text:
(*
    File:     Group_Relations.thy
    Author:   Joseph Thommes, TU München
*)
section \<open>Group relations\<close>

theory Group_Relations
  imports Finite_Product_Extend
begin

text \<open>We introduce the notion of a relation of a set of elements: a way to express the neutral
element by using only powers of said elements. The following predicate describes the set of all the
relations that one can construct from a set of elements.\<close>

definition (in comm_group) relations :: "'a set \<Rightarrow> ('a \<Rightarrow> int) set" where
  "relations A = {f. finprod G (\<lambda>a. a [^] f a) A = \<one>} \<inter> extensional A"

text \<open>Now some basic lemmas about relations.\<close>

lemma (in comm_group) in_relationsI[intro]:
  assumes "finprod G (\<lambda>a. a [^] f a) A = \<one>" "f \<in> extensional A"
  shows "f \<in> relations A"
  unfolding relations_def using assms by blast

lemma (in comm_group) triv_rel:
  "restrict (\<lambda>_. 0::int) A \<in> relations A"
proof
  show "(\<Otimes>a\<in>A. a [^] (\<lambda>_\<in>A. 0::int) a) = \<one>" by (intro finprod_one_eqI, simp)
qed simp

lemma (in comm_group) not_triv_relI:
  assumes "a \<in> A" "f a \<noteq> (0::int)"
  shows "f \<noteq> (\<lambda>_\<in>A. 0::int)"
  using assms by auto

lemma (in comm_group) rel_in_carr:
  assumes "A \<subseteq> carrier G" "r \<in> relations A"
  shows "(\<lambda>a. a [^] r a) \<in> A \<rightarrow> carrier G"
  by (meson Pi_I assms(1) int_pow_closed subsetD)

text \<open>The following lemmas are of importance when proving the fundamental theorem of finitely
generated abelian groups in the case that there is just the trivial relation between a set of
generators. They all build up to the last lemma that then is actually used in the proof.\<close>

lemma (in comm_group) relations_zero_imp_pow_not_one:
  assumes "a \<in> A" "\<forall>f\<in>(relations A). f a = 0"
  shows "\<forall>z::int \<noteq> 0. a [^] z \<noteq> \<one>"
proof (rule ccontr; safe)
  fix z::int
  assume z: "z \<noteq> 0" "a [^] z = \<one>"
  have "restrict ((\<lambda>x. 0)(a := z)) A \<in> relations A"
    by (intro in_relationsI finprod_one_eqI, use z in auto)
  thus False using z assms by auto
qed

lemma (in comm_group) relations_zero_imp_ord_zero:
  assumes "a \<in> A" "\<forall>f\<in>(relations A). f a = 0"
  and "a \<in> carrier G"
  shows "ord a = 0"
  using assms relations_zero_imp_pow_not_one[OF assms(1, 2)]
  by (meson finite_cyclic_subgroup_int infinite_cyclic_subgroup_order)

lemma (in comm_group) finprod_relations_triv_harder_better_stronger:
  assumes "A \<subseteq> carrier G" "relations A = {(\<lambda>_\<in>A. 0::int)}"
  shows "\<forall>f \<in> Pi\<^sub>E A (\<lambda>a. generate G {a}). finprod G f A = \<one> \<longrightarrow> (\<forall>a\<in>A. f a = \<one>)"
proof(rule, rule)
  fix f
  assume f: "f \<in> (\<Pi>\<^sub>E a\<in>A. generate G {a})" "finprod G f A = \<one>"
  with generate_pow assms(1) have "\<forall>a\<in>A. \<exists>k::int. f a = a [^] k" by blast
  then obtain r::"'a \<Rightarrow> int" where r: "\<forall>a\<in>A. f a = a [^] r a" by metis
  have "restrict r A \<in> relations A"
  proof(intro in_relationsI)
    have "(\<Otimes>a\<in>A. a [^] restrict r A a) = finprod G f A"
      by (intro finprod_cong, use assms r in auto)
    thus "(\<Otimes>a\<in>A. a [^] restrict r A a) = \<one>" using f by simp
  qed simp
  with assms(2) have z: "restrict r A = (\<lambda>_\<in>A. 0)" by blast
  have "(restrict r A) a = r a" if "a \<in> A" for a using that by auto
  with r z show "\<forall>a\<in>A. f a = \<one>" by auto
qed

lemma (in comm_group) stronger_PiE_finprod_imp:
  assumes "A \<subseteq> carrier G" "\<forall>f \<in> Pi\<^sub>E A (\<lambda>a. generate G {a}). finprod G f A = \<one> \<longrightarrow> (\<forall>a\<in>A. f a = \<one>)"
  shows "\<forall>f \<in> Pi\<^sub>E ((\<lambda>a. generate G {a}) ` A) id.
         finprod G f ((\<lambda>a. generate G {a}) ` A) = \<one> \<longrightarrow> (\<forall>H\<in> (\<lambda>a. generate G {a}) ` A. f H = \<one>)"
proof(rule, rule)
  fix f
  assume f: "f \<in> Pi\<^sub>E ((\<lambda>a. generate G {a}) ` A) id" "finprod G f ((\<lambda>a. generate G {a}) ` A) = \<one>"
  define B where "B = inv_into A (\<lambda>a. generate G {a}) ` ((\<lambda>a. generate G {a}) ` A)"
  have Bs: "B \<subseteq> A"
  proof
    fix x
    assume x: "x \<in> B"
    then obtain C where C: "C \<in> ((\<lambda>a. generate G {a}) ` A)" "x = inv_into A (\<lambda>a. generate G {a}) C"
      unfolding B_def by blast
    then obtain c where c: "C = generate G {c}" "c \<in> A" by blast
    with C someI_ex[of "\<lambda>y. y \<in> A \<and> generate G {y} = C"] show "x \<in> A"
      unfolding inv_into_def by blast
  qed
  have sI: "(\<lambda>x. generate G {x}) ` B = (\<lambda>x. generate G {x}) ` A"
  proof
    show "(\<lambda>x. generate G {x}) ` B \<subseteq> (\<lambda>x. generate G {x}) ` A" using Bs by blast
    show "(\<lambda>x. generate G {x}) ` A \<subseteq> (\<lambda>x. generate G {x}) ` B"
    proof
      fix C
      assume C: "C \<in> (\<lambda>x. generate G {x}) ` A"
      then obtain x where x: "x = inv_into A (\<lambda>a. generate G {a}) C" unfolding B_def by blast
      then obtain c where c: "C = generate G {c}" "c \<in> A" using C by blast
      with C x someI_ex[of "\<lambda>y. y \<in> A \<and> generate G {y} = C"] have "generate G {x} = C"
        unfolding inv_into_def by blast
      with x C show "C \<in> (\<lambda>x. generate G {x}) ` B" unfolding B_def by blast
    qed
  qed
  have fBc: "f (generate G {b}) \<in> carrier G" if "b \<in> B" for b
  proof -
    have "f (generate G {b}) \<in> generate G {b}" using f(1)
      by (subst (asm) sI[symmetric], use that in fastforce)
    moreover have "generate G {b} \<subseteq> carrier G" using assms(1) that Bs generate_incl by blast
    ultimately show ?thesis by blast
  qed
  let ?r = "restrict (\<lambda>a. if a\<in>B then f (generate G {a}) else \<one>) A"
  have "?r \<in> Pi\<^sub>E A (\<lambda>a. generate G {a})"
  proof
    show "?r x = undefined" if "x \<notin> A" for x using that by simp
    show "?r x \<in> generate G {x}" if "x \<in> A" for x using that generate.one B_def f(1) by auto
  qed
  moreover have "finprod G ?r A = \<one>"
  proof (cases "finite A")
    case True
    have "A = B \<union> (A - B)" using Bs by auto
    then have "finprod G ?r A = finprod G ?r (B\<union>(A-B))" by auto
    moreover have "\<dots> = finprod G ?r B \<otimes> finprod G ?r (A - B)"
    proof(intro finprod_Un_disjoint)
      from True Bs finite_subset show "finite B" "finite (A - B)" "B \<inter> (A - B) = {}" by auto
      show "(\<lambda>a\<in>A. if a \<in> B then f (generate G {a}) else \<one>) \<in> A - B \<rightarrow> carrier G" using Bs by simp
      from fBc show "(\<lambda>a\<in>A. if a \<in> B then f (generate G {a}) else \<one>) \<in> B \<rightarrow> carrier G"
        using Bs by auto
    qed
    moreover have "finprod G ?r B = \<one>"
    proof -
      have "finprod G ?r B = finprod G (f \<circ> (\<lambda>a. generate G {a})) B"
      proof(intro finprod_cong')
        show "?r b = (f \<circ> (\<lambda>a. generate G {a})) b" if "b \<in> B" for b using that Bs by auto
        show "f \<circ> (\<lambda>a. generate G {a}) \<in> B \<rightarrow> carrier G" using fBc by simp
      qed simp
      also have "\<dots> = finprod G f ((\<lambda>a. generate G {a}) ` B)"
      proof(intro finprod_comp[symmetric])
        show "(f \<circ> (\<lambda>a. generate G {a})) ` B \<subseteq> carrier G" using fBc by auto
        show "inj_on (\<lambda>a. generate G {a}) B"
          by (intro inj_onI, unfold B_def, metis (no_types, lifting) f_inv_into_f inv_into_into)
      qed
      also have "\<dots> = finprod G f ((\<lambda>a. generate G {a}) ` A)" using sI by argo
      finally show ?thesis using f(2) by argo
    qed
    moreover have "finprod G ?r (A - B) = \<one>" by(intro finprod_one_eqI, simp)
    ultimately show ?thesis by fastforce
  next
    case False
    then show ?thesis unfolding finprod_def by simp
  qed
  ultimately have a: "\<forall>a\<in>A. ?r a = \<one>" using assms(2) by blast
  then have BA: "\<forall>a\<in>B\<inter>A. ?r a = \<one>" by blast
  from Bs sI have "\<forall>a\<in>A. (generate G {a}) \<in> ((\<lambda>x. generate G {x}) ` B)" by simp
  then have "\<forall>a\<in>A. \<exists>b\<in>B. f (generate G {a}) = f (generate G {b})" by force
  thus "\<forall>H\<in>(\<lambda>a. generate G {a}) ` A. f H = \<one>" using a BA Bs by fastforce
qed

lemma (in comm_group) finprod_relations_triv:
  assumes "A \<subseteq> carrier G" "relations A = {(\<lambda>_\<in>A. 0::int)}"
  shows "\<forall>f \<in> Pi\<^sub>E ((\<lambda>a. generate G {a}) ` A) id.
         finprod G f ((\<lambda>a. generate G {a}) ` A) = \<one> \<longrightarrow> (\<forall>H\<in> (\<lambda>a. generate G {a}) ` A. f H = \<one>)"
  using assms finprod_relations_triv_harder_better_stronger stronger_PiE_finprod_imp by presburger

lemma (in comm_group) ord_zero_strong_imp_rel_triv:
  assumes "A \<subseteq> carrier G" "\<forall>a \<in> A. ord a = 0"
  and "\<forall>f \<in> Pi\<^sub>E A (\<lambda>a. generate G {a}). finprod G f A = \<one> \<longrightarrow> (\<forall>a\<in>A. f a = \<one>)"
  shows "relations A = {(\<lambda>_\<in>A. 0::int)}"
proof -
  have "\<And>r. r \<in> relations A \<Longrightarrow> r = (\<lambda>_\<in>A. 0::int)"
  proof
    fix r x
    assume r: "r \<in> relations A"
    show "r x = (\<lambda>_\<in>A. 0::int) x"
    proof (cases "x \<in> A")
      case True
      let ?r = "restrict (\<lambda>a. a [^] r a) A"
      have rp: "?r \<in> Pi\<^sub>E A (\<lambda>a. generate G {a})"
      proof -
        have "?r \<in> extensional A" by blast
        moreover have "?r \<in> Pi A (\<lambda>a. generate G {a})"
        proof
          fix a
          assume a: "a \<in> A"
          then have sga: "subgroup (generate G {a}) G" using generate_is_subgroup assms(1) by auto
          show "a [^] r a \<in> generate G {a}"
            using generate.incl[of a "{a}" G] subgroup_int_pow_closed[OF sga] by simp
        qed
        ultimately show ?thesis unfolding PiE_def by blast
      qed
      have "finprod G ?r A = (\<Otimes>a\<in>A. a [^] r a)" by(intro finprod_cong, use assms(1) in auto)
      with r have "finprod G ?r A = \<one>" unfolding relations_def by simp
      with assms(3) rp have "\<forall>a\<in>A. ?r a = \<one>" by fast
      then have "\<forall>a\<in>A. a [^] r a = \<one>" by simp
      with assms(1, 2) True have "r x = 0"
        using finite_cyclic_subgroup_int infinite_cyclic_subgroup_order by blast
      thus ?thesis using True by simp
    next
      case False
      thus ?thesis using r unfolding relations_def extensional_def by simp
    qed
  qed
  thus ?thesis using triv_rel by blast
qed

lemma (in comm_group) compl_fam_iff_relations_triv:
  assumes "finite gs" "gs \<subseteq> carrier G" "\<forall>g\<in>gs. ord g = 0"
  shows "relations gs = {(\<lambda>_\<in>gs. 0::int)} \<longleftrightarrow> compl_fam (\<lambda>g. generate G {g}) gs"
  using triv_finprod_iff_compl_fam_PiE[of _ "\<lambda>g. generate G {g}", OF assms(1) generate_is_subgroup]
        ord_zero_strong_imp_rel_triv[OF assms(2, 3)]
        finprod_relations_triv_harder_better_stronger[OF assms(2)] assms by blast

end

subset_name: curated


TITLE: Sketching the graph and finding the function by using the given.
QUESTION [3 upvotes]: A function f that is defined and continuous for $x\neq 0$ satisfies the following conditions:

f($2-{\sqrt 2}$) = $\sqrt[3]{1-1/\sqrt 2}$, f($\frac{2}{3}$) = $0$,  f(2) = $\sqrt[3]{2}$, f($2+{\sqrt 2}$) = $\sqrt[3]{1+1/\sqrt 2}$
$\lim_{x \to 0^-}f(x)$ = $-\infty$, $\lim_{x \to 0^+}f(x)$ = $\infty$, $\lim_{x \to -\infty}f(x)$ = $0$, $\lim_{x \to \infty}f(x)$ = $0$
$f^\prime$($x$) $<$ $0$ for $x$ $<$ $\frac{2}{3}$ and $x\neq 0$, and for $x$ $>$ $2$; $f^\prime$($x$) $>$ $0$ for $\frac{2}{3}$ $<$ x $<$ $2$
$\lim_{x \to \frac{2}{3}^-}f^\prime(x)$ = $ -\infty$,  $\lim_{x \to \frac{2}{3}^+}f^\prime(x)$ = $\infty$
$f^{\prime\prime}(x)$ $<$ $0$ for $x$ $<$ $0$ and for $2-{\sqrt 2}$ $<$ $x$ $<$ $2+{\sqrt 2}$ and $x\neq \frac{2}{3}$;$f^{\prime\prime}(x)$ $>$ $0$ for $0$ $<$ $x$ $<$ $2-{\sqrt 2}$ and for $x$ $>$ $2+{\sqrt 2}$

The question asks to come up with function $f$($x$)$ = $ (a$x$ $+$ b)$^{c}$$x$$^{d}$ that satisfies 1-5 conditions. What are a, b, c and d?
I drew the graphand found that d is obviously -1 however I couldn't find a, b, and c. Please help.

REPLY [1 votes]: First, since the left and right limits at zero are unequal (in fact, do not exist), $f$ is discontinuous at $0$, so that $d<0$. Since $f\left(\frac{2}{3}\right)=0$, the factor $(ax+b)^c$ must vanish there, so that $c>0$ and $ax+b = r(3x-2)$ for some constant $r$. So rewriting, we get
$$f(x) = r^c(3x-2)^c x^d.$$
Next, $f(2)>0$ shows that $r^c>0$; assume going forward that in fact $r=1$, and we will try to fix things up later if that proves not to work. This gives
$$f(x) = (3x-2)^c x^d.$$
Then $f(2) = 2^{2c+d} = 2^{1/3}$, so that $2c+d = \frac{1}{3}$. Next,
\begin{align*}
f(2-\sqrt{2})f(2+\sqrt{2}) &= ((4-3\sqrt{2})(4+3\sqrt{2}))^c((2-\sqrt{2})(2+\sqrt{2}))^d \\
&= (-2)^c2^{d} = (-1)^c2^{c+d}\\
\left(1-\frac{1}{\sqrt{2}}\right)^{1/3}\left(1-\frac{1}{\sqrt{2}}\right)^{1/3}
&= \left(\frac{1}{2}\right)^{1/3} = 2^{-1/3}.
\end{align*}
Since these two expressions must be equal, we see that $(-1)^c=1$ and $c+d = -\frac{1}{3}$.
Solving
\begin{align*}
  2c+d &= \phantom{-}\frac{1}{3}\\
\phantom{2}c+d &= -\frac{1}{3}
\end{align*}
gives $c=\frac{2}{3}$ and $d=-1$, so that
$$f(x) = (2x-3)^{2/3}x^{-1}.$$
Although we did not use the various conditions on the derivatives, this function is easily seen to satisfy all of them.
A plot is shown below:


TITLE: Definitions of inner and outer regular measures
QUESTION [8 upvotes]: The definitions can be found here.
I am trying to figure out why it's defined specifically this way. More precisely- Why compact sets are used to define inner regularity while open sets are used for outer regularity?
I thought that maybe it has something to do with the facts that the union of open sets is open and that (in a Hausdorff space) the intersection of compact sets is compact, but I dont see how these matter.
Another thought I had is that maybe the motive comes from Riesz–Markov–Kakutani representation theorem.

REPLY [5 votes]: Terminology differs among authors. Another def'n is that a measure $m$ is Radon-regular iff each measurable set $S$ satisfies $m(S)=\sup \{m(T):S\supset T \land T$ compact $\},$ and that inner-regular means $m(S)=\sup \{m(U):S\supset U=\bar U\}.$ In many cases, Radon-regular and inner-measure co-incide.
The motivation is that we can say much more about regular measures because regularity is a relation between the measure  and the topology of the measure space.
For example a set $S$ is Lebesgue-measurable iff there exists a $G_{\delta}$ set $ S^i$ and an $F_{\sigma}$ set $S^o$ with $S^i\subset S \subset S^o$ and $m^o (S^o$  \ $S^i)=0.$
One of the motives of Lebesgue measure was attempts to extend calculus to  a much broader class of functions,  which resulted in extending the notion of length or volume. ...Note the outer measure involves  intersections of subsets of a (usually infinite ) family of open sets, and the inner measure involves  unions of closed sets.


TITLE: Let $\mathbf{a}$, $\mathbf{b}$, and $\mathbf{c}$ be unit vectors, with $\mathbf{a+b+c=0}$. The angle between any two of these vectors is $120^\circ$.
QUESTION [1 upvotes]: Let $\mathbf{a}$,  $\mathbf{b}$, and $\mathbf{c}$ be unit vectors, such that  $\mathbf{a}+\mathbf{b}+\mathbf{c} = \mathbf{0}$. Show that the angle between any two of these vectors is $120^\circ$.
Hi,
I have been having some trouble with this problem.
I have tried to assign variables to the vectors, and creating various equations with them.  However, I can't figure out how to use the $\mathbf{a+b+c=0}$ information.
Thanks!

REPLY [0 votes]: Geometrically, $\mathbf a + \mathbf b + \mathbf c = \mathbf 0$ means we can form a triangle $\triangle XYZ$ whose oriented sides $\overrightarrow{XY}, \overrightarrow{YZ}, \overrightarrow{ZX}$ are parallel copies of $\mathbf a$, $\mathbf b$, $\mathbf c$.
Since $\mathbf a$, $\mathbf b$, $\mathbf c$ are unit vectors, $\triangle XYZ$ is equilateral, so $\angle X = \angle Y = \angle Z = 60^\circ$.
But $\angle X$ is the angle between $\overrightarrow{XY}$ and $\overrightarrow{XZ}$, or between $\mathbf a$ and $-\mathbf c$. If the angle between $\mathbf a$ and $-\mathbf c$ is $60^\circ$, then the angle between $\mathbf a$ and $\mathbf c$ is its supplementary angle, $120^\circ$.
Similarly, we conclude from $\angle Y = \angle Z = 60^\circ$ that the angle between $\mathbf a$ and $\mathbf b$, and the angle between $\mathbf b$ and $\mathbf c$, are both $120^\circ$.


TITLE: Are set operations same as lambdas?
QUESTION [1 upvotes]: I'm teaching myself algebra (programmer by profession), and I wanted to clear a constant doubt that I'm having.
My doubt arose from the self-study of algebraic structures- Groups, Fields and Rings and these SE hits:
What are the differences between rings, groups, and fields?
Can anyone explain the difference between a ring, a group, and a field in a way so that your average 15 year old can understand?
Let's use the partial definition of a group as:

A group is a set of elements $E$ with an operation $\circ{}$ to combine the elements of $E$....

Most of the answers above use addition, subtraction, multiplication, and division as the operation $\circ$ to explain these concepts. Is there some special reason behind this other than illustration purposes? My programmer brain wonders why can't I use any lambda with binary arity instead, of course, as long as it follows the definitional requirements of the particular structure? Say, lambda x,y: math.log(x,y).
That leads to the last related question- an integer ring $Z_m$ consists of:

The set $Z_m$ = $\{0, 1, ..., m-1\}$
Two operations $+$, and $\times$ such that ....

By now, I know that the inverse of operation $+$ exists, but not necessarily for $\times$ for an integer ring. Then, can I assume that there are rings out there that could be working with different sets and operations that are not even remotely close to the arithmetic operations of addition and multiplication?

REPLY [1 votes]: Ultimately, I would say that it depends on what you would like to do with your binary operation $\circ.$ Given a set $S$ with a binary operation $\circ,$ you have a magma. Going one step further, you could make sure that $\circ$ is associative, and you'd wind up with a semigroup. If you desire an identity element $e_S,$ then you are dealing with a monoid. Last but not least, requiring your elements to have inverses under $\circ$ will give you a group.
Given that $S$ is a commutative group under $\circ,$ an additional binary operation $*$ will make $S$ into a ring if
1.) $S$ is a monoid under $*$ and
2.) the operation $*$ is distributive, i.e., $r * (s \circ t) = (r * s) \circ (r * t)$ and $(s \circ t) * r = (s * r) \circ (t * r).$
Consequently, the answer to your last question is probably no because the operation $\circ$ acts in an analogous manner to addition of integers, and the operation $*$ acts similarly to multiplication of $n \times n$ matrices.


TITLE: Find distribution function
QUESTION [0 upvotes]: Let $X$ be random variable such that $$\Pr(X = 0) = 0.8 \\ \Pr(X = 1) = 0.1 \\ \forall x \in (0,1) f_X(x) = 0.1$$
Find distribution function $F_X(x)$.
I have no idea how solve it. Maybe I need some theory, but I don't know where should I search. I will grateful for any help.

REPLY [1 votes]: This problem can be solved by using the definition of the cumulative distribution function $F_X(x)$. Recall that
$$F_X(x)=\Pr(X\le x).$$
We will have to pay careful attention to the inequality sign: $\le x$ is not the same thing as $\lt x$.
First let $x\lt 0$. Then $\Pr(X\le x)=0$. There is no "stuff" to the left of $0$.
Next let $x=0$. Note that $\Pr(X\le 0)=\Pr(X=0)=0.8$. There is a "point mass" of $0.8$ at $x=0$.
More generally (and this includes the case previously dealt with),
suppose  that $0\le x\lt 1$. The mass up to $x$ is the point mass at $0$, plus the contribution from the interval $(0,x]$. This contribution is $(0.1)(x-0)$.  Alternately, we could use the equivalent $\int_0^x (0.1)\,dt$.  Thus if $0\le x\lt 1$, then $F_X(x)=0.8+(0.1)x$.
Now let $x= 1$. Note that $\Pr(X\le 1)=0.8+\int_0^1 (0.1)\,dt +0.1=1$. More generally, if $x\ge 1$ then $F_X(x)=1$.
Remark: Intuition about "weights" is often more accurate than intuition about probabilities. What we have here is a point mass of $0.8$ at $x=0$, a point mass of $0.1$ at $x=1$, and a wire of density $0.1$ joining $0$ and $1$. Everywhere before $x=0$, and after $x=1$, we have "wire" of density $0$.
Then $F_X(x)$ is the total mass from "$-\infty$" all the way up to and including $x$.


TITLE: word problem involving interest
QUESTION [1 upvotes]: Upon entering college, Meagan borrowed the limit of $5000 on her credit card to help pay for expenses. The credit company charges 19.95 % interest compounded continuously. How much will Meagan owe when she graduates in four years ?
I wanted to use A(t)=A(0)e^rt, and r=19.95%, t=4, A(0)=5000
so I was thinking
A(t) = A(0)e^(rt)
A = (5,000)e^(.1995)(4)

Am I doing this correctly ? how do I calculate the rest? Is this is all I need ?
thanks!

REPLY [1 votes]: A = p(1+r/n)^nt

P = principal amount (the initial amount you borrow or deposit)
r  = annual rate of interest (as a decimal)
t  = number of years the amount is deposited or borrowed for.
A = amount of money accumulated after n years, including interest.
n  =  number of times the interest is compounded per year
A = (5,000)(1+.1995)^(4)


TITLE: How many positive two-digit integers have exactly 8 positive factors?
QUESTION [3 upvotes]: I solved this problem by listing all two-digit integers and going through each one. Is there an easy way to solve the problem?

How many positive two-digit integers have exactly 8 positive factors?

REPLY [1 votes]: If $n$ has prime power factorization $n=p^aq^br^c$ then the number of factors for $n$ is $(a+1)(b+1)(c+1)$ ( A number with 4 or more distinct prime factors will exceed 100). For numbers of the type $p^aq^b$ the count is $(a+1)(b+1)$.
Now see how to get $8$ as product of the above type $(a+1)(b+1)(c+1)$, or $(a+1)(b+1)$ then choose primes $p,q,r$ to ensure the number is $<100$.


TITLE: Direction of EMF and Induced current through the circuit
QUESTION [0 upvotes]: When the circuit shown above is moved with a constant velocity to the left I know that B is going to have the higher potential.
And I believe that the current flows clockwise that is in the sense of AEDCB.
But my tutor says that the current flows in the sense ABCDE,saying that it is similar to having a battery across BD with the positive terminal at B instead of the half loop BCD with the magnetic field..
I am very confused with this matter.
Am I correct or is my tutor correct?

REPLY [0 votes]: Before starting the answer one you can replace BCD with BC to easily understand ,let for understanding the direction we will apply lenz law that is current induced in a coil in such fashion that it will oppose the cause ,as flux in the coil is decreasing as it is move toward left so current will be in clockwise direction to oppose decrease in current it mean your TUTOR IS RIGHT now coming to your question that B should be at higher potential yiu can simply understand it by lorentz force equation F=QVB there is cross product between V and B that is when your teacher talk about battery yiu can simply derive polarity use right hand palm rule put right hand finger in direction of flux your thumb will indicate velocity and and point that faces palm will be at higher potential YOUR TEACHER IS is simply talking about electron because in this equation q replace by e which is counter or anti clockwise


TITLE: Property of representing a convex function as the supermum of affine functions
QUESTION [2 upvotes]: It is well-known that a convex function can be represented as the pointwise supremum of some affine functions. Hence let $f:\mathbb{R}\to \mathbb{R}$ be a convex function, we know there exists a set $\mathcal{A}$ such that
$$
f(x)=\sup_{\alpha\in \mathcal{A}}a(\alpha)x+b(\alpha).
$$
My question is, with suitable assumptions on $f$, is it possible to deduce certain properties of this representation? For example, $\mathcal{A}$ is compact, or $a$ and $b$ are continuous in $\alpha$?

REPLY [1 votes]: you can use the fact that the double conjugate is the function itself (since your $f$ is convex and closed), so you can use $\mathcal{A} = \mathbb{R}$, $a(\alpha)=\alpha$ and $b(\alpha) = -f^*(\alpha)$. If your function is polyhedral, then so is $b$.


TITLE: To find the equation of the ellipse with foci $\pm3i$ that goes through the point $8-3i$.
QUESTION [2 upvotes]: To find the equation of the ellipse with foci $\pm3i$ that goes through the point $8-3i$.
The standard equation of ellipse given foci is $|z-f|+|z+f|=2 \mu |f|$, where $\pm f$ is foci and $z$ is a point on the ellipse.
So, $|8-3i-3i|+|8-3i+3i|=2 \mu |3i| \implies 18=2 \times3 \mu \implies\mu=3.$
Thus the equation of ellipse is: $|z-3i|+|z+3i|=6|3i| = 18.$
Is the solution correct?

REPLY [2 votes]: More simply: the sum of the distances from a point of the ellipse to the foci should be constant: since
$$
|8-3i-3i|+|8-3i+3i|=18
$$
the equation is
$$
|z-3i|+|z+3i|=18
$$
No need to compute $\mu$.


TITLE: Why direction of Ampere's Law follows right hand rule?
QUESTION [0 upvotes]: Why direction of Ampere's Law follows right hand rule?

REPLY [0 votes]: Because we conventionally use a right-handed coordinate system.
In a left-handed coordinate system, it would follow a left-hand rule.


TITLE: Special case of Leibniz formula
QUESTION [1 upvotes]: I am wondering if I understand it right. Considering the following partial derivative:
$$
\partial_tF(t, x) = \partial_t\int_{a(t)}^{b(t)} f(t-s, x)ds
$$
It looks to me as Leibniz should be used. So we should get
$$
\partial_t\int_{a(t)}^{b(t)} f(t-s, x)ds \\ = \int_{a(t)}^{b(t)} \partial_t f(t-s, x)ds - f(t-a(t), x)(1-a'(t)) + f(t-b(t), x)(1-b'(t))
$$
Is this correct, or did I get something wrong?

REPLY [1 votes]: Applying the Leibniz integral rule we obtain
\begin{align*}
\color{blue}{\frac{d}{dt}\left(\int_{a(t)}^{b(t)}f\left(u(t,s),x\right)\,ds\right)}
&\color{blue}{=f\left(u(t,b(t)),x\right)\,\frac{d}{dt}b(t)}\\
&\qquad\color{blue}{-f(u(t,a(t)),x)\,\frac{d}{dt}a(t)}\\
&\qquad\color{blue}{+\int_{a(t)}^{b(t)}\frac{\partial}{\partial t}f\left(u(t,s),x\right)\,ds}\tag{1}
\end{align*}
provided the functions are appropriately differentiable.

Example: We look at a small example.
\begin{align*}
u(t,s)&=t-s\qquad\qquad a(t)=4t\qquad b(t)=3t^2\\
f(u(t,s),x)&=u(t,s) x\\
&=(t-s)x
\end{align*}

The left-hand side of (1) gives
\begin{align*}
\frac{d}{dt}\left(\int_{a(t)}^{b(t)}f(u(t,s),x)\,ds\right)&=\frac{d}{dt}\int_{s=4t}^{3t^2}(t-s)x\,ds\\
&=x\frac{d}{dt}\left(ts-\frac{1}{2}s^2\bigg|_{s=4t}^{3t^2}\right)\\
&=x\frac{d}{dt}\left(3t^3-\frac{9}{2}t^4-4t^2+8t^2\right)\\
&\,\,\color{blue}{=x\left(-18t^3+9t^2+8t\right)}
\end{align*}
The right-hand side of (1) gives
\begin{align*}
&f\left(u(t,b(t)),x\right)\,\frac{d}{dt}b(t)-f(u(t,a(t)),x)\,\frac{d}{dt}a(t)
+\int_{a(t)}^{b(t)}\frac{\partial}{\partial t}f\left(u(t,s),x\right)\,ds\\
&\qquad=\left(t-\left(3t^2\right)\right)x\cdot 6t-\left(t-(4t)\right)x\cdot 4
+\int_{4t}^{3t^2}\frac{d}{dt}(t-s)x\,ds\\
&\qquad=\left(6t^2x-18t^3x\right)+12tx+x\int_{4t}^{3t^2}\,ds\\
&\qquad=\left(6t^2-18t^3\right)x+12tx+\left(3t^2-4t\right)x\\
&\qquad\,\,\color{blue}{=x\left(-18t^3+9t^2+8t\right)}
\end{align*}
and we see according to (1) the blue marked results coincide as expected.


TITLE: Show that at least one of the intervals is contained in the union of the other two.
QUESTION [1 upvotes]: Question: Let $I_1,I_2,I_3$ be three open intervals of $\mathbb{R}$ such that none is contained in another. If $I_1\cap I_2\cap I_3$ is non-empty, then show that at least one of these intervals is contained in the union of the other two.

My approach: Let us assume the contrary for the sake of contradiction, that is we have $$I_1\nsubseteq I_2\cup I_3, \\I_2\nsubseteq I_3\cup I_1\text{ and} \\I_3\nsubseteq I_1\cup I_2.$$
Now, I can observe that if we assume so, then $\exists i,j$ such that $i\neq j$ and $I_i\subseteq I_j$, which is a contradiction to the statement which states that none of the intervals $I_i$ is contained in another interval $I_j$.
But how to formally prove the same?

REPLY [1 votes]: Let $I_i=(a_i,b_i), i=1,2,3.$ Now since $I_1\cap I_2\cap I_3\neq \phi\implies \exists x\in\mathbb{R}$, such that $x\in I_1\cap I_2\cap I_3$.
Again since $x\in I_1\cap I_2\cap I_3$, implies that $x\in I_i\cap I_j, \forall i\neq j.$
Now let $a'=\min\{a_1,a_2,a_3\}$ and let that $b'=\{b_1,b_2,b_3\}$.
Now since none of $I_i$ is contained in $I_j$, where $i\neq j$, implies that, we have $$(a_i,b_i)\nsubseteq (a_j,b_j), \forall i\neq j.$$
We also have that $x\in I_i\cap I_j\implies I_i\cap I_j\neq \phi.$ Therefore, either we have $a_i<a_j<b_i<b_j$ or we have $a_j<a_i<b_j<b_i$.
This implies that $\forall i,j,$ we have $a_i<b_j$.
The conclusion $a_i<a_j<b_i<b_j$ or $a_j<a_i<b_j<b_i, \forall i\neq j$, helps us in concluding that if $a'=a_k$ and $b'=b_l$, then we must have $k\neq l$.
This implies that $a_k<a_l<b_k<b_l$. Now let the other set be $(a_m,b_m)$.
Therefore, we have $a_k<a_m<b_k<b_m\implies a_k<a_m<b_k<b_m<b_l$.
Now two cases are possible, either we will have $a_k<a_m<a_l<b_k<b_m<b_l$ or we will have $a_k<a_l<a_m<b_k<b_m<b_l$.
Observe that if any of the above mentioned cases is true then we can conclude that $(a_m,b_m)\subseteq(a_k,b_k)\cup (a_l,b_l)$.
This implies that at least one of the given sets is contained in the union of the other two.


TITLE: Diagonalization of real symmetric matrices with symplectic matrices
QUESTION [4 upvotes]: Consider the following real symmetric matrix
$M=\left[\begin{array}{ccc} A & B\\ B^T & D \end{array}\right]$
Both A and D are real symmetric $n\times n$ matrices. B is a real $n\times n$ matrix but not necessarily symmetric. I am interested in diagonalizing the matrix with a symplectic matrix R satisfying
$R^TJR=J$
where
$J=\left[\begin{array}{ccc} 0 & 1\\ -1 &0 \end{array}\right]$
such that $R^TMR$ is block-diagonal. In particular, I am interested in an algorithm that can be coded using, say Matlab. I am aware of the following similar posts.
1.  This post discussed a similar question for Hamiltonian matrix. However, in the above case, matrix M may not anti-commute with J. I am thinking along the line that if it is possible to similar transform M into a Hamiltonian matrix. So I guess the question boils down to how to transform a real symmetric matrix into a Hamiltonian matrix, if possible.
2. This post gives criteria that if JM is diagonalizable, then the above procedure exists. However a clear procedure is still lacking, say to conjugate JM into the Cartan subalgebra as suggested, if we assume JM is diagonalizable. Any help is appreciated.
3. I am aware of the Willianmson's theorem from This post. But in this case, M may not be positive definite.
Any help is appreciated, and thanks in advance.

REPLY [1 votes]: I'm aware I'm answering 4 years after you asked, but if you still want to know, check out Appendix B of this https://arxiv.org/abs/0902.1502 or this paper https://arxiv.org/abs/2108.05364. Implementations are available here: https://github.com/softquanta/symplectic_decomposition.


TITLE: Devise a combinatorial problem in which it is easier to solve via probability theory than by counting methods
QUESTION [1 upvotes]: We easily find probabilities by counting the number of ways we can meet a condition and dividing it by the number of total possible outcomes.  This is using combinatorics to solve probabilities.
Im looking for a scenario, or a type of complicated combinatorial problem, in which the approach at counting the number of ways of meeting a condition is actually easier by multiplying the probability of meeting the condition (determined in other ways) by the total possible number of outcomes.
I realize this may be an absurd question, but its been in the back of my mind for some time.  Can anyone imagine up such a problem?  This is for the sake of education, so that we may all learn of new problem solving techniques.

REPLY [3 votes]: There is a very nice proof of Hook Length Formula. Its basic idea is to create an algorithm that generates random instances with uniform (discrete) distribution  and use $\frac{1}{p}$ for counting. Take a look at

C. Greene, A. Nijenhuis, H. S. Wilf, “A probabilistic proof of a formula for
  the number of Young tableaux of a given shape”, Adv. in Math. 31
  (1979), 104-109,

and

Jean-Christophe Novelli, Igor Pak, Alexander V. Stoyanovskii, “A direct
  bijective proof of the hook-length formula”.

I hope it is an example of what you asking for ;-)


TITLE: Is the coordinate ring of SL2 a UFD?
QUESTION [18 upvotes]: Is the ring $K[a,b,c,d]/(ad-bc-1)$ a unique factorization domain?

I think this is a regular ring, so all of its localizations are UFDs by the Auslander–Buchsbaum theorem. However, I know there are Dedekind domains (which are regular; every local ring is a PID, so definitely  UFD) that are not UFDs, so being a regular ring need not imply the ring is a UFD.
With the non-UFD Dedekind domains (at least the number rings), I can usually spot a non-unique factorization, but I don't see any here in this higher dimensional example.

REPLY [8 votes]: Let $R=K[X,Y,Z,T]/(XY+ZT-1)$. It's easily seen that $R$ is an integral domain.
In the following we denote by $x,y,z,t$ the residue classes of $X,Y,Z,T$ modulo the ideal $(XY+ZT-1)$.
First note that $x$ is prime: $R/xR\simeq K[Z,Z^{-1}][Y]$. Then observe that $R[x^{-1}]=K[x,z,t][x^{-1}]$ and $x$, $z$, $t$ are algebraically independent over $K$. This shows that $R[x^{-1}]$ is a UFD and from Nagata's criterion we get that $R$ is a UFD.


TITLE: Prime powers that divide a factorial
QUESTION [3 upvotes]: If we have some prime $p$ and a natural number $k$, is there a formula for the largest natural number $n_k$ such that $p^{n_k} | k!$.
This came up while doing an unrelated homework problem, but it is not itself homework. I haven't had any good ideas yet worth putting here.
The motivation came from trying to figure out what power of a prime you can factor out of a binomial coefficient.  Like $\binom{p^m}{k}$.

REPLY [7 votes]: This follows from a famous result of Kummer:
Theorem. (Kummer, 1854) The highest power of $p$ that divides the binomial coefficient $\binom{m+n}{n}$ is equal to the number of "carries" when adding $m$ and $n$ in base $p$.
Equivalently, the highest power of $p$ that divides $\binom{m}{n}$, with $0\leq n\leq m$ is the number of carries when you add $m-n$ and $n$ in base $p$.
As a corollary, you get
Corollary. For a positivie integer $r$ and a prime $p$, let $[r]_p$ denote the exact $p$-divisor of $r$; that is, we say $[r]_p=a$ if $p^a|r$ but $p^{a+1}$ does not divide $r$. If $0\lt k\leq p^n$, then
$$\left[\binom{p^n}{k}\right]_p = n - [k]_p.$$
Proof. To get a proof, assuming Kummer's Theorem: when you add $p^n - k$ and $k$ in base $p$, you get a $1$ followed by $n$ zeros. You start getting a carry as soon as you don't have both numbers in the column equal to $0$, which is as soon as you hit the first nonzero digit of $k$ in base $p$ (counting from right to left). So you really just need to figure out what is the first nonzero digit of $k$ in base $p$, from right to left. This is exactly the $([k]_p+1)$th digit. So the number of carries will be $(n+1)-([k]_p+1) = n-[k]_p$, hence this is the highest power of $p$ that divides $\binom{p^n}{k}$, as claimed. $\Box$


TITLE: If $m_n + n$ is true, then prove that $m_{n+1} + n + 1$ is true. Algebraic breakdown help
QUESTION [1 upvotes]: I'm a 46 year old Discrete Math student, and with all of the gaps in my math education, remembering the algebra to do the last step of my Induction proofs have been the hardest part for me. How do I deal with the subscript and where do I even start to break the right side of this down to prove that it is equal to the left.  Looking forward to learning this. I have added some example pics of the program that I have created to run the Truth Table examples.  It sums up the total of each row and proves that my theory is correct.
Let $m_n$ be the number of true $p_i$ in the chained bi-conditional $(p_1 \leftrightarrow p_2 ... \leftrightarrow p_n)$
Claim: The chained bi-conditional is true if and only if $m_n + n$ is even for all positive integers. This formula covers both odd and even cases.
$m_n + n$ $\rightarrow$ $m_{n+1} + n + 1$

REPLY [1 votes]: So if I understand correctly you want to see the truth of the biconditional
$$P \leftrightarrow p_{n+1}$$ where $P$ itself is a complex chain of smaller statements with arrows between them. The beauty of the situation is that you discovered the a different way of describing the truth or falsehood of $P$: $P$ is true if $m_n + n$ is even and false if it is odd.
This is great. Now we don't have to worry about the inner structure of $P$ and just role with this.
We know that there are just 4 possible situations:

$P$ is true and $p_{n+1}$ is true
$P$ is true and $p_{n+1}$ is false
$P$ is false and $p_{n+1}$ is true
$P$ is false and $p_{n+1}$ is false

In each of the four situations you can compute two things:

Whether the 'big' statement $P \leftrightarrow p_{n+1}$ is true or not.
Whether $(n+1) + m_{n+1}$ is even or odd

Once you have done that you can check if the relation between truth/falsehood of the big statment and even/oddness (parity) of $(n+1) + m_{n+1}$ that you conjectured holds in all four cases and if it does you are done.
Now I think you don't have any trouble with item 1. So let's talk about 2.
First: why $(n+1) + m_{n+1}$? Well because there are $n + 1$ terms in the big statement $P \leftrightarrow p_{n+1}$
As I said you want to compute this quantity in four situations. I'll do the first, you can do the other three.
$P$ is true so $n + m_n$ is even.
Let's distinguish two more cases: $n$ even and $n$ odd.
If $n$ is even then $m_n$, the number of true statements among $p_1, \ldots, p_n$ must be even as well. Now we add the aditional true (IN THIS CASE) statement $p_n$ so we get that $m_{n+1} = m_n + 1$: the total number of true statments is one more so $m_{n+1}$ is odd. On the other hand, since $n$ is even, $n+1$ is odd as well and hence $(n+1) + m_{n+1}$ is odd + odd = even.
If $n$ is odd then, since $n + m_n$ is even, $m_n$ is odd as well. We add one more true statement ($p_n$) so we find (IN THIS CASE) that $m_{n+1}$ is even. But $n + 1$ is even as well (since $n$ is odd) so again $(n+1) + m_{n+1}$ is even.
Conclusion: in the scenario of the first bullet point we have that $(n+1) + m_{n+1}$ is even. One down, three to go.


TITLE: A nice variety without a smooth model
QUESTION [8 upvotes]: Is there a simple example of a smooth proper variety $X$ over $K=\mathbb{Q}_p$ ($p$ prime) such that
--- $X(K)\neq\emptyset$,
--- the $l$-adic étale cohomology $H^i(X\times_K\bar K,\mathbb{Q}_l)$ is unramified for every prime $l\neq p$ and every $i\in\mathbb{N}$,
--- the $p$-adic étale cohomology $H^i(X\times_K\bar K,\mathbb{Q}_p)$ is crystalline for every $i\in\mathbb{N}$,
and yet,
--- $X$ is not the generic fibre of any smooth proper $\mathbb{Z}_p$-scheme ?
My example of a Châtelet surface with these properties is simple enough, but can one do better ?

REPLY [3 votes]: As Minhyong suggests, a curve $C$ (with $C(\mathbb{Q}_p)\neq0$) which has bad reduction but whose jacobian $J$ has good reduction would do the affair.  This works because the cohomology of $C$ is essentially the same as that of $J$, and because an abelian $\mathbb{Q}_p$-variety has good reduction if and only if its $l$-adic étale cohomology is unramified for some (and hence for every) prime $l\neq p$ (Néron-Ogg-Shafarevich) or its $p$-adic étale cohomology is crystalline (Fontaine-Coleman-Iovita).
I asked Qing Liu for explicit examples.  He suggested the curve
$$
y^2=(x^3+1)(x^3+ap^6)\qquad  (a\in\mathbb{Z}_p^\times)
$$
when $p\neq2,3$, and $y^2=(x^3+x+1)(x^3+a3^4x+b3^6)$, with $a,b\in\mathbb{Z}_3^\times$, for $p=3$.
He refers to Proposition 10.3.44 in his book for computing the stable reduction of these $C$, and to Bosch-Lütkebohmert-Raynaud,
Néron models, Chapter 9, for showing that $J$ has good reduction.
I "accept" this answer as coming from Minhyong Kim and Qing Liu.

REPLY [1 votes]: Chandan, can you tell me the numbers of the theorems in Bosch-Lütkebohmert-Raynaud?


TITLE: Magnetic lines of force , from south to north
QUESTION [0 upvotes]: I have understand that even inside of a bar magnet magnetic lines move from north to south from this que below.
Why is the direction of magnetic field from South to North Pole inside a magnet?.
But my doubt is that , when a current is passed in upward ⬆ direction through a mettalic conductor /wire , the magnetic lines of force are produced like concentric circles in anticlockwise direction. Then south pole is produced at front side of the wire while north pole is produced at back side of wire. We know that magnetic lines of force are directed from north to south pole ,
But in this scenario , though the lines of force are directed from north to south but at the same time it is also directed from south to north because it forms concentric circles . So pls explain me how can these lines are directed from south to north
If you have got my point , then pls answer me

REPLY [0 votes]: We know that magnetic lines of force are directed from north to south
pole

Yes. In the case of the Earth magnetic field, the magnetic South pole is close to the geographic North pole and vice versa. The North pole of a compass needle points to that direction, to the Earth magnetic South pole.
If a horizontal conductor has a conventional current flowing South to North, and a compass is placed over it, the needle deflects to East. That indicates that the $B$ field lines generated by the current points to East over the wire.
Here, thinking about South and North poles are confusing because the lines form circles around the wire, there is no poles. But anyway the convention that the lines are from North to South works. The vectorial sum of the Earth and the wire B-fields now points somewhere between North and East. And the needle aligns with that new direction.
If the compass is placed under the wire, the deflection is to the west, and the same reasoning applies.


TITLE: Using $\varepsilon$-$\delta$, prove that $\lim\limits_{(x,y) \rightarrow (3,2)} (xy^2+\frac{3}{x})=13$
QUESTION [1 upvotes]: I started working on factoring $|xy^2+\frac{3}{x}-13|$ and got to $|xy^2-12-\frac{x-3}{x}|$. I have to get to $(y-2)$ somehow, but am confused how to get that from $xy^2-12$. Can anyone give even a slight hint? :)

REPLY [0 votes]: Let $\varepsilon>0$ be fixed.
We can write
\begin{align}
\big|xy^2+\frac 3x-13\big| &= \big|xy^2-3\cdot2^2+\frac3x-1\big| =\\[1.5ex]
&= \big|(x-3)\,4+x(y^2-2^2)+\frac {3-x}x\big|\leq\\[1.5ex]
&\leq \big|x-3\big|\,4+\big|x\big|\cdot\big|y-2\big|\cdot\big|y+2\big|+\frac{\big|3-x\big|}{\big|x\big|}\leq\ldots\\[1.5ex]
\end{align}
$\Big[$Because $(x,y)$ converges to $(3,2)$, we can suppose that $2\leq x\leq 4,\quad 1\leq y\leq 3$, i.e. $(x,y)$ belongs to the closed square $S\Big((3,2),1\Big)$ centered in$(3,2)$ with half-side $1\Big]$
\begin{align}
\phantom{ll}\ldots&\leq\big|x-3\big|\,4+4\,\big|y-2\big|\,5+\frac12\,\big|3-x\big|=\\[1.5ex]
&=\frac92\,\big|x-3\big|+20\,\big|y-2\big|\leq\\[1.5ex]
\end{align}
\begin{align}
&\leq 20\Big(\big|x-3\big|+\big|y-2\big|\Big)\leq\ldots\\[1.5ex]
\end{align}
$\bigg[$ By using that $\;a,b\in\mathbb R \implies a+b\leq\sqrt2\sqrt{a^2+b^2}\bigg]$
\begin{align}
\phantom{l}\ldots&\leq 20\,\sqrt2\,\sqrt{\big|x-3\big|^2+\big|y-2\big|^2}\leq\varepsilon
\end{align}
when
$$ (x,y)\in B\Big((3,2), \frac{\varepsilon}{20\sqrt2}\Big)\cap S\Big((3,2),1\Big), $$
where $B\Big((\xi,\eta),r\Big)$ denote the closed disk centered in $(\xi,\eta)$ and with radius $r$.
Hence a possible choice for $\delta$ is
$$\delta=\min\Big(\frac\varepsilon{20\sqrt2},1\Big). $$


TITLE: Deep theorem with trivial proof
QUESTION [47 upvotes]: It is the snobbishness of the young to suppose that a theorem is trivial because the proof is trivial.
-- Henry Whitehead

I have been awestruck by the beauty of this quote.
What is in your opinion a good contender to exemplefy the meaning intended by Whitehead?
Off the top of my head I am thinking of Langrange's theorem in Group Theory, which is rather simple to prove but provides a very useful insight.

REPLY [6 votes]: 1.The Hahn-Banach theorem for real Banach spaces. The proof is a few lines of calculation, which are then baked by Zorn's Lemma.

There is no largest prime (or an infinite set of primes, depending on your choice of axioms). The first chapter of the book Prime Number Records gives about 23 different proofs. One due to Prof. Leo Morse:"It suffices to give a strictly increasing unbounded sequence of pair-wise co-prime natural numbers. For example, the Fermat numbers."
With some modern machinery (Liouville's theorem that a bounded entire function is a constant), the Fundamental Theorem of Algebra is proved as follows: If p is a  non-zero polynomial and p(z) is never 0 then 1/p is a  bounded  entire function.


TITLE: What is the General procedure for graphing heavidside functions?
QUESTION [0 upvotes]: I was given an example of a second order differential equation with U1(t)-U(3t) as the forcing function. I was asked to graph the forcing function and the answer is that the function is 1 when t is between 1 and 3. I understand that heavidside functions are like a switch that turn on to be one. I thought that the interval would be one when t was less than one and 1 when t was greater than one. Is there a general procedure when asked to graph heaviside functions in a similar fashion  to this problem? What if there was a constant infront of the heviside or what if they were added instead of subtracted?

REPLY [0 votes]: (I'm understanding here that your $U_1(t)$ here is a Heaviside function that transitions at $t=1$.)
It's true to say that Heaviside functions are "like a switch" -- they start at one value, 0, and then they abruptly transition to the value of 1. You probably understand how to graph continuous functions, like $f(t) = t$, very well, and so only the discontinuous part makes it tricky, but you are correct that it changes at $t=1$ and $t=3$. If you can find these transition points, break the function up into parts where it is continuous, and graph those separately.
In this case, you correctly found that the transitions points are $1$ and $3$. In general, these will occur whenever the argument of $U_1$ passes through $1$. For instance, if I had $U_1(x^2-3)$ in my function, then my transition would be at $-2$ and $2$.
Take one point between each transition point, and consider what the function looks like there. Since $U_1$ is "locally constant", you can replace it with a constant as long as you're thinking about that interval. Here, your three intervals are $t < 1$, $1 < t < 3$, and $3 < t$. So just plug in values from each: evaluate your forcing function at $t=0$, $t=2$, and $t=4$. You know how $U_1$ will act with certain values plugged in, so then you can just see that you'll get $0, 1, 0$. Since it's constant there, that fully describes your function.
This procedure will also work for other things you mentioned, like adding, subtracting, leading constants, or other coefficients. But it also extends to many other things! If you want to plot a complicated function like
$$f(t) = e^t U_0(-t) + t U_0(t) U_0(\sin t)$$
Notice that the first two $U$'s would transition when $t=0$, and the last transitions whenever $\sin t = 0$, that is, at $t = n \pi$. We want to see what happens on either side of $t=0$, so plug in $t = -1$ (only to the $U$s!), and we get
$$f(t) = e^t * 1 + t * 0 * U(\sin t) = e^t$$
So we know that everywhere up to $t=0$, it's just an exponential function. Then we plug in $t=1$, and we get
$$f(t) = e^t * 0 + t * 1 * U(\sin t) = t U(\sin t)$$
and this narrows it down. Then we know it's positive for $t$ after an even multiple of $\pi$, so it's just going to keep turning on and off there, with a steady ramp of $t$ the whole way.
I hope this illustrates how to tackle these kinds of problems in general!


text:
\begin{document}

\title[Topological conjugacy]{Topological conjugacy for unimodal nonautonomous discrete dynamical systems}

\author{Ermerson Araujo}

\date{\today}
\keywords{Combinatorial equivalence, kneading sequences, unimodal maps}
\subjclass[2010]{37B10, 37E05}

\address{Ermerson Araujo, Departamento de Matem\'atica, Universidade Federal do Cear\'a (UFC),
Campus do Pici, Bloco 914, CEP 60440-900, Fortaleza -- CE, Brasil}
\email{ermersonaraujo@gmail.com}

\begin{abstract}

The goal of this article is to study how combinatorial equivalence implies
topological conjugacy. For that, we introduce the concept of kneading sequences for nonautonomous
discrete dynamical systems and show that these sequences are a complete invariant for
topological conjugacy classes.

\end{abstract}

\maketitle

\section{Introduction}

A {\it nonautonomous discrete dynamical system} (short NDS) is a pair $(X,\mathcal{F})$,
where $X$ is a metric space and
$\mathcal{F}=(f_n)_{n\geq1}$ is a sequence of continuous maps
$f_n:X\to X$. The orbits of the system are described by the
maps $f^\ell_n:X\to X$, defined by
\[
f^\ell_n(x):=(f_{n+\ell-1}\circ\cdots\circ f_n)(x)\; \text{for each}\; n,\ell\in\N \;\text{and}\; x\in X,
\]
\[
f_n^0:=\text{id}_{X}\; \text{for each}\; n\in\N.
\]
The classical autonomous setting is obtained by letting $f_n=f$, for every $n\geq1$.
Furthemore, we define $f_n^{-\ell}:=(f^\ell_n)^{-1}$, which is only applied to sets. (We do not
assume that the maps $f_n$ are invertible.)

Nonautonomous discrete dynamical systems were introduced by S. Kolyada and
L. Snoha \cite{ks} motivated by the desire to understand better the
topological entropy of skew products.

In recent years, a large number of papers have been devoted to dynamical
properties in nonautonomous discrete systems.

Huang \etal \cite{hwzz} introduced and studied
topological pressure for nonautonomous discrete dynamical systems.
Metric entropy of NDS has been studied in \cite{kawan11} and \cite{kawan22}.
The notion of chaos was extended to NDS setting by many authors
(\eg \cite{tichen,shi,wuzhu,zhshsh}).
So, although recognizably distinct from classical autonomous dynamical systems, the theory
of the nonautonomuos discrete dynamical systems
has developed into a highly active field of research.
In this way, it is natural to search for ways to classify the NDS
in classes with similar dynamical behavior.

The easiest way to see when two NDS have the same dynamical behavior
is when there exists a topological conjugacy between them.
For example, topological entropy for NDS is invariant by topological conjugacy \cite[Sec. 5]{ks}.
Let $(X, \mathcal{F})=(X, (f_n)_{n\geq1})$ and $(Y, \mathcal{G})=(Y, (g_n)_{n\geq1})$
be two nonautonomous discrete dynamical systems. We say that
$(X, \mathcal{F})$ and $(Y, \mathcal{G})$
are topologically conjugate if there exists a sequence
$(h_n)_{n\geq1}$ of homeomorphisms from $X$ into $Y$ such that
both families $(h_n)_{n\geq1}$ and $(h_n^{-1})_{n\geq1}$ are equicontinuous and
$h_{n+1}\circ f_n=g_n\circ h_n$ for every $n\geq1$.
If $h_n$ is a continuous surjective map for all $n\geq1$, then we say that
$(X, \mathcal{F})$ and $(Y, \mathcal{G})$
are topologically semi-conjugate. When $X, Y$ are intervals we also require
that all homeomorphisms $h_n$ are order preserving.

We can not remove the equicontinuity condition of the family $(h_n)_{n\geq1}$
as this would imply that all NDS $(f_n)_{n\geq1}$, with $f_n$ homeomorphism
for each $n\geq1$, are topologically conjugate to a trivial NDS, see \cite[Prop. 2.1]{AF}.

Thus, inspired by this we ask the following.

\medskip
\noindent
{\bf Problem 1:} Let $(X, \mathcal{F})$ and $(Y, \mathcal{G})$ be two
nonautonomous discrete dynamical systems.
Under which conditions $(X, \mathcal{F})$ and $(Y, \mathcal{G})$
are topologically (semi-)conjugate?

\medskip
In this short work, we introduce the notion of combinatorial equivalence for NDS
on the particular case where $X$ is an interval and $f_n$ is a unimodal
map for every $n\geq1$. We can give the following answer to Problem 1.

\begin{theorem}\label{thmA}
Let $(J^\F, \F)$ and $(J^\G, \G)$ be two unimodal nonautonomous
discrete dynamical systems and assume that
both satisfy the limit property. Then $(J^\F,\F)$ and $(J^\G,\G)$ are topologically
conjugate if and only if they have the same kneading sequence.
\end{theorem}

To prove this theorem, we will construct kneading sequences in a similar way as
Milnor and Thurston made in their famous paper \cite{MT}.

\section{Kneading sequences for UNDS}

Let $(J^\mathcal{F}, \mathcal{F})$ be the NDS defined as follows:
let $J^\F=[a^\F, b^\F]$ be an interval, and
$f_n:J^\F\to J^\F$ be a continuous map satisfying
$f_n(a^\F)=f_n(b^\F)=a^\F$ for each $n\geq1$.
Besides that, there is $c_n^\F\in(a^\F,b^\F)$
such that $f_n\restriction_{[a^\F, c_n^\F]}$ is strictly increasing
and $f_n\restriction_{[c_n^\F, b^\F]}$ is strictly decreasing.
The points $\{c_n^\F:\; n\geq1\}$ are called {\it turning points} of the NDS.
We call the NDS defined above {\it unimodal nonautonomous discrete dynamical system} (short UNDS)

Let's proceed by constructing the symbolic space for UNDS.
Consider the alphabet $\mathcal{A}_\F=\{L, c_n^\F, R: n\geq1\}$.

\medskip
\noindent
{\sc Address of a point:} Let $n\geq1$. The {\em address} of a point $x\in J^\F$
on the level $n$ is the letter $i_{\F,n}(x)\in\mathcal A_\F$ defined by
$$
i_{\F,n}(x)= \left\{
\begin{array}{ll}
L    & \textrm{, if }  x\in[a^\F,c_n^\F)     \\
c_n^\F  & \textrm{, if }  x=c_n^\F \\
R    & \textrm{, if }  x\in(c_n^\F,b^\F].
\end{array}
\right.
$$

\medskip
\noindent
{\sc Itinerary of a point:} Let $n\geq1$. The {\em itinerary} of a point $x\in J^\F$
on the level $n$ is the sequence $I_{\mathcal{F},n}(x)\in\mathcal A_\F^{\{0,1,2,\ldots\}}$ defined by
\[
I_{\mathcal{F},n}(x)=(i_{\mathcal{F},n}(x),i_{\mathcal{F},n+1}(f_n^1(x)),\ldots,
i_{\F,n+\ell}(f_n^\ell(x)),\ldots).
\]

\medskip
\noindent
{\sc Kneading Sequence:} The {\em kneading sequence} of $(J^\F,\F)$
is the sequence $\V(\F)=\{\V_n^\F\}_{n\geq1}$, where $\V_n^\F:=I_{\F,n}(c_n^\F)$.

\medskip
The proposition below ensures that the kneading sequences
are preserved by topological conjugacy.

\begin{proposition}\label{prop1}
Let $(J^\F,\F)$ and $(J^\mathcal{G},\mathcal{G})$ be two UNDS.
If $(J^\F,\F)$ and $(J^\mathcal{G},\mathcal{G})$ are topologically conjugate
then $\V(\F)=\V(\mathcal{G})$.
\end{proposition}

Before going to the proof, we observe that $\V(\F)=\V(\mathcal{G})$ means
that we are identifying $\mathcal{A}_\F$ with $\mathcal{A}_\G$.

\begin{proof}
We have shown that $I_{\F,n}(c_n^\F)=I_{\G,n}(c_n^\G)$ for all $n\geq1$.
Let $(h_n)_{n\geq1}$ be a conjugacy between $(J^\F,\F)$ and $(J^\G,\G)$.
Since $h_n:J^\F\to J^\G$ is an order preserving homeomorphism,
$h_n$ sends the increasing (decreasing) interval of $f_n$ to the increasing (decreasing)
interval of $g_n$. Thus, for all $n\geq1$ we have $h_n(c_n^\F)=c_n^\G$ and
\[
(h_{n+\ell}\circ f_n^\ell)(c_n^\F)=(g_n^\ell\circ h_n)(c_n^\F)=g_n^\ell(c_n^\G),\ \forall\ell\geq 1.
\]
Once $h_{n+\ell}$ is ordering preserving on $J^\F$ for all $\ell\geq1$,
\[
i_{\F,n+\ell}(f_n^\ell(c_n^\F))=i_{\G,n+\ell}(g_n^\ell(c_n^\G)).
\]
Therefore $I_{\F,n}(c_n^\F)=I_{\G,n}(c_n^\G)$.
\end{proof}

\section{Combinatorial equivalence for UNDS}

The purpose of this section is to introduce the concept of
combinatorial equivalence between two UNDS.
We will prove that equality between kneading sequences
is sufficient to ensure combinatorial equivalence.

For each $n\geq1$ set
\[
\cc_n(\F)=\left\{x\in J^\F;\;\exists\; \ell\geq0 \;
\textrm{such that}\; f_n^\ell(x)=c_{n+\ell}^\F\right\}.
\]

\begin{definition}
We say that two UNDS $(J^\F,\F)$ and $(J^\mathcal{G},\mathcal{G})$ with
turning points $\{c^\F_n;\;n\geq1\}$ respectively $\{c^\mathcal{G}_n;\;n\geq1\}$
are {\it combinatorially equivalent} if there exists a family of order preserving bijections
$h_n:\cc_n(\F)\to\cc_n(\mathcal{G})$ such that $h_{n+1}\circ f_n=g_n\circ h_n$
on $\cc_n(\F)\backslash\{c_n^\F\}$ for all $n\geq1$.
\end{definition}

In the next section we prove that, with the right hypotheses (limit property), the
equality between kneading sequences implies topological conjugacy.
The following theorem is fundamental for that.

\begin{theorem}\label{thm1}
Let $(J^\F,\F)$ and $(J^\mathcal{G},\mathcal{G})$ be two UNDS with
kneading sequences $\V(\F)$ and $\V(\mathcal{G})$. If
$\V(\F)=\V(\G)$, then $(J^\F,\F)$ and $(J^\G,\mathcal{G})$
are combinatorially equivalent.
\end{theorem}

The proof of Theorem \ref{thm1} is inspired by \cite[Thm. 1]{Rand}.
Before proving the theorem, let us introduce some notations.

Given an interval $J$, let $\partial J$ denote its boundary.
For each $k,\ell\geq1$, the functions $f_k$ and $f_k^\ell$
will be denoted by $f(k)$ and $f(k,\ell)$ respectively.
In the same way we denote $g_k$ and $g_k^\ell$ by $g(k)$ and $g(k,\ell)$ respectively.

Fix $n\geq1$. For each $k\geq1$ let $\cc^k_n(\F)=\{x\in J^\F: f(n,\ell)(x)=
c_{n+\ell}^\F\; \textrm{for some}\; 0\leq\ell\leq k-1\}$. Note that
$\cc_n(\F)=\bigcup_{k\geq 1}\cc^k_n(\F)$.

Since $\cc_n^{k+1}(\F)=\cc_n^k(\F)\cup f(n,k)^{-1}(c_{n+k}^\F)$,
we have that $\cc_n^k(\F)\subset\cc_n^{k+1}(\F)$. Let
$\pp_n^k(\F)=\{J_n^\F(\ell)\subset J^\F:\; \partial J_n^\F(\ell)
\subset\cc_n^{k}(\F)\cup\{a^\F,b^\F\}
\; \textrm{and}\; 1\leq\ell\leq\theta^k_n(\F)\}$,
where the increasing order of the index of
$J_n^\F(\ell)$ is the same order as the intervals are placed in $J^\F$
and $\theta^k_n(\F):=\#\pp_n^{k}(\F)$.
By definition, $J_n^\F(\ell)$ is a maximal
monotonicity closed interval of $f(n,k)$ for all $1\leq\ell\leq \theta^k_n(\F)\}$.

We denote $f(k)\restriction_{[a^\F,c_k^\F]}$ and
$f(k)\restriction_{[c_k^\F,b^\F]}$
by $f_-(k)$ and $f_+(k)$ respectively. Furthermore, given a sequence
$j=j_1j_2\ldots j_k \in\{-, +\}^k$ put
\[
f_j(n,k):=f_{j_k}(n+k-1)\circ\cdots\circ f_{j_2}(n+1)\circ f_{j_1}(n).
\]
Given $x\in\cc_n^k(\F)\backslash\{c_n^\F\}$ we
consider $j=j_1j_2\ldots j_\ell \in\{-, +\}^\ell$, with $1\leq\ell\leq k-1$,
the minimal sequence such that $f_j(n,\ell)(x)=c_{n+\ell}^\F$.
We denote $x$ by $x_j^\F(n)$. Consider the set
\[
\aaa_n^k(\F):=\biggl\{j=j_1\ldots j_\ell\in\{-, +\}^\ell:
\begin{array}{c}
\exists x\in\cc_n^k(\F)\textrm{ such that }\\
x=x^\F_j(n)\textrm{ for some }0\leq\ell\leq k-1
\end{array}
\biggr\},
\]
where $c_n^\F=x_j^\F(n)$ when $j=\emptyset$ ($\ell=0$). Thus
\[
\cc^k_n(\F)=\left\{x_j^\F(n):\; j\in\{-,+\}^\ell\; \textrm{with}\; 0\leq\ell\leq k-1\right\}.
\]

Finally, note that for each $J_n^\F(\ell)\in\pp_{n}^k(\F)$ there is a unique sequence
$j=j_1\ldots j_k\in\{-,+\}^k$ such that
$f_{j}(n,k)$ is strictly monotone on $J_n^\F(\ell)$. So
\[
\partial J_n^\F(\ell)\in\left\{a^\F, b^\F, c_n^\F, x^\F_{j_1\ldots j_m}(n)\right\},
\]
for $1\leq m\leq k-1$.

In the same way we can define $\cc_{n}^k(\G)$, $\pp_{n}^k(\G)$ and $\aaa_n^k(\G)$.

So the main idea in the proof is to show that for each $k\geq1$ there exists a strictly
increasing bijection map $h_n^k:\cc_n^k(\F)\to\cc_n^k(\G)$ (see Figure \ref{fig1}) such that:
\begin{enumerate}[$\circ$]
\item $h_{n+1}^k\circ f(n)=g(n)\circ h_n^k$ on $\cc_n^k(\F)\backslash\{c_n^\F\}$,
\item $h_n^k\restriction_{\cc_n^{k-1}(\F)}=h_n^{k-1}$.
\end{enumerate}

\begin{figure}[H]\centering
\begin{tikzpicture}[scale=2]
      \draw[line width=1pt] (1,-1) -- (-1,-1);
			\draw[line width=1pt] (1,1) -- (1,-1);
			\draw[line width=1pt] (1,1) -- (-1,1);
			\draw[line width=1pt] (-1,1) -- (-1,-1);
			\draw[|->] (1.4,-1.15) -- (1.8,-1.15) ;
			\fill[black] (1.6,-1.15) circle (0.mm) node[above] {$h_n^2$};
			\draw[dashed] (0.54,0.2) -- (-1,0.2) node[left]{$c_{n+1}^\F$};
      \draw[dashed] (0,0) -- (0,-1) node[below] {$c_n^\F$};
			\draw[dashed] (-1,-1) -- (1,1);
			\draw[dashed] (-0.54,0.2) -- (-0.54,-1) node[below]{$x_-^\F(n)$};
			\draw[dashed] (0.54,0.2) -- (0.54,-1) node[below]{$x_+^\F(n)$};
  \draw[color=red,line width=1.5pt] (0,0.7) parabola (-1,-1) ;
	\draw[color=red,line width=1.5pt] (0,0.7) parabola (1,-1) ;
	\fill[red,thick] (0.74,0.18) circle (0.mm) node[above] {$f_n$};

      \draw[line width=1pt] (4.2,-1) -- (2.2,-1);
			\draw[line width=1pt] (4.2,1) -- (4.2,-1);
			\draw[line width=1pt] (4.2,1) -- (2.2,1);
			\draw[line width=1pt] (2.2,1) -- (2.2,-1);
			\draw[dashed] (2.2,-1) -- (4.2,1);
			\draw[dashed] (3.92,-0.2) -- (2.2,-0.2) node[left]{$c_{n+1}^\G$};
      \draw[dashed] (3.2,0) -- (3.2,-1) node[below] {$c_n^\G$};
			\draw[dashed] (2.47,-0.2) -- (2.47,-1) node[below]{$x_-^\G(n)$};
			\draw[dashed] (3.92,-0.2) -- (3.92,-1) node[below]{$x_+^\G(n)$};
			\draw[color=blue,line width=1.5pt] (3.2,0.7) parabola (2.2,-1) ;
	    \draw[color=blue,line width=1.5pt] (3.2,0.7) parabola (4.2,-1) ;
  \fill[blue,thick] (3.94,0.18) circle (0.mm) node[above] {$g_n$};
\end{tikzpicture}
\caption{Construction of $h_n^2$.}
\label{fig1}
\end{figure}

We proceed by induction on $k$.

\begin{proof}[Proof of Theorem \ref{thm1}.]
Fix $n\geq1$. It is clear that we can define $h_n^1:\cc_n^1(\F)\to\cc_n^1(\G)$
as $h_n^1(c_n^\F):=c_n^\G$. Suppose by induction that there exists
$h_n^k:\cc_n^k(\F)\to\cc_n^k(\G)$ strictly increasing bijection such that
$h_{n+1}^k\circ f(n)=g(n)\circ h_n^k$ on $\cc_n^k(\F)\backslash\{c_n^\F\}$
and $h_n^k\restriction_{\cc_n^{k-1}(\F)}=h_n^{k-1}$. This implies that
$\#\pp_{n}^k(\F)=\#\pp_{n}^k(\G)$ and $\aaa_n^k(\F)=\aaa_n^k(\G)$.

Let $J_n^\F(i)\in\pp_{n}^k(\F)$ and
$j=j_1\ldots j_k\in\{-,+\}^k$ such that
\[
J_n^\F(i)=\left[ x_{j_1\ldots j_m}^\F(n), x_{j_1\ldots j_\ell}^\F(n) \right],
\]
with $1\leq m\neq\ell\leq k-1$. By the manner that we order the intervals
on $\pp_{n}^k(\F)$ and $\pp_{n}^k(\G)$ we have
\[
h_n^k(x_{j_1\ldots j_{m}}^\F(n))=x_{j_1\ldots j_{m}}^\G(n)  \; \textrm{and} \;
h_n^k(x_{j_1\ldots j_{\ell}}^\F(n))=x_{j_1\ldots j_{\ell}}^\G(n),
\]
where
\[
J_n^\G(i)=\left[ x^\G_{j_1\ldots j_m}(n), x_{j_1\ldots j_\ell}^\G(n) \right].
\]
Furthermore,
\begin{eqnarray*}
f_j(n,k)(J_n^\F(i))&=&[f_{j_k}(n+k-1)\circ\cdots\circ f_{j_{m+1}}(n+m)(c_{n+m}^\F), \\
                   & & \qquad \qquad \qquad \qquad
										   f_{j_k}(n+k-1)\circ\cdots\circ f_{j_{\ell+1}}(n+\ell)(c_{n+\ell}^\F)]
\end{eqnarray*}
and
\begin{eqnarray*}
g_j(n,k)(J_n^\G(i))&=&[g_{j_k}(n+k-1)\circ\cdots\circ g_{j_{m+1}}(n+m)(c_{n+m}^\G), \\
                   & & \qquad \qquad \qquad \qquad
										   g_{j_k}(n+k-1)\circ\cdots\circ g_{j_{\ell+1}}(n+\ell)(c_{n+\ell}^\G)].
\end{eqnarray*}

Since $\V(\F)=\V(\G)$,
\[
f_j(n,k)(J_n^\F(i))\cap\{c_{n+k}^\F\}\neq\emptyset
\]
if, and only if
\[
g_j(n,k)(J_n^\G(i))\cap\{c_{n+k}^\G\}\neq\emptyset.
\]

If $f_j(n,k)(J_n^\F(i))\cap\{c_{n+k}^\F\}=\emptyset$, then
$J_n^\F(i)\in\pp_n^{k+1}(\F)$.
On the other hand, if $\partial\left(f_j(n,k)(J_n^\F(i))\right)\cap\{c_{n+k}^\F\}\neq\emptyset$
we get $J_n^\F(i)\in\pp_n^{k+1}(\F)$.
In any case, we also have $J_n^k(\G)\in\pp_n^{k+1}(\G)$.
Thus, $x_{j_1\ldots j_m}^\F(n)$ and $x_{j_1\ldots j_\ell}^\F(n)$ are the only points
of $\cc_n^{k+1}(\F)$ on $J_n^\F(i)$. The same thing holds on $J_n^\G(i)$.

Now, if $\textrm{int}\left(f_j(n,k)(J_n^\F(i))\right)\cap\{c_{n+k}^\F\}\neq\emptyset$,
then there exists a unique
\[
x_j^\F(n):=f_j(n,k)^{-1}(c_{n+k}^\F)\in\textrm{int}(J_n^\F(i)).
\]
Hence, there also exists a unique
\[
x_j^\G(n):=g_j(n,k)^{-1}(c_{n+k}^\G)\in\textrm{int}(J_n^\G(i)).
\]
We define
\[
h_n^{k+1}(x_j^\F(n)):=x_j^\G(n),
\]
where $x_j^\F(n)\in\cc_n^{k+1}(\F)\backslash\cc_{n}^k(\F)$
and $x_j^\G(n)\in\cc_n^{k+1}(\G)\backslash\cc_{n}^k(\G)$.
Moreover, if $x\in\cc_n^k(\F)$, then we put $h_n^{k+1}(x)=h_n^k(x)$.

The cases where  $J_n^\F(i)=[a^\F, x_{j_1\ldots j_\ell}^\F(n)]$,
$J_n^\F(i)=[x^\F_{j_1\ldots j_\ell}(n), b^\F]$,
or $J_n^\F(i)=[x^\F_{j_1\ldots j_\ell}(n), c_n^\F]$, with $1\leq\ell\leq k-1$, are similar. Thus
we have that $h_n^{k+1}:\cc_n^{k+1}(\F)\to\cc_n^{k+1}(\G)$ is
a strictly increasing bijection such that:
\begin{enumerate}[$\circ$]
\item $h_{n+1}^{k+1}\circ f(n)=g(n)\circ h_n^{k+1}$ on $\cc_n^{k+1}(\F)\backslash\{c_n^\F\}$, and
\item $h_n^{k+1}\restriction_{\cc_n^k(\F)}=h_n^k$.
\end{enumerate}

Therefore, for all $n\geq1$ we can define
\[
\begin{array}{cccl}
h_n \ : & \! \cc_n(\F) & \! \longrightarrow  & \! \cc_n(\G) \\
        & \! w & \! \longmapsto      & \! h_n(w)=h_n^k(w),
\end{array}
\]
where $k$ is such that $w\in\cc_n^k(\F)$. Note that
$h_{n+1}\circ f(n)=g(n)\circ h_n$ on $\cc_n(\F)\backslash\{c_n^\F\}$.
This concludes the proof that $(J^\F,\F)$ is combinatorially equivalent
to $(J^\mathcal{G},\mathcal{G})$.
\end{proof}

\section{Topological conjugacy for UNDS}

Once that combinatorial equivalence has been established by the equivalence
between kneading sequences, our goal now is to extend each bijection $h_n$
to the entire interval $J^\F$ to get a topological conjugacy.

As a consequence of Theorem \ref{thm1}, we have the following

\begin{proposition}\label{prop2}
Let $(J^\F,\F)$ and $(J^\G,\G)$ be two UNDS with
kneading sequences $\V(\F)$ and $\V(\mathcal{G})$.
Assume that $\cc_n(\mathcal{G})$ is dense in $J^\G$ for all $n\geq1$.
If $\V(\F)=\V(\G)$, then $(J^\F,\F)$ and $(J^\G,\mathcal{G})$
are combinatorially equivalent and $h_n:\cc_n(\F)\to\cc_n(\G)$ can be extended
continuously to $J^\F$ as an order preserving continuous surjective map
such that $h_{n+1}\circ f_n=g_n\circ h_n$ for each $n\geq1$.
\end{proposition}

\begin{proof}
We only need to prove that each $h_n$ can be extended to $J^\F$.
Fix $n\geq1$ and let $h_n:\cc_n(\F)\to\cc_n(\G)$.

We claim that we can extend continuously $h_n$ to $\overline{\cc_n(\F)}$.
In fact, take $w\in\overline{\cc_n(\F)}\backslash\cc_n(\F)$. Suppose that there
exist $w_k^j\in\cc_n(\F)$, $j=1,2$, such that $w_k^1\uparrow w$
and $w_k^2\downarrow w$. The cases where we have only $w_k\uparrow w$ or
$w_k\downarrow w$, with $w_k\in\cc_n(\F)$, are similar.

Since $h_n$ is strictly increasing there are unique

\[
h_n^1(w):=\displaystyle\lim_k h_n(w_k^1) \ \ \textrm{and} \ \
h_n^2(w):=\displaystyle\lim_k h_n(w_k^2).
\]
Note that $h_n^j(w)$ does not depend on the sequence $w^j_k$ converging to $w$, and
$h_n^1(w)\leq h_n^2(w)$. Since $\cc_n(\G)$ is dense in $J^\G$ and $h_n$ is a strictly monotone map

we get $h_n^1(w)=h_n^2(w)$.
Therefore, for $w\in\overline{\cc_n(\F)}\backslash\cc_n(\F)$ we can define
\[
h_n(w):=\displaystyle\lim_k h_n(w_k),
\]
where $w_k\in\cc_n(\F)$ is some (any) sequence such that $w_k\to w$.
Whence $h_n:\overline{\cc_n(\F)}\to J^\G$ is a strictly increasing continuous
surjective map.

Now, let $J=(\alpha, \beta)$ be a connected component
of $J^\F\backslash\overline{\cc_n(\F)}$. Once $h_n(\alpha)=h_n(\beta)$,
we can extend $h_n$ to $J$ as $h_n(w):=h_n(\alpha)$ for all $w\in J$.
Observe that $h_n(a^\F)=a^\F$ and $h_n(b^\F)=b^\F$.

\medskip
\noindent
{\sc Claim:} $h_{n+1}\circ f_n=g_n\circ h_n$ on $J^\F$ for all $n\geq1$.

\medskip
\noindent
{\em Proof of the claim.} From Theorem \ref{thm1}, we have
$h_{n+1}\circ f_n=g_n\circ h_n$ on $\cc_n(\F)\backslash\{c_n^\F\}$.
Take $w\in J^\F\backslash\cc_n(\F)$ so that there is a sequence
$(w_k)_{k\geq1}\in\cc_n(\F)\backslash\{c_n^\F\}$ satisfying $\displaystyle\lim_k w_k=w$. Thus,
\[
(h_{n+1}\circ f_n)(w)=\displaystyle\lim_k (h_{n+1}\circ f_n)(w_k)
=\displaystyle\lim_k (g_n\circ h_n)(w_k)=(g_n\circ h_n)(w).
\]

Let $J=(\alpha,\beta)$ be a connected component of $J^\F\backslash\overline{\cc_n(\F)}$.
Observe that $f_n(J)$ also is a connected component of $J^\F\backslash\overline{\cc_{n+1}(\F)}$.
Without loss of generality we may assume that
$\alpha\in\overline{\cc_n(\F)\backslash\{c_n^\F\}}$ or $\alpha\in\{a^\F, b^\F\}$,

and so $h_{n+1}(f_n(\alpha))=g_n(h_n(\alpha))$. Thus
\[
h_{n+1}(f_n(w))=h_{n+1}(f_n(\alpha))=g_n(h_n(\alpha))=g_n(h_n(w)),
\]

for all $w\in J$. It remains to prove that $(h_{n+1}\circ f_n)(c_n^\F)=(g_n\circ h_n)(c_n^\F)$.
If $c_n^\F\in\overline{\cc_n(\F)\backslash\{c_n^\F\}}$, then equality
follows from the first part. Otherwise, there exists $\alpha\in J^\F$ so that
$(\alpha, c_n^\F)\subset J^\F\backslash\overline{\cc_n(\F)}$ and
$\alpha\in\overline{\cc_n(\F)\backslash\{c_n^\F\}}$ or $\alpha\in\{a^\F, b^\F\}$,
then equality follows from the second one. This proves the claim.

\medskip
The proof of the proposition is finished.
\end{proof}

Notice that if we also assume that $\cc_n(\F)$ is dense in $J^\F$ for all
$n\geq1$ on the proposition above, then $h_n:J^\F\to J^\G$ is a
homeomorphism for all $n\geq1$.

\medskip
\noindent
{\sc Limit Property:} Let $(J^\F, \F)$ be a UNDS. We say that $(J^\F,\F)$ satisfies the
{\em limit property} if there is a unimodal map $f:J^\F\to J^\F$
so that the turning point $c^f$ of $f$ is not periodic, and such that:
\begin{enumerate}[$\circ$]
\item $f_n$ converges uniformly to $f$;
\item $\cc_n(\F)$ is dense in $J^\F$, $\forall n\geq1$;
\item $\cc(f)$ is dense in $J^\F$.
\end{enumerate}

\medskip

Now we are ready to prove Theorem \ref{thmA}.

\begin{proof}[Proof of Theorem \ref{thmA}.]
Because of Proposition \ref{prop1} we only need to prove the reverse
implication. From Proposition \ref{prop2}, it remains to prove that the
families $(h_n)$ and $(h_n^{-1})$ are equicontinuous. By symmetry, it is enough
to prove that $(h_n)$ is equicontinuous. We will use the same notation
to the one used in the proof of Theorem \ref{thm1}. Let $\varepsilon>0$.
Since $(J^\G,\G)$ has the limit property, there exist a unimodal map
$g:J^\G\to J^\G$ and $k\geq1$ so that $\cc^k(g)$ is
$\tfrac{\varepsilon}{8}$--dense in $J^\G$. Since $c^g$ is not periodic, for each $m=1, \ldots, k-1$
we have that if $x_\alpha(g)$ and $x_\beta(g)$ are two consecutive elements
of $\cc^m(g)$ on $J^\G$, then either $c^g\notin[g^m(x_\alpha(g)),g^m(x_\beta(g))]$
or $c^g\in(g^m(x_\alpha(g)),g^m(x_\beta(g)))$.
Therefore, using that $g(n+m-1)\circ\cdots\circ g(n)$ converges uniformly
to $g^m$ for each $m\geq1$

, and proceeding as \cite[Lemma 3.1]{EA},
there is $N_1\geq1$ (see Figure \ref{fig2}) such that $\aaa_n^k(\G)=\aaa^k(g)$ and
\[
d(x_j(g),x_j^\G(n))<\tfrac{\varepsilon}{8} \; \text{for each}\; j\in\aaa^k(g) \;\text{and}\; n\geq N_1.
\]
This implies that $\cc_n^k(\G)$ is $\tfrac{\varepsilon}{4}$--dense in $J^\G$ for all $n\geq N_1$.

\begin{figure}[H]\centering
\begin{tikzpicture}[scale=1.4]
      \draw[line width=1pt] (2,-2) -- (-2,-2);
			\draw[line width=1pt] (2,2) -- (2,-2);
			\draw[line width=1pt] (2,2) -- (-2,2);
			\draw[line width=1pt] (-2,2) -- (-2,-2);
			\draw[dashed] (-1.1,-0.8) -- (-1.1,-2);
			\fill[black] (-1.15,-2) circle (0.mm) node[below]{$a$};
			\draw[dashed] (-1,-0.7) -- (-1,-2);
			\fill[black] (-1,-1.95) circle (0.mm) node[below]{$b$};
			\draw[dashed] (0.4,1.2) -- (0.4,-2);
			\fill[black] (0.35,-2) circle (0.mm) node[below]{$c$};
      \draw[dashed] (0.5,1.2) -- (0.5,-2);
			\fill[black] (0.5,-1.95) circle (0.mm) node[below]{$d$};
			\draw[dashed] (-2,0.2) -- (-0.6,0.2);
			\draw[dashed] (-2,0.3) -- (-0.45,0.3);
			\draw[dashed] (-0.58,0.2) -- (-0.58,-2);
			\fill[black] (-0.57,-2) circle (0.mm) node[below]{$e$};
			\draw[dashed] (-0.45,0.3) -- (-0.45,-2);
			\fill[black] (-0.45,-1.95) circle (0.mm) node[below]{$f$};
			\coordinate (A) at (-1.1,-0.8);
      \coordinate (B) at (0.4,1.2);
	    \draw[color=blue,line width=1.5pt]  (A) to [bend left=15] (B);
			\coordinate (C) at (-1,-0.7);
      \coordinate (D) at (0.5,1.2);
	    \draw[color=red,line width=1.5pt]  (C) to [bend left=15] (D);
			\fill[black] (-2,0.1) circle (0.mm) node[left]{$c^g$};
			\fill[black] (-2,0.4) circle (0.mm) node[left]{$c_{n+m}^\G$};
			\fill[black] (0.7,1.8) circle (0.mm) node[right]{$a=x_\alpha(g)$};
			\fill[black] (0.7,1.4) circle (0.mm) node[right]{$b=x_\alpha^\G(n)$};
			\fill[black] (0.7,1) circle (0.mm) node[right]{$c=x_\beta(g)$};
			\fill[black] (0.7,0.6) circle (0.mm) node[right]{$d=x_\beta^\G(n)$};
			\fill[black] (0.7,0.2) circle (0.mm) node[right]{$e=x_j(g)$};
			\fill[black] (0.7,-0.2) circle (0.mm) node[right]{$f=x_j^\G(n)$};
			\fill[black] (0.6,-0.6) circle (0.mm) node[right]{$d(e,f)<\tfrac{\varepsilon}{8}$};
			\fill[blue] (-1.8,1.8) circle (0.mm) node[right]{$g^m$};
			\fill[black] (-1.5,1.8) circle (0.mm) node[right]{,\;$1\leq m\leq k-1$};
			\fill[red] (-1.8,1.4) circle (0.mm) node[right]{$g(n,m)$};
			\fill[black] (-1,1.4) circle (0.mm) node[right]{,\;$n\geq N_1$};

\end{tikzpicture}
\caption{Case $c^g\in(g^m(x_\alpha(g)),g^m(x_\beta(g)))$. In the other
case, $x_j(g)$ and $x_j^\G(n)$ do not exist.}
\label{fig2}
\end{figure}

\medskip
\noindent
{\sc Claim 1:} For any $n\geq N_1$, let $x_j^\G(n)$ and $x_{\ell}^\G(n)$ be two consecutive elements
of $\cc_n^k(\G)$ on $J^\G$. Then $d(x_j^\G(n), x_{\ell}^\G(n))<\tfrac{\varepsilon}{2}$.

\medskip
\noindent
{\em Proof of the claim.} Suppose by contradiction this claim is false.
Let $x$ be the midpoint of the interval $[x_j^\G(n), x_\ell^\G(n)]$.
Hence $d(x, x_j^\G(n))\geq\tfrac{\varepsilon}{4}$ and $d(x, x_{\ell}^\G(n))\geq\tfrac{\varepsilon}{4}$.
This contradicts the $\tfrac{\varepsilon}{4}$--density of $\cc_n^k(\G)$
and finishes the proof of the claim.

\medskip
By the limit property of $(J^\F,\F)$ and $\V(\F)=\V(\G)$, there exists $N_2\geq1$
such that $\aaa_n^k(\F)=\aaa^k(f)=\aaa^k(g)$ and
\[
d(x_j(f),x_j^\F(n))<\tfrac{M}{4} \; \text{for each}\; j\in\aaa^k(f) \;\text{and}\; n\geq N_2,
\]
where
\[
M=\displaystyle\min_{\substack{j,\ell\in\aaa^k(f)\\j\neq\ell}}\left\{d(x_j(f), x_{\ell}(f)),
d(x_j(f), a^\F), d(x_j(f), b^\F)\right\}.
\]

\medskip
\noindent
{\sc Claim 2:} For any $n\geq N_2$, let $x_j^\F(n)$ and $x_{\ell}^\F(n)$ be two consecutive elements
of $\cc_n^k(\F)$ on $J^\F$. Then $d(x_j^\F(n), x_{\ell}^\F(n))\geq\tfrac{M}{2}$.

\medskip
\noindent
{\em Proof of the claim.} Suppose that $\alpha,\beta\in\aaa^k(f)$
are such that $x_\alpha(f)$ and $x_\beta(f)$ are consecutive on $J^\F$ and
$d(x_\alpha(f), x_\beta(f))=M$. The other possibilities $d(x_\alpha(f), a^\F)=M$,
and $d(x_\alpha(f), b^\F)=M$, are treated similarly.

For any two consecutive elements $x_j^\F(n)$ and $x_{\ell}^\F(n)$ of $\cc_n^k(\F)$
we have that $d(x_j^\F(n), x_{\ell}^\F(n))\geq d(x_j(f), x_\ell(f))-2\tfrac{M}{4}$.
Hence $d(x_j^\F(n), x_{\ell}^\F(n))\geq d(x_\alpha(f), x_\beta(f))-2\tfrac{M}{4}=\tfrac{M}{2}$
and the proof of the claim is finished.

\medskip
Take $0<\delta_1<\tfrac{M}{2}$ and $N=\max\{N_1, N_2\}$. Let $x,y\in J^\F$
such that $d(x, y)<\delta_1$ and $n\geq N$. Without loss of generality,
we may assume that $x\leq y$. We have two cases to consider:
\begin{enumerate}[$\circ $]
\item There are $x_\alpha^\F(n)$ and $x_\beta^\F(n)$ two consecutive elements
of $\cc_n^k(\F)$ so that $x,y\in[x_\alpha^\F(n), x_\beta^\F(n)]$. By construction
of $h_n$, $x_\alpha^\G(n)$ and $x_\beta^\G(n)$ are two consecutive elements
of $\cc_n^k(\G)$ and $h_n(x),h_n(y)\in[x_\alpha^\G(n), x_\beta^\G(n)]$.
Claim $1$ implies that $d(h_n(x), h_n(y))<\tfrac{\varepsilon}{2}$.
\item There are $x_\alpha^\F(n)$, $x_\beta^\F(n)$ and $x_\gamma^\F(n)$ three
consecutive elements of $\cc_n^k(\F)$ so that $x\in[x_\alpha^\F(n), x_\beta^\F(n)]$
and $y\in[x_\beta^\F(n), x_\gamma^\F(n)]$ (Claim $2$). Again by construction of $h_n$,
$x_\alpha^\G(n)$, $x_\beta^\G(n)$ and $x_\gamma^\G(n)$ are three consecutive elements
of $\cc_n^k(\G)$ and $h_n(x)\in[x_\alpha^\G(n), x_\beta^\G(n)]$ and
$h_n(y)\in[x_\beta^\G(n), x_\gamma^\G(n)]$. Consequently $d(h_n(x), h_n(y))<\varepsilon$.
\end{enumerate}

Now, by continuity there exists $\delta_2>0$ such that $x,y\in J^\F$ with
$d(x,y)<\delta_2$ implies $d(h_n(x),h_n(y))<\varepsilon$ for each $n=1, 2, \ldots, N-1$.
Hence taking $\delta=\min\{\delta_1,\delta_2\}$ the theorem follows.
\end{proof}

Notice that the same proof above can be used to prove a variant of Theorem \ref{thmA}:
Let $(J^\F,\F)$ and $(J^\G,\G)$ be two unimodal nonautonomous
discrete dynamical systems, and assume that
$(J^\G,\G)$ satisfies the limit property. If $(J^\F,\F)$ and $(J^\G,\G)$
have the same kneading sequences and there exists a unimodal map $f:J^\F\to J^\F$
such that the turning point of $f$ is not periodic, $\overline{\cc(f)}=J^\F$ and $f_n\to f$,
then they are topologically semi-conjugate.

\begin{example}
For each $n\geq1$, let $f_n,g_n,q_n:[0,1]\to[0,1]$ be three unimodal maps defined by
$$
f_n(x)=
\left\{
\begin{array}{ll}
(\frac{2n+4}{4+n})x  &\textrm{, if } x\in[0, \frac{4+n}{2n+4}] \\
(\frac{2n+4}{n})(1-x)&\textrm{, if } x\in[\frac{4+n}{2n+4},1]
\end{array}
\right.,
$$
$$
g_n(x)= \left\{
\begin{array}{ll}
(\frac{4n+4}{3n-1})x   &\textrm{, if } x\in[0, \frac{3n-1}{4n+4}] \\
(\frac{4n+4}{n+5})(1-x)&\textrm{, if } x\in[\frac{3n-1}{4n+4}, 1]
\end{array}
\right.,
$$
$$
q_n(x)= \left\{
\begin{array}{ll}
2(\frac{n+4}{n+5})x   &\textrm{, if } x\in[0, \frac{1}{2}] \\
2(\frac{n+4}{n+5})(1-x)&\textrm{, if } x\in[\frac{1}{2}, 1].
\end{array}
\right.
$$
Observe that $f_n, q_n$ converges uniformly to $f$ and $g_n$ converges uniformly to
$g$, where
$$
f(x)= \left\{
\begin{array}{ll}
2x   &\textrm{, if } x\in[0, \frac{1}{2}] \\
2(1-x)&\textrm{, if } x\in[\frac{1}{2}, 1]
\end{array}
\right.,
\ \ g(x)= \left\{
\begin{array}{ll}
\frac{4}{3}x   &\textrm{, if } x\in[0, \frac{3}{4}] \\
4(1-x)&\textrm{, if } x\in[\frac{3}{4}, 1].
\end{array}
\right.
$$
It is easy to see that $([0,1], (f_n))$, $([0,1], (g_n))$, and
$([0,1], (q_n))$ satisfy the limit property. From Theorem \ref{thmA},
it follows that $([0,1], (f_n))$ is topologically conjugate to $([0,1], (g_n))$.
On the other hand, $([0,1], (f_n))$ is not topologically conjugate to $([0,1], (q_n))$,
since $q_n$ is not surjective for any $n\geq1$. More generally, we can construct a
family $\mathscr{M}$ of UNDS satisfying the limit property containing
conjugated and non-conjugated UNDS. For that, let $\mathscr{U}$ be the set of
unimodal maps $f:[0,1]\to[0,1]$ so that $|f'|\geq\theta(f)>1$.
Let $\mathscr{M}:=\{([0,1], \F): f_n\to f \text{ with } f,f_n\in\mathscr{U}
\text{ and } f \text{ has no periodic turning point}\}$.
Observe that the three UNDS constructed above belong to $\mathscr{M}$.
\end{example}

We finish this section remarking that many concepts used in classical dynamics
such as periodicity, recurrence and wandering domains have generalizations
for nonautonomous systems even though none work easily, while other concepts
such as entropy can be generalized. As the notion of attracting periodic points
and wandering intervals are the main obstructions to the density of critical set
(preimages of turning points) on the autonomous setting, it is natural we ask:

\medskip
\noindent
{\bf Problem 2:} Let $(J^\F,\mathcal{F})$ be a unimodal nonautonomous
discrete dynamical systems. Under which conditions we have that
$\cc_n(\F)$ is dense in $J^\F$ for all $n\geq1$?

\medskip

\appendix

\section{The multimodal case}

In this appendix, we work with NDS where each map $f_n:J^\F\to J^\F$ is a multimodal map.
Let $(J^\mathcal{F}, \mathcal{F})$ be the NDS defined as follows:
let $J^\F=[a^\F, b^\F]$ be an interval, and
$f_n:J^\F\to J^\F$ be a  piecewise monotone continuous map such that
either $f_n(a^\F)=a^\F$ for each $n\geq1$ or $f_n(a^\F)=b^\F$ for each $n\geq1$
and either $f_n(b^\F)=a^\F$ for each $n\geq1$ or $f_n(b^\F)=b^\F$ for each $n\geq1$.
Furthermore, there exists $\ell\geq1$ so that for each $n\geq1$
there are $a^\F<c_n^\F(1)<\cdots<c_n^\F(\ell)<b^\F$
such that the intervals $I_n^\F(1)=\left[a^\F, c_n^\F(1)\right),
I_n^\F(2)=\left(c_n^\F(1), c_n^\F(2)\right), \ldots, I_n^\F(\ell+1)=\left(c_n^\F(\ell), b^\F\right]$
are the largest intervals in which $f_n$ is strictly monotone.
The points $\{c_n^\F(1), \ldots, c_n^\F(\ell):\; n\geq1\}$ are the {\it turning points} of the NDS.
We call the NDS defined above $\ell$-{\it modal nonautonomous discrete dynamical system} (short $\ell$-MNDS)

Consider the alphabet $\mathcal{A}_\F=\{I_n^\F(1), c_n^\F(1), I_n^\F(2),
\ldots, c_n^\F(\ell), I_n^\F(\ell+1):\; n\geq1\}$.

\medskip
\noindent
{\sc Address of a point:} Let $n\geq1$. The {\em address} of a point $x\in J^\F$
on the level $n$ is the letter $i_{\F,n}(x)\in\mathcal A_\F$ defined by
$$
i_{\F,n}(x)= \left\{
\begin{array}{ll}
I_n^\F(j)    & \textrm{, if }  x\in I_n^\F(j)     \\
c_n^\F(j)  & \textrm{, if }  x=c_n^\F(j).
\end{array}
\right.
$$

\medskip
\noindent
{\sc Itinerary of a point:} Let $n\geq1$. The {\em itinerary} of a point $x\in J^\F$
on the level $n$ is the sequence $I_{\mathcal{F},n}(x)\in\mathcal A_\F^{\{0,1,2,\ldots\}}$ defined by
\[
I_{\mathcal{F},n}(x)=(i_{\mathcal{F},n}(x),i_{\mathcal{F},n+1}(f_n^1(x)),\ldots,
i_{\F,n+\ell}(f_n^\ell(x)),\ldots).
\]

\medskip
\noindent
{\sc Kneading Sequences:} The {\em kneading sequences} of $(J^\F,\F)$
are the sequences $\V^\F(j)=\{\V_n^\F(j)\}_{n\geq1}$,
where $\V_n^\F(j):=I_{\F,n}(c_n^\F(j))$ and $j=1, \ldots, \ell$.

\medskip
\begin{definition}\label{defMNDS}
Let $(J^\F,\F)$ and $(J^\G,\G)$ be two $\ell$-MNDS. We say that

\begin{enumerate}[\rm (1)]
\item {\em $(J^\F,\F)$ and $(J^\G,\G)$ have the same kneading sequences} if
\[
\V^\F(j)=\V^\G(j),\ \textrm{for each } j=1, \ldots, \ell.
\]
\item $(J^\F,\F)$ is {\em monotonically equivalent}
to $(J^\G,\G)$ if for all $n\geq1$ and each $j=1, \ldots, \ell+1$:
\begin{enumerate}[$\circ$]
\item either $f_n\restriction_{I_n^\F(j)}$ and $g_n\restriction_{I_n^\G(j)}$ are
strictly increasing;
\item or $f_n\restriction_{I_n^\F(j)}$ and $g_n\restriction_{I_n^\G(j)}$
are strictly decreasing.
\end{enumerate}
\end{enumerate}
\end{definition}

The limit property for $\ell$-MNDS has a little change: if $f$ is the
$\ell$-modal map such that $f_n$ converges uniformly to $f$, then
$c^f(i)$ is not periodic and $c^f(i)\notin\{f^{-m}(c^f(j)): m\geq1\}$
for all $i,j=1,\ldots,\ell$. The other items remain the same.
With these more general definitions, we have the following.

\begin{theorem}\label{thmA2}
Let $(J^\F,\F)$ and $(J^\G,\G)$ be two monotonically equivalent $\ell$-modal
nonautonomous discrete dynamical systems, and assume that both satisfy the
limit property. Then $(J^\F,\F)$ and $(J^\G,\G)$ are topologically
conjugate if and only if they have the same kneading sequences.
\end{theorem}

The proof of Theorem \ref{thmA2} follows exactly the same ideas and arguments
as in the proof of Theorem \ref{thmA}, and we leave it to the reader.

\bibliography{NADS}
\bibliographystyle{alpha}

\end{document}

config: arxiv


TITLE: Propierties of the submersion
QUESTION [0 upvotes]: Let $p:M\rightarrow N$ a surjective submersion of manifolds. Let $Y\subset N$ I want to prove that

$Y$ is a submanifold of $N$ if and only if $p^{-1}(Y)$ is a submanifold $M$.

I could prove that if $Y$ is a submanifold then $p^{-1}(Y)$ using the regular point theorem, but I could not prove the converse.
Any hint?

REPLY [0 votes]: You can directly show that something is a manifold by exhibiting a system of charts for it with the right transition maps. Can you find a way of taking the system of charts on $p^{-1}(Y)$ and moving them over to give a system of charts on $Y$?


TITLE: What is the remainder of dividing $(116+17^{17})^{21}$ by $8$?
QUESTION [3 upvotes]: What is the remainder of dividing $(116+17^{17})^{21}$ by $8$?

How to solve this? Solving the congruence (find the value of $a$) in $$(116+17^{17})^{21}\equiv a\pmod8$$ and how is this done?

REPLY [4 votes]: As $17\equiv1\pmod 8,17^{17}\equiv1\pmod 8$ and $116\equiv4\pmod 8$
$\displaystyle\implies 116+17^{17}\equiv 4+1\pmod8\equiv5$
$\displaystyle\implies(116+17^{17})^{21}\equiv 5^{21}\pmod8 $
Now, $\displaystyle5^2=25\equiv1\pmod 8\implies 5^{21}=5\cdot(5^2)^{10}\equiv5\cdot1^{10}\pmod 8\equiv5$


text:
import pseudo_normed_group.system_of_complexes
import Lbar.pseudo_normed_group
import breen_deligne.homotopy

.

open_locale nnreal

open category_theory ProFiltPseuNormGrpWithTinv opposite

variables (r r' : ℝ≥0) [fact (0 < r)] [fact (0 < r')] [fact (r < r')] [fact (r' < 1)]
variables (BD : breen_deligne.package) (κ : ℕ → ℝ≥0)
variables [BD.data.very_suitable r r' κ] [∀ (i : ℕ), fact (0 < κ i)]

include r r' BD κ

@[nolint unused_arguments]
def first_target_stmt : Prop :=
  ∀ m : ℕ,
  ∃ (k K : ℝ≥0) (hk : fact (1 ≤ k)),
  ∃ c₀ : ℝ≥0,
  ∀ (S : Type) [fintype S],
  ∀ (V : SemiNormedGroup.{0}) [normed_with_aut r V],
    ((BD.data.system κ r V r').obj (op $ of r' (Lbar r' S))).is_weak_bounded_exact k K m c₀


TITLE: Capped Geometric Distribution
QUESTION [0 upvotes]: The Geometric Distribution is defined as the number of trials it takes for the first success to appear in a sequence of Bernoulli trials.
My question is what happens when the number of trials is capped at a number $N$? That is what would the distribution of $Y$ be if we define it as
$$Y = min(X, N) \quad X \sim Geom(p), \quad N \in \mathbb{N}$$
I found one article that references this type of problem (group of people playing a dice game, $Y$ determines the winner) but the main difference is that $Y$ gets reset to the support of $X$ and not set equal to $N$.

REPLY [1 votes]: The only difference in truncated and original geometric distribution is in "last value": $X$ can take values $1,2,\ldots$ and $Y$ can take values $1,2,\ldots, N$ only.  And $Y=N$ iff $X\geq N$, so
$$
\mathbb P(Y=N)=\mathbb P(X\geq N) = \mathbb P(N-1 \text{ unsuccesses})=(1-p)^{N-1}.
$$
All the other values of $Y$ have the same probabilities as for $X$: for $k=1,2,\ldots,N-1$,
$$
\mathbb P(Y=k)=\mathbb P(X=k) = p(1-p)^{k-1}.
$$


TITLE: Is it known if the ground state of an arbitrary electronic many-body problem has antisymmetric spatial part?
QUESTION [1 upvotes]: For non-interacting electrons, Hund's rule (#1) says that the state with the highest total spin has the lowest total energy. For two independent electrons, it seems then that the electrons should be in a triplet configuration, i.e. that the spatial part be antisymmetric.
In complete generality: Is it true that the spatial part of the ground state of interacting electrons in an external potential is anti-/symmetric? If so, can you explain why, or provide a reference?

REPLY [0 votes]: It is known that this is not a rule. One counterexample is enough, $H_2$. Molecules with an even number of electrons and nonzero electronic spin in the ground state are rare. $O_2$ is an exception. See however https://en.wikipedia.org/wiki/Single-molecule_magnet.


TITLE: nonorthogonal diagonalization and quadratic forms
QUESTION [2 upvotes]: Consider a quadratic form $q(x)=x^TAx$ where $A$ is a real $n\times n$ symmetric matrix. If we diagonalize $A$, using a nonorthogonal matrix $P$, i.e. $A=PDP^{-1}$ where $D$ is a diagonal matrix of eigenvalues, then does the quadratic equation $x^TDx+c=0$ represent the same geometric shape as $x^TAx+c=0$? In other words, does a nonorthogonal change of coordinates necessarily not alter the basic shape? So does an ellipse remains an ellipse albeit it becomes distorted? If so, my question is why?
I think this is because the coordinate transformation is invertible but I cannot formally explain how that helps.

REPLY [0 votes]: The short answer is no, the quadric’s basic shape might not be preserved.
The overall shape of the quadric (hyper)surface defined by the implicit equation $q(\mathbf x)=\mathbf x^TA\mathbf x=1$ is determined by its spectrum: the number of positive, negative and zero eigenvalues of the symmetric matrix $A$. In the plane, $++$ and $--$ correspond to ellipses, $+-$ to parabolas, $+0$ and $-0$ to parabolas, and $00$ to various degenerate conics. Note that for the nondegenerate conics, the spectrum is completely determined by the sign of $\det(A)$.
If you have a linear transformation represented by the matrix $M$, so that $\mathbf x'=M\mathbf x$, then $$\mathbf x^TA\mathbf x = (M^{-1}\mathbf x)^TA(M^{-1}\mathbf x) = \mathbf x^T(M^{-T}AM^{-1})\mathbf x.$$ What this means is that, when interpreting $A$ as the matrix of a quadratic form, a change of basis takes the form $P^TAP$, not $P^{-1}AP$ as it does when $A$ is the matrix of a linear transformation. This is the correct expression to use for a quadratic form whether or not $P$ is orthogonal. The Spectral Theorem assures us that such linear transformations don’t change the overall shape of the quadric. So, in particular, the image of an ellipse under an invertible linear transformation is another ellipse, and so on.
On the other hand, the expression that you’re asking about, $P^{-1}AP$, doesn’t represent a linear change of coordinates of the quadratic form (unless, of course, $P$ is orthogonal). Now, $\det(A)=\det(P^{-1}AP)$, so one might think that at least in the two-dimensional case the overall shape of the two conics is the same. Unfortunately, we’re not really interested in the eigenvalues of $B=P^{-1}AP$: only the symmetric part of a matrix contributes to the quadratic form, so we’re really interested in the spectrum of $B+B^T$. In general, though, there’s no systematic relationship between the eigenvalues of a matrix and its symmetric part. Indeed, it’s not too hard to construct a $2\times2$ example for which the the signs of $\det(B)$ and $\det(B+B^T)$ differ.
For example, let $$B=\begin{bmatrix}2&3+\sqrt6\\3-\sqrt6&2\end{bmatrix}.$$ Then $\det(B)=1$, but the determinant of its symmetric part is $-5$. We have $$B = \begin{bmatrix}-\sqrt3 & \sqrt3 \\ 3-\sqrt6 & 3-\sqrt6\end{bmatrix} \begin{bmatrix}2-\sqrt3&0\\0&2+\sqrt3\end{bmatrix} \begin{bmatrix}-\sqrt3 & \sqrt3 \\ 3-\sqrt6 & 3-\sqrt6\end{bmatrix}^{-1},$$ so taking $D$ to be the diagonal matrix in the middle and $P$ the matrix on the left, $\mathbf x^TD\mathbf x=1$ is the ellipse $(2-\sqrt3)x^2+(2+\sqrt3)y^2=1$, but $\mathbf x^T(PDP^{-1})\mathbf x=1$ is the hyperbola $2x^2+6xy+2y^2=1$. At the risk of belaboring the obvious, this demonstrates that if you have a diagonal matrix $D$ and invertible $P$, then the basic shapes of the curves $\mathbf x^TD\mathbf x=1$ and $\mathbf x^T(PDP^{-1})\mathbf x=1$ aren’t necessarily the same.

Actually, we can do a bit better that this in the $2\times2$ case. Consider the symmetric matrix $$Q = \begin{bmatrix}a&b\\b&d\end{bmatrix}$$ and let $$M = Q+\begin{bmatrix}0&c\\-c&0\end{bmatrix}.$$ Then $\det(Q)=ad-b^2$ and $\det(M)=ad-b^2+c^2\ge\det(Q)$. So in particular, if $\det(Q)\gt0$ then $\det(M)\gt0$, so if you end up with an ellipse via $M=PDP^{-1}$, then you must also have started with an ellipse. I don’t know if this holds in higher dimensions, but it seems an interesting problem in its own right.


TITLE: Dark standard model linked though Higgs
QUESTION [1 upvotes]: is it possible to have a dark std model (ie image of the std model) linked to the std model only through the Higgs? That would allow dark matter condensation, dark galaxies, stars, planets and biology.
Is there any astronomical evidence for (or search for) large condensed dark matter structures though gravitational lensing? "distant Light bending around nothing?"
all fanciful, but fun to think about.

REPLY [1 votes]: Yes. Actually, it is also possible to have such a Dark Standard Model that is linked to the Standard Model only through gravity.
Original theories for the nature of dark matter assumed all of the dark matter to be made of a single particle. WIMPs and Axions are the two prototype examples. Since these have not yet been found, it is perhaps no surprise that also more complex "Dark Sectors" are being looked into. A simple extension beyond a single dark matter particle for example includes a fifth force that lives in that dark sector. "Dark Photons" are one particular and currently rather popular model of that class. The problem with all these more complex models is though that the experimental constraints are minimal or non-existent, and one enters philosophical thin ice when developing such complex models that have no foundation in any experimental data.
Entirely dark galaxies are unlikely. The science of structure formation in combination with extensive gravitational lensing studies and analysis of the cosmic microwave background rule out the hypothesis that most of dark matter would be in clumps exceeding a few solar masses.
Concerning lower-mass clumps, there are also significant constraints, again from gravitational lensing. The corresponding observations typically are referred to "MACHO searches", for MAssive Compact Halo Objects. It is thus not yet ruled out completely that there could be dark stars or dark planets, but indeed it is rather unlikely, given the strong constraints from these searches.
To summarize, while it is allowed by current data that there could be dark aliens wandering around their planets, existing constraints make this quite unlikely.


TITLE: Correctness of the sweep line algorithm for line segment intersection in the plane
QUESTION [0 upvotes]: Suppose we are given a finite set $S$ of line segments in the plane and

the intersection between two segments is empty or a single point in the interior of both segments
at most two segments intersect in a single point
all endpoints of segments and intersection points have pairwise different $x$-coordinate

Let's imagine that we sweep an imaginary vertical line $\ell$ rightwards over the plane, starting from a position left of all segments. Let

$T$ be the set of segments $s\in S$ which intersect $\ell$ at a fixed time (i.e. position on the $x$-coordinate)
$y_s$ be the $y$-coordinate of the intersection of $s\in T$ with $\ell$ and define $$s<t:\Leftrightarrow y_s<y_t\;\;\;\text{for }s,t\in T\tag 1$$

I want to show that if $s,t\in S$ with $s\cap t=\left\{p\right\}$, then $s,t$ eventually become neighbors (wrt to $<$) in T.

That's what I've tried (but I think it's too complicated): By assumption (2.), there is some $\varepsilon>0$ with $$B_\varepsilon(p)\cap u=\emptyset\;\;\;\text{for all }u\in S\setminus\left\{s,t\right\}\tag 2$$ and there is some point $v_u$ with $v_u.x<p.x$ and $$v_u\in\partial B_\varepsilon(p)\cap u$$ for $u\in\left\{s,t\right\}$. We can assume that $v_s.x>v_t.x$.

Now, the idea is to show that $s,t$ are neighbors in $T$ in the whole time interval $I_x:=(v_s.x,p.x)$, i.e. there is no $u\in s$ whose orthogonal projection onto the $x$-coordinate contains some $x\in I_x$ such that $$s<u<t\text{ or }t<u<s\;\;\;\text{in }T\tag 3$$ at time $x$.

What's the easiest way to prove this?

My idea was to let $$I_y:=\begin{cases}(v_s.y,v_t.y)&\text{, if }v_s.y<v_t.y\\ (v_t.y,v_s.y)&\text{, if }v_t.y<v_s.y\end{cases}$$ and show that $I_x\times I_y\subseteq B_\varepsilon(x)$. By $(2)$ it would immediately follow there can't be any $u\in S\setminus\left\{s,t\right\}$ with $(3)$.
I've tried to prove this in the following way: Let $q\in I_x\times I_y$. Then,
\begin{equation}
\begin{split}
\left\|p-q\right\|_2^2&<|p.x-v_s.x|^2+\max\left(\left|p.y-v_s.y\right|^2,\left|p.y-v_t.y\right|^2\right)\\
&\le\left\|p-v_s\right\|_2^2+\left\|p-v_t\right\|_2^2=2\varepsilon^2
\end{split}
\end{equation}
but that estimate is to weak. So, how can we prove the statement?

REPLY [1 votes]: I think you can simplify your proof after (2). You already know that
\begin{align}
  B_{\epsilon}(p) \cap u = \emptyset \tag{2}
\end{align}
for all $u \in S \setminus \{s, t\}$. Furthermore, there must be an $x$ such that there exist points $v_s \in s \cap B_{\epsilon}(p)$ and $v_t \in t \cap B_{\epsilon}(p)$ with $v_s.x = v_t.x$. Without loss of generality let $v_s.y < v_t.y$. By (2) you know that there exists no $u \in S$ with $v_u.x = v_s.x$ and $v_s.y < v_u.y < v_t.y$ (This would imply $v_u \in B_{\epsilon}(p)$ and, thus, $B_{\epsilon}(p) \cap u \neq \emptyset$). Therefore, $s$ and $t$ must be neighbors.