问:
How to prove to be an irrational number? Like $\sqrt{2}$, $\sqrt{3}$, or $\sum\limits_{k=1}^{\infty} \frac{1}{n^2}=\pi^2/6$
As we know $\sqrt{2},\sqrt{3}$ are irrational numbers. And I see some proofs on the net.

So I doubt that how $e,\pi$ or already known irrational numbers are proved to be irrational.

In fact, I got interested in Riemann zeta function
$$\zeta(s)=\sum_{n=0}^{\infty} \frac{1}{n^s},$$

we know $\zeta(2)=\pi^2/6$ from Euler, 1737.

One mathematician (sorry to forgot his name) proved $\zeta(3)$ to be also irrational 40 years before.

Can somebody explain how he could do with that? To understand Apéry's theorem, is it very hard?

An question raises that could one real number make up of two different irrationals (for example: $e,\pi$), $e\pi$, or others can be rational?

答:
Some combinations of irrationals can be proved to be irrational in a more or less systematic way.  For example, $(\sqrt{2})^\pi$ is irrational (and in fact transcendental) by Gelfond's theorem (see http://en.wikipedia.org/wiki/Gelfond%E2%80%93Schneider_theorem).

One approach to proving the irrationality of a number is to show that it is approximated "too well" by rational numbers $p/q$ where "too well" is assigned a specific meaning in terms of the denominator $q$.  From this point of view, $e$ is irrational because the terms in the familiar power series defining it: $e=\sum 1/n!$ tend to zero rapidly, so that the partial sums give extremely good approximations of $e$.

To get more details, see Spivak's book "Calculus" where he proves irrationality of $\pi$ on page 307, and that $e$ is transcendental on page 409.


问:
Is this an error in the book or am I missing something? Proof that an imaginary number z is equal to it's conjugate (no mention of b = 0).
Ok, so here's what I have in the textbook:

Prove: For any complex number $z\in C$, $\overline{z} = z$.

Suppose that $z = a + bi$ where $a,b \in R$.

The proof:

$\overline{z} = a+bi = \overline{\overline{a-bi}} = a-(-b)i = a+bi = z$

Isn't the solution already assuming that the conjugate of z is equal to z by stating that $\overline{z}= a +bi$ when they defined $z = a+bi$ ?

I know this is true when $b = 0$, but there's no reference to $b=0$ in the book, I wrote all there is.

Edit: The book is Schaum's outlines Linear Algebra, 5th edition.

Edit 2: I checked in the 6th edition, to see if it was a typo that had been fixed. It is exactly the same in the 6th edition as in the 5th edition.

Edit 3: I put the parts of the book in quotes. They are not direct quotes as there are 2 other problems in the book, but the 2 others are not relevant to this one. If you want the original thing here is a picture of the whole thing (I highlighted the parts that I wrote in the question):

答:
There's a typo, the item (iii) is really wrong. It may be $\overline{\overline{z}}=z$. So, the proof is good:

$$\overline{\overline{z}}=\overline{a-bi}=a-(-b)i=a+bi=z$$

Good studies!

ContentLicense: CC BY-SA 4.0


问:
Will all the models of this FOL theory represent only bipartite graphs ? And viceversa?
Will all the models of the following formula be bipartite graphs, with edges being represented by $R$ and $A$ and $\neg A$ being two disjoint groups of vertices ?

$$\forall x \forall y R(x,y) \rightarrow (\neg A(x) \leftrightarrow A(y))$$

Is this encoding right ?

答:
All models will be bipartite graphs. However, "are all bipartite graphs models?" is complicated to answer.

You've demonstrated that bipartitions are expressible in first-order logic. That is, given a graph (specified by a relation $R$) and a predicate $A$, we get a partition of the vertex set into two parts: those for which $A$ holds, and those for which $A$ does not hold. Then we can express the statement "This partition is a bipartition of the graph" in first-order logic, which is essentially what you've done.

This is different from the statement "The graph is bipartite" or "The graph has some bipartition." To build from the axioms in the question to such a statement, we'd need to begin with "There exists a predicate $A$ such that ..." which is not allowed in first-order logic.

We can get closer. A graph is bipartite if and only if it does not have any odd cycles. For any fixed $n$, we can express the statement "The graph does not have any cycles of length $2n+1$" in first-order logic. So with infinitely many axioms, we can express "A graph is bipartite" just by combining these statements for $n=1,2,3,\dots$.

With finitely many axioms we can't get there. I don't have the technical know-how to explain why, but the intuitive idea is that any finite set of axioms only works on some "local" scale. For large enough $n$, $C_{2n+1}$ will be indistinguishable from a bipartite graph on that scale, because you have to go all the way around the very long odd cycle to notice that the graph is not bipartite.


问:
Confirming a easy proof: the product of two consecutive numbers is always even.
Can someone confirm if my prove is right?

Theorem. The product of two consecutive integers is always even.

Proof. Define a number $n$ such that $n:=2k$ where $k\in\mathbb{Z}$, this ensures that $n$ is an even number. Define a second number $p$ such that $p:=n+1=2k+1$ this ensures that $p$ is an odd number and the numbers $n$ and $p$ are consecutive because they differ by $1$. There product is given by:

$$np=2k(2k+1)=2(2k^2+k)=4k^2+2k\tag1$$

Which is clearly a multiple of $2$.

Define a number $h$ such that $h:=2m+1$ where $m\in\mathbb{Z}$, this ensures that $h$ is an odd number. Define a second number $z$ such that $z:=n-1=2m$ this ensures that $z$ is an even number and the numbers $h$ and $z$ are consecutive because they differ by $1$. There product is given by:

$$hz=2m(2m+1)=2(2m^2+m)=4m^2+2m\tag2$$

Which is clearly a multiple of $2$, this proves that $hz$ is an even number.

This proves this theorem.

If I am wrong how can I make my proof valid?

答:
Let $a$ and $b$ be two consecutive integers, with $a<b$. If $a$ is even, then $b$ is odd. Hence,
$$
ab=(2n) \cdot (2n+1)=4n^2+2n=2(n^2+n) \, ,
$$
where $n \in \mathbb{Z}$ is the unique integer such that $2n=a$ and $2n+1=b$. If $a$ is odd, then $b$ is even. Can you finish the proof?


问:
Does $y = f(x) = ax+b$ actually have two mappings inside it?
I’m just a high school student, so I may be somewhat logically flawed in understanding this.

According to wikipedia, the definition of function requires an input $x$ with its domain $X$ and an output $y$ with its domain $Y$, and the function $f$ maps $x$ to $y$.

But how about $f(x)$? I often see syntaxes such as $f(1) = 0$  in my text book. Doesn’t that mean it is $f(x)$ being first assigned a value and then transfer the value into $y$? So, there must be two transitions/mappings between the input $x$ and the output $y$ right?

My conceptual model of function is like this: A definition of function requires an input $x$ with its domain $X$, a forwarder $f(x)$ with its domain $F$ and an output $y$ with its domain $Y$. The function $f$ first maps $x$ to $f(x)$ then maps $f(x)$ to $y$.

These two definitions are not quite the same.

On 2022.6.29: The picture below had solved my confusion.

答:
I'm the OP, and I now kind of know what was causing me confusion back then.

I was actually thinking of function in a more computer science way. When we use the notation $y = f(x)$ in programming languages, it actually functions in the same way as I described in the question above. It first calculates $f(x)$ with the input $x$ and then stores the result into a temporary variable, and it then assigns the value of the temporary variable to the variable $y$. It does have two transitions in the process.

Yes, of course one can argue that function in mathematics doesn't need to have the same definition as function in programming languages. I agree, even though it is really weird to have two different definitions describing the same concept. But it is actually worthwhile to talk about which of them is more appropriate and more intuitive.

If we redefine function in mathematics in a more computer science way. We can define a function as $f: I → O$ , where $f$ maps the input $x$ in the domain $I$ to the output $f(x)$ in the codomain $O$. Defining function in this way makes a much more elegant expression, which would get rid of the redundant usage of $y$ which just means the same as $f(x)$. Confusion in choosing which notation to use happens all the time, not only in choosing between $f(x) = ax+b$ or $y = f(x) = ax+b$ but also $dy/dx$ or $df(x)/dx$.

Moreover, if we look back to the definition of function in programming languages, this kind of redundancy just doesn’t exist at all, and it never caused any problem. The variable $y$ is not pre-defined in the function itself, and is actually only needed when we want to “store” the final result, not to “use” the result for further calculation. The “real” output of the function $f$ is $f(x)$, not $y$.

The variable y can still be used with function in mathematics, but now it should no longer be seen as part of the definition of function. The expression $y = f(x) = ax+b$ thus should be understood alternatively as simultaneous equations of $y=f(x)$ and $f(x) = ax+b$.

To conclude, I think the original definition is obviously outdated and flawed compared to the more modern one used in computer science. It deserves a refurbished definition as it has been a quite fundamental and frequently used concept in mathematics, hopefully urgently.


问:
Transverse submanifolds
Take $M$ to be a manifold and $N_{1},N_{2} \subset M$ to be two submanifolds such that $dimN_{1}+dimN_{2}=dimM$. Then $N_{1}$ and $N_{2}$ are transverse if and only if their intersection is infinite discrete (i.e. contains either finitely or countably infinite many elements) and $TN_{1} \bigoplus TN_{2} \cong TM $ at any point $p \in N_{1} \cap N_{2}$.

Recall that any two $N_{1},N_{2}$ are transverse submanifolds of $M$ if $T_{x}N_{1} + T_{x}N_{2}=T_{x}M$ for all $x \in N_{1} \cap N_{2}$.

Suppose that $M=\mathbb{R}^{3}$ and take $N_{1}$ and $N_{2}$ to be such that $dimN_{1}+dimN_{2}=3$ but they are not transverse.

More precisely I would like to see an example where $TN_{1} \bigoplus TN_{2} \cong TM$ fails.

Is there a concrete example of such situation, maybe $S^{2}$ and the circle $S^{1}$ touching at one unique point?

Thanks in advance!

答:
One way to do this is to start in $\mathbb{R}^{2}$ and then move one dimension up.

In $\mathbb{R}^{2}$, consider the line $\ell = \{(x,y) \, \mid \, y = 0\}$. Here are two curves $\gamma$ that intersect it at a single point, but not transversely: $\gamma_{1} = \{(x,x^{2}) \, \mid \, x \in \mathbb{R}\}$ and $\gamma_{2} = \{(x,x^{3}) \, \mid \, x \in \mathbb{R}\}$.

Now move to $\mathbb{R}^{3}$ in the following way.  Let $P$ be the plane $P = \{(x,y,z) \, \mid \, y = 0\} = \pi^{-1}(\ell)$.  (Here $\pi : \mathbb{R}^{3} \to \mathbb{R}^{2}$ is the projection $\pi(x,y,z) = (x,y)$.)  Next, let $\tilde{\gamma}$ be the curve $\tilde{\gamma} = \{(x,y,0) \, \mid (x,y) \in \gamma\}$.

We have $\text{dim} \, P + \text{dim} \, \tilde{\gamma} = 3$.  However, they do not intersect transversely.

Title: Transverse submanifolds


问:
$f$ is integrable in $\mathbb{R}^n$ iff $r \mapsto r^{n-1} g(r)$ is integrable in $\mathbb{R}_{\geq 0}$
Let $g: \mathbb{R}_{\geq 0} \to \mathbb{R}$ be a function and let $f: \mathbb{R}^n \to \mathbb{R}$ be $f(x)=g(|x|)$.

Show that $f$ is integrable in $\mathbb{R}^n$ iff $r \mapsto r^{n-1} g(r)$ is integrable in $\mathbb{R}_{\geq 0}$.

Can someone help me with this?

答:
There is nothing to prove actually if $n=1.$ So we assume $n\ge 2.$ First of all note that every $0\neq x\in\mathbb{R}^n$ can be uniquely written as $x=r\omega$ where $r>0$ and $\omega\in S^{n-1}.$ The rest is essentially a change of variable. Note that
$$\int_{\mathbb{R}^n} f(x)dx=\int g(|x|)dx=\int_{S^n} \int_{0}^{\infty}g(r)r^{n-1}drd\omega.$$

Now note that $\int_{S^n}d\omega=C_n$ is just the surface measure of sphere which is a constant (depending upon the dimension). The Fubini’s theorem therefore gives you the desired the conclusion.


答:
Style and sufficiency of proof generally depends on where you take the "floor of certainty" to be - what facts and statements can be taken as true and what needs to be demonstrated.

With a simple theorem like this you might say that we should set the floor a little lower, that is provide an exposition of even simple steps in the proof. For example, the fact that alternating integers are odd and even might need to be supported - or not.

Commenting on your proof, then, in the knowledge that this is a little subjective:

Breaking into the cases of lower number even and lower number odd looks good, and you could do this more clearly.

You don't need to define a separate number for $n{+}1$. It doesn't add clarity and could easily confuse the matter.

You can continue to use $n$ and $n{+}1$ as the successive numbers in the two cases - again limit your new defined numbers, for example to where they will interact with each other; there's no such interaction between your cases.

So using your method I would write:

Theorem. The product of two consecutive integers is always even.

Proof. Consider successive integers $n$ and $n{+}1$ with product $n(n{+}1)$ in two cases, (1) where $n$ is even and (2) where $n$ is odd.

Case 1, $n$ even:

Find $k\in\mathbb{Z}$ such that $n=2k$. Now the desired product is given by:

$$n(n+1)=2k(2k+1)=2(k(2k+1))\tag1$$

And since $k(2k+1)\in\mathbb{Z}$ this is thus a multiple of $2$ and even as required.

Case 2, $n$ odd:

Find $k\in\mathbb{Z}$ such that $n=2k+1$. Now the desired product is given by:

$$n(n+1)=(2k+1)(2k+2)=2(2k+1)(k+1)\tag 2$$

And since $(2k+1)(k+1)\in\mathbb{Z}$ the product is again a multiple of $2$ and even as required.

Since both cases demonstrate an even product, this proves this theorem.

Title: Confirming a easy proof: the product of two consecutive numbers is always even.

Body:
Can someone confirm if my prove is right?

Theorem. The product of two consecutive integers is always even.

Proof. Define a number $n$ such that $n:=2k$ where $k\in\mathbb{Z}$, this ensures that $n$ is an even number. Define a second number $p$ such that $p:=n+1=2k+1$ this ensures that $p$ is an odd number and the numbers $n$ and $p$ are consecutive because they differ by $1$. There product is given by:

$$np=2k(2k+1)=2(2k^2+k)=4k^2+2k\tag1$$

Which is clearly a multiple of $2$.

Define a number $h$ such that $h:=2m+1$ where $m\in\mathbb{Z}$, this ensures that $h$ is an odd number. Define a second number $z$ such that $z:=n-1=2m$ this ensures that $z$ is an even number and the numbers $h$ and $z$ are consecutive because they differ by $1$. There product is given by:

$$hz=2m(2m+1)=2(2m^2+m)=4m^2+2m\tag2$$

Which is clearly a multiple of $2$, this proves that $hz$ is an even number.

This proves this theorem.

If I am wrong how can I make my proof valid?


问:
Truncated exponential map from $\mathfrak{gl}_n$ to $GL_n$
Let $k$ be a field of characteristic $p>0$.  If $A$ is a nilpotent matrix in $\mathfrak{gl}_n(k)$, with $p>n$, then we can define the unipotent matrix:

$$\exp(A)=\sum_{i=0}^{p-1}\frac{A^i}{i!}$$

If $f_p(x,y)=\frac{1}{p}((x+y)^p-x^p-y^p)\in k[x,y]$, and if $AB=BA$, we have the formula

$$\exp(A)\exp(B)=\exp(A+B-f_p(A,B))\tag{$*$}$$

Notice the expansion of $f_p$ is well-defined in $k[x,y]$.  I would like to be able to prove the following:

If $A_1,\ldots,A_k$ are linearly independent, pairwise commuting, nilpotent matrices, then $\prod_{i=1}^k\exp(A_i)^{m_i}=I_n$ if and only if $m_i=0$ for $i=1,2,\ldots,k$.

One step I've made is to see that $f_p(x,x)=0$ implies that $\exp(A)^m=\exp(mA)$.  As I try to proceed further, the $f_p(x,y)$ terms get messy.

答:
Yes, this is true, assuming that by $m_i=0$ you mean $m_i \equiv 0 \mod p$.

If the $A_i$ commute, then $\prod \exp(A_i)^{m_i} = \exp \left( \sum m_i A_i \right)$. Since the $A_i$ are commuting and nilpotent, the sum $\sum m_i A_i$ is nilpotent. The exponential map is injective on nilpotent matrices (with inverse $\log$) so $\exp \left(  m_i A_i \right)=I_n$ implies that $\sum m_i A_i=0$. Since the $A_i$ are linear independent, this means that all the $m_i$ are $0$.

In the above, I am using the following Lemma: If $p > n$, and $A$ and $B$ are commuting $n \times n$ nilpotent matrices, then $\exp(A+B) = \exp(A) \exp(B)$. The key to this is to recall that, if $A$ and $B$ are commuting nilpotent matrices, then we can choose a basis in which they are both strictly upper triangular. Using this, if $A$ and $B$ are commuting nilpotent matrices, then $A^i B^j=0$ for any $(i,j)$ with $i+j \geq n$. Thus
$$\exp(A+B) = \sum_{m < n} \frac{(A+B)^m}{m!} = \sum_{i+j < n} \frac{A^i B^j}{i! j!} = \sum_{i < n} \sum_{j < n} \frac{A^i}{i!} \frac{B^j}{j!} = \exp(A) \exp(B).$$

It's worth pointing out that this is not true if you only assume that $A_i^p=0$ and not $p>n$. For example, take $p=2$ and
$$A = \begin{pmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \end{pmatrix} \quad B = \begin{pmatrix} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix} \quad C = \begin{pmatrix} 0 & 1 & 1 & 1 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \end{pmatrix}.$$

If I haven't made any errors, then $A^2=B^2=C^2=0$, $A$, $B$ and $C$ commute and $A$, $B$ and $C$ are linearly independent, but $\exp(A) \exp(B) \exp(C) = (1+A)(1+B)(1+C) = 1$. I can explain how I found this if there is a need.

ContentLicense: CC BY-SA 3.0


问:
Inequality for definite integral.
In a proof in my class we used the following inequality: $$\left| \int_0^{\varepsilon}te^{ut}\,du\right|\le |\varepsilon| |t|(e^{\varepsilon_0t}+e^{-\varepsilon_0t})\quad \forall |\varepsilon| <\varepsilon_0.$$However, I'm not quite sure how we got that inequality so I need some help.

答:
Since the exponential function is strictly increasing then for any $|\epsilon|<\epsilon_0$ we have that
$$e^{ut}\leq e^{|ut|}\leq e^{|\epsilon| |t|}\leq e^{\epsilon_0 |t|} = \max\{e^{\epsilon_0t},e^{-\epsilon_0t}\}\leq e^{\epsilon_0 t}+e^{-\epsilon_0t} $$
Hence
\begin{align*}\Big|\int_0^\epsilon te^{tu} du\Big| &\leq sign(\epsilon)\int_0^\epsilon |t|e^{ut}du\leq sign(\epsilon)\int_0^\epsilon |t|(e^{\epsilon_0 t}+e^{-\epsilon_0t})du\\
&= [|t|(e^{\epsilon_0 t}+e^{-\epsilon_0t})] sign(\epsilon)\int_0^\epsilon du\\
&= |t|(e^{\epsilon_0 t}+e^{-\epsilon_0t}) sign(\epsilon)\epsilon = |t||\epsilon|(e^{\epsilon_0 t}+e^{-\epsilon_0t})
\end{align*}

Title: Inequality for definite integral.


问:
Finding Pythagorean triple where the squares of sides are weighted
Suppose $a, b, c \in \mathbb{Z}_{>0}$. The following is a variation of a Pythagorean triple, but with weighted squares:

$$a^2 + 3 b^2 = 4 c^2$$

Can I find something like Euclid's formula that can generate all solutions for this formula? I'm not sure how to approach this problem.

答:
Given Euclid's formula $ \quad A=m^2-k^2,\quad B=2mk,\quad C=m^2+k^2\quad$ we can adapt it to yours where

$$(2mk)^2=3B^2\implies 2mk=\sqrt{3}B\implies B=\frac{2mk}{\sqrt{3}}$$

$$(m^2+k^2)^2=4C^2\implies (m^2+k^2)=2C\implies C=\frac{m^2+k^2}{2}$$

None of the  $B$-values or $C$-values will be integers for primitive triples and $B$ will not even be rational unless $A=B=C=1$. That triple cannot be generated by the formula using integers $(m,k)$.


问:
Calculate $\lim _{x\to 0^+}\frac{\left(e^x-e^{-x}\right)}{\sin x}$
***How to find the this limit without using L'Hopital's Rule.

$\lim _{x\to 0^+}\frac{\left(e^x-e^{-x}\right)}{\sin x}$

答: Two famous limits give it as$$\lim_{x\to0^+}\frac{x}{\sin x}\left(\frac{e^x-1}{x}+\frac{e^{-x}-1}{-x}\right)=1(1+1)=2.$$


问:
Covering image of a connected CW-complex need not be a CW-complex.

Problem: Let $X$ be a connected CW-complex, and $Y$ be a connected topological space. Suppose $p: X\to Y$ is a covering map. Does there
exist a CW-structure on $Y$? More generally, is $Y$ homotopically equivalent to a CW-complex?

Note that the other direction is clear: Any covering of a connected CW-complex can always be given a CW-structure, lifting the characteristics map of cells of base space such that the covering map is a cellular map.

I believe that the answer to the above problem is no, but I have no counterexample.

$\bullet$ Notice that $Y$ is locally path-connected as the covering map is a local homeomorphism, hence $Y$ is path-connected also. So, we can not consider spaces $\{0\}\cup\left\{\frac{1}{n}:n\in\Bbb N\right\}$ or Topologist Sine Curve as $Y$. Notice that both $\{0\}\cup\left\{\frac{1}{n}:n\in\Bbb N\right\}$ or Topologist Sine curve are not homotopically equivalent to a CW-complex.

$\bullet$ Similarly, we can not consider the Hawaiian Earring(this is not semi-locally simply connected) as $Y$: The connected CW-complex $X$ has the universal cover so that $X$ is semi-locally simply connected, but the property "semi-locally simply connected" is preserved under a local homeomorphism.

So, I am run out of examples. Any help will be appreciated. Thanks in advance.

答:
Andrew Ranicki conjectured (or at least asked) if a compact, 4 manifold has a CW structure, if and only if, it is smoothable.

As far as I know this is open, but assuming the conjecture is true the answer to your question is that a CW complex may cover a non CW complex. Let $M$ be a nonsmoothable 4-manifold with infinite fundamental group, these are known to exist. Let $\widetilde{M}$ denote its universal cover. $\widetilde{M}$ is noncompact, and it is known that noncompact 4-manifolds are triangulated. Hence, $\widetilde{M}$ has a CW structure and it covers $M$ which is not a CW complex, if Ranicki's conjecture is true.

In this case, the base of this cover is homotopy equivalent to a CW complex since all compact manifolds are homotopy equivalent to CW complexes.


问:
Good book on topological group theory?
I'm looking for a good introduction to the theory of locally compact groups and their representations. It may assume the reader to be familiar with basic group theory, topology and measure theory. </p>

答:
There is a classical Lev Pontrjagin’s book “Continuous groups” or “Topological groups” (original is in Russian, but there exists an English translation too). Also I often encountered references to
“Abstract Harmonic Analysis” by E.Hewitt and K.Ross it this context, but I never saw this book. :-) </p>

Title: Good book on topological group theory?


问:
Defining the heavy-tail index

Consider the following two results, https://arxiv.org/pdf/0907.2261.pdf (Theorem 1.8) and https://arxiv.org/pdf/2006.04740.pdf (Theorem 1). In the former a proof of a random vector $\vec{w} \in \mathbb{R}^n$ to be heavy-tailed with index $\alpha$ is obtained by showing that  $\lim_{t\to\infty} t^{\alpha} \Pr(\Vert \vec{w} \Vert  > t) > 0$. In the later it is obtained by showing that $\exists \vec{u} \in S^{n-1}$ and a positive continuous function $f_\alpha$ on $S^{n-1}$ s.t we have $\lim_{t\to\infty} t^{\alpha} \Pr(\langle \vec{u}, \vec{w} \rangle  > t) = f_\alpha (\vec{u})$

Are these two definitions of the heavy tail index equivalent?

In works like https://link.springer.com/article/10.1007/s00184-014-0515-7 we see the following definition of ``strictly $\alpha-$stably" distributed random vector $\vec{w}$ i.e for any $m$ independent copies of it say $\vec{w}_1,\ldots,\vec{w}_m$ we must have,
$$\sum_{i=1}^m \vec{w}_i \overset{\text{in distribution}}{=} m^{\frac {1}{\alpha}} \vec{w}$$

This a priori looks like another new definition of the heavy tail index $\alpha$.

Is this equivalent to any of the above 2 definitions and if yes, how?

If the ch.f. of $X$ is $\phi_X(t) = \exp(-\sigma^{\alpha}\vert t\vert^{\alpha})$ for some $\sigma >0$ then it is said to be a ``Symmetric $\alpha-$Stable $(\sigma)$". Its a separate question as to how this is a special case of any of the above $3$ definitions - assuming that they are equivalent!

答:
Note that for a unit vector $u$, one has that $\langle u, w \rangle$ is the projection of $w$ in the direction of $u$. Of course this projection is maximized when $u$ is the unit vector in the direction of $w$, in which case $\langle u, w \rangle = |w|$ exactly when $u$ is parallel to $w$. Thus, for any $u$, if one has $\langle u, w \rangle > t$ then
$ t < \langle u, w\rangle \leq \langle w/|w|, w\rangle = |w|$ so $|w| > t$ as well.

Thus $t^\alpha P(\langle u, w\rangle > t) \leq t^\alpha P(|w| > t)$ and taking the liminf we find $0 < f_\alpha(u) \leq \liminf t^\alpha P(|w| > t)$ for all $u$, showing $w$ has a tail at least as heavy as $\alpha$.

On the other hand, if $0 < \lim t^\alpha P(|w| > t)$ this doesn't tell us about the existence of $f_\alpha(u)>0$. Indeed, if $w$ is concentrated on a line, then $P(\langle u, w\rangle > t) = 0$ when $u$ is perpendicular to this line, so the first condition does not imply the second.

Title: Defining the heavy-tail index


问:
Restrictions on the number of edges in a graph
I have been wondering, can you prove that if we suppose in a graph (connected or not) there are no cycles of length 3 (so at least 4) and each vertex has a degree 4 or less, then the number of edges will be less than $\frac{3}{2}*n - 2$ where n is the number of vertices.

I tried using Euler's Formula, namely $v - e + f = k + 1$. If we then take into consideration the restriction on cycle length we get $2f < m$ so we can substitute and get $m < 2n - 4$, however that is not enough and also is not using the fact that each vertex has a degree 4 or less.

I'm sorry, I forgot to tell the graph has to be planar.

答:
A non-planar counterexample to your hypothesis. the Folkman graph is quartic (4-regular, all vertices have degree 4) and has girth (minimum cycle length) of 4. This has 20 vertices and 40 edges.

Adding in the planar requirement, an infinite square grid would obviously not meet your formula, so a large square grid with the corners looped around to each other should also fail... let's try $6\times 6$, which has $36$ vertices and $64$ edges breaks your proposed limit of $\frac{3}{2}*n - 2$:

... and I could sneak in four more edges to get to $2n{-}4$ in fact, by joining the nodes adjacent to the corners also along each edge.

It's pretty clear that this construction allows a $2n{-}4$ edged triangle-free graph on $n$ nodes for a grid that is $2p\times 2q$ with $p,q>1$, nearly $4$-regular except for $8$ degree-$3$ nodes. The smallest is thus on $16$ vertices with $28$ edges:


问:
Coordinate transformation formula for a pinhole camera model
This is a pinhole camera model

(I don't get, is there [R t], or (R, t)). This formula is used to model the projection from a space point M to an image point m. Projection drawing:

Tilde over vector, means that "1" is added to that vector as the element.
M is the coordinate of the point in the 3d space, and m is the coordinate of the point in the picture, f is the focal length of the camera, and s is the pixel aspect ratio. (R, t) describes the 3D transformation between the world coordinate system, in which the rectangle is described, and the camera coordinate system.

It is unclear to me, what does "t" letter and (R, t) after A mean, and how, by inserting the 3D coordinates (pixel aspect ratio = 1) of the rectangle corners to the formula we get this:
$$
\begin{aligned}
\lambda_1\tilde{\mathbf{m}}_1&=\mathbf{A}\mathbf{t} \\
\lambda_2\tilde{\mathbf{m}}_2&=w\mathbf{A}\mathbf{r}_1+\mathbf{A}\mathbf{t} \\
\lambda_3\tilde{\mathbf{m}}_3&=h\mathbf{A}\mathbf{r}_2+\mathbf{A}\mathbf{t} \\
\lambda_4\tilde{\mathbf{m}}_4&=w\mathbf{A}\mathbf{r}_1+h\mathbf{A}\mathbf{r}_2+\mathbf{A}\mathbf{t}
\end{aligned}
$$

I found this formula in this document (page 13).

答: [R,t] is a 3x4 matrix obtained by juxtaposing the 3x3 matrix R representing the orientation of the camera and the 3x1 vector t representing its position. To be more precise, the rows or R are the axes of the camera (Y up, X left, Z toward the scene) in the world reference frame, and the centre  of the camera is $-R^t t$ in world reference frame.</p>


问:
Matrix times vector vs vector times vector
I can memorize this operation/logic easily. (3x3 * 3x1 = 3x1) And I understand it's a Matrix times vector.

But I can't operate a column vector * a column vector.

a * b =14 + 25 + 3*6

It's logic is different from Matrix. I thought vector is a matrix. I knew I misunderstand somewhere. Can anyone help?

答:
The dot product in matrix notation would be $\mathbf{a}^T\mathbf{b}$, assuming $\mathbf{a}$ and $\mathbf{b}$ are column vectors (i.e., matrices with a single column). If they are row vectors (matrices with a single row) then it would be $\mathbf{a}\mathbf{b}^T$, with $^T$ denoting transposition.

So that could be one way for you to remember it. It's just the case that matrix notation is not always used for describing the dot product of two vectors.


问:
ranking a matrix in an equation
I've been given a equation where I need to find $C$ (matrix) where I know what $A$ and $B$ are (also matrices), the equation is :

$$ 13I= -0.2*A*C+B $$

I want to canonize $B$ with elementary row operations to make it easier - can I do that? if I can does it affect the value of $B$? if so, how can I negate that?

答:
It would help to know the sizes of $A,B$, and $C$. Assuming they are all square matrices, then you can rewrite your equation as $-5(13I - B) = AC$. Then you can solve this system simultaneously by considering the augmented matrix
$$[A \vert 5B - 65I].$$
Then you need to do Gaussian elimination (row operations) to put this into reduced row echelon form. If this system only has one solution, then the reduced row echelon form will be of the following form
$$[I \vert C].$$
Otherwise, there will be some free parameters or there is no solution.

Title: ranking a matrix in an equation


问:
Adding decreasing power of the same number
Simpler ways of writing for example $a^{n}$ + $a^{n-1}$ + $a^ {n-2}$ + ... + a=

what is this even called ?

答:
This is a (finite) geometric series, which can be given by the formula
$$
\sum_{k=0}^{n}a^k= \frac{a(1-a^n)}{1-a} \, .
$$


问:
How do I integrate $\ln(x)/(x-1)$
How do I integrate $\ln(x)/(x-1)$?

\begin{align}
f(x)&=\int\frac{\ln(x)}{x-1} \,dx\\
&=\ln(x)\ln|x-1|-\int\frac{\ln|x-1|}{x}\, dx\\
&=\ln(x)\ln|x-1|-\left(\ln|x-1|\ln(x)-\int\frac{\ln(x)}{x-1}\, dx\right)\\
&=\ln(x)\ln|x-1|-(\ln|x-1|\ln(x)-f(x))\\
2f(x)&=0\\
f(x)&=0
\end{align}
I end up getting this but I know its incorrect. How do I do solve this

答:
As was mentioned in the comments $$ \int \frac{\ln(x)}{x-1}\,dx=-\operatorname{Li}_{2}(1-x)+C$$ where $\operatorname{Li}_{2}(x)$ is the polylogarithm function.  There isn't a way of writing this in terms of elementary functions.  Your error was in the line
\begin{align}
f(x) &= \ln(x)\ln|x-1|-(\ln|x-1|\ln(x)-f(x))\\
\implies 2f(x) &= 0.
\end{align}
You dropped a minus sign, so you should have
\begin{align}
f(x) &= \ln(x)\ln|x-1|-(\ln|x-1|\ln(x)-f(x))\\
\implies f(x) &= \ln(x)\ln|x-1|-\ln|x-1|\ln(x)+f(x)\\
\implies 0 &= 0
\end{align}
i.e. you haven't actually simplified the original expression.

Title: How do I integrate $\ln(x)/(x-1)$


问:
If $R=\mathbb{Z}[\sqrt{p}]$ with $p$ an odd prime and $I=(2,1+\sqrt{p})$, then $I^2=(2)$
I'm trying to show that if $R=\mathbb{Z}[\sqrt{p}]$ with $p$ an odd prime such that $p \equiv 3 \hspace{1mm} (\operatorname{mod} 4)$ and $I$ is the ideal generated by $2$ and $1+\sqrt(p)$ then $I^2$ is the ideal generated by $2$

I know that
$$I^2=(4,2+2\sqrt{p},1+p+2\sqrt{p})$$
and that $4,2+2\sqrt{p},1+p+2\sqrt{p} \in (2)$, so I just need to prove that $2 \in I^2$ but I cannot prove it.

Any ideas?

Body: If $ p \equiv 3 \pmod{4} $, then $ (4, 2 + 2 \sqrt{p}, 1 + p + 2 \sqrt{p}) = (4, 2 + 2 \sqrt{p}, p - 1) $ and since $ \gcd(p-1, 4) = 2 $ this is equal to $ (2, 2 + 2 \sqrt{p}) = (2) $.


问:
$\lim_{x\to \pi/2} \;\frac 1{\sec x+ \tan x}$
how to solve it answer is $0$, but $\frac 1{\infty + \infty}$ is indeterminate form

$$\lim_{x \to \pi/2} \frac 1{\sec x + \tan x}$$

答:
$$\lim_{x \to \frac\pi2} \frac 1{\sec x + \tan x}$$

$$=\lim_{y \to 0} \frac 1{\sec \left(\frac\pi2-y\right) + \tan \left(\frac\pi2-y\right)}$$ (Putting $\frac\pi2-x=y,$ as $x \to \frac\pi2,y\to0$)

$$=\lim_{y \to 0} \frac 1{\csc y + \cot y}$$

$$=\lim_{y \to 0} \frac {\sin y}{1 + \cos y}$$

$$=\lim_{y \to 0} \frac {2\sin \frac y2\cos\frac y2}{2\cos^2\frac y2}$$

$$=\lim_{y \to 0} \tan  \frac y2$$ as $\cos\frac y2\ne0$

$$=0$$


问:
Why principal ideal ring has Krull dimension at most 1?
We know PID which is not has a Krull dimension at most one.
But can we say the same thing with PIR（which is not necessarily integral domain）？

We cannot say in PIR that every nonzero prime ideal is maximal, so I wonder we can say the statement.

答:
A PIR is a Noetherian ring, so Krull's height theorem holds.

Let $M$ be a maximal ideal in the ring: since it is principal, Krull's theorem says it is of height at most $1$. Therefore the whole ring can only be Krull dimension at most $1$.


问:
If Jacobi iterations converges to $Ax=b$ then $A$ is invertible.

Prove or Disprove:

If Jacobi iterations converges to $Ax=b$ then $A$ is invertible.

I think this is true but not sure how to connect between $A$ and $B_J$

答: With $B_J:=I-D^{-1}A$, assume there is an $x\neq 0$ such that $Ax=0$. It means that $B_Jx=x$. Hence $\rho(B_J)\geq 1$ and the Jacobi method is not convergent.


答:
That's not matrix multiplication but the inner product of vectors in some inner product space.

In general, you can only multiply two matrices when they have compatible sizes. In other words, you can only do multiplication to get $MN$ when the number of columns of $M$ is the same as that of rows of $N$.

Title: Matrix times vector vs vector times vector

Body:
I can memorize this operation/logic easily. (3x3 * 3x1 = 3x1) And I understand it's a Matrix times vector.

But I can't operate a column vector * a column vector.

a * b =14 + 25 + 3*6

It's logic is different from Matrix. I thought vector is a matrix. I knew I misunderstand somewhere. Can anyone help?


问:
The graph of $x^x$
I have a question about the graph of $f(x) = x^x$. How come the graph doesn't extend into the negative domain? Because, it is not as if the graph is undefined when $x=-5$. But according to the graph, that seems to be the case. Can someone please explain this?

Thanks

答: Well, for a START, what would happen if your exponents were to be -0.5, -1.5, -2.5, -3.5 etc? Convert that into radical notation and try to evaluate for those negative x-values. (And there is lots more going on for negative x-values)</p>

Title: The graph of $x^x$


问:
What is the simplest and most intuitive way of computing the mixed strategy of a matrix game?
I have learned how to compute the mixed Nash equilibrium (NE) of a game many times, however, whenever I am faced with a matrix, I just cannot recall the exact way/logic of going about computing it.

I think the main problem is that the formulation process is just too long. I need to either pose it as a linear program (but who can remember how to formulate the objectives) or use a graphical method such as the best response curve. But I just keep on forgetting them because I lack the intuition.

For example, I am given the matrix,

I can visually check that (Fink, Fink) or $(2,2)$ is the pure NE. But I cannot remember how exactly to compute the mixed NE.

I have some rules of thumbs, which are unreliable, for example: by inspection I can see that the players probably want to earn a higher payoff than $1$ sometimes. So the players probably use a mixed strategy $(2/3, 1/3)$ of sorts to switch between Quiet and Fink. As I said, totally unreliable.

What is the simplest and most intuitive method that you use to compute the NE for matrix games (and does it generalize to larger matrices)?

答:
This game has no mixed-strategy NE. The only NE is (Fink, Fink) as playing fink is a strictly dominant strategy.

In mixed-strategy NEs players are indifferent between the strategies they play with positive probability. This gives a simple algebraic solution.

In your game let $q$ be the probability of playing Quiet. Note that the game is symmetric so it suffices to look at the indifference condition for one player.

In a mixed strategy NE player 1 is indifferent if $2q=3q+1(1-q)$. Rearranging would give the mixed-strategy NE if there was one. In this case we get $0=1$ and there is no mixed-strategy-NE.

You can also derive mixed-strategy NEs as the intersection of best-response functions.


问:
What are the math processes to get from $1-a$ to $1 - \frac{.1a}{1-.9a}$?
What are the math processes to get the following equation?
$$\begin{align}
&\;1 - a \\[4pt]
=&\; 1 -.9a-.1a \\[4pt]
=&\; 1 - \frac{.1a}{1-.9a}
\end{align}$$
I understand the second line, splitting the $a$ into $.9a$ and $.1a$. I just have no idea how they managed to get the final form.

答:
$$1-0.9a-0.1a$$
$$=\frac{1-0.9a-0.1a}{1-0.9a}\times(1-0.9a)$$
$$=\biggl(\frac{1-0.9a}{1-0.9a}-\frac{0.1a}{1-0.9a}\biggr)\times(1-0.9a)$$
$$=\biggl(1-\frac{0.1a}{1-0.9a}\biggr)\times(1-0.9a)$$

But the final form in your post lacks the term $(1-0.9a)$ multiplied to it. So, your final form is either wrong
$$(OR)$$

Otherwise, it is true for certain values of $a$. So we need to solve for $a$ in your equation.
$$1-a=1-\frac{0.1a}{1-0.9a}$$
$$-a=-\frac{0.1a}{1-0.9a}$$
$$0=a-\frac{0.1a}{1-0.9a}$$
$$\Rightarrow a-0.9a^2-0.1a=0$$
$$\Rightarrow 0.9a-0.9a^2=0$$
$$\Rightarrow 0.9a(1-a)=0$$
$$\Rightarrow a=1,a=0$$

So, the final form holds good if $a=0$ or $a=1$. It is wrong For other values of $a$.


问:
Parametric function

$$\begin{cases}
x(t)=\sin ^2 t\\
y(t)=\cot ^2 t\\
\end{cases}
$$

I need to explore this function, but I can't find enough information in my language. I don't know the steps of exploring the parametric function and building the graphic. I'd like to learn how to explore such a function in parametric form.

答:
$$x=\sin ^2 t,y=\cot ^2 t$$
Eliminate $t$
$$y=\frac{\cos^2 t}{\sin^2 t}=\frac{1-\sin^2 t}{\sin^2 t}\to y=\frac{1-x}{x}$$
the rectangular (cartesian) form of the curve is
$$y=\frac{1}{x}-1$$

Hope this is useful

Title: Parametric function


问:
Double summation: how to solve?
I have this summation that I need to solve:

$$\sum_{i=2}^{n}\sum_{j=i}^{n} 3i$$

Can someone please help? I have no idea how to start this

答:
Just write it out explicitly to see what happens:

\begin{align}
\sum_{i=2}^{n}\sum_{j=i}^{n} 3i
&=\sum_{i=2}^{n} \underbrace{(3i+3i+\cdots +3i)}_{n-i+1 \textrm{ terms}}\\
&=\sum_{i=2}^n (n-i+1)(3i)\\
&=\sum_{i=2}^n 3n \cdot i - 3i^2+3i\\
&=\bigg((3n+3)\sum_{i=2}^ni\bigg)-3\bigg(\sum_{i=2}^ni^2\bigg)
\end{align}

Now work on the following:
$$
\sum_{i=2}^n i,\quad \sum_{i=2}^ni^2
$$

Title: Double summation: how to solve?


问:
How to calculate $\lim\limits_{x\to1^+}\frac{1}{(x-1)^2} \int\limits_{1}^{x} \sqrt{1+\cos(\pi t)}\,\mathrm dt$
Can anyone help me by calculating this limit?

I know that I need L'Hôpital but how?

$$
\lim_{x \to 1^+}\frac{1}{(x-1)^2} \int_{1}^{x} \sqrt{1+\cos(\pi t)} \,\mathrm dt
$$

Thank you very much!!

答:
Since my hints were not helpful, here a is my solution
$$
\begin{align}
\lim\limits_{x\to 1^+}\frac{1}{(x-1)^2}\int\limits_{1}^{x}\sqrt{1+\cos(\pi t)}dt
&=\lim\limits_{x\to 1^+}\frac{\left(\int\limits_{1}^{x}\sqrt{1+\cos(\pi t)}dt\right)'}{((x-1)^2)'}&\text{L'Hopitale rule}\\
&=\lim\limits_{x\to 1^+}\frac{\sqrt{1+\cos(\pi x)}}{2(x-1)}& y\to x-1\\
&=\lim\limits_{y\to 0^+}\frac{\sqrt{1+\cos(\pi (y+1))}}{2y}&\cos(\pi+\alpha)=-\cos\alpha\\
&=\lim\limits_{y\to 0^+}\frac{\sqrt{1-\cos(\pi y)}}{2y}&1-\cos\alpha=2\sin^2\frac{\alpha}{2}\\
&=\lim\limits_{y\to 0^+}\frac{\sqrt{2}|\sin\frac{\pi y}{2}|}{2y}&y>0\\
&=\lim\limits_{y\to 0^+}\frac{\sqrt{2}\sin\frac{\pi y}{2}}{2y}&\sin\alpha\sim\alpha\\
&=\lim\limits_{y\to 0^+}\frac{\sqrt{2}\frac{\pi y}{2}}{2y}\\
&=\frac{\pi\sqrt{2}}{4}
\end{align}
$$</p>


问:
Prove a mod b = b mod a iff a=b
I'm just starting to learn the three types of proofs and I came across this question.

```<code>                 a mod b = b mod a iff a=b
</code>```

I tried looking for solution to prove this but couldn't find it. Most examples I have are of the same divisor like: a mod n = b mod n. But i couldn't find anything on this.

I assumed it would be by using a contradiction proof but what got me confused is the if and only if condition.

Thanks for any help you send this way!!

答:
Since this is an if and only if proof, we have to do it in two directions. As is often the case in if and only if proofs, one direction is much easier to do than the other. We will do the easy direction first.

Proof.

⇐ We assume a=b and prove a mod b = b mod a. But this is clearly true, since x mod x= 0 for all x, and a and b are equal.

⇒ Here we have to prove that if a mod b = b mod a, then a=b.  Try to find a direct proof, and then give up and do the contrapositive:  assume a ≠ b, and prove that a mod b ≠ b mod a. There are two possible cases: a < b and b < a (we’ve already assumed a≠b).

Case 1: a  b Then b < a, and the proof is the same as above with a replaced by b and b replaced by a.

□

Source

Title: Prove a mod b = b mod a iff a=b

Body:
Since this is an if and only if proof, we have to do it in two directions. As is often the case in if and only if proofs, one direction is much easier to do than the other. We will do the easy direction first.

Proof.

⇐ We assume a=b and prove a mod b = b mod a. But this is clearly true, since x mod x= 0 for all x, and a and b are equal.

⇒ Here we have to prove that if a mod b = b mod a, then a=b.  Try to find a direct proof, and then give up and do the contrapositive:  assume a ≠ b, and prove that a mod b ≠ b mod a. There are two possible cases: a < b and b < a (we’ve already assumed a≠b).

Case 1: a < b Then a mod b=a.  But b mod a, as we noted, is strictly less than a, so it cannot be equal to a and hence can’t be equal to a mod b.

Case 2: a > b Then b < a, and the proof is the same as above with a replaced by b and b replaced by a.

□

Source


问:
Sum and product problem
How can I find the result of:

$\sum\limits_{i=1}^{n}\prod\limits_{j=1}^{2} ij$

I know that $\prod\limits_{j=1}^{2} ij = 2i^2$, so I should simply do the summation as $\sum\limits_{i=1}^n2i^2$?

Title: Sum and product problem

Body:
This should be easy:

$$ \sum_{i=1}^n 2i^2 = 2 \sum_{i=1}^n i^2 = 2 \frac{n(n+1)(2n+1)}{6} = \frac{n(n+1)(2n+1)}{3}$$


问:
Does the inclusion map satisfy Unique (CP) Extension Property?
Consider the $C^\star$ algebra $C(\mathbb{T}^2)$ of complex-valued continuous functions on the 2-torus with the sup norm. Let $S$ be the operator system that is the (norm) closure of the linear span of $\{z^m w^n| m,n \in \mathbb{Z}, mn \geq 0\}$. Let $i:S\to C(\mathbb{T}^2)$ denote the (canonical) inclusion map of $S$ in $C(\mathbb{T}^2)$. $i$ is completely positive (CP), and by Arveson's Extension Theorem, it must have CP extension(s) $\tilde{i}: C(\mathbb{T}^2) \to C(\mathbb{T}^2)$, such that $\tilde{i}|_S=i$. One such CP extension is, $\tilde{i}=\mathbb{I}_{C(\mathbb{T}^2)}$, which is, extending as the identity map on $C(\mathbb{T}^2)$. My question is, is it the only CP extension? Or, in other words, does the inclusion map have the unique extension property (CP) in this case? If not, what would be another example of a CP extension?

Perhaps an interesting point to note: the trivial extension doesn't work, as aptly demonstrated in the answer to a
previous question here.

答:
Yes,  the unique CP extension applies to the present situation.  It is in fact a consequence of the following more
general fact:

Theorem. Let $A$ be a unital C$^*$-algebra and let $\mathscr U$ be a set of unitaries in $A$,
generating $A$ as a C$^*$-algebra,  and  such that $1\in \mathscr U$.
Given a unital representation $\pi :A\to B(H)$, suppose that $\varphi :A\to B(H)$ is a completely positive map such that
$$
  \varphi (u)=\pi (u), \quad\forall u\in \mathscr U.
  \tag 1
  $$
Then $\varphi =\pi $.

Proof.  By Stinespring Theorem,  there exists a Hilbert space $\tilde H$, a  unital representation  $\tilde\pi:A\to B(\tilde H)$,
and a bounded operator $V:H\to \tilde H$, such that
$$
  \varphi (a)=V^*\tilde \pi (a)V, \quad\forall a\in A.
  \tag 2
  $$
Observing that both $\varphi $ and $\pi $ are unital maps, we deduce that $V^*V=1$, which is to say that $V$ is an isometry.

Notice that, for each $u$ in $\mathscr U$ we have
$$
  \pi (u) = \varphi (u) = V^*\tilde \pi (u)V,
  $$
so $\tilde \pi (u)$ commutes with $VV^*$ by Lemma (1), below.

Since the commutant of $VV^*$ is a closed $^*$-subalgebra, and since $A$ is generated as a C$^*$-algebra by $\mathscr U$, it
follows that $\tilde \pi (a)$ commutes with $VV^*$ for every $a$ in $A$.

From (2) we then deduce  that $\varphi $ is a $^*$-homomorphism, and then (1) implies that $\varphi =\pi $.   $ \tag*{$\square$}$

Lemma 1.  If $V:H\to\tilde H$ is an isometry, and $U$ is a unitary operator on $\tilde H$ such that  $V^*UV$
is unitary, then $U$ commutes with $VV^*$.

Proof.  Identifying  $H$ with  a subspace of $\tilde H$ via $V$,  we may write $\tilde H = H \oplus  H^\perp$, and hence $U$ may be
represented as a matrix
$$
  U = \pmatrix {A & B \cr C & D},
  $$
where $A = V^* UV$, hence $A$ is unitary.  It follows that
$$
  \pmatrix {1 & 0 \cr 0 & 1} =   \pmatrix {A & B \cr C & D}   \pmatrix {A^* & C^* \cr B^* & D^*} = \pmatrix {AA^* + BB^*
&\cdots  \cr \cdots & \cdots},
  $$
so $B=0$, and a similar argument proves that $C=0$.  This implies that $U$ commutes with
$$
  \pmatrix {1 & 0 \cr 0 & 0} = VV^*.
  \tag*{$\square$}
  $$


问:
SVD decomposition of orthonormal matrix Issue
If i have a matrix $E \in \Re^{nxm}$ with orthonormal columns (e.g $E^TE = I_m$) then how to demonstrate (through Singular Value decomposition on $E$) that :

$I_n-EE^T = U(I_n -\Sigma\Sigma^T)U^T$ ?

PS : through SVD i establish the relation : $E^TE = VV^T$

Thanks for the help

答:
If you have the SVD: $E=U\Sigma V$ then

\begin{align}
I_n-EE^T
&=I_n-U\Sigma V(U\Sigma V)^T\\
&=I_n-U\Sigma (VV^T)\Sigma^TU^T\\
&=UI_nU^T-U\Sigma (VV^T)\Sigma^TU^T\\
&= U(I_n -\Sigma\Sigma^T)U^T
\end{align}


问:
Finding maximum value of periodic function.
In my question, It is asked to find the maximum value of the equation of motion such as u_t=(11/210)sin(10t)-(1/21)sin(t11). I have found that when t=2pi*n/21 it reached its maximum value. But it is obvious from its graph and calculations that the value of function increases from n=1 to n=10 and equals to 0.0997 when n=11. After that, it decreases. My question is there any way to calculate it more easily or it is what it is?

答:
The graph of a function has slope 0 at it's maximum point.

Slope of a function can be determined by dy/dx that is its derivative.

So, by equating it's derivative to 0 we get,

dy/dx = {(11/210) * 10 * cos(10t)} - {(1/21) * 11 * cos(11t)} = 0

= (11/21) * [cos(10t) - cos(11t)] = 0     (since, cos(c)-cos(d) we know)

0 = 2 * sin(21t/2) * sin(t/2)

so, we get two cases

sin(21t/2) = 0 or sin(t/2) = 0

Since we know, sin^-1(0) = n*pi

therefore t = 2 * n * pi /21  or t = 2 * n * pi.

and the maximum value by putting the above t value of the function will be 0.099717665

Check this graph solution for the problem


问:
If $A=B+C$, $BC=CB$ and $AD=DA$, does $BD=DB$?
If $A$, $B$, $C$ and $D$ are $n\times n$ matrices s.t. $A=B+C$, $B$ and $C$ commute and $A$ and $D$ commute, do $B$ and $D$ have to commute? Saw this used in another proof and it sounded like this was trivial but I tried proving it and I couldn't. Thanks.

答:
Just repeating Jack's comment but it's too big for a comment:

Start with $A = B + C$ and multiply through by $D$ on the left to get $$DA = DB + DC \:.$$ Now repeat but multiply through by $D$ on the right to get $$AD = BD + CD \:.$$ Subtract one result from the other: $$\left(DA - AD\right) = \left(DB - BD\right) + \left(DC - CD\right)$$

The left side is given as zero, and the right side is only zero if we simultaneously have: $$DB = BD \hspace{2.54cm} DC = CD$$

... what has not been explicitly used yet is $BC=CB$.


问:
find surface area using double integral of: $\sqrt{(x^2+y^2)}+z=1$ and plane $x+2z+1=0$
So I have to find surface area using double integral of: $\sqrt{(x^2+y^2)}+z=1$ above the plane $x+2z+1=0$. I drew the picture and I see that an ellipse is the intersection between the cone and plane.

Now, $z(x,y)=1-\sqrt{(x^2+y^2)}$ and from $z$ I will find partial derivatives by $x$ and $y$ and put them in a standard formula. Now what is giving me trouble is when I have to project into the x-y plane. I can't seem to understand over which area should I integrate.. Is it the ellipse?

------------------------EDIT: based on Math Lover comment------------------------

So I expressed $z$'s from cone and plane equation and set them equal in order to get the ellipse equation. At the end I got $\frac{(\sqrt{3}x-9)^2}{18}+\frac{y^2}{\frac{9}{2}}=1$. Then I performed the change of variables $x=\frac{9+\sqrt{18}u}{\sqrt{3}}$, $y=\frac{\sqrt{9}}{\sqrt{2}}v$ in order to get unit circle in $u-v $ coordinate system. Also I have found the Jacobian $\frac{\partial(x,y)}{\partial(u,v)}$. Finally at the end I used polar coordinates and that's it. Is this the right way to do so?

答:
Something went wrong in one of your steps. I get a different equation for ellipse.

Parametrization of the cone, $ \ c(t) = (r \cos t, r \sin t, 1 - r)$.

So you can easily find $|c'_{t} \times c'_{r}| = r \sqrt2$

So surface area of the cone $S = \sqrt2 \iint_D r \ dr \ dt$

Where $D$ is the projection of the intersection of the cone and the plane in $XY$ plane. So our next step is to find the area of the intersection.

Intersection of cone and the plane is given by,

$1 + \frac{1+x}{2} = \sqrt{x^2 + y^2} \implies 3x^2 - 6x + 4y^2 = 9$

So equation of ellipse is $ \ \frac{(x-1)^2}{4} + \frac{y^2}{3} = 1$

We know the area of the ellipse with semi-major and semi-minor axes lengths of $a$ and $b$ is $\pi ab = 2 \sqrt3 \ \pi$

But if you need to solve this by integral then,

Using substitution, $x = 1 + 2 \ r \cos t, y = \sqrt3 \ r \sin t$, our new domain is a circle of unit radius.

$|J| = 2 \sqrt3 \ r$

So the area of ellipse $\displaystyle \iint_D dx \ dy = \int_0^{2\pi}\int_0^1 2 \sqrt3 \ r \ dr \ dt = 2 \pi \sqrt3$

So surface area of the cone $S = 2 \pi \sqrt6$


答:
Integrate the equivalent, valid when $t\gt1$, $t\to1$,
$$
\sqrt{1+\cos(\pi t)}=\sqrt2\,\sin\left(\frac\pi2(t-1)\right)\sim\frac\pi{\sqrt2}\,(t-1).
$$
This yields, for $x\to1$, $x\gt1$,
$$
\int_1^x\sqrt{1+\cos(\pi t)}\mathrm dt\sim\frac\pi{\sqrt2}\int_1^x(t-1)\mathrm dt=\frac\pi{2\sqrt2}(x-1)^2.
$$</p>

Title: How to calculate $\lim\limits_{x\to1^+}\frac{1}{(x-1)^2} \int\limits_{1}^{x} \sqrt{1+\cos(\pi t)}\,\mathrm dt$

Body:
Can anyone help me by calculating this limit?

I know that I need L'Hôpital but how?

$$
\lim_{x \to 1^+}\frac{1}{(x-1)^2} \int_{1}^{x} \sqrt{1+\cos(\pi t)} \,\mathrm dt
$$

Thank you very much!!


问:
Is changing limits in this fashion okay? What is the justification for it?
I am currently working through chapter $5$ on limits of Spivaks calculus book. I have come to exercise $15$ix. I feel like I can answer this question correctly, but there is one small step I know to be true but can't seem to show it concisely.

I would like to note, I am aware similar questions have been asked, but I am querying a specific part of this question.

The question:

Assume that: $$\lim_{x\to{0}}{}\frac{\sin(x)}{x}=\alpha.$$

Find, in terms of $\alpha$, the following limit:

$$\lim_{x\to1}\frac{\sin(x^2-1)}{x-1}.$$

My attempt:

Firstly, multiplying bottom and top by $x+1$ yields:

$$\lim_{x\to1}\frac{\sin(x^2-1)(x+1)}{(x-1)(x+1)}=\lim_{x\to1}\frac{\sin(x^2-1)(x+1)}{x^2-1}$$

Now, IF $(*):=\lim_{x\to1}\frac{\sin(x^2-1)}{x^2-1}$ exists, we can split the above into:

$$\lim_{x\to1}\frac{\sin(x^2-1)}{x^2-1}\times{}\lim_{x\to1}{x+1},$$

using algebra of limits. Which gives us:

$$\lim_{x\to1}\frac{\sin(x^2-1)}{x^2-1}\times{}2.$$

So I'm essentially left to show $(*)$ exists and is equal to $\alpha$. Which means the original limit is equivalent to $2\alpha$.

Now clearly as $x\to{1}$ we have that $(x^2-1)\to{0},$ so can we write this limit as:

$$\lim_{x\to0}\frac{\sin(x)}{x}\text{ or }\lim_{x^2-1\to0}\frac{\sin(x^2-1)}{x^2-1}?$$

If so, why exactly? It just feels a bit imprecise.

I guess what I'm asking exactly is:

$$\lim_{x\to0}\frac{\sin(x)}{x}\overset{?}{=}\lim_{x\to1}\frac{\sin(x^2-1)}{x^2-1}$$

答:
There are several exercises in this chapter that require steps that aren't justified fully in the published solutions.

Problem 5-1.(vi) uses $\lim_{h \to 0}\sqrt{a+h} = \sqrt{a}$, without any justification.

5-2.(ii) and (iii) involve $\lim_{x \to 0}\sqrt{1-x^2}$.

5-14 and 5-15 use $\lim_{x \to 0} \cos{x} = 1$ and $\lim_{x \to 0} \sin{x} = 0$ without justification (This comes from the continuity of $\sin$ and $\cos$, which is assumed in the next chapter but not proven until these functions are defined formally in chapter 15 [3rd ed.]).

5-15(ix), the exercise asked about here, uses the fact that if $\lim_{x \to 0} f(x) = \ell$, then $\lim_{x \to a} f(x^2 - a^2) = \ell$.

This fact is not yet justified and is worth proving to yourself, using an $\varepsilon$-$\delta$ argument.

IMO it's also worth proving that for $a > 0$, $\lim_{x\to a} \sqrt{x} = \sqrt{a}$ (or prove this statement for $a + h$ with $h \to 0$, if you prefer.)

The trig ones, don't worry about for now :)

Spoiler Alert: Below are proofs for the two postulates mentioned above.

Postulate 1 If $\lim_{x \to 0} f(x) = \ell$, show that $\lim_{x \to a} f(x^2 - a^2) = \ell$.

For the latter limit to be true we need to show that for any $\varepsilon > 0$, there exists some $\delta > 0$ such that for all $x$, if
$$0 < |x - a| < \delta \text{ then }|f(x^2 - a^2) - \ell| < \varepsilon$$

From the first limit, we have for $\varepsilon$, there exists a $\delta_1 > 0 $ such that for all $x$ if

$$0 < |x| < \delta_1 \text{ then } |f(x) - \ell| < \varepsilon$$

Therefore, if $0 < |x^2 - a^2| < \delta_1$, then $|f(x^2 - a^2) - \ell| < \varepsilon$.

So we want to find a $\delta$ such that if $0 < |x - a| < \delta$ then $0 < |x^2 - a^2| < \delta_1$.

Taking the requirement
$$0 < |x^2 - a^2| < \delta_1$$

and looking just at the left-hand side, we have

$$0 < |x^2 - a^2|$$

which is met by requiring $x \neq \pm a$.

We can guarantee this with the requirement

$$0 < |x - a| < 2|a|$$

Returning to our requirement $0 < |x^2 - a^2| < \delta_1$, the right-hand side has

$$|x^2 - a^2| < \delta_1$$
$$|x - a||x + a| < \delta_1$$

If you recall Spivak's proof that $\lim_{x \to a} x^2 = a^2$, you can see that the above requirement is similar, only with $\delta_1$ substituted for $\varepsilon$. Using the tricks employed in that proof, we find that

$|x - a| < \delta_2 = \min(1,\frac{\delta_1}{1+2|a|})$, will guarantee $|x^2 - a^2| < \delta_1$

Combining all our $\delta$-requirements, we have for all $x$ if
$$0 < |x - a| < \delta \text{ then } |f(x^2 - a^2) - \ell| < \varepsilon$$

with $\delta = \min(1, 2|a|, \frac{\delta_1}{1+2|a|})$, where $\delta_1$ is given by the existence of the first limit $\lim_{x \to 0} f(x)$.

Postulate 2 Show that for $a > 0$, $\lim_{x\to a} \sqrt{x} = \sqrt{a}$.

Suppose the statement is true. Then for any $\varepsilon > 0$ there exists some $\delta > 0$ such that for all $x$ if

$$0 < |x-a| < \delta \text{ then } |\sqrt{x} - \sqrt{a}| < \varepsilon$$

Let's take the $\varepsilon$ requirement and work backwards to find a suitable $\delta$:

$$|\sqrt{x} - \sqrt{a}| < \varepsilon$$
Multiplying both sides by $|\sqrt{x} + \sqrt{a}|$ (which equals $\sqrt{x} + \sqrt{a}$ since both terms are nonnegative) we have
$$(\sqrt{x} + \sqrt{a}) \cdot |\sqrt{x} - \sqrt{a}| < (\sqrt{x} + \sqrt{a}) \cdot \varepsilon$$

$$|x - a| < (\sqrt{x} + \sqrt{a}) \cdot \varepsilon$$

If we require $|x - a| < \sqrt{a} \varepsilon$, this should suffice, since $\sqrt{x} + \sqrt{a} \geq \sqrt{a}$. We also want to prevent $x$ from being negative, which we can do by requiring $|x - a| < a$.

Combining these requirements we have $\delta = \min(a,\sqrt{a}\varepsilon)$

Checking that this $\delta$ works, if $\varepsilon > 0$ then for all $x$ with

$$0 < |x-a| < \sqrt{a}\varepsilon$$
$$|\sqrt{x} - \sqrt{a}| \cdot (\sqrt{x} + \sqrt{a}) < \sqrt{a}\varepsilon$$
$$|\sqrt{x} - \sqrt{a}| < \frac{\sqrt{a}\varepsilon}{(\sqrt{x} + \sqrt{a})} \leq \varepsilon$$

You can combine these 2 postulates along with the other properties of limits to fill in some of the gaps in the chapter exercises.

Edit
I suspect that exercise 5-15(ix) and the others mentioned were overlooked by Spivak. The published solutions involve steps that are completely without justification.

When I first did this chapter I was pretty discouraged by these lapses. I feared I might have to skip these things and just hope they didn't create flaws in my understanding. (I wasn't on MSE at the time, or I would have come here for answers!)

However, I was eventually able to bridge the gaps myself by using $\varepsilon$-$\delta$ arguments, which are at the heart of the chapter. This came as a huge relief.

Upon seeing kam's question, my instinct was to reassure them "no, it's not just you" and to steer them towards what I did.

In so doing, I failed to account for a few things. First, IIRC back when I first did the problems in this chapter I probably skipped the parts I wasn't sure of, or maybe even just accepted them and moved on. It wasn't until finishing the other exercises that I came back and attacked $\varepsilon$/$\delta$-Style. Sorry, kam. I should have encouraged you to come back to these later after finishing the rest of the problems.

Second, Paramanand Singh is right. Though it's maybe not something Spivak intended (I reckon these extra problematic problems were an oversight), this is a good reason to think more generally about limit substitutions, and that's maybe closer to what kam was asking about anyway.

Furthermore, general ideas on such substitutions aren't really in the book aside from a few exercises (6-12, maybe others?) but they are worth being aware of. (Perhaps they show up after Chapter 12, where I am? Seems unlikely.) Update: the proof of an important theorem in chapter 12 (Inverse Functions) uses one of these substitutions, again without quite giving a complete explanation.

Limit Substitutions without continuity, basic gist
Suppose we have two functions $f$ and $g$ both with limits

$$\lim_{x \to a} g(x) = b$$
$$\lim_{x \to b} f(x) = \ell$$

Suppose also that there exists an interval around (possibly excluding) $a$ such that $g$ in this interval is never equal to $b$. In other words, there exists some $\delta_a > 0$ such that for all $x$ if
$$0 < |x -  a| < \delta_a \text{ then } g(x) \neq b$$

Then the limit of composite function  $\lim_{x \to a} f(g(x)) = \ell$.
Intuitively, you can see that this should be true. As $x$ approaches $a$, $g(x)$ gets close to $b$, so $f(g(x))$ approaches $\ell$, its limit at $b$.

The condition that $g$ never actually reach $b$ within some interval around $a$ is to ensure $g$ properly "mimics" the behavior of $x$ approaching $b$. That is, we know only the behavior of $f(x)$ for $x$ near but not equal to $b$. Without this restriction, we don't know if $f(g(x))$ approaches $\ell$ or not. It might but we can't be certain.

Consider the functions $f$ and $g$ with
$$f(x) = 1 \text{  (for $x \neq 0$)}$$
$$f(x) = 50 \text{  (for $x = 0$)}$$
$$g(x) = 0$$

Here, $\lim_{x \to 0} f(x) = 1$, $\lim_{x \to a} g(x) = 0$ (for any $a$), but $\lim_{x \to a} f(g(x)) = 50$. A silly example but it shows the point.

Proof:
By definition, for any $\varepsilon > 0$ there exists some $\delta_f > 0$ such that for all $y$ if

$$0 < |y -  b| < \delta_f \text{ then } |f(y) - \ell| < \varepsilon$$

For $\delta_f$ there is some $\delta_g$ such that for all $x$ if

$$0< |x-a| < \delta_g \text{ then } |g(x) - b| < \delta_f$$

Therefore there is a $\delta = \min(\delta_a, \delta_g)$ such that for all x if

$$0< |x-a| < \delta \text{ then } 0<|g(x) - b| < \delta_f$$

Thus, for any $\varepsilon > 0$ there exists some $\delta$ such that for all x if

$$0< |x-a| < \delta \text{ then } |f(g(x)) - \ell| < \varepsilon$$

or,
$$\lim_{x \to a} f(g(x)) = \ell = \lim_{x \to b} f(x) = \ell$$

The consequence of this is that we can sometimes swap out more complicated limits and replace them with simpler ones.

Problem 5-15(ix) presents us with $\lim_{x \to 1} \frac{\sin(x^2 - 1)}{x^2-1}$ which is $\lim_{x \to 1}f(g(x))$ with $f(x) = \frac{\sin x}{x}$ and $g(x) = x^2 - 1$.

We know
$$\lim_{x \to 1} x^2 - 1 = 0$$
$$\lim_{x \to 0} \frac{\sin x}{x} = \alpha$$

Furthermore, you can see that as $x$ approaches $1$, $ x^2 - 1$ gets close to but never equal to $0$ (Where does it equal $0$?). In this way it "mimics" the behavior of $x\to 0$, so we are ok replacing $g(x)$ in the limit.

$$\lim_{x \to 1} \frac{\sin(x^2 - 1)}{x^2-1} = \lim_{x \to 0} \frac{\sin x}{x} = \alpha$$

Finally, note that there are other criteria for making similar substitutions with limits. One thing to look at is the "continuity" of the outer function $f$. This is examined in the next chapter of the book.


问:
Linear functional on LCS is continuous iff it is bounded by finite linear combination of topology-defining semi-norms
Consider a linear functional f on a TVS, whose topology is generated by a family of semi-norms $\mathcal{P}$, such that the topology is Hausdorff. In functional analysis by Conway, it is said that f is continuous if and only if $$|f(x)|\leq\sum_{k=1}^{n}\alpha_np_n(x)$$
For all x. My question is, why can this sum only include a finite number of terms from $\mathcal{P}$? I know that f is continuous if and only if the map $x\rightarrow|f(x)|$ is a continuous semi-norm, and I know that semi-norms are continuous iff they are bounded by some continuous semi-norm. I can't seem to find the missing piece though.

答:
Since $f$ is continuous, there exists an open neighborhood $V$ of $0$, such that
$$
  x\in V \Rightarrow |f(x)|<1.
  $$
By definition of the topology defined by semi-norms, there exists a finite set $\{p_1,p_2,\ldots,p_n\}\subseteq \mathcal P$, and $\epsilon>0$, such that
$$
  (\forall i)\ p_i(x)<\epsilon \Rightarrow x\in V.
  $$
Can you now finish the argument?


问:
Is this particle always equilibrium vertically that is $R\sin(\alpha)=mg$ where $R$ is reaction on particle
A thin smooth straight tube $OA$ is constrained to rotate with constant angular velocity $ω$ about a fixed vertical axis through $O$, and a particle is free to move in the tube. The angle between $OA$ and the upward vertical is a fixed acute angle $α$ and describe a fixed horizontal circle of radius $a$,
While the particle is in a state of steady motion the angular velocity of the tube is suddenly reduced to $\dfrac{\omega}{2}$ and is then maintained constant at the new value. Find
the time the particle takes to reach O,

My  problem is

Is this particle always equilibrium vertically that is $Rsin(\alpha)=mg$ where $R$ is reaction on particle

Edit:

$\underline{Attempt}$

I got $r$ is distance from vertical axis through $O$ which is perpendicular to the this axis then I used velocity and accelerataion  in polar coordinates

So $f=ma$ along the $r$

$$-R\cos(\alpha)=m(\ddot{r}- \frac{r{\omega^2}}{4})$$

and $f=ma$ perpendicular to the $r$

$$R\sin({\alpha})=mg$$

so then $-g\cot(\alpha)=\ddot{r}- \dfrac{r{\omega^2}}{4}$

Is that correct?

答:
Analyzing the movement along the rod and defining $\delta(t)$ the distance from $O$, in the absence of friction, the dynamics are given by

$$
m\ddot\delta(t) = \frac 14m\omega^2(\delta(t)\sin\alpha)\sin\alpha-m g\cos\alpha
$$

or

$$
\ddot\delta(t) = \frac 14\omega^2\delta(t)\sin^2\alpha-g\cos\alpha
$$

Solving this ODE for $\delta(0) = \frac{g\cos\alpha}{\omega^2\sin^2\alpha},\ \ \dot\delta(0)=0$. Note that the rotation will be reduced from $\omega$ to $\omega/2$ so we have

$$
\delta(t) = \frac{g \cot (\alpha ) \csc (\alpha ) \left(4-3 \cosh \left(\frac{1}{2} \omega  t \sin (\alpha )\right)\right)}{\omega ^2}
$$

solving now $\delta(t^*)=0$ we obtain

$$
t^*=\frac{2\cosh ^{-1}\left(\frac{4}{3}\right) \csc (\alpha )}{\omega }
$$


问:
Using the sinc integral to estimate $\pi$
We know that
$$
\frac{\pi}{2} = \int_0^\infty \frac{\sin(x)}{x}dx
$$
By separating our integral and using a substitution we obtain:
\begin{align}
\int_0^\infty \frac{\sin(x)}{x}dx &= \int_0^1 \frac{\sin(x)}{x}dx + \int_1^\infty \frac{\sin(x)}{x}dx \\
&= \int_0^1 \frac{\sin(x)}{x}dx \, + \int_0^1 \frac{\sin(\frac{1}{x})}{x}dx
\end{align}
Using taylor series:
$$
\frac{\pi}{2} = \int_0^1 \left[\sum_{n=0}^\infty\frac{(-1)^nx^{2n}}{(2n+1)!} + \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!x^{2n+2}} \right]
$$
Evaluating:
$$
\frac{\pi}{2} = \sum_{n=0}^\infty \frac{(-1)^n-(-1)^n}{(2n+1)(2n+1)!}
$$
Meaning that
$$
\frac{\pi}{2} = 0\,?
$$
Where did I go wrong. Thank you!

答:
The reason why it is wrong is that Taylor series for $\sin\left(\frac{1}{x}\right)$ is not uniformly convergent on $(0, 1)$.

Analogy:

You could say that

$$1 + 2 + 3 + 4 + ... = -\frac{1}{12}$$

if $1+2+3+4+...$ is convergent, but it is not, so the equality does not hold.


问:
Let a, b and c be positive real numbers such that abc=1. Let d = {a^(b+c)} * {b^(c+a)} * {c^(a+b)}. Find the maximum possible value of a.
Let a, b and c be positive real numbers such that abc=1. Let d = {a^(b+c)} * {b^(c+a)} * {c^(a+b)}.
Find the maximum possible value of d.

I found that the answer is 1 by using some hacks/fakesolve but I am unable to prove it.

答:
Using the simplification given by @lulu, we have to minimise $aln(a)+bln(b)+cln(c)$ subject to $abc=1$.

Using the method of Lagrange multipliers we have $$ln(a)+1=\lambda /a,ln(b)+1=\lambda /b,ln(c)+1=\lambda /c.$$

Then $$aln(a)+a=bln(b)+b=cln(c)+c$$

Since $xln(x)+x$ is an increasing function we have $a=b=c$. These variables and $d$ must therefore be $1$.


问:
Vector calculus for ellipse in polar coordinates
I'm having trouble with this question, can somebody please help me with it! I'll thanks/like your comment if help me =)

I know that for a ellipse the parametric is $x=a\sin t$ , $b= b \cos t$, $t\in (0, 2\pi)$ (?)

For part a) I drew up the graph but not sure if it's right. The circle has radius 1 and for the ellipse I'm able to find $x= 2 $.
$$
(x-1)^2 =9  \\
\sqrt{(x-1)^2}=\sqrt 9\\
x-1=3\\
x=4$$

For $y=2\sqrt 2$
$$
y^2=8\\
\sqrt y=\sqrt 8\\
y=\sqrt 4 \sqrt 2\\
y= 2\sqrt 2 \text{ or } 2.8 $$

So the region should be the circle? since the ellipse covers the whole circle?
Thank you very much for helping!
Cheers.

答:
It's not enough to compute the four apexes of the ellipse, since there might be some intersection in between. Argue instead as follows:

When $x^2+y^2\leq 1$ then $|x-1|\leq 2$ and therefore
$${(x-1)^2\over 9}+{y^2\over8}\leq {4\over9}+{1\over8}<1\ .$$

This shows that the unit circle is completely contained in the interior of the given ellipse.


问:
Example of increasing function such that $f(x+y)=f(x)+f(y)$
Is there a strictly increasing continuous function $f:[0,m]\rightarrow [0,2000]$ such that $f(0)=0$, $f(m)=2000$ (where $m>2000$) and $f(x+y)=f(x)+f(y)$? If so, can someone give me an example?

What I'm trying to do is to apply a function to a bar chart that has values like 300, 40000, 10000, 30000, 12345 etc in order to 'normalize' the chart (make the chart look more friendly, without seeing a very big bar, then a tiny one). I applied the function $f(x)=\frac{x}{300/40000*x + 1}$, but the chart started to look weird because $f(40000)  \ne f(10000)+f(30000)$. This is why I'm searching for a function that satisfies $f(x+y)=f(x)+f(y)$.

答:
When @J.W.Tanner gave an example, he basically gave the prototype for all such functions.  $$f(x+y)=f(x)+f(y)$$ is linear, and with your other conditions $$f(x)=ax, \quad \text{with }0<a<1.$$

UPDATE

Why is the function necessarily linear?  We reasoned in a comment above that $f(kx)=kf(x)$ and $f(x/k)=f(x)/k$ when $k$ is an integer.  Combining these, we have $f(px/q) = pf(x)/q,$ so it holds for rationals.  A continuity argument gives $f(a x)=af(x)$ for $a \in \text{Reals},$ which implies $f(x) = ax.$


答: The function $f(x) = \frac{2000x}{m}$ will do, as SenJen noted in the comments.

Title: Example of increasing function such that $f(x+y)=f(x)+f(y)$

Body:
Is there a strictly increasing continuous function $f:[0,m]\rightarrow [0,2000]$ such that $f(0)=0$, $f(m)=2000$ (where $m>2000$) and $f(x+y)=f(x)+f(y)$? If so, can someone give me an example?

What I'm trying to do is to apply a function to a bar chart that has values like 300, 40000, 10000, 30000, 12345 etc in order to 'normalize' the chart (make the chart look more friendly, without seeing a very big bar, then a tiny one). I applied the function $f(x)=\frac{x}{300/40000*x + 1}$, but the chart started to look weird because $f(40000)  \ne f(10000)+f(30000)$. This is why I'm searching for a function that satisfies $f(x+y)=f(x)+f(y)$.


问:
Example of a zero sequence $x_n$, such that $\sum_{k=1}^\infty \frac{x_n}{n}$ diverges.
As stated in the title:
Let $x_n$ be a zero sequence, such that $\sum_{k=1}^\infty \frac{x_n}{n}$ diverges.

either: Find a sequence $x_n$ for which the condition is met.

or: Prove that such a sequence cannot exist.

So far I have tried several known zero sequences: $\frac{1}{n},(\sqrt[n]{n}-1),\frac{n!}{n^n}$ but none seemed to work.
I also did not manage to prove that the statement is wrong. (I only managed to prove that for every zero sequence $x_n$ the series $\sum_{k=1}^\infty \frac{x_n}{n^2}$ converges.)

I don't think blindly trying several zero sequences is the best idea, so I would like to know how to approach this problem.

答:
What about
$$x_n = \frac{1}{\ln(n+1)} \quad ?$$


答: Consider $A:=I$ and $B$ and $D$ being arbitrary matrices that do not commute.  Then $C=I-B$.  Clearly, $A$ and $D$ commute, and it can easily be seen that $B(I-B)=(I-B)B$, so $B$ and $C$ commute.  Since the hypotheses are true and the result is false, the implication must be false.

Title: If $A=B+C$, $BC=CB$ and $AD=DA$, does $BD=DB$?

Body: If $A$, $B$, $C$ and $D$ are $n\times n$ matrices s.t. $A=B+C$, $B$ and $C$ commute and $A$ and $D$ commute, do $B$ and $D$ have to commute? Saw this used in another proof and it sounded like this was trivial but I tried proving it and I couldn't. Thanks.


问:
Good book on topological group theory?
I'm looking for a good introduction to the theory of locally compact groups and their representations. It may assume the reader to be familiar with basic group theory, topology and measure theory. </p>

答: You may consider G. B. Folland's book A Course in Abstract Harmonic Analysis.</p>

Title: Good book on topological group theory?


问:
Ball with radius that tends to zero equal the point
Is it correct that in a metric space, for $B[x,r]$ an open ball centered in x with radiuis r:

$\lim_{r\rightarrow0}B[x,r]=\{x\}$ or it is =$\emptyset$ ?

Thank you for the answers to this very basic question

答: By definition $$B[x,r] = \{a : |x-a|< r \}$$ which means sending $r$ to $0$, we get $|x-a| \le 0$. This is true iff $x=a$.


问:
Get vector coordinates
I'm currently working on a ray tracer, more specifically on ray-cylinder intersection. The cylinder is defined by a position (that I've called $coord$ and is the position vector of one of the caps, $cap1$), a height (which is a float) and an orientation (a vector of which its coordinates range from $-1.0$ to $1.0$). Based on this, I should define another cap, $cap2$.

This is a representation in 2D of what it should look like.

My issue is I can't quite figure out how to get the coordinates for the second cap. I've tried doing such things :

$cap2 = (orient + coord) * height$
 $cap2 = (orient + coord) - coord$
 or $cap2 = coord + (posit * height)$ without much success.
It should be noted that $orient$ and the vector going to $cap2$ are parallel.

Help much appreciated.

答:
Usually, a right circular cylinder in 3D is defined using the following properties:

Cylinder base, $\vec{b}$.  This is the center of the circular disc at one end of the cylinder.

Cylinder axis, $\vec{a}$.  This is the direction and length of the cylinder axis.  Sometimes the axis is split into an unit direction vector $\hat{a}$ ($\lVert \hat{a} \rVert = 1$) and height $h$, so that $\vec{a} = h \hat{a}$.

Cylinder radius, $r$.

Since $\vec{b}$ is one end of the cylinder, the other end is at $\vec{b} + \vec{a} = \vec{b} + h \hat{a}$.

The volume enclosed by this cylinder consists of points $\vec{p}$ that fulfill
$$\left\lbrace \; \begin{aligned}
\left\lVert \hat{a} \times (\vec{p} - \vec{b}) \right\rVert & \le r \\
0 \le \hat{a} \cdot (\vec{p} - \vec{b}) &\le h \\
\end{aligned} \right.$$
where $\times$ denotes vector cross product, $\cdot$ denotes vector dot product, and $\lVert \; \rVert$ denotes Euclidean norm.

Title: Get vector coordinates


答:
I'm not sure what the limit alone means.

I think it's helpful to reframe the limit a bit more rigorously as:
$$ \lim_{r \to 0} B[x, r] = \bigcap_{n=1}^\infty B[x, \tfrac{1}{n}]. $$

Then, we can see that $x \in B[x, \tfrac{1}{n}] $ for all $n$.
It's not hard from here to show that if $ y \ne x$ implies that there is an $n$ such that $y \notin B[x, \tfrac{1}{n} ]$.

This shows that the limit is just the point.

Title: Ball with radius that tends to zero equal the point

Body:
Is it correct that in a metric space, for $B[x,r]$ an open ball centered in x with radiuis r:

$\lim_{r\rightarrow0}B[x,r]=\{x\}$ or it is =$\emptyset$ ?

Thank you for the answers to this very basic question


问:
Convexity and supporting lines
While studying the properties of the Legendre-Fenchel transform I encountered this concept of convexity of a function: given $f:[a,b]\rightarrow \mathbb{R}$ we say that it admits a supporting line in $x\in[a,b]$ if $\exists \alpha\in\mathbb{R}$ such that $$f(y)\ge f(x)+\alpha(y-x)\quad \quad \forall y\in[a,b]$$
My book calls convex functions that admit supporting lines. My question is: how can I prove (if it's even true) that if the function admits a supporting line $\forall x\in[a,b]$ then it is convex in $[a,b]$ in the usual way, i.e. $\forall x_1,x_2\in[a,b]$ and $\forall c\in[0,1]$ $$cf(x_1)+(1-c)f(x_2)\ge f(cx_1+(1-c)x_2)$$?

It's intuitive that it's true if I plot an example, but I wasn't able to prove it.

答:
Hint: Let $D = \{(t, \mu)\in [a, b]\times {\mathbb R},\forall x\in [a,b], \mu\ge f(x)+\alpha_x(t-x)\}$.

Show that $D$ is a convex set as an intersection of convex sets

Show that $D$ is the epigraph of $f$

Conclude that $f$ is convex.

Title: Convexity and supporting lines


问:
Proving that $||v - \text{proj}_F^\perp (v)|| \le ||v - u||,\forall u \in F $
I'm having some trouble proving the following proposition:

Let $V$ be a finite-dimensional inner product real vector space: For each $v\in V$ it's orthogonal projection over $F \le V$ (denoted by: $\text{proj}_F^\perp
(v)$) is the closest vector of $F$ to $v$, in other words: $$||v - \text{proj}_F^\perp
(v)|| \le ||v - u||,\forall u \in F $$

I tried a direct proof, I tried a proof by contradiction, but I wan's able to prove this proposition. How can this proof be done?

答: If $v\in V$, and if $f\in F$ and $f'\in F^\perp$ are such that $v=f+f'$, then $f=\operatorname{proj}F^\perp(v)$ and, for any $u\in F$,\begin{align}\|v-u\|^2&=\bigl\|(v-f)+(f-u)\bigr\|^2\\&=\|v-f\|^2+\|f-u\|^2\text{ (because $f-u\in F$ and $v-f\in F^\perp$)}\\&\geqslant\|v-f\|^2.\end{align}


问:
Quaternions as rotations: Notation
I am reading through this notes on quaternions. I am trying to understand how they work as rotations as long as their norm is always 1.

First, in page 4 above the figure, the quaternion rotation operator is defined as follows

$$q = \cos(\theta) + u\sin(\theta)$$

But later, in equation 4 it is defined with half the angle:
$$q = \cos(\frac{\theta}{2}) + u\sin(\frac{\theta}{2})$$
Is this a mistake or are they telling the same?

答:
It doesn't matter if your quaternion has norm $1$ or not. It always induces a rotation on the subset of pure quaternions.

This is because if $q$ is nonzero, $qxq^{-1}=\frac{|q|}{|q|}qxq^{-1}=\frac{q}{|q|}x\left(\frac{q}{|q|}\right)^{-1}$ and of course $\frac{q}{|q|}$ is a unit quaternion.

So saying "it's a rotation as long as it's a unit quaternion" does not really carry any weight.

But choosing a unit-length representative reduces the number of quaternions being considered all the way down to two (instead of uncountably many.) In fact one can go a step further and insist that the real part is positive to get a single representative for a nontrivial rotation.

Strictly speaking, there is no mistake, but there is an awkwardness being caused by using a $\theta$ in both places.  The question is "how is $\theta$ related to the angle of rotation caused by the unit quaternion?"

One could write $q=\cos(\theta)+u\sin(\theta)$ with $\theta\in [0,\pi/2]$ and rightly say "it rotates by an angle of $2\theta$".

OR one could have written $q=\cos(\theta/2)+u\sin(\theta/2)$ with $\theta\in [0,2\pi]$ to begin with and said "this rotates by an angle of $\theta$.

Title: Quaternions as rotations: Notation


问:
How to show in a clean way that $z^4 + (x^2 + y^2 - 1)(2x^2 + 3y^2-1) = 0$ is a torus?
How to show in a clean way that the zero-locus of $$z^4 + (x^2 + y^2 - 1)(2x^2 + 3y^2-1) = 0$$ is a torus? </p>

答:
If one happens to know some Morse theory the following is another possible solution:

Consider the function $\Phi:\mathbb{R^3}\rightarrow\mathbb{R}$ defined by $$\Phi(x,y,z)=z^4+(x^2+y^2-1)(2x^2+3y^2-1).$$ You want to prove that $\Phi^{-1}(0)=:S$ is a torus. Computing $d_{(x,y,z)}\Phi$ you can check that $0$ is a regular value and hence $S$ is indeed a smooth surface embedded in $\mathbb{R}^3$. In particular $S$ is orientable.

Define now the function $$f:S\rightarrow\mathbb{R},\ (x,y,z)\mapsto y.$$ Using that $T_{(x,y,z)}S=\ker d_{(x,y,z)}\Phi$ it is not difficult to check that $f$ is a Morse function with critical points $(0,-1,0),(0,-\frac{1}{\sqrt{3}},0),(0,\frac{1}{\sqrt{3}},0),(0,1,0)$. Now use the Hessian of $f$ at these points to show that their indices are $0,1,1$ and $2$, respectively. Using the Morse inequalities you get that $\chi(S)=0$.

Since $S$ is an orientable surface with Euler characteristic equal to $0$, it must be a torus.


问:
Is there a general formula for the eccentricity $e$ of ellipsoid?
For an ellipse of eccentricity $e$ the formulas are:

${x^2 \over a^2} + {y^2 \over b^2} = 1 \\ e = \sqrt {1-\left({b \over a} \right)^2}$

what about the "3D case"?

答:
There is no 3D analogue of eccentricity, in the sense that there isn't a single real parameter that uniquely characterizes the shape up to scale. An $n$-dimensional ellipsoid's shape can be characterized up to rotations and reflections by $n$ real parameters, the lengths of its principal axes. These generalize the semimajor and semiminor axes, and for an ellipsoid with equation

$$\sum_{i=1}^n \left( \frac{x_i}{\sigma_i} \right)^2 = 1$$

are given by the denominators $\sigma_i$. More generally, for an ellipsoid with equation $x^T M x = 0$ where $x = \left[ \begin{array}{c} x_1 \\ \vdots \\ x_n \end{array} \right]$ and $M$ is a positive-definite symmetric matrix, the principal axes are the lines spanned by the eigenvectors of $M$ and the lengths of the principal axes are the inverses of the corresponding eigenvalues.

An ellipsoid is characterized up to rotations, reflections, and scaling by the ratios $\frac{\sigma_i}{\sigma_j}$ among the lengths of its principal axes; it suffices to consider the $n-1$ ratios $\frac{\sigma_i}{\sigma_{i+1}}$, arranging the lengths in order $\sigma_1 \ge \sigma_2 \ge \dots \ge \sigma_n$. Only when $n = 2$ does this boil down to a single number.

Quadrics which are not ellipses also have principal axes although the corresponding eigenvalues may be negative or zero, so the corresponding "lengths" may be negative or infinite.


问:
Solving for $g(y)$ in the following integral equation
Given the following integral equation where $f, h$ are known:

$$ f(t) = \int_{t-1}^t{g(y)\cdot h(y)\ \textrm{d}y}$$

Is it possible to solve this for $g(t)$, i.e. get an equation of the form $g(t) = \dots$ (supposing $g, h$ are suitably nice)? I have no idea where to start. I acknowledge that an analytical solution may not even exist in the general case, but I remain hopeful!

答:
If $ g $ and $ h $ are continuous at $ \Bbb R $ , then $ f $ will be differentiable at $ \Bbb R $ and, by chain rule,

$$f'(t)=g(t)h(t)-g(t-1)h(t-1)$$

So,

$$g(t)=g(t-1)\frac{h(t-1)}{h(t)}+\frac{f'(t)}{h(t)}$$

which is a kind of recursive relation.


问:
Geometric interpretation of FINDING eigenvectors
I have a pretty good idea for the geometric interpretation of what eigenvectors are I think.

When following the steps:
\begin{equation}
Ax = \lambda x
\\
(A - \lambda I)x = 0
\end{equation}

Obviously it works because subtracting $\lambda x$ from $Ax$ will only be zero when they are equal.  However, I don't quite intuitively understand.

Subtracting by $\lambda$ in $R^2$ seems to move $\hat i $ in the x-direction by $\lambda$ and $\hat j $ in the y-direction by $\lambda$ until they are on the same line thus meaning $det(A - \lambda I) = 0 $.  I still don't completely understand why this gives us the eigenvector of our original transformation $A$ though.

Hope this has not been asked before, and makes sense. Thanks for reading.

答:
The operator $\lambda I$ scales every vector by $\lambda$. You want to find the particular vectors $x$ that $A$ does that to (that's the definition of an eigenvalue/eigenvector pair). So you solve for $x$ in
$$
Ax = \lambda Ix .
$$

In operator terms that's
$$
(A - \lambda I)x = 0.
$$

One way to find out whether that equation has any solutions other than the $0$ vector is to look at the determinant of $A-\lambda I$.


问:
Regular functions and morphisms of varieties for field not algebraically closed
Let us work over a field that is NOT algebraically closed. Specifically, $K=\mathbb R$.

We can think of the affine line $\mathbb A_K^1$ as a ringed space $(\mathbb A_K^1,\mathcal O_{\mathbb A^1_K})$, where for an open set $U\subseteq\mathbb A^1_K$, we have $\mathcal O_{\mathbb A^1_K}(U)=(\text{the ring of regular functions over $U$})=\{\frac fg\mid f,g\in K[x] \,\&\,(\forall P\in\mathbb A^1_K)(g(P)\ne 0)\}$.

I think that according to this, $\frac1{1+x^2}$ should be a regular function over the entire $\mathbb A^1_K$. Moreover, it seems to me that $f:\mathbb A_K^1\to\mathbb A_K^1$ defined as $f(x)=\frac1{1+x^2}$ is a morphism from the ringed space $(\mathbb A^1_K,\mathcal O_{\mathbb A^1_K})$ to itself (in the sense that it pulls back regular functions to regular functions). Am I right?

I am probably confused by either different sources using different definitions or me seeing statements that only hold for $K$ algebraically closed. E.g. the Wikipedia article Morphism of algebraic varieties states

The regular functions on $\boldsymbol A^n$ are exactly the polynomials in $n$ variables [...]

,
which I only agree with for $K$ algebraically closed and not in the case I've shown above.

答:
The definition of a regular map needs to be modified substantially when you work over a field that isn't algebraically closed. There are different ways to set things up at different levels of sophistication; one way to say it is that regular maps should have the property that they remain regular maps after extension of scalars, and your proposed map does not remain regular after extension of scalars to $\mathbb{C}$ because of the poles at $x = \pm i$.

You are implicitly thinking only in terms of what are called the $K$-points of a variety; this is only safe to do when $K$ is algebraically closed. For example if $K = \mathbb{F}_q$ is a finite field then the polynomial $f(x) = x^q - x$ vanishes on $K$-points but is not considered to be identically zero as a regular map, because it does not remain identically zero after extension of scalars, e.g. to the algebraic closure $\overline{\mathbb{F}_q}$. We also have issues such as varieties, for example $\text{Spec } \mathbb{R}[x, y]/(x^2 + y^2 + 1)$, which have no $K$-points (here $K = \mathbb{R}$) but which it would be incorrect to consider empty; here again the issue is that points exist after extension of scalars.

Here I am implicitly taking a functor of points approach which I personally find conceptually cleanest. The standard approach you will find in most textbooks is based on the Zariski spectrum of all prime ideals (which capture points over extensions of scalars but somewhat indirectly), not just maximal ideals corresponding to $K$-points.

On the other hand, there is a somewhat separate field of real algebraic geometry in which we only consider $\mathbb{R}$-points, and which is mostly not studied by algebraic geometers but by model theorists or something like that.


答:
First of all note that any function with given condition must be $f(x)=ax$, $a\in\mathbb R$ (real numbers).
In our case $a=2000/m$.
So required function is $f(x)=2000x/m$.
Please verify this function follow all your given conditions.


问:
Find a delta given an interval
So the question is the following:

We consider $f(x) = x^2$ on the interval $[0,100]$. Let $\epsilon > 0$ be given. Find a number $\delta > 0$ so that if the length of an interval $I \in [0, 100]$ is less than delta, then $f(u) - f(l) < \epsilon$, where $f(u), f(l)$ are max and min of $f$ on $I$, respectively.

It's probably really simple but I just can't seem to wrap my head around it. Thanks!

答:
So this question is about showing the (uniform) continuity of a function on an interval. Since this seems like a homework problem, here's a somewhat simpler problem that will hopefully give some intuition:

Consider the function $f(x) = 10 x$ on $[0, 10]$.
Then, given $\varepsilon > 0$, we want to find $\delta > 0$ such that any interval of width less than $\delta$ has $f(u) - f(\ell) < \varepsilon$.

A first guess might be $\delta := \varepsilon$.
However, this doesn't work. If $\varepsilon = 1$, then $\delta = 1$, so an interval might be $[9, 10]$.
However, $f(u) - f(\ell) = f(10) - f(9) = 10 > 1 = \varepsilon$. (Why is $u = 10$, $\ell = 9$?)

Setting $\delta$ to be something less than $\frac{1}{10}\varepsilon$, will make this work out though.

In your example, you'll need to figure out where the "least well behaved" part of the function is and choose your $\delta$ based around that (hint: the steeper the function, the smaller $\delta$ has to be relative to $\varepsilon$).

Title: Find a delta given an interval


问:
How can we represent the following case in limit notation?
Suppose, we have the following sequence:

$$ 1, 2, 3,..., n $$

This sequence doesn't converge as it extends to $ \infty $.

How can we represent this case using limit notation?

Is the following correct?

$$\lim_{n\to\infty}n=\infty$$

答: Yes, combining the meanings of $\lim_{n\to\infty}$ and $\lim_{x\to c}f(x)=\infty$ discussed here.


问:
An intuitive explanation for the ecological fallacy
I believe that it is called the ecological fallacy. People say that one cannot apply population-wide statistics to individuals of that population. So, just because some trait exists in a higher proportion in population A than it does in B, that doesn't mean that an individual from pop. A is more likely to have that trait than an individual from pop. B.

Now, this doesn't make sense to me. An individual coming from a population where a certain trait is more common, is more likely to have that trait than an individual coming from a population where said trait is less common, right?

I guess this depends on how one mathematically defines "common". The mean and the average are a bit scary statistical quantifications. I see how "the average" is problematic, but a weighted average is surely quite illuminatory of the real likelihoods?

答:
There is no one ecological fallacy; rather, it is a category of fallacies that share the trait you describe: incorrectly extrapolating a group property to an individual property.

Very often, it is because some correlation is being ignored.  You are correct that if Group A contains a higher proportion of, say, color-blind people than Group B, then a uniformly randomly selected person from Group A is more likely to be color-blind than one drawn from Group B.

But suppose the person from Group A is female.  Females are rather less likely to be color-blind than males.  With this additional information, the conclusion that this (possibly still otherwise randomly selected) person is more likely to be color-blind is no longer reliable.  This presumption that the conclusion is robust to such considerations marks it as an example of an ecological fallacy.


问:
In how many ways can the 10 vacant positions be filled?
There are 10 vacant positions in the central cabinet of ministers. There is a total of 350 MPs in the ruling party. A cabinet post can be given to any of these MPs and an MP can take more than one cabinet position. In how many ways can the 10 vacant positions be filled?</p>

答: Since each post has 350 possibilities the answer is $350^{10}$.


问:
Well ordered set
(ZFC) Let X ⊆ R is a well ordered form the standart order in R. Prove that X is either finite or countable.

My attempt: Since X is well ordered then if L is a chain where every countable subset of X is well-ordered, L is well ordered as well. In other words: then L contains a finite descending sequence
a0 > a1 > a2 > . . . with a minimal element an, hence X is finite;

If I manage to find a function which maps every element of my set into a unique ordinal I think that this is satisfying. Still I know that well-ordered set is order-isomorphic to a unique ordinal. However, I do not how to make use of it.

I hope that someone will help me. Thanks!

答:
I might be missing something, but it seems to me that since the well-ordering comes from $\mathbb{R}$, we can use the countability of the rationals:

For each $a_i$ (with $i \in I$, your indexing set), we can assign a rational $q_i \in (a_i, a_{i+1} )$.
This $q_i$ is only mapped to by $a_i$, so we have an injection, $I \to \mathbb{Q}$.
Thus, $|I| \le | \mathbb{Q} |$, so it is either countable or finite.

Title: Well ordered set


问:
What is the distribution of sampling top $p$ proportion of overlapping normal distributions?
I have a tricky distribution question. Say I have two normal distributions (call them $n_1$ and $n_2$) described by $\mu_1, \sigma_1$ and $\mu_2, \sigma_2$ such that $\mu_2 >\mu_1$ and the distributions are overlapping. Now suppose I have a random variable $\rho$ with unknown distribution $\phi_{\rho}$ such that $0<\rho<1$ (i.e. $\phi_{\rho}(x)=0$ $\forall x<0$ and $\forall x>1$).

I now have 1000 observations. $\rho*1000$ of them are distributed according to $n_1$ and the other $(1-\rho)*1000$ of them are distributed according to $n_2$. If $q\in [0,1]$ and I grab the greatest $q * 1000$ observations from my $1000$ total observations, how many of them are likely to come from $n_2$ (as a function of $\rho$)?

I know I am interested in the expected value of this but I am not sure exactly how to think about the distribution I am describing here (or what its expected value is). Intuitively I am thinking for instance that if $\rho$ is $0$, then they all come from $n_2$. If $\rho$ is $\frac{1}{1000}$, then only $1$ observation comes from $n_1$ so by taking, say, the $100$ largest observations it is highly highly unlikely I will get the one observation from $n_1$. But this becomes much less clear for less extreme values of $\rho$.

I know this is very general and theoretical but I appreciate any thoughts or comments about similar problems that might be helpful.

答:
I don't have a full solution but I can tell you "how to think about this question". The way is to use priors and the Bayes theorem:

Suppose you knew  the conditional probability to come from each distribution, given $x$ :
$P({\bf 1}|x)$ and $P({\bf 2}|x)$, where ${\bf 1};{\bf 2}$ represent which distribution was used. This is a useful quantity and you can ask questions like what the probability than (say) $n_1$ observations were above $x_0$ and were all from $\bf 1$, while the other $n_2$ were below it and from $\bf 2$, e.g. $$\begin{equation}{n_1+n_2 \choose n_1}\left(\int_{x_0} dxP({\bf 1}|x)\right)^{n_1}\left(\int^{x_0} dxP({\bf 2}|x)\right)^{n_2}\end{equation}\tag{1}$$
This is very close to what you want, and the best I can do.

In order to compute this, we will use Bayes's theorem several times:
Suppose we knew the joint probability $P(x,{\bf 1} )$, that is the probability to both observe $x$ and be in distribution $\bf 1$. Then, Bayes's theorem states that $ P({\bf 1}|x)= {P(x,{\bf 1} )\over P(x)}$

To compute the joint probabilities $P(x,{\bf 1} );P(x,{\bf 2} )$ let us start with the conditional probability to pick the value $x$ given  the observation was from distribution $\bf 1$. Thus $$P(x|{\bf 1})=N(x;\mu_1,\sigma_1)$$ (where $N$ is the normal distribution.  Similarly $$P(x|{\bf 2})=N(x;\mu_2,\sigma_2)$$ is the probability to pick $x$ given that the observation was from distribution $\bf 2$. The joint probabilities can be deduced ( again via Bayes's theorem):
$$P(x,{\bf 1})=P(x|{\bf 1})P({\bf 1})=P(x|{\bf 1})\rho \tag{2}$$
$$P(x,{\bf 2})=P(x|{\bf 2})P({\bf 2})=P(x|{\bf 2})(1-\rho) \tag{3}$$

So, we have $P(x,{\bf \{1,2\}})$, we are lacking $P(x)$, but this can be deduced by normalization: $P(x)=P(x|{\bf 1})P({\bf 1})+P(x|{\bf 2})P({\bf 2})$ ( the probability to see $x$ at all).

Finally, we can recover $$P({\bf 1}|x)={P(x|{\bf 1})\rho \over P(x|{\bf 1})\rho+P(x|{\bf 2})(1-\rho)}\tag{4}$$


问:
What does it mean by $f$ stays in a bounded set in $C^{k+1}(X)$?
I am trying to understand the statement of Theorem 7.7.1 in The analysis of linear partial differential operator I by Hormander. What exactly is meant by  $f$ stays in a bounded set in $C^{k}(X)$? I assume it is with respect to some norm on this space but it isn't lear to me what the norm is. Any clarification appreciated. Thank you.

Notation. $X$ is an open subset of $\mathbb{R}^n$ and $C^{k}(X)$ is the space of real valued functions defined on $X$ which are $k$ times continuously differentiable.

答:
Theorem 7.7.1 asserts that if $B \subset C^{k+1}(X)$ is bounded, then there exists a constant $C>0$ such that the claimed estimate holds for all $u \in C_0^k(K)$ and $f \in B$ such that $\operatorname{Im} f \geq 0$ in $X.$

What does it mean for $B$ to be bounded? It appears as though Hörmander does not explicitly define this, but the space $C^k(X)$ is equipped with the seminorms $\lVert \cdot \rVert_{C^k(K)}$ for all $K \subset X$ compact, which gives it the structure of a locally convex space. We say $B$ is bounded if it is bounded with respect to all of these seminorms, that is $\sup_{f\in B} \lVert f \rVert_{C^k(K)} < \infty$ for all $K \subset X$ compact (see Definition 9 here).


答: That looks correct, but it is much easier to see that\begin{align}4(n+1)^2>4n^2&\implies(4(n+1)^2)!>(4n^2)!\\&\implies\frac1{(4(n+1)^2)!}<\frac1{(4n^2)!}.\end{align}

Title: Solution verification of $\frac{1}{(4n^2)!} >\frac{1}{(4(n+1)^2)!}$ inequality

Body:
\begin{align}
\frac{1}{(4n^2)!} &>\frac{1}{(4(n+1)^2)!}\\
\frac{(4(n+1)^2)!}{(4n^2)!} &>\frac{1}{1}\\
\frac{(4(n^2+2n+1))!}{(4n^2)!} &>1\\
\frac{(4n^2+8n+4)!}{(4n^2)!} &>1\\
\iff \frac{(4n^2+8n+4)}{(4n^2)}&>1\\
4n^2+8n+4&>4n^2\\
8n &> -4\\
n&>-\frac{1}{2}
\end{align}
This implies that the inequality holds for all natural numbers, thus also the primary inequality $$\frac{1}{(4n^2)!} >\frac{1}{(4(n+1)^2)!}$$ holds.


问:
Geometric Progressions: Sum to Infitity
Good evening, would anyone be able to help with the question reading:

Find the sum to infinity of the geometric series:
$$ 3 + \frac{3}{4} + \frac{3}{16} + \cdots $$
Find also the sum of the first n terms of this series. Show that however large $n$ may be, the sum cannot exceed $4$.

I'm still trying to wrap my head around the topic of Progressions as I'm studying it at school, any and all help would be appreciated.

答: So you see ratio is 1/4 and first term=3 so using formula a/(1-r) where a=first term r=ratio you have the sum is 3/(1-1/4)=3/(3/4)=4. So the infinite sum is 4, and because it is always increasing, the first n terms can't go over 4. So formula for sum of first n terms is S_n=(a(1-r^n))/(1-r) where a is first term r=ratio and S_n is sum of first n terms so stated before a=3 and r=1/4 so S_n=(3(1-(1/4)^n)/(1-1/4)=(3(1-(1/4)^n))/(3/4)=4(1-(1/4)^n).</p>

Title: Geometric Progressions: Sum to Infitity


问:
Question on Contractions
Let $S = \{x \in \mathbb{R}^n ;  \|x\| \le 1  \}$ and $f: S \to S$ be a contraction. Determine one can have $f(S) = S$.

I really need some help with this question. In advance I wanted to give all the necessary definitions needed for this question.

Definition:

Let $(X, P)$, $(Y, P')$ be metric spaces and $f: X \rightarrow Y$ a function. Let $M \in \mathbb{R}$ with $M > 0$. We say that $f$ satisfies a Lipschitz  condition with Lipschitz constant $M$ if $\forall x, y \in X: P'(f(x), f(y)) \le M P(x,y)$

Let $(X, P)$, $(Y, P)$ be metric spaces and $f: X \rightarrow Y$ a function. We say that $f$ is a contraction (mapping) if $f$ satisfies the  Lipschitz  condition on $X$ with some Lipschitz constant $M \in (0,1)$.

答:
Let $\xi \in S$ by the fixed point of $f$ which exists by the Banach fixed point theorem. Now the function $d \colon S \to \mathbb R$, $x \mapsto \|x-\xi\|$ attains its maximum on $S$, as $S$ is compact. Let $\eta \in S$ such that $d(\eta) \ge d(x)$ for all $x \in S$. Let $q \in [0,1)$ by a Lipschitz constant for $f$, then for all $x \in S$:
$$
  d(f(x)) = \|f(x) - \xi\| = \|f(x) - f(\xi)\| \le q \|x-\xi\| = qd(x) \le qd(\eta) < d(\eta).
$$
Hence $f(x) \ne \eta$ and $f$ isn't onto.</p>

Title: Question on Contractions


问:
Find a real-valued function which satisfied Cauchy-Riemann equations
Question: Find a real-valued function $v(x,y)$ such that $v(0,0)=1$ and together, $u$ and $v$ satisfy the Cauchy-Riemann equations in the entire complex plane.

Let $v(x,y)=M(x,y)+i.0$ which is a real-valued function.

CR equations for $v(x,y)$,

$$M_x=0$$

$$M_y=0$$

But it seems I am going to a wrong direction. Any solution or hints will be great helpful for me.

答: You can take $u(x,y)=x$ and $v(x,y)=y+1$. Then, $u_x=v_y=1$ and $u_y=-v_x=0$. Besides, $v(0,0)=1$.


答:
If $ (v_n)$ is a geometric sequence, then there is a constant $ q$, such that

$$(\forall n\ge 0)\;\; v_n=v_0.q^n$$

in your case, $ v_0=3 $ and $ q=\frac 14$.

The sum of termes of $ (v_n) $ is
given by

$$S_N=v_0+v_1+v_2+...+v_N=v_0\frac{1-q^{N+1}}{1-q}$$

$$=3\frac{1-\frac{1}{4^{N+1}}}{1-\frac 14}$$

$$=4-\frac{1}{4^N}$$

Title: Geometric Progressions: Sum to Infitity

Body:
Good evening, would anyone be able to help with the question reading:

Find the sum to infinity of the geometric series:
$$ 3 + \frac{3}{4} + \frac{3}{16} + \cdots $$
Find also the sum of the first n terms of this series. Show that however large $n$ may be, the sum cannot exceed $4$.

I'm still trying to wrap my head around the topic of Progressions as I'm studying it at school, any and all help would be appreciated.