问:
Linear transformation $T$ such that for every extension $\overline{T}$, $\|\overline{T}\|>\|T\|$.
Let $E$ and $F$ be normed spaces such that $\dim F < \infty$, $G$ a subspace of $E$ and $T:G\rightarrow F$ a continuous linear map. I know that there exists a continuous linear extension $\overline{T}:E\rightarrow F$. Also, if $E$ is a Hilbert space, then $\overline{T}$ can be chosen in the way that $\|\overline{T}\|=\|T\|$.

Problem: Find an example of $E$, $F$, $G$ and $T$ (like above) such that every continuous linear extension $\overline{T}$ has a greater norm, i.e. $\|\overline{T}\|>\|T\|$.

Now, $F$ must be at least a 2-dimensional space, otherwise I could use Hahn-Banach to find an extension with equal norm.  My professor told me it could be done with $E$ of finite dimension. Of course, I tried to come up with an example of $E$ with a norm that doesn't satisfy the parallelogram law. For example, $E=\left(\mathbb{R}^3,\|\cdot\|_{\infty}\right)$ and $F=\left(\mathbb{R}^2,\|\cdot\|_1\right)$. But I couldn't prove that it works with any example I tried using those spaces.

Can somebody help me to find an example and assure that it really has that property?

EDIT:
Apparently, it can't be done with $E$ of finite dimension nor with $F$ equipped with the $\sup$ norm, as @Hamza proved below.

答:
Let $E=\mathbb R^3$, equipped with the sup norm, namely
$$
  \|(x,y,z)\|_\infty  = \max\{|x|, |y|, |z| \},
  $$
and let
$$
  F=G=\{(x, y, z)\in E : x+y+z=0\},
  $$
equipped with the induced norm from $E$.  Also let
$$
  T:G\to F
  $$
be the identity function.  Then clearly $\|T\|=1$, but any extension $\bar T:E\to F$ has norm strictly bigger than
one.
The reason is that any such  $\bar T$ is necessarily  a projection from $E$ to $G$ and there is no
such projection with norm $1$.
The best way to convince oneself of this fact  is to make a cardboard
model of this cube

cut it along the red line, place one of the two halves on top of the table with the red hexagon down,

and attempt to shine a flashlight so that the shadow is restricted to within the hexagon.  It is impossible!

This is based on another answer I recently gave to this question.


问:
Ordering of triples question
I have read this question in an exercise book:

"Consider the following strict ordering (denoted by >)...

a > b > c > d

Now consider in what way the following triples are related in >^3

(c, d, a),(d, a, d),(d, d, d),(a, a, a),(d, c, d)."

I am unsure what exactly the question is asking. Is it perhaps asking what properties of relation it has? e.g reflexive, irreflexive, total, linear, transitive etc. Any help would be great.

答: If I’m interpreting the question correctly, it misuses the expression $>^3$ to mean the relation on $\{a,b,c,d\}^3$ defined by setting $\langle u,v,w\rangle>^3\langle x,y,z\rangle$ if and only if $u>x$, $v>y$, and $w>z$. Thus, for instance, $\langle a,a,a\rangle>^3\langle d,d,d\rangle$, because $a>d$, but $\langle c,d,a\rangle\not>^3\langle d,d,d\rangle$, because even though $c>d$ and $a>d$, in the middle $d\not>d$. It appears that for each pair of the listed triples you are to determine which member of the pair (if either) is $>^3$ the other.

ContentLicense: CC BY-SA 4.0

Title: Ordering of triples question


问:
Quadratic residue modulo $p$ if quadratic residue module $p^k$
I'm trying to prove this statement. I've already checked here in StackExchange but I can only find the proof of the converse.

In Wikipedia, it is stated: "A number $a$ relatively prime to an odd prime $p$ is a residue modulo any power of $p$ if and only if it is a residue modulo $p$", and I want to prove the only if part.

This problem appeared to me when I wanted to solve the following problem:

Check that the only solutions, up to congruence, of $X^2 \equiv 25 $ (mod $p^k)$, for any $k$ and for any prime $p \neq 2$, $5$, are $X \equiv \pm 5 $  (mod $p^k)$.

Any help would be appreciated.

答:
You indicated that the problem

Let $k$ be a positive integer and let $p$ be a prime other than $2$ or $5$.

Show that the only solutions, up to congruence, of $x^2 \equiv 25\;(\text{mod}\;p^k)$ are $x \equiv \pm 5\;(\text{mod}\;p^k)$.

is the underlying problem which you are trying to solve.

For the above problem, there is no need to consider quadratic residues.

Instead, we can argue as follows . . .

Suppose $x$ is an integer such that $x^2 \equiv 25\;(\text{mod}\;p^k)$.

Note that $x+5$ and $x-5$ can't both be divisible by $p$, else their difference
$$
(x+5)-(x-5)=10
$$
would be divisible by $p$, contrary to $p\ne 2,5$.
\begin{align*}
\text{Then}\;\;&
x^2 \equiv 25\;(\text{mod}\;p^k)
\\[4pt]
\implies\;&
p^k{\,\mid\,}x^2-25
\\[4pt]
\implies\;&
p^k{\,\mid\,}(x+5)(x-5)
\\[4pt]
\implies\;&
p{\,\mid\,}(x+5)(x-5)
\\[4pt]
\implies\;&
p{\,\mid\,}(x+5)\;\text{or}\;p{\,\mid\,}(x-5)\;\text{but not both}
\\[4pt]
\end{align*}
Then since $p^k{\,\mid\,}(x+5)(x-5)$ and exactly one of $x+5,x-5$ is divisible by $p$, it follows that exactly one of $x+5,x-5$ is divisible by $p^k$.

Therefore $x \equiv \pm 5\;(\text{mod}\;p^k)$, as was to be shown.


问:
Supremum Norm Identity
Let $f,g : [0,1] → \mathbb{R}$ be bounded functions. Prove that $ ∥fg∥  ≤  ∥f ∥∥g ∥$

Where we define the sup norm of a bounded function $f : D → \mathbb{R}$  as   $∥f∥ := sup${|f(x)| : x ∈ D}$ $.

Any ideas of where to start with this proof or theorems that are useful would be helpful.

答: $$\|f \cdot g\|_\infty=\sup_{x\in D} |f(x) \cdot g(x)|= \sup_{x\in D}(|f(x)|| g(x)|)\leq \sup_{x\in D}( \|f\|_\infty\cdot |g(x)|) = \|f\|_\infty \cdot \|g\|_\infty$$

Title: Supremum Norm Identity


问:
Finding Expectation Given Conditional Expectation
Suppose random variable $ Y_{il} $ has the expectation:

$E(Y_{il}) = \int E(Y_{il} | Z_i= z)P^{Z_i}(dz)$

I am given that

$ E(Y_{il} | Z_i= z) = 1-exp(-\lambda_izB) $

Thus we get end up with the following equation for the Expectation.

$E(Y_{il}) = \int (1-exp(-\lambda_izB)P^{Z_i}(dz))$

Skipping over the constants $\lambda_i, B$ because they are irrelevant to the question.

Suppose that $Zi$ ~ $\gamma(\theta,\frac{1}{\theta}) $, and $P^{Z_i} $ is the probability distribution for Zi.

My Question: Because in the conditional expectation we condition on $Z_i= z$, Does this mean that our probability distribution also in terms of $Z_i= z$ or just a generic random variable $Z_i$?

Apologies if this is a stupid question.

答:
z should be lowercase, because you want to integrate over all its possible values.

You have $$E(Y)=\int E(Y|Z=z)p(Z=z)dz \text { integral rule}\\
=\int(1-e^{-\lambda Bz})p(Z=z)dz\\
=\int_{-\infty}^\infty(1-e^{-\lambda Bz})\frac{(1/\theta)^\theta}{\Gamma(\theta)}z^{\theta-1}e^{-\frac z \theta}dz$$

Indeed in your very first line you should write $P^{Z=z}$ not $P^Z$ otherwise that equality is not technically true.


Title: Extension Galois field, multiplication of elements

Body:
Consider the extension field $GF(2^4)$ and the primitive polynomial $P(x)=x^4+x^3+1$. How would I find the result of multiplication of elements $1011$ and $1100$?

My work:

$1011$ corresponds to $\{1x^3 + 0x^2 + 1x^1 + 1x^0\} = x^3 + x + 1$

$1100$ corresponds to $\{1x^3 + 1x^2 + 0x^1 + 0x^0\} = x^3 + x^2$

Multiplying:
$(x^3 + x + 1 ) * ( x^3 + x^2 )\\
= x^6 + x^ 5 + x^4 + 2x^3 + x^2\\
= x^2(x^3+1) + x(x^3+1) + (x^3+ 1) + 2x^3 + x^2\\
= x^5 +  x^2 + x^4 + x + x^3 + 1 + 2x^3 + x^2 \\
= x(x^3 + 1) +x^2 + (x^3 + 1) + x + x^3 + 1 + 2x^3 + x^2 \\
= x^4 + x + x^2 + x^3 + 1 + x + x^3 + 1 + 2x^3 + x^2 \\
= (x^3 +1) + x^2 + x^3 + 1 + x + x^3 + 1 + 2x^3 + x^2 \\
= 5x^3 + 2x^2 + x + 3
= x^3 + x + 3 \\
 $

Binary representation = 1011

But, I'm not sure if this is correct.

Body:
You could do this in binary, using long hand (carryless - xor instead of add) multiplication and (borrowless - xor instead of subtract) division:

Multiply

    1011
    1100
    ----
    0000
   0000
  1011
 1011
 -------
 1110100  == x^6 + x^5 + x^4 + x^2

Divide

            101
      ---------
11001 | 1110100
        11001
        -----
         01000
         00000
         -----
          10000
          11001
          -----
           1001   == x^3 + 1

Or following the method in the original question, noting that x^i + x^i == 0:

$(x^3 + x + 1 ) * ( x^3 + x^2 )\\
= (x^6 + x^4 + x^3) + (x^5 + x^3 + x^2) \\
= x^6 + x^5 + x^4 + x^2 \\
= x^2(x^3 + 1) + x(x^3 + 1) + (x^3 + 1) + x^2 \\
= (x^5 + x^2) + (x^4 + x) + x^3 + 1 + x^2 \\
= x^5 + x^4 + x^3 + x + 1 \\
= x(x^3 + 1) + (x^3 + 1) + x^3 + x + 1 \\
= (x^4 + x) + (x^3 + 1) + x^3 + x + 1 \\
= x^4 \\
= x^3 + 1
$


问:
How to compute $\int^{\infty}_{0} t^{(\frac1n-1)}\cos t \,\mathrm{d}t$?
How to calculate the below integral?
$$
\int^{\infty}_{0} \frac{\cos t}{t^{1-\frac{1}{n}}} \textrm{d}t = \frac{\pi}{2\sin(\frac{\pi}{2n})\Gamma(1-\frac{1}{n})}
$$
where $n\in \mathbb{N}$.</p>

答:
Write $t^{n^{-1}-1}\cos t$ as
$$\frac12t^{n^{-1}-1}e^{it}+\frac12t^{n^{-1}-1}e^{-it}.$$

Consider the integral as complex contour integral and

Turn the contour of integration for $\frac12t^{n^{-1}-1}e^{it}$ by $\frac{\pi}{2}$ counterclockwise (so that it becomes $t=is$, $s\in\mathbb{R}_{>0}$), then
$$\int_0^{\infty}\frac12t^{n^{-1}-1}e^{it}dt=\frac{e^{\frac{i\pi}{2}n^{-1}}}{2}\int_0^{\infty}s^{n^{-1}-1}e^{-s}ds=\frac{e^{\frac{i\pi}{2}n^{-1}}}{2}\Gamma(n^{-1}).$$

Similarly, turn the contour of integration for $\frac12t^{n^{-1}-1}e^{it}$ by $\frac{\pi}{2}$ clockwise, then
$$\int_0^{\infty}\frac12t^{n^{-1}-1}e^{-it}dt=\frac{e^{-\frac{i\pi}{2}n^{-1}}}{2}\int_0^{\infty}s^{n^{-1}-1}e^{-s}ds=\frac{e^{-\frac{i\pi}{2}n^{-1}}}{2}\Gamma(n^{-1}).$$

The sum of two integrals is equal to $\Gamma(n^{-1})\cos\frac{\pi}{2n}$ and it is an easy exercise (using Euler reflection formula $\Gamma(z)\Gamma(1-z)=\frac{\pi}{\sin\pi z}$) to show that this equals to the right side of your identity.

ContentLicense: CC BY-SA 3.0


问:
Generalization of $\sqrt{n}+\sqrt{n+2005}=m$
I was solving the math Olympiad of Belgium the year $2005$, and the last problem was :

If $n$ is an integer then find all values for $n$ for which $\sqrt{n}+\sqrt{n+2005}$ is an integer as well.

My question is just to generalize this problem, so instead of $2005$ , we can solve this over all $x\in \mathbb{N}$.

My Attempt:

If you have solved the problem you will realized that this equation :
$$\sqrt{n}+\sqrt{n+x}=m$$
Has solution if and only if $m\mid x$. But I am not sure if there are some other restrictions in this equation to has a solution.

答:
By squaring you get $$2\sqrt{n(n+x)} = m^2-2n-x$$

On the other hand we have, by multiplying both sides of starting equation with $\sqrt{n+x}-\sqrt{n}$:  $$(n+x)-n= m(\sqrt{n+x}-\sqrt{n})$$ and if we square this again we get $$x^2=m^2 (4n+2x-m^2)\implies m^2\mid x^2\implies m\mid x$$


问:
alphabet $\{0, 1\}$, the set of non-empty words whose first letter occurs at least two other times
Build finite automata that accept the language: the alphabet $\{0, 1\}$, the set of non-empty words whose first letter occurs at least two other times. Automata can be non-deterministic if that makes it easier for you. However, you cannot use transitions labeled with a regular expression or the empty word $\epsilon$.

I made this finite automata, but it seems to be wrong. Can you tell me what would be a good automata for this exercise?

EDIT

答:
It is indeed wrong: it accepts every non-empty word over the alphabet $\{0,1\}$. It’s actually just as easy to make the automaton deterministic. (By the way, automata is plural: the singular is automaton.) Like your attempt, it will have two independent tracks, one for a first letter $0$ and one for a first letter $1$.

The $0$ track will have states $q_1,q_2$, and $q_3$. Starting at $q_0$ there should be a $0$ transition to $q_1$ to start the track. A $1$ input at $q_1$ should just loop at $q_1$, and the $0$ transition should go to $q_2$. (Note that the subscript shows how many zeroes the automaton has read so far.) A $1$ input at $q_2$ should simply loop at $q_2$, and the $0$ transition should go to $q_3$, which should be an acceptor state. Both possible inputs should simply loop at $q_3$.

The $1$ track is very similar.


问:
How to formalize this proof?
I'm trying to prove the following claim, and I believe I have a proof.

Claim: If the system $Ax\leq b$ has a solution, then $b$ can be written as $u+v$, where $u \in \text{Im}(A)$ and $v \geq 0$.

Proof: Consider the following decomposition of $b$ into the basis vectors of $\mathbb{R}^m$, scaled by the vector $c$: $b=c_1e_1+c_2e_2 + \dots c_me_m$. This can be rewritten using the matrix $I_m$.

$$
   b=  \begin{bmatrix}
    1 & & \\
    & \ddots & \\
    & & 1
   \end{bmatrix}
  \begin{bmatrix}
   c_1 \\
   \vdots \\
   c_m
  \end{bmatrix}
$$

Now, let $c^{+}$ be the vector obtained by replacing any negative coordinate in $c$ with $0$, and $c^-$ be the vector obtained by replacing any positive coordinate in $c$ with $0$. Clearly, $c^{+}+c^-=c$. Now, let $u=I_mc^-$ and $v=I_mc^{+}$. We have $u+v=b$, since $I_mc^-+I_mc^{+}=I_m(c^-+c^+)=I_mc=b$. By construction, $v$ is strictly greater than $0$, so all that is left to show is that $u \in \text{Im}(A)$. Since $Ax\leq b$, $b$ cannot be negative in the direction of any basis vector $e_i$ if $e_i \notin \text{Im}(A)$, because if $e_i$ were not in the image of $A$, then the smallest magnitude the vector $Ax$ could have in the direction of $e_i$ is $0$. Thus, if $b$ is negative in the direction of $e_i$, and $Ax \leq b$, then we must have $e_i \in \text{Im}(A)$. Since $c^-$ captures only the negative magnitudes in the basis-vector decomposition of $b$, it must be that $u \in \text{Im}(A)$.
QED

This proof seems pretty hand-wavy to me, specifically the argument I make about $b$ having negative magnitude in the "direction" of basis vectors. Does anyone have any thoughts about how I could formalize this notion and approach it in a more structured, algebraic, way? Thanks!

答:
The matrix $A$ does not necessarily have anything to do with the standard basis $\{e_i\}$. Indeed, it is possible for none of the standard basis vectors to be in the image of $A$ even if that image has dimension $m-1$ (the image might be the set of vectors orthogonal to $(1,\dots,1)$, for example).

To solve the problem, it suffices to let $x_0$ be a solution to $Ax\le b$, and consider $Ax_0 + (b-Ax_0)$.

In general, I find it more effective to avoid specifying bases (even the standard basis) whenever possible; and when I do need to specify a basis, I'd rather specify a basis having to do with the matrix involved (its eigenvectors, for example).

Title: How to formalize this proof?


问:
Irreducible polynomial with $\alpha^l=\beta^l \in \mathbb{Q}$ for any two roots
Let $f$ be an irreducible polynomial in $\mathbb{Q}[x]$ with $\alpha^l=\beta^l \in \mathbb{Q}$ for any two roots $\alpha,\beta \in \mathbb{C}$ of $f$. Can I follow $f | (x^l-\alpha^l)$ from this? If $\deg f \leq l$, the statement is obvious. But is $\deg f > l$ possible?

答: Let $c=\alpha^l$ and $g(X)=X^l-c$. As $f$ is irreducible, $\gcd(f,g)$ is either $1$ or $f$ itself. In the latter case, of course $f\mid g$ and $\deg f\le \deg g=l$. In the first case, $f$ has no root in common with $g$, whereas all roots of $f$ must be roots of $g$; then $f$ has no roots at all and is constant.


问:
Does $t\delta(t) = 0$?
In middle of solving a problem I encountered terms looking like this:
$$t\delta(t)$$
Or in a more general form:
$$(t-t_{0})\delta(t-t_{0})$$
It appears that in the textbook I am reading "Linear Systems and Siganls by B. P. Lathi and Roger Green" this value is considered to equal zero although it is not explicitly mentioned (probably to avoid some complex math).
Now from the properties of the impulse function:
$$t\delta(t) = 0\times\delta(t)$$
Is this value $0$ or it is undefined? similar to $0\times\infty$? If the value is not $0$, does this mean that the unit step function $u(t)$ is not the derivative of the unit ramp function $tu(t)$ (as shown bellow)?
$$\frac{d(tu(t))}{dt}=t\delta(t)+u(t)$$

答:
As a distribution, yes this is true. Let your distribution be $T$, then for Schwartz $f$,

$$T(f) = \delta(tf) = 0\cdot f(0) = 0.$$

The shifted result also holds because any Schwartz function is a translation of another.

Title: Does $t\delta(t) = 0$?


问:
How does one compute the sign of a permutation?
The sign of a permutation $\sigma\in \mathfrak{S}_n$, written ${\rm sgn}(\sigma)$, is defined to be +1 if the permutation is even and -1 if it is odd, and is given by the formula

$${\rm sgn}(\sigma) = (-1)^m$$

where $m$ is the number of transpositions in the permutation when written as a product of transpositions.

Alternatively the sign is -1 if, when we express $\sigma$ as a product of disjoint cycles, the result contains an odd number of even-length cycles. Otherwise the sign is +1.

My permutations are expressed as tuples $(\sigma_1,\dots,\sigma_n)$, so neither the expression of the tuple as a product of disjoint cycles nor as a product of transpositions are immediately available to me.

This suggests two high-algorithms to compute the sign of a permutation:

Express the permutation as a product of transpositions and count the number of transpositions.

Express the permutation as a product of disjoint cycles and count the number of even-length cycles.

What are typical algorithms for accomplishing these tasks, and how do they vary in running type depending on $n$? Are there more efficient algorithms for calculating the sign of a permutation?

Additional info

The motivation is to quickly decide whether instances of the fifteen puzzle are solvable. I want to generate a large number of solvable instances of the fifteen puzzle for testing some search algorithms. At the moment I generate a random instance, and test whether it is solvable by trying to solve it with depth-first search, which is fairly slow going, and won't generalize well to larger puzzles (24 puzzle, 35 puzzle...) due to time and memory limitations. Since solvable instances of the fifteen puzzle are in 1-1 correspondence with even elements of $\mathfrak{S}_{16}$, I figure that there must be a faster way of generating solvable instances.

It has just occurred to me that a better way of generating solvable instances of the puzzle might be to generate an even number of transpositions and multiply them together to generate an even permutation. I'd prefer an algorithm that was guaranteed to return an even distribution over $\mathfrak{S}_n$ though, and in fact I'm now sufficiently interested in the answer to this question in its own right.

答:
Welcome to $O(n)$'s world. Python code

```<code>def decomposition(n, sigma):
   lista = list()
   for k in range (1, n + 1):
      boo = False
      for i in range (0, n):
         if sigma[i] == k:
            boo = True
            break
      if not boo:
         print("Error")
         return

   contador = 0
   for i in range (0, n - 1):
      if sigma[i] != i + 1:
         k = sigma[i] - 1
         sigma[i], sigma[k] = i + 1, k + 1
         s = "(" + str(i + 1) + ", " + str(k + 1) + ")"
         lista.insert(0, s)
         contador = contador + 1

   s = ""
   for cadaUm in lista:
     s = s + cadaUm
   print("sigma =", s)
   print("size =", contador)
   return
</code>```

Body:
Welcome to $O(n)$'s world. Python code

def decomposition(n, sigma):
   lista = list()
   for k in range (1, n + 1):
      boo = False
      for i in range (0, n):
         if sigma[i] == k:
            boo = True
            break
      if not boo:
         print("Error")
         return

   contador = 0
   for i in range (0, n - 1):
      if sigma[i] != i + 1:
         k = sigma[i] - 1
         sigma[i], sigma[k] = i + 1, k + 1
         s = "(" + str(i + 1) + ", " + str(k + 1) + ")"
         lista.insert(0, s)
         contador = contador + 1

   s = ""
   for cadaUm in lista:
     s = s + cadaUm
   print("sigma =", s)
   print("size =", contador)
   return


问:
Density of smooth functions in $L^p$ space on Cantor Set
Let $\mu$ be the Cantor measure on the Cantor set $C$ in $[0,1]$. Is the space of functions obtained by restricting smooth compactly supported functions to $C$ dense in $L^p(\mu)$ for $1 \leq p < \infty$?

答:
Let $\mu$ be a regular Borel measure on $\mathbf{R}^d$. Then the Riesz Kakutani representation theorem implies that $C_c(\mathbf{R}^d)$ is dense in $L^p(\mu)$ for $1 \leq p < \infty$. If $\{ \phi_\varepsilon \}$ is a mollifier on $\mathbf{R}^d$ and $f \in C_c(\mathbf{R}^d)$, then a simple application of uniform continuity shows that the smooth, compactly supported functions $f * \phi_\varepsilon$ converge to $f$ uniformly.

If $\mu$ is a finite measure, this implies that $f * \phi_\varepsilon$ converges to $f$ in $L^p(\mu)$ for all $1 \leq p \leq \infty$. On the other hand, since we may choose $\phi_\varepsilon$ to be supported on a ball of radius $\varepsilon$, all the functions $f * \phi_\varepsilon$ are supported on $\text{supp}(f) + B$, where $B$ is the unit ball at the origin, and the regularity of $\mu$ implies $\mu|_B$ is a finite regular Borel measure, so there is no loss of generality in assuming that $\mu$ is finite.


问:
theorem Let A be a set. Then ∪P(A)=A
prove this theorem. Let A be a set. Then ∪P(A)=A.

I've tried to do a similar problem but I am unsure how to solve this problem.

答: Just use the definitions to show that $\bigcup\wp(A)\subseteq A$ and $A\subseteq\bigcup\wp(A)$. I’ll start the first one for you to get you started. Let $x\in\bigcup\wp(A)$; then by the definition of union there is an $S\in\wp(A)$ such that $x\in S$. But then $x\in S\subseteq A$, so … ?

Title: theorem Let A be a set. Then ∪P(A)=A


问:
Monotone Convergence Theorem - Bartle Elements of Real Analysis Book
I am not sure how the author went from this $x - l \leq x_n \leq x + l$ to this $x - l \leq sup\{x_n\} \leq x + l$, I boulded this part on the theorem for clarity.

Let $X = (x_n)$ be a sequence of real numbers which is monotone increasing in the sense that $ x_1 \leq x_2 ...\leq x_n$

Then the sequence $X$ converges if and only if it is bounded, in which case $\lim(x_n) = \sup \{x_n\}$.

Proof. If  $x=\lim(x_n)$ and $l > 0$, then there exists a natural number $K(l)$ such that if $K(l)\leq$ n then

$\textbf{x - l $\leq$ x_n $\leq$ x + l}$

Since x is monotone, this relation yields

$x - l \leq\sup\{x_n\} \leq x + l$

....

答:
Let $(x_{n})_{n=0}^{\infty}$ be an increasing sequence of real numbers.

Let us also suppose that it is bounded.

Since the set $\{x_{n}\in\mathbb{R}\mid n\in\mathbb{N}\}$ is not empty and bounded, it admits a supremum, which we denote by $s$.

This means that for every $\varepsilon > 0$, there corresponds $n_{\varepsilon}\in\mathbb{N}$ s.t.
\begin{align*}
n\geq n_{\varepsilon} \Rightarrow s - \varepsilon < x_{n_{\varepsilon}} \leq x_{n} \leq s < s + \varepsilon \Rightarrow |x_{n} - s| < \varepsilon
\end{align*}
whence we conclude that $x_{n}\to s$.

Similar reasoning applies if the sequence $(x_{n})_{n=0}^{\infty}$ is decreasing and bounded below.

In such case, the sequence $(x_{n})_{n=0}^{\infty}$ converges to $\inf\{x_{n}\in\mathbb{R}\mid n\in\mathbb{N}\}$.

On the other hand, if the sequence converges to some $L\in\mathbb{R}$, then it is bounded.

Indeed, according to the definition of convergence, let $\varepsilon = 1$.

Then there corresponds $n_{1}\in\mathbb{N}$ s.t.
\begin{align*}
n\geq n_{1} \Rightarrow |x_{n} - L| < 1 \Rightarrow |x_{n}| < |L| + 1
\end{align*}

Then it suffices to take $M = \max\{|x_{0}|,|x_{1}|,\ldots,|x_{n_{1}}|,|L| + 1\}$, and we are done.

Hopefully this helps!


答:
Let $U=\sup\{x_n\}$. Since $(x_n)$ is monotone increasing, then for $l>0$, there exists an $N\geq 1$, such that $x_N > U - l$. If this were not the case, then it would imply that $x_N\leq U-l < U$ for all $N\geq 1$, which contradicts that $U$ is the least upper bound on $(x_n)$. Thus, we have that
$$
n\geq N \implies U - l < X_N \leq x_n < U + l \implies \lvert x_n - U \rvert \leq l.
$$
So $x_n\to U = x$ as $n\to \infty$.

Title: Monotone Convergence Theorem - Bartle Elements of Real Analysis Book

Body:
I am not sure how the author went from this $x - l \leq x_n \leq x + l$ to this $x - l \leq sup\{x_n\} \leq x + l$, I boulded this part on the theorem for clarity.

Let $X = (x_n)$ be a sequence of real numbers which is monotone increasing in the sense that $ x_1 \leq x_2 ...\leq x_n$

Then the sequence $X$ converges if and only if it is bounded, in which case $\lim(x_n) = \sup \{x_n\}$.

Proof. If  $x=\lim(x_n)$ and $l > 0$, then there exists a natural number $K(l)$ such that if $K(l)\leq$ n then

$\textbf{x - l $\leq$ x_n $\leq$ x + l}$

Since x is monotone, this relation yields

$x - l \leq\sup\{x_n\} \leq x + l$

....


问:
Confusion with vector spaces
If we consider $\mathbb{R}^2$ space with cartesian coordinates, then $\vec{V_{c1}}+\vec{V_{c2}} = \overrightarrow{V_{c1}+V_{c2}}$ implies we add components of the vectors. This idea of adding vectors should be invariant to the basis used but if we consider adding vectors using polar coordinates the way we add components is completely different compared to using cartesian coordinates.

Therefore, does this mean when we define a vector space it's basis specific and we must specify how actual addition/scaling operations are implemented under the basis OR are the implementations of the operations overlooked and by specifying that these operations exist we merely mean that we can put the "tail" of one vector on the "head" of another.

答:
You got a good answer already, but here's another way of putting it.

If two vectors, say $v$ and $w$, are a basis for $\mathbb R^2$ (for example, vectors $\hat i$ and $\hat j$),

then any vectors $a$ and $b$ in $\mathbb R^2$ can be expressed as a linear combination of $v$ and $w$,

say $a=a_vv+a_ww$ and $b=b_vv+b_ww$.  Because of the properties of vector spaces,

then $a+b=a_vv+a_ww+b_vv+b_w=(a_v+b_v)v+(a_w+b_w)w$,

so we add vectors component-wise.

On the other hand, when we express vectors using polar coordinates $r$ and $\theta$,

the vectors are not expressed as linear combinations of two vectors,

but rather as $(r\cos\theta,r\sin\theta)$, which is not linear in $\theta$,

so addition of vectors is not so convenient in polar coordinates,

though other operations may be more convenient in polar coordinates.

Title: Confusion with vector spaces


问:
Trigonometry/Geometry triangle help
In right triangle $ABC$, ($\angle BAC = 90$), $D$ is found on side $AC$, $BD$ is the angle bisector of $\angle ABC$ and $BC= \sqrt3 BD$.

Find the value of $\angle CBD$.

With regards to the point $E$ (found on $AC$), establish whether the areas of the circles which circumscribe the triangles $BDE$ and $BCE$ are equivalent.

(Not connected to number 2): Given that $CE=2DE$, find the value of the angle $\angle BEC$.

Note: I tried to call $\angle ABC =\alpha$ and thereby bisect it. I found different trigonometric relationships, but I could not work out how to solve for either the values of the sides nor the angles.

答:
1.$AB=BD\cos(\alpha)\hspace 1cmAC=BC\sin(2\alpha)=\sqrt3BD\sin(2\alpha)\hspace 1cm BC=\sqrt3 BD$

this implies $2\sqrt3\cos(\alpha)^2-\cos(\alpha)-\sqrt3=0\Rightarrow \cos(\alpha)=\dfrac{\sqrt3}{2}$. Then $\angle{CBD}=30^{\circ}$

2.The two respective radius should be equal. From the two areas
$$A=\frac{BD\cdot DE\cdot BE}{4R}\\A'=\frac{BE\cdot EC\cdot BC}{4R'}$$ we have the condition $$R=R'\iff(BD\cdot DE)A'=(BC\cdot EC)A$$ This can be set straightforward but it is some tedious.

Note that $\triangle{BAD}$ is the half of an equilateral triangle so $\angle{BDC}=120^{\circ}$ and $\angle{BCD}=30^{\circ}$, etc.

You can take my answer as a simple comment if you consider it as incomplete.

Title: Trigonometry/Geometry triangle help


问:
Sequential Criteria of Limit Points in Hausdorff Spcaes
By the sequential criteria, I convey that :
A point p of set S is limit point of S if there exists a sequence of distinct points in S that converge to p.

I understand that if the space is metriziable, then this holds.
But for general spaces I haven't been able to find a counter example.
Any help would be appreciated.

答:
Here’s a simple example that does not use ordinals. Let $Y=\Bbb R\setminus\{0\}$, and let $\tau$ be the following topology on $\Bbb R$

$$\tau=\wp(Y)\cup\{\Bbb R\setminus F:F\subseteq Y\text{ is countable}\}\,;$$

that is, points of $Y$ are isolated, and a set $U\subseteq\Bbb R$ is an open nbhd of $0$ if and only if $0\in U$ and $\Bbb R\setminus U$ is countable. It’s easy to check that $\langle\Bbb R,\tau\rangle$ is Hausdorff and that $0\in\operatorname{cl}_\tau Y$.

However, if $\langle y_n:n\in\Bbb N\rangle$ is any sequence in $Y$, the set $\Bbb R\setminus\{y_n:n\in\Bbb N\}$ is an open nbhd of $0$ that contains no points of the sequence, so no sequence in $Y$ converges to $0$.


问:
How can I verify that these two calculations for a derivative are equal?
I am working through Calculus with Analytic Geometry by Simmons. Section 3.6 (Derivatives of Higher Order) page 107 question #25 has me calculating the derivative of the following function..

$$
(\frac{x^3+8}{x^2})^{3/4}
$$

The answer in the back of the book is
$$
\frac{3(x^3 - 16)}{4x^3} \sqrt[4]{\frac{x^2}{x^3 + 8}}
$$

The answer I get when I use the deriv feature on wolfram alpha is
$$
\frac{3(x^3-16)}{4x^3\sqrt[4]{\frac{8}{x^2} + x}}
$$

And when I calculate it myself (still working on how to multiply these two values..)

$$
\frac{3}{4\sqrt[4]{\frac{8+x^3}{x^2}}} \cdot (1 - \frac{16}{x^3})
$$

How can I verify that these answers are actually the same value?

答:
HINT

\begin{align*}
\sqrt[4]{\frac{a}{b}} = \frac{1}{\sqrt[4]{\dfrac{b}{a}}}
\end{align*}


问:
Knowing that $X$ independent of itself and $ \mathbb{E}[X^2] < \infty $, how to show that $X$ is a constant?
I have a random variable $X$ that is independent of itself.
How knowing that $$ \mathbb{E}[X^2] < \infty $$ is suppose to help me finding that $X$ is a constant.

I already found out that if $X$ is independent, $ \mathbb{P}(X \in A) = 0  $ or $ \mathbb{P}(X \in A) = 1 $. And I just pick $x \in \mathbb{R}$ and I have $\mathbb{P}(X \le x)^2 = \mathbb{P}(X \le x) \ge \mathbb{P}(X \le x) = 0  $ or $ \mathbb{P}(X \le x) = 1$

But how is it related to that expected value ? I really tried but I do not come with any ideas.

Many thanks ! If the question is ambiguous or unclear, please edit it :)

答: As I said in the comments, by the independence, we have that  $E(X^2)=E(X)E(X)$ and thus $Var(X)=0$. As a result, $X=E(X)$ (constant) almost surely.


Title: How can I verify that these two calculations for a derivative are equal?

Body:
I am working through Calculus with Analytic Geometry by Simmons. Section 3.6 (Derivatives of Higher Order) page 107 question #25 has me calculating the derivative of the following function..

$$
(\frac{x^3+8}{x^2})^{3/4}
$$

The answer in the back of the book is
$$
\frac{3(x^3 - 16)}{4x^3} \sqrt[4]{\frac{x^2}{x^3 + 8}}
$$

The answer I get when I use the deriv feature on wolfram alpha is
$$
\frac{3(x^3-16)}{4x^3\sqrt[4]{\frac{8}{x^2} + x}}
$$

And when I calculate it myself (still working on how to multiply these two values..)

$$
\frac{3}{4\sqrt[4]{\frac{8+x^3}{x^2}}} \cdot (1 - \frac{16}{x^3})
$$

How can I verify that these answers are actually the same value?

Body:
$$\frac{3(x^3-16)}{4x^3\sqrt[4]{\frac{8}{x^2} + x}}= \frac{3(x^3-16)}{4x^3}\cdot\frac{1}{\sqrt[4]{\frac{8+x^3}{x^2}}}= \frac{3(x^3 - 16)}{4x^3} \sqrt[4]{\frac{x^2}{x^3 + 8}}$$

Also, the same way,

$$\frac{3}{4\sqrt[4]{\frac{8+x^3}{x^2}}} \cdot (1 - \frac{16}{x^3})=\frac{3}{4\sqrt[4]{\frac{8+x^3}{x^2}}} \cdot ({\frac{x^3-16}{x^3}})= \frac{3(x^3-16)}{4x^3\sqrt[4]{\frac{8}{x^2} + x}}= \frac{3(x^3-16)}{4x^3}\cdot\frac{1}{\sqrt[4]{\frac{8+x^3}{x^2}}}= \frac{3(x^3 - 16)}{4x^3} \sqrt[4]{\frac{x^2}{x^3 + 8}}$$


问:
For what $x$ values does the integral have a minimum

Question

My Understanding

I'm really confused because I flipped the function to solve it (turned the integral negative and made lower value upper value...) and I got $2$ as the answer, but when I checked it, it was wrong. I know it can't be $0$ either. So it's $-2$? Or can there be multiple minimum points? Because $-2$ is a minimum when we don't flip the function. But $2$ is the minimum when we flip it. I'm really confused.

答:
If we let $G^\prime(x)=g(x)$ then we have $f(x)=G(1)-G(x)$. Then $f^\prime(x)=-G^\prime(x)=-g(x)$.

We know that $f$ will have a minimum value at the value $x_0$ where $f^\prime(x_0)=0$ and where $f^\prime(x)<0$ on some interval $(a,x_0)$ and $f^\prime(x)>0$ on some interval $(x_0,b)$. This is because $f$ will be decreasing to the left of $x_0$ and increasing to the right of $x_0$.

Since $f^\prime(x)=-g(x)$ you can find $x_0$ by inspecting the graph of $y=-g(x)$.

Turn the graph of $g$ "upside down" and we see that the conditions are met for $x_0=2$.


问:
Is a prime number still a prime when in a different base?
Is a prime number in the decimal system still a prime when converted to a different base?</p>

答:
Alas, the accepted answer is misleading (and arguably incorrect). The reason that primality (or any other purely arithmetic property) is preserved in radix representation is simply that such representation faithfully preserves all of the arithmetic operations on integers. More precisely, $\:$ if $\rm\;n\to r(n)\;$ is the map from $\rm\:n\:$ to its radix $\rm\:d\;$ representation, then it preserves addition $\rm\;r(m+n) = r(m) + r(n),\;$ and multiplication $\rm\;r(mn) = r(m)\ r(n),\;$ and $\rm\;r\;$ has an inverse $\rm\;s\;$ that similarly also preserves addition and multiplication (technically: $\rm\:r\:$ is a ring isomorphism). This readily implies that the relation  of divisibility is faithfully preserved in radix representation, because the relation of divisibility can be expressed as an equation involving only arithmetic (ring) operations (namely multiplication), and such equations are necessarily preserved by the maps $\rm\;r\;$ and $\rm\;s\;$ - indeed these maps are defined precisely so to preserve these fundamental operations.

It's instructive to examine more closely the preservation of divisibility. First, we recall the standard notation $\rm\;a\mid b\; := \: a\;$ divides $\rm\:b,\;\;$ i.e. $\rm\:\exists \:n\in\mathbb Z : \ an = b,\;$ i.e. there exists an integer $\rm\:n\:$ such that $\rm\;an = b\;$.

Lemma $\rm\ \ \ a\mid b \iff r(a)\mid r(b)\quad\quad$ (Divisibility Preservation by Isomorphisms)

Proof: $\rm\;(\;\Rightarrow\;)\quad a\mid b \;\Rightarrow\; \exists\:n\in\mathbb Z: \: an = b \;\Rightarrow\; r(a)\ r(n) = r(an) = r(b) \;\Rightarrow\; r(a)\mid r(b)$

$\rm\;(\Leftarrow)\quad r(a)\mid r(b) \;\Rightarrow\;\exists\:c\in r(\mathbb Z): \: r(a)\: c = r(b)\;\Rightarrow\; r(a)\ r(n) = r(b) \;\Rightarrow\; a n = b\;$

The final $\;\Rightarrow\;$ above is by applying $\rm\;r\:$'s inverse $\rm\:s\:$ so to cancel the $\rm\;r\:$'s using $\rm\;\: sr = 1 =\;$ identity map.

Note: this employs $\rm\:s\:$'s preservation of multiplication, $\:$ viz. $\rm\:s(r(a)\: r(n)) \:=\: sr(a)\: sr(n) \:=\: a\: n\;$

As a corollary, we conclude that primes (i.e. irreducibles) are also preserved, since they are definable purely in terms of divisibilty, viz. $\rm\,p\,$ is prime $\, :=\ \rm p = ab \Rightarrow p\mid a\;$ or $\rm\;p\mid b\;$ and $\rm\;p\;$ is not a unit, i.e. not $\rm\;p\nmid 1\;$.

So your question reduces to the more fundamental  why is radix representation a ring isomorphism, i.e. why really does radix representation preserve the addition and multiplication operations? This is a very good question that deserves a thoughtful answer. It's a serious pedagogical oversight that this topic is rarely discussed in algebra textbooks. Although most students understand this fact subconsciously, many have difficulty providing a rigorous proof (or, worse, they overlook the fact that it does require a rigorous proof). Your question would attract much more interest and receive much more interesting replies if you rephrase it in this manner. Thus I propose the following:

Suggestion: Edit the title of your question to say "Why is radix representation a ring isomorphism, i.e. why does it preserve addition and multiplication?". Include in your question a brief description of your knowledge level so to help set the proper level for replies, e.g. do you already know any abstract algebra or elementary number theory?
Unaccept the currently accepted answer so that the software will continue to bump up the question till it gets a number of good answers. IMPORTANT: Wait at least a week before accepting any answer so that everyone has a chance to see the question. This helps serve to maximize its exposure and hence maximize your potential of receiving insightful answers. If all goes well this should result in an interesting thread that will help serve as a future reference for many similar frequently asked questions.

Note to much more experienced readers: this problem is not as trivial as you might think at first glance (and certainly less trivial for novices). For example, the analogous problem for real numbers (or p-adics) is the subject of a famous paper [1]. For an introduction see e.g. here. Its closing line is quite apropos:

This is very much in keeping with Rota’s thinking that mathematics is not just a quest to solve problems, it is
  also a quest to understand the mathematical universe as
  clearly and as deeply as possible

[1] F. Faltin, N. Metropolis, B. Ross, G.-C. Rota,

The real numbers as a wreath product.

Advances in Math. 16 (1975), 278-304

ContentLicense: CC BY-SA 2.5


问:
Random walk probability non-symmetric steps
I currently have a probability class tutorial question that I have no idea where to begin. At first instinct, I thought it may have been a CTMC question or branching question, but now I have no idea, any help would be greatly appreciated!

Question:

A node goes down $1$ step with probability $p$ and up $t$ steps with probability $1-p$.

Let $X_n=Y_1+\ldots +Y_n$ where $(Y_n)$ is i.i.d with $P(Y_1 = -1) = p$ and $P(Y_1 = t ) = 1-p$.

Assuming that at $n=0$ the node is at $0$,

a) find $P(X_n = 0)$.

b) find a simple sequence $(u(n))$ such that $P(X_{(t+1)n} = 0) \sim u(n)$.

I know that if $p=0$ or $p=1$ then the answer to a) is 0. I have trouble with the case $0<p<1$.
I've also worked out that if the node takes $t$ steps up, then it must take $t$ steps back as well to get to the initial point.
And I am not sure as well, but I think we can use the $\phi (u)$ formula.

答:
That $X_n=0$ means that there was $i$ steps $+t$ and $n-i$ steps $-1$ with $+t\cdot i-1\cdot(n-i)=0$, that is, $i=n/(t+1)$. Hence $P[X_n=0]=0$ if $n$ is not a multiple of $t+1$ and, for every $n\geqslant0$,
$$
P[X_{(t+1)n}=0]={(t+1)n\choose n}(1-p)^np^{tn}.
$$
Stirling's formula yields that, when $n\to\infty$,
$$
P[X_{(t+1)n}=0]\sim\frac{c}{\sqrt{n}}\vartheta^n,
$$
with
$$
c=\sqrt{2\pi}\sqrt{\frac{t+1}t},\qquad
\vartheta=(t+1)^{t+1}t^{-t}(1-p)p^t.
$$
Note that $\vartheta=1$ if $p=t/(t+1)$ (this is the zero drift case) and $\vartheta\lt1$ otherwise.</p>


问:
Is it true that $\ker(\operatorname{coker}\varphi)$ satisfies the universal property of images in a pre-abelian category?
Let $\varphi:A\to B$ be a morphism in a pre-abelian category (that is, an additive category where every morphism has a kernel and a cokernel) $\sf{C}$. If $\sf{C}$ is indeed abelian, then I know that $i=\ker(\operatorname{coker}\varphi):K\to B$ satisfies the following universal property: $\varphi$ factors through $i$ and if $L\to B$ is another monomorphism through which $\varphi$ also factors, then it exists a unique morphism $K\to L$ such that the diagram

commutes. I want to know if $\ker(\operatorname{coker}\varphi)$ satisfies this universal property even if $\sf{C}$ is only pre-abelian.

I don't think this is the case since this universal property "gives" a morphism $\operatorname{im} \varphi\to\operatorname{coim}\varphi$ (given that we define $\operatorname{im}\varphi$ as the source of $\ker(\operatorname{coker}\varphi)$ and $\operatorname{coim}\varphi$ as the target of $\operatorname{coker}(\ker\varphi)$) and not the other way around. This is ok if $\operatorname{im}\varphi$ and $\operatorname{coim}\varphi$ are isomorphic (which happens in an abelian category) but is strange otherwise. Also, I only know how to prove that $\ker(\operatorname{coker}\varphi)$ satisfies this universal property whenever $\sf{C}$ is abelian. Of course these are nothing but heuristics for me to think that $\ker(\operatorname{coker}\varphi)$ does not satisfy the universal property above.

答: I think we get a map $\operatorname{coker} (\ker \varphi) \to \ker(\operatorname{coker} \varphi)$, see Tag 0107; for an example where this map is not an isomorphism, see Tag 0108 (in this example, both (the source of) $\ker \varphi$ and (the target of) $\operatorname{coker} \varphi$ are $0$).


问:
Is there a formal name for an "undecorated" variable?
That is, is there a term to specifically distinguish the type of thing x is, as juxtaposed with x', x'', etc?

The wikipedia page on the prime symbol mentions that "The prime is said to "decorate" the letter to which it applies.", which gives me that idea that perhaps "undecorated variable" is the term I'm looking for, but I haven't found any examples of people using it. Is there a standard term for this?

答:
No, no such name exists … besides "undecorated".

 I'm posting this CW answer so that users who confidently concur that the answer to the question is "No" have an answer to vote on, and so this question doesn't stagnate in the Unanswered Questions Queue. If however anyone has an affirmative response to the question, please downvote this and post your answer.


问:
How many ways can $10$ different kids be distributed to $4$ distinct classes?
I want to find the number ways to distribute 10 different kids to 4 different classes, such that each class has at least 2 kids.

My attempt: First distribute 8 kids to 4 classes such that each class has exactly 2 kids. Then there are 16 option to split the remaining 2 kids.

This is my result: $\binom{10}{2}\binom{8}{2}\binom{6}{2}\binom{4}{2}*16$ which is wrong. I don't have exact answer but I know it's either: $226800, 75600, 44100$.

Thanks.

答:
I think the answer is 226800.

The sizes of the classes can be $(2,2,2,4)$ (+ any permutation) or $(2,2,3,3)$ (+ any permutation).

$$\binom{10}{4}\binom{6}{2}\binom{4}{2}\cdot\binom{4}{1} + \binom{10}{3}\binom{7}{3}\binom{4}{2}\cdot\binom{4}{2} = 226800$$


答: Note that there are 2 different size types for classes, $3,3,2,2$ or $4,2,2,2$. For the first one, choose firstly which classes will have $3$ kids, $\binom{4}{2}$, then distribute the kids, $\binom{10}{3} \cdot \binom{7}{3} \cdot \binom{4}{2} \cdot \binom{2}{2}$. Do the same for the other size type and you will arrive at $ \binom{4}{2} \cdot \binom{10}{3} \cdot \binom{7}{3} \cdot \binom{4}{2} \cdot \binom{2}{2} + \binom{4}{1} \cdot \binom{10}{4} \cdot \binom{6}{2} \cdot \binom{4}{2} \cdot \binom{2}{2} = 226800$

Title: How many ways can $10$ different kids be distributed to $4$ distinct classes?

Body:
I want to find the number ways to distribute 10 different kids to 4 different classes, such that each class has at least 2 kids.

My attempt: First distribute 8 kids to 4 classes such that each class has exactly 2 kids. Then there are 16 option to split the remaining 2 kids.

This is my result: $\binom{10}{2}\binom{8}{2}\binom{6}{2}\binom{4}{2}*16$ which is wrong. I don't have exact answer but I know it's either: $226800, 75600, 44100$.

Thanks.


问:
Drawing from an urn - probability of drawing maximum 2 of each color.
I have an urn with 100 balls; 10 colors with each 10 balls.

Now I draw 10 balls from the urn without replacement - I want to calculate the probability of getting 3 or more of any color.

How do I calculate this? My problem is that for example the event "more than 3 greens" and "more than 3 yellows" are not exclusive, there are possible draws which fit into both categories. So I cant the hypergeometric distribution and multiply it by 10.

I feel like this question is school level.... I got a math degree, so the answer doesnt need to be super basic, I just suck at stochastics. Help greatly appreciated.

答:
There are $\frac{100!}{90!}$ possible combinations.  To get an answer compute the number of combinations $(C)$where there are at most $2$ of any color. The combinations can be broken down into $6$ sets depending on how many balls appear twice.  These are $A_k=90^k\times 10^{10-2k}$ and The number of permutations for each $k$ are $B_k=\frac{10!}{(k!)^2(10-2k)!}$  $C=\sum_{k=0}^5 A_kB_k$, giving $Q=\frac{90!C}{100!}$ as the probability of no more than $2$ of any color and the desired result is $P=1-Q$

Calculating $Q$ i s tedious but not impossible.


问:
Probability of no matches taking $N$ cards out of $rN$ cards at random and matching them against a target deck
In Feller's introduction to probability the answer to this problem is given by:

$ $$\tag1 p = \sum_{v={2}}^{N} (-1)^{v}{N \choose v}r^{v}{\frac{(rN-v)!}{rN!}}$

Acording to the means of the chapter from which this problem comes from, the probability of at least one of the cards being matched is given by:

$ $$\tag1 P_1 = \sum_{v=1}^{N}(-1)^{v+1}S_v $

Where $S_v$ is the sum of probabilities of the intersection of exactly $v$ cases where there are $v$ matches.

Yet i have some problems with the answer given, because to begin, if you are choosing $N$ cards out of $2N$ at random, should't the factor ${2N \choose N}$ be in the denominator? And also because you are choosing $N$ cards to work with, in the sum, it should't be just $N$ instead $rN$?

答:
Feller has simplified some intermediate results.

The solution is by application of inclusion/exclusion.  We consider all the cards to be distinct, including those that have the same rank, so the $N$ cards out of $rN$ can be selected in $\binom{rN}{N}$ ways and then ordered in $N!$ ways, resulting in $\binom{rN}{N} N!$ possible arrangements, all of which are equally likely.

Using Feller's notation, let $S_{\nu}$ be the probability (with deliberate over-counting) that $\nu$ of the cards in the target deck are matched.  We want to count the number of ways that this can happen.  The matched ranks can be chosen in $\binom{N}{\nu}$ ways, and each of the $\nu$ ranks can be matched by any of $r$ cards. The remaining cards in the sample of $N$ can be selected in $\binom{rN-\nu}{N-\nu}$ ways, and then ordered in $(N-\nu)!$ ways. So the total number of arrangements is
$$\binom{N}{\nu} r^{\nu} \binom{rN-\nu}{N-\nu} (N-\nu)!$$
Therefore
$$S_{\nu} = \frac{\binom{N}{\nu} r^{\nu} \binom{rN-\nu}{N-\nu} (N-\nu)!}{\binom{rN}{N} N!}$$
Simplifying,
$$\begin{align}
S_{\nu} &= \frac{\binom{N}{\nu} r^{\nu} (rN-\nu)! / N!}{(rN)! / N!} \\
&= \frac{\binom{N}{\nu} r^{\nu} (rN-\nu)!}{(rN)!}
\end{align}$$
which is Feller's result.

Feller makes one more simplification.  The probability of no match is, by inclusion / exclusion,
$$P_0 = \sum_{\nu=0}^N (-1)^{\nu} S_{\nu}$$
but $S_0 = S_1$, so the first two terms cancel, and
$$P_0 = \sum_{\nu=2}^N (-1)^{\nu} S_{\nu}$$


问:
What does Cohen independence theorem say?
In his celebrated paper, the independence of the continuum hypothesis, P. Cohen proved that there is a model of Z-F in which the continuum hypothesis fails. As a corollary, continuum hypothesis is independent to Z-F. This is indeed a famous story in math -- perhaps most graduate students know about it and can roughly recite the heuristic.

However, thinking a little more led me wondering what this statement really means. In particular, I'm not comfortable with the meaning of the term "model". Whatever a model is, suppose there are two models of Z-F, $M_1$ and $M_2$, in which CH holds and fails respectively. Since the $M_i$'s are models of the set theory, what is a set depends on $i$. That is to say, a set in the sense of $M_1$ is not necessarily a set in the sense of $M_2$. However, all definitions I've seen for the term "model" depend on the notion of a set.

Thus my first question: What is a model in this context? If you'd like to use the term "set" in your definition, then in which sense does that mean?

The answer, as I guess, could be that either the term "model" or "set" isn't formally defined here. That's fine, as in the end there must be some undefined terms, and this could be one of them. However, if that's the case, then in what sense can we say that such model exists? That is, we might not know what a model formally is, but at least we should know what qualifies a mathematical work to be a legit construction of a model. This is my second question.

EDIT I was benefited much from the comments, answers, and discussions below, and am grateful for it. What follows is my understanding.

By a "model" we mean a set in ZF theory. By a "set" in ZF theory we mean a class $A$ that exists in the von Neumann universe $V$, or simply denoted by $A \in V$. By "$V$" we mean the syntactic expression $\{x | x = x\}$, and by "exists in $V$" (or $A \in V$) we mean the expression given here.

Therefore, a "model" is really just a syntactic expression derivable from the ZF axioms that satisfy a bunch of syntactic requirements.. but we rather think of it as a real set that we call "set". It is the "set" that we talk about in usual mathematical context (e.g. in a finite group theory course). This answers my first question, and deprecates the second.

答:
Cohen showed that if $M$ is a countable transitive model of $V=L$,1 then there is another, larger model, $M[G]$, which is also countable and transitive, has the same ordinals as $M$, and in $M[G]$ the Continuum Hypothesis fails. Indeed, a model is simply a set with a binary relation, and in Cohen's case, it is a [countable] transitive set in some ambient universe of $\sf ZF$, with the relation that is the real $\in$ of that ambient universe (restricted to the model).

By restricting to transitive models, Cohen ensures that the notion of sethood does not change from one model to the next.

Of course, that raises the obvious question: can we do forcing without countable transitive models? Indeed, without any models of $\sf ZF$? The answer is yes, and from a technical standpoint there is little to no difference: start with a countable transitive model, look at the proofs from $\sf ZF$ that the forcing theorem holds for whatever statement we wish to prove, and extract a finite fragment $\sf ZF^*$ which is sufficient for this proof; next given any finite extension of this fragment, we can find a countable transitive model of this fragment and force over that; finally, by a meta-theoretic argument we get that $\sf ZF$ cannot prove $\sf AC$, and $\sf ZFC$ cannot prove $\sf CH$, etc.

But it really is simpler to just take countable transitive models on the chin and move on. Finally the main counterpoint to your issue with "different sethood" is that $M$ is a submodel of $M[G]$, so that $M[G]$ and $M$ agree on the sets of $M$, and their elements, and their elements' elements, and so on.

In fact, in Cohen's case, the forcing does not even change cofinalities of ordinals. So $M$ and $M[G]$ agree on which ordinal is $\omega_1$ and which one is $\omega_2$, etc. The thing they disagree on, for example, is to what extent subsets of $\omega$ are in the model: in $M[G]$ there are more subsets of $\omega$ than in $M$, to the point that there is no longer a way to match them with $\omega_1$ (of $M$ or $M[G]$) inside $M[G]$ itself. And again, because these models are transitive, they also agree with $V$, with the "real universe", as to what are subsets of $\omega$. It's simply that they don't know all the subsets.

Cohen does not assume transitivity, but just that the $\in$ relation is the correct one, but these are all isomorphic to a transitive model, and so he also assumes this.


问:
Prove equality using sum of squares and determinants
Let $a$, $b$, $c$, $d$ be four positive real numbers satisfying that $a^2+b^2 = S = c^2+d^2$. I want to show that, under that hypotheses, it also holds:
$$ (ab+cd)S = (ac+bd)(ad+bc) $$

It seems to me that this equality can be deduced using suitable determinants of matrices. At a certain point one should use that $a^2+b^2$ and $c^2+d^2$ coincide.

答:
You can simply calculate this algebraically, no need to consider matrices:

$$\begin{align*}(ac+bd)(ad+bc) &= a^2cd + abc^2 + abd^2 + b^2cd \\
&=  abc^2 + abd^2 +a^2cd + b^2cd \\ &= ab(c^2+d^2) + cd(a^2+b^2)\\
&= abS+cdS \\
&= (ab+cd)S\end{align*}$$


问:
If $[X,Y]=0$ then $X=0$
Let M be a manifold and $X \in \mathfrak{X}(M)$ show that if for any $Y \in \mathfrak{X}(M)$ , $[X,Y]=0$ then $X=0$.

I think we can use following theorem

Theorem: $[X, Y]=0$ iff the flows of $X$ and $Y$ commute locally, meaning $\left(\Phi_{t}^{Y} \Phi_{s}^{X}\right)(x)=\left(\Phi_{s}^{X} \Phi_{t}^{Y}\right)(x)$ for all $x \in M$ and
sufficiently small s, t.
This is a special case of the Frobenius integrability
theorem.

答:
Suppose $[X,Y] = 0$ for all vector fields $Y$. Then, for all $f \in \mathcal{C}^{\infty}(M)$ and all $Y$, we have $0=[X,fY] = \left(X\cdot f\right) Y + f [X,Y] = \left(X \cdot f\right) Y$.

Fix a point $p \in M$ and choose a vector field that does not vanish in a neighbourhood of $p$, say $Y$. Then for all function $f$, we have $(X\cdot f) Y= 0$. Thus in a neighbourhood of $p$, we have $X\cdot f = 0$. Note that this is true for all $p$.

For now, we have shown that $X$ is a vector field satisfying
$$
\forall f \in \mathcal{C}^{\infty}(M),~ X\cdot f = 0.
$$
The exercice now reduces to show that there is a unique vector field $X$ that acts trivially on the set $\mathcal{C}^{\infty}(M)$, that is the zero vector field. If you know the different equivalent definitions for vector fields, this should be pretty straightforward from there.

Title: If $[X,Y]=0$ then $X=0$


问:
Size of the image (rank) in $T_n$
For $\alpha, \beta \in T_n$ (full transformation semigroup/monoid - set of all maps from $\{1,2,\ldots, n\}$ to itself), show that $|\text{Im}(\alpha\beta)| \leqslant |\text{Im}(\alpha)|$ and $|\text{Im}(\alpha\beta)| \leqslant |\text{Im}(\beta)|$.

How do you show this (apparently it's easy)?

答:
I haven't read Sammy Black's proof in detail, but I think an answer consisting mainly of words might be more illuminating.

Let $X = \{1,\ldots,n\}$.  Let $k = |\operatorname{Im}(\alpha)|$ and $m = |\operatorname{Im}(\beta)|$.

You didn't say whether your transformations are acting on the left or the right.  I'll assume they act on the right.

To obtain $\operatorname{Im}(\alpha\beta)$, we act on $X$ first by $\alpha$ and then by $\beta$.  Once we have acted by $\alpha$, we have $k$ points left to be acted on by $\beta$.  Since $\beta$ is a function, this can produce at most $k$ points.  But also, since $\beta$ maps the whole of $X$ to $m$ points, it certainly maps $\operatorname{Im}(\alpha)$ (which is a subset of $X$) to at most $m$ points.

So $|\operatorname{Im}(\alpha\beta)|$ is bounded above by both $k = |\operatorname{Im}(\alpha)|$ and $m = |\operatorname{Im}(\beta)|$.

Title: Size of the image (rank) in $T_n$


问:
summing over distribution of occurence of letters in english language
i have this structure, representing the percentwise distribution of the usage of letters in the english language:

```<code>letterFrequency = {
'E' : 12.0,
'T' : 9.10,
'A' : 8.12,
'O' : 7.68,
'I' : 7.31,
'N' : 6.95,
'S' : 6.28,
'R' : 6.02,
'H' : 5.92,
'D' : 4.32,
'L' : 3.98,
'U' : 2.88,
'C' : 2.71,
'M' : 2.61,
'F' : 2.30,
'Y' : 2.11,
'W' : 2.09,
'G' : 2.03,
'P' : 1.82,
'B' : 1.49,
'V' : 1.11,
'K' : 0.69,
'X' : 0.17,
'Q' : 0.11,
'J' : 0.10,
'Z' : 0.07 }
</code>```

Fairly simple. The thing that I am having difficulty understanding, is that in this book (https://eclass.uniwa.gr/modules/document/file.php/CSCYB105/Reading%20Material/[Jonathan_Katz%2C_Yehuda_Lindell]_Introduction_to_Mo(2nd).pdf?fbclid=IwAR1hf1OTKAhf4ZHvswERpcZ3ZVDQMxHuP2FWRg2tvlo3-tUMSdFIPLWZR_8) [introduction to modern cryptography page 11] the following claim about this distribution is made.

```<code>Let p_i,with 0 ≤ p_i ≤ 1, denote thefrequency of theith letter in normal English text Calculation
    using Figure 1.3 gives:
</code>```

$ \sum_{i=0}^{25}p_i^2\approx0.065 $

This makes absolutely no sense at all to me, when i do the calculation, and sum over every frequency lifted to a power of two I get 646.6717. What am I doing wrong?

答:
The table gives percents, so as probabilities each must be divided by $100.$ So if you don't do that first, the sum of squares that way being $646.6717,$ then you still need to divide that by $10000$ to get $.06466717$ which rounds to their number $.065.$

[since your raw numbers are $100$ times the right ones, their squares are $10000$ times the right frequency squares.]


问:
Number Theory: Möbius Function

Background

Definition:

If $S$ with partial order ≤ is locally finite with minimal element, then the generalized Möbius function $\mu$: S $\times$ S $\rightarrow$ $\mathbb{N}$ by

(a) $\forall$ s $\in$S, $\mu$ (s, s)=1

(b) for r< s, $\mu$ (r, s) = - $\sum_{t \in [r, s) } μ(r, t)$\

Theorem to be proved:

Let S $\neq$ ∅ and define if and only if a | b, then a ≤ b. If {d, n}$\mathbb{Z}$, μ(d, n)= $\frac{n}{d}$

Source:

Number Theory with Applications (James A. Anderson, James M. Bell)

My Question

Sorry that I have tried to prove this theorem for many times but I fail. Could somebody help me?

答:
To avoid any possible confusion, I’ll use $\mu'$ for the number theoretic function.

You can do it by induction on $n$. Assume that

$$\mu(d,m)=\mu'\left(\frac{m}d\right)$$

whenever $1\le d\mid m<n$, and let $1\le d\mid n$. If $d=n$, we have

$$\mu(d,n)=\mu(n,n)=1=\mu'(1)=\mu'\left(\frac{n}d\right)\,,$$

so assume that $d<n$.

Let $\prod_{k=1}^\ell p_k^{\alpha_k}$ be the prime factorization of $\frac{n}d$, where each $\alpha_k\ge 1$. then $m\in[d,n)$ iff $\frac{m}d=\prod_{k\in S}p_k^{\beta_k}$, where $S\subseteq\{1,\ldots,\ell\}$, $1\le\beta_k\le\alpha_k$ for $k\in S$, and either $S\ne\{1,\ldots,\ell\}$, or there is a $k\in S$ such that $\beta_k<\alpha_k$.

Suppose first that $\alpha_k=1$ for $k=1,\ldots,\ell$, so that $\mu'\left(\frac{n}d\right)=(-1)^\ell$. Then $m\in[d,n)$ iff there is an $S\subsetneqq\{1,\ldots,\ell\}$ such that $\frac{m}d=\prod_{k\in S}p_k$, and therefore

$$\begin{align*}
\mu(d,n)&=-\sum_{m\in[d,n)}\mu(d,m)\\
&=-\sum_{m\in[d,n)}\mu'\left(\frac{m}d\right)\\
&=-\sum_{S\subsetneqq\{1,\ldots,\ell\}}(-1)^{|S|}\\
&\overset{(1)}=-\sum_{k=0}^{\ell-1}(-1)^k\binom{\ell}k\\
&\overset{(2)}=-\left((1-1)^\ell-(-1)^\ell\right)\\
&\overset{(3)}=(-1)^\ell\\
&=\mu'\left(\frac{n}d\right)\,.
\end{align*}$$

Here $(1)$ is because $\{1,\ldots,\ell\}$ has $\binom{\ell}k$ subsets of cardinality $k$, $(2)$ follows from the binomial theorem, and $(3)$ holds because $0^\ell=0$, since $\ell\ge 1$.

The other possibility is that $\alpha_k>1$ for some $k\in\{1,\ldots,\ell\}$, so that $\mu'\left(\frac{n}d\right)=0$. See if you can carry out the induction step in this case: the computation is actually very similar to the one that I just did, though for some $m\in[d,n)$ you’ll have $\mu'\left(\frac{m}d\right)=0$.

Added: If $\alpha_k>1$ for some $k\in\{1,\ldots,\ell\}$, we have to split the $m\in[d,n)$ into those that will actually contribute to $\mu(d,n)$ and those that contribute $0$. For each $m\in[d,n)$ let $\frac{m}d=\prod_{k=1}^\ell p_k^{\beta_k(m)}$, where $0\le\beta_k(m)\le\alpha_k$ for each $k\in\{1,\ldots,\ell\}$. Let

$$S_m=\{k\in\{1,\ldots,\ell\}:\beta_k(m)=1\}$$

and

$$T_m=\{k\in\{1,\ldots,\ell\}:\beta_k(m)>1\}\,.$$

If $T_m\ne\varnothing$, then $\mu(d,m)=\mu'\left(\frac{m}d\right)=0$. Moreover, for each $S\subseteq\{1,\ldots,\ell\}$ there is exactly one $m\in[d,n)$ such that $S_m=S$ and $T_m=\varnothing$, and each of these is a proper divisor of $n$, so

$$\begin{align*}
\mu(d,n)&=-\sum_{m\in[d,n)}\mu(d,m)\\
&=-\sum_{m\in[d,n)}\mu'\left(\frac{m}d\right)\\
&=-\sum_{\substack{m\in[d,n)\\T_m=\varnothing}}\mu'\left(\frac{m}d\right)\\
&=-\sum_{S\subseteq\{1,\ldots,\ell\}}(-1)^{|S|}\\
&=-\sum_{k=0}^\ell(-1)^k\binom{\ell}k\\
&=-(1-1)^\ell\\
&=0\\
&=\mu'\left(\frac{n}d\right)\,.
\end{align*}$$

Title: Number Theory: Möbius Function


问:
Normal Distribution Dog Weight
Suppose the weights of full grown female beagles follow a normal distribution with a mean weight of $20$ pounds and a standard deviation of $3$ pounds.

A beagle was brought in for an exam and found to weigh $16.1$ pounds. Based on the provided weight model, approximately what percentage of beagles weigh less than $16.1$ pounds?

Based on the given distribution, the heaviest $15$% of full grown female beagles weigh at least how many pounds?

Let $Y=$ the number of full grown female beagles who weigh at least $20$ pounds in a random sample of $60$.  What is the exact model of $Y$?

a) Normal$(30, 3.873)$

b) Binomial$(60, 0.20)$

c) Binomial$(60, 0.50)$

d) Normal$(20, 3)$

My answers:

$9.68%$%

$23.12$ lbs

I am stuck and need help. I think the answer is B but I am not sure. Any help?

答:
You can get answers to (1) and (2) by standardizing and
using printed normal tables. But using printed tables can require rounding, which may introduce slight errors.

Exact computations in R, where pnorm and qnorm are normal CDF and quantile functions, respectively, can give
more exact answers. [The second parameter in these R procedures is the normal population standard deviation.]
Thank you for showing your work; your answers are in good
agreement with results from R.

```<code>pnorm(16.1, 20, 3)
[1] 0.09680048
qnorm(.85, 20, 3)
[1] 23.1093
</code>```

Also, I agree with @patricio's comment.

Below is a graph of the density function of $\mathsf{Norm}(\mu=20, \sigma=3),$ with vertical lines (dotted red, and dashed blue) relevant to parts (1) and (2), respectively.

R code for figure:

```<code>curve(dnorm(x, 20, 3), 10, 30, lwd=2, ylab="PDF",
      main = "Density of NORM920,3)")
 abline(h = 0, col="green2")
 abline(v = 16.1, col="red", lwd=2, lty="dotted")
 abline(v = 23.1, col="blue", lwd=2, lty="dashed")
</code>```

Title: Normal Distribution Dog Weight

Body:
You can get answers to (1) and (2) by standardizing and
using printed normal tables. But using printed tables can require rounding, which may introduce slight errors.

Exact computations in R, where pnorm and qnorm are normal CDF and quantile functions, respectively, can give
more exact answers. [The second parameter in these R procedures is the normal population standard deviation.]
Thank you for showing your work; your answers are in good
agreement with results from R.

pnorm(16.1, 20, 3)
[1] 0.09680048
qnorm(.85, 20, 3)
[1] 23.1093

Also, I agree with @patricio's comment.

Below is a graph of the density function of $\mathsf{Norm}(\mu=20, \sigma=3),$ with vertical lines (dotted red, and dashed blue) relevant to parts (1) and (2), respectively.

R code for figure:

curve(dnorm(x, 20, 3), 10, 30, lwd=2, ylab="PDF",
      main = "Density of NORM920,3)")
 abline(h = 0, col="green2")
 abline(v = 16.1, col="red", lwd=2, lty="dotted")
 abline(v = 23.1, col="blue", lwd=2, lty="dashed")


问:
Problem by proving a statement by induction
I want to prove the following statement by induction for $n\leq 1$:
\begin{align*}
\sum_{k=n}^{2n-1}\frac{1}{k}= \sum_{k=1}^{2n-1}\frac{(-1)^{k+1}}{k}
\end{align*}

For $n=1$ its easy to see:
\begin{align*}
\sum_{k=1}^{2-1}\frac{1}{k}= \sum_{k=1}^{1}\frac{1}{k}= 1 = \sum_{k=1}^{1}\frac{(-1)^{2}}{1}
\end{align*}
For $n\mapsto n+1$
\begin{align*}
\sum_{k=n+1}^{2(n+1)-1}\frac{1}{k}= \sum_{k=n+1}^{2n+1}\frac{1}{k} = \sum_{k=n-1}^{2n-1}\frac{1} {k}+\frac{1}{n+1}+\frac{1}{n}\dots \ ?
\end{align*}
I am not really sure how to deal with the $k=n+1$ in the sum, to use the case that the statement holds for n. Could anybody help me? Thank you very much!

答:
Call the left sum $L(n)$ and the right sum $R(n)$. Then we have

$$L(n + 1) - L(n) = \frac{1}{2n}+\frac{1}{2n+1} - \frac{1}{n},$$

and

\begin{align}
R(n + 1) - R(n) &= \frac{(-1)^{2n}}{2n+1} + \frac{(-1)^{2n+2}}{2n+1} \\
&= - \frac{1}{2n} + \frac{1}{2n+1}.
\end{align}

So by comparison, we need to check that

$$\frac{1}{2n} - \frac{1}{n} = - \frac{1}{2n}, $$

which is easy enough.

Therefore $L(n+1) - L(n) = R(n + 1) - R(n)$. So if $L(n) = R(n)$ then $L(n + 1) = R(n + 1)$.


问:
Cardinality of the set of divergent sequences
How to find the cardinality of the set of divergent sequences? Let's name this set $A$.

I know that cardinality of the set of sequences equals $2^{\lvert \mathbb{N}\rvert}$, so $\lvert A\rvert\le 2^{\lvert \mathbb{N}\rvert}$.
How to strictly prove to the other side?

Thank you for your time.

答: Hint: Show that the set of unbounded increasing sequences has cardinality $\mathfrak{c}$.


问:
Isomorphism of Lie algebras $\mathfrak{sl}(4, \mathbb{C}) \cong \mathfrak{so}(\bigwedge^2 \mathbb{C}^4,B)$ via the differential of a representation
As in a previous question, letting $V = \mathbb{C}^4$ with basis $\{e_1, \dots, e_4\}$, and define the representation $\rho$ of $G = \operatorname{SL}_4(\mathbb{C})$ on $\bigwedge^4 V$ given by $g \cdot (v_1 \wedge v_2) = gv_1 \wedge gv_2$.
Set $\Omega = e_1 \wedge e_2 \wedge e_3 \wedge e_4$, and let $B$ be the non-degenerate symmetric bilinear form on $\bigwedge^2 V$ defined by $u \wedge v = B(u,v)\Omega$.

I know that $B(\rho(g) a, \rho(g)b) = B(a,b)$ and $B(d\rho(X)a,b) + B(a,d\rho(X)b) = 0$ for all $g \in G$, $X \in \mathfrak{sl}(4,\mathbb{C})$, and $a, b \in \bigwedge^2 V$.
I need to use the differential $d\rho$ to obtain a Lie algebra isomorphism $\mathfrak{sl}(4, \mathbb{C}) \cong \mathfrak{so}(\bigwedge^2 \mathbb{C}^4,B)$, using the fact that $\mathfrak{sl}(4, \mathbb{C})$ is a simple Lie algebra (a Lie algebra with no proper $\text{ad}(\mathfrak{g})$-invariant subspaces), and without using any knowledge of the general classification of simple Lie algebras/their Dynkin diagrams.

I am not sure how this hint is helpful though.
In trying to find some non-trivial $\text{ad}(\mathfrak{g})$-invariant subspace, I am not going anywhere.
The two identities involving $B$ only tell me that $B$ is invariant under $G$ and $\mathfrak{g}$, but I can draw no further conclusions.

答:
The differential of any Lie group homomorphism is a lie algebra homomorphism.

You already have a Lie group homomorphism $\rho:\mathrm{SL}_4\mathbb{C}\to\mathrm{SO}(\Lambda^2\mathbb{C}^4)$, which means the differential must be a lie algebra homomorphism $\mathrm{d}\rho:\mathfrak{sl}_4\mathbb{C}\to\mathfrak{so}(\Lambda^2\mathbb{C}^4)$. You can check it is a nonzero map, so the kernel is proper, but $\mathfrak{sl}_4\mathbb{C}$ is simple so the kernel must be trivial. Then check dimensions to conclude this is a lie algebra isomorphism (after all, $\mathrm{d}\rho$ is a linear map!).


问:
Inverse complex tangent in two forms
I have one doubt, I'm studying complex trigonometric functions and in the books I've seen there are two versions of the inverse tangent. I've seen: $$\tan^{-1}(z) = \frac{1}{2i} \log \frac{1+zi}{1-zi}$$

and, as well: $$\tan^{-1}(z) =  \frac{i}{2} \log \frac{i+z}{i-z} $$

How are these two the same? Thanks in advance.

答:
firstly $\frac{1}{2i} =-\frac{i}{2}$

Multiplying the expression inside the log by $i\over i$ ...
$$ \frac{1+zi}{1-zi} \times {i\over i}= \frac{i-z}{i+z}  $$
so $$\log \frac{1+zi}{1-zi} = - \log \frac{i+z}{i-z} $$

Title: Inverse complex tangent in two forms


问:
In how many ways can we split 8 animals into four groups with these requirements...
So we have $2$ cats, $2$ dogs, $2$ lions, $2$ tigers, and we want to split them into $4$ groups such that each group has $2$ none similar animals.

In how many ways we can do that?

I'm trying to use the inclusion exclusion principle, since I tried to approach it normally and got stuck, and here's what I reached:

numbers of ways with $1$ group with $2$ similar animals : $8 \choose 2$$\frac {6!}{2!2!2!}$ .

number of ways with $2$ groups with $2$ similar animals: $8 \choose 4$$\frac{4!}{2!2!}$.

number of ways with $3$ groups with $2$ similar animals: $8 \choose 6$.

number of ways with $4$ groups with $2$ similar animals: $8 \choose 8$.

So my answer will be $\frac {8!}{2!^4}$ - $4 \choose 1$$8 \choose 2$$\frac {6!}{2!2!2!}$ + $4 \choose 2$$8 \choose 4$$\frac{4!}{2!2!}$ - $4 \choose 3$$8 \choose 6$ +$1$.

$=5040 - 10080 + 2520 - ...= (-)$.

So basically I got it all wrong since I got a minus in the end, would be happy to see where are my mistakes and how to solve this. Thanks in advance.

答:
For applying P.I.E, the correct way should have been

$\displaystyle 4! - {4 \choose 1} 3! + {4 \choose 2} 2! - {4 \choose 3} 1! + {4 \choose 4} 0! = 9$

which is nothing but the derangement of $4$ i.e. $!4$.

But here is the problem. It is not a straightforward application of derangement either. That is because you are pairing animals that are of same kind (cat, dog, lion and tiger with cat, dog, lion and tiger).

Consider for example, a group $\{C \ D, D \ L, L \ T, T \ C\}$ is no different than $ \{C \ T, D \ C, L \ D, T \ L \}$. The derangement will count them as two different. The number of such repetitions should be ${3 \choose 2} = 3$ which is choosing two animals out of remaining $3$ to pair with a given animal. Once we fix two pairs, other two are automatically fixed. Here are the other two that repeat -

$\{C \ D, D \ T, L \ C, T \ L\}$ and $ \{C \ L, D \ C, L \ T, T \ D \}$

$\{C \ L, D \ T, L \ D, T \ C\}$ and $ \{C \ T, D \ L, L \ C, T \ D \}$

So the right answer is $ = !4 - 3 = 6$.

By the way, writing all permissible arrangements is a very viable option here.


问:
Why is a triangle a rigid shape?
Triangles are used in structures because they are the most rigid shape. And they are many descriptions of this on network and, if I am not mistaken, this all comes down to the law of cosines.But why this holds? Is it maybe because if we connect three random points on a planes their internal angles will add to half revolution? Is there a more fundamental reason for the rigidness of triangles?

Engineer here, please don't mind the naive question

答:
The triangle is the only polygon that cannot change shape without changing the length of one or more of its sides. For all other polygons, the shape can change by merely changing four or more angles. For instance, a rectangle can become a parallelogram just by increasing two opposite angles and decreasing the other two angles.

A triangle can change shape by "breaking" one of its sides into two parts and making an angle with them (like putting a dent in the base of an equilateral triangle). This turns it into a quadrilateral... or maybe it was 4-sided before but two of the sides were in a straight line.,

But (and I know this is repeating) unless you change the length of one or more sides of a triangle, you cannot change its shape.

Title: Why is a triangle a rigid shape?


答:
There are essentially two different patterns: either you have two species in two groups (forcing the choice for the other two groups) or you do not (in which case each group is distinct and with four groups you must have a cycle):

(a) AB AB CD CD

(b) AB BC CD DA

But the next issue is what counts as a different split.

If the species are distinguishable, but the groups and individual animals are not then (a) has $3$ possibilities (the cats can be with one other species) and (b) has $3$ possibilities (the cats can not be with one other species), making $6$ in total

If the species and individual animals are distinguishable, but the groups are not then (a) has $3\times 2^2$ possibilities (since you can only split a pair of species two ways, but have two sets of pairs) and (b) has $3\times 2^4$ possibilities (since going round the cycle you get two choices for each species), making $60$ in total

Title: In how many ways can we split 8 animals into four groups with these requirements...

Body:
So we have $2$ cats, $2$ dogs, $2$ lions, $2$ tigers, and we want to split them into $4$ groups such that each group has $2$ none similar animals.

In how many ways we can do that?

I'm trying to use the inclusion exclusion principle, since I tried to approach it normally and got stuck, and here's what I reached:

numbers of ways with $1$ group with $2$ similar animals : $8 \choose 2$$\frac {6!}{2!2!2!}$ .

number of ways with $2$ groups with $2$ similar animals: $8 \choose 4$$\frac{4!}{2!2!}$.

number of ways with $3$ groups with $2$ similar animals: $8 \choose 6$.

number of ways with $4$ groups with $2$ similar animals: $8 \choose 8$.

So my answer will be $\frac {8!}{2!^4}$ - $4 \choose 1$$8 \choose 2$$\frac {6!}{2!2!2!}$ + $4 \choose 2$$8 \choose 4$$\frac{4!}{2!2!}$ - $4 \choose 3$$8 \choose 6$ +$1$.

$=5040 - 10080 + 2520 - ...= (-)$.

So basically I got it all wrong since I got a minus in the end, would be happy to see where are my mistakes and how to solve this. Thanks in advance.


问:
Am looking to prove following conjecture:
Am looking to prove that the tangent to the cubic $y = x^3 + 7x^2 + 4x + 28$ at any point midway between any two roots will itself cut the x-axis at the remaining root. The cubic factorizes as $y = (x+2i)(x-2i)(x+7)$. And here's my problem: complex roots. How can I solve the above?</p>

答:
The only way I can see you can try in this case is taking all the roots as points in $\,\Bbb C\cong\Bbb R^2\,$ , so that the roots are $\,(-7,0)\,,\,(0,2)\,,\,(0,-2)\,$ , but then you have a new problem (or perhaps you only need a good definition): is the midpoint of the line segments between any two of these points a point on the function's graph? No, it's not. For example, the midpoint between points $\,2\,,\,3\,$ is the origin $\,(0,0)\,$ , which is not part of the graph, or the midpoint between points $\,1\,,\,2\,$ , which is $\,\displaystyle{\left(-\frac72\,,\,1\right)}\;$, and it is also not on the graph.

Thus, the conjecture is false under the above assumption. If you have other ones write them down.


问:
How to use undetermined coefficients to solve an equation?
For this question, I'm having trouble solving this equation. Here is what I have so far. Can anyone please help me out?

$$u'' + u' + 2u = \sin^2t$$

$$ u'' + u' + 2u = 0$$

$$ m^2 + m + 2 = 0$$

$$ m = \frac{-1 +- \sqrt{1-4(1)(2)}}{2}$$

$$ m = \frac{-1}{2} + \frac{\sqrt{7}{i}}{2}$$

$$ u = e^\frac{-t}{2}(c_1\cos\sqrt{7}t + c_2\sin\sqrt{7}t)$$

答:
For the complementary solution it should be

$$u_c(t)=e^{-\frac{t}{2}}\big((c_1\cos\big(\frac{\sqrt{7}}{\color\red{2}}t\big)+c_2\sin\big(\frac{\sqrt{7}}{\color\red{2}}t\big)\big)$$

Then using the identity $\cos(2t)=1-2\sin^2(t)$ we have  $$u''+u'+2u=\sin^2(t)=\frac{1}{2}-\frac{\cos(2t)}{2}$$ So for the particular integral try $u_{p}(t)=A\cos(2t)+B\sin(2t)+C.$

 You should get $$u(t)=e^{-\frac{t}{2}}\big((c_1\cos\big(\frac{\sqrt{7}}{\color\red{2}}t\big)+c_2\sin\big(\frac{\sqrt{7}}{\color\red{2}}t\big)\big)+\frac{1}{8}\big(\cos(2t)-\sin(2t)\big) +\frac{1}{4}.$$

We have $$u_{p}=A\cos(2t)+B\sin(2t)+C$$
$$u_{p}'=-2A\sin(2t)+2B\cos(2t)$$
$$u_{p}''=-4A\cos(2t)-4B\sin(2t)$$

Substituting, we obtain

$$u_{p}''+u_{p}'+2u_{p}=-4A\cos(2t)-4B\sin(2t)$$
$$-2A\sin(2t)+2B\cos(2t)$$
$$+2A\cos(2t)+2B\sin(2t)+2C$$
$$=-2A\cos(2t)-2B\sin(2t)+2B\cos(2t)-2A\sin(2t)+2C$$
$$=(-2A+2B)\cos(2t)-(2A+2B)\sin(2t)+2C=\frac{1}{2}-\frac{1}{2}\cos(2t)$$

And comparing coefficients we have $$-2A+2B=-\frac{1}{2}$$
$$2A+2B=0$$
$$2C=\frac{1}{2}$$

$$\implies A=\frac{1}{8}, \space B=-\frac{1}{8}, \space C=\frac{1}{4}$$


答:
Use Euler's formula to express $\sin^2(t)$ in terms of exponentials

$$\sin^2(t) = -\frac 14 (e^{2it} -2 + e^{-2it}) $$

now try a solution of the form $Ae^{2it} + B + C e^{-2it}$

Title: How to use undetermined coefficients to solve an equation?

Body:
For this question, I'm having trouble solving this equation. Here is what I have so far. Can anyone please help me out?

$$u'' + u' + 2u = \sin^2t$$

$$ u'' + u' + 2u = 0$$

$$ m^2 + m + 2 = 0$$

$$ m = \frac{-1 +- \sqrt{1-4(1)(2)}}{2}$$

$$ m = \frac{-1}{2} + \frac{\sqrt{7}{i}}{2}$$

$$ u = e^\frac{-t}{2}(c_1\cos\sqrt{7}t + c_2\sin\sqrt{7}t)$$


问:
How do I compute this integral with a Dirac's delta?
While studying probability I encountered this integral
$$I=\int_{\mathbb{R}^2}\exp\left({-\frac{x_1^2+x_2^2}{2}}\right)\delta\left(r-\sqrt{x_1^2+x_2^2}\right)\,dx_1\,dx_2$$
If I compute this in polar coordinates i get
$$I=\int_0^{2\pi}\,d\theta \int_0^{+\infty}\exp\left(-\dfrac{\rho^2}{2} \right)\rho\delta(r-\rho)\,d\rho=2\pi r\exp\left(-\dfrac{r^2}{2}\right)$$
but in cartesian coordinates I only get
$$I=\exp\left(-\frac{r^2}{2}\right)$$
I don't understand why. I just thougth that I was using the Dirac's delta's properties in both cases. I think the first result is the correct one and there is something I don't know about Dirac's delta with more than one variable.

Which result is correct and why?

答:
In Cartesian coordinates, you have
\begin{align}
I=&\ \int dx\ \exp\left(-\frac{x_1^2+x_2^2}{2} \right)\delta(r-\sqrt{x_1^2+x_2^2})\\
 =&\ \int_{r=\sqrt{x_1^2+x_2^2}} d\sigma\
\exp\left(-\frac{x_1^2+x_2^2}{2} \right)
\end{align}
since
\begin{align}
|\nabla (r- \sqrt{x_1^2+x_2^2})| = 1.
\end{align}
Hence, it follows that
\begin{align}
I = \int_{r=\sqrt{x_1^2+x_2^2}} d\sigma\
\exp\left(-\frac{r^2}{2} \right) = 2\pi r\exp\left(-\frac{r^2}{2} \right).
\end{align}


问:
Possible error in Wikipedia article for $(p,q)$ tensor space notation
I am currently studying tensor space theory (over a field) and was reading through the generic Wikipedia article for tensors when I cam across the following:

A type $(p,q)$ tensor $T$ is a multilinear map, $T: \underbrace{V^*\times...\times V^*}_\text{p-copies}\times\underbrace{V\times...\times V}_\text{q-copies} \to K$ where $K$ is the underlying field.

Where the author(s) then goes onto define the notation for the  tensor product as

$\underbrace{V\otimes...\otimes V}_\text{p-copies} \otimes \underbrace {V^* \otimes...\otimes V^*}_\text{q-copies}$

My question: why did they flip the usage of the $p$'s and $q$'s? Namely, they first called copies of the dual space $p$ and copies of the vector space $q$ $-$ then went on to reverse this in the notation for the tensor product (calling copies of the vector space $p$ and copies of the dual $q$). Is this an error that should be corrected? I also noticed this in a lecture by Frederic P. Schuller on tensor space theory. He did the exact same thing when defining what a tensor is and then the tensor product (at minute $29:23$).

Please correct me if I am wrong, as I am new to differential geometry and tensors.

答:
Note that a $(0,1)$-tensor is a linear map $V \to K$, so $V^*$ is the space of such tensors. Similarly, a $(1,0)$-tensor is a linear map $V^* \to K$, so $V^{**}$ (which can be canonically identified with $V$ if we are in finite dimensions) is the space of such tensors.

Now, if $v \in V$ and $f \in V^*$, define $v \otimes f : V^* \times V \to K$ by $(v \otimes f)(g,w) = g(v)f(w)$. Note then that $v \otimes f \in V \otimes V^*$, not $v \otimes f \in V^* \otimes V$. Thus $$v \in V,\, f \in V^* \implies v \otimes f \in V \otimes V^*$$ as naturally one would like it to happen.


答: No, this is correct. The asterisk denotes the dual, i.e. if $V$ is a $K$-vector space, $V^*$ is the vector space of linear maps $V \to K$. Intuitively, in the first case, you have something very similar (except with multilinear maps). In the second case, you are missing this step (the “$\to K$” part), so you have to dualize everything in the domain, which gives you the swapped asterisks (as $V^{**} = V$ for finite-dimensional vector spaces).

Title: Possible error in Wikipedia article for $(p,q)$ tensor space notation

Body:
I am currently studying tensor space theory (over a field) and was reading through the generic Wikipedia article for tensors when I cam across the following:

A type $(p,q)$ tensor $T$ is a multilinear map, $T: \underbrace{V^*\times...\times V^*}_\text{p-copies}\times\underbrace{V\times...\times V}_\text{q-copies} \to K$ where $K$ is the underlying field.

Where the author(s) then goes onto define the notation for the  tensor product as

$\underbrace{V\otimes...\otimes V}_\text{p-copies} \otimes \underbrace {V^* \otimes...\otimes V^*}_\text{q-copies}$

My question: why did they flip the usage of the $p$'s and $q$'s? Namely, they first called copies of the dual space $p$ and copies of the vector space $q$ $-$ then went on to reverse this in the notation for the tensor product (calling copies of the vector space $p$ and copies of the dual $q$). Is this an error that should be corrected? I also noticed this in a lecture by Frederic P. Schuller on tensor space theory. He did the exact same thing when defining what a tensor is and then the tensor product (at minute $29:23$).

Please correct me if I am wrong, as I am new to differential geometry and tensors.


答:
First, read WW1's comment on the question.  I agree that his interpretation of the question is most likely what is intended.

Now, since no two animals of the same type in are to be in the same group, by the pigeon hole principle, there must be a group with a cat, a second group, with a dog, yet a third group with a tiger, and a fourth group with a lion.  This is what I meant by my first sentence.

We concentrate on the remaining $4$ animals.  We can arrange them in $4!=24$ ways, put the first animal in the first group, the second animal in the second groups, and so on.  Not all these permutations are admissible.  Any of the $3!=6$ permutations in which the cat comes first is invalid, because we will end up with two cats in the first group. Similarly, we must subtract the $6$ permutations where the dog comes second, and so on.  Since we have to do this for each of the animals, we are left with $24-\cdot6=0$ permutations.

Wait, though; we've done some double counting.  If the cat comes first and the dog second in the permutation, then we subtracted it twice.  We have to add these back in.  There are $\binom42=6$ ways to choose the two animals, and $2$ ways to permute the other $2$ animals, so we have to add $12$.

We're not done yet.  Suppose $3$ of the animals match the animal already in the group.  Then of course, so does the fourth, and there's only one such permutation.  How many times have we counted it?  We counted it once to begin with, subtracted it $4$ times (once for each animal), and added it back $6$ times (once for each pair of animals).  In all, we counted it $1-4+6=3$ times.  And of course, we don't want to count it all all, so we subtract $3$, giving a final answer of $9$.

This is equal to the number of derangements of $4$ objects, where a derangement is a a permutation that leaves no element fixed. That is $\sigma(x)\neq x$ for every $x$ in the set.  I hope the relation of derangements to this problem is clearer now.


问:
Drawing from an urn - probability of drawing maximum 2 of each color.
I have an urn with 100 balls; 10 colors with each 10 balls.

Now I draw 10 balls from the urn without replacement - I want to calculate the probability of getting 3 or more of any color.

How do I calculate this? My problem is that for example the event "more than 3 greens" and "more than 3 yellows" are not exclusive, there are possible draws which fit into both categories. So I cant the hypergeometric distribution and multiply it by 10.

I feel like this question is school level.... I got a math degree, so the answer doesnt need to be super basic, I just suck at stochastics. Help greatly appreciated.

答:
The probability you are looking for can be found by inversion: it is one minus the probability of drawing two or less of each color, which is much easier to calculate since a draw has to satisfy the condition for each color rather than at least for some.

What combinations can you have that would count in? There can be 10 balls of 10 different colors $(1,1,\ldots,1)$, or 2 of the same color + 8 uniquely colored ones $(2,1,\ldots,1,0)$, or $(2,2,1,\ldots,1,0,0)$, etc. Write down the probability of each case, denoting each by the number $i \in \{0, \ldots, 5\}$ of same-color couples:
\begin{align}
p_i &= \frac{90!}{100!} \cdot 10^{10-i} \cdot 9^{i} \cdot (10-i)! \cdot \frac{10!}{2^i}.
\end{align}

Here the first term is the probability of each particular 10-ball draw (with the order of drawing kept important), the second term reflects the probability of drawing a new color the first time, and the third term, the second time; the fourth term is for all the ways to assign a particular color the number of occurrences in the draw, and the rest account for the fact that the order of drawing is actually irrelevant.

Important: I haven't done combinatorics in a while, so my exact formula for $p_i$ may be flawed, but I stand for the overall logic.

Then the probability you're looking for is $1 - \sum_{i=0}^{5} p_i$.


问:
Nonsingular affine curve which is not unmixed
Let $C$ be any nonsingular curve in $A^3_{\mathbb C}$. Can a point be an irreducible component of $C$? I am not able to find an example of such $C$.</p>

答:
First, note that a point is an irreducible component of a variety if and only if its complement is closed, i.e. if and only if the point is isolated. So any example would be a disjoint union of another non-singular curve and a point.

The following equations cut out the disjoint union of a line and a point in the affine plane:
\begin{align}
x y & = 0 \\
x^2 - x & = 0
\end{align}
Thus it is seen that the coordinate ring of this affine variety is isomorphic to $\mathbb{C}[y] \times \mathbb{C}$, which is certainly of dimension 1. It is a regular variety in the sense that every local ring is regular, but perhaps you would not count it as non-singular, because the tangent space at the isolated point is 0-dimensional instead of 1-dimensional.


问:
Is it possible for the graph of a multivariable function to be a curve instead of a surface?
The graph of a function, let's say for example $f: \mathbb{R}^2  \rightarrow \mathbb{R} $ is a subset of the Cartesian product of the domain and the codomain, in this case of $\mathbb{R}^3$. When speaking about the graph of functions like $f$ the term that is always associated is that one of surface. However i was wondering if there is a possibility that the graph of a function of this type is instead a curve, that can be represented as a vector function $\vec{f}$. For example, a function $F(x, y)$ with the property that the condition of each level set is only satisfied by one point.

$$ F(x, y) = c$$

I imagine that the previous example is not possible, nonetheless i haven't been able to prove it. I don't know if the question is rather trivial, and i have a misconception about the concept of the  graph of a function.

答:
I answer here your question in my comment. If we add continuity we cannot construct such a function (the one with level sets consisting of just one point). Let $F:\mathbb{R}^2\longrightarrow\mathbb{R}$ be a continuous function, and let $(x_1,y_1),(x_2,y_2)\in\mathbb{R}^2$. Applying Intermediate Value Theorem to the single variable functions:
$$g_1(t):=\begin{cases}F((x_1,y_1)+2t\cdot(x_2-x_1,0)) &\text{ if }t\in[0,\frac{1}{2}] \\F((x_1,y_1)+(2t-1)\cdot(0,y_2-y_1)) &\text{ if } t\in(\frac{1}{2},1]\end{cases}\\
g_2(t):=\begin{cases}F((x_1,y_1)+2t\cdot(0,y_2-y_1)) &\text{ if }t\in[0,\frac{1}{2}] \\F((x_1,y_2)+(2t-1)\cdot(x_2-x_1,0)) &\text{ if } t\in(\frac{1}{2},1]\end{cases}$$

We get that for all $y_1\in(g_1(0),g_1(1))$ there exists $c_1\in(0,1)$ such that $g_1(c_1)=y_1$, and for all $y_2\in (g_2(0),g_2(1))$ there exists $c_2\in(0,1)$ such that $g_2(c_2)=y_2$. Note that $g_1(0)=g_2(0)$ and $g_1(1)=g_2(1)$ because of how we defined them, then for every: $$y\in(g_1(0),g_1(1))=(g_2(0),g_2(1))=(F(x_1,y_1),F(x_2,y_2))$$ there exists $c_1,c_2\in(0,1)$ such that $g_1(c_1)=g_2(c_2)=y$. The existence of this implies that there are: $$(a,b)\in L((x_1,x_2),(x_2,y_1))\cup L((x_2,y_1),(x_2,y_2)) \\ (c,d)\in L((x_1,y_1),(x_1,y_2))\cup L((x_1,y_2),(x_1,y_2))$$
(where $L((u,v),(w,z)):=\{(u,v)+t\cdot (w-y,z-v) : t\in[0,1]\}$ is the segment joining two points).

Such that $F(a,b)=y=F(c,d)$. This makes impossible what we want.


问:
Finding area based on only the sides
I came across this simple looking question:

The perimeter of a triangle is $42$ cm. One side of a triangle is $8$ cm longer than the smallest side and the third side is $1$ cm less than $3$ times the smallest side. Find the area of the triangle.

At first it seemed easy enough. The smallest side is $7$ (which can be derived from:$\frac{42-8+1}{5}$, where subtracting $8$ takes care of the long side and now we have $2$ units of the shortest side, then adding one takes care of the third side now we have $5$ units of the third side and dividing by $5$ we can get the length of the smallest side, $7$) and with that knowledge the length of the first and third side is $15$ and $20$ respectively.

That's when I got stuck. The question gives us no indication of the angles in the triangle. I thought we could use the Pythagorean theorem to see if the triangle is right-angled, but it's not. Is there a rule we can use to find the area? Yes, I'm familiar with trigonometry and $sin()$, $cos()$ and $tan()$ so you can use those in your answer.

If you don't mind can you give me the rule to finding the area, not the answer to the problem? Because I want to solve this question myself. Many thanks to whoever answer's this!

答:
Here is a proof for Heron's formula, as requested.

We begin with the Cosine rule:
$$a^2=b^2+c^2-2bc\cos A\iff\cos A=\frac{b^2+c^2-a^2}{2bc}$$
Recall the identity $$\sin^2x+\cos^2x\equiv 1\iff\sin^2x\equiv1-\cos^2x$$
This means that $$\begin{align}
\sin^2A&=1-\cos^2A\\
&=1-\frac{(b^2+c^2-a^2)^2}{(2bc)^2}\\
&=\frac{(2bc)^2-(b^2+c^2-a^2)^2}{(2bc)^2}\\
&=\frac{(2bc)^2-(a^2-b^2-c^2)^2}{(2bc)^2}\\
&=\frac{(a^2-(b^2-2bc+c^2))((b^2+2bc+c^2)-a^2)}{(2bc)^2}\\
&=\frac{(a^2-(b-c)^2)((b+c)^2-a^2)}{(2bc)^2}\\
&=\frac{(a+b-c)(a-b+c)(a+b+c)(-a+b+c)}{(2bc)^2}
\end{align}$$
So that factors really nicely! (Just for clarification: I have used the formulae $(x+y)^2=x^2+2xy+y^2$ and $x^2-y^2=(x+y)(x-y)$ quite a few times in the manipulations above.)

Now consider the expression for the 'semi-perimeter' $s$:
$$s=\frac{a+b+c}{2}$$
This means that
$$s-b=\frac{a+b+c}{2}-\frac{2b}{2}=\frac{a-b+c}{2}$$
and so on for $a$ and $c$ as well. Thus, we have
$$\begin{align}
\sin^2A&=\frac{2s \times 2(s-a)\times 2(s-b)\times 2(s-c)}{(2bc)^2}\\
&=\frac{4^2s(s-a)(s-b)(s-c)}{(2bc)^2}
\end{align}$$

Now, in a triangle, the $\sin$ of any angle will be positive. This is because any angle $\theta$ in a triangle must satisfy $0<\theta<180^\circ$, and for $0<\theta<180^\circ$ we know that $\sin\theta$ is positive. Hence when we now square root both sides we can (and must) take the positive square root. So we have
$$\begin{align}
\sin A&=\frac{4}{2bc}\sqrt{s(s-a)(s-b)(s-c)}\\
&=\frac{2}{bc}\sqrt{s(s-a)(s-b)(s-c)}\\
\end{align}$$
But wait! We know that the area of a triangle, let's call it $S$, is equal to $\frac{1}{2}bc\sin A$!! This means that
$$\begin{align}
S&=\frac{1}{2}bc\sin A\\
&=\frac{1}{2}bc\times\frac{2}{bc}\sqrt{s(s-a)(s-b)(s-c)}\\
&=\sqrt{s(s-a)(s-b)(s-c)}
\end{align}$$
So we finally have our truly remarkable endpoint:

$$\mathbf{S=\sqrt{s(s-a)(s-b)(s-c)}}$$

as required!!

I hope you enjoyed that as much as I enjoyed writing it! :) If you have any questions please don't hesitate to ask.

Title: Finding area based on only the sides


问:
Probability of getting one correct multiple choice question out of 2 with different number of choices?
Say you have two questions, one with 5 choices and one with 6 (only 1 choice each question is correct). What is the probability of getting Exactly one question correct out of the two. I didn't specify if it is the one with 5 choices or the one with 6.
I know that getting both correct would be 1/5 * 1/6. and if both had the same number of choices (say 4) getting only one correct would be 1/4 * 3/4. but how do I approach the above problem? Thanks

答:
There are 5x6 ways to pick two answers. There are 5 ways to pick the first right and the second wrong, and 4 ways to pick the first wrong and the second right, that’s nine ways to get exactly one right. The chances are 9 in 30 or 3 in 10.

You could also count 4x5 ways to get both wrong, 1 way to get both right, and the 30-20-1 = 9 remaining ways are exactly one right.


问:
Decreasing functions $f:\mathbb{N} \rightarrow \mathbb{N}$
Let $P=\{f:\mathbb{N} \rightarrow \mathbb{N}; f(i+1)\leq f(i) \ \forall \ i \in \mathbb{N}\}.$

I need to show 2 elements of $P$ and prove that $P$ is countable (or not).

I think that $P$ only have constant functions, and if this is right, i can prove that $P$ is countable. There is some counter-example for this?

答: It is not true that every $f\in P$ is constant. Take, say,$$f(n)=\begin{cases}2&\text{ if }n=1\\1&\text{ otherwise.}\end{cases}$$However, it is true that, if $f\in P$, then there is some $N\in\Bbb N$ such that $n\geqslant N\implies f(n)=f(N)$. And it is not hard to deduce from this that $P$ is indeed countable.


问:
if A is compact and B is closed, there exist $u$ in $A$ and $v$ in $B$ such that $\|u-v\|=\inf\{\|a-b\|\mid a\text{ in }A\text{ and }b\text{ in }B\}$
I can't prove this theorem. Could you help me? Thanks for your helps.

$A$ and $B$ are subsets of an $n$-dimensional Euclidean space. Prove that, if $A$ is compact and $B$ is closed, there exist $u$ in $A$ and $v$ in $B$ such that $\|u-v\|=\inf\{\|a-b\| \mid a \in A\text{ and }b \in B\}$.

答:
You may assume that $B$ is compact by embedding $B$ inside a big enough cube (the intersection of this cube with $B$ is closed and bounded, so it's compact). You have to be a bit careful about how you choose this cube, but I'll let you think about that.

Then, $A\times B$ is compact so the function $(a,b) \mapsto \|a-b\|$ maps onto a compact set (image of compact is compact) and reaches its bounds (compact subsets of $\mathbb{R}$ are closed and bounded).


问:
Number of points chosen form a polygon to have no isosceles and equilateral triangles.

Let $\Omega$ be a regular polygon with $n$ sides. Let's choose $\Gamma$ a set of vertices, for which any triangle with the vertices in $\Gamma$ is neither isosceles, nor equilateral. Find $\max |\Gamma|$. (Bulgarian Team Selection Test)

Attempt:

I have tried to prove for small cases: $3$, $4$, $5$, $6$, $7$ etc., but no rule confirms, to be proven by induction. And, technically, induction is not possible in this case, because, polygon $\Omega \setminus A$, where $A$ is a vertex, won't be regular.

答:
Upfront warning: this is not really an answer to the question asked, more like a report on my search for sources that couldn't possibly fit in a comment.

Witness the fact that most comments reference integer indices, it is clear that this problem has very little to do with geometry and polygons. The question asked is about avoiding arithmetic progressions in cyclic groups. The triangle aspect of the original question is important though, since it prescribes whether $\{1,6\}$ in $\mathbb{Z}_{10}$ is to be considered an arithmetic progression or not.

Halbeisen (2005) is an example of a treatment that doesn't match our context, but it is still informative, at least for groups with odd cardinalities which do not have this issue. In contrast, Lev (2004) is an example of the opposite choice. At any rate, there seems to be ongoing research on the topic, as can be seen in section 2.2 of this 2011 review by Croot & Lev. I'll quote an interesting excerpt:

(...) it is easy to construct a progression-free set $A \subseteq \mathbb{F}_q$ such that $\vert A \vert = 2^r$: just fix arbitrarily a basis $\{e_1,\dots,e_r\}$ of $\mathbb{F}_q$ over $\mathbb{F}_3$ and let $A := \{\epsilon_1 e_1+\cdots + \epsilon_r e_r:  \epsilon_1, \dots,\epsilon_r \in \{0,1\}\}$

Which is a rigorous mathematical formulation of the "greedy algorithm" used in this related math-contest question. Given the fact that the question mentions "Bulgarian Team Selection Test", I'm inclined to think that the expected answer is simply an application of the "greedy algorithm" and would be something along the lines of

$$ 3^{m-1} \leqslant n \leqslant 3^m  \implies 2^{m-1} \leqslant \max \vert \Gamma \vert \leqslant 2^m $$

Halbeisen, Lorenz; Halbeisen, Stephanie, Avoiding arithmetic progressions in cyclic groups, Elem. Math. 60, No. 3, 114-123 (2005). ZBL1161.11307.

Lev, Vsevolod F., Progression-free sets in finite abelian groups Journal of Number Theory,
Volume 104, Issue 1, 162-169 (2004).


问:
Is every Riemannian metric "locally" flat?
I have the following cuestion: Is every Riemannian metric "locally" flat?, i.e., for every poit $p\in M$ there is a smooth chart around $p$ such that the Christoffel symbols are all zero.

If this is true, how can i obtain such chart?

Thanks in advance.

答:
This is not true as the curvature is $C^2$-invariant and is determined by the Christoffel symbols.

It is possible to prescribe the metric at a point (this is a matter of linear algebra), but not in an open neighbourhood.


问:
Bijection between $ A ^ {B \cup C} $ and $ A ^ B \times A ^ C $, with $B\cap C=\varnothing$.
Let us consider $ f | _B $ the function restricted to $ B $ and analogously $ f | _C $. The formula for both is $ f | _B (b) = f (b) $ and $ f | _C (c) = f (c) $, for all $ b, c \ in B, C $ respectively.

Let's define $ F: A ^ {B \cup C} \to A ^ B \times A ^ C $, given by $$ F (f) = (f | _B, f | _C) $$.

To demonstrate injectivity, let $ f, f '\in A ^ {B \cup C} $, such that $ F (f) = F (f') $, that is $ (f | _B, f | _C) = (f '| _B, f' | _C) $. Hence, it follows that $ f (b) = f | _B = f '| _B = f' (b) $ and $ f (c) = f | _C = f '| _C = f' (c) $. This is that $ f = f '$. That way $ F $ is injective.

Some help demonstrating surjectivity.

答: In general, given sets $B$ and $C$, we can form the disjoint union $B \sqcup C$ by choosing disjoint copies $B'$ of $B$ and $C'$ of $C$, and then $B \sqcup C = B' \cup C'$; for example $$B \sqcup C := (B \times \{0\}) \cup (C \times \{1\}).$$ Now, we claim that there is a bijection between $A^{B \sqcup C}$ and $A^B \times A^C$ (this is the reason of why the disjoint union $B \sqcup C$ is also denoted by $B + C$, or $B \oplus C$). Indeed, given functions $g: B \to A$ and $h : C \to A$, define $g \sqcup h : B \sqcup C \to A$ by $$(g \sqcup h) \circ j_B = g \textrm{ and } (g \sqcup h) \circ j_C = h$$ where $j_Z : Z \to B \sqcup C$ (for $Z \in \{B,C\}$) sends each element in $Z$ to its corresponding copy in $Z' \subseteq B \sqcup C$. Thus, the function $$A^B \times A^C \to A^{B \sqcup C}; \quad (g,h) \mapsto g \sqcup h$$ is a bijection, with inverse $$A^{B \sqcup C} \to A^B \times A^C; \quad f \mapsto (f \circ j_B,f \circ j_C).$$