问:
Determine if relation $P: R^+ \rightarrow R$ is a function, $(x, y) \in P$ means that $x = y^2$
I am self studying discrete mathematics using the book Discrete Mathematics with Applications (Epp, 4e international version) and I have been stuck trying to figure out the answer to the below question (pg. 22, question 11)

Define a relation $P$ from $\mathbb{R^+}$ to $\mathbb{R}$ as follows: For all real numbers $x$ and $y$ with $x > 0$, $(x,y) \in P$ means that $x = y^2$

Is $P$ a function?

I reasoned that $P$ is a function as it satisfies the properties of a function:

For every element $x$ in $\mathbb{R^+}$ there is an element $y$ in $\mathbb{R}$

No two distinct ordered pairs in $P$ have the same first element because, the domain $\mathbb{R^+}$ which means that $x$ is always positive, so y = $\sqrt x$ would also be a positive real number.

However, the solution provided for the problem states: $P$ is not a function because, for example, $(4, 2) \in P$ and $(4, -2) \in P$ but $2 \ne -2$

I can't work out why it would not be function given that $P$ explicitly means $x = y^2$, so using the example provided, $(4, -2) \not\in P$ because $\sqrt 4 \ne -2$.

Can anyone give a hint that might help me to get unstuck or should I re-read this chapter as I've missed something fundamental?

答: Your mistake is that you assumed that, for $x>0$: $x=y^2$ is equivalent with $y=\sqrt{x}$. This is not true. $x=y^2$ is actually equivalent with: $$y=\sqrt{x}\text{ or }y=-\sqrt{x}$$


问:
Find the coordinates of the points where the line $4y = 7x + 1$ intersects the curve $x^2 - 2x + y^2 - 4y = 60$
I have tried equating the two equations to each other and then solving for x; however, I believe this is not correct.  I also tried factorizing the curve equation but I could not see a way through.

Thank you for your help.

答: Your first approach is fine. Since$$4y=7x+1\iff y=\frac74x+\frac14$$you can replace $y$ with $\frac74x+\frac14$ in the second equation, thereby getting $\frac{5}{16} \left(13 x^2-26 x-3\right)$. And now you solve the equation$$\frac{5}{16} \left(13 x^2-26 x-3\right)=60,$$which has two solutions: $-3$ and $5$. So, the intersection points are $(-3,-5)$ and $(5,9)$.

ContentLicense: CC BY-SA 4.0


问:
Example of a locally Noetherian space $X$ and a point $x$ such that $\dim_xX=\infty$
There is a well-known example that a Noetherian ring $A$ with infinite Krull dimension due to Nagata, which gives a Noetherian space $\operatorname{Spec}A$ with infinite Krull dimension.

Given the definition $\dim_xX=\inf_{x\in U}\dim U$ where $U$ takes over all open neighborhood of $x$, then does there exist a locally Noetherian space $X$(not necessary a scheme) with a point $x\in X$ such that $\dim_xX=\infty$?

Edit. User leoli1 has given an example, could someone give an example of scheme?

答: Consider the space $X=\{1,2,3\dots\}\cup\{\infty\}=\Bbb N\cup\{\infty\}$. Define the topology on $X$ so that the open sets are $\emptyset$ and sets of the form $[n,\infty]$ with $n<\infty$. Then $X$ is noetherian and $\dim X=\infty$. Now any neighborhood of $x=\infty$ is of the form $[n,\infty]$ and hence homeomorphic to $X$, so $\dim_xX=\infty$.


答:
$\newcommand{\R}{\mathbb R}$

Although there might be some ambiguity in the way $P$ is defined (using the verb "mean" in "$(x, y) \in P$ means that ..." is not very precise. Is this is an equivalence or an implication ?!), we can safely assume that what was intended is the formal definition

$$\forall (x, y) \in \R^+ \times \R, (x, y) \in P \Leftrightarrow (x = y^2) $$

Using this definition, we have indeed that both $(4, 2)$ and $(4, -2)$ are in $P$, so $P$ can not be a function.

Your second point is false: $x$ is indeed non-negative, but it does not imply that $y \in \R^+$; as we just saw, it can very much be negative, as in $(4, -2) \in P$.

Title: Determine if relation $P: R^+ \rightarrow R$ is a function, $(x, y) \in P$ means that $x = y^2$

Body:
I am self studying discrete mathematics using the book Discrete Mathematics with Applications (Epp, 4e international version) and I have been stuck trying to figure out the answer to the below question (pg. 22, question 11)

Define a relation $P$ from $\mathbb{R^+}$ to $\mathbb{R}$ as follows: For all real numbers $x$ and $y$ with $x > 0$, $(x,y) \in P$ means that $x = y^2$

Is $P$ a function?

I reasoned that $P$ is a function as it satisfies the properties of a function:

For every element $x$ in $\mathbb{R^+}$ there is an element $y$ in $\mathbb{R}$

No two distinct ordered pairs in $P$ have the same first element because, the domain $\mathbb{R^+}$ which means that $x$ is always positive, so y = $\sqrt x$ would also be a positive real number.

However, the solution provided for the problem states: $P$ is not a function because, for example, $(4, 2) \in P$ and $(4, -2) \in P$ but $2 \ne -2$

I can't work out why it would not be function given that $P$ explicitly means $x = y^2$, so using the example provided, $(4, -2) \not\in P$ because $\sqrt 4 \ne -2$.

Can anyone give a hint that might help me to get unstuck or should I re-read this chapter as I've missed something fundamental?


问:
Can someone explain to me what are these 2 statements talking about?
I have to prove that these 2 statements are equivalent, but I can't even understand them.

There exist $\epsilon_0>0$ such that for all $k\in\mathbb N$, there exist $n_k\in\mathbb N$ such that $n_k\geq k$ and $|x_{n_k}-x|\geq\epsilon_0$?

There exist $\epsilon_0>0$ and a subsequence $(x_{n_k})$ of sequence $(x_n)$ such that $|x_{n_k}-x|\geq\epsilon_0$ for all $k\in\mathbb N$.

Thank you.

答:
We have a sequence $\sigma=\langle x_n:n\in\Bbb N\rangle$ of real numbers. (1) and (2) are both statements about that sequence. The second is a little simpler, so I’ll start with it.

(2) says that there is some positive distance $\epsilon_0$ such that a whole subsequence of $\sigma$ consists of terms that are at least $\epsilon_0$ units away from the number $x$. That is, there is an increasing sequence $n_0<n_1<n_2<\ldots$ of natural numbers such that every one of the terms $x_{n_k}$ of the original sequence $\sigma$ is at least $\epsilon_0$ units distant from the number $x$. Equivalently, each of these terms $x_{n_k}$ lies outside the open interval $(x-\epsilon_0,x+\epsilon_0)$. We might say that $\sigma$ has a subsequence whose terms are bounded away from $x$ by a distance $\epsilon_0$. (You should note, by the way, that this just says that infinitely many terms of $\sigma$ are outside the interval $(x-\epsilon_0,x+\epsilon_0)$: once you have infinitely many such terms, you can certainly list them in order of increasing subscripts to get a subsequence of $\sigma$.)

(1) also says that there is a positive distance $\epsilon_0$ such that certain terms of the sequence $\sigma$ are at least $\epsilon_0$ units distance from $x$, i.e, outside the open interval $(x-\epsilon_0,x+\epsilon_0)$. Specifically, there is a function $\varphi:\Bbb N\to\Bbb N$ such that for each $k\in\Bbb N$, $\varphi(k)\ge k$ and the point $x_{\varphi(k)}$ is at least $\epsilon_0$ units distant from $x$, i.e., not in the open interval $(x-\epsilon_0,x+\epsilon_0)$.

The problem statement uses the notation $n_k$ for this integer that I’ve called $\varphi(k)$, but that’s potentially confusing, since it makes $\langle x_{n_k}:k\in\Bbb N\rangle$, my $\langle x_{\varphi(k)}:k\in\Bbb N\rangle$, look like a subsequence of $\sigma$, which it technically is not, since the sequence of indices $\langle n_0,n_1,n_2,\dots\rangle$, or in my notation $\langle\varphi(0),\varphi(1),\varphi(2),\dots\rangle$, need not be increasing.

Proving that (2) implies (1) is fairly straightforward, once you notice that if $\langle n_0,n_1,n_2,\dots\rangle$ is an increasing sequence of natural numbers, then $n_k\ge k$ for each $k\in\Bbb N$. This is easily proved by induction on $k$.

Proving that (1) implies (2) is a little harder. You’ll want to assume that you have an $\epsilon_0$ and a function $\varphi$ as in (my version of) (1), and use them to construct a subsequence $\langle x_{n_k}:k\in\Bbb N\rangle$ of $\sigma$ as in (2). This construction will be recursive. You can start by letting $n_0=\varphi(0)$: then $n_0\ge 0$ (which of course it was bound to be anyway!), and $x_{n_0}\notin(x-\epsilon_0,x+\epsilon_0)$. Now you need an $n_1$ such that $n_1>n_0$ and $x_{n_1}\notin(x-\epsilon_0,x+\epsilon_0)$. Any $\varphi(k)$ has the property that $x_{\varphi(k)}\notin(x-\epsilon_0,x+\epsilon_0)$, so we could get the desired separation of $x_{n_1}$ from $x$ with any $x_{\varphi(k)}$, but we also want $n_1=\varphi(k)>n_0$. This is easily managed when we recall that $\varphi(k)\ge k$ for every $k\in\Bbb N$: just let $n_1=\varphi(n_0+1)$. All that remains is to generalize this to explain how to get $n_{k+1}$ once you have $n_k$.

ContentLicense: CC BY-SA 3.0


问:
Suppose $V$ is an $n$-dimensional space on the field $\mathbb F_p$ and $W$ is an $m$-dimensional subspace of $V$
Suppose $V$ is an $n$-dimensional space on the field $\mathbb F_p$ and $W$ is an $m$-dimensional subspace of $V$ where $m <n$.

Show that $p^{n − m}$ sets can be found, each of which is $ a_j + W = \{a_j + w: w ∈ W\} $, so that

$$
V = \bigcup_{i=1}^{p^{n-m}}a_i + W
$$

答: Since $\dim(V/W)=\dim(V)-\dim(W)$, we have $\mid V/W | = p^{n-m}$. Since the elements of $V/W$ are equivalence classes, their union is $V$.


问:
Unlike Jensen's Inequality, can we upper bound $\log \sum_{i}{u_i \exp(x_i)}$?
According to Jensen's Inequality, it is not hard to derive the lower bound for $\log \sum_{i}{u_i \exp(x_i)}$ due to the convexity of $\log(\cdot)$:

$\log \sum_{i}{u_i \exp(x_i)} \geq \sum_{i}{u_i x_i}$.

However, my question is, can we somehow upper bound the quantity. The idea is that the set $\{x_i\}$ is bounded by $\min x_i$ and $\max_i$ and the second-order derivative of $\log(\cdot)$ is also bounded in the interval $[\min x_i, \max x_i]$. Therefore, we can upper bound it but the question is to what simplicity the bound could be.

I have tried some combinations under some numerical test, and found this bound might work:

$\log \sum_{i}{u_i \exp(x_i)} \leq \sum_{i}{u_i x_i}+\frac{1}{8}\Big(\max x_i\Big)^2$.

But I do not have a way to prove it. Any suggestions? Thanks.

答:
I guess there is some typo in the previous answer.

If  $f(x) = \log(x)$(a concave function)then you will get the upperbound for log sum as:
$$ \log \left(\sum_{i}p_{i} x_{i}\right) \leq \sum_{i} p_{i} \log \left(x_{i}\right) - \log (a)  - \log (b) + 2 \times \log \left(\frac{a+b}{2}\right)$$

By the way, all variable $x_i$ shall vary from a to b.

Moreover, this inequality is not proposed by SLAVKO SIMIC.


问:
How can there be multiple irreducible representations of a group each having distinct dimension?
The matrix elements for the $(2l+1)$-dimensional irreducible representation of SO(3) are given by:

$D_{m',m}^l(\phi_1,\Phi,\phi_2)=[i^{m'-m}\sqrt{(l+m')!(l-m')!(l+m)!(l-m)!}$
$\times(\frac{1+\cos\Phi}{2})^l (\frac{1-\cos\Phi}{1+\cos\Phi})^{(m'-m)/2}$
$\times\sum\limits_k \frac{(-1)^k}{(l+m-k)!(l-m'-k)!(m'-m+k)!k!}$
$\times(\frac{1-\cos\Phi}{1+\cos\Phi})^k] e^{im'\phi_2}e^{im\phi_1}$

My question is, how can there be multiple irreducible representations of SO(3) each of which has a different dimension?  I thought that by definition an irreducible representation is one which has no non-trivial invariant subspace.  I know that SO(3) can be represented by the group of 3x3 symmetric matrices of determinant = 1 (I thought this had something to do with the fact that it takes 3 parameters to describe a 3D rotation).  If I have a larger matrix (e.g. 5x5) doesn't it have to have a non-trivial invariant subspace (e.g. of dimension 2), and therefore be reducible?

The application here is the expansion of functions using generalized spherical harmonics, which is accomplished (per the Peter-Weyl theorem) using an infinite linear combination of the matrix elements of the irreducible representationS (emphasis on the S) of SO(3).

答: That's the most confusing way to write down the irreducible representations of $\text{SO}(3)$ I've ever seen. Here's a much less confusing one: the irreducible representation of $\text{SO}(3)$ of dimension $2n+1$ is precisely its action on the space of harmonic polynomials in three variables $x, y, z$ of degree $n$ (given by extending its action on $x, y, z$ multiplicatively). Maybe this will help you play with these representations and see why they don't behave the way you think they do. </p>


问:
The successor of a set
Definition : The successor of a set $x$ is the set $S(x) = x \bigcup \{x\}$

Prove that $x \subseteq S(x)$ and there is no $z$ such that $  x \subset z \subset S(x)$

I really battle with proofs :(

Here is what I have:

let $ y \in x $

then $y \in S(x)$

therefore $ x \subseteq S(x) $

I think I have over simplified something here and am not really proving anything :(

The second thing to prove is that there is no $z$ such that $ x \subset z \subset S(x) $

I really don't know how to approach this one.
Would it be by contradiction?
Assume that such $z$ exists, and then show that it can't exist?

I know that the statement $ x \subset z \subset S(x) $ means there is at least one element in $z$ that is not in $x$ and at least one element in $S(x)$ that is not in $z$ which implies that there are elements in $S(x)$ which are not in $x$.  Doesn't this contradict the definition of $S(x)$?

答:
Straightforward proof of 2nd question

Suppose $x\subset z\subset S$ (1)

then $z=x\cup A$ where $A$ is a set such that $A \cap x =\emptyset$

Suppose $A \neq \emptyset$ for non triviality purposes

By definition of $S$ and using (1) $$ (x\cup A) \subset (x\cup \{x\})$$

Since $A \cap x =\emptyset$ , $A \subset \{x\}$

But $\{x\}$ is a single-element set

Hence $A=\{x\}$

Hence $z=S$

Title: The successor of a set


问:
Find $\varlimsup_{x \rightarrow + \infty} (\cos\sqrt{x-2015}-\cos\sqrt{x+2015})$
Find $$\varlimsup_{x \rightarrow + \infty} (\cos\sqrt{x-2015}-\cos\sqrt{x+2015})$$

I don't understand what the overline means. I think about upper limit of the sequence, but I'm not sure that's right for function.
Thank you.

答:
It's right for functions. You can find the definition there https://encyclopediaofmath.org/wiki/Upper_and_lower_limits It's very natural:

Let $f$ be defined on $E \subset \mathbb{R}$ and there exists a sequence $\{y_n\} \subset E$ converging to $x_0$ (i.e. $x_0 \in \overline{E\setminus \{x_0\}}$. Then
$$
\overline{\lim_{x \to x_0}}f(x) = \inf_{\varepsilon>0} \sup\{f(x) : 0 <|x-x_0| < \varepsilon, \, x \in E\}.
$$

When $x_0 = +\infty$, we have a definition
$$
\overline{\lim_{x \to +\infty}}f(x) = \inf_{r>0} \sup\{f(x) : r < x, \, x \in E\},
$$
which is correct if there exists a sequence $\{y_n\} \subset E$ s.t. $y_n \to +\infty$ when $n \to \infty$. For $x_0 = -\infty$ we have analogous definition.


问:
How to prove that a particular form of numbers achieve all remainders modulo $n$?
(Note: Here numbers mean positive integers, all variables used are positive integers)

This query is regarding a particular question where I need to prove that a particular form of numbers ("particular form of numbers", for example squares, cubes, $4n+1$ etc.) can attain every remainder modulo $n$ for some positive integer $n$.

So my question is, what is the general process to approach this type of problem?

For example suppose we need to find the type of numbers $k$ for which the sequence $1^2, 2^2, 3^2 , \cdots $ contains all the remainders modulo $k$.
(That is , the equation
$r^2 = x \pmod k$ has a solution $r$ for all $x=1,2, \cdots k$).

So far, I have not been able to make any rigourous proof. So please give some general method to approach these type of question.

答:
Partial progress :
Here's the proof that $n$ power numbers for any even $n>1$ (i.e. squares, $4$ th powers etc.)  can't achieve all remainders modulo $k$ for $k \gt 2$.

Define sets $A$ and $B$ as
$A=\{ km, km+1, \cdots , km+k-1 \} $
$B=\{ 0,1,2 \cdots , k-1 \}$.
See that $|A|=|B|$.

Define a map $p : A \mapsto B$  such that $p(a)=r$ where $ a^n \equiv r \pmod k$.

For any $a \in A$,
Note that
$a^n \equiv (k-a)^n \pmod k$.  (Since $n$ is even, it is trivial by binomial theorem)

$\implies p $ is not a one one map and because $|A|=|B|$, $p$ can not be onto.


问:
The distance of hyperbola to a point
Let the hyperbola equation is
$$ \frac{x^2}{a^2}-\frac{y^2}{b^2}=1$$
I want to know the distance of the hyperbola curve to any point. Since there are symmetrical branches of hyperbola, I don't know how to derive the equation. Thank you for your help.

答:
Hyperbola: $\displaystyle \frac{x^2}{a^2} - \frac{y2}{b^2} = 1$

Say you have a point $P (h,k)$ which is not on the hyperbola and you want to find its minimum distance to the hyperbola curve.

The square of distance of any point $(x,y)$ on the hyperbola to point $P$ can be written as

$d^2 = (x-h)^2 + (y-k)^2$ ...(i)

Now from the equation of hyperbola,

$\displaystyle y = \pm \frac{b}{a} \sqrt{x^2-a^2}$

Plugging this in $(i)$,

$d^2 = f(x) = (x-h)^2 + (\pm \frac{b}{a} \sqrt{x^2-a^2} - k)^2$

Now we take derivative of this for both plus and minus values of $y$, equate to zero to find critical points and check which point minimizes the square of distance $f(x)$.

$\displaystyle f'(x) = 0 = (x-h) +  \frac{b^2}{a^2} x \mp \frac{b}{a} \frac{kx}{\sqrt{x^2-a^2}}$

I hope you can take it from here based on the given values.

Title: The distance of hyperbola to a point


问:
Clarifying the notion of basic/free variables in a system of equations
I have a fundamental confusion with regards to the notion of free/basic variables.

Consider the following linear system
$$
\begin{pmatrix}
1 & 2 & -1 \\
2 & -4 & 0
\end{pmatrix}\begin{pmatrix}
x_1\\
x_2 \\
x_3
\end{pmatrix}=\begin{pmatrix}
4\\
5
\end{pmatrix}
$$
This system has augmented matrix
$$
\begin{pmatrix}
1 & 2 & -1 & 4 \\
2 & -4 & 0  & 5
\end{pmatrix}
$$
which row reduces to
$$
\begin{pmatrix}
1 & 2 & -1 & 4 \\
0 & -8 & 2 & -3
\end{pmatrix}
$$
The last matrix is in echelon form. Columns 1 and 2 are pivot columns, so $x_1,x_2$ are basic variables. The third column is not a
pivot column, so $x_3$ is a free variable.   Finally, the last column is not a pivot column, so the system is consistent.

The above analysis tells me that $x_3$ is a free variable. I interpret this as saying "$x_1,x_2$ can be expressed as a function of $x_3$, while $x_3$ can be set equal to any number". For example, I can set $x_3=200$. In turn,
$$
\begin{cases}
x_1+2x_2-200=4 \\
2x_1-4x_2=5
\end{cases}\Rightarrow x_2=\frac{403}{8}, x_1=\frac{413}{4}
$$

Nevertheless, just by doing naive computations, I realised that we could also "express $x_1$, $x_3$ as a function of $x_2$, while setting $x_2$ equal to any number". For example, I can set $x_2=7$. In turn,
$$
\begin{cases}
x_1+14-x_3=4\\
2x_1-28=5
\end{cases}\Rightarrow x_1=\frac{33}{2}, x_3=\frac{53}{2}
$$
Alternatively, we could also "express $x_2$, $x_3$ as a function of $x_1$, while setting $x_1$ equal to any number". For example, I can set $x_1=0$. In turn,
$$
\begin{cases}
2x_2-x_3=4\\
-4x_2=5
\end{cases}\Rightarrow x_2=-\frac{13}{2}, x_3=-\frac{5}{4}
$$

Hence, my question: what is the relation between

the fact that $x_3$ is a free variable and $x_1,x_2$ are basic variables (from the row-reduced matrix)

and the fact that in the system above I am free to fix the value of anyone among $x_1,x_2,x_3$ and I will always be able to solve for the other two unrestricted variables

答:
You've got three column vectors, and any pair of them are independent and thus span $\Bbb R^2$. The solution set is one-dimensional (affine) (aka a line) so of the form $v+tw$, where $v,w$ are some vectors  and $t$ is a free real variable. We can choose this to be $x_3$, or any of the other variables if you prefer.

You can then transform the equations to $$\begin{align}x_1+2x_2 &= 4+x_3\\
x_1 - 4x_2 &= 5\end{align}$$

and solve these in terms of $x_3$ uniquely. We get (add twice the first equation to the second and we get $3x_1 = 13+2x_3$ so $x_1 = \frac{1}{3}(13+2x_3)$, and subtraction the equations: $6x_2 = x_3-1$ and hence $x_2 = \frac{1}{6}(x_3 - 1)$ (we could also do another elimination step on this new system) so we can write the solution line as

$$\begin{pmatrix} \frac{13}{3}\\ -\frac16\end{pmatrix} + x_3  \begin{pmatrix} \frac{2}{3}\\ \frac16\end{pmatrix}$$


问:
If $\mathcal V:=\{\vec v_i\}$ and $\mathcal U:=\{\vec u_i\}$ are bases that is $\mathcal V_i:=\{\vec v_1,...,\vec u_i,...\vec v_n\}$ a base?
So let be $V$ a vector space of finite dimension and let be $\mathcal V:=\{\vec v_i:i=1,...,n\}$ and $\mathcal U:=\{\vec u_i:i=1,...,n\}$ two different bases so that there exist $i=1,...,n$ such that $u_i\neq v_j$ for each $j=1,...,n$. Therefore I asked if the collection $\mathcal V_i:=\{\vec v_1,...,\vec v_{i-1},\vec u_i,\vec v_{i+1},...\vec v_n\}$ is a base too and I answered to me that it could be not because it seems to me that it is possible that the vector $\vec u_i$ lies in the subspace spanned by the vector $\vec v_1,...,\vec v_{i-1},\vec v_{i+1},...,\vec v_n$ but unfortunately I did not able to find a counterexample. So could someone clarify if $\mathcal U$ is effectively a base, please?

答: Suppose that $V=\Bbb R^3$, that$$\mathcal U=\{(0,1,1),(1,0,1),(1,1,0)\}\quad\text{and that}\quad\mathcal V=\{(1,0,0),(0,1,0),(0,0,1)\}.$$Then $\mathcal U\cap\mathcal V=\emptyset$, but if the first element of $\mathcal V$ gets replaced by the first element of $\mathcal U$, then you don't have a basis anymore.


答:
Any relation from $\mathbb R^+\rightarrow\mathbb R$ is a subset of $\mathbb R^+\times\mathbb R$ which contains the distinct ordered pairs of type $(a^2,a) $ as well as $(a^2,-a)$ for some $a\ne 0(say)$. Now it is upto you which type of pair you choose exclusively in order to have a well defined function.

Infact, defining the function like $x\mapsto y$ by rule $x=y^2$ doesn't work as it enforces you to pick up both the types of pairs in $f$. In order to have a positive answer define something like $y=\sqrt{x}$ or $y=-\sqrt{x}$.


问:
The preopen set
A subset $S$ of $(X, \tau)$ is said to be "preopen" if $S \subset \text{int}(\text{cl}(S))$.

The family of all preopen subsets of $(X, \tau)$ is denoted by $\text{PO}(X, \tau)$.

The complement of a preopen set in $(X, \tau) $ is called preclosed.

The intersection of all members of the family of all preclosed sets in $(X, \tau)$ containing a set $S$ is called the "preclosure" of $S$ and is denoted by $\text{pcl}_{\tau}(S)$.

I can prove that, if $S$ is preclosed, $\text{pcl}_{\tau}(S) = S$, but I cannot prove when $\text{pcl}_{\tau}(S) = \text{cl}_{\tau}(S)$. Except that $S$ is closed, is there any other condition that $\text{pcl}_{\tau}(S) = \text{cl}_{\tau}(S)$? I mean, if $S$ is preopen, preclosed, or both of them, or $X$ is extremally disconnected, or any other condition, is $\text{pcl}_{\tau}(S) = \text{cl}_{\tau}(S)$?

Can you help me?

($S$ is preclosed is $\text{cl}(\text{int}(S)) \subset S$.)

An extremally disconnected space is a topological space in which the closure of every open set is open.

答:
Some ideas: note that not only a preclosed set is equal to its preclosure, but also the other way: a set equal to its preclosure is preclosed. In other words, any intersection of preclosed sets is still preclosed.

You may want to look at the classical Kuratowski closure complement problem to get some intuition on how a set can look like with respect to its Kuratowski operations.

You have proposed some conditions – if you can't find a proof, try to find a counterexample.

Title: The preopen set


问:
Cardinality of the set of surjective functions on $\mathbb{N}$?
I know that the set of all surjective mappings of ℕ onto ℕ (lets name this set as F) should have cardinality |ℝ|.

How to strictly prove that?

From the fact that cardinality of every possible function is |ℝ|, |F| <= |ℝ|.
Saw similar question on this site, but I need strictly defined function that shows that |F|>=|ℝ|.

Thank you for your time.

答:
HINT: Let $M=\{2n+1:n\in\Bbb N\}$. For each $A\subseteq M$ define

$$f_A:\Bbb N\to\Bbb N:n\mapsto\begin{cases}
n/2,&\text{if }n\in\Bbb N\setminus M\\
1,&\text{if }n\in A\\
0,&\text{if }n\in M\setminus A\;.
\end{cases}$$


问:
Angle between tangents to circle given by $x^2 + y^2 -12x -16y+75=0$?

Given the circle: $C(x,y)=x^2 + y^2 -12x -16y+75=0$, find the two tangents from origin

First, I get the line which passes through point of contact of tangents from origin using result here which is :

$$ -12x-16y-2 \cdot 75 = 0$$

Or,

$$ 6x + 8y -75 =0 $$

Now, I use the result discussed in this answer, which says that pair of straight from point $(P)$ to conic is given as:

$$ C(0,0) C(x,y) =  (6x+8y-75)^2$$

This leads to:

$$ 75 (x^2 + y^2 -12x-16y+75) = (6x+8y-75)^2$$

$$ 0 = 75(x^2 +y^2 - 12 x - 16y +75) - (6x+8y-75)^2$$

For applying the result in this answer, then the formula

$$ a= 75 - 36,  b=75-64 , h= - \frac{(6 \cdot 8 \cdot 2 )}{2}$$

Or,

$$ a = 39, b= 11 , h=-48$$

$$ \tan \theta = \frac{2 \sqrt{(-48)^2-39 \cdot 11)}}{39+11} = \frac{2 \sqrt{5^4 \cdot 3}}{25} = 2 \sqrt{3}$$

However, the intended answer was:

$$ \tan \theta = \frac{1}{\sqrt{3} } $$

Where have I gone wrong?

答:
Found the problem! To understand why it fails , refer back to the derivation here, it was assumed that:

$$C(P) + \lambda T(P)^2 = 0$$

Where we are subbing in the coordinate of point $P$ into equation of circle, and we assumed that $T(P) = C(P)$ , now this is not true if we directly scale up the equation of tangents (read Jocahimsthal's notation for understanding), now suppose we say,:

$$ T'(x,y) = \lambda T(x,y)$$

Then,

$$ \frac{T'(P)}{\phi} = T(x,y)$$

Or,

$$C(P) = \frac{T'(P)}{\phi} \tag{1} $$

And if we used scaled tangent equation in original equation

$$ C(P) +  \lambda (T'(P))^2=0 \tag{2}$$

From (1) and (2),

$$ C(P) + \lambda ( \phi C(P) )^2 = 0$$

Or,

$$ -\frac{1}{\phi^2 C(P)} = \lambda$$

Which changes our tangent equation to:

$$ C(x,y) C(P) = \frac{T'(x,y) )}{\phi^2}$$


问:
l'Hospitals rule for multivariate limit
I have the following bivariate function where $\alpha,\beta,\gamma,\delta$ are constants, $a,b,c,d$ are integers and $x,y\in [0,1]$
$$
f(x,y) = \frac{\alpha(1-x^ay^b) + \beta(1-x^cy^d)}{\gamma(1-x^2)+\delta(1-y^3)}
$$
which I would like to examine the limit of as $x=y=1$ from below. My instinct was to apply l'Hospital's rule by setting $x=y$ and converting to a univariate function.
$$
f(x,x) = \frac{\alpha(1-x^{(a+b)}) + \beta(1-x^{(c+d)})}{\gamma(1-x^2)+\delta(1-x^3)}
$$
from which we can gather the limit is
$$
\lim_{x\rightarrow 1}f(x,x) = \frac{\alpha(a+b)+\beta(c+d)}{2\gamma +3\delta}
$$
However, am also aware that l'Hospital's rule is strictly for univariate functions. In this particular case, how can I be certain that my limit is correct?

答:
Your approach is incomplete. If the bidimensional limit exists, the univariate one that you presented also exists and has the same value. However, the computation of the univariate limit only shows that if the bidimensional exists, it must have the same value.

What you have computed, by taking $y=x$, is called a directional limit. You could have considered, more generally, that $y = 1 + m(x-1)$. In this case, the directional limit (here you may use L'Hopital's rule) is given by
$$
\lim_{x\to 1} f(x, 1+m(x-1))=
\frac{\alpha (a + b m) + \beta (c + d m)}{3 \delta  m + 2 \gamma}.
$$

The dependence in $m$ proves that the bidimensional limit does not exist.

Title: l'Hospitals rule for multivariate limit


问:
Is hypothesis space a random variable?
I am reading where it states:

The previous statement is true for a given training set $D$. However, remember that $D$ itself is drawn from $P^{n},$ and is therefore a random variable. Further, $h_{D}$ is a function of $D,$ and is therefore also a random variable. And we can of course compute its expectation:

The definition of hypothesis space is given in the section Loss Functions.

There are typically two steps involved in learning a hypothesis function $h() .$ First, we select the type of machine learning algorithm that we think is appropriate for this particular learning problem. This defines the hypothesis class $\mathcal{H},$ i.e. the set of functions we can possibly learn. The second step is to find the best function within this class, $h \in \mathcal{H}$. This second step is the actual learning process and often, but not always, involves an optimization problem. Essentially, we try to find a function $h$ within the hypothesis class that makes the fewest mistakes within our training data. (If there is not a single function we typically try to choose the "simplest" by some notion of simplicity - but we will cover this in more detail in a later class.) How can we find the best function? For this we need some way to evaluate what it means for one function to be better than another. This is where the loss function (aka risk function) comes in. A loss function evaluates a hypothesis $h \in \mathcal{H}$ on our training data and tells us how bad it is. The higher the loss, the worse it is - a loss of zero means it makes perfect predictions. It is common practice to normalize the loss by the total number of training samples, $\mathrm{n},$ so that the output can be interpreted as the average loss per sample (and is independent of $\mathrm{n}$ ).

This is how I would rewrite the document:

We have a table where the last column are targets (predictions). This can be represented as a multivariate random variable or random vector $X$ and a target random variable $Y$. The data distribution of the random variable $D$ representing our dataset is: $P(X,Y)$. When concrete dataset $\mathcal D$ is drawn from sample space $P^n$ we get our dataset $\mathcal D$ with $n$ rows.

But I am not sure of $h$? It is inside the "Loss Function" paragraph where actually to me it reflects the machine learning algorithm function. One of the infinitely many. Further hypothesis class may be a random variable $\mathcal H$, and $h$ just a realization.

1. How can dataset $\mathcal D$ be a random variable?

In here when they refer to a dataset they use $D$.

2. How can $h_{D}$ be a random variable if this is a function associated to a concrete dataset, and most likely it means a function tryout (one of the many) that translate the input to the output?

答:
I will summarise you what is a learning task. There is a set of features $X$ which you can observe and a set of features $Y$ which you have to guess. And there is a probability distribution $P(X,Y)$ which is constant but unknown. We can make some assumptions about this distribution but it's not important now.

You are looking for a function $h:X\to Y$ which guesses the feature $y$ only based on the feature $x$. There are many such functions and you are looking for the function $h$ with minimal loss. You can either search in all possible functions or you restrict yourself for functions in a so-called hypothesis class $H$.

If your function $h$ guesses $y$ from $x$ wrongly, then the bigger $P(x, y)$ is, the bigger impact it has on the loss. To calculate the loss, you actually need $P(X,Y)$, which you don't have. You can only estimate it from the dataset. However, the loss you calculate is also only an estimation.

The doubles $(x,y)$ taken independently from $P(X,Y)$ are random variables. So a list of these doubles, which is called dataset $D$, is also a random variable. The functions in the hypothesis space $H$ are not random variables. However, their estimations of losses on the dataset $D$ are random variables. And when you are looking for the minimum of random variables, it is also a random variable.

And the argument of minimum, the function $h_D$ you choose, may not be the same function as you would choose if you knew the distribution $P(X,Y)$. So it is actually also a random variable.

Is it more clear now? And if not, can you please tell me which paragraph you don't understand?

Title: Is hypothesis space a random variable?


问:
Correlation coefficient between X and Y, linear regression.
The formula that I know for correlation coefficient r:$\frac{\sum (x_i - \bar{x}) (Y_i - \bar{Y})}{\sqrt{\sum(x_i-\bar{x})^2\sum(Y_i-\bar{Y})^2}}$

If the only given values I have are $\sum x_i, \sum x_i^2, \sum y_i, \sum x_iy_i$ is it even possible to compute the correlation coefficient?

答:
Apparently you are referring Pearson's sample correlation coefficient. In that case, one of the known alternative formulas is

$ r_{xy} = \frac{\sum x_i y_i - \sum xi \sum yi}{\sqrt{ n \sum x_i^2 - \left ( \sum x_i \right )^2}\sqrt{ n \sum y_i^2 - \left ( \sum y_i \right )^2}} $

So I believe that you would still need to have at least $\sum y_i^2$ and $n$ to compute it.


问:
How do mathematically I represent data binding between a set and a list?
Suppose I have a set of entities e.g.

$$S = \{apple, cat, hat, fruit\}$$

and I have their corresponding similarity scores

$$ E = \{0.8, 0.1, 0.9, 1.0\} $$

How do I represent the S in decreasing order of their corresponding similarity scores in E in another list?

This is what I have done

We are using scores in E and sorted the list R in decreasing order
$$R = [s_i, s_{i+1}, s_{i+2}, ... , s_n] : s \varepsilon S$$ where $$\{s_i > s_{i+1} > s_{i+2} ... > s_n\} : s \varepsilon E$$

I apologize in advance if I have come up with an unintuitive solution but after trying for hours, this is what I could only think of. Is it correct and how can we represent it in a better way?

答:
We could create a mapping $f: S \to E$ such that $f(apple)=0.8,\ f(cat)=1.0, f(hat)=0.9$ and $f(fruit)=1.0$.

Then we could define our set $R$ as follows:

$$R=\{s_1,\cdots,s_{|S|} |\ f(s_i) \geq f(s_j),\ i < j \leq |S|, s_i,s_j \in S \}$$


问:
How to find inverse function of function with parameter?
I'm trying to find the inverse function of this function:
$$
f_a(x) = \frac{x^a}{x^a + (1-x)^a}
$$

$a$ - parameter

$x$ - variable

Is it possible to find if there is a parameter here?

Hope for your help! Thanks in advance!

答: $$f_a(x) = \frac{x^a}{x^a + (1-x)^a} \implies \frac 1{f_a(x)}=1+\frac{(1-x)^a } {x^a}\implies $$ $$\frac 1{f_a(x)}-1=\Bigg[\frac {1-x}x\Bigg]^a\implies\Bigg[\frac 1{f_a(x)}-1\Bigg]^{\frac 1a}=\frac {1-x}x\implies x=???$$


问:
Circle through point of contact of tangents from external point of another cirlce
In this truly amazing answer to a question I had asked this morning, a following construction is done:

Consider the  a vector from origin to a point on the $(x,y) $ plane, also a circle $C$ having center $(a,b)$. The vector from the circle $C$ center to this point is $(x-a,y-b)$ , now it seems that the set of points for which these two vectors are perpendicular are another circle:

$$ (x,y) \cdot (x-a,y-b) = 0$$

$$ x^2 - ax + y^2 - by = 0$$

But I just can't understand why this set of points satisfying above two criteria must be a circle, could someone shed some intuition on why we should expect this? You can find an example of the above discussion here

答:
The locus of points subtending a right angle on the segment $OC$ is a circle with diameter $OC$ and center its midpoint.

The above equation can be rewritten more clearly as
$$\left(x-\frac{a}{2} \right)^2+\left(y-\frac{b}{2} \right)^2=\left(\frac{\sqrt{a^2+b^2}}{2} \right)^2=\left(\frac{OC}{2} \right)^2$$


问:
$\frac 21 \times \frac 43 \times \frac 65 \times \frac 87 \times \cdots \times \frac{9998}{9997} \times \frac {10000}{9999} > 115$
Prove that
$$x = \frac 21 \times \frac 43 \times \frac 65 \times \frac 87 \times \cdots \times \frac{9998}{9997} \times \frac {10000}{9999} > 115$$

saw some similar problems like
show $\frac{1}{15}< \frac{1}{2}\times\frac{3}{4}\times\cdots\times\frac{99}{100}<\frac{1}{10}$ is true

but didn't manage to get 115. I could get a weaker conclusion of $x>100$ though.

\begin{align}
x^2 &= \left(\frac 21 \times \frac 21\right) \times \left(\frac 43 \times \frac 43\right) \times \cdots \times \left(\frac{10000}{9999} \times \frac {10000}{9999}\right) \\
&\ge \left(\frac 21 \times \frac 32\right) \times \left(\frac 43 \times \frac 54\right) \times \cdots \times \left(\frac{10000}{9999} \times \frac {10001}{10000}\right) \\
&= 10001
\end{align}

so $x > 100$

Body:
We have
$$\dfrac{2}{1}\cdot \dfrac{4}{3}\cdots \dfrac{10000}{9999}= \dfrac{10000!!}{9999!!}= \dfrac{2^{5000}\cdot 5000!}{\dfrac{10000!}{2^{5000}\cdot 5000!}}= 2^{10000}\cdot \dfrac{\left ( 5000! \right )^{2}}{10000!}$$
Using Sterling's approximation $n!= \sqrt{2n\pi}\left ( \dfrac{n}{e} \right )^{n}$ then we have
$$x\cong 2^{10000}\cdot \dfrac{\left ( \sqrt{10000\pi}\left ( \dfrac{5000}{e} \right )^{5000} \right )^{2}}{\sqrt{20000\pi}\left ( \dfrac{10000}{e} \right )^{10000}}= 2^{10000}\cdot \dfrac{50\sqrt{2\pi}}{2^{10000}}= 50\sqrt{2\pi}$$
Another solution is
$$x= \frac{2^{10000}}{\dbinom{10000}{5000}}= \frac{2^{10000}}{\dfrac{2^{10001}}{\pi}\int_{0}^{\infty}\dfrac{{\rm d}x}{\left ( x^{2}+ 1 \right )^{5001}}}= \frac{\pi}{2\int_{0}^{\infty}\dfrac{{\rm d}x}{\left ( x^{2}+ 1 \right )^{5001}}}$$
Then by the transformation $\int_{0}^{\infty}f\left ( x \right ){\rm d}x= \int_{0}^{1}\dfrac{f\left ( \dfrac{x}{1- x} \right )}{\left ( 1- x \right )^{2}}{\rm d}x$ we have
$$\int_{0}^{\infty}\frac{{\rm d}x}{\left ( x^{2}+ 1 \right )^{5001}}= \int_{0}^{1}\frac{{\rm d}x}{\left ( \left ( \dfrac{x}{1- x} \right )^{2}+ 1 \right )^{5001}\left ( 1- x \right )^{2}}= \int_{0}^{1}\frac{\left ( x- 1 \right )^{10000}}{\left ( 2x^{2}- 2x+ 1 \right )^{5001}}{\rm d}x=$$
$$= \int_{0}^{1}\frac{\left ( \dfrac{x^{2}- 2x+ 1}{2x^{2}- 2x+ 1} \right )^{5000}}{2x^{2}- 2x+ 1}{\rm d}x$$
We can doublecheck the result then in Wolfram|Alpha or Mathematica.


问:
Joint density of uniform variables on different support
I have two random variables $X$ and $Y$ where $X$ is uniformly distributed in $[1/3,4/3]$ and $Y$ is uniformly distributed in $[0,1]$. I would like to get the probability that $X$ is greater than $Y$.  I assume I need the joint density first which should be :

$$f(x,y) = \begin{cases}1, & (x,y)\in[1/3,4/3]\times[0,1]\\ 0, & \mathrm{else} \end{cases} $$

According to this, I assume the $Pr(X>Y)$ should be equal to $2/3$ but I am not sure if this is correct.

答: Uniform distribution makes this problem much simpler, as probability is proportional to area of domain.  Graph the square that is your domain. Then graph y=x. The region below/to the right of that line has area equal to the probability you want.  In particular, the area not included is the triangle's 1/2 2/3 2/3 = 2/9, so the answer is 7/9.</p>


问:
Find a density function $g$ that verify the following integrale convergence and divergence.
For a given real $a>0$, I'm looking for a measurable function $g : [0,a] \rightarrow \mathbb{R}^+ \in L^1([0,a])$  (i.e $g$ is a density for a random variable valued in $[0,a]$) such that :

$\int_0^a |\ln x| g(x) \ dx $ is a divergent integrale.

$\int_0^a |\ln y|  g(y) \left( \int_0^y  g(x)\ dx \right) \ dy$ is a convergent integrale.

I actually have no idea if such a function exists, do you have any thoughts on this ?

答:
I'm gonna choose $a<1$ to avoid sign issue on the log. The function :

$$g : x \mapsto \frac{1}{x |\ln(x)|^2}$$

should definitely work using Bertrand integrale convergence. To verify the second hypothesis we just have to note that :

$$\int_0^y g(x) dx = - \frac{1}{\ln(y) }$$
for $y < a < 1$.


问:
natural numbers and successors
I hope this question is not a repeat.

Given there is no $k \in N $ such that $ n < k < n+1 $ for $n \in N$

Prove for all $ m, n \in N $ if $ m < n $ then $ m+1 \leq n$

Conclude $ m < n $ implies $ m+1 < n + 1 $  and therefore $S(n) = n+1 $ defines a one - to - one function on $ N $.

NOTE $ N = \text{  natural numbers}$

Ok, this is what I have :

from the given, the first number greater than $n$ is $n+1$.

therefore if $m < n$ then $n \geq m+1 $

therefore $m+1 \leq n $

Is this correct?

I am not sure how to proceed with the second part, where I need to make the conclusions.

since $m < n$ I know from the proof that $ m + 1 \leq n$ and since $ n < n+1 $ then $ m + 1  < n + 1$

is this correct?

showing if S(n) is one to one:

assume $S(a) = S(b)$ for some $a, b \in N$
then $ a + 1 = b + 1 $
then $ a = b $
therefore $ S(n) $ is a one to one function.

答:
The first conclusion is almost immediate. Suppose, seeking a contradiction, that there were $m, n \in \mathbb{N}$ such that $m< n$ but $m + 1 > n$. But that would mean that there was an $n \in \mathbb{N}$ such that $m < n < m+ 1$, a contradiction. So we must have $m < n \implies m+1 \leq n$. It immediately follows that $m+1 < n+1$, as you have shown. </p>

<p>To show it's injective, I would again use proof by contradiction. Suppose there were $a, b \in \mathbb{N}$ such that $a+1 = b+1$, but $a \neq b$. Then, WLOG, assume $a<b$. But then we have $a+1 < b+1$, a contradiction by what we've just proved. So we're done. </p>

Title: natural numbers and successors

Body:
The first conclusion is almost immediate. Suppose, seeking a contradiction, that there were $m, n \in \mathbb{N}$ such that $m< n$ but $m + 1 > n$. But that would mean that there was an $n \in \mathbb{N}$ such that $m < n < m+ 1$, a contradiction. So we must have $m < n \implies m+1 \leq n$. It immediately follows that $m+1 < n+1$, as you have shown.

To show it's injective, I would again use proof by contradiction. Suppose there were $a, b \in \mathbb{N}$ such that $a+1 = b+1$, but $a \neq b$. Then, WLOG, assume $a<b$. But then we have $a+1 < b+1$, a contradiction by what we've just proved. So we're done.


问:
A number N in base 10, is 503 in base b and 305 in base b + 2. What is the product of the digits of N?
Pls help me with A number N in base 10, is 503 in base b and 305 in base b + 2. What is the product of the digits of N? I amen't solve it. Help needed</p>

答:
A number $\overline{xyz}$ in base $b$ is equal to $x\cdot b^2 + y\cdot b + z$ and so on for larger numbers continuing the pattern increasing the power of $b$ as we continue to the left.  For instance, in base 10 we have $35013$ is equal to $3\cdot 10^4 + 5\cdot 10^3 + 0\cdot 10^2 + 1\cdot 10 + 3$.

So, we are told that $N = 5\cdot b^2 + 0\cdot b + 3$ and also that $N = 3\cdot (b+2)^2 + 0\cdot (b+2) + 5$

Setting these equal, we can find $b$.

Now, armed with the knowledge of what $b$ is we can find $N$.

Now, armed with the knowledge of what $N$ is in decimal, we can find the product of the digits of $N$.


问:
Coordinates of a point after rotating a line from one end
I am working on a project in javascript and p5.js and am stuck at this point. I have to rotate a line from one of its ends with a certain angle in Radians/Degrees. Here's what I want to get-the image

Sorry for my bad drawing. So I want a formula to get C(x,y) to use in my code.

What should be the values of x and y in terms of t-ratios and x1,y1,x2,y2

Sorry,flip the positions of A and B.

答: A \theta counterclockwise rotation matrix about the origin takes an (x,y) input and gives an output $(x cos \theta - y sin \theta, x sin \theta + y cos \theta) $. So you can shift your angle vertex to the origin, perform the rotation, and then shift back; if A is the vertex with coordinates (x1,y1) then subtract that from B, transform, and then add A back on to get C(x,y).


问:
Fenchel-Young inequality for matrices
Background: For $f:D\subset\mathbb{R}^d \to \mathbb{R}$ we call
$$f^*:D^*\subset\mathbb{R}^d \to \mathbb{R}, ~ y\mapsto \underset{x\in D}{\sup}(\langle y, x \rangle - f(x))$$
the conjugate of $f$, where $D^*:=\{y\in\mathbb{R}^d | \underset{x\in D}{\sup}(\langle y, x \rangle - f(x)) < \infty \}$. From the definition we directly get the Fenchel inequality:
$$
\langle y, x \rangle \leq f(x) + f^*(y)
$$
If we set $D=\mathbb{R}^d, f(x)=\lVert x \rVert$ for some vector norm $\lVert \cdot \rVert$ and consider the dual norm $ \lVert y \rVert^* = \underset{\lVert x \rVert=1}{\sup}\langle y, x \rangle$ (which in fact is a dual norm thanks to Riesz' representation theorem) we get
$$
f^*(y) = \begin{cases} 0, &\lVert y \rVert^* \leq 1 \\
                       \infty, &\mathrm{otherwise},
 \end{cases}
$$
and therefore $\langle y, x \rangle \leq \lVert x \rVert$ whenever $\lVert y \rVert^* \leq 1$.

My questions: The author of a paper I'm reading defines something he calls a matrix dual norm by
$$
\lVert W \rVert^* := \underset{\lVert u \rVert = \lVert v \rVert = 1}{\sup} u^TWv
$$
for some vector norm $\lVert \cdot \rVert$ and $W\in\mathbb{R}^{d\times d}$. He then claims that
$$
\langle A, B \rangle_{\mathrm{F}} \leq \lVert A \rVert\qquad (*)
$$
by the Fenchel-Young inequality whenever $\lVert B \rVert^* \leq 1$, where $\langle \cdot, \cdot \rangle_{\mathrm{F}}$ is the Frobenius SP and I'm guessing (this is not clearly defined) $\lVert \cdot \rVert$ is the matrix norm induced by the aforementioned vector norm $\Vert \cdot \Vert$.

How is the defined matrix norm $\lVert \cdot \rVert^*$ a dual norm? I'm only familar with $\mathbb{R}^{d\times d}$ as a Hilbert space when equipped with the Frob. SP, but the dual norm defined here does not seem to coincide with the dual norm induced by the Frobenius SP.

Since the author does not use the Frobenius norm as dual norm, can we still apply Fenchel-Young to achieve the inequality (*)?

Thanks a lot!

Edit:
I just realized that it might be somewhat exaggerated to use Fenchel-Young here, since for all $x \neq 0, y \in \mathbb{R}^d$ with $\lVert y \Vert^* \leq 1$ we have
$$
\langle x, y \rangle = \langle x/\lVert x \rVert, y \rangle \lVert x \rVert \leq \lVert y \Vert^* \lVert x \rVert \leq \lVert x \rVert.
$$
My second question therefore reduces to whether the following is true:
$$
\langle A, B \rangle_F = \langle A/\lVert A \rVert, B \rangle_F \lVert A \rVert \overset{\mathrm{?}}{\leq} \lVert B \Vert^* \lVert A \rVert.
$$
I'm pretty sure it is not since by choosing $\lVert \cdot \lVert = \lVert \cdot \lVert_1$ we have
$$
\lVert B \Vert^* = \underset{\lVert u \rVert_1 = \lVert v \rVert_1 = 1}{\sup} u^TBv = \underset{i,j}{\max}|b_{i,j}|
$$
but choosing $A$ appropriately (one entry per column +/-1, otherwise zero s.t.  $\lVert A \rVert_1$ = 1)
$$
\langle A, B \rangle_F = \sum_{j=1}^d \underset{i=1, ..., d}{\max} |b_{ij}| > \lVert B \Vert^*
$$
Could someone confirm?

答:
If we define the dual norm on $\mathbb{R}^{m\times n}$ as
$$
\lVert X \rVert^* = \sup\{\operatorname{Tr}(X^\top Y)~\vert~\lVert Y \rVert \leq 1\} = \sup\{\lvert\operatorname{Tr}(X^\top Y)\rvert~\vert~\lVert Y \rVert \leq 1\},
$$
thus
$$
\langle X,Y\rangle_F = \langle X,Y/\lVert Y\rVert\rangle_F\lVert Y\rVert\leq \lVert X \rVert^*\lVert Y\rVert
$$

Furthermore, we have an induced dual matrix norm $\lVert A \rVert^*$ equivalently defined as
$$
\lVert A \rVert^* = \sup_{\lVert x\rVert^* = 1,\lVert y\rVert=1}x^\top Ay.
$$

Proof:
From the definition of dual norm on vectors:
$$
\lVert z\rVert^* = \sup_{\lVert y \rVert=1}\lvert y^\top z\rvert.
$$

Thus
$$
\lVert A\rVert^* = \sup_{\lVert x\rVert^*=1}\lVert Ax\rVert^* = \sup_{\lVert x\rVert^*=1,\lVert y\rVert=1}\lvert y^\top Ax\rvert.
$$

EDIT 1: To complement the answer a bit, we also have from Fenchel-Young on vectors
$$
x^\top A^\top By\leq \lVert Ax\rVert^*\lVert By\rVert \leq \lVert A\rVert^*\lVert B\rVert \lVert x\rVert^*\lVert y\rVert.
$$

The dual norm of the spectral norm $\lVert A \rVert_2$ is not itself. It is in fact the so-called nuclear norm, and a proof is available here: https://math.stackexchange.com/a/1145246/443030

$\lVert A\rVert_1$ is the dual of $\lVert A\rVert_\infty$, for matrices, and the inequality still holds.

Title: Fenchel-Young inequality for matrices


答:
Starting with your corrected sketch:

HINT: Let point $D$ be a point on $\overline{AB}$ such that angle $\angle{ADC}$ is a right angle. Using trig functions, we can then calculate the side lengths of $\triangle ADC$. Do you see how those side lengths would be enough to find $(x,y)$?

Title: Coordinates of a point after rotating a line from one end

Body:
I am working on a project in javascript and p5.js and am stuck at this point. I have to rotate a line from one of its ends with a certain angle in Radians/Degrees. Here's what I want to get-the image

Sorry for my bad drawing. So I want a formula to get C(x,y) to use in my code.

What should be the values of x and y in terms of t-ratios and x1,y1,x2,y2

Sorry,flip the positions of A and B.


答:
Using complex numbers, translate both points to bring the rotation center to the origin, rotate by multiplying by $e^{i\theta}$ and translate back to the original center.

$$q'=(q-p)e^{i\theta}+p.$$

You can rewrite this formula using real numbers only.


问:
Prove that a non-abelian group of order $pq$ ($p<q$) has a nonnormal subgroup of index $q$
So I've come up with a proof for the following question, and I'd like to know if it's correct (as I couldn't find anything online along the lines of what I did).

Question Let $p$ and $q$ be primes with $p<q$.  Prove that a non-abelian group of order $pq$ has a nonnormal subgroup of index $q$, so there there eixists and injective homomorphism into $S_q$

First I prove a Lemma:

If $H\triangleleft G$, then for every conjugacy class $\mathcal{K} \subseteq H$ or $\mathcal{K} \cap H = \emptyset$.

proof:

If $x\in K\cap H$, then $gxg^{-1} \in gHg^{-1}$ for all $g\in G$.  Since $H$ is normal $gHg^{-1}=H$, so that $H$ contains all the conjugates of $x$,i.e., $\mathcal{K} \subseteq H$ (Dummit and Foot, 4ed, pg 127).

Here's my attempt on the question:

proof:

Let $G$ be a group such that $|G|=pq$ and $p<q$.   Then $G \cong \mathbb{Z}_{pq}$, and so $G$ has two subgroups, say $H$ and $K$, of orders $p$ and $q$, respectively.  'Suppose that $H$ is a subgroup in $G$ of order $p$; a subgroup $K$ of order $q$ exists in $G$ by Cauchy (Nicky Hekster).'  Also, by Lagrange, $H=\langle x \rangle$ and $K=\langle y \rangle$ for some $y,x \in G$.  Since $p$ is the smallest prime that divides the order of $G$ and $|G:K|=p$, then $K\triangleleft G$ (by some theorem).

Suppose, by contradiction, that $H\triangleleft G$.  Then for $x^i \in H$, $gx^ig^{-1}=x^k \in H$ for all $g\in G$.  So $H\subseteq \mathcal{O}_{x^i}$ (the conjugacy class of $x_i$).  Since $|x^i|$ divides $p$, then $x_i\not \in K$.  Thus $\mathcal{O}_{x_i} \not \subseteq K$.  Also, since $e=x^p$ and $gx^pg^{-1}=e$, then $e\in \mathcal{O}_{x_i}$.  Thus $\mathcal{O}_{x_i} \cap K \neq \emptyset$.  Therefore $K$ is not normal in $G$, which is absurd.

Now let $\phi: G \rightarrow S_q$ be the permutations representation afforded by the action of $G$ on $A=G/H$ by left multiplication -- say $\pi_g:A\rightarrow A$ defined by $aH\mapsto gaH$ for all $g\in G, \; aH\in A$.  We know that $\ker \pi_g \leq H$ and $\ker \pi_g \triangleleft G$.  Since $|H|=p$, then $|\ker \pi_g|=p$ or $|\ker \pi_g|=1$.  If $|\ker \pi_g|=p$, then $\ker \pi_g$ is not normal in $G$.  Therefore, $|\ker \pi_g|=1$.  Since $\ker \pi_g=\ker \phi$, then $\phi$ must be injective.

Sorry for any typos (and a HUGE 'thank you' to anyone who actually reads this).

答: Since $G$ is non-Abelian the center $Z(G)$ must be trivial. Therefore, every conjugacy class except $\langle 1\rangle$ must be of order $p$ or $q$. If a subgroup of index $q$ (and therefore order $p$) is normal it must be the union of conjugacy classes. So we must have $p = 1+mp+nq$ for some non-negative integers $m$ and $n$ which is not possible.


问:
Relation between different types of elliptic functions
Disclamer: all formulas in this post are intended to be schematic, coefficients signs etc. are mostly wrong.

I do various physics-motivated computations involving elliptic functions and the more sources I look at the bigger grows my confusion. It seems there are many different parametrizations of elliptic functions, for example

Elliptic integrals such as $K(x)=\int_0^1\frac{dt}{\sqrt{(1-t^2)(1-xt^2)}}$.

Jacobi theta functions like $\vartheta_3=\sum_{n\in\mathbb{Z}}q^{n^2}$

Double-periodic functions such as Weierstrass function $\wp(z)\sim\sum_{n,m}\frac{1}{(z-n-m\tau)^2}$

Modular forms like $E_2(q)=1-24\sum_n\frac{nq^n}{1-q^n}$ or $\eta=q^{1/24}\prod_n(1-q^n)$

For example, some relations between them are

$K(x)=\frac{\pi}{2}\vartheta_3^2(q)$ for $q=\exp\Big(-2\pi\frac{K(1-x)}{K(x)}\Big)$

$\vartheta_1'=\eta^3$

$\wp(z)\sim \frac1{z^2}+z^2E_2+z^4E_4+\dots$

Is it possible to give a concise summary of how different types of functions are related to each other? Make these relationships less surprising? For example I've known for a long time that $\vartheta_3^2(q)=K(x)$ but still learning that $\vartheta_2^2(q)=xK(x)$ surprised me.

Of course a deeper theory behind all these objects exist but learning it seems quite an investment. Are there some exposition intermediate between full-blown modern algebraic geometry and simple statement of definitions and formulas relating them? Especially I'm interested in some resource that would connect many of these thigs. For example Mumford's lectures on theta-functions are great but they mostly don't touch other types of functions mentioned above.

答:
One great book that I've found is Elliptic Curves
Function Theory, Geometry, Arithmetic by Henry McKean and Victor Moll.


问:
Can the concept of field extensions be applied equally well to UFDs?
In a nutshell, a field extension is where you take a polynomial $p(x)$ that is irreducible in some field $F$, then define $\alpha$ as a root of $p$, then add $\alpha$ to $F$, then add the minimum number of additional elements to $f$ to maintain its field structure.

Can this still be done if $F$ is merely a UFD rather than a field?  Does something fundamental break down in the construction of a "UFD extension" that goes through in a field extension?  If not, why do people talk about field extensions when "UFD extension" seems more general?

答:
Given a ring $R$ and an irreducible polynomial $p(x)\in R[x]$, you can construct the ring $R[\alpha]$ in the same way as you would a field extension of the form $F(\alpha)$, namely $R[\alpha]\cong R[x]/(p(x)).$

Even if $R$ is a UFD, there's no guarantee that $R[\alpha]$ will be a UFD. For example, take $R=\mathbb{Z}$ and $p(x) = x^2+5$. Then the resulting ring is $\mathbb{Z}[i\sqrt{5}]$, which is not a UFD: in this ring, $6$ factors as both $2\cdot 3$ and as $(1+i\sqrt{5})(1-i\sqrt{5})$.


问:
Is this a contradiction?
I have proven that a group has one and only one element $a$ such that $a=a^2$, that is $a$ is the square of itself.

However, then I encounter this question about finding $2\times 2$ matrices such that $A=A^2$. It turned out that there are more than one of them.

Could anyone explain why that is the case? I understand that these "special" matrices are in a group of matrices.

The three matrices I have found so far are:
\begin{bmatrix}
1&0\\
0&1
\end{bmatrix}
,\begin{bmatrix}
0&1\\
1&0
\end{bmatrix}
and the zero matrix.

Thank you very much.

Nick

答:
$a=a^2\implies a^{-1}a=a^{-1}a^2\implies a=e$(identity element) holds.

The validity of above implication is due to the guaranteed existence of $a^{-1}$ in group as you pre-multiplied by this inverse element. Such existence can not be promised in case of matrices so $A=A^2\not\implies A=I$(identity matrix) unless you are dealing with a multiplicative group of invertible matrices.

Title: Is this a contradiction?


问:
$a+\frac{1}{2M}\le\int^b_a xf(x)\,dx\le b-\frac{1}{2M}$

Let $f:[a,b] \to (0, +\infty)$ be a continuous function such that $\int^b_a f(x)\,dx =1 $ and $M=\max_{x\in[a,b]} f(x)$. Prove that
$$a+\frac{1}{2M}\le\int^b_a xf(x)\,dx\le b-\frac{1}{2M}.$$

This is the problem where I thought I could use that  $\int^b_a xf(x)\,dx =\frac{a+b}{2}$ because from the mean value theorem exists $c\in(a,b)$ such that $f(x)=(b-a)f(c)=1$ and also $f(c)<M$, so $1/(b-a)<M$, so
$$
\frac{1}{2M}+a< \frac{a+b}{2}<b-\frac{1}{2M}
$$
It is useful?It is another way I could prove this $$a+\frac{1}{2M}\le\int^b_a xf(x)\,dx\le b-\frac{1}{2M}.$$ ?

答:
hint

make the substitution $$t=(a+b)-x$$

the integral becomes
$$I=\int_a^bxf(x)dx=$$
$$\int_a^b(a+b-t)f(a+b-t)dt=$$
$$(a+b)\int_a^bf(x)dx-\int_a^btf(a+b-t)dt$$


问:
inequality of a continous function for which $f(a)=f(b)=0$
Let $f:\Bbb R \to \Bbb R$ a twice diferrentiable function with its derivate continous and $f(a)=f(b)=0$. Prove that
$$2\int^b_a (f'(x))^2\,dx\le\int^b_a (f(x))^2\,dx +\int^b_a (f''(x))^2\,dx$$
The only idea that I have is to separate de parts maybe like this
$$\int^b_a (f'(x))^2\,dx-\int^b_a (f(x))^2\,dx\le \int^b_a (f"(x))^2\,dx-\int^b_a (f'(x))^2\,dx$$ but from I do not know what to do.

答: First integrate $\displaystyle \int_a^b f'(x) f'(x)\, dx$ by parts. Then apply an inequality of the form $ab \le \dfrac{a^2+b^2}{2}$.


问:
For which $a$ is the matrix negative definite?
Find the number of integer numbers of $a$ such that the matrix $$\begin{pmatrix}-1& 0 & 5 \\ 0 & a-50 & 0 \\ 5 & 0 & 6-a\end{pmatrix}$$ is negative definite.

For that, the eigenvalues must be negative, right?

I calculated the eigenvalues using Wolfram and we get these.

So all these have to be negative and we get a system of inequalities :

$$a-50<0 \\ \frac{1}{2}\left (-\sqrt{a^2-14a+149}-a+5\right ) <0 \\ \frac{1}{2}\left (\sqrt{a^2-14a+149}-a+5 \right )<0$$ So we get from the first $a<50$, from the second $a\in \mathbb{R}$ and from the last one $a>31$.

So the integer values for $a$ are from $32$ to $49$, so there are $49-32+1=18$ values, right?

答:
Your approach is correct, but it could be slightly simpler: with the help of the rule of Sarrus or by other means, it's easy to find the characteristic polynomial is:

$$(\lambda-a+50)(\lambda^2+(a-5)\lambda+a-31)$$

The eigenvalues must all be negative, so, for the first factor, that means $a<50$. For the second factor, instead of computing the roots, note that the sum of the roots must be negative, while the product of the roots must be positive. With Vieta's formulas, that means $a>31$ and $a>5$. So the condition is indeed $31<a<50$.


问:
l'Hospitals rule for multivariate limit
I have the following bivariate function where $\alpha,\beta,\gamma,\delta$ are constants, $a,b,c,d$ are integers and $x,y\in [0,1]$
$$
f(x,y) = \frac{\alpha(1-x^ay^b) + \beta(1-x^cy^d)}{\gamma(1-x^2)+\delta(1-y^3)}
$$
which I would like to examine the limit of as $x=y=1$ from below. My instinct was to apply l'Hospital's rule by setting $x=y$ and converting to a univariate function.
$$
f(x,x) = \frac{\alpha(1-x^{(a+b)}) + \beta(1-x^{(c+d)})}{\gamma(1-x^2)+\delta(1-x^3)}
$$
from which we can gather the limit is
$$
\lim_{x\rightarrow 1}f(x,x) = \frac{\alpha(a+b)+\beta(c+d)}{2\gamma +3\delta}
$$
However, am also aware that l'Hospital's rule is strictly for univariate functions. In this particular case, how can I be certain that my limit is correct?

答:
If you set $y=1$, the expression turns to

$$\lim_{x\to 1}\frac{\alpha(1-x^a)+\beta(1-x^c)}{\gamma(1-x^2)}$$ and by L'Hospital,

$$=\lim_{x\to 1}\frac{\alpha ax^{a-1}+\beta cx^{c-1}}{\gamma2x}=\frac{\alpha a+\beta c}{\gamma2}.$$

But with $x=1$, you will get a very different expression.

Title: l'Hospitals rule for multivariate limit


问:
Limit points of $\frac{p}{2^k}$
Find all limit points of the set $S = \{\frac{p}{2^k} : p \in \mathbb{Z}, k \in \mathbb{N}\}$.

I could prove that all elements of $S$ are limit points of $S$. However, answer is all real numbers. Any help?

答: Take an arbitrary point on the reals, and write it in binary notation.  For example, x = 101.111100001010...Then x can be approached by 101, 101.1, 101.11, 101.111, 101.1111, 101.11110, ...each of which is a fraction with denominator a power of 2.</p>

Title: Limit points of $\frac{p}{2^k}$


问:
How to transform this system to $\dot{X}=AX+F$ form and find A, F matrices?
How to transform this system to $\dot{X}=AX+F$ form and find A, F matrices?
$$
a)\ddot{x}+\dot{x}=x-y+t+sint$$
$$\ddot{y}+\dot{y}=x+y+e^t+t^2$$
$$
b)\dddot{x}+\dot{x}+2x=t$$
$$\dddot{y}-\dot{y}-2y=t^2$$

my tryings are :
$x_1=x\\x_2=\dot{x_1}=\dot{x}\\x_3=y\\x_4=\dot{x_3}=\dot{y}\\ \\$

then; $\dot{x_1}=x_2\\\dot{x_2}=x_1-x_3+t+sint-x_2\\\dot{x_3}=x_4\\\dot{x_4}=x_1-x_4+x_3+e^t+t^2$

and
$\dot{X}=$$
    \begin{pmatrix}
     \dot{x_1}\\
    \dot{x_2} \\
    \dot{x_3}\\
\dot{x_4}\\
    \end{pmatrix}
$=$
    \begin{pmatrix}
     0&1&0&0\\
    1&0&-1&0 \\
    0&0&0&1\\
1&0&1&-1\\
    \end{pmatrix}$$
    \begin{pmatrix}
     x_1\\
    x_2 \\
    x_3\\
x_4\\
    \end{pmatrix}$+$
    \begin{pmatrix}
     0\\
    t+sint \\
    0\\
e^t+t^2\\
    \end{pmatrix}$

is it true like this way? If it is, for the second question do I need to find $x_1,x_2,x_3,x_4,x_5,x_6$ and therefor 6x6  matrix?

答: Yes, your reasoning is correct. You just introduce new variables, corresponding to derivatives of the existente ones, to prevent the occurence of derivatives with order other than one.</p>


问:
Strong mathematical induction: Prove Inequality for a provided recurrence relation $a_n$

The sequence $a_1,a_2,a_3,\dots$ is defined by: $a_1=1$, $a_2=1$, and $a_n=a_{n-1}+a_{n-2}-n+4$ for all integers $n\ge 3$. Prove using strong mathematical induction that $a_n\ge n$ for all integers $n\ge 3$.

I'm comfortable solving questions that require mathematical induction but I always struggle with strong mathematical induction. I've read the steps provided in my textbook and looked at examples but I still don't get it

答:
Take a $m>1\in N$. In the strong form of mathematical induction you consider that $\forall n<m,n\in N$ the given proposition is true (of course after proving it for first few initial cases though it is enough to prove it for the first case) and then you go on to prove it for $m$.

Similarly in this case.

$a_3=1+1-3+4=3\ge 3$

Now you consider it to be true for $n\in N.n\le k$. Using this hypothesis we will prove it for $n=k$.

$a_k=a_{k-1}+a_{k-2}-k+4\ge (k-1)+(k-2)-k+4=k+1>k$(using the fact that $a_{k-1}\ge k-1 ,a_{k-2}\ge k-2$. Note that these are from the hypothesis)


问:
Compute $\int_{0}^{\infty}\lambda^{k+\alpha-1}\text{e}^{-(\beta+1)\lambda} \, d\lambda$
This definite integral relates to Bayesian statistics and is a problem that needs to be solved in order to show that the $\text{Gamma}(\alpha,\beta)$ distribution is a (prior) conjugate family for the $\text{Poisson}(\lambda)$ distribution. Now, I know there are ways to prove this by means of proportionality arguments, and then identify the posterior distribution to make the necessary conclusion, but I want to do it the hard (but more intuitive) way. So I have,

$$
\int_0^\infty \lambda^{k+\alpha-1}\text{e}^{-(\beta+1) \lambda} \, d\lambda
$$

where $k$ is an integer $\geq0$ and $\alpha, \beta > 0$. In a "normal" situation where $\lambda$ is raised to the power of an integer, I would be able to solve it by integration of parts, but know I am not so sure. Any clues?

Edit

I use the hints user NHL provided:

\begin{align}
\int_0^\infty \lambda^{k+\alpha-1}\text{e}^{-(\beta+1)\lambda} \, d\lambda &= \big\{\mu = (\beta+1)\lambda,\, d\mu = (\beta+1) \, d\lambda \big\} \\[3mm]&= \frac{1}{(\beta + 1)^{k+\alpha}} \int_0^\infty \mu^{k+\alpha-1} \text{e}^{-\mu} \, d\mu \\[3mm] &= \big\{ \Gamma(z)\triangleq \int_{0}^{\infty}x^{z-1}\text{e}^{-x} \, dx\big\}\\[3mm] &= \frac{\Gamma(k+\alpha)}{(\beta + 1)^{k+\alpha}}.
\end{align}

答:
I use the hints user NHL provided:

\begin{align}
\int_{0}^{\infty}\lambda^{k+\alpha-1}\text{e}^{-(\beta+1)\lambda}d\lambda &= \big\{\mu = (\beta+1)\lambda,\; d\mu = (\beta+1)d\lambda\big\} \\[3mm]&= \frac{1}{(\beta + 1)^{k+\alpha}}\int_{0}^{\infty}\mu^{k+\alpha-1}\text{e}^{-\mu}d\mu \\[3mm] &= \big\{ \Gamma(z)\triangleq \int_{0}^{\infty}x^{z-1}\text{e}^{-x}dx\big\}\\[3mm] &= \frac{\Gamma(k+\alpha)}{(\beta + 1)^{k+\alpha}}.
\end{align}


问:
Matrix derivative (chain rule application)
Let $x$, $y$ by vectors s.t. $x=f(y)$ and let $B$ be a constant matrix.

What is $\frac{\partial x'Bx}{\partial y}$?

The partial derivative $\frac{\partial x'Bx}{\partial x}=2Bx$ and we need to use the chain rule using the matrix $J=\frac{\partial x}{\partial y}$ but I'm not sure how.

Is it simply $2(Bx)^T J$?

Thanks.

答:
$\def\p#1#2{\frac{\partial #1}{\partial #2}}$The
scenario is the following:

You successfully calculated the gradient of a certain function
$$\phi = x^TBx \quad\implies\quad g = \p{\phi}{x} \doteq 2Bx$$
Sometime later you discover that $x$ is not an independent vector,
but is actually dependent on a more fundamental vector for which
you know the Jacobian, i.e.
$$x = f(y) \quad\implies\quad J=\p{x}{y}$$
How do you use this new information to calculate the gradient of the original function?

Instead of the Chain Rule,
you can substitute the differentials of the various quantities
$$\eqalign{
d\phi &= g^Tdx \quad\qquad dx = J\,dy
}$$
to effect a Change of Variables from $x\to y\;$ like so
$$\eqalign{
d\phi &= g^Tdx \\&= g^T(J\,dy) \\
  &= (J^Tg)^Tdy \\
  &= (2J^TBx)^Tdy \\
\p{\phi}{y} &= 2J^TBx \\
}$$


问:
How is the normal vector to the parabola $r(t)=(t,t^2)$ calculated?
I think the parametrisation of a parabola ought to be like so:

$r(t)=(t,t^2)$

When I compute $r'(t)$ I get $r'(t)=(1,2t)$, and when I compute $r''(t)$ I get $r''(t)=(0,2)$. At each point $0i + 2j$ isn't orthogonal to the tangent everywhere. What is the correct normal vector calculation?

答:
There is a fact that if $r(t)$ is a vector such that $\| r(t) \|$ is constant, then $r'(t)$ is orthogonal to $r(t)$. So, to fix your calculations you can first normalize $r'(t)$ to have constant magnitude, and then calculate the second derivative.

Edit: the start of the first step is to calculate
$$\| r'(t) \| = \sqrt{1^2 + (2t)^2 } = \sqrt{1 + 4t^2}, $$
so our unit tangent vector is
$$T(t) = \left\langle \frac{1}{\sqrt{1 + 4t^2}}, \frac{2t}{\sqrt{1 + 4t^2}} \right\rangle.$$


问:
Show that $x^{(k)} \not\to x$ in the $l_\infty$ norm

Given a fixed element $x\in l_1$, show that $x^{(k)} = (x_1,x_2,\ldots,x_k,0,0,\ldots,0)$ converges to $x$ in the $l_1$ norm. Show that $x^{(k)} \to x$ in the $l_2$ norm as well. Produce a counterexample to show that $x^{(k)}\not\to x$ in the $l_\infty$ norm.

I'm struggling with the last part, i.e. the counterexample. I am showing my work for the remaining parts:

$$y_k = \|x-x^{(k)}\|_1 = \sum_{i=k+1}^\infty |x_i - x^{(k)}_i|$$
and
$$z_k = \|x-x^{(k)}\|_2 = \left(\sum_{i=k+1}^\infty |x_i - x^{(k)}_i|^2 \right)^{1/2}$$

Clearly $y_k,z_k$ are decreasing and non-negative, so $y_k \stackrel{k\to\infty}{\longrightarrow} 0$ and $z_k \stackrel{k\to\infty}{\longrightarrow} 0$.

I don't know how to proceed in the $l_\infty$ norm case.

Thanks a lot!

答:
hint: maybe you can get $x^{(k)}$ not to converge at all and therefore it especially can't converge to $x$.

possible solution:

Take $x = (1,1,1,\ldots) \in \ell^\infty$. Now
 \begin{align*} \|x^{(N)}-x^{(M)}\|_\infty = 1 \end{align*}
 for all $M,N \in \mathbb{N}$ with $M > N$. This means $x^{(k)}$ is not a cauchy sequence and therefore it cannot converge.

edit:
If you had $x=(x_n)_{n\in\mathbb N}\in\ell^{1}$ you would get $|x_n| \to 0$ as $n\to\infty$ and therefore
$$
\|x-x^{(k)}\|_\infty = \sup_{n\in\mathbb{N},\ n>k} |x_n| \to 0
$$
as $k\to\infty$ so you wouldn't be able to find a counterexample. Phrased in other words your statement also holds for the space $(c_0, \|\cdot\|_\infty)$ of sequences that converge to 0.

Regarding your proofs of the other statements: A real sequence being decreasing and non-negative isn't enough for convergence to 0 (take for example $1+\frac 1 n$), but I think you know that and just phrased it wrong. It probably becomes clearer if you just take a power:
$$
\|x-x^{(k)}\|_{\ell^p}^p = \sum_{n=k+1}^\infty |x_n|^p \to 0
$$
as $k\to\infty$ for $p\in [1,\infty[$.


问:
C*-Algebras and Operator Theory - Gerald J. Murphy Theorem 4.1.2
I'm stuck On the theoremTheorem 4.1.2, from   C*-Algebras and Operator Theory by Gerald J. Murphy

4.1.2. Theorem. Suppose that $(p_\lambda)_{\lambda\in\Lambda}$ is a net of projections on a Hilbert space $H$.

(1) If $(p_\lambda)$ is increasing, then it is strongly convergent to the projection of $H$ onto the closed vector subspace $(\cup_{\lambda}p_\lambda(H))^-$.

(2) If $(p_\lambda)$ is decreasing, then it is strongly convergent to the projection of $H$ onto $\cap_\lambda p_\lambda(H)$.

I have no idea how to deal with. Any hints or suggestions
would be welcome!

答:
As established earlier in Murphy, an increasing net of projections that is bounded above by a self-adjoint element converges strongly to its supremum. Hence, for $(1)$ we must check that if $p$ is the projection onto $K= \overline{\bigcup_\lambda p_\lambda H}$, then $p$ is the supremum of the net $\{p_\lambda\}$.

This is rather easy, since $p \geq p_\lambda$ for all $\lambda$ because $pH \supseteq p_\lambda H$ by definition of $p$. Thus $p$ is an upper bound for this net.

Moreover, if $q \geq p_\lambda$ for all $\lambda$, then clearly $q \geq p$ and
$$(p-q)p_\lambda = pp_\lambda-q p_\lambda = p_\lambda- p_\lambda = 0$$
so that $p-q$ annihilates the subspace $K$. Hence,
$$0 = (p-q)p = p-qp= p-q$$
and thus $q=p$. Thus, $p$ is the least upperbound. Can you now do $(2)$ yourself?


答:
Given $x \in \Bbb R$, consider the sequence $$x_n := \dfrac{\lfloor 2^n x\rfloor}{2^n}.$$
Show that $\{x_n\} \subset S$ and $x_n \to x$. Conclude from this.

Title: Limit points of $\frac{p}{2^k}$

Body:
Find all limit points of the set $S = \{\frac{p}{2^k} : p \in \mathbb{Z}, k \in \mathbb{N}\}$.

I could prove that all elements of $S$ are limit points of $S$. However, answer is all real numbers. Any help?


问:
TP and TQ are the two tangents to a circle with centre O so that ∠POQ = 110°
If TP and TQ are the two tangents to a circle with centre O so that ∠POQ = 110°, then ∠PTQ is equal to

My attempt:

答:
$ OP ⊥ PT and TQ ⊥ OQ $

$ ∴∠OPT = ∠OQT = 90° $

Now, in the quadrilateral POQT, we know that the sum of the interior angles is 360°

So, $ ∠PTQ+∠POQ+∠OPT+∠OQT = 360° $

Now, by putting the respective values we get,

$ ∠PTQ +90°+110°+90° = 360° $

$ ∠PTQ = 70° $


答:
Rotating a vector $(a,b)$ by 90 degrees counterclockwise yields $(-b,a)$. The unit tangent vector to $r(t)=(t,t^2)$ is $$T(t)=\frac{r'(t)}{\|r'(t)\|}=\frac{(1,2t)}{\sqrt{1+4t^2}}.$$Rotate to obtain $$N(t)=\frac{(-2t,1)}{\sqrt{1+4t^2}}.$$

Bonus: plane curves can be studied with complex numbers. Then, rotation by 90 degrees counterclockwise amounts to multiplying by $\rm i$. So if $r(t)=t+t^2{\rm i}$, divide $r'(t)$ and ${\rm i}r'(t)$ by $|r'(t)|$ to reobtain $T(t)$ and $N(t)$.

Title: How is the normal vector to the parabola $r(t)=(t,t^2)$ calculated?

Body:
I think the parametrisation of a parabola ought to be like so:

$r(t)=(t,t^2)$

When I compute $r'(t)$ I get $r'(t)=(1,2t)$, and when I compute $r''(t)$ I get $r''(t)=(0,2)$. At each point $0i + 2j$ isn't orthogonal to the tangent everywhere. What is the correct normal vector calculation?


问:
Every first contable locally convex space has a countable neighborhood basis of balanced and convex sets
Terminology: By a neighborhood of a point $x$ on a topological space, I mean any subset $V$ which contains an open set containing $x$. A set $B$ in a vector space $X$ is called balanced if $\lambda B \subseteq B$ for every $|\lambda| \le 1$.

Let $X$ be locally convex and suppose it is first countable. In particular, $X$ has a countable neighborhood basis of the origin. Because $X$ is locally convex, we know that (a) every element of such basis is balanced and (b) $X$ has a neighborhood basis of the origin consisting of balanced and convex sets.

Question: How can we find a countable neighborhood basis of the origin consisting of balanced and convex sets?

答:
Let $(U_n)_n$ be any countable local base at $0$ for $X$.

As you stated under (b), for each $n$ we can find a convex balanced neighbourhood $V_n$ of $0$ such that $V_n \subseteq U_n$, as $X$ is locally convex.

It’s immediate that $(V_n)_n$ is your required local countable base at $0$.


问:
Angle between tangents to circle given by $x^2 + y^2 -12x -16y+75=0$?

Given the circle: $C(x,y)=x^2 + y^2 -12x -16y+75=0$, find the two tangents from origin

First, I get the line which passes through point of contact of tangents from origin using result here which is :

$$ -12x-16y-2 \cdot 75 = 0$$

Or,

$$ 6x + 8y -75 =0 $$

Now, I use the result discussed in this answer, which says that pair of straight from point $(P)$ to conic is given as:

$$ C(0,0) C(x,y) =  (6x+8y-75)^2$$

This leads to:

$$ 75 (x^2 + y^2 -12x-16y+75) = (6x+8y-75)^2$$

$$ 0 = 75(x^2 +y^2 - 12 x - 16y +75) - (6x+8y-75)^2$$

For applying the result in this answer, then the formula

$$ a= 75 - 36,  b=75-64 , h= - \frac{(6 \cdot 8 \cdot 2 )}{2}$$

Or,

$$ a = 39, b= 11 , h=-48$$

$$ \tan \theta = \frac{2 \sqrt{(-48)^2-39 \cdot 11)}}{39+11} = \frac{2 \sqrt{5^4 \cdot 3}}{25} = 2 \sqrt{3}$$

However, the intended answer was:

$$ \tan \theta = \frac{1}{\sqrt{3} } $$

Where have I gone wrong?

答:
Expansion on Math777's answer, see here first, the rest of the answer's idea can be understood in the following way:

Reference picture:

Fact: From any external point there exists two tangents to the circle, therefore there must be two tangents from the point $(0,0)$ to our circle since this point is external to our circle.

Now, the vector from origin to point of contact of tangent is perpendicular to radius vector because tangent and radius are perpendicular, there are two such points like this. These two points satisfy equation of the bigger diametric circle, hence these two points are the intersection of big diameteric circle and original one (See the linked post to understand the diameteric circle idea)


答:
Here's some intuition to get you started, in order to understand why your answer is wrong and the correct answer is all real numbers.

Look at an ordinary ruler marked in inches (using the imperial system of units). You will see each inch divided in half, each $1/2$ inch divided in half to give fourths, each $1/4$ inch divided in half to give eighths, each $1/8$ inch divided in half to give sixteenths... In the abstract, you can continue this forever, using arbitrarily short subdivision lengths $1/2^n$ as $n$ goes to $\infty$ (although I don't think I've ever seen an actual physical ruler go any shorter than $1/2^5$, and even that is rare).

Perhaps from this it will become evident to you that inside the 1 inch long interval from inch mark number $N$ to inch mark number $N+1$, the subdivision points are dense, and every real number in that interval is a limit point of the set of subdivision points. And then, letting $N$ vary over $\mathbb Z$, you can see that the set of subdivision points over the entire real line is dense.

Once you understand this picture, turning it into a mathematical proof is not actually very hard. The key is to notice that for each $r > 0$, if you choose a natural number $n$ so large that $2^n > \frac{1}{r}$, then $\frac{1}{2^n} < r$. Each real number $x$ is contained in one of the subdivision intervals $\left[\frac{k}{2^n},\frac{k+1}{2^n}\right]$ (i.e. the union of those intervals is the entire real line), and therefore $x$ has distance less than $r$ from either of the two subdivision points at the ends of that interval.

Title: Limit points of $\frac{p}{2^k}$


问:
How to calculate the limit of the following function?
How can I calculate the limit of the following function?

$$\lim_{(x,y)\to(0,0)}\frac{-|x+y|}{e^{x^2}{^{+2xy+y^2}}}$$

Body:
Hint.

$(x+y)^2=x^2+2xy+y^2$ and $a^2=|a|^2$.


问:
Proving an identity regarding the Cauchy problem (using convolutions)
Given $u_0 \in C_c(\mathbb{R}^n)$, consider the solution of the Cauchy problem $$u(x,t) = \int_{\mathbb{R}^n} \Gamma (x - y,t)u_0(y) dy \qquad x \in \mathbb{R}^n,t>0\, \, .$$ Given $0<s<t$ , define $ v:\mathbb{R}^n \to \mathbb{R}$ as $$v(x) = \int_{\mathbb{R}^n} \Gamma(x-z,t-s)u(z,s)dz \, \, .$$ How would you prove that $v(x) = u(x,t)$?

In the help sheet it says notice notice that $$v(x) = \Gamma(t-s)\star u(s) = \Gamma(t-s)\star (\Gamma(s) \star u_0)$$ $$ = (\Gamma(t-s) \star \Gamma(s) )\star u_0$$ $$= \Gamma(t) \star u_0$$

but I don't understand how they get from the second last like to the last line? Could someone please explain?

答:
I assume $u$ is a solution of the heat equation and $\Gamma$ is its fundamental solution.
The property $\Gamma(x,t)=\Gamma(\cdot,t-s) \star \Gamma(\cdot,s)\,$, $0<s<t\,$, can be verified directly by computing the integral for convolution. Or it can be done using the uniqueness theorem for bounded solutions of the Cauchy problem. Namely, function $\Gamma(x,t)$ is evidently a solution of
$$
u_t-\Delta u=0,\ \  x\in \mathbb R,\ t>s,\quad u|_{t=s}=\Gamma(x,s).
$$
And
$$
u(x,t) = \int_{\mathbb{R}^n} \Gamma (x - y,t-s)\Gamma(y,s)\,dy
$$
is also a solution.</p>


问:
Gradient of squared dot product
Let $x,y \in \mathbb{R}^d$. What is

$$\nabla_x\langle x,y\rangle^2$$

I know that we have

$$
\nabla_x\langle x,x\rangle^2 = \|x\|^2 = 2x
$$
but how this changes for the case $\langle x,y\rangle^2$?

答:
We have
$$\begin{align}d_p\langle x,y\rangle^2&=2\langle x,y\rangle\langle p,y\rangle\\
&=\langle2\langle x,y\rangle y,p\rangle,
\end{align}
$$
hence the gradient is $2\langle x,y\rangle y.$

Title: Gradient of squared dot product


答:
The only difference is that strong induction gives you more to work with at the induction step: you get to assume that the result is true for all smaller values of $n$, not just for the immediately preceding value.

Here your induction hypothesis should be that $n>4$ and that $a_k\ge k$ for all integers $k$ such that $3\le k<n$, and the induction step is then to prove from this that $a_n\ge n$. As it happens, you don’t even need the full strength of the induction hypothesis: you need only that $a_{n-1}\ge n-1$ and $a_{n-2}\ge n-2$. From these you can argue that

$$a_n=a_{n-1}+a_{n-2}-n+4\ge(n-1)+(n-2)-n+4=n+1>n\;.$$

This is actually more than you were asked to show, but that’s okay: if $a_n>n$, then certainly $a_n\ge n$.

Note that the induction step does require knowing the result for the next two smaller values, $n-1$ and $n-2$, so you have to start with an $n$ that’s at least $5$, and your basis step has to verify two things, namely, that $a_3\ge 3$ and that $a_4\ge 4$.

Title: Strong mathematical induction: Prove Inequality for a provided recurrence relation $a_n$

Body:
The sequence $a_1,a_2,a_3,\dots$ is defined by: $a_1=1$, $a_2=1$, and $a_n=a_{n-1}+a_{n-2}-n+4$ for all integers $n\ge 3$. Prove using strong mathematical induction that $a_n\ge n$ for all integers $n\ge 3$.

I'm comfortable solving questions that require mathematical induction but I always struggle with strong mathematical induction. I've read the steps provided in my textbook and looked at examples but I still don't get it


问:
When are we free in setting the free/basic variables in a linear system of equations
Consider the following two linear system of equations of the form $b=Ax$

case 1:
$$
\begin{pmatrix}
4\\
5
\end{pmatrix}= \begin{pmatrix}
1& 2 & -1\\
2 & -4 & 0
\end{pmatrix}\begin{pmatrix}
x_1\\
x_2\\
x_3
\end{pmatrix}
$$

case 2:
$$
\begin{pmatrix}
0\\
5
\end{pmatrix}= \begin{pmatrix}
0 & 0\\
0 & 1
\end{pmatrix}\begin{pmatrix}
x_1\\
x_2\\
\end{pmatrix}
$$

The reduced row echelon form  of case 1 is
$$
\begin{pmatrix}
1 & 0 & -0.5 & 3.25\\
0 & 1 & -0.25 & 0.3750
\end{pmatrix}
$$
Hence, $x_1,x_2$ are basic variables, $x_3$ is free variable.

The reduced row echelon form  of case 2 is
$$
\begin{pmatrix}
0 & 1 & 5\\
0 & 0 & 0
\end{pmatrix}
$$
Hence, $x_2$ is basic variable, $x_1$ is free variable.

Question: There is a key difference between case 1 and case 2.

In case 1, I am free to decide which two variables among $\{x_1,x_2,x_3\}$ are set as basic and which one is set as free. The reduced row echelon suggests to set $x_1,x_2$ as basic variables and $x_3$ as free. However, I could equivalently set $x_1,x_3$ as basic variables and $x_2$ as free. Or, I could also set $x_3,x_2$ as basic variables and $x_1$ as free. This can be seen by doing naive calculations from the system of equations.

In case 2, I am not free to decide which variable between $\{x_1,x_2\}$ should be set as basic and which one should be set as free. Again, this can be seen by doing naive calculations from the system of equations.

More generally (and for more complicated systems) I am wondering whether there is an easy way to detect when I am (not) free in setting the free/basic variables from just looking at the reduced row echelon, without the need of doing other calculations. Is this a matter of zero rows or is there something more subtle?

答:
It depends on the rang of the row echelon matrix , say the rang of the row echelon matrix matches the dimension of the $x$ vecotr in your case, then you cannnot choose any of the $x_1,x_2\ldots x_n $ as free variables. In case 2 you can choose either $x_1$ or $x_2$ as a free variable , lets say $t\in \mathbb{R}$ and calculate the rest accordingly.
The amount of free varibales you can chosse is given by '$rows(x)-rang(A)$'. The varibale you choose to be free is up to you.


问:
Closedness of a set
I am trying to understand closed sets.

Is the set
$$
A =\{x \in \mathbb{R}^2: \lVert x \rVert_2 \leq 1 \} -\{0\}
$$
closed?

I think it is not, because if we take the sequence $(\frac{1}{n})_{n=1}^{\infty}$ we can see it lies in A, but its limit is zero which is not in the set A. And because of that, A is not a closed set.

Is this right? Thank you.

答: It is almost correct, but you missed the fact that $(\forall n\in\Bbb N):\frac1n\notin\Bbb R^2$. Take the sequence $\left(\left(\frac1n,0\right)\right)_{n\in\Bbb N}$ instead.

Title: Closedness of a set


问:
Similar triangles question
If I have a right triangle with sides $a$.$b$, and $c$ with $a$ being the hypotenuse and another right triangle with sides $d$, $e$, and $f$ with $d$ being the hypotenuse and $d$ has a length $x$ times that of $a$ with $x \in \mathbb R$, is it necessary for $e$ and $f$ to have a length $x$ times that of $b$ and $c$ respectively?

EDIT: The corresponding non-right angles of both triangles are the same.

答:
As the comments say, no

Think about a triangle $ABC$ with hypotenuse $A$ and then another triangle $DEF$ with hypotenuse $D$. We know that there's a relationship between $A$ and $D$, so it's as if $D$ is given to us. You can imagine extending (or retracting) one of the legs on $DEF$ to some factor of the corresponding leg on $ABC$ and easily find another leg to complete the triangle.

Now let's say you wanted to keep the triangles as right triangles. Then it's clear that $A^2 = B^2 + C^2$ and that $D^2 = E^2 + F^2$. We know that $D = xA$ so we know $x^2A^2 = E^2 + F^2 = x^2 \left( B^2 + C^2 \right)$ so it follows that $\frac{E^2 + F^2}{x^2} = B^2 + C^2$. Now if we are initially given the $ABC$ triangle then we know $B , C$. It then follows that $\frac{E^2 + F^2}{x^2} = B^2 + C^2$ has many different solutions for a given $x$ (at first glance it appears to be an infinite number but I wouldn't quote me on that).

Hope this helps!

EDIT:

With the addition to the original question that all angles are the same then the answer is yes, all sides will be scaled equally. Let's call $B$ the side corresponding to $E$ and $C$ to $F$. Next let's call the angle between $A,B$ as $\theta$ (and similarly the angle between $D,E$). It must be true that $\cos \theta = \frac{B}{A} = \frac{E}{D}$. We take note that $D = xA$ so $\frac{B}{A} = \frac{E}{xA} \iff xB = E$. The same argument can be applied to side $C$ and $F$ using the angle between $A,C$ (and $D,F$).

Again I hope this helps!

Title: Similar triangles question


问:
One mysterious step in proof of Lemma 3 Devroye 83
Can someone help me with one step of the following proof of a multinomial concentration inequality taken from Lemma 3 of "The equivalence of weak, strong and complete convergence in l1 for kernel density estimate" (Devroye 83).

Let $p_1, p_2, ..., p_k$ be the parameter of a multinomial distribution ($\sum p_i = 1$). In $n$ experiments for some fixed $n$, let $X_i$ be the number of times that $i$ appears. $\sum_{i=1}^k X_i = n$

Now, let $N \sim $Poisson(n). After another $N$ experiments (completely independent of the previous n experiments), let $X'_i$ be the number of times that $i$ appears in these $N$ experiments. $\sum_{i=1}^k X'_i = N$. This is also known as Poissonization and each $X'_i$ is actually independent Poisson variable with parameter $np_i$.

I understand and can reproduce every other step except for the following inequality:

$$ P(\sum_{i=1}^k \frac{1}{n} |X_i - np_i| \geq \epsilon ) \leq P(|N-n| \geq \frac{2n\epsilon}{5}) + P(\sum_{i=1}^k\frac{1}{n} |X_i'- np_i| \geq \frac{3n\epsilon}{5}) $$

I'm not sure how he obtained this inequality. Also note how there seems to be an extra "n" in the second term. Inequality (2) (Which I think he used for this step) from the proof is the following simple observation:
$$ \frac{1}{n} \sum_{i=1}^k |X_i - np_i| \leq \frac{1}{n}\sum_{i=1}^k |X_i-X'_i| + \frac{1}{n}\sum_{i=1}^k |X'_i - np_i| $$

Would greatly appreciate any help/suggestions/pointers to similar proofs. The full proof is pasted below for context:

答:
Turns out that I needed to think about Poissonization in the right way to see this.

Fixing an n, $ U_1, U_2, ..., U_n $ are drawn from the multinomial distribution. Now imagine that these draws are snap-shots taken from another random process where we first pick the number of experiments according to $N \sim Poi(n)$, and then take $N$ draw from the multinomial distribution like before.

So there are two cases: if $ n \geq N$, we have something like:
$$ U_1=U'_1, ..., U_N=U'_N, U_{N+1},..., U_n $$
On the other hand, if $N > n$:
$$ U_1=U'_1,..., U_n=U'_n, U'_{n+1},..., U_N $$

So we have something rather surprising (to me at least) that is $|N-n| = \sum_{i=1}^k |X_i - X'_i|$. Combining with (2) in the proof and union bound argument, we get

$$ p(\sum_{i=1}^k \frac{1}{n} |X_i-np_i| \geq \epsilon) \leq p(\frac{1}{n}\sum_{i=1}^k |X_i-X'_i| + |X'_i - npi| \geq \epsilon) $$
$$ \leq p(\sum_{i=1}^k \frac{1}{n} |X_i-X'_i| \geq \frac{2\epsilon}{5}) + p(\sum_{i=1}^k \frac{1}{n} |X'_i - npi| \geq \frac{3\epsilon}{5}) $$
$$ = p(|N-n|\geq \frac{2n\epsilon}{5}) + p(|X'_i - npi| \geq \frac{3n\epsilon}{5}) $$

So there is actually an extra $\frac{1}{n}$ in the manuscript.


问:
Answer to an algebra question

The parabola $y=x^2-3bx-5$, in the $x-y$ plane, has its vertex at
$(3,4)$. What is the value of $b$?

I can't answer it. There is no value for $b$ such that, when multiplied by 3, gives 5 and also makes the equation equal 4 when $x$ is 3 (to get the vertex). I'm pretty stumped all around. The problem is from a private SAT prep company. I would appreciate any help.

答:
Hint.

What does "vertex" mean for a parabola? In general,
$$
y=a(x-h)^2+k\tag{1}
$$
has the vertex $(h,k)$.

Now write the equation of your parabola in the form of (1) by completing the square.

Alternatively, your parabola can be written as
$$
y=a(x-3)^2+4\tag{2}
$$
for some real number $a$. Expand (2) and compare it with your equation to find $b$.

Following the hint above, you will find that the number $b$ that satisfies the assumptions does not exist.

Title: Answer to an algebra question


答: Another way to see it is via the fact that closed sets are (perhaps defined as, depends on your definition of open set) the complements of open sets. The complement of $A$ is not open because there is no open ball of positive radius contained in $A^\complement$ around $(0,0)$

Title: Closedness of a set

Body:
I am trying to understand closed sets.

Is the set
$$
A =\{x \in \mathbb{R}^2: \lVert x \rVert_2 \leq 1 \} -\{0\}
$$
closed?

I think it is not, because if we take the sequence $(\frac{1}{n})_{n=1}^{\infty}$ we can see it lies in A, but its limit is zero which is not in the set A. And because of that, A is not a closed set.

Is this right? Thank you.


答:
Triangles are similar if and only if their angles match up.
For the triangles to be similar then, the right angle of the one triangle must match up with the right angle of the other (why?).
This means that side $d$ pairs with side $a$.

But there are still two possibilities: $b$ and $c$ can pair up with $e$ and $f$, respectively, or they can pair up with $f$ and $e$, respectively.
Your conjecture assumes $e = xb$ and $f = xc$, which is the case when $b, c$ pair up with $e, f$; but what about the other case when $b,c$ pair up with $f,e$?

Title: Similar triangles question

Body:
If I have a right triangle with sides $a$.$b$, and $c$ with $a$ being the hypotenuse and another right triangle with sides $d$, $e$, and $f$ with $d$ being the hypotenuse and $d$ has a length $x$ times that of $a$ with $x \in \mathbb R$, is it necessary for $e$ and $f$ to have a length $x$ times that of $b$ and $c$ respectively?

EDIT: The corresponding non-right angles of both triangles are the same.


答:
the exponential function is continuos so you can pull the limit inside and thus get $$\exp\left({\lim_{(x,y)\to0}\frac{-|x+y|}{|x+y|^2}}\right) .$$
I presume you can go on from here by yourself

Title: How to calculate the limit of the following function?

Body:
How can I calculate the limit of the following function?

$$\lim_{(x,y)\to(0,0)}\frac{-|x+y|}{e^{x^2}{^{+2xy+y^2}}}$$


答:
In these sort of cases, the solution is a line or plane or something similar, embedded within the input space.  When describing this solution we can choose any variable that is not orthogonal to it as a free variable, and then once we have enough to fully describe a projection of the solution with the same number of dimensions, we can say any other variables we would need to get back to the solution set have you are "basic", forced by the selection of free variables to take on a particular value.

In case 1, the solution is $x = (\frac{109}{42}, \frac{1}{21}, -\frac{55}{42}) + (2,1,4)t$, and since none of $(1,0,0)$, $(0,1,0)$, and $(0,0,1)$ are orthogonal to $2,1,4$, we can pick any one of $x_1$, $x_2$, and $x_3$ to be the free variable.

In case 2, the solution can be described as the line $x = (0,5) + (1,0)t$, where it is obvious that $(0,1)$ is orthogonal to the solution line, and so $x_2$ cannot be considered free.

I've done a bit of searching to figure out how to detect this in the RREF.  here's an example.

$$
\begin{pmatrix}
1\\
4
\end{pmatrix}= \begin{pmatrix}
4& -6 & 3\\
-2 & 3 & 4
\end{pmatrix}\begin{pmatrix}
x_1\\
x_2\\
x_3
\end{pmatrix}
$$

This has RREF

$$
\left(\begin{array}{ccc|c}
1 & -\frac{3}{2} & 0 & -\frac{4}{11}\\
0 & 0 & 1 & \frac{9}{11}
\end{array}\right)
$$

And it turns out that $x_3 = \frac{9}{11}$ all the time.  The solution set is $(-\frac{2}{11}, \frac{4}{33}, \frac{9}{11}) + (2,3,0)t$; this is clearly orthogonal to $(0,0,1)$ and so $x_3$ is fixed.

This provides guidance to the answer The second row of the RREF contains only columns from $x_3$ and the solution vector, so this one must be fixed; I can select from $x_1$ and $x_2$ to be free because they share a row.

Applying this wisdom to case 1: selecting one of $x_1$ or $x_2$ to be free fixes $x_3$ by the action of one of the rows, which then (since everything else in the other row is fixed) fixes the other one, and picking $x_3$ to be free fixes $x_1$ and $x_2$ by the respective actions of the two rows in themselves.

Applying similar wisdom to case 2: $x_1$ gets to be free because it affects absolutely nothing; $x_2$ does not because it is surrounded by already-fixed results.

So it looks like it goes something like this: if a variable is in a position in one of the rows where it can affect only variables that are either fixed or already chosen as free, then it must be fixed.  Otherwise, you may choose it to be free.

Title: When are we free in setting the free/basic variables in a linear system of equations

Body:
Consider the following two linear system of equations of the form $b=Ax$

case 1:
$$
\begin{pmatrix}
4\\
5
\end{pmatrix}= \begin{pmatrix}
1& 2 & -1\\
2 & -4 & 0
\end{pmatrix}\begin{pmatrix}
x_1\\
x_2\\
x_3
\end{pmatrix}
$$

case 2:
$$
\begin{pmatrix}
0\\
5
\end{pmatrix}= \begin{pmatrix}
0 & 0\\
0 & 1
\end{pmatrix}\begin{pmatrix}
x_1\\
x_2\\
\end{pmatrix}
$$

The reduced row echelon form  of case 1 is
$$
\begin{pmatrix}
1 & 0 & -0.5 & 3.25\\
0 & 1 & -0.25 & 0.3750
\end{pmatrix}
$$
Hence, $x_1,x_2$ are basic variables, $x_3$ is free variable.

The reduced row echelon form  of case 2 is
$$
\begin{pmatrix}
0 & 1 & 5\\
0 & 0 & 0
\end{pmatrix}
$$
Hence, $x_2$ is basic variable, $x_1$ is free variable.

Question: There is a key difference between case 1 and case 2.

In case 1, I am free to decide which two variables among $\{x_1,x_2,x_3\}$ are set as basic and which one is set as free. The reduced row echelon suggests to set $x_1,x_2$ as basic variables and $x_3$ as free. However, I could equivalently set $x_1,x_3$ as basic variables and $x_2$ as free. Or, I could also set $x_3,x_2$ as basic variables and $x_1$ as free. This can be seen by doing naive calculations from the system of equations.

In case 2, I am not free to decide which variable between $\{x_1,x_2\}$ should be set as basic and which one should be set as free. Again, this can be seen by doing naive calculations from the system of equations.

More generally (and for more complicated systems) I am wondering whether there is an easy way to detect when I am (not) free in setting the free/basic variables from just looking at the reduced row echelon, without the need of doing other calculations. Is this a matter of zero rows or is there something more subtle?


问:
Expected Value for mutually independent random variables
Let $X_1, ..., X_n$ be random variables. Show that if for any functions from $\mathbb{R}$ to $\mathbb{R}$  $\beta_1, ... , \beta_n$, that
$$E[\beta_1(X_1) \times ... \times \beta_n(X_n)] = E[\beta_1(X_1)] \times ... \times E[\beta_n(X_n)] \implies X_1, ..., X_n \text{ are mutually independent} $$

This is where I'm at with the question so far:

From the definition of the probability density function and mutual in-dependency, we have that
$$X_1, ..., X_n \text{ are mutually independent} \iff f_{X_1, ..., X_n}(x_1, ..., x_n) = f_{X_1}(x_1) \times ... \times f_{X_n}(x_n) $$
And by the definition of expected value:
$$E[\beta_1(X_1) \times ... \times \beta_n(X_n)] = \int_{-\infty}^{+\infty} ... \int_{-\infty}^{+\infty}\beta_1(x_1)... \beta_n(x_n)f_{X_1, ..., X_n}(x_1, ... x_n)dx_1 ... dn_x$$
$$E[\beta_1(X_1)] \times ... \times E[\beta_n(X_n)] = \int_{-\infty}^{+\infty}\beta_1(x_1)f_{X_1}(x_1)dx_1 ... \int_{-\infty}^{+\infty}\beta_n(x_n)f_{X_n}(x_n)dx_n$$
Somehow I need to show that those two expected values above implies $f_{X_1, ..., X_n}(x_1, ..., x_n) = f_{X_1}(x_1) \times ... \times f_{X_n}(x_n)$, but I'm not really sure how to do so.

答:
For simplicity consider the case $n = 2$ (the general case goes the same). Let $\beta_1$ be the indicator function on some interval $[x_1,x_2]$ and $\beta_2$ the indicator function on $[y_1,y_2]$. Then
$$E(\beta_1 X_1 \times \beta_2 X_2) = \int_{x_1}^{x_2} \int_{y_1}^{y_2} f_{1,2}(x,y) \,dx\,dy$$
whereas
$$E(\beta_1 X_1) \times E(\beta_2 X_2) = \int_{x_1}^{x_2} \int_{y_1}^{y_2} f_1(x)\,f_2(y) \,dx\,dy$$
So what we see is that we have two functions in $2$ variables, namely $f_{1,2}(x,y)$ and $f_1(x)\,f_2(y)$, whose double integrals are the same on every square.

I claim that this is sufficient to conclude that $f_{1,2}(x,y) = f_1(x) \,f_2(y)$ almost everywhere (i.e. everywhere if your functions are continuous). If your functions are continuous, you can make the following argument. Suppose that at some point $(x,y)$ they differ, say $f_{1,2}(x,y) < f_1(x) \,f_2(y)$, then this inequality remains true on a small neighbourhood of $(x,y)$, and we can just choose our square sufficiently small to ensure that the two integrals are unequal as well. But more generally, the conclusion remains valid if the functions are merely measurable, see e.g. this answer.


答:
I don't think there is any value of $b$ for which it can be the vertex.

Proof:

$4=9-9b-5$ (Vertex lies on the parabola)

$\implies b=0$

But the slope of tangent at the vertex should be zero,
that is $\frac{dy}{dx}=0$ at the vertex.

$\implies 2x-3b=0$ at $x=3$

$\implies b=2$ so there isn't any value of $b$ which satisfies that.

Title: Answer to an algebra question

Body:
The parabola $y=x^2-3bx-5$, in the $x-y$ plane, has its vertex at
$(3,4)$. What is the value of $b$?

I can't answer it. There is no value for $b$ such that, when multiplied by 3, gives 5 and also makes the equation equal 4 when $x$ is 3 (to get the vertex). I'm pretty stumped all around. The problem is from a private SAT prep company. I would appreciate any help.


问:
Prove that a linear mapping is not continuous by a counterexample
Let consider $\mathcal{C}([0,1], \mathbb{R})$. Let $x_0 \in [0, 1]$ and $L$ a linear mapping such that $ \forall f \in \mathcal{C}([0,1], \mathbb{R}), L(f)=f(x_0)$. Prove that $L$ is not continuous when $\mathcal{C}([0,1], \mathbb{R})$ is endowed with the $l_1$ norm.

I tried to consider a piecewise continuous function, but I can't manage to organize the proof.

答:
Let $f_n: [0, 1] \rightarrow \mathbb{R}$ be defined as
$$
f_n(x) :=
\begin{cases}
\left(\frac{x}{x_0} \right)^n & ,~ x<x_0 \\
1 &,~ x=x_0 \\
\left(\frac{x_0}{x} \right)^n &,~ x>x_0
\end{cases}.
$$
for $n \in \mathbb{N}$. It is not hard at all to check continuity of the $f_n$. We have (for $n>1$):
$$
\lVert f_n \rVert_{L^1([0, 1])} = \int_{0}^{x_0} \left(\frac{x}{x_0} \right)^n~\mathrm{d}x + \int_{x_0}^1 \left(\frac{x_0}{x} \right)^n~\mathrm{d}x = \frac{x_0}{n+1} + \frac{x_0 - x_0^n}{n-1} \overset{n \rightarrow \infty}{\longrightarrow} 0
$$
But $\lvert Lf_n \rvert = 1$ for all $n \in \mathbb{N}$.
This means that $L$ is discontinuous.