How prove this $x+y=0$ if $\left(\sqrt{y^2-x^3}-x\right)\left(\sqrt{x^2+y^3}-y\right)=y^3$ Question:

let $x,y$ are real numbers,and such

  $$\left(\sqrt{y^2-x^3}-x\right)\left(\sqrt{x^2+y^3}-y\right)=y^3$$
  show that
  $$x+y=0\tag{1}$$

before I have solve following problem:

if

  $$(x+\sqrt{x^2+1})(y+\sqrt{y^2+1})=1,\Longrightarrow x+y=0\tag{2}$$
  solution:we have
  $$y+\sqrt{y^2+1}=\sqrt{x^2+1}-x\tag{3}$$
  $$x+\sqrt{x^2+1}=\sqrt{y^2+1}-y\tag{4}$$
  $(3)+(4)$
  then  $x+y=0$

But for $(1)$ we can't use this methods to solve  it.maybe can have other nice methods, Thank you

Below some characteristics of the function $f(x,y) = \left(\sqrt{y^2-x^3}-x\right)\left(\sqrt{x^2+y^3}-y\right)-y^3$ are shown. The area where $f(x,y)$ is undefined is in gray: $y^2 - x^3 < 0$ or $x^2 + y^3 < 0 $. Therefore the domain of the function is mainly restricted to $x\le 0$ and $y\ge 0$. The line $x+y=0$ is in red, some contour lines of $f(x,y)$ are in blue. The function is somewhat symmetrical around $x+y=0$, but not quite. Shouldn't the restriction $x\le 0$ and $y\ge 0$ be added to the problem in the first place?It's easy to prove that $x+y=0$ is a solution of $f(x,y)=0$, but proving the reverse seems to be tedious; at least I see no "nice" way to do it.

We'll replace $x$ on $-x$ and have the following problem.

Let $x$ and $y$ be real numbers such that $$\left(\sqrt{y^2+x^3}+x\right)\left(\sqrt{x^2+y^3}-y\right)=y^3.$$
Prove that
$$x=y.$$
Indeed, we need to prove that
$$\left(\sqrt{y^2+x^3}+x\right)\left(\sqrt{x^2+y^3}-y\right)=\left(\sqrt{x^2+y^3}+x\right)\left(\sqrt{x^2+y^3}-x\right)$$ or
$$\left(\sqrt{y^2+x^3}+x\right)\left(\sqrt{x^2+y^3}-y\right)-\left(\sqrt{x^2+y^3}+x\right)\left(\sqrt{x^2+y^3}-y\right)+$$
$$+\left(\sqrt{x^2+y^3}+x\right)\left(\sqrt{x^2+y^3}-y\right)-\left(\sqrt{x^2+y^3}+x\right)\left(\sqrt{x^2+y^3}-x\right)=0$$ or
$$\left(\sqrt{y^2+x^3}-\sqrt{x^2+y^3}\right)\left(\sqrt{x^2+y^3}-y\right)+(x-y)\left(\sqrt{x^2+y^3}+x\right)=0,$$ which gives $$x=y$$ or
$$\frac{(x^2+xy+y^2-x-y)\left(\sqrt{x^2+y^3}-y\right)}{\sqrt{y^2+x^3}+\sqrt{x^2+y^3}}+\sqrt{x^2+y^3}+x=0,$$ which is
$$\sqrt{(x^2+y^3)(y^2+x^3)}+x\sqrt{y^2+x^3}+(x^2+xy+y^2-y)\sqrt{x^2+y^3}+x^2+xy+y^2-x^2y-xy^2=0.$$
We'll prove that
$$\sqrt{(x^2+y^3)(y^2+x^3)}+x\sqrt{y^2+x^3}+(x^2+xy+y^2-y)\sqrt{x^2+y^3}+x^2+xy+y^2-x^2y-xy^2\geq0,$$ for which we consider four cases.

$x\geq0$ and $y\geq0.$

We see that $$\sqrt{(x^2+y^3)(y^2+x^3)}-y\sqrt{x^2+y^3}=\sqrt{x^2+y^3}\left(\sqrt{y^2+x^3}-y\right)\geq0$$ and
$$(x^2+xy+y^2)\sqrt{x^2+y^3}-x^2y-xy^2\geq x(x^2+xy+y^2)-x^2y-xy^2=x^3\geq0.$$
The equality occurs for $x=y=0$ only.

$x\geq0,$ $y\leq0$.

After replacing $y$ on $-y$ we need to prove that
$$\sqrt{(x^2-y^3)(y^2+x^3)}+x\sqrt{y^2+x^3}+(x^2-xy+y^2+y)\sqrt{x^2-y^3}+x^2-xy+y^2+x^2y-xy^2\geq0,$$ where $x$ and $y$ are non-negatives.

We see that $x^2\geq y^3$, which gives
$$x\sqrt{y^2+x^3}\geq x\sqrt{y^2+y^{4.5}}\geq xy^2.$$
The equality occurs only for $x=y=0$ again.

$x\leq0$ and $y\geq0.$

We'll replace $x$ on $-x$ and we need to prove that:
$$\sqrt{(x^2+y^3)(y^2-x^3)}-x\sqrt{y^2-x^3}+(x^2-xy+y^2-y)\sqrt{x^2+y^3}+x^2-xy+y^2-x^2y+xy^2\geq0,$$
where $x$ and $y$ are non-negatives now.

We see that
$$\sqrt{(x^2+y^3)(y^2-x^3)}-x\sqrt{y^2-x^3}=\sqrt{y^2-x^3}\left(\sqrt{x^2+y^3}-x\right)\geq0.$$
Thus, it's enough to prove that
$$(x^2-xy+y^2-y)\sqrt{x^2+y^3}+x^2-xy+y^2-x^2y+xy^2\geq0,$$ which is true, but my proof of this statement is still very ugly.

$x\le0$ and $y\leq0$.

After replacing $x$ on $-x$ and $y$ on $-y$ we need to prove that:

$$\sqrt{(x^2-y^3)(y^2-x^3)}-x\sqrt{y^2-x^3}+(x^2-xy+y^2+y)\sqrt{x^2-y^3}+x^2-xy+y^2+x^2y+xy^2\geq0,$$
where $x$ and $y$ are non-negative, for which it's enough to prove that
$$x^2-xy+y^2\geq x\sqrt{y^2-x^3}.$$
Now, by AM-GM $$x\sqrt{y^2-x^3}\leq\frac{1}{2}(x^2+y^2-x^3).$$
Id est, it's enough to prove that
$$x^2-xy+y^2\geq\frac{1}{2}(x^2+y^2-x^3)$$ or
$$(x-y)^2+x^3\geq0,$$ which is obvious.

The equality occurs for $x=y=0.$


Finding $a_{1996}$ if $\prod_{n=1}^{1996}(1+nx^{3^n})= 1+a_1x^{k_1} + a_2x^{k_2} + \cdots + a_mx^{k_m}$ I need to find the coefficient $a_{1996}$

$$\prod_{n=1}^{1996}(1+nx^{3^n})= 1+a_1x^{k_1} + a_2x^{k_2} + \cdots + a_mx^{k_m}$$

$a_1, a_2, ... , a_m$ are non zero.

$k_1 < k_2 <... < k_m$

So if $x=1$ you can find the sum of all coefficients, but I am not sure how to use this fact

Do I need to simplify the product somehow plug different values of x and check if something can be approximated that way, or would that be a waste of time?

Thanks,

What powers of $x$ appear as a term in the expression? Of course $x^0$. And $x^3$. More generally, any positive integer that can be built from summing powers of $3$ with no repeats.
$$x^0,\overset{1\text{st}}{x^3},\overset{2\text{nd}}{x^9},\overset{3\text{rd}}{x^{12}},\overset{\cdots}{x^{27}},x^{30},x^{36},x^{39},\ldots$$
In base 3, these are numbers that use only 0 and 1 for their digits, with a 0 in the "ones" place.
$$x^0,x^{10},x^{100},x^{110},x^{1000},x^{1010},x^{1100},x^{1110},\ldots$$
So what would be the 1996th term's power of $x$? In binary, 1996 is $$11111001100$$
So we are looking for $x^{111110011000}$ but the exponent is in base 3. It is built from multiplying $$x^{3^{11}}x^{3^{10}}x^{3^{9}}x^{3^{8}}x^{3^{7}}x^{3^{4}}x^{3^{3}}$$
So its coefficient is $$11(10)(9)(8)(7)(4)(3)=665280$$

All the non-zero coefficients are of the form $a_n$ with $n=\sum_k3^{i_k}$, $i_k\ge1$. As a result, $n$ must be divisible by $3$ if $a_n\ne 0$. Since $1996$ is not divisible by $3$, its coefficient is $0$.


What is the $3$D equation for a damped sine wave? This is the equation for a $2$D damped sine wave, what is the $3$D equivalent?
For reference, a $2$D sine wave would be of form $y=\sin x$, and a 3D sine wave would be of form $z=\sin(\sqrt{x^2 + y^2})$.

As in the comments, the equation
$$z=e^{-\sqrt{x^2+y^2}}\sin(\sqrt{x^2+y^2})$$
or else phase-shifted and scaled as
$$z=e^{-\sqrt{x^2+y^2}}\cos(2\pi\sqrt{x^2+y^2})$$
expresses damping along any ray $ax+by=0$ extending from the origin, and this was achieved by substitution of $\sqrt{x^2+y^2}$ in place of $t$ in the Wikipedia listing at https://en.m.wikipedia.org/wiki/Damped_sine_wave.


Evaluate $\lim_{x\to 0}\frac{(\tan x)^{2008}-(\arctan x)^{2008}}{x^{2009}}$ without using Taylor series. Evaluate $$\lim_{x\to 0}\frac{(\tan x)^{2008}-(\arctan x)^{2008}}{x^{2009}}$$ without using Taylor series.

I have a solution using  $\lim_{x\to 0}\frac{\tan x-\arctan x}{x^2}=0$, but I would really like to see a different idea.

You can define

$$f(x) = \left(\frac{\tan(x)}{x}\right)^{2008}\qquad g(x) = \left(\frac{\arctan(x)}{x}\right)^{2008}$$

then use
$$\frac{f(x)-g(x)}{x} \longrightarrow f^\prime(0) - g^\prime(0)$$

Remark

The following useful inequalities hold

$$|\tan(x) -x| = \int_0^{|x|}\tan^2(t)dt\le \tan^2(x)\int_0^{|x|}dt\le |x| \tan^2(x) \quad \text{for }|x|<\frac{\pi}{2}$$

and
$$|\arctan(x) - x| = \int_0^{|x|}\frac{t^2}{1+t^2}dt\le
\int_0^{|x|}t^2dt\le \frac{|x|^3}{3}\quad\text{for } x\in{\mathbb R}$$
They imply that $\frac{\tan(x)}{x} - 1 = o(x)$ and $\frac{\arctan(x)}{x}-1 = o(x)$, hence $f^\prime(0)$ and $g^\prime(0)$ exist and are $0$.


Bioinformatics Probability Question I have the following problem from my homework set:
The human genome is 3 billion bp in length, if we only consider one strand. The complement strand has the same information content, so for many purposes this is a reasonable and very common simplification. When looking for motifs and short sequence matches, however, we need to consider both strands of DNA, so here the genome is 6 billion bp long.
For this problem we will assume that the frequency of all four nucleotides is the same $(p(A)=p(T)=p(C)=p(G)=0.25)$.

A) What is the probability that the nucleotide at a random position i is a C? (1 point)

B) If you have the sequence AATTG (a "5mer"), what is the probability that you will find that sequence starting at the same randomly chosen position i? (2 points)

C) What is the probability of NOT finding this sequence at all in the entire human genome? (5 points)

D) You are given sequences ranging from 5bp to 20bp in length. For each of these sequences, what is the probability of finding that sequence once and only once in the human genome (use the same assumptions about genome composition)? (15 points)

I have been able to find the answers for the first three parts.

A) $.25$

B) $.25^5$

C) $(1-.25^5)^{6,000,000,000}$

I am having difficulty determining how to find a kmer once and only once. I thought I might use the binompdf command on my Ti-89 but I don't know if that would be correct. Any help is appreciated, thank you.

You have the right idea using a binomial distribution:

For each block of five nucleotides, flip a coin that comes up heads (success) with probability $p$. We define success as finding our specific sequence, so $p=0.25^5$. We require that we only have a single success in our $ 6\cdot10^9-5+1$ flips, so the probability of this is
$$\binom{6\cdot10^9-5+1}{1}\times p(1-p)^{6\cdot10^9-1}\approx 6\cdot10^9 \times p(1-p)^{6\cdot10^9}\approx 0.$$


Exist a linear combination for full rank matrix? I met a problem this morning.

Suppose that there exists a full rank matrix $X = \begin{bmatrix} a & b & c \\ b & c & b\\ c & d & a  \end{bmatrix} \in \mathbb{R}^{3\times3}$. From the definition of linear independence we know that if the matrix is not a full rank matrix, a relationship would exist: $k_1X_1 + k_2X_2 = X_3$, where $X_1$, $X_2$ and $X_3$ are corresponding columns of $X$, and $k_1$, $k_2 \in \mathbb{R}\backslash0$.

If $k_1$ and $k_2$ exist, then:
$$\begin{bmatrix} X_1 & X_2\end{bmatrix} \begin{bmatrix} k_1 \\ k_2\end{bmatrix}  = \begin{bmatrix} X_3 \end{bmatrix}$$
$$ \begin{bmatrix} k_1 \\ k_2\end{bmatrix}  = pinv (\begin{bmatrix} X_1 & X_2\end{bmatrix}) \begin{bmatrix} X_3 \end{bmatrix}$$
$$ \begin{bmatrix} k_1 \\ k_2\end{bmatrix}  =  (\begin{bmatrix} X_1 \\X_2\end{bmatrix} \begin{bmatrix} X_1 & X_2\end{bmatrix})^{-1}\begin{bmatrix} X_1\\ X_2\end{bmatrix} \begin{bmatrix} X_3 \end{bmatrix}$$

And I try some simulations in MATLAB, the $k_1$ and $k_2$ exist. But the first equation is not satisfied.

So I am very confused. I must be wrong. But where? Could anyone tell me?

It seems you have solved a least square problem finding the projection of $X_3$ onto $span(X_1,X_2)$.

Refer to the related Is the pseudoinverse matrix the solution to the least squares problem?

Your characterization of linear dependence is incorrect. The columns of $X$ are linearly dependent iff there is a nontrivial linear combination $k_1X_1+k_2X_2+k_3X_3$ that vanishes, but this does not let you conclude that $k_1X_1+k_2X_2=X_3$. Consider, for example, $$X=\begin{bmatrix}1&0&0\\0&0&0\\0&0&1\end{bmatrix}.$$ The columns are linearly dependent (every set that contains the zero vector is), but the third column is obviously not a scalar multiple of the first.

If the columns are linearly dependent, then you know that at least one of them is a linear combination of the other two, but you can’t a priori decide which one that is.


Minimum of a three variable function (2) This question is related to the question here.The referred question was about finding mininum/maximum of $\frac{x^2+y^2+z^2}{xyz}$.

The answers were clear. But one of the hint given by @Marconius is given below

"Also try to exploit the symmetry of the function, i.e. if you can prove that a minimum can only occur when $x=y=z$, then you can reduce the problem to a single-variable one"

My question is that how such a proof can be derived. Could anyone provide me right directions. For example, can we prove that $\frac{x^2+y^2+z^2}{xyz}$ has a minimum when $x=y=z$ (for positive real values of $x,y,z$)?

The graph of the function $\tfrac{x^2+y^2+z^2}{xyz}$ is the subset of $\Bbb{R}^4$ given by
$$\left\{(x,y,z,w)\in\Bbb{R}^4\mid\ w=\tfrac{x^2+y^2+z^2}{xyz}\right\}.$$
If you can prove that a minimum/maximum can only occur if $x=y=z$, then any minimum/maximum must be in the subset
$$\left\{(x,y,z,w)\in\Bbb{R}^4\mid\ w=\tfrac{x^2+y^2+z^2}{xyz}\text{ and }x=y=z\right\}=\left\{(x,x,x,w)\in\Bbb{R}^4\mid\ w=\tfrac{3}{x}\right\}.$$
In particular $x$ is then a minimum/maximum of the function $\tfrac{3}{x}$. This reduces the problem to a single-variable problem.

In this particular case the function $\tfrac{3}{x}$ has no minimum/maximum, so the original function has no minimum/maximum with $x=y=z$, and the comment is of no use here.

In general, if you can prove that for a minimum/maximum $m=(x_1,\ldots,x_n)$ of a multivariate function $f(x_1,\ldots,x_n)$ we have an explicit relation $x_n=g(x_1,\ldots,x_{n-1})$, then we can substitute $x_n$ to look for a minimum/maximum of
$$f(x_1,\ldots,x_{n-1},g(x_1,\ldots,x_{n-1})),$$
making it a problem in one variable less. In this way every relation we find reduces the number of variables by one. In this particular case we are looking at
$$f(x,y,z)=\frac{x^2+y^2+z^2}{xyz},$$
and the hint given by @Marconius suggests your prove that the relations
$$z=y\qquad\text{ and }\qquad y=x,$$
both hold. This reduces the problem to finding a minimum/maximum of $f(x,x,x)$, which is a single-variable function. However, the hint is misleading because your function has no minima or maxima at all.


What do you call an equation that is impossible to algebraically solve? For example...

$$3.5=\sin{(x)}+\frac{\cos{(x)}}{\ln{(x)}+\sin{(x)}}$$

So your options to solving such a thing are iteration and other numerical methods. Does an equation that's not solvable algebraically have a specific name?

These are called transcendental equations. When you have a polynomial equation that you cannot solve, then you say that the equation "is not solvable by radicals."


Origin of the Power Rule Proof: Who first proved the power rule? Who was the first person to prove the power rule for derivatives? The person could have proved the power rule using limits and the binomial theorem or difference of two nth powers, or the implicit differentiation method. I have no preference as to how the person proved it--as long as the person proved it correctly--I'm just wondering: who did it in the first place? Who first proved the power rule?

The power rule was already in Fermat, Hudde, Wallis, and Barrow in the 17th century, a few decades before the invention of the calculus by Newton and Leibniz, and two centuries before Cauchy's work in the 19th century (for those who are curious, here is Cauchy's 1821 definition of a continuous function: $f$ is continuous if a change in $x$ by an infinitesimal $\alpha$ necessarily produces an infinitesimal change $f(x+\alpha)-f(x)$ in $y$). Fermat did it using a technique anticipating the calculus called adequality. More details can be found in this recent study.

Since there's a general unease about infinitesimal methods, which was probably how the power rule was first proven, here's a proof I recently found in an old textbook. If $f(x) = ax^n$, where $a$ is a constant, then:
$$f'(x) = \frac{a(x_1^n - x_0^n)}{x_1 - x_0}$$
$$ = a(x_1^{n-1} + x_0x_1^{n-2} + x_0^2x_1^{n-3} + ... + x_0^{n-2}x_1 + x_0^{n-1})$$
Letting $x_1 → x_0$ in value we obtain:
$$f'(x) ⇒ nax^{n-1}$$
This approach is usually credited to Caratheodory, but according to the Brian M Scott answer here Cauchy got there first.

EDIT But, the underlying logic was strongly implied by Leibniz:

  For instead of the infinite or the infinitely small, one takes quantities as large, or as small, as necessary in order that the error be smaller than the given error, so that one differs from Archimedes' style only in the expressions, which are more direct in our method and conform more to the art of invention.
  The Calculus in the Eighteenth Century, p56, HJM Bos.


Probability that 20 sided die beats 12 sided die with reroll Alice rolls a 12 sided die (the faces labeled 1 through 12) and Bob rolls a 20 sided die (the faces labeled 1 through 20). After seeing their roll (but not the other person's roll), each person can choose to roll again. Bob wins if his final roll is strictly greater than Alice's final roll. What is the probability that Bob wins?

I'm assuming that Alice's optimal strategy is to reroll if she gets a roll that is less than 7, and Bob's optimal strategy is to reroll if he gets less than 11. Is this correct?

What I will do to solve that question :

Consider the 13 cases for Alice and 21 cases for Bob (from reroll always, to never reroll, through reroll on 1, reroll on 2 or less, reroll on 3 or less...) and compute the probability of winning for each cases (use a computer program for the $21\times13$ cases).

Then I'll compute the nash equilibrium and the minmax. If they are unique and the same, you have your best moves for Alice and Bob, if not, then it depends of what's in the head of the other player.

HINTS

[more hints added, assuming downvote means my first attempt does not make it easy enough]

I am assuming that A's die is 1-12 and B's is 1-20 (not stated in the question as I write this).

An elementary point to get started. If your die has 12 sides, you have a half-chance of getting 1-6. So you might think it was best to reroll iff you got that. If you follow that strategy, you have a half chance of getting 7-12 on the first roll, and a half chance of rerolling. So you have $\frac{1}{24}$ for each of 1-6, and $\frac{3}{24}$ for each of 7-12, expected value 8.

Now consider B's strategy. If B gets 13 or better he is certain to win. If he leaves a 9 he expects to win 50% of the time. So suppose he rolls a 10 (which if he followed the analogous strategy to A would mean a reroll). It is easy to check that A's chance of leaving a 10 or better (on her assumed strategy) is $\frac{9}{24}$. In other words, if B does not reroll, he expects to win $\frac{15}{24}$ of the time. So should he reroll? It is easy to check that he should not.

But how do we avoid a kind of infinite regress? How can either A or B decide whether to make a second roll without knowing the other's strategy? Do we need some kind of minimax approach?

Note the asymmetry. With a single roll, the game would clearly favour B.


Number of letters required to make three letter names If a monster has 63 children and he wants to keep 3 letter names for each of them so that they are distinct,but with the condition that you can use the same letter more than once,how many letters at minimum does the monster need to name it's children?-source-BdMO 2011question.I have done a bit of listing and I think it is 5.
As others pointed out,my listing was wrong.

Let $n$ be the number of letters we need to form $63$ distinct three-letter names.

So, we have $n$ choices for the first letter, $n$ choices for the second letter and $n$ choices for the third letter: This means we have $n\times n\times n = n^3$ possible combinations of distinct names.

We need $n^3 \geq 63$ where $n$ is an integer. $n = 4$ gives us $4^3 = 64$ distinct names, and this is indeed the least integer giving us at least $63$ distinct names. (For $n = 3$, we have only $3^3 = 27$ possible distinct names, and for $n = 5$, we have as many as $5^3 = 125$ distinct three letter names.

Hence, the monster needs minimally $4$ distinct letters to create 63 distinct three-letter names.

The answers listed above have attached an unspecified paradigm - i.e.visual form of the names. The Question states that this is a monster, not a man, and the criteria is "distinction". Why do we seek an answer in the written form? Is this a known characteristic of monsters?

Webster - DISTINCTION : a difference that you can see, hear, smell, feel, etc. : a noticeable difference between things or people

Wouldn't 3 letters, pronounced at different octaves, or in ascending tones, or descending tones, etc. meet the test of distinction? and since letters may be repeated..........

My answer is 1 letter


What is the difference between an identity, an equation and a conditional equation? What is the difference between an identity, an equation and a conditional equation?

Thank you?

Equation means equality. They are both related to the word equal. If such an equality is true for all values of the variable, it is called an identity, e.g., $\sin^2x+\cos^2x=1$ is true for all x. If however the equation in question only holds for some values, which one is supposed to determine, then it's called conditional, and its variable is termed an unknown.

Equation is a mathematical description which is equal only for one variable. But identity is a mathematical description which is always equal for any variable.


Twenty students in a class when they are required to choose a partner The question consists of two subquestions:
There are 20 students in the class.
A. The students are required to divide into 10 pairs. In how many ways can they are divided?
Attempt: First of all, we know that there are 20 students so in the first stage we can arrange them in upline so we get 20! for that, In the second stage, we determine two students in the line will be pair and the change between them makes the calculation $2^n$, the last stage we can see of each pair can be move in the line together which gives us the $10!$.
Final solution: $\frac{20!}{10!\cdot 2^{10}}$.
This solution correct?
B. The students are required to divide into pairs again, however, they cannot be with the same partners as in subquestion - A.. In how many ways can they do so?
Attempt:  I tried to use the inclusion-exclusion principle without any success.

In your attempt on section A, the naïve approach of lining up the students and pairing them off leads to $20!$ scenarios, which we can then assess to find which other scenarios will produce the same ten pairs. We can rearrange the pairs in $10!$ ways, and we can rearrange each pair of students, $2$ options per pair. These changes are independent of each other - these rearrangements can't cancel each other out. So for any set of pairs, there are $10!\cdot 2^{10}$ scenarios that will generate those same ten pairs. So overall we have $\dfrac{20!}{10!\cdot 2^{10}}$ different pairings here.
For section B,  we'll call the section A result $\mathcal P_{20}$, for making pairings from $20$ students. Now for reducing this total, we can consider how many different pairings exist with $18$ students (assuming a pairing carried over from the first time) - call this $\mathcal P_{18}$, calculated in a similar way to section A. Then also calculate how many for $16$ students (two pairings carried over), $\mathcal P_{16}$, etc. Now combine all these results using inclusion-exclusion to remove over- and under-counts from each adjustment based on choosing pairs from the first pairing:
$$ \mathcal P_{20} - \binom{10}{1}\mathcal P_{18} + \binom{10}{2}\mathcal P_{16} - \binom{10}{3}\mathcal P_{14}  + \binom{10}{4}\mathcal P_{12} - \binom{10}{5}\mathcal P_{10} \\ + \binom{10}{6}\mathcal P_{8} - \binom{10}{7}\mathcal P_{6} + \binom{10}{8}\mathcal P_{4} - \binom{10}{9}\mathcal P_{2}+ \binom{10}{10}\mathcal P_{0}$$ (noting that $\mathcal P_{2}=\mathcal P_{0}=1$)

Part A:
Label each student $1,\dots,20$. Pair student $1$ off ($19$ ways to do this). Regroup all the remaining students, and relabel $1,\dots,18$. Pair student $1$ off ($17$ ways to do this), regroup, etc...
This gives $19!!=19\cdot17\cdots3\cdot1$ ways to pair the students of.
Which is the same as $\frac{20!}{10!2^{10}}=\frac{20\cdot19\cdots1}{2\cdot10\cdot2\cdot9\cdots2\cdot1}$, just a different way of counting them.


What is the name of this kind of integral called? $$f_n(x) = \frac{1}{n}\int_x^{x+n} f(t) dt$$ What is this kind of integral called? Thanks!

I've heard it referred to as an integral with respect to a "dummy" variable.


The Hole in One Pizza In a recent issue of Crux, at the end of the editorial (which is public), it appears the following very nice problem by Peter Liljedahl.

I couldn't resist sharing it with the MSE community. Enjoy!

Nice riddle! My solution would be to cut along a line through the center of the circle and the center of the rectangle.

Proof.

A cut through the center of a circle divides it into pices of equal size. The same holds for rectangles. Therefore everyone gets the same amount of pizza minus the same "amount of hole". $\square$

$\qquad$

It amazed me that this works for pizzas and holes of even stranger shapes as long as they are point-symmetric. In this way one can make the riddle even more interesting, e.g. an elliptic pizza with a hole in the shape of a 6-armed star.

Lay both pizza's one on top of the other and cut through the whole so that it sliced in half.
Each person get a slice from the pizza with the whole and the remainder of the other pizza.
So the exterior area of the pizzas are equal and the share of the whole is equal.


In how many ways, Miyad can do the Trick? The numbers 1 to 6 are written on six balls. Miyad took four balls from there such that if he multiplies the numbers of the balls and then say it to Mun, then Mun can’t say the number of the balls surely. In how many ways, Miyad can do this?

This is a question of Math Olympiad and I'm slightly confused in line 2. Can someone please help?

The prime $5$ occurs in only one of the numbers, so Mun can tell whether it’s one of the $4$ balls, and if it is, divide it out of the product. Thus, we can split the problem into two cases: Choose $4$ numbers without $5$ such that their product is not unique, or choose $3$ numbers in addition to $5$ such that their product is not unique.

The first case is impossible, since only one of the other $5$ numbers is omitted, so Mun can find that number by dividing the product of all $5$ other numbers by the given product.

For the second case, if we choose two different triples out of $5$ numbers, they have at least one number in common. They can’t have the same product if they differ only in one number; so they must have exactly one number in common. The only two pairs of numbers up to $6$ that have the same product are $2\cdot3=1\cdot6$ and $2\cdot6=3\cdot4$.

Thus, there are exactly four ways in which Miyad can choose the numbers such that their product doesn’t uniquely identify them, namely $\{1,4,5,6\}$ and $\{2,3,4,5\}$ (for $2\cdot3=1\cdot6$) and $\{1,2,5,6\}$ and $\{1,3,4,5\}$ (for $2\cdot6=3\cdot4$).

In general there are $\displaystyle{\binom{6}{4}=15}$ ways for Miyad to choose the balls.

For instance $\{1,2,3,4\}$ giving product $24$, or $\{2,3,5,6\}$, giving product 180.

If the product is unique to one selection, the balls can be identified with certainty, if the same product occurs with more than one selection, they can't be identified with certainty.


Quotient space of unit sphere is Hausdorff Let $S^1=\{e^{2\pi it}|t\in\mathbb{R}\}$ be the unit sphere. Define $\sim$ on $S^1$ where two points are identified if $t_1-t_2=k$, for some $k\in\mathbb{Z}$. It must be shown that $S^1/\sim$ is Hausdorff.

 The quotient space of a topological space is Hausdorff if its graph is closed. That is, if $R=\{(x,y)\in S^1\times S^1|x \sim y \}$ is closed, then $S^1/\sim$  is Hausdorff. I am struggling to show that $R$ is closed.

$R=\{(x,y)\in S^1\times S^1|x \sim y \}=\{(e^{2\pi it_1},e^{2\pi it_2})\in S^1\times S^1|t_1-t_2=k,k \in\mathbb{Z}\}$. Easiest way I think, is to show that $S^1\times S^1-R$ is open. If $(x,y)\in S^1\times S^1-R$ where $x=e^{2\pi it_1},y=e^{2\pi it_2}$ $\implies$ $t_1-t_2=c,c\notin \mathbb{Z}$. But I do not know how to proceed from here. If this approach is too complicated, do suggest a different method. Thank you in advance. i changed the r to a t

Define $f:\mathbb S^1\times\mathbb S^1\to\mathbb S^1$ by $f(x,y)=xy^{-1}$, where if $y^{-1}:=e^{-ia}$ if $y=e^{ia}$.  Since $f$ is continuous, the preimage $f^{-1}(\{1\})$ is closed.  But this is precisely $R$.

Your question is  misstated  .if t_1 - t_2 = k then e^2(pi)it_1 =e^2(pi)it_2 so the equivalence relation is the identity on S^1 and S^1 as a subspace of R^2  is clearly Hausdorff -It's a metric space . Your question is probably about the isomorphism between S^1 and R/Z induced by the map t--> e^2(pi)it which has kernel Z,


Is it possible to find the criminal with graph-theoretic methods? I've been presented to a problem:

  Someone commited a crime. When interrogated, the people, named $G,m,M,J,D$ argued:

  $G:$ It wasn't $D$; It was $M$.
  $m:$ It wasn't $M$; It wasn't $D$
  $M:$ It was $D$; It wasn't $G$.
  $J:$ It was $M$; It was $m$.
  $D:$ It was $J$; It wasn't $G$.

  Knowing that each one of them told a lie, who was the criminal?

I'm pretty sure that graph theoretic methods could be used to solve this problem. I tried to write a graph matrix with $a_{ij}=-1$ if it is said that wasn't him and $a_{ij}=1$ if someone said it was him and $0$ otherwise (I left the $0$ entries null for simplicity), then:

$$\begin{vmatrix}
{}&{G}&{m}&{M}&{J}&{D}\\
{G}&{}&{1}&{}&{}&{-1}\\
{m}&{}&{}&{-1}&{}&{-1}\\
{M}&{-1}&{}&{}&{}&{1}\\
{J}&{}&{1}&{1}&{}&{}\\
{D}&{-1}&{}&{}&{1}&{}
\end{vmatrix}$$

But I'm unable to find some kind of pattern to solve this problem with this matrix. So, Is there something that could be used to find the criminous with graph-theoretic methods?

Besides this, I've tried to think about matrix permutations assuming that the first proposition of each one is a lie and the second is truth, then the first is truth, then the second is a lie, for example:

$$\begin{vmatrix}
{}&{G}&{m}&{M}&{J}&{D}\\
{G}&{}&{1_L}&{}&{}&{-1_T}\\
{m}&{}&{}&{-1_L}&{}&{-1}\\
{M}&{-1_L}&{}&{}&{}&{1}\\
{J}&{}&{1_L}&{1}&{}&{}\\
{D}&{-1_L}&{}&{}&{1}&{}
\end{vmatrix}$$

And then permute the $L$ies and $T$ruths to obtain a set of matrices and then find some property on one of these matrices that could indicate who's the criminous. But I also can't see some pattern here.

The problem has no solution. One of the statements It was M and It wasn't M must be true, so It wasn't D must be false. Therefore $D$ is guilty. But then both of $M$'s statements are true.

I didn't use any of this matrix stuff. I just simply assumed every statement they made was a lie and for G I constructed the set {D,M'} where M' means M didn't do it. Then for m I have (M, D}. Continuing I have {D', G}, {M', m'}, and {J', G} and conclude that G is the one none of them said did do it therefore G did it.


Simplify $(\neg p \wedge q \wedge r) \vee \neg(q\vee \neg r)$ Anyone can help me  simplify $$(\neg p \wedge q \wedge r) \vee \neg(q\vee \neg r)$$ by using algebraic propositions laws? I'm stuck.
According to a program I'm using the result is: $(\neg p \wedge r) \vee (\neg q \wedge r)$.

Does anyone know a software or webpage that shows you the steps?

$$(\neg p \wedge q \wedge r) \vee \neg(q\vee \neg r) \Leftrightarrow \text{ (DeMorgan)}$$

$$(\neg p \wedge q \wedge r) \vee (\neg q \land \neg \neg r) \Leftrightarrow \text{ (Double Negation)}$$

$$(\neg p \wedge q \wedge r) \vee (\neg q \land r) \Leftrightarrow \text{ (Distribution)}$$

$$((\neg p \wedge q ) \vee \neg q) \land r \Leftrightarrow \text{ (Distribution)}$$

$$((\neg p \vee \neg q) \wedge (q  \vee \neg q)) \land r \Leftrightarrow \text{ (Complement)}$$

$$((\neg p \vee \neg q) \wedge \top) \land r \Leftrightarrow \text{ (Identity)}$$

$$(\neg p \vee \neg q) \land r \Leftrightarrow \text{ (Distribution)}$$

$$(\neg p  \land r) \vee (\neg q \land r) $$

There are some advanced equivalence principles that can do this a bit more quickly. For example, if you have:

Reduction

$p \land (\neg p \lor q) \Leftrightarrow p \land q$

$p \lor (\neg p \land q) \Leftrightarrow p \lor q$

then you could go from line 4 immediately to line 7

In fact, the generalized form of Reduction says:

Reduction (generalized)

$(r \lor p) \land (r \lor \neg p \lor q) \Leftrightarrow (r \lor p) \land (r \lor q)$

$(r \land p) \lor (r \land \neg p \land q) \Leftrightarrow (r \land p) \lor (r \land q)$

and with that you can go from line 3 to line 8 in 1 step!

HINT: Check this:

Wolfram Alpha

https://www.wolframalpha.com/input/?i=(not+p+AND+q+AND+r)+OR+not+(q+OR+not+r)


Describe the idempotents in a polynomial ring $R[x]$. In general, given that $R$ is a commutative ring, how to identify the idempotent elements in $R[x]$?

I tried by solving equalities of polynomials but nothing works...

So how can I see that in which form can we get some idempotents in $R[x]$?

Thanks in advance!

$f^2 = f\,\Rightarrow f(0)^2\! = f(0) := a.\,$ If $\,f\!-\!a\ne 0\,$ it has order $\,k\ge 1,\,$ say $\,f = a + bx^k+\cdots,\,b\ne 0.\,$ Then $\,f^2\! = f\,\Rightarrow 2ab = b,\,$ times $\,a\,\Rightarrow\,2ab = ab\,\Rightarrow\,ab=0\,\Rightarrow\,b = 2ab = 0,\,$ contradiction. Therefore $\,f = a = f(0)\,$ is constant.

CASE I:$R$ is an integral domain .Then $p(x)^2=p(x)\implies \deg p(x)=0\implies p(x) \text{is a constant}$

So the idempotents in $R[x]$ are only the idempotents in $R$.

CASE II:$R$ is not an integral domain .Here though the idempotents in $R$ will be idempotents in $R[x]$ it can't be said whether they will be the only idempotents or not.


Linear combination on $\mathbb R_{1} \left[ x \right ]$ Given the canonical basis of $\mathbb R_{1} [ x ]$,  $\{x,1\}$, I tryed to write each vector as a linear combination of the basis $B=\{3x+6,2x+10\}$.

So, I started by $x$.

$x=\lambda_{1}(3x+6)+\lambda_{2}(2x+10)=(3\lambda_{1}+2\lambda_{2})x+6\lambda_{1}+10\lambda_{2}$

Then I solved the system by a matrix,

$\begin{bmatrix}
 3&2\\6&10
\end{bmatrix}$
$\begin{bmatrix}
 \lambda_{1}\\\lambda_{2}
\end{bmatrix}$=
$\begin{bmatrix}
 1\\0
\end{bmatrix}$

The result was $\lambda_{1}=\frac{5}{9}$ and $\lambda_{2}=-\frac{1}{3}$.

But now I don't know how to define the coordinates of $x$ on the basis $B$. Thanks for the help

If I understand your question properly, then $x$ written in the basis $B$ simply has the form
$$ \begin{pmatrix} \frac{5}{9} & -\frac{1}{3} \end{pmatrix}. $$
This is analogous to the fact that the basis vectors of $B$ have the form
$$ \begin{pmatrix} 3 & 6 \end{pmatrix} \quad\text{and}\quad \begin{pmatrix} 2 & 10 \end{pmatrix} $$
when written in the canonical base $(\,x,\, 1\,)$.

The coordinates of $x$ should be your $\lambda_1, \lambda_2$. You have already changed its basis to be written in terms of vectors in $B$ rather than vectors in the standard basis.

$$
[x]_{\{x,1\}} = (1, 0) \\
[x]_{B} = (\lambda_1, \lambda_2)
$$


Finding the order of permutations in $S_8$ Consider the group $S_8$. What is the order of $\sigma = (4,5)(2,3,7)$ and $\tau = (1,4)(3,5,7,8)$?

My book says I should just use a trick by the order of a permutation expressed as a product of disjoint cycles is the least common multiple of the lengths of the cycles.

I don't understand the trick very well and would like to see how the counting process here actually works. I can count for non disjoint ones like say $(1,2,3,8)$ has order 4

I don't know how to count $\sigma = \begin{pmatrix}
1 &2  &3  &4  &5  &6  &7  &8 \\
1 &3  &7  &5  &4  &6  &2  &8
\end{pmatrix}$

EDIT Just so I am clear, the answer for the order $\sigma$ is 6 and $\tau$ is 4. I am still confused as to how they got this

I have an extremely poor understanding of this subject so please help me! Thank you

So take $\sigma = (4,5)(2,3,7)$. The order, by definition, is the the smallest natural number $n$ such that $\sigma^n = (1)$ (i.e. the identity element in the group, i.e. the element that sends every number to itself). Since the cycles $(4,5)$ and $(2,3,7)$ are disjoint you have

$$\begin{align}
\sigma^n &= (4,5)(2,3,7)(4,5)(2,3,7)\dots (4,5)(2,3,7)(4,5)(2,3,7)\\
&= (4,5)(4,5)\dots (4,5)(4,5)(2,3,7)(2,3,7)\dots (2,3,7)(2,3,7)\\
&= (4,5)^n(2,3,7)^n
\end{align}$$

(note that the two elements $(4,5)$ and $(2,3,7)$ commute).

So the order of $\sigma$ is exactly the smallest natural number $n$ such that $(4,5)^n = (1)$ and $(2,3,7)^n = (1)$ (think about this fact for a moment).

But what is the order of a each of $(4,5)$ and $(2,3,7)$?

Well, the order of $(4,5)$ is two exactly because $(4,5)^2 = (4,5)(4,5) = (1)$. The order of $(2,3,7)$ is $3$ because
$$\begin{align}
(2,3,7)^1 &= (2,3,7) \\
(2,3,7)^2 &= (2,7,3) \\
(2,3,7)^3 &= (1)
\end{align}
$$

Now it is not to hard to see that the order of $\sigma$ is exactly the least common multiple of $2$ and $3$ (since we need both $(4,5)^m = (1)$ and $(2,3,7)^m = (1)$ and the smallest $m$ where this happens is exactly the least common multiple). Hence the final answer is $6$.

Addendum: I just wanted to add a bit about orders of these elements. First note that for example the element $(4,5)$ is just the element
$$
(4,5) = \begin{pmatrix}
1 & 2 &3 & 4 & 5 & 6 & 7 & 8 \\ 1 & 2 & 3 &5 & 4& 6 & 7 & 8
\end{pmatrix}.
$$
(Hence $4$ maps to $5$ and $5$ to $4$). So when you compose (multiply) the element with itself, then you get
$$
\begin{align}
(4,5)(4,5) &= \begin{pmatrix}
1 & 2 &3 & 4 & 5 & 6 & 7 & 8\\ 1 & 2 & 3 &5 & 4& 6 & 7 & 8
\end{pmatrix}\begin{pmatrix}
1 & 2 &3 & 4 & 5 & 6 & 7 & 8\\ 1 & 2 & 3 &5 & 4& 6 & 7 & 8
\end{pmatrix} \\ &= \begin{pmatrix}
1 & 2 &3 & 4 & 5 & 6 & 7 & 8\\ 1 & 2 & 3 &4 & 5& 6 & 7 & 8
\end{pmatrix} = (1)
\end{align}
$$
(I usually write the identity as $(1)$).

This means that the order of $(4,5)$ is $2$. Likewise you find that
$$
(2,3,7) = \begin{pmatrix}
1 & 2 &3 & 4 & 5 & 6 & 7 & 8\\ 1 & 3 & 7 &4 & 5& 6 & 2 & 8
\end{pmatrix}.
$$

Prove that easy lemma: the order of a permutation expressed as a product of disjoint cycles is the minimal common multiple of the lengths of the cycles.

To prove the above you may want to use the fact (also easily provable) that two disjoint cycles commute...


Probability of 4 of a kind in 7 cards, if first dealt 2 different cards? Question: In Texas Hold 'Em, you choose your hand from among seven cards: two that are dealt to you, plus five more that are available to all players.

Suppose the two cards dealt to you are of different denominations. What is the probability that you'll be able to make four of a kind out of all seven cards available to you?

I'm trying to use Bayes conditional to solve this:

Prob (4 of a kind in 7 cards after first being dealt 2 different denominations ) = ( Prob(Being dealt 2 different denominations first (in 2 cards), given 4 of a kind in 7 cards) * Prob(4 of a kind in 7 cards) ) / Prob (Being dealt 2 different denominations first (in 2 cards))

I'm looking for clarity on how to think about this problem (and, I believe, how I should avoid overthinking it as I've spent too much time pondering "what ifs").

I can't solve "Prob (Being dealt 2 different denominations first (in 2 cards), given 4 of a kind in 7 cards)", so I'm not sure if I'm stating the possibilities properly. What if I'm dealt Queen, King, and 3 more Kings in the 5 cards. Or what if there are 4 Jacks in the 5 cards and the 4-of-a-kind is made completely in the 5? The location of the 4 of a kind (if linked with one of the initial 2 cards or not), along with how to separate the probabilities of the initial 2 cards from the 5 is confusing me.

I would recommend doing this directly rather than using $Pr(A\mid B)=\frac{Pr(A\cap B)}{Pr(B)}$

You are dealt two cards of different ranks.  Without loss of generality, suppose that the first two cards dealt to you are the Ace of spades and the King of spades.

Now... in order to have a four of a kind, among the next five cards one of the following three situations must occur:

Among the next five cards, the remaining three aces appear
Among the next five cards, the remaining three kings appear
Among the next five cards, four of the cards are all of the same rank (not an ace and not a king)

Note that these cases do not in fact overlap.

There are $\binom{50}{5}$ possible ways to deal the next five cards (where order doesn't matter).  $\binom{47}{2}$ of those ways have all three remaining aces.  Similarly $\binom{47}{2}$ have all three remaining kings.

As for counting the final case, first pick which rank it is that appears four times.  There are $11$ options.  Then, pick which the final remaining card is.  There are $46$ options here.  This gives $11\cdot 46$ such choices.

All in all, this gives the probability as:

$$\dfrac{\binom{47}{2}+\binom{47}{2}+11\cdot 46}{\binom{50}{5}}$$

If you insist on using $Pr(A\mid B)=\frac{Pr(A\cap B)}{Pr(B)}$ where $A$ represents getting a four of a kind and $B$ represents the first two cards are of different ranks, I recommend using the sample space whose elements are of the form $(\text{first card},\text{second card},\text{set of remaining cards})$, which we recognize being equiprobable so we can use counting techniques.

You will get $Pr(B)=\dfrac{52\cdot 48\cdot \binom{50}{5}}{52\cdot 51\cdot \binom{50}{5}}=\dfrac{48}{51}$

Then, in calculating $Pr(A\cap B)$, we do so similarly to the first method, except having to deal with the extra step of deciding the specific cards used in the first two slots:

$Pr(A\cap B)=\dfrac{52\cdot 48\cdot \binom{47}{2}+52\cdot 48\cdot \binom{47}{2}+52\cdot 48\cdot 11\cdot 46}{52\cdot 51\cdot \binom{50}{5}}$

Completing the arithmetic arrives at the same answer as before.

Let's say the first two cards are "A" and "B".  To get "four of a kind", you the next three cards must be of type "A" or of type "B".  There are 50 cards left to draw from, 3 of type "A" and 3 of type "B".  The probability the third card matches either "A" or "B" is 6/50= 3/25.  After that we must match the third card drawn.  On the fourth draw there are 49 cards left, 2 of the correct type.  The probability the fourth card matches the third card is 2/49.  There are then 48 cards left, just 1 of the correct type.  The probability the fifth card matches the third and fourth cards is 1/48.

The probability of four of a kind given that the first two cards are different is (3/25)(2/49)(1/48).


How do i prove there exists a function $F(n+1)=f_n(F(1),...,F(n))$? Let $X$ be a set and $f_n:X^n \rightarrow X$ be a function for all $n\in \mathbb{N}$ and $c\in X$.

How do i prove that there exists $F:\mathbb{N} \rightarrow X$ such that $F(n+1)=f_n(F(1),...,F(n))$ and $F(1)=c$?

Plus, is this $F$ unique?

The quick brute-force method would be to use Zorn's lemma.

Consider the set of all functions $A\to X$ where $A$ is an initial sequence of $\mathbb N$ and the function satisfies your equation. Order the functions by set inclusion. Zorn's lemma says there is a maximal element. The domain of this maximal element cannot have a largest element, because then you would easily construct a function that went one higher. Therefore the domain must be $\mathbb N$ itself, and the maximal element is your $F$.

(It can be done without Zorn or choice, but there will be more thinking involved then).

For uniqueness, suppose there are $F$ and $G$ that both satisfy your equation. Prove by (long) induction on $n$ that $F(n)=G(n)$ for all $n$, and therefore $F=G$.


Lagrange vs Eulerian coordinates I have to solve this exercise about the Eulerian and Langrangian approach and I would like if you can take a look on my answers:

In the evening, at $t=0$ say, the temperature increases southwards in Sweden at a rate of $0.1$ K/km. A wind from the south at $3.0$ m/s brings warm air with it. However, net radiation from the air (into the clear sky) cools the air such that material air particles, travelling with the wind at speed $3.0$ m/s, decrease their temperature with time at a given constant rate of $−0.36$ K/hour.
a) What is the time derivative of temperature in [K/hour] for an observer standing still on the ground?
b) In what direction, and at what speed in [km/hour] should an observer move in order to experience a constant temperature?

These are my answers

a)

$$\frac{\partial T}{\partial t} = -0.36 K/hour$$

b)

$$\frac{D T}{D t} = \frac{\partial T}{\partial t}  + u \frac{\partial T}{\partial x} = 0$$

$$-0.36 + u (-0.1) = 0$$

$$u(-0.1) = 0.36$$

$$u = \frac{0.36}{-0.1}$$

$$u = -3.6 km/h$$

I have some doubts about the second because I don't know if I have to include the velocity of the fluid and calculate the observer velocity relative to that.
Note I assume the positive direction in the same direction of the wind velocity.

Thank you for help.

Edit* I think you need to consider the observer's movement, as the temperature of the particle changes over a fixed distance, but that rate is subject to both the speed of the wind (3m/s) and the perceived rate of change would thus be dependent on the speed of the observer.

Also I presume you're in SG2214 this semester, if you'd like to work on problems together I'd be happy to have a work buddy :) my email is (shsng@kth.se)


Stock behaviour probability I found this question in a financial mathematics course exam, could anyone please help with a solution and some explanation? Thanks in advance :)

  A stock has beta of $2.0$ and stock specific daily volatility of $0.02$.
  Suppose that yesterday’s closing price was $100$ and today the market
  goes up by $1$%. What’s the probability of today’s closing price being
  at least $103$?

Expectation of the stock price is $100*(1+2*0.01)=102$

standard deviation is $2\%$, which is $2$

$103$ is $0.5$ standard deviation away,

$1-N(0.5)=31\%$

Assuming normality, vola being specified as standard deviation and assuming a risk free rate (r_risk_free) of zero the following reasoning could be applied:

1) From CAPM we see that the expected return of the stock (E(r)=r_risk_free+beta*(r_market-r_risk_free) here E(r)=0+2.0*.01=0.02

2) From a casual definition of beta we know that it relates stock specific vola to market vola (sd_stock_total=beta*sd_stock_specific) here sd_stock_total=2*0.02=.04

3) Since we have data on returns only, we transfer the value of interest (103) into returns space which is r_of_interest=(103-100)/100=0.03

4) Now we are looking for the probability that the return of interest occurs (P(x>=0.03)) given the probability distribution defined as N(0.02,0.04) or written differently P(x>=0.03)~N(0.02,0.04)

5) In R you could write 1-(pnorm(.03,mean=.02,sd=.04)) where 1-pnorm(...) is necessary because pnorm() returns P(X<=x) of the cumulative distribution function (CDF) and you are interested in P(x>=0.03) - more pnorm() hints.


Distributive Property - A Converse Theorem Suppose that $\ast$ distributes over $+$, where $0$ is the additive identity.  We can conclude the following.

$a \ast b = a \ast \left( b + 0\right)$

$a \ast b = a \ast b + a \ast 0$

$\therefore a \ast 0 = 0$

In other words, the additive identity and multiplicative 'annihilator' become equivalent under the distributive property.  However, the converse is not true.  Equating the two $0$'s is a necessary but insufficient condition to establish distributivity.

My question: suppose we know that the following two statements are true for any element $a$.

$a+0=a$

$a \ast 0 = 0$

What other conditions (in addition to the above) are sufficient to imply a distributive property?

Listen, the converse of the distributive property does work. It is common in algebra for you to use the converse of the distributive property ex.) 2x + 3x = 5x or x(2 + 3).


Inferior limit of a sequence of measurable functions Suppose $f_n$ is a sequence of non-negative measurable functions that converges in measure to a measurable function $f$. I am trying to show that

$$\int f\leq\liminf_n\int f_n$$

Using Fatou's lemma we know that

$$\int\liminf_n f_n\leq\liminf_n\int f_n$$

so it would be enough to show that
$$f\leq\liminf_n f_n$$
but, is this even true? am I going in the right direction?

Hints:

There exists a subsequence $(f_{n_k})_k$ of $(f_n)_n$ such that $$\liminf_{n \to \infty} \int f_n = \lim_{k \to \infty} \int f_{n_k}.$$
Check that $f_{n_k} \to f$ in measure as $k \to \infty$.
There exists a further subsequence, say $(g_j)_j$, of $(f_{n_k})_k$ such that $g_j \to f$ almost everywhere. Applying Fatou's lemma yields, $$\int f \leq \liminf_{j \to \infty} \int g_j= \liminf_{n \to \infty} \int f_n.$$

Consider $g_n(x)=\inf\limits_{k \geq n}f_k(x)$, if so, we obtain $g_1 \leq g_2 \leq \ldots \leq g_n \leq \ldots$ and $\lim\inf f_n(x) \leq \lim\limits_{n \to \infty} g_n(x)$. By monotone convergence $\int\limits_X f(x)dx = \lim\limits_{n \to \infty} \int\limits_X g_n(x)dx$, which gives $g_n(x) \leq f_n(x)$, so $\int\limits_X g_n(x) dx \leq \int\limits_X f_n(x)dx$.


Black and white beads on a circle There are $n$ beads placed on a circle, $n\ge 3$. They are numbered in random order as viewed clockwise. Beads for which the number of the previous bead is less than the number of a next bead are painted in white color,and others - in black.

Two colourations that can be made equal by rotation are considered identical. How many different colourations can occur?

I've write a programm and for $n=3...11$ I've got answers $2 , 1 , 6 , 7 ,  18 , 25 , 58 , 93 , 186$

What follows  is not  an answer but  a conjecture and  some additional
material. We  start by working  with permutations. I assumed  that the
bit  pattern  /  black-white  pattern  at position  $q$  reflects  the
less-than  greater-than relation  at position  $q$ in  the permutation
(whether the  pair at $q$  and $q+1$ is  an ascent or a  descent, with
circular wrap-around).

I  wrote   a  Perl  program  to  investigate   these  necklaces  (Perl
aficionados are invited to verify  and improve this work, e.g. the bit
strings can be produced with pack/unpack). The program
iterates  over  all bit  patterns  and  attempts  to find  a  circular
permutation  that fits  the  pattern using  a backtracking  algorithm,
which I hope I've implemented correctly.

The program produced the following table:

$ time ./blg2.pl 12 2>/dev/null
3: 4 2/2
4: 6 4/2
5: 8 6/2
6: 14 12/2
7: 20 18/2
8: 36 34/2
9: 60 58/2
10: 108 106/2
11: 188 186/2
12: 352 350/2

real    7m9.188s
user    7m7.926s
sys     0m0.045s

This table shows the total number of circular colorations obtained and
indicates furthermore that for all  $n$ all possible bit patterns were
realized except two. The totals give the sequence

$$4,6,8,14,20,36,60,108,188,352,\ldots $$

which is  OEIS A00031 which  is simply the
substituted cycle index of the cyclic group
$$\left.Z(C_n)(A+B)\right|_{A=1,B=1},$$
and can be  taken as evidence that the program is  working since it is
the  correct  answer  (total  number  of circular  patterns  with  two
colors.)

Observe that the exact values which are
$$2,4,6,12,18,34,58,106,186,350,\ldots $$
also have an OEIS entry namely
OEIS A052823
which  reflects  the underlying  necklace  combinatorics  but not  the
permutation aspect.

The   program  can  also   output  permutations   that  fit   a  given
pattern. Here are the examples for $n=3,4,5$ and $n=6$:

$ time ./blg2.pl 5
100 3-1-2
110 3-2-1
111 FAIL
000 FAIL
3: 4 2/2
1010 3-1-4-2
1000 4-1-2-3
1110 4-3-2-1
1100 4-2-1-3
1111 FAIL
0000 FAIL
4: 6 4/2
11100 5-3-2-1-4
11110 5-4-3-2-1
10100 4-1-5-2-3
11010 4-2-1-5-3
11000 5-2-1-3-4
10000 5-1-2-3-4
00000 FAIL
11111 FAIL
5: 8 6/2
100000 6-1-2-3-4-5
111000 6-3-2-1-4-5
101000 5-1-6-2-3-4
110100 5-2-1-6-3-4
100100 4-1-5-6-2-3
111110 6-5-4-3-2-1
110000 6-2-1-3-4-5
110110 4-2-1-6-5-3
111100 6-4-3-2-1-5
101010 3-1-5-4-6-2
111010 5-3-2-1-6-4
101100 4-1-6-5-2-3
111111 FAIL
000000 FAIL
6: 14 12/2

Studying the  examples we immediately  have a conjecture,  namely that
all black-white  patterns can be  realized except the  monochrome ones
(this is obvious  as the circularity makes it  impossible to have just
one run in a circular permutation since not all $n$ elements can be to
the left  of a larger/smaller  element).  The conjectured  formula is
thus

$$-2 + \left.Z(C_n)(A+B)\right|_{A=1,B=1}.$$
This is
$$-2 +
\left.\frac{1}{n} \sum_{d|n} \varphi(d) (A^d+B^d)^{n/d}\right|
_{A=1, B=1}.$$

This gives the following
CONJECTURE:
$$\large{\color{#0A0}{
-2 + \frac{1}{n} \sum_{d|n} \varphi(d) 2^{n/d}}}.$$

Concluding  remark.  With  the  amount  of data  and  context  now
available it should not be difficult to produce a combinatorial proof,
which I suspect will turn out to be simple.

This is the Perl code that was used to compute the above data.

#! /usr/bin/perl -w
#

sub search {
    my($n, $bits, $q, $avail, $key, $seen, $sofar) = @_;

    if($q==$n){
        if(($bits->[$n-1] == 0 &&
            $sofar->[$n-1] < $sofar->[0]) ||
           ($bits->[$n-1] == 1 &&
            $sofar->[$n-1] > $sofar->[0])){
            $seen->{$key} = join('-', @$sofar);
        }

        return;
    }

    my $pos = 0;
    while(!exists($seen->{$key}) && $pos<$n-$q){
        my $nxttry = $avail->[$pos];

        if($q==0 ||
           (($bits->[$q-1] == 0 &&
             $sofar->[$q-1] < $nxttry) ||
            ($bits->[$q-1] == 1 &&
             $sofar->[$q-1] > $nxttry))){
            push @$sofar, $nxttry;
            splice @$avail, $pos, 1;

            search($n, $bits, $q+1,
                   $avail, $key, $seen, $sofar);

            splice @$avail, $pos, 0, $nxttry;
            pop @$sofar;
        }

        $pos++;
    }
}

 MAIN: {
     my $mx = shift || 6;

     for(my $n=3; $n<=$mx; $n++){
         my $seen = {};  my $failed = {};

         for(my $ind=0; $ind<2**$n; $ind++){
             my $bits = [];

             for(my ($pos, $indx)=(0, $ind);
                 $pos<$n; $pos++){
                 push @$bits, ($indx %2);
                 $indx = ($indx-$bits->[-1])/2;
             }

             my $rot;
             for($rot=0; $rot<$n; $rot++){
                 my @rotbits =
                     (@$bits[$rot..($n-1)],
                      @$bits[0..($rot-1)]);
                 my $rotkey = join('', @rotbits);

                 last if
                     exists($seen->{$rotkey}) ||
                     exists($failed->{$rotkey});
             }

             if($rot==$n){
                 my $key = join('', @$bits);
                 search($n, $bits, 0, [1..$n], $key, $seen, []);

                 $failed->{$key} = 'FAIL'
                     if !exists($seen->{$key});
             }
         }

         my $total = scalar(keys %$seen) + scalar(keys %$failed);

         foreach my $pat (keys %$seen){
             print STDERR "$pat " . $seen->{$pat} . "\n";
         }
         foreach my $pat (keys %$failed){
             print STDERR "$pat " . $failed->{$pat} . "\n";
         }

         print "$n: $total " . scalar(keys(%$seen)) . "/" .
             scalar(keys(%$failed)) . "\n";
     }
}

There is a radically simplified  version of the above program which is
not  as fast,  however. Instead  of  iterating over  bit patterns  and
backtracking  to  find a  matching  permutation  we  iterate over  all
permutations and collect the  bit patterns / black-white patterns that
appear. The slow-down in the speed  when solving the case of $n=10$ is
on the order of a factor  of $60$. This code uses the factorial number
system to iterate over permutations  and never allocates more than one
permutation at a time.

#! /usr/bin/perl -w
#

sub fact {
    my ($n) = @_;

    return 1 if ($n == 0 || $n == 1);
    return $n*fact($n-1);
}

 MAIN: {
     my $mx = shift || 6;

     for(my $n=3; $n<=$mx; $n++){
         my $seen = {};

         for(my $ind=0; $ind<fact($n); $ind++){
             my @perm = (1..$n);

             for(my ($pos, $indx) = ($n-1, $ind);
                 $pos > 0; $pos--){
                 my $targ = $indx % ($pos+1);
                 $indx = ($indx-$targ)/($pos+1);

                 my $tmp = $perm[$pos];
                 $perm[$pos] = $perm[$targ];
                 $perm[$targ] = $tmp;
             }

             my @bits = ();

             for(my $pos=0; $pos<$n-1; $pos++){
                 my $bit =
                     ($perm[$pos] < $perm[$pos+1] ? 1 : 0);
                 push @bits, $bit;
             }
             push @bits, ($perm[$n-1] < $perm[0] ? 1 : 0);

             my $rot;
             for($rot=0; $rot<$n; $rot++){
                 my @rotbits =
                     (@bits[$rot..($n-1)],
                      @bits[0..($rot-1)]);
                 my $rotkey = join('', @rotbits);

                 last if exists($seen->{$rotkey});
             }

             if($rot==$n){
                 my $key = join('', @bits);
                 $seen->{$key} = join('-', @perm);
             }
         }

         foreach my $pat (keys %$seen){
             print STDERR "$pat " . $seen->{$pat} . "\n";
         }

         print "$n: " . scalar(keys(%$seen)) . "\n";
     }
}

Let $a_n$ be the number of configurations for this problem
and let $b_n$ be the number of two-colored necklaces with $n$ beads (no flips allowed).
It is well-known that $b_n=\frac 1n\sum_{d\mid n}\phi(d)2^\frac nd$ (this is OEIS A000031).

Computer runs suggest that for odd $n$ all these patterns also occur for this problem,
except the two monochromatic ones, so $a_n=b_n-2$.

They also suggest that for even $n$ all patterns occur except the ones where one color occurs every second spot.
In this case the number of necklaces where black occurs every second spot is $b_{\frac n2}$.
We get the same number for the patterns where white occurs every second spot,
and then we have double counted the one pattern where black and white alternate all around the necklace,
so we get $a_n=b_n-2b_{\frac n2}+1$.


Why is compactness so important? I've read many times that 'compactness' is such an extremely important and useful concept, though it's still not very apparent why. The only theorems I've seen concerning it are the Heine-Borel theorem, and a proof continuous functions on R from closed subintervals of R are bounded. It seems like such a strange thing to define; why would the fact every open cover admits a finite refinement be so useful? Especially as stating "for every" open cover makes compactness a concept that must be very difficult thing to prove in general - what makes it worth the effort?

If it helps answering, I am about to enter my third year of my undergraduate degree, and came to wonder this upon preliminary reading of introductory topology, where I first found the definition of compactness.

As many have said, compactness is sort of a topological generalization of finiteness. And this is true in a deep sense, because topology deals with open sets, and this means that we often "care about how something behaves on an open set", and for compact spaces this means that there are only finitely many possible behaviors.

But why finiteness is important? Well, finiteness allows us to construct things "by hand" and constructive results are a lot deeper, and to some extent useful to us. Moreover finite objects are well-behaved ones, so while compactness is not exactly finiteness, it does preserve a lot of this behavior (because it behaves "like a finite set" for important topological properties) and this means that we can actually work with compact spaces.

The point we often miss is that given an arbitrary topological space on an infinite set $X$, the well-behaved structures which we can actually work with are the pathologies and the rare instances. This is throughout most of mathematics. It's far less likely that a function from $\Bbb R$ to $\Bbb R$ is continuous, differentiable, continuously differentiable, and so on and so forth. And yet, we work so much with these properties. Why? Because those are well-behaved properties, and we can control these constructions and prove interesting things about them. Compact spaces, being "pseudo-finite" in their nature are also well-behaved and we can prove interesting things about them. So they end up being useful for that reason.

Every continuous function is Riemann integrable-uses Heine-Borel theorem. Since there are a lot of theorems in real and complex analysis that uses Heine-Borel theorem, so the idea of compactness is too important.


Conditional probability - a formal discussion This is a rather philosophical question.

P[B|A] is formally defined as P[B and A]/P[B] where A and B are events in a sigma algebra and P is a probability mass function.

That is, P[B|A] is just a division of two numbers. If so, how come there are problems where we find it hard to calculate P[B and A] as well as P[B], but it is easy for us to reason about P[B|A] and so we assign a value to P[B|A] immediately without going through the division? (I can't think of an example for this, but I surely recall there are such cases. Can anyone share an example?)

To be more concrete, I'd be happy to see an example where it's hard\impossible to calculate P[A and B] or P[B] but it is easy to reason about P[A|B] on intuitive levels along with a justification for this reasoning (I'm talking about sample space and probability function definitions).

I think you mean $P(A|B)$ rather than $P(B|A)$; I'll assume that.

It might happen that event $B$, if it happens, controls the conditions for $A$ to happen, which does not imply that one has any idea of how probable $B$ is. As an extreme case, $B$ might logically imply $A$, in which case $P(A|B)=1$ regardless. Another example is if someone tosses a coin but I have no idea whether the coin is fair; for the events $A$: I win the toss, and $B$: the coin is fair, I know by definition that $P(A|B)=0.5$, even though I know nothing about $P(B)$ or $P(A\cap B)$.

Role of Conditional Probability - My thoughts:

A need for representing event in the presence of prior knowledge: Consider the probability of drawing a king of heart randomly from a standard deck of 52 cards. The probability of this event without any prior knowledge is 1/52. However, if one learns that the card drawn is red, then the probability of getting a king of heart becomes 1/26. Similarly, if one gathers knowledge that the card drawn is a face card (ace, king, queen or jack) then the probability gets shifted to 1/16. So we see that representing an event in the presence of prior knowledge is important and conditional event representation of (A|H) is the most adopted representation to solve this need. No matter what representation we adopt, we can agree that conditional event solves an unique concern of representing conditional knowledge.

What is most elemental - unconditional Vs. conditional: The debate whether conditional probability is more elemental than (unconditional) probability remains as an enticing subject for many statistician [1] as well as philosophers [2]. While the most adopted notation of conditional probability and its ratio representation viz. P(A|H)=P(AH)/P(H) where P(H)>0 indicates (unconditional) probability is more elemental; the other school of thoughts has their logic too. For, them, when we say probability of getting face value of 2 in a random throw of a fair dice is 1/6, we apply prior knowledge that all throw will lands perfectly on a face such that a face will be visible unambiguously, or that the dice will not break into pieces when rolled and so on. Therefore we apply a prior knowledge in order to determine a sample space of six face values. No matter what kind of probability is the most elemental, following the notation of conditional probability, we can agree that we speak of (unconditional) probability when we’ve accepted a sample space as the super most population and we’re not willing to get astray by adding further sample points to this space.  Similarly, we speak of conditional probability when we focus on an event with respect to the sub-population of the super-most (absolute in this sense) population.

Is there any case which can be solved only by conditional probability: Once again, as long as we accept the ratio representation of the conditional probability, we see that conditional probability can be expressed in terms of unconditional probability. Thus, conceptually, any problem where conditional probability is used, can also be solved without use of conditional probability as well. However, we must appreciate that for cases where population and sub-population are not part of the same experiment, the use of conditional probability is really useful (not necessarily inevitable). To explain this further, in case of finding probability of a king of heart given that the card is red, we don’t really need conditional probability because the population of 52 cards and sub-population of 26 red cards are very clear to us. However, for cases such as applying a medicinal test on a cow to determine if it has mad-cow-disease, if we know false positive and false negative probabilities of the test, then to find out probability that a cow has disease given that it has tested positive, conditional probability can be used with great effect. If I may bring an analogy of ‘plus’ and ‘multiplication’ symbols of mathematics, we all know that any problem that uses multiplication symbol, can also be solved without it by mere use of ‘plus’ symbol. Similarly, in terms of solving problems, conditional probability can be avoided altogether just like multiplication symbol in mathematics. Still, we can appreciate the usefulness of conditional probability just like we can appreciate the use of multiplication in mathematics.

----------Bibliography------------
[1] H. Nguyen and C. Walker, “A history and introduction to the algebra of conditional events and probability logic,” Systems, Man and Cybernetics, IEEE Transactions on
   (Volume:24 , Issue: 12 ), pp. 1671 - 1675, Dec 1994.

[2] A. Hájek, “Conditional Probability,” Handbook of the Philosophy of Science. Volume 7: Philosophy of Statistics., vol. 7, p. 99, 2011.


Do matrices have a "to the power of" operator? Well I was sure that saying "$A^3$"  (where $A$ is an $n\times n$ matrix) is nonsense. Sure one could do $(A\cdot A) A$  But that contains different operators etc. So what did my prof mean by the following statement:

show that $A^{25}\mathbf{x} = \mathbf{0}$ has only the trivial solution? (We're also given the determinant of A).

I know the proof will probably end with stating: "This means that $A^{25}$ is invertible, so $A^{25}\mathbf{x} = \mathbf{0}$ has only the trivial solution. And well I could state that $\det(A^{25}) = 5^{25} \neq 0$.

But then again: I really wonder what the "to the power of" operator means? Or did my prof make a mistake here?

If you want a formal definition of matrix exponentiation for non-negative integer values, just define $A^n = A^{n-1}\cdot A$ and $A^0 = I$. Since matrix multiplication is associative, we won't have any ambiguity there.

Edit: As Tobias Kildetoft points out below, it might be wiser to define the base case as  $A^1=A$ instead of $A^0=I$, so as to not have to worry about how $\det(A^n)=\det(A)^n$ for $\det(A)=0$ would imply $0^0=1$. Which isn't false, depending on how we want to define it, but is something we might not want to worry about for the purposes of defining matrix powers.

But matrix multiplication is not commutative, $A^{2}\cdot A^{3} \neq A^{3}\cdot A^{2}$ except in some particular cases.
$$
    A = \pmatrix{0 & 3 \\ 1&2}; A^2 = \pmatrix{3 & 6 \\ 2 & 7}; A^3 = \pmatrix{6&  21 \\ 9 & 20}
$$
$$A^2·A^3 = \pmatrix{72 & 183\\ 75 & 182}; A^3·A^2 = \pmatrix{60 & 183 \\ 61 & 182}
$$
So $A^5$ has more than 1 solution, at least in this case.


Probability that X is less than Y directly from joint CDF? Suppose X and Y are arbitrary random variables with joint cdf F. Is it possible to find $P(X<Y)$ or $P(X\le Y)$ directly from F? I think it's easier to think in terms of the equivalent $P(X-Y<0)$, but I'm not sure how to go from there. I know that if X and Y are continuous, we can differentiate the joint cdf to get the joint pdf and we can integrate over that to solve it. But is there a more general way to get it directly from the joint cdf? e.g., what if X or Y is not continuous or discrete?

$\displaystyle \mathbb{P}(X\leq Y)= \int_{-\infty}^{\infty}\int_{-\infty}^{y} f(x,y)dxdy$. Assuming continuous r.v pdf $f(x,y)$


Blow up of a solution What exactly does blow up mean, when people say, for example, that a solution (to a pde (say)) blows up.

Thanks.

The meaning is, of course, context-dependent...

In the context of differential equations, that a solution to an equation with a "time" variable blows up usually means that the maximal domain for which it is defined is finite, so that at the endpoint of that interval something `bad' happens: either the solution goes to infinity, or it stops being smooth (in a way that makes the differential equation stop having sense, maybe), or something. This is an important phenomenon, one which causes trouble. A couple of examples:

Perelman's solution of the Poincaré conjecture—in a very vague sense—consists of a way to `work around' the fact that certain solutions of a (very complicated non-linear) PDE blow up;
the third `Millenium' Clay problem is (very roughly) the question «do the solutions of the Navier-Stokes equation blow up?».

Consider, as a very simple example, the equation $$\frac{\mathrm dx}{\mathrm dt}=x^2.$$ This equation makes sense and satisfies the conditions for existence and uniqueness of solutions on all of the $(t,x)$-plane, but if you solve it (which is easy to do explicitely, as it has the two variabls separated) you'll see that all of its solutions have a maximal interval which is a half-line (which is bounded on the left or on the right, depending on the initial condition) and that at the finite end of that interval the solutions become unbounded. We thus say that all solutions of our equation blow up in finite time.

There are also equations which have some solutions which blow up and some which live forever. One example is $$\frac{\mathrm dx}{\mathrm dt}=\begin{cases}x^2&\text{if $x\geq0$}\\0&\text{if $x\leq0$}\end{cases}$$ and you'll surely find lots of fun in trying to concoct examples where even more interesting phenomena occur.

"Blows up" means it goes to infinity.

1/x "blows up" at x = 0.


Summation of an infinite series The sum is as follows:

$$
\sum_{n=1}^{\infty} n \left ( \frac{1}{6}\right ) \left ( \frac{5}{6} \right )^{n-1}\\
$$

This is how I started:

$$
= \frac{1}{6}\sum_{n=1}^{\infty} n  \left ( \frac{5}{6} \right )^{n-1} \\
= \frac{1}{5}\sum_{n=1}^{\infty} n  \left ( \frac{5}{6} \right )^{n}\\\\
 = \frac{1}{5}S\\
S = \frac{5}{6} + 2\left (\frac{5}{6}\right)^2 + 3\left (\frac{5}{6}\right)^3 + ...
$$

I don't know how to group these in to partial sums and get the result. I also tried considering it as a finite sum (sum from 1 to n) and applying the limit, but that it didn't get me anywhere!

PS: I am not looking for the calculus method.

I tried to do it directly in the form of the accepted answer,

$$
\textrm{if} \ x= \frac{5}{6},\\
S = x + 2x^2 + 3x^3 + ...\\
Sx = x^2 + 2x^3 + 3x^4 + ...\\
S(1-x) = x + x^2 + x^3 + ...\\
\textrm{for x < 1},\ \  \sum_{n=1}^{\infty}x^n = -\frac{x}{x-1}\  (\textrm{I looked up this eqn})\\
S = \frac{x}{(1-x)^2}\\
\therefore S = 30\\
\textrm{Hence the sum} \sum_{n=1}^{\infty} n \left ( \frac{1}{6}\right ) \left ( \frac{5}{6} \right )^{n-1} = \frac{30}{5} = 6
$$

Letting $a = d = 1/6$ and $r = 5/6$, our sum is:
$$
S = a + (a + d)r + (a + 2d)r^2 + (a + 3d)r^3 + \cdots
$$
Scaling by $r$, we find that:
$$
rS = ar + (a + d)r^2 + (a + 2d)r^3 + \cdots
$$
Subtracting the two equations (by collecting like powers of $r$), we obtain:
$$
(1 - r)S = a + dr + dr^2 + dr^3 + \cdots = a + dr(1 + r + r^2 + \cdots) = a + \frac{dr}{1 - r}
$$
Hence, we conclude that:
$$
S = \frac{a}{1 - r} + \frac{dr}{(1 - r)^2} = \frac{1/6}{1 - 5/6} + \frac{(1/6)(5/6)}{(1 - 5/6)^2} = \frac{1}{6 - 5} + \frac{(1)(5)}{(6 - 5)^2} = 6
$$

hint: differentiate  the identity

$$\sum_{k=0}^{\infty} x^k = \frac{1}{1-x} $$


Limit comparison test; negative outcome It says the outcome must be a positive finite answer, what happens if it's a negative finite answer? Is it inconclusive as if it were 0 or infinity?

If $\lim |a_n/b_n|=0$ and $\sum |b_n|$ converges, then $\sum |a_n|$ converges.  If $\lim |a_n/b_n|=\infty$ and $\sum |a_n|$ diverges, then $\sum |b_n|$ diverges.

I'm not sure how the above answer relates, but in my calculus 3 class our teacher, when he finds a negative result ex: -1/10 when using the limit comparison theorem, he simply combines the theorem with the absolute value theorem that says if the absolute value converges so does the series, so then you get 1/10, which is positive. Combining theorems I believe is the complicated idea here.


Topology on the space of test functions I try to read into the theory of distributions and there is one thing which bothers me. I read that a distribution is a linear, continuous functional from the space of test functions, which, depending on the author, is sometimes defined as the Schwartz space and sometimes as $C^\infty_K$ (smooth functions with compact support).

For starters, where does this ambiguity come from and does it matter?

Now for the case $C^\infty_K$, I found two definitions of the associated topology.  One as the final topology induced by the inclusion maps from $C^\infty(K)$ (I already understand how the topology on those spaces is defined) and one as the final locally convex topology induced by those maps, i.e. the finest locally convex topology which makes those maps continuous.

I couldn't proof their equivalence, i.e. how do I show that the final topology induced by those maps is already linear and locally convex?

OK, I think I got a counterexample now, i.e. the final topology w.r.t. to the inclusion maps is not linear.  I only worked it through on ${\bf R}$ though.

The only assumption that I need is that for given $\epsilon>0$ and $k,n\in{\bf N}$ we find some $f\in C^\infty([-1,1])$ such that $\|f^{(j)}\|_\infty<\epsilon$ for $j<k$ and $f^{(k)}(0)>n$, which I think is easily constructable.

Now the set $$A:=\{f\in C^\infty_K:\forall j>0.f(j)<1/f^{(j)}(0)\}$$ (where we set $1/0=\infty$) is open in the final topology, i.e. $A\cap C^\infty([-i,i])$ is open for each $i$.  Yet, if the final topology were linear, we would find an neighbourhood $B$ of $0$ such that $B+B\subseteq A$.  But any such $B$ contains a basic open neighbourhood from $C^\infty([-i,i])$ of the form $$U_i=\{f\in C^\infty([-i,i]):\forall j<k(i).\|f^{(j)}\|_\infty<\epsilon(i)\}$$ for each $i$.  Now we can construct $f\in U_1$ such that $f^{(k(i))}(0)$ is large enough so $f+U_{k(i)+1}\subseteq B+B$ is no longer a subset of $A$.

So the majority of books and scripts I read (e.g. Rudin) got it right, when they defined the topology on the space of test functions as the limit topology in the category of locally convex spaces, i.e. the finest locally convex topology which makes the inclusion maps continuous.

Let $K_n$ be the unit ball of radius $n$, so your $C_K^\infty = \bigcup_n C^\infty(K_n)$ and let $f_n:C^\infty(K_n)\longmapsto C_K^\infty$ be the natural embeddings. The system
$$
\bigl\{ \bigcup_n f_n[U_n]\, ;\, U_n \text{ is a $0$-neighbourhood in } C^\infty(K_n) \bigr\}
$$
is a filter base in $C_K^\infty$, immediately to verify (using $U_n = f_n^{-1}(f_n[U_n])$) that it is a filter base for a for a fundamental system of $0$-neighbourhoods of a linear topology.

An absolutely convex set U is $0$-neighbourhood in $C_K^\infty$ for the topology described above, iff $f_n^{-1}[U]$ is a $0$-neighbourhood in $C^\infty(K_n)$ for each $n$. But the latter is precisely the characterization for a $0$-neighbourhood in the final topology on $C_K^\infty$.


Arc length of the squircle The squircle is given by the equation $x^4+y^4=r^4$. Apparently, its circumference or arc length $c$ is given by

$$c=-\frac{\sqrt[4]{3} r G_{5,5}^{5,5}\left(1\left|
\begin{array}{c}
 \frac{1}{3},\frac{2}{3},\frac{5}{6},1,\frac{4}{3} \\
 \frac{1}{12},\frac{5}{12},\frac{7}{12},\frac{3}{4},\frac{13}{12} \\
\end{array}
\right.\right)}{16 \sqrt{2} \pi ^{7/2} \Gamma \left(\frac{5}{4}\right)}$$

Where $G$ is the Meijer $G$ function. Where can I find the derivation of this result? Searching for any combination of squircle and arc length or circumference has led to nowhere.

You can do this using this empirical formula to find perimeter of general superellipse

$L=a+b*(((2.5/(n+0.5))^{1/n})*b+a*(n-1)*0.566/n^2)/(b+a*(4.5/(0.5+n^2)))$
This mentioned in my research

By your definition, $\mathcal{C} = \{(x,y) \in \mathbb{R}^{2}: x^4 + y^4 = r^4\}$. Which can be parametrized as
\begin{align}
\mathcal{C} =
\begin{cases}
\left(+\sqrt{\cos (\theta )},+\sqrt{\sin (\theta )} \right)r\\
\left(+\sqrt{\cos (\theta )},-\sqrt{\sin (\theta )} \right)r\\
\left(-\sqrt{\cos (\theta )},+\sqrt{\sin (\theta )} \right)r\\
\left(-\sqrt{\cos (\theta )},-\sqrt{\sin (\theta )} \right)r
\end{cases}
, \qquad 0 \leq \theta \leq \frac{\pi}{2}, \, 0<r
\end{align}

Now, look at this curve in $\mathbb{R}^{2}_{+}$ as $y = \sqrt[4]{r^4-x^4}$, then observe that symmetry with both axis. It yields the arc length is just:
$$c = 4 \int_{0}^{r} \sqrt{1+\left(\dfrac{d}{dx}\sqrt[4]{r^4-x^4}\right)^2} \,dx = 4 \int_{0}^{r} \sqrt{1+\frac{x^6}{\left(r^4-x^4\right)^{3/2}}} \,dx$$


Find a confidence interval using as pivotal quantity a function of the MLE Let $X_1,\ldots,X_n$ be a random sample from $f(x\mid\theta)=\theta x^{\theta -1}$ for $0<x<1$. Find a confidence interval for $\theta$ using as pivotal quantity a function of the maximum likelihood estimator for $\theta$.

Well, using the logaritmic version of the likelihood function, I got that the MLE of $\theta$ is $T_1=\frac{-n}{\sum\limits_{i=1}^n \log x_i}$. Is this correct? And how could I use a function of it as a pivotal quantity?

I know that a pivotal quantity is a function of the sample and the parameter, whose distribution doesn't depend on $\theta$.

Assuming you want a confidence interval for $\theta$, you may use the CLT or a known chi distribution for variances and some integrals to get an interval for the parameter. First, we have that

$$
\hat{\theta}=-\frac{n}{\sum ln\ x_i}.
$$

For the expected value we have the integral

$$
E[X] = \int_0^1 \theta x^{(\theta-1)}x\ \text{d}\theta = \frac{\theta}{1+\theta}
$$

And for the second moment

$$
\mu_2 = \int_0^1 \theta x^{\theta-1}x^2\ \text{d}\theta = \frac{\theta}{2+\theta}
$$

So the variance is

$$
Var(X) = \frac{\theta}{2+\theta} - \left( \frac{\theta}{1+\theta} \right)^2 = \\
= \frac{\theta}{(\theta + 1)^2(\theta+2)}.
$$

Now then, we know that the following quotient has a chi-squared distribution, so in this case

$$
C = \frac{(n-1)S^2}{\sigma^2} \sim \chi^2_{n-1}
\text{ and } C = \frac{(n-1)S^2}{\frac{\theta}{(\theta + 1)^2(\theta+2)}}
$$

where $S^2 = \frac{1}{n-1}\sum(X_i-\overline{X})^2$. Finally we can work with the random variable $C$ distributed as a chi squared. We want a confidence interval of $(1-\alpha)$. Then

$$
P \left( \chi^2_{n-1}(\alpha/2) < C < \chi^2_{n-1}(1-\alpha/2)\right) = 1-\alpha
$$

denoting that $\chi^2_{n-1}(\beta)$ is the $\beta$ percentile of the $\chi^2$ which may be found in reference tables.

From here we work with the interval only substituting the percentiles with $\frac{1}{D_l}$ and $\frac{1}{D_s}$, lower and superior.

$$
\frac{1}{D_l} < C < \frac{1}{D_s} \implies
\frac{1}{D_l} < \frac{(n-1)S^2}{\frac{\theta}{(\theta + 1)^2(\theta+2)}} < \frac{1}{D_s} \\\ \\\ \\\
D_l > \frac{\frac{\theta}{(\theta + 1)^2(\theta+2)}}{(n-1)S^2} > D_s \\\ \\\ \\\
D_l((n-1)S^2) > \frac{\theta}{(\theta + 1)^2(\theta+2)} > D_s((n-1)S^2) \\\ \\\ \\\
(D_l((n-1)S^2)) > \frac{\theta}{(\theta + 1)^2(\theta+2)} > (D_s((n-1)S^2)) \\\ \\\ \\\
(\hat{\theta} + 1)^2(\hat{\theta}+2)\left( D_l((n-1)S^2) \right)
> \theta >
(\hat{\theta}+ 1)^2(\hat{\theta}+2)\left( D_s((n-1)S^2) \right) \\\ \\\ \\\
$$

where, taking the before definition and derivation, $\hat{\theta} = -\frac{n}{\sum ln\ x_i}$ and $S^2 = \frac{1}{n-1}\sum(X_i-\overline{X})^2$. When we sent the $\theta$ to the other side, we simply used its estimate to work around the problem. (However a mathematician should confirm this is valid!) Either way, if you find a better known distribution that may include $\theta$, then try it and tell us if it is simpler!

Hope it helps!

Note: made some edits due to an error :/


Relation between eigenvalues of $A$ and $B=(f(a_{i,j}))$, where $f$ is concave/convex Let $A$ be an $n\times n$ real and symmetric matrix with unit diagonal. Furthermore, let the entries $0\leq  a_{ij} \leq 1, i\neq j$.

Now, define a function $f:[0,1] \mapsto [0,1]$ which is concave. For instance we can take $f(x)=2\sin\left(\frac{\pi x}{6}\right)$ which is the transformation of rank correlation coefficients to linear correlation coefficients in a Gaussian copula.

This defines  $B=(f(a_{i,j}))$, which is still symmetric with unit diagonal and $b_{ij}\geq a_{ij}$.

I have three related questions:

1) What is the relationship between the eigenvalues of $A$ and $B$ in this case? Does a general result exist when $f$ is concave or convex?

2) Imagine that $A$ has $t$ negative eigenvalues, how about $B$?

3) If $B$ is positive semi-definit is $A$?

The question arised when i implemented an algortihm to find the nearest (rank) correlation matrix to $A$ (where a requirement is non-negative eigenvalues) which worked as intended. Afterwards, I converted the rank-coefficients to linear correlation coefficients and this correlation matrix had some negative eigenvalues. This made me curious about the relation between the eigenvalues as the function $f$ possesses very nice properties (monotone, concave, infinitely differentiable). Perhaps a general result exists in this case?

Some quick observations. Presumably $n>1$.

(2) Since your $A$ always has nonnegative entries and your $f$ is a function on $[0,1]$, $B$ must have a nonnegative trace. It follows that $B$ must possess at least one non-negative eigenvalue, but the number of non-negative entries may vary wildly.

Example: let $A$ be indefinite, then:

when $f(x)=x$, $B=A$ is indefinite;
when $f(x)=1$, all eigenvalues of $B=ee^T$ (where $e$ is the all-one vector) are non-negative.

(3) No. In the above example with $f(x)=1$, $B$ is positive semidefinite but $A$ is indefinite.


On the idealizer of the set of elementary wedge products of two vectors in $K^4$, for a field $K$ Let $K$ be a field of characteristic zero. Consider $V=K^4$ with standard basis vectors $e_1,e_2,e_3,e_4$. We can consider the second exterior product $\bigwedge^2 V $ of $V=K^4$ with  a basis given by $\{e_1\wedge e_2,e_1\wedge e_3,e_1\wedge e_4, e_2\wedge e_3, e_2\wedge e_4, e_3\wedge e_4\}$. For $\hat x=x_1e_1+x_2e_2+x_3e_3+x_4e_4 \in V$ and $\hat y=y_1e_1+y_2e_2+y_3e_3+y_4e_4 \in V$, we have the elementary wedge product

$\hat x \wedge \hat y=\sum_{i<j}(x_iy_j-x_jy_i) e_i \wedge e_j$.

Now let $R=K[T_1,...,T_6]$ and   Consider  $$I :=\{f \in R : f(a_1b_2-a_2b_1,a_1b_3-a_3b_1,a_1b_4-a_4b_1,a_2b_3-a_3b_2,a_2b_4-a_4b_2,a_3b_4-a_4b_3) = 0, \forall (a_1,...,a_4); (b_1,...,b_4)\in K^4\}$$.

Then $I $ is an ideal of $R$. My question is :

Is $I$ always a principal, radical ideal ?

(May assume $K$ is algebraically closed if need be)

First thing's first: $I$ is the ideal corresponding to the closed embedding $G_k(n)\hookrightarrow \Bbb P(\bigwedge^k K^n)$ of the Grassmanian of $k$-planes in $n$-space into projective space. This enables us to make many determinations about $I$ from knowing what's going on with these geometric objects.

As $n,k$ vary, $I$ is not always principal: if $I$ were principal, then $G_k(n)$ would be a subvariety of codimension at most one in $\Bbb P(\bigwedge^k K^n)$ by Krull's Height Theorem. But the dimension of $G_k(n)$ is $k(n-k)$ while the dimension of the projective space in question is $\binom{n}{k}-1$. It is not hard to see that $k(n-k)+1$ is not always greater than $\binom{n}{k}-1$: for instance, choosing $n=5,k=2$ gives that the former is $7$ while the latter is $9$.

In your specific case of $k=2,n=4$, the ideal is principal and is generated by $p_{23}p_{14} - p_{13}p_{24} + p_{12}p_{34}$, which may be verified by examining the relations between the the coordinates you have written (we write $p_{ij}$ for the determinant of the minor composed of rows $i$ and $j$, or equivalently the coefficient of $e_i\wedge e_j$). This equation may be seen to generate a radical without too much trouble. In general, the equations which generate $I$ are called the Plucker relations, and it's known how to construct all of them (see wikipedia, for instance).

In general, as $n,k$ vary, $I$ is always radical, since the Grassmanian over a field is reduced. This is in fact a special case of a more general theorem: determinental ideals are radical (see Are the determinantal ideals prime?) for several links and sources.

Once you fix $x$ and $y$, your question is equivalent to determining the ideal vanishing at a point in $K^6$. Such an ideal is nothing, but the maximal ideal corresponding to the point.


Sample space for die throwing experiment I am throwing $8$ non-identical dice, and I want to find the probability of getting a sum of the numbers on the dice equal to $30$.

The number of ways of getting the sum equal to $30$ is $125588$ (using multinomial theorem).

Is this the number of favorable cases?

What will be the sample space?

I am a little confused, please help me out...

Answer:

Number of favorable cases $ = (-1)^0 {8\choose0}{29\choose7}+(-1)^1 {8\choose1}{23\choose7}+(-1)^2 {8\choose2}{17\choose7}+(-1)^3 {8\choose3}{11\choose7} = 1560780 - 8*245157 + 28*19448-56*330 =125588$

In general, the formula for finding the distribution of sum s in throwing n dice with x sides goes like this

$$\sum_{k = 0}^{[(s-n)/x]} -1^k {n\choose k}{(s-1-xk)\choose (x-1)}$$


How to find the gradient of norm square How to find the gradient of the function $f(x) = ||g(x)||^2$

where $x \in \Bbb R^d$ and $g: \Bbb R^d \to \Bbb R$.

I will assume $g:\mathbb R^d \to \mathbb R^n$.

It's nice to avoid using components.  Notice that
\begin{equation}
f(x) = h(g(x))
\end{equation}
where $h(x) = \|x\|^2$.
The chain rule tells us that
\begin{align}
f'(x) &= h'(g(x)) g'(x) \\
&= 2 \underbrace{g(x)^T}_{1 \times n} \underbrace{g'(x)}_{n \times d}.
\end{align}

If we use the convention that $\nabla f(x)$ is a column vector, rather than a row vector, then
\begin{equation}
\nabla f(x) = f'(x)^T = 2 g'(x)^T g(x).
\end{equation}


Why does this series converge? My question is:

  Why does the series $$ \sum_{j,k=1}^\infty \frac{1}{j^4+k^4} $$ converge?

I tested the convergence with Mathematica and Octave, but I can't find an analytical proof. In fact, numerical computations suggest that the value of the series is $<1$.

One obvious thing to do would be to use the generalized harmonic series to see that \begin{align} \sum_{j,k=1}^\infty \frac{1}{j^4+k^4} &= \sum_{k=1}^\infty \frac{1}{2k^4} + \sum_{j,k=1; j\neq k} \frac{1}{j^4+k^4} \\ &= \frac{\pi^4}{180} + \sum_{j=1}^\infty \sum_{k=1}^{j-1} \frac{1}{j^4+k^4} + \sum_{j=1}^\infty \sum_{k=j+1}^{\infty} \frac{1}{j^4+k^4}\\ &\leq \frac{\pi^4}{180} + \sum_{j=1}^\infty \sum_{k=1}^{j-1} \frac{1}{(j-k)^4} + \sum_{j=1}^\infty \sum_{k=j+1}^{\infty} \frac{1}{(j-k)^4} \end{align} but unfortunately the last two (double-)series do not converge.

The problem arises when one tries to estimate the Hilbert-Schmidt norm of the Laplacian in $H^2(\mathbb{T}_\pi^2)$.

You can use the inequality $2xy\le x^2+y^2$ (seen by expanding $(x-y)^2\ge0$) to get $2j^2k^2\le j^4+k^4$:
$$\sum_{j,k=1}^\infty\frac1{j^4+k^4}\le\frac12\sum_{j=1}^\infty\sum_{k=1}^\infty\frac1{j^2}\frac1{k^2}=\frac12\Bigl(\sum_{j=1}^\infty\frac1{j^2}\Bigr)^2<\infty.$$

Straightforward application of comparison test:
$$
\sum_{j,k=1}^\infty \frac{1}{j^4+k^4} < \sum_{j=1}^\infty \frac{1}{j^4} = \frac{\pi^4}{90}
$$


Distance between two skew lines I have 2 skew lines $L_A$ and $L_B$ and 2 parallel planes $H_A$ and $H_B$.

The line $L_A$ lies in $H_A$ and $L_B$ in $H_B$. If the equations of $H_A$ and $H_B$ are given like this:

$x+y+z = 0$ (for $H_A$)

$x+y+z = 5$ (for $H_B$)

Can I just simply say that the distance between two lines $L_A$ and $L_B$ is 5 since there the two planes they lies are separated apart by 5?

No. The distance between the two planes is not 5 in the first place. However, if you find the correct distance between the two planes, then your answer may still be wrong if the lines are parallel. If they are not parallel, then it happens to be correct. You gave the condition of skew lines, but I mention these two cases because it shows that it is not at all trivial why the distance should be as claimed, and there is something crucial about the lines being skew.

Yes,the shortest distance between the skew lines will equal the distance between the planes.


Find three real orthogonal matrices of order $3$ having all integer entries. Find three real orthogonal matrices of order $3$ having all integer entries.

I have no idea to solve the problem. I don't know how to start. If $A$ be such matrix then $AA^T=A^TA=I_3$. Please help me.

Let $A$ be such a matrix, and $a_1$, $a_2$ and $a_3$ its columns. Given that $A$ is orthogonal means that
$$\langle a_i,a_j\rangle =\left\{\begin{array}{ll}1&\text{ if } i=j\\0&\text{ if }i\neq j\end{array}\right.$$
So in particular $\langle a_i,a_i\rangle=1$. Because the $a_i$ have all integer coefficients, it follows that
$$a_1,a_2,a_3\in\{(1,0,0),(-1,0,0),(0,1,0),(0,-1,0),(0,0,1),(0,0,-1)\},$$
leaving six options for each of $a_1$, $a_2$ and $a_3$. For any given value of $a_1$ there are precisely $4$ values of $a_2$ such that $\langle a_1,a_2\rangle=0$. For any given values of $a_1$ and $a_2$ there are precisely $2$ values of $a_3$ such that $\langle a_1,a_3\rangle=\langle a_2,a_3\rangle=0$. Hence there are $6\times4\times2=48$ such matrices in total.

If $a,b,c$ are the entries of the first row of a matrix, then $a^2+b^2+c^2$ should be equal to one. this equation has the following integer solutions : (1,0,0),(0,1,0),(0,0,1),(-1,0,0),(0,-1,0),(0,0,-1). Thus, each row of the desired matrix should be filled with exactly one of these solutions. So, you have 120 possible matrices.


the definition of autocorrelation I find the definition of autocorrelation from wiki:
$$
R(s,t)=\frac{E\left[ \left( {{X}_{t}}-{{\mu }_{t}} \right)\left( {{X}_{s}}-{{\mu }_{s}} \right) \right]}{{{\sigma }_{t}}{{\sigma }_{s}}}
$$
But I also find the definition of autocorrelation as below somewhere:
$$
R(s,t)=E\left[ X(s)X(t) \right]
$$

Are these two definitions equal? Can anyone provide the proof?

The bottom definition is the one I often use, i.e. without subtracting the mean and without normalizing. When you take the expectation with the mean subtracted it is often termed the Autocovariance, i.e.:

$C_{xx}(\tau) = E[(x(t)-\mu)(x(t+\tau)-\mu)]$

Having said that both definitions you provide get (confusingly) used, depending on the field of application. But just to make it clear they are not equivalent.

The first definition is the true definition for the autocorrelation.

The second equation you got from "somewhere" just looks like the autocorrelation function for a Gaussian R.V. with $\mu = 0$ and $\sigma^2= 1.$


Let A, B, and C be events such that A and B are both subsets of C. Also, let P(A) = 0.3, P(B) = 0.4, and P(C) = 0.6. Then, P(A|B) could be what Let A, B, and C be events such that A and B are both subsets of C. Also, let P(A) = 0.3, P(B) = 0.4, and P(C) = 0.6. Then, P(A|B) could be what, there is hint saying that there is not only one answer and I do not get it.

$$0.6\geq P(A\cup B)=0.3+0.4-P(A\cap B).$$

Thus, $$P\left(A|B\right)=\frac{P(A\cap B)}{P(B)}\geq\frac{0.1}{0.4}\geq\frac{1}{4}.$$
Also, use that $P(A\cap B)\leq P(A).$


Min value of cos(sinx)+sin(cosx) The max value of $f(x) = \sin(\cos x) + \cos(\sin x)$ is equal to $1 + \sin 1$.  What is the minimum value of $f(x)$?

I found the max value using that range $\sin(\cos x)$ is $[- \sin 1, \sin 1]$ and range of $\cos(\sin x)$ is $[ \cos 1, 1]$ and the maximum of both occurs at $x=0$.  Although I can't find the min value using this approach. Please try to find this  without using graphing calculators.

Look at this graph of your function to see the multiple minima, essential for understanding and solving your problem.


Can anyone make me understand in simple language why the second condition for being an Euclidean domain is superfluous? Can anyone make me understand in simple language why the second condition for being an Euclidean domain is superfluous ?

Why $v(a)  \leq v(ab)$ is not needed?  How we can deduce from the first one?

Suppose $V$ satisfies only the first Euclidean property, i.e. for all $\,a,b\in D\,$ if $\,b\neq 0\,$ then there are $\,q,r\in D\,$ such that $\, a = qb +r\,$ with $\, V(r) < V(b),\,$ where $V$ maps $D$ into  (well-ordered)  $\,\Bbb N.\,$ We show how to construct from $V$ another Euclidean function $v$ that satisfies $\, v(a) \le v(ab)\,$ if $\,ab\neq 0$.
Derive $\,v\,$ from $\,V\,$ as follows
$$\begin{align}
v(0) &= V(0)\\
v(a) &= {\rm min}\{ V(b)\ :\ b\in aD\backslash 0\}
\end{align}$$
Note $\,v(a)\le V(a)\,$ since it is clear if $\,a = 0,\,$ else  it follows by $\, a\in aD\backslash 0$
$v$ is also a Euclidean function: if $\,a,b\in D\,$ and $\,b\neq 0\,$ then $\,v(b) = V(bc)\,$ for $\,0\neq c\in D.\,$  Since $\,V\,$ is a Euclidean function there are $\,q,r\in D\,$ such that $\, a = qbc + r\,$ and $\,V(r) < V(bc) = v(b).\,$ But by above we know $\,v(r)\le V(r)\,$ thus $\,v(r) < v(b),\,$ so $\,v\,$ is a Euclidean function.
Note $\, v(a) \le v(ab)\,$ if $\,ab\neq 0\,$ since $\,aD\backslash 0\supseteq abD\backslash 0$ $\,\Rightarrow\,{\rm min}\,V(aD\backslash 0) \le {\rm min}\,  V(abD\backslash 0)$
Remark $ $ See the paper cited here by Agargun & Fletcher for a comprehensive study of the logical relationships between various common definitions of Euclidean domains and rings.

The second property is superfluous because only the first one is needed to prove that every ideal of a Euclidean domain is principal.


Proof Involving Rational Numbers I asked this same question last night and got some answers but still can't make sense of this, normally I'd move on but since I know how to do everything else for the test I'm going to try to get this down. Anyway .....

Stuck on a tutorial question trying to study for a test. The question is :

Consider the following statement: "Between any two different rational numbers, there are at least two different rational numbers."
(a) Write this statement as a logical expression. The universe is all numbers. Use Q to denote the set of rational numbers.

(b) Prove or disprove this statement.

Thanks, proofs are what I'm having the hardest time with. There are actually other parts to the question but I know how to do those, can someone tell me what they'd consider the full answer?? Our prof gives us little to no examples so I have nothing to go on, plus I learn best from looking at example

For the proof you can just give some explicit construction that works. E.g., note that if $x < y$, then $x < x + \frac{1}{3}(y - x) < x + \frac{2}{3}(y - x) < y$. Then argue that if $x, y \in\mathbb{Q}$, then the two new numbers are rationals, too.

You can just use the fact that rationals numbers are dense.  So given $a$ and $b$, there is a rational number in $(a,b)$ and there is a rational number in both of $(a,x)$ and $(x,b)$, where $x$ is any number in $(a,b)$.


equation of circle tangent to line with radius Find the equation of a circle tangent to line $3x + y - 2 = 0$ at $(-1,5)$ and with radius $\sqrt{10}$. I've no idea on how to do this.

Now as per the question the circle touches the line only at one point so it is a tangent and there can be two circles touching the given line and passing through (-1,5).

so if the given line is a tangent to the circle then the centre lies on a line perpendicular to the given line, passing through (-1,5) and at a distance of $\sqrt(10)$.

slope of line perpendicular to it is $$m1*m2=-1$$ where m1=-3 which implies m2 =1/3.

let the line be y=(1/3)x+c. Now this passes through (-1,5) substitute and get c value

The resulting equation which is perpendicular to the given line is $$3y=x+16---(1)$$

so to find the circle equation we need to find the centre of the circle which can be found with the equation shown which is nothing but the distance formula$$(x-(-1))^2+(y-5)^2=(\sqrt10)^2   ---(2)$$

since the circle has a radius of $\sqrt(10)$ units which implies the centre is at a distance of $\sqrt(10)$ units from it.

solving (1) and (2) we get $$x=-4,2$$ and $$y=4,6$$

therefore the equations of two circles are $$(x-(-4))^2+(y-4)^2=(\sqrt10)^2$$ and $$(x-2)^2+(y-6)^2=(\sqrt10)^2$$

hope this helped.

Let the equation be $$10=(x-a)^2+(y-b)^2$$
$$\implies10=(a+1)^2+(b-5)^2\ \ \ \  (1)$$
Again like equation of circle tangent to line with radius,
$$\dfrac{3a+b-2}{\sqrt{3^2+1^2}}=\pm\sqrt{10}$$
$$3a+b-2=\pm10$$
Considering '+' sign, $$3a+b-2=10\iff b=12-3a\ \ \ \  (2)$$
Use $(1),(2)$ find $a$ and hence $b$
Similarly consider '-' sign


How to solve for $N$ in $\prod\limits_{k=1}^{N}\left(1-\frac{k}{b}\right) = P$ As the title states, I'm trying to solve for $N$ in

$$\displaystyle \prod_{k=1}^{N}\left(1-\frac{k}{b}\right) = P$$

where $0 < P < 1$ and $b > N$ and is an integer.

I'm trying to solve this for extremely large values of $b$ and $N$ ($\sim 10^{50}$) so a brute-force numerical approach won't work in any reasonable time. I also can't make approximations for very small values of $P$. I'd like $P$ to be able to take on any value between $0$ and $1$.

Can this be solved? If not are there certain approximations that can be made?

If $k$ is fixed, and $b$ is large, then the product is approximately equal to $e^{k N/b},$ so that
$$\frac{N}{b} = \frac{P}{k},$$
so $$N = \frac{Pb}{k}.$$

Rewrite your LHS as

$$LHS = \prod_{k=1}^N \frac{b-k}{b} = \frac{(b-1)!}{(b-N-1)! b^N}.$$

Now use Stirling's approximation, to get

$$LHS\approx \frac{\sqrt{2\pi(b-1)} (b-1)^{b-1}/e^{b-1}}{b^N (b-1-N)^{b-N-1}/e^{b-N-1}} = \sqrt{\frac{b-1}{b-N-1}}\frac{(b-1)^{b-1} e^N}{b^N (b-N-1)^{b-N-1}}.$$ Now, you have to decide whether $b$ is a lot or a little bigger than $N.$ The first option is easier.


Possible definitions of exponential function I was wondering how many definitions of exponential functions can we think of. The basic ones could be:

$$e^x:=\sum_{k=0}^{\infty}\frac{x^k}{k!}$$
also
$$e^x:=\lim_{n\to\infty}\bigg(1+\frac{x}{n}\bigg)^n$$
or this one:
Define $e^x:\mathbb{R}\rightarrow\mathbb{R}\\$ as unique function satisfying:
\begin{align}
e^x\geq x+1\\
\forall x,y\in\mathbb{R}:e^{x+y}=e^xe^y
\end{align}
Can anyone come up with something unusual? (Possibly with some explanation or references).

The exponential function is the unique solution of the initial value problem

$y'(x)=y(x) , \quad y(0)=1$.

Here's an "unusual" one: $e$ is the positive real number such that

$$
\sqrt{6\log\left(e\sqrt[4]{e}\sqrt[9]{e}\sqrt[16]{e}\sqrt[25]{e}\ldots\right)} = \pi.
$$


Matrix inequality after taking inverse Let A and B be Positive definite Matrices with $ A\leq B$ in the sense that $B-A$ is positive definite. Is it true that $A^{-1} \geq B^{-1} $?

Consider $B-A\geq 0$ using Schur complement, this is equivalent to
$$\begin{bmatrix}B&I\\I&A^{-1}\end{bmatrix}\geq 0, \quad B>0$$
Since $A^{-1}>0$, now apply the Schur complement one more time to obtain
$$A^{-1}-I(B)^{-1}I=A^{-1}-B^{-1}\geq  0$$
therefore we have $A^{-1}\geq B^{-1}$.

It is true. If you have matrices with this conditions ,then you can start from
$A≤B$.

Now multiply with $A^{-1}$  on the left
$$A^{-1} A ≤ A^{-1} B$$
Now you have the identity Matrix on the left. Multiply with $B^{-1}$ on the right and you will get
 $$B^{-1} ≤ A^{-1}$$