Prove that the set $\{\sin(x),\cos(x)\}$ is linearly independent using derivative. Prove that the set $\{\sin(x),\cos(x)\}$ is linearly independent using matricies.

Method 1

Let $r\sin(x)+s\cos(x)=0$

If $x=0$, this means that $s=0$

If $x=\pi/2$, this means that $r=0$.

Therefore the set is linearly independent.

Method 2 - using derivatives

$$r\sin(x)+s\cos(x)=0 \tag 0$$
Differentiating once,

$$r\cos(x)-s\sin(x)=0 \tag 1$$
Differentiating again,

$$-r\sin(x)-s\cos(x)=0 \tag 2$$

Take equation $0$ and $2$ in a matrix.

\begin{pmatrix}
1 & 1\\
-1 & -1
\end{pmatrix} RREF
\begin{pmatrix}
1 & 1\\
0 & 0
\end{pmatrix}

So this still brings us back to our original $r\sin(x)+s\cos(x)=0$. Does this mean that it cannot be solved with a matrix, and we must solve it manually?

The first equation and the differentiated equation give a nontrivial element $(\sin(x),\cos(x))$ of the kernel of the matrix

$$\begin{bmatrix} r & s \\ -s & r \end{bmatrix}.$$
Note that the determinant of this matrix is $r^2+s^2=0$, so $r=s=0$. Note that this method proves more: that there is no interval on which the sine and cosine functions are constant multiples of one another.

Following your argument, you have two equations:
$$r\sin(x)+s\cos(x)=0 \tag 0$$
$$r\cos(x)-s\sin(x)=0 \tag 1$$
In matrix notation:
$$\begin{pmatrix}r & s\\ -s & r\end{pmatrix}
\begin{pmatrix}
\sin x \\
\cos x
\end{pmatrix}=\begin{pmatrix}
0 \\
0
\end{pmatrix}$$
for all $x$. This implies that
$$\begin{pmatrix}r & s\\ -s & r\end{pmatrix}
\begin{pmatrix}
x \\
y
\end{pmatrix}=\begin{pmatrix}
0 \\
0
\end{pmatrix}$$
for all x and y. It follows that
$$\begin{pmatrix}r & s\\ -s & r\end{pmatrix}=0$$


Finding domain and range I just started learning engineering functions and I'm now at Domain and range ..

I was trying this question -

$$\frac{x+\sqrt{x+1}}{2x-1} $$

I found the domain to be

$x$ less than $1/2$
$x$ more than $1/2$
$x$ more than or equal to $-1$

Now about finding the range ,

I'm having huge difficulty...

I thought spitting the fractions will help but it didn't . -

Is there any way to see how to get the range easily ? Thanks

Let $t=\sqrt{x+1}\geq 0$ so $x=t^2-1$. Thus we have to find all $b\in \mathbb{R}$ that satisfy equation:

$$\frac{t^2 +t-1}{2t^2 -3} =b$$

If you draw a graph of rational function $f(t) = \frac{t^2 +t-1}{2t^2 -3}$, for $t\geq 0$ you will see that the range is $(-\infty,{1\over 3}]\cup ({1\over 2},\infty)$.

The domain is $[-1,+\infty)\setminus\{\frac{1}{2}\}$.

$f$ is continuous function on $\left(\frac{1}{2},+\infty\right)$, $\lim\limits_{x\rightarrow\frac{1}{2}^+}f(x)=+\infty,$ $\lim\limits_{x\rightarrow+\infty}f(x)=\frac{1}{2}$

and easy to see that the equation $f(x)=\frac{1}{2}$ has no solutions.

Indeed, $$f(x)=\frac{1}{2}$$ it's
$$\frac{x+\sqrt{x+1}}{2x-1}=\frac{1}{2}$$ or
$$2x+2\sqrt{x+1}=2x-1$$ or
$$2\sqrt{x+1}=-1,$$ which is impossible.

Thus, $$f:\left(\frac{1}{2},+\infty\right)\rightarrow\left(\frac{1}{2},+\infty\right).$$

The case $x\in\left[-1,\frac{1}{2}\right)$ is the similar:

$f$ is a continuous function on $\left[-1,\frac{1}{2}\right)$, $\lim\limits_{x\rightarrow\frac{1}{2}^-}f(x)=-\infty$  and $f(-1)=\frac{1}{3}.$

Thus, $$f:\left[-1,+\frac{1}{2}\right)\rightarrow\left(-\infty,\frac{1}{3}\right).$$
Id est, the range is:
$$\left(-\infty,\frac{1}{3}\right]\cup \left(\frac{1}{2},\infty\right).$$
Done!


Is it possible when multiplying two polynomials that, after collecting similar terms, all terms vanish? Algebra by Gelfand poses this question with the remarkably unhelpful answer of:

No. Probably this problem seems silly; it is clear that it cannot happen. If you think so, please reconsider the problem several years from now.

I'm sure the mathematical wit is just lost on me, but since you can lose some terms from multiplying, it doesn't seem too far fetched through some mathematical wizardry that there are cases where they all vanish. So why is this?

If you are working with polynomials over a ring with zero divisors, such as $\mathbb{Z}/4\mathbb{Z}$, then it is possible for the product of two polynomials to vanish.  This may be what Gelfand is coyly alluding to.  But in the ordinary sense of polynomials with rational, real, or complex coefficients, the degree of the product is the sum of the degrees of the polynomials being multiplied together.

Sure, if the polynomials are piecewise defined.  Let p(x)={x^2, x>0; 0,x<0}, q(x)={x^3,x<=0;0,x>0}.  But these wouldn't be polynomials in the strict definition.


Easy proof that $\{(x,y) \in \mathbb{R}^2 : y=\tan(x) \}$ is a closed set. I'd like to know whether there's an "easy" proof that $$A:= \{(x,y) \in \mathbb{R}^2 : y=\tan(x) \} .$$
I've tried to prove that its complement is open, but given an $(x,y)$ such that $\tan(x) \neq y$, it's a bit of a grind to find (in the general case) an open set that contains $(x,y)$ and does not intersect $A$. Is there a simpler argument using basic euclidean topology?
Some of the comments here gave me an idea: If $U:=\{{(x,y) \in \mathbb{R}^2: cos(x) \neq 0}\}$ and $f:U -> \mathbb{R}$ is such that $f(x,y)=y-tan(x)$, then$A=f^{-1}(\{0\})$, therefore A is closed in U. Does this imply that A is closed in \mathbb{R}^2, though?

If $p\in A^c:=\Bbb R^2\setminus A$, let the perpendicular from $p$ to $A$ have length $r$: then the open neighbourhood of $p$ of radius $r$ is a subset of $A^c$, so $A^c$ is open. You can formalize the result that $r>0$ with $A$'s radius of curvature.

You know that
Introducing $A_k=] -\pi/2+k\pi,\pi/2+k\pi[$
$$ f(x, y) =y-tan(x) $$
$f$ is continous on each set$G_k={ A_k} \times \mathbb{R} $ where sit the zeros.
$$ A=f^{-1}(\{0\}) $$
because $f$ is continous on each$G_k$ where sit the zeros and the reciprocal image of a closed set by a continous function is closed hence
A is closed in $\cup G_k$
In order to get to the slides take $(x_n,y_n)=((\pi/2+1/n) ,\tan(1/n-\pi/2)) $
$$y_n-\tan(x_n) = \dfrac{\sin(\pi)} {\cos(\pi/2 +1/n)\cos(\pi/2-1/n)} =0$$
So you can complete the missing lines...


How can one determine whether the following series converges or diverges $$
\sum_{n = 2}^{\infty}\frac{(-1)^{n}}{\sqrt{n} + (-1)^{n}}
$$

Wolfram Alpha returns nothing useful, except that the ratio test was inconclusive.

One way of doing it, which is not necessarily the most sophisticated way to go about it, is to look at the partial sums and see if those trend toward some limit. Let's, for the sake of clear example, use the sum "$1 + 1/2 + 1/4 + 1/8 + ...$" if we add the first two terms, our partial sum is 1.5. For the first three it's 1.75. And for the first 4 it's 1.875. As you can seem the partial sums are trending toward two and if we had a function which gave us these partial sums it's limit as x approaches infinity would be two as well.


Prove $4(x^2+1)^3 \leq 5(x^3+2)^2$ I need to prove that $4(x^2+1)^3 \leq 5(x^3+2)^2$ ,  for all $x\geq 0$, in the context of an introductory analysis course. I thought I could somehow use convex functions to prove this, but I simply cannot get it to work. Hints anyone?


Continuity of a probability measure I am studying Measure Theory and I am stuck in some concepts about continuity of a measure.

Let $(S_{1}, \Sigma,\mu)$ be a measurable space, where $\mu$ is a probability mesure such that $\mu (S_{1}) = 1.$ Let also $S_{\delta} \subset \Sigma, 0\leq \delta \leq 1,$ be a family of sets such that $S_{\delta_{1}} \subseteq S_{\delta_{2}}$ for $0 \leq \delta_{1}\leq \delta_{2} \leq 1.$

Theorem- Continuity from bellow:
 If $E_{j} \subset E_{j+1}$ is an increasing sequence of mesurable sets, then
$$
 \mu\left( \bigcup_{j=1}^{\infty} E_j \right)
  = \lim_{j\rightarrow \infty}\mu(E_{j}).
$$

Define $E_{j} = S_{1-1/j}$, for $j \in\{1,2,\ldots\}$.  My question is the following:

If $\mu(S_{\delta}) = c$, with $c < 1$, for $0 \leq \delta < 1,$ how the theorem holds at $\delta =1$ (or $ j\rightarrow \infty$)?

I think I missed something. Thanks in advance!

I see no problem in this.

By the theorem you have $\mu(\bigcup_{j=1}^\infty E_j) = c$ with $\bigcup_{j=1}^\infty E_j = \bigcup_{j=1}^\infty S_{1-1/j} \subseteq S_1$. That $c < 1 = \mu(S_1)$ just tells us that $\bigcup_{0 \leq \delta < 1} S_\delta = \bigcup_{j=1}^\infty S_{1-1/j} \subsetneq S_1$ is a proper subset. (This is what you missed.)

Take for example the unit interval $S_1 = [0,1]$ with the Lebesgue measure $\lambda$ and set $S_\delta = \emptyset$ for every $0 \leq \delta < 1$ (i.e. $c = 0$).

If you define $\mu(S_\delta)=c$ for $c<1$ then $\mu(E_j)=c$ for all $j$, but then $\mu(S_1)=c$ which contradicts with the condition $\mu(S_1)=1$.


semi ellipsoid and cylinder parametrize the curve the curve of intersection of the semi ellipsoid $x^2 + y^2 +4z^2 = 4, y >=0 $ (i.e. the half of the ellipsoid with positive y coordinates), and the cylinder $ x^2 + z^2 =1 $.
Consider the polar angle in the xz-plane with the positive z axis (see figure). Use this coordinate as the parameter to parametrize the curve with the counterclockwise direction looking into the y axis. Sketch the curve (hint: simplify the expression for y(t) so that it is given by only one
term).

figure :http://i.imgur.com/m1uENAg.jpg

How do I solve this? Solution if possible.

Plug $z=\cos t, x=\sin t$ into the first equation:
$$\sin^2t+y^2+4\cos^2t=4$$
We get $y^2=3-3\cos^2t=3\sin^2t$, since $y>0$, $y=\sqrt{3}|\sin t|$.

The projection of the curve C of intersection onto the $xz$ -plane is the circle
$x^2+z^2 = 1, y =0$, so we can write $x=\cos t, z=\sin t, 0\le t \le 2\pi$.

$C$ also lies on the surface $x^2+y^2+4z^2 =4$ and since $y\ge 0$ we can write

$$y=\sqrt{4-x^2-4z^2}=\sqrt{4-\cos^2t-4\sin^2t}=\sqrt(4-\cos^2t-4(1-\cos^2t)=\sqrt(3\cos^2t)$$

Hence the parametric equation
$x=cost y=\sqrt(3)|\cos t|,z=\sin t,0\le t\le 2\pi$ and the corresponding vector
$$r(t)=\cos t i +\sqrt(3)|\cos t|j +\sin t k,0\le t\le 2\pi$$


Sum of points' powers to circles is constant Given 2 circles $\omega_1, \omega_2$, find the locus of all points $P$ such that $\mathcal{P}ow(P, \omega_1) + \mathcal{P}ow(P, \omega_2) = k$ (i.e: sum of powers of point $P$ with respect to the two circles $\omega_1, \omega_2$ is constant).

UPDATE: I found out two solutions:

1. Analytical geometry approach:

let $O_1=(0, 0), r_1, O_2=(z, 0), r_2$ be the centers and radiuses of $\omega_1, \omega_2$ respectively. Denote by $d_1, d_2$ to be the distances between $P$ and $O_1, O_2$. Let $P=(x, y)$. By the definition of power of a point: $d_1^2 -r_1^2+d_2^2-r_2^2=k\Leftrightarrow d_1^2 +d_2^2 =k+r_1^2 +r_2^2$
$$
\Leftrightarrow x^2 + y^2 + (z-x)^2 +y^2=k+r_1^2+r_2^2\\
\Leftrightarrow (x-\frac{z}{2})^2+y^2=\frac{k+r_1^2 +r_2^2-z^2}{2} +\frac{z^2}{4}
$$

Which is a circle centered at the midpoint of $O_1, O_2$

2. Euclidean geometry: using Apollonius theorem:

Consider the midpoint $M$ of $O_1O_2$, by applying Apollonius theorem in $\triangle PO_1O_2$: $d_1^2 +d_2^2 = PM^2 +\frac{z^2}{4}$ This means that $PM$ is constant. Therefore, $P$ is on the circle centered at $M$ with radius $\sqrt{k+r_1^2 +r_2^2-\frac{z^2}{4}}$.

$Pow(P, \omega_1) = d(P, O_1)^2 - r_1 ^2$
$Pow(P, \omega_1) = d(P, O_2)^2 - r_2 ^2$

So, we want to find the set of points $P$ such that

$d(P, O_1)^2 + d(P, O_2 )^2 = k + r_1^2 + r_2^2$

What is the shape of the graph?

   It is a circle, whose center is the midpoint of $O_1O_2$.
 To see this, set $P = (x,y), O_1 = (x_1, y_1), O_2 = (x_2, y_2)$, expand out the equation and simplify.

See here: https://artofproblemsolving.com/wiki/index.php/Radical_axis

I believe this has something to do with something along the lines of "Newton's Lemma"


Proof of Eckart-Young-Mirsky theorem Could someone please explain why in this Wiki page one says "we know  that $\exists(k+1)$ dimension space $(v_1,v_2, \dots, v_n)$" ?

One needs to show that if $\mathrm{rank}(B)=k$, then $\|A-B\|_2\geq\|A-A_k\|_2$. This can be done as follows.
Since $\mathrm{rank}(B)=k$, $\dim\mathcal{N}(B)=n-k$ and from $$\dim\mathcal{N}(B)+\dim\mathcal{R}(V_{k+1})=n-k+k+1=n+1$$ (where $V_{k+1}=[v_1,\ldots,v_{k+1}]$ is the matrix of right singular vectors associated with the first $k+1$ singular values in the descending order), we have that there exists an
$$x\in\mathcal{N}(B)\cap\mathcal{R}(V_{k+1}), \quad \|x\|_2=1.$$
Hence
$$
\|A-B\|_2^2\geq\|(A-B)x\|_2^2=\|Ax\|_2^2=\sum_{i=1}^{k+1}\sigma_i^2|v_i^*x|^2\geq\sigma_{k+1}^2\sum_{i=1}^{k+1}|v_i^*x|^2=\sigma_{k+1}^2.
$$
From $\|A-A_k\|_2=\sigma_{k+1}$, one gets hence $\|A-B\|_2\geq\|A-A_k\|_2$. No contradiction required, Quite Easily Done.

EDIT An alternative proof, which works for both the spectral and Frobenius norms, is based on the Weyl's theorem for eigenvalues (or more precisely, its alternative for singular values): if $X$ and $Y$ are $m\times n$ ($m\geq n$) and (as above) the singular values are ordered in the descreasing order, we have
$$\tag{1}
\sigma_{i+j-1}(X+Y)\leq\sigma_i(X)+\sigma_j(Y) \quad\text{for $1\leq i,j\leq n, \; i+j-1\leq n$}
$$
(this follows from the variational characterization of eigen/singular values; see, e.g., Theorem 3.3.16 here).
If $B$ has rank $k$, $\sigma_{k+1}(B)=0$.
Setting $j=k+1$, $Y:=B$, and $X:=A-B$, in (1) gives
$$
\sigma_{i+k}(A)\leq\sigma_i(A-B) \quad \text{for $1\leq i\leq n-k$}.
$$
For the spectral norm, it is sufficient to take $i=1$. For the Frobenius norm, this gives
$$
\|A-B\|_F^2\geq\sum_{i=1}^{n-k}\sigma_i^2(A-B)\geq\sum_{i=k+1}^n\sigma_i^2(A)
$$
with the equality attained, again, by $B=A_k$.

The original statement of Eckart-Young-Mirsky theorem on wiki is based on Frobenius norm, but the proof is based on 2-norm. Though Eckart-Young-Mirsky theorem holds for all norms invariant to orthogonal transforms, I think it is necessary to add a proof purely based on Frobenius norm since it is even easier to prove than that based on 2-norm.

The following proof is replicated from these notes.

Since $||M-X_k||_F = ||U\Sigma V^\intercal - X_k||_F = ||\Sigma - U^\intercal X_k V ||_F$, denoting $N = U^\intercal X_k V$, an $m \times n$ matrix of rank $k$, a direct calculation gives
\begin{equation}
  ||\Sigma-N||_F^2 = \sum_{i,j} |\Sigma_{i,j} - N_{i,j}|^2 = \sum_{i=1}^r |\sigma_i-N_{ii}|^2+\sum_{i>r}|N_{ii}|^2+\sum_{i\neq j} |N_{i,j}|^2
\end{equation}
which is minimal when all the non diagonal terms of $N$ equal to zero, and so are all  diagonal terms with $i > r$. Obviously, the minimum of the terms left is attained when $N_{ii} = \sigma_i$ for $i = 1,2,\cdots,k$ and all other $N_{ii}$ are zero.


Is $\{x : x \in \mathbb{R} \land (x \ge 0 \lor x < 0) \} \neq \mathbb{R}$ provable in a constructive setting? The main thing I'm trying to figure is whether (101) itself is a theorem in a constructive setting, whether it is a meta-theorem, or something else.

Andrej Bauer's lecture The Five Stages of Accepting Constructive Mathematics gives the following example of a counterintuitive result in a generic constructive setting. I'm guessing based on the content of his lecture that this generic constructive setting is similar to the usual ZFC but without the axiom of choice, and without the law of the excluded middle. This is only my guess though, I don't know the exact framework that Bauer is assuming for the purposes of this example.

One example that he gave near the start of the lecture is the following (101). Discussion of this example begins at the offset 7:17.

$$ \left\{ x : x \in \mathbb{R} \;\land\; \left( x < 0 \lor x \ge 0 \right) \right\} \subsetneq \mathbb{R} \tag{101} $$

Bauer also claims that the related statement (102) is true.

$$ \mathbb{R} \setminus \left\{x : x \in \mathbb{R} \;\land\; \left(x < 0 \lor x \ge 0\right)\right\} \tag{102} = \{\} $$

(201) and (202) are my two guesses as to what (101) might mean.

(201): (101) is a theorem

I think this just means we can construct a term that witnesses (251).

$$ \left(\left\{ x : x \in \mathbb{R} \land \left(x \ge 0 \lor x < 0\right) \right\} = \mathbb{R}\right) \to \bot \tag{251} $$

(202): (102) is a metatheorem

I interpret this as meaning that, in any model of constructive mathematics, it is not the case that (252).

$$ \left[\left\{ x : x \in \mathbb{R} \land \left(x \ge 0 \lor x < 0\right) \right\}\right] = \left[ \mathbb{R} \right] \tag{252} $$

Where $\left[\psi\right]$ refers to the element of the domain in the model that corresponds to a well-formed formula $\psi$ and $=$ refers to the inherent equality that the model itself is equipped with. Equivalently, $\left[\cdot\right]$ sends closed terms to their interpretation, which I'm pretty sure is a legitimate thing to do.

Disclaimer: There are many foundational theories of constructive mathematics: to answer questions like the one you ask, one has to be explicit about which particular foundation is used. What I say below holds if we choose constructive set theory CZF as our foundational theory. CZF is the theory implicitly used in Bauer's talk, and is the common setting for Bishop's constructive mathematics. Just like in classical mathematics, other foundations are possible, and what I say here does not hold in some of those alternate foundations.

Constructive mathematics is what you get if you remove the law of excluded middle and other principles that imply it from classical mathematics. If you add back the law of excluded middle, you get classical mathematics again (indeed, CZF Set Theory extended with the law of excluded middle is the same as ordinary classical Zermelo-Fraenkel set theory).

This means that you won't be able to prove theorems that contradict established theorems of classical mathematics: if you could prove such a theorem, adding the law of excluded middle would yield an inconsistent theory!

Consequently, (101) is not a theorem of constructive mathematics. For if $S$ denote the set $\{ x \in \mathbb{R} \:|\: x < 0 \vee x \geq 0 \}$, then your (101) can be rephrased as $S \subseteq \mathbb{R} \wedge S \neq \mathbb{R}$. If we could prove $S \neq \mathbb{R}$ in constructive mathematics, then the same proof would work in classical mathematics. But in classical mathematics we can prove $S = \mathbb{R}$, so in that case we could prove both $S \neq \mathbb{R}$ and $S = \mathbb{R}$ in classical mathematics, which would lead to a contradiction.

I'm not an expert so take what I say with a grain of salt.

First, if we are working with say ZF without choice or excluded middle and the set $\mathbb{R}$ and the ordering $<$ is defined as in classical mathematics, then $\{x \in \mathbb{R} \mid x < 0 \lor x\geq 0 \} \neq \mathbb{R}$ cannot possibly be true as it would also have to be true in ZFC, since our system is just a strict subsystem of ZFC.

So how can this statement be true? There are a few ways. If suppose you are working in ZF minus choice and EM, then you could have a different construction of $\mathbb{R}$ and $<$ (this is what more-or-less happens in Bishop and Bridges, Constructive Analysis). The other thing that could happen is you are working with a different system that is not a subsystem of ZFC.


What does it mean for a function to be continuous on its domain? I never understood the phrase "continuous on its domain."

Isn't everything continuous on its own domain, since the domain are all the $x$ values that we can plug into $f(x)$ and get a defined $y$ value back? i.e. doesn't the domain by definition tell you where the function is continuous? Why would the domain ever include something not continuous / not defined?

Not necessarily. The domain of a function tells you over what values the function $f(x)$ exists, not where it is continuous. Take the piecewise function:

$$f(x) =
\begin{cases}
1 & x<0\\
2 & x\geq0
\end{cases}$$

This function is defined for all $x\in\mathbb{R}$, but is not continuous at $x=0$. It still has a valid value: $f(0)=2$, but that doesn't make it continuous at that point.

For a function to be continuous at a point, its limit must be the same regardless of what direction of approach. In this case, $\lim\limits_{x\to0^-}{f(x)}=1$ while $\lim\limits_{x\to0^+}{f(x)}=2$, making it discontinuous at that point.

A function is said to be continuous if it continues at each point. This means that over the domain. Functions that are not continuous do not exist for every x value over the domain. For example if a function is defined near an open interval (the circle that is not shaded on a graph) then the function is discontinuous. However, if a function is defined near a closed circle (the shaded circle of a graph) then that function is continuous. This problem usually occurs when trying to find the Domain and Range of a function or if a problem ask to graph the Domain and Range of a function.


Point reflection over a line I'm having trouble understanding the solution presented below (It's from a textbook).
I tried to get something similar, but to no avail. Help me find the way he derived those a,b,x2 and y2 expressions.

Point p is a point which we want to reflect (It's a simple structure with two integers x and y representing coordinates on a 2d plane); (x0,y0) and (x1,y1) are coordinates of endpoints of a line.
Expected result is a new point with reflected coordinates.

Point mirror(Point p, int x0, int y0, int x1, int y1)
{

   double dx,dy,a,b;
   long x2,y2;
   Point p1; //reflected point to be returned

   dx  = (double) (x1 - x0);
   dy  = (double (y1 - y0);

   a   = (dx * dx - dy * dy) / (dx * dx + dy*dy);
   b   = 2 * dx * dy / (dx*dx + dy*dy);

   x2  = Math.round(a * (p.x - x0) + b*(p.y - y0) + x0);
   y2  = Math.round(b * (p.x - x0) - a*(p.y - y0) + y0);

   p1 = Point((int)x2,(int)y2);

   return p1;

}

Here's an explanation. It is easier to follow, if you draw a picture. Let $\vec{u}=(dx,dy)$
be the vector from the point $P_0=(x_0,y_0)$ to the point $P_1=(x_1,y_1)$, i.e. a vector pointing in the direction of the mirror line. Then $\vec{n}=(-dy,dx)$ is perpendicular to it. Let's name the vector from $P_0$ to $P$ $\vec{v}=(p.x-x_0,p.y-y_0)$. The projection of the vector $\vec{v}$ along the normal $\vec{n}$ is then
$$\vec{p}=\frac{\vec{n}\cdot\vec{v}}{\vec{n}\cdot\vec{n}}\vec{n}=\frac{-(p.x-x_0)dy+(p.y-y_0)dx}{dx^2+dy^2}\vec{n}.$$

We compute the mirror image point $P'=(x_2,y_2)$ by comparing the representations of the vectors $\vec{P_0P}=\vec{v}$ and $\vec{P_0P'}=\vec{v'}$ in the (orthogonal) basis $\{\vec{u},\vec{n}\}$. The reflection maps $\vec{u}$ to itself, but the vector perpendicular to mirror is mapped to its negative: $\vec{n}\mapsto -\vec{n}$. Therefore the reflection maps
$$
\vec{v}\mapsto\vec{v}'=\vec{v}-2\vec{p}.
$$
The rest amounts to just plugging in the numbers. Write $N=dx^2+dy^2$ for short. We get
$$
\begin{align}
\vec{v}'&=\frac1N\left(N(p.x-x_0)-2dy^2(p.x-x_0)+2dy\,dx(p.y-y_0)\right)\vec{i}\\
&+\frac1N\left(N(p.y-y_0)+2dy\,dx(p.x-x_0)-2dx^2(p.y-y_0)\right)\vec{j}\\
&=\left(a(p.x-x_0)+b(p.y-y_0),b(p.x-x_0)-a(p.y-y_0)\right).
\end{align}
$$
Your formula comes from the fact that the above vector $\vec{v'}$ is the separation vector from $P_0$ to $P'$. Therefore we need to add the coordinates of $P_0$ to the result.

Edit: A crude image here. The points $P,P_0,P_1,P'$ are marked with the dots, and the vectors are $u,n,v,p,v',-2p$. Sorry about the missing arrows on top of those letters - can't do any better :-(

It worked for me in C++.
Where p is the point to reflect,
and P0(x0, y0) P1(x1, y1) is the mirror line,
and p1 is the point reflected.
See the graph again

struct Point{
      int x;
      int y;
  };

  Point mirror(Point p, int x0, int y0, int x1, int y1){

     double dx,dy,a,b;
     long x2,y2;
     Point p1; //reflected point to be returned

     dx  = (double) (x1 - x0);
     dy  = (double) (y1 - y0);

     a   = (dx * dx - dy * dy) / (dx * dx + dy*dy);
     b   = 2 * dx * dy / (dx*dx + dy*dy);

     x2  = round(a * (p.x - x0) + b*(p.y - y0) + x0);
     y2  = round(b * (p.x - x0) - a*(p.y - y0) + y0);

     p1.x = (int)x2;
     p1.y = (int)y2;

     return p1;
  }


Convergence of an exponential martingale Let $Y_i$ be a sequence of independent and identically distributed non-degenerate variables, $X_n$ their n-th partial sum, $\phi(u)=\log E[\exp(uY_1)]$ and define $M_n(u)=\exp(uX_n-n\phi(u))$.
Then $M_n(u)$ for u such that $\phi(u)\in\mathbb{R}$ is a martingale, converges almost surely, and for $u\ne0$ it converges almost surely to zero.

Verifying that $M_n(u)$ is a martingale is just a matter of plugging things in and calculating. In addition, $M_n(u)$ converges almost surely by the martingale convergence theorem since it is $L^1$-bounded (their expectation is $1$).
Now I am having trouble verifying the convergence to zero. I found that, noting that with $Z_i:=uY_i-\phi(u)$ we can write $\ln M_n(u)$ as the $n$-th partial sum of the $Z_i$'s. Noting the independence of the $Z_i$'s and that $EZ_i=EuY_i-\phi(u)\lt 0$ by Jensen, we have, by the law of large numbers, almost sure convergence of $\ln M_n(u)/n$ to $EZ_1\lt 0$. Now I feel like that this should enable me to show that the $M_n(u)$ must converge to zero, but I can not see how. Any help would be very much appreciated.

Jensen is the right tack, but apply it to the exponential, not the exponent: The limit $M_\infty(u)$ is a.s. equal to $\exp(uX_1-\phi(u))\cdot M'_\infty(u)$, where $M'_\infty(u):=\lim_n \exp((X_n-X_1)-(n-1)\phi(u))$, and the two factors are independent. Take square roots and then expectations. Either $\Bbb E[\sqrt{M_\infty(u)}]=0$ (in which case you are done) or $\Bbb E[\sqrt{\exp(uX_1-\phi(u)}]=1$. This latter equality would violate the strict concavity of the square root function unless $X_1$ were degenerate, which you have assumed it not to be.


What are the cosets of $\mathbb{Q_8} / Z(\mathbb{Q_8})$? $\mathbb{Q_8} = \{\pm 1, \pm i, \pm j, \pm k\}$ where 1 is the identity and $\mathbb{Q_8}$ is a non-abelian group under ordinary multiplication

Let it be given that $i^2=j^2=k^2=ijk=-1$

I already calculated $Z(\mathbb{Q_8}) = \{-1, 1 \}$ by constructing a Cayley table

I want to find $\mathbb{Q_8} / Z(\mathbb{Q_8})$

Now, I know Lagrange's theorem tells us that $\mid \mathbb{Q_8} / Z(\mathbb{Q_8}) \mid = 8/2 =4$. So there must be 4 distinct cosets

Where do I go from here?

The congruence classes modulo a a normal subgroup are obtained multiplying each element of the group by all elements of the subgroup. Therefore, it is not hard to check that you obtain the four classes
$$\{1,-1\},\; \{i,-i\}, \; \{j,-j\}, \; \{k,-k\}.$$
Furthermore, the quotient $Q_8/Z$  is abelian, and each class has order $2$, so $Q_8/Z$ is isomorphic to Klein's Vierer Gruppe.

There are only two groups of order $4$.  Check the number of elements of order $2$ (or $4$) to decide which one.


Finding powers of 2 and 3 in modular arithmetic Find all the powers of $2$ and $3$ modulo $17$.

How would you solve this question and explain the steps please!

If you want all the powers explicitly there's no other way but computing them. If you want some beforehand information on about how many computations you have to do you can reason along the following lines.

You know that $2^4=16\equiv-1\bmod17$. Thus $2^8=(2^4)^2\equiv(-1)^2=1\bmod17$ and so there are just $8$ different powers of $2\bmod17$ to compute.
$3^2=9\equiv-8=-2^3\bmod17$, so up to a sign the even powers of $3$ are powers of $2$ which you already have computed. In fact $3^8=(3^2)^4\equiv(-2^3)^4=(2^4)^3\equiv-1\bmod17$ because of the previous computation so that the cycle of the powers of $3\bmod 17$ has length $16$ thus including all non-zero classes in $\Bbb Z_{17}$.

Just do it.

As $17$ is prime we know by Fermat's Little Theorem that at most there will be $17-1=16$ of them, but there might be fewer if $a^k \equiv 1\pmod{17}$ and $k\mid 16$.

$2^1 \equiv 2 \pmod {17}$

$2^2 \equiv 2*2\equiv 4 \pmod {17}$

$2^3 \equiv 2*4 \equiv 8 \pmod {17}$

$2^4 \equiv 2*8\equiv 16 \equiv -1 \pmod {17}$.  COOL!  $2^8\equiv (2^4)^2 \equiv (-1)^2 \equiv 1 \pmod{17}$ so there will only be $8$ of them!

$2^5 \equiv 2*(-1) \equiv -2 \equiv 15 \pmod{17}$

$2^6 \equiv 2*(-2) \equiv -4 \equiv 13 \pmod{17}$

$2^7 \equiv 2*(-4) \equiv -8 \equiv 9 \pmod {17}$

$2^8 \equiv 2*9\equiv 18 \equiv 1 \pmod{17}$

So that's it for $2$s.

....

$3$s.

$3^1 \equiv 3\pmod{17}$

$3^2 \equiv 3*3 \equiv 9 \pmod{17}$

$3^3 \equiv 3*9 \equiv 27 \equiv 10\equiv -7 \pmod {17}$

$3^4 \equiv 3*(-7)\equiv -21\equiv -4 \equiv 13 \pmod {17}$

$3^5 \equiv 3*(-4) \equiv -12\equiv 5 \pmod{17}$.

$3^6 \equiv 3*5 \equiv 15\equiv -2 \pmod {17}$

$3^7 \equiv 3*(-2) \equiv -6 \equiv 11 \pmod {17}$.

$3^8 \equiv 3*(-6) \equiv -18 \equiv -1 \pmod {17}$

As $3^8 \equiv -1$ we can do the rest by taking negatives

$3^9 \equiv -3^1 \equiv -3 \equiv 14 \pmod{17}$.

$3^{10} \equiv -3^2 \equiv -9 \equiv 8 \pmod {17}$

$3^{11} \equiv -3^3 \equiv 7\pmod {17}$

$3^{12} \equiv -3^4 \equiv 4 \pmod {17}$

$3^{13} \equiv - 3^5 \equiv 12 \pmod {17}$

$3^{14} \equiv -3^6 \equiv 2 \pmod {17}$

$3^{15} \equiv -3^7 \equiv 6 \pmod{17}$

$3^{16} \equiv -3^8 \equiv 1 \pmod {17}$.

And that's it for the $3$s.


solving differential equations of the following form I understand that the questions are from the 'Differential Equations' part of the syllabus. It also seems that p means $dy/dx$ in these types. But I am not sure about the rest of the notations and also what exactly in 'differential equations' topic do these relate to. Most of the questions I have encountered do not have other letters apart from p,x and y. Any simple reference would be much appreciated.

$px+qy=pq$
$pt-qs=q^3$
$pxy+pq+qy=yz$
$pq=x(ps-qr)$

HINT to begin:

When $z=f(x,y)$, partial derivatives of z are
$$ p=\dfrac{\partial z}{\partial x} ,\,q=\dfrac{\partial z}{\partial y}  ,\, r=\dfrac{\partial^2 z}{\partial x^2} ,\,t=\dfrac{\partial^2 z}{\partial y^2} ,\,s=\dfrac{\partial^2 z}{\partial x\, \partial y} . $$


Prove $\lim\limits_{h\to0}\int_a^b\left|f(x+h)-f(x)\right|\,\mathrm dx=0$ Let $f(x)$ be a Riemann integrable function on any finite interval $I\subset\mathbb{R}$. Supposing $f(x)=0$ for $x\notin[a,b]$, show that $$\lim\limits_{h\to0}\int_a^b\left|f(x+h)-f(x)\right|\,\mathrm{d}x=0.$$
How to prove that if not familiar with the fact that Riemann integrable function is almost everywhere continuous?

Brutal force: Let $\epsilon >0$. Then since $f$ is Riemann integrable, there is a $\delta >0$ so that any partition $P$ of
$$a-1= x_0<x_1 < x_2 < \cdots x_n= b+1$$
with $x_{i+1} - x_i < \delta$ will have

$$ U(f, P)- L(f, P)<\epsilon.$$

Now consider the partition

$$ a < a+ h < a+ 2h < \cdots <a+ kh < b$$

where $a+ (k+1)h \ge b$ and $h < \delta/2$. Then

\begin{equation}
\begin{split}
\int_a^b |f(x+ h) - f(x)| dx &= \sum_i \int_{a+ ih}^{a+ (i+1)h} |f(x+ h) - f(x)|dx \\
&\le \sum_i h \left( \sup_{x\in [a+ih,a+ (i+2) h]} f(x) -  \inf_{x\in [a+ih,a+ (i+2) h]} f(x)\right) \\
&\le  U(f, \hat P ) - L(f, \hat P) < \epsilon
\end{split}
\end{equation}

where $\hat P$ is the partition

$$ a< a+2h< a+ 4h < \cdots < a +\ell (2h) < b,$$

where $a + (\ell +1) 2h \ge b$. Thus

$$\int_a^b |f(x+ h) - f(x)| dx < \epsilon$$

whenever $h < \delta/2$ and we are done.

Here is a cleaner argument. We know that $C_c(\mathbb{R})$, namely the continuous functions with compact support are dense in $L^1(\mathbb{R})$. With this, denote $f_h \triangleq f(x+h)$. Take a $g \in C_c(\mathbb{R})$ such that
$$
||f-g||_{L^1(\mathbb{R})}<\frac{\epsilon}{2}.
$$
Now note by triangle inequality that,
$$
||f_h-f||_{L^1(\mathbb{R})} \leq \underbrace{||f_h-g_h||_{L^1(\mathbb{R})}}_{\leq \frac{\epsilon}{2}} + \underbrace{||g_h-g||_{L^1(\mathbb{R})}}_{\to 0,\ \text{as} \ h\to 0} + \underbrace{||g-f||_{L^1(\mathbb{R})}}_{<\frac{\epsilon}{2}}
$$
where the middle convergence follows since, as $|g(x+h)-g(x)|\to 0$ as $h\to0$, due to continuity of $g(\cdot)$ and since $|g(x+h)-g(x)|\leq 2|g(x)|\in L^1$, we have by dominated convergence that $\int |g(x+h)-g(x)|\to 0$.

Hence, for every $\epsilon>0$,
$$
\lim_{h\to 0}||f_h-f||_{L^1(\mathbb{R})} < \epsilon.
$$
Since $\epsilon>0$ is arbitrary, we conclude.


Global sections of Proj In constructing the relative Proj for a graded ring $S$, one inevitably has that the distinguished opens, $D_+ (f)$, where $f \in S_+$ gives rise to a scheme that is isomorphic to $\textrm{Spec } S_{(f)}$, the latter denoting the degree zero elements in the localization $S_f$.

In all constructions I have seen of this, including Hartshorne and Liu, they usually leave it at that with no mention of global sections. I was tempted to handle this by saying $f = 1$ then $D_+ (f) = \textrm{Proj } S$ is isomorphic to $\textrm{Spec } S_{(1)} = \textrm{Spec } S_0$ (degree 0 elements of $S$). The Wikipedia article mentions this without proof or justification under the "Twisting Sheaf of Serre" section (https://en.wikipedia.org/wiki/Proj_construction#The_twisting_sheaf_of_Serre) lending credence to this fact.

However, the condition that $f \in S_+$, where $1$ typically does not reside (take the natural grading on the polynomial ring for example) seems to invalidate this proof. What justification should I use instead?

Background: I am interested in justifying that for $X = \textrm{Proj } A[x_0,\dots,x_n]$, that $\mathcal{O}_X (1)$ can be generated by the global sections $x_0, \dots, x_n$.

Let $A$ be a commutative unital ring and let $E:=A\{e_0,..,e_n\}$ be the free $A$-module of rank $n+1$ on the elements $e_i$. Let $E^*:=A\{x_0,..,x_n \}$ be the dual of $E$ and let $S:=Sym_A(E^*)\cong A[x_0,..,x_n]$. Let $\mathbb{P}(\mathcal{E}^*):=Proj(S)$ be the projective space bundle of $\mathcal{E}$ (the sheafification of $E$). There is a canonical exact sequence of graded $S$-modules
$\phi: \oplus_{i=0}^n S x_i \rightarrow S(1) \rightarrow 0$
defined by
$\phi( f_0,f_1,..,f_n):= \sum_{i=0}^n f_i x_i .$
Let $\pi:\mathbb{P}(\mathcal{E}^*) \rightarrow Y$ (here $Y:=Spec(A)$) be the projection map. When sheafifying the map $\phi$ we get the tautological sequence
$\pi^*\mathcal{E}^* \rightarrow^{\phi} \mathcal{O}(1) \rightarrow 0.$
Since $\pi^*\mathcal{E}^*$ is a free $\mathcal{O}$-module it follows this proves the invertible sheaf $\mathcal{O}(1)$ is generated by the global sections $s_i:=\phi(x_i)$.
Lemma: Let $X$ be a scheme. For a surjection $\phi: \mathcal{O}_X \{y_1,..,y_l\} \rightarrow \mathcal{L}$ where $\mathcal{L}$ is an invertible sheaf on $X$ and $\mathcal{O}_X\{y_1,..,y_l\}$ is the free sheaf of rank $l$ on the elements $y_i$, it follows $\phi(y_i)$ are global sections generating $\mathcal{L}$. Conversely given a set of global sections $s_1,..,s_l$ of $\mathcal{L}$ that generated $\mathcal{L}$ it follows there is a canonical surjective map of $\mathcal{O}_X$-modules
$\phi: \mathcal{O}_X\{y_1,..,y_l\}\rightarrow \mathcal{L}$ defined on an open set $U$ by $\phi(\sum_i u_iy_i):=\sum_i u_i(s_i)_U$ where $(s_i)_U$ is the restriction of $s_i$ to $U$. The details may be found in Hartshorne's book  Chapter II.7.


Any two groups of three elements are isomorphic - Fraleigh p. 47 4.25(b) The answer has no details. Hence maybe the answer is supposed to be quick. But I can't see it?

Hence I took two groups. Call them $G_1 = \{a, b, c\}, G_2 = \{d, e, f\}$.
Then because every group has an identity, I know $G_1, G_2$ has one each.
Hence WLOG pick $c$ as the identity in $G_1$. I want to match letters so pick $e$  as the identity in $G_2$.
Now we have $G_1 = \{a, b, \color{magenta}{c}\}, G_2 = \{d, \color{magenta}{e}, f\}$.

I know every group has an inverse. But how do I apply this to $G_1, G_2$ to simplify them?

And to prove $G_1, G_2$ are isomorphic, how do I envisage and envision what the isomorphism is?

Update Dec. 25, 2013 (1). Answer from B.S. Why does $ab = b$ fail?
$\begin{align} ab & = b \\ & = bc \end{align}$. What now?

(2.) Do I have to do all the algebra work for $G_1$ for $G_2$? Is there some smart answer?

Update Jan. 8, 2014 (1.) I'm confounded by drhab's comment on Dec. 30 2013. Is drhab saying: Even if the domain's identity is the $\color{green}{third}$ letter $\color{magenta}{c}$  but the codomain's identity is the second letter $\color{magenta}{e}$, $d^{\huge{\color{green}{3}}} = e$ anyways? Hence I should've chosen the $\color{green}{third}$  letter in the codomain as the identity too?

What else is drhab saying about this?

(2.) I don't understand drhab's comment on Dec. 28 2013. Why refer to commutativity? It's not a group axiom? And what are the binary operations?

Update: I didn't realize this before, but by dint of Martin Sleziak's comment, this question is just a sepcial case of Fraleigh p. 63 Theorem 6.10 = Pinter p. 109-111 Theorem 11.1.

Let's work on $G_1$ and assume $c=e_{G_1}$. So $$ac=a, bc=b,cc=c$$ But what about $ab$. It is either $a$, $b$ or $c$. If $ab=a=ac$, since $G_1$ is a group so we can cancel $a$ to have $b=c$ which is wrong because $b\neq c$. The same short story is when we assume $ab=b$, so we just have $ab=c$. This means that $$a=b^{-1}$$ and vise versa. Now what about $aa$? Is it equal to $a$, $b$ or $c$? We have $$aa=a\to a=c\\ aa=b\to aa=a^{-1}\to a^3=c\\ aa=c\to a=a^{-1}=b$$ The first one and the last one is clearly wrong so we just get $a^3=c$. Hence our group $G$ is change to the following form: $$G_1=\{c,a,a^{-1}\}$$ in which $a^3=c$. Now do the same way for $G_2$. It gives us: $$G_2=\{e,d,d^{-1}\}$$ What are the differences between these two groups? Just changing the alphabet $a\to d$ and vice versa? So, there are no differences between them and they are really the same things.

Write down a map of the sets, then prove that it is an isomorphism.  Bijectivity should follow from your construction.  So prove that it is a homomorphism.  Hint:  Order of the elements should help.


Domain and Range Notation I'm doing homework for my Number Theory class and the professor uses a notation that I am not familiar with.

An example problem says:
Consider the function $f(x)=x^2-1$. Express the function as a relation, and find the domain and range.

The answer to this question is $\{(x,x^2-1)|x\in\Bbb R\}$

I have to use the same notation to express the domain and range of $f(x)=\sqrt{2x-1}+1$.

By the domain convention, I know that $2x$ must be greater than or equal to $1$ for the domain, and that the range is defined by where $x$ is defined. How would I write this using this notation?

I propose the frequently used notation: the domain is denoted by $\text{dom}$ while the range is $\text{rng}$. For the 2nd function we have $\text{dom}(f)=\left(\frac{1}{2},+\infty\right)$ and $\text{rng}(f)=\left[0,+\infty\right)$.


On the Cramér-Granville Conjecture and finding prime pairs whose difference is 666 Questions

If $p= \text{NextPrime}[q]$ (the smallest prime greater than $p$), and $p-q = 666,$ what are $p$ and $q$?

(There may be multiple choices. I am interested in finding one.)

  Cramér-Granville Conjecture:
  Defining $p_0=2$, and $p_n$ as the nth odd prime, and the nth prime gap as
  $g_n=p_{n+1}-p_n$, then  $g_n< M \log(p_n)^2$ for some $M>1$.

Reference link: http://mathworld.wolfram.com/Cramer-GranvilleConjecture.html

If $g_n =666$,  what are generally good estimates for $M$ and $p_n$ according to the
Cramér-Granville Conjecture? Can we make use of current sieve technology and probabilistic modeling to solve our problem efficiently?

Commentary & Previous Work

If $g_n = 666$ and $p_n = 18691113008663$, then the conjecture is satisfied for any $M>1$.

We let  $\text{primegap}_{avg} = x/\pi(x) = g_n = 666$.
If $\pi(x) = x/(li(x) + sqrt(x) * log(x)/(8\pi))$, then we have  $x/\pi(x) = 666$ which implies $x = 4.73231\times10^{289}$ and $\pi(x) =7.10558\times10^{286}$.  So our upper bound estimate of $p_n$ will be less than $4.73231\times10^{289}$.

Therefore, we shall focus our attention on finding consecutive prime pairs whose difference is $666$ in the open interval, $(18691113009329, 4.73231*10^{289})$. And we should expect to find approximately $(c/333)(\pi(4.73231×10^{289})-\pi(18691113009329)) = c *7.10558×10^{286}/333$, or $c * 2.133808*10^{284}$ consecutive prime pairs whose difference is $666$ where
$0.5 < c < 1$.

Note:  $c \to 1$ as $x\to\infty$ according to the Polignac Conjecture.

Furthermore, we also expect to discover sufficiently many prime gaps greater than $666$ in the open interval, $(18691113009329, 4.73231×10^{289})$, so that the prime gap density or average of $666$ is maintained.  And according to the Cramér-Granville Conjecture, the maximum prime gap of $\log(4.73231\times10^{289})^2 = 444892$, more or less, exists in the open interval.

Reference links

https://terrytao.wordpress.com/2009/08/18/the-least-quadratic-nonresidue-and-the-square-root-barrier/#comment-472548;

http://www.ams.org/journals/mcom/1989-52-185/S0025-5718-1989-0947470-1/S0025-5718-1989-0947470-1.pdf/;

https://en.wikipedia.org/wiki/Polignac%27s_conjecture;

https://www.quora.com/What-great-conjectures-in-mathematics-combine-additive-theory-of-numbers-with-the-multiplicative-theory-of-numbers/answer/David-Cole-146;

https://terrytao.files.wordpress.com/2009/08/prime-number-theory1.pdf;

Boeyens, Jan C. A.; Levendis, Demetrius C. (2008), Number Theory and the Periodicity of Matter, Berlin: Springer-Verlag, ISBN 978-1-4020-6659-7;

'A Primer in Density Functional Theory', http://link.springer.com/book/10.1007%2F3-540-37072-2;

https://www.wired.com/2016/11/physicists-uncover-strange-numbers-particle-collisions/.

“Repetition and growth of prime gaps are essential for the efficient generation of the integers.”

According to the table compiled by Thomas Nicely, $q=18{,}691{,}113{,}008{,}663$ is the first occurrence of $666$ as a prime gap.  Nicely attributes the value to a 1989 paper by Jeff Young and Aaron Potler.

You can also use prime number theorem to get an estimation. Something like:
$$\text{Find }x\text{ such that}\\\;\\ \frac{x+666}{\ln(x+666)}=\frac{x}{\ln x}+1$$
Edit:

Thanks to the online Wolfram Cloud, I came up with something like this:

gap[x_,d_]:=(x+d)/Log[x+d]-x/Log[x]-1;
FindRoot[gap[x,666],{x,1000}]

which resulted in $x=3.8013\times 10^{17}$


sufficient vs. minimally sufficient vs. complete statistic and MLE for uniform distribution I think I have some confusion going on when it comes to sufficient vs. minimally sufficient vs. complete statistic. I have read some posts here (ex: Finding complete sufficient statistic) but I am still not exactly sure I understand. So I summarized all the questions I have from all the different posts I have read here into one problem/question.

Let $\theta > 0$ be a parameter and let $X_1, ..., X_n$ be a random sample with pdf $f(x|\theta)=\frac{1}{2\theta}$ if $-\theta \leq x \leq \theta$ and $0$ otherwise.
a) Find the MLE of $\theta$

$L(\theta|X)=\frac{1}{(2\theta)^n}I(\theta)_{(max(-X_{(1)},X_{(n)}), \infty)}=\frac{1}{(2\theta)^n}I(|X|_{(n)})_{(0,\theta)}$.
So my question: How do you know if the MLE is $max(-X_{(1)},X_{(n)})$ or $|X|_{(n)}?$.

b) Is the MLE sufficient for $\theta$?

From a), regardless of which is the MLE, both $max(-X_{(1)},X_{(n)})$ and $|X|_{(n)}$ are sufficient by the factorization theorem. $(X_{(1)},X_{(n)})$ would also be sufficient because of the factorization theorem as well.

c) Is the MLE minimally sufficient for $\theta$?

Regardless of which one is the MLE, wouldn't all $max(-X_{(1)},X_{(n)})$, $|X|_{(n)}$, and $(X_{(1)},X_{(n)})$ be minimally sufficient?
From Casella and Berger's book, a sufficient statistic $T(X)$ is minimally sufficient if for every two sample points $x,y$, the ratio $f(x|\theta)/f(y|\theta)$ is free of $\theta$ iff $T(x)=T(y).$
For $max(-X_{(1)},X_{(n)}):$ $\frac{f(x|\theta)}{f(y|\theta)} = \frac{\frac{1}{(2\theta)^n}I(\theta)_{(max(-X_{(1)},X_{(n)}), \infty)}}{\frac{1}{(2\theta)^n}I(\theta)_{(max(-Y_{(1)},Y_{(n)}), \infty)}}$, which is free of $\theta$ iff $max(-X_{(1)},X_{(n)})=max(-Y_{(1)},Y_{(n)})$
For $|X|_{(n)}$: $\frac{f(x|\theta)}{f(y|\theta)} = \frac{\frac{1}{(2\theta)^n}I(|X|_{(n)})_{(0,\theta)}}{\frac{1}{(2\theta)^n}I(|Y|_{(n)})_{(0,\theta)}}$, which is free of $\theta$ iff $|X|_{(n)}=|Y|_{(n)}$
For $(X_{(1)},X_{(n)})$: $ \frac{f(x|\theta)}{f(y|\theta)}=\frac{\frac{1}{(2\theta)^n}I(X_{(1)})_{(-\theta, \infty)}I(X_{(n)})_{(-\infty, \theta)}}{\frac{1}{(2\theta)^n}I(Y_{(1)})_{(-\theta, \infty)}I(Y_{(n)})_{(-\infty, \theta)}}$, which is free of $\theta$ iff $(X_{(1)},X_{(n)})=(Y_{(1)},Y_{(n)})$

d) Is the MLE complete for $\theta$?

Regardless of which is the MLE, which of $max(-X_{(1)},X_{(n)})$, $|X|_{(n)}$, and $(X_{(1)},X_{(n)})$ would be complete?

As is possible to see in the comment section, all questions in the post have been answered before. Take a look at the links provided in the comment section. They  summarize everything well.


What are all the concordant forms $n$ such that $a^2+b^2 = c^2,\,a^2+nb^2=d^2$ for $n<1000$? Part I. The list of congruent numbers $n<10^4$ such that the system,

$$a^2-nb^2 = c^2$$
$$a^2+nb^2 = d^2$$

has a solution in the positive integers is known (A003273)

$$n = 5, 6, 7, 13, 14, 15, 20, 21, 22, 23, 24, 28, 29, 30, 31, 34,\dots$$

Part II. Strangely, for the similar concordant forms/numbers $n$ such that,

$$a^2+b^2 = c^2$$
$$a^2+nb^2 = d^2$$

is not even in the OEIS,

$$n = 1,7,10,11,17,20,22,23,24,27,30,31,34,\dots$$

The list of $104$ prime $n<10^3$ is known (by Kevin Brown, David Einstein (hm?), and Allan MacLeod) though several troublesome primes were excluded assuming the Birch/Swinnerton-Dyer conjecture.

  Question: Anybody knows how to generate the list of all concordant forms $n<1000$? (Elkies describes a method here.)

P.S. Incidentally, the special case $n=52$ appears in equal sums of like powers. Let,

$$a^2+b^2 = c^2$$
$$a^2+52b^2 = d^2$$

Then for $k=1,2,4,6,8,10,$

$$(8b)^k + (5a-4b)^k + (-a-2d)^k +  (a-2d)^k + (-5a-4b)^k  + (-12b+4c)^k + (12b+4c)^k =\\
(4a+8b)^k + (3a-2d)^k + (-3a-2d)^k + (-4a+8b)^k + (-16b)^k + (a+4c)^k + (-a+4c)^k$$

found by J. Wroblewski and yours truly. An initial primitive solution is $a,b,c,d = 3,4,5,29$ and an infinite more.


Use comparison test to show that $\sum^{+\infty}_{k=1} \frac{1}{k(k+1)(k+2)}$ converges and find its limit Use comparison test to show that $\sum^{+\infty}_{k=1} \frac{1}{k(k+1)(k+2)}$ converges and find its limit

I tried expanding out the denominator, and then using the comparison test with $\frac{1}{k^3}$ but I think this is an incorrect use of the comparison test as I get divergence. I know that the limit is $\frac{1}{4}$ but I am not sure how to use the comparison test to then apply the limit.

Yes you compare with $\frac{1}{k^3}$, as follows,
$$\frac{k^3}{k(k+1)(k+2)}\rightarrow 1$$ so the series converges, because $\sum \frac{1}{k^3}$ converges.


Trouble with gradient intuition I'm currently learning about gradients, and I thought khanacademy could help me acquiring some intuition. The actual computation is clear to me, however I'm having trouble understand the intuition.

This is the video I'm talking about, and the part I have a part with is 6:11. He says the gradient vector shows the direction you have to travel in the x,y-plane in order to get a maximum slope in the z-direction.

This sounds like gibberish to me. How do you have to 'travel' it? Why do you get a maximum slope? I just have no clue.

At each point in the $xy$-plane the function you have gives you a value. Normally this value is visualized as altitude, that is, a $z$-value, but it can just as easily be anything else. Examples that come to mind are grayscale tones (if the $xy$-plane is a black-and-white picture), density (if the $xy$-plane is a plate made from varying materials), or, maybe the most important one, potential in some force field (often gravitational or electric fields). Of course, in practice, these examples rarely gives you a function that is nice to work with, but in textbook examples it usually works out.

Now, imagine that you are walking around in the $xy$-plane, sniffing or measuring that value, whatever it represents. At some point in time you're standing at the point $(5, 3)$, and the gradient of your function at that point is $(-1, 2)$. That means if you turn around so that you're facing the direction $(-1, 2)$ at the point $(5, 3)$ (that is, you're looking directly at the point $(4, 5)$), then that is the direction where (at least for the first gazillionths of a meter you travel) your value will increase the fastest, out of all the directions you can pick.

I can do even better. I can tell you how much it will grow for that gazillionths of a meter. Since the amplitude of the gradient is $\sqrt 5$, your function value will grow approximately  $\dfrac{\sqrt{5}}{\text{gazillion}}$ of whatever unit it's measured in. (Approximately because even over that short distance, the gradient might change a little bit. Studying how these changes affect the end result is what partial differential equations, and especially using computers to solve them, are all about. But that's another story entirely)

Traveling around in the $xy$-plane is something we mathematicians do all the time. When you finally get some intuition, it's much easier (and more fun) to play around with that intuition if you imagine yourself being at a point in the plane and walk around, rather than just imagining a point moving around.

Also, as for why the gradient gives you a maximal slope, say again that you have a gradient of $(-1, 2)$ at the point $(5, 3)$. That means if you go from that point, and move directly parallel to the $x$-axis (for a very short distance), in the positive direction (towards $(6, 3)$), then your function value will decrease at the speed of $1$ per meter travelled. This is, after all, how the $-1$ in $(-1, 2)$ came to be in the first place. So if you want the value to increase, you're better off moving in the negative $x$-direction.

Likewise, if you move in the positive $y$-direction (towards $(5, 4)$), your function value will increase by $2$ per meter travelled. So you can see that the function grows faster in the positive $y$-direction than in the negative $x$-direction. However, the direction that will give you the fastest growth is a balanced mix of the two. Hence both the directions $(-1, 0)$ and $(0, 1)$ will give you growth, but the fastest growth is the direction $(-1, 2)$. Twice as much in the $y$-direction because the function grows twice as fast that way as in the $x$-direction.

Gradient intuition.
I had problems too trying to visualize intuitively the gradient in R 3.
I want to share how I finally got it.
For simplicity let us consider a function valued at x=0, y=0 that takes a value z=0 at that point. The shape of this function does not matter because we will look at the tangent plane at x=0, y=0, z=0. Let us also assume that the partial derivative with respect to x at x=0 is 1 and that the partial derivative with respect to y is also 1.
Let us now look at the tangent plane defined by the two partial derivatives.
(The tangent lines).
The partial derivative with respect to x will bisect the z x plane with a slope of 45 degrees. Similarly, the partial derivative with respect to y will bisect the z y plane with a slope of 45 degrees too.
So these two lines define our tangent plane. This tangent touches our z function at z=0.
Now imagine to project a plane, parallel to the x y plane that that intersects the z axis at z=1.
This plane parallel to the x y plane will intersect the tangent plane ( the one mentioned above defined by the two partial derivatives). The intersection is a line.
This line passes through the points  x=0, y=1, z=1 and x=1, y=0, z=1.
Now let us project this line on the x y plane. This projection is the contour line for z=1 on the x y plane.
This contour line (A) passes through the points x=1, y=0, z=0 and x=0, y=1, z=0.
Now, what is the shortest direction to take from the origin (the point where the tangent plane touches our function) to reach on the the x y plane the contour line for z=1 described above (A)?
It is obvious that the direction is 45 degrees, or the components of the the vector describing the direction are <1,1>. Same as the partial derivatives.
One can see that this vector is perpendicular to the contour line on the x y plane.
Now assume that the partial derivative with respect to y is 2, while the partial derivative with respect to x remains 1.
The tangent plane will shift. The plane parallel to the x y plane and with z= 1 will intersect the new tangent plane at points
x=1, y=0, z=1 and x=0, y=1/2, z=1.
The projection of this line on the x y will give us the new contour line for z=1.
This new contour line (B) will pass through the points
x=1, y=0, z=0 and x=0, y=1/2 z=0.
What is the direction to take now to reach from the origin this new contour line (B) on the x y plane?
The vector must be perpendicular to the new contour line (B) and so has components <1,2>, same numbers as the partial derivatives.


Taking the negative of a continued fraction If I have a continued fraction for an irrational number $z= \langle a_0;a_1,a_2,a_3,\ldots\rangle$ it seems that $(-1)*z = \langle-a_0;-a_1,-a_2,-a_3,\ldots\rangle$. Is this true?

In general, if you have the continued fraction representation for $y$ and $z$ can you say something about the continued fraction representation of $y*z$?

While evaluating the continued fraction of the negative coefficients of the continued fraction expansion of $z$ does indeed evaluate to $-z$, your formula $[-a_0; -a_1, \dots, ]$ is not regarded as "the continued fraction" of $-z$, which is usually defined using the Euclidean algorithm resulting in non-negative coefficients after the first. In general, for $z \in \mathbb{R}$ and $z= [a_1; a_2, a_3 \dots]$, then
\begin{align}
-z= [-a_1-1;1,a_2-1, a_3, \dots],
\end{align}
where the terms in the ellipses are identical. For example, $\frac{4}{3} = [1,3]$, while $-\frac{4}{3} = [-2,1,2]$. To address your second question, there are formulas to compute the continued fraction expansion of $\frac{az+b}{cz+d}$, where $a, b, c, d \in \mathbb{Z}$, relying only on the continued fraction expansion of $z$ and certain $2 \times 2$ matrices defined using the coefficients. See An Introduction to Continued Fractions by van der Poorten.

If continued fraction
$$
a_0+\frac{1}{\displaystyle a_1+\frac{1}{a_2+\ddots}}
$$
converges to $z$, where all $a_k$ and $z$ are complex numbers, then continued fraction
$$
-a_0+\frac{1}{\displaystyle -a_1+\frac{1}{-a_2+\ddots}}
$$
converges to $-z$.

Is that what you mean?  You should be able to prove it!


Tricky, Interval Proof, is my Proof correct? Prove that if  x is in $[0,b]$, then $x=tb$ for some $t$ where  $0 \le t \le 1$

My Attempt at a proof:

Interval Midpoint: $ x = \frac{b}{2} $

$  \frac{b}{2} = tb $
$  \frac{1}{2} = t $

This is implies that value of $t$ is equal to  $ \frac{1}{2} $  when $x$ is the mid point of  $[0,b]$.

Thus for the first element in the interval: $  t = \frac{1}{2}- \frac{1}{2} = 0  \rightarrow x= 0b = 0 $

Thus for the last element in the interval: : $  t = \frac{1}{2}+ \frac{1}{2} = 1 \rightarrow x= 1b = b $

Q.E.D

Is this a correct proof? is my reasoning correct? I am self learning maths but not sure if this rigorous enough.

You're given that $0\le x\le b$ Suppose $b>0$. Then
$$
0\le\frac{x}{b}\le1
$$
and, setting $t=\frac{x}{b}$, you have $x=tb$ as required.

If $b=0$, then $x=0$ and $t=1$ is good.

Your proof is not enough: that's just for three points, but when $b>0$ there are infinitely many points in $[0,b]$, so you can't do it by cases.

HINT: solve the inequality $$0\le tb\le b$$
since $b\geq 0$ we get $t\geq 0$ and $tb\le b$ gives $$b(1-t)\geq 0$$ and since $b\geq 0$ we get $t\le 1$


Solve two equations for $a$ and $b$ \begin{cases}
c_2=\dfrac{c_1}{a} \left( \left(\dfrac{c_3}{b}\right)^3 - 1 \right) \\[2ex]
b^2 = a^2 + c_3^2 + 2(a)\, (c_3)\, (c_4)  \\
\end{cases}

I am stuck at this point. Not sure on how to move forward.  ( A small change made)

The system \begin{cases}
c_2=\dfrac{c_1}{a} \left( \left(\dfrac{c_3}{b}\right)^3 - 1 \right) \\[2ex]
b^2 = a^2 + c_3^2 + 2ac_3c_4  \\
\end{cases}

is equivalent to

\begin{cases}
a=c_{1}\dfrac{c_{3}^{3}-x^{3}}{c_{2}x^{3}}\\[2ex]
b=x ,
\end{cases}

where $x$ is a solution of the following septic equation I've obtained in SWP:

\begin{eqnarray*}
0 &=&c_{2}^{2}x^{7}+c_{3}c_{2}^{2}x^{6}+\left(
-c_{1}^{2}+2c_{1}c_{3}c_{4}c_{2}\right) x^{5}+\left(
2c_{3}^{2}c_{1}c_{4}c_{2}-c_{3}c_{1}^{2}\right) x^{4} \\[2ex]
&&+\left( 2c_{3}^{3}c_{1}c_{4}c_{2}-c_{1}^{2}c_{3}^{2}\right)
x^{3}+c_{1}^{2}c_{3}^{3}x^{2}+c_{1}^{2}c_{3}^{4}x+c_{1}^{2}c_{3}^{5}.
\end{eqnarray*}

  I am stuck at this point.

The general septic equation cannot be solved algebraically.

Note: In the present form of the system $a$ should be different from $0$. So $b=c_{3},a=0$
is no longer a solution.

sub your equational value for a into your second line (b squared etc)
ie sub c1c2((c3b)3−1) into b2=a2+c23+2(a)(c3)(c4)
Hope this helps you :)


ordinary differential equation solving I have a diffeq:

I have a nonlinear Diffeq:

$$\frac{d^2x}{dt^2}+\beta \frac{dx}{dt}+\varepsilon e^{- \lambda x} = f(t) $$

where $f(t)$ is a function that is known, and $\beta$ and $\lambda$ are constants that are known. Also, we know that $\epsilon$ is a constant parameter that is small.

I first need to obtain the zero order solution $x_0$, before finding the first order solution $x_1$

The first thing that I need to do is to use asymptotic expansions to obtain solutions of order $\epsilon=0$ and (TYPO)

Note that general solution for f(t) that will have two unknown constants.

UPDATE: After the first order term is solve, it needs to be plugged back in. The exponential needs to be linearized and things should start cancelling out. I am not sure how to do this, I just know this is what needs to be done.

UPDATE2: Correction, $\epsilon = 1$ was a typo. It should be $\epsilon^1$
I need to find a solution in the form:

$$x(t)=x_0(t)+\epsilon^1x_1(t)+\epsilon^2x_2 (t) + ... $$

So initially, $\epsilon$ needs to be set to 0 in order to obtain $x_0$. To find $x_1$, I need $\epsilon^1$

UPDATE3: I know now that I need to plug:

$$x=x_0+\epsilon_1x_1 $$ back into the original equation

Thus:

$$\frac{d^2}{dt^2}(x_0+\epsilon_1x_1) + \beta\frac{d}{dt}(x_0+\epsilon_1x_1)+\epsilon \times exp(-\lambda(x_0+\epsilon_1x_1))  $$

Then

$$\frac{d^2}{dt^2}x_0+\frac{d^2}{dt^2}\epsilon_1x_1+\beta \frac{d}{dt}x_0 +\beta \frac{d}{dt}\epsilon_1 x_1+\epsilon \times exp(-\lambda x_0))+\epsilon \times exp(-\lambda \epsilon_1 x_1)$$

I think then the $x_0$ terms may cancel with f(t) or something like that? It may be some sort of approximation.

I still need to linearize the exponential. Any help is appreciated.

Update4: Taking the solution a little but further...

We know that:

$$\frac{d^2x_0}{dt^2}+\beta \frac{dx_0}{dt} = f(t) $$

So, those terms all cancel. And now we have:
$$\frac{d^2}{dt^2}\epsilon_1x_1 +\beta \frac{d}{dt}\epsilon_1 x_1+\epsilon \times exp(-\lambda(x_0+\epsilon_1x_1))=0$$

But we dont want $\epsilon^2$ terms, to part of the exponential goes away as well.

We are left with:
$$\frac{d^2}{dt^2}\epsilon x_1 +\beta \frac{d}{dt}\epsilon x_1+\epsilon \times exp(-\lambda x_0)=0$$

Where we know $x_0$. This now means that the exponential is no longer a function of arbitrary x.

$$\frac{d^2x_1}{dt^2} +\beta \frac{dx_1}{dt}+ e^{-\lambda x_0}=0$$

Where we know $x_0$. Should I use method of undetermined coeffs?

solving this 2nd order ODE

you solved this kind of 2nd order diff eqn before for x0

consider e^-ramda xo as a f(t)

the solution will be similar


What independent events actually means? A bin has $2$ balls, one is black and one is white. Every round a uniformly chosen ball is drawn from the bin. If the color of the ball is white, then the ball is returned to the bin with an additional white ball. If the ball is black the experiment is over. Let $X$ be the number of rounds in the above experiment.

  Compute the distribution of $X$.

$$P(X=k)= ?$$

I have 3 questions regarding my (partial) solution.

My (partial) solution:

For $1<=i<=k$, I defined $X_i$ to be the number of white balls drawn in the $i$th round.

Is it valid to conclude without any explanation or proof that (and everything I did so far):
$$ P(X=k)=P(X_1=1,X_2=1,....X_{k-1}=1,X_k=0) $$
for every $1<=i<j<=k$, $X_i$ and $X_j$ seems to be dependent because if I draw a white ball then in the next round, the probability to draw a white ball is smaller. Also, the only way to go to the next round is if we drew a white ball. Then, logically, why they are independent? (in the following section I'm trying to prove that they are independent mathematically).

I was trying to prove that for every $1<=i<j<=k-1$, $X_i$ and $X_j$ are independent (Eventually, I will prove for $1<=i<j<=k$).

$$ P(X_i=1,X_j=1)=P(X_j=1 | X_i=1)*P(X_i=1)=\frac{j}{j+1} * \frac{i}{i+1} $$
The above is true because $i<j$, so we can conclude that because $P(X_i=1)$ then the experiment isn't over so the event $X_i=1$ doesn't have any infloence over $P(X_j=1)$.

But the opposite seems to me not true:
$$ P(X_i=1,X_j=1)=P(X_i=1 | X_j=1)*P(X_j=1)= 1 * \frac{j}{j+1} $$

The above is true because $i<j$ so the experiment wasn't over at the round number $i$, so at the round number $i$ we must have drawn a white ball - and there is no other option.

$ P(X_i=1,X_j=1)=P(X_j=1 | X_i=1)*P(X_i=1)=P(X_i=1 | X_j=1)*P(X_j=1)$. Why am I wrong?

A difficulty encountered by your argument is that you do not have a clear definition of $X_j$ from the problem statement.

You seem to be tempted to say $X_j = 0$ if the experiment ends before the $j$th drawing.
But if there is no $j$th drawing, is "the number of white balls in the $j$th drawing" even defined?

One way to resolve this is to make it part of the definition of $X_j$ that $X_j = 0$ if the experiment ends before $j$ balls are drawn.
Then $X_i$ and $X_j$ clearly are not independent.
In particular, because $X_2 = 1 \implies X_1 = 1,$
$$P(X_1 = 1 \cap X_2 = 1) = P(X_2 = 1) \neq P(X_1 = 1)P(X_2 = 1).$$

Both of your attempts to evaluate $P(X_1 = 1 \cap X_2 = 1)$
by means of conditional probability were incorrect,
because you assumed that in general $P(X_i = 1) = \frac{i}{i+1}$
for every positive integer $i.$
In fact, according to the way you defined $X_i,$
$P(X_i = 1) < \frac{i}{i+1}$ whenever $i > 1.$

Here is a way out of this difficulty. The value of $X_j$ is irrelevant to the outcome of the experiment when there is no $j$th drawing.
So you can define $X_j$ any way you like in that case.
Define $X_j$ as follows for any positive integer $j$:
$$
X_j = \begin{cases}
1 & \text{at least $j$ balls are drawn and the $j$th ball is white,} \\
0 & \text{at least $j$ balls are drawn and the $j$th ball is black,} \\
1 & \text{with probability $\frac{j}{j+1}$ if fewer than $j$ balls are drawn,}\\
0 & \text{otherwise.}\\
\end{cases}
$$

When you define $X_j$ this way, it turns out that indeed
$P(X_j = 1) = \frac{j}{j+1},$
and $X_j$ is independent of $X_i$ whenever $i \neq j.$

Now you can proceed to compute $P(X = k)$ without getting hung up on the dependence of $X_j$ on $X_i$ or on whether $X_j$ is even defined.

Yes, $P(X=k)=P(1, 1,\cdots1, 0)$, by definition.
Of course they are dependent, because $X_i=0\implies P(X_j=0)=P(X_j=1)=0$. And conversely $X_j=0\implies P(X_i=1)=1$, $X_j=1\implies P(X_i=1)=1$.

In this exercise, we have an unusual situation, as $X_j=0,X_j=1$ are not the only cases. $X_j\text{ not drawn}$ is also possible, with some probability.


What is the difference between stochastic process and random variable? I am having a hard time grasping the core difference between a random variable and a stochastic process.

A random variable assigns a number to every outcome of an experiment.
A random process assigns a function of time to every outcome of an experiment.
But the values of this function of time can be represented with ONE SINGLE random variable as well. So what is the point in having a stochastic process when you can represent an experiment with only random variables? Could somebody make one or two examples where the difference is clear?

Appreciate it

A stochastic process is a family of random variables indexed by some set, usually $\mathbb{Z}^{n}$ or $\mathbb{R}^{n}$.
It's additional structure over random variables that let you establish notions of trajectories, association over a space and other interesting properties.
In empirical studies, the set may represent discrete time ($\mathbb{Z}$), continuous time ($\mathbb{R}$), geographical location in a map on a given discrete time ($\mathbb{R}^2 \times \mathbb{Z}$) and et cetera.

A random process is a function, for instance a random walk over the plane.
A random variable has a value, for instance the average distance of the random walk after a specified number of steps.


Limit of a sequence $\sqrt{1+\sqrt{\underbrace{1+ \ldots +\sqrt{1}}_{n}}}$ $\sqrt{1+\sqrt{\underbrace{1+ \ldots +\sqrt{1}}_{n}}}$

I think the limit is $2$ for intuitive reasons.
Now, to prove it, I need to use the definition of a sequence's limit.
So, I need to solve the following inequality

$|\sqrt{1+\sqrt{\underbrace{1+ \ldots +\sqrt{1}}_{n}}}-2| < \epsilon$

for an arbitrary $\epsilon$.
But I'm not sure it is possible.
One implication of the inequality is

$\sqrt{1+\sqrt{\underbrace{1+ \ldots +\sqrt{1}}_{n}}} < \epsilon + 2$

What I found out is the same as

$ \underbrace{( \ldots ((}_{n-1}\epsilon + 2\underbrace{)^2 - 1)^2 \ldots )^2}_{n-1} - 2 > 0$

But I'm not sure it is solvable in this form.

Hints: Suppose the limit exists, and denote it by $L$. Then notice that $L=\sqrt {1+L}$, from which you get $L^2-L-1=0$ and you can find the roots to obtain candidates for the limit. If you can then argue why the limit does in fact exist, and you can eliminate one of the limits, then you will have found the answer. In any case, $2$ does not solve the equation, so the limit is certainly not $L=2$. (There are better ways of showing that a sequence converges than playing with $\varepsilon $. Ask yourself, is the sequence monotone? is it bounded?)

Based on the statement on the problem and my interpretation of the "$\dots$," it appears that all of the previous answers given are not correct.  The answers seem to assume an embedded square root, but I don't see that here.  The sequence as written seems to be nothing more than $a_n=\sqrt{1+\sqrt{n}}$, which clearly diverges.


Importance of Representation Theory Representation theory is a subject I want to like (it can be fun finding the representations of a group), but it's hard for me to see it as a subject that arises naturally or why it is important. I can think of two mathematical reasons for studying it:

The character table of a group is packs a lot of information about the group and is concise.
It is practically/computationally nice to have explicit matrices that model a group.

But there must certainly be deeper things that I am missing. I can understand why one would want to study group actions (the axioms for a group beg you to think of elements as operators), but why look at group actions on vector spaces? Is it because linear algebra is so easy/well-known (when compared to just modules, say)?

I am also told that representation theory is important in quantum mechanics. For example, physics should be $\mathrm{SO}(3)$ invariant and when we represent this on a Hilbert space of wave-functions, we are led to information about angular momentum. But this seems to only trivially invoke representation theory since we already start with a subgroup of $\mathrm{GL}(n)$ and then extend it to act on wave functions by $\psi(x,t) \mapsto \psi(Ax,t)$ for $A$ in $\mathrm{SO}(n)$.

This Wikipedia article on particle physics and representation theory claims that if our physical system has $G$ as a symmetry group, then there is a correspondence between particles and representations of $G$. I'm not sure if I understand this correspondence since it seems to be saying that if we act an element of G on a state that corresponds to some particle, then this new state also corresponds to the same particle. So a particle is an orbit of the $G$ action? Anyone know of good sources that talk about this?

One comment about your sentence "this seems to only trivially invoke representation theory". It might be surprising, but such obvious representations are actually the source of interesting mathematics, and a lot of effort of representation theorists is devoted to studying them.

More precisely: start with a group (in your example $SO(n)$) acting on a space $X$ (in your example $\mathbb R^n$), and look at the space of functions on $X$ (let me write it $\mathcal F(X)$; in a careful treatment, one would have to think about whether we wanted continuous, smooth, $L^2$, or some other kind of functions, but I will suppress that kind of technical consideration).

Then, as you observe, there is a natural representation of $G$ on $\mathcal F(X)$.

You are right that from a certain point of view this seems trivial, because the representation is obvious.  Unlike when one first learns rep'n theory of finite groups, where one devotes a lot of effort to constructing reps., in this context, the rep. stares you in the face.

So how can this be interesting?

Well, the representation $\mathcal F(X)$ will almost never be irreducible. How does this representation decompose?

Suddenly we are looking at a hard representation theoretic problem.

First, we have to work out the list of irreps of $G$ (which is much
like what one does in a first course on rep'n theory of finite groups).
Second, we have to figure out how $\mathcal F(X)$ decomposes, which involves representation theory (among other things, you have to develop methods for investigating this sort of question), and also often a lot of analysis (because typically $\mathcal F(X)$ will be infinite dimensional, and may be a Hilbert space, or have some other similar sort of topological vector space structure which should be incorporated into the picture).

I don't think I should say too much more here, but I will just give some illustrative examples:

If $ X = G = S^1$ (the circle group, say thought of as $\mathbb R/\mathbb Z$)
acting on itself by addition, then the solution to the problem of decomposing
$\mathcal F(S^1)$ is the theory of Fourier series.  (Note that a function on $S^1$ is the
same as a periodic function on $\mathbb R$.)
If $ X = G = \mathbb R$, with $G$ acting on itself by addition, then
the solution to the above question (how does $\mathcal F(\mathbb R)$ decompose under the
action of $\mathbb R$) is the theory of the Fourier transform.
If $ X = S^2$ and $G = SO(3)$ acting on $X$ via rotations, then decomposing $\mathcal F(S^2)$ into irreducible representations gives the theory of spherical harmonics.
(This is an important example in quantum mechanics; it comes up for example in the theory of the hydrogen atom, when one has a spherical symmetry because the electron orbits the nucleus, which one thinks of as the centre of the sphere.)
If $ X = SL_2(\mathbb R)/SL_2(\mathbb Z)$ (this is the quotient of a Lie group by a discrete subgroup, so is naturally a manifold, in this case of dimension 3),
with $G = SL_2(\mathbb R)$ acting by left multiplication, then the problem of decomposing $\mathcal F(X)$ leads to the theory of modular forms and Maass forms, and is the first example in the more general theory of automorphic forms.

Added: Looking over the other answers, I see that this is an elaboration on  AD.'s answer.

Although many years later, for the sake of completeness -regarding the last question on references about uses in physics- i believe one might find lots of interest on the classic text by J. Baez and J. Huerta:

The Algebra of Grand Unified Theories

Lots of the mathematical physics developed during the last decades of the 20th century is reviewed there, in a particularly didactic way.


Deal combinations in solitaire game 'Klondike' It seems that the solitaire card game 'klondike' presents some head scatching when trying to work out odds for winnable games etc, but I would image a 'simpler' question is 'how many different deal combinations are there in the 'Klondike' card game?

Assuing that you're referring to the standard version of the game as described at Wikipedia, and that the order of the cards in the deck is part of what distinguishes deals, the answer is $52!$, the number of different permutations of $52$ cards. If you disregard the order of the cards in the deck and are only interested in the different ways of dealing the $28$ cards on the table, the answer is $\displaystyle\frac{52!}{(52-28)!}$: There are $52$ choices for the first card to deal, $51$ for the next one, ..., and $52-27$ for the last one.

If you are looking for the number based on any card in any possible position then the correct answer would be somewhere north of $8\cdot 10^{67}$
(As there are $52!$ ways to deal a pack of cards)


Converse to a result of convergence of $\lim\frac{1}{x}\sum_{n\leq x}a_n=k$ In this post it is proved that, if $(a_n)$ is a sequence in $\mathbb C$:
$$
\lim_{x\to\infty}\frac{1}{x}\sum_{n\leq x}a_n=k\implies\lim_{x\to\infty}\frac{1}{\log x}\sum_{n\leq x}\frac{a_n}{n}=k.
$$
Is the converse to this statement true? In other words, is there a sequence $(a_n)$ with
$$
\lim_{x\to\infty}\frac{1}{\log x}\sum_{n\leq x}\frac{a_n}{n}=k
$$
but $\lim_{x\to\infty}\frac1x\sum_{n\leq x}a_n$ converges to some other limit, or diverges? Thanks.

If $a_n=(-1)^n n$, then:
$\displaystyle \lim_{x \rightarrow +\infty} \dfrac{1}{\log x} \sum_{n\leqslant x} \dfrac{a_n}{n} = \lim_{x \rightarrow +\infty} \dfrac{1}{\log x} \sum_{n\leqslant x} (-1)^n = 0 $
$\displaystyle \lim_{x \rightarrow +\infty} \dfrac{1}{ x} \sum_{n\leqslant x}a_n$ doesn't exist.


Find the exhaustive values of "a", given the condition If the equation $2^{2x} + a*2^{x+1} + a + 1=0$ has roots of opposite sign then the exhaustive values of a are?

I tried taking $2^x = t$. But then didn't know what to do.

The equation became, $t^2 + 2at + a +1 =0$. But then, what conditions should I impose?

Since one root is negative and the other positive, the only conclusion that I could draw was that $x<0$ for the first condition and $x>0$ for the second condition. But, I don't know how do I proceed from here?

Any help would be appreciated.

So far, so good! Now: if $x>0$, then $t=2^x>1$; and if $x<0$, then $t=2^x<1$. So for your new equation $t^2+2at+a+1=0$ you want it to have two roots $t$ of which one is greater and one is less than $1$. Comparing roots with $1$ seems a bit difficult; comparing with $0$ would be easier… So let's make another change of variables: $t$ is greater than or less than $1$ if and only if $y=t-1$ is greater than or less than $0$ respectively. Substituting $y=t-1$, i.e. $t=y+1$, the quadratic equations transforms into
$$\begin{split}
t^2+2at+a+1=0 &\iff (y+1)^2+2a(y+1)+a+1=0\\
&\iff y^2+2(a+1)y+(3a+2)=0.
\end{split}$$

Now, we have to satisfy two conditions:

A quadratic equation (with real coefficients and the leading term of $1$, as is the case here) has roots of opposite signs if and only if its constant term is negative, so we need $3a+2<0$.
But also don't forget that $t=2^x$ can't be negative! In other words, $t>0$, and therefore $y=t-1>-1$. For a parabola $f(y)=y^2+2(a+1)y+(3a+2)$ opening up with one positive and one negative root, we need to guarantee that even the negative root is greater than $-1$, or in other words, that $-1$ lies to the left of both roots. This conditions will be satisfied if $f(-1)>0$, i.e. $(-1)^2+2(a+1)\cdot(-1)+(3a+2)>0$, which simplifies to $a+1>0$.


The Eigenvalues of a block matrix It is known that all $B$, $C$ and $D$ are $3 \times 3$ matrices. And the eigenvalues of $B$ are $1, 2, 3$; $C$ are $4, 5, 6$; and $D$ are $7, 8, 9$. What are the eigenvalues of the $6 \times 6$ matrix
$$\begin{pmatrix}
B & C\\0 & D
\end{pmatrix}$$
where $0$ is the $3 \times 3$ matrix whose entries are all $0$.
From my intuition, I think the eigenvalues of the new $6 \times 6$ matrix are the eigenvalues of $B$ and $D$. But how can I show that?

By definition, an eigenvalue $\lambda$ of the block matrix $A$ satisfies

$$\det \begin{pmatrix} B-\lambda I & C \\ 0 & D-\lambda I \end{pmatrix} = 0.$$

Using a property of block matrix determinants, we have

$$\det \begin{pmatrix} B-\lambda I & C \\ 0 & D-\lambda I \end{pmatrix} = \det(B-\lambda I)\det(D-\lambda I) = 0$$

Thus the eigenvalues of $B,D$ are also the eigenvalues of $A$.

$BE=-CD^{-1}$
By your assumptions $B$ is invertible, hence we have:
$-B^{-1}CD^{-1}$


What is the probability that the system is operational? Consider a system of components in which there are 4 independent components, each of which possesses an operational probability of 0.9. The system does have a redundancy built in such that it does not fail if 3 out of the 4 components are operational. What is the probability that the total system is operational?

I have no idea how to approach this. Can someone help?

The system fails when two, or more, components fail. If each component falls out independently (which is probably true) then we have a binomial distribution $Bin(0.9)$. Now it is easy to calculate the probability of system failure.  $$P( \text{failure}) = 1 - P(\text{operational}) = 1 - \left(P(\text{#fallout} = 0)+ P(\text{#fallout} = 1)\right)$$

$$ = 1-\left(\binom{4}{0}0.9^4 + \binom{4}{1}0.9^3\cdot0.1\right)$$


Calculate $E[a^N]$ for Poisson random variable $N$ Calculate $E[a^N]$ for a Poisson random variable $N$ with parameter $\lambda \cdot T$, where $a>0$ is a constant independent of $N$.

My answer:

\begin{align}
E[a^N] &= \sum_{n=0}^{\infty} \frac{e^{-\lambda T}(\lambda T)^n}{n!} a ^n\\
&= \sum_{n=0}^{\infty} \frac{e^{-\lambda T}(a\lambda T)^n}{n!} \\
&= \sum_{n=0}^{\infty} e^{-\lambda T} e^{a\lambda T}\frac{e^{-a\lambda T}(a\lambda T)^n}{n!} \\
       &= \sum_{n=0}^{\infty} {e^{-\lambda T(1-a)}} \frac{e^{-a\lambda T}(a \lambda T)^n}{n!}\\
       &= {e^{-\lambda T(1-a)}}  \sum_{n=0}^{\infty} \frac{e^{-a\lambda T}(a \lambda T)^n}{n!}\\
       &= {e^{-\lambda T(1-a)}}\\
\end{align}

Is this right?

Yes, this is correct.  In general, if $X$ is a random variable (typically a discrete random variable) then $\mathbb{E}[z^X]$ is referred to as the probability generating function (PGF for short) of $X$.  Your answer for the Poisson random variable of mean $\lambda T$ is correct, as can be verified on Wikipedia.

\begin{align}
\operatorname{E} \Big[a^N\Big] &= \sum_{n=0}^\infty \frac{e^{-\lambda T}(\lambda T)^n}{n!} a^n \\[10pt]
&= \sum_{n=0}^\infty e^{-\lambda T(1-a)} \frac{e^{-a\lambda T}(a \lambda T)^n}{n!} \longleftarrow \text{I'm not sure what you did here.}
\end{align}

I would do this:
\begin{align}
\operatorname{E} \Big[a^N\Big] &= \sum_{n=0}^\infty \frac{e^{-\lambda T}(\lambda T)^n}{n!} a^n \\[10pt]
&= e^{-\lambda T} \sum_{n=0}^\infty \frac{(\lambda T)^n}{n!} a^n & \longleftarrow & \text{ This can be done because $e^{-\lambda T}$ does not} \\
& & & \text{ change as $n$ goes from $0$ to $\infty$.} \\[10pt]
& = e^{-\lambda T} \sum_{n=0}^\infty \frac{(a\lambda T)^n}{n!} \\[10pt]
& = e^{-\lambda T} e^{a\lambda T} = e^{-\lambda T(1-a)}.
\end{align}


Convert Riemann sum to definite integral $ \lim_{n\to\infty}\sum_{i=1}^{n} \frac{21 \cdot \frac{3 i}{n} + 18}{n} $ The limit

$\quad\quad \displaystyle \lim_{n\to\infty}\sum_{i=1}^{n} \frac{21 \cdot \frac{3 i}{n} + 18}{n} $

is the limit of a Riemann sum for a certain definite integral

$\quad\quad \displaystyle \int_a^b f(x)\, dx $

a = ?

b = ?

f(x) = ?

Attempt at solution:

So I know:

$ \displaystyle \sum_{i=1}^{n} f(a + i dx) dx $ = $ \int_a^b f(x)\, dx $

and that $x_i = a + i \cdot \frac {b-a}{n}$

So I will try to rewrite my riemann sum as convenient:

$\frac {3}{n} (7 \cdot \frac {3i}{n} + 6) $

so my $dx$ is $\frac {3}{n}$

since +6 is supposed to be my a, b is therefore 9 because $dx = \frac {b-a}{n}$

but that 7 inside the parenthesis spoils everything... so I can't say what's inside the parenthesis is equal to $x_i$

So what's next?

Divide through by 63. We then find that $f(a)=\frac{18}{63}$ by plugging in $i=0$. But then $f(x)=\frac{18}{63}+x$ because $f(a+idx)=f(a)+idx$. Hence $a=0$ and $b=1$. Therefore our integral is $63\int_{0}^{1} x+\frac{18}{63}dx$

You don't need to adjust your boundaries for the integral. The easiest way is to choose arbitrarily $a=0$  and $b=1$  so that $\Delta x = 1/n$ and $x^{\ast}$ is $i/n$. See the picture:


Basis of 4th degree polynomial linear space that doesn't contain 2nd or 3rd degree polynomials In K4[x], the linear space of the polynomials of degree 4 or less with coefficients in in K, determine:

(a) A basis that doesn't contain 2nd or 3rd degree polynomials

(b) If there exists a basis that doesn't contain 4th degree polynomials

(c) A supplementary subspace of the one generated by the polynomials
p(x)=−1+2x+x^2 y q(x)=1+x+x^2

Suggestion: use coordinate row (or column) matrices

The basis for a 4th-degree polynomial is {1,x,x^2,x^3,x^4}. So I think the answer to part (a) is {1,x,x^4}.
For part (b) there is not.


two definitions of arc length Let $f:[a,b]\rightarrow\mathbb{R}$ be an absolute continuous function with $-\infty <a <b <\infty. $ Define the length $L $ be the total variation of the graph $g (x)=(x,f (x)) $ on $[a,b] $. Show that $L=\displaystyle\int^b_a \sqrt {1+f'^2(x)} dx. $

I think $L\le\displaystyle\int^b_a \sqrt {1+f'^2(x)} dx $ is easily provable but the converse doesn't seem easy.. can anyone help me?

Yes, one half is trivial. Since $f$ is AC it follows that $g$ is AC, hence $g$ is the integral of its derivative. So $$|g(y)-g(x)|
=\left|\int_x^yg'(t)\,dt\right|\le\int_x^y|g'(t)|\,dt=\int_x^y(1+f'(t)^2)^{1/2}\,dt.$$This shows that $$L\le\int_a^b(1+f'^2)^{1/2}.$$

The other inequality follows from a certain lemma. I'll let you consider why the result follows from the lemma (take $F=g'$).

Lemma If $F\in L^1([0,1])$ then $$\lim_{n\to\infty}\sum_{j=1}^n\left|\int_{(j-1)/n}^{j/n}F\right|=\int_0^1|F|.$$

Proof. Say the left side is $\lim_{n\to\infty}I_n(F)$. Note that $|I_n(F)-I_n(G)|\le||F-G||_1$. Let $\epsilon>0$. Choose $G\in C([0,1])$ with $||F-G||_1<\epsilon$.

The lemma holds for continuous functions, hence with $G$ in place of $F$ (another exercise). So if $n$ is large enough we have $$|I_n(F)-||F||_1|
\le|I_n(F)-I_n(G)|+|I_n(G)-||G||_1|+|||G||_1-||F||_1|<\epsilon+\epsilon+\epsilon.$$


Implication and equivalence arrows, when to use them? In my course book we have something called implication arrows $\Rightarrow$ and equivalence arrows $\Leftrightarrow$ and I have never managed to understand them.

When do I know which to use and how do I know that I'm correct when I use them?

$A \Rightarrow B$ has the meaning $B$ follows from $A,$ but this does not necessarily hold the other way around. For instance, from $x = 13$, it follows that $x$ is prime. But you can't argue the other way around. So if $\mathbb P$ is the set of prime numbers, you can write $x = 13 \Rightarrow x\in\mathbb P$, but not $x\in\mathbb P\Rightarrow x=13.$

$A\Leftrightarrow B$ has the meaning that $A$ and $B$ are the same statement, just transformed.  If $A\Leftrightarrow B$, then both $A\Rightarrow B$ and $B\Rightarrow A.$  For instance, consider an integer $x$.  You can say that if $x$ is even, then $x$ is not odd.  This certainly also holds the other way around, so you can write $\mathrm{even}(x)\Leftrightarrow\lnot\;\mathrm{odd}(x).$

I find the implication of logical implication really interesting. Starting off with the truth tables for the two, note only one subtle difference:

+---+---+-------+
| P | Q | P ⇒ Q |
+---+---+-------+
| T | T |     T |
| T | F |     F |
| F | T |     T |
| F | F |     T |
+---+---+-------+

+---+---+-------+
| P | Q | P ⇔ Q |
+---+---+-------+
| T | T |     T |
| T | F |     F |
| F | T |     F |
| F | F |     T |
+---+---+-------+

There is a common misconception that P ⇒ Q means that if P is true, Q is true.  What the truth table shows is that P ⇒ Q is always true when P is false, which I think has profoundly interesting implications.

In some ways the truth table is simply a formal version of the old computer cliche "garbage in, garbage out": P actually tells us nothing about Q.

Q, on the other hand, if we know it is false, tells us if P is true or not provided we know if P ⇒ Q is true or false. But if Q is true, it gives us no clue as to whether P is true or false.

It is a lesson that we can only advance knowledge by working back from known wrong conclusions. There is no way to link known correct conclusions to what we think causes them with logical certainty.

This may seem obvious to lawyers and detectives, but it is very counterintuitive to us computer programmers who think of P as a pattern to be matched by an if statement, and Q as the resulting action.

My view is logical implication creeps up in other fields such as the old statistics cliche "correlation does not imply causation" and Karl Popper's scientific method of falsifiability. But I've never really seen this articulated anywhere.

Through deduction we can whittle down the choice of possible causes, Ps, for a given Q until the answer seems obvious, but even then it is never 100% certain.

Logical implication encourages an open mind which remembers “there are more things in heaven and earth, Horatio, than are dreamt of in your philosophy”. Many zealots out there could have been cured if our basic education system taught logical implication at an early age.


Diophantine equation $a^m + b^m = c^n$ ($m, n$ coprime) Arising from this recent question, and in particular the answer by Gerry Myerson, it occurs to me that, if $m$ and $n$ are coprime integers, non-trivial solutions can be found to any Diophantine equation of this form (or with more powers of $m$ on the left hand side):
$$a^m + b^m = c^n$$
The method is to take any integers d and e, find $f = d^m + e^m$, then find an integer $r$ such that $r$ is congruent to $0 \mod m$ and to $-1 \mod n$. The Chinese Remainder Theorem tells us that such an integer can be found. Then:
$$(d^m)(f^r) + (e^m)(f^r) = f(f^r)$$
has powers of m on the left and a power of n on the right.
Is this correct, or am I missing something? (I realise that to find solutions with no common divisor is much harder or perhaps impossible.)

This question has been answered in the above comments by Gerry Myerson and Will Jagy.  The method in the question is correct.  There is also a variant possible if $f$ is divisible by an $n$-th power.  Solutions obtained by this method always have $a,b,c$ with a common divisor.  It is much harder to find solutions with $a, b, c$ coprime, especially if it also required that $2m^{-1} + n^{-1} < 1$.

(This answer is posted according to guidance in this question on meta.)


Differentiate ln$(100|x|)$ I asked to find $\frac{\text{dy}}{\text{dx}}$ for $y = \ln(100|x|)$.
\begin{eqnarray}
\frac{\text{d}\,\text{ln}(100|x|)}{\text{dx}}  &=& \frac{\text{d ln(100|x|)}}{\text{d 100|x|}} \cdot \frac{\text{d (100|x|)}}{\text{dx}}\\
&=&\frac{1}{\text{100|x|}}\cdot \frac{\text{d (100|x|)}}{\text{dx}}
\end{eqnarray}

I'm not sure how to calculate the derivative of 100$|x|$. Any hints would be appreciated.

Notice that

$$\ln(100|x|)=\ln(100)+\ln|x|$$

Thus, derivative is simply given as

$$\frac d{dx}\ln(100|x|)=\frac1x$$

The derivative of ln functions is simply ln(f(x))=f'(x)/f(x)


Inequality and sequence of functions I am trying to prove that the sequence of functions $f_n(x)=\sqrt{n}\cos(nx)-\frac{\sin(nx)}{2n\sqrt{n}}$ does not converge for any $x\in\mathbb{R}$.
My idea was to fix $x\in\mathbb{R}$ and then obtain a bound like $|f_n(x)|\geq $ constant $\times$ (function of $n$ which goes to infinity as $n\to\infty$) so $|f_n(x)|=|\sqrt{n}\cos(nx)-\frac{\sin(nx)}{2n\sqrt{n}}|\geq ||\sqrt{n}\cos(nx)|-|\frac{\sin(nx)}{2n\sqrt{n}}||\geq \sqrt{n}||\cos(nx)|-\frac{1}{2n^2}|$ which is almost what I wanted but I don't see how to go further so I would appreciate an hint. Thank you.

As the sequence of functions $\{g_n\}$ with
$$g_n(x)=\frac{\sin(nx)}{2n\sqrt{n}}$$ converges to $0$ for all $x \in \mathbb R$, it is enough to prove that $\{h_n\}$ does not converge for any $x \in \mathbb R$ with
$$h_n(x) = \sqrt{n}\cos(nx).$$
And this is clear as $\cos nx$ never converges to zero which would be required for $\{h_n(x)\}$ to converge as $\lim\limits_{n \to \infty} \sqrt n = \infty$.
Proof that $\{\cos nx\}$ never converges to zero:
As $$\cos(n+1)x = \cos nx \cos x - \sin nx \sin x,$$ if $\{ \cos nx \}$was converging to zero, we would have
$$\lim\limits_{n \to \infty} \sin x \sin nx = 0$$
which implies either $\sin x = 0$, i.e. $x \in {\pi \mathbb Z}$ and a contradiction as $\{\cos np\pi\}$ doesn't converge to zero. Or $\lim\limits_{n \to \infty} \sin nx = 0$ and another contradiction as
$$\cos^2 nx + \sin^2 nx = 1.$$

$f_n(x)=\sqrt{n}\cos(nx)-\frac{\sin(nx)}{2n\sqrt{n}}$
$\frac{\sin(nx)}{2n\sqrt{n}} \rightarrow 0$ when $n\rightarrow +\infty$.
And $\sqrt{n} \cos (nx)$ diverges (we can easily extract a subsequence that diverges to $+\infty$)


Show that $n$ is a prime number Let $n\in\mathbb{N}$, $n\geq2$ and $(G,.)$ an abelian group of order $n$, with the property that for any $a\in G$, the number of endomorphisms of $G$ with $f(a)=a$ is equal to $n+1-ord(a)$, where $ord(a)$ is the order of $a$. Show that $n$ is a prime number.

My approach is to suppose that $n$ is not prime, so there is a decomposition for $n=p_1p_2...p_k$, where $p_1,\dots ,p_k$ are prime numbers and from Cauchy's theorem there is an $a\in G$ such  that $ord(a)=p_i$. Now $f(e)=f(a^{p_i})=(f(a))^{p_i}=e$, and so the number of endomorphisms $f$ is $n+1-1=n$.

However, from the hypothesis, the number of $f$ is $n+1-p_i$, and so $n+1-p_i=n$ $\Rightarrow p_i=1$. As we chosen $p_i$ arbitrarily, we get that all the numbers $p_i,i\in{0,1,...,k}$ are $1\Rightarrow n=1$, false with $n\geq2$.

  Is my proof correct?

You haven't used the hypothesis that the group is abelian.

I corrected some points of grammar & punctuation, though, too, like not starting a sentence with a mathematical symbol, adding a space after each comma, etc.


Remainder Theorem: $f(x) = x^3+1$ divided by $\operatorname {d}(x)= x$ Take for example $\operatorname {f}(x) = x^3+1$ divided by $\operatorname {d}(x)= x-1$ $$\frac {x^3+1}{x-1}$$
The Remainder Theorem tells us that the remainder will be $$\operatorname {f}(zero-of-\operatorname {d}(x))$$
In this case the remainder should be $$\operatorname {f}(1) = 1^3+1 = 2$$
so the remainder should be 2 for any value of $x$
But for $x=3$
$$f(3) = 3^3+1$$
when this is divided by $x-1$ which is $3-1$ which equals to $2$, $$\frac {3^3+1}{3-1}$$ $$= \frac {27+1}{2}$$ $$= \frac {28}{2}$$ $$= 14, remainder = 0 ≠ 2$$
Please explain me why this happened

There is a distinction between the statements :

The division of the polynomial $f(x)$ by the polynomial $x-a$, yields the remainder as the constant polynomial $f(a)$.

and

For each $y$, if we divide the number $f(y)$ by the number $y-a$, the remainder is always $f(a)$.

The difference is simple : one refers to polynomial division, the other to integer division. You seem to think that these statements are both implied by one another. The second, is in fact NOT TRUE.

Why? Well, note that $f(y) - f(a)$ is always a multiple of $y-a$, so what is true is this :

For each $y$, the numbers $f(y)$ and $f(a)$ leave the same remainder when divided by $y-a$.

Now, what if $f(a) \geq y-a$? Then it cannot be a possible remainder when something is divided by $y-a$.
That's what is happening in your example : you have $f(1) = 2$, and $y=3$ so $y-a = 2$ as well. So if you divide $f(3)$ by $2$, you get $0$ as the remainder, which is the same as when you divide $f(1)$ by $2$. But $f(1)$ itself can't be the remainder because the remainder upon division by $2$ can only be $1$ or $0$, which $f(1)$ is not.

Basically, if you want to find the remainder when $f(y)$ is divided by $y-a$, the answer is NOT $f(a)$ , but rather the remainder when $f(a)$ is divided by $y-a$. That number in your case is $2 \pmod 2 = 0$.

$$\frac{x^3+1}{x-1}=x^2+x+1+\frac{2}{x-1}$$
Here $f(x)=x^3+1$, so the remainder when $f(x)$ is divided by the linear factor $x-1$ is a constant 2 as seen above and which is nothing but $f(1)=2$.
If the divisor is quadratic, the remainder is linear function of $x$ or a constant.


Solve $x + y + z = xyz$ such that $x , y , z \neq0$ I came across the equation $x+y+z=xyz$ such that $x , y , z \neq 0$.

I set $x=1, y=2, z=3$ but how can i reach formal mathematical solution without " guessing " the answer ? Thank you

you can write $y+z=x(yz-1)$ and if $yz\ne1$ you will get $$x=\frac{y+z}{yz-1}$$
if $yz=1$ we get $y+z=0$ or $y^2+1=0$ which gives us complex solutions.

Solve for one variable. As the problem is symmetric, wlog solve for $z$:
$$z = \frac{x+y}{xy-1}.$$
Now you can choose any real values for $x$ and $y$, as long as $xy-1\ne0$. For your choice of $x=1$, $y=2$, you can see it gives the correct result.

If you need a solution in $\mathbb Z$, things get more complicated as the value of $z$ could be a rational for most choices of $x$ and $y$.


Four-potential, electric and magnetic field generated Given a four-potential (coordinates are written in Einstein's notation)
  $A=(x^1, 2x^1, x^3 - 2x^2, 2x^3)$ which electric and magnetic field are generated?

  $E=(-c,0,2c), B=(0,1,-2)$
  $E=(c,0,2c), B=(1,0,-2)$
  $E=(-c,-2c,2c), B=(1,0,1)$
  $E=(-c,0,0), B=(1,0,0)$
  $E=(c,0,0), B=(0,-2,-2)$

I compute

$\dfrac{E^1}c = \dfrac{\partial A^0}{\partial x^1} - \dfrac{\partial A^1}{\partial x^0} = 1$ and so $E^1=c$, $E^2=0 , E^3=0$.

So intuitively the answer is $5$. but components of the magnetic field are not the same I found, can someone help me?

The magnetic field is not zero, as $A^2$ depends on $x^3$, yielding $B_1=1$.

You didn't say whether the components you specified for $A$ are $A^\mu$ or $A_\mu$. The notation $E^1$ doesn't make sense, since $E$ is a three-vector with normal indices, not a four-vector with covariant and contravariant indices. Also,

$$
\frac{\partial A^0}{\partial x^1} - \frac{\partial A^1}{\partial x^0}
$$

doesn't make sense, since this is $\partial_1A^0-\partial_0A^1$, that is, a mixture of terms from $F_1^0$ and $F_0^1$. You need

$$
\partial^1A^0-\partial^0A^1=\frac{\partial A^0}{\partial x_1} - \frac{\partial A^1}{\partial x_0}\;.
$$

I suspect if you sort all these things out, you'll get a different sign for $E_1$, in agreement with answer $4)$.

Using
$$ F^{\mu \nu} = \partial^\mu A^\nu - \partial^\nu A^\mu = \begin{pmatrix} 0 & -E_1/c & -E_2/c & -E_3/c \\ E_1/c & 0 & -B_3 & B_2 \\ E_2/c & B_3 & 0 & -B_1 \\ E_3/c & -B_2 & B_1 & 0 \end{pmatrix} $$
I find that all three components of the magnetic field are zero, so either we both made a mistake, or the question is wrong. Is that the same you are getting?


Irreducible Polynomial over a Field I am studying elementary Field Theory. I have a problem that I have been wrestling with for a bit:

Let $p(x)$ be an irreducible polynomial over a field $F$. Suppose that $p(x)$ divides $f_1(x)....f_n(x)$ in $F[x]$. Prove that $p(x)$ divides $f_i(x)$ for some $i$ in {$1,...,n$}.

I'm assuming we can use induction to prove this but I'm not really sure how to go about it.

Thanks!

The base case of the induction is obvious.

Assume the inductive hypothesis $p\mid f_1f_2\cdots f_n$ implies $p\mid f_i$ for some $i\in\{1,\ldots,n\}$

Now, consider $p\mid f_1f_2\cdots f_nf_{n+1}$. If $p\mid f_1f_2\cdots f_n$, we're done by the inductive hypothesis. Suppose not, then $p\mid f_1f_2\cdots f_{n+1}$ and $p\not\mid f_1f_2\cdots f_n$ implies $p\mid f_{n+1}$ since $p$ is irreducible which completes the inductive step.

Addendum: This is basically a generalization of Euclid's lemma for the Euclidean domain of the ring of polynomials in $x$

You have to prove that if $p$ divides $fg$, then $p$ divides $f$ or $p$ divides $g$, then you can proceed easily by induction. The argument is pretty much the same as for integers, you can look it up in any book.


What does it mean to say that an event is determined by certain random variables? So I was reading Moser and Tardos' paper on their algorithmic proof for the Lovász Local Lemma and came across the following:

  Let $\mathcal{P}$ be a finite collection of mutually independent random variables in a fixed probability space $\Omega$. We will consider events $A$ that are determined by the values of some subset $S \subseteq \mathcal{P}$ of these variables.

(This excerpt was taken from the second paragraph on page 2 of the above mentioned paper.)

What does it mean to say that an event is determined by (the values of) certain random variables? I think it would make sense, for instance, to say that the event $A$ = [$X \neq Y$] is determined by the r.v.s $X$ and $Y$, but I don't know how one would formally write the definition in general.

The event is a function of the random variables.

$A = f(X, Y, Z ...)$ with $X, Y, Z \in S$


About connected Lie Groups How can I prove that a connected Lie Group is generated by any  neighborhood of the identity?

The result is almost trivial for $R^n$ but I tried using the open subgroup generated by this neighborhood.

An open subgroup $H$ of a topological group $G$ is closed because
$$
G \smallsetminus H = \bigcup_{g \notin H} gH
$$
is open as union of the open sets $gH$.

Now take your neighborhood $U$ of the identity, let $H = \bigcup_{n \in \mathbb{Z}} U^{n}$ and check that $H$ is an open (hence closed) subgroup of $G$. By connectedness $G = H$.

Is the question is similar to: Prove that any open subgroup of a connected lie group is the whole group then here is my answer, if not then please ignore it.

Take any $g\in G$ define $l_g:G\rightarrow G, l_g(k)=gk$, Let $H$ be any open subgroup of $G$,clearly as $l_g$ is a homeomorphism, and $H$ is open so $l_g(H)=gH$ is also open, but $G$ is connected so $H=G$


Round of Eight tournament pairing combinations I notice recently that in a Starcraft tournament in the round of eight, there were 3 Zerg players, 3 Protoss players and 2 Terran players, yet none of the four pairings had a Terran versus Zerg matchup. What are the chances of that?

I computed an answer to this problem by writing a computer program that enumerated all the possibilities and counted those pairings in which a TvZ occurred, but is there a mathematical way to compute the odds without enumerating every possibility manually?

For example, lets say the players are Z1, Z2, Z3, P1, P2, P3, T1 and T2. One possible pairing might be:

Z1 vs P1
Z2 vs P2
Z3 vs P3
T1 vs T2

In this set of pairings there are no TvZ matches. What is the odds of this occurring?

I don't know the game, so will assume that anyone can be matched with anyone else. I will also assume all pairings are equally likely. Let the Terrans be A and B. Choose A's opponent first. Maybe it is B, in which case no Terran is matched with a Zerg. This has probability $\frac{1}{7}$.

But maybe A is matched with someone else. The probability this is a non-Zerg is $\frac{3}{7}$. And if that has happened, the probability B is matched with a non-Zerg is $\frac{2}{5}$. So the required probability is
$$\frac{1}{7}+\frac{3}{7}\cdot \frac{2}{5}.$$
This turns out to be $\frac{11}{35}$. In the language of odds, which is little used in probability theory, the odds are $11$ to $24$.

Total ways of pairing from a pool of $n = (n-1)\cdot(n-3). . .\cdot3\cdot1 = (n-1)!!$

Total ways of pairing from two groups with $m$ and $n = mn$

Pairings when both T's together = $5!!$

Pairings between T's and P's and between remaining pool of $4 = 2\cdot3\cdot3!!$

$$Pr = \frac{5!! + 2\cdot3\cdot3!!}{7!!} = \frac{33}{105} = \frac{11}{35}$$

Odds are usually used for betting as odds against

Odds against = $24:11$