Solve Group Equation Given Two Equations Given the group equations $x^{2}=a^{2}$ and $x^{5}=e$, I attempted to solve for $a$ by substituting $a^{2}x^{3}$ into $x^{5}=e$.

Then I solved for $x$ by noting $a^{2}x^{3} = e \implies (a^{2}x^{3})x^{-3} = (e)x^{-3}\implies a^{2}=x^{-3}$.

Is this correct? If so, can it be broken down further? Why/why not?

If you are trying to find the structure of the group, you might proceed as follows:

$$a^{10}=(x^2)^5=(x^5)^2=e$$

You have $x^2=a^2$ whence $x^4=a^4$ and $x^6=a^6$ but also $x^6=x^5x=x$ so that $x=a^6$, and $x^3=a^8$.

You can check that if $a^{10}=e$ and $x=a^6$ the cyclic group of order $10$ generated by $a$ satisfies the given relations. No larger group can do so, but a smaller group might - for example the trivial group with $a=x=e$. I'll leave you to explore the other cases which may arise.

Note: within the cyclic group of order $10$ the powers of $x$ are all even powers of $a$, so you don't get a nice equation $a=x^n$ with $n$ an integer.

We have, $x^2= a^2$ and $x^5 = e$
Now,

$X^5= e$

or, $x^2. x^2. x= e$

or, $a^2 . a^2 . x =e$ ( substituting from 1st equation)

or, $a^4 . x= e$

Operating both sides by $a^{-4}$ on the left. We get,

$(a^{-4}. a^4). x= a^{-4}. e$

or, $e. x = a^{-4}$
Therefore, $x= a^{-4}$


If $b \equiv 0 \pmod a$ and $c \equiv 0 \pmod b$, then $c \equiv 0 \pmod a$ The question is If $b \equiv 0 \pmod a$ and $c \equiv 0 \pmod b$, then $c \equiv 0 \pmod a$.

My attempt is that $b \equiv 0 \pmod a$ can be written $a\mid b-0 = a\mid b$
and the same with $c \equiv 0 \pmod b$ can be written $b\mid c-0 = b\mid c$
then the $c \equiv 0 \pmod a$ can also be written $a\mid c-0 = a\mid c$.

Now all I have to show is that $a\mid b$ and $b\mid c$ then $a\mid c$ and to show $a\mid c$ I can write
$c= ak$ for some integer $k$.

Now I start my proof
let $a\mid b$ be $b=a\cdot r$ for some integer $r$ (1)
let $b\mid c$ be $c=b\cdot s$ for some integer $s$ (2)

Sub (1) into (2) and get
$c=(ar)\cdot s$
$c=a\cdot(r\cdot s)$
so $c=a\cdot k$
therefore the statement $b \equiv 0 \pmod a$ and $c \equiv 0 \pmod b$, then $c \equiv 0 \pmod a$ is true.

I'm wondering if this is a good/correct proof?
any input would be appreciated.

You did not define what $k$ is, so no, your proof is not correct. You simply state

  So $c=ak$.

What is $k$? How did you prove that $c=ak$?


Is $(-\infty, 0]$ closed in $\mathbb{R}$? Using Rudin's definition that a closed set contains all of its limit points, can I say that the interval  $(-\infty, 0]$ is closed in $\mathbb{R}$ (using the regular metric $d(x,y) = |x-y|$ for $x,y \in \mathbb{R}$)?

Let $A=(-\infty,0]$, so $A^c=(0,\infty)$, which is clearly open. Hence $A$ is closed.

To show that this definition of a closed set is equivalent to your definition:

Here is the proof that $B$ closed if and only if $B^c$ open.

Suppose $B$ is closed (with your definition of closed). Then choose $x\in B^c$. That is, you choose $x\notin B$. Since $x\notin B$, we know that $x$ is not  a limit point. Since it's not a limit point, then we know that there exists some a $\epsilon$-neighborhood for which $N_{\epsilon}(x)\cap B=\emptyset$ where $N_{\epsilon}(x)\subset B^c$. Hence we have that $x$ is an interior point of $B^c$ and consequently $B^c$ is open.

For the reverse direction:

Suppose that $B^c$ is open, so every point in $B^c$ is an interior point. Let $x$ be a limit point of $B$. Then $\forall\epsilon>0$ we have $N_{\epsilon}(x)\cap B\neq \emptyset$, so $x$ is not an interior point of $B^c$. Hence $x\in B$ and so every limit point of $B$ is a point in $B$ and hence $B$ is closed.

Hence $B$ is closed if and only if its complement $B^c$ is open.


Explanation of ln derivative... So we know that for example :

I'd like to know how do we get to this result based on:

I've been trying to do it for myself but I really have no idea how to get to the result starting.

I'll show you an alternative way of evaluating deriving that $\frac{d}{dx}\ln(x)=\frac{1}{x}$.

We know that if $f$ is differentiable in an interval $I$, with $f'(x) < 0$ or $f'(x) > 0$ for all $x \in I$, then $f$ has an inverse funcion $f^{-1}$which is defined and is differentiable on $f(I)$.

So let $f(x) = e^x$. Then $f$ is defined and differentiable in $\mathbb R$ and $\forall x \in \mathbb R,f'(x) > 0$, so there exists an inverse function defined and differentiable in $f(\mathbb R) = \mathbb R^+$, which is $f^{-1}(x) = \ln(x)$
We know that $f(f^{-1}(x)) = x$. If we differentiate from both sides we get:
$$\begin{align}
&(f^{-1})'(x)f'(f^{-1}(x)) = 1 \iff
\\
\\
&(f^{-1})'(x) = \frac{1}{f'(f^{-1}(x))}
\end{align}$$
We know that $f'(x) = e^x$ so $f'(f^{-1}(x)) = x$. So we conclude that
$$\begin{align}
&(f^{-1})'(x)=\frac{1}{x} \iff
\\
\\
&\frac{d}{dx}\ln(x)=\frac{1}{x}
\end{align}$$
Using then the chain rule you can get the derivative you want.
This is a common way of finding out derivatives for inverse functions.


Indented Path Integration The goal is to show that
$$\int_0^\infty \frac{x^{1/3}\log(x)}{x^2 + 1}dx = \frac{\pi^2}{6}$$
and that
$$\int_0^\infty \frac{x^{1/3}}{x^2 + 1}dx = \frac{\pi}{\sqrt{3}}.$$

So, we start with the function
$$f(z) = \frac{z^{1/3}\log(z)}{z^2 + 1}.$$

Let $c_r$ be the upper semicircle with radius r such that $r< 1$ and let $c_R$ be the upper semicircle with radius R such that $R>1$. Let $L_1$ be the interval $(r,R)$ on the reals and let $L_2$ be the interval $(-R,r)$ on the reals.

Then
$$\int_{c_r} f + \int_{c_R} f + \int_{L_1} f + \int_{L_2} f$$
taken counterclockwise is equal to
$$2\pi i\text{ res}(f,i) = -\frac{\pi^2  e^{\frac{\pi i}{6}}}{2}$$

The integrals along $c_r$ and $c_R$ should go to $0$ as $r$ goes to $0$ and $R$ goes to infinity.

So:
$$\int_{L_1} + \int_{L_2}= \int_0^{\infty}\frac{x^{1/3}\log(x)}{x^2 + 1}dx - \int_{0}^{\infty} \frac{x^{1/3}e^{i\pi/3}(ln(x) + i\pi) }{x^2 + 1}dx$$
where x is the polar radius.

Let $A =  \int_0^\infty \frac{x^{1/3}\log(x)}{x^2 + 1}dx$
and let $B = \int_0^\infty \frac{x^{1/3}}{x^2 + 1}dx $.

Then we have $A*(1-e^{\frac{i\pi}{3}}) - \pi i e^{\frac{i\pi}{3}}B =  $-$\frac{\pi^2  e^{\frac{\pi i}{6}}}{2}$.

This doesn't mesh with the given values for $A$ and $B$, so I'm wondering where I went wrong. Thanks in advance.

In this case, I would use a keyhole contour about the positive real axis.  Consider the contour integral

$$\oint_C dz \frac{z^{1/3}}{1+z^2} \log{z}$$

where $C$ is the keyhole contour.  As it is clear that the integral about the circular arcs, large and small, vanish as their respective radii go to infinity and zero, the contour integral is equal to

$$\int_0^{\infty} dx \frac{x^{1/3}}{1+x^2} \log{x} - e^{i 2 \pi/3} \int_0^{\infty} dx \frac{x^{1/3}}{1+x^2} (\log{x}+i 2 \pi) $$

which is equal to

$$\left (1-e^{i 2 \pi/3} \right )\int_0^{\infty} dx \frac{x^{1/3}}{1+x^2} \log{x} - i 2 \pi e^{i 2 \pi/3} \int_0^{\infty} dx \frac{x^{1/3}}{1+x^2}$$

By the residue theorem, the contour integral is $i 2 \pi$ times the sum of the residues at the poles $z=e^{i \pi/2}$ and $z=e^{i 3 \pi/2}$, or

$$i 2 \pi \frac{e^{i \pi/6}}{2 i} i \frac{\pi}{2} + i 2 \pi \frac{e^{i \pi/2}}{-2 i} i \frac{3 \pi}{2} = \frac{5 \pi^2}{4} + i \frac{\sqrt{3} \pi^2}{4}$$

As you did above, let $A$ be the first integral and $B$ be the second.  Then

$$\left (1-e^{i 2 \pi/3} \right ) A - i 2 \pi e^{i 2 \pi/3} B = \frac{5 \pi^2}{4} + i \frac{\sqrt{3} \pi^2}{4}$$

So we equate real and imaginary parts.  Rearranging, we get two equations and two unknowns:

$$\frac{3}{2} A + \sqrt{3} \pi B = \frac{5 \pi^2}{4} $$
$$-\frac{\sqrt{3}}{2} A + \pi B = \frac{\sqrt{3} \pi^2}{4}$$

From this, you may easily verify that

$$A = \int_0^{\infty}dx \frac{x^{1/3}}{1+x^2} \log{x} = \frac{\pi^2}{6}$$

$$B = \int_0^{\infty}dx \frac{x^{1/3}}{1+x^2} = \frac{\pi}{\sqrt{3}}$$

Indented contours pick up half a residue. So the integral along $c_r$ is not zero.


Consistency of a System of linear equations Test the consistency of the system of linear equations

$$\begin{align} 4x-5y+z & =2 \\ 3x+y-2z& = 9 \\ x+4y+z& =5\end{align}$$

We have the system of linear equations given by: $$\begin{align} 4x-5y+z & =2\\ \\ 3x+y-2z& = 9 \\ \\ x+4y+z& =5\end{align}$$

This system can be represented by either of the following augmented coefficient matrices:

$$\begin{pmatrix} 4 & -5 & 1 &\mid 2 \\ 3 & 1 & -2 &\mid  9 \\ 1 & 4 & 1 &\mid 5\end{pmatrix}\to \quad \text{Swap Row 1 with Row 3}\to \quad \begin{pmatrix} 1 & 4 & 1 & \mid 5 \\ 3 & 1 & -2 & \mid 9 \\ 4 & -5 & 1 & \mid 2 \end{pmatrix}$$

On the second matrix, we can continue to perform elementary row operations to row reduce the matrix (i.e. using Gaussian Elimination) in order to check for consistency. Does this sound familiar?

If we row reduce to obtain a row, e.g., of $(0 \; 0 \; 0 \; c)$ where $c$ is non-zero, we have an inconsistent system. Else, the system is consistent. Consistency simplyy means: Can all equations simultaneously be true? If so, then the system is consistent. If not, then it is inconsistent. When a system is inconsistent, no solution can possibly exist. When it is consistent, either a unique solution exists, or infinitely many solutions exist.

If you row reduce correctly, to reduced row echelon form, you should obtain the matrix:

$$\begin{pmatrix} 1 & 0 & 0 &\mid 2\\ 0 & 1 & 0 & \mid 1 \\ 0 & 0 & 1 & \mid -1 \end{pmatrix}$$

Now, what can you say about the system's consistency?

Examine the consistency for following system and solve if consistent.
$$3x + y +2z = 3,$$      $$2x – 3y – z = -3,$$      $$x + 2y + z = 4$$


Why does this series have a different sum when its terms are rearranged? The problem is:

Give an example of a convergent series such that, when the terms are rearranged, the series sums to a different value.

A solution is:

Although everything makes sense in this solution, I don't get how what it claims is possible.  How can a series have two different sums?

Is this caused by the fact that subtraction is not commutative or associative?  Hence, in fact, we really have two different series here?

No, that's not the reason.

When we sum a series, the order by which we sum is essential. That is, we first take $a_1$, then $a_1+a_2$, then $a_1+a_2+a_3$ and so on. Then we see which limit we get.

If, instead, we sum it as$$a_1,a_1+a_3,a_1+a_3+a_2,a_1+a_3+a_2+a_5,a_1+a_3+a_2+a_5+a_7,a_1+a_3+a_2+a_5+a_7+a_4,\ldots$$then we may well get a different limit. However, if the series $|a_1|+|a_2|+|a_3|+\cdots$ converges, this cannot happen. That is, the limit will always be the same.

In order to know more, read about Riemann's rearrangement theorem.

I think that main (and only) reasons why sum of series that is not absolutely convergent is not invariant with respect to permutations are that the operation of subtraction is neither commutative nor associative in general.

Because, if the terms are all positive we obtain always the same sum, no matter how the series is rearranged, because of commutativity and associativity of addition over the positive numbers.

So, I tend to believe that you are on the right track when viewing different sums as a consequunces of non-commutativity (and non-associativity) of subtraction.

Because of non-commutativity and non-associativity of subraction it is natural to have different sums when we permute the series, but it can come as a surprise that we can obtain any value we want with different permutations.


Perfect Shuffle Card Permutation Problem I'm given the problem where one can perform perfect shuffles (i.e. you split the deck into halves and then interweave them) on a deck of $52$ cards (both in and out shuffles) and I am supposed to determine whether all $52!$ possible deck orderings are possible through a composition of such shuffles. I know that given only in or out shuffles you cannot do so since they are cyclic and of order $8$ and $52$ but I really have no idea how to even begin to tackle this problem of composing them. Was hoping for any hints or thoughts on as to how I should attempt this problem? Thanks!

EDIT: An out shuffle is when you interweave leaving the top card on the top while an in shuffle is when you interweave by putting the top card of the bottom half on top of the whole deck.

Not an answer, just a suggestion.

You have two permutations, and you want to find out if they generate the entire group of all permutations.

If $a$ is one permutation, and $b$ is the other, then $ab^{-1}=(1\,27)(2\,28)(3\,29)\cdots(26\,52)$. We can see that $a$ and $c=ab^{-1}$ generates the same subgroup as $a$ and $b$.

$c=ab^{-1}$ is probably easier to deal with, since its order is $2$.

Given an initial configuration of a deck of $52$ cards, perfectly shuffling them $52$ times will take you through exactly $52$ of the $52!$ permutations on the deck. In order to hit every one of them, you must perfectly shuffle the deck and then switch the top two cards before perfectly shuffling it again. Of course you'd be doing this for several lifetimes. In the same vein, perfectly out-shuffling $52$ times, and then in-shuffling once before out-shuffling again $52$ times will indeed have the same effect as composing (or switching just the top two cards) the permutations. But again we could never experience this in hundreds of lifetimes. A computer simulation could demonstrate it perhaps. The (n-1) k-cycle has $8$ elements in its one-orbit for $n=52$, whereas the (n+1) k-cycle has $52$. This is how order is determined. There is no multiplication that can be performed on these orders to determine whether they span the full symmetric set. We can see the composition. Go through all $52$ configurations stemming from the initial position, make one alteration, repeat. Time permitting, we will see every member of S sub $52$.


How can I check that a hexagon intersects a circle? If I know the center positions and radius of a circle and hexagon, how can I determine whether are they touches or intersects each other?

Update:
Also the orientation of the hexagon is know:

This is not easy because there are several configurations to consider.  You could have an intersection along the line connecting the centers, but you can have intersections at other points as well.  Think of shortening $D$ in the bottom figure, moving the hexagon nearer the circle.  The corner below the line between the centers will hit the circle before the point on the line between the centers.  You can't rely on the corners, either, because the hexagon could be much larger so the corner misses the circle but the edge hits.

My approach would be to find the side of the hexagon that crosses the line between the centers, then see if that segment intersects the circle.  It is easier if we put the origin at the center of the hexagon with the $x$ axis running through a vertex.  This corresponds to taking $\theta=0$.  You can find the center of the circle in the new coordinate system first by subtracting the center of the hexagon to get the center as $(x_c',y_c')$, then rotating the system by $-\theta$ to get $x_c=x'_c \cos \theta+y'_c\sin \theta, y_c=-x_c \sin \theta+y'_c \cos \theta$

Now find the angle the center of the circle is at using $\arctan2(y_c,x_c)$.  If it is in the range $0$ to $\frac \pi 3$ the side of interest will be the one that starts from the $x$ axis and goes up.  Your diagram is like this.  The equation of the line is $y=-\sqrt 3(x-R)$.  The line segment extends from $x=R$ to $x=\frac R2$.  Now you can check whether the two endpoints are inside the circle, plus find the perpendicular from the line segment to the center and find the intersection with the line the side is on.  If that intersection is between the corners, you need to check that point, too.

Hint:

There is an intersection if the center of the circle belongs to the Minkowski sum of the hexagon inside and the disk. This sum is a larger hexagon with rounded corners.

To determine if a point lies inside it, find the tangency points between the circles and the sides. This will allow you to split the plane in twelve sectors, each bounded by a straight edge or an arc.

For a straight boundary, it is an easy matter to check insideness. For a curved boundary, you can check the position with respect to the chord and then with respect to the arc.


Finding Angles Given Two Triangles with Equal Perimeters The two right triangles shown below have equal perimeters

The hypotenuse of
the orange triangle is one leg of the green triangle stacked on top of it. If the smallest angle of the orange triangle is 20 degrees, what are the angles of the green right triangle?

I think I have a solution but it will be difficult to explain. I will use the following variables:

$$
\begin{split} a&=\text{base of orange triangle} \\ b&=\text{height of orange triangle} \\ c&= \text{"left" side of green triangle} \end{split}
$$

Then the hypotenuse of the orange triangle is $\sqrt{a^2+b^2}$ and the hypotenuse of the green triangle is $\sqrt{a^2+b^2+c^2}$. Since the perimeters of both triangles are equal we must have:

$$
\begin{split} &&a+b=c+\sqrt{a^2+b^2+c^2} \\ &\implies& a+b-c=\sqrt{a^2+b^2+c^2} \\ &\implies& (a+b-c)^2=a^2+b^2+c^2 \\ &\implies& a^2+b^2+c^2+2ab-2ac-2bc=a^2+b^2+c^2 \\ &\implies& ab=ac+bc \\ &\implies& c=\frac{ab}{a+b} \end{split}
$$

So now we have all sides in terms of $a$ and $b$. But we can relate $a$ and $b$ using the angle we are given. So:

$$
\tan(20)=\frac{b}{a} \implies b=a\tan(20)
$$

Using this we get:

$$
\begin{split} a^2+b^2=a^2+a^2\tan^2(20)=a^2(1+\tan^2(20))=\frac{a^2}{\sin^2(20)}  \end{split}
$$

Now we can get an expression for $a^2+b^2+c^2$ purely in terms of $a$:

$$
\begin{split} a^2+b^2+c^2&= \frac{a^2}{\sin^2(20)}+\frac{a^2b^2}{\frac{a^2}{\sin^2(20)}+2ab}\\ &=\frac{a^2}{\sin^2(20)}+\frac{a^4\tan^2(20)}{\frac{a^2}{\sin^2(20)}+2a^2\tan(20)} \\ &=\frac{a^2}{\sin^2(20)}+\frac{a^4\tan^2(20)\sin^2(20)}{a^2+2a^2\tan(20)\sin^2(20)} \\ &=\frac{a^2}{\sin^2(20)}+\frac{a^2\tan^2(20)\sin^2(20)}{1+2\tan(20)\sin^2(20)} \\ &=a^2\left[ \frac{1}{\sin^2(20)}+\frac{\tan^2(20)\sin^2(20)}{1+2\tan(20)\sin^2(20)} \right] \\ &= a^2K^2 \end{split}
$$

where $K^2$ is just the number in the square brackets (the only reason I made it $K^2$ instead of $K$ is because we will need to take square roots). Finally, letting $\theta$ be the smallest angle of the green triangle, we get:

$$
\begin{split} \cos(\theta)&=\frac{\sqrt{a^2+b^2}}{\sqrt{a^2+b^2+c^2}} \\ &=\frac{\frac{a}{\sin(20)}}{aK} \\ &=\frac{1}{K\sin(20)} \end{split}
$$

You can work out that $K\approx 2.9262$ and then plugging this in gives us $\theta\approx 2.34^{\circ}$.

I went through the same process as the previous answer, but discovered a mistake:

$$a^2 + b^2 = a^2 + a^2 \tan^2(20) = a^2 (1 + \tan^2 (20)) = a^2 / \cos^2 (20) \textrm{, not } \sin^2 (20)$$


$x_1,x_2,...,x_n$ are positive real numbers, $\sum_{i=1}^n x_i^2 = 1$, to find the minimum value of: $$\sum_{i=1}^n \frac{x_i^5}{s - x_i}$$

with $$s = \sum_{i=1}^{n}x_i$$

I used Cauchy Schwarz inequality:

$$(\sum_{i=1}^n \frac{x_i^5}{s - x_i})(\sum_{i=1}^{n}\frac{s-x_i}{x_i}) \geq 1$$

$$(\sum_{i=1}^n \frac{x_i^5}{s - x_i})\geq \frac1{\sum_{i=1}^{n}\frac{s-x_i}{x_i} }$$

the denominator is $$s \sum_{i=1}^n \frac1{x_i} - n$$

i don't know a good way to limit it above

Hint:  Using CS and Power Mean inequalities
$$LHS \cdot \sum x_i(s-x_i)\ge \left(\sum x_i^3\right)^2\ge \frac1n$$
Further $\sum x_i(s-x_i)=s^2-1\le n-1$, so LHS $\ge \dfrac1{n(n-1)}$, with equality when all the $x_i=\frac1{\sqrt n}$.

Ok, let's use the LM

$$F(x_1,...,x_n,\lambda) = (x_1 + ... + x_n)(\frac{1}{x_1} + ... + \frac{1}{x_n}) - \lambda(x_1^2 + ... + x_n^2 - 1)$$

we must have $$\nabla F = \vec 0$$

$$\frac{\partial F}{\partial x_i} = (\frac{1}{x_1} + ... + \frac{1}{x_n}) + (x_1 + ... + x_n)\frac{-1}{x_i^2} - 2\lambda x_i = 0$$

I am convinced that this is not a good aproach


Satisfying explanation of Aristotle's Wheel Paradox. The paradox:

We have a circle and there is another circle with smaller radius. They are co-centeric.

If circle make full turn without sliding, both smaller and bigger circle make full turn too. If we assume that the passed road is equal to the circumference of circles. We have got smaller circle's radius is equal to bigger one's.

Unsatisfying Solutions I found:

"Do not assume that smaller circle's circumference is equal to passed road since the surface that contacts to the ground is bigger one. " // Okey but it does not explain the paradox, it explains just what is the wrong assumption (even does not explain why it is a wrong assumption.)
It is undeniable that every point on both smaller and bigger circle will contact exactly one and only one point on their path. Therefore we can think that this is a bijective maps and smaller circle is isomorphic to bigger one. (Okey but ....)

Question: What is the true answer? What is wrong with the definition of circumference of a circle and relationship with its taken path.

The velocity of any point $P$ on a wheel can be written as the sum of two velocities: the velocity $\vec V$ of the center $O$ and the velocity $\vec\omega\times\vec{OP}$ of rotation about the center, where $\vec\omega$ is angular velocity (perpendicular to the plane of the wheel).

A wheel turns without sliding with respect to a given path if the velocity of the contact point between wheel and path vanishes. Let then $C$ and $C'$ be the contact points of the two wheels. We have
$$
\vec v_C=\vec V+\vec\omega\times\vec{OC}
\quad\text{and}\quad
\vec v_{C'}=\vec V+\vec\omega\times\vec{OC'}
$$
If $\vec v_C=0$ then $\vec V=-\vec\omega\times\vec{OC}$ and
$$
\vec v_{C'}=-\vec\omega\times\vec{OC}+\vec\omega\times\vec{OC'}
=\vec\omega\times(\vec{OC'}-\vec{OC})=\vec\omega\times(\vec{CC'}).
$$
This cannot vanish, unless $C=C'$. So the assumption that both circles turn without sliding is false.

Its really quite simple and doesnt even need math to show..
Take any given point on the outside  larger circle as it moves along..an do the same for the smaller one...neither one travels in a straight line from point A to point B....and they dont travel the same distance. Its seems like they do but they dont..Thats they key..
.
For proof..stick a pencil  in the outermost part of the larger circle and one in the smaller one..rotate the circle and notice the paths traced out..
They will not be straight lines but curved lines..and the curved line not the same length for both..Bigger the circle, the longer the path traced out from A to B.


In what sense of "structure" do group homomorphisms "preserve structure"? It is commonly said that group homomorphisms "preserve the structure of the group", e.g., from Wikipedia:

  The purpose of defining a group homomorphism as it is, is to create functions that preserve the algebraic structure.

Now, my default notion of structure is the one that holds that isomorphisms are structural identities---two isomorphic mathematical objects have the same structure. But that notion of structure requires preserving not just the operations on the elements of the domain but also requires the two sets of objects be equipollent.

But homomorphisms aren't, in general, reversible---or else they'd be isomorphisms. Since homomorphisms don't preserve size, in what sense does it preserve "structure"? Is there any more to the notion of structure here than the simple definition of algebraic structure as a set endowed with certain operations?

Toy example illustrating the problem: suppose you have a group homomorphism between groups $G$ and $H$ (but no isomorphism). Suppose also that you have an isomorphism between $G$ and some further group $J$. Since there is a homomorphism between $G$ and $H$, the structure in $G$ is preserved in $H$. Since there is an isomorphism between $G$ and $J$ they have the same structure. But presumably if $H$ preserves the structure of $G$ then they have the same structure. That can't be true, though, since they aren't isomorphic. It seems then that either "structure" means two different things when discussing isomorphisms and homomorphisms, or "preserves" doesn't mean that the homomorphic sets have the same structure (and then I have no idea what "preserves" actually means in this context).

I'm clearly confused about something. I'm just not sure what.

TL;DR What is the difference between the notions of "structure" when you say that isomorphic structures have the same structure and when you say that a group homomorphism preserves the group structure?

The "structure"  preserved is how the elements relate to each other under the operations of the algebraic structure. For example, for commutative rings, if $\ a = b^2 - c^2 = (b-c)(b+c)\ $ is a difference of squares then applying a ring hom $\,h\, :\, r\mapsto \bar r\,$ shows that it remains a difference of squares in the image ring, viz. $\ \bar a = \bar b^2-\bar c^2  = (\bar b - \bar c)(\bar b +\bar c),\,$ so this particular ring-theoretic structure is preserved. Similarly preserved are polynomial expressions, i.e. expressions composed of basic ring operations (addition, multiplication) and constants $\,0,1.\,$ For example,  a root of a polynomial remains a root of the image of the polynomial. So we can prove that a polynomial has no roots by showing it has no roots in a simpler homomorphic image, e.g. a ring with size so small that we can easily test all of its elements to see if they are roots, e.g. common parity arguments: $\,f(x) = x(x\!+\!1)+2n\!+\!1\,$ has no integer roots since $\,x(x\!+\!1)\,$ is even, so adding $2n\!+\!1$ yields an odd hence  $\ne 0;\,$ equivalently $\!\bmod 2\!:\ f(0)\equiv 1\equiv f(1),\,$ i.e. $\,f\,$ has no roots mod $\,2,\,$ hence it has no integer roots.  This idea generalizes as follows

Parity Root Test $\ $ A polynomial $\,f(x)\,$ with integer coefficients
has no integer roots when its $\rm\,\color{#0a0}{constant\,\ coefficient}\,$ and $\,\rm\color{#c00}{coefficient\,\ sum}\,$ are both odd.

Proof $\ $ If so then $\  \color{#0a0}{f(0)} \equiv  1\equiv \color{#c00}{f(1)}\,\pmod 2,\ $ i.e. $\:f\:$ has no roots in $\,\Bbb Z/2 = $ integers mod $\,2,\,$ therefore $\,f\,$ has no integer roots. $\ $ QED

In the same way, we can often reduce problems to "smaller", simpler problems in modular images. Because  homs preserve the ambient algebraic structure, as above, we can often deduce information about the original problems from information gleaned in the simpler modular images. Such problem solving by modular reduction is an algebraic way of "dividing and conquering".

Functors are the most general case of structure preserving function. A functor is a function that sends objects to objects and arrows to arrows in such a way that the structure is preserved which means 3 things:
1)an identity arrow is sent into an identity arrow
2)given an arrow between two objects, the image of the arrow is another arrow between the images of the objects
3)operation of arrow is preserved (composition is the general case of an associative operation)

A group is a one object category, so that we really only have to pay attention to 1) and 3) which means that the function preserves unit and operation between elements. The last condition means that the function ¨commutes¨ with the operation in the sense that you get the same result from operating two elements and mapping the result, or if you map the original elements and then operate them in the image structure.

One thing that seems to confuse also is the fact that a structure preserving function does not have to bijective. For example, if the morphism $\phi:G\rightarrow H$ is not surjective this means we are mapping the group $G$ into a subgroup of $H$ (that is why the structure is preserved even if we dont cover all tof the image). When the morphism is not injective this means we dont preserve distinctness (there will be at least one pair of different elements in G mapped into the same element of H).


Proof by Induction, is it correct? For $n \in N$, we define a set of successors of $n$ through

$A_n := \{s(n), s(s(n)), ...\} := \{ \varphi_m(n), \text{ where } m \in N \setminus\{0\} \}$

$\varphi_m : N \to N$ is the addition function, as in m + n

I'm trying to prove through induction that $n \notin A_n$ for every $n \in N$

I started by proving that $0 \notin A_n$ based on Peano Axioms and the definition of addition function: $\varphi_m(0) = m$.

Assuming that $n \notin A_n$ for every $n \in N$, I have to prove that $s(n) \notin A_{s(n)}$

Perhaps it would work by defining $A_{s(n)}$ in terms of $A_{n}$ and using the induction assumption, but I am not sure how to do that. Am I on the right track?

Recall (from Peano's axioms) that the successor function is injective. Let $s^k$ denote $s\circ s \circ \cdots \circ s$ ($k$ times).

If $s(n) \in A_{s(n)}$, then there exists $k \ge 2$ such that $s(n) = s^k(n) = s(s^{k-1}(n))$ hence by injectivity $n = s^{k-1}(n)$, so $n \in A_n$. A contradiction.

Proof by contradiction. Assume $s(n) \in A_{s(n)}$. Then there exists some $m \in \mathbb{N} \setminus \{ 0 \}$ such that $\varphi_m(s(n)) = s(n)$, i.e. $m + s(n) = s(n)$. Given that $m \in \mathbb{N} \setminus \{ 0 \}$, it must be the case that $m = s(k)$ for some $k$, and thus $s(k) + s(n) = s(n)$.

Now, by the Peano Axiom that says:

$\forall x \forall y \ x + s(y) = s(x + y)$

we thus have that

$s(s(k) + n) = s(n)$

And by the Peano Axiom that says:

$\forall x \forall y \ (s(x) = s(y) \rightarrow x = y)$

We thus have that:

$s(k) + n = n$

But that means that there is some $m \in \mathbb{N} \setminus \{ 0 \}$ (namely, $s(k)$) such that $m + n = n$, and thus $n \in A_n$.  But that contradicts the inductive hypothesis that $n \not \in A_n$.  Hence our assumption that $s(n) \in A_{s(n)}$ must be rejected, and thus $s(n) \not \in A_{s(n)}$


How many 3 character combinations can be made using letters AND numbers? I'm wondering how many possible 3 character combinations can be made using the 26 letters of the alphabet, and 0-9. I've seen that with just the alphabet you can create somewhere around 17k different combinations of 3 letters, but I assume adding in 0-9 must increase that number substantially.

Well, there are 36 combinations for each, and they're all independent. 36x36x36 is 46656.

It would have to be :
$999×26= ?$  $(? × 3) =?? (?? × 26) =$ and I believe there would be approximately  $2.023$ million combinations.


How many people have X and Y? I have a problem for which I know the answer, but have no idea why it works.

Of 40 people, 30 have a condition. 15 have condition X and 25 have condition Y. How many people have both X and Y?

I know the answer is 10 (thanks to the answers sheet) but why? When I simulate the question by giving random people X's and Y's, and count the people with both, I get answers ranging from 8 to 13. Is my simulation off? Is my thinking wrong?

Why would it be 10, all the time?

Only theory is 15+25=40, 40-30=10, but this doesn't seem logical. (25-15=10, but that's just coincidence right?)

Use the law $n(X \cup Y)=n(X)+n(Y)-n(X \cap Y),$ where $n(A)$ means the number of elements in set $A$. In your case, you know $n(X \cup Y)$ and also know $n(X),n(Y).$

EDIT: Note that the 40 in "Of 40 people..." has no effect. The same answer of 10 would result if this phrase were replaced by "Of 1000 people." Only the people having a condition can count toward those having both conditions X and Y.

I misunderstood the question earlier. By condition you mean symptom - which means that there could be people with a disease who don't have either symptom X or Y. I am posting a new answer:

Total no. of people = 40 = number of people with only symptom X + Number of people with only symptom Y + number of people with both symptom X and Y + number of people with neither symptom X or Y.

number of people with neither symptom X or Y = total population - people with either symptom = 40 - 30 = 10

Number of people without symptom X = total population - people with symptom X = 40 - 15 = 25

number of people with only symptom Y = people without symptom X - people with neither symptom = 25 - 10 = 15

number of people without symptom Y = total population - people with symptom Y = 40 -25 = 15

number of people with only symptom X = people without symptom Y - people with neither symptom = 15 - 10 = 5

therefore total number of people with only X or only Y symptom  = 15 + 5 = 20

therefore total number of people with both X and Y = Total number of people with either symptom X or Y - total number of people with only X or only Y symptom = 30 - 20 = 10


Use Green’s Theorem to evaluate $\int_C \cdot dr $ Use Green’s Theorem to evaluate

$\int_CF\cdot dr$, if $F(x,y)=<\sqrt{x}+y^3, x^2+\sqrt{y}>$ and $C$ consists of the arc of the curve $y =sin(x)$ from $(0, 0)$ to $(\pi, 0)$ and the line segment from $(\pi, 0)$ to $(0, 0)$.

Trying to get the hang of the coding, so forgive if their are multiple edits. There is a photo tagged to the title if need be. Thanks

You should provide some of your attempts. Otherwise it is unclear where your confusion is. I assume that it is converting the line integral into a integral over a region. If your vector field is  $F = (P,Q)$, then Green's Theorem says that
$$
\int_C F\cdot dr = \iint_D \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = \iint_D 2x - 3y^2 dA.
$$


Is there any way to solve integral of $\sqrt{8-x^{2}}$ without using $\sin$ or $\cos$ formulas? I was thinking about the following integral if I could solve it without using trigonometric formulas. If there is no other way to solve it, could you please explain me why do we replace $x$ with $2\sqrt 2 \sin(t)$? I'm really confused about these types of integrals.

$$\int \sqrt{8 - x^2} dx$$

By rescaling the variable, let us replace the constant $8$ by $1$, for convenience.

The equation $y=\sqrt{1-x^2}$ represents the upper-half of the unit circle, and the integral

$$\int_{t=0}^x\sqrt{1-t^2}dt$$ is the area of a vertical "slice" between the abscissas $0$ and $x$. You can compute it as the area of a sector of aperture $\theta$ such that $\sin(\theta)=x$, plus a triangle of base $x$ and height $\sqrt{1-x^2}$.

Hence,

$$A=\frac12\theta+\frac12x\sqrt{1-x^2}=\frac12\arcsin(x)+\frac12x\sqrt{1-x^2}.$$

This is how a trigonometric function appears, and you can't avoid it because it belongs to the final solution.

You also see the connection by taking the derivative

$$(\arcsin(x))'=\frac1{\sqrt{1-x^2}}.$$
The trigonometric function disappears and is replaced by a rational expression.

A similar phenomenon occurs with the logarithm,

$$(\ln(x))'=\frac1x,$$

and this is why you will see logarithms appear now and then in antiderivatives.


$F[x]/(x)^n\simeq F[y]/(ay+b)^n$? Let $F$ be a field. Then $F[x]/(x)^n\simeq F[y]/(ay+b)^n$ for $a,b\in F\setminus\{0\}$

This is a question related to this question. Is the statement true? I can't apply the method mentioned in the comment there. Could you give any hints?

This kind of questions is sometimes confusing for beginners, as they use the letter $x$ on both sides.
It would be much clearer if the question is stated like this:

Show that $F[x]/(x)^n \simeq F[y]/(ay + b)^n$.

It is then easy to construct homomorphisms in both directions.
$\rightarrow$: send $x$ to $ay + b$.
$\leftarrow$: send $y$ to $a^{-1}(x - b)$.
And then easy to check that they are well-defined and are inverses of each other.


$p_1^2 + p_2^2 = q_1^2 + q_2^2 = r_1^2 + r_2^2 = 2$, $p_1^2 + q_1^2 + r_1^2 = 3$, positive distinct rational solution I would like to know if equations

$p_1^2 + p_2^2 = 2$

$q_1^2 + q_2^2 = 2$

$r_1^2 + r_2^2 = 2$

$p_1^2 + q_1^2 + r_1^2 = 3$

have a solution where $p_1,p_2,q_1,q_2,r_1,r_2$ are pairwise distinct and positive rational numbers.

I will appreciate any help.

Yes, the possible solutions in the rationals are $x,y = \pm1, \pm1$, so

$$p = [1,1]$$
$$q = [-1,1]$$
$$r = [-1,-1]$$

is a solution

Edit:

Based on comments.

We seek a solution $x,y$ such that $x^2 +y^2=2$ where $x,y \in \mathbb{Q}$. To fulfill this (and without loss of generality within the specified domain) we require that $0 \le x \le 1$ and $1\le y\le 2$. Let's then define two new variables $a,b$ and require $0 \le a,b \le 1$ and reexpress our equation as

$$(1+a)^2 + (1-b)^2 = 2$$

Expanding this equation

$$ a^2 + 2 a + 1 + b^2 - 2b + 1 = 2$$

Simplifying,

$$ b^2 - 2b + a^2 + 2 a= 0$$

Therefore,

$$ a^2 + b^2 = 2 ( b - a )$$

so $b \ge a$

If $b = a + \epsilon$, for some non-negative $\epsilon$

$$ a^2 + (a+\epsilon)^2 = 2 ( a+\epsilon - a )$$

$$ 2 a^2 + 2a\epsilon + \epsilon^2 = 2 \epsilon $$

Solving for $a$

$$a = \frac{-2 \epsilon \pm \sqrt{(2 \epsilon)^2 - 4 (2) (\epsilon^2 - 2 \epsilon ) }}{2 \cdot 2}$$

$$a = \frac{-\epsilon \pm \sqrt{4 \epsilon - \epsilon^2}}{2}$$

so for $a$ to be real, $0 \le \epsilon \le 4$. And for $a$ to be rational, there must be some coprime integers $m,n$ such that

$$4 \epsilon - \epsilon^2 = \left( \frac{m}{n} \right)^2$$

Or equivalently

$$ n^2( 4 \epsilon - \epsilon^2 ) = m^2$$

The only way for this to be the case is if $4 \epsilon - \epsilon^2$ is a square number, which within the domain limits on $\epsilon$ means $\epsilon=0,4$

This corresponds to $a=0$ and $b=0$ or $b=2$, thus the solutions are $x,y= 1,\pm1$ and cannot be distinct and positive.

Note:

Because of the definition of $a$, we lose the $[-1,-1]$ solution


Geometric construction showing independency of constant By means of a geometric construction show that

$$\frac{1-\cos t + k\sin t}{\sin t+ k \ (1+\cos t)} $$

does not depend on $k$

To be honest, I don't think this question is of very much meaning.

Note that $\tan(\frac{x}{2})=\frac{1-\cos t}{\sin t}=\frac{k \sin t}{k(1+\cos t)}=\frac{1-\cos t+k \sin t}{\sin t+k(1-\cos t)}$.

So it suffices to make a geometric construction to show $tan\left(\frac{x}{2}\right)=\frac{1-\cos t}{\sin t}$.

The following image shows the construction.

$$\frac{1-\cos x + k\sin x}{\sin x + k(1+\cos x)}=\\
\frac{2\sin^2(\frac x2) + k2\sin (\frac x2)\cos (\frac x2)}{2\sin (\frac x2)\cos (\frac x2) + 2k\cos^2(\frac x2)}=\\
\frac22.\frac{\sin (\frac x2)}{\cos (\frac x2)}.\frac{\sin(\frac x2) + k\cos (\frac x2)}{\sin (\frac x2) + k\cos(\frac x2)}=1.\tan(\frac{x}{2}).1=\\\tan(\frac{x}{2})$$


Find k , if k = $\lim \limits_{n \to \infty} n[(n^3+3n^2+2n+1)^{1/3}+(n^2-2n+3)^{1/2}-2n]$ Find k , if k = $\lim \limits_{n \to \infty} n[(n^3+3n^2+2n+1)^{1/3}+(n^2-2n+3)^{1/2}-2n]$

I converted it to $\lim \limits_{t \to 0}$ and tried using L'Hospital's rule and I got it after differentiating twice. It looks pretty bad. Please give an elegant solution. Thank you.

Start with $$\lim \limits_{n \to \infty} n[(n^3+3n^2+2n+1)^{1/3}+(n^2-2n+3)^{1/2}-2n] = \lim \limits_{n \to \infty} n^2\left(\left(1+\frac 3n+\frac 2{n^2}+\frac 1{n^3}\right)^{1/3}+\left(1- \frac 2n+ \frac 3{n^2}\right)^{1/2}-2\right), $$

then use the expansion $(1+x)^a = 1 + a x + \frac{a(a-1)}2 x^2 + o(x^2)$ to expand the two remaining terms


CG Optimisation Function I need to prove the following statement:

  Given the unique minimiser $\overline{x}$ of the function $f(x) := \frac{1}{2}<Ax,x> + <b,x> + c$ for some $c \in \mathbb{R}, b \in \mathbb{R}^n$ and a symmetric positive definite matrix $A$. Prove that $f(x) = f(\overline{x}) + \frac{1}{2}\vert\vert x-\overline{x} \vert\vert_A^2$ holds for all $x \in \mathbb{R}^n$

So far I got to this point:
$\begin{align*}
                f(\overline{x}) + \frac{1}{2}\vert\vert{x - \overline{x}}\vert\vert_A^2 &= \frac{1}{2}\overline{x}^TA\overline{x} + \overline{x}^Tb + c + \frac{1}{2}((x-\overline{x})^TA(x-\overline{x})) \\
                &= \frac{1}{2}\overline{x}^TA\overline{x} + \overline{x}^Tb + c + \frac{1}{2}(x^TAx - x^TA\overline{x} - \overline{x}^TAx + \overline{x}^TA\overline{x}) \\
                &= \frac{1}{2}\overline{x}^TA\overline{x} + \overline{x}^Tb + c + \frac{1}{2}x^TAx - x^Tb + \frac{1}{2}\overline{x}^TA\overline{x} \\
                &= \overline{x}^TA\overline{x} + \overline{x}^Tb + c + \frac{1}{2}x^TAx - x^Tb \\
                &= \frac{1}{2}x^TAx - x^Tb + c + 2\overline{x}^Tb
            \end{align*}$

How can I get $<b,x>$ from $2\overline{x}^Tb - x^Tb$?

There is probably an error in your third equality, because you cannot get from $2\overline{x}^Tb - x^Tb$ to $<b,x>$ (since $\overline{x}$ depends on $A$). It goes well until the expression right before that:

$$\overline{x}^TA\overline{x} + \overline{x}^Tb + c + \frac{1}{2}x^TAx - x^TA\overline{x},$$

but I do not see what the next step was that you took. If you plug in the expression for $\overline{x}$ (you can explicitly write the minimizer in terms of $A$ and $b$), you should arrive at the solution.

For the function $f(x) = \frac{1}{2}< Ax,x > + < b,x > + c$ the equality $f(x) = f(\overline{x}) + \frac{1}{2}\vert\vert x - \overline{x}\vert\vert_A^2$ can not hold. It holds only for the function $f(x) = \frac{1}{2} < Ax,x > - < b,x > + c$.


Problem with two improper integral problems. In the first one I am supposed to just tell if the improper integral
$$\int\limits_1^\infty \frac{\cos^2(x)}{x^2}\,dx$$
is convergent or divergent. I am not supposed to calculate the integral.

The best I can do is show that $$\int\limits_1^\infty \frac{1}{x^2}\,dx$$ is convergent and since that $$\int\limits_1^\infty \frac{\cos^2(x)}{x^2}\,dx\lt\int\limits_1^\infty \frac{1}{x^2}\,dx$$ and is there for also convergent. But I have a feeling that I am completely wrong on this one.

The other problem is: Is it possible that $$\int\limits_{-5}^{10} \cos(\tanh(x^2))\,dx =10\pi.$$
The teacher then gave a hint that don't even try to calculate the integral. On this problem I have no idea were to start.
Any help is greatly appreciated.

For the first one, you are right. But you should elaborate your proof.

Remember, there is a theorem that tells you that if $0<f(x)<g(x)$ for all $x>a$, then if the integral $$\int_a^\infty g(x)$$ converges, the integral $$\int_a^\infty f(x)$$ also converges.

For the second, a hint:

$\cos$ is bounded by $1$, and the length of the integral is $15$.

$10\cdot \pi$ is greater than $30$, on the other hand.

Let $f$ defined by

$f(x)=\frac{cos^2(x)}{x^2}$.

$f$ is continuous at $ [1,+\infty)$

and

$(\forall x\geq 1) \;   0\leq f(x) \leq\frac{1}{x^2}$

but

$\int_1^{+\infty} \frac{1}{x^2}$

is convergent , so

$\int_1^{+\infty}f(x)dx$

is also convergent.


Finding out the number of solutions If $a,b,c$ are three distinct real numbers for which there exists $x,y,z$ such that $$x+y+z=1$$$$ax+by+cz=t$$$$a^2x+b^2y+c^2z=t^2$$then $a^3x+b^3y+c^3z-t^3=P(t)$,be a polynomial in $t$,then the question is to find out the number of solutions to P(t)=0 .

My attempt at the solution

We are given three equations.From there we can solve for $x,y,z$ using Cramer's rule.by solving I got $$x=\frac{(t-b)(b-c)(c-t)}{(a-b)(b-c)(c-a)}$$  $$y=\frac{(a-t)(c-a)(t-c)}{(a-b)(b-c)(c-a)}$$    $$z=\frac{(a-b)(b-t)(t-a)}{(a-b)(b-c)(c-a)}$$  however plugging it in the equation I got $$P(t)=a^3( t-b)(b-c)(c-t)+b^3 (a-t)(c-a)(t-c)+c^3 (a-b)(b-t)(t-a)-t^3(a-b)(b-c)(c-a)=0$$ from here I couldn't proceed.please help me in this regard.thanks.

you are given $a^{3}x+b^{3}y+c^{3}z−t^{3}=P(t)$.Equate $P(t)=0$.Then multiply $(a+b+c)$ on both sides and then simplify to get
$t^{3}-(a+b+c)t^{2}+(ab+bc+ca)t-abc=0 $ which shows that the roots are $a,b,c$.

Now just simply observe that $P(a)=0$ and $P(b)=0$ and $P(c)=0$

Well just plug in $x,y,z$ respectively...

So when you calculate $P(a)$, you will find that $y,z$ simply vanish and $x=1$.
Hence it is a root...


Conditional probability of IIDs $\{X_i\}$ are i.i.d. standard Normal variables, with mean 0 and standard deviation 1, $$S={1\over{n}}\sum_{i=1}^nX_i$$
What is the conditional probability $P(X_1|S\ge1)$?

Call $\varphi$ the standard normal PDF and $\Phi$ its CDF. For each $n\geqslant2$, the density $g_n$ of the desired conditional distribution is such that
  $$
g_n(x)=\frac{\varphi(x)\,\Phi\left(\frac{x-n}{\sqrt{n-1}}\right)}{\Phi(-\sqrt{n})}.
$$

To see this, fix a bounded measurable function $u$ and compute
$$
E(u(X_1); S\geqslant 1)=E(u(X_1);X_1+\sqrt{n-1}Y\geqslant n),
$$
where $Y=\frac1{\sqrt{n-1}}(X_2+\cdots+X_n)$ is standard normal and independent of $X_1$. Thus,
$$
E(u(X_1); S\geqslant 1)=\iint u(x)\mathbf 1_{x+\sqrt{n-1}y\geqslant n}\varphi(x)\varphi(y)\mathrm dx\mathrm dy.
$$
Note that, for every fixed $x$,
$$
\int\mathbf 1_{x+\sqrt{n-1}y\geqslant n}\varphi(y)\mathrm dy=1-\Phi\left(\frac{n-x}{\sqrt{n-1}}\right)=\Phi\left(\frac{x-n}{\sqrt{n-1}}\right),
$$
hence
$$
E(u(X_1)\mid S\geqslant 1)=\frac1{P(S\geqslant1)}\int u(x)\varphi(x)\Phi\left(\frac{x-n}{\sqrt{n-1}}\right)\mathrm dx.
$$
Finally, $S$ is centered normal with variance $n$ hence $P(S\geqslant1)=1-\Phi(\sqrt{n})=\Phi(-\sqrt{n})$, and the expression of $E(u(X_1)\mid S\geqslant 1)$ computed above, holds for every bounded measurable function $u$, hence the density $g_n$ follows.

As $N$ goes to infinity, this conditional probability does not make sense, because the probability of the event $\hat{S} \geq 1$ goes to 0 by the LLN.


How to find antilog with simple calculator? I know how to find log to base $10$ using simple calculator:

say if you want to find log of $12$ you can do as blow:

Step 1: $13$ times $\sqrt{\star} \implies 1.00030338$;
Step 2: subtract 1: $1.00030338 - 1 = 0.00030338$;
Step 3: Multiply by $3558 = 1.07942$.

Can I find $\operatorname{antilog}$ too?

By simple calculator i mean this:

Using your calculator's square root key, you can approximate any antilog as closely as you like.  Say you want $10^{1.234}$.  You start by writing $$1.234 \approx  1 + \frac18 + \frac1{16} + \frac1{32} + \frac1{128}$$ which you can find by any of several straightforward methods.  (If this isn't clear, leave a comment and I will explain it further.)

Then you can calculate $$\begin{align}10^{1/8} &= \sqrt{\sqrt{\sqrt{10}}}\\
10^{1/16} &= \sqrt{10^{1/8}}\\
10^{1/32} &= \sqrt{10^{1/16}}\\
10^{1/128} &= \sqrt{\sqrt{10^{1/32}}} \\
\end{align}$$

and so on.  Then $$10^{1.234} \approx 10\cdot 10^{1/8}\cdot 10^{1/16}\cdot 10^{1/32} \cdot 10^{1/128}.$$

You can be a little more clever than this.  $1.234$ is almost, but not quite, $1 + \frac18 + \frac1{16} + \frac1{32} + \frac1{64}$, so you can get a much better approximationby writing $$1.234\approx 1 + \frac18 + \frac1{16} + \frac1{32} + \frac1{64} \color{red}{- \frac1{512}}$$

and then $$10^{1.234}\approx 10\cdot 10^{1/8}\cdot 10^{1/16}\cdot 10^{1/32}\cdot 10^{1/64}\color{red}{\div 10^{1/512}}.$$

Just do the reverse of what u did. Like if u want to find antilog of 2.345
Step-1 Take numbers after decimal : 0.345
Step-2 Divide it by 3558: 0.00009696459
Step-3 Add 1: 1.00009696459
Step-4 Square it 13 times: Press * and = 13 times: 2.218 (approx)
Step 5- Now the number before decimal is used as power of 10 and multiplied to the answer: 2.218 *10^2
Final answer= 221.8
(You may also take the whole number -2.345 and use it in the method. But answer may vary!)


Is the conditional density of $X$ given $Y=y$ the density of $X$ with respect to some probability measure $P(\cdot \mid Y=y)$? Let $(\Omega,\Sigma,P)$ be a probability space and let $X:\Omega \to \mathbb R$ and $Y:\Omega \to \mathbb R$ be jointly continuous random variables with density functions $f_X$ and $f_Y$, respectively, and joint density function $f_{X,Y}$. The book Introduction to Probability by Bertsekas and Tsitsiklis defines
$$
f_{X \mid Y}(x \mid y) = \frac{f_{X,Y}(x,y)}{f_Y(y)}.
$$
Question: For a given $y \in \mathbb R$ such that $f_Y(y) > 0$, does there exist a probability measure $Q$ such that the distribution of $X$ with respect to $Q$ has density function $x \mapsto f_{X\mid Y}(x \mid y)$?

In other words, does there exist a probability measure $Q$ such that
$$
Q(X \in S) = \int_S f_{X \mid Y}(x \mid y) \, dx
$$
for all measurable subsets $S$ of $\mathbb R$?

If such a measure $Q$ exists, I might call it $P(\cdot \mid Y = y)$ rather than $Q$. The book seems to sidestep and avoid the thorny issue of how to define $P(\cdot \mid Y = y)$, but it feels to me that $P(\cdot \mid Y = y)$ is a vital concept that is missing from the exposition.

I believe that you have precisely defined the probability measure that you are interested in.

Intuitively, when we have the joint distribution of $X$ and $Y$, we can plot the contours of their joint density in $\mathbb{R}^2$. If we then fix $Y = y$, we get the conditional density $f_{X|Y}(x|y)$, which is precisely the density of $X$ on this "slice" of $\mathbb{R}^2$ where $Y = y$. That is to say, conditioning on $Y$, we are simply converting to a one-dimensional density, in a particular subspace of $\mathbb{R}^2$. Hence, any usual rules that would apply to a univariate density will apply here.


Reducing Spaces: Domain Problem

Given a Hilbert space $\mathcal{H}$.

Consider normal operators:
$$N:\mathcal{D}(N)\subseteq\mathcal{H}\to\mathcal{H}:\quad N^*N=NN^*$$

Denote for readability:
$$\mathcal{D}:=\mathcal{D}(N)=\mathcal{D}(N^*)$$

Regard a closed space:
$$\mathcal{S}\leq\mathcal{H}:\quad\mathcal{H}=\mathcal{S}\oplus\mathcal{S}^\perp$$

Suppose invariance:
$$N(\mathcal{S}\cap\mathcal{D})\subseteq\mathcal{S}\quad N(\mathcal{S}^\perp\cap\mathcal{D})\subseteq\mathcal{S}^\perp$$

  Then it may still happen:
  $$\mathcal{D}\supsetneq\mathcal{D}\cap\mathcal{S}+\mathcal{D}\cap\mathcal{S}^\perp$$

Does someone have a nonexample?

Reference

For a preexample: Preliminary

If $PT = TP$, then for any $x\in Z$
$$
Tx = TP(x) = PT(x) \in Z
$$
and so $T(Z) \subset Z$. Now since
$$
(I-P)T = T-TP = T-PT = T(I-P)
$$
it follows that $T(Z^{\perp}) \subset Z^{\perp}$.

Conversely, if $Z$ is reducing, then for any $x\in H$, write $x = Px + (I-P)x$, then
$$
Tx = P(Tx) + (I-P)Tx = T(Px) + T(I-P)x
$$
and so
$$
P(Tx) - T(Px) = T(I-P)x - (I-P)Tx \in Z\cap Z^{\perp} = \{0\}
$$
and so $TP = PT$.


What are some different approaches to scoring an old Stock price prediction? I want to compare predictions of 2 different Stock Analysts and would like to assign a score to each prediction made by each of them in a way that they can be compared to decide who is the best Analyst.

The factors to take into account would be:

How earlier was the prediction made; a correct prediction of certain Stock price being at $100 is worth more if it was made a year before the target date than if it was made a month before.
How close to the actual price was the prediction; someone predicting $110 would score higher than someone predicting 120.

When we mix both of them, someone predicting 120 a year earlier scores higher than someone predicting 110 a month before.

I understand how I could do this to compare the two analysts on one single stock, by adding different weights to each factor, but what should I do if I'm comparing stocks that cost anywhere from a couple of pennies to $100.000?

I'm looking for anything that can point me in the right direction.

How about this approach: Fit the parameters of a stochastic process to the history of the stock price in question, such as geometric Brownian motion (GBM). GBM is used to model stock prices in the Black-Scholes model and is the most widely used model of stock price behavior. GBM has two parameters $\mu$ and $\sigma$ representing the percentage drift and percentage volatility, which can be easily estimated (through maximum likelihood) from historical data.

Fit a GBM process to the stock price data from a year ago to today. Find the probability density of such a GBM process ending at $120 today. (Notice the probability will not be entirely concentrated at \$100 because this is a stochastic process. Repeating the process will yield different outcomes even if the parameters are unchanged. A stock price with higher volatility will have a correspondingly more "spread out" price distribution at today. Moreover, this spread increases the longer the GBM random walk is run, again reflecting the fact that it's easier to attain an error margin over a shorter time frame).
Fit a GBM process to the stock price data from a month ago to today. Find the probability density of such a GBM process ending at $110 today.

Compare analysts 1 and 2 on the basis of the probabilities that were obtained from the above fitted stochastic processes, weighing them according to the entropy of the GBM-induced present price distribution (which is higher for the more spread out distribution, so its not unfairly penalized). These are the probabilities that they would be "correct" if history were "re-run", assuming such a history can be approximately modelled by the GBM process.

Here's an example of what samples from a GBM process with a given drift and volatility look like:

EDIT: It occurred to me that it makes more sense to do the above but backwards. Have each analyst decide what they think $\mu$ and $\sigma$ are . Their belief about $\mu$ can be obtained from their prediction about what the price will be today, i.e. just pick the $\mu$ which maximizes the probability density of their predicted price:

\begin{align*}
\mu_1 &= \operatorname*{argmax}_{\mu_1 \in \mathbb{R}^+} \mathrm{P}(S_\text{today} = \text{price prediction 1} \mid S_\text{a year ago}, S \sim \text{GBM}(\mu_1, \sigma_1)) \\
\mu_2 &= \operatorname*{argmax}_{\mu_2 \in \mathbb{R}^+} \mathrm{P}(S_\text{today} = \text{price prediction 2} \mid S_\text{a month ago}, S \sim \text{GBM}(\mu_2, \sigma_2))
\end{align*}

Alternatively it could represent what they think is the expected price:
\begin{align*}
\mu_1 &: \mathrm{E}(S_\text{today} \mid S_\text{a year ago}, S \sim \text{GBM}(\mu_1, \sigma_1)) = \text{price prediction 1} \\
\mu_2 &: \mathrm{E}(S_\text{today} \mid S_\text{a month ago}, S \sim \text{GBM}(\mu_2, \sigma_2)) = \text{price prediction 2}
\end{align*}

Then find the probability that the price would be what it actually is today under each analysts' prediction, i.e.

\begin{align*}
L_1 &= \mathrm{P}(S_\text{today} = \$100 \mid S_\text{a year ago}, S \sim \text{GBM}(\mu_1, \sigma_1)) \\
L_2 &= \mathrm{P}(S_\text{today} = \$100 \mid S_\text{a month ago}, S \sim \text{GBM}(\mu_2, \sigma_2))
\end{align*}

$L_i$ is the likelihood of analyst $i$'s beliefs given the actual outcome, or equivalently, the probability of the actual outcome given each analyst $i$'s beliefs. Use $L_i$ as the score for analyst $i$.

Implicitly what you're asking each analyst to do is to give you the parameters for a distribution reflecting what they think today's stock price will be, and you're rewarding them with the likelihood of the parameters they gave you.

As some of the other answers/comments have pointed out, there are plenty of ways you could mix the two criteria to satisfy both earlier predictions being better and more accurate predictions being better.

If $t$ represents the length of time between the prediction and the event (how early it is) and $d$ represents the distance between the prediction and the actual value obtained, then we want to maximize $t$ and minimize $d$, and any score $s(t,d)$ with $\frac{\partial s}{\partial t}>0$ and $\frac{\partial s}{\partial d}<0$ satisfies the qualitative properties you want. For the record, there are a LOT of functions satisfying those properties.

In practice, you probably want a score that carries more meaning. Scores like profit (or profit less opportunity cost) are impossible because of your requirement that some more profitable predictions receive lower scores (e.g. \$120 instead of \$110 for a \$100 security). Since your focus seems to be on evaluating the accuracy of the analyst, you might try comparing their predictions to the best possible predictions.

For each prediction the analysts made, you know the lead time and the distance from the true price (which should probably be normalized to allow a comparison across securities) and consider only the lower-right portion of that hull. See below on simple simulated data.

Each prediction is represented by one of those points, and the best predictions are those with large times and low distances represented by the black line. Your score for any given prediction is simply how far it is from the black line.

This doesn't quite agree with your 1st constraint, but it arguably better captures the information you care about. Since you're penalizing some predictions that would yield high profits, business impact is not the dominating factor. You explicitly mentioned that you're trying to rank analysts. If that ranking is only internal, comparing their performance to the best possible performance they could hope to have is a logical, interpretable metric.

In most industries, boiling people down to a single aggregate metric causes problems. In software it causes people to write meaningless comments to have lots of lines of code, to split small problems into tiny ones to have more "commits" to make it seem like more work is being done, and so on. Combining multivariate problems into single variable problems causes information to be lost; some good performers will be rated as bad and vice versa. That isn't even counting the HR nightmare and morale drop that accompany draconian, numbers-driven policies. Knowing their performance is valuable, but be very careful how that information is used and communicated.


How to prove the that no formula can be represented in the form (F . G), "No formula can be represented in the form (F . G), where F and G are
formulas and  is a binary connective, in more than one way.

By representing a formula in the form ¬F or (F . G) we start “parsing” it. The
assertion of the previous problem shows that a formula can be parsed in only one way.
From now on, we will abbreviate formulas of the form (F . G) by dropping the
outermost parentheses in them. We will also agree that ≡ has a lower binding power
than the other binary connectives. For instance,

p ∨ q ≡ p ⊃ r

will be viewed as shorthand for

((p ∨ q) ≡ (p ⊃ r)).

Finally, for any formulas F1, F2, . . . , Fn (n > 2),

F1 ∧ F2 ∧ · · · ∧ Fn

will stand for

(· · ·(F1 ∧ F2) ∧ · · · ∧ Fn).

The abbreviation F1 ∨ F2 ∨ · · · ∨ Fn will be understood in a similar way."

I understood it ,but i could not prove it with form of induction.

I know that,
In another useful form of induction, we check that all atoms and 0-place connectives
have property P, and that the property is preserved when a new formula is formed
using a unary or binary connective. More precisely, we show that

• every atom has property P,

• both 0-place connectives have property P,

• if a formula F has property P then so does ¬F,

• for any binary connective , if formulas F and G have property P then so does

(F . G).

Then we can conclude that property P holds for all formulas.

Do F and G, if they are formulas and not abbreviations, both have the property that they can get represented in only one way?

If you can't tell, what about all of the subformulas {F$_1$ ... F$_n$} of F and {G$_1$ ... G$_n$} of G?  Can they get represented in only one way?

If you can't tell that, can you tell that all of the subformulas of F$_1$, all the subformulas of F$_2$ ... all the subformulas of F$_n$ and similarly the subformulas of G$_1$ ... those of G$_n$ can only get represented in one way?

Do you eventually end up with letters if you repeat the above?  And if you do, then can you build up F and G via the formation rules?

Remember, the only objects which qualify as formulas consist of strings which satisfies the formation rules (for that language).  Nothing else is a formula.


How to prove $14^{11}\pmod {23}$ is same as $22 \pmod {23}$ How do we simplify the exponent in:

$$14^{11}\pmod {23} \rightarrow 22 \pmod {23} ?$$

We have that

$$14^{2}\equiv 196 \equiv 12 \mod 23$$

$$14^{4}\equiv 144 \equiv 6 \mod 23$$

and therefore

$$14^{10}\equiv 12\cdot 6^2 \equiv 432 \equiv -5 \mod 23$$

hence

$$14^{11}\equiv -70 \equiv -1 \mod 23$$


Solution to underdetermined linear equations I have a set of numbers $x_i$ and I know sums of certain subsets $y_i=\sum x_{\sigma_k}$. All $x_i>0$ and I'm looking for a simple solution.

With some internet research I found that this might be related to problems in signal processing. So basically I have given a vector $\mathbf{y}$ and a matrix $\mathbf{A}$ with $y_i>0$ and $A_{ij}\in\{0,1\}$. I'm looking for a solution to the vector $\mathbf{x}$ ($x_i\geq 0$) with

$\mathbf{A}\mathbf{x}=\mathbf{y}$

where this linear equation is underdetermined.

Apparently to complete this problem several norms to minimize on $\mathbf{x}$ are possible. For my particular task it's not clear whether I need L0, L1 or L2 norm, so any solution will do - as long as it's simple. Approximate solution like iterative approaches are also fine.

Can you suggest a way to solve this problem?
I'm looking for a reference to an algorithm which I can understand as a non-mathematician. Even better would be an open source implementation that I can download. And it would be perfect if it were a Python solution.

You can solve the system in a least-squares sense:

$$\mathbf{Ax}=\mathbf{y}$$
$$\mathbf{A^{T}Ax}=\mathbf{A^{T}y}$$
$$\mathbf{Jx}=\mathbf{r}$$
$$\mathbf{x}=\mathbf{J^{-1}r}$$

where $\mathbf{J=A^{T}A}$ and $\mathbf{r=A^{T}y}$.

Note that $\mathbf{J^{-1}r=(\mathbf{A^{T}A)^{-1}A^{T}y}}$ which is application of left pseudoinverse of $\mathbf{A}$ - this obtain the least-squares solution in $\mathbf{x}$ if $\mathbf{A}$ is overdetermined or have full rank - but this may no be our case.

The $\mathbf{J}$ is $n\times n$ and is possibly rank-deficient (underdetermined solution).

The SVD of $\mathbf{J}$ is then

$$\mathbf{J}=USV^{T}$$

where

$U$ is $n\times n$ orthogonal.

$V$ is $n\times n$ orthogonal.

$S$ is $n\times n$ diagonal, with diagonal elements $\sigma_{1} \geq \sigma_{2} \geq \cdots \geq \sigma_{n} > 0$.

The solution of your linear system is given by

$$\mathbf{x}=\mathbf{J^{-1}r}=\left(USV^{T}\right)^{-1}=VS^{-1}U^{T}\mathbf{y}$$

or more specifically:

$$\mathbf{x}=\sum_{i=1}^{n}\frac{u_{i}^{T}\mathbf{r}}{\sigma_{i}}v_{i}$$

where $u_{i}\in \mathbb{R}^{m}$ and $v_{i}\in \mathbb{R}^{n}$ are i-th columns of $U$ and $V$, respectively.

We can extend the above sum for rank-deficient cases:

$$\mathbf{x}=\sum_{\sigma_{i}\neq 0}\frac{u_{i}^{T}\mathbf{r}}{\sigma_{i}}v_{i}+\sum_{\sigma_{i}=0}\tau_{i}v_{i}$$

where $\tau_{i}$ are arbitrary coefficients (any choice of $\tau_{i}$ satisfies your linear system).

It should be noted that by choosing $\tau_{i}=0$ yield minimum-norm solution, which is usually the most desirable one in undetermined and ill-conditioned systems (where singular values are almost zero).

Source: Nocedal, Wright: "Numerical Optimization, Second Edition", chapter 10.2 Linear Least-Squares Problems, p. 250

If you have MATLAB simply use the "\" or "/" operator will do the trick. It solves the equation Ax = y in a least squared sense if it's under-determined. It determines the solution by solving |Ax-y|^2 = min. If you don't have Matlab you can do it manually by defining your error E = (Ax-y) * (Ax-y)' and then calculating the partial differentials dE/dyi setting them to zero. This give and system of linear equations that matches the number of elements and can therefore be solved using a regular linear equation solver.


Compute marginal probabilities for a given substitution I am trying to compute the marginal probability for this:

$$f_{XY}(x,y) =  \begin{cases}
      2e^{-y} & \text{if  } -y< x < y \ \text{ and }\ 0 < y < \infty ,\\
      0 & \text{otherwise.}   \end{cases} $$

Is my following solution correct?

$$f_X(x) = \int_{y}^{-y}\,2e^{-y}\,\mathrm{d}y = -2e^{-y} + 2e^y$$

and

$$f_Y(y) = \int_{0}^{\infty}\,2e^{-y}\,\mathrm{d}x = \infty$$

Your joint pdf is not correctly normalized:
$$
\int_0^\infty dy\int_{-y}^y dx\ 2 e^{-y}=4
$$
instead of $1$. So, assuming that your jpdf is instead $(1/2)e^{-y}$ over the same $(x,y)$-domain, the marginal pdf of $X$ is obtained by integrating over $y$ (be careful about the integration range, though!)
$$
f_X(x)=\Theta(-x)\int_{-x}^\infty dy (1/2)e^{-y}+\Theta(x)\int_{x}^\infty dy (1/2)e^{-y}=\Theta(-x) e^{x}/2+\Theta(x)e^{-x}/2\ ,
$$
where $\Theta(x)$ is the Heaviside step function. The integration limits are obtained by resolving the constraints $-y<x<y$ together with $y>0$ for $y$. Note that this marginal pdf is correctly normalized over the support $x\in (-\infty,\infty)$.

For the other marginal
$$
f_Y(y)=\Theta(y)\int_{-y}^y dx (1/2)e^{-y}=y e^{-y}\Theta(y)\ ,
$$
correctly normalized over $y\in (0,\infty)$.

Nope your answer is not correct.
The fact you need the pdf of X shoul prompt you to integrate y.which you have done but the limits of integration are incorrect..limits should be from $ 0 to  \infty $

For your second part the limit should be

$ -y to y $.


Non-trigonometric Proof for values of $\sin(\frac{\pi}{6})$ and $\cos(\frac{\pi}{6})$ I'm looking for non-trigometric (also, purely real analysis) proofs for the following facts. (For reference, I'm working with the series definitions for sine and cosine.)

$\sin(\frac{\pi}{6})= \cos(\frac{\pi}{3})=\frac{1}{2}$.

$\cos(\frac{\pi}{6})= \sin(\frac{\pi}{3})=\frac{\sqrt{3}}{2}$.

I've proven the values of $\sin$ and $\cos$ at $0, \pi, \frac{\pi}{2}, 2\pi $and $
\frac{\pi}{4}$ as well as the the standard summation / double-angle formulas. But I'm still having trouble. My only strategy so far has been to write something like the following and perhaps expand out with the summation formulas.

$\cos(2(\frac{\pi}{3})+\frac{\pi}{3})= \cos(\pi)=-1$ and $\sin(2(\frac{\pi}{3})+\frac{\pi}{3})=\sin(\pi)=0$.

Since I obviously don't know the value of these functions at $\frac{\pi}{3}$, I'm not sure if this will get me anywhere.
Could someone explain to me how to solve this problem?

Hint:

from $\cos(2(\frac{\pi}{3})+\frac{\pi}{3})= \cos(\pi)=-1$, using summation and double-angle formulas we have:
$$
\left(2\cos^2(\pi/3)-1 \right)\cos(\pi/3)-2\left(1-\cos^2(\pi/3)\right)\cos(\pi/3)+1=0
$$
that for $\cos(\pi/3)=y$ becomes:
$$
4y^3-3y+1=(y+1)(2y-1)^2=0
$$

In $\triangle ABC$, $AB=BD=DA$, $AC=CD$

$$\sin\frac{\pi}6=\frac{BC}{AB}=\frac{a}{2a}=\frac12$$


Showing that $\sqrt[105]{105}>\sqrt[106]{106}$ How can one prove that

$$\sqrt[105]{105}>\sqrt[106]{106}\text{ ?}$$
Induction on the statement $$\sqrt[n]{n}>\sqrt[n+1]{n+1} \text{ for } n \in \mathbb{N}| n>2$$ would yield $\sqrt[3]{3}>\sqrt[4]{4}$ at the base step $n=3$, which we cannot assume.

So, based on the properties of powers and square roots alone, can we prove the first statement?

EDIT: No calculus, no functions, not even logs.

It's more of a riddle that anything else.

One way to do this is to simply note that the function $x^{1/x}$ is decreasing over a range including $105$ and $106$.

To this end, it suffices to show that $\ln[x^{1/x}] = \frac{\ln(x)}{x}$ is decreasing over such a range.

We calculate:
$$
\frac{d}{dx}\left( \frac{\ln(x)}{x} \right) =
\frac{1 - \ln(x)}{x^2}
$$
So, this function is decreasing when $1 - \ln(x) < 0 \implies x > e$.  So, the function is decreasing over the range $(e,\infty)$, which suffices for our purposes.

For that base step, should you want to prove this without calculus:
$$
3^{1/3} > 4^{1/4} \iff\\
3^4 > 4^3 \iff \\
81 > 64 \quad \checkmark
$$

Graphing $y=x^{\frac{1}{x}}$ , Y has a maximum value at x = e.

It's all asymptotic to 1 as x approaches infinity

Updated with graph to show exact area of interest. When properly scaled, it's clear that Y is dropping like a rock as X increases. The good answers here still don't adhere to Op's request for "powers and square roots alone" and his edit of "No calculus, no functions, not even logs."
The highest rated answer? ln x from the start.


To clarify my doubt over my wrong approach in combination question The question is:

  A box contains 5 different red and 6 different white balls. In how many ways can 6 balls be selected so that there are at least two balls of each colour.

I know this an easy question and we can easily do this by taking different cases and adding them like:-  $ { 5 \choose 2 }×{6\choose4}+  { 5 \choose 3}× { 6 \choose 3 }+ { 5 \choose 4 }× { 6 \choose 2 } =425$

But i want to know that why my first approach is wrong that is since we want two red balls and two white balls so we take  $ { 5 \choose 2 }×{6\choose2}$ and now the remaining two balls could be taken from the remaining 7 balls (11-4) so the final answer according to me -
$ { 5 \choose 2 }×{6\choose2}×{7\choose2}$

Please tell what is wrong with my approach I am very confused (the correct answer is 425).

Let's say the balls are numbered. Red balls from $1$ to $5$, and white balls from $1$ to $6$.

In your approach, you first pick two red balls and two white balls, then two balls freely. That means that you count

  Red ball 1 and 2, then white ball 1 and 2, then red ball 3 and white ball 3

as a different choice from

  Red ball 1 and 3, then white ball 2 and 3, then red ball 2 and white ball 1

and there are several other combinations that ought to be counted as the same choice as this one, but your method considers them different. In fact, there are 9 of them in total. If the final choice had 2 whites or 2 reds, that number would instead be 6.

This is why you get an answer that is much too large (somewhere between 6 and 9 times too large).

Let red balls are named as R1,R2,R3...R6 and white are named as W1 , W2, W3......W6.
In selection of 2 Red and 2 white let us say you have first got R1,R2,W1,W2 and later in next two you got R3 , W3 i.e you have now R1,R2,R3,W1,W2,W3 in 1800 ways.
This also includes when in first 4 you withdraw R1,R3,W1,W2 and next two are R2 , W3 now again you can see it is same set of balls but counted twice like wise there are many repetition of cases.


Does l'Hopital rule work for -inf/inf? If you have an indeterminate form:

$\frac{-\infty}\infty$
$\frac\infty{-\infty}$
$\frac{-\infty}{-\infty}$

does l'Hôpital's rule also apply?

Yes, but you should first make sure whether or not you really need L'Hopital's rule! If you apply it to stuff which doesn't require you to, weird things will happen.

Take for example the following limit:

  $\displaystyle\lim_{x \to 0^+}x \ln x = \lim_{x \to 0^+}\dfrac{\ln x}{1/x}$.

  Blindly applying L'Hopital's Rule repeatedly gives:
  $\displaystyle\lim_{x \to 0^+}x \ln x = \lim_{x \to 0^+}\dfrac{\ln x}{1/x} = \lim_{x \to 0^+}\dfrac{1/x}{-1/x^2} = \lim_{x \to 0^+}\dfrac{-1/x^2}{2/x^3} = \lim_{x \to 0^+}\dfrac{2/x^3}{-6/x^4} = \cdots$.

  But it we stop after applying L'Hopital's Rule once and simplify stuff, we get:
  $\displaystyle\lim_{x \to 0^+}x \ln x = \lim_{x \to 0^+}\dfrac{\ln x}{1/x} = \lim_{x \to 0^+}\dfrac{1/x}{-1/x^2} = \lim_{x \to 0^+} -x = 0$.

Yes, it does apply,

Consider taking limits of these, by using l.Hopital

$1$) $\lim_ {x- \to \infty} \dfrac{-x}{x}$

$2$)  $\lim_ {x- \to \infty} \dfrac{x}{-x}$

$3$) $\lim_ {x- \to \infty} \dfrac{-x}{-x}$


moments of gaussian distribution by taylor series expansion I have calculated characteristic function of the normal distribution
$$f_X(k)=\exp\left(ika-\frac{\sigma ^{2}k^{2}}{2}\right)$$ and now I would like to find the moments, so I know that you could expand characteristic function by Taylor series

$$
\begin{split}
f_X(k)
  &= \left(1 + \frac{1}{1!}\left(ika - \frac{\sigma^2k^2}{2}\right)
                 + \frac{1}{2!}\left(ika - \frac{\sigma^2k^2}{2}\right)^2
                 + \frac{1}{3!}\left(ika - \frac{\sigma^2k^2}{2}\right)^3
                 + \ldots \right)\\
  &= \left(1+\frac{(ik)}{1!}\left \langle X^1 \right \rangle+\frac{(ik)^2}{2!}\left \langle X^2 \right \rangle+\frac{(ik)^3}{3!}\left \langle X^3 \right \rangle+...\right)
\end{split}
$$

and the moments will be
$\left \langle X^n \right \rangle$

Now the problem is that I completely forgot how to evaluate Taylor series.
Could you be so kind and help me to calculate for example second moment? I know what the answer should be, but I couldn't get it right.

Any help would be appreciated!

I'm not sure we need to use Taylor series to compute moments given the moment generating function.

Let $X$ be RV with MGF:

$$M_X(t) := E[e^{Xt}]$$

We have

$$M_X^{'}(0) = E[Xe^{X(0)}], M_X^{''}(0) = E[X^2e^{X(0)}]$$

Now if you wanted to get the mgf given the moments $E[X^n]$ do this:

$$E[e^{Xt}] = E[\sum_{n=0}^{\infty} \frac{X^n}{n!}]$$
$$= \sum_{n=0}^{\infty} E[\frac{X^n}{n!}]$$
$$ = \sum_{n=0}^{\infty} \frac{1}{n!} E[X^n]$$


Understand and an algorythm to Maximize number of triangles from a set of points on XY plane Given: Set of points (x, y)
Looking to: Maximize count of triangles that can be formed.
Each triangle which is enclosed within another (with/without shared edge) will be counted again.
Specifics on which triangle will be counted:
  Order of aplying edges matters.
  Direction of edges matters.
  Edges cannot cross aother edges.
  Points already surrounded completely with triangles cannot be connected to any point anymore, but may be connected from another point that is not completely surrounded by triangles.
  From each point, a maximum of 8 edges can originate.

Say, triangle ABC with another point D inside of it.
  Connecting each point to all others creates 3 inner triangles + 1 outer.  If the count is 3 or 4 is explained as follows:
  If outer triangle was formed first, D cannot be connected to A, B or C, howeve, D can be connected from A, B, and C.
  Best solution: triangle ABD, then ABC, then edge CD (not DC) crates 2 more, CDA and CDB = count of 4 triangles.
  If CDA and CDB are formed first, then edge AB or BA is applied, only one more triangle is counted, namely: ABC.  ABD is NOT counted as it is SMALLER than ABC.

Another example: Say, points L(eft) and R(ight) are aligned horizontaly.  above the would-be-edge are a set of points U(p)1-Un.  Connecting U1 to U2, U2 to U3, ..., Un-1 to Un, and then L to each U and R to each U will result in 2*(n-1) triangles.  Then connectine L to R will only count one more triangle, namely: LRUn, with total count of 2*(n-1)+1.  Insted, connecting LRU1, then LRU2, then U2 to U1, and repeating (LRU3, U3 to U2) until lastly connecting LRUn, Un to Un-1, will result in 3*(n-1)+1 triangles.

I am looking for an idea of how to approach the matter in order to create an algorythm that is given set of point and returns an ordered list of edges (from which point to which point) that will give the maximum count of triangles.
I am familiar with recursion.

Thank you!
Oren. :)

Sounds like someone is looking for an algorithm to maximize field count in Ingress :)

Having thought about the problem for a bit, I think the task can be split in a few parts:

1) Find a set of links (edges) that create the maximum possible number of triangles. At this stage, the only restriction is that the edges cannot cross. There is often more than one solution for this. I'm sure brute force can do this given enough time. However, a LOT of time may be required.

2) Once a solution is chosen/found, determine the sequence of links in order to comply with the other requirements (a new link can create triangles inside pre-existing triangles, but not a new triangle inside another new triangle). This part can also be done by brute force, but again may take too long.

I have a matlab script that tests a given sequence of links and calculates how many valid triangles are formed (and how many triangles are "lost"). For a set of 25 nodes, a sequence (about 50 links and 50 triangles) takes about 0.3 seconds to run and validate, which is too long if we're trying to brute-force all possible solutions...


If $x,y \in R$ and $xy^2$ then: If $x,y \in R$ and $x<y\implies x^2>y^2$ then:

a). $x>0$

b). $x<0$

c). $y>0$

d). $y<0$

My Attempt:

Here, $x<y \implies x^2>y^2$ but How?
As for example: Let $x=0.001$ and $y=0.1$.
So,
$$0.001<0.1$$
Squaring Both Sides, We get:
$$0.000001<0.01$$
??

As you noticed, the implication is false when x is positive. So x must be negative and the answer is (b).

Only possibility of holding inequality is that y must be negative. I.e option (d) is correct.
We check this by an example
-2<-1 imply 4>1 (square of both sides)


Checking if 8 points in the projective plane lie on a singular cubic. I need to check if 8 points of $\mathbb{P}^2$ (over a finite field) lie on a singular cubic with one of them a double point.
I know that to check if a point is singular we suffice to compute the partial derivatives in it and look if they are all 0.
I also know I can use the Veronese embedding to prove that 10 points of $\mathbb{P}^2$ are independent with regard to cubics, i.e. don't lie in one, if and only if the matrix that has as rows the evaluation of monomials of degree 3 in a given point has rank 10.
However I'm not sure how to unite the two concepts to find an easy (as in "easy to feed to a computer to check") condition to check if there is a singular cubic passing through the 8 points and having a singularity in one of them.
My idea is to use a matrix that has the following rows:

rows 1-7 evaluation of monomials of degree 3 in each of 7 points
rows 8-10 the partial derivatives to such monomials evaluated in the 8-th point

Repete 8 times changing the point we want to check is singular.
But I can't find a proper proof that this does actually work.

At the end the method I proposed actually works


"If A then B" in Venn (or Euler) Diagrams How can I represent "If A then B" in a diagram?
I thought it would be a simple subset like $A ⊂ B$. However this material says that

If $A$ then $B$ $=$ $A^c ∪ B$.

Now I am confused.

You want to construct the set $\;\{x\mid x\in A \to x\in B\}\;$.

Then by implication equivalence this is $\;\{x\mid x\not\in A \vee x\in B\}\;$.

Which is simply $\;A^\complement \cup B\;$.

This is the set of all elements that, if they're in A then they're in B

The statement $A\subseteq B$ is not a set.  It is a relation.  It is the statement that $y\in A \implies y\in B$.

In the specific case that  $A$ is a subset of $B$, then there is no element that is not in $A^\complement \cup B$.

So if you wanted to represent the statement "if $A$ then $B$", you could have $A$ as a subset of $B$.

But if you wanted to represent all elements that "if in $A$ then in $B$" you would use the union: $A^\complement\cup B$.

It is quite simple - the external shape is B and A is fully included


Are these two statement equivalent? $\forall x \exists y P(x,y)$

$\exists x \forall y P(x,y)$

where P(x,y) means x is smaller than y.

I believe that they mean the same thing.

Assuming you mean $\exists x \forall y P(x,y)$ for your second one, no these are not the same.

Put them into words and it will become clearer:

The first one says:

" Every $x$ is smaller than some $y$. "

The second one says:

" There is some $x$ which is smaller than every $y$. "

These are certainly not saying the same thing. For instance the first one is true in $\mathbb{R}$ but the second one is not.

The second condition asserts a form of uniformity.


If a sequence $\{a_n\}$ satisfies the Inequality $a_{n+1} < ka_{n}$, then show that $ \lim\limits_{n \to \infty} a_n =0$ where $0< k , a_n< 1$ I know one solution.

Consider $\sum a_n$

Then use ratio test to show that the series converges, hence the sequence.

Any other Ideass !

I will suppose $a_n \ge 0$, as otherwise the statement is wrong.

By induction, we have $a_n < k^na_0$, hence $0 \le a_n \le k^n a_0$. As $0 \to 0$ and $k^n a_0 \to 0$ (we have $k \in (0,1)$), by the squeeze theorem $a_n \to 0$.

we have a theorem if a sequence is bounded and descend this sequence ins convergence and this condition is true for your question .and i think we have $a_n\geq 0 $


Solving an equation including $\ln$ $$8x^2+2\ln x-6=0$$

How should I approach this? Should I do $8x^2=e^{\ln{x^2}}$?

The real solution:

$$8x^2+2\ln(x)-6=0\Longleftrightarrow$$
$$8x^2+2\ln(x)=6\Longleftrightarrow$$
$$2\left(4x^2+\ln(x)\right)=6\Longleftrightarrow$$
$$4x^2+\ln(x)=3\Longleftrightarrow$$
$$e^{4x^2+\ln(x)}=e^{3}\Longleftrightarrow$$
$$e^{4x^2}\cdot e^{\ln(x)}=e^{3}\Longleftrightarrow$$
$$e^{4x^2}\cdot x=e^{3}\Longleftrightarrow$$
$$xe^{4x^2}=e^{3}\Longleftrightarrow$$
$$x=\frac{e^{3}}{e^{4x^2}}\Longleftrightarrow$$
$$x=e^{3-4x^2}\Longleftrightarrow$$
$$x^2=\left(e^{3-4x^2}\right)^2\Longleftrightarrow$$
$$x^2=e^{6-8x^2}\Longleftrightarrow$$
$$x=\sqrt{\frac{\text{W}\left(8e^6\right)}{4}}\Longleftrightarrow$$
$$x=\frac{1}{2}\sqrt{\frac{\text{W}\left(8e^6\right)}{2}}$$

With $\text{W}(z)$ is te product log function

The best method to solve these types of transcendental equations is with graphs.


Prove $a_n =3\cdot2^{n-1} +2(-1)^n$ $a_1=1, a_2=8$ and $a_n=a_{n-1} +2a_{n-2}$ for $n\ge3$

Prove that

$$a_n =3\cdot2^{n-1} +2(-1)^n$$
for $n\in N$

Base Case: n=1

$a_1= 3\cdot2^0+2(-1)^1, a_1 =1$, the base case holds true

I.H: Suppose its true for $$a_k =3\cdot2^{k-1} +2(-1)^k$$

How do i proceed for induction step, am i suppose to use $a_n=a_{n-1} +2a_{n-2}$ in the induction step?

NOTE: For this problem, you need two base cases, since the recursive sequence is defined in terms of the previous two terms. I'll leave that part up to you.

INDUCTION STEP: Assume that both
$$a_n=3(2^{n-1})+2(-1)^n$$
and
$$a_{n+1}=3(2^{n})+2(-1)^{n+1}$$
for some $n\ge 3$. Then, by hypothesis
$$a_{n+2}=a_{n+1}+2a_n$$
so substituting the values of $\,a_{n+1}\,$ and $\,a_{n}$ from our induction hypothesis yields
$$\begin{align} a_{n+2}&=3(2^{n})+2(-1)^{n+1}+2(3(2^{n-1})+2(-1)^n)\\
&=3(2^{n})+2(-1)^{n+1}+3(2^{n})+4(-1)^n\\
&=6(2^{n})+2(-1)^{n}\\
&=3(2^{n+1})+2(-1)^{n+2}
\end{align}$$
which completes the proof of the inductive step.

hint

Let for $n\ge 3$,
$$b_n=a_n+a_{n-1} $$

then
$$b_n=2b_{n-1} $$
Geometric.


Proving an inequality using the divergence theorem I have a question that reads

  $$\frac{\partial}{\partial t}w=\nabla^2w$$
  for $w(t,\vec x)$ with initial condition $w(0,\vec x) = w(0,x)$ in $V$ , boundary condition $w(t,\vec x) = f(\vec x)$ on $S$ for all $t > 0$.

  Show that
  $$\frac{\partial}{\partial t}\iiint |\nabla w|^2 dV\le0$$

So far I have

$$\frac{\partial}{\partial t}\iiint |\nabla w|^2 =
\iiint\frac{\partial}{\partial t}(\nabla.(w\nabla w)-w\nabla^2w)dV$$
$$= \iint\frac{\partial}{\partial t}(w\nabla w).d\vec S-\iiint\frac{\partial}{\partial t}(w\nabla^2 w)dV$$
(divergence theorem)
$$=-\iiint\frac{\partial}{\partial t}(w\nabla^2 w) dV$$
($w$ constant on $S$)
$$=-\iiint(\frac{\partial w}{\partial t})^2+(w\frac{\partial^2}{\partial^2 t}w) dV$$
(using the PDE given).

The first term is then clearly negative, but I'm not sure about the second?

You could try integration by parts on $w \frac{\partial^2 w}{\partial^2 t}$, as this would then lead to a square term.


Prove if $a\mid b$ and $b \mid (ap)$ where $p$ is prime, then $b=a$ or $b=ap$ Let $a,b,p \in \mathbb{N}$ such that $a\mid b$ and $b\mid (ap)$ where $p$ is prime. Show that $b=a$ or $b=ap$

So far I know that $b=a(x)$ and $(ap)=b(y)$ where $x,y \in \mathbb{Z}$. Also fro that you can conclude that $a \mid (ap)$ so $ap=a(z)=b(y)$ where $z \in \mathbb{Z}$. Now I don't really see the next steps to use or where the idea that $p$ is prime comes into play. I only have the definition of divide and prime right now also to use.

Hint $\,\ b/a\mid p\,\Rightarrow\,\underbrace{ b/a = 1}_{\large b\ =\ a}\,$ or $\ \underbrace{b/a = p}_{\large b\ =\ ap}$

By the fundamental theorem of arithmetic, every positive rational number $x$ can be uniquely written as

$$x = \prod\limits_p p^{\textrm{ord}_p x}$$

where $p$ runs through all the prime numbers, $\textrm{ord}_p x$ is an integer, and $\textrm{ord}_p x = 0$ for all but finitely many $p$.    The number $x$ is an integer if and only if none of the numbers $\textrm{ord}_p x$ are negative.

Let $q$ be a particular prime number, and suppose that $a$ divides $b$ and $b$ divides $qa$.  This means that $b/a$ and $qa/b$ are both integers. Thus all the exponents in

$$b/a = \prod\limits_p p^{\textrm{ord}_p b - \textrm{ord}_p a}$$

are nonnegative, and the same for

$$aq/b = \prod\limits_p p^{\textrm{ord}_p a + \textrm{ord}_p q - \textrm{ord}_p b}$$

Since $\textrm{ord}_p q$ is $0$ or $1$, depending on whether $p \neq q$ or $p = q$, this tells us that $\textrm{ord}_p b = \textrm{ord}_p a$ for all $p \neq q$, and $\textrm{ord}_q b - \textrm{ord}_q a$ and $\textrm{ord}_q a - \textrm{ord}_q b + 1$ are both nonnegative.

The only way this last thing can happen is if either $\textrm{ord}_q a = \textrm{ord}_q b$ or $\textrm{ord}_q a = \textrm{ord}_q b - 1$.  In other words, either $a = b$ or $aq = b$.


Is it true that $A \in A$? I defined the set $A$ as follow:
\begin{align}
A_0 & =\varnothing \\
A_1 & =\{A_0\}=\{\varnothing\} \\
A_2 & =\{A_1\}=\{\{\varnothing\}\} \\
A_3 & =\{A_2\}=\{\{\{\varnothing\}\}\} \\
& {}\ \ \vdots \\
A & =A_\infty= \{\{\{\cdots\{\varnothing\}\cdots\}\}\}
\end{align}

And now this is clear that $A\in A$.

Is there any mistake in my conclusion?

I think there are two separate questions here:

Why does the sequence of sets $\{\}, \{\{\}\}, \{\{\{\}\}\}$, . . . not "approach a limit"?
Is it possible to have a set $A$ such that $A\in A$?

The second one is more easily dealt with, since it a precise question. The usual axioms of set theory, ZFC, rule out any such set; specifically, the axiom of Foundation (or Regularity) prevents the existence of such sets. However, if we remove the axiom of Foundation from ZFC, then such sets can exist, and there are even set theories - Quine's NF comes to mind, as well as its better-behaved variant NFU - which require such sets to exist.

For the former, I think you would find it helpful to try to define in what sense the sequence of sets $\{\}, \{\{\}\}, \{\{\{\}\}\}$, . . . does (in your mind) have a limit, and then try to see what it would take to prove that. Note that there is a clear notion of limit for a monotonic sequence of sets: e.g., if $\alpha>\beta$ implies $A_\alpha\supset A_\beta$ for $\alpha<\kappa$, then it makes sense to define $$\lim_{\alpha\rightarrow\kappa}A_\alpha=\bigcup_{\alpha\in\kappa} A_\alpha,$$ and similarly for descending sequences of sets. But your sequence of sets isn't monotonic, so there's no natural notion of what its "limit" should be. And, even if there were, you would need to argue that properties of terms of the sequence induce properties of the limit, which depends on the properties in question.

Set theory forbids the existence of a set $A$ such that $A \in A$, as that kind of construction easily leads to paradoxes, such as Russell's paradox.

In ZFC, the axiom of regularity has this immediate consequence.


Can we represent all algebraic structures in First-Order logic? Can we represent all algebraic structures in First-Order logic?

The subject of universal algebra is the study of algebraic structures as classes of first-order structures in languages having only function symbols as opposed to relation symbols. This subject is regarded by many as much closer to algebra than the rest of model theory, allowing one to combine methods from model theory with a stronger category-theoretic perspective. A central concern are the classes of algebras, called varieties, defined by a system of algebraic equations.

For example, Birkoff's theorem states that a class of algebraic structures is a variety of algebras if and only if it is closed under homomorphisms, subalgebras and products.

A similar corresponding theorem in first order logic is due to Kiesler and Shelah: two structures are elementary equivalent if and only if they have a common ultrapower, and one can use this to characterize which collections of algebras form an elementary class.

No, you cannot.  Fuzzy subalgebras cannot get represented in first-order logic since the background logic for a fuzzy subalgebra consists of a multivalued or infinite-valued logic.


Compute the unit digit of ${{43}^{43}}^{43}$ I am trying to find the unit digit of ${{43}^{43}}^{43}$. Normally I would try to break up the exponent to make it easier though 43 is prime so I can't do that. So instead I first considered $43^{43}$ and tried finding a pattern for the unit digits by looking at the first few values:

$43^0=1$

$43^1=43$

$43^2=1849$

$43^3=79507$

$43^4=3418801$

$43^5=147008443$

...

I then noticed that they repeated every 4 values with a unit digit pattern of {1, 3, 9, 7}.

This is where I am stuck, for a few reasons:

I know I need to prove that this unit digit pattern continues but I'm not quite sure how to do this.
I know the unit digit is 1, 3, 7 or 9, but how do I know which one it is?
After finding the unit digit of $43^{43}$, how do I then use this to find the unit digit of ${{43}^{43}}^{43}$? Lets say $x$ is the unit digit of $43^{43}$ so would I then need to find the unit digit of $43^x$?

Can someone help me with where I am confused, let me know if the proccess I wrote is incorrect, or explain an easier approach to this problem?


Can $\frac{\csc \alpha +\cos \alpha}{\cos \alpha - \tan \alpha - \sec \alpha}$ be simplified? I am trying to simplify the following but I cannot.
$$
\frac{\csc \alpha +\cos \alpha}{\cos \alpha - \tan \alpha - \sec \alpha}
$$
Can it be simplified?
Edit
My last result is
$$
- \frac{\cos \alpha \left( 1 + \sin \alpha \cos \alpha\right)}
{\sin^2 \alpha \left(1+\sin \alpha\right)}
$$
I am wondering it might be a wrong question given by my student's teacher.

If you multiply the numerator and denominator of your last expression by $4(1-\sin\alpha)$ and use the identity $\sin2\alpha=2\sin\alpha\cos\alpha$, you can rewrite your result as $$(1-\csc\alpha)(2\csc2\alpha+1).$$ This reduces the original 24 symbols to 17.

First, let us state of the fundamentals.
$$\csc \alpha = \frac{1}{\sin \alpha};\ \tan \alpha = \frac{\sin \alpha}{\cos \alpha};\ \cot \alpha = \frac{\cos\alpha}{\sin \alpha};\ \sec \alpha = \frac{1}{\cos \alpha};\ \sin^2\alpha + \cos^2\alpha = 1;$$

So,
$$\frac{\csc \alpha +\cos \alpha}{\cos \alpha - \tan \alpha - \sec \alpha} \cdot 1= \frac{\frac{1}{\sin \alpha} + \cos \alpha}{\cos \alpha - \frac{\sin\alpha}{\cos\alpha} - \frac{1}{\cos\alpha}} \cdot\frac{\cos \alpha}{\cos \alpha} = \frac{\cot \alpha + \cos^2\alpha}{\cos^2\alpha - \sin\alpha - 1} = \frac{\cot\alpha + 1 -\sin^2\alpha}{1-\sin^2\alpha - \sin\alpha - 1} = \frac{\cot\alpha + 1 -\sin^2\alpha}{\left(0-\sin\alpha\right)\cdot\left(1+\sin\alpha\right)}\cdot\frac{\sin\alpha}{\sin\alpha} = \frac{\cos\alpha + \sin\alpha - \sin^3\alpha}{- \sin^2\alpha\left(1+\sin\alpha\right)} = -\frac{\cos\alpha + \sin\alpha\cos^2\alpha}{\sin^2\alpha\left(1+\sin\alpha\right)} = - \frac{\cos\alpha\left(1 + \sin\alpha\cos\alpha\right)}{\sin^2\alpha\left(1+\sin\alpha\right)}$$
Alright, a great number of further simplifications are possible from here, but what do you think will eliminate the denominator and bring the order down to one without changing the inner $\alpha$. There are 4 terms and various manipulations left.

You'll see this is actually the simplest you can get it to. It is not trivial to prove this though.


Bound the norm of the partial trace of an operator on a Hilbert space Let $H=H_1 \otimes H_2$ a composite Hilbert space and let $A, B$ bounded linear operators on $H$, and we can assume they are trace class. Let $A_2$ we denote the operator on $H_2$ obtained by taking the partial trace on $H_1$, i.e. $A_2 = \mathrm{tr}_1 (A) \in B(H_2)$. Assume that we know there is some $\varepsilon > 0$ so that $$ ||A-B||_{B(H)} < \varepsilon. $$ Can we deduce a similar bound for $$ ||A_2 - B_2 ||_{B(H_2)} ? $$

Notes:
1) The finite dimensional case $\dim (H_1) = n_1$, $\dim (H_2) = n_2$, $\dim (H) = n$ all finite would be fine (but if you have a hint or source for an answer in the general case I'd appreciate!).

2) (My approach) Choose the Frobenius (or Hilbert-Schmidt) norm, $$ ||A||_{B(H)} = \sqrt{\mathrm{tr}(A^\dagger A)}$$ and let $C=A-B$. We know that $ ||I_1 \otimes C_2 ||_{B(H)} = ||I_1||_{B(H_1)} ||C_2||_{B(H_2)}$ (cf. Reed Simon Vol. 1 page 299), then
$$
\begin{array}{rcl}
||C_2||_{B(H_2)} &=& \frac{1}{\sqrt{n_1}} ||I_1 \otimes C_2||_{B(H)} \\
&=& \frac{1}{\sqrt{n_1}} ||C - (I_1 \otimes C_2) - C ||_{B(H)} \\
&\leq& \frac{\varepsilon}{\sqrt{n_1}} + \frac{1}{\sqrt{n_1}} ||C - (I_1 \otimes C_2)||_{B(H)}
\end{array}
$$
and expect to bound the term on the right by $\alpha ||C||_{B(H)}$. To see what is going on I put
$$
C = C^{(1)}_1 \otimes C^{(2)}_1 + C^{(1)}_2 \otimes C^{(2)}_2
$$
(taking enough terms of these form should be fine to represent an arbitrary $C$). A calculation gives
$$
||C - (I_1 \otimes C_2)||_{B(H)} = ||(C^{(1)}_1 - (\mathrm{tr} C^{(1)}_1) I_1)\otimes C_1^{(2)} + (C^{(1)}_2 - (\mathrm{tr} C^{(1)}_2) I_1)\otimes C_2^{(2)}||_{B(H)}
$$
how close this is to $||C||_{B(H)}$?

There is a bound on the norm of partial trace of operator that depends on the type of norm and the size of the Hilbert space you traced out. For the Schatten $p$-norm, which is defined for $p\ge1$ as
$$\|A\|_p=\left[\text{Tr}(|A|^p)\right]^{1/p}$$
one can show a bound, due to Rastegin, where
$$\|\text{Tr}_1(A)\|_p \le [\dim(H_1)]^{(p-1)/p} \|A\|_p$$
This bound can be made dimension-independent by choosing $p=1$, in which case the norm is the trace norm. But if you choose $p=2$, which gives the Frobenius norm, you'll have a dimension-dependent bound.

(a colleague just helped me to arrive at the answer. Moreover, it seems to me that it carries over to the infinite dimensional space so long as the partial trace is well defined).

Let $ C = \sum_{ij} \alpha_{ij} e_i \otimes e_j,$ where $e_i$ is a matrix with only one non-zero entry equal to one, so we have this basis for the complete space. We calculate
$$
\begin{array}{rcl}
||C_2||_{(B(H_2),||\cdot||_F)}^2 &=& ||\sum_{ij} \alpha_{ij} \mathrm{Tr}(e_i) e_j ||_{(B(H_2),||\cdot||_F)}^2 \\
&=& \sum_{ij} |\alpha_{ij}|^2  |\mathrm{Tr}(e_i)|^2 \\
&\leq& \sum_{ij} |\alpha_{ij}|^2 = ||C||_{(B(H),F)}^2
\end{array}
$$
where we used that $\mathrm{Tr}(e_i)$ is either one or zero. We conclude that $||C_2|| \leq \varepsilon$.


Is the set $A$ of anti-symmetric relations over $\mathbb{N}$ countable? I think this set is uncountable, but I don't really know how to prove this. Maybe somehow find an injection from $\mathcal{P}( \mathbb{N} \times  \mathbb{N}) \rightarrow A$?

For any $X\subseteq \mathbb N$, the relation $R_X = \{(x,x) | x \in X\}$ is anti-symmetric.

For any $S \subset \mathbb{N}$ take $\mathcal{R}_S = \bigcup_{s \in S} \{(s,s+1)\}$. It is clear that $\mathcal{R}$ is anti-symmetric, as it's a subset of anti-symmetric relation $\mathcal{R}=\{(x,y):x<y\}$, and that this transformation is injective, so there are at least $\# \mathcal{P}(\mathbb{N})$ anti-symmetric relations on $\mathbb{N}$, thus their number is uncountable.


Do All Primitive Triples Belong to the Plato, Pythagoras and Fermat Families? After running into this terminology on the "Formulas for generating Pythagorean triples" Wikipedia page, I was curious whether all triples fit into these categories.

The article states:

  Plato: c - b = 1, Pythagoras: c - a = 2, Fermat: |a - b| = 1

Another page lists these as the first primitive triples where c<100:

(3, 4, 5)       (5, 12, 13)     (8, 15, 17)     (7, 24, 25)
(20, 21, 29)    (12, 35, 37)    (9, 40, 41)     (28, 45, 53)
(11, 60, 61)    (16, 63, 65)    (33, 56, 65)    (48, 55, 73)
(13, 84, 85)    (36, 77, 85)    (39, 80, 89)    (65, 72, 97)

However, I noticed that some sets like the (36, 77, 85) triple have a very large difference between each integer. Does this mean that not all pythagorean triples are part of one of those families? And if so, is there a formula for generating all the pythagorean triples?

The views of these people only apply to a small number of triples and sometimes depend on which of $A or B$ is odd or even. Below are original functions gleaned from 8 million spreadsheet formulas.

These generate all triples where GCD(A,B,C) is the square of an odd number. $\mathbf {\text{This includes all primitives}}$ and excludes all non-odd-square multiples of primitives. In the following sample of sets of triplets ($Set_1, Set_2, Set_3, \text{ and }Set_{25}$), we can see a great variation in the differences between $A,B,C$. We can also see that $\mathbf {(C-B) \text{ is the }n^{th} \text{odd square}}$. In the example: $C_1-B_1=1^2, C_2-B_2=3^2, C_3-B_3=5^2\text{ and }C_{25}-B_{25}=49^2=2401$.

$$\begin{array}{c|c|c|c|c|}
 \text{$Set_n$}& \text{$Triplet_1$} & \text{$Triplet_2$} & \text{$Triplet_3$} & \text{$Triplet_4$}\\ \hline
\text{$Set_1$} & 3,4,5 & 5,12,13& 7,24,25& 9,40,41\\ \hline
\text{$Set_2$} & 15,8,17 & 21,20,29 &27,36,45 &33,56,65\\ \hline
\text{$Set_3$} & 35,12,37 & 45,28,53 &55,48,73 &65,72,97 \\ \hline
\text{$Set_{25}$} &2499,100,2501 &2597,204,2605  &2695,312,2713 &2793,424,2825\\ \hline
\end{array}$$
Note: These triples can be generated by a variation of Euclid's formula where $A=(2m-1+n)^2-n^2\quad B=2(2m-1+n)n\quad C=(2m-1+n)^2+n^2\quad$ but we will use a formula developed from observations of these sets because it is easier to see that $n$ is a set number and $k$ is a member number.

Theorem:
$$\forall n,k \in \mathbb{N}, \exists A,B,C\in \mathbb{N}:A^2+B^2=C^2 \iff A=(2n-1)^2+2(2n-1)k$$

Proof: Let $$A=(2n-1)^2+2(2n-1)k$$

Solving $A^2+B^2=C^2$ for $B$ and $C$, respectively, and substituting $A$,  we find that
$$B=2(2n-1)k+2 k^2$$and$$C=(2n-1)^2+2(2n-1)k+2k^2$$
We can then show that
$$A^2=(2n-1)^4+4(2n-1)^3 k+4(2n-1)^2 k^2$$
$$B^2=4(2n-1)^2 k^2+8(2n-1) k^3+4k^4$$
$$C^2=(2n-1)^4+4(2n-1)^3 k+8(2n-1)^2 k^2+8(2n-1) k^3+4k^4$$
 $$A^2+B^2=(2n-1)^4+4(2n-1)^3 k+8(2n-1)^2 k^2+8(2n-1) k^3+4k^4=C^2$$

$\therefore \forall n,k \in \mathbb{N},\exists A,B,C\in \mathbb{N}:A^2+B^2=C^2 \iff A=(2n-1)^2+2(2n-1)k\text{ } \blacksquare$

The Plato family $C-B=1$ applies only to members of $Set_1$ as we can see in the sample above. The Pythagoras family $C-A=2$ applies only to the first members of all sets. I haven't heard of the Fermat family $| A-B |=1$ and it seems to apply at random as in these examples of $f(n,k)=A,B,C$.
$$f(1,1)=3,4,5$$
$$f(2,2)=21,20,29$$
$$f(4,5)=119,120,169$$
$$f(9,12)=697,696,985$$
$$f(21,29)=4059,4060,5741$$
$$f(50,70)=23661,23660,33461$$
I stopped checking at $Set_{50}$ and I don't see a pattern, do you? Do note that the first two families are nothing more than the members of $Set_1$: f(1,k) or the first members of all sets: f(n,1). I hope I have shown you there is much more to explore than the families mentioned.

If you wanted to explore other configurations, I have another set of equations for side $A$ even:$$A=2n^2+2{N}n+4kn-4n\text{, }B=2n(2k-1)+(2k-1)^2\text{, }C=2n^2+2n(2k-1)+(2k-1)^2$$ and it generates the same triplets rotated $90$ degrees in the example with $A$ and $B$ reversed.

Update: I have recently inferred, but not proven, that triples where $GCD(A,B,C)\ne 2$ and $GCD(A,B,C)\ne x^2,x\in\mathbb{N}$ cannot be generated by Euclid's formula $(A=m^2-n^2\quad B=2mn\quad C=M^2+n^2)$ alone without a multiplier, as shown here.

Yes,

For $~a~$ is an even number then $~b= \frac{1}{2i}(a^2-i^2)~$ , where $~i~$ is an positive even integer such that $~i<[a/2]~$, $~b~$ belong to $~N~$ , and $~c= b+i~$
e.g., $~a=36~$

For $~i=~$ even positive number $~2n^2; ~~~n~$ belong to $~N~$

$~~~~~~= 2,8<[36/2]$

$i=2,~~~~ (36,323,325)~$

$i=8, ~~~~(36,77,85)~$

From this it is clear that for even number a, we can find more than one primitive triples, it may have Plato, pythagorean , and Fermat family triples....this is the generating formula for even primitive triples....

For odd generating primitive Pythagorean Triples, I will send it later on....
Regards,
Shailesh