问:
Question about SIS model
Modify the following SIS model

```<code>S´= -bSI + aI
I´ = bSI - aI
</code>```

to the situation in which there are two competing strains of the same disease, generating two infective classes I1 , I2 under the assumption that coinfections are not possible. Does the model predict coexistence of the two strains or competitive exclusion?

I believe the modification should look like this:

```<code>S´= -b1SI1 -b2SI2 + a1I1 + a2I2
I1´ = b1SI1 - a1I1
I2´= b2SI2 -a2I2
</code>```

But the book I am reading doesn’t really talk about coexistence of strains or exclusions so I don’t really know where to go from here. The basic reproduction number R0 is pretty obvious but what do I need to check regarding it?

Thanks in advance

答:
I think that the exclusion of the possibility of getting both diseases means that you don't have to have any $I_1$ term in your $I_2'$ equation, and no $I_2'$ term in the $I_1'$ equation.  Likewise you don't have to have a separate $I_1I_2$ equation either. Your modification looks good.

$a_1$ is the recovery rate for people sick with strain $1$.
People become sick when they catch strain $1$ or strain $2$.
It looks like reinfection is possible in this model, as those who recover are returned to the "susceptible" category.

So long as the recovery rates are high enough that there is no possibility of everyone being infected at once, I don't see why either would die out, as there will always be more available susceptible hosts for either strain. I'm not sure how to prove it though.

Title: Question about SIS model


问:
What's the condition required in terms of roots for getting exact three integral negative values for this quadratic?
The number of integral values of a for which $ax^2 - (4-2a)x - 8<0$ for exactly three integral value of x .what i did was find the roots of this equation lets say $x_1$ ,$x_2$ ,will this condition satisfy the criteria: $2<|x_1 -x_2|<4$ and a>0? Or is there some more elegant way to get the required condition in terms of roots?

答:
The general quadratic
$Ax^2 + Bx + C = 0$

can be converted to
$\left(x + \frac{B}{2A}\right)^2 -
\frac{B^2 - 4AC}{(2A)^2} = 0.$

So, a given value of $x$ will satisfy
$Ax^2 + Bx + C < 0 \iff
\left(x + \frac{B}{2A}\right)^2 < \frac{B^2 - 4AC}{(2A)^2}.$

With this query's quadratic, you have that
$A,B,C = a, (2a-4), (-8)$ respectively.

This means that the values of $x$ that will satisfy the
inequality are
$$\left(x + \frac{2a-4}{2a}\right)^2 < \frac{(2a-4)^2 + 32a}{(2a)^2}
= \frac{(2a+4)^2}{(2a)^2}.\tag1$$

Since it is given that $a > 0$, (1) Can be re-expressed as

$$- \frac{2a+4}{2a} < x + \frac{2a-4}{2a} < \frac{2a+4}{2a}$$

which implies that

$$-1 - \frac{2}{a} < x + 1 - \frac{2}{a} < 1 + \frac{2}{a} $$

which implies that

$$-2 < x < \frac{4}{a}. \tag2$$

This means that the positive integer values of $a$ must be found such that inequality (2) above will have exactly 3 satisfying integer values of $x$.

It is immediate that inequality (2) above will be satisfied by
$\{-1,0\}$, regardless of the value of $a$.  Therefore, the problem reduces to identifying for which values of $a$,
$\frac{4}{a} > 1.$

Edit

Thanks to heropup for noticing a mistake of mine.

The actual constraint should be

$1 < \frac{4}{a} \leq 2$, since $x \geq 2$ must be prevented.

Noting that $a$ is required to be a positive integer, this means that $a \in \{2,3\}.$  Therefore, there are exactly 2 satisfying positive integer values for $a$.

ContentLicense: CC BY-SA 4.0


问:
Find the GCD of a polynomial using the extended Euclidean algorithm and express it in the form $a(x)f(x)+b(x)g(x)$
I am working on a problem that I can not seem to finish.

Find the gcd of $f(x)=x^7+1$ and $g(x)=x^6+x^5+x^3+1$, and express it in the form $a(x)f(x)+b(x)g(x)$ using the extended euclidean algorithm.

I have performed the extended euclidean algorithm to find the following
$$f(x)=(x-1)g(x)+(x^5-x^4+x^3-x+2)$$
$$g(x)=(x+2)(x^5-x^4+x^3-x+2)+(x^4-x^3+x^2-3)$$
$$x^5-x^4+x^3-x+2=(x)(x^4-x^3+x^2-3)+(2x+2)$$
$$x^4-x^3+x^2-3=(\frac{1}{2}x^3-x^2+\frac{3}{2}x-\frac{3}{2})(2x+2)+0$$
This should give us the GCD as following
$$\gcd(f,g)=\gcd(x^5-x^4+x^3-x+2, 2x+2)=2x+2$$
And the monic GCD then becomes
$$\gcd(f,g)=x+1$$

This is where I get stuck. Now that I have this I am not sure how to algebraically get back to the form $a(x)f(x)+b(x)g(x)$

The answer as provided in the solutions is
$$\gcd(f(x),g(x))=x^4+x^3+x^2+1=xf(x)+(x^2+x+1)g(x)$$

I do not understand how to proceed and get to the answer. I have tried going backward but I am not getting anywhere. How do I go from $x+1$ to the correct solution?

答: There is an interesting way to find the solutions for "Bezout's equation" $ax+by=mdc(a,b)$ using continued fractions. For instance, for numbers: find $(x,y) \in \mathbb Z^2$ s.t. $11x + 7y = 1$. The continued fraction of $11/7$ is $[1;1,1,3]$. By removing the last digit, $[1;1,1] = 3/2$, we see that $11 \cdot 2 + 7 \cdot (-3) = 1$, therefore $(x,y) = (2,-3)$ satisfy the equation $11x + 7y = 1$. The thing is: the algorithm is exactly the same for polynomials!


问:
Prove that if $A|B$ and $B|A$, then $A=B$?
Attempt:

Prove that if $A|B$ and $B|A$, then $A=B$? Both $A$ AND $B$ are positive integers.

I tried to substitute with the fact that $A|B$ means $\exists c$ s.t $Ac_2=B$ and same $Bc_1=A$ when $B|A$, but I did not get anywhere.

答: If $A=c_1 B$ and $B= c_2A$, then we get $A = c_1B = c_1 c_2 A$, and if all these are non negative integers, it follows that $c_1 =c_2 =1$.


问:
Need help understanding Erdős' proof about divergence of $\sum\frac1p$
I'm looking at proofs from Proofs from the Book (Martin Aigner, Günter M. Ziegler). The proof I'm having trouble is the sixth proof of the infinitude of the primes they give (on page 5; although I'll be reconstructing it here). I'll be highlighting the areas I'm having trouble with. They set out to prove two things; namely, and letting $P$ be the set of all primes:

$P$ is infinite.

$\sum_{p \in P} {1 \over p}$ diverges.

They begin by assuming the sum in 2. is convergent and letting $P = \{p_1, p_2, ...\}$ The first problem comes up here. By assuming the given form for $P$, haven't they already assumed that it is infinite? Then, by assumption,

$$\sum_{i > k} {N \over p_i} < {N \over 2}, \ \  (N \in \mathbb{Z}^+). \tag{1}$$

They then define $p_1, p_2, ..., p_k$ to be the small primes, and $p_{k + 1}, p_{k + 2}, ...$ to be the large primes, and define $N_b$ to be the number of positive integers $n \leq N$ which are divisible by at least one of the big primes; and, similarly, $N_s$ the number of positive integers $n \leq N$ which only have small prime divisors. By definition, $N_b + N_s = N$, and the intended contradiction will hinge on showing $N_b + N_s < N$.

Note, since $\left \lfloor {N \over p_i} \right \rfloor$ gives the number of positive integers $n \leq N$ which are multiples of $p_i$,

$$N_b \leq \sum_{i > k} \left \lfloor {N \over p_i} \right \rfloor < {N \over 2}. \tag{2}$$

Then, they let $n = a_n b_n^2$, where $n \leq N$ such that it has only small prime divisors. Here $a_n$ is the square free part. Then, every $a_n$ is a product of different small primes. Assuming that by this it is simply meant that every factor of $a_n$ is distinct, they assert there are exactly $2^k$ different square free parts. Notice, also, $b_n \leq \sqrt{n} \leq \sqrt{N} \implies b_n \leq \sqrt{N}$, and as such,

$$N_s \leq 2^k \sqrt{N}. \tag{3}$$

Here, is the second problem for me. I can see intuitively how $(3)$ holds; but, I'm not really sure, and I'd like to see it through a more rigorous approach, and also, why it's relevant. I'm not really sure, and I'd like a more rigorous approach.

They conclude the proof by showing that, since $(2)$ holds for any $N$, $2^k \sqrt{N} \leq {N \over 2}$ holds for $N = 2^{2k + 2}$. This is the third area I'm having trouble. I'm not really sure why this is relevant and how the contradiction $N_s + N_b < N$ follows from it. Maybe I missed it, but I also didn't see a proof of the infinitude of primes here.

Finally, I have a question as well. Looking back, how could one have guessed that this approach will lead us to a proof of the two claims above? In other words, what's the insight behind taking this particular approach?

Thanks to all in advance!

Note: I wasn't really sure what tags relating to "Proofs" would be appropriate here, so I've only used tags concerning number theory and primes. Please feel free to tag away.

答:
First Problem

Initially, no assumption is made about $P = \{p_1, p_2 \dots \}$ being finite or infinite. The aim of the proof is to show that $\sum_{p \in P} 1/p$ diverges.  Once this is established, it follows immediately that $P$ is infinite, since otherwise, $\sum_{p \in P} 1/p$ would be a finite sum.

Second Problem

Every element $n \le N$ with only small prime divisors is of the form $a_nb_n^2$.  The numer of distinct $a_n$ is given by the number of subsets of $\{p_1, \dots, p_k\}$ which is exactly $2^k$. The inequality, $b_n \le \sqrt{N}$ implies there are at most $\sqrt{N}$ distinct $b_n$.  So putting the two together, we get

$$N_s \le (\text{# of choices of } a_n) \times (\text{# of choices of } b_n) \le 2^k\sqrt{N}$$

Third problem

By letting $N = 2^{2k + 2}$, we have the bound $N_s \le 2^k\sqrt{N} \le N/2$.  Since $N_b < N/2$,

$$N_s + N_b < \frac{N}{2} + \frac{N}{2} = N$$

However, by construction $N_b + N_s = N$, and so we have reached a contradiction.
Therefore, $\sum_{p \in P} 1/p$ must diverge.

ContentLicense: CC BY-SA 3.0


问:
Definition of direct sum of Hilbert spaces
Let $\mathcal{H}=(\mathcal{H},(\cdot, \cdot)_{\mathcal{H}})$ be a Hilbert space and $\mathcal{H}_1, \mathcal{H}_2 \subset \mathcal{H}$ subspaces. When I write $\mathcal{H}=\mathcal{H}_1\oplus\mathcal{H}_2$ this means that $\mathcal{H}_1$ and $\mathcal{H}_2$ are closed subspaces of $\mathcal{H}$ such that

$(1)$ for every $z\in\mathcal{H}$ there exist $x\in\mathcal{H}_1$ and $y\in\mathcal{H}_2$ such that $z=x+y$.

$(2)$ The elements  $x\in\mathcal{H}_1$ and $y\in\mathcal{H}_2$ in $(1)$ are unique.

Question. The item $(2)$ is equivalent to saying that $(x,y)_{\mathcal{H}}=0$?

In the second answer of this question it is said that $(x,y)_{\mathcal{H}}=0$ implies that $x$ an $y$ are uniquely determined. So, the converse holds?

答:
The converse doesn't hold.

As an example, take $H = \Bbb R^2$ equipped with the Euclidean norm, i.e. $|(x, y)|^2 = x^2  + y^2$.

Take $H_1 = \Bbb R\cdot (1, 0)$ and $H_2 = \Bbb R\cdot (1, 1)$.

It is clear that they satisfy the conditions (1) and (2) in your question. These conditions in fact requires that $H = H_1 \oplus H_2$ just as vector spaces.

However this is not a direct sum of Hilbert spaces, as $v = (1, 0) \in H_1$ and $w = (1, 1)\in H_2$ are not orthogonal.


问:
Questions about SVD, Singular Value Decomposition
I am not a mathematician, so I need to understand what SVD does and WHY more than how it works exactly from the math perspective. (I understand at least what is the decomposition though).

This guy on youtube gave the only human explanation of SVD saying, that the U matrix maps "user to concept correlation" Sigma matrix defines the strength of each concept, and V maps "movie to concept correlation" given that initial matrix M has users in the rows, and movie (ratings) in the columns.

He also mentioned two concept specifically "sci fi" and "romance" movies. See the picture below.

My questions are:

How SVD knows the number of concepts. He as human mentioned two - sci fi, and romance, but in reality in resulting matrices are 3 concepts. (for example matrix U - that one with blue titles - has 3 columns not 2).

How SVD knows what is the concept after all. I mean, what If i shuffle the columns randomly how SVD then knows what is sci fi, what is romance. I mean, I suppose there is no rule, group the concepts together in the column order. What if scifi movie is the first and last one? and not first 3 columns in the initial matrix M?

What is the practical usage of either U, Sigma or V matrices? (Except that you can multiply them to get the initial matrix M)

Is there also any other possible human explanation of SVD than the guy up provided, or it is the only one possible function? Matrices of correlations.

答:
A matrix, $A \in \mathbb{R}^{M \times N}$, maps vectors in $\mathbb{R}^N$ to vectors in $\mathbb{R}^M$ via matrix multiplication,
$$v \mapsto A v$$
If you multiply the matrix with every vector on the unit sphere in $\mathbb{R}^N$, the resulting vectors you get out will form an ellipsoid in $\mathbb{R}^M$. The primary axes of the ellipsoid in $\mathbb{R}^M$ are orthogonal.
Also, it is possible to prove that the input vectors that get mapped to those primary axes are orthogonal in $\mathbb{R}^N$.

The singular value decomposition $A = U \Sigma V^T$, is an encoding of the sphere and the ellipsoid.

The left singular vectors, $u_i$, are primary axes of the ellipsoid, except scaled to be unit length.

The singular values, $\sigma_i$, are the lengths of the primary axes of the ellipsoid

The right singular vectors, $v_i$, are the unit vectors on the sphere that get mapped to the primary axes of the ellipsoid.

This was all assuming the dimensions of the input and output are the same. But the idea extends to the case where dimensions of the input and output are different ($M \neq N$).

If the input dimension is larger than the output dimension ($N > M$), then there are $N-M$ orthogonal vectors on the unit sphere that get squashed to zero, but in the other $M$ complementary directions everything is the same as described above.

If the output dimension is larger than the input dimension ($M > N$), then the output is still an ellipsoid, but it is contained in a $N$-dimensional hyperplane, and is totally flat in the $M-N$ directions complementary to the hyperplane.


问:
Evaluating $\lim_{x\to-1}\frac{2x^2+5x+3}{2-\sqrt{2+\sqrt{3-x}}}$

What is the value of the limit:
$$\lim_{x\to-1}\frac{2x^2+5x+3}{2-\sqrt{2+\sqrt{3-x}}}$$
$1)8\quad\quad\quad\quad\quad2)12\quad\quad\quad\quad\quad3)16\quad\quad\quad\quad\quad4)24$

I evaluated the limit by L'Hopital rule:
$$\lim_{x\to-1}\dfrac{4x+5}{\dfrac1{2\sqrt{3-x}\times2\sqrt{2+\sqrt{3-x}}}}=\dfrac{1}{\dfrac{1}{16}}=16$$
But I wonder is there other approach to evaluate the limit?

答:
HINT

What about multiplying by the conjugate twice?

\begin{align*}
\frac{1}{2 - \sqrt{2+\sqrt{3-x}}} = \frac{2 + \sqrt{2+\sqrt{3-x}}}{2 - \sqrt{3-x}}
\end{align*}

Similarly, we have
\begin{align*}
\frac{1}{2-\sqrt{3-x}} = \frac{2 + \sqrt{3-x}}{1 + x}
\end{align*}

Finally, notice that
\begin{align*}
2x^{2} + 5x + 3 & = (2x^{2} + 2x) + (3x + 3)\\\\
& = 2x(x+1) + 3(x+1)\\\\
& = (2x+3)(x+1)
\end{align*}

Can you take it from here?


问:
Question on multivariable chain rule (voltage-current equation of a capacitor with time-, voltage- and current-dependent capacitance)
This is not homework. I want to obtain the equation that relates the voltage across a capacitor with the current through that capacitor, where the capacitance is a function of the voltage, current and time; and where the voltage and current are a function of time. To explicitly indicate that, I'm going to denote those quantities as $C(i,v,t)$, $v(t)$ and $i(t)$. We don't know the specific expressions for $v(t)$, $i(t)$ or $C(i,v,t)$, because I want to obtain a general formula, we just know the dependencies I said. Also, I want to write the equation such that $i(t)$ is in terms of $C(i,v,t)$ and $v(t)$ (and possibly of the charge $q(t)$).

First, we begin with the definition of capacitance of a two-terminal ideal capacitor, which as we know it's:

$C(i,v,t) = \dfrac{q(t)}{v(t)} \tag 1$

Multiplying both sides by $v(t)$, we get:

$C(i,v,t) \, v(t) = q(t)  \tag 2$

Differentiating both sides with respect to time, we get:

$\dfrac{\mathrm dq(t)}{\mathrm dt} = \dfrac{\mathrm d}{\mathrm dt} [C(i,v,t) \, v(t)] \tag 3$

Applying the definition of current ($i(t) = \mathrm dq(t)/\mathrm dt$) in left-hand side of the previous equation, we get:

$i(t) = \dfrac{\mathrm d}{\mathrm dt} [C(i,v,t) \, v(t)] \tag 4$

Notice if the capacitance was constant, then we could take it out of the derivative and we'd end up with the typical current-voltage equation of a capacitor:

$i(t) = C \, \dfrac{\mathrm dv(t)}{\mathrm dt} \tag 5$

But we're assuming the capacitance is a function of voltage, current and time, so we can't apply that property, so equation (5) is not valid in our case.

So far I don't have questions. If I made an error, please tell me. Going back to equation (4), I'm not exactly sure how to decompose the derivative of the right-hand side. In my Calculus 3 class, I was taught that if $w = w(x,y)$, where at the same time $x = x(t)$ and $y = y(t)$, then:

$\dfrac{\mathrm dw}{\mathrm dt} = \dfrac{\partial w}{\partial x} \dfrac{\mathrm dx}{\mathrm dt} + \dfrac{\partial w}{\partial y} \dfrac{\mathrm dy}{\mathrm dt} \tag 6$

However, in equation (4), we identify $x$ as $C$ and $y$ as $v$, so we have $w = w(C,v) = C \, v$, where at the same time $C = C(i,v,t)$ and $v = v(t)$ and $i = i(t)$. Notice the difference between this and the previous example is that in the previous example, $x$ was a function of one variable, while in my case it's of three variables. I think the following is true:

$\begin{align} \dfrac{\mathrm dw}{\mathrm dt} &= \dfrac{\partial w}{\partial C} \dfrac{\mathrm dC}{\mathrm dt} + \dfrac{\partial w}{\partial v} \dfrac{\mathrm dv}{\mathrm dt} \\ &= \dfrac{\partial w}{\partial C} \dfrac{\partial C}{\partial t} + \dfrac{\partial w}{\partial v} \dfrac{\mathrm dv}{\mathrm dt} && \text{$C$ is a function of three variables so we use $\partial$} \\ &= \dfrac{\partial [C \, v]}{\partial C} \dfrac{\partial C}{\partial t} + \dfrac{\partial [C \, v]}{\partial v} \dfrac{\mathrm dv}{\mathrm dt} && \text{Substituting the expression for $w$} \\ &= v \dfrac{\partial C}{\partial t} + C \dfrac{\mathrm dv}{\mathrm dt} && \text{Evaluating} \\ \implies \dfrac{\mathrm d}{\mathrm dt} [C \, v] &= v \dfrac{\partial C}{\partial t} + C \dfrac{\mathrm dv}{\mathrm dt} \tag 7 \end{align}$

Is equation (7) true/correct? Because if so, finally we substitute its right-hand side into the right-hand side of equation (4) to get:

$\boxed{i(t) = v(t) \dfrac{\partial C(i,v,t)}{\partial t} + C(i,v,t) \dfrac{\mathrm dv(t)}{\mathrm dt}} \tag 8$

Based on the textbook Network analysis by Mac E. Van Valkenburg, it looks like equation (8) is correct at least when the capacitance is a function of time but not of voltage nor current (notice in the following image he used a complete differential instead of a partial one when differentiating $C(t)$), and still assuming the voltage and current are a function of time, since in such case equation (8) reduces to:

$i(t) = v(t) \dfrac{\mathrm dC(t)}{\mathrm dt} + C(t) \dfrac{\mathrm dv(t)}{\mathrm dt} \tag 9$

Here's what the author says:

The following example from that textbook illustrates the equation (1-21) when the voltage is constant (notice he writes with an uppercase $V$, which is common when voltage is a constant):

答:
$$i(t) = \frac{d}{dt} [C(i(t),v(t),t) \, v(t)]$$

By the product rule,

$$i(t) = v(t) \, \frac{d}{dt} [C(i(t),v(t),t)] + C(i(t),v(t),t) \, \frac{dv}{dt}$$

By the chain rule for total derivatives,

$$\frac{d}{dt} [C(i(t),v(t),t)] = \frac{\partial C}{\partial t} + \frac{\partial C}{\partial i} \frac{d i}{d t} + \frac{\partial C}{\partial v} \frac{d v}{d t}$$


问:
finite union of connected dense subsets of $\Bbb R^2$
For each $i\in\{1,2,\dots ,n\}$ define a set $$A_i:= \bigg\{(x,y)\colon x\in(k_i,k_i+1) , y\neq 0\bigg\}$$, to be dense and connected in $(k_i,k_i+1)\times\Bbb R$ where the $k_i$ are integer numbers . Put$$ M:=\bigg\{(x,0)\colon x\in\Bbb R\setminus\bigcup_{i=1}^{n}(k_i,k_i+1)\bigg\}.$$ Now, $$D:= \bigcup_{i=1}^{n} A_i \cup M$$ I am sure that $D$ is still a connected subset of $\Bbb R^2$ since the density of $A_i$ will ensure that they will have a limit point in common, By using the well known fact that says :If $B\subset\Bbb R^2$ is connected and $x$ is a limit of of $B$, then $B\cup \{x\}$ is still connected.

My question I need to see if my explanation is strong enough to show that $D$ is connected subset $\Bbb R^2$,  Thank you in advance.

Edit: The original was missing an important assumption for $A_i$ which is $y\neq 0$

答:
1) this is the answer before you added the condition that $y \neq 0$ to the definition of the $A_i$:

Your statements about density and limit points aren't really the important things here. To see that $D$ is connected, show that it contains the $x$-axis, $X$ say, which meets each of the connected sets $A_i$. As the $x$-axis is connected this means that $D$ is connected, because you can write $D$ as:
$$
D = \bigcup_{i=1}^n (A_i \cup X)
$$
which displays $D$ as the union of a family of connected sets $X_i = A_i \cup M$ such that the intersections $X_i \cap X_j$ are all non-empty.

2) now you have added the condition that $y \neq 0$ to the definition of the $A_i$, the set $D$ remains connected:

To see this, let $D'$ be $D$ as it was without the restriction that $y\neq 0$ in the definition of the $A_i$. Then

$$
D = D' \setminus \bigcup_{i=1}^n (k_i, k_i + 1)\times \{0\}
$$

Removing these open line segments from $D'$ does not disconnect it, in the same way that removing a rung from a ladder (with at least two rungs) does not disconnect it).

To give a more formal proof, observe that you can replace each $A_i$ in the definition by:
$$
B_i:= \left\{(x,y)\colon x\in(k_i,k_i+1) , y\neq 0 \mbox{ or } x \in \{k_i, k_i+1\}\right\}
$$
since the points I have added to $A_i$ to form $B_i$ are in $M$. It is easy to see that the $B_i$ are path-connected (or you can use your argument about limit points to display $B_i$ as the union of two intersecting connected sets). The closed line segments in the set $M$ then join each $B_i$ to its east and west neighbours. To be even more formal, you can show by induction that $D$ can be constructed by repeatedly taking the union of pairs of intersecting connected sets.


问:
Prove $C(E)$ is a Banach space with E a compact topological space
Let be E a compact topologycal space. I know that $C(E)\subset B(E)$ with $C(E)=\{f:E\rightarrow \mathbb C \quad$continuous$\}$ and $B(E)=\{f: E\rightarrow \mathbb C \quad$continuous and bounded$ \}$.

$B(E)$ is a Banach space so $C(E)$ is a Banach space because subset of a Banach space?

答:
You can verify, without too many difficulties, that $C(E)$ is a vector space (just like $B(E)$), according to the definition of vector space. Furthermore, with the supremum norm, $C(E)$ is also a normed vector space (as $B(E)$). As $\mathbb{C}$ is a Banach space, with the norm sup $C(E)$ it is also a Banach space (just like $B(E)$ is also).

In addition, you can prove the following proposition (with a 3-$\epsilon$ argument, like in tis post):

Proposition. The $B(E)$ space is a closed subspace of $C(E)$, that is, the uniform limit of continuous bounded functions is continuous.

As the E space is compact, so the reciprocal of the proposition is valid, in the sense that we also have $C(E)\subset B(E)$.


问:
Definability in hereditarily finite and countable sets
Suppose a set $X\subseteq H(\omega_1)$ of hereditarily countable sets is parameter-free definable in the structure $(H(\omega_1),\in)$ by a formula in the language of set theory, say $\phi(v)$, i.e.
$$(H(\omega_1),\in)\models\phi[x]\text{ iff }x\in X$$
for all $x\in H(\omega_1)$.

Can we then define the set $X\cap H(\omega)$, i.e. the subset of all hereditarily finite sets in $X$ in the structure $(H(\omega),\in)$ of hereditarily finite sets, i.e. is there a $\psi$ such that
$$(H(\omega),\in)\models\psi[x]\text{ iff }x\in X\cap H(\omega)$$
for all $x\in H(\omega)$?

答: No. For example, the set of (Gödel numbers of) sentences true in $(H(\omega),\in)$ is definable in $(H(\omega_1),\in)$ but not in $H((\omega),\in)$.


问:
How do I find the asymptote of $y= -\tan(x+\pi/4)$?
$$y= -\tan(x+\pi/4).$$
How do I find the asymptote of this?

答:
We need to find asymptotes of $$y = -\tan \left (x + \frac{\pi}{4} \right).$$

It is well-known that the $\tan$ function is undefined whenever $\cos(x) = 0$, which happens at $x = \frac{\pi}{2} + \pi k$ for integer $k$.

Hence, we must have: $$x + \frac{ \pi}{4} = \frac{\pi}{2} + \pi k.$$

Simplifying, we get: $$x = \pi k + \frac{ \pi}{4}.$$

Thus, the asymptotes are $$x = \pi k + \frac{\pi}{4}$$ for integer $k$.


问:
A modified secretary problem where time is of essence
The famed secretary problem asks when you should decide to hire, given that you cannot postpone the decision, and that you can objectively rank the candidates. I've seen that the solution for getting the best expected value candidate is to stop after $\sqrt{N}$ candidates, for a given $N$ total candidates.

I'm wondering if there are any closed solutions for the following modification:

There are an infinite amount of secretaries, but the rate they apply is constant, say $f$.

Time has value - so that one is willing to compromise in exchange for less time without a secretary. That is, one is interested in maximizing the expected value of $X_t - g(t)$, where $g(t)$ is some increasing function of time.

I feel this problem is a more realistic model of the real world problem, but I haven’t seen a solution for it. Any ideas?

答:
This kind of problem is known as "optimal stopping," and there's an entire theory of optimal stopping with lots of textbooks and courses available.  Some of the variants you've described above are analyzed here, for instance:

http://www.math.ucla.edu/~tom/Stopping/sr6.pdf


问:
Invariant factors of a cyclic $K[X]$-module and of its dual
Let $Q\in\mathbb{K}[X]$ be a non zero polynomial of degree $m$. ($\mathbb{K}$ is a field.)

We consider the $\mathbb{K}[X]$-module $M_Q = \mathbb{K}[X] / (Q) $.

What are the invariant factors of $M_Q$ and ${M_Q}^*$ (the dual space of $M_Q $)?

Does there exist an example of a $\mathbb{K}[X]$-module which is of finite dimension over $\mathbb{K}$ such that $M$ and $M^*$ are not isomorphic?

Does there exist a square matrix $A \in M_n(\mathbb{K}) $ which is not similar to its transpose $ A^T $ ?

So most examples of invariants factors I see are with modules over $ \mathbb{Z} $ or $ \mathbb{Z}[i] $ but here we are looking at a module over a polynomial ring.

I have been trying to find the presentation matrix of $M_Q$ but I'm not sure how to proceed...

As $ M_Q = \left< \overline{1}, \overline{X}, \overline{X^2}, ... , \overline{X^{m-1}} \right> $ first we need to solve $ \ker(\varphi) $ with

$ \varphi :  \mathbb{K}[X]^m \rightarrow M_Q $

$  (r_i)_{i=0,1,\dots,m-1} \mapsto \sum_{i=0}^{m-1} r_i \overline{X^i} $

which here is just a surjective homomorphism as $ M_Q$ may not be a free module.

Would $ r_i = QX^k $ with $ 0 \leq k \leq m-1 $ fit for the kernel?

Thanks in advance for any insights given on the matter.

答:
For the question 2.

$M_Q^*= Hom_K(M_Q,K)$ with its natural $K[X]$-module structure $(X\cdot f)(a)=f(X  a)$.

Factorize $Q=\prod_j P_j^{e_j}$, let $R_i=P_i^{e_i-1}\prod_{j\ne i} P_j^{e_j}\in K[X]/(Q)$ and take $f\in M_Q^*$ such that $f(R_i)\ne 0$ for each $i$.

Then $$\ker(K[X]\to K[X]\cdot f )= (Q)$$
Therefore $\dim_K(K[X]\cdot f)= \dim_K(M_Q)=\dim_K(M_Q^*)$ so that as $K[X]$-modules $$ M_Q^*=K[X]\cdot f \cong K[X]/(Q)=M_Q$$
Such a $f$ exists because we can take $f_i\in Hom_K(K[X]/(P_i^{e_i}),K), f_i(R_i)=1$ and $f=\sum_i f_i$.


问:
$\space \lim_{n \to \infty} s_n(x)=0$ and $ \lim_{n \to \infty} \int_{0}^{1}s_n(x)\,dx=1.$ Possible for step functions?
Question : Show that there exists step functions

$\space$ {$ \space {s_n : [0,1] \to R }\space  $}${_{n=1}^{\infty}}$ such that for any $x\in(0,1)$ we have $\space \lim_{n \to \infty} s_n(x)=0$  and $\space \lim_{n \to \infty} ({ \int_{0}^{1}{(s_n(x)\,dx)} }=1.$

Corrected approach for the Question : for all $ x \in (0,1)$

Let $f:[0,1]\to\mathbb{R}$ be such that $\int_{0}^{1}f(x)\,dx=1$. For each $n\in\mathbb{N}$, define $f_{n}:[0,1]\to\mathbb{R}$ by
$$S_{n}(x)=
\begin{cases}
    n-nx & 0 \leq x \leq \tfrac{1}{n} \\
    0      & \tfrac{1}{n} < x \leq 1
\end{cases}$$

as I need to prove such that for any $x\in(0,1)$ we have $\space \lim_{n \to \infty} s_n(x)=0$  and $\space \lim_{n \to \infty} ({ \int_{0}^{1}{(s_n(x)\,dx)} }=1.$

$x \in (0,1)$ = $x \in (0, \frac{1}{n}] \space U \space (\frac{1}{n}, 1)$

Case 1 : for $\frac{1}{n}<x<1$

as $\space {n \to \infty},  \space  \frac{1}{n} \to 0$ and $\space \lim_{n \to \infty} S_n(x) = \lim_{n \to \infty} 0=0$

Case 2 : for $0<x \le \frac{1}{n}$

as $\space {n \to \infty},  \space  \frac{1}{n} \to 0$ and $\space \lim_{n \to \infty} S_n(x) = \lim_{n \to \infty} n-nx $

but here’s an observation through number line for case 2 :

as $\space {n \to \infty} \space  \frac{1}{n} \to 0     $ for a large value of n,  $\space \frac{1}{n}<x$ . $\space$  So , $\space \lim_{n \to \infty} S_n(x) = \lim_{n \to \infty} 0=0$

Integrating step :

$\lim_{n \to \infty}  $$\int_{0}^{1}S_{n}(x)\,dx = \int_{0}^{1/n}(n-nx)\,dx + \int_{1/n}^{1}0\,dx$

$\space \space$= $\lim_{n \to \infty}[{nx}-\frac {nx^2}{2}]_{0}^{1/n} + (\lim_{n \to \infty} 0 = 0)$

$\lim_{n \to \infty}  $$\int_{0}^{1}S_{n}(x)\,dx $ = $\lim_{n \to \infty} 1-\frac{1}{2n}= 1 - 0$

$\lim_{n \to \infty} $$\int_{0}^{1}S_{n}(x)\,dx $ = $\lim_{n \to \infty} 1 = 1 $

Thanks for the response guys, I changed and took an  approach through your suggestion guys. Do correct if needed

答:
First, remember that step functions are piecewise constant functions with finitely many pieces. So $g(x)=x^{n}$ is not a step function, but something like
$$g(x)=
\begin{cases}
    1 & 0 \leq x \leq \frac{1}{3} \\
    3 & \frac{1}{3} < x \leq \frac{2}{3} \\
    6 & \frac{2}{3} < x \leq 1
\end{cases}
$$
is a step function.

Here is an idea for you to think about. Let $f:[0,1]\to\mathbb{R}$ be such that $\int_{0}^{1}f(x)\,dx=1$. For each $n\in\mathbb{N}$, define $f_{n}:[0,1]\to\mathbb{R}$ by
$$f_{n}(x)=
\begin{cases}
    nf(nx) & 0 \leq x \leq \tfrac{1}{n} \\
    0      & \tfrac{1}{n} < x \leq 1
\end{cases}$$
What is $\int_{0}^{1}f_{n}(x)\,dx$? For a given $x\in(0,1)$, what is $\lim_{n\to\infty}f_{n}(x)$? Can you apply this idea to make the desired sequence of step functions?


问:
Probability: Expected Value definition question
Can anyone explain how the following equality is true?

a)
$$
E[X + Y] = E[X] + E[Y]
$$

I considered the following definitions:

b)
$$
E[X] = \sum_{i=1}^k c_i p_X(c_i)
$$
c)
$$
p_X(z) = P(\{s \in \Omega : X(s) = z\}) = P(X = z)
$$

d)
$$
E[X + Y] = \sum_{i=1}^k c'_i f_{X+Y}(c'_i)
$$

e)
$$
p_{X+Y}(z) = P(\{s \in \Omega : X(s) + Y(s) = z\}) = P(X + Y = z)
$$

For example if form (c) has mutually exclusive sets of elementary events that satisfy the random variable equality.  $p_X(z) = P(s_1 \subset \Omega)$ and $p_Y(z) = P(s_2 \subset \Omega)$ and $s_1 \cap s_2 = 0$ then how can this be true, $E[X] + E[Y] = E[X+Y]$?

答:
$X$ and $Y$ don't have to have the same sample space/support, i.e. $X$ can have sample space $\Omega_1$ and $Y$ sample space $\Omega_2$. Further, if $z\in\Omega_1\cap\Omega_2$, then in the discrete case $f_X(z)=Pr(s_1\subset \Omega_1)=Pr(\{z\}\subset\Omega_1)$ and $f_Y(z)=Pr(s_2\subset \Omega_2)=Pr(\{z\}\subset\Omega_2)$ and so $s_1=s_2=\{z\}$ and so $s_1\cap s_2\ne 0$.

But it is not necessary that $\Omega_1\cap\Omega_2\ne0$. From the definitions above, $E(X)+E(Y)=E(X+Y)$ is $\sum_{i=1}^k c_i f_X(c_i)+\sum_{i=1}^{k''}c_i''f_Y(c_i'')=\sum_{i=1}^{k'}c_i'f_{X+Y}(c_i')$.

Example:

$X$ is $1$ with probability $1$, $Y$ is $2$ with probability $1$. Then $X+Y=3$ with probability 1. We have $E(X)+E(Y)=1\cdot 1+2\cdot 1=3$ and $E(X+Y)=3\cdot 1=3$. But $\left(\Omega_1=\{1\}\right)\cap\left(\Omega_2=\{2\}\right)=0$. Suppose instead that $Y=1$ with probability $1$. Then $X+Y=2$ with probability 1 and $f_X(1)=Pr(\{1\}\subset\Omega_1)$ and $f_Y(1)=Pr(\{1\}\subset\Omega_2)$ and so it is not possible that $s_1\cap s_2=0$ as $s_1=s_2$.


问:
Why this equation is able to be said as the condition for composing a constant function?
$a,b,c,d:=$non zero real number constants.

$t\geq0$

$f(t):=\frac{a+bt}{c+dt}$

The thing which I want to do is determine the values of the above $4$ constants to make $f(t)$ as a constant function.

The textbook gave the condition for it as below.

$\frac{a}{c}=\frac{b}{d}$

Why this equation is able to be said as the condition for composing a constant function?

答:
We want $ f(t) = \frac{a + bt}{c + dt} = e$, where $e$ is some constant

This should hold for all values of t. So taking $t = 0$

$ \frac{a}{c} = e \Rightarrow a = ec$

If $t = 1$:

$\frac{a+b}{c+d} = e \Rightarrow a + b = (c + d)e = ec + ed = a + ed$

$\Rightarrow b = ed \Rightarrow e = \frac{b}{d}$

$\Rightarrow \frac{a}{c} = \frac{b}{d}$


问:
How do I show that $P_{range(A)}$ is an orthogonal projector onto A?
I have the first part of the proof, which is that P is indeed an orthogonal projector ($P = P^2 = P^T$). Now, how do I show that $Px \in range(A) \forall x \in R^n$? The linear algebra class I took before numerical was... less than ideal, so I don't know much about the properties of the range.

答:
$P_{\text{range}(A)}  (x) = A \big( (A^T A)^{-1} A^T x \big) = Ay$, for $y = (A^T A)^{-1} A^Tx $.

$P_{\text{range}(A)}  (x)$ is therefore in the range of $A$, because it is equal to $Ay$ for some $y$.

Side note: You are assuming that $(A^TA)$ is invertible, which means that the columns of $A$ are linearly independent. That is not necessarily the case in general.


问:
Showing that $a^{12}\equiv 1\pmod{210}$ if $a$ is coprime to $210$
I ran into this next question:

Show that if $a$ is coprime to $210$:

$a^{12}\equiv1 \pmod{210}$.

This is not a homework question.

Thank you very much in advance,

Yaron

答: hints:first find phi(210)=the number of positive integers less than 210 and prime to 210....then apply Euler's theorem : a^phi(210) congruence to 1 mod(210)...</p>


问:
Von Neumann Algebra built from the bilateral shift
Let $T: \ell^2(\mathbb{Z}) \rightarrow \ell^2(\mathbb{Z})$ be the bilateral shift, and let $S= \{ I, T, T^* \}$.

Why do we have $S'= S'' = \{ W\in B(\ell^2(\mathbb{Z})) : \langle e_n, We_k \rangle = \langle e_{n-k}, We_0 \rangle\} $?

I know that if we think of it in terms of matrices then this is $w_{{n+j},{k+j}} = w_{n,k}$. But I still don't really see what's going on here. I wonder if there is way to see this with out the use of matrices.

Thank you in advance!

答:
A more suggestive way of describing the von Neumann algebra you are dealing with might be "all the bounded operators whose matrices with respect to the standard basis of $\ell^2(\mathbb{Z})$ are constant along diagonals". Note that the matrix of the bilateral shift itself has this property; it has one diagonal consisting of $1$s and the rest of the diagonals consist only of $0$s.

To avoid working with matrices, you can apply the Fourier transform. Then you are working on $L^2(\mathbb{T})$, where $\mathbb{T} \subseteq \mathbb{C}$ is the unit circle, and the generators become the coordinate functions $z$ and $\overline z$. The $C^*$-algebra generated by those functions is the commutative algebra $C(\mathbb{T})$ of continuous functions on the circle, represented as multiplication operators. The weak-star closure must also be commutative and indeed you get $L^\infty(\mathbb{T})$, likewise represented as multiplication operators.

The dual perspective is probably not that important at the moment because one is mainly interested in the von Neumann algebras of noncommutative groups, and the harmonic analysis of noncommutative groups is much more complicated. The matrix perspective is the right place to start, I think.


问:
Sized Biased Picking Distribution
I am having trouble understanding the following proof on sized biased picking. We have the following situation:

Let $ X_1, \cdots , X_n $ be i.i.d. and positive, and $S_i = X_1 + \cdots + X_i$ for $ 1 \leq i \leq n $. The values of $S_i/S_n$ are used to partition the interval $[0,1]$, each sub-interval has size $Y_i = X_i/S_n$. Suppose $U$ is an independent uniform r.v. on $(0,1)$, and let $\hat{Y}$ denote the length of the sub-interval containing $U$. We aim to calculate the distribution of $\hat{Y}$.

The claimed result is $ \mathbb{P}(\hat{Y} \in dy) = n y \, \mathbb{P}(Y \in dy) $, where the notation means $\mathbb{P}(\hat{Y} \in A) = \int_A n y \mu(y) \, dy$ with $\mu$ the law of $Y$ (I think perhaps they mean the density function $f_Y(y)$).

The proof given is
\begin{align*}
\mathbb{P}(\hat{Y} \in dy) &  = \sum_{i=1}^n \mathbb{P}\left(\hat{Y} \in dy, \frac{S_{i-1}}{s_n} \leq U < \frac{S_i}{S_n}\right) \\
& = \sum_{i=1}^n \mathbb{P}\left(\frac{X_i}{S_n} \in dy, \frac{S_{i-1}}{S_n} \leq U < \frac{S_i}{S_n}\right) \\
&  =  \sum_{i=1}^n \mathbb{E}\left[\frac{X_i}{S_n} \, 1\left(\frac{X_i}{S_n} \in dy\right)\right] \tag{$\ast $} \\
& = \sum_{i=1}^n y \, \mathbb{P}\left(\frac{X_i}{X_n} \in dy \right) \tag{$\ast  \ast $} \\
& =ny \, \mathbb{P}(Y \in dy) \\
\end{align*}

I do not understand the equalities $(\ast)$ and $(\ast \ast)$, and do not fully understand the notation given in the proof.

答:
I'm not sure I understand the notation in the proof, but the result is quick with a continuous version of Bayes' rule. Let $A_i$ be the event that $U$ lands in the $i^{\text{th}}$ interval. By symmetry, the unconditional probabilities $\Pr[A_1], \dots, \Pr[A_n]$ are all $\frac1n$.

Then
$$
   f_{Y_i \mid A_i}(y) = \frac{\Pr[A_i \mid Y_i = y] f_{Y_i}(y)}{\Pr[A_i]} = \frac{y f_{Y_i}(y)}{1/n} = n y f_{Y_i}(y) = ny f_Y(y).
$$
The conditioned random variable $Y_i \mid A_i$ is not literally the same thing as $\hat Y$. However, by the law of total probability, we have $$f_{\hat Y}(y) = \sum_{i=1}^n f_{\hat Y \mid A_i} (y) \Pr[A_i] = \sum_{i=1}^n f_{Y_i \mid A_i}(y) \Pr[A_i].$$ This is just an average of $f_{Y_i \mid A_i}(y)$ over $i=1,\dots,n$, and since they're all $ny f_Y(y)$, we conclude that $f_{\hat Y}(y) = ny f_Y(y)$.

Moreover, the intuition for the result comes from Bayes' rule. We are saying that the distribution of the size of the picked interval is scaled up in proportion to the probability that an interval of that size would be picked (turning $f_Y(y)$ into $y f_Y(y)$). The constant factor of $n$ is just there to make the whole thing integrate to $1$ again.

Title: Sized Biased Picking Distribution


问:
Find the GCD of a polynomial using the extended Euclidean algorithm and express it in the form $a(x)f(x)+b(x)g(x)$
I am working on a problem that I can not seem to finish.

Find the gcd of $f(x)=x^7+1$ and $g(x)=x^6+x^5+x^3+1$, and express it in the form $a(x)f(x)+b(x)g(x)$ using the extended euclidean algorithm.

I have performed the extended euclidean algorithm to find the following
$$f(x)=(x-1)g(x)+(x^5-x^4+x^3-x+2)$$
$$g(x)=(x+2)(x^5-x^4+x^3-x+2)+(x^4-x^3+x^2-3)$$
$$x^5-x^4+x^3-x+2=(x)(x^4-x^3+x^2-3)+(2x+2)$$
$$x^4-x^3+x^2-3=(\frac{1}{2}x^3-x^2+\frac{3}{2}x-\frac{3}{2})(2x+2)+0$$
This should give us the GCD as following
$$\gcd(f,g)=\gcd(x^5-x^4+x^3-x+2, 2x+2)=2x+2$$
And the monic GCD then becomes
$$\gcd(f,g)=x+1$$

This is where I get stuck. Now that I have this I am not sure how to algebraically get back to the form $a(x)f(x)+b(x)g(x)$

The answer as provided in the solutions is
$$\gcd(f(x),g(x))=x^4+x^3+x^2+1=xf(x)+(x^2+x+1)g(x)$$

I do not understand how to proceed and get to the answer. I have tried going backward but I am not getting anywhere. How do I go from $x+1$ to the correct solution?

Body:
$$  \left(   x^{7}  + 1 \right)  $$

$$  \left(   x^{6}  +  x^{5}  +  x^{3}  + 1 \right)  $$

$$  \left(   x^{7}  + 1 \right)  =  \left(   x^{6}  +  x^{5}  +  x^{3}  + 1 \right)  \cdot \color{magenta}{  \left(   x  - 1 \right) } +  \left(   x^{5}  -  x^{4}  +  x^{3}  -  x  + 2 \right)  $$
$$  \left(   x^{6}  +  x^{5}  +  x^{3}  + 1 \right)  =  \left(   x^{5}  -  x^{4}  +  x^{3}  -  x  + 2 \right)  \cdot \color{magenta}{  \left(   x  + 2 \right) } +  \left(   x^{4}  -  x^{3}  +  x^{2}  - 3 \right)  $$
$$  \left(   x^{5}  -  x^{4}  +  x^{3}  -  x  + 2 \right)  =  \left(   x^{4}  -  x^{3}  +  x^{2}  - 3 \right)  \cdot \color{magenta}{  \left(   x  \right) } +  \left(  2 x  + 2 \right)  $$
$$  \left(   x^{4}  -  x^{3}  +  x^{2}  - 3 \right)  =  \left(  2 x  + 2 \right)  \cdot \color{magenta}{  \left(   \frac{  x^{3}  - 2 x^{2}  + 3 x  - 3 }{ 2 }  \right) } +  \left( 0 \right)  $$
$$ \frac{ 0}{1} $$
$$ \frac{ 1}{0} $$
$$ \color{magenta}{  \left(   x  - 1 \right) }  \Longrightarrow  \Longrightarrow  \frac{  \left(   x  - 1 \right) }{ \left( 1  \right) } $$
$$ \color{magenta}{  \left(   x  + 2 \right) }  \Longrightarrow  \Longrightarrow  \frac{  \left(   x^{2}  +  x  - 1 \right) }{ \left(   x  + 2 \right) } $$
$$ \color{magenta}{  \left(   x  \right) }  \Longrightarrow  \Longrightarrow  \frac{  \left(   x^{3}  +  x^{2}  - 1 \right) }{ \left(   x^{2}  + 2 x  + 1 \right) } $$
$$ \color{magenta}{  \left(   \frac{  x^{3}  - 2 x^{2}  + 3 x  - 3 }{ 2 }  \right) }  \Longrightarrow  \Longrightarrow  \frac{  \left(   \frac{  x^{6}  -  x^{5}  +  x^{4}  -  x^{3}  +  x^{2}  -  x  + 1 }{ 2 }  \right) }{ \left(   \frac{  x^{5}  +  x^{2}  -  x  + 1 }{ 2 }  \right) } $$
$$  \left(   x^{6}  -  x^{5}  +  x^{4}  -  x^{3}  +  x^{2}  -  x  + 1 \right)  \left(   \frac{  x^{2}  + 2 x  + 1 }{ 2 }  \right)  -  \left(   x^{5}  +  x^{2}  -  x  + 1 \right)  \left(   \frac{  x^{3}  +  x^{2}  - 1 }{ 2 }  \right)  =  \left( 1  \right)  $$
$$  \left(   x^{7}  + 1 \right)  =  \left(   x^{6}  -  x^{5}  +  x^{4}  -  x^{3}  +  x^{2}  -  x  + 1 \right)  \cdot \color{magenta}{  \left(   x  + 1 \right) } +  \left( 0 \right)  $$
$$  \left(   x^{6}  +  x^{5}  +  x^{3}  + 1 \right)  =  \left(   x^{5}  +  x^{2}  -  x  + 1 \right)  \cdot \color{magenta}{  \left(   x  + 1 \right) } +  \left( 0 \right)  $$
$$  \mbox{GCD} =   \color{magenta}{  \left(   x  + 1 \right) }   $$
$$  \left(   x^{7}  + 1 \right)  \left(   \frac{  x^{2}  + 2 x  + 1 }{ 2 }  \right)  -  \left(   x^{6}  +  x^{5}  +  x^{3}  + 1 \right)  \left(   \frac{  x^{3}  +  x^{2}  - 1 }{ 2 }  \right)  =  \left(   x  + 1 \right)  $$

.......


问:
Prove $\pi=\lim_{n\to\infty}2^{4n}\frac{\Gamma ^4(n+3/4)}{\Gamma ^2(2n+1)}$
How could it be proved that
$$\pi=\lim_{n\to\infty}2^{4n}\frac{\Gamma ^4(n+3/4)}{\Gamma ^2(2n+1)}?$$

What I tried

Let
$$L=\lim_{n\to\infty}2^{4n}\frac{\Gamma ^4(n+3/4)}{\Gamma ^2(2n+1)}.$$
Unwinding $\Gamma (n+3/4)$ into a product gives
$$\Gamma \left(n+\frac{3}{4}\right)=\Gamma\left(\frac{3}{4}\right)\prod_{k=0}^{n-1}\left(k+\frac{3}{4}\right).$$
Then
$$\lim_{n\to\infty}\frac{(2n)!}{4^n}\prod_{k=0}^{n-1}\frac{16}{(3+4k)^2}=\frac{\Gamma ^2(3/4)}{\sqrt{L}}.$$
Since
$$\frac{(2n)!}{4^n}\prod_{k=0}^{n-1}\frac{16}{(3+4k)^2}=\prod_{k=1}^n \frac{4k(4k-2)}{(4k-1)^2}$$
for all $n\in\mathbb{N}$, it follows that
$$\prod_{k=1}^\infty \frac{4k(4k-2)}{(4k-1)^2}=\frac{\Gamma ^2(3/4)}{\sqrt{L}}.$$
But note that this actually gives an interesting Wallis-like product:
$$\frac{2\cdot 4\cdot 6\cdot 8\cdot 10\cdot 12\cdots}{3\cdot 3\cdot 7\cdot 7\cdot 11\cdot 11\cdots}=\frac{\Gamma ^2(3/4)}{\sqrt{L}}.$$

I'm stuck at the Wallis-like product, though.

答:
I suppose you could do it the cheap way and use Stirling's approximation:

$$n! \sim \sqrt{2\pi n} (n/e)^n$$ implies $$\Gamma^4(n+3/4) \sim 4\pi^2 \frac{(n-1/4)^{4n+1}}{e^{4n-1}},$$ and $$\Gamma^2(2n+1) \sim 2\pi \frac{(2n)^{4n+1}}{e^{4n}};$$ hence $$2^{4n} \frac{\Gamma^4(n+3/4)}{\Gamma^2(2n+1)} \sim \pi \left(1 - \frac{1}{4n}\right)^{4n+1} e,$$
and the rest is straightforward.


问:
How to uncurl $\mathbf{B}=\mathbf{\nabla\times A}$ if we only know $\mathbf{B}$ at $z=z_0$?
Let
$$\mathbf{B} = \mathbf{\nabla\times A},$$
where
$$\mathbf{B}(x,y,z) = B_x(x,y,z)\mathbf{\hat{x}} + B_y(x,y,z)\mathbf{\hat{y}} + B_z(x,y,z)\mathbf{\hat{z}},$$
$$\mathbf{A}(x,y,z) = A_x(x,y,z)\mathbf{\hat{x}} + A_y(x,y,z)\mathbf{\hat{y}} + A_z(x,y,z)\mathbf{\hat{z}},$$
Assume that
$$\mathbf{\nabla \cdot A} = 0.$$
This ensures that $\mathbf{A}$ is uniquely defined by the top equation.

Is it possible to calculate $A_x(x,y,z_0)$, $A_y(x,y,z_0)$ if we only know $\mathbf{B}(x,y,z)$ at $z=z_0$?

Here is my naive attempt.
Assume that
$$A_x(x,y,z_0) = \frac{\partial \phi}{\partial y},$$
$$A_y(x,y,z_0) = -\frac{\partial \phi}{\partial x},$$
where
$$\phi=\phi(x,y).$$
Substituing this into the $z$-component of the top equation gives the following 2D Poisson equation
$$\left(\frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2}\right)\phi(x,y) = -B_z(x,y,z_0).$$
We can solve this for $\phi(x,y)$ which we can use to calculate $A_x(x,y,z_0)$ and $A_y(x,y,z_0)$. Is this a valid solution? There seems to be problems. For example, since
$$\mathbf{\nabla\cdot A} = 0,$$
this implies that
$$\left.\frac{\partial A_z}{\partial z}\right|_{z=z_0}=0,$$
but $z_0$ is arbitrary so $\partial A_z / \partial z = 0$ $\forall z$?

答:
It doesn't appear so since we can, for example, add a constant field $\mathbf{A}_0$ to $\mathbf{A}$ so that $$\mathbf{A}_2=\mathbf{A}+\mathbf{A}_0$$
We now have that if $\mathbf{A}$ satisfies
$$\nabla \cdot \mathbf{A}=0$$
then also $\mathbf{A}_2$  satisfies
$$\nabla \cdot \mathbf{A}_2=0$$
since
$$\nabla \cdot \mathbf{A}_2=\nabla \cdot (\mathbf{A}+\mathbf{A}_0)=\nabla \cdot \mathbf{A}+\nabla \cdot \mathbf{A}_0=0+0=0$$
So, you cannot "see" the value of the constant field $\mathbf{A}_0$ from $$\mathbf{B}=\nabla \times \mathbf{A}$$
since also
$$\mathbf{B}=\nabla \times \mathbf{A}_2$$
due to
$$\mathbf{B}=\nabla \times \mathbf{A}_2=\nabla \times \mathbf{A}+\nabla \times \mathbf{A}_0=\nabla \times \mathbf{A}+0=\nabla \times \mathbf{A}$$


问:
$x,y\in \mathbb Z^+$ Show that $P(x)=P(y) \implies x=y$

Let $x,y\in \mathbb Z^+$, we define $P(a)$ as the product of all the divisors of $a$ including $1$ And $a$, Show that $$P(x)=P(y)\implies x=y.$$

Im gonna show my Attempt, which is based on cases.

Case 1: if  $x,y$ Are primes.
$$x,y\in \mathbb P\implies \cases{P(x)=x\\P(y)=y}\implies (P(x)=P(y) \iff x=y)$$
Where $\mathbb P$ Is the set of prime numbers.

Case 2: if $x,y$ are some powers of prime.
$$\cases{x=p^n\\ y=q^m } \implies \cases{P(x)=1\cdot p\cdot p^2...p^n=p^{\frac{n(n+1)}{2}}\\P(y)= 1\cdot q\cdot q^2...q^m= q^{\frac{m(m+1)}{2}}} $$ So to prove the statement We set $$p^{\frac{n(n+1)}{2}} =q^{\frac{m(m+1)}{2}}$$
The next step is to show that this statement implies $x=y$, I have no idea how to do that. (By the way the next case is to consider $x,y$ As a product of primes.)

答:
Deep breath.

So every positive integer is going to have a unique prime factorization.  Each factor is going to be a product of the these primes to some power and so $P(x)$ is going to be a product of these factors is going to be a product of these primes to powers.

So if $x = \prod p_i^{k_i}$ is the unique prime factorization of $x$ then $P(x) = \prod p_i^{m_i}$ where $\{m_i\}$ is a distinct set of natural numbers based on some manipulation of the values of $k_i$ and possibly, but probably not, the values of $p_i$.  Just what this manipulation is ... well, we'll figure that out later.

So if $P(x) = P(y) = M = \prod p_i^{m_i}$ then it must be that $x$ and $y$ have the exact same prime factors, namely $\{p_i\}$ and $x = \prod p_i^{k_i}$ and $y = \prod p_i^{w_i}$ and that somehow the set of $\{k_i\}$ and the set of $\{w_i\}$ both get manipulated to the same set $\{m_i\}$.

We have to show nothing more or less than if two set's of $\{k_i\}$ and $\{w_i\}$ both get manipulated to the same set of $\{n_i\}$ then that would imply that each $k_i$ is equal $w_i$ respectively.

Okay... so just what is the manipulation[1]?

......

Well.  Consider $x = \prod p_i^{k_i}$ and lets consider one particular prime designated $Q$ (with a capital) so that $x = Q^K \prod_{p_i\ne Q} p_i^{k_i}$ and let $N = \frac x{Q^K}$ and so that $p\not \mid N$.

Okay.... Let $F = \{$ factors of $N\}$ and so each factor of $x$ will be expressible as $Q^jf$ for some $f\in F$ and some power $j$ where $w = 0,...., K$.

So $P(x) = \prod_{j= 0}^K [\prod_{f\in F} Q^j f]=$

$(\prod_{j=0}^K Q^j)^{|F|}\prod_{f\in F} f$

Okay... $\prod_{f\in F} f = P(N) = P(\frac x{Q^K})$ and $\prod_{j=0}^K Q^j= Q^{\sum_{j=0}^K j} =Q^{\frac {K(K+1)}2}$ and $|F|$....

well.... if a number has $\prod p_i^{k_i}$ as it's prime factorization then a factor $f$ is of the form $\prod p_i^{a_i}$ where $0 \le a_i \le k_i$.  There are $k_i +1$ options for $a_i$ so there are $\prod (k_i + 1)$ factors.

So $|F| =\prod_{p_i \ne Q} (k_i + 1)$.

So we have $P(x) = Q^{\frac K2\cdot (K+1)\prod_{p_i \ne Q} (k_i + 1)}\cdot P(\frac x{Q^K})$ but we can put that $(K+1)$ inside the product so $P(x) = Q^{\frac K2\prod(k_i + 1)}\cdot P(\frac x{Q^K})$

And so recursively we have

$P(x) = \prod p_j^{\frac {k_j}2 \prod(k_i + 1)}$.

........

So if $P(x) = P(y) = \prod p_j^{m_j}$ then $x = \prod p_j^{k_j}$ and $y = \prod p_j^{w_j}$ and we have for each $j$ than

$m_j = \frac {k_j}2(\prod (k_i +1)) =\frac {w_j}2(\prod (w_i +1))$ and our task is simply to show that implies each $k_j = w_j$.

....Phew.....

Okay so we know that

each $k_j = w_j \cdot \frac {\prod(w_i+1)}{\prod(k_i+ 1)}$

but $\frac {\prod(w_i+1)}{\prod(k_i+ 1)}=\prod \frac {w_i + 1}{k_i + 1}$ is a positive constant.

If we call it $C$ we have:

$C = \prod \frac {w_i + 1}{k_i + 1} = =\prod \frac {w_i + 1}{Cw_i + 1}$

Now we just have to show that is only possible if $C = 1$.

And if $C < 1$ then $\frac {w_i + 1}{Cw_i + 1} > 1$ and $C = \prod \frac {w_i + 1}{Cw_i + 1} > 1$.  A contradiction.

And if $C > 1$ then $\frac {w_i + 1}{Cw_i + 1} < 1$ and $C = \prod \frac {w_i + 1}{Cw_i + 1} < 1$.  A contradiction.

======

[1] Credit where credit is due.  Calvin Lin had a for more elegant calculation and expression of $P(N)$ that escaped me.

Let $x =\prod p_i^{k_i}$ and let $\tau(x) = \prod(k_i + 1)=$ the number of factors of $x$.  Now if $d|x\iff\frac xd|x$ obviously. So....

$P^2(x) = (\prod_{d|x}) d\cdot (\prod_{d|x})\frac xd = \prod_{d|x} N= N^{\tau(x)} = N^{\prod(k_i + 1)}$

So $P(x) = N^{\frac {\prod(k_i + 1}2}$....

Which as $\sqrt{N} = \prod p_i^{\frac {k_i}2}$ is the same result as mine.


Title: $\space \lim_{n \to \infty} s_n(x)=0$ and $ \lim_{n \to \infty} \int_{0}^{1}s_n(x)\,dx=1.$ Possible for step functions?

Body:
Question : Show that there exists step functions

$\space$ {$ \space {s_n : [0,1] \to R }\space  $}${_{n=1}^{\infty}}$ such that for any $x\in(0,1)$ we have $\space \lim_{n \to \infty} s_n(x)=0$  and $\space \lim_{n \to \infty} ({ \int_{0}^{1}{(s_n(x)\,dx)} }=1.$

Corrected approach for the Question : for all $ x \in (0,1)$

Let $f:[0,1]\to\mathbb{R}$ be such that $\int_{0}^{1}f(x)\,dx=1$. For each $n\in\mathbb{N}$, define $f_{n}:[0,1]\to\mathbb{R}$ by
$$S_{n}(x)=
\begin{cases}
    n-nx & 0 \leq x \leq \tfrac{1}{n} \\
    0      & \tfrac{1}{n} < x \leq 1
\end{cases}$$

as I need to prove such that for any $x\in(0,1)$ we have $\space \lim_{n \to \infty} s_n(x)=0$  and $\space \lim_{n \to \infty} ({ \int_{0}^{1}{(s_n(x)\,dx)} }=1.$

$x \in (0,1)$ = $x \in (0, \frac{1}{n}] \space U \space (\frac{1}{n}, 1)$

Case 1 : for $\frac{1}{n}<x<1$

as $\space {n \to \infty},  \space  \frac{1}{n} \to 0$ and $\space \lim_{n \to \infty} S_n(x) = \lim_{n \to \infty} 0=0$

Case 2 : for $0<x \le \frac{1}{n}$

as $\space {n \to \infty},  \space  \frac{1}{n} \to 0$ and $\space \lim_{n \to \infty} S_n(x) = \lim_{n \to \infty} n-nx $

but here’s an observation through number line for case 2 :

as $\space {n \to \infty} \space  \frac{1}{n} \to 0     $ for a large value of n,  $\space \frac{1}{n}<x$ . $\space$  So , $\space \lim_{n \to \infty} S_n(x) = \lim_{n \to \infty} 0=0$

Integrating step :

$\lim_{n \to \infty}  $$\int_{0}^{1}S_{n}(x)\,dx = \int_{0}^{1/n}(n-nx)\,dx + \int_{1/n}^{1}0\,dx$

$\space \space$= $\lim_{n \to \infty}[{nx}-\frac {nx^2}{2}]_{0}^{1/n} + (\lim_{n \to \infty} 0 = 0)$

$\lim_{n \to \infty}  $$\int_{0}^{1}S_{n}(x)\,dx $ = $\lim_{n \to \infty} 1-\frac{1}{2n}= 1 - 0$

$\lim_{n \to \infty} $$\int_{0}^{1}S_{n}(x)\,dx $ = $\lim_{n \to \infty} 1 = 1 $

Thanks for the response guys, I changed and took an  approach through your suggestion guys. Do correct if needed

Body: Let $s_n = n\chi_{(0,1/n)}.\,\,$


问:
difference of either or, neither nor, both or and either and
I am answering a compound proposition exercise. I need to translate a phrase to logical connectives (only ~, ^,v) but I am confused with both or and either and as there is a note  on the questionnaire that "both" goes with "and" while "either" goes with "or" does that mean that I will disregard the or in the first question and and in the second?

```<code>Example
1.both p or q and r
2. either p and q or r

My answer
1.p^q^r
2.pVqvr
</code>```

I am also confused with either or and neither nor.

```<code>Example
1. either -6 is a positive number or 3 is a positive number
2.  neither -6 is a positive number nor 3 is a positive number

Where
: -6 is a positive number
: 3 is a positive number

My answer
1.pvq
2.~pv~q
</code>```

Is my answer to those examples right? If wrong can you explain it to me? Thanks!

答:
The point of the "both" and "either" here is just to disambiguate the parentheses.  If you just say "p or q and r", that's ambiguous: do you mean "(p or q) and r" or "p or (q and r)"?  But if you put "both" at the start, that tells you it can't be the second option: "both p or (q and r)" doesn't make sense because you can't have "both x or y", only "both x and y".

So "both p or q and r" just means "(p or q) and r", or $(p\vee q)\wedge r$ in symbols.  Similarly, "either p and q or r" just means "(p and q) or r".

For your second set of examples, you are right about the first one but the second one is wrong.  "Neither p nor q" means that p and q are both false, whereas your proposed $\neg p\vee\neg q$ means only that at least one of p and q is false (since it says "p is false or q is false").


问:
Question about SIS model
Modify the following SIS model

```<code>S´= -bSI + aI
I´ = bSI - aI
</code>```

to the situation in which there are two competing strains of the same disease, generating two infective classes I1 , I2 under the assumption that coinfections are not possible. Does the model predict coexistence of the two strains or competitive exclusion?

I believe the modification should look like this:

```<code>S´= -b1SI1 -b2SI2 + a1I1 + a2I2
I1´ = b1SI1 - a1I1
I2´= b2SI2 -a2I2
</code>```

But the book I am reading doesn’t really talk about coexistence of strains or exclusions so I don’t really know where to go from here. The basic reproduction number R0 is pretty obvious but what do I need to check regarding it?

Thanks in advance

答:
Since the total population $N = S + I_1 + I_2$ is constant in this model, we can substitute $S = N - I_1 - I_2$
and get two equations for $I_1$ and $I_2$:
$$ \eqalign{\dfrac{dI_1}{dt}&= b1 I_1 (N - I_1 - I_2) - a_1 I_1 = (b_1 N - a_1) I_1 - b_1 I_1^2 - b_1 I_1 I_2\cr
    \dfrac{dI_2}{dt} &= b_2 I_2 (N - I_1 - I_2) - a_2 I_2 = (b_2 N - a_2) I_2 - b_2 I_2^2 - b_2 I_1 I_2\cr}$$

If $a_1 b_2 \ne a_2 b_1$, there are three equilibrium points:

$I_1 = I_2 = 0$, $S = N$

$I_1 = 0$, $I_2 = N - a_2/b_2$, $S = a_2/b_2$

$I_2 = 0$, $I_1 = N - a_1/b_1$, $S = a_1/b_1$.

Of course if $a_2/b_2 > N$ or $a_1/b_1 > N$, these are not really possible because you can't have a negative population.

The Jacobian matrix at equilibrium point (1.) is
$$ \pmatrix{b_1 N - a_1 & 0\cr 0 & b_2 N - a_2\cr}$$
The Jacobian matrix at equilibrium point (2.) is
$$ \pmatrix{ (a_2 b_1 - a_1 b_2)/b_2 & 0 \cr
            a_2 - N b_2 & b_2 N - a_2\cr} $$
and the Jacobian matrix at equilibrium point (3.) is
$$ \pmatrix{ (a_1 b_2 - a_2 b_1)/b_1 & 0 \cr
            a_1 - N b_1 & b_1 N - a_1\cr} $$
Let's say $a_1/b_1 < a_2/b_2$.  There are essentially three
possibilities.

If $N < a_1/b_1 < a_2/b_2 $, the only "real" equilibrium point is (1.), and it is stable.  The infections always die out as $t \to \infty$.

If $a_1/b_1 < N < a_2/b_2$, the "real" equilibrium points are (1.) and (3.) . (1.) is unstable and (3.) is stable.
Strain 2 dies out as $t \to \infty$, and  we end up with  strain 1 endemic (unless we started with $I_1 = 0$).

If $a_1/b_1 < a_2/b_2 < N$, we have all three equilibrium points, but (1.) and (2.) are unstable and (3.) is stable.
Again we end up with  strain 1 endemic (unless we started with $I_1 = 0$).

The marginal cases where two or more of $a_1/b_1$, $a_2/b_2$ and $N$ are equal are left as an exercise.

Title: Question about SIS model


问:
how to solve vectors in 3D?

We had our examination a few days ago. I was not able to answer this question. So I decided to remember it and answer at home but still I can't answer this on my own.

Please help me to answer this so that if this type of question will be asked again in our incoming major exam, I can answer it.

答: First use trigonometry to find the x, y and z components of each vector.</p>

Title: how to solve vectors in 3D?


问:
Rejecting null hypothesis using a t stat
Suppose $X_1,…, X_n$ (where n is large) are modeled as iid normal. The T statistic is calculated to be $−1.7$.

Do we have enough evidence to reject $H_0:μ≥0$ at the $5\%$ level?

I am wondering wow  the t statistic give us information regarding whether to reject or not.

答:
If n is large, you don't need a $T$ statistic, because a $Z$ statistic will do the job.  A one-sided hypothesis test at $\alpha = 0.05$ has a critical $Z = 1.645$.  So yes.

If $n$ were small, you would need a $T$ statistic, and the critical value would depend on $n$, but in general would be larger than the critical $z$ value.  So if it wasn't enough to reject with $Z$, it definitely would not be enough to reject with $T$.


问:
Looking for alternative ways to solve $3^{2^x}=2^{3^x}$

If $\log2=m$ and $ \log3=n$, What is the answer of
$\large3^{2^x}=\large2^{3^x}$? $1)\log
 mn\quad\quad\quad\quad\quad\quad2)\frac{m-n}{\log m-\log
 n}\quad\quad\quad\quad\quad\quad3)\frac{\log \frac
 mn}{m-n}\quad\quad\quad\quad\quad\quad4)\frac{m+n}{\log mn}$

Here is my approach:

To solve the equation I took logarithm in base $3$ of both sides of the equation:

$$2^x=\log_32^{3^x}=3^x\log_32=3^x\times\frac{m}{n}$$
So we have $(\frac23)^x=\frac mn$. by taking logarithm in base $\frac23$ we have:

$$x=\log_{\large\frac23}(\frac mn)=\frac{\log(\frac mn)}{\log(\frac23)}=\frac{\log \frac mn}{m-n}$$

My questin: Is there other approach to solve this problem?

答:
You could take natural log. It will reduce the change of base steps

$\Rightarrow 2^x\log 3 = 3^x\log 2 \\\Rightarrow (\frac23)^x =\frac mn \\\Rightarrow x\log(\frac23) = x(m -n) = \log(\frac mn)\\
\Rightarrow \boxed{x = \frac{\log \frac mn}{m-n}}  $


问:
If $f(x)=x^3-\lambda x$ then $\Big\{ x\in\mathbb R : \Big| \lim_{n\to \infty}f^n(x)\Big|<+\infty \Big\}$ is a Cantor set.

This problem is from  An introduction to chaotic dynamical systems  Robert L. Devaney

I've already  prove (a) and (b) but i'm a bit confused about (c).

For (c) we can assume that $\lambda>4$, I have seen a similar post here but it did not helped me at all, i was thinking of a different approach.

$f$ has $3$ fixed points since $f(x)=x\iff x_0=0 $ or $x_1=-\sqrt{1+\lambda}$, or $x_2=\sqrt{1+\lambda}$, if we define
$$ A=\Big\{ x\in\mathbb R : \Big| \lim_{n\to \infty}f^n(x)\Big|<+\infty \Big\}$$
then $A\subseteq [x_1,x_2]\Rightarrow A=\Big\{ x\in[x_1,x_2] : f^n(x)\nrightarrow \pm \infty \Big\}$ ,

But how can i continue from this point on?

Any suggestions?

答:
Note that for all $x>\sqrt{\lambda+1}, f(x) > x$, and so $\lim_{n\to\infty}f^n(x)=\infty$. Similarly for $x < -\sqrt{\lambda+1}$.

So the interesting thing is the behaviour of $x\in[-\sqrt{\lambda+1}, \sqrt{\lambda+1}]$. The critical question is if $x$ starting in $[-\sqrt{\lambda+1}, \sqrt{\lambda+1}]$ ever leaves that interval. We first note that the interval contains a local min and a local max, located at $x=\pm\sqrt{\lambda/3}$ the minimum and maximum value reached evaluate to $\pm\lambda^{3/2}\frac{2}{3\sqrt3}$. So $\lambda$ will be large enough to make $x$ jump out of the interval for some value of $x$ if:

$$
\frac{2\lambda^{3/2}}{3\sqrt3} > \sqrt{\lambda+1}
$$

This gives a critical value of $\lambda=3$. For larger values of $\lambda$, subintervals around $x=\pm\sqrt{\lambda/3}$ get mapped outside of the range $[-\sqrt{\lambda+1}, \sqrt{\lambda+1}]$. If we remove these 2 subintervals, since they will eventually go to $\pm\infty$, the 3 remaining subintervals all have $f$ increasing or decreasing monotonically between $-\sqrt{\lambda+1}$, and  $\sqrt{\lambda+1}$. This is just our original interval again, so applying $f$ another time will remove another 2 subintervals from each of our 3 remaining intervals. Applying this process ad infinitum produces a Cantor set.


问:
Define the mass function
Consider the following experiment: A coin is tossed with probability $\alpha \in (0,1)$ of head and a dice is rolled at the same time. Define the following random variable:

\begin{equation*}
        X = \begin{cases}
        \text{number in the face of the dice} \ , \ \text{if heads} \\
        \text{number in the face of the dice + 2} \ , \text{if tails}
        \end{cases}
\end{equation*}

Define the probability mass function of $X$.

Well, i think that the elements of the random variable are something like $(1,H),(2,H),\dots,(6,H),(3,T),\dots, (9,T)$. The probability mass function would be something like $$P(X=x)=P(X=(n,H))=\alpha\cdot \frac{1}{6}$$ for $n \in \{1,2,\dots,6\}$. This makes sense? I'm really lost i'm this question.

答: These are the outcomes, with entries in the first column having chance $\frac 1 {6}\alpha$ of occurring and entries in the second column having pure luck $\frac 1 6(1-\alpha)$, except four of the outcomes occur twice each, with probability $\frac \alpha 6+\frac{1-\alpha} 6 =\frac 1 6$ of occurring.

Title: Define the mass function


答:
Start  with $\large3^{2^x}=\large2^{3^x}$,

Take logarithm 1st time:

\begin{align}2^x\ln 3=3^x\ln2,  \frac{\ln 3}{\ln 2}= (\frac{3}{2})^x \end{align}

Take logarithm 2nd  time:
\begin{align}\ln (\frac{\ln 3}{\ln 2})=x(\ln 3-\ln 2),x=\frac{\ln (\frac{\ln 3}{\ln 2})}{\ln 3-\ln 2}\end{align}

Replace $\ln 3$  and $\ln 2$ with m and n:

\begin{align} x=\frac{\ln (\frac{m}{n})}{m-n}\end{align}

Title: Looking for alternative ways to solve $3^{2^x}=2^{3^x}$

Body:
If $\log2=m$ and $ \log3=n$, What is the answer of
$\large3^{2^x}=\large2^{3^x}$? $1)\log
 mn\quad\quad\quad\quad\quad\quad2)\frac{m-n}{\log m-\log
 n}\quad\quad\quad\quad\quad\quad3)\frac{\log \frac
 mn}{m-n}\quad\quad\quad\quad\quad\quad4)\frac{m+n}{\log mn}$

Here is my approach:

To solve the equation I took logarithm in base $3$ of both sides of the equation:

$$2^x=\log_32^{3^x}=3^x\log_32=3^x\times\frac{m}{n}$$
So we have $(\frac23)^x=\frac mn$. by taking logarithm in base $\frac23$ we have:

$$x=\log_{\large\frac23}(\frac mn)=\frac{\log(\frac mn)}{\log(\frac23)}=\frac{\log \frac mn}{m-n}$$

My questin: Is there other approach to solve this problem?


问:
Permutation and Combinations with conditions
Hallo :) This is a question about permutations but with conditions.

2 boys and 4 girls are to be arranged in a straight line.
In how many ways can this be done if the two boys must be separated? (The order matters)

Thank You.

答:
Think about it this way:

How many ways are there of arranging all six people without restriction?

How many arrangements have the two boys together?

Therefore how many arrangements have the two boys separated?


问:
Limit of $\int_0^1\left(\frac {2}{\sqrt {(1-t^2)(1-xt^2)}}-\frac{x}{1-xt}\right)\,dt$ as $x\to 1^{-}$
While going through this question I was reminded of one of my earlier questions and I found that there is some unfinished business which needs some further exploration.

Let $$F(x) =\int_0^1\left(\frac{2}{\sqrt{(1-t^2)(1-xt^2)}} -\frac{x}{1-xt}\right)\, dt=\int_0^1 f(x, t) \, dt\tag{1}$$ for $x\in[0,1]$.

Let's observe that $$\lim_{x\to 1^-}f(x,t)=\frac{1}{1+t}\tag{2}$$ and hence it is natural to expect that $\lim_{x\to 1^-}F(x)$ should equal $\int_0^1 dt/(1+t)=\log 2$ but numerical evidence as well as some amount of elliptic function theory tells (see one of my questions linked earlier for details) us that this particular limit is $4\log 2$.

This suggests that there is some weird behavior of integrand as $x\to 1^{-}$ (in particular the convergence is not uniform).

I would like to have this limit evaluated using some analysis related to convergence of integrand as $x\to 1^-$. Any help in this direction would be appreciated.

Note: I have asked a new question instead of bumping an old one. The old question is more about solution verification and is related to elliptic integrals. I wanted to have a different perspective which involves general issues of uniform convergence to handle the limit of this integral.

答:
Remark: @metamorphy gave a nice solution. I constructed something slightly different
from @metamorphy's $G(x, t)$ due to the term $\sigma_x$.
I hope my answer is helpful as a supplement.

Let (replace $x$ with $x^4$ for convenience)
\begin{align*}
&G(x)\\
=\ & \int_0^1 \left(\frac{2}{\sqrt{(1 - t^2)(1 - x^4 t^2)}}  - \frac{x^4}{1 - x^4t}\right) \mathrm{d} t\\
=\ & \int_0^1 \left(\frac{2}{\sqrt{(1 + t)(1 + x^2t)}\sqrt{(1 - t)(1 - x^2 t)}}  - \frac{x^4}{1 - x^4t}\right) \mathrm{d} t\\
=\ & \int_0^1 \left(\frac{2}{\sqrt{(1 + t)(1 + t)}\sqrt{(1 - t)(1 - x^2 t)}}  - \frac{x^4}{1 - x^4t}\right) \mathrm{d} t\\
&\quad + \int_0^1 g(x, t)\mathrm{d} t \tag{1}
\end{align*}
where
$$g(x, t) = \frac{2}{\sqrt{1 + t}\sqrt{(1 - t)(1 - x^2 t)}}
\left(\frac{1}{\sqrt{1 + x^2t}} - \frac{1}{\sqrt{1 + t}}\right).$$

First, we deal with the second integral in (1), i.e., $\int_0^1 g(x, t)\mathrm{d} t$.
Clearly, $g(x, t) \ge 0$ for all $x$ in $(0, 1)$ and all $t$ in $[0, 1)$. Also, we have,
for all $x$ in $(0, 1)$ and all $t$ in $[0, 1)$,
\begin{align*}
&g(x, t) \\
\le\ & \frac{2}{\sqrt{(1 - t)(1 - x^2)}}
\left(\frac{1}{\sqrt{1 + x^2t}} - \frac{1}{\sqrt{1 + t}}\right)\\
=\ & \frac{2}{\sqrt{(1 - t)(1 - x^2)}}\frac{(1 - x^2)t}{(1 + x^2t)(1 + t)}\Big(\frac{1}{\sqrt{1 + x^2t}} + \frac{1}{\sqrt{1 + t}}\Big)^{-1}\\
\le \ & \frac{2}{\sqrt{(1 - t)(1 - x^2)}}(1-x^2)\\
=\ &  \frac{2\sqrt{1 - x^2}}{\sqrt{1 - t}}.
\end{align*}
Since $\int_0^1 \frac{2\sqrt{1 - x^2}}{\sqrt{1 - t}} \mathrm{d} t = 4\sqrt{1 - x^2}$,
we have $\lim_{x\to 1^{-}} \int_0^1 g(x, t)\mathrm{d} t = 0$.

Second, we deal with the first integral in (1), denoted by $H(x)$.
With the substitution $\frac{1 - x^2 t}{1 - t} = u^2$ (correspondingly $t = \frac{u^2 - 1}{u^2 - x^2}$),
we have (after doing the partial fraction decomposition for the integrand)
\begin{align*}
&H(x)\\
=\ & \int_1^\infty \left(\frac{\frac{2}{\sqrt{2 + 2x^2}}}{u - \sqrt{\frac{1 + x^2}{2}}}
- \frac{\frac{2}{\sqrt{2 + 2x^2}}}{u + \sqrt{\frac{1 + x^2}{2}}}
 + \frac{2(1 + x^2)u}{(1 + x^2)u^2 - x^2} - \frac{2u}{u^2 - x^2}\right)  \mathrm{d} u\\
=\ &\left(\frac{2}{\sqrt{2 + 2x^2}}
\ln \frac{u - \sqrt{\frac{1 + x^2}{2}}}{u + \sqrt{\frac{1 + x^2}{2}}}
 + \ln\frac{(1 + x^2)u^2 - x^2}{u^2 - x^2}\right)\Big\vert_1^\infty\\
=\ & \ln(1 + x^2) - \frac{2}{\sqrt{2 + 2x^2}}
\ln \frac{1 - \sqrt{\frac{1 + x^2}{2}}}{1 + \sqrt{\frac{1 + x^2}{2}}} - \ln\frac{1}{1 - x^2}\\
=\ & \ln(1 + x^2) + \frac{2}{\sqrt{2 + 2x^2}}\ln 2 + \frac{4}{\sqrt{2 + 2x^2}}
\ln \left(1 + \sqrt{\frac{1 + x^2}{2}}\right)\\
&\quad + \left(1 - \frac{2}{\sqrt{2 + 2x^2}}\right)\ln(1 - x^2).
\end{align*}
It is easy to obtain $\lim_{x\to 1^{-}} H(x) = 4\ln 2$.

Thus, we have $\lim_{x\to 1^{-}} G(x) = 4\ln 2$.

We are done.


问:
Why is the graph of the reciprocal function $f(x)=\frac{1}{x}$ not a one-to-one function?
The horizontal line test seems to show that no horizontal line intersects the graph of $f$ at more than one point.  However, the textbook tells me it's not one-to-one.  It is related to the asymptotes, I believe, but I'm not sure how.

答:
If the domain  of $f$ is taken to be the entire real line $\mathbb{R}$, then $f(0)$ is not defined. Likewise there is no value of $x$ that gives $f(x)=0$.

However, $f(x)$ is one-to-one if the domain is restricted to $\mathbb{R}\setminus\{0\}$.

ETA: In fact, on $f$'s domain $\mathbb{R}\setminus \{0\}$, not only if $f$ one-to-one, but it is its own inverse!


问:
Computing the dimension of $O(n)$ - why is this matrix invertible?
I am looking at Lee's Smooth Manifolds second edition and on page 167 he computes the dimension of $O(n)$. Now there he defines a map
$$\begin{eqnarray*}\Phi : &\text{GL}(n,\Bbb{R})& \to M(n,\Bbb{R})  \\
&A& \mapsto A^T A \end{eqnarray*}$$

and wants to compute the rank of $\Phi$. Now he says for any $B \in T_I\text{GL}(n,\Bbb{R})$ let $\gamma : (-\epsilon,\epsilon) \to \text{GL}(n,\Bbb{R})$ be the smooth curve $\gamma(t) = I + tB$ and so on.

My question is: Why should the matrix $I + tB$ be an invertible matrix for all $t \in (-\epsilon,\epsilon)$? My guess is this: at $t = 0$ the determinant of $I + (0\cdot B)$ is $1$. Now consider the map $$f(t)\stackrel{\text{def}}{\equiv} \det \gamma(t) : \Bbb{R} \to \Bbb{R}.$$ Now $f(0) = 1$ and because $\Bbb{R} - \{0\}$ is open I can choose a $\varepsilon > 0 $ so that $B_\varepsilon(1) \subseteq \Bbb{R} - \{0\}$. Then by continuity there exists $\delta > 0$ so that $f(B_\delta(0)) \subseteq B_\varepsilon(1)$, that is to say I can choose an $\delta > 0$ so that $\gamma(t)$ is invertible for all $t \in (-\delta,\delta)$.

答:
You are correct.

The curve $\gamma$ you defined is continuous from $\mathbb{R}$ to $M(n, \mathbb{R})$ and the determinant function is continuous from $M(n, \mathbb{R})$ to $\mathbb{R}$.

So, the map $t \mapsto \operatorname{det}(\gamma(t))$ is continuous on $\mathbb{R}$, taking the value $1$ at $t = 0$. Therefore there is a small interval around $0$ on which $\operatorname{det}(\gamma(t))$ is nonzero.


答:
The textbook might have some nitpick in mind like "$f$ is not a one-to-one function because it's not a function: it's not defined at $0$".

But the horizontal line test tells us that there can't be two inputs which $f$ sends to the same output. This means if we restrict our domain to anything where $f$ is a function (such as the nonzero real numbers) then $f$ will be one-to-one. So you're basically right.

Title: Why is the graph of the reciprocal function $f(x)=\frac{1}{x}$ not a one-to-one function?

Body: The horizontal line test seems to show that no horizontal line intersects the graph of $f$ at more than one point.  However, the textbook tells me it's not one-to-one.  It is related to the asymptotes, I believe, but I'm not sure how.


问:
How do I find the asymptote of $y= -\tan(x+\pi/4)$?
$$y= -\tan(x+\pi/4).$$
How do I find the asymptote of this?

答:
Normally, I wouldn't answer, because the OP didn't show work.  But since another answer has already been given ...

Alternative approach

$$y= -\tan(x+\pi/4).$$
How do I find the asymptote of this?

Use that

$\tan(a+b) = \frac{\tan(a) + \tan(b)}{1 - \tan(a)\tan(b)},$

$\tan(\pi/4) = 1$

$\tan(x) = 1 \iff x \in \{\pi/4, 5\pi/4\}$, within a modulus of $2\pi$.

$$y = - \tan(x + \pi/4) = - \frac{\tan(x) + 1}{1 - \tan(x)}.$$

For an asymptote, it is sufficient that the denominator $= 0$ and the numerator $\neq 0$.  Sufficiency refers to the fact that when the numerator and denominator are both $(0)$, then the situation is unclear.  In this problem, no such complication is present.

Therefore, it is immediate that (within a modulus of $2\pi$), the asymptotes occur at $\{\pi/4, 5\pi/4\}.$


问:
Intuition behind Group Action
I am going over the group action and unfortunately I am not sure if I understand it very well. So for this reason, I have a few questions that possibly could help me to understand this concept better. A group action is defined as: Group $G$ acts on a set $X$ if there is a homomorphism $\sigma:G \rightarrow S_X$. I am aware group action has other definitions but for me this one is easier to understand.

Now $S_X$ is the group of permutations of $X$. It is not crystal clear what is meant by this to me. Is $S_X$ the same thing as $S_{|X|}$? When we talk about $S_n$, I have a clear mental picture; $n$ is an integer and I think of all the permutations with $n$ elements. But in $S_X$, $X$ is not an integer but a set. So suppose $X=\{1,2,3\}$. Is $S_X$ the same thing as $S_3$ in this case?

The other part of my question is about what the definition implies: it says "... if there is a homomorphism". Is it possible that homomorphism does not exist? Is the homomorphism unique? Or is it possible to have more than one homomorphism? Once we specify a group and a set, how do we find such homomorphism(s)? Is there a requirement for the size of the group and the set to make this definition work, for instance $|G| \geq |X|$?

Please let me know if you would like me to clarify my question. Any help with making a better intuition is appreciated!

答:
For your first question, $S_X$ is defined as the group of bijections from $X$ to $X$.  The group operation is function composition.  If $X$ is finite, say with $n$ elements, then the groups $S_X$ and $S_n$ are obviously (noncanonically) isomorphic.

There is always at least one group homomorphism from $G$ to $S_X$, namely the one which sends everything in $G$ to the identity map in $S_X$.  This is the "stupid group action" which doesn't do anything ($g.x = x$ for all $g \in G$ and $x \in X$).

There are typically many homomorphisms from $G$ to $S_X$.  Therefore, there are typically many group actions of $G$ on $X$.

"Once we specify a group and a set, how do we find such homomorphisms?"

Good question.  In many settings, the group action comes up naturally.  It's not like people are taking random sets $X$ and $G$ and asking, "I wonder how many different actions of $G$ I can find on $X$."  I mean maybe some people are doing this, but usually the group  action is a convenient way to describe some existing phenomenon they are trying to study.

Example: Let $X = \{ z \in \mathbb C: \operatorname{Im}(z) > 0\}$ be the upper half plane, and let $G = \operatorname{SL}_2(\mathbb Z)$ be the group of integer matrices with determinant $\pm 1$.  There is a natural group action of $G$ on $X$ by

$$\begin{pmatrix} a & b \\  c & d \end{pmatrix}.z = \frac{az+b}{cz+d}.$$

This came up (I believe) when people were working out problems like describing all holomorphic bijections from the Riemann sphere to itself, and it turns out that they all look like the above (where the matrix can be anything in $\operatorname{GL}_2(\mathbb R)$).

One example where people are taking certain groups $G$ and certain sets $X$, and asking what are the ways $G$ can act on $X$, is when $X$ is a vector space.  The bijections from $X$ to itself coming from the elements of $G$ are required to also be linear transformations on $X$.  Such group actions are called representations.  Representation theory includes the problem of describing all representations of a given group on a given vector space, and this turns out to be an extremely difficult problem in general.  Finding such group actions is no simple matter.

Title: Intuition behind Group Action


答:
There may not be such a non-trivial group action, to answer your question, because for instance if the groups are finite, we need the orders of the groups to satisfy certain conditions.  For instance there's no non-trivial homomorphism from $\mathbb Z_m\to\mathbb Z_n$, when $(m,n)=1$.

For example this prevents us from forming certain non-trivial semi-direct products, since we need a group action to do so.  So, take the case when $p\nmid q-1, q\gt p\gt2$.  Then we can't do anything but the standard direct product $\mathbb Z_q\times\mathbb Z_p\cong\mathbb Z_{pq}$.

Also we do have $|G|\ge|X|$ when the action is transitive.  That's because in that case there is only one orbit.  The orbit-stablizer theorem says the index of the stabilizer is the size of the orbit (of an element).  We can easily say that each orbit has size less than or equal to that of $G$, since the orbit is $Gx$.

Title: Intuition behind Group Action

Body:
I am going over the group action and unfortunately I am not sure if I understand it very well. So for this reason, I have a few questions that possibly could help me to understand this concept better. A group action is defined as: Group $G$ acts on a set $X$ if there is a homomorphism $\sigma:G \rightarrow S_X$. I am aware group action has other definitions but for me this one is easier to understand.

Now $S_X$ is the group of permutations of $X$. It is not crystal clear what is meant by this to me. Is $S_X$ the same thing as $S_{|X|}$? When we talk about $S_n$, I have a clear mental picture; $n$ is an integer and I think of all the permutations with $n$ elements. But in $S_X$, $X$ is not an integer but a set. So suppose $X=\{1,2,3\}$. Is $S_X$ the same thing as $S_3$ in this case?

The other part of my question is about what the definition implies: it says "... if there is a homomorphism". Is it possible that homomorphism does not exist? Is the homomorphism unique? Or is it possible to have more than one homomorphism? Once we specify a group and a set, how do we find such homomorphism(s)? Is there a requirement for the size of the group and the set to make this definition work, for instance $|G| \geq |X|$?

Please let me know if you would like me to clarify my question. Any help with making a better intuition is appreciated!


问:
Why/How was Hardy impressed with Ramanujan’s work prior to seeing any proofs of their validity?
Hardy’s famous correspondence with Ramanujan highlights an interesting aspect of the mathematics discipline; the appreciation of a mathematical expression prior to knowing its validity.

Why/how was Hardy impressed with Ramanujan’s early work on continued fractions if he didn’t even know they were true?

There is the proclamation by Hardy:

Must be true, because, if they were not true, no one would have the
imagination to invent them

— G. H. Hardy, 1913

...which suggests there is some aesthetic to the expression that is noticed prior to any proof. What is this aesthetic? Is it the apparent symmetry? Some immediate familiarity to other known expressions?

Or is it simply plugging in numbers into the RHS and being able to immediately validate that it “must/should” be true?

答:
I quote a few excerpts from the foreword (by C. P. Snow) to A Mathematician's Apology (by G. H. Hardy) (emphasis in italics mine) :

One morning early in 1913, he found among the
letters on his breakfast table, a large untidy envelope decorated with Indian stamps. When he
opened it, he found sheets of paper by no means fresh, on which, in a non-English holograph, were line after line of symbols. Hardy glanced at them without enthusiasm.

So Hardy felt, more than anything, bored. He glanced at the letter, written in halting English,
signed by an unknown Indian, asking him to give an opinion of these mathematical discoveries.
The script appeared to consist of theorems, most of them wild or fantastic looking, one or two
already well-known, laid out as though they were original. There were no proofs of any kind.
Hardy was not only bored, but irritated. It seemed like a curious kind of fraud.

That particular day, though, while the timetable wasn’t altered, internally the things were not going according to plan. At the back of his mind, getting in the way of his complete pleasure in his game, the Indian manuscript nagged away. Wild
theorems. Theorems such as he had never seen before, nor imagined. A fraud of genius? A
question was forming in his mind. As it was Hardy’s mind, the question was forming itself with
epigrammatic clarity: is a fraud of genius more probable than an unknown mathematician of
genius? Clearly the answer was no. Back in his rooms in Trinity, he had another look at the script.
He sent word to Littlewood that they must have a
discussion after hall.

Anyway, by nine o’clock or so they were in one of Hardy’s rooms, with the manuscript stretched out in front of them. Apparently it did not take them long. Before midnight they knew, and knew for certain. The writer of these manuscripts was a man of genius. That was as much as they could judge, that night.

Fron the above it should be clear that at first Hardy got really bored and irritated with Ramanujan's letter. But somehow the theorems of the letter got registered in his mind and this was probably due to extreme nature of the theorems (wild, not seen before, couldn't be imagined by Hardy).

When Hardy and Littlewood met together they probably tried to verify some of the theorems and concluded that there was some genuine mathematics involved. In particular the theorems on Rogers-Ramanujan continued fraction were the most difficult. Hardy remarked that "they defeated him completely". The quote by Hardy in your question is also related to the same Rogers-Ramanujan continued fraction formulas.

I have shared some of my thoughts on Ramanujan in a blog post and I have mentioned there something about the nature of Ramanujan's formulas:

meaning of the formula could be understood by anyone with basic knowledge of algebra and calculus

focus was on special cases of general formulas with actual numbers rather than general formula itself

minimal use of symbolism and wherever possible indicate a pattern by exhibiting it numerically or by writing about the pattern in English rather than describing pattern via a formula

each formula had a certain unexpectedness providing a shock treatment to the reader

most of the formulas had deep theories behind them

avoiding use of $\sum$ and $\prod$ symbols to represent infinite series and products

I think these qualities are reasonable enough to impress some people. It is also well known that Ramanujan sent the same letter to H. F. Baker and E. W. Hobson and they did not give any feedback. So probably the formulas couldn't impress everyone.


问:
Investigating the generalised Hamming distance of some code and information transmission of a binary code.
Letting $A$ be an alphabet of size $a$ and $S = A^{n}$. Meaning the set of all worlds of length $n$ with bits chosen from the alphabet $A$. We now take $w ∈ S$.

Assume $n = 14$. How many words in $S$ are of Hamming distance less than or equal to $4$ from $w?$

Secondly, assume  $n = 14$ and $a = 2$ and that $w ∈ S$ is transmitted through a Binary Symmetric Channel with probability $p$ of correct transmission for each individual bit. How would you generalise a formula for the probability that $5$ or fewer errors will occur during transmission of $w?$

The idea I've had so far is generalising some form of binomial probability from zero errors up to 5. Something like:

$p^{14} + 14p^{13}(1-p)^{1}+...$

Appreciate any of your time/thoughts in advance.

答:
To have Hamming distance $\le4$, we pick those positions that are intended to be different from $w$ and purposely choose them to be different from $w$ at that position, of which we have $a-1$ choices, hence the desired answer is

$$\sum_{i=0}^4 \binom{14}{i}(a-1)^i.$$

Similarly, for the case where $n=14$, $a=2$, and $p$ being the probability of success, we choose the position of mistakes. At those positions, we multiply by probability $1-p$ and at the correct positions, we multiply by $p$. Hence the expression:

$$\sum_{i=0}^5 \binom{14}{i}p^{14-i}(1-p)^i.$$

That is your idea was right, just write down $6$ terms.


问:
If an orthogonal matrix $O$ has characteristic polynomial $p=(t-1)^3$ why must it be $I_3$?
I came across this statement in my linear algebra class:

If $p(t)=(t-1)^3$ for an orthogonal matrix $O$, then the matrix is $I_3$ and $O\in SO(3,\mathbb{R})$.

I know that this gives that the eigenvalues are all equal to $1$, but I am having difficulty understanding why the rest of the statement is true. Here is what I have tried:

Used the generalized eigenspace decomposition to write $O$ as an upper-triangular matrix,

Used the property that $\langle v_1, v_2\rangle =\langle O(v_1), O(v_2)\rangle$.

However I am unable to ascertain why this is the case despite repeated attempts. Is there any proof as to why this is true?

答: Since orthogonal matrices are diagonalisable, we can write $O = PDP^{-1}$ with $D$ a diagonal matrix whose entries are the eigenvalues of $O$, i.e. $D = I_3$. But then $O = P P^{-1} = I_3$.


问:
On a specific C*-algebra
How to describe the C*-algebra $C^*(\eta, \eta^*)/<\eta^*\eta=\mu \eta \eta^*>$ where $\mu$ is a positive real number less than $1$? What happens if $\mu$ is replaced by a complex number? Any hint/suggestion/reference will be highly appreciated.

答:
If a C*-algebra $A$ is generated by an element $\eta $ satisfying $\eta^*\eta=\mu \eta \eta^*$,  where $\mu$ is a positive
real number less than $1$,  then $\eta =0$, and hence
$A=\{0\}$.  The reason is that
$$
  \|\eta \|^2 = \|\eta ^*\eta \| = \mu \|\eta \eta^*\| = \mu \|\eta\|^2,
  $$
which imply that $\|\eta \|=0$.

If $\mu $ is any complex number not equal to $1$, one also deduces that $\eta =0$, because of comparison  of spectra:
$$
  \sigma (\eta \eta ^*) \cup \{0\} =
  \sigma (\eta ^*\eta ) \cup \{0\} =
  \sigma (\mu \eta \eta ^*) \cup \{0\} =
  \mu \sigma (\eta \eta ^*) \cup \{0\},
  $$
while the only nonempty subset $S\subseteq {\mathbb R}_+$ such that $\mu S=S$ is the set $S=\{0\}$.

Title: On a specific C*-algebra


问:
What is the correct answer to this infamously ambiguous arithemetic problem 6÷2(1+2)?
I am going to make the case that $6÷2(1+2) = 1$. Many people often quote the PEMDAS rule and get the answer $9$ but look at this way:

$$ 6 \div 2(1+2)= \frac{6}{2(1+2)}$$

Algebraically, we can treat the division operator(obelus or solidus) as a fraction. Everything to the left of the operator is in the numerator and everything to the right is in the denominator. So then you apply the usual PEMDAS rule to simplify the denominator:

$$\frac{6}{2(1+2)}=\frac{6}{2(3)}=\frac{6}{2\ast 3}=\frac{6}{6}=1$$

A lot of engineering types just work left to right and end up with $9$ because they don't appreciate how division lacks 2 properties of multiplication: commutativity and distributivity. But if there is a good mathematical reason why I'm wrong, I'd like to hear it!

答:
Actually, no, you cannot treat $\div$ as a fraction line, it should be treated as dividing the thing that comes right after, not ALL the things that come right after. There is no really good mathematical reason you are wrong, that's just it, $\div$ doesn't work like a fraction line. Thus, by PEMDAS, you should actually get

$6 \div 2(1+2) = 6 \div2(3) =3(3) = 9$

If you're still confused, try the mnemonic PEMoDAoS, which stands for parantheses, exponents, (multiplication or division), (addition or subtraction). Multiplication and division have the same priority, so you have to evaluate from left to right.

Hope this helps


答:
In order to get $\frac{6}{ 2(1+2)}$, we would need the original problem to have: $$6 \div (2 (1+2)).$$

Because there are no extra parentheses, we cannot put the $1+2$ under the fraction bar as well. You MUST work from left to right and get $6 \div 2 \cdot (1+2) = 3 \cdot (1+2) = 9.$

Title: What is the correct answer to this infamously ambiguous arithemetic problem 6÷2(1+2)?

Body:
I am going to make the case that $6÷2(1+2) = 1$. Many people often quote the PEMDAS rule and get the answer $9$ but look at this way:

$$ 6 \div 2(1+2)= \frac{6}{2(1+2)}$$

Algebraically, we can treat the division operator(obelus or solidus) as a fraction. Everything to the left of the operator is in the numerator and everything to the right is in the denominator. So then you apply the usual PEMDAS rule to simplify the denominator:

$$\frac{6}{2(1+2)}=\frac{6}{2(3)}=\frac{6}{2\ast 3}=\frac{6}{6}=1$$

A lot of engineering types just work left to right and end up with $9$ because they don't appreciate how division lacks 2 properties of multiplication: commutativity and distributivity. But if there is a good mathematical reason why I'm wrong, I'd like to hear it!


问:
Construction of triangle
I don't know how to prove or disprove the following problem:
How to construct triangle if elements $a$, $b$, $\beta-\gamma$ are given?
Is it constructible (if not, how to prove it)? Any help is welcome.</p>

答:
I don't think that such a triangle can be constructed. A triangle construction needs 3 values known. These three values should be valid Congruence rule (such as SAS, SSS, ASA etc.).

Here, we are given 2 edges and difference of angles opposite to those sides. The two edges are 2 values and for third, we have no usable means to deduce. Just one relation between two angles is not enough.

Title: Construction of triangle


问:
Unitary Operators in Hilbert's space are arcwise connected using functional calculus
This is a follow up question on: unitary operators in Hilbert's space are arcwise connected.

I see that Theorem 2.12 of Murphy's book shows that if $\sigma(u) \ne \mathbb{T}$, then there exists $a \in A_{sa}$ such that $u =e^{ia}$ using functional calculus.

I wonder how to continue the argument and show that the unitary operators are arcwise connected as https://math.stackexchange.com/a/3348044/836719 has suggested.

Thank you!

答:
The function
$$
  \text{lg}:S^1\to[0,2\pi),
  $$
defined by $\text{lg}(e^{i\theta})=\theta$, for all $\theta \in  [0,2\pi)$, is Borel measurable on $\sigma (u)$, so  the Borel
functional calculus provides a meaning to $\text{lg}(u)$, and one has that
$$
  e^{i\, \text{lg}(u)}=u.
  $$


答:
The 'ambiguity' here comes from the fact that we use parenthesis as a multiplication operator as well. We can agree that $2(1+2) = 2 \times (1+2)$

The problem is then:
$ 6 \div 2 \times (1+2)$

By convention, multiplication and division have the same hierarchy and when this happens, we always go from left to right. Conventions are followed to avoid ambiguous situations such as this.

If you mean to use the division operator as you did, then you're not following the usual convention. That is, you're using the same symbols differently than what most mathematicians would interpret it.

Obviously, if you're using symbols differently, then anything can happen.

How calculators evaluate the input all depends on how it was programmed, hence why there is the 'ambiguity'


问:
Kronecker Zeta function
If we define the Kronecker symbol K(a,n) as at Wikipedia, can we define

$$\zeta_K(s,a) = \sum_n \dfrac{K(a,n)}{n^s}$$? If so, what does it equal?

答:
Although not a complete answer, I will list some observations.

First of all, the series is certainly defined for $\text{Re}(s)>1$, since the absolute value of $\left(\frac an\right)$ is bounded ($\le1$).

Now, since $K(a,n)$, by definition, is completely multiplicative with respect to $n$, we have the following Euler product:

$$\zeta_K(s,a)=\prod_{p:\text{prime}}\left(1-\left(\frac ap\right)p^{-s}\right)^{-1}.$$

Moreover, in the case $a$ is of the form $2^k(4m+1)$, because of quadratic reciprocity we have

$$\zeta_K(s,a)=\sum_{n=1}^\infty\left(\frac na\right)n^{-s},$$

where clearly $\left(\frac na\right)$ is a Dirichlet character modulo $a$. Thus, we have $\zeta_K(s,a)=L(s,\chi)$ with $\chi(n)=\left(\frac n a\right)$.

Title: Kronecker Zeta function


问:
Is it possible to identify a unique square within a grid given its vertices?
For example, I have a grid where the x-axis goes from -1.5 to 1.5, with an interval of 0.3 (i.e. -1.5 would go to -1.2 then -0.9 and so on).

The y-axis goes from 3.0 to -3.0 with an interval of 0.3 as well.

How can I identify a unique square within this grid just by its vertices?

I tried to sum up the vertices which make up the square, but the sum of the vertices is not unique and can apply to a few other squares within the grid as well.

EDIT FOR CLARIFICATION:

Is there a way to use vertices like these to help me identify the square in the graph? I am designing a game where I want to be able to check if the box is currently occupied. I only have access to the vertices surrounding the box.

答:
You can find the center of a square by finding the mid-point of it's diagonal:
$$(h,k) = \left( \frac {x_2+x_1} {2}, \frac {y_2+y_1} {2} \right)$$
Where $(h,k)$ represents the center. You can then get the side length by finding the distance between any two adjacent vertices:
$$\text{side length (r)} = \sqrt{(x_2-x_1) + (y_2-y_1)}$$
And then you can use this formula to identify the square:
$$\left|(x - h) + (y - k)  \right|+\left| (x - h)  - (y - k)  \right|=r$$


问:
Accessible explanation of the link between the distribution of primes and the Riemann zeta function.
I am looking for an accessible explanation of the link between the distribution of the primes and the Riemann zeta function.

I have read the related questions and answers here (eg this), and also via internet search, and also key recommended books (popular maths, like Prime Obsession, texts like Apostol, etc) - and they either gloss over the key points, or are aimed at readers with university level maths.

I would appreciate answers here, or pointers to explanations elsewhere.

My students understand the Euler product formula, complex functions, calculus.

Note - this is a repeat of a previous question (now deleted) as it was marked negative without explanation.

答:
$\frac{-\zeta'(s)}{s\zeta(s)}$ is the Laplace transform of $\psi(e^u)$.

If $\psi(e^u)-e^u=O(e^{\sigma u})$ then $\frac{-\zeta'(s)}{s\zeta(s)}-\frac1{s-1}$ is analytic for $\Re(s)>\sigma$. The converse is the purpose of the Tauberian theorems, highly non-trivial, the topic of several textbooks.

The main results are

the PNT $\psi(e^u)-e^u=o(e^{u})$ (following from that $\frac{-\zeta'(s)}{s\zeta(s)}-\frac1{s-1}$ is analytic on $\Re(s)\ge 1$ and shifting the inverse Laplace transform integral to the left),

that under the RH $\psi(e^u)-e^u=O(e^{u/2} u^2)$ (shifting the inverse Laplace transform integral even more to the left after proving a lot of non-trivial facts on $\zeta(s)$)

and the Riemann explicit formula.


答:
In addition to the other excellent answers, I wanted to add another example of group actions. Let $S$ be a square inside a plane and let $G$ be the set of all rigid transformations $f$ of the plane (i.e.  rotations, translations, reflections, or their combination), such that $f(S)=S$. Turn $G$ into a group by equipping it with composition. Now notice that $G$ acts in a natural way not only on $S$, but also on the set of vertices of $S$, as well as on the set of sides of $S$, the set of diagonals of $S$, and so on...

Of course you can do the same thing starting with any geometric object in any space with a properly selected set of transformations (bijections of the space that preserve/have some desired properties).

Title: Intuition behind Group Action


问:
Quick Confirmation of Fourier series using trigonometric identities
The Fourier series expansion  for $f(x) = \sin 5x \sin x$ is $\dfrac{\cos 4x - \cos 6x}{2}$? This makes sense as $f(x) = \sin 5x \sin x$ is made up of the product of two odd functions which equals an even function and hence why there are are no sine functions in the answer.

Cheers

答:
Note that for $x,y \in \mathbb R$, we have by the addition formulae for the cosine:
$$ \cos(x \pm y) = \cos x \cos y \mp \sin x \sin y
$$
Subtracting, we get
$$ \cos(x+y) - \cos(x-y) = -2 \sin x \sin y $$
Now let $y = 5x$, to obtain
$$ \cos(6x) - \cos(4x) =- 2 \sin x \sin 5x $$</p>


问:
Is the following a Bijective Function?
In a bijective function is it necessary that all the elements of the domain correspond to a value in the range? Like for example can the following be a bijective function -

If not then what type of function is it?

答:
As per the diagram, it's not a function. A function assigns to every value $x$ in domain (which from the diagram seems to be the set $\{1,2,3,4,5,6\}$)  a unique value in Co-domain (which seems to be equal to range from the diagram). Therefore, in your case, $f:\{1,2,3,4,5,6\}\to \{a_1,a_2,a_3,a_4\}$ is  not a function.

However, if you consider restriction of $\{1,2,3,4,5,6\}$ to $\{1,2,3,4\}$, then yes, $g:\{1,2,3,4\}\to \{a_1,a_2,a_3,a_4\}$ is a bijection in the way your diagram shows.

Title: Is the following a Bijective Function?


Title: Is the following a Bijective Function?

Body:
In a bijective function is it necessary that all the elements of the domain correspond to a value in the range? Like for example can the following be a bijective function -

If not then what type of function is it?

Body:
This is a bijection from the set $\{1,2,3,4\}$ to the set $\{a_1,a_2,a_3,a_4\}.$

It is NOT a bijection from $\{1,2,3,4,5,6\}$ to $\{a_1,a_2,a_3,a_4\}.$


问:
Finding the area of my lake
Last year I took AP Calculus BC and learned some basic information on areas of irregular objects (Riemann sums and thats about it). I thought it was very interesting and this year we were assigned a math project on the topic of our choice.

I have a lake behind my house that I decided to find the area of for the project. I went into google maps and clicked on some points, but I'm not sure what to do from here. I've been reading online and saw some interesting methods like Green's theorem and Planimetry, but I'm not sure how to utilize those in the context of my lake since I don't have an equation. I assumed I would have to plot these points on a graph to start off with, but I'm not sure where or how to do that using the information from my Google Map. Can anyone help me mathematically go about this? For reference, an image of the lake is attached below:

(I know it shows the area, I just want to learn - and show - how to do that)

Heres a link to the lake for anyone that wants to experiment with the 'points' on a google map. You just right click and press 'measure distance'.

https://goo.gl/maps/M8WH6wPYGyxMP97z9

答:
Something that might interest you is the shoelace formula for the area of a polygon given the cartesian coordinates of the vertices. According to the wikipedia page:

[The shoelace formula] can be verified by dividing the polygon into triangles, and can be considered to be a special case of Green's theorem.

Mathologer has a youtube video on this topic. His videos are a delight to watch and I always learn something new when watching them, so I recommend checking that out.

As for applying the formula to your situation, we would like to choose an origin and set up a cartesian coordinate system. One way to do this is using a program like photoshop which displays the cartesian coordinates of the cursor in terms of pixels. Then using the shoelace formula you could find an estimate $A$ of the area of the lake in square pixels. After that you could convert pixels to metres by having a reference length in the image (e.g. by using the length scale provided by google) and measuring that in terms of pixels again using a photo editor. From this we could find the value of $x$ in the conversion $1 \ \text{pixel} = x \ \text{metres}$, then our estimate of the area of the lake in square metres is $x^2 A$.

As pointed out in the comment section of the question, the calculation of the area won't be exact, so ideally we would want to add an uncertainty $\delta$ in the measurement of the area of the lake so that we are confident that the true area of the lake in square metres lies in the interval $[x^2 A - \delta, x^2 A + \delta]$.

There are various ways to do this, one being to find the area of a polygon that envelopes the lake and finding the area of a polygon lying inside the lake. These will give you upper and lower bounds for the true area, allowing for a suitable choice of $\delta$.

Title: Finding the area of my lake