TITLE: Expressing a wave on a string using Fourier series
QUESTION [1 upvotes]: Where does time function of wave on string go when expressed in the Fourier series?
A Standing wave on string of length $L,$ fixed at its ends $x=0$ and $x=L$ is: $\quad y(x, t)=A \sin (k x) \cos \left(\omega t+\phi_{0}\right) \quad$
Where: $k=\frac{n \pi}{L}$
A periodic function $f(x)$ with period $P$ is represented by the Fourier series:
$f(x)=\frac{1}{2} a_{0}+\sum_{n=1}^{\infty} a_{n} \cos \left(n \frac{2 \pi}{p} x\right)+\sum_{n=1}^{\infty} b_{n} \sin \left(n \frac{2 \pi}{p} x\right)$
Where:
$$
a_{0}=\frac{2}{p} \int_{-P / 2}^{P / 2} f(x) d x \quad a_{n}=\frac{2}{p} \int_{-P / 2}^{P / 2} f(x) \cos \left(\frac{2 \pi}{p} n x\right) d x \quad b_{n}=\frac{2}{p} \int_{-P / 2}^{P / 2} f(x) \sin \left(\frac{2 \pi}{P} n x\right) d x
$$
For question where a guitar is played and the string is put into motion by plucking it. If we want to write $y(x)$ as a sum of the basis function, $y_{n}(x)$ we write:
$$
y(x, 0)=\sum_{n=1}^{\infty} a_{n} \sin \left(k_{n} x\right) \quad \rightarrow \quad y(x, t)=\sum_{n=1}^{\infty} a_{n} \sin \left(k_{n} x\right) \cos \left(\omega_{n} t\right)
$$
[since the wave function is usually odd, so the $a_n$ function will be eliminated)
Also in the case where the wave is neither an odd or even function when we have values for $a_{0}, a_{n}, b_{n}$ (not just 0).
And the periodic function is given by $f(x)=\frac{1}{2} a_{0}+\sum_{n=1}^{\infty} a_{n} \cos \left(n \frac{2 \pi}{p} x\right)+\sum_{n=1}^{\infty} b_{n} \sin \left(n \frac{2 \pi}{p} x\right)$.
Where do we add the $\cos \left(\omega_{n} t\right)$ part?
Does the equation look like this: $y(x, t)=\frac{1}{2} a_{0} \cos \left(\omega_{n} t\right)+\sum_{n=1}^{\infty} a_{n} \cos \left(n \frac{2 \pi}{p} x\right) \cos \left(\omega_{n} t\right)+\sum_{n=1}^{\infty} b_{n} \sin \left(n \frac{2 \pi}{p} x\right) \cos \left(\omega_{n} t\right)$

REPLY [0 votes]: For each time $t$, there is a different Fourier series. The $t$-dependence is incorporated via the Fourier coefficients. For a function $y(x,t)$ that is always zero at $x = 0$ and at $x = L$, the Fourier series is
\begin{equation}
\sum_{n=1}^{\infty}b_n(t)\sin\left(\frac{2\pi}{L}x\right).
\end{equation}
There are no cosine terms because of the boundary conditions. More generally, the Fourier series would be
\begin{equation}
\frac{1}{2}a_o(t) + \sum_{n=1}^{\infty}\left(a_n(t)\cos\left(\frac{2\pi}{L}x\right) + b_n(t)\sin\left(\frac{2\pi}{L}x\right)\right).
\end{equation}
If your function of $t$ and $x$ is
\begin{equation}
y(x,t) = A\sin\left(\frac{2\pi}{L}x\right)\cos(\omega t + \phi),
\end{equation}
then your Fourier expansion with $t$-dependent Fourier coefficients is
\begin{equation}
\underbrace{A\cos(\omega t + \phi)}_{b_n(t)}\sin\left(\frac{2\pi}{L}x\right).
\end{equation}
All $a_m(t)$ and all other $b_m(t)$ ($m\neq n$) are identically 0.


TITLE: Example of a series of functions that converges for Leibniz rule but is not uniformly convergent?
QUESTION [2 upvotes]: Is there a series of functions $\sum (-1)^n u_n(x)$ that is convergent for Leibniz rule (alternating series test) in some interval but it is not uniformly convergent?
In particular the series must be (for using Leibniz rule)

positive $(u_n(x) \geq 0)$
decreasing $(u_{n+1}(x)\leq u_n(x))$
such that $\lim_{n \to \infty} u_n(x)=0$

Edit: I refer to the following definition: a series of function is uniformly convergent in a subset $S$ if
$$\lim_{n \to \infty} \sup_{x \in S} |\sum_{n \geq 0} u_n(x) -\sum_{n=0}^{N} u_n(x)| =0$$

REPLY [2 votes]: Let $$u_n(x)=\frac{x^n}{n}, \quad n\in\mathbb N.$$ Then $u_n: [0,1]\to\mathbb R$ is a decreasing sequence of non-negative functions.
Clearly
$$
\sum_{n=1}^\infty u_n(1)=\sum_{n=1}^\infty\frac{1}{n}=\infty,
$$
while
$$
\sum_{n=0}^\infty (-1)^nu_n(x)=\sum_{n=0}^\infty \frac{(-1)^nx^n}{n}=-\log(1+x),
\quad\text{for all $x\in [0,1]$.}
$$
In fact,
$$
\sum_{k=0}^n (-1)^ku_k(x)=-\log(1+x)-\frac{(-1)^{n+1}\xi^{n+1}}{n+1},
\quad \text{for some $\xi\in (0,x)$},
$$
and thus
$$
\bigg|\sum_{k=0}^n (-1)^ku_k(x)+\log(1+x)\,\bigg|\le\frac{1}{n+1},
\quad \text{for all $x\in [0,1]$},
$$
i.e., the series $\,\sum (-1)^nu_n\,$ converges uniformly in $[0,1]$.
Note. That the advantage of this example is that $[0,1]$ is compact.


TITLE: Interpreting inequalities
QUESTION [0 upvotes]: Two conditions are met for $x$:
$x \geq 4$ and $x \geq 1$
What's the final value of $x$? I believe it is $x \geq 4$, by simple logic if $x$ is said to be greater or equal than $1$ and greater or equal than $4$, then it means it is the greatest one..
However, what's the justification behind this? I cannot just put 'by looking at it' on my notebook.

REPLY [1 votes]: $x\geq4$ is a subset of $x\geq1$.
Since the condition mentioned is $x\geq4$ 'and' $x\geq1$, the intersection of the two solutions must be taken which is subset of the two.
And hence the final solution is $x\geq4$.


text:
/-
Copyright (c) 2021 Johan Commelin. All rights reserved.
Released under Apache 2.0 license as described in the file LICENSE.
Authors: Aaron Anderson, Alex J. Best, Johan Commelin, Eric Rodriguez, Ruben Van de Velde
-/

import algebra.char_p.algebra
import field_theory.finite.basic
import field_theory.galois

/-!
# Galois fields

If `p` is a prime number, and `n` a natural number,
then `galois_field p n` is defined as the splitting field of `X^(p^n) - X` over `zmod p`.
It is a finite field with `p ^ n` elements.

## Main definition

* `galois_field p n` is a field with `p ^ n` elements

## Main Results

- `galois_field.alg_equiv_galois_field`: Any finite field is isomorphic to some Galois field
- `finite_field.alg_equiv_of_card_eq`: Uniqueness of finite fields : algebra isomorphism
- `finite_field.ring_equiv_of_card_eq`: Uniqueness of finite fields : ring isomorphism

-/

noncomputable theory

open polynomial
open_locale polynomial

lemma galois_poly_separable {K : Type*} [field K] (p q : ℕ) [char_p K p] (h : p ∣ q) :
  separable (X ^ q - X : K[X]) :=
begin
  use [1, (X ^ q - X - 1)],
  rw [← char_p.cast_eq_zero_iff K[X] p] at h,
  rw [derivative_sub, derivative_pow, derivative_X, h],
  ring,
end

/-- A finite field with `p ^ n` elements.
Every field with the same cardinality is (non-canonically)
isomorphic to this field. -/
@[derive field]
def galois_field (p : ℕ) [fact p.prime] (n : ℕ) :=
splitting_field (X^(p^n) - X : (zmod p)[X])

instance : inhabited (@galois_field 2 (fact.mk nat.prime_two) 1) :=
⟨37⟩

namespace galois_field
variables (p : ℕ) [fact p.prime] (n : ℕ)

instance : algebra (zmod p) (galois_field p n) :=
splitting_field.algebra _

instance : is_splitting_field (zmod p) (galois_field p n) (X^(p^n) - X) :=
polynomial.is_splitting_field.splitting_field _

instance : char_p (galois_field p n) p :=
(algebra.char_p_iff (zmod p) (galois_field p n) p).mp (by apply_instance)

instance : fintype (galois_field p n) := by {dsimp only [galois_field],
  exact finite_dimensional.fintype_of_fintype (zmod p) (galois_field p n) }

lemma finrank {n} (h : n ≠ 0) : finite_dimensional.finrank (zmod p) (galois_field p n) = n :=
begin
  set g_poly := (X^(p^n) - X : (zmod p)[X]),
  have hp : 1 < p := (fact.out (nat.prime p)).one_lt,
  have aux : g_poly ≠ 0 := finite_field.X_pow_card_pow_sub_X_ne_zero _ h hp,
  have key : fintype.card ((g_poly).root_set (galois_field p n)) = (g_poly).nat_degree :=
    card_root_set_eq_nat_degree (galois_poly_separable p _ (dvd_pow (dvd_refl p) h))
    (splitting_field.splits g_poly),
  have nat_degree_eq : (g_poly).nat_degree = p ^ n :=
    finite_field.X_pow_card_pow_sub_X_nat_degree_eq _ h hp,
  rw nat_degree_eq at key,
  suffices : (g_poly).root_set (galois_field p n) = set.univ,
  { simp_rw [this, ←fintype.of_equiv_card (equiv.set.univ _)] at key,
    rw [@card_eq_pow_finrank (zmod p), zmod.card] at key,
    exact nat.pow_right_injective ((nat.prime.one_lt' p).out) key },
  rw set.eq_univ_iff_forall,
  suffices : ∀ x (hx : x ∈ (⊤ : subalgebra (zmod p) (galois_field p n))),
    x ∈ (X ^ p ^ n - X : (zmod p)[X]).root_set (galois_field p n),
  { simpa, },
  rw ← splitting_field.adjoin_root_set,
  simp_rw algebra.mem_adjoin_iff,
  intros x hx,
  -- We discharge the `p = 0` separately, to avoid typeclass issues on `zmod p`.
  unfreezingI { cases p, cases hp, },
  apply subring.closure_induction hx; clear_dependent x; simp_rw mem_root_set aux,
  { rintros x (⟨r, rfl⟩ | hx),
    { simp only [aeval_X_pow, aeval_X, alg_hom.map_sub],
      rw [← map_pow, zmod.pow_card_pow, sub_self], },
    { dsimp only [galois_field] at hx,
      rwa mem_root_set aux at hx, }, },
  { dsimp only [g_poly],
    rw [← coeff_zero_eq_aeval_zero'],
    simp only [coeff_X_pow, coeff_X_zero, sub_zero, ring_hom.map_eq_zero, ite_eq_right_iff,
      one_ne_zero, coeff_sub],
    intro hn,
    exact nat.not_lt_zero 1 (pow_eq_zero hn.symm ▸ hp), },
  { simp, },
  { simp only [aeval_X_pow, aeval_X, alg_hom.map_sub, add_pow_char_pow, sub_eq_zero],
    intros x y hx hy,
    rw [hx, hy], },
  { intros x hx,
    simp only [sub_eq_zero, aeval_X_pow, aeval_X, alg_hom.map_sub, sub_neg_eq_add] at *,
    rw [neg_pow, hx, char_p.neg_one_pow_char_pow],
    simp, },
  { simp only [aeval_X_pow, aeval_X, alg_hom.map_sub, mul_pow, sub_eq_zero],
    intros x y hx hy,
    rw [hx, hy], },
end

lemma card (h : n ≠ 0) : fintype.card (galois_field p n) = p ^ n :=
begin
  let b := is_noetherian.finset_basis (zmod p) (galois_field p n),
  rw [module.card_fintype b, ← finite_dimensional.finrank_eq_card_basis b, zmod.card, finrank p h],
end

theorem splits_zmod_X_pow_sub_X : splits (ring_hom.id (zmod p)) (X ^ p - X) :=
begin
  have hp : 1 < p := (fact.out (nat.prime p)).one_lt,
  have h1 : roots (X ^ p - X : (zmod p)[X]) = finset.univ.val,
  { convert finite_field.roots_X_pow_card_sub_X _,
    exact (zmod.card p).symm },
  have h2 := finite_field.X_pow_card_sub_X_nat_degree_eq (zmod p) hp,
  -- We discharge the `p = 0` separately, to avoid typeclass issues on `zmod p`.
  unfreezingI { cases p, cases hp, },
  rw [splits_iff_card_roots, h1, ←finset.card_def, finset.card_univ, h2, zmod.card],
end

/-- A Galois field with exponent 1 is equivalent to `zmod` -/
def equiv_zmod_p : galois_field p 1 ≃ₐ[zmod p] (zmod p) :=
have h : (X ^ p ^ 1 : (zmod p)[X]) = X ^ (fintype.card (zmod p)),
  by rw [pow_one, zmod.card p],
have inst : is_splitting_field (zmod p) (zmod p) (X ^ p ^ 1 - X),
  by { rw h, apply_instance },
by exactI (is_splitting_field.alg_equiv (zmod p) (X ^ (p ^ 1) - X : (zmod p)[X])).symm

variables {K : Type*} [field K] [fintype K] [algebra (zmod p) K]

theorem splits_X_pow_card_sub_X : splits (algebra_map (zmod p) K) (X ^ fintype.card K - X) :=
(finite_field.has_sub.sub.polynomial.is_splitting_field K (zmod p)).splits

lemma is_splitting_field_of_card_eq (h : fintype.card K = p ^ n) :
  is_splitting_field (zmod p) K (X ^ (p ^ n) - X) :=
h ▸ finite_field.has_sub.sub.polynomial.is_splitting_field K (zmod p)

@[priority 100]
instance {K K' : Type*} [field K] [field K'] [fintype K'] [algebra K K'] : is_galois K K' :=
begin
  obtain ⟨p, hp⟩ := char_p.exists K,
  haveI : char_p K p := hp,
  haveI : char_p K' p := char_p_of_injective_algebra_map' K K' p,
  exact is_galois.of_separable_splitting_field (galois_poly_separable p (fintype.card K')
    (let ⟨n, hp, hn⟩ := finite_field.card K' p in hn.symm ▸ dvd_pow_self p n.ne_zero)),
end

/-- Any finite field is (possibly non canonically) isomorphic to some Galois field. -/
def alg_equiv_galois_field (h : fintype.card K = p ^ n) :
  K ≃ₐ[zmod p] galois_field p n :=
by haveI := is_splitting_field_of_card_eq _ _ h; exact is_splitting_field.alg_equiv _ _

end galois_field

namespace finite_field

variables {K : Type*} [field K] [fintype K] {K' : Type*} [field K'] [fintype K']

/-- Uniqueness of finite fields:
  Any two finite fields of the same cardinality are (possibly non canonically) isomorphic-/
def alg_equiv_of_card_eq (p : ℕ) [fact p.prime] [algebra (zmod p) K] [algebra (zmod p) K']
  (hKK' : fintype.card K = fintype.card K') :
  K ≃ₐ[zmod p] K' :=
begin
  haveI : char_p K p,
  { rw ← algebra.char_p_iff (zmod p) K p, exact zmod.char_p p, },
  haveI : char_p K' p,
  { rw ← algebra.char_p_iff (zmod p) K' p, exact zmod.char_p p, },
  choose n a hK using finite_field.card K p,
  choose n' a' hK' using finite_field.card K' p,
  rw [hK,hK'] at hKK',
  have hGalK := galois_field.alg_equiv_galois_field p n hK,
  have hK'Gal := (galois_field.alg_equiv_galois_field p n' hK').symm,
  rw (nat.pow_right_injective (fact.out (nat.prime p)).one_lt hKK') at *,
  use alg_equiv.trans hGalK hK'Gal,
end

/-- Uniqueness of finite fields:
  Any two finite fields of the same cardinality are (possibly non canonically) isomorphic-/
def ring_equiv_of_card_eq (hKK' : fintype.card K = fintype.card K') : K ≃+* K' :=
begin
  choose p _char_p_K using char_p.exists K,
  choose p' _char_p'_K' using char_p.exists K',
  resetI,
  choose n hp hK using finite_field.card K p,
  choose n' hp' hK' using finite_field.card K' p',
  have hpp' : p = p', -- := eq_prime_of_eq_prime_pow
  { by_contra hne,
    have h2 := nat.coprime_pow_primes n n' hp hp' hne,
    rw [(eq.congr hK hK').mp hKK', nat.coprime_self, pow_eq_one_iff (pnat.ne_zero n')] at h2,
    exact nat.prime.ne_one hp' h2,
    all_goals {apply_instance}, },
  rw ← hpp' at *,
  haveI := fact_iff.2 hp,
  exact alg_equiv_of_card_eq p hKK',
end

end finite_field

subset_name: curated


TITLE: Rank of a Decomposable Tensor
QUESTION [3 upvotes]: I'm independently studying Stephen Roman's Advanced Linear Algebra, and I came across a line of reasoning that appears obvious but that I don't understand, and was hoping someone might help me clarify.
'$\textbf{The Rank of a Decomposable Tensor}$
If $\{u_i \; | \; i \in I\}$ is a basis for $U$ and $\{v_j \; | \; j \in J\}$ is a basis for $V$, then any decomposable vector has the form $$ u \otimes v = \sum_{i,j} r_i s_j (u_i \otimes v_j)$$ Hence, the rank of a decomposable vector is $1$, since the rank of a matrix whose $(i, j)$th entry is $r_i s_j$ is $1$.'
I do not understand how a matrix having $(i, j)$th entry $r_i s_j$ implies that it is rank $1$.

REPLY [3 votes]: The rank is one provided that $u\otimes v \neq 0$. In particular, $u, v \neq 0$.
Consider the matrix
$$A = \left[\begin{array}{cccc}
r_1s_1 & r_1s_2 & \dots &r_1s_n\\
r_2s_1 & r_2s_2 & \dots & r_2s_n\\
\vdots & \vdots & \ddots&\vdots\\
r_ms_1 & r_ms_2 & \dots & r_ms_n
\end{array}\right].$$
Note that
\begin{align*}
\operatorname{rank}(A) &= \dim\operatorname{Col}A\\
&= \dim\operatorname{span}\left\{\left[\begin{array}{c} r_1s_1\\ \vdots\\ r_ms_1\end{array}\right], \dots, \left[\begin{array}{c} r_1s_n\\ \vdots\\ r_ms_n\end{array}\right]\right\}\\
&= \dim\operatorname{span}\left\{s_1\left[\begin{array}{c} r_1\\ \vdots\\ r_m\end{array}\right], \dots, s_n\left[\begin{array}{c} r_1\\ \vdots\\ r_m\end{array}\right]\right\}\\
&= \dim\operatorname{span}\left\{\left[\begin{array}{c} r_1\\ \vdots\\ r_m\end{array}\right]\right\}\\
&= 1
\end{align*}
as $u = \sum r_iu_i$ where $u_i$ form a basis and $u \neq 0$, so $r_i \neq 0$ for some $i$.
Another way of seeing it uses the notion of outer product (see here and here). Let $v^T = [r_1\ \dots\ r_m]$ and $w^T = [s_1\ \dots\ s_n]$, then the outer product of $v$ and $w$ is
$$vw^T = \left[\begin{array}{c} r_1\\ \vdots\\ r_m\end{array}\right][s_1\ \dots\ s_n] = \left[\begin{array}{cccc}
r_1s_1 & r_1s_2 & \dots &r_1s_n\\
r_2s_1 & r_2s_2 & \dots & r_2s_n\\
\vdots & \vdots & \ddots&\vdots\\
r_ms_1 & r_ms_2 & \dots & r_ms_n
\end{array}\right] = A.$$
As $A$ can be written as the sum of one outer product and no fewer (because $A$ is not the zero matrix), $A$ has rank one.

REPLY [0 votes]: Let $r$ be the column vector with entries $r_i$ and let $s$ be the column vector with entries $s_j$. Then if $M=rs^T$, we have $M_{i,j}=r_is_j$. Clearly $M$ has rank one, since its column are all multiples of $r$.


TITLE: On a better solution for $\large {\int_{-x}^x\frac{dx}{\sqrt {1-x^2} e^{bx^2+ax}}}$
QUESTION [6 upvotes]: In

this user’s question

I was able to find out the following. It is for a bigger physics research problem having to do with comparing a Wigner function corresponding to a state of light for the user named @Jayanth Jayakumar. Although, I know almost nothing on this subject, I still find the result very interesting. In case you do not believe the results, here is graphical proof for each step. I would like to support both of us, but I am more interested in finding out a way to significantly simplify the result below:
$$\mathrm{\int_{-x}^x\frac{dx}{\sqrt {1-x^2} e^{bx^2+ax}}=\quad \frac12 \sum_{n=0}^\infty\sum_{m=0}^n\frac{b^ma^{n-m}sgn\left(x^{m+n+1}\right)\left[(-1)^m+(-1)^n\right]B_{x^2}\left(\frac{m+n+1}{2},\frac12\right)}{m!(n-m)!}=\quad \sum_{n=0}^\infty\sum_{m=0}^n\frac{b^ma^{n-m}k^{m+n+1}\left[(-1)^m+(-1)^n\right]\,_2F_1\left(\frac12,\frac{m+n+1}{2}, \frac{m+n+3}{2},k^2\right)}{m!(n-m)!(m+n+1)}=\quad \sum_{n=0}^\infty\sum_{m=0}^n\frac{b^{2m}a^{2(n-m)}sgn\left(x^{2(m+n)+1}\right)B_{x^2}\left(\frac{2(m+n)+1}{2},\frac12\right)}{(2m)!(2(n-m))!-}\quad -\sum_{n=0}^\infty\sum_{m=0}^n\frac{b^{2m+1}a^{2(n-m)}sgn\left(x^{2(m+n)+3}\right)B_{x^2}\left(\frac{2(m+n)+3}{2},\frac12\right)}{(2m+1)!(2(n-m))!}, }$$
For the full area under the function, set $x=\pm1$:
$$\mathrm{\int_{-1}^1 \frac{e^{-bx^2-ax}}{\sqrt{1-x^2}}dx=\quad\frac{\sqrt\pi}{2}\sum_{n=0}^\infty\sum_{m=0}^n\frac{b^ma^{n-m}sgn\left(x^{m+n+1}\right)[(-1)^m+(-1)^n]}{m!(n-m)!} \left(\frac{m+n-1}2\right)^{\left(-\frac12\right)},B_1\left(\frac{m+n+1}2,\frac12\right)= B\left(\frac{m+n+1}2,\frac12\right)= \frac{\sqrt\pi\left(\frac{m+n-1}2\right)!}{\left(\frac{m+n}2\right)!}= \sqrt\pi\left(\frac{m+n -1}2\right)^{\left(-\frac12\right)}}$$
This problem uses a double series, the link has a way to turn it into a single sum, Incomplete Beta function properties, hypergeomtric function values, falling factorial $x^{(y)}$ and the Wigner function. An exact solution is needed, so how can one simplify or evaluate this series? I said before, that the special case was a goal, but now the main goal is simplifying the general case or integrating another way. Please correct me and give me feedback!
Here is a single sum solution inspired by @Harry Peter and @James Arathoon. It uses the Lower Incomplete Fox-Wright function and Central Binomial Coefficient:
$$\displaystyle\begin{align}\displaystyle\int_{-t}^t\frac{dx}{\sqrt {1-t^2} e^{bt^2+at}}=\sum _{n=0}^{\infty } \frac{(2 n)! }{2^{2 n} (n!)^2}\sum _{m=0}^{\infty } \frac{a^{2 m} }{(2 m)!} b^{-\big(m+n+\frac{1}{2}\big)} \gamma\left(m+n+\frac{1}{2},bt^2\right)=\\ b^{-\frac12}\sum _{n=0}^{\infty } \binom {2n}n (4b)^{-n}\,_2Ψ^{(\gamma)}_1\left[\,^{\left(1,n+\frac12,bt^2\right),(1,1)}_{\quad \quad (1,2)}\ \frac{a^2}b\right]\end{align}$$
which is super close to this would be closed form of equation $2.27$ in this Korea Science article. If the coefficients of the factorials of $n$ were the same in the denominator and numerator, also for $m$, then we would technically have a closed form.

REPLY [3 votes]: Since $e^{-a x}=\cosh (a x)-\sinh (a x)$ and $\sinh (a x)$ is an odd function it follows that
$$\int_{-t}^t \frac{e^{-b x^2} \sinh (a x)}{\sqrt{1-x^2}} \, dx=0 $$
and
$$I(t)=\int_{-t}^t \frac{e^{-a x-b x^2}}{\sqrt{1-x^2}} \, dx=\int_{-t}^t \frac{e^{-b x^2} \cosh (a x)}{\sqrt{1-x^2}} \, dx$$
Given this
$$I(t)=\sum _{n=0}^{\infty } \frac{(2 n)! }{2^{2 n} (n!)^2}\sum _{m=0}^{\infty } \frac{a^{2 m} }{(2 m)!}\int_{-t}^t e^{-b x^2} x^{2 (m+n)} \, dx$$
and with Mathematica giving
$$\int_{-t}^t e^{-b x^2} x^{2 (m+n)} \, dx=  b^{-m-n-\frac{1}{2}} \left(\Gamma \left(m+n+\frac{1}{2}\right)-\Gamma \left(m+n+\frac{1}{2},b t^2\right)\right)$$
At the suggestion of @TymaGaidash, this can be simplified using the lower incomplete gamma function, $\gamma(s,x) = \int_0^x t^{s-1} e^{-t} dt$, as
$$\gamma\left(m+n+\frac{1}{2},bt^2\right)=\left(\Gamma \left(m+n+\frac{1}{2}\right)-\Gamma \left(m+n+\frac{1}{2},b t^2\right)\right)$$
Thus
$$\int_{-t}^t e^{-b x^2} x^{2 (m+n)} \, dx=  b^{-m-n-\frac{1}{2}} \gamma\left(m+n+\frac{1}{2},bt^2\right)$$
and finally
$$I(t)=\sum _{n=0}^{\infty } \frac{(2 n)! }{2^{2 n} (n!)^2}\sum _{m=0}^{\infty } \frac{a^{2 m} }{(2 m)!} b^{-(m+n+\frac{1}{2})} \gamma\left(m+n+\frac{1}{2},bt^2\right)$$
Update: Also maybe worth looking at
$$I(t)=\sum _{n=0}^{\infty } \frac{(2 n)!}{2^{2 n} (n!)^2} \int_{-t}^t e^{-b x^2} x^{2 n} \cosh (a x) \, dx$$
for approximations as Mathematica can calculate the integral when $n$ is fixed, for example the largest $n=0$ term for which
$$I(t)\approx \frac{\sqrt{\pi } e^{\frac{a^2}{4 b}} \left(\text{erf}\left(\frac{a+2 b t}{2 \sqrt{b}}\right)-\text{erf}\left(\frac{a-2 b t}{2 \sqrt{b}}\right)\right)}{2 \sqrt{b}}$$
where $\text{erf}(x)$ is the error function


TITLE: Monadic second order theory of "backwards tree" -- is it decidable?
QUESTION [4 upvotes]: A famous result by Rabin states that the monadic second order theory of the binary tree is decidable. By the binary tree, understand the free monoid $\{0,1\}^*$ of words, with operations $S_0(w)=w0$ and $S_1(w)=w1$, or the set $\{1,2,\dots\}=\mathbb N$ of naturals with operations $S_i(n)=2n+i$.
I would like to know if the same is true for a kind of "backwards-growing" tree: the union of all height-$s$ binary trees with the root moving away to infinity. Formally, it can be described as the set $\mathbb N$ of naturals with partially-defined operations $S_0(n)=n-2^{s-1}$ and $S_1(n)=n+2^{s-1}$ if $n=2^s(2m+1)$ (so they're defined only if $s>0$). The span of $2^s(2m+1)$ is a height-$s$ binary tree, with leaves the odd numbers, reached from numbers $\equiv2\pmod4$, reached from numbers $\equiv4\pmod8$, etc.
Certainly the theory is not the same as that of the binary tree --- even the first-order logic is different, since odd numbers have no successor. However, I suspect that, just as for the usual tree, the monadic second-order theory is decidable.
Good pointers to the literature would also be most welcome!

REPLY [3 votes]: I was kindly informed by Bruno Courcelle that the theory is decidable. Starting from vertex $1$, the vertices of the "backwards tree" are reached by prefixes of the context-free language $\bigcup_{s\ge0}S_0^{-s-1}S_1(S_0+S_1)^s$, so is the set of reachable total states of a deterministic pushdown automaton.


TITLE: Calculating winding number
QUESTION [1 upvotes]: Let
$$
\begin{align}
\gamma= \gamma_1 +\gamma_2+\gamma_3,\\
\gamma_1(t)=e^{it}, t\in[0,2\pi] \\
\gamma_2(t)=-1+2e^{-2it}, t\in [0,2\pi]\\
\gamma_3(t)=1-i+e^{it},t\in [\frac{\pi}{2},\frac{9\pi}{2}]
\end{align}
$$
Calculate the value of $n(\gamma,z)$ as $z$ takes its value in $\mathbb{C}\backslash\gamma.$
$\gamma$ is the image of the closed curve.
as $\gamma$ is a closed and smooth by parts, we have that
$$
n(\gamma,z)= \frac{1}{2\pi i}\int\limits_\gamma\frac{dc}{c-z} = \frac{1}{2\pi i} \left[ \int\limits_{\gamma_1}\frac{dc}{c-z}+ \int\limits_{\gamma_2}\frac{dc}{c-z} + \int\limits_{\gamma_3}\frac{dc}{c-z}\right]
$$
as  $\gamma_1'(t) = ie^{it}$, $\gamma_2'(t) = -4ie^{-2it}$ and $\gamma_3'(t) = ie^{it}$
i can compute those integrals as
$$
n(\gamma,z)=\frac{1}{2\pi i} \left[ \int\limits_{0}^{2\pi}\frac{ie^{it}dt}{e^{it}-z}+ \int\limits_{0}^{2\pi}\frac{-4ie^{-2it}dt}{-1+2e^{-2it}-z} + \int\limits_{\pi /2}^{9\pi /2}\frac{ie^{it}}{1-i+e^{it}-z}\right]
$$
but when i compute then for any point i am finding $0$, did i made any mistake or am i making one when computing those integrals?

REPLY [2 votes]: The value depends on $z$. Note that $\gamma_1$ runs once counterclockwise around the circle with center $0$ and radius $1$, $\gamma_2$ runs twice clockwise around the circle with center $-1$ and radius $2$, $\gamma_3$ runs twice counterclockwise around the circle with center $1-i$ and radius $1$. Let $D_i$ be the open disk bounded by $\gamma_i$. Note that $D_1 \subset D_2$. Drawing a picture is helpful. For a point $z \in \mathbb C \setminus \gamma$ we get $n(\gamma,z) = n(\gamma_1,z) + n(\gamma_2,z) + n(\gamma_3,z)$. We have $n(\gamma_1,z) = 1$ for $z \in D_1$, $n(\gamma_1,z) = 0$ for $z \notin D_1$, $n(\gamma_2,z) = -2$ for $z \in D_2$, $n(\gamma_2,z) = 0$ for $z \notin D_2$, $n(\gamma_3,z) = 2$ for $z \in D_3$, $n(\gamma_3,z) = 0$ for $z \notin D_3$. Thus:

If $z \in D_1 \cap D_3$, then $n(\gamma,z) = 1 -2 + 2 = 1$.
If $z \in D_1 \setminus D_3$, then $n(\gamma,z) = 1- 2 + 0 = -1$.
If $z \in D_2 \setminus (D_1 \cup D_3)$, then $n(\gamma,z) = 0 -2 + 0 = -2$.
If $z \in (D_2 \cap D_3) \setminus D_1 $, then $n(\gamma,z) = 0 -2 + 2 = 0$.
If $z \in D_3 \setminus D_2$, then $n(\gamma,z) = 0 + 0 + 2 = 2$.
If $z \notin D_1  \cup D_2 \cup D_3$, then $n(\gamma,z) = 0 + 0 + 0 = 0$.


TITLE: How to prove this trigonometric limit?
QUESTION [1 upvotes]: How to prove that
$$\lim _{x\to 0}\frac{\sin x^{\circ} }{x }=\frac{\pi}{180}$$
We cant separate limits in nummerator and denominator as denominator limit is zero so how do we proceed ?

REPLY [0 votes]: If you change the unit to arc degree, then it is equivalent to
$$\lim_{x\rightarrow 0}\frac{\sin x}{x}=1.$$
Then there are lots of reference and questions regarded, such as
How to prove that $\lim\limits_{x\to0}\frac{\sin x}x=1$?


TITLE: Density of primes among the first $k$ numbers formed by the digits of $\pi$?
QUESTION [3 upvotes]: Consider the digits numbers formed by the first $k$ digits in the decimal representation of $\pi, k \ge 1$
$$
3\\31\\ 314 \\3141 \\31415\\314159 \\ 3141592 \\31415926 \\314159265 \\ ...
$$
Out of the first $10^4$ such numbers $4000$ (approximately $40\%$) end in $1,3,7$ or $9$. Since all primes $> 5$ end in one of these four digits I checked how many of these $4000$ numbers were primes and I could find only corresponding to $k = 2,6,38$ which is much lower than what I anticipated.

Question: In general, assuming $0 < \alpha < 1$ to be a normal in base $10$, what is the expected density of primes among the first $k$ numbers formed by the digits of
  $\alpha$ as explained above?

REPLY [3 votes]: The Prime Number Theorem states, essentially, that the number of primes less than $N$ is approximately $\frac{N}{\ln{N}}$. That means (roughly) that the probability of $N$ itself being prime is about $\frac{1}{\ln{N}}$. Remember, though, that $N$ is the number itself, not the number of digits it has. Since $\ln{N}$ is roughly equal to the number of digits of $N$ multiplied by $\ln{10}$, and since a number's index in your sequence is just the number of digits it has, that suggests that the probability that the $k$th number is prime should be about $\frac{1}{k\ln{10}}$. Assuming that whether each number is prime is independent of the primality of the ones before it, the expected number of primes should be
$$\frac{1}{\ln{10}}\left(1 + \frac12 + \frac13 + \frac14 + \cdots + \frac1k\right)$$
By a result of Euler and Mascheroni, this happens to be almost exactly $\frac{\ln{k}}{\ln{10}} = \log_{10}k$. $\log_{10}{10000} = 4$, which is exactly what your data indicates!
This is a very cool question - I've never had a density estimate work out this perfectly!


TITLE: A non constant complex valued function with zero derivative
QUESTION [1 upvotes]: I need an example of a non constant complex valued function whose derivative is zero at every point in in the complex plane.

REPLY [3 votes]: A complex valued function on a connected set with derivative $0$ is a constant. (Refer Stein - Shakarachi Complex analysis Chapter 1 for proof.)
If the function is defined on disconnected regions then its derivative is zero need not imply function is constant.
For example take $f$ to be $1$ on $|z|<1$ and $2$ on $|z-3|<1$. Then $f$ is non-constant but $f'(z)=0$.


TITLE: If X has a dense subset, does that imply that X is closed?
QUESTION [0 upvotes]: Assume X has a dense subset Y. Then every point in X is a limit point in Y, and so the union of Y and it's closure contains X. Therefore X must be closed. Is this correct?

REPLY [1 votes]: the union of Y and its closure contains X

Sure (incidentally, the closure of a set contains the set itself already). But so what? All that means is that we have $$Y\subseteq X\subseteq cl(Y).$$ But unless we know $Y=cl(Y)$ we can't "close the gap" to argue $X=cl(Y)$ and hence be closed.
On the other hand, the above does show that a set with a closed dense subset is itself closed.


text:
\section{Fibrations in 2-categories}\label{2catsec}

The notions of fibration and two-sided discrete fibration internal to a 2-category are due to \cite{streetfibrationsyoneda}; a good summary of the main results can be found in \cite[\S 2]{weberyoneda}. In order to perform desired constructions, we work in \emph{finitely complete} 2-categories $\K$, i.e., a 2-category that admits finite conical limits and cotensors with the ``walking arrow'' category $\2$ --- though the weaker hypothesis that $\K$ admits only PIE-limits would also suffice. In particular:

\begin{lem} A finitely complete 2-category $\K$ has all comma objects.
\end{lem}
\begin{proof}
First note that
\[
\xymatrix@=10pt{ & A^\mathbb{2} \ar[dl]_c \ar[dr]^d \ar@{}[dd]|{\Leftarrow} \\ A \ar@{=}[dr] & & A \ar@{=}[dl] \\ & A}
\] is a comma object, where $d$ and $c$ are induced by the domain and codomain inclusions $\mathbb{1} \rightrightarrows \2$. To see this, recall that comma objects, like all weighted limits, are defined \emph{representably}, meaning in this case that the comma object $\comma{A}{A}$ of the depicted opspan must induce isomorphisms of categories \[ \K(X,\comma{A}{A}) \cong \comma{\K(X,A)}{\K(X,A)}\] for all $X \in \K$, where the right hand side denotes the comma category for the pair of identity functors on $\K(X,A)$. But we know that in $\Cat$, this comma category is $\K(X,A)^\2$. Hence, $\comma{A}{A}$ must induce isomorphisms of categories \[\K(X,\comma{A}{A}) \cong \K(X,A)^{\2}\]  which is the defining universal property of the cotensor of $A \in \K$ by $\2$.

Given an opspan $A \sr{g}{\ra} C \sr{f}{\la} B$ in $\K$, its comma object is the wide 2-pullback
\[
\xymatrix@=10pt{
	& A \ar[dr]^g\ar@{=}[dl]&& C^{\2} \ar[dr]^d\ar[dl]_c&& B\ar@{=}[dr]\ar[dl]_f \\
	A && C && C && B.
}
\]
with 2-cell defined by whiskering the 2-cell of the comma object $C^\2$. This can be proven directly in $\Cat$, implying the result for a generic 2-category $\K$ by the representability of weighted limits.
\end{proof}

Another proof of the previous lemma, left as an exercise to the reader, uses the pasting lemma for comma squares.

\begin{lem}\label{pastinglem} Given a diagram in a 2-category $\K$ such that the right-hand square is a comma square \[\xymatrix{ \cdot \ar[d] \ar[r] & \cdot \ar[d] \ar[r] & \cdot \ar[d] \\ \cdot \ar[r] & \cdot \ar[r] \ar@{}[ur]|{\Leftarrow} & \cdot}\] the whole diagram is a comma square if and only if the left-hand square is a 2-pullback.
\end{lem}
\begin{proof} Analogous to the pasting lemma for ordinary pullbacks.
\end{proof}

\subsection{Fibrations}

Fibrations in a 2-category are defined representably.

\begin{defn} A 1-cell $p\colon E \ra B$ in a 2-category $\K$ is a \emph{fibration} iff $\K(X,p)$ is a fibration  for all $X \in \K$ and if \[\xymatrix{ \K(X,E) \ar[d]_{\K(X,p)} \ar[r]^{\K(x,E)} & \K(Y,E) \ar[d]^{\K(Y,p)} \\ \K(X,B) \ar[r]_{\K(x,B)} & \K(Y,B)}\] is a cartesian functor for all $x \colon Y \ra X$ in $\K$.
\end{defn}

Unpacking this definition, $p \colon E \ra B$ is a \emph{fibration} if every 2-cell \[\xymatrix{ X \ar[r]^e \ar[dr]_b & E \ar[d]^p \\ \ar@{}[ur]|(.7){\Rightarrow \beta } & B}\] has a $p$-cartesian lift $\a \colon e' \Ra e$ so that $p\a = \beta $. A 2-cell
\[\xymatrix{ X \ar@/^/[rr]^{e'} \ar@{}[rr]|{\Downarrow \a} \ar@/_/[rr]_e & &E}\] is $p$-\emph{cartesian} when for all $x \colon Y \ra X$, $\a x$ is a $\K(Y,p)$-cartesian arrow in $\K(Y,E)$. This means that for all 2-cells \[\xymatrix{ Y \ar[rr]^{e''} \ar[dr]_x & \ar@{}[d]|{\Downarrow \xi}& E & Y \ar[d]_x \ar[rr]^{e''} &  \ar@{}[d]|{\Downarrow \gamma} & E \ar[d]^p \\ & X \ar[ur]_e &  & X \ar[r]_{e'} & E \ar[r]_p & B}\] such that $p\xi = p\a x \cdot \gamma$, then there is a unique 2-cell $\zeta\colon e'' \Ra e'x$ such that $\xi =  \a x \cdot \zeta$ and $p\zeta = \gamma$.

Note this definition did not require any hypotheses on the 2-category $\K$, but to prove the results that follow whose statements refer to certain finite 2-limits in $\K$, we do require something like the hypothesis of finite completeness to guarantee that these exist.

\begin{thm} In any finitely complete 2-category $\K$ \begin{enumroman} \item the composite of fibrations is a fibration \item the pullback of a fibration is a fibration \end{enumroman}
\end{thm}
\begin{proof} Follows from Theorem \ref{catstabthm}.
\end{proof}

\begin{thm} Let $\K$ be a finitely complete 2-category, $p \colon E \ra B$ a 1-cell. The following are equivalent: \begin{enumroman} \item $p$ is a fibration. \item For all $b\colon X \ra B$, the map $i\colon X \times_B E \ra \comma{b}{p}$ has a right adjoint in $\K/X$ \[\xymatrix@R=10pt@C=10pt{ X \times_B E \ar@{-->}[dr]_i \ar@/_/[dddr] \ar@/^/[drrr] \\ & \comma{b}{p} \ar[rr] \ar[dd] & & E \ar[dd]^p \\  & & &  \\ & X \ar[rr]_b \ar@{}[uurr]|{\Rightarrow}  & & B}\]

 \item The map $E \ra \comma{B}{p}$ has a right adjoint in $\K/B$. \item the canonical arrow $E^\2 \ra \comma{B}{p}$ has a right adjoint right inverse. \end{enumroman}
\end{thm}
\begin{proof}
(iii) is (ii) with $b = 1_B$. (iii) implies (ii) by the pasting lemma \ref{pastinglem}. Equivalence with (i) requires some cleverness; see \cite[2.7]{weberyoneda}. (i) $\Leftrightarrow$ (iv) is analogous to Theorem \ref{catadjthm}; see \cite[9]{streetfibrationsyoneda}.
\end{proof}

An \emph{opfibration} in $\K$ is a fibration in $\K^{\co}$. It follows from characterization (iii) above that any 2-functor between finitely complete 2-categories that preserves comma objects preserves fibrations and opfibrations. We briefly mention the very simplest examples.

\begin{ex} \quad
\begin{enumroman}
\item The fibrations internal to the 2-category $\Cat$ are exactly the fibrations introduced in Definition \ref{defn:1-fib}.
\item A fibration internal to the 2-category $\Cat_{/A}$ is a functor $p\colon E \ra B$ such that arrows $b \ra pe$ in the fiber over an identity in $A$ have $p$-cartesian lifts. If the functors $E \ra A$ and $B \ra A$ are fibrations in $\Cat$ and $p$ preserves cartesian arrows, then $p$ is a fibration in $\Cat$ if and only if it is a fibration in $\Cat_{/A}$. In general, the notion of fibration in $\Cat_{/A}$ is weaker than the notion of fibration in $\Cat$.
\end{enumroman}
\end{ex}

Discrete fibrations in a 2-category $\K$ with cotensors by $\2$ can either be defined representably or in analogy with Definition \ref{internaldefn} and these definitions are equivalent.

\subsection{Two-sided discrete fibrations} In a 2-category $\K$, we write $\Span(\K)$ for the bicategory of spans in $\K$, whose objects and 1-cells coincide with those in the 1-category underlying $\K$, and whose 2-cells are defined fiberwise. If $\K$ has binary products, the hom-categories $\Span(\K)(A,B)$ are isomorphic to the comma categories $\K_{/A \times B}$; hence, they are actually 2-categories.

\begin{defn} A span $A \sr{q}{\la} E \sr{p}{\ra} B$ is a \emph{two-sided discrete fibration} if and only if it is representably so, i.e., if for all $X \in \K$, \[ \xymatrix{ \K(X,A) & \K(X,E) \ar[l]_-{\K(X,q)} \ar[r]^-{\K(X,p)} & \K(X,B)}\] is a two-sided discrete fibration.
\end{defn}

As in $\Cat$, comma objects provide examples of two-sided discrete fibrations.

\begin{thm} Given $f \colon A \ra C$ and $g \colon B \ra C$, the span $A \la \comma{f}{g} \ra B$ is a two-sided discrete fibration.
\end{thm}
\begin{proof} Because weighted limits are also defined representably, it suffices to prove when $\K= \Cat$. See Theorem \ref{catcommathm}.
\end{proof}

\begin{thm} If $A \sr{q}{\la} E \sr{p}{\ra} B$ is a \emph{two-sided discrete fibration}, then $p$ is a fibration and $q$ is an opfibration.
\end{thm}
\begin{proof} The proof is technical, but  at least the two parts are dual, by interpreting the two-sided discrete fibration in  $\K^{\co}$.
\end{proof}
\begin{rmk}\label{fibinbic}
In his original paper, Street defines fibrations, opfibrations, and two-sided discrete fibrations to be pseudo-algebras for certain 2-monads on the appropriate hom-2-category of $\Span(K)$. For instance, the 2-monad on $\Span(K)(A,B)$ for two-sided discrete fibrations sends a span $A \sr{q}{\la} E \sr{p}{\ra} B$ to the 2-pullback of
\[ \xymatrix@=10pt{ & A^\2 \ar[dl]_{d} \ar[dr]^c & & E \ar[dl]_q \ar[dr]^p & & B^\2 \ar[dl]_d \ar[dr]^{c} \\ A & & A & & B & & B}\]
See \cite{streetfibrationsyoneda} for details.
\end{rmk}
\subsection{Yoneda lemma} Part of the motivation for defining two-sided discrete fibrations internally to a 2-category was to state and prove a Yoneda lemma in this context. While this is peripheral to our discussion, we nonetheless take a brief detour to give the statement.

\begin{thm} Let $\K$ be finitely complete 2-category, $A \sr{q}{\la} E \sr{p}{\ra} B$ a two-sided discrete fibration, and $f\colon A \ra B$ a 1-cell. The identity 2-cell $\id_f$ induces a canonical arrow $i \colon A \ra \comma{B}{f}$ from the 2-pullback of $f$ along the identity at $B$ to the comma object. Precomposition with $i$ induces a bijection between arrows of spans $\comma{B}{f} \ra E$ and arrows of spans $B \ra E$.
\end{thm}
\begin{proof}[Proof sketch] \eadd{Rather than reproduce the full proof from \cite[16]{streetfibrationsyoneda} or \cite[2.12]{weberyoneda} here, we sketch the main ideas to outline a nice exercise in categorical yoga.}

Unwinding the definition, the 1-cell $i$ is induced by the 2-universal property of the comma object:
\[
\begin{tikzcd}[sep=1.25em]
& A \arrow[ddr, bend left, "f"] \arrow[ddl, bend right, equals] \arrow[d, dashed, "i"] \\ & \comma{B}{f} \arrow[dr, "d"] \arrow[dl, "c"'] \arrow[dd, phantom, "\scriptstyle\Leftarrow"] \\ A \arrow[dr, "f"'] & & B \arrow[dl, equals] \\ & B
\end{tikzcd}
\qquad = \qquad \begin{tikzcd}[sep=1.25em] & A \arrow[dr, "f"] \arrow[dl, equals] \arrow[dd, phantom, "\scriptstyle=\ \id_f"] \\ A \arrow[dr, "f"'] & & B \arrow[dl, equals] \\ & B
 \end{tikzcd}
 \]
The Yoneda lemma asserts that restriction along $i$ induces a bijection between the maps of spans:
	\[
\left\{
\vcenter{
\xymatrix@R=3mm@C=3mm{
& A \ar@{.>}[dd]\ar@{=}[dl]\ar[dr]^f & \\
A && B \\
& E\ar[ur]_p\ar[ul]^q &
}}
\right\}
\leftrightarrows
\left\{
\vcenter{
\xymatrix@R=3mm@C=3mm{
& \comma{B}{f} \ar@{.>}[dd]\ar[dl]_c\ar[dr]^d & \\
A && B \\
& E\ar[ur]_p\ar[ul]^q &
}}
\right\}
	\]

 The classical Yoneda lemma is what we get from this theorem when $\K=\Cat$ and $A=1$ is the terminal category; the other leg $p\colon E \to B$ is then a discrete fibration. In this case,  the above correspondence reduces to
	\[
	\left\{
\vcenter{
\xymatrix@R=4mm@C=4mm{
1 \ar[dr]_b \ar[rr] && E \ar[dl]^p\\
&B &
}}\right\}
\qquad
\leftrightarrows
\qquad
\left\{
\vcenter{
\xymatrix@R=4mm@C=4mm{
\comma{B}{b} \ar[dr]_d\ar[rr] && E \ar[dl]^p\\
&B  &
}}\right\}
	\]
	Note that the left-hand side is isomorphic to the fiber $E_b$. Without loss of generality $p\colon E \to B$ can be thought of $\Sigma\colon\comma{*}{F} \to B$ for some $F \colon B^\op\to \Set$. Now via Theorem \ref{thm:discrete-equiv},  the Yoneda lemma asserts that there is a bijection between functors $\comma{B}{b} \to E$ over $B$ and elements of the set $Fb$.
\end{proof}

config: arxiv


TITLE: Concern about fair coin toss probability
QUESTION [0 upvotes]: Suppose we're not sure if we want to go out or not.
We decide to test our luck and we start tossing a fair coin.
We set this ground rules :
If in the string composed by H's and T's TT appears before HT we decide to go, else we stay home.
What is the probability that we're going out ?
I really thought hard but with some friends we settled that the probability is $\frac{1}{4}$, by thinking that every coin toss i independent , thus the probability of going out is P(T)*P(T).
But this doens't feel right to me because of two reasons:

nothing assures me that the first coin toss is an H.
even if the first toss would be an H then no matter how long we wait eventually a T will come up.
So it seems "more frequent" HT than TT.

Anyone can explain me what kind of argument is right for this situation ?

REPLY [1 votes]: If you roll a $\sf T$ first then either you roll a $\sf T$ second (and have $\sf TT$) or you roll a $\sf H$ second.   Now whether you roll that, or a $\sf H$ first, this will be followed by an indefinite sequence of $\sf H$ until you roll a $\sf T$.
Thus you will roll $\sf\ldots TT$ before $\sf\ldots HT$ only if you roll tails on the first two rolls, which occurs with a probability of $1/4$.


TITLE: Strong deformation retraction $\mathbb{I}\!\times\!\mathbb{B}^l \longrightarrow \{1\}\!\times\!\mathbb{B}^l\cup \mathbb{I}\!\times\!\{0\}$?
QUESTION [1 upvotes]: In Shastri's Elements of Differential Topology, p. 225, there is written:

I don't understand this map $R$. Why is $\theta\!\in\!\mathbb{S}^{l-1}$, shouldn't we have $\theta\!\in\!\mathbb{B}^l$? Furthermore, how do I construct such a map myself? I see that the idea is for each $(s,x)\!\in\!\mathbb{I}\!\times\!\mathbb{B}^l$ to construct a ray starting in $(0,\frac{x}{\|x\|})\!\in\!\{0\}\!\times\!\mathbb{S}^{l-1}$ and going through $(s,x)$, i.e. the ray
$$\mathbb{I}\!\times\!\mathbb{B}^l\!\times\![0,\infty)\!\rightarrow\!\mathbb{I}\!\times\!\mathbb{B}^l,\;\; (s,x,t)\mapsto
\begin{cases}
(0,\frac{x}{\|x\|})+t\big((s,x)-(0,\frac{x}{\|x\|})\big)& x\!\neq\!0\\
(s,x) & x\!=\!0
\end{cases},$$
and then for each $(s,x)$ to choose the right $t_0\!\in\![0,\infty)$ where that ray hits $\{1\}\!\times\!\mathbb{B}^l\cup \mathbb{I}\!\times\!\{0\}$, divide that ray by $t_0$ so that the ray becomes a line parametrized by $\mathbb{I}$. I do not see a way to do this.

REPLY [0 votes]: I've figured it out. Turns out I misunderstood what the text was suggesting.
Claim: $\mathbb{B}^l\!\times\!\mathbb{I}$ strongly deformation retracts onto $\mathbb{B}^l\!\times\!\{1\} \cup \{0\}\!\times\!\mathbb{I}$, and $\mathbb{B}^l\!\times\!\{0\}$ goes to $(0,0)$.
Proof: Each element of $\mathbb{B}^k$ is written in polar coordinates as $(\theta,r)\in\mathbb{S}^{k-1}\!\!\times\!\mathbb{I}$, since $\frac{\mathbb{S}^{k-1}\!\times\mathbb{I}}{\mathbb{S}^{k-1}\!\times\{0\}}\approx\mathbb{B}^k$. The strong deformation retraction $h$ is obtained by not changing $\theta$ (i.e. $h$ is the same on each slice $\{(\theta,r,s)\!\in\!\mathbb{S}^{k-1}\!\!\times\!\mathbb{I}\!\times\!\mathbb{I};\,\theta\!=\!\text{const.}\}\!\approx\!\mathbb{I}\!\times\!\mathbb{I}$), on the lower half of the cylinder, $\mathbb{B}^l\!\times\![0,\frac{1}{2}]$, the outer line $\{(\theta_0,1)\} \!\times\! [0,\frac{1}{2}]$ is pushed to the middle line $\{(\theta_0,0)\} \!\times\! [0,1]$, on the upper half of the cylinder, $\mathbb{B}^l\!\times\![\frac{1}{2},1]$, the outer line $[\frac{1}{2},1] \!\times\!\{(\theta_0,1)\}$ is pushed to the top radial line$\{\theta_0\}\!\times\mathbb{I}\times\!\{1\}$, where the point $(1,\theta_0,1)$ does not move. Let us write $(r,s)$ instead of $(\theta_0,r,s)$. For $0\!\leq\!s\!\leq\!\frac{1}{2}$ the point $(1,s)$ therefore travels to $(0,2s)$, i.e. we have a straight path $t \mapsto (1,s) + t\big((0,2s)\!-\! (1,s)\big) \!=\!\big(1\!-\!t, s(1\!+\!t)\big)$. For $\frac{1}{2}\!\leq\!s\!\leq\!1$ the point $(1,s)$ travels to $(2(s\!-\!\frac{1}{2}),1)$, i.e. we have a path $t \mapsto (1,s) +t\big((2s\!-\!1,1) \!-\!(1,s)\big) \!=\!\big(1\!+\!2t(s\!-\!1),s\!+\!t(1\!-\!s)\big)$. Let us construct the homotopy for the remaining points as well (those from the whole square $\mathbb{I}\!\times\!\mathbb{I}$). The interval $0\!\leq\!s\!\leq\!\frac{1}{2}\!+\!\frac{1-r}{2} \!=\!\frac{2-r}{2}$ will be mapped at the end of the homotopy to $0\!\leq\!s\!\leq\!1$, hence each point $(r,s)$ travels to $(0,\frac{2}{2-r}s)$, i.e. we have a path $t \mapsto (r,s) +t\big((0,\frac{2}{2-r}s)\!-\! (r,s)\big) \!=\!\big(r\!-\!rt,s\!+\!t(\frac{2}{2-r}s\!-\!s)\big) \!=\!\big(r(1\!-\!t),s\!+\!st\frac{r}{2-r}\big)$. Now we must deal with the remaining points $(r,s)$  from the triangle  $\frac{2-r}{2}\!\leq\!s\!\leq\!1$. The line  through $(0,1)$ and $(1,\frac{1}{2})$ is $y\!-\!1\!=\!\frac{1/2-1}{1-0}(x\!-\!0)$, i.e. $y\!=\!-\frac{1}{2}x\!+\!1$. The line through $(1,1)$ in $(r,s)$ is $y\!-\!1\!=\!\frac{s-1}{r-1}(x\!-\!1)$. These two lines intersect in $(\frac{2s-2}{r+2s-3},\frac{r+s-2}{r+2s-3})$. The ratio of the distances is $\|-(1,1)\!+\!(r,s)\| \,/\, \|-(1,1)\!+\!(\frac{2s-2}{r+2s-3}, \frac{r+s-2}{r+2s-3})\|\!=\!\ldots\!=\!|r\!+\!2s\!-\!3|\!=\!3\!-\!r\!-\!2s\in[0,1]$. Therefore the point $(r,s)$ is mapped to $\big(1\!-\!(3\!-\!r\!-\!2s),1\big)\!=\! (r\!+\!2s\!-\!2,1)$, i.e. we have a path $t\mapsto (r,s)\!+\!t\big( (r\!+\!2s\!-\!2,1) \!-\!(r,s)\big) \!=\!\big(r\!+\!2t(s\!-\!1),s\!+\!t(1\!-\!s)\big)$. The desired strong deformation retraction is therefore
$$\;\;\;\;\;\;\;h\!:\;\;\mathbb{B}^l\!\times\!\mathbb{I}\! \times\!\mathbb{I}\!\longrightarrow\!\mathbb{B}^l\!\times\!\mathbb{I},\;\;\; (\theta,r,s,t)\mapsto
\begin{cases}
   \displaystyle \big(\theta,r(1\!-\!t),s\!+\!st{\textstyle\frac{r}{2-r}})\big)
   & \hspace{-8pt} ;\;\; \displaystyle r\!+\!2s\!\leq\!2,\\
   \displaystyle \big(\theta,r\!+\!2t(s\!-\!1),s\!+\!t(1\!-\!s)\big)
   &\hspace{-8pt} ;\;\; \displaystyle r\!+\!2s\!\geq\!2. \;\;\;\;\;\;\;\;\;\blacksquare
\end{cases}$$


text:
\section{Introduction}
\par Mathematical modelling to understand the development of tumour cells and their dynamics is of great importance as it in turn helps in devising appropriate treatment methods. In this study, we introduce fractional time derivatives in a tumour growth model with mechanical coupling. The fractional derivatives have the role of accounting for anomalous diffusion, more precisely subdiffusion, seen in tumour growth.

The tumour microenvironment has a strong influence on tumour cell proliferation and migration \cite{Balkwill12,wang17,yuan16}. Depending on the environment of the surrounding host tissue, tumour not only migrates using typical Fickian diffusion, but it also migrates more generally using subdiffusion, superdiffusion, and even ballistic diffusion. Haptotaxis and chemotaxis, which are initiated by extracellular matrix and nutrient supply respectively, and cell-cell adhesion all drastically affect a tumour's diffusion mode when a tumour invades its surrounding host tissue and proliferates. In particular, experimental results by \cite{jiang2014anomalous} both from \textsl{in vitro} and \textsl{in vivo} show evidence of anomalous diffusion in cancer growth. Taking the average radius of the tumour to be an indicator of the root-mean-squared displacement of the cells, they observed anomalous diffusion in \textsl{in vitro} experiments of growing cultured cells from the breast line, and in the clinical data from patients with adrenal tumour and liver tumour.

Anomalous diffusion is a diffusion process with a nonlinear relation between mean squared displacement and time, unlike the normal diffusion process where the relation is linear. In the microscopic setting, the diffusion processes are presented by the continuous time random walk (CTRW) model \cite{tahir1983correlated}, wherein the particle jumps in random directions, and the waiting time before the next jump and jump lengths are given by random processes. We have the following three relevant examples of CTRW. When the mean of the probability density function (PDF) of the waiting time (first moment) and the variance of the PDF of the jump length (second moment) are finite, in the long-time limit we have a behaviour described by an integer-order diffusion equation. In this case, the solution for a point initial condition is a Gaussian PDF, and the mean square displacement (MSD) has a linear dependence on time. A PDF of the waiting time $\sim t^{-1-\alpha}$ as $t\rightarrow \infty$ with $0<\alpha<1$, results, in the continuum limit, in a time fractional diffusion equation, represented by a power-law dependence of MSD on time of the form $<x^2(t)>\sim t^{\alpha}$ leading to subdiffusive behaviour. A PDF of jump length $\sim |\vec{x}|^{-1-\beta}$ as $|\vec{x}|\rightarrow\infty$ with $0<\beta<2$ gives us in the long-time limit a behaviour described by fractional diffusion equations in space, leading to superdiffusive behaviour. Fractional differential equations were obtained from the CTRW formulation in \cite{compte1996stochastic} and \cite{metzler2000random}. The description of reactions that take place in systems with anomalous diffusion is discussed in \cite{henry2006anomalous,seki2003recombination, yuste2004reaction,nepomnyashchy2016mathematical}. These examples of CTRW discussed above are adapted to cancer modelling by the migration proliferation dichotomy observed in the development of cancer cells in \cite{iomin2005superdiffusion,iomin2005fractional,iomin2007fractional,fedotov2007migration}. We consider a subdiffusion limited reaction equation for the density of tumour cells, in contrast to the normal reaction diffusion for nutrients and chemotherapeutic density, to take into account the memory effects of cells. This involves the introduction of the Riemann--Lioville fractional derivative, which has the memory kernel in its definition, in the flux and reaction terms in the equation concerning tumour density, as seen in \cite{iomin2015continuous}, resulting in a multi-order system of fractional differential equations. The model can be modified to the one with Caputo fractional derivative assuming sufficient regularity as seen in \cite{yuste2004reaction}.

\par It is important to incorporate mechanical effects in tumour growth model since the growth of the tumour increases mechanical stress due to the surrounding host tissues, which in turn impede the further growth of the tumour. Experimental evidence can be seen in \cite{helmlinger1997solid}, where multi-cellular spheroids were grown in agar gel concentrations ranging from 0\% to 1\% and increasing the agar concentration resulted in the inhibited expansion of the spheroid as the substrate stiffness increased. In the literature, reaction-diffusion models with mechanical coupling are seen in \cite{lima2016selection,lima2017selection,faghihi2020coupled,hormuth2018mechanically} for modelling tumour growth. In our model, we incorporate the mechanical effects in a similar way to the aforementioned papers.

\par After having introduced the mathematical model, we proceed with analysing existence and uniqueness of a weak solution. We remark that, while the mathematical analysis of Cahn--Hilliard equations with mechanical effects is well addressed in the literature see, e.g. \cite{miranville2001long,carrive1999cahn, garcke2003cahn, garcke2005mechanical, garcke2019phase}, the analysis of reaction-diffusion equations with mechanical coupling is not straightforward. The traditional Caputo derivative, which is valid for absolutely continuous functions, is extended to a wider class of functions through various generalisations in the study of weak solutions to fractional differential equations. For instance, some generalisations of the Caputo derivative in the literature are given in \cite{kilbas2006theory,allen2016parabolic,gorenflo2015time,li2018generalized,akilandeeswari2017solvability}, and they all reduce to the traditional one under the assumption of sufficient regularity of the function. In the analysis, we use the one in \cite{kilbas2006theory,diethelm2010analysis}, which relies on Riemann--Liouville derivatives and is, in contrast to the traditional one, also valid for some functions that do not necessarily have the first derivative. Using Galerkin methods for showing the existence of weak solution to partial fractional differential equations is quite popular and it is seen for instance in \cite{djilali2018galerkin,ouedjedi2019galerkin,zacher2009weak,zacher2019time,mclean2019well,lean2020regularity,li2018some,manimaran2019numerical} and for tumour growth models in \cite{garcke2017well,fritz2018unsteady,fritz2019local}. The key variations from that of integer-order in the analysis are the fractional Gronwall Lemma \cite{lean2020regularity}, some special estimates due to the lack of chain rule for fractional derivatives \cite{vergara2008lyapunov}, and compactness theorems similar to the Aubin--Lions theorem  \cite{li2018some}. The multi-order ordinary fractional differential system is well addressed \cite{diethelm2010analysis}, however there is not much in the literature on the multi-order partial differential system.

\par The main novelties of our work can be summarised as follows: (a) we consider a nonlinear reaction-diffusion system with a fractional time derivative, capable of modelling subdiffusion in tumour progression, and we illustrate, by numerical simulations, its flexibility in describing the tumour dynamics by varying the fractional exponent; (b) we provide a rigorous mathematical analysis of the existence and uniqueness of a weak solution to this model, the original aspects consisting in the treatment of the fractional time derivative and of the mechanical coupling.

\par This exposition has the following structure:
Section \ref{sec:mathematical_modelling} gives the mathematical modelling of the tumour growth with mechanical effects and fractional derivative. In Section \ref{sec:preliminaries}, we introduce the notations and preliminary results needed in the later sections. The mathematical analysis of the model giving existence and uniqueness of the weak solution using Galerkin methods is worked out in Section \ref{sec:mathematical_analysis}. The numerical discretisation of the model using finite element method for space and finite differences in time with a convolution quadrature formula for the fractional derivative is given in Section \ref{sec:numerical_discretization}. The results of the numerical experiments in Section \ref{sec:numerical_simulation} show the effects of the fractional derivative and the mechanical coupling in the model.


text:
\section{Distances -- Upper bound}
\label{upProofSec}

For the lower bound on the distances in Section \ref{lowBound}, it was sufficient to control the deviation of a single walker. Establishing the upper bound is substantially more involved, as we need to understand the joint movements of the left and the right walker.  To lighten notation, we omit the torus size $N$ in the quantities $\Xln$, $\Xrn$, $\Fln$ and $\Frn$.

\ssec{Proof for $\de < 1$}
	\label{heavUpSec}
	We begin by discussing the heavy-tailed setting, as the argument is particularly short. The reason herefore lies in the rapid growth of the fitnesses following from recursion \eqref{hDiffEq}. In particular, the upper bound $N$ for the fitnesses becomes absorbing: after reaching it, it remains there for a long period of time.

	\bel[Absorbing upper fitness bound]
		\label{absorbLem}For $\de<1$,
		$$\lim_{N \to \i}\P\bi(\vp(\Fl_h) \ge N > \vp( \Fl_{h + 1}) \text{ for some $h \le N$}\bi) = 0.$$
	\enl
	\bep
		Although the recursion leading to the bound \eqref{hDiffEq} refers to the limiting model where the torus is replaced by the integers, we obtain a finite-volume version by the same arguments together with an additional truncation:
		\begin{equation}\label{eqFlRestricted}
		\Fl_{h + 1} = \bi((\vp(\Fl_h) \wedge N)F_{h + 1}\bi)^{1/\de}.
		\end{equation}
		Thus,
		$$\P\bi(\vp(\Fl_h) \ge  N  > \vp(\Fl_{h + 1})\bi) \le \P(F_{h + 1} <  N^{\de - 1}) = \exp\big(- N^{1 - \de}\big),$$
		so that invoking the union bound over $h \le N$ concludes the proof.
	\enp

	Equipped with this auxiliary result, we now establish the tightness asserted in Theorem \ref{distThm}.
	\bep[Proof of Theorem \ref{distThm}; upper bound; $\de < 1$]
		First, the walkers coalesce with certainty once the scopes reach $N$. Hence, by Lemma \ref{absorbLem}, it suffices to show that
		$$\lim_{K \to \i}\ls_{N \to \i}\P\Bi( \sup_{h \le \ll(N) + K}\vp(\Fl_h) < N\Bi) = 0.$$
		Now, as long as $\vp(\Fl_h) \le N$, the fitness in the torus model coincides with the one in the infinite limit. Next, by Lemma \ref{heavTightLem}, the fitnesses in the limit model grow as $\mu^{Z\de^{-h}}$ for a positive random variable $Z$. We conclude by noting that
		$$ \lim_{K \to \i}\sup_{N \ge 1}\P\Bi(\mu^{Z\de^{-(\ll(N) + K)}} < N\Bi) = \lim_{K \to \i} \P\bi(Z\de^{- K} < 1\bi)  = 0.$$
	\enp

\ssec{Proof for $\de = 1$}
	\label{critUpSec}
To prove the upper bound for $\de = 1$, we first show that, with high probability, the walker's fitness is close to $N$ after at most
$$\t_0 := \t_0(N) := \lm(N) + \lm(N)^{7/8}$$
steps.

\bel[Lower bound on fitnesses]
	\label{t0Prop}
	For $\de=1$,
	$$\lim_{N \to \i}\P\Bi(\inf\bi\{\Fl_h:\,{ \t_0(N) \le h \le 2\lm(N)}\bi\} \ge N\exp(-\lm(N)^{3/4})\Bi) = 1.$$
\enl

The second ingredient is a specifically constructed coupling between $\bi\{\xff\bi\}_{h \ge 0}$ and independent walkers $\bi\{\xfii\bi\}_{h \ge 0}$, the latter moving w.r.t.\ two independent copies of the fitnesses $(F_{i,h})_{i,h}$.
We write
$$\Esu_h := \bi\{\xff = \xfii\bi\},$$
and
$$\Efa_h := \bi\{\xff \ne \xfii\bi\}$$
for the events that the coupling succeeds, respectively fails at level $h$. Moreover, let
$$\FFco_h := \s\bi(\Xl_i,  \Xr_i, \Xli_i, \Xri_i, \Fl_i, \Fr_i,\Fli_i, \Fri_i\bi)_{i \le h}$$
denote the $\s$-algebra of information on the coupled walkers up to level $h$.
Note that $\Fl_h = \Fr_h$ if coalescence occurs at level $h$, whereas $\Fli_h \ne \Fri_h$ almost surely,  by absolute continuity of the fitnesses.  Therefore,
$\{\Xl_h = \Xr_h\} \subset  \Efa_h.$
The crux of the coupling is that, whenever it fails, the walkers coalesce with probability at least $1/4$.

\bel[Coupling with independent walkers]
	\label{coupProp}
	There is a coupling between the true walkers $\bi\{\xff\bi\}_{h \ge 0}$ and independent walkers $\bi\{\xfii\bi\}_{h \ge 0}$ such that almost surely on the event $\Esu_h$,
	$$\P(\Xl_{h + 1} = \Xr_{h + 1}\,|\, \FFco_h)  \ge \tfrac14 \P(\Efa_{h + 1},|\, \FFco_h)  .$$
\enl

Finally, we show that for the independent walkers starting from fitnesses at least as large as $N \exp(-\lm(N)^{3/4})$,  with high probability $\Fl_h = \Fr_h = N$ for some $h \le \lm(N)^{7/8}$.

\bel[Absence of long excursions]
	\label{noExcProp}
	For $\delta=1$ and every $\e > 0$ there exists $\Ne = \Ne(\e)$ such that if $N \ge \Ne$, then
	$$\P\bi(\Fli_h = \Fri_h = N \text{ for some $h \le \lm(N)^{7/8}$}\,|\, \Fri_0, \Fli_0\bi) \ge 1 - \e$$
	holds almost surely on the event
	$$\Ek = \Ek(N) = \bi\{\min\bi\{\Fri_0 ,  \Fli_0 \bi\} \ge N \exp(-\lm(N)^{3/4})\bi\}.$$
\enl

Before establishing Lemmas \ref{t0Prop}, \ref{coupProp} and \ref{noExcProp}, we complete the proof of Theorem \ref{distThm}.
\bep[Proof of Theorem \ref{distThm}, upper bound; $\de = 1$]
	 By Lemma \ref{t0Prop}, we may assume that $\min\{\Fl_h, \Fr_h\} \ge N\exp(-\lm(N)^{3/4})$ for all $\t_0 \le h \le 2\lm(N)$. Now, if the coupling fails without coalescence at time $h \ge \t_0$, then we restart it by initializing $(\Xli_h, \Xri_h)$ at time $h + 1$ with the values of the true system $(\Xl_h, \Xr_h)$. Proceeding recursively, this produces stopping times $\{\te_i\}_{i \ge 1}$ encoding the sequence of coupling failures, where we impose that $\te_0 = \t_0$. To ease notation, we let the sequence $\te_i$ be constant after the index $i_0$ where $\te_{i_0} = H_N$, i.e., where coalescence occurs.

	In particular,
	$$
		\P\bi(H_N \ge \lm(N) + \Ke\lm(N)^{7/8}\bi)
		\le \P(\te_{\Ke} \ne H_N) +  \sum_{i \le \Ke}   \P\bi(\te_i - \te_{i - 1}\ge  \lm(N)^{7/8}\bi).
		$$
Now, fixing $\e > 0$ and $\Ke$ with $4^{-\Ke} \le \e$, applying Lemmas \ref{coupProp} and \ref{noExcProp} concludes the proof.
\enp

Next, we prove Lemma \ref{t0Prop}.
\bep[Proof of Lemma \ref{t0Prop}]
	First,
	$$\P\bi(F_h \le  1/(2\log\log(N))\bi) = \exp\bi(-2\log\log(N)\bi) = \log(N)^{-2},$$
	so that we may assume $F_h \ge 1/(2\log\log(N))$ for all $h \le 2\lm(N)$.

	By Lemma \ref{critTightLem}, the event  $\bi\{\sup_{h \le \t_0}\vp(\Fl_h) \ge N\bi\}$ occurs with high probability. Hence, it remains to show that
	$$\lim_{N \to \i}\P\Bi(\inf_{h, h'\le 2\lm(N)} (\Fl_{h + h'} / \Fl_h) \le \exp(-\lm(N)^{3/4}/2)\Bi) = 0.$$
	To that end, we fix $h, h' \le 2\lm(N)$ and apply the union bound afterwards.
	First, by \eqref{cDiffEq} in companion with \eqref{eqFlRestricted},
	$$\P\Bi(\Fl_{h + h'} / \Fl_h \le \exp(-\lm(N)^{3/4}/2)\Bi) \le \P\Bi(\prod_{h < i \le h + h'}(2F_i) \le \exp(-\lm(N)^{3/4}/2)\Bi).$$
	If $h' \le \log(N)^{2/3}$, then
	$$\prod_{h < i \le h + h'}(2F_i) \ge (\log\log(N))^{-h'} > \exp(- \lm(N)^{3/4}/2).$$
	On the other hand, if $h' \ge \log(N)^{2/3}$, then using moderate deviations \cite[Theorem 11.2]{mdp} for $\{\lm(2F_i)\}_{i \ge 1}$ implies that
	$$\P\Bi(\prod_{h < i \le h + h'}(2F_i) \le 1\Bi) \le \exp\bi(-\log(N)^{1/2}\bi),$$
	as asserted.
\enp

To construct the coupling, we write
$$\ccl_h := \{i \in \{0, \dots, N - 1\}:\, d_N(i , \Xl_h) \le \lc \Fl_h\rc\}$$
for the scope of $\Xl_h$ and define $\ccr_h$ accordingly. We also put
$$\cca_h := \ccl_h \cap \ccr_h, \qquad \text{ and } \qquad \ccu_h := \ccl_h \cup \ccr_h,$$
for the intersection and union of the scopes of the two walkers at time $h$.

\bep[Proof of Lemma \ref{coupProp}]

 Write $\nl_h, \nr_h, \nca_h, \ncu_h \ge 0$ for the cardinalities of the scopes $\ccl_h, \ccr_h, \cca_h, \ccu_h$, respectively. Then, the coalescence event $\Xl_{h + 1} = \Xr_{h + 1}$ means that the maximum of all fitnesses in $\ccu_h$ is contained in $\cca_h$. This event is of probability $\nca_h/\ncu_h$. To prove the assertion, we therefore need to construct a coupling such that $\P(\Efa_{h + 1}\,|\,\FFco_h) \le {4\nca_h}/{\ncu_h}$ holds under $\Esu_h$.

	 To that end, we proceed inductively and assume the coupling to be constructed until step $h$ and work under the event of coupling success $\Esu_h$. In order to define the true process $\bi((\Xl_{h + 1}, \Fl_{h + 1}\bi), \bi(\Xr_{h + 1}, \Fr_{h + 1})\bi)$, we let $F_1, \ld, F_{\ncu_h}$ be a sequence of i.i.d.\ fitnesses. Here, we think of $F_1, \ld, F_{\nl_h}$ to be in the scope of the left walker and $F_{\ncu_h - \nr_h + 1}, \ld, F_{\ncu_h}$ to be in the scope of the right walker. See Figure \ref{figCoupling}. To recover the true process, each of the walkers selects the maximum fitness within its scope. Note that $\cca_h$ is empty if $d_N(\Xl_h, \Xr_h) > \lc\Fl_h\rc + \lc\Fr_h\rc$.

\begin{figure}[ht]
	\begin{center}
	\input{coupling}
	\end{center}
\caption{An illustration of the coupling in the proof of Lemma \ref{t0Prop}.}
\label{figCoupling}\end{figure}

To construct the coupling, we may w.l.o.g.~assume that $\nl_h \le \nr_h$. Otherwise, reverse the roles of $\ms L$ and $\ms R$ in the following argument.
In order to define the independent process $\bi((\Xli_{h + 1}, \Fli_{h + 1}), (\Xri_{h + 1}, \Fri_{h + 1})\bi)$, we need to remove the dependence coming from the observation that in the true process, both the left and the right walker query the fitnesses $F_{\ncu_h - \nr_h + 1}, \ld, F_{\nl_h}$. To this end, we introduce a further copy of $\nca_h$ independent uniform fitnesses $F_{\ncu_h + 1}, \ld, F_{\ncu_h + \nca_h}$. Then, the left walker selects the maximum fitness among $F_1, \ld, F_{\nl_h}$ and the right walker selects the maximum fitness among $F_{\nl_h + 1}, \ld, F_{\nl_h + \nr_h}$. Hence, under $\Esu_h$, we may express the probability of a coupling failure succinctly as
\begin{align*}
	\P(\Efa_{h + 1}\,|\,\FFco_h) = \P\bi(\{\mr \in \cca_h\} \cup \{\mrs \ge \ncu_h + 1\}\bi),
	\end{align*}
	where $\mr \in \cca_h = \{\ncu_h - \nr_h + 1, \ld, {\ncu_h}\}$ and $\mrs \in \{\nl_h + 1, \ld, \nl_h + \nr _h\}$ denote the locations of the maxima of $F_{\ncu_h - \nr_h + 1}, \ld, F_{\ncu_h}$ and $F_{\nl_h + 1}, \ld, F_{\nl_h + \nr_h}$, respectively. Since both $\mr$ and $\mrs$ are uniformly distributed in their domains of size $\nr_h$, we deduce that
	$$\P\bi(\{\mr \in \cca_h\} \cup \{\mrs \ge \ncu_h + 1\}\bi) \le \P(\mr \in \cca_h) + \P(\mrs \ge \ncu_h + 1) = \frac{2\nca_h}{\nr_h} \le \frac{4\nca_h}{\ncu_h},$$
	the last step relying on the assumption that $\nl_h \le \nr_h$.
\enp

Finally, we establish Lemma \ref{noExcProp}.
\bep[Proof of Lemma \ref{noExcProp}]
	Recalling the recursion formula \eqref{eqFlRestricted} for $\de=1$, we introduce a modified fitness by setting $\Flm_0 = (2\Fli_0 \wedge N) / N$ and then $\Flm_{h + 1} = ((2\Flm_hF_{h + 1})\wedge 1)  $. Then, by induction, $\Flm_h \le \vp(\Fli_h) / N$ for all $h \ge 0$. Defining $\Frm_h$ similarly, it therefore suffices to prove the assertion with $\vp(\Fli_h)$ and $\vp(\Fri_h)$ replaced by $N\Flm_h$ and $N\Frm_h$.

	Now, $\Glm_h := \lm(\Flm_h)$ is a random walk in $(-\i, 0]$ truncated at 0 with drift to the right. Hence, there exists $K > 0$ such that for $h_0 =\lm(N)^{7/8} / 2$ we have
	$\P(\Glm_{h_0} \le -K) \le \e.$
	In particular, $\P\bi((\Glm_{h_0}, \Grm_{h_0}) \in [-K, 0]^2\bi) \ge 1 - 2\e$. Finally, note that the set $[-K, 0]^2$ has finite expected return time and that there is a positive probability $p(K) > 0$ such that almost surely on the event $(\Glm_h, \Grm_h) \in [-K, 0]^2$ we have
	$$\P\bi(\Glm_{h + 1} = \Grm_{h + 1} = 0 \,|\, \Glm_h, \Grm_h\bi) \ge p(K).$$
	This gives domination by a geometric random variable and thereby concludes the proof.
\enp

\ssec{Proof for $\de \in (1, 2) \cup (2, \ff)$}
\label{lightUpSec}
For $\de > 2$ we think of $\{\Xl_h - \Xr_h\}_{h \ge 0}$ as a centered random walk whose step size admits a finite second moment, so that we obtain convergence to Brownian motion just as in Lemma \ref{invPrinLem}. For $\de \in (1, 2)$ we proceed similarly, but now with the stable limit law (Lemma \ref{stabPrinLem}) rather then the invariance principle of Lemma \ref{invPrinLem}.  Since $\{\Xl_h\}_{h \ge 0}$ and $\{\Xr_h\}_{h \ge 0}$ are independent, we conclude from Lemma \ref{stabPrinLem} that also $\big\{N^{-1/\de}(\Xl_{Nt} - \Xr_{Nt}) \big\}_{t \le 1}$ converges in distribution to a $\de$-stable process.
\bep[Proof of Theorem \ref{distThm}; upper bound; $\de \in (1, 2) \cup (2, \ff)$]
	We write
	$$T_N := \{h \ge 0:\, \Xr_h \le \Xl_h\}$$
	for the first time, where the left walker moves past the right one. Note that when the left walker moves past the right one, then necessarily the scopes must intersect and there is a positive probability of coalescence.	Thus, it suffices to derive bounds on $T_N$. To that end, note that the distance $|\Xl_0 - \Xr_0|$ of the initial locations is at most $N$.

	Now, we write $T$ for the first time that Brownian motion, respectively a symmetric $\de$-stable process exceeds 1, so that
	$$\limsup_{N \to \ff}\P(T_N \ge KN^{2 \wedge \de}) \le \P(T \ge K).$$
	Finally, we leverage that 	hitting times are almost surely finite, which is true not only  for Brownian motion but also for the symmetric $\de$-stable process because it is recurrent when $\de \in (1,2)$, see \cite[Theorem I.17]{bertoin}. Therefore, the right-hand side tends to 0 as $K \to \infty$.
\enp


TITLE: Show that $\frac3{16}\sum\limits_{n\ge1}\frac{(-1)^n-8}{n^3}=-\frac{105}{64}\zeta(3)$
QUESTION [2 upvotes]: Consider the series

$$\frac34\sum_{n\ge1}\left[\frac{(-1)^n}{(2n)^3}-\frac2{n^3}\right]=\frac3{16}\sum_{n\ge1}\frac{(-1)^n-8}{n^3}$$

I have come across this sum while evaluating the integral
$$\int_{0}^{\pi/4}\frac{x^3}{\sin^2x}\,{\rm d}x=\frac{3\pi}{4}G-\frac{\pi^3}{64}+\frac{3\pi^2}{32}\log2-\frac{105}{64}\zeta(3)$$
where $G$ denotes Catalan's Constant and $\zeta(z)$ the Riemann Zeta Function. To do so I followed this approach (Solution to Problem $18$ given by ysharifi)

Integration by parts with $x^3=u$ and $\frac{{\rm d}x}{\sin^2 x}={\rm d}v$ reduces the problem to $\int_0^{\pi/4}x^2\cot x\,{\rm d}x$. Again, integration by parts with $x^2=u$ and $\cot x\,{\rm d}x={\rm d}v$ reduces the problem to $\int_0^{\pi/4}x\ln \sin x\,{\rm d}x$ and this one can by easily found using the identity $$\ln \sin x =-\ln 2-\sum_{n=1}^{\infty}\frac{\cos(2nx)}{n}$$ which holds for $x\in(0,\pi)$

The sum I failed to express in terms of the Riemann Zeta Function occured within the last step of calculation. To put in a nutshell I do not know how to show

$$\frac3{16}\sum_{n\ge1}\frac{(-1)^n-8}{n^3}=-\frac{105}{64}\sum_{n\ge1}\frac1{n^3}=-\frac{105}{64}\zeta(3)$$

As this is the last step of evaluating the integral I would be glad if someone could explain to me how to show the equalitiy of these two sums.
Thanks in advance!

REPLY [4 votes]: $$
\begin{align}
\color{red}{S} & =+\frac{3}{16}\sum_{n=1}^{\infty}\frac{(-1)^n-8}{n^3} =-\frac{3}{16}\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^3}-\frac{3}{16}\sum_{n=1}^{\infty}\frac{8}{n^3} \\[2mm]
& =-\frac{3}{16}\eta(3)-\frac{3}{2}\zeta(3) =\left(-\frac{3}{16}\left(1-2^{1-3}\right)-\frac{3}{2}\right)\zeta(3) =\color{red}{-\frac{105}{64}\zeta(3)}
\end{align}
$$


TITLE: What angle does minute hand and hour hand clock inclined to each other at $2:25$
QUESTION [2 upvotes]: What angle does minute hand and hour hand clock inclined to each other at $2:25$
My Attempt:
The hour hand of a clock rotate throufh an angle of $30°$ in $1$ hour,
So,
At $2$o'clock, $2$ hrs$\equiv 60°$
Again,
The minute hand of a clock rotate through an angle of $6°$ in $1$ minute.
So, At $2:25$, i.e, $25$min$\equiv 150°$
What should I do further?

REPLY [2 votes]: I'd envision it like so:

The blue arc represents a rotation of $2/12$ of a full rotation, or $60^{\circ}$ since a full rotation is $360^{\circ}$.
The green segment represents a rotation in the opposite direction of $(1/12)-(25/60)\times(1/12)$ of a full rotation or $17.5^{\circ}$; the $1/12$ represents the rotation from the three o'clock hour back to the two o'clock hour, while the $(25/60)\times(1/12)$ represents the proportion that the hour hand moves as the time changes from 2:00 to 2:25.
Taken together, the angle between the two is $60^{\circ}+17.5^{\circ}$ or $77.5^{\circ}$.


TITLE: The Tate conjecture for abelian varieties
QUESTION [12 upvotes]: Let $k$ be a number field. Recall that Faltings proved the famous Tate conjecture, which states that for any abelian variety $X$ over $k$ and any prime $\ell$, the natural map
$$\mathrm{End}(X) \otimes \mathbb{Q}_\ell \to \mathrm{End}(V_\ell(X))^{\mathrm{Gal}(\bar k/k)},\qquad (*)$$
is an isomorphism, where $V_\ell(X)$ denotes the usual Tate module of $X$ tensored with $\mathbb{Q}_\ell$.
If I understand it correctly however, this is but one part of the Tate conjecture, which in general states that the cycle class map
$$A^r(X) \otimes \mathbb{Q}_\ell \to H^{2r}(\bar X, \mathbb{Q}_\ell)^{\mathrm{Gal}(\bar k/k)},$$
is an isomorphism for all $r$. The "Tate conjecture" as I state it in $(*)$ is but the case $r=1$ of this conjecture.

Is the full Tate conjecture still open for abelian varieties? Are there any interesting special cases where it is known?

REPLY [16 votes]: Let me put it this way, Tate's conjecture for abelian varieties is known to imply the Hodge conjecture for abelian varieties, and the last is very much open for this class. For the implication, see the article by Deligne and Milne on "Hodge cycles on abelian varieties" (you can get a copy off of Milne's website).
There are lot's of interesting cases known for Tate. People like Zarhin, who seems to be around and has done a lot of this, can give more precise statements.


TITLE: solve the integral
QUESTION [1 upvotes]: Could anyone help me on how to go about this integral.
Calculate:  $$\int_0^{\pi/6}\left(\sin2x-\frac{\cos 3x}{3}+\frac{x^2}{2}\right)dx$$
This is how I've done and not coming up with right answer:
I get a primitive function
$$\left[-\frac{\cos 2x}{2}-\frac{1}{3}\frac{\sin3x}{3}+\frac{x^3}{6}\right]_0^{\frac{\pi}{6}}$$
and then I go ahead substituting in the respective $\pi/6$ och 0 boundary values which lead to $$\frac{1}{4}-\frac{1}{9\sqrt{2}}+\frac{\pi^3}{216*6}$$ and it's a wrong answer.
Any suggestions are highly appreciated!

REPLY [0 votes]: $$\int_{0}^{\frac{\pi}{6}}\left(\sin(2x)-\frac{\cos(3x)}{3}+\frac{x^2}{2}\right)\space\text{d}x=$$
$$\int_{0}^{\frac{\pi}{6}}\sin(2x)\space\text{d}x-\frac{1}{3}\int_{0}^{\frac{\pi}{6}}\cos(3x)\space\text{d}x+\frac{1}{2}\int_{0}^{\frac{\pi}{6}}x^2\space\text{d}x=$$

Substitute $u=2x$ and $\text{d}u=2\space\text{d}x$.
This gives a new lower bound $u=2\cdot0=0$ and upper bound $u=2\cdot\frac{\pi}{6}=\frac{\pi}{3}$:

$$\frac{1}{2}\int_{0}^{\frac{\pi}{3}}\sin(u)\space\text{d}u-\frac{1}{3}\int_{0}^{\frac{\pi}{6}}\cos(3x)\space\text{d}x+\frac{1}{2}\int_{0}^{\frac{\pi}{6}}x^2\space\text{d}x=$$

Substitute $s=3x$ and $\text{d}s=3\space\text{d}x$.
This gives a new lower bound $s=3\cdot0=0$ and upper bound $s=3\cdot\frac{\pi}{6}=\frac{\pi}{2}$:

$$\frac{1}{2}\int_{0}^{\frac{\pi}{3}}\sin(u)\space\text{d}u-\frac{1}{9}\int_{0}^{\frac{\pi}{2}}\cos(s)\space\text{d}s+\frac{1}{2}\int_{0}^{\frac{\pi}{6}}x^2\space\text{d}x=$$
$$\frac{1}{2}\left[-\cos\left(u\right)\right]_{0}^{\frac{\pi}{3}}-\frac{1}{9}\left[\sin\left(s\right)\right]_{0}^{\frac{\pi}{2}}+\frac{1}{2}\left[\frac{x^3}{3}\right]_{0}^{\frac{\pi}{6}}=$$
$$-\frac{1}{2}\left(\cos\left(\frac{\pi}{3}\right)-\cos\left(0\right)\right)-\frac{1}{9}\left(\sin\left(\frac{\pi}{2}\right)-\sin\left(0\right)\right)+\frac{1}{6}\left(\left(\frac{\pi}{6}\right)^3-0^3\right)=$$
$$-\frac{1}{2}\left(\frac{1}{2}-1\right)-\frac{1}{9}\left(1-0\right)+\frac{1}{6}\left(\frac{\pi^3}{216}-0\right)=\frac{5}{36}+\frac{\pi^3}{1296}$$