Normal distribution tricky problem The question goes like this:

The amount of regular petrol purchased every week at a petrol pump in a city follows the normal distribution with mean 50000 litres and standard deviation 10000 litres. The starting supply of petrol is  74000 litres, and there is scheduled weekly delivery of 47000 litres. Find the probability that, after 11 weeks, the supply of petrol will be below 20000 gallons.

Without that weekly information the problem is easy. But, with that information I am not sure which quantity changes. I believe mean must change, but I am not sure about the standard deviation.

The change of supply from week to week is normally distributed, with mean -3000 litres, and a standard deviation of 10000 litres.

The sum of $11$ independent normal variables all with mean $m$ and standard deviation $s$ is again a normal variable with mean $11m$ and standard deviation $\sqrt{11}s$ (this is also true for any other number in place of $11$). Assuming the tank is allowed to go into negative supply (or, more likely, the restocking can be expedited if needed), you can tell from there how much the supply has changed with what probability.

First, take into account the things that are constant: starting with $74000 L$ and gaining $47000 L$ $11$ times.

$$74000 + 11*47000 = 591000 L$$

In order for the station to have $20000 L$ or less, they need to sell at least $571000 L$ over those $11$ weeks. Based on the information we were given, we expect that they will sell $11*50000 = 550000 L$ in $11$ weeks with a standard deviation of $11*10000 = 110000 L$.

From this, we calculate a $Z-$score of about $.19$. Thus the probability of this happening is about $.4247$.


Number of possible inequalities for $n$ variables. Let's say there are two variables $x$ and $y$. Here are the possible inequalities:
(i) $x<y$
(ii) $x>y$
(iii) $x=y$
So, there are $3$ possible inequalities. But let's say there are $n$ variables $x_1,x_2,x_3,x_4,\cdots,x_n$. How many possible distinct inequalities are there?
I assume this problem requires combinatorics , but I am not sure how to apply them in this problem.
Edit: Maybe this problem can be related with coin arranging (point me if I am wrong)
First of all , we can arrange coins (where order matters).But ,we will also allow for the coins to stack on top of each other.In the stacks (There can be multiple stacks) , order doesn't matter.Stacks will be represented with parenthesis.
For $1$ coin , there is $1$ way to arrange them.

$1$ $[x_1]$

For $2$ coins , there are $3$ ways to arrange them.

$12$ $[x_1 < x_2]$
$21$ $[x_1 > x_2]$
$(12)$ $[x_1 = x_2]$

For $3$ coins , there are $13$ ways to arrange them.

$123$ $[x_1 < x_2,x_2 < x_3,x_3 > x_1]$
$132$ $[x_1 < x_2,x_2 > x_3,x_3 > x_1]$
$213$ $[x_1 > x_2,x_2 < x_3,x_3 > x_1]$
$231$ $[x_1 > x_2,x_2 < x_3,x_3 < x_1]$
$312$ $[x_1 < x_2,x_2 > x_3,x_3 < x_1]$
$321$ $[x_1 > x_2,x_2 > x_3,x_3 < x_1]$
$(12)3$ $[x_1 = x_2,x_2 < x_3,x_3 > x_1]$
$3(12)$ $[x_1 = x_2,x_2 > x_3,x_3 < x_1]$
$1(23)$ $[x_1 < x_2,x_2 = x_3,x_3 > x_1]$
$(23)1$ $[x_1 > x_2,x_2 = x_3,x_3 < x_1]$
$2(13)$ $[x_1 > x_2,x_2 < x_3,x_3 = x_1]$
$(13)2$ $[x_1 < x_2,x_2 > x_3,x_3 = x_1]$
$(123)$ $[x_1 = x_2,x_2 = x_3,x_3 = x_1]$

I thought of thinking of stacks as variables that are equal.And a variable being left/right to another variable corresponding to a varialbe less than/ greater than another variable.So my thought is..
number of ways to arrange $n$ coins = number of distinct inequalities for $n$ variables.(Hope this helps)

As Mike Earnest commented, these are ordered Bell numbers or Fubini numbers, and OEIS A000670 shows many ways to calculate them, though not a closed form.
My preference would be with a recursion, letting $f(n,k)$ be the number of orderings of $n$ variables involving $k$ distinct values. When you introduce the $n$th variable, you can

either introduce it to an ordering of $n-1$ variables with $k$ distinct values by matching it to one of the existing distinct values

or introduce it to an ordering of $n-1$ variables with $k-1$ distinct values by matching it to one of the $k$ gaps between, before or after the existing distinct values

So $$f(n,k)=k\big(f(n-1,k) + f(n-1,k-1)\big)$$
and you can start with $f(1,1)=1$ and $f(1,k)=0$ for $k\not=1$, or from $f(0,0)=1$ and $f(0,k)=0$ for $k\not=0$.
You then get a table like this
$$\begin{matrix}
 & k& 0& 1& 2& 3& 4& 5& 6& 7& 8& & \text{Sum}\\
n& & & & & & & & & & & & \\
0& & 1& 0& 0& 0& 0& 0& 0& 0& 0& & 0\\
1& & 0& 1& 0& 0& 0& 0& 0& 0& 0& & 1\\
2& & 0& 1& 2& 0& 0& 0& 0& 0& 0& & 3\\
3& & 0& 1& 6& 6& 0& 0& 0& 0& 0& & 13\\
4& & 0& 1& 14& 36& 24& 0& 0& 0& 0& & 75\\
5& & 0& 1& 30& 150& 240& 120& 0& 0& 0& & 541\\
6& & 0& 1& 62& 540& 1560& 1800& 720& 0& 0& & 4683\\
7& & 0& 1& 126& 1806& 8400& 16800& 15120& 5040& 0& & 47293\\
8& & 0& 1& 254& 5796& 40824& 126000& 191520& 141120& 40320& & 545835\\
\end{matrix}$$
with the final column giving the result you want

To make an inequality, you need two of the variables and one of three operators, so $n (n-1) 3.$ Notice that the order of the variables matters, so you do not divide by two.


One sided calculus limit I am trying to find the one sided limits numerically for this equation:

$$\lim_{x\to0_\pm} \frac{x−\sin\left|x\right|}{x^3}$$

I know the answer for $\lim_{x\to0_{-}}$ is $\infty$, and $\lim_{x\to0_{+}}$ is $\frac16$ but I am having the hardest time arriving at these conclusions.

Thank you all for the help.

If the question says numerically, that means that it wants you to do something like this:

If $f(x) = \frac{x - \sin |x|}{x^3}$, then the numerical limit would be calculated like this:

The limit as $x$ approaches the left: $f(x-h)$ for very small $h$. (i.e. Make an $h$ like $.0001$ and plug it into a calculator)

Limit as $x$ approaches from the right: $f(x+h)$ for every small $h$.

Perhaps its useful to know that $\sin(x) \approx x-\frac{x^3}{3!}$ for $x \approx 0.$


Exactly one in Predicate Logic Could anyone tell me how to translate the following sentence into predicate logic.

E : the set of elephants

A : the set of animals

G(x) : x is green

E(x) : x is an elephant

N(x; y) : name of x is y

There is exactly one green elephant and his name is James.

For example
$$\exists j\;\Bigl( E(j)\land G(j)\land\forall x\; \bigl((E(x)\land G(x))\to (x=j\land N(x,\text{James}))\bigr)\Bigr) $$
("There exists an object $j$ that is both an elephant and green and any (other) object that is also an elephant and green is this $j$ and is named James")
or
$$\exists j\;\Bigl( E(j)\land G(j)\land N(j,\text{James})\land\forall x\; \bigl((E(x)\land G(x))\to x=j\bigr)\Bigr) $$
("There exists a green elephant named James and any green elephant equals this")
or if $\exists !$ is available
$$ \bigl(\exists!x\,E(x)\land G(x)\bigr)\land \forall x\,\bigl((E(x)\land G(x))\to N(x,\text{James})\bigr)$$
("There exists exactly one green elephant and all a green elephants are named James").

Note that
$$ \exists !j\;\bigl(E(j)\land G(j)\land N(j,\text{James})\bigr)$$
is not correct


Question on Arnoldi Iteration Let $p_n$ be a monic polynomial of degree n (a polynomial of degree n with the coefficient of the term of degree n equal to 1).

Let A be a square matrix and $A=Q^*HQ$ where H is a Hessenberg matrix.
Given a vector b, why can the polynomial $p(A)b$ be written as $A^nb - Qy$ for some vector y?

This is the first step in proving the following property of the arnoldi iteration, which I am trying to understand.

"The matrix Hn can be characterized by the following optimality condition. The characteristic polynomial of Hn minimizes ||p(A)q1||2 among all monic polynomials of degree n. This optimality problem has a unique solution if and only if the Arnoldi iteration does not break down."

The question should be edited in this way: any monic polynomial p(A) can be written as $A^n-Qy$ where y is an arbitrary vector and Q is the orthonormal basis for $K_n$ (The $n_{th}$ Krylov subspace of A). Hence, the answer to this question becomes straightforward: The term with the highest degree of the monic polynomial would be by definition $A^n$ and the rest of polynomial can be formed by Q which is the orthonormal basis for $K_n:=<b,Ab,A^2b,...,A^{n-1}b>$
For a more comprehensive elaboration on Arnoldi iteration and Krylov subspaces you can look into numerical linear algebra handbooks e.g. Chapter 34 "Numerical Linear Algebra" by Lloyd N. Trefethen.


Express multiple conditional probability in terms of single conditional probabilities I have to calculate/approximate the following multiple conditional probability:

$$P(X|C_1,C_2,C_3)$$

with only $P(X|C_1)$, $P(X|C_2)$ and $P(X|C_3)$ given. Is it possible to express the above in terms of these single conditional probabilities only? What independence assumptions would I need to make?

Edit: Concretely, I have an algorithm that can calculate single conditional probabilities $P(X|C_i)$ directly, and I want to use these to calculate multiple conditional probabilities. I'm fairly certain it is not possible to calculate this exactly, but perhaps I can make some independence assumptions so some terms can be dropped?


Prove that for $n \gt 6$, there is a number $1 \lt k \lt n/2$ that does not divide $n$ My nine year old asked this question at lunch today: Is there a number that is divisible by everything that is half or less than the number?

I immediately answered, "No. I mean, 6. But not for any number bigger than 6."

So I tried to think why that was true, and my first efforts didn't quite work. I did come up with a proof, but it isn't as elegant as I hoped.

How would you prove this?

if $n\geq 7$ and $n$ is divisible by all numbers less then $\frac{n}{2}$, so $n$ is even and we can deduce easily that $\frac{n}{2}(\frac{n}{2}-1)$ divides $n$ (because $\gcd(\frac{n}{2},(\frac{n}{2}-1))=1$) and the most important:
$$\frac{n}{2}(\frac{n}{2}-1)\leq n$$

whence : $n-2=2(\frac{n}{2}-1)\leq 4$ so $n\leq 6$

if n is odd, $f=\frac{n-1}2$ is an integer less than $\frac n2$. $2f=n-1$, so if $f\gt1$ f is not a factor of n. if $f\gt1$ then $n\gt3$, so for all odd $n\gt3$ n there is a number f which is not a factor.
if n is even, $f=\frac{n-2}2$ is an integer less than $\frac n2$. $2f=n-2$, so if $f\gt2$ f is not a factor of n. if $f\gt2$ then $n\gt6$, so for all even $n\gt6$ n there is a number f which is not a factor.


Proving $\gcd \left(\frac{a}{\gcd (a,b)},\frac{b}{\gcd (a,b)}\right)=1$ How would you go about proving that $$\gcd \left(\frac{a}{\gcd (a,b)},\frac{b}{\gcd (a,b)}\right)=1$$

for any two integers $a$ and $b$?

Intuitively it is true because when you divide $a$ and $b$ by $\gcd(a,b)$ you cancel out any common factors between them resulting in them becoming coprime. However, how would you prove this rigorously and mathematically?

Let $d=\gcd(a,b)$. Let $a=md$ and $b=nd$. If some $k\gt 1$ divides $m$ and $n$, then $kd$ divides $a$ and $kd$ divides $b$, contradicting the fact that $d$ is the greatest common divisor of $a$ and $b$.

Suppose $d \mid \dfrac a{\gcd(a,b)}$ and $d\mid \dfrac b{\gcd(a,b)}$.

Then $nd = \dfrac a {\gcd(a,b)}$ and $md\mid \dfrac b {\gcd(a,b)}$.

So $nd\cdot\gcd(a,b) = a$ and $md\cdot\gcd(a,b) = b$.

So $d\cdot\gcd(a,b)$ is a common divisor of $a$ and $b$.

Since $\gcd(a,b)$, is the greatest of those, we must have $d\not>1$.


Lebesgue measure/Measurable sets Question :
Let $f,g$ be measurable real valued functions on $\mathbb{R}$ such that  :
$$\int_{-\infty}^{\infty} (f(x)^2+g(x)^2)dx=2\int_{-\infty}^{\infty} f(x)g(x)dx$$
Let $E=\{x\in \mathbb{R} : f(x)\neq g(x)\}$ . Which of the followng statements are necessarily true?

$E$ is empty set
$E$ is measurable
$E$ has lebesgue measure $0$
For almost all $x\in \mathbb{R}$ we have $f(x)=0$ and $g(x)=0$

Explanation: What all I could see is that second bullet and third bullet are probably correct. Because :
$$\int_{-\infty}^{\infty} (f(x)^2+g(x)^2)dx=2\int_{-\infty}^{\infty} f(x)g(x)dx$$
i.e.,
$$\int_{-\infty}^{\infty} (f(x)^2+g(x)^2)dx-2\int_{-\infty}^{\infty} f(x)g(x)dx=0$$
i.e.,
$$\int_{-\infty}^{\infty}(f(x)-g(x))^2dx=0$$
Though I have negative limits my function $(f(x)-g(x))^2$ is positive
So, I would see that $E=\{x\in \mathbb{R} : f(x)\neq g(x)\}$ is measurable and has measure $0$
Please tell me if what I have done is sufficient/clear.

Yes, I think what you have done is right, provided that f and g are square integrable. And the last statement that f and g are zero is trivially false.

OP's explanation for second and third bullets are no-doubt correct but by giving a counter-example, we shall also have to explain why first and fourth bullets are false. Here I am going to provide it.
For the first bullet, let $~f(x)=5,~~g(x)=\begin{cases}
      5 &\text{if}~~~~ x\ne 0 \\
      1 &\text{if}~~~~ x= 0
         \end{cases}$
Clearly $~\int_{-\infty}^{\infty} (f(x)^2+g(x)^2)dx=2\int_{-\infty}^{\infty} f(x)g(x)dx~$ and $~0\in E\implies E=\phi~.$hence first bullet is not correct.
For the fourth bullet, from the above example, as neither $~f(x)=0~$ nor $~g(x)=0~,$  we can conclude that the fourth bullet is also not correct.

Note: I know that I am attempting to answer an old question. But I think my answer will also useful for solving the problem to the future reader. So if you downvote my answer, please tell me, where I am wrong. Thank you.


Find the minimum value of $a^4+b^4+\sqrt {ab}$ if $a+b=1$. Find the minimum value of $a^4+b^4+\sqrt {ab}$ if $a+b=1$.

My attempt:
If we put $a=b=\dfrac {1}{2}$, we get $\dfrac {5}{8}$ which is the minimum value. But are there any other methods.?

Note that:

$$a+b=1 \Rightarrow a^2+b^2=1-2ab \Rightarrow a^4+b^4=1+2a^2b^2-4ab.$$

If we denote $\sqrt{ab}=t$, the LHS function will be:

$$f(t)=2t^4-4t^2+t+1.$$

Noting that $0\le \sqrt{ab} \le 0.5,$ minimum will be $f(0.5)=0.625$.

To prove that the minimum of the function is at $\frac{5}{8}$, let us assume that we have a positive real number $k$, that is the absolute value of difference between $a$ or $b$ and $\frac{1}{2}$. Any other pairs of solutions can be represented as $(0.5+k)^4 + (0.5-k)^4 + \sqrt{(0.5+k)(0.5-k)}$.

Using the binomial theorem, we have:
$$(0.5+k)^4 = [(1)(0.5)^4 (k)^0 + (4)(0.5)^3(k)^1 + (6)(0.5)^2(k)^2 + (1)(0.5)^1 (k)^3$$

$$(0.5-k)^4 = [(1)(0.5)^4 (k)^0 + (4)(0.5)^3(-k)^1 + (6)(0.5)^2(-k)^2 + (1)(0.5)^1 (-k)^3 + (4)(0.5)^0(-k)^4]$$

The terms that have $1$st, $3$rd powers cancel, leaving:

$$(0.5+k)^4+(0.5-k)^4 = [(2)(0.5)^4 (k)^0+ (12)(0.5)^2(k)^2 + (2)(0.5)^0(-k)^4]$$
$$=2*\frac{1}{16} + 12*\frac{1}{4}*k^2 + 2*k^4$$
$$=2k^4+3k^2+\frac{1}{8}$$

The last term $\sqrt{(0.5+k)(0.5-k)}$ can be simplified into $\sqrt{-k^2+0.25}$ by the difference of squares.

Adding the terms together gives:
$$=(2k^4+3k^2+0.125) + (\sqrt{-k^2+0.25})$$

Since the derivative of $2k^4+3k^2+0.125$ is larger than $-k^2+0.25$, it follows that $2k^4+3k^2+0.125$ is larger than $\sqrt{-k^2+0.25}$, so the first function grows faster.

The first term has a minimum of $0.125$ when $k=0$, and also when $k=0$ the function $\sqrt{-k^2+0.25}$ evaluates to $0.5$, $0.5 + 0.125 = 0.625$, so the function is at is minimum when $a,b $ = $0.5+0$, or when $a,b = 0.5$.


Is the projection from $\mathbb R^n$ locally isometric to the flat torus? I’m reading a problem asking to introduce a Riemannian metric on $T^n$ in such a way that the projection $\pi(x_1,...,x_n)=(\exp(ix_1),...,\exp(ix_n))$ from $\mathbb R^n$ to $T^n$ is a local isometry. Does the flat metric apply here?

By definition, flat torus is defined as the product of n copies of the Riemannian manifold $S^1$ with its metric induced by $\mathbb R$.

Yes; both of these spaces are "flat" in the sense that the curvature of the Riemannian metric is identically zero.

Notice that the natural projection written above maps a point of $R^n$ to a point in $T^n$. This may seem odd at first but notice how the coordinate curves of $R^n$ get mapped to $T^n$. Because $e^{ix}$ is a point in a complex circle, the projection essentially wraps each axis of $R^n$ into a circle of the torus. This maps coordinate curves into coordinate curves. Since the projection is evidently the cartesian product of a number of complex circles (which are the same as $S^1$) you can introduce the product metric the same as if you were working with real numbers exclusively. Since you can take the vectors to be coordinate curves locally, the product metric is a local isometry.


Projective Plane for F3 I have to calculate the points of a projective Plane on $\mathbb{Z}_3^2$. I thing I understood the way how to do this for the Fano-Plane but I am not sure how to do this here because I have more than one nonzero scalar?

As in all projective planes over fields, the points are $[x_1,x_2,x_3]$, where $(x_1,x_2,x_3)\neq (0,0,0)$ and $[x_1,x_2,x_3]=[kx_1,kx_2,kx_3]$ for all $k\neq 0$.

In this case the points are
$A=[0,0,1]$, $B=[0,1,0]$, $C=[1,0,0]$, $D=[0,1,1]$, $E=[1,0,1]$, $F=[1,1,0]$, $G=[0,1,2]$, $H=[1,0,2]$, $I=[1,2,0]$, $J=[1,1,1]$, $K=[1,1,2]$, $L=[1,2,1]$, $M=[2,1,1]$.

The interesting part is giving the lines. You have to select every two points, for example $A$ and $B$, and the other two points on their line are given by every possible linear combinations of $A$ and $B$. In this case they are $A+B$ and $A+2B$.

So the lines are
$(A,B,D,G)$ $(A,C,E,H)$, $(A,F,J,K)$, $(A,I,L,M)$, $(B,C,F,I)$, $(B,E,J,L)$, $(B,H,K,M)$, $(C,D,J,M)$, $(C,G,K,L)$, $(D,E,K,I)$, $(D,F,L,H)$, $(E,F,M,G)$, $(G,H,J,I)$.

Assuming that by calculating points of projective plane on $\mathbb Z^2_3$ you mean finite projective plane of the order $9 = 3^2$, then as the first you should generate Galois field $GF(3^2)$. You can use Wikipedia article $GF(p^2)$ for an odd prime. After having multiplicative group of the Galois Field $GF(3^2)$, use it to rotate permutation sub-matrices $C_{ij}$ in incidence matrix in canonical form (See Paige L.J., Wexler Ch., A Canonical Form for Incidence Matrices of Projective Planes...., In Portugalie Mathematica, vol 12, fasc 3, 1953). This is incidence matrix of points and lines and it defines the projective plane.


How to compute infinite limits of summations (Riemann sums) when there are i's in the denominator? I'm really lost here. I'm trying to use a right Riemann sum to compute:
$$\int_0^1{\frac{x}{x^4+2x^2+1}dx}$$
Eventually I get here:
$$\lim_{n\to\infty}\frac{1}{n}\sum_{i=1}^n\frac{n^3i}{n^4+2n^2i^2+i^4}$$
Using $\Delta{x}=1/n$ and $x_i=a+i\Delta{x}=0+i/n=i/n$
But now I'm stuck since the only formulas I know to manipulate the summation and get rid of the i's are like $\sum_{i=1}^ni^2$. But since I have i's in the denominator, how can I proceed? Thanks!

I hope you are OK with the substitution $x^2+1=u$ to get the integral as $$\int_{1}^{2}\frac{du}{2u^2}$$ A standard technique now is to use a Riemann sum where the partition points are in geometric progression (in contrast to the more common arithmetic progression).
Let $x_i=2^{i/n}$ be the points of partition and we form the Riemann sum $$\sum_{i=1}^{n}f(x_i)(x_i-x_{i-1})$$ where $f(x) =1/(2x^2)$. The computation of the limit is not difficult and should give you the answer $1/4$.

If on the other hand you wish to use the partition points in arithmetic progression as $x_i=1+(i/n)$ then you need some approximation technique for the Riemann sum. Thus we have $$f(x_i) \Delta x_i=\frac{1}{2n(1+i/n)^2}$$ and $$\frac{1}{2}\left(\frac {1}{1+i/n}-\frac{1}{1+(i+1)/n}\right) \leq f(x_i)\Delta x_i\leq \frac{1}{2}\left(\frac{1}{1+(i-1)/n}-\frac{1}{1+i/n}\right)$$ Adding these inequalities for $i=1,2,\dots, n$ we get $$\frac{1}{2}\left(\frac{1}{1+1/n}-\frac{1}{1+(n+1)/n}\right)\leq \sum_{i=1}^{n}f(x_i)\Delta x_i\leq \frac {1}{2}\left(1-\frac{1}{2}\right)$$ By squeeze theorem the limit of Riemann sum as $n\to \infty $ is $1/4$.
The approximation used above can be applied to the original integral also (without the substitution $u=1+x^2$) but it is slightly more complicated to deal with.

As per @Aryadeva:
$$I=\int_{1}^{2} \frac{du}{2u^2}=\frac{1}{2}\lim_{n \to \infty} \sum_{k=n}^{2n} \frac{n^2}{k^2} \frac{1}{n}= \lim_{n \to \infty} \frac{n}{2} \sum_{k=n}^{2n} \frac{1}{k^2}=\lim_{n \to \infty}\frac{n}{2}[H^{(2)}_{2n}-H^{(2)}_n]=\lim_{n \to \infty}\frac{n}{2}[\psi^1(n)-\psi^1(2n+1)]~~~~~~(1)$$
Here $H^{(2)}_n$ are harmonic numbers and $\psi^m(x)$ are called polygamma functions.
Using the bounds on $\psi^m(z)$ as
$$ \frac{(m-1)!}{x^m}+\frac{m!}{2x^{m+1}} \le \psi^m(x) \le \frac{(m-1)!}{x^m}+\frac{m!}{x^{m+1}}~~~~(2) $$
we get $$ 1+\frac{1}{2n} \le n~\psi^1(n) \le  1+\frac{1}{n}$$
So by sandwich theorem (squeez law) $$\implies \lim_{n \to \infty} n~ \psi^1(n)=1~~~~(3)$$
and
$$ \frac{n}{2n+1} +\frac{n}{2(2n+1)^2}\le  n \psi^1(2n+1) \le \frac{n}{2n+1}+\frac{n}{(2n+1)^2}$$
Again $$\lim_{n \to \infty} n \psi^1(2n+1) =\frac{1}{2}~~~~~~(4)$$
Fimally using (3) and (4) in (1), we get
$$I=\frac{1}{4}.$$
For polygamma functions see
https://en.wikipedia.org/wiki/Polygamma_function


Are all disjoint events dependent? This is a basic question about probability theory.

My reasoning goes as follows:

If $A$ and $B$ are independent events, the probability they both happen is their multiplication:
$$\Pr(A \text{ and } B) = \Pr(A) \times \Pr(B)$$
If their marginal probability is not impossible, also their product is non-zero:
$$\Pr(A) > 0,\, \Pr(B) > 0 \implies \Pr(A \text{ and } B) > 0$$
Hence, independent events cannot be disjoint
Hence, only dependent events can be disjoint
Hence, all disjoint events are dependent.

Can you help me point out the error in my argument?

Yes, as long as A and B are mutually exclusive and sum up to 1 or 100%,  P(A) + P(B) = 1.
Example : You are either dead or alive. The chance of being alive or dead always sums up tp 1 or 100%, no matter what condition you live in.


Show that $n-n\max\{X_i\}$ equals exponential distribution for $X_i$ i.i.d. and uniformly distributed on $[0,1]$ Say, we have a sequence of random variables $(X_i)_{1\leq i \leq n}$ that are i.i.d. and distributed according to a uniform distribution defined on interval $[0,1]$.

Why is $n-n\cdot\max\{X_1,\cdots,X_n\}$ distributed like an exponential distribution for real and positive $n$?

I know that $\lim_{n\rightarrow\infty}e(x)=(1+\frac{x}{n})^n$...

I also know that $\max\{X_1,\cdots,X_n\}=P(X_1)\cdot\cdots\cdot P(X_n)$ for independent variables...

I have:

\begin{align}
F(x) & =P(n-n\cdot\max\{X_1,\cdots,X_n\}<x)=P(\max\{\ldots\} \\[10pt]
& \geq1-x/n)=1-P(\max\{\ldots\}<1-x/n).
\end{align}

Now, due to the cdf of $U(0,1)$ which is just $x$, this is equivalent to $1-(1-x/n)^n$.

But how do I proceed now? Or are there any mistakes so far?

Let $M_n = \max_{1\leqslant i\leqslant n}X_i$ for each positive integer $n$. For $0<t<1$ we have $$\{M_n \leqslant t\} = \bigcap_{i=1}^n \{X_i\leqslant t\},$$
so by independence it follows that
$$
\mathbb P(M_n\leqslant t) = t^n.
$$
Let $Y_n = n(1-M_n)$. Let $t>0$ and choose $N>1/t$. Then for $n\geqslant N$ we have
\begin{align}
\mathbb P(Y_n\leqslant t) &= \mathbb P(n(1-M_n)\leqslant t)\\
&=\mathbb P(1-M_n\leqslant t/n)\\
&=\mathbb P(M_n\geqslant 1-t/n)\\
&=1-\mathbb P(M_n\leqslant 1-t/n)\\
&=1 - (1-t/n)^n.
\end{align}
Since $$\lim_{n\to\infty}1 - (1-t/n)^n  = 1 - e^{-t},$$
it follows that $Y_n$ converges in distribution to the exponential distribution with mean $1$.

You are right in your calculations. There are two points and you are done:

$1.$ From uniform to exponential you have the condition $0 \leq 1-x/n \leq 1$

$2.$ You have to consider $\lim n\rightarrow\infty$. Then you get $\exp(-x)$. Your formula about this is also correct.

So in conclusion you get an exponential distribution with parameter $+1$ and it is $exp(-1)$ safely enough it doesnt blow up.


Sheaf of $1$-forms on affine opens First some definitions:

Let $k$ be algebraically closed. Let $X$ be a topological space. We define an algebraic variety $X$ to be a $k$-space $(X,\mathcal O_X)$, such that for each $x\in X$, we have an open $U\subset X$ with $x\in U$, such that $(U,\mathcal O_X\vert_U)\cong (Z,\mathcal O_Z)$, where $Z\subset\mathbb A^n$ is an (embedded) affine variety, and $\mathcal O_Z$ is its sheaf of regular functions. In short: $X$ is locally affine.

An (embedded) affine variety $Z\subset\mathbb A^n$ is a closed subset of $\mathbb A^n$ (with the Zariski topology), and for open $U\subset Z$, $\mathcal O_Z(U)$ is the $k$-algebra consisting of functions $U\to k$ that locally can be written as $f/g$ with $f,g\in k[x_1,\dots,x_n]$.

Given an algebraic variety $X$ we have the so-called sheaf of 1-forms $\Omega^1_X$, which is defined to be the sheafification of the presheaf $U\mapsto \Omega^1_{\mathcal O_X(U)}$, where $\Omega^1_{\mathcal O_X(U)}$ is the module of Kähler differentials of $\mathcal O_X(U)$ (the regular functions on $U$).

Now, my lecture notes state the following two results without proof:

For an affine open $U\subset X$, we have $\Omega^1_X(U)\cong\Omega^1_{\mathcal O_X(U)}$.

For affine opens $V\subset U\subset X$, we have $\Omega^1_{\mathcal O_X(V)}\cong\Omega^1_{\mathcal O_X(U)}\otimes_{\mathcal O_X(U)}\mathcal O_X(V)$, and furthermore, the restriction map $\Omega^1_X(U)\to\Omega^1_X(V)$ is given by $\omega\mapsto\omega\otimes 1$.

When they say the restriction map is given this way, do they mean that we have the following commutative diagram:

Here, the restriction mapping $\Omega^1_{\mathcal O_X(U)}\to\Omega^1_{\mathcal O_X(V)}$ is the one induced from $\iota^*$ (exactly such that this diagram commutes).

Your question: "For an affine open $U \subseteq X$, we have
$$\text{O1}. \text{   }\Omega^1_X(U) \cong \Omega^1_{\mathcal{O}_X(U)}.$$
For affine open sets  $V\subseteq U \subseteq X$ we have
$$\text{O2}. \text{   }  \Omega^1_{\mathcal{O}_X(V)} \cong \Omega^1_{\mathcal{O}_X(U)}\otimes_{\mathcal{O}_X(U)} \mathcal{O}_X(V),$$
and furthermore, the restriction map $\Omega^1_X(U) \rightarrow \Omega^1_X(V)$ is given by $ω \rightarrow ω\otimes 1$.
It seems to me the following is true:
Question O1. Since your sheaf $\Omega^1_X$ is defined as the sheafification $F^{+}$ of the pre-sheaf $F(U):= \Omega^1_{\mathcal{O}_X(U)}$, it seems to me $F^{+}$ and $F$ should have she same sections on affine open subsets $U:=Spec(A) \subseteq X$. Hence
$$F(U)=F^{+}(U) \cong \Omega^1_{A/k}:=\Omega^1_{\mathcal{O}_X(U)}.$$
Question O2. The construction of the module of differentials in functorial in the sense that for any open subscheme $U \subseteq X$ it follows $i^*(\Omega^1_{X/k}) \cong \Omega^1_{U/k}$ where $i: U \rightarrow X$ is the inclusion  map.
Moreover for any pair of maps of commutative rings $A \rightarrow B$ and $A \rightarrow S$
let $B_S :=S\otimes_A B$. There is (Matsumura's book "Commutative ring theory", Exercise 25.4 page 198) a canonical isomorphism
$$ B_S \otimes_B \Omega^1_{B/A} \cong \Omega^1_{B_S/S}.$$
Hence for any inclusion of basic open sets $D(f) \subseteq D(g) \subseteq U:=Spec(A)$
you should get the following:
$$(\Omega^1_{D(g)/k})_{D(f)}:= \Omega^1_{A_g/k} \otimes_{A_g} A_f \cong \Omega^1_{A_{fg}/k}\cong \Omega^1_{A_f/k}.$$
By $(\Omega^1_{D(g)/k})_{D(f)}$ I mean the restriction of $\Omega^1_{D(g)/k}$
to the open set $D(f)$. Here we have used that $D(f)\cap D(g):=D(fg) =D(f)$ since $D(f) \subseteq D(g)$.
The restriction map
$$\rho: \Omega^1_{D(g)/k}\rightarrow (\Omega^1_{D(g)/k})_{D(f)} := \Omega^1_{D(g)/k}\otimes_{A_g} A_f\cong \Omega^1_{D(f)/k}$$
is the canonical map $\rho(\omega):=\omega \otimes 1$ where $1\in A_f$ is the multiplicative identity. Formula O2 should follow since $U$ and $V$ have open covers on the form $D(f)$.


$\det(I+A) = 1 + tr(A) + \det(A)$ for $n=2$ and for $n>2$? Let $I$ the identity matrix and $A$ another general square matrix. In the case $n=2$ one can easily  verifies that
\begin{equation}
\det(I+A) = 1 + tr(A) + \det(A)
\end{equation}
or
\begin{equation}
\det(I+tA) = 1 +  t\  tr(A) + t^2\det(A)
\end{equation}
for some scalar $t \in \mathbb{R}$.

I have tried to see if there exists a similar formula for $n>3$. This is a natural question. But the calculations are very big and difficulty to see. Then I do the answer. Is there a similar formula for $n>2$?

If $a_1,a_2,\ldots,a_n$ are the eigenvalues of $A$ then the characteristic polynomial of $A$ is
$$
\det(tI-A)=(t-a_1)(t-a_2)\cdots(t-a_n).$$
Therefore

$\det(I+A)=(1+a_1)(1+a_2)\cdots(1+a_n) = 1 + \operatorname{tr}A+ \ldots + \det A$

and

$\det(I+tA)=(1+ta_1)(1+ta_2)\cdots(1+ta_n)= 1 + t\operatorname{tr}A+ \ldots + t^n\det A$.

it is like the development of the characteristic polynomial (each term of the polynomial is a combination of symmetric sub-determinants of the initial matrix), I think you can easily find a proof of this fact on the internet.
Or you can even prove it by yourself, using the multi-linearity of the determinant, and the fact that it is alternate.


For a compact K, prove that there exists $m > 0$ such that $f(x) ≥ m$ for all $x$ in K Pasted image for time's sake...

I'm not sure how to approach this besides using the case $f(x) = x$ on the set $(0,1)$. Would I use 1/m?

By Weierstrass' theorem, $f$ attains a global minimum in $K$, say $f(x_m)=\min_{x \in K} f(x)$. By assumption, $f(x_m)>0$, and hence $f(x) \geq f(x_m)=m>0$ for every $x \in K$.

In general, if $K$ is compact and $f:K\rightarrow Y$ is continuous then $f\left(K\right)$
is a compact subspace of $Y$ (not difficult to verify just by the definition of compact). Applying this here, we find that $f\left(K\right)\subset\left(0,\infty\right)\subset\mathbb{R}$
is compact. Then it is closed too (since compact sets in a Hausdorff
space are closed) and we have $0\notin f\left(K\right)$. Then some
$m>0$ exists such that $f\left(K\right)\cap\left(-m,m\right)=\emptyset$. Combined with $f\left(K\right)\subset\left(0,\infty\right)$ this
leads to $f\left(x\right)\geq m$ for each $x\in K$.


Solving a $1^\infty$ indeterminate form. I'm preparing for my calculus exam and I can't solve this limit:

$$\lim_{x\rightarrow\infty}\left(\frac{1+\tan(1/x)}{1-\tan(1/x)}\right)^x$$

The limit tends to $1^\infty$, which is indeterminate.
I've tried several things and I couldn't solve it.

Any idea? Thanks in advance.

Note that
$$\tag{1}\lim_{x\rightarrow\infty}\left(\frac{1+\tan(1/x)}{1-\tan(1/x)}\right)^x=\lim_{x\rightarrow\infty}e^{\displaystyle x\ln\left(\frac{1+\tan(1/x)}{1-\tan(1/x)}\right)}=e^{\displaystyle\lim_{x\rightarrow\infty}x\ln\left(\frac{1+\tan(1/x)}{1-\tan(1/x)}\right)}$$
since $e^x$ is a continuous function.

Note that
$$\lim_{x\rightarrow\infty}x\ln\left(\frac{1+\tan(1/x)}{1-\tan(1/x)}\right)=
\lim_{x\rightarrow\infty}\frac{\ln\left(\frac{1+\tan(1/x)}{1-\tan(1/x)}\right)}{\frac{1}{x}}  \cdot \left(\frac{0}{0}\right)$$
We can apply the L'Hospital rule to the previous limit. Since
$$\frac{d}{dx}\ln\left(\frac{1+\tan(1/x)}{1-\tan(1/x)}\right)=\frac{1-\tan(1/x)}{1+\tan(1/x)}\cdot \frac{d}{dx}\left(\frac{1+\tan(1/x)}{1-\tan(1/x)}\right)
$$
$$=\frac{1-\tan(1/x)}{1+\tan(1/x)}\cdot \frac{d}{dx}\left(\frac{2}{1-\tan(1/x)}-1\right)=\frac{1-\tan(1/x)}{1+\tan(1/x)}\cdot \frac{2\sec^2(1/x)\cdot(-\frac{1}{x^2})}{(1-\tan(1/x))^2},$$
we have
$$\lim_{x\rightarrow\infty}\frac{\ln\left(\frac{1+\tan(1/x)}{1-\tan(1/x)}\right)}{\frac{1}{x}}=
\lim_{x\rightarrow\infty}\frac{\frac{1-\tan(1/x)}{1+\tan(1/x)}\cdot \frac{2\sec^2(1/x)\cdot(-\frac{1}{x^2})}{(1-\tan(1/x))^2}}{-\frac{1}{x^2}}$$
$$\tag{2}=\lim_{x\rightarrow\infty}\frac{1-\tan(1/x)}{1+\tan(1/x)}\cdot \frac{2\sec^2(1/x)}{(1-\tan(1/x))^2}=2.$$

Combining $(1)$ and $(2)$, we have
$\lim_{x\rightarrow\infty}\left(\frac{1+\tan(1/x)}{1-\tan(1/x)}\right)^x=e^2.$

EDIT:
you can write your expression as
$$
\bigg(1+\frac{2\tan(1/x)}{1-\tan(1/x)}\bigg)^x \sim \bigg(1+\frac{2}{x}\bigg)^x \rightarrow e^2
$$
when $x\rightarrow \infty$.


Given $u_1,u_2,u_3,u_4$ independent vectors in $R^5$. $v_1,v_2,v_3$ in $R^5$ too. Given $u_1,u_2,u_3,u_4$ linearly independent vectors in $R^5$. $v_1,v_2,v_3$ $\in R^5$ .

They are defined like that:

$v_1 = u_1 + u_2 - u_4$

$v_2 = au_1 - u_3 + u_4$

$v_3 = u_2 + au_3 - 6u_4$

How can I find for which $a$ , vectors $v_1,v_2,v_3$ will be linearly dependent?

We show that $v_1,v_2,v_3\in \mathbb{R}^5$ are always linearly independent. First suppose $a=0$. Now let $\lambda_1,\lambda_2,\lambda_3\in \mathbb{R}$ such that $\lambda_1v_1+\lambda_2v_2+\lambda_3v_3=0$. Then $$\lambda_1u_1+2(\lambda_1+\lambda_3)u_2-\lambda_2u_3-6(\lambda_1+\lambda_2+\lambda_3)u_4=0  \implies \lambda_1=\lambda_2=\lambda_3=0$$
Now suppose $a\not=0$. We express the $v_i$ as column vectors now and perform elementary matrix operations on
\begin{bmatrix}
1 & a & 0 \\
1 & 0 & 1 \\
0 & -1 & a \\
-1 & 1 & -6 \\
0 & 0 & 0
\end{bmatrix}
to obtain
\begin{bmatrix}
a & 0 & 0 \\
0 & a & 0 \\
0 & 0 & 1-\frac{1}{a^2} \\
0 & 0 & -1 -\frac{1}{a}+\frac{6}{a^2} \\
0 & 0 & 0
\end{bmatrix}
Note that since $a=1,-1$ both do not yield the zero vector in the third column, we can conclude that $v_1,v_2,v_3\in \mathbb{R}^5$ are always linearly independent

$v_1, v_2, v_3$ will be linearly dependent if there are $\alpha_1, \alpha_2, \alpha_3$ not all $0$ such that $\alpha_1v_1 + \alpha_2v_2 + \alpha_3v_3 = 0$

For your $v_1, v_2$ and $v_3$ we have:

$0 = \alpha_1v_1 + \alpha_2v_2 + \alpha_3v_3 = \alpha_1(u_1 + u_2 - u_4) + \alpha_2(au_1 - u_3 + u_4) + \alpha_3(u_2 + au_3 - 6u_4) = u_1(\alpha_1 + a\alpha_2) + u_2(\alpha_1 + \alpha_3) + u_3(-\alpha_2 + a\alpha_3) + u_4(-\alpha_1 + \alpha_2 -6\alpha_3)$

As you have that $u_1, u_2, u_3, u_4$ are linearly independent, you have that

\begin{cases}
\alpha_1 + a\alpha_2 = 0 \\
\alpha_1 + \alpha_3 = 0\\
-\alpha_2 + a\alpha_3 = 0 \\
-\alpha_1 + \alpha_2 -6\alpha_3 = 0
\end{cases}

$\implies$

\begin{cases}
a = \frac{-\alpha_1}{\alpha_2} \\
\alpha_1 = -\alpha_3\\
a = \frac{\alpha_2}{\alpha_3} \\
-\alpha_1 + \alpha_2 -6\alpha_3 = 0
\end{cases}


Given a surface how can I get an $\epsilon$-sample? Few definitions and then my question:

  Definition 1.1. The medial axis $M$ of a curve (surface) $\Sigma \in \mathbb{R}^k$ is the closure of the set of points in $\mathbb{R}^k$ that have at least two closest points in $\Sigma$.

  Definition 1.2. The local feature size at a point $x \in \Sigma$ is the value of a function
  $$
f(x) : \Sigma \to \mathbb{R} \;\;\text{where} \; f(x) = d(x,M)
$$
  In other words $f(x)$ is the distance of $x \in \Sigma$ to the medial axis $M$.

  Definition 1.3. A sample $P$ of $\Sigma$ is a $\epsilon$-sample if each point $x \in \Sigma$ has a sample point $p \in P$ such that $\left\|x - p \right\| \leq \epsilon f(x)$

These definition are taken from chapter 1 in Curve and Surface Reconstraction : Algorithms and Mathematical analysis.

My question is : Suppose $\Sigma$ is some regular surface given, is there a known algorithm to generate a sample $P$ which is a $\epsilon$-sample?

I would imagine just a simple sampling of the surface wouldn't really work, what I want is to generate random samples which are also $\epsilon$ samples.

I guess the question should be broken down into :
1. Known algorithms to construct the medial axis $M$
2. Known algorithms to perform $\epsilon$-sampling.

I'm asking because I'd like to implement such an algorithm and I don't have a clue where to start.

I think the difficulty of this depends on how 'nice' your surface is. If the surface is smooth with a global curvature bound, one can prove that $f(x)$ is bounded away from zero. So any sufficiently fine/dense sample of points on $\Sigma$ would work. Although I think generating such a sample for an arbitrary surface still looks hard.

If the surface is not smooth, a general procedure seems even harder. Consider $\Sigma$ a triangle in $\mathbb{R}^2$, then the corners need to part of your sample as they are part of the medial axis.


Two tangent circles inscribed in a rectangle (Compute the area) Consider two circles with a diameter equal to $a$, externally tangent to each other, and whose centers are at the same height. Those circles are inscribed inside a rectangle of length $2a$ and height $a$. This is a sketch I made for this problem (please, forgive my unsteady handwriting):

I am asked to calculate the shaded area.
I can do it using:

Symmetry. This is the easiest way in my opinion, since we know the area of each circle ($\pi a^2/4$) and the area of the rectangle ($2a^2$), which gives us: $\boxed{A_{\text{shaded}}=\dfrac{4-\pi}{4} a^2}$
Mathematical functions. We can set the origin at the bottom left corner, calculate each of the circles' analytical functions (as well as that of the line), compute the intersections, and make use of definite integrals to compute the final area. It will yield the exact same result as the above, yet the process to achieve it would be much longer.

However, I'm not interested in any of these 2 methods (as they look pretty easy). I'm interested in finding a pure geometrical way to solve it. No functions. No symmetry as I used above. But pure geometric relationships.

I thought about drawing some lines from the center of each circle to each intersection, like this:

This would give us 4 areas hopefully easy to solve for ($S_1$, $S_1^{\prime\prime}$ and $S_2$, $S_2^{\prime\prime}$). Of course, $S_1=S_1^{\prime\prime}$ and $S_2=S_2^{\prime\prime}$ but as I said, I don't want to make use of symmetry (I'm a bit masochistic after all). So I thought about this solving scheme:

$A_{\text{shaded}}=\frac12A_{\text{rectangle}}-S_2-(A_{\text{circle}}-S_2^{\prime\prime})$
$A_{\text{sector}}=S_1+S_2$
Since $S_1$ is a triangle, we could use some trigonometric relations in it to solve for $S_1$, and since the area of a circular sector is known, we could solve for $S_2$ (and $S_2^{\prime\prime}$, and very innocently find that actually $S_2=S_2^{\prime\prime}$)

The main problem I have is that I don't know the inner angle of the circle sectors, because of the (seemingly randomly placed) intersection points near the borders of the rectangle.

IF ONLY I were able to locate those intersection points using pure trigonometry and geometric relationships/theorems, the problem could be solved.

Any hints or ideas?

PD: Please, don't question my (foolish) decision of not making use of symmetry. I want to take this problem as a personal challenge. Obviously, if this were to be solved quickly I wouldn't scratch my head too much and go for the easy solution.

From the lower left-hand corner of the rectangle (let's call that point $A$), consider the diagonal line as a secant line of the left-hand circle,
intersecting the circle at points $B$ and $C$, where $B$ is between $A$ and $C$.
Also consider one of the edges of the rectangle adjacent to $A$
as a tangent line touching the circle at $D$.

Then by a theorem about tangent and secant lines from a point outside a circle, we have the following relationship of the lengths of the segments
from $A$ to each of the three points $B$, $C$, and $D$:
$$
(AD)^2 = AB \times AC. \tag1
$$

It is easy to find that $AD = \frac12a$.
Now let $E$ be the midpoint of the bottom side of the rectangle;
then $AC$ is the hypotenuse of right triangle $\triangle AEC$,
which has legs $a$ and $\frac12a$, and therefore
$AC = (\frac12\sqrt5)a$.

We can then use Equation $(1)$ to find the length $AB$,
so we can find the length of the chord $BC$; from that chord and
the radius of the circle we can get the angle of $S_1$ at the
center of the circle.

I think you can consider the full rectangle area 2a^2 subtracting the areas of two circles then devide the result by 2.


Solving $u_y + (1-2u)\cdot u_x = 0$ using characteristic equations I need to solve the following partial differential equation:
$$F(x, y, u, p, q) = u_y + (1-2u)\cdot u_x = 0$$
with $$u(x, 0) = \left\{\begin{array}{cc} \frac{1}{4} & x < 0 \\ \frac{3}{4} & x>0 \end{array}\right.$$
Here we have $p = u_x$ and $q = u_y$. $u(x,y)$ is a continously differentiable function and $-\infty < x < \infty$ and $0 < y < \infty$.

I use the characteristic differential equations to get a solution. These are
$$
\left\{\begin{array}{ll}
\frac{dx}{ds} = F_p = 1-2u \\
\frac{dy}{ds} = F_q = 1\\
\frac{du}{ds} = pF_p+qF_q = p(1-2u) + q\\
\frac{dp}{ds} = -F_x - pF_u = 2p^2\\
\frac{dq}{ds} = -F_y - qF_u = 2pq \end{array}\right.
$$

The initial conditions are, if $s=0$, that $x = \lambda, y = 0,$
$$
u = \left\{\begin{array}{cc}
\frac{1}{4} & \lambda < 0\\
\frac{3}{4} & \lambda > 0\end{array}\right.
$$
and, using that $$0 = \frac{du}{d\lambda} = p(\lambda)\frac{dx}{d\lambda} + q(\lambda)\frac{du}{d\lambda} = p(\lambda)\cdot 1 + 0$$
and $$ 0 = q + (1-2u)p = q + (1-2)\cdot 0 = q$$
we get $p=q=0$.

From this, it is easily seen that $y = s$. But from here, I'm stuck. The general solution of $\frac{dp}{ds} = 2p^2$ is $p(s) = \frac{1}{c_1 - 2s}$. It impossible for this function to sattisfy the initial condition, so does that mean that this partial differential equation doesn't have a solution?

Another thing is that the initial condition for $u(x, 0)$ isn't continous, so how can the solution of this problem be continously differentiable?

The PDE $u_y + (1-2u)u_x = 0$ is the Lighthill-Whitham-Richards traffic flow equation. The proposed initial-value problem may be interpreted as a one-way road with the car density $u=1/4$ before the abscissa $x=0$ (fast cars), and the car density $u=3/4$ after the abscissa $x=0$ (slow cars). In mathematical words, this is a Riemann problem, which cannot be solved by using the method of characteristics alone: we should consider weak solutions. The Lax entropy condition tells that the solution is a shock wave. Its speed is given by the Rankine-Hugoniot condition $s = 1- \frac{1}{4} - \frac{3}{4} = 0$. Therefore, the solution is a static shock wave
$$
u(x,y) = \left\lbrace
\begin{aligned}
&1/4 &&\text{if}\quad x<0\\
&3/4 &&\text{if}\quad x>0
\end{aligned}\right.
$$

Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:

$\dfrac{dy}{dt}=1$ , letting $y(0)=0$ , we have $y=t$

$\dfrac{du}{dt}=0$ , letting $u(0)=u_0$ , we have $u=u_0$

$\dfrac{dx}{dt}=1-2u=1-2u_0$ , letting $x(0)=f(u_0)$ , we have $x=(1-2u_0)t+f(u_0)=(1-2u)y+f(u)$ , i.e. $u=F(x+(2u-1)y)$

$u(x,0)=\begin{cases}\dfrac{1}{4}&x<0\\\dfrac{3}{4}&x>0\end{cases}$ :

$F(x)=\begin{cases}\dfrac{1}{4}&x<0\\\dfrac{3}{4}&x>0\end{cases}$

$\therefore u=\begin{cases}\dfrac{1}{4}&x+(2u-1)y<0\\\dfrac{3}{4}&x+(2u-1)y>0\end{cases}=\begin{cases}\dfrac{1}{4}&x-\dfrac{y}{2}<0\\\dfrac{3}{4}&x+\dfrac{y}{2}>0\end{cases}$

Hence $u(x,y)=\begin{cases}\dfrac{1}{4}&x<\dfrac{y}{2}\\\dfrac{3}{4}&x>-\dfrac{y}{2}\\c&\text{otherwise}\end{cases}$


$\sum_{n=1}^\infty \frac{1}{n^{1+\frac{1}{n}}}$ Divergent? Is this thorough enough?

$$\frac{1}{n^{1+\frac{1}{n}}} = \frac{1}{n \cdot n^{\frac{1}{n}}} \ge \frac{1}{n} \quad as\ n\to \infty$$

Therefore by comparison w/ $\frac{1}{n}$, $\sum_{n=1}^\infty \frac{1}{n^{1+\frac{1}{n}}}$ is divergent.

Thanks!

What you have written is incorrect, since it is not true that $\frac{1}{n \cdot n^{\frac{1}{n}}} \ge \frac{1}{n}$.  Indeed, $\frac{1}{n \cdot n^{\frac{1}{n}}} < \frac{1}{n}$ for all $n>1$.  You seem to be trying to claim that your inequality holds "in the limit", but it is unclear what you mean by this precisely or how you are claiming this would imply your sum diverges.

To solve the problem correctly, I would suggest trying to compare $\frac{1}{n \cdot n^{\frac{1}{n}}}$ to $\frac{1}{2n}$ instead.

$\sum_{n=1}^{\infty} \frac{1}{n^p}$ diverges if and only if $s < 1$. Clearly you have that $1+ \frac{1}{n} \geq 1$ and for $n\to\infty$ you have that $1+ \frac{1}{n} \to 1$


Is the row echelon form of a system of equations unique? I'm talking about the row echelon form, not the reduced row echelon form. If it isn't can you give me some examples?

This matrix is already in row echelon form:
$$
\begin{bmatrix}
1 & 1 \\
0 & 1 \\
\end{bmatrix}
$$
but we can apply the row operation $R_1 \gets R_1 - R_2$ which gives another row echelon form
$$
\begin{bmatrix}
1 & 0 \\
0 & 1 \\
\end{bmatrix}.
$$

Reduced row echelon forms are unique, however.

Reduced row-echelon form of a matrix is unique, but row-echelon is not.
You may have different forms of the matrix and all are in row-echelon forms.


Why Do The Axioms of Euclidean Geometry Not Need To Include the Definition of Space? EDIT: update, I found that Euclid's axioms are not considered rigorous.  David Hilbert did a full axiomatization of Euclidean Geometry (1899 in his book Grundlagen der Geometrie--tr. The Foundations of Geometry).  To do so, he required 6 primitive terms which were undefined, including points, lines, and planes.  Lines and planes are both spaces.  Therefore, a true axiomatization of Euclidean Geometry does in fact require space.

source: http://userpages.umbc.edu/~rcampbel/Math306/Axioms/Hilbert.html
http://en.wikipedia.org/wiki/Hilbert%27s_axioms#The_axioms

I left the rest of the post as it was below, but I believe ^^this is the answer.
...........

  aside: Here's idea: Euclidean geometry is done using points and lines (0 and 1 dimensional spaces embedded in 2 dimensional space.)  Why stop there?  Why not do geometry with plane, line, point constructions embedded in 3 space?  Or 3-space constructions in 4-space?  The intersection of two lines is a point, the intersection of two planes would be a line, and the intersection of two "non parallel" 3-spaces would be a plane.

Here are the axioms of Euclidean Geometry:

A straight line segment can be drawn joining any two points.
Any straight line segment can be extended indefinitely in a straight line.
Given any straight line segment, a circle can be drawn having the segment as radius and one endpoint as center.
All right angles are congruent.

That's without even including the parallel postulate.

Here are the axioms: http://mathworld.wolfram.com/EuclidsPostulates.html

  EDIT: here's a possible first postulate that I came up with. Timax's postulate: There exists a 2-dimensional space which has distance-preserving maps.  (maybe some linear algebra or topology would be necessary to define it fully)

But how can you even say "points exist" without first saying "space exists"? If points are defined as being mathematically represented by co-ordinates, (eg (23, 1, 9) is a point in 3-space,) then it seems the first axiom would have to be "3-space exists", and "real numbers exist" in order to create co-ordinates.

It seems pretty obvious that space must come before points.

I guess you could say a point exists, but to get beyond a zero dimensional space, and to get more than 1 point, you would need to establish that space exists.  You can't have 2 points without at least 1 dimension.  To get more than 1 line, you have to have 2 dimensions, to get more than 1 plane, you have to have 3 dimensions.

This is probably why the concept of "space" always seems so mysterious, BECAUSE THE EXISTENCE OF SPACE IS AN AXIOM THAT WAS NEVER STATED!

Here's another idea: Do Euclid's Elements-style constructions in 3 dimensional space or even N dimensional space, instead of only 2 dimensional space.  Has that been done?

2015 M. Wanzek

Why would you have to say “space exist” before you can say “points exist”? In order to have something that contains all those points? But then by the same reasoning, you can't say “space exist” without first postulating the existence of some entity that can contain spaces. Before you know it, you require an infinite regress of existence statments.

In short, “points exist” is a fine start. It was good at Euclid's time, and it's good now.

You've got to remember that this was done thousands of years ago. Euclid probably didn't realize the need for such an axiom. Euclid's Elements is one of the first attempts at making math logically rigorous. That took mathematicians hundreds of years to do and only really got going around the 1600s.


Unimodular Matrix Inverse Proof Confusion Regarding the proof of Lemma 2: Matrix $A$ is totally unimodular if and only if the matrix $[A |I]$ is TU, I do not understand the first step on permuting square submatrix of B to the desired form.

Proof of Lemma 2: Let $A$ be totally unimodular. Any square submatrix $B$ of $\begin{bmatrix} A & I\end{bmatrix}$ can be permuted to the form
$$
\tilde B = \begin{bmatrix}
  A_1 & 0 \\
  A_2 & I_k \\
\end{bmatrix}.
$$
where $A_1$ is a square sub-matrix of $A$, and so $\det(A_1) \in \{-1,0,1\}$.

We have

$$ \det(B) = \pm\det(\tilde B) = \det(A_1) \in \{-1,0,1\}.$$

Take a submatrix $B$ of $\begin{bmatrix} A & I\end{bmatrix}$,

Suppose you pick $k$ columns from $I$.

Note that row swapping doesn't change the magnitude of the matrix. Hence when you examine the TU property, you can permute the rows. Just move those zero rows to the top. That would result in the form that they describe.

This proof is easy piece.
Okay, so for a matrix $A$ to be $TU$ we need to have that any submatrix $A'$ in $A$ must verify:
$$det(A') \in \{-1,0,1\}$$
Let's say that $B$ is then $[A~~ I]$ and that $B'$ is a submatrix of $B$ then $B'$ can be written as $[A'~~ I']$
By the Determinant Expansion by Minors from the columns in $I'$ we have:
$$det(B') = \pm det(B'') = \ldots = \pm det(B'''') = \pm det(A'''')$$
$$rank(I') = rank(I'') + 1 = rank(I''') + 2 = \ldots$$


How to prove that a sequence is unbounded I want to ask, how to start a proof that shows a sequence to be unbounded

Define a sequence $\{X_n\}$ by

$$X_1 = 1 ,\quad X_{n+1} = X_n + \sqrt{X_n}  \quad \text{for}\ n \geq 1$$

Prove that $\{X_n\}$ is unbounded.

First,
$x_{n+1} > x_n \ge 1$
for all $n$.

Second,
$x_{n+1} \ge x_n+1
$.

Therefore,
$x_n \ge n$,
so $x_n$ is unbounded.

In this case, you can show that $\lim_{n\to\infty} a_n=\infty$, which implies that the sequence is unbounded.  In other words, one shows that:
$$\text{For any $M>0$, there exists $N>0$ such that $n\geq N$ implies $|a_n|>M$}$$
In this case, what you want to note is that $X_1 =1$, and $\sqrt{x}\geq 1$ for any $x\geq 1$.  With $X_{n+1} := X_{n+1} + \sqrt{X_{n}}$, prove that $X_n \geq n$ for all $n$, and conclude that $X_n$ is unbounded.


If $G$ is a non-cyclic group of order $n^2$, then $G$ is isomorphic to $\mathbb{Z_n} \oplus \mathbb{Z_n}$ I've independently come up with a question (I know it's been asked before, but I can't find the question online) involving the external direct product, non-cyclic groups and isomorphisms.  So, is the following statement true?

Claim:  "If $G$ is a non-cyclic group of order $n^2$, then $G$ is isomorphic to $\mathbb{Z_n} \oplus \mathbb{Z_n}$."
Where $\mathbb{Z_n} \oplus \mathbb{Z_n}$ is the external direct product of $\mathbb{Z_n}$ and $\mathbb{Z_n}$.

I've been thinking about this for a few hours, but really can't figure out a good place to start (or a counter-example).

First, $\mathbb{Z}_n \oplus \mathbb{Z}_n$ is abelian, while there are many non-cyclic groups that are non-abelian (take $S_3$ for example), so the answer to your question as written is immediately no.

However, what if we only consider abelian non-cyclic groups? Then $\mathbb{Z}_2 \oplus \mathbb{Z}_6$ and $\mathbb{Z}_2 \oplus \mathbb{Z}_2 \oplus \mathbb{Z}_2$ are two counterexamples you might consider. [After OP's edit: a counterexample is $\mathbb{Z}_2 \oplus \mathbb{Z}_2 \oplus \mathbb{Z}_2 \oplus \mathbb{Z}_2$, which has order $16$ but is not $\mathbb{Z}_4 \oplus \mathbb{Z}_4$.]

[Note: this is a relevant result that you might already know: if $m$ and $n$ are coprime, then $\mathbb{Z}_m \oplus \mathbb{Z}_n \cong \mathbb{Z}_{mn}$.]

What you might then ask is if every abelian group can be written as the direct product of cyclic groups, and this is true, but not obvious: Classification of finitely generated abelian groups.

The non cyclic-group $G=Z_2\times Z_2$ is of order $4$, but is not isomorphic to $Z_4\oplus Z_4$.


Limits and While Loops Question: Consider the following program. Does $f(1)=\infty$?

  \begin{align*}
f(i):=&|\text{while } \frac{1}{i}>0\\
&||i\leftarrow i+1\\
&|i
\end{align*}

I would say that $f(1)=\infty$ is a true statement. The program does not terminate, but one could consider the sequence of points $\{x_i\}_{i\in \mathbb{N}}$ given by $x_i:=\frac{1}{i}$, so that  $\{x_i\}_{i\in \mathbb{N}}$ converges to the limit $0$ which makes $\underbrace{\frac{1}{i}}_{=0}>0$ false which means $f(1)=\infty$. However, I could be wrong.

The only thing we can say here is that $f(1)$ is not defined, or that the program does not terminate for the input $1$.

We cannot say that $f(1)=0$.


Equation of straight lines from a point making equal angles with given two lines. Given two lines $3x = 4y +7$ and $5y =12x + 6$. Find the angle between them and also the equations to the two striaght lines which pass through point $(4,5)$ and make equal angles with two lines.

I can get the angles easily as $\pi - \tan^{-1}(\frac{33}{56})$ and $\tan^{-1}(\frac{33}{56})$, But the problem is the second part.

    I know that if a point is given to me as $(x^{\prime} , y^\prime)$ and a line $y =mx + c$, then if we consider the line as the base of a isosceles triangle(it is the unequal side of that isosceles triangle) and the given point is the opposite to the line, then the slope of other two sides are $$ m^\prime = {\tan \alpha \pm m \over 1 \mp m\tan\alpha }$$
    where $\alpha$ is equal angle of the triangle.

So to solve that question I supposed that the given two lines are the two equal sides of a isosceles triangle. With $\displaystyle\tan \alpha =  \tan\left({\pi - \tan^{-1}(\frac{33}{56}) \over 2}\right),  \tan\left({\tan^{-1}(\frac{33}{56}) \over 2}\right) = {11\over 3}, {3\over11}$

Since I know $m^\prime$ in this situation (I assume it to be ${12\over5}$ and ${3\over 4}$(Slopes of the given lines)).

so putting all this in the equation

$$ \left({12\over5} \text{ or } {3\over 4}\right)= {\left({11\over 3} \text { or } {3\over11}\right) \pm m \over 1 \mp m\left({11\over 3} \text { or } {3\over11}\right) }$$

Solving for $m$ in the above equation gives $6$ different values of $m$ which directly implies $6$ different striaght line equations; Of which only two are correct.

How can I find these two out from these $6$ equations ?
Any easier way of doing this question ?

Here dotted lines are the desired equations and solid lines are the given equations.

(Click on the picture for the link to graphing calculator)

Let $y=m(x-4)+5$ be the equation of a generic line through $P=(4,5)$. Compute the intersection points $A$ and $B$ of that line with the two given lines (as a function of $m$). Find the squared distances of $A$ and $B$ from $O=(-59/33,-34/11)$, the common point of the given lines. Finally, set $AO^2=BO^2$ to find the values of $m$ which give an isosceles triangle.

I make you clear that I'm writing equation first as

12y = 5x +6   ----(1)  instead of 5y =12x +6

And 2nd eqⁿ is 4y = 3x -7    ------(2)

Here
m₁ = 5/12, m₂ =3/4.

Let m is slop of desire line, then

y = m(x-4) +5   ------(3)

Now angle between (1)&(3) will be

Θ₁ = |(12m -5)/(12 +5m)|

Similarly angle between (2)&(3) will be

Θ₂ = |(4m -3)/(4 +3m)|

Now since Θ₁=θ₂ given,then

(12m -5)/(12 +5m)
           = ± (4m -3)/(4 +3m)

After cross multiplication,we get

48m+36m² -20-15m

= ± (48m-36+20m²-15m)

On Taking +ve sign

16m² = -16, which is unacceptable.

On taking -ve sign

96m+56m² -30m-56=0

⇒56m² +66m -56=0

⇒28m² +33m -28=0

⇒m = (-33±65)/56

    =32/56 or -98/56

    =4/7 or -7/4

Thus desired line would be

7y = 4x -11 & 4y = -7x +33.


Alternate proof of prime avoidance lemma. All the ideals mentioned below are in a commutative ring $A$ with identity. Let me start with a special case. Note that the ideals in Lemma 1 below do not need to be prime.

  Lemma 1. If $\mathfrak a\subset \mathfrak b\cup\mathfrak c$, then $\mathfrak a\subset \mathfrak b$ or $\mathfrak a\subset \mathfrak c$.

Here is my proof: If $\mathfrak a\subset \mathfrak b\cup\mathfrak c$, then it’s not hard to see that $\mathfrak a=(\mathfrak a\cap\mathfrak b)\cup(\mathfrak a\cap\mathfrak c)$. The right hand side union is an ideal thus one of them is contained in another, i.e. $\mathfrak a= \mathfrak a\cap\mathfrak b$ or $\mathfrak a= \mathfrak a\cap\mathfrak c$.

The above result is not true in general for three or more unions. Now I am trying to give a proof without involving argument about elements for Prime Avoidance Lemma:

  If an ideal contained in a finite union of prime ideals, say $\mathfrak p_1,...,\mathfrak p_n$, then it is contained in at least one $\mathfrak p_i$ for some $i$.

Attempt for a proof:

  Lemma 2. An ideal $\mathfrak p$ is prime if and only if $\mathfrak{ab}\subset\mathfrak p$ implies that $\mathfrak a\subset \mathfrak p$ or $\mathfrak b\subset\mathfrak p$.

Proof is omitted for Lemma 2. Let $\mathfrak a$ be an ideal contained in $\bigcup\limits_{i=1}^n\mathfrak p_i\subset \mathfrak p_n\cup\sum\limits_{i=1}^{n-1} \mathfrak p_i$. By applying Lemma $1$ we can reduce the question to the following:

  Prove that if $\mathfrak a\subset \sum\limits_{i=1}^{n-1} \mathfrak p_i$, then $\mathfrak a\subset \mathfrak p_i$ for some $i$.

By the induction hypothesis if $\mathfrak a\subset \bigcup\limits_{i=1}^{n-1}\mathfrak p_i$ then we are done. Thus by applying the minimality of the sum of ideals we have restricted our attention to the following two cases:

(1). $\mathfrak a=\sum\limits_{i=1}^{n-1}\mathfrak p_i$.

(2). $\mathfrak a$ is not contained in $\bigcup\limits_{i=1}^{n-1}\mathfrak p_i$ and is properly contained in $\sum\limits_{i=1}^{n-1}\mathfrak p_i$.

How do I finish the proof? Any other different proof is also welcomed. Thank you.

Note first that for a ring with three distinct maximal ideals $\mathfrak{a},\mathfrak{b},\mathfrak{c}$, then $\mathfrak{a} \subset \mathfrak{b}+\mathfrak{c}$.

So we use indeed induction: assume that $\mathfrak{a} \subset \bigcup_{i=1}^n{\mathfrak{p}_i}$.

If any smaller reunion of the $\mathfrak{p}_i$ contains $\mathfrak{a}$, we are done by induction hypothesis. So in particular, we have some $x_i \in \mathfrak{p}_i \backslash \mathfrak{p}_n$. We also have some $a \in \mathfrak{a}$ not in any $\mathfrak{p}_k$ (induction hypothesis) with $k <n$, and some $b \in \mathfrak{a} \backslash \mathfrak{p}_n$.

If $a \notin \mathfrak{p}_n$, we are done. Otherwise, consider $a+bx_1 \ldots x_{n-1}$.


Construct a field of 25 elements. We need to construct a field of 25 elements.

By using a result : For a prime p and a monic irreducible polynomial $p(x)$ in $\Bbb F_p [x]$ of degree $n$ , then the ring $(\Bbb F_p [x] / <p(x)>)$ is a field of order $p^{n}$.

So , for the question problem above , we need a irreducible polynomial of degree $2$ in $\Bbb Z_5 [x]$.

Solution says , $x^{2} + x +1$ does our job , since it's irreducible in $\Bbb Z_5 [x]$.

My question : Can't we take $ x^{2} + 2$ , it has no zeroes in $\Bbb Z_5$ , and is irreducible. In every solution I've gone through , the polynomial $x^{2} + x+1$ is taken. Can't we consider $ x^{2} + 2$ ?

Yes, you can take $x^2+2$: any degree two irreducible polynomial would do.

If you take $K=\mathbb{F}_p[x]/\langle x^2+2\rangle$ and denote by $u$ an element such that $u^2=3$, you can find in it a root of $x^2+x+1$:
$$
(a+bu)^2+(a+bu)+1=0
$$
is the same as
$$
a^2+3b^2+a+1=0,\qquad 2ab+b=0
$$
so we must take $a=-1/2=2$ and so $3b^2=3$; thus $2+u$ and $2-u$ are the solutions.

there is a theorem that says that if you have a field $F$ and a polynomial $f(x)$ of degree 2 or 3 in $F[x]$ that does not have zeros in $F$, then $f(x)$ is irreducible in $F$, therefore $x^2+2$ is irreducible in $\mathbb{Z}_5$.


Which of the numbers $99^{100}$ and $100^{99}$ is the larger one? Which of the numbers $99^{100}$ & $100^{99}$ is the larger? Solve without using logarithms.

Note that
$$\begin{align}
99^{100} > 100^{99} &\iff 99 \cdot 99^{99} > 100^{99}
\\
&\iff 99 > (100/99)^{99}
\\
&\iff 99 > \left( 1 + \frac{1}{99}\right)^{99}
\end{align}$$

Since $(1 + \frac{1}{n})^n < 3$ for all integers $n$, the above inequalities are all true.  Thus, $99^{100} > 100^{99}$.  In general, you should expect that $x^y > y^x$, whenever $y > x$.

From experimenting with small numbers:

scala> (0 to 5).map (x=> (math.pow (x, x+1) - math.pow (x+1, x))).mkString ("; ")
res18: String = -1.0; -1.0; -1.0; 17.0; 399.0; 7849.0

scala> (0 to 5).map (x=> (math.pow (x, x+1), math.pow (x+1, x))).mkString ("; ")
res19: String = (0.0,1.0); (1.0,2.0); (8.0,9.0); (81.0,64.0); (1024.0,625.0); (15625.0,7776.0)

you can conclude, that the first one is growing faster than the second. Of course this is only an indication.


Every infinite set has an infinite countable subset? As the title says, that's all my question. Let me state it again:

  Is it true that every infinite set has an infinite countable subset?

It seems so trivial, my thought goes like this: pick an arbitrary element and denote it as $x_1$; pick the next one and denote it $x_2$, and so on.

Is my proof correct? Since it seems so simple, I'm not sure of it.

To avoid any further confusion, the definitions used are:

Finite: In bijection with $\{1\ldots n\}$ for some $n$.
Infinite: Not finite.
Countably infinite: In bijection with $\Bbb N$.
Countable: Finite or countably infinite.

You are on the right track, however you cannot use the phrase "the next one".  The idea is:Let $S$ be an infinte set. Pick an element $x_1\in S$, then since $\{x_1\}$ is finite,
$S\setminus\{x_1\}\ne\emptyset$, so pick an element $x_2\in S\setminus\{x_1\}$ and so on.

Note the Axiom of Choice is involved here.

If you are using Dedekind's definition of infinity, then there must exist an injective, but not surjective function $f: S \to S$, where $S$ is your infinite set. Then there must also exist a countable $n\subset S$ that is order-isomorphic to $\mathbb {N}$ with the function $f$ serving as the successor function and any element outside the range of $f$ serving as the first element $0$ (or $1$).

For details and formal proofs, see my posting "Infinity: The Story So Far" at my math blog.


Prove there is sequence $(t_n)$ such that: i) $t_n →\infty $, ii)$ f'(t_n) =0$, and iii) $ \lim f(t_n) = 1$. Let $f : [0, \infty) \rightarrow \mathbb{R} $ be a bounded, differentiable function satisfying $\underset{ t \to \infty } \lim \ inf \ f(t) = 0$ and $\underset{ t→∞ } \lim \sup \ f(t) = 1$. Prove that there is a sequence $(t_n)^∞ _{n=1}$ such that: i) $t_n →\infty \ as \  n \to \infty$, ii)$ f'(t_n) =0$ for all $n \in N$, and iii) $ \underset{n \to \infty}\lim f(t_n) = 1$.
(i) Since, every bounded sequence in $R^k$ contains a convergent subsequence, there exists a sequence $(t_n)$ such that $t_n \to \infty$ as $n \to \infty$.
(iii) Since $\underset{ t→∞ } \lim \sup \ f(t) = 1$, we have $ \underset{n \to \infty}\lim f(t_n) = 1$.
(ii) To show $ f'(t_n) =0$ , how to use  $\underset{ t \to \infty } \lim \ inf \ f(t) = 0$ and $\underset{ t→∞ } \lim \sup \ f(t) = 1$.

If $f:[0,\infty)\to\mathbb{R}$ is bounded then $|f(x)|<M$ for all $x\in[0,\infty)$. You are given that $\liminf_{t\to\infty}f(t)=0$ and $\limsup_{t\to\infty}f(t)=1$, so $f$ must be bouncing back and forth in some manner, think of oscillations. The assumption that $f$ is differentiable, and hence continuous, is important now because it tells us that as $f$ is oscillating, it does so continuously and so it must have infinitely many turning points.
Can you use this now to find $(t_{n})$?

To get condition (ii), you are going to have to explicitely define the $t_n$ to satisfy this condition, it can't stem from a property about the limits.
Here is a way to do it: first define a sequence $s_n$ such that:

$0<f(s_n)<1$

$s_n$ is increasing

$s_n \to \infty$

$\forall n, \exists x \in ]s_n, s_{n+1}[:\ f(x) > \max\big(f(s_n), f(s_{n+1}),  1 - \frac{1}{n}\big)$

This can easily be constructed by induction: take any $s_1$ such that $0<f(s_1)<1$. Once you have $s_n$, since $f(s_n)<1$ and $\lim \sup f = 1$, there exists $x_0 > s_n+1$ such that $f(x_0) > f(s_n)$ and $f(x_0)>1-\frac{1}{n}$. And then since $\lim \inf f = 0$, you can choose $s_{n+1} > x_0$ such that $f(s_{n+1}) < f(x_0)$ and $f(s_{n+1}) < 1$. Simply note that $s_{n+1} > x_0 > s_n+1$, so $(s_n)$ is increasing and $s_n \to +\infty$. The fourth point is obvious by construction.
$ $
Once we have this sequence, just define $t_n = \mbox{argmax}_{t \in [s_n, s_{n+1}]} f(t)$. Using the fourth point, $t_n \in ]s_n, s_{n+1}[$,and since $f$ is differentiable, $f'(t_n) = 0$. Also, there exists $x_0 \in ]s_n, s_{n+1}[$ such that $f(x_0) > 1 - \frac{1}{n}$ so $f(t_n) \ge f(x_0) > 1 - \frac{1}{n}$. Thus, $f(t_n) \to 1$.
Finally, you have $t_n \ge s_n$, so $t_n \to +\infty$.


Evaluate $ \int_{0}^{1} \sqrt{1+\sqrt[3] {x}}\,dx,$ Evaluate
  $ \int_{0}^{1} \sqrt{1+\sqrt[3] {x}} \,dx,$

So,i've tried doing the substitution $ u=\sqrt[3] {x}$( this was a suggestion by wolfram), but frankly i am reaching a point where it doesn't work . Could help me out ?

If $u = \sqrt{1+\sqrt[3]{x}}$, then the integral becomes
$$\int_0^1 \sqrt{1+\sqrt[3]{x}} \ dx = 6 \int_1^\sqrt{2} u^2 \, (u^2-1)^2du.$$
Now, expand the integrand and integrate a polynomial.


Trying to show that $\sum_{n=1}^{\infty}{\frac{1}{n^2}}$ converges by using the direct comparison test I know that if $\forall n \in \mathbb{N}$, $ 0 \le a_n \le b_n$. Then if $\sum b_n$ converges then $\sum a_n$ converges.

I am having trouble finding a convergent $\sum b_n$ for which
\begin{align*}
 \sum_{n=1}^{\infty}{\frac{1}{n^2}} \le \sum b_n
\end{align*}
Is it possible to do it with comparison test, or do I need another method?

You can compare, as said in the comments, by
$$
\frac1{n^2}<\frac1{n(n-1)}=\frac1{n-1}-\frac1n
\text{ so that } ~~
\sum_{n=1}^\infty\frac1{n^2}\le1+\sum_{n=2}^\infty \frac1{n-1}-\frac1n=2$$
by telescoping the upper bound. A closer bound is
$$
\frac1{n^2}<\frac1{(n+1)(n-1)}=\frac1{2(n-1)}-\frac1{2(n+1)}
\text{ so that } ~~
\sum_{n=1}^\infty\frac1{n^2}\le1+\frac12\sum_{n=2}^\infty \frac1{n-1}-\frac1{n+1}=\frac74
$$
where the telescoping "leap-frogs".

You could also use the construction of the Cauchy condensation test where
$$
\frac1{n^2}<\frac1{4^k}~~\text{ for }~~ 2^k\le n < 2^{k+1} ~~
\text{ so that } ~~
\sum_{n=1}^\infty\frac1{n^2}\le\sum_{k=0}^\infty2^k\cdot \frac1{4^k}=\sum_{k=0}^\infty2^{-k}=2$$

We know $$n^2 > (n - 1)^2$$

Inverting the fractions the sign changes and we get

$$n^2 > (n - 1)^2 \implies \frac{1}{n^2} < \frac{1}{(n - 1)^2}$$

Take $b_n = \frac{1}{(n - 1)^2}$

Hence $a_n < b_n$ is satisfied.

P-series are of the form $$f(x) = \frac{1}{x^p}$$

And when $p > 1$ the series converges; $p \le 1$ the series diverges.

Since $b_n$ is a p-series with $p > 1 (= 2)$, $b_n$ converges.

Since $b_n$ converges and $b_n > a_n$; $a_n$ converges as well.


What is the probability that a run of $n$ consecutive successes occurs before a run of $m$ consecutive failures? This example comes from the book "A First Course in Probability" by Sheldon Ross, Chapter 3, Section 3.5, Example 5C, Page 95. The example is given a solution but there is a part that I don't really understand. I hope someone can explain the logic to me.

The original question is as follow:

Independent trials, each resulting in a success with probability $p$ or a failure with probability $q = (1-p)$, are performed. We are interested in computing the probability that a run of $n$ consecutive successes occurs and before a run of $m$ consecutive failures. We are given the solution as follow:

Solution.: Let E be the event that a run of $n$ consecutive successes occurs before a run of $m$ consecutive failures. To obtain P(E), we start by conditioning on the outcomes of the first trial. That is letting H denote the event that the first trial results in a success, we obtain $$P(E) = pP(E|H)+qP(E|\bar H)$$ Now, given that the first trial was successful, one way we can get a run of $n$ successes before a run of $m$ failures would be to have the next $n-1$ trials all result in successes. So, let us condition on whether or not that occurs. That is, letting F be the event that trials 2 through n all are successes, we obtain $$P(E|H) = P(E|FH)P(F|H)+ P(E|\bar{F}H)P(\bar{F}|H)$$ On the one hand, clearly, P(E|FH) = 1; on the other hand, if the event $\bar{F}H$ occurs, then the first trial would result in a success, but there would be a failure some time during the next $n-1$ trials. However, when this failure occurs, it would wipe out all of the previous successes, and the situation would be exactly as if we started out with a failure. Hence, $$P(E|\bar{F}H) = P(E|\bar{H})$$
Pause !! This is the part where I don't really understand. I agree that when this failure occurs, it would wipe out all of the previous successes. But the situation for E to occur shouldn't be exactly the same as if we started out with a failure because if $F$ is the event that trials $2$ to $n$ are successes, then $\bar{F}$  is the event where not all $2$ to $n$ trials are successes. That being said, the last next $n-1$ trials could be successes. As an example, if the last two of $n-1$ trials is a failure then a success; then the situation for E to occur should be exactly the same as $P(E|H)$. So, can someone explain to me why $P(E|\bar{F}H) = P(E|\bar{H})$ ?

Since we are looking at an infinite number of trials, after removing a finite prefix of trials, we still have an infinite number of trials. Therefore, we can simply ignore this prefix for the probability of $E$ (even the conditions for $E$ would already be fulfilled by the prefix) and therefore $P(E) = P(E \mid B)$ for any event $B$ that restricts the outcomes of the trials only for a finite prefix.

This hopefully answers your question, but the partial solution you gave above seems rather complicated to me. Thankfully, I was able to find a pdf version of the book via Google. There, the author gives a rather complicated probability for $P(E)$ (not included in your question). That surprised me, because intuitively I thought, $P(E) = 1$.

Thinking about that in more detail, I am now convinced that, indeed, $P(E) = 1$: Let $A$ be the event, that we consecutively have $n$ successes and $m$ failures in $n + m$ trials. Clearly, $P(A) ≠ 0$, because $P(A) = p^n ⋅ q^m$ (assuming $0 < p < 1$).

Now let us iterate these $n + m$ trials and let $E'$ be the event that event $A$ occurs in the first block of $n + m$ trials, or in the second block of $n + m$ trials, or in the third block, etc. Since $P(A) ≠ 0$ and therefore $P(\overline{A}) ≠ 1$ it is easy to see that $P(\overline{E'}) = 0$, since we are looking at an infinite number of trials. Therefore we have $P(E') = 1$.

Since $E' ⊆ E$, it follows that also $P(E) = 1$.

For short: In an infinite sequence of trials you are able to find any finite pattern with probability $1$.

Therefore I suspect that there is a mistake in the solution in the book – unless I made a mistake myself or misunderstood the problem.

There are several errors which suggest a fundamental misunderstanding of probability on the behalf of the author of this textbook. For starters, this problem is not properly defined - how many trials are there in total? Are we talking about precisely $n+m$ trials, or perhaps some finite number $N>n+m$, or an infinite sequence? The worked solution is inconsistent in which assumption it is using, and arrives at a conclusion which is incorrect regardless of which assumption we use. Let's go through it:

"One the one hand, clearly, $P(E|FH)=1$" - nope. $E$ is the event that we have $n$ successes followed by $m$ failures; $H$ is the event that the first trial is a success; $F$ is the event that the second through $n^\text{th}$ trials are successes. If we are assuming precisely $n+m$ trials, then given $F,H$, the event $E$ occurs if and only if trial $n+1$ through $m$ are failures. This clearly has probability less than $1$. Similarly, this will be less than $1$ (but larger than $P(E)$) if we have $N>n+m$ trials. If we have an infinite number, then yes, this probability will be $1$; but this is no more "clear" than the original problem.
"Hence, $P(E|F^cH)=P(E|H^c)$" - what?! You gave a very simple counterexample which shows this is not the case. The instance of the first failure indeed "resets" the count in a sense, but we might have only a single failure on the second trial, in which case we still have $n-2$ successes in a row. If we are assuming $n+m$ trials, the probability is already zero; if we are assuming $N>n+m+1$, this situation leaves us with a positive probability (whereas a later failure may make the probability zero) and if we are assuming infinitely many trials, then as before, each side is equal to $1$. (As Tony Dietze points out, since $E$ is an event only interested in the existence of a certain string, $P(E|C)=1$ for any event $C$ which depends only on finitely many trials.)
"Now, $GH^c$ is the event that the first $m$ trials all result in failures, so $P(E|GH^c) = 0$" - why? This is true if we have only $n+m$ trials, but then, $P(E|C)=0$ for any event $C$ which includes a failure in the first $n$ trials. In general, this statement is only true if we have $N<n+2m$ total trials. The event $G$ is also an extremely strange event to condition on - why should we care if all the first $m$ trials are failures? Isn't one enough? Additionally, the author makes the same weird error as before, namely asserting that $G^cH^c$ involves a success wiping out all the previous failures and so claiming that $P(E|G^cH^c)=P(E|H)$. Absolute nonsense.
"Thus, $P(E)=\ldots=\frac{p^{n-1}(1-q^m)}{p^{n-1}+q^{m-1}-p^{n-1}q^{m-1}}$" - clearly not true. If there $n+m$ trials, then $E$ is precisely the event where the first $n$ are success, and the rest are failures, so $P(E)=p^nq^m$. If there are infinitely many trials, the reasoning Tony Dietze provided shows that $P(E)=1$ (intuitively speaking, if we run trials for long enough, eventually any finite string will appear). If there are $N>n+m$ trials, $P(E)=P_N(E)$ must depend on $N$ with $P_N(E)\to1$ as $N\to\infty$.

It really blows my mind that such bad mathematics could appear in a textbook. I have not read any other part of this book, but if this example is anything to go by, you should get rid of it immediately and find a new textbook on probability.


Do you prove all theorems whilst studying? When you come across a new theorem, do you always try to prove it first before reading the proof within the text? I'm a CS undergrad with a bit of an interest in maths. I've not gone very far in my studies -- sequence two of Calculus -- but what I'm trying to understand right now, though, is how one actually goes about studying so that when finished with a good text, there's more of an intuitive understanding than superficial.

After reading "The Art Of Problem Solving" from the Final Perspectives section of part eight in 'The Princeton Companion to Mathematics', it seems to hint at approaching studying in that very way. A quote in particular, from  Eisenstein, that caught my attention was the following -- I'm not going to paraphrase much:

  The method used by the director was as follows: each student had to
  prove the theorems consecutively. No lecture took place at all. No one
  was allowed to tell his solutions to anybody else and each student
  received the next theorem to prove, independent of the other students,
  as soon as he had proved the preceding one correctly, and as long as
  he had understood the reasoning. This was a completely new activity
  for me, and one which I grasped with incredible enthusiasm and an
  eagerness for knowledge. Already, with the first theorem, I was far
  ahead of the others, and while my peers were still struggling with the
  eleventh or twelfth, I had already proved the hundredth. There was
  only one young fellow, now a medicine student, who could come close to
  me. While this method is very good, strengthening, as it does, the
  powers of deduction and encouraging autonomous thinking and
  competition among students, generally speaking, it can probably not be
  adapted. For as much as I can see its advantages, one must admit that
  it isolates a certain strength, and one does not obtain an overview of
  the whole subject, which can only be achieved by a good lecture. Once
  one has acquired a great variety of material through [...] For
  students, this method is practicable only if it deals with small
  fields of easily, understandable knowledge, especially geometric
  theorems, which do not require new insights and ideas.

I feel that this type of environment is something you don't often see, especially in the US -- perhaps that's why so many of our greats are foreign born. As I understand it, he does go on to say that he wouldn't particularly recommend that method of study for higher mathematics, though.

A similar question was posed to mathoverflow where Tim Gowers (Fields Medal) went on to say that he recommended similar methods to study: link

I'm not quite certain that I understood the context of it all, though. Upon asking a few people whose opinion mattered to me, I was told that it if time were precious to me, it would be a waste going about studying mathematics in that way, so I'd like to get some perspective from you math.stackexchange. How do you go about studying your texts?

Edit: Broken link added fixed.

There is a continuum in the way one understands a theorem.
At one end  of the spectrum mathematicians  just try to understand the statement and use it as a black box .
At the other end they understand the theorem so well that they improve on it: this is called research.

An important thing to keep in mind is that your attitude toward a result is not fixed for ever: you may first consider  it as a black box and solve exercises by blindly using it, then see how it is quoted in proving corollaries or other theorems and finally come back to it and realize that it is actually quite natural.

Professors  have  the advantage that they really have to understand a theorem if they want to teach it well and answer the students' questions.
One of the great aspects of this site is that everybody can be a teacher:
 I strongly advise you to try and answer questions here.  They are at all possible levels and I am sure you can find some that you will  answer very competently.

A paradoxical way of expressing what it means to have understood a theorem is to say that ideally you have to reach the stage where you consider that all its proofs in the literature are "wrong": it is a patently absurd statement but it conveys the idea that the theorem is now yours because you have integrated it into your own mathematical world.

Edit
Since Neal asks about this in his comment, let me emphasize that when I say that proofs in the literature are "wrong" I mean that, although they are technically 100% correct, they don't correspond to the subjective way one has organized one's understanding of the subject.

For example, the definition I like for a finite  field extension $K/k$ to be separable is that it is étale i.e. that the tensor product with an algebraic closure of $k$ is split: $K\otimes_k\bar k\cong \overline {k} ^n$ .
I know this is rather idiosyncratic and of course I know the equivalence with the usual definition, but then I feel that long proofs that $\mathbb C\otimes _\mathbb R\mathbb C$ is not a field are "wrong" since I know,  by the  definition of separable  I have interiorized, that  $\mathbb C\otimes _\mathbb R\mathbb C=\mathbb C^2$.
Let me emphasize  that all this a completely personal and secret [till today :-)] attitude within myself and that I absolutely don't advocate that other mathematicians should change their definition of separable.

See my question on mathoverflow for books that are designed for this sort of study:

https://mathoverflow.net/questions/12709/are-there-any-books-that-take-a-theorems-as-problems-approach

Also, from a short bio of S.Ramanujan:

  At sixteen, Ramanujan borrowed the English text "Synopsis of Pure Mathematics". This work was to prove a deep influence on Ramanujan's development as a mathematician, for it offered mathematical theorems without accompanying proofs, thereby prompting Ramanujan to prove the material by his own mathematical cunning.

Full text here: http://myhero.com/go/hero.asp?hero=s_ramanujan


Compute $(\sin4^\circ)^2 +(\sin8^\circ)^2+(\sin12^\circ)^2+\cdots+(\sin176^\circ)^2$ Angle of sine is in degrees, can anyone show me an easy soln to this? This was question was given to us for 1minute without calcu.

I know that $\sin4^\circ=\sin176^\circ$, $\sin8^\circ=\sin172^\circ$.....
But dont know how to proceed

We may add $\sin^2(0)=0$, too.

In that way, one is summing $180/4=45$ terms of the form $\sin^2(x)$ where $x$ is uniformly and rather densely distributed over the angles (because the periodicity is $180$ degrees), so one basically calculates $45$ times the average value of $\sin^2(x)$. The average value of $\sin^2(x)$ is $1/2$ so the result is approximately $45/2=22.5$.

One may explicitly verify that this result $22.5$ is exact by using
$$\sin^2(x) = \frac{1-\cos(2x)}{2}$$
Now, sum the right hand side for $x$ going from 0 degrees to 176 degrees, with the step 4 degrees. The first term gives us $45/2=22.5$.

The second term, the sum of $-\cos(2x)/2$ for $x$ from 0 degrees to 176 degrees with the step 4 degrees, is zero. It's because the argument $2x$ of the cosine goes from 0 to 352 degrees, with the step of 8 degrees.

The cosine $\cos A$ is the real part of $\exp(iA)$. But the sum of $\exp(iA)$ over $A$ from 0 to 352 degrees with the step of 8 degrees vanishes because these complex numbers $\exp(iA)$ are uniformly distributed along the unit circle. By the ${\mathbb Z}_{45}$ symmetry of the polygon with 45 vertices, the sum is zero. Schematically,
$$\sum \cos(2x) = {\rm Re}\sum \exp(2ix),\\
\sum\exp(2ix) = \exp(i\cdot 8^\circ) \sum\exp(2ix)=0 $$
The sum of the phases has to vanish because it is equal to itself times a number different from one (the phase that just cyclically permutes the 45 terms).

So the result is $22.5$.

A simple trigo way as you noted continuing from there we can write it as $$2(\sin^2(4)\ldots\sin^2(88))$$ then we can use $$2\sin(\theta)\sin(\theta)=\cos(0)-\cos(2\theta)$$ thus we get $$22-(\cos(8)+\ldots +\cos(176))$$ now as terms are in ap ie angles we use A.P. formula for $\cos$ i.e. $$\frac{\sin(a+(n-1)b/2)}{\sin(b/2)}\cdot \cos((c+d)/2)$$ where a is first term,b is common difference and c,d are first,last angles respectively which yields result as $-0.500000$ thus sum is $22-(-0.500000) = 22.50000$


Intermediate value property for derivatives I'm going over a proof of a special case of the Intermediate value theorem for derivatives.

It says:

  Consider $a, b\in I$ with $a<b$.

  Suppose first that $f'(a)<0<f'(b)$. We will show $\exists x\in(a, b)$
  such that $f'(x)=0$.

  Since $f'$ exists on $[a, b]$, we must have $f$ continuous on $[a, b]$, so the  Extreme Value Theorem implies that $f$ attains its minimum
  at some point $x\in[a, b]$. This minimum point cannot be an endpoint
  of $[a, b]$.

I just don't understand this last sentence. Why can't the minimum point be an endpoint of $[a, b]$?

Since $$f'(a)=\lim\limits_{h\rightarrow0^+} {f(a+h)-f(a)\over h }<0,$$ it follows that  ${f(a+h)-f(a)\over h }<0$ for $h>0$ sufficiently small (with $a+h<b$). From this, it follows that  ${f(a+h)-f(a)  }<0$ for $h>0$ sufficiently small; thus, the minimum can't occur at $x=a$.

A similar argument will show the minimum can't occur at $x=b$.


Prove that $\sqrt{8}$ is irrational in different method I tried to prove that $\sqrt{8}$ is irrational. I said let $\sqrt{8}$ be rational then $\sqrt{8}$ = $a/b$ where $a$ and $b$ are relatively prime.
Then $2\sqrt{2}=a/b$ , and $\sqrt{2} =a/(2b)$. it is obvious that $RHS$ is rational and $LHS$ is irrational (assumed that $\sqrt{2}$ is proved). So there is a contradiction and proof done.
My question is that is there other ways to prove that $\sqrt{8}$ is irrational?

You could use the equation $x^2-8=0$ and apply the Rational Root Theorem.

let $\sqrt{8}$ is equal to $\frac {a}{b}$ where $(a,b)=1$
Then $8 = \frac {(a^2)}{(b^2)}$ , so $8(b^2)= a^2$
Because of the fact that $(a,b)$ are relatively prime $b$ cannot divide $a$.Hence $8|a^2$.
Then ,let's say $a=2k$ where $k$ is a positive integer.
Then $8(b^2)=(2k)^2$. Then,by simplification $2(b^2)=k^2$ .It is obvious that  $k$ is an even integer, so k=2m where m is positive integer.
so $2(b^2)=(k^2)=(2m)^2$.then $b=8m b^2 = 2(m^2)$.As a result $2|b$ .
we concluded that a and b have common divisor which is $2$. it is a contradiction.


Number of Dyck paths with k returns and b peaks The number of Dyck paths from the origin to $(2n,0)$ which touch the $x$-axis $k+1$ times ($k$ internal touches) is given by
$$\frac{k}{2n-k}{2n-k \choose n}.$$

The number of Dyck paths from the origin to $(2n,0)$ which have $b$ peaks (i.e. local maxima) is given by the Narayana numbers
$$\frac{1}{n}{n \choose b}{n \choose b-1}.$$

My questions is: What is the number of Dyck paths from the origin to $(2n,0)$ which have both $k$ returns and $b$ peaks?

Thanks!

If $n$=semi length of Dyck Path, $p$=peaks and $r$=returns then:

$$F(n,p,r)=\frac{r(n-1)!(n-r-1)!}{p!(p-r)!(n-p)!(n-p-1)!}$$

If $n=p=r$ then $F(n,p,r)=1$.

Example:
$$F(7,4,3)=\frac{3(7-1)!(7-3-1)!}{4!(4-3)!(7-4)!(7-4-1)!}=45$$
There are 45 Dyck Paths with a semi length of 7, 4 peaks and 3 returns.

See Online Encyclopedia of Integer Sequences A253938:

  A pyramid $F(n,p,r)$ of successive triangular arrays read by rows, relating Dyck path peaks and returns to the $x$ axis ($n = $ semilength of Dyck paths, $p =$ number of peaks, $r =$ returns to the $x$ axis).


The simplification of divided difference of cosine function What is the following limit?

$$\lim_{h \to 0}\frac{\cos(\pi/2+h)-\cos(\pi/2)}{h}$$

Why when simplified do you get $(-\sin(h))/h)$?

A cosine wave is just a sine wave shifted to the left by $\frac{\pi}{2}$, so use the following trig identity:

$$\cos(\theta + \frac{\pi}{2}) = \sin(\theta + \frac{\pi}{2} + \frac{\pi}{2}) = \sin(\theta + \pi) = -\sin(\theta)$$

you know :$$cos(a+b)=cos a cos b -sin a sin b\\cos(\frac{\pi}{2}+h)=cos(\frac{\pi}{2})cos h - sin (\frac{\pi}{2}) sin h=0 -1* sin h$$
use this to find that limit
$$\lim_{h\to 0}\frac{cos(\frac{\pi}{2}+h)-cos \frac{\pi}{2}}{h}=\\\lim_{h\to 0}\frac{-1*sin h-cos \frac{\pi}{2}}{h}=\\\lim_{h\to 0}\frac{-1*sin h-0}{h}$$


Finding ideal fluid flow with a boundary I need to find the stream lines of an ideal fluid flow with the given potential of $F(z)=z^2=x^2-y^2+i2xy$ on the upper half of the Cartesian plane.

I have determined that the velocity of the fluid must be $\overline{F'(z)}=2x-i2y$ however this has does not fit the constraint because the fluid has a non-zero imaginary component at the line $\Im(z)=0$ and thus violates the boundary.  It seems like you cannot have this potential and this boundary condition because the fluid would be forced to flow through a boundary.

How do I make the fluid fit the boundary?  How does one do this for a general potential and boundary condition?

Ok, I figured it out, so heres an answer for anyone confused like me in the future (probably still me).

$F'(z)$ is the derivative with respects to $z$ so the velocity is $\overline{F'(z)} = 2\bar{z}$.  This fits our constraint.

Essentially when asked to find the fluid flow of a potential with a constraint, the constraint should not have any impact on the solution other than to provide a domain.  Either the potential fits the constraint or it does not.  You cannot make a potential fit an arbitrary constraint (unless the potential is zero everywhere), which is where the confusion I had was coming from.

The streamlines should look like as the following:

  which belongs to a family of curves $$2xy=c$$


Angles between two vertices on a dodecahedron Say $20$ points are placed across a spherical planet, and they are all spaced evenly, forming the vertices of a dodecahedron. I would like to calculate the distances between the points, but that requires me to find out the angles between the vertices.

From the origin of the dodecahedron, how would I find the angle between two adjacent vertices on the same face, and the angle between two vertices on the same face but not connected by an edge?

As noted in Wikipedia's "Dodecahedron" entry, if $s$ is the length of an edge of a dodecahedron, and $r$ the radius of its circumsphere, then

$$r = s \frac{\sqrt{3}}{4}\left( 1 + \sqrt{5} \right)$$

So, if two points $A$ and $B$ are joined by and edge, and $O$ is the center of the dodecahedron, then $\triangle AOB$ is isosceles with legs $r$ and base $s$; applying the Law of Cosines to its vertex angle, we have ...

$$s^2 = r^2 + r^2 - 2 r\cdot r\cos\angle AOB = 2 r^2 \left( 1 - \cos\angle AOB \right) = 4 r^2 \sin^2\frac{1}{2}\angle AOB$$

so that

$$\sin\frac{1}{2}\angle AOB = \frac{s}{2r} = \frac{2s}{s\sqrt{3}\left(1+\sqrt{5}\right)} = \frac{\sqrt{3}\left(\sqrt{5}-1\right)}{6}$$

whence

$$\angle AOB = 2 \arcsin \frac{\sqrt{3}\left(\sqrt{5}-1\right)}{6} = 41.8103\dots^\circ$$

If $A$ and $C$ are non-adjacent vertices on a face, then $d := |AC|$ is a diagonal of a regular pentagon with side length $s$. Thus,

$$d = \frac{s}{2}\left( 1 + \sqrt{5} \right)$$

Just as above, we can compute

$$\sin\frac{1}{2}\angle AOC = \frac{d}{2r} = \frac{s\left(1+\sqrt{5}\right)}{s\sqrt{3}\left(1+\sqrt{5}\right)} = \frac{\sqrt{3}}{3}$$

whence

$$\angle AOC = 2 \arcsin \frac{\sqrt{3}}{3} = 70.5288\dots^\circ$$

(You may recognize this as the central angle between adjacent vertices of a cube. It's often helpful to realize that a dodecahedron's face diagonals form the edges of a family of cubes, as shown in the Wikipedia entry. Moreover, one can think of constructing a dodecahedron by taking a cube and pitching a pup-tent on each face, where a triangular tent face and a quadrilateral tent face form a regular pentagon.)

There are 3 angles to consider in a polygon involving the origin point. the circumsphere radius (vertices) what you are seeking, the midsphere (center of edges, aka. 1/2 dihedral angle), and the insphere (right angle perpendicular to the face).

Formulas for finding these three radii are:

C = a/4 * √3 * (1 + √5)

M = a/4 * (3 + √5)

I = a/2 * √([25 + {11*√5}]/10)

Also useful to know the radii formulas for the pentagon.

c = a/10 * √(50 + [10*√5])

i = a/10 * √(25 + [10*√5])

Then you just solve for the right triangle OIC, where you have the three sides and the right angle, height=I, base=c, and hypotenuse=C, just have to find the other two angles.


Let $a,b,c\in \Bbb R^+$ such that $(1+a+b+c)(1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c})=16$. Find $(a+b+c)$ Let $a,b,c\in \Bbb R^+$ such that $(1+a+b+c)(1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c})=16$.

Find $(a+b+c)$.

I computed the whole product ;If  $(a+b+c)=x\implies (1+x)(1+\frac{bc+ca+ab}{abc})=16$. Unable to view how to proceed further.

Please help.

By $AM \ge GM$ inequality,$$(1+a+b+c)\left(1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge \left(4\sqrt[4]{abc}\right)\left(4\sqrt[4]{\frac{1}{abc}}\right)=16$$and equality holds when $1=a=b=c$.

By applying Caushy Schwartz inequality we get
( 1 + a + b + c ) ( 1 + 1/a + 1/b + 1/c ) >= 16
Equality holds when a^2 = b^2 = c^2 = 1
Therefore
a = b = c = 1


Example of Sufficient and Insufficient Statistic? I am having trouble understanding the concept of a sufficient statistic.  I have read What is a sufficient statistic? and Sufficient Statistic (Wikipedia)

Can someone please give an example of:

a simple (but non-trivial) statistical model
a sufficient statistic of that model
an insufficient statistic of that model
how you identified 2 & 3 as having and lacking, respectively, the sufficiency property

$\def\E{\mathrm{E}}$Consider samples $X = (X_1,X_2)$ from a normally distributed population $N(\mu,1)$ with unknown mean.

Then the statistic $T(X)=X_1$ is an unbiased estimator of the mean, since $\E(X_1)=\mu$. However, it is not a sufficient statistic - there is additional information in the sample that we could use to determine the mean.

How can we tell that $T$ is insufficient for $\mu$? By going to the definition. We know that $T$ is sufficient for a parameter iff, given the value of the statistic, the probability of a given value of $X$ is independent of the parameter, i.e. if

$$P(X=x|T=t,\mu)=P(X=x|T=t)$$

But we can compute this:

$$P(X=(x_1,x_2) | X_1=t,\mu) =
\begin{cases}
0 & \mbox{if }t\neq x_1 \\
\tfrac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}(x_2-\mu)^2} & \mbox{if }t=x_1
\end{cases}$$

which is certainly not independent of $\mu$.

On the other hand, consider $T'(X) = X_1+X_2$. Then we have

$$P(X=(x_1,x_2) | X_1+X_2=t, \mu) = \frac{1}{2\pi}\int_{-\infty}^{\infty}e^{-\frac{1}{2}(s-\mu)^2 - \frac{1}{2}(t-s-\mu)^2}ds$$

and you can complete the square and show that this is independent of $\mu$, and hence $T'$ is a sufficient statistic for the mean $\mu$.

Here is my attempt to answer part 4 of the question.  If you have a parameteric family of distributions with a parameter theta (theta can be K-dimensional for K>=1) then a statistic is sufficient if all the information about theta is contained in it.  That means that given the parametric family the conditional distribution of the data given the sufficient statistic is independent of the parameter theta.  For many problem there are several functions T of the data that are sufficient.  Sufficiency usually carries data reduction with it.  For example say the parametric family is N(m,1) (m is the mean and 1 is the variance).  To estimate m you take a sample of size n=10.  The data involves ten values but the sufficient statistic (the sample mean) is just a single number.  That is the least amount of information possible for a sufficient statistic.  This type of sufficient statistc is better than others in data reduction and is called minimal sufficient.  To determine that a candidate statistic is sufficient you can use the factorization theorem to test it.  If the candidate T is sufficent it will factor the density as follows f(x|theta)= g(T(x)|theta) h(x) where f is the density for the data given theta, g is the density for T(x) given theta and the term h(x) does not depend on theta.  So you can check sufficiency by calculating f and g, divide f by g and T will be sufficient iff the resulting function h doesn't depend on (involve) theta.  Besides sufficiency and minimal sufficiency there is a concept called complete sufficiency.  All the cocepts are important in the parametric theory of point estimation.  This is nicely covered in Casella and Berger's book Statistical Inference 2e Chapter 6.


Use the PMI to prove the following $4^{k}-1$ is divisible by 3. $4^{n}-1$ is divisible by $3.$

(i) Basis Step: $P(1)$is true because $4^1 -1=3 $ and $3\mid3$.

(ii) Suppose $P(k)$ is true for induction hypothesis $3\mid4^{k}-1$.

Show $(k+1)$ is true $3\mid4^{k+1}$

Now here is where things gets tricky.

$\Rightarrow 4^{k+1} -1 = \left(4^{k}*4 \right)-1 =$

I cannot seem to figure out how one factors this any further. I just cannot seem to make it divisible by 3. Could someone show me how to do this?

Inductive assumption: $3 \mid (4^k-1)$, and therefore $(4^k-1) = 3m$ for some $m$.

$4(4^k-1) = 4^{k+1} -4 = 4^{k+1}-3-1$

so $4^{k+1}-1= 4(4^k-1)+3 = 12m+3$ which is divisible by $3$, completing the induction.


Use the definition of continuity to prove a function is defined at every nonnegative real number Use the definition of continuity to prove that the function f defined by f(x) = (x)^(½) is continuous at every nonnegative real number.

Our definition of continuity: Let I be an interval, let f:I-R, and let c(element of)I. The function f is continuous at c if for each (epsilon)>0 there exists (delta)>0 such that |f(x)-f(c)| < (epsilon) for all x (element of) I that satisfy |x-c|<(delta). The function f is continuous on I if f is continuous at each point of I.

I'm not sure how to the nonnegative real number comes into play here in terms of proving the continuity.

Continuity at $c=0$ is continuity from the right, so one wants to show given $\varepsilon$ there is $\delta$ for which $0 \le x < \delta$ implies $0 \le \sqrt{x} < \varepsilon$. The natural choice here is $\delta=\varepsilon^2.$

Otherwise $c>0$ and we initially restrict to $\delta<c/2$ (and throw that in at the end via the usual "min" definition.) From this assumption we have $c/2<x<3c/2$ and so
$$\sqrt{c/2}<\sqrt{x}<\sqrt{3c/2}.$$

Now we consider that
$$|\sqrt{x}-\sqrt{c}|=\frac{|x-c|}{\sqrt{x}+\sqrt{c}}. \tag{1}$$
Given our bounds for $\sqrt{x}$ the right side of $(1)$ is at most $|x-c|/(\sqrt{c/2}+\sqrt{c}),$ so that if we finally define $\delta$ as the minimum of the two numbers $c/2$ and $\varepsilon \cdot (\sqrt{c/2}+\sqrt{c})$ we'll have $|x-c|<\delta$ implies $|\sqrt{x}-\sqrt{c}|<\varepsilon.$

The function g : x->x is continuous, the function h : x->x^(1/2) is also continuous, so the function f is also continuous, because it is equal to h(g(x)) which is continuous as a composed of two continuous function.


Counting process which is not a Poisson process Please construct a counting process N, whose r.v. N(t) are distributed as Poisson(λt) but the process N itself is not a Poisson process.

This is an assignment in our Stochastic Process class. So I suppose this counting process N should meet all but one of a Poisson's process conditions. 1) N(0)=0 2) independent increments 3) At any given time t N(t) ~ Poiss( λt). So making the increments dependent should probably be the way to go. However, I've no idea how that could be done.

Let $\{N_t, t\geq 0\}$ be a homogeneous Poisson process with intensity $\lambda$, it holds that $N_{t+s}-N_s\sim Pois(\lambda t)$. Thus, let $Z_t=N_{t+s}-N_{s}$ for any $t,s\geq 0$. Then clearly $\{Z_t, t\geq 0\}$ is a counting process, $Z_t\sim Pois(\lambda t)$, but most importantly - its increments are dependent, e.g. let $m>s\geq 0,\,t_2>t_1\geq 0$ and then $Z_{t_2}-Z_{t_1}=N_{t_2+s}-N_{t_1+s}$ and $Z_{t_1}-Z_{0}=N_{t_1+m}-N_{m}$, intervals $\left(t_1,t_2\right]$ and $(0,t_1]$ do not overlap, but $N_{t_2+s}-N_{t_1+s}$ and $N_{t_1+m}-N_{m}$ are dependent since $m>s$.

Let $$N(t) \sim \text{Poisson}(\lambda t) \quad\forall t > 0$$

Let's examine $T_n$ which is the sum of the inter event times of a Poisson Process.
We know that $$T_n \sim \text{sum of n i.i.d } exp(\lambda)$$

$$\{T_n \leq t\} = \{N(t) \geq n\}$$

We also know that the sum of i.i.d exponentials follows the Erlang-n distribution.
Let $F_n(t) = \mathbb{P}\{T_n \leq t\} \sim \text{Erlang}(n,\lambda)$ and let $U \sim unif(0,1)$.

$$F_n^{-1}(U) \sim \text{Erlang}(n,\lambda)$$

Now if we define $$t_i = F_i^{-1}(U)$$ and examine the counting process on $t_i$.

$N(t) \sim \text{Poisson}(\lambda t)$ but all the times are dependent. If we know any one value we know the entire counting process and the independent increments property is lost!