问:
How to generate vertex-transitive tilings?
It is trivial to construct vertex-transitive polytopes: choose any finite matrix group $\Gamma\subseteq\mathrm O(\Bbb R^d)$ and some point $v\in\Bbb R^d$. Take the convex hull of the orbit $\Gamma v$, then this defined the orbit polytope

$$\mathrm{Orb}(\Gamma,v):=\mathrm{conv}(\Gamma v),$$

which is vertex-transitive (i.e. for any two vertices there is a symmetry of the polytope mapping the first vertex to the second).

Question: Is there a simiar way to generate an abundance of vertex-transitive tilings of $\Bbb R^d$?

By tiling I mean a face-to-face tiling of $\Bbb R^d$ by convex polytopes.

Update

I just had the following idea: take any lattice $\Lambda\subseteq\Bbb R^d$ and consider the Vornoi decomposition of $\Bbb R^d$ w.r.t this lattice. It is cell-transitive. So we compute its dual, which should then be vertex-transitive. Do we get the dual as the Vornoi decomposition of $\Bbb R^d$ w.r.t the vertices of the first Vornoi decomposition? I am not sure.

答:
As you noted in the comments, the problem with $\mathbb R^d$ is that it's flat, and the convex hull of a set of points "forgets about" the interior points.

Consider the mapping from $\mathbb R^d$ to a paraboloid in $\mathbb R^{d+1}$ defined by

$$x=x_1e_1+\cdots+x_de_d\mapsto$$

$$x^\smile=x+\tfrac12x^2e_{d+1}$$

$$=x_1e_1+\cdots+x_de_d+\tfrac12(x_1^2+\cdots+x_d^2)e_{d+1}$$

where $e_i$ are basis vectors. This gives us a curved model of Euclidean space. It's not intrinsic curvature, though; if we use a degenerate dot product on $\mathbb R^{d+1}$, with $e_i\cdot e_j=(1-\delta_{i,d+1})\delta_{i,j}$, then the resulting geometry on the paraboloid is ordinary Euclidean geometry.

Furthermore, translations, rotations, reflections, etc. can be induced by affine transformations on $\mathbb R^{d+1}$. We need only consider reflections, since other Euclidean transformations are compositions of these. Let $y\in\mathbb R^d$ be any point, and $u\in\mathbb R^d$ any unit vector, normal to a hyperplane passing through $y$. Then the reflection of $x\in\mathbb R^d$ across this hyperplane is

$$x\mapsto$$

$$x-2\big((x-y)\cdot u\big)u.$$

Now consider this affine transformation on $\mathbb R^{d+1}$:

$$x+x_{d+1}e_{d+1}\mapsto$$

$$x+x_{d+1}e_{d+1}-2\big((x-y)\cdot u\big)\big(u+(y\cdot u)e_{d+1}\big).$$

(Notice that $u+(y\cdot u)e_{d+1}=u_y^\smile$ is a tangent vector to the paraboloid at the point $y^\smile$, and that factor in the middle is $(x-y)\cdot u=(x+x_{d+1}e_{d+1}-y^\smile)\cdot u_y^\smile$ using the degenerate dot product.)

Its restriction to the paraboloid is

$$x^\smile\mapsto$$

$$x+\tfrac12x^2e_{d+1}-2\big((x-y)\cdot u\big)\big(u+(y\cdot u)e_{d+1}\big)$$

$$=\Big(x-2\big((x-y)\cdot u\big)u\Big)+\tfrac12\Big(x-2\big((x-y)\cdot u\big)u\Big)^2e_{d+1}$$

$$=\Big(x-2\big((x-y)\cdot u\big)u\Big)^\smile$$

which has the same effect as the reflection on $\mathbb R^d$.

Using this, you should be able to express any Euclidean isometry group in terms of these non-Euclidean reflections on $\mathbb R^{d+1}$. Then you can take the convex hull of the orbit of any point $x^\smile$, find the hull's faces, and project these back onto $\mathbb R^d$.

Note that the convex hull of a set $S$ has $F(\text{conv}(S))=\text{conv}(F(S))$ for any affine transformation $F$, so it doesn't matter where the origin is in $\mathbb R^d$, or where the vertex of the paraboloid is.

Since the paraboloid is strictly convex, no points are "forgotten" (absorbed into the convex hulls of other points).

Any face of the convex hull which is unbounded must be vertical (its covector must have no $e^{d+1}$ component). Equivalently, any non-vertical face is bounded. For, if a face is defined by $\alpha_1x_1+\cdots+\alpha_dx_d+x_{d+1}=b$, since the convex hull is above the paraboloid, we have $x_{d+1}\geq\tfrac12x^2$ and thus $\alpha_1x_1+\cdots+\alpha_dx_d+\tfrac12x^2\leq b$, which describes a ball in $\mathbb R^d$. So $x_1,\cdots,x_d$ are bounded, and it follows that $x_{d+1}=b-\alpha_1x_1-\cdots-\alpha_dx_d$ is also bounded.

With a vertical face, all of the points in $\mathbb R^d$ are on one side of a hyperplane. So if your chosen group is cocompact, then any point's orbit crosses any hyperplane, and each face must be bounded.

If your chosen group is discrete, then each bounded face must be the convex hull of a finite set of points. (Otherwise the points would accumulate and not be discrete.)


问:
Existence of a continuous and non-constant function $f: X \to [0,1]$
Let $(X, \mathcal{T})$ a normal space and $T_1$ that has more than one point. Prove that exist a continuous and non-constant function $f:(X, \mathcal{T}) \to ([0,1], \mathcal{T}_{us})$ .

My idea If $X$ has only two points then $X=\{a,b\}$, since $X$ is $T_1$ then $\{a\}$ and $\{b \}$ are closed, and are disjoint. By the Urysohn's Lemma exists a continuos fuction $f:(X, \mathcal{T}) \to ([0,1], \mathcal{T}_{us})$ such that $f(\{a\})=0$ and $f(\{b\})=1$ and $f$ wouldn't be constant.
But I don't know how to do if $X$ is infinite

答: $\{a\}$ and $\{b\}$ are closed (by $T_1$-ness) and disjoint. So Urysohn's lemma applies to these two closed sets (regardless of the size of $X$; it holds for any normal space).

ContentLicense: CC BY-SA 4.0


问:
Countability of $\mathbb Q\cap[0,1]$
I do not understand why this set is countable
$$\mathbb Q\cap[0,1]$$

答: It's well known that the intersection between a countable set and an uncountable set is countable. Intuitively, this makes sense because the intersection between two sets has size equal to at most the minimum size of the two sets that you are taking an intersection of. In other words, the size of $\mathbb{Q} \cap [0, 1]$ is at most the size of $\mathbb{Q}$, which is known to be countable.

Title: Countability of $\mathbb Q\cap[0,1]$


问:
Minimum value of $a\cdot b+2a\cdot c+b\cdot c$

find the minimum value of $$S=\vec{a}\cdot\vec{b}+2 \vec{a}\cdot\vec{c}+\vec{b}\cdot\vec{c}$$ if $|a|=1,|b|=2,|c|=3$

The solution is pretty easy: $$|2\vec{a} +2\vec{c}+\vec{b}|^2\ge 0$$ $$\iff 4+36+4+4S\ge 0$$ $$\iff S\ge -11$$ However if i want to solve this like this it becomes difficult

Let $p,q,r$ be the angles btw the vectors $(a,b),(b,c),(c,a)$ respectively.Then we have to minimise $$S=2\cos p+6\cos q+6\cos r$$ We cannot take $p=q=r=-\pi$ as that would mean all vectors make an angle of $-\pi$ with each other which is insane!

How would i go about solving from this method??

UpdateAs Calvin Lin pointed out My first meethod is wrong

答:
You need the additional constraint that $ p + q + r \leq 2 \pi $ (where we take the angle in the range $ [ 0, \pi ] $).

This is because the angles at the vertex/origin sum up to at most $2 \pi$.

Furthermore, because the coefficients are positive, we will always want to increase $p/q/r$, so the minimum will occur when $ p + q + r = 2 \pi$.

At this point, we can use Lagrange Multipliers to calculate the minimum.


问:
Prove that if $R$ is a preorder over a set A, then $S_R$ is a strict order.
$aS_Rb$ if $aRb \cap bꞦa$

My working:

Theorem: If $R$ is a preorder over a set $A$, then $S_R$ is a strict order.

Example relation:

R = { {a,b}, {b,c}, {c,c}, {f,c}, {f,e}, {e,f}, {e,b}, {b,e}, {d,e}, {d,d}, {d,a}, {a,d} }

Therefore $S_R$ = { {a,d}, {a,b}, {d,e}, {a,e}, {b,e}, {b,c}, {e,f}, {c,f} }

Lemma 1: The binary relation $aS_Rb$ is irreflexive.

Proof: Consider an arbitrary $a \in S_R$. We need to prove $S_R$ is irreflexive. Consider an element $b=a$. If $aRb$ is false we can conclude that the relation is irreflexive.

Lemma 2: The binary relation $aS_Rb$ is transitive.

Proof: Consider an arbitrary $a, b, c, \in S_R$. We need to prove that $S_R$ is transitive. To do so, begin by checking if $aRb$ and $bRc$ is true. After that check the relation $aRc$. If $aRb \cap aRc \implies aRc$ then we can conclude that $R$ is transitive. $\blacksquare$

I was wondering if someone can check this proof, while i was writing it something didn't seem right about it i.e. it seemed incredibly small in length and i felt like i regurgitated the definitions of irreflexivity and transitivity.

答:
Seems that you made some incorrect assumptions in your proof.

First of all, when you want to prove that a relation has or doesn't have some properties (reflexivity, transitivity, asymmetry etc.) it's often better to employ the definitions of those properties.

Second of all, you need to learn how to use these definitions correctly, this will prevent you from wrong assumptions which might make your proof incorrect.

So, let's go back to your proof.

Lemma 1: The binary relation $S_R$ is irreflexive.

As far as I'm concerned your statement doesn't prove anything. It only uses some vague assumptions:

If $aRb$ is false we can conclude that the relation is irreflexive.

And note you can't prove statements by saying "if something is true we can conclude". To really prove the statement you need to show that "something is really true".

Let's see what the definition of irreflexivity:

A binary relation $R$ on set $A$ is called irreflexive if for each $a \in A$ $a \not R a$.

So as it turned out we have to show that for an arbitrary $a \in A$, we have that $a \not S_R a$. There's also an interesting point here that we have to prove a negation, so we need to either negate a definition of $S_R$ to proceed with the prove or employ proof by contradiction.

Proof: So consider an arbitrary $a \in A$, we will show that $a \not S_R a$ (see how I use the definition of irreflexivity here). For the sake of contradiction, let's assume that $a S_R a$ which means that $aRa \land a\not Ra $ (see how I use the definition of $S_R$). But these two statements are impossible, because they contradict each other, so we have reached the contradiction and our assumption that $a S_R a$ is wrong, therefore $a \not S_R a$. $\blacksquare$

Lemma 2: The binary relation $aS_Rb$ is transitive.

In this proof there is the same mistake, it doesn't prove anything, it only assumes that if something were true, then what I'm trying to prove would be also true. But it's important to understand that proofs don't work that way, you shouldn't assume, you should prove, your reader shouldn't even have any doubts about what you're saying in your proofs.

It's a very good point that you're using the definition of transitivity in this proof. But be careful, when using definitions like the transitivity which is below:

A binary relation $R$ on set $A$ is called transitive if for each $a, b, c \in A$ if $a R b \land b R c$ then $a R c$.

Here you don't need to prove that $a R b$ and $b R c$, you can assume this for free, because you already selected them from the set, you intentionally select such $a, b, c$, where $a R b \land b R c$. You ignore elements which doesn't meet this condition (this how implication works). And after you select such elements you need to prove that $a R c$. This is what you need to prove.

Proof: Consider arbitrary $a, b, c \in A$ such that $a S_R b \land b S_R c$. We will prove that $a S_R c$, by showing that $a R c \land c \not R a$. Since $a S_R b$ we know that $a R b \land b \not R a$. Since $b S_R c$ we know that $b R c \land c \not R b$. Since $R$ is a preorder, it's transitive and since $a R b \land b R c$, consequently we have that $a R c$.

Now we need to show that $c \not R a$. For the sake of contradiction, let's assume that $c R a$. Then since $R$ is a preorder, it's transitive and since $c R a \land a R b$, we have that $c R b$. But this contradicts the fact that $b R c \land c \not R b$, therefore our assumption is wrong and $c \not R a$. Since $a R c \land c \not R a$, we know that $a S_R c$ as required. $\blacksquare$


问:
Euler circuits for the graph
How many different Euler circuits are there in the graph that begins at $A$ and then visits $D$ as the immediate next vertex?

My answer is $16$, and I suppose I have counted all possible such Euler circuits. I wonder whether I have missed additionally possible paths, and would be welcomed to know that.

答: I think $16$ seems to be the correct value. I also counted and obtained $16$ as an answer.

Title: Euler circuits for the graph


问:
What's the difference between arccos(x) and sec(x)
My question might sound dumb, but I don't really see why the graphics of $\arccos(x)$ and $\sec(x)$ are different.

As far as I know $\arccos(x)$ is the inverse cosine function $(\cos(x)^{-1})$ and $\sec(x)$ = $\frac{1}{\cos(x)}$ (source Wolfram|Alpha).

So why aren't they equal?

Thanks in advance.

答:
$\arccos x$ means the angle $\alpha$ in the interval $[0,\pi]$ such that

$$
\cos\alpha=x
$$

while

$$
\sec x=\frac{1}{\cos x}
$$

So $\arccos0=\pi/2$, while $\sec0=1$.

I believe you're misled by the $\cos^{-1}$ notation somebody uses for $\arccos$.

From a "function theory" point of view, the notation $\cos^2 x$ is inappropriate, because it should mean $\cos(\cos x)$; however, $\cos^2x=(\cos x)^2$ has been used for centuries because it's practical and in many formulas you need the cosine squared, while the cosine of the cosine is rarely needed.

Mixing the two notations, that is using $\cos^{-1}$ for the "inverse function of the cosine" is, at the least, confusing.

ContentLicense: CC BY-SA 3.0


问:
Is function $f(x,y)=x(4-y^2)$ injective / surjective? Sketch $f^{-1}[B]$, where $b = [0 ; +\infty)$.
There is a fuction $f: \mathbb{R}^2 \implies \mathbb{R} $  given by:
$$f(x,y)=x(4-y^2)$$

Is the function injective?

Is the function surjective?

find and sketch $f^{-1}[B]$, where $b = [0 ; +\infty)$

Here is an sketch:
sketch (I am not allowed to share images)

As we can see, function $f$ can be described as a linear function with all it's rotations besides those that would cover the blue part of cartesian coordinate system. The function gets most vertival for $y = 0$.

Therefore, I would say that the function is injective.

However, I am not sure if it's surjective since for $y = -2$ the function is constant and returns only the value $0$.

And finally, the last part. I know that $f^{-1}[B]$, for $b = [0 ; +\infty)$ is an counterimage. The problem is, that in that case I have 2 separate types of argements. I would say that the answer is:

{$n \in (-2 ; 0)\cup(0;2)\}\cap $ {$x \in (0; +\infty)$}

{$n \in (-\infty ; -2)\cup(2;\infty)\}\cap $ {$x \in (-\infty;0)$}

答:
This function is not injective.  For all $x,y \in \Bbb{R}$, $f(x,y) = f(x,-y)$, so most points in the image have at least two preimages.  (In fact, most points have infinitely many preimages.)

This function is easily seen to be surjective.  Let $u \in \Bbb{R}$ be an arbitrary point in the codomain.  Then $f(u,\sqrt{3}) = u$.

For $f(x,y) \geq 0$: the product of two terms is nonnegative when both are nonnegative or both are nonpositive.  So,
$$  \{(x,y) \mid x \geq 0 \text{ and } 4 - y^2 \geq 0\} \\ \quad \cup \{(x,y) \mid x \leq 0 \text{ and } 4 - y^2 \leq 0\}  \text{.}  $$
Easy simplifications: \begin{align*}
& \{(x,y) \mid x \geq 0 \text{ and } 4 - y^2 \geq 0\}  \\
    &\quad {} \cup \{(x,y) \mid x \leq 0 \text{ and } 4 - y^2 \leq 0\}  \\
    &\quad {} = \{(x,y) \mid x \geq 0 \text{ and } 4 \geq y^2\}  \\
        &\qquad {} \cup \{(x,y) \mid x \leq 0 \text{ and } 4 \leq y^2\}  \\
    &\quad {} = \{(x,y) \mid x \geq 0 \text{ and } -2 \leq y \leq 2\}  \\
        &\qquad {} \cup \{(x,y) \mid x \leq 0 \text{ and } ( y \leq -2 \text{ or } 2 \leq y)\}
\end{align*}

And this is easy to plot.


答:
For any $a \in \mathbb{R}$, $f(\frac{a}{4},0)=a$. So surjective.

$f(x,2)=0$ for any $x$. So not injective.

For the image of $f^{-1}[B]$, use case analysis method:

$|y|>2 \longrightarrow$ $x \in (-\infty,0]$

$|y|=2 \longrightarrow$ $x$ is any real number.

$|y|<2 \longrightarrow$ $x \in [0,\infty)$

Here is the image

Title: Is function $f(x,y)=x(4-y^2)$ injective / surjective? Sketch $f^{-1}[B]$, where $b = [0 ; +\infty)$.

Body:
There is a fuction $f: \mathbb{R}^2 \implies \mathbb{R} $  given by:
$$f(x,y)=x(4-y^2)$$

Is the function injective?

Is the function surjective?

find and sketch $f^{-1}[B]$, where $b = [0 ; +\infty)$

Here is an sketch:
sketch (I am not allowed to share images)

As we can see, function $f$ can be described as a linear function with all it's rotations besides those that would cover the blue part of cartesian coordinate system. The function gets most vertival for $y = 0$.

Therefore, I would say that the function is injective.

However, I am not sure if it's surjective since for $y = -2$ the function is constant and returns only the value $0$.

And finally, the last part. I know that $f^{-1}[B]$, for $b = [0 ; +\infty)$ is an counterimage. The problem is, that in that case I have 2 separate types of argements. I would say that the answer is:

{$n \in (-2 ; 0)\cup(0;2)\}\cap $ {$x \in (0; +\infty)$}

{$n \in (-\infty ; -2)\cup(2;\infty)\}\cap $ {$x \in (-\infty;0)$}


问:
Using $\int_{2}^{\infty } \frac{dx}{x^{n+1}} =\frac{1}{n 2^{n}}$
find the sums of the series
$$\frac{1}{1\cdot 2}+\frac{1}{2\cdot 2^{2}}+\cdots +\frac{1}{n 2^{n}}$$

I solved this below, but did not know what my mistake is
But now the page asks me to provide additional context
Solution.

$$\sum_{n=1}^{\infty }\frac{1}{n 2^{n}}=
\sum_{n=1}^{\infty }\int_{2}^{\infty }\frac{dx}{x^{n+1}}=\int_{2}^{\infty }\sum_{n=1}^{\infty }\frac{dx}{x^{n+1}}=\int_{2}^{\infty }\frac{dx}{x(x-1)}$$

$$\int_{2}^{\infty }\frac{dx}{x(x-1)}=\lim_{b\to \infty }  \int_{2}^{b }(\frac{1}{x-1}-\frac{1}{x})dx=\lim_{b\to \infty } \left [  \ln(b-1 ) - \ln(b)+\ln(2)  \right  ] =\ln(2)$$


问:
Show that $\lim_{n \to \infty} \prod_{k=1}^{k=n} (1+ \frac{1}{n}f(\frac{k}{n})) = e^{\int_{0}^{1}f(x) dx}$
Let $f:[0,1] \to \mathbb{R}$ be a continuous function. Show that $\lim_{n \to \infty} \prod_{k=1}^{k=n} (1+ \frac{1}{n}f(\frac{k}{n})) = e^{\int_{0}^{1}f(x) dx}$.

Attempt:  let $\lim \prod_{k=1}^{k=n} (1+ \frac{1}{n}f(\frac{k}{n})) = l$ then $\lim_{n \to \infty} \log(\prod_{k=1}^{k=n} (1+ \frac{1}{n}f(\frac{k}{n})))= \log l \implies \lim \sum_{k=1}^{k=n} \log(1+ \frac{1}{n}f(\frac{k}{n})) = \log l$.

and working backwards we should have $\log l = \int_{0}^{1} f(x)dx = \lim_{n \to \infty} \sum_{k=1}^{k=n} \frac{1}{n} f(\frac{k}{n})$. So here I'm trying to find the link between $\lim \sum_{k=1}^{k=n} \log(1+ \frac{1}{n}f(\frac{k}{n}))$ and $\lim_{n \to \infty} \sum_{k=1}^{k=n} \frac{1}{n} f(\frac{k}{n})$.

but $\log(1+x) \approx x$ for $x <1$, so in order to use this approximation I have to prove that $\frac{1}{n}f(\frac{k}{n})< 1$ (or maybe I can use the fact that f is continuous and prove that f attains supremum at some point in $[0,1]$) after some terms and I'm stuck .  Is there any other approach which avoids this approximation?

答:
$$ \sum_{k=1}^n\log\left(1+\frac{1}{n}f\left(\frac{k}{n}\right)\right)=\frac{1}{n}\sum_{k=1}^n f\left(\frac{k}{n}\right)+\mathcal{O}\left(\frac{1}{n^2}\sum_{k=1}^n f\left(\frac{k}{n}\right)^2\right) $$
Since $\lim\limits_{n\rightarrow +\infty}\frac{1}{n}\sum_{k=1}^n f\left(\frac{k}{n}\right)^2=\int_0^1f(x)^2dx$, then $\lim\limits_{n\rightarrow +\infty}\frac{1}{n^2}\sum_{k=1}^n f\left(\frac{k}{n}\right)^2=0$ and thus
$$ \lim\limits_{n\rightarrow +\infty}\sum_{k=1}^n\log\left(1+\frac{1}{n}f\left(\frac{k}{n}\right)\right)=\int_0^1 f(x)dx $$


问:
What does it mean when two Groups are isomorphic?
I'm not asking for the formal definition I know it. An isomorphism is a bijective homomorphism. In my book it's indicated many times when two groups are isomorphic, and I don't understand what's the reason for that. What can we "do" when we know that 2 groups are isomorphic? What does it really mean when 2 groups are isomorphic? </p>

答:
We'll try to build up to the definition of group isomorphism.

Suppose $ (G, \cdot) $ is a group, and say we have a bijection $ \Phi : G \rightarrow \Phi(G) $ ( which can be thought of as a relabelling $ g \mapsto \Phi(g) $ of elements of $ G $ ). A natural question arises : Can we endow $ \Phi(G) $ with a group operation $ \ast $ such that groups $ (G, \cdot)$ and $ (\Phi(G), \ast) $ "become copies of one another", i.e. are such that $ ( x \cdot y = z \text{ in group } G ) \iff ( \Phi(x) \ast \Phi(y) = \Phi(z) \text{ in group } \Phi(G) ) $ ?

It's clear $ \Phi(x) \ast \Phi(y) := \Phi(x \cdot y) $ is the only potential candidate for the definition of $ \ast $ , and a simple check shows this definition actually works : It makes $ (\Phi(G), \ast) $ a group and also a copy of $ (G, \cdot) $ in the above sense.

This makes us arrive at the formal definition :
Groups $ (G, \cdot) $ and $ (G', \ast) $ are called isomorphic if there is a bijection $ \phi : G \rightarrow G' $ such that $ \phi(x \cdot y) = \phi(x) \ast \phi(y) $.


问:
Parity of all possible sums

The numbers $1, 2, 3, \dots, 1024$ are written on a blackboard. They are divided into pairs. Then each pair is wiped off the board and non-negative difference of its numbers is written on the board instead. $512$ numbers obtained in this way are divided into pairs and so on. One number remains on the blackboard after ten such operations. Determine all its possible values.

In this soution (1st one), how is it clear that all possible values must have same parity ?

I think I am missing something obvious, pls someone help...

Thankyou

答:
Any integer and it's negative is of same parity.  $$a \equiv -a \pmod 2$$

We see $$\pm a_1 \pm a_2 \pm a_3 \ldots \pm a_n \equiv a_1 + a_2 + a_3 + \ldots a_n \pmod 2$$

Any combination of addition, subtraction results in same parity as all additions.

Title: Parity of all possible sums


问:
Evaluation of $\lim_{x\to 0^+}\frac{(1+x)^x - 1}{x^2}$
I'm interested in evaluating mentioned limit without using l'Hopital's rule, if possible.
What I'm thinking of is doing the following:

Let $$f(x) = \frac{(1+x)^x - 1}{x^2} = \frac{\left((1+x)^\frac{1}{x}\right)^{x^2} - 1}{x^2}$$

Now if we define:
$$g(x) = (1+x)^\frac{1}{x} \Rightarrow \lim_{x\to 0^+}g(x) = e > 0$$
and $$h(x) = x^2 \Rightarrow \lim_{x\to 0^+}h(x) = 0$$
then $f$ becomes a compound function $f(x) = \frac{g(x)^{h(x)} - 1}{h(x)}$

Now I'd like to argue that since $g(x) \xrightarrow{x\to 0} e > 0$, then $$\lim_{x\to 0^+}f(x) = \lim_{x\to 0^+}\frac{g(x)^{h(x)} - 1}{h(x)} \stackrel{?}{=} \lim_{x\to 0^+}\frac{\Big(\displaystyle\lim_{x\to 0^+}g(x)\Big)^{h(x)} - 1}{h(x)} = \lim_{x\to 0^+}\frac{e^{h(x)} - 1}{h(x)} = 1$$

However, the step with the question mark is troublesome for me and I struggled to prove it. Which got me thinking, is such statement even true or possible in this case? Or were such cases namely the very reason why L'Hopital's rule was discovered and is applicable?

Any clarification is greatly appreciated! Thanks a lot in advance!

Body:
$$\lim_{x\to 0} \frac{(1+x)^x-1}{x^2}=\lim_{x \to 0} \frac{(1+x.x)-1}{x^2}=1$$
Use binomial approximation: $(1+z)^k=1+kz+o(z^2)$, if $|z|<<1$


答:
If Taylor for $e^t$ and the standard limit $\lim_{x\to 0}\frac{\ln (1+x)}{x}=1$ are allowed, you have

$$\frac{(1+x)^x-1}{x^2} = \frac{e^{x\ln (1+x)}-1}{x^2}= \frac{x\ln (1+x) + O(x^2\ln^2 (1+x))}{x^2}$$$$=\frac{\ln (1+x)}{x} + O(\ln^2(1+x))\stackrel{x\to 0}{\longrightarrow}1$$

Adressing your question:

Taking "partial limits" is not a rule although it sometimes may give correct results.

Just consider $(1+x)^{\frac 1x}$ for $x\to 0$. If you applied your "rule" you would get $1^{\frac 1x}\stackrel{x\to 0}{\longrightarrow}1$, which is wrong, since $\lim_{x\to 0}(1+x)^{\frac 1x}=e$.

Title: Evaluation of $\lim_{x\to 0^+}\frac{(1+x)^x - 1}{x^2}$

Body:
I'm interested in evaluating mentioned limit without using l'Hopital's rule, if possible.
What I'm thinking of is doing the following:

Let $$f(x) = \frac{(1+x)^x - 1}{x^2} = \frac{\left((1+x)^\frac{1}{x}\right)^{x^2} - 1}{x^2}$$

Now if we define:
$$g(x) = (1+x)^\frac{1}{x} \Rightarrow \lim_{x\to 0^+}g(x) = e > 0$$
and $$h(x) = x^2 \Rightarrow \lim_{x\to 0^+}h(x) = 0$$
then $f$ becomes a compound function $f(x) = \frac{g(x)^{h(x)} - 1}{h(x)}$

Now I'd like to argue that since $g(x) \xrightarrow{x\to 0} e > 0$, then $$\lim_{x\to 0^+}f(x) = \lim_{x\to 0^+}\frac{g(x)^{h(x)} - 1}{h(x)} \stackrel{?}{=} \lim_{x\to 0^+}\frac{\Big(\displaystyle\lim_{x\to 0^+}g(x)\Big)^{h(x)} - 1}{h(x)} = \lim_{x\to 0^+}\frac{e^{h(x)} - 1}{h(x)} = 1$$

However, the step with the question mark is troublesome for me and I struggled to prove it. Which got me thinking, is such statement even true or possible in this case? Or were such cases namely the very reason why L'Hopital's rule was discovered and is applicable?

Any clarification is greatly appreciated! Thanks a lot in advance!


问:
Let A be an mxn matrix of rank 1. Using the definition of rank, show that there exists an mx1 matrix X and an 1xn matrix Y such that A=XY
How do I approach this question? I know the X and Y can be the basis for the columnspace of A and the basis of rowspace of A, but how can it be? Thanks for help!</p>

答:
If the rank of $A$ is equal to $1$, then its columns are all in the vectorial line generated by a non zero vector $Y$ of dimension $m$.

That means that the columns of $A$ are
$$x_1 Y, x_2 Y, \dots , x_nY.$$ If you denote $X^T=(x_1, \dots , x_n)$, you get the desired result

$$A=X^T Y.$$


问:
Evaluating the sum $\sum_{n=1}^{\infty}\frac{1}{F_n}$
Let $F_n$ be the $n^\text{th}$ Fibonacci number. I wanted to  calculate
$$\sum_{n\ge1}\frac{1}{F_n}$$
I simplified it to
$$\sqrt{5}\sum_{n\ge1}\frac{1}{\varphi^n-\phi^n}$$
But this didn't seem to help. The value is approximately $3.35988566624317755317201130291892...$. The OEIS entry of this constant is this. Mathematica gives the answer $$\frac{\sqrt{5}}{4} \left(\frac{\log 5+2\psi_{\phi^{-4}}(1)-4\psi_{\phi^{-2}}(1)}{2\log\phi}+\vartheta_{2}^{2}\left(\phi^{-2}\right)\right)$$
Where $\vartheta$ is the Elliptic Theta function and $\psi$ is the q-polygamma function. This looks very hard to derive. But, this can be calculated, which means that there is surely a method to do this. So, my question is,

How can the closed form be derived?

If the derivation is very long, an article containing the proof can be mentioned, but a derivation in the answers would be more helpful.

Body:
$$F_n=\frac{1}{\sqrt{5} }\left({\varphi^n-\phi^n}\right)=\frac{\varphi^n}{\sqrt{5} }\left(1-\left(\frac{\phi }{\varphi }\right)^n\right)$$
$$\frac 1{F_n}=\frac{\sqrt{5} }{\varphi^n} \frac 1{1-\left(\frac{\phi }{\varphi }\right)^n }$$

$$\sum_{n=1}^{\infty}\frac{1}{F_n}=\sqrt{5}\sum_{n=1}^{\infty}\frac{1 }{\varphi^n} \frac 1{1-\left(\frac{\phi }{\varphi }\right)^n }=\sqrt{5}\sum_{n=1}^{\infty}\frac{1 }{\varphi^n}\sum_{k=0}^{\infty}\left(\frac{\phi }{\varphi }\right)^{k n}$$

It is now for the second portion of the summand where the special functions are comming from.

You can have very good approximations writing
$$\sum_{n=1}^{\infty}\frac{1}{F_n}\sim \sum_{n=1}^{p}\frac{1}{F_n}+\sqrt{5}\sum_{n=p+1}^{\infty}\frac{1 }{\varphi^n}=\sum_{n=1}^{p}\frac{1}{F_n}+\frac{5+3 \sqrt{5}}{2}  \varphi ^{-(p+1)} $$
$$\left(
\begin{array}{cc}
p & \text{approximation} \\
 5 & 3.35957125458259727076544117389 \\
 10 & 3.35988589577114994666863433486 \\
 15 & 3.35988566607489618438018003508 \\
 20 & 3.35988566624330092650853900324 \\
 25 & 3.35988566624317746272239950104 \\
 30 & 3.35988566624317755323832329939 \\
 35 & 3.35988566624317755317196268711 \\
 40 & 3.35988566624317755317201133856 \\
 45 & 3.35988566624317755317201130289 \\
 50 & 3.35988566624317755317201130292
\end{array}
\right)$$


问:
Show that the integral $\int_{0}^\infty\frac{1}{x\sqrt{1+x^2}}dx$ is divergent
I have some trouble with this integral. Actually, I know when the interval ranges from 1 to $\infty$, this integral is convergent. But the difficulty I have encountered is how to show the divergence where the case of interval ranges from 0 to 1.

答:
By Taylor expansion around $x=0$ $$\frac{1}{x\sqrt{1+x^2}}=\frac{1}{x}-\frac{x}{2}+O\left(x^3\right)$$ This is more than sufficient.

Still faster, using equivalents, if $x$ is small then $x^2\ll 1$ and
$$\frac{1}{x\sqrt{1+x^2}}\sim \frac 1x$$


问:
What are some examples of "visually beautiful" math texts?
Are there good examples of math textbooks that seamlessly combines art, cartoon drawing or computer graphics with the texts? I know that there are a few books on the market, which are not text-book per se, that are geared towards general audience, where there are plenty of artistic figures naturally generated by the phenomena under study, usually things of geometric nature, Escher tiling, or things like chaos.

Wolfram's a New Kind of Science comes to mind. What are some other examples of "visually beautifu" or "artistic" math textbooks or texts.

答: A good example is, in my opinion, Islamic Geometric Patterns by Jay Bonner.


问:
Find the average of the number $n \sin n^\circ$ for $n=2,4,6\cdots,180$
I have been asked in a exam to find the average of the number: $$n \sin n^\circ$$ for $n$=$2,4,6,\cdots,180$

I have tried a lot basically with sum product, or pairing the inputs...but at the end don't able to find any way to solve it, can someone help me with the approach?

答:
\begin{align}
\sum_{r=1}^{90}2r\sin\left(\dfrac{2r\pi}{180}\right)&=2\sum_{r=1}^{45}r\sin\left(\dfrac{r\pi}{90}\right)+2\sum_{r=1}^{45}(90-r)\sin\left(\pi-\dfrac{r\pi}{90}\right)\\
&=2\sum_{r=1}^{45}r\sin\left(\dfrac{r\pi}{90}\right)+180\sum_{r=1}^{45}\sin\left(\dfrac{r\pi}{90}\right)-2\sum_{r=1}^{45}r\sin\left(\dfrac{r\pi}{90}\right)\\
&=180\times\sum_{r=1}^{45}\sin\left(\dfrac{r\pi}{90}\right)
\end{align}
Now apply sum of sine of AP formula and you're done!


问:
If $X_i \sim \text{Exp}(\theta)$ what is the distribution of $\frac{n}{X_1 + \dots + X_n}$?
If $X_1, \dots X_n$ are independent and Exponentially distributed with parameter $\theta$, then by examining the Characteristic function (or Moment generating function) of the sum
$$
X_1 + \dots + X_n
$$
one sees that the sum follows a Gamma distribution with parameters $n$ and $\theta$.

I wonder if similarly one may derive the distribution of $$
\frac{n}{X_1 + \dots + X_n} \quad ? $$

Most grateful for any help provided!

答:
Very easy:

if (as it is) $Y=\Sigma_i X_i\sim Gamma(n;\theta)$ then $\frac{1}{Y}\sim \text{Inverse Gamma}$

thus the law of $\frac{n}{Y}$ can be derived immediately by a simple transformation


问:
Prove inequality with powers of $n$
Prove that $(a+b)^n \le a^n+b((a+1)^n-a^n)$,
n in N,
$a \ge 0, 0<b<1$,
$a,b$ in $R$.,

With the little I remember from high school and college, I have tried the binomial expansion of $(a+1)^n$ and $(a+b)^n$.

Also tried to move $a^n$ at the LHS and use the formula:
$x^n-y^n=(x-y)\left(x^{n-1}+x^{n-2}y+x^{n-3}y^2+\dots+x^2y^{n-3}+xy^{n-2}+y^{n-1}\right)$ but I am not getting anywhere.

FYI this is not homework or anything. Just trying to keep my brain alive, to minimize the risk of developing dementia :)
Thank you very much in advance!!

答:
Let $f(x) = x^n$. Since this is convex, so
$$\dfrac{f(x+b)-f(x)}{b}\le \dfrac{f(x+1)-f(x)}{1}\qquad \forall b\in(0,1)$$
Hence the result.
Consider the following image:

For convex red curve, the slope of blue secant is more than that of the green one. In general, slope of secant joining points $x$ and $x+y$ increases with $y$, where $y>0$.

Title: Prove inequality with powers of $n$


问:
Combinatorics question in field of characteristic 2...
I'm trying to figure out the number of solutions to a particular system of equations. The variables here are $A_1,A_2,A_3,B_1,B_2,B_3,C_1,C_2,C_3$, all in $\mathbb{F}_2$, constraint to

(a) $B_1^2 = A_1 C_1$,

(b) $A_1 C_2 + A_2 C_1 = 0$, and

(c) $A_1 C_3 + A_3 C_1 = 0$.

This problem actually relates to something I'm doing in algebraic geometry right now--I just suck at counting, so forgive me if they I have been approaching this isn't optimal. Here's what I've got so far:

$B_2, B_3$ are free, so that is 4 options.

I split the problem into two main cases, case 1 being when $B_1 = 1$ and when $B_1 = 0$.

Case $B_1 = 1$: Then $A_1 = C_1 = 1$ by (a). By (b), we must then have $A_2 + C_2 = 0$, and since we are in $\mathbb{F}_2$, this means that $A_2 = C_2$, so we have 2 options here. Likewise we must have $A_3 = C_3$ by (c), giving us 2 more options, independent from what we did in (b). So for this case, we have

(4 [from $B_2, B_3$]) $\times$ (2 [from $A_2, C_2$]) $\times$ (2 [from $A_3, C_3$]) = 16.

So it is just the case when $B_1^2 = 0$ that I am having difficulty with, as this implies either $A_1 = 0$ or $C_1 = 0$ (or both), splitting the problem into many sub-cases whhich might intersect.

答:
I don't think the $B_1=0$ subcases are that bad. If $A_1=C_1=0$, then $A_2,A_3,C_2,C_3$ can be whatever they want. If $A_1=0$ but $C_1=1$, then $A_2=A_3=0$ but $C_2,C_3$ can be whatever. The case $C_1=0$ but $A_1=1$ is similar.

(If you really just care about this one specific problem, you can check all $2^9=512$ options by brute force!)


答: The book of Douglas Hofstadter, ''Gödel, Escher, Bach - an eternal golden braid'', is an excellent example how lively mathematics can be. The pics of Escher are impressive.</p>

Title: What are some examples of "visually beautiful" math texts?

Body:
Are there good examples of math textbooks that seamlessly combines art, cartoon drawing or computer graphics with the texts? I know that there are a few books on the market, which are not text-book per se, that are geared towards general audience, where there are plenty of artistic figures naturally generated by the phenomena under study, usually things of geometric nature, Escher tiling, or things like chaos.

Wolfram's a New Kind of Science comes to mind. What are some other examples of "visually beautifu" or "artistic" math textbooks or texts.


问:
Convergence of independent random variables with different distributions
Let $Y_1,Y_2,...,$ be independent r.v. with distribution $P(Y_k=k)=P(Y_k=−k)=1/2$ for all k. Here, $S_n=Y_1+Y_2+...+Y_n \ \ ∀k>0$.

Does ${S_n/n^{3/2}}$ converge in distribution? If yes, please write the limit distribution.
Note: each Yi has a different distribution

I know that the Law of Large Numbers requires i.i.d. random variables but I don't know how to start given this condition.

答:
You can apply the Lindeberg central limit theorem to the sequence $(Y_n)$. The $i$th $Y_i$ has mean $0$ and variance $i^2$ so the variance of the sum $S_n:=Y_1+\cdots+Y_n$ is
$$
s_n^2 := \operatorname{Var}(S_n)=1^2+\cdots+n^2=\frac 16n(n+1)(2n+1)\sim \frac{n^3}3.\tag1$$
Check that the Lindeberg condition holds: For every $\epsilon>0$,
$$
\lim_{n\to\infty} \frac1{s_n^2}\sum_{i=1}^n E\left[Y_i^2 I\left(|Y_i|>\epsilon s_n\right)\right]=0.\tag2
$$
Indeed, observe for every $i=1,\ldots,n$:
$$
Y_i^2 = i^2 \le n^2=n^3\frac1n\le 3s_n^2\frac1n=\frac 3ns_n^2.\tag3$$
In view of (3) we see that given $\epsilon>0$ the condition $|Y_i|>\epsilon s_n$ is never true when $n$ is sufficiently large (namely, when $\frac3n<\epsilon^2$), hence for $n$ sufficiently large each term in the sum in (2) is zero, and the Lindeberg condition holds.

The conclusion of the Lindeberg CLT is that $S_n/s_n$ converges in distribution to standard normal, hence the limit distribution for $S_n/n^{3/2}$ is $N(0,\frac13)$.


问:
Showing that every automorphism of $S_3$ is a conjugation without using the orders of the elements
I'd like to demonstrate that every automorphism of $S_3$ is a conjugation but without looking at the order of the elements. My attempt is the following :

I've already proved that there is an injective homomorphism $f : Aut(S_3) \hookrightarrow S_3$ and also that the conjugation map $Ad$ is an automorphism.

Now let's consider $g : S_3 \longrightarrow Aut(S_3)$ such as $g(\sigma) = Ad(\sigma)$. It is easy to prove that $g$ is an injective homomorphism.

Let $\psi$ be in $Aut(S_3)$ and $\sigma$ be in $S_3$. Thus we have :

$$g \circ f : Aut(S3) \hookrightarrow S_3 \hookrightarrow Aut(S_3)$$
$$ (g \circ f)(\psi) = g(\sigma) = Ad(\sigma)$$

Because of the two injections, we can deduce that $S_3 \cong Aut(S_3)$ and thus $g \circ f$ is an automorphism. I think it also proves that there is as much elements in $Aut(S_3)$ than conjugations by the elements of $S_3$. Therefore, as the set of conjugations by the elements of $S_3$ is a subset of $Aut(S_3)$ we can deduce that every automorphism of $S_3$ is a conjugation.

I'm not quite sure about what I said. Could you please help me ?

Thanks a lot.

答:
I'd attempt the following geometric argument.

Consider 2-simplex (an equilateral triangle on the plane). Its symmetry group is precisely $S_3.$ Moreover, $S_3$ acts simply transitively on the "flags" (sequences of faces ordered by inclusion) of faces of the triangle. In other words: it acts simply transitively on the poset of faces $\mathcal{P}(S_3)$ of $\Delta^2.$ Thus we can identify the poset of faces with the group $S_3$ as sets.

Now consider an arbitrary automorphism $\varphi:S_3\to S_3.$ It induces an automorphism of $\mathcal{P}(S_3).$ Geometric realization of this poset=barycentric subdivision of $\Delta^2.$ Any symmetry of $\Delta^2$ clearly induces an automorhism of the barycentric subdivision and vice versa, hence automorphisms of $S_3$ is the group $S_3$ itself.


答:
Since $\sin(180^\circ - \theta) = \sin(\theta)$, $\sin{90^\circ} = 1$, and $\sin{180^\circ} = 0$, we can write the sum as
$$
(2 \sin{2^\circ} + 178 \sin{2^\circ}) + (4 \sin{4^\circ} + 176 \sin{4^\circ}) + \ldots + (88 \sin{88^\circ} + 92 \sin{88^\circ}) + 90\text.
$$

To get the average, divide by the number of terms, $90$, and get
$$
2 \sin{2^\circ} + 2 \sin{4^\circ} + \ldots + 2 \sin{88^\circ} + 1\text.\tag{*}
$$

Now, $\cos(\theta - 1^\circ) - \cos(\theta + 1^\circ) = 2 \sin\theta \sin 1^\circ$.  Therefore,
$$
2\sin\theta = \frac{\cos(\theta - 1^\circ) - \cos(\theta + 1^\circ)}{\sin{1^\circ}}\text.\tag{**}
$$

When you plug $\text{(**)}$ into $\text{(*)}$, most of the $\cos$ terms cancel out and you are left with
$$
\frac{\cos{1^\circ} - \cos{89^\circ}}{\sin{1^\circ}} + 1 = \frac{\cos{1^\circ} - \sin{1^\circ}}{\sin{1^\circ}} + 1 = \color{red}{\cot{1^\circ}}\text.
$$

Title: Find the average of the number $n \sin n^\circ$ for $n=2,4,6\cdots,180$

Body:
I have been asked in a exam to find the average of the number: $$n \sin n^\circ$$ for $n$=$2,4,6,\cdots,180$

I have tried a lot basically with sum product, or pairing the inputs...but at the end don't able to find any way to solve it, can someone help me with the approach?


问:
Find how many solutions the equation $x^3\equiv 5 \pmod {143}$ has.

Find how many solutions the following equation has:
$x^3\equiv 5 \pmod {143}$

Using Euler's criterion I found that the equation $x^3\equiv 5 \pmod {11}$  has one solution, and $x^3\equiv 5 \pmod {13}$ has 3 solutions.

How can I conclude from that how many solutions the original equation has? I thought maybe to use the CRT.

答:
If you have $m$ solutions to $f(x)\equiv0\bmod p$ and $n$ to $f(x)\equiv0\bmod q$, every pair of solutions to the sub-congruences combines by the CRT into a distinct solution modulo $pq$, so there are $mn$ solutions to the mod-$pq$ congruence.

Thus, in your case, there are $3×1=3$ solutions.


问:
How to show if $G$ is a finite group and $H$, $K$ normal subgroups of $G$, then $G = HK$ if and only if $G / (H \cap K) \cong G / H \times G / K$?
Let $G$ be a finite group and $H$, $K$ normal subgroups of $G$. Prove that $G=HK$ if and only if $G/(H\cap K)$ is isomorphic to $G/H\times G/K$.

For first part i think I have to take the function $\phi\colon G\rightarrow G/H\times G/K$ defined by $\phi(g)=(gH,gK)$ then the kernel would be $H\cap K$ but I am confusion how to show this function is onto and how to use the fact $G=HK$?

For converse part since H and K be normal then $HK$ be a subgroup of $G$ so clearly $HK\subset G$ but now how to show that $G\subset HK$?

It will be enough if I get a proper hint for both part.Thank you.

答:
HINT: For the first part, we know that
$$|HK| = \frac{|H|\cdot |K|}{|H \cap K|}$$
So, what can we say about $|H \cap K|$ if $G = HK$ considering $HK \le G$? Also, the function $\phi$ you defined is not surjective unfortunately. Here, using $G = HK$, we can write any element of $g \in G$ as $g = hk$ for some $h \in H$ and $k \in K$ uniquely. Now, you may try to define it as $\phi: G \to G/H \times G/K$ such that $\phi(g) = (kH, hK)$. Then, $\phi$ is well-defined since $g = hk$ is a unique way to write $g$. Now, you can either check $\phi$ is a surjective homomorphism or check $\phi$ is injective and $|G| = |G/H \times G/K|$ (Since $|G|$ is finite, these two together imply $\phi$ is surjective).

For the second part, we can write
$$\frac{|G|}{|H \cap K|} = \frac{|G|}{|H|}\cdot \frac{|G|}{|K|}$$
Then, what can we say about $|HK|$ considering the above hint for the first part?


问:
Viewing tensors as multilinear maps, what is the difference between a vector and a covector?
I am currently taking a module on Differential Geometry, and as such I have been mostly looking at tensors as objects following the usual transformation rule. However I would like to understand how that is equivalent to viewing tensors as multilinear maps from direct products of Vector and their Dual spaces to the corresponding field of scalars.

Here is my understanding of the distinction between a covector and a vector(as multilinear maps):

Let $V$ be a vector space over $\mathbb{R}$. The dual space of $V$, denoted $V^*$, is defined as $V^* =\{ϕ:V→\mathbb{R}∣ϕ$ is linear$\}$. $V^*$ is an n-dimensional vector space over the same field. Thus, $V$ is isomorphic to $V^*$. Thus they are, as vector spaces, identical. But we consider them to be different by considering elements of $V^*$(covectors) to act on vectors as linear maps.

My confusion:

I have seen it said that you can consider vectors to also be linear a maps of covectors. Doesn't this contradict the distinction of $V$ and $V^*$?

My attempted explanation for this:

Either $V$ and $V^*$ can be considered the dual space of the other, the choice is arbitrary

It seems to me that when viewing vectors and covectors as objects obeying their respective transformation rule, this confusion is not an issue, since we simply define the operation of contraction to combine a vector and a covector into a scalar.

I really can't make intuitive sense of this, so any insights would be appreciated.

答:
This mainly is a question on linear algebra, and it touches on a subtle point. As you say, two (finite dimensional) vector spaces of the same dimension are isomorphic. But this does not mean that they are (or can be considered as) identical. Indeed, there is a big freedom of choice for the isomorphism between two spaces of the same dimension. In particular, if you look at an abstract vector space $V$, you know that $V^*$ has the same dimension, but there is no "natural" way to associate to $v\in V$ a linear functional in $V^*$. To get such an association you need to make additional choices, e.g. a basis for $V$ or an inner product on $V$. This is related to the situation you meet in differential geometry, where you know that all tangent spaces have the same dimension, but there is no natural way to identify different tangent spaces without making additional choices (say a chart). So you cannot simply consider them as "identical" although they are isomorphic. In the same spirit, you have to distinguish between a tangent space and its dual space.

Still, there is a certain symmetry in the situation between $V$ and $V^*$. This comes from the fact that you can also consider the dual $(V^*)^*$ of $V^*$. Of course, this has again the same dimension as $V^*$ and hence as $V$, but here much more is true: There is a natural way to identify the space $(V^*)^*$ with $V$. (But there is no natural way to identify $(V^*)^*$ with $V^*$.) The natural identification sends a vector $v\in V$ to the linear map $V^*\to\mathbb R$ given by $\phi\mapsto\phi(v)$. This means that in a way you can swap the roles of $V$ and $V^*$ (and this is used frequently when dealing with tensor fields in differential geometry), but this does not mean that you have a way to identify $V$ with $V^*$.

This does not mean that there is no difference between vectors and covectors. From a differential geometry point of view, this is because smooth maps (and diffeomorphisms) act on them differently. Consequently, there is a diffeomorphism invariant linear operation on covectors (the exterior derivative) and no such operation exists on vectors. But you can also interpret vectors in terms of linear functionals.


问:
Decomposing matrix $A$ into $A^{\|}$ and $A^{\perp}$ where $A^{\|}$ and $A^{\perp}$ are perpendicular to each other
For a given matrix $V$, I am trying to decompose a matrix $A$ into $A^{\|}$ and $A^{\perp}$ such that

$AV = (A^{\|} +A^{\perp})V = A^{\|}V + A^{\perp}V = \mathbf{0} + A^{\perp}V$,
where $A^{\|}$ and $A^{\perp}$ are orthogonal to each other.

So from what I gather, to do this task, $A^{\|}$ is in the left null space of $V$, so I can first find the null space of $V^T$ by doing SVD for instance (SVD would give orthogonal basis of the null space of $V^T$, which is equal to the left null space of $V$). Then I can project $A$ onto the left null space of $V$ to obtain $A^{\|}$, so I will somehow have to construct the matrix $X$ from the orthogonal basis of the left null space of $V$ so that I can do $A^{\|}= \frac{Tr(A^TX)}{Tr(X^TX)}X$.

The problem I am facing here is I don't know how to construct the matrix $X$ from the orthogonal basis of the left null space of $V$ in such a way that $A^{\|}= \frac{Tr(A^TX)}{Tr(X^TX)}X$ would be orthogonal to $A^{\perp}=A-A^{\|}$ .

How can I do this?

Thank you,

答:
You can take any matrix $X$ whose columns form basis of the left null space of $V$ and define
$$A^\parallel = \frac{\operatorname{tr}(A^TX)}{\operatorname{tr}(X^TX)}X,
\qquad\text{and}\qquad
A^\perp = A-A^\parallel.$$
Then $A=A^\parallel + A^\perp$, $A^\parallel V = 0$ and
\begin{align}
  \operatorname{tr}((A^\parallel)^TA^\perp)
    &= \operatorname{tr}((A^\parallel)^TA)
     - \operatorname{tr}((A^\parallel)^TA^\parallel)\\
    &= \frac{\operatorname{tr}(A^TX)}{\operatorname{tr}(X^TX)}
       \operatorname{tr}(X^TA)
     - \frac{\operatorname{tr}(A^TX)^2}{\operatorname{tr}(X^TX)^2}
       \operatorname{tr}(X^TX)
     = 0,
\end{align}
meaning that $A^\parallel$ and $A^\perp$ are orthogonal.


问:
What does it mean when two Groups are isomorphic?
I'm not asking for the formal definition I know it. An isomorphism is a bijective homomorphism. In my book it's indicated many times when two groups are isomorphic, and I don't understand what's the reason for that. What can we "do" when we know that 2 groups are isomorphic? What does it really mean when 2 groups are isomorphic? </p>

答:
I think that the point is to settle the definition of "isomorphic groups" on another basis than the one invoking the isomorphism (which sounds like swapping the effect with the cause), and then to show that the "operation-preserving property" of (necessarily) a bijection between the two groups is indeed the enabling property to allow the two groups to be isomorphic (according to the said different definition). Hereafter is an attempt to do so.

A group $G$ shows its structure as soon as we allow the internal operation to fully deploy its effects. Accordingly, we can reasonably state the following:

Definition 1. The structure of a group $G$ is the set $\theta_G:=\{\theta_a, a\in G\}\subseteq \operatorname{Sym}(G)$, where $\theta_a$ is the bijection on $G$ defined by: $g\mapsto \theta_a(g):=ag$.

Here a problem arises, if we want to determine whether two groups, $G$ and $\tilde G$, "have the same structure", since in general $\operatorname{Sym}(G)\cap\operatorname{Sym}(\tilde G)=\emptyset$, and then any attempt to "compare by overlapping" the structures $\theta_G$ and $\tilde\theta_\tilde G$ is doomed to fail. We can overcome this issue "by transporting" the structure of $G$ in $\operatorname{Sym}(\tilde G)$, and see whether we can make the "transported $\theta_G$" to overlap with $\tilde\theta_{\tilde G}$. If we succeed, then we can rightly say that $G$ and $\tilde G$ are isomorphic, since we have been able to bring the structure of one onto precisely that of the other. So, with reference to the following diagram:

$$
\newcommand{\ra}[1]{\kern-1.5ex\xrightarrow{\ \ #1\ \ }\phantom{}\kern-1.5ex}
\newcommand{\ras}[1]{\kern-1.5ex\xrightarrow{\ \ \smash{#1}\ \ }\phantom{}\kern-1.5ex}
\newcommand{\las}[1]{\kern-1.5ex\xleftarrow{\ \ \ \smash{#1}\ \ }\phantom{}\kern-1.5ex}
\newcommand{\da}[1]{\bigg\downarrow\raise.5ex\rlap{\scriptstyle#1}}
\begin{array}{c}
 &  & \\
G & \ras{\space\space\space f \space\space\space}   &   \tilde G \\
\da{\theta} &  &  \da{\tilde\theta}  \\
\operatorname{Sym}(G) & \ras{\varphi^{(f)}} & \operatorname{Sym}(\tilde G) \\
\end{array}
$$

we set forth this other:

Definition 2. Two groups, $G$ and $\tilde G$, are isomorphic if there is a bijection $f\colon G\to \tilde G$ such that the above diagram commutes, namely:

$$\tilde\theta=\varphi^{(f)}\theta f^{-1}\tag 1$$

where $\varphi^{(f)}\colon \operatorname{Sym}(G)\to \operatorname{Sym}(\tilde G)$ is the "structure transporting" bijection defined by $\sigma\mapsto f\sigma f^{-1}$.

Now, as a caracterization of such an "enabling" bijection $f$, the following holds:

Claim. Two groups $G$ and $\tilde G$ are isomorphic (as per the Definition 2) if and only if there is a bijection $\psi\colon \tilde G \to G$ such that:

$$\psi(\tilde a\tilde g)=\psi(\tilde a)\psi(\tilde g), \space\space\forall \tilde a,\tilde g\in \tilde G \tag 2$$

Proof.

\begin{alignat}{1}
&\tilde\theta=\varphi^{(f)}\theta f^{-1} &\iff \\
&\tilde\theta_\tilde a(\tilde g)=(\varphi^{(f)}\theta f^{-1})(\tilde a)(\tilde g), \space\space\forall \tilde a,\tilde g\in \tilde G &\iff \\
&\tilde a\tilde g=(\varphi^{(f)}\theta f^{-1})(\tilde a)(\tilde g), \space\space\forall \tilde a,\tilde g\in \tilde G &\iff \\
&\tilde a\tilde g=(\varphi^{(f)}(\theta_{f^{-1}(\tilde a)}))(\tilde g), \space\space\forall \tilde a,\tilde g\in \tilde G &\iff \\
&\tilde a\tilde g=(f\theta_{f^{-1}(\tilde a)}f^{-1})(\tilde g), \space\space\forall \tilde a,\tilde g\in \tilde G &\iff \\
&\tilde a\tilde g=f(\theta_{f^{-1}(\tilde a)}(f^{-1}(\tilde g)), \space\space\forall \tilde a,\tilde g\in \tilde G &\iff \\
&\tilde a\tilde g=f(f^{-1}(\tilde a)f^{-1}(\tilde g)), \space\space\forall \tilde a,\tilde g\in \tilde G &\iff \\
&f^{-1}(\tilde a\tilde g)=f^{-1}(\tilde a)f^{-1}(\tilde g), \space\space\forall \tilde a,\tilde g\in \tilde G\\
\tag 3
\end{alignat}

So, $(1)\Longrightarrow (2)$, by setting $\psi:=f^{-1}$, and $(2)\Longrightarrow (1)$, by setting $f:=\psi^{-1}$. $\space\space\Box$

Therefore, a bijection between two groups with the property $(2)$ is rightly called isomorphism.


问:
What's the correct null and alternative hypothesis?
For the scenario below, write down the most appropriate null and alternative hypotheses that you would use for statistical inference on the population mean. Write down the critical value that you would use to run your test. Choose a significance level for the test of 5%.

Mercury levels higher than two parts per billion in drinking water are unsafe. You are responsible for determining whether drinking water in a given city is safe, based on 100 water samples that have been taken from different areas of that city.

My idea is that the null is that Mercury level is three or higher than three, which means that drinking water is unsafe. The alternative is that Mercury level is less than three, which means that water is safe for drinking. If we find evidence against null, we can reject null and conclude that water is safe for drinking. However, in group discussions, my classmates get different ideas from me.

答:
Mercury levels higher than two parts per billion in drinking water are unsafe.

Reading the statement, the obvious system of hypothesis to be verified is the following

$$\begin{cases}
\mathcal{H}_0,  & \mu=2 \\
\mathcal{H}_1,  & \mu>2
\end{cases}$$

as the test is One Tailed, and considering that $n=100$ is greater enough, you will use the following statistic

$$Z_{stat}=\frac{\overline{X}_{100}-2}{S}\sqrt{100}\dot{\sim}N(0;1)$$

and you will Reject $\mathcal{H}_0$ (EDIT: $H_0 \leq2$ means the water is safe) iff

$$Z_{stat}\geq 1.64$$

or, equivalently

$$\overline{X}_{100}\geq 2+1.64\frac{S}{10}
$$

($S$ is the sample standard deviation)


答:
Working with $\phi$ and the first isomorphism theorem is indeed a proper way.

Hints:

If $x,y\in KH=HK$ with $x=kh,\,y=h'k'$, then $\phi(kh')=(xH,yK)$.

If $\phi$ is surjective, then $\forall x\in G\,\exists g\in G:\phi(g)=(xH,K)$.

Title: How to show if $G$ is a finite group and $H$, $K$ normal subgroups of $G$, then $G = HK$ if and only if $G / (H \cap K) \cong G / H \times G / K$?

Body:
Let $G$ be a finite group and $H$, $K$ normal subgroups of $G$. Prove that $G=HK$ if and only if $G/(H\cap K)$ is isomorphic to $G/H\times G/K$.

For first part i think I have to take the function $\phi\colon G\rightarrow G/H\times G/K$ defined by $\phi(g)=(gH,gK)$ then the kernel would be $H\cap K$ but I am confusion how to show this function is onto and how to use the fact $G=HK$?

For converse part since H and K be normal then $HK$ be a subgroup of $G$ so clearly $HK\subset G$ but now how to show that $G\subset HK$?

It will be enough if I get a proper hint for both part.Thank you.


问:
Piecewise function derivative at the point x = 0
I can't find the derivative of xsin (pi / x) in this question.
$$
f(x) =
\begin{cases}
x \sin \frac{\pi}{x} & x \neq 0\\
0 & x = 0
\end{cases}.
\quad \text{Compute }f'(0).
$$

There must also be a differentiability rule for f '(x) to exist.

How do I solve it?

答:
$$f'(x)=\lim_{h \to 0} \frac{h \sin (\pi/h)-0}{h}= \lim_{h \to 0} \sin (\pi/h)=\text{does not exist}.$$

$L=\lim_{x \to 0} \sin(\pi/x)$ does not exist for two sequeces $x_n=\frac{1}{n}$ and $x'_n=\frac{1}{(n+1/2)}$, let $g(x)=\sin(\pi/x)$, then $g(x_n)=0, g(x'_n)=1$ these being different there are two accumulation points. So, the limit $L$ does not exist. Hence, $f(x)$ is not differentiable at $x=0$.


问:
About Normal Distribution Probability Density Function Notation
The equation for probability density function is:

$$f(x|\mu, \sigma^{2})=(1/(2\pi \sigma^2 ))^{1/2}exp[-(x-\mu)^{2}/2\sigma^{2}]$$

What does the notation $f(x|\mu, \sigma^{2})$ mean? Does it have the same meaning as in conditional probability? And sometimes I'll see $\mu|\sigma^2$ being omitted becoming $f(x)$, or only $\sigma^2$ being omitted becoming $f(x|\mu)$. Does they mean the same?

And what about something like $f(x|a,b,\sigma^2)$ and $f(x|a,b)$?

答:
Anything written after the pipe $\mid$ indicates parameters of the function that are usually held constant for each instance of the function – they can be inferred from context when omitted.

The same notation is found with elliptic integrals and functions: A&S uses $F(\varphi\mid m),\operatorname{sn}(u\mid m)$ and so on, where $m$ is called… the parameter.


问:
Space of bounded continuous functions from $\mathbb R$ to $\mathbb C$
Let $V$ denote the space of bounded continuous functions from $\mathbb R$ to $\mathbb C$. Enumerate the rationals by $r_1,r_2,\ldots$. Define the inner product on $V$ by
$$\langle f,g \rangle = \sum \frac{1}{2^k}f(r_k) \overline{g(r_k)}$$

The question is whether $V$ endowed with this inner product becomes a Hilbert space.

It is fairly easy to check the above operation is indeed an inner product, but I got stuck about the completeness part. Any help is appreciated.

答:
This is not a Hilbert space. Define
$$
f_n(x) = \max(0, \min(nx,1)).
$$
Then $f_n(x) \to 1$ for $x>0$ and $f_n(x)=0$ for all $x<0$. However $(f_n)$ is a Cauchy sequence in the given norm. Let $n>m\ge1$. Then
$$
\|f_n-f_m\|^2 = \sum_{k: \ r_k \in (0,\frac1m)} 2^{-k} |f_n(r_k)-f_m(r_k)|^2
\le \sum_{k: \ r_k \in (0,\frac1m)} 2^{-k}.
$$
It remains to show that the right-hand side converges to zero for $m\to\infty$. Take $g$ with $g(x)=1$ for all $x$. Then
$$
1 = \|g\|^2 \ge  \sum_{k: \ r_k \in (0,1)} 2^{-k} = \sum_{i=1}^\infty \sum_{k: \ r_k \in [\frac1{i+1},\frac1i)} 2^{-k}.
$$
The sum $\sum_{k: \ r_k \in (0,\frac1m)} 2^{-k}$ is a tail of this converging series, hence $\sum_{k: \ r_k \in (0,\frac1m)} 2^{-k} \to 0$ for $m\to\infty$.
Thus, $(f_n)$ is Cauchy. The sequence converges pointwise on the rationals to a function that cannot be extended to be a continuous function.


问:
Finding the values of $t$ such that the limit is zero.

Let a real valued sequence ${x_n}$  such that

$$\lim_{n\to\infty} n\cdot (x_n) = 0$$

Find all possible real numbers $t$ such that
$\lim_{n\to\infty} x_n \cdot (\log n) ^ t = 0$

Source

My approach was to use the Epsilon-Delta definition which results in
$|n x_n| <  \varepsilon$

This does not help in any way...a rigorous proof is needed.

答: For any real number $t$, $|x_n(\log n)^t|=|nx_n|\left|\frac{(\log n)^t}{n}\right|$ and since $\lim\limits_{n\rightarrow +\infty}nx_n=0$ and $\lim\limits_{n\rightarrow +\infty}\frac{(\log n)^t}{n}=0$, we have $\lim\limits_{n\rightarrow +\infty}x_n(\log n)^t=0$ so any $t$ fits.


问:
If $F_{\mu_n}(x) \longmapsto F_{\mu}(x) \hspace{0.1cm}\forall x \in D$ countable dense then $\mu_n \longmapsto \mu$ vaguely
Given $\mu$ a finite measure and $(\mu_n)_{n=1}^{+\infty}$ measure of probability on $(\mathbb{R},\mathcal{B}(\mathbb{R}))$ then the following statements holds :

Theorem : If $F_{\mu_n}(x) \longmapsto F_{\mu}(x) \hspace{0.1cm}\forall x \in D$ countable dense then $\mu_n \longmapsto \mu$ vaguely.

I remember the definition of vague convergence : $\mu_n \longmapsto \mu$ vaguely if $\lim\limits_{n \to +\infty} \int f d\mu_n = \int f d\mu \hspace{0.1cm} \forall f \hspace{0.1cm}\in C_c(\mathbb{R})$.

The proof of the theorem is an approximation argument, noting that if $f = \sum\limits_{k=1}^n a_i \mathbb{1}_{(x_i,x_{i+1}]}$ with $x_i \in D$ then $$\int f d\mu_n = \sum\limits_{k=1}^n a_i(F_{\mu_n}(x_{i+1})-F_{\mu_n}(x_{i})) \overset{n}{\longmapsto} \sum\limits_{k=1}^n a_i(F_{\mu}(x_{i+1})-F_{\mu}(x_{i}))= \int f d\mu$$

So the problem is to approximate every $f \in C_c(\mathbb{R})$ with functions belonging to the class $\sum\limits_{k=1}^n a_i \mathbb{1}_{(x_i,x_{i+1}]} : x_i \in D$.

I know that if $ f\in C_c(\mathbb{R})$ then $ \exists\hspace{0.1cm} K$ compact such that $f|{_{\mathbb{R}-K}} = 0$ so $f$ is uniformely continuos on $K$, but how to procede from here ?

Any help or reference would be appreciated.

答: Let $\epsilon >0$. There exists $\delta >0$ sch that $|f(x)-f(y)| <\epsilon$ if $|x-y| \leq \delta$. There exists $N$ such that $f(x)=0$ if $|x|>N$. Let $\{x_i\}$ be a partition of $[-N,N]$ such that $x_{i+1}-x_i <\delta$ for each $i$. Let $g =\sum a_i1_{(x_i,x_{i+1}]}$ where $a_i=f(x_i)$. Then $|f(x)-g(x)| <\epsilon$ for all $x$.


答:
Despite what some comments have claimed, such expressions are common in Bayesian probability and statistics and do have meaning in terms of conditional densities with respect to parameters and hyperparameters.

For instance, we could define a hierarchical model in which $$X \mid \mu, \sigma \sim \operatorname{Normal}(\mu, \sigma^2),$$ and $\mu$ is itself a random variable:  $$\mu \sim \operatorname{Normal}(\mu_0, \sigma_0^2).$$  In this model, $\sigma$ is fixed and known; $\mu_0$ and $\sigma_0$ are hyperparameters representing the prior distribution of $\mu$.  Then based on an observed sample $\boldsymbol X = (X_1, X_2, \ldots, X_n)$, we can compute a posterior density for $\mu$ given the data $\boldsymbol X$.  We write this as $$f(\mu \mid \boldsymbol X, \sigma, \mu_0, \sigma_0) = \frac{f(\boldsymbol X \mid \mu, \sigma)f(\mu \mid \mu_0, \sigma_0)}{f(\boldsymbol X)},$$ which is essentially Bayes' theorem.  Here, $f(\mu)$ is the prior density, and the denominator is the marginal density of the joint distribution of the sample $\boldsymbol X$.

The takeaway is that the symbols that are written after the $|$ sign are intended to represent those quantities that are fixed with respect to those symbols written before the $|$ sign.  Note that if we wanted to, we could write $$f(\boldsymbol X \mid \mu, \sigma) = f(X_1, X_2, \ldots, X_n \mid \mu, \sigma),$$ showing explicitly that the joint density is a mapping $f : \mathbb R^n \to \mathbb R$.

Title: About Normal Distribution Probability Density Function Notation

Body:
The equation for probability density function is:

$$f(x|\mu, \sigma^{2})=(1/(2\pi \sigma^2 ))^{1/2}exp[-(x-\mu)^{2}/2\sigma^{2}]$$

What does the notation $f(x|\mu, \sigma^{2})$ mean? Does it have the same meaning as in conditional probability? And sometimes I'll see $\mu|\sigma^2$ being omitted becoming $f(x)$, or only $\sigma^2$ being omitted becoming $f(x|\mu)$. Does they mean the same?

And what about something like $f(x|a,b,\sigma^2)$ and $f(x|a,b)$?


问:
Real function that has a taylor series of zero convergence radius in some point
Is there any example of a real valued function $f$ defined on an open interval $I$(possibly the real line) that satisfies the following property?

There is an $a\in I$ such that the Taylor series expansion of $f$ relative to $a$ has zero radius of convergence. (The function doesn't have to be equal to its Taylor series in some interval, just thinking about the convergence radius of the Taylor series itself.)

答: By Borel's theorem, every sequence of real numbers may be the coefficients of the Taylor series of some function. If the coefficients are made to be some sufficiently fast-increasing function, the Taylor series diverges at all non-zero values.</p>


问:
Definition of invariant bilinear form
Given a bilinear form $\beta $ on a Lie algebra $ \mathfrak{g} $, what does it mean for $ \beta $ to be $ \mathfrak{g} $-invariant?

答:
It means $\beta([x, y], z) = \beta(x, [y, z])$.

It is not at all obvious that this is the correct definition. It's written to look nice but you may rightfully ask why this condition deserves the name "invariant." It may make a little more sense to rewrite it as

$$\beta(\text{ad}_y(x), z) + \beta(x, \text{ad}_y(z)) = 0;$$

this reveals that $\mathfrak{g}$-invariant means invariant with respect to the tensor action of $\mathfrak{g}$ on $\mathfrak{g} \otimes \mathfrak{g}$, where $\mathfrak{g}$ carries the adjoint representation. (Invariant with respect to a Lie algebra action means annihilated.) If you've never done this calculation, it would be good to check that

the tensor action of a Lie algebra $X \mapsto X \otimes 1 + 1 \otimes X$ exponentiates to the tensor action of a Lie group $g \mapsto g \otimes g$, and

exponentiation identifies "invariant with respect to $\mathfrak{g}$" and "invariant with respect to $G$" in the following sense: a vector $v \in V$ in a $\mathfrak{g}$-representation is $X$-invariant in the sense that $Xv = 0$ iff it is $\exp(tX)$-invariant in the sense that $\exp(tX) v = v$ for all $t \in \mathbb{R}$.

This means that a bilinear form on $\mathfrak{g}$ is $\mathfrak{g}$-invariant iff it is $\text{Inn}(\mathfrak{g})$-invariant in the obvious sense ($\beta(gx, gy) = \beta(x, y)$), where $\text{Inn}(\mathfrak{g})$ is the connected Lie subgroup of $\text{Aut}(\mathfrak{g})$ with Lie algebra the Lie algebra of inner derivations $\text{ad}_y(-)$.

Title: Definition of invariant bilinear form


问:
What's the difference between arccos(x) and sec(x)
My question might sound dumb, but I don't really see why the graphics of $\arccos(x)$ and $\sec(x)$ are different.

As far as I know $\arccos(x)$ is the inverse cosine function $(\cos(x)^{-1})$ and $\sec(x)$ = $\frac{1}{\cos(x)}$ (source Wolfram|Alpha).

So why aren't they equal?

Thanks in advance.

答:
That's a problem of notation and probably a lack of definitions. We define $\sec x$ as the multiplicative inverse of $\cos x$, in other words, fixed $a \in \mathbb{R}$, $\sec a$ is the number such that $\sec a \cos a = 1$. Now $\arccos x$ is a little different thing: it's the inverse function of $\cos x$.

I don't know if you've learned this but the formal definition of a function is that of a collection of ordered pairs. In other words, since a function from a set $A$ to a set $B$ should be a rule assigning for each $a \in A$ some $b \in B$ we can simply define a function as the set of all ordered pairs of elements in $a$ together with related elements in $b$. However, we require the additional property that if $(a,b) \in f$ and if $(a,c)\in f$ then $b = c$ and this is just the formal way to state the "vertical line rule". Since the second element in each pair is unique we give it a name: if $(a,b) \in f$ then $b = f(a)$. Also to state starting and ending sets we write functions from $A$ to $B$ as $f: A \to B$.

Now, if you have a function you have a collection of ordered pairs right? So, you can create a new set of ordered pairs by reversing the pairs. So if $f : A \to B$ is a function from $A$ to $B$ we define the inverse $f^{-1}$ by the property that $(a,b) \in f^{-1}$ when $(b,a)\in f$. Now it's not at all clear when $f^{-1}$ is a function. Just to show you that consider the following function that maps naturals to naturals:

$$f = \{(1,2), (3,2), (4,1)\}\subset \mathbb{N}\times \mathbb{N}$$

This is a function by our definition. Now the inverse is $f^{-1} = \{(2,1), (2,3), (1,4)\}$, now this isn't a function because $(2,1)\in f^{-1}$ and $(2,3)\in f^{-1}$. So $f^{-1}$ will be a function if the original function also satisfies $f(x) = f(y)$ implying $x = y$. This kind of function is called one-one, and so if $f$ is one-one, $f^{-1}$ will be a function called then inverse function.

Also, if $f: \mathbb{R} \to \mathbb{R}$ has an inverse function $f^{-1}:\mathbb{R} \to \mathbb{R}$ then $f(f^{-1}(x)) = x$ and $f^{-1}(f(x)) = x$. So $\arccos$ is defined precisely this way: fixing one interval where $\cos$ is one-one, you define $\arccos$ in that interval by the property that $\arccos x$ is the number $y$ such that $\cos y = x,$ in other words, it returns you the value of angle whose cosine is $x$.

Just a reference to finish: you can find treatments like this in books like Spivak's Calculus or Apostol's Calculus Vol. 1. I hope the way I exposed this helps you a little. Good luck!

EDIT: The problem of notation I've mentioned and forgot talking about is that both the multiplicative inverse and the inverse function are in some contexts denoted by $\cos^{-1}$ and this usually happens to all trigonometric functions. So to avoid confusion, I recommend writing $\arccos$, $\arcsin$ and so on for the inverse functions.


问:
Theorem concerning Linear Cauchy functional equation $ f ( a x + b y + c ) = p f ( x ) + q f ( y ) + r $
I was reading a book on functional equation by Titu Anderescu and came across the following functional equation.

Linear Cauchy Equation

This functional equation is an extension of both Cauchy's and Jensen's equation. It has the form
$$ f ( a x + b y + c ) = p f ( x ) + q f ( y ) + r \text , \tag {*} \label {eqn} $$
where $ a $, $ b $, $ c $, $ p $, $ q $ and $ r $ are real constants and $ a b \ne 0 $.

Then we have the following Theorem.

Theorem. Let $ f : \mathbb R \to \mathbb R $ be a nonconstant continuous function satisfying \eqref{eqn}. Then $ f ( x ) = s x + t $ for some constants $ s \ne 0 , t \in \mathbb R $ and $ p = a $, $ q = b $, $ r = s c + t ( 1 - a - b ) $.

Proof. setting
$$ ( x , y ) = \left( - \frac c a , 0 \right) , \left( \frac { u - c } a , 0 \right) , \left( - \frac c a , \frac v b \right) , \left( \frac { u - c } a , \frac v b \right) $$
in \eqref{eqn} gives respectively
$$ f ( 0 ) = p f \left( - \frac c a \right) + q f ( 0 ) + r \text , $$
$$ f ( u ) = p f \left( \frac { u - c } a \right) + q f ( 0 ) + r \text , $$
$$ f ( v ) = p f \left( - \frac c a \right) + q f \left( \frac v b \right) + r \text , $$
$$ f ( u + v ) = p f \left( \frac { u - c } a \right) + q f \left( \frac v b \right) + r \text . $$
Hence $ \require {color} \color {red} f ( u + v ) = f ( u ) + f ( v ) - f ( 0 ) $ for all $ u , v \in \mathbb R $.

Thus $ g ( u ) = f ( u ) - f ( 0 ) $ is a continuous additive function and we conclude that $ g ( x ) = s x $ for some constant $ s \in \mathbb R $, i.e. $ f ( x ) = s x + t $, where $ t = f ( 0 ) $ and $ s \ne 0 $. Now plugging $ f ( x ) $ in \eqref{eqn} shows that $ p = a $, $ q = b $ and $ r = s c + t ( 1 - a - b ) $.

I am unable to understand that how the author has concluded $ f ( u + v ) = f ( v ) + f ( u ) + f ( 0 ) $, as for that to happen, $ 2 \Big( p f \left( - \frac  c a \right) + q f (0) + r \Big) = 2 f ( 0 ) $ should be equal to $ 0 $, but the author has not proved anything like that in the theorem.

It would be highly helpful if I am provided with a pedantic proof that how $ f ( 0 ) = 0 $.

答: Actually you get $f(u+v)=f(u)+f(v)+tf(0)$ for some $t$ [ $t=1-2q$]. Put $u=v=0$ to get $(2+t)f(0)=f(0)$ If $f(0) \neq 0$ then this gives $t=-1$ so   $f(u+v)=f(u)+f(v)-f(0)$. If $g(u)=f(u)-g(0)$ then  we get $g(u+v)=g(u)+g(v)$. So in both the cases $f(0)=0$ and $f(0) \neq 0$ we can reduce the given functional equation to Cauchy's equation.


问:
Differentiation on $(C[0,1], d_\infty)$ and $(C[0,1], d_1)$
Define $F:C[0,1]\rightarrow \mathbb{R}$ as $F(u)=\int_{0}^{1} (u(x))^2 dx$.

We have verified that directional derivative of $F$ at $u$ in direction $g$ is $D_gF(u)=\int_0^12u(x)g(x)dx$.

Now there is a claim that the differential of $F$ at $u$ is the same map $(F'(u))(v)= \int_0^12u(x)v(x)dx$, under both $d_1$ and $d_\infty$.

I use definition of differentiation, after several steps, I get

$\frac{|F(u+h)-F(u)-F'(u)(h)|}{\|h\|} = \frac{\int_0^1 (h(x))^2dx}{\|h\|}$

But downstairs is either a sup (in $d_\infty$) or an integral (in $d_1$), and I don't know how to continue to bound the expression to $0$ (as $h\rightarrow0$) in either case..

答:
Note that
\begin{align}
\|h\|_2 \le \|h\|_\infty^\frac{1}{2}\|h\|_1^\frac{1}{2}
\end{align}
then it follows that
\begin{align}
\frac{\int^1_0 h(x)^2\ dx}{\|h\|_\infty} \le \|h\|_1
\end{align}
and likewise
\begin{align}
\frac{\int^1_0 h(x)^2\ dx}{\|h\|_1} \le \|h\|_\infty.
\end{align}

It remains to prove the first inequality. Observe
\begin{align}
\int^1_0 h(x)^2\ dx \le \|h\|_\infty \int^1_0 |h(x)|\ dx .
\end{align}


答:
As I have commented below the answers Bamboo and Ethan Bolker, ismomorphism between two groups  does not mean "they are one and the same".  They are same only theoretically, i.e., algebraically, but not practically. Actually they could be differnet practically is the reason isomorphisms are useful!

Isomorphism only means what it says, a homomorphism which is bijective. As a consequence two isomorphic groups share many properties, number of elements of a specific order, being abelian, being solvable/nilpotent etc.

Actually isomorphism means two vastly different groups can be understood easily in terms of each other. For example in those days when there were no calculators (for get computers) one used logarithm tables to carry out multiplication. The reason that adding the logarithms and then finding the antilogarithm produces the product of two numbers is because: logarithm gives an isomorphism of the group of positive real numbers under multiplication with the group of all real numbers under addition.

Knowing a newly found  group is isomorphic to a better understood known group  gives us a handle to work with the former.

Many theorems are celebrated because they show two groups occurring in different contexts are isomorphic. (de Rham's theorem in differential geometry/topology, Artin's Reciprocity Law in Algebraic Number Theory)

Easier example: All polynomials with integer coefficients in a single variable is a group under usual addition of polynomials  and is isomorphic to the group of all POSITIVE RATIONAL numbers under usual multiplication of numbers.

Title: What does it mean when two Groups are isomorphic?

Body: I'm not asking for the formal definition I know it. An isomorphism is a bijective homomorphism. In my book it's indicated many times when two groups are isomorphic, and I don't understand what's the reason for that. What can we "do" when we know that 2 groups are isomorphic? What does it really mean when 2 groups are isomorphic? </p>


问:
Show this inequality for a normally distributed random variable
Let $X$ be normally distributed with mean $\mu=0$ and variance $\sigma^2>0$.

I want to show that given  any real number $a \geq 0$ we have $\mathbb{P}(\mid X \mid \leq a) \leq \frac{a}{\sigma}$
How can I show this inequality in a simple way? I tried doing it just by computing it with the probability density function, but then you get an annoying calculation with the error function. Any help would be appreciated.

答: $\frac 1  {\sqrt {2 \pi} \sigma} \int_{-a}^{a} e^{-x^{2}/2\sigma^{2}} dx \leq \frac1  {\sqrt 2 \pi \sigma } \int_{-a}^{a}  dx <\frac a {\sigma}$ since $\sqrt {\frac2  {\pi}} <1$.


答:
We need to show that $\log^t(n)=o(n)$, which is fairly obvious for $t\le 0$.

Now equivalently, for $t>0$,

$$\log(n)=o(\sqrt[t]n)$$

and

$$\log(m^t)=o(m)$$

and

$$\log(m)=o(m).$$

Title: Finding the values of $t$ such that the limit is zero.

Body:
Let a real valued sequence ${x_n}$  such that

$$\lim_{n\to\infty} n\cdot (x_n) = 0$$

Find all possible real numbers $t$ such that
$\lim_{n\to\infty} x_n \cdot (\log n) ^ t = 0$

Source

My approach was to use the Epsilon-Delta definition which results in
$|n x_n| <  \varepsilon$

This does not help in any way...a rigorous proof is needed.


答:
The choice of null and alternative hypothesis depends on which claim requires the burden of proof.

A commonly used analogy comes from the legal doctrine of "innocent until proven guilty."  In a jury trial, defendants are presumed innocent, and are only convicted if there is guilt "beyond a reasonable doubt."  This means the evidence weighing in favor of guilt must be so overwhelming that it is highly implausible to a reasonable person that the defendant could still be innocent in light of the evidence presented.

The rationale for this doctrine is that it is a far greater miscarriage of justice to wrongly convict an innocent person than it is to fail to convict the guilty on the basis of inadequate evidence of guilt.  Consequently, such a legal system would prefer to reduce the likelihood of the former than the latter.

In hypothesis testing, a similar notion applies to the null and alternative hypotheses, with the null being "innocence" and the alternative being "guilt."  Therefore, which claim is the null and which is the alternative depends on which one, if true, requires the data to show with a high degree of confidence.  Correspondingly, the only conclusions that are possible in a statistical hypothesis test are:

Reject the null hypothesis (i.e., accept the alternative)

Fail to reject the null hypothesis (i.e., the evidence is inconclusive).

It is a common misconception to characterize the second conclusion as being equivalent to "accepting the null hypothesis."  This is wrong because the entire test is performed under the presumption of the null, just as a jury trial is held under the presumption of innocence:  the prosecutor will argue "if the defendant is innocent, then how could all the evidence point to their guilt?"

To further extend the analogy, a conclusion in which we reject the null hypothesis in error is analogous to a defendant being wrongfully convicted.  This error is called Type I error, and we design the test in such a way as to limit the probability of making such an error to not exceed some predefined value called $\alpha$, the significance level of the test.  Conversely, the failure to reject the null when the alternative is true, is called a Type II error, and is analogous to the failure to convict a guilty defendant.

With all of this in mind, we now turn our attention to your specific question.  Here, you are presented with a scenario in which we very clearly want to conclusively demonstrate that the water is safe--in other words, the penalty for erroneously claiming the water is safe when it is not, is quite a lot higher than the cost of claiming the water is unsafe when it is.  Consider that many people could be severely poisoned or permanently harmed in the former case.  So the former error is the one that must be tightly controlled.  Consequently, the claim that requires the burden of proof is the claim that the water is safe.  This must be your alternative hypothesis.  Therefore, the correct structure of the hypothesis test is

$$H_0 : \mu \ge 2 \quad \text{vs.} \quad H_1 : \mu < 2.$$

Under the assumption of the null being true, the choice of test statistic will be based on a null mean of $\mu_0 = 2$, since this is the choice that will ensure that the Type I error of the test will not exceed the significance level--i.e., it is the most conservative and ensures that if we reject $H_0$, we do so in error with probability at most $\alpha$.

Title: What's the correct null and alternative hypothesis?

Body:
For the scenario below, write down the most appropriate null and alternative hypotheses that you would use for statistical inference on the population mean. Write down the critical value that you would use to run your test. Choose a significance level for the test of 5%.

Mercury levels higher than two parts per billion in drinking water are unsafe. You are responsible for determining whether drinking water in a given city is safe, based on 100 water samples that have been taken from different areas of that city.

My idea is that the null is that Mercury level is three or higher than three, which means that drinking water is unsafe. The alternative is that Mercury level is less than three, which means that water is safe for drinking. If we find evidence against null, we can reject null and conclude that water is safe for drinking. However, in group discussions, my classmates get different ideas from me.


答:
Thus they are, as vector spaces, identical.

Be careful. It's dangerous to consider mathematical objects to be identical just because there happens to be an isomorphism between them. The problem is that isomorphisms that fail to be "natural" generalize poorly. For example, if $V$ carries a representation of a group $G$, $V^{\ast}$ also carries a representation of $G$, and it will generally be a non-isomorphic representation. There are many other contexts like this, for example in differential geometry, where you'll get very confused if you carelessly identify a vector space with its dual. Tangent vectors are not cotangent vectors and vector fields are not differential $1$-forms.

I have seen it said that you can consider vectors to also be linear a maps of covectors. Doesn't this contradict the distinction of $V$ and $V^*$?

This is a point that is very often poorly explained. The resolution is that the term "covector" is relative, and what it's relative to is a choice of "base" vector space to act as the "vectors." That is, the grammar of the term "covector" is "an object is a covector with respect to $V$," which means that it's an element of the dual space $V^{\ast}$.

So it's true that $V^{\ast}$ consists of the covectors with respect to $V$, and it's also true that $V$ consists of the covectors with respect to $V^{\ast}$. Formally what this means is that there is a natural double dual isomorphism $V \to (V^{\ast})^{\ast}$ (when $V$ is finite-dimensional); this time it is actually mostly harmless to consider these as identical, due to naturality. For example the double dual of a $G$-representation is a naturally isomorphic $G$-representation.

Title: Viewing tensors as multilinear maps, what is the difference between a vector and a covector?

Body:
I am currently taking a module on Differential Geometry, and as such I have been mostly looking at tensors as objects following the usual transformation rule. However I would like to understand how that is equivalent to viewing tensors as multilinear maps from direct products of Vector and their Dual spaces to the corresponding field of scalars.

Here is my understanding of the distinction between a covector and a vector(as multilinear maps):

Let $V$ be a vector space over $\mathbb{R}$. The dual space of $V$, denoted $V^*$, is defined as $V^* =\{ϕ:V→\mathbb{R}∣ϕ$ is linear$\}$. $V^*$ is an n-dimensional vector space over the same field. Thus, $V$ is isomorphic to $V^*$. Thus they are, as vector spaces, identical. But we consider them to be different by considering elements of $V^*$(covectors) to act on vectors as linear maps.

My confusion:

I have seen it said that you can consider vectors to also be linear a maps of covectors. Doesn't this contradict the distinction of $V$ and $V^*$?

My attempted explanation for this:

Either $V$ and $V^*$ can be considered the dual space of the other, the choice is arbitrary

It seems to me that when viewing vectors and covectors as objects obeying their respective transformation rule, this confusion is not an issue, since we simply define the operation of contraction to combine a vector and a covector into a scalar.

I really can't make intuitive sense of this, so any insights would be appreciated.


问:
X and Y are independent iff $P(a < X \leq b, c < Y \leq d) = P(a < X \leq b) P(c < Y\leq d)$ proof
To phrase it another way, how would one show that these two expressions are equivalent?

$P(X \leq x, Y \leq y) = P(X \leq x) P (Y \leq y)$

$P(a < X \leq b, c < Y \leq d) = P(a < X \leq b) P(c < Y\leq d) \quad$ for all $a < b, c < d$

I tried to do it by writing the events as some sort of disjoint unions, but didn't really get anywhere. Any help would be appreciated! Thanks!

答:
Hint

If 2. holds

$$
\begin{align}
\mathbb P(X\leq x, Y\leq y)
&=\mathbb P(\bigcup_{n,m\geq 1}\{-n<X\leq x\}\cap\{ -m<Y\leq y\})
\\&=\lim_{n\to \infty }\lim_{m\to \infty }\mathbb P(-n<X\leq x, -m<Y\leq y)
\\&=\lim_{n\to \infty }\mathbb P(-n<X\leq x)\lim_{m\to \infty }\mathbb P(-m<Y\leq y)
\\&=\mathbb P(X\leq x)\mathbb P(Y\leq y).
\end{align}
$$

Conversely, if 1. holds, try a similar argument.


问:
Positional notation: proof of relations for different base but same digits
I have the following situation:

Be $n_1, n_2, b_1, b_2 \in \mathbb{N}$ with the notation of $n_1$ in base $b_1$ and $n_2$ in $b_2$ and the digits $z_{l-1}z_{l-2}...z_0$

Furthermore, be $b_1 > b_2$ and $m \in \mathbb{N}$

Proof (or refute) that there exists an $N \in \mathbb{N}$ so that if $l > N$ and $z_{l-1} \neq 0$, then $n_1 > mn_2$.

I think that this implication is correct but I struggle with how to proof it. Does anyone have a tip how to go about this?

答:
For a $l$-digit number $n$ in base $b$, we have the following limits: $(\underbrace{100\ldots 0}_l)_b\le n\lt (\underbrace{100\ldots 0}_{l+1})_b$, i.e. $b^{l-1}\le n \lt b^l$.

Now, if we can have $mb_2^l\le b_1^{l-1}$ for a sufficiently large $l$, then, for any $l$-digit number $n_1$ in $b_1$ and any $l$-digit number $n_2$ in $b_2$ we will have:

$$mn_2<mb_2^l\le b_1^{l-1}\le n_1$$

All that is left is to prove that $mb_2^l\le b_1^{l-1}$ is achievable for sufficiently large $l$. However, this inequality is equivalent with $mb_1\le\left(\frac{b_1}{b_2}\right)^l$, and this is certainly true for "large enough" $l$ because $b_1>b_2$, and so we have that the right-hand side diverges to $+\infty$ (when $l\to\infty$), while the left-hand side is a constant. More explicitly, you may take $l\ge N=\lceil\log_{b_1/b_2}(mb_1)\rceil$.

(Note we have proven the inequality not only in the case $n_1$ and $n_2$ have the same digits, but in the case of them having arbitrary digits, as long as both are with the same number of digits and the leading digit is nonzero in both.)


问:
How do I implement the Euler scheme for this SDE?
I have the following problem:

Consider the system of (Ito) SDE's for the price $S_{t}$ of a stock, the stochastic and past-dependent
volatility $\sigma_{z}$ and the long-term averaged volatility $\xi_{t}:$

\begin{array}{l}
d S_{t}=\alpha S_{t} d t+\sigma_{i} S_{t} d w_{t}^{1} \\
d \sigma_{t}=-\left(\sigma_{t}-\xi_{t}\right) d t+p \sigma_{t} d w_{t}^{2} \\
d \xi_{t}=\frac{1}{\alpha}\left(\sigma_{z}-\xi_{i}\right) d t
\end{array}with $S_{0}=€ 50, \quad \sigma_{0}=0.20, \quad \xi_{0}=0.20 \quad$ and $\alpha=0.10$

For $p=0($ and $\alpha \neq 0)$ the well-known Black-Scholes model is obtained:
$$
d S_{t}=\alpha S_{i} d t+\sigma_{0} S_{i} d w_{i}^{1}
$$

Question: How do I implement the Euler scheme for this SDE?

Nadine :)

Additional Question after answer of   :

Unfortunately my knowledge is not sufficient to follow your answer. I found the following derivation on page 2 of: http://people.math.gatech.edu/~meyer/MA6635/chap4.pdf

We can now state Itô's lemma:
Let $X$ satisfy
$$
\begin{aligned}
d X(t) &=a(X, t) d t+b(X, t) d W \\
X(0) &=X
\end{aligned}
$$

or in the case of my specific SDE:

$d S_{t}=\alpha S_{t} d t+\sigma_{i} S_{t} d w_{t}^{1} \\$

with $a(X,t) = \alpha S_{t}$ and $b(X,t) = \sigma_i S_t$

Then the following is stated:

"Assume that u(x, t) is a smooth function of the independent variables x and t. Then"

$d u(t)=\left[\frac{\partial u}{\partial t}+a(X, t) \frac{\partial u}{\partial x}+\frac{1}{2} \frac{\partial^{2} u}{\partial x^{2}} b(X, t)^{2}\right] d t+b(X, t) \frac{\partial u}{\partial x} d W$

I don't understand why only the $a(X,t)$ term gets replaced by this îto solution.

The second part states that:

If we apply Itô's lemma to $u=\ln s$ where
$$
d S(t)=\mu S d t+\sigma S d W
$$
then (with $X=S, a(S, t)=\mu S, b(S, t)=\sigma S)$ we find
$$
\frac{\partial u}{\partial s}=\frac{1}{s}, \quad \frac{\partial^{2} u}{\partial s^{2}}=-\frac{1}{s^{2}}, \quad \frac{\partial u}{\partial t}=0
$$

I see where the derivatives come from, but I don't see why u = ln(s).

The document states that:

$d u(t)=\left[\mu-\sigma^{2} / 2\right] d t+\sigma d W$

But find that really strange since there are no $S$ terms or $a(X,t)$ or $b(X,t)$ terms are observed.

Thank you very much for your first reply, but it went way too quick. Since I can't go to univeristy due covid, this forum is the only forum of instruction I have.

Would you mind to elaborate the answers with more steps based on the additional questions?

Thankyou,
Nadine  :)

答:
The Euler method is a rather crude method. The same goes for Euler-Maruyama. However, there is no trivial extension of ODE methods to SDE methods of the same order. The Milshtein method, even in its derivative-free variation, raises the strong approximation order to one. The next expansion terms of the exact solution contain integrals over the Brownian motion that can not be reduced to values at the sample points. This is an obstacle for higher strong order, it is slightly easier to construct methods with higher weak order, see Kloeden, Platen "Numerical Solution of Stochastic Differential Equations", Springer, 1992.

So it would be best to make the equation as simple as possible before starting with the numerics. Here one can do what is also done for the simple geometric Brownian motion and set $S_t=\exp(X_t)$ or $X_t=\log(S_t)$. Then the Ito formula gives
$$
dX_t = \frac{dS_t}{S_t}-\frac12\frac{d\langle S_t\rangle}{S_t^2}
=α dt+σ_tdW^1_t-\frac12σ_t^2dt
$$
For the numerical step you now translate all differentials into differences and use their known properties
$$\newcommand{\D}{\mathit{\Delta}}
\begin{align}
X_{t+\D t}&=X_t+(α-\tfrac12σ_t^2)\D t+σ_tz^1_t\sqrt{\D t}\\
σ_{t+\D t}&=σ_t−(σ_t−ξ_t)\D t+pσ_tz^2_t\sqrt{\D t}\\
ξ_{t+\D t}&=ξ_t+\frac1α(σ_t−ξ_t)\D t
\end{align}
$$
where $z^1_t$, $z^2_t$ are random numbers from a standard normal distribution. They get computed anew in every step. Alternatively, if for instance you want to compare the "strong" numerical convergence for different step sizes, you can generate the Brownian motions $W^k_t$ for the smallest step size and use the appropriate differences for larger step sizes. (To do that in the other direction "constructively", you would need to fill the gaps with Brownian bridges.)

```<code>t = linspace(0,T,N+1); dt = t[1]-t[0]
W = cumsum(randn([2,N+1])*dt**0.5)
# for an integration test with step size M*dt use
do_integration_test(t[::M],W[:,::M],...)
</code>```

As the Brownian motion can be considered as approximately a random walk with very small step size, simplified numerical methods that might be somewhat faster are possible, for details and more see the always recommendable
P. Forsyth: "An introduction to Computational Finance without Agonizing Pain".

As to the more general situation, when translating to the given equation you get $X_t=X(t)=S_t$ and the coefficients as you describe. If necessary, which it here is not, you could translate back, $u=\ln s\implies s=exp(u)$. With constant coefficients, there is no more to do, you can integrate directly $U_t=U_0+[μ−σ^2/2]t+σW_t$, so that $S_t=S_0\exp([μ−σ^2/2]t+σW_t)$, the equation for the geometric Brownian motion.

Body:
The Euler method is a rather crude method. The same goes for Euler-Maruyama. However, there is no trivial extension of ODE methods to SDE methods of the same order. The Milshtein method, even in its derivative-free variation, raises the strong approximation order to one. The next expansion terms of the exact solution contain integrals over the Brownian motion that can not be reduced to values at the sample points. This is an obstacle for higher strong order, it is slightly easier to construct methods with higher weak order, see Kloeden, Platen "Numerical Solution of Stochastic Differential Equations", Springer, 1992.

So it would be best to make the equation as simple as possible before starting with the numerics. Here one can do what is also done for the simple geometric Brownian motion and set $S_t=\exp(X_t)$ or $X_t=\log(S_t)$. Then the Ito formula gives
$$
dX_t = \frac{dS_t}{S_t}-\frac12\frac{d\langle S_t\rangle}{S_t^2}
=α dt+σ_tdW^1_t-\frac12σ_t^2dt
$$
For the numerical step you now translate all differentials into differences and use their known properties
$$\newcommand{\D}{\mathit{\Delta}}
\begin{align}
X_{t+\D t}&=X_t+(α-\tfrac12σ_t^2)\D t+σ_tz^1_t\sqrt{\D t}\\
σ_{t+\D t}&=σ_t−(σ_t−ξ_t)\D t+pσ_tz^2_t\sqrt{\D t}\\
ξ_{t+\D t}&=ξ_t+\frac1α(σ_t−ξ_t)\D t
\end{align}
$$
where $z^1_t$, $z^2_t$ are random numbers from a standard normal distribution. They get computed anew in every step. Alternatively, if for instance you want to compare the "strong" numerical convergence for different step sizes, you can generate the Brownian motions $W^k_t$ for the smallest step size and use the appropriate differences for larger step sizes. (To do that in the other direction "constructively", you would need to fill the gaps with Brownian bridges.)

t = linspace(0,T,N+1); dt = t[1]-t[0]
W = cumsum(randn([2,N+1])*dt**0.5)
# for an integration test with step size M*dt use
do_integration_test(t[::M],W[:,::M],...)

As the Brownian motion can be considered as approximately a random walk with very small step size, simplified numerical methods that might be somewhat faster are possible, for details and more see the always recommendable
P. Forsyth: "An introduction to Computational Finance without Agonizing Pain".

As to the more general situation, when translating to the given equation you get $X_t=X(t)=S_t$ and the coefficients as you describe. If necessary, which it here is not, you could translate back, $u=\ln s\implies s=exp(u)$. With constant coefficients, there is no more to do, you can integrate directly $U_t=U_0+[μ−σ^2/2]t+σW_t$, so that $S_t=S_0\exp([μ−σ^2/2]t+σW_t)$, the equation for the geometric Brownian motion.


问:
Finding a function which is distance $1$ from $x^2$ along its normals
So I recently thought of a problem, which I have tried to solve many ways but couldn't. It's very simple to describe...

Find the function which is distance $1$ from parabola $y=x^2$ along its normals. Visual of what I mean.

This one of the things I've tried...

This function would get the corresponding $x$ value on the function we want to find from an $x$ value on curve $y=x^2$. The inverse would give the corresponding $x $ value on $x^2$ for a given $x$ value which would make it trivial to determine the $y$ value.
$$
f(x)=x+\sin(\arctan(2x))
$$
This can be resolved to this...
$$
f(x)=x+\frac{2x\sqrt{1+4x^2}}{1+4x^2}
$$
However I don't know how to get the inverse of this function.

答:
$(f-x)=2x/\sqrt{1+4x^2}$
so
$(f-x)^2(1+4x^2)=4x^2$
or, according to Wolfy,
$4 f^2 x^2 + f^2 - 8 f x^3 - 2 f x + 4 x^4 - 3 x^2 = 0$.

This is a quartic in $x$
which can be solved but is incredibly messy as expected.


答: The word "inverse" is context-sensitive.  There are additive, multiplicative and compositional inverses.  The $\arccos$ function is a compositional inverse.  The secant function is a multiplicative inverse to the cosine function, which is defined off the zeroes of the cosine function.  </p>

Title: What's the difference between arccos(x) and sec(x)

Body:
My question might sound dumb, but I don't really see why the graphics of $\arccos(x)$ and $\sec(x)$ are different.

As far as I know $\arccos(x)$ is the inverse cosine function $(\cos(x)^{-1})$ and $\sec(x)$ = $\frac{1}{\cos(x)}$ (source Wolfram|Alpha).

So why aren't they equal?

Thanks in advance.


问:
Hall's Marriage Theorem
I am aware that Hall's Marriage theorem for complete matching goes like "A bipartite graph $G$ with bipartition $(V_1, V_2)$ has a complete matching from $V_1$ to $V_2$ if and only if $$ |N(A)| \geq |A|, \forall A \subseteq V_1$$

I want to know in which cases does an equality hold, i.e. $$ |N(A)| = |A|, \forall A \subseteq V_1 $$

Any help is greatly appreciated.

答:
This can only hold if every vertex of $V_1$ has degree $1$ exactly (so the graph is a disjoint union of edges).

Why? Consider each vertex $u\in V_1$ as the singleton set $A=\{u\}$. If it has any number of neighbors not equal to one, you've already found a set breaking the equality.

Title: Hall's Marriage Theorem


问:
How do I estimate $\int_{2}^{\infty} \frac{|\log \log t|}{t\, (\log t)^2} \, dt$
Question: How do I estimate the following integral? $$\int_{2}^{\infty} \frac{|\log \log t|}{t\, (\log t)^2} \, dt$$

Attempt: Setting $u=\log t$, we get

$$\int_{2}^{\infty} \frac{|\log \log t|}{t\, (\log t)^2} \, dt = \int_{\log 2}^{\infty} \frac{|\log u|}{u^2}\, du$$ and then I am stuck.

答:
Observe that
\begin{align}
\int^\infty_2 \frac{|\log\log t|}{t(\log t)^2}\ dt = \int^\infty_{\log\log 2} |u|e^{-u}\ du.
\end{align}

You can even evaluate this by hand.


问:
is the $0$ vector of $V$ also in $W$? How do I verify this?
Why is W not a subspace of V here?

If we let $w(t)=p(t)+q(t)$, then $w(0)=2$ which is not equal to 1. Does this mean that W is not closed under vector addition?

is the $0$ vector of $V$ also in $W$? How do I verify this?

答: Not_tiger you know $W$ is not empty. So let $v \in W$. Assume $W$ is a subspace. So $-v$ is also in $W$. $v$ and $-v$ are in $V$. The zero vector is $0=v+(-v)$. The zero vector of $V$ is also zero vector of $W$. But in $W$ there is no zero vector. If there is a zero vector, we must have $a=b=c=0$ but this contradicts with $p(0)=1$.


问:
What does it mean for a matrix to be orthogonally diagonalizable?
I'm a little confused as to when a matrix is orthogonally diagonalizable.

I understand that if symmetric, it's always orthogonally diagonalizable, but in what other cases can you orthogonally diagonalize a matrix?

答:
I assume that by $A$ being orthogonally diagonalizable, you mean that there's an orthogonal matrix U and a diagonal matrix $D$ such that

$$A = UDU^{-1} = UDU^T.$$

A must then be symmetric, since (note that since $D$ is diagonal, $D^T = D$!)
$$A^T = \left(UDU^T\right)^T = \left(DU^T\right)^TU^T = UD^TU^T = UDU^T = A \text{.}$$


问:
Finding the linear span of a polynomial with a condition
I'm trying to find the linear span of the following set:

$M = \{p(x)∈ \mathbb R_4 [x] : p(x)=p(x-1),  x∈\mathbb R\} $

I've tried taking a vector from $M$ and applying the condition on it. Like that:

Let:
$p(x) ∈ M$

$p(x) = p(x-1)$

$a + bx + cx^2 + dx^3 = a + b(x-1) + c(x-1)^2 + d(x-1)^3 $

$b(1) + c(2x-1) + d(3x^2 -3x +1) = 0$

$(b-c+d) + (2c-3d)x + (3d)x^2 = 0$

Now, if I open an equation system based on the variables of the polynomial, I get that b, c and d are equal to 0.
What does that mean in terms of the linear span?
Am I trying to find it wrong?

答: Equate  the coefficients to $0$ and you will get $b=c=d=0$. Since $a$ is arbitrary the span is simply the set of all constant polynomials.


问:
When can we inscribed sphere in tetrahedron?
I'm wondering is there any conditions to do that. I believe propably there are but what ?</p>

答:
Here a "physical" proof that any tetrahedron has an inscribed sphere.

Let $\Delta$ be a tetrahedron.

Consider a vertex $v$ and the three planes containing the faces meeting at $v$. Consider the triangular infinite cone they form at $v$. Think of having it in a vertical position. Now put a unit sphere inside the cone. The sphere will fall until it touches the three faces.

Now consider the fourth face of $\Delta$, the one opposite to $v$, and consider the family of planes parallel to it. The higher planes won't touch the sphere, wile planes closer to the vertex will slice a circle on the sphere.
By continuity there is one of such plane $P$, which touches the sphere in one point (and having the sphere below $P$). It determines a tetrahedron where the unit sphere is inscribed. Since $P$ is parallel to the original face of $\Delta$, then the obtained tetrahedron is similar to $\Delta$. So by scaling you obtain a sphere inscribed in $\Delta$.


答:
Suppose a polynomial satisfies $p(x)=p(x-1), \forall x \in \mathbb{R}$.

Consider $q(x)= p(x)-p(0)$, then $0$ is a root of $q$, and every integer is a root of $q$.

Hence $q$ is the $0$ function and we conclude that $p$ is a constant function.

In general, we have shown that a periodic polynomial is a constant function.

Title: Finding the linear span of a polynomial with a condition

Body:
I'm trying to find the linear span of the following set:

$M = \{p(x)∈ \mathbb R_4 [x] : p(x)=p(x-1),  x∈\mathbb R\} $

I've tried taking a vector from $M$ and applying the condition on it. Like that:

Let:
$p(x) ∈ M$

$p(x) = p(x-1)$

$a + bx + cx^2 + dx^3 = a + b(x-1) + c(x-1)^2 + d(x-1)^3 $

$b(1) + c(2x-1) + d(3x^2 -3x +1) = 0$

$(b-c+d) + (2c-3d)x + (3d)x^2 = 0$

Now, if I open an equation system based on the variables of the polynomial, I get that b, c and d are equal to 0.
What does that mean in terms of the linear span?
Am I trying to find it wrong?