答:
To add on leoli1's answer.

Take $L=\Bbb{Q}(\alpha)$ a complex root of the irreducible polynomial $f=x^4+3x+1$.

If $L\cap \Bbb{R}$ is larger than $\Bbb{Q}$ then $[L:L\cap \Bbb{R}]=2$ thus $L$ contains some $L\cap \Bbb{R}$-conjugate of $\alpha$, its complex conjugate.

Factorize $f=(x-\alpha)(x-\overline{\alpha})(x-b)(x-c)$.

$(x-\alpha)(x-\overline{\alpha})\in L[x]$ means that $(x-b)(x-c)\in L[x]$.

Therefore, the splitting field of $f$ is $L(b)$ having degree at most $8$ over $\Bbb{Q}$.

This is a contradiction because the splitting field of $f$ has degree $24$.

http://magma.maths.usyd.edu.au/calc/

```<code>   P<x>:=PolynomialRing(Integers());
   f := x^4+3*x+1;
   Degree(SplittingField(f));
</code>```

Body:
To add on leoli1's answer.

Take $L=\Bbb{Q}(\alpha)$ a complex root of the irreducible polynomial $f=x^4+3x+1$.

If $L\cap \Bbb{R}$ is larger than $\Bbb{Q}$ then $[L:L\cap \Bbb{R}]=2$ thus $L$ contains some $L\cap \Bbb{R}$-conjugate of $\alpha$, its complex conjugate.

Factorize $f=(x-\alpha)(x-\overline{\alpha})(x-b)(x-c)$.

$(x-\alpha)(x-\overline{\alpha})\in L[x]$ means that $(x-b)(x-c)\in L[x]$.

Therefore, the splitting field of $f$ is $L(b)$ having degree at most $8$ over $\Bbb{Q}$.

This is a contradiction because the splitting field of $f$ has degree $24$.

http://magma.maths.usyd.edu.au/calc/

   P<x>:=PolynomialRing(Integers());
   f := x^4+3*x+1;
   Degree(SplittingField(f));


问:
When computing the Taylor series of $(\cos x)^2$ how does the slide jump to concluding it is $1-(\sin x)^2$?
In the following slide it shows how the taylor series of $(\cos x)^2$ is computed:

On the first line they simply take the taylor series of cosx and write it out twice, which makes sense.  However, the order in which they multiply out the 1st line does not make sense to me.

For example, how did they get the 3rd term on the 2nd line: $(x^2/2!)^2$ ?

When I multiply out the 1st line I'm doing a straightforward distribution and I'm getting terms like this:

$$1 - (x^2/2!) + (x^4/4!) - (x^6/6!) - (x^2/2!) + (x^6/2!4!)  + (x^{10}/2!6!) + (x^4/4!) - (x^6/4!2!) + (x^8/4!4!) - (x^{10}/4!6!)$$

When I combine the like terms above I do not end up with the 3rd line below and I was wondering if someone could point me in the right direction.

ContentLicense: CC BY-SA 3.0

Body:
The terms are arranged as follows:
$$
\begin{align}
\cos^2 x &= (a_1 + a_2 + a_3 + a_4\ldots)(a_1 + a_2 + a_3 + a_4\ldots)\\
&= a_1^2 + 2a_1a_2 + a_2^2 + 2a_1a_3 + 2a_2a_3 + 2a_1a_4 + \ldots
\end{align}
$$


问:
Show the given relation related to partial derivatives
Assume $z=f(x,y)$ with $x= \rho \cos(\theta)$ and $y= \rho \sin(\theta)$,then show

$$(\frac{\partial z}{\partial x})^2+(\frac{\partial z}{\partial y})^2=(\frac{\partial z}{\partial \rho})^2+\frac{1}{\rho^2}(\frac{\partial z}{\partial \theta})^2$$

$$\frac{\partial z}{\partial \theta}=\frac{\partial z}{\partial x}\frac{\partial x}{\partial \theta}+\frac{\partial z}{\partial y}\frac{\partial y}{\partial \theta}$$

and $$\frac{\partial z}{\partial \rho}=\frac{\partial z}{\partial x}\frac{\partial x}{\partial \rho}+\frac{\partial z}{\partial y}\frac{\partial y}{\partial \rho}$$

But I don't know how to compute $\frac{\partial z}{\partial x}$ or $\frac{\partial z}{\partial y}$,can someone help me?

答:
Just write down the expressions:

$$\left(z_\rho\right)^2=\left(z_xx_\rho+z_yy_\rho\right)^2=\left(z_x\cos t+z_y\sin t\right)^2=z_x^2\cos^2t+2z_xz_y\cos t\sin t+z_y^2\sin^2t$$

and

$$\frac1{\rho^2}(z_t)^2=\frac1{\rho^2}\left(z_x(-\rho\sin t)+z_y(\rho\cos t)\right)^2=$$

$$=\frac1{\rho^2}\left(z_x^2\rho^2\sin^2 t-2z_xz_y\rho^2\cos t\sin t+z_y^2\rho^2\cos^2t\right)$$

Well, now just add the above two expressions and see what you get...

ContentLicense: CC BY-SA 4.0


问:
prove or disprove in $\mathbb{Z}[i]$ decomposition of real integer to 2 elements with the same norm is unique up to multiplying by unit
let $a,b,c\in\mathbb{Z}[i]$such that a=bc and say $|b|=B$ and $|c|=C$ . prove or disprove, there is no other decomposition such: $a=de ,|d|=B,|e|=C$
up to multiplying b and c by unity.

Edit: this claim is not true for complex number, but is it for reals? prove it!

答:
We can easily find a counterexample by finding two non-associates with the same norm, such as $1+7i$ and $5+5i$. Then, all we need is $$(1+7i)(x) = (5+5i)(y)$$
This yields $$\frac{y}{x} =\frac{1+7i}{5+5i} = \frac{4+3i}{5}$$
so by choosing $x = 5$ and $y = 4+3i$, we see that $$(1+7i)(5) = (5+5i)(4+3i)$$ which serves as a counterexample, since $|1+7i| = |5+5i|$ and $|5| = |4+3i|$, yet neither are associates.


问:
Poincaré-Bendixson theorem and the Sinai Billiard
One of the conclusion of the Poincaré-Bendixson Theorem is that in planar dynamical systems chaotic motion could not arise.
But in the Sinai Billiard trajectories in fact are chaotic...how is this possible?</p>

答:
Alessandro's remark is a red herring. True, Sinai billiard flows are not differentiable, but that's not the main issue.

The phase space of the billiard flow is the space of couples $(x,v)$, where $x$ belongs to the billiard and $v$ is a unit speed vector. Hence you have $3$ degrees of freedom, so the billiard flow acts on a $3$-dimensional space. Obviously Poincaré-Bendixon does not apply.

(there are further issues about the differentiability of the system and the topology of the phase space, but they seem secondary to me).


问:
Markov inequality with $a$ lower than mean
In markov's inequality we have $P(X \geq a) \leq \displaystyle\frac{\mathbb{E}[X]}{a}$ but isn't it possible to have $a < \mathbb{E}[X]$ giving a probability lower or equal than a number greater than 1?

答:
You're right, and then the inequality is still true but trivial. Hence, the Markov inequality is only useful for $a > E[X]$.

Analogously, the Chebyshev's inequality is only useful when the distance from the mean is greater than one standard deviation.


问:
How to find the sum of $\sum_{n=0}^{\infty} \frac{z^{3 n}}{(3 n) !}$?
I'm really stuck with this problem and I hope some of you could give me a hint.

Consider the functions: $$f(z)=\sum_{n=0}^{\infty} \frac{z^{3 n}}{(3 n) !}, \quad f^{\prime}(z)=\sum_{n=0}^{\infty} \frac{z^{3 n+2}}{(3 n+2) !}, \quad f^{\prime \prime}(z)=\sum_{n=0}^{\infty} \frac{z^{3 n+1}}{(3 n+1) !}, \quad f^{\prime \prime \prime}(z)=f(z)$$
I need to find the sum of each of these series.

My first thought was to show that:
$$f(z)+f^{\prime}(z)+f^{\prime \prime}(z)=\sum_{n=0}^{\infty} \frac{z^{3 n}}{(3 n) !}+\sum_{n=0}^{\infty} \frac{z^{3 n+2}}{(3 n+2) !}+\sum_{n=0}^{\infty} \frac{z^{3 n+1}}{(3 n+1) !}=\sum_{n=0}^{\infty} \frac{z^{n}}{n !}=e^{z}$$
And then solve this differential equation.

However, officially I haven't had a course in differential equations. So is there another way I can solve this using basic Complex Analysis? I hope some of you can guide me in the right direction, what theorems could be useful?

答:
Let $\omega$ be the cube root of unity then denote $f_1$ by $e^{\omega x}$,$f_2$ by $e^{\omega^2 x}$ and $f_3$ by $e^x$

$$\begin{align*}f_1 &= \sum_{n\ge0} \left(\dfrac{x^{3n}}{(3n)!} + \omega \dfrac{x^{3n+1}}{(3n+1)!} + \omega^2 \dfrac{x^{3n}}{(3n+2)!}  \right)\\
f_2 &= \sum_{n\ge0} \left(\dfrac{x^{3n}}{(3n)!} + \omega^2 \dfrac{x^{3n+1}}{(3n+1)!} + \omega \dfrac{x^{3n+2}}{(3n+2)!}  \right)\\
f_3 &= \sum_{n\ge0} \left(\dfrac{x^{3n}}{(3n)!} + \dfrac{x^{3n+1}}{(3n+1)!} +  \dfrac{x^{3n}}{(3n+2)!}  \right) \end{align*} $$

Add all of them and get your result


问:
Defining a triangle with vertices $a$, $b$, $c$ by $t_1a+t_2b+t_3c$, where $0\leq t_1, t_2, t_3\leq 1$ and $1=t_1+t_2+t_3$. Why does this work?
For $a$, $b$, $c$ in $\mathbb{R}^2$, I know that the set
$$A=t_1a+t_2b+t_3c$$
with $0\leq t_1,t_2,t_3\leq 1$ and $1=t_1+t_2+t_3$
is a triangle with vertices $a$, $b$, $c$.

Could someone explain to me how we came up with those parameters geometrically? How could you explain it and why do they work?

Thank you in advance.

答:
You forgot to add that $a,b,c\in\Bbb R^2$. Clearly, $a,b,c\in A$; for intance, $a=1\times a+0\times b+0\times c$. Furthermor, the line segment that goies from $a$ to $b$ is a subset of $A$, since any $p$ from that line segment can be written as $(1-t)a+tb$, for some $t\in[0,1]$, and therefore $p=(1-t)a+tb+0\times c$. For the same reason, $A$ contains the line segments joining $a$ and $c$ and joining $b$ and $c$.

Let $p$ be an element of $A$. If $t_1,t_2,t_3\in[0,1]$ are such that $t_1+t_2+t_3=0$, and if $p=t_1a+t_2+t_3c$, then\begin{align}p&=t_1a+t_2b+\bigl(1-(t_1+t_2)\bigr)c\\&=(t_1+t_2)\left(\frac{t_1}{t_1+t_2}a+\frac{t_2}{t_1+t_2}b\right)+\bigl(1-(t_1+t_2)\bigr)c.\end{align}Since$$\frac{t_1}{t_1+t_2}+\frac{t_2}{t_1+t_2}=1\quad\text{and}\quad\frac{t_1}{t_1+t_2},\frac{t_2}{t_1+t_2}\geqslant0,$$the point$$\frac{t_1}{t_1+t_2}a+\frac{t_2}{t_1+t_2}b\tag1$$belongs to the line segment joining $a$ to $b$, and therefore $p$ belongs the line segment joining $(1)$ to $c$; so, $p$ belongs to the triangle.

This works in the opposite direction too: if $p$ belongs to the trangle, find the point $q$ of the line segment joining $a$ to $b$ such that $q$ $p$ and $c$ are collinear. Then $q=(1-t)a+tb$, for some $t\in[0,1]$, and $p=(1-t')q+t'c$ for some $t'\in[0,1]$. So\begin{align}p&=(1-t')\bigl((1-t)a+tb\bigr)+t'c\\&=\bigl((1-t')(1-t)\bigr)a+(1-t')tb+t'c\\&\in A.\end{align}


问:
Can Dijkstra's algorithm be exponential?

Teacher say:  Given a directed a-cyclic graph $G = (V , E)$, with
negative edges weights, now We can claim that, Dijkstra's relaxation
function do $\Omega(2^n)$ relaxation  edges, but every times i think,
i get stuck to prove it. In addition, I found reference of  This problem come from "Jeff
Erickson, Algorithms text book".

Anyone can give me a hint to construct an example to prove it?

答:
The standard implementation of Dijkstra does not work with negative edge weights. This is due to its greedy nature. Here is an example:

Consider the directed graph $G=(V,E)$ with $E=\{(a,b, 4), (a, c, 2), (b, c, -3)\}$ and assume that we want to find the shortest path from $a$ to $c$.
Dijkstra's algorithm will extract vertex $c$ from the queue before it will extract $b$ and report $2$ as the shortest path from $a$ to $c$. But the correct answer would have been $1$ ($a \to b$, $b \to c$).

However, if we restrict $G$ to the acyclic case, we can make Dijsktra’s algorithm work for negative weights but we must allow visiting vertices more than once (meaning, we do not mark vertex $c$ as closed).

However, this leads to calculating each possible path from $a$ to $c$ and in the worst case there are $2^{n-2}$ possible paths in a directed acyclic graph on $n$ vertices (checkout this question and its answers), thus $\Omega(2^n)$.


问:
Finding the formula for T from Newton's Law of Cooling
I think I got a wrong answer because I skipped a particular step which seemed optional. I'm still not too sure what happened though and would appreciate your help...

Background:

Newton’s law of cooling says that the rate of change of temperature is proportional to the temperature difference. In symbols, if a body is at a temperature $T$ at time $t$ and the surrounding region is at a constant temperature $T_e$ ($e$ for external), then the rate of change of $T$ is given by
$$dT/dt=k(T_e −T)$$.
The constant $k > 0$ is a constant of proportionality that depends properties of the
body like specific heat and surface area.

Problem:

Find the formula for $T$ if the initial temperature at time $t = 0$ is $T_0$.

My Working:

$\frac{dT}{dt} = k(T_e-T)$

$\int\frac{1}{(T_e-T)}dT = \int k dt$

$ln|T_e-T| = kt + c$

$|T_e-T| = e^{kt+c} $

$T_e-T = \pm e^{kt+c}$

$T=T_e-(\pm e^{kt}\cdot e^c) = T_e - (\pm e^c \cdot e^{kt})$

When $t=0$ , $T=T_0$. Hence,

$ln|T_e-T_0| = c$

$T_e - T_0 = \pm e^c$

Thus,

$T = T_e - ((T_e - T_0)\cdot e^{kt}) = T_e + (T_0 - T_e)e^{kt}$ (correct answer should be $T_e + (T_0 - T_e)e^{-kt}$)

Working from Answer Key:

Separate variables and integrate.

$(T −T_e)^{−1}dT =−kdt \Rightarrow ln|T −Te|=−kt+c$

$T − T_e = \pm e^ce^{−kt} = Ae^{−kt}$

The initial condition $T(0) = T_0$ implies $A = T_0 − T_e$ . Thus

$T =T_e +(T_0 −T_e)e^{−kt}$

My Comment:

It seems to me that I did not get the right answer just because I did not remove $-1$ from $(T_e-T)$ to change $k(T_e-T)$ into $-k(T-T_e)$ right at the start. But shouldn't this step be an optional step? Why, then, am I unable to get the correct answer just because I omited this step? I suspect there's something I've missed but I just can't find out what's wrong with my working!

Please help me and thank you in advance!

(Note: This problem is from MIT OCW's Single Variable Calculus (18.01) course.)

答: The issue is in the integral $$\int \frac{1}{T_e - T} \, \mathrm{d}T$$ In order to integrate this, we need the integrand in the form $\frac{f'(x)}{f(x)}$, where here $f'(x) = -1$, which means we need an extra minus sign on the numerator. That is,  $$\int \frac{1}{T_e - T} \, \mathrm{d}T = -\int \frac{-1}{T_e - T} \, \mathrm{d}T = - \ln|T_e -T| $$ which recovers the missing minus sign.


问:
What is the pullback of an alternating multilinear map
I'm trying to understand the proof of the statement which says that up to multiplication by a nonzero scalar, there is a unique left invariant volume form on a lie group G. What I don't understand in this proof  is what is $L_g^*\omega_e$, Where $L_g$ is the left  translation by $g \in G$ and $\omega_e$ is $\wedge^n(T_eG)^*$. I know what is the pullback of differential form by a map but in this case $\omega_e$ is not a differential form on G it's an alternating multilinear map. So in general if V is a vector space of dimension n  and $\phi : V \rightarrow V $ is a map and $T \in \wedge^n V^*$  what is the definition of $\phi^*(T)$?

Thank you!

答:
The pullback of $T \in \bigwedge^nW^*$ by a linear map $\phi : V \to W$ is defined by

$$(\phi^*T)(v_1, \dots, v_n) = T(\phi(v_1), \dots, \phi(v_n)).$$

Once you have this definition, you can define the pullback of differential forms pointwise. That is, if $f : M \to N$ is a smooth map, we obtain a linear map $(f_*)_p : T_pM \to T_{f(p)}N$, so if $\alpha \in \Omega^n(N)$, we define $(f^*\alpha)_p := (f_*)^*_p\alpha_{f(p)}$, i.e.

$$(f^*\alpha)_p(v_1, \dots, v_n) := ((f_*)^*_p\alpha_{f(p)})(v_1, \dots, v_n) = \alpha_{f(p)}((f_*)_pv_1, \dots, (f_*)_pv_n)$$

for all $v_1, \dots, v_n \in T_pM$.


问:
adjunction space $\mathbf{S}^{n-1}\times\mathbf{D}^{n}\cup\mathbf{D}^{n}\times\mathbf{S}^{n-1}$ is homeomorphic to $\mathbf{S}^{2n-1}$
Consider the adjunction space $\mathbf{S}^{n-1}\times\mathbf{D}^{n}\cup\mathbf{D}^{n}\times\mathbf{S}^{n-1}$ by gluing $\mathbf{S}^{n-1}\times \mathbf{S}^{n-1}$ together, is there a good way to prove that $\mathbf{S}^{n-1}\times\mathbf{D}^{n}\cup\mathbf{D}^{n}\times\mathbf{S}^{n-1}\cong\mathbf{S}^{2n-1}$? When $n=1$, I can check manually that it is true (attach four lines together by glueing their end points, constructing a square that is homeomorphic to $\mathbf{S}^{1}$), but I don't know how to prove it in general.
Thanks!

答:
First observe that you are not really gluing something. Both $\mathbf{S}^{n-1}\times\mathbf{D}^{n}$ and $\mathbf{D}^{n}\times\mathbf{S}^{n-1}$ are genuine subspaces of $\mathbf{D}^{n}\times\mathbf{D}^{n}$ and $B := \mathbf{S}^{n-1}\times\mathbf{D}^{n}\cup\mathbf{D}^{n}\times\mathbf{S}^{n-1}$ is their ordinary union. But in fact $\mathbf{S}^{n-1}\times\mathbf{D}^{n}\cap\mathbf{D}^{n}\times\mathbf{S}^{n-1} = \mathbf{S}^{n-1}\times \mathbf{S}^{n-1}$. You can of course regard $B$ as the adjunction space $X \cup_f Y$ with $X = \mathbf{S}^{n-1}\times\mathbf{D}^{n}, Y = \mathbf{D}^{n}\times\mathbf{S}^{n-1}$ along the inclusion map $X \supset \mathbf{S}^{n-1}\times \mathbf{S}^{n-1} \stackrel{f}{\hookrightarrow} Y$. But this is unnecessary here.

Also note that the topologies on $\mathbf{S}^{n-1}$ and $\mathbf{D}^{n}$ are the subspace topologies inherited from $\mathbb R^n$. The topology on $\mathbf{D}^{n}\times\mathbf{D}^{n}$ is the product topology which agrees with the subspace topology inherited from $\mathbb R^{n} \times \mathbb R^{n} = \mathbb R^{2n}$. Thus also $B$ carries the subspace topology inherited from $\mathbb R^{2n}$.

Let $\lVert - \rVert$ denote the Euclidean norm on $\mathbb R^{m}$. On $\mathbb R^{2n} = \mathbb R^{n} \times \mathbb R^{n}$ define
$$\lVert (x,y) \rVert' = \max(\lVert x \rVert, \lVert y \rVert) .$$
This is also a norm on $\mathbb R^{2n}$ and it is equivalent to $\lVert - \rVert$. In fact, all norms on $\mathbb R^{m}$ are equivalent. But note that we do not need this general theorem. We have
$$\max(\lVert x \rVert, \lVert y \rVert) = \max\left(\sqrt{\sum_{i=1}^n x_i^2}, \sqrt{\sum_{i=1}^n y_i^2} \right) \le \sqrt{\sum_{i=1}^n x_i^2 + \sum_{i=1}^n y_i^2} = \lVert (x,y) \rVert,$$
$$\lVert (x,y) \rVert^2=  \sum_{i=1}^n x_i^2 + \sum_{i=1}^n y_i^2 \le 2 \max\left(\sum_{i=1}^n x_i^2, \sum_{i=1}^n y_i^2 \right) = 2 \max(\lVert x \rVert^2,\lVert y \rVert^2 ) \\ = 2 \max(\lVert x \rVert,\lVert y \rVert)^2  = 2 (\lVert (x,y) \rVert')^2 .$$

Let $S  = \{(x,y) \in \mathbb R^{2n} \mid \lVert (x,y) \rVert' = 1\}$ denote the unit sphere with respect to $\lVert - \rVert'$. We have $S = B$. In fact, we have $(x,y) \in S$ iff both $\lVert x \rVert, \lVert y \rVert \le 1$ and at least one of them is $1$, in other words iff $(x,y) \in \mathbf{S}^{n-1}\times\mathbf{D}^{n}$ or $(x,y) \in \mathbf{D}^{n}\times\mathbf{S}^{n-1}$.

Define

$$\phi: S \to \mathbf{S}^{2n-1}, \phi(z) = \frac{z}{\lVert z \rVert } ,$$
$$\psi: \mathbf{S}^{2n-1} \to S, \psi(z) = \frac{z}{\lVert z \rVert' } .$$
These maps are continuous with respect to the standard subspace topologies on $\mathbf{S}^{2n-1}$ and $S$ which are inherited from the "usual" Euclidean topology on $\mathbb R^{2n}$. This topology is by definition induced by the Euclidean norm $\lVert - \rVert$, but is also induced by the norm $\lVert - \rVert'$ because it is equivalent to $\lVert - \rVert$. Thus the continuity of $\phi, \psi$ follows from the fact that $\lVert - \rVert$ and $\lVert - \rVert'$ are continuous real-valued functions on $\mathbb R^{2n}$.

Since $\psi \circ \phi = id$ and $\phi \circ \psi = id$, we see that they are homeomorphisms which are inverse to each other.


问:
If $\beta$ is an $\mathcal{L}$ formula and $\beta_P$ is a tautology then $\beta $ is valid
I am reading A Friendly Introduction to Mathematical Logic by Christopher Leary and Lars Kristiansen. Let $\beta$ be a $\mathcal{L}$-formula in first order language $\mathcal{L}$. In the first line of Page 52, the authors claim that if $\beta_P$, a formula in propositional logic, is a tautology then $\beta$ is valid.

$\beta_P$ is obtained from $\beta$ in the following fashion (which is written is Page 51):

Find all subformulas of $\beta$ of the form $\forall x\alpha$ that are not in the scope of another quantifier. Replace them with propositional variables in a systematic fashion. This means that if $\forall yQ(y,c)$ appears twice in $\beta$, it is replaced by same letter both times, and distinct subformulas are replaced with distinct letters.

Find all atomic formulas that remain, and replace them systematically with new propositional variables.

At this point, $\beta$ will have been replaced with a propositional formula $\beta_{P}$.

I was trying to prove this by induction on the structure of formula $\beta$ but I got stuck in the base case. Here's my attempt: Let $\mathfrak{U}$ be any $\mathcal{L}$-structure and $s$ be any variable assignment function to $\mathfrak{U}$. Suppose that $\beta\equiv=t_1t_2$ (written in Polish notation) where $t_1 , t_2$ are terms of the language $\mathcal{L}$. Then $\beta_P$ is $A$ where $A$ is a propositional variable. Let's assume that $\beta _P$ is a tautology. We need to prove that $\beta$ is valid, i.e., $\mathfrak{U}\models \beta [s]$.

Here's my question: Since $\beta_P$ is a tautology, are not allowed to make the substitution $A=F$ because if we do, then $\beta_P$ would not longer be a tautology. How do we remedy this situation?

答:
We need a preliminary Lemma in order to define a "procedure" that assigns to every interpretation $\mathfrak A$ and variable assignment function $s$ a boolean valuation $v$ such that:

(1) $\mathfrak A,s \vDash \beta \text {  iff  } v(\beta_{\text P})= \text T$.

The idea is to consider the rule for producing $\beta_{\text P}$ from $\beta$ above and let $P_i$ the propositional variable corresponding to subformula $Q_i$, i.e. an atom or subformula $\forall x \alpha$.

We define:

$v(P_i)= \text T \text { when } \mathfrak A,s \vDash Q_i, \text { and } v(P_i)= \text F$ otherwise.

And then we use the clauses for the satisfaction relation [Def.1.7.4, page 29] to prove (1) by induction, considering that we have three "atomic" cases: (1) $=t_1t_2$, (2) $Rt_1 \ldots t_n$ and (3) $∀xα$, and only two compound cases: (4) $¬ \varphi$ and $\varphi ∨ \psi$.

Having proved the Lemma, we can use it to prove the main result.

Let $\beta$ a formula and let $\mathfrak A$ and $s$ whatever.

We compute the corresponding $v$. But $\beta_{\text P}$ is a tautology; thus: $v(\beta_{\text P})=\text T$, that implies: $\mathfrak A,s \vDash \beta$.


问:
Reduction modulo $p$ in number fields

For every prime number $p$, there exist a map
  $$f:\mathbb{P}^n(\mathbb{Q})\to\mathbb{P}^n(\mathbb{F}_p)$$ defined
  by: for $P\in \mathbb{P}^n(\mathbb{Q})$, we can find a unique tuple
  $(x_1,\dots,x_n)\in\mathbb{Z}^n$ of coprime integers such that
  $P=[x_1,\dots,x_n]$. Then
  $f(P):=[\overline{x_1},\dots,\overline{x_n}]\in \mathbb{P}^n(\mathbb{F}_p)$.

Now, replace $\mathbb{Q}$ by a number field $K$ and $p$ by a prime ideal $\mathfrak{p}$ of $\mathcal{O}_K$. A paper I read speaks of a map
$$f:\mathbb{P}^n(K)\to\mathbb{P}^n(\mathcal{O}/\mathfrak{p}).$$
But how is this defined ? For $P\in\mathbb{P}^n(K)$, we can find $(x_1,\dots,x_n)\in\mathcal{O}_K^n$ such that $P=[x_1,\dots,x_n]$, however, if $\mathcal{O}_K$ is not a UFD, we cannot necessarily choose them such that $x_i\not\in\mathfrak{p}$ for some $i$ as in the rational case...

答:
We cannot necessarily choose them such that $x_i \notin \mathfrak{p}$ for some $i$ as in the rational case...

Right, but we don't need to.  In order for the reduction map to be defined we don't need $x_i \in \mathcal{O}_K$ for all $i$; we need $v_{\mathfrak{p}}(x_i) \geq 0$ for all $i$ and $v_{\mathfrak{p}}(x_i) = 0$ for at least one $i$.  In other words, we can define the reduction map by replacing $R$ with its localization $R_{\mathfrak{p}}$, which has the same fraction field $K$, and is a DVR hence a UFD.

Note that this works for a nonzero prime ideal $\mathfrak{p}$ in any Dedekind domain $R$ with fraction field $K$.

Title: Reduction modulo $p$ in number fields


答:
In general, it is possible for us to extract components of power series by playing with roots of unity. Particularly, let's say

$$
f(z)=\sum_{n=0}^\infty c_nz^n
$$

converges absolutely for some $z\in\mathbb C$. If we set $\zeta=e^{2\pi i/q}$ then

$$
\frac1q\sum_{r=0}^{q-1}\zeta^{-ra}f(\zeta^rz)=\sum_{k=0}^\infty c_{qk+a}z^{qk+a}\tag1
$$

Consequently, we get

$$
\sum_{k=0}^\infty{z^{3k+1}\over(3k+1)!}=\frac13\sum_{r=0}^2 e^{-2\pi ir/3}\exp(e^{2\pi ir/3}z)
$$

Proof for (1):

Given $f$ converges absolutely at $z$, we can plug in the power series expansion for $f(z)$ to get

$$
\begin{aligned}
\sum_{r=0}^{q-1}\zeta^{-ra}\sum_{n=0}^\infty c_n\zeta^{rn}z^n
&=\sum_{n=0}^\infty c_nz^{n}\left(\sum_{r=0}^{q-1}\zeta^{r(n-a)}\right)
\end{aligned}
$$

If $n\equiv0\pmod q$, then

$$
\sum_{r=0}^{q-1}\zeta^{r(n-a)}=\sum_{r=0}^{q-1}1=q
$$

Otherwise we have $\zeta^{n-a}\ne1$:

$$
\sum_{r=0}^{q-1}\zeta^{r(n-a)}={\zeta^{q(n-a)}-1\over\zeta^{n-a}-1}=0
$$

Consequently, the purpose of sum of roots of unity in (1) is to extract coefficients from power series.

Title: How to find the sum of $\sum_{n=0}^{\infty} \frac{z^{3 n}}{(3 n) !}$?

Body:
I'm really stuck with this problem and I hope some of you could give me a hint.

Consider the functions: $$f(z)=\sum_{n=0}^{\infty} \frac{z^{3 n}}{(3 n) !}, \quad f^{\prime}(z)=\sum_{n=0}^{\infty} \frac{z^{3 n+2}}{(3 n+2) !}, \quad f^{\prime \prime}(z)=\sum_{n=0}^{\infty} \frac{z^{3 n+1}}{(3 n+1) !}, \quad f^{\prime \prime \prime}(z)=f(z)$$
I need to find the sum of each of these series.

My first thought was to show that:
$$f(z)+f^{\prime}(z)+f^{\prime \prime}(z)=\sum_{n=0}^{\infty} \frac{z^{3 n}}{(3 n) !}+\sum_{n=0}^{\infty} \frac{z^{3 n+2}}{(3 n+2) !}+\sum_{n=0}^{\infty} \frac{z^{3 n+1}}{(3 n+1) !}=\sum_{n=0}^{\infty} \frac{z^{n}}{n !}=e^{z}$$
And then solve this differential equation.

However, officially I haven't had a course in differential equations. So is there another way I can solve this using basic Complex Analysis? I hope some of you can guide me in the right direction, what theorems could be useful?


问:
Integrability of Thomae function and value of its integral
I am trying to solve the following exercise (from Axler's Measure, Integration and Real Analysis) and I would like to have some help in finishing my proof.

"Define $f:[0,1]\to\mathbb{R}$ as follows:
$
  f(a) =
  \begin{cases}
                                   0 & \text{if $a$ is irrational} \\
                                   \frac{1}{n} & \text{if $a\in\mathbb{Q}$ and $n$ is the smallest positive integer such that $a=\frac{m}{n}$ for some integer $m$}
  \end{cases}
$.

Show that $f$ is Riemann integrable and compute $\int_{0}^{1}f$."

My solution: (NOTE: THIS SOLUTION IS WRONG, I HAVE WRITTEN A CORRECT ONE IN THE ANSWER TO THIS QUESTION BELOW)

We immediately note that for any partition $P$ of $[0,1]$ we have $L(f,P,[0,1])=\sum_{j=1}^{n}(x_j-x_{j-1})\inf_{[x_{j-1},x_j]}f=\sum_{j=1}^{n}(x_j-x_{j-1})\cdot 0=0$ since in any sub-interval of $[0,1]$ there will be an irrational number.

Now we let $\varepsilon>0$ and consider the size of the set $D_{\varepsilon}:=\{x\in [0,1]:f(x)\geq \varepsilon\}\subset\{x\in [0,1]:f(x)\geq\frac{1}{n}\}=:D_{1/n}$, where $\varepsilon\geq\frac{1}{n},\ n>1+\frac{2}{\varepsilon}$ fixed (such an $n$ must exist, since $\frac{1}{n}\overset{n\to\infty}{\to}0$).

We note that if $\varepsilon>1$ we have $D_{\varepsilon}=\emptyset\Rightarrow \#(D_{\varepsilon})=0$,

if $\varepsilon = 1$ we have  $D_{\varepsilon}=\{0,1\}\Rightarrow \#(D_{\varepsilon})=2$,

if $\frac{1}{2}\leq\varepsilon\leq 1$ we have $D_{\varepsilon}\subset D_{1/2}=\{0,1,\frac{1}{2}\}\Rightarrow \#(D_{\varepsilon})\leq \#(D_{1/2})=3$,

if $\frac{1}{3}\leq\varepsilon\leq \frac{1}{2}$ we have $D_{\varepsilon}\subset D_{1/3}=\{0,1,\frac{1}{2},\frac{1}{3},\frac{2}{3}\}\Rightarrow \#(D_{\varepsilon})\leq \#(D_{1/3})=5$,

if $\frac{1}{4}\leq\varepsilon\leq \frac{1}{3}$ we have $D_{\varepsilon}\subset D_{1/4}=\{0,1,\frac{1}{2},\frac{1}{3},\frac{2}{3},\frac{1}{4},\frac{2}{4},\frac{3}{4}\}=\{0,1,\frac{1}{2},\frac{1}{3},\frac{2}{3},\frac{1}{4},\frac{3}{4}\}\Rightarrow \#(D_{\varepsilon})\leq \#(D_{1/4})=7$,

if $\frac{1}{5}\leq\varepsilon\leq \frac{1}{4}$ we have $D_{\varepsilon}\subset D_{1/5}=\{0,1,\frac{1}{2},\frac{1}{3},\frac{2}{3},\frac{1}{4},\frac{3}{4},\frac{1}{5},\frac{2}{5},\frac{3}{5},\frac{4}{5}\}\Rightarrow \#(D_{\varepsilon})\leq \#(D_{1/5})=11$,

$\dots$,

so we see that in each case $D_{1/N},\ (N\geq 2)$ contains $0,1$ and the $k-1$ elements $\frac{1}{k},...,\frac{k-1}{k}$ for $2\leq k\leq N$ (this is an overestimate for the size of $D_{1/N}$ since some of the elements in this set might be present also in $D_{\frac{1}{N-1}}$ - see for example $D_{1/4}$) hence an upper bound for the size of $D_{1/N}$ is $$2+\sum_{k=2}^{N}(k-1)=2+\sum_{k=1}^{N}k-\sum_{k=1}^{N}1=1+\frac{N(N+1)}{2}-(N-1)=\frac{N^2-N+4}{2}$$
which in turn implies that $\#(D_{\varepsilon})\leq \frac{N^2-N+4}{2}$.

So, we consider again now the aformentioned $n$ such that $\frac{1}{n}\leq\varepsilon$, set $M:=\frac{n^2-n+4}{2}$ and consider the partition $P_{\varepsilon}$ of $[0,1]$ such that
$$P_{\varepsilon}:=\{0,n_1-\frac{\varepsilon}{4M}, n_1+\frac{\varepsilon}{4M},\dots, n_M-\frac{\varepsilon}{4M}, n_M+\frac{\varepsilon}{4M},1\}.$$

We have that $$U(f,P_{\varepsilon},[0,1])\leq (n_1-\frac{\varepsilon}{4M})\cdot 1+\frac{\varepsilon}{2M}\cdot 1+(n_2-n_1-\frac{\varepsilon}{2M})\cdot\frac{1}{n}+\frac{\varepsilon}{2M}\cdot 1+(n_3-n_2-\frac{\varepsilon}{2M})\frac{1}{n} +\dots +\frac{\varepsilon}{2M}\cdot 1+ (1-n_M-\frac{\varepsilon}{4M})\cdot 1= M\cdot\frac{\varepsilon}{2M}+(1-n_M-(\frac{\varepsilon}{4M}+\frac{\varepsilon}{4M}+M\frac{\varepsilon}{2M}))\frac{1}{n}=\frac{\varepsilon}{2}+(1-n_M-\frac{\varepsilon}{2M}+\frac{\varepsilon}{2})\frac{1}{n}<\frac{\varepsilon}{2}+(1+\frac{\varepsilon}{2})\frac{1}{n}<\varepsilon$$ so $U(f,P_{\varepsilon},[0,1])-L(f,P_{\varepsilon},[0,1])=U(f,P_{\varepsilon},[0,1])<\varepsilon$.

We have thus shown that for each $\varepsilon>0$ there exists a partition of $[0,1]$ such that $U(f,P_{\varepsilon},[0,1])-L(f,P_{\varepsilon},[0,1])<\varepsilon$ so $f$ is Riemann integrable on $[0,1]$, and $\int_{0}^{1}f=U(f,[0,1])=L(f,[0,1])=0$.

答:
Let $\varepsilon>0$, set $n=\frac{2}{\varepsilon}$, $N=n(n+1)$ and consider the partition $P$ of $[0,1]$ such that $P=\{0,\frac{1}{N},\frac{2}{N},\dots,\frac{N-1}{N},1\}$. Since $f$ takes up the value $1$ in two places (in $0$ and $1$) and for $n\geq 2$ it takes up the value $\frac{1}{n}$ in at most $n-1$ places ($\frac{1}{n}, \frac{2}{n},\dots,\frac{n-1}{n}$ - note that if $\frac{j}{n},\ 1\leq j \leq N-1$ is not in lowest terms it could take on a higher value) we have  $$U(f,P,[0,1])=\sum_{j=1}^{N}\frac{1}{N}\sup_{[\frac{j-1}{N},\frac{j}{N}]}f=\frac{1}{N}(2\cdot 1+1\cdot\frac{1}{2}+2\cdot\frac{1}{3}+\dots +n\cdot\frac{1}{n+1})\leq\frac{1}{N}(2\cdot 1 + 2\cdot\frac{1}{2}+3\cdot\frac{1}{3}+\dots + (n+1)\cdot\frac{1}{n+1})=\frac{2+n}{N}=\frac{2+n}{n(n+2)}=\frac{1}{n}=\frac{\varepsilon}{2}<\varepsilon.$$ Being $L(f,P,[0,1])=\sum_{j=1}^{N}\frac{1}{N}\inf_{[\frac{j-1}{N},\frac{j}{N}]}f=\sum_{j=1}^{N}\frac{1}{N}\cdot 0=0$ (since in any subinterval of $[0,1]$ there is an irrational number) it follows that $U(f,P,[0,1])-L(f,P,[0,1])=U(f,P,[0,1])<\varepsilon$.

Thus, for arbitrary $\varepsilon>0$ we have found a partition $P$ of $[0,1]$ such that $U(f,P,[0,1])-L(f,P,[0,1])<\varepsilon$ hence $f$ is integrable on $[0,1]$.

By definition of Riemann integral we know that $\int_{0}^{1}f=U(f,[0,1])=L(f,[0,1])$ and since $L(f,[0,1])=\sup_{P} L(f,P,[0,1])=\sup_{P} \{0\}=0$ we conclude that $\int_{0}^{1}f=0.$


问:
Unbounded sequences.
I was hoping to see if anyone could help me with a thought that I have stumbled upon.

Is there a sequence $a_n$ which is unbounded and its inverse is also unbounded?

I can't think of one myself and I am just looking for some help.

Thanks in advance

答:
As Kavi is correctly suggesting in one comment, we can consider

$$a_n=\begin{cases}  n  &\text{ if } n\text{ is even }\\
          \frac{1}{n}  &\text{ if } n \text{ is odd.}\end{cases}$$

Clearly it is unbounded since the even terms are pushing it to infinity.

Its inverse, that we denote $b_n:=\frac{1}{a_n}$, is the same sequence but exchanging the parity of $n$ in the definition. More formally, we have that:

$$b_n=\begin{cases}  n  &\text{ if } n\text{ is odd }\\
          \frac{1}{n}  &\text{ if } n \text{ is even.}\end{cases}$$

Hence it is also unbounded.

Title: Unbounded sequences.


问:
Algebraic expression for $\frac{x^z}{z} + \frac{x^{\bar{z}}}{\bar{z}}$ using only real variables
UPDATED:

(I originally asked for a proof that the expression was real; I meant to ask for the real algebraic expression that the sum equates to - i.e., an expression in terms of $x$, $\sigma$ and $t$ but not $i$.)

Let $z = \sigma + i t$ be a complex number and let $\bar{z} = \sigma - i t$ be its conjugate. Let $x$ be a positive real number. I want to find the real expression corresponding to

$$\frac{x^z}{z} + \frac{x^{\bar{z}}}{\bar{z}}$$

(I am generalising from an expression in von Mangoldt's explicit formula for the Chebyshev $\psi$ function here.)

I realise that since $\frac{x^{\bar{z}}}{\bar{z}}$ is the conjugate of $\frac{x^z}{z}$, their sum must be real - but what form does it take?

This should involve a fairly simple piece of algebra, but I can't get there. I begin as follows:

$$
\begin{aligned}
\frac{x^z}{z} + \frac{x^{\bar{z}}}{\bar{z}} &= \frac{x^{\sigma +i t}}{\sigma +i t}+\frac{x^{\sigma -i t}}{\sigma -i t}
\\ &= \frac{e^{(\sigma +i t) \log (x)}}{\sigma +i t}+\frac{e^{(\sigma -i t) \log (x)}}{\sigma -i t}
\\ &= \frac{(\sigma +i t) e^{(\sigma -i t) \log (x)}+(\sigma -i t) e^{(\sigma +i t) \log (x)}}{\sigma ^2+t^2}
\end{aligned}
$$

But I don't know how to proceed. I'd like to apply Euler's identity, but as far as I'm aware, the identity $e^{a+ib} = e^a e^{ib}$ does not apply because the original exponent is complex, so I cannot split out the complex component of $e^{(\sigma \pm i t) \log (x)}$. And even if it is valid to write $e^{a+ib} = e^a e^{ib}$, I still end up with a factor of $-2 i \sin (t \log (x))$ in the numerator.

Could someone show me how to arrive at the purely real expression that $\frac{x^z}{z} + \frac{x^{\bar{z}}}{\bar{z}}$ resolves to?

答:
The complex conjugate of the first term is exactly the second term. Hence their sum  is real.

$x^{z}=e^{z \ln x}$ and the conjugate of $e^{c}$ is $e^{\overline c}$ as seen using power series.


问:
Proving a result for cohomology for real projective plane
I am trying to show that the cup product of any two elements in $H^{1}(\mathbb{R}P^{2};\mathbb{Z}/4\mathbb{Z})$ is zero.

I have computed the cohomology groups $H^{*}(\mathbb{R}P^{2},\mathbb{Z}/4\mathbb{Z})$ using the singular homology

$ \displaystyle H_{n}(\mathbb{R}P^{2}) \cong  \begin{cases} \mathbb{Z} & n =0 \\ \mathbb{Z}/2\mathbb{Z} & n=1 \\ 0 & \text{otherwise} \end{cases}$

and the Universal Coefficient Theorem $H^{n}(\mathbb{R}P^{2};\mathbb{Z}/4\mathbb{Z})\cong \operatorname{Hom}(H_{n}(\mathbb{R}P^{2}),\mathbb{Z}/4\mathbb{Z})\oplus \operatorname{Ext}(H_{n-1}(\mathbb{R}P^{2}),\mathbb{Z}/4\mathbb{Z})$ to get that

$ \displaystyle H^{n}(\mathbb{R}P^{2};\mathbb{Z}/4\mathbb{Z}) \cong  \begin{cases} \mathbb{Z}/4\mathbb{Z} & n =0 \\ \mathbb{Z}/2\mathbb{Z} & n=1,2 \\ 0 & \text{otherwise} \end{cases}$.

For any to elements of $H^{1}(\mathbb{R}P^{2};\mathbb{Z}/4\mathbb{Z})$ say $a,b$ we have that $a\smile b=0$. My issue is I can't really see where to get started. We have $\smile:H^{1}(\mathbb{R}P^{2};\mathbb{Z}/4\mathbb{Z})\times H^{1}(\mathbb{R}P^{2};\mathbb{Z}/4\mathbb{Z}) \to H^{2}(\mathbb{R}P^{2};\mathbb{Z}/4\mathbb{Z})$ which one could view as $\smile:\mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z} \to \mathbb{Z}/2\mathbb{Z}$... but alas I am at a loss for ideas.

答:
$\require{AMScd}$
Consider the cellular cochain maps induced by $\phi:\Bbb Z/4\to \Bbb Z/2$.
\begin{CD}
0 @<{}<<\Bbb Z/4 @<{2}<<\Bbb Z/4 @<{0}<<\Bbb Z/4@<{}<<0\\
\text{ } @VVV @VVV @VVV\\
0@<{}<< \Bbb Z/2 @<{0}<< \Bbb Z/2@<{0}<<\Bbb Z/2 @<{}<< 0
\end{CD}
We see that the induced homomorphism is trivial for $H^1$ and bijective for $H^2$.
Consider the induced homomorphism on the cohomology rings, i.e.,
$$g:H^*(\Bbb RP^2;\Bbb Z/4)\to H^*(\Bbb RP^2;\Bbb Z/2)$$

We have $g(\alpha\smile\alpha)=g(\alpha)\smile g(\alpha)=0$ if $\alpha$ generates $H^1(\Bbb RP^2;\Bbb Z/4)$. Also, as $\alpha\smile\alpha\in H^2(\Bbb RP^2;\Bbb Z/4)$, we know that $\alpha\smile\alpha=0$ as the induced ring homomorphism is bijective in dimension $2$.


问:
Which trigonometric identity is applicable to show $\sin\left(\frac{(2n+1)\pi}{2a}(x + a)\right) = \cos\left(\frac{(2n+1)\pi x}{2a} + n\pi \right)$?
I'm confused with regards to the following and want to know which trigonometric identity is applicable here:

$$\sin\left(\frac{(2n+1)\pi}{2a}(x + a)\right) = \cos\left(\frac{(2n+1)\pi x}{2a} + n\pi \right)$$
where $a$ is a constant and $n \in \mathbb{Z}$.

Furthermore, I want to know why the above expression is equal to $$(-1)^n\cos\bigg(\frac{(2n+1)\pi x} {2a}\bigg).$$

答:
Your first equation implies that
$$\sin\left(\phi+\frac{\pi}{2}\right)=\cos(\phi)$$
for any angle $\phi$, which is a defining identity of the trigonometric functions $\sin$ and $\cos$. Furthermore, you can easily see that both hand sides of your last equation are equal to
$$(-1)^n\cos(\phi)\,,$$
since $\cos(a+b)=\cos(a) \cos{b}-\sin(a) \sin{(b)}$.


问:
Obtain the coefficient of $x^2$ in the expansion of $1+\frac{6}{2x+1}+\frac{5}{2-3x}$
Hello so this is a 2 part question and I managed to express that praction as a partial fraction which was equaled to $$1+\frac{6}{2x+1}+\frac{5}{2-3x}$$ I will add my work below I tried lot to Obtain the coefficient of $x^2$ but cant get to the answer of 237/8 and help is much appreciated. Thank you!

答:
Recall that
\begin{align}
\frac1{a-b z} &=\frac1a\frac1{1-\left(\frac b a z\right)}\\
&=\frac1a\sum_{n=0}^{\infty}\left(\frac b a z\right)^n
\end{align}

We have $$\frac1{1+2x}=\frac1{1-(-2x)}=1-2x+4x^2-8x^2\pm \ldots$$ and $$\frac1{2-3x}=\frac12\frac1{1-\frac32x}=\frac12\times\left(1+\frac32x+\frac94x^2\pm\ldots\right)$$

Now you have $1+6\times\frac1{1+2x} + 5 \times \frac1{2-3x}$, $\color{red}{\mbox{and you can substitute the latter expansions and obtain the result you are after.}}$


问:
What does the equation $f(A)=A$ tell about the function $f$?
Given a function $f: A \to A$ what does $f(A)=A$ mean? Does it mean $f$ is bijective on $A$? I could see it being bijective on $A$ if $A$ is a finite set, but I'm not sure what happens if $A$ is infinite? $f(A)=A$ surely means $f$ is surjective whether or not $A$ is finite. But I don't know how to go about proving the injectivity of $f$?

答: The assertion $f(A)=A$ does not tell you that $f$ is injective. For instance, if $f(x)=\sin(2\pi x)$, then $f([-1,1])=[-1,1]$, but $f$ is not injective on $[-1,1]$.


问:
The Cardinality of the Surreal Numbers and Games
I've been learning about John H Conway's surreal numbers, which are a rich superset of the reals. It's clear that the real numbers are dense in the middle section of the surreals, with the surreals extending much further toward the left and right.

I'm wondering about the cardinality of the surreals. Can we put the surreal numbers into a relationship with the power set of the reals, for instance?

The surreals form a proper subset of another JH Conway discovery, the games. Is anyone aware of the cardinality of these games?

答: They have no "cardinality" as they form a proper class: every ordinal number is also a surreal number. So they are not in 1-1 correspondence to any set, in ZFC. If you have a set theory with proper classes, the surreal numbers are probably as large as the whole universe.</p>


问:
Distributing quantifiers across implication
I've been given a series of questions about the distribution of quantification across logical operators. For example:

Is $\;\forall x \in S \;(p(x) \; or \; q(x))\;$ the same as
$\;(\forall x \in S \; p(x)) \; or \; (\forall x \in S \; q(x))$?

Most of these make pretty good intuitive sense to me (in the above, for example, obviously these are not "the same" as for the first it is sufficient for a member have either $p$ or $q$, but for the latter it is necessary that all members have either $p$ or all members have $q$). I'm stuck, however, on "distribution" across implication:

Is $\;\forall x \in S \;(p(x) \; \implies \; q(x))\;$ the same as
$\;(\forall x \in S \; p(x)) \; \implies \; (\forall x \in S \; q(x))$?

I ask that you not provide the answer, but a method for dissecting this. It seems like they should be the same, however this is just a "gut feeling" that I've failed to quantify. I've been given the following example situation to use:

Let $S$ be a collection of two coins, one with an $H$ on top and a $T$
on bottom, and one with an $H$ on top and a $Q$ on bottom. Let $p(x)$
denote "$x$ has an $H$ on top" and $q(x)$ denote "$x$ has a $Z$ on
bottom".

This hasn't helped much, though.

答:
So, to get an intuition for logical formulae I can't quite get my head around, I usually try to rewrite them into something that's easier for me to reason about.

If you're using classical FOL, then you could apply
$(\varphi \Rightarrow \psi) \Leftrightarrow (\neg \varphi \vee \psi)$
and
$\neg (\forall x \in A. \phi) \Leftrightarrow (\exists x \in A. \neg\phi)$
to obtain

"Is $\forall x \in A. \neg P(x) \vee Q(x)$ the same as $(\exists x \in A. \neg P(x)) \vee (\forall x \in A. Q(x))$?"

and see if that is more intuitive.

Also, if you know a method of proving formulae, you could also just try to directly show $$\big(\forall x \in A. \neg P(x) \Rightarrow Q(x)\big)
\Leftrightarrow
\big((\forall x \in A. \neg P(x)) \Rightarrow (\forall x \in A. Q(x))\big)$$
and
$$\big(\forall x \in A. \neg P(x) \Rightarrow Q(x)\big)
\not\Leftrightarrow
\big((\forall x \in A. \neg P(x)) \Rightarrow (\forall x \in A. Q(x))\big)$$
and see if one works. If a proof fails, that can help you come up with a counterexample model.


Title: Non-trivial zero(s) of Akiyama-Tanigawa triangle

Body:
Introduced in 1997, the Akiyama-Tanigawa triangle is a doubly-indexed recursion that encodes the Bernoulli numbers, among other sequences. It is defined as follows: let $a:\mathbb{N^0}\times\mathbb{N^+}\to \mathbb{R}$ (this indexing is to agree with that of the Bernoulli numbers) be given by:
$$
a_{0,j}=\frac{1}{j}\qquad a_{i,j} = j(a_{i-1,j}-a_{i-1,j+1})
$$Here is a table for $1\le i+1,j\le10$:
$$
\begin{array}{c|cccccccccc}
  & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\
\hline 0 & 1 & \frac{1}{2} & \frac{1}{3} & \frac{1}{4} & \frac{1}{5} & \frac{1}{6} & \frac{1}{7} &
   \frac{1}{8} & \frac{1}{9} & \frac{1}{10} \\
 1 & \frac{1}{2} & \frac{1}{3} & \frac{1}{4} & \frac{1}{5} & \frac{1}{6} & \frac{1}{7} &
   \frac{1}{8} & \frac{1}{9} & \frac{1}{10} & \frac{1}{11} \\
 2 & \frac{1}{6} & \frac{1}{6} & \frac{3}{20} & \frac{2}{15} & \frac{5}{42} & \frac{3}{28} &
   \frac{7}{72} & \frac{4}{45} & \frac{9}{110} & \frac{5}{66} \\
 3 & 0 & \frac{1}{30} & \frac{1}{20} & \frac{2}{35} & \frac{5}{84} & \frac{5}{84} &
   \frac{7}{120} & \frac{28}{495} & \frac{3}{55} & \frac{15}{286} \\
 4 & -\frac{1}{30} & -\frac{1}{30} & -\frac{3}{140} & -\frac{1}{105} & \color{red}{0} & \frac{1}{140} &
   \frac{49}{3960} & \frac{8}{495} & \frac{27}{1430} & \frac{125}{6006} \\
 5 & 0 & -\frac{1}{42} & -\frac{1}{28} & -\frac{4}{105} & -\frac{1}{28} & -\frac{29}{924} &
   -\frac{7}{264} & -\frac{28}{1287} & -\frac{87}{5005} & -\frac{27}{2002} \\
 6 & \frac{1}{42} & \frac{1}{42} & \frac{1}{140} & -\frac{1}{105} & -\frac{5}{231} &
   -\frac{9}{308} & -\frac{343}{10296} & -\frac{1576}{45045} & -\frac{27}{770} &
   -\frac{205}{6006} \\
 7 & 0 & \frac{1}{30} & \frac{1}{20} & \frac{8}{165} & \frac{5}{132} & \frac{295}{12012} &
   \frac{67}{5720} & \frac{4}{6435} & -\frac{6}{715} & -\frac{75}{4862} \\
 8 & -\frac{1}{30} & -\frac{1}{30} & \frac{1}{220} & \frac{7}{165} & \frac{200}{3003} &
   \frac{1543}{20020} & \frac{3997}{51480} & \frac{464}{6435} & \frac{1539}{24310} &
   \frac{775}{14586} \\
 9 & 0 & -\frac{5}{66} & -\frac{5}{44} & -\frac{44}{455} & -\frac{629}{12012} &
   -\frac{41}{12012} & \frac{133}{3432} & \frac{140}{1989} & \frac{1113}{12155} &
   \frac{9597}{92378} \\
\end{array}
$$As claimed (I could provide a proof, if there's interest, but it's not relevant to my question), $a_{i,1}=B_i$, the $i^{th}$ Bernoulli number, with $B_1=1/2$. As such, $a_{2i+1,1}=0$ for $i>1$; call these trivial zeros. Note $a_{4,5}=0$ as well: my conjecture is that this is the only such non-trivial zero.

I have verified this claim for $a_{i,j}$ for $1\le i+1,j\le 100$. I also believe that for fixed $j$, $a_{n,j}>0$ for some $n=n(j)$. Additionally, for several fixed $j$ there is a closed-form by using the recursion. However, I'm not sure which of any of these lines is a feasible plan-of-attack, or if miraculously a closed-form for $a_{i,j}$ exists.

Body:
Here I obtain a closed form of the recurrence relation with a view to later proving your conjecture, if possible. Using Mathematica's FindSequenceFunction I inferred the possible ansatz $$a_{i,j}=\frac{p_i(j)}{(j)_{i+1}},$$ where $(x)_n$ is the Pochhammer symbol and $p_i(j)$ is some polynomial in $j$.

Now using

a[0, j_] := 1/j
a[i_, j_] := j (a[i - 1, j] - a[i - 1, j + 1])
i := 6
FindSequenceFunction[Table[a[i, j] Pochhammer[j, i + 1], {j, 1, 10}],
  j]

I was able to find the following numerator polynomials $p_i(j)$:

$$\begin{array}{l}
p_0(j)=1\\
p_1(j)=j\\
p_2(j)=j^2\\
p_3(j)=(-1 + j) j^2\\
p_4(j)=-5 j^3 + j^4\\
p_5(j)=(-1 + j) (-4 j^2 - 15 j^3 + j^4)\\
p_6(j)=42 j^3 + 119 j^4 - 42 j^5 + j^6\\
p_7(j)=(-1 + j) (120 j^2 + 518 j^3 + 659 j^4 - 98 j^5 + j^6)\\
p_8(j)=-2160 j^3 - 7250 j^4 - 6189 j^5 + 3721 j^6 - 219 j^7 + j^8\\
p_9(j)=(-1 + j) (-12096 j^2 - 57720 j^3 - 98906 j^4 - 57735 j^5 +
   15241 j^6 - 465 j^7 + j^8)\\
\end{array}
$$

My first port of call was the OEIS but the coefficients of each polynomial do not appear to be listed.

Let's try working out generating functions for each row $(a_{i,j})_{j=1}^\infty$ instead:

$$P_i(x):=\sum_{j=1}^\infty\frac{p_i(j)}{(j)_{i+1}}x^j.$$

Mathematica tells me the following:

$$\begin{array}{l}
P_0(x)=-\log (1-x)\\
P_1(x)=-\frac{x+\log (1-x)}{x}\\
P_2(x)=\frac{(x-2) \log (1-x)-2 x}{x^2}\\
P_3(x)=\frac{3 (x-2) x-\left(x^2-6 x+6\right) \log (1-x)}{x^3}\\
P_4(x)=\frac{\left(x^3-14 x^2+36 x-24\right) \log (1-x)-4 x \left(x^2-6 x+6\right)}{x^4}\\
P_5(x)=\frac{5 x \left(x^3-14 x^2+36 x-24\right)-\left(x^4-30 x^3+150 x^2-240 x+120\right) \log (1-x)}{x^5}\\
P_6(x)=\frac{\left(x^5-62 x^4+540 x^3-1560 x^2+1800 x-720\right) \log (1-x)-6 x \left(x^4-30 x^3+150 x^2-240 x+120\right)}{x^6}\\
P_7(x)=\frac{7 x \left(x^5-62 x^4+540 x^3-1560 x^2+1800 x-720\right)-\left(x^6-126 x^5+1806 x^4-8400 x^3+16800 x^2-15120 x+5040\right) \log (1-x)}{x^7}\\
P_8(x)=\frac{\left(x^7-254 x^6+5796 x^5-40824 x^4+126000 x^3-191520 x^2+141120 x-40320\right) \log (1-x)-8 x \left(x^6-126 x^5+1806 x^4-8400 x^3+16800 x^2-15120 x+5040\right)}{x^8}\\
P_9(x)=\frac{9 x \left(x^7-254 x^6+5796 x^5-40824 x^4+126000 x^3-191520 x^2+141120 x-40320\right)-\left(x^8-510 x^7+18150 x^6-186480 x^5+834120 x^4-1905120 x^3+2328480 x^2-1451520 x+362880\right) \log (1-x)}{x^9}\\
\end{array}$$
The coefficients of the polynomials appear to be related to the sequence A090582 and on further investigation we find that
$$P_i(x)=(-1)^{i+1}\frac{ixu_{i-1}(x)-u_i(x)\log(1-x)}{x^i},$$
where
$$u_i(x)=\sum _{k=1}^i (-1)^{k+1} k! \mathcal{S}_i^{(k)} x^{i-k},$$
where $\mathcal{S}_i^{(k)}$ is the Stirling number of the second kind.

Now we'd like to find out when the coefficient of $x^j$ in $P_i(x)$ is zero, where $j>1$. The case $j=1$ relates to the Bernoulli numbers in the first column and their trivial zeros, defined by the OP.

We can now use Cauchy's integral formula,

$$P_i^{(j)}(0)=\frac{j!}{2\pi\sqrt{-1}}\oint\frac{P_i(z)}{z^{j+1}}dz,$$

to compute the entries in your original matrix (note I used $\sqrt{-1}$ to avoid confusion with the index $i$), e.g. $a_{4,5}=0$ corresponds to
$$\oint\frac{P_4(z)}{z^6}dz=\oint\frac{\left(z^3-14 z^2+36 z-24\right) \log (1-z)-4 z \left(z^2-6 z+6\right)}{z^{10}}=0.$$
Expanding the general integrand we have
$$\frac{(-1)^{i+1}iu_{i-1}(z)}{z^{j+i}}+(-1)^{i+1}u_i(z)\sum_{k=1}^\infty\frac{z^{k-(j+i+1)}}{k}.$$
Substituting $u_i(x)$ we have
$$(-1)^{i+1}i\sum _{k=1}^{i-1} (-1)^{k+1} k! \mathcal{S}_{i-1}^{(k)} \frac{1}{z^{j+k+1}}+(-1)^{i+1}\sum _{\ell=1}^i (-1)^{\ell+1} \ell! \mathcal{S}_i^{(\ell)}\sum_{k=1}^\infty\frac{1}{k}\frac{1}{z^{j+1-k+\ell}}.$$
Since the first sum has no $1/z$ term that just leaves the double sum.

Now $i$ and $j$ are effectively fixed (they correspond to the row and column in your original matrix) and since $1\leq\ell\leq i$ then the coefficients of $1/z$ occur when $j+1-k+\ell=1\implies k=j+\ell$, hence the sum of coefficients gives you the closed form you were wondering about,

$$a_{i,j}=(-1)^i\sum _{\ell=1}^i (-1)^{\ell} \ell! \mathcal{S}_i^{(\ell)}\frac{1}{j+\ell},$$
which again passes the $a_{4,5}=0$ test. Now we just need to prove $a_{i,j}=0$ if and only if $i=4$ and $1<j=5$, but I've not been able to do this yet...

It's worth taking a look at the list of polynomials $p_i(j)$. Notice how every odd polynomial from $p_3(j)$ onward has a factor of $(j-1)$ which corresponds to $B_n=0$ (odd Bernoulli numbers) when you plug in $j=1$ !

You can also see why $a_{4,5}=0$ since $-5j^3+j^4$ has a root at $j=5$.

If your conjecture is true then these polynomials must be irreducible over $\mathbb{Z}$, excluding the $(j-1)$ factor of the odd polynomials. Factorising the polynomials,
$$\begin{array}{l}
p_0(j)=1\\
p_1(j)=j\\
p_2(j)=j^2\\
p_3(j)=(-1 + j) j^2\\
p_4(j)=j^3(-5 + j)\\
p_5(j)=(-1 + j) j^2(-4  - 15 j + j^2)\\
p_6(j)=j^3(42 + 119 j - 42 j^2 + j^3)\\
p_7(j)=(-1 + j) j^2(120 + 518 j + 659 j^2 - 98 j^3 + j^4)\\
p_8(j)=j^3(-2160  - 7250 j - 6189 j^2 + 3721 j^3 - 219 j^4 + j^5)\\
p_9(j)=(-1 + j) j^2(-12096  - 57720 j - 98906 j^2 - 57735 j^3 +
   15241 j^4 - 465 j^5 + j^6)\\
\end{array}
$$

Proving $p_0,p_1,p_2,p_3\neq 0$ is easy for $j>1$. It's also easy to see that $p_4=0$ if and only if $1<j=5$.

Let's try to use the Rational Root Theorem (RRT) to see if each polynomial is irreducible. First of all notice that the coefficient of the highest $j$ power in each un-factored term is $1$ so any rational root $r=n/d$ with $\gcd(n,d)=1$ must have be such that $d\mid 1$, so the denominator can only every be $1$ which implies $r\in\mathbb{Z}$. RRT also tells us that $n\mid c_0$ where $c_0$ is the constant term of each un-factored polynomial.

So for $p_5$ the candidate roots are $r\in\pm\{1,2,4\}$ and since $p_5(r)\neq 0$ then $p_5$ is irreducible, and you will never find a zero.

For $p_6$ the candidate roots are $r\in\pm\{1,2,3,7,42\}$ and since $p_6(r)\neq 0$ then $p_6$ is irreducible, and you will never find a zero.

For $p_7$ the candidate roots are $r\in\pm\{1,2,4,8,3,5,120\}$. Again no zeros.

For $p_8$ the candidate roots are $r\in\pm\{1,2,4,8,16,3,9,27,5,2160\}$. Again no zeros.

For $p_9$ the candidate roots are $r\in\pm\{1,2,4,8,16,32,64,3,9,27,7,12096\}$. Again no zeros.

So in general after fully factoring the polynomial as far as possible, you may be able to find a proof based on an analysis of the constant term and whether or not it's $\pm$ divisors are roots of the polynomial factor.

Another perspective is this: the un-factored polynomial terms of each $p_i(j)$ are $O(j^{i-3})$ so there always exists an $N(j)$ such that $a_{i,j}\neq 0$ for all $j>N(j)$, so that covers an infinite number of cases for each row $i$.


问:
entailment relationships in FOL

I was doing this question, but it tells me one of the option I selected was wrong.

The second option I thought it means "all of the things in A is not in B" indicates that "there exists something in A that is not in B"

The last option I thought means "not all things in A are in B" indicates that "there exists something in A that is not in B"

Both make sense to me, please help explain which one I selected is wrong. Also is there any that I haven't selected is correct?

答:
Only the last implication is correct.

As you have observed, you can think of the predicates $A$ and $B$ as sets. So, knowing the first three don't hold, we should try to construct a counter-model.

Both the first and second fail on the model $A = \emptyset, B = \emptyset$,
and the third fails on the model $A = \{o_1,o_2\}, B = \{o_2\}$.

Where you went wrong is assuming $A$ will contain an element. Indeed
$$\exists x( A(x) ), \forall x( A(x) \Rightarrow B(x) ) \vDash \exists x( A(x) \wedge \neg B(x) )$$
actually holds.

So, beware cases where you just have $\forall$ assumptions, because they don't guarantee that objects will actually exist in your set.

Title: entailment relationships in FOL


问:
Surface inside cylinder $y_1^2+y_2^2=r^2$ using parametrisation
The question is

Let $r>0$. Let S⊂ $R^3$ the surface given by $y_1y_2-ry_3=0$. Determine the area of the portion of S inside the cylinder given by $y_1^2+y_2^2=r^2$.

In class we saw a formula to calculate the area of parametric surfaces (the one with the cross product). The problem is, I don't have a parametrisation of S. And to be very honest, I have no idea how it's done. I thought maybe cylindrical coordinates would help? I also tried isolating r to see if it could help in any way. How do I approach a question like this? We didn't see any examples in class and I am very confused.

答:
You need to find surface area of the hyperbolic paraboloid $z = \frac{xy}{r}$, inside the cylinder $x^2+y^2 = r^2$.

The formula for the surface area is given by,

$S = \displaystyle \iint_A \sqrt{1+ \bigg(\frac{\partial z}{\partial x}\bigg)^2 + \bigg(\frac{\partial z}{\partial y}\bigg)^2} \ dA$

$ = \displaystyle \iint_A \sqrt{1+ \frac{x^2}{r^2} + \frac{y^2}{r^2}} \ dA$

where $A$ is the projection of the surface in $XY$ plane and $dA = dx \ dy$.

Converting to polar coordinates,

$x = \rho \cos \theta, y = \rho \sin \theta, 0 \leq \rho \leq r, 0 \leq \theta \leq 2\pi$.

$S = \displaystyle \frac{1}{r}\int_0^{2\pi} \int_0^r \rho \sqrt{\rho^2 + r^2} \ d\rho \ d\theta = \displaystyle \frac{2 \pi}{r} \int_0^r \rho \sqrt{\rho^2 + r^2} \ d\rho $

To evaluate the integral, you can substitute $\rho^2 + r^2 = t, \rho d\rho = \frac{1}{2} dt$.

Can you take it from here?


问:
Use proposition 2 to prove proposition 1. Would you correct my proof?
PROPOSITION 1:  If $f:U\subset \mathbb{R}^2\rightarrow \mathbb{R}$ is a differentiable map, where $U$ is an open subset of $\mathbb{R}^2$, then a graphic of $f$, given by $(x, y, f(x,y))$, where $(x,y)\in U$ is a regular surface.

PROPOSITION 2: If $f:U\subset \mathbb{R}^3\rightarrow \mathbb{R}$ is a differentiable map and $a\in f(U)$ is a regular value of $f$, then $f^{-1}(a)$ is a regular surface in $\mathbb{R}^3$.

DEFINITION: Given a differentiable map $F:U\subset \mathbb{R}^n \rightarrow \mathbb{R}^m$ defined in an open set $U$ of $\mathbb{R}^n$ we say that $p\in U$ is a critical point of $F$ if the differential $dF_p:\mathbb{R}^n\rightarrow \mathbb{R}^m$ is not a surjective (or onto) mapping. The image $F(p)\in \mathbb{R}^m$ of a critical point is called a critical value of $F$. A point of $\mathbb{R}^m$ which is not a critical value is called a regular value of $F$.

MY ATTEMPT:
Define $h:U\subset\mathbb{R}^3\rightarrow \mathbb{R}$ given by $h(x, y, z)= f(x, y)-z$ where $f$ is differentiable. Note that $$\nabla h = (h_x(x, y,z), h_y(x, y,z), h_z(x, y,z))= (f_x, f_y, f_z)=(f_x, f_y, -1) \implies \nabla h(p) \neq 0,\; \forall p.$$
Therefore, $(0,0,0)$ is a regular value of $h$. Now, note that

\begin{align*}
h^{-1}(0)&=\{(x, y, z); h(x, y, z)=0\}\\
&=\{(x, y, z); f(x, y)=z\}\\
&=\{(x, y, f(x, y))\}:=\text{graphic of}\; f
\end{align*}

Thus, by proposition 2, graphics of differentiable functions are regular surfaces.


问:
Pigeonhole principle application sums and differences
Let $A \subset \{1,2,...,99\}$, prove or disprove the following:

a. For $|A| = 27$

b. For $|A| = 26$

There are $2$ different numbers in $A$ that their sum or their difference can be divided with $50$.

I am sure the question points me to use Pigeonhole Principle, but I'm not quite sure which are the pigeons and which are the holes because the question deals with or so proving that a sum (or a difference) can be divided with $50$ closes the proof without handling the other case.

I've started with saying that there are $(99+98)-(1+2)+1=195$ different sums,
and $(99-1)-(1-99)+1=197$ differences (including negatives) in $\{1,2,...,99\}$.

Also, when dividing with $50$ you can get maximum 50 remainders (let $r \in \{0,1,...,49\}$ mark the reminders)

Can anyone hint me with how to proceed from here?

Body:
Hint: Consider the sets:
$\{1, 49, 51, 99 \},
\{2, 48, 52, 98 \}, \{3, 47, 53, 97 \}, \ldots, \{24, 26, 74, 76 \},\{25, 75 \}, \{50 \}$.


问:
Convolutional inverse of distribution (Van Dijk)
I am going over the problems in van Dijk's distribution theory book and trying to find the (convolution) inverse of the following distributions:

$$S=\delta''+H,\quad T=He^x +\delta'$$
using Heaviside calculus but I'm not sure the solutions make sense. First (using Heaviside symbolic calculus)

$$S=H+\delta''=\frac{1}{\delta'}+\delta''=\frac{\delta'\delta''+1}{\delta'}=\frac{\delta'''+1}{\delta'}$$
so the inverse of $S$ is given by
$$\delta' (\delta'''+1)^{-1}=\delta' H Z$$
where $Z$ solves $(d^3+1)Z=0$
but this doesn't seem correct. For $T$ I get using $(He^x)^{-1}=\delta'-\delta$
$$T=He^x + \delta' = \frac{1}{\delta'-\delta}+\delta'=\frac{\delta''-\delta'}{\delta'-\delta}$$
and so $T^{-1}=(\delta'-\delta)(\delta''-\delta')^{-1}$
Solving the ODE $(d^2-d)Z=0$ with $Z(0)=0,Z'(0)=1$ I get the solution
$$(\delta''-\delta')^{-1} =H(e^x-1)=He^x-H$$
and so
$$T^{-1}=(\delta'-\delta)(e^x H-H)=H+e\delta-\delta'$$
which again, I'm not sure is correct!
Hints/references/Feedback appreciated.

答:
Not sure what you mean with $Z$ solves $(d^3+1)Z=0$.

The Laplace transform of $\delta'''+\delta$ is $1+s^3$ and $\frac1{1+s^3} = \frac1{3\zeta_6^2}\frac{1}{s-\zeta_6}+\frac1{3\zeta_6^6}\frac{1}{s-\zeta_6^3}+\frac1{3\zeta_6^{10}}\frac{1}{s-\zeta_6^5}$ whose (causal) inverse Laplace transform is $U=\frac1{3\zeta_6^2} e^{\zeta_6 t}+\frac1{3\zeta_6^{6}} e^{\zeta_6^3 t}+\frac1{3\zeta_6^{10}} e^{\zeta_6^5 t})1_{t>0}$ which is the convolution inverse of $\delta'''+\delta= S'$ so the convolution inverse of $S$ is $U'$


问:
Show that $\|x\|_1 = 1/2 \inf_{y > 0}(\sum_{i=1}^{m}\frac{x_i^2}{y_i} + 1^Ty)$

Show that $\|x\|_1 = 1/2 \inf_{y > 0}(\sum_{i=1}^{m}\frac{x_i^2}{y_i} + 1^Ty)$ for $x \in R^{m}$.

I think an equivalent problem is:
minimize $f(y) = 1/2 (\sum_{i=1}^{m}\frac{x_i^2}{y_i} + 1^Ty)$.

$1/y$ is a convex function on $y > 0$, $1^Ty$ is a convex function so, $1/y + 1^Ty$ is a convex function. I remember something like that the $\inf$ of a convex function over a convex set is convex.

$df(y)/dy_i = 1/2(-\frac{x_i^2}{y_i^2} + 1) = 0, y_i^2 = x_i^2, y_i = |x_i|$

The function value at $y_i$ becomes $m/2 + 1^Ty = m/2 + 1^T|x|$ which is not equal to the $\|x\|_1 = 1^T|x|$.

Body:
We have for $y>0$
\begin{align}
\sum_{i=1}^{m}\frac{x_i^2}{y_i} + 1^Ty &= \sum_{i=1}^{m}(\frac{x_i^2}{y_i}+y_i) \\
&= \sum_{i=1}^{m}(\frac{x_i^2}{y_i}+y_i-2|x_i|) + 2\sum_{i=1}^{m}|x_i|\\
&= \sum_{i=1}^{m}(\frac{|x_i|}{\sqrt{y_i}}-\sqrt{y_i})^2 + 2\sum_{i=1}^{m}|x_i| \ge  2\sum_{i=1}^{m}|x_i| = 2||x||_1\\
\end{align}

Hence, $\|x\|_1 = 1/2 \inf_{y > 0}(\sum_{i=1}^{m}\frac{x_i^2}{y_i} + 1^Ty)$ for $x \in R^{m}$


问:
How prove this inequality $\left(\int_{0}^{1}f(x)dx\right)^2\le\frac{1}{12}\int_{0}^{1}|f'(x)|^2dx$

Let $f\in C^{1}[0,1]$ such that $f(0)=f(1)=0$. Show that
  $$\left(\int_{0}^{1}f(x)dx\right)^2\le\dfrac{1}{12}\int_{0}^{1}|f'(x)|^2dx.$$

I think we must use Cauchy-Schwarz inequality
$$\int_{0}^{1}|f'(x)|^2dx\ge \left(\int_{0}^{1}f(x)dx\right)^2$$
but this maybe is not useful to problem.

The coefficient $\dfrac{1}{12}$ is strange. How find it ?

答:
By integration by parts,
\begin{align*}
\int_0^1f(x)\mathrm{d}x&=\int_0^1f(x)\mathrm{d}(x+C)\\
&=(x+C)f(x)\bigg|_0^1-\int_0^1(x+C)f'(x)\mathrm{d}x\\
&=-\int_0^1(x+C)f'(x)\mathrm{d}x
\end{align*}
Hence by Cauchy-Schwarz inequality
$$\left(\int_0^1f(x)\mathrm{d}x\right)^2\leqslant\int_0^1(x+C)^2\mathrm{d}x\int_0^1(f'(x))^2\mathrm{d}x$$
and,
$$\int_0^1(x+C)^2\mathrm{d}x=\frac{1}{3}(3C^2-3C+1)=g(C)$$
we can easily get $\min g(C)=g(\frac{1}{2})=\frac{1}{12}$. Therefore,
$$\left(\int_0^1f(x)\mathrm{d}x\right)^2\leqslant\frac{1}{12}\int_0^1(f'(x))^2\mathrm{d}x$$


问:
Show that $\sigma(n)$ is odd if $n$ is a power of two.
Show that $\sigma(n)$ is odd if $n$ is a power of two.

What I know: Since $n$ is a power of two we can say $n=2^x$. So $\sigma(n)=\sigma(2^x)=2^{x+1}-1$ which is clearly odd. This identity was given and proved by the reader in Elementary Number Theory by Dudley. Is this correct? It seems too obvious.

Also $\sigma(n)$ is the sum of all the divisors of $n$.

答: That is correct, yes, but you can prove it easily. If $n=2^k$, then the divisors of $n$ are $1,2,2^2,\ldots,2^k$ and so$$\sigma(n)=\sum_{j=0}^k 2^j=\frac{2^{k+1}-1}{2-1}=2^{k+1}-1,$$which is odd, as you noticed.


问:
Parabola equation (sign of coefficients)
Consider the equation
$$
y=a^2x +bx
$$

Which signs of $a,b$ can deliver the picture below? I suspect that I need $a>0$ and $b<0$ with $|b|$ larger enough than $a$. Is that correct?

答: None. If $y=ax^2+bx$, then $(0,0)$ belongs to the parabola. So, you will not get a graph like that picture.


答:
There is a lemma that states, for $p$ prime, $\sigma(p^k) = \frac{p^{k+1}-1}{p-1}$. As long as you can use that, your proof would look like this:

Let $n = 2^k$, for $k \in \mathbb{Z}$. Then we have that $\sigma(n) = \sigma(2^k) = \frac{2^{k+1} - 1}{2 - 1} = 2^{k+1} - 1$, which is odd.

Body:
Show that $\sigma(n)$ is odd if $n$ is a power of two.

What I know: Since $n$ is a power of two we can say $n=2^x$. So $\sigma(n)=\sigma(2^x)=2^{x+1}-1$ which is clearly odd. This identity was given and proved by the reader in Elementary Number Theory by Dudley. Is this correct? It seems too obvious.

Also $\sigma(n)$ is the sum of all the divisors of $n$.


问:
Difficult Continuity Multivariable Question
WIll you help me understand the following?

$ f(x,y)=\begin{cases}
\sin(y-x) & \text{for} & y>|x| \\  \\
0 & \text{for} & y=|x| \\  \\
\frac{x-y}{\sqrt{x^2 + y^2}} & \text{for} & y<|x|
\end{cases}
$

I need to check differentiability and continuity.
I tried substituting $x= \frac{1}{2} (u-v)   , y=\frac{1}{2} (u+v)$  but it doesn't help me...
Will you help me figure this thing out?

Thanks !

答:
Hint:

Calculate $\lim\limits_{x\to{0}}{f(x,\,y)}$ along lines $y=kx$ for different $k.$


问:
Algebraic expression for $\frac{x^z}{z} + \frac{x^{\bar{z}}}{\bar{z}}$ using only real variables
UPDATED:

(I originally asked for a proof that the expression was real; I meant to ask for the real algebraic expression that the sum equates to - i.e., an expression in terms of $x$, $\sigma$ and $t$ but not $i$.)

Let $z = \sigma + i t$ be a complex number and let $\bar{z} = \sigma - i t$ be its conjugate. Let $x$ be a positive real number. I want to find the real expression corresponding to

$$\frac{x^z}{z} + \frac{x^{\bar{z}}}{\bar{z}}$$

(I am generalising from an expression in von Mangoldt's explicit formula for the Chebyshev $\psi$ function here.)

I realise that since $\frac{x^{\bar{z}}}{\bar{z}}$ is the conjugate of $\frac{x^z}{z}$, their sum must be real - but what form does it take?

This should involve a fairly simple piece of algebra, but I can't get there. I begin as follows:

$$
\begin{aligned}
\frac{x^z}{z} + \frac{x^{\bar{z}}}{\bar{z}} &= \frac{x^{\sigma +i t}}{\sigma +i t}+\frac{x^{\sigma -i t}}{\sigma -i t}
\\ &= \frac{e^{(\sigma +i t) \log (x)}}{\sigma +i t}+\frac{e^{(\sigma -i t) \log (x)}}{\sigma -i t}
\\ &= \frac{(\sigma +i t) e^{(\sigma -i t) \log (x)}+(\sigma -i t) e^{(\sigma +i t) \log (x)}}{\sigma ^2+t^2}
\end{aligned}
$$

But I don't know how to proceed. I'd like to apply Euler's identity, but as far as I'm aware, the identity $e^{a+ib} = e^a e^{ib}$ does not apply because the original exponent is complex, so I cannot split out the complex component of $e^{(\sigma \pm i t) \log (x)}$. And even if it is valid to write $e^{a+ib} = e^a e^{ib}$, I still end up with a factor of $-2 i \sin (t \log (x))$ in the numerator.

Could someone show me how to arrive at the purely real expression that $\frac{x^z}{z} + \frac{x^{\bar{z}}}{\bar{z}}$ resolves to?

答:
Your calculation is almost done.
$$
2\,{\frac {{{\rm e}^{\sigma\,\ln  x }} \big( \sigma\cos
 \left( t\ln  x   \right)+t \sin \left( t\ln
 x   \right) \big) }{{\sigma}^{2}+{t}^{2}}}
$$


问:
Convergence rate of maximum interval of Poisson process
Let $\{t_0,t_1,\ldots,t_{N}\}\subset [0,1]$ be a collection of points from a Poisson process with intensity $\lambda$.
Here, for convenience, assume that $t_0=0$.
Then what is the convergence rate of maximum interval of these points,

\begin{align*}
 \max_{i:t_i\in(0,1]} (t_i-t_{i-1}) = O_p(?)
\end{align*}

as $\lambda\rightarrow \infty$?

Here, what I have done.

Let $\max \Delta t_i = \max_{i:t_i\in(0,1]} (t_i-t_{i-1})$.

We have $N\sim \operatorname{Pois}(\lambda)$, and "without considering" the restriction $t_N \le 1$, each interval follows
\begin{align*}
  t_i - t_{i-1} \sim \operatorname{Exp}(\lambda),
\end{align*}
independently.

So, for some fixed $n$, we have
\begin{align*}
\def\bbP{{\mathbb{P}}}
\def\eps{{\varepsilon}}
  \bbP\{ \max_{i\le n} \Delta t_i > \eps \} &= \bbP \left\{ \Delta t_1 > \eps, \ldots, \Delta t_n > \eps\right\}  \\
&= \prod_{i=1}^n e^{-\lambda \eps} \\
&= (e^{-\lambda \eps})^n.
\end{align*}
Since $N$ is not a constant, but a random, we have to sum up with probability $\bbP\{N=n\}$ as follows:
\begin{align*}
  \bbP\{ \max_{i\le N} \Delta t_i > \eps \} &= \sum_{n=0}^{\infty} (e^{-\lambda \eps})^n \bbP\{N=n\}  \\
&= \sum_{n=0}^{\infty} (e^{-\lambda \eps})^n \frac{\lambda^n e^{-\lambda}}{n!}  \\
&= e^{-\lambda} \sum_{n=0}^{\infty}  \frac{\left(e^{-\lambda \eps}\lambda\right)^n}{n!}  \\
&= e^{e^{-\lambda \eps}\lambda - \lambda}  \\
&= e^{\lambda \left( e^{-\lambda \eps} - 1\right)}.
\end{align*}

Then we have $\max_{i\le n} \Delta t_i \xrightarrow{\bbP} 0$ by
\begin{align*}
  \lim_{\lambda \rightarrow \infty} \lambda \left( e^{-\lambda \eps} - 1\right) = \lim_{\lambda \rightarrow \infty} -\lambda=-\infty,
\end{align*}
and the definition of in probability convergence:
\begin{align*}
  \bbP\{ \max_{i\le n} \Delta t_i > \eps \} \rightarrow  \lim_{\lambda \rightarrow \infty} e^{\lambda \left(e^{-\lambda \eps}-1\right)} = e^{-\infty} = 0.
\end{align*}
Thus, for some small $c>0$,
\begin{align*}
  e^{\lambda \left( e^{-\lambda \eps} - 1\right)} &= e^{-c}, \\
   \lambda \left( e^{-\lambda \eps} - 1\right) &= -c, \\
   e^{-\lambda \eps} &= 1- \frac{c}{\lambda}, \\
  \lambda \eps &= \log \left( \frac{\lambda}{\lambda-c} \right),  \\
  \eps &= \frac{\log \frac{\lambda}{\lambda-c}}{\lambda},
\end{align*}
and we have the convergence rate (exactly) as
\begin{align*}
  \max_{i:t_i\in(0,1]} (t_i-t_{i-1}) = O_p\left(\frac{\log \frac{\lambda}{\lambda-c}}{\lambda}\right).
\end{align*}

However, I have no idea how to handle the restriction $t_N \le 1$.

Of course the convergence rate may be similar to this, but I don't know how to rigorously prove it.

Thanks,

答:
I don't have a solution, but I have a couple of ideas:

1.

Conditional on $N$, the distribution of $(t_1,...,t_N)$ is the same as the distribution of the order statistics of a random sample of size $N$ from the Unif(0,1) distribution.

Now, let $M_N$ be the maximum spacing from a random sample from the Unif(0,1) distribution of size $N$.

$$P[N \times M_N -\log{N}<x|N]\rightarrow e^{-e^x}$$
$$E[N \times M_N -\log{N}|N]\rightarrow \gamma =0.5772157..$$
$$Var[N \times M_N|N]\rightarrow \frac{\pi^2}{6}$$

So, the unconditional mean is
$$E[M]\approx  E\left[\frac{\gamma+\log{N}}{N}\right]\approx \frac{\gamma+\log{\lambda}}{\lambda}$$

and the unconditional variance is

$$Var[M] \approx \frac{\pi^2}{6}E\left[\frac{1}{N^2}\right]+Var\left[\frac{\gamma+\log{N}}{N}\right]
\approx \frac{\pi^2}{6}\frac{1}{\lambda^2}+\frac{\gamma+\log{\lambda}-1}{\lambda^3}$$
Devroye, Luc. "Laws of the iterated logarithm for order statistics of uniform spacings." The Annals of Probability (1981): 860-867.

2.
This seems like a way to eliminate the restriction that $t_N<1$.

Let $N^*=ceiling(\lambda+\sqrt{\lambda}\log{\lambda})$

Then,

$$P\left[\max_{i\le N}\Delta t_i>x\right]$$
$$=P\left[\max_{i\le N}\Delta t_i>x,N<N^*\right]+P\left[\max_{i\le N}\Delta t_i>x,N\ge N^*\right]$$
$$\le P\left[\max_{i\le N^*}\Delta t_i>x,N<N^*\right]+P\left[\max_{i\le N}\Delta t_i>x,N\ge N^*\right]$$
$$\le P\left[\max_{i\le N^*}\Delta t_i>x\right]+P\left[N\ge N^*\right]$$

$N$ has mean and variance $\lambda$ and is asymptotically normal, so $P[N \ge N^*]\approx 1-\Phi(\log{\lambda})\rightarrow 0$ where $\Phi$ is the standard normal cdf.

For large $\lambda$, $\lambda-\sqrt{\lambda}\log{\lambda}<N<\lambda+\sqrt{\lambda}\log{\lambda}$ with probability approaching 1.

Therefore, $P\left[\max_{i\le N}\Delta t_i>x\right]$ is bounded between the similar probabilities with these lower and upper bounds replacing the $N$.


问:
What is the best way to prove general differentiation product rule w/o going back to limit interpretation?
I am trying  to put some flow in how concepts and rules are established in basic differentiation: which comes first and which comes second.  Particularly, my question is what the best way is  to prove product rule in general w/o going to the very definition of differentiation. Here is my approach.

The 4 basic foundations needed are below.

Definition of differentiation: $f' (x)=\lim\limits_{h\rightarrow 0} \frac{f(x+h)-f(x)}{h}$

Chain rule: $f'(x) = \frac {d f} {dx}
= \frac {df_1} {df_2} \frac {df_2} {df_3} \frac {df_3} {dx}$

Derivative of $ e^x:   \frac{d}{dx} e^x = e^x$

Derivative of $\ln x: \frac{d}{dx}\ln x = \frac{1}{x}$

Then we prove  product rule : $(uv)'=u'v+uv'$ in the following steps.

Let $y=u\cdot v$

$ln(y) = \ln(u)+\ln(v) $            - property of logarithm

$\frac{1}{y}y'=\frac{1} {u}u'+\frac{1} {v}u'$ - differentiation on both sides  with  chain rule, $\frac{d}{dy} ln (y) =\frac{1}{y} $

$y'=y\frac{1} {u}u'+y\frac{1} {v}u'$ - multiply both sides by $y$

$(uv)'=u'v+uv'$ - replace $y$ with $uv$

The above process can be easily applied to  $(uvw)'=u'vw+uv'w+uvw^\prime$ and so on...

Any one has suggestions for  different and better approaches?

答:
Making johnnyb's approach more rigorous. My answer goes to definitions (against the request to avoid this), but it is just to show why johnnyb's approach actually works.

By the definition of a derivative
$$f(x+h)=f(x)+f'(x)h+o(|h|),\ g(x+h)=g(x)+g'(x)h+o(|h|)$$
$$f(x+h)g(x+h)=f(x)g(x)+(f(x)g(x))'h+o(|h|)$$
but also from the first line
$$f(x+h)g(x+h)=(f(x)+f'(x)h+o(|h|))(g(x)+g'(x)h+o(|h|))=$$
$$=f(x)g(x)+[f'(x)g(x)+f(x)g'(x)]h+o(|h|)$$
and this holds as $g'(x)f'(x)h^2+g(x)o(|h|)+g'(x)o(|h|)h+o(|h|)(f(x)+f'(x)h+o(|h|))\in o(|h|)$ because
$$\lim|\frac{g'(x)f'(x)h^2+g(x)o(|h|)+g'(x)o(|h|)h+o(|h|)(f(x)+f'(x)h+o(|h|))}{h}|=0$$
So from the uniqueness of derivative we have
$$(f(x)g(x))'=f'(x)g(x)+f(x)g'(x)$$

There is also a definition of derivative that simplifies this kind of derivations (and complicates other kinds), but i can't recall it now.


问:
Rolling a die until the sum of all numbers rolled is greater than 6
I'm trying to solve a problem in which I'm asked the probability mass function of $X$ where $X$ is the number of times a dice is rolled until a sum greater than six is encountered.
I was able to deduce that the possible values that $X$ can take are $2,3,4,5,6,7$ (at most we keep gettings $1$s each time), however deducing how often does each of them occurs wasn't easy for me to conclude without drawing the sample space tree, but the problem is that it gets out of control (in terms of size) quickly and I'm not able to find the pattern that would make me solve it quicker otherwise.

Thank you, have a nice day.

答:
Many parts of the tree are the same.  There are only seven probability mass functions you need:  If you have already reached $n$, how many more rolls will you need?

Work backwards,, starting with $n=6$ and ending with $n=0$.


问:
Explanation of where this trig identity comes from
I'm working on a problem but it's been a while since I last saw trig identities so I'd love some help or being pointed in the right direction.

Basically, I'd like to understand where this identity comes from;

$$\tan(2t) = \dfrac{2\tan(t)}{1 - \tan^2(t)}$$

Thanks for any help you can give - if it's useful to know the context of the problem, I'm writing a bit of code that converges on $\pi$ faster than Leibniz's series. (Please don't give too much away about the rest of the problem though, I'd like to get there myself :) )

答: If you know that $\sin(a+b)=\sin a\cos b+\cos a\sin b$ and $\cos(a+b)=\cos a\cos b-\sin a\sin b$, you can apply that to $\tan(a+b)=\dfrac{\sin(a+b)}{\cos(a+b)}$ as a function of $\tan a$ and $\tan b$.  Once you've got that, consider the case where $a$ and $b$ are both the same number, and you've got it.</p>


Title: Rolling a die until the sum of all numbers rolled is greater than 6

Body:
I'm trying to solve a problem in which I'm asked the probability mass function of $X$ where $X$ is the number of times a dice is rolled until a sum greater than six is encountered.
I was able to deduce that the possible values that $X$ can take are $2,3,4,5,6,7$ (at most we keep gettings $1$s each time), however deducing how often does each of them occurs wasn't easy for me to conclude without drawing the sample space tree, but the problem is that it gets out of control (in terms of size) quickly and I'm not able to find the pattern that would make me solve it quicker otherwise.

Thank you, have a nice day.

Body:
Recursion will do:

If $f(x,n)$ is the probability you need $x$ throws until you exceed $n$, then

$f(0,n)=1$ for $n<0$ and $f(x,n)=1$ for $x \not=0$ and $n<0$

$f(x,n)=\frac16(f(x-1,n-1)$ $+f(x-1,n-2)$ $+f(x-1,n-3)$ $+f(x-1,n-4)$ $+f(x-1,n-5)$ $+f(x-1,n-6))$ for $n\ge 0$

giving the table

   x  0   1    2      3      4       5       6       7
n
<0    1
0     0  6/6
1     0  5/6  6/36
2     0  4/6 11/36  6/216
3     0  3/6 15/36 17/216   6/1296
4     0  2/6 18/36 32/216  23/1296  6/7776
5     0  1/6 20/36 50/216  55/1296 29/7776  6/46656
6     0   0  21/36 70/216 105/1296 84/7776 35/46656 6/279936

and you want the bottom row


答:
Let $P (X) $ be the probability that the sum does not exceed 6 after $X $ throws. The number of such events is the number of positive integer solutions to the inequality:
$$k_1+k_2+\cdots+k_X\le6,
$$
which by stars and bars is
$$
\binom 6X.
$$

Hence $$P (X)=\frac1 {6^X}\binom6X.$$

Now observe that the probability  you are looking for is
$$P (X-1)-P (X).
$$


问:
Weak stationary white noise sequence that are identically distributed but not independent
Let $\{Z_t\}$ be a sequence of random variables that are weakly stationary, i.e., $\mathbb{E}[Z_t]=0$ and $Var[Z_t]=\sigma^2$ and $Z_t$'s are uncorrelated, i.e., $cov(Z_s,Z_t)=0$ for all $t\neq s$. Can we create such a sequence of $Z_t$ that are not independent? Also, can we have the other way around, that is, $Z_t$'s are independent but not identically distributed?

My try:
For the first question, I am thinking of $Z_t = (-1)^tX_t$ where $X_t \sim N(0,1)$ but do not know how to show they are dependent.

答:
For the second question, note the weak stationarity only puts restrictions on first two moments, but knowing first two moments is not enough to determine a general distribution.

Here's one construction. Let $f_{0}\left(z\right)=\left(2\pi\right)^{-1/2}z^{-1}\exp\left(-\left(\log z\right)^{2}/2\right)$ be the lognormal density. For $t\geq1$, let $f_{t}\left(z\right)=f_{0}\left(z\right)\left(1+\sin\left(2\pi\log z\right)/t\right)$ which is still a density. Let $\left\{Z_{t}\right\}_{t\geq0}$ be a sequence of random variables. Apply product measure method, they can be constructed independent, and each follows a distribution with density $f_{t}\left(z\right)$. Clearly they have distinct distributions. However, it follows Heyde (1963) that they have the same moments $\mathbb{E}\left[Z_{t}^{k}\right]=\exp\left(k^{2}/2\right)$ for all $k\geq1$, hence weakly stationary.


问:
Routine question about derivatives of automorphic forms being L^2
I consider Automorphic forms on $G = SL_2 (\mathbb{R})$, which are $\Gamma$-invariant, $K$-finite, $Z(g)$ finite, and of moderate growth. If I have such an automorphic form, which happens to be in $L^2(\Gamma \backslash G)$, is it true that if we will act on it by some element of the Lie algebra, the result will also be in $L^2$?

Thank you,
Sasha

答: There is a claim (Borel, Automorphic forms on $SL_2(\mathbb{R})$, 2.14) that if $f$ is $K$-finite on the right and $Z(g)$-finite, then it can be represented as convolution of itself with a smooth function $\phi$ with compact support. Then it is quite clear that if it is in $L^2$, if we take derivative, and "throw" it into the $\phi$ place, we will get a convolution of $L^2$ with a smooth function with compact support, hence again something in $L^2$.</p>


问:
Example for a faithful and finite dimensional representation over a sovable and finite dimensional Lie-Algebra
I am new to StackExchange.  I am learning about Lie-Algebras and I was wondering whether somebody can give me an example for a finite dimensional and faithful representation of a sovable and finite dimensional Lie-Algebra $\mathcal{L}$ over a Vektorspace V (finite dimensional): $\rho: \mathcal{L} \longrightarrow gl(V)$, such that if $\mathcal{N}$ is the Nil-Radical of $\mathcal{L}$,  $~ \rho(\mathcal{N})$ is nilpotent.

Such a representation should exists (see: Theorem 9 of https://terrytao.wordpress.com/2011/05/10/ados-theorem/). I can think of one for a nilpotent Lie-Algebra, but I have trouble finding one for a sovable Lie-Algebra.

Thank you!  :)

答: What about the natural action of the Lie algebra of upper triangular $n\times n$ matrices on $k^n$? The nilradical consists of strictly upper triangular matrices, which are nilpotent.


答:
The best way is to go back to the limit definition, as this only involves elementary means.

(When using alternatives approaches, make sure that you don't invoke the product rule indirectly - unknowingly - as this would make a circular argument. So you should check the definitions of the exponential and the logarithms, and how their derivatives are established, as well as how the chain rule is obtained.)

This said, you ought to know the process named as the logarithmic derivative,

$$\frac{(tuvw)'}{tuvw}=(\log(tuvw))'=(\log(t)+\log(u)+\log(v)+\log(w))'=\frac{t'}t+\frac{u'}u+\frac{v'}v+\frac{w'}w$$

and

$$(tuvw)'=tuvw\left(\frac{t'}t+\frac{u'}u+\frac{v'}v+\frac{w'}w\right).$$

Title: What is the best way to prove general differentiation product rule w/o going back to limit interpretation?

Body:
I am trying  to put some flow in how concepts and rules are established in basic differentiation: which comes first and which comes second.  Particularly, my question is what the best way is  to prove product rule in general w/o going to the very definition of differentiation. Here is my approach.

The 4 basic foundations needed are below.

Definition of differentiation: $f' (x)=\lim\limits_{h\rightarrow 0} \frac{f(x+h)-f(x)}{h}$

Chain rule: $f'(x) = \frac {d f} {dx}
= \frac {df_1} {df_2} \frac {df_2} {df_3} \frac {df_3} {dx}$

Derivative of $ e^x:   \frac{d}{dx} e^x = e^x$

Derivative of $\ln x: \frac{d}{dx}\ln x = \frac{1}{x}$

Then we prove  product rule : $(uv)'=u'v+uv'$ in the following steps.

Let $y=u\cdot v$

$ln(y) = \ln(u)+\ln(v) $            - property of logarithm

$\frac{1}{y}y'=\frac{1} {u}u'+\frac{1} {v}u'$ - differentiation on both sides  with  chain rule, $\frac{d}{dy} ln (y) =\frac{1}{y} $

$y'=y\frac{1} {u}u'+y\frac{1} {v}u'$ - multiply both sides by $y$

$(uv)'=u'v+uv'$ - replace $y$ with $uv$

The above process can be easily applied to  $(uvw)'=u'vw+uv'w+uvw^\prime$ and so on...

Any one has suggestions for  different and better approaches?


问:
Logarithmic Equations
How does one go about solving:

$(5x+2)^{\frac{4}{3}} = 16$

I'm confused as how to parse through the equation to solve it using logs.

Title: Logarithmic Equations

Body:
\begin{align}
&(5x+2)^{\frac{4}{3}} = 16\\\implies
&5x+2 = (16)^{\frac{3}{4}}\\\implies
&5x+2 = (2^4)^{\frac{3}{4}}\\\implies
&5x+2 = 2^3\\\implies
&5x+2 = 8\\\implies
&5x= 6\\\implies
&x=\frac{6}{5}
\end{align}


问:
$yB$ is not a prime ideal in $B.$
Let $B = \mathbb R[x,y]$ where $x^2 + y^2 = 1$ which is called the coordinate ring of the unit circle.

I am trying to prove that $yB$ is not a prime ideal in $B.$

I have the following information about $B$:

1- $B$ is an integral domain.

And I know that in an integral domain a prime ideal is defined as an ideal $I \neq R$ such that if $ab \in I,$ then either $a \in I$ or $b \in I.$

Still I do not know how to use that definition to prove that $yB$ is not a prime ideal.

Any help will be appreciated!

答: Let $B= \frac{\mathbb{R}[X,Y]}{(X^2+Y^2-1)}$ and $yB=\frac{(Y,X^2+Y^2-1)}{(X^2+Y^2-1)}$ when $y$ is the image of $Y$ in $B$. Then $\frac{B}{yB}=\frac{\mathbb{R}[X,Y]}{(Y,X^2+Y^2-1)}=\frac{\frac{\mathbb{R}[X,Y]}{(Y)}}{\frac{(Y,X^2+Y^2-1)}{(Y)}}=\frac{\mathbb{R}[X]}{(X^2-1)}$ which is not a domain. Hence, it is not a prime ideal.

Title: $yB$ is not a prime ideal in $B.$


答:
Let $\mathfrak{r}_2(K)$ be the non-abelian Lie algebra of dimension $2$ over $K$ with basis $(e_1,e_2)$ and $[e_1,e_2]=e_1$. Then the adjoint representation is a faithful linear representation of $\mathfrak{r}_2(K)$ into $\mathfrak{gl}_2(K)$ and the nilradical $\mathfrak{n}=\langle e_1\rangle $ is mapped to the nilpotent matrix
$$
{\rm ad}(e_1)=\begin{pmatrix} 0 & 1 \cr 0 & 0 \end{pmatrix}.
$$
The adjoint representation is faithful in this case, because its kernel, the center of the Lie algebra, is trivial. This is never the case for non-abelian nilpotent Lie algebras.

Title: Example for a faithful and finite dimensional representation over a sovable and finite dimensional Lie-Algebra

Body:
I am new to StackExchange.  I am learning about Lie-Algebras and I was wondering whether somebody can give me an example for a finite dimensional and faithful representation of a sovable and finite dimensional Lie-Algebra $\mathcal{L}$ over a Vektorspace V (finite dimensional): $\rho: \mathcal{L} \longrightarrow gl(V)$, such that if $\mathcal{N}$ is the Nil-Radical of $\mathcal{L}$,  $~ \rho(\mathcal{N})$ is nilpotent.

Such a representation should exists (see: Theorem 9 of https://terrytao.wordpress.com/2011/05/10/ados-theorem/). I can think of one for a nilpotent Lie-Algebra, but I have trouble finding one for a sovable Lie-Algebra.

Thank you!  :)


问:
Pathological vector fields
Is it true that any smooth curve is an integral curve for some smooth vector field?

More concretely, consider the curve $\gamma:\mathbb{R}\rightarrow \mathbb{R}^2$ defined by $\gamma(t)=(t,t^3sin(1/t))$ if $t\neq 0$ and by $\gamma(0)=(0,0)$. It is a $C^1$ curve, so i was wondering if it could exist a $C^1$ vector field $v=v(x,y,t)$ such that $\gamma$ in an integral curve.

Of course i'm interested in the somehow pathological behaviour near $(0,0)$, more generally i was wondering if this kind of oscillation phenomena could happen with integral curves of vector fields (also in the $C^k$, $k\in \mathbb{N}\cup\{\infty\}$ case).

Thank you for reading.

答:
I was thinking about this, and actually it seems that it is true, at least locally.

It is actually the same construction for the canonic form of flows near non singular points (for an autonomous vector field).

Say that you have a smooth curve $t\rightarrow \gamma(t)\in \mathbb{R}^2$ for $t\in I=(-\epsilon, \epsilon)$ with $\gamma(0)=(0,0)$ and $\gamma'(0,0)=(1,0)$. Define the function $F:I\times\mathbb{R}\rightarrow \mathbb{R}^2$ by $F(x,y)=\gamma(x)+(0,y)$. We have that $F(x,0)=\gamma(x)$ and $DF(0,0)=I_2$. So $F$ is a diffeomorfism between two open sets $U\subseteq I\times\mathbb{R}$ and $V\subseteq\mathbb{R}^2$ both containing $(0,0)$. Now if we take the constant vector field $v(x,y)=(1,0)$ on $U$, we note that tha curve $\alpha(t)=(t,0)$ is an integral curve for $v$, so $F\circ\alpha$ is an integral curve for the push forward vector field $F_*v$ on $V$, but $F\circ\alpha=\gamma$.

Title: Pathological vector fields


问:
Process of proving that an odd and even recursive sequence is converging
We have recursively given sequence:
$$
a_1 = 0\\
a_{2n}=\frac{1}{2}a_{2n-1}\\
a_{2n+1}=a_{2n}+\frac{1}{2}
$$
What I did:
I separated the two analysis of two sequences into even and odd sequence convergence tests, if they are equal, then the whole sequence converges.

$$a_{2n}=\frac{1}{2}a_{2n-1}=\frac{1}{2}(a_{2n-2}+\frac{1}{2})$$
We now prove that the sequence is increasing:
$$a_{2(n+1)}\geq a_{2n}$$
Base:
$$a_2 = 0, a_4 = \frac{1}{4}\implies a_4 \geq a_2$$
Induction step:
$$a_{2n}\geq a_{2n-2}\\
\frac{1}{2}(a_{2n-2}+\frac{1}{2})\geq a_{2n-2}\\
\vdots\\
\frac{-1}{2}a_{2n-2}\geq\frac{-1}{4} \iff a_{2n-2}\leq \frac{1}{2}
$$
So now we know that the even sequence increases if the terms are bellow $\frac{1}{2}$.
Let's see if that is true:
$$a_{2n}\leq \frac{1}{2}\\
a_0=0\leq\frac{1}{2}\\
a_{2n+2}=\frac{1}{2}(a_{2n}+\frac{1}{2})\leq \frac{1}{2}(\frac{1}{2}+\frac{1}{2})=\frac{1}{2}
$$
Now we know that the sequence is increasing and bounded above, so the limit must exist:
$$
\lim_{n\rightarrow \infty}a_{2n+2} = \frac{1}{2}(\lim_{n\rightarrow \infty}a_{2n}+\frac{1}{2})\\
\vdots\\
2a = a + \frac{1}{2}\\
a = \frac{1}{2} = L_1
$$
Proof ended for even sequences. Then I would proof for odd sequences in the same way and see if $L_1$ is equal to $L_2$, which I already know it isn't since I already solved the whole problem.

But my question is, is my process of proving OK? Are there any mistakes ?

(the dots and other trivial calculations that I skipped would be of course sufficiently done)

答:
What you have done is correct. You can even show that $a_{2n} = 1/2 - (1/2)^n$. This implies that $a_{2n-1} = 2a_{2n} = 1 - (1/2)^{n-1}$. Hence $a_{2n} \to 1/2$ and $a_{2n-1} \to 1$.

But is immediately clear, without the above formulae, that $(a_n)$ does not converge. If it would, then $a = \lim_{n \to \infty} a_{2n+1} = \lim_{n \to \infty} (a_{2n} + 1/2) = a +1/2$ which is absurd.


问:
Expected number of coin tosses until same consecutive output
I looked a lot for the answer to this question online, with no success.

The coin used is not fair, and has: $\displaystyle P( H) =\frac{2}{3}$. $\displaystyle P( T) =\frac{1}{3}$.

I wrote a solution that makes sense to me, but the overall answer is kinda weird, so I would love if anyone could solve/review my answer.

My solution:

Let $\displaystyle W=number\ of\ tosses\ until\ same\ consecutive\ output\ ( including)$.

Let's look at the following complementary events:

$\displaystyle HH,\ HT,\ TT,\ TH$.

From Total Law of Examplaction:

$\displaystyle  \begin{array}{{>{\displaystyle}l}}
E[ W] =P( HH) \cdotp E[ W|HH] +P( TT) \cdotp E[ W|TT] +P( HT) \cdotp E[ W|HT] +P( TH) \cdotp E[ W|TH]\\
=\frac{4}{9} \cdotp 2+\frac{1}{9} \cdotp 2+\frac{2}{9} \cdotp ( 2+E[ W]) +\frac{2}{9} \cdotp ( 2+E[ W])\\
=2+\frac{4}{9} \cdotp E[ W]\\
\Longrightarrow \\
\frac{5}{9} E[ W] =2\\
E[ W] =\frac{18}{5}
\end{array}$

Which rounds up to $\displaystyle 4$.

Is the logic behind even correct? How am I not getting an integer as anwer?

答:
(N.B.  this question is a duplicate, but I think the simple recursive argument presented here is worth recording).

Suppose the coin comes up $H$ with probability $p$.

Let $E_H$ (resp. $E_T$) denote the expected number of tosses to get a match, given that the prior toss was $H$ (resp. $T$).

Then we see that $$E_H=p\times 1 +(1-p)\times (E_T+1)\quad \& \quad E_T=(1-p)\times 1+p\times (1+E_H)$$

Which yields $$E_H=\frac {2-p}{p^2-p+1}\quad \&\quad E_T=\frac {p+1}{p^2-p+1}$$

Letting $E$ be the expectation from the start (prior to the first toss) we see that $$E=p\times (E_H+1)+(1-p)\times (E_T+1)$$

and combining the results completes the computation:  $$E=\frac 3{p^2-p+1}-1$$

Sanity check:  Note that this correctly yields $3$ when $p=\frac 12$

When $p=\frac 23$ as in your specific question, we get $E=\frac {20}7$.  As mentioned in the comments, there is no reason to expect the answer to be an integer.


问:
Number of messages per node in a fully connected graph
I have a fully connected graph $G$. Each node in the graph sends 1 message per second to another randomly selected node. The message is not forwarded directly to the destination node, but instead, the sender node selects 3 random intermediate nodes through which the message should go before reaching the destination (e.g, if node A wants to send a message to node B, A selects three intermediate nodes $X_1, X_2, X_3$ and sends the message as A->$X_1$->$X_2$->$X_3$->B). How can I calculate what is the average throughput of a single node, i.e., how many messages per second a single node receives?

答:
Think about the starting node for each message actually sending $4$ small messages, each of which are received by a separate node (intermediate or end). Thus, if the number of nodes is equal to $n$, there are $4n$ small messages sent per second. However, since each sent message is received, the rate at which messages are received must be the rate at which messages are sent. Thus, there are also $4n$ small messages received per second on average, so the number per node is $\frac{4n}{n} = \boxed{4\text{ messages per second.}}$

Another argument:

Note that the probability of any given node receiving a message sent by another given node is $\frac{4}{n-1}$, because the sending node must choose $4$ out of the $n-1$ remaining nodes to route its message through. Then, since the rate of messages received is the same as the rate sent, namely $1$ per second, the node gets an expected value of $\frac{4}{n-1}$ messages per second from the other given node. Multiplying this by the $n-1$ possible sending nodes, the node receives $(n-1)\frac{4}{n-1} = \boxed{4\text{ messages per second.}}$


问:
Pythagors Theorem : Geometrical interpretation
If x,y,z are the 3 sides of a right angled triangle,
where say, x = hypo, y=vertical, z=base then
I learnt that the
area of a square on hypo = area of square on vertical + area of square on base
i.e $$x^2 = y^2 + z^2$$
That's the usual picture students are taught at elementary math.
I never thought that this also means :
area of a circle(x=dia) on hypo = area of circle(z=dia) on  base + area of circle(y=dia) on base

Question :

Is it true for any other regular polygons ?

Am I right when I say x,y,z can be rational or irrational?

Is there any connection between Pythagoras Theorem and FLT ?

答:
First a right triangle is NOT  a "regular polygon" itself so asking if it is true for "other regular polygons" makes no sense.  Second, yes, that is true whether the side lengths are rational or irrational.  For example $3^2+ 4^2= 25= 5^2$ where 3, 4, and 5 are rational (in fact integer) while $\pi^2+ \pi^2= (\sqrt{2}\pi)^2$ where $\pi$ and $\sqrt{2}\pi$ are irrational.

It is easy to see that the equation $x^2+ y^2= z^2$ has integer solutions, x= 3, y= 4, z= 5, above.  "FTL" says that the equation $x^n+ y^n= z^n$, with n a positive integer larger than 2, has NO integer solutions.


Title: Logarithmic Equations

Body:
How does one go about solving:

$(5x+2)^{\frac{4}{3}} = 16$

I'm confused as how to parse through the equation to solve it using logs.

Body:
$$(5x+2)^{\frac{4}{3}} = 16$$
$$\left((5x+2)^{\frac{4}{3}}\right)^{\frac{3}{4}} = 16^{\frac{3}{4}}$$
$$(5x+2)^{\frac{4}{3}\cdot \frac{3}{4}}=2^{4\cdot \frac{3}{4}}$$
$$5x+2=2^3$$
$$5x+2=8$$
$$x=\frac{6}{5}$$


问:
Find $\frac{\partial (x,y)}{\partial (u,v)}$
Given $$x=\frac{u+v}{1-uv}$$ and $$y=\arctan\left(u\right)+\arctan\left(v\right)$$

Then find $$\frac{\partial (x,y)}{\partial (u,v)}$$

I don't understand the notation $(x,y)$ or $(u,v)$,since I've always seen a function instead of $(x,y)$ and a variable instead of $(u,v)$,can someone help?

答:
I don't understand the notation $(x,y)$ or $(u,v)$,since I've always seen a function instead of $(x,y)$ and a variable instead of $(u,v)$,can someone help?

It could be confusing if you're thinking about the standard notation for a partial derivative of a function of several variables, but it's not that different: $x$ and $y$ are in fact functions of $u$ and $v$.

The notation
$$\frac{\partial (x,y)}{\partial (u,v)} \tag{1}$$
refers to the Jacobian (determinant) which is, in this case:
$$
\det\begin{bmatrix} \dfrac{\partial x}{\partial u} & \dfrac{\partial x}{\partial v} \\[4pt] \dfrac{\partial y}{\partial u} & \dfrac{\partial y}{\partial v} \end{bmatrix}\tag{2}
$$
So you should remember $(1)$ as a (compact) notation for $(2)$.

I would expect your text(book) to introduce this notation, if you're giving this kind of exercise.


答:
For your first question, Pythagoras theorem works for any object as long as their side lengths are similar

Reference here:

https://betterexplained.com/articles/surprising-uses-of-the-pythagorean-theorem/#:~:text=The%20Pythagorean%20Theorem%20can%20be,formula%20that%20squares%20a%20number.

For your second question, Yes you are right a simple example of this is:

$x = \sqrt{\sqrt{2}}$, $y = \sqrt{\sqrt{3}}$

$\sqrt{2} + \sqrt{3} = c^2$

$c = \sqrt{\sqrt{2} + \sqrt{3}}$

All numbers are irrational.

For your last question on FLT there is a connection with Pythagorean triplets:

https://www.math.mun.ca/~drideout/pytrip06.pdf

Title: Pythagors Theorem : Geometrical interpretation

Body:
If x,y,z are the 3 sides of a right angled triangle,
where say, x = hypo, y=vertical, z=base then
I learnt that the
area of a square on hypo = area of square on vertical + area of square on base
i.e $$x^2 = y^2 + z^2$$
That's the usual picture students are taught at elementary math.
I never thought that this also means :
area of a circle(x=dia) on hypo = area of circle(z=dia) on  base + area of circle(y=dia) on base

Question :

Is it true for any other regular polygons ?

Am I right when I say x,y,z can be rational or irrational?

Is there any connection between Pythagoras Theorem and FLT ?


问:
If $f(a) \leq 0$, $f(b) < 0$ and $\int_{a}^b f(x)\text{dx} \geq 0$ for $f: [a,b] \to \mathbb{R}$, then $f'(x_0) = 0$.
Let $a,b \in \mathbb{R}$ with $a < b$ and $f : [a,b] \to \mathbb{R}$ be differentiable with
$$
f(a) \leq 0, \quad f(b) < 0 \quad \text{and} \quad \int_{a}^b f(x)\text{dx} \geq 0.
$$
I have to show that there exists a $x_0 \in (a,b)$ with $f'(x_0) = 0$.

Here is my attempt: From the mean-value theorem I know that there exists a point $c \in (a,b)$ such that
$$
f'(c) = \frac{f(b) - f(a)}{b-a}
$$

and from the mean value theorem for definite integrals I know that there exists some point $\xi \in [a,b]$ such that
$$
\int_{a}^b f(x)\text{dx} = f(\xi)(b-a).
$$
From here I don't know what to do next. Can someone give me a tip how I can continue? The case $f (a) = f (b)$ is clear, so that we can assume that $f (a) \neq f(b)$.

答:
well if:
$$\int_a^bf(x)dx\ge0$$
then that means that the mean value of $f$ over $[a,b]$ is also greater than or equal to zero i.e. $\bar{f}_{a,b}\ge0$
this means there exists a domain $[c,d]$ where $f>0$ where $a<c<d<b$. Now since $f(b)<0$ there has to be a maximum in $[c,d]$, which means there is also a maximum in $[a,b]$ i.e. a point where $f'(x_0)=0$


问:
If $X\subset \mathbb{R}$ is bounded and a infinity set, then $X \cap X^\prime \not = \emptyset$.
First of all, let us define $X^\prime = \{a \in \mathbb{R} \mid \forall\epsilon>0, ((a - \epsilon, a + \epsilon)-\{a\}) \cap X \not = \emptyset\}$. The result is true when $X$ is an uncountable set: if $X \cap X^\prime = \emptyset$, then for all $x \in X$ there is an $\epsilon_x>0$ such that $((x - \epsilon_x, x + \epsilon_x)-\{x\}) \cap X = \emptyset$. It implies that all points of $X$ are isolated points. Then, $X$ should be a countable set (a contradition). So, $X \cap X^\prime \not = \emptyset$.
But, what happens when $X$ is a countable set?

答: Consider $X = \left\{\frac{1}{n} : n \in \Bbb N\right\}.$ Then, $X' = \{0\}$ and $X \cap X' = \varnothing$.


答:
As far as I understand, $$\frac{\partial (x,y)}{\partial (u,v)}$$ means, that you have to find the determinant of a Jacobian matrix, given $x(u,v)$ and $y(u,v)$, i.e.:

$$\frac{\partial (x,y)}{\partial (u,v)} =
\begin{vmatrix}
\frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\
\frac{\partial y}{\partial u}  & \frac{\partial y}{\partial v}
\end{vmatrix}.$$
Taking partial derivatives of your given $x(u,v)$ and $y(u,v)$.

Title: Find $\frac{\partial (x,y)}{\partial (u,v)}$

Body:
Given $$x=\frac{u+v}{1-uv}$$ and $$y=\arctan\left(u\right)+\arctan\left(v\right)$$

Then find $$\frac{\partial (x,y)}{\partial (u,v)}$$

I don't understand the notation $(x,y)$ or $(u,v)$,since I've always seen a function instead of $(x,y)$ and a variable instead of $(u,v)$,can someone help?


问:
what is the formula for this function?
I've been thinking about the differences in numbers so for example:

$\begin{array}{ccccccc} &&&0&&&\\ &&1&&1&&\\&1&&2&&3&\\0&&1&&3&&6
\end{array}$

or with absolute differences:

$\begin{array}{ccccccccccc}
&&&\vdots&&\vdots&&\vdots\\
&&1&&0&&2&&2\\
&1&&2&&2&&4&&2\\
2&&3&&5&&7&&11&&13
\end{array}$

so I found this and I wanted to know if this has an actual mathematical formula:

$\begin{array}{cccccccc}
&&&&\vdots\\
&&&3&\cdots&\vdots\\
&&2&&5&\cdots&\vdots\\
&1&&3&&8&\cdots&\vdots\\
0&&1&&4&&12&\cdots
\end{array}$

and the sequence continues  0, 1, 4, 12, 32, 80, 192, 448, 1024, 2304, 5120, 11264, 24576, 53248, 114688, 245760,... If you couldn't tell what I was doing to generate them the first diagonal going up from the 0 increased by one. I wasn't able to find the formula, but I have found some other diagonals have properties like, the second (1,3,5...) is the odd numbers, all of those diagonals are linear.

Title: what is the formula for this function?

Body:
It seems to be the number of edges in N-dimensional cube:

4 = 1 * 2 + 2 (square)
12 = 4 * 2 + 4 (ordinary cube)
32 = 12 * 2 + 8 (4-D cube)
80 = 32 * 2 + 16
192 = 80 * 2 + 32
448 = 192 * 2 + 64
1024 = 448 * 2 + 128
2304 = 1024 * 2 + 256
5120 = 2304 * 2 + 512
11264 = 5120 * 2 + 1024
24576 = 11264 * 2 + 2048
53248 = 24576 * 2 + 4096
114688 = 53248 * 2 + 8192
245760 = 114688 * 2 + 16384
524288 = 245760 * 2 + 32768


问:
Best book of topology for beginner?
I am a graduate student of math right now but I was not able to get a topology subject in my undergrad... I just would like to know if you guys know the best one..</p>

答: Topology For Beginners. Dr Steve Warner</p>

Title: Best book of topology for beginner?

Body: Topology For Beginners. Dr Steve Warner


问:
show that the set of orthogonal matrices is bounded
I'm trying to demonstrate that the set of orthonormal matrices $\mathcal{O}(n,m) =\{ A : A^TA = I_m\} $ is bounded but sort of lost

答: Note that the columns must form an orthonormal basis, so no entry can have absolute value greater than 1.</p>


问:
Is this type of a transformation possible in propositional logic?
Let $\Phi_1(A,B)$ be a propositional logic formula consisting only of propositions $A$ and $B$,
similarly we have $\Phi_2(C,D)$. Now lets say we have a formula as follows:
$$(M \leftrightarrow \Phi_1(A,B)) \land (N \leftrightarrow \Phi_2(C,D)) $$

Where $M$ and $N$ are propositional variables as well.
Is there any transformation possible which converts this form into a formula of the following form :
$$\Phi_1'(M,N) \land \Phi_2'(A,B,C,D)$$

The transformed formula must be equivalent to the original formula.
Where $\Phi_1'(M,N)$ consists only of $M$ and $N$ and similarly for $ \Phi_2'(A,B,C,D)$.
No new propositional variables should be introduced.

Corrections made in the problem, thanks to @Alex Kruckman's comment

答:
No such transformation is possible unless $\Phi_1$ and $\Phi_2$ are constant. WLoG, let $A_1, B_1$ be such truth values that $\Phi_1(A_1, B_1)$ is true, and $A_2, B_2$ be such truth values that $\Phi_1(A_2, B_2)$ is false. Moreover, fix $N, C, D$ such that $N \leftrightarrow \Phi_2(C, D)$.

Then, the following are true:
$$(\top \leftrightarrow \Phi_1(A_1, B_1)) \land (N \leftrightarrow \Phi_2(C, D))$$
$$(\bot \leftrightarrow \Phi_1(A_2, B_2)) \land (N \leftrightarrow \Phi_2(C, D))$$

By the equivalence you're asking for, the following are also true:
$$\Phi_1'(\top, N) \land \Phi_2'(A_1, B_1, C, D)$$
$$\Phi_1'(\bot, N) \land \Phi_2'(A_2, B_2, C, D)$$

Thus all of the conjuncts are true and so is:
$$\Phi_1'(\bot, N) \land \Phi_2'(A_1, B_1, C, D)$$

By the equivalence again,
$$(\bot \leftrightarrow \Phi_1(A_1, B_1)) \land (N \leftrightarrow \Phi_2(C, D))$$

But, by assumption, we have $\top \leftrightarrow \Phi_1(A_1, B_1)$, so $\top \leftrightarrow \bot$, which is a contradiction.