TITLE: ""Proof""check for an inequality involving the exponential function
QUESTION [1 upvotes]: This is a trivial question but I am having trouble to see what wehther my argumentation is valid or if I did some sign error. I suspect I messed somethin up again.
I start with the inequality
$$
1- \exp(-f(t)^2) \leq Ct \leq \frac{1}{2}
$$
for $t \leq \frac{1}{2C}$, $C>0$ and $f$ continuous. What I want to get is some inequality like
$$
f(t)^2 \leq 2Ct.
$$
on some reasonable interval $(0,t_0)$.
To this end, I subtract 1 on both sides of the first equation, divide by -1 and take the logarithm on both sides. Then I get
$$
-f(t)^2 \geq \ln(1-Ct).
$$
again dividing by -1 gives
$$
f(t)^2 \leq -\ln(1-Ct).
$$
Now wolframalpha says that
$$
-\ln(1-Ct)<2 Ct
$$
whenever
$$
0<x<1/2 (W(-2/e^2) + 2).
$$
I don't know what the function $W$ is but it works for me. Can I now conclude that
$$
f(t)^2 \leq 2Ct
$$
on the interval $(0,min(\frac{1}{2C},1/2 (W(-2/e^2) + 2))$? Also if it is true, is there some inequality to justify wolframalphas findings? Is it just the Taylor expansion of $\ln$?

REPLY [1 votes]: You dropped $C$ somewhere in your calculations that you should not have. There is no way that in a function where $t$ only shows up multiplied by $C$, any upper limit is not going to involve $C$ as well. Yet your expression involving $W$ has no $C$ in it.

Let's look carefully at where the inequality $- \ln(1 - Ct) < 2Ct$ holds. First, we can rewrite it as $$0 < \ln(1 - Ct) + 2Ct$$
Let $g(t) = \ln(1 - Ct) + 2Ct$. Then $$g'(t) = 2C - \dfrac{C}{1-Ct}$$
Note that $g(0) = 0$ and $g'(0) = C > 0$. Further, setting $g'(0) = 0$ gives $t = \frac 1{2C}$, exactly the upper limit you had set. Thus $g'(t) > 0$ on the interval $\left(0, \frac 1{2C}\right)$, and so $g$ is strictly increasing on this interval. Since $g(0) = 0$, we have that $g(t) > 0$ on all of $\left(0, \frac 1{2C}\right)$. And since $g$ is still non-decreasing at $\frac 1{2C}$, we also have $$g\left(\frac 1{2C}\right) \ge g\left(\frac 1{4C}\right) > 0$$
Which means
$$0 < \ln(1−Ct)+2Ct\\ -\ln(1−Ct) < 2Ct\\f(t)^2 < 2Ct$$ on
$\left(0, \frac 1{2C}\right]$.


TITLE: Show up to equivalence that there is only one group action of $\mathbb{Z}_{3}$ on $X=\left \{ 1,2,3 \right \}$.
QUESTION [1 upvotes]: Show up to equivalence that there is only one group action of $\mathbb{Z}_{3}$ on $X=\left \{ 1,2,3 \right \}$.

I am not sure where to begin with such a proof. I would assume it is proof by contradiction and I use the definition of equivalent actions? Is there an easier way, because I am not very knowledgable on equivalent actions and how to apply this to the proof! Any help and guidance is appreciated.

REPLY [1 votes]: Remember that $\mathbb{Z}_3$ is cyclic generated by some element g. So any action on X is determined by what g$\cdot$X is, with the added condition that g$^3$ = $e$. Also, I think you mean there's one non-trivial action.
To show that two actions $\phi_1,\phi_2$ are equivalent, you will have to show  that there is some bijection f:X$\rightarrow$X such that $\phi_1$(g,x)=$\phi_2$(g,f(x)) $\forall g \in G$.

REPLY [1 votes]: This is false although close to truth. There are two possible group actions on $\{1,2,3\}$.
By $G$-set I will understand $(X,\cdot)$ pair where $X$ is a set and $\cdot:G\times X\to X$ is a group action.
First of all note that any orbit is isomorphic to $G/H$ as a $G$-set for some subgroup $H$. Since $G=\mathbb{Z}_3$ has only two possible subgroups, namely $G$ and $0$ then there are only two possible orbits: $G/0$ and $G/G$. Non-isomorphic because of different cardinality.
Now since every $G$-set is a disjoint union of orbits (up to $G$-isomorphism) than $X=\{1,2,3\}$ has two possible $G$-set structures: namely $X\simeq G/0$ or the trivial one $X\simeq G/G\sqcup G/G\sqcup G/G$.