问:
probability of selecting two roses of different colors when two roses are drawn at random from $2$ red roses, $4$ yellow roses and $6$ pink roses
A bag contains $2$ red roses, $4$ yellow roses and $6$ pink roses. Two roses are drawn at random. What is the probability that they are not of same color?

This problem can be easily solved with $$\frac{2C1 \cdot 4C1 + 4C1 \cdot 6C1 + 6C1 \cdot 2C1}{12C2} = \frac{2}{3}$$

But I want to do this using the ratio method and here is my attempt:
RY + YP + PR =
$$\frac{2}{12} \cdot \frac{4}{12} + \frac{4}{12} \cdot \frac{6}{12} + \frac{6}{12} \cdot \frac{2}{12} = \frac{11}{36}$$
which is way off, what am I doing wrong? and also how does one decide when to use the ratio method and when to use the Combinatorics method.

Thank you for any help.

答:
As I said in my comment, you must also count $YR,PY$, and $RP$. Note that when you are picking the roses for $R$, once you pick a rose for $Y$, you must use $11$ in the denominator as one rose has already been picked out.

So then we have the following:

$\frac{2}{12}\cdot \frac {4}{11} + \frac{4}{12} \cdot \frac {6}{11} + \frac{6}{12} \cdot \frac{2}{11} = \frac{11}{33}$.

Then we have $2\cdot\frac{11}{33} = \frac {22}{33} = \frac{2}{3}$ after scaling by two when taking into account the other three cases listed above as the resulting value is clearly the same from the commutativity of multiplication.


答:
Your computation using direct probabilities has two types of errors:

(a) you have forgotten that each duo can occur in two sequences.

(b) The denominator for any duo can't remain the same !

So the correct computation presented taking into account the above simply, would be

$\frac{2}{12\cdot 11}(2\cdot4 + 4\cdot 6 + 6\cdot 2) $

The two answers must necessarily tally, and you need to decide which of the two methods is easier for the particular problem. In general, direct use of probabilities is simpler only when there is a fixed order, else beginners tend to make errors.

ContentLicense: CC BY-SA 4.0

Title: probability of selecting two roses of different colors when two roses are drawn at random from $2$ red roses, $4$ yellow roses and $6$ pink roses

Body:
A bag contains $2$ red roses, $4$ yellow roses and $6$ pink roses. Two roses are drawn at random. What is the probability that they are not of same color?

This problem can be easily solved with $$\frac{2C1 \cdot 4C1 + 4C1 \cdot 6C1 + 6C1 \cdot 2C1}{12C2} = \frac{2}{3}$$

But I want to do this using the ratio method and here is my attempt:
RY + YP + PR =
$$\frac{2}{12} \cdot \frac{4}{12} + \frac{4}{12} \cdot \frac{6}{12} + \frac{6}{12} \cdot \frac{2}{12} = \frac{11}{36}$$
which is way off, what am I doing wrong? and also how does one decide when to use the ratio method and when to use the Combinatorics method.

Thank you for any help.


问:
Multivariable Function. Differentiable $\Rightarrow$ Partial Differentiable.
I cannot understand why this proposition holds. Let

$U$ be an open set in $\Bbb R^{n+m}(=\Bbb R^n\times \Bbb R^m)$,

$f$ : $U\to \mathbb{R}^d$ be a function,

$x_0 \in \Bbb R^n, y_0 \in \Bbb R^m,$ be two points such that $(x_0,y_0)\in U$.

Then we can pick up a neighbourhood $V$ of $x_0$, and define
$g : V \to \mathbb{R}^d$ as

$$
g(x):=f(x,y_0).
$$
If $f$ is differentiable at $(x_0,y_0)(=:z_0)$, i.e.
\begin{equation}
\lim_{h\to 0} \dfrac{\|f(z_0+h)-f(z_0)-Df(z_0)h\|}{\|h\|}=0
\end{equation}
holds,
then $g$ is differentiable, i.e.
\begin{equation}
\exists A \ (\text{a $d\times n$  matrix}) \ ; \
\lim_{h\to 0} \dfrac{\|g(x+h)-g(x)-Ah\|}{\|h\|}=0
\end{equation}
holds (In other words, $f$ is partial differentiable about $x$ at $(x_0,y_0)$).

I have no idea to prove this.

Maybe I have to evaluate the quantity
$$\dfrac{\|g(x+h)-g(x)-\square h\|}{\|h\|}
$$
but I don't know what $\square$ is.

I would like you to tell me some ideas.

答:
You have to specialize your $h$. Choose $h=(k_1,k_2,\dots k_n,0,0,\dots 0)$ where there are $m$ zeroes. Call $k=(k_1,k_2,\dots k_n)$. Observe that $g(x)=f(x,y_0)$ and $g(x_0+k)=f(z_0+h)$. Using the definition of differentiability of $f$ at $z_0$ you get
$$
\frac{\|g(x_0+k)-g(x_0)-Ak\|}{\|k\|}\to 0
$$
where $k$ is arbitrary and $A$ is the sub matrix of $D$ consisting of the first $n$ columns of the matrix $D$. Therefore, $g$ is differentiable at the point $x_0$ and its Jacobian matrix is $A$.


问:
Evaluating $\lim_{x \to 0} x\sin\left(\frac{1}{x}\right)$ or showing it does not exist

Does the following limit exist?
$$\lim_{x \to 0} x\sin\left(\frac{1}{x}\right)$$

If yes, then I have to find it; if no, then I need to give reason why.

I couldn’t figure out how to prove this formally, can someone please help me out.

答: You can let $$z=\frac{1}{x}$$ so then the limit becomes $$\lim_{z \rightarrow \infty} \frac{1}{z} \sin z \:,$$ which clearly pulls down to zero.


问:
On the proof of Thm.3.50 in Baby Rudin
I have a question on the proof of Thm.3.50 in Rudin's Principles of Mathematical Analysis. This theorem is about the convergence of the Cauchy product of two infinite series. For convenience, I write the necessary details of the theorem and proof.

Theorem

Suppose

(a) $\sum_{n = 0}^{\infty}a_n$ converges absolutely,

(b) $\sum_{n = 0}^{\infty}a_n = A$,

(c) $\sum_{n = 0}^{\infty}b_n = B$

(d) $c_n = \sum_{k = 0}^{n}a_kb_{n - k}.$

Then,
\begin{align*}
\sum_{n = 0}^{\infty}c_n = AB.
\end{align*}

Proof

Put
\begin{align*}
A_n = \sum_{k = 0}^{n}a_k, \ \  B_n = \sum_{k = 0}^{n}b_k, \ \ C_n = \sum_{k = 0}^{n}c_k, \ \  \beta_n = B_n - B.
\end{align*}
Then we can rewtrite $C_n$ as
\begin{align*}
C_n = A_nB + a_0\beta_n + a_1\beta_{n-1} + \cdots + a_n\beta_0.
\end{align*}
Put
\begin{align*}
\gamma_n = a_0\beta_n + a_1\beta_{n-1} + \cdots + a_n\beta_0 \\
\end{align*}
so that
\begin{align*}
C_n = A_nB + \gamma_n.
\end{align*}
Our aim here is to show $C_n \to AB$. Since $A_nB \to AB$, it suffices to show
\begin{align*}
\gamma_n \to 0.
\end{align*}
Put
\begin{align*}
\alpha = \sum_{n = 0}^{\infty}|a_n|.
\end{align*}
By (c), we have that $\beta_n \to 0$, or that for any $\varepsilon > 0$, there is some integer $N$ such that $|\beta_n| \leq \varepsilon$ for $n \geq N$. Hence, for sufficiently large n
\begin{align}
|\gamma_n| &\leq |\beta_0a_n + \cdots + \beta_Na_{n - N}| + |\beta_{N + 1}a_{n - N - 1} + \cdots + \beta_na_0| \\
&\leq |\beta_0a_n + \cdots + \beta_Na_{n - N}| + \varepsilon\alpha. \tag{1}
\end{align}
Keeping $N$ fixed, and letting $n \to \infty$, we get
\begin{align}
\limsup_{n \to \infty} |\gamma_n| \leq \varepsilon\alpha, \tag{2}
\end{align}
since $a_n \to 0$ as $n \to \infty$. Since $\varepsilon$ is arbitrary, we obtain $\lim_{n \to \infty}\gamma_n = 0.$ (end)

Question

The quetion I have here is about (2). I see that, by triangular inequality and the fact that $|a_n| \to 0$, the term $|\beta_0a_n + \cdots + \beta_Na_{n - N}|$ goes to zero with $N$ fixed. Because we take the limit of (1), it seems to me that we obtain
\begin{align}
\lim_{n \to \infty} |\gamma_n| \leq \varepsilon\alpha \tag{3}
\end{align}
instead of (2). I notice that we get (3) if (2) holds since $|\gamma_n|$ is nonnegative, but I feel we can derive (3) directly from taking the limit of (1). Am I wrong in some points?

Any comment would be appriciated. Thank you!

答:
If $|t_n| \leq s_n$ and $\lim s_n=s$ we cannot say that $\lim t_n$ exists. What we can say is $\lim \sup |t_n| \leq \lim \sup s_n=\lim s_n=s$ since $\lim \sup$ and $\lim $ are the same for  a convergent sequence $(s_n)$.

Example: Let $t_n=0$ for $n$ even and $t_n=1$ for $n$ odd. Let $s_n=1$ for all $n$. Then $\lim t_n$ does not exist even thoug $0 \leq t_n \leq s_n$ and $s_n \to 1$.

Title: On the proof of Thm.3.50 in Baby Rudin


答:
First Approach

To begin with, notice that
\begin{align*}
\left|x\sin\left(\frac{1}{x}\right)\right| \leq |x| < \varepsilon
\end{align*}

Thus, for every $\varepsilon > 0$, there corresponds a $\delta = \varepsilon$ s.t.
\begin{align*}
0 < |x - 0| < \delta \Rightarrow \left|x\sin\left(\frac{1}{x}\right) - 0\right| < \varepsilon
\end{align*}

whence we conclude the given function converges to zero as $x$ approaches zero.

Second Approach

The limit exists and converges to zero due to the sandwich theorem.

That is because $x\to 0$ and the $\sin$ function is bounded.

Hopefully this helps!

Title: Evaluating $\lim_{x \to 0} x\sin\left(\frac{1}{x}\right)$ or showing it does not exist

Body:
Does the following limit exist?
$$\lim_{x \to 0} x\sin\left(\frac{1}{x}\right)$$

If yes, then I have to find it; if no, then I need to give reason why.

I couldn’t figure out how to prove this formally, can someone please help me out.


答:
Yes, the limit there exist. Note that the funcion $f(x)=\sin(x)$ is a bounded function, with   $$|\sin(x)|\leq 1 \iff -1\leq \sin(x)\leq 1$$
Therefore, $$-1\leq \sin\left(\frac{1}{x}\right)\leq 1$$
$$\lim_{x\to 0} -x \leq \lim_{x\to 0} x \sin\left(\frac{1}{x}\right)\leq \lim_{x\to 0}x$$
$$ 0\leq \lim_{x\to 0} x\sin\left(\frac{1}{x}\right)\leq 0$$
Finally, by the well-know Squeeze theorem we can conclude that $$\color{blue}{\boxed{\lim_{x\to 0}x\sin\left(\frac{1}{x}\right)=0}}$$


问:
Showing that degree of polynomial reduced modulo $p$ divides degree of splitting field
Let $f$ be a polynomial of degree $n$ with integer coefficients and $p$ a prime for which $f$, considered modulo $p$, is a degree-$k$ irreducible polynomial over $\mathbb{F}_p$. Show that $k$ divides the degree of the splitting field of $f$ over $\mathbb{Q}$.

(this is Miklos Schweitzer 2020, Problem 10)

My miscellaneous thoughts: We know that for a polynomial of degree $n$, the degree of its splitting field is at most $n!$ from induction on the Ring Tower Theorem. Additionally, since $f(x)$ is irreducible over $\mathbb{F}_p$, then the field $\mathbb{F}_p[x]/(f(x))$ has a root of $f(x)$. Now suppose $k = 2$. Then $f(x) = a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0 \in \mathbb{Z}[x]$ reduces to $f(x) \equiv b_2 x^2 + b_1x + b_0 \in \mathbb{F}_p [x]$ where $b_i \equiv a_i \pmod{p}$ where applicable. I'm really not sure if these things are particularly useful or how to combine them.

This is really hard!

答:
I'll start with a hint/sketch as Quinn does in his answer, if someone wants to try this before reading the details: factor $f(x)$ in $\mathbf Z_p[x]$  (not just in $\mathbf Q_p[x]$) using Hensel's lemma or Newton polygons to get a factor of degree $k$ in $\mathbf Z_p[x]$ that has an irreducible reduction mod $p$ of degree $k$. Then think of the Galois group of the splitting field of $f(x)$ over $\mathbf Q_p$ as a decomposition group at a prime ideal over $p$ in the integers of the splitting field of $f(x)$ over $\mathbf Q$.

We will use the decomposition group at a prime ideal, as Quinn does in the other solution posted here, but we'll interpret decomposition groups as Galois groups of $\mathfrak p$-adic completions.  This has the advantage that a decomposition group is genuinely a Galois group over $\mathbf Q_p$ even in the case of ramified primes, whereas if you work with localizations then you can say $D(\mathfrak p|p)/I(\mathfrak p|p)$ is a Galois group over $\mathbf F_p$ but you can't say that about $D(\mathfrak p|p)$ itself for all $p$.

The first step is to factor $f(x)$ in $\mathbf Z_p[x]$ into a factor of degree $k$ and a complementary factor, where the first factor reduces mod $p$ to an irreducible polynomial in $\mathbf F_p[x]$.  I will show two ways to do this: by Hensel's lemma and by Newton polygons.

Method 1: Factoring $f(x)$ in $\mathbf Z_p[x]$ by Hensel's lemma.

We are going to use a version of Hensel's lemma in $\mathbf Z_p[x]$ that allows for polynomials whose leading coefficients are not $p$-adic units. Here is the version I have in mind.

Theorem:  Let $f(x) \in \mathbf Z_p[x]$ with at least one coefficient being in $\mathbf Z_p^\times$, and assume the reduced polynomial $\overline{f}(x)$ in $\mathbf F_p[x]$ can be written as $\alpha(x)\beta(x)$ where $\alpha(x)$ and $\beta(x)$ are relatively prime polynomials in $\mathbf F_p[x]$. Then in $\mathbf Z_p[x]$ we can write $f(x) = a(x)b(x)$ where (i) $\overline{a}(x) = \alpha(x)$ and $\overline{b}(x) = \beta(x)$ in $\mathbf F_p[x]$ and (ii) $\deg a(x) = \deg \alpha(x)$.

Proof: See Gouvea, "$p$-Adic Numbers: an Introduction" (2nd ed.) Theorem 3.4.6, p. 72 or Borevich and Shafarevich, "Number Theory," Theorem 2, p. 275 for a proof.

Remark. I am using the hypotheses as they appear in Borevich and Shafarevich. Gouvea instead essentially assumes $\alpha(x)$ is monic and says in the conclusion that $a(x)$ is monic, but that is morally the same kind of condition as $\deg a(x) = \deg \alpha(x)$ in the theorem above.

Note $\overline{f}(x)$ is not $0$ in $\mathbf F_p[x]$ since some coefficient of $f(x)$ (not necessarily the leading coefficient!) is in $\mathbf Z_p^\times$.

We are not requiring $\deg b(x) = \deg \beta(x)$, and the following example illustrates that.

Example. In $\mathbf Z_2[x]$ let $f(x) = 6x^5 + 4x^3 + x^2 + x + 3$. Then in $\mathbf F_2[x]$, $\overline{f}(x) = x^2+x+1$, which is irreducible and the degree has dropped from $5$ down to $2$. We can write $\overline{f}(x)$ in $\mathbf F_2[x]$ as a product of relatively prime factors $\alpha(x)\beta(x)$ in two ways, and each lifts to a corresponding factorization of $f(x)$ as $a(x)b(x)$ in $\mathbf Z_2[x]$ where $\deg a(x) = \deg \alpha(x)$.

First way: $\alpha(x) = 1$ and $\beta(x) = x^2+x+1$. Then we can use $a(x) = 1$ and $b(x) = f(x)$. Indeed, $f(x) = 1 \cdot f(x)$, $\overline{1}(x) = \alpha(x)$, $\overline{f}(x) = \beta(x)$, and $\deg a(x) = \deg \alpha(x)$ (but $\deg b(x) = 6 \not= \deg \beta(x))$. This is how the theorem looks when we use $\alpha(x) = 1$. It is silly, but it is valid.

Second way: $\alpha(x) = x^2+x+1$ and $\beta(x) = 1$. Then the theorem says $f(x) = a(x)b(x)$ in $\mathbf Z_2[x]$ where $\overline{a}(x) = x^2+x+1$, $\overline{b}(x) = 1$, and $\deg a(x) = 2$ (but $\deg b(x) = 3 \not= \deg \beta(x)$). Using PARI,
$$
a(x) = x^2 + (1 + 2 + 2^5 + 2^{15} + 2^{16} + \cdots)x + (1 + 2^2 + 2^5 + 2^6 + 2^{10} + 2^{13} + 2^{16} + \cdots)
$$
and
$$
b(x) = 6x^3 + (2 + 2^2 + 2^3 + 2^5 + 2^8 + 2^9 + 2^{10} + 2^{11} + 2^{12} + 2^{13} + 2^{14} + 2^{15} + \cdots)x^2 + (2^2 + 2^3 + 2^4 + 2^6 + 2^9 + 2^{14} + 2^{16} + \cdots)x + (1 + 2 + 2^2 + 2^6 + 2^{11} + 2^{14} + 2^{15} + \cdots).
$$
Notice $\overline{a}(x) = x^2+x+1$ and $\overline{b}(x) = 1$ in $\mathbf F_2[x]$.  This concludes the example.

We'll now apply Hensel's lemma to the contest problem. We are told $f(x) \in \mathbf Z[x]$ has degree $n$ and $\overline{f}(x) = \pi(x)$ in $\mathbf F_p[x]$ is irreducible of degree $k$, so $1 \leq k \leq n$. Use Hensel's lemma with $\alpha(x) = \pi(x)$ and $\beta(x) = 1$, as in the "second way" of the example above.  We obtain $f(x) = a(x)b(x)$ in $\mathbf Z_p[x]$ where $\overline{a}(x) = \pi(x)$, $\overline{b}(x) = 1$, and $\deg a(x) = \deg \pi(x) = k$. That means $a(x)$ has a leading coefficient in $\mathbf Z_p^\times$, since its degree did not drop under reduction mod $p$. So $a(x)$ is a polynomial of degree $k$ in $\mathbf Z_p[x]$ whose reduction mod $p$ is irreducible of degree $k$.

Method 2:  Factoring $f(x)$ in $\mathbf Z_p[x]$ by Newton polygons.

By the hypotheses of the problem, the constant term $f(0)$ is not divisible by $p$, the coefficient of $x^k$ in $f(x)$ is not divisible by $p$, coefficients of $f(x)$ of degree strictly between $0$ and $k$ may or may not be divisible by $p$ (but at least they are integral) and the coefficients of $f(x)$ of degree greater than $k$ are all divisible by $p$. Therefore the first segment of the $p$-adic Newton polygon of $f(x)$ is the segment on the $x$-axis from $(0,0)$ to $(k,0)$. Further segments of the $p$-adic Newton polygon of $f(x)$ have positive slope, so $f(x)$ has $k$ roots in $\overline{\mathbf Q_p}$ with $p$-adic absolute value $1$ ($p$-adic valuation $0$) and its other roots in $\overline{\mathbf Q_p}$ have $p$-adic absolute value greater than $1$.

Numbers algebraic over $\mathbf Q_p$ that are roots of the same irreducible polynomial in $\mathbf Q_p[x]$ have the same $p$-adic absolute value, so the $k$ roots of $p$-adic absolute value $1$ are $\mathbf Q_p$-conjugate only to other roots of $p$-adic absolute value $1$. Therefore if we let $a(x)$ be the product of the monic irreducible factors of $f(x)$ in $\mathbf Q_p[x]$  whose roots have $p$-adic absolute value $1$,  then $\deg a(x) = k$, $a(x)$ is in $\mathbf Z_p[x]$ (not just $\mathbf Q_p[x]$!) and $f(x) = a(x)b(x)$ where $b(x) \in \mathbf Q_p[x]$.  We want to show $b(x) \in \mathbf Z_p[x]$. Since $a(x)$ is monic in $\mathbf Z_p[x]$ and $f(x) \in \mathbf Z_p[x]$, we can apply the division algorithm in $\mathbf Z_p[x]$ to write $f(x) = a(x)q(x) + r(x)$ in $\mathbf Z_p[x]$ where $r(x) = 0$ or $\deg r(x) < k$.
Viewing this in $\mathbf Q_p[x]$ and comparing it to $f(x) = a(x)b(x)$ in $\mathbf Q_p[x]$, the uniqueness of the division algorithm  in $\mathbf Q_p[x]$  tells us that $q(x) = b(x)$ and $r(x) = 0$. Therefore $b(x) \in \mathbf Z_p[x]$, so the factorization $f(x) = a(x)b(x)$ is in $\mathbf Z_p[x]$. Why is $a(x) \bmod p$ irreducible in $\mathbf F_p[x]$?

Reduce the equation $f(x) = a(x)b(x)$ modulo $p$: in $\mathbf F_p[x]$, $\overline{f}(x) = \overline{a}(x)\overline{b}(x)$. Since $a(x)$ is monic of degree $k$, $\overline{a}(x)$ has degree $k$ in $\mathbf F_p[x]$. By the hypothesis of the problem, $\overline{f}(x)$ is irreducible of degree $k$.    Therefore by unique factorization in $\mathbf F_p[x]$, $\overline{a}(x)$ is irreducible and $\overline{b}(x)$ is a nonzero constant.

Now we continue with the rest of the argument, knowing (by two methods) that $f(x)$ factors as $a(x)b(x)$ in $\mathbf Z_p[x]$ where $a(x) \bmod p$ is irreducible of degree $k$.  By the reduction mod $p$ test, $a(x)$ is irreducible in $\mathbf Q_p[x]$.

Let $r_1, \ldots, r_n$ be a full set of roots of $f(x)$ in an extension of $\mathbf Q_p$.  Then $\mathbf Q_p(r_1, \ldots, r_n)$ is a splitting field of $f(x)$ over $\mathbf Q_p$ and $K := \mathbf Q(r_1, \ldots, r_n)$ is a splitting field of $f(x)$ over $\mathbf Q$.  We want to show $k \mid [K:\mathbf Q]$.

Some $r_i$ is a root of $a(x)$, so $[\mathbf Q_p(r_i):\mathbf Q_p] = \deg a(x) = k$ on account of irreducibility of $a(x)$ over $\mathbf Q_p$. Therefore $k$ divides $[\mathbf Q_p(r_1,\ldots,r_n):\mathbf Q_p]$.

Let $\mathfrak p$ be a prime ideal lying over $p$ in $\mathcal O_K$. Then the $\mathfrak p$-adic completion $K_{\mathfrak p}$ is isomorphic to $\mathbf Q_p(r_1,\ldots, r_n)$. (Since $K/\mathbf Q$ is Galois, its completions at different prime ideals lying over $p$ are isomorphic as extensions of $\mathbf Q_p$, so the specific choice of $\mathfrak p$ lying over $p$ is not important; any one is as good as any other.) The extension $K_{\mathfrak p}/\mathbf Q_p$ is Galois and ${\rm Gal}(K_{\mathfrak p}/\mathbf Q_p)$ is isomorphic to the decomposition group $D(\mathfrak p|p)$ in ${\rm Gal}(K/\mathbf Q)$. From the "$efg = n$" formula for Galois extensions of number fields, $|{\rm Gal}(K/\mathbf Q)|$ is divisible by $|D(\mathfrak p|p)|$, so $[K:\mathbf Q]$ is divisible by $|{\rm Gal}(K_{\mathfrak p}/\mathbf Q_p)|$, which equals $[K_{\mathfrak p}:\mathbf Q_p]$, and that is divisible by $k$. Thus $k \mid [K:\mathbf Q]$.

In fact, since $a(x)$ has an irreducible reduction mod $p$, each root of $r$ of $a(x)$ generates an unramified extension of $\mathbf Q_p$, so $[\mathbf Q_p(r):\mathbf Q_p]$ divides the degree of the residue field extension for $K_{\mathfrak p}/\mathbf Q_p$. Residue field degrees do not change under completion, so $k$ divides $f(\mathfrak p|p)$, which is also what Quinn found at the end of his answer ($k \mid [(\mathcal O_K)_{\mathfrak p}/\mathfrak p:\mathbf F_p]$).

Боже мой, какая замечательная задача!


问:
Quadrilateral $ABCD$ with chord of inscribed circle meeting $AC$ at $P$. Find ratio $AP$ : $PC$
In a quadrilateral $ABCD$, there is an inscribed circle centered at $O$. Let $F,N,E,M$ be the points on the circle that touch the quadrilateral, such that $F$ is on $AB$, $N$ is on $BC$, and so on. It is known that $AF=5$ and $EC=3$. Let $P$ be the intersection of $AC$ and $MN$. Find the ratio $AP:PC$.

I know that $AM=AF=5$ and $CN=CE=3.$ The answer is equivalent to the ratio of the areas $[ADP]:[DPC]$. I cannot continue on from this point. Would anyone please help?

答:
COMMENT:

As can be seen ratio $\frac{AP}{PC}$ depends on the positions of C related to A. So more constrains must be included, for example " A, O(center of circle) and C are colinear, otherwise there can be infinite solutions.


问:
Trigonometric Identity Problem - Cos Tan and Sin
I have been going through my lecture notes for a structures question (the solution of a 2nd order ode for a buckling problem) when I came across a very weird trigonometric simplification which I just cannot get my head around. Could anyone shed any light on this?

Further down the page, there is another similar simplification, but this one seems to be incorrect:

Testing some random values, the simplification is incorrect unless I replace the C/2 with a C.

答:
This simplification is correct as

$$\cos t+\tan 2t\sin t=\cos t+\frac{\sin 2t}{\cos 2t}\sin t=\frac{\cos t\cos 2t+\sin t\sin 2t}{\cos 2t}=\frac{\cos(2t-t)}{\cos2t} $$

Similarly, $$-\sin t+\tan2t\cos t=-\sin t+\frac{\sin2t\cos t}{\cos2t}=\frac{\sin2t\cos t-\sin t\cos2t}{\cos2t}=\frac{\sin(2t-t)}{\cos2t}$$

Put $t=\frac{\mu'L}4$

ContentLicense: CC BY-SA 3.0


答:
$$-1 \le \sin {x} \le 1$$

$$-|x| \le x \sin \frac{1}{x} \le |x|$$

$$\lim_{x \to 0} -|x| \le \lim_{x \to 0} x \sin \frac{1}{x} \le \lim_{x \to 0} |x|$$

$$0 \le \lim_{x \to 0} x \sin \frac{1}{x} \le 0$$

This means a limit does exist and is 0:

$$\lim_{x \to 0} x \sin \frac{1}{x} = 0$$


问:
Are there infinitely many primes which are factors of no terms in the sequence $a_0 = 4$, $a_{n}=$ $a_{n-1}^{2}-a_{n-1}$?
Problem statement:

Let $a_{0}=4$ and define a sequence of terms using the formula $a_{n}=$ $a_{n-1}^{2}-a_{n-1}$ for each positive integer $n$

a) Prove that there are infinitely many prime numbers which are factors of at least one term in the sequence;

b) Are there infinitely many prime numbers which are factors of no term in the sequence?

My attempted proof is as follows:

a) Write $a_n = a_{n-1} (a_{n-1} - 1)$. If a prime $p$ divides $a_{n-1}$ then $p$ cannot divide $a_{n-1} - 1$, else $p$ would divide their difference, $1$. So there must exist another prime $p_1 \ne p$ such that $p_1 \mid (a_{n-1} - 1)$ which implies $p_1 \mid a_n$. Similarly, since $p$ and $p_1$ both divide $a_n$, neither can divide $a_n - 1$, so $\exists p_2 \ne p,p_1$ such that $p_2 \mid a_{n+1}$. Continuing in this way we generate an infinite sequence of distinct primes $(p_k)_k$ where $p_k \mid a_{n+k-1}$.

b) I noticed that if $a_N \cong 2 \pmod p$ for some $N$, then $a_n \cong 2 \pmod p$ for all $n \ge N$ by the recurrence relation, since $2\times(2-1) = 2$. Clearly such a prime cannot be a divisor of any term in the sequence, for if it were, each term in the sequence thereafter would be congruent to $0 \pmod p$. It thus suffices to find infinitely many primes $p$ such that $a_n \cong 2 \pmod p$ for at least one term in the sequence. If we define a new sequence $b_n = a_n - 2$ then this is equivalent to finding infinitely many primes $p$ which are factors of at least one term in the sequence $(b_n)_n$. This suggests that we could perhaps proceed with a similar argument to that made in part a). The new recurrence relation becomes $b_n + 2 = (b_{n-1} + 2) (b_{n-1} + 1)$, or simplifying: $b_n = b_{n-1}(b_{n-1} + 3)$. The argument should now be almost identical to that in part a), with the observation that $p \ne 3$ becomes the sequence $(b_n)_n$ modulo $3$ is $2, 1, 1, 1, \ldots$.

Is this argument correct? I would be interested in seeing other solutions too, in any case.

答:
the numbers in the sequence are divisible by $2$ and $3$ but are not  divisible by any larger prime $q$ with Legendre symbol $(q,5) = -1.$

For $j\ge1$ We have a sequence $$ t_j = 1, 5, 65, 8645, ...$$
which is $t_{j+1} = t_j (2 t_j + 3) $
Your sequence numbers are $a_j =4 t^2 + 6 t + 2$ and the new multipliers are
$$a_{j+1}/a_j = a_j-1 = 4 t^2 + 6t + 1 = (3t+1)^2 - 5 t^2 $$
As this $a_j - 1$ is primitively represented by $x^2 - 5 y^2,$  it cannot be divisible by any prime $q$ where $q \equiv 2,3 \pmod 5$. Thereby no prime factors $\equiv 2,3\bmod 5$ divide any $a_j$ except the factors $2$ and $3$ of $a_1=12$.

The first few such ``multipliers'' $a_{j+1}/ a_j$ are
$$  11 $$
$$  131  $$
$$ 17291 $$
$$  298995971  $$
$$   89398590973228811  =   8779  \cdot       10079  \cdot     1010341471  $$
$$   7992108067998667938125889533702531 =                  29  \cdot                    59  \cdot                     241  \cdot             2511683491  \cdot       7716660340023314591 $$
$$ 63873791370569400659097694858350356285036046451665934814399129508491 =    109  \cdot    2309  \cdot     1307641  \cdot     1312951  \cdot     4627751  \cdot   19075520521  \cdot   1674514206139655462590242193707851  $$

Notice how each prime factor $p$ has last (decimal) digit $1$ or $9$, and are thereby $p \equiv\pm 1 \pmod{10}$   so  $p \equiv\pm 1 \pmod{5}$

Thus no prime $q > 3$ appears as a factor when   $q \equiv\pm 3 \pmod{5},$   that is $q \equiv\pm 3 \pmod{10},$ so that the last decimal digit of this $q$ is $3$ or $7.$  These ratios are not divisible by $7, 13, 17, 23, 37, 43, 47...$


问:
Determinants of matrices defined by the minimum/maximum indices of their entries
$
\newcommand{\m}[1]{\left( \begin{matrix} #1 \end{matrix} \right)}
$
The Problem: Consider the matrices $A_n := \left(a^{(n)}_{i,j}\right)_{1 \le i,j \le n},B_n := \left(b^{(n)}_{i,j}\right)_{1 \le i,j \le n} \in M_{n \times n}(\Bbb R)$ with entries defined as follows:

$$a^{(n)}_{i,j} := \min \{i,j\} \qquad b^{(n)}_{i,j} := \max \{i,j\}$$

(Examples of these matrices follow momentarily, to show how the structure of each "evolves" and emerges as $n$ increases.)

What are $\det(A_n)$ and $\det(B_n)$ in the general $n \times n$ case, for each $n \in \Bbb Z^+$?

Background/Context: I happened to have a question similar to this on my linear algebra final a few weeks ago. (In full disclosure: final grades have already been determined and all that, so this isn't any longer an "active" test question.) However, that problem used a matrix similar to $A_n$, without specific numbers, and only the $4 \times 4$ case.

I then recently found a video by Dr. Peyam which considered the $5 \times 5$ cases for both $A_n$ and $B_n$, and hinted that these possibly came from a Putnam exam. I'm not sure if it was, though. I looked back through all of the Putnam problems back to $1985$ and didn't see this exact one, though one was similar: $A2$ from $2014$, where $A_n$ is instead defined by $a^{(n)}_{i,j} = 1/\min\{i,j\}$. (Maybe I'll look at that some other time.)

I imagine the definition for $B_n$ is just considered a natural generalization of this problem. In any event, I couldn't find this problem handled in its generality on MSE, so I figured I'd try to figure it out, and post my own solution.

Of course, if you have your own solutions, they're greatly welcomed! Each can provide their own insights.

Small Cases for the Matrices: To help show the structure of the matrices, $A_n$ and $B_n$ look like, in the cases for $n=1,2,\cdots,6$, as follows. Determinants are included, but calculated by Wolfram.

\begin{alignat*}{3}
&\boxed{n=1} &&\qquad A_1 = \m{
1  }  &&\qquad \det(A_1) = 1\\
& &&\qquad B_1 = \m{
1 }
&&\qquad \det(B_1) = 1\\
&\boxed{n=2} &&\qquad A_2 = \m{
1 & 1 \\
1 & 2  }
&&\qquad \det(A_2) = 1\\
& &&\qquad B_2 = \m{
1 & 2    \\
2 & 2   }
&&\qquad \det(B_2) = -2\\
&\boxed{n=3} &&\qquad A_3 = \m{
1 & 1 & 1 \\
1 & 2 & 2 \\
1 & 2 & 3 }
&&\qquad \det(A_3) = 1\\
& &&\qquad B_3 = \m{
1 & 2 & 3   \\
2 & 2 & 3  \\
3 & 3 & 3  }
&&\qquad \det(B_3) = 3\\
&\boxed{n=4} &&\qquad A_4 = \m{
1 & 1 & 1 & 1 \\
1 & 2 & 2 & 2 \\
1 & 2 & 3 & 3 \\
1 & 2 & 3 & 4 }
&&\qquad \det(A_4) = 1\\
& &&\qquad B_4 = \m{
1 & 2 & 3 & 4  \\
2 & 2 & 3 & 4 \\
3 & 3 & 3 & 4   \\
4 & 4 & 4 & 4  }
&&\qquad \det(B_4) = -4\\
&\boxed{n=5} &&\qquad A_5 = \m{
1 & 1 & 1 & 1 & 1 \\
1 & 2 & 2 & 2 & 2 \\
1 & 2 & 3 & 3 & 3 \\
1 & 2 & 3 & 4 & 4 \\
1 & 2 & 3 & 4 & 5 }
&&\qquad \det(A_5) = 1\\
& &&\qquad B_5 = \m{
1 & 2 & 3 & 4 & 5  \\
2 & 2 & 3 & 4 & 5 \\
3 & 3 & 3 & 4 & 5  \\
4 & 4 & 4 & 4 & 5 \\
5 & 5 & 5 & 5 & 5 }
&&\qquad \det(B_5) = 5\\
&\boxed{n=6} &&\qquad A_6 = \m{
1 & 1 & 1 & 1 & 1 & 1\\
1 & 2 & 2 & 2 & 2 & 2\\
1 & 2 & 3 & 3 & 3 & 3\\
1 & 2 & 3 & 4 & 4 & 4\\
1 & 2 & 3 & 4 & 5 & 5\\
1 & 2 & 3 & 4 & 5 & 6}
&&\qquad \det(A_6) = 1\\
& &&\qquad B_6 = \m{
1 & 2 & 3 & 4 & 5 & 6 \\
2 & 2 & 3 & 4 & 5 & 6\\
3 & 3 & 3 & 4 & 5 & 6 \\
4 & 4 & 4 & 4 & 5 & 6 \\
5 & 5 & 5 & 5 & 5 & 6 \\
6 & 6 & 6 & 6 & 6 & 6}
&&\qquad \det(B_6) = -6
\end{alignat*}

One reasonably conjectures that $\det(A_n) = 1$ for all $n$ and $\det(B_n) = (-1)^{n-1} n$ for all $n$.

答:
Here's an alternative way to think about $A_n$ using probability.

Let $X_i$ be a collection of independent unit gaussians, and let $S_i:=\sum_{k\leq i} X_k$ be the running sum. Define $Y_n=(S_1, \dots, S_n)$. Then $Y_n$ is multivariate gaussian, and it is easy to check that the covariance matrix is $A_n$.

The key observation is that $S_n-S_{n-1}$ is independent of $Y_{n-1}$. This is because $S_n-S_{n-1}=X_n$, while $Y_n$ depends only on $\{X_i\}_{i<n}$. Therefore, we can expand the probability density as follows:

\begin{align*}
P(Y_n)
&=P(Y_n \mid Y_{n-1})P(Y_{n-1})\\
&=P(S_n-S_{n-1} \mid Y_{n-1})P(Y_{n-1})\\
&=(2\pi)^{-1/2}e^{-(S_n-S_{n-1})^2/2}P(Y_{n-1})
\end{align*}

On the other hand, $P(Y_n)=(2\pi)^{-n/2}\det(A_n)^{-1/2}e^{-Y_n^tA_n^{-1}Y_n/2}$ so

$$e^{-Y_n^tA_n^{-1}Y_n/2}\det(A_n)^{-1/2}=e^{-(S_n-S_{n-1})^2/2}e^{-Y_{n-1}^tA_{n-1}^{-1}Y_{n-1}/2}\det(A_{n-1})^{-1/2}$$

where we have cancelled the common factor of $(2\pi)^{-n/2}$.

Thus formula holds for any value of $Y_n$, so set $Y_n=(0,\dots,0)$.  Plugging into the above formula we conclude that $\det(A_n)=\det(A_{n-1})$. Obviously $\det(A_1)=1$ so we are done.


问:
Total Derivative & Maximum Directional Derivative & dx
Is it correct to say that the total derivative is the maximum directional derivative? I am studying electromagnetism this semester, and this equation appeared: $$dT = \nabla T \cdot dr$$ where $dr$ is the infinitesimal displacement vector. The right hand side is the equation for the directional derivative. Before we get this equation, however, we have this equation: $$dT = (\frac{\partial T}{\partial x}) dx + (\frac{\partial T}{\partial y}) dy + (\frac{\partial T}{\partial z}) dz$$ I bring this second equation up because it is the result of my ultimate confusion. It has been years since I have taken vector calculus, and even when I did, I did not feel comfortable with gradients and such. I know that $\frac{\partial T}{\partial x}$ is the partial derivative of T with respect to x, while keeping y and z constant -- and I can easily calculate these, I just need an intuitive approach. My main confusion comes from the meaning of adding the $dx, dy,$ and $dz$ after the partials. Is this just notation? Is it just a way to make the gradient (a vector) into the scalar function $dT$?

Thank you! :)

答:
This is a good question and it's actually pretty complicated. Mostly, the symbol $d$ is difficult to deal with. There is always a very precise mathematical meaning to it but it is hard to describe, especially in higher dimensions. It involves some multivariable calculus and linear algebra. Below, I'll let $d$ denote the total derivative rather than an infinitesimal displacement vector. For $dx$, $dy$, and $dz$, they can be regarded as both as I'll show below. I'll introduce the general idea first but it's much simpler in your case.

We can regard $T$ as a differentiable function $T: \mathbb{R}^3 \to \mathbb{R}$. If $f: \mathbb{R}^n \to \mathbb{R}^m$ is differentiable, then we usually regard the Total Derivative $df$ to be the linear map associated with the matrix $$df = \begin{bmatrix}
\frac{\partial f_1}{\partial x_1} & \frac{\partial f_1}{\partial x_2}  & \dots &\frac{\partial f_1}{\partial x_n}\\
\frac{\partial f_2}{\partial x_1} & \frac{\partial f_2}{\partial x_2}  & \dots &\frac{\partial f_2}{\partial x_n}\\
\vdots & &  \ddots & \vdots\\
\frac{\partial f_m}{\partial x_1} & \frac{\partial f_m}{\partial x_2}  & \dots &\frac{\partial f_m}{\partial x_n}
\end{bmatrix}$$ called the Jacobian of $f$. Here we take the $f_i$ to be defined by $$f(x_1, \dots, x_n) = (f_1(x_1, \dots, x_n), f_2(x_1, \dots, x_n), \dots, f_m(x_1, \dots, x_n))$$ which is just $f$ expanded into the coordinates of $\mathbb{R^m}$. This is found by taking the best linear approximation of $f$ at any point and is then, a streightforward generalization of the usual derivative.

Then, we can do some calcuations. Let $x: \mathbb{R}^3 \to \mathbb{R}$ be the function which takes $(x_1,x_2,x_3) \mapsto x_1$ and we define $y$ and $z$ similarly. $y$ takes $(x_1,x_2,x_3) \mapsto x_2$ and $z$ takes $(x_1,x_2,x_3) \mapsto x_3$. Then, we find that the total derivative of $x$ is $$dx = \begin{bmatrix}
\frac{\partial x}{\partial x_1} & \frac{\partial x}{\partial x_2} & \frac{\partial x}{\partial x_3}
\end{bmatrix}= \begin{bmatrix} 1 & 0 & 0 \end{bmatrix}$$ by the definition of $x$. Similarly, we have $$dy = \begin{bmatrix} 0 & 1 & 0 \end{bmatrix} \;\text{and}\;dz = \begin{bmatrix} 0 & 0 & 1 \end{bmatrix}.$$

In the case of $T$, the Jacobian is given by $$dT = \begin{bmatrix}
\frac{\partial T}{\partial x} & \frac{\partial T}{\partial y} &\frac{\partial T}{\partial z}
\end{bmatrix}.$$ Then, we note that given this $dT(x,y,z)$ is not a real number: it's a linear map from $\mathbb{R^3}$ ro $\mathbb{R}.$  Because of that, it would not make sense to write $dT$ as the sum of partial derivatives. It wouldn't take into account the different coordinates of the input. Using the linear properties of matrices, though, we can obtain the next best thing. \begin{align*} dT &= \begin{bmatrix}
\frac{\partial T}{\partial x} & \frac{\partial T}{\partial y} &\frac{\partial T}{\partial z}
\end{bmatrix} \\ &= \frac{\partial T}{\partial x} \begin{bmatrix}
1 & 0 & 0
\end{bmatrix} + \frac{\partial T}{\partial y} \begin{bmatrix}
0 & 1 & 0
\end{bmatrix} +\frac{\partial T}{\partial z} \begin{bmatrix}
0 & 0 & 1
\end{bmatrix} \\&= \frac{\partial T}{\partial x_1} dx + \frac{\partial T}{\partial x_2} dy +\frac{\partial T}{\partial x_3}dz. \end{align*}

This is the second formula you mentioned.

Essentially, the total derivative is a little different from partial derivatives in the sense it gives a linear map rather than a scalar function. In multivariable calculus and differential geometry, there are numerous "derivatives" and they always have some intersection with each other. In the above case, the Total Derivative and its relation to the Partial Derivative is brought to light. I use these two derivatives much more often than the gradient, but that may be due to my lack of experience in fields that don't. A lot of this notation can be converted to gradients but I find it more intuitive to work with just partial derivatives.


问:
Confusion in representation of complex numbers .

To begin, in the above image, $z$ has been taken to to be $x+iy$, allowing a multiplication by $i$ to result in: $iz=-y+ix$. In the diagram, they have actually shown the 2nd quadrant than the 4th quadrant. If the point given by $z$ was in the 4th quadrant, then wouldn't the angle be $-90^{\circ}$?

Finally, when we have $x+iy$, then it means that the y axis is the imaginary axis. Only, here we see the x axis as said imaginary axis. How is this possible?

答:
The notation is misleading. It would be easier to understand, if they used following notation:

Coordinates of the point A: z1 = x1 + iy1

Coordinates of the point B: z2 = iz1 = x2 + iy2 = -y1 + ix2


问:
Splitting of prime ideals in algebraic extensions
I'm reading Algebraic Number Theory by Jurgen Neukirch. I have some problems with some of the exercises in Section 9 of Chapter 1.

They are:

1) If $L / K$ is a Galois extension of algebraic number fields with non-cyclic
Galois group, then there are at most finitely many non-split prime ideals of $K$.

2) Let $L / K$ be a finite (not necessarily Galois) extension of algebraic number
fields and $N / K$ the normal closure of $L / K$. Show that a prime ideal $p$ of $K$ is totally
split in $L$ if and only if it is totally split in $N$.

I have worked on them for a long time but couldn't get any idea of it.

Can you please help? Thank you!

答:
for question1,what about using the following lemma:

```<code>    If L / K is separable, then there are only finitely many prime
    ideals of K which are ramified in L.
</code>```

we can suppose the extension L/K is finite.And we know a prime splits totally iff it is unramified,so we can say that when a prime ideal is nonsplit ,it must be ramified...so we can say 1) is right.

Body:
for question1,what about using the following lemma:

    If L / K is separable, then there are only finitely many prime
    ideals of K which are ramified in L.

we can suppose the extension L/K is finite.And we know a prime splits totally iff it is unramified,so we can say that when a prime ideal is nonsplit ,it must be ramified...so we can say 1) is right.


问:
Evaluating variables for constrained Lagrange
So if we had the equation bound by a constant such that

$(L(x, y, λ) = 9x^2 - y^2 - λ(9x^2 + y^2 - 81)$

I can determine that

$\frac{∂L}{∂x} = 18x - 18λ$

$\frac{∂L}{∂y} = -2y - 2λ$

$\frac{∂L}{∂λ} = -(9x^2 + y^2 - 81)$

Assuming this is correct, when we equate each partial derivative to $0$ it may be clear that for $f_x, x = 1, λ = 1$ and presumably $y = 0$ but I have an issue with the constraint equation. I understand there are two answers and at first I assumed that $±9 - y = 9$ was the answer as $x = ±1$. However, I realised that there is a $-$ sign and wondered if that still made the application of the square root possible. Also, the $y$ is squared, so doesn't that mean that there is a $±y$ value as well? Is the value of $λ = 0$? Just not sure if I have equated the values correctly.

答:
Use Lagrange Multiplier method for this problem only if you have not been given a choice. Now coming to your working, there is a mistake.

$L(x, y, λ) = 9x^2 - y^2 - λ(9x^2 + y^2 - 81)$. Equating partial derivatives to zero give you,

$18x - 18\lambda x = 0$

$-2y - 2\lambda y = 0$

$9x^2 + y^2 = 81$

From first, either $x = 0$ or $\lambda = 1 \,$ for $x \ne 0$.

If $\lambda = 1, y = 0$ from the second equation. So $x = \pm3$ given the constraint.

Similarly if $x = 0, y \ne 0,$ (due to constraint). Hence $\lambda = - 1 $ from second equation and $y = \pm 9$ given the constraint.

So you get critical points as $(\pm 3, 0), (0, \pm 9)$.


答:
The coefficient of $i$ gives us the vertical position.

$x+iy$ corresponds to the coordinate $(x,y)$. When $x, y $ are both positive, it is a point in the first quadrant.

$iz=ix-y = -y+ ix$ corresponds to the point $(-y, x)$. If $x, y$ are both positive, the first coordinate $-y$ is negative and the second coordinate $x$ is positive and hence we get a point in the second quadrant.

In general $iz$ rotate $z$ by $90^\circ$ anticlockwise.

Title: Confusion in representation of complex numbers .

Body:
To begin, in the above image, $z$ has been taken to to be $x+iy$, allowing a multiplication by $i$ to result in: $iz=-y+ix$. In the diagram, they have actually shown the 2nd quadrant than the 4th quadrant. If the point given by $z$ was in the 4th quadrant, then wouldn't the angle be $-90^{\circ}$?

Finally, when we have $x+iy$, then it means that the y axis is the imaginary axis. Only, here we see the x axis as said imaginary axis. How is this possible?


答:
You have two axes: the axis in the "$1$" direction and the axis in the "$\mathrm{i}$" direction.  Any point of the plane $(a,b)$ can be written as having coordinate $a$ in the $1$ direction and coordinate $b$ in the $\mathrm{i}$ direction:
$$  1 \cdot a + \mathrm{i} \cdot b  \text{.}  $$
Of course, no one ever writes down the "$1\cdot {}$", so we actually write and see
$$  a + \mathrm{i} b  \text{.}  $$

There is a potential ambiguity in the uses of $x$ and $y$.  $x$ is being used as both the label for the real axis and for the first coordinate of the point $A$.  $y$ is being used as the label for the imaginary axis and for the second coordinate of the point $A$.  It would have been better to not use the axis labels as coordinates of the point $A$, but it was done.  You may be better served (mentally) replacing the label on the horizontal axis with "Re" and the label on the vertical axis with "Im".

So you start with point $A$ having coordinates $(x,y)$, corresponding to the complex number $x + \mathrm{i} y$.  Notice that $x$ and $y$ are positive real numbers, so $A$ is in the first quadrant.  Then you multiply by $\mathrm{i}$, obtaining $\mathrm{i} x - y$.  That is, you have
$$  1 \cdot (-y) + \mathrm{i} \cdot x  \text{.}  $$
This is the point with coordinates $(-y,x)$ which has negative first coordinate (left half-plane) and positive second coordinate (upper half-plane), so is in quadrant II, as labelled by the point $B$.


答:
You’re letting the letters $x$ and $y$ confuse you. Let’s start with a complex number $u+iv$ instead, where $u$ and $v$ are positive real numbers. This complex number has real part $u$ and imaginary part $v$. (Remember, the imaginary part is the coefficient of $i$.) It is plotted at the point $\langle u,v\rangle$ in the first quadrant, because we normally plot the real part on the $x$-axis and the imaginary part on the $y$-axis.

When you multiply the number by $i$, you get $ui-v$, or $-v+ui$. Now the real part, i.e., the part without the factor of $i$, is $-v$, and the imaginary part is $u$, so we plot the complex number at the point $\langle -v,u\rangle$, which is in the second quadrant.

It’s fine to use the terms $x$-coordinate and $y$-coordinate instead of real coordinate and imaginary coordinate, or abscissa and ordinate, but only if you remember that the $x$ and $y$ in these terms have nothing to do with the symbols used to denote the actual coordinate numbers. This is true even when you’re just dealing with points in the plane: the $x$-coordinate of the point $\langle y,x\rangle$ is $y$, not $x$.


问:
What's the exact/closest approximate to the inverse of getting 70%?
To clarify: I'm creating a computer program and I'm trying to find the actual mathematical way of approaching my problem. Given a number x, a is 0.7x. After doing some testing, it seems that to get back up to x from a, I can multiply by 1.42857143, but sometimes this isn't enough precision and gets a number that is very wrong.

Is there a blatantly obvious solution to this that I can't see? I want to find a better way to solve this without repeatedly adding 1 until it's right.

Thanks!

答:
Precision of operations is limited on  computers.

What data type do you use? Float? Double?

To get better precision, you may want to use the types like BigDecimal.

If this is not sufficient, use other tools for programming like Mathematica.


问:
Show that there does not exist a polynomial $f \in \mathbb{R}[x]$ such that $f^2(x) = f(x) \cdot f(x) = 1 + x + x^3$.
Show that there does not exist a polynomial $f \in \mathbb{R}[x]$ such that $f^2(x) = f(x) \cdot f(x) = 1 + x + x^3$.

I really have no idea where to begin and would appreciate all help I can get to solve this.

i tried to use the information from this link Is there a polynomial $f\in \mathbb Q[x]$ such that $f(x)^2=g(x)^2(x^2+1)$ - because it resembles my doubt. But I couldn’t do it.

答:
Hint:

If the degree of $f$ is $n$, what is the degree of $f^2$?


问:
How to show the existence, or lack thereof, for $\lim_{x \to 0} \frac{1}{x}\sin\left(\frac{1}{x}\right)$

Does the following limit exist?
$$\lim_{x \to 0} \frac{1}{x}\sin\left(\frac{1}{x}\right)$$

If yes, then I have to find it; if no, then I need to give reason why.

I couldn’t figure out how to prove this formally, can someone please help me out.

Edit: please note that sin is multiplied by 1/x here and not x.

答:
Define the sequences $(x_n)$ and $(y_n)$ by
$$x_n = \frac{1}{n\pi}\: \text{ and } \: y_n = \frac{1}{\left(2n+\frac{1}{2}\right)\pi}$$
for each $n\in\mathbb{N}$. Clearly, $x_n,y_n\to0$ as $n\to\infty$. Now, it is easy to see that
$$\lim_{n\to\infty}\frac{1}{x_n}\sin\left(\frac{1}{x_n}\right)=0\:\text{ and }\:\lim_{n\to\infty}\frac{1}{y_n}\sin\left(\frac{1}{y_n}\right)=\infty$$
Thus, the limit does not exist.


问:
A proof that the constant-coefficient in $U_{n}$ is $0$ when $n$ is odd.
Let $(U_{n}(x))_{n≥0}$ be the Chebyshev polynomials of the second kind (https://mathworld.wolfram.com/ChebyshevPolynomialoftheSecondKind.html#:~:text=A%20modified%20set%20of%20Chebyshev,of%20the%20Gegenbauer%20polynomial%20with%20.).
We know that the constant coefficient in $U_{n}$ is $(-1)^{n/2}$ when $n$ is even. It seems from the examples of the above link that the constant-coefficient in $U_{n}$ is $0$ when $n$ is odd. But I not able to find proof for this.

答:
I believe this follows from the defining formula
$$ U_n( \cos \theta ) \sin(\theta) = \sin ( (n + 1)\theta). $$
If $x = \cos \theta$ and $n$ is odd, then $-x = \cos (\pi + \theta)$ so that
$$ U_n( -x) = \frac{ \sin( (n+1)(\theta + \pi))  }{\sin(\theta + \pi) } = \frac{\sin((n+1)\theta)}{-\sin(\theta)} = -U_n(x). $$
(when $\sin(\theta) \neq 0$; if $\sin(\theta) = 0$, then $U_n(x) = 0$). That is, $U_n$ is an odd function of $x$ when $n$ is odd. Since $U_n$ is a polynomial, this implies that the constant term is zero.

EDIT: There is a much simpler way of proving this. Since $\cos(\pi/2) = 0$, then
$$ U_n(0) \sin(\pi/2) = \sin( (n+1)\pi/2), $$
i.e.
$$ U_n(0) \cdot 1 = \sin\bigg( \frac{(n+1)\pi}{2} \bigg). $$
If $n$ is odd, then $n+ 1$ is even so that
$$U_n(0) = \sin k\pi$$
for $n$ odd. Therefore $U_n(0) = 0$ and so the constant term of $U_n$ is $0$ when $n$ is odd.


问:
Finding the standard deviation from data grouped by intervals

Starting Monthly Salary
Number of Graduates

1,001 - 1,400
1

1,401 - 1,800
11

1,801 - 2,200
14

2,201 - 2,600
38

2,601 - 3,000
36

Total
100

What I have done, is averaged out each value in Monthly Salary column and continuing from there by calculating the deviation at the end

Calculating all 100 of these values is tedious:

(1001 + 1400) / 2 = 1200.50

(1401 + 1800) / 2 = 1600.50

(1401 + 1800) / 2 = 1600.50

(1401 + 1800) / 2 = 1600.50

Continued for all 100 values to the calculate the deviation for each...

Then calculating the standard deviation with σ=√((Σ(x-µ)^2)/2). This formula once everything has been entered in will look really nasty. Is there an easier (or cleaner) solution to finding the (approximate) standard deviation of the monthly starting salaries above?

答:
Your grouped data have midpoints $m_i$ with respective
frequencies $f_i.$

```<code>m = c(12,16,20,24,28)*100
f = c(1,11,14,38,36)
</code>```

The sample mean is approximately
$A =\bar X = \frac 1n\sum_{i=1}^5 f_im_i =2388,$ where
$n = \sum_{i=1}^5 f_i = 100.$  Using R as a calculator:

```<code>n = sum(f); n
[1] 100
a = sum(f*m)/n; a
[1] 2388
</code>```

The sample variance
$S^2 \approx \sum_{i=1}^5 \frac{1}{n-1}f_i(m_i-\bar X)^2 =166\,319.2$ and
the sample standard deviation is $S =\sqrt{S^2} = 407.82.$

```<code>v = sum(f*(m-a)^2)/(n-1);  v
[1] 166319.2
s = sqrt(v);  s
[1] 407.8225
</code>```

If you are using some kind of spreadsheet, there might be
a built-in function for finding the mean and variance of
a column of $n = 100$ numbers. If so, you could find
exact values of the sample mean and standard deviation.
(Some information is lost when data are put into groups and summarized.)

```<code> x = rep(m, times=f)  # 'data' reconstructed from m & f
 mean(x);  sd(x)
 [1] 2388
 [1] 407.8225

 cutp=seq(1000, 3000, by=400)
  hist(x, br=cutp, ylim=c(0,45),
       col="skyblue2", label=T)
</code>```

Body:
Your grouped data have midpoints $m_i$ with respective
frequencies $f_i.$

m = c(12,16,20,24,28)*100
f = c(1,11,14,38,36)

The sample mean is approximately
$A =\bar X = \frac 1n\sum_{i=1}^5 f_im_i =2388,$ where
$n = \sum_{i=1}^5 f_i = 100.$  Using R as a calculator:

n = sum(f); n
[1] 100
a = sum(f*m)/n; a
[1] 2388

The sample variance
$S^2 \approx \sum_{i=1}^5 \frac{1}{n-1}f_i(m_i-\bar X)^2 =166\,319.2$ and
the sample standard deviation is $S =\sqrt{S^2} = 407.82.$

v = sum(f*(m-a)^2)/(n-1);  v
[1] 166319.2
s = sqrt(v);  s
[1] 407.8225

If you are using some kind of spreadsheet, there might be
a built-in function for finding the mean and variance of
a column of $n = 100$ numbers. If so, you could find
exact values of the sample mean and standard deviation.
(Some information is lost when data are put into groups and summarized.)

 x = rep(m, times=f)  # 'data' reconstructed from m & f
 mean(x);  sd(x)
 [1] 2388
 [1] 407.8225

 cutp=seq(1000, 3000, by=400)
  hist(x, br=cutp, ylim=c(0,45),
       col="skyblue2", label=T)


问:
Reduce the Cauchy-Euler equation to a smaller expression?
Given the Cauchy-Euler second order differential equation:
$$ \left( x^2\frac{d^2}{d x^2} + x\frac{d}{d x} -1\right)y(x)=0 $$
is it possible to find a function $u(x)$ , which satisfies the following simple form:?
$$ \frac{d^2 y(u)}{d u^2}=0 $$
***notice that $t=\ln(x)$ yields:
$$ \frac{d^2 y(t)}{d t^2}-1=0 $$

答:
notice that t=ln(x) yields:
$$\dfrac {d^2y(t)}{dt^2}−1=0$$

$$
\left( x^2\frac{d^2}{d x^2} + x\frac{d}{d x} -1\right)y(x)=0$$
It yields:
$$\dfrac {d^2y(t)}{dt^2}−\color {red}{y(t)}=0$$
That you can rewrite as:
$$y''+y'-y'-y=0$$
$$(ze^{-t})'=0$$
Where $z=y'+y$.
$$((ye^t)'e^{-2t})'=0$$

Note that:
$$\frac{d^2 y(u)}{d u^2}=0$$
Has solution:
$$y(u)=c_1u+c_2$$
But $c_2$ is not a solution of the original equation:
$$ x^2\frac{d^2y}{d x^2} + x\frac{dy}{d x} -y=0$$
So it 's not possible to find such $u=u(x)$.


答:
Alternative approach:

$f(x)^2$ should be non-negative for all $x\in\mathbb R$.

What about $1+x+x^3$, particularly when $x$ is a negative number with large magnitude?

Body:
Show that there does not exist a polynomial $f \in \mathbb{R}[x]$ such that $f^2(x) = f(x) \cdot f(x) = 1 + x + x^3$.

I really have no idea where to begin and would appreciate all help I can get to solve this.

i tried to use the information from this link Is there a polynomial $f\in \mathbb Q[x]$ such that $f(x)^2=g(x)^2(x^2+1)$ - because it resembles my doubt. But I couldn’t do it.


答:
$$\cos \dfrac{A}{2}+\tan A \sin \dfrac{A}{2}$$

$$=\cos\dfrac{A}{2}+\dfrac{\sin A}{\cos A} \sin \dfrac{A}{2}$$

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

$$=\cos \dfrac{A}{2}(\dfrac{\cos A+ 2 \sin^2 \frac{A}{2}}{\cos A})$$

$\cos A=1-\sin^2 \dfrac{A}{2} \implies(\dfrac{\cos A+ 2 \sin^2 \frac{A}{2}}{\cos A})=\dfrac{1}{\cos A}$

And the final expression becomes: $\cos \dfrac{A}{2} \cdot \dfrac{1}{\cos A}=\sec A \cdot \cos \dfrac{A}{2}$

Title: Trigonometric Identity Problem - Cos Tan and Sin

Body:
I have been going through my lecture notes for a structures question (the solution of a 2nd order ode for a buckling problem) when I came across a very weird trigonometric simplification which I just cannot get my head around. Could anyone shed any light on this?

Further down the page, there is another similar simplification, but this one seems to be incorrect:

Testing some random values, the simplification is incorrect unless I replace the C/2 with a C.


答:
Any

$f(x) \in F[x], \tag 1$

where $F$ is any field, may be written

$f(x) = \displaystyle \sum_0^n f_i x^i = f_nx^n + \sum_0^{n - 1}f_ix^i, \; f_i \in F; \tag 2$

then

$f(x) \cdot f(x) = f_n^2x^{2n} + \text{lesser degree terms}; \tag 3$

it follows that $f(x) \cdot f(x)$ is always of even degree; but

$\deg \; x^3 + x + 1 = 3, \; \text{odd}, \tag 4$

so . . .

Note that

$F = \Bbb R \tag 5$

needn't hold to attain the above result.

Title: Show that there does not exist a polynomial $f \in \mathbb{R}[x]$ such that $f^2(x) = f(x) \cdot f(x) = 1 + x + x^3$.


问:
If $f$ is $\mathcal{S}$-measurable, and $f$ is also $\mathcal{S}'$-measurable, then $\mathcal{S} \subseteq \mathcal{S}'$.
Let $f: \mathbb{R} \to \mathbb{R}$ where $f(x) = x^2$. Let $\mathcal{S} = \{A \in \mathcal{B}(\mathbb{R}): A = -A\}$, where $\mathcal{B}(\mathbb{R})$ are the Borel subsets of $\mathbb{R}$, and $-A = \{-a : a \in A\}$. (In other words, $\mathcal{S}$ is the set of all Borel subsets that are equal to their negative). We have previously shown that $\mathcal{S}$ is a $\sigma$-algebra and that $f$ is $\mathcal{S}$-measurable.

Show that if $\mathcal{S}'$ is another $\sigma$-algebra on $\mathbb{R}$ for which $f$ is $\mathcal{S}'$-measurable, then $\mathcal{S} \subseteq \mathcal{S}'$.

As mentioned below in the comments, this means showing that $\mathcal{S}$ is the smallest $\sigma$-algebra for which $f$ is measurable, but I am not sure how to do that.

答:
Consider a symmetric Borel set $A=-A$. Now the function $x^2$ is a homeomorphism of $[0,\infty)$ onto $[0,\infty)$ ( $\sqrt{x}$ is a continuous inverse). Since, $A\cap [0,\infty)$ is a borel set and $x^2$ a homeomorphism on $[0,\infty)$ it follows that the image
$$B=A^{2}\cap [0,\infty)$$
of $A$ through the function $x^2$ is a Borel set too. We claim that $f^{-1}(B)=A$. Indeed,
if $x\in A$ then $f(x)=x^2 \in A^2\cap [0,\infty)=B$. Hence, $A\subseteq f^{-1}(B)$. Now, if $x\in f^{-1}(B)$ then $f(x)=x^{2}\in A^{2}\cap [0,\infty)$. In particular, $x^2\in A^2$. Hence, $x\in A$ or $x\in -A$. But since $A$ is symmetric it follows that $x\in A$. So, $f^{-1}(B)=A$. Now, since $B$ is a Borel set it follows that $f^{-1}(B)\in \mathcal{S'}$. But, $f^{-1}(B)=A\in \mathcal{S'}$. Hence, $\mathcal{S}\subseteq \mathcal{S'}$.


问:
Convergence of quadratic variations
This may be a fundamental question on a martingale theory.

Let $n \in \mathbb{N}$ and $M^n=(M^{n,1},\ldots,M^{n,d})$ be a $d$-dimensional square integrable martingale on a probability space with probability measure $P_n$. Each $M^n$ may not be continuous.

We denote by $E_n$ the expectation under $P_n$. We assume that for any $t \in [0,1]$
\begin{align*}
\lim_{n \to \infty}E_n\left[\left|[M^{n,i},M^{n,j}]_t - \delta_{i,j}t\right|\right]=0.
\end{align*}
Then, can we show that the law of $M^n$ converges weakly to that of a $d$-dimensional Brownian motion in $D([0,1])$? Here, $D([0,1])$ is the space of right continuous functions with finite left limits with Skorohod topology.

This may be true. However, I do not the proof.

Please tell me a reference.

答:
Adding a bit to the previous answer, if one formulates the question in terms of the angle bracket (a more natural question in some contexts) instead of the square bracket, the statement is also not true. Not even when $M_n = M$ for all $n$ and $\langle M \rangle_t = t$.

A martingale $M$ with cadlag paths and quadratic variation $\langle M \rangle_t = t$ need not be a standard Brownian motion---take $M_t$ to be the compensated Poisson process
$N_t - t$ where $N_t$ is Poisson process with intensity $1$.
However, considering this counterexample would lead to a sufficient condition for when the statement holds, under weak convergence on $D[0,1]$.

The issue is whether $M$ has same finite dimensional distribution as Brownian motion, since Kolmogorov's continuity criterion gives a version with continuous sample paths once
this is satisfied.
Having the right first moment and quadratic variation/"infinitesmal second moment" is not enough to guarantee having the
same finite dimensional distribution as Brownian motion.

Suppose $\langle M \rangle_t = t$ and that $M$ has independent increments---e.g. $M_t = N_t - t$. (The general martingale case is not that different; one needs to replace unconditional means by conditional means.)

$M$ has the same finite dimensional distribution as Brownian motion if and only if
$$
\phi_{\lambda}(t) = E[e^{i \lambda M_t}] = e^{-\frac12 \lambda^2 t},
$$ or

$$
\frac{d}{dt} \phi_{\lambda}(t) = -\frac12 \lambda^2 \phi_{\lambda}(t), \; \phi_{\lambda}(0) = 1.
$$
This ODE means that, for a small increment $\Delta_h = M_{t+h} - M_t$, we must have
$$
E[e^{i \lambda \Delta_h}] = E[ 1 +  i \lambda \Delta_h - \frac12 \lambda^2 \Delta_h^2 + \cdots ] = 1 - \frac12 \lambda^2 h + r(h), \quad \quad (*)
$$
where the remainder term $r(h) = o(h)$, i.e. it must vanish faster than $h$.
This is not true in general. E.g. for the process $N_t - t$, one can calculate directly
$$
E[e^{i \lambda \Delta_h}] = e^{-\frac12 \lambda^2 h + O(h)} = 1 - \frac12 \lambda^2 h + O(h).
$$
The remainder term is $O(h)$ and not $o(h)$, and $N_t - t$ does not have the same finite dimensional distribution as Brownian motion.

A standard sufficient condition that the remainder $r(h)$ in $(*)$ to be $o(h)$ is that, for all $t$,
$$
\lim_{\alpha \rightarrow \infty} \limsup_{h \rightarrow 0} E[  \frac{ \Delta_h^2 }{h} \cdot 1_{ \{ \frac{ \Delta_h^2 }{h} > \alpha \} }  ] = 0. \quad \quad (**)
$$
The uniform integrability condition $(**)$ is a kind of an infinitesmal version of the Lindeberg condition for CLT's.

Couple Comments:

Every local martingale $M$ with continuous paths and $\langle M \rangle_t = t$ must be a standard Brownian motion (Levy's Theorem).
In the context of this discussion, this say that continuity of sample path guarantees that $r(h)$ in $(*)$ is $o(h)$---indeed, this is Ito's lemma for
continuous local martingales. The requires the martingale property, namely that $\int d \langle M \rangle_t$ over an interval can be approximated by sum of $\Delta_h^2$'s as $h \rightarrow 0$ when paths are continuous.

Quoting Ito's lemma again, $(*)$ can be recast in terms of the infinitesmal generator. A martingale is Brownian motion if and only if its infinitesmal generator is $\frac{d^2}{dx^2}$.

With a sequence $\{ M_n, n \geq 1\}$---first, there is no need or reason to assume each $M_n$ is a martingale. See, for example, various versions of Functional Central Limit Theorem where partial sums of dependent sequences (strong-mixing/mixingales/etc) converge weakly to the Brownian motion.
The martingale condition only needs to hold "in the limit".

Second, tightness must be considered. Separate assumption needs to be made so that  $\{ M_n, n \geq 1\}$ is tight on $D$.
Then a standard sufficient condition that ensures that the weak limit is Brownian motion is again $(**)$, extended appropriately to sequences of random elements on $D[0,1]$.
A detailed discussion can be found in Section 19 of Convergence of Probability Measures by Billingsley.

Title: Convergence of quadratic variations


问:
Infinite Series question

The first one, the effective resistance is $2R$, then $5R/3$ then $13R/8$ etc....
My job is to find the pattern/equation so I can find the total resistance when $20$ resistors are connected. Of course I can do the long way by using this formula I created
$R_n = (R_{n-1} \times R)/(R_{n-1} +R) +R$. Can you guys help me find an equation where I can just input the value of $n$ and simply get an answer. Thanks!

The second question is: What would I get if I had an infinite number of resistance.

Help is very much appreciated :)

答:
It is a fact of life that homographic transforms like
$$
u:x\mapsto R+\frac{Rx}{R+x},
$$
are conjugate to affine transforms. The first step to make use of this fact for the iterates of $u$ is to determine a fixed point of $u$, here
$$
x^*=R\frac{1+\sqrt5}2,
$$
and to rewrite everything in terms of $1/(x-x^*)$, here
$$
\frac1{u(x)-x^*}=b+\frac{a}{x-x^*},
$$
for some $(a,b)$ you will compute, hence
$$
\frac1{u(x)-x^*}-c=a\left(\frac1{x-x^*}-c\right),
$$
for some $c$ you will compute. In particular,
$$
\frac1{u^n(x)-x^*}-c=a^n\left(\frac1{x-x^*}-c\right).
$$
This yields an explicit formula for $R_n=u^{n-1}(2R)=u^n(\infty)$, namely
$$
R_n=x^*+\frac1{c(1-a^n)}.
$$
Since $x^*$ is known, note that one can skip everything above and identify $a$ and $c$ through the values of $R_1=2R$ and $R_2=\frac53R$, using
$$
c(1-a)=\frac1{R_1-x^*}=\frac1R\frac{3+\sqrt5}2,\qquad
c(1-a^2)=\frac1{R_2-x^*}=\frac3R\frac{7+3\sqrt5}2.
$$
Better still, use $a=1/u'(x^*)$ and the expression of $c(1-a)$ above, to get, for every $n\geqslant1$,
$$
R_n=(s-1)R+\frac1s\frac{s^2-1}{s^{2n}-1}R,\qquad s=\frac{3+\sqrt5}2.
$$</p>

Title: Infinite Series question


问:
A lot of confusion in this question of finding purely real values of theta?

For purely imaginary , it should have been
$\dfrac{8i\sin\theta}{1+4\sin^2\theta}$

Why written $3-4\sin^2\theta$.

Also , I read that the range of theta has to be between $-\pi$ and $\pi$. Why written $n\pi$ here?

I couldn’t find the homework question tag so posted it here. Please help in understanding where am I wrong.

答:
Yes, for purely imaginary, $z=\dfrac{8\sin \theta}{1+4\sin^2\theta}$, but for that to happen, it's real part must be $0$, shouldn't it? and that's why the real part that is,  $\Re(z)=\dfrac{3-4\sin^2\theta}{1+4\sin^2\theta}=0$.

As for, the $n\pi$ part, $\theta$ can take infinite values as $\sin \theta$ is defined for all $\theta$, (yes, even complex values too! but that is not relevant here), so the general value of $\theta$ that satisfies the final equation is $\theta=n\pi \pm \dfrac{\pi}3$, you can put it back in the equation and verify that it satisfies the equation or not.


问:
Top 10 math mnemonics
If you study undergraduate medicine, mnemonics are almost indispensable - there is so much factual material to learn. I was never given any mnemonics in my time as a maths undegraduate. But Robert Israel just mentioned "iciacids" to help remember the important features when looking at a function plot Function - Main Features?  I would certainly have forgotten some of them.

So maybe mnemonics are useful for maths too. Maybe also for tricky proofs? Or for areas knee-deep in definitions, like topology? Does anyone know of a source for math mnemonics? Eg a top 10 type website? Or a kind of successor to sci.math FAQs (although I do not remember any mnemonics there)? Or a book?

答:
Here is the E E E rule for signs of permutations:

Even number of Even cycles makes an Even permutation.

Yes, the sign is $(-1)^ {\sum_{c \textrm{ cycle}} (|c|-1) }$, but I remember it better with the triple E rule.

Title: Top 10 math mnemonics


答: The complex number $a+b i$ [where $a,b$ are real numbers] is purely imaginary iff $a=0.$ So write your number in $a+b i$ form and see that $a$ has numerator $3-4 \sin^2 \theta$ and a strictly positive denominator.

Title: A lot of confusion in this question of finding purely real values of theta?

Body:
For purely imaginary , it should have been
$\dfrac{8i\sin\theta}{1+4\sin^2\theta}$

Why written $3-4\sin^2\theta$.

Also , I read that the range of theta has to be between $-\pi$ and $\pi$. Why written $n\pi$ here?

I couldn’t find the homework question tag so posted it here. Please help in understanding where am I wrong.


问:
Method to partial fractions.
For Example:

$$\frac{ax^2 + bx+c}{(dx+e)(fx^2+g)}\equiv\frac{A}{dx+e}+\frac{Bx+C}{fx^2+g}$$

and

$$\frac{ax^4 + bx^3+cx^2+dx+e}{(x+f)(x^2+g)}\equiv Ax+B+\frac{C}{x+f}+\frac{Dx+E}{x^2+g}$$

How do you know how to format the right hand side, in the equation below in partial fractions? (what is the 'pattern'?)

$$\frac{2ax}{(x-2a)(x^2+a^2)}$$
where a is a non zero constant

Could you look at it as the top of the fraction being a simple derivative of the bottom? If so how would you know to add the Ax+B in the latter example?

答:
Using Partial Fraction Decomposition,
$$\frac{2ax}{(x-2a)(x^2+a^2)}=\frac A{(x-2a)}+\frac {Bx+C}{(x^2+a^2)} $$</p>

Title: Method to partial fractions.


问:
Show that this family is equicontinuous at $0$
Let $E$ be a normed vector space, $$p_K(\varphi):=\sup_{x\in E}|\varphi(x)|\;\;\;\text{for }\varphi\in E'$$ for compact $K\subseteq E$ and $\sigma_c(E',E)$ denote the initial topology with respect to $(p_K,K\subseteq E\text{ is compact})$, i.e. the subspace topology on $E'$ inherited from the topology of compact convergence on $C(K)$.

Let $\mathcal C\subseteq C(E')$ be uniformly $\sigma_c(E',E)$-equicontinuous.

Why can we conclude that $$\forall\varepsilon>0:\exists\delta>0:\forall\varphi\in E':\left\|\varphi\right\|_{E'}<\delta\Rightarrow\sup_{f\in\mathcal C}\left|f(0)-f(\varphi)\right|<\varepsilon?\tag1$$

Most probably the desired claim is trivial to obtain, but I'm not able to see how due to the rather complicated setting.

$(1)$ is obviously some kind of equicontinuity at $0$. I'm not sure if it's relevant, but by the Banach-Alaoglu theorem $\{\varphi\in E':\left\|\varphi\right\|_{E'}\le\delta\}$ is $\sigma_c(E',E)$-compact for all $\delta>0$.

答:
Recall the definition of uniform equicontinuity of $\mathcal{C}$ as a set of maps $(E',\sigma_c(E',E)) \to \Bbb{R}$:

For every neighbourhood $V \subseteq \Bbb{R}$ of $O$ there is a neighbourhood $U$ of $0$ in $(E',\sigma_c(E',E))$ such that $$\varphi,\psi \in V \implies f(\varphi)-f(\psi) \in V, \, \text{for all }f \in \mathcal{C}.$$

Now for $\psi = 0$ and $V = \left\langle-\frac\varepsilon2, \frac\varepsilon2\right\rangle$, we get a neighbourhood $U$ of $0$ such that $$\varphi \in U \implies |f(\varphi)-f(0)|<\frac\varepsilon2, \, \text{for all }f \in \mathcal{C} \implies \sup_{f \in \mathcal{C}} |f(\varphi)-f(0)|\le\frac\varepsilon2<\varepsilon$$
$U$ being a neighbourhood of $0$ contains an intersection of finitely many open balls around the origin of radii $\delta_1, \ldots, \delta_k$ with respect to the seminorms of compact sets $K_1, \ldots, K_n \subseteq E$:
$$\bigcap_{k=1}^n \{\phi \in E' : p_{K_k}(\phi) < \delta_k\} \subseteq U.$$
Sets $K_k$ are bounded in norm by some $M_k > 0$ so if we set $$\delta := \min_{1 \le k \le n}\frac{\delta_k}{M_k}$$ then for any $\varphi \in E'$ we have
$$\|\varphi\|_{E'} < \delta \implies p_{K_k}(\varphi) = \sup_{x \in K_k}\|\varphi(x)\| \le \|\varphi\|_E'\sup_{x \in K_k}\|x\| < \delta M_k \le \delta_k$$
for all $k=1, \ldots, n$ so
$$\|\varphi\|_{E'} < \delta \implies \varphi \in \bigcap_{k=1}^n \{\phi \in E' : p_{K_k}(\phi) < \delta_k\} \subseteq U \implies \sup_{f \in \mathcal{C}} |f(\varphi)-f(0)|<\varepsilon.$$


问:
If $\mu$ is a probability measure s.t. the characterizing function of $\mu$ maps a closed ball to $\{1\}$, then $\mu=\delta_0$
Let $\mu$ be a probability measure on the Borel $\sigma$-algebra $\mathcal B(E)$ of a normed vector space $E$, $B_{E'}$ denote the closed unit ball in $E'$ and $\hat\mu$ denote the characteristic function of $\mu$.

Assume there is a $\delta>0$ with $$\tag1\hat\mu(\delta B_{E'})=\{1\}.$$ Why are we able to conclude $$\hat\mu(E')=\{1\}?\tag2$$

The claim should follow from the inequality $$\left|\hat\mu(\varphi_1)-\hat\mu(\varphi_2)\right|^2\le2\hat\mu(0)\left(\hat\mu(0)-\Re\hat\mu(\varphi_1-\varphi_2)\right);\;\;\text{for all }\varphi_i\in E'.\tag3$$ But how exactly?

答:
The first claim follows from (3) by induction: Let's show that $\hat \mu(2^n \delta B_{E'})=\{1\}$ for all $n\in \mathbb{N}$. For $n=0$ this is true by assumption. For the induction step take $\phi\in 2^{n+1}\delta B_{E'}$. Since $\frac 1 2 \phi\in 2^{n}\delta B_{E'}$, we have $\hat \mu(2^{-1}\phi)=1$ by induction hypothesis. Therefore
$$
|\hat \mu(\phi)-1|^2=|\hat \mu(\phi)-\hat\mu(2^{-1}\phi)|^2\leq 2(1-\hat \mu(2^{-1}\phi))=0
$$
by (3).


问:
Find $\sup _\limits{Q \in M_{4\times 2} (\mathbb{R}), Q^{T} Q=I_{2}} \operatorname{tr}\left(Q^{T} A Q\right)$
Let
$$
A:=\left[\begin{array}{llll}
3 & 1 & 0 & 0 \\
1 & 3 & 0 & 0 \\
0 & 0 & 6 & 2 \\
0 & 0 & 2 & 6
\end{array}\right]
$$
Find $\sup _\limits{Q \in M_{4\times 2} (\mathbb{R}), Q^{T} Q=I_{2}} \operatorname{tr}\left(Q^{T} A Q\right)$, where $M_{4 \times 2}(\mathbb{R})$ represents the set of all matrices of size $4\times 2$.

I know that $\mathrm{tr}A=\sum _i A_{ii}$, but how can we deal with this upper bound? It is obvious that $Q^T AQ$ is a $2\times 2$ matrix, but I don't know how does the condition $Q^TQ=I_2$ help. Also, are there any backgrounds for this problem? I seldom see (linear algebra) problems asking the uppper bound for a trace and I hope I could get further information about these kind of problems (if possible).

答:
$A$ is positive definite and its four eigenvalues are $2,4,4,8$. Von Neumann's trace inequality gives
$$
\operatorname{tr}(Q^TAQ)\le\sum_{i=1}^2\sigma_i(Q^T)\sigma_i(AQ)=\sum_{i=1}^2\sigma_i(A)=\sum_{i=1}^2\lambda_i^\downarrow(A)=8+4=12.
$$
Alternatively, note that $Q^TAQ$ is a principal submatrix of $U^TAU$ for some orthogonal matrix $U$. By Cauchy's interlacing inequality for bordered submatrices of Hermitian matrices or by Courant-Fischer minimax inequality, we have $\lambda_i^\downarrow(Q^TAQ)\le\lambda_i^\downarrow(U^TAU)=\lambda_i^\downarrow(A)$. Therefore $\operatorname{tr}(Q^TAQ)=\sum_{i=1}^2\lambda_i^\downarrow(Q^TAQ)\le\sum_{i=1}^2\lambda_i^\downarrow(A)=12$.

Obviously, equalities hold in the above when the two columns of $Q$ are two unit eigenvectors corresponding to the eigenvalues $8$ and $4$ respectively.


问:
Learning roadmap for Knot theory
I want to study Knot theory in the summer break. What are some good resources on this topic? My background is a a first course in Analysis and Linear Algebra, and Hatcher's notes on point-set topology. Also, what should I study next, and from where, to be sufficiently prepared for learning Knot theory? I am interested in learning some Algebraic Topology as well.</p>

答:
I'm a big fan of Justin Roberts excellent set of notes titled "knots knotes" which is available on his web page.

There are other excellent references such as Rolfsen's Knots and Links, Lickorish's Introduction to Knot Theory, and Adam's Knot Book. Related to this field is 3-manifold topology, which is also a very interesting subject to read. Hatcher has some online notes.

I also highly recommend some passing familiarity with homotopies, isotopies, and what have you (basic algebraic topology via Hatcher or Fulton or Bredon etc.) and some differential topology (Milnor or Guellemin-Pollack). If you're completely new to manifolds, Spivak's Calculus on Manifolds is a great short text; Milnor's Differential Topology is even shorter! Below is my take on a few texts.

Roberts - Very approachable, very Modern, easy to read, fun exercises.

Rolfsen - A little dated, but beautiful diagrams and great exercises. Lots of good stuff here.

Lickorish - Seems to be a little more sophisticated than the others. I have no personal experience with this text.

Schulten's Intro to 3 Manifold Topology - Difficult text, but very good.

Have fun reading!

Title: Learning roadmap for Knot theory


问:
Cohomology $H^ {i}(\mathbb{R}P^n, \mathbb{Z}_2)$
I have to compute $H^{i}(\mathbb{R}P^2,\mathbb{Z}_2)$. I know that is $\mathbb{Z}_2$ for $i=0,1,2$ but I'm looking for a proof without universal coefficient theorem. Have you some ideas?</p>

答:
If you look at Hatcher page 144 he writes down the cellular chain complex for $\Bbb{R}P^n$ for both even and odd $n$. Consider the case now when $n$ is even and our coefficient theory is in $\Bbb{Z}/2\Bbb{Z}$. Then your cellular chain complex is looking like

$$0 \to \Bbb{Z}/2\Bbb{Z} \stackrel{2\cdot}{\longrightarrow} \Bbb{Z}/2\Bbb{Z} \stackrel{0}{\longrightarrow} \ldots \stackrel{0}{\longrightarrow} \Bbb{Z}/2\Bbb{Z} \to 0.$$

However because our chain groups are now $\Bbb{Z}/2\Bbb{Z}$ the multiplication by two map is the zero map. If you now apply the functor $\hom(-,\Bbb{Z}/2\Bbb{Z})$ you will get the exact same chain complex but with arrows going the other way. Now what happens when you take the homology of a chain complex where all the maps are the zero maps?


问:
What is this symmetric-like property of $n$-ary relations?
Let's consider that there is a set $S$ for which $R \subseteq S \times S \times S$ is a relation on that set. With $x,y,z \in S$, imagine that the following holds:

$$R(x, y, z) \iff R(x, z, y) \iff R(y, x, z) \iff R(y, z, x) \iff R(z, x, y) \iff R(z, y, x)$$

This 3-ary relation has a property that reminds me of symmetric relations in which the order of the elements doesn't matter. This property can hold non-trivially for arities $n>1$, and still hold for $n=0$ or $n=1$ in a trivial way, making it something rather general. What is this property called of an $n$-ary relation?

答:
Well, there is the notion of symmetric function.

An $n$-ary function $f(x_1,\ldots,x_n)$ is symmetric if its value is the same no matter the order of its arguments.

Symmetric relations can be defined similarly - after all, functions are special relations.


问:
Proof of $\left(K_{1} \cap K_{2}\right)^{*}=K_{1}^{*}+K_{2}^{*}:$ the dual of intersection of convex cones is the sum of their duals
I'm trying to prove the following:

Let $K_{1}$ and $K_{2}$ be pointed closed convex cone in $E$ such that $K_{1 }∩ K_2 \neq \emptyset$.

Show that
  $\left(K_{1} \cap K_{2}\right)^{*}=K_{1}^{*}+K_{2}^{*}$

under an appropriate assumption on $K_{1}$ and $K_{2}$.

I have managed to show $K_{1}^{*}+K_{2}^{*}  \subseteq \left(K_{1} \cap K_{2}\right)^{*} $ without making any assumptions, but I have a hard time showing $\left(K_{1} \cap K_{2}\right)^{*}\subseteq K_{1}^{*}+K_{2}^{*}$.

would appreciate some help

答:
The proof only needs a basic property of dual cone:
$$
\begin{align}
K_1 \subseteq K_2 \implies K_1^* \supseteq K_2^*, \tag{1}
\end{align}
$$
which is easy to verify.

Apply (1) to $K_1 \cap K_2$, and $K_1,K_2$, we have
\begin{align}
    \left.\begin{array}{l}
    (K_1 \cap K_2) \subseteq K_1 \implies
    (K_1 \cap K_2)^* \supseteq K_1^* \\
    (K_1 \cap K_2) \subseteq K_2 \implies
    (K_1 \cap K_2)^* \supseteq K_2^* \\
    \end{array}\right\} \implies
    (K_1 \cap K_2)^* \supseteq (K_1^* \cup K_2^*). \tag{2}
\end{align}
Take convex hull of both side of (2). Left side won't change, and right side becomes $K_1^*+K_2^*$:
\begin{equation}
    (K_1 \cap K_2)^* \supseteq K_1^* + K_2^*. \tag{3}
\end{equation}

Apply (1) to $K_1^* \cup K_2^*$, and $K_1^*,K_2^*$ similarly, we have
\begin{align}
    \left.\begin{array}{l}
    K_1^* \subseteq (K_1^* \cup K_2^*) \implies
    K_1^{**} \supseteq (K_1^* \cup K_2^*)^* \\
    K_2^* \subseteq (K_1^* \cup K_2^*) \implies
    K_2^{**} \supseteq (K_1^* \cup K_2^*)^*
    \end{array}\right\} \implies
    (K_1^{**} \cap K_2^{**}) \supseteq (K_1^* \cup K_2^*)^*. \tag{4}
\end{align}
Apply (1) to (4):
\begin{align}
             (K_1^{**} \cap K_2^{**})^* & \subseteq (K_1^* \cup K_2^*)^{**} \\
\end{align}
For any set $K$, dual cone of dual cone is the convex hull of origin set, i.e., $K^{**}=\operatorname{ConvexHull}(K)$, so
\begin{align}
    \implies (K_1 \cap K_2)^* & \subseteq \operatorname{ConvexHull}(K_1^* \cup K_2^*) \\
    \implies (K_1 \cap K_2)^* & \subseteq K_1^* + K_2^* \tag{5}
\end{align}

(3)(5) implies $$(K_1 \cap K_2)^* = K_1^* + K_2^*$$

Actually, this holds not only for closed convex cone $K_1, K_2$, but also holds for arbitrary set $K_1,K_2$.


问:
Not understanding the meaning of a logic question.
I am trying to understand the meaning of this exercise question in a logic textbook.

For each of the following statement forms,find a statement form that is logically equivalent to its negation and in which negation signs apply only to statement letters.

i. $A \rightarrow ( B \leftrightarrow \lnot C )$

ii. $\lnot A \lor ( B \rightarrow  C )$

iii. $A \land (B \lor \lnot C)$

But I keep failing. I don't understand  the part "find a statement form that is logically equivalent to its negation and in which negation signs apply only to statement letters". So I am unable to solve it.Can someone tell me what that means?

答: It means you are only using $\neg$ in front of $A$ $B$ or $C$. So instead of writing $\neg(A \lor B)$, for example, you would write $\neg A \land \neg B$


答:
Here is a more elementary solution.

Let $Q=\begin{bmatrix}Q_1\\Q_2\end{bmatrix}$. Then $$I_{2\times2}=Q^TQ=\begin{bmatrix}Q_1^T Q_2^T\end{bmatrix}\begin{bmatrix}Q_1\\Q_2\end{bmatrix}=Q_1^TQ_1+Q_2^TQ_2$$
$$Q^TAQ=\begin{bmatrix}Q_1^T Q_2^T\end{bmatrix}\begin{bmatrix}B&0\\0&2B\end{bmatrix}\begin{bmatrix}Q_1\\Q_2\end{bmatrix}=Q_1^TBQ_1+2Q_2^TBQ_2$$
Note that $Q_1^TQ_1$ and $Q_2^TQ_2$ are simultaneously diagonalizable, with non-negative eigenvalues that add to one, i.e., $Q_1^TQ_1=PD_1P^T$, $Q_2^TQ_2=P(I-D_1)P^T$ with $P$ orthogonal.

Since the problem is to maximize the trace of the sum, both terms of which are similar to each other, it is optimal to choose $Q_1=0$. Then $Q_2$ is orthogonal and the maximum trace is $2\mathrm{tr}B=2\times6=12$.

Body:
Let
$$
A:=\left[\begin{array}{llll}
3 & 1 & 0 & 0 \\
1 & 3 & 0 & 0 \\
0 & 0 & 6 & 2 \\
0 & 0 & 2 & 6
\end{array}\right]
$$
Find $\sup _\limits{Q \in M_{4\times 2} (\mathbb{R}), Q^{T} Q=I_{2}} \operatorname{tr}\left(Q^{T} A Q\right)$, where $M_{4 \times 2}(\mathbb{R})$ represents the set of all matrices of size $4\times 2$.

I know that $\mathrm{tr}A=\sum _i A_{ii}$, but how can we deal with this upper bound? It is obvious that $Q^T AQ$ is a $2\times 2$ matrix, but I don't know how does the condition $Q^TQ=I_2$ help. Also, are there any backgrounds for this problem? I seldom see (linear algebra) problems asking the uppper bound for a trace and I hope I could get further information about these kind of problems (if possible).


答:
In each problem you have a statement form $\varphi$; you are to start with its negation $\neg\varphi$ and use logical equivalences to transform that into a statement form in which no compound expression is negated: the only expressions that are preceded by a negation are individual statement letters. As Couchy said in the comments, you cannot have $\neg(A\land B)$, because in it the compound expression $A\land B$ is negated, but you can apply De Morgan’s laws to convert that to $\neg A\lor\neg B$, in which the only things negated are the statement letters $A$ and $B$. I’ll work the first one in gory detail as an example.

$$\begin{align*}
\neg\big(A\to(B\leftrightarrow\neg C)\big)&\equiv\neg\big(\neg A\lor(B\leftrightarrow\neg C)\big)\\
&\equiv\neg(\neg A)\land\neg(B\leftrightarrow\neg C)\\
&\equiv A\land\neg(B\leftrightarrow\neg C)\\
&\equiv A\land\neg\big((B\land\neg C)\lor(\neg B\land \neg(\neg C))\big)\\
&\equiv A\land\neg\big((B\land\neg C)\lor(\neg B\land C)\big)\\
&\equiv A\land\big(\neg(B\land\neg C)\land\neg(\neg B\land C)\big)\\
&\equiv A\land\neg(B\land\neg C)\land\neg(\neg B\land C)\\
&\equiv A\land\big(\neg B\lor\neg(\neg C)\big)\land\big(\neg(\neg B)\lor\neg C\big)\\
&\equiv A\land(\neg B\lor C)\land(B\lor\neg C)
\end{align*}$$

Title: Not understanding the meaning of a logic question.

Body:
I am trying to understand the meaning of this exercise question in a logic textbook.

For each of the following statement forms,find a statement form that is logically equivalent to its negation and in which negation signs apply only to statement letters.

i. $A \rightarrow ( B \leftrightarrow \lnot C )$

ii. $\lnot A \lor ( B \rightarrow  C )$

iii. $A \land (B \lor \lnot C)$

But I keep failing. I don't understand  the part "find a statement form that is logically equivalent to its negation and in which negation signs apply only to statement letters". So I am unable to solve it.Can someone tell me what that means?


问:
How good is the approximation $e^{-x} = \left(1- \frac{x}{n}\right)^n$?
We know that for $n\rightarrow \infty$, the above converges. But how good of an approximation is it for non arbitrarily large $n$ (i.e. $n << \infty$)?

答:
By the Taylor series of the $\log$ and $\exp$ functions,
\begin{align*}
\left( {1 - \frac{x}{n}} \right)^n & = \exp \left( {n\log \left( {1 - \frac{x}{n}} \right)} \right) = \exp \left( {n\left( { - \frac{x}{n} + \mathcal{O}\!\left( {\frac{{x^2 }}{{n^2 }}} \right)} \right)} \right)
\\ & = e^{ - x} \exp \left( {\mathcal{O}\!\left( {\frac{{x^2 }}{n}} \right)} \right) = e^{ - x} \left( {1 + \mathcal{O}\!\left( {\frac{{x^2 }}{n}} \right)} \right),
\end{align*}
as $n\to +\infty$, provided $|x| \ll n$.


问:
Post-Uni Calculus/Probabilities Book Suggestion
I have a Computer Science Background, recently graduated and I would like to refresh/improve my knowledge about probabilities and statistics (also calculus). The priority is probabilities and statistics and then calculus. Can you suggest me any good books?

Also, is the * For Dummies series any good? They have Statistics and Probabilities books, and I do not know which one will be more useful for my programming abilities. Also I like calculus but the books I found were for high school level.

Any suggestions please? Thanks! (also I am looking at a price range of < 30 pounds.)

答:
I would suggest Probability Theory and its Application Volume I by William Feller
and Calculus Volume I by Tom M Apostol.

Feller is an excellent book for Probability.


问:
Probability of picking 4 red balls one by one?

There are 3 red balls and 2 blue balls in a jar. You pick any 4 balls
without looking in the jar. What is the probability that you would be
having 4 red balls in hand?

Note that you are picking balls one by one means after picking one you put that ball in the jar and again pick another ball without looking and by this way you have to pick 4 red balls

答: The answer would be: $(\frac{3}{5})^4$ since each time the probability of picking a red ball is $\frac{3}{5}$ and you need this to happen $4$ times while the events are independent since you will put back the ball you pick each time.


问:
For multiplication with conditional probabilities, why can't $A_1 \cap A_2 \cap ... \cap A_n = \emptyset$?
My book says this for the formula of multiplication with conditional probabilities:

For any events $A_1, A_2, ..., A_n$ with $A_1 \cap A_2 \cap ... \cap A_n \neq \emptyset$

$P(A_1 \cap A_2 \cap ... \cap A_n) = P(A_1)P(A_2 |A_1)P(A_3|A_1\cap A_2)...P(A_n|A_1\cap A_2  \cap ...\cap A_{n-1})$

This makes intitutive sense to me as $P(A_1 \cap A_2 \cap ... \cap A_n)$ is just equivalent to $P(A_1)$ occurs and then given that $A_1$ occurs, we need $A_2$ to occur too, and then,
given that $A_1$ and $A_2$ occur, we need $A_3$ to occur too, and then,
given that $A_1$ and $A_2$ and $A_3$ occur, we need $A_4$ to occur too and so on till we reach $P(A_n|A_1\cap A_2  \cap ...\cap A_{n-1})$ and stop because then all the events in the intersection in the LHS occurred!

My question is, why isn't $A_1 \cap A_2 \cap ... \cap A_n \neq \emptyset$ allowed? Wouldn't that just make $P(A_1 \cap A_2 \cap ... \cap A_n) = 0$ which is a completely legimately value for probability? Is any mathematical law broken with $A_1 \cap A_2 \cap ... \cap A_n = \emptyset$ that makes it not allowed?

Thank You!

答:
The right condition you want is $P(A_1 \cap \ldots \cap A_{n-1}) \neq 0$, however, this is not implied by $A_1 \cap \ldots \cap A_{n-1} \neq \emptyset$ (unless maybe your book considers discrete probability spaces and requires no $0$-sets).

If $A_1 \cap \ldots \cap A_{n-1} = \emptyset$ (or even $P(A_1 \cap \ldots \cap A_{n-1}) = 0$), then the last probability would not be defined since

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

However, if $P(A_1 \cap \ldots \cap A_{n}) = \emptyset$ and $P(A_1 \cap \ldots \cap A_{n-1}) \neq 0$ then the formula is well defined and true since the LHS is $0$ and the last enumerator in the last factor is also $0$.


问:
Not understanding how to solve this question?

If $z=x+iy$ lies in the third quadrant, then prove that $\frac{\bar{z}}{z}$ also lies in the third quadrant when $y<x<0$.

My attempt:  I got equation $$\frac{x^2 - y^2 - 2ixy}{x^2+y^2}$$

Now , I got $x^2 - y^2<0 $ and $-2xy<0$ , $x^2 < y ^2$. Therefore $y$ is greater than $x$. I got $y<x$ this I did not get.

答: Well, how did you get $x^2-y^2<0$ if you didn't assume the inequality $y<x<0$? You can only get that if you assume $y<x<0$ because that would imply $|y|>|x|$ hence $y^2>x^2$.


问:
When is a weak limit of martingales a martingale
Suppose I have a sequence $\{X_t^n,{\cal F}_t^n\}$ of martingales (with respect to different filtrations), and suppose that $X_n$ converge weakly to a continuous process $X$. What are the conditions for $X$ to be a martingale with respect to its natural filtration ${\cal F}_t=\sigma(X_s,0\leq s\leq t)$?

In the problem I have in my research I can show that the $\{X^n\}$ are uniform integrable. So, in order to show $E[C_{t+u}1_A]=E[C_t1_A]$ for all $A\in{\cal F}_t$, I can show first
$$E[C_{t+u}g(C_s,0\leq s\leq t)]=E[C_{t}g(C_s,0\leq s\leq t)]$$
for all continuous and bounded $g$.

Then, I'm using the functional monotone class theorem (link); but I'm missing one step, which is to show that the $\sigma$-field generated by all bounded and continuous functions of $\{C_s,0\leq s\leq t\}$ is ${\cal F}_t$.
I'm pretty sure this is true but I don't know how to prove this. I think this will answer my question.

I hope this question makes sense...

Thanks

答:
In the book:  "J. Jacod and  A. N. Shiryayev, Limit Theory for Stochastic Processes, 2ed. Springer, 2003'',
There is following result(p.522 Propsition 1.1): Assume that $ (M^n) $ is a sequence of martingales converging in law to a limit process $ M $, and that $ |M^n|\le b $
identically for some constant $ b $. Then $ M $ is a martingale with respect to the filtration that it generates.


问:
Give a counterexample of the compactness property
I want to refute that if $\Omega$ is complete metric space, $F$ is Borel $\sigma$-algebra $\Omega$, then for any positive $\varepsilon$ there is a compact set $K$ such that $\mathbb{P}(K)>1−\varepsilon$.

I know that for this limitation to be true, I need need separability of $\Omega$ or at least separable support of $\mathbb P$. Therefore, I need to give a counterexample, but unfortunately I cannot come up with one. I would be very grateful for help

答:
One particularly simple example we might hope for is if we let $\Omega$ be a discrete metric space. Then the compact sets are all finite, so as long as finite sets are measure-zero, we're done. Of course the simplification to the compact sets comes with a price elsewhere: now the Borel algebra is the full power set and thus we need a (countably-additive) probability measure that is defined on the whole power set and such that finite sets have measure zero. This is the measure problem. The nonexistence of such a set and measure is consistent with ZFC, and the existence of one is consistent with ZFC if and only if a measurable cardinal is consistent. So we have a "trivial" counterexample, but at a high consistency strength.

But it turns out any counterexample requires a measurable cardinal, so the discrete counterexample isn't suboptimal in terms of consistency strength.

Addition

Since the reference given to this second claim is an encyclopedic 900-page volume that doesn't take a particularly direct route toward the result, I'll include a more streamlined version of Fremlin's proof here. This is my first pass at understanding this theorem, so I may have introduced some errors and there may well be some additional simplifications I missed.

First, a definition.

A cardinal $\kappa$ is measure-free if it does not admit a non-zero finite measure on all of $P(\kappa)$ such that finite sets have measure zero.

If a there is a cardinal that is not measure-free, the least such cardinal is real-valued measurable, and the existence of a real-valued measurable cardinal is equiconsistent with the existence of a measurable cardinal. (See, e.g. Jech Set Theory 3rd edition, chapter 10 and the beginning of chapter 22.) The main theorem is

Let $\Omega$ be a complete metric space with weight $w(\Omega)=\kappa$ (i.e. $\kappa$ is the minimum cardinality of a base for $\Omega$'s topology). If $\kappa$ is measure-free, then for any finite Borel measure $(\Omega, \mathcal B(\Omega), \mu)$ and $\epsilon > 0$, there is a compact $K\subseteq \Omega$ with $\mu(\Omega\setminus K) < \epsilon.$

First, note that if we can find a closed, separable $\Omega_0\subseteq \Omega$ such that $\mu(\Omega\setminus\Omega_0)=0,$ we're done: $\Omega_0$ is a complete and separable metric space and the restriction of $\mu$ is a finite Borel measure on $\Omega_0.$ Thus, by the result OP cites that additionally assumes separability, there is a compact set $K\subseteq \Omega_0$ with $\mu(\Omega_0\setminus K)<\epsilon,$ and since $\mu(\Omega\setminus\Omega_0)=0,$ we have $\mu(\Omega\setminus K)<\epsilon.$

So, it suffices to prove

If $\Omega$ is a complete metric space, $w(\Omega)$ is measure-free, and $\mu$ is a finite Borel measure on $\Omega$, then there is a closed, separable $\Omega_0\subseteq \Omega$ such that $\mu(\Omega\setminus \Omega_0)=0.$

First a couple lemmas:

Lemma 1. Let $\Omega$ be a metric space with $w(\Omega)$ measure-free, $\mu$ a finite Borel measure on $\Omega,$ and let $\mathcal F$ be some collection of disjoint open subsets of $\Omega$. Write $$\mathcal F^0 := \{U\in\mathcal F: \mu(U)=0\}\\\mathcal F^{>0}:=\{U\in \mathcal F: \mu(U)> 0\}.$$ Then $\mathcal F^{>0}$ is countable and $\mu\left(\bigcup \mathcal F^{0}\right) =0.$

Proof. It's clear that $\mathcal F^{>0}$ is countable since the measure is finite and the sets in $\mathcal F$ are disjoint.

For the other claim, say that to the contrary, $\mu\left(\bigcup \mathcal F^{0}\right) > 0.$ We will show this means $w(\Omega)$ is not measure-free. Write $\mathcal F^0$ as an indexed set $\mathcal F^0 = \{U_i: i\in I\}.$ Then, we can define a non-zero finite measure space $(I, P(I), \nu)$ by letting $\nu(J) = \mu\left(\bigcup_{i\in J} U_i\right)$ for $J\subseteq I.$ Since the $U_i$ are disjoint, it's easy to verify countable additivity and show that $\nu$ is a measure on the full power set of $I$ with $\nu(\{i\}) = \mu(U_i) = 0.$ Thus $|I|$ is not measure-free.

Since the $U_i$ are disjoint, for any base for $\Omega$'s topology, there will be some distinct basis element inside each $U_i.$ Thus, we have $|I|\le w(\Omega).$ But it's well-known that if a cardinal $\kappa$ is not measure-free, then neither is any $\lambda>\kappa$: we can extend a witnessing measure on $\nu$ $P(\kappa)$ to one on $P(\lambda)$ by letting $\nu'(X) = \nu(X\cap\kappa).$ So $w(\Omega)$ is not measure-free.

Lemma 2. If $\Omega$ is a metric space, there is a base $\mathcal B$ for its topology that is a countable union of disjoint collections. In other words, $\mathcal B = \bigcup_{n=1}^\infty \mathcal B_n$ where each $\mathcal B_n$ is a disjoint collection of open sets.

Proof Sketch. Let $(q_n,q_n')$ be an enumeration of $\{(q,q')\in\mathbb Q^2: 0<q<q'\}.$ Enumerate $\Omega=\{x_\xi: \xi < \kappa\}$ where $\kappa=|\Omega|.$ Let $$G_{n\xi} = \{x\in \Omega: d(x,x_\xi) < q_n\land \inf_{\eta < \xi}d(x,x_\eta)>q_n'\}$$ and then take $\mathcal B_n=\{G_{n\xi}: \xi<\kappa\}.$  and verify the $G_{n\xi}$ are open sets, disjoint for fixed $n,$ and that $\bigcup_n \mathcal B_n$ is a base. For more info, see Fremlin's Measure Theory, Volume 4, section 4A2L(g)(ii)

Then we can put it together as follows. Let $\mathcal B=\bigcup_n\mathcal B_n$ be a base as in the second lemma. For each $n,$ similarly to the first lemma, let $\mathcal B_n^{>0}$ denote the collection of positive-measure sets in $\mathcal B_n$ and $\mathcal B_n^0$ the null sets. By the first lemma, each $\mathcal B_n^{>0}$ is countable and the union of the sets in $\mathcal B_n^0$ is null. Define $\Omega_0 = \Omega\setminus \bigcup_n \left(\bigcup \mathcal B_n^0\right).$ $\Omega_0$ is clearly closed with $\mu(\Omega\setminus \Omega_0)=0,$ so we just must show that it is separable.

Claim: If $U\in\mathcal B$ and $U\cap\Omega_0\ne \emptyset,$ then $U\in \bigcup_n\mathcal B_n^{>0}.$

Proof. Let $x\in U\cap\Omega_0.$ $U\in \mathcal B,$ so $U\in \mathcal B_n$ for some $n$. Since $x\in \Omega_0,$ it follows that $U\notin \mathcal B_n^0$ (or we'd have $x\in \bigcup \mathcal B_n^0$), and thus $U\in \mathcal B_n^{>0}.$

So, noting that $\bigcup_n \mathcal B^{>0}_n$ is countable, $\{U\cap \Omega_0:U\in\mathcal B\}$ is a countable base for $\Omega_0,$ so $\Omega_0$ is separable.


问:
Why is this not a equivalence relation in the $\mathbb{Z}$ set?

The relation $xRy$ on the set $\mathbb{Z}$ if $x + y$ is divisible by $3$.

$xRx = x + x = 2x$. In order to $2x$ be divisible by $3$, $x$ itself need to be divisible by $3$, it is not a problem is it?

$xRy = yRx$

$xRy = yRz = xRz$

I think the last two are immediate, so probably my error is in the first.

答:
Note that $1R2$ ($1+2$ is divisible by $3$) and  $2R4$ ($2+4$ is divisible by $3$) do not imply  $1R4$ ($1+4$ is not divisible by $3$). Hence transitive property does not hold.

Also reflexive property $xRx$ is not satisfied when $x\equiv 1\pmod{3}$ or $x\equiv 2\pmod{3}$.


答:
You have

$$\frac{\bar z}z=\frac{x-iy}{x+iy}=\frac{x^2-y^2-2xyi}{x^2+y^2}\,,$$

with real part $\frac{x^2-y^2}{x^2+y^2}$ and imaginary part $-2xy$. You seem to have got that far just fine.

You’re given that $y<x<0$: it’s not something that you have to prove or discover. That implies that $y^2>x^2$, so $x^2-y^2<0$. Moreover, $x^2+y^2>0$, so $\frac{x^2-y^2}{x^2+y^2}<0$: that shows that the real part is negative. Finally, $x$ and $y$ are negative, so $xy>0$, and $-2xy<0$: that shows that the imaginary part is also negative. And that puts $\frac{\bar z}z$ in the third quadrant.

Note that $y^2>x^2$ does not imply that $y>x$: it implies only that $|y|>|x|$. And when $x$ and $y$ are negative, as is the case here, that means that $y<x$.

Title: Not understanding how to solve this question?

Body:
If $z=x+iy$ lies in the third quadrant, then prove that $\frac{\bar{z}}{z}$ also lies in the third quadrant when $y<x<0$.

My attempt:  I got equation $$\frac{x^2 - y^2 - 2ixy}{x^2+y^2}$$

Now , I got $x^2 - y^2<0 $ and $-2xy<0$ , $x^2 < y ^2$. Therefore $y$ is greater than $x$. I got $y<x$ this I did not get.


问:
Asymptotic behaviour of quantity related to an ODE
Consider the differential equation

$$u''(t) + u(t) + u^3(t) = f(t) u(t) + g(t),\,\, u(0) = u_0, \,u'(0) = u_1$$

where $u :\mathbb{R}^+ \to \mathbb{R}$ is a bounded solution  and $f, g: \mathbb{R}^+ \to \mathbb{R}$ are given continuous functions such that $$\lim_{t\to \infty} f(t) = \lim_{t\to \infty} g(t) = 0.$$
Define the quantity $E(t) := \frac 12 (u(t))^2 +\frac 12 (u'(t))^2 + \frac 14 (u(t))^4.$
I want to study the asymptotic of $E$ when $t \to \infty.$  First, I considered many examples for $f, g$ using mathematica (See the attached images where I plotted $t\mapsto E(t)$). I'm interested in the case where
$\liminf_{t\to \infty}E(t) >0$ or roughly $\lim_{t\to \infty} E(t) >0.$ I wonder if this is the general case or it depends on the initial values $u_0, u_1$ and $f,g.$ In all the examples that I tested numerically, I found that $\lim_{t\to \infty} E(t) >0.$
My attempt started by differentiating $E$ with respect to $t.$ It seems that in the case where
$u'$ is bounded, we have $$\lim_{t\to \infty} E'(t) =0,$$
since $E' = (f u + g) u'.$ But this does not imply necessarily that $\lim_{t\to \infty}E(t) >0.$
The limit might be zero.

Thank you for any hint.

答:
If $\;t\to\infty,\;$ then the given ODE takes the form of
$$\ddot u+u+u^3 =0,\tag1$$
which does not contain variable $\;t.\;$

Substitution $\;P=\dot u,\;\ddot u = P'_u\, \dot u = P\,\dfrac{\text dP}{\text du},\;$ presents $(1)$ in the form of
$$P\,\text dP = -(u^3+u)\,\text du,$$
$$P=\dfrac12\sqrt{(a^2+1)^2-(u^2+1)^2},$$
$$\dfrac{\text du}{\sqrt{a^2-u^2}} = \sqrt{2+a^2+u^2},\tag2$$
and after substitutiion
$$v=\arcsin \dfrac ua$$
one can get
$$\text dv = \sqrt{2+a^2(1+\sin^2 v)}\,\text dt,$$
with the solutions
$$v(t) = \text{am}\left(\sqrt{a^2+2}\,(t+b)\bigg|\dfrac{a^2}{2+a^2}\right)$$
and
$$u(t) = a \text{ sn}\left(\sqrt{a^2+2}\,(t+b)\bigg|\dfrac{a^2}{2+a^2}\right),\tag3$$
where $\;\text{am}\;$ and $\;\text{sn}\;$ are the Jacobi functions.

The Jacobi functions looks too heavy for applying of the constant variations method.

Taking in account derivatives
\begin{cases}
\left(\text{sn}(x|m)\right)' = \text{cn}(x|m)\text{ dn}(x|m)\\[4pt]
\left(\text{cn}(x|m)\right)' = -\text{sn}(x|m)\text{ dn}(x|m)\\[4pt]
\left(\text{dn}(x|m)\right)' = -m\text{ sn}(x|m)\text{ cn}(x|m),
\end{cases}

looks possible to find parameters $\;a\;$ and $\;b\;$ for the given functions $\;f(t)\;$ and $\;g(t)\;$ numerically, using the next statement:

$E(t) = \dfrac14(2u'^2+u^4+2u^2) \ge 0,$

$E(0) = \dfrac14(2u_1^2+u_0^4+2u_0^2),$

$E'(\infty) = 0,$

$E(\infty) = E(0)+\int\limits_0^\infty(uf+g)\,\text dt.$


问:
Can $\lim_{n\to\infty}(a_n-a_{2n})=0$ ensure $\lim_{n\to\infty}a_n=0$?
Can $\lim_{n\to\infty}(a_n-a_{2n})=0$ ensure $\lim_{n\to\infty}a_n=0$? It seems difficult to find a counterexample. The often used  $\sum \frac{1}{n}$ is not such one.

答: $a_n=1$ for all $n$ is a counterexample:  $\lim\limits_{n\to\infty}(a_n-a_{2n})=0,$ but $\lim\limits_{n\to\infty} a_n=1$.


答:
Let $y>x>0$ and $z=-x-iy$ which is in the third quadrant since $-y<-x<0$, then
$$w=\frac{\bar{z}}{z}=\frac{-x+iy}{-x-iy}=\frac{-x+iy}{-x-iy}\frac{-x+iy}{-x+iy}=\frac{x^2-y^2-2ixy}{x^2+y^2}.$$
Since $y>x>0$ then $x^2-y^2<0$ and $-2xy<0$ and $w$ is in the third quadrant.


答:
Every probability measure verifies that $P(\emptyset)=0$

So $A_1 \cap A_2 \cap ... \cap A_n = \emptyset$ $\implies$ $P(A_1 \cap A_2 \cap ... \cap A_n) = P(\emptyset)=0$ is trivial

The most important property that probability measures have (that other measures doesn't necesarily verify) is that if two events A, B are independent then $P(A \cap B) = P(A) P(B)$

Title: For multiplication with conditional probabilities, why can't $A_1 \cap A_2 \cap ... \cap A_n = \emptyset$?

Body:
My book says this for the formula of multiplication with conditional probabilities:

For any events $A_1, A_2, ..., A_n$ with $A_1 \cap A_2 \cap ... \cap A_n \neq \emptyset$

$P(A_1 \cap A_2 \cap ... \cap A_n) = P(A_1)P(A_2 |A_1)P(A_3|A_1\cap A_2)...P(A_n|A_1\cap A_2  \cap ...\cap A_{n-1})$

This makes intitutive sense to me as $P(A_1 \cap A_2 \cap ... \cap A_n)$ is just equivalent to $P(A_1)$ occurs and then given that $A_1$ occurs, we need $A_2$ to occur too, and then,
given that $A_1$ and $A_2$ occur, we need $A_3$ to occur too, and then,
given that $A_1$ and $A_2$ and $A_3$ occur, we need $A_4$ to occur too and so on till we reach $P(A_n|A_1\cap A_2  \cap ...\cap A_{n-1})$ and stop because then all the events in the intersection in the LHS occurred!

My question is, why isn't $A_1 \cap A_2 \cap ... \cap A_n \neq \emptyset$ allowed? Wouldn't that just make $P(A_1 \cap A_2 \cap ... \cap A_n) = 0$ which is a completely legimately value for probability? Is any mathematical law broken with $A_1 \cap A_2 \cap ... \cap A_n = \emptyset$ that makes it not allowed?

Thank You!


问:
Find the sum of the series $1+(\frac 12 +\frac 13)\frac 14 +(\frac 14 +\frac 15)\frac{1}{4^2}.....$ upto $\infty$
$$S = \frac 12 .\frac 14 + \frac 14.\frac{1}{4^2} + \frac 16 . \frac{1}{4^3} ....... + \frac 13. \frac 14 + \frac 15 . \frac{1}{4^2}.....$$

The first part of the series is a standard logarithmic series shoes sum is $\frac {-1}{2} (\log \frac 32 +\log \frac 12)$, but I don’t not know how to deal with the second part, which I also believe is a logarithmic expansion

答:
The series given in your title is the following with $x=1/4$:
$$\begin{align}
S(x)&=1+\sum_{k=1}^{\infty}\left(\frac{1}{2k}+\frac{1}{2k+1}\right)x^k
=\frac{1}{2}\sum_{k=1}^{\infty}\frac{x^k}{k}+\sum_{k=0}^{\infty}\frac{x^k}{2k+1}\\
&=\frac{1}{2}\sum_{k=1}^{\infty}\frac{x^k}{k}+\frac{1}{z}\sum_{k=0}^{\infty}\frac{z^{2k+1}}{2k+1}\\
&=-\frac{\log(1-x)}{2}+\frac{-\log(1-z)+\log(1+z)}{2z}
\end{align}$$
where $z=\sqrt{x}$.
Finally, we get
$$S(1/4)=\frac{\log(3)}{2}+\log(2).$$