TITLE: Very different starting conditions into similar outcomes within Chaos Theory.
QUESTION [1 upvotes]: Chaos Theory demonstrates how systems with very similar starting conditions can end up in very different states, but can it demonstrate whether systems with very different starting conditions can end up in very similar states?

REPLY [3 votes]: The exponential divergence of initial conditions only holds if they are very close to each other (and only in average).
The trajectories emerging from two distant initial conditions may very well converge to each other (only to eventually separate again due to the positive Lyapunov exponent of the system).
To see that this must happen, consider the evolution of phase-space volumes¹:

If we consider a small bounded volume of initial conditions (e.g., a cube), the volume covered by the states arising from the evolution of these points blows up over time, until the entire attractor is covered.

On the other hand, the attractor itself is an invariant set (under the dynamics), i.e., it is unchanged by temporal evolution and so is its volume.

So, if we cover the entire neighbourhood of the attractor in small cubes, the future of any of these phase-space volumes is again a (smaller) neighbourhood of the same attractor. Therefore some initial conditions from different (distant) cubes must be evolved to be close to each other.
This property is called topological mixing and one of the defining features of chaos.
A simple example that illustrates this is the bit-shift map $T$: If we represent the state of the system in binary, applying the map corresponds to removing the first digit after the radix point. Suppose we choose two arbitrary irrational numbers $x$ and $y$ as initial conditions, and denote their $i$th digits with $x_i$ and $y_i$ respectively.
The distance after $n$ iterations ($δ:=|T^n(x)-T^n(y)|$) depends primarily on how many of their digits following the $n$th one are identical, i.e., what is the largest $p$ such that $x_{n+i}=y_{n+i} ∀ i ∈ {0,…,p}$.
Now, unless you carefully craft your initial conditions to avoid this, you will eventually find a block of $p$ matching digits for any $p$, and thus at some time in the future, $δ$ will become arbitrarily small.
¹ assuming for simplicity that there is only one attractor, whose basin of attraction is the entire phase space


TITLE: How to determine at which point a polynomial function with even and odd x-exponents is symmetrical, algebraically?
QUESTION [5 upvotes]: Say you have a polynomial function like $f(x) = 2x^3 - 6x^2 + 8$. This function won't be point symmetrical over $(0, 0)$, nor will it be symmetrical over the y-axis, but after plotting it, you can see that it's point symmetrical over the point $(1, 4)$.
For a polynomial function like $g(x) = (x+1)^3$, you can tell that it was shifted by one unit to the left on the x-axis, $h(x) = x^3$ is symmetrical over the point $(0, 0)$, so $g(x)$ will be symmetrical over the point $(-1, 0)$.
I don't really see how you could use this approach for $f(x)$ yet though.
One approach I found that works for $f(x)$ is to take the second derivative and set it to $0$.
$$f'(x) = 6x^2 - 12x$$ $$f''(x) = 12x - 12$$ $$f''(x) = 0$$ $$12x = 12$$ $$\fbox{x = 1}$$
Now to find the corresponding y value, plug in $x = 1$ into $f(x)$.
This second derivative approach doesn't seem to work for all functions though, there could be a point where the graph changes concavity but still isn't point symmetrical over.
Now my question is, is there a way to determine over which point any function like this will be symmetrical over without plotting it?

In the comments, it was stated that for a point of symmetry to exist, all the derivatives of the function must be (anti-)symmetric with respect to a line x = a. How can you check if a function is (anti-)symmetric to this line algebraically?

REPLY [4 votes]: If $f(x) = a_n x^n + a_{n-1} x^{n-1} +...$, the only possible abscissa for a symmetry point is the roots mean, i.e. $-\frac {a_{n-1}} {n \, a_n}$.
First proof
As was said in the comments, if $f$ is symmetric or anti-symmetric, all its derivatives are symmetric or anti-symmetric around the same abscissa. This alternates: even degree polynomials are symmetric (this is because of the behavior on $\infty$), odd degree polynomials are anti-symmetric.
So we can find this abscissa $a$ with the $n-1$th derivative: $f^{(n-1)}(x) = a_n \, n! \, x + (n-1)! \, a_{n-1} = 0$
$x=-\frac {a_{n-1}} {n \, a_n}$.
Tentative of a second proof
Another possible proof would proceed like this.
If there is a symmetry or anti-symmetry point $(a,b)$, we can vertically translate $f$ by subtracting $b$: $g(x)=f(x)-b$.
Now $g$ is (anti-)symmetric around $(a,0)$. So its real roots are symmetric around $a$.
If $g$ has all its $n$ roots real, because of the symmetry, the roots mean will be $a$, so $a=-\frac {a_{n-1}} {n \, a_n}$.
Now how to deal with the case where $g$ has some non-real complex roots?
One possible way would be to show that we can build a polynomial $h$ such that $h-g$ is a polynomial with degree at most $n-2$, and $h$ is (anti-)symmetric around a point with abscissa $a$ such as $g$. We would then be able to apply $a=-\frac {a_{n-1}} {n \, a_n}$: as we have only added an $n-2$ degree polynomial, $h$ has same first two coefficients $a_n$ and $a_{n-1}$ as $f$ and $g$.
Unfortunately I don't succeed in proving that.
Third proof, which gives the complete list of conditions
The method here is to directly expand the (anti-)symmetry equation.
If $n$ is odd, $f$ is anti-symmetric, so
$f(a+x)+f(a-x)=2b$
$\sum_{k=0}^n a_k(a+x)^k + \sum_{k=0}^n a_k(a-x)^k=2b$
$\sum_{k=0}^n a_k \sum_{j=0}^k {k \choose j} x^j a^{k-j} + \sum_{k=0}^n a_k \sum_{j=0}^k {k \choose j} (-1)^j x^j a^{k-j}=2b$
This can be split into separate equations, one for each power of $x$, i.e. one for each $j$. Equations for $j$ odd give $0=0$. Equations for $j$ even and $\ge 2$ give the following:
$\forall j$ even and $\ge 2$, $\sum_{k=j}^n a_k {k \choose j} a^{k-j} + \sum_{k=j}^n a_k {k \choose j} (-1)^j a^{k-j} = 0$
$\sum_{k=j}^n a_k {k \choose j} a^{k-j}=0$
For $j=n-1$, we get
$\sum_{k=n-1}^n a_k {k \choose {n-1}} a^{k-(n-1)} = 0$
$a_{n-1} a^0 + a_n \, n \, a^1 = 0$
$a = - \frac {a_{n-1}} {n \, a_n}$
The other conditions are obtained with the other even values for $j$. The value for $b$ can be computed from $j=0$ (but that's not much different from computing $b=f(a)$). Unfortunately, the conditions grow ugly when $j$ decreases; but it is the necessary and sufficient list and there does not seem to be a simpler way.
As example, the next condition, for $j=n-3$, expands into:
$3n^2a_n^2a_{n-3} - 3n(n-2)a_na_{n-1}a_{n-2}+(n-1)(n-2)a_{n-1}^3=0$
If $n$ is even, $f$ is symmetric, so
$f(a+x)-f(a-x)=0$
$\sum_{k=0}^n a_k(a+x)^k - \sum_{k=0}^n a_k(a-x)^k=0$
$\sum_{k=0}^n a_k \sum_{j=0}^k {k \choose j} x^j a^{k-j} - \sum_{k=0}^n a_k \sum_{j=0}^k {k \choose j} (-1)^j x^j a^{k-j}=0$
As above, this splits into separate equations, one for each $j$. Equations for $j$ even give $0=0$. Equations for $j$ odd give:
$\forall j$ odd and $\ge 1$, $\sum_{k=j}^n a_k {k \choose j} a^{k-j} - \sum_{k=j}^n a_k {k \choose j} (-1)^j a^{k-j} = 0$
so $\sum_{k=j}^n a_k {k \choose j} a^{k-j}=0$ as above.
For $j=n-1$, we get $a = - \frac {a_{n-1}} {n \, a_n}$ as before.
The two cases (n even; n odd) can be summarized in the following conditions:
$f(x) = a_n x^n + a_{n-1} x^{n-1} +...$ is symmetric (if $n$ even) or anti-symmetric (if $n$ odd) around $a=-\frac {a_{n-1}} {n \, a_n}$ (and $b=f(a)$, for anti-symmetry) iff:
$\fbox {$\forall j=n-1, n-3, ... \text{ 2 or 1, }\sum_{k=j}^n a_k {k \choose j} a^{k-j}=0$}$
Example
Using $f(x)=(x-4)(x-2)(2x+1)(x+3)(x+5)-209$
which means we have an anti-symmetry around $(a=-\frac 1 2,b=-209)$
$f(x)=2x^5+5x^4-48x^3-77x^2+214x-89$
(The reason why I don't use the example in the question post, is because I would like to have no null coefficient, and $a_n \ne 1$).
$a=-\frac {a_{n-1}} {n\,a_n}=-\frac {5} {5 \times 2}=-\frac 1 2$
$3 n^2 a_n^2 a_{n-3} - 3n(n-2)a_n a_{n-1} a_{n-2}+(n-1)(n-2)a_{n-1}^3 =$
$3 \times 5^2 \times 2^2 \times (-77) - 3 \times 5 \times 3 \times 2 \times 5 \times (-48) + 4 \times 3 \times 5^3=-23100+21600+1500=0$
There is no other condition (the one above is $\sum_{k=j}^n a_k {k \choose j} a^{k-j}=0$ for $j=n-3=2$), so the polynomial is anti-symmetric around $a=-\frac 1 2$ and $b=f(a)=-209$.


TITLE: Spectral resolutions of a second-order orthogonal tensor
QUESTION [0 upvotes]: Page 37 of Continuum mechanics by C. S. Jog lists the following formulae as the "spectral resolutions" of an orthogonal tensor $\bf R$ as having the eigenvectors ${\bf e} , \, {\bf n} , \, {\bf \hat{n}}$ and the corresponding eigenvalues $1, \lambda , \hat{\lambda}$, where ${\bf n}$ and ${\bf \hat{n}}$ are complex conjugates.
$$
\begin{equation}\begin{aligned}
{\bf I} & = {\bf e} \otimes {\bf e} +  {\bf n} \otimes {\bf \hat{n}} + {\bf \hat{n}} \otimes {\bf n} \\
{\bf R} & = {\bf e} \otimes {\bf e} +  \lambda \; {\bf n} \otimes {\bf \hat{n}} + \hat{\lambda} \; {\bf \hat{n}} \otimes {\bf n} \\
{\bf R}^2 & = {\bf e} \otimes {\bf e} +  \lambda^2 \; {\bf n} \otimes {\bf \hat{n}} + \; \hat{\lambda}^2 {\bf \hat{n}} \otimes {\bf n} \\
\end{aligned}\end{equation}
$$
I'd appreciate help deriving these "spectral resolutions".

REPLY [0 votes]: I finally have the answer with help from the author. What follows is a sketch of the proof.
First, because the eigenvalues $1$, $\lambda$ and $\hat{\lambda}$ are distinct they are linearly independent (see a simple proof here). Next because the eigenvalues of $\bf R$ are linearly independent it can be expressed as the product of three matrices
$$
{\bf R} = {\bf P} \, {\bf \Lambda} \, {\bf P}^{-1}
$$
Proof given here, where $\bf P$ is the the matrix whose columns consist of the eigenvectors $\bf e$, $\bf n$ and $\bf \hat{n}$ and $\bf \Lambda$ is the diagonal matrix whose diagonal elements are the eigenvalues $1$, $\lambda$ and $\hat{\lambda}$ i.e.
$$
{\bf P} = [ \, {\bf e} \, , \,  {\bf n} \, , \, {\bf \hat{n}} \, ]
$$
$$
\Lambda =
\begin{pmatrix}
1 & 0 & 0 \\
0 & \lambda & 0 \\
0 & 0 & \hat{\lambda}
\end{pmatrix}
$$
Because
$${\bf P} = e_1 \otimes {\bf e} \, + \,e_2 \otimes {\bf n} \, + \, e_3 \otimes {\bf \hat{n}}$$
where $e_1 = [1 \, , \, 0 \, , \,  0]^T$, $e_2 = [0 \, , \, 1 \, , \,  0]^T$ and $e_3 = [0 \, , \, 0 \, , \,  1]^T$.
The transpose conjugate is
$${\bf P}^T = {\bf e} \otimes e_1 \, + {\bf \hat{n}} \otimes \,e_2 \, + \, {\bf n}\otimes e_3 $$
Note that the conjugate is taken because $\bf P$ is complex valued. It follows that:
$${\bf P}^T {\bf P} \, = \, {\bf P} \, {\bf P}^T =  \, {\bf I} = \, {\bf e} \otimes {\bf e} \, + \,{\bf n} \otimes {\bf \hat{n}} \, + \,  {\bf \hat{n}} \otimes {\bf n}^T $$
Therefore ${\bf P}^T$ is the inverse of ${\bf P}$, i.e.
$${\bf P}^T = \, {\bf P}^{-1} = \, [ \, {\bf e}^T \, , \,  {\bf n}^T \, , \, {\bf \hat{n}}^T \, ]$$
Therefore
$$
\begin{equation}\begin{aligned}
{\bf R} &= {\bf P} \, {\bf \Lambda} \, {\bf P}^T \\
        &= {\bf e} \, {\bf e}^T \, + \, \lambda \, {\bf n}\, {\bf \hat{n}}^T \, + \, \hat{\lambda}\, {\bf \hat{n}} \, {\bf n} \\
        &= {\bf e} \otimes {\bf e} \, + \, \lambda \, {\bf n} \otimes {\bf \hat{n}} \, + \, \hat{\lambda}\, {\bf \hat{n}} \otimes {\bf n}^T
\end{aligned}\end{equation}
$$


text:
(* Copyright 2021 (C) Mihails Milehins *)

section\<open>Utilities\<close>
theory CZH_Utilities
  imports Main
  keywords "lemmas_with" :: thy_decl
begin

text\<open>
Then command \<^text>\<open>lemmas_with\<close> is a copy (with minor amendments to formatting)
of the command with the identical name that was introduced by Ondřej Kunčar in
\<^text>\<open>HOL-Types_To_Sets.Prerequisites\<close>. A copy of the function was produced,
primarily, to avoid the unnecessary dependency of this work on the
axioms associated with the framework \<open>Types-To-Sets\<close>.
\<close>

ML\<open>
val _ =
  Outer_Syntax.local_theory'
    \<^command_keyword>\<open>lemmas_with\<close>
    "note theorems with (the same) attributes"
    (
      Parse.attribs --| \<^keyword>\<open>:\<close> --
      Parse_Spec.name_facts --
      Parse.for_fixes >>
      (
        fn (((attrs),facts), fixes) =>
          #2 oo Specification.theorems_cmd Thm.theoremK
          (map (apsnd (map (apsnd (fn xs => attrs@xs)))) facts) fixes
      )
    )
\<close>

text\<open>\newpage\<close>

end

subset_name: curated


TITLE: Double Fourier Series coefficient derivation
QUESTION [1 upvotes]: We are looking at the rectangular membrane problem.
We're assuming that $f(x,y)$ is a smooth function defined on the rectangle $\{(x,y) : 0 \leq x \leq a, 0 \leq y \leq b \}$. We want to expand $f$ as
$$f(x,y) = \sum_{m \geq 1} \sum_{n \geq 1}a_{mn}\sin( \frac{m\pi x}{a}) \sin (\frac{n\pi y}{b}) \text{ (1)}$$
We make use of the identity:
$$\int_0^a \sin( \frac{m'\pi x}{a})\sin( \frac{m\pi x}{a}) \mathrm{d}x=\frac{a}{2}\delta_{mm'},$$
where $m,m' \in \mathbb{Z}$.
For the derivation we are assuming that all the conditions are satisfied to interchange the integral and sum and we get
$$\int_0^b \int_0^a f(x,y)\sin( \frac{m'\pi x}{a})\sin( \frac{n'\pi x}{b})\mathrm{d}x\mathrm{d}y= \frac{ab}{4}a_{m'n'} \text{ (2)}$$
at this point assuming that $(1)$ is true to find out what the coefficients $a_{mn}$ have to be.
In the textbook once we get to line $(2)$ which I can do the derivation of, they then go on to say that this implies that
$$a_{mn}= \frac{4}{ab}\int_0^b \int_0^a f(x,y)\sin( \frac{m\pi x}{a})\sin( \frac{n\pi x}{b})\mathrm{d}x\mathrm{d}y,$$
which is what I am having trouble with.
From what I understand $(2)$ is the inner product of $(f,\sin( \frac{m'\pi x}{a})\sin( \frac{n'\pi x}{b}))$ and then this somehow implies that the $a_{mn}$ must be as above.
I am missing something here =/
EDIT Bah, I am thinking it may just be simple algebra now that I look at it. -_-
I still don't know, any help is appreciated if it isn't too much to ask ;)

REPLY [1 votes]: First of all I think in (1), (2) and last formula in the second sine you mean $y$ instead of $x$, e.i. it should be $\sin\left(\frac{n\pi y}{b}\right)$ but not $\sin\left(\frac{n\pi x}{b}\right)$. Under that assumption all your formulas would be correct but I will add some small explanation.
Having (1) you can multiply it by $\sin\left(\frac{m'\pi x}{a}\right)\sin\left(\frac{n'\pi x}{b}\right)$ and using integral identity will get the last equality.


text:
\begin{document}

\title[]{Harnack Estimate for \\ the Endangered Species Equation}

\author[]{Xiaodong Cao}
\address[]{ Department of Mathematics, Cornell University\\ Ithaca, NY 14853-4201, USA}
\email{cao@math.cornell.edu}

\date{\today}

\author[]{Mark Cerenzia}
\address[]{ Department of Mathematics, Cornell University\\ Ithaca, NY 14853-4201, USA}

\curraddr{Department of Operations Research and Financial Engineering, Princeton University, Princeton, NJ 08544}

\email{ cerenzia@princeton.edu}

\author[]{Demetre Kazaras}
\address[]{ Department of Mathematics, University of Oregon\\ Eugene, OR 97403-1222, USA}
\email{demetre@uoregon.edu}

\date{\today}

\keywords{differential Harnack inequality; endangered species equation.}

\maketitle

\begin{abstract}
We prove a differential Harnack inequality for  the \emph{Endangered Species Equation}, which is a nonlinear parabolic equation. Our derivation relies on an idea related to the parabolic maximum principle. As an application of this inequality, we will show that positive solutions to this equation must blowup in finite time.
\end{abstract}

\section{ Introduction}\label{intro}
We consider positive smooth solutions $f(x,t) : \R^n\times [0,\infty)\to \mathbb{R}$ to the following Cauchy problem:
\begin{equation} \label{gese}
\begin{cases}
\frac{\partial }{\partial t} f=\Delta f + f^p,\\
f(x,0)=f_0(x),
\end{cases}
\end{equation}
for $p>1$. It is well known that solutions for (\ref{gese}) may blow up in finite time. There are some early studies in \cite{kaplan63, friedman65, ito66} for certain quasi-linear parabolic equations that show finite time blow up under suitable conditions. H. Fujita \cite{fujita66} studied equation \eqref{gese} as an example of a general quasi-linear parabolic partial differential equation whose failure for long-time existence depends on both the spacial dimension $n$ and the power $p$,  but not on the (positive) initial value $f_0$, provided that $0<n(p-1)<2$. In \cite{hamilton11}, R. Hamilton  labeled \eqref{gese} as the  \emph{Endangered Species Equation (ESE)} since one may think of $f(x,t)$ as the population density of a certain species evolving in time t. The population evolves according to diffusion  (the term $\Delta f$), but the equation also incorporates an additional change (the term $f^p$) in population that results from a pair's meeting.  \\

Our main result Theorem \ref{maintheorem} is a differential Harnack inequality (also known as a Li-Yau type estimate). As an application, we find that \emph{any} positive solution with positive initial condition becomes unbounded in finite time. So the population becomes arbitrarily large no matter how endangered initially. Our approach thus provides a new derivation of Theorem 1 in Fujita \cite{fujita66} (our Theorem \ref{mainthm}). In addition, upon integrating this inequality along a space-time path, we can recover a classical Harnack inequality (Corollary \ref{cor.class}).

The importance of parabolic Harnack inequalities, introduced in \cite{moser64, moser67}, is well-established.  Classical applications include deriving Holder continuity, obtaining Gaussian bounds for the heat kernel, and drawing many other conclusions about the underlying geometry of the space. While the study of differential Harnack inequalities and applications  originated in P. Li and S.-T. Yau \cite{ly86} (see also D. Aronson and P. B\'{e}nilan \cite{ab79} for a precursory form), this method was later brought into the study of geometric flows by Hamilton and played an important role  in the field, especially for the study of the Ricci flow (see for example, \cite{hamilton93}).   A more sophisticated use of such estimates for applications in geometry can be found in the details of the program for three dimensional geometrization.  This leads to G. Perelman's differential Harnack inequality  \cite[Corollary 9.3]{perelman1}, which is a crucial step in his solution to the Poincar\'e Conjecture. Since then, a systematic method to find Harnack inequality for geometric evolution equations was developed  in \cite{cao08, ch09}.\\

One of the main motivations in writing this paper is to suggest that the method developed in geometric flows can also be used for the study of  long-time existence (or non-existence)  for nonlinear parabolic equations, especially for finite-time blow-ups. In \cite{ab79}, the estimate was used to prove existence of solutions. For a probabilistic analysis of the Dirichlet problem associated to the endangered species equation with $1<p\leq 2$, see E.B. Dynkin \cite{dynkin92, dynkin93}. In his work, the solution $f$ of \eqref{gese} appears in the expression for the Laplace functional of a certain measure-valued Markov process, a so-called superprocess. Another goal of this paper is to generalize the estimate in \cite{hamilton11}, which also partially answers a question of Hamilton.\\

\paragraph{\textbf{Acknowledgements:}}  X. CaoÕs research was partially supported by NSF grant DMS 0904432. M. Cerenzia and D. Kazaras's research was supported by NSF grant DMS 0648208, through the Research Experience for Undergraduates Program at Cornell University. Note that any opinions, findings, and conclusions or recommendations expressed in this material are those of the authors and do not necessarily reflect the views of the National Science Foundation.

The authors would like to thank the referee and Dr. Hao Jia for their helpful comments and suggestions on an earlier version of this article. They would also like to thank Professor Robert Strichartz for his encouragement.

\section{Harnack Estimate}

In this section, we shall first derive our differential Harnack estimate.
Let $f(x,t) \in C^{\infty}(\mathbb{R}^n\times[0,\infty))$ be a positive solution to \eqref{gese} and  $u:=\log f$, then we have
$$
u_t = \Delta u + \left \vert \nabla u \right \vert ^2 + e^{u(p-1)}.
$$
The main technical result, Theorem 2.2, relies upon calculation of the evolution of a Harnack quantity and use of  the parabolic maximum principle. The main object of our study is the following Harnack quantity
\begin{equation}\label{hq1}
H:= \alpha \Delta u + \beta \left \vert \nabla u \right \vert ^2 + c e^{u(p-1)} + \phi,
\end{equation}
where $\alpha, \beta, c\in\mathbb{R}$ and $\phi:\mathbb{R}^n\times[0,\infty)\to[0,\infty)$ will be chosen suitably later.  Our first task is to compute the evolution of $H$.

\begin{lemma}\label{mainlemma}
Suppose $f(x,t)$ is a positive solution to \eqref{gese}, $u=\log f$ and $H$ is defined as in \eqref{hq1}. Then we have
\begin{align}\label{mainevolu}\nonumber
H_t = & \Delta H + 2 \nabla H \cdot \nabla u + (p-1)e^{u(p-1)}H + 2(\alpha - \beta) \left \vert \nabla \nabla u \right \vert ^2 \\ \nonumber
& +\left ( \alpha (p-1) + \beta - cp  \right ) (p-1)e^{u(p-1)} \mid \nabla u \mid^2 \\
&- (p-1)e^{u(p-1)} \phi + \phi_t - \Delta  \phi - 2 \nabla  \phi \cdot \nabla u .
\end{align}
\end{lemma}

\begin{proof}The proof follows from direct calculation.  First note the evolution equations:
$$
\partial_t (\mid \nabla u \mid^2) = \Delta \mid \nabla u \mid^2 - 2\mid \nabla \nabla u \mid^2 + 2 \nabla \mid \nabla u \mid^2 \cdot \nabla u + 2 (p-1) e^{u(p-1)} \mid \nabla u \mid^2,
$$
and
$$
\partial_t (\Delta u) =  \Delta (\Delta u) + \Delta \mid \nabla u \mid^2 + (p-1) e^{u(p-1)} \Delta u + (p-1)^2 e^{u(p-1)} \mid \nabla u \mid^2.
$$
In the above, we use
\begin{equation}
\label{ricci2}
\Delta \mid \nabla u \mid^2 =  2\nabla u \cdot \nabla \Delta u +  2 \mid \nabla \nabla u \mid^2.
\end{equation}
Hence we  have
\begin{align}\nonumber
H_t = &
\alpha \left [ \Delta (\Delta u) + \Delta\left \vert \nabla u \right \vert ^2  + (p-1) e^{u(p-1)} \Delta u + (p-1)^2 e^{u(p-1)} \left \vert \nabla u \right \vert ^2  \right ] \\ \nonumber
& + \beta \left [ \Delta \left \vert \nabla u \right \vert ^2  - 2\left \vert \nabla \nabla u \right \vert ^2  + 2 \nabla \left \vert \nabla u \right \vert ^2 \cdot \nabla u + 2 (p-1) e^{u(p-1)} \left \vert \nabla u \right \vert ^2  \right ] \\ \nonumber
&+c(p-1)e^{u(p-1)}\left [ \Delta u + \left \vert \nabla u \right \vert ^2 + e^{u(p-1)}\right ]  +\phi_t .
\end{align}
Using (\ref{ricci2}) again, we arrive at
\begin{align}\nonumber
H_t = &\Delta H + 2 \nabla H \cdot \nabla u
+ \alpha [  2|\nabla \nabla u | ^2  + (p-1) e^{u(p-1)} \Delta u
+ (p-1)^2 e^{u(p-1)} \left \vert \nabla u \right \vert ^2  ]\\ \nonumber
&+ \beta \left [2 (p-1) e^{u(p-1)}\left \vert \nabla u \right \vert ^2 - 2\left \vert \nabla \nabla u \right \vert ^2   \right ]
+ c(p-1)e^{2u(p-1)}\\ \nonumber & - c(p-1)e^{u(p-1)}\left \vert \nabla u \right \vert ^2 - c(p-1)^2e^{u(p-1)}\left \vert \nabla u \right \vert ^2  + \phi_t - \Delta  \phi - 2 \nabla  \phi \cdot \nabla u.
\end{align}
The lemma then follows upon expanding and reordering.

\end{proof}

\begin{thm}\label{maintheorem}
Suppose $f(x,t)$ is a positive solution to $\eqref{gese}$ and  $u=\log f$.   If $\alpha, \beta, a$ and $c$  satisfy
\begin{equation}\label{cond}
\alpha > \beta\geq 0, \ \ \frac{\alpha(p-1)+2\beta}{p}\geq c \geq \frac{(p-1)n\alpha^2}{4(\alpha-\beta)},
\end{equation} and
\begin{equation}\label{conda}
a\geq \frac{n\alpha^2}{2(\alpha-\beta)}>0,
\end{equation}
then we have
\begin{equation}
\label{harnack}
H_0\equiv \alpha \Delta u + \beta \left \vert \nabla u \right \vert ^2 + c e^{u(p-1)} + \frac{a}{t} \geq0
\end{equation}
 for all $t$.
\end{thm}

\begin{remark}
Our conditions \eqref{cond} allow us to choose $\beta=0$ in \eqref{harnack}. However, while the argument in the proof below requires that $\beta>0$ initially, we can let $\beta \to 0$ at the end. \end{remark}

\begin{remark}
We will see that the inequality (\ref{condnp}) in the proof below exhibits a restriction on the power $p$ with respect to the spacial dimension $n$ similar to the primary condition of  Theorem 1 in Fujita \cite{fujita66}.
\end{remark}

\begin{remark}
This result also gives a partial answer to Question 4 in Hamilton \cite{hamilton11}.
\end{remark}

\begin{proof}
Choose $\phi_R(x,t)$ so that it is defined on the n-rectangle $R \subset \R^n$ made up of a cartesian product of $n$ intervals $[p_i, q_i]$ so that $\phi_R \rightarrow \infty$ if $x_i \rightarrow p_i, \ q_i $ or if $t \rightarrow 0$. More explicitly, we may take
\begin{equation}\label{defphi}
\phi_R(x,t) = \frac{a}{t} + \sum_{k=1}^n \left ( \frac{b}{(x_k - p_k)^2} + \frac{b}{(q_k - x_k)^2}    \right ),
\end{equation}
for $t>0$ and $x = (x_1, \ldots, x_n) \in R = \Pi_1^n [p_i, q_i]$, and extend it to be $\infty$ elsewhere. The corresponding Harnack quantity is
$$H_R= \alpha \Delta u + \beta \left \vert \nabla u \right \vert ^2 + c e^{u(p-1)} + \phi_R(x,t).$$
Note that $H_R \to H_0$ as the rectangle $R= \Pi_1^n [p_i, q_i]$  exhausts $\R^n$, and  $H_R>0$ for small $t$ since $\phi_R \to \infty$ as $t\to0$.

For the sake of contradiction, assume that there exists a first time $t_0$ and point $x_0 \in R$ where $H_R(x_0, t_0) = 0$. At $(x_0,t_0)$, we have
$$
(H_R)_t \leq 0, \ \ \ \nabla H_R = 0, \ \ \ \Delta H_R \geq  0,
$$
and
$$
\Delta u = -\frac{1}{\alpha} (\beta \left \vert \nabla u \right \vert ^2  + c e^{u(p-1)} + \phi_R).
$$
Applying Lemma 2.1 and Cauchy-Schwarz in the form $\left \vert \nabla \nabla u \right \vert ^2 \geq \frac{1}{n} (\Delta u)^2$ yields that
\begin{align} \label{expression}\nonumber
0  \geq &\frac{2(\alpha - \beta)}{n\alpha^2} [\beta \left \vert \nabla u \right \vert ^2  + c e^{u(p-1)} + \phi_R]^2   - (p-1)e^{u(p-1)} \phi_R \\
& + \left [ \alpha (p-1) + \beta - cp  \right ] (p-1)e^{u(p-1)} \mid \nabla u \mid^2 + (\phi_R)_t - \Delta  \phi_R - 2 \nabla  \phi_R \cdot \nabla u.
\end{align}
Set $X = e^{u(p-1)}$ and $Y= \left \vert \nabla u \right \vert^2 $. Expanding and combining terms gives
\begin{align}\label{eq6} \nonumber
0 \geq &\frac{2(\alpha-\beta)}{n \alpha^2}(c^2 X^2+\beta^2 Y^2) + \left [\alpha (p-1) - c p + \beta + \frac{4(\alpha-\beta)\beta c}{n \alpha^2 (p-1)} \right](p-1) XY  \\
 &+ \left [ \frac{4(\alpha-\beta)c}{n \alpha^2} - (p-1) \right]\phi_R X + \frac{4(\alpha-\beta)\beta }{n \alpha^2} \phi_R Y \\\nonumber
 &+ (\phi_R)_t  - \Delta  \phi_R - 2 \nabla  \phi_R \cdot \nabla u+ \frac{2(\alpha-\beta)}{n \alpha^2}\phi_R^2.
\end{align}
We now claim that the right hand side is in fact positive, which will give us a contradiction. First note the quadratic inequality
$$
\frac{4(\alpha-\beta) \beta }{n\alpha^2}  \phi_R Y - 2 \nabla  \phi_R \cdot \nabla u \geq \frac{-n \alpha^2 \mid  \nabla \phi_R \mid^2}{4 (\alpha-\beta) \beta \phi_R}.
$$
Given (\ref{cond}), it follows that
\begin{equation}\label{cond1}
\alpha (p-1) - c p + \beta + \frac{4(\alpha-\beta)\beta c}{n \alpha^2 (p-1)} \geq 0,\quad
\frac{4(\alpha-\beta)c}{n \alpha^2} - (p-1)\geq 0.
\end{equation}
Dropping several nonnegative terms in the right hand side of (\ref{eq6}), we  arrive at

$$
0 \geq (\phi_R)_t  - \Delta  \phi_R - \frac{n \alpha^2 \mid  \nabla \phi_R \mid^2}{4 (\alpha-\beta) \beta \phi_R} + \frac{2(\alpha-\beta)}{n \alpha^2}\phi_R^2.
$$

We then compute
$$
\Delta \phi_R = \sum_{i=1}^n \left ( \frac{6b}{(x_k-p_k)^4}+  \frac{6b}{(q_k-x_k)^4} \right ),
$$
$$
|\nabla \phi_R |^2 =  \sum_{k=1}^n \left( -  \frac{2b}{(x_k-p_k)^3}+\frac{2b}{(q_k-x_k)^3}  \right )^2,
$$
and observe that
\begin{align}\label{gradest}\nonumber
\frac{|\nabla \phi_R|^2}{\phi_R} & = \sum_{k=1}^n \left(  - \frac{2b}{(x_k-p_k)^3 \sqrt{\phi_R}}+\frac{2b}{(q_k-x_k)^3 \sqrt{\phi_R}}  \right )^2 \\
& \leq \sum_{k=1}^n \left(   \frac{2\sqrt{b}}{(x_k-p_k)^2 }+\frac{2\sqrt{b}}{(q_k-x_k)^2 }  \right )^2.
\end{align}
Set
$$
 A :=  \frac{2(\alpha-\beta)}{n \alpha^2}>0 ,\ \ \ \ B:= \frac{n\alpha^2}{4(\alpha-\beta)\beta}>0.
$$
To arrive at a contradiction, it suffices to show that
\begin{equation}\label{eq12}
A \phi_R^2 - \Delta \phi_R - B \frac{ \mid \nabla \phi_R \mid^2}{\phi_R} + (\phi_R)_t>0.
\end{equation}
Now we choose $a$ as in (\ref{conda}), so that $Aa^2-a\geq 0$. Next, plugging (\ref{gradest}) and   (\ref{defphi}) into (\ref{eq12}), we conclude that it is sufficient to have  $b>0$ and
$$
Ab^2-b(6+4B) >0 ,
$$
which reduces to
$$
b>\frac{1}{A}(6+4B).
$$
In summary, the conditions on $a$ and $b$ are
$$
a\geq  \frac{n\alpha^2}{2(\alpha-\beta)}, \ \ \ b> \frac{n\alpha^2}{2(\alpha-\beta)}\left[6+\frac{n \alpha^2} {(\alpha-\beta)\beta}\right].
$$
Recall that our constants $\alpha, \beta, c$ must satisfy $\alpha>\beta>0$ along with (\ref{cond1}):
$$
\alpha (p-1) - c p + \beta + \frac{4(\alpha-\beta)\beta c}{n \alpha^2 (p-1)} \geq 0,\ \ \ \  \frac{4(\alpha-\beta)c}{n \alpha^2} - (p-1) \geq 0.
$$
These latter two inequalities can be satisfied as long as we choose $c$ such that
$$
\frac{\alpha(p-1)+2\beta}{p}\geq c \geq \frac{(p-1)n\alpha^2}{4(\alpha-\beta)}.
$$
For given $n,p$, we may choose such $c$ as long as we have
\begin{equation}\label{condnp}
\frac{4(\alpha(p-1)+2\beta)(\alpha-\beta)}{\alpha^2} \geq p(p-1)n.
\end{equation}

Our choice of constants $\alpha, \beta, a, c$ now implies  that the right hand side of \eqref{eq6} is in fact positive, which is a contradiction.  Assuming the solution exists in all of space $\R^n$, we can let $R \to \R^n$ so that $\phi_R \to a/t$.
This completes the proof of Theorem \ref{maintheorem}.
\end{proof}

\section{Applications}

In this section, we shall give a few applications of Theorem \ref{maintheorem}. We first recover and improve estimate in \cite{hamilton11}; then we use it to study long time existence problem of (\ref{gese}); finally we integrate along space-time curve to derive a classical Harnack inequality.

\subsection{Hamilton's result}
In this subsection, we study the case of $n=1$ and $p=2$, which was studied  in  \cite{hamilton11} by Hamilton. In particular, we apply Theorem \ref{maintheorem} with $n=1$ and $p=2$  by picking $\alpha=1$, $\beta=0$, $c=\frac{1}{2}$, and $a=\frac{2}{3}$, to conclude
$$
\Delta u +\frac{1}{2}e^{u} +\frac{2}{3t} \geq 0.
$$
Recalling that $u=\log f$, we  recover \cite[Theorem 5.1]{hamilton11}:

\begin{thm}[Hamilton \cite{hamilton11}]
Let $f$ be a positive solution to \eqref{gese} with $n=1$, $p=2$. Then
\begin{equation}\label{hharnack}
f_t + \frac{2f}{3t} \geq \frac{f_x^2}{f} + \frac{f^2}{2}.
\end{equation}
\end{thm}
Furthermore, our proof in fact shows that \eqref{hharnack} can be improved by picking $a=\frac12$ and $c=\frac14$ to get
$$
\Delta u +\frac{1}{4}e^{u} +\frac{1}{2t} \geq 0,
$$
yielding
\begin{equation}\label{hharnack2}
f_t + \frac{f}{2t} \geq \frac{f_x^2}{f} + \frac{3f^2}{4}.
\end{equation}

\begin{remark}In \cite{hamilton11}, Hamilton asks if it is possible to improve (\ref{hharnack}).
The above (\ref{hharnack2}) gives an affirmative answer to that question.
\end{remark}

If the dimension $n=2$, and $p=2$, a similar Harnack estimate can be derived  by picking $\alpha=1$, $\beta=0$, $a=1$ and $c=\frac12$, which we state as our next theorem.
\begin{thm}
Let $f$ be a positive solution to \eqref{gese} with $n=2$, $p=2$. Then
\begin{equation}\label{hharnack3}
f_t + \frac{f}{2t} \geq \frac{f_x^2}{f} + \frac{3f^2}{4}.
\end{equation}
\end{thm}

\subsection{Finite-time Blow Up}

In this subsection, we reprove Fujita's result \cite[Theorem 1]{fujita66}, which states that if $0<n(p-1)<2$, then any positive solution $f$ to \eqref{gese} will blow up in finite time, however small the positive initial value may be. We remark that Fujita also shows in the same paper that the condition $n(p-1)>2$ implies there exists some small positive initial data such that the solution exists for all time. For blow up in the case $n(p-1)=2$, see V. Galaktionov \cite{galak94}.\\

We in fact start by proving the following weaker version. Recall that $p>1$, so the lower bound for $n(p-1)$ in the next statement is always satisfied.

\bp
Suppose that $f$ is a positive solution to $\eqref{gese}$, and $c$ is a constant satisfying $0<n(p-1)\leq c < 2$. Then $f$ blows up in finite time provided that
\begin{equation} \label{lowbound}
f(x_0, t_0) \geq \left (\frac{4n}{2-c} \right )^{1/(p-1)}
\end{equation}
 at some point $(x_0, t_0)$.
\ep
\begin{proof}
Picking $\alpha=2$, $\beta=1$, $a=2n$ and $c$ such that $(p-1)n \leq c < 2$ in Theorem \ref{maintheorem} yields

$$
2 \Delta f - \frac{| \nabla f |^2}{f}  + c f^p +\frac{2n}{t}f \geq 0.
$$
Since $f_t=\Delta f+f^p$, we have
$$
2f_t - \frac{|\nabla f|^2}{f} + \frac{2nf}{t} \geq  (2-c)f^p.
$$
Hence
$$
2f_t + \frac{2nf}{t} \geq (2-c)f^p.
$$
The above inequality implies that
\begin{equation}\label{inverse}
2\frac{\partial}{\partial t} \left ( \frac{1}{f} \right ) \leq  \frac{1}{f} \left  ( \frac{2n}{t} - (2-c)f^{p-1}  \right ) =\frac{1}{f^{2-p}} \left  (\frac{2n}{t f^{p-1}} - (2-c)  \right ).
\end{equation}
Without loss of generality, we may assume that $f\geq (\frac{4n}{2-c})^{1/(p-1)}$ at the origin $x_0=0$ for $t_0=1$. This assumption together with (\ref{inverse}) gives
$$
2\frac{\partial}{\partial t} \left ( \frac{1}{f} \right )(0,t)\leq \frac{2-c}{f^{2-p}(0,t)} \left (\frac{1}{2t} - 1 \right )< 0,
$$
so that $f(0, t)$ is strictly increasing for $t\geq 1$ provided that $f(0, t)$ is finite.

Now if $p>2$, then $f^{p-2}(0,t) \geq f^{p-2}(0,1)$ for $t \geq 1$ and (\ref{inverse}) becomes
$$
2\frac{\partial}{\partial t} \left ( \frac{1}{f} \right )(0,t) \leq \frac{2n}{tf(0,1)}-(2-c)f^{p-2}(0,1).
$$
On the other hand, if $1<p\leq 2$, manipulation of (\ref{inverse}) yields
\begin{equation*}\label{pl2}
\frac{2}{p-1} \frac{\partial }{\partial t} \left [ \left ( \frac{1}{f} \right )^{p-1} \right ](0,t)=2f^{2-p} \frac{\partial}{\partial t} \left ( \frac{1}{f} \right ) (0,t) \leq \frac{2n}{t f^{p-1}(0,1)} - (2-c) .
\end{equation*}
In both cases, there exists  $\delta>0$, when $t$ is large enough, the right hand sides are less than $-\delta<0$,  so that $\frac{1}{f} \to 0$ in finite time. This proves our proposition.
\end{proof}

Intuitively, the above result says that if the population is ever large enough somewhere, then it will become unbounded in finite time. Through a parabolic rescaling argument, without loss of generality, we can always assume a given positive solution $f$ satisfies the condition \eqref{lowbound} of the previous proposition at some point. To see this, let $\lambda >0$ and $\delta \in \R$, define a new function $\tilde{f}(\tilde{x},\tilde{t}) := \lambda^\delta f(x, t)$, where $\tilde{x} := \lambda x$, $\tilde{t} := \lambda^2 t$. Then we can see that with the choice of $\delta := -\frac{2}{p-1}$, $\tilde{f}$ also satisfies equation \eqref{gese}:
\begin{equation}
\label{rescale}
\frac{\partial}{\partial \tilde{t}}\tilde{f} = \lambda^{\delta-2} \left (\frac{\partial}{\partial t}f \right ) =\lambda^{\delta-2} (\Delta f)+ \frac{\lambda^{\delta-2}}{\lambda^{\delta p}} (\lambda^\delta f)^p = \tilde{\Delta} \tilde{f} + \tilde{f}^p.
\end{equation}
In particular, $\lambda>0$ is arbitrary and can be chosen so that condition \eqref{lowbound} is met by $\tilde{f}$. Once $\lambda$ is chosen and fixed,  $\tilde{f}$ remains bounded in finite time if and only if $f$ does. We have thus proved the following theorem, which says that no matter how small the initial population is, it will become unbounded in finite time.

\begin{thm}[Fujita \cite{fujita66}] \label{mainthm}
Let $0<n(p-1) < 2$. Then any positive solution $f$ to the equation $\eqref{gese}$ blows up in finite time.
\end{thm}

\subsection{Classical Harnack Inequality}

In this subsection,  we shall integrate our differential Harnack (\ref{harnack}) along a space-time path to derive  a classical Harnack type inequality, which provides a comparison of values
of positive solutions at different points in space-time.

\bc \label{cor.class}
Let $f$ be a positive solution to the generalized endangered species equation \eqref{gese} and $u = \log f$.  Let $\gamma(t) = (x(t),t)$, $t \in [t_1,t_2]$, be a space-time curve joining  two given points $(x_1,t_1), (x_2,t_2) \in \R^n \times [0, \infty)$ with $0 < t_1<t_2$. Assume further that $\alpha \geq 2 \beta$ so  $c \leq \alpha$, and  $a = \frac{n \alpha^2}{2(\alpha-\beta)}\leq n \alpha$. Then we have
\begin{equation}\label{intharnack}
f(x_1,t_1) \leq f(x_2,t_2) \left( \frac{t_2}{t_1} \right)^n \exp{\left[\frac{|x_2-x_1|^2}{2(t_2-t_1)}\right]}.
\end{equation}

\ec
\bpf
Recall the evolution equation for $u = \log f$ is
$$
u_t = \Delta u + | \nabla u |^2+e^{u(p-1)}.
$$
By our differential Harnack inequality (\ref{harnack}), we have $H\geq0$, this yields that
$$
\begin{aligned}
\Delta u
\geq \alpha^{-1} (- \beta | \nabla u |^2 - c e^{u(p-1)} - \frac{a}{t}).
\end{aligned}
$$
We then compute the evolution of $u$ along $\gamma$:
$$
\begin{aligned}
\frac{d}{d t} [u(x(t),t)] &=\nabla u \cdot \dot{x} + u_t  \\
& = \nabla u \cdot \dot{x} + \Delta u + | \nabla u |^2+e^{u(p-1)} \\
& \geq | \nabla u |^2 \left (1-\frac{\beta}{\alpha} \right ) + \nabla u \cdot \dot{x} - \frac{a}{\alpha t} + e^{u(p-1)}\left (1-\frac{c}{\alpha} \right ) \\
& \geq | \nabla u |^2 \left (\frac{1}{2}-\frac{\beta}{\alpha} \right ) - \frac{1}{2} | \dot{x}|^2 - \frac{a}{\alpha t} + e^{u(p-1)}\left (1-\frac{c}{\alpha} \right ) \\
& \geq  - \frac{1}{2} | \dot{x}|^2- \frac{a}{\alpha t}, \\
\end{aligned}
$$
where we have used the assumption $\alpha \geq 2 \beta$ with  $c \leq \alpha$ for the last inequality. Hence, we have
\begin{equation}\label{curve}
\frac{d}{d t} [-u(x(t),t)] \leq  \frac{1}{2} | \dot{x}|^2+ \frac{n}{t}.
\end{equation}
Integrating the above inequality \eqref{curve} along $\gamma$, and taking the infimum over all such space-time paths yields
$$
u(x_1,t_1) - u(x_2,t_2) \leq \inf_{\gamma(t) = (x(t),t)} \int_{t_1}^{t_2} \left[ \frac{1}{2} | \dot{x}|^2+ \frac{n}{ t} \right] .
$$
Recalling that $u = \log f$, we arrive at \eqref{intharnack}.
\epf

\def\cprime{$'$}
\bibliographystyle{plain}

config: arxiv


TITLE: Why is it necessary $n\geq 2$ in this problem about boundary and isolated point?
QUESTION [2 upvotes]: Let $A\subset\mathbb{R}^n$ be an open set, where $n\geq2$. Prove that given $a\in\mathbb{R}^n-A$, are equivalent:
(i) the set $A\cup\{a\}$ is open;
(ii) $a$ is an isolated point of the boundary of $A$;
(iii) There exists $r>0$ such that $B(a;r)-\{a\}\subset A$.
Why is it necessary $n\geq 2$ in this problem? I'm confusing because I belive that I've proved (i)$\Leftrightarrow$(iii) and (iii) $\Rightarrow$(ii)  without using this hypothesis.
Furthermore, I would like to know how to prove (ii) $\Rightarrow$(i) (or (ii) $\Rightarrow$(iii)).
Thanks.

REPLY [2 votes]: You're right, you don't need $n \geqslant 2$ for the equivalence $(i) \iff (iii)$ or the implication $(iii) \Rightarrow (ii)$.
What you need $n \geqslant 2$ for is the implication $(ii) \Rightarrow (iii)$. Let's look at the example $A = (0,\,1) \subset \mathbb{R}^1$ to see where it breaks down for $n = 1$. Let's take $a = 0$. $a$ is an isolated boundary point, for example the $\frac12$-neighbourhood of $0$ doesn't contain another boundary point. Now, any $\varepsilon$-neighbourhood of $0$, for $0 < \varepsilon < 1$, consists of three parts

$(0,\, \varepsilon) \subset A$,
$\{0\} \subset \partial A$,
$(-\varepsilon,\,0) \subset (\mathbb{R}\setminus A)^\circ$.

The part inside $A$ and the part "outside" $A$ are separated by the part on the boundary of $A$. That is generally so, for each $M \subset X$, where $X$ is a topological space, you have a disjoint union
$$X = \overset{\circ}{M} \,\dot{\cup}\, \partial M\, \dot{\cup}\, (X\setminus M)^\circ.$$
Now, if you have an isolated boundary point $p$ of $M$, a small enough punctured open neighbourhood of $p$ (that is, a $V \setminus \{p\}$, where $V$ is an open neighbourhood of $p$), splits into two disjoint open parts, one in the interior of $M$, the other in the exterior of $M$, since
$$V = (V \cap \overset{\circ}{M})\, \dot{\cup}\, (V \cap \partial M)\,\dot{\cup}\, \bigl(V \cap (X\setminus M)^\circ\bigr)$$
and $V\cap \partial M = \{p\}$ if $V$ is small enough.
For $X = \mathbb{R}^n$, we have that an $\varepsilon$-neighbourhood minus the centre point is connected when $n \geqslant 2$, but not connected when $n = 1$.
So in the case $n \geqslant 2$, either $(B_\varepsilon(a) \setminus \{a\}) \cap A$ or $(B_\varepsilon(a)\setminus \{a\}) \cap (\mathbb{R}^n \setminus \overline{A})$ must be empty, since by definition a connected set cannot be written as the disjoint union of two nonempty open parts. $(B_\varepsilon(a) \setminus \{a\}) \cap A$ cannot be empty, since $a \in \partial A$ and $a \notin A$, hence $(B_\varepsilon(a) \setminus \{a\}) \subset A$.


TITLE: How do we prove $|\cos x-\cos y|\le|x-y|$ for all $x,y\in\mathbb{R}$?
QUESTION [7 upvotes]: My thought:
Fix any point $y_0\in\mathbb{R}$, since we know $\cos' x=-\sin x$ and $|-\sin|\le1$, thus $|\cos' x|=\lim_{x\to y_0}\frac{|\cos x-\cos y_0|}{|x-y_0|}=...\le 1$. I think my issue is how to translate $\lim_{x\to y_0}\frac{|\cos x-\cos y_0|}{|x-y_0|}$ into $\frac{\cos x-\cos y}{x-y}$. Could someone give some insight?

REPLY [0 votes]: We will prove this for all $0<x,\ y <\frac{\pi}{2}$
If $\pi :\mathbb{R}^2\rightarrow \mathbb{R},\ \pi (a,b)=a$, then $$
|\pi (P)-\pi (Q)| \leq |P-Q|$$
If $$P=(\cos\ x,\sin\ x),\ Q=(\cos\ y,\sin\ y),$$ then $$ |x-y| \geq
|P-Q| \geq |\pi(P)-\pi (Q) |$$


TITLE: Show that the set $D_b$ is a Dynkin-System
QUESTION [1 upvotes]: let $X \neq \emptyset$, ${D_B} := \{Q \subseteq X: Q \cap B \in \sigma(\mathcal{E}) \}$, where $\sigma(\mathcal{E}):= \bigcup_{\mathcal{E} \subseteq \mathcal{D_{i}}} D_i$ and $D_i$ is a family of Dynkin systems $ \forall i \in I$.
That is, $\emptyset \neq X, D \subseteq \mathcal{P}(X)$ is a Dynkin system when:

$\emptyset, X \in D$.
$A \in D \to X\setminus A \in D$.
For every sequence $(A_{n})_{n\in\mathbb{N}}$ of pairwise disjoint sets $A_i$ in $D$, the union $\bigcup_{n\in\mathbb{N}} A_n \in D$.

Moreover, $B  \in \sigma(\mathcal{E})$ and $ \mathcal{E} \subseteq \mathcal{P}(X)$. What I want to prove is that $D_B$ is indeed a Dynkin System. I struggle with the second and third condition and would appreciate some help.

REPLY [2 votes]: It seems that the Dynkin system generated by $\mathcal E$ is the intersection of all the Dynkin systems containing $\mathcal E$. Denoting by $\mathcal D$ the collection of the Dynkin systems containing $\mathcal E$, this means that $\sigma\left(\mathcal E\right)=\bigcap_{D\in\mathcal D}D$. With these notations,
$$
D_B=\left\{Q\subseteq X: Q\cap B\in \bigcap_{D\in\mathcal D}D\right\}
=\bigcap_{D\in\mathcal D} \left\{Q\subseteq X: Q\cap B\in D\right\}.
$$
To conclude, we need the following two facts:

An arbitrary intersection of Dynkin systems is a Dynkin system.
For each Dynkin system $D$, the collection $\mathcal A_D=\left\{Q\subseteq X: Q\cap B\in D\right\}$ is a Dynkin system.

Let us show the second fact. Observe that for all $D\in\mathcal D$, $B\in\mathcal D$ hence $X\in \mathcal A_D$. Let $Q\in\mathcal A_D$. Write
$$
\left(X\setminus Q\right)\cap B=B\cap Q^c=\left(B^c\cup \left(Q\cap B\right) \right)^c.
$$
Since $D$ is Dynkin system, $B^c\in  D$ and by assumption $Q\cap B\in D$ hence by the third point of the definition, the disjoint union $B^c\cup \left(Q\cap B\right) $ belongs to $D$. Using again the second point of the definition shows that $X\setminus Q\in \mathcal A_D$.
For the third condition, let $\left(Q_n\right)_{n\geqslant 1}$ be a pairwise  disjoint sequence of elements of $\mathcal A_D$. We know that $Q_n\cap B\in D$ for all $n$; since the sequence $\left(Q_n\cap B\right)_{n\geqslant 1}$ is disjoint and consists of elements of $D$, we know that $\bigcup_{n\geqslant 1}\left(Q_n\cap B\right)=
\left(\bigcup_{n\geqslant 1}Q_n\right)\cap B$ is an element of $D$ hence so is $\bigcup_{n\geqslant 1}Q_n$.


TITLE: Prove that the following function $f(x)=\log_{2}\left(x+{\sqrt{x^2+1}}\right)$ is invertible on the whole number line.
QUESTION [0 upvotes]: I think that it would help to show that this function is odd. But how can I show that the function $$\log_{2}\left(x+{\sqrt{x^2+1}}\right)$$ is invertible?

REPLY [2 votes]: Check that $x+\sqrt {x^2+1}$ maps $\mathbb {R}$ bijectively onto $(0,\infty).$ Then recall $\log_2 x$ maps $(0,\infty)$ bijectively onto $\mathbb {R}.$ It follows that the composition is a bijection of $\mathbb {R}$ onto $\mathbb {R},$ which gives the conclusion.


TITLE: into function vs injective function
QUESTION [5 upvotes]: In many mathematical books that I have read and from lectures from professors, the words 'into' and 'injective' were used interchangeably, but in Patrick Suppes book Axiomatic Set Theory he gives a percise definition of what it means for a function to be 'into':
$f$ is a function from $\:A$ into $\:B \leftrightarrow\: f$ is a function & $D(f) = A$ & $R(f) \subseteq B$
where $D(f)$ is the Domain of $f$ and $R(f)$ is the Range of $f$.
Is the definition given by Suppes the correct meaning of 'into', or is 'into' simply a synonym for 'injective'

REPLY [9 votes]: Into is not a synonym for "injective". There is, however, another way of referring to an injective function: such a function is sometimes said to be "one-to-one function", which is not to be mistaken with a "one-to-one correspondence"/bijective function.
Even though we do refer to a surjective function as being "onto", it does not follow that an injective function is therefore "into."


TITLE: Why in professional spectroscopes diffraction gratings are used instead of prisms?
QUESTION [4 upvotes]: Why in professional spectroscopes diffraction gratings are used instead of prisms?
I guess transparency is not an issue, as we need optics anyway.

REPLY [0 votes]: In addition, glass prism will experience its resonance frequency absorption at certain wavelength (eg: some glasses absorb UV). That why it is not a brilliant choice to disperse the light spectrum in most of the professional spectroscopy.


text:
\begin{document}
\title{An optimal algorithm for stochastic strongly-convex optimization}
\author{ Elad Hazan\\
IBM Almaden\\
{\tt elad.hazan@gmail.com}
\and
Satyen Kale \\
Yahoo! Research \\
{\tt skale@yahoo-inc.com} }
\date{}
\maketitle

\begin{abstract}
We consider stochastic convex optimization with a strongly convex (but not necessarily smooth) objective. We give an algorithm which performs only gradient updates with optimal rate of convergence.
\end{abstract}

\section{Setup}

Consider the problem of minimizing a convex function on a convex domain $\K$:
$$ \min_{\x \in \K} f(\x). $$
Assume that we have an upper bound on the values of $f$, i.e. a number $M > 0$ such that for any $\bx_1, \bx_2 \in \K$, we have $|f(\bx_1) - f(\bx_2)| \leq M$. Also, assume we can compute an unbiased estimator of a subgradient of $f$ at any point $\x$, with $L_2$ norm bounded by some known value $G$. Assume that the domain $\K$ is endowed with a projection operator $\prod_{\K}(\y) = \arg \min_{\x \in \K} \|\x-\y\|$. Finally, we assume that $f$ satisfies the following inequality
$$ f(\x) - f(\x^*) \geq \lambda \|\x-\x^*\|^2 $$
where $\x^*$ is the point in $\K$ on which $f$ is minimized. This property holds, for example, if $f$ is $\lambda$-strongly-convex.
The canonical example of such an optimization problem is
support-vector-machine training.

\section{The algorithm}
The algorithm is a straightforward extension of stochastic gradient descent.
The new feature is the introduction of ``epochs" inside of which standard
stochastic gradient descent is used, but in each consecutive epoch the learning
rate decreases exponentially.

\begin{algorithm}[h!]
  \caption{\label{alg:online-variance-updates} Epoch-GD}
  \begin{algorithmic}[1]
    \STATE Input: parameters $M$,$G$, $\lambda$, $\eta_k$, $T_k$ and error tolerance  $\varepsilon$. Initialize $\x_1^1 \in \K $ arbitrarily.
    \FOR{$k  = 1$ to $\lceil \log_2 \frac{M}{\varepsilon}\rceil$}
    \STATE Set $V_k = \frac{M}{2^{k-1}}$, and start an epoch as follows
    \FOR{$t = 1$ to $T_k$}
    \STATE Let the estimated subgradient of $f$ at $\x_t^k$ be $\hatg_t$
    \STATE Update
    $$
    \x_{t+1}^k = \prod_{\K}\left\{\x_t^k - {\eta_k} \hatg_t  \right\}
    $$ \label{alg:update}
    \ENDFOR
    \STATE Set $\x_1^{k+1} = \frac{1}{T_k} \sum_{t=1}^{T_k} \x_t^{k} $ (or pick one iterate at random).
    \ENDFOR
    \RETURN $\x_1^{k+1}$.
  \end{algorithmic}
\end{algorithm}

\section{Analysis}

Our main result is the following Theorem
\begin{Thm} \label{thm:main}
The final point $\x_1^{k+1}$ returned by the {\it Epoch-GD} algorithm, with
parameters $T_k = \left\lceil\frac{16G^2 }{\lambda V_k}\right\rceil$ and
$\eta_k = \frac{{V_k}}{4G^2}$,  has the property that $\E[f(\x_1^{k+1})] -
f(\x^\star) \leq \varepsilon$. The total number of gradient updates is
$O(\frac{G^2}{\lambda \varepsilon})$.
\end{Thm}

The inter-epoch use of standard gradient decent is analyzed using the following Lemma from \cite{Zinkevich03}:
\begin{Lem}[Zinkevich \cite{Zinkevich03}] \label{lem:zink}
Let $D = \|\x^* - \x_1\|_2$ and $\|\hatg_t\|\leq G$. Apply $T$ iterations of
the update $\x_{t+1} = \prod_{\K}\left\{\x_t - {\eta}  \hatg_t \right\}$. Then
$$\sum_{t=1}^T \hatg_t\cdot(\x_t - \x^*)  \leq  \eta  G^2 + \frac{D^2 }{\eta T}. $$
\end{Lem}

Now, if we set $\hatg_t$ to be the unbiased estimator of a subgradient $\g_t$
of $f$ at $\x_t$, then by the convexity of $f$, we get
$$f(\x_t) - f(\x^*)\ \leq\ \g_t\cdot (\x_t - \x^*) = \E_{t-1}[\hatg_t\cdot (\x_t - \x^*)],$$
where $\E_{t-1}[\cdot]$ denotes expectation conditioned on all the randomness
up to round $t-1$. This immediately implies the following:

\begin{Lem} \label{thm:batchzink}
Let $D = \|\x^* - \x_1\|_2$. Apply $T$ iterations of the update $\x_{t+1} =
\prod_{\K}\left\{\x_t - {\eta}  \hatg_t \right\} $, where $\hatg_t$ is an
unbiased estimator for the (sub)gradient of $f$ at $\x_t$ satisfying
$\|\hatg_t\|\leq G$. Then
$$ \frac{1}{T} \E[ \sum_{t=1}^T f({\x}_t)] - f(\x^*)  \leq  \eta  G^2 + \frac{D^2 }{\eta T } . $$
By convexity of $f$, we have the same bound for $ \E[ f(\bar{\x})] - f(\x^*)  $, where $\bar{\x} = \frac{1}{T} \sum_{t=1}^T \x_t$.
\end{Lem}
Define $\Delta_k = f({\x}_1^k) - f(\x^*)$. Using Theorem~\ref{thm:batchzink} we
prove the following key lemma:
\begin{Lem} \label{lem:boundOnQ}
For any $k$, we have $\E[\Delta_k] \leq V_k$.
\end{Lem}
\begin{proof}
We prove this by induction on $k$. The claim is true for $k = 1$ since $\Delta_k \leq M$. Assume that $\E[\Delta_k] \leq V_k$ for some $k \geq 1$ and now we prove it for $k+1$. For a random variable $X$ measurable w.r.t. the randomness defined up to epoch $k+1$, let $\E_k[X]$ denote its expectation conditioned on all the randomness up to phase $k$.
By Lemma \ref{thm:batchzink} we have
\begin{eqnarray*}
\E_k[ f(\x_1^{k+1})] - f(\x^*) & \leq  \eta_k  G^2 + \frac{ \|\x_1^k - \x^* \|^2}{\eta_k T_k}  \\
 & \leq  \eta_k  G^2 + \frac{ \Delta_k }{\eta_k  T_k  \lambda}  &\mbox{(by $\lambda$-strong convexity)}
\end{eqnarray*}
and hence,
$$\E[\Delta_{k+1}]\ \leq\  \eta_k G^2 + \frac{ \E[\Delta_k] }{\eta_k T_k \lambda} \ \leq\ \eta_k  G^2 + \frac{ V_k }{\eta_k T_k \lambda} \ \leq\ \frac{V_k}{2}\ =\ V_{k+1},$$
as required. The second inequality uses the induction hypothesis, and the last
inequality and equality use the definition of $V_k$ and the values  $\eta_k = \frac{{V_k}}{4 G^2} $ and $T_k = \lceil\frac{ 16 G^2 }{\lambda V_k}\rceil$.

\end{proof}

We can now prove our main theorem:
\begin{proof}[Proof of Theorem~\ref{thm:main}.]
By the previous claim, taking $k = \lceil \log_2 \frac{M}{\varepsilon}\rceil$
we have
$$ \E[ f(\x^{k+1}_1)] - f(\x^*)\ =\ \E[ \Delta_{k+1}]\ \leq\ V_{k+1}\ =\ \frac{M}{2^k}\ \leq\ \varepsilon,$$
as claimed.

To compute the total number of gradient updates, we sum up along the epochs: in
each epoch $k$ we have $T_k=\lceil \frac{ 16 G^2   }{\lambda V_k}\rceil$ gradient updates,
for a total of
$$ \sum_{k=1}^{\lceil \log_2 \frac{M}{\varepsilon} \rceil}  \left\lceil\frac{16G^2}{\lambda V_k}\right\rceil\ \leq\ \sum_{k=1}^{\lceil \log_2\frac{M}{\varepsilon} \rceil}  \frac{16G^2 \cdot 2^{k-1}}{\lambda M} + 1\ \leq\ \frac{20G^2}{\lambda
\epsilon},$$
assuming that $\lceil \log_2 \frac{M}{\varepsilon} \rceil \leq
\frac{2G^2}{\lambda \varepsilon}$.
\end{proof}

\section{Conclusions}

Extension of the above result to stochastic optimization of strongly convex functions with respect to norms other than the Euclidean norm are straightforward via standard online learning techniques. A factor two speedup can be obtained by stoping the epoch at a random point.

We thank Nati Srebro for bringing the problem of deriving an efficient attention algorithm for stochastic strongly-convex optimization to our attention.

\bibliographystyle{plain}
\bibliography{svm}
\appendix

\section{High probability bounds}

We briefly sketch how using essentially the same algorithm with slightly more
iterations, we can get a high probability guarantee on the quality of the
solution. The update in line 6 requires a projection onto a smaller set, and
becomes
$$ \text{6: Update }\ \x_{t+1}^k = \prod_{\K \cap B_{V_k} (\x_1^k)}\left\{\x_t^k - {\eta_k} \hatg_t  \right\} $$
Here $B_r(\x)$ denotes the $L_2$ ball of radius $r$ around the point $x$. We
assume that such a projection can be computed very efficiently. In particular,
if $\K = \mathbb{R}^n$, then the projection is simply a scaling down of the
vector towards the center of the ball.

We prove:
\begin{Thm} \label{thm:main2}
The final point $\x_1^{k+1}$ returned by the modified {\it Epoch-GD} algorithm,
with parameters $T_k = \left\lceil\frac{100 G^2 \ln(1/\deltil)}{\lambda
V_k}\right\rceil$, $\eta_k = \frac{V_k}{10 G^2}$, where $\deltil = \frac{\delta }{4 \log \frac{M}{\varepsilon}}$, has the property that $f(\x_1^{k+1})
- f(\x^\star) \leq \varepsilon$ with probability at least $1-\delta$. The total
number of gradient updates is $O(\frac{G^2\log(1/\deltil)}{\lambda \varepsilon})$.
\end{Thm}

The following Lemma is analogous to Lemma \ref{thm:batchzink}, but provides a
high probability guarantee.
\begin{Lem} \label{lem:highprobGD}
Let $D = \|\x^* - \x_1\|_2$. Apply $T$ iterations of the update $\x_{t+1} =
\prod_{\K \cap B_{D}(\x_1) }\left\{\x_t - {\eta} \hatg_t \right\} $, where
$\hatg_t$ is an unbiased estimator for the (sub)gradient of $f$ at $\x_t$
satisfying $\|\hatg_t\|\leq G$. Then with probability at least $1-\deltil$
$$\frac{1}{T}\sum_{t=1}^T f(\x_t) - f(\x^*)\  \leq\ \eta G^2 +  \frac{D^2 }{\eta T }  + \frac{8GD \sqrt{\ln(1/\deltil)}}{\sqrt{T}}.$$
By the convexity of $f$, the same bound holds for $f(\bar{\x}) - f(\x^*)$,
where $\bar{\x} = \frac{1}{T} \sum_{t=1}^T \x_t$.
\end{Lem}
\begin{proof}
Let $\g_t = \E_{t-1}[\hatg_t]$, a subgradient of $f$ at $\x_t$, where
$\E_{t-1}[\cdot]$ denotes the expectation conditioned on all randomness up to
round $t-1$. Consider the martingale difference sequence given by
$$X_t\ =\ \g_t\cdot (\x_t - \x^*) - \hatg_t \cdot (\x_t - \x^*).$$
We can bound $|X_t|$ as follows:
$$|X_t| \leq \| \hatg_t \| \| \x_t - \x^*\| + \E_{t-1}[\| \hatg_t \|] \| \x_t - \x^*\|\ \leq\ 4 G D,$$
where the last inequality uses the fact that $\x_t \in B_{D}(\x_1)$, and
hence by the triangle inequality $\|\x_t - \x^*\| \leq \| \x_t - \x_1 \| +
\|\x_1 - \x^*\| \leq 2D$.

By Azuma's inequality (see Lemma~\ref{lem:azuma}), with probability at least
$1-\deltil$, the following holds:
\begin{equation} \label{eq:azuma}
\frac{1}{T}\sum_{t=1}^T  \g_t\cdot (\x_t -
\x^*) - \frac{1}{T}  \sum_{t=1}^T \hatg_t\cdot (\x_t - \x^*)\  \leq\  \frac{8GD
\sqrt{\ln(1/\deltil)}}{\sqrt{T}}.
\end{equation}
 Note that by the convexity of $f$, we have
$f(\x_t) - f(\x^*) \leq \g_t\cdot(\x_t - \x^*)$. Then, by using
Lemma~\ref{lem:zink} and inequality~(\ref{eq:azuma}), we get the claimed
bound.
\end{proof}

We can now proceed along the same lines as Theorem \ref{thm:main} and prove the same result with high probability, the derivation is completely analoguous.
\begin{Lem} \label{lem:hiprobboundOnQ}
For an appropriate choice of $\eta_k, T_k$, the following holds. For any $k$,
with probability $(1-\deltil)^k$ we have $\Delta_k \leq V_k$.
\end{Lem}
\begin{proof}
We prove this by induction on $k$. The claim is true for $k = 1$ since
$\Delta_k \leq M$. Assume that $\Delta_k \leq V_k$ for some $k \geq 1$ with
probability at least $(1 - \deltil)^k$ and now we prove it for $k+1$. We
condition on the event that $\Delta_k \leq V_k$. By Lemma \ref{lem:highprobGD},
we have with probability at least $1-\deltil$,
\begin{align*}
\Delta_{k+1} & = f(\x_1^{k+1}) - f(\x^*) \\
& \leq  \eta_k  G^2 + \frac{ \|\x_1^k - \x^* \|^2}{\eta_k T_k} + \frac{ 8G \|\x_1^k - \x^*\| \sqrt{\ln(1/\deltil)}}{\sqrt{T_k}} & \mbox{(by Lemma \ref{lem:highprobGD})}\\
 & \leq  \eta_k  G^2 + \frac{ \Delta_k }{\eta_k  T_k  \lambda} + \frac{ 8G \sqrt{\Delta_k} \sqrt{\ln(1/\deltil)}}{\sqrt{\lambda T_k}} &\mbox{(by $\lambda$-strong convexity)} \\
 & \leq  \eta_k  G^2 + \frac{ V_k }{\eta_k  T_k  \lambda} + \frac{ 8G \sqrt{V_k} \sqrt{\ln(1/\deltil)}}{\sqrt{\lambda T_k}} &\mbox{(by the conditioning)}
 \end{align*}
Let $T_k = \left\lceil\frac{100 G^2 \ln(1/\deltil) }{\lambda V_k}\right\rceil$,
and we get
$$f(\x_1^{k+1}) - f(\x^*)\ \leq \ \eta_k  G^2 +  \frac{1}{\eta_k} \frac{ V_k^2
}{100 G^2  \ln(1/\deltil)} +  \frac {V_k}{10}$$ Next set $\eta_k = \frac{V_k}{10
G^2}$, and we get that
$$\Delta_{k+1} = f(\x_1^{k+1}) - f(\x^*)\ \leq\ \frac{V_k}{10} + \frac{ V_k }{10 \ln(1/\deltil)} +  \frac {V_k}{10} \leq \frac{V_k}{2} =
V_{k+1}.$$ Factoring in the conditioned event, which happens with probability
at least $(1-\deltil)^k$, overall, we get that $\Delta_{k+1} \leq V_{k+1}$ with
probability at least $(1-\deltil)^{k+1}$.
\end{proof}

We can now prove our high probability theorem:
\begin{proof}[Proof of Theorem \ref{thm:main2}]
By the previous claim, taking $k = \lceil \log_2 \frac{M}{\varepsilon}\rceil$
we have with probability at least $(1-\tilde{\delta})^k$ that
$$  f(\x^{k+1}_1) - f(\x^*)\ =\  \Delta_{k+1}\ \leq\ V_{k+1}\ =\ \frac{M}{2^k}\ \leq\ \varepsilon,$$
Since $\tilde{\delta} = \frac{\delta}{4k}$, and hence $(1-\tilde{\delta})^k \geq 1 - \delta$ as needed.

To compute the total number of gradient updates, we sum up along the epochs: in
each epoch $k$ we have $T_k=O(\frac{ G^2 \log(1/\tilde{\delta}) }{\lambda V_k})
$ gradient updates, for a total of
\begin{align*}
\sum_{k=1}^{\lceil \log_2 \frac{M}{\varepsilon} \rceil}  O\left(\frac{ G^2 \log(1/\deltil)}{\lambda V_k}\right)
&= \sum_{k=1}^{\lceil \log_2\frac{M}{\varepsilon} \rceil}  O\left(\frac{ 2^{k-1}G^2 \log(1/\deltil)}{\lambda M}\right)\\
&= O\left( \frac{ G^2 \log({1}/{\deltil})}{\lambda \varepsilon}\right).
\end{align*}
\end{proof}

\subsection{Martingale concentration lemma}
The following inequality is standard in obtaining high probability regret bounds:
\begin{Lem}[Azuma's inequality] \label{lem:azuma}
Let $X_1, \ldots, X_T$ be a martingale difference sequence. Suppose that $|X_t|
\leq b$. Then, for $\delta > 0$, we have
$$ \Pr\left[\sum_{t=1}^T X_t \geq \sqrt{2 b^2 T \ln(1/\delta)}   \right]\ \leq\ \delta.$$
\end{Lem}

\end{document}


TITLE: limit 0 times infinity, rewrite to find the limit
QUESTION [1 upvotes]: I need some help with: $\lim_{x\to 0+} x^3\cdot e^{1/x}$. How to start?
I've tried substitution $(y=1/x)$ without any luck.
I would prefer not to use L'Hopitals rule and apologizes for a bad title line.

REPLY [1 votes]: You can carry on with your substitution. In the case $y = 1/x$, then as $x \to 0^+$, $y \to \infty$, and you want to look at the limit of
$$
\lim_{y\to\infty} (1/y^3) e^y = \lim_{y\to\infty} \frac{e^y}{y^3}.
$$
If you know, for instance, that the exponential grows faster than any polynomial, you can avoid L'Hopital's rule.

REPLY [1 votes]: Substitute $x=\frac1t$. Then this is the same as $$\lim_{t\to+\infty}\frac{e^t}{t^3}$$
Using the wellknown inequality $e^x\ge 1+x$, we have $e^t=(e^{t/4})^4\ge (1+\frac t4)^4$, so for $t>0$
$$ \frac{e^t}{t^3}\ge\frac{(1+\frac t4)^4}{t^3}>\frac1{256}t.$$
(In short: Exponential always wins over polynomial)


TITLE: probability - Difference between two statements
QUESTION [1 upvotes]: I have the following scenario.
There are a number of people on an island. Each person has a chance of 0.015 of dying on any given day, independent of other people on the island.
I have two different probabilities for two statements that were calculated by the professor, but I can't see the difference between the meaning of the two statements.
Statement 1: Person A survived the sixth day.
Probability of statement 1:
$P=(1-0.015)^5\cdot 0.015=0.0139$ (probability person A died on the 6-th day)
$P(Statement 1)=1-0.0139=0.986$
Statement 2: Person A survived for 6 straight days.
Probability of statement 2:
$P(Statement 2)=0.985^6=0.913$
I understand how each probability was calculated, but I don't see the difference between each statement logically. Please explain what is the difference, or give an example when statement 1 occurs but statement 2 doesn't, or vice versa.
Thank you.

REPLY [0 votes]: Given a different event (not dying and survivinf) they would be different. But since the event of surviving the 6th day induces surviving previous days, they have the same probability. If event i.e. was being healthy, probabilities would be different imo