For which $n\geq2$ does there exist $\sigma \in A_n$ with $|\sigma|&gt;n$?  I have found this same question , But I have not been able to understand how it works. On my own I came to the conclusion that for $n$ less than $8$ the result is not satisfied, but for $n = 8$, I found the permutation $\sigma$ formed by a $5$-cycle and a $3$-cycle. Which is in $A_n$ and $|\sigma|=15&gt;8$. In particular, I cannot understand how the proof follows, to conclude that the result is true for all $n\geq 8$. Could you help me? If there is an easier way to see this, please, I would appreciate if you shared it.
For which $n≥2$ does there exist $\sigma\in A_n$ with $|\sigma| &gt; n$? I have a proof that I'd like to be verified. My proof that there exists $\sigma\in A_n$ with $|\sigma| &gt; n$ if and only if $n\geq 8$: For $n&lt;8$, one can manually verify for each $n$ that there is no $\sigma\in A_n$ with order greater than $n$. Suppose $n\geq 8$. Case 1: $n$ is even and $3\not\mid n$. Then the permutation $(1,2,3)(4,\dots,n)$ is even and has order $\operatorname{lcm}(3,n-3)=3n-9\geq n+16-9\geq n + 7$. Case 2: $n$ is even and $3\mid n$. Then $n\geq 12$, so $n-2\geq 8$ is even and is not divisible by 3. Applying Case 1, we see that $(1,2,3)(4,\dots,n-2)\in A_n$ has order at least $n+5$. Case 3: $n$ is odd. Then apply one of the previous two cases with $n-1$ to get an even permutation with order at least $n+4$. $\blacksquare$ Is my proof correct? It doesn't seem very elegant. Is there some slicker proof?
Finishing a proof: $f$ is injective if and only if it has a left inverse I've already done a lot of searching (in particular: ) to try to prove this statement:      $f: A \to B$ is injective if and only if it has a left inverse. [I'm going to also assume $A$ and $B$ are nonempty.]   My attempt so far:     $\Leftarrow$ Suppose $f$ has a left inverse and $f(a_1) = f(a_2)$ for    $a_1, a_2 \in A$. Then there is a function $g: B \to A$ such that   $g\left(f(a_1)\right) = a_1$ and $g\left(f(a_2)\right) = a_2$. Since   $g$ is well-defined, it follows that $g\left(f(a_1)\right) =  g\left(f(a_2)\right)$, and hence $a_1 = a_2$.         $\Rightarrow$Now   suppose $f$ is injective. Then $f(a_1) = f(a_2) \implies a_1 = a_2$.   Furthermore, since $A$ is nonempty, there is an element $x \in A$.   Define $g: B \to A$ by  \begin{equation*} g(b) = \begin{cases}  f^{-1}(b), &amp; b \in \text{Im}(f) \\ x, &amp; b \in A \setminus  \text{Im}(f)\text{.} \end{cases} \end{equation*} This mapping is   well-defined, since if $f$ is injective, $f^{-1}(b)$ is unique.   My main question: does this imply that if $f: A \to B$ is injective that any mapping $g: \text{Im}(f) \to A$ is bijective? Otherwise, I don't understand why we can even use $f^{-1}$ on an element. [Please read the link above for more details - in proof 1.] Why is there a clear mirror image off a rough surface at shallow angles? The mirrors we use everyday have a highly reflective smooth surface, so that we can see a clear image.  But I found the same happens on a rough (not so smooth) surface, though only at shallow angles.  For example    this is the logo of physics stackexchange, reflected from the back of an ipad.  Considering that aluminum may contributed to reflection, I tried a different material:    this time a piece of printer paper.  Though not as clear, there is still some visible pattern resembling the logo.  The same effect can be observed on highways, where distant vehicles have a mirror image off the road; I do not have a picture at hand however.  So, why is this happening, and why only on shallow angles? Intellisense in IntelliJ no longer working This question has been asked before, at least twice by people who were trying to cheat on the PROMYS Europe 2020 Application, and once in a  because the OP showed no effort. I haven't been able to solve it, so I'm going to to ask it again.  PROMYS applications have closed, and I hope the effort shown below is enough to keep this question from being closed.     The Mathematical Forest is grown in a two-dimensional plane, where   trees can only grow on points with integer coordinates. To start with,   there are no trees at all. The foresters plant the first tree at   $(0,0)$. Each year, they carry out tree planting according to the   following rule. If there is a tree on the point (𝑚,𝑛) but there are   no trees on the points $(m+1,n)$ and $(m,n+1)$, then they can choose   to remove the tree on $(m,n)$ and plant new trees on the points   $(m,n+1)$ and $(m+1,n)$. For an integer $k\geq1$, the $k$th diagonal   consists of all points $(m,n)$ with $m+n=k−1$. Is it possible for the   foresters to arrange their planting so that eventually there are no   trees on the first $2$ diagonals? What about the first $3$ diagonals?    $4$ diagonals? Can you generalize?   It's easy to get to a position with no trees on the first two diagonals.  (It only takes $4$ plantings.)  I can prove it's impossible to reach a position with no trees on the first $4$ diagonals, and I believe it's impossible to reach a position with no trees on the first $3$ diagonals, but I can't prove it.  For the $k=4$ case, I used a pagoda function, as in a peg solitaire problem.  For $k=1,2,3,\dots$ define the potential of a tree on the $k$th diagonal as $\frac1{2^{k-1}}$, and the potential of the forest as the sum of the potentials of all the trees in it.  When a tree of potential $\frac1{2^{k-1}}$ is removed, it is replaced by $2$ trees of potential $\frac1{2^k}$ so the potential of the forest never changes.  Initially, the potential of the forest is $1$.  Since there are $k$ trees on diagonal $k$, the potential of all points beyond the fourth diagonal is $$\sum_{k=5}^\infty\frac k{2^{k-1}}=\frac34&lt;1,$$ so it is impossible that there are no trees on the first $4$ diagonals.  The potential of the fourth diagonal is $\frac12$, so this argument doesn't show that it's impossible to have no trees on the first $3$ diagonals.  I've done a lot of experimenting, though, and I believe that the statement is true.  (I even wrote a little computer game so I could experiment quickly.)  The best I've been able to do is to get down to one man on the third diagonal.   The black circles are trees eligible for removal, and the gray circles are other trees.  You can see that there is a "traffic jam" in front of the lone tree on diagonal $3$, and there seems to be no way to clear it.  I've been trying to come up with an argument that reflects this traffic jam, but I haven't been at all successful. We know that after $n$ removals there are $n+1$ trees, and I've been trying to prove somehow that they can't all be sufficiently far from the origin to allow all trees to get past diagonal $3$, but I haven't come close.   In the diagram above, if the tree at $(1,1)$ were moved to $(0,2)$, and the tree at $(1,2)$ were moved to $(2,0)$, then it would be possible to remove the tree at $(0,2)$, emptying the second diagonal.  In the fictitious position, we still have one tree on diagonal $2$ and two on diagonal $3$.   So, I think an argument along the lines I was trying cannot just be concerned with the distance from the origin; it must somehow take into account how the trees interfere with one another.  I haven't been able to do this.    I'd be grateful for solutions, hints, or counterexamples.  I wasn't really sure what tags to attach.  Please add any that seem appropriate. Prove there's a simple path of length $k$ in a simple graph $G$ where all the vertices have degree of at least $k$.   Relevant definitions:  $G$ is a simple graph that consists of a vertex set $V(G) = \{v_1, v_2, ..., v_n\}$ and an edge set $E(G) = \{e_1, e_2, ..., e_m\}$ where each edge is an ordered pair of vertices. The edge $\{u,v\}$ is denoted $uv$.  A walk of length $k$ is a sequence $v_0,e_1,v_1,e_2,...,e_k,v_k$ of vertices and edges such that $e_i=v_{i-1}v_i$ for all $i$. A path is a walk with no repeated vertex.   My attempt:  Induction, for $k=1$ it's obvious.  Suppose for $k-1$ and we'll prove for $k$.  Let $G$ be a simple graph such that $\forall v \in V : d(v)\ge k$.  We can assume that there are at least $k+1$ vertices since there are $k$ neighbours to every vertex.  Let $v_0$ be some vertex, it has $k$ neighbours, we'll move to one of its neighbours say $v_1$ and remove $v_0$. So now there are $k$ vertices with a degree of at least $k-1$ and from the induction hypothesis we'll have a path of length $k$.  Is using only $k$ vertices from $n$ right? Does it keep the generality when using only all the neighbours of $v_0$? Comparison between integrals Let's say - $F(x) &gt; G(x)$ for all $x$ in $E$.  So is it true that ?  $$\int F(x)\,dx  &gt; \int G(x)\,dx$$ or I can use this statement only when the two functions are non-negative functions.  Thanks! Radius of convergence of $\sum_{n=0}^\infty n!x^{n^2}$ The question is to calculate the radius of convergence of $\sum\limits_{n=0}^\infty n!x^{n^2}$.  I dont know how to calculate R.O.C for such type of questions involving $n^2$ as power of x. How do we do this? Any help appreciated. Short sci-fi story about Earth blockaded by other human colonies Like the subject: short sci-fi story about Earth blockaded by other human colonies.  The best humans have colonized some stellar systems. But they begin to treat humans of earth like inferiors. A war began, colonies won the war and blockaded the earth, but despite this the earth won. Who is the author? Lets say I have a function defined as $f(x) = \sqrt {x^2}$. Common knowledge of square roots tells you to simplify to $f(x) = x$ (we'll call that $g(x)$) which may be the same problem, but it isn't the same equation. For example, say I put $-1$ into them:  $\begin{align} f(x) &amp;= \sqrt {x^2} \\ f(-1) &amp;= \sqrt {(-1)^2} \\ f(-1) &amp;= \sqrt {1} \\ f(-1) &amp;= 1 \end{align}$  $\begin{align} g(x) &amp;= x \\ g(-1) &amp;= -1 \end{align}$  thereby, we conclude that $f(x)$ and $g(x)$ do not produce the same results even though they are mathematically the same. This is also shown when we try to graph the functions:  :    :    :    From this, we can see that given $f(x) = \sqrt {x^2}$, when simplified is not the same as $f(x) = x$. So, is there any way to prove that $y = \sqrt {x^2}$ is not the same as $y = x$ for negative values, but is infact the same as $y = \mid x \mid$? This is a  about Hosting multiple SSL websites on the same IP.   I was under the impression that each SSL Certificate required it's own unique IP Address/Port combination. But the  is at odds with this claim.  Using information from that Question, I was able to get multiple SSL certificates to work on the same IP address and on port 443. I am very confused as to why this works given the assumption above and reinforced by others that each SSL domain website on the same server requires its own IP/Port.  I am suspicious that I did something wrong.  Can multiple SSL Certificates be used this way? How to keep rice in a biryani from burning and sticking together? The bottom part of my  Biryani always  gets burnt, rice always gets sticky but I want each grain separated. How can I prevent burning &amp; stickiness of rice? Joining non-spatial CSV file to spatial data (shapefile) using QGIS? I have a CSV file which has postcode data in a field, but it is non-spatial e.g. no Easting and Northing.   I have a shapefile with postcode in a field and some additional data.  I want to join the CSV with the shapefile based on attributes (postcode fields). I know I can do this in ArcMap, but how can I achieve the same thing in QGIS? I'm trying to run an interactive command through paramiko.  The cmd execution tries to prompt for a password but I do not know how to supply the password through paramiko's exec_command and the execution hangs.  Is there a way to send values to the terminal if a cmd execution expects input interactively?  ssh = paramiko.SSHClient() ssh.connect(server, username=username, password=password) ssh_stdin, ssh_stdout, ssh_stderr = ssh.exec_command("psql -U factory -d factory -f /tmp/data.sql")   Does anyone know how this can addressed? Thank you. How to show similar display of codes of different programming languages in LaTeX I am very new to LaTeX and I am writing my thesis. In a chapter, I want to show my codes, which are in Mathematica.   I use the listings package as follows:  \documentclass{article}   \usepackage{listings}   \begin{document}   \lstset{language=Mathematica}    \begin{lstlisting}   Do[i^j,{i,1,10},{j,1,10}]   \end{lstlisting}    \end{document}   However, the output is much different to what the code looks like in the Mathematica IDE (even if I use other languages in \lstset{language=&lt;name&gt;}):    I read some questions on the site but, in some of them, the font or color of the codes were not the same. Could you please help me to display something like the picture below in LaTeX?  If it is important, I am using the BiDi package. In the simple linear regression case $y=\beta_0+\beta_1x$, you can derive the least square estimator $\hat\beta_1=\frac{\sum(x_i-\bar x)(y_i-\bar y)}{\sum(x_i-\bar x)^2}$ such that you don't have to know $\hat\beta_0$ to estimate $\hat\beta_1$  Suppose I have $y=\beta_1x_1+\beta_2x_2$, how do I derive $\hat\beta_1$ without estimating $\hat\beta_2$? or is this not possible? Do stone quarries respawn in Hearthfire? XPath - Get node with no child of specific type Possible cardinality of power sets What cardinals can provably not occur as the cardinality of a power set? I know that $\mathbb N$ and natural numbers that are not powers of two are such cardinals. What else is out there? Evaluating $ \lim_{x \rightarrow 0}\left (\frac{\sin(x)}{x}\right)^\frac{1}{x^2}$ Command with percent symbols not running in crontab The following crontab does not work:  # TEST LINE DOES DOT RUN */1 * * * * /bin/echo 'test '`/bin/date +%Y-%m-%d` &gt;&gt; /tmp/test   I also tried starting it with:  SHELL=/bin/bash   Update: I thought the backtick characters ` were the villains, but as the answer below clarifies the percentage % was the culprit!