TITLE: Proof that there will always be data within 1 SD
QUESTION [0 upvotes]: So we just started stats at school, and our teacher told us that no matter the data, no matter how distorted or weird it is, there will always be data within 1 standard deviation of the mean. Is this true?? Any way to prove this?

REPLY [0 votes]: The "most extreme" distributions would be ones containing $2N$ data points , half of which have a value of $a$ and the other half a value of $b$
The mean of such a distribution is $\mu = \frac12(a+b)$
the standard deviation is $\sigma = \frac12 |b-a|$
So for this distribution every point is exactly one standard deviation from the mean


TITLE: Expected Value of a sum of a sequence
QUESTION [0 upvotes]: I have encountered this problem and don't know even where to start.
Let $ M = \{s_1,s_2,s_3...s_k : s_i \in \{1,...,i\}\} \subseteq \{1,...,k\}^k $
We select a sequence from M in a uniform distribution.
What is the expected value of the sum of elements in the chosen sequence?
Besides that I know that I have to define a Random Variable such that $ f(a) = sum$, I am pretty stuck.
Thanks in advance.

REPLY [1 votes]: You want to find $E[\Sigma_{i=1}^ks_i]$. Note that, by the linearity of expectation, you have $E[\Sigma_{i=1}^ks_i] = \Sigma_{i=1}^kE[s_i]$. If $s_i$ are randomly selected, then $s_i\sim Uniform(a,b)$ and $E[s_i]=\frac{b-a}{2}$. Thus,
\begin{eqnarray*}
E[\Sigma_{i=1}^ks_i] &=& \Sigma_{i=1}^kE[s_i] \\
&=& \Sigma_{i=1}^k \frac{b-a}{2} \\
&=& \frac{k(b-a)}{2}
\end{eqnarray*}


TITLE: Simple systems are the "smallest"
QUESTION [1 upvotes]: Let $\Phi$ be a root system of a finite reflection group $W$. Let $\Delta$ be a simple system in $\Phi$. I want to prove that $\Delta$ is "the smallest" set which generates $W$, more precisely: There is no proper subset $A$ of $\Delta$ which generates $W$.
I start with the following: suppose well, then we can find $\alpha\in\Delta\setminus A$ such that $s_{\alpha}$ is not needed as a generator. But now? Is the goal to prove that $A=\Delta$ or how to get the contradiction?
Thank you for help.

REPLY [0 votes]: Assume $\Delta$ irreducible, otherwise take components.
The elements of $\Delta$ that are combinations of basic roots from $A$ form a root system   $\Phi'\subset \Phi$. The reflection $(s_{\alpha})_{\alpha \in A}$ generate its Weyl group $W'$. Let $\beta \ne \alpha$, $\beta \in \Delta'$. We have $W' \cdot \beta \subset \Phi'$. Note that $\Phi' \ne \Phi$ since $\alpha$ is not an integral combination of the other basic roots. However, $W\cdot \beta = \Phi$ since $\Delta$ is irreducible. We conclude that $W'\ne W$.


TITLE: What's wrong when a particle is accelerating?
QUESTION [2 upvotes]: This is surely a dumb question, calculus isn't probably inconsistent - but I can't guess the flaw.
Suppose a ball is at rest. Except position, all other derivatives of it is zero.  When I hit it, the position is changing, so there is a velocity; the velocity has gone from zero to nonzero, so there is acceleration; and crazily enough the acceleration has gone from zero to nonzero, so it is changing too (ad infinitum).
One portion is a physics question, that there should be no terms higher than acceleration, I am not asking that. I am asking if I expand the position from its taylor series just before the hit, everything is zero all the time, but it is actually not. Why?

REPLY [5 votes]: Your "mistake" is here:

if I expand the position from its taylor series

The function that describes the movement of the ball is zero for times, smaller than $t_0$, and nonzero for times larger than $t_0$. This is only possible if the function s not analytic, which means that the function either does not have, or is not equal to, its Taylor series.

There's nothing strange about this. For example, you can look at the function $$f(x)=\begin{cases}e^{-\frac1{t}} & x>0\\ 0& x\leq 0\end{cases}$$
which is a continuous, infinitely differentiable function. The Taylor series for this function around $0$ is equal to $0$, but the function itself is not equal to $0$.


TITLE: How to show that a set is convex?
QUESTION [2 upvotes]: My problem is as follows:
Let f(x,y) be a general concave function (i.e. any concave function), and let S be points (x,y) such that f(x,y) greater than or equal to zero.
I understand that this means S is the set of points on or below f(x,y) and >0, and I can show how it is convex in a sketch/graph. I just don't understand how you show it mathematically in a sufficient way using the definition of a convex set.
Any help would be greatly appreciated.

REPLY [2 votes]: If $\xi_0=(x_0,y_0)$ and $\xi_1=(x_1,y_1)$ are two points in $S$ then we must have $f(\xi_0)\geq 0$ and $f(\xi_1)\geq 0$. Let $0\leq t\leq 1$ and consider $\xi_t=(1-t) \xi_0+t\xi_1$. By concavity of $f$ we must have:
$$ f(\xi_t) \geq (1-t) f(\xi_0) + t f(\xi_1) \geq 0$$
so also $\xi_t\in S$.


TITLE: Final step in solving differential equation
QUESTION [4 upvotes]: I can't figure out the last step of the following question:
Show that the general solution of the differential equation: $dy/dx + y\tan x=1$ is given by: $y\sec x=\ln(\sec x + \tan x) + C$
This is what I've worked out so far:
To make things easier let $u(x)=\exp(\int\! \tan(x)\,\mathrm dx)=\sec(x)$
Multyply both sides by $u(x): \sec(x)\frac {dy(x)} {dx} + {\sec(x)}{\tan(x)}y(x)=\sec(x)$
Then substitute $\sec(x)\tan(x)= \frac {d\sec(x)} {dx}$
$\sec(x) \dfrac {dy(x)}{dx}+ \dfrac{d \sec(x)} {dx}y(x)= \sec(x)$
Using the reverse product rule of the LHS:
$\displaystyle\frac d {dx} (\sec(x)y(x))dx= \int \sec(x)\,dx$
then integrating both sides with respect to x
$\int \frac d {dx} (\sec(x)y(x))dx= \int \sec(x)dx$
Therefore:
$\sec(x)y= -\log(\cos(\tfrac x 2)- \sin(\tfrac x 2))+\log(\cos(\tfrac x 2) + \sin(\tfrac x 2)) + C$
Where do I go from here to find the solution (given in the question)? Thanks in advance!

REPLY [4 votes]: You have a slightly unusual but essentially correct version of an integral of $\sec x$. There is an issue of missing absolute value signs, which are also missing in the problem's version, which should have been $\log(|\sec x+\tan x|$).
To see it is (apart from negativity issues) the same as the standard one used in the problem, we show that
$$\frac{\cos(x/2)+\sin(x/2)}{\cos(x/2)-\sin(x/2)}=\sec x+\tan x.$$
To see that this is enough, note that $\log a-\log b=\log(a/b)$.
Multiply top and bottom by $\cos(x/2)+\sin(x/2)$.
At the bottom we get $\cos^2(x/2)-\sin^2(x/2)$, which by a double angle formula is equal to $\cos x$. On top we get $\cos^2(x/2)+\sin^2(x/2)+2\cos(x/2)\sin(x/2)$, which is $1+\sin x$.
So we end up with $\frac{1+\sin x}{\cos x}$, which is $\sec x+\tan x$.

REPLY [1 votes]: Since they're handing you a solution,
$$
y = \frac{\ln|\sec x+\tan x| + C}{\sec x} = (\cos x) \ln|\sec x+\tan x| + C\cos x,
$$
you can diffentiate, finding $dy/dx$, and plug it into the equation in place of $y$, and see if the two sides are equal.
In that way, you verify that it's a solution.  The next question is whether it's general, i.e. there are no other solutions.  That might follow from a theorem stated in a section of a textbook that you recently read.