问:
Condition on $u$ so that $ |\frac{x \cdot u } {u \cdot u}u - x |< |x|$
If $u,x \in \mathbb{R}^n$ then am I right to assert that the above inequality will hold if $u$ makes an angle $\alpha \in (0,\pi/2)$ with span$\{x\}^{\perp}$?

When $n=3$ it seems quite apparent that $|\frac{x \cdot u } {u \cdot u}u - x | = \cos(\alpha)|x|<|x|.$ I think it's exactly the same in higher dimensions.

答:
The expression only makes sense when $u \neq 0$, and then $\hat{u} := \frac{x \cdot u}{u \cdot u} u$ is the projection of $x$ onto $u$, i.e. $\hat{u}$ is parallel to $u$ and $x = v + \hat{u}$ such that $v \cdot u = v \cdot \hat{u} =0$.

Consequently, $|\frac{x \cdot u}{u \cdot u} u - x| = | v | < |\hat{u} +v| = |x|$ for all $x$.


问:
Natural Deduction Proof for Modus Tollens
Suppose I would like to proof modus tollens, i.e.  $P\ \to Q,\ \lnot Q\ \vdash\lnot P$, based on Gentzen-style natural deduction for classical logic. Using rules of inference for NK as given in Katalin Bimbo's Proof Theory: Sequent Calculi and Related Formalism (p.270), I made the derivation below:

$\dfrac{\begin{matrix}\dfrac{\left(P\rightarrow Q\right)\land\lnot Q}{\lnot Q}\ \land\! E_1&\dfrac{\begin{matrix}\dfrac{\left(P\rightarrow Q\right)\land\lnot Q}{P\rightarrow Q}\land\! E_2&\begin{matrix}\\P\\\end{matrix}\\\end{matrix}}{Q}\ \to\!E\\\end{matrix}}{\dfrac{\bot}{\lnot P}\bot E}\bot I$

I have a few questions:

Is my NK proof above correct? In its linear variant, I have seen some proofs that include $\neg Q \to \neg P$ as the last step after $\neg  P$ (using $\to I$). Is that last step necessary? I mean, did we not already derive $\neg P$ (which is the right side of $\vdash$) from $P \to Q, \neg Q$?

As I understand it, in this proof the only assumption I make is $P$ (to the left of $\to E$), whereas $P \to Q \wedge\neg Q$ is not an assumption but a premise (keeping in mind that they appear before $\vdash$ in the proposition, and so must be taken as given, not an arbitrary assumption). Is my reasoning correct?

How to differentiate between theorem and true proposition in NK proof system? Semantically speaking, a theorem is a proposition that is true in all interpretations, whereas a true proposition need not be true on all interpretations. In truth table, this is easy to show. But in NK, I don't see anything like that. Is theorem a proposition which can be proved without assumption (only using premise), whereas true proposition (not a theorem) can only be proved with assumption? Is this related to the place of $\vdash$ in the proposition?

Lastly, is NK proof for $P\ \to Q,\ \lnot Q\ \vdash\lnot P$ is the same as NK proof for $\vdash P\ \to Q \wedge \lnot Q\ \to \lnot P$ ? I mean, what is the difference between $\vdash$ and $\to$, on the one hand, and the difference between comma (,) and $\wedge$? Is it not just the difference between object-language and metalanguage that we must aware at all times yet, in the actual labour of proof, have no impact whatsoever in the way we operates on those symbols (both truth functionally/semantically and proof theoretically/syntactically)? I know that we must be mindful of tiny differences between levels of language so that we don't succumb to the dreadful self-referential paradoxes, such as Russel's or Gödel sentence. But, really, we don't stumble into Gödel sentence everyday, do we? So, back to the point, what is the actual, operational, difference between using $\vdash$ and $\to$ and between using comma (,) and $\wedge$ in natural deduction (or any other proof systems) other than healthy precautionary attitude of anticipating over something that happens once in a blue moon?

答:
Natural Deduction Proof for Modus Tollens

Using a simple form of natural deduction...

ContentLicense: CC BY-SA 4.0


问:
Combo problem, complementary counting
You have 5 blue nails in a column and 3 red nails in another column. You can attach a string between any red nail and blue nail. How many ways can we attach these strings such that every nail has at least 1 string attached to it?

For example, drawing all possible strings would work. Or connecting the first 3 blue with first 3 red, then connecting the last 2 blue to any red would work.

I approached this using Principle of Inclusion-Exclusion (if there is a simpler solution, let me know), but I am stuck. All I have is

$(2^5-1)^3$-something

$(2^5-1)^3$ is the number of non-empty subsets of blue nails for the red nails to connect to, cubed because 3 red nails. I don't know what to subtract from this though

答:
Represent the connected pairs of nails in a $5\times3$ grid, where an $S$ in row $i$ and column $j$ represents a string between the $i$-th blue nail and the $j$-th red nail.

For example, here’s collection that you would want to count, because every nail has at least one string connected to it.

So if I understood your question correctly, you want to count the number of ways to place $S$’s in the grid so that no row or column is $S$-less. Equivalently, you want to count the number of $5\times3$ matrices containing $0$s and $1$s with no row or column that is all zeroes.

That question has been answered in general here, and as you and others recognize, it relies on the inclusion-exclusion principle.

Number of $(0,1)$ $m\times n$ matrices with no empty rows or columns

Title: Combo problem, complementary counting


问:
Other ways to evaluate $\lim_{x \to 0} \frac 1x \left [ \sqrt[3]{\frac{1 - \sqrt{1 - x}}{\sqrt{1 + x} - 1}} - 1\right ]$?
Using the facts that:
$$\begin{align}
\sqrt{1 + x} &= 1 + x/2 - x^2/8 + \mathcal{o}(x^2)\\
\sqrt{1 - x} &= 1 - x/2 - x^2/8 + \mathcal{o}(x^2)\\
\sqrt[3]{1 + x} &= 1 + x/3 + \mathcal{o}(x)
\end{align}$$

I was able to evaluate the limit as follows:
$$\begin{align}
\lim_{x \to 0} \frac 1x \left [ \sqrt[3]{\frac{1 - \sqrt{1 - x}}{\sqrt{1 + x} - 1}} - 1\right ] &\sim \lim_{x \to 0} \frac 1x \left [ \sqrt[3]{\frac{\dfrac x2 + \dfrac{x^2} 8}{\dfrac x2 + \dfrac{x^2} 8}} - 1\right ] =\\
&= \lim_{x \to 0} \frac 1x \left [ \sqrt[3]{1 + \frac{2x^2}{4x - x^2}} - 1\right ] \sim\\
&\sim \lim_{x \to 0} \frac{2x^2}{12x^2 - 3x^3} = \frac 16
\end{align}$$

What are other ways to evaluate it? Maybe pure algebraically? I tried to rationalize the denominator, but got stuck at some point...

ContentLicense: CC BY-SA 3.0

Body:
Let us start by noting that:

$\sqrt[3]{1-\sqrt{1-x}}\cdot \sqrt[3]{1+\sqrt{1-x}}=\sqrt[3]{x}$

$\sqrt[3]{\sqrt{1+x}-1}\cdot \sqrt[3]{\sqrt{1+x}+1}=\sqrt[3]{x}$

$$\sqrt[3]{\frac{(1-\sqrt{1-x})(1+\sqrt{1-x})}{(\sqrt{1+x}-1)(\sqrt{1+x}+1)}}=1$$

Using these algebraic facts, we can write that: $$\begin{align} \lim_{x\to 0} \frac{1}{x} \cdot \Biggl[\sqrt[3]{\frac{1-\sqrt{1-x}}{\sqrt{1+x}-1}}-1\Biggl]
&= \lim_{x\to 0}\frac{1}{x} \cdot \Biggl[\sqrt[3]{\frac{1-\sqrt{1-x}}{\sqrt{1+x}-1}}-\sqrt[3]{\frac{(1-\sqrt{1-x})(1+\sqrt{1-x})}{(\sqrt{1+x}-1)(\sqrt{1+x}+1)}}\Biggl] \\
&= \lim_{x\to 0}\frac{1}{x} \cdot \Biggl[\sqrt[3]{\frac{(1-\sqrt{1-x})(\sqrt{1+x}+1)}{(\sqrt{1+x}-1)(\sqrt{1+x}+1)}}-\sqrt[3]{\frac{(1-\sqrt{1-x})(1+\sqrt{1-x})}{(\sqrt{1+x}-1)(\sqrt{1+x}+1)}}\Biggl] \\
&= \lim_{x\to 0}\Biggl[\frac{\sqrt[3]{\sqrt{1+x}+1}-\sqrt[3]{1+\sqrt{1-x}}}{x}\Biggl] \cdot \Biggl[\sqrt[3]{\frac{1-\sqrt{1-x}}{x}}\Biggl] \\
&= \lim_{x\to 0}\frac{\sqrt[3]{\sqrt{1+x}+1}-\sqrt[3]{1+\sqrt{1-x}}}{x} \cdot \lim_{x\to 0} \sqrt[3]{\frac{1-\sqrt{1-x}}{x}} \\ \end{align}$$

If we set $-\sqrt{1-x}$ and $\sqrt[3]{\sqrt{1+x}+1}$ to be $f(x)$ and $g(x)$ respectively, it can be written that: $$\begin{align} \lim_{x\to 0} \frac{g(x)-g(-x)}{x} \cdot\lim_{x\to 0}\sqrt[3]{\frac{f(x)-f(0)}{x}} \tag{1}\label{eq1}
&= \lim_{x\to 0} 2\Biggl[\frac{g(x)-g(-x)}{2x}\Biggl]\cdot \lim_{x\to 0}\sqrt[3]{\frac{f(x)-f(0)}{x}} \\
&= 2g'(0)\cdot \sqrt[3]{f'(0)} \tag{2}\label{eq2} \\
&= \Biggl[\frac{(\sqrt{1+x}+1)^{-\frac{2}{3}}}{3\sqrt{1+x}}\Biggl]_{x=0}\cdot \Biggl[\sqrt[3]{\frac{1}{2\sqrt{1-x}}}\Biggl]_{x=0} \tag{3}\label{eq3} \\
&= \frac{1}{3\sqrt[3]{4}}\cdot \frac{1}{\sqrt[3]{2}} \\
&= \frac{1}{6} \\   \end{align}$$

$(2)$: limit definition of the derivative

Hope it helps!


问:
Composite of relation and its opposite
I have a relation set $R = \{(a, b), (a, c), (b, c), (b, d), (c, d), (c, a)\}$ included in set $\{a, b, c, d\}\times\{a, b, c, d\}$.

Its opposite of relation is $R^{−1} = \{(b, a), (c, a), (c, b), (d, b), (d, c), (a, c)\}$.

I'm asked to find the composite of $R\circ R^{−1}$ as a set, and I came up with:

$R\circ R^{−1} = \{(a, a), (a, d), (b, b), (b, c), (c, b), (c, c), (d, c), (d, d), (d, a)\}$.

However, the practice exam's answer says it is:

$R\circ R^{−1} = \{(a, a), (b, b), (c, c), (d, d), (a, d), (d, c)\}$

After that, I'm asked to find $R^{−1}\circ R$, which I came up with to be:

$R^{−1}\circ R = \{(a, c), (a, a), (a, b), (b, a), (b, b), (b, c), (c, c), (c, b)\}$

Whereas again the practice exam's solution tells me:

$R^{−1}\circ R = \{(a, a), (b, b), (c, c), (d, b), (d, c)\}$

Who is correct? I came to my answers by drawing out arrow diagrams and followed the each member of $X$ to its $Y$, but I seem to be getting way more relations than the solution says I should be.

答: I think the text is wrong. Whoever wrote the question might have been working under the assumption that the relation was injective (one-to-one), in which case you get back the identity relation when you compose $R$ with its inverse (this is what you called the "opposite" relation). However, clearly, $R$ is not injective as $\left(a,b\right)$ and $\left(a,c\right)$ are in $R$.</p>

Title: Composite of relation and its opposite


问:
Images under the transformation of inversion
Consider the transformation of inversion $Tz = 1/z$. Find the image of under $T$ of the circle $x^2+y^2=ax$ where $a \in \mathbb{R}$.

I guess I am just a bit confused about these types of problems. I do know that Mobius transformations take extended circles to extended circles. So since we are given a circle, to begin with, I expect the image to be a line or circle. I tried converting the equation of the circle to "complex" form so that we would have $|z| = \sqrt{2a(z+\bar{z})}$, whenever $z = x+iy$. However, I don't think this really leads to anything.

Edit: Okay first off thank you for the comments I got lots of insight into this problem. So finding three points on the circle, computing the images, and determining the circle that goes through these images is going to be the image of our circle $x^2+y^2=ax$. Clearly the points $z = 0,a,\frac{a}{2}(1+i)$ all lies on this circle. Computing these images yields:
$$0 \rightarrow \infty \\
a \rightarrow 1/a \\
\frac{a}{2}(1+i) \rightarrow \frac{1}{a}(1-i).$$
The only extended circle that can go through these three points must be a line.

答:
How about completing the square to get $(x-a/2)^2+y^2=a^2/4$.  Then since you know you're going to get a circle or a line, check three points.  After all, this is a Mobius transformation.  Or we could express it as $|z-a/2|=|a|/2$.

So $a\mapsto 1/a, 0\mapsto\infty$, and we know we have a line, through $1/a$.

Finally $a/2+a/2i\mapsto (1-i)/a$, and we conclude that it's vertical.


问:
Quot scheme on open subset
Let $X$ be a scheme, and $U$ an open subset of $X$. If $X$ is projective and $F$ some coherent sheaf on it, then the Quot scheme $Quot_{X}(F)$ exists, and $Quot_{U}(F|_{U})$ too. How can one describe the embedding of $Quot_{U}(F|_{U})$ into $Quot_{X}(F)$. I.e., given a quotient $F|_{U}\rightarrow Q$, how do we get a quotient of $F$? I think we should take the composite $F\rightarrow i_{\ast} i^{\ast} F\rightarrow i_{\ast} Q$, but why should this be surjective? Since $i:U\rightarrow X$ is an open immersion, the pushforward is not exact.

答:
Your intuition is correct. The map $p\colon F \to i_\ast i^\ast F \to i_\ast Q$ is still surjective. In short, this is because you can check this after replacing $i\colon U\hookrightarrow X$ with $i|_{\textrm{Supp}(Q)}\colon \textrm{Supp}(Q) \,\,\widetilde{\to}\,\, \textrm{Supp}(i_\ast Q)$.

First of all, $\alpha\colon i^\ast i_\ast Q \,\,\widetilde{\to}\,\, Q$ is an isomorphism because $i$ is an immersion. The surjection $F|_U \twoheadrightarrow Q$ you started with is recovered as $\alpha\circ p|_U \colon F|_U \to i^\ast i_\ast Q \,\,\widetilde{\to}\,\, Q$, thus $p|_U \colon F|_U \to i^\ast i_\ast Q$ is surjective. But $i$ is an isomorphism around the support of $Q$ and our calculation takes place precisely there; thus we may assume $i$ is surjective, so $i^\ast$ is in particular faithfully flat, hence $i^\ast$ reflects epimorphisms, hence $p$ is surjective.

See Lemma A.1 here for a generalisation.

Title: Quot scheme on open subset


问:
Picture of the generators of fundamental group of genus g surface
It's a standard result that the fundamental group of a genus g surface has presentation

$$\pi_1(\Sigma_g)=\langle a_1,b_1 \ldots a_g,b_g | \prod_i [a_i,b_i] =1 \rangle.$$

Does anyone have a picture of say a genus 3 surface where these generating loops $a_i,b_i$ are explicitly drawn onto the surface?

For example I have see pictures like the one in this answer
the fundamental group of a Riemann surface with n points removed
or
https://www.researchgate.net/figure/Fundamental-group-of-a-genus-3-surface_fig1_226810380
but those sets of 6 loops are different and it's not clear to me which one is correct for the above presentation, or which pairs are $a_1,b_1$ or $a_2,b_2$ or $a_3,b_3$.

Thanks in advance for any help.

答:
That genus 3 picture does, indeed, give your presentation, up to an appropriate bijection $\{a,b,c,d,e,f\} \leftrightarrow \{a_1,b_1,a_2,b_2,a_3,b_3\}$.

Maybe this picture is a little more standard, but I don't know if the presentation is going to be any easier to read from it.


问:
Method of Lagrange multipliers for multiple constraints
I'm having trouble with a past exam question regarding the use of Lagrange multipliers for multiple constraints. The question is:

Using the method of Lagrange multipliers for multiple constraints, find the absolute maximum and minimum of
$$f(x,y,z) = xy + 2z$$
On the intersection of
$$x + y + z =0$$ And
$$x^2 + y^2 + z^2 = 24$$
I'm kinda unsure where to start, I haven't really come across many of these types of questions.

Any help would be appreciated.

答:
It seemed like this is a good problem for illustrating the solution of an extremization using variable "elimination" and a single Lagrange multiplier versus the use of two multipliers.  Solving the resulting systems proved to be a bit "tricky" in either case.

The function for which we seek extrema, $ \ f(x,y,z) \ = \ xy  \ + \ 2z \ , $  has a symmetry under exchanging the coordinates $ \ x \ $ and $ \ y \ $ , so we may expect critical points to be located symmetrically about the plane $ \ y \ = \ x \ $ .  The constraint curve is an inclined "great circle" on the surface of a sphere centered on the origin, so this may also produce a limited amount of symmetry.

$$ $$

In the two-variable, one-multiplier system Claude Leibovici describes, the Lagrange equations are

$$ y \ - \ 2 \ \ = \ \ \lambda · (2x \ + \ y) \ \ \ , \ \ \ x \ - \ 2 \ \ = \ \ \lambda · (2y \ + \ x) \ \ . $$

We will not be able to factor simply the equations we produce by moving all the terms to one side; instead, we consider solving the equations for $ \ \lambda \ : $

$$ \lambda \ \ = \ \ \frac{x \ - \ 2}{2y \ + \ x} \ \ = \ \  \frac{y \ - \ 2}{2x \ + \ y} \ \ . $$

Provided the denominators are not zero (we'll see that they aren't), we "cross-multiply" the ratio to find

$$ 2x^2 \ + \  xy \ - \ 4x \ - \ 2y \ = \ 2y^2 \ + \ xy \ - \ 4y \ - \ 2x \ \ \Rightarrow  \ \ x^2   -  x \  = \ y^2   -   y  \ \ . $$

(The challenge in solving systems of non-linear equations is in finding an evident path to a solution -- when there is such a path; I'll admit that it wasn't clear to me right away how to continue. In fact, I didn't see it until I'd already solved the two-multiplier system.)

$$ \Rightarrow  \ \ y^2   -  x^2 \  = \ y    -   x  \ \ \Rightarrow \ \ (y + x) · (y - x) \ \ = \ \ y - x \ \ \Rightarrow \ \  (y + x - 1)·(y - x) \ \ = \ \ 0 \ \ . $$

[PROTIP:  Resist the urge to simply "cancel" like factors on either side of an equation; you may neglect possible solutions to the equation or create false ones in doing so.]

So we have the two cases, $ \ y \ = \ x \ $ (as anticipated) and $ \ y \ = \ 1 - x \ . $ With the constraint equation written as $ \ x^2  \ + \ xy \ + \ y^2 \ = \ 12 \ , $ we find

$$ \ \mathbf{y = x :  } \quad x^2 \ + \ x·x \ + \ x^2 \ = \ 12 \ \ \Rightarrow \ \ 3x^2 \ = \ 12 \ \ \Rightarrow \ \ x \ = \ y \ = \ \pm 2 \ \ , \ \ z \ = \ -(x+y) \ = \ \mp 4 \ \ ; $$

$$ \ \mathbf{y = 1-x :  } \quad x^2 \ + \ x·(1-x) \ + \ (1-x)^2 \ = \ x^2 \ - \ x \ + \ 1 \ = \ 12 \ \ \Rightarrow \ \ x^2 \ - \ x \ - \ 11 \ = \ 0 $$ $$ \Rightarrow \ \ x \ = \ \frac{ 1 \ \pm \ \sqrt{ \ 1 \ - \ 4·1·(-11)}}{2} \ = \ \frac{1 \ \pm \ \sqrt{45}}{2} \ \ \text{or} \ \ \frac{1 \ \pm  \ 3\sqrt{5}}{2} \ \ , \ \ y \ = \ \frac{1 \ \mp  \ 3\sqrt{5}}{2} \ \ , $$
$$  z \ = \ -  \left( \frac{1 \ \pm  \ 3\sqrt{5}}{2} \ + \ \frac{1 \ \mp  \ 3\sqrt{5}}{2} \right)   \ = \ -1 \ \ . $$

The two extremal values from the $ \ y = x \ $ case yield  $ \ f( \ 2 \ ,\ 2 \ , \ -4 \ ) \ \  = \ \ 2·2  \ + \ 2·(- 4)  \  =  -4 \ $  and

$$ \ f( \ -2 \ ,\ -2 \ , \ +4 \ ) \ \  = \ \ (-2)·(-2)  \ + \ 2·4  \ \ = \ \ 12 \ \ , $$

the maximum value for our function.  From the second case, we obtain the minimum value,

$$ \ f \left( \frac{1 \ \pm  \ 3\sqrt{5}}{2} \ , \ \frac{1 \ \mp  \ 3\sqrt{5}}{2} \ , \ -1 \right) \ \  = \ \ \left( \frac{1 \ \pm  \ 3\sqrt{5}}{2}  \right)·\left( \frac{1 \ \pm  \ 3\sqrt{5}}{2}  \right)  \ + \ 2·(-1) $$

$$  = \ \ \frac{1}{4} \ - \ \frac{45}{4} \ - \ 2 \ \ = \ \ \frac{-52}{4} \ \ = \ \ -13 \ \ . $$

$$ $$

We can also work with all three coordinate variables and two multipliers to produce the system of equations

$$ y \ = \  \lambda · 1 \ + \ \mu · 2x \ \ , \ \ x \ = \  \lambda · 1 \ + \ \mu · 2y \ \ , \ \ 2 \ = \  \lambda · 1 \ + \ \mu · 2z \ \ . $$

Solving each of these for $ \ \lambda \ $ gives us

$$ \lambda \ \ = \ \ y \ - \ 2 \mu x \ \ = \ \ x \ - \ 2 \mu y \ \ = \ \ 2 · (1 \ - \ \mu z) \ \ . $$

The first pair-equation can be re-written as $ \ y \ + \ 2 \mu y \ - \ x \ - \ 2 \mu x \ = \ 0 \ \ \Rightarrow \ \ (y - x) · ( 1 \ + \ 2 \mu) \ = \ 0 \ \ , $
leading us to the two cases $ \ y \ = \ x \ $ and $ \ \mu \ = \ -\frac{1}{2} \ . $

For $ \ y \ = \ x \ , $ we obtain $ \ z \ = \ -(x+y) \ = \ -2x \ \ \Rightarrow \ \ z^2 \  = \ 4x^2 \ = \ 4y^2  \ . $  Putting this into the spherical constraint yields
$$ \ x^2 \ + \ x^2 \ + \ 4x^2 \ = \ 24 \ \ \Rightarrow \ \ x^2 \ = \ y^2 \ = \ 4 \ \ , \ \ z^2 \ = \ 16 $$

and the relations here imply $ \ x \ = \ y \ = \ \pm 2 \ \ \ , \ \ z \ = \ \mp 4 \ , \ $ as above.

Using $ \ \mu \ = \ -\frac{1}{2} \   $  in the $ \lambda-$ equation produces $ \ \lambda \   = \  x \ + \  y \  = \  2 \ + \ z \ \ . $  Inserting this into the planar constraint then gives us $ \ x \ + \ y \ + \ z \ = \ ( 2 + z) \ + \ z \ = \ 0 \ \Rightarrow \ z \ = \ -1 \ , \ $ and consequently, $ \ x \ + \ y \ = \ 1 \ \Rightarrow \ y \ = \ 1 \ - \ x \ . \ $  The spherical constraint equation becomes

$$ x^2 \ + \ (1 \ - \ x)^2 \ + \ (-1)^2 \ \ = \ \ 24 \ \ \Rightarrow \ \ x^2 \ + \ 1 \ - \ 2x \ + \ x^2 \ + \ 1 \ \ = \ \ 24 $$ $$ \Rightarrow \ \ 2x^2 \ - \ 2x \ - \ 22 \ \ = \ \ 0 \ \ , $$

which is essentially the quadratic equation we found earlier.


问:
Span of an empty list
Its a simple question . But still I thought of asking it coz I don't see the logic behind it. Recently I read a statement "the Span of empty list ( ) equals {0}" .What we know is an empty list has no elements in it,so how can we even talk of a span of it as span is a linear combination of all the vectors in a list and in empty list no vector is present at all. So, is it kind of just a declaration that we make just like that?? i guess there is no logic behind it and we just assume it like that. But why? What's the reason of doing so? And why do we need to declare a span of an empty list anyway?? Sorry I am not getting the logic behind it and that's why I thought of asking you guys. Thanks.</p>

答:
You can just assume it (or more correctly define it) that way.

That being said the span of a set $S \subset V$ in a vector space $V$ is just the smallest subspace of $V$ containing $S$, as such $\{0\}$ is the span of $\emptyset$. (Note that any subspace must contain $0$ by definition).

In terms of linear combinations you could think of the span of a set $S \subset V$ as the set $\{\lambda_1v_1 + ...\lambda_nv_n + 0 : n \in \mathbb N , v_1,...,v_n \in S, \lambda_1,...,\lambda_n \in \mathbb R\}$ (assuming $\mathbb R$ is the underlying field). This definition accommodates the possibility that the set $S$ is empty, as well as the possibility that the set $S$ is infinite.

Title: Span of an empty list


问:
Countable Union of an Infinite Set Equivalence
Assuming the link: Infinite set as union of disjoint countable sets. implies the following proposition:

Prop: If $\{A_i:i\in \mathbb{N}\}$ is a collection of disjoint infinite sets of the same cardinality, then
$$|\cup_i A_i|=|A_1|.$$
I would like to show it implies the next claim below. Given two sets $A$ and $B$, we define
$$|A|\cdot |B|:=|A\times B|.$$
Claim: If $T$ is an infinite set, then
$$\aleph_0\cdot |T|=|T|.$$

答: $\aleph_0 \times T$ is a countable collection of disjoint sets of the same cardinality, ( $A_i = \{i\} \times T$ ).


问:
Why does an exponential function eventually get bigger than a quadratic
I have seen the answer to this question and this one.

My $7$th grade son has this question on his homework:

How do you know an exponential expression will eventually be larger than any quadratic expression?

I can explain to him for any particular example such as $3^x$ vs. $10 x^2$ that he can just try different integer values of $x$ until he finds one, e.g. $x=6$. But, how can a $7$th grader understand that it will always be true, even $1.0001^x$ will eventually by greater than $1000 x^2$? They obviously do not know the Binomial Theorem, derivatives, Taylor series, L'Hopital's rule, Limits, etc,

Note: that is the way the problem is stated, it does not say that the base of the exponential expression has to be greater than $1$. Although for base between $0$ and $1$, it is still true that there exists some $x$ where the exponential is larger than the quadratic, the phrase "eventually" makes it sound like there is some $M$ where it is larger for all $x>M$. So, I don't like the way the question is written.

答:
I'd use a geometric argument combined with the principle of induction to convince your son. A simple, but fairly rigorous proof that is probably understandable to a middle schooler taking algebra.

Have your son imagine a square sheet of paper with side length $b$, where $b$ is very large, then by increasing its side length by $1$ we get a strip of paper added to the top and side of square, but increasing its dimension by $1$ will add far more material, since the strip of paper can clearly be cut and fit completely inside the cube with plenty of room to spare.

Therefore, $(b + 1)^2 < b^{2 + 1}$ (again, reminding your son that $b$ is very large).

In fact, if you like, you can just exhibit such a $b$ without giving the argument above, but I think it may help it sink in that increasing the variable in the exponential increases the volume of the square by an entirely new dimension while increasing the variable in a square just adds a tiny 2d slit of paper in comparison.

Now we just "repeat" the argument (you can use the word 'induction' if you like, but I think 'repeat' gets the idea across). Suppose that we know

$$(b + n)^2 < b^{2 + n}$$

for some $n$. We want to show that

$$(b + n + 1)^2 < b^{2 + n + 1}$$

we can rewrite this as

$$(b + n)^2 + 2(b + n) + 1 < b^{2 + n} b$$
The LHS rewrite can be seen geometrically. The side and top strips have length $(b + n)$ and height 1, and  the corner has area 1, a diagram can be drawn to see this.

(Alternatively, not sure if they use this term in school anymore, but this way of rewriting the LHS was called "FOILing" when I was in middle school)

This inequality is definitely true, since

$$b^{2 + n} b = b^{2 + n}  + \dots + b^{2 + n} \text{ ($b$ times)} $$,

This can also be argued geometrically, but it's trickier to visualize in dimensions higher than 3.

By our assumption,  $$(b + n)^2 < b^{2 + n}$$

Therefore, $$(b + x)^2 < b^{2 + x}$$ for all natural numbers $x$ greater than or equal to 1.

If your son is interested in these types of things, then thinking about 'higher' dimensions will probably add an air of intrigue to the whole thing. The above induction step can be explained geometrically, and even still remain rigorous if a bit more work is put into it.

The idea is that, when you add 1 to the length of a square, you are adding only a 2 dimensional amount of stuff, but when you add 1 to the dimension of a square, you are adding a 3 dimensional amount of stuff.

When you add two to the length of the square, you are still adding only a 2 dimensional amount of stuff. But when you add two to the dimension of a square, not only are you adding the 3 dimensional amount of stuff, but you are also adding 4 dimensional amount of stuff now as well, and so on.


答: It makes sense if you think about the span of a set of vectors as the smallest vector space containing the set of vectors. What is the smallest vector space containing the empty set? It's the smallest vector space since all vector spaces contain the empty set. Therefore, the span of the empty set is the zero vector space.</p>

Title: Span of an empty list

Body: Its a simple question . But still I thought of asking it coz I don't see the logic behind it. Recently I read a statement "the Span of empty list ( ) equals {0}" .What we know is an empty list has no elements in it,so how can we even talk of a span of it as span is a linear combination of all the vectors in a list and in empty list no vector is present at all. So, is it kind of just a declaration that we make just like that?? i guess there is no logic behind it and we just assume it like that. But why? What's the reason of doing so? And why do we need to declare a span of an empty list anyway?? Sorry I am not getting the logic behind it and that's why I thought of asking you guys. Thanks.</p>


问:
Find the point on the graph of$ z=x^2−y^2$ at which vector $\vec{n}=⟨−8,12,−2⟩$ is normal to the tangent plane $P$.
The normal vector to a tangent plane is the $∇F(x,y,z)$, hence $\langle 2x,-2y,-1 \rangle $ from $F(x,y,z)=x^2-y^2-z$.
So, if the normal vector is equal to $\langle -8,12,2 \rangle$, then $-8=2x, 12=-2y$ and $-1=-2$ (?)
It is wrong, but I don't see how.
Any help is appreciated.
Thank you

答: You need to have:  $\vec{n} \parallel \nabla{F}$ at $(x,y,z) \in C, C: z= x^2-y^2$. This means: $\langle -8,12,-2 \rangle = k\langle 2x,-2y,-1\rangle \implies 2kx=-8, -2ky=12, -k = -2\implies k = 2, x = -2, y = -3, z = (-2)^2 - (-3)^2 = 4 - 9 = -5.$ Thus the point you are solving for is: $(-2,-3,-5)$.


问:
Two definitions of representation of a topological group
Please compare the following two definitions:

Definition 1

A representation of a topological group $G$ in a vector space ${\mathbb{V}}$ over complex numbers is a continuous map
$$
  A\, :\quad G\times {\mathbb{V}}\longrightarrow{\mathbb{V}}\;\,,\qquad (g\,,\;v)\longmapsto A(g)\,v\;\,,\;\;\;\;\; g\in G\,,\;\;v\in{\mathbb{V}}\;\; \label{1}\tag{1}
  $$
with the following properties:

(1) $\;\; A\,$ is a group action;

(2) $\;\; A\,$ is linear, i.e.
$$
  A(g)\,(\alpha\, v+\beta)=\alpha\, A(g)\, v+ A(g)\, w\quad \mbox{for}\quad\forall g\in G\,,\;\; v,\; w\in {\mathbb{V}}\,,\;\; \alpha\in{\mathbb{C}}\,\;.
  $$

Definition 2

A representation of a topological group $G$ in a vector space ${\mathbb{V}}$ is a homomorphism of $ G$ into the group of invertible linear transformations of ${\mathbb{V}}$:
$$
 A\,:\quad G\longrightarrow GL({\mathbb{V}})\;\,. \label{2}\tag{2}
 $$

$$
\,
$$
Definition 2 $\;\;\Longrightarrow\;\;$ Definition 1,

Indeed, the $A$ from Definition 2 is linear and ensures $\,(g\,,\;v)\longmapsto A(g)\,v\,$.

Also, Definition 2 says that $A$ is a homomorphism $-$ which guarantees the continuity of $\,(g\,,\;v)\longmapsto A(g)\,v\;$ in Definition 1.

QED

To show that Definition 1 $\;\Longrightarrow\;$ Definition 2, we must demonstrate that Definition 1 ensures \eqref{2} being a homomorphism.

How to do this?

答:
Actually, a group action by $G$ on a set $X$ is equivalent to a group homomorphism $\psi: G\to \operatorname{Aut}(X)$, where in this case automorphisms are bijections of the set with itself. Indeed, the group action can be thought of as $\alpha:G\times X\to X$ such that $\alpha(hg,x)=\alpha(h,\alpha(g,x))$ etc. This is equivalent to $\psi(hg)=\psi(h)\circ \psi(g)$ where $\psi(g)(x)=\alpha(g,x)$. $\alpha(e_G,x)=x$ for all $x\in X$ is equivalent to $\psi(e_G)=\operatorname{Id}_X$. These are good facts to check carefully yourself as an exercise in the definitions (also usually in any algebra text).

So, definition (1) already gives us a group homomorphism $G\to \operatorname{Aut}^{\operatorname{Sets}}(V)$. A priori, the codomain on the right is self-bijections. However, the requirement that $G$ act linearly means that $G$ maps into $\operatorname{GL}(V)\subsetneq \operatorname{Aut}^{\operatorname{Sets}}(V)$. The continuity of the action stipulated in definition (1) then implies that as a map $G\to \operatorname{GL}(V)$ it is continuous, where $\operatorname{GL}(V)$ is viewed as a topological group.


问:
Relation between Geometric and Exponential random variables
Let $Y_n$ be a random variable following the exponential distribution with $\lambda$ as a parameter, $Y_n \sim \mathcal E (\lambda)$

For $\theta > 0$, we define $X_n$ as following :

$$X_n = [\theta Yn]$$
Where $[\theta Y_n]$ denotes the superior whole part :

$$[x]=min\{k\in \mathbb Z, k\geq x\}$$

How to find $\theta$ such that $X_n$ follows a geometric distribution ? $X_n \sim \mathcal G (p)$ ?

$\textbf{My attempt : }$
\begin{align}
\mathbb P (X_n=j)
&=\mathbb P(min\{k\in \mathbb Z, \theta Y_n \leq k\}=j)\\
&=\mathbb P(\theta Y_n\leq j)\\
&=\mathbb P(Y_n\leq \frac{j}{\theta})
\end{align}
Since $Y_n \sim \mathcal E (\lambda)$, then:
$$\large \mathbb P(Y_n\leq \frac{j}{\theta})=1-e^{-\lambda\frac{j}{\theta}}$$
And that's where I got. I didn't quite understant the part of the min.

答:
You define $$[x] = \min\{k \in \mathbb Z, k \ge x\}.$$  This means, $[x]$ is the smallest integer $k$ such that $k \ge x$.  So for instance, $[3.2] = 4$, because $3 < 3.2 < 4$.

In the mathematical literature, what you define as $[x]$ is more commonly written $\lceil x \rceil$, called the ceiling of $x$.  I will use this notation but the meaning is the same.

What does it mean when we say $$X_n = \lceil \theta Y_n \rceil?$$  Specifically, what is the relationship between $\Pr[X_n = x]$, the probability that $X_n = x$, and the corresponding values of $Y_n$?  By definition, $$\Pr[X_n = x] = \Pr[\lceil \theta Y_n \rceil = x].$$  But what values of $Y_n$ satisfy $\lceil \theta Y_n \rceil = x$?  For example suppose we are interested in the case $x = 4$.  Then $\theta Y_n$ must be strictly greater than $3$, but less than or equal to $4$.  Since if $\theta Y_n = 3$, then $\lceil \theta Y_n \rceil = 3$; and if $\theta Y_n = 4.00001$, then $\lceil \theta Y_n \rceil = 5$--we still have to "round up" to the nearest larger integer.  So $$\Pr[\lceil \theta Y_n \rceil = x] = \Pr[x-1 < \theta Y_n \le x], \tag{1}$$ where we remind ourselves that $x$ must be an integer.

Now, we can apply the fact that $Y_n$ is exponential with rate $\lambda$:  $$\begin{align}
\Pr[x - 1 < \theta Y_n \le x] &= \Pr\left[\frac{x-1}{\theta} < Y_n \le \frac{x}{\theta}\right] \\
&= F_{Y_n}(x/\theta) - F_{Y_n}((x-1)/\theta) \\
&= (1 - e^{-\lambda x/\theta}) - (1 - e^{-\lambda (x-1)/\theta}) \\
&= e^{-\lambda (x-1)/\theta} - e^{-\lambda x/\theta} \\
&= (1 - e^{-\lambda/\theta}) e^{-\lambda (x-1)/\theta}. \tag{2} \end{align}$$

where $F_{Y_n}(y) = 1 - e^{-\lambda y}$ is the cumulative distribution function for an exponential random variable with rate $\lambda$.  Notice that we've written the probability as two factors, the first of which is independent of $x$, depending only on fixed constants $\lambda$ and $\theta$.

What does a geometric distribution look like?  Well, suppose $W$ is geometric with parameter $p$; e.g., $$\Pr[W = w] = (1-p)^{w-1} p, \quad w \in \{1, 2, 3, \ldots \}. \tag{3}$$  Notice that I have chosen a parametrization such that $W$ has positive integer support; some parametrizations include $0$, but because $\lceil x \rceil$ for $x > 0$ is never $0$, we won't use such a parametrization.

Then all that is left is to compare the geometric PMF in Equation $(3)$ against the result we got in Equation $(2)$.  Here, $x$ and $w$ must play the same role.  What is $p$?  Clearly, we must have $$(1-p)^{w-1} = e^{-\lambda (x-1)/\theta} = (e^{-\lambda/\theta})^{x-1},$$ so $$p = 1 - e^{-\lambda/\theta},$$ and we write
$$\Pr[X_n = x] = (1-p)^{x-1} p, \quad p = 1 - e^{-\lambda/\theta}, \quad x \in \{1, 2, 3, \ldots\}.$$

What this tells us is that for any $\theta > 0$, the distribution of $\lceil \theta Y_n\rceil$ is geometric with parameter $p = 1 - e^{-\lambda/\theta}$.  If you want a specific $p$, then the value of $\theta$ that gives you a geometric distribution with such a $p$ is $$\theta = -\frac{\log(1-p)}{\lambda}.$$


问:
Find $P(X_{(1)} < \mu < X_{(4)})$
Let $X_1, X_2, X3, X4$ be iid random variables from a normal distribution with a mean of $\mu$.

Find $P(X_{(1)} < \mu < X_{(4)})$

My try:

$P(X_{(1)} < \mu < X_{(4)})$ = $P(\mu < X_{(4)})$ - $P(\mu < X_{(1)})$

And I know how to get the distributions of $X_{(1)}$ and $X_{(4)}$. But I think there must be a simpler way to find $P(X_{(1)} < \mu < X_{(4)})$. Any suggestions would be great!

Title: Find $P(X_{(1)} < \mu < X_{(4)})$

Body:
\begin{align}
P(X_{(1)} < \mu < X_{(4)})&=1-P(X_{(4)}<\mu)-P(\mu<X_{(1)})\\
&=1-\left(\dfrac{1}{2}\right)^4-\left(\dfrac{1}{2}\right)^4\\
&=\boxed{\dfrac78}
\end{align}


问:
Calculation in Routh's theorem
The proof of Routh's theorem concludes with showing$$1-\frac{x}{zx+x+1}-\frac{y}{xy+y+1}-\frac{z}{yz+z+1}=\frac{(xyz-1)^2}{(xz+x+1)(xy+y+1)(yz+z+1)}.$$I seek an elegant "proof from the book" of this, rather than one that involves tedious, potentially error-prone algebra. In particular, it feels like symmetries, degree-counting etc. should make this obvious, rather than an accident where the numerator happens to have a nice factorization. My best approaches are these two:

Option 1

The LHS's first two, three and four terms have respective sums $\frac{zx+1}{zx+x+1}$,$$\frac{(zx+1)(xy+y+1)-y(zx+x+1)}{(zx+x+1)(xy+y+1)}=\frac{x^2yz+zx+1}{(zx+x+1)(xy+y+1)}$$and$$\frac{x^2yz+xz+1}{(zx+x+1)(xy+y+1)}-\frac{z}{yz+z+1}=\frac{(x^2yz+xz+1)(yz+z+1)-z(zx+x+1)(xy+y+1)}{(zx+x+1)(xy+y+1)(yz+z+1)}.$$The numerator is nine monic terms of degree $0$ to $6$ minus nine monic terms of degree $1$ to $5$, so the terms of degree $0$ and $6$ will survive as $(xyz)^2+1$, and any other surviving term(s) will have coefficients summing to $-2$. The problem's symmetries mandate $-2xyz$ to finish the job.

That's quite nice, but the third partial sum probably can't be done in one's head. What one can say, however, is the third partial sum's numerator will be six terms of degrees $0$ to $4$ minus three of $1$ to $3$, so a $0$ and a $4$ survives, but it's harder to deduce without calculation that the third uncancelled term will be of degree $2$.

Option 2

This one looks like it might end up more elegant at first, but it looks like it ultimately requires some of Option 1's techniques to finish.

The case $x=\tfrac{1}{yz}$ has left-hand side$$1-\frac{1}{yz+z+1}-\frac{yz}{yz+z+1}-\frac{z}{yz+z+1}=0,$$so the general case's numerator must be divisible by $xyz-1$. In the special case $x=y=z$, the left-hand side is$$1-\frac{3x}{x^2+x+1}=\frac{(x-1)^2}{x^2+x+1}=\frac{(x^3-1)^2}{(x^2+x+1)^3}.$$The most obvious generalization with appropriate symmetries and denominator is $\frac{(xyz-1)^2}{(xz+x+1)(xy+y+1)(yz+z+1)}$, as desired. The most general numerator is of the form$$(xyz-1)(xyz+1+p(x,\,y,\,z)),$$where $p$ is invariant under a cyclic permutation of $x,\,y,\,z$, with $p(x,\,x,\,x)=0$ and $p\left(\tfrac{1}{yz},\,y,\,z\right)=0$.

The first constraint makes $p$ a polynomial in $a:=x+y+z,\,b:=xy+yz+zx,\,c:=xyz$; the second ensures that polynomial vanishes when $a=3x,\,b=3x^2,\,c=x^3$. These are achievable with a factor such as $a^2-3b$, $a^3-27c$, $ab-9c$ or $b^3-27c^2$. The third constraint only adds one requirement, divisibility by $c-1$. Ultimately, some careful degree-counting is needed to prove $p=0$.

答:
The question asks to find an elegant proof of
$$ 1-\frac{x}{zx\!+\!x\!+\!1}-\frac{y}{xy\!+\!y\!+\!1}
 -\frac{z}{yz\!+\!z\!+\!1}= \\
\frac{(xyz-1)^2}
 {(zx\!+\!x\!+\!1)(xy\!+\!y\!+\!1)(yz\!+\!z\!+\!1)}. \tag{1} $$
In order to simplify algebraic manipulation define
$$ X:= zx+x+1,\quad Y:= xy+y+1,\quad Z:= yz+z+1. \tag{2}$$
Move all terms to the same side of the equation and
eliminate all denominators to get
$$ 0 = -X\,Y Z+x\, Y Z+X\,y\, Z+X\,Y z+(1-xyz)^2. \tag{3} $$
Define the polynomial expression
$$ A := X\,Y Z-x\, Y Z-X\,y\, Z-X\,Y z. \tag{4} $$
Proving equation $(3)$ and equation $(1)$ is equivalent to
proving $$ A = (1-xyz)^2.\tag{5} $$

One possible proof is to identity the homogeneous parts of
the degree six polynomial $\,A.\,$

For degrees $0$ and $6$ the only contribution is from the
first term of $\,A\,$ and thus
$$ A_0 = 1, \qquad A_6 = (xyz)^2.  $$
For similar reasons,
$$ A_1 = (x+y+z)-x-y-z=0.  $$

$$ A_5 = (xyz)^2(1/x+1/y+1/z) - (xyz)^2(1/z+1/x+1/y) = 0.$$

$$ A_2 = 2(zx+xy+yz)-x(y+z)-y(x+z)-z(x+y)=0.  $$

$$A_4 = 2xyz (x+y+z)-xyz ((x+y)+(y+z)+(z+x))=0. $$

$$ A_3 \!=\! (4xyz\!+\!z^2x\!+\!x^2y\!+\!y^2z) \\
\!-\!xy(2z\!+\!x)\!-\!yz(2x\!+\!y)\!-\!zx(2y\!+\!z)
\!=\! -2xyz. $$
Putting all the parts together proves equation $(5)$.

This is essentially expanding the polynomial expression
$\,A\,$ and doing the same with $\,(1-xyz)^2.$

Another possible proof is to note that $\,A\,$ is a
polynomial in $\,x,y,z\,$ with maximum degree of each
variable being $2$. There are $3^3=27$ possible
monomials in $\,A.\,$ If equation $(5)$ can be proved
for at least $27$ generic values of $\,x,y,z\,$ then
the equation holds in general. Consider
$$ x=a/b,\;\; y=b/c,\;\; z=c/a,\;\;
x y z = 1, \\ d:=a+b+c,\quad (X,Y,Z) =
d\Big(\frac1b,\;\frac1c,\;\frac1a\Big), \\
A = \frac{d^2}{abc}(d-a-b-c) = 0 = (1-1)^2. $$
We can choose any nonzero values for $\,a,b,c\,$
which gives more than enough values of $\,x,y,z\,$
which proves equation $(5)$.

A simpler variation of this proof uses
$$ x = y = z =: w,\quad x y z = w^3,\quad X=Y=Z=1+w+w^2 =: W,
\\ A = W^3-3wW^2 = W^2(W-3w) = (W(1-w))^2 = (1-w^3)^2. $$

Title: Calculation in Routh's theorem


问:
Is the "derivative limit theorem" true  in the higher dimension?
<s>I am wondering if the following statement is a canonical theorem in real analysis. Does anybody here know the exact reference for it? It maybe a corollary of the mean value theorem, I think. Motivated by the answer to this question, I curious about that if this statement is still true in the higher dimension case.</s>

The following statement is a corollary of the Mean Value Theorem.

Let $x_0\in{\bf R}$ and $U(x_0)$ be
  the neighborhood of $x_0$.

$$f:U(x_0)\to{\bf R}$$

is continuous on $U(x_0)$ and
  differentiable on $U(x_0)\setminus\{x_0\}$.
   If the limit

$$\lim_{x\to x_0;x\in
U(x_0)\setminus\{x_0\}}f'(x)$$

exists, then $f$ is differentiable at
  $x_0$ and

$$f'(x_0)=\lim_{x\to x_0;x\in
U(x_0)\setminus\{x_0\}}f'(x)$$

Here is my question:

Is this statement still true for the higher dimension if
$$f:U(x_0)\subset{\mathbb R}^n\to{\mathbb R}^m ?$$

Since I don't know any generalization of the MVT in the higher dimension, I think one may need to approach it directly by definition. On the other hand, if one can deal with it "component-wise", one may be able to reduce it to the one dimension case.

答:
It also follows from L'Hospital's rule applied to $\frac{f(x)-f(x_0)}{x-x_0}$.

Added (to clarify for the ones who downvoted the answer)

Just to clarify a little, actually in this case using L'Hospital is really the same way as using Patrick proof implicitely.

L'H is proven using Cauchy's mean value Theorem (which by the way doesn't ask for $f$ not be constant). But in our case, $g(x)=x-x_0$, and hence Cauchy's Mean Value Theorem is exactly the standard Mean Value Theorem.

Last but not least, if you look over Patrick's proof he basically proves uusing the MVT the following: If $f(x), x-x_0$ are continuous on $U$, differentiable on $U-\{ x_0 \}$ and $\lim_{x-x_0} \frac{f'(x)}{1} =L$ then $\lim_{x-x_0} \frac{f(x)-f(x_0)}{x-x_0} =L$.

While his proof is easier and much cleaner than simply using L'H, this is just a particular case on L'H... And L'H Theorem doesn't say we cannot use it in the trivial case, is just that we usually don't because in that case it is overkill. But it is not wrong....


问:
Kalman filter: observability
A standard state-space for Kalman filter is:
\begin{align}
x_{t+1}&= F x_{t} + Gw_t\\
y_t&= Hx_t + v_t.
\end{align}

We know that there exists a similarity transform (i.e., an invertible matrix) such that an equivalent input-output relation can be described by the state-space
\begin{align}
x_{t+1}&= \begin{pmatrix}F_{obs}&0 \\ F_{mixed}& F_{Non-Obs}
\end{pmatrix} x_{t} + \tilde{G} w_t\\
y_t&= \begin{pmatrix}H_{obs} &0 \end{pmatrix}x_t + v_t.
\end{align}
The objective is to construct an optimal estimator of $x_{t+1}$ based on $y_1,\dots,y_{t}$.

I have two questions:

Why not omitting the unobservable part of the state from the state-space? Since the unobservable part does not effect the measurements $y_t$, their best estimation is independent of the measurements.

My guess is that, in most cases, the state vector has some physical interpretation so that people do not want to omit this part from the problem.

In the more general setup, we can estimate a linear combination of the state. If we aim to estimate $Hx_t$ rather than $x_t$, is it reasonable to assume that the system is fully observable?

答:
For convenience for further use I first define the equivalent system as

\begin{align}
\begin{bmatrix}
x_{1,t+1} \\ x_{2,t+1}
\end{bmatrix} &=
\begin{bmatrix}
F_1 & 0 \\ F_2 & F_3
\end{bmatrix}
\begin{bmatrix}
x_{1,t} \\ x_{2,t}
\end{bmatrix} +
\begin{bmatrix}
G_1 \\ G_2
\end{bmatrix} w_t, \\
y_t &=
\begin{bmatrix}
H_1 & 0
\end{bmatrix}
\begin{bmatrix}
x_{1,t} \\ x_{2,t}
\end{bmatrix} + v_t.
\end{align}

The prediction and correction step of the Kalman filter can be written as

\begin{align}
\begin{bmatrix}
\hat{x}_{1,t+1|t} \\ \hat{x}_{2,t+1|t}
\end{bmatrix} &=
\begin{bmatrix}
F_1 & 0 \\ F_2 & F_3
\end{bmatrix}
\begin{bmatrix}
\hat{x}_{1,t|t} \\ \hat{x}_{2,t|t}
\end{bmatrix}, \\
\begin{bmatrix}
\hat{x}_{1,t+1|t+1} \\ \hat{x}_{2,t+1|t+1}
\end{bmatrix} &=
\begin{bmatrix}
\hat{x}_{1,t+1|t} \\ \hat{x}_{2,t+1|t}
\end{bmatrix} +
\begin{bmatrix}
K_1 \\ K_2
\end{bmatrix}
\left(y_{t+1} -
\begin{bmatrix}
H_1 & 0
\end{bmatrix}
\begin{bmatrix}
\hat{x}_{1,t+1|t} \\ \hat{x}_{2,t+1|t}
\end{bmatrix}\right).
\end{align}

Here the correction gains $K_1$ and $K_2$ might be time varying. Substituting the prediction step in the correction step and splitting the expressions for $\hat{x}_1$ and $\hat{x}_2$ yields

\begin{align}
\hat{x}_{1,t+1|t+1} &= (I - K_1\,H_1)\,F_1\,\hat{x}_{1,t|t} + K_1\,y_{t+1}, \\
\hat{x}_{2,t+1|t+1} &= (F_2 - K_2\,H_1\,F_1)\,\hat{x}_{1,t|t} + F_3\,\hat{x}_{2,t|t} + K_2\,y_{t+1}.
\end{align}

From this it can be observed that the observable part of the state, $\hat{x}_1$, is only affected by the previous estimate of the observable part of the state and the measured output. Therefore, the best estimate of $H_1\,x_{1,t}$ would be $H_1\,\hat{x}_{1,t|t}$, whose dynamics thus doesn't depend on $\hat{x}_2$. So using a reduced observable model by omitting $\hat{x}_2$ from the state of the initial model would yield the same estimating performance regarding $\hat{x_1}$ and thus $H_1\,\hat{x}_{1,t|t}$.

The unobservable part of the state, $x_2$, might however still be of interest in some cases. For example when that part of the state is controllable and you do want to drive it to some desired reference. Even though that part of the state is not observable, it can still be detectable if $F_3$ is stable/Schur. Therefore, it can still be possible to drive $x_2$ (close) to such desired reference, but usually when a state is not observable its associated estimation error covariance matrix can be quite large and thus such tracking performance of the desired reference might not be great.

Title: Kalman filter: observability


答: The expression under limit can be written as $$\frac{y^{1/3}-1}{x}=\frac{y^{1/3}-1}{y-1}\cdot\frac{y-1}{x}\tag{1}$$ where $$y=\frac{1-\sqrt{1-x}}{\sqrt{1+x}-1}=\frac{x}{1+\sqrt{1-x}}\cdot\frac{\sqrt{1+x}+1}{x}=\frac{\sqrt{1+x}+1}{1+\sqrt{1-x}}\tag{2}$$ Since $y\to 1$ as $x\to 0$ we can see that the first fraction in $(1)$ tends to $1/3 $ via the standard formula $$\lim_{t\to a} \frac{t^n-a^n}{t-a}=na^{n-1}\tag{3}$$ Hence the desired limit is equal to the limit of $$\frac{1}{3}\cdot\frac{y-1}{x}=\frac{\sqrt {1+x}-\sqrt{1-x}}{3x(1+\sqrt{1-x})}=\frac{2x}{3x}\cdot\frac{1}{(\sqrt{1+x}+\sqrt{1-x})(1+\sqrt{1-x})}$$ It follows that the desired limit is $$\frac{2}{3}\cdot\frac{1}{2\cdot 2}=\frac{1}{6}$$ In general avoid rationalization of expressions containing radicals beyond square roots and instead use the limit formula $(3)$. This saves a lot of typing effort and reduces visual clutter.

Title: Other ways to evaluate $\lim_{x \to 0} \frac 1x \left [ \sqrt[3]{\frac{1 - \sqrt{1 - x}}{\sqrt{1 + x} - 1}} - 1\right ]$?

Body:
Using the facts that:
$$\begin{align}
\sqrt{1 + x} &= 1 + x/2 - x^2/8 + \mathcal{o}(x^2)\\
\sqrt{1 - x} &= 1 - x/2 - x^2/8 + \mathcal{o}(x^2)\\
\sqrt[3]{1 + x} &= 1 + x/3 + \mathcal{o}(x)
\end{align}$$

I was able to evaluate the limit as follows:
$$\begin{align}
\lim_{x \to 0} \frac 1x \left [ \sqrt[3]{\frac{1 - \sqrt{1 - x}}{\sqrt{1 + x} - 1}} - 1\right ] &\sim \lim_{x \to 0} \frac 1x \left [ \sqrt[3]{\frac{\dfrac x2 + \dfrac{x^2} 8}{\dfrac x2 + \dfrac{x^2} 8}} - 1\right ] =\\
&= \lim_{x \to 0} \frac 1x \left [ \sqrt[3]{1 + \frac{2x^2}{4x - x^2}} - 1\right ] \sim\\
&\sim \lim_{x \to 0} \frac{2x^2}{12x^2 - 3x^3} = \frac 16
\end{align}$$

What are other ways to evaluate it? Maybe pure algebraically? I tried to rationalize the denominator, but got stuck at some point...


问:
Approximating nonlinear first order recurrence with differential equation
I have been investigating the following recurrence that pops out in describing the probability a particular Monte Carlo algorithm for min cuts succeeds:

$$P_k = P_{k-1} - P_{k-1}^2 / 4$$

where $P_0 = c$ for some $0 < c < 1$. In my case, all I need is a lower bound for $P_k$ but ultimately I want it to be asymptotically as tight as I can get because this success probability plays a role in how many times I will re-run the algorithm to make it much less likely the algorithm fails. A naive approach uses the fact that $0 \leq P_{k-1} \leq 1$ to make the recurrence become

$$P_{k} \geq P_{k-1}\left(1 - 1/4\right) = \frac{3}{4} P_{k-1} \geq \left(3/4\right)^{k} P_{0}$$

The depth of the recursion tree in the algorithm is $d = c \log(n)$ for some constant $c$ which makes the above bound state that $P_{d} = \Omega(1/n^b)$ for some constant $b$. This is not great because it can be shown that $P_{d} = \Omega(1/\log(n))$. From the original paper on the min cut algorithm, they tackled this recurrence by making $P_k = 4/Z_k$ for some new variable $Z_k$. With this, one can then find that $P_k = \Omega(1/k)$ and so the desired result falls out.

I asked a question recently here to try and get some intuition into this transformation because it was not intuitive to me. One of the simpler approaches mentioned in the answer effectively treated the recurrence as a finite difference approximation to some differential equation and used that idea to spit out a differential equation we could work on instead. By this approach, one can find that the related diffeq is $\dot{p} = -p/4$ with $p(0) = c$ and it gives us that $p(k) = 4/(4c^{-1} + k)$. Interestingly, this $p(k)$ is asymptotically correct when you compare to the $P_k = \Omega(1/k)$ result.

My question is, is there any rigorous way to relate the answer to a recurrence of the form $x_{k} = x_{k-1} + f(x_{k-1})$ for some function $f$ to the solution you get from the differential equation $\dot{x}(t) = f(x(t))$? In particular, are there any well known results that give inequalities that relate the two so I can find lower bounds or upper bounds of a recurrence using the corresponding differential equation (and vice versa)?

Edit

More specifically, if we have the recurrence $X_{k} = X_{k-1} + f(X_{k-1})$ with $X_{0} = c$ and differential equation $\dot{x}(t) = f(x(t))$ with $x(0) = c$, do we have any bound of the form $K_1 x(k) + d_1 \leq X_{k} \leq K_2 x(k) + d_2$ for all $k \geq 0$ with constants $K_1, d_1, K_2, d_2$?

答:
The calculus argument is that you have $\frac{dx}{dk}\approx -\frac{x^2}{4}$ suggesting $\int -\frac1{x^2}\, dx \approx \int \frac14\, dk$  i.e. $\frac{1}{x}\approx \frac{k}{4}+B$ for some constant $B$, i.e. $x\approx\frac{1}{\frac{k}{4}+B}$.  But here you have a discrete recursion so may need some adjustment.

When $0 < P_0=c \le 1$ you have the bounds $$\dfrac{1}{\frac k{3} + \frac1c} \le P_k \le \dfrac{1}{\frac k4 + \frac1c}$$ or if you dislike fractions of fractions, $$\dfrac{3c}{kc + 3} \le P_k \le \dfrac{4c}{kc + 4}$$ and this can be shown by induction, since $x-\frac{x^2}{4}$ is a monotone function on $[0,1]$.

You had already found the upper bound.  You can tighten the lower bound when $c<1$ so as to have $P_1$ exactly, with  $$\dfrac{1-\frac{c}4}{\frac{(k-1)}4 + \frac1c} \le P_k \le \dfrac{1}{\frac {k}4 + \frac1c}$$


问:
Interchanging expectations of log likelihood
I see in papers (here in eq. 3 or here on page 4, for example) that it can be done like this using Fubini
$$\mathbb{E}_x\mathbb{E}_\theta\log f(x|\theta)=\mathbb{E}_\theta\mathbb{E}_x\log f(x|\theta)$$
Where $f(x|\theta)$ is a conditional probability density.

I don't immediately see why $\mathbb{E}_\theta\mathbb{E}_x|\log f(x|\theta)|\leq\infty$. Is it actually the case? Or are some additional conditions on $f$ needed?

答:
The identities
$$\mathbb{E}_x\mathbb{E}_\theta \bigl[{\bf 1}_{\{f(x|\theta)>1\}}\log f(x|\theta)  \bigr]=\mathbb{E}_\theta\mathbb{E}_x\bigl[{\bf 1}_{\{f(x|\theta)>1\}}\log f(x|\theta)  \bigr]\quad \; \; (*)$$
and
$$\mathbb{E}_x\mathbb{E}_\theta \bigl[{\bf 1}_{\{f(x|\theta)\le 1\}}\log f(x|\theta)  \bigr]=\mathbb{E}_\theta\mathbb{E}_x\bigl[{\bf 1}_{\{f(x|\theta)\le 1\}}\log f(x|\theta)  \bigr]\quad \; \; (**)$$
always hold by Tonelli's_theorem [1]. Adding them will give the desired formula,  provided at least one of them is finite.

On the other hand, if $(*)$ yields $+\infty$ and  $(**)$ yields $-\infty$, then the original formula might be undefined. For example,
consider the densities
$$f(x|\theta)=\frac{C}{|x-\theta |\, [1+(\log |x-\theta|)^2]} \,,$$
where $C$ is chosen to ensure $\int f(x|\theta) \, dx=1$.
(One can give an even simpler degenerate example, replacing $\theta$ on the RHS of  this formula by $0$.)
To verify the example, the integration fact needed is that $$\int_1^\infty \frac{dx}{x (1 +(\log  x)^\beta)} $$
is finite for $\beta>1$ but infinite for $\beta=1$. The change of variable $u =\log x$ transforms this to the standard integrals
$$\int_0^\infty \frac{du}{  (1 +u^\beta )} \,. $$
Similarly,
$$\int_0^1 \frac{dx}{x (1 +|\log x|^\beta )} $$
is finite for $\beta>1$ but infinite for $\beta=1$.

[1]
https://en.wikipedia.org/wiki/Fubini%27s_theorem#Tonelli's_theorem_for_non-negative_measurable_functions


问:
Finding tangent planes using differential geometry "language"
I'm studying differential geometry and I have a couple of questions (which are related).

I'm supposed to find the tangent planes to the surface given by $x^2+y^2-z^2=1$ at points $(x,y,0)$.

Is it enough to consider only the "upper" part of this hyperboloid, by using $\varphi(x,y)=(x,y,\sqrt{x^2+y^2-1})$ and work it as I would with the graph of functions?

If so, why? My reasoning was that even if I had to "complete" the surface with the lower part of the hyperboloid,  $\tilde{\varphi}(x,y)=(x,y,-\sqrt{x^2+y^2-1})$, the tangent planes would have to agree at $(x,y,0)$ so I can choose either one. I guess this would imply that it is sufficient to investigate tangent planes in any local parametrization that contains the points of interest.

Supose I can proceed as stated. Then I know how to find the tangent plane as I used to do in multivariable calculus, but I was also given the following deffinition:

Given a local parametrization $\varphi: U \subset \mathbb{R}^2 \to S$ where $S$ is a regular surface and a point $p \in \varphi(U)$. The tangent plane to this point is :

$$T_pS=D\varphi(\varphi^{-1}(p))(\mathbb{R}^2)$$

Which leads me to my next problem:

I'm having a hard time understanding the differential notation.

I tried to apply this to $\varphi(x,y)=(x,y,\sqrt{x^2+y^2-1})$ at points $p=(x_0,y_0,0)$.

As far as I'm concerned, $\varphi^{-1}(p)=(x_0,y_0)$ and

$$
D\varphi((x_0,y_0))=\begin{bmatrix}
1 &0 \\ 0& 1 \\ \dfrac{x_0}{\sqrt{x_0^2+y_0^2-1}} &\dfrac{y_0}{\sqrt{x_0^2+y_0^2-1}}\\
\end{bmatrix}
$$

I don't know what to do next. What does the $\mathbb{R^2}$ has to do in the definition of the tangent plane, and how do I get to the tangent plane expression from here?

This was a long question but I would really appreciate anyone taking the time to help me here :´)

答:
1- I don't think its a good idea to use that parametrization. In differential geometry, parametrizations of regular surfaces are defined at open sets, so all the properties are well defined. That parametrization would not work for that, at least not in $z=0$

2-The two vectors on the columns are the basis of the tangent plane at that point. So any vector on the plane can be espressed by a linear combination of those vectors. Let's say you wanna find the tangent of a curve $\alpha (u(t), v(t))$ at a point $t=0$ (This curve is at your surface parametrized by $u$ and $v$) If $D_u$ and $D_v$ are the column vectors, so the tangent would be $D_u u'(0)+ D_v v'(0)$. In other words, it would be the matrix multiplication $D * (u'(0), v'(0))^T$
I'm kind new to differential geometry, but I have a test about it after tomorrow, so I've been studying. Hope this helps!


问:
If $a+b=2$ so $2+\frac{ab(a-b)^2}{4}\leq a^{1+\sqrt{b}}+b^{1+\sqrt{a}}\leq2+\frac{(a-b)^2}{4}$

Let $a$ and $b$ be non-negative numbers such that $a+b=2$. Prove that:
  $$2+\frac{ab(a-b)^2}{4}\leq a^{1+\sqrt{b}}+b^{1+\sqrt{a}}\leq2+\frac{(a-b)^2}{4}$$

In the left inequality the equality occurs also for $a=2$ and $b=0$, which makes additional problems.

For $b=2-a$ we get inequalities of one variable, but derivatives are very ugly and I think it can't help (I mean here $f'(x)=0$...).

Since the equality occurs for $a=b=1$, we can try the Taylor series, but I  failed in this way.

Thank you!

Body:
(Partial solution) For the right inequality:

WLOG, assume $a \le b$. Then $0 < a \le 1 \le b < 2$.
We have $\sqrt{b} = 1 + \frac{b - 1}{\sqrt{b} + 1} \ge 1 + \frac{b - 1}{\frac{b + 1}{2} + 1} = \frac{3b + 1}{b + 3}$
and $\sqrt{a} = 1 - \frac{1 - a}{\sqrt{a} + 1} \le 1 - \frac{1 - a}{(a + 1)/2 + 1} = \frac{3a + 1}{a + 3}$.
It suffices to prove that
$$a^{1 + \frac{3b + 1}{b + 3}} + b^{1 + \frac{3a + 1}{a + 3}}\leq 2+\frac{(a-b)^2}{4}. \tag{1}$$

Let $a = 1 - x, b = 1 + x$ for $0 \le x < 1$. (1) becomes
$$(1 - x)^2 \cdot (1 - x)^{\frac{2x}{4 + x}} + (1 + x)^2 \cdot (1 + x)^{\frac{-2x}{4 - x}} \le 2 + x^2.$$

Fact 1: $(1 + u)^{\frac{-2u}{4 - u}} \le 1 - \frac{1}{2}\,{u}^{2} + \frac{1}{8}\,{u}^{3}-{\frac {1}{96}}\,{u}^{4}+{\frac {11}{384}}
\,{u}^{5} \triangleq F(u)$ for all $u$ in $(-1, 1)$. (Hint: Take logarithm and then take derivatives.)

By Fact 1, it suffices to prove that
$$(1 - x)^2F(-x) + (1 + x)^2F(x) \le 2 + x^2$$
which is written as $\frac{1}{96}x^4(50 - 9x^2) \ge 0$. True.

We are done.


问:
Show the Mean Squared Error of the Sample Mean
I am trying to find the mean squared error of the sample mean $\frac{Y_1 + \dots + Y_n}{n}$, but I am a little stuck. Here is what I've done so far.

$$MSE[\hat{\theta}] = E\bigg[ \bigg(\frac{Y_1 + \dots + Y_n}{n} - \theta^2 \bigg)  \bigg] = E\bigg[ \bigg(\frac{Y_1 + \dots + Y_n}{n}  \bigg)^2 \bigg] - 2\mu E\bigg[ \frac{Y_1 + \dots + Y_n}{n}  \bigg] + \mu^2 $$

I tried factoring (as shown above), but I don't know how to simplify everything to get the mean squared error of the sample mean. Any suggestions?

答:
Suppose your estimator for $\mu$ is $\bar Y$. Write $$\begin{split}MSE(\bar Y)&=E(\bar Y^2)-2\mu E(\bar Y)+\mu^2\\&=Var(\bar Y)+[E(\bar Y)]^2-2\mu E(\bar Y)+\mu^2\end{split}$$

If each of the Y's has expected value $\mu$ and variance $\sigma^2$, we have $\frac{\sigma^2}{n}+\mu^2-2\mu^2+\mu^2=\frac{\sigma^2}{n}$

This is in the form of $Var(\bar Y)+Bias(\bar Y, \mu)=\frac{\sigma^2}{n}+0$ as $\bar Y$ is an unbiased estimator of $\mu$.


问:
Proof: A bijective linear map cannot send a line to a point
Let $T\colon\mathbb R^3 \rightarrow\mathbb R^3$ be a bijective linear map. Prove that $T$ can not send a line to a point.

I started off thinking that $T(t\vec{m}+\vec{b}) = \vec{x}$ should have infinitely many solutions, but then I don't know how to go on with the equation.

答: Assume that $T(x_0 + tx_1) = x_2$ for all $t\in\mathbb R$ with some $x_1\neq 0$ (otherwise this wouldn't be a line). By linearity, $x_2 = Tx_0 + tTx_1$, that is, $t\cdot Tx_1 = x_2 - Tx_0$ for all $t\in\mathbb R$. Setting $t=0$ yields $x_2 - Tx_0 = 0$ and then $t=1$ shows that $Tx_1 = 0$ and so $x_1 = T^{-1}(0) = 0$.  Contradiction!


问:
Computing continued fraction expansions
My question concerns the numerical accuracy of a continued fraction expansion.  A typical algorithm for computing a continued fraction can be written in Python as :

```<code>x0 = sqrt(2)

N = 40
a = [0]*N
u = [0]*N

x = x0
for k in range(N):
    a[k] = int(x//1)
    u[k] = x % 1      # Often replaced with x - a[k]  ???
    x = 1/u[k]

print(a)
</code>```

For $x=\sqrt{2}$, this produces

```<code>[1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 1, 1, 1, 3, 3, 1, 3, 1, 1, 2, 1809, 1, 2, 5, 2, 2, 1, 2, 1]
</code>```

Or for $x = \pi$, I get :

```<code>[3, 7, 15, 1, 292, 1, 1, 1, 2, 1, 3, 1, 14, 3, 3, 23, 1, 1, 7, 4, 35, 1, 1, 1, 2, 3, 3, 3, 3, 1, 1, 14, 6, 4, 5, 1, 7, 1, 5, 1]
</code>```

One observation is that round-off error has clearly crept into the calculation.  The expansion for $\sqrt{2}$ should be repeating 2s, and the approximation for $\pi$ is accurate up to the first appearance of 14 (12 terms after the leading $a_0=3$ term).

What I also found is that while round off error does creep in, the number of correct terms in the expansion is exactly what is needed to get convergents that agree with $x$ up to machine precision.  So for example,  convergent 20 of $\sqrt{2}$  is given by

```<code>54608393 / 38613965
</code>```

This convergent approximates $\sqrt{2}$ to $4.44 \times 10^{-16}$.  Convergent 21 agrees with $\sqrt{2}$ exactly (in finite precision arithmetic).  Interestingly, the expansion has exactly 20 correct terms (i.e. 2s) after the leading $a_0 = 1$.

Similarly, for $\pi$, the convergent 12 is given by

```<code>80143857 / 25510582
</code>```

which agrees with $\pi$ to within an error of $4.44 \times 10^{-16}$.  Convergent 13 agrees with $\pi$ exactly (in finite precision arithmetic).

In both cases, the numerators for the convergents agree with the OEIS. See A001333 and A002485.

The above observations led me to these questions :

Is there a standard stopping criteria that can be used to determine when the expansion has converged to a desired tolerance?

Is it always the case that one will have enough correct terms in the expansion to approximate the desired number to machine precision?

Is it possible to detect whether a continued fraction is periodic?  Or, if one knows it will be periodic (i.e. $x$ is a root of a quadratic with integer coefficients), is it possible to get the periodic sequence exactly?

It has also occurred to me that nobody would think of computing continued fraction expansions using finite precision arithmetic!  I would like to do this problem as an exercise in a beginning computational math course, and was hoping not to go into variable precision arithmetic (an area which is really outside of my wheel house).

答:
Here is a method for $\sqrt n$  that is likely to be what Fermat used in hand computations.

$$  \sqrt { 5} = 2 +     \frac{  \sqrt {5} - 2 }{ 1 }  $$
$$    \frac{ 1 }{   \sqrt {5} - 2 }  =  \frac{    \sqrt {5} + 2 }{1 } = 4 +  \frac{    \sqrt {5} - 2 }{1 } $$

Simple continued fraction tableau:

$$
 \begin{array}{cccccccc}
 & & 2 & & 4 & & 4 & \\
 \\
  \frac{ 0 }{ 1 }   &   \frac{ 1 }{ 0 }   & &   \frac{ 2 }{ 1 }   & &   \frac{ 9 }{ 4 }    \\
  \\
 & 1 & & -1 & & 1
 \end{array}
 $$

$$
 \begin{array}{cccc}
  \frac{ 1 }{ 0 }   & 1^2 - 5 \cdot 0^2 = 1 &     \mbox{digit}  &  2  \\
  \frac{ 2 }{ 1 }   & 2^2 - 5 \cdot 1^2 = -1 &     \mbox{digit}  &  4  \\
  \frac{ 9 }{ 4 }   & 9^2 - 5 \cdot 4^2 = 1 &     \mbox{digit}  &  4  \\
 \end{array}
 $$

$$\bigcirc  \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc \bigcirc\bigcirc  \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc  \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc \bigcirc $$

$$  \sqrt { 13} = 3 +     \frac{  \sqrt {13} - 3 }{ 1 }  $$
$$    \frac{ 1 }{   \sqrt {13} - 3 }  =  \frac{    \sqrt {13} + 3 }{4 } = 1 +  \frac{    \sqrt {13} - 1 }{4 } $$
$$    \frac{ 4 }{   \sqrt {13} - 1 }  =  \frac{    \sqrt {13} + 1 }{3 } = 1 +  \frac{    \sqrt {13} - 2 }{3 } $$
$$    \frac{ 3 }{   \sqrt {13} - 2 }  =  \frac{    \sqrt {13} + 2 }{3 } = 1 +  \frac{    \sqrt {13} - 1 }{3 } $$
$$    \frac{ 3 }{   \sqrt {13} - 1 }  =  \frac{    \sqrt {13} + 1 }{4 } = 1 +  \frac{    \sqrt {13} - 3 }{4 } $$
$$    \frac{ 4 }{   \sqrt {13} - 3 }  =  \frac{    \sqrt {13} + 3 }{1 } = 6 +  \frac{    \sqrt {13} - 3 }{1 } $$

Simple continued fraction tableau:

$$
 \begin{array}{cccccccccccccccccccccccc}
 & & 3 & & 1 & & 1 & & 1 & & 1 & & 6 & & 1 & & 1 & & 1 & & 1 & & 6 & \\
 \\
  \frac{ 0 }{ 1 }   &   \frac{ 1 }{ 0 }   & &   \frac{ 3 }{ 1 }   & &   \frac{ 4 }{ 1 }   & &   \frac{ 7 }{ 2 }   & &   \frac{ 11 }{ 3 }   & &   \frac{ 18 }{ 5 }   & &   \frac{ 119 }{ 33 }   & &   \frac{ 137 }{ 38 }   & &   \frac{ 256 }{ 71 }   & &   \frac{ 393 }{ 109 }   & &   \frac{ 649 }{ 180 }    \\
  \\
 & 1 & & -4 & & 3 & & -3 & & 4 & & -1 & & 4 & & -3 & & 3 & & -4 & & 1
 \end{array}
 $$

$$
 \begin{array}{cccc}
  \frac{ 1 }{ 0 }   & 1^2 - 13 \cdot 0^2 = 1 &     \mbox{digit}  &  3  \\
  \frac{ 3 }{ 1 }   & 3^2 - 13 \cdot 1^2 = -4 &     \mbox{digit}  &  1  \\
  \frac{ 4 }{ 1 }   & 4^2 - 13 \cdot 1^2 = 3 &     \mbox{digit}  &  1  \\
  \frac{ 7 }{ 2 }   & 7^2 - 13 \cdot 2^2 = -3 &     \mbox{digit}  &  1  \\
  \frac{ 11 }{ 3 }   & 11^2 - 13 \cdot 3^2 = 4 &     \mbox{digit}  &  1  \\
  \frac{ 18 }{ 5 }   & 18^2 - 13 \cdot 5^2 = -1 &     \mbox{digit}  &  6  \\
  \frac{ 119 }{ 33 }   & 119^2 - 13 \cdot 33^2 = 4 &     \mbox{digit}  &  1  \\
  \frac{ 137 }{ 38 }   & 137^2 - 13 \cdot 38^2 = -3 &     \mbox{digit}  &  1  \\
  \frac{ 256 }{ 71 }   & 256^2 - 13 \cdot 71^2 = 3 &     \mbox{digit}  &  1  \\
  \frac{ 393 }{ 109 }   & 393^2 - 13 \cdot 109^2 = -4 &     \mbox{digit}  &  1  \\
  \frac{ 649 }{ 180 }   & 649^2 - 13 \cdot 180^2 = 1 &     \mbox{digit}  &  6  \\
 \end{array}
 $$

$$\bigcirc  \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc \bigcirc\bigcirc  \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc  \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc \bigcirc $$

Title: Computing continued fraction expansions


问:
A basic question on conditional expectation and variance
How calculate variance of $X$ from the following statements : </p>

<p>$$ E[X\mid Y=1] = 2 + E[X]$$
$$ E[X\mid Y=2] = 3 + E[X]$$
$$ E[X\mid Y=3] =0$$
$$ P(Y=1)=0.5, P(Y=2)=0.3 ~~\text{and}~~ P(Y=3) =0.2 $$</p>

答:
I think you need to give more information. You can easily find
$$
  \mathbb{E}[X]
= \mathbb{E}[\mathbb{E}[X\mid Y]]
= 0.5(2 + \mathbb{E}[X]) + 0.3(3 + \mathbb{E}[X])
= \frac{19}{10} + \frac{4}{5}\mathbb{E}[X]
$$
so that $\mathbb{E}[X] = \frac{19}{2}$, but to find the variance you will need to know
$\mathbb{E}[X^2]$ or at least $\mathbb{E}[X^2\mid Y]$. Otherwise, there may be two different random variables with the properties you list yet having different variances. For example, the condition you gave is the same as
$$
  \mathbb{E}[X \mid Y ]
= -2Y^2 + 7Y - 3 + \frac{19}{2} \chi\{ Y \neq 3 \},
$$
so let
$$
  X_1
= -2Y^2 + 7Y - 3 + \frac{19}{2} \chi\{ Y \neq 3\}
$$
and let
$$
  X_2
= -2 N(Y^2,10) + 7U(0,2Y) - 3 + \frac{19}{2}\chi\{ Y \neq 3\},
$$
where $N(Y^2,10)$ is a normal random variable with mean $Y^2$ and variance $10$, and $U(0,2Y)$ is a uniform random variable on the interval $(0,2Y)$.
Both satisfy the given conditions, yet obviously have different variances.</p>


问:
Binomial Distribution Word Problem
A student gets at least 8 hours of sleep 45% of the nights; the sleeping schedule is independent from night to night. Let Xi indicate whether the student gets at least 8 hours of sleep during the next 4 night respectively for i = 1, 2, 3, and 4. Let X = X1 + X2 + X3 + X4. Find the variance of X.

I'm struggling to understand what kind of distribution this problem is follow. I think it would be binomial. In that case wouldn't the variance of having x successes in 4 nights (trials) be modelled by the binomial. Hence the variance of X is (npq)^0.5 ? However, I feel like the question wants me to do some RV algebra. But a bit stuck.

答:
There are four requirements for a binomial distribution:

Fixed number of trials

Only two possible outcomes per trial

Same success probability for every trial

Trials are independent of each other.

All four conditions are satisfied. $n = 4, p = 0.45$

Title: Binomial Distribution Word Problem


答:
This is the Gauss-Lagrange method of neighboring reduced indefinite forms. I learned this in Binary Quadratic Forms by D. A. Buell. It is also in Dickson (1929) Introduction to the Theory of Numbers.

The output below says that a root of $7x^2 + 3 xy - 7 y^2$  has purely periodic fraction 1,4,4,1. (You just take the absolute values of my "delta" numbers to make the continued fraction). Let me think about which root that might be. Alright, got it. It is actually the ratio  $r=y/x$ which solves $7+3r-7r^2,$  the relevant root being  $$ \frac{3+ \sqrt{205}}{14}  $$

```<code>=========================================
jagy@phobeusjunior:~/old drive/home/jagy/Cplusplus$ ./indefCycle 7 3 -7

  0  form              7           3          -7

           1           0
           0           1

To Return
           1           0
           0           1

0  form   7 3 -7   delta  -1
1  form   -7 11 3   delta  4
2  form   3 13 -3   delta  -4
3  form   -3 11 7   delta  1
4  form   7 3 -7

  form   7 x^2  + 3 x y  -7 y^2

minimum was   3rep   x = -1   y = -1 disc 205 dSqrt 14  M_Ratio  4
Automorph, written on right of Gram matrix:
17  21
21  26
=========================================
</code>```

This one is for $\frac{13 + \sqrt{1313}}{44}$

```<code>jagy@phobeusjunior:~/old drive/home/jagy/Cplusplus$ ./indefCycle 13 13 -22

  0  form             13          13         -22

           1           0
           0           1

To Return
           1           0
           0           1

0  form   13 13 -22   delta  -1     ambiguous
1  form   -22 31 4   delta  8
2  form   4 33 -14   delta  -2
3  form   -14 23 14   delta  2
4  form   14 33 -4   delta  -8
5  form   -4 31 22   delta  1
6  form   22 13 -13   delta  -1
7  form   -13 13 22   delta  1     ambiguous            -1 composed with form zero
8  form   22 31 -4   delta  -8
9  form   -4 33 14   delta  2
10  form   14 23 -14   delta  -2
11  form   -14 33 4   delta  8
12  form   4 31 -22   delta  -1
13  form   -22 13 13   delta  1
14  form   13 13 -22

  form   13 x^2  + 13 x y  -22 y^2

minimum was   4rep   x = -1   y = -1 disc 1313 dSqrt 36  M_Ratio  7.668639
Automorph, written on right of Gram matrix:
-486641  -921536
-544544  -1031185
=========================================
</code>```

Title: Computing continued fraction expansions

Body:
My question concerns the numerical accuracy of a continued fraction expansion.  A typical algorithm for computing a continued fraction can be written in Python as :

```<code>x0 = sqrt(2)

N = 40
a = [0]*N
u = [0]*N

x = x0
for k in range(N):
    a[k] = int(x//1)
    u[k] = x % 1      # Often replaced with x - a[k]  ???
    x = 1/u[k]

print(a)
</code>```

For $x=\sqrt{2}$, this produces

```<code>[1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 1, 1, 1, 3, 3, 1, 3, 1, 1, 2, 1809, 1, 2, 5, 2, 2, 1, 2, 1]
</code>```

Or for $x = \pi$, I get :

```<code>[3, 7, 15, 1, 292, 1, 1, 1, 2, 1, 3, 1, 14, 3, 3, 23, 1, 1, 7, 4, 35, 1, 1, 1, 2, 3, 3, 3, 3, 1, 1, 14, 6, 4, 5, 1, 7, 1, 5, 1]
</code>```

One observation is that round-off error has clearly crept into the calculation.  The expansion for $\sqrt{2}$ should be repeating 2s, and the approximation for $\pi$ is accurate up to the first appearance of 14 (12 terms after the leading $a_0=3$ term).

What I also found is that while round off error does creep in, the number of correct terms in the expansion is exactly what is needed to get convergents that agree with $x$ up to machine precision.  So for example,  convergent 20 of $\sqrt{2}$  is given by

```<code>54608393 / 38613965
</code>```

This convergent approximates $\sqrt{2}$ to $4.44 \times 10^{-16}$.  Convergent 21 agrees with $\sqrt{2}$ exactly (in finite precision arithmetic).  Interestingly, the expansion has exactly 20 correct terms (i.e. 2s) after the leading $a_0 = 1$.

Similarly, for $\pi$, the convergent 12 is given by

```<code>80143857 / 25510582
</code>```

which agrees with $\pi$ to within an error of $4.44 \times 10^{-16}$.  Convergent 13 agrees with $\pi$ exactly (in finite precision arithmetic).

In both cases, the numerators for the convergents agree with the OEIS. See A001333 and A002485.

The above observations led me to these questions :

Is there a standard stopping criteria that can be used to determine when the expansion has converged to a desired tolerance?

Is it always the case that one will have enough correct terms in the expansion to approximate the desired number to machine precision?

Is it possible to detect whether a continued fraction is periodic?  Or, if one knows it will be periodic (i.e. $x$ is a root of a quadratic with integer coefficients), is it possible to get the periodic sequence exactly?

It has also occurred to me that nobody would think of computing continued fraction expansions using finite precision arithmetic!  I would like to do this problem as an exercise in a beginning computational math course, and was hoping not to go into variable precision arithmetic (an area which is really outside of my wheel house).

Body:
This is the Gauss-Lagrange method of neighboring reduced indefinite forms. I learned this in Binary Quadratic Forms by D. A. Buell. It is also in Dickson (1929) Introduction to the Theory of Numbers.

The output below says that a root of $7x^2 + 3 xy - 7 y^2$  has purely periodic fraction 1,4,4,1. (You just take the absolute values of my "delta" numbers to make the continued fraction). Let me think about which root that might be. Alright, got it. It is actually the ratio  $r=y/x$ which solves $7+3r-7r^2,$  the relevant root being  $$ \frac{3+ \sqrt{205}}{14}  $$

=========================================
jagy@phobeusjunior:~/old drive/home/jagy/Cplusplus$ ./indefCycle 7 3 -7

  0  form              7           3          -7

           1           0
           0           1

To Return
           1           0
           0           1

0  form   7 3 -7   delta  -1
1  form   -7 11 3   delta  4
2  form   3 13 -3   delta  -4
3  form   -3 11 7   delta  1
4  form   7 3 -7

  form   7 x^2  + 3 x y  -7 y^2

minimum was   3rep   x = -1   y = -1 disc 205 dSqrt 14  M_Ratio  4
Automorph, written on right of Gram matrix:
17  21
21  26
=========================================

This one is for $\frac{13 + \sqrt{1313}}{44}$

jagy@phobeusjunior:~/old drive/home/jagy/Cplusplus$ ./indefCycle 13 13 -22

  0  form             13          13         -22

           1           0
           0           1

To Return
           1           0
           0           1

0  form   13 13 -22   delta  -1     ambiguous
1  form   -22 31 4   delta  8
2  form   4 33 -14   delta  -2
3  form   -14 23 14   delta  2
4  form   14 33 -4   delta  -8
5  form   -4 31 22   delta  1
6  form   22 13 -13   delta  -1
7  form   -13 13 22   delta  1     ambiguous            -1 composed with form zero
8  form   22 31 -4   delta  -8
9  form   -4 33 14   delta  2
10  form   14 23 -14   delta  -2
11  form   -14 33 4   delta  8
12  form   4 31 -22   delta  -1
13  form   -22 13 13   delta  1
14  form   13 13 -22

  form   13 x^2  + 13 x y  -22 y^2

minimum was   4rep   x = -1   y = -1 disc 1313 dSqrt 36  M_Ratio  7.668639
Automorph, written on right of Gram matrix:
-486641  -921536
-544544  -1031185
=========================================


问:
Closure and limit of a sequence
Let $E$ be a subset of a metric space $(S,d)$. I'm trying to show that an element is in $\overline{E}$ if and only if it is the limit of some sequence of points in $E$.

Suppose there is a sequence $(p_n) \subseteq E$. Then because $E \subseteq \overline{E}$, I know that $(p_n) \subseteq \overline{E}$. Since $\overline{E}$ is closed, $\overline{E}$ contains the limit of every convergent sequence of points in $\overline{E}$. So $p \in \overline{E}.$

This is how I approached the converse: I assumed $p \in \overline{E}$ and that there does not exist a sequence $(p_n) \subseteq E$ such that $p_n \to p$. Then I concluded that there exists  $\varepsilon > 0$ such that $d(p, x) \geq \varepsilon \ \forall x \in E$.

Then there exists an open ball centered at $p$ that does not contain a point of $E$ (I wasn't sure how to make this statement rigorous). So $p \notin \overline{E}$.

Could I get some feedback on my proof?

答:
The result is always true in the first countable space. Note that every metric space is first countable.

Let $X$ be first countable and $E$ is a subset of $X$. If $p∈\bar{E}$ iff there  exists a sequence $(p_n)⊆E$ such that $p_n→p$.

Title: Closure and limit of a sequence


问:
limit $\lim_{x \to \infty} \frac{\sqrt{x+2} - \sqrt{x+1}}{\sqrt{x+1} - \sqrt{x}}$
How can I find the limit to infinity of this function? As this is a $0/0$ equation, I tried using the L'Hôpital's rule in this but ended up making it more complex. I've also tried rationalising the denominator but it didn't lead to anywhere.

$$\lim_{x \to \infty} \frac{\sqrt{x+2} - \sqrt{x+1}}{\sqrt{x+1} - \sqrt{x}} $$

答: Assume WLOG $x\geq0$. Thus, $$\frac{\sqrt{x+2}-\sqrt{x+1}}{\sqrt{x+1}-\sqrt x}\cdot\frac{\sqrt{x+1}+\sqrt x}{\sqrt{x+1}+\sqrt x}$$$$=(\sqrt{x+1}+\sqrt x)(\sqrt{x+2}-\sqrt{x+1})\cdot\frac{\sqrt{x+2}+\sqrt{x+1}}{\sqrt{x+2}+\sqrt{x+1}}$$$$=\frac{\sqrt{x+1}+\sqrt x}{\sqrt{x+2}+\sqrt{x+1}}=1-\frac{\sqrt{x+2}-\sqrt x}{\sqrt{x+2}+\sqrt{x+1}}$$


问:
Finding the general solution of a differential equation using two methods.
Can somebody please help me in finding the general solution of $2x^3dy=y(y^2+3x^2)dx$ using another method?

My attempt for the first method:

$2x^3y^{-3}dy-3x^2y^{-2}dx=dx$

Let $v=y^{-2}$; $dv=-2y^{-3}dy$

The equation now becomes:

$-x^3dv-3x^2vdx=dx$

$\frac{dv}{dx}+\frac{3v}{x}=-\frac{1}{x^3}$

$P(x)=\frac{3}{x}$, I.F.$=e^{3\int x^{-1}dx}=x^3$

$vx^3=-\int \frac{x^3}{x^3}dx$

$vx^3=-x+C$

$x^3=y^2(C-x)$

答:
\begin{gather*}
\frac{dy}{dx} =\frac{y^{3}}{2x^{3}} +\frac{3y}{2x}\\
\\
Let\ y=tx\\
\frac{dy}{dx} =t+x\frac{dt}{dx}\\
t+x\frac{dt}{dx} =\frac{t^{3}}{2} +\frac{3t}{2}\\
x\frac{dt}{dx} =\frac{t^{3} +t}{2}\\
\\
\int \frac{2dt}{t^{3} +t} =\int \frac{dx}{x}\\
\\
Now,\ \int \frac{2dt}{t^{3} +t} =\int \frac{2dt}{t\left( t^{2} +1\right)} =\int \frac{2tdt}{t^{2}\left( t^{2} +1\right)}\\
\\
Let\ t^{2} =u\\
2tdt=du\\
\\
\int \frac{2dt}{t^{3} +t} =\int \frac{du}{u( u+1)} =\ln\frac{u}{u+1} =\ln\frac{t^{2}}{t^{2} +1} +C\\
\\
So,\ form\ the\ original\ differential\ equation,\\
\ln\frac{t^{2}}{t^{2} +1} +C_{1} =\ln x+C_{2}\\
\ln\frac{y^{2}}{y^{2} +x^{2}} -\ln x=C\\
\\
So,\ the\ general\ solution\ is\ \\
y^{2} =e^{c} x\left( x^{2} +y^{2}\right)
\end{gather*}
Hope this answers your question!


问:
Infinity in ZFC vs infinity in the metatheory
I was reading about different ways to formalize the notion of infinity in ZFC. The classic axiom is of course to have the existence of an inductive set, but you can also assert the existence of a Dedekind infinite set, and it turns out it is equivalent. (This question shows this: How can I define $\mathbb{N}$ if I postulate existence of a Dedekind-infinite set rather than existence of an inductive set?)

They both give a notion of infinity internal to the theory. However, I am not quite sure how much this coincides with a "metatheoretic" understanding of infinity. For instance, let $L=\{=,\in\}$ and let $T$ be the $L$-theory consisting of all axioms of ZFC except for the axiom of infinity. Define a new language $L'=L\cup\{c\}$ with a single additional constant symbol, and define a new theory $$T'=T\cup\{\exists x_1\dots\exists x_n\bigwedge_{1\leq i<j\leq n}x_i\neq x_j\wedge\bigwedge_{i=1}^n x_i\in c:n\in\mathbb{N}\}.$$ This seems to be a very natural way to capture the notion of infinity, but I don't know whether we could prove that the realization of $c$ in a model of $T'$ is Dedekind-infinite in the sense of ZFC. Therefore, in light of the question I put a link to, my question is: is $T'$ equivalent to ZFC?

答:
Not in any sense. $T'$ is in fact a conservative extension of $\mathsf{ZF}$-$\mathsf{Inf}$; this is a consequence of the compactness theorem.

Now any well-founded model (or even $\omega$-model) of $T'$ restricts to a model of $\mathsf{ZF}$, but in the absence of such a semantic restriction $T'$ is no better than $\mathsf{ZF}$.


答: No; in fact, $T'$ is conservative over $T$ (that is, it does not prove any theorems in $L$ that $T$ does not prove).  Indeed, every finite fragment of $T'$ is satisfied in every model of $T$, by letting $c$ denote some set with enough elements to satisfy all of the finitely many axioms you include.  Every theorem of $T'$ is proved from only finitely many of its axioms, so this means it can't prove anything over $L$ that isn't true in every model of $T$.

Title: Infinity in ZFC vs infinity in the metatheory

Body:
I was reading about different ways to formalize the notion of infinity in ZFC. The classic axiom is of course to have the existence of an inductive set, but you can also assert the existence of a Dedekind infinite set, and it turns out it is equivalent. (This question shows this: How can I define $\mathbb{N}$ if I postulate existence of a Dedekind-infinite set rather than existence of an inductive set?)

They both give a notion of infinity internal to the theory. However, I am not quite sure how much this coincides with a "metatheoretic" understanding of infinity. For instance, let $L=\{=,\in\}$ and let $T$ be the $L$-theory consisting of all axioms of ZFC except for the axiom of infinity. Define a new language $L'=L\cup\{c\}$ with a single additional constant symbol, and define a new theory $$T'=T\cup\{\exists x_1\dots\exists x_n\bigwedge_{1\leq i<j\leq n}x_i\neq x_j\wedge\bigwedge_{i=1}^n x_i\in c:n\in\mathbb{N}\}.$$ This seems to be a very natural way to capture the notion of infinity, but I don't know whether we could prove that the realization of $c$ in a model of $T'$ is Dedekind-infinite in the sense of ZFC. Therefore, in light of the question I put a link to, my question is: is $T'$ equivalent to ZFC?


问:
Triple integral for a pyramid (calculating volume)
Fix $H>0$ and let $P\subset\mathbb{R}^3$ be the solid pyramid with vertices at $(0,0,0),(1,0,0),(1,2,0),(0,2,0)$ and $(0,0,H)$. Calculate the volume $V_P$ of $P$.

I know that I must set up a triple integral to solve this problem, but I'm unsure of how to set this up as I have never done this before. How do I create the limits of the integrals and how do I set up the function within the integral?

答:
You can do this using two integrals, one for each triangular prism. Begin by drawing the xy plane and the four points of the base, as well as a diagonal from $(1,2)$ to $(0,0)$. For the lower right triangle, the plane formed by the side of the pyramid contains vectors $(0,0,H)-(1,0,0)=(-1,0,H)$ and $(0,0,H)-(1,2,0)=(-1,-2,H)$. Take the cross product to get $(2H, 0, 2)$, a normal vector to the plane. The equation of the plane can be found by the dot product $(2H, 0, 2) \cdot (x-0, y-0, z-H)=0$, giving $z=H(1-x)$. Then the limits of integration are $0<z<H(1-x), y/2<x<1, and 0<y<2$.

For the other half, the same thing happens. Find two vectors in the plane, they can be $(0,0,H)-(0,2,0)=(0,-2,H)$ and $(0,0,H)-(1,2,0)=(-1,-2,H)$ with a cross product of $(0, H, 2)$. Thus an equation of the plane is $(0,H,2)\cdot (x-0,y-0,z-H)=0$ giving us $z=H(1-y/2)$. The limits of integration are therefore $0<z<H\left(1-\frac y2\right),0<x<\frac y 2,0<y<2$.

The volume of the pyramid is given by $$\int_0^2\int_{\frac y 2}^1\int_0^{H(1-x)}dzdxdy+\int_0^2\int_0^{\frac y 2}\int_0^{H\left(1-\frac y 2\right)}dzdxdy$$

It turns out that the volume of the two pieces are the same.


问:
Isomorphisms between the induced topologies on countable spaces
A topological space $T$ is called canonical if all induced topologies on its infinite subsets are homeomorphic to $T$.
Is it true that every topology on an infinite countable set has a subset upon which the induced topology is canonical?

Thanks in advance.

答:
The following paper by Ginsburg and Sands offers a beautiful solution to this question.

https://www.jstor.org/stable/2320588?seq=1#metadata_info_tab_contents

Edit: The statement is correct, and this theorem allows one to prove that the only canonical topological spaces are those with

Discrete topology.

Indiscrete topology.

Cofinite topology.

Initial segment topology.

Final segment topology.

A trivial consequence is that any infinite topological space contains (at least) one of the above topological subspaces.

The proof essentially classifies the types of canonical subspaces that a space $T$ may contain by considering first the sets $cl\{a\}$ for each point $a\in T$. The induced topology on each $cl\{a\}$ is the indiscrete topology, hence if any of the $cl\{a\}$ are infinite we are done. Otherwise, all $cl\{a\}$ are finite and in particular we have infinitely many of them so we may replace the space $T$ by a space $T_0$ by taking a single member of the closure of each element, the new space $T_0$ has the separation property, namely that for all $a,b\in T_0$, $cl\{a\} \neq cl\{b\}$. One may define a partial ordering of the space by $a\le b$ if $a\in cl\{b\}$ and use a classification theorem of infinite graphs by Ramsey and Whaley saying that an infinite graph must contain an infinite chain, an infinite antichain, or one other thing that is harder to describe.

If $T_0$ contains an infinite chain, i.e. it has elements $a_1 < a_2 < ...$, then if $A$ is a non-empty open subset of $T_0$ intersecting the chain, then it has a minimal element, say $a_k$, and since an open set containing $a_k$ must contain $a_n$ for all $n\ge k$, by considering the subspace induced by the chain we obtain a subspace with the final segment topology. One continues by case analysis in the same spirit.


问:
Closed form for $\int_1^\infty\int_0^1\frac{\mathrm dy\,\mathrm dx}{\sqrt{x^2-1}\sqrt{1-y^2}\sqrt{1-y^2+4\,x^2y^2}}$
Consider the following integral:
$$\mathcal{I}=\int_1^\infty\int_0^1\frac{\mathrm dy\,\mathrm dx}{\sqrt{x^2-1}\sqrt{1-y^2}\sqrt{1-y^2+4\,x^2y^2}}.$$
It can be represented as
$$\mathcal{I}=\int_1^\infty\frac{K(\sqrt{1-4\,x^2})}{\sqrt{x^2-1}}\mathrm dx=\Re\int_0^1\frac{K\left(\sqrt\frac{1+3\,y^2}{1-y^2}\right)}{1-y^2}\mathrm dy,$$
where $K(x)$ is the complete elliptic integral of the 1st kind.

I was not able to further simplify any of these integrals, but numerical integration suggests the following conjectural closed form:
$$\mathcal{I}\stackrel{?}{=}\frac{3\,\Gamma(1/3)^6}{2^{17/3}\pi^2}.$$
Can you suggest a proof of this conjecture?

Using Mathematica I obtained a closed form for $\mathcal{I}$ in terms of the Meijer G-function:
$$\mathcal{I}=\frac{\pi^{3/2}}{4}G_{3,3}^{2,1}\left(\frac{1}{4}\left|\begin{array}{c}1,\ \ 1,\ \ 1\\\frac{1}{2},\frac{1}{2},\frac{1}{2}\\\end{array}\right.\right),$$
but I have no manual proof for that and do not know how to reduce it to $\frac{3\,\Gamma(1/3)^6}{2^{17/3}\pi^2}.$

答:
Edited It turns out this integral is done in http://arxiv.org/abs/0801.0891, which O.L. linked to in a comment to another question. This isn't much of an answer, really, as for all the closed forms I refer you to that paper.

Write the integral (by change of variable) as
$$ \frac12 \int_0^\infty K(\sqrt{-1-2\cosh \theta})\,d\theta, $$
then expand this into a double integral from the definition of $K$, and integrate over $\theta$ to get
$$ \int_0^1 \frac{dt}{\sqrt{1-t^2}\sqrt{1+3t^2}} K\left( \sqrt{\frac{1-t^2}{1+3t^2}} \right). $$
Another change of variable
$$ t^2 = \frac{1}{v^2+1} $$
will bring it to the form
$$ \frac1\pi\overline V_3(1,1,1) = \int_0^\infty \frac{dv}{\sqrt{(v^2+4)(v^2+1)}}K\left(\sqrt{\frac{v^2}{v^2+4}}\right), $$
which is equation (32) in that paper. It is evaluated by considering an integral of a product of three modified Bessel functions, and that integral was done in two papers by W. N. Bailey in 1934 ("Some indefinite integrals involving Bessel functions"). As a consequence, the integral equals
$$ \Re\left( K(e^{-\pi i/6})^2\right). $$
In that paper, (equation 39) they also give the equivalent form (I don't know a proof why this is equivalent to the above)
$$ \frac12 K(\sin(\pi/12))K(\cos(\pi/12)) = \frac{\sqrt3}2 K(\sin(\pi/12))^2, $$
and the closed form exists because $\sin\frac\pi{12}$ is a singular value (see mathworld) of the elliptic integral:
$$ \frac{3\Gamma(\frac13)^6}{32\pi^2 2^{2/3}}. $$


答:
\begin{gather*}
\lim _{x\rightarrow \infty }\frac{\sqrt{x+2} -\sqrt{x+1}}{\sqrt{x+1} -\sqrt{x}}\\
=\lim _{x\rightarrow \infty }\frac{\sqrt{1+\frac{2}{x}} -\sqrt{1+\frac{1}{x}}}{\sqrt{1+\frac{1}{x}} -1}\\
\end{gather*}
(Taking $\sqrt{x}$ common from both numerator and denominator)

Now, from the binomial series expansion, $(1+ay)^{b} =1+aby$, when $y\to 0$

\begin{gather*}
\sqrt{1+\frac{2}{x}} =1+\frac{1}{2} \times \frac{2}{x}\\
\sqrt{1+\frac{1}{x}} =1+\frac{1}{2} \times \frac{1}{x}\\
\lim _{x\rightarrow \infty }\frac{\sqrt{1+\frac{2}{x}} -\sqrt{1+\frac{1}{x}}}{\sqrt{1+\frac{1}{x}} -1} =\lim _{x\rightarrow \infty }\frac{\left( 1+\frac{1}{2} \times \frac{2}{x}\right) -\left( 1+\frac{1}{2} \times \frac{1}{x}\right)}{\left( 1+\frac{1}{2} \times \frac{1}{x}\right) -1} =1
\end{gather*}

Title: limit $\lim_{x \to \infty} \frac{\sqrt{x+2} - \sqrt{x+1}}{\sqrt{x+1} - \sqrt{x}}$

Body:
How can I find the limit to infinity of this function? As this is a $0/0$ equation, I tried using the L'Hôpital's rule in this but ended up making it more complex. I've also tried rationalising the denominator but it didn't lead to anywhere.

$$\lim_{x \to \infty} \frac{\sqrt{x+2} - \sqrt{x+1}}{\sqrt{x+1} - \sqrt{x}} $$


问:
if $p_L=\sum_{i=1}^{L}|a_i|\cdot \sum_{j=1}^{L} |b_j|$ is a Cauchy sequence then $\sum_{i\text{ or }j>L}|a_i|\cdot |b_j|<\varepsilon$
I don't understand how inequality $(1)$ is derived in the following proof (where $\{a_n\}$ and $\{b_n\}$ are assumed to converge absolutely).

PROOF

Notice that the sequence
$$p_L=\sum_{i=1}^{L}|a_i|\cdot \sum_{j=1}^{L} |b_j|$$
converges, since $\{a_n\}$ and $\{b_n\}$ are absolutely convergent, and since the limt of a product is the product of the limits. So $\{p_L\}$ is a Cauchy sequence, which means that for any $\varepsilon>0$, if $L$ and $L'$ are large enough, then
$$\left| \sum_{i=1}^{L'}|a_i|\cdot \sum_{j=1}^{L'} |b_j|-\sum_{i=1}^{L}|a_i|\cdot \sum_{j=1}^{L} |b_j|\right|<\frac{\varepsilon}{2}.$$
It follow that
$$\sum_{i\text{ or }j>L}|a_i|\cdot |b_j|\leq\frac{\varepsilon}{2}<\varepsilon. \tag{1}$$
$[\dots]$

(Calculus 3rd edition, Michael Spivak. Theorem 23-9, PROOF.)

I must be missing something very simple but can't seem to understand how $(1)$ follows from what came before it.

Body:
WLOG assume $L' > L$. Writing as a double sum and splitting into four blocks,

$$\sum_{i=1}^{L'}|a_i| \sum_{j=1}^{L'}|b_j|= \sum_{i=1}^{L'}\sum_{j=1}^{L'}|a_i||b_j| \\=  \sum_{i=1}^{L}\sum_{j=1}^{L}|a_i||b_j|+ \sum_{i=1}^{L}\sum_{j=L+1}^{L'}|a_i||b_j|+ \sum_{i=L+1}^{L'}\sum_{j=1}^{L}|a_i||b_j|+ \sum_{i=L+1}^{L'}\sum_{j=L+1}^{L'}|a_i||b_j|, $$

Hence,

$$\sum_{i=1}^{L'}|a_i| \sum_{j=1}^{L'}|b_j|- \sum_{i=1}^{L}|a_i| \sum_{j=1}^{L}|b_j|\\ =  \underbrace{\sum_{i=1}^{L}\sum_{j=L+1}^{L'}|a_i||b_j|+ \sum_{i=L+1}^{L'}\sum_{j=1}^{L}|a_i||b_j|+ \sum_{i=L+1}^{L'}\sum_{j=L+1}^{L'}|a_i||b_j|}_{i \text{ or }j > L}$$


问:
There's 8 black balls and 7 white balls. 3 of the balls are drawn at random. Probability of drawing 2 of one color and 1 of the other color?
A bin has 8 black balls and 7 white balls. 3 of the balls are drawn at random. What is the probability of drawing 2 of one color and 1 of the other color?

Here's what I tried:

Case 1: 2 black balls and 1 white ball

8/15 * 7/14 * 7/13 = 392/2730 = 28/195

Case 2: 2 white balls, 1 black ball

7/15 * 6/14 * 8/13 = 336/2730 = 24/195

28/195 + 24/195 = 52/195 = 4/15

My teacher said it was wrong and I don't get why. She didn't give me the solutions either and just told me the answer was 4/5. Can someone tell me what I'm doing wrong? Thanks!

答:
The answer follows from the story of the Hypergeometric distribution! A random variable that counts the number of successes (in this case, let's call a success drawing a white ball) from a fixed number of draws (3, in our case) without replacement is said to have a Hypergeometric distribution. (If the scenario were instead with replacement, we'd have a Binomially distributed random variable on our hands.)

Let's let $X$ be this particular random variable. We know that at a minimum $X = 0$, and at a maximum, $X = 3$, but we're interested in the probability that $X = 2$ (2 white balls, 1 black ball) or $X = 1$ (1 white ball, 2 black balls). Since these cases are disjoint, we can simply add the probabilities of these two cases: $P(X=1) + P(X=2)$.

There are $\binom{15}{3}$ ways to pick 3 balls. To get exactly 2 white balls (and therefore, 1 black ball), we need to choose 2 white balls from the 7 possible white balls, and 1 black ball from the 8 possible black balls. The same logic can be used to evaluate the other case:

$$
\begin{align}
P(X=1) &= \frac{\binom{7}{1}\binom{8}{2}}{\binom{15}{3}}\\
P(X=2) &= \frac{\binom{7}{2}\binom{8}{1}}{\binom{15}{3}}
\end{align}
$$

Adding the probabilities together, we arrive at:

$$
P(X=1 \cup X=2) = \frac{4}{5}
$$


问:
How exactly are polynomials related to their evaluation maps?
My abstract algebra textbook* has taken great care in the last few sections to stress that, "polynomial are NOT functions to just 'plug' values of x into". To fully 'get' this, and to 'get' what polynomials really are then, I read through the proofs and explanations in the appendix that build up the idea of what a polynomial rings really is, and how we know a polynomial ring does exist for all rings.

In the current section of the book however, the book simply stated that "every polynomial induces a function from R to R", and then started to use polynomials as functions again, and the way it's doing so has me confused- what does it mean that it "induces" a function? The book will use the polynomial as function in a proof, and then make a conclusion about the polynomial itself- why, really, is that step valid?

The easiest example of this is it's proof for the polynomial remainder theorem.

The remainder theorem for polynomials.

The proof.

I get that, by the division algorithm for polynomials and the fact that (x - a) is degree 1, the following is a true statement: $\exists q(x),r(x)\in R[x] \; s.t \; f(x) = (x - a)q(x) + r(x)$ and $r=0 _r$ or $degree(r) = 0$

Isn't the next statement, namely that $f(a) = (a - a)q(a) + r(x) $, just a statement about some elements of in the ring R? Why can we say that the r(x) from the first statement is the same r(x) in the second statement? In other words, why is this second statement true?

/***********************/

I've worked ahead on some of the more complicated and involved stuff, but I have this nagging frustration in some part of my brain, as I don't "get" how these polynomials are really related with their evaluation maps, and when/why we can/cannot use a polynomial's evaluation map to make a conclusion about the polynomial itself.

*Here are some related threads I looked at that I thought about for a bit:

Why is it important that the evaluation map is a homomorphism?

What is the exact definition of polynomial functions?

*I feel like there is something very simple I'm missing.. I've spent too much time thinking about this and I think I've muddled my brain. Also, my bad if I've formatted this poorly, this is my first post. Thanks a lot for the help.

Edit:
*My textbook: Abstract Algebra, An Introduction - Third Edition, by Thomas W Hungerford and David Leep

答: If $p(x)$ is irreducible over the field $\Bbb F$, then the evaluation homomorphism at any root $\alpha$ of $p$ is an isomorphism between $\Bbb F[x]/(p)$ and $\Bbb F(\alpha)$, the field you get by adjoining the root to $\Bbb F$.


问:
Probability problem where I remove 4 balls without reposition
One box has 7 white balls, 3 black balls and 3 red balls. We remove four balls without replacement.

What is the probability that there are two white, one black and one red.

答:
Following @Numericallyilliterate's Comment or @John's Answer (+1), the answer is
$\frac{{7\choose 2}{3\choose 1}{3\choose 1}}{{13\choose 4}} = 0.2643,$
as computed using R:

```<code>choose(7,2)*choose(3,1)*choose(3,1)/choose(13,4)
[1] 0.2643357
</code>```

By simulation, letting white be worth 100, black be worth 10, and red be worth 1, we seek total score $211.$ Simulating four draws
a million times, we can expect two (maybe three) place accuracy. We have the following:

```<code>set.seed(2021)
box = c(rep(100,7), rep(10,3), rep(1,3))
box
[1] 100 100 100 100 100 100 100  10  10  10   1   1   1

x = replicate(10^6, sum(sample(box,4)))
mean(x == 211)
[1] 0.263744      # aprx 0.2643357
2*sd(x==211)/1000
[1] 0.0008813247  # 95% margin of simulation error
</code>```

Body:
Following @Numericallyilliterate's Comment or @John's Answer (+1), the answer is
$\frac{{7\choose 2}{3\choose 1}{3\choose 1}}{{13\choose 4}} = 0.2643,$
as computed using R:

choose(7,2)*choose(3,1)*choose(3,1)/choose(13,4)
[1] 0.2643357

By simulation, letting white be worth 100, black be worth 10, and red be worth 1, we seek total score $211.$ Simulating four draws
a million times, we can expect two (maybe three) place accuracy. We have the following:

set.seed(2021)
box = c(rep(100,7), rep(10,3), rep(1,3))
box
[1] 100 100 100 100 100 100 100  10  10  10   1   1   1

x = replicate(10^6, sum(sample(box,4)))
mean(x == 211)
[1] 0.263744      # aprx 0.2643357
2*sd(x==211)/1000
[1] 0.0008813247  # 95% margin of simulation error


问:
When to use these fractions properly?
I'm self-taught in mathematics, however, I have missed many basic ideas of arithmetic, where I'll usually use symbolab to help me get through some simplification of equations. although, I'm starting to get annoyed by this, as its impeding my learning. I would really like your support in helping me understand the distinction in fractions likeso:

$$[a] \space \cfrac{x}{1-x\over x}$$ when this could be simplified as either:

$$[1] \space\frac{x}{(1-x)x} $$
or $$[2] \space \frac{x^2}{1-x}$$

Though, I've also used symbolab and it's given me fractions likeso:

$$[b] \space \cfrac{x \over1+x}{1-x\over x}$$

Where it will be split like the first fraction into either (1) or (2). Are there rules that help me decide if its either (1) or (2) given equation (a), and when is it acceptable to use these for equation (b).

Or does it not matter for (b), and that the result should ensue likeso:

$$[b] \space \cfrac{x \over1+x}{1-x\over x} = \frac{x}{1+x}\frac{1-x}{x}$$

Thank you for the support! this should help me from relying on symbolab all the time for this.

答:
In my opinion, making the reader figure out how to parse an expression is, at best, an aggravating speed bump. Order of operations is for idiotic questions on facebook and computer programmers. The expression $\frac{\frac{x}{1-x}}{x}$ could mean

$$\text{either $\frac{\left( \frac{x}{1-x} \right)}{x}$
            or $\frac{x}{\left(\frac{1-x}{x}\right)}$}.$$

If you have to check yourself what an expression means, then some readers are going to have to do the same thing. So try not to do that. I can't say don't do that because there are always exceptions. But you should always strive for clarity and ease of reading.

Title: When to use these fractions properly?


问:
Given a scalene triangle $ABC$ with $H$ orthocenter, prove that two lines are parallel
Given a scalene triangle $\triangle ABC$ with $H$ the orthocenter of the triangle. The internal bisector of the angle $\angle BAC$ intersects the lines $BH$ and $CH$ at the points $Λ$ and $Θ$ correspondingly. If (c) is the circumcircle of the triangle $\triangle ΗΛΘ$ and (e) is the tangent of (c) at $Λ$. The perpendicular line from $H$ towards (e) is the line (e) at the point $N$. If $T$ is the point where the perpendicular line from $Λ$ towards $ΗΘ$ intersects the line $ΗΘ$, prove that $NT$ and $ΛΘ$ are parallel.

My thoughts are the following:

$\angle ΛΤΗ+\angle ΗΝΛ=90^{o}$

Hence $ΤΗΝΛ$ is inscribable.

Hence we have that $\angle ΤΛΝ+ \angle ΤΗΝ=180^{o}$

And also $\angle ΛΝΤ=\angle ΛΗΤ$

Hence it is enough to prove that $\angle ΛΗΘ = \angle ΛΘΗ$.

Moreover it seems like $\triangle ΛΘΗ$ is equilateral, but I can't prove that. Could you please explain to me how to solve this question?

答:
I will call the points where the angle bisector of $\angle BAC$ intersects $BH$ and $CH$ as $P$ and $Q$ respectively.  Let $S$ be some arbitrary point on the right[Since $P$, $Q$ are on the right of $H$ ] of point $P$ on the tangent to the  circumcircle of $\triangle HPQ$.

Observe that,  $\angle SPQ=\angle PHQ=\angle PHT=\angle PNT=\angle SNT$ and thereafter $PQ\parallel NT$.

$\triangle HPQ$ is not necessarily equilateral but isosceles since $\angle HPQ=\angle HQP=90-\frac {\angle A}{2}$.