TITLE: equation for irregular shapes and determining their similarity
QUESTION [4 upvotes]: I am a biologist interested in studying cell shape. Cells comes in all different shapes and forms. Some are circular and oval but most have a shape that could be asymmetrical or irregular. A circle and an oval have specific equations. Is it possible to create an equation of a shape like for example the fifth one in the image?
I would also like to know if different cell shapes are similar, by that I mean if they could change into one another just by stretching, squeezing, twisting or bending or they are dissimilar, which mean one would need to do something else?
In addition, cells change their shape through time, a circle can become an oval and back to circle as the cell moves. Is there a part of math that study changes of shape through time?
Sorry that my question is not precise, I'd appreciate any help, for example, which books or branch of mathematics I should look into and try to find experts in, to find answer for such questions?
Thank you!

REPLY [3 votes]: I find this question extremely interesting, but at the same time it's evident that it's also very complicated. On the other hand, I believe that the winning strategy is always to start from something simple and then, once consolidated, generalize it at will until you reach your goals.
So, assuming that I'm not an expert in the field, what I can say is that it often happens that we've a set of points to which we want to associate a function in one or more variables that describes them in the best possible way. A first question to ask is whether we want a function that passes absolutely through these points (interpolation) or if we want a function that passes us as close as possible, not necessarily that touches all the points (fitting).
In both cases, then, it's necessary to establish the basic functions to be used, usually polynomial, rational or trigonometric. Furthermore, it must be established whether this function is to be defined elementary or piecewise and in the latter case it's necessary to establish how to connect these sections, usually something smooth is desired, etc etc.
The answer to this list of questions depends on many factors, first of all the context in which one operates and the objectives we have set. Specifically, since the starting data may be affected by measurement errors, by noise, I would say that a curve fitting is preferable and in particular I thought that it was advantageous to rely on the Discrete Fourier Transform to quickly find a least squares trigonometric fit to the data.

In light of all this, by defining the following routine in Mathematica:
TrigFit[v_, m_, {x_, x0_, x1_}] := Module[{fc, fs, i, imax, j, n, p, t},
  n = Length[v]; imax = Min[m, Floor[(n - 1) / 2]]; t = 2 π (x - x0)/(x1 - x0);
  fc = Table[Sum[v[[j + 1]] Cos[2 π i j / n] / n, {j, 0, n - 1}], {i, 0, imax}];
  fs = Table[Sum[v[[j + 1]] Sin[2 π i j / n] / n, {j, 0, n - 1}], {i, 0, imax}];
  p = fc[[1]] + 2 Sum[fc[[i + 1]] Cos[i t] + fs[[i + 1]] Sin[i t], {i, 1, imax}];
  Expand@Rationalize[p, 0.01]]

we can begin with an equally spaced distribution of points of which we wish to determine a function in a variable with at most $m$ modes and defined on the period considered:
pts = Table[N[Cos[x] / (5/4 + Sin[x]) + Sin[x] / (5/4 - Cos[x])],
            {x, 0, 2 π - 2 π / 11, 2 π / 11}];

f[x_] = TrigFit[pts, 4, {x, 0, 2 π}];

Show[Plot[f[x], {x, 0, 2 π}],
     ListPlot[pts, DataRange -> {0, 2 π - 2 π / 11}, PlotStyle -> Red],
     AxesLabel -> {"x", "y"}, GridLines -> Automatic]

TeXForm@f[x]

$\quad\quad\quad\quad\quad\quad$

$\sin (x)+\frac{1}{4} \sin (3 x)+\frac{1}{4} \sin (4 x)+\cos (x)-\frac{1}{4} \cos (3 x)$

Therefore, we can raise the bar by considering a distribution of points in the $\langle x,\,y \rangle$ plane of which we wish to determine a $m$ modes parameterization of the closed curve that best describes them:
pts = Table[N[{JacobiP[5, 1, 2, Cos[t]], JacobiP[5, 1, 2, Sin[t]]}],
            {t, 0, 2 π, π / 25}];

{x[t_], y[t_]} = TrigFit[#, 3, {t, 0, 2 π}]& /@ Transpose[pts];

Show[ParametricPlot[{x[t], y[t]}, {t, 0, 2 π}],
     ListPlot[pts, PlotStyle -> Red],
     AxesLabel -> {"x", "y"}, GridLines -> Automatic]

TeXForm@{x[t], y[t]}

$\quad\quad\quad\quad\quad\quad$

$\left\{-\frac{2 \sin (t)}{5}+\frac{2}{5} \sin (2 t)-\frac{8}{9} \sin (3 t)+\frac{25 \cos (t)}{4}-\frac{13}{4} \cos (2 t)+\frac{34}{7} \cos (3 t)-\frac{17}{9},\frac{584 \sin (t)}{97}-\frac{3}{4} \sin (2 t)-\frac{60}{13} \sin (3 t)+\frac{6 \cos (t)}{13}+\frac{44}{13} \cos (2 t)-\frac{8}{7} \cos (3 t)-\frac{101}{50}\right\}$

It's not all. The power of this approach is that it extends even to closed three-dimensional curves, everything always depends on having a set of sufficiently accurate coordinates. For example, in Mathematica we can find the coordinates of some knots:
pts = First@@KnotData[{8, 3}, "ImageData"];

{x[t_], y[t_], z[t_]} = TrigFit[#, 10, {t, 0, 2 π}]& /@ Transpose[pts];

ParametricPlot3D[{x[t], y[t], z[t]}, {t, 0, 2 π}, Axes -> False,
                 Boxed -> False, ImageSize -> 400, PlotRange -> All,
                 PlotStyle -> Directive[Orange, Specularity[White, 70]]
                ] /. Line[pts_, rest___] :> Tube[pts, 1/25, rest]

TeXForm@{x[t], y[t], z[t]}

$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad$

$\left\{-\frac{\sin (t)}{9}+\frac{2}{9} \sin (3 t)+\frac{1}{17} \sin (5 t)+\frac{1}{14} \sin (7 t)-\frac{\cos (t)}{8}-\frac{2}{81} \cos (2
   t)-\frac{2}{11} \cos (3 t)+\frac{2}{23} \cos (5 t)-\frac{2}{43} \cos (7 t)+\frac{2}{71} \cos (10 t),\frac{\sin (t)}{9}-\frac{4}{11} \sin (2
   t)-\frac{1}{8} \sin (3 t)+\frac{2}{57} \sin (5 t)-\frac{2}{57} \sin (6 t)-\frac{2 \cos (t)}{19}-\frac{2}{75} \cos (2 t)-\frac{2}{13} \cos (3 t)-\frac{2}{21} \cos (4 t)-\frac{1}{39} \cos (5 t)+\frac{1}{8},-\frac{1}{44} \sin (5 t)-\frac{1}{11} \sin (8 t)-\frac{1}{28} \cos (5 t)-\frac{2}{77} \cos (8 t)\right\}$

At this point, once we understand the academic examples, we can finally put them into practice by considering something similar to what you mentioned in the question. In particular, considering a two-dimensional figure with sufficiently well defined and closed edges, i.e. something like this:
$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad$
we can import it into Mathematica to extract the edge points:
img = Binarize@Import["https://i.stack.imgur.com/RpSbo.png"];
lines = Cases[Normal@ListContourPlot[Reverse@ImageData[img], Contours -> 1], _Line, -1];
pts = Reverse@SortBy[lines[[All, 1]], Length];

and consequently, by selecting the points of the three main edges, we obtain what is desired:
p = Table[TrigFit[#, 10, {t, 0, 2 π}]& /@ Transpose[pts[[i]]], {i, 3}];
ParametricPlot[p, {t, 0, 2 π}, AxesLabel -> {"x", "y"}, GridLines -> Automatic]

$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad$
where, as an example, a parameterization of the orange edge is:
TeXForm@p[[2]]

$\left\{-\frac{432 \sin (t)}{13}-\frac{29}{5} \sin (2 t)-\frac{14}{9} \sin (3 t)-\frac{1}{6} \sin (4 t)+\frac{8}{5} \sin (5 t)-\frac{76}{25} \sin (6 t)+\frac{13}{5} \sin (7 t)+\frac{19}{6} \sin (8 t)-\frac{3}{2} \sin (9 t)-\frac{2}{11} \sin (10 t)+\frac{48 \cos (t)}{5}-\frac{13}{4} \cos (2 t)+\frac{38}{13} \cos (3 t)-\frac{8}{5} \cos (4 t)-\frac{44}{15} \cos (5 t)-\frac{50}{11} \cos (6 t)+\frac{8}{13} \cos (7 t)+\frac{12}{13} \cos (8 t)+\frac{11}{7} \cos (9 t)-\frac{5}{6} \cos (10 t)+73,-\frac{41 \sin (t)}{4}+\frac{36}{7} \sin (2 t)-\frac{4}{3} \sin (3 t)-\frac{6}{5} \sin (4 t)-\frac{4}{7} \sin (5 t)+\frac{2}{3} \sin (6 t)+\frac{1}{7} \sin (7 t)+\frac{3}{5} \sin (8 t)-\frac{14}{11} \sin (9 t)+\frac{22}{21} \sin (10 t)-\frac{155 \cos (t)}{3}-\frac{48}{11} \cos (2 t)-\frac{22}{3} \cos (3 t)-\frac{26}{9} \cos (4 t)-\frac{4}{3} \cos (5 t)-\frac{2}{5} \cos (6 t)-\frac{52}{11} \cos (7 t)+\frac{5}{4} \cos (8 t)-\frac{4}{9} \cos (9 t)-\frac{1}{3} \cos (10 t)+\frac{1168}{9}\right\}$

Once acceptable parameterizations have been obtained for each edge of interest, we can proceed to work on the change of shape and in this regard I had spent a fairly detailed answer. Of course, this will haven't solved all the problems, but I hope it's a good starting point. ^_^


TITLE: Dual of linear map into a tensor product
QUESTION [3 upvotes]: Suppose we have (finite dimensional) vector spaces $U$, $V$ and $W$ and a linear map
$f: U\otimes V \to W; u\otimes v\mapsto f(u\otimes v)$.
Is there a standard way to write the dual linear map? I mean, it is of course
$f^*: W^* \to (U\otimes V)^*; w^* \mapsto f^*(w^*)$
but since $(U\otimes V)^*\simeq U^* \otimes V^*$ in this case, can we somehow put emphasis on the fact that $f^*(w^*)$ is a linear sum of tensor components? Like
$f^*(w^*)=\sum_{something-I-dont-see} f^*_{?}(w^*)\otimes f^*_{?}(w^*)$

REPLY [2 votes]: First, there is some major problem with your question. Notice that there is no chance to write
$$f^*(w^*)=\sum_{something-I-dont-see} f^*_{?}(w^*)\otimes f^*_{?}(w^*),$$
because the left hand side is linear in $w^*$ and the right hand side is quadratic!
But something can be done!
You said that $(U\otimes V)^*\simeq U^* \otimes V^*.$ However, it is only true, because $U$ and $V$ have finite dimensions. In general, for $U,V$ with arbitrary dimensions we only have an injective map $\phi:U^* \otimes V^*\to(U\otimes V)^*$ given by formula
$$\phi\left(\sum_i\alpha_i\otimes\beta_i\right)\left(\sum_jx_j\otimes y_j\right)=\sum_{i,j}\alpha_i(x_j)\beta_i(y_j).$$
Here you have a related discussion about this problem.
So your question can be reformulated as

Problem. Let $f:U\otimes V\to W$ be given and let $f^*:W^*\to(U\otimes V)^*$ denote the dual map. Is there a map $\bar{f}:W^*\to U^*\otimes V^*$ such that
  $$f^*=\phi\circ \bar{f}?$$

Answer. If $U,V$ have finite dimension, then there is such $\bar{f}$ but not defined in canonical manner (it depends on basis of $U$ and $V$).
In fact, if $u_1,\dots,u_n$ and $v_1,\dots,v_m$ are basis in $U$ and $V$ respectively, then we have a dual basis $\omega_1,\dots,\omega_n$ and $\eta_1,\dots,\eta_m$ in $U^*$ and $V^*$ respectively. As a result $\lbrace \omega_i\otimes\eta_j\rbrace_{i,j}$ form a base in $U^*\otimes V^*$ and $\lbrace \phi(\omega_i\otimes\eta_j)\rbrace_{i,j}$ form a base in $(U\otimes V)^*$. Thus, for any $w^*\in W^*$ we have that
$$f^*(w^*)=\sum_{i,j}a_{i,j}(w^*)\phi(\omega_i\otimes\eta_j).$$
Putting $\bar{f}=\phi^{-1}\circ f^*$ we obtain that
$$\bar{f}(w^*)=\sum_{i,j}a_{i,j}(w^*)\omega_i\otimes\eta_j.$$
This is all I can tell for certain! Notice also that we do not need $W$ to have finite dimension.


TITLE: Show the sum is equal to a product of six primes
QUESTION [0 upvotes]: On a set of math challenges, one of them is to prove that $$145678+456781+567814+678145+781456+814567$$ is the product of six different primes.  This sounds like number theory to me, but I have no background in the subject, so I was wondering if anyone can indicate how the proof is supposed to go?

REPLY [1 votes]: Taken from here,

Write $145678=1\cdot 10^5+4\cdot 10^4+5\cdot 10^3+6\cdot 10^2+7\cdot 10+8$, and do the same for all other numbers. Each of the digits appears exactly once at each of the six place-value positions, giving the result of addition as
  $$(1+4+5+6+7+8)\cdot(10^5+10^4+10^3+10^2+1)=\\
31\cdot 111\cdot1001=31\cdot3\cdot37\cdot11\cdot7\cdot13$$

REPLY [1 votes]: HINT: Have you noticed that each separate digit appears exactly once in each position. Try adding without carrying and see what happens.
[it took me less than a minute to write down the six primes - so you should give a bit more effort before expecting a solution - sometimes the process of solving it yourself is valuable]


TITLE: Derivative of $\sin(-x) = -\cos(x)$?
QUESTION [1 upvotes]: Doesn't derivative of $\sin(-x)$ equal:
$f'(x) = \cos(-x)(-1)$
$f'(x)= -\cos(-x)$
how do you get $-\cos x$?

REPLY [1 votes]: $\cos(x)$ is an even function which means $\cos(-x)=\cos(x)$. Using this we have $-\cos(-x)=-\cos(x)$


TITLE: Why does the code not work?
QUESTION [0 upvotes]: I want to write a program which gives me all prime ideals in the ring of integers $\mathcal{O}_K $,  $K= \mathbb{Q}(\sqrt{-5})$ . They should have the property to be principal.  My problem is that the condition if(J.gens()[0]-1 in L):
does not really work. By this I want that the generator of the principal ideal minus $1$ is contained in the ideal L.
   sage: K.<a> = QuadraticField(-5)
             cl = K.class_group()
              g = cl.gens()[0]
               L=K.fractional_ideal(-a-1)
    sage: for norm, ideals in K.ideals_of_bdd_norm(10).items():
             for J in ideals:
                 if (J.is_prime()):
                   if  (cl(J)==g^4):
                     if(J.gens()[0]-1 in L):
                         print norm,J

REPLY [1 votes]: I answered this on Ask SageMath:

What you are looking for is J.gens_reduced() instead of J.gens().

See the documentation of gens_reduced:

Express this ideal in terms of at most two generators, and one if possible.

By contrast, gens() are just the generators which were given when the ideal was constructed (in this case, by Sage).


text:
\section{\texorpdfstring{$\mathbb{F}_{\infty}$}{Foo}-modules}

In this section, we introduce our motivating example, the category of finitely generated $\F_\infty$-modules. Our main reference for this section is $\cite{habjuh2018}$.

We start with the definition of the generalized ring, $\F_\infty$ (see \cite{durov2007}, \textbf{5.1.16}).

\begin{definition}
The \textbf{generalized field} or \textbf{field} $\F_\infty$ is defined as the set $\{-1,0,1\}$ equipped with three operations: a unary $-$, and a binary $\plus$ and $\cdot$, with the following properties:
\begin{itemize}
\item $(a\plus b)\plus c=a\plus (b\plus c)$, $a\plus b=b\plus a$;
\item $0\plus a=a\plus 0=0$, hence $0$ is an \textbf{absorbing element};
\item $a\plus a=a$;
\item $a\plus(-a)=0$;
\item $-(1)=-1$, $-(-1)=1$, $-0=0$;
\item $(a\cdot b)\cdot c=a\cdot (b\cdot c)$, $a\cdot b=b\cdot a$;
\item $1\cdot a=a\cdot 1=a$;
\item $(-1)\cdot(-1)=1$.
\end{itemize}
\end{definition}

We remark that the operator $\plus$ is unlike the ordinary addition: the element $0$ is an absorbing element ($0\plus a=0$); and the operation $-$ is not the additive inverse, for instance, $(a\plus a)\plus (-a)$ is in general not $a$, but rather $a\plus (a\plus (-a))=a\plus 0=0$.

This discussion leads us to the definition of a module over $\F_\infty$ (see \cite{durov2007}, \textbf{4.3.7}).

\begin{definition}
\label{def:module}
An \textbf{$\F_\infty$-module} is a structure $(V,0,-,\plus)$ such that:
\begin{itemize}
\item $(a\plus b)\plus c=a\plus (b\plus c)$, $a\plus b=b\plus a$;
\item $a\plus a=a$, $a\plus (-a)=0$;
\item $-(a\plus b)=(-a)\plus (-b)$; $-(-a)=a$.
\end{itemize}
Again, the notions of \textbf{submodule}, \textbf{congruence}, \textbf{quotient module}, \textbf{module homomorphism} and \textbf{finitely generated modules} are defined as usual.
\end{definition}

We remark that contrary to usual conventions, $0$ is not a neutral element, and $-a$ is not an additive inverse. In fact, an $\F_\infty$-module structure on a set $V$ gives rise to a natural partial order on $V$ given by $a\leq b$ if $a+b=b$; in this partial order $0$ is the \textbf{maximal} element.

We, again, consider projective $\F_\infty$-modules, modules with the usual lifting property. Similarly as with usual modules, a finitely generated $\F_\infty$-module $P$ is projective if and only if there exists a splittable surjection $F\rightarrow P$ from a free $\F_\infty$-module $F$ to $P$.

\begin{Example}\label{ex:f}
We modify Example \ref{ex:b} to obtain projective $\F_\infty$-modules. Let $D_n$ be the projective $\B$-module as in Example \ref{ex:b}. For each element $a_{ij}$ we add another element $-a_{ij}$. The addition $\plus$ is defined in an opposite fashion as before, explicitly, on the $a_{ij}$, we have
\[a_{ij}\plus a_{kl}=a_{\min(i,k), \min(j,l)},\]
on the $-a_{ij}$, we have
\[(-a_{ij})\plus (-a_{kl})=-a_{\min(i,k), \min(j,l)},\]
and finally
\[a_{ij}\plus (-a_{kl})=0.\]
In this fashion, we obtain an $\F_\infty$-module, which we call $E_n$ (the figure below illustrates the module $E_3$).

Again, there exists a surjection $g:\F_\infty[A_1,A_2,...,A_{2n-1}]\ra E_n$ generated by the images
\[g(A_1)=a_{nn}, g(A_2)=a_{n,n-1}, g(A_3)=a_{n-1,n},...\]
\[g(A_{2i})=a_{n-i,n-i+2}, g(A_{2i+1})=a_{n-i+1,n-i}.\]
This surjection splits, the map $h:E_n\ra \F_\infty[A_1,A_2,...,A_{2n-1}]$ given by the images
\[h(a_{nn})=A_1, h(a_{n,n-1})=A_1\plus A_2, h(a_{n-1,n})=A_1\plus A_3,...\]
\[h(a_{n-i,n-i+2})=A_1\plus A_2\plus ...\plus A_{2i-2}\plus A_{2i},h(a_{n-i+1,n-1})=A_1\plus ...\plus A_{2i-1}\plus A_{2i+1}\]
provides a right inverse.
\begin{center}
    \definecolor{xdxdff}{rgb}{0,0,0}
\definecolor{ududff}{rgb}{0,0,0}
\definecolor{cqcqcq}{rgb}{0,0,0}
\begin{tikzpicture}[line cap=round,line join=round,>=triangle 45,x=0.5cm,y=0.5cm]
\clip(6.673223054618417,-5.137551424556686) rectangle (17.17897535725542,18.24760788413266);
\draw [line width=2.pt] (16.,12.)-- (12.,8.);
\draw [line width=2.pt] (10.,14.)-- (14.,10.);
\draw [line width=2.pt] (10.,10.)-- (12.,8.);
\draw [line width=2.pt] (12.,16.)-- (16.,12.);
\draw [line width=2.pt] (12.,16.)-- (10.,14.);
\draw [line width=2.pt] (10.,10.)-- (14.,14.);
\draw [line width=2.pt] (12.,8.)-- (12.,4.);
\draw [line width=2.pt] (12.,4.)-- (8.,0.);
\draw [line width=2.pt] (8.,0.)-- (12.,-4.);
\draw [line width=2.pt] (12.,4.)-- (14.,2.);
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\draw [line width=2.pt] (10.,2.)-- (14.,-2.);
\draw [line width=2.pt] (12.,-4.)-- (14.,-2.);
\begin{scriptsize}
\draw [fill=ududff] (12.,8.) circle (2.5pt);
\draw[color=ududff] (12.79700685660759,7.763624437166554) node {$a_{11}$};
\draw [fill=ududff] (14.,10.) circle (2.5pt);
\draw[color=ududff] (14.834755945813466,10.014645006989227) node {$a_{12}$};
\draw [fill=ududff] (12.,12.) circle (2.5pt);
\draw[color=ududff] (12.79560306398616,12.023797888816526) node {$a_{22}$};
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\draw[color=ududff] (9.124304156939111,9.681187968222842) node {$a_{21}$};
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\draw[color=ududff] (14.793073815967668,14.182857991569035) node {$a_{23}$};
\draw [fill=ududff] (16.,12.) circle (2.5pt);
\draw[color=ududff] (16.376994750108,12.765665576811902) node {$a_{13}$};
\draw [fill=ududff] (10.,14.) circle (2.5pt);
\draw[color=ududff] (9.249350546476506,14.683043549718613) node {$a_{32}$};
\draw [fill=ududff] (12.,16.) circle (2.5pt);
\draw[color=ududff] (12.62560306398616,16.266964483858942) node {$a_{33}$};
\draw[color=black] (13.625974180285318,10.848287603905188) node {$$};
\draw[color=black] (11.625231947687004,11.89034085005014) node {$$};
\draw[color=black] (10.624860831387847,8.889227501152678) node {$$};
\draw[color=black] (13.625974180285318,13.89108308264845) node {$$};
\draw[color=black] (10.624860831387847,15.85014318540096) node {$$};
\draw[color=black] (12.583920934140362,11.89034085005014) node {$$};
\draw [fill=ududff] (12.,6.) circle (2.5pt);
\draw[color=ududff] (12.792146025219775,6.263438879016976) node {$0$};
\draw [fill=ududff] (12.,4.) circle (2.5pt);
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\draw [fill=ududff] (10.,2.) circle (2.5pt);
\draw[color=ududff] (9.79140379262146,2.76195441382036) node {$-a_{12}$};
\draw [fill=ududff] (14.,2.) circle (2.5pt);
\draw[color=ududff] (14.29288825781809,2.76195441382036) node {$-a_{21}$};
\draw [fill=xdxdff] (12.,0.) circle (2.5pt);
\draw[color=xdxdff] (11.892146025219775,0.8612121812220518) node {$-a_{22}$};
\draw [fill=xdxdff] (8.,0.) circle (2.5pt);
\draw[color=xdxdff] (7.790661560023146,0.7612121812220518) node {$-a_{13}$};
\draw [fill=ududff] (10.,-2.) circle (2.5pt);
\draw[color=ududff] (9.89140379262146,-1.1395300513762564) node {$-a_{23}$};
\draw [fill=ududff] (12.,-4.) circle (2.5pt);
\draw[color=ududff] (11.892146025219775,-3.240272283974565) node {$-a_{33}$};
\draw [fill=ududff] (14.,-2.) circle (2.5pt);
\draw[color=ududff] (14.29288825781809,-1.2395300513762564) node {$-a_{32}$};
\draw[color=black] (11.416821298458013,6.388299710404794) node {$$};
\draw[color=black] (9.62448971508869,2.845318673511956) node {$$};
\draw[color=black] (9.62448971508869,-2.1148547781380165) node {$$};
\draw[color=black] (12.62560306398616,2.8870008033577546) node {$$};
\draw[color=black] (12.583920934140362,-0.11411254553970801) node {$$};
\draw[color=black] (11.625231947687004,-0.11411254553970801) node {$$};
\draw[color=black] (13.584292050439519,-3.1152258944371702) node {$$};
\end{scriptsize}
\end{tikzpicture}
\end{center}
\end{Example}

As before, we have the following property.

\begin{proposition}\label{prop:fnonnot}
Let $E_n$ and $E_m$ be $\F_\infty$-modules as in Example \ref{ex:f}, with elements $a_{11}$, ..., $a_{nn}$, and $b_{11}$, ..., $b_{mm}$ respectively. Then, any injective $\F_\infty$-module morphism $f:E_n\ra E_m$ satisfying
\[f(a_{11})=b_{11}, f(a_{12})= b_{12}, f(a_{21})= b_{21}, f(a_{22})= b_{22}, \mbox{and}\]
\[f(a_{n-1,n-1})=b_{m-1,m-1}, f(a_{n-1,n})= b_{m-1,m}, f(a_{n, n-1})=b_{m, m-1}, f(a_{nn})= b_{mm}\]
is the identity morphism (and hence $n=m$). \qed
\end{proposition}

We omit the proof of the proposition above, it follows the same strategy as in Proposition \ref{prop:bnonnot}, but in this case, the recursive process begins with $a_{n-2,n}$ instead of $a_{13}$.

\begin{Example}
As in Example \ref{ex:bbmod} we define an $\F_\infty$-module of 17 elements depicted below. We denote this module by $\E_0$. For every $n>3$, we have an injection $\E_0\ra E_n$ defined as
\[k(A_{11})=a_{11}, k(A_{12})=a_{12},k(A_{21})=a_{21}, k(A_{22})=a_{22},\]
\[k(A_{33})=a_{n-1,n-1}, k(A_{34})=a_{n-1,n}, k(A_{43})=a_{n,n-1}, k(A_{44})=a_{nn}\]
and $k(-A_{ij})=-k(A_{ij})$. This injection is a splittable injection as in Lemma \ref{lem:splinjb}, we omit the proof of this statement, it is completely parallel (reversed).

\begin{center}
\definecolor{xdxdff}{rgb}{0,0,0}
\definecolor{ududff}{rgb}{0,0,0}
\definecolor{cqcqcq}{rgb}{0,0,0}
\begin{tikzpicture}[line cap=round,line join=round,>=triangle 45,x=1.0cm,y=1.0cm]
\clip(-0.72,-4.3) rectangle (10,8.52);
\draw [line width=2.pt] (5.,8.)-- (4.,7.);
\draw [line width=2.pt] (4.,7.)-- (5.,6.);
\draw [line width=2.pt] (5.,8.)-- (6.,7.);
\draw [line width=2.pt] (6.,7.)-- (5.,6.);
\draw [line width=2.pt] (5.,6.)-- (5.,5.);
\draw [line width=2.pt] (5.,5.)-- (4.,4.);
\draw [line width=2.pt] (4.,4.)-- (5.,3.);
\draw [line width=2.pt] (5.,5.)-- (6.,4.);
\draw [line width=2.pt] (6.,4.)-- (5.,3.);
\draw [line width=2.pt] (5.,3.)-- (5.,1.);
\draw [line width=2.pt] (5.,1.)-- (4.,0.);
\draw [line width=2.pt] (4.,0.)-- (5.,-1.);
\draw [line width=2.pt] (5.,1.)-- (6.,0.);
\draw [line width=2.pt] (6.,0.)-- (5.,-1.);
\draw [line width=2.pt] (5.,-1.)-- (5.,-2.);
\draw [line width=2.pt] (5.,-2.)-- (4.,-3.);
\draw [line width=2.pt] (4.,-3.)-- (5.,-4.);
\draw [line width=2.pt] (5.,-4.)-- (6.,-3.);
\draw [line width=2.pt] (6.,-3.)-- (5.,-2.);
\begin{scriptsize}
\draw [fill=ududff] (4.,7.) circle (2.5pt);
\draw[color=ududff] (4.56,7.11) node {$A_{43}$};
\draw [fill=ududff] (5.,8.) circle (2.5pt);
\draw[color=ududff] (5.14,8.37) node {$A_{44}$};
\draw [fill=ududff] (6.,7.) circle (2.5pt);
\draw[color=ududff] (6.4,7.09) node {$A_{34}$};
\draw [fill=ududff] (5.,6.) circle (2.5pt);
\draw[color=ududff] (5.4,6.00) node {$A_{33}$};
\draw [fill=ududff] (5.,5.) circle (2.5pt);
\draw[color=ududff] (5.38,5.11) node {$A_{22}$};
\draw [fill=ududff] (4.,4.) circle (2.5pt);
\draw[color=ududff] (4.42,4.11) node {$A_{21}$};
\draw [fill=ududff] (5.,3.) circle (2.5pt);
\draw[color=ududff] (5.42,3.09) node {$A_{11}$};
\draw [fill=ududff] (6.,4.) circle (2.5pt);
\draw[color=ududff] (6.38,4.05) node {$A_{12}$};
\draw [fill=ududff] (5.,2.) circle (2.5pt);
\draw[color=ududff] (5.36,2.15) node {$0$};
\draw [fill=ududff] (5.,1.) circle (2.5pt);
\draw[color=ududff] (5.4,1.31) node {$-A_{11}$};
\draw [fill=xdxdff] (4.,0.) circle (2.5pt);
\draw[color=xdxdff] (4.5,0.07) node {$-A_{12}$};
\draw [fill=ududff] (6.,0.) circle (2.5pt);
\draw[color=ududff] (6.48,0.11) node {$-A_{21}$};
\draw [fill=ududff] (5.,-1.) circle (2.5pt);
\draw[color=ududff] (5.38,-0.91) node {$-A_{22}$};
\draw [fill=ududff] (5.,-2.) circle (2.5pt);
\draw[color=ududff] (5.38,-1.93) node {$-A_{33}$};
\draw [fill=ududff] (4.,-3.) circle (2.5pt);
\draw[color=ududff] (4.46,-2.87) node {$-A_{34}$};
\draw [fill=ududff] (6.,-3.) circle (2.5pt);
\draw[color=ududff] (6.42,-2.95) node {$-A_{43}$};
\draw [fill=ududff] (5.,-4.) circle (2.5pt);
\draw[color=ududff] (5.5,-3.93) node {$-A_{44}$};
\end{scriptsize}
\end{tikzpicture}
\end{center}
\end{Example}

config: arxiv


TITLE: How many numbers are there where $x \mod 818 = 610$ and...
QUESTION [0 upvotes]: How many numbers are there, where the next two conditions are met, how many such conditions are required to make it just one option for $x$?

$x \mod 818 = 610$
$x \mod 819 = 615$

REPLY [3 votes]: An infinity of numbers, they are exactly the numbers of the form $n(818\cdot819)+666462$, with $n\in\mathbb Z$
The solutions to a system of conguence equations in $\mathbb Z$ are $0$ or infinite. This is because the solution will be determined $\bmod lcm(M_1,M_2\dots M_n$). Where the system is:
$x\equiv a_1\bmod M_1$
$x\equiv a_2\bmod M_2$
$\dots$
$x\equiv a_n\bmod M_n$
For more information see the Chinese Remainder Theorem.

REPLY [1 votes]: Since $819-818=1$ the solutions are:
$$x\equiv 610\cdot819-615\cdot818=-3480\mod 818\cdot 819\quad\text{i.e.}\quad x\equiv666\,462\mod 669\,942.$$


TITLE: How do I show that the following is a definition of an alternating form?
QUESTION [3 upvotes]: Suppose $E$ is an open set in $\mathbb{R}^n$. A differential form of order
$k$ in $E$ (briefly, a k-form in E) is a function $\omega$, which assigns to each $k$ surface $\Phi$ in $E$ a number $\omega(\Phi) = \int_{\Phi} \omega$
$$\int_{\Phi}\omega = \int_{D} \sum a_{i_1 \cdots i_k}(\Phi(u)) \frac{\partial(x_{i_1},\cdots,x_{i_k})}{\partial(u_{i_1},\cdots,u_{i_k})}du  \cdots \cdots \cdots (1)$$
where $\omega$ is symbollically representated as $\omega = \sum a_{i_1 \cdots i_k}(x) dx_{i_1} \wedge \cdots \wedge dx_{i_k}$
The definition of differential forms that was given to us in class is that $\omega(x)$ is an alternating $k$ tesnor. How are these two definitions related?How do I show that $(1)$ is an alternating $k$ tensor?

REPLY [0 votes]: It is clear that when a differenial form
$$\alpha = \sum_{i_1< \cdots > i_k} \alpha_{i_1\cdots i_k} dx^{i_1} \wedge \cdots \wedge dx^{i_k}$$
is given, one uses it to define integration for any $k$ surface $\Phi$ using (1). One can show that any continuous differential form $\alpha$ can be recovered (by continuous I mean that $\alpha_{i_1\cdots i_k}$ are continuous functions), if we know $\int_\Phi \alpha$ for all $k$-surface $\Phi$.
To see this, let $x = (x^1, \cdots, x^n) \in E$ and any $i_1 < \cdots < i_k$. Let $R>0$ so that $B^n_R(x) \subset E$. Let $\Phi$ be the parametrized surface given by
$$\Phi : B^k_R(0) \to E, \ \ \Phi(u^1, \cdots, u^k) = (x^1 (u) , \cdots, x^n(u)),$$
where
$$ x^j (u) = \begin{cases} x^{i_l} + u^{i_l}, & \text{ if } j = i_l, l =1, \cdots, k, \\
x^j & \text{ otherwise}. \end{cases}$$
For any $0<r<R$, let $\Phi_r = \Phi|_{B^k_r(0)}$. Then
$$ \int_{\Phi_r} \alpha = \int_{B^k_r(0)} \alpha_{i_1\cdots i_k} (\Phi (u)) du$$
and by continuity of $\alpha_{i_1\cdots i_k}$,
$$ \alpha_{i_1\cdots i_k} (x) = \lim_{r\to 0} \frac{1}{|B^k_r(0)|} \int_{\Phi_r} \alpha.$$


TITLE: Simulator for electrostatics
QUESTION [2 upvotes]: I am searching for a tool that visualizes the electric field/potential for simple electrostatic configurations like

conducting/polarized sphere in an uniform electric field
conducting/polarized sphere in a field generated by some simple charge density (like e.g. finite line charge)

I did a search on the internet but did not find anything useful. Does someone know such a tool (preferably for Linux). Thanks a lot.

REPLY [4 votes]: I've used Paul Falstad's applets for classes I've taught. He has some very impressive visualizations there including some that you don't see often (like the magnetic vector potential). He always links to the Java source code, so if you can program, it'll give you a good base to start with.
Check out http://www.falstad.com/vector3de/ for an example.

REPLY [2 votes]: There are lot of visualizations available on mathematica demonstration site.

All you need is free wolfram player(downloadable from the same website) to run these free to use demonstrations. This application is platform independent.
    I hope they help !!
Check this link:
Wolfram electrostatic demos


TITLE: Ideals in $C[0,1]$
QUESTION [3 upvotes]: Let $C[0,1]$ be the ring of continuous real-valued functions on $[0,1]$, with addition and multiplication defined pointwise.
  For any subset $S$ of $C[0,1]$ let $Z(S)=\{f\in C[0,1]: f(x)=0 \text{ for all }x\in S\}$. Then which of the following statements are true?
  (a) If $Z(S)$ is an ideal in $C[0,1]$ then $S$ is closed in $[0,1]$.
  (b) If $Z(S)$ is a maximal ideal then $S$ has only one point.
  (c) If $S$ has only one point then $Z(S)$ is a maximal ideal.

(a) Not necessary: if I take $S=(1/2,1/3)$ still $Z(S)$ is an ideal.
(b) I know that maximal ideals in $C[0,1]$ come in this way (I don't know the proof rigorously), i.e. $C_a=\{f\in C[0,1]:f(a)=0\}$ so $S$ may be finite or countable set? So I guess $(b)$ is a true statement and for the same reason $(c)$ is also true. But I will be happy if someone can explain me a bit about (b) and (c). Thank you.

REPLY [2 votes]: Another approach to proving maximality:  If $S, T$ are two nonempty subsets of $[0,1]$ with $T \subset S$, then $Z(S) \subset Z(T)$.   So $Z(S)$ is maximal if and only if it does not contain any proper subsets -- i.e., it must be a one-point set.


TITLE: Z-Transform Solving
QUESTION [0 upvotes]: i was trying to solve this exercise with Z-Transform:
$   \begin{cases}
   x(n+2)-x(n+1)=a(n) & n\ge0 \\ x(0)=x(1)=0
   \end{cases}  $
where:
$ a(n)=\begin{cases}
   n^2-n & n & even \\ n^2+n & n & odd
   \end{cases}
$
I usually solve them using the geometric series of $ w $, that converges in
$\frac{1}{1-w} $ , but i can't manage to solve it.

REPLY [1 votes]: Just a hint (but I'm not sure if this is easier than AlexSilva's answer): Consider $b[n]=a[n]-a[n-1]$. Then
$$b[n]= \cases{ 0& if $n$ is even
\\4 n-2 & if $n$ is odd
}$$
(you can also assume that it's zero for negative $n$). Then you can obtain $B(z)$ (knowing the transform of $u[n]$ and $n u[n]$), by a composition with the decimation operator and shifting). And $X(Z) (z^2-z) = A(z)$ and $B(z)= A(z) [1-z^{-1}]$