Is it possible to place 26 points inside a rectangle that is 20 cm by 15 cm so that the distance between every pair of points is greater than 5 cm? I need help to answer the following question:

  Is it possible to place 26 points inside a rectangle that is $20\, cm$ by
  $15\,cm$ so that the distance between every pair of points is greater
  than $5\, cm$?

I haven't learned any mathematical ways to find a solution; whether it maybe yes or no, to a problem like this so it would be very helpful if you could help me with this question.

No, it is not. If we assume that $P_1,P_2,\ldots,P_{26}$ are $26$ distinct points inside the given rectangle, such that $d(P_i,P_j)\geq 5\,cm$ for any $i\neq j$, we may consider $\Gamma_1,\Gamma_2,\ldots,\Gamma_{26}$ as the circles centered at $P_1,P_2,\ldots,P_{26}$ with radius $2.5\,cm$. We have that such circles are disjoint and fit inside a $25\,cm \times 20\,cm$ rectangle. That is impossible, since the total area of $\Gamma_1,\Gamma_2,\ldots,\Gamma_{26}$ exceeds $500\,cm^2$.

Highly non-trivial improvement: it is impossible to fit $25$ points inside a $20\,cm\times 15\,cm$ in such a way that distinct points are separated by a distance $\geq 5\,cm$.

Proof: the original rectangle can be covered by $24$ hexagons with diameter $(5-\varepsilon)\,cm$. Assuming is it possible to place $25$ points according to the given constraints, by the pigeonhole principle / Dirichlet's box principle at least two distinct points inside the rectangle lie in the same hexagon, so they have a distance $\leq (5-\varepsilon)\,cm$, contradiction.

Further improvement:

the depicted partitioning of a $15\,cm\times 20\,cm$ rectangle $R$ in $22$ parts with diameter $(5-\varepsilon)\,cm$ proves that we may fit at most $\color{red}{22}$ points in $R$ in such a way that they are $\geq 5\,cm$ from each other.

Although the conclusion is correct, the reasoning in this answer is not.  The premise I started from does not apply to the question asked.

(After not finding any apparent community standard for what to do with this sort of incorrect answer, I decided to leave it in place - so that no-one else will duplicate it, and to preserve whatever value it may have, - but with the above note to prevent confusion.)

tl-dr; No.

This can be thought of as a circle packing question: treat each point as the center of a circle, and the distance no greater than constraint becomes a circles don't overlap constraint.  (It's a little different from the usual "how many circles will fit", because this question is "how many circle-centers will fit" with the rest of each circle allowed to stick out of the rectangle.)

The densest packing of circles that don't overlap is the hexagonal tiling, where each circle is contained by (or inscribed in) a hexagon. (The description where center of each circle is also the center of a hexagon, and the centers of the nearest neighboring circles are the vertices of that hexagon, gives the same arrangement of circles but with larger hexagons; The following may be easier to understand with non-overlapping hexagons.)

The question can be restated as: At a 5cm spacing, will 26 hexagon centers fit inside a 20×15cm rectangle?

We can start by putting one in a corner; this might not be necessary, but it covers the minimum area within the rectangle so there's no better starting point.

Since the hex tiling has straight lines at minimum spacing, we can put a line of points along an edge of the rectangle - along the 20cm edge we can put 5 points (the last of which is on another corner).

Alongside that first line of hexagons we can put another line of hexagons, also spaced 5cm from eachother, which will be $\frac{5\cdot\sqrt{3}}{2} \approx 4.33$cm away from the first line.  (Because that's the height of an equilateral triangle with 5cm sides, and the points in the second line are necessarily positioned to make equilaterial triangles with the points in the first line.)

With that line spacing, after the initial line, $\left \lfloor\frac{15\cdot2}{5\cdot\sqrt{3}}\right\rfloor=\left\lfloor\frac{15}{4.33}\right\rfloor=\left\lfloor3.36\right\rfloor=3$ more lines of hexagons will have some centers within the rectangle.  Because of the positions of the centers along the lines, they will alternate having 5 and 4 centers within the rectangle.

So the total number of centers within the rectangle will be $5+4+5+4=18$.  But that might not be a maximum; it might be possible to get more with a different orientation.

The obvious next orientation to try is with the first line along the 15cm side:  $\left\lfloor\frac{20}{4.33}\right\rfloor=\left\lfloor4.62\right\rfloor=4$, so the total number of centers within the rectangle this way will be $4+3+4+3+4=18$.

Is there another orientation that could fit more?  The diagonal of the rectangle is $\sqrt{20^2+15^2}=25$, just enough to fit 6 centers along the diagonal.  The height of the 15-20-25 triangles on each side is 12cm, so there is room for 2 lines on either side of the diagonal.  The first line on each side partitions off a similar triangle with $\frac{12-4.33}{12}\approx0.64$ of the height, so it has a base of $0.64\times25\approx15.98$, so at most 4 centers can be on each of those lines.  The second line on each side partitions off a similar triangle with $\frac{12-2\cdot4.33}{12}\approx0.28$ of the (original) height, so it has a base of $0.28\times25\approx6.96$, so at most 2 centers can be on each of those lines.  Without considering alignment this gives an upper bound, the maximum number of centers within the rectangle this way is $2+4+6+4+2=18$.

This is not a rigorous proof, but it's enough to convince me.  With a minimum distance of exactly 5cm, only 18 points can fit into the rectangle, so with all distances strictly greater than 5cm it is not possible to place 26 points.


Solving the functional equation $ T (n) = 3 ~ T(\frac{n}{2}) + n^{2} $. I’m trying to solve the following recurrence relations for run-time $ T(n) $. Assuming $ T(n) $ is constant for $ n \leq 2 $, solve
$$
T(n) = 3 ~ T \! \left( \frac{n}{2} \right) + n^{2}.
$$

For real $n$ there are exact solutions $A|n|^b + cn^2$ where $A$ is a free parameter and $b,c$ are determined by the recurrence.  The recurrence for integer $n$ will approximate that after defining what $T(n/2)$ means.

The exponent $b$ is the same as the fractional dimension of the Sierpinski triangle which is made of $3$ copies of itself at $1/2$ the scale, a form of the same recurrence (without the $n^2$ term).

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

  $\ds{\mrm{T}\pars{n} = 3\,\mrm{T}\pars{n \over 2} + n^{2}:\ ?\,,\qquad
\left.\vphantom{\large A}\mrm{T}\pars{n}\right\vert_{\ n\ \leq\ 2} = \mbox{'initial' constant}}$.

Lets $\ds{\,\mrm{T}\pars{n} \equiv \tilde{\mrm{T}}\pars{n}n^{2}\implies
\tilde{\mrm{T}}\pars{n} = {3 \over 4}\,\tilde{\mrm{T}}\pars{n \over 2} + 1
\implies\tilde{\mrm{T}}\pars{n} - 4 =
{3 \over 4}\bracks{\tilde{\mrm{T}}\pars{n \over 2} - 4}}$ which leads to:
\begin{align}
\tilde{\mrm{T}}\pars{n} - 4 & =
{3 \over 4}\,{3 \over 4}\bracks{\tilde{\mrm{T}}\pars{n \over 2^{2}} - 4} =
{3 \over 4}\,{3 \over 4}\,{3 \over 4}
\bracks{\tilde{\mrm{T}}\pars{n \over 2^{3}} - 4} = \cdots =
\pars{3 \over 4}^{k}
\bracks{\tilde{\mrm{T}}\pars{n \over 2^{k}} - 4}
\end{align}
which is equivalente to
$$
\mrm{T}\pars{n} = 4n^{2} +
\pars{3 \over 4}^{k}n^{2}
\bracks{{\mrm{T}\pars{n/2^{k}} \over \pars{n/2^{k}}^{2}} - 4}
$$

$$\bbox[8px,#efe,border:0.1em groove navy]{%
\mrm{T}\pars{n} = 4n^{2} +
\pars{3 \over 4}^{k}
\bracks{2^{2k}\,\mrm{T}\pars{n \over 2^{k}} - 4n^{2}}}
$$

  The final solution is found with a value of $\ds{k}$ which satisfies
  $\ds{{n \over 2^{k}} \leq 2 \implies k \geq {\ln\pars{n} \over \ln\pars{2}} - 1}$

$$\bbox[8px,#efe,border:0.1em groove navy]{%
\mrm{T}\pars{n} = 4n^{2} +
\pars{3 \over 4}^{k_{c}}
\bracks{2^{2k_{c}}\,\mrm{T}\pars{n \over 2^{k_{c}}} - 4n^{2}}}\,,\qquad
\left\{\begin{array}{rcl}
\ds{k_{c}} & \ds{\equiv} &
\ds{\left\lceil{\ln\pars{n} \over \ln\pars{2}} - 1\right\rceil}
\\[2mm]
\ds{\mrm{T}\pars{n \over 2^{k_{c}}}} & \ds{=} & \mbox{'initial' constant}
\end{array}\right.
$$


Solve logarithmic inequality I need help solving $-\log_2(x-1)*\log_2(3x-4)>0$

The logarithms have a base number of $2$.

Set $u=\log_2(x-1)$ and $v=\log_2(3x-4)$. Then you need to find when $-uv>0$, that is, $uv<0$. This becomes

  either $u>0$ and $v<0$ or $u<0$ and $v>0$.

Now replace $u$ and $v$ by their definition and solve the inequalities.

For instance, the first case becomes
$$
\begin{cases}
\log_2(x-1)>0 \\
\log_2(3x-4)<0
\end{cases}
$$
so you get $x-1>1$ from the first and $0<3x-4<1$ from the second. Can you go on?

Not a mathematician, but I can solve this using grade 11 math. All you need is to find when you get a positive number as a total. It does not matter what the base of the logarithm is.

$\log(x)$ is negative when $0 < x < 1$, $\log (x)$ when $x > 1$ is positive no matter what the base of the log is.

A negative number times a negative number is positive
A negative number times a positive number is negative
A positive number times a positive number is positive

$0 < x - 1 < 1$ when $1 < x < 2$
$0 < 3x - 4 < 1$ when $3/4 < x < 1$

your inequality is therefore true when $1 < x < 2$


If $\vec{\nabla} \times \langle P,Q,Q \rangle=\vec{0}$ iff $Pdx+Qdy+Rdz$ is exact differential form. If $\vec{\nabla} \times \langle P,Q,R\rangle=\vec{0}$ iff $Pdx+Qdy+Rdz$ is an exact differential form.

My attempt:-

If $Pdx+Qdy+Rdz$ is an exact differential form. Then there exists $U(x,y,z): dU=Pdx+Qdy+Rdz$. From the mixed partial theorem, It is easy to prove that $curl \langle P,Q, R\rangle=\vec{0}$.

If $\vec{\nabla} \times \langle P,Q,R \rangle=\vec{0}$

$\implies R_y=Q_z,R_x=P_z $ and $Q_x=P_y$. How do I prove the existence of scalar function $U$:$dU=Pdx+Qdy+Rdz$?

This is quite standard. However, you need an additional assumption to prove your result. The easiest is to assume that the domain on which $dU$ is defined is a star domain.

You can then find $U$ by simply integrating $dU$ from $\vec{r}_0= (x_0,y_0,z_0)$ to $\vec{r}=(x,y,z)$ along the path $$\vec{\gamma(}t) = t\vec{r}_0+ (1-t)\vec{r},\qquad t\in[0,1].$$
In particular, we set
$$U(\vec{r}) = \int_{\gamma}(Pdx+Qdy+Rdz)\,. \tag{1} $$

Now, what is left to show is that $dU$ is indeed $(P,Q,R)$. The relevant property thereby is that in (1) the value of $U$ is independent of he path choses (and only depends on the endpoints). This fact is due to Stokes theorem and the vanishing of the curl. The result then follows from the fundamental theorem of calculus.

In general we know that
$$dU = U_x dx + U_y dy + U_z dz $$
then you want to find $U$ such that
$$P = U_x, Q = U_y, Z = U_z.$$
Note that from the curl equation you get that $U_{xy} = U_{yx}$ and $U_{xz}=U_{zx}$ and $U_{yz}=U_{zy}.$ From the first equation you get that:
$$U = \int P(x,y,z)dx + \phi(y,z)$$
From the second equation you get that:
$$Q = U_y = \frac{\partial }{\partial y}\int Pdx + \phi_y \implies \phi_y =Q - \frac{\partial }{\partial y}\int Pdx.$$
Since $\phi_y$ is a function of $y$ and $z$ we have that $\phi_{yx} = 0$ and thus
$$\frac{\partial }{\partial x}\left(Q -\frac{\partial }{\partial y}\int Pdx\right)=0.$$
Thus
$$\phi (y,z) = \int Q(x,y,z)dy-\int \frac{\partial }{\partial t}\left(\int P(x,y,z)dx\right)dt + h(z).$$
Thus our current expression for $U$ is
$$U = \int P(x,y,z)dx + \int Q(x,y,z)dy-\int \frac{\partial }{\partial t}\left(\int P(x,y,z)dx\right)dt + h(z).$$
Now recall that
$$Z = U_z = \frac{\partial }{\partial z}\int P(x,y,z)dx + \frac{\partial }{\partial z}\int Q(x,y,z)dy-\frac{\partial }{\partial z}\int \frac{\partial }{\partial t}\left(\int P(x,y,z)dx\right)dt + h_z(z).$$
Thus
$$\frac{dh}{dz} = Z - \frac{\partial }{\partial z}\int P(x,y,z)dx - \frac{\partial }{\partial z}\int Q(x,y,z)dy+\frac{\partial }{\partial z}\int \frac{\partial }{\partial t}\left(\int P(x,y,z)dx\right)dt.$$
And so
$$h(z) = \int Z(x,y,z) dz - \int \frac{\partial }{\partial t}\left(\int P(x,y,z)dx\right) dt - \int \frac{\partial }{\partial t}\left(\int Q(x,y,z)dy\right)dt+
\int \frac{\partial }{\partial m}\left[  \int\frac{\partial }{\partial t}\left(\int P(x,y,z)dx\right)dt \right]dm.$$
Thus we have
$$U = \int P(x,y,z)dx + \int Q(x,y,z)dy-\int \frac{\partial }{\partial t}\left(\int P(x,y,z)dx\right)dt + \int Z(x,y,z) dz - \int \frac{\partial }{\partial t}\left(\int P(x,y,z)dx\right) dt - \int \frac{\partial }{\partial t}\left(\int Q(x,y,z)dy\right)dt+
\int \frac{\partial }{\partial m}\left[  \int\frac{\partial }{\partial t}\left(\int P(x,y,z)dx\right)dt \right]dm$$


How can we define the set $\{\{\}, \{\{\}\}, \{\{\{\}\}\}, ...\}$? How do you define the set of elements which can be made by repeatedly wrapping the null set in another set: $\{\{\}, \{\{\}\}, \{\{\{\}\}\}, ...\}$?

I've tried $A = \{x\ |\ x=\emptyset \lor \exists y\in A : x = \{y\}\}$, but that would give us a recursive formula, because $y\in A$ is replaced by that same formula.

Is this set definable, and if yes, how?

How about:
$$\{x\mid\forall A,(\{\}\in A\land(\forall a\in A,\\\{a\}\in A))\Rightarrow x\in A\}$$
To make that easier to read, let me define the following. A set $A$ is upwards if $a\in A$ implies $\{a\}\in A$. (I made up this word just now.) A set $A$ is called totally upwards if it's upwards and it contains $\{\}$. (I also just made this up.) Then the above reads:
$$\{x\mid\forall A,\\A\text{ is totally upwards }\Rightarrow x\in A\}$$

Yes, because there's an axiom that says you can do exactly that, because you can't derive it from the intuitively more "basic" axioms. It is called the axiom of infinity, and states that there is a non-empty set $S$ such that whenever $X \in S$, then $(\{X\} \cup X) \in S$ as well.


Nilpotent blocks of matrices I'm trying to read through Hirsch and Smale's "Differential Equations, Dynamical Systems, and Linear Algebra", and I don't understand how this theorem follows from this other theorem.

The first theorem says

  Theorem 1. Let $N$ be a nilpotent operator on a real or complex vector space $E$.  Then $E$ has a basis giving $N$ a matrix of the form $$A = diag\{A_1, \dots, A_r\}$$ where $A_j$ is an elementary nilpotent block, and the size of $A_k$ is a nonincreasing function of $k$.  The matrices $A_1, \dots, A_r$ are uniquely determined by the operator $N$.

They go on to say that if $A$ is an elementary nilpotent matrix, then the dimension of $Ker A = 1$, since the rank of $A$ is $n-1$.  That part I understand.

What I don't understand is how Theorem 2 follows from Theorem 1:

  Theorem 2. In Theorem 1 the number $r$ of blocks is equal to the dimension of $Ker A$.

Can you help me?

Thanks!

Each elementary nilpotent block has rank $n-1$ and the rank the matrix is the sum of the ranks of the block diagonal matrices. Equivalently the nullity of the matrix is the sum of the nullities of the block diagonal matrices. If the nullity of $A$ is $k$ it follows that there must be precisely $k$ blocks, each contributing $1$ to the nullity.

Also, the form that the book is talking about is much more general than just for nilpotent operators. This type of decomposition is called the Jordan Normal Form. The fact that the geometric multiplicity of an eigenvalue is the number of Jordan blocks is a fundamental fact of the Jordan Normal Form and your result is an immediate consequence.

i found answer of your question  in matrix analysis and applied linear algebra (carld.meyer)


Find all solutions (if there are any) to the following congruences Find all solutions (if there are any):

$x^2 \equiv -1 \pmod{7}$

I found $x$ $\equiv$ $\sqrt6 \pmod{7}$

Is this the answer?

Can I use the same procedure for $x^2 \equiv -1 \pmod{13}$

I think the question requires you to find integer solutions, and there are none.

In fact, there are no solutions to $x^{2} \equiv -1 \pmod p$ if $p$ is prime and satisfies $p \equiv 3 \pmod 4$.

$$x^2 \equiv -1 \pmod{p} \implies x^4 \equiv 1 \pmod{p}$$

We have $p = 4m+3$ which gives us

$$x^{p-1} = x^{4m+2} = x^2x^{4m} \equiv-1\pmod{p}$$

This would be a contradiction to Fermat's little theorem, and hence has no solutions, for $x \in \mathbb{Z}$.

You haven't described your procedure, so how should we be able to answer that?

Modular arithmetic is normally only interesting in $\mathbb N\not\ni \sqrt{6}$, so chances are that you haven't found what was searched for, which would suggest you need a different approach.


$X,Y$ distributes geometrically, calculate $P(X+Y=n)$? I have this question (Answer provided).

I'm asking only about $P(X+Y=n)$

My calculation is:
$$\sum_{k=0}^n P(X = k, Y = n-k) $$
And when calculated like this, the final answer is: $(n+1) p^2 (1-p)^n$

I've searched online and found that in a lot of places $(n-1)$ will recieved if the sigma is: $$\sum_{k=1}^{n-1}$$ So it might be the key to the answer.

BUT, I don't have any constraints in the question so why should it be like this?
Thanks

There are two different definitions of Geometric distribution (check wikipedia or anything else). https://en.m.wikipedia.org/wiki/Geometric_distribution

You use the one which starts from zero, authors of this problem use the one that starts from 1.

The geometric distribution has probability mass zero on $0$.  Consequently,
$$ P(X = 0, Y = n) = P(X = n, Y = 0) = 0  \text{.}  $$

Therefore, your
$$ \sum_{k=0}^n P(X = k, Y = n-k) = 0 + \left( \sum_{k=1}^{n-1} P(X = k, Y = n-k) \right)  + 0  \text{.}  $$


Is $A$ \ $B$ ~ $A$ true? 1) If $A$ is infinite and uncountable, and $B$ is finite or countable, is $A$ \ $B$ ~ $A$ true?

2) If $A$ is infinite uncountable and $B \subset A$ is finite or countable, is $A$ \ $B$ ~ $A$ true?

1)I assume it is true, because of the cardinality of the continuum?
2)Also true for the same reason?

1) follows from 2), because in all cases $A \setminus B = A \setminus (A \cap B)$, and if $B$ is finite or countably infinite, so is $A \cap B$.

To prove 2), if $B$ is finite or countable, it is not all of $A$. So there exists $x \in A \setminus B$. Now you can apply the same argument to $A$ and $B \cup \{x\}$. Repeat. You will be able to find a countable subset $C$ of $A$, disjoint from $B$. Now: $A \setminus B = (A \setminus (B \cup C))\cup C$ and $A = (A \setminus(B\cup C)) \cup (B \cup C)$. The "outer" unions are disjoint unions in both cases.

Now compare the RHS of the two equations: The first term of the union is the same, and the second is a countably infinite set in both cases. There is a bijection between $C$ and $B\cup C$, and you can extend this with the identity map on $A\setminus(B\cup C)$ to get a bijection from $A\setminus B$ to $A$.

They are indeed both true.
Assume that $A$ is uncountable, $B$ is countable, and $A - B$ is countable. Since the union of two countable sets is countable, then $(A - B) \cup B$ is countable. But $A \subset (A - B) \cup A;$ since a subset of a countable set is itself countable, we conclude that $A$ has to be countable. This contradiction proves that $A - B$ is uncountable.


This can be proven easily by taking advantage of the AM-GM inequality:

$$\sum_{i=1}^{n}{a_i^n } \geq \sum_{i=1}^{n}{b_i^n } \Leftrightarrow \sum_{i=1}^{n}{a_i^n } \geq n\cdot \sqrt[n]{\prod_{i=1}^{n}b_{i}^{n}} |\cdot \frac{1}{n} \overset{\prod_{i=1}^{n}a_{i}}{\rightarrow}\frac{1}{n}\cdot \sum_{i=1}^{n}{a_i^n } \geq \sqrt[n]{\prod_{i=1}^{n}a_{i}^{n}}$$


Probability that $A \cup B$ = S and $A \cap B = \phi $ Let $S$ be a set containing $n$ elements and we select two subsets: $A$ and $B$ at random then the probability that $A \cup B$ = S and $A \cap B = \varnothing $ is?

My attempt

Total number of cases= $3^n$ as each element in set $S$ has three option: Go to $A$ or $B$ or to neither of $A$ or $B$

For favourable cases: Each element has two options: Either go to $A$ or to $B$ which gives $2^n$ favourable cases.

Is my approach correct?

Pick any subset $A$, and there is only one subset $B$, namely $S \setminus A$  which satisfies $A \cup B = S$ and $A \cap B = \emptyset $.

There are $2^n$ subsets to choose from so the probability of selecting such a pair is $1/2^n$.

(Or, $1/(2^n - 1)$ if one constrains that $A \ne B$).

Your proposed solution assumes A and B are disjoint in all possible cases.  The question appears to intend A and B are not necessarily disjoint, and requires them to be disjoint for success.

Also you don't state whether it's equally likely to choose e.g. 3 or 1 elements since the natural scenario in the real world is often a binomial distribution, with ~n/2 being the most likely number of elements chosen each time.  But let's assume that's not the case here.

Each element has 4 options: Only A, Only B, both or neither so there are $4^n$ cases.

The cases not fitting the first condition are those in which one element was distributed to neither, the probability being $1-(3/4)^n$.

The cases not fitting the 2nd condition are those in which one element was distributed to both, the probability again being $1-(3/4)^n$.

The combination of the two is a conditional probability so these cannot simply be multiplied.  The quickest way is to deduct the two probabilities of failure from 1 so the answer is $(1/4)^n+(1/4)^n$.


Wieferich primes in base $47$ Wieferich primes are defined as prime numbers $p$ such that $p^2$ divides $2^{p − 1} − 1$. While reading about such primes, I came upon the following curious conjecture on the Wikipedia page of "unsolved problems in number theory":

  Are there any Wieferich primes in base $47$?

Since no explanation is given for this strange question, I find myself puzzled by the importance of the number $47$ within this context. What role does this base in particular have in the context of Wieferich primes and why would it be important to solve this problem in particular, instead of another number base?

After some research on the internet, it indeed seems that Peter was right in the comment section, and $47$ is the smallest base for which no Wieferich primes are known.

Weiferich primes exist only in base 2. He wrote his paper in 1908, but the question has been around since the time of Euler.  Dickson's history of mathematics devotes 12 pages to this question, but affords Weiferich only five lines in a paragraph on the nineth page.  It is misleading to use this term.

Weiferich did not discover sevenites, but he noted that a particular solution to fermat's last proposition exists only for binary sevenites.

There is no particular reason for 47 not to have any particular sevenites.  A similar situation exists with there being no known iso-sevenites for 3, (the Fibonacci series) which means that no instance of where if $p \mid F_n$ then also $p^2 \mid F_n$.  However, the octagon-series and the Heron series, which correspond to isobases 6 and 4, do have sevenites.

Thus, in the series of Heron triangles, (triangles of sides e-1, e, e+1 and integer area), if 103 divides a side, so does $103^2$.

47 has 2 as a sevenite (or 'weiferich prime').  The two-place period of 2 supposes only 8, as can be seen in 11, 13 and 45.  Here 32 divides 47^2-1, and thus it has two as a sevenite.

EDIT:

The table of 'sevenites' for particular bases, up to b=14400 and p=2000000, do not produce a list of primes longer than 80 digits, except in one or two cases.  The number is quite small.  Sort of in the $\sum 1/p$ range.  There are a good scattering of unfilled rows, 47 is the first.

In the early versions of the tables I produced 6 was the first unfilled row.  But then 61661 came along.


Need help in solving the differentiation I want to differentiate the following expression with respect to n:
$\frac{d}{dn}[1-{(1-p)^n}^d]$
I do not know which formula to use to differentiate this term with respect to n. Please can anybody help me?
I have tried like this:
$\frac {d}{dn}(1-{(1-p)^n}^d)= -\frac{d}{dn}[{(1-p)^n}^d]=-\frac{d}{dn}[e^{n^dln(1-p)}]=-e^{n^dln(1-p)}.ln(1-p).dn^{d-1}$   Is my answer correct?

Assuming the $d$ in the exponent is a real number and that $p\leq 1$:
$$\frac{d}{dn}[1-(1-p)^{n^d}] = -\frac{d}{dn}(1-p)^{n^d}
= - \ln(1-p) \times (1-p)^{nd} \times \frac{d}{dn} (n^d) \\ = - n^{d-1}d(1-p)^{nd}\ln(1-p)$$
Which I believe is the same as your answer, although simplified.


How to solve this congruence $17x \equiv 1 \pmod{23}$? Given $17x \equiv  1 \pmod{23}$

How to solve this linear congruence?
All hints are welcome.

edit:
I know the Euclidean Algorithm and know how to solve the equation $17m+23n=1$
but I don't know how to compute x with the use of m or n.

Algorithmically one can employ the Extended Euclidean Algorithm to invert $\,17\,$ modulo $\,23\,$ (possible since they are coprime). But for small numbers it is usually quicker to use other methods, e.g. rewriting  $\,17^{-1} \equiv \frac{1}{17}$ as a fraction, then adding $\pm$modulus to the numerator and denominator until one obtains an exact quotient. Do this $\rm\color{#0a0}{first}$ to minimize the absolute value of the denominator (to increase the probability of an exact quotient), and  $\rm\color{#c00}{then}$ do the same to the numerator, aiming to obtain a multiple of the denominator. For example

$${\rm mod}\ 23\!:\,\ 17x\equiv 1\,\Rightarrow\,x \,\equiv\, \dfrac{1}{\color{#0a0}{17}} \,\equiv\, \dfrac{\color{#c00}1}{\color{#0a0}{-6}}\,\color{#c00}\equiv\,\dfrac{\color{#c00}{24}}{-6} \,\equiv\, -4 \,\equiv\, 19\qquad\qquad$$

We subtracted $23$ from the denominator to get the minimal integer $\,\color{#0a0}{-6 \equiv 17}\pmod {23},\,$ then since $\,23\equiv -\color{#c00}{1}\pmod 6,\,$ we add it the numerator $(= \color{#c00}1)$  to make it be divisible by $6$.

Beware $\ $ For non-prime moduli one must be sure to restrict to denominators that are coprime to the modulus (so invertible), so to avoid the modular analog of dividing by zero.

In this example the described method ends up being essentially a special case of said algorithm, reinterpreted in fractional form. But that is not always so. The advantage of the fractional form is that it allows one to use well-known fractional arithmetic (subject to said restriction), e.g. cancelling common factors. One can find many worked examples is my prior posts.

simplest  answer is x=1/17 because 17x=1mod 23 17

17x=1mod23,
23 divede (17x-1)
if x=1/17,
17*1/17=1,
1=1mod23,
since 23 divide (1-1),
23 divide 0


How many leading principal minors are there for a $4 \times 4$ matrix? How many leading principal minors are there for a $4 \times 4$ matrix?

Please explain in detail. I know for a $3 \times 3$ matrix.

The kth order leading principal minor is the determinant of the kth order principal
submatrix formed by deleting the last n − k rows and columns.

I think you are referring to leading principal submatrix rather than leading principal minors, which is a bit broad. If you can find for 3*3 matrix and you know how to do it, you can use the same method to find for 4*4 matrix.


Is there an elementary proof that next prime number is guaranteed to be relatively nearby, that does not involve prime number theorem-related maths? I asked this question yesterday, perhaps a bit too hastily:
Does the prime number theorem tell us that the next prime number is guaranteed to be relatively nearby?
I think I bit off more than I can chew with this question, as I know little about the maths surrounding the prime number theorem. However, I asked the question assuming that any proof that
$$ \displaystyle\lim_{n\to\infty}\ \frac{p_n}{p_{n+1}} = 1\quad \text{where}\ p_n\ \text{is the n-th prime number}?$$
will require knowledge of the PNT, especially the proof of the PNT itself, which I do not have. Is there an elementary proof of the above statement, that does not require knowledge of the PNT or related facts, or difficult number theory?
For example, the proof that there are infinitely many primes is not too difficult to understand, and doesn't require any advanced number theory. So I am wondering if there is a proof that the primes are eventually relatively close with a similar level of difficulty to the proof that there are infinitely many primes. Sorry for not providing this information sooner.

The first elementary proof of the PNT famously happened when Erdos proved that $\lim_{n\to\infty}\frac{p_n}{p_{n+1}}=1$, and then Selberg combined that with some of his other works to derive the PNT in an elementary fashion. If you wish to call Erdos's proof of the $\lim_{n\to\infty}\frac{p_{n}}{p_{n+1}}=1$ an "Elementary proof without prime number theorem related maths" then you have your answer, but seeing as this was the crucial step missing for a full elementary PNT proof I'd call that a stretch. I don't think that there is any more complete answer to your question one can give, short of transcribing a rendition of Erdos's proof.
An account of the first elementary proof of the PNT is here. Be warned, however. What Erdos and Selberg call elementary might not look so elementary to you.


What is the number of consignments that would be packed in the initial $113$ hours? QUESTION: In a factory, $20$ workers start working on a project of packing consignments. They need exactly $5$ hours to pack one consignment. Every hour $4$ new workers join the existing workforce. It is mandatory to relieve a worker after $10$ hours. Then what is the number of consignments that would be packed in the initial $113$ hours?

MY APPROACH: It seems a bit complicated to me.. Firstly, every hour $4$ new workers join the workforce.. I don't know how to handle that.. and above that I have to subtract workers who have worked for $10$ hours.. I am sorry that this time I could not provide any working as I could not do anything as such.. I hoped to learn this here but noone has asked it before.. so I am forced to ask it.. one more thing, I do not know which tag is appropriate for this type of problem, so suggestions on that too are welcome..

Any help will be much appreciated. Thank you so much.

I tried it too but my approach was different .
Let as per condition after 113 hours we have 441 workers they can do one consignment in 100/441 hrs ..now this thing when multiplied by some number should yield 10 cuz every 10 hours a worker is relieved which makes the total number of workers not equal to 441 therefore the greatest integer is 45 which is multiplied to give 10.20 something.
For better understanding see this video
https://youtu.be/3mJ0U-yIXz8


If $G$ is an inﬁnite simple group then any proper subgroup of $G$ has inﬁnite index. If $G$ is an inﬁnite simple group then any proper subgroup of $G$ has inﬁnite index.

This question's hint is use the $n!$-theorem but i dont understand how i use it for answer.

$n!$-theorem: Let $G$ be a group and $H$ be a subgroup of $G$ of finite index, say $|G:H|=n$. Then there is a normal subgroup N of $G$ such that $N\subseteq H$ and $G/N$ is isomorphic to a subgroup of $S_n$ and so $|G/N|$ divides $n!$. Indeed, ${\rm core}_G(H)$ is such a normal subgroup of $G$.

Assume for contradiction that $G$ is an infinite simple group and has a finite index subgroup $H$ such that $|G:H|>1$.  Then, $\mathrm{Core}_G(H)$ is a finite index normal subgroup because the intersection of finite index subgroups is finite and by definition $\mathrm{Core}_G(H)$ is invariant under conjugation by $G$.  In particular, $G$ has a non-trivial proper normal subgroup and therefore cannot be simple.  Thus, $G$ does not have any proper finite index subgroups.

From a stylistic perspective, quoting a "theorem" here this is unnecessarily complicated and makes things harder to understand.  The proof of this statement relies on two basic facts; and these are facts that everyone who studies groups should know anyways.  1. There are only finitely many subgroups of a given finite index.  2. The intersection of two finite index subgroups has finite index.  I will take these for granted; each is an easy exercise.

Now given a finite index subgroup $H$, it has only finitely many conjugates by 1.  So intersect all of them.  This subgroup has finite index by 2.  It is clearly normal because conjugating the intersection amounts to intersecting the conjugates, but that just permutes the things we're intersecting and we end up intersecting the same finite set of subgroups.


"implied multiplication" operator precedence? I hold a masters in computer science from one of the worlds top universities and until today I thought I more or less know basic math.

I'm sure you guys all know these click-bait simple "90% of people can't solve this equation" posts on facebook where everyone starts to argue over a simple equation (and I DEFINITELY don't want to kick off one of those - but I'd like to discuss the roots of this confusion).

So today there was another one of those:

6/2(1+2)

Based on what I learned and applied throughout all my years in university, this equals 9, since brackets are evaluated first and then it's left to right, since division and multiplication have the same operator precedence.

Google agrees with that:

Wolfram Alpha agrees with that:

and my texas instruments agrees with that too. So it's 9, right?

Well today I came across a claim I hadn't heard before, which is "implied multiplication takes precedence over both explicit multiplication and division" - so by that rule it would not be left to right in the above example, but the implied multiplication would be evaluated before the division, which would mean that

6/2(1+2) == 1 != 6/2*(1+2)

So, are google, wolfram alpha and my calculator all wrong (they by the way also yield 9 if ÷ is used instead of /)?

The only thing i found on the issue so far is this statement on Wikipedia (https://en.wikipedia.org/wiki/Order_of_operations):

  Mixed division and multiplication:
  Similarly, there can be ambiguity in the use of the slash symbol / in expressions such as 1/2x.[5] If one rewrites this expression as 1 ÷ 2x and then interprets the division symbol as indicating multiplication by the reciprocal, this becomes:

1 ÷ 2 × x = 1 × 1/2 × x = 1/2 × x.

  With this interpretation 1 ÷ 2x is equal to (1 ÷ 2)x.1[6] However, in some of the academic literature, multiplication denoted by juxtaposition (also known as implied multiplication) is interpreted as having higher precedence than division, so that 1 ÷ 2x equals 1 ÷ (2x), not (1 ÷ 2)x.
  For example, the manuscript submission instructions for the Physical Review journals state that multiplication is of higher precedence than division with a slash,[7] and this is also the convention observed in prominent physics textbooks such as the Course of Theoretical Physics by Landau and Lifshitz and the Feynman Lectures on Physics.[a]

So one thing this tells me is clearly AVOID IMPLIED MULTIPLICATION

but what is internationally actually 'more correct and less wrong'?
Also, I don't fully see how ÷ vs. / is relevant to this question?

By the way, some calculators do seem to evaluate to 1:

In your picture of the $2$ calculators the one with the $9$ answer was only programed for order of operations. The calculator with the answer of $1$ was also programed to consider the distributive property. A calculator is only as good as its programming and only as intelligent as its user.
When I went to school we were taught implied multiplication took precidence because of the distributive property.  And the order of operations was a suggestion not concrete.


Strategies for solving second order DEs with variable coefficients Does anyone have any good videos/ article links for techniques to solve second order DEs with variable coefficients?

I mean, just as an example, a DE like:

$$ x \frac{d^{2}y}{dx^{2}} - (2x-1) \frac{dy}{dx} + (x-1)y = 0$$

(this one is fairly simple to solve but the one I have in mind is tricker (for me anyway)).

If anyone's curious, the DE I have is:

$$\frac{d^2 y}{dx^2} + \frac{2(x^2 - 1)}{x^3} \frac{dy}{dx} + \frac{x^2 + 1}{x^6} y = 0$$

Are there any techniques to systematically find a general solution for DEs like this?

Thanks very much

The best technique is a long, hard stare at the answer given by Wolfram Alpha:
$$y(x)=\left(\frac{C_1}{x}+C_2\right)\exp\left\{-\frac{1}{2x^2}\right\}.$$

More seriously, when you have a linear 2nd order equation with rational coefficients, the first thing you need to look at is the structure of its singular points on the Riemann sphere. For example,

if there are $3$ regular singularities, we are sure that the equation can be solved with Gauss hypergeometric functions,
if there are $>3$ singular points, it is (almost) certain that the solutions cannot be written in a closed form.
when we have $\leq 2$ singularities (like in your example), some of which are irregular (here $x=0$), then one should look in more details at the solution asymptotics to decide if there are any chances to solve the equation in terms of elementary or special functions.

In your case the only (irregular) singular point is $x=0$. To investigate the asymptotics, it is convenient to send it to infinity by the change of variables $t=1/x$, $u(t)=y(x)$ and then look for the asymptotic solution in the form $u(t)=t^{a}\exp P(t^b)$, where $P$ is a polynomial. But in your case this already gives the exact answer.

The ODE that you have gets dramatically simplified if you convert it into standard form: $$z''(x)+v(x)z(x)=0.$$ In fact, for your example, $v(x)=0$, so that $z=c_1+c_2x$, and since $$y(x)=\frac{z}{x}\exp\{-\frac{1}{2x^2}\},$$ you get the answer.

For converting to standard form, see
Second-Order Ordinary Differential Equation on MathWorld.


Is the class of algebraic extensions distinguished? In paragraph V.1 of Algebra proposition 1.7 Lang claims that the class of algebraic extensions is distinguished. I know that if $F/k$ and $E/F$ are algebraic extensions than so is the $E/k$ - that is easy to prove. However, the second required property is that if $E/k$ is algebraic and $F/k$ is arbitrary, then (assuming the compositum is defined) $EF/F$ is algebraic. Lang more or less skips the proof of this, only saying that "an element remains algebraic under lifting, and hence does the extension."

However, there are no finiteness conditions here and no element given in advance. Although $EF$ is generated over $F$ by a set of elements which are algebraic over $F$ (namely, all the elements of $E$), there are no facts known (up to this point in the book at least) about the infinitely generated extensions.

What am I missing?

Thank you.

In Lang's definition of distinguished, both conditions (2)-(3) require that $\,k\subset E,F\,$ and both $\,E,F\,$ are contained in some object $\,R\in\mathcal C\,$ , so $\,F/k\,$ cannot be "arbitrary":

$\,\Bbb Q/\Bbb Q\,$ is algebraic and $\,\Bbb Q(\pi)/\Bbb Q\,$ is any extension, but $\,\Bbb Q\cdot\Bbb Q(\pi)/\Bbb Q=\Bbb Q(\pi)/\Bbb Q\,$ is not algebraic.

Added: The elements of $\,EF\,$ can be seen as rational functions in the elements of $\,E\,$ with coefficients of $\, F\, $ (and the other way around, of course).

Let $\,x\in EF\,$. Then there exists a finite number of elements in $\,E\,$ , say $\,e_1,...,e_r\,$ which actually appear in the above rational expression of $\,x\,$, and thus $\,x\in F(e_1,...,e_n)\subset EF\,$.

Since clearly $\,F(e_1,...,e_n)/F\,$ is algebraic we're done.

If $E/F$ is an algebraic extension, $K/F$ is an extension. Prove that $EK/K$ is algebraic.

Assume $H=\{a\in EK|a \text{ is algebraic over } K\}$. It is easy to see that $H$ is a field and $K<H<EK$. $E/F$ is algebraic, so all the elements in $E$ are algebraic over $F$, also over $K$; so $E\subseteq H$; then we have $EK\subseteq H$, which means $EK=H$. $EK$ is an algebraic extension over $K$. ■


Which is the greatest possible natural number that divides $(p+3)(p-7)$, where $p$ is a prime number greater than $3$? Which is the greatest possible natural number that definitely divides
  $(p+3)(p-7)$, where $p$ is a prime number greater than $3$?

This one is from my module, comes as a fill in the blanks with no answer. I have a feeling that there is something wrong with this question, since for $p=5$, one has $(p+3)(p-7)=-16$.

ADDED:

As mentioned, in one answer I tried substituting $p=5,7,9,11,13,17,19,...$ and spotted that the greatest number is $8$. I am just wondering .. is there any other way (probably using modulus) to directly find this number without actually substituting ?

I have a feeling that there is something wrong with this question

The "greater than $3$" condition is a bit strange, because the answer is unchanged if $p=3$ is allowed. And if they wanted to exclude negative values of $(p+3)(p-7)$, they should say "greater than 7". But the question as it stands is technically correct. It would perhaps be better like this:

  Which is the greatest natural number that divides $(p+3)(p−7)$ for every odd prime number $p$?

If $p$ is odd, then $p = 2k+1$ for some $k$. So $(p+3)(p-7) = (2k+4)(2k-6) = 4(k+2)(k-3)$, which is divisible by $8$ since one of $k+2, k-3$ must be even. So $8$ divides $(p+3)(p−7)$ for every odd prime number $p$.

That's half the question. The other half is to show that no natural number $n > 8$ divides $(p+3)(p−7)$ for every odd prime number $p$. Note that if $n$ has this property, then so does every factor of $n$. So it is enough to show that (i) $16$ doesn't have this property; and (ii) no odd prime $q$ has this property.

To show that $16$ doesn't have this property, just put $p=11$. So suppose now that $q$ is an odd prime. We need to produce another odd prime $p$ that is not equal to $q-3 \pmod q$ and not equal to $7 \pmod q$. Then we can be sure that $q$ doesn't divide $(p+3)(p-7)$. It seems a bit like overkill to me, but we can use Dirichlet's theorem on arithmetic progressions for this: for coprime $a$ and $q$, the arithmetic progression $a + rq$  $(r = 0, 1, 2,...)$ contains infinitely many primes. So we just have to choose $a$ with $1 \le a \le q-1$ that is not equal to $q-3 \pmod q$ and not equal to $7 \pmod q$. If $q=3$, we can choose $a=2$; and if $q \ne 3$, we can choose $a=3$.

And we're done.

The question is sloppily phrased, because to me it reads more like "Which is the greatest possible natural number that definitely divides $(p+3)(p−7)$, where $p$ is some given prime greater than 3?", in which case the answer is trivially $(p+3)(p-7)$ !

But I imagine (as does everyone else by the look of it) the OP meant "Which is the greatest possible natural number that definitely divides $(p+3)(p−7)$ for every prime $p$ greater than 3?"


A limit related to super-root (tetration inverse). Recall that tetration ${^n}x$ for $n\in\mathbb N$ is defined recursively: ${^1}x=x,\,{^{n+1}}x=x^{({^n}x)}$.

Its inverse function with respect to $x$ is called super-root and denoted $\sqrt[n]y_s$ (the index $_s$ is not a variable, but is part of the notation — it stands for "super"). For $y>1, \sqrt[n]y_s=x$, where $x$ is the unique solution of ${^n}x=y$ satisfying $x>1$. It is known that $\lim\limits_{n\to\infty}\sqrt[n]2_s=\sqrt{2}$. We are interested in the convergence speed. It appears that the following limit exists and is positive:
$$\mathcal L=\lim\limits_{n\to\infty}\frac{\sqrt[n]2_s-\sqrt2}{(\ln2)^n}\tag1$$
Numerically,
$$\mathcal L\approx0.06857565981132910397655331141550655423...\tag2$$

Can we prove that the limit $(1)$ exists and is positive? Can we prove that the digits given in $(2)$ are correct? Can we find a closed form for $\mathcal L$ or at least a series or integral representation for it?

I have some observations, from where someone more experienced might be able to derive the proof - maybe this is helpful)

Let the iterated functional root ("superroot of order") $B(z,n)$  (which finds the " B "ase of the powertower) be defined as
$$ \;^n b = z  \qquad \to \qquad  B(z,n) = b  $$
For the following let us always denote $u$ for the $\log(z)$ for notational convenience.
Let us then define the generalization of the Lambert-W-function to higher iterates as a simple conjugate of the base-finding $B()$-function:
$$ W(u,n) = W(\log(z),n)  = \log( B(\exp(u),n)) \tag 1$$
where we can find a formal power series for $W(u,n)$ from the Lagrange series reversion of $u \cdot \exp(u)$, such that
$$ \begin{array}{} \mathcal {\text{ Taylor:} } & W^{-1}(u,1) = u  \\
 \mathcal {\text{ Taylor:} }& W^{-1}(u,2) = u \cdot \exp(u) \\
 \mathcal {\text{ Taylor:} }& W^{-1}(u,3) = u \cdot \exp(u \cdot \exp(u) ) \\
\end{array} \tag 2$$
Then your limit for a general $z$ and $u = \log(z)$ reads
$$ \lim_{n \to \infty} {B(z,n)-z^{1/z} \over \log(z)^n} \qquad \text{or}
\qquad \lim_{n \to \infty} {\exp(W(u,n))-\exp(u \cdot \exp(-u)) \over u^n} \tag 3 $$
Actually we use $z=2$ (and thus $u=\log(2)$) and this reads a bit more friendly:
$$ \lim_{n \to \infty} {\exp(W(u,n))- \sqrt 2 \over u^n} \tag 4$$

Now, if we look at the formal power series for $W(u,n)$ we find a striking pattern, in that with increasing n the coefficients of the equivalently longer leading part become constant, which can be written like the following
$$ \begin{array} {}
W(u,n) &=& u \Large \left(1-u+{ u^2 \over 2!}-{ u^3 \over 3!} + ... \right. \\
&& \qquad +  u^{n-1} \Large \left(  {a_{n,1} u \over 1!} -{a_{n,2} u^2 \over 2!} + ...   \right) \\
 && \left. \quad \Large \right)\end{array} \tag 5$$
Let us denote the part in the inner parenthese as $R(u,n)$
$$ R(u,n) = {a_{n,1} u \over 1!} -{a_{n,2} u^2 \over 2!} - ...  \tag 6
$$
which makes then
$$ \begin{array} {rrl}
 W(u,n) &=& u \cdot \exp(-u) + u^n \cdot R(u,n) )\\
 B(z,n) &=& \exp(u \cdot \exp(-u) + u^n \cdot R(u,n)) \\
        &=& \exp(u \cdot \exp(-u)) \cdot \exp( u^n \cdot R(u,n)) \\
        &=& \sqrt 2\cdot \exp(  u^n \cdot   R(u,n)) \\
\end{array} \tag 7$$

Something on convergence: because the first line in fomula (5) contains an exponential series in $u$ and such a series is entire, the consideration of range of convergence of the whole construct focuses on $R(u,n)$ - other than with $z=2$ and $u=\log(2)$ I did not check it, but for this settings it seems to converge to something $\lim_{n \to \infty} R(u,n) =r_\infty \approx 0.0484903140769$ just by numerically searching (binary search) $b_n=B(2,n)$ and computing backwards. Interestingly the coefficients of the power series of $R(u,n)$ have the same scheme to become constant for the leading part which extends when n increases and has thus also a limiting power series. Pari/GP gives me
$$ R(u,12)= u - 5 \cdot u^2/2! + 13 \cdot u^3/3! - 19 \cdot u^4/4! + 1  \cdot u^5/5! + 231 \cdot u^6/6! - ... $$
and this is the same for all values n greater than 12, so we might assume this constant also for the case of $n \to \infty$.
I did not find anything related in OEIS yet, perhaps I can later find some pattern...

We see in the last formula in (7) the squareroot occuring, so we get the initial formula as
$$ \lim_{n \to \infty}\sqrt2 \cdot {  \exp(u^n  R(u,n)) -1  \over u^n} \tag 8 $$
where the exponentialseries (divided by $u^n$) has no constant term and the linear term is $ R(u,n)$ and the following terms are powers of it, multiplied by n'th powers of $u=\log(2) \approx 0.693 \lt 1$.

These are so far only some (organized) observations, a real proof of the existence of a finite limit must be provided by analysis of the power series of $R(u,n)$ (but besides the heuristics I don't have the proof yet)

I got a message from Tommy1729.

He considered nonzero $T$ such that :

$$T = \lim \frac{A^n}{(f(n,2) - \sqrt 2 - L \ln 2 ^n)}$$.

Where $f(n,2)$ is the nth superroot of $2$ , $L$ is the constant from the Op and $A$ is a constant.

If the limit $T$ does not exist at least the best fitting $A$ is considered.
In other words :

$$A^n $$

~

$$\frac{1}{(f(n,2) - \sqrt 2 - L \ln 2 ^n)}$$

Generalizing the Op is easy.
And the problem was not new to him.

These generalizations seem even harder and he did not provide formal proofs.

However what amazed me was this :

$$L \ln2 = r = 0.0475...$$

Where $ r $ is conjectured the radius of Gottfried's answer !

Sorry too add questions instead of answers. But this is too much for a comment.


Why is it that solutions $x_1 \text{ and } x_2 \text{ to a differential equation are linearly dependent } \iff W(x_1,x_2)(t_0)=0$ Can someone explain  from where the aforementioned fact is derived? By the way $W = \text { Wronskian determinant}.$

For the non-trivial direction, suppose $x_1$ and $x_2$ are solutions to $$x''+bx'+cx=0,$$define $$W=x_1x_2'-x_1'x_2,$$and assume $$W(t_0)=0.$$Define $$y(t)=x_2'(t_0)x_1(t)-x_1'(t_0)x_2(t).$$Of course $y$ is a solution to the original DE, being a linear combination of $x_1$ and $x_2$. It's clear that $y'(t_0)=0$, and $W(t_0)=0$ shows that $y(t_0)=0$. So uniqueness shows that $$y(t)=0,$$which says that $x_1$ and $x_2$ are dependent.


Direct Proof that $1 + 3 + 5 + \cdots+ (2n - 1) = n\cdot n$ Prove that for all integers $n$, $n \geq 1$,
  $$1 + 3 + 5 + \cdots + (2n - 1) = n\cdot n$$

How would I go about proving this?

Hint:  When $n=5$, $$\boldsymbol{1+{\color{red}3}+{\color{green}5}+{\color{purple}7}+{\color{blue}9}=} $$ $$\begin{array}{ccccc}
\blacksquare & {\color{red}\blacksquare} & {\color{green}\blacksquare} & {\color{purple}\blacksquare} & {\color{blue}\blacksquare}\\
{\color{red}\blacksquare} & {\color{red}\blacksquare} & {\color{green}\blacksquare} & {\color{purple}\blacksquare} & {\color{blue}\blacksquare}\\
{\color{green}\blacksquare} & {\color{green}\blacksquare} & {\color{green}\blacksquare} & {\color{purple}\blacksquare} & {\color{blue}\blacksquare}\\
{\color{purple}\blacksquare} & {\color{purple}\blacksquare} & {\color{purple}\blacksquare} & {\color{purple}\blacksquare} & {\color{blue}\blacksquare}\\
{\color{blue}\blacksquare} & {\color{blue}\blacksquare} & {\color{blue}\blacksquare} & {\color{blue}\blacksquare} & {\color{blue}\blacksquare}
\end{array}  $$

$$\boldsymbol{=5^2}.$$

Turn this into a general proof.

you can see this link in generally solution  :

http://en.wikipedia.org/wiki/Arithmetic_progression

This section is about Finite arithmetic series. For Infinite arithmetic series, see Infinite arithmetic series.

The sum of the members of a finite arithmetic progression is called an arithmetic series.

Expressing the arithmetic series in two different ways:

$    S_n=a_1+(a_1+d)+(a_1+2d)+\cdots+(a_1+(n-2)d)+(a_1+(n-1)d)\\
    S_n=(a_n-(n-1)d)+(a_n-(n-2)d)+\cdots+(a_n-2d)+(a_n-d)+a_n.$

Adding both sides of the two equations, all terms involving d cancel:

$    \ 2S_n=n(a_1+a_n).$

Dividing both sides by 2 produces a common form of the equation:

$    S_n=\frac{n}{2}( a_1 + a_n).$

An alternate form results from re-inserting the substitution:$ a_n = a_1 + (n-1)d:$

$    S_n=\frac{n}{2}(2a_1 + (n-1)d). $


Why $\mathbb{Q}$ is not a projective $\mathbb{Z}$-module? From the fact that $\operatorname{Hom}_{\mathbb{Z}}(\mathbb{Q},\mathbb{Z})=0$, how do we conclude that $\mathbb{Q}$ is not a projective $\mathbb{Z}$-module?

Alternatively, another fast way to prove by some Theorems is that
“Any modules over a PID,
it is a projective module
if and only if
it is a free module.”

So by contradiction,
suppose Q is a projective Z-module.
Because Z is a PID,
Q is also a free Z-module
But It's not.

Because for all submodules of Q \ {0},
they are not linearly independent over Z.
And thus the only independent submodule of Q is {0},
which cannot span the whole Q.
So Q cannot find a basis of Z-module.
Then this will complete the proof.

We can simplify the
linearly independence clarification above to show that

“For all q/p and n/m belongs to Q \ {0},
where p, m are not 0,
q/p and n/m are not linearly independent over Z.”

Suppose a(q/p)+b(n/m)=0 for some a, b belong to Z.
Take a=pn and b=-mq belong to Z.
Then a(q/p)+b(n/m)
=(pn)(q/p)+(-mq)(n/m)
=qn - qn =0.
Therefore, q/p and n/m
are not linearly independent over Z.

In sum, because Z is a PID
and Q is not a free Z-module,
then Q is not a projective Z-module, too.

I try to create a counterexample but later to find out there is some mistake in the condition I've made. Maybe someone can help to fix it or try to find another better counterexamples??
Anyway, here's my (false) counterexample I'd tried to find.

//
Consider A→B→0 with Z-module the exact sequence
with A=Z, B=Z,
g:A→B is an epimorphism defined by
g(z)=z+1 for all z belongs to Z.

Because f:Q→B
can only be the zero map
and h:Q→A as well.
yet gh(q)=g(0)=1=/=0=f(q)
for all q belongs to Q.

Thus there isn't exist any
(actually it's because we've only have that one for us to check and however it doesn't fulfill.)
homomorphism h:Q→A such that gh=f
for this exact sequence to be commutative.

Therefore, Q is not an projective Z-module.
//

The mistake is...even though g:A→B is onto in the example above, it does not form a module map...：(
Maybe we need to try to find another g such that it is a Z-module epimorphism pair of hom. f and epi. g to make the counterexample exist.


The sum of two positive numbers is 1. The sum of their cubes is a maximum. What are the numbers? I set this up and end up finding the minimum (the two numbers would both be $1/2$).  To find a maximum value, I could reflect the functions and use $y^3-x^3$ but I still end up finding $1/2$ as the two numbers.  It does make sense that two two values to produce a max would be 0 and 1 but I can't figure out how to set up the problem from the start.  What I have is...

$x+y=1 \\
x^3 + y^3 = max$

Subsitution...

$x^3 - (1-x)^3 = max \\
x^3 - 1 + 3x - 3x^2 + x^3 = max \\
2x^3 - 3x^2 + 3x - 1 = max \\
6x^2 - 6x + 3 = 0 \\
x = 1/2 \text{ which makes }y = 1/2 \\
$

That is where my issue is.  How do I set it up to find the sum of the cubes to be a max?

For a function defined on a closed interval. The maximum and minimum are found where the derivative is zero, or at the boundaries of the interval.

You can take derivative, but you also need to check for the case $x=0$ and $x=1$. By the way your original question does not have an answer. I suppose you should mean "nonnegative".

Your answer gives the MINIMUM. As explained above, the diff is not zero at the ends of the range, where the answer is. So that answer is (1-€)x€.

(€=epsilon)


Sylow p-subgroup is p-group Let P be a prime. A p-group is a group of order $P^{r} \forall r\geq 0$.
If G is a group of order $P^{r}m \forall m \in \mathbb{Z}$ where p does no divide m then a Sylow p-subgroup of G is a subgroup of order $p^{r}$.

From this, would readers be valid in making the deduction that a Sylow p-subgroup is a p-group?
Thanks in advance.

The deduction would be valid, since you seems to allow for the possibility of $r=0$ for both p-groups and Sylow-p-subgroups.

However, my textbook (Artin) defines p-groups such that $r\geq1$. If we defined Sylow-p-subgroups such that trivial Sylow-p-subgroups would exist (like my professor does), then not all Sylow-p-subgroups would be p-subgroups if we combine Artin's definition with my professor's definition.


Look at the following infinite sequence: 1, 10, 100, 1000, 10000, . . .. What is the proof that there are 2 numbers in this sequence that differ by a multiple of 12345678987654321?

Let $n = 12345678987654321$.  Consider the first $n+1$ elements of the powers-of-10 sequenc modulo $n$.  According to the pigeonhole principle,  there are at least two of these elements that have the same residue modulo $n$.  Their difference is a multiple of $n$.

Take the digit 1 from the 10 and take the digit 1 from the 100.  They differ by 0 which is a multiple of 12345678987654321.


Calculating permutations of a certain type I've some troubles calculating the total numbers of permutations of a specific type:

  "How many permutations in $\ S_{10}$ are of type (3,5) and how many of type (4,4)?"

For the type, I assume permutations $\omega_i \in \ S_{10}$ of type (3,5) are made by one 3-cycle, one 5-cycle and two 1-cycle, something like:  $\omega_1$ = (1 2 3)(4 5 6 7 8)(9)(10) = (1 2 3)(4 5 6 7 8).

Same for permutations $\sigma_i \in \ S_{10}$ of type (4,4) : something like $\sigma_1$ = (1 2 3 4)(5 6 7 8)(9)(10) = (1 2 3 4)(5 6 7 8)

I'm not sure if I'm correctly interpretating the definition of type of a permutation.

Latest, for the total number of permutations, I'm calculating in this way:

Permutations $\omega_i \in \ S_{10}$ of type (3,5) : $\frac{10!}{3 * 5 * 1^2}$
Permutations $\sigma_i \in \ S_{10}$ of type (4,4) : $\frac{10!}{4^2 * 1^2 * 2!}$ , where 2! is added because we have two 4-cycles.

If I should calculate the number of permutations $\delta_i\in \ S_{50}$ of type (2,5,5,6,6,6,10,10) with the same technique: $\frac{50!}{2 * 5^2 * 6^3 * 10^2 * 2! * 3! * 2!}$ where 2!*3!*2! are the exponents, which are the "repeated" cycles of the same length.

Is this the correct way? Thanks you!

Concerning "interpretation", well that's a standard way
to classify permutations, based on their cycle structure,
so yours is a plausible interpretation.

We can represent a given permutation through its cycles, as in one of your example
$$
\left( {1,2,3} \right)\left( {4,5,6,7,8} \right)\left( 9 \right)\left( {10} \right)
$$
However, that same permutation cold be represented as well by
$$
\left( {4,5,6,7,8} \right)\left( {3,1,2} \right)\left( 9 \right)\left( {10} \right)
$$
that is, the cycles can be permuted between them, and each cycle
can be internally cyclically permuted.

The representation can be made univocal if
 - each cycle is taken to start with its (e.g.) lowest element;
 - the cycles are ordered (e.g. increasingly) according to their first element;
which is the example you gave.

Another univocal representation is possible, and is more convenient for the calculations
you are requiring, which is:
 - each cycle is taken to start with its (e.g.) lowest element;
 - the cycles are ordered firstly by their length and secondly by their first element;
in this way our example becomes
$$
\left( 9 \right)\left( {10} \right)\left( {1,2,3} \right)\left( {4,5,6,7,8} \right)
$$

Permutations can be partitioned according to what is called the cycle structure,
which corresponds to a vector $C$ where each component $C_k$ represents
the number of cycles of length $k$, and clearly
$$
\sum\limits_{1\, \le \,k\, \le \,n} {k\,C_{\,k} }  = n
$$
with $n$ being the number of elements of the permutation.
Our example will correspond to
$$
C = \left( {2,0,1,0,1,0,0,0,0,0} \right)
$$

So, as interpreted, the problem is asking the number of permutations
corresponding to a given cycle structure $C$.

The number of ways to compose a $k$-cycle from $m$ available elements
is the same as that of composing a $k$-subset from a $m$-set, multiplied
by the $(k-1)!$ number of ways to permute the elements beyond the first,
which is assumed to be the lowest. So
$$
N_{\,k} (m) = \left( {k - 1} \right)!\left( \matrix{
  m \cr
  m - k \cr}  \right) = \left[ {k \le m} \right]{{m!} \over {k\,\left( {m - k} \right)!}}
$$
where $[P]$ denotes the Iverson bracket.

To compose $C_k$ $k$-cycles from $m$ available elements, the number of ways
will be
$$
\eqalign{
  & N_{\,k} (m)N_{\,k} (m - k)\, \cdots \,N_{\,k} (m - k\,C_{\,k} )/C_{\,k} ! =   \cr
  &  = \left[ {k\,C_{\,k}  \le m} \right]{1 \over {k^{\,C_{\,k} } \,C_{\,k} !}}{{m!} \over {\,\left( {m - k\,C_{\,k} } \right)!}} \cr}
$$
where the division by $C_k!$ is because the cycles order is fixed.

Thus the number of permutations of $n$ elements with a given cycle structure
$C$ is:
$$
N(C) = \;{{n!} \over {\prod\limits_{1\, \le \,k\, \le \,n} {k^{\,C_{\,k} } C_{\,k} !} }} =
\;{{\left( {\sum\limits_k {k\,C_{\,k} } } \right)!} \over {\prod\limits_k {k^{\,C_{\,k} } C_{\,k} !} }}
$$
refer for instance to this paper.

Thus your calculation is correct.

not it is not correct.
for the first type: $\frac{10!}{(3! \cdot 5!\cdot 1^3\cdot 2^5}$,
for the second type: $\frac{10!}{(4!)^2\cdot 1^4\cdot 2^4}$

in general:

$$\frac{n!}{a_1!\cdot a_2!\cdots a_n!\cdot 1^{a_1}\cdot 2^{a_2}\cdots n^{a_n}}$$


Show that a number is not prime? Show that for any integer $n>1$, all the numbers $n!+2, n!+3, \ldots, n!+n$ are composite (i.e. not prime).

Hint:
Try to show, that if for two numbers $a$ and $b$, $a$ is divisible by $d$ and $b$ is divisible by $d$, then so is their sum. Then go looking for such a common divisor in your sums.

Take an arbitrary number $n! + k$, where $n > 1, 1 < k \leq n$.

The following is the definition of the factorial function:

$$n! = \prod_{i=1}^n i$$

Hence every i where $1 \leq i \leq n$ is a factor of $n!$. Since the range k covers is within the range of i, this is clearly true for all k.

We know that k is a factor of $n!$, so let us say $n! = qk, k \in \mathbb{Z}$. Hence $n! + k = qk + k = (q + 1)k$, which implies q + 1 and k are integral factors of $n! + k$. QED

This is a reasonably rigorous proof. When you become more familiar with the concepts of divisibility and factorials the result should be fairly apparent from first sight.


Angle between tangents and angle subtended by radii are supplementary Using the result that the length of the tangents draw from an external point to a circle are equal, prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line segment joining the points of contact at the center.

I can easily prove this by the following:

$\angle OQP = \angle ORP =  90^{\circ}$
So in quadrilateral $OQPR$ $ \angle QPR + \angle QOR = 180^{\circ}$ and therefore these angles are supplementary.

But I have not used what is said in the question. How should I do it?

Hint: Try to use congruence. If you can prove that the two triangles are congruent then it is a kite. Now, you can prove it easily.

To prove:  $BPA + BOA = 180^\circ$

Solution: In quadrilateral $AOBP$, $PBO = 90^\circ$ (as the angle between the radius and the tangent to the circle is $90^\circ$). Similarly, $PAO = 90^\circ$. $PAO + PBO + BOA + BPA = 360^\circ$ (the sum of the angles of a quadrilateral), therefore $90^\circ + 90^\circ + BOA + BPA = 360^\circ$, so (since we have proved that $PAO = 90^\circ$ and $PBO = 90^\circ$) $BOA + BPA = 360^\circ - 180^\circ = 180^\circ$.


Where does the Normal Distribution of Error terms come from, and why? In a Simple or Multiple Linear Regression Model, we are given the assumption that the error terms, $\epsilon$, are distributed like so:
$ε ∼ N(0, I_nσ^2).$
Can anybody explain where the mathematical logic comes from to derive this assumption, or what the intuition is behind it?

It comes mainly from the requirements of the classical statistical theory. For example, in econometrics you usually assume only uncorrelated errors with constant variance and zero expectation. These assumptions are suffice to derive the best linear unbiased estimator (Gauss-Markov theorem). However, this is not enough for the likelihood-based statistical inference that relies on a specified probability distribution. The most straightforward distribution that satisfies the aforementioned conditions is multivariate normal. In such a case, the uncorrelated error terms become independent and thus the estimation and inference becomes very simple. Before the computer era, mathematical simplicity was crucial, since non-analytical derivations were very hard and even impossible, hence closed-form solutions were much desirable.
In practice, normality assumed merely as approximation, if assumed at all, and much of the inference relies on large-sample theory, i.e., the asymptotic distributions of the estimators. However, note that at least one assumption is really essential for the very analysis. The finite variance assumption is crucial, otherwise your estimators (either OLS or ML) will not converge to finite values. The zero expectation assumption should not bother you, since it is mere technical. If you have non-zero constant expectation you can add it to the intercept term, if it depends on the explanatory variable, you should change your model (conditional expectation of $Y$ given $X$). In both cases, you can obtain the zero expectation after a proper modification. The uncorrelated errors assumption can be dropped by using GLS or GLM.
To sum it up, normal distribution is straightforward model to work with that is usually plausible to assume where linear regression models are applicable. However, you should always bear in mind that as more assumptions you have, the more mistakes you prone to do. Thus, you have various tests of model assumptions and model-misspecifcation, that checks the validity of your assumptions and quantifies the cost of error.

Some of the reasons for this choice, I would say, include

Intuitively reasonable: the Normal distribution is very nice, in that it is mean zero, symmetric (i.e. there is no bias direction in the error), has a memorable bell-shape, etc. All these properties make it a very "plausible" assumption for how errors would be distributed.

Mathematically convenient: this is essentially the backbone for a lot of statistical inference, including hypothesis tests, models, etc. For example, this allows you to deduce the distributions of different test statistics, sums, etc.

Of course, it is not the only possibility; after all, it is only an assumption. For example, generalisations such as GLMs allow for other error distributions of the response variable.


What is the geometric meaning of singular matrix Could anyone help explain what is the geometric meaning of singular matrix ? What's the difference between singular and non-singular matrix ? I know the definition, but couldn't understand it very well.

Thanks.

If you are in $\Bbb{R}^3$, say you have a matrix like

$$\left[\begin{array}{ccc} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\a_{31} & a_{32} & a_{33} \end{array}\right].$$

Now you can think of the columns of this matrix to be the "vectors" corresponding to the sides of a parallelepiped. If this matrix is singular i.e. has determinant zero, then this corresponds to the parallelepiped being completely squashed, a line or just a point.

In my opinion, zero determinant of a matrix with nonzero entries shows the balance like the resultant of two equal forces in opposite direction is zero or equilibrium state.


What operations can be performed on three variables $x$, $y$, and $z$ and produce one unique sum? I'm having a programming problem, but I feel this would be more relevant here. I have three unique variables, $x$, $y$, and $z$. What operations can I perform on these variables to produce a unique value? Are there any approaches I could take?

Thanks!

EDIT: To be more specific, I want some function $f(x, y, z)$ that produces a unique output for each set of inputs (assuming uniqueness in inputs, $1$, $2$, $3$ does not need to output something different than $3$, $2$, $1$). I realize that a generalized case of this is probably not possible, but I would definitely be able to co-opt similar solutions

$f(x,y,z) = x\sqrt{2} + y\sqrt{5} + z\sqrt{7}$

This function would increase by a similar order of magnitude to its inputs. Moreover it's possible to prove that for $x,y,z \in \mathbb{Z}, f(x,y,z)$ must be unique $\forall x,y,z$.

Edit: Proof:

Suppose for contradiction that $\exists x_{1},y_{1},z_{1},x_{2},y_{2},z_{2} \in \mathbb{Z}$:

$f(x_{1},y_{1},z_{1}) = f(x_{2},y_{2},z_{2})$ with at least one of the following being true: $x_{1} \neq x_{2}, y_{1} \neq y_{2} , z_{1} \neq z_{2}$

Then we have $x_{1}\sqrt{2} + y_{1}\sqrt{5} + z_{1}\sqrt{7} = x_{2}\sqrt{2} + y_{2}\sqrt{5} + z_{2}\sqrt{7}$

$\Rightarrow (x_{1} - x_{2})\sqrt{2} + (y_{1} - y_{2})\sqrt{5} + (z_{1} - z_{2})\sqrt{7} = 0$

In the case that $x_{1} = x_{2}$ and $y_{1} \neq y_{2}$ and $z_{1} \neq z_{2}$:

$\frac{\sqrt{5}}{\sqrt{7}} = \frac{z_{2} - z_{1}}{y_{1}-y_{2}}$

This would imply that $\frac{\sqrt{5}}{\sqrt{7}} \in \mathbb{Q}$ which leads to a contradiction (distinct irrational divided by distinct irrational must also be irrational - simple to prove)

A similar proof applies to the case where $y_{1} = y_{2}, x_{1} \neq x_{2}$ and  $z_{1} \neq z_{2}$ and the case when $z_{1} = z_{2}, x_{1} \neq x_{2}$ and $y_{1} \neq y_{2}$

If we have $x_{1} = x_{2}$ and $y_{1} = y_{2}$, then we get $z_{1} = z_{2}$ which is a contradiction. Therefore we have proved the following conditions:

$x_{1} \neq x_{2}, y_{1} \neq y_{2} , z_{1} \neq z_{2}$

let $a = x_{1} - x_{2}, b = -(y_{1}-y_{2}), c = -(z_{1} - z_{2})$

Then we have $a\sqrt{2} = b\sqrt{5} + c\sqrt{7}$

$\Rightarrow \sqrt{2} = \frac{b\sqrt{5} + c\sqrt{7}}{a}$

let $b_{1} = \frac{b}{a}, c_{1} = \frac{c}{a}$

$\Rightarrow b_{1}, c_{1} \in \mathbb{R}$

$2 = (b_{1}\sqrt{5} + c_{1}\sqrt{7})^{2}$

$\sqrt{35} = 2 - 5b_{1}^{2} - 7c_{1}^{2}$

$\Rightarrow \sqrt{35} \in \mathbb{Q}$ which is a contradiction.

This implies that $\forall x,y,z \in \mathbb{Z}, f(x,y,z)$ is unique.

You can use the concept of binary expression.for example, (1,2,3) goes to the binary expression of 123.


What is the domain of this function in 3 variables? Consider the function
$$f(x, y, z)=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+xyz.$$
Which is the domain of this function?
Certainly, we have to ask $(x, y, z)\neq(0, 0, 0)$ but also the points of the type
$$ (0, y, z), \ (x, 0, z), \ (x, y, 0), \ (0, 0, z), \ (x, 0, 0), \ (0, y,0)$$
are not allowed. How to write in an "easy" way?
I would say
$$D(f)=\{(x, y, z)\in\mathbb{R}^3 : (x, y, z)\neq (0, 0, 0), ..\}$$
but I don't know how to continue. Could anyone please help me?
Finally, all these points to be excluded, consitute a straight line, a plane or something else?
Thank you in advance!

$\newcommand{\R}{\mathbb{R}}$
You've already found an answer by yourself. You can just set the domain of your function $f:X \rightarrow Y$ to
$$
\begin{align}
X
&= \{(x,y,z) \in \R^3 ~ | ~ x \neq 0 \land y \neq 0 \land z \neq 0 \}\\
&= \R^3 \setminus \{(x,y,z) \in \R^3 ~ | ~ x = 0 \lor y = 0 \lor z = 0 \}\\
&= \R^3 \setminus \left[\left\{(x,y,z) \in \R^3 ~ | ~ x = 0\} \cup \{(x,y,z) \in \R^3 ~ | ~ y = 0\} \cup \{(x,y,z) \in \R^3 ~ | ~ z = 0\right\}\right]
\end{align}
$$
It constitutes an entire three-dimensional Euclidean space with the $x-y$, $y-z$, and $z-x$ planes being removed.
Meanwhile, the codomain should be $Y = \R$.

For your first question consider labeling the domain:
$D(f)=\{(x, y, z)\in\mathbb{R}^3 : x,y,z \neq 0\}$.
The following is a response to your second question from the bottom of your post:
As $\mathbb{R^{3}}$ represents all of $3$-dimensional space, $D(f)$ is $\mathbb{R^{3}}$ without the points previously mentioned. The origin $(0,0,0)$ is removed, WLOG points of the form $(0,0,z)$ lie on one of the $3$ axes, so each axis is removed. WLOG points of the form $(0,y,z)$ represent an arbitrary point on the $xy$ plane. Altogether, we end up with $\mathbb{R^{3}}$ without the $xy$,$xz$,and $yz$ planes.


Proof that ratio of vertices to edges in an infinite (square) grid is 1:2? I'm making the statement that the ratio of vertices to edges in an infinite square grid is $1:2$. I need this fact for deducing further theorems specific for my problem, however I can't find any theory on infinite grids.

I would prefer to cite some literature on that (maybe also for hexagonal and triangular grids).

If no formal literature exists that covers that issue I would want to write the proof. My thinking so far is each vertex has $4$ edges connected to it and each edge $2$ vertexes, which makes for a $2$ to $4$ ratio $-> 1:2$. However I feel this is not formal enough and the proof should involve infinity of the grid.

Please post possible literature or a formal proof if you think mine isn't sufficient.

If we're being "careful with $\infty$", then let's consider an $n \times n$ square grid (with $n^2$ vertices). In this grid, there are $(n-2)^2$ interior vertices, which have $4$ edges out of them; $4(n-2)$ edge vertices, which have $3$ edges out of them; $4$ corner vertices, which have $2$ edges out of them.

By the degree sum formula, adding these numbers gives twice the number of edges (because each edge is counted twice, once by each of its endpoints). If we call the number of edges $m$, then
$$
   2m = 4(n-2)^2 + 3(4n-8) + 2(4) = 4n^2-4n
$$
so there are $2n^2-2n$ edges. The ratio $\frac mn$ of edges over vertices is $\frac{2n^2-2n}{n^2} = 2 - \frac 2n$, which approaches $2$ as $n \to \infty$.

We can be fully rigorous and still do less work with asymptotic analysis. If we were to claim that each vertex has degree $4$, and therefore that $2m = 4n^2$, we would be wrong only for edges around the border. But there are $O(n)$ edges along each border, whose degree differs from $4$ by a constant, so we have $2m = 4n^2 + O(n)$, and therefore $\frac mn = \frac{2n^2 + O(n)}{n^2} = 2 + O(\frac1n)$. Once again, as $n \to \infty$, this approaches $0$.

Also, we can consider a finite toroidal square grid, which "wraps around" at the borders, Pac-Man style. Then every single vertex has degree $4$, and the ratio of vertices to edges is $1:2$ with no approximation.

Consider that any vertex has four edges connected to it. Pick two edges that are adjacent, that is, are adjacent sides of a square. The vertex and the two edges can be translated to every other vertex and this partitions all vertices and edges. Thus the ratio of vertices to edges is $1:2$. What is important is the local connection between vertices and edges despite the fact that there are many different ways of pairing edges and vertices. There is only one way that is translation invariant.

A similar proof applies to an infinite one dimensional grid where the ratio is $1:1$.

A similar proof applies to an infinite (triangular) grid where the ratio is $1:3$.

In general, if you have an infinite homogeneous grid with the degree of each vertex being a constant $D$, then since each edge connects exactly two vertices, the ratio of vertices to edges is $2:D$. This applies to infinite homogeneous trees as a special case.

So in the case of an infinite hexagonal grid the ratio is $2:3$.

The same proofs apply to a finite cycle, infinite cylinder, finite torus by formally identifying edges of infinite grids. It is clear that for a finite cycle there is a $1:1$ ratio of vertices to edges, and this also applies in the limit of an infinite cycle. The infinite total number of vertices or edges makes no difference. It is the local connections between them that counts.

If you want to be really careful, consider the lattice of points in the plane with integer coordinates as vertices with horizontal edges $\{(n,y)\}$ where $m\le y<m+1$, and vertical edges $\{(x,m)\}$ where $n\le x<n+1$. Each period square is $\{(x,y)\}$ where $n\le x<n+1, m\le y<m+1$ and each contains one lattice point $(n,m)$, one horizontal edge, and one vertical edge. The ratio is $1:2$ in any period square. Similar proofs apply to other cases.

Read the Wikipedia article Fundamental pair of periods for the related idea of fundamental period parallelogram which is very important in elliptic function theory.


Homotopy composition Hatcher exercise Show that composition of paths satisfies the following cancellation property: if $f_0 \cdot g_0 \simeq f_1 \cdot g_1 $ and $g_0 \simeq g_1$, then $f_0 \simeq f_1$.

So I have two homotopies.

So say $g_0,g_1: X \rightarrow Y$ and $f_0,f_1:Y \rightarrow Z$.

Then we know that $G:X \times I \rightarrow Z$ s.t. $G(x,0)=f_0 \cdot g_0(x)$ and $G(x,1)=f_1 \cdot g_1 (x)$. Also, $H:X \times I \rightarrow Y$ s.t. $H(x,0)=g_0(x)$ and $H(x,1)=g_1(x)$.

I was wondering how do you construct the homotopy for f? or is there a simplier way.

I would think you construct a homotopy, but can't see how to.

It helps if you first show that inverses of homotopic paths are homotopic, which is relatively easy to do from the definitions.  That is, show that if $f_1 \simeq f_2$, then $\bar{f_1} \simeq \bar{f_2}$.  Once you have that, start with $f_0 \simeq f_0 \cdot (g_0 \cdot \bar{g_0}) \simeq (f_0 \cdot g_0) \cdot \bar{g_0}$, and notice that there's a nice copy of $f_0 \cdot g_0$ in there.

Since $f \cdot g(s) = f(2s){\kern 1pt} {\kern 1pt} {\kern 1pt} for{\kern 1pt} 0 \le s \le \frac{1}{2}{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} ;g(2s - 1){\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} for{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} \frac{1}{2} \le s \le 1
$ and $G$ is a homotopy between $f_0 \cdot g_0
$and $f_1 \cdot g_1$.we could find a homotopy $F$ between $f_0$ and $f_1$ by $F(s)=G(1/2 s)$


determinant of a a linear operator I understand that matrix of $T$ will be $6\times 6$ and will consists of $1,0$ but not able to find out what will be the determimant.

Here the space of matrices is viewed as a 6-dimensional space.  The 2 by 3 matrix structure is irrelevant.  Therefore it is preferable to view this map as $T:\mathbb{R}^6\to \mathbb{R}^6$ given by $(x_1,x_2,x_3,x_4,x_5,x_6)\mapsto(x_6,x_4,x_1,x_3,x_5,x_2)$.  Then one immediately recongnizes a permutation matrix.  Hence its determinant is either $1$ or $-1$.  Now all you have to do is determine whether the permutation is even or odd, for example by decomposing it into cycles.

I just noticed the minus sign on $x_6$, so this is not exactly a permutation matrix.  That changes the sign of the determinant to its opposite.

I think that for the transformation to be defined, $T$ must be a $2x2$ matrix.After which you will have a system of equations. Then you may solve for the entries of $T$.


Let $X_1,X_2,...$ be i.i.d. RVs and let N be another independent RV which takes values from nonnegative integers. Let $Y=\sum_{k=1}^N X_k$ Using the fact that $M_Y(t)=M_N(logM_{X_1}(t))$, derive $$E[Y]=E[N]E[X_1]$$

I don't quite see how to use the hint to prove this. Any ideas?

There is a simpler way to do this BTW, just for the record.

$$E(Y) = E(E(Y|N))$$

$$E(Y|N) = E(\sum_{k=1}^{N} X_{k})$$

$$=\sum_{k=1}^{N} E(X_{k})$$
$$=N\times E(X_1)$$

Therefore,

$$E(E(Y|N)) = E(X_1)\times E(N)$$


$\int_{-\infty}^{+\infty} e^{-x^2} dx$ with complex analysis Inspired by this recently closed question, I'm curious whether there's a way to do the Gaussian integral using techniques in complex analysis such as contour integrals.

I am aware of the calculation using polar coordinates and have seen other derivations. But I don't think I've ever seen it done with methods from complex analysis. I am ignorant enough about complex analysis to believe it can somehow be done without knowing how it would be done.

What follows is a list of solutions that I enjoy, and use complex analysis either implicitly or explicitly.  I will update the list as I come up with more. (Note:  Solution 4 is my favorite, and is completely complex analysis oriented.  I also quite like Solution 6.)

First, let $u=x^{2}$, $du=2xdx$. Then our integral becomes $$\int_{-\infty}^\infty e^{-x^2}dx=\int_{0}^{\infty}u^{-\frac{1}{2}}e^{-u}du=\Gamma\left(\frac{1}{2}\right).$$  where $\Gamma(s)$ is the Gamma function.

Solution 1:
Since $$\Gamma(1-s)\Gamma(s)=\frac{\pi}{\sin\pi s}$$ for all complex $s$, we conclude $$\Gamma\left(\frac{1}{2}\right)=\sqrt{\pi}.$$

Solution 2: Recall  the Beta function, $$\text{B}(x,y)=\int_{0}^{1}t^{x-1}(1-t)^{y-1}dt=\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}.$$ Setting $x=y=\frac{1}{2}$ we have

$$ \left(\Gamma\left(\frac{1}{2}\right)\right)^{2}=\int_{0}^{1}\frac{1}{\sqrt{t(1-t)}}dt.$$

To evaluate this, set $t=\sin^{2}(x)$ to find $$\left(\Gamma\left(\frac{1}{2}\right)\right)^{2}=\int_{0}^{\frac{\pi}{2}}\frac{2\sin x\cos x}{\sin x\cos x}dt=\pi.$$
Alternatively, we could evaluate the last integral by choosing branch's such that the integrand is analytic on $\mathbb{C}-[0,1]$ and then integrating around this cut.  (The residue then comes from the residue at infinity)

Solution 3: Setting $s=\frac{1}{2}$ in the duplication formula, $$\Gamma(s)\Gamma\left(s+\frac{1}{2}\right)=\sqrt{\pi}2^{1-2s}\Gamma(2s),$$ yields $$\Gamma\left(\frac{1}{2}\right)=\sqrt{\pi}.$$

Solution 4:  My personal favorite:
Recall the functional equation for the zeta function, namely that $$\pi^{-\frac{z}{2}}\Gamma\left(\frac{z}{2}\right)\zeta(z)=\pi^{-\frac{1-z}{2}}\Gamma\left(\frac{1-z}{2}\right)\zeta(1-z).$$ Taking the limit as $z\rightarrow1$, we know that $\zeta(z)\sim\frac{1}{z-1}$ and $\Gamma\left(\frac{1-z}{2}\right)\sim2\frac{1}{\left(z-1\right)}$ so that we must have the equality $$\pi^{-\frac{1}{2}}\Gamma\left(\frac{1}{2}\right)=2\zeta(0).$$ By taking the limit in the right half plane as $s\rightarrow0$ using the identity $$\zeta(s)=\frac{s}{s-1}-s\int_{1}^{\infty}\{u\}u^{-s}du,$$ which holds for $\sigma>0$, we can find that $\zeta(0)=\frac{1}{2}.$ (notice the pole/zero cancellation). Consequently $$\Gamma\left(\frac{1}{2}\right)=\sqrt{\pi}.$$

Solution 5: From complex integration, for $a,b>0$ we have the identity $$\int_{-\infty}^{\infty}(1-ix)^{-a}(1+ix)^{-b}dx=\frac{2^{2-a-b}\pi\Gamma(a+b-1)}{\Gamma(a)\Gamma(b)}.$$ Set $a=\frac{1}{2},b=\frac{3}{2}$ to find that
$$
\int_{-\infty}^{\infty}\frac{1-ix}{\left(1+x^{2}\right)^{\frac{3}{2}}}dx=\frac{2\pi}{\Gamma\left(\frac{1}{2}\right)^{2}}.$$ Hence $$\int_{0}^{\infty}\frac{1}{\left(1+x^{2}\right)^{\frac{3}{2}}}dx=\frac{\pi}{\Gamma\left(\frac{1}{2}\right)^{2}}.$$ Since the integrand on left hand side has anti derivative $\frac{x}{\sqrt{x^{2}+1}}+C$, it follows that the integral is $1$ and hence$$\Gamma\left(\frac{1}{2}\right)=\sqrt{\pi}.$$

Solution 6: More with the Beta function.  Consider the Mellin Transform $$\mathcal{M}\left(\frac{1}{\left(1+t\right)^{a}}\right)(z):=\int_{0}^{\infty}\frac{t^{z-1}}{(1+t)^{a}}dt=\text{B}(a-z,z).$$ The last equality follows by substituting $v=\frac{1}{1+t}$, and then rewriting the integral as $\int_{0}^{1}v^{a-z-1}(1-v)^{z-1}dv.$ Now, plug in $a=1$ and $z=\frac{1}{2}$ to get $$\int_{0}^{\infty}\frac{1}{\sqrt{t}(1+t)}dt=\Gamma\left(\frac{1}{2}\right)^{2}$$ and then let $t=x^{2}$ to find
$$
2\int_{0}^{\infty}\frac{1}{1+x^{2}}dx=\pi=\Gamma\left(\frac{1}{2}\right)^{2}.
$$

Solution 7:  We can also prove the result by using Stirling's formula. Admittedly, this isn't really using complex analysis, but I find it interesting.

Since $z\Gamma(z)=\Gamma(z+1)$ we see that
$$
\Gamma\left(n+\frac{1}{2}\right)=\Gamma\left(\frac{1}{2}\right)\cdot\left(\frac{1}{2}\right)\left(\frac{3}{2}\right)\cdots\left(\frac{2n-1}{2}\right)=\Gamma\left(\frac{1}{2}\right)\left(\frac{(2n)!}{n!4^{n}}\right)=\Gamma\left(\frac{1}{2}\right)\binom{2n}{n}\frac{n!}{4^{n}}.
$$
By Stirling's formula,
$$
\binom{2n}{n}\frac{1}{4^{n}}\sim\frac{1}{\sqrt{\pi n}}\ \text{as}\ n\rightarrow\infty
$$
and
$$
\frac{\Gamma\left(n+\frac{1}{2}\right)}{n!}\sim\frac{1}{\sqrt{ne}}\frac{\left(n+\frac{1}{2}\right)^{n}}{n^{n}}.
$$
Using the fact that $\lim_{n\rightarrow\infty}\left(1+\frac{a}{n}\right)^{n}=e^{a}$, it then follows that
$$
\frac{\Gamma\left(n+\frac{1}{2}\right)}{n!}\sim\frac{1}{\sqrt{n}}.
$$
Consequently, taking the limit as $n\rightarrow\infty$ in the formula $$\Gamma\left(\frac{1}{2}\right)=\frac{\Gamma\left(n+\frac{1}{2}\right)}{n!}\frac{4^{n}}{\binom{2n}{n}}$$ yields $$\Gamma\left(\frac{1}{2}\right)=\sqrt{\pi}.$$

Hope that helps,

Remark: All of the formulas used here can be proven without use the fact that $\Gamma(1/2)=\sqrt{\pi}$, so that none of these are cyclic.  This is mainly worth pointing out for $4$.

Edit:  I put what were solutions 2 and 3 together since they were not different.

$\text{MR}1641980\, (99\text{h}:26002)$

Desbrow, Darrell.

On Evaluating $\displaystyle\int_{-\infty}^\infty e^{ax(x-2b)}\, dx$ by Contour Integration Round a Parallelogram.

Amer. Math. Monthly $105\, (1998)$, no. $8,\, 726–731$.


The definition of Arithmetic Function Consider following functions (written on Page 120 of the book "Summing It Up" by Avner Ash and Robert Gross, 2016):
$$
a(n)= \begin{cases}
      0 & \text{if $n$ is not prime and} \\
      1 &  \text{if $n$ is prime.}
\end{cases}
$$
... It turns out that $a(n)$ is not a very easy function to work with. Numbers like to be multiplied, so our next guess is to use
$$
b(n) = \begin{cases}
      0 & \text{ if n is not a power of a prime and} \\
      1 &  \text{if n is a power of a prime.}
      \end{cases}
$$
This is better but still not so great. The problem is that $b(n)$ does not depend on which prime $n$ is a power of, if indeed $n$ is a prime power. Finally, we try
$$
\Lambda(n)=\begin{cases}0,&\text{if $n$ is not a power of a prime and}\\ \log p,&\text{if $n = p^m$ for some prime $p$.}\end{cases}
$$
QUESTION:
What is the meaning of

$a(n)$ is not a very easy function to work with. Numbers like to be multiplied (how this affect $a(n)$)

The problem is that $b(n)$ does not depend on which prime $n$ is a power of, if indeed $n$ is a prime power (what  $\Lambda(n)$ does different besides the output).

There's a saying in mathematics: "A mathematical object is what it does." The authors of your textbook are trying to find a function that does something. What that "something" is is hard to say without reading the textbook and seeing what the function $\Lambda$ does in the proof or example that the textbook outlines. But to answer your questions directly:

$a(n)$ is a fine function to work with. Its values on the positive integers are $0,1,1,0,1,0,1,0,0,0,1,0,1,0,0,\dots$; finding the value of $a(n)$ for an arbitrary $n$ is a question of open research: see Primality Test - Wikipedia for more information. However, we usually don't just define functions for the fun of it. We define functions for a purpose, be that educational or to help in a proof. The quote "numbers like to be multiplied" probably means that the proof or example in question has to do with multiplying numbers. The function $a$ is not very useful in that regard; in general we don't have $a(nm)=a(n)a(m)$ or $a(nm)=a(n)+a(m)$, or any other identities like that. In fact, unless $n$ or $m$ is equal to $1$, we have that $a(nm)=0$, so $a$ isn't very useful when it comes to multiplying numbers. $b$ doesn't have any such property either, but it does have a somewhat weaker property, that $b(n^k)=(b(n))^k$ for any natural numbers $n,k$. This makes it a bit more useful when it comes to multiplication, but it's hard to know why that function was chosen without knowing more about the application. In general, when we talk about multiplying numbers, we like to choose a multiplicative function; the page I linked shows many common examples of such functions.
You ask what $\Lambda$ does differently besides the output. That's a bit of a complicated question because what is the difference between any two functions with the same domain, except for the outputs? In this case, in the proof or example that your textbook is going over, it's important that if a number is a prime power, we know which prime we are raising to a power. I can't tell you why that is important unless you tell me what your textbook is doing with $\Lambda$. Usually, in number theory, logarithms are usually useful for converting products into sums, using the property that $\log(mn)=\log m+\log n$. This benefit is a little diluted when we set $\Lambda(n)=0$ whenever $n$ is not a prime power, but it is still beneficial. Maybe you're testing whether two numbers $m,n$ are powers of the same prime, which is only true when $\Lambda(mn)\neq 0$. Apparently, the value of the prime is important here, which is why we use $\Lambda$ instead of $b$.

I hope this answers your question. In general, numbers don't like to be multiplied unless we care about multiplying them. And we don't care about which prime $n$ is a power of, unless there's a reason we should. $\Lambda$ is what it does, and clearly we care about these things for some yet-unknown purpose. I'm sorry I can't give you more information, but if you want to leave a comment about this I can update the answer :)


Prove $a^{n}*a^{m}=a^{n+m}$ $a^{n}*a^{m}=a^{n+m}$

Induccion over $a$

Being $a=0$ $$0^{n}*0^{m}=0^{n+m}$$
            $$0*0=0$$
it works!

Hip. $$a^{n}*a^{m}=a^{n+m}$$

Now let $S(n)$ be the succesor of $n$
We have to show the following: $$S(a)^{n}*S(a)^{m}=S(a)^{n+m}$$

I been trying to develop the left side of the equality, but seems a little obious to me the conection with the right side. Does any suggestion of whats could go in the middle?

Induct on $m$ instead. The case $m=0$ is trivial by $a^0=1$. If the case $m=k$ works, $a^na^{k+1}=a^naa^k=a^{n+1}a^k=a^{n+1+k}=a^{n+k+1}$.

Induction over $a$ is terrible, if doable at all.

Easy over $m$:

$$a^na^1=a^{n+1}$$ by definition of the integer powers, and

$$a^{n}a^{m+1}=a^n(a^ma)=(a^na^m)a=a^{n+m}a=a^{n+m+1}=a^{n+(m+1)}.$$


Isomorphism between quotient rings of $\mathbb{Z}[x,y]$ I need to find the condition on $m,n\in\mathbb{Z}^+$ under which the following ring isomorphism holds:
  $$
\mathbb{Z}[x,y]/(x^2-y^n)\cong\mathbb{Z}[x,y]/(x^2-y^m).
$$

My strategy is to first find a homomorphism
$$
h:\mathbb{Z}[x,y]\rightarrow\mathbb{Z}[x,y]/(x^2-y^m)
$$
and then calculate the kernel of $h$.

To achieve this, I furthermore try to identify the isomorphism between $\mathbb{Z}[x,y]$ and itself, which I guess is
$$
f:p(x,y)\mapsto p(ax+by,cx+dy)
$$

where $ad-bc=\pm 1,a,b,c,d\in\mathbb{Z}$.

Then $f$ induces a homomorphism $h$. But from here I failed to move on.

I believe there is some better idea, can anyone help?

Updated:

It should be isomorphism between quotient rings, not groups. Very sorry for such mistake.

Well, the additive group of $\mathbb{Z}[x,y]/(x^2 - y^n)$ is just a free abelian group on the generators $1,x, y, xy, xy^2, xy^3,\ldots$ - ie its the free abelian group on countably many generators, which is independent of $n$.

Hence, $m,n$ can be anything.

Basically, the $\mathbb{Z}$-module $\mathbb{Z}[x,y]$ is generated by all monomials with coefficient 1, ie, $1, x, y, x^2, y^2, xy, x^3, y^3, x^2y, xy^2,\ldots$. The relation $x^2 - y^n$ just allows you to replace any $x^k$ you see with $x^{k-2}y^n$ (for $k\ge 2$). However, in both cases you still end up having countably infinitely many generators (and hence countably infinite rank), and since a free abelian group is determined up to isomorphism by its rank, they're isomorphic.

If you're talking about ring isomorphisms, then Potato is right - if $n\ne m$, then the two rings are not isomorphic.

Let $R_n = \mathbb{Z}[x,y]/(x^2 - y^n)$.

As to why they're not isomorphic as rings, this seems to me to be a rather deep question, and I feel like the best explanation is through algebraic geometry. Essentially, the polynomial $x^2 - y^n$ defines a curve $C_n$ in the plane (namely the set of points (a,b) where $a^2 - b^n = 0$). These curves $C_n$ are birationally defined by their function fields, which in this case is just the quotient field of your ring $R_n$. If the rings $R_n,R_m$ are isomorphic, then their quotient fields must be isomorphic as well, and so the curves $C_n,C_m$ they define must be birationally equivalent. However, it can be computed via the Riemann-Hurwitz formula on the coordinate function $y$, viewed as a function from your curve to $\mathbb{P}^1$ that the curve associated to $R_n$ has geometric genus $(n-1)(n-2)/2$ (as long as $n\ge 1$, see exercise 2.7 in Silverman's book "The Arithmetic of Elliptic Curves"), which being a birational invariant, tells you that the function fields for your curves $C_n,C_m$ are not isomorphic for $n\ne m$, and hence $R_n, R_m$ could not be isomorphic either.

Finally it's easy to see that $R_0$ is not isomorphic to $R_n$ for any $n\ge 1$ since $R_0$ has nilpotent elements, and $R_n$ for $n\ge 1$ does not.

I can think of some other proof ideas, but essentially they all rely on some form of algebraic geometry. Many of these ideas I could phrase purely ring-theoretically, but it would seem complicated and completely unmotivated without explaining the connection to geometry.

We can write $\;x^2=y^k\;$ in the ring $\,R_k:=\Bbb Z[x,y]/(x^2-y^k)\;$ , so that any polynomial in $\;\Bbb Z[x,y]\;$ is mapped to a polynomial with $\,y-$ degree at most $\,k\,$ , for example in $\,R_3\,$ :

$$3xy^2 - xy^4-y^3+2x^3y\mapsto3xy^2-x(yx^2)-x^2+2x^3y=2x^3y-x^2+3xy^2$$

Now, if $\,n\neq m\,$ , say WLOG $\,n<m\,$ , we have

$$x^2-y^m\neq0\;\;\text{in}\;\;R_n$$

since otherwise we'd have a polynomial in $\;R_n\;$ with $\,y-$degree higher than $\,n\,$ ...


Intuitive view on $H_2(K)$ where $K$ is the M$\ddot{o}$bius strip Take a Torus $T^2$ and compute its homology group $H_2(T^2)$. It can be deduced that $H_2(T^2) \cong \mathbb{Z}$, and this is correspondent to the intuitive view that the only 2-simplex that is a cycle (boundaryless) and is not a boundary of 3-simplex is the surface itself.

Simiarly, consider a Möbius strip $K$. It can be computed that $H_2(K)\cong\{0\}$. However, in the intuitive view, the surface of $K$ itself is boundaryless, but not a boundary of a 3-simplex, and hence $H_2(K)\cong \mathbb{Z}$.

Is this correct? If it is wrong, which assumption is wrong or which assumption did I miss?

For shure it is something coming from orientability. Applying results of De Rham cohomology you can procide as follows. Retract by a deformation retraction the general Möbius strip to a Möbius strip with fibre $[0,1]$, in this case cohomology with compact support coincides with De Rham cohomology by compactness. Now an application of non-orientability implies that $H_{c}^n(K)$, the cohomology with compact support of $K$, has dimension $0$.  These cohomologies coincides with the singular one in these cases.

Perhaps these arguments are not so satisfying because you need to deal with manifolds and homology theories on manifolds to get a meaning of your computation. However, you can look at the first chapter of "Elements of Algebraic Topology" by Munkres in which you can find a computational approach with triangulations and then prove that the generators of the simplicial homology vanish. As above, in compact topological spaces, singular and simplicial homology are the same.


What happens to a random walk when we increase the probabilities of going right? Consider a random walk on the integers where the probability of transitioning from $n$ to $n+1$ is $p_n$ (and of course, the probability of transitioning from $n$ to  $n-1$ is $1-p_n$); we assume all $p_n$ are strictly less than $1$. Suppose we know that this random walk is pretty well concentrated; for example, let us assume that we know that
$$P(|X(t) - (1/3)t| \geq  c \sqrt{t}) \leq e^{-c^2}$$ where $X(t)$ is the state of the walk after $t$ steps.

Now suppose we increase every $p_n$ by $\epsilon$ (and correspondingly decrease the probability of transitioning from $n$ to $n-1$ by $\epsilon$), where $\epsilon$ is some number such that $p_n + \epsilon < 1$ for all $n$. Let $Y(t)$ be the state of the new chain after $t$ steps. My question is: does a similar concentration result hold for $Y(t)$?

It seems very intuitive that $Y(t)$ should concentrate around $(1/3)t + 2 \epsilon t$.

Write $Z(t) = Y(t) - X(t)$. Then
\[
Y(t)-(1/3 + 2\epsilon)t = \Big(X(t) - (1/3)t\Big) + \Big(E[Z(t)]-2\epsilon t\Big) + \Big(Z(t) - E[Z(t)]\Big).
\]
The concentration of the first term is given by your assumption.
For the third term, $Z(t)$ is the sum of $t$ iid bounded random variables,
so $P(|Z(t) -E[Z(t)] | \ge x \sqrt{t} ) \le 2 e^{-C x^2}$ for some constant $C>0$ by Hoeffding's inequality; this takes care of the third term above.

For the middle term, $E[Z(t)]-2\epsilon t=0$ by noticing that $Z(t)-2\epsilon t$ is a Martingale with zero mean at time 0.

By the union bound, with probability at least $1-2e^{-C x^2} - e^{-c^2}$ we get
\[
|Y(t)-(1/3 + 2\epsilon)t| \le (c+x)\sqrt t.
\]

NOTE: This is totally wrong - see comments.

Let $E(t)$ be the one-way random walk that adds one to its current state with probability $\epsilon$ and remains in its current state with probability $1-\epsilon$. Is it not true that $Y(t) = X(t) + E(t)$ exactly? - that is, the distributions of the random variables $Y(t)$ and $X(t) + E(t)$ are exactly the same?

Assuming I'm right that they are the same, then a concentration result for $Y$ follows from the hypothesized concentration result for $X$ and the standard concentration that can be calculated for $E(t)$: if $Y(t) - (\frac13+\epsilon)t$ exceeds $c\sqrt t$, then one of $X(t) - \frac13t$ or $E(t) - \epsilon t$ must exceed $\frac c2\sqrt t$.


Let $n\in \Bbb N$. Is there a prime $p>n$ for which $\gcd(n,p-1)=1$? For which $n\in \Bbb N$, is there a prime number $p>n$ for which $\gcd(n,p-1)=1$?

When $n$ is odd, Dirichlet's theorem implies there are infinitely many primes in the sequence
$$2,\quad 2+n,\quad 2+2n,\quad\ldots$$
and taking such a prime $p=2+kn$ (other than $2$), we have
$$\gcd(p-1,n)=\gcd(1+kn,n)=1$$
When $n$ is even, it is impossible to have $\gcd(p-1,n)=1$ for a prime $p>n$ because any prime greater than $n$ is odd, so $\gcd(p-1,n)\geq 2$.

Yes, because if n is even then it is of the form 2c where c is some integer, and it is a classical result of number theory that any integer can be factored into a unique product of a finite number of primes. Thus to find a number m such that (n,m)=2  you find a number of the form 2p where p is some prime that does not divide c, which is always possible because of an infinite number of primes. Now you need merely to prove that there are an infinite number of primes a such that a=(b-1)/2, where b is also prime, which is relatively easy.