问:
Combinations' Problem
A person has six friends and during a certain vacation, he met them during several dinners. He found that he dined:-

with all the six exactly on one day,

with every five of them on $2$ days,

with every four of them on $3$ days,

with every three of them on $4$ days, and,

with every two of them on $5$ days.

Further every friend was present at $7$ dinners and every friend was absent at seven dinners.
How many dinners did he have alone?

I got puzzled with so much of information and couldn't get the answer. Please help.

答:
You can solve the problem via integer linear programming as follows.  Let $F=\{0,\dots,5\}$ be the set of friends. For each subset $S \subseteq F$, let nonnegative integer decision variable $x_S$ indicate the number of times that subset is the guest list for a dinner.  The constraints are:
\begin{align}
\sum_{S \subseteq F:\ T \subseteq S} x_S &= 7-k &&\text{for $k \in \{2,\dots,6\}$ and $T \subseteq F$ such that $|T|=k$} \tag1 \\
\sum_{S \subseteq F:\ i \in S} x_S &= 7 &&\text{for $i\in F$} \tag2 \\
\sum_{S \subseteq F:\ i \notin S} x_S &= 7 &&\text{for $i\in F$} \tag3
\end{align}
Constraint $(1)$ forces each $k$-subset to appear the correct number of times.
Constraint $(2)$ forces each friend to be present for $7$ dinners.
Constraint $(3)$ forces each friend to be absent for $7$ dinners.

There are $2^6=64$ variables and $\sum_{k=2}^6\binom{6}{k}+6+6=69$ constraints, and the solution turns out to be unique, even if you do not force $x_S$ to be a nonnegative integer.  Explicitly, here are the constraints, where each $S$ is expressed in binary notation (for example, $x_3 = x_{\{0,1\}}$):
$$
x_{3} + x_{7} + x_{11} + x_{15} + x_{19} + x_{23} + x_{27} + x_{31} + x_{35} +
x_{39} + x_{43} + x_{47} + x_{51} + x_{55} + x_{59} + x_{63} = 5 \\
x_{5} + x_{7} + x_{13} + x_{15} + x_{21} + x_{23} + x_{29} + x_{31} + x_{37} +
x_{39} + x_{45} + x_{47} + x_{53} + x_{55} + x_{61} + x_{63} = 5 \\
x_{6} + x_{7} + x_{14} + x_{15} + x_{22} + x_{23} + x_{30} + x_{31} + x_{38} +
x_{39} + x_{46} + x_{47} + x_{54} + x_{55} + x_{62} + x_{63} = 5 \\
x_{9} + x_{11} + x_{13} + x_{15} + x_{25} + x_{27} + x_{29} + x_{31} + x_{41} +
x_{43} + x_{45} + x_{47} + x_{57} + x_{59} + x_{61} + x_{63} = 5 \\
x_{10} + x_{11} + x_{14} + x_{15} + x_{26} + x_{27} + x_{30} + x_{31} + x_{42} +
x_{43} + x_{46} + x_{47} + x_{58} + x_{59} + x_{62} + x_{63} = 5 \\
x_{12} + x_{13} + x_{14} + x_{15} + x_{28} + x_{29} + x_{30} + x_{31} + x_{44} +
x_{45} + x_{46} + x_{47} + x_{60} + x_{61} + x_{62} + x_{63} = 5 \\
x_{17} + x_{19} + x_{21} + x_{23} + x_{25} + x_{27} + x_{29} + x_{31} + x_{49} +
x_{51} + x_{53} + x_{55} + x_{57} + x_{59} + x_{61} + x_{63} = 5 \\
x_{18} + x_{19} + x_{22} + x_{23} + x_{26} + x_{27} + x_{30} + x_{31} + x_{50} +
x_{51} + x_{54} + x_{55} + x_{58} + x_{59} + x_{62} + x_{63} = 5 \\
x_{20} + x_{21} + x_{22} + x_{23} + x_{28} + x_{29} + x_{30} + x_{31} + x_{52} +
x_{53} + x_{54} + x_{55} + x_{60} + x_{61} + x_{62} + x_{63} = 5 \\
x_{24} + x_{25} + x_{26} + x_{27} + x_{28} + x_{29} + x_{30} + x_{31} + x_{56} +
x_{57} + x_{58} + x_{59} + x_{60} + x_{61} + x_{62} + x_{63} = 5 \\
x_{33} + x_{35} + x_{37} + x_{39} + x_{41} + x_{43} + x_{45} + x_{47} + x_{49} +
x_{51} + x_{53} + x_{55} + x_{57} + x_{59} + x_{61} + x_{63} = 5 \\
x_{34} + x_{35} + x_{38} + x_{39} + x_{42} + x_{43} + x_{46} + x_{47} + x_{50} +
x_{51} + x_{54} + x_{55} + x_{58} + x_{59} + x_{62} + x_{63} = 5 \\
x_{36} + x_{37} + x_{38} + x_{39} + x_{44} + x_{45} + x_{46} + x_{47} + x_{52} +
x_{53} + x_{54} + x_{55} + x_{60} + x_{61} + x_{62} + x_{63} = 5 \\
x_{40} + x_{41} + x_{42} + x_{43} + x_{44} + x_{45} + x_{46} + x_{47} + x_{56} +
x_{57} + x_{58} + x_{59} + x_{60} + x_{61} + x_{62} + x_{63} = 5 \\
x_{48} + x_{49} + x_{50} + x_{51} + x_{52} + x_{53} + x_{54} + x_{55} + x_{56} +
x_{57} + x_{58} + x_{59} + x_{60} + x_{61} + x_{62} + x_{63} = 5 \\
x_{7} + x_{15} + x_{23} + x_{31} + x_{39} + x_{47} + x_{55} + x_{63} = 4 \\
x_{11} + x_{15} + x_{27} + x_{31} + x_{43} + x_{47} + x_{59} + x_{63} = 4 \\
x_{13} + x_{15} + x_{29} + x_{31} + x_{45} + x_{47} + x_{61} + x_{63} = 4 \\
x_{14} + x_{15} + x_{30} + x_{31} + x_{46} + x_{47} + x_{62} + x_{63} = 4 \\
x_{19} + x_{23} + x_{27} + x_{31} + x_{51} + x_{55} + x_{59} + x_{63} = 4 \\
x_{21} + x_{23} + x_{29} + x_{31} + x_{53} + x_{55} + x_{61} + x_{63} = 4 \\
x_{22} + x_{23} + x_{30} + x_{31} + x_{54} + x_{55} + x_{62} + x_{63} = 4 \\
x_{25} + x_{27} + x_{29} + x_{31} + x_{57} + x_{59} + x_{61} + x_{63} = 4 \\
x_{26} + x_{27} + x_{30} + x_{31} + x_{58} + x_{59} + x_{62} + x_{63} = 4 \\
x_{28} + x_{29} + x_{30} + x_{31} + x_{60} + x_{61} + x_{62} + x_{63} = 4 \\
x_{35} + x_{39} + x_{43} + x_{47} + x_{51} + x_{55} + x_{59} + x_{63} = 4 \\
x_{37} + x_{39} + x_{45} + x_{47} + x_{53} + x_{55} + x_{61} + x_{63} = 4 \\
x_{38} + x_{39} + x_{46} + x_{47} + x_{54} + x_{55} + x_{62} + x_{63} = 4 \\
x_{41} + x_{43} + x_{45} + x_{47} + x_{57} + x_{59} + x_{61} + x_{63} = 4 \\
x_{42} + x_{43} + x_{46} + x_{47} + x_{58} + x_{59} + x_{62} + x_{63} = 4 \\
x_{44} + x_{45} + x_{46} + x_{47} + x_{60} + x_{61} + x_{62} + x_{63} = 4 \\
x_{49} + x_{51} + x_{53} + x_{55} + x_{57} + x_{59} + x_{61} + x_{63} = 4 \\
x_{50} + x_{51} + x_{54} + x_{55} + x_{58} + x_{59} + x_{62} + x_{63} = 4 \\
x_{52} + x_{53} + x_{54} + x_{55} + x_{60} + x_{61} + x_{62} + x_{63} = 4 \\
x_{56} + x_{57} + x_{58} + x_{59} + x_{60} + x_{61} + x_{62} + x_{63} = 4 \\
x_{15} + x_{31} + x_{47} + x_{63} = 3 \\
x_{23} + x_{31} + x_{55} + x_{63} = 3 \\
x_{27} + x_{31} + x_{59} + x_{63} = 3 \\
x_{29} + x_{31} + x_{61} + x_{63} = 3 \\
x_{30} + x_{31} + x_{62} + x_{63} = 3 \\
x_{39} + x_{47} + x_{55} + x_{63} = 3 \\
x_{43} + x_{47} + x_{59} + x_{63} = 3 \\
x_{45} + x_{47} + x_{61} + x_{63} = 3 \\
x_{46} + x_{47} + x_{62} + x_{63} = 3 \\
x_{51} + x_{55} + x_{59} + x_{63} = 3 \\
x_{53} + x_{55} + x_{61} + x_{63} = 3 \\
x_{54} + x_{55} + x_{62} + x_{63} = 3 \\
x_{57} + x_{59} + x_{61} + x_{63} = 3 \\
x_{58} + x_{59} + x_{62} + x_{63} = 3 \\
x_{60} + x_{61} + x_{62} + x_{63} = 3 \\
x_{31} + x_{63} = 2 \\
x_{47} + x_{63} = 2 \\
x_{55} + x_{63} = 2 \\
x_{59} + x_{63} = 2 \\
x_{61} + x_{63} = 2 \\
x_{62} + x_{63} = 2 \\
x_{63} = 1 \\
x_{1} + x_{3} + x_{5} + x_{7} + x_{9} + x_{11} + x_{13} + x_{15} + x_{17} + x_{19}
+ x_{21} + x_{23} + x_{25} + x_{27} + x_{29} + x_{31} + x_{33} + x_{35} + x_{37} + x_{39} + x_{41} + x_{43}
+ x_{45} + x_{47} + x_{49} + x_{51} + x_{53} + x_{55} + x_{57} + x_{59} + x_{61} + x_{63} = 7 \\
x_{2} + x_{3} + x_{6} + x_{7} + x_{10} + x_{11} + x_{14} + x_{15} + x_{18} +
x_{19} + x_{22} + x_{23} + x_{26} + x_{27} + x_{30} + x_{31} + x_{34} + x_{35} + x_{38} + x_{39} + x_{42} +
x_{43} + x_{46} + x_{47} + x_{50} + x_{51} + x_{54} + x_{55} + x_{58} + x_{59} + x_{62} + x_{63} = 7 \\
x_{4} + x_{5} + x_{6} + x_{7} + x_{12} + x_{13} + x_{14} + x_{15} + x_{20} +
x_{21} + x_{22} + x_{23} + x_{28} + x_{29} + x_{30} + x_{31} + x_{36} + x_{37} + x_{38} + x_{39} + x_{44} +
x_{45} + x_{46} + x_{47} + x_{52} + x_{53} + x_{54} + x_{55} + x_{60} + x_{61} + x_{62} + x_{63} = 7 \\
x_{8} + x_{9} + x_{10} + x_{11} + x_{12} + x_{13} + x_{14} + x_{15} + x_{24} +
x_{25} + x_{26} + x_{27} + x_{28} + x_{29} + x_{30} + x_{31} + x_{40} + x_{41} + x_{42} + x_{43} + x_{44} +
x_{45} + x_{46} + x_{47} + x_{56} + x_{57} + x_{58} + x_{59} + x_{60} + x_{61} + x_{62} + x_{63} = 7 \\
x_{16} + x_{17} + x_{18} + x_{19} + x_{20} + x_{21} + x_{22} + x_{23} + x_{24} +
x_{25} + x_{26} + x_{27} + x_{28} + x_{29} + x_{30} + x_{31} + x_{48} + x_{49} + x_{50} + x_{51} + x_{52} +
x_{53} + x_{54} + x_{55} + x_{56} + x_{57} + x_{58} + x_{59} + x_{60} + x_{61} + x_{62} + x_{63} = 7 \\
x_{32} + x_{33} + x_{34} + x_{35} + x_{36} + x_{37} + x_{38} + x_{39} + x_{40} +
x_{41} + x_{42} + x_{43} + x_{44} + x_{45} + x_{46} + x_{47} + x_{48} + x_{49} + x_{50} + x_{51} + x_{52} +
x_{53} + x_{54} + x_{55} + x_{56} + x_{57} + x_{58} + x_{59} + x_{60} + x_{61} + x_{62} + x_{63} = 7 \\
x_{0} + x_{2} + x_{4} + x_{6} + x_{8} + x_{10} + x_{12} + x_{14} + x_{16} + x_{18}
+ x_{20} + x_{22} + x_{24} + x_{26} + x_{28} + x_{30} + x_{32} + x_{34} + x_{36} + x_{38} + x_{40} + x_{42}
+ x_{44} + x_{46} + x_{48} + x_{50} + x_{52} + x_{54} + x_{56} + x_{58} + x_{60} + x_{62} = 7 \\
x_{0} + x_{1} + x_{4} + x_{5} + x_{8} + x_{9} + x_{12} + x_{13} + x_{16} + x_{17} +
x_{20} + x_{21} + x_{24} + x_{25} + x_{28} + x_{29} + x_{32} + x_{33} + x_{36} + x_{37} + x_{40} + x_{41} +
x_{44} + x_{45} + x_{48} + x_{49} + x_{52} + x_{53} + x_{56} + x_{57} + x_{60} + x_{61} = 7 \\
x_{0} + x_{1} + x_{2} + x_{3} + x_{8} + x_{9} + x_{10} + x_{11} + x_{16} + x_{17} +
x_{18} + x_{19} + x_{24} + x_{25} + x_{26} + x_{27} + x_{32} + x_{33} + x_{34} + x_{35} + x_{40} + x_{41} +
x_{42} + x_{43} + x_{48} + x_{49} + x_{50} + x_{51} + x_{56} + x_{57} + x_{58} + x_{59} = 7 \\
x_{0} + x_{1} + x_{2} + x_{3} + x_{4} + x_{5} + x_{6} + x_{7} + x_{16} + x_{17} +
x_{18} + x_{19} + x_{20} + x_{21} + x_{22} + x_{23} + x_{32} + x_{33} + x_{34} + x_{35} + x_{36} + x_{37} +
x_{38} + x_{39} + x_{48} + x_{49} + x_{50} + x_{51} + x_{52} + x_{53} + x_{54} + x_{55} = 7 \\
x_{0} + x_{1} + x_{2} + x_{3} + x_{4} + x_{5} + x_{6} + x_{7} + x_{8} + x_{9} +
x_{10} + x_{11} + x_{12} + x_{13} + x_{14} + x_{15} + x_{32} + x_{33} + x_{34} + x_{35} + x_{36} + x_{37} +
x_{38} + x_{39} + x_{40} + x_{41} + x_{42} + x_{43} + x_{44} + x_{45} + x_{46} + x_{47} = 7 \\
x_{0} + x_{1} + x_{2} + x_{3} + x_{4} + x_{5} + x_{6} + x_{7} + x_{8} + x_{9} +
x_{10} + x_{11} + x_{12} + x_{13} + x_{14} + x_{15} + x_{16} + x_{17} + x_{18} + x_{19} + x_{20} + x_{21} +
x_{22} + x_{23} + x_{24} + x_{25} + x_{26} + x_{27} + x_{28} + x_{29} + x_{30} + x_{31} = 7
$$
You can easily solve this linear system of equations by hand by starting from the trivial constraint $x_{63}=1$ and substituting this value in the other constraints, some of which then become trivial.  Repeat until all variables are determined, or just solve for $x_0$.  The unique solution is
$$x_{0}=x_{1}=x_{2}=x_{4}=x_{8}=x_{16}=x_{31}=x_{32}=x_{47}=x_{55}=x_{59}=x_{61}=x_{62}=x_{63} = 1$$ with all other $x_S=0$.  In particular, $x_0=1$, so only one dinner is eaten with no friends.

Title: Combinations' Problem


问:
Exercise on Venn diagrams.
At an international conference on environmental pollution, they attend

90 specialists, of which 33 speak English, 52 Portuguese and 45 Spanish;

28 of them speak Portuguese and English; 22 English and Spanish; 15 Portuguese and

Spanish. How many attendees speak the three languages?

Who help me, with this. Please.

答:
$|E \cup P \cup S| =90$, as there $90$ people in total.

We know that $$|E \cup P \cup S| = |E| + |P| + |S| - |E \cap P| - |E \cap S| -|P \cap S| + |E \cap P \cap S|$$

by the inclusion-exclusion formula. E.g. see here.

Plugging in your data:

$$90 = 33 + 52 + 45 - 28 -22 -15  + |E \cap P \cap S|$$

so  $$90 = 65  + |E \cap P \cap S|$$

so $25 = 90 - 65$ speak all three. Venn diagrams are not needed.

ContentLicense: CC BY-SA 4.0

Title: Exercise on Venn diagrams.


问:
Evaluating the congruence $(1+195) \pmod 7$
$$(1 + 195) \pmod 7 \equiv \quad ?$$

How would I get the answer? Because when I divide $196 / 7$ I get $28$ which is not a decimal to multiply by $7$.

答:
195 mod 7 is 6, and 1 mod 7 is 1

which means the sum is 7 mod 7 = 0

you recieve 28 which is 0 mod 7, which is the answer

ContentLicense: CC BY-SA 3.0


问:
How do we prove that given a set S such that |S|=|SxS| then |P(S)|=|P(S)xP(S)| where P(S) is the power set of S?
Given a bijection f from S to SxS, I can't seem to find an appropriate mapping from P(S) to P(S)xP(S). I don't know if I am missing something obvious but I have tried making a connection through P(SxS) as well but to no avail.
Edit: I'm especially interested in the case of S being the natural numbers as I want to use this and prove that |R|=|R^2|.</p>

答:
You can define a bijection between $\Bbb{P}(\Bbb{N})$ and $\Bbb{P}(\Bbb{N}) \times \Bbb{P}(\Bbb{N})$ by:
$$
A \mapsto (\{ i \in \Bbb{N} \mid 2i \in A\}, \{ j \in \Bbb{N} \mid 2j+1 \in A\})
$$
and that will give you that $|\Bbb{R}| = |\Bbb{R}^2|$. Space filling curves give another (and perhaps more entertaining) take on the latter fact.


问:
analytical function
Could someone please provide a step by step explanation of the following exercise?  Determine the largest open set where f is analytic.$ f(z)=|x^2−y^2|+2i|xy|$

this right? Please I need help to finish it, I don't know what to do?

$f(z) = x^2 - y^2 + 2i |xy| $
$ u = x^2 - y^2 $
$ v = 2i $ $xy $

$\frac{2u}{2x}= 2x $ , $ \frac{2v}{2y} = 2x$

$\frac{2u}{2y}= -2y$

$\frac{2v}{2x} = 2y $

$\frac{2u}{2x} = \frac{2v}{2y}$

$\frac{2u}{2y} = -\frac{2v}{2x}$
Is correct ?

答:
$\textbf{Hint}$: We can write the function piecewise by recognizing that it can take on four different forms

$$f(z) = \begin{cases}z^2 \\ \bar{z}^2 \\ -z^2 \\ -\bar{z}^2\end{cases}$$

depending on which "slice" of the complex plane we are on. The absolute values divide the plane into eight equal slices, and the only ones that the function can be analytic on are the ones where the function is either $z^2$ or $-z^2$. Can you take it from here?

Title: analytical function


问:
Conclude integral is zero without calculation
I have the following two integrals:
$$\int_0^{2\pi} \frac{\cos{\theta}d\theta}{\sqrt{a^2\cos^2{\theta}+b^2\sin^2{\theta}+c^2}},$$
$$\int_0^{2\pi} \frac{\sin{\theta}d\theta}{\sqrt{a^2\cos^2{\theta}+b^2\sin^2{\theta}+c^2}},$$
and I know both are zero. I can argue why the second one is so:
$$\int_0^{2\pi} \frac{\sin{\theta}d\theta}{\sqrt{a^2\cos^2{\theta}+b^2\sin^2{\theta}+c^2}}=\int_{-\pi}^{\pi} \frac{\sin{\theta}d\theta}{\sqrt{a^2\cos^2{\theta}+b^2\sin^2{\theta}+c^2}}.$$
Therefore,
$$\int_0^{2\pi} \frac{\sin{\theta}d\theta}{\sqrt{a^2\cos^2{\theta}+b^2\sin^2{\theta}+c^2}}=\int_{-\pi}^{0} \frac{\sin{\theta}d\theta}{\sqrt{a^2\cos^2{\theta}+b^2\sin^2{\theta}+c^2}}+\int_{0}^{\pi} \frac{\sin{\theta}d\theta}{\sqrt{a^2\cos^2{\theta}+b^2\sin^2{\theta}+c^2}}.$$
With the variable change $\phi=-\theta$ I get $\sin{\theta}d\theta=\sin{\phi}d\phi$ because sine is an odd function (also, the denominator does not change), so the first term becomes
$$\int_{-\pi}^{0} \frac{\sin{\theta}d\theta}{\sqrt{a^2\cos^2{\theta}+b^2\sin^2{\theta}+c^2}}=\int_{\pi}^{0} \frac{\sin{\phi}d\phi}{\sqrt{a^2\cos^2{\phi}+b^2\sin^2{\phi}+c^2}}=-\int_{0}^{\pi} \frac{\sin{\phi}d\phi}{\sqrt{a^2\cos^2{\phi}+b^2\sin^2{\phi}+c^2}},$$
and it cancels the second term so the sine integral is zero.

Is there a similar procedure for the cosine function? Since it is an even funcion the argument can't be exactly the same. Any help will be appreciated!

答: Hint: Use the substitution $\phi = \pi/2 - \theta$.


问:
General Topology- If $f:X \to Y$ is a map between spaces and if $X$ is separable, then the subspace $f(X) \subseteq Y$ is separable
I'm having a hard time constructing the proof for this question. Here's what I have

Let $X$ be separable. Then $f(X) \subseteq Y$ is a subspace of $Y$. Without loss of generality, let $f$ be an open map. Since $X \subseteq X$ is open, $f(X) \subseteq Y$ is also open. By proposition there exists a countable dense subset $A \subseteq X$.

From here I just stated to choose $A=X$ and incorrectly closed the proof. Where should I go from here? Should I utilize the definition of the $f(X)$ subspace somehow?

There is a part 2 to this question involving 2nd countability with basically the same premise, so I'd like to understand the separability proposition first.

Thanks :)

答:
Let $D$ be countable dense in $X$. Then $f[D]$ is dense in $f[X]$: let $O$ be non-empty open in $f[X]$ so we can write $O= f[X] \cap O'$ where $O$ is open in $Y$.

So $f^{-1}[O']$ is open in $X$ as $f$ is continuous and it's non-empty too: pick $y \in O$ so $y \in f[X]$ so for some $x \in X$ we have $f(x)=y \in O \subseteq O'$ showing $x \in f^{-1}[O]$. So as $D$ is dense in $X$, we have some $d \in f^{-1}[O'] \cap D$ and so $f(d) \in f[D] \cap O$ and so $f[D]$ intersects $O$. Done.

You surely cannot assume $f$ is open. "WLOG", it's a huge loss of generality...


问:
How does a column of zeros in a matrix influence its right singular vectors?
Here is the SVD decomposition calculated in Julia:

Why is the 4th column of $V^{T}$ also zero? How is that connected with columns of $A$?

Appreciate any help!

答:
Note that if $A$ has singular value decomposition $A = U \Sigma V^T$, then the columns of $V$ (the rows of $V^T$) form a basis for the row space of $A$. Because every row of $A$ is orthogonal to $(0,0,0,1,0)$, this must hold for the columns of $V$ as well.

Equivalently, the nullspace of $V^T$ must be equal to the nullspace of $A$, and the nullspace of $A$ contains the vector $(0,0,0,1,0)$.


问:
Classic example of a non exact form
Let $\dfrac{xdy-ydx}{x^2+y^2}$ be a 1-form defined in $\mathbb{R}^2\backslash\{0\}$.
Where can I find a detailed proof that it is not exact? I would prefer a proof that doesn't use results about conservative vector fields (I know that way). I have searched Jefree Lee, Do Carmo & some others, but they all skip this part.  Thanks a lot.</p>

答:
I'm not sure whether this is acceptable or not, but the simplest thing to do (since the form is closed) is to attempt to integrate it to find supposed $\phi$ such that your form $\omega$ is $\mathrm d \phi$.

One can derive a contradiction most straightforwardly by going around a circle, say $x^2+y^2=1$. We use the path $x(t)=\cos t,y(t)=\sin t$.
$$\int_0^{2\pi} \omega = \int_0^{2\pi} \frac{\cos^2 t +\sin^2 t}{1}\mathrm d t = 2\pi$$

But if $\omega=\mathrm d \phi$ then the expression on the left is $\phi((1,0))-\phi((1,0))=0$, a contradiction.

In response to your complaints about this method: The definition of exact is that $\omega=\mathrm d \phi$ for some $\phi$. The only result we've used is that integrals of derivatives are given by boundary values (a very mini version of Stokes's theorem), i.e.
$$\int_{\mathbf x}^{\mathbf y} \mathrm d \phi = \left[\int_{t_x}^{t_y} \frac{\mathrm d \phi}{\mathrm d t} \mathrm d t = \right]\phi({\mathbf y})-\phi({\mathbf x})$$
which I don't think is really a mysterious result. If there's something you're not happy with do just say.

Title: Classic example of a non exact form


问:
Solve constrained equation of vector product
I am not really well experienced in linear algebra, so please apologize when the solution to my question is either trivial or unfeasible. I already tried to find a solution using Lagrange multipliers (see below), but my mathematical impotence keeps from from solving the problem.

Let $\vec{w}$ be a known d-dimensional real column vector with norm $1$ ($\vec{w}^T\vec{w} = 1$) and $c$ be a known real-valued scalar. I want to find the solution for:

$$\vec{w}^T\vec{x} = c$$ subject to the constraint $$\vec{x}^T\vec{x}=1$$ (so $\vec{x}$ should be a unit norm vector).

I came up with the Lagrangian minimzation problem

$$\min f(x, \lambda) = \frac{1}{2}(\vec{w}^T\vec{x} - c)^2 - \lambda(\vec{x}^T\vec{x} - 1)$$.

Finding the partial derivatives w.r.t. $\vec{x}$ and $\lambda$ leads to (if I am correct):

$$\frac{\partial f}{\partial \vec{x}} = \vec{w}^T(\vec{w}^T\vec{x} - c) - 2\lambda\vec{x} = \vec{0}$$
and
$$\frac{\partial f}{\partial \lambda} = -\vec{x}^T\vec{x} + 1 = 0$$

But from here, I really don't know how to proceed. Maybe someone could help out?

Thanks a lot in advance!

答:
Consider what happens when $d = 3$. The set of $\vec x$ for which $\vec w^T \vec x = c$ is a plane, and the set of $\vec x$ for which $\vec x^T \vec x = 1$. By drawing yourself a picture, you should be able to see that there are 3 possibilities:

The sphere does not intersect the plane

The plane "cuts" through the sphere so that their intersection is a circle

The plane is tangent to the sphere so that their intersection is a single point.

Whether we are in situation 1,2, or 3 depends on the vector $\vec w$ and the constant $c$. In situation 1, there is no solution. In situation 2, there are infinitely many solutions. In situation 3, there is exactly one solution.

With a bit more work, we can actually come to the following conclusion: if $c = 0$, then we are in situation 2 (and the plane cuts through the center of the sphere). For $c \neq 0$, situation 1 occurs when $\|\vec w\| < c$, situation 2 occurs when $\|\vec w\| > c$, and situation 3 occurs when $\|\vec w\| = c$. This holds in all dimensions $d \geq 3$.


问:
Is the following polynomial reducible?
I need to check whether the following polynomial $g(x)=x^2+t^2−1∈\Bbb{C}(t)$ (where $\Bbb{C}(t)$ is the field of all rational expressions over $\Bbb{C}$) is reducible. My attempt: I assume by contradiction that it's reducible, and I write it as:
$$(x+p(t))(x+q(t))=x^2+t^2−1$$

So we get that:
$$p(t)+q(t)=0$$
$$p(t)q(t)=t^2−1$$

Since $\Bbb{C}(t)$ is a field, it's also a UFD. So either $p(t)q(t)=(t−1)(t+1)$ or WLOG, $p(t)=λ(t^2−1)$ for some $0≠λ∈\Bbb{C}$. Either way, we get a contradiction to the first condition over $p(t),q(t)$. Is my approach correct or am I missing something?

答:
You have mostly put together a correct argument, but let me fill in a few details.

You have a polynomial $g(x) = x^2 + t^2 - 1$ with coefficients taken from $\mathbb C(t)$, the field of rational functions in $t$ over the complex numbers.  We ask whether $g(x)$ is irreducible in $\mathbb C(t)[x]$.

As you pointed out, to factor this monic polynomial $g(x)$ would mean:

$$ g(x) = (x + p(t))(x + q(t)) = x^2 + t^2 - 1 $$

and thus $p(t)+q(t) = 0$ and $p(t)q(t) = t^2 - 1$ for some rational functions $p(t),q(t)\in \mathbb C(t)$.

Since the first of these relations gives $q(t) = -p(t)$, we can dispense with $q(t)$ and focus on the second relation in terms only of $p=p(t)$, namely:

$$ p^2 + t^2 -1 = 0 $$

It is pretty obvious that no polynomial solution $p \in \mathbb C[t]$ is possible.  But the next thing to ask is whether a rational solution $p \in \mathbb C(t)$ is possible.

Here we use the observation that $\mathbb C[t]$ is a very nice UFD (in fact a Euclidean domain), and a UFD is integrally closed.  That means that if a solution to the equation above for $p$ exists in the field of quotients $\mathbb C(t)$ of $\mathbb C[t]$, then that solution would belong already to the base ring of polynomials $\mathbb C[t]$.  Done!

Title: Is the following polynomial reducible?


问:
How many lines are determined by $n$ coplanar points (no three of which are collinear)?
I have a problem in which I need to find how many lines there would be if there are n points. No 3 points are collinear. All points are coplanar. I have figured out this much:

Lines
Points

0
1

1
2

3
3

6
4

10
5

15
6

So on...

I have noticed that the increment between the number of lines is one.

So:

0 + 1 = 1

1 + 2 = 3

3 + 3 = 6

6 + 4 = 10

10 + 5 = 15

Thank you!

答: @JonathanMiller has a website that has the answer. The formula is: $$ \frac{x(x-1)}{2} $$


问:
Cambridge Admissions Exam Statistics 2006
The question I am trying to answer is

My work:

I understand that there are three cases. One where the two points are on the diameter, one where they are both on the circumference of the half circle, and one where one is on the diameter and one is on the circumference. The area in the first case is obviously $0$, however, I do not know how to proceed from here.

I have tried a couple of methods non of which seemed to work. The very first thing I tried was to assign angles to the two points, say $\theta$ and $\alpha$. The area is then given by
$$\frac{\sin (\theta - \alpha)}{2}$$

I thought of now defining a new angle $\phi=\theta - \alpha$ and then to take the expectancy of the area, however, this seems to go nowhere as I first don't know the distribution of this angle, and moreover, I am not sure you can take expectancy over $\sin$
.

Then I thought about using Heron's Formula for the area of a triangle and consider the distribution of the distance between the two points. Again, that led me to nowhere as I do not the distribution of the distance between them.

There are published answers for this question :
https://www.thestudentroom.co.uk/showthread.php?t=894936&page=4#post32939428
https://pmt.physicsandmathstutor.com/download/Maths/STEP/Solutions-and-Reports/2006%20Hints%20and%20Answers.pdf (page 28)

however I do not understand them. They assume that the points are in some regions and then they assume that the distance between them is just the distance between these regions. I am not following the logic of the whole argument. If anybody could help me understand this problem and help me answer it I would very greatful.

答:
Not a full answer but I don't the the rights to comment or the time to give a full answer.  The points are presumably selected uniformly at random with the length of the perimeter the basis for two identical probability spaces.  This is a 2d-probability space with the positions of the two points the two dimensions.

You simply need to integrate across the range of point positions a function for the area of the triangle then divide by the total space.

We can assign the entire probability space $S=(2+\pi)^2$ in size.

An area of probability $\dfrac {2^2}S$ will return a triangle of size zero as you correctly observed, leaving $\dfrac{4\pi + \pi^2}S$ to be assessed.

To get the one point on diameter, one on perimeter area space, you need to parametrise the two points. The intuition they're testing is your ability to discern, is that you need to parametrize the point on the diameter based on its linear co-ordinate and the circumference point based upon its radial co-ordinate, because these parameters "move at constant speed" along the perimeter.

But bear in mind there are two such cases - $a$ on perimeter, $b$ on diameter and the converse so that gets doubled.  This is a double-integral, first between $-1,1$ and then between $-\pi$ and $0$

Then you need to parametrise both points being on the perimeter, this time, both radially, so a double integral with both going from $-\pi$ to $0$.


问:
Is it possible to isolate p in $"a = 2bpq + 2apq + a(p^2) + 2aqq + bqq"?$
Is it possible to simplify this equation so that $p$ is isolated on one side instead of $a$? I tried factoring out $p$ on the right side, but I get stuck with the $ap^2$.

$$a = 2bpq + 2apq + ap^2 + 2aq^2 + bq^2$$

答:
$$a = 2bpq + 2apq + ap^2 + 2aq^2 + bq^2$$
$$ap^2+p(2bq + 2aq) + (2aq^2 + bq^2-a)=0$$
It is  a quadratic equation of $p$.

for $ax^2+bx+c=0\;$

$x=\dfrac{-b\pm\sqrt{b^2-4\cdot a\cdot c}}{2a}$

similarly:

$$p=\dfrac{-(2bq + 2aq)\pm\sqrt{(2bq + 2aq)^2-4\cdot a\cdot (2aq^2 + bq^2-a)}}{2a}$$

I hope you can solve from here


答:
First, we need to consider the probabilities of observing each of the three cases.  To do this, we consider the probability $p$ that a randomly selected point occurs on the diameter:  this is simply $$p = \frac{2}{2+\pi},$$ since the diameter has length $2$ and the semicircle's circumference has length $\pi$.  So the probability of the diameter-diameter ($DD$) case is $$\Pr[DD] = p^2 = \frac{4}{(2+\pi)^2};$$ the circumference-diameter ($CD$) case has probability $$\Pr[CD] = 2p(1-p) = \frac{4\pi}{(2+\pi)^2};$$ and the circumference-circumference ($CC$) case has probability $$\Pr[CC] = (1-p)^2 = \frac{\pi^2}{(2+\pi)^2}.$$

So the expected area is $$\operatorname{E}[A] = \operatorname{E}[A \mid DD]\Pr[DD] + \operatorname{E}[A \mid CD]\Pr[CD] + \operatorname{E}[A \mid CC]\Pr[CC],$$ and you noted already that $A \mid DD = 0$.  We want to figure out the other two conditional expectations.  Let's work with the circumference-diameter case.  In this situation, we are given that one point is on the diameter and the other is on the circumference.  So each point is placed independently and uniformly at random on their respective segments.  Therefore, if we place the semicircle on a coordinate plane with the midpoint of the diameter at the origin and the circumference corresponding to the curve $y = \sqrt{1-x^2}$, then the diameter point has some random $x$-coordinate, say $(X, 0)$ where $X \sim \operatorname{Uniform}(-1,1)$.  The point on the circumference needs to be parametrized by the counterclockwise angle it makes with the positive $x$-axis, i.e., $(\cos \theta, \sin \theta)$ for some $\theta \sim \operatorname{Uniform}(0, \pi)$.  Then the conditional area random variable $A \mid CD$ is a function of $X$ and $\theta$ as follows:
$$A \mid CD = \frac{1}{2} |X| \cdot \sin \theta.$$  The absolute value is needed since $|X|$ represents the absolute unsigned distance between $X$ and the origin.  Since $X$ and $\theta$ are independent, it follows that $$\operatorname{E}[A \mid CD] = \frac{1}{2} \operatorname{E}[|X|] \operatorname{E}[\sin \theta] = \frac{1}{2} \int_{x=-1}^1 |x| \cdot \frac{1}{2} \, dx \int_{\theta=0}^\pi \sin \theta \cdot \frac{1}{\pi} \, d\theta = \frac{1}{2\pi}.$$

Now we do the same thing for the $CC$ case.  Here, we have two angles, say $\theta$ and $\varphi$, where both are IID uniform on $(0,\pi)$.  The tricky part is that the conditional area uses a different formula, from $$|\triangle ABC| = \frac{1}{2}a b \sin C,$$ where $a = b = 1$ because the radius is $1$, and $C = |\theta - \varphi|$ is the magnitude of the difference in the two angles between the sides of unit length.  Consequently,
$$\operatorname{E}[A \mid CC] = \frac{1}{2} \int_{\theta = 0}^\pi \int_{\varphi = 0}^\pi \sin |\theta - \varphi| \frac{1}{\pi^2} \, d\varphi \, d\theta.$$  We can use a symmetry argument to write this double integral as twice an integral over half the domain:
$$\begin{align}
\operatorname{E}[A \mid CC]
&= \frac{1}{\pi^2} \int_{\theta = 0}^\pi \int_{\varphi = 0}^\theta \sin (\theta - \varphi) \, d\varphi \, d\theta \\
&= \frac{1}{\pi^2} \int_{\theta = 0}^\pi \bigl[\cos (\theta - \varphi)\bigr]_{\varphi=0}^\theta \, d\theta \\
&= \frac{1}{\pi^2} \int_{\theta = 0}^\pi (1 - \cos \theta) \, d\theta \\
&= \frac{1}{\pi^2} \bigl[ \theta - \sin \theta \bigr]_{\theta = 0}^\pi \\
&= \frac{1}{\pi}.
\end{align}$$

Now we put everything together:

$$\operatorname{E}[A] = \frac{1}{2\pi} \cdot \frac{4\pi}{(2+\pi)^2} + \frac{1}{\pi} \cdot \frac{\pi^2}{(2+\pi)^2} = \frac{1}{2+\pi},$$ again remembering that the diameter-diameter case is zero.  This matches the claimed answer.

In case there is some confusion about the symmetry argument in the $CC$ case, note that we can explicitly write for any bounded function $f$
$$\begin{align}
\int_{\theta = 0}^\pi \int_{\varphi = 0}^\pi f(\theta, \varphi) \, d\varphi \, d\theta
&= \int_{\theta = 0}^\pi \int_{\varphi = 0}^\theta f(\theta, \varphi) \, d\varphi \, d\theta + \int_{\theta = 0}^\pi \int_{\varphi = \theta}^\pi f(\theta, \varphi) \, d\varphi \, d\theta \\
&= \int_{\theta = 0}^\pi \int_{\varphi = 0}^\theta f(\theta, \varphi) \, d\varphi \, d\theta + \int_{\varphi = 0}^\pi \int_{\theta = 0}^\varphi f(\theta, \varphi) \, d\theta \, d\varphi.\end{align}$$  The double integral in the second term simply has its order of integration switched via Fubini's theorem.  Now we can exploit the aforementioned symmetry, with the choice $f(\theta, \varphi) = \sin |\theta - \varphi| = \sin |\varphi - \theta| = f(\varphi, \theta)$, from which it follows that the interchanged double integral is identical to the first term.

Title: Cambridge Admissions Exam Statistics 2006

Body:
The question I am trying to answer is

My work:

I understand that there are three cases. One where the two points are on the diameter, one where they are both on the circumference of the half circle, and one where one is on the diameter and one is on the circumference. The area in the first case is obviously $0$, however, I do not know how to proceed from here.

I have tried a couple of methods non of which seemed to work. The very first thing I tried was to assign angles to the two points, say $\theta$ and $\alpha$. The area is then given by
$$\frac{\sin (\theta - \alpha)}{2}$$

I thought of now defining a new angle $\phi=\theta - \alpha$ and then to take the expectancy of the area, however, this seems to go nowhere as I first don't know the distribution of this angle, and moreover, I am not sure you can take expectancy over $\sin$
.

Then I thought about using Heron's Formula for the area of a triangle and consider the distribution of the distance between the two points. Again, that led me to nowhere as I do not the distribution of the distance between them.

There are published answers for this question :
https://www.thestudentroom.co.uk/showthread.php?t=894936&page=4#post32939428
https://pmt.physicsandmathstutor.com/download/Maths/STEP/Solutions-and-Reports/2006%20Hints%20and%20Answers.pdf (page 28)

however I do not understand them. They assume that the points are in some regions and then they assume that the distance between them is just the distance between these regions. I am not following the logic of the whole argument. If anybody could help me understand this problem and help me answer it I would very greatful.


问:
if a set S is convex then $\left\langle x-y,z\right\rangle <\left\langle x-y,\frac{x}{2}+\frac{y}{2}\right\rangle $ for any z in the set
let S be a convex set, and let $x\notin S;y,z\in S$.
such y is the closest to x in S.
prove:
$\left\langle x-y,z\right\rangle <\left\langle x-y,\frac{x}{2}+\frac{y}{2}\right\rangle $.
The inner product here is of course the euclidean one. and the distance is with the regular Euclidean distance.
I tried using Cauchy Schwartz, but it has not been successful.

答:
Here is geometric proof.  Let $t={1\over 2}(x+y).$ (We shall only use that
$t$ is strictly between $x$ and $y$ on the line segment $xy.$) Let $B$ be the closed ball of radius
|x-y| and center $x.$  Observe that $S\cap B=\{y\}.$ Let $H$ be the hyperplane through
$y$ with normal $x-y.$ If $z$ were on the same side of $H$ as $t,$ then the line segment $yz$
would intersect $B$ at points other than $y,$ a contradiction. Thus, $z$ is on $H$ or on the opposite
side from $t.$ Therefore, $\langle x-y,t-z\rangle\gt 0.$ In other words,
$$\left\langle x-y,z\right\rangle <\left\langle x-y,t\right\rangle.$$


问:
Negation of $\forall x T(x) \implies \exists y L(y)$
I am trying to figure out the negation of $\forall xT(x) \implies \exists y L(y)$

I have two choices in mind

$\exists x T(x) \land \forall y \neg L(y)$

$\forall x T(x) \land \forall y \neg L(y)$

Not sure which is correct. I know in general the negation of $p \implies q$ is $p \land \neg q$ but here how to handle the quantifiers is confusing me.

For instance if the bracketing was $\forall x(T(x) \implies \exists y L(y))$, then I believe the negation would be $\exists x T(x) \land \forall y \neg L(y)$, but in this case since $\forall x T(x)$ is a premise, I think that $2$ is correct.

答: You're nearly there. Just note that $$\lnot\exists yLy\equiv \forall y\lnot Ly.$$


答: The traditional convention is that quantifiers bind tighter than the binary connectives: so you should read your formula as  $(\forall x T(x)) \implies (\exists y(L(y))$, which makes your proposed answer 2 the correct one. (Your idea about $\forall x(T(x) \implies \exists y(L(y))$ isn't right: $\implies$ should still turn into $\land$ when you negate it. This would lead to answer 1.)

Title: Negation of $\forall x T(x) \implies \exists y L(y)$

Body:
I am trying to figure out the negation of $\forall xT(x) \implies \exists y L(y)$

I have two choices in mind

$\exists x T(x) \land \forall y \neg L(y)$

$\forall x T(x) \land \forall y \neg L(y)$

Not sure which is correct. I know in general the negation of $p \implies q$ is $p \land \neg q$ but here how to handle the quantifiers is confusing me.

For instance if the bracketing was $\forall x(T(x) \implies \exists y L(y))$, then I believe the negation would be $\exists x T(x) \land \forall y \neg L(y)$, but in this case since $\forall x T(x)$ is a premise, I think that $2$ is correct.


问:
If fundamental group of a compact path connected space is finite then its universal cover is compact
Let $X$ be a compact and path connected space and let $p:\tilde X\rightarrow X$ be its universal cover. I can show that if $\tilde X$ is compact then $\pi_{1}(X)$ is finite:

$\pi_{1}(X, x_{0})$ acts on $p^{-1}(x_{0})$  via $[\gamma].y=\tilde\gamma(1)$ where $\tilde\gamma$ is the unique lifting strating at y. Another observation is that
Stab(y)={$[\gamma]$$\in\pi_{1}(X, x_{0})$ : $\tilde\gamma$ is a closed loop}=$p_{*}(\pi_{1}(\tilde X, \tilde x_{0}))$ where $p_{*}$ is the induced homomorphism of $p$. Since this action is transitive,  there is a bijection between  $p^{-1}(x_{0})$ and $\pi_{1}(X, x_{0})$/$p_{*}(\pi_{1}(\tilde X, \tilde x_{0}))$. In our case $\tilde X$ is simply connected. So $p^{-1}(x_{0})$ and $\pi_{1}(X, x_{0})$ has the same cardinality. But $p^{-1}(x_{0})$ is closed and discrete subspace of compact space $\tilde X$. so it is finite, as desired.

But I couldn't show the converse. I am new in algebraic topology and I follow Hatcher's book chapter 1. Can any one give a hint or answer? Thanks!

答:
Let $\{ \tilde{U}_i : i \in I \}$ be an open cover of $\tilde{X}$. For each point $\tilde{x}$ in the fibre $p^{-1} \{ x \}$, choose an open neighbourhood $\tilde{U}_{\tilde{x}}$ of $\tilde{x}$. Since $p$ is a local homeomorphism, we may choose an open neighbourhood $\tilde{V}_{\tilde{x}} \subseteq \tilde{U}_{\tilde{x}}$ of each $\tilde{x}$ that $p$ maps homeomorphically onto an open neighbourhood $V_{\tilde{x}}$ of $x$, and then $V_x = \bigcap_{\tilde{x} \in p^{-1} \{ x \}} V_{\tilde{x}}$ is an open neighbourhood of $x$ such that $p^{-1} V_x$ is a disjoint union of homeomorphic copies of $V_x$, each contained in some $\tilde{U}_i$.

Observe that $p$ is surjective, so $\{ V_x : x \in X \}$ is an open cover of $X$, so has a finite subcover. Consider $\{ p^{-1} V_x : x \in X \}$. This is an open cover of $\tilde{X}$, but not necessarily subordinate to $\{ \tilde{U}_i \}$. Nonetheless, by construction, there is a finite refinement of $\{ p^{-1} V_x : x \in X \}$ that is subordinate to $\{ \tilde{U}_i \}$, and this implies the existence of the desired finite subcover. I leave the details to you.


问:
Reference: Covariance operator is compact
I am looking for a reference for the following theorem:

Let $(X,\Vert \cdot \Vert)$ be a separable Banach space and $C: X^{\ast} \rightarrow X \subseteq X^{\ast \ast}, f \mapsto \int_X f(x) \cdot - ~ d\mu(x) =: q(f, -)$ be the covariance operator of a Gaussian measure on $X$. Then $X$ is compact, where $X^{\ast}$ carries the operator norm topology and $X$ the norm topology.

For finite dimension this is trivial.

For a separable Hilbert spaces one can even show that $C$ is trace class by choosing an orthonormal basis $\{e_n \}_{n \in \mathbb{N}}$ and

$$\int_H \Vert x \Vert_H^2 d \mu(x) = \sum_{n = 1}^{\infty} \int_H \langle x, e_n \rangle^2 d \mu(x) = \sum_{n = 1}^{\infty} q(e_n, e_n) = \sum_{n = 1}^{\infty} \langle C e_n, e_n \rangle = \text{tr} ~ C$$

In his notes on SPDE, Martin Hairer gives the following hint:

Proceed by contradiction by first showing that if $C$ wasn’t
compact, then it would be possible to find a constant $c > 0$ and a sequence of elements $(f_n)_{n \in \mathbb{N}}$ such that $\Vert f_n \Vert = 1$, $q(f_k, f_n) = 0$ for any $k \neq n$ and $q(f_n, f_n) \geq c$ for every $n \geq 1$.

Conclude that if this was the case, then the law of large numbers applied to the sequence of random variables $(f_n)_{n \in \mathbb{N}}$ would imply that $\sup_{n \in \mathbb{N}} f_n(x) = \infty$ for $\mu$-almost every $x \in X$, thus obtaining a contradiction with the fact that $\sup_{n \in \mathbb{N}} \vert f_n(x) \vert \leq \Vert x \Vert < \infty$ for $\mu$-almost every $x \in X$.

答:
One approach is to note that $C$ is essentially the embedding of $X^*$ into the Cameron-Martin space $H \subset X$.  Now the result follows from the fact that $H$ is compactly embedded into $X$.  This can be found as:

Lemma I.4.5 of Kuo, Hui-Hsiung, Gaussian measures in Banach spaces, Lecture Notes in Mathematics. 463. Berlin-Heidelberg-New York: Springer-Verlag. VI, 224 p. DM 23.00 (1975). ZBL0306.28010.

Corollary 3.2.4 of Bogachev, Vladimir I., Gaussian measures. Transl. from the Russian by the author, Mathematical Surveys and Monographs. 62. Providence, RI: American Mathematical Society (AMS). xii, 433 p. (1998). ZBL0913.60035.

Another approach, similar to what I did in Proposition  4.16 of these lecture notes, is to first prove that $\int_X \|x\|^2\,d\mu(x) < \infty$, e.g. as a corollary of the much stronger Skorokhod or Fernique theorems.  Now suppose we have a sequence $f_n$ in the unit ball of $X^*$. By Alaoglu, we can pass to a subsequence converging weak-* to some $f$.  Now since the $f_n$ have norm at most one, then as functions on $X$, they are dominated by the square-integrable function $\|\cdot\|$.  Hence by dominated convergence, they converge to $f$ strongly in $L^2(\mu)$.  Now Cauchy-Schwarz gives, for any $g \in X^*$ with $\|g\| \le 1$,
$$|q(f_n - f, g)| \le \|f_n - f\|_{L^2} \|g\|_{L^2} \le \|f_n - f\|_{L^2} \left(\int \|x\|^2\mu(dx)\right)^{1/2}$$
where the right side approaches 0 uniformly in $g$, so we have $q(f_n - f, \cdot) \to 0$ in $X^{**}$.


问:
Proof of whether or not a set P is an open set
Let $P=\{(x_1, x_2,0) :x_1, x_2 \in \mathbb{R}\}$ be the $x_1$-$x_2$ plane in $\mathbb{R}^3$. Is $P$ an open subset of $\mathbb{R}^3$?

I know that $P$ is an open set if for each points $x_1, x_2$ in $P$, there exists a radius $r>0$ such that $B_r(x) \subseteq P$. I think that $P$ is an open set because it is just a circle in the $\mathbb{R}^3$ plane. This is assuming the $0$ doesn't change in the definition of open considering it is not a variable. I am not sure how to prove the set is open from here.

答:
Hint: it isn't open.

Negate the definition of an open set. Choose the origin (edit: tbc, the origin in $\mathbb{R}^3$). Let $r>0$ be given. Show that there exists a point within distance $r$ of the origin, yet which isn't in $P$. Conclude that the origin is not an interior point of $P$.


问:
Does $\int_0^{+\infty}\cos(x^3)\,\mathrm{d}x$ converge absolutely or conditionally?
$$\int\limits_0^{+\infty}\cos(x^3)\,\mathrm{d}x$$

I think that the integral absolutely does not converge, since the function does not tend to zero, but conditionally it converges, since if you look at the graph of the function, the areas of the cosine humps decrease, and the signs alternate, and then the Leibniz integral converges, but I think this solution is not strict and I would like to get a normal one, but there are no ideas.

答:
The integral converges, but you should look in an integral table (or perform computer algebra) to find:

$$\frac{\Gamma \left(\frac{1}{3}\right)}{2 \sqrt{3}} \approx 0.773343$$

Here's a graph of the integral up to $t$:


问:
Splitting the representation of $\mathbb{R}$ on $L^2(\mathbb{R})$ into irreducibles.
On p2 of these notes, the author writes that the translation action of $\mathbb{R}$ on $L^2(\mathbb{R})$ is a reducible representation which does not contain any irreducible subrepresentations. Presumably by $L^2(\mathbb{R})$ he means the square integrable functions $f : \mathbb{R}\rightarrow\mathbb{C}$.

"Indeed, for any measurable set $S\subset\mathbb{R}$, the space of functions $f$ for which the Fourier transform $\hat{f}$ has support in $S$ is a proper closed invariant subspace."

This space is certainly invariant - how would one see that it is closed?

"By Schur's lemma, any irreducible subrepresentation $W\subset L^2(\mathbb{R})$ would have dimension 1, ie $W = \langle F\rangle$ where $F: \mathbb{R}\rightarrow\mathbb{C}$ is a function such that $F(x+r)$ is proportional to $F(x)$. Thus, $F(x)$ is proportional to $e^{\lambda x}$ for some $\lambda\in\mathbb{C}$, but no such function lies in $L^2$"

No issues with this.

"However, a representation "decomposes" into a (possibly uncountable) number of irreducibles. This is the theory of "unitary disintegration""

This is my main question - what does he mean by this? Googling "unitary disintegration" did not yield any relevant results. Furthermore, how can a representation decompose into irreducibles, but have no irreducible subrepresentations?

References would be appreciated.

答:
Let $G = \mathbb R$, and $V = L^2(\mathbb R)$, which is a unitary representation of $G$ with the translation action.  It is true (more generally of unitary representations of locally compact groups) that if $W \subset V$ is a subrepresentation, then there exists another subrepresentation $W'$ of $V$ such that $V = W \oplus W'$.  Consequently, $V$ is semisimple as an abstract representation of $G$: there exist irreducible subrepresentations $W_i$ of $V$ such that

$$V = \bigoplus\limits_i W_i.$$

This is an algebraic direct sum, which means for every $v \in V$, there exist unique $w_i \in W_i$, all but finitely many of which are $0$, such that $v = \sum\limits w_i$.

This is not what representation theorists who study Hilbert space representations are interested in.  They are interested, in decomposing representations into a direct sum of irreducibles in a different sense: that is, in the best case scenario they are interested in finding closed irreducible representations $W_i$ of $V$ (that is, closed subspaces $W_i$ of $V$ which are subrepresentations of $G$, and which themselves admit no nonzero, proper, closed  subrepresentations $W_i'$ of $W_i$) such that the $W_i$ are mutually orthogonal, and such that

$$V = \hat{\bigoplus\limits_i} W_i \tag{Hilbert space direct sum}$$
This last condition means that for every $v \in V$, there exist $w_i \in W_i$ such that $v = \sum\limits_i w_i$ (as an absolutely convergent sum in the norm topology).

This turns out to be impossible in this case, since $V = L^2(\mathbb R)$ does not have any closed, irreducible subrepresentations: if $W$ were such a subrepresentation, it would have to be one-dimensional and correspond to a unitary character of $\mathbb R$.  Then there would have to exist a function $f: \mathbb R \rightarrow \mathbb C$ and a real number $t$, such that

$$\int\limits_{\mathbb R} |f(x)|^2 dx < \infty$$

$$ f(x+y) = e^{ 2 \pi i ty} f(x) \tag{$x,y \in \mathbb R$}$$
and this is impossible.


问:
How to approximate a sharply peaked function with deltas?
I have the function

$$
f(x)=\cases{\frac{1}{\sqrt{\sinh^2(a)-\sinh^2(ax)}} & $-1 \leq x \leq1$ \\ 0 & otherwise}
$$

Where $a \in \mathbb{R}^+$. Here is a plot of $f$:

It would be very convenient if I could justify (in any way: maybe distributionally?) that

$$
f(x)=A\left[ \delta(x-1)+\delta(x+1) \right]+B+\mathcal{O}(?) \quad ; \quad a \to \infty \tag{1}
$$

For some $x$ independent constants $A$ and $B$. I am uncertain of how to do this, or how to estimate the error.

What I've tried

My thought was to look at the Fourier series of $f$, and associate terms with the Fourier series of (2). By expanding (1) around $x= \pm1$, I can find a Laurent series in $(x \pm 1)$, and perform the Fourier integral term by term. The result is messy, and worse: does not allow easy comparison to the Fourier series of (2). This may not be a fruitful approach.

By looking at the wiki page about mollifiers, it seems I need to claim something like: $a f(x) \to \delta (x \pm1), \ a \to \infty$. Superficially, this appears to work, but I'm not sure how to make it concrete, or if there should be any other factors next to $af(x)$ (other than a factor to ensure normalization to unity).

Questions

How can I estimate the error and find $A$ and $B$ in (1)?

If $A$ and $B$ cannot be found, how can I find the ratio $A/B$?

Background

$f(x)$ will eventually appear beneath an integral with a Green's function $G$. I'd like to formulate this for arbitrary $G$. We can take some liberties about the 'niceness' of $G$, as necessary. In particular, it has a well behaved Fourier transform.

答:
This answer is intended to provide a more complete mathematical account of the heuristic approach used by @Han de Bruijn in a previous answer.

The function as prescribed above does not converge to a delta-function in the limit $a\to\infty$. However it seems to converge to one with some appropriate modifications. Consider the modification

$$f_a(x)=N(a)\left(\frac{1}{\sqrt{\sinh^2 a-\sinh^2ax}}-\frac{1}{\sinh a}\right)$$

So far the normalization factor in front is undetermined but we will see that it is necessary for the convergence towards a distribution with the desired properties. Consider now an integral over a test function and perform the change of variables $\sinh a \tanh t=\sinh ax$

$$\int_{-1}^{1}dx\phi(x)f_a(x)=\frac{N(a)}{a}\int_{-\infty}^{\infty}\phi\left(\frac{1}{a}\sinh^{-1}(\sinh a\tanh t)\right)\frac{\cosh t-1}{\cosh^2 t\sqrt{1+\sinh^2a\tanh^2 t}}$$

Note that $\lim_{a\to\infty}\frac{1}{a}\sinh^{-1}(\sinh a\tanh t)=\text{sgn}(t)$. In order for the limit of the integrand to exist we need to assume that

$$\lim_{a\to\infty}\frac{N(a)}{a\sinh a}=L\in\mathbb{R}$$

With this determined, it is easy to see that we can use the dominated convergence theorem since the integrand is dominated by an integrable function

$$|\phi|\frac{\sinh a}{\sqrt{1+\sinh^2 a\tanh^2 t}}\frac{\cosh t -1}{\cosh^2 t}\leq M_\phi\frac{\cosh t -1}{\cosh t|\sinh t|} $$

where $M_\phi$ is the maximum of $|\phi(x)|,x\in(-1,1)$  (here we have assumed that $\phi$ is an appropriate test function).

Whenever the condition on the normalization factor is satisfied we readily see that

$$\lim_{a\to\infty}\int_{-1}^1dx \phi(x)f_a(x)=L(\phi(1)+\phi(-1))\int_0^{\infty}dt\frac{\cosh t-1}{\cosh t\sinh t}$$

where the integral can be exactly evaluated to be

$$\int_0^{\infty}dt\frac{\cosh t-1}{\cosh t\sinh t}=\log2$$

It can be shown that normalizing $f_a(x)$ by the area under it's curve naturally chooses $L=(2\log 2)^{-1}$ and is enough to guarantee the convergence to exactly $\frac{\delta(x-1)+\delta(x+1)}{2}$, as required by the fact that the area of the normalized curve is $1$.

$$$$


问:
How to understand the reduced cost in simplex method?
I’m reading a note on the simplex method. The author mentions a quantity called “reduced cost”, yet no interpretation of it is provided.

Here’s the setup: $c \in \mathbb{R}^n$ is the cost vector, $A \in \mathbb{R}^{m \times n}$ is the constraint matrix in our system of $m$ equations in $n$ variables, $I \in [n]$ the index set of non-basic variables, and $I^C$ its complement. We start at a vertex $v$ of the associated polyhedron and try to move along an “interesting” direction $d$ (defined below). The reduced cost is defined as:

$$c - ((c_{I^C})^T (A_{I^C})^{-1}A))^T$$

where $c_{I^C}$ and $A_{I^C}$ are $c$ and $A$ restricted to the columns with indices in the subscripts, respectively.

I have two questions concerning this notion.

What does this vector tell us?

A bit further down, in showing that the simplex method is correct, the author argues in the following way. Let $v$ be a vertex outputted by the simplex method. Then for any $I \subseteq [n]$ “related to $v$” (defined below), and for any $i^* \in I$:

  $\bullet$ either $(c - ((c_{I^C})^T (A_{I^C})^{-1}A))^T)_{i^*} \leq 0$, or

  $\bullet$ there exists index $j$ such that $(v_{I^C})_j = 0$ and $((-A_{I^C})^{-1}A_{i^*})_j < 0$, and so $t = 0$.

The comment is that the first case corresponds to the situation where moving along the direction $d$ makes the objective function worse, and the second case the situation where doing so makes the solution infeasible.

I hope to understand the former case by reading answers to question $1$ above. As for the latter, I’m confused as to why this case can even exist. From what I understand, moving along a direction $d$ from $v$ is actually the same thing as moving along a $1$-face of the associated polyhedron. Since we start moving from a vertex with $(v_{I^C})_j = 0$, $v$ should already be a point of the polyhedron with the smallest $j$-value. So how can a direction $d$ that leads to another vertex with even smaller (in fact, negative) $j$-value exist?

I should note that the notations that I’m reading are rather unconventional. The author actually doesn’t use the words “non-basic variables”, but specifies $I$ (and $d$) as follows:

Given a vertex $v$ of the polyhedral set $Ax=b, x \geq 0$, and a subset $J$ of size $n - m$ of $[n]$, such that $v$ is the unique solution of the system of equations $Ax = b, x_j = 0$ for $j \in J$, we find:

  $\bullet$ a subset $I$ of $[n]$ of size $n - m$,

  $\bullet$ an index $i^* \in I$,

  $\bullet$ and a direction $d$,

such that:

  (i) $v$ is the unique solution of $Ax=b, x_{i}=0$ for $i \in I$

  (ii) $A d=0, d_{i}=0$ for $i \in I \setminus \{i^*\}, d_{i^*} > 0$

  (iii) $c^Td>0$

  (iv) the solution $v+t d$ is a feasible solution for some $t>0$.

答:
The fundamental idea is that when $A\mathbf x = \mathbf b$ holds, we can express the objective function $\mathbf c^{\mathsf T} \mathbf x$ in several equivalent ways. (For example, if $x_1 + x_2 = 1$ holds, we can express $2x_1 + x_2$ as $x_1 + (x_1 + x_2) = x_1 + 1$.)

In particular, if we're at a solution where only the basic variables indexed by $B \subseteq [n]$ are nonzero, and the nonbasic variables indexed by $N = [n] - B$ are all zero, it makes sense to express the objective function in terms of the nonbasic variables only. That will let us see which nonbasic variables are good to bring into the basis. To do this, solve $A\mathbf x = \mathbf b$ for $\mathbf x_B$ in terms of $\mathbf x_N$:
$$
   A \mathbf x = \mathbf b \implies A_B \mathbf x_B + A_N \mathbf x_N = \mathbf b \implies \mathbf x_B = A_B^{-1}(\mathbf b - A_N \mathbf x_N).
$$
Then substitute this into the objective function
$$
   \mathbf c^{\mathsf T}\mathbf x = \mathbf c_B^{\mathsf T} \mathbf x_B + \mathbf c_N^{\mathsf T} \mathbf x_N = \mathbf c_B^{\mathsf T}A_B^{-1}(\mathbf b - A_N \mathbf x_N) + \mathbf c_N^{\mathsf T} \mathbf x_N.
$$
This has a constant component $\mathbf c_B^{\mathsf T}A_B^{-1}\mathbf b$ (which is the value of the objective function at the current basic solution) and a variable component $(\mathbf c_N^{\mathsf T} - \mathbf c_B^{\mathsf T}A_B^{-1}A_N) \mathbf x_N$. We might as well extend the variable component to $(\mathbf c^{\mathsf T} - \mathbf c_B^{\mathsf T}A_B^{-1}A) \mathbf x$, which includes all variables (the basic variables are included with coefficient $0$).

The coefficient vector of $\mathbf x$ in the variable component is exactly the reduced cost. It is the cost vector, re-expressed in terms of the nonbasic variables only.

The vector $\mathbf v$ "related to $N$" is the basic solution with basic variables $B$: it has $\mathbf v_B = A_B^{-1} \mathbf b$ and $\mathbf v_N = \mathbf 0$.

We've expressed the cost function already as $\mathbf c_B^{\mathsf T}A_B^{-1}\mathbf b +  (\mathbf c^{\mathsf T} - \mathbf c_B^{\mathsf T}A_B^{-1}A) \mathbf x$. Let $\mathbf r^{\mathsf T} = \mathbf c^{\mathsf T} - \mathbf c_B^{\mathsf T}A_B^{-1}A$ be the reduced cost, noting that $\mathbf r_B = \mathbf 0$. We spot that $\mathbf c_B^{\mathsf T}A_B^{-1}\mathbf b = \mathbf c_B^{\mathsf T} \mathbf v_B = \mathbf c^{\mathsf T} \mathbf v$, the objective value at $\mathbf v$.

So we can write the objective function at any point $\mathbf x$ as
$$\mathbf c^{\mathsf T} \mathbf x = \mathbf c^{\mathsf T} \mathbf v + \mathbf r^{\mathsf T} \mathbf x.$$ In particular:

Suppose $\mathbf r \le \mathbf 0$. Then because $\mathbf x \ge \mathbf 0$ for every feasible $\mathbf x$, we have $\mathbf r^{\mathsf T} \mathbf x \le 0$, so $\mathbf c^{\mathsf T} \mathbf x \le \mathbf c^{\mathsf T} \mathbf v$. In other words, $\mathbf v$ is optimal.

Suppose $r_j > 0$ for some $j \in N$. If we can pivot to a solution $\mathbf x$ where $x_j > 0$, but still $x_k = 0$ for all $k \in N - \{j\}$, then its objective value will be $\mathbf c^{\mathsf T} \mathbf v + \mathbf r^{\mathsf T} \mathbf x = \mathbf c^{\mathsf T} \mathbf v + r_j x_j > \mathbf c^{\mathsf T} \mathbf v$. In other words, the objective value will increase.


问:
Divisibility by 37 .
Let the sum of two three-digit numbers be divisible by 37. Prove that the six-digit number obtained by concatenating the digits of those numbers is also divisible by 37.

$\overline {abc}$ + $\overline {def}$ is divisible by 37. Prove $$\overline{abcdef}$$ is divisible by 37.

$$\overline {abc} = 100a + 10b + c$$
$$\overline {def} = 100d + 10e + f$$
then we have
$$\overline {abc}+ \overline {def} = 100a + 10b + c + 100d + 10e + f = 100(a+d) + 10(b+e) + c + f $$
And I'm stuck here. Can anyone help me?

答:
Since $37 \mid 111$, then $37 \mid 999$. Hence, for really large numbers, first cast out $999s$ untill you get a $3$ or less digit number. Then check that number for divisibility by  $37$.

Example: $N = 285566$.

$285566 \to 285 + 566 = 851 \to |851 - 888| = 37$

Hence $37 \mid 285566$

Title: Divisibility by 37 .


答:
$(0, 0, 0) \in P$, but for any $r > 0$, $B_r((0, 0, 0))$ contains the point $(0, 0, r/2)$ which is not in $P$. So $P$ is not open. (Your description of what it means for $P$ to be open is nearly right, but you only need to talk about one $x$, not $x_1$ and $x_2$.)

Another way of seeing this is to note that $P$ is the inverse image of $\{0\}$ under the continuous function from $\Bbb{R}^3$ to $\Bbb{R}$ which maps a point $(x, y, z)$ to $z$. As this function is continuous, the inverse image $P$ of the closed set $\{0\}$ must also be closed. But the only subsets of $\Bbb{R}^3$ that are both open and closed are $\emptyset$ and $\Bbb{R}^3$, and, as $P$ is neither of these, it is not open. This approach may involve ideas about continuity and connectedness that you haven't studied yet, but when you have learnt those ideas, it can give you an "at a glance" way of seeing whether a set defined by a system of one or more constraints is open or closed. (The constraint in your example is $z = 0$. If you changed it to $z > 0$ you would get an open set.)

Title: Proof of whether or not a set P is an open set

Body:
Let $P=\{(x_1, x_2,0) :x_1, x_2 \in \mathbb{R}\}$ be the $x_1$-$x_2$ plane in $\mathbb{R}^3$. Is $P$ an open subset of $\mathbb{R}^3$?

I know that $P$ is an open set if for each points $x_1, x_2$ in $P$, there exists a radius $r>0$ such that $B_r(x) \subseteq P$. I think that $P$ is an open set because it is just a circle in the $\mathbb{R}^3$ plane. This is assuming the $0$ doesn't change in the definition of open considering it is not a variable. I am not sure how to prove the set is open from here.


问:
Given $\textrm{gcd}(p,q)=1$, in what cases do positive integers $k$ and $a$ exist for which $k^2 + (2q)^2 = (aq+p)^2$?
Given $\textrm{gcd}(p,q)=1$, in what cases do positive integers $k$ and $a$ exist for which $k^2 + (2q)^2 = (aq+p)^2$?

I have tried to approach this problem using Euclid's parameterization of Pythagorean triples, obtaining for even $q$:

$aq+p = u^2+v^2$

$q = uv$

$k=u^2-v^2$

However, I am unable to find the cases in which there is a way to select such $u$ and $v$. Any form of help would be greatly appreciated.

答:
(Elevating my comment to an answer)

Given $q$, write any factorization $q=uv$ with $u>v>0$ and $\gcd(u,v)=1$; choose any $a$ such that $aq<u^2+v^2$; let $p=u^2+v^2-aq$; let $k=u^2-v^2$, and you have a solution.


问:
How to compute the argmin over vectors?
Can I just differentiate to find the argmin if I'm looking for a vector and if yes which differentiation is used for such questions as below?  $f$ is differentiable.
$v_0 = argmin_{v\in \mathbb{R^n}} f(v)$
And further can anyone give me a source with useful differentiation rules if I have vectors, e.g. how to differentiate $\|v-x\|_2^2.$

答:
Your question is not too clear. So I make some assumptions.

Yes, you can use ArgMin to get the argument of the minimum.

Concerning your vectors, I assume that one vector b is a given constant and the other a  is a position vector, With this, the derivatives are:

```<code>a = {x, y};
b = {x0, y0};
D[(a - b).(a - b), x]
(* 2 (x - x0 *)

D[(a - b).(a - b), y]
(* 2 (y - y0) *)
</code>```

OwnerDisplayName: guest

Title: How to compute the argmin over vectors?

Body:
Your question is not too clear. So I make some assumptions.

Yes, you can use ArgMin to get the argument of the minimum.

Concerning your vectors, I assume that one vector b is a given constant and the other a  is a position vector, With this, the derivatives are:

a = {x, y};
b = {x0, y0};
D[(a - b).(a - b), x]
(* 2 (x - x0 *)

D[(a - b).(a - b), y]
(* 2 (y - y0) *)

OwnerDisplayName: Daniel Huber


答:
But $\mathbb R^3$ is not a plane.

It's a three-dimensional space.  And the open set $P$ is not  a circle.  It is a sphere.... a $3D$ sphere... and a sphere can't lie entirely in a plane.

Any open ball $B_r(w_1, w_2, 0) = \{(x,y,z)\in \mathbb R^3| d((w_1,w_2,0), (x,y,z)) < r\}$ will include the point $(w_1,w_2, \frac 12 r)$.  (because $d((w_1,w_2,0),(w_1,w_2,\frac 12r)) = \sqrt {(w_1-w_1)^2 + (w_2-w_2)^2 + (\frac 12 r- 0)^2} =\frac 12 r < r$).  And $(w_1, w_2, \frac 12 r) \not \in \{(x_1, x_2, 0)|x_1,x_2 \in \mathbb R\}$.

So no, a 2-dimensional plane existing in a 3-dimensional space is not all open set.


问:
Relationships between $2$ mutually tangent circles in a semi-circle
In the attempt to answer this question, I was brought to another one, connected to a certain family of circles.

Consider the semicircle made of the upper half of the unit circle and the horizontal diameter with endpoints $A(-1,0)$ and $B(1,0)$.

Let us now consider the family $(F)$ of circles inscribed into this semicircle as can be seen on Fig. 1.

Fig. 1: Certain circles of family $(F).$

They can be described in a parametric way as circles

$$\text{with center} \ (x_c=-\cos a, y_c=\frac12 \sin^2a) \ \text{and radius} \ y_c, \ a \in (0,\pi) \tag{1}$$

or equivalently as circles $(C_T)$

$$\text{with center} \ \left(x_c=\frac{T^2-1}{(1+T)^2}, y_c=\frac{2T}{(1+T)^2}\right) \ \text{and radius} \ y_c, \ T \in (0,+\infty) \tag{2}$$

(Connection between the two parameterizations: $T=t^2$ where $t:=\tan \tfrac{a}{2}$).

The locus of these centers is parabola $(P)$ with cartesian equation:

$$y=\tfrac12(1-x^2),\tag{3}$$

represented on Fig. 2 as the light blue curve.

Each circle $(C_T) \in F$ is associated with a unique circle $(C_{T'}) \in F$ which is tangent to $(C)$ as depicted on Fig. 2.

Fig. 2: Two circles of family $(F)$ are mutually tangent if their parameters $T$ and $T'$ comply with relationship (4).

Let $I$ be the point of tangency of $(C_T)$ and $(C_{T'})$.

Question: how can we describe the locus of $I$?

(sketched as the magenta curve on Fig. 2.)

My work: I have obtained through tedious calculations an unexpected linear relationship between parameters $T$ and $T'$ which is:

$$T=(3-2 \sqrt{2})T'\tag{4}$$

Using (4), I have tried to obtain the locus of $I$ as the envelope of chords defined by the line segment joining the centers of circles $C_T,C_{T'}$ using homographies but to no avail.

Edit: Nice solution using inversion (a little different from the solution given in a comment by Blue).

'Refer to Fig. 3). Consider inversion with pole $S(0,-1)$ and power $2$ meaning that $M$ is exchanged with $M'$ iff $S,M,M'$ are aligned and $SM.SM'=2$. Its circle of inversion is  the black circle (center $S$, radius $\sqrt{2}$). It is easy to see that any circle of family (F) is globally invariant. This is in particular the case for the two blue circles. As the image by inversion of two tangent curves in a certain point $I$ is two tangent curves with contact point $I'$ where $M'$ is the image of $M$, point $I$ is invariant, establishing that $I$ belongs to the circle of inversion.$\square$

As a consequence, tangency points $T_1$ and $U_1$, $T_2$ and $U_2$, are exchanged, meaning that $S,T_1,U_1$ and $S,T_2,U_2$ are aligned.

Fig. 3.

Remark: Two properties of circles of family (F) deserving to be known:

http://www.gogeometry.com/problem/p311_circle_inscribed_semicircle_chord.htm

http://www.gogeometry.com/problem/p310_semicircle_circle_tangent_angle.htm

Body:
$$(x_1,y_1)=\left(\frac{T-1}{T+1},\frac{2T}{(T+1)^2}\right)\\
(x_2,y_2)=\left(\frac{U-1}{U+1},\frac{2U}{(U+1)^2}\right)\\
(x_1-x_2)^2+(y_1-y_2)^2=(y_1+y_2)^2\\
\frac{4(T-U)^2}{(T+1)^2(U+1)^2}=4y_1y_2=\frac{16TU}{(T+1)^2(U+1)^2}\\
T^2-6TU+U^2=0$$
The tangent point is $(x,y)=$

$$(y_2(x_1,y_1)+y_1(x_2,y_2))/(y_1+y_2)\\
=\left(\frac{U(T^2-1)+T(U^2-1)}{U(T+1)^2+T(U+1)^2},\frac{4TU}{U(T+1)^2+T(U+1)^2}\right)\\
\frac{4x}y=T-\frac1T+U-\frac1U\\
\frac4y=T+2+\frac1T+U+2+\frac1U\\
\left(\frac4y-4\right)^2=\left(\frac{4x}y\right)^2+4(T+U)(\frac1T+\frac1U)\\
(1-y)^2=x^2+2y^2\\x^2+(y+1)^2=2$$


问:
Find the coordinates of the centroid
I'm having a problem setting this problem up and solving it for the $X$ and $Y$;

Set up the integrals (but do not evaluate them) to find the coordinates of the centroid of region A ( The curve is $y=\sin(e^x-1)$ shown on the interval $0 \leq x \leq 1$)

enter image description here

答:
Assuming I read your statement correctly, here's the region $A$:

Its area is $$A = \int\limits_{x=0}^1 f(x)\ dx$$

and center of mass has coordinates:

$$\bar{x} = \frac{\int\limits_{x=0}^1 x f(x)\ dx}{A}$$

and

$$\bar{y} = \frac{\int\limits_{y=0}^{f(1)} y (1 - f^{-1}(y))\ dy}{A}$$

and $f^{-1}(y) = 1+ \ln (\pi - \sin^{-1} y)$

Title: Find the coordinates of the centroid


答:
The integral is convergent (conditionally) as shown here.

The integral is not absolutely convergent. We have

$$\int_0^R|\cos x^3|\, dx \geqslant \int_0^R\cos^2 x^3 \, dx = \int_0^R\left(\frac{1}{2}\cos 2x^3 + \frac{1}{2}\right)\, dx \\ =  \frac{1}{2}\int_0^R\cos 2x^3 \, dx + \frac{R}{2},$$

and the RHS diverges to $+\infty$ as $R \to +\infty$ since $\int_0^\infty\cos 2x^3 \, dx = 2^{-1/3}\int_0^\infty \cos u^3 \, du$ is convergent.

Title: Does $\int_0^{+\infty}\cos(x^3)\,\mathrm{d}x$ converge absolutely or conditionally?

Body:
$$\int\limits_0^{+\infty}\cos(x^3)\,\mathrm{d}x$$

I think that the integral absolutely does not converge, since the function does not tend to zero, but conditionally it converges, since if you look at the graph of the function, the areas of the cosine humps decrease, and the signs alternate, and then the Leibniz integral converges, but I think this solution is not strict and I would like to get a normal one, but there are no ideas.


问:
How to get the area of the triangle in a trigonometric circumpherence when there's a negative angle?
The problem is as follows:

The figure from below represents a trigonometric circle or unit circle. Assume $PQ=|\sin\theta|$. Find the area of the region represented by $ABC$.

The choices given are as follows:

$\begin{array}{ll}
1.&\frac{\sin\theta}{2}\\
2.&1\\
3.&\frac{1}{2}\\
4.&-\frac{\sin\theta}{2}\\
\end{array}$

Well, I'm totally lost in this problem. As I don't know how to understand the negative value which it has been given in the figure.

The image has been reconstructed from the original source which isn't very clear, so I did to my best of drawing abilities to put all the information but it is not very clear if the angle starts on point which lies on $\textrm{x-axis}$ or in point $B$ so, please consider this in your answer, so it may match the answers given.

In short my biggest complain is that weird arrow how should I understand it?. Can you please address this doubt of mine in the heading of your answer?

Other than that what I could notice is that catch here is how to get the value of the base of the triangle and the height for $\triangle ABC$. The thing is that I'm confused how to use the fact that it confuses me that $PQ$ is given as an absolute value.

Since this was found in an old trigonometry practice sheet the sort of strategy which I'm looking to get is by relying in the concepts of trigonometric circle.

As an extension to this I also notice that the radius here is one unit.

Thus can someone guide me here on exactly what to do, and what should I do to get one of the answers given in the choices?.

Since I'm not very sure with this topic I appreciate that the answer given would include the necessary steps to get to that answer.

答:
Given  the configuration, |OA|=1, |OB|=1, $\angle$ GOB=$\theta-90^o$

|OC|=$\frac{|OQ|}{\cos (\theta-90^o)}$=$\frac{(|OP|+|OQ|)}{ \cos (\theta-90^o)}$=$\frac{(1+\sin \theta)}{\sin \theta}$

Apply the area of triangle formula based upon two adjacent  sides and the angle in between,

$S_\Delta ABC$ =$S_\Delta AOC$ -$S_\Delta AOB$

= $\frac{1}{2}|OA||OC|\sin(\theta)$-$\frac{1}{2}|OA||OB|\sin(\theta)$

= $\frac{1}{2}\sin\theta$ $\frac{(1+\sin \theta)}{\sin \theta}$-$\frac{1}{2}\sin\theta$

= $\frac{1}{2}$


问:
Is there a way to write $1/(x-y) = f(x) g(y)$?
For $x>0$ and $y>0$, do there exist some single-variable functions $f$ and $g$ for which
$$
\frac{1}{x-y} = f(x) g(y) \ ?
$$

答:
There isn't.  Let $d=x-y$.  Then $$\frac1d = f(x)g(x-d).$$

Fix $x$ and let $y$ approach $x$ so that $d$ approaches zero.  The left side goes to infinity.  On the right side, the factor $f(x)$ is constant.  For the product to go to infinity, we must have $$\lim_{d\to 0} g(x-d) = \infty.$$

But this last equation holds for any fixed value of $x$.  This is impossible.

The argument for $\frac1{x+y}$ will be essentially the same.


问:
Can a Field lie "strictly" inside itself?
Is there a field $F$ and a non-zero ring homomorphism $f:F \rightarrow F$ such that $f$ is not surjective?

In other words, is it possible for a field to strictly lie inside itself? I couldn't come up with an example, but I don't see this couldn't be the case. Clearly this is impossible for finite fields and for $\mathbb{Q}$.

答:
$\mathbb{C} \rightarrow \overline{\mathbb{C}(t)}$ are both algebraically closed fields, and have transcendence degree $2^{\aleph_0}$ over $\mathbb{Q}$.

Therefore, $\mathbb{C} \simeq \overline{\mathbb{C}(t)}$ (see here), but the inclusion is not surjective.


问:
How to verify whether or not there exists a vector $x$ such that $Ax > 0$?
Given a matrix $A$ in $\mathbb R^{m\times n}$, if I want to know whether or not there is an $x \in \mathbb R^{n}$ such that
$$
Ax > 0,
$$
(meaning all elements in $Ax$ are positive). Is there some easy way to verify whether or not such $x$ exists?</p>

答:
This is an elaboration on LinAlg's comment on my answer.

Using the simplex method, solve the optimization problem of maximizing $t \in \mathbb R$ subject to $$A \mathbf x - t \mathbf 1 \ge \mathbf 0$$ (where $\mathbf 0, \mathbf 1$ are the constant vectors). I say "maximize" but actually you can stop as soon as $t>0$.

If there is a solution with $t>0$, then $A \mathbf x > \mathbf 0$ at that solution, because $(A \mathbf x)_i \ge t > 0$ for all $i$.

Conversely, if there is any $\mathbf x \in \mathbb R^n$ with $A \mathbf x > \mathbf 0$, then by setting $t$ to be the smallest component of the vector $A\mathbf x$, we get a solution to $A \mathbf x - t \mathbf 1 \ge \mathbf 0$ with $t > 0$.


问:
Set of Infinite Sets Contradiction?
Consider the set $S$ that contains all sets $X$ such that $|X|\geq |\Bbb N|$.

The whole reason ZFC was made was because of Russel’s paradox right? And so we know that we can't have a set that contains itself. So because of this: $S \notin S$

And so we have that $|S| < |\Bbb N|$.(i.e: that there are only finitely many sets of infinite cardinality).

But clearly this is not the case! One way to see this is to note that all of:

$\{x \in \Bbb{Z}$ $s.t. x \equiv 1$ (mod 2)}, $\{x \in \Bbb{Z}$ $s.t. x \equiv 1$ (mod 3)}, $\{x \in \Bbb{Z}$ $s.t. x \equiv 1$ (mod 4)}, ... $\in S$

So clearly $|S| \geq |\Bbb N|$

This brings about a contradiction. And one would determine that it was irrational to consider the set $S$.

My question, essentially, is: What's going on here? Surely there is some problem with the logic above, as it should be rational to consider the collection of sets with infinite cardinality, right?

I'm afraid my ZFC knowledge isn't quite up to par (at all) so I'm not able to work out exactly what is wrong with my argument.

In my search for answers I came across some other posts on here, but none of them seemed to be directly comparable to my question.

Thank you in advance!!

答:
The thing is that you define $S$ to be the "set" of $X$ that satisfy some property. If you do not tell where you pick $X$ from, the object $S$ you're creating might not even be a set. Some of the sets that are constructed this way and is axiomatized to be a set are the power set $P(A)=\{X\mid X\subseteq A\}$, and the union $\bigcup C=\{x\mid \exists A\in C : x\in A\}$. If $S$ is not a set, some properties we are familiar with might not be true.

See the term "class" for further information.

Title: Set of Infinite Sets Contradiction?


问:
Understanding sequent calculus

Are the following rules correct?

$(i)$ $\dfrac{\Phi \Rightarrow \Delta}{\Phi, \psi \Rightarrow \Delta}$

$(ii)$ $\dfrac{\Phi, \psi \Rightarrow \Delta}{\Phi \Rightarrow \Delta}$

Intuitively, I would've said that rule $i$ is incorrect and rule $ii$ is correct. We know that a rule is correct when the validity of the upper sequence(s) implies the validity of the lower sequence(s). So for $i$, if $\Phi \Rightarrow \Delta$ is valid, then $\Phi, \psi \Rightarrow \Delta$ must also be valid, but we can pick $\psi$ to be the empty set, making $\Phi, \psi$ unsatisfiable. The same argument can be used in the different direction for $ii$.

However, my textbook gives the exact opposite answer: $i$ is correct and $ii$ is incorrect. Can someone explain to me why this is the case?

答:
The book is correct. (i) is a valid inference rule and (ii) is not valid. Also, the capital Greek letters $\Phi$ and $\Delta$ refer to sets of well-formed formulas, but $\psi$ refers to a single well-formed formula.

Let's consider the two sequents $\Phi \vdash \Delta$ and $\Phi,\psi \vdash \Delta$ and consider when they are true.

Also, let's define $\Phi''$ as $\lnot\Phi_1, \cdots, \lnot\Phi_n$. The double prime is not standard notation. I will use $\bigvee S$ to denote a disjunction over a finite set $S$.

$$ \Phi \vdash \Delta \; \text{holds} \;\; \text{if and only if} \;\;\bigvee \Phi'' \cup \Delta \;\text{holds} $$

$$ \Phi, \psi \vdash \Delta \; \text{holds} \;\; \text{if and only if} \;\;\bigvee \Phi'' \cup \{\lnot\psi\} \cup \Delta \;\text{holds} $$

$\Phi'' \cup \Delta$ is a subset of $\Phi'' \cup \{\lnot \psi\} \cup \Delta$, therefore the truth of the former implies the truth of the latter and thus

$$ \frac{\Phi \vdash \Delta}{\Phi, \psi \vdash \Delta} $$

Similarly, the converse is invalid because of the case where $\bigvee \Phi'' \cup \Delta$ is false, but $\lnot\psi$ is true.

Title: Understanding sequent calculus


问:
Method for estimating number of fish in my Aquarium
Context: I have an aquarium (50 Liters) with lots of fishes from the same specie (Guppy) and i have noticed that recently their number as grown exponentially due to their reproductive cycle. I would like to estimate their number.

Problem: It's not possible for me to count them individually, and i cannot dry the aquarium due to the fact that the newborns are translucid and have sizes bellow 4 milimeters, it would be an impossible task to distinguish them from the grains of smalls rocks in my aquarium.

Pre-Condition: Due to the fact that newborns stay relatively close to the ground and as they grow they will occupy (swim) in places increasingly closer to the surface level, we can consider that the fishes are uniformly distributed in the aquarium. Fish Tank

Possible method: Take a cup of water (for example 150 milliliters) consistently from the same place of the aquarium, and count the number of fishes from inside the cup. By repeating this method multiple time i could in theory estimate the total number of fishes in my aquarium.

Question: Is this a good method? Are there any big flaws in my lines of thinking? Could you think of a better, more pratical method?

Thanks in advance for your help :)

答: You can try the capture-recapture method which uses the expectation of the hyper-geometric distribution.</p>


问:
Prove that for every positive whole number $n$, there exists a number $m_{n}$
Use mathematical induction for the following.
Prove that for every positive whole number $n$, there exists a number $m_{n}$ such that  $m_{n}$ has $n$ digits in its decimal representation, all of its digits belong to $\left \{ 1, 2 \right \}$ and  $m_{n}$ is divisible by $2^{n}$.

答:
Let $R_n= m_n \cdot 2^{-n}$

The assertion is equivalent to saying that for every $n$ there exists an $m_n$ such that $R_n$ is an integer.

This is clearly true for $n=2$ for which $m_2=12\;\text{ and }\;R_2=3$

for $n=k$ assume that $R_k$ is an integer

Then we can form $m_{k+1}$ in two ways ...

$$m^{(1)}_{k+1} = 10^k + m_k
\\[1em] = 2^k 5^k + 2^kR_k = 2^k(5^k + R_k)$$
Since $5^k$ is always odd, $m^{(1)}_{k+1}$  will be  divisible by $2^{k+1}$ if and only if $R_k$ is odd
$$m^{(2)}_{k+1} = 2\cdot10^k + m_k
\\[1em] = 2^{k+1} 5^k + 2^kR_k = 2^k(2\cdot5^k + R_k)$$
Since $2 \cdot 5^k$ is always even, $m^{(2)}_{k+1}$  will be
divisible by $2^{k+1}$ if and only if $R_k$ is even.

Since $R_k$ is an integer it must be either even or odd so either $m^{(1)}_{k+1}$  or $m^{(2)}_{k+1}$  will be divisible by $2^{k+1}$


答: Start with an arbitrary field $k$, and let $F=k(x)$ be the field of rational functions in one variable $x$ over $k$. There is a homomorphism $f:k(x)\to k(x)$ that acts on any rational function by substituting $x^2$ for $x$ (leaving the coefficients unchanged). It's clearly a non-zero homomorphism, and $x$ is not in its range.

Title: Can a Field lie "strictly" inside itself?

Body:
Is there a field $F$ and a non-zero ring homomorphism $f:F \rightarrow F$ such that $f$ is not surjective?

In other words, is it possible for a field to strictly lie inside itself? I couldn't come up with an example, but I don't see this couldn't be the case. Clearly this is impossible for finite fields and for $\mathbb{Q}$.


问:
Proving Jensen's inequality for the general case starting from the finite case
For a finite convex combination, Jensen’s inequality is given by: $$f\left(\sum_i^n a_i x_i\right) \leq \sum_i^n a_i f(x_i)$$ for a convex $f$. Proving this is not so bad starting from the definition of a convex function (iteratively break the convex combination into a convex combination between just two points and apply the definition of a convex function).

For the case of general probability measures: $$f(\mathbb E X) \leq \mathbb E f(X)$$ where $X$ is a real valued random variable, the approach I know is pretty different (show that a tight affine lower bound of $f$ exists then apply linearity of expectation) but I feel that the method for the finite case could be made to work here.

Here is my attempt, drawing inspiration from the construction of the Lebesgue integral:

Given a real valued random variable $X$, let $X^+,X^-$ be such that $X = X^+ - X^-$ so that it suffices to show the claim for non-negative random variables (apply the definition of convexity to the convex combination $\mathbb P [X>0] X^+ + \left(1-\mathbb P[X>0]\right) X^-$). Let $X_j$ be a sequence of simple random variables (defined as a finite sum of indicator functions on the sample space) dominated by $X$ such that $$\mathbb E X_j \rightarrow \sup \left\{\mathbb E Y: Y \leq X,\, Y\text{ is simple}\right\}$$

Now we have:
$$f\left(\mathbb E X\right) = f\left(\mathbb E X_j + o(1)\right) = f\left(\mathbb E X_n\right) + o(1) \leq \mathbb E f\left( X_j\right) + o(1)$$

Where the first equality is by definition of the sequence $X_j$, the second equality is from the fact that $f$ is convex and so continuous on the interior of its domain, the inequality is from the finite case of Jensen's inequality.

Next, I would like to claim that $X_j \rightarrow ^p X$ and apply the continuous mapping theorem to complete the argument, but I'm not sure how to justify that claim.

I could use feedback on my reasoning as well as some help justifying the last claim.

EDIT: there is a simpler argument. Instead of defining $X_j$ as I have, I could instead define it as a sequence of simple random variables increasing towards $X$ pointwise (for every event in the sample space). I had not realize that we could define such a sequence in general, I am not sure how to show that this is true. In that case we have:

$$\mathbb E f\left( X_j\right) = \mathbb E f\left( X\right) + o(1)$$

by the continuous mapping theorem.

Any feedback on this followup approach would be very appreciated.

答:
I think your argument goes through fine if you choose the $X_j$ in a nice way. Don't just take any $X_j$, take $X_j$ nonnegative, simple, and increasing to $X$ with $X_j \to X$ a.s. (and hence in probability as well, but the need for the continuous mapping theorem goes away since monotone convergence applies.) The existence of such an approximating sequence is usually part of the process of defining the Lebesgue integral. Split $\mathbb{R}$ into multiples of $2^{-k}$ from $0$ to $k$ and take $X_k$ to be the largest multiple of $2^{-k}$ that is less than $X$ and $k$. Then $0 \leq X_k \leq \min(X, k)$ and $X_k$ increases to $X$ pointwise.

I would like to note, however, that proving the general case of Jensen directly is somewhat simpler than your extension of the finite case. For any line $L \leq f$, one has
$L(E[X]) = E[L(X)] \leq E[f(X)]$, now take the sup over lines (with rational slope and intercept, say, so the sup is countable) to get $f(E[X]) \leq E[f(X)]$.


问:
Integration by parts of $\int_0^\infty \! n^2 \ln(1-e^{-an}) \, \mathrm{d} n$
Im trying to do the following integration by hand,

$\int_0^\infty \! n^2 \ln(1-e^{-an}) \, \mathrm{d} n$

I have tried to use integration by parts and substitution, but each time it gets to complicated and messy (polylog terms etc...), is there any straight forward way to solve this?

答:
This is a generalization. By expanding the logarithm in its Taylor series (the same method in the other answers) you'll get that
$$ \int_0^\infty x^{n-1}\log(1-e^{-\alpha x})dx=-\frac{1}{\alpha^n}\zeta(n+1)\Gamma(n)$$
Your case is $n=3$</p>


问:
Property of Kulkarni-Nomizu product
I am reading about Weyl tensor in Introduction to Riemannian manifolds by John M. Lee. In Lema 7.22 (e), there is the next assertion.

Let $V$ be an $n$-dimensional vector space endowed with a scalar product $g$, let $h$ and $k$ be symmetric $2$-tensors on V, let T be and algebraic curvatura tensor on $V$, and let tr_g denote the trace on the first and las indices. Then $g(T,h \mathbin{\bigcirc\mspace{-15mu}\wedge\mspace{3mu}} g)=4g(tr_g T,h)$.

$\mathbin{\bigcirc\mspace{-15mu}\wedge\mspace{3mu}}$ denotes de Kulkarni-Nomizu product.
I am a bit lost in my try of the proof, what is the meaning for the metric applied on $(0,4)$ covariant tensors?

答:
When I did that problem, I furthermore chose a basis $ \left(b_i \right)$ for $V$ with respect to which the matrix for $g$ consists of entries of the form $\pm\delta_{ij}$. It made the computation easier since the product $\langle T,h\mathbin{\bigcirc\mspace{-15mu}\wedge\mspace{3mu}} g\rangle _g$ can then be written in coordinates as
$$\langle T,h\mathbin{\bigcirc\mspace{-15mu}\wedge\mspace{3mu}} g\rangle _g=\sum_{i,j,l,m=1}^{n}{T_{ijlm} \left(h_{im}g_{jl}+h_{jl}g_{im}-h_{il}g_{jm}-h_{jm}g_{il} \right)}g^{ii}g^{jj}g^{ll}g^{mm}.$$
Now distribute and show that each of the four resultant sums equals $\langle {\rm tr}_g{T},h\rangle _g$.

Title: Property of Kulkarni-Nomizu product


问:
How do I intuitively make sense of "logically implies" in first order logic
I am reading Friendly Introduction to Mathematical Logic by Christopher Leary and Lars Kristiansen. In Definition 1.9.1., the authors define logically implies for two sets of $\mathcal{L}$-sentences $\Delta$ and  $\Gamma$ as follows:

We will say $\Delta$ logically implies $\Gamma$ and write $\Delta \models\Gamma$ if for each $\mathcal{L}$-structure $\mathfrak{U}$, if $\mathfrak{U}\models\Delta$ then $\mathfrak{U}\models \Gamma$.

The definition seems okay but it would really help me to understand as to why we are making such a definition and what the motivation really is behind this definition. I would like to know why we are demanding "for each model of $\Delta$" and not discussing truth in a particular model of $\Delta$.

答:
This is just the basic concept of logical implication: Something (call it $X$) logically implies something else (call it $Y$) if and only if $Y$ is true whenever $X$ is true.

Consider this. In our world, grass is green, and snow is white. Now, would you sat that grass being green logically implies that snow it white just because there is a world in which both statements are true?  Of course not. And of course grass being green does not logically imply snow being white: you can imagine a logically possible world where grass is green, but snow is not white.

In other words, $X$ does not imply $Y$ if there is some world where $X$ is true but $Y$ is false.  But that means that $X$ does imply $Y$ when there is no such world ... which is to say that in every world where $X$ is true, $Y$ will also be true.


问:
How to get the area of the triangle inscribed in the unit circle when a slope is tangent?
The problem is as follows:

The circle shown in the figure from below is a unit circle. $PT$ is a line tangent to that circle on point $T$. Angle alpha measured in radians is an acute angle. Using this information, find the area of the orange shaded region which is comprised $\triangle TQR$. Assume that the length of the radius of the unit circle is of 1 decameter.

The choices given in my book are as follows:

$\begin{array}{ll}
1.&200\cos\alpha\sin\left(\frac{\pi}{6}-\frac{\alpha}{2}\right)\\
2.&100\sqrt{2}\cos\frac{\alpha}{2}\sin\left(\frac{\pi}{4}-\frac{\alpha}{2}\right)\\
3.&50\sin\alpha\sin\left(\frac{\pi}{4}-\frac{\alpha}{2}\right)\\
4.&100\sqrt{2}\sin\frac{\alpha}{2}\sin(45^\circ+\alpha)\\
\end{array}$

I'm not sure exactly how to solve this question, it seems kind of difficult to my abilities.

I tried all sorts of manipulations on the image but I can't find a proper relation to get the base of the triangle. I also considered the oportunity to use the tangent secant theorem but this did not helped me much in the solution.

This half angle thing which is on the third quadrant it is making me feel confused. Needless to say that it is more than thirty minutes and I don't think I am getting anywhere with this problem.

Can someone help me here?.

I am totally lost here. Since I'm not good at spotting things it would help a lot that a drawing could accompany the solution as I guess that it is needed some labels to see where should additional or rearranging the angles would be needed to solve this problem.

答:
Add lines OT and OR. Notice that $\triangle OTA$ is a right triangle. Then you have $$|\sphericalangle TOA| = 90^\circ - \alpha$$
$$|\sphericalangle TQA| = \frac12|\sphericalangle TOA| = 45^\circ - \frac{\alpha}{2}$$
Then you should be able to find all the angles in the picture. Eventually you should get
$$|\sphericalangle TOQ| = 90^\circ+\alpha$$
$$|\sphericalangle TOR| = 90^\circ$$
$$|\sphericalangle QOR| = 180^\circ-\alpha$$
We have then
\begin{align} S(\triangle TQR) &= S(\triangle TOQ)+ S(\triangle TOR) + S(\triangle QOR) = \\
&= \frac12 10^2(\sin|\sphericalangle TOQ| + \sin|\sphericalangle TOR| + \sin|\sphericalangle QOR|) = \\
&= 50\big(\sin(90^\circ+\alpha) + 1 + \sin(180^\circ-\alpha)\big) \end{align}

After some trigonometry you can show that it's equal to
$$ 100\sqrt{2}\cos\frac{\alpha}{2}\sin\big(45^\circ+\frac{\alpha}{2}\big) $$

Strangely enough, it's neither of the answers you can choose from, but you can also find
$$|\sphericalangle TQR| = 45^\circ$$
$$|\sphericalangle QRT| = 45^\circ+\frac{\alpha}{2}$$
$$|\sphericalangle RTQ| = 90^\circ-\frac{\alpha}{2}$$
There is a well-known formula that links the radius of the circumscribed circle to the lengths of the sides and angles of the triangle:
$$ 2R = \frac{|TR|}{\sin|\sphericalangle TQR|} = \frac{|QT|}{\sin|\sphericalangle QRT|} = \frac{|RQ|}{\sin|\sphericalangle RTQ|}  $$
We have then
$$ |TR| = 2R \sin |\sphericalangle TQR|$$
$$ |QT| = 2R \sin |\sphericalangle QRT|$$
\begin{align} S(\triangle TQR) &= \frac12 R^2\, |TR| |QT| \sin|\sphericalangle RTQ| = \\
&= 2R^2 \sin|\sphericalangle TQR| \sin|\sphericalangle QRT| \sin|\sphericalangle RTQ| = \\
&= 200 \cdot \frac{\sqrt{2}}{2} \sin\big(45^\circ+\frac{\alpha}{2}\big) \sin\big(90^\circ-\frac{\alpha}{2}\big) = \\
&= 100\sqrt{2} \sin\big(45^\circ+\frac{\alpha}{2}\big) \cos\frac{\alpha}{2}\end{align}
and we get the same result.


答:
It might help if we think about when $\Delta \models \Gamma$ is false. Let's consider the contrapositive of the forall definition and call it (I).

$$ \textbf{(I)}\;\;\; \Delta \not\models \Gamma \;\;\text{iff}\;\; \text{there exists a model $M$ such that $M \models \Delta$ and $M \not\models \Gamma$ } $$

If we only looked at a single not-fixed model for $\models$, then the contrapositive would be as follows

$$  \textbf{(II)}\; \textbf{BAD} \;\;\; \Delta \not\models \Gamma \;\;\text{iff} \;\; \text{for all models $M$ where $M \models \Delta$, $M \not\models \Gamma$ holds} $$

The statement (II) is not compatible with the fact that implication only requires a single counterexample to be false.

If we look at a single fixed model $M$, then we're not talking about a relationship between $\Gamma$ and $\Delta$ that always holds, but rather a relationship between them in $M$ alone.

Title: How do I intuitively make sense of "logically implies" in first order logic

Body:
I am reading Friendly Introduction to Mathematical Logic by Christopher Leary and Lars Kristiansen. In Definition 1.9.1., the authors define logically implies for two sets of $\mathcal{L}$-sentences $\Delta$ and  $\Gamma$ as follows:

We will say $\Delta$ logically implies $\Gamma$ and write $\Delta \models\Gamma$ if for each $\mathcal{L}$-structure $\mathfrak{U}$, if $\mathfrak{U}\models\Delta$ then $\mathfrak{U}\models \Gamma$.

The definition seems okay but it would really help me to understand as to why we are making such a definition and what the motivation really is behind this definition. I would like to know why we are demanding "for each model of $\Delta$" and not discussing truth in a particular model of $\Delta$.


问:
Did I sketch this polar curve correctly?
The equation is:

$r^2=-4 \sin(2\theta)$

I first made a reference graph in cartesian coordinates using values $\displaystyle \frac{\pi}{4}$, $\displaystyle \frac{\pi}{2}$, $\displaystyle \frac{3 \pi}{4}$, $\displaystyle \pi$. Then from that I formed this:

Something seems off about that though. Should it be across the other axis instead?

答:
You can let WolframAlpha plot this by rewriting it in Cartesian coordinates:

$$r^2=-4\sin2\theta=-2\sin\theta\cos\theta\;,$$
$$r^4=-2\sin\theta r\cos\theta r=-2xy\;.$$

Concerning your own plot: It seems it's not the axes you got mixed up, but the sine and cosine.


问:
If I have three points, is there an easy way to tell if they are collinear?
Points $(a,b)$, $(m,n)$, and $(x,y)$ are selected at random. What is the quickest/easiest way to tell if they are collinear? At first I thought it was a matter of comparing slopes but that doesn't appear to be enough.</p>

答:
The area is zero, in formula $\det \pmatrix{1 &a &b\\
               1 &m &n\\
               1 &x &y}=0$</p>


问:
$[A^n,B]$ in terms of nested commutators
For a project I am working on, it would be helpful to have an expression for $[A^n,B]$ for $[A,B] = AB-BA \neq 0$, in terms of nested commutators $[A,[A,B]]$ and so forth, with any $A$'s all moved to the left. This is because the nested commutators are trivial for me to calculate in my case. So far I have been able to derive:

$$[A^2,B] = 2A[A,B] - [A,[A,B]]$$

and that,

$$[A^3,B] = 3A^2[A,B] - 3A[A,[A,B]] + [A,[A,[A,B]]].$$

It is in theory easy to compute $[A^n, B]$ for a specific $n$, you write down all nestings up to $n$ nestings, and for each term you include the necessary factors of $A$ to the left, so that you have $n$ factors of $A$ in every term. However this is tedious, and does not provide a general formula. Is there a general expression for $[A^n,B]$ in the form I have derived above?

Update

As a comment pointed out, the first two examples hinted at binomial coefficients, and I am conjecturing that,

$$[A^n,B] = \sum_{k=1}^n (-1)^{k-1} \binom{n}{k} A^{n-k} \underbrace{[A,[A,\dots]]}_{k\; \text{times}}.$$

Here $k=1$ nestings means the commutator $[A,B]$. I have checked this for $n=4$ by hand, and it continues to hold as well as for the other examples of course. I do not know how to prove it though.

答:
Let $\text{ad}_A(X)=[A,X]$.
We can easily find a recursive formula for $[A^n,B]$:
$$[A^n,B]=A^nB-BA^n$$
$$=A^{n-1}BA-BA^n+A^{n-1}[A,B]$$
$$=[A^{n-1},B]A+A^{n-1}[A,B]$$
$$=A[A^{n-1},B]-[A,[A^{n-1},B]]+A^{n-1}[A,B]$$
$$=A[A^{n-1},B]-\text{ad}_A[A^{n-1},B]+A^{n-1}[A,B]$$
It is straightforward to check that your conjectured closed formula
$$[A^n,B] = \sum_{k=1}^n (-1)^{k-1} \binom{n}{k} A^{n-k} \text{ad}_A^k(B)$$
indeed satisfies this recursion, which proves it.


问:
Minimum of $|z_1| + |z_2|+ \dots + |z_n|$ when $z_1 + \dots + z_n = 1$ and all the arguments of $z_m$ are prescribed.
Let $-\frac{\pi}{2} \le \phi_1 < \dots < \phi_k < 0 < \phi_{k+1} < \dots < \phi_N < \frac{\pi}{2}$ be $N$ prescribed phases. Let $a_1, \dots, a_N$ be $N$ real numbers such that
$$
a_1 e^{i \phi_1} + \dots + a_N e^{i \phi_N} = 1.
$$
What is the minimum of $|a_1|+\dots+|a_N|$?

The minimum is achieved when $a_m=0$ for $m \ne k, k+1$. The proof can be easily understood in a geometrically point of view, as kindly suggested in the following answer. Many thanks to all the answers and comments.

答:
Now we have one proof very rigorous and another one more based on intuition. I'd like to post my own proof that somewhat lies between them.

Denote $R=\{ z_1, \dots, z_N \}$, $R_1 = \{ z \in R: \text{Im}  z > 0 \}$, $R_2 = \{ z \in R: \text{Im}  z < 0 \}$. Both $R_1$ and $R_2$ must be non-empty. Denote
$$
z^+ = \sum_{z \in R_1} z = |z^+| e^{i \phi_+}, \text{ and } z^- = \sum_{z \in R_2} z = |z^-| e^{i \phi_-}.
$$
There must be $\pi> \phi_+ \ge \phi_{k+1}$ and $-\pi < \phi_- \le \phi_k$.
It is easy to see that
$$
|z_1| + \dots + |z_N| \ge |z^+| + |z^-|.
$$

Because $z^+ + z^- = 1$, we have
$$
|z^+| + |z^-| = \frac{\sin \phi_+ - \sin \phi_-}{\sin (\phi_+ - \phi_-)}
$$
which is monotonically increasing with $\phi_+$ and decreasing with $\phi_-$ (just examine the sign of derivatives, or directly seen by plotting triangles with the bottom being 1). Thus, the minimum of $|z^+| + |z^-|$ is attained when $\phi_+ = \phi_{k+1}$ and $ \phi_- = \phi_k$, or $z_m=0$ for $m \ne k, k+1$.


答:
You write in particular:

I would like to know why we are demanding "for each model of $\Delta$" and not discussing truth in a particular model of $\Delta$.

Often logic is introduced in the context of understanding some fixed structure such as the semiring of natural numbers $\mathbb{N}$ or (stretching the term "structure" a bit) the universe of set theory $V$. However, that's not something we have to do. The apparatus of first-order logic provides us with a broad collection of sentences which make sense (whether true or false) in arbitrary structures, and we care right at the outset about the general behavior of this apparatus. The definition of logical entailment reflects this level of generality. We may later consider modifications of the entailment relation (although spoiler alert, none of them seem to have the same ultimate staying power), but to start with we should at least understand the "no-unnecessary-restrictions" situation.

It may help at this point to look at a particular mathematical claim in this context. Take for example the statement "If every element has order $2$, then every pair of elements commutes" in the context of group theory. The class of groups is axiomatizable: there is a set $\Delta_{grp}$ of first-order sentences such that the models of $\Delta_{grp}$ are exactly the groups. The fact that the statement above is true in every group amounts to the following logical entailment: $$\Delta_{grp}\models[\forall x(x*x=e)\rightarrow\forall x,y(x*y=y*x)].$$ Note that it's important that we range over all $\mathfrak{U}$s here; we're really trying to make a statement about arbitrary models of $\Delta_{grp}$ here.

Ultimately, I consider it a mistake to think about model theory as being oriented towards individual structures at all; model theory, in my opinion, is better thought of as the study of certain classes of, operations on, dividing lines amongst, and other things around structures in general.


答: Consider the two operators $L_A(B)=AB$ and $R_A(B)=BA$.  These two operators commute with each other, so by the binomial theorem, $$R_A^n=(L_A-(L_A-R_A))^n=\sum_{k=0}^n (-1)^k \binom{n}{k} L_A^{n-k}(L_A-R_A)^k,$$ and thus $$L_A^n-R_A^n=\sum_{k=1}^n(-1)^{k-1}\binom{n}{k} L_A^{n-k}(L_A-R_A)^k.$$ This is exactly your conjectured formula since $L_A-R_A$ is the commutator operator $[A,-]$ and $L_A^n-R_A^n$ is $[A^n,-]$.

Title: $[A^n,B]$ in terms of nested commutators

Body:
For a project I am working on, it would be helpful to have an expression for $[A^n,B]$ for $[A,B] = AB-BA \neq 0$, in terms of nested commutators $[A,[A,B]]$ and so forth, with any $A$'s all moved to the left. This is because the nested commutators are trivial for me to calculate in my case. So far I have been able to derive:

$$[A^2,B] = 2A[A,B] - [A,[A,B]]$$

and that,

$$[A^3,B] = 3A^2[A,B] - 3A[A,[A,B]] + [A,[A,[A,B]]].$$

It is in theory easy to compute $[A^n, B]$ for a specific $n$, you write down all nestings up to $n$ nestings, and for each term you include the necessary factors of $A$ to the left, so that you have $n$ factors of $A$ in every term. However this is tedious, and does not provide a general formula. Is there a general expression for $[A^n,B]$ in the form I have derived above?

Update

As a comment pointed out, the first two examples hinted at binomial coefficients, and I am conjecturing that,

$$[A^n,B] = \sum_{k=1}^n (-1)^{k-1} \binom{n}{k} A^{n-k} \underbrace{[A,[A,\dots]]}_{k\; \text{times}}.$$

Here $k=1$ nestings means the commutator $[A,B]$. I have checked this for $n=4$ by hand, and it continues to hold as well as for the other examples of course. I do not know how to prove it though.


问:
Prove that the set of Turing-decidable languages are closed under union.
I.e., prove if both $L_1$ and $L_2$ are Turing-decidable, then $L_1∪L_2$ is Turing-decidable. I know that the statement is true, but struggling with how to prove it formally.

答:
Create a Turing machine which simulates the machine for $L_1$ except that if the simulation enters the REJECT state, it then simulates the machine for $L_2$.

A more interesting approach is to simulate both machines interlaced. (That is, do one step on machine 1 and then one step on machine 2, repeatedly.) If one machine enters the REJECT state, continue by just simulating yhe other machine. If one machine enters the ACCEPT state or the second machine enters REJECT, that's where you end up. The interlaced machine show that Turing recognisability is closed under union.

The interesting part is showing that these machine transformations are possible, but that's not too hard.