问:
3- regular connected graph $G$ has independent subgraph $H$ such that $G-H$ is still connected.

There are $1998$ cities in Dussia, each being connected (in both directions) by flights to three other cities. Any city can be reached by any other city by a sequence of flights. The DSS (Dussia secret service) plans to close off $200$ cities, no two joined by a single flight. Show that this can be done so that any open city can be reached from any other open city by a sequence of flights only passing through open cities.

I don't know how to find such $H$, probably it should be constructed with some algorithm or perhaps take some extrem versions of $H$ (say with minimal or maximal number of neighbour set $N(H)$). Any hint what to do?

While writing this problem, this problem pop out as similar question: Every 3-connected graph has a cycle C such that G-V(C) is connected

Ohh, I can't read, it is 3-connected in that link not 3 regular.

答:
A solution to this problem follows from establishing the following result--which is actually a decent amount stronger, we don't have to settle for merely 200 vertices, we can remove 250 vertices:

Theorem A: Let $G$ be a 3-regular graph on $n$ vertices. Then there is an independent set $S$ of vertices in $G$ satisfying $|S|= \frac{n}{8}$ so that $G \setminus S$ is connected.

Idea of proof: The idea that gives a bound of a bit better than
$\frac{n}{10}$ is that you end up with a connected graph $G'$ where the   cycles are all vertex-disjoint and  so the upper bound on the number of edges is $\frac{4n(G')}{3}$ where $n(G')$ is the number of vertices  in $G'$, and this upper-bound is achieved when all the    cycles have length 3. [The smaller you can bound the average degree on $G'$,
the larger you can guarantee the size of the set of vertices that you can remove.] In   fact, we establish that the connected graph $G'$ that we achieve the cycles remain vertex disjoint, and $G'$ is guaranteed to have either many cycles of length 4 or larger, or many vertices not in any cycle.

The proof I gave last would establish a bound of $\frac{n}{10}$
but not $\frac{n}{8}$. The problem was as this: If you contract all degree-2 vertices
as in Lemma 3 to an edge then what may result is a multigraph with parallel edges
between two vertices and so you cannot assume that the length of the smallest cycle
is 3 anymore.

We first make the following observation:

Claim 1: If $v$ is a cut-vertex in a graph of maximum degree-3, then at
least one edge in that graph incident to $v$ is a cut-edge.

Proof of Claim 1: Removing $v$ makes at least 2 connected components. And so as
$v$ has degree-3, there is at least one such component where exactly
one edge $e$ incident to $v$ has its endpoint in that component. Then
$e$ is a cut edge. $\surd$

We now establish this theorem. Let us suppose that we have already removed
an independent set from $G$ so that the resulting graph is still
connected. Now let $v$ be a vertex in the remaining graph where $v$
still has degree-3 in the remaining graph and $v$ is not a
cut-vertex in the remaining graph. Then if we remove $v$,
then the resulting set of vertices that we have removed from $G$ remains
independent, and the resulting graph is still connected. Thus we do
the following:

(I) If there is an available vertex $v$ that
is in a triangle [on the graph that is left], pick a vertex $v'$ that
is in a triangle to remove. Otherwise
pick another available vertex to remove. STOP when there are no more
available vertices to remove. Write as $G'_{int}$ the resulting graph.
Then the set of vertices in $G \setminus G'_{int}$ is an independent set
of vertices in $G$. Furthermore, $G$ and $G'_{int}$ together have
the property:

(A) Let $C$ be a triangle in $G'_{int}$ and let $w'$ be a vertex
not in $G \setminus G'_{int}$ such that $w'$ is adjacent in $G$ to a
vertex in $C$. Then $w'$ is in a triangle in the original graph $G$.

Indeed, for $w'$ to have been taken out in the first place,
by Claim 1 above $w'$ is not incident to any cut edges at the time. So let
$v$ be the vertex in $C$ adjacent to $w'$ in $G$. Then $vw'$ was
not a cut-edge at the time. As $v$ is in a cycle $C$, neither
are the other 2 edges incident to $v$. So $v$ was available.
As $v$ is in a triangle, by the rule (I) $w'$ had to have been in a
triangle as well.

Then, from $G'_{int}$, do the following if and so long as
there is a triangle $C$ in $G'_{int}$ that
satisfies the following: There is a vertex $w'$ in $G \setminus G'_{int}$ such
that both:
(a) The vertex $w'$ is adjacent in $G$ to a vertex $v$ in $C$, and $w'$
is in a triangle in $G$.
(b) A triangle $w'x'y'$ that $w'$ is in, in $G$, satisfies
the additional property that $x'y'$ is in a cycle in $G'_{int}$.

Then do the following:

Add back in $w'$ and so the edges $vw'$, $w'x',w'y'$.

Then check that both $v$ and $x'$ together are available, are not
adjacent to each other, and are each in triangles, so remove
both $x'$ and $v$.

Then resetting the resulting graph to $G'_{int}$, note that $G'_{int}$
remains connected, and has one fewer triangle than
before.

Then $G'_{int}$ satisfies the invariant (A) after each of the above
iterations as well. Indeed, the set of vertices in a triangle after each
iteration is a subset of the set of vertices that were in a triangle
before. And every vertex $v$ taken out of $G'_{int}$
was in a triangle in $G'_{int}$ before so $v$ is also in a triangle in
$G$.

When there are no more vertices satisfying (a) and (b) above,
the resulting graph $G'$ also satisfies (B):

(B) Let $C$ be a triangle in $G'$ and let $w'$ be a vertex
not in $G \setminus G'$ such that $w'$ is adjacent in $G$ to a
vertex in $C$. Then $w'$ is in a triangle $w'x'y'$
in the original graph $G$, where the additional property is satisfied
that both $x',y'$ are not in any cycles in $G'$.

Claim 2: The cycles in $G'$ are vertex-disjoint.

Proof of Claim 2: In $G'$ there are no more available vertices. So all vertices of
degree-3 are cut-vertices. Thus by Claim 1,
every vertex $v$ of degree-3 is incident to a cut edge. So let $M$ be
the set of cut edges. Each such $e \in M$ is not in a cycle, and as every
vertex of degree-3 in $G'$ is incident to a cut edge, it follows that
$G \setminus M$ has maximum degree-2 and is a collection of
vertex-disjoint cycles. As every edge in $M$ is a cut-edge and is in no cycle in $G'$, it follows
that the set of cycles in $G' \setminus M$ is the same as the set
of cycles in $G' \setminus M$, and these are vertex-disjoint. So the
claim follows. $\surd$

Now let $\cal{C}$ be the collection of cycles in $G'$ and for each integer $k$
let us write as $\cal{C}_k$ the set of cycles of length $k$.
We now make some observations:

As the cycles are vertex-disjoint,
let us collapse each cycle $C \in \cal{C}$ in $G'$ to a vertex $v_C$;
the resulting graph $G"_1$ is a tree. Furthermore, every vertex in $G''_1$
is either in $\{v_C; C \in \cal{C}\}$ or in $A$ where
$A$ $\doteq \{v \in G'$; $v$ is not in any cycle in $G'\}$.
Now let $W'$ be the set of vertices $W'\doteq \{w';$
$w' \in G\setminus G';$ $w'$ is adjacent in $G$ to a vertex in
a triangle $C \in \cal{C}_3\}$. By the fact that $G''_1$ is a
tree gives the following:

Claim 3:
$$|{\cal{C}}_4|+|A| \ge 1+\left[\sum_{C \in \cal{C}_3}d_{G''_1}(v_C)\right]
-2|\cal{C}_3|.$$

Furthermore, the following is also true:

Claim 4:
$$W' \ge \sum_{C \in \cal{C}_3} [3-d_{G''_1}(v_C)].$$

Proof: As Invariant (A) is maintained,
for each cycle $C \in \cal{C}_3$ there are
$3-d_{G''_1}(v_C)$ vertices $w'$ in $G \setminus G'$
such that $w'$ is in a triangle in $G$, and $w'$ is adjacent in
$G$ to a vertex in $C$. So $w'$ is in $W'$.

Thus, to establish this, it suffices to show that each $w' \in W$
is adjacent in $G$ to exactly one vertex $v$ that is any
$C \in \cal{C}_3$. Let $w'$ be a vertex in $W'$, and let
$w'x'y'$ be a triangle
that $w'$ is in, in the original graph $G$, and let $v$ be the
vertices adjacent in $G$ to $w'$ so that $v$ is in a triangle in $G'$.
Then $v,x',y'$ are all of $w$'s neighbors in $G$ because $w'$ has
degree-3, and furthermore, as (B) is satisfied, $x'$ and $y'$ cannot
be in any triangles in $G'$.  So the only vertex that $w'$ can be
adjacent to that is in a triangle in $G'$, is $v$. Thus indeed,
each $w' \in W'$ is adjacent in $G$ to exactly one vertex $v$ that is any
$C \in \cal{C}_3$, and so $|W'|$ must be at least
$\sum_{C \in \cal{C}_3} [3-d_{G''_1}(v)]$.

So putting Claims 3 and Claims 4 together yields:

Claim 5: $$|{\cal{C}}_4|+|A|+|W'| \ge |\cal{C}_3|+1$$

We now partition the set $A$ of vertices not in a cycle
into 2 sets $A_1$ and $A_{2}$.

The set $A_1$ is defined to be the set of vertices
$v \in A$ such that either $v$ has degree-3 in $G''$,
or $v$ has degree 2 or 1 but $v$ has no neighbor
in $G'' \cap A$ that has degree 2 or 1.

The set $A_{2}$ is the set of vertices $v \in A \setminus A_1$;
equivalently, the set of vertices $v \in A$ such that both $v$ has
degree-2 or 1, and $v$ is adjacent in $G'$ to another vertex in $A$
that has degree-2 or 1.

Claim 6: Then the inequality
$$|{\cal{C}}_4|+|A|+|W'| \ge |\cal{C}_3|+1$$
from Claim 5 can be strengthened to
$$|{\cal{C}}_4|+|A_1|+\frac{|A_2|}{2}+|W'| \ge |\cal{C}_3|+1.$$

Proof of Claim 6: From $G''_1$ contract each path of 2 or more vertices
in $A$ of degree 2 or less to a single vertex and let
$G''_2$. Then let $A'$ be the set of vertices in $G''_2$
that correspond to vertices in $A$ in $G''_1$. Then
$|A_1| + \frac{|A_2|}{2} \ge |A'|$ on the one hand, and on the
other hand, $G''_2$ remains a tree, the line of reasoning used to
establish Claim 3 above gives:
$$|{\cal{C}}_4|+|A'| \ge 1+
\left[\sum_{C \in \cal{C}_3}d_{G''_2}(v_C)\right]-2|\cal{C}_3|.$$

However, $d_{G''_1}(v_C)=d_{G''_2}(v_C)$
for each $C \in \cal{C}$.  So this gives

$$|{\cal{C}}_4|+|A'| \ge 1+
\left[\sum_{C \in \cal{C}_3}d_{G''_1}(v_C)\right]-2|\cal{C}_3|.$$

Plugging this together with Claim 4, and then
noting $|A_1|+\frac{|A|}{2} \ge |A'|$  yields Claim 6.
$\surd$

We also make the following claim:

Claim 7: $|A| \ge \max\{2|W'|+|A_1|, |A_2|\}$.

Proof of Claim 7: We first show the following: Let $w'$ be a vertex
in $W'$ and let $w'x'y'$ be a triangle. We claim that the only
vertex in $W'$ that $x'$ and $y'$ can be adjacent to is $w'$.

[Indeed, then $x'$ and $y'$ are both in
$G'$ and so is the edge $x'y'$. If on the one hand there is
another vertex $w'' \in W'$ such
that $w''x'y'$ is a triangle then
that leaves $D_{G'}(x')=d_{G'}(y')=1$ with the edge $x'y'$ in $G'$
so $x'$ and $y'$ form their own connected component, a
contradiction. If on the other hand there is another
vertex $w'' \in W'$ such that
$w''x'x''$ form a triangle for a vertex $x'' \not = y'$, then
$x''$ and $y'$ are 2 distinct vertices in $G'$ and $w'$ and $w''$ are 2
distinct vertices in $G \setminus G'$, all adjacent to $x'$
which gives $d_G(x') \ge 4$.]

We next claim that $w'$ is not adjacent in $G$ to any vertex in
$A_1$. Indeed, both $x'$ and $y'$ as above
are in $A$ because invariant (B) is satisfied, but both $x'$ and $y'$
have degree no greater than 2 and are adjacent to each other so $x'$ and
$y'$ are both not in $A_1$. So Claim 7 follows. $\surd$

Claim 6 and Claim 7 together imply Claim 8:

Claim 8: $$|A| \ge |A|_1+\frac{|A_2|}{2}+|W'|$$, and so
$$|{\cal{C}}_4|+|A| \ge |\cal{C}_3|+1.$$

Claim 8 implies Claim 9:

Claim 9: Let $n$ be the number of vertices in the original graph $G$ and
let $R$ be the number of vertices in $G \setminus G'$ and let
$n(G')=n-R$ be the number of vertices in $G'$ and let
$m(G')$ be the number of edges in $G'$. Then
$m(G') \le \frac{9n(G')}{7}$, and this is when every vertex is in either a 3-cycle
in $G'$ or a 4-cycle.

Then putting this together and doing algebra on the inequality
$$\frac{9(n-R)}{7} \ge \frac{3n}{2}-3R$$
yields $R \ge \frac{n}{8}$
and thus the theorem. $\surd$

And this is essentially the best you can do, you cannot guarantee a set larger than $\frac{n}{8}+O(1)$: First let $T_1$ and $T_2$ be
2 complete binary trees of the same size where the roots of $T_1$ and $T_2$ have degree-2 and every other interior vertex of each of $T_1$ and $T_2$ has degree-3. Then

Put an edge between the roots of $T_1$ and $T_2$ and call the resulting graph $T$.

Replace each vertex $v$ of $T$ with a cycle $c_v$, where $c_v$ is a triangle if $v$ is an interior vertex, and $c_v$ is a 4-cycle if $v$ is a leaf.

Then connect the $c_v$s as follows: (a) If $v$ and $w$ are adjacent vertices in $T$ then there is exactly one edge from a vertex in $c_v$ to a vertex in $c_w$; if $v$ and $w$ are non-adjacent distinct vertices in $T$ then there are no edges from $c_v$ to $c_w$. (b) If $v$ is an interior vertex of $T$, then each of the exactly 3 vertices in $c_v$ is adjacent to exactly 1 vertex outside of $c_v$ each. If $v$ is a leaf vertex, then exactly 1 of the exactly 4 vertices in $c_v$ is adjacent to exactly one vertex outside of $c_v$.

Then, for each leaf vertex $v$ of $T$, add an additional vertex $i_v$ and put an edge from $i_v$ to the 3 of the vertices in $c_v$ that have degree exactly 2.

Call the resulting graph $G$. Then if $T$ has $n$ vertices, then $G$ has
$5 \times \left(\frac{n}{2}\right)+ 3 \times \left(\frac{n}{2}\right)$ vertices, plus or minus $O(1)$. The largest independent set of vertices that we can remove from $G$ so the resulting graph is connected is $\{i_v; v$ and leaf vertex in $T\}$, and this set has cardinality $\frac{n}{2} = \frac{|V(G)|}{8}+O(1)$. $\surd$


问:
Joint distribution of multiple binomial distributions
In the picture below, how do they arrive at the joint density function? I understand how Binomial distributions work, but have never seen the joint distribution of them.

The original file can be found here: Link

答:
If you have $N$ independent random variables with densities $f_1,\ldots,f_N$, then the joint density is simply $$
  f(x_1,\ldots,x_N) = f_1(x_1)\cdot\ldots\cdot f_N(x_N)
$$

The join density of $N$ independent random variables with $X_i \sim \textrm{Bin}(m,p)$ is thus $$
  f(x_1,\ldots,x_N) = \prod_{i=1}^N \underbrace{\binom{m}{x_i} p^{x_i}(1-p)^{m-x_i}}_{=\mathbb{P}(X_i=x_i)}
  = p^{\sum_{i=1}^N x_i} (1-p)^{\sum_{i=1}^N (m-x_i)} \prod_{i=1}^N \binom{m}{x_i} \text{.}
$$

ContentLicense: CC BY-SA 3.0


问:
are two the following functors isomorphic
Consider the following two functors that go $C^{op}\times C\to Set$, where $X$ is an object in $C$

$(A,B)\mapsto Set(C(X,A)\times A,B)$

$(A,B)\mapsto Set(C(X,A),Set(A,B))$

Are these two functors isomorphic? (i.e. is there an natural transformation between them that is an isomorphism?)
I want to think that they are due to $Set(A\times B,C)\cong Set(A,Set(B,C))$, but I haven't grasped all the details.

答:
Yes, in the end the two functors are isomorphic, but you need some care in specifying in what sense $Set(X,Y)\times Z$ is an object of $\cal C$ for every $Z$ (instead of $Set(C(X,A)\times A,B)$, you want to write ${\cal C(\cal C}(X,A)\times A,B)$, or the only way to make $Set(C(X,A)\times A,B)$ type-check is to impose ${\cal C} = Set$).

The correct statement involves tensors over the category of sets.

Now, the adjunction given by the fact that $\cal C$ is tensored
$$
{\cal C}({\cal C}(X,A)\times A,B)\cong Set({\cal C}(X,A), {\cal C}(A,B))
$$
is a natural isomorphism in all arguments, and the claim follows.

ContentLicense: CC BY-SA 4.0


问:
General form for this integer-factor-shifting bijection?
Let $f(k,s)$ be a function that takes all the factors in an integer $k$ and increases their prime indices by $s$.

e.g.

$$f(28,2)=f(2^2 \cdot 7,2)=f(p_1^2 p_4,2)=p_3^2 p_6 =5^2 \cdot 13=325.$$

I've determined that for any positive integer $n$,

$$\left\vert\{1 \leq k \leq n : f(k,1)>2n\}\right\vert=\left\vert\{n+1 \leq k \leq 2n : f(k,1)\leq 2n\}\right\vert.$$

I understand why this happens; $x\iff f(x,1)$ is a bijection in $\mathbb{N}$, so symmetry requires it.

Through experimentation, it also seems that

$$\left\vert\{1 \leq k \leq n : f(k,2)>3n\}\right\vert=\left\vert\{n+1 \leq k \leq 3n : f(k,2)\leq 3n\}\right\vert.$$

I'm less clear on why this works, particularly because there is no obvious candidate for $x$ in situations like $f(x,2)=2y$ that maintains a 1-to-1 mapping.

More to the point, I haven't been able to extend this pattern any further, and am wondering whether there is a good reason for that. If anyone can shed some insight on whether this general form can or cannot be extended for values of $s>2$, I'll consider my question answered.

答:
First of all, the map $x \mapsto f(x, 1)$ isn't a bijection from $\mathbb N$ to $\mathbb N$; it's a bijection from $\mathbb N$ to the set of positive odd integers.  Similarly, $x \mapsto f(x, s)$ is a bijection from $\mathbb N$ to the set of positive integers not divisible by any of $p_1 = 2, p_2 = 3, \dots, p_s$.

Your observation about $f(k, 1)$ can be rephrased as follows:  if you add to each side the quantity
$$
|\{1 \leq k \leq n: f(k, 1) \leq 2n\}|,
$$
you get
$$
n = |\{1 \leq k \leq 2n: f(k, 1) \leq 2n\}|.
$$
This is easy to prove:  There are exactly $n$ odd numbers below $2n$, each of them equals $f(k, 1)$ for exactly one choice of $k$, and this $k$ is less than $2n$ because $x \leq f(x, 1)$ for all $x$.

You can restate your claim about $f(k, 2)$ similarly:  adding
$$
|\{1 \leq k \leq n: f(k, 2) \leq 3n\}|
$$
to each side gives
$$
n = |\{1 \leq k \leq 3n: f(k, 2) \leq 3n\}|.
$$
To prove this, it suffices to show that exactly $n$ of the numbers $1, \dots, 3n$ are divisible by neither 2 nor 3.  You can check this by casework on the parity of $n$.

If you want to generalize this to study $f(k, s)$ for $s > 2$, the trick of adding a term to each side will still work; the only thing that will change is the problem of counting integers below $p_s n$ that are not divisible by a prime $\leq p_s$.  For example, when $s = 3$, there are approximately $\frac 12 \cdot \frac 23 \cdot \frac 45 \cdot 5n = \frac{4n}{3}$ integers below $5n$ that are not divisible by 2, 3, or 5; but the exact number will fluctuate a little depending on congruence conditions on $n$.


问:
line at infinity intersected with a line from points of the parabola
Suppose we are looking at a parabola as a conic in $PG(2,\mathbb{K})$ with the line at infinity denoted $\ell_\infty$. I am still working on a previous problem I posted here.

Suppose we have two points $A(a,a^2)$ and $B(b,b^2)$ on the parabola. The slope of the line $AB$ is $b+a$. How does this line intersect the line at infinity? The idea of the exercise is to get the intersection, $L_{AB}$ of $AB$ and $\ell_\infty$, and then take a line from $L_{AB}$ to the origin and see where it intersects the conic. In this case I think that I always get the origin as the result, but I feel like this should not be the case. I know that the line at infinity is tangent to the parabola, does this mean that the new line will be parallel to $AB$?

答:
If you are allowed to use homogeneous projective coordinates, then
some calculations are enough to identify the objects involved.
Suppose we have two field elements $\,a,b,\,$ and let $\,c:=a+b.\,$
The point coordinates for the points $\,A,B,L_{AB}\,$ are:
$$ A = (a, a\,a, 1), \qquad
   B = (b, b\,b, 1) \qquad  \text{ and }\qquad
  L_{AB} = (1, c, 0). \tag{1}$$
The line coordinates for the lines $\,AB, \ell_{\infty}, \ell_2\,$ ($\ell_2$
is line through origin and $L_{AB}$) are:
$$ AB = (-c, 1, a\,b), \qquad
   \ell_{\infty} = (0, 0, 1) \qquad \text{ and }\qquad
  \ell_2 = (c, -1, 0). \tag{2}$$
Define the point $\,C := (c, c\,c, 1)\,$ on the parabola
and on the line $\,\ell_2.\,$
Check that all of the other incidence relations hold. For
example, line $\,AB\,$ contains point $\,A\,$ since
$$ -c(a) + 1(a\,a) + a\,b(1) = -(a+b)a + a\,a + a\,b = 0. \tag{3}$$

Projectively, this is equivalent to a circle with a point
$\,O\,$ on it regarded as the origin and then two points $\,A\,$
and $\,B\,$ on the circle determine a third $\,C\,$ so that the
line $\,AB\,$ is parallel to $\,OC.\,$ This process defines a
group operation on the circle points.


问:
Verify function solves heat equation $\Delta f = \partial_t f$
So the notorious function is written as $f(x,t) = t^{-n/2}\,e^{\frac{-x^2}{4\,t}} $

It strikes me as that I have to show that the first derivative with respect to t equals the sum of both second derivatives. But on the other side this seems irritating because the first derivative looks already complicated:

$$\partial_tf(x,t) = -\frac{n}{2} \,t^{-n/2-1} \,e^{\frac{-x^2}{4\,t}}+ t^{-n/2}\,\frac{x^2}{4\,t^2}\, e^{\frac{-x^2}{4\,t}}.$$ Should this really equal $\Delta f = \partial_x^2f +\partial_t^2 f$?

I'm not really confident whether this might be the task or is it?

答:
It is a little annoying, but that's just how it goes sometimes. First of all, since you have it written as $x^2,$ you're clearly working with $n=1.$ Otherwise, this should be $\|x\|^2.$

Calculate that

$$\partial_x f=-\frac{1}{2}t^{-3/2} xe^{-x^2/4t},$$ and $$\partial_{xx} f=-\frac{1}{2}t^{-3/2} e^{-x^2/4t}+\frac{1}{4}t^{-5/2}x^2e^{-x^2/4t}.$$ When $n=1,$ this is $\Delta f$ and is exactly the same as what you computed for $\partial_t f$ (modulo a sign error on the power of $t$ in the first term: it should read $t^{-n/2-1}$).


问:
Understanding Direct Sums and Product (Vector Spaces)
On p.21 of A Course in Homological Algebra - Hilton and Stammbach. there is a proof for a (possibly infinite) collection of modules that there is an isomorphism $\eta$

$\eta: \text{Hom} (\bigoplus_{j \in J} A_j, B) \to \prod_{j \in J} \text{Hom} (A_i, B)$

For $\psi: \bigoplus A_j \to  B$ define  $\eta(\psi) = (\psi i_j)_{j \in J}$
where the $i_j$ are the injections  $i_j: A_j \to \bigoplus A_j$

I'm fine with seeing this as a bijection, and with its linearity.

The proof continues to describe the projections associated with the product, saying the projections $\pi_j : \prod_{j \in J} \text{Hom} (A_i, B) \to \text{Hom} (A_i, B)$
are given by $\pi_j \eta = \text{Hom} (i_j, B) $

I don't see this last part at all (it appears to equate a map with a collection of maps), nor do I have an idea what it should be. Assistance would be appreciated.

答:
Recall that $\hom$ is a functor $\mathcal A^{op}\times\mathcal A\to\mathcal Ab$ when $\mathcal A$ is an Abelian category (for arbitrary categories its codomain is $\mathcal Set$), acting by pre- and post composition.

In particular, $\hom(i_j, B)$ is the map $\hom(\bigoplus_kA_k,\,B)\to\hom(A_j,B)$ assigning $f\mapsto f\,i_j$.

On the other hand, for any $f\in\hom(\bigoplus_kA_k,\,B)$, by definition of $\eta$, the $j$th coordinate of $\eta(f)\in\prod_k\hom(A_k,B)$ is the same: $f\,i_j$.


问:
Uniform Topology on $\mathbb{R}^{\Gamma} $ is different from Product Topology on $\mathbb{R}^{\Gamma}$
This question is from My topology assignment and i was unable to solve it and unfortunately I have no intution on how this question should be approached.

Question:If $\Gamma$ is an infinite set then prove that Uniform Topology on $\mathbb{R}^{\Gamma} $ is different from Product   Topology on $\mathbb{R}^{\Gamma}$.

Definition of uniform Topology: Let $\Gamma$ be a non empty set and let $\bar{d}$ be the standard bounded metric on $\mathbb{R}$ . Define a metric $\bar{\mathbb{\rho}}$ on $\mathbb{R}^{\Gamma}$ by  $\bar{\mathbb{\rho}}(x,y)$= {lub${\bar{d}(x_{\alpha} ,y_{\alpha})}: \alpha \in \Gamma$}.Then $ \bar{\mathbb{\rho}}$ is a metric called uniform metric and topology induced by $\bar{\mathbb{\rho}}$ is called uniform topology on $\mathbb{R}^{\Gamma}$.

Can you please tell me on how should this question should be solved.  Also, I am really sorry that I would not be able to Get any attemp as I don't have anything worthy to show.

Thank you!

答:
For each $\gamma\in\Gamma$, define $f_\gamma\colon\Gamma\longrightarrow\Bbb R$ as$$f_\gamma(\alpha)=\begin{cases}1&\text{ if }\alpha=\gamma\\0&\text{ otherwise.}\end{cases}$$Let $0\in\Bbb R^\Gamma$ be the null function. Then, for each $\gamma\in\Gamma$, $\overline\rho(f_\gamma,0)=\overline d(0,1)>0$. So, $0\notin\overline{\{f_\gamma\mid\gamma\in\Gamma\}}$.

However, with respect to the product topology on $\Bbb R^\Gamma$ we do have$$0\in\overline{\{f_\gamma\mid\gamma\in\Gamma\}}.\tag1$$In fact, if $A$ is an open set of $\Bbb R^\Gamma$ with respect to that topology and if $0\in A$, then, for some finite subset $F$ of $\gamma$, and for a set $\{(-\varepsilon_\gamma,\varepsilon_\gamma)\mid\gamma\in F\}$ of intervals of $\Bbb R$ (with each $\varepsilon_\gamma$ greater than $0$), $A$ contains the set$$\left\{f\in\Bbb R^\Gamma\,\middle|\,(\forall\gamma\in F):f(\gamma)\in(-\varepsilon_\gamma,\varepsilon_\gamma)\right\}.$$In particular, if $\gamma\in\Gamma\setminus F$, then $f_\gamma\in A$. So, since $\Gamma$ is infinite, $A$ contains some element of $\{f_\gamma\mid\gamma\in\Gamma\}$ (actually, infinitely many of them), and so we do have $(1)$.


问:
Landmarks of subjects of mathematics
In order to have a good view of the whole mathematical landscape one might want to know a deep theorem from the main subjects (I think my view is too narrow so I want to extend it).

For example in natural number theory it is good to know quadratic reciprocity and in linear algebra it's good to know the Cayley-Hamilton theorem (to give two examples).

So, what is one (per post) deep and representative theorem of each subject that one can spend a couple of months or so to learn about? (In Combinatorics, Graph theory, Real Analysis, Logic, Differential Geometry, etc.)

答: Spectral theorem in Functional Analysis.

Title: Landmarks of subjects of mathematics


问:
Find the MLE of a parameter for a random variable with a given density.
Let $X$ be a random variable with density:

$    f_{X}(x) = \begin{cases}
          1 - \theta & \frac{-1}{2} < x < 0 \\
          1 + \theta & 0 \leq x < \frac{1}{2} \\
          0 & otherwise
       \end{cases}
$

Where $\theta \in (-1,1)$ is an unkown value. We have a random sample $X_{1},...,X_{n}$.

Find the maximun likelihood estimator for $\theta$.

This was an excersive on a test I took a few weeks ago and while I passed the test I still can't figure out what is the correct approach to this problem. I went through my notes and haven't seen a problem like this before.

I tried the usual way of writing $L(\theta, X_{1},...,X_{n})$ and trying to maximize it but that got me nowhere useful.

Any help would be greatly appreacited, thanks!

答: Let $a=\sum_{i=1}^n\mathbf 1_{(-\frac{1}{2},0)}(X_i)$ be the number of observations between $-\frac{1}{2}$ and $0$. Then the number of observations in $\left[0,\frac{1}{2}\right)$ is $n-a$. The likelihood function is $$L(\theta)=(1-\theta)^a(1+\theta)^{n-a}$$ and the log-likelihood function is $$l(\theta)=a\log (1-\theta)+(n-a)\log (1+\theta)$$ Differentiating $l(\theta)$ and equating it to $0$, $$l'(\theta)=-\frac{a}{1-\theta}+\frac{n-a}{1+\theta }=\frac{-a(1+\theta)+(n-a)(1-\theta)}{1-\theta^2}=\frac{-n\theta -2a+n}{1-\theta ^2}$$ This gives our desired MLE: $$\theta_{\text{MLE}}=\frac{n-2a}{n}=\frac{n-2\sum_{i=1}^n\mathbf 1_{(-\frac{1}{2},0)}(X_i)}{n}$$


问:
$3^{1234}$ can be written as $abcdef...qr$. What is the value of $q+r$?
It was possible to find $3^{15} ≡ 7\pmod{100}$. Knowing that $7^{4k}≡1\pmod{100}, (3^{15})^{80}≡3^{1200}≡(7)^{80}≡1\pmod{100}$.
$(3^{15})(3^{15})3^{1200}≡1\cdot 7\cdot 7\pmod{100}, 3^{1230}\cdot 3^{4}≡49\cdot 81\pmod{100}≡69\pmod{100}$ to get $6+9=15$. I felt that there must be a better way to solve this problem. What ideas should I keep in mind?

答:
The Carmichael function value for $100$, $\lambda(100)=\text{lcm}(\lambda(25),\lambda(4)) = \text{lcm}(20,2) = 20$ gives us $1234 \equiv 14 \bmod \lambda(100) \implies 3^{1234} \equiv 3^{14} \bmod 100$.

Probably the easiest way to calculate $3^{14} \bmod 100$ is as $81^3 \cdot 9$ discarding higher digits, so $81^2 \equiv 61$ and $81\cdot 61 \equiv 41$ leading to $3^{1234} \equiv 3^{14} \equiv 69 \bmod100$.

Starting alternatively from your observation that $3^{15} \equiv 7 \bmod 100$, we could find $3^{14} \equiv 7\cdot 3^{-1} \equiv 7\cdot 67 \equiv 469 \equiv 69 \bmod 100$.


问:
Modular Arithmetic Operations..
I am studying 7th and I am having hard time with this problem. Need help.
How to find the value of B in the following equation?

A=B^e mod C

Thanks in advance :-)

答:
One particular method works if $C$ is prime and $A$ is coprime to $C$. Then all nonzero remainders modulo $C$ make up a cyclic group, i.e. there is one generator (call it $G$) such that $1, G, G^2,\ldots,G^{C-2}$ is the set of all of those remainders, with $G^{C-1}=1$. This number is called "primitive root modulo $C$" (Cf. Wikipedia article.) There is no known algorithm for finding $G$, but it is usually found fairly quickly via trial-and-error.

Now, $A=G^i$ for some $i$, $0\le i\le C-2$ and finding $B=G^j$ such that $B^e=A$ is equivalent to finding $j$ such that $j\cdot e=i\pmod{C-1}$. Surely the bigger problem here is finding $i$, which is generally hard. However, it may happen that $A$ is already a primitive root modulo $C$, so you can take $G=A, i=1$ and the whole problem reduces to inverting $e$ modulo $C-1$.

(Side note: this method is analogous to calculating $B=A^{\frac{1}{e}}$ as $B=G^{\frac{1}{e}\log_G A}$ in $\mathbb R$, for a nicely chosen base $G$ ... sort of.)

Apart from $C$ being prime, there are other cases where the group of all invertible remainders $\pmod C$ is cyclic: when $C=2, C=4, C=p^k$ or $C=2p^k$ for an odd prime $p$ and an exponent $k\ge 1$. One can use a similar approach in those cases.

Title: Modular Arithmetic Operations..


问:
A question on The Prime Factorization of a Polynomial
I just proved the following theorem:

Let $F$ be a field. A non-scalar monic polynomial in $F[x]$ can be factored as a product of monic prime polynomials in $F[x]$ in one and only one way except for order.

In some problems, I see that this theorem is used for polynomials which are not monic and still the factorization is assumed to be unique up to the order.  My question is that will this statement still be true if I remove the word "monic" from in front of "polynomial" and "prime polynomial"?

答:
A ring $F[x]$ is a Unique Factorization Domain (UFD) (proof below) so you can write any non-zero non-unit element as a product of primes up to order and up to units.

Now, look what are the units in $F[x]$. If $a\neq 0$ and there is $b$ such that $ba=1$ then necessarily those are both degree zero polynomials (or in your terminology scalars). Note that this is not true for polynomials over rings with zero divisors. But for polynomials over domains, this is clear.
Observe that opposite inclusion is also true. Take a scalar, ie. element of $F$ then it is invertible as $F$ is a field.

So if we lose the condition of polynomials being monic we still have a unique decomposition but up to units (or up to association in different terminology).

Now to show that $F[x]$ is UFD. One can either use Theorem that polynomials over UFD are UFD. Or one can go directly. I am not sure what definition of UFDs you used in your course  of algebra but this one should be rather well-known:

1 GCD always exists (unique up to association) - you can easily check that division with remainder works for polynomials (with the remainder having lesser degree then divisor)

2 Ascending Chain Condition on Principle Ideals - this one can be more abstract but one can rephrased it as follows: there exists no infinite chain of divisors $... |r_n|r_{n-1}| ... |r_2|r_1$ where each $r_i/r_{i+1}$ is non-unit. Now clearly if one polynomial divides the other and the result of the division is non-unit (not a scalar) then they have a different degree.


问:
Coordinates for a neighborhood of a totally real submanifold
Motivation: I'm interested in which submanifolds of (half dimension) a complex manifold can be expressed locally in coordinates as the real part of the complex coordinates, and was wondering if the name "totally real" had any connection to this.

Definition: Let $M$ be a $2n$-dimensional real manifold with an almost complex structure $J$. A submanifold $L \subset M$ is called totally real if it is of half the dimension of $M$ and $T_q L \cap J T_q L= \{0\}$ for $q$ in $L$.

Question: I have two questions, but one is a generalization of the other.

Question 1: If $M$ is actually a complex manifold (i.e., $J$ is integrable) and $L \subset M$ is a totally real submanifold, does this imply that there exists holomorphic coordinates $(z^k= x^k + iy^k)$ on some neighborhood $U$ in $M$ such that $U \cap L = \{y^k =0\}$?

Question 2: Same as Question 1, but where $M=T^*L$ and $L$ is viewed as zero section, and it is assumed that $T^*M$ is complex manifold.

答: I do not understand the 2nd question (I think, there are some assumptions missing here). Formally speaking the answer to Question 1 is negative since the existence of such a holomorphic change of variables implies that $L$ is a real-analytic submanifold (and not every totally real submanifold is real-analytic as you can see already in the case of real curves in ${\mathbb C}$). However, if you assume, additionally, that $L$ is real-analytic, then the answer is positive, see Lemma in my answer to this Mathoverflow question.


问:
What 'complete' means in Hilbert Space
From my Professor's Lectures, a Hilbert Space is defined as an inner product space $\bigoplus$ complete. In the same lecture, 'complete' here means, a system $\{\varphi_j\}_{j=1}^\infty \subset V$ is complete if $\nexists g \in V\setminus\{0\} $ such that $g$ is perpendicular to $\varphi_j, \forall j $.

It is also explicitly stated that 'complete' here doesn't mean 'complete space' where all Cauchy sequences are convergent. However, in the proof for a lemma that states:

An orthonormal system $\{\varphi_j\}_{j=1}^\infty \subset V$ that is complete is a basis in $V$,

We came to a point where we proved that $S_n(f) = \sum_{k=1}^{\infty} C_k\varphi_k$ is a Cauchy sequence, and then my Professor said that because we're in a Hilbert Space (complete), thus $S_n(f)$ is convergent. Is there something I'm missing here? Are both definition of 'complete' true in Hilbert Spaces?

答:
There's a notion of completeness of a metric space and a notion of completeness of a basis, and they're not the same thing. Hilbert spaces are defined to be, in particular, complete metric spaces. Completeness of a basis means something different. It means what you said, and another way of stating it is that the span of the given system is dense in the Hilbert space. So, one is an intrinsic property of the space itself and the other is a property of particular subspaces.

For the record, I think a more clear way to state completeness of a system than you did is that if $\langle g,\varphi_j\rangle=0$ for all $j$, then $g=0$.

Title: What 'complete' means in Hilbert Space


问:
Construction of non-empty perfect set in $\mathbb{R}^1$ which contains no rational points

We say a set is perfect if it is closed and every point is limit point.

The question is from baby Rudin$(2.18)$. I understood the construction given in solution manual by J. D. Taylor. I have constructed two examples and wanted to verify if they are correct.

$1.$ Before giving first example, I introduce one terminology.

Let $S\subset \mathbb{R}$. Let $x\in S$ but is not a limit point of
$S$. So we can find $\epsilon>0$ such that $B(x,\epsilon) \cap S=x$.
Then we choose $n_1\in \mathbb{N}$ big enough so that
$x+\frac{1}{n_1}\in B(x,\epsilon)$. Then $\forall k>n_1, k\in
 \mathbb{N}$, we have  $x+\frac{1}{k}\in B(x,\epsilon)$. These elements
$\{x+\frac{1}{k}:k\geq n_1\}$, we will call neighbouring elements of
x and this process of 'choosing appropriate $\epsilon, n_1$ for
such $x$ and finding it's neighbouring elements', we will call
"setting up neighbourhoood for x with respect to S".

Now let $S_1=\sqrt{2}\cup \{\sqrt{2}+\frac{1}{n}:n\in \mathbb{N}\}$. Only $\sqrt2$ is limit point of $S_1$. So we 'set up neighbourhood with respect to $S_1$' for all other elements and let denote that entire collection(i.e. $S_1$ with all those neighbouring elements) by $S_2$. Now, all numbers present in $S_1$ are limit points of $S_2$. But we have whole new bunch of numbers in $S_2$ which are not limit points of $S_2$. So we do same with $S_2$ what we did for $S_1$(but here we set up neighbourhood with respect to $S_2$)and call it $S_3$. Let
$S=\cup_{n=1}^{\infty}S_n$. All numbers in $S$ are irrationals as they all are of the form $\sqrt2 +r$, for some rational $r$. Each number will be limit point because if that number is in $S_i$, in the next step, i.e. in $S_{i+1}$, we set up neighbourhood for it. But I am bit skeptical about the closeness of $S$. Actually inspecting my construction closely, I think that each $S_i$ is countable, and thus $S$ is countable but we know that perfect set is uncountable, so I think my this example won't work. But I would like to know your thoughts on this. Like can I somehow make this perfect doing some small changes. And if $S$ is not closed, I tried to find it's limit points which are not in $S$, but I am not seeing it straightaway.

$2$. The idea behind this construction is using previous exercise. Let $E$ be the set of all $x\in[0,1]$, whose decimal expansion contains only the digits $4$ and $7$. I know this set is perfect. The only problem here is that $E$ contains rationals too. Suppose we translate $E$ by some irrational $y$ in such a way that $y+E$ contains irrationals only, then we are done I guess. We can find such $y$, one such is $y\in(0,1)$ having only $0$ and $1$ in it's decimal expansion. Am I correct?

Thanks.

答:
Let's tweak the $4,7$ example a bit. Consider decimal expansions of the form

$$.\text{_}0\text{_}00\text{_}000\text{_}0000\text{_}00000\text{_} \dots$$

where each $\text{_}$ is either a $4$ or $7.$ Denote the set of all numbers having such a decimal expansion by $E.$ Then $E$ is perfect.

Let $x\in E.$ If $x$ were rational, then its decimal expansion would be repeating after some point. But that can't happen because of the arbitrarily large stretches of $0$'s that occur in the expansion of $x.$

Therefore $E$ is a perfect set with no rationals.


答:
In the context of Hilbert spaces, the term 'complete' is used in both ways.

A metric space is complete if every Cauchy sequence in this space is convergent.

An orthonormal system $M:=\{\varphi_{\alpha}\}_{\alpha}$ in a Hilber space $\mathcal{H}$ is complete if $M^{\perp}=\{0\}$, that is, if $\langle\varphi, \varphi_{\alpha}\rangle = 0$ for every $\alpha$ implies $\varphi = 0$

Title: What 'complete' means in Hilbert Space

Body:
From my Professor's Lectures, a Hilbert Space is defined as an inner product space $\bigoplus$ complete. In the same lecture, 'complete' here means, a system $\{\varphi_j\}_{j=1}^\infty \subset V$ is complete if $\nexists g \in V\setminus\{0\} $ such that $g$ is perpendicular to $\varphi_j, \forall j $.

It is also explicitly stated that 'complete' here doesn't mean 'complete space' where all Cauchy sequences are convergent. However, in the proof for a lemma that states:

An orthonormal system $\{\varphi_j\}_{j=1}^\infty \subset V$ that is complete is a basis in $V$,

We came to a point where we proved that $S_n(f) = \sum_{k=1}^{\infty} C_k\varphi_k$ is a Cauchy sequence, and then my Professor said that because we're in a Hilbert Space (complete), thus $S_n(f)$ is convergent. Is there something I'm missing here? Are both definition of 'complete' true in Hilbert Spaces?


问:
Proving that $ka \equiv kb \pmod{m} \iff a \equiv b \pmod{m/\gcd(k,m)}$
I'm having some trouble proving the following proposition:

Let $m \in \mathbb N$ and $a,b,k \in \mathbb Z$, then:
$$ka \equiv kb \pmod{m} \iff a \equiv b \pmod{\frac{m}{\gcd(k,m)}}$$

I began by trying to prove that $a \equiv b \pmod{\frac{m}{\gcd(k,m)}} \implies ka \equiv kb \pmod{m}$ but I got stuck with:

$$a\gcd(k,m) \equiv b \gcd(k,m) \pmod{n}$$

Then I also tried to prove that $ka \equiv kb \pmod{m} \implies a \equiv b \pmod{\frac{m}{\gcd(k,m)}}$ But I also was not able to conclude anything. This is my first time learning modular arithmetic and I'm still confused with the concepts and have no intuition with this at all, this is perhaps why I'm having some trouble solving these types of questions. How can I prove this?

答:
Lemma. Let $k,m\in\mathbb N$ and $n\in\mathbb N_0$. Then $mn \equiv 0\pmod k$ is equivalent to $\operatorname{gcd}(k,m)n \equiv 0\pmod k$.

Proof. The direction $\impliedby$ is clear. For $\implies$, write $m=v_m\operatorname{gcd}(k,m)$, $k=v_k\operatorname{gcd}(k,m)$ and $mn = Ck$ for some $v_m, v_k\in\mathbb N$, $C\in\mathbb N_0$.

Then $$v_m\operatorname{gcd}(k,m) n = mn=C v_k \operatorname{gcd}(k,m),$$ so $$n=\frac{C}{v_m} v_k,$$ which, since $v_m$ and $v_k$ are co-prime by definition of GCD, implies that $C$ is divisible by $v_m$ (exercise: fully formulate this argument*).

Hence $$\operatorname{gcd}(k,m)\cdot n=\frac{C}{v_m} k \equiv 0\pmod k.$$
$\square$

Now assume that $ka\equiv kb\pmod m$, then WLOG $ka = kb + m n$ for some $n\in\mathbb N_0$ so that $$a=b+\frac{mn}k=b+\frac{m}{\operatorname{gcd}(k,m)}\frac{n\operatorname{gcd}(k,m)}k,$$ and the Lemma can be used to conclude since $\frac{mn}k\in\mathbb Z\implies \frac{n\operatorname{gcd}(k,m)}k \in\mathbb Z$.

The other direction is easier: If $$a=b+\frac{m n}{\operatorname{gcd}(k,m)}$$ for some $n\in\mathbb Z$, then $$ka=kb+m\frac{nk}{\operatorname{gcd}(k,m)}$$ and you can conclude with $\frac{nk}{\operatorname{gcd}(k,m)}\in\mathbb N$.

* In fact, my current formal proof would rely on the fundamental Theorem of arithmetic but I am convinced that there is a much easier way to show this.


问:
Residue field of composite of two field

[Question]

I know that $K'\cdot K''$ is an unramified extension of $K$ but I don't know why $K'\cdot K''$ have a residue field $k'$.

is it always true that $K_1\cdot K_2$ have a residue field $k_1 \cdot k_2$?
(where $k_1,k_2$ are residue fields of $K_1, K_2$)

I think that if we prove the proposition 7.50 , then we can use " $K_1\cdot K_2$ have a residue field $k_1 \cdot k_2$" in this situation.

However, we can't use that fact while proving this proposition.

How can I prove this?

Thank you for your attention.

reference(J.S. Milne's Algebraic Number Theory) and this post
1: Strange reasoning of unramified extensions having same residue fields are the same.

答:
For $K/\Bbb{Q}_p$ a finite extension then $F/K$ is unramified iff $F=K(\zeta_n)=K(\zeta_{q-1})$ with $p\nmid n$ and $q= |O_F/(\pi_F)|$. This is the main application of Hensel lemma.

When $E/K,E'/K$ are ramified then it is not always the case that the residue field of $EE'$  is the smallest field contained those of $E,E'$, try with $E=\Bbb{Q}_2(2^{1/3}),E'=\Bbb{Q}_2(\zeta_3 2^{1/3})$.

When $E'/K$ is unramified then $EE'=E(\zeta_{q-1})$ has residue field $O_E/(\pi_E)(\zeta_{q-1})$.

Title: Residue field of composite of two field


问:
Proving that $f=x\cdot g(x)$ is differentiable at $0$ if and only if $\lim_{x\to0}g(x)$ exists.

Let $f$, $g$ two functions that are defined on a neighborhood of $0$ such that $f(x)=x\cdot g(x)$ for every $x$ in that neighborhood. Then $f$ is differentiable at $0$ if and only if $\lim_{x\to0}g(x)$ exists.

Okay so I'm stuck in the second direction. If I assume $g(x)$ has a limit, where does it get me?
The first direction is quite straightforward.

Assume $f$ is differentiable at $0$, then the limit of $\lim _{x\to 0}\frac{\left(f\left(x\right)-f\left(0\right)\right)}{x-0}$ exist, and is equal to some $L ∈ R$.

Then we have

$\lim_{x\to 0}\frac{\left(f\left(x\right)-f\left(0\right)\right)}{x}=\lim _{x\to 0}\frac{\left(x\cdot g\left(x\right)-0\cdot g\left(0\right)\right)}{x}=\lim_{x\to 0}\frac{\left(x\cdot g\left(x\right)\right)}{x}=\lim_{x\to \:0}g(x)=L$

Therefore $\lim_{x\to \:0}g(x)$ exist, and is equal to $L$.

Assume $\lim _{x\to 0}\left(g\left(x\right)\right)$ exists, and is equal to some $L ∈ R$.

But this can also be written as

$\lim _{x\to 0}\left(g\left(x\right)\right)=\lim _{x\to 0}\frac{\left(g\left(x\right)\right)\cdot x}{x}=\lim \:_{x\to \:0}\frac{\left(f\left(x\right)\right)}{x}=L$

Now we check if $f$ is differentiable at $0$

$\lim _{x\to 0}\frac{\left(f\left(x\right)-f\left(0\right)\right)}{x-0}=\lim \:_{x\to 0}\frac{\left(f\left(x\right)-0\cdot g\left(0\right)\right)}{x}=\lim \:_{x\to \:0}\frac{\left(f\left(x\right)\right)}{x}=L$

Therefore the limit exists and $f$ is differentiable at $0$.

答:
If $f(x)=xg(x)$ (in a neigbourhood of $0$) then $f(0)=0.$ Thus we have (for any $x\ne 0$)

$$\dfrac{f(x)-f(0)}{x-0}=\dfrac{f(x)}{x}=g(x),$$ from where

$$\lim_{x\to 0}\dfrac{f(x)-f(0)}{x-0}=\lim_{x\to 0}\dfrac{f(x)}{x}=\lim_{x\to 0}g(x).$$

In other words, $f$ is differentiable at $0$ if and only if $\lim_{x\to 0}g(x)$ exists.

EDIT

If $f$ is differentiable at $0$ then there exists $f'(0)=\lim_{x\to 0}\dfrac{f(x)-f(0)}{x-0}.$ Thus there exists $\lim_{x\to 0}g(x)$ and it takes the value $f'(0).$

Conversely, assume that $\lim_{x\to 0}g(x)$ exits. Since $\lim_{x\to 0}\dfrac{f(x)-f(0)}{x-0}=\lim_{x\to 0}g(x)$ we can say that $f$ is differentiable at $0$ and moreover $f'(0)=\lim_{x\to 0}g(x).$


答:
The following may or may not be the exact definitions your professor is using. However, even when the terminology is slightly different, the versions in use have more or less the same nuances.

A Hilbert space is an inner product space such that the distance $\sqrt{\langle x-y,x-y\rangle}$ makes it a complete metric space, i.e. a metric space where sequences are convergent if and only if they are Cauchy. In that sense it is said to be complete.

A sequence $\{x_i\}_{i\in I}$ in an inner product space $(E,\langle\cdot,\cdot\rangle)$ is said to be orthogonal if $\langle x_i,x_j\rangle=0$ for all $i\ne j$.

A sequence $\{x_i\}_{i\in I}$ is said to be orthonormal if it is orthogonal and $\langle x_i,x_i\rangle=1$ for all $i\in I$.

A sequence in a Hilbert space $H$ is said to be a complete orthonormal system (or a Hilbert basis) if it is orthonormal and $\{x\in H\,:\, \forall i\in I, \langle x_i, x\rangle=0\}=\{0\}$. Equivalently, a Hilbert basis is an orthonormal sequence with dense linear span.

Title: What 'complete' means in Hilbert Space


问:
Prove this field equality.
Let $A$ be an $n×n$ matrix over $\Bbb C$ (complex) and $F(A)$ the field of values. Let $U$ be an $n×n$ unitary matrix.

(i) Show that $F(U^*AU) = F(A)$.

I am not sure how to deal with equality of fields. How do we show that both fields are equal?
Showing that if one element belong to one field, it belongs to the other field too?

答:
First thing don't worry about field equality as this leads to unnecessarily abstract concepts. Both fields are subsets of complex numbers so you can treat this just as a question about equality of subsets. Sou you just prove two inclusions.

I think that for the complete proof you need these two observations:

1 Multiplication by unitary matrix does not change the norm of a vector
see $(Ux)^*(Ux)=x^*U^*Ux=x^*Ix=x^*x$ where $I$ is the identity matrix.

2 Unitary matrix is regular. In other words, it is surjective as a linear operator on a vector space.

Now the equation $x^*U^*AUx=(Ux)^*A(Ux)$ will allow you to complete the proof. Just for each value in the one field of values find a vector that gives you this value and try to prove the existence of a vector that gives you the same value in the other field of values.

Title: Prove this field equality.


问:
Logarithm question involving different base
Calculate the values of $z$ for which $\log_3 z = 4\log_z3$.</p>

答: Hint : use the formula for the change of base $\log_b(x) = \frac{\log_k(x)}{\log_k(b)}$ to change the base $z$ of the expression on the right.</p>


问:
Finding the coefficients in a vector space of numbers
Consider the vector space $V=\operatorname{span}_{\mathbb Q}(\sqrt2,\sqrt 3)$ of all rational linear combinations of $\sqrt 2$ and $\sqrt 3$ over $\mathbb Q$. If it makes things easier, take the module $\operatorname{span}_{\mathbb Z}(\sqrt 2,\sqrt 3)$.

My question is:

Given $\alpha = a\sqrt 2+b\sqrt 3$, how can we express the coefficients $a$ and $b$ in terms of $\alpha$?

Clearly by a simple linear algebra argument, the coefficients are uniquely determined for any $\alpha \in V$, but it's not obvious how to determine them given access to $\alpha$ alone (e.g. say we have access to arbitrarily many decimal digits $d_i\in\{0,\dots,9\}$ such that $\alpha = \sum_{i\in\mathbb Z}10^id_i$).

I suppose one can find them if we can define an inner product on $V$, but its not obvious how to define an inner product explicitly in terms of $\alpha$. For instance, I can easily define
$$\langle a\sqrt2+b\sqrt3,c\sqrt2+d\sqrt3\rangle = ac+bd,$$
and then we would get $a=\langle \alpha,\sqrt2\rangle$ and $b=\langle\alpha,\sqrt 3\rangle$, but obviously this is cheating.

I've tried messing around with different combinations of $\alpha$, $\alpha^2$, etc., but I can't seem to get anywhere. I appreciate any help with this.

Disclaimer

I suspect this is impossible, because it feels like if it were possible, then we could use something similar to express $\mathfrak{Re}(z)$ in terms of $z$ alone (without using $\bar z$ or anything like that), and I suspect that if that were possible, I would surely have come across it somewhere.

If it is impossible, a proof of impossibility would be nice, though I have no idea how one would go about such a thing.

答:
$\newcommand{\ZZ}{\mathbb Z}\newcommand{floor}[1]{\left\lfloor#1\right\rfloor}$As alluded to by Gerry Myerson, there are algorithms which take algebraic integers like $\alpha=a\sqrt2+b\sqrt3$ of arbitrary (say, decimal) precision and can output good (and small) $a,b\in\ZZ$ which approximate $\alpha.$

Even stronger, there are algorithms which can produce minimal polynomials: given a good enough approximation of a real number $\alpha$ which is the root of a polynomial $p(x)\in\ZZ[x]$ and given the degree $d:=\deg p,$ we can provide a good $p(x).$ Plugging in your $\alpha$ with $d=4$ into the algorithm would do what you want.

I will outline one way to do this. The idea is to use lattice basis reduction. Pick some very large integer $N$ representing precision of $\alpha$ and consider the set of $d$ vectors
$$\left\langle 1, 0, \cdots, 0, \floor{N\alpha^0}\right\rangle, \\
\left\langle 0, 1, \cdots, 0, \floor{N\alpha^1}\right\rangle, \\
\vdots \\
\left\langle 0, 0, \cdots, 1, \floor{N\alpha^{d-1}}\right\rangle.$$
These vectors define a lattice in $\mathbb R^{d+1}.$ After lattice basis reduction (say, with Lenstra–Lenstra–Lovász), we will have small vectors like
$$v=\left\langle a_0,a_1,\cdots,a_d,\sum_{k=0}^{d-1}a_k\floor{N\alpha^k}\right\rangle.$$
If the last component of $v$ is small enough, then we guess that it's supposed to be $0$ if we replaced the approximations $\floor{N\alpha^k}$ with actual $\alpha^k$s. In other words, we hope
$$\sum_{k=0}^{d-1}a_k\alpha^k=0.$$
Thus, we can read the minimal polynomial for $\alpha$ straight off the other coefficients $a_0,\ldots,a_d$ of $v.$


问:
Rank $ (a) {\aleph_4}^{\aleph_4}, (b) {\aleph_0}^{\aleph_4}, (c) {\aleph_4}^{\aleph_0}, (d) {\aleph_4}, (e) {\aleph_5}, (f) {\aleph_5}^{\aleph_0} $
If we work in a model of set theory where $2^{\aleph_0} = \aleph_5$, then Rank in ascending order the following cardinals:

$$
(a) {\aleph_4}^{\aleph_4}, (b) {\aleph_0}^{\aleph_4}, (c) {\aleph_4}^{\aleph_0}, (d) {\aleph_4}, (e) {\aleph_5}, (f) {\aleph_5}^{\aleph_0}
$$

If possible, specify whether there is a strict equality or inequality

My attempt:

the only one I could figure out was: (f) $ {\aleph_5}^{\aleph_0} = (2^{\aleph_0})^{\aleph_0} = 2^{\aleph_0} $.

What will be $\aleph_4$ here?

答: Since $2^{\aleph_4}\le\aleph_4^{\aleph_4}\le(2^{\aleph_4})^{\aleph_4}=2^{\aleph_4}$ etc. (a) and (b) are both $2^{\aleph_4}\ge\aleph_5$ (both (e) and (f), which exceed (d)). Analyze (c) using $\aleph_5=2^{\aleph_0}\le\aleph_4^{\aleph_0}\le2^{\aleph_4\aleph_0}=2^{\aleph_4}$, so dropping the brackets $a=b\ge c\ge e=f>d$. But $e=\aleph_5^{\aleph_0}\ge c$, co the second $\ge$ becomes $=$. Since $b=c$ is equivalent to $2^{\aleph_4}=2^{\aleph_0}$, it's undecidable as @spaceisdarkgreen notes (indeed, @tomasz notes Easton's theorem shows this).


问:
Find the probability that exactly $12$ customers arrive from $9$A.M to $11$ A.M
Customer arrivals at a payment counter in a with an average of nine per hour

Assume that the number of customer arrivals at a counter in each hour behaves independently. Find the probability that exactly $12$ customers arrive from $9$A.M to $11$ A.M

Parameter $\lambda=9$ per hour so there $2$ hours between  $9$A.M to $11$ A.M therefore $\lambda=18$

$P(X=12)=\dfrac{e^{-18} 18^{12}}{12!}$

Is that correct?

答: Yes, that is correct.  For a Poisson process with rate $\lambda$ per unit of time, the random number of events per $t$ units of time is a Poisson random variable $X(t)$ with rate $\lambda t$, with probability mass function $$\Pr[X(t) = x] = e^{-\lambda t} \frac{(\lambda t)^x}{x!}, \quad x \in \{0, 1, 2, \ldots\}.$$  In your case, $\lambda = 9$, your units of time is $1$-hour increments, and $t = 2$ hour increments.  Thus the probability that there are $12$ arrivals during a $2$-hour increment is $\Pr[X(2) = 12]$, which is what you wrote.


问:
Why does the probability function of a Bernoulli variable $X\sim Bern(p)$ have the form $f(x|p)=p^x(1-p)^{1-x}$?
Why does the probability function of a Bernoulli variable $X\sim Bern(p)$ have the form $f(x|p)=p^x(1-p)^{1-x}$?

As I understand, the density function $f(x|p)$ for $X\sim Bern(p)$ should satisfy
$$f(0|p)=1-p,\ f(1|p)=p$$
Such functions are certainly not unique. In general, I would also require that $f(x|p)=0$ for $x\neq0,1$. So I would define it as $f(x|p)=(1-p)I_{\{0\}}(x)+pI_{\{1\}}(x)$. But if we only require that the conditions hold at $0$ and $1$, $f(x|p)=(2p-1)x+1-p$ is also a choice.

Why is there only $f(x|p)=p^x(1-p)^{1-x}$ that is used, especially when maximum likelihood estimation problems are involved? Is there a theory behind this? Can it be generalized to other discrete distributions?

答:
Suppose we have a sample $X_1,\cdots ,X_n$, in which there are $a=\sum_{i=1}^nx_i$ successes and $n-a$ failures. We want to estimate the probability of success. Given a probability $p$, the probability of getting $a$ successes and $n-a$ failures (in the particular order of the given sample) is $$p^a(1-p)^{n-a}$$
The logic behind maximum likelihood estimation requires you to maximize the likelihood function (which is the probability of getting the given sample given the value of $p$ in this case) by varying the value of $p$. If the pmf of a Bernoulli r.v. is chosen to be $f(x|p)=p^x(1-p)^{1-x}$, then we can come up with the likelihood function $$L(\theta)=\prod_{i=1}^np^{x_i}(1-p)^{1-x_i}=p^a(1-p)^{n-a}$$
which is our desired probability. Other choices of the p.m.f. like $f(x|p)=(2p−1)x+1−p  $ would give $$L(\theta)=\prod_{i=1}^n((2p−1)x_i+1−p)$$ which would not give a convenient result.


问:
Continuity of $f(x)=\sum_{n=0}^{\infty} (-1)^n \frac{1}{x+n}$
I am trying to solve the following exercise and I would like to have a hint about the continuity part:

Let $f(x)=\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+2}-\frac{1}{x+3}+\frac{1}{x+4}-\dots$.

Show $f$ is defined for all $x>0$. Is $f$ continuous on $(0,\infty)$? How about differentiable?

My solution (NOTE: completed the exercise):

f defined We can rewrite $f$ as $f(x)=\sum_{n=0}^{\infty} (-1)^n \frac{1}{x+n}$ which converges by the Alternating Series Test so the function is surely defined for all $x>0$.

f continuous Now, let $\varepsilon>0$: then if we take $N>\frac{1}{\varepsilon}$ we have, for all $n,m\geq N, n>m$ and $x>0$ $|\frac{(-1)^{m+1}}{x+(m+1)}+\frac{(-1)^{m+2}}{x+(m+2)}+\dots+\frac{(-1)^n}{x+n}|\leq\frac{1}{x+(m+1)}<\frac{1}{x+m}<\frac{1}{m}<\varepsilon$
(note that wheter $m$ is even or odd makes no difference since $|-x|=|x|$)
so the series converges uniformly Cauchy Criterion for Uniform Convergence of Series and $f$ is thus continuous by Term-by-Term Continuity Theorem.

f differentiable  $|f'_n(x)|=|\frac{(-1)^{n+1}}{(x+n)^2}|=\frac{1}{(x+n)^2}\leq\frac{1}{n^2}$ and $\sum_{n=0}^{\infty}\frac{1}{n^2}$ is convergent so the series of derivatives $\sum_{n=0}^{\infty}\frac{(-1)^{n+1}}{(x+n)^2}$ is uniformly convergent on $(0,\infty)$ by Weierstrass M-Test and in particular on any interval $[a,b]\subset (0,\infty)$; also, since for any $x_0\in [a,b]$, $\sum_{n=0}^{\infty}f_n(x_0)$ is convergent by the Alternating Series Test we can conclude, by the Term-by-Term Differentiability Theorem, that on any interval $[a,b]\subset (0,\infty)$, $\sum_{n=0}^{\infty}f_n(x)$ converges uniformly to a differentiable function $f(x)=\sum_{n=0}^{\infty}f_n(x)=\sum_{n=0}^{\infty} (-1)^n \frac{1}{x+n}$ such that $f'(x)=\sum_{n=0}^{\infty}f'_n(x)=\sum_{n=0}^{\infty}\frac{(-1)^{n+1}}{(x+n)^2}$.

Thank you.

答:
Note for $x>0$,
$$
0< \frac{1}{x+n}-\frac{1}{x+n+1}+\frac{1}{x+n+2}-... \, ... \, ... <\frac{1}{x+n},
$$
so
$$
\Big|\frac{1}{x+n}-\frac{1}{x+n+1}+\frac{1}{x+n+2}-... \, ... \, ... \Big|<\frac{1}{x+n}\leq \frac 1n.
$$
So this series converges uniformly for $0\leq x<\infty$, so the sum is continuous.


问:
Edge case with sampling and reconstruction.
I know I had been dabbling around this question before, here and here, but does anyone have in their bag of tricks the most simpliest and concise proof that:

$$\sum_{n=-\infty}^{\infty} (-1)^n \, \operatorname{sinc}(t-n) =  \cos(\pi t) $$

where

$$ \operatorname{sinc}(x) \triangleq \begin{cases}
\frac{\sin(\pi x)}{\pi x} \qquad & x \ne 0 \\
\\
1 & x = 0 \\
\end{cases} $$

and $t\in\mathbb{R}$ and $n\in\mathbb{Z}$ ?

I can show that both sides are an even function in $t$ and that both sides have agreement when $t$ is an integer.  But what is the simplest way to show equality for all real $t$ ?

This is something that I want to put together for us Neanderthal electrical engineers.  (and thank you.)

答:
This answer is largely based on this (very concise) answer to a related question of the OP.

Note that for $t\in\mathbb{Z}$ the equality is straightforward to show. The interesting case is when $t$ is not an integer. The derivation below is valid for non-integer real values of $t$.

Using $\cos(x)\sin(y)=\frac12\big[\sin(x+y)-\sin(x-y)\big]$ we can write

$$\begin{align}\sum_{n=-\infty}^{\infty}(-1)^n\textrm{sinc}(t-n)&=\sum_{n=-\infty}^{\infty}\cos(n\pi)\frac{\sin[\pi(t-n)]}{\pi(t-n)}\\&=\frac{\sin(\pi t)}{\pi}\sum_{n=-\infty}^{\infty}\frac{1}{t-n}\\&=\frac{\sin(\pi t)}{\pi}\left[\frac{1}{t}+\sum_{n=1}^{\infty}\left(\frac{1}{t-n}+\frac{1}{t+n}\right)\right]\\&=\frac{\sin(\pi t)}{\pi}\left[\frac{1}{t}+\sum_{n=1}^{\infty}\frac{2t}{t^2-n^2}\right]\tag{1}\end{align}$$

Now we need the following result:

$$\frac{1}{t}+\sum_{n=1}^{\infty}\frac{2t}{t^2-n^2}=\pi\cot(\pi t)\tag{2}$$

which can be found here, here and here, and which can be derived from the well-known infinite product representation of the sinc function

$$\frac{\sin(\pi t)}{\pi t}=\prod_{n=1}^{\infty}\left(1-\frac{t^2}{n^2}\right)\tag{3}$$

Combining $(1)$ and $(2)$ yields the desired result.


答:
Let $d:=\gcd(m,k)$.

If $\frac{m}{d}\mid a-b$ then $a-b=l(\frac{m}{d})$ for some $l\in\mathbb{Z}$ and so $$k(a-b)=\underbrace{l\left(\frac{k}{d}\right)}_{\in\mathbb{Z}} m.$$
This proves $m\mid k(a-b)$.
Conversely, if $m\mid k(a-b)$ then $k(a-b)=nm$ for some $n\in\mathbb{Z}$ and so
$$\frac{k}{d}(a-b)=n\cdot\frac{m}{d}.$$
Now note that $k/d$ and $m/d$ are coprime. Can you conclude from here?

Title: Proving that $ka \equiv kb \pmod{m} \iff a \equiv b \pmod{m/\gcd(k,m)}$

Body:
I'm having some trouble proving the following proposition:

Let $m \in \mathbb N$ and $a,b,k \in \mathbb Z$, then:
$$ka \equiv kb \pmod{m} \iff a \equiv b \pmod{\frac{m}{\gcd(k,m)}}$$

I began by trying to prove that $a \equiv b \pmod{\frac{m}{\gcd(k,m)}} \implies ka \equiv kb \pmod{m}$ but I got stuck with:

$$a\gcd(k,m) \equiv b \gcd(k,m) \pmod{n}$$

Then I also tried to prove that $ka \equiv kb \pmod{m} \implies a \equiv b \pmod{\frac{m}{\gcd(k,m)}}$ But I also was not able to conclude anything. This is my first time learning modular arithmetic and I'm still confused with the concepts and have no intuition with this at all, this is perhaps why I'm having some trouble solving these types of questions. How can I prove this?


问:
Any two non-separating curves on a surface are equivalent

Problem: Let $\Sigma$ be a orientable surface possibly non-compact with boundary. Let $C,$ and  $D$ be two simple-closed curves(smooth embedding) on
$\Sigma\backslash \partial \Sigma$ such that both $\Sigma\backslash
 C$, and $\Sigma\backslash D$ are connected. Does there always exist a
homeomorphism $f:\Sigma\to \Sigma$ such that $f(C)=D$.

This is clear if our surface $\Sigma$ is compact: considering tubular-neighborhoods we have two compact subsurface $M:=\Sigma\backslash\big((-1,1)\times C\big)$, and $N:=\Sigma\backslash\big((-1,1)\times D\big)$ with $\chi(M)=\chi(\Sigma)=\chi(N)$. Note that we are using the following fact: If a finite CW-complex $X$ can be written as union of two subcomplexes $X=Y\cup Z$, then we have the inclusion-exclusion formula: $\chi(X)=\chi(Y)+\chi(Z)-\chi(Y\cap Z)$. Now, we have the

Classification theorem(compact surfaces): Two compact orientable surfaces having the same number of boundary components and same Euler-characteristic, are
homeomorphic.

Now, after boundary rearrangement(which is again a variation of classification theorem for compact surfaces), we have a homeomorphism $\widetilde f:M\to N$ with $\widetilde f(-1\times C)=-1\times D$, and $\widetilde f(1\times C)=1\times D$.  Now, extends the $\widetilde f$ in the whole tubular-neighborhoods.

So, the crucial step is to use Euler-characteristic, and for the non-compact case we have compactly supported Euler-characteristics(considering the rank of compactly supported cohomology), and the inclusion-exclusion formula does hold.

But, the problem is in the classification theorem(compact surface) and its relation with compactly supported cohomology. Also, the boundary rearrangement of a compact surface is another one.

Any help will be appreciated.

答:
Here are the steps of the proof.

Suppose that $S$ is a compact connected oriented surface with boundary, $C_1, C_2$ are distinct boundary components of $S$. Then there exists a pair of pants $P\subset S$ such that $\partial P= C_1\cup C_2\cup C$, where $C$ is a loop in the interior of $S$.

Given a pair of pants $P$ as in 1, there is an orientation-preserving homeomorphism $h: P\to P$ which swaps $C_1$ and $C_2$.

For a homeomorphism $h$ as in 2, there exists an extension of $h$ to the rest of $S$, $\tilde{h}: S\to S$, which preserves all the boundary components of $S$ except for $C_1, C_2$, of course.

If $F$ is a connected oriented surface with boundary, $A_1, A_2$ are simple nonseparating oriented loops in the interior of $F$. Then there exists an orientation-preserving homeomorphism $f: F\to F$ which sends $A_1$ to $A_2$ and preserves each boundary component of $F$.

Suppose that $\Sigma$ is a connected oriented surface, possibly with boundary and $A_1, A_2$  are oriented simple loops in the interior of $\Sigma$. Then there exists a compact subsurface $F\subset int(\Sigma)$ containing $A_1\cup A_2$.

Suppose that $F$ is compact subsurface of the interior of a surface $\Sigma$ and $f: F\to F$ is an orientation-preserving homeomorphism preserving each boundary component. Then $f$ extends to an (orientation-preserving) homeomorphism $\Sigma\to \Sigma$.

Combining steps 4, 5, and 6, you obtain the desired result.


问:
Help to understand proof for any positive integer to be used as a base.
Help needed for alternate proof, & understand proof for any positive integer to be used as a base.

Problem statement:

Let $g$ be a positive integer $\gt 1$. Then every positive $a$ can be written uniquely in the form $$a = c_ng^n+\ldots+c_g+c_0, -(1)$$
where $n\ge 0$, $c_i$ is an integer, $0\le c_i \lt g$, $c_n\ne 0$. $g$ is called the base of $a,$ which is denoted by $(c_nc_{n-1}\ldots c_1c_0)_g$.

Proof: Using induction on $a$. When $a=1$, have $n=0$ & $c_0=1$. So true for $a=1$.

Now assume that the theorem is true for any integer less than $a$. Since $g\gt 1, a\gt 0, a$ must lie between two certain consecutive numbers of the following sequence:
$$g^0,g^1,g^2,\ldots,g^n,\ldots$$
More explicitly, there is a unique integer $n,$  s.t. $$g^n\le a\lt g^{n+1}$$
By divisibility theorem, can state: $$a=c_ng^n+r, 0\le r\lt g^n,$$
obviously, $g\gt c_n \gt 0$. If $r=0$, then $$a=c_ng^n+0g^{n-1}+\ldots +0g+0.$$
If $r\ne 0$, by induction hypothesis $$r=b_tg^t+\ldots+b_g+b_0, t\lt n,$$
where $0\le b_i \lt g.$ Thus $$a=c_ng^n+b_tg^t+\ldots+b_1g+b_0,$$ and the given statement $(1)$ is true.

To prove the uniqueness, assume there is another representation $$a=d_mg^m+\ldots+d_1g+d_0,-(2)$$
with $m\ge 0, 0\le d_i\lt g.$ If $c_i$ & $d_i$ are not all equal, by subtracting $(1)$ from $(2)$, get $$0=e_sg^s+\ldots +e_1g+e_0,$$
where $s$ is the largest value of $i$ for which $c_i\ne d_i,$ so that $e_s\ne 0$. If $s=0$, then $c_1=c_0=0$, which is a contradiction. If $s\gt 0$, $$|e_i|=|c_i-d_i|\lt g-1, i=0,\ldots, s-1,$$
and $$e_sg^s=-(e_{s-1}g^{s-1}+\ldots+e_0),$$
so that $$g^s\lt |e_sg^s|=|e_{s-1}g^{s-1}+\ldots+e_0| \lt (g-1)(g^{s-1}+\ldots+g+1)= g^s -1,$$
which is also a contradiction. Thus conclude $$n=m, c_i=d_i, i=0,1,\ldots,n,$$ as well as the representation is unique.

I feel that the inductive proof is unclear on how it has used the second step (i.e., is true for any value for $n=n'$) of weak-induction to get to the final step.

Also, the proof part for showing uniqueness is inelegant, as per me.

I request an alternate proof (or its source), & have found only one other post on topic on MSE. That post did not address the uniqueness aspect also, although was non-inductive.

答:
Here is a sketch of an alternative proof.

Let $\beta \ge 2$ and define $D \subset \Bbb N$ by writing

$\quad D = \{0,1\} \text{ when } \beta = 2 \quad \text{else } D = \{0,1,\dots,\beta-1\}$

Let the integer $a$ have a $\text{base-}\beta$ representation,

$\tag 1 \displaystyle \large a = \sum_{k = 0}^{+\infty} a_k \, \beta^{k} \quad \text{where } a_k \in D$

Let $j \ge 0$ and write as true

$\quad a_j + 1 = r \text{ where } 0 \le r \lt \beta \quad \text{XOR} \quad a_j + 1 = \beta + s \text{ where } 0 \le s \lt \beta$

Define the subscripted family $\large ({a'}_k)_{k \ge 0}$

$\quad {a'}_j = r \; \text{ XOR } \; {a'}_j = s \quad \text{AND}  \quad {a'}_k = a_k \text{ when } k \ne j$

It is elementary to write as true

$\quad \displaystyle \large  \beta^j  + a = \sum_{k = 0}^{+\infty} {a'}_k \, \beta^{k} \; \text{ (the term is 'absorbed')}\quad \text{ XOR } \quad \beta^j  + a  = \beta^{j+1} + \sum_{k = 0}^{+\infty} {a'}_k \, \beta^{k}  \; \text{ (we have to 'carry' the $1$ to the next exponent)}$

Since the integer $\large  \beta^j  + a$ is bounded above, we conclude that it must also have a representation

(eventually the '$\beta \text{ to a power}$' term can be 'folded' into the representation).

Since any $\text{base-}\beta$ representation can be expressed as a sum of these primitive $\beta^j$ terms,

we've demonstrated the following result.

Proposition 1: The sum of any two $\text{base-}\beta$ representations also has a representation.

Using proposition 1,

Proposition 2: There exist a $\text{base-}\beta$ representations for every natural number.

Proof

This is immediate since any submonoid of $(\Bbb N,+)$ containing $1$ contains all the natural numbers. $\quad \blacksquare$

If $a$ satisfies $a_k = 0$ for $k \ge j$ then it is not difficult to show that

$\quad a \lt \beta^j$

Using this fact and a proof by infinite descent, one also get the uniqueness.

Proposition 3: Any two $\text{base-}\beta$ representations of a natural number are identical.


Title: Logarithm question involving different base

Body: Calculate the values of $z$ for which $\log_3 z = 4\log_z3$.</p>

Body:
hint :

$$\frac{\log (z)}{\log (3)}=4\frac{\log (3)}{\log (z)}$$

$$(\log (z))^2=4(\log (3))^2$$


问:
Does this Make Sense ? Riemannian Manifold Directional Derivative.
Take a Riemannian Manifold $(M,g)$, where the tangent space at $p\in M$ is denoted $T_p M$. Let $f:M\to\mathbb{R}$.

Does it make sense to ask for the directional derivative of $f$ at $p$ in direction $v\notin T_p M$?

I'm guessing no?

答:
This only makes sense if $f$ is a function on some manifold $N$, for which $M$ is a submanifold, but at that point, you don't even need to reference the space $M$ so it doesn't really make too much sense.

If $f$ were some function only on $M$ there is a non-unique extension of $f$ to some neighborhood of $M$ in $N$ for which you could certainly talk about a directional derivative in a direction non-tangent to $M$ but tangent to $N$, however, the non-uniqueness of this extension likely makes the result non-consistent based on your choice of extension.


问:
In triangle $ABC$, $AA_1$, $BB_1$, $CC_1$ divide sides in ratio of $1: 2$ and meet at $M$, $K$, $L$. Find area relation of $KLM$ and $ABC$

Points $A_1$, $B_1$, $C_1$ divide sides $BC$, $CA$, $AB$ equilateral triangle $ABC$ in a ratio of $1: 2$.
The line segments $AA_1$, $BB_1$, $CC_1$ determine the triangle $KLM$.

Is the triangle $KLM$ also an equilateral side?
In what relation are the area of triangle $KLM$ the area of triangle$ ABC$?

My attempt:

I can see that $KLM$ is an equilateral triangle.  But, why the fraction $\frac{1}{7}$ is the area ratio of the triangle $KLM$ to the triangle $ABC$?

答:
According to Routh’s theorem
the ratio of the areas in general case is

\begin{align}
\frac{S_{KLM}}{S_{ABC}}
&=
\frac{(xyz-1)^2}{(xy+y+1)(yz+z+1)(zx+x+1)}
.
\end{align}

The question is a special case, for which
$x=y=z=\tfrac12$, so

\begin{align}
\frac{S_{KLM}}{S_{ABC}}
&=
\frac{(\tfrac18-1)^2}{(\tfrac14+\tfrac12+1)^3}
=\tfrac17
.
\end{align}


问:
Which convex shapes are the hardest to bind together with a rubber band?
Suppose I have a convex set $S\subset \mathbb{R}^2$ of unit area. In fact, I have two congruent copies of $S$ which I would like to bundle together with a rubber band, i.e. take the convex hull $C$ of a disjoint union of these two copies. Assuming I do this in an area-minimizing way, what is the worst case scenario for the area $A$ of the excess space $C\setminus(S\sqcup S')$? Which shapes attain or approach this upper bound?

I can prove that one always has $A \le 1$, by inscribing the set in a rectangle with its diameter spanning the long axis, placing the two rectangles side-to-side, and shaving the excess off of the outer end of each rectangle.

It's easy to see that a circle forces $A=\frac{4-\pi}{\pi}\approx 0.2732$. I believe that a hexagon yields $A=1/3$, as realized by either of these configurations:

Are there shapes that do worse? Better upper bounds on $A$? In the event that this question is resolved, what about the case of three dimensions or higher?

In the case when both copies have the same orientation, and are simply translations of one another, I can prove $A$ is equal to the excess area of the smallest parallelogram containing $S$. (As a consequence, better upper bounds on this excess area provide bounds on $A$.)

These questions seem adjacent to questions of packing density, as they approach such behavior in the limit as the number of copies ($2$ here) goes to infinity.

Update: Regular pentagons sharing an edge yield $A=\frac3{\sqrt{5}}-1\approx0.3416$. I haven't proved this is optimal, though it's superior to meeting at a vertex; it's possible some better pentagon packing yields a lower $A$ (would love to see one if so!).

Update 2020-10-28: After writing some code to compute excess areas for arbitrary convex polygons, I have become more optimistic that the pentagon (and some affine transformations thereof) are maximal for this problem; at the least, I do not think there are any local variations to the shape which make it harder to pack with itself. (I can also use this code to try various families of convex regions, and see if any exceed $0.3416$ - suggestions welcome.)

答:
By a 1973 result of G. D. Chakerian (PDF link), we know that any convex set $S$ of unit area is contained in a convex quadrilateral of area at most $\sqrt{2}$. It is easy to see that in any convex quadrilateral, there is a side with two adjacent angles whose sum is at most $180$ degrees. (In fact, there are always at least two such sides.)

Then, by taking this quadrilateral and its $180^\circ$ rotation about the midpoint of such a side, we obtain a convex set of area $2\sqrt{2}$ containing two disjoint copies of $S$. So the excess area of the two copies in this configuration is at most $2\sqrt{2}-2\approx 0.8284$.

It is conjectured by W. Kuperberg that the minimal quadrilateral can have area at most $\frac{3}{\sqrt{5}}$, attained by the regular pentagon; if so, this would reduce the bound to $\frac{6}{\sqrt{5}}-2\approx 0.6833$.

I suspect that at least in the $\sqrt{2}$ case, one may be able to refine this result slightly and show that any convex shape taking up at least $\frac1{\sqrt{2}}-\epsilon$ of the quadrilateral can be packed more efficiently somehow, but my attempts thus far haven't proven fruitful.


答:
Hint

$$\log_3 z = 4\log_z3 \Rightarrow \dfrac{\log z}{\log 3}=4\dfrac{\log 3}{\log z}$$

Answer (don't look until you try the hint)
$$\color{lightgrey}{\dfrac{\log z}{\log 3}=4\dfrac{\log 3}{\log z}\Rightarrow (\log z )^2=4(\log 3)^2 \Rightarrow \log z =2\log 3 \Rightarrow \log z =\log 3^2 \Rightarrow z = 3^2}$$

$\color{lightgrey}{\text{Therefore} \ z=9.}$


问:
Integral from infinite sum
I am trying to find if there is a way of turning the following sum into an integral:
$$\lim_{x\to\infty} \sum_{k=1}^{x}\operatorname{tr}\left(\frac{C^{k}}{x^2+x}\right)$$
I have looked into Riemann sums however I wasn't able to do it with this sum, however this could well be I didn't understand them fully.

答:
This is a geometric series. Assuming that $\|C\|<1$,
$$\begin{split}
\lim_{x\rightarrow+\infty}\sum_{k=1}^{x}\operatorname{tr}\left(\frac{C^{k}}{x^2+x}\right) &= \lim_{x\rightarrow+\infty}\frac 1 {x^2 + x}\operatorname{tr}\left(\sum_{k=1}^{x}C^{k}\right)\\
&= \lim_{x\rightarrow+\infty}\frac 1 {x^2 + x}\operatorname{tr}\left(C(I-C^x)(I-C)^{-1}\right)\\
&= \operatorname{tr}\left(C(I-C)^{-1}\right)\lim_{x\rightarrow+\infty}\frac 1 {x^2+x}\\
&=0
\end{split}$$

Title: Integral from infinite sum


问:
Calculating the order of permutation subgroup generated by 3 elements
I’m calculating the automorphism group of $\mathrm{GP}(8,3)$ (generalized Petersen graph GP(8,3), but it doesn’t matter for this question). Let $A := \operatorname{Aut}(\mathrm{GP}(8,3))$.

I know $A \leq S_{16}$, $|A| = 96$, and that $\alpha, \beta, \gamma \in A$.

$\alpha = (1, 7, 8)(2, 15, 13)(3, 12, 5)(6, 11, 9)$

$\beta = (7, 1)(6, 2)(5, 3)(15, 9)(14, 10)(13, 11)$

$\gamma = (0, 1, 2, 3, 4, 5, 6, 7)(8, 9, 10 ,11, 12 ,13, 14, 15)$

I can see why $|\langle \alpha, \beta, \gamma \rangle| \geq 24$, since $\alpha^3 = \beta^2 = \gamma^8 = 1$.

Is there any way to show that $|\langle \alpha, \beta, \gamma \rangle| = 96$, which would mean $\langle \alpha, \beta, \gamma \rangle = A$ ?

答:
It is easy to see that the group $G=\langle\alpha,\beta,\gamma\rangle$ acts transitively on 16 points. Now note that
$$\delta=\alpha\beta=(2,9)(3,12)(6,13)(7,8)(10,14)(11,15)
$$
and
$$
\epsilon=\alpha\gamma^2=(2,9,16)(3,14,8)(4,6,13)(7,10,12)
$$
(I multiply permutations left to right, otherwise you need to reverse the product order) are both in the stabilizer of $1$, and have orders $2$, respectively $3$. Thus the stabilizer has at least order $6$, and the whole group at least order $6\cdot 16=96$.


问:
Convergence of alternating series involving a squared sine
I need to determine whether the following series converges or not:

$\sum_{n=1}^\infty((-1)^{n}(\sin(\frac{\pi}{n}))^2 $

I thought to use the Alternating Series Test, but ran in to two problems:

After rewriting the series as $\sum_{n=2}^\infty((-1)^{n}(\sin(\frac{\pi}{n}))^2 $ to ensure the sine squared sequence was monotonically decreasing, I had trouble proving that it was actually decreasing. I thought of using a sketch but I don't think that would be rigorous enough.

I had a problem showing the sine squared sequence was null - it seems obvious that:

$\lim_{n\to \infty} (\sin(\frac{\pi}{n}))^2$         $= (\sin(\lim_{n\to \infty} \frac{\pi}{n}))^2 $ = $0$

but I am not sure how to justify moving the limit inside the sine.

答: First of all, note that$$(\forall n\in\Bbb N):\left((-1)^n\sin\left(\frac\pi n\right)\right)^2=\sin^2\left(\frac\pi n\right).$$So, your series is not alternating. And, since\begin{align}\lim_{n\to\infty}\frac{\sin^2\left(\frac\pi n\right)}{\frac1{n^2}}&=\pi^2\lim_{n\to\infty}\frac{\sin^2\left(\frac\pi n\right)}{\frac{\pi^2}{n^2}}\\&=\pi^2\lim_{n\to\infty}\left(\frac{\sin\left(\frac\pi n\right)}{\frac\pi n}\right)^2\\&=\pi^2\left(\lim_{n\to\infty}\frac{\sin\left(\frac\pi n\right)}{\frac\pi n}\right)^2\\&=\pi^2\left(\lim_{x\to0}\frac{\sin x}x\right)^2\\&=\pi^2\end{align}and since the series $\sum_{n=1}^\infty\frac1{n^2}$ converges, your series converges too.


答:
2: You can move the limit inside $\sin$ because it is continuous, as is $x \mapsto x^2$.

1: Note that $\frac{\pi}{n}$ is "obviously" decreasing, so it suffices to show that $\sin(x)^2$ is increasing on some interval containing all but finitely many terms of the sequence $\frac{\pi}{n}$.

That being said, if there is no typo, then the alternating series test doesn't actually apply because $(-1)^n$ is getting squared too. To deal with that, you'll want to use a comparison test.

Title: Convergence of alternating series involving a squared sine

Body:
I need to determine whether the following series converges or not:

$\sum_{n=1}^\infty((-1)^{n}(\sin(\frac{\pi}{n}))^2 $

I thought to use the Alternating Series Test, but ran in to two problems:

After rewriting the series as $\sum_{n=2}^\infty((-1)^{n}(\sin(\frac{\pi}{n}))^2 $ to ensure the sine squared sequence was monotonically decreasing, I had trouble proving that it was actually decreasing. I thought of using a sketch but I don't think that would be rigorous enough.

I had a problem showing the sine squared sequence was null - it seems obvious that:

$\lim_{n\to \infty} (\sin(\frac{\pi}{n}))^2$         $= (\sin(\lim_{n\to \infty} \frac{\pi}{n}))^2 $ = $0$

but I am not sure how to justify moving the limit inside the sine.


问:
Is AC required to demonstrate that every collection of disjoint open intervals in R is countable?
I understand that to prove this, one considers that each open interval contains a rational number and since the open intervals are disjoint, there exists an injection from the collection of intervals to the rational numbers, which are countable, hence the collection of open intervals is countable.

I noticed, however, that this requires arbitrarily many simultaneous, non-defined, choices from the intersection of the rational numbers with each interval in the collection. So, my question is, does this proof require AC?

答: No, we don't need the axiom of choice here. Take any bijection $f:\mathbb{Q}\to\mathbb{N}$, and for each interval in your collection choose the rational number $q$ in the interval for which $f(q)$ has the smallest value. This is a very specific choice function.


问:
When do we consider a boolean expression "fully simplified"? (Eg, is $B'+ BA'C$ the simplest form of $AB'C + A'BC+ A'B'C+A'B'C'+AB'C'$?)
Edit:

I have simplified a boolean expression (see below) and there still some terms who share the same element (it is very hard to simplify more). So,

When do we consider a boolean expression to be "fully simplified" (can't be simplified anymore)?

Here is an example:

$$AB'C + A'BC+ A'B'C+A'B'C'+AB'C'$$
where ($'$) means "not". For this example, I think the expression can not be simplified more than this: $$B'+ BA'C$$

So, is there any way to make sure the expression can't be simplified any more?

Thanks in advance

答:
Here is a Vietch Diagram supporting the logic of reduction to   $\quad\textbf{B'+A'C}$

Note how I labeled the (A,A), (B,B'), (C,C') cells outside the big rectangle so they overlap. The result you seek comes from taking the largest rectangular area(s) found after entering the data.

Note the rectangle for terms $(5,1,3,4)\implies B')\quad$ and $\quad (2,3)\implies A'C$.

I did this in EXCEL, entering, for example $\quad 1(AB'C)\quad$ because it let me keep track of which terms had been entered, and verify they afterward. Note, If $A'C$ had been in the original terms, you would have had  to enter it in the cells currently occupied by $2$ and $3$. You'll have to develop your own means of keeping track of stuff. This just works for me and I'm $70$.  I hope this helps explain the usage of Vietch diagrams.


问:
If we have a minimal sufficient statistic for a parameter, is this also a minimal sufficient statistic for any function of the parameter?
If we know that $\theta^{*}$ is a minimal sufficient statistic for a certain parameter $\theta$ , is it true that $\theta^{*}$ will also be a minimal sufficient statistic for any one-to-one function of $\theta$, $f(\theta)$?</p>

答:
Denote $\eta = f(\theta)$. $f$ is one-one.

For any sufficient statistic $T$ of $\eta$ we have

\begin{align*}
T(x)=T(y)\implies\delta_\eta(x)&=\delta_\eta(y) \\
\implies\delta_{f(\theta)}(x)&=\delta_{f(\theta)}(y)\\
\implies\delta_\theta(x)&=\delta_\theta(y)\\
\end{align*}

(The last $\implies$ holds because we can treat $\delta$ a function of $\theta$)

Thus $T$ is sufficient for $\theta$. Similarly, $\theta^*$ is sufficient for $\eta$.

Since $\theta^*$ is a minimal sufficient statistic of $\theta$ and $T$ sufficient for $\theta$, $\theta^*$ is a function of $T$.

This means for any $T$ sufficient for $\eta$, and $\theta^*$, being sufficient for $\eta$, is a function of $T$, thus $\theta^*$ is minimal sufficient for $\eta$.


问:
Show convergence for this sequence only by using the definition
I need to prove convergence for
$(b_n)_{n ∈ ℕ}=\left(\frac{(-1)^nn}{2n+1}\right)_{n∈ℕ}$ and also show the limit.

I may only use the following definition:  $∀ɛ > 0∃n_0∈ℕ∀n≥n_0:|a_n-a|< ɛ$.

So far we've been "guessing" $a$ as the limit (which should be in this case$\left(\frac{1}{2}\right)$) an then doing $|a_n-a|< ɛ$ for $a_n$ and $a$ as we had both of them.

My problem is, that I'm afraid it won't end well doing it like this. First, because it's an alternating sequence and I'm not even sure if the limit is simply $1/2$ or $+-1/2$ in which case I would be even more clueless on how to do it.
How should I start and how proceed? Do I maybe need to use Triangle inequality?

答:
The sequence clearly diverges, as already noted. To show this "using the definition", we need to find an $\varepsilon > 0$ such that for every $n_0$ there is an $n \ge n_0$ such that $|a_n - a| > \varepsilon$ no matter what $a$ we choose. (Check to see that you understand why.)

I'll split the argument into three cases. Assume $a > 0$ and take $\varepsilon = a$. Then for all odd values of $n$, $a_n < 0$, so $|a_n - a| > a = \varepsilon$. Similarly, if $a < 0$, we can choose even values of $n$.

The only remaining possibility is $a = 0$, but
$$\left|\frac{(-1)^n n}{2n+1} \right| = \frac{n}{2n+2} > \frac{n}{2n+n} = \frac13$$
if $n > 2$. This shows that we can take $\varepsilon = \frac13$ to see that $a$ can't be $0$ either.


答:
Because
$$\left| \sum\limits_{k=1}^{n} (-1)^k \right |\leqslant 1$$
and functional sequence $\frac{1}{x+n}$ uniformly with respect to $x$ tends to zero, then according to Dirichlet test series is uniformly converged, which gives continuity for every $x \in (0, +\infty)$.

Then, as formal derivative
$$\sum\limits_{k=1}^{n} \frac{(-1)^{k+1}}{(x+n)^2} $$
also is converged uniformly by $\left| \frac{(-1)^{k+1}}{(x+n)^2}\right | \leqslant \frac{1}{n^2}$, then we can state that $f$ have derivative for every $x \in (0, +\infty)$ and holds
$$f'(x)=\sum\limits_{k=1}^{\infty} \frac{(-1)^{k+1}}{(x+n)^2}$$
(btw last gives continuity as well).

Title: Continuity of $f(x)=\sum_{n=0}^{\infty} (-1)^n \frac{1}{x+n}$

Body:
I am trying to solve the following exercise and I would like to have a hint about the continuity part:

Let $f(x)=\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+2}-\frac{1}{x+3}+\frac{1}{x+4}-\dots$.

Show $f$ is defined for all $x>0$. Is $f$ continuous on $(0,\infty)$? How about differentiable?

My solution (NOTE: completed the exercise):

f defined We can rewrite $f$ as $f(x)=\sum_{n=0}^{\infty} (-1)^n \frac{1}{x+n}$ which converges by the Alternating Series Test so the function is surely defined for all $x>0$.

f continuous Now, let $\varepsilon>0$: then if we take $N>\frac{1}{\varepsilon}$ we have, for all $n,m\geq N, n>m$ and $x>0$ $|\frac{(-1)^{m+1}}{x+(m+1)}+\frac{(-1)^{m+2}}{x+(m+2)}+\dots+\frac{(-1)^n}{x+n}|\leq\frac{1}{x+(m+1)}<\frac{1}{x+m}<\frac{1}{m}<\varepsilon$
(note that wheter $m$ is even or odd makes no difference since $|-x|=|x|$)
so the series converges uniformly Cauchy Criterion for Uniform Convergence of Series and $f$ is thus continuous by Term-by-Term Continuity Theorem.

f differentiable  $|f'_n(x)|=|\frac{(-1)^{n+1}}{(x+n)^2}|=\frac{1}{(x+n)^2}\leq\frac{1}{n^2}$ and $\sum_{n=0}^{\infty}\frac{1}{n^2}$ is convergent so the series of derivatives $\sum_{n=0}^{\infty}\frac{(-1)^{n+1}}{(x+n)^2}$ is uniformly convergent on $(0,\infty)$ by Weierstrass M-Test and in particular on any interval $[a,b]\subset (0,\infty)$; also, since for any $x_0\in [a,b]$, $\sum_{n=0}^{\infty}f_n(x_0)$ is convergent by the Alternating Series Test we can conclude, by the Term-by-Term Differentiability Theorem, that on any interval $[a,b]\subset (0,\infty)$, $\sum_{n=0}^{\infty}f_n(x)$ converges uniformly to a differentiable function $f(x)=\sum_{n=0}^{\infty}f_n(x)=\sum_{n=0}^{\infty} (-1)^n \frac{1}{x+n}$ such that $f'(x)=\sum_{n=0}^{\infty}f'_n(x)=\sum_{n=0}^{\infty}\frac{(-1)^{n+1}}{(x+n)^2}$.

Thank you.


问:
Formally assigning probabilities to sample spaces that change over time
As mentioned in the title, how are probabilities assigned to changing sample spaces?

To make my question more clear, take the classic Polya's Urn Scheme, where we have an urn containing $b$ black balls and $r$ red balls. After each draw of a ball from the urn, the ball is replaced with $c$ more balls of the same color back into the urn. Regarding the first draw, I'm clear..since each ball is equally likely to be picked, a uniform probability space $(\Omega, \mathcal{F}, \mathbb{P})$ does the job once $\Omega$ is taken to be the set of all balls in the urn.

After the first draw, or any draw for that matter, the set of balls in the urn is no longer the same as $\Omega$. But any solution to this problem that I've seen does not mention this..and the answer is computed directly. Of course, its intuitively clear what's going on..but is there a more formal way of depicting this? Do we move to a new sample space $(\Omega', \mathcal{F'}, \mathbb{P}')$ after each draw?

What is the general way of working with sample spaces that change over time / after the occurrence of some events?

答:
I think the typical way to do this would be to set $\Omega := \{B,R \}^{\mathbb{N}}$, i.e. $\Omega$ is the set of all sequences of black and red ball draws.  Then $\mathcal F_n$ would be the filtration generated by the first $n$ draws, and $\displaystyle \mathcal F = \mathcal F_\infty = \bigvee_{n=1}^{\infty} F_n$.

Defining the probability measure $\mathbb{P}$ is slightly trickier, and relies on the Kolmogorov extension theorem.  Essentially we define $\mathbb{P}_n$ on $(\Omega, \mathcal F_n)$ to be the correct marginal distribution for the first $n$ tosses, so if $\omega = \omega_1\omega_2 \cdots \omega_n \cdots \in \Omega$ then $\mathbb{P}_n$ is the probability of getting the sequence $\omega_1\omega_2\cdots \omega_n$ in the first $n$ tosses.  That probability is probably slightly difficult to compute, but can be done recursively.  Then the Kolmogorov extension theorem guarantees the existence of a probability measure $\mathbb P$ on $(\Omega, \mathcal F)$ that agrees with the $\mathbb{P}_n$s on each $\mathcal F_n$.  Specifically, if $A \in \mathcal F_n$ then $\mathbb P(A) = \mathbb P_n(A)$.

So our probability space is $(\Omega, \mathcal F, \mathbb{P})$, and we keep the same $\Omega$ throughout.  This way we don't really have the probability space changing, because we think of the probability space as the entire sequence of draws.  The way we keep track of what we intuitively might think of as the "probability space changing" is with the filtrations $\mathcal F_n$ and the finite marginal probabilities $\mathbb{P}_n$.  The $\mathcal F_n$ represent the information we have after seeing $n$ draws, and the $\mathbb{P}_n$ assign probabilities to events that are observable within the first $n$ draws.  I think part of the reason this is confusing is because in order to actually do a Polya's Urn Scheme in the real world you would probably have to be changing what you're drawing from each time, but that's different from the way it is modeled.  That's because it would be quite complicated to model a randomly changing probability space, and we would have to do a lot more work to use theorems like Fatou's lemma or the dominated convergence theorem.

EDIT: For a simple example, suppose we only draw from the urn twice and we start with $1$ black ball and $2$ red balls and add $1$ ball at each step.  Then $\Omega = \{BB,BR,RB,RR\}$ is the set of possible sequences and serves as our sample space, our $\sigma$-algebras $\mathcal F_n$ (basically the elements we can tell apart after observing the first $n$ draws) are

\begin{align*}
\mathcal F_0 &= \{\emptyset, \Omega\} \\
\mathcal F_1 &= \{\emptyset, \Omega, \{BR,BB\}, \{RB,RR\} \} \\
\mathcal F_2 &= 2^\Omega.
\end{align*}

Our probability measure $\mathbb P$ is found by finding the probability of each sequence, so for example $\mathbb P(\{BB\}) = \frac 13 \cdot \frac 12 = \frac 16$.  If you wanted to, you could also define a probability measure $\mathbb{P}_1$ on $\mathcal F_1$ corresponding to the probability of the first draw, so

\begin{align*}
\mathbb{P}_1(\{BR,BB\}) &= \frac 13 \\
\mathbb{P}_1(\{RB,RR\}) &= \frac 23.
\end{align*}


问:
How to find global Maximum of this function.
This question which I am asking is to be used in another question and I am unable to do this. So, I am asking for help.

Consider the function$s_n(x)=\frac{2}{\pi} \int_{0}^{x} \frac{\sin (2nt)} {\sin (t)} dt $ . I proved that this function has Local Maxima at $x_1 ,x_3,x_5,...,x_{2n-1}$ where $x_m=\frac{m \pi } {2n}$.

But I am not able to prove that $s_n(\frac{\pi}{2n}) $is global maximium.

The method I thought was to compare all the local maximum but I am not able to compute the integral involving $s_n(x)$ .

I think that it has to be done in some other way( using some clever way and indirectly) and I am not able to figure out that .

So, Can you please help with that?

答:
Above Figure illustrates the graph of $f(t):=\frac{\sin(2nt)}{\sin(t)}$ on $[0, \pi]$.

Idea.
The idea to achieve the goal is very simple
thank to the following geometric observations:

Obs1. $f_n$ is symmetric with center $x=\pi/2$.

Obs2. Obs1 implies that $s_n$ is symmetric with axis $x=\pi/2$. Thus, it is sufficient to prove that $s_n( \frac{\pi}{2n} )$ is the maximum of $s_n$ on $[0, \frac{\pi}{2}]$, i.e. prove that it is the maximum of the set $\{ s_n( \frac{i\pi}{2n}) : i=1, ... n\}$.

Obs3. The area of $f_n$ over $[\frac{i\pi}{2n}, \frac{(i+1)\pi}{2n}]$, denoted by $s_n^i$, is decreasing for $i=0, ..., n-1$.

Obs4.
\begin{align}
s_n( \frac{\pi}{2n} )  & = s_n^1\\
s_n( \frac{2\pi}{2n} ) & = s_n^1-s_n^2\leq s_n^1\\
s_n( \frac{3\pi}{2n} ) & = s_n^1-(s_n^2-s_n^3) \leq s_n^1\\
s_n( \frac{4\pi}{2n} ) & = s_n^1-(s_n^2-s_n^3) -s_n^4 \leq s_n^1\\
.....
\end{align}

Hence we conclude that $s_n^1$ is the global maximum and $\frac{\pi}{2n}$ is global maxima on $[0,\pi]$.

Based on geometric properties of $f_n$, similar arguments can be applied to prove that $\frac{\pi}{2n}$ is global maxima on $[0,2\pi]$. Since$f_n$ is periodic $2\pi$, thus $\frac{\pi}{2n}$ is global maxima on $\mathbb R$.

We now move into the details.

Proof of Obs1. We need to show $f_n(x)=-f_n(\pi-x)$. Indeed,
\begin{align}
f_n(\pi-t)     & = \frac{\sin(2n(\pi-t))}{\sin(\pi-t)}\\
           & = -\frac{\sin(2nt)}{\sin(t)}\\
           & = -f_n(t).
\end{align}

Proof of Obs2.
We need to show $s_n(x)=s_n(\pi-x)$.
Indeed,
\begin{align}
s_n(x)     & = \int_0^{x}  f_n(t) dt\\
           & = \int_0^{x} -f_n(\pi-t) dt\\
           & = \int_{\pi-x}^\pi -f_n(t) dt\\
           & = \int_0^{\pi-x} f_n(t) dt\\
           & = s_n(\pi-x).
\end{align}
Here the fourth equation follows from the fact that $\int_0^\pi f_n(t)dt=0$ as $f_n$ is symmetric with center $x=\frac{\pi}{2}$.

Proof of Obs3.
Let $x_i=\frac{i\pi}{2n}$ for $i=1, n$.
\begin{align}
s^i_n      & = \int_{x_i} ^{x_{i+1}} |f_n(t)| dt\\
           & = \int_{x_i} ^{x_{i+1}}  \frac{|\sin(2nt)|}{\sin(t)} dt\\
           & = \int_{x_i+x_1} ^{x_{i+1}+x_1}  \frac{|\sin(2n(t+x_1))|}{\sin(t+x_1)} dt\\
           & = \int_{x_{i+1}} ^{x_{i+2}}  \frac{|\sin(2nt)|}{\sin(t+x_1)} dt\\
           & < \int_{x_{i+1}} ^{x_{i+2}}  \frac{|\sin(2nt)|}{\sin(t)} dt\\
           & = s^{i+1}_n
\end{align}
Here we have used $\sin(t+x_1)>\sin(t)$ on $[0, \frac{\pi}{2}]$.

Proof of Obs4. This is easy because we just need to investigate the sign of $f_n$. Note that we are working on $[0, \frac{\pi}{2}]$ and $\sin(t)$ is positive now.


问:
Is my solution correct? Generating functions question: How many non-negative solutions does the equation $x_1+x_2+x_3+x_4+x_5+x_6=12$ have?
so we began studying this subject, and I tried solving this question:  How many non-negative and whole ($\in \Bbb Z$) solutions does the equation $x_1+x_2+x_3+x_4+x_5+x_6=12$ have?

I would like to know if my solution is correct.

For non negative solutions it means $0\leq x_1, x_2, x_3, x_4, x_5, x_6$, so I built the next function:

$A(x)=(1+x^1+x^2+x^3+...)^6$, and due to the formula of the sum of an infinite geometric sequence:

$A(x)= {1 \over (1-x)^6}= \sum_{n=0}^∞ {n+6-1 \choose 6-1}x^n$.

Since we need the sum of 12, we need the coefficient of $x^{12}$, so I place n=12, and got the number of solutions to be 6188.

Now this seemed a bit weird to me. Is it so?

Thanks for any input!

答:
Your method of solving this is correct. It comes down basically to remembering the formula for the coefficients of $(1-x)^{-n}$. If you just call the coefficient of $x^k$ in that power series $(-1)^k\binom{-n}k$ (after all the coefficient of $x^k$ in $(1-x)^m$ is $(-1)^k\binom mk$ for $m\geq0$ too), then these new coefficients continue to satisfy
$$
  \binom nk=\frac{n\times(n-1)\times\cdots\times(n-k+2)\times(n-k+1)}
    {k\times(k-1)\times\cdots2\times\times1}
\qquad\text{for all $n$,}
$$
and therefore in particular
$$
\binom {-n}k=\frac{-n\times(-n-1)\times\cdots\times(-n-k+2)\times(-n-k+1)}
    {k\times(k-1)\times\cdots2\times\times1}
=(-1)^k\frac{n\times(n+1)\times\cdots\times(n+k-2)\times(n+k-1)}
    {k\times(k-1)\times\cdots2\times\times1}
=(-1)^k\binom {k+n-1}k.
$$
Therefore $(-1)^k\binom{-n}k=\binom {k+n-1}k=\binom {k+n-1}{n-1}$ which is (probably) your formula for the coefficients. (Be warned that although the first transformation is always valid, the symmetry of binomial coefficients used in the second step is only valid if the top index, here $k+n-1$, is non-negative! So there was no possibility if applying symmetry before the first transformation.)

What I'm saying is basically that you can remember the identity for all $n$ and for $k\in\mathbf N$ $\binom {-n}k=(-1)^k\binom {k+n-1}k$ (or if you prefer it in the form $\binom nk=(-1)^k\binom {k-n-1}k$) instead of explicitly remembering the formula for the coefficients of $(1-x)^n$ (the first identity, and more generally the use of binomial coefficients with "strange" upper indices, is a very useful tool, but there is no harm in remembering the latter as well, as it comes in handy rather often.).


问:
How many N-digit numbers have the sum of odd places digits equal the sum of even places digits?
We define a N-digit number as a sequence $d_{1}d_{2}d_3...d_N$, where $d_k\in\{0,1,2,3,4,5,6,7,8,9\}$. So the first digits can be equal to zero. N is a positive even integer.
The problem is to find, how many N-digit numbers are there, which have their sum of the digits on odd places equal to the sum of digits on the even places. Example number would be 2563 for 4-digit numbers as 2 + 6 = 5 + 3.

My thought was to choose the odd digits, we can do that in $10^{N/2}$ ways and then for each of that combinations make even digits a permutation of them ($\frac{N}{2}!$), but this, if correct at all, is only a small fraction of all the possibilities. Why? Because the digits do not have to be the same. Only the sums have to be equal.

Then my second though was that each of such numbers consists of digits from other (shorter) numbers that satisfy the requirement, but it is not true since only the sums have to be equal.

We can for each sum of odd/even digits calculate how many ways we can express it on N/2 digits and then sum the squares of such numbers of combinations. However that leads to very unfriendly calculations.

After you find a nice way, please provide some example calculations for N=4 and N=6.

答:
Hint:
The key to solve this is to know what the digits add up to say a number $k$. Iterate over it and use stars and bars with inclusion exclusion to get the number of compositions of $k$ where the parts of the compositions are bounded by $9.$ Call this number $\mathcal{A}_k,$ then the answer is
$$\sum _{k=0}^{9\cdot N}\mathcal{A}_k^2.$$
To compute $\mathcal{A}_k$ notice first that if we allow "digits" greater than $9$ then it would be $\mathcal{B}_k=\binom{k+N/2-1}{k}.$ So consider set $B_{k,i}$ where the $i-th$ digit is greater than $9$ then you want to compute
$$\left |\mathcal{B}_k\setminus \bigcup _{i=1}^{N/2}B_{k,i}\right |.$$
Use the principle there to conclude with a not very nice formula.