TITLE: Sequence: Sup, Inf, LimSup and LimInf
QUESTION [0 upvotes]: Given the sequence: $ a_n = (n^2+3n+1)^{1/n}-1 $
. Find $ \sup(a_n) $ , $ \inf(a_n) $, $ \lim[\sup(a_n)] $ and $ \lim[\inf(a_n)] $ .
So at first I tried to make the set puting some numbers $ n = 1,2,3,4,... $ to see how the set of numbers $ a_n $ behaves:
$ Set{(a_n)} = \{4, \sqrt{11}-1, \sqrt[3]{19}-1, \sqrt[4]{29}-1, \dots\} $
It is easy to see that $ \sup(a_n) = 4 $ and $ \inf(a_n) = 0 $ but I cannot find a way to prove it
For instance I tried the Supremum Property:
$ \sup(a_n) = 4 \iff \forallε>0 : \sup(a_n) - ε < a_n \iff 4 - ε < (n^2+3n+1)^{1/n}-1 \iff (5-ε)^n < n^2+3n+1 \iff (1+(4 - ε))^n < n^2+3n+1 $
Then in the left part of the above inequality I tried Bernoulli's inequality which working it a little bit gave me that: $ n^2+(ε-1)n \ge 0 $
I am not sure though how to continue. Also I am not sure about whether $ \sup(a_n) = \lim[\sup(a_n)] $ and same for the $ \inf(a_n) $. Can someone help me ?

REPLY [0 votes]: For each $n\in\Bbb N\setminus\{1\}$, you have $1\leqslant n^2+3n+1\leqslant n^3$, and therefore$$0=\sqrt[n]1^2-1\leqslant\sqrt[n]{n^2+3n+1}-1\leqslant\sqrt[n]n^3-1.$$So, since $\lim_{n\to\infty}\sqrt[n]n=1$, you have$$\lim_{n\to\infty}\sqrt[n]{n^2+3n+1}-1=0.$$So,$$\limsup_{n\in\Bbb N}\sqrt[n]{n^2+3n+1}-1=\liminf_{n\in\Bbb N}\sqrt[n]{n^2+3n+1}-1=0$$and, since it's a succession of non-negative numbers, $\inf\left\{\sqrt[n]{n^2+3n+1}-1\,\middle|\,n\in\Bbb N\right\}=0$. And it's not hard to see that $\sup\left\{\sqrt[n]{n^2+3n+1}-1\,\middle|\,n\in\Bbb N\right\}=5-1=4$.


TITLE: Picking up children from kindergarten - Conditional probability
QUESTION [2 upvotes]: In the evening, every child in a kindergarten is independently picked up by its parents. The mother will pick up the child with a probability of $40\%$ and the father will pick up the child with a probability of $60\%$. If the mother decides to pick up the child, there is a $15\%$ chance that she will be late. If the father decides to pick up the child, there is a $20\%$ chance that he will be late.
(a) What is the probability that a child will be picked up late from kindergarten?
(b) If a particular child is not picked up from kindergarten too late, what is the probability  of its father picking it up?
(c) Given that two different children are picked up by their fathers, what is the probability that both children will be picked up late?

We define the events:
$M:$ Child is picked up by its mother.
$F:$ Child is picked up by its father.
$L:$ Child is picked up late.
(a) $P(L) = P(M) \cdot P(L | M) \cdot P(F) \cdot P(L | F) = 0.4 \cdot 0.15 + 0.6 \cdot 0.2 = 0.18.$
(b) $P(F | L^c) = \dfrac{P(F) \cdot P(L^c | F)}{1-P(L)} = 0.585.$
(c) Since each child is picked up independently from other children: $P(L | F)^2 = (0.6 \cdot 0.2)^2 = 0.0144.$

My book however gives $0.04$ as an answer for $c$. What's wrong with my solution?

REPLY [1 votes]: You write that:
"(c) Since each child is picked up independently from other children: $P(L|F)^2=(0.6⋅0.2)^2=0.0144$."
What's wrong: It is given that the $2$ children are picked up by their fathers. You do not need to multiply by $0.6$, the probability that a child is picked up by their father.
So the answer is $0.2^2 = 0.04$.


TITLE: How to write the inverse matrix $A^{-1}$ as a polynomial in $A$?
QUESTION [6 upvotes]: I came a cross a question that gives a matrix $A$, and asks to write $A^{-1}$ as a polynomial in $A$ with real coefficients.
I don't know what this means, and googling didn't clarify very much. I found the inverse $A$, but I do not know what it means to write it as a polynomial in $A$.

REPLY [4 votes]: This can be done for any invertible $A$ using the Cayley–Hamilton theorem that any matrix satisfies its own characteristic equation. If $a_i$ are the coefficients of the characteristic polynomial then $a_0I + a_1A + \cdots + a_nA^n=0$ with $a_0=\det A$. Hence
$$A^{-1}=-\frac1{a_0}(a_1I + \cdots + a_nA^{n-1}).$$
You can find the characteristic polynomial by computing the determinant $\det(A-\lambda I)=a_0 + a_1\lambda + \cdots + a_n\lambda^n$.


TITLE: How to deal with a sum in the denominator
QUESTION [1 upvotes]: In a book, I saw the following inequality:
$$e^{-\frac{\lambda^2}{2 (\lambda + \Delta)}}\leq e^{- \min\{\lambda, \frac{\lambda^2}{\Delta}\}/4},$$
for some (specific) $\lambda, \Delta \geq 0$
Does this hold for any such $\lambda, \Delta$ or is this a special property of how these two quantities are defined?
In general, how can I work with such quantities, when I have a sum in the denominator and I would like to get only the dominant term (exactly) and the remaining part possibly approximated?

REPLY [1 votes]: Rearranging the original inequality:
$${-\frac{\lambda^2}{2 (\lambda + \Delta)}}\le {- \min\left\{\lambda, \frac{\lambda^2}{\Delta}\right\}\over 4}\\{2\frac{\lambda^2}{(\lambda + \Delta)}}\ge {\min\left\{\lambda, \frac{\lambda^2}{\Delta}\right\}}$$
If $\lambda, \Delta > 0$:
$$\begin{array}{cl}\lambda < \Delta \implies& {2\frac{\lambda^2}{(\lambda + \Delta)}} > {2\frac{\lambda^2}{2\Delta}} = \frac{\lambda^2}{\Delta} = \min\left\{\lambda, \frac{\lambda^2}{\Delta}\right\}\\
\lambda > \Delta \implies& {2\frac{\lambda^2}{(\lambda + \Delta)}} > {2\frac{\lambda^2}{2\lambda}} = \lambda =\min\left\{\lambda, \frac{\lambda^2}{\Delta}\right\}\\
\lambda = \Delta \implies& {2\frac{\lambda^2}{(\lambda + \Delta)}} = \lambda =\min\left\{\lambda, \frac{\lambda^2}{\Delta}\right\}\end{array}$$
The inequality holds for $\lambda, \Delta > 0$.


TITLE: Statistic - different question (person's breathing )
QUESTION [0 upvotes]: Research psychologists are interested in finding out whether a person's breathing patterns are affected by a particular experimental treatment. To determine the general respiratory patterns of the $n=30$ people in the study, the researchers collected some baseline measurements - the total ventilation in liters of air per minute adjusted for body size - for each person before the treatment. The data are shown here:
$$\begin{matrix}
5.23 & 7.79 & 5.83 & 5.37 & 4.35 & 5.54 & 6.04 & 5.48 & 6.58 & 4.82 \\
5.92 & 5.38 & 6.34 & 5.12 & 5.14 & 4.72 & 5.17 & 4.99 & 4.51 & 5.70 \\
4.67 & 5.77 & 5.84 & 6.19 & 5.58 & 5.72 & 5.16 & 5.32 & 4.96 & 5.63
\end{matrix}$$
I want to find the sample mean, median and standard deviation, then I want to construct a stem-and-leaf plot. How could I describe the shape of the distribution? How large or small does a ventilation measurement have to be before it is considered unusual?
thanks for your help :)
I use Mathematica and I obtained :
$$\begin{array}{c|c}
\mbox{Stem} & \mbox{Leaves} \\ \hline
4 & 45778\\
5 & 00112223445566778889\\
6 & 0236\\
7 & 8
\end{array}$$
but my teacher said is not ok!
can you help me please:)

REPLY [1 votes]: The mean/average is simply: $\dfrac{\Sigma x}{n}$
To get the median, you need to sort the data from smallest to largest and find the $\dfrac{n+1}{2}th$ term. Since you will get $15.5th$ term, find the mean/average of the $15th$ and $16th$ term.
The standard deviation is given by: $\sqrt{\dfrac{\Sigma x^2}{n} - \left(\dfrac{\Sigma x}{n}\right)^2}$.
As for the stem & leaf diagram, I would make the units as the stem, and the decimals as the leaves.
For instance, the first row will be like this:
4 | 45 82
5 | 23 37 48 54 83
6 | 04 58
7 | 79

How to fill it:

The first number is 5.23, so on the branch with "5", put "23" next to it.
Next comes 7.79, put "79" next to "7".
Third comes 5.83. Now, you will need to put this in the branch "5", and there, you already have "23". Put it after "23" since $83>23$.
Next is 5.37. This goes in the branch "5" as well, and since $23<37<83$, you will put "37" between "23" and "83".

On a computer it's easy to fill in it, but on paper, you will have to sort the sample from smallest to largest, so that you don't miss smaller amounts. In the above case on paper, you would have to cut/erase 83 to make room for 37.

REPLY [1 votes]: For Sample mean, just add up all the observations and divide the result by the total number of observations.
For Sample median, sort the data and then find the average of $\dfrac{n}{2}$th term and $\left(\dfrac{n}2{+1}\right)$th term (Since the number of observations is even).
For Sample standard deviation, apply the formula,
$$S=\sqrt{\frac{\sum_{i=1}^n(X_i-\bar{X})^2}{n-1}}$$
Where $S$ is the Sample standard deviation, $X_i$ is the $i$th observation, $\bar{X}$ is sample mean and $n$ is the number of observations.
For the shape of the distribution, you can find the measure of skewness and kurtosis.
The last part is $\text{Sample Mean}\pm\text{Sample Standard Deviation}$.