问:
polar decomposition of a square A when A is not full rank/invertible
I am on the proof of the polar decomposition w/o using SVD:

Let $A$ be a real $n \times n$ matrix. Show that there exists $Q$: orthogonal, $P$: positive semi-definite symmetric matrix such that $A = QP$.

Consider the product $A^*A$, it is a self-adjoint and positive semi-definite (as we are in $\mathbb{R}$) symmetric matrix, so it has a decomposition $EDE^{-1}$ where $D$ is diagonal and $E$ is orthonormal and contains eigenvectors of $A^*A$. Now let
$$
  P = \sqrt{A^*A} =  E\sqrt{D}E^{-1}
  $$
where $\sqrt{D}$ has diagonal elements being the squareroot of $D$. Notice that $||Px|| = ||Ax||,  $
so we can expect the product $AP^{-1}$ to be orthogonal. Provided that $P^{-1}$ exists (i.e. $P$ is positive definite), we have $||AP^{-1}x||=||x||$,
so we do find our desired decomposition by letting $Q = AP^{-1}$.

The problem is that if $A$ is not invertible, then $P$ will also not invertible, regarding the whole proof useless. How should I fix it in this case?

答:
A standard workaround for singularities is to use analytic techniques.   The below is an existence argument, not a uniqueness argument.  Note Polar Decomposition is not unique when $\det\big(A\big)=0$.

$A_k := A + \frac{\delta}{k}I$ for some small enough $\delta \gt 0$. Then for all $k=1,2,3,...$

$A_kP_k^{-1}=Q_k \in O_n\big(\mathbb R\big)$

by compactness of $O_n\big(\mathbb R\big)$, the $Q_k$ form a bounded sequence and hence there is (at least one) convergent subsequence $k_1\lt k_2\lt...$ whose limit is given by $\lim_{i\to\infty}Q_{k_i}=Q \in  O_n\big(\mathbb R\big)$  .

Similarly

$P_{k_i} = Q_{k_i}^TA_{k_i}$

whose limit exists because each term on the RHS does. And for avoidance of doubt:

1.) $P_{k_i}$ is symmetric for all $i$ hence the limit is too.

2.) by topological continuity of eigenvalues, since $P_{k_i}$ has all positive eigenvalues for all $i$, its limit has all real non-negative eigenvalues, i.e. it is PSD.


问:
Show that there exist a set $R$ and functions $β : S → R$ and $γ : R → T$ such that $β$ is onto, $γ$ is one-to-one and $α = γβ.$
Let $S$ and $T$ be sets and $α : S → T$ a function.  Show that there exist a set $R$ and functions $β : S → R$ and $γ : R → T$ such that $β$ is onto, $γ$ is one-to-one and $α = γβ.$

My proof:

Define $α : S → T$ via $\alpha(s)=t$. Assume that there exist a set $R$ and functions $β : S → R$ and $γ : R → T$. Since $\beta$ is onto, then for all $r\in R$, there is an element $s\in S$ such that $\beta(s)=r$. Also, by assuming problem $\gamma$ is one-to-one, so if $\gamma(r)=t$ and $\gamma(u)=t$ implies $r=u$. Thus $\gamma \beta(s)=\gamma(r)=t=\alpha(s)$ as desired.

Is my proof correct? I have a clear idea of why it works but I'm not very convinced of what I did.

答:
Your proof is doing it backwards to some extent. Define the set $R$ as some set which has $\big|R\big|=\big|\alpha(S)\big|$, i.e. $R$ is the same size as the range of $\alpha$. The easiest way to do this is to let $R=\alpha(S)$; note that this makes $R\subseteq T$ because $T$ is the codomain of $\alpha$.

We can define $\gamma:R\to T$ as the function $\gamma(r)=r$, which is well-defined because $R=\alpha(S)\subseteq T$. Clearly, $\gamma$ is injective because every distinct element in $R$ is mapped to itself (as an element of $T$). You can think of $\gamma$ as an identity map where the codomain is (possibly) larger than the domain.

Then, just define $\beta: S\to R$ as $s\mapsto \alpha(s)$. Remember that we defined $R=\alpha(S)$, and so $\beta$ is well-defined and it is clearly surjective as well. (The proof of this is left as an exercise for the reader.)

Your answer gets the general idea of what to do, but make sure you aren't "assuming" the requirements of the problem. We don't "assume" that $\gamma$ is one-to-one, we define $\gamma$ and then we show that it is one-to-one. It's a subtle distinction, but it's critically important to understanding the essence of the problem.

ContentLicense: CC BY-SA 4.0


问:
To prove $EA = FB$ or that $CQ'$ is radical axis
Given disjoint circles $c_1 = \odot(P,PA), c_2 = \odot(O,OB)$ such that $B$ and $A$ are in the same half-plane wrt $OP$ and that $PA \parallel OB \perp OP$.

Line $CDQ$ is the perpendicular bisector of $AB$, $D \in AB,Q \in OP$.

Point $Q'$ is the reflection of $Q$ wrt to $M$, the midpoint of $OP$.

Point $C$ on line perpendicular to $OP$ in $Q'$.

$E = CA \cap c_1, F = CB \cap c_2$.

I want to prove $EA = FB$.

We know that $DC = DQ$ because $DM$ is midsegment of trapezium $OBAP$.

答:
Since $BQ=AQ$ (red segments) we have $$r_1^2+(m-x)^2 = r_2^2+(m+x)^2$$ so
$$p(Q',c_1)- p(Q',c_2) = \big((m+x)^2-r_1^2\big)- \big((m-x)^2-r_2^2\big) =0$$ we see that $Q'$ has equal power with respect to both circles so it lies on a radical axis which is perpendicular to $OP$ so it is $CQ'$.

Now $C$ has also equal power to both circles so $$CB\cdot CF = CA\cdot CE$$
Since $CA= CB$ we have now $CF = CE$ and we are done.


问:
It can be shown that the multiplicative group $(\mathbb{Z}_{10})^\times$ is cyclic. To which group is it isomorphic?
It can be show that the multiplicative group $(\mathbb{Z}_{10})^\times$ is cyclic. To which group is it isomorphic?

What I have tried:

I have calculated the order of $(\mathbb{Z}_{10})^\times$ to be 4. Hence I think it is isomorphic to $\mathbb{Z}_{4}$.

Would someone be able to verify if my answer is correct?

Thank you for taking the time to respond to my question.

答:
The order of $\Bbb Z_n^\times$ is $\varphi(n)$, where $\varphi$ is the totient function.  $\varphi(10)=4$.

You can check that $3$ has order four in $\Bbb Z_{10}^\times$.  Because $3^2\equiv9, 3^3\equiv7\bmod10$.  That is, $3$ is a primitive element $\bmod10$.

Thus we have the cyclic group of order $4$.


问:
Dominant morphism between affine varieties induces injection on coordinate rings?
Here are the definitions that we use for this problem:

A morphism $\varphi : X \to Y$ between two varieties is said to be dominant if the image of $\varphi$ is dense in $Y$ (c.f. Hartshorne exercise 1.3.17)

We say $X$ is an affine variety if it is an irreducible closed subset of $\Bbb{A}^n$. (see the definition after example 1.1.4 of Hartshorne)

Now let $X$ and $Y$ be affine varieties and $\varphi : X \to Y$ a dominant morphism. Now this means that for all $p \in X$, $\varphi^\ast_p : \mathcal{O}_{\varphi(p),Y} \to \mathcal{O}_{p,X}$ is injective (c.f. Hartshorne exercise 1.3.3 (c)). Because we have assumed that $X,Y$ are affine varieties we identify $\mathcal{O}_{p,X} \cong A(X)_{\mathfrak{m}_p}$ and similarly for $Y$ to get that

$$(\varphi^\ast)_p : A(Y)_{\mathfrak{m}_{\varphi(p)}} \to A(X)_{\mathfrak{m}_p}$$

is injective for all $p\in X$.

My question is: Is it true that $\varphi^\ast : A(Y) \to A(X)$ is injective?

Here are some partial results I can get. Choose a maximal ideal $\mathfrak{m}_q \subseteq A(Y)$. If $\ker \varphi^\ast$ is completely contained in $\mathfrak{m}_q$ then

$$(\mathfrak{m}_q)_{\mathfrak{m}_q} (\ker \varphi^\ast)_{\mathfrak{m}_q} = (\ker \varphi^\ast)_{\mathfrak{m}_q}.$$

The ring $A(Y)_{\mathfrak{m}_q}$ is local and Noetherian so $(\ker \varphi^\ast)_{\mathfrak{m}_q}$ is finitely generated as a module over this ring. By Nakayama's lemma it follows $(\ker \varphi^\ast)_{\mathfrak{m}_q} =0$.

Edit: Adeel's answer below shows that under my assumptions we have $\ker \varphi \subseteq \text{nilrad}  A(Y)$, so my proof is complete.

答:
The usual definition of dominant would be that the image of $\varphi$ is dense, or, equivalently, contains a non-empty open subset of the target.

[The equivalence presumes that we are talking about irreducible varieties, so that
non-empty open sets are automatically dense.  The definition in terms of density of the image makes sense, and can be used, even if the source and target are reducible.]

Now this is equivalent to asking that the morphism on coordinate rings be injective, essentially because restricting to a dense subset is injective for continuous functions.

Again presuming that we are talking about irreducible varieties, this is equivalent to injectivity on stalks, as in the OP, although I've never seen this used as the definition of dominant.  Note that if the source is reducible, than this requirement is different to having dense image; it is equivalent to asking that if $P$ is a point of $X$, then the union of the components through $P$ has dense image in the union of components through $\varphi(P)$.  [So it would count the inclusion of the punctured line into two crossed lines --- the puncture being at the crossing-point --- as dominant, but wouldn't count the inclusion of the whole line into two crossed lines as dominant.]

All this being said, I don't understand the proof given in the OP: where does the claim that $(\mathfrak m_q)_{\mathfrak m_q} (ker \varphi^*)_{\mathfrak m_q} = ker \varphi^*$ come from?  As noted above, the claim that this implies injectivity on coordiante rings is only valid when $Y$ is irreducible, so somewhere in the proof you have to use that $A(Y)$ is a domain. (Actually the way the argment is written it is not even clear how the assumption of injectivity on stalks is being used at all.)

My suggestion would be that you draw some picures to think about this, including the cases of a line/punctured line mapping into two crossed lines that I mentioned above, and some other morphisms between affine algebraic sets, some having dense image and some not, and thinking about how the algebra will reflect the geometry.  The way the OP is written right now, it suggests that you are not thinking about the geometric meaning much at all, but more just manipulating the algebraic formalism.

Added: Here are some exercises to prove the equivalence b/w dominant and injectivity on coordinate rings, for morphisms $\varphi: X \to Y$ between affine algebraic sets over an algebraically closed field $k$.  I realize that a proof of this has already been given in another answer, but these exercises outline a different argument which is more geometric, in the spirit of the rant of the preceding paragraph.

Show that if $\varphi: X \to Y$ is a continuous map of top. spaces with dense image, then the induced map $\varphi^*: \mathcal C(Y,k) \to \mathcal C(X,k)$ is injective.  Here $\mathcal C(S,k)$ is the space of continuous morphisms from a top. space $S$ to the field $k$, where $k$ is given its Zariski topology (i.e. identified as a topological space with $\mathbb A^1$).

Note that since $A(X) \subset \mathcal C(X,k)$ when $X$ is an affine algebraic set, and similarly with $A(Y)$, having dense image implies
that $\varphi^*: A(Y) \to A(X)$ is injective.

Show that if $Z$ is any proper closed subset of $Y$, then there exists
a non-zero function $f \in A(Y)$ such that $f$ vanishes on all the points in $Z$.

Show that if $\varphi: X\to Y$ does not have dense image, then $A(Y) \to A(X)$ has non-trivial kernel, hence is not injective.

The point is that one doesn't need commutative algebra to prove the desired equivalence.  All it uses is the basic relationship between $A(X)$ and the Zariski topology on $A(X)$ for an affine algebraic set $X$, i.e. that elements of $A(X)$ are Zariski continuous, and that closed subsets can be cut out as the zero loci of elements of $A(X)$.

It's good to learn to think this way, i.e. to argue with geometry or topolology without translating everything into commutative algebra all the time, otherwise it becomes hard to see the geometry of what is going on (the more so as your study of algebraic geometry progresses).

ContentLicense: CC BY-SA 3.0


问:
Factoring with integer coefficients vs factoring with real coefficients
What is the difference between factoring with integer coefficients and factoring with real coefficients? Although the answer to this question probably sounds pretty obvious, I'm not sure how my answer changes in the question below.

If I was to factor $$3x^5-x^4-3x^3+x^2-6x+2$$
with integer coefficients, the answer is $$(3x-1)(x^2-2)(x^2+1)$$
How does my answer change if I am factoring with real (not necessarily integer) coefficients?

答:
Factoring over the integers, we see that $$3x^5 - x^4 - 3x^3 + x^2 - 6x + 2 = (3x - 1)(x^2 - 2)(x^2 + 1).$$ Factoring over the reals, we see that $$3x^5 - x^4 - 3x^3 + x^2 - 6x + 2 = 3(x - 1/3)(x - \sqrt{2})(x + \sqrt{2})(x^2 + 1).$$
Factoring over the complex numbers, we have $$3x^5 - x^4 - 3x^3 + x^2 - 6x + 2 = 3(x - 1/3)(x - \sqrt{2})(x + \sqrt{2})(x + i)(x - i).$$


答:
Your method works fine, however it's rather algebra-heavy which can lead to accidental mistakes. I would note that $$z+\frac 1z \in \Bbb R \iff \mathfrak{Im}(z+\frac 1z)=\mathfrak{Im}(z)+\mathfrak{Im}(\frac 1z)=0$$

Then, it is easily seen that $$\mathfrak{Im}(\frac 1z)=\frac{-\mathfrak{Im}(z)}{|z|^2}$$ and so we are solving the equation:
$$\mathfrak{Im}(z)\bigg(1-\frac{1}{|z|^2}\bigg)=0$$
which has solutions when $\mathfrak{Im}(z)=0$ (i.e. $z\in\Bbb R)$ or $$|z|^2=1\implies |z|=1\implies \mathfrak{Im}^2(z)=1-\mathfrak{Re}^2(z)\to z=t\pm i\sqrt{1-t^2}$$ which is what you got.

Title: Find z s.t. $z+\frac{1}{z}$ is real. Is my solution good?

Body:
I must find all z s.t. $z+\frac{1}{z}$ is real. I know that $z = a + bi$ is real when the Imaginary part is 0. So, there we go:

$$z+\frac{1}{z}= \frac{z^2+1}{z}=\frac{(a+bi)^2+1}{a+bi}=\frac{[(a+bi)^2+1](a-bi)}{(a+bi)(a-bi)}$$
$$\frac{[(a+bi)^2+1](a-bi)}{(a+bi)(a-bi)} = \frac{(a^2 +2abi-b^2+1)(a-bi)}{(a^2+b^2)}$$
$$\frac{(a^2 +2abi-b^2+1)(a-bi)}{(a^2+b^2)}=\frac{a^3 +2a^2bi-b^2a+a-ba^2i+2ab^2+b^3i-bi}{(a^2+b^2)}$$
$$2a^2b-ba^2+b^3-b=0$$
$$a^2b+b^3-b=0$$
$$b_1=0$$
$$a^2+b^2-1=0$$
$$b^2=1-a^2$$
$$b_2=\sqrt{1-a^2}$$
$$b_3=-\sqrt{1-a^2}$$

so, the solutions are:

$$z_1=a+0i$$
$$z_1=a+\sqrt{1-a^2}i$$
$$z_1=a-\sqrt{1-a^2}i$$

I am not sure if that should be done like I did it.


问:
Finding the $x^n$ coefficient of the power series $\sum\limits_{n=0}^\infty\frac{x^{2n+3}}{n!}$
I have a practice test question that asks:

Given the following Maclaurin series representation, $$\sum\limits_{n=0}^\infty\frac{x^{2n+3}}{n!}$$ what is the coefficient of $x^n$?

I have the correct solution, but I do not know how to derive it. How would I go about doing this?

答:
Let $\mathbb{N}_0$ denote the nonnegative integers, and $a_n$ the coefficient desired.

$$a_n=\begin{cases} 0& \frac{n-3}{2}\notin \mathbb{N}_0\\ \frac{1}{\left(\frac{n-3}{2}\right)!} & \frac{n-3}{2}\in \mathbb{N}_0\end{cases}$$

Details added upon request.  If $n=2m+3$, then $m=\frac{n-3}{2}$.  If $m$ is an integer, we have a coefficient; otherwise we don't (and the coefficient is 0).


问:
How to solve ODE: $\frac{dy}{dt} - 2ty = -3t^2e^{t^2}$?
I am trying to solve first order ODE: $\frac{dy}{dt} - 2ty = -3t^2e^{t^2}$. I tried using substitution $t^{2} = u$, but after substitution I get $\frac{dy}{dt} - y\frac{du}{dt} = -3ue^u$, which I am unable to solve.</p>

答:
Integration Factor

To begin to answer your question one can observe that this problem is easily solved using an integration factor.

\begin{equation}\frac{dy}{dt} - 2ty = -3t^2e^{t^2} \end{equation}
Then I would state the following: \begin{align}\mu(x)&=e^{P(x)}\\ &=e^{\int -2t\ dt}\\&= e^{-t^2} \end{align}
Then using this integrating factor I would multiply through the DE to solve it, and observe something neat happens.
\begin{align}e^{-t^2}\frac{dy}{dt}-2te^{t^2}y&=-3t^2\\ \frac{dy}{dt}\Bigg[ye^{-t^2}\Bigg]&=-3t^2\\ ye^{-t^2}&=-t^3+C \\ y&=Ce^{t^2}-t^3e^{t^2}\end{align}


问:
Relation between the spectrum of a Banach algebra and the spectrum of its unitization.
Let $A$ be any Banach algebra. We can always consider its unitization $A_{1}:=A\oplus\mathbb{C}$, even if $A$ is already unital. The multiplication on $A_{1}$ is defined as $$(a,\lambda)(b,\mu):=(ab+\lambda a+\mu b,\lambda\mu).$$ One can easily prove that a character $\phi$ on a unital algebra $A$ satisfies $\phi(1)=1$. The spectrum $\Omega(A)$ of $A$ is defined as the set of all non-zero homomorphisms $A\to\mathbb{C}$. We endow $\Omega(A)$ with the topology of pointwise convergence.

Note that $\hat{0}\colon A_{1}\to\mathbb{C}$, $(a,\lambda)\mapsto\lambda$ is a  character on $A_{1}$.

I am trying to prove that $\Omega(A)$ is homeomorphic to $\Omega(A_{1})\setminus\{\hat{0}\}$. Also, I want to prove this directly from the definitions, i.e. without using (local) compactness or Hausdorffness.

Here is what I tried:

For $\phi\in\Omega(A)$ I define $\hat{\phi}\colon A_{1}\to\mathbb{C}$ by $\hat{\phi}(a,\lambda):=\phi(a)+\lambda$. Now I tried to prove that $\phi\mapsto\hat{\phi}$ is a homeomorphism.

Injectivity: Assume that $\hat{\phi}=\hat{\psi}$. Then $$\phi(a)=\hat{\phi}(a,0)=\hat{\psi}(a,0)=\psi(a)$$ for all $a\in A$, thus $\phi=\psi$.

Surjectivity: Let $\alpha\in\Omega(A)\setminus\{\hat{0}\}$ be given. Define $\phi\colon A\to\mathbb{C}$ by $\phi(a):=\alpha(a,0)$. Then $\phi\neq0$, otherwise $\hat{\phi}=\hat{0}$. Also, it is easy to see that $\phi$ is a homomorphism, thus $\phi\in\Omega(A)$. Now observe that $$\hat{\phi}(a,\lambda)=\phi(a)+\lambda=\alpha(a,0)+\lambda\alpha(0,1)=\alpha(a,\lambda).$$

Continuity: Let $(\phi_{i})$ be a net in $\Omega(A)$ that converges to a character $\phi\in\Omega(A)$. Since $\phi_{i}(a)\to\phi(a)$ for all $a\in A$, it is easy to see that $\hat{\phi}_{i}(a,\lambda)\to\hat{\phi}(a,\lambda)$ for all $(a,\lambda)\in A_{1}$, so $\hat{\phi}_{i}\to\hat{\phi}$ in $\Omega(A)\setminus\{\hat{0}\}$.

But I dont know how to conclude that $\phi\mapsto\hat{\phi}$ is a homeomorphism, i.e. that its inverse is also continuous. Any help would be greatly appreciated!

答: When you proved surjectivity, you already showed how to construct an explicit inverse to your map (i.e., the inverse is restriction to $A$). This map is obviously continuous.


问:
Show this function is injective
I want to show that the curve $x^a=y^b$ is irreducible if and only if $\gcd(a,b)=1$.

This is easy to prove if I consider the next function:

$$f:\mathbb{C}[x,y]/\langle x^a-y^b\rangle\rightarrow\mathbb{C}[t]$$ such that $x\rightarrow t^b$ and $y\rightarrow t^a$, and show that $f$ is injective if $a$ and $b$ are coprime.
The first implication in our initial problem is easy to do by contradiction and the binomial formula on $(x^{a/d})^d - (y^{b/d})^d$ with $d=\gcd(a,b)$.
The other implication I have to use the function $f$ and the statement above, and then see its kernel is trivial (because I supposed it is injective) and then play with that.
I have all the problem done except the part that $f$ is injective if $a$ and $b$ are coprime, I can’t find the solution.

答: I claim that for any $p\in \Bbb C[x,y]/(x^a-y^b)$, we can write $p$ as $\sum_{i=0}^{a-1} x^ip_i(y)$: just use the relation $x^a=y^b$ to write $x^{a+i}=x^iy^b$ as many times as necessary. Now suppose that $f(\sum_{i=0}^{a-1} x^ip_i(y))=0$: substituting in $x=t^b$ and $y=t^a$, we get that $\sum_{i=0}^{a-1} t^{ib}p_i(t^a)=0$. Now look at the coefficients of each monomial term: since $\{0,b,2b,\dots,(a-1)b\}$ form a complete set of residue classes modulo $a$, we can't have any terms in common between $t^{ib}p_i(t^a)$ and $t^{jb}p_j(t^a)$ if $j\neq i$ (if we did, we'd have that $ib+na=jb+na$, or that $ib=jb$ modulo $a$). This implies that $p$ was zero to begin with, or that $f$ is injective.

Title: Show this function is injective


问:
Derivative of Trig Functions (Intuition Help?)
Looking for some intuition help here.

I have the following exercise and these are the steps I take:
$$
y = \sin\left(\frac{1}{x}\right)
$$

$$
u=\frac{1}{x}
$$

$$
y = \sin u,\;\;\frac{dy}{du} = \cos u= \cos\left(\frac{1}{x}\right)
$$

$$
u=x^{-1};\;\frac{du}{dx} =-x^{-2}=-\frac{1}{x^2}
$$

$$
\frac{dy}{dx}=\frac{dy}{du}\times\frac{du}{dx}=cos\frac{1}{x}\times-\frac{1}{x^2}
$$
This is incorrect but intuitively I want to multiply it this way.
$$
cos\frac{1}{x}\times-\frac{1}{x^2}=cos-\frac{1}{x^3}
$$

But the correct answer is:
$$
-\frac{cos-\frac{1}{x}}{x^2}
$$

Help me absorb the why so I can intuitively solve problems like these.

答:
First, you are taking short cuts when you are writing your equations that simply confuse things.  Also, LaTeX has \sin and \cos. Start over.

$$
\begin{aligned}
\text{Let}\quad f(x) &= \sin\frac{1}{x}.\\
\text{Let}\quad u &= \frac{1}{x}.\\
f(x) &= \sin u\\
\frac{df}{dx} &= \frac{df}{du} \frac{du}{dx} \quad\text{by the Chain rule}\\
\frac{df}{du} &= \cos u\\
\frac{du}{dx} &= -\frac{1}{x^2} \\
\frac{df}{dx} &= \frac{df}{du} \frac{du}{dx} \\
&= \cos u \cdot [-\frac{1}{x^2}] \\
&=  \cos \frac{1}{x} \cdot [-\frac{1}{x^2}]\\
&=  -\frac{1}{x^2} \cos \frac{1}{x}.
\end{aligned}
$$

By the way, you are panicking and falling into the "Universal Distributive Law of Freshmen"; you think that all operations distribute over each other, and all operations commute.  This is what has caused legions of students to replace $\sqrt{1+x^2}$ with $1 + x$.  You wrote:

$$
\cos\frac1x \cdot \frac{1}{x^2} = \cos\frac{1}{x^3}.
$$

Even though $\cos x$ looks like a product, it isn't.  Had you been using a programming language, you'd see that COS(X)*Y is not the same as COS(X*Y).  Try any of these cases with actual numbers and a pocket calculator; you'll see what I mean.

When you panic like that, always take a deep breath and start over.  And, never write a function next to its derivative and equate them.


答:
The inverse is given by
$$\Omega(A_1) \setminus\{\hat{0\}}\to \Omega(A): \chi \mapsto \chi\vert_A$$

Clearly this is weak$^*$-continuous, because if $\chi_i\to \chi$ pointwise on $A_1$, then also on $A$ since $A$ embeds in $A_1$.

Title: Relation between the spectrum of a Banach algebra and the spectrum of its unitization.

Body:
Let $A$ be any Banach algebra. We can always consider its unitization $A_{1}:=A\oplus\mathbb{C}$, even if $A$ is already unital. The multiplication on $A_{1}$ is defined as $$(a,\lambda)(b,\mu):=(ab+\lambda a+\mu b,\lambda\mu).$$ One can easily prove that a character $\phi$ on a unital algebra $A$ satisfies $\phi(1)=1$. The spectrum $\Omega(A)$ of $A$ is defined as the set of all non-zero homomorphisms $A\to\mathbb{C}$. We endow $\Omega(A)$ with the topology of pointwise convergence.

Note that $\hat{0}\colon A_{1}\to\mathbb{C}$, $(a,\lambda)\mapsto\lambda$ is a  character on $A_{1}$.

I am trying to prove that $\Omega(A)$ is homeomorphic to $\Omega(A_{1})\setminus\{\hat{0}\}$. Also, I want to prove this directly from the definitions, i.e. without using (local) compactness or Hausdorffness.

Here is what I tried:

For $\phi\in\Omega(A)$ I define $\hat{\phi}\colon A_{1}\to\mathbb{C}$ by $\hat{\phi}(a,\lambda):=\phi(a)+\lambda$. Now I tried to prove that $\phi\mapsto\hat{\phi}$ is a homeomorphism.

Injectivity: Assume that $\hat{\phi}=\hat{\psi}$. Then $$\phi(a)=\hat{\phi}(a,0)=\hat{\psi}(a,0)=\psi(a)$$ for all $a\in A$, thus $\phi=\psi$.

Surjectivity: Let $\alpha\in\Omega(A)\setminus\{\hat{0}\}$ be given. Define $\phi\colon A\to\mathbb{C}$ by $\phi(a):=\alpha(a,0)$. Then $\phi\neq0$, otherwise $\hat{\phi}=\hat{0}$. Also, it is easy to see that $\phi$ is a homomorphism, thus $\phi\in\Omega(A)$. Now observe that $$\hat{\phi}(a,\lambda)=\phi(a)+\lambda=\alpha(a,0)+\lambda\alpha(0,1)=\alpha(a,\lambda).$$

Continuity: Let $(\phi_{i})$ be a net in $\Omega(A)$ that converges to a character $\phi\in\Omega(A)$. Since $\phi_{i}(a)\to\phi(a)$ for all $a\in A$, it is easy to see that $\hat{\phi}_{i}(a,\lambda)\to\hat{\phi}(a,\lambda)$ for all $(a,\lambda)\in A_{1}$, so $\hat{\phi}_{i}\to\hat{\phi}$ in $\Omega(A)\setminus\{\hat{0}\}$.

But I dont know how to conclude that $\phi\mapsto\hat{\phi}$ is a homeomorphism, i.e. that its inverse is also continuous. Any help would be greatly appreciated!


问:
drawing balls from a bin but the balls are changing over time
I am doing this problem:

An urn initially has w white balls and k black balls. The balls are randomly drawn and replaced in the bin. Whenever a white ball is selected, it is painted black before it is replaced. What is the probability that the ball selected on the (n+1)st draw is white?

What I am thinking is to conditional on there are j times that a white ball is selected in the first n draws. So

$$P(the\ (n+1)st\ draw\ is\ white) = \sum_{j=0}^{n}P(the\ (n+1)st\ draw\ is\ white\ |\ j\ whites\ in\ the\ first\ n\ draws)P(j\ whites\ in\ the\ first\ n\ draws) $$
And I know

$$P(the\ (n+1)st\ draw\ is\ white\ |\ j\ whites\ in\ the\ first\ n\ draws) = \frac{w-j}{k+w}$$

But I am wondering how to calculate the probability that there are j whites in the first n draws. I think it should be $\frac{w(w-1)...(w-j+1)}{(w+k)^j}$, but I am not sure if it is correct.

Can anyone give me a hint? Thank you.

答:
let $q:=\frac{1}{w+k}$
$$\#\text{whites at n-th draw}= x, \\
p(\text{white at n-th draw}) = xq,
\\ \#\text{whites at n-th+1 draw}= x - 1 * p(\text{white at n-th draw}) = x(1-q), \\
p(\text{white at n-th+1 draw}) = \#\text{whites at n-th+1 draw}* q = (x - p(\text{white at n-th draw}))*q = xq(1-q) = p(\text{white at n-th draw})(1-q)$$
so from now we have:
$$p(\text{white at 1 draw}) = wq, \\
p(\text{white at n+1 draw}) = wq(1-q)^n, $$


问:
Complex log problem. How do I find the answer?
I have this problem and I'm trying to solve and find all values:

$$(1+i)^{2-i}$$

So far I have

$$1 + i = \sqrt{2} e^{i(\frac{\pi}{4})}$$ I got the above by converting 1+i into polar form. Also:

$$1+i = e^{ln(1+i)}$$

$$(1+i)^{2-i} = e^{(2-i)*ln(1+i)}$$

Now what?

答:
Let $w=1+i$ and $z=2-i$. Then, as you noted, for any choice of log (there are countably many), we have $w^z=e^{z\log w}$. As you noted, in polar coordinates, $w=\sqrt{2}e^{i\frac{\pi}{4}}$. Therefore, the set of logarithms is the family
\begin{align}
\log w&=\ln r + i(Arg(w)+2k\pi), \quad k\in \mathbb{Z}\\
&=\ln (\sqrt{2}) + i\left(\frac{\pi}{4}+2k\pi \right), \quad k\in \mathbb{Z}
\end{align}
In general, taking complex bases to complex powers yields varying results. The point of the exercise is to see how many truly distinct values of $w^z$ are possible.

For example, suppose $z=n$, for some positive integer $n$. Then
\begin{align}
e^{z\log w}&=e^{n\left[\ln (\sqrt{2}) + i\left(\frac{\pi}{4}+2k\pi \right)\right]} \\
&=e^{n\left[\ln (\sqrt{2}) + i\frac{\pi}{4}\right]}\underbrace{e^{2kn\pi i}}_{=1}
\end{align}
so even though we can choose any log, there is always one output.

In your case, $z=2-i$. Multiply that through the logarithm set and see if any of the terms of the form $e$ to an even integer times $\pi$ times $i$ pop out.


问:
Suppose |$A$| = |$B$| and let $f: A \to B$ be a surjective function. Can we/ should we define the ratio of size of set $A$ : set $B$ wrt function $f$?
I've not seen this idea anywhere before, but I'm not a set theory buff either.

Let $A$ and $B$ be sets with the same cardinality and let $f: A \to B$ be a surjective function. Suppose, for every $y \in B$, there are exactly $n \in \mathbb{N}$ members of $A$ such that $f$ maps these members of $A$ to $y$.

Then define the Rubinson-ratio of the size of set $A$ : set $B$ wrt function $f$ to be $n:1$.

...and perhaps we may say, wrt function $f$, $A$ is $n$ times Rubinson-larger than $B$.

We can probably extend this idea in a few ways, but let's leave it as is for now.

Is this a known idea in set theory? Else, may I claim the name for it?

答:
This idea comes up frequently in combinatorics. It is called a $k$-to-$1$ correspondence. It allows you to relate the cardinality of the domain and range via the $k$ factor, which can allow for counting one of the sets more conveniently than it would otherwise be possible directly. Also, I don't think that people can "claim" names like that. One can be the first discoverer, but names must be accepted by the relevant community. On a more humorous note, see Stigler's law of eponomy.

As requested by the OP in the comments, here is a link to a reference that uses this term in the finite setting. (However, a k-to-1 correspondence is also applicable to an infinite setting where the two sets $A$ and $B$ share the same cardinality $\aleph_i$)


问:
Prove $S_n/s_n\rightarrow N(0,1)$ in distribution
Given $B_1, B_2,\ldots$ are independent and bounded variables with $E(B_i) = 0$ for all $i=1,2,\ldots$. Define $S_n = B_1+ B_2+\ldots + B_n$ with variance $s_n^2\rightarrow \infty$. Prove that $\frac{S_n}{s_n}$ has a central limit.

My attempt: Due to the given condition, without i.i.d property, I try to prove that this sequence satisfies the Lindeberg condition and then applying Lindeberg-Feller theorem, we're done. So for every $\epsilon>0$ , I need to show for each positive $\varepsilon$,
$$\lim_{n\to +\infty}   \frac 1{\sum_{j=1}^n\sigma_j^2 }\sum_{j=1}^n\mathbb E\left[B_j^2\mathbf 1\left\{ \left|B_j\right|^2\gt \varepsilon\sum_{i=1}^n   \sigma_i^2\right\}\right]=0 .$$

Since $\sigma_i^{2} = E(B_i^2)$ for all $\ i=1,2,\ldots$ and $B_is$ are bounded variables, $B_j^2\leq M=max(|B_1|,|B_2|,\ldots)$. Thus, $$\lim_{n\to +\infty}   \frac 1{\sum_{j=1}^n\sigma_j^2 }\sum_{j=1}^n\mathbb E\left[B_j^2\mathbf 1\left\{ \left|B_j\right|^2\gt \varepsilon\sum_{i=1}^n   \sigma_i^2\right\}\right]\leq \lim_{n\to +\infty} \frac 1{\sum_{j=1}^n\sigma_j^2 } M^2\sum_{j=1}^n \mathbb E\left[\mathbf 1\left\{ \left|B_j\right|^2\gt \varepsilon\sum_{i=1}^n   \sigma_i^2\right\}\right] .$$ As $n\rightarrow \infty$, $s_n^2\rightarrow \infty$, so all the sets $\mathbf 1\left\{ \left|B_j\right|^2\gt \varepsilon\sum_{i=1}^n \sigma_i^2\right\}\rightarrow 0$. And we would be done if we could show that $\lim_{n\rightarrow \infty} \sum_{j=1}^n \mathbb E\left[\mathbf 1\left\{ \left|B_j\right|^2\gt \varepsilon\sum_{i=1}^n   \sigma_i^2\right\}\right]\rightarrow 0$. But this might not be true (counter example is the harmonic series with $p=1$) unless there is something that I was missing.

My question: Could someone please help me overcome this last step? In case I was on the wrong track, please let me know as well.

答:
Since each $X_k$ is bounded then for any $\epsilon>0$ there exists $N\in\mathbb N$ such that

$$\mathbf{1}\left(|X_k|>\epsilon s_n\right)=0$$

for all $k$ and all $n\geq N$. Hence

$$\lim_{n\rightarrow\infty}\sum_{k=1}^n\mathbb E\left[\mathbf{1}\left(|X_k|>\epsilon s_n\right)\right]=0$$

Thus taking $M$ as the bound for each $|X_k|$ we have

\begin{align*}
\lim_{n\rightarrow\infty}\frac{1}{s_n^2}\sum_{k=1}^n\int_{|X_k|>\epsilon s_n}X_k^2dP
&=\lim_{n\rightarrow\infty}\frac{1}{s_n^2}\sum_{k=1}^n\mathbb E\left[X_k^2\mathbf{1}\left(|X_k|>\epsilon s_n\right)\right]\\\\
&\leq\lim_{n\rightarrow\infty}\underbrace{\frac{1}{s_n^2}}_{\rightarrow0}M^2\underbrace{\sum_{k=1}^n\mathbb E\left[\mathbf{1}\left(|X_k|>\epsilon s_n\right)\right]}_{\rightarrow0}\\\\
&\rightarrow 0
\end{align*}

Therefore the Lindeberg condition is met and we have that $\frac{S_n}{s_n}$ has a central limit.


问:
Prove that $g(r, \alpha, \beta) = (r\cos(\alpha)\cos(\beta), r\sin(\alpha)\cos(\beta), r\sin(\beta))$ is an injection.
Prove that

$g: (0,∞)×(0,2π)×(−π/2,π/2)→R^3,$

$g(r, \alpha, \beta) = (r\cos(\alpha)\cos(\beta), r\sin(\alpha)\cos(\beta), r\sin(\beta))$

is an injection.

Exactly what it says on the tin. I haven't gone beyond assuming $g(a,b,c) = g(c,d,e)$. After constructing the relevant system of equations I am stuck trying to equate the variables from there. I might have forgotten too much from trigonometry. This is homework, so I would appreciate if someone could walk me through the process, but not the solution.

Body:
hint

Put
$$g(a,b,c)=(A,B,C)$$
$$g(d,e,f)=(D,E,F)$$

then

$$g(a,b,c)=g(d,e,f) \implies$$

$$(A,B,C)=(D,E,F)\implies$$

$$A^2+B^2+C^2=D^2+E^2+F^2\implies$$
$$a^2=d^2\implies a=d$$


答:
As mentioned in the comments:

In order for draw number $(n+1)$  to be white it must first have been white originally, probability $\frac w{b+w}$ and, secondly, it must never have been drawn before, probability  $\left(\frac {b+w-1}{b+w}\right)^n$.  These two events are independent (being drawn, or not, in the past has nothing to do with being drawn on the current round).  Thus the answer is the product $$\boxed {\frac w{b+w}\times \left(\frac {b+w-1}{b+w}\right)^n}$$

Note:  I have used $w$ for the initial number of white balls and $b$ for the initial number of black balls.

Title: drawing balls from a bin but the balls are changing over time

Body:
I am doing this problem:

An urn initially has w white balls and k black balls. The balls are randomly drawn and replaced in the bin. Whenever a white ball is selected, it is painted black before it is replaced. What is the probability that the ball selected on the (n+1)st draw is white?

What I am thinking is to conditional on there are j times that a white ball is selected in the first n draws. So

$$P(the\ (n+1)st\ draw\ is\ white) = \sum_{j=0}^{n}P(the\ (n+1)st\ draw\ is\ white\ |\ j\ whites\ in\ the\ first\ n\ draws)P(j\ whites\ in\ the\ first\ n\ draws) $$
And I know

$$P(the\ (n+1)st\ draw\ is\ white\ |\ j\ whites\ in\ the\ first\ n\ draws) = \frac{w-j}{k+w}$$

But I am wondering how to calculate the probability that there are j whites in the first n draws. I think it should be $\frac{w(w-1)...(w-j+1)}{(w+k)^j}$, but I am not sure if it is correct.

Can anyone give me a hint? Thank you.


问:
Preservation of completeness under isomorphisms between Boolean algebras

A Boolean algebra $\mathcal{B}:=(B,\leq,\lor,\land,^c,0,1)$ is said to be complete if every non-empty subset of $B$ has a greatest lower bound (g.l.b). Prove that a Boolean algebra that is isomorphic to a complete Boolean algebra is complete.

Let us start with a complete Boolean algebra, $\mathcal{B}:=(B,\leq,\lor,\land,^c,0,1)$ which is isomorphic to $\mathcal{B'}:=(B',\leq,\lor,\land,^c,0,1)$. Let $h:B\to B'$ be an isomorphism between the two.

For a non-empty subset $X' \subseteq B'$, $\exists X\in B$ such that $h(X)=B$. Is this correct, or does it need more justification? I think this is fine because: for every element $x' \in X'$ we have $x \in X$ such that $h(x) = x'$ since $h$ is an isomorphism. We can construct $X$ such that $h(X)= X'$ by $X = \{x\in B:x = h^{-1}(x') \forall x' \in X'\}$.

Since $\mathcal{B}$ is complete, we know that $X$ has a g.l.b, say $x_g \in X$. This means that for all $x\in X$, $x_g\leq x$. Since isomorphisms between Boolean algebras preserve partial orders, it follows that $h(x_g)\leq h(x)$ for all $x\in X$. Since $X'=\{x'\in B':x'=h(x) \forall x\in X\}$, $h(x_g)$ is in fact the g.l.b of $X'$. Is this fine?

This works for all $X'\subseteq B'$, hence $\mathcal{B}'$ is complete.

I'm new to proof-writing, and I tend to make mistakes or leave out things that need to be stated in order to complete a proof. Hence, it would be helpful if someone could just go through this and let me know if the aforementioned steps are complete, or some gaps need to be filled in? I'd be happy to know about ways to make this proof more rigorous.

答: What you’ve done is correct, but the fact that $h(x_g)\le x'$ for each $x'\in X'$ isn’t enough to show that $h(x_g)=\bigwedge X'$: it shows that $h(x_g)$ is a lower bound for $X'$, but it doesn’t show that $h(x_g)$ is the greatest lower bound for $X'$. For that you need to verify that $b'\le h(x_g)$ whenever $b'$ is a lower bound for $X'$. But that’s straightforward: if $b'\le x'$ for each $x'\in X'$, then $h^{-1}(b')\le h^{-1}(x')$ for each $x'\in X'$, so $h^{-1}(b')\le x_g$, and therefore $b'\le h(x_g)$.


问:
For every integer a, b, c, if 3c is not divisible by a, then b is not divisible by a or 3c is not divisible by b
I can't seem to figure this proof out. How would I prove this by contradiction and contraposition?

I tried doing it by saying $3c=bk=(ak)k=a(kk)$
since $b=ak$, $3c=bk$ for some interger $k$

$3c=a(kk)$ contradicts "3c is not divisible by a"
contraposition would follow suit since we prove that $3c=a(kk)$

What am I doing wrong?

答: If b is divisible by a and 3c is divisible by b then 3c is divisible by a, a contradiction . Thus one of them must be true.</p>


答: Hint : The contraposition of this proposition is "if $b$ is dividible by $a$ and $3c$ is dividible by $b$, then $3c$ is dividible by $a$".

Title: For every integer a, b, c, if 3c is not divisible by a, then b is not divisible by a or 3c is not divisible by b

Body:
I can't seem to figure this proof out. How would I prove this by contradiction and contraposition?

I tried doing it by saying $3c=bk=(ak)k=a(kk)$
since $b=ak$, $3c=bk$ for some interger $k$

$3c=a(kk)$ contradicts "3c is not divisible by a"
contraposition would follow suit since we prove that $3c=a(kk)$

What am I doing wrong?


问:
Failure of exchange of integral and limit of function $f_n(x)=ne^{-nx}$
Suppose you have a sequence of functions indexed by $n\in\mathbb{N}$, for $x\in[0,1]$, defined by $f_n(x)=ne^{-nx}$. Is it true that
$$\lim_{n\to\infty}\int_0^1f_n(x)dx= \int_0^1\lim_{n\to\infty}f_n(x)dx.$$

The answer is no. I post the question as a teaching example for people unsure of what could go wrong, especially because $f_n(x)$ is an expression that is fairly common in econometrics/economics and I've seen people just recklessly exchange limit and integral.

答:
The answer is that it cannot be exchanged because $f_n$ is neither monotonic (decreasing or increasing) nor there exists a dominant function such that $|f_n(x)|\leq g(x)$ for all $n$ and $x$ ($\{f_n\}_{n\in\mathbb{N}}$ is not uniformly integrable, either). To see it:

Monotonicity: One could be tempted to say that, for each $x\in[0,1)$ ($x=1$ is of measure zero and thus irrelevant), there is $n_x$ such that if $n\geq n_x$, $0\leq f_n(x)<f_{n+1}(x)$, and then try to apply this version of Monotone Convergence Theorem on the tail of the sequence. However, $n_x$ grows unboundedly in $x$, namely, as $x\to 1$, $n_x\to\infty$, which means that there is no $n^*$ such that if $n\geq n^*$, $0\leq f_n(x)<f_{n+1}(x)$ for every $x\in[0,1)$. So, no hope of using MCT.

Dominance: no hope of finding $g$ such that $|f_n(x)|\leq g(x)$ for all $n$ and $x$, because $f_n(0)\to\infty$ as $n\to \infty$. So no hope of using DCT either.

What happens is that we have $\int_0^1f_n(x)dx=1-e^{-n}$ and $\lim_{n\to\infty}f_n(x)=0$, so

$$\lim_{n\to\infty}\int_0^1f_n(x)dx=1\neq 0=\int_0^1\lim_{n\to\infty}f_n(x)dx.$$


问:
Ambiguity in a grammar and getting rid of it
I'm trying to figure out how I can change my grammar to get rid of the ambiguity but I'm having a hard time figuring it out, unfortunately I haven't been able to find an example that's close enough to my problem either.

The grammar:

$S\rightarrow A|C$

$A\rightarrow aaaB$

$B \rightarrow bBc | bc$

$C\rightarrow Dccc$

$D\rightarrow aDb|ab$

I believe this grammar is ambiguous because I can take two different paths and still get the string w='aaabbbccc'.

I'm not too sure where to start here, the grammar above is a representation of the union of two different grammars represented by the following

{$aaab^nc^n$ | $n>=1$} U {$a^mb^mccc$ | $m>=1$}

I don't have any work to show so far as I'm still trying to figure out the right course of action for getting rid of the ambiguity.

答:
HINT: There are probably better ways, but since the word $a^3b^3c^3$ is the only one that’s problematic, one very simple way is to use the same basic idea to generate the language

$$\{a^3b^nc^n:n\ge 4\}\cup\{a^nb^nc^3:n\ge 4\}$$

and to generate each of the $5$ words $a^3b^nc^n$ and $a^nb^nc^3$ with $n\in\{1,2,3\}$ individually, directly from $S$.


答:
let's talk about the Rubinson ratio!

First of all, just to answer your question directly, I have never heard of a name for this concept in the context of infinite sets, although there are a few related concepts that I'll get into shortly.

But I've also never seen this concept used (other than as a part of the related concepts). There's an saying that "a mathematical object is what it does." What does this concept do? What sorts of proofs can we construct with it? You can call this the "Rubinson ratio," and you can say that that's the name of it, but until and unless you do something with it, it's not really yours. The person who gets to name a concept generally isn't the first person who comes up with it, but the first person to do something with it.

In my (albeit limited) encounters with set theory, it's been very rare that I'm using a function like you describe, between two sets of equal cardinality, where the size of the preimage of each element in the range is equal. However, I think you're essentially trying to define a way to think about the "size" of one infinite set compared to another, and in that case we do have a few examples:

First, if we have a subset $A\subseteq\mathbb{N}$ of the natural numbers, we can define the Natural density of $A$ as follows: First, we define $f_A(n)=\big|A\cap \mathbb{N}^{\leq n}\big|$ be the number of elements of $A$ that are $\leq n$. Then the natural density $d(A)=\lim_{n\to\infty}\frac{f_A(n)}n$. This works similarly to how you would expect: the density of the even numbers is $\frac12$, as is the density of the odd numbers. The density of the multiples of $3$ is $\frac13$, etc. The primes, the squares, and all finite sets have natural density $0$.

There's also a related concept in group theory: If we have a group $G$ and a subgroup $H$, then we can define $[G:H]$ (read as "G over H") to be the number of left cosets of $H$ in $G$. When $G$ and $H$ are finite, $[G:H]=\frac{|G|}{|H|}$. We also have $[\mathbb{Z}:2\mathbb{Z}]=2$ and more generally $[\mathbb{Z}:n\mathbb{Z}]=n$. In general, the function $f:G\to (G/H)$ which maps an element of $G$ to the coset to which it belongs maps exactly $|H|$ elements of $G$ to each coset.

There are other concepts like this in other fields (@Favst gives a good example from combinatorics), but I'm not aware of a general term that applies across fields of mathematics. If you ever publish a paper that uses this concept, try naming it the Rubinson ratio! Maybe it'll be in a textbook someday.

Title: Suppose |$A$| = |$B$| and let $f: A \to B$ be a surjective function. Can we/ should we define the ratio of size of set $A$ : set $B$ wrt function $f$?

Body:
I've not seen this idea anywhere before, but I'm not a set theory buff either.

Let $A$ and $B$ be sets with the same cardinality and let $f: A \to B$ be a surjective function. Suppose, for every $y \in B$, there are exactly $n \in \mathbb{N}$ members of $A$ such that $f$ maps these members of $A$ to $y$.

Then define the Rubinson-ratio of the size of set $A$ : set $B$ wrt function $f$ to be $n:1$.

...and perhaps we may say, wrt function $f$, $A$ is $n$ times Rubinson-larger than $B$.

We can probably extend this idea in a few ways, but let's leave it as is for now.

Is this a known idea in set theory? Else, may I claim the name for it?


问:
Prove that $u(x)=\log\log\left(1+\frac{1}{\vert x\vert}\right)\in H^1(B(0,1))$
Problem: Let $\Omega\subset\mathbb R^2$ denote the open unit ball in $\mathbb R^2$. Prove that the unbounded function
$$f(x)=\log\log\left(1+\frac{1}{\vert x\vert}\right)$$
belongs to $H^1(\Omega).$

My Attempt: Let $\{\varepsilon_n\}_{n=1}^\infty\subset[0,1]$ such that $\varepsilon_n\searrow0$ as $n\to\infty$. Put $\Omega_n=B(0,\varepsilon_n)$. Define the sequence of functions
$$f_n(x)=\begin{cases}f(x)&\text{if }x\in\Omega\setminus\Omega_n\\0&\text{otherwise.}\end{cases}$$
Note that $\vert f_n\vert^2\nearrow\vert f\vert^2$ as $n\to\infty$, so by the monotone convergence theorem we have $\|f_n\|_{L^2(\Omega)}^2\to\|f\|_{L^2(\Omega)}^2$ as $n\to\infty$. Using integration in polar coordinates, as shown in Folland's Real Analysis text, we have that
\begin{align*}
\|f_n\|_{L^2(\Omega)}^2
&=\int_{\Omega\setminus\Omega_n} \vert f(x)\vert^2\,dx=\int_{\Omega\setminus\Omega_n}\left\vert\log\log\left(1+\frac{1}{\vert x\vert}\right)\right\vert^2\,dx\\
&=2\pi\int_{\varepsilon_n}^1 r\left\vert\log\log\left(1+\frac1r\right)\right\vert^2\,dr\\
&\leq2\pi\int_{\varepsilon_n}^1 e^r\,dr\\
&\leq2\pi\int_0^1e^r\,dr\\
&=2\pi e\\
&<\infty.
\end{align*}
Since the bound above does not depend on $n$, letting $n\to\infty$ shows that $f\in L^2(\Omega)$, by the monotone convergence theorem.

Next, observe that
$$\nabla f(x)=\left(-\frac{x_1}{\log\left(1+\frac{1}{\vert x\vert}\right)(1+\vert x\vert)\vert x\vert^2},-\frac{x_2}{\log\left(1+\frac{1}{\vert x\vert}\right)(1+\vert x\vert)\vert x\vert^2}\right),$$
so that
$$\vert\nabla f(x)\vert^2=\frac{1}{\log\left(1+\frac1{\vert x\vert}\right)^2(1+\vert x\vert)^2\vert x\vert^2}.$$
Using the same method as above we have that $\|\nabla f_n\|_{L^2(\Omega)}^2\to\|\nabla f\|_{L^2(\Omega)}^2$ by the monotone convergence theorem. Then, integrating in polar coordinates once again, we have
\begin{align*}
\|\nabla f_n\|_{L^2(\Omega)}^2
&=\int_{\Omega\setminus\Omega_n}\vert\nabla f(x)\vert^2\,dx=\int_{\Omega\setminus\Omega_n}\frac{1}{\log\left(1+\frac1{\vert x\vert}\right)^2(1+\vert x\vert)^2\vert x\vert^2}\,dx\\
&=2\pi\int_{\varepsilon_n}^1\frac{1}{\log\left(1+\frac1{r}\right)^2(1+r)^2r^2}\,dr\\
&\to\infty\quad\text{as }n\to\infty.
\end{align*}
It follows that $f\notin H^1(\Omega)$ since $\vert\nabla f\vert\notin L^2(\Omega)$.

Do you agree with my proof above? I am not sure that I fully understood and applied the definition of the Sobolev Space $H^1(\Omega)$, especially in the second part of the proof. Any clarification if I am in the wrong would be much appreciated.

Thank you for your time and valuable feedback.

答:
In the computation of $\int_{\Omega}|\nabla f|^2 \,dx$ you forgot $r\,dr$.

Hence the integral converges since $\frac{1}{\log(1+1/r)^2 r(1+r)^2}$ is integrable close to $r=0$ and we get $f\in H^1(\Omega)$.


问:
Find the coefficient of $x^n$ in the generating functions: $g(x) = \frac{x^3}{(1+x)^5 (1−x)^6}$
One of the problems in my Discrete Math course states that we need to find the coefficient of $x^n$ in generating function $g(x) = \frac{x^3} {(1+x)^5 (1−x)^6}$

I separated $g(x) = \frac{x^3} {(1+x)^5 (1−x)^6}$ into $x^3$ * $\frac{1}{(1+x)^5}$ * $\frac{1}{(1-x)^6}$

Then got $f(x) = x^3 * \sum_{k=0}^\infty \binom{n+4}{4}(-1)^kx^k * \sum_{k=0}^\infty \binom{n+5}{5}x^k$

I dont know what to do from here, can you tell me what is the next step?

答:
$$
\begin{align}
\frac{x^3}{(1+x)^5 (1−x)^6}
&=\frac{x^3+x^4}{\left(1-x^2\right)^6}\tag1\\
&=\left(x^3+x^4\right)\sum_{k=0}^\infty(-1)^k\binom{-6}{k}x^{2k}\tag2\\
&=\left(x^3+x^4\right)\sum_{k=0}^\infty\binom{k+5}{5}x^{2k}\tag3\\
&=\sum_{k=0}^\infty\binom{k+5}{5}\left(x^{2k+3}+x^{2k+4}\right)\tag4\\
&=\sum_{k=2}^\infty\binom{k+3}{5}\left(x^{2k-1}+x^{2k}\right)\tag5
\end{align}
$$
Explanation:

$(1)$: multiply by $\frac{1+x}{1+x}$

$(2)$: use the series for $(1+x)^{-6}$

$(3)$: $\binom{-6}{k}=(-1)^k\binom{k+5}{5}$ when $k\ge0$

$\phantom{\text{(3):}}$ (negative binomial coefficients)

$(4)$: distribute the $x^3+x^4$

$(5)$: substitute $k\mapsto k-2$

Thus, if $n=2k-1$ or $n=2k$, then the coefficient of $x^n$ is $\binom{k+3}{5}$.

This can be stated as the coefficient of $x^n$ is $\binom{\left\lfloor\frac{n+7}2\right\rfloor}{5}$


问:
How can you compute a set of extensions up to isomorphism from Ext?
Given $A$ and $B$ in an abelian $\mathbf{k}$-linear category, let $\mathrm{E_{A,B}}$ denote the set of all isomorphism classes of objects that fit in the middle of a short exact sequence $B \hookrightarrow \_\_ \twoheadrightarrow A$. Can we explicitly describe the set $\mathrm{E_{A,B}}$ if we know $\mathrm{Ext}^1_\mathbf{k}(A,B)$? I'm hoping that if $\{X_1, \dotsc, X_n\}$ is a basis for $\mathrm{Ext}^1_\mathbf{k}(A,B)$ as a $\mathbf{k}$-vector space, then we can say
$$
\mathrm{E_{A,B}} \;\;=\;\; \big\{[X_1], \dotsc, [X_n], [A\oplus B]\big\}\,,
$$
but this seems too hopeful.

答:
No.  As Roland commented, there is a simple counterexample in the category of chain complexes of $k$-vector spaces: let $B$ be $k^n$ concentrated in degree $1$ and $A$ be $k^m$ concentrated in degree $0$.  An extension of $A$ by $B$ is then just a chain complex of the form $0\to k^n\to k^m\to 0$ and so $\operatorname{Ext}^1(A,B)\cong \operatorname{Hom}(k^n,k^m)\cong k^{mn}$.  Up to isomorphism, though, such a chain complex is determined by the rank of the map $k^n\to k^m$, and so there are $\min(m,n)+1$ isomorphism classes.  Since $\min(m,n)+1$ is not determined by the product $mn$, the cardinality of $\mathrm{E_{A,B}}$ is not determined by $\operatorname{Ext}^1(A,B)$ up to isomorphism.

Here are some things you can say.  The automorphism groups of $A$ and $B$ each act on $\operatorname{Ext}^1(A,B)$ as isomorphisms of the middle objects, so $\mathrm{E_{A,B}}$ is no larger than the quotient of $\operatorname{Ext}^1(A,B)$ by these actions.  However, it may be even smaller, since there can be extensions whose middle objects are isomorphic but such an isomorphism cannot preserve the subobject $B$.  For instance, in the category of $k[x]$-modules, consider $A=B=C=(k[x]/(x))^{\oplus \mathbb{N}}\oplus (k[x]/(x^2))^{\oplus \mathbb{N}}$.  Then there are lots of short exact sequences $0\to B\to C\to A\to 0$ which have different images of $B\to C$ even up to automorphisms of $C$, since you can have different numbers of $k[x]/(x)$ summands that map into a $k[x]/(x^2)$ summand to form a nontrivial extension.

Note moreover that $\mathrm{E_{A,B}}$ can also be larger than your guess $\{[X_1], \dotsc, [X_n], [A\oplus B]\}$.  For instance, in the category of $k[x,y]$-modules, let $A=B=k[x,y]/(x,y)$.  Then $\operatorname{Ext}^1(A,B)\cong k^2$ is finite-dimensional, but if $k$ is infinite, then $\mathrm{E_{A,B}}$ is infinite.  Indeed, an element $(a,b)\in k^2$ corresponds to the extension with $k$-basis $\{e_1,e_2\}$ in which $xe_1=ae_2$ and $ye_1=be_2$ (and $x$ and $y$ both annihilate $e_2$).  For $(a,b)\neq (0,0)$, the annihilator of this module is the ideal generated by $bx-ay$.  In particular, such modules can only be isomorphic when their $(a,b)$'s are scalar multiples of each other.  So in this case, $\mathrm{E_{A,B}}$ is actually the projectivization of $\operatorname{Ext}^1(A,B)$ together with the trivial extension, which is larger than just a basis together with the trivial extension.

(Note that in general, $k^\times$ acts by automorphisms of $A$ and $B$ and this induces the scalar action of $k^\times$ on $\operatorname{Ext}^1(A,B)$, so $\mathrm{E_{A,B}}$ will always be no larger than the projectivization together with the trivial extension.)


问:
Can someone check if I'm finding marginal pmf correctly?
So I was given this question and chart:

My first thought is using the equation from the book.

to get the answer
$$P_Y(x)=\begin{cases}1/6 & x = 0,1,2,3,4,5\\0 & otherwise\end{cases}$$

but I also learned in class that P_X(x) is just summing up everything in the column, so I'm thinking that the other answer might also be
$$P_X(0)=6/21, P_X(1)=5/21, P_X(2)=4/21, P_X(3)=3/21, P_X(4)=2/21, P_X(5)=1/21$$

So which way is the correct way of doing it? Thanks in advance.

答:
You are not given that $Y$ is uniformly distributed so your first method does not work. It is $(X,Y)$ which is uniformly distributed over $21$ points of equal probability.

Indeed, the second method works and by symmetry, you can also conclude that $Y$ will be  distributed identically to $X$ i.e. will have the same PMF.


问:
Find the locus of $z$ such that $\arg \frac{z-z_1}{z-z_2} = \alpha$: confused when $\alpha = 0$.

Find the locus of $z$ such that $\arg \frac{z-z_1}{z-z_2} = \alpha$.
Use and draw $w = \frac{z-z_1}{z-z_2}$.

This exercise was discussed many times -- 1, 2, 3, 4 -- but I was unable to find answers to my problem with $0$ there.

I believe I understand where the arcs came from, here's my work:

If I understand correctly, for $\alpha = \pm\pi$, the locus would be the segment connecting $z_2$ and $z_1$, not including the points themselves.

I can not understand what is happening when $\alpha = 0$.

$\alpha = 0 = \arg \frac{z-z_1}{z-z_2} = \arg w \Longrightarrow \frac{z-z_1}{z-z_2} = k \in \mathbb{R}, \frac{z-z_1}{z-z_2} = k\frac{z-z_2}{z-z_2}.$

Solving this for $z$, $z = \frac{x_1-kx_2}{1-k} +i\frac{y_1-ky_2}{1-k}$. I am having trouble understanding the locus of this $z$. The textbook says it should be 'two segments with end points in $z_1$ and $z_2$, and one of this segments contains an infinitely distant point'. How to understand why is this answer right, and how to draw it? It seems the infinitely distant point matches $k=1$, but why should it lie in the 'direction' of the line passing through $z_1$ and $z_2$?

My class notes are messy. Why is $(0, 1)$ special on $w$ plane?

Thank you.

Body:
Let $z=x+iy;\;z_1=x_1+iy_1;\;z_2=x_2+iy_2$
$$\frac{z-z_1}{z-z_2}= \frac{x^2-x (x_1+x_2)+x_1 x_2+(y-y_1) (y-y_2)}{(x-x_2)^2+(y-y_2)^2}+i\frac{x (y_2-y_1)+x_1 (y-y_2)+x_2 (y_1-y)}{(x-x_2)^2+(y-y_2)^2}$$
$$\text{arg}\left(\frac{z-z_1}{z-z_2}\right)=\arctan\frac{{\frac{x (y_2-y_1)+x_1 (y-y_2)+x_2 (y_1-y)}{(x-x_2)^2+(y-y_2)^2}}}{{\frac{x^2-x (x_1+x_2)+x_1 x_2+(y-y_1) (y-y_2)}{(x-x_2)^2+(y-y_2)^2}}}=\\=\arctan\frac{x (y_2-y_1)+x_1 (y-y_2)+x_2 (y_1-y)}{x^2-x (x_1+x_2)+x_1 x_2+(y-y_1) (y-y_2)}$$

$$\text{arg}\left(\frac{z-z_1}{z-z_2}\right)=0\to x (y_2-y_1)+x_1 (y-y_2)+x_2 (y_1-y)=0$$
Rearrange
$$x (y_2-y_1)+y (x_1-x_2)+x_2y_1-x_1y_2=0$$
which is the equation of a line.

In the general case, let $a=\cot\alpha$

$$x^2+y^2+x (-x_1-x_2+y_1 a -y_2 a )+y (-x_1 a +x_2 a -y_1-y_2)+x_1 x_2+x_1 y_2 a -x_2 y_1 a +y_1 y_2=0$$
$$x^2+y^2+px+qy+r=0$$
we get a circle.

Hope this can be useful.


答:
First consider $f(x,y,z)=x^2+y^2+z^2$ and observe that
$$
f(g(r,\alpha,\beta))=r^2\cos^2\alpha\cos^2\beta+r^2\sin^2\alpha\cos^2\beta+r^2\sin^2\beta=r^2
$$
Therefore if $g(r,\alpha,\beta)=g(s,\gamma,\delta)$, then applying $f$ to both sides yields $r^2=s^2$, hence $r=s$.

Now the equations are simpler:
$$
\begin{cases}
\cos\alpha\cos\beta=\cos\gamma\cos\delta \\[1ex]
\sin\alpha\cos\beta=\sin\gamma\cos\delta \\[1ex]
\sin\beta=\sin\delta
\end{cases}
$$
Square the first two equations and sum them, getting $\cos^2\beta=\cos^2\delta$. Since $\beta,\delta\in(-\pi/2,\pi/2)$, we obtain $\cos\beta=\cos\delta$ and now with the third equation we can conclude that $\beta=\delta$.

Finally, from $\cos\beta=\cos\delta\ne0$, we obtain $\cos\alpha=\cos\gamma$ and $\sin\alpha=\sin\gamma$, so $\alpha=\gamma$.

Title: Prove that $g(r, \alpha, \beta) = (r\cos(\alpha)\cos(\beta), r\sin(\alpha)\cos(\beta), r\sin(\beta))$ is an injection.

Body:
Prove that

$g: (0,∞)×(0,2π)×(−π/2,π/2)→R^3,$

$g(r, \alpha, \beta) = (r\cos(\alpha)\cos(\beta), r\sin(\alpha)\cos(\beta), r\sin(\beta))$

is an injection.

Exactly what it says on the tin. I haven't gone beyond assuming $g(a,b,c) = g(c,d,e)$. After constructing the relevant system of equations I am stuck trying to equate the variables from there. I might have forgotten too much from trigonometry. This is homework, so I would appreciate if someone could walk me through the process, but not the solution.


问:
Finding the $x^n$ coefficient of the power series $\sum\limits_{n=0}^\infty\frac{x^{2n+3}}{n!}$
I have a practice test question that asks:

Given the following Maclaurin series representation, $$\sum\limits_{n=0}^\infty\frac{x^{2n+3}}{n!}$$ what is the coefficient of $x^n$?

I have the correct solution, but I do not know how to derive it. How would I go about doing this?

答:
Remember, the $n$ in the sum $\sum_{n=0}^{\infty} \frac{x^{2n+3}}{n!}$ is a dummy variable; it doesn't matter what you call it.  The confusion is that you were given $n$ used in a different way.  So, change the index variable in the sum.

$$
\sum_{n=0}^{\infty} \frac{x^{2n+3}}{n!} = \sum_{k=0}^{\infty} \frac{x^{2k+3}}{k!}
$$

So, $x^n$ has a nonzero coefficient only if $n = 2k + 3$ for some $k$, which happens if $n$ is odd and $n\ge3$. And then, $k=\frac{n - 3}{2}$.


问:
How to solve this type of function
This is an image of the question being asked

On top of that this is the question shown below:

```<code>                f[n]=((0.45)^n)*u[n-3]
                g[n]=f[n-4]
</code>```

Solve for g[n]?

My question is can g[n] be represented as g[n] = f[n] + f[-4]
And then solved for, or am I off?

Title: How to solve this type of function

Body: $$g(n) = f(n-4) = 0,45^{n-4}. u(n-7)$$


问:
Prove that $\vdash (p \rightarrow q) \wedge (p \wedge q \rightarrow r) \rightarrow (p \rightarrow r)$.
I keep trying to prove the propositional logic theorem:

$$\vdash (p \rightarrow q) \wedge (p \wedge q \rightarrow r) \rightarrow (p \rightarrow r)$$

using the theorem of deduction and its reverse (and obviously using the axioms and modus ponens), but I get nowhere. I've tried every possibility that I could come up with, I'm starting to lose my mind. The fact that I have $2$ conjunctions within the formula ruins everything. So I know I need to use the reverse of the theorem of deduction to simplify things (so we successively move the left sides of implications to the left of the deduction symbol $\vdash$):

$$ (p \rightarrow q) \wedge (p \wedge q \rightarrow r) \vdash p \rightarrow r$$

$$(p \rightarrow q) \wedge (p \wedge q \rightarrow r), p \vdash r$$

And from here I keep getting stuck. I tried to transform that conjunction into an implication using:

$$U \wedge V \equiv \neg (\neg U \lor \neg V) \equiv \neg(U \rightarrow \neg V)$$

And I use this twice, for both conjunctions. But the formula gets so complicated that I can't see anything and there seems to be no place to use modus ponens on. And if I use the above on just one conjunction I still don't get anything that I can use.

答:
Use conditional proof inference twice. Statement $$ (p \rightarrow q) \wedge (p \wedge q \rightarrow r) \vdash p \rightarrow r$$

is equivalent to $$ (p \rightarrow q),(p \wedge q \rightarrow r) \vdash p \rightarrow r$$

and then $$(p \rightarrow q),(p \wedge q \rightarrow r),p \vdash r$$

which is obviously valid statment.


问:
Tell me which of $691$ and $697$ is prime (no calculator)
For two numbers, $691$ and $697$, tell me which (or both) is prime.

Without using calculator, what is the strategy for tackling this? I tried all the numbers up to 8 which are easy to test. How do you deal with better numbers?

答:
$697=625+72(\to 25^2+2×6^2)=49+648(\to 7^2+2×18^2)$.  Both are of the form $a^2+2b^2$, and this nonuniqueness implies proper prime factors $\in\{1,3\}\bmod 8$.  Actual factors are $17$ and $41$.

Alternatively, we can set out to prove that $691$ is prime.  As this number $\equiv 3\bmod 8$, it is certified prime if it has a unique representation as $a^2+2b^2$ and the squares in this representation are relatively prime.

To prove uniqueness efficiently we need to cut down on trials.  To effect this, first note that $a^2$ must be odd and can't end in $5$ ($b^2$ can't end in $3$ or $8$).  Then if $a^2$ ends in $1$ it must end in $41$ or $91$ because $b$ would be a multiple of $5$ and thus $2b^2$ would have to be a multiple of $50$.  We rule out an ending of $91$, however, because this fails $\bmod 4$.

That leaves only six values of $a^2$ ending with the proper digits and less than $691$, and we try those.  Thus $691$ equals each of the following:

$9+682(\to b^2=341)$

$49+642(\to b^2=321)$

$169+522(\to b^2=261)$

$289+402(\to b^2=201)$

$441+250(\to b^2=125)$

$\color{blue}{529+162(\to b^2=81,\therefore a=23,b=9)}$

The existence of a unique sum hits on our very last chance!


Title: Tell me which of $691$ and $697$ is prime (no calculator)

Body:
For two numbers, $691$ and $697$, tell me which (or both) is prime.

Without using calculator, what is the strategy for tackling this? I tried all the numbers up to 8 which are easy to test. How do you deal with better numbers?

Body:
The only nontrivial trial divisions can all be done uniformly easily mentally as follows:

$\bmod 13\!:\ \color{#c00}{40\!\equiv\! 1}\Rightarrow\,691={17}(\color{#c00}{40})\!+\!11\equiv\,\ 4\,(\color{#c00}1)+11\equiv 2,\,$ so $\ 691\!+\!6\equiv 8$

$\bmod 17\!:\ \color{#c00}{17\!\equiv\! 0}\Rightarrow\,691=\color{#c00}{17}(40)\!+\!11 \equiv\,\ \color{#c00}{0}\,(40)\!+\!11\!\equiv\! 11,\,$ so  $\ 691\!+\!6\color{#0a0}{\equiv 0}$

$\bmod 19\!:\ \color{#c00}{40\!\equiv\! 2}\Rightarrow\,691=17(\color{#c00}{40})\!+\!11\equiv\, (-2)\color{#c00}2\!+\!11\equiv 7,\,$ so $\ 691\!+\!6\equiv 11$

$\bmod 23\!:\ \color{#c00}{17,40\equiv -6}\  \Rightarrow \ \color{#c00}{17(40})\!+\!11\equiv \color{#c00}{-6(-6)}\!+\!11\!\equiv\! 1,\,$ so $\ 691\!+\!6\equiv 7$

More generally see the universal divisibility test, e.g. see here and  here.


问:
drawing balls from a bin but the balls are changing over time
I am doing this problem:

An urn initially has w white balls and k black balls. The balls are randomly drawn and replaced in the bin. Whenever a white ball is selected, it is painted black before it is replaced. What is the probability that the ball selected on the (n+1)st draw is white?

What I am thinking is to conditional on there are j times that a white ball is selected in the first n draws. So

$$P(the\ (n+1)st\ draw\ is\ white) = \sum_{j=0}^{n}P(the\ (n+1)st\ draw\ is\ white\ |\ j\ whites\ in\ the\ first\ n\ draws)P(j\ whites\ in\ the\ first\ n\ draws) $$
And I know

$$P(the\ (n+1)st\ draw\ is\ white\ |\ j\ whites\ in\ the\ first\ n\ draws) = \frac{w-j}{k+w}$$

But I am wondering how to calculate the probability that there are j whites in the first n draws. I think it should be $\frac{w(w-1)...(w-j+1)}{(w+k)^j}$, but I am not sure if it is correct.

Can anyone give me a hint? Thank you.

答:
Label the $w$ white balls with the numbers $1,2,...,w$.

Define $$_rA_q = \{\text{the $q$-th white ball is drawn at the $r$-th draw}\}$$
$$_rB_q = \{\text{the $q$-th white ball is drawn at a previous draw than the $r$-th}\}$$

So that

$$P(\text{$r$-th draw is white}) = $$
$$P(\ _{r}A_1\cup  \  _{r}A_2 \cup \dots \cup \ \ _{r}A_w) = $$
$$P(_rA_1)+P(_rA_2)+\dots +P(_rA_w) = w P(_rA_1)$$

but

$$P(_rA_1) = P(_rA_1|_rB_1) P(_rB_1) +P(_rA_1|_rB_1^c)P(_rB_1^c) = $$
$$ 0 \cdot P(_rB_1)+ \frac{1}{w+k}\cdot (\frac{w+k-1}{w+k})^{r-1} $$

then, when $r = n+1$ you have

$$P(\text{$n+1$-th draw is white}) = \frac{w(w+k-1)^n}{(w+k)^{n+1}}$$


问:
Is there a name for "inverted ball"?
Is there a name for "inverted ball", that is a body filling the whole Euclidean space with a ball cut off?</p>

答: I have found the use of "exterior of a ball" or naturally "complement of the ball". I have never found one-word name for it.</p>

Title: Is there a name for "inverted ball"?


问:
What could be better than base 10?
Most people use base 10; it's obviously the common notation in the modern world.

However, if we could change what became the common notation, would there be a better choice?

I'm aware that it very well may be that there is no intrinsically superior base, but for the purposes of humans, is there a better one?

I've heard from sources such as this and this that base 12 is better, from here that base 8 is better, and, being into computer science, I would say that base 16 is the most handy.

Base 12 does seem to be the most supported non-base 10 number system, mainly due to the following reason pointed out by George Dvorsky:

First and foremost, 12 is a highly composite number — the smallest
  number with exactly four divisors: 2, 3, 4, and 6 (six if you count 1
  and 12). As noted, 10 has only two. Consequently, 12 is much more
  practical when using fractions — it's easier to divide units of
  weights and measures into 12 parts, namely halves, thirds, and
  quarters.

And, on top of that, previous societies considered very advanced used other systems, such as the Mayans using base 20, and the Babylonians using base 60.

So, summarized, my question is: Is there an intrinsically superior base? If not, is there one that would be best for society's purposes? Or does the best base depend on the context it is being used in?

答: I think base $6$ would make counting on our hands particularly convenient, we would have a $1$'s hand and a $6$'s hand and would be able to count up to $35$.</p>

Title: What could be better than base 10?


问:
meaning of $\leftrightarrows$
What is the meaning of $\leftrightarrows\,$?

It has been suggested that it is a synonym for $\leftrightarrow$ or $\Leftrightarrow$ or both, but in MSE posts it appears to have some other meaning associated with category theory.

答:
I'm answering my own question in order to summarise people's useful comments.  Thanks everyone!

The symbol can be used to indicate a bijection, as here.

In category theory: morphisms going both ways/adjunction/split epimorphisms.

Title: meaning of $\leftrightarrows$


问:
Uniform Hölder continuity also holds for smaller exponent $c' \in (0,c]$
Let $(S,\rho)$ and $(T,d)$ be metric spaces. Let $(S,\rho)$ be complete and $D\subseteq T$ be dense. Furthermore let $f:D\to S$ be uniformly Hölder continuous with exponent $c>0$ in the sense of

There exists a constant $K \in \mathbb R_+$ such that $$\rho(f(s),f(t))\leq K(d(s,t))^c,\ \ \ s,t\in D\ \mathrm{with}\ d(s,t)<1$$

Why does this inequality also hold for every $c' \in (0,c]$?

Perhaps these help:

link 1
link 2

Thanks!

答: If $0\leq d<1$ and $c'<c$ we have $d^c<d^{c'}$. Think about how the graph of $x^a$ for $x\in [0,1]$ changes by changing $a$.


问:
Name and layperson's explanation for an E8 group diagram.
I am taking a risk here, but hoping it will not ignite wrath in the reader. In trying to get an intuition of Lie theory this diagram is all but impossible to ignore:

Unfortunately, there are many youtube videos on the group E8 and its applications to subatomic particles and such. Yet, the actual diagram is not clearly addressed.

So, knowing fully well that the math behind it is at this point beyond my level, I would like to ask for:

What is the specific name of this diagram? Is it a Dynkin diagram? Does it have any other names?

What is the name and/or layperson's idea of what the color coding, nodes and edges represent?

答:
The diagram you posted is not a Dynkin diagram, but is the projection of the convex hull of the root system $E_8$ in the Coxeter plane, i.e. the invariant plane for the action of the Coxeter element on the Euclidean space where you embed the Root System of type $E_8$.

Moreover, the different colors of the edges in the picture correspond to the ordering relations between the roots, w.r.t. the ordering induced by the choice of a set of simple roots.

Similar pictures, but probably less astonishing, can be obtained in the same way for all the other irreducible root systems.

Here you can find a complete reference here.

EDIT: About Root Systems.
What is a root?
Suppose $\mathfrak{g}$ is simple.
If you consider a maximal Cartan subalgebra $\mathfrak{h}$, i.e. a maximal abelian subalgebra of your Lie algebra $\mathfrak{g}$, the adjoint action of $\mathfrak{h}$ is diagonal and each eigenspace is indexed by a linear functional in
$\mathfrak{h}^*$.
The set of these functionals is a root system for $\mathfrak{g}$ and it has many nice rigidity properties.
(See Humphrey's book, for example, for an exhaustive review of representations). However, the eigenspaces are not stable for the action of nilpotent elements and their permutations can be described in a nice way by combinatorics.
In other words, a root system for $\mathfrak{g}$ is the set of weights for the adjoint representation of $\mathfrak{g}$.

Edit 2: How to classify the irreducible root systems? Well, one can consider a suitable subset of roots that spans the ambient space and construct the Dynkin diaagrams, that classify completely the irreducible finite dimensional root systems and then the complex simple Lie algebras.

Edit3: Weyl Group and symmetries.
Let's suppose that your root system is embedded in an Euclidean space $E$.
You can consider the group of transformations of $E$ that preserve your Root System.
Such a group is called the Weyl group $W$ attached to the Root System you are considering.
In the case of a complex simple Lie algebra it is a finite reflection group.
It is easy to see that, because W permutes the roots, this action induces a permutation of the vertices of the polytope obtained as convex hull of the (long) roots.
In this sense you can see the symmetry of the root system of type $E_8$ that is object of the thread.
Moreover,  you were asking for eigenspace decomposition in general representations.
If you consider the lattice of weights of a finite dimensional complex representation, this lattice has a symmetry that is in some sense analogous: it comes from the fact that the lattice of weights of a finite dimensional representation must be invariant for the action of W!

By this picture come the nice polygons that you found described in the online lesson you posted in the comments.


问:
Proving $6+12+18+24+...+6n=3n(n+1)$ by induction
I'm learning prove by induction, but for some reason I just can't figure out the following example. I'll just work out the example and tell you guys where I'm stuck.

prove the following by induction $$6+12+18+24+...+6n=3n(n+1)$$

prove basis step (n=1).
$$
6 = 6*1 = 3*1(1+1) = 6
$$

induction step, assume Sk is true, n = k
$$
6 + 12 + 18 + 24 + ... + 6k = 3k(k+1)
$$

induction step n = k + 1

$$
6 + 12 + 18 + 24 + ... + 6(k+1) = 3(k+1)(k+2)
$$

bring Sk to both sides
$$
3k(k+1)+6(k+1) = 3(k+1)(k+2)
$$

Simplify, I really have no idea how these equal eachother, what are the simplifying steps taken in this example? Any help would be greatly appreciated!

答:
I'll write out how I would prove this.

Prove $$6 + 12 + \cdots + 6n = 3n(n + 1).$$

Proof. We will first prove that $$1 + 2 + \cdots + n = \frac{n(n+1)}{2} \tag{1}$$ for all $n \in \mathbb{N}.$ Note that $$1 = \frac{1(1+1)}{2}.$$ Suppose $(1)$ holds when $n = k.$ If that is the case, then $$1 + 2 + \cdots + k = \frac{k(k + 1)}{2}. \tag{2}$$ We now add $k + 1$ to both sides to $(2)$ to get $$1 + 2 + \cdots + k + k + 1 = \frac{k(k + 1)}{2} + k + 1. \tag{3}$$ Doing some algebra on the right hand side of $(3),$ we see that $$\frac{k(k + 1)}{2} + k + 1 = \frac{k(k+1) + 2(k + 1)}{2} = \frac{(k + 1)(k + 2)}{2},$$ which implies that we have proved that $(1)$ holds for all $n \in \mathbb{N}.$ We now multiply both sides of $(1)$ by $6$ to get $$6 + 12 + \cdots + 6n = 3n(n + 1).$$


答:
Steps 3 and 4 don't look right. You can't just assume:

$6 + 12 + 18 + 24 + ... + 6k + 6(k+1) = 3(k+1)(k+2)$ and manipulate it.

You need to prove that from the inductive assumption. So you need to use the following:

$6 + 12 + 18 + 24 + ... + 6k = 3k(k+1)$ to try and prove:

$6 + 12 + 18 + 24 + ... + 6k + 6(k+1) = 3(k+1)(k+2)$

So what you can do is manipulate the left hand side, and show that it comes out to the right hand side

$6 + 12 + 18 + 24 + ... + 6k + 6(k+1)$

$=3k(k+1) + 6(k+1)$ (this is substitution from our inductive assumption)

$=(3k+6)(k+1)$

$=3(k+2)(k+1)$

$=3(k+1)(k+2)$

So that shows the left hand side equals the right hand side. And that completes the proof.

Title: Proving $6+12+18+24+...+6n=3n(n+1)$ by induction

Body:
I'm learning prove by induction, but for some reason I just can't figure out the following example. I'll just work out the example and tell you guys where I'm stuck.

prove the following by induction $$6+12+18+24+...+6n=3n(n+1)$$

prove basis step (n=1).
$$
6 = 6*1 = 3*1(1+1) = 6
$$

induction step, assume Sk is true, n = k
$$
6 + 12 + 18 + 24 + ... + 6k = 3k(k+1)
$$

induction step n = k + 1

$$
6 + 12 + 18 + 24 + ... + 6(k+1) = 3(k+1)(k+2)
$$

bring Sk to both sides
$$
3k(k+1)+6(k+1) = 3(k+1)(k+2)
$$

Simplify, I really have no idea how these equal eachother, what are the simplifying steps taken in this example? Any help would be greatly appreciated!


问:
Discrete Math: Multiplication Principle Word Problem
Assuming that there will never be more than 500 million people in the United States before the year 2050 AD, how many alphanumeric-upper digits are necessary to provide every person with a personal zip code?  (Notice that businesses and institutions are being neglected in this count.)

If the answer is 9 digits, how can I write this in the form ${n \choose x}$?

答:
Let $r ~=~ 500000000$.

You want the smallest positive integer $n$ such that

$$36^n \geq r.$$

Taking logarithms of both sides:

$$n \times \log ~36 \geq \log ~r \implies n \geq \frac{\log ~r}{\log ~36}.$$

By the way, this can actually be approximately solved without a calculator, by using logarithms, base $10$, rather than base $e$, if you happen to know that

$$\log_{10} ~2 ~\text{approximately} ~=~ 0.301
~~\text{and} ~~ \log_{10} ~3 ~\text{approximately} ~=~ 0.477.$$

Then, you have that

$$\log_{10} ~36 ~\text{approximately} ~=~
2 \times \left(0.301 + 0.477\right)
$$

and

$$\log_{10} r ~\text{approximately} ~=~
8.699.
$$

Therefore, $n$ is the smallest positive integer such that

$$n \geq \frac{8.699}{2 \times \left(0.301 + 0.477\right)}.$$


答:
Here is how I would do the inductive step in this specific case:

Inductive hypothesis: for some $k$, we have $\:6 + 12 + 18 + 24 + ... + 6k = 3k(k+1)$.

We have to deduce that $\:6 + 12 + 18 + 24 + ... + 6(k+1) = 3(k+1)(k+2)$.

Now, we can group the terms in the l.h.s. and use the inductive hypothesis:
$$(6 + 12 + 18 + 24 + ... + 6k)+6(k+1) = 3k(k+1)+6(k+1)=3(k+1)(k+2).$$


问:
The set of differences for a set of positive Lebesgue measure
Quite a while ago, I heard about a statement in measure theory, that goes as follows:

Let $A \subset \mathbb R^n$ be a Lebesgue-measurable set of positive measure. Then we follow that $A-A = \{ x-y \mid x,y\in A\}$ is a neighborhood of zero, i.e. contains an open ball around zero.

I now got reminded of that statement as I have the homework problem (Kolmogorov, Introductory Real Analysis, p. 268, Problem 5):

Prove that every set of positive measure in the interval $[0,1]$ contains a pair of points whose distance apart is a rational number.

The above statement would obviously prove the homework problem and I would like to prove the more general statement. I think that assuming the opposite and taking a sequence $\{x_n\}$ converging to zero such that none of the elements are contained in $A$, we might be able to define an ascending/descending chain $A_n$ such that the union/intersection is $A$ but the limit of its measures zero. I am in lack of ideas for the definition on those $A_n$.

I am asking specifically not for an answer but a hint on the problem. Especially if my idea turns out to be fruitful for somebody, a notice would be great. Or if another well-known theorem is needed, I surely would want to know. Thank you for your help.

答:
Assume $A$ is the set contained in [0,1] with positive measure, say $m(A) > 0$ with $m$ the Lebesgue measure on $[0,1]$. Let Q be the set of all rational numbers in $[0,1]$ Since $\mathbb{Q}$ is countable, it can be presented as

$$\mathbb{Q}=\{p_1, p_2, ..., p_n, ...\}$$

Let

$$A_n = A+ p_n = \{x+p_n\mid x\in A\}.$$

If there exists a pair of integers $n$ and $m$ such that $A_n$ and $A_m$ intersect, then the claim of this proposition is proved. If no such pair exists, then the set of $\{A_n\}$ are all disjoint. Since the union of this family of sets is contained in $[0,2]$ and since $m(A) > 0$, we have

$$2 = m([0,2]) \geq m( \bigcup A_n ) = \sum\limits_{n \in \mathbb{N}}  m(A_n)  =\infty\cdot m(A) = \infty .$$


问:
Dominant morphism between affine varieties induces injection on coordinate rings?
Here are the definitions that we use for this problem:

A morphism $\varphi : X \to Y$ between two varieties is said to be dominant if the image of $\varphi$ is dense in $Y$ (c.f. Hartshorne exercise 1.3.17)

We say $X$ is an affine variety if it is an irreducible closed subset of $\Bbb{A}^n$. (see the definition after example 1.1.4 of Hartshorne)

Now let $X$ and $Y$ be affine varieties and $\varphi : X \to Y$ a dominant morphism. Now this means that for all $p \in X$, $\varphi^\ast_p : \mathcal{O}_{\varphi(p),Y} \to \mathcal{O}_{p,X}$ is injective (c.f. Hartshorne exercise 1.3.3 (c)). Because we have assumed that $X,Y$ are affine varieties we identify $\mathcal{O}_{p,X} \cong A(X)_{\mathfrak{m}_p}$ and similarly for $Y$ to get that

$$(\varphi^\ast)_p : A(Y)_{\mathfrak{m}_{\varphi(p)}} \to A(X)_{\mathfrak{m}_p}$$

is injective for all $p\in X$.

My question is: Is it true that $\varphi^\ast : A(Y) \to A(X)$ is injective?

Here are some partial results I can get. Choose a maximal ideal $\mathfrak{m}_q \subseteq A(Y)$. If $\ker \varphi^\ast$ is completely contained in $\mathfrak{m}_q$ then

$$(\mathfrak{m}_q)_{\mathfrak{m}_q} (\ker \varphi^\ast)_{\mathfrak{m}_q} = (\ker \varphi^\ast)_{\mathfrak{m}_q}.$$

The ring $A(Y)_{\mathfrak{m}_q}$ is local and Noetherian so $(\ker \varphi^\ast)_{\mathfrak{m}_q}$ is finitely generated as a module over this ring. By Nakayama's lemma it follows $(\ker \varphi^\ast)_{\mathfrak{m}_q} =0$.

Edit: Adeel's answer below shows that under my assumptions we have $\ker \varphi \subseteq \text{nilrad}  A(Y)$, so my proof is complete.

答:
Proposition: If $X = \mathrm{Spec}(A)$ and $Y = \mathrm{Spec}(B)$ are affine schemes, a morphism $f : X \to Y$ is dominant if and only if the kernel of the corresponding homomorphism $\varphi : B \to A$ is contained in the nilradical of $B$.

Proof: Recall that there is an equality
  $$ V(\varphi^{-1}(I)) = \overline{f(V(I))} $$
for ideals $I \subset A$; in particular,
  $$ V(\varphi^{-1}(0)) = \overline{f(X)} = Y $$
if and only if $f$ is dominant.  This is equivalent to the ideal $\varphi^{-1}(0) = \ker(\varphi)$ being contained in every prime ideal, i.e. in the intersection of all the prime ideals; but this intersection is precisely the nilradical of $B$.

In particular when Y is reduced (e.g. a variety), the homomorphism is injective.


问:
Ratio of circumradius to inradius of a regular tetrahedron
The circumradius of a set in $\mathbb{R}^n$ is the radius of the smallest sphere enclosing the set.  Similarly, the inradius is the radius of the largest sphere fitting inside a set.

Question:  What is the ratio $\frac{\text{circumradius}}{\text{inradius}}$ for a regular tetrahedron?

This is one of my all time favorite questions because it admits at least three very different solutions, two of which fit anyone's definition of "elegant".  I'm posting my favorite solution, but I am keen to see what solutions others can come up with.

答:
Referring to the diagram in Yly's answer, we see that r is the
the center's altitude.  Altitude of an average is average of the altitudes (this can be seen by choosing a cartesian coordinate system such that the altitude direction is one of the coordinate axes); therefore r = average of the 4 vertices' altitudes = (apex altitude + 0 + 0 + 0)/4 = (apex altitude)/4.

The circumradius R, on the other hand, is (apex altitude) - (center altitude) = (apex altitude) - (apex altitude)/4 = 3/4 apex altitude.

So their ratio R/r is (3/4 apex altitude) / (1/4 apex altitude) = 3.


问:
Find the area under one arch of the cycloid $x=6(t-\sin(t))$, $y=6(1-\cos(t))$
Find the area under one arch of the cycloid $x=6(t-\sin(t))$, $y=6(1-\cos(t))$

I'm trying to figure this out using calculus. the first cycle of this cycloid will achieve a maximum height of $y=12$ and will go from $x=0$ to $x=12\pi$

So i set up the integral:

$\int_0^{12\pi}ydx=\int_0^{12\pi}6(1-\cos(t))(6-6\cos(t))dt=36\int_0^{12\pi}1+\cos^2(t)-12\cos(t)dt$

$=36\int_0^{12\pi}1+(\frac{1+\cos(2t)}{2})-12\cos(t)dt$

$=36\int_0^{12\pi}\frac{3}{2}+\frac{\cos(2t)}{2}-12\cos(t)dt$

$=36[\frac{3x}{2}+\frac{\sin(2t)}{4}-12\sin(t)]|_0^{12\pi}$

$=18(12\pi)$

Which is not correct... I'm supposed to get $108\pi$. I would appreciate some help!

答: Everything is fine, except the limits of integration for $t$. Those should be $0$ to $2\pi$. Then your next to last line should be $$36\left[\frac{3t}{2}+\frac{\sin(2t)}{4}-12\sin(t)\right]\bigg|_0^{2\pi}=36\cdot 3\pi=108\pi$$


问:
Find two integers $x$ and $y$ such that $15x+28y=1$
I tried to solve it using the Euclidean algorithm but after finding the gcd$(15, 28)=$ $1$, and reversing the steps of the Euclidean algorithm I get the following expression:

These are the reversal steps:
$$13=2\times6+1$$
$$1=13-(2\times6)$$
$$1=13-6\left(15-13\right)$$
$$1=28-15-6\left(15-13\right)$$
$$1=28-15\left(1+6\right)+6\text{·}13$$

The expected result should be some expression of the form
$$1=28\left(y\right)-15\left(x\right)$$

答:
You already replaced a 13 by
$$13=28-15$$

Why don't you replace the last 13 also by this?

It would have been faster if you combined the 13's together before replacing.


答:
The two points i and -i are distance 2 apart.
The origin is distance one from one of them and one from the other so it satisfies the equation.

All the points on the line between those two points satisfy the equation.

Any other point got on the line between these two has the sum of the distances to those two points being greater than 2. Think of the theorem in euclidean geometry that the sum of the length of two sides of a triangle is greater than the length of the third side.

Therefore the only solutions are all the points between the two points.

In general the solutions to the equation
$|z-a|+|z-b| = c$
where $c$ is greater than the distance between
$a$ and $b$
is the eclipse with foci at a and b
and the sum of the distances to a and b equal to c.

If c is equal to the distance between a and b,
the solution is the points on the line segment joining a and b.

If c is less than the distance between a and b,
there is no solution.

Title: Need help with complex equation: |z−i|+|z+i|=2

Body:
I am trying to solve this equation: |z−i|+|z+i|=2 and don't know how to do it. This what I have:
$$\sqrt{(x+1)^2+y^2}+ \sqrt{(x-1)^2+y^2} = 2 /^2$$
$$(x+1)^2+y^2+(x-1)^2+y^2 + 2\sqrt{[(x+1)^2+y^2][(x-1)^2+y^2]} = 4$$
$$x^2 +2x + 1+y^2+x^2-2x + 1+y^2 + 2\sqrt{[(x^2 +2x + 1)+y^2][(x^2-2x + 1)+y^2]}=4$$
$$2x^2+2y^2 + 2 + 2\sqrt{[(x^2 +2x + 1)+y^2][(x^2-2x + 1)+y^2]}=4$$
$$2x^2+2y^2 + 2 + 2\sqrt{(x^2 +2x + 1)(x^2-2x + 1)+x^2y^2-2xy^2+y^2+x^2y^2-2xy^2+y^2+y^4}=4$$
$$2x^2+2y^2 + 2 + 2\sqrt{(x^2 +2x + 1)(x^2-2x + 1)+2x^2y^2-4xy^2+2y^2+y^4}=4$$
$$2x^2+2y^2 + 2 + 2\sqrt{x^4+2x^3+x^2-2x^3+4x^2-2x+x^2+2x+1+2x^2y^2-4xy^2+2y^2+y^4}=4$$
$$2x^2+2y^2 + 2 + 2\sqrt{x^4+6x^2+1+2x^2y^2-4xy^2+2y^2+y^4}=4$$

And I have no idea what to do next.
Any help would be much appreciated


问:
Prove that quadratic field Q(√2) is closed
For the quadratic field Q(√2), how can I prove that it is closed? This means that any addition, subtraction, multiplication, or division between any a+b√2 and c+d√2 will still be from the set Q(√2)?

Addition and subtraction are pretty trivial, so I won't include it here. For multiplication, I did the following:

$$(a+b√2)(c+d√2)$$
$$=ac+bc√2+ad√2+2bd$$
$$=(ac+2bd)+(bc+ad)√2$$

Therefore Q(√2) is closed for multiplication.

For division, I did the following:

$$\frac{a+b√2}{c+d√2}$$
$$=\frac{(a+b√2)(c-d√2)}{(c+d√2)(c-d√2)}$$
$$=\frac{ac-ad√2+bc√2-2bd}{c^2-2d^2}$$
$$=\frac{(ac-2bd)+(bc-ad)√2}{c^2-2d^2}$$

How do I prove that this value is of the set Q(√2)?

Body:
$$\frac{(ac-2bd) + (bc-ad)\sqrt{2}}{c^2 - 2d^2} = \frac{ac-2bd}{c^2 - 2d^2} + \frac{bc-ad}{c^2 - 2d^2}\sqrt{2}$$

and $a,b,c,d \in \Bbb Q.$


答:
You're almost there.

Is $\dfrac{ac-2bd}{c^2-2d^2}\in\mathbb Q$?

Is $\dfrac{bc-ad}{c^2-2d^2}$ ?

Then $\dfrac{(ac-2bd)+(bc-ad)\sqrt2}{c^2-2d^2}\in\mathbb Q(\sqrt2)$.

(Note:  you should exclude $c=d=0$.)

Title: Prove that quadratic field Q(√2) is closed

Body:
For the quadratic field Q(√2), how can I prove that it is closed? This means that any addition, subtraction, multiplication, or division between any a+b√2 and c+d√2 will still be from the set Q(√2)?

Addition and subtraction are pretty trivial, so I won't include it here. For multiplication, I did the following:

$$(a+b√2)(c+d√2)$$
$$=ac+bc√2+ad√2+2bd$$
$$=(ac+2bd)+(bc+ad)√2$$

Therefore Q(√2) is closed for multiplication.

For division, I did the following:

$$\frac{a+b√2}{c+d√2}$$
$$=\frac{(a+b√2)(c-d√2)}{(c+d√2)(c-d√2)}$$
$$=\frac{ac-ad√2+bc√2-2bd}{c^2-2d^2}$$
$$=\frac{(ac-2bd)+(bc-ad)√2}{c^2-2d^2}$$

How do I prove that this value is of the set Q(√2)?


答:
From the beginning, move one of the absolute values to the other side. It will simplify the calculations:
$$|z+i|=2-|z-i|$$
Square this and you get $$|z+i|^2=(2-|z-i|)^2=4-4|z-i|+|z-i|^2$$
$$x^2+(y+1)^2=4-4|z-i|+x^2+(y-1)^2$$
$$x^2+y^2+2y+1=4-4|z-i|+y^2-2y+1$$
$$4y-4=-4|z-i|$$
$$y-1=-|z-1|$$
$$(y-1)^2=x^2+(y-1)^2$$
The only solution is $x=0$. This is independent of $y$. Next is to check if by squaring we add extra solutions. Not all $y$'s might be valid.
$$|y-1|+|y+1|=2$$
If $y\ge1$, the equation is $$2y=2$$So $y=1$. If $y\le -1$ then $$-(y-1)-(y+1)=2$$, meaning $y=-1$.In between, $$-(y-1)+y+1=2$$or $2=2$. The final answer is $z=\alpha i$, for $\alpha\in[-1,1]$


问:
A matrix norm optimization problem involving matrix inversion
I am trying to characterize the the matrix $X^*$ that solves the following optimization (that is an analytical or semi-analytical solution form).

\begin{equation}
X^*=\arg\min_X \| (A+X)^{-1}\|_F \quad\mbox{s.t.}\quad \|X\|_p=r , X \succeq 0
\end{equation}

$A$ and $X$ are both symmetric square matrices. In fact, $A$ is positive definite and $X$ is positive semi-definite. Here $\|.\|_F$ and $\|.\|_p$ are Frobenius and p-Schatten norms.

I am interested in $X^*$ when $p=1$ and $p=2$ (but even knowing the solution in one of these cases is still very helpful).

The problem statement itself looks simple, but I have not been able to characterize its solution. Any thoughts are greatly appreciated.

Thanks!

Golabi

答:
Presuming (as discussed in the question comments), the constraint is $\|X\|_p \le r$ rather than $\|X\|_p=r$, this can be numerically solved in CVX.

Formulation is described in http://ask.cvxr.com/t/represent-schatten-p-norm-in-cvx/649 .


问:
How to compute derive a complex number from a square root of another complex number?
In plane wave in a lossy medium, the complex propagative constant is given to be

$$\gamma = \alpha\ + j\beta $$
$$= j \omega \sqrt {\mu \epsilon_c}$$
$$= j \omega \sqrt { \mu \left(\epsilon - j\frac{\sigma}{\omega}\right) }$$

Which will eventually result in a square root of a complex number like

$$\gamma = j\omega \sqrt {X+jY}$$

how do I proceed on to find $\gamma$ in terms of $\alpha$ and $\beta$ ?

答:
Let $z$ be the complex radicand and render

$(z+|z|)^2=z^2+2z|z|+|z|^2=z^2+2z|z|+z\overline z=z(2|z|+2\Re z)$

Thus for $z$ not a negative real number so that $|z|+\Re z>0$:

$z+|z|=\pm(\sqrt z)(\sqrt{2(|z|+\Re z)})$

$\color{blue}{\sqrt z=\pm\dfrac{z+|z|}{\sqrt{2(|z|+\Re z)}}}$


Title: Find two integers $x$ and $y$ such that $15x+28y=1$

Body:
I tried to solve it using the Euclidean algorithm but after finding the gcd$(15, 28)=$ $1$, and reversing the steps of the Euclidean algorithm I get the following expression:

These are the reversal steps:
$$13=2\times6+1$$
$$1=13-(2\times6)$$
$$1=13-6\left(15-13\right)$$
$$1=28-15-6\left(15-13\right)$$
$$1=28-15\left(1+6\right)+6\text{·}13$$

The expected result should be some expression of the form
$$1=28\left(y\right)-15\left(x\right)$$

Body:
$1=13-6\times2=13-6(15-13)=7\times13-6\times15=7\times(28-15)-6\times15$

$=7\times28-13\times15$


答:
The fast method uses the extended Euclidean algorithm, based  on the observation that all reminders in the Euclidean algorithm can be written as a linear combination of $15$ and $28$, with a recursive relation deduced from the relation step $i$:
$$ r_{i-1}=q_i r_i+r_{i+1}\iff r_{i+1}= r_{i-1}-q_i r_i,$$
from which we deduce the relations (with obvious notations)
$$  x_{i+1}= x_{i-1}-q_i x_i,\qquad  y_{i+1}= y_{i-1}-q_i y_i.$$
As it is a recursion of order $2$, we have to set the initial values $r_0=15$, $r_{-1}=28$, whence the following layout:

\begin{array}{rrrl}
r_i &x_i&y_i&q_i \\ \hline
28 & 0 & 1 \\
15 & 1 & 0 & 1 \\ \hline
13 & -1 & 1 & 1 \\
2 & 2 & -1 & 6 \\
\color{red}1 & \color{red}{-13} & \color{red}{7} \\
\hline
\end{array}


问:
$||E[X]|| \leq E[||X||]$
Let $X$ be a r.v. taking values in $\mathbb{R}^d,$ such that $X \in L^1.$ $||.||$ is an arbitrary norm on $\mathbb{R}^d.$ Prove that $$||E[X]||\leq E[||X||].$$

When $d=1,$ then letting $\theta=\frac{E[X]}{|E[X]|},$ $|E[X]|=\frac{1}{\theta}E[X] \leq E[\frac{1}{|\theta|}|X|]\leq E [|X|].$

Having problems with $d>1,$ what do suggest to do? Is there a way to prove it using the above argument?

答:
Hint: If you want to prove this statement - you'll have to prove 2 fairly easy things: (a) that norms are convex functions and (b) jensen's inequality for convex functions.
Both proofs are very fun - esp. when you arrive at them by yourself, but also easily google-able.</p>

Title: $||E[X]|| \leq E[||X||]$


答:
The set of solutions is the segment $\{\alpha i\mid-1\le\alpha\le1\}$.

This follows from the following geometrical argument. $|z-i|$ is the distance from $z$ to $i$; $|z+i|$ is the distance from $z$ to $-i$. Which point $z\in\mathbb C$ can have the sum of distances to $i$ and to $-i$ equal to $2$, knowing that the distance between $i$ and $-i$ is already $2$? It must be colinear with $i$ and $-i$ (or otherwise the LHS would be $\gt 2$ due to triangle inequality) and it must lie in the actual segment with endpoints $i,-i$, or otherwise one of the moduli $|z-i|,|z+i|$ would alone be $\gt 2$.


答:
Applying the Euclidean algorithm we have $$28=1\cdot15+13$$
$$15=1\cdot13+2$$
$$13=6\cdot2+1$$
$$2=2\cdot1+0$$

So we have $\gcd(13,28)=1$ and for the reversal steps $$1=13-6\cdot2$$ and now substituting the second $$2=15-1\cdot 13$$ we have $$1=13-6\cdot2=13-6\cdot(15-1\cdot13)=7\cdot13-6\cdot15$$ and finally substituting the first $13=28-1\cdot15$ we obtain $$1=7\cdot13-6\cdot15=7\cdot(28-1\cdot15)-6\cdot15$$
$$=7\cdot28-13\cdot15$$

Thus $$7\cdot28-13\cdot15=1$$


问:
Computing a 3D quadrilateral inscribed sphere at its center?
For my game, I need to know when the player is over a powerup and for this I use a sphere collider:

3rd side view:

Legend:

orange, the powerup outline, a quad for which I know its centroid

green, sphere collider outline, its center is quad centroid but radius is random for now

Radius for circumscribed circle is easy since I know center and vertices it should encompass:

Radius for inscribed circle is what I'd like to achieve instead:

I guess I am close with the following links but unsure as on how to do it for 3D points:

https://en.wikipedia.org/wiki/Tangential_quadrilateral

https://en.wikipedia.org/wiki/Bicentric_quadrilateral

So, knowing a 3D quadrilateral and its centroid, how can I find the radius of an inscribed sphere at its center?

答:
(Comment, but not enough rep for commenting.)

For a square quadrilateral with inradius $r$ and outradius $R$, $R = r \sqrt{2}$.

For a generic quadrilateral, I would suggest half of some sort of mean (or minimum or maximum) of the two diagonals; for example,
$$\begin{aligned}
d_{sum} &= \sqrt{ (x_3 - x_1)^2 + (y_3 - y_1)^2 + (z_3 - z_1)^2 } \\
~       &+ \sqrt{ (x_4 - x_2)^2 + (y_4 - y_2)^2 + (z_4 - z_2)^2 }\\
r &= \frac{d_{sum}}{4\sqrt{2}}
\end{aligned}$$
Depending on the shapes, you could use half the root-mean-square of the diagonals divided by square root of two,
$$r = \frac{1}{4}\sqrt{ (x_3 - x_1)^2 + (y_3 - y_1)^2 + (z_3 - z_1)^2
                      + (x_4 - x_2)^2 + (y_4 - y_2)^2 + (z_4 - z_2)^2 }$$


问:
Weight function and orthogonalization
If we have the weight function $w(x) = e^{-x}$, then what are the constant and linear polynomials $p_0(x)$, $p_1(x)$, that are orthogonal on $[0,\infty)$ w.r.t. $w(x)$?

And if $f(x) =\cos(x)$, what is the best linear approximation to $f(x)$ for the given weight?

答:
I think you are interested in the Laguerre polynomials $L_n$, which form an orthonormal basis for the weighted space $L^2((0,\infty), e^{-x}\,dx)$. In particular, $L_0(x)=1$ and $L_1(x)=-x+1$.

To approximate $\cos x$, just compute its coefficients as
$$c_n=\int_0^\infty \cos x L_n(x) e^{-x}\,dx$$
For example, I get $1/2, 1/2, 1/4, 0, -1/8, -1/8$ for $n=0,\dots,5$. The corresponding sum
$$\sum_{n=0}^5 c_n L_n(x)= 1+\frac{1}8 x-\frac{7}{8}x^2+\frac{7}{24} x^3-\frac{1}{32} x^4+\frac{1}{960}x^5$$ gives this approximation:

This isn't picture-perfect but what can you expect from degree 5 polynomial. The weight $e^{-x}$ makes sure the approximation is okay at the beginning, but it does not care much about what happens when $x$ is large.

Title: Weight function and orthogonalization


问:
Is there a practical formula for quantization on arbitrary subsets of $\mathbb{R}^n$
On $\mathbb{R}^n$, there is a convenient formula for quantization in terms of the Fourier transform. More precisely, given a (sufficiently nice) function $a :\mathbb{R}^{2n}\to\mathbb{C}$,
$$
\operatorname{Op}_h(a)u(x) = \frac{1}{(2\pi h)^n}\int_{\mathbb{R}^n}\int_{\mathbb{R}^n}e^{\frac{i}{h}\langle x-y, \xi\rangle}a(x,\xi)u(y)\,\mathrm{d}y\mathrm{d}\xi
$$
Quantization can also be defined on subsets of $\mathbb{R}^n$. On such subsets, is there still a convenient formula that one can work with?

答:
In fact, there is. If it's okay, I'll do away with the semi-classical formulation and talk microlocally.

If $\Omega\subset\mathbb{R}^n$ is open, then we can define a local version of the Kohn-Nirenberg symbol class $S^m_{loc}(\Omega\times\mathbb{R}^n)$ as the set of all $a\in C^\infty(\Omega\times\mathbb{R}^n)$ so that for any compact $K\subset \Omega$ and every multi-indices $\alpha,\beta,$ there exists $C_{K,\alpha,\beta}$ so that $$|\partial_\xi^\beta\partial_x^\alpha a(x,\xi)|\leq C_{K,\alpha,\beta}\langle\xi\rangle^{m-|\alpha|}$$ for all $(x,\xi)\in K\times\mathbb{R}^n.$ One can prove that if we take such a symbol, then the corresponding pseudodifferential operator sends

$C_c^\infty(\Omega)\rightarrow C^\infty(\Omega)$

$\mathcal{E}'(\Omega)\rightarrow\mathcal{D}'(\Omega)$

$H^{s+m}_{comp}(\Omega)\rightarrow H^s_{loc}(\Omega)$ for all real $s$

There's a bit of an issue, in that it doesn't generate an operator algebra. To get around this, one can utilize a locally-finite partition of unity on $\Omega$ (conjugate by cutoffs and sum over all cutoffs with overlapping support). This gives us a properly-supported pseudo-differential operator, and the symbol will differ from the original by a symbol which corresponds to a smoothing operator. This class of operators actually sends all of the relevant spaces to themselves, and we can generate a nice algebra here (e.g. we can compose in the same way).


问:
Find the locus of $z$ such that $\arg \frac{z-z_1}{z-z_2} = \alpha$: confused when $\alpha = 0$.

Find the locus of $z$ such that $\arg \frac{z-z_1}{z-z_2} = \alpha$.
Use and draw $w = \frac{z-z_1}{z-z_2}$.

This exercise was discussed many times -- 1, 2, 3, 4 -- but I was unable to find answers to my problem with $0$ there.

I believe I understand where the arcs came from, here's my work:

If I understand correctly, for $\alpha = \pm\pi$, the locus would be the segment connecting $z_2$ and $z_1$, not including the points themselves.

I can not understand what is happening when $\alpha = 0$.

$\alpha = 0 = \arg \frac{z-z_1}{z-z_2} = \arg w \Longrightarrow \frac{z-z_1}{z-z_2} = k \in \mathbb{R}, \frac{z-z_1}{z-z_2} = k\frac{z-z_2}{z-z_2}.$

Solving this for $z$, $z = \frac{x_1-kx_2}{1-k} +i\frac{y_1-ky_2}{1-k}$. I am having trouble understanding the locus of this $z$. The textbook says it should be 'two segments with end points in $z_1$ and $z_2$, and one of this segments contains an infinitely distant point'. How to understand why is this answer right, and how to draw it? It seems the infinitely distant point matches $k=1$, but why should it lie in the 'direction' of the line passing through $z_1$ and $z_2$?

My class notes are messy. Why is $(0, 1)$ special on $w$ plane?

Thank you.

答:
You know that (within a modulus of $2\pi$),

$\arg \left(w_1 \times w_2\right) ~=~ \arg(w_1) + \arg(w_2).$

Therefore, $\arg \left(\frac{w_1}{w_2}\right) ~=~ \arg(w_1) - \arg(w_2).$

I can not understand what is happening when $\alpha = 0.$

In this situation, you have that

$$0 = \alpha ~=~ \arg \left(\frac{z - z_1}{z - z_2}\right)
~=~ \arg(z - z_1) - \arg(z - z_2).$$

Imagine the infinite line that passes through $z_1$ and $z_2$.

Note, that the $\arg$ function is not defined on the complex number $(0 + i[0]).$

Therefore, $z$ is not allowed to equal either $z_1$ or $z_2$.

There are 3 possibilities:

$\underline{\text{case 1} ~z ~\text{is not on this line}}$

Then,

$$\arg(z - z_1) \neq  \arg(z - z_2).$$

Therefore, this possibility must be excluded from the locus of satisfying points.

$\underline{\text{case 2} ~z ~\text{is on this line}, ~\textbf{but between}
~z_1 ~\text{and} ~z_2}$

Then,

$$\arg(z - z_1) ~=~  \arg(z - z_2) ~\pm ~\pi.$$

Therefore, this possibility must also be excluded from the locus of satisfying points.

$\underline{\text{case 3} ~z ~\text{is on this line}, ~\textbf{but not between}
~z_1 ~\text{and} ~z_2}$

Then, regardless of whether the point $z$ is closer to $z_1$ or closer to $z_2$,

$$\arg(z - z_1) ~=~  \arg(z - z_2).$$

Therefore, this possibility represents the locus of all satisfying points.

Thus, the locus of all satisfying points, when $\alpha = 0,$ is all $z$ that are
on the line formed by $z_1$ and $z_2$, but are not between $z_1$ and $z_2$.


答:
The integrand should be
$36(1-\cos(t))^2
= 36(1-2\cos(t)+\cos^2(t))
= 36(1-2\cos(t)+(1+\cos(2t))/2)
$

and then integrate from 0 to $\pi$.

Body:
Find the area under one arch of the cycloid $x=6(t-\sin(t))$, $y=6(1-\cos(t))$

I'm trying to figure this out using calculus. the first cycle of this cycloid will achieve a maximum height of $y=12$ and will go from $x=0$ to $x=12\pi$

So i set up the integral:

$\int_0^{12\pi}ydx=\int_0^{12\pi}6(1-\cos(t))(6-6\cos(t))dt=36\int_0^{12\pi}1+\cos^2(t)-12\cos(t)dt$

$=36\int_0^{12\pi}1+(\frac{1+\cos(2t)}{2})-12\cos(t)dt$

$=36\int_0^{12\pi}\frac{3}{2}+\frac{\cos(2t)}{2}-12\cos(t)dt$

$=36[\frac{3x}{2}+\frac{\sin(2t)}{4}-12\sin(t)]|_0^{12\pi}$

$=18(12\pi)$

Which is not correct... I'm supposed to get $108\pi$. I would appreciate some help!


问:
Doubt - Hausdorff Space
I am solving exercise 3-17 in Lee's Topological Manifolds.

I'm having problems with question b). I can't imagine the diagonal.

Can someone help me. Thanks

答:
There is no need to look at the diagonal in $X\times X$: just show that any two distinct points of $X$ have disjoint open nbhds. There are three cases; I give them here with a couple of hints.

Both points are in $(0,1)\times(0,1)$: that case is very easy, since this subspace of $X$ just has its usual topology.

One point is in $(0,1)\times(0,1)$, and the other is either $(0,1)$ or $(1,0)$; that takes just a little work. If $(x,y)$ is the point in $(0,1)\times(0,1)$, pay particular attention to $y$.

The points are $(0,0)$ and $(1,0)$; that one is very easy.

Title: Doubt - Hausdorff Space