答:
Well, first of all you mean two maximal $k$-connected subgraphs $G_1$ and $G_2$;
as in $G_1$ and $G_2$ are $k$-connected but there is no subgraph $G'_1 \supset G_1$ of $G_1$ nor is there a subgraph $G'_2 \supset G_2$ of $G_2$ that is $k$-connected.

The result you want is Claim 1 below:

Claim 1: Suppose that $G_1$ and $G_2$ share $k$ or more vertices. Then $G_1 \cup G_2$ is $k$-connected, which contradicts that $G_1$ and $G_2$ are maximal $k$-connected graphs.

Indeed, we show that $G_1 \cup G_2$ is still connected even after removing any set $S$ of $k-1$ vertices if $G_1$ and $G_2$ intersect on a set $U$ of $k$ or more vertices. Indeed, let $S$ be any set of $k-1$ vertices of $G_1 \cup G_2$. Then $G_1 \setminus S$ is still connected and $G_2 \setminus S$ is still connected [why is this]. And so for any vertex $w$ in $U \setminus S$ then there is a path in $G_1 \setminus S$ from $w$ to any vertex in $G_1 \setminus S$, and there is a path in $G_2 \setminus S$ from $w$ to any vertex in $G_2 \setminus S$, so $(G_1 \cup G_2) \setminus S$ is still connected [make sure you see why]. Thus indeed $G_1 \cup G_2$ is still connected even after removing any set $S$ of $k-1$ vertices if $G_1$ and $G_2$ intersect on a set $U$ of $k$ or more vertices.

Title: length of common cycle in two k-connected subgraphs of a graph

Body:
In a given graph $G$, $G_1$ and $G_2$ are maximal k-connected subgraphs, I have to prove that the length of the common cycle is atmost $K$ i.e. they have atmost $k-1$ common nodes.

I am unsure on how to get started in the proof, any help and related links are appreciated.


问:
Show the existence and evaluate $F'(t)$ knowing $F(t)$
We are given the following expression :

$$F_1(t) = \int_{0}^{\pi}{\frac{\cos(x)dx}{(1+t\cos(x))^2}}$$
$$F_2(t) = \int_{0}^{1}{\frac{\log(1+tx)}{1+x^2}}dx$$

Show the existence and evaluate $F_1'(t)$ and $F_2'(t)$

I don't know how I can prove the existence of both  $F_1'(t)$ and $F_2'(t)$.

To evaluate $F_1'(t)$ and $F_2'(t)$, I thought that using the partial fraction decomposition method could help us but I'm not sure.

Any help would be a lot appreciated, thanks in advance.

答:
As the integrand is $\mathcal{C}^1([0,1])$, if $t>-1$ we can use the Leibniz integral rule and get
$$F'_2(t)=\int_0^1 \frac{\partial \frac{\log (t x+1)}{x^2+1}}{\partial t} \, dx=\int_0^1 \frac{x}{\left(x^2+1\right) (t x+1)} \, dx=\frac{\pi  t-4 \log (t+1)+\log (4)}{4 t^2+4}$$

ContentLicense: CC BY-SA 4.0


问:
How many vertices of degree 1 in a tree?
How many vertices of degree 1 are there in a tree with no vertices of degree more than 4?
The only thing that I have right now is that the number of edges in a tree is n-1 where n is the number of vertices. Any help would be appreciated. Thanks in advance.

ContentLicense: CC BY-SA 3.0

Body:
If there are $n_i$ vertices of degree $i$, $1\le i\le 4$, then we must  have $$n_1+2n_2+3n_3+4n_4=2(n-1)=2(n_1+n_2+n_3+n_4-1),$$
hence $$n_1=n_3+2n_4+2. $$
Especially,
$$ 2\le n_1\le \frac{2n+2}3.$$
(The latter from $3n_1= 2(n_1+n3+n_4)+2-n_3\le 2n+2$).


问:
Convolution of the delta distribution, basic property
I am trying to prove that  for all $\phi \in S(\mathbb{R}), \phi*\delta = \phi$. I tried with a direct calculation of $ \phi*\delta(\psi)$. At the very end, I got (expanding convolution and exchange the order of the integral, where $\tilde{\phi}(y-x) = \phi(x-y)$
$$
\int \delta(y)(\tilde{\phi} * \psi)(y) dy
$$
But I am not sure if it's legitimate to write the above integral as
$$
T(\tilde{\phi} * \psi) = T_{\phi}(\psi)
$$
Or is there another way to show this property?

答:
You are nearly finished. Note that by definition of the delta distribution it holds that
$$\langle \phi*\delta,\psi\rangle = \int (\phi*\delta)(x) \psi(x) dx = \int \int \underbrace{\phi(x-y)}_{\tilde{\phi}(y-x):=} \delta(y) \psi(x) dy dx = \int \delta(y)(\tilde{\phi}*\psi)(y)dy =\langle \delta, \tilde{\phi}*\psi\rangle =  (\tilde{\phi}*\psi)(0) = \int \phi(x) \psi(x) dx = \langle \phi,\psi\rangle,$$
where the second to last step follows since inserting 0 just gives you this integral. Note that I use $\langle .,.\rangle$ as a notation for the integral. Since this equality holds for all test functions $\psi$, you get $\phi*\delta=\phi$.


问:
Reference request for functorial approach to classifying spaces
In the introduction to Bott and Tu's Differential Forms in Algebraic Topology, there was mention that

...there was no time left within the scope of our book to explain the functorial approach to classifying spaces in general and to make the connection with the Eilenberg--MacLane spaces.

Is there a good reference that has covered this material?

答: One source is Algebraic Topology by Tammo Tom Dieck, section 14.4.


问:
Why is "If every proper subgraph of G is bipartite, then G is bipartite." False
"If every proper subgraph of G is bipartite, then G is bipartite" is apparently false.

I can't think of a justification in my head. I've been going over this, sketching out multiple non-bipartite graphs with all proper subgraphs being Bipartite, but the main graph not. I haven't come up with anything. Can someone help me?

答: You can consider any odd cycle, $C_{2n+1}$. Removing an edge you have a $P_{2n+1}$ and removing a vertex you have a $P_{2n}$. Both are bipartite.


问:
Showing that an increasing and continuous function's inverse is increasing and continuous
Show that if $f:[a,b] \to \mathbb{R}$ is continuous and increasing, then the inverse function $f^{-1}$ exists and is also continuous and increasing on the interval on which it is defined.

Proof

Show that $f$ is one-to-one

Since $f$ is strictly increasing on $[a,b]$, for each $x_n \in [a,b]$ with $n \in \mathbb{N}$, $f(x_n)<f(x_{n+1})$

Then if $f(x_n)<f(x_{n+1}), \Rightarrow x_n<x_{n+1}$

So when $f(x_n)\neq f(x_{n+1}), \Rightarrow x_n \neq x_{n+1}$. Thus, $f$ is one-to-one

Show that $f$ is onto

Take $y \in f([a,b])$

Since $f$ is continuous on $[a,b],$ we can find an $x \in [a,b]$ such that $f(x)=y$ for all $y \in f([a,b])$

Thus, $f$ is onto

Since $f$ is one-to-one and onto, $f$ is bijective, which means $f^{-1}$ exists

Now, since $f$ is defined on $[a,b]$, $f^{-1}$ is defined on the interval $[f(a),f(b)]$

We know that $(f^{-1})'(x)=\frac{1}{f'(f^{-1}(x))}$ by the inverse function theorem which tells us that if $f'>0$ on an interval, then $(f^{-1})'>0$ on that interval.

Since $f$ is increasing on $[a,b], f'(x)>0$ on $[a,b]$. Then $(f^{-1}){'}(x)>0$ on $[a,b]$

Thus, $f^{-1}$ is increasing.

I can't figure out how to show that $f^{-1}$ is continuous based off of all this.

I feel like this proof is full of mistakes so feel free to show me where I'm going wrong. I'm having trouble seeing all this.

答:
You have correctly established that $f^{-1}$ exists.

Now, use the definition of "increasing" to conclude that $f^{-1}$ is increasing.

From there, use the definition of continuity. Consider any $y \in (f(a),f(b))$. Let $\epsilon > 0$. We want to show that there is a $\delta > 0$ for which
$$
|y - y'| < \delta \implies |f^{-1}(y) - f^{-1}(y')| < \epsilon.
$$
Let $x = f^{-1}(y)$. Define $y_1 = f(x- \epsilon)$ and $y_2 = f(x + \epsilon)$. Let $\delta = \min\{y - y_1, y_2 - y\}$. We note that for any $y'$ with $|y - y'| < \delta$, we have
$$
y_1 < y' < y_2 \implies f^{-1}(y_1) < f^{-1}(y') < f^{-1}(y_2)
\\\implies f^{-1}(y) - \epsilon < f^{-1}(y')
< f^{-1}(y) + \epsilon\\
\implies |f^{-1}(y') - f^{-1}(y)| < \epsilon,
$$
which is what we wanted.

The proof where $y$ is equal to either endpoint is similar.


问:
Deterministic sequence of probability measures
Suppose $\mu_0$ is a probability measure on the reals with finite first moment $\int_{-\infty}^\infty |x| \, \mu_0(dx) < \infty$ so that $m:= \int_{-\infty}^\infty x \, \mu_0(dx) \in \mathbb R$ is well defined: If $X_0\sim\mu_0$ is a random variable with law $\mu_0$, it has expectation $\mathbb E[X_0]=m$. Now, given $\mu_n$ for $n\in\mathbb N_0$, we construct the probability measure $\mu_{n+1}$ deterministically as follows. Taking $(U_n, V_n, W_n)$ i.i.d. each distributed according to $\mu_n$, let $\mu_{n+1}$ the law of $$ U_n + V_n-\ln(1+e^{V_n})+\ln(1+e^{W_n}). $$ In other words, $$ \mu_{n+1} := \mu_n \ast \mu_n \circ(x\mapsto x-\ln(1+e^x))^{-1} \ast\mu_n\circ (x\mapsto \ln(1+e^x))^{-1} ,\quad n\in\mathbb N $$ where $\ast$ denotes convolution of measures.

If $X_n\sim \mu_n$, it holds $\mathbb E[X_n]=2^n\, m$, so I wonder: If $m>0$, is it (generally) true that $F(t):=\mu_n((-\infty,t])\to 0$ as $n\to\infty$ for any $t\in\mathbb R$? Also, in the case $m=0$, can anything be said about the asymptotic behavior of $\mu_n$ as $n\to\infty$? Clearly, if $\mu_0 =\delta_a$ is the point measure at $a\in\mathbb R$, it holds $\mu_n=\delta_{2^n a}$. Otherwise, the variance $\text{Var}(X_{n+1})>\text{Var}(X_n)$ of $X_{n+1}\sim\mu_{n+1}$ is strictly larger than the variance of $X_n\sim\mu_n$.

答:
Let $A_n := \frac{1}{2^n}U_n$

Though the following result can be generalized, I'll stay with the condition that $\mu_0 \in L^2$ for the simplicity

Theorem 1

If $\mu_0 \in L^2$, we have
$$ \mathcal{L}( 2^{cn}( A_n-m )  ) \longrightarrow \delta_{0}$$
for all $0<c<\frac{1}{2} $, where $\mathcal{L}(X)$ is defined as the law of the random variable $X \square $.

Demonstration

We have:
$$\begin{array}{lcl}\text{Var}(A_{n+1})&\underbrace{=}_{\text{def of } \mu_{n+1}}& \dfrac{1}{4^{n+1}} \left[ \text{Var}(U_n)+\text{Var}\left\{ \ln( e^{W_n}+1 )-\ln( e^{-V_n}+1 ) \right\} \right]\\
&\le&\dfrac{1}{4^{n+1}} \left[ \text{Var}(U_n)+ \text{Var}\left\{ \ln( e^{W_n}+1 )-\ln( e^{-W_n}+1 ) \right\} \right]\\
&=& \dfrac{1}{4^{n+1}} \left[ \text{Var}(U_n)+\text{Var}(W_n) \right]= \frac{1}{2}\text{Var}(A_n)
\end{array}
$$
Where in the second line, I have used a variation of Chebyshev's inequality, which is:
$$
\text{Var}( f(X)-g(Y)) \le \text{Var}( f(X)-g(X))
$$
for any two random variables $X, Y$ such that $X \sim Y$ and $f$ is increasing while $g$ is decreasing.

Thus $$\text{Var}(A_n) \le \dfrac{1}{2^n}\text{Var}(A_0)$$
Hence the conclusion $\square$.

Corollary 2

If $m>0$ and $\mu_0 \in L^2$, we have:
$$F_{\mu_n}(t) \xrightarrow{ n \rightarrow +\infty} 0 $$
for all $t$.
$\square$

Corollary 3

If $m=0$, $\mu_0 \in L^2$, then:
$$\mu_{n}( 2^{cn}a, 2^{cn}b  ) \rightarrow 1 $$
for $a<0<b$ and $c>1/2$
$\square$

Comment:

If the condition $L^2$ is relaxed, the same result can be found but more subtle treatment is needed, and some optimal transport inequalities might be involved.

I would be nice if OP can help me know where did you come up with this question.


问:
Recursive function that outputs its own code
This problem is probably a rather trivial one, since I have the impression, that it is a textbook-style one, but nonetheless somehow it won't give in. Here it is:

I have to show that there exists a (unary) recursive function, that has code $c$ and also takes the constant value $c$ (i.e., it outputs its own code).

I am pretty sure, I have to use Kleene's (second) recursion theorem, that says that for a given (total) recursive function $f:\mathbb{N} \rightarrow \mathbb{N}$ there is a number/code $c$ such that $\phi_c=\phi_{f(c)}$ (where $\phi_a$ is the partial recursive function that has code $a$), but I can't figure out how...

答: Just let $f(x)$ be a function that, on input $x$, creates and returns a program $P_x$ such that $P_x$ ignores its input and returns the number $x$. </p>


问:
Are all biregular graphs connected?
a simple question that I can't really seem to prove.
I have a biregular graph (as in a graph with n vertices and each vertex has a degree of 'd') and it is also bipartite. I'd like to prove it is connected, as in there is a path between every two vertices.

Any advice?

答:
It isn’t necessarily true:</p>
```<code>                    *       *
                    |       |
                    *       *
</code>```

Title: Are all biregular graphs connected?

Body:
It isn’t necessarily true:

                    *       *
                    |       |
                    *       *


问:
Testing for isomorphism between normal subgroup lattices
The question is whether we can use GAP to construct the normal subgroup lattice of a finite group $G$, call it $\mathcal{N}(G)$, and to then test whether $\mathcal{N}(G) \cong \mathcal{N}(H)$, where $G, H$ are non-isomorphic finite groups. An easy example of a pair $(G,H)$ such that $\mathcal{N}(G) \cong \mathcal{N}(H)$ and $|G|=|H|$ is the obvious $(Q_8,D_8)$, but for groups of larger size the task becomes unmanageable to carry out by hand.

I will provide here a little background on how and where this question comes up.

Assume that for a finite group $G$ we are given the following information: the normal subgroup lattice $\mathcal{N}(G)$, where each node representing a normal subgroup displays the order of that subgroup. The bigger question to answer here is which characteristic subgroups of $G$ can be located in $\mathcal{N}(G)$ by using only the information above.

It is clear that some subgroups can be located and sometimes this can be done with virtually no effort. For example, we can easily locate $\operatorname{F}(G)$ because the various $\operatorname{O}_p(G)$'s are directly visible in the lattice.
Some other subgroups can still be located, but that requires more effort on our part. In case $G$ is soluble for instance, we can detect $\operatorname{\Phi}(G)$, the Frattini subgroup of $G$. (If $G$ is not soluble, then it is not possible to locate $\operatorname{\Phi}(G)$ and to see this one needs only look at the two perfect groups of order $1344$.)
We can also locate $G'$ because there is a systematic way to tell if a group is abelian by looking at the normal subgroup lattice with orders.

The question I would like to answer is whether $\operatorname{Z}(G)$ can be located if we assume that $G$ is soluble. I suspect that the answer is "no" in general and that $p$-groups alone should suffice as sources of counterexamples. (Interestingly, if $G$ is soluble and we detect that $\operatorname{\Phi}(G) =1$ then it is possible to locate $\operatorname{Z}(G)$.)

答:
Here is a very basic implementation that takes two lists of sets (e.g. lists of normal subgroups if the two groups) and tries to find an inclusion and order preserving map between them, returning fail if no such map can be found. It arranges the subgroups according to orders, calculates minimal inclusion relations, and then tries to map the one set of relations to the other one. It should do better than hand, but likely will fail if there are dozens of subgroups of a given order.

```<code># take two lists of subsets, inclusion+order-preserving bijection, if any
InclusionOrderPreservingBijection:=function(l,m)
local inclusionSet,lsz,msz,lp,mp,lpair,mpair,c,G,i,f,e;

  # compute minimal inclusion releations
  inclusionSet:=function(a)
  local ai,i,j,apair;
    ai:=[];
    for i in [1..Length(a)] do
      ai[i]:=[];
      for j in [i+1..Length(a)] do
        if Size(a[j])>Size(a[i]) and IsSubset(a[j],a[i]) then
          AddSet(ai[i],j);
        fi;
      od;
    od;

    # minimal inclusions
    apair:=[];
    for i in [1..Length(a)] do
      for j in ShallowCopy(ai[i]) do
        ai[i]:=Difference(ai[i],ai[j]);
      od;
      for j in ai[i] do
        AddSet(apair,Immutable([i,j]));
      od;
    od;
    return apair;
  end;

  lsz:=List(l,Size);
  msz:=List(m,Size);

  # normalize arrangement
  lp:=Sortex(lsz);
  l:=Permuted(l,lp);
  mp:=Sortex(msz);
  m:=Permuted(m,mp);

  if lsz<>msz then
    return fail; # different orders
  fi;

  # inclusion sets
  lpair:=inclusionSet(l);
  mpair:=inclusionSet(m);

  if Length(lpair)<>Length(mpair) then
    return fail; # different inclusions
  fi;

  # mapping group
  c:=List(Collected(lsz),x->x[2]); # count how many of same sizes
  G:=[];
  f:=1;
  for i in c do
    Append(G,GeneratorsOfGroup(SymmetricGroup([f..f+i-1])));
    f:=f+i;
  od;
  G:=Group(G);

  # now G is the direct product of symmetric groups on the different sizes

  e:=RepresentativeAction(G,lpair,mpair,OnSetsSets);

  if e=fail then return fail;
  else
    return lp*e/mp;
  fi;
end;
</code>```

Body:
Here is a very basic implementation that takes two lists of sets (e.g. lists of normal subgroups if the two groups) and tries to find an inclusion and order preserving map between them, returning fail if no such map can be found. It arranges the subgroups according to orders, calculates minimal inclusion relations, and then tries to map the one set of relations to the other one. It should do better than hand, but likely will fail if there are dozens of subgroups of a given order.

# take two lists of subsets, inclusion+order-preserving bijection, if any
InclusionOrderPreservingBijection:=function(l,m)
local inclusionSet,lsz,msz,lp,mp,lpair,mpair,c,G,i,f,e;

  # compute minimal inclusion releations
  inclusionSet:=function(a)
  local ai,i,j,apair;
    ai:=[];
    for i in [1..Length(a)] do
      ai[i]:=[];
      for j in [i+1..Length(a)] do
        if Size(a[j])>Size(a[i]) and IsSubset(a[j],a[i]) then
          AddSet(ai[i],j);
        fi;
      od;
    od;

    # minimal inclusions
    apair:=[];
    for i in [1..Length(a)] do
      for j in ShallowCopy(ai[i]) do
        ai[i]:=Difference(ai[i],ai[j]);
      od;
      for j in ai[i] do
        AddSet(apair,Immutable([i,j]));
      od;
    od;
    return apair;
  end;

  lsz:=List(l,Size);
  msz:=List(m,Size);

  # normalize arrangement
  lp:=Sortex(lsz);
  l:=Permuted(l,lp);
  mp:=Sortex(msz);
  m:=Permuted(m,mp);

  if lsz<>msz then
    return fail; # different orders
  fi;

  # inclusion sets
  lpair:=inclusionSet(l);
  mpair:=inclusionSet(m);

  if Length(lpair)<>Length(mpair) then
    return fail; # different inclusions
  fi;

  # mapping group
  c:=List(Collected(lsz),x->x[2]); # count how many of same sizes
  G:=[];
  f:=1;
  for i in c do
    Append(G,GeneratorsOfGroup(SymmetricGroup([f..f+i-1])));
    f:=f+i;
  od;
  G:=Group(G);

  # now G is the direct product of symmetric groups on the different sizes

  e:=RepresentativeAction(G,lpair,mpair,OnSetsSets);

  if e=fail then return fail;
  else
    return lp*e/mp;
  fi;
end;


问:
complex integral, help.
Proof that, if $f$ is holomorphic in the ring $U = {r <| z - a | <R}$, then the integral of $f$, along the circumference with center $a$ and radius $ρ$, $\int_{| z - a |=ρ}f(z)dz$, has the same value, for all $r <ρ <R$.

idk how to start :(

答:
We can define a closed contour in four pieces; one around $|z-a|=R$  One around $|z-a| =p$  and two line segments between these two circles along the positive real line.  Let's call this contour $\gamma$

$\int_{|z-a|= p} f(z)\ dz + \int_{\gamma} f(z)\ dz = \int_{|z-a| = R} f(z)\ dz$

As $f$ is holoporphic everywhere inside the contour of $\gamma, \int_{\gamma} f(z)\ dz = 0$

$\int_{|z-a|= p} f(z)\ dz = \int_{|z-a| = R} f(z)\ dz$ for all $p$ such thata $r<p<R$

Title: complex integral, help.


问:
Transform algebraic curves
For $i=1,2,3$ let $P_i$ be real polynomials such that $P_3$ is positive. Define the mapping
\begin{align*}
f(x,y) = \left(x+\frac{P_1(x,y)}{1+\sqrt{P_3(x,y)}},y+\frac{P_2(x,y)}{1+\sqrt{P_3(x,y)}}\right), x,y \in \mathbb{R}.
\end{align*}

Does $f$ map algebraic curves to algebraic curves?

Note: Numerically, I investigated many non-trivial examples and they all looked algebraic again.

答:
Explaining my comment:

Assume that the original algebraic curve is given in parametric form
$$x(t)=r_1(t), \ \ \  y(t)=r_2(t),$$
where $r_i$ are rational functions. Applying $f$ to it gives
$$f(x(t),y(t))=(u,v)=\left(r_1(t) + \frac{r_3(t)}{1+\sqrt{r_5(t)}},r_2(t) + \frac{r_4(t)}{1+\sqrt{r_5(t)}}\right).$$

This can be rewritten as a polynomial equation system in $t$. Computation of the determinant $\det$ of its Sylvester matrix gives a polynomial equation $\det(u,v)=0$, such that $(u,v)$ is again an algebraic curve.

Problem here: As far as I know, an algebraic curve is rationally parameterizable if and only if its genus is equal to zero.

Title: Transform algebraic curves


问:
Rings | Homomorphisms | Units

Question

Show that if $f :R\rightarrow S$ is a homomorphism, and if $a$ is a unit of $R$, then $f(a)$ is
a unit of $S$. Show, in fact, that $f(a^{−1}) = f(a)^{−1}$ for any unit $a$ of $R$.

Attempt

Let $f:R\rightarrow S$ be a homomorphism. Then

$$ f(a)=f(a\cdot 1_R)=f(a)f(1_R),$$

so

$$ f(1_R)=\frac{f(a)}{f(a)}=1_S, $$

which is me trying to show why $(6)$ holds:

Consequently, we then have that,

$$ f(a\cdot a^{-1})=f(a)f(a^{-1})=f(1_R)=1_S, $$

so it is then easy to see that because $f(a)f(a^{-1})=1_S$, then $f(a)$ is a unit in $S$. Furthermore, by the same equality

$$ f(a^{-1})=\frac{1_S}{f(a)}=f(a)^{-1}, $$

as needed.

答:
You've got something of the right idea, here, but are off in a few details. I assume that your rings are abelian groups under addition, monoids under multiplication, with multiplication being left- and right-distributive over addition. I further assume your definition of a ring homomorphism $R\to S$ requires $1_R\mapsto 1_S.$ Let me know if my assumptions are correct.

Now, given those assumptions, we know for any unit $a\in R$ that $$1_S=f(1_R)=f(aa^{-1})=f(a)f(a^{-1}),$$ and likewise, $$1_S=f(a^{-1})f(a).$$ Thus, indeed, we have that $f(a)$ is a unit of $S$ and $f(a)^{-1}=f(a^{-1}).$

Added: The assumptions that I made above are actually necessary. Without them, the proposition is false.

On the one hand, if we only require a ring homomorphism $f:R\to S$ to satisfy $$f(x+y)=f(x)+f(y)\quad\text{and}\quad f(xy)=f(x)f(y)\tag{$\clubsuit$}$$ for all $x,y\in R$, then let's consider $R=S=\Bbb Z$, and let $f:R\to S$ be given by $f(n)=0$ for all $n.$ Then $-1$ is a unit of $R$, but $f(-1)=0$ is not a unit of $S$, but $f$ is a ring homomorphism by this definition. This shows that without requiring $1_R\mapsto 1_S$, the proposition fails.

Now, suppose we don't require that a ring be a monoid under multiplication. (The even integers are then a ring, for example.) The most we can require in general, then, is that ring homomorphisms $f$ on $R$ satisfy $(\clubsuit)$, which we already saw wasn't enough.

Title: Rings | Homomorphisms | Units


问:
Points in plane with every pair having at least two equidistant points?
I was given this question in person by a fellow trainee at the downtime of an IMO training session, which made me think this problem is Olympiad related. I am interested in the solution as much as the origin!

$\textbf{Problem:}$Does there exist a non-empty finite set of points $P$ in $\mathbb{R}^2$ such that for each pair of points $x$, $y$ in $P$ there exists at least two distinct points $a$ and $b$ in $P$ such that both $a$ and $b$ are equidistant from $x$ and $y$.

To be honest, in over a decade and getting a degree in Mathematics, I have not made much headway into this problem. I tried to do some counting based on the number of pairs and pigeonhole to get the conclusion that there must be a point that is the circumcenter of a triangle made up of points in $P$. Assuming $P$ is a minimal solution (in terms of size of $P$) does not seem to help that much.

EDIT: originally the question left open whether it was in the plane or space, but the space case is trivial.

EDIT EDIT: added a note to say the set should be non-empty. Note that we do not need at least two points, because when we say "pairs of points" we don't say "distinct pair of points"!

答:
Let $P$ be the finite set of points above mentioned.
We have, by definition, that for every pair of points $x$ and $y$, there is another different pair $a$ and $b$ that are located in the line that passes through the midpoint of $x$ and $y$ and is perpendicular to the line segment $\overline{\rm xy}$ connecting such points.

Let $L$ be the set of all lines that can be constructed by passing through each pair of points in $P$ (this is a much bigger set, but still finite). Now consider the pair of lines $l_1,l_2\in L$ which present the smallest nonzero angle between them (there can be more than one, but it's enough to consider one pair). Then there is a pair of points $x_1\in l_1$ and $x_2\in l_2$ which doesn't satisfy the condition, i.e., there is no other pair $a,b$ equidistant from $x_1$ and $x_2$, which leads to a contradiction.

Note that if such pair $a,b$ existed, the condition "$l_1$ and $l_2$ having the smallest nonzero angle between them" would not hold, as the angle between $l_1$ or $l_2$ and the line perpendicular to the segment $\overline{\rm x_1x_2}$ (the line passing through $a$ and $b$) would be smaller.

Therefore, $P$ and $L$ must be both infinite sets.


问:
How to prove $ ((S\times_A T)(1))_{(f\otimes g)}\cong S(1)_{(f)}\otimes_{S_{(f)}}(S\times_A T)_{(f\otimes g)}\otimes_{T_{(g)}} T(1)_{(g)}$?
I'm currently trying to solve Exercise 5.11 in chapter 2 of Hartshorne: Let $S,T$ be $\mathbb{Z}_{\geq 0}$-graded rings with $S_0=T_0=A$, and define their Cartesian product $S\times_A T$ to be
$$
S\times_A T=\bigoplus_{d\geq 0}S_d\otimes_A T_d.
$$
I already showed that $\operatorname{Proj}(S\times_A T)\cong \operatorname{Proj}(S)\times_{\operatorname{Spec}(A)}\operatorname{Proj}(T)$. Now I am trying to show that
$$
\mathcal{O}_{\operatorname{Proj}(S\times_A T)}\cong p_1^{*}\mathcal{O}_{\operatorname{Proj}(S)}\otimes p_2^{*}\mathcal{O}_{\operatorname{Proj}(T)}
$$
where $p_1$ and $p_2$ are the structure maps making $\operatorname{Proj}(S\times_A T)$ the product of $\operatorname{Proj}(S)$ and $\operatorname{Proj}(T)$. I'm stuck at the following very specific part of the proof:

Is it true that
$$
((S\times_A T)(1))_{(f\otimes g)}\cong S(1)_{(f)}\otimes_{S_{(f)}}(S\times_A T)_{(f\otimes g)}\otimes_{T_{(g)}} T(1)_{(g)}
$$
For all $f\in S_d$ and $g\in T_d$ where $d\in\mathbb{Z}_{\geq 1}$ is arbitrary? I constructed a map going from right to left by mapping $s/f^n\otimes[(s'\otimes t')/(f\otimes g)^n]\otimes t/g^n$ to $((ss')\otimes (tt'))/(f\otimes g)^{2n}$ and showed that it is surjective. However, I'm struggling to prove that the map is injective.

Edit: As suggested by KReiser, I tried to directly construct an inverse. However, I'm only able to do this under the extra assumption that $S$ and $T$ are generated by $S_1$ resp. $T_1$ as $A$-algebras, so that we can assume that $f$ and $g$ are of degree $1$. Here's why.

To justify the well-definedness of the map
$$
((S\times_A T)(1))_{(f\otimes g)}\to S(1)_{(f)}\otimes_{S_{(f)}}(S\times_A T)_{(f\otimes g)}\otimes_{T_{(g)}}T(1)_{(g)}\\
(s\otimes t)/(f\otimes g)^n\mapsto (s/f^n)\otimes 1\otimes (t/g^n)
$$
we would like to start by defining it on the homogeneous pieces of $(S\times_A T)(1)$ and then pass to the localisation. However, for this to work, we would have to relax the target to $S(1)_{f}\otimes_{S_{(f)}}(S\times_A T)_{f\otimes g}\otimes_{T_{(g)}}T(1)_{g}$, and only afterwards show that the image is contained in $S(1)_{(f)}\otimes_{S_{(f)}}(S\times_A T)_{(f\otimes g)}\otimes_{T_{(g)}}T(1)_{(g)}$, because otherwise, the map
$$
s\otimes t\in (S\times_A T)(1)\mapsto (s/1)\otimes 1\otimes (t/1)
$$
wouldn't be well-defined. So at this point we have a map of abelian groups $(S\times_A T)(1)\to S(1)_{f}\otimes_{S_{(f)}}(S\times_A T)_{f\otimes g}\otimes_{T_{(g)}}T(1)_{g}$. Now to pass to the localisation on the source, as we have only a morphism of abelian groups at this point, we would have to justify by hand why it is well defined. That is, we want to show that if $(s\otimes t)/(f\otimes g)^n$ is equal to $0$ inside $(S\times_A T)(1)_{f\otimes g}$, then $(s/1)\otimes 1\otimes (t/1)$ is $0$ inside $S(1)_{f}\otimes_{S_{(f)}}(S\times_A T)_{f\otimes g}\otimes_{T_{(g)}}T(1)_{g}$. If we now suppose that $f$ and $g$ are of degree $1$, then this can be done: if $e$ is the degree of $s$ and $t$, then
$$
(s/1)\otimes 1\otimes (t/1)=((s/f^{e+1})sf^{e+1})\otimes 1\otimes ((t/g^{e+1})tg^{e+1})=\\
=(sf^{e+1}/1)\otimes ((s\otimes t)/(f\otimes g)^{e+1})\otimes (tg^{e+1}/1)=0
$$
because of the middle term. However, I don't see how to do this without assuming that $f$ and $g$ are of degree $1$.

答:
How about constructing an inverse directly? If $(s\otimes t)/(f\otimes g)^d$ is an element of $(S\times_A T)(1)_{(f\otimes g)}$, send it to $s/f^d \otimes 1 \otimes t/g^d$. To see that this is mutually inverse with your map, we just need to do a little calculation. The first composition is relatively straightforward:

$$(s\otimes t)/(f\otimes g)^d \mapsto s/f^d \otimes 1 \otimes t/g^d = s/f^d \otimes [(f\otimes g)^d/(f\otimes g)^d] \otimes t/g^d \\ \mapsto (sf^d\otimes tg^d)/(f\otimes g)^{2d} = (s\otimes t)/(f\otimes g)^d.$$

The second direction requires a small tensor manipulation:

$$s/f^n \otimes [(s'\otimes t')/(f\otimes g)^n] \otimes t/g^n \mapsto (ss'\otimes tt')/(f\otimes g)^{2n} \\\mapsto ss'/f^{2n} \otimes 1 \otimes tt'/g^{2n} = s/f^n \otimes [(s'/f^n \otimes 1)\cdot (1\otimes t'/g^n)] \otimes t/g^n$$

where we used that $s'/f^n$ and $t'/g^n$ are actually elements of $S_{(f)}$ and $T_{(g)}$ so they can be pulled across the tensor product.

The lesson here is that sometimes it can be easier to write down an inverse map than to check bijectivity directly.

Edit: Since your troubles aren't quite resolved, let's write down a more explicit construction and check we don't need $f,g$ of degree one, just that they're the same degree.

Start with the map $S_d\times T_d \to S(1)_f \otimes_{S_{(f)}} (S\times_A T)_{f\otimes g} \otimes_{T_{(g)}} T(1)_g$ defined by $(s,t)\mapsto s\otimes 1 \otimes t$. This is $A$-bilinear, as $A$ sits naturally inside the degree zero parts of $S_{(f)}$, $T_{(g)}$, and $(S\times_A T)_{f\otimes g}$, and the map clearly respects addition. So we get that our map factors through $S_d\otimes_A T_d$. Taking direct sums, we get a map $S\times_A T\to S(1)_f \otimes_{S_{(f)}} (S\times_A T)_{f\otimes g} \otimes_{T_{(g)}} T(1)_g$ defined on pure tensors by $s\otimes t\mapsto s\otimes 1 \otimes t$.

Next, we check that the action of $f\otimes g$ on $S\times_A T$ is invertible on the target: $f\otimes 1 \otimes g$ has inverse $1/f\otimes 1 \otimes 1/g$ in the target, so our map factors through $(S\times_A T)_{f\otimes g}$ and we get a map $$(S\times_A T)_{f\otimes g} \to S(1)_f \otimes_{S_{(f)}} (S\times_A T)_{f\otimes g} \otimes_{T_{(g)}} T(1)_g.$$

It remains to check that the homogeneous elements of total degree one in $(S\times_A T)_{f\otimes g}$ map to homogeneous elements of the appropriate degree in $S(1)_f \otimes_{S_{(f)}} (S\times_A T)_{f\otimes g} \otimes_{T_{(g)}} T(1)_{g}$. If $(s\otimes t)/(f\otimes g)^n$ is of homogeneous degree one, then both $s/f^n$ and $t/g^n$ are too, so $s/f^n\otimes 1 \otimes t/g^n$ lies in $S(1)_f \otimes_{S_{(f)}} (S\times_A T)_{(f\otimes g)} \otimes_{T_{(g)}} T(1)_{(g)}$ and we've won.


答: Proof by contradiction. Assume that every non-bipartite graph has a non-bipartite proper subgraph. Then, starting with any non-bipartite finite graph $G_0$, say $G_0=K_5$,  we can get an infinite decreasing sequence of nom-bipartite graphs $G_0\supset G_1\supset G_2\supset G_3\supset\cdots$ such that $G_{n+1}$ is a proper subgraph of $G_n$ for each $n$. This is impossible, seeing as the finite graph $G_0$ has only a finite number of subgraphs.

Title: Why is "If every proper subgraph of G is bipartite, then G is bipartite." False

Body:
"If every proper subgraph of G is bipartite, then G is bipartite" is apparently false.

I can't think of a justification in my head. I've been going over this, sketching out multiple non-bipartite graphs with all proper subgraphs being Bipartite, but the main graph not. I haven't come up with anything. Can someone help me?


问:
Show that matrices are not similar
I have to show that the following matrices are not similar:

$$A = \left[\begin{matrix} 1 & 3 & -3 \\ -3 & 7 & -3 \\ -6 & 6 & -2\end{matrix}\right]$$

and $$A' = \left[\begin{matrix} 5 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 3\end{matrix}\right]$$

I know that 2 matrices $A$ and $A'$ are similar if there exists an invertible matrix $B$, such that $$A'= B^{-1}AB$$

According to Wikipedia, similar matrices share some properties (for example they have the same eigen values), but I don't know how to start, since I have missed the last lectures of my linear algebra course, unfortunetely. Unfortuntely, life has not been so completely fair with me :(

Should I just check if they have the same eigen values?

答: Does the above method always work? For example, if you have 2 matrices in Jordan form, the determinants can both be 0 (say, all 0 along their diagonals). In this case, the 2 matrices can be NOT similar but still share the same determinant, making the above proof not necessarily true for all cases?</p>

Title: Show that matrices are not similar


问:
Tangent plane to $f(x, y)= x^3 - x \, \ln y + y^2 + 3 y \ln x + 5$ in $(1,1,z_0)$
Let be $f(x, y)= x^3 - x \, \ln y + y^2 + 3 y \ln x + 5$

Calculate  $z_0=f(1,1)$

I calculate that and gives me $7$ I mean $f(1,1)=7$

and  give the tangent plane to $f(x, y)= x^3 - x \, \ln y + y^2 + 3 y \ln x + 5$ in $(1,1,z_0)$

But I think the point (1,1,7) is not in $f(x, y)= x^3 - x \, \ln y + y^2 + 3 y \ln x + 5$

Then how can I calculate the tangent plane?

Body:
The equation of the plane tangent to $z=f(x,y)$ at point $(x_0,y_0,z_0)$ is$$z  = f\left( {{x_0},{y_0}} \right) + {f_x}\left( {{x_0},{y_0}} \right)\left( {x - {x_0}} \right) + {f_y}\left( {{x_0},{y_0}} \right)\left( {y - {y_0}} \right)$$
$$f_x(x,y)=3 x^2+\frac{3 y}{x}-\log y;\;f_y(x,y)=-\frac{x}{y}+3 \log x+2 y$$
$$z=7+6(x-1)+(y-1)\to z=6x+y$$


问:
commutes with a nilpotent matrix and invertible
The exercise is:

Given $V$ a vector space, $\dim V<\infty$ and $T, N\in \mathcal{L}(V)$, where $N$ is nilpotent and $TN=NT$. Prove that: $T$ is invertible iff $T+N$ is invertible. Furthermore, $\det(T)=\det(T+N)$ and $p_T(t)=p_{T+N}(t)$.

I need a hint about how can I start this kind of exercise, which involves a nilpotent matrix.

答:
For a rather different approach one can directly prove $p_T(t)=p_{T+N}(t)$ which then gives you the equality of determinants and in turn the equivalence of invertibility.  I assume $\dim V = m$ and assume the field has characterisitic zero (or if one wants to contemplate positive characteristic fields, then $\text{char}(\mathbb F)\gt m$)

lemma 1:

by nilpotence, $N^m = \mathbf 0$ and with commutativity this means you can apply the binomial theorem, so $\big(T+N\big)^j= T^j +\Big(\sum_{k=1}^{j-1} \binom{j}{k}T^kN^{j-k}\Big) + N^j$

lemma 2:

$T^k N^r$ for $r\geq 1$ is nilpotent, because

$(T^k N^r)^m =  T^{km} N^{rm}=  T^{km}\mathbf 0 = \mathbf 0$

and recall that since nilpotent matrices are similar to strictly upper triangular matrices, they necessarily have trace zero

main argument:

$\text{trace}\Big(T+N\Big)= \text{trace}\Big(T\Big)+\text{trace}\Big(N\Big)= \text{trace}\Big(T\big) + 0 $

and for $j\in\big\{1,2,...,m\big\}$

$\text{trace}\Big(\big(T+N\big)^j\Big)= \text{trace}\Big(T^j\Big) + \sum_{k=0}^{j-1} \binom{j}{k}\cdot\text{trace}\Big(T^kN^{j-k}\Big)= \text{trace}\Big(T^j\Big) + 0$

Thus by application of Newton's Identities $\Big(T+N\Big)$ and  $\Big(T\Big)$ have the same characteristic polynomial.


问:
Identity of inverse matrix
How can I prove that

$$(C_{N}^{-1} + W_{N})^{-1} = C_{N}(I + W_{N}C_{N})^{-1}$$

I tried to use the Woodbury identity:

$$(A + BD^{-1}C)^{-1} = A^{-1}-A^{-1}B(D+CA^{-1}B)^{-1}CA^{-1}$$

which seems to be useful in this case, but I can't simplify enough to get the result.

Suggestions?

答:
Think about what happens if they were just numbers (but don't forget that it's not commutative):

Let's call $A:=C_N$ and $B:=W_N$ for simplicity. If these were numbers, you are looking for $\displaystyle\frac1{1/A\,+B}$. Now we want to pull out $\displaystyle\frac1{1/A}$ (alias $A$) from the left, so that, because of the inverse, we have to pull out $1/A$ from the right:
$$1/A+B=(1+BA)\cdot 1/A\\
\frac1{1/A\,+B}=\frac1{1/A}\cdot\frac1{1+BA}\,.$$
Writing it with inverses, we get the same line:

As $A^{-1}+B=(I+BA)A^{-1}$, we have
  $$(A^{-1}+B)^{-1}= A(I+BA)^{-1}\,.$$

Title: Identity of inverse matrix


问:
Satisfying Hypotheses For L'Hospital's Rule
I have that the following hypothesis need to be satisfied in order to use L'Hospital's rule for,
\begin{equation}
   \lim_{x\to a}\frac{f(x)}{g(x)}
\end{equation}

$f(x)$ and $g(x)$ have to be functions differentiable near $a$.

$\lim_{x\to a}\frac{f(x)}{g(x)}$ has to have the indeterminate form $\frac{0}{0}$ or $\frac{\pm\infty}{\pm\infty}$.

$\lim_{x\to a}\frac{f'(x)}{g'(x)}$ has to exist.

If possible, find the following limit using L’Hospital’s rule.  If not possible, explain why,
\begin{equation}
   \lim_{x\to0^+}\frac{\ln{x}}{\frac{1}{\sin{x}}}
\end{equation}
How do I check whether the hypothesis are satisfied and whether I can use L’Hospital’s rule for this or not.

I tried differentiating $\ln(x)$ and $\frac{1}{\sin(x)}$ at $0$ and saw that the derivative of $\ln(x)$ does not exist at $0$. From this, I came to the conclusion that L'Hospital's rule cannot be used. I tried using L'Hospital's anyway and still arrived at the correct answer, $0$, which I verified using Desmos. What am I doing wrong?

答:
Hypothesis 1 only says that the functions need to be differentiable near $a$, not differentiable at $a$. Using symbols, they need only be differentiable on some deleted neighbourhood $(a - \delta, a) \cup (a, a + \delta)$ for some $\delta > 0$.

This is similar to what we think of for limits: $\lim_{x \to a} F(x)$ only depends on the behaviour of $F(x)$ as $x$ approaches $a$, and doesn't depend on $F(a)$ itself.


问:
Lexicographic ordering
Let <lex denote the lexicographic ordering by their co-ordinate vertices...i.e. X <lex Y
if $x_1=y_1, x_2=y_2 ... x_i< y_i.$ where X =($x_1,x_2...x_d$) and Y =($y_1,y_2...y_d$) and i ∈ [d].
Prove that for any y ∈ $R^d$ the sets {x ∈ $R^d$: x ≤lex y} and {x ∈ $R^d$: y ≤lex x} are convex.

So for proving convexity usually one assumes two points and check if tx + (1-t)y (which is basically a line segment) satisfies the given condition where t ∈ [0,1] but in lexicographic ordering, they individually should (as mentioned the coordinates) satisfy the given condition, not a line. So how can I prove them when I am not allowed to take a line segment?

答:
Let $y=\langle y_1,\ldots,y_d\rangle$, and suppose that $x=\langle x_1,\ldots,x_d\rangle<_{\text{lex}}y$. There is a $k\in\{1,\ldots,d\}$ such that $x_i=y_i$ for $1\le i<k$, and $x_k<y_k$. Now let $t\in[0,1]$;

$$tx+(1-t)y=\big\langle tx_1+(1-t)y_1,\ldots,tx_d+(1-t)y_d\big\rangle\,,$$

and we want to show that $tx+(1-t)y\le_{\text{lex}}y$, so we consider its coordinates.

If $1\le i<k$, then $tx_i+(1-t)y_i=ty_i+(1-t)y_i=y_i$.

If $t\ne 0$, $tx_k+(1-t)y_k<ty_k+(1-t)y_k=y_k$.

If $t=0$, then $tx+(1-t)y=y$.

The first two bullet points are exactly what is needed to show that if $t\ne 0$, then $tx+(1-t)y<_{\text{lex}}y$, while the third shows that $tx+(1-t)y=y$ when $t=0$, so $\overline{xy}\subseteq\left\{z\in\Bbb R^d:z\le_{\text{lex}}y\right\}$.

The argument for $\left\{z\in\Bbb R^d:y\le_{\text{lex}}z\right\}$ is very similar.

Title: Lexicographic ordering


问:
Simple inequality question for infimum of $A = \{n \in\text{integers}:\frac{n}{2|n|+1} \}$
$$\beta >-\frac{1}{2}$$
There exists $a$ in $A = \{n \in\text{integers}:\frac{n}{2|n|+1} \}$ such that $a < \beta$
$$\beta > \frac{n}{2|n|+1}$$
$$n < \beta (2|n|+1) $$
$$n - 2|n|\beta<\beta $$

I don't know how to solve this, because there is an absolute value.

答:
If $ \beta>0 $ take $ n=0$.
If $ \beta =0 $, take $ n=-1$.

If $ -\frac 12<\beta<0 $ then

$$2\beta+1>0$$ and $ n $ should be such that $ n<0$.

So, we look for $ n<0 $ satisfying

$$\frac{n}{1-2n}<\beta $$
or

$$n<\beta-2n\beta$$
$$\iff n(2\beta+1)<\beta$$
$$\iff n<\frac{\beta}{2\beta+1}$$

thus, you can take $$n=\lfloor \frac{\beta}{2\beta+1}\rfloor-2$$


问:
Proving $ab^T$ is diagonalizable $\iff a^T b \neq 0$
Assuming $a$ and $b$ as two $n$-element vectors, how can we show that $$a b^T$$ is diagonalizable, if and only if their dot product is nonzero ( = they are not orthogonal)?

I've tried to prove $ab^T$ has $n$ linearly independent eigenvectors and then, automatically as a result, the beginning statement would be true, but I've not been successful in proving it.

Any help would be appreciated.

答:
Let me give very simple solution. Let $A=ab^T$. Since $Ax=ab^Tx=(b^Tx)a$,  $ImA=span\{a\}$. So rank of $A$ is 1 and by rank nullity, dimension of the null space of $A$ is $n-1$. What is kernel of $A$?

Kernel of $A$ is obviously $b^\perp$. So if $a$ and $b$ are orthogonal, then $a$ is also in kernel. So $ImA \subset KerA$. Thus the only eigenvalue is zero and since the zero eigenspace (kernel) is $n-1$ dimensional, $A$ is not diagonalizable.

Assume that $a$ and $b$ are NOT orthogonal. So $a$ is not in kernel. In this case, eigenvalues are $0$ and $b^Ta$. Since the zero eigenspace $n-1$ dimensional and $b^Ta$ eigenspace 1 dimensional and their sum is $n$, $A$ is diagonalizable.


问:
Strange Fourier series with all coefficients being null
I needed to calculate the Fourier series expansion of the function
$$
f(x) = \sin(x)\cos(x)\ ,
$$
so I started computing the coefficients of the series
$$
f(x) = A_0 + \sum_{n=1}^\infty A_n\cos(nx) + B_n\sin(nx)\ .
$$

The $A_0$ and $A_n$ coefficients became zero because of the parity of the function $f(x)$, so only the $B_n$ coefficients should survive. However, I got surprised when entering the following integral into the Maple software,
$$
B_n = \dfrac{1}{\pi} \int_{-\pi}^\pi \sin(x)\cos(x)\sin(nx)\ dx\ ,
$$
which returned the result
$$
B_n = \dfrac{2\sin(n\pi)}{\pi(n^2-4)}\ ,
$$
that seems to be always null since $\sin(n\pi) = 0$ when $n\in\mathbb{N}$.

This would mean that all the coefficients of the series are zero, obviously not matching with the original behaviour of $f(x)$. Where did I make a mistake? What is happening here?

答:
Note that $B_n$ is undefined when $n=2$ (a zero appears in the denominator), so you have to calculate that separately. Then

$$B_2 = \frac 1 \pi \int_{-\pi}^\pi \sin(x)\cos(x)\sin(2x) \, \mathrm{d}x$$

You can show that $B_2 = 1/2$. Then you have a single-term Fourier series showing that

$$\sin(x)\cos(x) = \frac 1 2 \sin(2x)$$


答:
When $n = 2$ both the numerator and denominator are zero. Taking the limit as $n \to 2$ reproduces the correct answer: we have, by l'Hopital's,

$$\lim_{n \to 2} \frac{2 \sin n\pi}{\pi(n^2 - 4)} = \lim_{n \to 2} \frac{2 \pi \cos n \pi}{2n \pi} = \frac{1}{2}$$

which correctly reproduces the double-angle formula $\sin x \cos x = \frac{1}{2} \sin 2x$ as already mentioned.

Title: Strange Fourier series with all coefficients being null

Body:
I needed to calculate the Fourier series expansion of the function
$$
f(x) = \sin(x)\cos(x)\ ,
$$
so I started computing the coefficients of the series
$$
f(x) = A_0 + \sum_{n=1}^\infty A_n\cos(nx) + B_n\sin(nx)\ .
$$

The $A_0$ and $A_n$ coefficients became zero because of the parity of the function $f(x)$, so only the $B_n$ coefficients should survive. However, I got surprised when entering the following integral into the Maple software,
$$
B_n = \dfrac{1}{\pi} \int_{-\pi}^\pi \sin(x)\cos(x)\sin(nx)\ dx\ ,
$$
which returned the result
$$
B_n = \dfrac{2\sin(n\pi)}{\pi(n^2-4)}\ ,
$$
that seems to be always null since $\sin(n\pi) = 0$ when $n\in\mathbb{N}$.

This would mean that all the coefficients of the series are zero, obviously not matching with the original behaviour of $f(x)$. Where did I make a mistake? What is happening here?


问:
$E(M_{n\wedge \tau}) = E(M_0)$ implies $M$ is a martingale
$(M_n)_{n\geq 0}$ is an adapted family such that $E|M_n|<\infty$, then, I need to show that $E(M_{n\wedge \tau}) = E(M_0)$ implies $M$ is a martingale, for every stopping $\tau$.

Honestly, I do not know how to start. Any hint? Thanks

答:
This is some kind of variation of no-arbitrage principle.

So, for any $k<n$, we define the following time:
$$\tau_k:=
    \begin{cases}
      k, & \text{if}\ M_{k} \ge \mathbb{E}(M_{k+1}|F_k) \\
      k+1, & \text{other wise}
    \end{cases}
$$
Clearly, $\tau_k$ is a stopping time.

On the other hand we have:

$$ \mathbb{E}(M_{k}) = \mathbb{E}(M_0)= \mathbb{E}( M_{n \wedge \tau_k})= \mathbb{E}( M_k \mathbb{1}_{\ M_{k} \ge \mathbb{E}(M_{k+1}|F_k)} +M_{k+1} \mathbb{1}_{\ M_{k} < \mathbb{E}(M_{k+1}|F_k)}  )   $$
Which implies $\mathbb{P}( M_k < \mathbb{E}(M_{k+1}|F_k) )=0$
Vice versa, we have:
$$\mathbb{P}( M_k > \mathbb{E}(M_{k+1}|F_k) )=0$$
Hence,
$$ \mathbb{E}(M_{k+1}|F_k)= M_k$$
Thus, the conclusion.


问:
Existence of a Hyperspace for any Vector Space.
Given any Vector space $V$(as mentioned by @GEdgar: $V \neq\{0\}$). It contains a Maximal-Proper-Subspace.

I thought to use Zorn's Lemma to prove this statement. I considered $\Sigma:=$ Set of all proper subspaces of $V$. Now taking a chain $\{V_i:i \in I\}$ in $\Sigma$, I want to obtain an upper-bound of that chain by considering $\cup_{i \in I} V_i$ which is a subspace of $V$. But I cannot show $\cup_{i \in I} V_i$ is a proper-subspace of $V$.

I also thought of proving this in another way, let $(x_i)_{i \in I}$ be a basis for $V$. I take out only one element from basis say $x_{i_0}$. Now I want to show that the subspace generated by $\{x_i:i \in I\} \setminus \{x_{i_0}\}$ is maximal. But I cannot show that.

Note: I think in the second method Zorn's Lemma is not needed.

答:
You are correct, once you use AC to get a Hamel basis, you do not need it again.  But we can do this with Zorn in almost the way you tried.

Note: your main assertion fails for $V = \{0\}$.  It has no proper subspace at all.

Now assume $V \ne \{0\}$.  Choose some $u \ne 0, u \in V$.  Then apply Zorn on Σ :=  Set of all subspaces of $V$ with $u  \notin V$.


问:
Expected Value of Exponential Function of Stochastic Process: $E[e^{X_t}]$
I am still getting used to expectations and stochastic processes. Supposed we have a stochastic process defined by
$$dX_t = \mu(t,X_t)dt+\sigma(t,X_t)dW_t$$
where $W_t$ is a Wiener process. The simple thing I need to do is evaluate the following expression for some time $T>t$.
$$E_{t,x}[e^{X_T}]$$
Would I be correct in assuming the following logic?
$$E_{t,x}[e^{X_T}]=e^{E_{t,x}[X_T]}$$
Which then we can use Itô's Lemma to find $E_{t,x}[X_T]$. First we define $f(t,x)=x$. Now we have the stochastic differential. Here we use the facts that $\frac{\partial f}{\partial t}=0$ and $\frac{\partial^2 f}{\partial x^2}=0$
$$df(t,X_t)=\mu(t,X_t)dt+\sigma(t,X_t)dW_t$$
Now we integrate
$$\int_t^T df(t,X_t)=\int_t^T\mu(t,X_t)dt+\int_t^T\sigma(t,X_t)dW_t$$
Which implies this (the expected value of the $dW_t$ integral is zero).
$$f(T,X_T)=f(t,X_t)+\int_t^T\mu(t,X_t)dt$$
Finally we take expectation.
$$E_{t,x}[f(T,X_T)]=E_{t,x}[X_T]=f(t,X_t)+ \int_t^T E_{t,x}[\mu(t,X_t)]dt$$
So then this leaves me with the final representation of the expectation that I am looking for.

$$E_{t,x}[e^{X_T}]=e^{E_{t,x}[X_T]}=e^{f(t,X_t)+ \int_t^T E_{t,x}[\mu(t,X_t)]dt}$$

I guess then we just hope that the $\mu$ function is easy enough to analyze. This just still doesn't feel so comfortable to me, I was wondering if I'm doing this right or if I'm missing something here.

Thanks.

答:
As pointed out in the comments above, it is not correct to assume that $\mathbb{E}[e^X]=e^{\mathbb{E}[X]}$.

The easiest counterexample is Normal vs. Lognormal distribution. If $X\sim N(\mu,\sigma)$, then $Y:=e^X$ is Lognormal by definition (i.e. see definition of Lognormal distribution).

The expected value of $Y$ is $\mathbb{E}[e^{X}]=e^{\mu+0.5\sigma^2}\neq e^{\mu}=e^{\mathbb{E}[X]}$.

You can either take the above as a "fact" that arises from the definition of Lognormally distributed random variable (i.e. you can memorize what the mean of a Lognormally distributed random variable looks like), or we can derive it by deriving the distribution of $Y$ from scratch:

$$F_Y(y)=\mathbb{P}(Y\leq y)=\mathbb{P}(e^{X}\leq y)=\mathbb{P}(X\leq ln(y))=\int_{-\infty}^{ln(y)}f_X(h)dh$$

Now:

$$f_Y(y):=\frac{\partial}{\partial y}\left(F_Y(y)\right)=\frac{\partial}{\partial y}\left(\mathbb{P}(X\leq ln(y))\right)=\frac{\partial}{\partial ln(y)}\left(\int_{-\infty}^{ln(y)}f_X(h)dh\right)\frac{\partial ln(y)}{\partial y}=\\=f_X(ln(y))\frac{1}{y}=\\=\frac{1}{\sqrt{2\pi}}e^{\frac{-(ln(y)-\mu)^2}{2\sigma^2}}\frac{1}{y}$$

Taking the expectation:

$$\mathbb{E}[Y]:=\int_{-\infty}^{\infty}\left(y f_Y(y)\right)dy=\\=\int_{-\infty}^{\infty}\left(\frac{1}{\sqrt{2\pi}}e^{\frac{-(ln(y)-\mu)^2}{2\sigma^2}}\right)dy$$

Taking the substitution $y=e^h$, we get:

$$\mathbb{E}[Y]=\int_{-\infty}^{\infty}\left(\frac{1}{\sqrt{2\pi}}e^{\frac{-(h-\mu)^2}{2\sigma^2}}\right)e^hdh$$

Now focusing on the exponents:

$$exp\left\{\frac{-(h-\mu)^2}{2\sigma^2}\right\}exp\left\{h\right\}=exp\left\{\frac{-h^2+2\mu h -\mu^2}{2\sigma^2}\right\}exp\left\{\frac{2\sigma^2h}{2 \sigma^2}\right\}=\\=exp\left\{\frac{-h^2+2h(\mu+\sigma^2) - (\mu+\sigma^2)^2 + (\mu+\sigma^2)^2 -\mu^2}{2\sigma^2}\right\}=\\=exp\left\{\frac{-(h-(\mu+\sigma^2))^2}{2\sigma^2}\right\}exp\left\{\frac{(\mu+\sigma^2)^2 -\mu^2}{2\sigma^2}\right\}=\\=exp\left\{\frac{-(h-(\mu+\sigma^2))^2}{2\sigma^2}\right\}exp\left\{\mu + 0.5\sigma^2\right\}$$

Plugging this back into the integral, we get:

$$\mathbb{E}[Y]=exp\left\{\mu + 0.5\sigma^2\right\}\int_{-\infty}^{\infty}\left(\frac{1}{\sqrt{2\pi}}e^{\frac{-(h-(\mu+\sigma^2))^2}{2\sigma^2}}\right)dh=\\=exp\left\{\mu + 0.5\sigma^2\right\}$$

Because the expression in the integral is the PDF of a Normally distributed random variable (with mean $\mu+\sigma^2$): which integrates to 1.

Now to your problem:

$$X_t=X_0+\int_{h=0}^{h=t}\mu(X_h,h)dh+\int_{h=0}^{h=t}\sigma(X_h,h)dW_h$$

If $\mu(X_h,h)=\mu X_h$ with $\mu$ being a constant and $\sigma(X_h,h)=\sigma X_h$ with $\sigma$ being a constant, then $X_t$ is Geometric Brownian Motion, which has the well known solution:

$$X_t=X_0exp\left\{\mu t -0.5 \sigma^2 t + \sigma W_t\right\}$$

From the above, we can see directly that $X_t$ is log-normally distributed. So that:

$\mathbb{E}[X_t]=X_0e^{\mu t}$

If you want to take the conditional expectation $\mathbb{E}_t[X_T]=\mathbb{E}[X_T|\mathcal{F}_t]$, then:

$$\mathbb{E}[X_T|\mathcal{F}_t]=\mathbb{E}\left[X_t exp\left\{\mu (T-t) -0.5 \sigma^2 (T-t) + \sigma W(T-t)\right\}\right]=\\=X_te^{\mu (T-t)}$$

Edit: if we cannot assume that the process is GBM (and therefore know its distribution), but we have to stick with the general $\mu(X_t,t)$ and $\sigma(X_t,t)$ functions (that, we must assume, are at least square-integrable, for $X_t$ to be an Ito Process), we can apply Ito's lemma as follows:

Ito's Lemma states that any twice differentiable function $F()$ of $X_t$ and $t$ follows the process:

$$F(X_t,t)= F(X_0,t_0) + \int_{h=0}^{h=t}\left(\frac{\partial F}{\partial t}+\frac{\partial F}{\partial X}\mu(X_h,h)+\frac{1}{2}\frac{\partial^2 F}{\partial X^2}\sigma(X_h,h)^2\right)dh + \int_{h=0}^{h=t}\left(\frac{\partial F}{\partial X}\sigma(X_h,h)\right)dW_h $$

We can apply the above formula to $e^{X_t}$ to get:

$$e^{X_t}=e^{X_0}+\int_{h=0}^{h=t}\left(0+e^{X_h}\mu(X_h,h)+\frac{1}{2}e^{X_h}\sigma(X_h,h)^2\right)dh + \int_{h=0}^{h=t}\left(e^{X_h}\sigma(X_h,h)\right)dW_h$$

Now we could attempt to take an expectation of the above: you are correct in your question to say that the expectation will "kill" the Ito Integral (because of the martingale property of the Ito integral, its expectation is equal to zero), but unless we know what the functions $\sigma(X_h,h)$ and $\mu(X_h,h)$ actually are, we won't be able to get any proper answer anyway, because we won't be able to simplify the expectation over the Riemann integral.


问:
In what kind of Banach algebras is 0 the only topological zero divisor?
On page 33 of http://math.aalto.fi/opetus/harmanal/pruju/calg04.pdf it is asked
in what kind of Banach algebras is 0 the only topological zero divisor.

What do they mean by kind of Banach algebras. Using part (b) of the exercise it
follows that the Banach algebras may not have invertible elements. Is
this a kind of Banach Algebras?

答:
Writing $G(A)$ to be the open subset of invertible elements, any $x \in \partial G(A)$ (the boundary) is a topological zero divisor.
To see this, just observe that $\frac{xx_n^{-1}}{\|x_n^{-1}\|}\to 0$ whenever $x_n\in G(A)$ converges to $x$.

If $A$ has no nonzero topological zero divisors then, by the above, $\{0\}\cup G(A)$ is closed and so the set of nonzero non invertible elements, call it $S(A)$, is open.
Further, if $x \in S(A)$, then the continuous function $f(\lambda)= (1-\lambda)x + \lambda e$ on $[0,1]$ has image in $S(A) \cup G(A)$, the union of two disjoint open sets. Contradiction. So the assumption that $S(A) \neq \emptyset$ was wrong.

Hence $A = \mathbb{C}$ by the Gelfand-Mazur theorem.


问:
Are Taylor series and power series the same "thing"?
I was just wondering in the lingo of Mathematics, are these two "ideas" the same? I know we have Taylor series, and their specialisation the Maclaurin series, but are power series a more general concept? How does either/all of these ideas relate to generating functions?</p>

答:
As others have noted, a power series is a series $\sum_{n=1}^{\infty} a_n x^n$ (or sometimes with $x$ translated by some $x_0$, to become $(x - x_0)$).  Normally
when one says Taylor series, one means the Taylor series of some particular smooth function $f$.  (So in mathematical speech, one wouldn't usually say "consider a Taylor series".  You might say "consider a power series", or
"consider the Taylor series of the function $f$".  At least, this is my experience.)

One complication in making too much of a distinction is that any power series (say with real coefficients) is the Taylor series of a smooth function (this is a theorem of Borel).  So the distinction is more terminological than logical.

Added: Borel's theorem is discussed here.


问:
Every finite dimensional Subspace of a Hausdorff toplogical vector space is closed
I'd like to proof the Theorem stated in the question Title.Which is

Let $(X,\tau )$ be a hausdorff topological vectorspace and $F\subseteq X$ be a finite dimensional Subspace. Show that F must be closed.

Now i am aware of the proof which uses an Isomorphism to Rn and uses local compactness of the preimage of the unit ball.
However i want to prove it in another more straight forward way.

The idea is the following:
$F\quad is\quad closed\quad <->\quad X\setminus F\quad is\quad open\quad <->\quad for\quad all\quad x\quad \in \quad X\setminus F\quad there\quad exists\quad an\quad open\quad neighbourhood\quad V\quad s.t\quad V\sqsubseteq X\setminus F\quad <->\quad Every\quad Point\quad in\quad X\setminus F\quad is\quad an\quad interior\quad Point\quad <->\quad X\setminus F\quad is\quad open$

Now my problem: To show the second part in the reasoning above we could show that for $x\in X\setminus F$ the set $F'\setminus F$ forms an open neighbourhood of x (in respect to the subspace topology of F') where $F'=Lin(F\bigcup  \left\{ x \right\} )$. Im struggeling to show this part.

My try: Let $x\in U$ and U be open in the subspace toplogy of F' i.e U is of the Form $U=O\bigcap  F'$ for some $O\in \tau $ so than we have

$O\bigcap  F'=O\bigcap  (Lin(F\bigcup  \left\{ x \right\} )\sqsubseteq (O\bigcap  Lin(F))\quad \bigcup  \quad (O\bigcap  \left\{ x \right\} )$

How do i continiue from here. I hope i made my problem clear. Thanks for any Help and Sorry for the bad editing!

答:
I think the result might not be true in general, if the field $\mathbb K$ over which your linear space is defined is not complete.

Maybe rather show directly that it is closed, instead of considering its complement.

If $F$ has dimension $1$, then $F=\{\lambda e:\lambda\in \mathbb K\}$, where $e$ is any fixed nonzero element of $F$. Then, show that the function that maps any element $\lambda e$ of $F$ to $\lambda\in\mathbb K$ (this map is well defined, since $\lambda$ is uniquely determined by $\lambda e$) is continuous. Hence, $F$ will automatically be closed.

Then, if $F$ has dimension $\geq 2$, you can maybe proceed by induction.


问:
Why does the cutting plane method for integer programming run in exponential time?
I am looking for a proof of the fact that the cutting plane algorithm for integer programming does not run in polynomial time. </p>

<p>The algorithm consists in adding constraints to the initial problem in order to approximate the convex hull of the polyhedron. I understand the exponential complexity comes from the fact that the number of such valid inequalities is exponential. But why? </p>

答:
Consider $\min \{ c^Tx : x \in P \cap\mathbb{N}\}$ for some polyhedron $P$. When you replace $P$ with $P'$ where all extreme points of $P'$ are integral and there is no integer vector in $P$ that is not in $P'$, then by solving $\min \{ c^Tx : x \in P'\}$ with a method that returns a cornerpoint (e.g., the simplex method), you obtain an integer solution to the original problem.

A cutting plane method constructs $P'$ from $P$ by iteratively 'slicing off' parts of $P$ that do not contain integer solutions.

The question assumes that all cutting plane methods run in exponential time. That is not true. It is possible to add a small number of cutting planes (equal to the number of decision variables) and obtain the optimal solution to an integer optimization problem. To do that, you need to add the cutting planes that are the facet defining inequalities for $P'$ at the optimal solution. In practice, you do not know what those cutting planes are, so you end up adding many more cutting planes than necessary. That number of cutting planes can be exponential, as the following example shows.

Consider the problem $\max_{x \in \mathbb{N}^2} \{ x_2 : 2kx_1 + x_2 \leq 2k$, $-2kx_1 + x_2 \leq 0 \}$ for $k\in\mathbb{N}$. (For example, take $k=4$ and draw along.) The feasible region is a triangle between $(0,0)$, $(1,0)$ and $(0.5, k)$ intersected with $\mathbb{N}^2$. The optimal solution to the relaxed problem is the top of the triangle. If you multiply the first inequality with $2k-1$, the second inequality with $2k+1$, add, and round, you get the cutting plane $-x_1 + x_2 \leq k-1$, which invalidates the first solution. Visually, you can deduce the following series of cuts (i.e., linear inequalities with integer coefficients and an integer right-hand side):
$$x_1 + x_2 \leq k-1$$
$$-x_1 + x_2 \leq k$$
$$x_1 + x_2 \leq k-2$$
$$-x_1 + x_2 \leq k-1$$
etc, until you have added the last three cuts $-x_1+x_2 \leq 0$, $x_1+x_2 \leq 1$, and $x_1 + x_2 \leq 0$. Thus, the number of cuts needed is $2k+1$, which is exponential in the input size, which is $\mathcal{O}(\log k)$.


问:
Random Index Central Limit Theorem (redux)
My question is similar to Random index central limit theorem.  I have a different proof of a similar theorem and would like to check if it is correct.  I am suspicious that I have made an oversight because my proof seems to be considerably simpler.

Let $X_1, X_2, ...$ be i.i.d. with $EX_i = 0$ and $EX_i^2 = 1$.  Let $a_n$ be an increasing sequence of natural numbers (I could just as well take $a_n = n$) such that $a_n \rightarrow \infty$ and $N_n$ be a random sequence of natural numbers with $\frac{a_n}{N_n} \rightarrow 1$ in probability.  I wish to show that $\frac{S_{N_n}}{\sqrt{N_n}} \implies \mathcal{N}(0, 1)$, i.e., converges in distribution.

proof: First, $\frac{S_{a_n}}{\sqrt{a_n}} \implies \mathcal{N}(0, 1)$ by the CLT and since a subsequence of a weakly converging sequence converges weakly.  Then, fix some $\epsilon > 0$ and let $A_n = \{|\frac{S_{N_n}}{\sqrt{N_n}} - \frac{S_{a_n}}{\sqrt{a_n}}| > \epsilon\}$.  Conditioning on the event $|N_n - a_n| \ge 1$:

\begin{align*}
P(A_n)
&= P(A_n\ \big|\ |N_n - a_n| \ge 1)P(|N_n - a_n| \ge 1) + P(A_n\ \big|\ |N_n - a_n| < 1) P(|N_n - a_n| < 1)\\
&= P(A_n\ \big|\ |N_n - a_n| \ge 1)P(|N_n - a_n| \ge 1)\\
&\le P(|N_n - a_n| \ge 1)\\
&\rightarrow 0
\end{align*}

where I've used $|N_n - a_n| < 1 \implies N_n = a_n$ for the second inequality and $P(\cdot) \le 1$ in the third, and that $N_n/a_n \rightarrow 1$ in probability in the final step.

Since $\epsilon$ is arbitrary, it follows that $|\frac{S_{N_n}}{\sqrt{N_n}} - \frac{S_{a_n}}{\sqrt{a_n}}| \rightarrow 0$ in probability and therefore $\frac{S_{N_n}}{\sqrt{N_n}} \implies \mathcal{N}(0, 1)$ by Slutsky's theorem. $\square$

Is this proof valid?

答: Normally, you don't have $\lim_{n \rightarrow +\infty} \mathbb{P}( |N_n-a_n| \ge 1 ) =0$ unless more conditions are specified.


问:
Does there exist an explicit formula for the result of polynomial division with non linear factors?
The remainder theorem states that, if a polynomial $P(x)$ is divided by a linear factor $x-b$ , then the remainder of the division is $ P(b)$. I am looking for a more general result of what would be the outcome if the dividing factor was some general factor instead of a linear one.. i.e:

$$ P(x) = R(x)  + Q(x) J(x)$$

Where, $ R(x)$ is the remainder, $J(x)$ is the dividing polynomial which has degree less then $P(x)$  and $Q(x)$ is the divisor. I have previously derived this result for factor of the form See here:

$$J(x) = (x-a)^n$$

but suppose I had some irreducible polynomial of the form( as an example):

$$J= x^2 + x +1$$

Then what would be the remainder?

答:
The remainder will be $0$ or a polynomial with degree less than $x^2+x+1$. For example, if we divide $x^4$ by $x^2+x+1$:
$$
\require{enclose}
\begin{array}{r}
\color{#C00}{1x^2}\color{#090}{-1x}\color{#00F}{+0}\\[-4pt]
x^2+x+1\enclose{longdiv}{x^4+0x^3+0x^2+0x+0}\\[-4pt]
\underline{\color{#C00}{x^4+1x^3+1x^2}}\phantom{+0x+0\ \ }\\[-4pt]
-1x^3-1x^2\phantom{+0x+0\ \ }\\[-4pt]
\underline{\color{#090}{-1x^3-1x^2-1x}}\phantom{+0\ \ }\\[-4pt]
1x+0\\[-4pt]
\underline{\color{#00F}{0x^2+0x+0}}\\[-4pt]
1x+0\\[-4pt]
\end{array}
$$
Thus,
$$
x^4=\left(x^2+x+1\right)\overbrace{\left(x^2-x\right)}^Q+\overset{\substack{R\\\downarrow\\{\phantom{A}}}}{x}
$$


答:
Consider $(x^3 + x^2 - x + 2):( x^2 + 1)$

The remainder has the form $R(x)=ax+b$. Divisor is $(x+i)(x-i)$

$x^3 + x^2 - x + 2=Q(x)(x+i)(x-i)+ax+b$

Plug $x=i$

$i^3+i^2-i+2=Q(i)(i+i)(i-i)+ai+b \to ai+b=1-2 i$

and $x=-i\to -ai+b=1+2 i$

we've got the system
$$
\begin{cases}
ai+b=1-2 i\\
-ai+b=1+2 i\\
\end{cases}
$$
$b=1,a=-2$

The remainder is $R(x)=-2x+1$

No general rule AFAIK

Title: Does there exist an explicit formula for the result of polynomial division with non linear factors?

Body:
The remainder theorem states that, if a polynomial $P(x)$ is divided by a linear factor $x-b$ , then the remainder of the division is $ P(b)$. I am looking for a more general result of what would be the outcome if the dividing factor was some general factor instead of a linear one.. i.e:

$$ P(x) = R(x)  + Q(x) J(x)$$

Where, $ R(x)$ is the remainder, $J(x)$ is the dividing polynomial which has degree less then $P(x)$  and $Q(x)$ is the divisor. I have previously derived this result for factor of the form See here:

$$J(x) = (x-a)^n$$

but suppose I had some irreducible polynomial of the form( as an example):

$$J= x^2 + x +1$$

Then what would be the remainder?


问:
Find distribution function and density of random variable
Point is selected uniformly from the shape $1 \leq |x| + |y| \leq 3, y > 0$.

Find distribution function and density of random variable $X = x$.

I find that $x \in [-3; 3]$ and $y \in (0,3]$, how can i find distribution function?

答:
HINT

In this uniformly drawn point, let $X$ denote the first coordinate and $Y$ denote the second one.

Plot the region on which $X,Y$ are defined.

Find the joint distribution function $f_{XY}$ of $X,Y$

You can compute the marginal distribution of $X$ as
$$
   f_X(x) = \int_\mathbb{R} f_{XY}(x,y)\ dy.
   $$


答:
A power series is a purely algebraic object, defined as a formal infinite sum $\sum_{n=0}^\infty a_nx^n$ where the $a_n$ are elements of some ring $R$ (for example, $R=\mathbb{R}$ or $\mathbb{C}$). You can choose any $a_n$ you like, and you still have a well-defined power series - one need not be concerned with questions of convergence. The set of all power series over a ring $R$ itself forms a ring $R[[x]]$. These power series do not in general define functions from $R$ to $R$; there is in general no way to make sense of an infinite sum of elements of $R$, so there is no sensible way to substitute an element of $R$ for $x$ in an arbitrary power series.

A Taylor series is a special kind of power series $T_{f,x_0}$ defined using a (real or complex) smooth function $f$ and a real/complex number $x_0$, as in Stahl's answer. Here we do have a sensible way to talk about convergence of limits in $\mathbb{R}$ and $\mathbb{C}$, and indeed we can interpret $T_f$ as giving us a function defined on a neighbourhood of $x_0$.

A generating function is a power series of the form $\sum_{n=0}^\infty a_nx^n$ where the coefficients $a_n$ are natural numbers. This is an algebraic object which encodes the sequence $\{a_n\}_{n=0}^\infty$ and does not in general define a real-valued function.

Title: Are Taylor series and power series the same "thing"?

Body: I was just wondering in the lingo of Mathematics, are these two "ideas" the same? I know we have Taylor series, and their specialisation the Maclaurin series, but are power series a more general concept? How does either/all of these ideas relate to generating functions?</p>


答:
To avoid case work you can simply define $c = \beta + \frac 12$ so $c>0$, $\beta=c-\frac 12$.

$$n - 2|n|\beta<\beta \iff n-2|n|c+|n|< c-\frac 12\\
\Leftarrow n<0, - 2|n|c < -\frac 12 < c-\frac 12 \Leftarrow |n| > \frac{1}{4c}
$$

Then if $n\le n_0 = - \lceil \frac{1}{4c} \rceil$ you have $\beta > \frac{n}{2|n|+1}$.

Title: Simple inequality question for infimum of $A = \{n \in\text{integers}:\frac{n}{2|n|+1} \}$

Body:
$$\beta >-\frac{1}{2}$$
There exists $a$ in $A = \{n \in\text{integers}:\frac{n}{2|n|+1} \}$ such that $a < \beta$
$$\beta > \frac{n}{2|n|+1}$$
$$n < \beta (2|n|+1) $$
$$n - 2|n|\beta<\beta $$

I don't know how to solve this, because there is an absolute value.


问:
disproving the set is connected
I am looking for a neat prove for If the closure of a set in $R$
is connected
too

Can you help ..

(concise explanition may help me)

答: $\mathbb{Q}^n$ seems good to me.  You just need to show that there's a point with rational coordinates arbitrarily close to any point in $\mathbb{R}^n$.  Let $x=(x_1,x_2,\dots,x_n)\in\mathbb{R}^n$.  Given $\varepsilon>0$, for each $1\leq k\leq n$ there is a $y_k\in\mathbb{Q}$ such that $\mid x_k-y_k\mid<\frac\varepsilon n$.  The let $=(y_1,y_2,\dots,y_n).$  We have $\mid x-y\mid<\varepsilon$.

Title: disproving the set is connected


问:
Multiple regression coefficients
A company is interested in examining the impact that negative reviews have on the sales of a product. So the Marketing department collected data for the 3 years of the monthly operation of the company and came up with the Regression model where Y is the monthly sales, X1 is the percentage of negative reviews the company received each month, X2 is the average monthly price of the product and X3 is the company's monthly expenses for advertising.

Assume that we are only interested in the impact that negative reviews have on monthly sales and that the negative reviews relate to a specific product feature.
So based to the following Excel output, how much money would you be willing to spend to improve that product feature so that to reset the negative reviews you receive monthly?

I would appreciate your quidance.

答:
I'm not sure im understading your question, I'm assuming you got that regression equation fitting some data or from some statistics book as an example. So you have 3 random variables $X_1$ , $X_2$ and $X_3$ , why do you want to force $X_1 = 0$?  What's the porpouse of the exercise you are doing? Maybe a elaborate a little more so I understand the problem.

Edit
 Ok, suppose you want to reset the negative impact the reviews have on your product. The assigment asks you, how much money will you spend to improve the feature of that product to reset the effect of the negative reviews? . We can't control the negative reviews, but we can control the price of the product and the marketing budget.
 Let's suppose that the inital price of the product is $X_2^0$ and the initial marketing budget is $X_3^0$. In the next month we decide lower the price of the product by an amount $\Delta X_2$, how are the sales affected? Using the coefficients that are given to you in the table:
 $$ Y = -0.030X_1 - 1.06(X_2^0 + \Delta X_2) - 0.0009  X_3^0 + 14.87$$
 $$ Y = -0.030X_1 -1.06\Delta X_2 - 1.06(X_2^0) - 0.0009 X_3^0 +  14.87$$ So setting $X_1$ to the amount of negative reviews of past month, we need to solve $$-0.030X_1 -1.06\Delta X_2 = 0$$ so:
$$ \Delta X_2 \approx -0.028X_1$$ That's no the complete answer because, this is how much the product price needs to be lowered such that it cancels the contribution of the negative reviews. So you would need to have additional information on how money spended on product improvement correlates to product price. Also just saying $\Delta X_2$ is not enough, because the coefficient in a linear fit have standard deviations. This means that we need to take into account error propagation, thats why I used the $\approx$ and not = in the result.  I suppose you have covered error propagation in your class if not check the wikipedia page to follow the reasoning: https://en.wikipedia.org/wiki/Propagation_of_uncertainty
 The correct presentation for the answer will be  $ \Delta X_2 = (-0.028 \pm \sigma)X_1$, where $\sigma$ is the standard error you estimate using the method in the link.

Moreover we haven't touched the marketing budget, maybe we can change that too! The problem is that the marketing budget seems negativility correlated to the sales even when you account the standard deviation, so we will leave that there.
 That's all the reasoning I can offer without more information, check with your TA or Professor if that's the correct line of reasoning to follow, if you are still stucked after talking to  them, make another edit and I see what can I do.

 (Also note that we have asume that all the variables are indepedent, in other words that changing $X_i$ doesn't affect $X_j$)

Title: Multiple regression coefficients


问:
Show that the set $U = \{(1,2), (1,3), (1,4)\}$ spans $\Bbb{R}^2$ but isn't a basis.
I really need help with that question, so I'd appreciate a solution with steps because this problem was given for practice and my exam is in a few days so I'd like to understand the steps to the solution. Thank you.</p>

答:
HINT

To spell out the comment hint a bit more, if $U$ spans $\mathbb{R}^2$, and $\vec{b} \in \mathbb{R}^2$, then there must exist $x,y,z \in \mathbb{R}$ such that
$$
\vec{b}
 = x\begin{bmatrix} 1\\2\end{bmatrix}
 + y\begin{bmatrix} 1\\3\end{bmatrix}
 + z\begin{bmatrix} 1\\4\end{bmatrix}
 = \begin{bmatrix} 1 & 1 & 1 \\2 & 3 & 4\end{bmatrix}
   \begin{bmatrix} x\\y\\z\end{bmatrix}
$$
So you have to prove that any $b$ can be generated that way...


问:
Prove (without quoting any theorems) that polynomials on [0,1] are continous
I'm confused as to go about this problem. I feel as if we have to show that

$P [0,1] \in C^{0}[0,1]$

by letting

$f = a_{n}x^{n} + a_{n-1}x^{n-1} + .... + a_{1}x^{1} + a_{0}$

We must show that if $f,g \in P([0,1])$, then $f+g \in P([0,1])$

Show that $a_{n}x^{n} \in C^{0}[0,1]$

Can anyone help me out where to go from here?

答:
Steps:

Prove that constant functions are continuous

Prove that the identity function $f(x)=x$ is continuous

Prove that if $f(x)$ and $g(x)$ are continuous, then so are $f+g$ and $f\cdot g$.

This suffices to prove that all polynomials are continuous.


问:
help with field notation.
I'm doing my undergraduate senior capstone on Schanuel's conjecture. Currently, I am researching James Ax's paper on the conjecture. In his paper he proves Schanuel's conjecture over the formal power series which says let $y_1, \dots, y_n \in  t\mathbb{C}[[t]]$ be $\mathbb{Q}$-linearly independent. Then $$\dim_{\mathbb{C}(t)}\mathbb{C}(t)(y_1,\dots,y_n,e^{y_1},\dots, e^{y_n}) \geq n $$ here the $\dim_{\mathbb{C}(t)}$ is the transcendence degree. My question here is two fold. First, What does $t\mathbb{C}[[t]]$ mean? I know $ \mathbb{C}[[t]]$ is the formal power series over the complex numbers, but that extra $t$ in front is throwing me. My other question is how to say all this notation out loud? Especially that long field notation in the conclusion of the theorem. I have a presentation on this and I am completely self taught through books so I don't know how to speak most of this stuff and I don't want to make a fool of myself at the presentation. If somebody could help I would appreciate it. Thank you.

答:
When $R$ is a ring, and $a$ is an element of $R$ then

$$ aR = \{ax : x \in R\} $$

Examples:

$2 \mathbb{Z} = \{2x : x \in \mathbb{Z}\} = $ the set of even integers

$t \mathbb{C}[[t]] = \{tf(t) : f(t) \in \mathbb{C}[[t]]\} = \{ \sum_{k \ge 1} a_k t^k : a_k \in \mathbb{C} \} = $ the set of power series which are divisible by t or which have no constant term

Alternative notations for $aR$ include:

$(a)$

$\langle a \rangle$

It is called the "principle ideal (of $R$) generated by $a$."

$\dim_{\mathbb{C}(t)}\mathbb{C}(t)(y_1,\dots,y_n,e^{y_1},\dots, e^{y_n}) \geq n$

I would say "the transcendence degree of this extension is at least $n$." You could give a name to "this extension" too, like define $K = \mathbb{C}(t)(y_1,\dots,y_n,e^{y_1},\dots, e^{y_n})$ or something.

$y_1,\dots,y_n \in t\mathbb{C}[[t]]$

Probably I would say "y one through y n are in t-c-t." You can also call $t\mathbb{C}[[t]]$ "the maximal ideal of $\mathbb{C}[[t]]$." Note: $\mathbb{C}[[t]]$ has only one maximal ideal.

Title: help with field notation.


问:
Suppose an acute triangle $ABC. D,E\in |BC|, |AD|=|CD|, |AE|=|BE|. F: FD||AB, FE||AC. $ Show that $|FB|=|FC|$.
It is clear that triangle $DFE$ is similar to $ABC$. I tried using inequalities such as $|AD|+|AE|\gt a$ to get information about the sides but with little success. Since $|FB|=|FC|\Longrightarrow \angle FCB = \angle FBC$ I attempted to divide the triangle into isosceles triangles and continue from there but I wasn't able to get the asnwer.

Thank you for any help.

答:
$D,E$ lie on the perpendicular bisectors of $AC, AB$ respectively.

Since $FD // AB$ and $FE//AC$, the perpendicular bisectors of $AC, AB$ are also perpendicular to $FE$ and $FD$, making them altitudes of $\triangle DEF$, and their intersection the orthocenter of $\triangle DEF$. This intersection is also the circumcenter of $\triangle ABC$.

The line joining the this intersection and the point $F$ is therefore also an altitude of $\triangle DEF$ with base $DE$, and is perpendicular to $BC$. The line passing through the circumenter of $\triangle ABC$ and perpendicular to $BC$ must be the perpendicular bisector of $BC$. Hence $|FB| = |FC|$.


问:
Composition of Riemann integrable function and C1 function is Riemann integrable
Let $f$ be a Riemann integrable function on $[a,b]$ and $g: [\alpha, \beta] \to [a, b]$ be a continuously differentiable function. Then how can I prove that $f \circ g$ is Riemann integrable? I think this is an easy task, but I can't get it. Maybe uniform continuity of $f$(+uniform continuity of $f'$) seems to be crucial, but I can't proceed any more.

There are no more assumptions about $f$ other than stated. According to the textbook containing this(Krantz, Real Analysis and Foundations, 4th ed, sec 7.3. It is an elementary real analysis textbook), this is an 'easier result' than proving that 'composition $\phi \circ f$ of continuous function $\phi$ on compact interval and Riemann integral function $f$ on $[a,b]$ is Riemann integrable'. I would appreciate answers not using measure theory if possible.

答:
This is true if in addition to $g \in C^1([\alpha,\beta]$),  the derivative $g'$ is non-zero. We then have an inverse function $g^{-1}$ which is continuously differentiable, mapping sets of measure zero into sets of measure zero.

Since $f$ is Riemann integrable it is bounded and the set of discontinuity points $D_f$ has measure zero. Thus, $f \circ g$ is Riemann integrable since it is bounded with discontinuity points in a set of measure zero, $D_{f \circ g} = g^{-1}(D_f)$.


问:
Distribution of Column Sums of Orthogonal Matrix?
Suppose I have a square real orthogonal matrix $A \in \mathbb{R}^D$, and I compute the element-wise sum of the $i$th column as $a_i := \sum_{d=1}^D A_{di}$. How can I describe the distribution of the $a_i$ values for the D columns?

I know that the maximum value $a_i$ can take is $\sqrt{D}$ and the minimum value is $-\sqrt{D}$. I'm wondering how fixing one $a_i$ affects the remaining $a_j$ for $j \neq i$.

Edit 1: I might be more interested in how the $a_i$ values decay if you order them such that $a_1 \geq a_2 \geq ... \geq a_D$. For instance, if $a_1 = \sqrt{D}$, what is the largest that $a_2$ can be?

答:
The possible vectors $(a_i)_{1\leq i \leq D}$ are exactly the vectors of $\mathbb{R}^n$ with norm $\sqrt{D}$. That is, we always have

$$ \sum_{i=1}^D a_i^2 = D $$

So to answer one question, if $a_1$ has the extreme value $a_1 = \pm \sqrt{D}$, then all other $a_i$ are exactly zero. This makes sense because the first column of $A$ must be $\pm D^{-1/2} [1, \ldots, 1]$, and taking the dot product with any other column must give zero, meaning the sum of that other column's elements is zero.

To prove the claim more generally, note that calling the unit basis $\{e_i\}$, we have $A_{ij} = e_i^T A\, e_j$, so

$$ a_i = \sum_{d=1}^D A_{di} = \sum_{d=1}^D (e_d^T A\, e_i) = \left[\sum_{d=1}^D e_d^T \right] A e_i $$

The left row vector is a vector with all elements $1$. So denoting $u = [1,\ldots,1]$, that gives

$$ a_i = u^T A \, e_i = (A^T u)^T e_i $$

That is, the numbers $a_i$ are the coordinates of the vector $A^T u$. Since $A$ is orthonormal, so is $A^T$, and therefore

$$\big| (a_i)_{1 \leq i \leq D} \big| = |A^T u| = |u| = \sqrt{\sum_{d=1}^D 1^2} = \sqrt{D} $$

One way of looking at this is that each value $(a_i / \sqrt{D})$ is the cosine of the angle between $A e_i$ (the image of an original basis vector) and the constant vector $u$.

For the converse, suppose we are given values $\alpha_i$ such that $\sum \alpha_i^2 = D$. Choose an orthonormal basis $\{x_i\}$ of $\mathbb{R}^D$ with $x_1 = D^{-1/2} (\alpha_i)_{1 \leq i \leq D}$, and choose another orthonormal basis $\{y_i\}$ of $\mathbb{R}^D$ with $y_1 = D^{-1/2} u$, for example by the Gram-Schmidt algorithm. If matrix $X$ has $\{x_i\}$ as columns and matrix $Y$ has $\{y_i\}$ as columns, then the matrix $A = Y X^T$ is one example of matrix with the values $a_i = \alpha_i$ as its column sums, since

$$ a_i = u^T A e_i = \sqrt{D} \cdot y_1^T Y X^T e_i = \sqrt{D} \cdot e_1^T X^T e_i = \sqrt{D} \cdot x_1^T e_i = \alpha_i $$

So for a probability distribution on vectors $(a_i)$, we could certainly use the usual distribution for the sphere $S^{D-1}$. I would expect this is the same distribution we would get from a sensible probability distribution on orthonormal matrices $A$ (however that would be defined), just as a principle of symmetry.


问:
how can I prove that if I have a regular language L, that L' is a regular language
how can I prove that if  I have a regular language L
and I create a new language L'

where L' = (L but with last letter repeated)
(i.e. if ab is in language L then abb is in language L')

that L' is a regular language

I tried solving the question but I have no way of being sure of
my solutions so I'll post it here:

L is a regular language therefor there is a regular expression for it,
we create a new language x where x =L*($\Sigma$),
$\Sigma$ is all the letters in language L
,x is a regular language because it is regular language chained with a letter
,and letters are a regular expression
the language L' is a subset of x,
there for x can be written as x=L' $\cup$ t (t=x/L'),
and therefor L' has to be a regular expression
because if it wasn't then x would not be a a regular language

答:
Since $L$ (over some alphabet $\Sigma$) is regular, there is a finite automaton $\mathcal{M}$ that recognises $L$. We now need to try and build a finite automaton $\mathcal{M}'$ that recognises $L'$. Let $Q_F$ be the collection of all state-symbol pairs of $\mathcal{M}$ that cause a transition into any of its accept states. More formally, $$Q_F := \{(q,s)\in Q\times \Sigma : \delta(q,s) \in F\},
$$ where $\delta\colon Q\times \Sigma \to Q$ is the transition function of $\mathcal{M}$, and $F$ is the set of accept states. We add a new state $A$ to $Q$ and make this the only accept state. Next, for every pair $(q,s)\in Q_F$, we add a new state $n_{(q,s)}$, and set $\delta^+(q,s) = \{n_{(q,s)}, \delta(q,s)\}$ and $\delta^+(n_{(q,s)},s) = A$. For $(q,s)\in (Q\times \Sigma) \setminus Q_F$ we leave $\delta^+(q,s) = \delta(q,s)$.

Note that the resulting automaton is non-deterministic, but this is fine since an equivalent DFA always exists. I now leave it up to you to show that this new NDFA $\mathcal{M}' := \{Q\cup A, \Sigma, \delta^+, q_o, A \}$ recognises $L'$.


答:
I think what this question is getting at is that, to be a basis of a vector space,

a set of vectors must span the vector space and be linearly independent.

If $\pmatrix{x\\y}$ is any element of $\mathbb R^2$, we can solve $\pmatrix{x\\y}=a\pmatrix{1\\2}+b\pmatrix{1\\3}$

to get $a=3x-y$ and $b=y-2x$,

so $\pmatrix{x\\y}$ is a linear combination of $\pmatrix{1\\2}$ and $\pmatrix{1\\3}$, so $\pmatrix{1\\2}$ and $\pmatrix{1\\3}$ span $\mathbb R^2$

(so certainly $\pmatrix{1\\2}$, $\pmatrix{1\\3},$ and $\pmatrix{1\\4}$ span $\mathbb R^2$).

Indeed, any two linearly independent vectors in $\mathbb R^2$ (which is a  $2$-dimensional vector space over $\mathbb R$)

will span $\mathbb R^2$ and be a basis.

Elements of a basis must be linearly independent, but $(1,4)=2(1,3)-(1,2)$,

so the set of three vectors given are not a basis.

Title: Show that the set $U = \{(1,2), (1,3), (1,4)\}$ spans $\Bbb{R}^2$ but isn't a basis.

Body: I really need help with that question, so I'd appreciate a solution with steps because this problem was given for practice and my exam is in a few days so I'd like to understand the steps to the solution. Thank you.</p>


问:
Confused about Cantor function and measure of Cantor set
OK, so I know that the Lebesgue measure of the ternary Cantor set is $0$.

However, in class the prof briefly mentioned that if we build a Lebesgue-Stieltjes measure $\mu$ out of the Cantor function, then $\mu(C) = 1$.

I don't understand why this is true, can someone please explain?

答:
The point here is that there are two different measures on the Cantor set $C$:

Viewing the Cantor set as a subset of $\mathbb{R}$, there is the Lebesgue measure $m_1$ that gives the Cantor set measure 0.

Viewing the Cantor set as the set $2^{\mathbb{N}}$ of all infinite binary sequences, the "fair coin" measure $m_2$ is defined so that for each $n \in \mathbb{N}$ and $i \in \{0,1\}$, the set  $\{ f \in 2^{\mathbb{N}} : f(n) = i\}$ has measure $1/2$. This measure is also called the Hausdorff measure of dimension $\log(2)/\log(3)$, where $\log(2)/\log(3)$ is the Hausdorff dimension of the Cantor set $C$.

The Cantor function $g(x)$ is defined such that $g(x) = m_2([0,x) \cap C)$. This is just the cumulative distribution function of $m_2$, but now we again view $C$ as a subset of the real line.


问:
Prove Matching of Size K
The problem is as follows: Given a graph with n vertices, each having degree >= 2k, prove that a matching of size >= k exists. (n > 2k > 0)

My initial approach has been that if we can just find a path within the graph of length 2k, then simply alternating the edges in that path would result in a matching of size k. But now I'm not sure how to prove that such a path of length 2k exists.

答: Hint: Suppose $(v_1, \ldots, v_m)$ is a path.  If $v_m$ has degree $\ge m$, there is some edge $(v_m, v_{m+1})$ so $(v_1, \ldots, v_{m+1})$ is a path.

Title: Prove Matching of Size K


问:
properties of a truncated tent maps on unit interval
given the following truncated tent map:

$T_h: [0,1] \rightarrow [0,1], x \mapsto min(h, 1-2|x -1/2|)$ for $0 \leq h \leq 1$

my script in dynamical systems states the following properties:

$T_1$ has only one periodic point (the fixed point 0) while the tent map $T_1$ has
a 3-cycle {2/7, 4/7, 6/7} and hence has all natural numbers as periods by the
Sharkovsky Forcing Theorem.

Any cycle $\mathcal{O} \subset [0, h)$ of $T_h$ is a cycle for $T_1$, and any cycle $\mathcal{O} \subset [0, h]$ of $T_1$ is a cycle for $T_h$.

Since the script does not provide any proofs on these claims I tried showing them myself, but got stuck at some point. Can someone help me?

Proof

My problem starts with the definition already. What is the difference between $T_1$ and tent map $T_1$? I thought $T_1$ was my tent map?
So, for the tent map: $T_1 = 1-2|x-1/2|$ for $0 \leq x \leq 1$. $T_1(0) = 0$, thus $x = 0$ is a fixed point and therefore of period one. Now, $T_1(2/3) = 2/3$, thus $x = 2/3$ is also a fixed point and of period one. What am I missing?

It is easy to check the second part, since $f(f(f(2/7))) = f(f(4/7)) = f(6/7) = 2/7$

I know how to show that the points $s^2/m (m,s \in \mathbb{N}, with 0<s\leq 2m)$ are eventually fixed points for T, s.d. $T^n(s/2^m) = p$. But I'm not sure that this is of any help here.

Any input is much appreciated!

答:
In case onyone ever comes across this question again, here are some answers:

There was a typing error in the script. It was supposed to say $T_0$ only has one fixed point. Which is obviously true, because $T_0 (x) = 0$. Thus $T_0(0) = 0$ is the only fixed point. The second part of the statement is correct.

Let $\mathrm(F)$ denote the set of fixed points for a function on an interval. Then, $$\mathrm(F)(T_h) \subset \mathrm(F)(T_1)$$ on the interval $[0,h)$. Notice that this only valid for the half open interval, since $T_h(h) = h$ is a fixed point for every $T_h$, but not for $T_1$. Likewise, $$\mathrm(F)(T_1) \subset (F)(T_h)$$ on $[0,h]$.


问:
Elementary substructure question
Suppose $\mathcal{M} \preccurlyeq \mathcal{N}$, where $M$ is the universe of $\mathcal{M}$, and $N$ is the universe of $\mathcal{N}$. Let $\phi(x,y)$ be an $L$-formula, where $x$ is n-ary and $y$ is m-ary. Suppose $\forall k \in N^m$, $\phi(x,k)$ defines a finite set in $N^n$.

Show that there exists $n\in \omega$ such that for all $k\in M^m$, $\phi(x,k)$ defines a set of size at most $n$ in $M^n$.

Attempt:

Since $\forall k \in N^m$, $\phi(x,k)$ defines a finite set in $N^n$. I thought about taking the maximum of the size of each $k$. But I'm not sure how to find some $n\in \omega$ such that for all $k\in M^m$ $\phi(x,k)$ defines a set of size almost $n$ in $M^n$.

I thought about using Tarski-Vaught test. But I couldn't get anywhere.

Conversely, suppose there exists $n\in \omega$ such that for all $k\in M^m$, $\phi(x,k)$ defines a set of size at most $n$ in $M^n$. Show if  $\mathcal{M} \preccurlyeq \mathcal{N}$, then $\forall k \in N^m$, $\phi(x,k)$ defines a finite set in $N^n$.

Attempt:

Suppose $\mathcal{M}\preccurlyeq \mathcal{N}$. Then we know $\mathcal{N} \models \phi(j(a))$ iff
$\mathcal{N} \models \phi(a)$, where $j$ is the containment map, $a \in M^n$. Hence, by definition of an elementary substructure. We get our desired result.

Could someone help me with the first part? And please let me know if my second part makes sense.

Thank you so much!

答:
This is false as written: take for example $\mathcal{M}=\mathcal{N}=(\mathbb{N}; <)$, and let $\varphi(x,y)\equiv x<y$. Clearly for each $k\in N$ there are only finitely many $a\in N$ such that $\mathcal{N}\models\varphi(a,k)$; however, it's also clear that there's no bound on the size of the sets picked out by $\varphi$ in $M$.

Here are a couple versions of the claim which are true:

Suppose $\mathcal{M}\preccurlyeq\mathcal{N}$, $\varphi(-,a)^\mathcal{N}$ is finite for each $a\in N$, and $\mathcal{N}$ is $\omega$-saturated. Then the conclusion holds.

Suppose $\mathcal{M}$ is such that every elementary extension $\mathcal{M}\preccurlyeq\mathcal{N}$ has $\varphi(-,a)^\mathcal{N}$ finite for each $a\in N$. Then the conclusion holds.

These two facts are really the same fact phrased two different ways, and the key to proving it/them is

 compactness.

Title: Elementary substructure question


问:
Why is $\hat y = y$ for me?
Assume a linear regression model $y=X\beta + \epsilon$ with $\epsilon \sim N(0,\sigma^2I)$ and $\hat y=Xb$ where $b=(X'X)^{-1}X'y$. Besides $H=X(X'X)^{-1}X'$ is the linear projection from the response space to the span of $X$, i.e., $\hat y=Hy$.

Now I have a really weird question.

If $\hat y=Hy$ then
\begin{align}
\hat y =& HX\beta + \epsilon \\
\hat y =& X\underbrace{(X'X)^{-1}X'X}_{\text{I}}\beta + \epsilon \\
\hat y =& X\beta + \epsilon \\
\hat y =& y
\end{align}

What's wrong here?

答:
The problem is that you are not multiplying the error term. It should be like that
\begin{equation*}
\begin{split}
\widehat{\mathbf{y}} & = \mathbf{H}\mathbf{y} = \mathbf{H}(\mathbf{X}\boldsymbol{\beta}+\boldsymbol{\varepsilon}) = \mathbf{H}\mathbf{X}\boldsymbol{\beta}+ \mathbf{H} \boldsymbol{\varepsilon}\end{split}
\end{equation*}

Title: Why is $\hat y = y$ for me?


问:
Number of Solutions for Congruency Equations
I'm leaning congruency equations, so for example:
$$
    ax \equiv b \pmod m
$$
I have that the number of solutions will be equal to $d$, where
$$
    d = \gcd(a, m).
$$
And the solutions ae:
$$
    x, x+m/d, x+2m/d, x+3m/d, \ldots , x+(d-1)m/d
$$
Now, having understood how I solve the equation, I am still not entirely sure as to why the number of solutions is given by $d$...

Anybody can calrify please?

Cheers! :)

答:
Let $d$ be a common divisor of $a$ and $m$. If the congruence $ax\equiv b\pmod{m}$ has a solution, then $ax-b$ is a multiple of $m$. Since $d$ divides both $m$ and $a$, it must divide $b$. In particular, if $d$ is the gcd of $a$ and $m$, the congruence has no solutions unless $d$ divides $b$. So from now on we suppose that $d$ divides $b$.  Let $a=a_1d$, $m=m_1d$, and $b=b_1d$.

The congruence $a_1dx\equiv b_1d\pmod{m_1d}$ holds if and only if $m_1d$ divides $a_1d x -b_1d$. This is the case if and only if $m_1$ divides $a_1 x-b_1$, that is, if and only if $a_1x\equiv b_1\pmod{m_1}$.

So now consider the congruence $a_1x\equiv b_1\pmod{m_1}$. Since $d$ is the greatest common divisor of $a$ and $m$, it follows that the numbers $a_1$ and $m_1$ are relatively prime.

Because $a_1$ and $m_1$ are relatively prime, the congruence $a_1x\equiv b_1\pmod{m_1}$ has a unique solution modulo $m$. For a proof of uniqueness, we can use the fact that $a_1$ has an inverse $c$ modulo $m_1$. Then $ax\equiv b\pmod{m_1}$ if and only if $ca_1x\equiv cb_1\pmod{m_1}$, that is, if and only if $x\equiv cb_1\pmod{m_1}$. (One can also give a direct proof of uniqueness without using the inverse.)

So now we ask: given the unique solution $x$ modulo $m_1$, which numbers are congruent to $x$ modulo $m$? Certainly the numbers $x+\frac{im}{d}$ are, where $i$ ranges from $0$ to $d-1$. For these are the numbers $x+im_1$, and they are all congruent to $x$ modulo $m_1$.

Are there any others? Suppose that $y\equiv x\pmod{m_1}$. Then $m_1$ divides $y-x$. So the possible remainders when $y-x$ is divided by $m$ are $0, m_1,2m_1,\dots, (d-1)m_1$. It follows that $y-x \equiv im_1 \pmod{m}$ for some $i$ where $0\le i\le d-1$. This just says that $y\equiv x+im_1 \pmod{m}$, that is, $y\equiv x+\frac{mi}{d}\pmod{m}$, for some $i$ ranging from $0$ to $d-1$.

Remark: The important part was the uniqueness modulo $m_1$. The rest is straightforward, and you know it well. To give a simple numerical example, suppose that we know that $x\equiv 3\pmod{16}$. What can we say about $x$ modulo $80$? We can say that $x$ is congruent to one of $3$, $19$, $35$, $51$, $67$.


问:
A problem about a rod breaking into $n$ pieces and probability theory

A rod of length $1$ is broken at random, which means that the remaining part has an even distribution at $(0,1)$. The remainder is broken in a similar way, and so on.

$1$.  Let $X_{n}$ be the length of the part left over after the rod has been broken $n$ times. Describe $X_{n}$ as a product.

$2$. Show that the sequence $\{\log(X_{n})/ n \}_{n\geq 1}$ almost sure converges, answering what the limit is.

$3.$  Obtain an approximation for the probability that $X_{36}\leq e^{-24}$

My attempt:

I think $\displaystyle X_{n}=\prod_{i=1}^{n}X_{i}$, where $i\in [0,n]$ with $x_{i}$ choosen uniformly from $(0,1)$. is it correct?

We need to prove that $$U_{n}=\frac{\log (X_{n})}{n} \overset{a.s}{\to} U$$
but $$\frac{\log(X_{n})}{n}=\frac{\log\left(\prod_{i=1}^{n}X_{i} \right)}{n}=\frac{\sum_{1\leq i\leq n}\log(X_{i})}{n}$$
but since that $X_{i}$ are i.i.d (I don't sure about it. Can you correct me if I'm wrong?) and since that $$\mathbb{E}[\log(X)]=\int_{0}^{1}1\cdot \log(x)dx=-1$$
where $f(x)=\frac{1}{1-0}$ is the density function of $X\sim U(0,1)$.
So, by Strong Law of Large Numbers we have $$U_{n} \overset{a.s}{\to}-1$$

I don´t know how to solve this option.

Can you teach me how can I solve this problem?

答:
$X_i$ are already defined in the problem, so it would be better to define a different random variable for your uniform random variables, e.g. $X_n = \prod_{i=1}^n Y_i$ for $Y_i \overset{\text{i.i.d}}{\sim} \text{Unif}(0,1)$.

You are correct that $\frac{1}{n} \log (X_n) = \frac{1}{n} \sum_{i=1}^n \log(Y_i) \overset{\text{a.s.}}{\to} E[\log(Y_1)]$, but $E[\log(Y_1)] = \int_0^1 \log x \, dx = -1$.

$P(X_{36} \le e^{-24}) = P(\log X_{36} \le -24) = P(\frac{1}{36}\sum_{i=1}^{36} \log(Y_i) \le -2/3)$. By a normal approximation (central limit theorem), you can approximate $\frac{1}{36}\sum_{i=1}^{36} \log (Y_i)$ with a $N(E[\log(Y_1)],  \text{Var}(\log(Y_1))/36)$ normal random variable to estimate the above probability.


问:
Generalized eigenvalue problem with symmetrical real matrices
I am investigating the generalized eigenvalue problem

$$(\lambda\,\boldsymbol{A}+\boldsymbol{B})\,\boldsymbol{x}=\boldsymbol{0}$$

where $\boldsymbol{A}$ and $\boldsymbol{B}$ are real-valued symmetrical matrices, $\lambda$ are the eigenvalues and $\boldsymbol{x}$ are the eigenvectors.

According to Wikipedia, the eigenvalues $\lambda$ are all real-valued if $\boldsymbol{A}$ is positiv definite. Is there a simple proof for that?

What happens if $\boldsymbol{A}$ is only positiv semidefinite? Are the eigenvalues are still all real-valued?

I assume, from a mechanical point of view, that for positiv definite matrices $\boldsymbol{A}$ and $\boldsymbol{B}$, the eigenvalues are all negative real-valued. Is there a simple proof for that statement? And again, what happens if the matrices (or one of them) is only positiv semidefinite?

答:
Assuming that $A$ is positive definite, then we have that it is also nonsingular. Thus, we can express the eigenvalue problem as follows:

$$
\lambda x = -A^{-1}Bx.
$$

But notice that since $A$ is PD, then so is $A^{-1}$, and so for all nonzero $x$:
$$
\lambda^2x^\top x =x^\top BA^{-2}Bx > 0,
$$
and thus $\lambda$ must be real, see https://math.stackexchange.com/a/354119/443030

EDIT: Regarding your second question, let $A=B=\begin{pmatrix}1&0 \\ 0&0\end{pmatrix}$ and $\lambda\in\mathbb{C}$. Clearly, $A$ is PSD, then

$$
\det(\lambda A+B) = \det\begin{pmatrix}\lambda+1&0\\0&0\end{pmatrix} = 0.
$$
It appears that if $A$ is PSD with $\ker A \cap\ker B \neq \{0\}$, then $\lambda$ is a free variable, i.e. we can choose any $\lambda\in\mathbb{C}$, so that
$$
(\lambda A + B)x = 0,
$$
for any $x = \begin{pmatrix}0\\ t\end{pmatrix},~ t\in\mathbb{C}$.

So yes, in this case, the generalized eigenvalue can be complex.


问:
Find the singularities and residues that correspond. Show details

Find all the singularities in the finite plane and the corresponding residues. Show the details.

$$f(z) = \frac{1}{1-e^z}$$

So I'm trying to find the singularities. So the singularities are located when the denominator = 0 and that occurs when $e^z = 1$ which occurs when $z = 2 \pi in$ when $n \in \mathbb{R}$

Now what? How can I find the residues?

so the residue formula that I know is:

$$\operatorname{Res}_{z = z_0}f(z) = \lim_{z \to z_0} (z-z_0) f(z)$$

That's what I'm trying to use. But how do I determine this when $z_0$ isn't fixed?

答: For each $n\in\Bbb Z$, you have\begin{align}\operatorname{res}_{z=2\pi in}\frac1{1-e^z}&=\lim_{z\to2\pi in}\frac{z-2\pi in}{1-e^z}\\&=-\frac1{\lim_{z\to2\pi in}\frac{e^z-e^{2\pi in}}{z-2\pi in}}\\&=-\frac1{\exp'(2\pi in)}\\&=-\frac1{e^{2\pi in}}\\&=-1.\end{align}


问:
Have no idea how to solve this task about probabilistic problem, just wonder what method should be applied
task:
From the box containing 7 black balls  and 3 red we select two.

What is the probability of getting two black balls if we know that one of the chosen balls is black ?

答:
Assuming you do not put back any of the chosen balls and that 10 is the total number of all balls in the box (thus there are no other colors in the box), then precede as follows:

first draw: P(drawing balck ball) = $\frac{7}{10}$
because you have 10 balls in total and seven of them are black.
Now, since you removed one ball, there are only 9 balls in the box left. Moreover, we know that the removed ball was black. Thus, there are now 6 black balls and three red balls in the box. Thence,

second draw: P(drawing black ball) = $\frac{6}{9}$.

Now, if you read your question carefully, you notice that the problem states a conditional probability. Notice the difference between these two statements:

What is the probability of drawing two balck balls?

What is the probability of drawing two black balls, knwoing that one of the balls is black.

The formula for conditional probability is given by:

$$P(A|B) = \frac{P(A \cap B)}{P(B)}$$

Thus,
P(second black ball, given we already have a black ball) = $\frac{\frac{7}{10} \cdot \frac{6}{9} }{\frac{7}{10}} = \frac{6}{9}.$


问:
probability of being crazy from jersy question
In a country, there are 2 states, Jersy and New Castle.
70% of the population are from Jersy, and 30% are from New Castle.
Out of the Jersy population, 6% are crazy, and out of New Castle, 3% are crazy.
if someone is crazy, what is the probability that he is from Jersy?

So I just said that $P(Jersy|Crazy)= \frac{P(Jersy\cap Crazy}{P(Crazy)}=\frac{0.7*0.06}{0.7*0.06+0.3*0.03} = 0.8235$

Is that correct?

答:
Yes, that is correct.

I would go into more detail than what you have shown, in order to be clear about how you performed the computation.

Let $J$ be the event that a randomly selected person is from Jersy and $N$ be the complementary event that a randomly selected person is from New Castle.  Let $C$ be the event that a randomly selected person is crazy.

Then we are asked to determine $$\Pr[J \mid C] = \frac{\Pr[C \mid J]\Pr[J]}{\Pr[C]}.$$  This is Bayes' rule.  Then the law of total probability gives for the denominator $$\Pr[C] = \Pr[C \mid J]\Pr[J] + \Pr[C \mid N]\Pr[N].$$  This is exactly what you have computed, with the following probabilities $$\Pr[J] = 0.7, \\ \Pr[C \mid J] = 0.06, \\ \Pr[N] = 1 - \Pr[J] = 0.3, \\ \Pr[C \mid N] = 0.03.$$