TITLE: Why Peirce's law implies law of excluded middle?
QUESTION [19 upvotes]: Why if in a formal system the Peirce's law
$((P\rightarrow Q)\rightarrow P) \rightarrow P$
 is true, the law of excluded middle $P \lor \neg P$ is true too?

REPLY [3 votes]: Why?  Because you can prove it.  A formal system has no (or minimal) reference to meaning, and consequently any logical theorem holds, because you can prove it given the axiom set and the rules of inference of the system.  I'll outline a formal system where you can prove the law of the excluded middle ApNp and Peirce's law CCCpqpp.  First formation rules:

All lower case letters of the Latin alphabet qualify as formulas.
If $\alpha$ qualifies as a formula, so dose N$\alpha$
If $\alpha$ and $\beta$ qualify as a formulas, so does C$\alpha$$\beta$, and A$\alpha$$\beta$.
The above rules come as sufficient for what follows.

The system (and almost all of the proofs below can get credited to the work of Jan Lukasiewicz and the publishers and translators of Lukasiewicz's Elements of Mathematical Logic) has three axioms:
1 CCpqCCqrCpr (hypothetical syllogism)
2 CCNppp
3 CpCNpq
In words you can translate the first axiom as saying "if the first (proposition) implies the second, if the second implies the third, then the first implies the third".  The second axiom can get translated into English as saying "if the negation of the first implies the first, then the first."  The third axiom can get translated as saying "if the first, then, if not the first, then the second."  The system has one definition
4' Apq:=CNpq
It has three inference rules
S: For any lower case letter $\lambda$ within any axiom we may uniformly substitute it with any other formula, provided that such substitution happens for all lower case letters equiform with $\lambda$ in that formula.  For instance, were it the case that CpCqp came as an axiom of our system,  we could substitute p with Cpp in CpCqp (substitutions will get abbreviated by the notation p/Cpp hereafter) and obtain C Cpp Cq Cpp, but not C Cpp C qp.  We can also apply such uniform substitution to any theorem we have derived already.
R: For any formula where Xpq, or some substitution instance of Xpq, we may replace Xpq with Yrs, where we have a definition such that we have definition Yrs:=Xpq, where the variables of Yrs contain only equiform variables with Xpq.  Such replacement does not have to happen uniformly for the formula.  It can happen only in one place within a formula.  In particular here this means that if we have some theorem like CCNpqCqCNpq, we may replace either instance of CNpq in CCNpqCqCNpq with Apq, to yield CApqCqCNpq.
D: If we have a theorem C$\alpha$$\beta$ and $\alpha$ qualifies as a theorem also, we may detach $\beta$ as a theorem also.
So, first we'll prove the law of the excluded middle, in symbols ApNp.  The notation "1, p/q" indicates that we will uniformly substitute formula p in the formula reference by the numeral 1 with q.  The "=" sign gets used to indicated that the formula on the left of the "=" sign comes as equiform with the formula on the right side of the "=" sign.  The notation C1-2 indicates that formula 1 comes as the antecedent or first formula which the C on the leftmost of the formula connects, 2 comes as the consequently or second formula which the C on the leftmost of the formula connects, and the "-" indicates that we will detach formula 2 as a theorem.  The notation 3 R = 4 indicates that we will replace formula 3, which comes as a substitution instance of CNpq, with the formula Apq (p and q remain equiform throught CNpq and Apq) and obtain theorem 4.
1 CCpqCCqrCpr axiom
2 CCNppp axiom
3 CpCNpq axiom
 1 q/CNpq = 4

4 C CpCNpq CCCNpqrCpr
 4 = C 3 - 5

5 CCCNpqrCpr
 5 q/p, r/p = 6

6 C CCNppp Cpp
 6 = C 3 - 7

7 Cpp
 7 p/Np = 8

8 CNpNp
 8 R = 9

9 ApNp
Next we'll prove prove CqCpq so that we can prove any two theorems from another theorem within this system.
  1 p/Cpq, q/CCqrCpr, r/s = 10

10 C CCpqCCqrCpr CCCCqrCprsCCpqs
  10 = C 1 - 11

11 CCCCqrCprsCCpqs
  11 q/Cqr, r/Csr, s/CCsqCpCsr = 12

12 C CCCCqrCsrCpCsrCCsqCpCsr CCpCqrCCsqCpCsr
  11 p/s, s/CpCsr = 13

13 CCCCqrCsrCpCsrCCsqCpCsr
  12 = C 13 - 14

14 CCpCqrCCsqCpCsr
  11 s/CCCprsCCqrs = 15

15 C CCCqrCprCCCprsCCqrs CCpqCCCprsCCqrs
  1 p/Cqr, q/Cpr, r/s = 16

16 CCCqrCprCCCprsCCqrs
  15 = C 16 - 17

17 CCpqCCCprsCCqrs
  14 p/Cpq, q/CCprs, r/CCqrs, s/t = 18

18 C CCpqCCCprsCCqrs CCtCCprsCCpqCtCCqrs
  18 = C17 - 19

19 CCtCCprsCCpqCtCCqrs
  5 r/CCCNpppCCqpp = 20

20 C CCNpqCCCNpppCCqpp CpCCCNpppCCqpp
  17 p/Np, r/p, s/p = 21

21 CCNpqCCCNpppCCqpp
  20 = C 21 - 22

22 CpCCCNpppCCqpp
  22 p/CCNppp = 23

23 C CCNppp CCCNCCNpppCCNpppCCNpppCCqCCNpppCCNppp
  23 = C 2 -24

24 C CCNCCNpppCCNpppCCNppp CCqCCNpppCCNppp
  2 p/CCNppp = 25

25 CCNCCNpppCCNpppCCNppp
  24 = C 25 - 26

26 CCqCCNpppCCNppp
  5 p/t, q/CCNppp, r/CCNppp = 27

27 C CCNtCCNpppCCNppp CtCCNppp
  26 q/Nt = 28

28 CCNtCCNpppCCNppp
  27 = C 28 - 29

29 CtCCNppp
  19 p/Np, r/p, s/p = 30

30 C CtCCNppp CCNpqCtCCqpp
  30 = C 29 - 31

31 CCNpqCtCCqpp
  1 p/CNpq, q/CtCCqpp = 32

32 C CCNpqCtCCqpp CCCtCCqpprCCNpqr
  32 = C 31 - 33

33 CCCtCCqpprCCNpqr
  33 t/NCCqpp, r/CCqpp = 34

34 C CCNCCqppCCqppCCqpp CCNpqCCqpp
  2 p/CCNqpp = 35

35 CCNCCNqppCCNqppCCNqpp
  35 = C 34 - 36

36 CCNpqCCqpp
  5 r/CCqpp = 37

37 C CCNpqCCqpp CpCCqpp
  37 = C 36 - 38

38 CpCCqpp
  38 p/q, q/Np = 39

39 CqCCNpqq
  14 p/q, q/CNpq, r/q, s/p = 40

40 C CqCCNpqq CCpCNpqCqCpq
  40 = C 39 - 41

41 C CpCNpq CqCpq
  41 = C 3 - 42

42 CqCpq
In words theorem 42 can get translated as "The first proposition implies that the second proposition implies the first proposition."  Or equivalently, "The first proposition implies that if the second proposition, then the first proposition."  On the strength of that you can show any proposition on the basis of any other proposition within this system (you might want to note exactly what the rule of substitution S says and what it does not say).  Now we'll prove Peirce's law CCCpqpp.
  1 p/q, q/Cpq = 43

43 C CqCpq CCCpqrCqr
  43 = C 42 - 44

44 CCCpqrCqr
  44 p/Nq, q/p, r/CCpqq = 45

45 C CCNqpCCpqq CpCCpqq
  36 p/q, q/p = 46

46 CCNqpCCpqq
  45 = C 45 - 47

47 CpCCpqq
  14 p/q, q/Cqr, s/p = 48

48 C CqCCqrr CCpCqrCqCpr
  47 p/q, q/r = 49

49 CqCCqrr
  48 = C 49 - 50

50 CCpCqrCqCpr
  1 p/CpCqr, q/CqCpr, r/s = 51

51 C CCpCqrCqCpr CCCqCprsCCpCqrs
  51 = C 50 - 52

52 CCCqCprsCCpCqrs
  52 q/Np, r/q, s/CCCpqpp = 53

53 C CCNpCpqCCCpqpp CCpCNpqCCCpqpp
  36 q/Cpq = 54

54 CCNpCpqCCCpqpp
  53 = C 54 - 55

55 CCpCNpqCCCpqpp
  55 = C 3 - 56

56 CCCpqpp
So, now we'll prove the law of the excluded middle ApNp from Peirce's law CCCpqpp in this system.
  42 q/ApNp, p/CCCpqpp

57 C ApNp CCCCpqppApNp
  57 = C 9 - 58

58 CCCCpqppApNp
  58 = C 56 - 59

59 ApNp
But, even here, there exists a catch...
What what if we only had Perice's law with the rules of inference S, D, and R above?  Could we prove the law of the excluded middle?  Prover9 gave me the following matrix (with condensed detachment as a rule of inference, but that won't end up mattering here):
 A|  0  1|C| 0 1|N
------------------
 0*| 1  0|0| 0 1|0
 1|| 0  0|1| 0 0|0

I've starred 0 to indicate it as the designated element here.  If you write out the truth tables for CCCpqpp and ApNp here, you can see that CCCpqpp always evaluates to the designated element 0.  However, when p=0, ApNp=A0N0=A00=1.  Also, and it does come as important to check this also, if Cpq=0, and p=0, then q=0, so rule D comes as valid for this system.  Rules R and D hold for this system, since the matrix above implies A, C, and N as truth-functional.  Thus, from Peirce's law "alone" (the rules mentioned above preferably will get understood as implicit in this statement), you simply cannot derive the law of the excluded middle.


TITLE: Chain rule (proof verification)
QUESTION [3 upvotes]: Hi everyone I'm asking two thinks is  this proof correct (my other idea was using limit of sequences)? and are there a simpler alternative than this using Newton's approximation? If someone could help me I'd be so thankful.
Proposition: Let $X,Y$ be subsets of $\mathbb{R}$, $f:X\rightarrow Y$ and $g:Y\rightarrow \mathbb{R}$ be functions, $x_0\in X$, $y_0\in Y$ are limit points of $X$ and $Y$ respectively, and $y_0=f(x_0)$. Then if  $f$ is differentiable at $x_0$ and $g$ is differentiable at $y_0$, then the composition is differentiable at $x_0$ and $(g\circ f)'(x_0)=g'(y_0)f'(x_0)$, i.e.,
$$\lim_{x\rightarrow x_0;x\in X\backslash\{x_0\}} \frac{(g\circ f)(x)-(g\circ f)(x_0)}{x-x_0}=g'(y_0)f'(x_0)$$
Proof: It will suffice to show that given a $\varepsilon>0$ there exists a $\delta>0$ such that $(g\circ f)(x)$ is $\varepsilon |x-x_0|$-close to $(g\circ f)(x_0)+g'(y_0)f'(x_0)(x-x_0)$ whenever $x\in X$ and $x$ is $\delta$-close to $x_0$.  After a little algebra this is equivalent to
$g(f(x))-g(f(x_0))-g'(y_0)f'(x_0)(x-x_0)=$
  \begin{align} \big[g(f(x))-g(f(x_0))-g'(f(x_0))(f(x)-f(x_0))\big]
+\big[ g'(f(x_0))(\,(f(x)-f(x_0)-f'(x_0)(x-x_0))\,) \big]\end{align}
Since $\big|f(x)-f(x_0)-f'(x_0)(x-x_0) \big|\le  |x-x_0| \varepsilon_1$ because $f$ is differentiable at $x_0$ by hypothesis. We set $\varepsilon_1=\varepsilon/2(| g'(f(x_0))|+1)$ and $\delta_1>0$ such that for all $\{x\in X: |x-x_0|\le \delta_1\}$  we have
$$\big|f(x)-f(x_0)-f'(x_0)(x-x_0) \big|\le  \frac{ |x-x_0| \varepsilon}{2(| g'(f(x_0))|+1)}$$
Now let $\varepsilon_2=\varepsilon/2(|f'(x_0)|+1))$. Since $|g(y)-g(y_0)-g'(y_0)(y-y_0)|\le |y -y_0|\varepsilon_2$ we can find some $r>0$ such that whenever $|y-y_0|\le r$ the above inequality holds. Since $f$ is differentiable it follows that is continuous at $x_0$ then there is $\delta_2$, $0<\delta_2\le \delta_1$, such that $|f(x)-f(x_0)|\le r$ and $|f(x)-f(x_0)|\le |x-x_0|(|f'(x_0)|+1)$ at the same time. Thus we have
\begin{align}|g(f(x))-g(f(x_0))-g'(f(x_0))(f(x)-f(x_0))|\le \frac{|f(x) -f(x_0)|\varepsilon}{2(|f'(x_0)|+1)}\\
\le|x-x_0|\varepsilon/2 \end{align}
For instance we have
$$\begin{align}|g(f(x))-g(f(x_0))-g'(y_0)f'(x_0)(x-x_0)|\le |x-x_0|\varepsilon /2+ |g'(f(x_0))|\frac{ |x-x_0| \varepsilon}{2(| g'(f(x_0))|+1)} \\
\le |x-x_0| \varepsilon\end{align}$$
whenever $x$ is $\delta_2$-close to $x_0$ as claimed.

REPLY [2 votes]: An idea for a much shorter and simpler proof using continuity, and using your same notation:
$$g\circ f(x_0):=\lim_{x\to x_0}\frac{g\circ f(x)-g\circ f(x_0)}{x-x_0}=$$
$$=\lim_{x\to x_0}\frac{g(f(x))-g(f(x_0))}{f(x)-f(x_0)}\cdot\frac{f(x)-f(x_0)}{x-x_0}=g'(f(x_0))\cdot f'(x_0)$$
since, by continuity, we have that
$$x\to x_0\implies f(x)\to f(x_0)$$


TITLE: Graph Part of a line
QUESTION [1 upvotes]: The formula and graph below get close to what I want. This integral shows the AREA of 0.1 to 0.7 in this formula, but I only want the line SEGMENT from 0.1 to 0.7
How do I do that?
Formula:$$ \int _{0.1}^{0.7}\frac{x}{1-x}dx$$
Integral of a formula graph

REPLY [1 votes]: To make life easier
$$\frac x {1-x}=\frac{x-1+1}{1-x}=-1+\frac1{1-x}$$
$$\int_a^b\frac x {1-x}\,dx=-\int_a^b dx+\int_a^b\frac {dx}{1-x}= ???$$


text:
\section{Steiner-like formulas for generators.}
\label{sec:Steiner-like}

In this paper, we consider the interior of a set instead of its exterior, as discussed in Section~\ref{sec:motivation}. However, our primary requirement of a generator is that it has a similar (inner) tube formula;
see Definition~\ref{def:Steiner-like} below and also Section~\ref{sec:motivation} for motivation of the nomenclature.

For a nonempty and bounded open set $\gen \ci\bR^d$, let $\genir=\gr(\gen)$ denote its \emph{inradius}; that is, the radius of the largest open ball
  contained in \gen. It is clear that $\genir$ is always positive and finite. In case \gen is the generator of a fractal spray $\sT$, we have
  \linenopax
  \begin{align}\label{eqn:gr(gen)=g}
    \gr(\gen^j)
    = \gr(\simm_j \gen)
    = \gr(\scale_j \gen)
    = \scale_j \gr(\gen)
    = \scale_j \genir
  \end{align}
  for the inradii of the components $\gen^j$ of $\sT$.

 \begin{comment}
 \begin{defn}[Inradius]\label{def:inradius}
  For any bounded open set $A \ci \bRd$, the \emph{inradius}
  $\gr(A)$ is given by the radius of the largest open ball
  contained in $A$:
  \linenopax
  \begin{equation}\label{eqn:def:inradius}
    \gr(A) := \sup\{ r>0 \suth B(x,r) \ci A, \text{ for some } x \in A\}.
  \end{equation}

  It is clear that $\gr(A)$ is always a well-defined finite real number, and that $\gr(A)>0$ whenever $A \neq \es$. When $A$ is a generator \gen, we use the notation $\genir := \gr(\gen)$ and get
  \end{defn}
\end{comment}
It will be useful to write the inner parallel volume $V(\gen,\ge)$ of the set $\gen \ci \bRd$ as a ``polynomial-like'' expansion in \ge of degree at most $d$. More precisely, we have the following definition.
\begin{defn}
  \label{def:Steiner-like}
  An \emph{(inner) Steiner-like formula} (or a \emph{Steiner-like representation} of the tube formula) for a nonempty and bounded open set $\gen \ci \bRd$ with inradius $\genir = \gr(\gen)$ is an expression for the volume of the inner \ge-parallel sets of \gen of the form
  \linenopax
  \begin{align}\label{eqn:def-prelim-Steiner-like-formula}
    V(\gen,\ge) = \sum_{k=0}^{d} \crv_k(\gen,\ge) \ge^{d-k},
    \qq\text{ for } 0 < \ge \leq \genir,
  \end{align}
  where for each $k=0,1,\dots,d$, the coefficient function $\crv_k(\gen,\cdot)$ is a real-valued function on $(0,\genir]$ that is bounded on $[\ge_0, \genir]$ for every given $\ge_0 \in (0,\genir]$.
\end{defn}

\begin{remark}[The choice of the coefficient functions $\crv_k(\gen,\ge)$]
  \label{rem:nonuniqueness-of-Steinerlike}
  Note that a representation of the form \eqref{eqn:def-prelim-Steiner-like-formula} always exists. For example, one can always take a trivial representation with $\crv_d(\gen, \ge) = V(\gen,\ge)$ and $\crv_0(\gen, \ge) = \dots = \crv_{d-1}(\gen, \ge) = 0$ on $(0,\genir]$. Another, slightly less trivial, representation is given by letting $\crv_k(\gen, \ge) = \frac1{d+1} V(\gen, \ge) \ge^{k-d}$ for $k=0,\dots,d$.
   For brevity, we may use the term ``Steiner-like generator/set'' to indicate that a fixed Steiner-like representation for the tube formula is intended, and write ``tube formula'' for  ``inner tube formula''.

  We have in mind nontrivial representations of the volume function, in which the coefficients allow interpretations in terms of curvature. Clearly, not every representation can have such an interpretation, and so some uniqueness condition will be needed to characterize the correct one for this purpose. However, this is not our aim here (this issue shall be addressed in a forthcoming paper by the same authors). For the main results of this paper, the tube formulas for fractal sprays (and tilings), we make no assumptions on the generators. In fact, our theorems provide many tube formulas for the same spray --- one for each choice of a Steiner-like representation for the generators. Our formulas should be seen as a general tool to transfer a given representation of the volume function of a generator into a tube formula for the generated sprays. We do not yet know what the canonical choice of the representation for the generator is, but our approach seems flexible enough to contain it. It seems that a reasonable strategy would be to choose the coefficients as ``constant as possible''. It is likely that some integrals of the support measures of \cite{HuLaWe} could provide the coefficients of some canonical representation.
\end{remark}

\begin{remark}[Monophase and pluriphase generators]
  \label{rem:monophase}\label{rem:monophase-and-pluriphase}
  As noted above, the coefficients in the expansion \eqref{eqn:def-prelim-Steiner-like-formula} are clearly not unique. However, if a set \gen has a Steiner-like representation with constant coefficients
\linenopax
\begin{align*}
  \crv_k(\gen,\ge) = \crv_k(\gen)
  \qq\text{for all }
  \ge \in (0,\genir] \text{ and } k=0,1,\dots,d,
\end{align*}
then such an expansion is unique, and the set is called \emph{monophase}.  More precisely, a bounded open set $\gen \ci \bRd$ is monophase iff its inner tube formula may be written in the form
  \linenopax
  \begin{align}\label{eqn:monophase}
    V(\gen,\ge) =

      \sum_{k=0}^{d} \crv_k(\gen) \ge^{d-k},
      \qq 0 < \ge \leq \genir.

  \end{align}

  For monophase sets, we always choose this canonical Steiner-like representation. In this case, one has $\gk_d(G)=0$, since otherwise $\lim_{\ge\to 0}\lambda_d(G_{-\ge})\neq 0$. The monophase case has been treated in \cite{TFCD}, at least from the distributional perspective. A variety of natural and classical examples of self-similar tilings in \bRd have monophase generators; see \cite[Section~9]{TFCD}. Furthermore, all geometric (or ordinary) fractal strings (i.e., 1-dimensional fractal sprays) also have monophase generators, since \gen is always an interval; see \cite[Section~8.2]{TFCD}. In general, however, it is rather restrictive to assume that the generator is monophase because many sets (including generators of self-similar tilings) do not have a polynomial expansion with constant coefficients; see Section~\ref{sec:Examples}.

  For monophase sets, the inner tube formula is a polynomial for $\ge \in (0,\genir]$
  and this is the reason for the nomenclature. More generally, as in \cite{TFCD,TFSST}, we say a bounded open set $\gen \ci \bRd$ is \emph{pluriphase} iff it has a Steiner-like tube formula with coefficient functions $\crv_k(\gen,\cdot)$  that are piecewise constant with respect to a finite partition of $[0,\genir]$.
  In short, the inner tube formula is piecewise polynomial, with finitely many pieces. (Such a representation is unique if it is assumed that one takes the partition to have as few components as possible.) We use the term \emph{general Steiner-like} (or \emph{Steiner-like with variable coefficients}) to emphasize the distinction from the special cases of monophase and pluriphase sets. It was conjectured in \cite{TFCD,TFSST} that all convex polyhedra are pluriphase.
\end{remark}

\begin{remark}\label{rem:Steinerlike-in-TFCD}
  The above definition of ``Steiner-like'' is slightly more general than in \cite[Definition~5.1]{TFCD}, where it was introduced: in particular, the local integrability of each coefficient function $\crv_k(\gen,\mydot)$ and the limit condition for $\lim_{\ge\to0^+} \crv_k(\gen,\ge)$ have both been removed.
  The content of Proposition~\ref{cor:V(Gj,ge)} was taken as a hypothesis in \cite{TFCD}, but is now seen to follow from the assumptions in Definition~\ref{def:Steiner-like}.
\end{remark}

No assumption is made on the uniqueness of the coefficients
$\crv_k(\gen,\ge)$ in Definition~\ref{def:Steiner-like} (as discussed in Remark~\ref{rem:nonuniqueness-of-Steinerlike}), but any
choice of coefficients for \gen satisfying
\eqref{eqn:def-prelim-Steiner-like-formula} gives rise to some coefficients $\crv_k(\gen ^j,\ge)$ for each set $\gen^j = \simm_j(\gen)$ by defining
   \linenopax
   \begin{equation}\label{eqn:scaling}
        \crv_k(\gen^j,\ge) := \scale_j^k \crv_k(\gen,\scale_j^{-1}\ge),
        \qq \text{for } \; 0 < \ge \leq \gr(\gen ^j) = \scale_j \genir,
      \end{equation}
as is seen in the following proposition. Here and henceforth, $\scale_j^k$ denotes the \kth power of $\scale_j$.

\begin{prop}\label{cor:V(Gj,ge)}
  Let $\sT=\{\gen ^j\}$ be a fractal spray on a generator \gen with a given Steiner-like representation as in \eqref{eqn:def-prelim-Steiner-like-formula}.
  Then the inner tube formula of each set $\gen^j$ has a Steiner-like representation in terms of the same coefficients:
  \linenopax
  \begin{equation}\label{eqn:V(Gj,ge)}
    V(\gen^j,\ge) = \sum_{k=0}^d \scale_j^k \crv_k(\gen,\scale_j^{-1} \ge) \ge^{d-k},
    \qq \text{for } 0<\ge \le \gr(\gen^j) = \scale_j \genir.
  \end{equation}
\end{prop}
\begin{proof}
   The motion invariance and homogeneity of Lebesgue measure implies that for each $j \geq 1$, $\lambda_d(\gen^j_{-\ge})=\lambda_d(\Psi_j (\gen_{-\ge/\scale_j}))=\scale_j^d \lambda_d(\gen_{-\ge/\scale_j})$, where $\Psi_j$ is the similarity transformation of \bRd described in Definition~\ref{def:fractal-spray}. Whence, by \eqref{eqn:def-prelim-Steiner-like-formula} and \eqref{eqn:scaling},
  \linenopax
  \begin{align*}
    V(\gen^j,\ge)
    = \scale_j^d V(\gen,\scale_j^{-1}\ge)
    = \scale_j^d \sum_{k=0}^{d} \crv_k(\gen, \scale_j^{-1}\ge) (\scale_j^{-1}\ge)^{d-k}
    = \sum_{k=0}^{d} \crv_k(\gen^j,\ge) \ge^{d-k}
  \end{align*}
  for $0 < \ge \leq \gr(\gen^j) = \scale_j\genir$, and the Steiner-like representation \eqref{eqn:V(Gj,ge)} follows. Note that the coefficients $\crv_k(\gen^j,\cdot)$ of $\gen^j$ clearly inherit the boundedness properties from the coefficients $\crv_k(\gen,\cdot)$ of $\gen$.
\end{proof}

In the sequel, we will always work with the coefficient functions of the sets $\gen^j$ chosen according to \eqref{eqn:scaling}.  Proposition~\ref{cor:V(Gj,ge)} ensures this choice is always possible.

\begin{comment}[Motion invariance and homogeneity]
  \label{prop:properties-of-kappas}
  For any Steiner-like representation for the tube formula of $\gen \ci \bRd$, the coefficients may be extended to all similar sets so as to be motion invariant and homogeneous. Then, for any rigid motion $T$ of \bRd  and any $r>0$, the sets $T \gen$ and $r \gen$ have a Steiner-like representation in terms of the same coefficients.
 \begin{proof}
  For $0 \leq \ge \leq \genir = \gr(\gen)$ and $k=0,1,\dots,d$, define
  \linenopax
  \begin{equation} \label{eqn:motion-inv}
    \crv_k(T(\gen),\ge) := \crv_k(\gen,\ge)
    \qq\text{and}\qq
    \crv_k(r \gen,r\ge) := r^k \crv_k(\gen,\ge),
  \end{equation}
  for any rigid motion $T$ and $r>0$.
  Then the invariance of the Lebesgue measure gives
  \linenopax
  \begin{equation*}
    V(T(\gen),\ge)
    = \gl_d(T(\gen)_{-\ge})
    = \gl_d(T(\gen_{-\ge}))
    = \gl_d(\gen_{-\ge})
    = V(\gen,\ge),
  \end{equation*}
  and the homogeneity of Lebesgue measure implies $V(r\gen,r\ge)=r^d V(\gen,\ge)$, whence
  \linenopax
  \begin{align*}
    V(r\gen,\ge)
    = r^d V(\gen,r^{-1}\ge)
    = r^d \sum_{k=0}^{d} \crv_k(\gen,r^{-1}\ge) (r^{-1}\ge)^{d-k}
    = \sum_{k=0}^{d} \crv_k(r\gen,\ge) \ge^{d-k}
  \end{align*}
  for $0 \leq \ge \leq \gr(r\gen) = r\genir$, and the Steiner-like representation follows; see Remark~\ref{rem:renormalize}.
  \end{proof}
\end{comment}

Up to this point, the coefficient functions $\crv_k(\gen,\cdot)$ in a Steiner-like formula for $\gen$ have been defined only for $0 < \ge \leq \genir = \gr(\gen)$. For $k=0,1,\dots,d$,
we define $\crv_k(\gen) := \crv_k(\gen, \genir)$ and then extend $\crv_k(\gen,\ge)$ to $\ge \in (\genir,\iy)$ as constant functions with this value:
\linenopax
\begin{equation}\label{eqn:coeff-ext}
  \crv_k(\gen,\ge) := \crv_k(\gen)
  \qq\text{for } \ge \geq \genir.
\end{equation}
Note that \eqref{eqn:def-prelim-Steiner-like-formula} need not hold for $\ge > \genir$ and so we have the freedom of the choice \eqref{eqn:coeff-ext}. We emphasize that this choice is vitally important for the tube formulas in Theorem~\ref{thm:pointwise-tube-formula} and its corollaries below to be correct.

Note that, for $\ge=g$, equation (\ref{eqn:def-prelim-Steiner-like-formula}) implies that

the $d$-dimensional Lebesgue measure of \gen satisfies
\begin{equation} \label{eqn:leb-coeff-rel}
  \gl_d(\gen)
  = V(\gen, \genir)
  = \sum_{k=0}^{d} \crv_k(\gen, \genir) \genir^{d-k}
  = \sum_{k=0}^{d} \crv_k(\gen) \genir^{d-k}.
\end{equation}

config: arxiv


TITLE: Dynamic Programming problem palindrome
QUESTION [3 upvotes]: I am stuck with the following problem:
Given a string of characters $w$, we want to know the minimum number of characters that we must add to $w$ in order to convert this string in a palindrome (note that we can add characters in any position of the string $w$)
This problem was given in one of my algorithms classes and the key was to solve it by dynamic programming. The suggestion given for this problem was to think of a function that takes two parameters $i,j$ to transform the string $a_i...a_j$.
I'll give a few examples of the problem at hand to illustrate:

$b$ is converted to the palindrome string $b$ by adding $0$ characters
$ab$ is converted to $aba$ by adding $1$ character
$adaf$ is converted to $fadaf$ by adding $1$ character
$abc$ is converted to $abcba$ by adding $2$ characters

I couldn't come up with any ideas to define a function $f(i,j)$ that helps to construct a dynamic programming solution, any suggestions with a dynamic programming approach towards the solution would be really appreciated.

REPLY [1 votes]: This is a well known problem with a dozen of variants all over the web. This page offers several solutions, including a dynamic one:
Minimum insertions to for a palindrome


TITLE: Homomorphism between $ S_n $ and the group of Automorphisms of a group of direct product
QUESTION [4 upvotes]: here I am referring to the Problem 8 of the Exercise for Section 5.1 from Dummit Foote. The problem refers to the previous problem which reads:

Let $ G_{1},\dots,G_{n} $ be groups and let $ \sigma\in S_{n} $ be fixed. Prove that the map $$ \varphi_{\sigma}:G_{1}\times \dots\times G_{n}\to G_{\sigma^{-1}(1)}\times \dots\times G_{\sigma^{-1}(n)} $$ defined by $$ \varphi_{\sigma}(g_1,\dots,g_n)=\big(g_{\sigma^{-1}(1)},\dots,g_{\sigma^{-1}(n)}\big) $$ is an isomorphism.

Now problem 8 reads as the following:

In the above exercise, let $ G_1=\dots=G_{n} $, and call $ G=G_1\times \dots\times G_{n} $. Then show that for every permutation $ \sigma\in S_{n} $, the map $ \varphi_{\sigma} $ is an automorphism on $ G $. Also show that the map $ \sigma\mapsto\varphi_{\sigma} $ is an injective homomorphism of $ S_{n} $ into $ \mathscr{A}(G) $, where $ \mathscr{A}(G) $ is the group of automorphisms of $ G $.

I am having trouble with the homomorphism part. I know that I am making a very silly mistake, but cannot find the same. My argument is the following:
For any arbitrary element $ (g_1,\dots,g_n)\in G $ and for arbitrary permutations $ \sigma,\tau\in S_n $, we have
$$\begin{align}
  \varphi_{\sigma\circ\tau}(g_1,\dots,g_n) &= \big(g_{(\sigma\circ\tau)^{-1}(1)}, \dots,g_{(\sigma\circ\tau)^{-1}(n)}\big)\\
  &= \big(g_{(\tau^{-1}\circ\sigma^{-1})(1)},\dots,g_{(\tau^{-1}\circ\sigma^{-1})(n)}\big)\\
  &= \big(g_{(\tau^{-1}(\sigma^{-1})(1))},\dots,g_{(\tau^{-1}(\sigma^{-1})(n))}\big)\\
  &= \varphi_{\tau}\big(g_{\sigma^{-1}(1)},\dots,g_{\sigma^{-1}(n)}\big)\\
  &= \varphi_{\tau}\big(\varphi_{\sigma}(g_1,\dots,g_n)\big)\\
  &= (\varphi_{\tau}\circ\varphi_{\sigma})(g_1,\dots,g_n).
 \end{align}$$
Therefore, $ \varphi_{\sigma\circ\tau}=\varphi_{\tau}\circ\varphi_{\sigma} $.
So, the order of the maps has reversed. I cannot seem to find a mistake. Please help.

REPLY [5 votes]: This is tricky, and the computation has to be done carefully. There is a bit of a surprise in it. Here we go:
The $j$th coordinate of $\varphi_{\rho}(g_1,\ldots,g_n)$ is the $\rho^{-1}(j)$th coordinate of $(g_1,\ldots,g_n)$.
So the $j$th coordinate of $\varphi_{\tau}(\varphi_{\sigma}(g_1,\ldots,g_n))$ is the $\tau^{-1}(j)$th coordinate of $\varphi_{\sigma}(g_1,\ldots,g_n)$.
The $\tau^{-1}(j)$th coordinate of $\varphi_{\sigma}(g_1,\ldots,g_n)$ is the $\sigma^{-1}(\tau^{-1}(j))$th coordinate of $(g_1,\ldots,g_n)$.
That is,
$$\begin{align*}
\varphi_{\tau}\circ\varphi_{\sigma}(g_1,\ldots,g_n) &=
\left(g_{\sigma^{-1}(\tau^{-1}(1))}, g_{\sigma^{-1}(\tau^{-1}(2))},\ldots,g_{\sigma^{-1}(\tau^{-1}(n))}\right)\\
&=\left(g_{\sigma^{-1}\circ\tau^{-1}(1)}, g_{\sigma^{-1}\circ\tau^{-1}(2)},\ldots, g_{\sigma^{-1}\circ\tau^{-1}(n)}\right)\\
&= \left( g_{(\tau\circ\sigma)^{-1}(1)},\ldots,g_{(\tau\circ\sigma)^{-1}(n)}\right)\\
&=\varphi_{\tau\circ\sigma}(g_1,\ldots,g_n).
\end{align*}$$
The error in your calculation is in line 4.


TITLE: Electromagnetic waves produced by a charged pith ball
QUESTION [6 upvotes]: Maxwell’s equations appear to have no limitation as to the length of an electromagnetic wave that can be produced by an accelerating electric charge. So in theory, if I use a charged rod to oscillate a charged pith ball at, say $1\, \mathrm{Hz}$, the wavelength of this wave is $3\times10^8\,\mathrm{m}$! Although the energy of this wave is very weak, it is an electromagnetic wave. I have two questions:
How would someone go about to detect such a weak wave, if possible?
Has anyone ever detected electromagnetic waves in nature (I am thinking outer space) with a comparable wavelength?

REPLY [5 votes]: While there does not exist currently any standardized technology for detecting such long-wavelength radiation, the technology does exist. In theory, one could build an extremely large antenna out of smaller antennae placed between here and the Moon (it would need to be that large to accommodate the long wavelength) and tune the receiver to that frequency. Or one could set up an omni-directional antenna that resonates at $1Hz$. However, the practicality of it is minimal. A long wavelength like that might easily pass though some objects, but the wave would have such low energy that massive amplification would be required for any transmitting or receiving apparatus. This would put the power consumption at some ridiculous level and any benefit of using a $1Hz$ signal would be lost.
As for detecting this frequency from nature, no. Everything emits blackbody radiation. The warmer something is, the higher the peak frequency and the reverse for colder things.  The coldest something could get in nature (meaning a system in equilibrium with its surroundings) is around $2.7K$.  The Cosmic Microwave Background radiation is emitted at approximately this temperature. That means the CMB should have one of the lowest peak frequencies of all natural bodies. The low-end peak frequency for the CMB is still, however, ~$160.2GHz$, a wavelength of ~$1.873 mm$.
That being said, all objects emit radiation at lower frequencies, including at the $1Hz$ range. However, when we observe these objects, we tend to look for frequencies that will have enough energy to be easily visible and using well known technology. Currently, we use Radio waves as the lowest range (Yes, I see the tautology there), specifically the mid to high-end radio waves. The reason being manifold; as wavelength goes up, your antenna dish needs to be on the same scale as it or larger. This means for even poor resolution, a $1Hz$ telescope would need to span almost to the Moon (as I mentioned earlier). Also, $1Hz$ is very far from the peak emitted frequency of any object, which means that the power at that frequency would be extremely low. In most cases, it is so low that one could approximate the amount of radiation at $1Hz$ as zero.
In summary, $1Hz$ is a very natural frequency, in fact cosmologists often talk about frequencies much lower; with wavelength on the order of the size of the visible universe. The technology does exists to be able to measure such a wave, but it is (in almost all cases) impractical to do so and thus, there is no standard equipment existing that can do it. All objects emit $1Hz$ radiation, however this is nowhere near the peak frequency of any natural body, so detecting it from a natural object would be pointless and require massive amounts of power.