TITLE: Does weak convergence in $L^2$ implies convergence almost everywhere along subsequence?
QUESTION [7 upvotes]: If I know $\int_{[0,1]} f_{n}(x) g(x) dx \rightarrow  \int_{[0,1]} f(x) g(x) dx$ as $n \rightarrow \infty$ for all $g \in L^2([0,1])$ (weak convergence in $L^2$)
and $|f_n(x)|_{L^2} <C$ (uniformly bounded in $L^2$-Norm),
does then exist a subsequence $f_{n_k}$ such that $\lim_{n \rightarrow \infty}f_{n_k}(x) = f(x)$ for almost every $x \in [0,1]$?
Edit: If uniform boundedness in $L^2$ norm is not enough would $f_n$ being uniformly bounded in $L^\infty$ norm help to get a subsequence?

REPLY [5 votes]: The example by David Mitra is perfectly valid, but it may be a little hard to see exactly what happens to $\sin nx$ as $n\to\infty$. Here's a slightly modified example: Rademacher functions
$$r_n(x) = \operatorname{sign}\sin (2^n \pi x), \quad x\in [0,1]$$
These converge to $0$ weakly in $L^2$, but $|r_n|=1$ a.e.
To prove weak convergence, first consider $\int r_n g$ when $g$ is continuous, putting the contributions of adjacent $\pm $ intervals against  each other. Then conclude by density (using the fact that $\|r_n\|_2=1$ and Hölder's inequality).

For complex-valued functions, $\exp(i nx)$ is a good counterexample.


TITLE: Are periodic tilings stable against defects outside of some region away from the defect?
QUESTION [1 upvotes]: Suppose that I have a set of Wang tiles on a 2D infinite grid, and that normally the tiling pattern is periodic. Assume it has period $p$ in both vertical and horizontal direction. Then at fixed points on the lattice, I remove the tile/introduce a defect. Is there any notion of the periodic tiling recovering its structure sufficiently far away from the defects?
There appears to be a similar result for types of aperiodic tilings. Here it's is shown that a Robinson tiling is stable with respect to defects: https://www.mimuw.edu.pl/~miekisz/stablejsp.pdf

REPLY [3 votes]: As I understand your question, you are asking:

Suppose we have a finite set of Wang tiles, which have a periodic tiling $T$ of the plane. Now, fix the colors on finitely many grid squares, such that the result can still be extended to a tiling of the plane. Can this extension can be chosen such that the tiling is identical to $T$ up to a translation, except for finitely many errors near the defect?

The answer to this question is "no". Consider the eight Wang tiles given by rotations of the following two tiles:

Observe that in any tiling of these, the green squares must be matched up, so we are using domino shapes with certain matching conditions. The resulting dominoes look like this, up to rotation:

where the grey circles indicate a side that can be either red or blue, whichever we choose.
After some thought, it should be clear that the only restriction this imposes on a domino tiling is that no two dominoes can be aligned edge-to-edge, with their union forming a $2\times 2$ square. Note that every arrangement of these Wang tiles uniquely determines a squarefree domino tiling of the plane, and every squarefree domino tiling yields at least one corresponding arrangement of the Wang tiles.
First, note that these tiles have a periodic arrangement:

(I've shown the domino tiling, but it should be obvious how to convert this into a periodic arrangement of the Wang tiles.)
On the other hand, suppose that we start off with a domino configuration like this:

Then it is not too hard to verify that the unique squarefree domino tiling containing this arrangement looks like:

But no translation of this latter tiling will ever line up with our periodic one! So we cannot place Wang tiles beyond those original eight in a way that reverts to our original periodic tiling. (In fact, we only need to fix three of them for this behavior to occur.)


TITLE: Union of two subsets is sum of the span?
QUESTION [3 upvotes]: Let $V$ be a Vector Space, and let $S_1, S_2 $ be subsets of V.
Prove that $$\text{span}(S_1 \cup S_2 )= \text{span}(S_1) +\text{span}(S_2)$$
I have no clue where to start.  Hints are greatly appreciated.

REPLY [3 votes]: Hints:
$$v\in\text{Span}\,(S_1\cup S_2)\implies \exists\,v_1,...,v_n\in S_1\,,\,w_1,...,w_m\in S_2\;\;s.t. $$
$$v=\sum_{i=1}^na_iv_i+\sum_{j=1}^mb_jw_j\in\text{Span}\,S_1+\text{Span}\,S_2$$
$$\text{So Span}\,(S_1 \cup S_2)\subseteq\text{Span}\,S_1+\text{Span}\,S_2 $$
Now you try the other direction...

REPLY [1 votes]: Hint: Show that span$\{S_1\} \subseteq span\{S_1 \cup S_2\}$ and similarly for span{$S_2$}.
Use that span of a set is a vector space to show RHS is contained in LHS.
Then show that $S_1 \cup S_2 \subseteq span\{S_1\}+span\{S_2\}$. Since RHS is a vector space, LHS is contained in RHS. Hence two sides are equal.

REPLY [0 votes]: Hint: $\text{span}(S)$ means a space which is the smallest in that contain $S$.

So we just need to prove:

First, $S_1\cup S_2$ is contained in $\text{span}(S_1)+\text{span}(S_2)$;
Second, for any $W$ is a subspace of $V$ which contains $S_1\cup S_2$, then we have that $W$ must contain $\text{span}(S_1)+\text{span}(S_2)$.


TITLE: differentiation of difficult function with unknown constants
QUESTION [0 upvotes]: The expression
$C = -A\frac{\frac{K1}{f1^2H}+2\frac{K1K2}{f1^2f2H^2}+3\frac{K1K2K3}{f1^3f3H^3}}{1+\frac{K1}{f1^2H}+\frac{K1K2}{f1^2f2H^2}+\frac{K1K2K3}{f1^3f3H^3}}$
where A is a positive constant, K1,K2,K3 are positive constants such that K1>K2>K3, and f1,f2,f3 are positive constants such that f1>f2>f3 shows the net negative charge on a trivalent acid with Ks being dissociation constants and fs being activity coefficients. When plotted with arbitrary values for Ks and fs the C is always increasing as H increases.
Now I have two problems, the first is whether I could show that C is always monotone increasing with H, e.g. that the
$\frac{dC}{dH}$
is always positive.
The second is if I can always know that if I have an arbitrary number of functions each monotone in the same direction, the sum of such functions will also be monotone.

REPLY [1 votes]: Actually you can simplify this function
$$
C(H) = -A\frac{\frac{K_1}{f_1^2H}+2\frac{K_1K_2}{f_1^2f_2H^2}+3\frac{K_1K_2K_3}{f_1^3f_3H^3}}{1+\frac{K_1}{f_1^2H}+\frac{K_1K_2}{f_1^2f_2H^2}+\frac{K_1K_2K_3}{f_1^3f_3H^3}}
$$
multiplying numerator and denominator by $f_1^3f_2f_3H^3$:
$$
C(H) = -A\frac{K_1f_1f_2f_3H^2+2K_1K_2f_1f_3H+3K_1K_2K_3f_2}{f_1^3f_2f_3H^3+K_1f_1f_2f_3H^2+K_1K_2f_1f_3H+K_1K_2K_3f_2}
$$
Then
$$
 \frac{dC}{dH} = {\frac {AK_{{1}}f_{{1}}f_{{2}}f_{{3}} \left( {f_{{1}}}^{3}{H}^{4}f_{{2}}f_{{3}}+4\,{H}^{3}K_{{2}}{f_{{1}}}^{3}f_{{3
}}+ \left( 9\,K_{{2}}K_{{3}}{f_{{1}}}^{2}f_{{2}}+K_{{1}}K_{{2}}f_{{1}}
f_{{3}} \right) {H}^{2}+4\,HK_{{1}}K_{{2}}K_{{3}}f_{{2}}+K_{{1}}{K_{{2
}}}^{2}K_{{3}}
 \right) }{ \left( {H
}^{3}{f_{{1}}}^{3}f_{{2}}f_{{3}}+{H}^{2}K_{{1}}f_{{1}}f_{{2}}f_{{3}}+H
K_{{1}}K_{{2}}f_{{1}}f_{{3}}+K_{{1}}K_{{2}}K_{{3}}f_{{2}} \right)^{2}
}}
$$
And you can verify it is positive for $H>0$.


TITLE: Where can I find the divisor class groups of du Val singularities?
QUESTION [9 upvotes]: The du Val singularities are the simplest type of surface singularities.  Each type of du Val singularity has a divisor class group.  Specifically, let $X$ be a surface with an isolated singularity at $P$; then the (analytic or étale) local ring at P depends only on the type of the singularity, and has a divisor class group.
The most familiar example is the quadric cone (A1 singularity), found in many algebraic geometry textbooks.  A line $L$ passing though the vertex of the cone is not locally principal, but $2L$ is, and we find that the divisor class group has order $2$.  (Note: in general an A1 singularity will be étale locally, but not Zariski locally, isomorphic to the vertex of the cone.  As far as I can see, there's no reason in general to expect the generator of the divisor class group to come from a divisor on the ambient surface; we may well have to pass to an étale (or analytic) neighbourhood.)
In a beautiful article, Lipman (Pub. Math. IHES 1969) studied these and computed the (finite) divisor class group of each du Val singularity.  However, as far as I can see, he does not give explicit generators like we have in the example of the quadric cone.
So:

Is there in the literature an explicit description (i.e. with explicit generators)
  of the divisor class groups of the du Val singularities?

REPLY [4 votes]: I do not know a reference, but here is my guess. I will use the notion of the wikipedia article. The order is: type, class group, the generator ideals. ($i^2=-1$, and I assume char. 0 for simplicity)
$A_n$, $\mathbb Z/(n+1)$, $(w+ix, y)$.
$D_n$ ($n$ even), $\mathbb Z/(2)\oplus \mathbb Z/(2)$, $(w,y), (w, x+iy^{(n-2)/2})$.
$D_n$ ($n$ odd), $\mathbb Z/(4)$, $(w,y)$.
$E_6$, $\mathbb Z/(3)$, $(x, w+iy^2)$.
$E_7$, $\mathbb Z/(2)$, $(w,x)$.
$E_8$, $0$.


TITLE: Finding coefficient of $x^k$ in summation of binomial series
QUESTION [0 upvotes]: I am not able to solve this problem
Find the coefficient of $x^k\;\;(k$ is greater than or equal to zero and lesser than or equal to $n$) in the expansion of $E = 1 + (1+x) + (1+x)^2 .... + (1+x)^n$
The final simplified answer is $(n+1)C(k+1)$
Any help would be thoroughly appreciated

REPLY [4 votes]: Since $E$ is just the sum of the geometric series, $E=\dfrac{(1+x)^{n+1}-1}{x}$ and thus the coefficient of $x^k$ is ${n+1 \choose k+1}$


TITLE: convergence of the series $\sum_{n≥ 1}\frac{(n+1)^n}{3^nn!}$
QUESTION [0 upvotes]: I want to determine the convergence of the series to do so, consider that $$\frac{(n+1)^n}{3^nn!}≤\frac{(n+1)^n}{3^n}$$
From where
$$\frac{(n+1)^n}{3^n}=\left({\frac{n}{3}}+{\frac{1}{3}}\right)^n$$
And how
$\lim_{n\to\infty}\sqrt[n]{\left({\frac{n}{3}}+{\frac{1}{3}}\right)^n}=\infty$ Then by the root criterion $\sum_{n≥ 1}\left({\frac{n}{3}}+{\frac{1}{3}}\right)^n=\sum_{n≥ 1}\frac{(n+1)^n}{3^n}$ diverges, and thus by the comparison criterion $\sum_{n≥ 1}\frac{(n+1)^n}{3^nn!}$ diverges
Am I right? or any suggestions? thank you.

REPLY [2 votes]: That is not correct. Yes,$$\sum_{n=1}^\infty\frac{(n+1)^n}{3^n}$$diverges, and, yes, you always have$$\frac{(n+1)^n}{3^nn!}\leqslant\frac{(n+1)^n}{3^n}.$$So what? It doesn't follow from this that$$\sum_{n=1}^\infty\frac{(n+1)^n}{3^nn!}$$diverges.
You can use the quotient test:\begin{align}\lim_{n\to\infty}\frac{\frac{(n+2)^{n+1}}{3^{n+1}(n+1)!}}{\frac{(n+1)^n}{3^nn!}}&=\lim_{n\to\infty}\frac{(n+2)^{n+1}}{3(n+1)^{n+1}}\\&=\lim_{n\to\infty}\frac13\left(1+\frac1{n+1}\right)^{n+1}\\&=\frac e3\\&<1,\end{align}and therefore your series converges.


TITLE: Sum of power series $\sum_{k=1}^{\infty}(-1)^{k-1}\frac{x^{k+1}}{k(k+1)}$
QUESTION [0 upvotes]: Calculate the sum of power series: $$\sum_{k=1}^{\infty}(-1)^{k-1}\frac{x^{k+1}}{k(k+1)}$$
I know that $$\ln x=\sum_{k=1}^{\infty}\frac{(-1)^{k-1}(x-1)^{k}}{k}$$ but I don't know if and how can I use it to calculate the given sum.

REPLY [0 votes]: An alternative to differentiation is simply to use $\frac{1}{k(k+1)}=\frac1k-\frac{1}{k+1}$, with $$\sum_{k=1}^{\infty}(-1)^{k-1}\frac{x^{k+1}}{k}=x\ln(1+x),\quad\sum_{k=1}^{\infty}(-1)^{k-1}\frac{x^{k+1}}{k+1}=x-\ln(1+x).$$


text:
(*
 * Copyright 2014, NICTA
 *
 * This software may be distributed and modified according to the terms of
 * the BSD 2-Clause license. Note that NO WARRANTY is provided.
 * See "LICENSE_BSD2.txt" for details.
 *
 * @TAG(NICTA_BSD)
 *)

section "64-Bit Machine Word Setup"

theory Machine_Word_64
imports Machine_Word_64_Basics More_Word Bit_Shifts_Infix_Syntax Rsplit
begin

context
  includes bit_operations_syntax
begin

type_synonym machine_word = \<open>machine_word_len word\<close>

lemma word_bits_len_of:
  \<open>LENGTH(machine_word_len) = word_bits\<close>
  by (simp add: word_bits_conv)

lemma word_bits_size:
  "size (w :: machine_word) = word_bits"
  by (simp add: word_bits_def word_size)

lemma word_bits_word_size_conv:
  \<open>word_bits = word_size * 8\<close>
  by (simp add: word_bits_def word_size_def)

lemma word_size_word_size_bits:
  \<open>word_size = (2 :: 'a :: semiring_1) ^ word_size_bits\<close>
  by (simp add: word_size_def word_size_bits_def)

lemma lt_word_bits_lt_pow:
  "sz < word_bits \<Longrightarrow> sz < 2 ^ word_bits"
  by (simp add: word_bits_conv)

lemma if_then_1_else_0:
  "((if P then 1 else 0) = (0 :: machine_word)) = (\<not> P)"
  by simp

lemma if_then_0_else_1:
  "((if P then 0 else 1) = (0 :: machine_word)) = (P)"
  by simp

lemmas if_then_simps = if_then_0_else_1 if_then_1_else_0

lemma bool_mask [simp]:
  \<open>0 < x AND 1 \<longleftrightarrow> x AND 1 = 1\<close> for x :: machine_word
  by (rule bool_mask') auto

lemma in_16_range:
  "0 \<in> S \<Longrightarrow> r \<in> (\<lambda>x. r + x * (16 :: machine_word)) ` S"
  "n - 1 \<in> S \<Longrightarrow> (r + (16 * n - 16)) \<in> (\<lambda>x :: machine_word. r + x * 16) ` S"
  by (clarsimp simp: image_def elim!: bexI[rotated])+

lemma le_step_down_word_3:
  fixes x :: machine_word
  shows "\<lbrakk>x \<le> y; x \<noteq> y\<rbrakk> \<Longrightarrow> x \<le> y - 1"
  by (fact le_step_down_word_2)

lemma shiftr_1:
  "(x::machine_word) >> 1 = 0 \<Longrightarrow> x < 2"
  apply transfer
  apply (simp add: take_bit_drop_bit)
  apply (simp add: drop_bit_Suc)
  done

lemma Suc_unat_mask_div:
  "Suc (unat (mask sz div word_size :: machine_word)) = 2 ^ (min sz word_bits - word_size_bits)"
  by (simp add: word_size_word_size_bits unat_drop_bit_eq unat_mask_eq drop_bit_mask_eq Suc_mask_eq_exp
    flip: drop_bit_eq_div word_bits_conv)

lemma ucast_not_helper:
  fixes a::"8 word"
  assumes a: "a \<noteq> 0xFF"
  shows "ucast a \<noteq> (0xFF::machine_word)"
proof
  assume "ucast a = (0xFF::machine_word)"
  also
  have "(0xFF::machine_word) = ucast (0xFF::8 word)" by simp
  finally
  show False using a
    apply -
    apply (drule up_ucast_inj, simp)
    apply simp
    done
qed

lemma unat_less_2p_word_bits:
  "unat (x :: machine_word) < 2 ^ word_bits"
  apply (simp only: word_bits_def)
  apply (rule unat_lt2p)
  done

lemma unat_less_word_bits:
  fixes y :: machine_word
  shows "x < unat y \<Longrightarrow> x < 2 ^ word_bits"
  unfolding word_bits_def
  by (rule order_less_trans [OF _ unat_lt2p])

lemma unat_mask_2_less_4:
  "unat (p AND mask 2 :: machine_word) < 4"
  by (rule unat_less_helper) (simp only: take_bit_eq_mod word_mod_less_divisor flip: take_bit_eq_mask, simp add: word_mod_less_divisor)

lemma unat_mult_simple:
  \<open>unat (x * y) = unat x * unat y\<close>
    if \<open>unat x * unat y < 2 ^ LENGTH(machine_word_len)\<close>
    for x y :: machine_word
  using that by (simp flip: unat_mult_lem)

lemma upto_2_helper:
  "{0..<2 :: machine_word} = {0, 1}"
  by (safe; simp) unat_arith

lemma word_ge_min:
  \<open>- (2 ^ (word_bits - 1)) \<le> sint x\<close> for x :: machine_word
  using sint_ge [of x] by (simp add: word_bits_def)

lemma word_rsplit_0:
  "word_rsplit (0 :: machine_word) = replicate (word_bits div 8) (0 :: 8 word)"
  by (simp add: word_rsplit_def bin_rsplit_def word_bits_def word_size_def Cons_replicate_eq)

lemma x_less_2_0_1:
  fixes x :: machine_word
  shows "x < 2 \<Longrightarrow> x = 0 \<or> x = 1"
  by (rule x_less_2_0_1') auto

end

end

subset_name: curated


TITLE: Help to define a sequence
QUESTION [1 upvotes]: Imagine a sequence of $N$terms where each term is $R$ percent larger then pair of following elements, and the sum of all terms is constant, say $1$?
For example:
$R = \dfrac{N_i}{N_{i+i} + N_{i+2}}$
$N_i = R \times (N_{i+1}+N_{i+2})$ if $i > N-2$
and
$\sum N_i = 1$
I wanted to create a function $f$ such as provided with number of elements in group $N$ , ratio between element and the sum of to following elements $R$ and index of an element $i$ would provide $i^{th}$ element:
$f(N, R, i) = N_i$
EDIT: Example:
For example a sequence (that doesn't add to 1), with $N = 5$, $R = 0.8$:
$7.68, 5.6, 4, 3, 2$
Note that $7.68 = 0.8 \times (5.6 + 4)$, $5.6 = 0.8 \times (4 + 3)$ and $4 = 0.8 \times (3+2)$
EDIT #2: Fixed first sentence to reflect what I wanted to know.
I'm a novice in math, and this question is totally over my head, so any help would be welcome.

REPLY [0 votes]: I am a little unsure what you are asking.
In your first sentence you say "each term is R percent larger then previous element", which leads to the recurrence $a_n = (1+R)a_{n-1}$ as DonAntonio and Arturo have analyzed.
But in your example each term is actually a constant factor larger than the sum of the previous two elements, which is $a_n = (1+R)(a_{n-1} + a_{n-2})$.  This is a more difficult problem.
For convenience write $a_1 = \alpha$, $a_2 = \beta$, and $(1+R) = r$.  Then the two-term recurrence gives
\begin{align*}
  a_1 &= \alpha \\
  a_2 &= \beta \\
  a_3 &= r(a_1 + a_2) = r\alpha + r\beta \\
  a_4 &= r(a_2 + a_3) = r^2\alpha + (r + r^2)\beta \\
  a_5 &= r(a_3 + a_4) = (r^2 + r^3)\alpha + (2r^2 + r^3)\beta \\
  a_6 &= r(a_4 + a_5) = (2r^3 + r^4)\alpha + (r^2 + 3r^3 + r^4)\beta
\end{align*}
and so on.  The corresponding sums $S_N = \sum_{i=1}^N a_i$ are
\begin{align*}
  S_1 &=\alpha \\
  S_2 &=\alpha + \beta \\
  S_3 &=(1+r)\alpha + (1+r)\beta \\
  S_4 &=(1+r+r^2)\alpha + (1 + 2r + r^2)\beta \\
  S_5 &=(1+r+2r^2+r^3)\alpha + (1+2r+3r^2+r^3)\beta \\
  S_6 &=(1+r+2r^2+3r^3+r^4)\alpha + (1+2r+4r^2+4r^3 + r^4)\beta
\end{align*}
It is possible that we can spot a pattern here.  Perhaps the coefficients in the expression before $\alpha$ are simply $1,1, 2, 3, ..., N-3, 1$ (though I don't see a nice pattern in the $\beta$ expression).  Then, if we are lucky, we could find a closed form expression for $S_N$, which we could use to choose $\alpha$ and $\beta$ to give us a sum of 1.
But I am not optimistic.
Edit My lack of optimism was misplaced.  If we take $\alpha = 2$ and $\beta = r$ (and index starting at zero rather than one) we have a (non-integer) Lucas sequence $V_n(r,-r)$:
\begin{align*}
  V_0(r, -r) &= 2, \\
  V_1(r, -r) &= r,\\
  V_n(r, -r) &= rV_{n-1}(r,-r) + rV_{n-2}(r,-r).
\end{align*}
The analysis in the Wikipedia page shows that $V_n = a^n + b^n$ where
$$a = \frac{r + \sqrt{\Delta}}{2}, \quad b = \frac{r - \sqrt{\Delta}}{2}, \quad \Delta=r^2 + 4r$$
and so the sum $S_N$ is
\begin{align*}
  S_N &= \sum_{n=0}^N a^n + b^n \\
      &= \frac{a^{N+1}-1}{a-1} + \frac{b^{N+1} -1}{b-1}.
\end{align*}
Then, by starting our sequence with $2/S_N$ and $r/S_N$ rather than with $2$ and $r$ we will get a sequence with a sum of 1.
Note, though, that this method doesn't allow us to choose $a_1/a_0$: the ratio must be $r/2$.


text:
\begin{document}

\vspace*{40mm}
\begin{center}
        {\Large {\bf{THE MODELLING AND ANALYSIS OF FRACTIONAL-ORDER
                     CONTROL SYSTEMS IN DISCRETE DOMAIN}}} \\
         \vspace{10mm}
         \normalsize
         {Ivo PETR\'A\v{S}, \softL{}ubom\'{\i}r DOR\v{C}\'AK, Imrich
	   KO\v{S}TIAL \\
               \vspace{1mm}
                {   Department of Informatics and Process Control  \\
                    BERG Faculty, Technical University of Ko\v{s}ice   \\
                    B. N\v{e}mcovej 3, 042 00 Ko\v{s}ice, Slovak Republic
		    \\
                    phone:      (+42195) 6025172                   \\
                    e-mail: {\it\{petras, dorcak, kostial\}@tuke.sk}}   \\
                  }

\end{center}
\vskip 5mm
\begin{abstract}
\hspace{-6.5mm}
This paper deals with fractional-order controlled systems and fractional-order controllers in the discrete domain. The mathematical description by the fractional difference equations and properties of these systems are presented. A practical example for modelling the fractional-order control loop is shown and obtained results are discussed in conclusion.

\end{abstract}

\hspace{2.5mm}
{\bf Key words:}\,discrete fractional calculus,
                  digital controller,  discrete fractional-order system.

\section*{1. INTRODUCTION}

Fractional calculus was used for modelling  of physical systems,
but we can find only few works dealing with the application of
this mathematical tool in control theory (e.g. \cite{Axtell, Dorcak,
Matignon, Petras1, Oustaloup, Podlubny}). These works used continuous
mathematical models of fractional order. The fractional-order
systems have a unlimited memory, being integer-order systems
cases in which the memory is limited.

It is necessary to realise that the case of discrete
fractional-order systems is very important for their description
to have a finite difference equation. Such an equation can be
obtained by numerical approximation and by the inverse Z-transform
of a~discrete transfer function.

The aim of this paper is to show, how by using the fractional
calculus, we can obtain a more general fractional-order models
of the controlled objects and general structure for the classical
$PID$ controller in the discrete domain.

\section*{2. DISCRETE FRACTIONAL CALCULUS}

The idea of fractional calculus has been known since the development
of the regular calculus, with the first reference
probably being associated with Leibniz and L'Hospital in 1695.

Fractional calculus is a generalisation of integration and differentiation
to non-integer order fundamental operator $_{a}D^{\alpha}_{t}$,
where $a$ and $t$ are the limits of the operation.
The continuous integro-differential operator is defined as
$$
 _aD^{\alpha}_{t} = \left \{
        \begin{array}{ll}
                \frac{d^{\alpha}}{dt^{\alpha}} & \mbox{$\Re(\alpha)>0$,} \\
                 1 & \mbox{$\Re(\alpha)=0$,} \\
                \int_{a}^{t} (d\tau)^{-\alpha} & \mbox{$\Re(\alpha)<0$.}
        \end{array}
        \right.
$$
The two definitions used for the general fractional differintegral
are the Gr\"unwald-Letnikov (GL) definition and the Riemann-Liouville (RL)
definition \cite{Oldham}. The GL is given here
\begin{equation}\label{GLD_d}
     _{a}D^{\alpha}_{t}f(t)=\lim_{h \to 0} h^{-\alpha}
     \sum_{j=0}^{[\frac{t-a}{h}]}(-1)^j {\alpha \choose j}f(t-jh),
\end{equation}
where $[x]$ means the integer part of $x$. The RL definition is given as
\begin{equation}\label{LRL}
   _{a}D_{t}^{\alpha}f(t)=
    \frac{1}{\Gamma (n -\alpha)}
    \frac{d^{n}}{dt^{n}}
    \int_{a}^{t}
    \frac{f(\tau)}{(t-\tau)^{\alpha - n + 1}}d\tau, \\
\end{equation}
for $(n-1 < \alpha <n)$ and
where $\Gamma (.)$ is the well known Euler's {\it Gamma}  function.

In general, the approximation for fractional operator of
order $\alpha$ can be expressed by generating function
$\omega(\xi^{-1})$, where $\xi^{-1}$ is the shift operator. This
generating function and its expansion determine both the form of
the approximation and the coefficients \cite{Lubich}.

For discrete approximation of the time derivative,  we can use
the generating function corresponding to the Z-transform of
backward difference rule,
\begin{equation}
   \omega(z^{-1}) = \frac{1-z^{-1}}{T},
\end{equation}
and performing the power series expansion (PSE) of $(1-z^{-1})^{\pm \alpha}$,
we obtain the Z version of the GL formula by using the short
memory principle \cite{Podlubny}, for the discrete equivalent
of the fractional-order integro-differential operator
$\omega(z^{-1})^{\pm \alpha}$,
\begin{equation}\label{DFD}
(\omega(z^{-1}))^{\pm \alpha} = T^{\mp \alpha} \sum_{j=0}^{[\frac{L}{T}]} (-1)^j
{\pm \alpha \choose j} z^{[\frac{L}{T}] - j},
\end{equation}
where $T$ is the sample period, $L$ is the memory length, $[x]$ is the
integer part of $x$ and $(-1)^j {\pm \alpha \choose j}$ are
a~binomial coefficients $c_j^{(\alpha)}, \,(j=0, 1, \dots)$.
For its calculation we can use the following expression:
\begin{equation}\label{b_k}
  c_0^{(\alpha)} = 1, \qquad c_j^{(\alpha)} = \left (1 -
                       \frac{1+ (\pm \alpha)}{j}\right)c_{j-1}^{(\alpha)}.
\end{equation}

Another possibility for the discrete approximation is the use of
the trapezoidal (Tustin) rule as a generating function for PSE.  For the discrete equivalent of the fractional-order integro-differential operator,
we can write a general formula \cite{Matignon}
\begin{equation}
(\omega(z^{-1}))^{\pm \alpha} = \left ( \frac{2}{T}
\frac{1-z^{-1}}{1+z^{-1}} \right )^{\pm \alpha}
\end{equation}
for obtaining the coefficients and the form of the approximation.

A detailed review of the approximation methods (Calson's,
Chareff's, CFE, Matsuda's, \dots) for continuous and
discrete fractional-order models was done in work \cite{Vinagre}.

\section*{3. FRACTIONAL-ORDER CONTROL CIRCUIT}

We will be studying the control system shown in Fig.1,
where $G_{c}(z)$ is the controller transfer function,
$G_{s}(z)$ is the controlled system transfer function,
$W(z)$ is an input, $E(z)$ is an error, $U(z)$ is the output
from controller and $Y(z)$ is the output from system \cite{Dorf}.

   \vspace*{4.1cm}
   \vspace{-20mm}
   \begin{picture}(110,42)
\setlength{\unitlength}{1mm}
   \put(26,27){$W(z) \ \ +$}
   \put(37,19){$ -$}
   \put(49.5,27){$E(z)$}
   \put(81.0,27){$U(z)$}
   \put(117.0,27){$Y(z)$}
   \put(26,25){\vector(1,0){15.9}}
   \put(45,10){\vector(0,1){11.7}}
   \put(45,25){\circle{6}}
   \put(42.9,27.0){\line(1,-1){4.0}}
   \put(42.9,23.0){\line(1, 1){4.0}}
   \put(48,25){\vector(1,0){12.2}}
   \put(45,10){\line(1,0){72.1}}
   \put(60,20){\framebox(20,10) [cc]{$G_c(z)$}}
   \put(80,25){\vector(1,0){10.0}}
   \put(90,20){\framebox(20,10) [cc]{$G_s(z)$}}
   \put(110,25){\vector(1,0){15.9}}
   \put(117,10){\line(0,1){15.1}}
\end{picture}
   \vspace{-10mm}
   \centerline{Figure 1: {\it Feed - back control loop}}
   \vspace{1mm}

\subsection*{3.1 Fractional-order controlled system}

The fractional-order controlled system will be represented with a fractional model with the fractional differential equation given by the
following expression $(_0D^{\mu}_{t} \equiv D^{\mu}_{t})$:
\begin{equation} \label{n-DR}
     a_{n}\, D^{\beta_{n}}_ty(t) + \ldots +
     a_{1}\, D^{\beta_{1}}_ty(t) +
     a_{0}\, D^{\beta_{0}}_ty(t) =
     b_{n}\, D^{\alpha_{n}}_tu(t) + \ldots +
     b_{1}\, D^{\alpha_{1}}_tu(t) +
     b_{0}\, D^{\alpha_{0}}_tu(t),
\end{equation}
where  $\beta_k, \alpha_k$  $(k = 0, 1, 2, \dots)$ are generally real
numbers, $\beta_{n} > \ldots > \beta_{1} > \beta_{0}$,
$\alpha_{m} > \ldots > \alpha_{1} > \alpha_{0}$ and
$a_k, b_k$ $(k= 0, 1, \dots)$ are arbitrary constants.

For obtaining a discrete model of the fractional-order system
(\ref{n-DR}), we have to use discrete approximations of the
fractional-order integro-differential operators and then we
obtain a general expression for the discrete transfer function of
the controlled system \cite{Vinagre}
\begin{equation}
  G_s(z)= \frac{b_m (\omega(z^{-1}))^{\alpha_m}+ \dots +
          b_1 (\omega(z^{-1}))^{\alpha_1}+b_0(\omega(z^{-1}))^{\alpha_0}}
               {a_n (\omega(z^{-1}))^{\beta_n}+ \dots +
          a_1 (\omega(z^{-1}))^{\beta_1}+a_0(\omega(z^{-1}))^{\beta_0}},
\end{equation}
where $(\omega(z^{-1}))$ denotes the discrete operator, expressed
as a function of the complex variable $z$ or the shift operator
$z^{-1}$.

For discrete time step $k$, according to relation (\ref{DFD})
and the inverse Z-transform of the difference equation (\ref{n-DR}),
we can write the fractional-order difference equation in the form
\begin{eqnarray}\label{r4_a}
  a_n\, T^{-\beta_n}\sum_{j=0}^{k}c_j^{(\beta_n)} y_{k-j} + \dots +
  a_1\, T^{-\beta_1}\sum_{j=0}^{k}c_j^{(\beta_1)} y_{k-j} +
  a_0\, T^{-\beta_0}\sum_{j=0}^{k}c_j^{(\beta_0)} y_{k-j} =
\nonumber \\
  b_m\, T^{-\alpha_m}\sum_{j=0}^{k}c_j^{(\alpha_m)} u_{k-j} + \dots +
  b_1\, T^{-\alpha_1}\sum_{j=0}^{k}c_j^{(\alpha_1)} u_{k-j} +
  b_0\, T^{-\alpha_0}\sum_{j=0}^{k}c_j^{(\alpha_0)} u_{k-j}.
\end{eqnarray}
From relation (\ref{r4_a}) the following form of
the difference equation can be obtained
\begin{equation}
y_k=\frac{\sum_{i=1}^{m}(b_i T^{-\alpha_i}\sum_{j=0}^{k}
          c_j^{(\alpha_i)}u_{k-j})
          - \sum_{i=1}^{n}(a_i T^{-\beta_i}\sum_{j=1}^{k}
          c_j^{(\beta_i)}y_{k-j})}
         {\sum_{i=0}^{n}a_i T^{-\beta_i}c_0^{(\beta_i)}},
\end{equation}
for $k=2, 3, \dots$, where $y_0=0$ and $y_1=0$.

\subsection*{3.2 Fractional order controller}

The fractional $PI^{\lambda}D^{\delta}$ controller will be represented
by discrete transfer function given in the following expression:
\begin{equation}\label{Gr}
      G_{c}(z)=\frac{U(z)}{E(z)}=
      K + \frac{T_i}{(\omega(z^{-1}))^{\lambda}}+T_d(\omega(z^{-1}))^{\delta},
\end{equation}
where $\lambda$ and $\delta$ are arbitrary real numbers
$(\lambda, \delta \geq 0)$, $K$ is the proportional constant,
$T_i$ is the integration constant and $T_d$ is the derivative
constant.

Taking $\lambda=1$ and $\delta=1$, we obtain a classical $PID$
controller. If $\lambda = 0$ and/or $T_i=0$, we obtain
a~$PD^{\delta}$ controller, etc.
All these types of controllers are particular cases of the
$PI^{\lambda}D^{\delta}$ controller, which is more flexible and
gives an opportunity to better adjust the dynamical properties
of the fractional-order control system.

Similar to methods for controlled system, they can be used for
obtaining the difference equation of the fractional-order controller
(\ref{Gr}).

\section*{4. ILLUSTRATIVE EXAMPLE}

We give in this section an example of modelling the
stable dynamical system by using fractional calculus in
discrete domain. The fractional-order control system consists of
the real controlled system with the coefficients:
\begin{equation}\label{sys}
a_2 = 0.8, \,\, a_1 = 0.5, \,\, a_0 = 1.0, \,\,
\beta_2 = 2.2, \,\, \beta_1 = 0.9, \,\, \beta_0 = 0, \,\,
b_0 = 1.0, \,\, \alpha_0 = 0
\end{equation}
and the fractional-order $PD^{\delta}$ controller, designed on the
stability measure $S_t=2.0$ and damping measure $\xi=0.4$, with
the coefficients:
\begin{equation}\label{cont}
K = 50.0, \,\, T_d = 5.326, \,\, \delta = 1.286.
\end{equation}
The fractional-order differential equation of the closed control loop
has the form $(_0D^{\mu}_{t} \equiv D^{\mu}_{t})$:
\begin{equation} \label{r_44}
  {a_2\, D^{\beta_2}_t y(t) + a_1\, D^{\beta_1}_t y(t) + T_d\, D^{\delta}_t y(t)
  + (a_0 + K) y(t)} = K~w(t) + T_d\,  D^{\delta}_t w(t).
\end{equation}
The resulting differential equation (\ref{r_44}) can be rewritten to the fractional difference equation of the closed control loop in general form
via inverse Z-transform of the relation (\ref{DFD}):
\begin{equation} \label{numer_PD}
       y_k =
       \frac{K w_k+
       T_d{T}^{-\delta}\sum\limits_{j=0}^{k}{c_j^{(\delta)} w_{k-j}}-
       a_2{T}^{-\beta_2}\sum\limits_{j=1}^{k}{c_j^{(\beta_2)} y_{k-j}}-
       a_1{T}^{-\beta_1}\sum\limits_{j=1}^{k}{c_j^{(\beta_1)} y_{k-j}}-
       T_d{T}^{-\delta}\sum\limits_{j=1}^{k}{c_j^{(\delta)} y_{k-j}}
            }
            {
       a_2{T}^{-\beta_2} c_0^{(\beta_2)} +
       a_1{T}^{-\beta_1} c_0^{(\beta_1)} +
       T_d{T}^{-\delta} c_0^{(\delta)} +
       (a_0 +K)
            },
\end{equation}
for $(k=1,2,\dots)$, where  $y_0 = 0,\,w_0 = 0,\, w_1 = 0$\,
and \, $w_k = 1$, for $(k=2,3,\dots)$. The binomial coefficients
were calculated according to relation (\ref{b_k}).

\section*{5. CONCLUSION}

The above methods make it possible to model and analyse
(simulate) fractional-order control systems in the discrete
domain and also to realise digital fractional-order controllers.
We have shown the survey of approximation methods
and Z-transform method as good
approximations of the fractional-order operator
$_aD^{\alpha}_{t}$.
These methods for discretisation of fractional calculus are
suitable for realisation and implementation of
fractional-order controllers, because this controllers
are more robust than classical one.

Stability investigation of discrete fractional-order control
system and practical simulation will be studied in further work.

\section*{ACKNOWLEDGEMENTS}

This work was partially supported by grant {\it \,Fractional-order Dynamical Systems and Controllers:
Discrete and Frequency-Domain Models and Algorithms} from
Austrian Institute for East and South-East Europe
and partially supported by grant VEGA 1/7098/20 from the
Slovak Agency for Science.

config: arxiv


TITLE: How to show that $z + \frac{1}{z}$ maps $\{z=x+iy|y>0\} \cap \{z||z|\leqslant1\}$ onto the lower half plane?
QUESTION [0 upvotes]: How to show that $z + \frac{1}{z}$ maps $\{z=x+iy|y>0\} \cap \{z||z|\leqslant 1\}$ onto the lower half plane?
Suppose $y$ belongs to the lower half plane, want to solve $z + \frac{1}{z}=y$. There must be two solutions(counting multiplicity) by fundamental theorem of algebra. But how do I guarantee that one solution must be in $\{z=ix+y|y>0\} \cap \{z||z|=1\}$?

REPLY [1 votes]: Let
$$S:=\{z=x+iy|y>0\} \cap \{z||z|\leq 1\} \\
T= \{w | Im(w) \leq 0 \} \\
f(z) =z+\frac{1}{z}$$
Then
$$f(z)=z+\frac{1}{z}=z+\frac{\bar{z}}{|z|^2}$$
Since $|z|\leq 1$ we have
$$
Im(f(z))=y-\frac{y}{|z|^2} \leq 0 \,.
$$
This shows that $f(S) \subset T$.
Next, to prove ONTO.
Let $w \in T$.
If $Im(w)=0$ then look for $z \in S$ of $|z|=1$ such that $f(z)=w$.
Recall that in this case you have
$$f(z)=z+\frac{\bar{z}}{|z|^2}=z+\bar{z}$$
If $Im(z)<0$ then you need
$$z+\frac{\bar{z}}{|z|^2}= w \Rightarrow \\
\bar{z}+\frac{z}{|z|^2}= \bar{w} \\
$$
Try adding and subtracting these two relations to get something about $x$ and $y$.


TITLE: Triangle law of vector addition
QUESTION [1 upvotes]: In the triangle law of vector addition, how can we prove that the resultant vector of a and b points in the direction c?

REPLY [1 votes]: Note we can prove it using physics . let the distance travelled by a particle be $a+b$ according to your figure. But the displacement is $c$ . now the triangle be $90$ at $B$ so we know magnitude if resultant is $\sqrt{a^2+b^2+2abcos(90)}=\sqrt{a^2+b^2}$ also magnitude of $c$ is $√c^2$ so if we square both we get $a^2+b^2,c^2$ but they are equal according to pythagoras theorem therefore not only resultant vector but magnitude of both vectors is the same. Also alternate way is complete the parallelogram and by parallelogram law of vectors the resultant of adjacent vectors  is diagonal . now as |d| which is parallel to $|b|$ are equal hence the result follows .


TITLE: Most efficient way to find spaces within a square on a 2D Cartesian grid?
QUESTION [0 upvotes]: The context here is that I'm writing a method to randomly place items on a game map and I want to be able to define an area around the items that is empty so they each have some breathing room. I'd still like to find a more efficient method if anyone has any ideas. I implemented this solution but it is slower than I'd like.
What's the most efficient way to find all spaces within a square with a given length centered on a given space in a 2D Cartesian grid? I get in over my head rapidly with math jargon so searching didn't help much.
Say I have a regular 2D grid, like graph paper. Pick a coordinate $c$ and make it the center of a square extending d distance from that point. So the square extends from $c-d$ to $c+d$ in both dimensions. Is there a clever/efficient way to get all the graph spaces falling within this square or am I overthinking it?

I mean get a list of all grid spaces in the square.

REPLY [0 votes]: If the coordinates of the centerpoint are $(c_1, c_2)$, then the coordinates of the cells are all given by $(c_1 + l, c_2 + k)$, where both $l$ and $k$ range from $-d$ to $d$.
An algorithm that outputs all coordinates would work like this:

Choose $l = -d$ and print $(c_1 - d, c_2 - d), (c_1 - d, c_2 - d + 1), \ldots, (c_1 - d, c_2 + d)$
Choose $l = -d + 1$ and print $(c_1 - d - 1, c_2 - d), (c_1 - d - 1, c_2 - d + 1), \ldots, (c_1 - d - 1, c_2 + d)$.
$\ldots$
Choose $l = d$ and print $(c_1 + d, c_2 - d), (c_1 + d, c_2 - d + 1), \ldots, (c_1 + d, c_2 + d)$.


TITLE: Does every continuous map from $\mathbb{Q}$ to $\mathbb{Q}$ extends continuously as a map from $\mathbb{R}$ to $\mathbb{R}$?
QUESTION [10 upvotes]: Given a continuous function $f:\mathbb{Q}\to\mathbb{Q}$ ,does there exist a continuous function  $g:\mathbb{R}\to\mathbb{R}$ such that   $g|_{\Bbb Q} = f$?
What I have no Idea about how to attempt this Question! Any suggestion will be very helpful.

REPLY [16 votes]: Not in general. Consider the function $f_{0}:\mathbb{Q}\rightarrow\mathbb{Q}$ defined by $$f_{0}(x)=\begin{cases}0 \hspace{4mm} \mbox{if } x^{2}<2 \\ 1 \hspace{4mm} \mbox{if } x^{2}>2\end{cases}.$$ It is easy to see that $f_{0}$ is continuous on $\mathbb{Q}$ (since $\sqrt{2}\notin\mathbb{Q}$), but if $g_{0}:\mathbb{R}\rightarrow\mathbb{R}$ is an extension of $f_{0}$ to $\mathbb{R}$, then $g_{0}$ is necessarily discontinuous at $\sqrt{2}$.
It is worth noting that if $f:\mathbb{Q}\rightarrow\mathbb{Q}$ can be extended to a continuous function $g:\mathbb{R}\rightarrow\mathbb{R}$, then such an extension is unique. That is, if $g:\mathbb{R}\rightarrow\mathbb{R}$ and $h:\mathbb{R}\rightarrow\mathbb{R}$ are continuous functions such that $g(x)=h(x)$ for all $x\in\mathbb{Q}$, then $g=h$. This is because $\mathbb{Q}$ is a dense subset of $\mathbb{R}$.


TITLE: Comparing energy band diagram for metals, semiconductors and insulators
QUESTION [0 upvotes]: The energy band diagram is a model for describing why an insulator is not a good conductor compared to a semiconductor: in the first case, electrons would need to pass a very large band gap in order to promote from the valence band to the conduction band. The energy required for this is simply too large (and would lead to too much heat in the material).
For the semiconductor, on the other hand, the energy is smaller and electrons can move from the valence band to the conduction band by thermal excitation.
Fine, but which of the following figures is "more" correct, the one where the conduction is of the same thickness, or the one where the conduction bands' upper levels are the same?
The first one is more often used, but that might not mean too much. Any thoughts or ideas?

REPLY [0 votes]: The upper bound of conduction band is $\infty$ so it doesn't matter which of the figure is correct and which one is incorrect.


TITLE: Proving $[A:B]=[A^f:B^f][A_f:B_f]$
QUESTION [2 upvotes]: I'm having a hard time showing an equality of indices holds. This is exercise 1.44 from Lang's Algebra.

Suppose $f\colon A\to A'$ is a homomorphism of abelian groups, and $B\leq A$. Denote by $A^f$ and $A_f$ the image and kernel of $f$ in $A$ respectively, and similarly for $B^f$ and $B_f$. Show that $[A:B]=[A^f:B^f][A_f:B_f]$ in the sense that if two of these three indices are finite, so is the third, and the stated equality holds.

My initial thought was to compose a bijection from $A/B$ to $A^f/B^f\times A_f/B_f$. I tried to map a coset $aB$ to $(f(a)B^f,\underline{\hspace{.75cm}})$ but couldn't think of a nice candidate for the right coordinate.
Does anyone have suggestions on the right approach here? Thanks for your input.

REPLY [0 votes]: Let $P$ be a representant system for cosets of $B^f$ in $A^f$. Let $Q$ be a representant system for cosets of $B_f$ in $A_f$. For every $p\in P$ choose an element $a_p\in A$ such that $f(a_p)=p$. We will prove that the indexed family $(a_p+q)_{(p,q)\in P\times Q}$ is a representant system for cosets of $B$ in $A$.
For $x\in A$ then $f(x)\in A^f$, so there is a $p\in P$ such that $f(x)\in p+B^f$, so $f(x)-f(a_p)\in B^f$, so there is a $b\in B$ such that $f(x)-f(a_p)=f(b)$, so $x-a_p-b\in A_f$, so there is a $q\in Q$ such that $x-a_p-b\in q+B$, so $x\in a_p+q+B$.
Also, if $a_p+q+B=a_{p'}+q'+B$ then $a_p+q-a_{p'}-q'\in B$, so $f(a_p)+f(q)-f(a_{p'})-f(q')\in B^f$, so $p-p'\in B^f$, so $p=p'$, so $q-q'\in B$, but $q-q'\in A_f$, so $q-q'\in B_f$, so $q=q'$.


TITLE: Calculate u in terms of time such that a particle maintains a constant speed following a parametric equation
QUESTION [2 upvotes]: I have a parametric equation given by:
$x=\cos(6u)$
$y=\sin(4u)$
And I understand that the speed of a particle at any given t is:
$\sqrt{\left(\dfrac{dx}{du}\right)^2 + \left(\dfrac {dy}{du} \right)^2} $
Which for my parametric equation would be:
$\sqrt{36 \sin^2(6u) + 16 \cos^2(4u)}$
I am trying to make a particle move at a constant speed along the parametric equation.
I have looked and seem to be unable to find information on how to do this, the only thing I did find is this paper on the subject.
Unfortunately given my rather basic grasp of calculus I wasn't able to make much sense of it.
To clarify, I am looking for $u$ in terms of time such that the particle maintains constant speed.
Thanks.

REPLY [2 votes]: As Yves Daoust explained, you can use one of the Runge-Kutta methods to numerically solve the differential equation for your parameter. However, if the step size is small (the speed is low) you can get by with simple Euler integration. In other words,
$\frac{\Delta u}{\Delta t} \approx \frac{du}{dt}$
so just let $\Delta u = \Delta t \cdot \frac{du}{dt}$
To optimize this calculation we can express the arc length differential in terms of $x$ and $y$. The equations below use $a$ and $b$ for the $x$ & $y$ frequencies, rather than your 6 & 4, in order to make the result apply to any Lissajous figure. They also use $\theta$ for the parameter; I figured that was appropriate because we're dealing with trigonometric functions, and I guess it looks a little more professional. :)
$$\begin{align}
\text{Let } x & = \cos (a \theta), \, y = \sin (b \theta) \\
\frac{dx}{d\theta} & = - a \sin (a \theta), \, \frac{dy}{d\theta} = b \cos (b \theta) \\
\\
\frac{ds}{d\theta}^2 & = \frac{dx}{d\theta}^2 + \frac{dy}{d\theta}^2 \\
& = a^2 \sin^2 (a \theta) + b^2 \cos^2 (b \theta) \\
& = a^2 (1 - \cos^2 (a \theta)) + b^2 (1 - \sin^2 (b \theta)) \\
& = a^2 (1 - x^2) + b^2 (1 - y^2) \\
\frac{ds}{d\theta} & = \sqrt{a^2 (1 - x^2) + b^2 (1 - y^2)}\\
\\
v & = \frac{ds}{dt} = \frac{ds}{d\theta} \cdot \frac{d\theta}{dt}\\
\\
\text{Therefore}\\
\frac{d\theta}{dt} & = \frac{v}{ds/d\theta}\\
\Delta\theta & \approx \frac{\Delta t \cdot v}{ds/d\theta}\\
& = \frac{\Delta t \cdot v}{\sqrt{a^2 (1 - x^2) + b^2 (1 - y^2)}}
\end{align}$$
Here's a small Python snippet to illustrate the algorithm:
def liss(a, b, speed, width, height):
    pix = bitmap(width, height)
    ox, oy = width // 2, height // 2
    rad = min(ox, oy) - 3

    #Precompute constants used in speed calculation
    r2 = rad * rad
    a2 = a * a
    b2 = b * b
    c2 = (a2 + b2) * r2

    maxtheta = 2. * pi
    theta = 0.0
    while theta < maxtheta:
        x = rad * cos(a * theta)
        y = rad * sin(b * theta)
        pix.put(int(.5 + ox + x), int(.5 + oy + y))
        dtheta = speed / (c2 - a2 * x*x - b2 * y*y) ** 0.5
        theta += dtheta

Here's the output of the above algorithm, with a=3, b=2, speed=10:

As you can see, it's fairly smooth. But we can make it better by following Yves' advice and using Runge-Kutta integration. Here's Python code that uses RK4:
def liss(a, b, speed, width, height):
    pix = bitmap(width, height)
    ox, oy = width // 2, height // 2
    rad = min(ox, oy) - 3

    def speedfunc(theta):
        s = (a * sin(a * theta))**2 + (b * cos(b * theta))**2
        return speed / (rad * s ** 0.5)

    maxtheta = 2. * pi
    theta = 0.0
    while theta < maxtheta:
        x = rad * cos(a * theta)
        y = rad * sin(b * theta)
        pix.put(int(.5 + ox + x), int(.5 + oy + y))

        #Calculate Runge-Kutta 4 increments
        k1 = speedfunc(theta)
        k2 = speedfunc(theta + 0.5 * k1)
        k3 = speedfunc(theta + 0.5 * k2)
        k4 = speedfunc(theta + k3)

        dtheta = (k1 + 2.0 * k2 + 2.0 * k3 + k4) / 6.0
        theta += dtheta

And its output, with the same parameters as above:

It's hard to see much difference here, but the RK4 algorithm is clearly superior for high speeds, especially in the regions of the curve where the acceleration is high.
In case you're interested, here's a Pastebin link to the full Python 2 program used to produce the RK4 version as a plain PBM file; you can easily adapt it to use the Euler algorithm. This program was tested on Python 2.6 but it probably runs on earlier versions; it will need minor modifications to run on Python 3.
LissRK4.py
Finally, I'd like to mention that although these constant speed algorithms make nice-looking plots, I have to agree with user86418 that the simple constant angle increment algorithm looks more natural in an animation; after all, this is simple harmonic motion. :)


TITLE: Chinese remaining theorem, help with surjectivity
QUESTION [0 upvotes]: Let $R$ be a ring and let $I_1, \dots, I_n \subset R$ be ideals, which are pairwise coprime. Let furthermore $\pi_j:R \to R/I_j,\; r \mapsto r + I_j$, $j=1, \dots, n$ be the projection to the $jth$ quotient ring.
I know that for all $j \in \{1, \dots, n\}$, the ideals $I_j$ and $\cap_{i \neq j}I_i$ are coprime. For all $j \in \{1, \dots, n\}$ we take $d_j \in I_j$ and $e_j \in \cap_{i \neq j}I_i$ s.t. $d_j + e_j=1$.
$$\Rightarrow \pi_i(e_j)=\pi_i(1-d_j)$$
This is now equal to $0$, if $i \neq j$ and equal to $1+I_j$, if $ i=j$. I don't see why we get this.
Many thanks for any help!

REPLY [1 votes]: Ideals $I_j$ and $\bigcap_{i\ne j} I_i$ are coprime, which means that $I_j + \bigcap_{i\ne j} I_i = R$. Thus for $1\in R$ there are $d_j\in I_j$ and $e_j\in \bigcap_{i\ne j} I_i$ with $d_j+e_j=1.$
Then $\pi_i(e_j) = 0$, since $e_j\in I_i$ for $i\ne j$.


TITLE: Computing norm for linear functional
QUESTION [1 upvotes]: Consider the space $\mathbb{E}:=\mathbb{R}^n$ to be equipped with the norm
$$||x||_p:=\Bigg(\sum_{k=1}^n |x_k|^p\Bigg)^{\frac{1}{p}},\,\,\,\,x=(x_1,...,x_n)^T\in\mathbb{E}$$
where $p>1$. Consider the space $\mathbb{E}^*$ and let $f\in\mathbb{E}^*$ be given by
$$\langle f,x\rangle:=\sum_{k=1}^n x_ky_k$$
Compute
$$||f||:=\sup_{||x||_p\le 1}\langle f,x\rangle$$
So far I did the following
BY uisng Holder's inequality we have
$$\langle f,x\rangle=\sum_{k=1}^n x_ky_k\le||x||_p ||y||_q$$
Therefore
$$||f||=\sup_{||x||_p\le 1}\langle f,x\rangle\le\sup_{||x||_p\le 1}||x||_p ||y||_q\le ||y||_q$$
I am not quite sure whether I am right so far, also I am stuck with the reverse way. Any help would be highly appreciated. Thanks in advance.

REPLY [2 votes]: Now, if we take $x_i=sig(y_i)|y_i|^{q-1}$, where $sig(y_i)$ is the sign of the real number $y_i$, we have that
$\lVert x \rVert_p=\left(\sum_{i=1}^n |y_i|^{qp-p}\right)^{\frac1p} = \left(\sum_{i=1}^n |y_i|^{q}\right)^{\frac1p}$. Now:
$$\langle f,x \rangle=\sum_{i=1}^n |y_i|^q = \left(\sum_{i=1}^n |y_i|^q \right)^{\frac 1p+\frac 1q}=\left(\sum_{i=1}^n |y_i|^q \right)^{\frac 1p}\left(\sum_{i=1}^n |y_i|^q \right)^{\frac 1q}=\lVert x\rVert_p \lVert y\rVert_q $$
So $\langle f,\frac{x}{\lVert x \rVert} \rangle=\lVert y\rVert_q$.


TITLE: Could the observable universe be bigger than the universe?
QUESTION [69 upvotes]: First of all, I'm a layman to cosmology. So please excuse the possibly oversimplified picture I have in mind.
I was wondering how we could know that the observable universe is only a fraction of the overall universe. If we imagine the universe like the surface of a balloon we could be seeing only a small part of the balloon

or we could be seeing around the whole balloon

so that one of the apparently distant galaxies is actually our own.
In the example with the balloon one could measure the curvature of spacetime to estimate the size of the overall universe, but one could also think about something like a cube with periodic boundary conditions.

Is it possible to tell the size of the overall universe?

Artistic image of the observable universe by Pablo Carlos Budassi.

REPLY [0 votes]: We could discover that the actual universe is larger than the observable universe when there is a non-zero constant curvature everwhere. Then, this would be like standing upon the earth and seeing as far as the horizon. In the picture drawn above what we can observe would be limited by a kind of cosmological horizon.


TITLE: What is $\mathbb{F}_7[X]$?
QUESTION [1 upvotes]: I do not understand what sets like these are. I know what something like $\mathbb{Z}_7$ is. It is the ring of integers modulo 7 so it is equal to ${0,1,2,3,4,5,6}$. But what is $\mathbb{F}_7[X]$ equal to. I don't understand. I have spent ages searching on the internet but can't find anything on it. All I know is that it has 7 elements but I just want to see a clear definition of it.

REPLY [4 votes]: Well, you know what the ring $\mathbb{Z}_{n}$ is for any integer $n$.  It turns out that if $n$ is a prime number (like $2$, $3$, $5$, $7$, etc.) (when $n$ is prime, it is usually replaced with the letter $p$) then $\mathbb{Z}_{p}$ is actually a field!  It's not just a ring, but it also has a multiplicative identity (if your definition of ring doesn't already come with this assumption), no zero divisors, and every nonzero element has a multiplicative inverse.
Usually, people write $\mathbb{F}_{p}$ instead of $\mathbb{Z}_{p}$ since this is a field.  Somehow, calling it $\mathbb{F}_{p}$ reminds us that it is the field of order $p$.
Now, what happens when we "adjoin $X$" to this field? That is, what is $\mathbb{F}_{p}[X]$?  It is simply the set $\{a_{0} + a_{1}X + a_{2}X^{2} + \dots + a_{n}X^{n} \mid a_{i} \in \mathbb{F}_{p}, n \in \mathbb{N} \}$.  In other words, $\mathbb{F}_{p}[X]$ is the set of polynomials whose coefficients come from $\mathbb{F}_{p}$.
So, for example, we know $\mathbb{F}_{7} = \mathbb{Z}_{7} = \{ [0], [1], [2], [3], [4], [5], [6] \}$.  Then some elements in $\mathbb{F}_{7}[X]$ are:
1) $[1] + [3]X^{2} + [5]X^{9}$
2) $[6]X^{2} + [4]X^{3}$
3) $[5]$ (a constant polynomial)
Any polynomial you can think of (remember that polynomials have finite degree, i.e., a finite highest power of $X$) with coefficients in $\mathbb{F}_{7}$ is in $\mathbb{F}_{7}[X]$.
$\mathbb{F}_{7}[X]$ is called the ring of polynomials with coefficients in $\mathbb{F}_{7}$, and this is actually a ring with the addition and multiplication you expect between two polynomials.


TITLE: Solution verification: $G$ and $G/H$ contain elements of same order
QUESTION [2 upvotes]: I just took my abstract algebra midterm, and was wondering if someone could confirm my solution to the following problem.

Problem: Let $G$ be a finite group and let $H$ be a normal subgroup of $G$. Let $n \geq 1$ be an integer and suppose there exists an element
  of order $n$ in $G/H$. Prove that there exists an element of order $n$
  in $G$.

My Solution: Since $H$ is a normal subgroup, there exists a surjective homomorphism $\varphi: G \to G$ such that $\ker(\varphi) = H$. By the First Isomorphism Theorem,
$$
G/\ker(\varphi) \cong \varphi(G).
$$
Since $\ker(\varphi) = H$ and $\varphi(G) = (G)$ (since $\varphi$ is surjective), this becomes
$$
G/H \cong G
$$
which implies that
$$
|G/H| = |G|.
$$
Since $G/H$ and $G$ are of the same order, if $G/H$ contains an element of order $n$, $G$ must as well.

REPLY [2 votes]: If $(gH)^m=H$, so $gH$ has order $m$ in G/H, then $g^m\in H$ (why?). If $g^m=1$, then we are done, otherwise $g^m\in H$ has some order, $n$ say. So $g^{mn}=1$. Consider $g^n$...
This "breaks" for infinite groups, as $g$ might have infinite order in $G$. For example, consider $G=\mathbb{Z}$ and $H$ any non-trivial subgroup.

Also, I should say that your concluding sentences incorrect, so your attempt is unfortunately wrong. A counter-example to the last line is the groups of order 4: The Klein 4-group has order 4, as does the cyclic group of order 4, but the Klein 4-group does not contain en element of order 4 while the cyclic group does.


TITLE: Recent Planck probe results: how unexpected?
QUESTION [5 upvotes]: The data from the Planck probe's observations are in, and according to the European Space Agency they show a "hemispheric asymmetry in the cosmic microwave background (CMB)". Quote:

an asymmetry in the average temperatures on opposite hemispheres of
  the sky [...] with slightly higher average temperatures in the
  southern ecliptic hemisphere and slightly lower average temperatures
  in the northern ecliptic hemisphere. This runs counter to the
  prediction made by the standard model that the Universe should be
  broadly similar in any direction we look.

How unexpected is this variance from the Standard Model and can it be quantified?
How certain is it that the data are accurate? For the recent discovery of the Higgs boson at the LHC, a five sigma result was considered sufficient to make the announcement. What is the sigma for the reported hemispherical asymmetry?
Yet the report of faster-than-light neutrinos, subsequently withdrawn due to equipment failures, was based on six sigma evidence. And in one of the backwaters of Wikipedia, we learn that:

Some anomalies in the background radiation have been reported which
  are aligned with the plane of the solar system, which contradicts the
  Copernican principle by suggesting that the solar system's alignment
  is special.[10] Land and Magueijo dubbed this alignment the "axis of
  evil" owing to the implications for current models of the cosmos,[11]
  although several later studies have shown systematic errors in the
  collection of that data and the way it is processed.[12][13][14]
  Various studies of the CMB anisotropy data either confirm the
  Copernican principle,[15] model the alignments in a non-homogeneous
  universe still consistent with the principle,[16] or attempt to
  explain them as local phenomena.[17] Some of these alternate
  explanations were discussed by Copi, et. al., who looked at data from
  the Planck satellite to resolve whether the preferred direction and
  alignments were spurious.[18][19]

(Wikipedia's main Planck probe article makes no mention of the hemispherical asymmetry.)
When can we expect this controversy to be resolved, and are more outcomes possible than (1) the Planck probe data are found to be in error or (2) the Standard Model must undergo major revision?

REPLY [6 votes]: One of the Planck papers discusses these anomalies in detail: Planck 2013 results. XXIII. Isotropy and Statistics of the CMB.

How unexpected is this variance from the Standard Model and can it be quantified? How certain is it that the data are accurate? For the recent discovery of the Higgs boson at the LHC, a five sigma result was considered sufficient to make the announcement. What is the sigma for the reported hemispherical asymmetry?

The anomalies found by Planck were already seen in the WMAP data, so they confirm the existence of these features. I can't find the specific sigma for the hemispherical asymmetry, but most of the anomalies are reported to have a significance of ~$3\sigma$. That's enough to raise eyebrows in the astronomical world, although particle physicists would be less impressed :-)
The question is if these anomalies are actually meaningful, and we can expect heated debates in the coming years between 'believers' and 'non-believers'. There are a few things to consider:

The CMB could be distorted by foreground objects, possibly by our local cluster, our own galaxy and even our solar system.
The CMB pervades the entire universe, but we can only observe a small part of it: we're seeing a cross-section in the form of a sphere, centred on us and with a certain radius (it took the photons 13.8 billion years to reach us). An observer in a different galaxy and/or at a different cosmic time would see a different part of the CMB.

These effects are sometimes called cosmic variance. Basically, it means that what we observe may not be truly representative of the entire  universe.
Also, we're dealing with statistical data, and flukes happen. We're biased at seeing patterns, even though they may not mean anything. For example, if you throw a dice 100 times in a row, then all sorts of apparent patterns may occur. For instance, it could contain the sequence 666666. That specific sequence may seem unlikely and significant, but it is just as likely as any other specific sequence, like e.g. 202020 or 675439.

When can we expect this controversy to be resolved

The Planck team has yet to release the polarisation data (probably next year), which may shed some light on the situation. But I think these issues will entertain the cosmologists for quite a few years. A promising field of research is the mapping of the large-scale distribution of (dark) matter, using gravitational lensing. This would help calculating the Integrated Sachs–Wolfe effect (which is the gravitational redshift/blueshift of CMB photons as they pass through the potential well of galaxy clusters) more accurately.

the Planck probe data are found to be in error

The data is reliable (the Planck data agrees with WMAP), it's all about the interpretation.

the Standard Model must undergo major revision?

It is possible that the explanation lies in a slight deviation from the Standard Model (but no major revision); after all, the Standard Model is an idealisation. Our universe may not be exactly homogeneous or isotropic at large scales. The Planck team actually tried fitting a non-standard model, a so-called Bianchi Model, with mixed results (Planck 2013 results. XXVI. Background geometry and topology of the Universe). Some others get rather carried away, speculating about the influence of 'other universes'.
As a final note, it is important to stress that, in the overall scheme, the impact of these anomalies is very small. The Standard Model fits the CMB almost perfectly, and is in agreement with studies of clusters of galaxies and supernovae.

(Source: Planck 2013 results. I. Overview of products and scientific results, Fig 19)
The fact that the Standard Model works so well and that the cosmological parameters are now known with such accuracy is a very remarkable feat. Thanks to the quality of the data and the success of the theories, cosmologists understand the overall picture with confidence, and can now focus on the details.
P.S. All the Planck papers can be found here: Planck 2013 Results Papers. The most important one is Paper XVI, in which the cosmological parameters are discussed.


TITLE: Sign convention for derivations in CDGAs
QUESTION [3 upvotes]: I'm trying to understand K\"ahler differentials for CDGAs (commutative differential graded algebras). A few minor things have been confusing me all afternoon. Pointers and references are welcome.
Let $A$ be a CDGA over a field of characteristic zero with differential of degree $1$ and with no cohomology in positive degrees. (If we made the differential of degree $-1$, then we might call these connective CDGAs), let $M$ be a left dg-module over $A$, and let $I=ker(A \otimes A \rightarrow A)$. Then I'd like to verify that $\delta:A \rightarrow I/I^{2}, a \mapsto 1 \otimes a - a \otimes 1$ is a derivation, just like for classical (not dg) commutative algebras.
For this to work, I seem to be forced to take the following convention for what I mean by derivation: $\delta(a \cdot b)=b \cdot \delta(a)+a\cdot \delta(b)$. On the other hand, some informal references I read give $\delta(a \cdot b)=\delta(a)\cdot b+a\cdot \delta(b)$, and use the convention that an action on the right $m \cdot x$ is defined by $(-1)^{|m||x|}x \cdot m$, where $x,m$ are homogeneous. Thus altogether
we would require $\delta(a \cdot b)=(-1)^{|a||b|+|b|}b \cdot \delta(a)+a\cdot \delta(b)$ for homogeneous $a,b$. But this seems to mess up the signs to make the given $\delta$ a derivation.
So the question is: what is the right notion of derivation $\delta: A \rightarrow M$ of degree $0$, and more generally of degree $k$?
My apologies for the triviality of the question.

REPLY [8 votes]: There are two steps to working smoothly with sign conventions.  The first is to do as much as possible categorically, where all choices of sign conventions are relegated to single black box.  The second step is only to unpack the black box when necessary, usually at the very end of a calculation.
Let $\mathcal C$ be a symmetric monoidal category.  I denote the monoidal structure by $\otimes$, the monoidal unit by $\mathbf 1$, and the symmetry by $\operatorname{flip}$.  I assume moreover that $\mathcal C$ is enriched in abelian groups (maybe because it is enriched in $k$-modules for some commutative ring $k$).  I assume also that $\mathcal C$ has kernels.
A commutative algebra in $\mathcal C$ is an object $A\in\mathcal C$ along with maps $m: A \otimes A \to A$ and $1: \mathbf 1 \to A$, such that: (commutativity) $m = m\circ \operatorname{flip}$, (associativity) $m\circ (m \otimes \operatorname{id}) = m \circ (\operatorname{id}\otimes m)$, and (units) $m \circ (1 \otimes \operatorname{id}) = \operatorname{id}$.  A (left) module for $A$ is an object $M \in \mathcal C$ and a map $m : A\otimes M \to M$ such that $m\circ (m \otimes \operatorname{id}) = m \circ (\operatorname{id}\otimes m)$.  Note that every left module is also a right module and in fact a bimodule by composing judiciously with $\operatorname{flip}$.
A morphism $\delta : A \to M$ is a derivation if $\delta \circ m = m\circ (\delta \otimes \operatorname{id}) + m\circ (\operatorname{id}\otimes \delta)$.  Then the set $\operatorname{der}(A,M)$ of derivations is naturally an abelian subgroup of $\hom(A,B)$.
Aside:  Suppose that $(\mathcal C,\otimes)$ is closed in the sense that for any object $X$, the functor $\otimes X : \mathcal C \to \mathcal C$ has a right adjoint $\underline{\hom}(X,-)$.  It follows that $\underline{\hom}(-,-)$ is contravariant in the first component.  Then in particular there is an object $\underline{\hom}(A,M) \in \mathcal C$.  It has a subobject $\underline{\operatorname{der}}(A,M) = \ker\bigl( \underline{\hom}(A,M) \rightrightarrows \underline{\hom}(A\otimes A,M)\bigr)$.  One of these maps corresponds to $m : A\otimes A \to A$ in the first component.  The second corresponds to $m : A\otimes M \to M$, and uses the adjunctions a few times in its definition.
Anyway, I believe that your claim that $I = \ker(m : A\otimes A \to A)$ comes equipped with a canonical derivation is satisfied in any symmetric monoidal linear category with kernels.  At worst, you need to include some more adjectives like "abelian" or "presentable".  But it is checkable without reaching inside to see how the objects are builts
Let me describe now a notation for doing such computations.  For any category $\mathcal C$ and any object $A\in \mathcal C$, a generalized element $x\in A$ is a morphism $X \to A$ for some $X\in \mathcal C$.  The generalized elements with domain $X$ are the $X$-points of $A$.  The Yoneda lemma implies that an object is determined by its generalized elements.  (In fact, in a sense $A$ is determined by its canonical $A$-point $\operatorname{id}:A \to A$.)  If $\mathcal C$ is symmetric monoidal, global element of $A \in \mathcal C$ is a morphism $\mathbf 1 \to A$.  Note that global elements do not determine the object.  Generalized elements in a symmetric monoidal category enriched over abelian groups can be manipulated like regular elements of regular abelian groups.  The requirement is that in any expression, there are no duplications or deletions: the expression should be homogeneous linear in each element.
For example, suppose that you have generalized elements $x: X \to A$ and $y : Y \to A$.  Then $xy = m(x\otimes y)$ is a generalized element with domain $X\otimes Y$.  On the other hand, $yx \in A$ has domain $Y \otimes X$.  There is a unique canonical isomorphism $\operatorname{flip}:X\otimes Y \to Y \otimes X$.  Then an equation like "$\forall x,y\in A$, $xy = yx$" has the interpretation: "for all pairs of generalized elements $x : X \to A$ and $y : Y \to A$, we have an equality of morphisms $m\circ (x\otimes y) = m\circ (y\otimes x)\circ \operatorname{flip} : X\otimes Y \to A$."  With this convention, you will never see signs.  For example, a morphism $\delta : A\to M$ is a derivation if for all generalized elements $a,b \in A$, we have $\delta(ab) = a\delta(b) + b\delta(a)$.
This is all well and good, but how do you unpack it?  The category of graded vector spaces is a monoidal category using the "convolution product" — this is what you're used to.  Its symmetric structure is determined by the requirement that $\operatorname{flip} : [1] \otimes [1] \to [1] \otimes [1]$ is minus the identity, where $[1]$ denotes the one-dimensional vector space in degree $1$ (or in degree $-1$, whichever is your convention); it and its inverse $\otimes$-generate the category of graded vector spaces.  You can skeletalize your category to decide that every object is a direct sum of various $[n] = [1]\otimes \dots \otimes [1]$.  With respect to this skeletalization, $[m+n] = [m]\otimes [n] \overset{\operatorname{flip}}\to [n]\otimes[m] = [n+m]$ is the map $(-1)^{mn}$, which is the usual Koszul rule.  This category has a distinguished Lie algebra, which is the abelian Lie algebra on the object $[1]$.  (By "abelian", I mean that the Lie bracket is $0$.)  The category of dg vector spaces is the category of representations of this Lie algebra (the Lie algebra action records the differential).  It is a symmetric monoidal category in the same way that the representation theory of any Lie algebra is a symmetric monoidal category.  In particular, forgetting from a dg vector space to a graded vector space is faithful and preserves the symmetric monoidal structure.  So a morphism in dgVect is a morphism in gVect that is equivariant for the Lie algebra action.
It's then important to remember that forgetting from a graded vector space to a regular vector space is monoidal, but does not preserve the symmetric structure.  So when working with generalized elements in gVect, you can say that a derivation should satisfy $\delta(ab) = a\delta(b) + b\delta(a)$.  But when working with elements of the underlying vector space, you must remember that there was an implicit $\operatorname{flip}$ when the $a$ and $b$ switched places; for that (worse) convention, we have $\delta(ab) = a\delta(b) + (-1)^{|a||b|}b\delta(a)$.
Finally, you asked about derivations of non-zero degree.  As in the aside, suppose that $\mathcal C$ is closed.  Then one always has $\hom(\mathbf 1,\underline{\hom}(A,M)) = \hom(A,M)$ as abelian groups, and also $\hom(\mathbf 1,\underline{\operatorname{der}}(A,M)) = \operatorname{der}(A,M)$.  When $\mathcal C$ is the category of graded or dg vector spaces, the monoidal unit is $\mathbf 1 = [0] = $ the one-dimensional vector space in degree $0$.  Up to a grading convention, a degree-$k$ differential is a morphism $[k] \to \underline{\operatorname{der}}(A,M)$.  It is not hard to unpack this definition: it says that $\delta(ab) = (-1)^{|a|k}a\delta(b) + (-1)^{|b|(|a|+k)}b\delta(a)$.  I think one of your formulas must have been about degree-odd derivations.
It is an unfortunate but common occurrence for writers to use worse conventions.  For example, remember that the canonical isomorphism $[k] \otimes X \cong X\otimes [k]$ is not the identity on underlying vector spaces.  In the conventions above, the adjunction implies that a map $[k] \to \underline{\operatorname{der}}(A,M)$ is a map $[k] \otimes A \to M$ satisfying some condition.  It is the same data as a map $A[k] = A\otimes [k] \to M$, but on underlying vector spaces those two maps are different.
The moral of the story is that you should never cut-and-paste sign-full formulas from anyone else.  Rather, pick a convention and stick to it, and as much as possible make no choices at all by using a categorical framework.


TITLE: Neutrinos detection. What mechanism for their interaction with matter?
QUESTION [0 upvotes]: As for the title, by which way neutrinos can interact with matter such as, e.g., pseudocumene or other scintillators?
Is it a merely mechanical collision?
Do neutrinos collide with nuclei and how "to see" the collision? Normally collisions involve electromagnetic interactions.
I didn't find any hint on Wikipedia articles about Borexino or other experiments.

REPLY [3 votes]: There are two ways to detect a neutrino:

The neutrino "collides" (i.e. interacts via the weak interaction) with a charged particle. In the interaction some of the neutrinos momentum is transferred to the charged particle.  if the neutrino is sufficiently energetic and the charged particle is sufficiently light (e.g. an electron) then the charged particle can be accelerated to relativistic speeds and will emit Cherenkov radiation as it interacts with other particles in the detector medium. This Cherenkov radiation can be detected.
The neutrino interacts with a heavier neutral particle, such as a neutron. The neutron is transformed into a proton, and to conserve charge an energetic charged particle such as an electron, a muon or a tau (depending on the flavour of the neutrino) is created. Again, the electromagnetic radiation given off by this energetic lepton can be detected.

Both of these detection methods depend on weak interactions which are very rare, so a high neutrino flux and a large detector volume is required to produce a useful number of interactions in the detector. Fortunately the flux of solar neutrinos is sufficient to produce useful detection rates. Detection of neutrinos that have originated outside of the solar system is still a very rare event.
The Wikipedia articles on neutrino detectors and neutrino astronomy provide more details.


TITLE: Change of summation index and Rewriting multiple summation as one sum - Induction Proof of Multinomial Theorem
QUESTION [2 upvotes]: Pictures from http://www.math.msu.edu/~dhand/MTH481FS08/Notes7.pdf
I checked out Changing Summation Index Question but I still can't figure this out.

I. First page of PDF. Substitute $a = n - k, b = k$ in the first sum to get the second sum. The first sum's index is $0 \le k \le n \iff$ Variable substitution  $ \color{blue}{n} = a + k = \color{blue}{a + b}, \, \color{green}{b = k}$ $\iff 0 \le \color{green}{b} \le \color{blue}{a -b}$.
But the PDF has $a + b = n$. How come?
II. Second page on PDF. What happened to the sum over $a_1 + ... + a_{k - 1} = b$? Did it get substituted into the first sum? Why? In https://math.stackexchange.com/a/448118/87058, James Cook recommends writing out the multiple sums. So we should not rewrite them as one sum.

REPLY [1 votes]: To the first question. This $a+b=n$ means that the sum is taking every pair $(a,b)$ such that
$$a\ge0, b\ge0, a+b=n.$$

REPLY [1 votes]: Manipulations of this sort are often manipulations of index sets in disguise.
\begin{eqnarray}
(x_1+...+x_k)^n &=& \sum_{i_k+j=n} { n! \over i_k! j! } (x_1+... x_{k-1})^j x_k^{i_k} \\
&=& \sum_{i_k+j=n} { n! \over i_k! j! } \left( \sum_{i_1+...i_{k-1}=j} { j! \over i_1! ... i_{k-1}!} x_1^{i_1}...x_{k-1}^{i_{k-1}} \right)  x_k^{i_k} \\
&=& \sum_{i_k+j=n} \sum_{i_1+...i_{k-1}=j} { n! \over i_1! ...  i_k!} x_1^{i_1}...  x_k^{i_k} \\
\end{eqnarray}
Now note that $\{(i_1,...,i_k) | i_1+...+i_k = n \} = \cup_{j=0}^n \{(i_1,...,i_{k-1}) | i_1+...i_{k-1}=j \} \times \{ (i_k) | i_k=n-j\}$ to get
\begin{eqnarray}
(x_1+...+x_k)^n &=&  \sum_{i_1+...i_{k}=n} { n! \over i_1! ... i_k!} x_1^{i_1}...  x_k^{i_k} \\
\end{eqnarray}


TITLE: About Descartes numbers
QUESTION [4 upvotes]: I just have a quick question regarding the definition of Descartes numbers, otherwise known as spoof odd perfect numbers.
In this Wikipedia page, Descartes numbers are defined as follows:

A Descartes number is defined as an odd number $n = mp$ where $m$ and $p$ are coprime and $2n = \sigma(m)\cdot(p+1)$.

My question is:  Is the divisibility constraint $\gcd(m,p)=1$ necessary?  I tried perusing Banks et. al.'s paper on the subject, but they do not seem to make that assumption.  I also checked Dittmer's "Spoof odd perfect numbers" and he does not make that supposition either.  Lastly, I tried searching for a page on Descartes numbers in MathWorld@Wolfram in the hopes of comparing the definition in Wikipedia, but there was none.
Can anybody verify and confirm if the definition of Descartes numbers, as stated in Wikipedia, is correct?

REPLY [1 votes]: Per an e-mail response from Professor Bill Banks:
From: Banks, William D.
Date: Sun, Mar 8, 2015 at 9:23 AM
Subject: RE: Your paper titled "Descartes Numbers"
To: Jose Arnaldo Dris
Dear Arnie,
To the best of my knowledge, the term "Descartes number" was first coined by me and first appeared in my paper with Guloglu, Nevans and Saidak.  However, I could easily imagine
that the terminology had been used before, though I had never seen it.
Because of that, I'd say the Wikipedia definition should be changed; the coprimality condition need not be included.  Another thing I dislike about the Wikipedia entry is the use of "$p$" in the definition, since $p$ may well be composite, as in Descartes'
example.
Cheers,
Bill
From: Jose Arnaldo Dris
Sent: Friday, March 06, 2015 1:05 PM
To: Banks, William D.
Subject: Re: Your paper titled "Descartes Numbers"
Hi again Professor Banks,
Additionally, I would like to refer you to a question that I've recently posted over at Math@StackExchange - About Descartes numbers.
Would you be able to confirm if the definition for Descartes numbers in Wikipedia is correct?
Thanks,
Arnie Dris


TITLE: On complemented copy of $c_{0}$ in projective tensor products
QUESTION [2 upvotes]: Suppose that the projective tensor product of $l_{\infty}$ and $X$ contains a complemented copy of $c_{0}$. Does it follow that $X$ contains a complemented copy of $c_{0}$?

REPLY [2 votes]: The answer is no. In fact, by the main theorem of G. Emmanuele,
 "On complemented copies of $c_{0}$ in $L_{X}^p$, $1 \leq p< \infty$",  Proc. Amer. Math. Soc. 104, 1988, 3, 785-786, it follows that the projective tensor product of $l_{\infty}$ and $L_{1}$ contains  a complemented copy of $c_{0}$.


TITLE: Is every field the field of fractions for some integral domain?
QUESTION [12 upvotes]: Given an integral domain $R$, one can construct its field of fractions (or quotients) $\operatorname{Quot}(R)$ which is of course a field. Does every field arise in this way? That is:

Given a field $\mathbb{F}$, does there exist an integral domain $R$ such that $\operatorname{Quot}(R) \cong \mathbb{F}$?

REPLY [8 votes]: To add to Qiaochu's answer, I'll point out that sometimes there is just one domain satisfying this (as in the case of finite fields) and sometimes there are many, many possible domains.  Take, for example, $$\mathbb{Q} = Frac(\mathbb{Z}) = Frac(\mathbb{Z}[1/2]) = Frac(\mathbb{Z}[m/n])$$ etc.  So the point I'm trying to make is that it's all up in the air.
If this is something that interests you, you might want to read up on commutative algebra, specifically localization, which abstracts the notion of a fraction field.  Eisenbud does a decent job of describing localization commutative algebra book.
Also related is the notion of an "algebraic integer".  The notion behind a ring of algebraic integers is this:
Suppose you add some element to $\mathbb{Q}$, and you end up with a bigger field $K$, (for example, $\mathbb{Q}(\sqrt{2})$).  Then, you look to find a domain $\mathcal{O} \subset \mathbb{Q}$ satisfying some basic properties such as:

$Frac(\mathcal{O}) =  K$
"Good" factorization (not necessarily unique)
$\mathcal{O}$ is Noetherian

And you end up with a generalization of $\mathcal{O} \subset K$ of  $\mathbb{Z} \subset \mathbb{Q}$


TITLE: Fixed points on complex function $T(z) = (1+i)z + 3-4i$
QUESTION [1 upvotes]: If $T(z) = (1+i)z + 3-4i$ is a complex function, and I got this fixed point: $z=4+3i$
Is this correct? Are there more fixed points?
$(1+i)z+3-4i=z$
$3-4i=z-(1+i)z$
$3-4i=-iz$
$z=-\frac{3-4i}{i}=4+3i$

REPLY [1 votes]: Your proof is correct and there are no other fixed points. You can see this abstractly by saying that the fixed point equation is a degree one polynomial and therefore has at must one solution. More concretely, each deduction you made was "reversible" so each implication was actually an equivalence.
You had
$$
\begin{align*}
(1+i)z+3-4i=z &\implies 3-4i=z-(1+i)z\\
&\implies 3-4i=-iz\\
&\implies z=-\frac{3-4i}{i}=4+3i
\end{align*}
$$
but at each step the reverse implication holds as well. For instance, in the last step you divided by $-i$, so to go the other way you multiply by $-i$. Hence, you really have
$$
\begin{align*}
(1+i)z+3-4i=z &\Longleftrightarrow 3-4i=z-(1+i)z\\
&\Longleftrightarrow 3-4i=-iz\\
&\Longleftrightarrow z=-\frac{3-4i}{i}=4+3i.
\end{align*}
$$
In other words, $z$ is a fixed point of $T$ if and only if $z = 4+3i$. Be warned that not all manipulations you will do are reversible, however. $(-1)^2 = 1^2$ but you cannot reverse this to get $-1=1$!


TITLE: Can we define near-rings as some kind of a monoid object in the category of groups?
QUESTION [2 upvotes]: I recently learned about the tensor product of Abelian groups, which can be used to define the concept 'ring.' In particular, a ring is just a monoid object in the monoidal category of Abelian groups, with tensor product for the monoidal product.
I'm wondering whether its possible to describe near-rings in a similar fashion. Here are my thoughts. Given groups $G$ and $I$, a set $H$, and a function $\varphi : G \times H \rightarrow I$, let us say that $\varphi$ is a levomorphism iff for every $h \in H$, the function $g \in G \mapsto \varphi(g,h)$ is a homomorphism. So $\varphi$ is required to be a homomorphism in the left argument, for any $h \in H$ that is put into the right argument.
Now given a group $G$ and a set $H$, by a tensor semiproduct of $G$ and $H$, let us mean a group $G \lhd H$ together with a levomorphism $$\varphi : G \times H \rightarrow G \lhd H$$
satisfying the obvious universal property. Namely that, that for any group $I$ and any levomorphism $\psi : G \times H \rightarrow I$, there exists a unique homomorphism $\psi' : G\lhd H \rightarrow I$ such that $\psi = \psi' \circ \varphi$.

Question. Do tensor semiproducts of groups actually exist, and if so, can we define near-rings as some kind of a monoid object in the category of groups?

REPLY [3 votes]: Yes, these tensor semiproducts exist, just let $$G \lhd H := \coprod_{h \in H} G.$$
Near-rings can be desribed via associative maps $G \lhd G \to G$. I have also explained this here.


TITLE: How to express using integrals the area of the region bounded by $xy=6, y=2, y=6, and x=6$ to find volume of region
QUESTION [0 upvotes]: Find the volume of the solid generated by revolving the region bounded by the above graphs of the equations around the line y=6.
My Attempt
$\int_1^6(6- 6/x)) ^2 dx$
Why would we need two integrals? The region is consistently defined by the curve y=6/x. My teacher wrote two integrals and summed them. She wrote $\int_1^3 (6-6/x))^2dx + \int_3^6 16 dx$
16 comes from $\pi r^2$.
She assumes the second region has a constant radius but that isn't true.

REPLY [0 votes]: For $1\le x\le3$, the radius $R=6-6/x$, but for $3\le x\le6$, the radius $R=4$; so
the volume of the solid is given by $\displaystyle V=\int_1^3\pi\big(6-\frac{6}{x}\big)^2dx+\int_3^6\pi(4)^2 dx$.
More simply, using the shell method gives $\displaystyle V=\int_2^62\pi(6-y)\big(6-\frac{6}{y}\big) dy$.


TITLE: A question about modified Newton’s Method on convex function
QUESTION [1 upvotes]: For a modified Newton’s Method, it simplify means multiplying a step size constant with the newton's direction such that
\begin{equation}
x_{k+1} = x_k -t_k \left(\nabla^2 f(x_k)\right)^{-1} \nabla f(x_k)
\end{equation}
So, for this modified Newton’s Method with a twice continuous differentiable function satisfying strong convexity condition $MI\succcurlyeq \nabla^2 f(x)\succcurlyeq mI$. If we suppose $\{x_k\}$ is the sequence generated by this modified Newton’s Method with step length $t_k = m/M$,  I want to prove that the following inequility hold:
\begin{align*}
f(x_k)  - f(x_{k+1}) \geq \frac{m}{2M} \nabla f(x_k)^T (\nabla^2 f(x_k))^{-1}\nabla f(x_k)
\end{align*}
What I have tried is directly substitute $x_{k+1} = x_k - \frac{m}{M}\left(\nabla^2 f(x_k)\right)^{-1} \nabla f(x_k)$ into the quadric expansion of $f(x)$:
\begin{align*}
f(x_{k+1}) &= f(x_k) + \nabla f(x_k)^T (x_{k+1} - x_{k}) + \frac{1}{2}(x_{k+1} - x_{k})^T \nabla f(x_k) (x_{k+1} - x_{k})\\
&\leq  f(x_k) + \nabla f(x_k)^T (x_{k+1} - x_{k}) + \frac{M}{2}\|x_{k+1} - x_{k}\|^2\\
& = f(x_k) - \frac{m}{M}\nabla f(x_k)^T (\nabla^2 f(x_{k}))^{-1}\nabla f(x_{k})) + \frac{M}{2} \frac{m^2}{M^2}\|\nabla^2 f(x_{k}))^{-1}\nabla f(x_{k})\|^2\\
&\leq f(x_k) - \frac{m}{L}\nabla f(x_k)^T (\nabla^2 f(x_{k}))^{-1}\nabla f(x_{k})) + \frac{M}{2} \frac{m^2}{M^2}\|\nabla^2 f(x_{k}))^{-1}\|^2 \|\nabla f(x_{k})\|^2\\
\end{align*}
Since the matrix 2 norm is just the largest eigen value and the eigenvalue of $(\nabla^2 f(x))^-1$ is equal to $1/\sigma(\nabla^2 f(x))$.
Then, use property $m \leq\sigma_{min}(\nabla^2 f(x))$ we have
$$\sigma_{max}((\nabla^2 f(x))^{-1})\leq \frac{1}{m} $$
\begin{align*}
f(x_{k+1})& \leq f(x_k) - \frac{m}{L}\nabla f(x_k)^T (\nabla^2 f(x_{k}))^{-1}\nabla f(x_{k})) + \frac{m^2}{2L} \frac{1}{m^2} \|\nabla f(x_{k})\|^2\\
& = f(x_k) - \frac{m}{L}\nabla f(x_k)^T (\nabla^2 f(x_{k}))^{-1}\nabla f(x_k)) + \frac{1}{2M}  \|\nabla f(x_k)\|^2\\
\end{align*}
Then, I am stacking at here , can anyone help one how could I go further to get the result? I think I am missing something simple. Thanks

REPLY [1 votes]: You stop one step earlier and use that
$$
\|v\|^2=v^TIv\le\frac1mv^T(∇^2f(x_k))v
$$
and apply that to $v=(∇^2f(x_k))^{-1}∇f(x_k)$ to get
$$
m\|(∇^2f(x_k))^{-1}∇f(x_k)\|^2\le ∇f(x_k)^T(∇^2f(x_k))^{-1}∇f(x_k).
$$


TITLE: Proving that a process is a Brownian motion by covariance and mean functions
QUESTION [0 upvotes]: The steps to showing that a process $(W_t)_{t \geq 0}$ is a Brownian motion (BM) are as follows:
(1)$W_0 = 0$
(2) $ \forall t ~~~W_t$ is continuous
(3)$W_t \sim N(0,t)$
(4)$W_{t+s}-W_{s} \sim N(0,t)$
(5)$W_{t+s}-W_{s} \bot \mathscr F_s=(W_u)_{u\in(0,s)} $
For example, if we take $X_t = W_{2t}-W{t}$ we can show that if $X$ satisfies these 5 steps then it is also a BM.
My question is that if a Gaussian process is fully characterized by its covariance and mean functions, is it enough to show that $X$ has the same covariance and mean function as $W$ instead of having to go through the 5 steps, or is there a contradictory case in which $X$ has the same mean and covariance function but does not satisfy some other condition such as continuity or independent increments? Answers should hopefully provide that contradiction.

REPLY [3 votes]: If $(X_t)_{t \geq 0}$ is a Gaussian process with the same mean and covariance function as a Brownian motion, then $(X_t)_{t \geq 0}$ satisfies (1),(3)-(5). Moreover, it follows from Kolmogorov's continuity theorem that $(X_t)_{t \geq 0}$ has a modification with (exclusively) continuous sample paths. However, the process $(X_t)_{t \geq 0}$ does not necessarily have continuous sample paths.


TITLE: How to turn into conditional statements the equations of physics? Looking for basic examples ( at the high school physics, college physics level).
QUESTION [0 upvotes]: My question is related to philosophy, but I do not ask for a philosophical answer. I would be interested in a  technical answer from a logician's / mathematician's point of view
In basic philosophy of science texts,  the name  of Karl Popper is often associated the claim that the basic form of a law of physics is
For all x ( if P(x) , then Q(x)).
Remark.- According to Popper, the fact that laws of physics are conditional statements is what makes them interesting, what gives them a cognitive value. That's what makes them risky statements, statements that can be tested.
What I am looking for is some examples of mathematical physics laws  that could actually be shown to have this conditional form. The problem ( for me) is that laws of physics are equations, I do not really know how to turn an equation into a conditional statement.
In basic texts, the examples given are oversimplistic, like :
For all x ( x is a raven --> x is black).
Remark.- I have no knowledge of advanced physics.

REPLY [2 votes]: If $O$ is an object that started at rest and has since been falling without resistance in a uniform gravitational field, if $g$ is the gravitational acceleration in that field, if $t$ is the time that $O$ has been falling, and if $s$ is the distance that $O$ has fallen, then $s = \frac12 gt^2.$
What might be missing from an oversimplified view of Popper's claim is that
$x$ might have many components (in the case above, we can identify at least
$O,$ $g,$ $t,$ and $s$) and that $P(x)$ might have a large number of clauses.
To put it another way, if you write down any equation of physics, an equation $Q(x)$ tells us nothing about the physical world until someone has said what each of the symbols in the equation means. The assignment of meaning to the symbols is $P(x),$ and it's what makes $Q(x)$ part of a physical law rather than just a random, meaningless equation.


TITLE: Sparser Bipartite graphs?
QUESTION [5 upvotes]: Maximal Planar Bipartite graphs are sparser than maximal planar graphs. For which other classes of graphs are maximal Bipartite members sparser than arbitrary maximal members.
Let $\mathcal{C}$ be a class of graphs and let $\mathcal{B}$ be the class of bipartite graphs. We say that $\mathcal{C}$ is biparted if $\rho(\mathcal{C}) > \rho(\mathcal{B}\cap\mathcal{C})$ (where $\rho(\mathcal{C})$ is defined below) i.e. the asymptotic density of arbitrary graphs in the class is larger than the asymptotic density of the bipartite graphs in the same class.
Examples of biparted graphs are:

Planar Graphs ($\rho(\mathcal{C}) = 6 > 4 = \rho(\mathcal{C}\cap \mathcal{B})$)
Graphs of finite genus (same parameters as above)
Subclasses of planar graphs e.g. outerplanar graphs

My question is:

Are there any other natural classes of biparted graphs?

For a graph $G$, let $\rho(G) = \frac{2|E(G)|}{|V(G)|}$ be the average degree of a vertex. We say that a class of graphs $\mathcal{C}$ is maximally $\rho$-sparse - written as $\rho = \rho(\mathcal{C})$ i.e. the maximum of average density (with maximum taken over all graphs from $\mathcal{C}$ with sufficiently many vertices) approaches $\rho$.
Note: Planar graphs can be proved to be biparted because Euler formula holds. The same is the case with bounded genus graphs. What about bounded tree-width graphs, graphs excluding a finite set of fixed minors,...?

REPLY [6 votes]: For an extreme example, chordal graphs can have as many as $\binom{n}{2}$ edges but chordal graphs that happen to also be bipartite can have only $n-1$ edges (they are forests). Or even more extremely, consider complete graphs versus (complete $\cap$ bipartite) graphs. But perhaps it makes sense to restrict your problem only to classes of graphs that are naturally sparse, as the planar and minor-closed graph families are.
It is not true that all minor-closed families are biparted; in particular, the graphs of treewidth $k$ are not. An $n$-vertex graph with treewidth $k$ has at most $\binom{k}{2}+k(n-k)$ edges, but (for large $n$ and small $k$) the complete bipartite graph $K_{n-k,k}$ has treewidth $k$ and is almost as dense with $k(n-k)$ edges.
One can extend the bounds from planar graphs to $k$-apex graphs (graphs such that you can delete $k$ vertices and get a planar graph), and their generalization to $(g,k)$-apex graphs for bounded genus $g$: these graphs have at most $(k+3)n+O(g)$ edges but their bipartite subclasses have at most $(k+2)n+O(g)$.
Moving beyond minor-closed properties but continuing with graphs related to planar graphs, the 1-planar graphs (graphs that can be drawn with at most one crossing per edge) can have at most $4n-8$ edges, but bipartite 1-planar graphs can have at most $3n-6$. See Czap and Hudák, "On drawings and decompositions of 1-planar graphs", Elect. J. Comb. 2013. And the graphs with book thickness $k$ have at most $(k+1)n-O(1)$ edges ($2n-3$ for the first page and $n-3$ for each subsequent page), but when they are bipartite the constant factor is lower because the first page is outerplanar. I suspect that similar things happen with other types of almost-planar graph.


TITLE: Proof by contradiction (2 ways of writing it)
QUESTION [1 upvotes]: I have a question regarding proof by contradiction (more like the way of writing it).
Let try to prove an easy proposition : For all integer $n$, if $n$ is even, then $n+1$ is odd.

Negate the whole proposition : There exists an integer $n$ such that $n$ is even and $n+1$ is even.
Since $n+1 = 2k$ for some integer $k$,
$n = 2k-1$, a contradiction.

Let $n$ be an integer. Assume that n is even.
We want to show that $n+1$ is odd.
Suppose to the contrary that $n+1$ is even.
Then $n+1=2k$ which gives $n = 2k-1$, a contradiction.

So the first way (1) negate the whole thing and get a contradiction. Along the way, there are steps setting up quantifier "there exists".
The second way, first set up the direct proof (there is a step doing for all quantifier). Then proceed to suppose a contradiction later.
The question is : Are they both a valid proof ? Are they both called a contradiction ? Or the first one is the contradiction ?(In this case, what is the second approach called ? Or it is NOT a valid proof ?)

REPLY [0 votes]: This is a useful question.  Allow me to streamline the contrast between the two setups:

Assume (to the contrary) that there exists an integer $n$ such
that $n$ is even and $n+1$ is even.
Let $n$ be an integer, supppose that $n$ is even, and assume (to the contrary) that $n+1$ is even.

When the two assumptions are negated, statements (1) and (2) become equivalent.


TITLE: Question related with Nakayama
QUESTION [3 upvotes]: Let $R$ be a commutative ring, $J$ denote the Jacobson radical of $R$. Let $M$ be an $R$-module, my question is:

Is $M=mM$ for all maximal ideal $m$ of $R$ if and only if $M=JM$.

It is clear that if $M=JM$, then $M=mM$ for all maximal ideal.
If $M$ is finite generated, suppose $M=mM$, for all maximal ideal. We know $M_m$ is finite generated $R_m$-module, hence we know $M_m=0$ by Nakayma since $M_m=mR_m M_m$. So we know $M=0$ since $M_m=0$ for all maximal ideal of $R$. So if $M$ is finite generated, the question is true.
Thank you in advance.

REPLY [2 votes]: No, this is false without some additional assumptions on $R$ or $M$.

For a simple counterexample, consider $\mathbb{Q}$ as a $\mathbb{Z}$-module.

Obviously $\mathbb{Q} = n \mathbb{Q}$ for any $0 \not= n \in \mathbb{Z}$, and in particular for any prime $p$, but the Jacobson radical of $\mathbb{Z}$ is $0$.
Below are some additional thoughts on your post.  For convenience I will label the asked-about implication
$$M = \mathfrak{m}M
   \; \forall \mathfrak{m} \in \operatorname{maxSpec}(R) \implies M = JM\tag{$*$}$$
To continue with the observation in your post about finitely generated modules, there are some more notable cases on $M$ or $R$ where $(*)$ is incidentally valid because $M = \mathfrak{m}M$ for all maximal ideals already implied that $M$ vanished.
For one, the behavior of projective modules is the same as what you noted about (locally) finite modules in your post:  for a projective module $M$ and a prime $\mathfrak{p}$, we again have $M_\mathfrak{p} = 0$ iff $M = \mathfrak{p}M$ (now the reasoning stems from $M_\mathfrak{p}$'s being free, as opposed to from NAK).  So again your condition would imply that projective modules vanish.  Contrast this to the counterexample I gave which featured an injective module (and even over a PID).
Another way this happens is when $R$ is Von Neumann Regular.  In this case $M=\mathfrak{p}M$ implies $M_\mathfrak{p}$ because $\mathfrak{p}R_\mathfrak{p} = 0$.
This last observation illuminates the unsurprising fact that

$(*)$ does hold whenever $\operatorname{maxSpec}(R)$ is compact with respect to the 'inverse' topology

(equivalently characterized as $R/J(R)$ being $0$-dimensional in the sense of Krull). Of course a special case of this is when $R$ is semi-local.