问:
How many five digit numbers can be formed that are both odd and less than $40,000$ with the set $\{2,3,4,5,6\}$
I have to find how many possible five digit numbers which are both less than 40,000 and odd can be made with the set $\{2,3,4,5,6\}$ by using each number once only .

So far I have noted that there are only two possible choices for the first digit; $2,3$

And only two possible choices for the last digit; $3,5$

Hence, by using dashes to represent the possible number of choices for each corresponding digit:

$$\underline{2} \space \underline{}\space \underline{}\space \underline{}\space \underline{2}$$

Filling in the remaining three middle digits:

$$\underline{2} \space \underline{3}\space \underline{2}\space \underline{1}\space \underline{2}$$
$$2*3*2*1*2=24$$
However I know this is incorrect, I am a bit confused on how to deal with the fact that both the first digit and the last digit have an overlapping choice of the number $3$, I have tried to split it into two cases with one case where the first digit is $2$ and the other case where the first digit is $3$ but it

How would I go about breaking down this problem logically and solving it?

答:
So you have find how many possible five digit numbers which are both less than 40,000 and odd can be made with the set {2,3,4,5,6}.Also considering that no numbers can be repeated.

So the possibilities for the first blank is 2 and 3, as they must be less than 40,000.

The second blank:4
The third blank:3
The fourth blank:2
The fifth blank has to be odd so it is 2.

So, you will get 243*2 =48 .


问:
Definition of open set/metric space
On Proof Wiki, the definition of an open set is stated as

Let $(M,d)$ be a metric space and let $U\subset M$, then $U$ is open iff for all $y\in U$, there exists $\epsilon \in \mathbb{R}_{>0}$ such that $B_M(y;\epsilon) \subset U$

Then there's a remark that states

It is important to note that, in general, the values of $\epsilon$ depend on $y$. That is, it is not required that $\exists \epsilon \in \mathbb{R}_{>0}$ s.t. $\forall y \in U$ s.t.  $B_M(y;\epsilon) \subset U$

I don't understand quite understand the remark, could someone please clearly explain this and perhaps provide an example? I believe I understand the definition, but the remark is throwing me off.

答: The statement is like uniform continuity (in terms of structure). All its saying is if you get really close to the "border" of an open set, you don't have mush wiggle room to stay in it ($\epsilon$ has to be small to stay in the set), versus if you were in the "middle" of it.</p>

ContentLicense: CC BY-SA 3.0

Title: Definition of open set/metric space


问:
Trivial Group Representation is Projective implies Semisimple
Let $G$ be a group and $\mathbb k$ a field. Then, let $\mathfrak 1$ be the trivial $\mathbb k G$-module.

According to Lorenz's A Tour of Representation Theory, he states "it turns out $\mathfrak 1$ has a great significance, if it is a projective as a $\mathbb kG$-module, then all $\mathbb kG$-module is projective" and goes on to reference Maschke's Theorem.

I can understand the following claim, that a ring $\mathbb kG$ is semisimple if and only if all its modules are projective, so Maschke's Theorem does indeed show that $\mathfrak 1$ is projective for a semisimple $\mathbb k G$. I can also see that if $\mathfrak 1$ is projective, then it is direct summand of some free module.

However, I don't understand how the converse implication (i.e. $\mathfrak 1$ projective implies $\mathbb kG$ semisimple) suggested by Lorenz's remark is shown. It seems to suggest that we want $\mathfrak 1$ to be the direct summand of $\mathbb kG$ as a $\mathbb kG$-module, but I don't quite see how this is done and even if it is, how the remark is proven.

答:
I'll write $k$ for the trivial module. There's a natural map $\varepsilon : k[G] \to k$ of $k[G]$-modules given by sending every $g \in G$ to $1$, and $k$ is projective iff this map splits as a map of $k[G]$-modules. A splitting of this map must send $1 \in k$ to a $G$-invariant element of $k[G]$, which must have the form $c \sum_{g \in G} g$ for some constant $c$; moreover, we must have

$$\varepsilon \left( c \sum_{g \in G} \right) = c |G| = 1$$

so a splitting exists iff $|G|$ is invertible in $k$, and by Maschke's theorem this condition is necessary and sufficient for $k[G]$ to be semisimple. (Maschke's theorem isn't always stated as an if-and-only-if but it is.)

ContentLicense: CC BY-SA 4.0


答:
One can see in the following way that every $G$-representation (over $\mathbb{k}$) is projective.

Step 1.
For every two representations $V$ and $W$ of $G$, the vector space $\operatorname{Hom}_{\mathbb{k}}(V,W)$ is again a representation of $G$ via
$$
  (g \cdot f)(v) = g \cdot f( g^{-1} \cdot v )
$$
for all $g \in G$, $f \in \operatorname{Hom}_{\mathbb{k}}(V, W)$ and $v \in V$.
A linear map $f$ from $V$ to $W$ is $G$-invariant if and only if it is a homomorphism of representations, i.e.
$$
  \operatorname{Hom}_{\mathbb{k}}(V,W)^G
  =
  \operatorname{Hom}_G(V,W) \,.
$$

Step 2.
We have for every $\mathbb{k}$-vector space $V$ the natural isomorphism of vector spaces
$$
  \operatorname{Hom}_{\mathbb{k}}(\mathbb{k}, V) \longrightarrow V \,,
  \quad
  f \longmapsto f(1) \,.
$$
Suppose now that $V$ is a representation of $G$.
The above isomorphism of vector spaces is then an isomorphism of $G$-representations.
It therefore restricts to a natural isomorphism of vector spaces
$$
  \operatorname{Hom}_G(\mathbb{k}, V)
  \longrightarrow
  V^G \,.
$$

Step 3.
Suppose now that the trivial representations $\mathbb{k}$ is projective.
This means that the functor
$$
  \operatorname{Hom}_G(\mathbb{k}, -)
  \colon
  \mathbf{Rep}(G)
  \longrightarrow
  \mathbf{Vect}(\mathbb{k})
$$
is exact.
It follows from the (natural) isomorphism $\operatorname{Hom}_G(\mathbb{k}, -) \cong (-)^G$ that the functor
$$
  (-)^G
  \colon
  \mathbf{Rep}(G)
  \longrightarrow
  \mathbf{Vect}(\mathbb{k})
$$is exact.

Step 4.
Let now $V$ be an arbitrary representations of $G$.
The functor $\operatorname{Hom}_G(V, -)$ equals the composition
$$
  \mathbf{Rep}(G)
  \xrightarrow{ \enspace \operatorname{Hom}_{\mathbb{k}}(V, -) \enspace }
  \mathbf{Rep}(G)
  \xrightarrow{ \enspace (-)^G \enspace }
  \mathbf{Vect}(\mathbb{k}) \,.
$$
Both of these functors are exact, whence $\operatorname{Hom}_G(V, -)$ is exact.
This shows that $V$ is projective.

There is also another way to look at this problem:
The group structure of $G$ gives its group algebra $\mathbb{k}G$ the structure of a Hopf algebra.

Partial Definition.
A Hopf algebra over $\mathbb{k}$ is a $\mathbb{k}$-algebra $H$ together with homomorphisms of $\mathbb{k}$-algebras
$$
    \Delta
    \colon
    H
    \to
    H \otimes H
    \,,
    \quad
    \varepsilon
    \colon
    H
    \to
    \mathbb{k}
  $$
such that certain conditions hold (which I won’t mention here).
The map $\varepsilon$ is the counit of $H$, and the map $\Delta$ (which we won’t care about in the following) is the comultiplication of $H$.

Given a Hopf algebra $H$, the ground field $\mathbb{k}$ becomes an $H$-module via
$$
  x \cdot y
  =
  \varepsilon(x) y
$$
for all $x \in H$ and $y \in \mathbb{k}$.
The counit map $\varepsilon$ becomes in this way a homomorphism of $H$-modules from $H$ to $\mathbb{k}$.

Example.
In the case of $H = \mathbb{k}G$ the maps $\Delta$ and $\varepsilon$ are given on the basis $G$ of $\mathbb{k}G$ by
$$
    \Delta(g)
    =
    g \otimes g \,,
    \quad
    \varepsilon(g)
    =
    1
  $$
for every $g \in G$.
The resulting $\mathbb{k}G$-module structure on $\mathbb{k}$ is given by
$$
    g \cdot y
    =
    \varepsilon(g) y
    =
    1 \cdot y
    =
    y
  $$
for all $g \in G$ and $y \in \mathbb{k}$.
This is precisely the $\mathbb{k}G$-module structure on $\mathbb{k}$ that corresponds to the trivial action of $G$ on $\mathbb{k}$.

There is now a generalization of Maschke’s theorem to Hopf algebras.

Theorem (Maschke’s theorem for Hopf algebras).
Let $H$ be a Hopf algebra.
The following conditions on $H$ are equivalent.

$H$ is semisimple.

There exists an element $a$ of $H$ such that

$x \cdot a = \varepsilon(x) a$ for all $x \in H$ and $a \in A$ (invariance), and

$\varepsilon(a) = 1$ (normalization).

There exists an $H$-submodule $A$ of $H$ with $H = A \oplus \ker(\varepsilon)$.

The kernel of $\varepsilon$ (which is an $H$-submodule of $H$ because $\varepsilon$ is a homomorphism of $H$-modules) is a direct summand of $H$.

The $H$-module $\mathbb{k}$ is projective.

One shows this by proving the implications
$$
  1 \implies 5 \implies 4 \implies 3 \implies 2 \implies 1 \,.
$$

Example.
Consider again $H = \mathbb{k}G$.
Let $a$ be an element of $H$ as in characterization 2 of the above theorem.
This element is given by a linear combination $a = \sum_{g \in G} a_g g$.
The condition $x \cdot a = \varepsilon(x) a$ holds for every $x \in \mathbb{k}G$ if and only if $g \cdot a = \varepsilon(g) a$ for every group element $g$ of $G$.
We have $\varepsilon(g) = 1$, so this condition is furthermore equivalent to $g \cdot a = a$ for every $g \in G$.
This means that all coefficients $a_g$ of $a$ have to be equal.

If $G$ is infinite, then this means that $a = 0$ (because $a$ can’t have infinitely many nonvanishing coefficients).
But then $\varepsilon(a) = 0$, contradicting $\varepsilon(a) = 1$.
So if $G$ is infinite, then no such element $a$ exist.

If $G$ is finite, then $a$ can be any element of the form $a = C \sum_{g \in G} g$ for some scalar $C$ in $\mathbb{k}$.
But we have
$$
    1
    =
    \varepsilon(a)
    =
    C \sum_{g \in G} \varepsilon(g)
    =
    C \cdot \lvert G \rvert.
  $$
We hence find that no such element $a$ exists if the characteristic of $\mathbb{k}$ divides $\lvert G \rvert$, and otherwise $C = 1 / \lvert G \rvert$ and thus
$$
    a
    =
    \frac{1}{\lvert G \rvert} \sum_{g \in G} g \,.
  $$
This is the “averaging element” (hence the letter $a$) used in the usual proof of Maschke’s theorem for groups.

By actually going through the full proof of the above theorem, one sees that when $\mathbb{k}$ is projective as an $H$-module, then the surjective homomorphism of $H$-modules $\varepsilon$ from $H$ to $\mathbb{k}$ splits.
Such a split gives us a homomorphism of $H$-modules from $\mathbb{k}$ to $H$, which in turn corresponds to an element of $H$.
This element is then the “averaging element” $a$.

Title: Trivial Group Representation is Projective implies Semisimple

Body:
Let $G$ be a group and $\mathbb k$ a field. Then, let $\mathfrak 1$ be the trivial $\mathbb k G$-module.

According to Lorenz's A Tour of Representation Theory, he states "it turns out $\mathfrak 1$ has a great significance, if it is a projective as a $\mathbb kG$-module, then all $\mathbb kG$-module is projective" and goes on to reference Maschke's Theorem.

I can understand the following claim, that a ring $\mathbb kG$ is semisimple if and only if all its modules are projective, so Maschke's Theorem does indeed show that $\mathfrak 1$ is projective for a semisimple $\mathbb k G$. I can also see that if $\mathfrak 1$ is projective, then it is direct summand of some free module.

However, I don't understand how the converse implication (i.e. $\mathfrak 1$ projective implies $\mathbb kG$ semisimple) suggested by Lorenz's remark is shown. It seems to suggest that we want $\mathfrak 1$ to be the direct summand of $\mathbb kG$ as a $\mathbb kG$-module, but I don't quite see how this is done and even if it is, how the remark is proven.


问:
Second order homogeneous differential equations: why do repeated roots modify the solution set?
Consider the following differential equation with constant coefficients:
$$
y'' + 2ky' + k^2y = 0 \, .
$$
The auxiliary equation is
\begin{align}
m^2 + 2km + k^2 &= 0 \\[4pt]
(m+k)^2 &= 0 \\[4pt]
m &= -k \, .
\end{align}
This means that the solutions of the differential equation are $Ae^{-kx}+B\color{red}{x}e^{-kx}$, where $A$ and $B$ are arbitrary constants. The $x$-term highlighted in red is what confuses me. Generally speaking, if the roots of the auxiliary equation are $\alpha$ and $\beta$, then the solutions to the differential equation are $Ae^{\alpha x} + Be^{\beta x}$. However, if $\alpha=\beta$, then the solutions are not $Ae^{\alpha x} + Be^{\alpha x}$—or at least, this solution is incomplete. I can verify that $y=xe^{-kx}$ is a solution of the differential equation by plugging it back in to the equation, but I'm looking for a deeper reason for why this occurs.

答:
Taking inspiration from Aryadeva's answer, I have found a direct method of solving the differential equation
$$
y'' + by + cy = 0
$$
which takes into account what happens when you have a repeated root. This equation has the auxiliary equation
$$
m^2+bm+c=0 \, .
$$
Say $\alpha$ and $\beta$ are the roots of the auxiliary equation. Then, the equation can be rewritten as
$$
(m-\alpha)(m-\beta)=0
$$
which implies that $b=-(\alpha + \beta)$ and $c=\alpha\beta$. Then, the original equation can be rewritten as
$$
y'' -(\alpha+\beta)y' + \alpha\beta y =0 \, .
$$
It can then be directly shown that
\begin{align}
y&=Ae^{\alpha x}+Be^{\beta x} \quad\text{if $\alpha\neq\beta$} \\[4pt]
y&=(A+Bx)e^{\alpha x} \quad\text{if $\alpha = \beta$} \, .
\end{align}
Here is how:
\begin{align}
y'' -(\alpha+\beta)y' + \alpha\beta y &= 0 \\[6pt]
y'' - \alpha y' - \beta y' + \alpha\beta y &= 0 \\[6pt]
y'' - \alpha y' - \beta (y' - \alpha y) &= 0 \\[6pt]
(y'' - \alpha y')e^{-\beta x} - \beta (y' - \alpha y)e^{-\beta x} &= 0 \\[6pt]
\left[(y' - \alpha y)e^{-\beta x}\right]' &= 0 \\[6pt]
(y' - \alpha y)e^{-\beta x} &= B \\[6pt]
y' - \alpha y &= Be^{\beta x} \\[6pt]
e^{-\alpha x}y' - \alpha e^{-\alpha x}y &= Be^{(\beta - \alpha)x} \\[6pt]
\left[e^{-\alpha x} y \right]' &= Be^{(\beta - \alpha)x}
\end{align}
Suppose that $\alpha \neq \beta$. Then, the following manipulations are justified:
\begin{align}
e^{-\alpha x}y &= \int Be^{(\beta - \alpha)x} \, dx = \frac{Be^{(\beta - \alpha)x}}{\beta - \alpha} + A\\[6pt]
y &= \frac{B}{\beta - \alpha}e^{\beta x} + Ae^{\alpha x} \, .
\end{align}
But then, since $\frac{B}{\beta - \alpha}$ is just some arbitrary constant, we may relabel it as $B$. We are left with
$$
\boxed{
\;\\[4pt]
\quad y=Ae^{\alpha x}+Be^{\beta x} \quad
\\
}
$$
as expected. If $\alpha = \beta$, then the above reasoning is invalid since we end up with division by zero. Instead,
$$
e^{-\alpha x}y = \int Be^{(\beta - \alpha)x} \, dx = Bx + A
$$
and therefore
$$
\boxed{
\;\\[4pt]
\quad y = (A+Bx)e^{\alpha x} \, . \quad
\\
}
$$


问:
Bound on determinant sum if some of rows and columns are zeros
Let $A$ be positive definite and $B$ be positive semidefinite. From the Minkowski inequality,

$$ \det [A + B] \geq \det [A] + \det [ B ] $$

Now, consider the case the $i$th row and column of $B$ are all zeros. Then, $\det [B] = 0$. Let $B’$ denote $B$ with the $i$th row and column deleted. Then, I believe we now have the following inequality:

$$ \det [ A + B ] \geq \det [ A ] + \det [ B’ ] $$

I am not sure how to prove this, but this seems true. My logic is that the following.

The determinant of a matrix is the volume spanned by the columns of the matrix. So, if we have a row and column of zeros, then adding the matrices together increases the volume of A by no more than the volume of $\det [B’]$. At least, this is my intuition. I’m struggling to prove this hypothesis.

So, any suggestion on proving this?

答:
As the other answer has pointed out, the inequality is not true, but in general, we do have
$$
\det(A+B)\ge\det(A)+a_{ii}\det(B').
$$
Without loss of generality, assume that $i=1$. Let
$$
A=\pmatrix{a_{11}&v^\ast\\ v&A'},\ B=\pmatrix{0&0\\ 0&B'}.
$$
Since $A$ is positive definite, the Schur complement $A/a_{11}=A'-\frac{1}{a_{11}}vv^\ast$ is positive definite. Therefore
\begin{aligned}
\det(A+B)
&=\det\pmatrix{a_{11}&v^\ast\\ v&A'+B'}\\
&=a_{11}\det\left(A'+B'-\frac{1}{a_{11}}vv^\ast\right)\\
&\ge a_{11}\left[\det\left(A'-\frac{1}{a_{11}}vv^\ast\right)+\det(B')\right]\\
&=\det(A)+a_{11}\det(B').
\end{aligned}


问:
Find the minimum of $q$.

Given $\displaystyle p,q\in\mathbb N, \frac p q=0.123456789...$(i.e. the first 9 digits after decimal point are $123456789$). Find the minimum of $q$.

I guessed the answer to be $111111111$ as $\displaystyle\frac {123456789} {999999999}=0.\dot{1}2345678\dot{9}$, but I'm not sure.

Thank you.

答:
Here is the magic algorithm which Hagen von Eitzen used to compute the answer.
This method is extremely elegant, and is derived from Farey sequences.

To shorten the computation, I will solve the analagous question of finding the smallest $q$ such that $\frac{p}{q}$ has decimal expansion $.789....$.

Let $\alpha = \frac{789}{1000}$; we are looking for rational numbers close to $\alpha$ with small denominators.  (In general, $\alpha$ can be any positive real number.)

Start with the two fractions $\frac01$ and $\frac10$.
Note that $\frac01 < \alpha < \frac10$.
Add the numerators and add the denominators to get the fraction
$\frac{0 + 1}{1 + 0} = \frac11$.  Of the three fractions we have listed so far ($\frac01, \frac11,$ and $\frac10$), take the two that are on either side of $\alpha$, i.e. $\frac01$ and $\frac11$.

We have the two fractions $\frac01$ and $\frac11$, with $\frac01 < \alpha < \frac11$.
Add the numerators and denominators to get $\frac{0 + 1}{1 + 1} = \frac12$.  Of the three fractions $\frac01$, $\frac12$, and $\frac11$, $\alpha$ is between $\frac12$ and $\frac11$, so take these two.

We have the two fractions $\frac12$ and $\frac11$.
Add the numerators and denominators to get $\frac{1 + 1}{2 + 1} = \frac23$.  Of the three fractions $\frac12$, $\frac23$, $\frac11$, $\alpha$ is between
$\frac23$ and $\frac11$, so take these two.

You can see where this is going.  Next we get the pairs of fractions $\frac34$ and $\frac11$, then $\frac34$ and $\frac45$, then $\frac79$ and $\frac45$, then $\frac{11}{14}$ and $\frac45$, then $\frac{15}{19}$ and $\frac{4}{5}$.
We stop because $\frac{15}{19}$ has decimal expansion $.789....$ (it is the first of the fractions we have listed).  This is the answer.  No matter what decimal expansion you want ($.123456789$ in your case), this method will always eventually give you the fration $\frac{p}{q}$ with minimum $q$.

Title: Find the minimum of $q$.


问:
How to calculate the range of $f$

Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be the cosine function, where the argument is given in radians.
Let $W=[-2 \pi,-\pi] .$ What is $f(W)$?

a. (0,1)

b. (-1,0)

c. (-1,1)

d. na

I think it should be $(-1,1)$ as based on $\sin$ graph. But it is given wrong. How is it possible?

答: Note that $f(-2\pi)=f(0)=1$. Then $\cos$ decreases on $\left[-2\pi,-\pi\right]$, until it reaches $-1$ on $-\pi$. So, the answer is $[-1,1]$.

Title: How to calculate the range of $f$


问:
How can I prove the identity $\dfrac{1}{1+\sin(x)}\equiv \dfrac{\sec^2\left(\frac{x}{2}\right)}{\left(\tan\left(\frac{x}{2}\right)+1\right)^2}$
I have to prove by working on the left hand side of the equation that

$$\dfrac{1}{1+\sin(x)}\equiv \dfrac{\sec^2\left(\frac{x}{2}\right)}{\left(\tan\left(\frac{x}{2}\right)+1\right)^2}$$
However I just cant seem to get it, I first started by rewriting the denominator using sin half angle formula

$$\frac{1}{1+2\sin(\frac{x}{2})\cos(\frac{x}{2})}=\frac{1}{\sin^2(\frac{x}{2})+\cos^2(\frac{x}{2})+2\sin(\frac{x}{2})\cos(\frac{x}{2})}$$

Then by simplifying further
$$=\dfrac{1}{(\sin(\frac{x}{2})+\cos(\frac{x}{2}))^2}$$
However I'm stumped here and do not know how to proceed, I even took an alternate path and tried rationalizing the fraction as such

$$\frac{1}{1+\sin(x)}*\frac{1-\sin(x)}{1-\sin(x)}=\frac{1-\sin(x)}{(1-\sin(x))^2}$$
And applying $\sin^2(x)+\cos^2(x)=1$
$$=\frac{1-\sin(x)}{\cos^2(x)}$$
$$=\sec^2(x)*(1-\sin(x))$$

But again I get stuck, is there some identity I am forgetting to implement or some other method to approach the left hand side with?

答:
Use $$\dfrac{\sec\dfrac x2}{\tan\dfrac x2+1}\equiv\dfrac1{\sin\dfrac x2+\cos\dfrac x2}$$

Now you have already identified $$\left(\sin\dfrac x2+\cos\dfrac x2\right)^2=?$$


答: Alternately, use half angle theorems: $$\sin(x) \equiv \frac{2\tan(x/2)}{1+\tan^2(x/2)} $$

Title: How can I prove the identity $\dfrac{1}{1+\sin(x)}\equiv \dfrac{\sec^2\left(\frac{x}{2}\right)}{\left(\tan\left(\frac{x}{2}\right)+1\right)^2}$

Body:
I have to prove by working on the left hand side of the equation that

$$\dfrac{1}{1+\sin(x)}\equiv \dfrac{\sec^2\left(\frac{x}{2}\right)}{\left(\tan\left(\frac{x}{2}\right)+1\right)^2}$$
However I just cant seem to get it, I first started by rewriting the denominator using sin half angle formula

$$\frac{1}{1+2\sin(\frac{x}{2})\cos(\frac{x}{2})}=\frac{1}{\sin^2(\frac{x}{2})+\cos^2(\frac{x}{2})+2\sin(\frac{x}{2})\cos(\frac{x}{2})}$$

Then by simplifying further
$$=\dfrac{1}{(\sin(\frac{x}{2})+\cos(\frac{x}{2}))^2}$$
However I'm stumped here and do not know how to proceed, I even took an alternate path and tried rationalizing the fraction as such

$$\frac{1}{1+\sin(x)}*\frac{1-\sin(x)}{1-\sin(x)}=\frac{1-\sin(x)}{(1-\sin(x))^2}$$
And applying $\sin^2(x)+\cos^2(x)=1$
$$=\frac{1-\sin(x)}{\cos^2(x)}$$
$$=\sec^2(x)*(1-\sin(x))$$

But again I get stuck, is there some identity I am forgetting to implement or some other method to approach the left hand side with?


问:
Question about test function
I'm reading this book and on page 66, the author gives this example about hypothesis testing:

I don't understand why the author has defined the test function in this way. Why did he choose $\phi(8)=67/75$? Why the size is exactly 0.05?

答:
considering the several question you are putting, I suppose that you are a little bit confused with this kind of problem. I try to explain you how to proceed in this case, trying to answer to all your questions.

First of all, considering the very well explanation of @heropup about uselessness of the randomized test, let's solve the following problem in a more commonly non - randomized test. If yuo want you can anyway randomize it...

The problem:
I played a "fair coin game" with a friend. I chose Head he Tail. We tossed "his fair coin" and bet $\$1,000$ ad any toss and these were the results

$$\{H-T-T-T-T-H-T-T-T-T\}$$

Result: in the time of a minute I loss 6 thousand dollars!!

The doubt:

Friend or not, I am wondering to understand if this was really a fair coin or it was biased in favour of Tail (he chose both Tail and the Coin)

The solution:

First of all I get a drawing of the binomial $Bin(10;0.5)$ that is the distribution of Tails under the Null hypothesis that my friend is not a robber...

As you can see, the probability of getting 8 or more tails in 10 tosses, is about $5.5\%$

This is the Type I error, the probability to reject $H_0$ in favour of a One tail hypothesis $H_1$ that the coin is biased in his favour.

You can randomize the test to get exactly 5% and not 5.47% as in the example but you do not get any plus-information. The test is "bordeline significative"  but it is not highly significative...if you state a Type I error of 1%, your pvalue of 5.47% is not enough to claim your money back...

Now suppose we fix an alternative hypotesis of $H_1: \theta=0.8$ and let's see what happens to our drawing

The yellow bars represent the probability to reject $H_0$ with the random sample above stated but under the alternative $H_1$: it's the Test Power:$68\%$.

The conclusion

The test is bordeline and not highly significative so I remain with my doubt because I'm not 99% sure of the robbery. Probably next time I will spend my $\$6,000$ for a Cruise to the Caribbean and not for gambling.

In a way I prefer to approach this problem (Bayesian approach), I assume a Uniform prior and evaluate

$$\frac{\int_0^{1/2}\theta^8(1-\theta)^2d\theta}{\int_{1/2}^1\theta^8(1-\theta)^2d\theta}=\frac{67}{1981}$$

Being the denominator greater than the numerator, my decision is in favour of the denominator, that is $\theta>\frac{1}{2}$

Title: Question about test function


问:
Find missing angle in triangle

In this triangle, we are given that $AB=AD=CE$, $BE=CD$ and angle  $ECD=2*AED=2a$.

We are asked to find $AED=a$.

I have seen several problems similar to this but can't follow any of the solution methods.
The only thing which is obvious to me is that since $AD+CD = BE+EC$, the triangle is isosceles.

Therefore its base angles (the ones I have drawn in green) are equal.
Also triangle ABD is isosceles, therefore its base angles are equal.

I don't see any other relation to use.

(FYI this is not homework or anything - just practice and self-motivation).
Thank you very much.

EDIT: If this is of any help, through Geogebra I get that $a=10^{\circ}$, therefore $ECD=20^{\circ}$ and we have a 80-80-20 triangle.

答:
First, from Euclid: an inscribed angle is half of the central angle that subtends the same arc on the circle.

Second, if you already have the angle 2a finding the angle a is trivial with a compass you can easily find the bisector, if you want to do it "pretty", if you are allowed to replicate C (with AD as a base instead of AB) and call that point F, and draw a circunference of center F and radius AF, the intersection with BC should give you point E.

Title: Find missing angle in triangle


问:
$\int_0^{+\infty}\frac{1}{e^{ax}+e^{bx}}dx$
Could you please give me some hints about evaluating this integration?

$$\int_0^{+\infty}\frac{1}{e^{ax}+e^{bx}}dx$$

where $a>0$, $b>0$ and $a\neq b$.

Thanks in advance.

答: If $a<b$,$$\begin{align}\int_0^\infty\tfrac{e^{-bx}dx}{1+e^{-(b-a)x}}&=\sum_{n\ge0}(-1)^n\int_0^\infty e^{-b-n(b-a)x}dx\\&=\tfrac{1}{b-a}\sum_{n\ge0}\tfrac{(-1)^n}{n+\tfrac{b}{b-a}}\\&=\tfrac{1}{b-a}\Phi(-1,\,1,\,\tfrac{b}{b-a})\end{align}$$in terms of the Lerch transcendent. If $a\ne b$, the result is $\tfrac{1}{|a-b|}\Phi\left(-1,\,1,\,\tfrac{\max\{a,\,b\}}{|a-b|}\right)$.


问:
Differents ways to evaluate the sum $\sqrt{12+\sqrt{12+\sqrt{12+\sqrt{12+\cdots}}}}$

Evaluate $$\sqrt{12+\sqrt{12+\sqrt{12+\sqrt{12+\cdots}}}}$$
My approach:
Let $$x=\sqrt{12+ \sqrt{12+\sqrt{12+\cdots}}}$$
so, we have that $$\sqrt{12+\sqrt{12+\sqrt{12+\sqrt{12+\cdots}}}}\iff \sqrt{12+x}=x \implies 12+x=x^{2} \iff (x+3)(x-4)=0$$
So, the answer is $\boxed{4}$.

Is correct my solution?
Can you show other ways for to solve this problem?
Can you suggest me any  textbooks with similar problems?
Thank you so much!

答:
Method 1 - contraction mapping theorem.

Let $g : [0,\infty) \to [0,\infty)$ be the map $x \mapsto \sqrt{x+12}$. For any $x, y \in [0,\infty)$, we have

$$\begin{align} & g(x) - g(y) = \sqrt{x+12} - \sqrt{y+12} = \frac{x - y}{\sqrt{x+12} + \sqrt{y+12}}\\
\implies & |g(x)-g(y)| \le \frac{|x-y|}{2\sqrt{12}}
\end{align}
$$
This means $g$ is a contraction mapping over $[0,\infty)$.

By Contraction mapping theorem, $g(x)$ have a unique fixed point over $[0,\infty)$. Furthermore, if one pick any $z \in [0,\infty)$ and construct a sequence $z_n$ by $$z_n = \begin{cases}z, &n = 0\\g(z) = \sqrt{z+12}, & n > 0\end{cases}$$
$z_n$ will converges to that fixed point.

Since $g(4) = 4$, that unique fixed point is $4$. By setting $z$ to $0$, we find

$$\begin{array}{rcl}
z_1 &=& \sqrt{12},\\
z_2 &=& \sqrt{12+\sqrt{12}},\\
z_3 &=& \sqrt{12+\sqrt{12+\sqrt{12}}}\\
&\vdots&
\end{array}$$
converges to $4$.

Method 2 - explicit bound.

Define $z_n$ as in method $1$ and let $y_n = 4 - z_n$ for $n \ge 1$.
Notice
$$y_{n+1} = 4 - z_{n+1} = 4 - \sqrt{12 + z_n}
= 4 - \sqrt{16-y_n}
= \frac{y_n}{4 + \sqrt{16-y_n}}\tag{*1}
$$
$y_{n+1}$ has same sign as $y_n$. Since $y_1 = 4 - \sqrt{12}> 0$, all $y_n$ are positive.

Notice $\sqrt{16-y_n} = \sqrt{12+z_n} \ge 12$, $(*1)$ implies

$$0 < y_{n+1} < \frac{y_n}{4+\sqrt{12}}$$

Replace $n$ by $1, 2, \ldots, m-1$ and combine the inequalities, we find for $m \ge 1$,

$$0 < y_m \le \frac{y_1}{(4+\sqrt{12})^{m-1}}\quad
\implies\quad 4 - \frac{4}{(4+\sqrt{12})^m}
\le z_m < 4$$
By squeezing,

$$\lim_{m\to\infty} z_m = \lim_{m\to\infty}
\underbrace{\sqrt{12 + \sqrt{12 + \sqrt{12 + \cdots}}}}_{m\text{ times}} = 4$$


问:
Are squares of uncorrelated Normal distributed random variables still uncorrelated?
If $X$ and $Y$ are two uncorrelated identically distributed random variables having the Normal distribution with $0$ mean, are $X^{2}$ and $Y^{2}$ also uncorrelated?

I know this holds for independent random variables, but I could not prove it for uncorrelated ones. I also know that this holds if $(X, Y)$ jointly follow the bivariate Normal distribution, but that is not known here.

答:
Hint:

Consider this frequently used example:

$X$ has a standard normal distribution with mean $0$ and variance $1$

$Z =+1$ or $-1$ each with probability $\frac12$, independent of $X$

$Y=XZ$

What is the distribution of $Y$?

What is the correlation between $X$ and $Y$?

What are the distributions of $X^2$ and $Y^2$?

What is the correlation between $X^2$ and $Y^2$?


问:
One way functions and P = NP?
How can I show that no one way functions exist under assumption of P = NP?</p>

答:
While we don't know whether $P\neq NP$ implies the existence of (cryptographic) one-way functions (OWFs), we do know that $P=NP$ implies their absence.

The standard argument goes like this: Assume there is a candidate OWF $f$, define the language $L_f:=\{(y,z)|\exists x:f(x)=y\land z\text{ is a prefix of }x\}$. Clearly $L_f$ is in NP as you can easily verify the condition on $(x,y,z)$ (as well as their syntactic correctness) in polynomial time, especially since we demand OWFs to be computable in polynomial time.

But given that $P=NP$ there is a polynomial-time algorithm to decide $L_f$. This then allows us to start with $z=\varepsilon$ - the empty string - and then query our polynomial-time algorithm with $z\|1$ and $z\|0$ and choosing the extension of $z$ as our new $z$ which yielded a positive result. If both give negative results we should be done and return $z$.

Title: One way functions and P = NP?


问:
Number of leaves in a tree with maximum degree $k$
Let $G$ be a tree with a maximum degree of the vertices equal to $k$.
At least how many vertices with a degree of $1$ can be in $G$ and why?

I think the answer must be $k$ but I don't know how to prove it.

答: HINT. Let $G$ be a smallest tree that meets the requirements. Could adding nodes to $G$ decrease the answer?


问:
Convergence of $\sum_{n=1}^{\infty}\frac{\frac{2}{3}(1+\frac{2}{3})\ldots(\frac{2}{3}+n)}{(1+\frac{3}{2})\ldots(\frac{3}{2}+n)}.$

Examine the convergence of the series:
$$\sum_{n=1}^{\infty}\frac{\frac{2}{3}(1+\frac{2}{3})\ldots(n+\frac{2}{3})}{(1+\frac{3}{2})\ldots(n+\frac{3}{2})}.$$

Attempt. Using ratio test for $a_n:=\frac{\frac{2}{3}(1+\frac{2}{3})\ldots(n+\frac{2}{3})}{(1+\frac{3}{2})\ldots(n+\frac{3}{2})}>0$, we have:

$$\frac{a_{n+1}}{a_n}=\frac{\frac{2}{3}+n+1}{\frac{3}{2}+n+1}\to 1,$$
so we do not conclude. Also $\sqrt[n]{(1+a)\ldots(n+a)}\to +\infty$ for $a>0$, so the root test also is inconclusive.

Thanks for the help.

答:
Let us use Gauss's test.
$$\frac{a_n}{a_{n+1}} = \frac{n+\frac52}{n+\frac53} = 1 + \frac{\frac56}{n(1+\frac{5}{3n})} = 1 + \frac{\alpha}{n} + \underline{O}(n^{-2})$$
with $\alpha = \frac{5}{6} < 1$. Hence $\sum_n a_n$ diverges.


问:
Name of the generalization of quadtree and octree?
What is the name of the equivalent of quadtrees and octrees in n-dimension ?</p>

答:
I have seen the generalization of octrees being called KD-tree-pyramids. A K-dimensional tree-pyramid (KD-TP) is a full tree where each node represents a K-dimensional subspace.

The most common instances of KD-TP are Full Binary Trees($K=1$), Quadtrees($K=2$), Octrees ($K=3$) and Hexadecatrees ($K=4$).

As Steven mentioned beyond 4D they don´t scale well and kd-trees are used the most.


问:
Closed form for n-th anti-derivative of $\log x$
Is it possible to write a closed-form expression with free variables $x, n$ representing the n-th anti-derivative of $\log x$?</p>

答:
$$\log^{(-n)}x=\frac{x^n}{n!}(\log x-H_n),$$
where $H_n$ is the harmonic number: $H_n=\sum_{k=1}^n k^{-1}=\,$$\gamma$$\,+\,$$\psi$$(n+1)=\gamma+\frac{\Gamma'(n+1)}{n!}$

Proof: By induction.


问:
Find all eigenvalue for $S$
Let $V$ be the vector space of all converging real sequences, that is

$$V = \lbrace (a_1,a_2,a_3,\ldots) |\forall i \in \mathbb {N}: a_i \in\ \mathbb{R}\;\mathrm{and}\; \exists\! \lim_{n\to\infty}a_n \rbrace$$

We are given this linear transformation:
$$ S:V \rightarrow  V,\; S(a_1,a_2,a_3,\ldots) = (a_2,a_3,a_4,\ldots)$$

Find all of the eigenvalues of $S$.

MY METHOD:
 In fact, $S$ "moves left". The linear transformation
deletes the first member in the sequence.
 I tried to find any
$\lambda$ $\in\ \mathbb{R}$, such that there is a sequence  with at
least one $a_i \neq 0$ that will satisfy:

$S(a_1,a_2,a_3,a_4,\ldots) = (\lambda a_1, \lambda a_2,
 \lambda a_3, \lambda a_4)$,

but without success to this moment

答:
You need to solve $$S(a_1,a_2,a_3,\ldots) = \lambda(a_1,a_2,a_3,\ldots)$$ i.e. $$(a_2,a_3,a_4,\ldots) = (\lambda a_1,\lambda a_2,\lambda a_3,\ldots).$$

Now equate each element of those two sequences. We need $a_2 = \lambda a_1$, $a_3 = \lambda a_2$, $a_4 = \lambda a_3$, ....

Substitute the first equation into the second to get $a_3 = \lambda a_2 = \lambda(\lambda a_1) = \lambda^2 a_1$.

Substitute the second equation into the third to get $a_4 = \lambda a_3 = \lambda(\lambda^2 a_1) = \lambda^3 a_1$.

So the first four terms of the eigensequence are $a_1$, $a_2 = \lambda a_1$, $a_3 = \lambda^2 a_1$, and $a_4 = \lambda^3 a_1$.

Can you figure out the rest of the eigensequence? Don't forget that $\lambda$ needs to be such that the eigensequence converges.

Title: Find all eigenvalue for $S$


答:
The equation $Sa = \lambda a$, for $a = (a_i)_{i \in \mathbb N} \in V$, implies that
$$\lambda a_i = a_{i+1}, \quad \forall i \in \mathbb N.$$
Therefore, $(1, \lambda, \lambda^2, \ldots)$ is an eigenvector if and only if $-1 < \lambda \le 1$.

Title: Find all eigenvalue for $S$

Body:
Let $V$ be the vector space of all converging real sequences, that is

$$V = \lbrace (a_1,a_2,a_3,\ldots) |\forall i \in \mathbb {N}: a_i \in\ \mathbb{R}\;\mathrm{and}\; \exists\! \lim_{n\to\infty}a_n \rbrace$$

We are given this linear transformation:
$$ S:V \rightarrow  V,\; S(a_1,a_2,a_3,\ldots) = (a_2,a_3,a_4,\ldots)$$

Find all of the eigenvalues of $S$.

MY METHOD:
 In fact, $S$ "moves left". The linear transformation
deletes the first member in the sequence.
 I tried to find any
$\lambda$ $\in\ \mathbb{R}$, such that there is a sequence  with at
least one $a_i \neq 0$ that will satisfy:

$S(a_1,a_2,a_3,a_4,\ldots) = (\lambda a_1, \lambda a_2,
 \lambda a_3, \lambda a_4)$,

but without success to this moment


问:
How to evaluate compounded annualised growth rate ("profit" growth rate) in case of losses?
A company makes a profit of \$50 in the first year, then another profit of \$50 in the next year and \$200 loss in the third year which translates to +50, +75 and -200.

What would be the Compounded Annualized "Growth" Rate of the profit if I compare 1st year profit vs 3rd year profit (+50 to -200)

I don't understand how do I calculate when present value is in negative.

If it were +50, +75,+25, and I compare  first year vs last, +50 to +25 (= a net of -25), I could've had calculated as 1-(present/past)^(1/years) = ~30%

If the (present/past) is -ve, wouldn't this give an imaginary number? Isn't this undefined if profit growth is 0? How to make sense of this and how to evaluate the Profit Growth metric in financial terms?

答:
Search for the Internal Rate of Return (IRR) and Net Present Value (NPV) formulas.
Still the stock market has more to do with art than with science.
Accounting books could be outdated with historical cost, cash flow can differ from earnings, there can be hidden liabilities, bad management... DYOR</p>


问:
Can an assumption be discharged without it being part of the tree?
Given the following formula, use natural deduction to prove that it holds.

The answer given by the professor was the following below:

I would like to understand how we can discharge the assumption of phi at the bracket 3 without actually deriving it from the tree itself like psi at the bracket 2?

答:
Can an assumption be discharged without it being part of the tree?

Yes.

See e.g. Dirk van Dalen (1997) "Logic and Structure", p. 34:

With respect to the cancellation of hypotheses, we note that one does not necessarily cancel all occurrences of a proposition $\psi$. This clearly is justified, as one feels that adding hypotheses does not make a proposition underivable (irrelevant information may always be added). It is a matter of prudence, however, to cancel as much as possible. Why carry more hypotheses than necessary?

Furthermore one may apply $(\to I)$ if there is no hypotheses available for cancellation e.g. $\dfrac{\phi}{\psi \to \phi}(\to I)$ is a correct derivation, using just $(\to I)$.

To sum it up: [...] striking out some (or all) occurrences, if any [...] .

The semantic justification for this is monotonicity (also known as weakening): We have that

If $\Gamma \vDash \phi$, then $\Gamma, \psi \vDash \phi$.

By the deduction theorem, it also follows that

If $\Gamma \vDash \phi$, then $\Gamma \vDash \psi \to \phi$.

If a conclusion can be established from a given set of premises, then it doesn't "get lost" by adding additional knowledge, so we can always add more premises or antecedents that are not actually being needed. This semantic idea transfers to derivations.

The same applies to all other rules that allow to discharge assumptions, i.e. $(\lor E)$, $(\neg I)$ and $(RAA)$.


问:
Definition of open set/metric space
On Proof Wiki, the definition of an open set is stated as

Let $(M,d)$ be a metric space and let $U\subset M$, then $U$ is open iff for all $y\in U$, there exists $\epsilon \in \mathbb{R}_{>0}$ such that $B_M(y;\epsilon) \subset U$

Then there's a remark that states

It is important to note that, in general, the values of $\epsilon$ depend on $y$. That is, it is not required that $\exists \epsilon \in \mathbb{R}_{>0}$ s.t. $\forall y \in U$ s.t.  $B_M(y;\epsilon) \subset U$

I don't understand quite understand the remark, could someone please clearly explain this and perhaps provide an example? I believe I understand the definition, but the remark is throwing me off.

答:
The subtlety is in $\exists \epsilon, \forall y$ and $\forall y,\exists \epsilon $. Do you know uniform continuity and continuity? You must be able to note the difference.

It should be clear why $\forall y,\exists \epsilon $ works. The point is same epsilon may not work for every y. Consider the interval $(1,2)$ for a point $1.2$, $\epsilon=0.1$ works, for $1.5$, $\epsilon=0.4$ works, The point is there need not be same epsilon which works for all.

Title: Definition of open set/metric space


问:
If $f(t)=\sum_{n=1}^{\infty} e^{-n^2t}t^a$ , then $ \lim_{t \to 0}f(t) = 0$
Let $$ f(t)=\sum_{n=1}^{\infty} e^{-n^2t}t^a , \hspace{5mm} t >0$$ for some $a>0$

Then show that $$ \lim_{t \to 0}f(t) = 0$$

[My attempt]

I think  $$ \lim_{t \to 0}f(t)=\lim_{t \to 0}\sum_{n=1}^{\infty} e^{-n^2t}t^a=\sum_{n=1}^{\infty}\lim_{t \to 0} e^{-n^2t}t^a=\sum_{n=1}^{\infty}0=0$$

However, for changing limit and series , we need to uniform convergence of $f$

I tried to use weierstrass M - test but it didn't work.

I'm stuck here, how that solve this?

答:
The limit is not always zero. It depends on $a$.

Let $\theta(t) := \sum_{n \in \mathbb{Z}}e^{-\pi  n^2 t}$ for $t >0$. It is well-known (and not hard to prove via Poisson summation) that
$$
\theta(t) = t^{-1/2}\theta(1/t)
$$
for all $t >0$. We can write
$$
f(t) = t^{a}\frac{\theta(t/\pi)-1}{2}  = t^{a}\frac{\pi^{1/2}t^{-1/2}\theta(\pi/t)-1}{2}.
$$
We have $\lim_{t \rightarrow 0}{\theta(\pi/t)} = 1$ and so
$$
f(t) \sim t^{a}\frac{\pi^{1/2}t^{-1/2}-1}{2} \sim  \frac{\pi^{1/2}}{2 }t^{a-1/2},
$$
where $g(t)\sim h(t)$ means that $g(t)/h(t) \rightarrow 1$ as $t \rightarrow 0$. In summary, for $a \in \mathbb{R}$,
$$
\lim_{t \rightarrow 0}{f(t)} = \begin{cases}
\infty & a < 1/2\\
\frac{\pi^{1/2}}{2 } &a =1/2\\
0 & a>1/2
\end{cases}
$$
The last case can also easily be shown via a geometric series estimate.


问:
How can I prove that this graph and $K_{3,3}$ are isomorphic?
If we have a simple graph with $6$ vertices, that does not have any three-vertex cycles and has $3$ edges going out of each vertex. How can I prove that this graph and $K_{3,3}$ are isomorphic?

答: Pick some vertex $v$ and consider its open neighborhood $N(v)=\{a,b,c\}$. Note that none of $a,b,c$ can be adjacent to another, since otherwise we would have a triangle. So each of them have the open neighborhood $\{u,v,w\}$, where $u,w$ are the remaining two vertices. This shows that the graph is 3-regular bipartite on 6 vertices with partitions of size 3, hence the $K_{3,3}$.


问:
ASK linear algebra

Concept of linear independent

 A. The linear system Ax=b has unique solution for b, then the columns of A are linearly independent

B. The column of the change-of-coordinate matrix P are linearly independent

 C. The subset of a linearly-dependent vector set is linearly-dependent

D. If A is diagonalization, then A has linearly-independent columns

 E. If AT*A(AT means transpose of A) is invertible, then A has linearly-independent columns

My answer is (A)(B)(E)

(A) A is invertible, so it is true

(B) true

(C) S={(1,0),(2,0)} and subset of S={(1,0)},so it is wrong

(D) A is diagonalization, does not mean A is invertible, so it is wrong

(E) determine of AT*A is not zero, so determine of A is not zero too

Is my answer correct?

答:
For (A), it depends on what the author means by "$Ax=b$ has unique solution (for $b$?)", but the point is that when $A$ is $m\times n$ with $m>n$, we may select $b\notin \operatorname{col}A$, in which case $[\forall x,\forall y,((Ax=b\land Ay=b)\rightarrow x=y)]$ is true and $[\exists z, Az=b]$ is false.

The rest seems fine.

Title: ASK linear algebra


答:
We claim that this graph $G$ has no $5$-cycles. To prove, suppose for a contradiction that such graph has a $5$-cycle.

Then, in the above graph, in order for vertex $1$ to have degree $3$, it must be adjacent to vertex $6$ since if it is adjacent to any other vertex, we would have a triangle. Similarly, vertex $5$ must be adjacent to vertex $6$ too. But then we have a $3$-cycle $(156)$, a contradiction. Thus, $G$ has no $3$-cycles and no $5$-cycles. But this mean $G$ cannot have an odd cycle, therefore is bipartite. Can you complete the argument now?

Body: If we have a simple graph with $6$ vertices, that does not have any three-vertex cycles and has $3$ edges going out of each vertex. How can I prove that this graph and $K_{3,3}$ are isomorphic?


问:
Derivation of Black-Scholes Equation: Why $d(\Delta S) = \Delta dS$
When deriving the Black-Scholes equation, a key step is to construct a portfolio
\begin{equation}
\Pi = V - \Delta S
\end{equation}
that contains a long position of the option and a short position of the stock. We then take derivative on both sides and get
\begin{equation}
d\Pi = dV - \Delta dS.
\end{equation}
By doing so we implicitly assume that $\Delta$ is a constant so that $d(\Delta S) = \Delta dS$.

However later on we choose $\Delta = \frac{\partial V}{\partial S}$ to eliminate the uncertainty. Since $V$ is a function of $S$ and $t$, so is $\Delta$. In that case, shouldn't we use
\begin{equation}
d(\Delta S) = \Delta dS + Sd\Delta
\end{equation}
and then apply Ito's lemma to $d\Delta$?

答:
In the standard derivation of the Black-Scholes model this portfolio is assumed to be self-financing that is there are no inflows or outflows of money. The formal definition of a self-financing portfolio in your case is that

$$d\Pi=dV-\Delta dS.$$

That is the self-financing assumption explains why you do not differentiate w.r.t. $\Delta$. For an intuitive explanation of the self-financing condition, see e.g. here Definition of self-financing strategy.


问:
Prove that $\Delta(x) = x \otimes1 + 1 \otimes x$ mod $I \otimes I $ for all $x \in I$.
I got stuck on this problem, so if anyone can give me a hint on this, I really appreciate.

Let $I$ be the augmentation ideal in Hopf algebra $A$. Prove that $\Delta(x) = x \otimes1 + 1 \otimes x$
mod $I \otimes I $ for all $x \in I$.

Because $I$ is a Hopf ideal, so $\Delta(I) \subset I \otimes A + A \otimes I$. So I can write $\Delta(x) = \sum a_i \otimes b_i + \sum c_i \otimes d_i$, here $a_i, d_i \in I$ and $b_i, c_i \in A$, and the sums are finite. I applied $\epsilon \otimes id$ to have a presentation $x = \sum \epsilon(c_i) \Delta(d_i)$. And I got stuck here. Thanks in advance for your help.

答:
Since the discussion in the chat is getting too long, let's write down our final conclusions in an answer. We first observe the following:

Since $(A,\Delta,\epsilon)$ is a coalgebra, the composites
\begin{aligned}
A &\xrightarrow{\Delta} A \otimes A \xrightarrow{\epsilon \otimes \operatorname{id}} k \otimes A \cong A \\
A &\xrightarrow{\Delta} A \otimes A \xrightarrow{\operatorname{id}\otimes \epsilon} A \otimes k \cong A
\end{aligned}
are the identity of $A$, i.e. we have
\begin{aligned}
(\epsilon \otimes \operatorname{id})(\Delta(x)) = 1 \otimes x \hspace{20pt} \text{and} \hspace{20pt} (\operatorname{id} \otimes \epsilon)(\Delta(x)) = x \otimes 1.
\end{aligned}

Since $\epsilon: A \to k$ is an algebra homomorphism, we have $\epsilon(1) = 1$.

It thus follows that
\begin{aligned}
(\epsilon \otimes \operatorname{id})(\Delta(x) - x \otimes 1 - 1 \otimes x) &=
1 \otimes x - 0 \otimes 1 - 1 \otimes x = 0 \\
(\operatorname{id} \otimes \epsilon)(\Delta(x) - x \otimes 1 - 1 \otimes x) &=
x \otimes 1 - 1 \otimes 0 - x \otimes 1 = 0
\end{aligned}
and thus $\Delta(x) - x \otimes 1 - 1 \otimes x \in \ker(\epsilon \otimes \operatorname{id}) \cap \ker(\operatorname{id}\otimes \epsilon)$. It thus remains to prove that
\begin{aligned}
\ker(\epsilon \otimes \operatorname{id}) \cap \ker(\operatorname{id}\otimes \epsilon) = I \otimes I.
\end{aligned}
Edit (updated after it turned out $k$ is not assumed to be a field.)
We claim that the short exact sequence
\begin{aligned}
0 \rightarrow I \rightarrow A \xrightarrow{\epsilon} k \rightarrow 0
\end{aligned}
splits. Indeed, the condition $\epsilon(1) = 1$ precisely tells us that the composition
\begin{aligned}
k \to A \xrightarrow{\epsilon} k
\end{aligned}
is the identity. Since split short exact sequences are closed under tensoring, we can tensor the sequence with $A$ to get a short exact sequence
\begin{aligned}
0 \rightarrow I \otimes A \rightarrow A \otimes A \xrightarrow{\epsilon \otimes \operatorname{id}} k \otimes A \rightarrow 0,
\end{aligned}
proving that $\ker(\epsilon \otimes \operatorname{id}) = I \otimes A$. Similarly we can tensor the above sequence on the left with $I$, giving a short exact sequence
\begin{aligned}
0 \rightarrow I \otimes I \rightarrow I \otimes A \xrightarrow{\operatorname{id} \otimes \epsilon} I \otimes k \rightarrow 0,
\end{aligned}
which shows that
\begin{aligned}
\ker(\epsilon \otimes \operatorname{id}) \cap \ker(\operatorname{id}\otimes \epsilon) = I \otimes I.
\end{aligned}


问:
a doubt with the series $ \sum_{n=0}^{\infty}e^{-nx} $
I know that the series is equal to

$$ \sum_{n=0}^{\infty}e^{-nx}= \frac{1}{1-e^{-x}}$$

However, if I expand each exponential term into a Taylor series I get

$$  \sum_{n=0}^{\infty}e^{-nx}= \sum_{n=0}^{\infty}\sum_{k=0}^{\infty}\frac{(-nx)^{k}}{k!} $$

Now if I use $ \sum_{n=0}^{\infty}n^{k}=\zeta (-k) $ I would get that

$$ \sum_{n=0}^{\infty}e^{-nx} = \sum_{k=0}^{\infty} \frac{(-1)^{k} \zeta (-k)}{k!}x^{k}$$

Is this last expression correct in the sense of Taylor series?

答:
6 years later, but the given answer is i feel incorrect/incomplete.

The relation with swapping sums was correct, and $\sum_n\sum_m a_{n,m}=\sum_m\sum_n a_{n,m}$ is valid for regularization, as long as the shape of the sums/whatever is regularized stays the same.

The question made the mistake assuming Taylor series have a the lower limit of 0. Just like most general series, this should be in reality minus infinity and positive infinity.

Let's take the example where we start at n=1, you can add yourself n=0.
Changed e^(-x) into c.

$$\sum_{n=1}^{\infty}c^{n}= \frac{c}{1-c}$$
$$\sum_{n=1}^{\infty}c^{n}= \sum_{n=1}^{\infty}\sum_{k \in \mathbb{Z}}\frac{(n\ln(c))^{k}}{k!}=$$
$$\sum_{k \in \mathbb{Z}}\frac{\zeta(-k)\ln^{k}(c)}{k!}=$$
if k<-2 it's 0. if k is -1, you need to look at the limit, which is $\frac{-1}{ln(c)}$.

$$\sum_{k=0}^{\infty}\frac{\zeta(-k)\ln^{k}(c)}{k!}=\frac{c}{1-c}+\frac{1}{ln(c)}$$
As can been seen writen in the answer above.

$$\sum_{k \in \mathbb{Z}}\frac{\zeta(-k)\ln^{k}(c)}{k!}=\frac{c}{1-c}$$


答:
In terms of the Gamma and Beta functions we are discussing
$$ \sum_{n\geq 1}\frac{\Gamma\left(n+\frac{5}{3}\right)/\Gamma\left(\frac{2}{3}\right)}{\Gamma\left(n+\frac{5}{2}\right)/\Gamma\left(\frac{5}{2}\right)}=\underbrace{\frac{\Gamma(5/2)}{\Gamma(2/3)\Gamma(5/6)}}_{K}\sum_{n\geq 1}B(n+5/3,5/6) $$
or
$$ K \sum_{n\geq 1}\int_{0}^{1} x^{n+2/3}(1-x)^{-1/6}\,dx=K\int_{0}^{1}x^{5/3}(1-x)^{-7/6}\,dx $$
which is clearly divergent.

Title: Convergence of $\sum_{n=1}^{\infty}\frac{\frac{2}{3}(1+\frac{2}{3})\ldots(\frac{2}{3}+n)}{(1+\frac{3}{2})\ldots(\frac{3}{2}+n)}.$

Body:
Examine the convergence of the series:
$$\sum_{n=1}^{\infty}\frac{\frac{2}{3}(1+\frac{2}{3})\ldots(n+\frac{2}{3})}{(1+\frac{3}{2})\ldots(n+\frac{3}{2})}.$$

Attempt. Using ratio test for $a_n:=\frac{\frac{2}{3}(1+\frac{2}{3})\ldots(n+\frac{2}{3})}{(1+\frac{3}{2})\ldots(n+\frac{3}{2})}>0$, we have:

$$\frac{a_{n+1}}{a_n}=\frac{\frac{2}{3}+n+1}{\frac{3}{2}+n+1}\to 1,$$
so we do not conclude. Also $\sqrt[n]{(1+a)\ldots(n+a)}\to +\infty$ for $a>0$, so the root test also is inconclusive.

Thanks for the help.


问:
$\vert G \vert < \infty$, $A,B,C \leq G$, $B \leq A \Rightarrow \vert A:B \vert \geq \vert C \cap A : C \cap B \vert$
I am stuck with the following problem. I am sure it cannot be that hard since it is intuitively true, but I can't find a way to prove it.

Let $A,B,C \leq G$ where $G$ is a finite group. Suppose moreover that $B \leq A$. Then $\vert A:B \vert \geq \vert C \cap A : C \cap B \vert$.

答:
Asked is to show that $\frac{|A|}{|B|} \geq \frac{|A\cap C|}{|C\cap B|}$ since everything is finite.

We may use that $|A|\cdot |C| = |A\cap C|\cdot |AC|$ and similarly for $B$ and $C$, this amounts to showing that $|AC| \geq |BC|$, which is obvious since $B\subseteq A$.


答:
The definition of open sets in terms of a metric states that for each point in an open set there'll be some open ball of radius $\epsilon > 0$ such that the ball is totally contained in the set. In other words, if $(M,d)$ is a metric space, a subset $U\subset M$ is open if for every $p \in M$ there's some $\epsilon >0$ with the property that $B(x; \epsilon)\subset U$. The remark says that the $\epsilon$ possibly depends on the point, so that you don't need to present some $\epsilon$ that works for every single point of $U$.

Since this can sound kind of abstract, I think an example in the plane can help a little. Consider $\mathbb{R}^2$ the euclidean plane with the usual euclidean metric. Given some subset $U\subset \mathbb{R}^2$ this set will be open if for every of it's points there's some open ball (which you can see for yourself in the plane will be discs without border). If the set $U$ is open, we can surround each point $p\in U$ with one of these balls so that all points sufficiently near $p$ will yet be in $U$. If you think for a while, points far from the border of the set $U$ can be surrounded by big discs, so that $\epsilon > 0$ is greater, and points nearer the border will need small $\epsilon$ to grant the disc still sits inside $U$.

Now, this is a good way to intuit that the $\epsilon$ depends on the point. I hope this helps you out. Good luck!

Title: Definition of open set/metric space

Body:
On Proof Wiki, the definition of an open set is stated as

Let $(M,d)$ be a metric space and let $U\subset M$, then $U$ is open iff for all $y\in U$, there exists $\epsilon \in \mathbb{R}_{>0}$ such that $B_M(y;\epsilon) \subset U$

Then there's a remark that states

It is important to note that, in general, the values of $\epsilon$ depend on $y$. That is, it is not required that $\exists \epsilon \in \mathbb{R}_{>0}$ s.t. $\forall y \in U$ s.t.  $B_M(y;\epsilon) \subset U$

I don't understand quite understand the remark, could someone please clearly explain this and perhaps provide an example? I believe I understand the definition, but the remark is throwing me off.


问:
Find the minimum and maximum value of $\frac{x^2-2x+5}{(x-1)^2}$
Find min,max of function $$A = \dfrac{x^2-2x+5}{(x-1)^2}.$$

We see that $A = 1 + \dfrac{4}{(x-1)^2}.$ However I have no next idea. How to solve this problem?

答: Since when $x\to 1$, $1/(x-1)$ approaches to $\infty$, thus $A$ doesn't have a maximum, or its maximum is $+\infty$. And notice when $x\to \infty$, $1/(x-1)^2\to 0$, it doesn't have a minimum either.


答: $A \geq 1$ but you can never have $A=1$. So the infimum is $1$ but $1$ is not a minimum. As $x \to 1$ we see that $A \to \infty$ so $A$ is not bounded above. So there is no maximum.

Title: Find the minimum and maximum value of $\frac{x^2-2x+5}{(x-1)^2}$

Body:
Find min,max of function $$A = \dfrac{x^2-2x+5}{(x-1)^2}.$$

We see that $A = 1 + \dfrac{4}{(x-1)^2}.$ However I have no next idea. How to solve this problem?


答: Any real number squared is non-negative: $$(x-1)^2 \ge 0 \implies 0\lt \frac{1}{(x-1)^2} \lt \infty \implies 0\lt \frac{4}{(x-1)^2} \lt \infty \\ \implies 1\lt 1+\frac{4}{(x-1)^2} \lt \infty $$


问:
last two digits of $9^{1500}$ (Dummit Foote -Abstract Algebra preliminaries $0.3.5$)
Question is to find last two digits of $9^{1500}$ (No Euler totient theorem please)

What i have done so far is :

$9^2\equiv 81\pmod{100}$

$9^4 \equiv  61\pmod{100}$

$9^8\equiv  21\pmod{100}$

$9^{16} \equiv  41\pmod{100}$

$9^{32} \equiv  81\pmod{100}$

higher powers of $9$ namely $64,128,256,512,1024$ will be in repeated pattern as above.

$9^{64} \equiv  61\pmod{100}$

$9^{128} \equiv  21\pmod{100}$

$9^{256} \equiv  41\pmod{100}$

$9^{512} \equiv  81\pmod{100}$

$9^{1024} \equiv  61\pmod{100}$

Now,I want to split power of $9$ i.e., $1500$ to powers which i have noted down above. i.e,

$9^{1500}=9^{1024}.9^{476}$

$9^{1500}=9^{1024}.9^{256}.9^{220}$

$9^{1500}=9^{1024}.9^{256}.9^{128}.9^{92}$

$9^{1500}=9^{1024}.9^{256}.9^{128}.9^{64}.9^{28}$

$9^{1500}=9^{1024}.9^{256}.9^{128}.9^{64}.9^{16}.9^{12}$

$9^{1500}=9^{1024}.9^{256}.9^{128}.9^{64}.9^{16}.9^{8}.9^4$

When you multiply two positive integers, the last digit in the product depends on those two integers only through their last digits.

So, I will look only for last two digits of $9$ in above powers.

$9^{1500}=9^{1024}.9^{256}.9^{128}.9^{64}.9^{16}.9^{8}.9^4$

$\equiv 61.41.21.61.41.21.61 \pmod{100} $

$\equiv (61.61).(21.21).(41.41).61\pmod{100}$

(we have already seen above $61.61\equiv 21 \text{mod}100$ and similarly for other cases). So, we would be left with :

$\equiv (21).(41).(81).(61)\pmod{100}$

$\equiv (61)(41)\pmod{100}$

$\equiv (01) \pmod{100}$
I would be happy if someone can verify the procedure I have done and I would be thankful if some one can help me to make this less laborious and more efficient.

答:
I saw the remaining answers, but I don't think that so much work is required.
$$9^{1500} = (10-1)^{1500} = 1 - 15000 + 10k = 1 + 10k'$$ where $k,k'\in\mathbb{Z}$, by binomial theorem.

So we have $$9^{1500} \equiv 1 + 10k' \bmod 100 $$
$$ 9^{1500} \equiv 1 \bmod 100$$
which is what we want!


问:
Number of leaves in a tree with maximum degree $k$
Let $G$ be a tree with a maximum degree of the vertices equal to $k$.
At least how many vertices with a degree of $1$ can be in $G$ and why?

I think the answer must be $k$ but I don't know how to prove it.

答: Hint Any tree on $\geq 2$ vertices has at least two leafs. Consider a tree with maximal degree $k$ and delete a vertex $v$ with degree $k$. You are left with a forest consisting of $k$ components, coming with a distinct neighbor $v_i$ of $v$ in $T$. Finish by making a case distinction on whether the components have 1 or more vertices and using the first sentence.


问:
Path between vertices with an odd degree

Is it true that in an undirected graph with exactly $2$ vertices with an odd degree there is a path that connects them and why?

It must be true but I cannot prove it.

答: Suppose there is no such path. Then the graph is not connected. Let $A$ and $B$ be connected components, $A$ with first vertex and $B$ with second. Then in each component we have, by Handshake lemma, at least two vertices with odd degree. So in total at least 4 vertices with odd degree. A contradiction.


问:
Affine transformation matrix for conversion between line and rectangle
Is there a way using an affine transformation matrix to convert between a rectangle of zero height (i.e. effectively having 2 different end/corner points) and a rectangle of > zero height (i.e. having 4 different end/corner points)?

If not, does another kind of transformation exist that can do the same?

Thanks in advance.

答:
The affine (linear) function

$$ f : (x,y) \mapsto \frac{r-s}{a-b}\,x + \frac{as-br}{a-b} $$

transform $[a,b]\times[c,d]$ into $[r,s]$, as it is easy to verify. Note however that $f$ is not bijective.

A bijective affine transformation can never transform $[a,b]\times[c,d]$ into $[r,s]$. In fact such a transformation $g$ would be a homeomorphism. And this is impossible: by removing an internal point from $[r,s]$ and the corresponding point (through $g$) from $[a,b]\times [c,d]$, we would again obtain, by restriction, a homeomorphisn between a connected space and a disconnected space, absurd.


问:
Second derivative test when function is non-analytic
When we are dealing with multivariate functions of type $f(\vec x)$ (which has a critical point at let's say $\vec x_o$), we find whether it's a minima or maxima by writing the Taylor's expansion of $f$ which is of the form:

$$f(\vec x) = f(\vec x_o) + \nabla f(\vec x_o) \cdot (\vec x - \vec x_o) + 1/2 (\vec x - \vec x_o)^TH_f(\vec x - \vec x_o)  + ...$$

Then, we say that if $H$ is positive definite, it has a minima. If $H$ is negative definite, it has a maxima. If $H$ is indefinite, it has a saddle point. Otherwise the test is inconclusive.

My question is what if $f$ is differentiable but non-analytic at $\vec x_o$? In that case, clearly, we can't write it's Taylor expansion, then how do we test? Or am I misunderstanding something?

答: To find the hermittian of $f,$ you need not write the series expansion. As long as all the required partial derivatives exist, you can always get $H.$


答:
The derivative is $-8/(x-1)^3$.
Since it's never zero it doesn't have a local max/min, then you have to look at the asymptotes(FFjet already done it).


问:
numbers 1 to 21 on circle with sum at least 33
if we write 1,2,3,....,21 on a circle. there are at least one three consecutive set with sum at least 33.

obviously the main idea is strictly increasing on sum of numbers. but i wanna know how to organize the solution.

答:
You have $21$ consecutive $3$ sets. Each number appears $3$ times in these $21$ sets. Moreover, the sum of numbers $1$ to $21$ is $ \dfrac{21 \times 22}{2} = 11 \times 21 $. So the sum of numbers of these sets will be:
$$ 3 \times \dfrac{21 \times 22}{2} = 33 \times 21. $$
Now, if all these sets have sum less than $33$, then the sum of numbers of these $21$ sets will be less than $33 \times 21$, a contradiction.


答:
As an example consider for instance $\mathbb R^2$ with its usual (Euclidean) metric structure (so distance look like what we are used to in the plane). Now consider the set $\{x\in \mathbb R^2\mid d(x,(0,0))<1\}$. This is the interior of the circle with centre the origin and unit radius. Now, this set is open, since for any point $p$ in this set, there is a radius $\epsilon>0$ such that a the ball with centre $p$ and radius $\epsilon$ is entirely contained in the set. But, the radius $\epsilon$ depends on the position of the point $p$. When $p$ is very close to the circumference of the circle the value of $\epsilon$ must be chosen to be very small.

It should be noted that it is actually extremely rare that in a metric space a set will be 'uniformly' open, in the sense that there will be a single $\epsilon$ that fits for all points in the set. As an extra exercise, try to find examples of such situations.

Title: Definition of open set/metric space


问:
$30$ Sided Dice Problem.
$4$ players roll a fair $30$-sided dice. Each of them rolls once and the $4$ numbers rolled are different. The $2$ individuals that rolled the biggest and small value will be in a team while the $2$ individuals who rolled the $2^{nd}$ and $3^{rd}$smallest value will be on the opposite team. The team with the largest average wins and the loser has to pay the winners' average number (for example, if the dice outcome is $27, 20, 15,$ and $11,$ then the people who get $27$ and $11$ get paid $19$).

Q) If you are the first to roll, what number would you prefer to maximize your payout?

答:
Here's a simple python program to find the answer:

```<code>from itertools import combinations
score=[0]*31
for x in combinations(range(1,31),4):
    outer=x[0]+x[3]
    inner=x[1]+x[2]
    if outer>inner:
        score[x[0]]+=outer
        score[x[1]]-=outer
        score[x[2]]-=outer
        score[x[3]]+=outer
    elif inner>outer:
        score[x[0]]-=inner
        score[x[1]]+=inner
        score[x[2]]+=inner
        score[x[3]]-=inner
for i in range(1,31):
    print(i,score[i],sep="\t")
</code>```

I assumed that if both totals are equal no-one wins or loses anything. To avoid floats I also doubled everything (so that you win or lose the sum, rather than the average). The highest value is best; here that is 24.

Title: $30$ Sided Dice Problem.

Body:
Here's a simple python program to find the answer:

from itertools import combinations
score=[0]*31
for x in combinations(range(1,31),4):
    outer=x[0]+x[3]
    inner=x[1]+x[2]
    if outer>inner:
        score[x[0]]+=outer
        score[x[1]]-=outer
        score[x[2]]-=outer
        score[x[3]]+=outer
    elif inner>outer:
        score[x[0]]-=inner
        score[x[1]]+=inner
        score[x[2]]+=inner
        score[x[3]]-=inner
for i in range(1,31):
    print(i,score[i],sep="\t")

I assumed that if both totals are equal no-one wins or loses anything. To avoid floats I also doubled everything (so that you win or lose the sum, rather than the average). The highest value is best; here that is 24.


问:
Intuition behind sums of sums of whole numbers
So I was playing around, and all this is just a curiosity and nothing serious.

Anyway, most readers probably know:
$$1+2+3+4+5+...+(n-1)+n=\frac{1}{2}n^{2}+\frac{1}{2}n=\binom{n+1}{n-1}$$

I started playing around, adding the individual sums of whole numbers instead of whole numbers alone. Words aren't very helpful to describe this process, instead consider the sum of sums for $n=4$, which we shall call $N_2(4)$ for simplicity:
$$\left ( 1+2+3+4 \right ) + \left ( 1+2+3 \right ) + \left ( 1+2 \right ) + \left ( 1 \right ) = 20$$

Remarkably, there is a simple formula (I did the math):
$$N_{2}(n)=\binom{n+2}{n-1}$$

Where $N_2(n)$ is the sum of sums as above. Formally, $N_2(n)=\sum_{1\leq i}^{n}\sum_{1\leq j\leq i}j$.

Now imagine going further, with sums of sums of sums, for instance:
$$N_3(4) = \left ( \left ( 1+2+3+4 \right ) + \left ( 1+2+3 \right ) + \left ( 1+2 \right ) + \left ( 1 \right ) \right ) + \left ( \left ( 1+2+3 \right ) + \left ( 1+2 \right ) + \left ( 1 \right ) \right ) + \left ( \left ( 1+2 \right ) + \left ( 1 \right ) \right ) + \left ( \left ( 1 \right ) \right ) = 35$$

Again, this seems to follow the pattern (I haven't explicitly checked):
$$N_3(n)=\binom{n+3}{n-1}$$

And we might conjecture:
$$N_k(n)=\binom{n+k}{n-1}$$

One angle of attack is this: realizing that the previous series always adds up to that of the differences between successive elements of the next series, and so verifying that:

$$\binom{n+k}{n-1} - \binom{(n-1)+k}{(n-1)-1}=\binom{n+(k-1)}{n-1}$$

I.e. that $N_{k}(n)-N_{k}(n-1)=N_{k-1}(n)$ for any suitable $n$ and $k$.

My question is if there's some intuition behind all this. Maybe an alternative way of looking at this, or proving it. Why are the sums so neatly expressible?

答:
Before I knew math, i had my own notation for such things but i never seen it elsewhere, Let the s'th sum be $$.^s\sum_{n}n$$ be the s'th "indefinite sum of n

indefinite sum of n $\sum_{n}=F(n)$ so that $\sum_{n=a}^{b} f(n)=F(b)-F(a-1)$.
The delta operator $\Delta$ is f(n)-f(n-1).

$$\Delta.^s\sum_{n}n=^{s-1}\sum_{n}n$$

$$.^s\sum_{n}n=\frac{(n+s)!}{(n-1)!(s+1)!}$$

$$\Delta.^s\sum_{n}n=\frac{(n+s)!}{(n-1)!(s+1)!}-\frac{(n-1+s)!}{(n-2)!(s+1)!}=\frac{(n+s-1)!(n+s))}{(n-1)!(s+1)!}-\frac{(n-1+s)!(n-1)}{(n-1)!(s+1)!}=$$

$$\frac{(n+s-1)!}{(n-1)!(s)!}=^{s-1}\sum_{n}n$$

Ps: in similair fashion: To figure sums similair to integrating, and if you combine this with the found sum in the question you can easily derive the bernouilli numbers aswell.

$.^s\sum_{n} n^{k+1} =n.^s\sum_{n} n^{k}-(s).^{s+1}\sum_{n} (n-1)^{k} $

Also quickly seen by delta'ing, officially the n-1 should be incorperate into the sum, as upper limit.


问:
Using the l'hopital's rule for a $1^∞$ case
I need to find the limit of:

$$\lim_{x \to 0} (1-2x)^{1/x}$$

I can see that this is a $1^∞$ case, so I will let $y =$ the equation and use the natural log on both sides:

$$\ln y = \lim_{x \to 0} \frac{1}{x}\ln(1-2x) $$

Now it is a $∞\times0$ case, so I can do:

$$\ln y = \lim_{x \to 0}\frac{\frac{1}{x}}{\frac{1}{\ln(1-2x)}}$$

and now it is a $\frac{∞}{∞}$ case. I think my mistake happens while I use the l'hopital's rule:

$$\ln y = \lim_{x \to 0}\frac{-x^{-2}}{-[\ln(1-2x)]^{-2}\times\frac{1}{1-2x}\times-2x}$$

$$\ln y = \lim_{x \to 0}\frac{-[\ln(1-2x)]^2}{\frac{2x^3}{1-2x}}$$

It is now a 0/0 case, and I have tried using the l'hopital's rule a few more times but to no such luck. Where did I make a mistake, and is there a better way to solve this?

答: To evaluate the limit, set $-2x=1/n,$ so that we now have $$\lim_{n\to\infty}\left(1+\frac1n\right)^{-2n}=\left[\lim_{n\to\infty}\frac{1}{\left(1+\frac1n\right)}\right]^2,$$ given that $x\mapsto x^2$ is continuous, so that the limiting value is $1/e^2.$


问:
Why is $\pi $ equal to $3.14159...$?
Wait before you dismiss this as a crank question :)

A friend of mine teaches school kids, and the book she uses states something to the following effect:

If you divide the circumference of any circle by its diameter, you get the same number, and this number is an irrational number which starts off as $3.14159... .$

One of the smarter kids in class now has the following doubt:

Why is this number equal to $3.14159....$? Why is it not some other irrational number?

My friend is in a fix as to how to answer this in a sensible manner. Could you help us with this?

I have the following idea about how to answer this: Show that the ratio must be greater than $3$. Now show that it must be less than $3.5$. Then show that it must be greater than $3.1$. And so on ... .

The trouble with this is that I don't know of any easy way of doing this, which would also be accessible to a school student.

Could you point me to some approximation argument of this form?

答:
I got this idea from a colleague in grad school. It's one of my favorite lessons.

Since my original posting of this answer, I've created an animation that appears on my trigonography site. The animation itself is too big to post here, but a frame of it appears below.

Take a rubber chicken ---yes, a rubber chicken--- to a chalkboard.

Press its feet together firmly to the board, hold a piece of chalk in its beak, and use the chicken like a compass to draw a nice, big circle.

Mark a point on the circle you've drawn.

Press the chicken's feet together at that point, bend the chicken around the circle, and mark where its beak lands.

Do that step again. And again. And again. And again. And again. And ag--- oh, wait ... we can't do it again: we've run out of circle! Hmmmm ...

Fold the chicken in half. Then fold it again. Voila! The double-folded chicken just fits into the left-over space on the circle.

Consequently, a circle with a radius of $1$ rubber chicken has a circumference of about $6.25$ rubber chickens ... an estimate that's within half a percent of the actual value (namely, $2\pi$)!

Now note: There's nothing special about the rubber chicken. We could've used a rubber cat, or a rubber python, or a rubber earthworm, or a rubber Godzilla, or a rubber band that extends from one end of the galaxy to the other. No matter what we object we use for a radius, we can fit just-about six-and-a-quarter of them around the circumference.

This doesn't explain anything about the irrationality of $\pi$ or that kind of thing, but it's a lesson the kids will never forget. :)

Title: Why is $\pi $ equal to $3.14159...$?


问:
If $f(t)=\sum_{n=1}^{\infty} e^{-n^2t}t^a$ , then $ \lim_{t \to 0}f(t) = 0$
Let $$ f(t)=\sum_{n=1}^{\infty} e^{-n^2t}t^a , \hspace{5mm} t >0$$ for some $a>0$

Then show that $$ \lim_{t \to 0}f(t) = 0$$

[My attempt]

I think  $$ \lim_{t \to 0}f(t)=\lim_{t \to 0}\sum_{n=1}^{\infty} e^{-n^2t}t^a=\sum_{n=1}^{\infty}\lim_{t \to 0} e^{-n^2t}t^a=\sum_{n=1}^{\infty}0=0$$

However, for changing limit and series , we need to uniform convergence of $f$

I tried to use weierstrass M - test but it didn't work.

I'm stuck here, how that solve this?

答:
The question relates to the Gauss circle problem. Indeed by considering
$$ \Theta(x)=\sum_{n\in\mathbb{Z}}e^{-n^2 x}$$
we have
$$ \Theta(-\log z)^2 = \sum_{N\geq 0} r_2(N) z^N $$
where $ r_2(N)=\left|\left\{(a,b)\in\mathbb{Z}^2:a^2+b^2=N\right\}\right| $. By Gauss geometric estimations we have
$$ \sum_{N=0}^{M}r_2(N) = \pi M + O(\sqrt{M}) $$
hence
$$ \frac{\Theta^2(-\log z)}{1-z}=\frac{\pi z}{(1-z)^2}+O\left(\sum_{n\geq 0}\sqrt{n}z^n\right) $$
and
$$ \Theta(-\log z)\sim\sqrt{\frac{\pi z}{1-z}} $$
as $z\to 1^-$. This implies
$$ \sum_{n\geq 0}e^{-n^2 x}\sim -\frac{1}{2}+\frac{1}{2}\sqrt{\frac{\pi e^{-x}}{1-e^{-x}}}\sim -\frac{1}{2}+\frac{\sqrt{\pi}}{2\sqrt{x}} $$
as $x\to 0^+$.


问:
Nim Game remove $1,3$ or $ 4$ matches
I was reading about variations of the nim game where 2 players remove matches from a pile of matchsticks.

In this variation players can remove $1,3$ or $4$ matches.

I can see that the player wins if the number of matches in the pile is:

```<code>{1, 3, 4, 5, 6, 8, 10, 11, 12, 13, 15, 17 ....}
</code>```

and the player loses if the number of matches in the pile is:

```<code>{0, 2, 7, 9, 14, 16 ...}
</code>```

It is not clear to me what is the pattern that indicates if the number of matches is a win.

I thought 4*2 = 8 (win) but also 4*2 + 1 = 9 (lose) so some multiple of the number of matches to remove is not a winning state.

What is the pattern here?

答:
Let $b_k=1$ when $k$ is a winning number and $b_k=0$ otherwise. Doing the analysis the OP did somewhat further one quickly conjectures that
$$(b_k)_{k\geq0}=(0,1,0,1,1,1,1,\ldots)\ ,$$
where the $\ldots$ suppress a periodic pattern with period $7$. That this pattern is true for $0\leq k\leq 6$ we know from looking at the computed table.

It remains to set up an induction proof: For each of the seven remainders mod $7$ we  go  back the necessary steps and verify that everything suits. The inductive hypothesis is: The pattern is true for all $k≤7j−1$. The induction is with respect to $j$, and the base case is $j=1$.

Title: Nim Game remove $1,3$ or $ 4$ matches


问:
Torsors and trivializations of a quasi-coherent sheaf
Greeting. As a part of my project i'm trying to understand the following paragraph of This article;

Here is what i understood so far:
Let $X$ be a scheme and $L$ a locally free sheaf of rank $g$ over $X$. Let $T_L$ be the functor from $(Sch/X)^{opp}$ to $Sets$ sending an $X$ scheme $S$ with structural morphism $f$ to $Isom_X(O_S^g,f^*L)$. This functor is representable by a scheme $\underline{T_L}$, moreover $GL_g$ acts on the right by $\omega\gamma=\omega \circ \gamma$. Let $\pi:T_L\rightarrow X$ the structural morphism. $Hom_X({T_L},{T_L})$ is naturally isomorphic to $Isom_X(O_{T_L},\pi^*L)$, taking the identity element we obtain a universal trivialization $\omega:O_{T_L}\simeq \pi^*L$.

What remains for me to understand is the following: How does the writer view $\pi_*O_{T_L}$ as a sheaf of sections on the form $f(\omega)$? What does $f$ stands for? I think $f$ is a function from $Isom_X(O_{T_L},\pi^*L)$ to $X$ but i can't see the relation with the sheaf $\pi_*O_{T_L}$. Thanks in advance!

答:
Let $u_X:X\rightarrow S$ be a scheme with $V:=Spec(R)\subseteq S$ an open subscheme with $u_X^{-1}(V)=U:=Spec(A)$, and let $\mathcal{L}_X$ be a locally trivial $\mathcal{O}_X$-module of rank $g$ where $\mathcal{L}_X(U)\cong L$.
Let $u:S:=Spec(A)\rightarrow T:=Spec(R)$ be the induced map and let $L:=A\{e_1,..,e_g\}$ be the free $A$-module of rank $g$. Let $\mathcal{L}$ be the sheafification of $L$. It seems in this situation there is an isomorphism $T_{\mathcal{L}}\cong Spec(B)$ where $B:=A[x_{ij},t]/(det(x_{ij})t-1)$, where $i,j=1,..,g$. Hence $A[x_{ij}]$ is the coordinate ring of the scheme $Mat_g(A)$ of square matrices of rank $g$ with coefficients in $A$, and $B$ is the coordinate ring of the group scheme $GL(g,A)$ of invertible $g\times g$-matrices $M:=(a_{ij})$ with $a_{ij}\in A$ and $det(M)\in A^*$ a unit. There is a canonical right action of $GL(g,A)$ on $B$ and a canonical structure morphism $\pi:T_{\mathcal{L}}\rightarrow Spec(A)$.

In this situation it follows the global sections of the  sheaf $\pi_*\mathcal{O}_{T_\mathcal{L}}$ is the ring $A[x_{ij},t]/(det(x_{ij})t-1)$ viewed as an $A$-algebra, and there is an inclusion of sheaves

I1. $\pi_*\mathcal{O}_{T_\mathcal{L}}[\kappa^*]\subseteq \pi_*\mathcal{O}_{T_\mathcal{L}}$ is a subsheaf

I have not read the paper, but it seems the subsheaf $\pi_*\mathcal{O}_{T_\mathcal{L}}[\kappa^*]$ is defined as the set of matrices $m$ in $B$ that are "semi-invariant" under the right action of the group $U$ of upper triangular matrices in $GL(g,A)$. Given $b\in U$ it follows $mb=\kappa^*(b)m$. You must figure out how $\kappa^*$ acts on elements $b\in U$. It seems the group $T$ acts on $U$.


答:
For the set $\{1,2,3,4,5,6,7\}$ the losers are $\{2,7\}.$

Inductively assume that for
$\{1,2,3, \cdots, (7n-2), (7n-1), 7n\}$ the losers are precisely those numbers that are congruent to either 2 or 7, mod 7.

Consider the numbers $\{(7n+1), (7n+2), \cdots, (7n + 7)\}.$

(7n+1) can take 1, leaving (7n), so (7n+1) is a winner.

(7n+2) must leave (7n+1), (7n-1), (7n-2), which are all inductively assumed to be winners.  Therefore, (7n+2) is a loser.

By very similar reasoning, (7n+3) through (7n+6) are winners and (7n+7) is a loser.

Title: Nim Game remove $1,3$ or $ 4$ matches

Body:
I was reading about variations of the nim game where 2 players remove matches from a pile of matchsticks.

In this variation players can remove $1,3$ or $4$ matches.

I can see that the player wins if the number of matches in the pile is:

```<code>{1, 3, 4, 5, 6, 8, 10, 11, 12, 13, 15, 17 ....}
</code>```

and the player loses if the number of matches in the pile is:

```<code>{0, 2, 7, 9, 14, 16 ...}
</code>```

It is not clear to me what is the pattern that indicates if the number of matches is a win.

I thought 4*2 = 8 (win) but also 4*2 + 1 = 9 (lose) so some multiple of the number of matches to remove is not a winning state.

What is the pattern here?


问:
Find missing angle in triangle

In this triangle, we are given that $AB=AD=CE$, $BE=CD$ and angle  $ECD=2*AED=2a$.

We are asked to find $AED=a$.

I have seen several problems similar to this but can't follow any of the solution methods.
The only thing which is obvious to me is that since $AD+CD = BE+EC$, the triangle is isosceles.

Therefore its base angles (the ones I have drawn in green) are equal.
Also triangle ABD is isosceles, therefore its base angles are equal.

I don't see any other relation to use.

(FYI this is not homework or anything - just practice and self-motivation).
Thank you very much.

EDIT: If this is of any help, through Geogebra I get that $a=10^{\circ}$, therefore $ECD=20^{\circ}$ and we have a 80-80-20 triangle.

答:
Choose $X$ so that $ADEX$ is a parallelogram. Then $XE=AD=CE$ and since $XE \parallel AC$, we get $\angle XEC = 180^\circ - 2a$. It follows that $a=\angle XCE$.

On the other hand, $\angle XAE = \angle DEA = a$. Therefore $\angle XCE = \angle XAE$ which shows that $AXEC$ is cyclic. Then we see that $\angle ACX = a = \angle XCE$, so the corresponding chords are equal: $AX=XE$. Also note that $X$ lies on the bisector of angle $ACB$ and since $ACB$ is isosceles, it follows that $AX=XB$.

Then we find $AX=XB=AB$ so $AXB$ is equilateral. Clearly $\angle CBX = XAC = 2a$. Writing down the sum of angles of triangle $ABC$ leads to $2a+2\cdot(2a+60^\circ)=180^\circ$ which yields $a=10^\circ$.

Below is another solution.

First we're going to show that $DE=AD$. To do so we exclude the possibilities that $DE>AD$ and $DE<AD$.

If $DE>AD$ then $\angle EAD > \angle DEA = a$. This means that $\angle EDC = a+\angle EAD > 2a = \angle DCE$ which implies that $CE>DE$ and we have a contradiction: $CE>DE>AD=CE$. If $DE<AD$ then all inequalities get reversed and we have a contradiction again. Hence $DE=AD$.

Then let $F$ be the circumcenter of $ADE$. Then $\angle DFA = 2\angle DEA = \angle ACB$, so the isosceles triangles $ADF$ and $BAC$ are similar. They are actually congruent because $AD=AB$. Since $AD=DE$, it also follows that $DEF$ is congruent to both $ADF$ and $BAC$. We can also see that $F$ lies on $AB$ because $\angle FAD = \angle BAC$.

Let $G$ be the point on $DF$ such that $DG=BE$. Then triangles $BAE$, $ADG$ and $EDG$ are congruent. In particular, $AE=EG=GA$, so triangle $AEG$ is equilateral.

What is left to do is some angle chasing. We have $\angle EGA = 60^\circ$, so $\angle DGA = 30^\circ$ as $DF$ is the axis of symmetry of $AEG$. Then $30^\circ = \angle DGA = \angle GFA + \angle FAG = \angle ACE + \angle EAC = 2a+\angle DEA = 2a+a=3a$, so $a=10^\circ$.

Title: Find missing angle in triangle


问:
conceptual doubt in finding $\lim \limits_{x \to \tfrac{\pi}{4}} \frac{4\sqrt 2 -\left ( \cos x + \sin x \right )^5}{1-\sin {2x}}$
While calculating the aforementioned limits I did the following.

$$\lim \limits_{x \to \tfrac{\pi}{4}} \frac{4\sqrt 2 -\left ( \cos x + \sin x \right )^5}{1-\sin {2x}}\\ =
\lim \limits_{x \to \tfrac{\pi}{4}} \frac{4\sqrt 2 -4\sqrt2 \left (\frac{1}{\sqrt 2} \cos x +\frac{1}{\sqrt 2} \sin x \right )^5}{1-\sin {2x}}\\=
\lim \limits_{x \to \tfrac{\pi}{4}} \frac{4\sqrt {2} \left ( 1 - \sin^5 (x + \frac{\pi}{4}) \right )}{1-\sin {2x}}\\
=\lim \limits_{x \to \tfrac{\pi}{4}} \frac{4\sqrt {2} \left ( 1 - \sin (x + \frac{\pi}{4}) \right ) \left(1 +\sin (x + \frac{\pi}{4}) +\sin^2 (x + \frac{\pi}{4}) +\sin^3 (x + \frac{\pi}{4}) +\sin^4 (x + \frac{\pi}{4}) \right)}{1-\sin {2x}}\\$$

As $x$ approaches $\frac{\pi}{4}$, the functions $\sin (x + \frac{\pi}{4})$ and $\sin 2x$ become close enough. So, I canceled them($1-\sin {2x}$ and $ 1 - \sin (x + \frac{\pi}{4})$ ) and the limit turns out to be $20\sqrt2$.

I made the above claim after comparing the graphs. But the fact that bothers me is L'Hospital's rule yielded the answer $5\sqrt2$ (and it is the correct answer to the problem)

Can anyone please explain?

答:
As $x$ approaches $\frac{\pi}{4}$ the functions $\sin (x + \frac{\pi}{4})$ and $\sin 2x$ become close enough

When we say "close enough", we are saying that $\lim_{x\to \pi/4} \dfrac{\sin (x + \frac{\pi}{4})}{\sin 2x}=1$. However, the limit
$\lim_{x\to \pi/4} \dfrac{1-\sin (x + \frac{\pi}{4})}{1-\sin 2x}=\dfrac{1}{4}\neq 1$

So we cannot cancel them simply.

We can view it from the perspective of Taylor Series:
$$
\frac{\sin \left(x+\frac{\pi }{4}\right)}{\sin (2 x)}=1+\frac{3}{2} \left(x-\frac{\pi }{4}\right)^2+\frac{19}{8} \left(x-\frac{\pi }{4}\right)^4+O\left(\left(x-\frac{\pi }{4}\right)^5\right) \\
\frac{1-\sin \left(x+\frac{\pi }{4}\right)}{1-\sin (2 x)}=\frac{1}{4}+\frac{1}{16} \left(x-\frac{\pi }{4}\right)^2+\frac{1}{96} \left(x-\frac{\pi }{4}\right)^4+O\left(\left(x-\frac{\pi }{4}\right)^5\right)
$$


问:
Showing that $\lim_{x \to 1} \left(\frac{23}{1-x^{23}}-\frac{11}{1-x^{11}} \right)=6$
How does one evaluate the following limit?
$$\lim_{x \to 1} \left(\frac{23}{1-x^{23}}-\frac{11}{1-x^{11}} \right)$$
The answer is $6$.

How does one justify this answer?

Edit: So it really was just combine the fraction and use L'hopital's rule twice (because function and its first derivative are of indeterminate form at $x=1$). This problem is more straightforward than it seems at first.

答:
This is definitely the best method to solve this question:

$$Let\space L = \lim_{x \to 1} \left(\frac{p}{1-x^{p}}-\frac{q}{1-x^{q}} \right)$$$\space$ $\space$ $\space$ $\space$ $\space$ $\space$ $\space$ $\space$ $\space$ $\space$ $\space$ $\space$ $\space$ $\space$ $\space$ $\space$       $\space$              Replace $x$ by $\frac{1}{x}$

$$\space L = \lim_{{x} \to 1} \left(\frac{p}{1-{\frac{1}{x}}^{p}}-\frac{q}{1-{\frac{1}{x}}^{q}} \right)$$
$$\space L = \lim_{{x} \to 1} \left(\frac{-px^p}{1-{{x}}^{p}}-\frac{-qx^q}{1-{{x}}^{q}} \right)$$
Adding both the equations we get $2L={p-q}$

And we are done


问:
Is it possible to write this rational sequence : $(u_n=\frac{2^{n-1}+1}{2^n})_{n \ge 1}$
using only the rational numbers : $1$; $\dfrac{1}{2}$; $\dfrac{1}{3}$; $\dfrac{1}{4}$ and the operation $+,-,\times$ ?

I get : $u_1 = 1 = \dfrac{1}{2} +\dfrac{1}{2}= \dfrac{1}{2}\left(1+1\right)=\dfrac{1}{3}\left(1+\dfrac{1}{2}\right)\left(1+\dfrac{1}{4}+\dfrac{1}{4}+\dfrac{1}{4}+\dfrac{1}{4}\right)$

$u_2 = \dfrac{3}{4} = \dfrac{1}{2} +\dfrac{1}{4}= \dfrac{1}{2}\left(1+\dfrac{1}{2}\right)= \dfrac{1}{3}\left(1+\dfrac{1}{2}\right)\left(1+\dfrac{1}{4}+\dfrac{1}{4}\right)$

$u_3 = \dfrac{5}{8}= \dfrac{1}{2} +\dfrac{1}{8}= \dfrac{1}{2}\left(1+\dfrac{1}{4}\right)=\dfrac{1}{3}\left(1+\dfrac{1}{2}\right)\left(1+\dfrac{1}{4}\right)$

$u_4=\dfrac{9}{16}=\dfrac{1}{2} +\dfrac{1}{16}=\dfrac{1}{2}\left(1+\dfrac{1}{8}\right)=\dfrac{1}{3}\left(1+\dfrac{1}{2}\right)\left(1+\dfrac{1}{8}\right)=\dfrac{1}{3}\left(1+\dfrac{1}{2}\right)\left(1+\dfrac{1}{2}\times \dfrac{1}{4}\right)$

etc...

$u_n = \dfrac{2^{n-1}+1}{2^n}= \dfrac{1}{2} +\dfrac{1}{2^{n}}=\dfrac{1}{2}\left(1+\dfrac{1}{2^{n-1}}\right)=\dfrac{1}{3}\left(1+\dfrac{1}{2}\right)\left(1+\dfrac{1}{2^{n-1}}\right)$ and following the value of $n$ we can make appear several products of $\dfrac{1}{2}$ and $\dfrac{1}{4}$.

Can we find a better form for $u_n$ ?

Moreover it seems possible to write the terms of this sequence (for $n>1$) using only once the rational numbers $1$; $\dfrac{1}{2}$; $\dfrac{1}{3}$; $\dfrac{1}{2^n}$ :

$u_2 = \dfrac{1}{3}\left(1+\dfrac{1}{2}\right)+\dfrac{1}{4}$

$u_3 = \dfrac{1}{3}\left(1+\dfrac{1}{2}\right)+\dfrac{1}{8}$

etc...

$u_n = \dfrac{2^{n-1}+1}{2^n}= \dfrac{1}{2} +\dfrac{1}{2^{n}}=\dfrac{1}{3}\left(1+\dfrac{1}{2}\right)+\dfrac{1}{2^{n}}$

Thanks in advance !

Body:
As mentioned in the comments, you can use a recurrence, which needs only $\frac{1}{2},+$ and $\times$:

Since $u_n=\frac{1}{2}+\frac{1}{2^n} \Rightarrow \frac{1}{2^n}=u_n-\frac{1}{2}$, we have $u_{n+1}=\frac{1}{2}+\frac{1}{2^{n+1}}=\frac{1}{2}(1+\frac{1}{2^n})$ and the recurrence relation $u_{n+1}=\frac{1}{2}(\frac{1}{2}+u_n)$. By $u_1=\frac{1}{2}+\frac{1}{2}$ we are finished.

The first n are:

$u_1=\frac{1}{2}+\frac{1}{2}$,

$u_2=\frac{1}{2}(\frac{1}{2}+u_1)=\frac{1}{2}(\frac{1}{2}+\frac{1}{2}+\frac{1}{2})$,

$u_3=\frac{1}{2}(\frac{1}{2}+u_2)=\frac{1}{2}(\frac{1}{2}+\frac{1}{2}(\frac{1}{2}+\frac{1}{2}+\frac{1}{2}))$.


问:
What's the condition for a group $G$ to be equal to the product of two normal subgroups
If $G$ is a group and $N,M$ are two normal subgroups. We know that the product $NM$ is normal subgroup of $G$, but when can I say that $G=NM$. What must be the conditions on $N,M$?

答:
Assuming all groups mentioned in these examples are finite.

An example: if $|G:M|$ and $|G:N|$ are coprime, then $G=NM$. Proof: $|G:NM| \mid |G:M|$ and $|G:NM| \mid |G:N|$.

Another example: if $|M|$ and $|N|$ are coprime and $|G|=|N| \cdot |M|$, then $G=NM$.

Yet another example: if $M$ is a maximal subgroup and $N \not \subseteq M$, then $G=NM$.

If you are familar with (ordinary) character theory of finite groups: if $\varphi$ is a character of $M$ and the induction-restriction $(\varphi^G)_N$ is irreducible, then $G=NM$.


问:
Is any compact metric totally disconnected space homeomorphic to a compact subspace of a Cantor space?
Every compact metric totally disconnected perfect space is homeomorphic to a Cantor space.

Is every compact metric totally disconnected space homeomorphic to a compact subspace of a Cantor space?

In other words, if you have a compact metric totally disconnected space can you embed it in a compact metric totally disconnected perfect space?

答:
Yes, first show that every zero-dimensional metric space is homeomorphic with a closed subset of the Baire space. Since the Baire space is a $G_\delta$ subset of the Cantor space, we have that every zero-dimensional metric space is a $G_\delta$ subspace of the Cantor space. Note that if our space was compact then the image of the homeomorphism is compact, as well.

The proof of this can be found in Kechris' Classical Descriptive Set Theory, Theorem 7.8.


问:
Why is $a^{-1}$ used instead of $\frac{1}{a}$ in Calculus by Spivak?
In Calculus by Michael Spivak, why did he use the form $a^{-1}$ instead of $\frac{1}{a}$ to explain the multiplicative inverse (reciprocal) property? Is there a conceptual reason why the former is preferred rather than the latter? I know they are the same thing but I can't help but feel those who are not strong on proofs would have an easier time seeing the solution to problem 1(i) if it was used in that way rather than what was given. Any insight as to why his style is used or preferred in analysis is greatly appreciated.

答: Because the notation $\frac1a$ suggests that this is the number that you get when you divide $1$ by $a$. And, yes, it is! But one should use that notation only after it was proved that $a^{-1}{}{}{}{}{}{}$ is indeed what you get when you divide $1$ by $a$.


答: I suspect this notation comes from abstract algebra, where we define the inverse of an element $a$ to be written as ${a}^{-1}$. You write "multiplicative inverse" in the problem description, and it agrees with my thoughts.

Title: Why is $a^{-1}$ used instead of $\frac{1}{a}$ in Calculus by Spivak?

Body: In Calculus by Michael Spivak, why did he use the form $a^{-1}$ instead of $\frac{1}{a}$ to explain the multiplicative inverse (reciprocal) property? Is there a conceptual reason why the former is preferred rather than the latter? I know they are the same thing but I can't help but feel those who are not strong on proofs would have an easier time seeing the solution to problem 1(i) if it was used in that way rather than what was given. Any insight as to why his style is used or preferred in analysis is greatly appreciated.


问:
Find the minimum and maximum value of $\frac{x^2-2x+5}{(x-1)^2}$
Find min,max of function $$A = \dfrac{x^2-2x+5}{(x-1)^2}.$$

We see that $A = 1 + \dfrac{4}{(x-1)^2}.$ However I have no next idea. How to solve this problem?

答:
The function $$A= \frac{x^2-2x+5}{(x-1)^2}=1+\frac{4}{(x-1)^2}\quad \forall x\in (-\infty,1)\cup(1,+\infty)$$

Hence, at $x=1$, the function is undefined. The minimum of the function however is when $x=\pm\infty, A=1$ while at $x=1$, the function achieves it maxima since
$$\lim_{x\to1}\frac{x^2-2x+5}{(x-1)^2}\to \infty$$
You can also check it through the graph of this function.

It is clear that the function tends towards infinity both from $x\to 1^-, x\to 1^+$


问:
Quadratic variation of Ito integral : Relationship between Ito Integral and Brownian Motion
This is theorem 4.2.3 from Shreve's Stochastic Calculus for finance II.

Statement of the theorem:

The quadratic variation accumulated up to time $t$ by the Ito integral is:
$$ [ I,I](t) = \int_{0}^{t} \Delta^{2}(u) du.$$

The proof revolves around computing quadratic variation accumulated by the Ito integral on one of the subintervals $[t_j, t_{j+1}]$ on which $\Delta(u)$ is constant. They choose the partition points:

$$t_{j} = s_{0} < s_{1} < \ldots < s_{m} = t_{j+1} $$

and compute

$$\sum_{i=0}^{m-1}[I(s_{i+1}) - I(s_{i})]^2 = \sum_{i=0}^{m-1}[\Delta(t_j)(W(s_{i+1}) - W(s_{i}))]^2 $$

I can't figure out how we got the aforementioned equality involving Ito integral and Brownian Motion. Any help appreciated.

Here we have the following definitions: if $t_{k} \leq t \leq t_{k+1}$ then
$$I(t) = \sum_{j=0}^{k-1}\Delta(t_{j})[W(t_{j+1} - W(t_j)] + \Delta(t_{k})[W(t) - W(t_{k})]$$
The process $I(t)$ is the Ito integral of the simple process $\Delta(t)$ that is written as
$I(t) = \int_{0}^{t}\Delta(u)dW(u).$

Suppose $\Delta(t)$ is an adapted stochastic process i.e. it is $\mathcal{F}(t)$ measurable for each $t \geq 0.$ Assume further that $\Delta(t)$ is constant in $t$ on each subinterval $[t_{j}, t_{j+1}).$ Then $\Delta(t)$ is a simple process.

Here u is just the dummy variable and W(t) is the Brownian Motion.

答:
It's not clear if $\Delta (u)$ is deterministic or not. I'll suppose it is, and if it's not, I let you adapt the proof. I denote $\sum_{j}$ for $\sum_{j=0}^{m-1}$. Then,

\begin{align*}
\mathbb E\left[\left(\sum_{j}(I(t_{j+1})-I(t_j))^2-\int_{0}^t\Delta (u)^2\,\mathrm d u\right)^2\right]&=\mathbb E\left[\left(\sum_{j}\big(\Delta (t_j)(W_{t_{j+1}}-W_{t_j})\big)^2-\Delta (t_j)^2(t_{j+1}-t_j)\right)^2\right]\\
&=\mathbb E\left[\left(\sum_{j}\Delta (t_j)^2\big((W_{t_{j+1}}-W_{t_j})^2-(t_{j+1}-t_j)\big)\right)^2\right]\\
&=\sum_{i,j}\mathbb E[\Delta (t_i)^2\big((W_{t_{i+1}}-W_{t_i})^2-(t_{i+1}-t_i)\big)\Delta (t_j)^2\big((W_{t_{j+1}}-W_{t_j})^2-(t_{j+1}-t_j)\big)]\\
&\underset{(*)}{=}2\sum_{i}\Delta (t_i)^4(t_{i+1}-t_i)^2\\
&\leq 2\max_{i}|t_{i+1}-t_i|\int_0^t\Delta (u)^4\,\mathrm d u\underset{m\to \infty }{\longrightarrow }0,
\end{align*}
where in $(*)$ we used the fact that Brownian motion has independent increments and that if $X\sim \mathcal N(0,1)$, then $\mathbb E[X^4]=3$.


问:
Show that no random variable exists which has a kurtosis which is strictly less than 1
Show that no random variable exists which has a kurtosis which is strictly less than 1.
The kurtosis has the following formula:$\frac{E[(X-\mu _{X})^4]]}{(E[(X-\mu _{X})^2])^2}$

Can anyone help me proceed?

答: Put $\xi = (X- \mu_X)^2$. We know that $D\xi \ge 0$ and hence $E\xi^2 \ge (E\xi)^2$. Q.e.d.


问:
An Olympiad geometry problem ; angle chasing / trig bashing / coordinate bashing
D is an inner point in equilateral triangle $ABC$ such that $\angle ADC=150^{\circ}$.Prove that the triangle formed by taking segments $AD$,$BD$ and $CD$ as its three sides is a right angled triangle. ...... [Source : Some European Olympiad in 2003]

This is the first time I am trying a problem like this (proving that these three lines make a right triangle).

I did manage to prove the special case when D $\in BM$ where M is midpoint of AC using simple sin rule, in which case I got $AD=CD=\frac{BD}{\sqrt2}=\frac{\sqrt3 - 1}{\sqrt 2}a$ where $a$ is the side length.

I did manage to find a solution online by AMAR_04 (Geo god on aops) (https://artofproblemsolving.com/community/c869463h1833925_build_up_your_understanding2_p21) but I am not very comfortable with rotation yet.

So, if anyone can please provide a different solution (trig bash/ coordinate bash or anything else except complex) OR explain the solution above, I would be really greatful!

Thanks!

答:
Simple solution: by analytic geometry, the locus of points $P$ such that $PC^2=PA^2+PB^2$ is a circle.

This circle clearly goes through $A$ and $B$ and through some point $Q$ on the $AM$ median.  Once proved that $\widehat{AQB}=150^\circ$ we are done. Without words:


问:
Relationship between three matrices
I think this might be an odd question, and a little vague.  But here goes.

This is related to coordinate transformations.  Three matrices are given: $G_1 , G_2$, and $\Lambda$.  $G_1$ and $G_2$ are symmetric.  (They are metrics, actually.)  $\Lambda$ is the matrix in question.  The three matrices satisfy

$\Lambda^{-1} = G_1^{-1} \Lambda^T G_2$

Is there anything we can say about the properties of $\Lambda$, such as it is symmetric, or it is either symteric or anti-symmetric, or ...?

Thanks.

答:
If you rewrite this as $$G_1 = \Lambda^T G_2 \Lambda\,,$$
this is just the change-of-basis formula for a symmetric bilinear form. That is, $\Lambda$ is the matrix that interpolates from the expression for the metric in one coordinate system to the expression in another coordinate system. As it stands, $\Lambda$ could be any invertible matrix.</p>

Title: Relationship between three matrices


问:
To show $|(\mathbb{Z}/n\mathbb{Z})^{\times}| = \varphi(n)$
I know this question has been asked before, but this is (hopefully) not a duplicate. Dummit and Foote's Algebra asks us to show $$|(\mathbb{Z}/n\mathbb{Z})^{\times}| = \varphi(n)$$
using the following definitions, and nothing else:

$(\mathbb{Z}/n\mathbb{Z})^{\times} := \{\bar a \in \mathbb{Z}/n\mathbb{Z} : \exists \bar c \in \mathbb{Z}/n\mathbb{Z}, \bar a \bar c = \bar 1\}$

Here, $\bar a$ denotes the equivalence class of $a$ in $\mathbb{Z}/n\mathbb{Z}$, namely one of $\bar 0,\bar 1,...,\overline{n-1}$. $\bar a\bar c$ refers to the usual multiplication, i.e. we take any elements from the equivalence classes of $\bar a$ and $\bar c$, multiply them, and then reduce the result $\bmod n$. We have shown this works in an earlier proposition, i.e. modular multiplication is well-defined. So, $(\mathbb{Z}/n\mathbb{Z})^{\times}$ is just the collection of residue classes of $\mathbb{Z}/n\mathbb{Z}$ which have a multiplicative inverse in $\mathbb{Z}/n\mathbb{Z}$.

$\varphi(n)$ is Euler's totient function, as is popularly known. It is not defined using $|(\mathbb{Z}/n\mathbb{Z})^{\times}| = \varphi(n)$ in this text. Rather it is defined as $\varphi(n) = |\{a\in\mathbb{N}: a\leq n, (a,n)=1 \}|$, i.e. number of positive integers less than and co-prime with $n$. $(a,n) = 1$ stands for $\operatorname{g.c.d}(a,n) = 1$.

We have not yet shown that

$(\mathbb{Z}/n\mathbb{Z})^{\times} = \{\bar a \in \mathbb{Z}/n\mathbb{Z} : (a,n) = 1\}$

so please do not use this in your proof. Knowing this, it is trivial, but in a later exercise, we are supposed to prove this using $|(\mathbb{Z}/n\mathbb{Z})^{\times}| = \varphi(n)$ - so let's avoid circular arguments.

Could someone please point me in the right direction? Thanks!

答:
HINT. Note that a solution of
$$
{\displaystyle ax\equiv 1{\pmod {m}}.}
$$
exists if and only if $\gcd(a, m) = 1$.

Proof: Let $x$ be the solution. This is the same thing as saying $ax + my = 1$ for some $y \in \mathbb{Z}$. But if $a$ and $m$ have a common factor $d$, then clearly $d$ divides $1$. This indicates that $d = \pm 1$, so the $\gcd$ of $a$ and $m$ must be $1$. More information here.


答:
There is a more general question, which has been studied intensively, namely when can we say that $G=AB$ for subgroups $A,B$? Such groups $G$ are called factorizable and there is a large literature about them.

There are some trivial conditions, for example, that $AB$ is a subgroup of $G$ if and only if $AB=BA$, see

Let $A,B$ be subgroups of a group $G$. Prove $AB$ is a subgroup of $G$ if and only if $AB=BA$

References on factorizable groups: for example Arad, and many papers by Amberg,B. Franciosi, S. and Degiovanni and others, also papers by Gorenstein, Herstein.

For more references see also this MO post.

Title: What's the condition for a group $G$ to be equal to the product of two normal subgroups

Body: If $G$ is a group and $N,M$ are two normal subgroups. We know that the product $NM$ is normal subgroup of $G$, but when can I say that $G=NM$. What must be the conditions on $N,M$?


问:
How to present cases when writing a proof.
I was writing a proof where I wanted to show a statement y is true and ended up with two different cases to consider which were essentially: $x = 1$ and $x \neq 1$. I showed the result I wanted is true in both cases and I was just wondering how I would phrase the resulting conclusion. Essentially I want to say "We have $x = 1$ or $x \neq 1$.  Since y is true if $x = 1$ and and y is true if $x \neq 1$, y must be true." What would be the best way of phrasing this so my intention is clear?

答:
I'd probably write it like this:
"At this point of the proof, we have two different cases: $x=1$ and $x \neq 1$.
First, if $x=1$, then ..., and $y$ is true.
Otherwise, we have ... and $y$ is also true.
Since it is true in both cases, $y$ must be true."


问:
If $a \not\equiv 0\ (\mod m\ )$ while m is prime, can I say that : $(a,m) = 1$?
It seems logical to me but I can't prove it.</p>

答:
This implication is only true if $m$ is prime.

If $m$ is composite number, then you can take $a$ such that $a$

is not divisible by $m$ but $a$ and $m$ share some common factor $d \gt 1$.

Example 1: $m = 20$, $a = 14$.

$d=2$ is a common factor but $a$ is not divisible by $m$

Example 2: $m = 11$.

Now if you pick an $a$ which is not divisible by $m$,

then for sure $a$ is also relatively prime with $m$.

That's because $m$ is prime in this case.


答:
Make life easier using $x=y+\frac{\pi}{4}$
$$\lim \limits_{x \to \tfrac{\pi}{4}} \frac{4\sqrt 2 -\left ( \cos (x) + \sin (x) \right )^5}{1-\sin ({2x})}$$ Combine
$$\cos (x) + \sin (x)=\sqrt{2} \cos (y)$$ So, in terms of $y$, the expression reduces to
$$\frac{4\sqrt 2 -4 \sqrt{2} \cos ^5(y) )}{2 \sin ^2(y)}=2\sqrt 2\,\frac{1-  \cos ^5(y) }{ \sin ^2(y)}$$

Now Taylor series
$$\cos(y)=1-\frac{y^2}{2}+\frac{y^4}{24}+O\left(y^6\right)$$
$$\cos^5(y)=1-\frac{5 y^2}{2}+\frac{65 y^4}{24}+O\left(y^6\right)$$

Just finish

Title: conceptual doubt in finding $\lim \limits_{x \to \tfrac{\pi}{4}} \frac{4\sqrt 2 -\left ( \cos x + \sin x \right )^5}{1-\sin {2x}}$

Body:
While calculating the aforementioned limits I did the following.

$$\lim \limits_{x \to \tfrac{\pi}{4}} \frac{4\sqrt 2 -\left ( \cos x + \sin x \right )^5}{1-\sin {2x}}\\ =
\lim \limits_{x \to \tfrac{\pi}{4}} \frac{4\sqrt 2 -4\sqrt2 \left (\frac{1}{\sqrt 2} \cos x +\frac{1}{\sqrt 2} \sin x \right )^5}{1-\sin {2x}}\\=
\lim \limits_{x \to \tfrac{\pi}{4}} \frac{4\sqrt {2} \left ( 1 - \sin^5 (x + \frac{\pi}{4}) \right )}{1-\sin {2x}}\\
=\lim \limits_{x \to \tfrac{\pi}{4}} \frac{4\sqrt {2} \left ( 1 - \sin (x + \frac{\pi}{4}) \right ) \left(1 +\sin (x + \frac{\pi}{4}) +\sin^2 (x + \frac{\pi}{4}) +\sin^3 (x + \frac{\pi}{4}) +\sin^4 (x + \frac{\pi}{4}) \right)}{1-\sin {2x}}\\$$

As $x$ approaches $\frac{\pi}{4}$, the functions $\sin (x + \frac{\pi}{4})$ and $\sin 2x$ become close enough. So, I canceled them($1-\sin {2x}$ and $ 1 - \sin (x + \frac{\pi}{4})$ ) and the limit turns out to be $20\sqrt2$.

I made the above claim after comparing the graphs. But the fact that bothers me is L'Hospital's rule yielded the answer $5\sqrt2$ (and it is the correct answer to the problem)

Can anyone please explain?


答:
You can say $(a,m)\neq m$. That's it. Any other divisor of $m$ is possible to achieve by simply choosing that to be $a$

$a\equiv 0\pmod m$ means $a$ is a multiple of $m$. And $(a,m)=m$ means $a$ is a multiple of $m$. So if one isn't true, neither is the other.

Title: If $a \not\equiv 0\ (\mod m\ )$ while m is prime, can I say that : $(a,m) = 1$?

Body: It seems logical to me but I can't prove it.</p>