问:
Is the posterior always a compromise between the prior and the data?
Suppose that we are interested in learning the proportion of the population $\theta$ with a particular property (for instance, the fraction of the population who are male). Suppose that we randomly sample $n$ members of this population (with replacement, to make things easier) and observe that $y$ of them have the property (so the fraction of the sample with the property is $y/n$). We start with a continuous prior $p(\theta)$ with full support $[0, 1]$ and update this using Bayes rule.

Question: does the expected value of the posterior always lie between the prior expectation and the sample fraction $y/n$?

Comments:
I know this is true in the case where my prior takes the form of a beta distribution (with parameters $\alpha$, $\beta$). In that case, we know that the prior expectation is
$$
\mathbb{E}[\theta] = \frac{\alpha}{\alpha + \beta}
$$

Due to random sampling, the probability that $y$ of the $n$ draws have the property is
$$
P(\text{data}\mid\theta) = \binom{n}{y} \theta^y (1 - \theta)^{n-y}
$$
which means (see, e.g., here) that the posterior $P(\theta\mid\text{data})$ is also beta distributed and has expected value
$$
\mathbb{E}[\theta\mid\text{data}] = \frac{\alpha + y}{\alpha + \beta + n}
$$
Moreover, it can be shown that
$$
\frac{\alpha + y}{\alpha + \beta + n} \in \left[\frac{y}{n}, \frac{\alpha}{\alpha + \beta}\right]
$$
where both inequalities are strict provided that $y/n \neq \alpha / (\alpha+\beta)$. So the posterior expectation does indeed lie between the prior expectation and the sample fraction in this instance.

To now examine the more general case, consider an arbitrary smooth prior $p(\theta)$ with full support on $[0, 1]$ and expected value $\mathbb{E[\theta]}$. By Bayes theorem, the posterior is
\begin{split}
p(\theta\mid\text{data}) &= \frac{p(\text{data}\mid\theta)P(\theta)}{P(\text{data})} \\
&= \frac{\binom{n}{y} \theta^y (1 - \theta)^{n-y}P(\theta)}{\int_0^1 P(\text{data}\mid\theta)P(\theta) d\theta}  \\
&= \frac{\binom{n}{y} \theta^y (1 - \theta)^{n-y}P(\theta)}{\int_0^1 \binom{n}{y} \theta^y (1 - \theta)^{n-y}P(\theta) \, d\theta} \\
&= \frac{\theta^y (1 - \theta)^{n-y}P(\theta)}{\int_0^1 \theta^y (1 - \theta)^{n-y}P(\theta) \, d\theta}
\end{split}

and so the posterior expectation is
\begin{split}
\mathbb{E[\theta}\mid\text{data}] &= \int_0^1 p(\theta\mid\text{data}) \theta \, d\theta \\
&= \int_0^1 \frac{\theta^y (1 - \theta)^{n-y}P(\theta)}{\int_0^1 \theta^y (1 - \theta)^{n-y}P(\theta) \, d\theta}  \theta \, d\theta \\
&= \frac{\int_0^1  \theta^y (1 - \theta)^{n-y}P(\theta) \theta \, d\theta}{\int_0^1 \theta^y (1 - \theta)^{n-y}P(\theta) \, d\theta}   \\
&= \frac{\mathbb{E}[\theta^{y+1} (1 - \theta)^{n-y}]}{\mathbb{E}[\theta^y (1 - \theta)^{n-y}]}
\end{split}

It remains to show that
$$
\frac{\mathbb{E}[\theta^{y+1} (1 - \theta)^{n-y}]}{\mathbb{E}[\theta^y (1 - \theta)^{n-y}]} \in \left[ \mathbb{E}[\theta], \frac{y}{n} \right];
$$
but this is where things begin to get tricky. Very heuristically (apologies for what is to come!), one might try to obtain one bound using an argument like
\begin{split}
\frac{\mathbb{E}[\theta^{y+1} (1 - \theta)^{n-y}]}{\mathbb{E}[\theta^y (1 - \theta)^{n-y}]} &\geq
\frac{\mathbb{E}[\theta^{y+1}] \mathbb{E}[(1 - \theta)^{n-y}]}{\mathbb{E}[\theta^y] \mathbb{E}[(1 - \theta)^{n-y}]} \\ &\geq
\frac{\mathbb{E}[\theta]^{y+1} \mathbb{E}[(1 - \theta)]^{n-y}}{\mathbb{E}[\theta]^y \mathbb{E}[(1 - \theta)]^{n-y}} \\ &=
\mathbb{E}[\theta]
\end{split}
...although, quite aside from the very dubious feel to this argument, it doesn't give us the $y/n$ bound. Any help would be much appreciated!

答:
Nice question! Sadly it is not true. I'll talk in terms of using coin flips to determine the bias $\theta$ of a weighted coin, where $\theta$ is the probability that the coin flips heads. Here's the idea: consider a prior in which we assign very high probability to values of $\theta$ very close to either $0$ or $1$ and very low probability otherwise, such that the prior expectation is $\frac{1}{2}$. Now suppose we see, say, $\frac{2}{3}$ heads in the sample. This is very unlikely if $\theta$ is close to $0$ and much more likely if it's close to $1$, so the posterior distribution is concentrated on values of $\theta$ close to $1$ and in particular the posterior mean is close to $1$, which makes it potentially larger than either the prior expectation $\frac{1}{2}$ or the sample fraction $\frac{2}{3}$.

You can see hints of this already in your beta distribution calculation: the inequalities you want assume that $\alpha$ and $\beta$ are positive, and are false for, say, $\alpha = 0, \beta = - \frac{1}{2}$. Of course the beta integral does not converge in this case but this gives us an idea of what to look for.

Formally, take $H$ (for "height") to be a large positive constant and $w$ (for "width") and $\epsilon$ to be small positive constants, and consider the "triangular" prior with probability density function

$$p(\theta) = \begin{cases} H \left( 1 - \frac{\theta}{w} \right) + \epsilon & \text{ if } 0 \le \theta \le w \\
 \epsilon & \text{ if } w \le \theta \le 1 - w \\
 H \left( 1 - \frac{1-\theta}{w} \right) + \epsilon & \text{ if } 1 - w \le \theta \le 1. \end{cases}.$$

We have $\int_0^1 p(\theta) \, d \theta = Hw + \epsilon$ so this is a pdf as long as $Hw + \epsilon = 1$. It's symmetric about $\frac{1}{2}$, so the prior expectation $\mathbb{E}(\theta)$ is $\frac{1}{2}$.

Now suppose we flip $3$ coins and $2$ of them are heads. Then the posterior density is the normalization of $\theta^2 (1 - \theta) p(\theta)$, and so the posterior expectation is $\frac{\mathbb{E}(\theta^3(1 - \theta))}{\mathbb{E}(\theta^2(1 - \theta))}$. This is a slightly tedious but doable calculation which I will punt to WolframAlpha; we get

$$\mathbb{E}(\theta^2(1 - \theta)) = \left( -\frac{w^3}{12} + \frac{w^2}{6}  \right) H + \frac{\epsilon}{12}$$
$$\mathbb{E}(\theta^3(1 - \theta)) = \left( - \frac{w^5}{15} + \frac{w^4}{5} - \frac{w^3}{4} + \frac{w^2}{6} \right) H + \frac{\epsilon}{20}$$

so the posterior expectation is their quotient. This is a bit annoying to write out in full so let's just talk about how it behaves asymptotically. If $w$ is small then the polynomials in $w$ above are dominated by their terms with smallest exponent, namely $\frac{w^2}{6}$, which in both cases comes from the portion of the integral corresponding to $\theta \approx 1$ and hence $\theta^n(1 - \theta) \approx 1 - \theta$; importantly this portion of the integral approximately does not depend on the exponent $n$ of $\theta$. We have $H = \frac{1 - \epsilon}{w}$ which gives, for both $w$ and $\epsilon$ small,

$$\mathbb{E}(\theta^2(1 - \theta)) = \frac{w}{6} + \frac{\epsilon}{12} + O(w^2 + \epsilon w)$$
$$\mathbb{E}(\theta^3(1 - \theta)) = \frac{w}{6} + \frac{\epsilon}{20} + O(w^2 + \epsilon w)$$

so we see that by taking $w$ to be small but $\epsilon$ to be much smaller we can arrange for the posterior expectation to be arbitrarily close to $1$, and in particular not in the interval $\left[ \frac{1}{2}, \frac{2}{3} \right]$, as expected. To be concrete we can take, say, $w = 0.01, \epsilon = 0.0001$.

If you're looking for a conceptual upshot, one conceptual upshot here is that when the prior is very "lopsided" like this, the prior mean is not a good summary of it, and when the prior is concentrated on two very different hypotheses, apparently small amounts of evidence can tilt the balance between them dramatically.

Edit: We can write down a simpler counterexample if we don't require the prior to have a continuous pdf or full support. Namely, we can take a discrete prior which assigns probability $\frac{1}{2}$ to either $\theta = w$ or $\theta = 1 - w$, where again $w$ is a small positive constant, and assigns $0$ probability otherwise. The prior mean is still $\frac{1}{2}$ by symmetry. Now if we flip $3$ coins and $2$ of them land heads the posterior mean is

$$\frac{w^3 (1 - w) + w (1 - w)^3}{w^2 (1 - w) + w (1 - w)^2} = \frac{w^2 + (1 - w)^2}{w + (1 - w)} = 1 - 2w + 2w^2$$

and we again see, very straightforwardly this time, that as $w \to 0$ the posterior mean gets arbitrarily close to $1$. Here we can actually determine explicitly  the values of $w$ for which the posterior mean is strictly larger than $\frac{2}{3}$: it occurs for all $0 < w < \frac{\sqrt{15} - 3}{6} \approx 0.145$.

Without an explicit calculation here we can at least check that the denominator and numerator are both $w + O(w^2)$ and so again have the same growth rate as $w \to 0$.


问:
Does the sequence $v_{n+1} = T(v_n)/|T(v_n)|$ converge to an eigenvector?
Let $\varphi:[-1,1]\times \mathbb{R}\to\mathbb{R}$ the map $\varphi(x,w)=2\omega +1$ and $\mathcal{C}^0([-1,1])=\{f:[-1,1]\to\mathbb{R};\ f \ \text{is continuous}\}$ . I am interested in studying the following bounded linear map
\begin{align*}
T:\left(\mathcal{C}^0([-1,1]),\|\cdot\|_\infty\right)&\to \left(\mathcal{C}^0([-1,1]),\|\cdot\|_\infty\right)\\
f&\mapsto \left(x\mapsto \frac{1}{2}\int_{-1}^1 f(2x+\omega)\cdot\mathbf{1}_{[-1,1]}(2x+\omega) \ \mathrm{d}\omega \right),
\end{align*}
where $\mathbf{1}_{[-1,1]}$ is the indicator function of the interval $[-1,1]$.

Using the same techniques used in the links Proving that there exists only one non-negative eigenfuction for the following operator., Is possible to show that the linear operator $T(\varphi)(x) = \int_{V_x\cap M} \varphi(y)\text{d}y$ has spectral radius $>0$., and Krein-Rutman theorem we can "easily" conclude that

$T$ is a compact linear map.

$T$ has positive radius (actually, $r(T) =1/2$).

$T$ has a unique eigenvector on the cone  $\mathcal{C}^0_+([-1,1]) =\{f:[-1,1]\to\mathbb{R};\ f \ \text{is continuous and }f\geq 0\}$ and its eigenvalue is equal to $1/2$.

I have done some numerical simulations and I think that the sequence the recursive sequence:
\begin{align*}
v_0 &= 1\\
v_n&= \frac{T(v_{n-1})}{\|T(v_{n-1})\|_{\infty}},
\end{align*}
converges to the unique eigenvector. This believe comes in light of some numerical simulations. In the picture below the red line represents the function $T(1)$, the green line $T^2(1)$, the orange one $T^3(1)$ and, finally, the blue one is $T^4(1)$.

My Question: Does anyone know if $\{v_n\}_{n\in\mathbb{N}}$ (or a subsequence of $\{v_n\}_{n\in\mathbb{N}}$)  converges to an eigenvalue? Moreover (this is just out of curiosity the important question is the first one), is there some well-known theorem that gives conditions to the sequence $\{v_n\}$ converges to an eigenvalue in the infinite-dimensional case?

It is worth mentioning that this paper gives us a result that under the "strong" Krein-Rutman theorem assumptions:

The sequence $\{v_n\}_n$ converges to an eigenvector. But, I was looking for a theorem that I could apply in my linear operator.

Can anyone please help me?

答:
Let $u \in \mathcal{C}_+^0([-1,1])$ be the normalized eigenvector of $T$ for the eigenvalue $r(T)$. One can indeed show the following result:

Theorem 1.
For every non-zero function $v_0$ in the positive cone, the iterates $\big(\frac{T}{r(T)}\big)^n v_0$ converge to a non-zero multiple of $u$ as $n \to \infty$.

Corollary 2.
Let $v_0$ be a non-zero function in the positive cone and define $v_n = \frac{Tv_{n-1}}{\|Tv_{n-1}\|}$ for each $n \ge 1$. Then $v_n$ converges to $u$ as $n \to \infty$.

Proof of Corollary 2.
Cearly, we have
$$
  v_n = \frac{\big(\frac{T}{r(T)}\big)^n v_0}{\left\| \big(\frac{T}{r(T)}\big)^n v_0 \right\|}
$$
for each $n \ge 1$.
We know from Theorem 1 that there exists a number $\alpha > 0$ such that $\big(\frac{T}{r(T)}\big)^n v_0 \to \alpha u$. Hence, the norm $\left\| \big(\frac{T}{r(T)}\big)^n v_0 \right\|$ converges to $\|\alpha u\| = \alpha$, so $v_n \to \frac{\alpha u}{\alpha} = u$. $\square$

In order to prove Theorem 1, one can use general results from the spectral theory of positive operators on Banach lattices. To get this machinery to work, the first thing to do is to observe that the function space $\mathcal{C}^0([-1,1])$ is, in a sense, "too large" for the operator. Indeed, for each $f$ from this space, the function $Tf$ vanishes at $-1$ and $-1$.

So let $\mathcal{C}_0((-1,1))$ denote that set of all functions from $\mathcal{C}^0([-1,1])$ that vanish at $-1$ and $1$. Then the range of $T$ is contained in $\mathcal{C}_0((-1,1))$, so it suffices to prove the result for the restriction of $T$ to $\mathcal{C}_0((-1,1))$. Hence, we will assume from now on that $T$ is defined on the slightly smaller space $\mathcal{C}_0((-1,1))$.

The space $\mathcal{C}_0((-1,1))$ is a Banach lattice (with respect to the usual pointwise ordering) and the operator $T$ on it is positive (i.e. it maps the cone into the cone).

Remark. The space $\mathcal{C}^0(([-1,1])$ is a Banach lattice, too, and $T$ is a positive operator on it, too; the reason why this space is not good to work with will be explained below.

In order to apply general results from the literature, we need the following terminology:

Definition 3.
Let $E$ be a Banach lattice (e.g., the space $\mathcal{C}_0((-1,1))$).

(a) A vector subspace $I$ of a Banach lattice $E$ is called an ideal if, for all $f,g \in E$ such that $g \in I$ and $\lvert f \rvert \le g$, we have $f \in I$.

(b) A positive linear operator $S$ on $E$ is called irreducible if the only closed $S$-invariant ideals in $E$ are $\{0\}$ and $E$. (Here, an ideal $I$ is said to be $S$-invariant if $SI \subseteq I$.)

In the Banach lattice $\mathcal{C}_0((-1,1))$ an explicit description of all closed ideals is known: they are precisely the sets of the form
$$
  \{f \in \mathcal{C}_0((-1,1)): \, f|_A = 0\},
$$
where $A$ is any closed subset of $(-1,1)$. This can be seen as follows:

Proof that each closed ideal can be described in this way: The fact that each such set is a closed ideal is obvious.

To see that the converse implication holds, one can, for instance, refer to [Mey91, Proposition 2.1.9]. There, a similar description of all closed ideals in the space $\mathcal{C}(K)$ for a compact Hausdorff space $K$ is given. We can simply apply this description to $\mathcal{C}^0([-1,1])$ (which would be denoted as $\mathcal{C}([-1,1])$ in the notation of [Mey91]). Since $\mathcal{C}_0((-1,1))$ is clearly a closed ideal in $\mathcal{C}^0([-1,1])$, it follows that each closed ideal $I$ in $\mathcal{C}_0((-1,1))$ is also a closed ideal in $\mathcal{C}^0([-1,1])$; hence, $I$ has the description claimed above. $\square$

With this description of closed ideals at hand, it is easy to show that our operator $T$ is irreducible. In fact, if $f$ is a non-zero function in the cone of $\mathcal{C}_0((-1,1))$, and if we use the notation
$$
  \{g > 0\} := \{x \in (-1,1): \, g(x) > 0\}
$$ for any real-valued function $g$ defined on $(-1,1)$, then we have
$$
  \{f > 0\} \subseteq \{Tf > 0\} \subseteq \{T^2f > 0\} \subseteq \{T^3f > 0\} \dots,
$$
and the (monotone) union of those sets equals the entire set $(-1,1)$. These observations are consequences of the definition of $T$.

Consequently, our operator $T$ has the following property:

Proposition 4. For each interger $n \ge 1$, the power $T^n$ is irreducible.

Remark. The fact that $T$ is irreducible and compact implies that $r(T) > 0$ by a general theorem of de Pagter [dPa86, Theorem 3]. However, I don't know whether the general theory is particularly helpful to determine the precise value of $r(T)$.

Now we apply the quite have machinery of infinite dimensional Perron-Frobenius theory, which yields the following result:

Proposition 5. Our operator $T$ has the following properties:

(a) The eigenspace $\ker(r(T) - T)$ is one-dimensional and spanned by a function $u$ which is strictly positive on $(-1,1)$.

(b) The dual eigenspace $\ker(r(T) - T)'$ is spannend by a strictly positive functional $\varphi$ in the dual space of $\mathcal{C}_0((-1,1))$ (where strictly positive means that $\langle \varphi, f\rangle > 0$ for every non-zero vector $f$ in the cone of $\mathcal{C}_0((-1,1))$).

(c) The eigenvalue $r(T)$ of $T$ is algebraically simple.

(d) Each eigenvalue of $T$ of modulus $r(T)$ is the product of $r(T)$ with a root of unity.

Proof.
Assertions (c) and (d) follow, for instance, from [Gro95, Theorem 5.3]. Assertion (b) now follows from [Gro95, Proposition 5.1]. Finally, assertion (a) follows from [Gro95, Theorem 5.2]. Note that, in order to translate the latter reference to assertion (a) above, one needs to observe that the quasi-interior points of the cone in $\mathcal{C}_0((-1,1))$ are precisely those functions $f$ that satisfy $f(x) > 0$ for each $x \in (-1,1)$. $\square$

Remark. The OP has already observed in the question that the eigenspace $\ker(r(T) - T)$ is one-dimensional. Part (a) of the previous proposition actually shows that this fact can also be derived from general infinite dimensional Perron-Frobenius theory.

For the previous proposition, we only needed $T$ itself to be irreducible, but not necessarily its powers. For the next result, we finally use the irreducibility of the powers of $T$:

Proposition 6.
The operator $T$ does not have any eigenvalue of modulus $r(T)$, except for $r(T)$ itself.

Proof.
Let us consider the operator $S := T/r(T)$. If $\lambda \not= 1$ is an eigenvalue of modulus $1$ of this operator, then it is a root of unity by Proposition 5(d). Hence, there exists an integer $n \ge 1$ such that $\lambda^n = 1$. But this implies that $1$ is an eigenvalue of $S^n$ of geometric multiplicity at least $2$. This is a contradiction: since $S$ is irreducible, the eigenspace $\ker(1-S)$ has dimension $1$ [Gro95, Theorem 5.2]. $\square$

Now, we can finally prove Theorem 1:

Proof of Theorem 1.
It suffices to prove the theorem for $v_0 \in \mathcal{C}_0((-1,1))$.

Set $S := T/r(T)$, and let $P$ denote the spectral projection of $S$ for the eigenvalue $1$. It follows from Proposition 5(a)-(c) that $P$ is a rank-$1$-projection that is given by $Pf = \langle \varphi, f \rangle u$ for all $f \in \mathcal{C}_0((-1,1))$ (provided that $\varphi$ is scaled such that $\langle \varphi, u\rangle = 1$).

According to Proposition 6, $S$ has no eigenvalues on the unit circle except for $1$. Hence, the powers $S^n$ converge to $P$ with respect to the operator norm as $n \to \infty$. In particular, for each $v_0 \in \mathcal{C}_0((-1,1))$, the sequence $(S^n v_0)$ converges to $Pf = \langle \varphi, v_0\rangle u$. $\square$

Remarks.

(a) The reason why we considered the space $\mathcal{C}_0((-1,1))$ rather than $\mathcal{C}^0([-1,1])$ is that the operator $T$ is not irreducible on $\mathcal{C}^0([-1,1])$ (in fact, the space $\mathcal{C}_0((-1,1))$ itself is a closed ideal in $\mathcal{C}^0([-1,1])$ which is invariant under $T$).

(b) The arguments shown above use quite general and abstract tools from operator theory on Banach lattices. For the concrete operator under consideration, it might be possible to find a more elementary proof (although I personally don't know one). Still, those general techniques have the advantage that they can be applied in a large variety of situations and that they yield rather strong qualitative results.

References:

[Gro95]: J. J. Grobler:  Spectral theory in Banach lattices, 1995. [link to zbMATH]

[Mey91] Peter Meyer-Nieberg: Banach lattices, 1991. [link to zbMATH]

[dPa86]: Ben de Pagter: Irreducible compact operators, 1986. [link to zbMATH]

ContentLicense: CC BY-SA 4.0


问:
Prove identity about divergence of vector fields
Let $f,g$ be scalar smooth functions and $\Phi:\mathbb R^2 \to \mathbb R^2$ be a smooth vector field. How do I prove that the following identity holds?
$$div\Bigg(g(x) \nabla h(f) - h(f) \nabla g(x) - g(x) h(f) \Phi \Bigg)= g(x)h''(f) |\nabla f|^2 - h(f)(\Delta g + \Phi \cdot \nabla g) + g h'(f) (\Delta f - \Phi \cdot \nabla f)$$

I tried to do the computations many times over, but I'm missing some pieces. The one $- h(f)(\Delta g + \Phi \cdot \nabla g)$ is the only I get consistently from my computations.

答:
You can try to use the dumb indexes technique (Einstein's summation technique, if I'm not mistaken): $div\vec{a}=\frac{\partial}{\partial{x_i}}a_i (or =\partial_i{a_i}), \Delta{g}=\partial_i\partial_ig$ (summation over repeated indices is implied), etc.

For instance, $div(\nabla{h(f)})=\partial_i\partial_i{h(f)}=\partial_i((\frac{d}{df}h)\partial_i{f})=(\frac{d^2}{df^2}h)\partial_i{f}\partial_i{f}+(\frac{d}{df}h)\partial_i\partial_i{f}=(\frac{d^2}{df^2}h)|\nabla{f}|^2+(\frac{d}{df}h)\Delta{f}$

$div(g(x)\vec{F})=\partial_i(g(x)F_i)=\partial_i{g(x)}F_i+g(x)\partial_iF_i=\nabla{g(x)}\vec{F}+g(x)div\vec{F}$, and so on.

Using this technique you can easily do any transformations of any complexity in a space of any dimension, not necessarily $R^2$.

Hopefully, this will help.


问:
If f has no local extrema, then it is monotonic
Let $f : \mathbb R \to \mathbb R$ continuous without local extremas. Show that $f$ is monotonic.

Here is my attempt, proving the contraposition :

If $f$ is not monotonic then $\exists a < b < c \text{ such that }  f(a) < f(b) \text{ and } f(b) > f(c)$ (or reverse sense).

Then I do not know what is the next right thing to say : maybe "since $f$ is continuous it admits a local extrema."

What do you think is the right thing to say to finish this proof ? Or maybe there is a direct proof you can give me ?

答:
This is more or less a direct application of the extreme value theorem :

First notice that constant functions have local extrema everywhere, so $f$ cannot be constant, nor locally constant for that matter.

Now suppose there exists $(a,b)\in\mathbb R^2$ such that $f(a)=f(b)$, since $f$ is not locally constant, there also exists $c\in(a,b)$ such that $f(a)\neq f(c)$.

By the extreme value theorem for $f$ continuous on interval $[a,b]$, $f$ reaches a min and  a max on $[a,b]$, but since $f(c)$ is distinct from $f(a),f(b)$ then these cannot be both $a$ and $b$, so either a min or a max exists on an interior point of $(a,b)$.

This contradicts the hypothesis that $f$ has no local extremum, therefore $f$ must be injective (i.e. refutation of $\exists\  f(a)=f(b)$).

We conclude invoking the equivalence between strictly monotonic functions and continuous injective functions.


问:
Extension of Field by two roots

Let $k \subseteq F$ be a extension, and let $a, b \in k$ be elements which are not squares, but such that $\sqrt{a}, \sqrt{b} \in F$, and are well defined and whatnot. Prove that, if the characteristic of $k$ is not $2$, then $k(\sqrt{a}, \sqrt{b}) = k(\sqrt{a} + \sqrt{b})$. Furthermore, show that the extension has degree 2 or 4 depending on if $ab$ is, or is not, a square in $k$, respectively.

Most of this question I have solved. I have shown the extensions to be equal, and that if $ab$ is a square, then the degree of the extension is $2$. The only part remaining is to show that the degree is $4$ if $ab$ is not a square.

I have made some progress in this direction. I have bounded the degree of the extension by $4$, both abstractly and by finding a monic degree $4$ polynomial with $\sqrt{a} + \sqrt{b}$ as a root. However, none of this is quite enough to show that the degree actually reaches $4$. There's a variety of options I can see, including showing that $\sqrt{ab} \notin k(\sqrt{a})$and showing that the polynomial I found is irreducible, but I can't see how to actually do that. Hints and solutions are greatly appreciated.

答:
Consider $k(\sqrt{ab})\subseteq k(\sqrt{a},\sqrt{b})$. Suppose $ab$ is not a square in $k$. Then either $k(\sqrt{ab})=k(\sqrt{a},\sqrt{b})$ (and the degree is $2$) or the degree is $4$.

Can we have equality? Suppose so. Then
$$
\sqrt{a}\in k(\sqrt{ab})=k(\sqrt{a},\sqrt{b})
$$
so there exist $p,q\in k$ such that
$$
\sqrt{a}=p+q\sqrt{ab}
$$
and so $2pq\sqrt{ab}=a-p^2-abq^2\in k$. This implies either $p=0$ or $q=0$.

Can you finish?

Title: Extension of Field by two roots


问:
Given $2$ randomly chosen integers $x,y$ what is $P(k=gcd(x,y))$?
Given $2$ randomly chosen integers $x,y$ what is the probability that a integer $k$ is the greatest common divisor of $x$ and $y$?

I know that the probability of $x,y$ being coprime is $\frac{1}{\zeta(2)} = \frac{6}{\pi^2}$ this could be a starting point. I think one could try to analyse $P(x/k $  and  $ y/k$ coprime $| k$ divides $x$ and $y)$ But then the distribution of $x$ and $y$ is no longer uniform. Maybe there is some helping hand here?

答:
What we will be computing is technically not a probability.  The most one can say is that it is a limit of probabilities.

Let $k$ be a fixed positive integer.  For any positive integer $N$, let $f_k(N)$ be the number of ordered pairs $(x,y)$ such that $1 \le x  \le N$, $1 \le y \le N$, and $\gcd(x,y)=k$.  We will (with the help of a major result that we only quote) show that
$$\lim_{N\to\infty} \frac{f_k(N)}{N^2}=\frac{6}{k^2\pi^2}$$

The proof is easy, but for the sake of completeness we do all the details.
Let $r(N)$ be the remainder when $N$ is divided by $k$.  Then $N=kM(N)+r(N)$ for some quotient $M(N)$.  For simplicity, we write $r$ for $r(N)$, and $M$ for $M(N)$, without forgetting their dependence on $N$.

Note that $f_k(N)=f_k(kM)$.  This is because if in the ordered pair $(x,y)$, one or both of $x$ and $y$ is in the interval $[kM+1, N]$, then that $x$ (or $y$) is not divisible by $k$.

Note also that if each of $x$, $y$ is in the interval $[1,kM]$, then $\gcd(x,y)=k$ if and only if $x=ku$, $y=kv$, where $\gcd(u,v)=1$.

It follows that
$$f_k(N)=f_k(kM)=f_1(M)$$
In less fancy language, the number of ordered pairs counted by $f_k(N)$ is the same as the number of ordered pairs $(u,v)$ such that $1 \le u\le M$, $1 \le v \le M$, and $\gcd(u,v)=1$.

We conclude that
$$\frac{f_k(N)}{N^2}=\frac{f_1(M)}{N^2}=\frac{f_1(M)}{(kM+r)^2}$$

A little manipulation shows that the quantity on the right is equal to
$$\frac{f_1(M)}{M^2}\left(\frac{1}{k^2}\right)\left(\frac{1}{1+\frac{2r}{kM}+\frac{r^2}{k^2M^2}}\right)$$

Now let $N \to \infty$.  Then $M=M(N)\to\infty$.  By a standard result about the number of ordered pairs of relatively prime integers up to $M$,
 $$\lim_{M\to\infty}\frac{f_1(M)}{M^2}=\frac{6}{\pi^2}$$

Note that as $M\to\infty$,
 $$1 +\frac{2r}{kM}+\frac{r^2}{k^2M^2} \to 1$$

and the result follows.

ContentLicense: CC BY-SA 3.0


问:
Hard Probability Inequality
I am new in this forum and I am happy to find it, because it seems a very precious place for asking questions.
My question is  about some probability inequality. I formulate this as following.
Let $(X_k)_{1\leq k \leq K}$, with $K$ a nonzero integer, be a discrete-random variables live in   $\{2, ..., n\}$ where $n \geq 2$ . Then, if  $(n_k)_{0\leq k \leq K}$, with $n_0 =1$, is a sequence of integers in  $\{1, ..., n\}$, one can prove that
$$ \mathbb{P}\big( \exists k\in \{1, ...., K\};\, X_k \leq n_{k-1}\big) \leq \sum_{k=1}^K 2^{k-1}\mathbb{P}\big(\max\{ 1\leq \ell \leq K; X_\ell \leq n_{\ell-1}\} = k\big).
$$

Thank you very much for your sugggestions,

Emera

答:
Do I overlook something? Can't we just argue that $\def\P{\mathbb P}$
$$ \P(\exists k; X_k \le n_{k-1}) = \sum_{k=1}^K \P(\max_{X_\ell \le n_{\ell -1}} \ell = k) $$
as if there is a $k$, there is exactly one maximal such. Now as $1 \le 2^{k-1}$ for $1 \le k \le K$ we are done.</p>

Title: Hard Probability Inequality


问:
The restriction of a quotient map to a saturated closed set is a quotient map
There are several similar questions on math.SE, but I couldn't find a resolution of this specific case. I apologize if it turns out this is duplicate.

I'm trying to prove the following property concerning quotient maps: the restriction of a quotient map to a saturated open or closed subset is a quotient map.

Here is my attempt for the case where the restricted domain is open.

Let $X,Y$ be topological spaces, $A$ be a saturated open subset of $X$, and $p:X \to Y$ be a quotient map. Define $q:A \to p(A)$ by $q = p|_A$. We prove that $q$ maps saturated open subsets to open subsets. Let $U \subset A$ be saturated w.r.t. $q$ and open in $A$. We prove it's saturated w.r.t. $p$ and open in $X$. If it weren't saturated, there would be some point in $p^{-1}(p(U))$ which does not lie in $U$. It must be contained in either $A$ or the complement of $A$ in $X$. It cannot lie in $A$, because $U$ is saturated w.r.t. $q$, which agrees with $p$ on $A$. And it cannot lie outside of $A$, because $A$ itself is saturated w.r.t. $p$, and our point is in $U \subset A$. So it must be in $U$. And $U$ is open by definition of the subspace topology, since $A$ is open. So $p$ takes $U$ to an open set in $Y$, and by definition of the subspace topology, $p(U) \cap p(A) = p(U)$ is open in $p(A)$. So $q$ is a quotient map.

But this finagling with the definition of the subspace topology doesn't work as nicely when $A$ is closed instead. Can anyone give me some pointers?

And my other question is: can anyone give a good counterexample for when $A$ is saturated but neither open nor closed?

答:
I figured it out with Daniel Fisher's hint in the comments. A quotient map sends saturated closed subsets to closed subsets. But then to use the same strategy, we want to show that a saturated (w.r.t. ) closed subset $U\subset A$ is saturated (w.r.t. ) and closed in , where  is a saturated (w.r.t. ) closed subset of . We can get that  is saturated w.r.t. $p$ by the same argument as above. We can see that it is closed in X via: Is it true that a subset that is closed in a closed subspace of a topological space is closed in the whole space?

Then $p$ sends $U$ to a closed subset $p(U) \subset p(A)$, since $p$ is quotient. We know $p(A)$ is closed in $Y$, so $p(U)$ is the intersection of a closed subset in $X$ (itself) with $p(A)$, hence closed in $p(A)$.


问:
Infinite subset of $\mathbb{N}$ is countable. Proof by math induction.
Proposition: Infinite subset of $\mathbb{N}$ is countable.

Proof: Suppose that $E$ is infinite subset of $\mathbb{N}$ and our goal to show that $E$ is countable, i.e. there is $f:\mathbb{N}\to E$ which is bijection.

Since $E\neq \varnothing$ then $\exists! e_1:=\min E$, then $E \backslash \{e_1\}\neq \varnothing$ then $\exists! e_2:=\min E \backslash \{e_1\}$. Also $\exists! e_3:=\min E \backslash \{e_1,e_2\}$ and we can continue this process by induction. The we define function $f:\mathbb{N}\to E$ in the following way: $f(n)=e_n$ for each $n\in \mathbb{N}$.

Intuitively I understand this very well. But how induction works here in a rigorous way?

Principle of mathematical induction is the following: If $S\subset \mathbb{N}$ with $1\in S$ and for each $n\in S$ we have $n+1\in S$ then it follows that $S=\mathbb{N}$.

How to write this set $S$ in our case in a rigorous way?

Would be very thankful for help!

答:
At the $n$-th iteration of the process you described, you essentially define an injection $f_n: [n] \to E$, where $[n] := \{1,2,\ldots,n\}$.

$f_{n+1}$ is defined as $f_n \cup (n+1, \min(E \setminus f_n([n]))$. In other words
$$f_{n+1}(i) := \begin{cases}
f_n(i) &, \;\; i \in \{1,\ldots,n\}\\
\min(E \setminus f_n([n]) &, \;\; i= n+1
\end{cases}.$$
The existence of these $f_n$ can be proven using induction. Moreover, these $f_n$ satisfy some additional properties that allow you to define $f: \mathbb N \to E$ as $\bigcup_{n \in \mathbb N} f_n$ and show it is bijective. You want to include those properties in your induction hypothesis as well.

So, given $n \in \mathbb N$ consider the property $$P(n)= \text{there are } f_1,f_2,\ldots,f_n \text{ such that [insert desired properties here]}$$ and the set $S = \{n \in \mathbb N \mid P(n) \text{ is true} \}$.
You can show by induction that $S = \mathbb N$, hence proving the existence of $f_1,f_2,\dots,$ with the desired properties.

Addendum

The desired properties are that $f_k$ is an injection $[k] \to E$, $f_k \subseteq f_{k+1}$ (this ensures that $f$ is well defined), and I think you'll need something like $\max(f_k([k])) < \min(E \setminus f_k([k]))$ to show that $f$ is surjective.


问:
In each cell of a $7\times 7$ grid is a number. The sum of all the numbers in each $2\times 2$ square and each $3\times 3$ square is zero.
In each cell of a $7\times 7$ grid is a number. The sum of all the numbers in each $2\times 2$ square and each $3\times 3$ square is zero. Prove that the sum of the numbers in the 24 bordering cells is $0$ too.

答:
You can split the $24$ bordering cells into 4 groups of $6\times 1$ rectangles
and then represent each rectangle as a set difference between an union of two $3\times 3$ squares and an union of three $2\times 2$ squares.

Sometimes a picture worth a thousand words....


问:
A question about the oriented angle of two lines
We know that if $d_1:a_1x+b_1y+c_1=0$ and $d_2:a_2x+b_2y+c_2=0$ then the angle between $d_1$ and $d_2$ is the angle between their normal vectors $\vec{N_1}=(a_1,b_1)$ and $\vec{N_2}=(a_2,b_2)$ and we can compute this angle by using the definition of the scalar product. But is there any way to find the oriented angle between $d_2$ and $d_1$? We are going to get that $\cos \alpha=\text{something}$ and the question is how to tell whether $\alpha$ is obtuse or not.

答: $\alpha$ is obtuse if $\alpha \gt 90 ° $. However, $cos(x)$, is negative for $x \gt 90 ° $. Therefore, $cos(\alpha) $ is negative if $\alpha$ is obtuse.


问:
Evaluate $\int\frac{x^4}{(x^2-1)^3}dx$
I tried:$$I=\int\frac{x^4}{(x^2-1)^3}dx=\int\frac{(x^2-1)(x^2+1)+1}{(x^2-1)^3}dx=\int\frac{x^2+1}{(x^2-1)^2}+\frac{1}{(x^2-1)^3}dx$$
For first fraction we can write it as $\frac{1}{x^2-1}+\frac{2}{(x^2-1)^2}$. therefor we have:

$$I=\frac12\ln\left|\frac{x-1}{x+1}\right|+\int\frac{2}{(x^2-1)^2}dx+\int\frac{1}{(x^2-1)^3}dx$$

I don't know how to evaluate remaining integrals.

答:
If the numerator is $x^{2r},$

$$x^{2r}=(1+x^2-1)^r=?$$

Had the numerator been $x^{2r+1},$

$x^{2r+1}=x(1+x^2-1)^r$ first substitute $x^2-1=y$

Now for $S=\dfrac{d(x/(x^2-1)^n)}{dx},$

$S=1/(x^2-1)^n-2(n+1)x^2(x^2-1)^{n+1}$

Write $x^2=x^2-1+1$ and integrate both sides to find

$$\dfrac x{(x^2-1)^n}=I(n)-2(n+1)(I (n) - I(n+1))$$ where $I(m)=\displaystyle\int\dfrac{dx}{(x^2-1)^m}$

Can you identify the values of  $n$ here?

Title: Evaluate $\int\frac{x^4}{(x^2-1)^3}dx$


问:
$7$ people sitting at a round table with $2$ next to each other
My question is:

$7$ people are sitting at a round table. $2$ of them must always be sitting next to each other. How many possible ways are there?

I thought it would be $(7-1)! \cdot 2!$ for the $2$ people but apparently the answer is $(6! \cdot 2!)/6 = 5! \cdot 2$.
Can someone explain why we divide by $6$ as well?

答: Looking at it another way, arrange the $5$ people that don't care where they sit, but leave two spaces open somewhere at the table next to each other. Then decide whether Buddy A sits to the left of Buddy B, or his right.


问:
How far would Rebecca walk if she walks 100 feet in a straight line and turns $165^{\circ}$ clockwise, repeating this pattern until...
If Rebecca walks 100 feet in a straight line. She turns $165^{\circ}$ clockwise and walks another 100 feet, and turns $165^{\circ}$ clockwise again. If she continues this pattern until she reaches the point where she started. How far did she walk?

I can't see how she would ever get back to her origin.
She turns $\frac{11}{24}$ of full rotation each time, this will not form a regular polygon.
It only makes sense she gets back to start if she turns for example $20^{\circ}$ - convenient factor of $360$, here I can use the fact the exterior angles of polygon add up $360^{\circ}$ to find the number of angles formed and derive an answer.

答:
Here is an approach using the representation of points in the plane as complex numbers.

For convenience, let's pick units of measurement so that $1$ unit = $100$ feet, and let Rebecca's location at step $n$ be $z_n$ for $n = 0, 1, 2, \dots$, with initial location $z_0 = 0$ and position after the first step $z_1 = 1$.  Then her position after moving $1$ unit of distance after a turn of $-11/12 \pi$ radians is $z_2 = 1 + a$, where we define $a = \exp(-(11/12) \pi i)$.  Her position after another step is $z_3 = 1 + a + a^2$, and her position at the $n$th step is
$$z_n = 1 + a + a^2 + \dots a^{n-1}$$
Applying the formula for the sum of a geometric series,
$$z_n = \frac{1 - a^n}{1-a}$$
So Rebecca gets back to her initial location when $a^n = 1$, i.e.
$$\exp \left(- \frac{11}{12} n \pi i \right) = 1$$
Since $\exp(z) = 1$ exactly when $z = 2m \pi i$ for some integer $m$, we know
$$\frac{11}{12} n \pi i = 2 m \pi i$$
so we need the smallest positive integer $n$ such that
$$11n = 24 m$$
for some positive integer $m$.  I'll leave the final step to the reader.


问:
Quick Volume with Cross Sections Question
So I was given the following prompt:

"Let $R$ be the region in the first quadrant bounded by the graph of $y=x^2$, the line $x=4$, and the $x$-axis. $R$ is the base of a solid whose cross sections perpendicular to the $x$-axis are equilateral triangles. What is the volume of the solid?"

I understand how to find the area of cross sections that might be squares or rectangles, but I haven't worked with equilateral triangles before. I also understand that the formula for the area of an equilateral triangle is: $\frac {a^2\sqrt3}{4}$, but I'm a bit lost over how I'd find the area or even the sides of the triangle in this case. Any help would be appreciated!

答:
Please note that the equilateral triangle cross-section is parallel to $YZ$ plane (perpendicular to $x-$axis). The base of the equilateral triangle is $x^2$ as it is between $y = 0$ and $y = x^2$.

So sides of equilateral triangle $ = x^2$ and height $ = \frac{\sqrt3}{2} x^2$

For a given value of $z$, $y$ will vary between $\frac{z}{\sqrt3}$ and $x^2-\frac{z}{\sqrt3}$.

The limits of $y$ come from the fact that at any given height ($z$) in the triangle if we drop perpendicular from two edges to the base, the distance from both vertices of the base to the foot of the perpendicular will be $ = z \cot 60^0= \frac{z}{\sqrt3} \ $.

So the integral to find volume should be

$\displaystyle \int_{0}^{4} \int_{0}^{\frac{\sqrt3}{2} x^2} \int_{\frac{z}{\sqrt3}}^{x^2- \frac{z}{\sqrt3}} dy \ dz \ dx$

If you integrate wrt $dz$ first, you will have to split the integral into two hence the choice of the order.


问:
Atiyah-Macdonald Exercise 2.6
I'm working on Exercise 2.6 from Atiyah-Macdonald's "Introduction to Commutative Algebra" and I had a question about the definition of bilinear maps.

Exercise 2.6: For any $A$-module $M$, let $M[x]$ denote the set of all polynomials in $x$ with coefficients in $M$, that is to say expressions of the form
$$ m_0 + m_1x + \dots + m_r x^r \qquad (m_i\in M)$$
Defining the product of an element of $A[x]$ and an element of $M[x]$ in the obvious way, show that $M[x]$ is an $A[x]$-module. Show that $M[x] \cong A[x]\otimes_A M$ (as $A[x]$-modules).

So I begin by trying to find explicit $A[x]$-linear maps from $M[x]\to A[x]\otimes_A M$ and then from $A[x]\otimes_A M \to M[x]$. I got a little confused about the definition of a bilinear map when trying to define the second map.

What is an $A[x]$-bilinear map on $A[x]\otimes_A M$? In particular, let $\phi: A[x]\times M \to M[x]$ be a map. Suppose I fix a first coordinate $f\in A[x]$, then what does it mean for $\phi(f,-) : M\to M[x]$ to be $A[x]$-linear? A priori, there is no defined multiplication of $A[x]$ onto $M$ right?

答:
First note that you want an $A[x]$-linear map from $A[x] \otimes_A M$ into $M[x]$, not an $A[x]$-bilinear map. This is because in the category of modules, the morphisms are the linear maps. Now in order to exploit the universal property of the tensor product in order to construct such a map, it is sufficient to start with a map on $A[x] \times M$ which is merely $A$-bilinear, because $A$ is the ring we take the tensor product over.

Such a map is given by
$$
(a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0, m) \mapsto a_n m x^n + a_{n-1} m x^{n-1} + \cdots + a_1 m x + a_0 m.
$$

Thus, the universal property of the tensor product implies that we get a map $A[x] \otimes_A M \to M[x]$, which a priori is only $A$-linear. But this map is in fact $A[x]$-linear because it's the inverse of a map which is $A[x]$-linear. Remember that the inverse of a module isomorphism is again a module isomorphism.

This $A[x]$-linear map is given by
$$
M[x] \to A[x] \otimes_A M, m_n x^n + \cdots + m_0 \mapsto x^n \otimes m_n + \cdots + 1 \otimes m_0.
$$

In order to show that these two maps are inverse to each other, it may be convenient to note that by linearity, it's sufficient to check this on convenient generating sets of each module.

Title: Atiyah-Macdonald Exercise 2.6


问:
How to show isomorphism of permutation groups?
Let $M = \{m_1,. . . , m_k\} ⊂ \{1, 2, 3,. . . , n\}$ and
$H = \{\sigma ∈ S_n | \forall m \in M, \; \sigma(m) \in M\}$.

Show that $H \subset S_n$ is a subgroup.

I tried the following:
$m \in M$ and $\sigma \in H$ implies that $\sigma(m) \in M$. If we take $\pi \in H$, then $\pi \sigma = \pi(\sigma(m)) \in M$ which in turn implies that $\pi\sigma \in H$(here the product is defined as a composition).

Update: I have shown that $H$ is nonempty, finite and closed under the operation of multiplication, hence I have shown that it is indeed a subgroup.

(Nonempty, because it contains at least the identity permutation, and finite because it can only have up to $n!$ elements which is a finite number.)

Now, to show that $H$ is isomorphic to $S_k × S_{n−k}$, I have to define a bijective homomorphism. I don't quite understand what this means.
I know that permutations are bijective, but how would this work for $S_k × S_{n−k}$?
What does $S_k × S_{n−k}$ actually mean?

答:
In Gallian's "Contemporary Abstract Algebra (Eighth Edition)", there is the following theorem.

Theorem [. . .] (Finite Subgroup Test)
Let $H$ be a nonempty finite subset of a group $G$. If $H$ is closed under the operation of $G$, then $H$ is a subgroup of $G$.

The proof is fairly straightforward.

So you are done.


问:
Convergence in $L^p$ and $L^q$ - multiplication
We have: $X_n \rightarrow X$ in $L^p$ and $Y_n \rightarrow Y$ in $L^q$. Moreover $p,q>1$ are such that $\frac{1}{p} + \frac{1}{q} =1$. Prove that $X_nY_n \rightarrow XY$ in $L^1$. Please, can you help?</p>

答:
Just use Hölder. We have $\def\norm#1{\left\|#1\right\|}$
\begin{align*}
  \norm{XY-X_nY_n}_1 &\le \norm{XY - XY_n}_1 + \norm{XY_n - X_nY_n}_1\\
      &\le \norm X_p\norm{Y-Y_n}_q + \norm{X-X_n}_p\norm{Y_n}_q
\end{align*}
Now $(Y_n)$ is bounded in $L_q$, hence both terms tend to zero. So we are done. </p>


问:
Estimate partial sum of binomial coefficient
$C^{m}_{n}=\frac{n!}{m!(n-m)!}$ are the binomial coefficients. Now for a fix $0<\alpha<1$, consider for any natural number $N$ the following sequence:
$$ T_N=\sum_{[\alpha N]\leq i\leq N}C^i_{N}$$
Could you help me with estimating the limit of $T_N/2^N$ when $N\to \infty$

答:
$$T(n)=\sum _{i=\lfloor n \alpha \rfloor }^n \binom{n}{i}=\frac{\Gamma (n+1) \, _2F_1(1,n (\alpha -1);n \alpha +1;-1)}{\Gamma (n-\alpha  n+1) \Gamma (n \alpha +1)}$$

From numerical results I got that
$$\underset{n\to \infty }{\text{lim}}\frac{T(n)}{2^n}=\begin{cases}
1; & 0<\alpha<\frac12\\
\frac12;& \alpha=\frac12\\
0; &\frac12<\alpha<1\\
\end{cases}$$


问:
Proof that $H \cap \operatorname{cl}(K) \subseteq \operatorname{cl}(H \cap K)$ for open $H$
This appears in W.A. Sutherland's "Introduction to Metric and Topological Spaces" (1st ed. 1975), chapter 3, exercise 3.9: 27 (iii).

"If $H$ is open in $T$ and $K \subseteq T$ then:
$$H \cap \operatorname{cl}(K) \subseteq \operatorname{cl}(H \cap K)$$"

(where $\operatorname{cl}$ denotes closure and $T$ is a topological space.)

It's obvious that $H \cap \operatorname{cl}(K) \subseteq \operatorname{cl}(H) \cap \operatorname{cl}(K)$ but then it does not follow that $\operatorname{cl}(H) \cap \operatorname{cl}(K) = \operatorname{cl}(H \cap K)$.

(In fact we have $\operatorname{cl}(H \cap K) \subseteq \operatorname{cl}(H) \cap \operatorname{cl}(K)$.)

So that line of approach does not look very fruitful.

So I drew it up as a Venn diagram to get inspiration, and I got:

$$H \cap \operatorname{cl}(K) = H \cap (K \cup \delta K)$$

where $\delta K$ is the boundary of $K$.

This leads us to:

$$H \cap \operatorname{cl}(K) = (H \cap K) \cup (H \cap \delta K) \subseteq (H \cap K) \cup (H \cap \delta K) \cup (\delta H \cap K)$$

and "all that remains to be done" is prove that $(H \cap \delta K) \cup (\delta H \cap K) \subseteq \delta (H \cap K)$ (because $(H \cap \delta K) \cup (\delta (H \cap K)) = \operatorname{cl}(H \cap K)$)

but it's starting to get messy and I'm unsure of my ground.

Besides, this does not make any use of the fact that $H$ is open.

I'm pretty sure there's an insight needed here which I have not been able to come up with.

答:
Suppose that $x \in H \cap \operatorname{cl}(K)$.  Then since $x \in \operatorname{cl}(K)$, there exists a net $y : I \to X$ with $y_i \in K$ for every $i \in I$ such that the net $y$ has limit $x$.  Now, since the limit point is in the open set $H$, there exists $i_0 \in I$ such that $x(i) \in H$ whenever $i_0 \le i$.  Therefore, the restriction of $y$ to $\{ i \in I \mid i_0 \le i \}$ is a net with limit $x$ where every element of the net is in $H \cap K$.  This implies that $x \in \operatorname{cl}(H \cap K)$.

If you are not familiar with nets, then you can think of the case where the topological space $X$ is first countable, so that closures are determined by limits of sequences, and then $y : \mathbb{N} \to X$ can be taken to be a sequence instead of a more general net.  The proof then says that for some $i_0$, the sequence $y_{i_0}, y_{i_0+1}, y_{i_0+2}, \ldots$ lies entirely within $H \cap K$ and still has limit $x$.


问:
Combinatorical permutation-cycles problem
How many permutations $\pi$ $\in$ $Sn$ ($n>2$) where $1$ and $2$ belong to the same cycle?

Here is what I have so far:

First I put $1$ and $2$ in the same cycle and make $j$ elements from the other $(n-2)$ join them in the same cycle: ${n-2 \choose j}$ with $0<j<n-2$

Then I must put the $(n-2-j)$ elements in other cycles. For that I use the Stirling numbers (which garantees that at least 1 element is put in a cycle) in  to distribuite those elements in the $k$ cycles, with $(1<k<j)$:

$S(n-2-j,k)$

My final answer is

$\sum_{j=0}^{\ n-2} {{n-2 \choose j} . \sum_ {k=0}^{\ j} {S(j,k)}}$

Is there something wrong with my resolution or a simpler way to solve it?

答:
There is a purely combinatorial approach as well. Let $\pi\in S_n$, and let $$\pi=\sigma_1\sigma_2\ldots\sigma_m\tag{1}$$ be its complete representation as a product of cycles, including the $1$-cycles for its fixed points. We can always take $(1)$ to be in the following standard form:

each cycle is written with its smallest element first, and

the cycles are arranged in descending order of their smallest elements.

Since $1$ is necessarily the smallest element of the cycle containing it, it must be the first element of $\sigma_m$.

Let $T$ be the set of $\pi\in S_n$ such that $1$ and $2$ are in different cycles of $\pi$. If

$$\pi=\sigma_1\sigma_2\ldots\sigma_m\in T\,,$$

then $2$ is the first element of $\sigma_{m-1}$. Let $\sigma'$ be the concatenation $\sigma_m\sigma_{m-1}$. That is, if $\sigma_{m-1}=(2\ldots s_k)$ and $\sigma_m=(1\ldots t_\ell)$, then $\sigma'=(1\ldots t_\ell 2\ldots s_k)$. (Note that this is not the product of $\sigma_m$ and $\sigma_{m-1}$.) Let $\hat\pi=\sigma_1\ldots\sigma_{m-2}\sigma$. Then $\hat\sigma\in S_n\setminus T$ and is written in the standard form $(1)$.

This transformation is invertible. If $\pi=\sigma_1\ldots\sigma_m\in S_n\setminus T$ is in standard form, then $\sigma_m$ has the form $(1\ldots t_\ell 2\ldots s_k)$. Let $\sigma_{m+1}'=(1\ldots t_\ell)$ and $\sigma_m'=(2\ldots s_k)$; then $\pi'=\sigma_1\ldots\sigma_{m-1}\sigma_m'\sigma_{m+1}'\in T$ is in the standard form $(1)$, and $\widehat{\pi'}=\pi$. Thus,

$$|S_n\setminus T|=|T|=\frac12|S_n|=\frac{n!}2\,.$$


问:
GCSE: The volume of frustum
Looking please for any help! The density of marble used for the pedestal shown in the figure below is $2750\text{ kg/m}^3$
. A crane can lift a maximum of 2 tons. Can the crane raise this pedestal?

My attempt was the following: To find the volume of the frustum: I use the formula of the trapezoid surface:  $$\frac{(1+0.4)\cdot 1.5}{2}=1.05\text{ m}^2$$
And then $1.05\times 1.5=1.575\text{ m}^3\longrightarrow$ the mass is $1.575\times 2750=4331\text{ kg}=4.3\text{ tons}$

And my answer was the crane can’t raise the pedestal.

I don’t know if this is the right answer! All help/solutions appreciated. Thanks

答:
I will, as an A Level student, try to show you exactly what you'd need to understand to get full marks in a similar GCSE question. Firstly, before I begin, you need to realise that you have not actually found the volume of the solid.

You have found the cross sectional area of this solid when looked at perpendicular to its base. Indeed, this area is equal to $1.05\text{m}$. But this is not relevant to the question. We are looking for volume, not area or surface area.

Now to begin:

Throughout this solution I will assume that the bases of this truncated pyramid are square, which would be a more normal GCSE question.

Think of the solid as a larger pyramid with its top, a smaller pyramid, truncated. I will avoid trigonometry in case you have not covered it yet.

Let the height of the larger pyramid (which currently has its top chopped off) be $h$. Then the height of the smaller pyramid (which is chopped off) is $h-1.5$. Using similar triangles, we see that

$$\frac{h}{0.5}=\frac{h-1.5}{0.2}\implies0.2h=0.5h-0.75\implies h=\frac{0.75}{0.3}=2.5$$
So the height of the smaller pyramid is $2.5-1.5=1$. Now, you should know that the volume of a square based pyramid with base length $a$ and perpendicular height $h$ is
$$\frac{1}{3}a^2h$$
Hence, the volume, $V_1$, of the smaller pyramid is $\frac{1}{3}(0.4^2)(1)=\frac{4}{75}$. Similarly, the volume of the larger pyramid, $V_2$, is $\frac{1}{3}(1^2)(2.5)=\frac{5}{6}$. Hence, the volume of the solid we are dealing with is
$$ V_2-V_1=\frac{5}{6}-\frac{4}{75}=0.78\text{m}^3$$
Using $$\text{density}=\frac{\text{mass}}{\text{volume}}$$
we can rearrange to find the required mass:
$$\text{mass}=\text{density}\times\text{volume}=0.78\times2750=2145\text{kg}$$
As $2145\text{kg}$ is larger than $2$ tonnes, we conclude that the crane cannot lift the pedestal.

I hope I have helped you. If you have any further queries, please don't hesitate to ask. I wish you luck in whatever assessments your school give you in place of GCSE's.

Please feel free to email me at mathshelper4you@gmail.com if you have any other problems. I will be glad to help you.

Title: GCSE: The volume of frustum


答:
Suppose the two people who sit next to each other are Alexandra and Benjamin.  Seat Alexandra.  It does not matter where we seat her since only the relative order of the people matters in a circular arrangement.  There are two ways we could seat Benjamin, to Alexandra's left or her right.  The remaining five people can be seated in the remaining five seats in $5!$ ways as we proceed clockwise around the table from Alexandra. Hence, there are indeed $2!5!$ admissible seating arrangements.

To address your question:  Treat Alexandra and Benjamin as a block.  Then we have six objects to arrange around the round table, the block and the other five people.  In a linear arrangement, the six objects could be arranged in $6!$ ways.  Since only the relative order matters in a circular arrangment, we divide by $6$ since there are six possible starting points for a linear arrangement corresponding to the same circular arrangement.  Hence, there are
$$\frac{6!}{6} = 5!$$
ways to arrange six distinct objects in a circle up to rotation.  Since the objects within the block can be arranged in $2!$ ways, there are
$$\frac{6!}{6} \cdot 2! = 5!2!$$
arrangements of the seven people at a round table if two particular people must be adjacent.

Title: $7$ people sitting at a round table with $2$ next to each other

Body:
My question is:

$7$ people are sitting at a round table. $2$ of them must always be sitting next to each other. How many possible ways are there?

I thought it would be $(7-1)! \cdot 2!$ for the $2$ people but apparently the answer is $(6! \cdot 2!)/6 = 5! \cdot 2$.
Can someone explain why we divide by $6$ as well?


问:
How accurate are most representations of pi?
I understand that  $\pi$  is the ratio of a circle's circumference to it's diameter and it is equal to about 3.14159265359(According to Google) but how accurate is this and most representations of $\pi$?</p>

答:
As you have been said alredy, no decimal representation will be exact, but we can know its error. If we extend those representations to any representation, we will have a perfect acuracy, but it won't be a numeric value. For example, you can represent it, by words, just as you said, by the ration of diameter/circunference, and that's exact.
About more mathematicals representations, there are a lot of them, in the form of infinite series, like the one made by Ramanujan, that converges to the actual value very fast:

$$\frac{1}{\pi}=\frac{2\sqrt2}{9801}\sum_{k=0}^\infty \frac{(4k)!(1103+26390k)}{(k!)^4396^{4k}}$$

Another exact one, by Leibniz:

$$\pi=4\sum_{n=0}^\infty \frac{(-1)^n}{2n+1}$$

Several values of the zeta function, specially with integers, have $\pi$ as the result (with a factor), like $\zeta(2)$ or $\zeta(4)$, which can be calculated by Fourier series.

And lots more, more info here: http://en.wikipedia.org/wiki/Pi


问:
Tangency points to a curve (with implicit differentiation)
I have the curve $x^2+xy+y^2=1$, whose graph is as follows:

I'm trying to find the points where the line tangent to the curve is parallel to the line $y = -x$. There are colleagues of mine who answered the points (0.5; 0.5) and (-0.5; -0.5).
I tried to solve by doing the implicit derivation and equating it to -1, which is the slope of the tangent line:

$$\frac{d}{dx}(x^2+xy+y^2)=\frac{d}{dx}(1)\leftrightarrow 2x+y+x\frac{dy}{dx}+2y\frac{dy}{dx}=0\leftrightarrow \frac{dy}{dx}=\frac{-2x-y}{x+2y}\leftrightarrow \frac{dy}{dx}=-\frac{2x+y}{x+2y}=-1\leftrightarrow \frac{2x+y}{x+2y}=1\leftrightarrow 2x+y=x+2y\leftrightarrow y=x$$

The problem is that I have two variables and an equation. I already thought of replacing $y$ with $x$ in the initial curve, but that doesn't respect the condition of the tangent line. How can I find the values of $x$ and $y$, given that I only have one equation?

答:
All the parallels to $y + x = 0$ have the representation $y+x=m$

Substituting $y = m-x$ into the ellipse equation we have

$$
x^2 + x(m-x)+(x+m)^2 = 1
$$

now solving for $x$

$$
x = \frac 12\left(m\pm \sqrt{4-3m^2}\right)
$$

but at tangency we should have $4-3m^2=0$ so the tangent lines are

$$
y+x=\pm\sqrt{\frac 43}
$$

at

$$
x =\frac 12 \left\{-\sqrt{\frac 43}, \sqrt{\frac 43}\right\}
$$


问:
Dividing both sides by 5, when solving $x = 5(\sin y) + 2$ for $y$?
How do I know which operations to do when solving an equation?
For example,
$$x = 5(\sin y) + 2 \tag1$$
If I wanted to make $y$ a function of $x$,
$$\begin{align}
x &= 5(\sin y) + 2 \tag2 \\[4pt]
x - 2 &= 5(\sin y) \tag3 \\[4pt]
\frac{x - 2}{5} &= \sin y \tag4 \\[4pt]
y &= \sin^{-1}\left(\frac{x - 2}{5}\right) \tag5
\end{align}$$
I believe this is the correct answer, but I was wondering why it isn't this way:

$$\begin{align}
x &= 5(\sin y) + 2 \tag6 \\
\frac{x}{5} &= (\sin y) + 2 \tag7 \\
\frac{x}{5} - 2 &= \sin y \tag8 \\
y &= \sin^{-1}\left(\frac{x}{5} - 2\right) \tag9
\end{align}$$

Why is equation $(9)$ wrong?

答:
@player3236's comment has already correctly pointed out the issue in your calculation: you simply make a mistake from (6) to (7) because you forgot to divide $2$ by $5$ as well.

In general, when you divide a nonzero constant into both sides of a equation, you need to do it for each term. To see a simpler example, if you have
$$
x=5y+2
$$
then, after dividing by $5$ on both sides, you get
$$
\frac15x=y+\frac25
$$
instead of
$$
\frac15x=y+2
$$

To fix your problem, check the following correct lines:

$$\begin{align}
x &= 5(\sin y) + 2 \tag6 \\
\frac{x}{5} &= (\sin y) + \color{red}{\frac{2}{5}} \tag{7'} \\
\frac{x}{5} - \color{red}{\frac{2}{5}} &= \sin y \tag{8'} \\
y &= \sin^{-1}\left(\frac{x}{5} - \color{red}{\frac{2}{5}}\right)
= \sin^{-1}\left(\color{red}{\frac{x-2}{5}}\right)
\tag{9'}
\end{align}$$


答: For visualization, Rebecca will walk the following path:

Title: How far would Rebecca walk if she walks 100 feet in a straight line and turns $165^{\circ}$ clockwise, repeating this pattern until...

Body:
If Rebecca walks 100 feet in a straight line. She turns $165^{\circ}$ clockwise and walks another 100 feet, and turns $165^{\circ}$ clockwise again. If she continues this pattern until she reaches the point where she started. How far did she walk?

I can't see how she would ever get back to her origin.
She turns $\frac{11}{24}$ of full rotation each time, this will not form a regular polygon.
It only makes sense she gets back to start if she turns for example $20^{\circ}$ - convenient factor of $360$, here I can use the fact the exterior angles of polygon add up $360^{\circ}$ to find the number of angles formed and derive an answer.


问:
Infinite subset of $\mathbb{N}$ is countable. Proof by math induction.
Proposition: Infinite subset of $\mathbb{N}$ is countable.

Proof: Suppose that $E$ is infinite subset of $\mathbb{N}$ and our goal to show that $E$ is countable, i.e. there is $f:\mathbb{N}\to E$ which is bijection.

Since $E\neq \varnothing$ then $\exists! e_1:=\min E$, then $E \backslash \{e_1\}\neq \varnothing$ then $\exists! e_2:=\min E \backslash \{e_1\}$. Also $\exists! e_3:=\min E \backslash \{e_1,e_2\}$ and we can continue this process by induction. The we define function $f:\mathbb{N}\to E$ in the following way: $f(n)=e_n$ for each $n\in \mathbb{N}$.

Intuitively I understand this very well. But how induction works here in a rigorous way?

Principle of mathematical induction is the following: If $S\subset \mathbb{N}$ with $1\in S$ and for each $n\in S$ we have $n+1\in S$ then it follows that $S=\mathbb{N}$.

How to write this set $S$ in our case in a rigorous way?

Would be very thankful for help!

答:
==== New Answer ====

First thing we need to be careful, exactly what where are proving by induction..

And what we are proving (by strong) induction is that the function described is well defined and true prove that by induction is to prove the statement:

$P(k):$ If $E_k= \{e_1,......,e_{k-1}\}$ is finite subset of of cardinality $k-1$ then $e_k := \min E\setminus E_k$ exists. And so $f(k) := e_k$ is well defined.

And that is done in a routine way:

Base case, $P(1)$:  $E_1 =\emptyset \subset E$.  $E\setminus \emptyset = E$ which is a nonempty subset of $E$.  So by well-ordering principal $\min E$ exists and so $f(1)=e_1 = \min E$ is a well-defined assignment.

Induction step:  If $E_{k}=\{e_1,....,e_{k-1}\}\subset E$ and $e_k =\min E\setminus E_k$ exists; then $e_k \in E$ and $e_k\not \in E_{k-1}$  and so $E_{k+1}=E_{k-1}\cup \{e_k\}=\{e_1,....,e_{k-1},e_k\}\subset E$. As $\{e_1,....,e_{k-1}\}$ is finite set of cardinality $k-1$ there exists a bijection $\phi_1: \{1,....,k-1\}\to \{e_1,....,e_{k-1}\}$.  so we can define $\phi_2:\{1,....,k-1\} \to \{e_1,....,e_k\}$ five if $n < k; \phi_2(n)= \phi_1(n)$ and $\phi_2(k) =e_k$.  $\phi_2$ is injective because: $\phi_2(k)=e_k\ne \phi_2(n)=e_n$ for all $n\ne k$ as $e_n \in E_k$ but $e_k\not \in E_k$. And for any $i\ne j; i\ne k;j\ne k$ then $\phi_2(i) =\phi_1(i)\ne \phi_2(j) = \phi_2(j)$ because $\phi_1$ is injective.

And $\phi_2$ is surjective as for all $w\in E_{k+1}$ where $w \ne e_k$ we have an $i< k$ where $\phi_1(i)=\phi_2(i) = w$ as $\phi_1$ is surjective.  And for $e_k \in E_{k+1}$ we have $\phi_2(k) = e_k$.

So $\phi_2$ is an injection and $E_{k+1}$ is a finite subset of $E$ with cardinality $k$.

.... Sheesh......

Okay... now $E_{k+1}$ is a  finite subset of $E$ and $E$ is infinite so $E_{k+1} \subsetneq E$.  so $E\setminus E_{k+1} \ne \emptyset$ and $E\setminus E_{k+1}$ is a proper subset of $\mathbb N$ so $\min E\setminus E_{k+1}$ exists and $f(k+1) = e_{k+1} = E\setminus \min E\setminus E_{k+1}$ is a valid assignment.

....

Dont worry.... No-one will ever expect you to go into that much detail....

....

And that proves that the function $f$ is well defined.

Now we just have to prove it is injective and surjective.

Injective is straightforward.  In the above proof $E_k=\{e_1,....., e_{k-1}\}=\{f(n)|n\in \mathbb N; n< k\}$ and $f(k) = e_k = \min E\setminus E_k$ were well defined.  For any $i\ne j; i< j$ we have $f(i)\in \{f(n)|n\in \mathbb N; n< k\}=E_j$ but $f(j) = \min E\setminus E_j \not \in E_j$ so $f(i) \ne f(j)$.

Proving $f$ is surjective is a little tricky and relies an the observation that it must be that $f(k) \ge k$.

To prove that by induction:

Base case:  $f(1) = \min E$.  As $E \subset \mathbb N$, $\min E \ge 1$.

Indcution.  If $f(k) \ge k$ then $\min E\setminus E_k \ge k$ and all $w\in E\setminus E_k$ are $w \ge k$ and if $w \ne e_k = \min E\setminus E_k$ then $w > e_k \ge k$ and $w \ge k+1$.  so $f(k+1) = \min E\setminus E_{k+1} = \min ( E\setminus E_k)\setminus \{e_k\}\ge k+1$.

...(ta-da.....)....

Foo, where were we... oh, yeah, proving surjective.

.... Hey! Watch this!

Let $w \in E$.  Let $U_w = \{n\in N| f(n) \ge w\}$.  As $f(w)\ge w$ we have $w\in U_w$ so $U_w$ is not empty.  So $\min U_w$ exists.  Claim: $f(\min U_w) = w$.

Pf:  Let $\omega = \min U_w$.  Then for all $k < \omega; k \not \in U_w$ so $f(k) < w$.  So $w \not \in E_\omega$ and $w \in E\setminus E_k$.

Now $f(\omega) = \min E\setminus E_k$ and $\omega \in U_w$ so $f(\omega)=\min E\setminus E_k \ge w$ and $w\in E\setminus E_k$.  So $w \ge \min E\setminus E_k = f(\omega)$.  So $w = f(\omega)$.

And that proves surjectivity.

===old answer ====

Yeah... this a matter of saying the right words in order at the right time.

Our Propostion:  $P(k)$ For every $i \le k$ we can map each $i\mapsto e_i$ to a distinct $e_i \in E$.

$P(1):$.  $E \subset \mathbb N$ so $\min E$ exists.  Let $1\mapsto e_1 = \min E$.

Induction step:  Assume $P(k)$. That is for $1,2,3,.....,k$ we have $E_k = \{e_1,e_2,....,e_k\}\subset E$ where the $e_i$ are distinct.

As $E$ is infinite and $E_k$ is finite $E_k \subsetneq E$ and $E\setminus E_k \ne \emptyset$. And as $E\setminus E_k\subset \mathbb N$ we know $\min E\setminus E_k$ exist.  And we know that as $E_k$ and $E\setminus E_k$ are disjoint $\min E\setminus E_k$ is distinct from all the other $e_i$.

So we let $k+1 \mapsto e_{k+1}:= \min E\setminus E_k$.  And thus $\{e_1,....,e_k,e_{k+1}\}$ has been a mapping of all $i \le k+1$ to distinct elements of $E$.

So we have proven $P(k+1)$.

So we conclude that the for any $n \in \mathbb N$, $P(n)$ is true.  And thus we have every $n \in \mathbb N$ will map to an $e_n$. And for all $i < n$ we know $e_i \ne e_n$  And for all $m > n$ we know $P(m)$ is true and $e_m \ne e_n$ so $e_n$ is distinct for all $e_i; i \ne n$.

And that's it.  Let $f(n) = e_n$ and that is an injective map.  Oh.... I guess we have to also prove it is surjective.

Well, that's a different step.


问:
Why is the magnitude of the vectors still not 1 after normalization?
I was reading that the vectors must have magnitude of $1$ to be called normalized; however, none of the problems I solved gave me a vector magnitude of $1$ after normalizing. I also thought that taking the norm of a vector produced the unit vector on the same line with that vector which also equals to $1$. What am I getting wrong here?

Note: I‘ve checked the other answers, but it didn‘t seem to click. Hope you can enlighten me.

答:
Why, all of the vectors are normalized. The magnitude of a vector is given by

$$ \|\textbf{v}\| = \sqrt{v_x^2 + v_y^2 + v_z^2}$$

The definition of a normalized vector is so that $\|\textbf{v}\|$, its magnitude is $1$.

Also, the way to normalize a vector is to scale the vector down by its magnitude. Let $\textbf{v}^\prime$ (bad notation, I don't know how to make the caret symbol in top) be the normalized vector. Then

$$ \textbf{v}^\prime = \frac{\textbf{v}}{\|\textbf{v}\|} $$

Reason why they are normalized:

$$ \sqrt{(1/\sqrt{2})^2 + 0^2 +(1/\sqrt{2})^2} = 1$$

$$ \sqrt{(1/2)^2 + (\sqrt{2}/2)^2 +(1/2)^2} = 1$$

$$ \sqrt{(1/2)^2 + (-\sqrt{2}/2)^2 +(1/2)^2} = 1$$

There is nothing fancy here. You can easily check that it is equal to $1$.


问:
$n$ divides $\phi(a^n -1)$ where $a, n$ are positive integer.

Let $n$ and $a$ be positive integers with $a > 1$. I need to show that $n$ divides $\phi(a^n -1)$.

Here, $\phi$ denotes the Euler totient function.

Could any one give me a hint?

答:
Hint:

$$a^n\equiv 1\pmod {a^n-1}$$
and $$a^d\not\equiv 1\pmod {a^n-1}$$
if $0<d<n$ (why?).


问:
Prove $f(x) = g(x)$ for all $x \in \overline{A}$.
Let $f,g : X \rightarrow Y$ be continuous functions from a topological space $X$ into a Hausdorff space $Y$. Let $A \subset X$ and suppose $f(x)=g(x)$ for all $x \in A$. Prove $f(x) = g(x)$ for all $x \in \overline{A}$.

$\textbf{Definition:}$ $X$ is $\textbf{Hausdorff}$ means for each distinct pair of $x_1,x_2 \in X$, $\exists \, \text{disjoint open} \, U,V \subset X$ with $x_1 \in U$ and $x_2 \in V$ such that $U \cup V = X$.

$\textbf{Definition:}$ A function $f: X \rightarrow Y$ where $X,Y$ are topological spaces is $\textbf{continuous}$ means $\forall \, \text{open} \, V \subset Y$, $f^{-1}(V)$ is open in $X$.

I am unsure how to start this problem off. I wrote the two given definitions to try and get a clearer view on the goal. Any hint on how to begin this would be appreciated.

答: Hint : What do you know about the diagonal of a an hausdorff space ?</p>


答: I would use the limit point definition of continuity:  $f$ is continuous iff $f$ preserves limits of all nets.

Body:
Let $f,g : X \rightarrow Y$ be continuous functions from a topological space $X$ into a Hausdorff space $Y$. Let $A \subset X$ and suppose $f(x)=g(x)$ for all $x \in A$. Prove $f(x) = g(x)$ for all $x \in \overline{A}$.

$\textbf{Definition:}$ $X$ is $\textbf{Hausdorff}$ means for each distinct pair of $x_1,x_2 \in X$, $\exists \, \text{disjoint open} \, U,V \subset X$ with $x_1 \in U$ and $x_2 \in V$ such that $U \cup V = X$.

$\textbf{Definition:}$ A function $f: X \rightarrow Y$ where $X,Y$ are topological spaces is $\textbf{continuous}$ means $\forall \, \text{open} \, V \subset Y$, $f^{-1}(V)$ is open in $X$.

I am unsure how to start this problem off. I wrote the two given definitions to try and get a clearer view on the goal. Any hint on how to begin this would be appreciated.


问:
How to prove that the stereographic projection of the $S^2$ is conformal
I read on Wikipedia that the stereographic projection of the $S^2$ sphere is conformal. However, there was no proof provided.
How can one show it?

答: In an infinitesimal neighbourhood of a point $p$ of $S^2$, the stereographic projection amounts to the reflection at the plane bisecting the angle between the tangent plane to $p$ and the projection plane.


问:
Ambiguity in the center of the locus of $|z-z_o|=r$
Let $z_o$ be a fixed complex number and $z$ be a variable complex number

I've been told that the locus of $z$ in  $|z-z_o|=r$ where $r\gt0$ is a circle centred at $z_o$

I have not been able to understand this . $z-z_o$ is a complex number originating at the origin so shouldn't the locus be centred at the origin ??

答:
Convince yourself by studying the real case $|x-5|=1$ for instance.

Similarly $x-5$ is a real originating at the origin, yet $x=4\text{ or }6$ both points at distance $1$ of center $5$.

The complex case is just the two dimensional extension of the real case.


答: Take $x\in\overline A$ and suppose that $f(x)\ne g(x)$. Since $Y$ is Hausdorff, there are neighborhoods $V_f$ and $V_g$ of $f(x)$ and of $g(x)$ respectively such that $V_f\cap V_g=\emptyset$. Since $f$ is continuous, $f^{-1}(V_f)$ and $g^{-1}(V_g)$ are neighborhoods of $x$. Therefore, $f^{-1}(V_f)\cap g^{-1}(V_g)$ is also a neighborhood of $x$ and so it contains some $a\in A$. But then $f(a)\in V_f$ and $g(a)\in V_g$, which is impossible, since $f(a)=g(a)$ and $V_f\cap V_g=\emptyset$.

Title: Prove $f(x) = g(x)$ for all $x \in \overline{A}$.


答:
Take a point $x$ such that $f(x)\neq g(x)$. Then because $Y$ is Hausdorff, there are open disjoint subsets $U\ni f(x)$ and $V\ni g(x)$ of $Y$.

Take the intersection $f^{-1}(U)\cap g^{-1}(V)$. It is open. It contains $x$. There is no point in this intersection where $f$ and $g$ agree. Is it possible for $x$ to lie in $\overline A$?


问:
Finding the maximum and minimum of $f$ on a set $Q$
I stumbled across a proposed task that I'm unable to solve.

We have a function $f:\mathbb{R^2}\to\mathbb{R}$ defined as:
$$
f(x,y):=x^2+xy+y^2+x+y+1
$$
The task is to "find the maximum and minimum" of $f$ on the set $Q$ which is defined as:
$$
Q:=\{(x,y)\in\mathbb{R^2}|\;\|(x,y)\|_\infty\le1\}
$$
Not knowing how to proceed, I decided to ignore $Q$ and simply calculated possible critical points and extrema. $f$ turns out to have a strict local minimum at $\left(-\frac13,-\frac13\right)$ which would fit into the constraints of $Q$ as $\|\left(-\frac13,-\frac13\right)\|_\infty$ is $\left(\frac13\right)\le1$.

How am I to proceed from this point on? Am I mistaken in thinking that the same strict local minimum is kept and that there is no maximum?

Cheers!

答:
There must be a maximum,as $Q$ is compact. Your method gives you all the local extrema in the interior of $Q$. So you need to check the values on the boundary of $Q$ next.

That means, you have to compute the extreme values of $x \mapsto f(x,\pm 1)$ and $y \mapsto f(\pm 1, y)$ on $[-1,1]$ and compare them to the inner extremum you allready found.


答:
Let $y_1=f(x)$ and $y_2=g(x)$ where $x\in\overline A$.

If $U_1\ni y_1$ and $U_2\ni y_2$ are any open neighbourhoods and $f,g$ are continuous, the set  $V=f^{-1}(U_1)\cap g^{-1}(U_2)$ is an open neighbourhood of $x$. As $x\in \overline A$, $V$ must intersect $A$. So let $a\in V\cap A$. Then $U_1\ni f(a)=g(a)\in U_2$, i.e., $U_1$ and $U_2$ are not disjoint.

Wec conclude that $y_1$ and $y_2$ cannot be separated in $Y$. As $Y$ is Hausdorff, this means that $y_1=y_2$.


问:
Iterated logarithm to the $n - 1$ of the auto tetration n^^n
I'm considering the sequence

$0 = \log(1)$

$\log\left(2^2\right)$

$\log\left(\log\left(3^{3^3}\right)\right)$

$\log\left(\log\left(\log\left(4^{4^{4^4}}\right)\right)\right)$

I tried solving this using Python and I get lots of $\log(x < 0)$ problems. I thought they were mostly fence post and recursion type errors, so I debugged my code but still the latest iteration has those problems for $n > 5$ or so.

I'm beginning to suspect that the iterated logs are too powerful, even when taken down a notch, and that I should do something to the argument to keep things finite. My first theory, on discovering a recursive solution, was that my sequence was increasing but I'm not at all sure anymore

Just to clarify: by iterated logarithm I mean $\log(\log(\log(... \text{Self tetration}, n^{n^{\dots}} )))$, is just $1$, $2^2$, $3^{3^3}$, ...

Iterated logs to the $n$ are provably overkill and iterated logs to the $n-2$ fail to tame the power tower.

答:
I'm assuming these are natural logarithms, although as it turns out the base of the logarithm doesn't matter much. Write $\log^k$ for the iterated (natural) logarithm and $n \upuparrows k$ for tetration. Note that with base-$n$ logarithms we clearly have

$$\log^k_n (n \upuparrows n) = (n \upuparrows (n-k)).$$

For $n \ge 3$ we have $\log n > 1$ and hence $\log_n x < \log x$, which gives

$$\log^{n-1}(n \upuparrows n) > \log \log_n^{n-2} (n \upuparrows n) = \log n^n = n \log n.$$

In particular there should in principle be no problems with taking logs of negative numbers, but I don't know what any particular piece of Python code might be doing to compute logs here.

But we can say more than this: it turns out that the base of practically every logarithm in the sequence except the outermost one doesn't matter, so the above is a pretty accurate approximation, which means

$$\boxed{ \log^{n-1} (n \upuparrows n) \approx \log n^n = n \log n }$$

so this sequence is increasing except for possibly the first few terms which can be checked by hand. To give an idea of how this works here's the $n = 4$ case. We get

$$\log 4^{4^{4^4}} = 4^{4^4} \log 4$$
$$\log \log 4^{4^{4^4}} = 4^4 \log 4 + \log \log 4.$$

Here $4^4 \log 4 \approx 355$ and $\log \log 4 \approx 0.326$ is much smaller in comparison so can be safely ignored when we take the next logarithm, hence

$$\log \log \log 4^{4^{4^4}} \approx 4 \log 4 + \log \log 4 \approx 4 \log 4.$$

In general we can argue by induction that $\log^k (n \upuparrows n) \approx (n \upuparrows (n-k)) \log n$. The error terms are quite tiny in comparison to how big the power towers are until we get to the outermost or maybe the second-to-outermost log. I can be more precise about this if you want.


问:
Product of Borel sigma algebra
Consider two generic topological spaces $(X,\mathcal{T}_1),(Y,\mathcal{T}_2)$.

We can then define the Borel measurable spaces $(X,\sigma(\mathcal{T}_1)),(Y,\sigma(\mathcal{T}_2))$, where $\sigma(\mathcal{T})$ is the smaller sigma algebra that cointains $\mathcal{T}$.

Now consider the product topological space $(X \times Y,\mathcal{T})$, where $\mathcal{T}$ is the product topology, namely the topology which has $\mathcal{B}=\{A_1 \times A_2 \mid A_1 \in \mathcal{T}_1, A_2 \in \mathcal{T}_2\}$ as a base.

We can now define a sigma algebra on $X \times Y$ in two different ways:

We can define the Borel product sigma algebra $\sigma(\mathcal{T}_1) \otimes \sigma(\mathcal{T}_2):=\sigma(\{A_1 \times A_2 \mid A_1 \in \sigma(\mathcal{T}_1), A_2 \in \sigma(\mathcal{T}_2)\})$;

We can instead define the Borel sigma algebra on product topology $\sigma(\mathcal{T})$.

I have read that it is always true that $\sigma(\mathcal{T}_1) \otimes \sigma(\mathcal{T}_2) \subseteq \sigma(\mathcal{T})$.

The converse ($\sigma(\mathcal{T}_1) \otimes \sigma(\mathcal{T}_2) \supseteq \sigma(\mathcal{T})$) is true for example when $(X,\mathcal{T}_1)$ and $(Y,\mathcal{T}_2)$ are second countable (namely they own a countable basis).

$ $

I want to proof these two relations. The second one is quite easy, and here is my attempt:

$X,Y$ are second countable, so there exist two countable basis $\mathcal{B}_1$ and $\mathcal{B}_2$ of these topological spaces.

We have that $\mathcal{B}:=\{B_1 \times B_2 \mid B_1 \in \mathcal{B}_1, B_2 \in \mathcal{B}_2\}$ is a countable base for $\mathcal{T}$, and then $\sigma(\mathcal{T})=\sigma(\mathcal{B})$.

Moreover $\mathcal{B} \subseteq \{A_1 \times A_2 \mid A_1 \in \mathcal{T}_1, A_2 \in \mathcal{T}_2\}=\{A_1 \times A_2 \mid A_1 \in \sigma(\mathcal{T}_1), A_2 \in \sigma(\mathcal{T}_2)\}$.

So $\sigma(\mathcal{T})=\sigma(\mathcal{B}) \subseteq \sigma(\mathcal{T}_1) \otimes \sigma(\mathcal{T}_2)$.

$ $

My question now is: how can I prove the first relation? (namely $\sigma(\mathcal{T}_1) \otimes \sigma(\mathcal{T}_2) \subseteq \sigma(\mathcal{T})$) I have searched a lot, but I didn't find a decent proof.

My attempt is the following:

$X,Y$ now are generic topological spaces (not second countable).

It is enough to prove that $\{A_1 \times A_2 \mid A_1 \in \sigma(\mathcal{T}_1), A_2 \in \sigma(\mathcal{T}_2)\} \subseteq \sigma(\mathcal{T})$.

However, I'm pretty stuck here. Any hint? Thank you so much!

答:
If $X,Y$ are second countable, so is $X \times Y$ and this implies that every open subset $O$ of $X \times Y$ is a countable union of basic open subsets (of the form $O_1 \times O_2$) which are by definition in $\sigma(\mathcal{T}_1) \otimes \sigma(\mathcal{T}_2)$. So all open subsets of the product are in $\sigma(\mathcal{T}_1) \otimes \sigma(\mathcal{T}_2)$, which is a $\sigma$-algebra and so contains $\sigma(\mathcal{T})$ by minimality.

OTOH, define $\mathcal{A} = \{A \subseteq X\mid: A \times Y \in \sigma(\mathcal{T})\}$ and show this is a $\sigma$-algebra that contains all open subsets of $X$, so $\sigma(\mathcal{T_1}) \subseteq \mathcal{A}$. Similarly, show that $$\sigma(\mathcal{T_2}) \subseteq \mathcal{B}:= \{B \subseteq Y\mid X \times B \in \sigma({T})\}$$ and note that from these facts it follows that for $A \in \sigma(\mathcal{T_1})$ and $B \in \sigma(\mathcal{T_2})$ we have that $A \times B = (A \times Y) \cap (X \times B) \in \sigma(\mathcal{T})$ and so the whole generating set of $\sigma(\mathcal{T}_1) \otimes \sigma(\mathcal{T}_2)$ is in $\sigma(\mathcal{T})$ which implies what you need:

$$\sigma(\mathcal{T}_1) \otimes \sigma(\mathcal{T}_2) \subseteq \sigma(\mathcal{T}) = \textrm{Bor}(X \times Y)$$

It can be shown that for a discrete uncountable space $X$ of size $|\Bbb R|$ we have that the diagonal $\Delta_X = \{(x,x) \mid x \in X\}$ is Borel in $X \times X$ but not in the product Borel algebra as defined under 1. Proofs can be found on this site.

Title: Product of Borel sigma algebra


问:
$\kappa$-tree with levels less then $\mu$ and $\mu^+ < \kappa$ has a cofinal branch
For the definitions used in this question I refer you to this previous question I asked.

Let $\kappa$ be an infinite regular cardinal and let $T$ be a $\kappa$ tree.

Prove that if there is some cardinal $\mu$ s.t $\mu^+ < \kappa$ and $|T_\alpha|\leq\mu$ for all $\alpha$ then $T$ has a cofinal branch.

I don't understand how to use the fact that $\mu^+<\kappa$ (I know that $\mu^+$ is regular as it is a successor cardinal).

We previously proved that we can look at a subtree $T'\subset T$ s.t   $T'$ is a $\kappa$-tree and $\forall \alpha<\beta<\kappa$ and $$\forall x\in T'_\alpha,\ \ \exists y\in T'_\beta,\ x \leq_T y$$

Moving to this $T'$ I tried building the branch inductively but I get stuck at the limit steps. My thought is that there is where I use the assumption.

答:
Let $L=\{\alpha<\kappa:\operatorname{cf}\alpha=\mu^+\}$, and for each $\alpha\in L$ let $x_\alpha\in T_\alpha$. For each $x\in T_\alpha\setminus\{x_\alpha\}$ there is a $\varphi_\alpha(x)<\alpha$ such that $x_\alpha$ and $x$ have different predecessors in $T_{\varphi_\alpha(x)}$. $|T_\alpha|\le\mu$, and $\operatorname{cf}\alpha=\mu^+>\mu$, so there is an $\eta(\alpha)<\alpha$ such that $\varphi_\alpha(x)<\eta(\alpha)$ for each $x\in T_\alpha\setminus\{x_\alpha\}$. Thus, the predecessor of $x_\alpha$ in $T_{\eta(\alpha)}$ is different from the predecessors in $T_{\eta(\alpha)}$ of all of the other members of $T_\alpha$.

The function $\eta:L\to\kappa$ is pressing-down (regressive) on the stationary set $L$, so the pressing-down lemma ensures that $\eta$ is constant on a cofinal subset of $L$. That is, there are a cofinal $C\subseteq L$ and a $\beta<\kappa$ such that $\eta(\alpha)=\beta$ for every $\alpha\in C$. $|T_\beta|\le\mu$, so there are a $y\in T_\beta$ and a cofinal $C_0\subseteq C$ such that $y<_Tx_\alpha$ for each $\alpha\in C_0$.

To finish the argument, use the definition of $\eta$ to show that $\{x_\alpha:\alpha\in C_0\}$ is a chain of length $\kappa$ in $T$ and therefore determines a cofinal branch through $T$.


问:
Analog to the identity theorem for holomorphic functions for functions of two complex variables
Consider a function $f:\mathbb{C}^2\rightarrow\mathbb{C}$ that is holomorphic and vanishes on a non empty bounded open set $S$. If $f$ were a function from $\mathbb{C}$, by the identity theorem, $f$ would be uniquely determined to be the zero function. Does a similar theorem hold when the domain is $\mathbb{C}^2$? If not, how can the vector space of all functions vanishing on $S$ be categorized or expressed? For example, I would be interested in listing them all if possible, or knowing the dimension of the vector space of all such functions.

答: Sure, you can deduce this from the single-variable case.  Pick a point $p\in S$, so $f$ vanishes on a neighborhood of $p$.  Now for any other $q\in\mathbb{C}^2$, consider the function $g(z)=f(zq+(1-z)p)$.  This is a holomorphic function $\mathbb{C}\to\mathbb{C}$, and is basically the restriction of $f$ to the (complex) line between $p$ and $q$.  Since $f$ vanishes in a neighborhood of $p$, $g$ vanishes in a neighborhood of $0$, so $g=0$.  In particular, $g(1)=0$ so $f(q)=0$.  Since $q$ was arbitrary, $f=0$.


答:
Indeed, by definition, a normalized vector is one with a magnitude of 1. Also by definition, taking the norm of a vector is actually taking the root of its scalar product (for a vector $v$, $|v| := \sqrt{\langle v,v\rangle}$). Thus, it is false to say that "taking the norm of a vector produc[e] the unit vector on the same line with that vector" : we clearly see that by definition, taking the norm of a vector $v$ output a scalar and not an other vector.

To obtain the vector $v_n$ of magnitude 1 that is "on the same line" as $v$ (the "normalized vector of $v$") we can simply take $\frac{v}{|v|}$ wich is what is done one the piece of mathematics that you have circled in red.

(you could easily prove that fact in $\mathbb{R^3}$ by letting $v := (a,b,c) \in \mathbb{R^3}$, calculating $\frac{v}{|v|}$ and taking the norm of $\frac{v}{|v|}$. You would see that it gives 1 for any value of $a$, $b$ and $c$)

Best of luck in your Linear Algebra journey,

LS

Title: Why is the magnitude of the vectors still not 1 after normalization?

Body:
I was reading that the vectors must have magnitude of $1$ to be called normalized; however, none of the problems I solved gave me a vector magnitude of $1$ after normalizing. I also thought that taking the norm of a vector produced the unit vector on the same line with that vector which also equals to $1$. What am I getting wrong here?

Note: I‘ve checked the other answers, but it didn‘t seem to click. Hope you can enlighten me.


问:
How to show isomorphism of permutation groups?
Let $M = \{m_1,. . . , m_k\} ⊂ \{1, 2, 3,. . . , n\}$ and
$H = \{\sigma ∈ S_n | \forall m \in M, \; \sigma(m) \in M\}$.

Show that $H \subset S_n$ is a subgroup.

I tried the following:
$m \in M$ and $\sigma \in H$ implies that $\sigma(m) \in M$. If we take $\pi \in H$, then $\pi \sigma = \pi(\sigma(m)) \in M$ which in turn implies that $\pi\sigma \in H$(here the product is defined as a composition).

Update: I have shown that $H$ is nonempty, finite and closed under the operation of multiplication, hence I have shown that it is indeed a subgroup.

(Nonempty, because it contains at least the identity permutation, and finite because it can only have up to $n!$ elements which is a finite number.)

Now, to show that $H$ is isomorphic to $S_k × S_{n−k}$, I have to define a bijective homomorphism. I don't quite understand what this means.
I know that permutations are bijective, but how would this work for $S_k × S_{n−k}$?
What does $S_k × S_{n−k}$ actually mean?

答:
Recall that a permutation of $\{1 \dots n\}$is a bijective map, then for a fixed $\sigma\in H$, the elements of $M$ are all of the form $\sigma(j)$ for $j \in M$.
Moreover for the inverse of $\sigma$ in $S_n$ the property
$\sigma(j)=j  $ holds and this implies  $\sigma^{-1} \in H$.

Now, a $\sigma \in H$, restricted to $M$, is a permutation of its $k$ elements.
Analogously $\sigma$ acts as a permutation on the complementary of $M$.

Consider the map $H \rightarrow S_k\times S_{n-k}$ that sends $\sigma$ to $(\sigma_{|M},\sigma_{|M^c})$ and try to prove that this is a group isomorphism.


问:
Number of binary strings without adjacent char more than $n$
$N$ is a natural number, let $0\leq\alpha\leq 1$ a fixed real number, I want to estimate the number of length-N binary strings which statisfy: the same character (i.e $0$ or $1$) could not occur repeatedly and subsequently more than $\alpha N$ times.

For example: substrings like $000....000$ (more than $\alpha N$ many $0$) could not exist.

I only need to know how the number $I(N)$ of these strings grows when $N\to \infty$.

答:
Partial answer: You can make (at least) somewhere around $2^{N-2/\alpha}$ different valid bitstrings (there is some rounding going on, but that's roughly the result you get). You do this by inserting $01$ every $\alpha N$ bits to break any potential consecutive substrings, and then choosing the remaining bits freely.

I don't know whether I am missing a significant amount of valid strings this way, though. It depends on what "significant" means.


问:
Newton's Theorem
For my Measure & Integration course, I've been asked to prove the following:

Let $g$ be a function taking values on $\Bbb{R_+}$, $f: \Bbb{R}^3 \to \Bbb{R}$ such that $f(x) = g(|x|)$. Suppose that $f \in L^1(\Bbb{R}^3, dx)$, and let

$$ \Phi(x) = \int_{\Bbb{R}^3} \frac{f(y)}{|x-y|}dy. $$

For all $r=|x| > 0$,

a) Show that $$\Phi(x) = \frac{4π}{r}\int_0^r g(s)s^2\ \mathrm ds+ 4\pi\int_r^{\infty}g(s)s\ \mathrm ds.$$

b) Deduce that $$\Phi(x) = \int_{\Bbb{R}^3} \frac{f(y)}{max\{|x|,|y|\}}\mathrm dy.$$

c) Show that $$\mu\{x: \Phi(x) > t\} < \infty\qquad\text{for all }t>0,$$
where $\mu$ denotes the Lebesgue measure

d) Conclude that $\Phi(x) = \Phi^*(x)$, where $\Phi^*(x)$ is the symmetric decreasing rearrangement of the function $\Phi(x)$.

After a lot of torture I managed to prove part a), but I'm completely stuck on the following parts. Any advice would be very very welcome!

答:
In part a, you showed that $\Phi$ itself is radial, it depends only on $r = |x|$.
Thus, $\Phi(x)$ is equal to its average value of the sphere of radius $|x|$.

I.e.

$$
\Phi(x) = \frac{1}{4\pi |x|^2}\int_{|u|=|x|} \Phi(u) du
$$
which by Fubini is
$$
= \frac{1}{4\pi |x|^2} \int_{\mathbb{R}^3}f(y) \int_{|u|=|x|} \frac{1}{|u-y|}du\, dy.
$$

This inner integral is a standard computation of a surface integral. You can see it's calculation e.g. here Surface integral of $f(x) = \frac{1}{ \Vert x -x_0 \Vert } $ over sphere.
Thus the last line is
$$
\frac{1}{4\pi |x|^2} \int_{\mathbb{R}^3}f(y) \frac{{4\pi |x|^2}}{\max(|x|, |y|)} dy
= \int_{\mathbb{R}^3}f(y) \frac{1}{\max(|x|, |y|)} dy
$$
as desired for part b.

By part b, we know that as $x \to \infty$ the integrand tends to $0$. Since the integrands with $|x|>1$ are dominated by the integrand with for a fixed $x_0$ on the unit sphere, we can apply dominated convergence to see that $\Phi \to 0$ as $|x| \to \infty$ (important that $f \in L^1$ here). This obviously implies part c as the measure in question is dominated by a that of a big ball.

For part d, just note that $\Phi$ is symmetric by b, and clearly decreasing in $|x|$ by c, so it is its own decreasing symmetric rearrangement.

Title: Newton's Theorem


问:
How to multiply 2 Binary Numbers?
Suppose I have two regular numbers $a$ and $b$ in base $10$ like this: (where $N$ is even)

$$a=a_{\frac N2}a_{\frac N2-1}\ldots a_1,\qquad b=b_{\frac N2}b_{\frac N2-1}\ldots b_1$$

So the result of their multiplication is:

$$a\cdot b=a_{\frac N2}a_{\frac N2-1}\ldots a\cdot b_{\frac N2}b_{\frac N2-1}\ldots b_1=\sum_{i=1}^{\frac N2}\sum_{j=1}^{\frac N2}(a_ib_j)10^{i+j-2}$$

Now suppose $a=a_Na_{N-1}...a_1$ and $b=b_Nb_{N-1}...b_1$ are (unsigned) binary numbers, such that each digit is actually 16 bits by its own. For example: $a_1$ can be: $1111010101000000$

So if $a$ has $N$ digits it's actually: $16N$ long binary number.

Given this fact how can I multiply them in a similar way to what is shown above?

Why I need this?

I want to multiply big numbers using the fact that I know only to multiply 16 bits binary numbers.

Small But Different Example:

if $a_M, b_M, a_L, b_L$ are all 16 bits binary numbers then I could use this fact to multiply 32 bits numbers like this:

答:
Replace $10$ by the base $b$,
either 2 or 16 or whatever to get

$a\cdot b=a_{\frac N2}a_{\frac N2-1}\ldots a\cdot b_{\frac N2}b_{\frac N2-1}\ldots b_1=\sum_{i=1}^{\frac N2}\sum_{j=1}^{\frac N2}(a_ib_j)b^{i+j-2}
$.

If $b=2$,
$b^{i+j-2}$ does
$i+j-2$ left shifts.

If $b=16$,
$b^{i+j-2}$ does
$4(i+j-2)$ left shifts.

Title: How to multiply 2 Binary Numbers?


问:
(In)equality between norm and scalar product
By the Cauchy-Schwarz inequality we always have $x=\sup_{||y||=1}|\langle x,y\rangle|$ for $x\in H$, where $H$ is a Hilbert space. Now let's narrow down our focus to $\mathbb{R}^{n}$ with the euclidean scalar product and suppose we have a symmetric matrix $A\in\mathbb{R}^{n \times n}$. Then we have
$$\sup_{||x||=||y||=1}\langle x,Ay \rangle =\sup_{||x||=1}||Ax||
 $$
by taking the suprema one at a time which is possible since $\langle x,Ay \rangle=\langle Ax,y\rangle$ by symmetry. Now what happens if we drop the symmetry assumption? A priori I don't think, I can use the line of argumentation here since taking the suprema in a different order will result in the supremum over either $||Ax||$ or $||A^Tx||$ which in general are not the same. If there is not, is there at least an inequality
$$ \sup_{||x||=1}||Ax||\le\sup_{||x||=1}\langle Ax,x\rangle
$$
for general $A$?

答: No, the inequality is not true. if $A$ is rotation by $90^{0}$ in $\mathbb R^{2}$ then  $\langle Ax, x \rangle=0$ for all $x$.


问:
Can there be exceptional points in the real axis?
Suppose a matrix $H$ that depends on some parameter $\lambda$ such that, when $\lambda=0$, the eigenvalues of $H$ are real (for instance you can assume that $H$ is Hermitian). By cranking up $\lambda$, the eigenvalues can be shifted into the complex plane (see for a $2\times 2$ matrix example. There are instances in which two eigenvalues coalesce and the matrix has degeneracy in the eigenvalues and the eigenvectors.
This is usually termed as an exceptional point "EP" in the literature.

I only have seen such EP where two eigenvalues coalesce in the complex plane. Can there be EP that occur in the real axis, when $H(\lambda=0)$ is Hermitian?

答:
You have asked two different questions: Can exceptional points (EPs) occur when the tuning parameter is real? And can EPs occur when the Hamiltonian is Hermitian?

For the first question, This reference answers with yes.

Since the Hamiltonians considered in this class can be non-Hermitian,
the EPs can occur for real parameter values.

For the second question, the answer is no because for a Hermitian matrix always has a set of eigenvectors that can span the whole Hilbert space. (EPs occur when eigenvectors coalesce a.k.a. when the geometric multiplicity of eigenvalues is less than the algebraic mulitplicity.)