问:
In what sense do isomorphisms "preserve all logical properties"?
My background in mathematical logic, model theory, etc. is patchy, so I'm looking for a clearer way to think about this. (Edit: I'm not asking what an isomorphism is, I'm asking how to formalize the idea of "preserving all logical properties" in order to state the principle described below in its full power.)

Motivating example: we know that $\Bbb N$ and $\Bbb Z$ are not order-isomorphic, because $\Bbb N$ has a least element whereas $\Bbb Z$ does not. The principle underlying this argument seems to be that an order isomorphism preserves the truth values of (first-order) sentences: if $(A,<_A)\cong(B,<_B)$ then $\operatorname{Th}(A,<_A)=\operatorname{Th}(B,<_B)$. Since the property "there is a least element" corresponds to the sentence $\exists x\forall y. x\le y$, the principle applies. The principle also seems to apply to higher-order properties. For example, we can conclude that $(\Bbb R\setminus\{0\},<)\not\cong(\Bbb R,<)$ by checking the least-upper-bound property, which is second-order.

More generally (if we think of set theory as modeling higher-order logic in first-order logic), I'm tempted to claim that the whole "set theory of <" (I'm not sure what this is exactly - a two-sorted first-order logic with relation symbols $<,\in$, maybe?) is preserved by isomorphisms. I think I can convince myself of this fact by observing that when $f:A\to B$ is an isomorphism, certain substitution rules involving $f$ in various syntactic contexts are valid, allowing us to prove equivalences between propositions about $<_A$ and the corresponding propositions about $<_B$.

I'm calling this a "principle" instead of a "theorem" because it seems easier to apply in specific cases (by reasoning directly about the elements and the isomorphism, without formalizing the notion of sentences) than to express in its full generality. My claim that "the set theory of $<$ is preserved" seems closely related to the fact that set theory is my metatheory (if that's the right word), so some kind of circularity or self-reference is appearing when I attempt to state the principle, and I wonder whether any formal version of the principle can fully eliminate the need to apply the principle directly.

So I guess my questions are:

In what sense (if any) is there a most general version of this principle, applying to all properties preserved by order isomorphisms?

Can it be expressed as a theorem in the deductive system where we want to apply it, or is there a theoretical obstacle to this?

Am I getting anything wrong? Or missing a clarifying perspective?

Thanks!

答: If $\mathbf{M_1}$ and $\mathbf{M_2}$ are structures for some signature $\Sigma$ (so $\Sigma = (<)$) in your example. Then, by definition, $\mathbf{M_1}$ and $\mathbf{M_2}$ are isomorphic iff there are functions $f : M_1 \to M_2$ and $g : M_2 \to M_1$ (where $M_i$ is the universe of discourse of $\mathbf{M_i}$), such that $g \circ f = 1_{M_1}$ and $g \circ f = 1_{M_2}$ and such that $f$ and $g$ are homomorphisms for the relations and functions in $\Sigma$ (so, in your example, $x < y$ iff $f(x) < g(y)$ and $u < v$ iff $g(u) < g(v)$). So if $\mathbf{M_1}$ and $\mathbf{M_2}$ are isomorphic, a logical language $\cal L$ (first-order or higher-order) over the signature $\Sigma$ cannot distinguish between $\mathbf{M_1}$ and $\mathbf{M_2}$, since $f$ induces a truth-preserving mapping from interpretations of $\cal L$ in  $\mathbf{M_1}$ and interpretations of $\cal L$ in $\mathbf{M_2}$ (and vice versa for $g$).


问:
prove f is separable if and only if f is relatively prime to its derivative.
Given the definition; a polynomial f in the complex numbers is separable if it has only simple roots.
My question is how do I prove f is separable if and only if f is relatively prime to its derivative.

Can someone help me go about doing this proof?

答:
Recall that a polynomial is separable iff it doesn't have duplicate roots in the algebraic closure. Hence it is enough to prove that: $f\in \mathbb{C}[X]$ has only simple roots iff $\gcd_{\mathbb{C}[X]}(f,f')=1$.

Proof: $\gcd(f,f')\neq 1$ iff there is an $a\in K$ such that $x-a|f$ and $x-a|f'$. That is, iff $f(a)=f'(a)=0$. This happens when the first two coefficients of
$$f(x)=f(a)+f'(a)(x-a)+\dots+\frac{f^{(k)}(a)}{k!}(x-a)^k$$
are $0$. Which is equivalent to $(x-a)^2|f(x)$.

As a remark, you can extend this proof to an algebraically closed field of positive characteristic by using the Hasse derivative in the expansion.

ContentLicense: CC BY-SA 4.0


问:
From 1 to 100 select 26 distinct numbers: pigeonhole principle
From 1 to 100 select 26 distinct numbers, there must exist one number which divides the product of the other 25 numbers.

This seems to be an easy application of the pigeonhole principle, but I could not figure out the exact argument. Can someone help?

答:
Hint to get you started:

How many primes numbers are there less than $100$?


问:
How to get $2700$ grams of corn from a bag of $6.8$ kilograms?
The problem is as follows:

A merchant has a kiosk and has one bag with $6.8$ kilograms of corn and a two pan scale and three weights. One measures $5$ kilograms, the other $2$ kilograms, and the other $1$ kilogram. In how many weighing trials at minimum could he serve an order of $2700$ grams of corn?

The choices given are:

$4$ trials

$3$ trials

$1$ trial

$2$ trials

I'm not sure exactly how to get to the right combination for this problem. It would be very nice if the answer could include the steps in a logical and very detailed way so I can understand.

To my best of ability I thought that I could arrange the weights as follows:

$5$, $2$, $1$, $6.8$

To get $2.7$ from these numbers is somewhat convoluted to me. Thus I need some help. What can I do?

The thing is either method which I've tried from subtracting this amount from $6.8$ does give me more than what I need, thus I am going in circles.

答:
Servaes has given a solution in two weighings.  I will show that it cannot be done in one weighing.  It cannot be done by weighing some amount of corn in only one of the pans, for the corn weighed would be an integral number of kilos, and the number of kilos of corn remaining in the sack would be congruent to $.8$ modulo $1$.  So, we must have corn in both pans, and unless we weigh all the corn, we have no way of knowing how much corn is in either pan.

Suppose one pan has $c$ kilos of corn, and $w_1$ kilos in weights, and the other pan had $6.8-c$ kilos of corn and $w_2$ kilos in weights.  We have $$\begin{align}c+w_1&=6.8-c+w_2\\c&=3.4-\frac{w_2-w_1}2\end{align}$$

Since $w_1$ and $w_2$ must be integers, we see that either $c\equiv.4\pmod1$, or $c\equiv.9\pmod1$.  In either case, we have $6.8-c\equiv c\pmod1$, and in no case do we have $2.7$ kilos of corn in one of the pans.


问:
Infinite Series Question, difference between infinities
Was just thinking about infinite series, what is wrong with the below?
$$s_1=\lim _{n\to\infty} \sum_{i=1}^n \frac{1}i =\infty $$
$$s_2=\lim _{n\to\infty} \sum_{i=1}^n i =\infty $$
$$\therefore \ s_1=s_2$$

Fixed the formula, I am not sure if this is wrong but every element of $s_2$ is larger than the corresponding element of $s_1$ (apart from i=1), is this just an oddity of infinity?

答:
The two series that you write diverges, so there is no real number that these sums approach as you take the limit. Sometimes one uses the symbol $\infty$ to denote the value of these sums, and in this sense, the two sums "converges" to the same limit -infinity. In my opinion, this is meaningless since this series diverges.

The example that you gave is good to see the differences between the infinities. The two sums have very different "rates of divergence". To see this, consider the partial sums
\begin{align}
a_n = \sum_{k=1}^{n} \frac{1}{k} \; , \\[6pt]
b_n = \sum_{k=1}^{n} k \; .
\end{align}
The sum of the first $n$ positive natural number is well known to give
$$ b_n = \frac{n(n+1)}{2} \; . $$
Using the integral test, we can see that the sum of the Harmonic numbers is in the limit
$$ \ln n  = \int_1^{n} \frac{1}{k} dk \; \leq a_n \leq  \; 1 +  \int_1^{n} \frac{1}{k} dk = 1 + \ln n \; . $$

Thus, for large $n$ the partial sums behave like
\begin{align}
a_n  \; &\sim \; \ln n \; , \\
b_n \;&  \sim \; \frac{1}{2} \, n^2 \; ,
\end{align}
which is enough to see that the two series diverge. However, more importantly, we can see that the Harmonic series diverges very slowly in a logarithmic way were as the sum of the natural number diverges quadratically.

To answer the question, there is nothing wrong with it,  it is just notation. I just wanted to explain that you are right when you compare one partial sum with the other and they have grown very differently. For finite $n$ we can easily see that $a_n < b_n$ (and even $ \text{Exp} \, a_n < b_n$). However, when you sum over infinite number $n\in \mathbb{N}$ the two sums goes to infinity. After you sum the information on the rate of grown of the function is lost and there is no meaning in "$\infty < \infty$", both diverges (are $\infty$).


问:
Nontrivial trivial integrals
Consider two propositions in geometry:

Circumscribe a right circular cylinder about a sphere.  The surface area of the cylinder between any two planes orthogonal to the cylinder's axis equals the surface area of the sphere between those two planes.  (Archimedes showed this.)

Let a curve running from the south pole to the north pole on a perfectly spherical earth meet every meridian of longitude at the same angle $\gamma$.  Then the length of the curve is proportional to $\sec\gamma$.  (In particular, it's infinite if $\gamma$ is a right angle, and one could even go beyond a right angle and consider it an oriented length, so that the oriented length is negative if the curve goes from the north pole to the south pole.  In that case, the infinite length would be the $\infty$ that's at both ends of the real line rather than $+\infty$ or $-\infty$, so the length depends continuously on $\gamma$.)

Both propositions admit the same kind of proof: For the first, consider what happens when the distance between the two planes is infinitesimal; then show that if it works then, then it also works for larger-than-infinitesimal distances.  For the second, show that the length of an infinitesimal increment of the curve is $\sec\gamma$ times the latitude-component of the distance, just by drawing an infinitesimal right triangle.

You might think that in both cases one is finding an integral, but it's an integral of a constant: just the constant itself times the length of the interval over which one integrates.  So it's as if you don't need to know about integrals and you certainly don't need to find any antiderivatives, but the method of proof doesn't work for larger-than-infinitesimal quantities.

One would like to say things like "energy is force times distance; it is only when the force is not constant that one must use integrals and multiply the force at an instant by the infinitesimal distance and then integrate".  But if proving the force is constant itself requires an infinitesimal viewpoint, then despite the triviality of computing the integral by multiplying two numbers, it seems necessary to the logic of the argument to view it as an integral.

So:

What other instances besides these two are known?  (I suspect there are many in geometry.)

What else of interest can be said about this phenomenon?

答:
Here's a problem that I think has a similar flavor:

Suppose that a bounded convex solid $S$ in $\mathbb{R}^3$ is positioned in a random orientation. Show that the expected area of its "shadow" onto the $xy$-plane is $1/4$ the surface area of $S$.

To solve this, we invert the problem: suppose that our set $S$ is fixed, and the angle of the light (and projected plane) are chosen at random.

Now, for each infinitesimal piece of boundary of $S$, say that its contribution to the shadow is half its projection onto the plane. Then, because there is a "near" and "far" portion of the convex set, the sum of all contributions across the surface of $S$ will be the area of its shadow, since we double-cover the shadow and divide by $2$. (This works because $S$ is convex, so every part of $S$ is either directly in the light, or directly on the dark side - every normal vector from the interior of the shadow passes through $S$ twice.)

So we just need to integrate, for each infinitesimal bit of surface, the expected size of its shadow when light is shone on it at a random angle. But this is just proportional to the surface area of that piece! So the answer must be linear in the surface area of $S$ when we integrate, and by considering the case of a spherical $S$ we see that the constant of proportionality ends up being $1/4$.

As a remark, this argument carries through in any dimension (we didn't use anything unique to $\mathbb{R}^3$ here), and in the two-dimensional case it gives us a constant of proportionality of $1/\pi$. When we apply this to the case of curves of constant width, we immediately recover Barbier's theorem. In fact, this proof shows us in a sense the "correct" generalization of curves of constant width into $n$ dimensions: they are the convex sets whose projections have constant $n-1$-dimensional measure! Barbier's theorem does generalize to such sets via the above proof, but when $n>2$ this is a different notion than "constant width" and so an analogous result does not hold.

ContentLicense: CC BY-SA 3.0

Title: Nontrivial trivial integrals


问:
Confidence intervals for a ratio of normal variances?
I was studying the F-distribution which says that given a pair of samples and their sizes and knowing the sample variances, we can compute the confidence interval of the ratio of true variances as the F statistic then follows the F distribution.

What I was wondering was does it make sense to know the ratio of the true variances and then estimate the "confidence interval" for the sample variances, i.e. that the ratio of the sample variance has a range centred at the ratio of true variances in which it lies 95% of the time for any samples of those sizes? Does such a computation make sense or have a meaning? Clearly if I put in the values then the sample variance can be associated with a range based on an F distribution, so it is possible. But does it make any sense?

答:
Let $X_{11}, X_{12}, \dots, X_{1,m}$ be a random sample of size $m$ from $\mathsf{Norm}(\mu_1, \sigma_1).$ and let
$X_{21}, X_{22}, \dots, X_{2.,n}$ be a random sample of size
$n$ from $\mathsf{Norm}(\mu-2, \sigma_2).$

Then $S_1^2 = \frac{1}{m-1}\sum_{i=1}^m (X_{1i} - \bar X_1)^2,$ where $\bar X_1 = \frac 1m\sum_{i=1}^m X_i,$ is an
unbiased estimator of $\sigma_1^2.$ Similarly for
$S_2^2$ and $\sigma_2^2.$

Then one can show that
$$\frac{S_2^2/\sigma_1^2}{S_2^2/\sigma_1^2}
= \frac{S_1^2/S_2^2}{\sigma_.^2/\sigma_2^2} = \frac{R}{\psi} \sim \mathsf{F}(m-1,n-1),$$
where $R = S_1^2/S_2^2$ and $\psi = \sigma_1^2/\sigma_2^2.$

Then $P(L < R/\psi < U) = 0.95,$ where $L$ and $U$ cut probability $0.025,$ respectively from the lower and upper tails of the distribution $\mathsf{F}(m-1, n-2).$
[Values $R$ and $L$ can be obtained from printed tables
of Snedecor's F distribution, but it is often more convenient to use software such as R or a statistical calculator.]

Upon pivoting (solving to isolate $\psi),$ we have
$(R/U < \psi < R/L) = 0.95,$ so that a 95% confidence
interval for $\psi = \sigma_1^2/\sigma_2^2$ is of the form
$(R/U, R/L) =
\left(\frac{S_1^2/S_2^3}{U},\, \frac{S_1^2/S_2^2}{L}\right).$

Example: Suppose we have samples of size $m = 10,n = 20$
from populations with $\sigma_1^2 = 25,\sigma_2^2 = 36,$ $\psi = 25/36 =0.6944.$

Then from the data with  $R = 0.4497,$ a 95% CI for $\psi$ is
$(0.156, 1.656),$ which does happen to cover the true value
$\psi = \sigma_1^s/\sigma_2^2 = 0.6944.$

```<code>set.seed(2021)
x1 = rnorm(10, 50, 5)
x2 = rnorm(20, 60, 6)
v1 = var(x1); v2 = var(x2)
v1; v2;  v1/v2
[1] 22.36833
[1] 49.7425
[1] 0.4496825

CI = (v1/v2) / qf(c(.975,.025), 9, 19);  CI
[1] 0.1561369 1.6563326
</code>```

Notes: (1) Even though $\psi < 1$ so that
$\sigma_1^2 < \sigma_2^2$ and $R = 0.4497 < 1,$ the sample sizes are not large enough to yield a CI short enough
to exclude $1.$

(2) The CI shown above is included in the R procedure var.test.

```<code> var.test(x1, x2)

        F test to compare two variances

 data:  x1 and x2
 F = 0.44968, num df = 9, denom df = 19, p-value = 0.2197
 alternative hypothesis: true ratio of variances is not equal to 1
 95 percent confidence interval:
   0.1561369 1.6563326
sample estimates:
ratio of variances
         0.4496825
</code>```

Body:
Let $X_{11}, X_{12}, \dots, X_{1,m}$ be a random sample of size $m$ from $\mathsf{Norm}(\mu_1, \sigma_1).$ and let
$X_{21}, X_{22}, \dots, X_{2.,n}$ be a random sample of size
$n$ from $\mathsf{Norm}(\mu-2, \sigma_2).$

Then $S_1^2 = \frac{1}{m-1}\sum_{i=1}^m (X_{1i} - \bar X_1)^2,$ where $\bar X_1 = \frac 1m\sum_{i=1}^m X_i,$ is an
unbiased estimator of $\sigma_1^2.$ Similarly for
$S_2^2$ and $\sigma_2^2.$

Then one can show that
$$\frac{S_2^2/\sigma_1^2}{S_2^2/\sigma_1^2}
= \frac{S_1^2/S_2^2}{\sigma_.^2/\sigma_2^2} = \frac{R}{\psi} \sim \mathsf{F}(m-1,n-1),$$
where $R = S_1^2/S_2^2$ and $\psi = \sigma_1^2/\sigma_2^2.$

Then $P(L < R/\psi < U) = 0.95,$ where $L$ and $U$ cut probability $0.025,$ respectively from the lower and upper tails of the distribution $\mathsf{F}(m-1, n-2).$
[Values $R$ and $L$ can be obtained from printed tables
of Snedecor's F distribution, but it is often more convenient to use software such as R or a statistical calculator.]

Upon pivoting (solving to isolate $\psi),$ we have
$(R/U < \psi < R/L) = 0.95,$ so that a 95% confidence
interval for $\psi = \sigma_1^2/\sigma_2^2$ is of the form
$(R/U, R/L) =
\left(\frac{S_1^2/S_2^3}{U},\, \frac{S_1^2/S_2^2}{L}\right).$

Example: Suppose we have samples of size $m = 10,n = 20$
from populations with $\sigma_1^2 = 25,\sigma_2^2 = 36,$ $\psi = 25/36 =0.6944.$

Then from the data with  $R = 0.4497,$ a 95% CI for $\psi$ is
$(0.156, 1.656),$ which does happen to cover the true value
$\psi = \sigma_1^s/\sigma_2^2 = 0.6944.$

set.seed(2021)
x1 = rnorm(10, 50, 5)
x2 = rnorm(20, 60, 6)
v1 = var(x1); v2 = var(x2)
v1; v2;  v1/v2
[1] 22.36833
[1] 49.7425
[1] 0.4496825

CI = (v1/v2) / qf(c(.975,.025), 9, 19);  CI
[1] 0.1561369 1.6563326

Notes: (1) Even though $\psi < 1$ so that
$\sigma_1^2 < \sigma_2^2$ and $R = 0.4497 < 1,$ the sample sizes are not large enough to yield a CI short enough
to exclude $1.$

(2) The CI shown above is included in the R procedure var.test.

 var.test(x1, x2)

        F test to compare two variances

 data:  x1 and x2
 F = 0.44968, num df = 9, denom df = 19, p-value = 0.2197
 alternative hypothesis: true ratio of variances is not equal to 1
 95 percent confidence interval:
   0.1561369 1.6563326
sample estimates:
ratio of variances
         0.4496825


问:
Transform into product: $\sin (3x)- \cos x - \sin x$
Can someone give an idea (an idea to start) about how to transform in product the following expression:
$$\sin(3x)-\cos(x)-\sin(x)$$
Mention that I tried to write $\sin(3x)$ as $\sin(2x+x)$, or $\cos(x)=\sin(\pi/2 + x)$ but I go back to the initial expression or it gets very complicated .
Thank you!

答:
I am agree totally with the user @Somos, because

$$\sin(3x)-\cos(x)-\sin(x) \iff \sin(3x)-(\cos(x)+\sin(x))$$
But
$$\cos(x)+\sin(x)=\sqrt2 \sin (x+\varphi)$$
(added angle method).

Hence
$$\sin(3x)-\cos(x)-\sin(x) \iff \sin(3x)-\sqrt2 \sin (x+\varphi)$$
and you can not apply the prostapheresis formulas for the presence of $\sqrt 2$.


问:
A branch cut problem
In Ahlfors' Complex Analysis text, chapter 3, section 4 the transformation $z=\zeta+\frac{1}{\zeta}$ is discussed. The author notes that for every $z$, there exists 2 solutions for $\zeta$ and they are inverses of each other. In order to get a unique $\zeta$ he suggests the restriction $|\zeta|<1$, and he says that in that case the interval $(-2,2)$ must be removed from the $z$ plane.

My question is: why must that interval be removed? In previous chapters it was shown that in order to get an analytic branch for $\sqrt{w}$ one might use a plane with a ray removed (say the negative real axis) - not an interval.

The inverse function is
$$\zeta=\frac{z \pm \sqrt{z^2-4}}{2}$$
and requiring that the argument of the square root is not negative doesn't get me to the interval $(-2,2)$.

答:
This function has two branch points at $\pm2$ as your inversion formula shows. By contrast, $\sqrt z$ alone has only one finite branch point at $0$; however, formally, it also has one at infinity. (Set $w=1/z$, or look at the Riemann sphere to make sense of this.)

Branch cuts always join two branch points, hence the line segment he chose. (An alternative would be two line segments from $2\to\infty$ and $-2\to-\infty$, which is secretly one line going via infinity.)

I can explain more if you like, but this is the reason.

Take $f(z)$ to be mostly two-valued, like $f=\sqrt{g(z)}$ for polynomial $z$. Now suppose you try to define $f$ by starting at some $z_0$ with one of the branches and smoothly (analytically) extending it. When does this go wrong? When you come back to a point $z_1$ you've seen before but with a different value.

Now imagine drawing the two curves you followed from $z_0$ to $z_1$; they form a loop. Start tightening it by bringing them closer together. Since it must be impossible for the two paths to be smoothly deformed onto each other (since they have different values at the $z_1$ end), there must be some point $z_\star$ (along the line you almost formed) such that you can't give a single consistent value to $f$ in any neighbourhood of $z_\star$. $z_\star$ is a branch point.

For example, with $\sqrt{z}$ this happens at the origin. Intuitively, if you go around it in an arbitrarily tiny circle, you get an inconsistent result.

It also happens at infinity; taking a small circle in the variable $1/z$ or a large circle in $z$ makes it clear that you have the same problem out here.

By contrast, $\sqrt{z^2-4}$ has two branch points at $\pm 2$. At infinity, for example, because $$z=Re^{i\theta} \implies (z^2-4)^{1/2}\sim (R^2e^{2i\theta})^{1/2}=Re^{i\theta}$$ there is no inconsistency on going round by a circle, as $\theta\sim\theta+2\pi$. (The $z^2$ goes round twice as fast to compensate for the way that $\sqrt z$ lags at half speed.)

Anyway, the point is that if you want to make it impossible to follow a pair of contradictory paths like those described above, then it seems you can't include any region containing a branch point, or indeed any region enclosing a branch point since the discontinuity will be felt further way. The exception is that when you enclose two branch points, it might be that going around both of them doesn't lead to a contradiction. In fact, we saw this for $\sqrt{z^2-4}$ above already. (It's an equivalent calculation in this case to seeing that infinity isn't a branch point.)

Thus is turns out that excluding the region $(-2,2)$, which prevents us from making any path which encircles just one of the branch points, is sufficient to allow a smooth definition everywhere. Hooray!

Title: A branch cut problem


问:
A question on a common surjetive map for proving uncountability of sets
With the aim of proving that something is uncountable, it's common one to utilize the Cantor diagonal method to achieve oneself goal. It's not hard to see that all of that arguments lies in proving the function given by
$$
f (\{\eta_k\}_{k\in\mathbb{N}}) = \sum^{\infty}_{k=1} \dfrac{\eta_k}{2^{k}},
$$ for all sequences $\{\eta_k\}_{k\in\mathbb{N}} \in\{0,1\}^{\mathbb{N}}$ is sujective on $[0,1]$. Is there a good way of proving that given an $x\in[0,1]$ is possible to find a sequence $\{\eta_k\}_{k\in\mathbb{N}}$ such that $f(\{\eta_k\}_{k\in\mathbb{N}})=x$, with each $n_k$ being 0 or 1?

答:
Let $\eta_1=0$ if $x<1/2,$ or $\eta_1=1$ if $x\ge 1/2.$ Let $x_1=\eta_1/2.$ Then $x_1\le x\le x_1+1/2.$

For $k\in\Bbb N,$ suppose that $x_k=\sum_{n=1}^k\eta_n/2^n$ and that  $x_k\le x\le x_k+1/2^k.$  Let $\eta_{k+1}=0$ if $x-x_k<1/2^{k+1},$ or $\eta_{k+1}=1$ if $x-x_k\ge 1/2^{k+1}.$ Let $x_{k+1}=x_k+\eta_{k+1}/2^{k+1}.$ Then we have $x_{k+1}\le x\le x_{k+1}+1/2^{k+1}.$

So by induction on $k\in\Bbb N$ we have $x_k\le x\le x_k+1/2^k$ for all $k\in \Bbb N.$ Therefore $$\sum_{k=1}^{\infty}\eta_k/2^k=\lim_{k\to\infty}x_k=x.$$ The construction (definition) of the sequence $(\eta_k)_{k\in \Bbb N}$  is sometimes called inductive but is more properly called a recursive construction.

If you want pedantic precision, append a sub-subscript "$x$" to each $\eta_k$.


问:
Coordinate vectors relative to an orthonormal basis
Let's say that we have $B$ = $(u$$1$$, ..., u$$n$$)$, which is an ordered orthonormal basis for $\mathbb{R}^n$. We then have an $n \times n$ matrix $U$ = $(u$$1$$, ..., u$$n$$)$. Then, for any $v$ in $\mathbb{R}^n$, would its coordinate vector relative to $B$ be $U^Tv$? (basically, would $v$$B$ = $U^Tv$?)

答:
Welcome to MSE!

If $\vec{v} \in \mathbb{R}^n$,

let $\vec{v}= c_1 \cdot \vec{u_1} +\cdots +c_n \cdot \vec{u_n} = U\begin{bmatrix}c_1 \\ \vdots \\ c_n\end{bmatrix} = U\cdot\vec{v_B}$

$\vec{v_B} = U^{-1}\cdot \vec{v}$

but $B$ is an orthonormal basis, $U^{-1} = U^T$, so yes.

You can see that $U^{-1} =U^T$ by

$$U^T \cdot U=\begin{bmatrix}\vec{u_1} \\ \vdots \\ \vec{u_n} \end{bmatrix} \begin{bmatrix}\vec{u_1} |& \vec{u_2}|& \cdots & |\vec{u_n}\end{bmatrix} = \begin{bmatrix} (\vec{u_1}\cdot\vec{u_1})& (\vec{u_1}\cdot\vec{u_2})&\cdots &( \vec{u_1}\cdot\vec{u_n} )\\(\vec{u_2}\cdot\vec{u_1})& (\vec{u_2}\cdot\vec{u_2})&\cdots &( \vec{u_2}\cdot\vec{u_n}) \\ \vdots& \vdots&\ddots & \vdots \\  (\vec{u_n}\cdot\vec{u_1})& (\vec{u_n}\cdot\vec{u_2})&\cdots & (\vec{u_n}\cdot\vec{u_n})\end{bmatrix}_{n\times n}$$
$$= I_{n\times n}$$

because by definition $\vec{u_i}\cdot\vec{u_j} =1$ if $i=j$, and $\vec{u_i}\cdot\vec{u_j} =0$ if $i\not=j$.


问:
Relation between diffeomorphism and immersion
Is it always true that a local diffeomorphism is an immersion? I was thinking about the specific case of the equivalence map:

$$ \rho : (x,y,z) \longrightarrow [(x,y,z)]$$

that goes from $S^2$ to $P(\mathbb{R^2})$, the latter being the projective space for $n=2$. We know that this is a local diffeomorphism and that the sphere is compact but does that allow us to conclude that it is also an immersion? Or is it not an immersion?

答:
This is trivial with certain definitions of "immersion" and "local diffeomorphism". Let $f$ be a smooth map between manifolds.

What it means for $f$ to be an immersion is that $df_x$ is injective for all $x$.

What it means for $f$ to be a local diffeomorphism is that $df_x$ is bijective for all $x$.

Thus, a local diffeomorphism is always an immersion.

Another definition of "local diffeomorphism" that you may have seen is "for all points $x$, there is an open neighborhood $U$ of $x$ such that $f|_U$ is a diffeomorphism onto its image". The inverse function theorem says precisely that this is equivalent to the definition of "local diffeomorphism" given above.


问:
Under which operations are singular values invariant?
Take a matrix A and construct B by a permutation of the columns of A. Apparently A and B have the same singular values, but I can't prove it for non 2x2 matrices.

Are there any other operations on matrices that preserve the singular values?

答:
Multiplying $A$ by an orthogonal matrix will necessarily preserve its singular values.

Here's an argument that proves this for square matrices. The singular values of $A$ are equal to the square roots eigenvalues of $A^TA$, which are equal to the square roots of the eigenvalues of $AA^T$. If we multiply $A$ by the left by an orthogonal matrix $U$, then we find that
$$
(UA)^T(UA) = A^TU^TUA = A^TA.
$$
Thus, the singular values of $UA$ are the square roots of the eigenvalues of $A^TA$, which are the singular values of $A$.  Similarly, if we multiply $A$ from the right by an orthogonal matrix $U$, we have
$$
(AU)(AU)^T = AUU^TA^T = AA^T.
$$
So, the singular values of $AU$ are the square roots of the eigenvalues of $AA^T$, which are the singular values of $A$.

If $U$ is a permutation matrix, then $AU$ is a matrix constructed by permuting the columns of $A$ and $UA$ is a matrix constructed by permuting the rows of $A$.


问:
What is the autocovariance of $y_t = \exp{(x_t)}$ where $x_t$ is a stationary normal process.

Let $x_t$ be a stationary normal process with mean $\mu_x$ and
autocovariance function $\gamma(h)$. Define the nonlinear time series
$y_t=\exp{(x_t)}$.

(a) Express the mean function $E(y_t)$ in terms of $\mu_x$ and
$\gamma(0)$. The moment generating function of a normal random
variable $x$ with mean $\mu$ and variance $\sigma^2$ is

$$M_x(\lambda)=E[\exp{(\lambda x)}]=\exp{(\mu \lambda +
> \frac{1}{2}\sigma^2 \lambda^2})$$

(b) Determine the autocovariance function of $y_t$. The sum of the two
normal randomvariables $x_{t+h}+x_t$ is still a normal random
variable.

EDIT:
I figured out the first part, but still struggling with part b.

First we not the MGF for a normal rv $x$ with mean $\mu_x$ and variance $\sigma_x^2$ is
$$E[\exp(xt)] = \exp\left[\mu_x t + \frac{\sigma_x^2 t^2}{2}\right]$$

so that $E[y_t] = E[\exp(x_t)]$ is the same as the normal MGF evaluated at $t=1$, which implies that the mean function of $y_t$ is

$$E[y_t] = E[\exp(x_t)]= \exp\left[\mu_x + \frac{\gamma_x(0)}{2}\right]$$

EDIT 5:  I think I finally see where the dependence of $x_t$ plays in.  How is this?

Note that since $x_t$ and $x_{t+h}$ are identically distributed normal but not independent then the mean and variance of $x_t+x_{t+h}$ are

$$\mu_{x_t+x_{t+h}} = 2\mu_x$$

$$\sigma^2_{x_t+x_{t+h}} = \gamma_x(0) + \gamma_x(0) + 2\rho_x(h)\gamma_x(0) = 2\gamma_x(0)(1+\rho_x(h))$$

Thus the MGF for $x_t+x_{t+h}$ is

$$M_{x_t+x_{t+h}}(\lambda) = E[\exp(\lambda(x_t+x_{t+h}))] = \exp\left[2\mu_x \lambda + \gamma_x(0)(1+\rho_x(h))\lambda^2]\right]$$

By the same argument above using the MGF evaluated at $\lambda = 1$ we have

$$E[y_ty_{t+h}] = E[\exp(x_t+x_{t+h})] = M_{x_t+x_{t+h}}(1) = \exp\left[2\mu_x + \gamma_x(0)(1+\rho_x(h))]\right]$$
And we can calculate the autocovariance:

$$\gamma_y(h) = E[y_ty_{t+h}] - \mu_y^2 = \exp[2\mu_x + \gamma_x(0) + \gamma_x(0)\rho_x(h)] - \exp[2\mu_x +\gamma_x(0)]$$

$$=\exp[2\mu_x + \gamma_x(0)]\exp[\gamma_x(0)\rho_x(h)] - \exp[2\mu_x + \gamma_x(0)]$$

$$=\exp[2\mu_x + \gamma_x(0)]\left(\exp[\gamma_x(0)\rho_x(h)] - 1\right)$$

答:
I figured it out so I'll answer my own question.

First we note the MGF for a normal rv $x$ with mean $\mu_x$ and variance $\sigma_x^2$ is
$$E[\exp(x\lambda)] = \exp\left[\mu_x \lambda + \frac{\sigma_x^2 \lambda^2}{2}\right]$$

so that $E[y_t] = E[\exp(x_t)]$ is the same as the normal MGF evaluated at $\lambda=1$, which implies that the mean function of $y_t$ is

$$E[y_t] = E[\exp(x_t)]= \exp\left[\mu_x + \frac{\gamma_x(0)}{2}\right]$$

Note that since $x_t$ and $x_{t+h}$ are identically distributed normal but not independent then the mean and variance of $x_t+x_{t+h}$ are

$$\mu_{x_t+x_{t+h}} = 2\mu_x$$

$$\sigma^2_{x_t+x_{t+h}} = \gamma_x(0) + \gamma_x(0) + 2\rho_x(h)\gamma_x(0) = 2\gamma_x(0)(1+\rho_x(h))$$

Thus the MGF for $x_t+x_{t+h}$ is

$$M_{x_t+x_{t+h}}(\lambda) = E[\exp(\lambda(x_t+x_{t+h}))] = \exp\left[2\mu_x \lambda + \gamma_x(0)(1+\rho_x(h))\lambda^2]\right]$$

By the same argument above using the MGF evaluated at $\lambda = 1$ we have

$$E[y_ty_{t+h}] = E[\exp(x_t+x_{t+h})] = M_{x_t+x_{t+h}}(1) = \exp\left[2\mu_x + \gamma_x(0)(1+\rho_x(h))]\right]$$
And we can calculate the autocovariance:

$$\gamma_y(h) = E[y_ty_{t+h}] - \mu_y^2 = \exp[2\mu_x + \gamma_x(0) + \gamma_x(0)\rho_x(h)] - \exp[2\mu_x +\gamma_x(0)]$$

$$=\exp[2\mu_x + \gamma_x(0)]\exp[\gamma_x(0)\rho_x(h)] - \exp[2\mu_x + \gamma_x(0)]$$

$$=\exp[2\mu_x + \gamma_x(0)]\left(\exp[\gamma_x(0)\rho_x(h)] - 1\right)$$


问:
Trouble understanding formula regarding random walks
Suppose we start with $n$ dollars and make a sequence of bets. For each bet, we win $1$
dollar with probability $p$ and lose $1$ dollar with probability $1−p$. We quit if either we go
broke, in which case we lose, or when we reach $n+m$ dollars, that is, when we win
$m$ dollars. From what I understand the probability we reach $n+m$ dollars before we go broke, given that we start with $n$ dollars, denoted by $P_n$, is given by

$$P_n = \Big( \frac{p}{1-p} \Big) ^m $$

However, $P_n > 1$ whenever $p > 0.5$. In other words, $P_n$ is not bounded by the interval $[0,1]$ as it should be. So how is this answer correct?

paper for reference: https://web.mit.edu/neboat/Public/6.042/randomwalks.pdf

答:
The exact formula is
$$
   P_n = \frac{\alpha^n - 1}{\alpha^{m+n}-1}
$$
where $\alpha = \frac{1-p}{p}$. When $\alpha > 1$ (so $p < 0.5$) and $n,m+n$ are large, $\alpha^n$ and $\alpha^{m+n}$ are much bigger than $1$, so $P_n \approx \frac{\alpha^n}{\alpha^{m+n}} = \alpha^{-m} = (\frac{p}{1-p})^m$, but this is only an approximation.

In general, the formula for $P_k$ must have the form $C_1 \alpha^k + C_2$ in order to satisfy the recurrence $P_k = p P_{k+1} + (1-p) P_{k+1}$, and satisfy the boundary conditions $P_0 = 0$, $P_{m+n}=1$. The boundary condition $P_0 = 0$ tells us that $C_1 + C_2 = 0$ or $C_2 = -C_1$, so $P_k$ is proportional to $\alpha^k - 1$. In order to get $1$ when $k=m+n$, the constant $C_1$ must be $\frac1{\alpha^{m+n}-1}$, giving us $P_k = \frac{\alpha^k-1}{\alpha^{m+n}-1}$.


问:
What should I learn next for pure math?
I can do basic single variable calculus which is essentially all you do at A Level in the UK. I also just read "What is mathematics?" by Richard Courant which I found very good. I would like to know where to go after I finish my proof writing and basic discrete maths books I am currently reading. Should I study real analysis? What book? Linear algebra?

I would appreciate advice on where to go next and what book would be good. I am currently 14 and my end goal would to become a pure mathematician.

The reason I ask this question is because I was trying to create a structure and I thought about the Gerard t'Hooft physics one or even the Pure mathematician or statiscian plan but they don't seem particularly specific in what you are meant to do where i.e they call a topic calculus and then a later one vector calculus and then analysis but as a pure mathematician I want to go straight into analysis, of reals obviously. In conclusion, I am asking this because I am going to go by steps now and choose the next topic as it comes.

Thank you in advance.

EDIT:

Would this work:

In order...

Elementary Discrete Maths,
Real Analysis,
Linear Algebra,
ODE's,
Probability,
Fourier Analysis,
Complex Analysis,
PDE's,
Graduate stuff which I will get to when needed.

I'm not sure if probability is necessary but I think it'll be interesting.

答:
The standard outline in many schools after single variable calculus is linear algebra, multivariable calculus, and differential equations, followed by several core classes, which often include analysis and abstract algebra but can include complex analysis (which does not have real analysis as a prerequisite necessarily), PDE's, numerical methods, etc. After that, people tend to take several electives, but in graduate school, you start over, often with topology, real analysis/measure theory, and abstract algebra again.

You can skip multivariable calculus and go into real analysis if you really want to, but multivariable calculus is beautiful, rigorous, fascinating, challenging, and studied and developed by Euler, Gauss, etc. Stewart Calculus has a great description of vector valued functions with the tangent, normal, and binormal, etc.

Analysis is not just more advanced calculus; it's a different emphasis, and tends to be less geometric than multivariable calculus.

This is all written from a limited perspective in the U.S.

Title: What should I learn next for pure math?


问:
Prove if $\lim_{k \rightarrow \infty} A_k = A$, then $\lim_{k \rightarrow \infty} A_k^{-1} = A^{-1}$
Let $\{A_k\}_{k=1}^{\infty}$ be a sequence of real $n \times n$ matrices. Suppose $A_k$ is invertible for all $k \in \mathbb{N}$, that $\lim_{k \rightarrow \infty} A_k = A$, and that $A$ is invertible. Is $\lim_{k \rightarrow \infty} A_k^{-1} = A^{-1}$?

This feels like it should be true but I'm having trouble coming up with a proof.

Also, I think defining $\lim_{k \rightarrow \infty} A_k = A$ as componentwise convergence makes the most sense for my problem but any other insights are welcome!

答:
Your question is asking whether $\Phi : GL(n, \mathbb{R}) \to GL(n, \mathbb{R})$ defined by
$$\Phi(A) = A^{-1}$$
is continuous. This is true. In fact, $\Phi$ is differentiable. Let $||\cdot||$ denote the operator norm on $M(n, \mathbb{R})$. A geometric series type argument yields that for $H \in M(n, \mathbb{R})$ with $||H|| < 1$,
\begin{align}
(I + H)^{-1} &= \sum_{k = 0}^{\infty}(-1)^kH^k \\
&= I - H + H^2 - \dots \\
&= I - H + O(||H||^2).
\end{align}
Now let $A \in GL(n, \mathbb{R})$ be arbitrary. For $H \in M(n, \mathbb{R})$ we have
$$(A + H) = A(I + A^{-1}H)$$
Hence if $||A^{-1}H|| < 1$, then $A + H$ is invertible. Since $||A^{-1}H|| \leq ||A^{-1}|| \cdot ||H||$, $A + H$ is invertible whenever $||H|| < \frac{1}{||A^{-1}||}$ and
\begin{align}
(A + H)^{-1} &= (A(I + A^{-1}H))^{-1} \\
&= (I + A^{-1}H)^{-1}A^{-1} \\
&= (I - A^{-1}H + O(||H||^2))A^{-1} \\
&= A^{-1} - A^{-1}HA^{-1} + O(||H||^2).
\end{align}
Hence $D\Phi(A)H = -A^{-1}HA^{-1}$.


问:
Trying to figure out the asymptotic density
Let $a>0$.

I have the following function $f: \mathbb{N} \to \mathbb{R} $ defined in the following way:

\begin{equation}\label{km relation}
 f(m) = m +\lfloor a  m -  \tfrac{a }{2} +\tfrac12 \rfloor.
\end{equation}

Now, define a `counting function' $A: \mathbb{R} \to \mathbb{R}$ by
\begin{equation}\label{}
 A(x) = \sum_{\substack{a \in f(\mathbb{N}) \\ 1 \le a \le x}} 1.
\end{equation}

where $f(\mathbb{N})$ is the image of $\mathbb{N}$ under $f$.

What  is $\lim_{x \to \infty} \tfrac{A(x)}{x}$?

I think that it should be $\frac{1}{a+1}$, but since $a$ can be irrational, I can't seem to quite prove it- I get stuck.

答:
Show that
$$\lim_{n\to \infty} \frac{f(n)}n=a+1,\qquad \lim_{x\to \infty} \frac{f(A(x))}{x}=1$$

$$\lim_{x\to \infty} \frac{f(A(x))}{x}=\lim_{x\to \infty} \frac{f(A(x))}{A(x)}\frac{A(x)}{x}=\lim_{x\to \infty} (a+1)\frac{A(x)}{x}$$


问:
For dual numbers, what does $\epsilon^\epsilon$ equal?
$$\epsilon^\epsilon=?$$
Where $\epsilon^2=0$, $\epsilon\notin\mathbb R$.
There is a formula for exponentiation of dual numbers, namely:
$$(a+b\epsilon)^{c+d\epsilon}=a^c+\epsilon(bca^{c-1}+da^c\ln a)$$
However, this formula breaks down in multiple places for $\epsilon^\epsilon$, yielding many undefined expressions like $0^0$ and $\ln 0$. So, here's my question: what is $\epsilon^\epsilon$ equal to for dual numbers?</p>

答:
Since $f(\varepsilon)=f(0)+f'(0)\varepsilon$,

formally we have

$\varepsilon^\varepsilon=1+\varepsilon+\varepsilon\ln 0$

The quantity $\ln 0$ is infinite with negative sign. We could stop here, but if you want you can write it as a divergent integral:

$\varepsilon^\varepsilon=1+\varepsilon-\varepsilon\int_0^1\frac{dx}x=1+\varepsilon-\varepsilon\gamma -\varepsilon\int_1^\infty\frac{dx}x$

I have no ideas where we can advance further.


问:
In how many ways can a pair be distributed over a set
Suppose a set $S$ of size $n$.
Let $P$ be all pairs of $S$ such that an element can only be paired with $k$ other elements. Hence, the size of $P$ is $\frac{nk}{2}$.

The question is: how many subsets of $S$, of size $g$, contains at least one element of $P$?

At first I tried
$$ \frac{\frac{nk}{2} \binom{n-2}{g-2}}{\binom{g}{2}} $$

And my reasoning was: I select a pair of $P$ and combine it with all subsets of $S$ of size $g-2$, that does not contains any of the two elements already selected, to create subset of size $g$. Then I divide by $\binom{g}{2}$ because that's the number of permutations of the pair in the created subset.

However I figured out it's wrong because $\binom{g}{2}$ does not represent the correct number of permutations of the pair, and now I don't really know how I should calculate this number.

Any help is appreciated. Thank you in advance

答:
There is not enough information to answer the question. For example, suppose that $n=6$, $k=2$, $g=3$. Then:

One possible choice of $P$ is $\{\{1,2\}, \{1,3\}, \{2,3\}, \{4,5\}, \{4,6\}, \{5,6\}\}$. In this case, out of the $\binom 63 = 20$ subsets $A \subseteq S$ of size $3$, all $20$ must contain an element of $P$: any $A$ contains either at least two of $\{1,2,3\}$ (and contains one of the first three elements of $P$) or at least two of $\{4,5,6\}$ (and contains one of the last three elements of $P$).

Another possible choice of $P$ is $\{\{1,2\}, \{2,3\}, \{3,4\},\{4,5\}, \{5,6\},\{1,6\}\}$. In this case, out of the $\binom 63 = 20$ subsets $A \subseteq S$ of size $3$, only $18$ contain an element of $P$: the exceptions are $\{1,3,5\}$ and $\{2,4,6\}$.

The answer depends on the structure of $P$, not just the size of $P$, or even on how many pairs contain each element.


问:
Proving that if $a^n=b^n$, then $a=b$, for positive $a$ and $b$, and natural $n$
In high school algebra, natural number exponents are defined as

$$\begin{aligned}a^1 &= a\\
a^{(n + 1)} &= a^n a\end{aligned}$$

With these I used induction to prove the 3 exponent laws. I also proved some ancillary theorems such as the nth power of a positive is positive, distribution of powers over fractions, and so on. Then I got to this statement:

For positive $a$, $b$ and natural number $n$,

$$\text{if }a^n = b^n\text{ then } a = b\text{.}$$

I have a proof, but it is very different from any of the other proofs. It requires $\Sigma$ notation, which none of the others require. However, I can't figure out how to do it without $\Sigma$. Is it possible?

Here's my proof:

$$a^n = b^n\\
  a^n - b^n = 0\\
  (a - b) \left(\sum_{k=1}^n \frac{a^n b^k}{a^k b}\right) = 0\\
  a = b  \quad\text{or}\quad \sum_{k=1}^n \frac{a^n b^k}{a^k b} = 0$$

As the summation has at least $1$ term and every term is positive, the summation must be positive, which is not equal to $0$. This leaves $a = b$.

Q.E.D.

答:
Prove the contrapositive: if $a\ne b$, $a^n\ne b^n$.

If $a\ne b$ then without loss of generality $a>b$. Now prove by induction $a^n>b^n$. The inductive step boils down to $a^{k+1}>a^kb>b^{k+1}$.


答:
Here's another approach.

Lemma If $a$ and $b$ are real numbers such that $0 < a < b$, and $n$ is a positive integer, then $0 < a^n < b^n$.

You should try to prove this by induction!

Claim If $a$ and $b$ are positive real numbers and $n$ is a positive integer such that $a^n = b^n$, then $a = b$.

Proof. Let $a$ and $b$ be positive real numbers, and let $n$ be a positive integer. We proceed by contrapositive: suppose $a \neq b$. Then $0 < a < b$ or $a > b > 0$. By the lemma, this yields $0 < a^n < b^n$ or $a^n > b^n > 0$. In both cases, $a^n \neq b^n$.

Title: Proving that if $a^n=b^n$, then $a=b$, for positive $a$ and $b$, and natural $n$

Body:
In high school algebra, natural number exponents are defined as

$$\begin{aligned}a^1 &= a\\
a^{(n + 1)} &= a^n a\end{aligned}$$

With these I used induction to prove the 3 exponent laws. I also proved some ancillary theorems such as the nth power of a positive is positive, distribution of powers over fractions, and so on. Then I got to this statement:

For positive $a$, $b$ and natural number $n$,

$$\text{if }a^n = b^n\text{ then } a = b\text{.}$$

I have a proof, but it is very different from any of the other proofs. It requires $\Sigma$ notation, which none of the others require. However, I can't figure out how to do it without $\Sigma$. Is it possible?

Here's my proof:

$$a^n = b^n\\
  a^n - b^n = 0\\
  (a - b) \left(\sum_{k=1}^n \frac{a^n b^k}{a^k b}\right) = 0\\
  a = b  \quad\text{or}\quad \sum_{k=1}^n \frac{a^n b^k}{a^k b} = 0$$

As the summation has at least $1$ term and every term is positive, the summation must be positive, which is not equal to $0$. This leaves $a = b$.

Q.E.D.


问:
imaginary numbers - how can I understand them - especially as they occur in 'roots' of polynomials?
In another question here, about roots of equations being imaginary, the accepted answer
said something interesting about "imaginary" (as a technical word in math) not meaning "not real".  I understand that imaginary numbers are "manipulable and usable to do all kinds of things".   And thus, "they exist" at least, in the realm of mathematics.

So my question is this. I was playing around in wxMaxima, and solving polynomials, to get their roots.  In maxima, $\sqrt{-1}$ is represented as %i.

Editor's note: all "%i"s have been replaced with $i$.

Here's an example of a polynomial, say $3x^6 + 4x^4$:

solve($3x^6 + 4x^4$,$x$);

Maxima's results:
[$x=\displaystyle -\frac{2i}{\sqrt{3}}$,$x=\displaystyle \frac{2i}{\sqrt{3}}$,$x=0$]

My exact question:

Is it common to have the same imaginary result twice with only the sign different (both q and -q are imaginary roots)?  (At first I thought it had reported the same root twice as I overlooked the negative sign.)

One can easily plot a real root and see how it intercepts the x axis.  How does one "visualize" or think of an imaginary root -- does it correspond to a local minimum that does not occur at the x axis?

What uses do imaginary roots of polynomials have in math?

答:
See this question for parts 2 and 3.

As for part 1, if you have a polynomial with real coefficients then for each complex root $x + iy$, its conjugate $x - iy$ is also a root.

The reason is that conjugation "commutes" with all the arithmetic operation, i.e.
$$ \overline{z+w} = \overline{z} + \overline{w}, \quad \overline{zw} = \overline{z} \overline{w}. $$
Here the line refers to conjugation. Since all coefficients are real, if you plug in your root and conjugate the entire expression, you end up with $\overline{0} = 0$ again, with the same polynomial, but with the conjugate root. So these roots come in pairs.


问:
Local fields and infinite extensions, basic questions
Notation throughout: Let $K$ be a discrete valuation field and $L/K$ an infinite (not necessarily Galois) extension of $K$.

1) How can/does one define a ramification index $e(L/K)$ for $L/K$? It seems that in order to first define an inertia group for $L/K$ one needs that $K$ be at least henselian and that $L/K$ be Galois (or else take a Galois closure in $L$); assuming these hypothesis, what is the definition of the ramification index $e(L/K)$? - is finiteness easy to see? aside: if $\pi_K$ is a uniformizer of $K$, (when) can we explicitly write down the uniformizer of $L$ in terms of $\pi_K$?

2) Separate question: let $\pi$ be a uniformizer of $K$ (not-necessarily henselian). We know that $1+\pi$ is a unit of $K$; how can one explicitly write down the inverse of $1+\pi$ in $K$? (we can't 'expand' as $K$ is not assumed, e.g., to be complete)

答:
1) You cannot, in general. The valuation on $L$ need no longer be discrete. For example, consider the extension $\mathbf Q_p(\zeta_{p^\infty})$ obtained by adding all $p$-power roots of $p$ to $\mathbf Q_p$. The only reasonable value for the ramification index of this extension is $e=+\infty$. As for writing down a uniformizer of $L$ in terms of $\pi_K$ : once again, this makes sense only if the valuation on $L$ is discrete. Assuming that it is, one way to do this would be to look at a finite sub-extension $L'/K$ which has the same ramification as the infinite extension $L$, and to find a uniformizer there (assuming you can solve the problem for a finite extension). (However, it is possible to define the ramification group of an infinite Galois extension.)

2) What kind of explicit expression are you looking for? Suppose $K = \mathbf C(z)$ with the discrete valuation at the origin. Then a uniformizer is $\pi = z$, and the inverse of $1+z$ is... $\frac{1}{1+z}$! How would you like to write $\frac{1}{1+z}$ more explicitly than that?


问:
For what values does the geothmetic meandian converge?
The geothmetic meandian, $G_{MDN}$ is defined in this XKCD as

$$F(x_1, x_2, ..., x_n) = \left(\frac{x_1 +x_2+\cdots+x_n}{n}, \sqrt[n]{x_1 x_2 \cdots x_n}, x_{\frac{n+1}{2}} \right)$$

$$G_{MDN}(x_1, x_2, \ldots, x_n) = F(F(F(\ldots F(x_1, x_2, \ldots, x_n)\ldots)))$$

The comic also (correctly) claims that $G_{MDN}(1, 1, 2, 3, 5) \approx 2.089$

There are two convergence questions I'm interested in:

For what values of $(x_1, x_2, \ldots, x_n)$ does $G_{MDN}$ converge to a single number?

For what values of $(x_1, x_2, \ldots, x_n)$ does $G_{MDN}$ converge, but not to a single number?

I've written up Python 3 code so that you can test out numbers yourself by changing the values at the bottom of the code. If the code never seems to stop running, then $G_{MDN}$ does not converge for your input. (In these situations, you can set verbose=True to see what's happening.)

```<code># assumes Python 3 because I assume that "/" always means float division
from typing import Iterable, Tuple, Union
from decimal import Decimal
import math
from functools import reduce  # Required in Python 3
import operator

# from https://stackoverflow.com/a/48648756
def prod(iterable):
    return reduce(operator.mul, iterable, 1)

def geothmetic_meandian(nums: Iterable[float], verbose=False) -> Tuple[bool, Tuple[float, float, float]]:
    def inner_geothmetic_meandian(nums: Iterable[float]) -> Tuple[float]:
        arithmetic_mean = sum(nums)/len(nums)
        geometric_mean = prod(nums)**(1 / len(nums))
        sorted_nums = sorted(list(nums))
        if len(nums) % 2 == 0:
            # even number of numbers
            higher_median_index = int(len(nums) / 2)
            lower_median_index = higher_median_index - 1
            median = (sorted_nums[higher_median_index] + sorted_nums[lower_median_index]) / 2
        else:
            # odd number of numbers
            median = sorted_nums[int((len(nums) - 1) / 2)]

        return (arithmetic_mean, geometric_mean, median)

    last_ans = None
    ans = inner_geothmetic_meandian(nums)
    converged = True
    while not (ans[0] == ans[1] == ans[2]):
        if ans == last_ans:
            converged = False
            break
        last_ans = ans
        ans = inner_geothmetic_meandian(ans)

        if verbose:
            print(ans)

    return converged, ans

if __name__ == "__main__":
    verbose = False
    values = (1, 1, 3, 2, 5)
    converged, results = geothmetic_meandian(values, verbose=verbose)
    if converged:
        print(f"The geothmetic meandian of {values} converged to: {results[0]}")
    else:
        print(f"The geothmetic meandian of {values} did not converge to a single value:\nArithmetic Mean: {results[0]}\nGeometric Mean:  {results[1]}\nMedian:          {results[2]}")
</code>```

I've tested this code to verify that $G_{MDN}(1, 1, 2, 3, 5) \approx 2.089$.

I have also not found any inputs that cause the program to not converge at all.

However, I have found that $G_{MDN}(1, 2, 3, 4, 5) = (2.8993696858822964, 2.899369685882296, 2.8993696858822964)$ which would mean that $(1, 2, 3, 4, 5)$ is in the second convergence class, where it converged, just not all to the same number. But I'm worried that this result is due to a rounding error in Python itself. (I quickly tried and failed to use the Decimal class due to the nth root operation.)

Thus, my hunch is that all inputs converge to a single number, but I have not been able to prove this yet. It looks like an epsilon-delta proof.

答:
Let $a_n,g_n$ and $m_n$ denote the three means computed at stage $n$, for $n\in \{1,2,3,\dots\}$. Let $$\begin{align}l_n&=\min(a_n,g_n,m_n)\\u_n&=\max(a_n,g_n,m_n)\end{align}$$ Since $u_n$ is a non-increasing sequence, and $l_n$ is a non-decreasing sequence, in order to prove convergence, it suffices to prove that $u_n-l_n\to 0$. We do this by showing $(u_{n+2}-l_{n+2})\le \frac79 (u_n-l_n).$

Note that $$a_{n+1}=\tfrac13(a_n+g_n+m_n)\le \tfrac{2}3u_n+\tfrac13 l_n,$$which follows by replacing the smallest of $\{a_n,g_n,m_n\}$ with $l_n$, and upper bounding the other two with $u_n$. Then, by the AM-GM inequality, $g_{n+1}\le \frac{2}3u_n+\frac13 l_n$ as well. The last two inequalities imply
$$
m_{n+2}\le \tfrac23u_n+\tfrac13l_n.
$$
We have given an upper bound for one of the means in the $(n+2)^{nd}$ stage. For the other two, note
$$
\begin{align}
g_{n+2}\le a_{n+2}
&=\frac{a_{n+1}+g_{n+1}+m_{n+1}}3
\\
&\le \frac{2a_{n+1}+m_{n+1}}3\tag{AM-GM}
\\
&=\frac{2a_n+2g_n+2m_n+3m_{n+1}}{9}
\\
&\le\frac{2a_n+2g_n+2m_n+3u_{n}}{9}\tag{$m_{n+1}\le u_n$}
\\
&\le\frac{7u_n+2l_{n}}{9}
\end{align}
$$
For the last step, we are replacing the smallest of $a_n,g_n,m_n$ with $l_n$, and upper bounding the other two with $u_n$.

We have shown that all three of $m_{n+2},g_{n+2}$, and $a_{n+2}$ are at most $\tfrac79u_n+\tfrac29u_n$, so we conclude that
$$
u_{n+2}\le \tfrac79u_n+\tfrac29l_n$$ which implies $$ u_{n+2}-l_{n+2}\le u_{n+2}-l_n\le \tfrac79(u_n-l_n)
$$

Note that this shows that $(u_n-l_n)\in O\left(c^n\right)$, where $c=\frac{\sqrt7}3$. It is actually possible to prove that for any $\epsilon>0$ that
$$
u_n-l_n\in O\left(\frac1{3^n} \right).
$$


问:
Why a point $P$ that maximizes $\angle OPQ$ must be a point of tangency of a circle with chord $OQ$?
I've seen the following problem with solution:

Can someone explain:

Why a point $P$ that maximizes $\angle OPQ$ must be a point of tangency of a circle with chord $OQ$?

How do we construct this circle? It seems he takes a circle with center $\left(\frac{1}{2}, k\right)$ but it's not clear to me where this chord $OQ$ must be constructed or where this circle must be constructed so that $OQ$ is a chord.

答:
You can prove the tangency by bashing easily which I assume is something you are more than capable of. But the geometric argument would b to argue by contradiction. Suppose $P$ on the line $l$ gives the maximum angle for $\angle OPQ.$ If $l$ is not tangent to the circumcircle of $\triangle OPQ$ at $P,$ then there is another intersection $P'$ and it is very easy to see that for any $P''$ on this segment $P'P$ gives a larger value for $\angle OP''Q,$ which is a contradiction.

I do not know what you are asking here. $OQ$ is fixed and the author is finding the unique solution for $P$ by simply coordinate-bashing. Do you mean how to construct this $P$ by compass and straightedge? Even if you did, you can still reverse-engineer this construction from the algebraic solution with ease.


问:
When is $(dy/dx)^{-1}=dx/dy$ for multivariable functions?
In general for a multivariable functions $y(m,n)$, $x(m,n)$,
$$\frac{dy}{dx}=\frac{\partial y}{\partial m}\frac{dm}{dx}+\frac{\partial y}{\partial n}\frac{dn}{dx}\tag{1}$$
and
$$\frac{dx}{dy}=\frac{\partial x}{\partial m}\frac{dm}{dy}+\frac{\partial x}{\partial n}\frac{dn}{dy}\tag{2}$$

so in general clearly $(dy/dx)^{-1}$ need not be equal to $dx/dy$.

My question is what is the precision Mathematical conditions on functions $y(m,n)$ and $x(m,n)$ under which this equality holds.

Clearly I could explicitly calculate $dy/dx$ and $dx/dy$ and check the equality, but is there a better way to see this equality from the point of view of the properties of the functions. Here, by functions I mean functions that commonly appear in physics (and definitely not some exotic function that has no physical origins - this is the reason I decided to post it in PhysicsSE and not MathSE).

If there exist any such properties, I would also like to have a mathematical proof that states that these properties are equivalent to proving the equality explicitly from the first principle i.e. using (1) and (2).

Edit: I have been notified of a possible ill definition of these particular chain rules, which would mean the question itself is invalid and I very likely might have misinterpreted something. To verify this possible ill definition I have posed a question in MathSE.

答:
You're allowed to do this when you're working with functions of a single variable.

If you check, you'll find that in all examples in physics where $dy / dx = (dx / dy)^{-1}$ is used, this is the case. Examples include the trajectory of a single particle, which can be parametrized as $x(t)$, and the potential in a problem with spherical symmetry, which can be parametrized as $V(r)$.

OwnerDisplayName: TheImperfectCrazy

OwnerDisplayName: knzhou


答:
Its only true when $y$ is a single variable function of $x$, so
$$
y = y(x).
$$
If $y$ is a function of multiple variables, say $y = y(m,n)$, then its derivative is really a vector.
$$
\nabla y = \left( \frac{\partial y}{\partial m}, \frac{\partial y}{\partial n}\right).
$$
Furthermore, the derivative of $x$ is also a vector, $\nabla x$.

Now, note that you can't even divide vectors even if you wanted to. In general, the expression
$$
\frac{\nabla y}{\nabla x}
$$
just makes no sense. So then, in the multivariable case, the expression
$$
\frac{dy}{dx}
$$
likewise makes no sense in general.

However, there is a special case in which it does make sense. That is when $y$ really can be written as a function of $x$, i.e.
$$
y(m,n) = f( x(m, n) )
$$
for some single variable function $f$. In this special case, $\nabla y$ is actually parallel to $\nabla x$, so it sort of does make sense to take the quotient $\nabla y / \nabla x$. And also, because $y = f(x)$, $y$ really is in a way just a single variable function of $x$, and so your identity will hold only in this case.

Also, as written now, your equations $(1)$ and $(2)$ are simply incorrect. $m$ and $n$ are not functions of $x$, so the expression $\partial m / \partial x$ doesn't make much sense. It's an incorrect application of the chain rule. Therefore you won't be able to prove anything from those starting points. To reiterate, the expression $dy/dx$ just has no meaning when $y$ and $x$ are general functions of $m$ and $n$, simply because a value of $x$ does not determine a unique value of $y$.

Edit:
As for your equations $(1)$ and $(2)$, here is the "meaning" of those equations. Each pair of functions of two variables defines a Jacobian which is a $2 \times 2$ matrix. So for the functions $(y(m,n), x(m,n))$ we have the Jacobian
$$
J_{x(m,n), y(m,n)} = \begin{pmatrix} \tfrac{\partial x}{\partial m} & \tfrac{\partial x}{\partial n} \\ \tfrac{\partial y}{\partial m} & \tfrac{\partial y}{\partial n} \end{pmatrix}
$$
and for the functions $(m(x,y), n(x,y))$ we have the Jacobian
$$
J_{m(x,y), n(x,y)} = \begin{pmatrix} \tfrac{\partial m}{\partial x} & \tfrac{\partial m}{\partial y} \\ \tfrac{\partial n}{\partial x} & \tfrac{\partial n}{\partial y} \end{pmatrix}.
$$
Because the maps here are inverse functions, one can prove that $J_{x(m,n), y(m,n)} J_{m(x,y), n(x,y)}= I$.

If we want to calculate the Jacobian for the composition of the two maps, which is $(x(m(x,y), n(x,y)), y(m(x,y), n(x,y) )$, the correct formula to use is to multiply the two Jacobians together. However, this is just $I$, which tautologically confirms that
$$
\begin{pmatrix} \tfrac{\partial x}{\partial x} & \tfrac{\partial x}{\partial y} \\ \tfrac{\partial y}{\partial x} & \tfrac{\partial y}{\partial y} \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}.
$$
However, you won't be able to draw any conclusions about the form of $x$ or $y$ from this calculation. All you can do is calculate the identity matrix tautologically.

Title: When is $(dy/dx)^{-1}=dx/dy$ for multivariable functions?

Body:
In general for a multivariable functions $y(m,n)$, $x(m,n)$,
$$\frac{dy}{dx}=\frac{\partial y}{\partial m}\frac{dm}{dx}+\frac{\partial y}{\partial n}\frac{dn}{dx}\tag{1}$$
and
$$\frac{dx}{dy}=\frac{\partial x}{\partial m}\frac{dm}{dy}+\frac{\partial x}{\partial n}\frac{dn}{dy}\tag{2}$$

so in general clearly $(dy/dx)^{-1}$ need not be equal to $dx/dy$.

My question is what is the precision Mathematical conditions on functions $y(m,n)$ and $x(m,n)$ under which this equality holds.

Clearly I could explicitly calculate $dy/dx$ and $dx/dy$ and check the equality, but is there a better way to see this equality from the point of view of the properties of the functions. Here, by functions I mean functions that commonly appear in physics (and definitely not some exotic function that has no physical origins - this is the reason I decided to post it in PhysicsSE and not MathSE).

If there exist any such properties, I would also like to have a mathematical proof that states that these properties are equivalent to proving the equality explicitly from the first principle i.e. using (1) and (2).

Edit: I have been notified of a possible ill definition of these particular chain rules, which would mean the question itself is invalid and I very likely might have misinterpreted something. To verify this possible ill definition I have posed a question in MathSE.


答:
As @knzhou has noted, in case of a function of a single variable
$$
\left[y'(x)\right]^{-1} = \frac{dx}{dy},
$$
since the derivative and the differential of a function of a single variable are equivalent.

This is not the case for the functions of several variables, see, e.g., my answers here and here.

OwnerDisplayName: Vadim


答:
It may not be obvious but you haven't specified all the information yet. Excuse me for changing your notation but define
\begin{align}
x(\vec u):\quad \mathbb R^n\rightarrow\mathbb R\\
y(\vec u):\quad \mathbb R^n\rightarrow\mathbb R
\end{align}
where I redefined $(m,n)$ as an n-dimensional point $\vec u$. Right now when I try to calculate $\frac{dy}{dx}$ I will get stuck because I'm missing information. You are differentiating $y(x)=y(\vec u(x))$ with respect to $x$ but we have never specified $\vec u(x)$. We can't just invert $x(\vec u)\rightarrow\vec u(x)$ because for a given value $x$ there's an infinite number of points $\vec u$ that give the same value $x$. To define a total derivative you implicitly have to specify a path in $\mathbb R^n$ because by definition a total derivative only depends on one variable. Once I have specified $\vec u(x)$ it is obvious that $dy/dx=(dx/dy)^{-1}$ because $y(\vec u(x))$ is just a function of one variable i.e. you can write it as $y(x)$. If $y(x)$ is invertible then
\begin{align}
\frac{dy}{dy}=\frac{d}{dy}y(x(y))&=\frac{dy(x(y))}{dx}\frac{dx(y)}{dy}\\
&=\frac{dy}{dx}\frac{dx}{dy}\overset !=1
\end{align}
So the mathematical conditions would be that $x(\vec u),\vec u(x),y(\vec u),\vec u(y)$ all exist and are continuous. I don't know much about function analysis so maybe there are better conditions.

OwnerDisplayName: AccidentalTaylorExpansion


答: Your idea will work if you get the summations right. As an alternative, you could prove by induction that, for positive $x$ and $y$, if $x > y$, then $x^n > y^n$. Given that, if $a^n = b^n$, then $a < b$ and $b < a$ are both impossible, so you must have $a = b$.


问:
Calculating the integral $\int_0^1 dx_1 dx_2 dx_3 \delta \left( x_1+x_2+x_3-1 \right) (x_1 x_2 x_3)^\alpha(x_2 x_3+x_1x_3+x_1x_2)^\beta.$
I need to calculate the integral

$$\int_0^1 \prod_{i=1}^N dx_i \delta \left( \sum_{j=1}^N x_j-1 \right) (x_1 x_2 \cdots x_N)^\alpha(x_2 x_3 \cdots x_{N} +x_1 x_3 \cdots x_{N} + \text{other terms with one } x_i \text{ absent})^\beta.$$

Here $\alpha$ and $\beta$ are real numbers. For instance, for $N=3$ we would have

$$\int_0^1 dx_1 dx_2 dx_3 \delta \left( x_1+x_2+x_3-1 \right) (x_1 x_2 x_3)^\alpha(x_2 x_3+x_1x_3+x_1x_2)^\beta.$$

Utilising the delta function and writing, say, $x_1$ in terms of $x_2, x_3$ does not help much.

Does anyone have any ideas about how to attack such a problem? If we can calculate the $N=3$ case, then I guess the generalisation will not be that difficult.

答:
Here is a method that produces an explicit evaluation of your integral if $\beta$ is a positive integer.

For simplicity, suppose that $N=3$.  With $x_3 = 1 - x_1 - x_2$, a simple calculation shows that
$$
x_2 x_3 + x_1 x_3 + x_1 x_2 = x_1 + x_2 - x_1 x_2 - x_1^2 - x_2^2.
$$
Raising this expression to the power $\beta$ and inserting it in the integrand, your integral becomes
$$
\int_\Delta x_1^\alpha x_2^\alpha (1-x_1-x_2)^\alpha (x_1 + x_2 - x_1 x_2 - x_1^2 - x_2^2)^\beta \, {\rm{d}}x_1 \, {\rm{d}}x_2,
$$
where the domain of integration is the simplex $\Delta = \{(x_1,x_2): x_1 > 0, x_2 > 0, x_1 + x_2 < 1\}$.

If $\beta$ is a positive integer then the term $(x_1 + x_2 - x_1 x_2 - x_1^2 - x_2^2)^\beta$ can be expanded {\it via} the multinomial theorem as a linear combination of monomials $x_1^j x_2^k$, where $j, k$ are nonnegative integers.  Then your integral is a linear combination of integrals of the form
$$
\int_\Delta x_1^{\alpha+j} x_2^{\alpha+k} (1-x_1-x_2)^\alpha \, {\rm{d}}x_1 \, {\rm{d}}x_2,
$$
and each of these integrals is a well-known \href{https://en.wikipedia.org/wiki/Dirichlet_distribution}{classical Dirichlet integral}.


问:
When does classical logic speed up arithmetic?
Let $HA$ denote the theory of Heyting Arithmetic, referring to the intuitionistic fragment of $PA$. Is there some infinite set of formulas $\Gamma$ such that both $HA \vdash \Gamma$ and $PA \vdash \Gamma$, yet the length of the shortest cut-free $HA$-proof of some formula $\phi \in \Gamma$ grows asymptotically faster than its shortest cut-free proof in $PA$?

In particular, are there sets of only $\Sigma_0$- or $\Sigma_1$-formulas such that this condition holds? Quantifier-free formulas can be ruled out, I think, since $HA$ decides equality on naturals and thus is equivalent to $PA$ on these formulas with only a multiplicative slowdown.

答:
I'm not sure if this is exactly what you want, but there's a fairly short argument along those lines. Given a natural number $n$, you can produce by Gödel style diagonalisation a sentence $\varphi$ that says "$\varphi$ is not provable by a derivation of length less than $n$". I believe this ends up being $\Delta_0$. By the usual diagonal tricks $\varphi$ is provable in $\mathbf{HA}$, but it is also true, and so has no derivation shorter than $n$. Then $\varphi \vee \neg \varphi$ is also provable in $\mathbf{HA}$ by first proving $\varphi$ and then applying $\vee$-introduction. Moreover, in cut free intuitionistic sequent calculus that is the only possible way to derive $\varphi \vee \neg \varphi$. On the other hand, in $\mathbf{PA}$ $\varphi \vee \neg \varphi$ is simply an instance of excluded middle and so has a trivial proof. For example, in the formulation of classical sequent calculus $\mathbf{LK}$ on the wikipedia page this can be done in 4 steps using $(I)$, $(\vee R_1)$, $(\vee R_2)$, and finally the right hand contraction rule, $(CR)$.

You might worried about the length of the formula $\varphi$, since it increases at least linearly with $n$. In that case you can instead consider $\varphi$ stating "$\varphi$ is not provable by a derivation of length less than $F(n)$" where $F$ is any primitive recursive function. That can ensure the shortest possible proof grows faster than the length of $\varphi$. I think this would make $\varphi$ a $\Sigma_1$ sentence.


答:
You could write it explicitly. Suppose $a,b$ are not-negative real numbers and $n$ is a positive natural number. Is easy to show that
\begin{align}
a^n - b^n
= (a-b)( a^{n-1} + a^{n-2}b^{1} + a^{n-3}b^{2} + ... + a^{1}b^{n-2} + b^{n-1})
\end{align}
This expression should be familiar from geometric sums and can be seen by distributing the first factor,  most of the term cancel out. From this, the second factor is not-negative,
$$  a^{n-1} + a^{n-2}b^{1} + a^{n-3}b^{2} + ... + a^{1}b^{n-2} + b^{n-1} \geq 0 \; , $$
since it is a sum of positive terms. So, the same way you proved,
\begin{align}
0 &= a^n - b^n \, \\[5pt] \Rightarrow \;  0&=a-b \\[5pt]
 \Rightarrow \; a & = b \; . \qquad \qquad \blacksquare
\end{align}

Note that you need $n$ to be is nonzero and that $a,b$ to be non-negative. For example $(2)^2=(-2)^2$ but $2 \neq -2$, well as $x^0 = 1 $ for any $x$.


问:
Calculate size of a windowed matrix
I was given the following problem:

Given a matrix $A \in \mathbb{R}^{n \times m}$, calculate the shape of a new matrix $B$ that is a sliding window view of $A \in \mathbb{R}^{48 \times 1536}$ with step $s$.

This is a little confusing by saying, but here's an example:

\begin{array}{|c|c|c|c|}
\hline
 1 & 2 & 3 & 4\\ \hline
 5 & 6 & 7 & 8\\ \hline
 9 & 10 & 11 & 12\\ \hline
 13 & 14 & 15 & 16\\ \hline
\end{array}
$$\bigg\downarrow$$
\begin{array}{|c|c|c|c|}
\hline
 1 & 2 \\ \hline
 2 & 3 \\ \hline
 3 & 4 \\ \hline
 5 & 6 \\ \hline
 6 & 7 \\ \hline
 7 & 8 \\ \hline
 9 & 10 \\ \hline
 10 & 11 \\ \hline
 11 & 12 \\ \hline
 13 & 14 \\ \hline
 14 & 15 \\ \hline
 15 & 16 \\ \hline
\end{array}

Original matrix has shape $4\times4$, and the transformed matrix with $s=2$ has shape $12\times2$. It is a little confusing for me to get the final number of rows in the transformed matrix because it doesn't seem to be related to the step size directly.

I've made some examples manually, like: matrix $[5\times4]$ with step size=$2$ generates shape $10\times2$, but same matrix with step size=$3$ generates $10\times3$ too. Some guidance is very welcome!

答: I'm not familiar with this sliding-window idea, but it appears the if the matrix has has shape $n\times m$, so that there are $m$ entries in a row, and the step size is $s$, then there will be $m-s+1$ steps per row, because the first element of the step can be in any of columns $1$ through $m-s+1$.  Then there will be $n(m-s+1)$ steps in all, and the resulting matrix will have shape $$n(m-s+1)\times s$$  For $n=5,m=4,s=3$ we get $10\times 3$ as in your example.

Title: Calculate size of a windowed matrix


问:
Measurability of function in product space.
I was wondering. Suppose that $(X,\Sigma_X)$ and $(Y,\Sigma_Y)$ are measure spaces and $f:X\to \Bbb R$ and $g:Y\to \Bbb R$ are measurable. Does this imply that $h:X\times Y\to \Bbb R $ defined
$h(x,y)=\min(f(x),g(y)) $
is measurable? Of course $X\times Y$ is equipped with product $\sigma$-algebra $\Sigma_X\otimes\Sigma_Y$ and $\Bbb R$ with the Borel $\sigma$-algebra.

I think to show this we need the measurability of the set $\{(x,y):f(x)\leq g(y)\}$, am I correct?

答:
I don't quite see where you would need that thing you said. For all $t\in\Bbb R$, $$\{(x,y)\in X\times Y\,:\, h(x,y)\ge t\}=\{x\in X\,:\, f(x)\ge t\} \times\{y\in Y\,:\, g(y)\ge t\}$$

and therefore $h$ is measurable.


问:
Measurable function that is constant on every subset of sigma algebra
I'm working on the following problem and got stuck:

Let $\Omega$ be a set, $(A_n)_{n\in \mathbb{N}}$ be a family of disjoint sets such that $\cup_{n\in \mathbb{N}} A_n = \Omega $ and  $\mathcal{A} = \big\{ \cup_{i\in I} A_i : I\subset \mathbb{N} \big\}$

Show that $\mathcal{A}$ is a $\sigma$-algebra.

Let $f:\Omega \to \mathbb{R}$, a $\mathcal{A}$-measurable function. Show that $f$ is constant on every $A_n$.

I managed to do the part 1.

$i) \Omega \in \mathcal{A}$

since for $I = \mathbb{N}$ we have $\cup_{i\in I} A_i = \Omega\in \mathcal{A}$

$ii) A\in \mathcal{A} \implies A^{c} \in \mathcal{A}:$

Let $A\in\mathcal{A}$. Then $A=\cup_{i\in I} A_i$. Thus
$A^{c} = \big(\cup_{i\in I} A_i\big)^{c} = \cup_{n\in \mathbb{N}} A_n \setminus \cup_{i\in I} A_i = \cup_{j\in J} A_j$

for some $J$ such that $J \cup I = \mathbb{N}$.

Similar for the countable unions.

For 2. I know that $f$ is measurable if:

$$\{f\geq t\}\in \mathcal{A}$$

I was wondering if I can show 2. knowing that $\{f\geq t\}\in \mathcal{A}$ is equivalent to $\{f\leq t\}\in \mathcal{A}$ and thus $f=t$?

Any help would be appreciated.

答: Let $x \in A_n$ and $y=f(x)$. Let $f_n$ be he restriction of $f$ to $A_n$.  Then $f_n^{-1}(y)$ is in $\mathcal A$ and conatisn $x$. It is also a subset of $A_n$. But the only subsets of $A_n$ in $\mathcal A$ are $\emptyset$ and $A_n$. Since $f_n^{-1}(y)$  is not empty, it must be $A_n$. This means $f=y$ on $A_n$.


答:
The key here is that $(A_n)_{n\in\mathbb{N}}$ is a disjoint family of sets. To prove the second part, fix $n\in \mathbb{N}$ and let $x,y\in A_n$. Aiming for a contradiction, assume that $f(x)\ne f(y)$.

Because $f$ is $\mathcal{A}$-measurable, we have that $f^{-1}\left(\left\{f(x)\right\}\right)\in\mathcal{A}$ and $f^{-1}\left(\left\{f(y)\right\}\right)\in\mathcal{A}$. By definition of $\mathcal{A}$, we have that $$f^{-1}\left(\left\{f(x)\right\}\right) = \bigcup_{i\in I_x} A_i\quad \text{and}\quad f^{-1}\left(\left\{f(y)\right\}\right) = \bigcup_{i\in I_y} A_i$$
for some sets $I_x,I_y\subset \mathbb{N}$. Obviously $x\in f^{-1}\left(\left\{f(x)\right\}\right)$ and $y\in f^{-1}\left(\left\{f(y)\right\}\right)$, so by disjointness of the family $(A_n)_{n\in\mathbb{N}}$, it follows that $n\in I_x\cap I_y$. By this and since $f(x)\ne f(y)$ we have $$A_n\subset f^{-1}\left(\left\{f(x)\right\}\right)\cap f^{-1}\left(\left\{f(y)\right\}\right)=\emptyset,$$
a contradiction.

Title: Measurable function that is constant on every subset of sigma algebra

Body:
I'm working on the following problem and got stuck:

Let $\Omega$ be a set, $(A_n)_{n\in \mathbb{N}}$ be a family of disjoint sets such that $\cup_{n\in \mathbb{N}} A_n = \Omega $ and  $\mathcal{A} = \big\{ \cup_{i\in I} A_i : I\subset \mathbb{N} \big\}$

Show that $\mathcal{A}$ is a $\sigma$-algebra.

Let $f:\Omega \to \mathbb{R}$, a $\mathcal{A}$-measurable function. Show that $f$ is constant on every $A_n$.

I managed to do the part 1.

$i) \Omega \in \mathcal{A}$

since for $I = \mathbb{N}$ we have $\cup_{i\in I} A_i = \Omega\in \mathcal{A}$

$ii) A\in \mathcal{A} \implies A^{c} \in \mathcal{A}:$

Let $A\in\mathcal{A}$. Then $A=\cup_{i\in I} A_i$. Thus
$A^{c} = \big(\cup_{i\in I} A_i\big)^{c} = \cup_{n\in \mathbb{N}} A_n \setminus \cup_{i\in I} A_i = \cup_{j\in J} A_j$

for some $J$ such that $J \cup I = \mathbb{N}$.

Similar for the countable unions.

For 2. I know that $f$ is measurable if:

$$\{f\geq t\}\in \mathcal{A}$$

I was wondering if I can show 2. knowing that $\{f\geq t\}\in \mathcal{A}$ is equivalent to $\{f\leq t\}\in \mathcal{A}$ and thus $f=t$?

Any help would be appreciated.


问:
Convergent sequences
If $l^p=\{\langle x_k \rangle \in R^n | \sum_{k=1}^{\infty} |x_k|^p < \infty \}$ and $1\leq p<q$, then $l^p\subset l^q$.

Prove.

I know that I have to show that if $x=\langle x_k \rangle\in l^p\rightarrow x\in l^q$ and I can use a fact that $l^p\subset c_0$, where $c_0=\{x=\langle x_k \rangle \in R^n | \displaystyle \lim_{k \to +\infty} x_k=0\}$.

答: One hint is that for a number $x$ less than 1, $x^q$ is less than $x^p$. And the numbers in the sequence are approaching 0.</p>

Title: Convergent sequences


问:
How calculate the FOV of the faces of a dodecahedron
Having a dodecahedron I want to calculate the field of view (FOV) in degrees of the faces of the shape. If each pentagon has a circunscribed circle, I want to know or how to calculate the angle from the center of the dodecahedron to both sides of the circle

答:
The radius of the circumscribing circle on your diagram can be found from the edge length of the pentagon $a$ and is $r=\frac{acosec(\pi/5)}{2}$

The radius R of the spherical surface passing through all 20 identical vertices of a dodecahedron is given by

${R=\frac{\sqrt{3}(\sqrt{5}+1)a}{4}}$

(see also What is the height of a Dodecahedron in terms of it's edge length and can the formula be simplified or generalised)

Then if $\alpha$ is half the angle you require $sin\alpha = \frac{r}{R}$

The answer is approximately 1.3047 rads or 74.755 degrees


问:
Prove that $G(B\cap C)=G(B)\cap G(C)$.
Let G be a graph and let B be a subclass of the domain of G. We use the symbol G(B) to designate
the class $$G(B)=\{y\mid \exists x\in B\ \text{such that} \ (x,y)\in G\} $$.

The one way containment is pretty straightforward.

But for the coming one, we take $ y \in G (B) \cap G (C) $, then $ y \in G (B) $ and $ y \in G (C) $. Then, there exists $ x \in B $ such that $ (x, y) \in G $, and there exists $ z \in C $, such that $ (z, y) \in G $, to get that $ y \in G (B \cap C) $ would have to guarantee that $ z = x $, but how do I do that?

Title: Prove that $G(B\cap C)=G(B)\cap G(C)$.

Body: This identity doesn't hold. Define $G = \{(0, 0), (1, 1), (2, 1)\}$. Then for $A = \{0, 1\}$ and $B = \{0, 2\}$, we see that $G(A) = G(B) = \{0, 1\}$ but $G(A \cap B) = G(\{0\}) = \{0\}$.


问:
Let $\{x_n\}$ be a sequence of real numbers that converges to infinity. Prove that $\{x_n\}$ is bounded from below.
I do not know how to start this proof.

I know that for a sequence to be bounded below that there is some number K such that $x_n \geq K$ for all $n \in \mathbb{N}$.
I also know that the sequence converges to infinity provided that for every $M >0$ there exists a $N \in \mathbb{N}$ such that $n \geq N \Rightarrow x_n > M$.

I am unsure how to use these two theorems to prove the sequence is bounded from below.

答:
The definition of

$\lim\limits_{n \to \infty} x_n = \infty$

is $\forall M > 0, \exists N \in \mathbb{N}, \forall n \geq N, x_n > M$.

So take, in particular, $M = 1$. Then take some $N$ such that for all $n \geq N, x_n > M$. Define $K = \inf(\{M\} \cup \{x_i | i < N\})$. This is valid because every finite non-empty set has an infinum.

Then $x \leq x_n$ for all $n$.


问:
Showing second-countable + T$_1$ + compact implies sequentially compact

Let $X$ be a second-countable T$_1$-space. If $X$ is compact then it is sequentially compact.

I know that compact implies countably compact. And under the condition that $X$ is first-countable: countably compact implies sequentially compact. Since $X$ is second-countable, thus first-countable, we have that $X$ is sequentially compact.

Is this correct? As the conditions of $X$ being a second-countable T$_1$-space weren't really used here, I suspect something is wrong.

However, those conditions may be needed to show that if $X$ is sequentially compact then it is compact.

答:
It is correct, and if one goes into a bit more detail, it is possible to eliminate the $T_1$ hypothesis.

Suppose that $X$ is compact and first countable, and let $\sigma=\langle x_n:n\in\Bbb N\rangle$ be a sequence in $X$. Let $A=\{x_n:n\in\Bbb N\}$. If $A$ is finite, $\sigma$ has a constant subsequence, which of course converges to a point of $X$. Otherwise, compactness of $X$ implies that the set $A$ has a limit point $p$.

We do not need $X$ to be $T_1$ for this to hold. If $A$ has no limit point, then each $x\in X$ has an open nbhd $U_x$ such that $U_x\cap A$ is finite. $\{U_x:x\in X\}$ is an open cover of $X$, so there is a finite $F\subseteq X$ such that $\{U_x:x\in F\}$ covers $X$. But then $A=\bigcup\{U_x\cap A:x\in F\}$ is a finite union of finite sets and so is finite, contrary to our assumption.

Now let $\{B_n:n\in\Bbb N\}$ be a countable local base at $p$; without loss of generality we may assume that $B_{n+1}\subseteq B_n$ for each $n\in\Bbb N$. (Why?) There is an $m_0\in\Bbb N$ such that $x_{m_0}\in B_0$. Given $m_k$ for some $k\in\Bbb N$, there is an $m_{k+1}\in\Bbb N$ such that $m_k<m_{k+1}$ and $x_{m_{k+1}}\in B_{k+1}$, since $B_{k+1}\cap A$ is infinite. In this way we can recursively construct a subsequence $\langle x_{m_k}:k\in\Bbb N\rangle$ of $\sigma$ such that $x_{m_k}\in B_\ell$ for all $k\ge \ell$, and clearly this subsequence converges to $p$. This shows that $X$ is sequentially compact.


问:
Uniform convergence of $f_n(x)=\frac{x^n}{n!}$
Let $f_n: [0,1]\rightarrow \mathbb{R}$ be a sequence of functions defined by $$f_n(x)=\frac{x^n}{n!}.$$
Prove that $f_n(x)$ converges uniformly to $0$.

We know that $$e^x = \sum_{n=0}^\infty \frac{x^n}{n!}$$ is this enought to prove that $f_n \rightarrow 0$?

答: Let $\varepsilon>0$: then $|f_n(x) - 0|=|\frac{x^n} {n!} |\leq \frac{1}{n!}<\varepsilon$ for $n>N:=\frac{1}{\varepsilon}$. Since $\varepsilon>0$ is arbitrary we have shown that $\forall\varepsilon>0$ there exists $N$ such that $|f_n(x) - 0|<\varepsilon\ \forall n>N$ thus $(f_n) $ converges uniformly to $0$ (on $[0,1]$) by definition of uniform convergence.


问:
Exponential simplification help
Why is it true that

$$
\frac{exp(x)}{1+exp(x)} = \frac{1}{exp(-x) + 1}
$$

Why did we negate $x$?

I tried to divide the left-hand side with $exp(x)$ and got
$$
\frac{exp(x)}{exp(x) + (exp(x))^2}
$$
but I cannot get forward.

Title: Exponential simplification help

Body: $$\dfrac{\exp(x)}{1+\exp(x)} = \dfrac{\frac1{\exp(x)}\exp(x)}{\frac1{\exp(x)}(1+\exp(x))} = \dfrac{1}{\frac1{\exp(x)}+1} =  \dfrac{1}{{\exp(-x)}+1}$$


答:
Using the two things you said you knew in your post, you can supply a number $K$ that is a lower bound for $\{x_n\}$ like this:

Take $N$ so that if $n\ge N$, $x_n \ge 100$.

We would be done with $K = 100$ if $x_n \ge 100$ for all $n$, but we only know $x_n \ge 100$ for $n\ge N$. So only $x_1,\dots,x_{N-1}$ could possibly be $< 100$.

Maybe even all of $x_1,\dots,x_{N-1}$ happen to actually be $\ge 100$, and then we are done with $K = 100$. However, it could transpire that some (or all) of $x_1,\dots,x_{N-1}$ are $< 100$. Fortunately, this is a finite list of numbers, so if we define $K = \min(x_1,\dots,x_{N-1}, 100)$, then $x_1,\dots, x_{N-1} \ge K$ (by definition), and $x_n \ge 100 \ge K$ for $n \ge  N$, so we have a lower bound $K$ which is not explicit, but is defined in terms of the sequence itself.

This kind of thing is pretty typical of many situations in analysis, where you have a good understanding of some "large" part of your problem, and "the rest" of the problem is hopefully smaller than the "large" part, and therefore ideally easier to understand.

Title: Let $\{x_n\}$ be a sequence of real numbers that converges to infinity. Prove that $\{x_n\}$ is bounded from below.

Body:
I do not know how to start this proof.

I know that for a sequence to be bounded below that there is some number K such that $x_n \geq K$ for all $n \in \mathbb{N}$.
I also know that the sequence converges to infinity provided that for every $M >0$ there exists a $N \in \mathbb{N}$ such that $n \geq N \Rightarrow x_n > M$.

I am unsure how to use these two theorems to prove the sequence is bounded from below.


问:
Proof verification: sequence $ a_n = (-1)^n $ does not converge
Theorem: Show that the sequence $ a_n = (-1)^n $ for all $ n \in \mathbb{N}, $ does not converge.

Proof: Suppose that there exists a limit $L$ such that $ a_n \rightarrow L  $. Specifically, for $ \epsilon = 1 $ there exists $ n_0 $ s.t. for all $ n > n_0$ then $|(-1)^n-L|<1$, from this we can also infer that $|(-1)^{n+1}-L|<1$ for all $ n > n_0 $.

Let $ n > n_0 $ be arbitrary, then $ 2 = | (-1)^n - (-1)^{n+1} | \leq | (-1)^n - L | + | L - (-1)^{n+1} | < 1 + 1 = 2  $, so since $ n > n_0 $ was arbitrary , hence for all $ n>n_0 $ we have 2<2 which is false, hence we get a contradiction.

My question: Regarding to when I wrote " $ n>n_0 $, we have 2<2 which is false, hence we get a contradiction. "

I'm trying to understand with respect to what there is a contradiction; we get that any statement of the form $ \forall n>n_0 (P(n) )$ is false ( since we got $ \forall n>n_0 . 2<2 ) $ [ and $ P(n) $ is a statement that depends on $ n $ ], and since we already have in the proof the statement " for all $ n > n_0$ then $|(-1)^n-L|<1$ " , then we get a contradiction. Is this correct?

Note: I know there are answers on this sequence's divergence, but they didn't really help me find an answer to my question.

答:
I say your statement $P(n)$ does not depend on $n$.

With respect to the known fact that $2=2$, you have deduced a contradiction. Hence your proof is good. (Although stop at the $2<2$ bit. You go on saying "so since $n$ was arbitrary", but that is not relevant.)

$a_n$ converges means that there is an $L$, so that for every positive epsilon, we can find a (big) index beyond which $a_n$ is epsilon-close to L.

$a_n$ does not converge means that for any $L$, there is a positive epsilon, so that for any index, we can find an even further index $t$, at which $a_t$ is not epsilon-close to L.

Maybe it will help you to see your proof from this second viewpoint.

Do you see how the "for anys" trade places with the "there exists at least one", when we contradict a complicated definition?

Try to step through your proof using the definition of "doesn't converge". It is in a way running your argument backwards.

So the task is: faced with any real $L$ candidate, to answer with a suitable epsilon.


问:
Does $\mathbb{E}[Y|X,Z] = \mathbb{E}[Y|Z]$ almost surely imply $Y \perp X | Z$
It is clear that $Y \perp X | Z \implies \mathbb{E}[Y|X,Z] = \mathbb{E}[Y|Z]$ almost surely,but does the reverse implication always hold?

答:
No.

$\mathbb{E}[Y\mid X] = \mathbb{E}[Y]$ does not imply $Y \perp X $

so conditioning on $Z$ is just as unsatisfactory

Consider the following equally likely sets of values each with probability $\frac18$

```<code>X   Y   Z
1   1   3
1   5   3
2   2   3
2   4   3
1   3   5
1   7   5
2   4   5
2   6   5
</code>```

You find $\mathbb{E}[Y\mid X,Z] = Z = \mathbb{E}[Y\mid Z]$, but $X$ determines the parity of $Y$ so they are not independent in any sense, conditional or otherwise

Body:
No.

$\mathbb{E}[Y\mid X] = \mathbb{E}[Y]$ does not imply $Y \perp X $

so conditioning on $Z$ is just as unsatisfactory

Consider the following equally likely sets of values each with probability $\frac18$

X   Y   Z
1   1   3
1   5   3
2   2   3
2   4   3
1   3   5
1   7   5
2   4   5
2   6   5

You find $\mathbb{E}[Y\mid X,Z] = Z = \mathbb{E}[Y\mid Z]$, but $X$ determines the parity of $Y$ so they are not independent in any sense, conditional or otherwise


问:
Proving that area of a triangle can be expressed with the determinant

I know that this might be a easy problem, but I'm stuck. I don't know how I'm supposed to use the information given in the task (the vectors)to prove that the area is equal to 1/2 det(P). I thought about starting with explaining why det(P) is equal to the area of a parallelogram and so divide by two. But still, I'm not sure how I'm supposed to use the vectors to do this. Really appreciate some help!

答:
Hint:

develop the determinant and see that it coincides with $r1 \times r_2$.


答:
As the other answer mentions, the condition does not suffice. However, if you assume $E[f(Y)|X,Z] = E[f(Y)|Z]$ for all bounded measurable $f$, that does suffice.

Write for any bounded measurable $f,g$ that $$
E[f(Y)g(X) | Z] = E[E[f(Y)g(X)| X, Z]|Z] = E[g(X) E[f(Y)|X,Z]|Z]
$$
$$
= E[g(X) E[f(Y)|Z] | Z]
= E[g(X) | Z] E[f(Y)|Z].
$$
Thus we have $Y\perp X$ conditional on $Z$.

Title: Does $\mathbb{E}[Y|X,Z] = \mathbb{E}[Y|Z]$ almost surely imply $Y \perp X | Z$

Body: It is clear that $Y \perp X | Z \implies \mathbb{E}[Y|X,Z] = \mathbb{E}[Y|Z]$ almost surely,but does the reverse implication always hold?


问:
Continuous QR decomposition
Let $F_n \rightarrow_{n\rightarrow \infty} 0$ and  $I+F_n=Q_nR_n$ be the QR decomposition with $r_{ii}>0$.

Then $Q_n \rightarrow I$. Why do we need the continuity of the mapping $A=QR\rightarrow Q$ for this conclusion?

答:
It is enough to show that any subsequence $(Q_{n_{k}})_{k}$ has a subsequence $(Q_{n_{k_{j}}})_{j}$ such that $Q_{n_{k_{j}}}\longrightarrow I$.

Since the columns of $Q_{n_{k}}$ are orthonormal for all $n$, then the sequence $(Q_{n_{k}})_{k\in\mathbb{N}}$ is bounded (you can consider any norm for our purposes, because all norms on a finite-dimensional space are equivalent). By compactness (of the closed balls in finite-dimensional normed spaces), there exists a convergent subsequence $(Q_{n_{k_{j}}})_{j}$, let's say, $Q_{n_{k_{j}}}\longrightarrow Q$. Therefore, $Q_{n_{k_{j}}}^{t}\longrightarrow Q^{t}$ (we are taking transposes)

Therefore $R_{n_{k_{j}}}=\underbrace{Q_{n_{k_{j}}}^{t}Q_{n_{k_{j}}}}_{=I}R_{n_{k_{j}}}=Q_{n_{k_{j}}}^{t}(I+F_{n_{k_{j}}})\longrightarrow Q^{t}$ as $j\rightarrow +\infty$

Hence $Q^{t}$ is upper triangular, and its diagonal consists of non-negative elements. But $Q^{t}$ is invertible (we are working with square matrices), because $I=Q_{n_{k_{j}}}^{t}Q_{n_{k_{j}}}\longrightarrow Q^{t}Q$, so $Q^{t}Q=I$. Hence, all the elements of the diagonal of $Q^{t}$ must be positive.

We have: $Q^{t}$ is upper triangular, all its diagonal consists of positive elements, and its rows are orthonormal. This implies that $Q^{t}=I$, and hence $Q=I$

Title: Continuous QR decomposition


问:
Definition of semimartingale and stochastic integration
Hey I found two definitions of semimartingale. One states that semimartingale is a sum of local martingale and FV process and the second one states that semimartingale is a sum of FV process and locally square integrable martingale. Are these definitions equivalent? I would like to define stochastic integral of caglad process with respect to semimartingale. Can I define this integral as the sum of Lebesgue-Stieltjes integral and integral w.r.t locally square integrable process? Or I have to define this integral using the first definition? What is the difference? Is this integral well defined? I would be thankful for any help.</p>

答:
Regarding your two definitions: I know the first one for classical semimartingales (or just semimartingales). The second one is the defintion of decomposable processes (e.g. in the book of Protter Ch.II,3).

The two definitions are equivalent: As every process which is locally a square integrable martingale is of course also a local martingale, all processes satisfying the second definition also satisfy the first definition. For the reverse you have to use the Fundamental Theorem of Local Martingales (e.g. in the book of Protter Ch.III,6,Th.25). This states that every local martingale can be written as the sum of two local martingales, whereas one of them is a FV process and the other has bounded jumps. The local martingale with bounded jumps is a locally bounded martingale und thus locally a square integrable martingale.

To your second question: As the FV part can be used for a Lebesgue-Stieltjes integral the answer is YES (for a FV process the stochastic and the Lebesgue-Stieltjes integral are indistinguishable, see Protter Ch.II,5,Th.17). Since you are only integrating càglàd processes the integral is well defined.


问:
Embedding into $\mathbb{R}^6$
I am trying to find a map that goes from $S^2$ to $\mathbb{R}^6$ such that it is an embedding. Im trying to use the fact that $S^2$ is compact and that I can find a map $\psi$ that is an injective immersion which would then make $\psi$ an embedding. Thinking about the Monge parametrization for the sphere i would get:

$$ \psi(x,y,z) = (x,y, \sqrt{1-x^2-y^2},0,0,0) $$

whose Jacobian matrix is:

$$  J(x,y,z) = \left(\begin{array}{cccc}
1 & 0\\
 0 & 1 \\
 \frac{-x}{\sqrt{1-x^2-y^2}} & \frac{-y}{\sqrt{1-x^2-y^2}}   \\
 0  & 0  \\
0 & 0 \\
0 & 0
\end{array}\right) $$

but this only covers the top hemisphere, I could do a composition with the negative root but I find it problematic. For instance I can't justify that the rank of the Jacobian is $3$ so I couldn't say its an immersion. Is there an injective immersion from $S^2$ to $\mathbb{R}^6$?

答: I don't understand the question. The standard realization of $S^2$ in $\mathbb{R}^3$ is an embedding. Adding three extra dimension does not make it any less of an embedding.

Title: Embedding into $\mathbb{R}^6$


问:
The 1-1 correspondence between sections of a principal G-bundle P and isomorphisms between the trivial principal G-bundle and P
Exercise: Show that a principal bundle admits a (global) section of and only if it is trivializable. In more detail: show that if $\sigma$ is a section of the principal G-bundle $\pi : P \rightarrow M$ then
$$ F_\sigma : M \times G \rightarrow P, F_\sigma (x,g) = \sigma(x).g $$
Is an isomorphism of pricinpal G-bundles; and then show that $\sigma \longleftrightarrow F_\sigma $ defines a 1-1 correspondence between sections of P and isomorphisms between the trivial principal G-bundle and P.

I know how to do the first part of this exercise, What it remains is how to construct a global section $\sigma $of P out of an isomorphism of trivial bundles : $ F: M \times G \rightarrow P $, such that $F= F_\sigma $.

Thanks!

答:
Let $F:M\times G\rightarrow P$ be an isomorphism and $e$ the neutral element of $G$. For every $x\in E$, write $\sigma(x)=F(x,e)$. The map $\sigma$ is a section of $\pi:P\rightarrow M$ since $F$ is a morphism of bundle.

To $\sigma$ is associated the isomorphism $F_\sigma: M\times G\rightarrow P$ defined by
$F_\sigma(x,g)=\sigma(x).g=F(x,e).g=F(x,g)$.


问:
Understanding an instance of Integration by Substitution w/the Dirac Delta Function
From Kanwal's Generalized Functions pg. 49:

Question: How does the author move from

$$
∫_{-∞}^∞ δ[f(x)]ϕ(x)dx = ∫_{-∞}^∞ δ(y)\frac{ϕ[x(y)]}{f'(x)}dy
$$

via Integration by Substitution at the start of the second equation? Specifically how does $f'(x)$ end up in the denominator (this makes no sense to me since $x$ is now interpreted as some fixed constant?)?  It seems to me that we should actually have

$$
∫_{-∞}^∞ δ[f(x)]ϕ(x)dx = ∫_{-∞}^∞ δ(y)ϕ[x(y)]f'(y)dy
$$

How am I mistaken?

答:
If your question is only about the change of variable you can read it backwards. We want to give meaning to $\delta(f(x))$, start with
$$
\int_{\mathbb{I}} \delta(f(x)) \, \phi(f(x))  \, f'(x) dx \; ,
$$
where $\mathbb{I}$ is some neighbourhood of the point $x_1$ where $f(x_1)=0$. And now change variables. Let $y=f(x)$, $dy = f'(x) dx$ , then
$$
\int_{\mathbb{I}} \delta(f(x)) \, \phi(f(x))  \, f'(x) dx \; =
\;  \int_{f(\mathbb{I})} \delta(y) \, \phi(y)  \, dx \; = \; \phi(0) .
$$
This is independent of the neighbourhood $\mathbb{I}$. In this region $f'(x)$ is nonzero by assumption. The distribution $\delta(f(x))$ must constraint $x=x_1$ and cancel the factor $ f'(x)$, thus the only possibility is
$$  \delta(f(x)) \; = \; \frac{ \delta(x-x_1) }{ f'(x_1) } \; . $$

Thus
\begin{align}
\int_{\mathbb{I}} \delta(f(x)) \, \phi(f(x))  \, f'(x) dx \; & =
\; \int_{\mathbb{I}} \delta(x-x_1) \, \phi(f(x))  \, \frac{f'(x)}{f'(x_1)}  \\
&= \; \phi(f(x_1))  \, \frac{f'(x_1)}{f'(x_1)}  \; = \; \phi(0)
\end{align}

A simple example

Calculate the integral  (for $x_0>0$)
$$
\int_{0}^{\infty} \delta(x^2 - x_0^2) \, \phi(x)   dx \; .
$$
From one side we could use $y=x^2-x_0^2$, $dy = 2x dx$. Let $x(y) = \sqrt{y+x_0^2}$ where $x(0) = x_0$, then
$$
\int_{0}^{\infty} \delta(x^2 - x_0^2) \, \phi(x)   dx \; = \;
\int_{-x_0^2}^{\infty} \delta(y) \frac{\phi[x(y)]}{2x(y)} dy \; = \; \frac{\phi(x_0)}{2x_0} \; .
$$
On the other hand, we could use the expression for the delta function. Let $f(x)=x^2-x_0^2$. In this region, $f$ has only one zero (that happens at $x=x_0$) and $f'(x)=2x > 0$. Thus
$$ \delta(x^2-x_0^2) = \frac{\delta(x-x_0)}{2x_0} \; , \text{ for } x> 0 \; . $$
Evidently, this gives the same answer for the integral.

Note that the statement of the book can be generalised to incorporate negative values to the derivative and more than one point where $f(x)=0$. In this example, for the general case where $f:\mathbb{R}\to\mathbb{R}$ with $f(x)=x^2-a^2$ the Dirac delta is
$$ (\delta \circ f)(x) \; = \;
\delta(x^2-a^2) \; = \; \frac{1}{2\vert a \vert } \, \Big( \,\delta(x-a) + \delta(x+a)   \,  \Big) \; .$$


问:
Proving that if $a^n=b^n$, then $a=b$, for positive $a$ and $b$, and natural $n$
In high school algebra, natural number exponents are defined as

$$\begin{aligned}a^1 &= a\\
a^{(n + 1)} &= a^n a\end{aligned}$$

With these I used induction to prove the 3 exponent laws. I also proved some ancillary theorems such as the nth power of a positive is positive, distribution of powers over fractions, and so on. Then I got to this statement:

For positive $a$, $b$ and natural number $n$,

$$\text{if }a^n = b^n\text{ then } a = b\text{.}$$

I have a proof, but it is very different from any of the other proofs. It requires $\Sigma$ notation, which none of the others require. However, I can't figure out how to do it without $\Sigma$. Is it possible?

Here's my proof:

$$a^n = b^n\\
  a^n - b^n = 0\\
  (a - b) \left(\sum_{k=1}^n \frac{a^n b^k}{a^k b}\right) = 0\\
  a = b  \quad\text{or}\quad \sum_{k=1}^n \frac{a^n b^k}{a^k b} = 0$$

As the summation has at least $1$ term and every term is positive, the summation must be positive, which is not equal to $0$. This leaves $a = b$.

Q.E.D.

答:
Direct Proof

Here, we will prove a stronger result:

If, $a^n=b^n$ and $a,b>0, n \in \mathbb {R} \setminus \left\{0\right\}$, then we have

$$\ln a^n =\ln b^n$$

$$\implies n \ln a=n \ln b$$

$$\implies \ln a=\ln b$$

$$\implies \ln a-\ln b=0$$

$$\implies \ln \left(\frac ab\right )=0$$

$$ \implies \dfrac ab =e^0=1$$

$$\color {gold}{\boxed {\color{black}{\dfrac ab=1 \Longrightarrow a=b.}}}$$


答:
$$\frac{dy}{-z+x}=\frac{dz}{3y^2}$$
You should use the constant $C_1$ instead. Since $C_1=x-z$. Your integration of this differential equation is  not correct.
$$3y^2{dy}=C_1{dz}$$
Integrate.
$$y^3=C_1z+C_2$$
$$y^3=(x-z)z+C_2$$
$$C_2=y^3-xz+z^2$$

This line seems not correct to me:
$$y^3=-\frac{z^2}{2}+zx+C_2$$

Title: Solving $3y^2\frac{\partial z}{\partial x}-\left(z-x\right)\frac{\partial z}{\partial y}=3y^2$, I keep getting the same wrong answer

Body:
For the conditions  $x\left(t\right)=t^{\frac{1}{2}},\:y\left(t\right)=t^{\frac{1}{3}},\:z\left(t\right)=0,\:t>0$

I get this:

$$\frac{dx}{3y^2}=\frac{dy}{-z+x}=\frac{dz}{3y^2}$$

$$(I)\:\:\: \frac{dx}{3y^2}=\frac{dz}{3y^2} $$

$$x=z+C_1$$

$$C_1 = x-z$$

$$(II)\:\:\: \frac{dy}{-z+x}=\frac{dz}{3y^2} $$

$$3y^2dy\:=\:\left(-z+x\right)dz$$

$$y^3=-\frac{z^2}{2}+zx+C_2$$

$$C_2=y^3+\frac{z^2}{2}-zx$$

Putting in conditions

$$C_1=t^{\frac{1}{2}}$$

$$C_2=t$$

$$C_2=C_1^2$$

This means that

$$y^3+\frac{z^2}{2}-zx = \left(x-z\right)^2$$

$$2y^3+z^2-2zx = 2 \left(x-z\right)^2$$

$$2y^3+\left(z^2-2zx+x^2-x^2\right)\:=\:2\:\left(x-z\right)^2$$

$$2y^3-x^2+\left(x-z\right)^2\:=\:2\:\left(x-z\right)^2$$

$$2y^3-x^2=\:\left(x-z\right)^2$$

$$\sqrt{2y^3-x^2}=x-z$$

$$x-\sqrt{2y^3-x^2}=z$$

But it's still wrong

So where's the mistake?


答:
This image should help to show the 'why' connecting the row-vectors and the determinant:

https://i.stack.imgur.com/a7QGB.png

Title: Proving that area of a triangle can be expressed with the determinant

Body: I know that this might be a easy problem, but I'm stuck. I don't know how I'm supposed to use the information given in the task (the vectors)to prove that the area is equal to 1/2 det(P). I thought about starting with explaining why det(P) is equal to the area of a parallelogram and so divide by two. But still, I'm not sure how I'm supposed to use the vectors to do this. Really appreciate some help!