION [1 upvotes]: Let $\emptyset \not = X\subseteq \Bbb{R}^n$ be convex and compact and let $\cal{A}$ be a commuting family of affine maps from $\Bbb{R}^n$ into $\Bbb{R}^n$ such that $X$ is invariant under each element of $\cal A$. How can I show that $\cal{A}$ has a common fixed point.
Remark I know this is a consequence of some stronger theorems such as Markov-Kakutani fixed point theorem. But I have no access to its proof. Besides I am looking for a much elementary proof (which I think must exists for this special case).

REPLY [2 votes]: For each $A \in \mathcal{A}$, let $K_A =\{ x \in X : A(x) = x\}$ the set of fixed points of $A$ in $X$. Since $A$ is continuous, $K_A$ is closed (hence compact), and since $A$ is affine, $K_A$ is convex. By - for example - Brouwer's fixed point theorem(1), $K_A \neq \varnothing$.
For $B \in \mathcal{A}$ and $x \in K_A$, we have
$$A(B(x)) = B(A(x)) = B(x),$$
i.e. $B(K_A) \subset K_A$, and since $K_A$ is convex and compact, $B$ has a fixed point in $K_A$.
From that, we can deduce that the family of convex compact sets $\mathcal{K}_\mathcal{A} = \{ K_A : A \in \mathcal{A}\}$ has the finite intersection property, $$\mathcal{F} \subset \mathcal{A} \text{ finite} \Rightarrow \bigcap_{A \in \mathcal{F}} K_A \neq \varnothing.$$
Therefore,
$$K := \bigcap_{A \in \mathcal{A}} K_A \neq \varnothing.$$
$K$ is the set of common fixed points of all $A \in \mathcal{A}$.

(1) Here, we can also prove directly that each $A$ has a fixed point in $X$. Let $\pi_r \colon \mathbb{R}^n \to \mathbb{R}$ the coordinate projection. Let $m_r(K) = \min \{ \pi_r(x) : x \in K\}$ and $M_r(K) = \max \{ \pi_r(x) : x \in K\}$ for any compact nonempty $K$. Let $X_0 = X$, and for $1 \leqslant r \leqslant n$, let $X_r = \{ x \in X_{r-1} : \pi_r(A(x)) = \pi_r(x)\}$. Since $A$ is continuous, each $X_r$ is closed in $X_{r-1}$, hence compact, and since $A$ is affine, each $X_r$ is convex. It remains to see that $X_{r-1} \neq \varnothing \Rightarrow X_r \neq \varnothing$.
Let $\varphi_r(x) = \pi_r(A(x)) - \pi_r(x)$. On $X_{r-1} \cap \pi_r^{-1}(m_r(X_{r-1}))$, we have $\varphi_r \geqslant 0$, and on $X_{r-1} \cap \pi_r^{-1}(M_r(X_{r-1}))$ we have $\varphi_r \leqslant 0$. Since $\varphi_r$ is continuous and $X_{r-1}$ connected, $X_r = X_{r-1} \cap \varphi_r^{-1}(0) \neq \varnothing$.


TITLE: Convergence in cdf and pdf
QUESTION [0 upvotes]: This question will probably be trivial to most. However I failed to find a clear answer on the internet.
When one says that n random variables converge "in distribution", does it mean that this random variables converge to the same cumulative distributive function (cdf) ? What does it say about their convergence in probability density function (pdf) ? Can two random variables have the same cdf but different pdf or the other way around ?

REPLY [2 votes]: Given a sequence of random variables, say $(X_n)_{n \in \mathbb{N}}$. We will say this sequence converges in distribution to the random variable $X$ if we take the sequence $(F_n(x))_{n \in \mathbb{N}}$ where $F_i(x)$ is the CDF of $X_i$ and have the following result

$\displaystyle \lim_{n \to \infty } F_n(x) = F(x)$ where $F(x) $ is the CDF of $X$.

Note: If we only have $m$ many random variables then we can form a sequence by taking $X_1,X_2, \ldots X_m$ and for each successive term after $X_m$, set $X_{m+j} = X_i$ where $j \in \mathbb{N}$ and $X_i \in \{ X_1, \ldots X_m \}$ and such a sequence trivially converges in distribution to $X_i$
If we have that $f_i(x)$ is the density of $X_i, \forall i$. That is if we have that the density exists for each random variable in our sequence then from Scheffé’s theorem we get that the convergence of the densities will imply convergence in distribution.
For the last question, say we have two continuous random variables of which both have a density function. Namely, say we have $X \sim F_X(z) $ and $Y \sim F_Y(z)$  then the denisties will be, respectively, $F'_X(z) = f_X(z)$ and $F'_Y(z) = f_Y(z)$.
Hence if $F_X(z) = F_Y(z)$ then $f_X(z) = f_Y(z)$. We can see the converse will hold in this case as well.


TITLE: Convergence of convolution of $L^p$ function with a sequence of distributions
QUESTION [3 upvotes]: let $h_n\in C_c^\infty (\mathbb{R}^d)$ s.t. $\int h_n dm = 1$ and $\operatorname{supp}(h_n)\to {0}$.
I've proven that $h_n\to\delta_0$ in $\mathcal{D}'(\mathbb{R}^d)$, now I'm trying to show that for any $f\in L^p(\mathbb{R}^d)$ where $0\le p<\infty$ I get that $f\star h_n\to f$ in $L^p$.
I tried just manually calculating $\|f\star h_n -f\|_p^p$ and showing that it is bound by something which vanishes as $n$ ascends, but not much luck there. I am not even sure if that is the right approach but I don't have any other ideas.
Any hint would be appreciated.

REPLY [2 votes]: Hint: for $p\in [1,\infty)$
1 - Prove that $C_c(\mathbb{R^d})$ is dense in $L^1(\mathbb{R}^d)$. You can find the proof here, in page 72, Theorem 1.3.20,
2 - By using item 1, prove that $C_c(\mathbb{R}^d)$ is dense in $L^p(\mathbb{R}^d)$,
3 - Prove that if $f\in C(\mathbb{R}^d)$, then $h_n\star f\rightarrow f$ uniformly on compact sets of $\mathbb{R}^d$,
4 - By using item 2 and 3, you can conclude.


TITLE: Series of Functions; Uniform Convergence
QUESTION [0 upvotes]: Problem:
Prove If the sequence of numbers $\{a_{n}{\}}_{n=1}^{\infty}$ satisfies $\sum_{n=1}^{\infty} \lvert a_n \rvert \lt \infty,$ then the series $\sum_{n=1}^{\infty}  a_ncos(nx)$ converges uniformly on $[0,2\pi]$. This means, the partial sums
$$
s_N(x) = \sum_{n=1}^N a_ncos(nx)
$$
define a sequence of functions $\{s_{N}{\}}_{n=1}^{\infty}$ that converges uniformly on $[0,2\pi]$. Hint: First show that the sequence is Cauchy w.r.t $\lVert  \cdot  \rVert_{\infty}$
$\\$
My thoughts:
I think I need to use the following property -
Property: If a sequence $\{f_{n}{\}}_{n=1}^{\infty}$ in $C([a,b])$ is Cauchy w.r.t $\lVert \cdot \rVert_{\infty}$ (uniformly cauchy) then it is uniformly convergent.
To be honest, the above (and hint) is the part I'm stuck on.
For just $g(x) = \sum_{n=1}^{\infty}  cos(nx)$ I feel like I could do this sort of proof.
W.L.O.G., let $k \lt L$
Now: $\lvert \sum_{n=1}^{L} g_n(x) - \sum_{n=1}^{k}  g_n(x) \rvert = \lvert \sum_{n=k+1}^{L} g_n(x) \rvert \le \int_k^L 1 \ dx \lt x \  \Big|_k^L$ = L - k
Anyway, I'm probably way off course at this point and I'm having a hard time actually figuring out how to go about this.

REPLY [0 votes]: This is just the Weierstrass $M$-test which states that $\sum f_n(x)$ converges uniformly on a set $E$ if $|f_n(x)|\le M_n$ for all $x\in E$ and $\sum M_n < \infty$. In this case, $|a_n\cos(nx)|\le |a_n|$ and $\sum |a_n| < \infty$.
To prove the $M$-test, you can prove that $\{s_N = \sum_{n=1}^Nf_n(x)\}_{N=1}^\infty$ is Cauchy by looking at $|\sum_{n=m}^p f_n(x)| \le \sum_{n=m}^p|f_n(x)| \le \sum_{n=m}^p M_n$, and use the fact that $\sum M_n < \infty$, so in particular $\{\sum_{n=1}^NM_n\}_{N=1}^\infty$ is Cauchy.


TITLE: Decide if the given endomorphism $F:V\rightarrow V$ is normal
QUESTION [0 upvotes]: Decide if the given endomorphism $F:V\rightarrow V$ is normal
Let $V := \mathbb{C}[x]_{\leq 2}$ with $ \langle g,h\rangle := \int_{-1}^{1}g(x)\overline{h(x)}dx$ and $F$ given by the derivative $F(g):=g'$

I would love to get some tipps on how to solve this one.
Any help is appreciated

REPLY [1 votes]: Tip: Your operator satisfies $F^3 = 0$ so $F$ is nilpotent. When can a nilpotent operator be normal?


TITLE: physics of the 1-dimensional wave equation
QUESTION [1 upvotes]: I was given the task to solve the 1-dimensional wave equation
$$
\partial_{tt}u-\partial_{xx}u=0
$$
with the conditions
\begin{align}
 u(x,0)&=f(x), \\
 \partial_tu(x,0)&=0,
\end{align}
where $\mathrm{supp}f\subseteq[-1,1]$.
By factorizing the equation into
$$
(\partial_t-\partial_x)(\partial_t+\partial_x)u=0
$$
and using new variables, and then applying the initial conditions, I get the general solution
$$
u(x,t)=\frac{1}{2}(f(x+t)+f(x-t)).
$$
Now I'm asked to interpret the solution the following way: If I'm sitting on the real line at point $x=10$, at what time will I be able to see the wave for the first time and how long am I able to observe it?
My guess would be that I don't see the wave at all, since $f\equiv0$ outside of $[-1,1]$ but maybe I'm misinterpreting that here.

REPLY [1 votes]: The wave is observable when at least one of $x\pm t$ is in $f$'s support, i.e. from $t=-11$ to $t=-9$, then again from $t=9$ to $t=11$.


TITLE: $n$-th derivative of $x^\alpha$ where $\alpha = m + 1/2$
QUESTION [3 upvotes]: It is well-known that, for any real $\alpha$ and nonnegative integer $n$
$$ \frac{d^n x^\alpha}{dx^n} = \alpha(\alpha-1)\cdots(\alpha - n + 1) x^{\alpha - n}  $$
I just found out that the coefficient is known as a falling factorial
$$ \alpha(\alpha-1)\cdots(\alpha - n + 1) =: (\alpha)_n $$
where $(\alpha)_n$ is Pochhammer's symbol.
It seems that in the case $\alpha = 1/2$ and more generally $\alpha = m + 1/2$ where $m$ is a nonnegative integer, it is possible to express $(\alpha)_n$ with powers of 2 and standard (integer) factorials.  For example:
$$ (1/2)_4 = \frac12\bigg(-\frac12\bigg)\bigg(-\frac32\bigg)\bigg(-\frac52\bigg) =- \frac{1\times3\times5}{2^4} = -\frac{5!}{2^4\times(2\times4)} =- \frac{5!}{2^4\times2^2\times 2!} $$
Would anyone know a good reference to such formulas?  I am guessing the cases $n>m$ and $n<m$ must be distinguished.
Thanks!
p.

REPLY [1 votes]: Notice that:
\begin{align}\left(m+\frac12\right)_n&=\prod_{k=0}^{n-1}\left(m+\frac12-k\right)\\&=\prod_{k=0}^{n-1}\frac{\color{#BB4466}{2m+1-2k}}2\frac{\color{#BB4466}{2m-2k}}{2\color{#66AA99}{(m-k)}}\\&=\frac1{4^n}\left(\prod_{k=0}^{n-1}\frac1{\color{#66AA99}{m-k}}\right)\left(\prod_{k=-1}^{2n-2}\color{#BB4466}{(2m-k)}\right)\\&=\frac1{4^n}\frac{(m-n)!}{m!}\frac{(2m+1)!}{(2m-2n-3)!}\end{align}
provided that $m\ge n$ to avoid $0/0$. If $m<n$, we then re-interpret the factorials using
$$\frac{(m-n)!}{(2m-2n-3)!}=\frac1{(2m-n-3)_{(m-3)}}$$
where $(a)_0=1$ and $(a)_{n+1}=(a)_n(a-n)$ recursively defines the falling factorial to negative values if $0\le m\le3$.


text:
\begin{definition}[Definition:Triangular Matrix/Upper Triangular Matrix]
An '''upper triangular matrix''' is a [[Definition:Matrix|matrix]] in which all the [[Definition:Lower Triangular Element|lower triangular elements]] are zero.
That is, all the non-zero [[Definition:Element of Matrix|elements]] are on the [[Definition:Main Diagonal|main diagonal]] or in the upper triangle.
That is, $\mathbf U$ is '''upper triangular''' {{iff}}:
:$\forall a_{ij} \in \mathbf U: i > j \implies a_{ij} = 0$
\end{definition}

config: wiki


TITLE: Finding maximal number of bad triplets
QUESTION [0 upvotes]: Let $a,b,c\in \mathbb{F}_{3^n}$. The summation of two vectors is done with modulo $3$. The elements of vectors are $0,1$ or $2$. We will say that $a,b,c$ form a bad triplet if $a\neq b,a\neq c,b\neq c$ and $a+b+c=0$.
Let $B_k$ be maximal number of bad triplets of set that has $k$ vectors. So we should add vectors to set so that number of total bad triplets in set is maximized.
I have calculated some values of $B_k$.
$B_1 = B_2 = 0,\ B_3 = B_4=1,\ B_5=2,\ B_6=3,\ B_7=5,\ B_8=8,\ B_9=12\ ...$
I need formula for $B_k$ or some evaluation of it. Tried here but no result.
For example (for $n = 2$) the set that produces $B_6=3$  may be $\{(0,0),(1,0),(2,0),(0,1),(1,1),(0,2)\}$
$(0,0)+(1,0)+(2,0) = (0,0)$
$(0,0)+(0,1)+(0,2) = (0,0)$
$(0,2)+(1,1)+(2,0) = (0,0)$
So we have $3$ bad triplets.
I believe that value of $B_k$ doesn't depend on $n$ (size of vectors).
I have already showed that $B_k \le$ $\frac{{k} \choose {2}}{3}$ but need more correct formula.

REPLY [1 votes]: Considering only the $j^{\operatorname{th}}$ coordinate, there are only two possibilities: either $a_j=b_j=c_j$, or all three are distinct.  In other words, you are counting the maximum number of Sets that may be made with $k$ cards and $n$ properties!
I know of a decent amount of research about the opposite problem, that of finding the largest possible subsets of $\mathbb{F}_3^n$ that contain no such triples.  Here is a post by Terence Tao on the subject, for example.  But I don't know offhand about research specifically on your problem.
Note that ${a,b,c}$ is a "bad" triplet if and only if it forms a line in $\mathbb{F}_3^n$.  This is an important geometric interpretation that might help in finding similar work.  Who knows, maybe somebody has applied algebraic geometry to this...


TITLE: How many increasing sequences of natural numbers are there that start with 1, end with 99, and have at least one even number? Prove your answer.
QUESTION [2 upvotes]: The first thought that came to mind was casework, but there are way too many cases, with the number of even integers ranging from 1 to 45. My second thought was to do complementary counting and count the number of sequences where there were no even integers, so there would be only odd. However, there are still tons of cases to consider. What do I use?

REPLY [4 votes]: The answer, ignoring the condition, is $2^{97}$; i.e. every subset of $\{2,3,\ldots, 98\}$.  We append this subset to $\{1,99\}$, and then order sequentially.
The answer, failing to meet the condition, is $2^{48}$, i.e. every subset of $\{3,5,7,\ldots, 97\}$.  We append this subset to $\{1,99\}$, and then order sequentially.
The answer you seek is the difference of the above.


TITLE: How to find the inverse cosine without a calculator
QUESTION [11 upvotes]: How to find the inverse of:
$$\cos(c)=\frac{1}{3}$$
In other words, i'm trying to solve for c and without a calculator. If it's hard or not possible, then how would you go about solving inverses in general. For example, lets say:
$$\cos(c)=\frac{3+1}{2\sqrt{2}}$$
how would we solve for c, which in this case is 105

REPLY [2 votes]: $$\arccos x =\dfrac{\pi}{2}-\displaystyle\sum_{n=0}^{\infty} \dfrac{(2n)!}{2^{2n}(n!)^2(2n+1)}x^{2n+1}=\dfrac{\pi}{2}-\biggl(x+\dfrac{1}{2}\dfrac{x^3}{3}+\dfrac{1}{2\cdot4}\dfrac{x^5}{5}+\underbrace{\cdots}_{o(x)}\biggr), \mid x\mid \lt 1$$

$$\arccos\biggl(\dfrac{1}{3}\biggr)\approx\dfrac{\pi}{2}-\biggl(\dfrac{1}{3}+\dfrac{1}{162}+\dfrac{1}{9720}+o\biggl(\dfrac{1}{3}\biggr)\biggr)\approx 1.231290=70.54773^{\circ}$$


TITLE: Equicontinuous set of functions.
QUESTION [3 upvotes]: This question comes up after going over Arzela-Ascoli theorems.
For a set of continuous functions $\mathbb F$ from $\mathbb R$ to $\mathbb R$  that is equicontinuous. How do I show that if sup{$|f(0)|:f \in F$} $< \infty$ , then $\mathbb F$ is pointwise bounded?
I know I need to show that since sup{$|f(0)|:f \in F$} = $M_0 < \infty$ then for each $x \in \mathbb R$ sup{$|f(x)|:f \in F$} = $M_x < \infty$.
What I have gotten so far is that I should fix $\epsilon$ and use the fact that the domain is $\mathbb R$.
I think I just need some help using equicontinuity and sup together.

REPLY [1 votes]: As an alternative to mookid's answer (which looks good), consider this one, which uses compactness (actually, mookid's answer reminds me of the proof of the compactness of $[0, 1]$, so maybe I'm just doing this at a less fundamental level).
Anyway, consider any $x \in \mathbb{R}$, and consider the set $[0, x]$ (we suppose WLOG that $x > 0$; if $x < 0$, we just examine $[-x, 0]$ and the rest of the proof is the same).
For $\varepsilon = 1$, for each $y \in [0, x]$, choose $\delta_y > 0$ so that $z \in (y - \delta_y, y + \delta_y) \Rightarrow |f(z)-f(y)| < 1$ for each $f \in F$; $\delta_y$'s existence is guaranteed by the equicontinuity of $F$.
Define $A_y = (y - \delta_y, y + \delta_y)$. Then $\{A_y\}_{y \in [0,x]}$ is an open cover of $[0, x]$, a compact set, and thus has a finite subcover, $\{A_{y_i}\}_{i=1}^n$.
Now, we know $M_0 < \infty$. If we order the $y_i$'s so that $y_1 < y_2 < \ldots < y_n$, then by the definition of equicontinuity of $F$, we know that $y \in A_{y_1} \Rightarrow |f(y)| \leq M_0 + 1$ for all $f \in F$, and more generally, $y \in A_{y_i} \Rightarrow |f(y)| \leq M_0 + i$ for all $f \in F$ (this might require a more formal proof--perhaps one using induction--but it shouldn't be too difficult). So, since $x \in A_{y_i}$ for some $i \in \{1, \ldots, n\}$, we know that $|f(x)| \leq M_0 + n$ for all $f \in F$, and therefore that $M_x \leq M_0 + n < \infty$.
I think explicitly using compactness makes the intuition a little more clear: because there is a bound for every function in $F$ at $0$, and because there is a bound on how much every function can change in a neighborhood of each point of $\mathbb{R}$, it can never be the case that some subset of $F$ increases locally at such a rate that it "breaks away" from the rest of the group toward infinity.


TITLE: number of ways to fill a 2D grid
QUESTION [0 upvotes]: We have a 2D grid with n rows and m columns, we can fill it with numbers between 1 and k (both inclusive). Only condition is that for each r such that 1<=r<=k ,no two rows must have exactly the same number of the number r. We have to find the number of ways in which we can fill the grid .

REPLY [0 votes]: Well, if no two rows can have the same number, you would be able to have a maximum of $k$ columns, since once you use the numbers from $1$ to $k$ in a row, you can't go any farther.
I think besides that there are no restrictions, so for a maximum size board you would have $nk!$ ways to do it.
If you mean that you can't have any numbers repeating in a row or column, then it would be a totally different story.


TITLE: is there anything intelligent to say about the following span of vectors?
QUESTION [1 upvotes]: Let's say I have a set of real vectors $v_1,\ldots,v_n$ such that $\sum_j v_{ij} = 1$ for all $i$ and $v_{ij} \ge 0$.
Now consider the set $\Gamma(n) = \{ \beta \mid \sum_i \beta_i = 1, \beta_i \ge 0 \}$, i.e. the set of vectors of dimension $n$ in the probability simplex.
Is there anything interesting to say about the span $\{ \sum_i \beta_i v_i \mid \beta \in \Gamma(n) \}$?
Under all kind of different conditions... let's say $v_i$ are independent, or that $n$ is larger than the length of each $v_i$, or anything at all. I am trying to see what properties I can have from such a span.
Thanks!

REPLY [1 votes]: The vectors $v_i$ are arbitrary points in the simplex of the unknown dimension of the space (since you write sums I will assume that we are in an $\mathbb R^m$).
The span consists of all convex combination of these points, so you get all possible convex sets contained in the standard simplex.
Certainly, the convex sets live at most in affine dimension $n-1$ and they have at most $n$ vertices, but they can be smaller and have less vertices if some vectors are in the span of other vectors.


TITLE: Do metals generally lose more heat by convection than non-metals?
QUESTION [1 upvotes]: I've been experimenting with radiative cooling lately and in my mind convection is always an enemy (especially forced convection since the modules are outdoors).
Looking at Newton's law of cooling, it doesn't seem to me that it accounts for surfaces made of different materials losing/gaining heat (can be wrong though, not uncommon especially when working with formulas).
Does it matter (much), that surfaces are made from different materials when we talk about convective heat gains/losses?
I'm only saying this because metal cooled more under radiation than a type of plastic I used, and to me that doesn't make much sense as the convective heat gain should, in my mind, be greater in the metal panel and therefore it shouldn't cool more than the plastic one (both had similar emissivity because of a very thin paint coating).
My experiments tell me that I'm most likely wrong in assuming the convective heat gains/losses will be greater in metals, but I thought it's most likely better to check with those who actually know what they're talking about, it may be some weird fluke in the tests.

REPLY [1 votes]: I'd expect metals to cool faster through convection because of a related heat property of theirs: conduction.  Metals are generally good heat conductors so as heat energy is removed from the surface of the metal by air, internal heat energy in the metal can quickly flow to the surface.
The rate heat flows between two systems is dependent on their temperature difference.  When air removes heat via convection, the surface of the object is cooler than the central portion of the object.  If heat in the object flows slowly from the center to the surface like in many plastics then the surface stays closer to air temperature and heat transfers to the surrounding air more slowly.  If the object is a metal, the heat removed at the surface is easily replaced by heat flowing from the center to the surface and the temperature delta between the surface and the air is greater so the object cools faster.


TITLE: Number of ways in which $5$ boys and $5$ girls are ordered in such a way that exactly $4$ girls stand consecutively in the queue
QUESTION [8 upvotes]: Let $n$ be the number of ways in which $5$ boys and $5$ girls can stand in a queue in such a way that all the girls stand consecutively in the queue. Let $m$ be the number of ways in which $5$ boys and $5$ girls in such a way that exactly $4$ girls stand consecutively in the queue. Then find $\frac{m}{n}$.

I could find $n$ correctly as $6!\times 5!$, but I am finding $m$ as $7!\times 4!\times \binom{5}{1}$, but the book says $m$ is $6!\times 4!\times 5\times \binom{5}{4}$. Where have I gone wrong? What is the proper logic to find $m$?

REPLY [5 votes]: You can line up the boys in $5!$ ways.
Next choose one of the $6$ gaps for the $4$ girls who are together.
Then choose the $4$ girls and arrange them in order, which can be done in $\binom{5}{4}4!=5!$ ways.
Finally, choose one of the $5$ gaps left for the remaining girl.

This gives $5!\cdot6\cdot5!\cdot5=5!\cdot6!\cdot5=432,000$ possibilities.


TITLE: Find impulse provided by the stopping object
QUESTION [0 upvotes]: I have tried to understand this problem, but I'm still struggling with it.
a 27g bullet hits a piece of wood and goes in 8.6 cm. if the bullet stopped in 0.44ms. what impulse did the wood provide?

REPLY [1 votes]: As the person commented on your post, we don't work homework problems like this for you. However, I can briefly explain what impulse is. The definition of impulse is $J = \int^{t_1}_{t_0} F(t) dt = (t_1-t_0)F_{avg} = \int^{t_1}_{t_0} \frac{dp}{dt}dt = p(t_1) - p(t_0)$ by mean value theorem and the integral theorem of calculus respectively. So you need to find either the average force applied to the bullet, or the momentum before it entered the wood. I'll leave it up to you to determine which is appropriate, as that is the point of this problem.


TITLE: Is there an infinitary sentence which is absolutely not second-order expressible?
QUESTION [13 upvotes]: This is a "forcing-absolute" followup to this question, whose answer was largely unsatisfying. The question is:

Suppose $V=L$. Is there an $\mathcal{L}_{\infty,\omega}$-sentence $\varphi$ such that in no forcing extension is $\varphi$ equivalent to a second-order sentence?

Throughout, by "forcing" I mean "set forcing," although the (tame) class forcing version also seems potentially interesting.
EDIT: I forgot to add that I'm restricting attention to infinite structures here (which is key to my comment below that every projectively-definable infinitary sentence is second-order expressible). As Fedor Pakhomov commented below, without this restriction the problem is trivial since second-order theories of finite structures can't be changed by forcing. I do not, however, want to restrict attention to countable structures.

I've decided to focus on models of $\mathsf{V=L}$ since that hypothesis seems to add interesting flavor to the question in a few ways:

It implies that such a $\varphi$ must not be in $\mathcal{L}_{\omega_1,\omega}$. This is because $(1)$ every constructible real is projective in some forcing extension and $(2)$ every projectively definable $\mathcal{L}_{\omega_1,\omega}$-sentence is equivalent to a second-order sentence. So a positive answer would have to crucially rely on uncountable Boolean combinations - which seems a bit odd, because each specific $\mathcal{L}_{\infty,\omega}$-sentence is equivalent to some $(\mathcal{L}_{\omega_1,\omega})^{V[G]}$-sentence in an appropriate forcing extension $V[G]$ (just collapse the size of the sentence), but isn't an obvious contradiction since the "potential projectivity" fact about $L$ doesn't seem to lift to arbitrary forcing extensions of $L$.

It rules out the "silly" solution provided by large cardinals. If large cardinals exist - specifically, enough to guarantee projective absoluteness - then any infinitary sentence which is not equivalent to a second-order sentence in $V$ remains so in all forcing extensions. (Note that this would give us an example in $\mathcal{L}_{\omega_1,\omega}$ for that matter.) But $\mathsf{V=L}$ breaks this "hammer," so that we seem to be forced to do some actual work.

If the answer is yes, there is in fact a definable example, namely the least such sentence with respect to the $L$-ordering. Of course this is silly, but it suggests that there might be canonical examples in a more interesting sense. By contrast I could imagine models of $\mathsf{ZFC+V\not=L}$ where the existence of such a sentence is guaranteed nonconstructively (e.g. by a more intricate counting argument), and so no canonical example need exist.

That said, since it seems plausible that the $\mathsf{V=L}$-situation is more difficult to attack than I'd hoped, I'm also interested in results for other extensions of $\mathsf{ZFC}$.

REPLY [5 votes]: For any given finite signature $\Omega$ there is a second-order sentence $\varphi$ of the signature $\Omega$ such that $\mathsf{ZFC}+V=L$ proves that for any $\mathcal{L}_{\infty,\omega}$-formula $\psi$ of the signature $\Omega$ there is a poset $P$ for which it is $\Vdash_P$-forced that for any infinite model $\mathfrak{M}$ we have $\mathfrak{M}\models \varphi$ iff $\mathfrak{M}\models \psi$.
$\varphi$ should be a second-order sentence expressing in all infinite models $\mathfrak{M}$ that the following holds (see below for a more explicit construction of $\varphi$):

there exists the greatest ordinal $\alpha$ such that the value $\aleph_\alpha^{L_\beta}$ is the same for all large enough countable limit ordinals $\beta$;
$\alpha$ is the position in $<_L$ of some $\mathcal{L}_{\omega_1,\omega}$-sentence $\psi$
$\psi$ is true in $\mathfrak{M}$.

Given $\mathcal{L}_{\infty,\omega}$-formula $\psi$ we consider $\alpha_0$ that is the position of $\psi$ in $<_L$. Let $P$ be the Levy collapse of $\aleph_{\alpha_0}$ onto $\omega$. Let us check that for this $\varphi,\psi$, and $P$ we have the desired equivalence.
Indeed let us reason inside $\Vdash_P$. Clearly, $\psi$ became an $\mathcal{L}_{\omega_1,\omega}$-formula.
Observe that $\alpha$ from 1. will always be equal to $\alpha_0$. Indeed, let us choose countable limit $\beta_0>\alpha_0$ such that all  ordinals $\gamma<\alpha_0$ that aren't $L$-cardinals aren't cardinals in $L_{\beta_0}$. Clearly, for any limit $\beta>\beta_0$, we have $\aleph_{\alpha}^{L_\beta}=\aleph_\alpha^L$. For any $\alpha'>\alpha_0$ since $\aleph_{\alpha'}^L$ isn't countable and any countable $\gamma$ we could find limit countable $\beta'$ such that $\aleph_{\alpha'}^{L_{\beta'}}$ is either undefined or has the value $>\gamma$. Hence $\alpha_0$ is the ordinal $\alpha$ from 1.
Using this the proof of semantical equivalence of $\varphi$ and $\psi$ in infinite models is trivial.

Now let me give more detailed description of $\varphi$.
Let $\mathsf{KPUL}_2(\in,\in_u)$ be a second-order formula depending on element variable $\alpha$ and binary predicate symbols $\in,\in_u$ asserting that we have $(\mathsf{KPU}-\mathsf{Foundation})+\textsf{Second-Order-Foundation}+L[\mathfrak{M}]=V$ for the structure where each element of the underlying domain simultaniously represent itself (as an urelement) and some set, $\in$ gives membership of sets in sets, $\in_u$ gives membership of urelements in sets. Formula $\mathsf{Emb}(R,\in,\in_u,\in',\in_u')$ expresses that $\mathsf{KPUL}_2(\in,\in_u)\land \mathsf{KPUL}_2(\in',\in_u')$  and the unary function $f$ gives an end embedding of the admissible set given by $(\in,\in_u)$ into the admissible set $(\in',\in_u')$. Formula $\mathsf{St}(\alpha,\beta,\kappa,\in,\in_u)$ expresses that $\mathsf{KPUL}_2(\in,\in_u)$ and in the corresponding admissible set $\alpha$ is a countable ordinal, $\beta$ is a countable limit ordinal, $\kappa$ is the ordinal that is $\alpha$-th cardinal according to $L_\beta$, and for any $f,\in',\in_u',\beta'$ if $\mathsf{Emb}(f,\in,\in_u,\in',\in_u')$ and $\beta'>f(\beta)$ is a countable limit ordinal according to the admissible set given by $(\in',\in_u')$, then there is $f(\alpha)$-th cardinal according to $L_{\beta'}$ and it is equal to $f(\kappa)$. Formula $\mathsf{MSt}(\alpha,\in,\in_u)$ expresses that $\mathsf{St}(\alpha,\in,\in_u)$ and we don't have $\mathsf{St}(\alpha+1,\in,\in_u)$. Formula $\mathsf{LC}(\alpha,\psi,\in,\in_u)$ expresses that in the admissible set given by $(\in,\in_u)$, $\alpha$ is an ordinal, $\psi$ is a constructible $\mathcal{L}_{\omega_1,\omega}$-sentence of the signature $\Omega$, and its place in $<_L$-order is $\alpha$. Formula $\mathsf{Tr}(\psi,\in,\in_u)$ expresses that in the admissible set given by $(\in,\in_u)$, $\psi$ is an $\mathcal{L}_{\omega_1,\omega}$-sentence of the signature $\Omega$ that is true in the underlying model $\mathfrak{M}$.
We put $\varphi$ to be:
$$\exists \in,\in_u,\alpha,\psi(\mathsf{KPUL}_2(\in,\in')\land \mathsf{MSt}(\alpha,\in,\in')\land \mathsf{LC}(\alpha,\psi,\in,\in')\land \mathsf{Tr}(\psi,\in,\in')).$$


TITLE: What correlation exists between the assumptions $a)$ and $b)$?
QUESTION [2 upvotes]: Let $F\in C^1(\mathbb{R}\setminus\{0\})$ and let $\varepsilon>0$ small. Consider the two assumptions:
$$a)\quad F(s)\ge \frac{1}{s^2}\quad\mbox{ for $0<s<\varepsilon$};$$
$$b)\quad sF^{\prime}(s)\ge \frac{1}{s}\quad\mbox{ for $0<s<\varepsilon$}.$$
As an exercise, I need to understand which is stronger between them; I mean, if $a)\implies b)$ or $b)\implies a)$. Or, in general, what kind of relation exists between them.
I tried this: conditions $b)$ is equivalent to $F^{\prime}(s)\ge \frac{1}{s^2}$ for $0<s<\varepsilon$. Thus
$$\int_s^{\varepsilon} F^{\prime}(v) dv \ge\int_s^{\varepsilon}\frac{1}{v^2} dv\iff F(\varepsilon)-F(s)\ge \frac{1}{s} -\frac{1}{\varepsilon}\iff F(s)\le C-\frac{1}{s},$$
for a suitable constant $C>0$.
Apparently, I do not see any relation beteween them.
Could someone please help me understand what I am doing wrong? Or please give some hints?
Thank you in advance!

REPLY [0 votes]: There is no implication between $a)$ and $b)$. Here are counterexamples to both implications :

the function $F : s \mapsto \dfrac{1}{s^2}$ obviously satisfies $a)$, but one has $F'(s) = - \dfrac{2}{s^3}$ which takes negative values, so $F$ does not satisfy $b)$.

conversely, the function $F : s \mapsto -\dfrac{1}{s}$ does not satisfy $a)$ because it takes negative values for $s > 0$, but one has $F'(s) = \dfrac{1}{s^2}$ do $F$ satisfies $b)$.


TITLE: If you have only 3 coins and 1 is a penny is it possible for you to have a Dollar?
QUESTION [0 upvotes]: If you have only 3 coins and 1 is a penny is it possible for you to have a Dollar?

REPLY [3 votes]: If it's a dollar or more, than Dominik has answered you. Yes there are dollar coins. However we can't have exactly $1$ dollar. If we had exactly $1$ dollar than that would require us to use coins lower than a dollar.
Coins with values:
$.01$
$.05$
$.1$
$.25$
$.50$
Since you already specified that one of the $3$ coins is a penny the other two should add up to $.99$.
We can't use two half dollars for an obvious reason.
And the next two coins of largest value would be $.5+.25$
But that isn't even enough to get to $.99$