Criteria for metric on a set Let $X$ be a set and $d: X \times X \to X$ be a function such that $d(a,b)=0$ if and only if $a=b$.

Suppose further that $d(a,b) ≤ d(z,a)+d(z,b)$ for all $a,b,z \in X$.

Show that $d$ is a metric on $X$.

Let $X$ be a set and $d: X \times X \to X$ be a function such that $$d(a,b)=0\text{ if and only if}\;\; a=b,\text{ and}\tag{1}$$ $$d(a,b) ≤ d(z,a)+d(z,b)\forall a,b,z \in X.\tag{2}$$

There's additional criterion that needs to be met for a function $d$ to be a metric on $X$:

You must have that $d(a, b) = d(b,a)$ for all $a, b \in X$ (symmetry).You can use the two properties you have been given to prove this.  $d(a,b)\leq d(b,a)+d(b,b)= d(b, a) + 0 = d(b,a)$ and vice versa, hence we get equality.
Having proven symmetry, you will then have that  $d(a,b) \leq d(z,a) + d(z, b) \iff d(a, b) \leq d(a, z) + d(z, b)$.
Finally, using the property immediately above, along with the $(1)$, you can establish that for all $a, b\in X$ such that $a\neq b$, we must have $d(a, b) > 0$.

Then you are done.

This must be the discrete metric.
1) The first condition follows by definition that d(a,b)=0 iff a=b;
2) Symmetry: this is trivial, because if a=b you have d(a,b)=0 and b=a gives you d(b,a)=0;
3) The triangle inequality: from the symmetry you can write d(z,a)=d(a,z).
Consider two cases:
 - if a=b, clear
 - if a is not equal to b, then either a not equal to z or z not equal to b. There you have $1\leqslant 1$ or $1\leqslant 2$.


Eisenstein integers and applications to Diophantine equations Solve the equation $7\times 13\times 19=a^2-ab+b^2$ for integers $a>b>0$. How many are there such solutions $(a,b)$?

I know that $a^2-ab+b^2$ is the norm of the Eisentein integer $z=a+b\omega$, but how can I make use of this? Thank you so much.

It is known that the Eisenstein integers $\mathbb{Z}[\omega]$ is an unique factorization domain and it has six units
$$\pm 1, \pm \omega, \pm \omega^2$$
Over $\mathbb{Z}[\omega]$, the numbers $7, 13, 19$ factorize into its prime factors as
$$\begin{cases}
7  &= (3 + \omega)(3 + \omega^2)\\
13 &= (4 + \omega)(4 + \omega^2)\\
19 &= (5 + 2\omega)(5 + 2\omega^2)
\end{cases}$$
This mean if we want to factorize $1729 = 7 \times 13 \times 19$ over $\mathbb{Z}[\omega]$ as
$$1729 = ( x + y\omega )(x + y\omega^2) = x^2 - xy + y^2
\quad x, y \in \mathbb{Z}
$$
the corresponding factor $x + y\omega$ must have the form

$$x + y\omega = u A B C\quad\text{ with }\quad
\begin{cases}
A &=  3 + \omega &\text{or}& 3 + \omega^2\\
B &=  4 + \omega &\text{or}& 4 + \omega^2\\
C &=  5 + 2\omega &\text{or}& 5 + 2\omega^2
\end{cases}
$$
and $u$ is one of above six units.

There are 8 possible choices of $A,B,C$. For each choice of $A,B,C$,
multiply by one of the six units allow one to obtain an pair of $x,y$
that satisfies $x \ge y \ge 0$:

$ABC = (3+\omega)(4+\omega)(5+2\omega) = 43+40\omega$.
$ABC = (3+\omega)(4+\omega)(5+2\omega^2) = 45+8\omega$.
$ABC = (3+\omega)(4+\omega^2)(5+2\omega) = 48+23\omega$.
$ABC = (3+\omega)(4+\omega^2)(5+2\omega^2) = 32-15\omega \implies -\omega^2 ABC = (47+32\omega)$
$ABC = (3+\omega^2)(4+\omega)(5+2\omega) = 47+15\omega$.
$ABC = (3+\omega^2)(4+\omega)(5+2\omega^2) = 25-23\omega \implies -\omega^2 ABC = 48+25\omega$
$ABC = (3+\omega^2)(4+\omega^2)(5+2\omega) = 37-8\omega \implies -\omega^2 ABC = 45+37\omega$
$ABC = (3+\omega^2)(4+\omega^2)(5+2\omega^2) = 3-40\omega \implies -\omega^2 ABC =
43+3\omega$

As a result, there are $8$ pairs of $(a,b)$ that solves the original problem:

$$(a,b) = (43, 3), (43,40), (45, 8), (45, 37), (47, 15), (47,32), (48,23), (48,25)$$

Plotting $7 \times 13 \times 19 = a^2 − ab + b^2$, You will get an ellipse like this:

.
But if we apply the condition $a$ & $b > 0$, both will be positive in only first Quadrant. And again $a > b$, we will end up with half of the ellipse in Quadrant 1 i.e. 1/8th of the total ellipse.

.
Set of all the points on this curve is our solution. Infinite number of Real solutions.

Below are the integer solutions:

(43,3), (43,40), (45,8), (45,37), (47,15),(47,32),(48,23),(48,25)


Can we differentiate a differential equation to determine its order and degree? It is known that we must need to convert the differential equation in polynomial equation of differential coefficients. But Can we differentiate a differential equation (whose degree is not defined) to determine its order and degree ?
Example to show my doubt clearly:
$y''=e^{y'}$
Above differential equation has its degree undefined. Differentiating it with respect to $x$
$y'''=(y'')^2$
So we may conclude that this is third order differential equation with degree $=1$
I know that on differentiating a differential equation number of arbitrary constants of solution equation increases hence we should not differentiate a differential equation in general. But I did not found a reference which states that we cannot differentiate a differential equation to determine its order and degree so I want to confirm my thoughts.

According to the link provided, a second order ODE can be said to have a degree if it is of the form
$$
(y'')^d+a_{d-1}(x,y,y')(y'')^{d-1}+...+a_1(x,y,y')y''+a_0(x,y,y')=0
$$
where the coefficient functions are sufficiently regular. The degree is then $d$.
So in the equation in question we have $d=1$ and $a_0(x,y,y')=-\ln(y')$, falling into the required pattern.

Note that assigning a degree to an ODE, and declaring what ODE can be assigned a degree, is a matter of opinion and circumstance. For example, in the intersection of differential equation and algebraic geometry a different definition might be more fruitful.


Find the horizontal asymptotes of $f(x) = \frac{x^3-2}{\lvert x\rvert^3+1}$ Taken from Thomas' Calculus 12e
Find the horizontal asymptote of the graph of:

$$f(x) = \frac{x^3-2}{\lvert x\rvert^3+1}$$

Solution: We calculate the limits as ${x \to \pm \infty}$

For $x\ge0$: $$\lim\limits_{x\to\infty}\frac{x^3-2}{\lvert x\rvert ^3+1} =\lim\limits_{x\to\infty}\frac{x^3-2}{x^3+1}=\lim\limits_{x\to\infty}\frac{1-(2/x^3)}{1+(1/x^3)}=1$$

I understand all except why the answer is one. I actually cannot find out how to get the answer.

As you find out, the limit as $x\to \infty$ is $1$ then $y=1$ is an horizontal asymptote.

For $x\to -\infty$ we have

$$\lim\limits_{x\to-\infty}\frac{x^3-2}{\lvert x\rvert ^3+1} =\lim\limits_{x\to-\infty}\frac{x^3-2}{-x^3+1}=\lim\limits_{x\to-\infty}\frac{1-(2/x^3)}{-1+(1/x^3)}=-1$$

then $y=-1$ is an horizontal asymptote for $x\to -\infty$.

I assume the function's domain is $\mathbb R$ except where $f$ is not defined namely $x=-1$.

Now if $\lim_{x \to \infty} f(x) = a$ for some $a \in \mathbb R$, then $y=a$ is a horizontal asymptote.

Similarly, if $\lim_{x \to -\infty} f(x) = b$, for some $b \in \mathbb R$, then $y=b$ is a horizontal asymptote.

For $x\to\infty$, we have

$$\lim\limits_{x\to\infty}\frac{x^3-2}{\lvert x\rvert ^3+1} =\lim\limits_{x\to\infty}\frac{x^3-2}{x^3+1}$$

$$=\lim\limits_{x\to\infty}\frac{x^3/x^3-(2/x^3)}{x^3/x^3+(1/x^3)}=\lim\limits_{x\to\infty}\frac{1-(2/x^3)}{1+(1/x^3)}$$

$$=\frac{\lim\limits_{x\to\infty} [1-(2/x^3)]}{\lim\limits_{x\to\infty} [1+(1/x^3)]}=\frac{ [\lim\limits_{x\to\infty}1-\lim\limits_{x\to\infty}(2/x^3)]}{ [\lim\limits_{x\to\infty}1+\lim\limits_{x\to\infty}(1/x^3)]}$$

$$=\frac{ [1-\lim\limits_{x\to\infty}(2/x^3)]}{ [1+\lim\limits_{x\to\infty}(1/x^3)]} = \frac{ [1-0]}{ [1+0]} = \frac11 = 1$$

Thus, $y=1$ is a vertical asymptote

Similarly, for $x\to-\infty$, we have

$$\lim\limits_{x\to-\infty}\frac{x^3-2}{\lvert x\rvert ^3+1} =\lim\limits_{x\to-\infty}\frac{x^3-2}{(-x)^3+1}$$

$$= \cdots$$

$$= \cdots$$

$$= \frac{ [1-\lim\limits_{x\to-\infty}(2/x^3)]}{ [-1+\lim\limits_{x\to-\infty}(1/x^3)]} = \frac{ [1-0]}{ [-1+0]} = \frac1{-1} = -1$$

Thus, $y=-1$ is a vertical asymptote

Soooo...do you know why

$$0 = \lim\limits_{x\to-\infty}(2/x^3) = \lim\limits_{x\to\infty}(2/x^3) = \lim\limits_{x\to-\infty}(1/x^3) = \lim\limits_{x\to\infty}(1/x^3)$$

?


distance between centres of two overlapping congruent circles If there are two overlapping congruent circles such that the area of intersection is 10% of the area of each circle, what is the distance between their centres in terms of the radius r cm?

Let $O$ and $O'$ be the centers and $P$ and $Q$ be the intersection points of both circles.

First, let's calculate the area of the circular segment $PQ$. We need the angle $\alpha=\angle POQ$.

Call $d=OO'/2$ and $r$ the radius of the circles. Then
$$\alpha=2\arccos\frac dr$$

The area of the circular segment is obtained substracting the triangle $OPQ$ to the circular sector $OPQ$, and is
$$\frac{\alpha r^2}{2}-d\sqrt{r^2-d^2}$$
being $\alpha$ expressed in radians (this makes the formulae clearer).

The area of intersection is twice the circular segment's. Therefore, you must solve this for $d$:

$$r^2\arccos\frac dr-d\sqrt{r^2-d^2}=0.05\pi r^2$$

I'd say that this is impossible to solve by algebraic methods, but any math program can do it with numerical methods, given the radius.

The equation becomes a bit cleaner if you call $k=d/r$ the ratio between $d$ and $r$. Dividing the equation by $r^2$ yields:
$$\arccos k-k\sqrt{1-k^2}=0.05\pi$$
Sadly it still can't be solved by algebraic methods.

PS: Don't forget that the arccos must be in radians, and that $d$ is the distance between centers halved.

Is it pi(rrr/5-rrrr/200)?
You assume one of the circles is xx+yy=rr while the other a circle with same radius but symmetric wrt x-axis. Assume one of the intersection points is (b,sqrt(rr-bb). Want b and then 2b is the required answer. The difference between the crcular sector  between the x-axis and the point (b,sqrt(rr-bb) and the triangle (0,0),(b,0),(b,sqrt(rr-bb)) is 1/40 .pi.rr.


What is worth of a stalk in red-blue Hackenbush?? I was studying about Red-Blue hackenbush from this link http://www.link.cs.cmu.edu/15859-s11/notes/Hackenbush.pdf

http://math.ucsd.edu/~wgarner/math168a/blueredhackenbush.htm this url shows a Red-Blue Hackenbush Calculator.

However I do not get the idea of worth of a stalk. What does worth of a stalk mean??

The worth (hereafter "value") of a Red-Blue Hackenbush stalk (sometimes known as a "Hackenstring", when written the RRBRB way) represents how many moves it gives to Blue (the player who can remove bLue edges is usually called "Left", but I won't follow that convention) when combined with other stalks. Since Hackenbush is a game you lose when it's your turn to move, the more moves you get the make, the safer you are.

The rest of this post is a long explanation of where the particular values all come from, which you can skip if you're not interested in how the above idea translates into all these weird numbers with denominators that are powers of 2 (called dyadic fractions/rationals).

Integer Values

For example, BBB is worth three moves to Blue in any combination. The reason is that Red (conventionally called "Right") can't touch this stalk, and Blue could move once in it to BB, then again to B, and then again to an empty stalk. Similarly, the value of an empty stalk is 0, and the value of n Bs in a row is just n.

Now, what about something like RRR? That's worth three moves to Red, and Blue isn't allowed to make a move on it. So should its value be 0? Well, RRR doesn't act like the empty stalk in combinations: If you play a two-stalk game with the empty stalk and BBB, then it's just like BBB, so the empty stalk doesn't change the fact that BBB gives Blue 3 moves. But if you play a two-stalk game with RRR and BBB (or a bigger game with those two stalks), then any move in one of those stalks can be mirrored in the other, so Blue's "extra moves" have been nullified.

If we use the plus sign to represent playing games with multiple stalks, then RRR+BBB is like an empty stalk (because of mirroring). If the value of BBB is 3, and the value of the empty stalk is 0, then we should have RRR+3=0, so that the value of RRR is -3. 3 moves for Red are like 3 negative moves for Blue.

In general, if a stalk is all one color, it has an integer value. n Rs is -n, the empty stalk is 0, and n Bs is n.

What is BR?

Let’s consider the game BR. If Red goes first, they can remove the R, leaving B=1. If Blue goes first, they can remove the B (and the R to the right of it) leaving “”=0. Since Blue can win by taking the B, this is better for Blue than the empty string - we should have BR>0. How does BR compare to B=1? Well, BR-B=BR+R. In that game, Red can move to B+R=1-1=0 (so Blue loses), and Blue can move to R=-1 (which Blue still loses). Thus, BR+R=BR-B is a game that Blue loses, so it's good for Red and we should have BR-B<0 so BR

We can get an even better understanding of the value of BR by comparing BR+BR with B=1. We must play BR+BR-B=BR+BR+R. If Blue goes first, they move to BR+R<0 and Blue loses. If Red goes first and takes the lone “R” then BR+BR>0 remains and Red loses. If Red goes first another R, then they leave BR+B+R=BR and Red loses. In any case, the person who goes first loses (if the other player plays perfectly). But when that happens, if you were to combine BR+BR+R with any other stalks, no one would want to move first in that piece, so it's like adding on the empty stalk. Therefore BR+BR+R=0, so BR+BR=B=1! These are pretty good reasons to call the value of BR 1/2, and that’s exactly what people do. (There's also a natural multiplication on Red-Blue stalks that I won't define, but it turns out that BB*X=X+X, so BB*BR=1, which is another reason to call it 1/2.)

Other fractions?

BR=1/2 and similarly, RB=-1/2. but what happens with more complicated stalks? Consider BRR; it’s clearly positive since Blue wins even if Red goes first, and it’s less than 1 since BRR+R is a win for Red. If you play BRR+BRR-BR=BRR+BRR+RB, then if Red goes first and they leave BRR+BR+RB=BRR>0, they lose. If Red goes first and they leave BRR+BRR>0, they lose. If Blue goes first and they leave BRR+RB then Red can respond with BR+RB=0 and they (Blue) lose. If Blue goes first and they leave BRR+BRR+R then Red can move to BRR+BR+R and Blue’s only moves are to BR+R (a win for Red) or BRR+R (also a win for Red). In all cases, the second player has a winning strategy, so BRR+BRR+RB is like the empty stalk, so BRR+BRR=BR. Thus, we can/should call BRR “1/4″. This pattern continues: BRRR=1/8, BRRRR=1/16, etc.

Let’s look at another family of positions, starting with BBR. Blue can take the middle B which the R is sitting on, but the bottom B intuitively seems independent of that top BR action. For this reason, I’d guess that BBR=B+BR. To test this, play BBR+R+RB. If Red goes first, they might take the first R, leaving BB+R+RB=B+RB>0, but they’d lose. If they take the third R, they’d leave BBR+R and Blue could respond with B+R=0 and Red would still lose. BBR+RB>BBR+R, so Blue would win if Red started by taking the other R, too. If Blue goes first, they can move to BBR+R+R

Finally, let’s look at one more family before I make a general claim. Consider BRB; this is potentially better for Blue than BR, but worse than B, so it’s somewhere in between 1/2 and 1. It actually turns out to be 3/4. We can prove that BRB+BRR=B by playing BRB+BRR+R. It’s easy to see (from previously known facts) that all of Blue’s starting moves result in negative games, so this is ≤0. If Red takes the first R, they leave B+BRR+R=BRR>0. If Red takes the third R (the second is worse) then Blue moves to 0 by leaving BR+BR+R. If Red takes the last R, Blue leaves BRB>0. (There is some advanced theory of games that helps to avoid these case analyses.) BRBB is even better for Blue, and equals 7/8, etc.

General Stalk Values

Since the negative of a basic stalk switches all of the Bs and Rs, and we know “”=0, it suffices to describe the values of strings starting with B. A string consisting only of n Bs just has the integer value n. Otherwise, suppose there are n Bs before the first R. Then the value turns out to be  $(n-1)+$the binary number you get by reading Bs as 1s and Rs as 0s after the first BR, with a 1 appended. This can be seen in: 10.110010 $1_{(2)}=$ BB BR BBRRBR.

  Note: The sections after "Integer Values" were hastily adapted from a
  blog post I wrote, so there may be typos/things that are unclear.

When you cut off a stalk, you remove the connection of other stalks to the ground.  The worth of a stalk is the value of that stalk and all those above it, which are dependent on that stalk.  The reason they start adding $\textbf{after}$ a change in color is because, as long as it is the same color, the opponent can cut from the bottom and benefit by $+n$.  However, with the addition of one of the opposite color, you can cut optimally cut from that stalk, making it a different value from the rest.  This value should be lower since you are removing less of the hackenbush game.  Like many numbers in game theory, the factor of $1/2$, to my knowledge, is arbitrary.


How to interpret $\nabla_x (f(u(x)))$? Is $\nabla_x (f(u(x)))= (\nabla f) (u(x)) \nabla u$ true? Let $u:\mathbb R^{n+1} \to \mathbb C$ be a function and $f:\mathbb C \to \mathbb C$ such that $f(z)=|z|^{\frac{4}{n-2}}z.$ Thus, composition of $f$ and $u$ is: $f(u)= |u|^{\frac{4}{n-2}}u.$

  My Question is: How should I interpret  $\nabla_x (f(u(x)))$? Is  $\nabla_x (f(u(x)))= (\nabla f) (u(x)) \nabla u$ true?

[Is this a standard notaions? My understnding for the gradient is: if $g:\mathbb R^n\to \mathbb C$ nice function then $\nabla g (x) = (\frac{\partial g(x)}{\partial x_1},..., \frac{\partial g(x)}{\partial x_n}), x=(x_1,...,x_n)\in \mathbb R^n$. Or am I missing something here?  But I do not knwo what is means $\nabla_x g$, which was used by the autors in the remark mentioned below]

Motivation: I have been trying to figure out  Remark 2.3 here See also this question

When you are starting with this work, it is best to look at the components.  Let $x= (x_1,x_2,...,x_n)$. $\nabla_x(f(u(x)))$ is an $n-$component vector.  The first component of $\nabla_x(f(u(x)))$ is $$\frac{\partial f}{\partial x_1}=\frac{d f}{du}\frac{\partial u}{\partial x_1}$$
The first derivative in this chain rule expression is not a partial derivative, since $f$ is a function of a scalar $u$ (so you won't have anything that looks like $\nabla f$).  Since the other components will be similar, you obtain
$$\nabla_x(f(u(x)))=(\frac{d f}{du}\frac{\partial u}{\partial x_1},\frac{d f}{du}\frac{\partial u}{\partial x_2},...,\frac{d f}{du}\frac{\partial u}{\partial x_n})$$
$$=\frac{df}{du} \nabla u$$

Try the chain rule for multivariate calculus: https://www.math.hmc.edu/calculus/tutorials/multichainrule/


Functional/Differential Equation In the midst of a calculation too long (and too irrelevant) to describe here, I've been forced to confront the following equation:

$$1-p-f(f(p))-f(p)f'(f(p))=0$$

Here $f$ is a differentiable function from the unit interval to itself, and the equation is supposed to hold identically in $p$.  How do I find all such functions $f$?

$\large\mbox{At least, two solutions.}$

Let's ${\rm f}\left(p\right) \equiv \mu p + \nu$.
$$
1 - p - \mu\left(\mu p + \nu\right) - \nu - \left(\mu p + \nu\right)\mu = 0
$$

$$
\left(1 - \mu\nu - \nu - \nu\mu\right)
+
\left(-1 - \mu^{2} - \mu^{2}\right)p
=
0
$$

$$
\mu = \pm\,{\sqrt{2} \over 2\,}\,{\rm i}\,,
\qquad
\nu
=
{1 \over 2\mu + 1}
=
{1 \over \pm\,\sqrt{2\,}\,{\rm i} + 1}
=
{1 \over 3}\,\left(1 \mp \sqrt{2\,}\,{\rm i}\right)
$$

$$
\color{#ff0000}{\large%
{\rm f}_{-}\left(p\right)
\color{#000000}{\ =\ }
-\,{\sqrt{2\,} \over 2}\,{\rm i}\,p
+
{1 \over 3}\,\left(1 + \sqrt{2\,}\,{\rm i}\right)\,,
\qquad
{\rm f}_{+}\left(p\right)
\color{#000000}{\ =\ }
{\sqrt{2\,} \over 2}\,{\rm i}\,p
+
{1 \over 3}\,\left(1 - \sqrt{2\,}\,{\rm i}\right)}
$$

$\large\mbox{Just trying to guess:}$

Let's $\quad x \equiv {\rm f}\left(p\right)$. Then,
$$
x{\rm f}'\left(x\right) + {\rm f}\left(x\right)
=
1 - p\left(x\right)\,,
\quad
x{\rm f}\left(x\right)
-
a{\rm f}\left(a\right)
=
x - a
-
\int_{a}^{x}p\left(x'\right)\,{\rm d}x'
$$

$$
{\rm f}\left(x\right)
=
{a\left[{\rm f}\left(a\right) - 1\right] \over x}
+
1
-
{1 \over x}\int_{a}^{x}p\left(x'\right)\,{\rm d}x'
$$


Refinement of triangles I have given the triangulation in black:

The task is to refine the two triangles on the top (left) in 4 congruent triangles each. I have done this in red colour.

In addition to that, there should not be any hanging nodes. I tried this using the violet colour. Is this correct, or is there any fault?

Thanks in advance for checking my draft.

Corrected version:

This is it maybe if I figured out your question correctly.


Special functions as representations of Lie Groups -The spherical harmonics $Y_{lm}$ are complete on $L^2(S^2)$.  They are also a representation of the (compact) Lie group $SO_3 (\mathbf{R})$.

-The functions $e^{i n x}$ are complete on $L^2([0,2\pi])$.  They are also a representation of the (compact) Lie group $U(1).$

My question is basically, how general is this phenomenon?  More specifically,

Bessel functions, Hermite polynomials, Legendre polynomials - do each of these represent some Lie group?  If so, what is it in each case?
Is there a nice example of some complete functions that represent a non-compact Lie group?
Is there a nice example of some complete functions that do not represent any Lie group at all?

Thanks!

The representations of $SO(3)$ are indexed by numbers $|\ell, m \rangle$ where $\ell \in \frac{1}{2}\mathbb{Z}$ and $|m| \leq \ell$ with $m - \ell \in \mathbb{Z}$.

The matrix elements of Wigner's small matrix give the Legendre polynomials:

$$ \langle \ell, 0 | e^{-i \theta J_z }|\ell, 0 \rangle = P_\ell(\cos \theta)$$

Here $J_z$ is a generator of the $\mathfrak{so}(3)$ Lie Algebra and the exponent is a rotation , $e^{-i \theta J_z } \in SO(3)$.


Inequality Involving $\int_V \left(\int_{B(x,\epsilon)} \eta_{\epsilon}(x-y) |f(y)|^p dy\right) dx $ I was reading the book "Partial Differential Equation" written by Lawrence C. Evans, coming up with a question.

On page 718, Evans wrote

  $$\int_V \left(\int_{B(x,\epsilon)} \eta_{\epsilon}(x-y) |f(y)|^p dy\right) dx \\ \leq \int_W |f(y)|^p \left(\int_{B(y,\epsilon)} \eta_{\epsilon}(x-y) dx\right) dy$$

Where $V,W,U$ are  open sets, $V\subset\subset W\subset \subset U$, and $\eta_{\epsilon}(x) := \frac{1}{\epsilon^n} \eta(\frac{x}{\epsilon})$, $\eta$ is the standard mollifier, $f \in L_{loc}^p(U)$

I want to ask why the inequality holds. I was trying to prove it so hard but I have no ideas.

This is an application of Fubini's theorem, followed by the fact that the integrand is positive and $V$ is a subset of $W$.

Since $V\subset \cup_{x\in V} B(x,\epsilon) \subset W$ if $\epsilon$ small enough,  we can show:
$\{(x,y): x\in V , y\in B(x,\epsilon)\} \subset \{(x,y):   y \in W, x\in B(y,\epsilon)\}.$

$\forall (x,y) \in \{(x,y): x\in V , y\in B(x,\epsilon)\}$, we know $x \in V$, then $y \in B(x,\epsilon)\subset \cup_{x\in V} B(x,\epsilon)\subset W$. And $y \in B(x,\epsilon)$ imples $x\in B(y,\epsilon)$. Therefore $(x,y) \in \{(x,y):   y \in W, x\in B(y,\epsilon)\}$, which proves "$\subset$".

Then by the above fact and the Fubini Thm,

$\int_V \int_{B(x,\epsilon)} (\cdot)dy\,dx = \int_{ \{(x,y): x\in V , y\in B(x,\epsilon)\}}(\cdot)dy\, dx \le  \int_{\{(x,y):   y \in W, x\in B(y,\epsilon)\}}(\cdot) dy\,dx = \int_{\{(x,y):   y \in W, x\in B(y,\epsilon)\}}(\cdot) dx\,dy = \int_W \int_{B(y,\epsilon)}(\cdot)dx\,dy$


If we know that the sequence of events $A_n$ converges, what can we say about the infinite union and intersection? Suppose that $\lim_{n\to\infty} A_n = A$. It is intuitive to me for the infinite intersection, that it is equal to $\bigcap_{n\geq1}A_n = A$. However, I am failing to see if there does indeed exist anything for the infinite union. Is there a nice result for that?

By $\lim_{n\to\infty}A_n=A$, I will assume you mean that the indicator functions ${\bf 1}_{A_n}$ converge pointwise to ${\bf 1}_A$. Then the statement
$$
\bigcap_{n\geq 1}A_n=A
$$
is not necessarily true: for example, you can make the intersection empty by setting $A_1$ to be the empty set and leaving all other terms in the limit unchanged (limits are unaffected by finitely many changes to the sequence).

$A \subseteq \bigcap_{n=1}^{\infty} A_n?$ No.

Suppose we have a measure space $(S, \Sigma, \mu)$ and measurable sets $A_1, A_2, ...$

'$\lim A_n = A$' is an abbreviation for

$$\limsup A_n = \liminf A_n = A$$

Definitions:

$$\limsup A_n = \bigcap_{m \ge 1} \bigcup_{n \ge m} A_n$$

$$\liminf A_n = \bigcup_{m \ge 1} \bigcap_{n \ge m} A_n$$

If $\omega \in \limsup A_n$, then $\forall m \ge 1, \exists n \ge m$ s.t.
$\omega \in A_n$ or:

$$\forall n \ge 1, \omega \in \bigcup_{m=n}^{\infty} A_n$$

If $\omega \in \liminf A_n$, then $\exists m \ge 1, \forall n \ge m$ s.t. $\omega \in A_n$ or:

$$\exists m \ge 1 s.t. \omega \in \bigcap_{n=m}^{\infty} A_n$$

In the latter case, we don't necessarily have $m=1$. Hence, it is not necessarily the case that

$$\omega \in \bigcap_{n=1}^{\infty} A_n \tag{*}$$

$A \supseteq \bigcap_{n=1}^{\infty} A_n?$ Yes.

Now obviously, if we have $(*)$, then we know that indeed that

$$\exists m \ge 1 s.t. \omega \in \bigcap_{n=m}^{\infty} A_n$$

namely $m = 1$ and hence

$$\omega \in \liminf A_n$$

Since $\liminf A_n \subseteq \limsup A_n$,

we have

$$\omega \in \limsup A_n$$

Hence

$$\omega \in \lim A_n = A$$


Shifting the index of $\sum_{n=2}^{\infty}a_n\cdot n\cdot (n-1)\cdot x^{n-2}$ I want to shift the index of
$$\sum_{n=2}^{\infty}a_n\cdot n\cdot (n-1)\cdot x^{n-2}$$
to start at $n=0$. If I expand this series I get:

$$a_2\cdot 2\cdot 1\cdot x^0+a_3\cdot 3\cdot 2\cdot x^1+\cdots$$

But if I start simply at $n=0$ without doing any modification, I get:

$$a_0\cdot 0\cdot (-1)\cdot x^{-2}+a_1\cdot 1\cdot0\cdot x^{-1}+a_2\cdot2\cdot1\cdot x^0+\cdots$$

which is the same as the series above, because the first $2$ terms are $0$, so I tought I could just write it like this:

$$\sum_{n=0}^{\infty}a_n\cdot n\cdot (n-1)\cdot x^{n-2}$$

but My book writes it as the following:

$$\sum_{n=0}^{\infty}(n+2)(n+1)a_{n+2}x^n$$

what's happening?

Both are correct. In the book, they just "shifted" the whole expression.

They just replaced $n$ with $n+2$.
So instead of starting with $n=2$ they start with $n=0$, but they use $n+2$ in the expression, so it is the same as just using $n=2$ in the original formula.

If you expand both series, you will see it is the same.

Nothing happens !

$\sum_{n=0}^{\infty}a_n\cdot n\cdot (n-1)\cdot x^{n-2}=\sum_{n=0}^{\infty}(n+2)(n+1)a_{n+2}x^n$


9th grade AMTI question $-$ $65$ bugs on a $9 \times 9$ board 65 bugs are placed at different squares of 9X9 square board. A bug in each moves to a horizontal or vertical adjacent square. No bug makes two horizontal or two vertical moves in succession. Show that after some moves, there will be at least two bugs in the same square.

Jack Frost's link contains a brief solution from v_Enhance at the Art of Problem Solving, but that answer was deleted by the owner. So here is an illustrated version:

There are $16$ red squares, so in four turns, at most $64$ bugs can visit a red square without colliding. But you should be able to convince yourself that each bug must visit a red square on every fourth turn. Hence, if there are no collisions, there can be at most $64$ bugs.

Remark: This argument shows that if there are $65$ bugs, then two bugs will collide after at most three moves; my comment to the OP shows that two moves are not enough.

This is a question based on topology of the bugs on the board.
So for that reason the bugs can be placed in any place on the board as the question mentions only that the bugs are placed on the different place on the board.
Now, Please refer the diagram.The Diagram of the board

In the diagram if one observes carefully. the board can be divided into boxes of 4X4.
The reason the boxes are of 4X4 is chosen is because 64+1=65. And 64 is a multiple of 4.
When the boxes are arranged in the 9X9 board as shown a corridor of 17 boxes are left in the corner perimeter. This is done to prove the question.
Now in all 4X4 boxes shown by mono color there are 4 bugs. These bugs will move in either clockwise or counter clockwise motion.

Just to give you an example. The Upper Left bug will come to the lower left, the lower left bug will move to the lower right, the lower right bug will go the upper right. and cycle will continue. This sort of ordered motion is done to symplify the problem
The bugs will never leave their individual boxes. This will not violate the rule of alternate successive motion.
Now we do some simple math to find out that only one bug is left. This bug can be placed in any place in the outer corridor. It is now easy to visualize that bug may move the first time. But the next time it cannot go to any empty box. (no horizontal motion if fist move is vertical or vice versa )
Therefore after some moves there will have to be at least 2 bugs in a block. In this case just after one move
Solution by Mechanical Engineering Student.


Help with understanding bond duration. I'm currently studying for a pure maths degree, and so have no background knowledge of bonds. I'm reading through some material regarding actuarial work, and came across the following definition of duration which I cannot wrap my head around:

The duration is the time before the average payments are made of the bonds held.

I've been googling the definition of duration, and understand it to be some sort of measure of the sensitivity of the bonds value to changes in the interest rates. Could someone give me a simple explanation of what the duration is, and the meaning of the above (what are the average payments?)
Thanks!

Consider a zero coupon bond-- it pays you 10 dollars next period.  What will you pay for it? It depends on the interest rate, or conversely, the interest rate is determined by the price of the bond.
$$P_1=\frac{10}{1+R}$$
If it pays you in 10 years,
$$P_{10}=\frac{10}{(1+R)^{10}}$$
So this bond has a maturity of 10 years, and note that $P_{10}$ is more sensitive to R than is $P_1$.  But if the bond has coupons, and along with paying you back 10  dollars in 10 years, it also pays you 1 dollar each year.  Duration is a way of saying that the 10 year coupon bond is less sensitive to changes in rates than a zero coupon bond, since
$$P_{c,10}=\frac{1}{1+R} +\frac{1}{(1+R)^2}+...\frac{10}{(1+R)^{10}}$$
It's like an average of the maturity of the different payments. That's how I've always considered it, anyways.

Duration is

One, how sensitive the price of bonds change relative to interest rate change
Suppose if we have a 3 year semi coupon bonds annual rate is 8% with a face value of $1000. If this bond has a duration of 2.5. It means that if there 1% interest rate increases, the price of bond will decrease 2.5%.

Two, it is also the time that you can collect your principals. Suppose you invest 1000 dollars for the bond above. It has maturity of 3 years. A duration of 2.5 means that in 2.5 years you can successfully collect ur principal $1000 back (because of the semi-annual coupons you get).


If $f(z)=(1+i)z+1$, what is $f^4$? Let $f(z)=(1+i)z+1$. Then $f(z)=\sqrt 2 e^{i\pi/4}z+1$ and thus $f=t\circ h\circ r$ where $t$ is the translation of vector $1$, $r$ the rotation of center $0$ and angle $\pi/4$ and $h$ the homothetic of parameter $\sqrt 2$. I found a fix point $z=\frac{-1}{2}+\frac{i}{2}$.

1) What if $f\circ f\circ f\circ f\circ f$ ? It looks to be an translation in the direction of $i$ and an homothetic of parameter 4, but is there an easy way to prove it (with out calculation) ?

2) Let $z_0=i$, $z_1=\sqrt 3+2i$ and $z_2=\sqrt 3-2i$, and consider the triangle $T=\Delta (z_0,z_1, z_3)$. Let $g_n=\underbrace{f\circ ...\circ f}_{n\ times}$. What is the smallest $n$ s.t. $$Area(g_n(T))\geq 100 Area(T)  \ \ ?$$

First, $Area(T)= 2\sqrt 3$. I think that $g_n$ is the same triangle with length of side bigger of $n\sqrt 2$, and thus, I would says that $$Area(g_n(T))=\frac{n4\sqrt 2\cdot n\sqrt 2\sqrt 3}{2}=4\sqrt 3n^2.$$
Therefore, $$4\sqrt 3n^2\geq 200\sqrt 3\implies n^2\geq 50\implies n\geq 8.$$
We conclude that the smallest $n$ is $n=8$. Is it correct ?

Hint: If $f(z)=az+b$, with $a\ne1$, then $f^n(z)=a^n z + b\dfrac{a^n-1}{a-1}$.

1) The linear part of $f^4$ is $-4z$, therefore, $f^4$ will be of the form $$f^4(z)=-4z+\tau,$$
therefore, it will be an homothetic transformation with scale factor $4$ composed by a point reflexion (you can also say that it's an homothetic transformation with scale factor $-4$) and a translation of vector $\tau$. To find $\tau$, you unfortunately need to do some calculation (i.e. compute explicitly $f^4$).

2) Yes, correct.


How can solve these system of differential equation? Assume that $U(\rho,z)$ and $V(\rho,z)$ functions are given.
We have the system of equations for these functions in the form
$$
\frac{\partial^2 U}{\partial\rho^2}+\frac{1}{\rho}\frac{\partial U}{\partial\rho}+\frac{\partial^2 U}{\partial z^2} = 0 \ ,
$$
and
$$
\frac{\partial V}{\partial\rho} = \rho\left[\left(\frac{\partial U}{\partial\rho}\right)^2+\left(\frac{\partial U}{\partial z}\right)^2 \right]\ ,
\qquad \frac{\partial V}{\partial z} = 2\rho\frac{\partial U}{\partial\rho}\frac{\partial U}{\partial z}\ ,
$$
Can we use separation of the variable in the first equation?
How about in the second and the third equations?

I will start the solution problem over here to show you exactly what I mean. The super-position principle, as you mentioned exists in linear systems. Hence, for the first part:

$ \frac{\partial^2 U}{\partial \rho^2}+\frac{ 1}{ \rho}\frac{\partial U}{\partial \rho}+\frac{\partial^2 U}{\partial z^2} = 0 $

with $U = R(\rho)Z(z)$

we get:

$ R''+(1/\rho) R' = -\lambda^2 R$

$ R = a_1 Y_0(\lambda \rho) + a_2 J_0(\lambda \rho) $

Here, $J_0$ and $Y_0$ are Bessel functions of the first and second kind, respectively.

$ Z'' = -\lambda^2 Z$

$ Z = b_1 \sin(\lambda z)+b_2 \cos(\lambda z)$

Next, as I tried to mention, $V$ depends only on $U$. There is no need of anything like the superposition principle:

$ \frac{\partial V}{\partial z} = 2\rho R'Z'(RZ)$

$ V = 2\rho R'R\big(\int Z'Z\partial z\big) +f(\rho) = \rho R'RZ^2 +f(\rho)$

$ \frac{\partial V}{\partial \rho} = \rho R''RZ^2 + R'RZ^2 + \rho (R')^2 Z^2 +f'(\rho)$

We also have:
$ \frac{\partial V}{\partial \rho} = \rho ((R')^2Z^2+R^2(Z')^2)$

Then you can use the quantities $a_1,a_2,b_1,b_2,\lambda$ and $f(\rho)$ to make sense of the equality:

$\rho R''RZ^2 + R'RZ^2 + \rho (R')^2 Z^2 +f'(\rho) = \rho ((R')^2Z^2+R^2(Z')^2)$

$\rho (-R'/\rho -\lambda^2 R)RZ^2 + R'RZ^2 + \rho (R')^2 Z^2 +f'(\rho) = \rho ((R')^2Z^2+R^2(Z')^2)$

$ f'(\rho) = \rho R^2(\lambda^2 Z^2+(Z')^2)$

Probably you should check again, but I believe that $ (\lambda^2 Z^2+(Z')^2) $ is always constant (i.e. it does not depend on $z$). Therefore, you will get:

$ f'(\rho) = \rho R^2(\text{some constant in }b_1,b_2,\lambda)$

and you can solve for $V$

Snifkes, Thank you very much for your solution. Actually, I have solved this problem in the prolate coordinates,
$$
\rho=\sqrt{(x^2-1)(1-y^2)} \ , z=xy
$$
and we get
$$
\left[\left(x^2-1\right)U_x\right]_x +
\left[\left(1-y^2\right)U_y\right]_y=0 \ ,
$$
$$
V_x = \frac{1-y^2}{x^2-y^2}\left[x(x^2-1)U_x^2-x(1-y^2)U_y^2
-2y(x^2-1)U_xU_y\right] \ ,
$$
$$
V_y = \frac{x^2-1}{x^2-y^2}\left[y(x^2-1)U_x^2-y(1-y^2)U_y^2
+2x(1-y^2)U_xU_y\right] \ ,
$$

Solution of the first equation is
$$
U(x,y)=\sum_{n=0}^{\infty}(a_n P_n(x)+b_n Q_n(x))(c_n P_n(x)+d_n Q_n(x))
$$
But my question was different: You are making separation of variable in the function $U(\rho,z)$, finally it depend on $\lambda$ which is separation constant and you need to sum over lambda like: $U= \sum_{\lambda} R_{\lambda}(\rho)Z_{\lambda}(z)$ - which is correct, but other two equations are not linear. As far as I know it is impossible to use summation over with separation constant for non linear equations. What do you think about it?


Show that $c(t)_{c(t)}$ and the tangent vector to $c$ at $t$ is perpendicular. Let $c(t)=(c_1(t),...,c_n(t)):[0,1] \to \Bbb R^n$ be a differentiable curve in $\Bbb R^n$ such that $|c(t)|=1$ $\forall t \in [0,1]$. Define tangent vector $v$ of $c$ at $t$ as $c_*((e_1)_t)=((c_1)'(t),...,(c_n)'(t))_{c(t)}$.

Show that $c(t)_{c(t)}$ and the tangent vector to $c$ at $t$ is perpendicular.

Now I was taking dot-product, $c(t).v=c_1(t)c_1'(t)+...+c_n(t)c_n'(t)$, so why it is zero? Can anyone help me with this arguement?

  Please don't tell $0=(T⋅T)′=T′⋅T+T⋅T′=2T⋅T′$.

I don't quite understand your last line. You know $|c(t)| = 1$ for all $t$, in other words $c(t) \cdot c(t) = 1$ for all $t$. The dot product is a bilinear map $\mu : \mathbb{R}^n \times \mathbb{R}^n \to \mathbb{R}$; for such a bilinear map, if $f, g : [0,1] \to \mathbb{R}^n$ are two differentiable maps, then the derivative of $[0,1] \to \mathbb{R}$, $t \mapsto \mu(f(t), g(t))$ is given by
$$\mu(f'(t), g(t)) + \mu(f(t), g'(t)).$$
I advise you to check this directly from the definition of the derivative and of bilinearity, if you're not convinced. Just compute $\frac{1}{h} \bigl( \mu(f(t+h), g(t+h)) - \mu(f(t), g(t)) \bigr)$ cleverly and take the limit as $h \to 0$.

So in your situation, the derivative of $t \mapsto c(t) \cdot c(t)$ is $c'(t) \cdot c(t) + c(t) \cdot c'(t) = 2 c(t) \cdot c'(t)$ by symmetry. But you know that $c(t) \cdot c(t) = 1$ for all $t$, so this derivative is zero and $c(t) \cdot c'(t) = 0$, i.e. $c(t)$ and $c'(t)$ are orthogonal.


convergence of Fourier coefficient Let $f(x)$ continuous function on $[-\pi,\pi]$ and $ c_n:=\int _{-\pi}^{\pi} f(x)\frac{e^{-inx}}{\sqrt{2\pi}}dx$. Suppose $f(-\pi)=f(\pi)$ and for all $x_0 \in [-\pi, \pi]$, there is $r>0$ s.t., $\forall x \in [-\pi, \pi]$ with $|x-x_0| \leq r, f(x)=\sum _{m=0}^{\infty} a_m(x-x_0)^m$

Then,  there exists $K >1$ and constant $C$ s.t., $|c_n|\leq \frac{C}{K^{|n|}}$

I guess I use Cauchy's integral expression, but I cannot take $K>1$.

Given the $2\pi$-periodicy of $f$ and the condition $f(0)=f(2\pi)$, it is well defined a continuous function $$F :S^1\rightarrow\mathbb{C}, e^{it}\mapsto f(t),$$ where $S^1$ is the unit circle in the complex plane. Then the condition $$\forall x_0\in[-\pi,\pi], \exists r>0, \exists(a_n)_{n\in\mathbb{N}}\subset\mathbb{C}, \forall x\in[-\pi,\pi],\left({\left(|x-x_0|\le r\right)\implies \left(f(x)=\sum_{m=0}^\infty a_m(x-x_0)^m\right)}\right) $$ can be worked out to $$\forall z_0\in S^1, \exists r>0, \exists(a_n)_{n\in\mathbb{N}}\subset\mathbb{C},  \forall z\in S^1\cap D_r(z_0), F(z)=\sum_{m=0}^\infty a_m(z-z_0)^m.$$

So, for each point of the unit circle, there exists a ball centered in that point where $F$ coincides with an holomorphic function. By the identity principle for holomorphic functions (or, say, analytic continuation), you can extend $F$ to an holomorphic function defined on an open set surrounding the unit circle. So there exists an annulus of convergence $A_{r,R}$ with $r<1<R$ where the Laurent series of $F$ centered in $0$ converges, say: $$\forall z\in A_{r,R}, F(z)=\sum_{n=-\infty}^\infty c_nz^n.$$ Then the series $$\sum_{n=0}^\infty c_nz^n$$ has a radius of convergence of at least $R$, while the series $$\sum_{n=0}^\infty c_{-n}z^n$$ has a radius of convergence of at least $\frac{1}{r}$. Define $H:=\min(R,\frac{1}{r})$. Then:
$$1\ge\limsup_{|n|\rightarrow\infty}(|c_n|H^{|n|})^{1/|n|}=H\limsup_{|n|\rightarrow\infty}|c_n|^{1/|n|}$$
So:
$$\limsup_{|n|\rightarrow\infty}|c_n|^{1/|n|}\le \frac{1}{H}.$$
Then, if $1<K<H$, easily follows that there exists a constant $C>0$ such that $$\forall n\in\mathbb{Z}, |c_n|\le\frac{C}{K^{|n|}}.$$
Finally, noticing that the coefficients $(c_n)_{n\in\mathbb{Z}}$ are also the Fourier coefficients of $f$, you have got the desidered result.


The graphs of $x^2+y^2+6x-24y+72=0$ and $x^2-y^2+6x+16y-46=0$ intersect at four points.Find the sum of the distances of these four points The graphs of $x^2+y^2+6x-24y+72=0$ and $x^2-y^2+6x+16y-46=0$ intersect at four points.Compute the sum of the distances of these four points from the point $(-3,2).$

$x^2+y^2+6x-24y+72=0$ is a circle and $x^2-y^2+6x+16y-46=0$ is a hyperbola.These cut at four points.
$x^2+y^2+6x-24y+72=0$ becomes $(x+3)^2+(y-12)=9^2$.
$x^2-y^2+6x+16y-46=0$ becomes $(x+3)^2-(y-8)^2=-9$.
When i solved these two equations to find the points of intesection,$y=10\pm \sqrt{41}$ and $x$ is $-3\pm\sqrt{36\pm 4\sqrt{41}}$
Now it is very difficult to find the sum of distances of these points from $(-3,2)$.Answer is $40$ given in the book.

We don't have to find the coordinates of the four points.

Since
$$(x^2+y^2+6x-24y+72)-(x^2-y^2+6x+16y-46)=0$$
$$\Rightarrow 2y^2-40y+118=0$$

we can set the four points as $(-3\pm\alpha,\beta),(-3\pm\gamma,\omega)$ where $$\beta+\omega=-(-40)/2=20,\qquad\beta\omega=118/2=59.$$

Now, noting that
$$(-3+\alpha)^2+\beta^2+6(-3+\alpha)-24\beta+72=0\Rightarrow \alpha^2+(\beta-2)^2=20\beta-59$$

the sum of the distances can be represented as
$$2\sqrt{\alpha^2+(\beta-2)^2}+2\sqrt{\gamma^2+(\omega-2)^2}=2\left(\sqrt{20\beta-59}+\sqrt{20\omega-59}\right)\tag1$$

Then,
$$\begin{align}&\left(\sqrt{20\beta-59}+\sqrt{20\omega-59}\right)^2\\&=20(\beta+\omega)-2\cdot 59+2\sqrt{20^2\beta\omega-20\cdot 59(\beta+\omega)+59^2}\\&=20\cdot 20-2\cdot 59+2\sqrt{20^2\cdot 59-20\cdot 59\cdot 20+59^2}\\&=20^2\end{align}$$

The result follows from $(1)$.

note: for any point of 4, the distance to $(-3,2)$ is $\sqrt{(-3\pm\sqrt{36\pm 4\sqrt{41}}-(-3))^2+(10\pm \sqrt{41}-2)^2}=\sqrt{36\pm 4\sqrt{41}+(8\pm \sqrt{41})^2}=\sqrt{(10\pm\sqrt{41})^2}=10\pm\sqrt{41}$

so it is easy to see the final answer.


Knowing pdf of a random variable, make a transformation, how to find the moments after the transformation? The random variable X has the exponential probability density $f_X(x)=\alpha e^{-\alpha x}, x\geq0$

A new random variable is given by $Y=e^{-X}$. How can I obtain an expression for the n-th moment of Y.

The $n$-th moment of $Y=e^{-X}$ is $\langle Y^n\rangle=\langle e^{-nX}\rangle$. So you can just do it in the distribution of $X$ without having to transform the distribution to $Y$. The moment is

$$\langle Y^n\rangle=\int_0^\infty\alpha e^{-(n+\alpha)x}dx=\frac{\alpha}{\alpha+n}.$$

The distribution of $Y$ can also be found. Since $X\in[0,\infty)$, $Y=e^{-X}\in(0,1]$. The probability

$$P(Y<y)=P(e^{-X}<y)=P(X>-\ln y)=y^\alpha,\quad y\in[0,1].$$

Then the density is $p(y)=\alpha y^{\alpha-1}$. From here one also obtains

$$\langle Y^n\rangle=\int_0^1\alpha y^{n+\alpha-1}dy=\frac{\alpha}{\alpha+n}.$$


Value of the integral $\int_{a}^{a}f(x)dx?$ I am confused about the value of the integral $\int_{a}^{a}f(x)dx,$ is it $0$ or not always? In Riemann integration case (i.e. $f$ is bounded ) its zero by definition. What about other cases for example $\int_{0}^{0}\frac{1}{x}? $ Is it zero? Please help. Thanks a lot.

It's always zero...except that there are some conventions in which this notation is used to denote something that's not zero.

So let me clarify "always": for the Riemann integral, and for the  Lesbegue integral with respect to the standard measure on the real line, and for functions $f: \mathbb R \to \mathbb R$, it's always zero. This property is very useful, for it helps ensure that
$$
\int_a^b + \int_b^c = \int_a^c
$$
in general.

So why do I say "except..."?

Well, for instance, if $f$ is a "delta function" (which isn't really a function, but instead a distribution), then this integral can denote something nonzero (or at least
$$
\int_{0^{-}}^{0^{+}} f(x) dx
$$
can be nonzero).

And if your measure "dx" on the real line happens to assign positive measure to the point $0$, then again the integral can denote something nonzero. But it'd be pretty odd to use "dx" to denote such a measure.

You can say like this $$\int_0^0{1\over x}dx=\lim_{c\to0+}\int_{-c}^{c}{1\over x}dx=\lim_{c\to0+}\left[\ln|x|\right]_{-c}^c=\lim_{c\to 0+}\{\ln|c|-\ln|-c|\}=0$$


Application of Unique Factorisation Theorem in Proof CONTEXT: Proof question made up by uni math lecturer

Suppose you have $x+y=2z$ (where $x$ and $y$ are consecutive odd primes) for some integer $z>1$, and that you need to prove that $x+y$ has at least three prime divisors (that don't have to be distinct).

Is it sufficient to say that, according to the unique factorisation theorem for integers, since $z$ can be expressed as a product of primes (and we already have the factor of $2$ which is prime), we know that $x+y$ has at least three prime divisors?

Or, would you need to do further working to show that $z=ab$ for some primes $a$ and $b$?

$z$ is strictly between $x$ and $y$, hence, not a prime.

Is it sufficient to say that, according to the unique factorisation
  theorem for integers, since $z$ can be expressed as a product of
  primes (and we already have the factor of $2$ which is prime), we know
  that $x+y$ has at least three prime divisors?

No, that's not enough. In fact, how do you know $z$ is not prime?

To prove $z$ is not prime, consider the fact that $y=x+2$, and therefore, $2z=x+x+2=2(x+1)$. Now, since $z=x+1$, can you conclude that $z$ is not prime?


The Four Fours puzzle using only $+$, $-$, $\times$, $\div$, $\sqrt{}$, and concatenation. Can we make $19$? I know the classic Four Fours puzzle, which asks to create whole numbers using exactly four "$4$"s combined via certain operations (arithmetic, concatenation, grouping, factorials, etc.), the exact roster of which can vary.

I have tried this specific version of it:
You can use only the basic arithmetic operations ($\times \div + - $), square root ($\sqrt{}$), and concatenation (eg, "$44$" is allowed).

  How far you can go? I cannot go further than $18$. I am stuck with $19$ and does not seem that there is a solution for it.

Let $\#$ denote concatenation. That is, $4\#3 = 43$. Then

$$(4 \# \sqrt 4) / \sqrt 4 - \sqrt4 = 42 / 2 - 2 = 19.$$

Not sure if you can use parentheses, but you would just have to perform concatenation first above all else.

According to Wikipedia, 19 has no solutions using only the operations you're describing.

Source: https://en.wikipedia.org/wiki/Four_fours#Solutions


A simpler method to prove $\log_{b}{a} \times \log_{c}{b} \times \log_{a}{c} = 1?$ This is the way I think about it:

$
1 = \log_aa = \log_bb = \log_cc \\~\\
\textbf{Using the ‘change of base rule':} \\
\log_{a}{b} = \frac{\log_{b}{b}}{\log_{b}{a}}, \hspace{0.3cm} \log_{b}{c} = \frac{\log_{c}{c}}{\log_{c}{b}}, \hspace{0.3cm} \log_{c}{a} = \frac{\log_{a}{a}}{\log_{a}{c}}\\~\\
\rightarrow \log_{b}{a} = \frac{\log_{b}{b}}{\log_{a}{b}}, \hspace{0.3cm} \log_{c}{b} = \frac{\log_{c}{c}}{\log_{b}{c}}, \hspace{0.3cm} \log_{a}{c} = \frac{\log_{a}{a}}{\log_{c}{a}}\\~\\
\rightarrow \large\log_{b}{a} \times \log_{c}{b} \times \log_{a}{c}\rightarrow\\~\\
\rightarrow\frac{\log_{b}{b}}{\log_{a}{b}} \times \frac{\log_{c}{c}}{\log_{b}{c}} \times \frac{\log_{a}{a}}{\log_{c}{a}} \overset? = 1\\
\rightarrow \frac{1}{\log_{a}{b} \times \log_{b}{c} \times \log_{c}{a}} \overset? = 1\\
\rightarrow \log_{a}{b} \times \log_{b}{c} \times \log_{c}{a} = 1\\
\rightarrow \log_{a}{b} \times \log_{b}{c} \times \log_{c}{a} \overset? = \log_{b}{a} \times \log_{c}{b} \times \log_{a}{c}\\
\rightarrow \frac{\log_{a}{b} \times \log_{b}{c} \times \log_{c}{a}}{\log_{b}{a} \times \log_{c}{b} \times \log_{a}{c}} = 1 \\
\rightarrow \frac{\log_{a}{b}}{\log_{b}{a}} \times \frac{\log_{b}{c}}{\log_{c}{b}} \times \frac{\log_{c}{a}}{\log_{a}{c}} = 1\\~\\
\small \text{------ Using the ‘change of base rule' again ------} \\~\\
\Large \therefore \hspace{0.2cm}\log_{b}{a} \times \log_{c}{b} \times \log_{a}{c} = 1
 $

Any other methods?

Thanks in advance.

Well it is much simpler. By definition
$$\log_b(a) = \frac{\ln(a)}{\ln(b)}, \quad \log_c(b) = \frac{\ln(b)}{\ln(c)}, \quad log_a(c) = \frac{\ln(c)}{\ln(a)}$$

where $\ln$ is the logarithm in base $e$.

Your identity directly follows.


Why does Euclid's HCF algorithm work? I saw this Khan Academy video on the visualization of Euclid's algorithm. The problem was to find the HCF of $32$ and $12$.

At $6:53$, the instructor reduced the original problem to  $12$ and $8$ as:
$$
\left\{
\begin{array}{c}
32 = 12 + 12 + 8 \\
12
\end{array}
\right.
$$
Why did he leave the numbers $9, 10$, and $11$ and directly jumped to $8$? How can I be sure that the numbers won't be a factor of $32$ and $12$?

He tried to express $32$ as a sum of $12$'s. This is why he "skipped" $9,10,11$. In addition to that note than none of $9,10,11$ divide 32. Hence, he reduced the problem to finding th HCF of 8 and 12, because it would work for 12 and 32 as well, since 12 is a divisor of 32.  So In fact, he didnt "jump" to 8, he just divided 32 with 12, and 8 was the remainder.
Indeed he writes $32 = 12 + 12 + 8$.
He says "if you found a break that can cover 12 and the you found another break for 8 with the condition it covers 12 too (5:45)". And that is of course, 4.

If you do some divisibility checks, you can see that because $3+2 = 5$ and $5 \nmid 9$. Additionally, neither of the numbers end in zero, or consist of the same digit (note this check only works for testing divisibility by eleven for numbers less than 99). I assume that's what's happened here.


Scaling the normal distribution? I might just be slow (or too drunk), but I'm seeing a conflict in the equations for adding  two normals and scaling a normal. According to page 2 of this, if $X_1 \sim N(\mu_1,\sigma_1^2)$ and $X_2 \sim N(\mu_2,\sigma_2^2)$, then $X_1 + X_2 \sim N(\mu_1 + \mu_2, \sigma_1^2 + \sigma_2^2)$, and for some $c \in \mathbb{R}$, $cX_1 = N(c\mu_1, c^2\sigma_1^2)$.

Then for $X \sim N(\mu,\sigma^2)$, we have $X + X = N(\mu + \mu,\sigma^2 + \sigma^2) = N(2\mu,2\sigma^2)$, but also $X + X = 2X = N(2\mu,2^2\sigma^2) = N(2\mu,4\sigma^2)$ ? Ie, the variances disagree.

edit: Oh, am I mistaken in saying that $2X = X + X$? Is the former "rolling" $X$ just once and doubling it while the latter "rolls" twice and adds them?

On the first page of the cited document, $X_1$ and $X_2$ were previously defined to be two (distinct) independent, identically distributed random variables. For your purposes, the "identically distributed" part is not important, but the "independent" part is.

On the second page, where $X_1$ and $X_2$ are considered to be normal variables, there's still the assumption that they're independent. Possibly this could have been stated more clearly, but in context this assumption makes sense.

When you consider $2X = X + X$, you are not dealing with two independent variables.
The two "copies" of $X$ are correlated (in fact, as correlated as any two variables can be).
The formula for the sum of two independent normal variables therefore does not apply.

You make a mistake assuming that $X+X = 2X$ like in arithmetic. This does not hold for random variables. As a simple counterexample take $X$ to be a Bernoulli trial with succes probability $0.5$. If $X+X=2X$ then you cannot flip two coins and get heads on one and tails on the other. Summing random variables is given by a convolution.


Pairs of integer pairs with same lcm, gcd and mean The problem is to find all pairs of two distinct pairs(up to permutation) of integer(!) numbers $(a, b)$ and $(c, d)$ s.t. $$\operatorname{lcm}(a, b) = \operatorname{lcm}(c, d)$$
$$\gcd(a, b) = \gcd(c, d)$$ and
$$\frac{a + b}{2} = \frac{c + d}{2}$$

It is easy to show that if both LCM and GCD are equal, then two pairs have the same product in absolute value and the same sum AND the same GCD.

There was a similar question on MathSE about natural numbers, answer to which is that such distinct pairs dont exist: Prove two distinct pairs of natural numbers with these properties do not exist

One such pair is ((-6; 35), (14;15)): their sums and LCMs and GCDs are all equal pairwise. How to find all such pairs?

Given $\operatorname{lcm}(x,y)\cdot\operatorname{gcd}(x,y)=|xy|$, we have $\operatorname{lcm}(a,b)\cdot\operatorname{gcd}(a,b)=\operatorname{lcm}(c,d)\cdot\operatorname{gcd}(c,d)$, so $|ab|=|cd|$. Given $\frac{a+b}{2}=\frac{c+d}{2}$, we have $c=(a+b)-d$. When substituted into $|ab|=|cd|$, this gives the two quadratics $d^2-(a+b)d\pm ab=0$. These have the two pairs of solutions $d=a,b$ (when $c,d$ are trivially a permutation of $a,b$) and $d=\frac{(a+b)\pm\sqrt{\Delta}}{2}$, where $\Delta$ is the discriminant $a^2+6ab+b^2$.

In the nontrivial case, $d$ is integral if $(a+b)\pm\sqrt{\Delta}$ is integral (and even), therefore $\Delta$ is a square. Hence, for some integer $k$, the triple $(a,b,k)$ is a solution of the 3-variable, 2nd-degree Diophantine equation $a^2+6ab+b^2=k^2$. This can be solved with a method analogous to finding Pythagorean triples: consider the intersections of the hyperbola $a^2+6ab+b^2=1$ with the line $a=m(b-1)$. The first intersection is $(1,0)$ and the second intersection is guaranteed to be rational when $m$ is integral, which paves the way for integer solutions when clearing denominators. The second solution is $b=\frac{(m+1)(m-1)}{m^2+6m+1},a=\frac{-2m(3m+1)}{m^2+6m+1}$, which we can substitute into $a^2+6ab+b^2=1$ and multiply through by $(m^2+6m+1)^2$, to give solutions for $(a,b,k)$, and by extension $(a,b,c,d)$ when substituted into the earlier equations. So the set of nontrivial solutions up to multiples and permutations is, for $m\in\mathbb{Z}$

$a=-2m(3m+1)\\
 b=(m+1)(m-1)\\
c=-(3m+1)(m+1)\\
d=2m(1-m)$

For example, $m=-5$ gives the solution $(-140,24,-56,-60)$, which is a multiple and permutation of the solution $(-6,35,14,15)$ mentioned by the asker. This encompasses all solutions.

Given $\gcd(a,b)=1$, since $ab=cd$, assume $$(-a,c)=x, (-a,d)=y, (b,c)=z, (b,d)=w$$
we have $$a=-xy, b=zw, c=xz,d=yw$$ so that $zw-xy=xz+yw$ or $(z-y)w=x(y+z)$
so
$x\mid z-y$ and $w\mid y+z$;
$z(w-x)=y(x+w)$
$z\mid x+w$ and
$y\mid w-x$.

Now let's assume that $w$ is largest (x,y,z,w are all positive integers) among x,y,z,w
since $w\mid y+z$, we must have $w=y+z$
so next we got $z=x+y$
So given any integer number pairs $(x,y)$, we have $$x=x,y=y,z=x+y,w=2y+x$$
$$a=-xy, b=(x+y)(x+2y), c=x(x+y), d=y(x+2y)$$ is a result.
And given any another integer $f$, we have
$$a=-xyf, b=(x+y)(x+2y)f, c=x(x+y)f, d=y(x+2y)f$$ is a result


Is this $0v=0$ proof correct? I'm taking my first linear algebra course at university and recently I've been introduced to vector spaces. Now, the teacher has asked us to prove that $0v=0$ for any $v$ using only the definition of a vector space. I've came up with a proof, but I'm unsure whether it's correct or not, as I'm not familiar with proofs in Math.

$$
\begin{align}(1+1)v &= 2v \\
(1+1)v-2v &= 2v-2v\\
[(1+1)-2]v &= 2(v-v) \\
0v &= 0
\end{align}
$$

If this is correct I can go on proving other properties of vector spaces. Any feedback appreciated!

EDIT: Ok, taking into account José's answer, I came up with this one
$$
(1+0)v=v \\
1v+0v=v \\
-v+v+0v=-v+v \\
0v=0
$$

Your proof has some problems. What is $-2v$? Is it $-(2v)$ or is it $(-2)v$? If it is $-(2v)$, then you could jump directly from $2v-2v$ to $0$, but then how do you know that $(1+1)v-2v=\bigl((1+1)-2\bigr)v$? And if $-2v$ is $(-2)v$, then how do you know that $2v-2v=2(v-v)$?

Here another method :
  where $θ$ is the zero vector and $a$ is a scaler
 We know that $av$ is a scaled vector and $v + θ = v$
So : $$av + θ = av $$
Hence :$$θ = (a-a)v$$
       $$θ = 0v$$


Stabilizers of permutations The problem is I need to list all elements in the stabilizer of the permutation $\sigma=(123)$ in $S_5$.

I know that the stabilizer of a permutation $\sigma$ is the subgroup $$Stab_{{S_n}}(\sigma) = \left\{ x\in S_n : x\sigma x^{-1}=\sigma \right\}$$

So that I need to find a permutation $x$ such that $$x\sigma x^{-1} = x(123) x^{-1} =(123)$$

Then I used the fact that $$x(123) x^{-1} =(x(1)x(2)x(3)) = (123)$$
Then I realize I'm just getting identities.. am I on the right track? Is it possible to have more than one value of x? Why is it so?

There are definitely more than one element stabilizing $(123)$.  Note that you can rewrite $(123) = (312) = (231)$, which should indicate more possible stabilizers using your method.

Furthermore, you should check that if $x$ fixes $(123)$ then $x \circ (45)$ also fixes $(123)$.

Hmm. $\text{Stab}(x) = \{g \in G: g \cdot x = x\}$. So we are letting elements in $S_5$ act on $\sigma=(123)$ by regular function composition. Let $\tau \in S_5$ then $\tau \cdot \sigma = \sigma \iff \tau(1)=1,\tau(2)=2$ and $\tau(3)=3$. It would seem as though we could then take any permutation of the form $\tau=(xy)$ where $x,y = 4$ or $5$. This is how you get $(45)$. Now do this for $(312)$ and $(231)$.


Bourbaki influence on the current mathematical research I know that the bourbakism had an influence in the academic world and an impact in secondary school and university. There are remarks in this Wikipedia dedicated to Nicolas Bourbaki.

  Question. Does Nicolas Bourbaki have an influence on the current mathematical  research? In what branchs/fields of mathematics, and why? Feel free if you need it to answer this as a reference request, and I try to find and read those facts. Many thanks.

I am asking from an informative point of view.

The main influence on current research of Bourbaki seems to be the Bourbaki Seminar. Here is a link to the program of the next seminar, on January 13, 2018.


Is this positive term series convergent? My series is:

$\frac{1}{1+2^{-1}}\ +\frac{1}{1+2^{-2}}\ +\ldots$

I see my $nth$ term is $\frac{1}{1+2^{-n}}$

How do I test for its convergence?

A necessary condition for the convergence of $\sum_n a_n$ is that $\lim_n a_n=0$. In your case, $\lim_n a_n=1$ so the series doesn't converge. In fact you can show that it diverges to $+\infty$ (every term in your sequence is greater than $\frac{1}{2}$).

Yes the series convergent to 1.
Series 1/2 power n divergent but 1/2 power n converges to infinity and 1/1+2 power n converges to 1.


Finding equivalent of an iterative sequence Define $(u_n)_{n\in\mathbb{N}^*}$:

$$u_1 =1 , u_{n+1}=1+\dfrac{n}{u_n}$$

Find an asymptotic equivalent of $u_n$ when $n\to+\infty$.

I guess that the answer should be $\sqrt{n}$, but I couldn't prove it...

You could try to show by induction that $\sqrt{n}<u_{n}<1+\sqrt{n}$ or something like that.


Infinite totally ordered set with all non-empty subsets containing both endpoints? I'm wondering if there are any infinite sets which can be totally ordered in such a way that all non-empty subsets contain a maximal and minimal element.

It is clear to me that this is impossible in discrete orderings and in the standard bounded dense orderings like $[0,1]$, but is it known that this is impossible in general?

This is indeed impossible. This is because any linear order $L$ has either an infinite increasing sequence or an infinite decreasing sequence (or both), and either provides a counterexample.

Technically, the observation above requires choice (think about a Dedekind-finite infinite set of reals). However, we do not in fact need choice for this problem:

Suppose $L$ is a linear order. Let

$$D=\{x\in L: \text{ there are finitely many points } <x\}$$

If $D$ has a maximal element $y$, then $\{z\in L: y<z\}$ is infinite (since $L$ is infinite) and has no least element, so $D$ can't have a maximal element. If $D$ is nonempty, then $D$ is a counterexample to the desired property.

So $D=\emptyset$. But then $L$ doesn't have a minimal element, so $L$ itself is a counterexample.

Concisely:

  Suppose $L$ is a linear order. If $L$ has a least element, then the set of elements with finitely many smaller elements has no maximum.

Similarly, if $L$ has a greatest element, then the set of elements with finitely many bigger elements has no minimum.

Incidentally, going back to the observation at the top of the answer: while this isn't the best way to solve this problem since it requires choice, it leads to a really interesting question. Namely, the set $\{\omega, \omega^*\}$ (where $L^*$ denotes the reverse of $L$) is a basis for the infinite linear orders: a linear order is infinite iff it contains a copy of $\omega$ or of $\omega^*$. Moreover, this is clearly the smallest basis for the infinite linear orderings. So we can ask:

  What about, say, uncountable linear orderings?

This turns out to be a very deep problem in set theory; the biggest positive result is due to Justin Moore:

  Suppose the Proper Forcing Axiom (PFA) holds. Then $\{\omega_1,\omega_1^*,S, C,C^*\}$ is a basis for the uncountable linear orders whenever $C$ is an arbitrary Countryman line and $S$ is an $\aleph_1$-dense set of real numbers.

(Here a Countryman line is a linear order of cardinality $\aleph_1$ whose square is a countable union of chains; the existence of such an object is not immediate, but provable in ZFC alone.)

Interestingly, the choice of Countryman line is meaningful - different Countryman lines can be non-isomorphic - but the choice of $S$ turns out not to be: assuming PFA, any two $\aleph_1$-dense sets of reals without endpoints are isomorphic (this follows from work of Baumgartner, I'm not sure where it was first explicitly stated).

Prove that a partial order with a top and bottom, for which
every not empty subset contains the top and the bottom, has
exactly one point.


Double negation vs. law of excluded middle? I think I am operating under classic logic here, or propositional logic? Not sure if these are the same thing or what all the options are here.

When I see how to prove double negation elimination, it uses law of excluded middle. When I see how to prove law of excluded middle, it uses double negation elimination.

Are these two things considered equivalent and axiomatic or is there some other proof that handles both? If it is an axiom, what other axioms exist?

In a propositional logic, the sentences are built out of primative abstract sentences $A_1,A_2,\ldots $ by combining them with connectives. The usual set of connectives are $\{\lnot,\land,\lor,\to, \leftrightarrow\},$ though sometimes a smaller set is used (or a smaller set is taken as primative and the others are defined off of them). Sometimes also there are other connectives too (e.g. in modal propositional logic), but since we're only really focusing on classical and intuitionistic here, which only use the ones I mentioned.

So things like $A_1\to A_2$ and $A_1\lor \lnot A_3$ would be sentences. Here we would write the law of excluded middle as $$ \mathcal A \lor \lnot\mathcal A$$ where it's understood that we can substitute any given sentence for $\mathcal A.$ Such a substitution is called an instance of LEM, so, for instance $ (A_1\land A_2)\lor \lnot(A_1\land A_2)$ is an instance of LEM.

The thing that makes the system "propositional" as opposed to, say, a predicate logic is that the basic building blocks are these primative abstract sentences $A_i.$ In predicate logic, the atomic sentences themselves have structure, and we have variables and constants representing individuals, and relation symbols representing relations between individuals (like equality), and quantifiers. For instance, you've probably seen things like $$\forall x_1 \forall x_2((x_1<x_2)\land (x_2<x_1)).$$ This is a sentence in predicate logic (that is also false in its usual interpretation as a sentence about numbers).

Predicate logic is related to propositional logic cause we can still write things like LEM $$ \mathcal A \lor \lnot\mathcal A,$$ where now we will substitute sentences of predicate calculus in for $\mathcal A$ to generate instances. So we can talk about the LEM in either context. From now on, I'll stick to propositional logic.

Now, we've talked about the grammar, but we still haven't said what these sentences are for. Of course we know that we intend $A_1\to A_2 $ to mean what "if $A_1$, then $A_2$" informally means and $A_1\land A_2$ to mean "$A_1$ and $A_2$", etc. We want is to be able to prove certain sentences valid, for instance, clearly we would like to be able to show somehow $A_1\to A_1$... this is clearly something that "should" be valid.

So we need a deductive system, which consists of rules for taking valid sentences and producing new valid sentences, as well as axioms, sentences we assume are valid. There are tons of different ways to do this: the main systems fall into three categories: Hilbert-style systems, natural deduction, and sequent calculus. No time to explain what these are, of course, other than to say a natural deduction system will be most convenient for our purposes here, so that's the category you're seeing here, though it will usually be presented much more formally (and carefully).

In this, we will take as a primative axiom scheme that $\mathcal A\to \mathcal A$ is valid where anything is substituted for $\mathcal A.$ This will be the only axiom, and everything else will be rules. (EDIT: I actually don't think this axiom scheme is even necessary since every instance of it follows from the $\to$ introduction rule described below.)

For instance, we have an introduction rule for $\lor$ that if a sentence $\phi$ is valid then $\phi\lor \psi$ and $\psi \lor \phi$ are valid for any sentence $\psi.$ Then we have an elimination rule, that if from the assumption of validity of a sentence $\phi$, we can derive that $\psi$ is valid, and separately, if from the assumption of the validity of a sentence $\phi',$ we can derive that same $\psi$ is valid, and $\phi\lor \phi'$ is valid, then $\psi$ is valid.

We can define similar introduction and elimination rules for $\land$... I won't mention what they are for lack of space: they're easier than the ones for $\lor$ and you can probably figure out what they are. Then there is an introduction rule for $\to$: if starting with a hypothetical assumption of $\phi,$ we can derive that $\psi$ is valid, then $\phi\to \psi $ is valid. And then the elimination rule is that if both $\phi$ and $\phi\to \psi$ are valid, then $\psi$ is valid.

Finally, we come to negation. First off, it is most convenient to not take negation as primative, but rather to take special primative sentence $\bot$ which intuitively represents a false statement or a contradiction. Then we define negation as $\lnot \mathcal \phi = \mathcal \phi\to \bot.$ Then we have an elimination rule for $\bot$ that from $\bot$ we can infer any sentence, which formalizes the notion that from an absurdity, we can derive anything. There is no introduction rule for $\bot.$

Notice this gels with our intuitive idea of contradiction, because if we have $\phi$ and $\lnot \phi,$ which recall is just an abbreviation for $\phi\to \bot,$ then by $\to$ elimination, we infer that $\bot$ is valid.

These rules give a version of the deductive system for intuitionistic propositional logic. Any statement whose validity can be proven by these rules is called intuitionistically valid. As I mentioned before, there are many different types of deductive systems that all produce the same valid sentences... I have only given one example.

Now, here's where we finally get down to answering your question. We can convince ourselves that we can neither prove $A_1 \lor \lnot A_1$ nor $\lnot\lnot A_1 \to A_1$ from the above rules. (It's a bit more complicated than this if we want to be careful, but the basic idea for LEM is that in order to get $A_1\lor\lnot A_1,$ we'd need to use the introduction rule for $\lor,$ which requires that we have either $A_1$ or $\lnot A_1$ valid. And it's clear, since $A_1$ is just some abstract sentence, there's no way of proving either of these things.)

So the schemes $\mathcal A \lor \lnot \mathcal A$ and $\lnot\lnot \mathcal A \to \mathcal A$ are not generally valid under intuitionistic logic. However, we can show that by taking either one of these schemes as an axiom, we can derive every instance of the other. So LEM and double negation elimination are equivalent over intuitionistic propositional logic. Adding either of these as an axiom results in classical propositional logic.

Two basic equivalences in classical logic are:

Double Negation

$$P \Leftrightarrow \neg \neg P$$

Implication

$$P \rightarrow Q \Leftrightarrow \neg P \lor Q$$

Thus, we have:

$$\neg P \lor P \Leftrightarrow P \rightarrow P \Leftrightarrow \neg \neg P \rightarrow P$$

Does that help?


What are Some Tricks to Remember Fatou's Lemma? For a sequence of non-negative measurable functions $f_n$, Fatou's lemma is a statement about the inequality

$$\int \liminf_{n\rightarrow \infty} f_n \mathrm{d}\mu \leq  \liminf_{n\rightarrow \infty}(\int f_n \mathrm{d} \mu)$$

or alternatively (for sequences of real functions dominated by some integrable function)

$$\limsup_{n\rightarrow \infty}(\int f_n \mathrm{d} \mu) \leq \int \limsup_{n\rightarrow \infty} f_n \mathrm{d}\mu$$

I keep forgetting the direction of these two inequalities. I know that using the concepts repeatedly is the best way to remember them.

But I am interested about learning intuitive tricks that people use to quickly remember them.

(For instance, to remember the direction of Jensen's inequality, I just picture a convex function and a line intersecting it.)

I like to think of the following pictures. The first two are $\int f_1$ and $\int f_2$ respectively, but even the smaller of these is larger than the area in the third picture, which is $\int \inf f_n$. Of course, Fatou's lemma is more subtle since we're talking about the limit infimum rather than just the minimum, but for the purpose of intuition this helps to make sure the inequalities go the right way.

Write the statement of the lemma on a credit-card sized piece of paper and carry it around for a month.


Bounds on $\sum\limits_{k=1}^n \frac{\sin(k)}{k}$ $\sum\limits_{k=1}^n \frac{\sin(k)}{k}$ converges as $n$ increases, to a limit of $\frac12(\pi-1) \approx 1.0708$
Empirically, it seems to be bounded by about $\frac12(\pi-1)  \pm \frac{1.043}{n}$, as shown in the chart below.
What is the precise value of this $1.043$ term?
Added: I think the three answers from Oliver Diaz, Gary, and Gabriel Romon, plus the comments, have demonstrated that the bounds are $$\frac12(\pi-1)  \pm \frac1{2\sin(1/2)}\frac{1}{n}$$ not only asymptotically but also as tight actual bounds; $\frac1{2\sin(1/2)}\approx 1.0429148214667441$

created in R with
n <- 1:11000
plot(n, cumsum(sin(n)/n), ylim=c(1.07,1.0715), xlim=c(1,10000), pch=46)
abline(h=(pi-1)/2, col="red")
curve((pi-1)/2 + 1.043/x, from=1, to=max(n), add=TRUE, col="red")
curve((pi-1)/2 - 1.043/x, from=1, to=max(n), add=TRUE, col="red")

This is a twice the bound of the OP:
Suppose $c_n\searrow0$.
Fix $M\in\mathbb{N}$ and for $n\geq N$ let
$$P_n(x)=\sum^n_{k=M}e^{ikx}=\frac{e^{iMx}-e^{i(n+1)}}{1-e^{ix}}=e^{iMx}\frac{1-e^{(n-M+1)ix}}{1-e^{ix}}=\frac{\sin\big((n-M+1)x/2\big)}{\sin(x/2)}e^{i(n-M)x/2}$$
From this, we get that
$$ S_n(x)=\sum^n_{k=M}\sin kx =\frac{\sin\big((n-M+1)x/2\big)}{\sin(x/2)}\sin\big((n-M)x/2\big)
$$
Then by summation by parts
$$\begin{align}
\sum^N_{n=M} c_n e^{inx} &=c_NP_N(x) +\sum^{N-1}_{n=M}(c_n-c_{n+1})P_n(x)\\
\sum^N_{n=M} c_n \sin nx & = c_NS_N(x) +\sum^{N-1}_{n=M}(c_n-c_{n+1})S_n(x)
\end{align}$$
Then
$$
\begin{align}
\Big|\sum^N_{n=M} c_n e^{inx}\Big| &\leq c_N|P_N(x)| +\sum^{N-1}_{n=M}(c_n-c_{n+1})|P_n(x)|\\
\Big|\sum^N_{n=M} c_n \sin nx\Big| &\leq c_N|S_N(x)| +\sum^{N-1}_{n=M}(c_n-c_{n+1})|S_n(x)|
\end{align}
$$
For $P_n$ and $S_n$ we have the estimates
$$\begin{align}
|P_n(x)|&= \big|\sin\big((n-M+1)x/2\big)\big|\csc(x/2)|\leq  |\csc(x/2)| \\
|S_n(x)|&=\Big|\sin\big((n-M+1)x/2\big)\sin\big((n-M)x/2\big)\Big||\csc(x/2)|\leq|\csc(x/2)|
\end{align}$$
for all $x\neq0$ and $n$,
\begin{align}
\Big|\sum^N_{n=M} c_n e^{inx}\Big|&\leq \Big(c_N +\sum^{N-1}_{n=M}(c_n-c_{n+1})\Big)|\csc(x/2)|\leq c_M |\csc(x/2)|\tag{1}\label{one}\\
\Big|\sum^N_{n=M} c_n \sin nx\Big|&\leq \Big(c_N +\sum^{N-1}_{n=M}(c_n-c_{n+1})\Big)|\csc(x/2)|\leq  c_M|\csc(x/2)|\tag{2}\label{two}
\end{align}
For the OP's series
$$
\sum^\infty_{n=1}\frac{\sin nx}{n}=\frac{1}{2i}\sum_{|n|\geq1}\operatorname{sign}(n)\frac{e^{inx}}{n}=\frac12(\pi-x),\qquad 0<x<2\pi
$$
From  \eqref{one} or \eqref{two} we obtain bound
$$\Big|\sum^{N-1}_{n=1}\frac{\sin nx}{n}-\frac{\pi-x}{2}\Big|=\Big|\sum^\infty_{n=N}\frac{\sin nx}{n}\Big|\leq \frac{1}{N}|\csc(x/2)|
$$
For $x=1$, we get twice the OP's bound. I ignore at the moment how to improve the bound in \eqref{one} by a factor of $1/2$ which is what the OP is observing.

Comment: The value of this sum is indeed (pi-1)/2. This is confirmed by noting that the function f(x) = pi-x on (0, pi), extended oddly to (-pi, 0), and then extended to all real numbers periodically and with the usual convention, has Fourier series = sum(sin(nx)/n; n=1,2,...). The extended function is continuous except at integer multiples of pi. Hence, we have equality at x=1, so the sum in question = f(1) = (pi-1)/2.
For more details, see:
Michael W. Ecker, Combining Mathematics and Technology to Sum Certain Series, Mathematics and Computer Education, Winter 2013, Vol. 47, No. 1, p. 19-23.


Question about open sets in the product topology in R^N Hi guys I have a statement that may be true or false. I believe it is false but I cannot prove or even show by example my claim.

If $U$ is open in $\Bbb R^\Bbb N$ then $U= \prod _{n=1} ^\infty U_n$ where each $U_n$ is open in $\Bbb R$.

I think this is false because I know that even for $X \times Y$ not every open set is of the form $U \times V$. Where $U$ and $V$ are open in $X$ and $Y$ respectively.Can someone help me here?

The statement is very false (for the product topology). Suppose the $U_n$ are non-empty open subsets of $\mathbb{R}$. Then $U = \prod_{n+1}^\infty U_n$ is open in the product topology iff for all but finitely many $n$, $U_n = \mathbb{R}$.

Proof: pick $(x_n) \in U$, which can be done (axiom of countable choice, formally) as all $U_n$ are non-empty. Then there is a finite subset $F$ of $\mathbb{N}$ and finitely many open set $O_n \subseteq \mathbb{R}$ for $n \in F$ such that $O = \prod_{n=1}^\infty O_n$ is a basic open product set (so $O_n = \mathbb{R}$ for $n \notin F$) and $x \in O \subseteq U$. Then for $n \notin F$, let $p \in \mathbb{R}$ be arbitrary. Then the point $y_m = x_m, m \neq n$, $y_n = p$ is in $O$ (as $x$ is, essentially, and we modify $x$ at a coordinate where $O_n = \mathbb{R}$), so $y \in U$, so in particular $y_n = p \in U_n$. As $p$ was arbitrary, $U_n = \mathbb{R}$, and this holds for all $n \in F$, so we are done.

There is another topology on $\mathbb{R}^\mathbb{N}$ that has exactly all those $U = \prod_n U_n$ (no restrictions of the all but countable type) as a base, which is, as we have seen now, stricly larger than the product topology, called the box topology. The product topology has much better properties (it's a connected metrisable space in this case, and separable etc.) while in the box topology this product is not even known to be normal in ZFC (IIRC). And it's certainly not connected, first countable, separable etc. in the box topology.

It's true that the statement as it stands already fails for finite products (the open unit circle is not of the form $U_1 \times U_2$ in the plane), but there at least we have that all open sets are (countable even) unions of such product open sets. In the infinite product topology, most of such sets are not even open. But again, if we apply the condition that all but finitely many $U_n$ equal the whole reals, we do get all usual product topology base elements, and we can write all open sets in $\mathbb{R}^\mathbb{N}$ as (countable) unions of those again.

You are correct, the statement is false. Consider some simple examples, such as the unit disc in $\mathbb{R}^2$; there is no way to express this as a product of two open sets in $\mathbb{R}$.

You can prove this explicitly by considering what sets you would have to pick in order to to include all the points on the disc. For example, you know that $(1, 0)$ and $(-1, 0)$ need to be included, so by considering the first entry, you have to have $U_1\subseteq (-1, 1)$, and similarly for $U_2$. However,  you immediately know that this won't work, as you are now including points like $(1, 1)$ in the product $U_1\times U_2$, even though this sits outside the unit circle.

There's a topological fact (sometimes given as a definition, sometimes a theorem, depending on the spaces you're looking at) that every open set in a product topology is a union of products of open sets - though this might be infinite!


Calculate the distance of the point $(1,1,1,1)$ from the subspace $W$. This is a duplicate question.

Let $W$=span$\left(\frac{1}{\sqrt{2}}(0,0,1,1),\frac{1}{\sqrt{2}}(1,-1,0,0)\right)$ be a subspace of the Euclidean space $\mathbb{R^4}$. Then the square of the distance from the point $(1,1,1,1)$ to the subspace $W$ is equal to_____?

So this question is already been asked and has an accepted answer here.
But I didn't understand the answer fully. To be honest, I didn't get the idea at all. The answer is talking about some distance $r_{AA_w}$ I don't know nothing how to get that.
So by this question, I want to know about the concept behind the distance of a point to a subspace. How to calculate this distance and what is the idea behind this? If someone can explain this to me then it would be really helpful. Thanks.

Start with simple cases: What’s the distance from a point to a line? How about to a plane? After a bit of experimentation, you might hit upon the idea that it’s the minimum distance between that fixed external point and any point on the line/plane. Simple geometric considerations show that this minimum distance is obtained when we measure it in a direction perpendicular to the line/plane, and so the nearest point to the fixed external point is the latter’s orthogonal projection onto the line/plane (labeled $\pi A$ in the diagram).

If we subtract this projection from the original point, we get a vector that’s orthogonal to the line/plane (called the orthogonal rejection from the line/plane), so the distance we’re looking for is equal to the length of this difference vector.

One can prove in various ways that this holds in general, namely, that the minimum distance from a point to an $m$-dimensional subspace lies along a direction that’s perpendicular to it. We’ll just take that as given. So, to compute this distance, you project the point onto the subspace and find the length of the difference between that projection and the original point.

There are many ways to construct this projection, depending on what you’re starting with. For this problem, you’ve got an orthonormal basis for $W$ (verify that for yourself), so you can use what you might recognize as a step in the Gram-Schmidt process, which computes successive orthogonal rejections from ever-growing subspaces. So, letting $\mathbf w_1$ and $\mathbf w_2$ be the two basis vectors of $W$, and $\mathbf v=(1,1,1,1)^T$, you would compute $\mathbf v-(\mathbf v\cdot \mathbf w_1)\mathbf w_1-(\mathbf v\cdot \mathbf w_2)\mathbf w_2$ and then compute the norm of this vector.

Here is the general idea behind this problem ...
Instead of working in $\mathbb{R^4}$, consider a more general euclidean vector space $E$ (that is a real vector space, equipped with an inner product (aka dot product)).
Let us denote by $\left<a\mid b\right>$ the scalar product of $a$ by $b$.
The norm of a vector is defined by $\Vert x\Vert=\sqrt{\left< x\mid x\right>}$.
The distance between two vectors $x$ and $y$ is $\Vert x-y\Vert$.
Consider a finite dimensional subspace $W$ of $E$ and $x\in E$.
The question is to minimize the distance from $x$ to the vectors of $W$.
It can be proved that :

there exist a unique $p\in W$ such that :
$$\Vert x-p\Vert=\inf\{\Vert x-w\Vert;\;w\in F\}$$

and that $p$ is obtained as the orthogonal projection of $x$ on $F$, that is :

$$p=\sum_{k=1}^n\left<x\mid e_k\right>\,e_k$$

where $(e_1,\cdots,e_n)$ denotes any orthonormal basis of $W$.
As an example, you could calculate the square of the distance of the map $x\mapsto x^2$ to $W=\mathrm{span}\left(x\mapsto 1,\,x\mapsto x\right)$. Here $E$ is the space of all continuous functions from $[0,1]$ to $\mathbb{R}$ and the inner product is defined by :
$$\forall (u,v)\in E^2,\left<u\mid v\right>=\int_0^1u(t)v(t)\,dt$$
The correct answer is $\frac{1}{180}$


Lebesgue measure of any line in $\mathbb{R^2}$. What is the Lebesgue measure of a line in $\mathbb R^2$? I am guessing that this zero. But I couldn't prove it rigorously. Please help.
From this can I conclude that any proper subspace of $\mathbb R^n$ has measure zero?

If the line is
$$
\ell=\{(x,0): x\in\mathbb R\},
$$
then for every $\varepsilon>0$, we have
$$
\ell\subset \bigcup_{k\in\mathbb Z}I_k^\varepsilon,
$$
where
$$
I_k^\varepsilon=[k,k+1]\times[2^{-|k|-2}\varepsilon,-2^{-|k|-2}\varepsilon].
$$
But
$$
m_2(\ell)\le\sum_{k\in\mathbb Z} m_2(I_k)=\sum_{k\in\mathbb Z} 2^{-|k|-1}\varepsilon=\varepsilon.
$$
Thus $m_2(\ell)<\varepsilon$, for every $\varepsilon>0$, and hence $m_2(\ell)=0$.

Any other straight line in $\mathbb R^2$ is obtained by a rigid motion of $\ell$, and rigid motion does not change the Lebesgue measure of a set.

Indeed, every proper linear subspace (or more generally, every proper hyperplane)
of $\mathbb R^n$ has zero $n-$dimensional Lebesgue measure.

Without loss of generality  we can suppose that over line $l$ has the form $l=\mathbb R \times 0$.

This follows from the translation- and rotation invariance of the lebesgue-measure.
(We can move our line such that it contains zero. Then we can rotate it)

We write $l=\cup_{k=0}^{\infty}([-k,k]\times 0)$

Now we obtain:

$\lambda_2(l)=\lambda_2(\cup_{k=0}^{\infty}([-k,k]\times 0))\leq\sum_{k=0}^{\infty}\lambda([-k,k]\times 0)=0 $


Prove $\int_1^\infty \frac{f(x)}{x}$ diverges Let $f$ be a continuous periodic function on $\mathbb{R},$ such that $0 \not\equiv f \geq 0.$
Prove $\int_1^\infty \frac{f(x)}{x}$ diverges.

My thoughts:
I tried applying the Limit Comparison Test with $\frac{1}{x}:$
$$\lim_{x\to \infty}\frac{\frac{f(x)}{x}}{\frac{1}{x}}=\lim_{x\to \infty} f(x)$$
Continuity and periodicity of $f$ implies $\lim_{x\to \infty}f(x)$ does not exist, so I got stuck.
Next I tried the Comparison Test with $\frac{1}{x},$ but couldn't manipulate the inequality to achieve $\frac{1}{x}\leq \frac{f(x)}{x}.$
Any help appreciated.

If $f$ has period $1$ then
$$\int_1^\infty\frac{f(x)}x\,dx=
\sum_{n=1}^\infty\int_n^{n+1}\frac{f(x)}x\,dx
=\sum_{n=1}^\infty\int_0^1\frac{f(t+1)}{n+t}\,dt
=\int_0^1 f(t+1)\sum_{n=1}^\infty\frac1{n+t+1}\,dt$$
and this is a divergent series. With rather more
fiddling around, one can do general period in a similar
fashion.

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

  Lets $\ds{T > 0}$ the $\ds{\mrm{f}}$-period.

With $\ds{R > 0}$:
\begin{align}
\int_{1}^{1 + R}{\mrm{f}\pars{x} \over x}\,\dd x & =
\int_{1}^{1 + \left\lfloor{R/T}\right\rfloor T + \braces{R/T}T}{\mrm{f}\pars{x} \over x}\,\dd x
\\[5mm] & =
\int_{1}^{1 + T}{\mrm{f}\pars{x} \over x}\,\dd x +
\int_{1 + T}^{1 + \left\lfloor{R/T}\right\rfloor T}{\mrm{f}\pars{x} \over x}\,\dd x +
\int_{1 + \left\lfloor{R/T}\right\rfloor T}^{1 + \left\lfloor{R/T}\right\rfloor T + \braces{R/T}T}{\mrm{f}\pars{x} \over x}\,\dd x
\\[5mm] & =
\bracks{\int_{1}^{1 + T}{\mrm{f}\pars{x} \over x}\,\dd x +
\int_{1}^{1 + \braces{R/T}T}{\mrm{f}\pars{x} \over x  + \left\lfloor{R/T}\right\rfloor T }\,\dd x} +
\bbox[15px,#ffe]{\ds{\int_{1 + T}^{1 + \left\lfloor{R/T}\right\rfloor T}{\mrm{f}\pars{x} \over x}\,\dd x}}
\end{align}

\begin{align}
\bbox[15px,#ffe]{\ds{\int_{1 + T}^{1 + \left\lfloor{R/T}\right\rfloor T}{\mrm{f}\pars{x} \over x}\,\dd x}} & =
\sum_{n = 1}^{\left\lfloor{R/T}\right\rfloor - 1}\int_{1 + nT}^{1 + nT + T}{\mrm{f}\pars{x} \over x}\,\dd x =
\sum_{n = 1}^{\left\lfloor{R/T}\right\rfloor - 1}\int_{1}^{1 + T}{\mrm{f}\pars{x} \over x + nT}\,\dd x
\\[5mm] & =
{1 \over T}\int_{1}^{1 + T}\mrm{f}\pars{x}
\sum_{n = 0}^{\left\lfloor{R/T}\right\rfloor - 2}{1 \over n + 1 + x/T}\,\dd x
\\[5mm] & =
{1 \over T}\int_{1}^{1 + T}\mrm{f}\pars{x}
\sum_{n = 0}^{\infty}\pars{%
{1 \over n + 1 + x/T} - {1 \over n + \left\lfloor{R/T}\right\rfloor + x/T}}
\,\dd x
\\[5mm] & =
{1 \over T}\int_{1}^{1 + T}\mrm{f}\pars{x}
\pars{H_{\left\lfloor{R/T}\right\rfloor + x/T - 1} - H_{x/T}}\,\dd x\qquad
\pars{~\substack{\ds{H_{z}:\ Harmonic}\\ \ds{Number}}~}
\end{align}

  Note that
  $\ds{H_{\left\lfloor{R/T}\right\rfloor + x/T - 1} \sim \ln\pars{\left\lfloor{R \over T}\right\rfloor + {x \over T} - 1}\quad\text{as}\quad R \to \infty}$. What do you conclude with the above relations ?.


uniqueness of solutions of $ax=b$ and $ya=b$ in a semigroup . Suppose $G$ is a semigroup in which every equation of the form $ax=b$ or $ya=b$ has a solution. Does this solution have to be unique?

This is the counter example @Rebecca noted. In fact, we see that in some cases we don't find any solutions and this happens cause there are more than one solutions in other case.

  gap> a:= Transformation([2, 3,3]);;
       b:= Transformation([ 3,2,3]);;
       s:= Semigroup(a, b);;
       e:=Elements(s);;
       for k in [1..3] do
           for j in [1..3] do
              for i in [1..3] do
              if e[j]*e[i]=e[k] then Print("k:=", k,"   ", "j:=",j, "   ",i,"\n");
              fi;
              od;
           od;
        od;

        k:=1   j:=1   2
        k:=2   j:=2   2
        k:=3   j:=1   1
        k:=3   j:=1   3
        k:=3   j:=2   1
        k:=3   j:=2   3
        k:=3   j:=3   1
        k:=3   j:=3   2
        k:=3   j:=3   3

$x$ has to be $a^{-1}b$ and $y$ has to be $ba^{-1}$. They are unique of course.


Proof of the affine property of normal distribution for a landscape matrix The widely used/mentioned/assumed affine property of multivariate normal distributions says that:

Given a random vector $x \in R^N$ with a multivariate normal distribution -- $x \sim N_x(\mu_x, \Sigma_x)$ -- then the random vector $y = Ax + b$ obtained by applying an affine/linear transformation to $x$ also has a normal distribution --> $y \sim N_y(A\mu_x+b, A\Sigma_x A^T)$

The above property is easy to prove if $A$ is an $N \times N$ matrix by writing $x = A^{-1}(y-b)$ and substituting it into $N_x(\mu, \Sigma)$ as shown below:

\begin{aligned}
p_y(y) & \propto p_x(A^{-1}(y-b)) \\
& \propto exp\{-0.5 \times (A^{-1}(y-b)-\mu_x)^T\Sigma_x^{-1}(A^{-1}(y-b)-\mu_x)\}\\
& = exp\{-0.5 \times (y - (A\mu_x + b))^T A^{-T}\Sigma_x ^{-1}A(y - (A\mu_x + b))\}\\
& = exp\{-0.5 \times (y - (A\mu_x + b))^T (A \Sigma_x A^T)^{-1}(y - (A\mu_x + b))\}\\
&\sim N_y(A\mu_x+b, A \Sigma_x A^T)
\end{aligned}

My questions are the following:

Does the affine propert hold true even if  A is a landscape $M \times N$ matrix with $M < N$ ? (most textbooks/lecture-notes say so and many papers assume this before deriving other things)
If the affine property is true, how do you prove it? because when A is a landscape $M \times N$ matrix with $M < N$ you cannot compute A^{-1} and hence you cannot express the random vector $x$ as  $x = A^{-1}(y-b)$

One should approach this through characteristic functions. Recall that $X$ is normal $N(\mu,\Sigma)$ if and only if, for every deterministic vector $t$ of size $N\times1$,
$$
E(\mathrm e^{\mathrm it'X})=\mathrm e^{\mathrm it'\mu-t'\Sigma t/2},
$$
where $t'$ denotes the transpose of $t$. For every $(A,b)$ of compatible sizes, if $Y=AX+b$, one gets
$$
E(\mathrm e^{\mathrm it'Y})=\mathrm e^{\mathrm it'b}E(\mathrm e^{\mathrm it'AX})=\mathrm e^{\mathrm it'b}E(\mathrm e^{\mathrm is'X}),
$$
where $s'=t'A$. Since $s=(t'A)'=A't$, one sees that
$$
E(\mathrm e^{\mathrm it'Y})=\mathrm e^{\mathrm it'b}\mathrm e^{\mathrm is'\mu-s'\Sigma s/2}=\mathrm e^{\mathrm it'b+\mathrm it'A\mu-t'A\Sigma A't/2}.
$$
The RHS is the characteristic function of the normal distribution $N(b+A\mu,A\Sigma A')$, and this is enough to identify the distribution of $Y$ as such. Note that no inverse or pseudo-inverse is involved and that this applies to $Y$ of any dimension.

Yes it does hold for a $M \times N$ matrix

You can use the generalized inverse in place of the normal inverse, where the original inverse is now the left pseudo-inverse, such that

$A^{-1}_{left}A=I$ (Property 1: Not that this inverse is not commutative)

Applying $A_{left}^{-1}$ to both sides of $y=Ax+b$, we get

$A^{-1}_{left}y = A^{-1}_{left}(Ax+b) =  A^{-1}_{left}Ax + A^{-1}_{left}b = x + A^{-1}_{left}b$

Isolating x then gives us

$x = A^{-1}_{left}(y-b)$

which is similar to the $A^{-1}(y-b)$ which you have used in your proof above.

If you were to trace through your proof (left as an exercise), now using the $A^{-1}_{left}$ instead of $A^{-1}$, along with property 1 above, you should arrive at the final form as you have derived.


Prove that if $f'(x) > g'(x)$, $f(a) \ge g(a)$ then $f(x) > g(x)$ in $(a,b]$ There's a similar question here: If $f(a) = g(a)$ and $f'(x) < g'(x)$ for all $x \in (a,b)$, then $f(b) < g(b)$ however in the original answer the conditions are a bit different as well as the solution uses Mean value theorem which we didn't learn.

  $f(x)$ and $g(x)$ are continuous in $[a,b]$ and differentiable in $(a,b)$. $\quad f(a) \ge g(a)$ and $f'(x) > g'(x)$ in $(a,b)$. Prove that $f(x) > g(x)$ in $(a, b]$.

Let $E(x) = f(x) - g(x)$.

If $E(a) = E(b)$ then by Rolle theorem there must be a $c$ such that $E'(c) = 0$.

But $E'(x) = f'(x) - g'(x) > 0$ for $a<x<b$ therefore $E'(x) \neq 0$. Therefore $E(a) \neq E(b)$.

Because $E(x)$ is continuous in $[a,b]$ there must be an absolute minimum and maximum for $E(x)$ in $[a,b]$. According to max/min theorem possible max/min can be on the edges of an interval, where the derivative is $0$ or where derivative doesn't exist. Because $E(x)>0$ and differentiable in $(a,b)$ the max/min has to be on the edges.

$E'(x) > 0 \Rightarrow$ $E(x)$ is ascending for $a<x<b$ therefore $E(a)$ = min, $E(b)$ = max.

Because $f(a) \ge g(a)$ then $E(a) \ge 0$. Because $E(a)$ is the absolute min from there $E(b) >E(x)>E(a)$ for $a<x<b$. Of course there can't be any $x_1, x_2$ such that $E(x_1)=E(x_2)$ because then again by Rolle $E'(x)$
 will have a zero but this contradicts the given conditions.

Therefore if $E(b) >E(x)>E(a)$ for $a<x<b$ then $f(x)>g(x)$ for $a<x<b$. Q.E.D.

Please let me know if there were parts I could've skipped in the proof.

The fact that $E$ is strictly increasing, is usually proven by MVT. Actually you don't need MVT.

Notice that for $x\in (a,b)$ we have
$$ 0 < E'(x) = \lim_{h\to 0^+} \underbrace{\frac1h}_{>0} (E(x+h) - E(x)). $$
That is, there exists some $h_x > 0$ such that for every $h\in (0,h_x)$ it follows $E(x+h) > E(x)$. Now let $y\in (x, b]$. If $E(y) > E(x)$ there is nothing to show. If $E(y) \le E(x)$ you obtain a contradiction by Rolle's theorem.

Then, by continuity of $E$ and for $t\in (0,1)$, $t\to 0$ we have
$$ E(y) > E(x) > E(a (1-t) + xt) \to E(a) \ge 0. $$

Everything is good except for one thing I believe, you said $f(x) > g(x) \in (a, b]$. Why is it one open one closed?


List of commonly occurring infinite sequences? I'm looking for an online list of commonly occurring convergent infinite sequence, such as

$n^a$ (converges to $0$ if $a < 0$, to $1$ if $a = 0$, and to $\infty$ if $a > 0$)
$\sqrt[n]{n}$ (converges to $1$)
$\sqrt[n]{n!}$ (converges to $\infty$)
$(1 + \frac{x}{n})^n$ (converges to $e^x$)

For comparison, here is a list of commonly occurring Maclaurin series, and here is a list of commonly occurring derivatives. I'm looking for a similar list, only for sequences.

$a^n$ to $0,1,\infty$ or does-not-exist is possibly the most important.

Consider $a_n=\frac{n}{\ln n\cdot \pi(n)}$. As usual $\pi(n)$ is the number of primes not exceeding $n$. It is known that $a_n\rightarrow 1$ and this fact is known as the Prime number theorem.


Proving a triangle inequality for some divergence The variational distance is defined by,
$$
V(P,Q)=\sum _{i}|p_{i} -q_{i} |
$$
where $P=(p_{1} ,...,p_{n})$ and $Q=(q_{1} ,...,q_{n} )$ are discrete distributions.

It is fairly easy to see that $V$ is a metric, and in particular that it satisfies the triangle inequality. I now introduce class prior, $0<\alpha<1$, which represents the relative 'size' of the distributions,
$$
V(\alpha ,P,Q)=\sum _{i}|\alpha p_{i} -(1-\alpha )q_{i} |
$$
and would like to prove that it again satisfies the triangle inequality,

$$V(\alpha ,P,Q)+V(\beta ,Q,R)\ge V(\gamma ,P,R) $$
where $R$ is another distribution, $\beta$ and $\gamma$ are respective priors, and $\gamma$ is naturally,
$$\gamma =\frac{\alpha \beta }{\alpha \beta +(1-\alpha )(1-\beta )}
$$

I have confidence (numerically) that this triangle inequality indeed holds, but would like a formal proof. Any ideas?

Actually, I've recently realized that the inequality is false. It is true only for equal class priors. Sorry about that.

But $V(\alpha, P, Q) != V(\alpha, Q, P)$ in general, therefore it would not be a metric, right? If that is true, what is the point of proving or disproving the triangle inequality?


Error update in matrix Kalman Filter Kalman filter represents a best pathway between prediction and measurement.  Since in its simplest form the model is purely Gaussian, it achieves this by creating a product Gaussian by multiplying together the pdfs of the prediction and measurement (from which the new mean and variance can be extracted).  In one dimension, the fused ${\sigma}_{fused}^2$ is given by:  $$ {\sigma}_{fused}^2={1\over{1/{\sigma}_{pred}^2+1/{\sigma}_{meas}^2}}$$
and I can see this formula in the one dimension error update.  In the multidimensional case, it becomes:  $${\Sigma}_{fused}=({\Sigma}_{pred}^{-1}+{\Sigma}_{meas}^{-1})^{-1}$$ There is probably some kind of algebraic simplification or other sleight of hand occurring but I cannot seem to find an expression that looks like this anywhere in the multidimensional update equations.  Perhaps there is someone out there that can shed light on what I am missing.  Thank you.

Short Answer:
The updated covariance matrix in the Kalman filter can be written concisely in a single equation as follows:
$$ P_{k+1} = \bigr( (FP_k F + Q)^{-1} + H^T R^{-1} H \bigr)^{-1}$$
where $P_{k+1}$ is the fused or updated covariance, $P_{pred} = FP_k F + Q$ is the predicted covariance, and $P_{meas} = \bigr( H^T R^{-1} H \bigr)^{-1}$ is the measurement information after projecting on the state space. If you have $H=I$ where $I$ is the identity matrix, then $H^T R^{-1} H = R^{-1}$.
This is equivalent to the equation you provided in the case that $H=I$. In other words, relating to your notation:
$$ P_{fused} = P_{k+1} = \bigr( (FP_k F + Q)^{-1} + H^T R^{-1} H \bigr)^{-1} = \bigr( P_{pred}^{-1} + P_{meas}^{-1} \bigr)^{-1}$$
Long Answer:
I am not sure why the downvotes (especially with no comments clarifying the reason...), but I will provide some additional information that will hopefully help clarify my answer.
Let $F$ be the process model, $H$ be the measurement model, $Q$ be the process covariance, and $R$ be the measurement covariance. See Wikipedia article for details on what these matrices represent. Furthermore, let $P_k^{-}$ and $P_k^{+}$ denote the covariance matrix at time step $k$ for the predict step (i.e., before acquiring a measurement) and the update step (i.e., after acquiring a measurement), respectively.
The covariance matrix computed by the predict step of the Kalman filter assuming additive noise is given by
$$P_{k+1}^{-} = FP_k^{+}F + Q$$
The covariance matrix computed by the update step of the Kalman filter assuming additive noise is given by
$$P_{k+1}^{+} = (I - P_{k+1}^{-} H^T (H P_{k+1}^{-} H^T + R)^{-1} H)P_{k+1}^{-} $$
where $I$ is the identity.
Using the Woodbury Matrix Identity as mentioned in the comments by @KwinvanderVeen, the information form of the updated covariance matrix is given by
$$ (P_{k+1}^+)^{-1} = (P_{k+1}^{-1})^{-1} + H^T R^{-1} H $$
where more details of this form referred to as the information filter can be found on the Wikipedia article. As an exercise, I suggestion attempting to apply the Woodbury Matrix Identity to obtain this equation.
Taking the inverse of both sides and substituting $P_{k+1}^{-1}$ we have the following:
$$ P_{k+1}^{+} = \bigr( (FP_k^{-} F + Q)^{-1} + H^T R^{-1} H \bigr)^{-1},$$
which is equivalent to the form you describe, which using your notation is given as
$$ P_{fused} = P_{k+1}^{+} = \bigr( (FP_k^{-} F + Q)^{-1} + H^T R^{-1} H \bigr)^{-1} = \bigr( P_{pred}^{-1} + P_{meas}^{-1} \bigr)^{-1}$$


Need to know what this theorem is asking Theorem
Let (Y,$\tau_1$) be a a Subspace of (X,$\tau$)
If (X $\tau$) satisfies Second Countable axiom
Show  (Y,$\tau_1$) does too
My question is this : is this theorem asking us
to prove the hereditary of the Second Countable axiom?

It is indeed asking to show that the property “$X$ is second countable “ is hereditary with respect to all subspaces.
Not that it helps with the proof.


Joint Default Probability Range There is a 50% probability that bond A will default next year and a
  30% probability that bond B will default. What is the probability
  range for event that at least one bond defaults?

The author then goes on with:

  To have the largest probability we can assume whenever A defaults, B
  does not default; and vice versa. A does not default -- so the maximum
  probability that at least one bond defaults is 50% + 30% = 80%.

OK, so he is talking about a scenario where A and B are independent.
$$p(\text{at least one defaults}) = 1 - p(\text{none default})$$
A and B are Bernoulli RVs, so:
$$p(\text{none default}) = 0.5 \cdot 0.7 = 0.35$$
$$p(\text{at least one defaults}) = 0.65$$

65% is nowhere close to 80%. What is going on?

"Whenever A defaults, B does not" is not independent at all.  The description explains how B's default depends on A's.

Let's talk about this with dice.  If we know that A is some event with a probability of 3/6 and B is some event with a probability of 2/6 but we don't know what those events are, what is the maximum probability that either A or B could happen?  Well, we could imagine an interpretation where A = "roll a 1, 2, or 3" and B = "roll a 4 or 5", so the probability of at least one occurring under this interpretation is 5/6.  It should be fairly evident that we could not do better.

The probability that either bond A or B defaults is
$$\Pr(A \text{ or } B) = \Pr(A) + \Pr(B) -\Pr(A \cap B)$$.
The maximum probability that at least one bond deafults is when we take $\Pr(A \cap B) = 0$.