问:
Upward Löwenheim-Skolem theorem without equality
I am confused about the following argument for a simplified proof of the upward Löwenheim-Skolem theorem for first-order logic without equality:

Let $\mathcal M$ be an infinite first-order structure. Given some element $a\in M$, we add a family of new elements $a_i$, $i\in I$ and stipulate that they behave as $a$ for every function and predicate symbol, creating a new structure $\mathcal M'$. Now, $\mathcal M$ and $\mathcal M'$ satisfy the same equality-free formulas if we allows variables assignments in $M$.

I don't quite understand how it is necessary that $\mathcal M$ is infinite here. But if the restriction would be dropped, then the result would be that any satisfiable set has models of arbitrary large cardinalities.

答:
Indeed, the requirement that $\mathcal{M}$ be infinite is superfluous, and the more general result you suggest holds: if $T$ is a satisfiable set of first-order sentences without equality, then $T$ has arbitrarily large models. The proof goes through unchanged.

In more detail, given $\mathcal{M},M,a,I$ as in the question, we define $\mathcal{M}'$ as follows:

The underlying set $M'$ of $\mathcal{M}'$ is $M\sqcup\{\langle a,i\rangle: i\in I\}$.

Let $i:M'\rightarrow M$ be the identity on $M$ and send $\langle a,i\rangle$ to $a$; basically, $i$ is the "collapse" map corresponding to the "blow-up" we're performing. If we were to "blow-up" more than one element of $\mathcal{M}$, which we could but don't need to, this $i$ would correspondingly be (slightly) more interesting.

For each $n$-ary relation symbol $R$ in the language of $\mathcal{M}$ we set $$R^\mathcal{M'}=\{(c_1,...,c_n)\in ((M')^n: (i(c_1),...,i(c_n))\in R^\mathcal{M}\}.$$

For each $n$-ary function symbol $f$ in the language of $\mathcal{M}$ we set $$f^\mathcal{M'}(c_1,...,c_n)=f^\mathcal{M}(i(c_1),...,i(c_n)).$$ (Note that $M\subseteq M'$: we've added new things corresponding to $a$, but we haven't gotten rid of $a$ itself. So the output of $f^\mathcal{M'}$ is in fact guaranteed to be an element of $\mathcal{M}'$.)

It's then easy to check that $\mathcal{M'}\equiv_{\mathsf{w/o=}}\mathcal{M}$.


问:
Using Inclusion-Exclusion, probability of occurrence of exactly $m$ events out of $\{A_1,\ldots,A_n\}$
Using Inclusion-Exclusion, prove that the probability of occurrence of exactly $m$ events out of $\{A_1,\ldots,A_n\} = \sum_{i=0}^{n-m} (-1)^i   \binom{m+i}{m} P_{m+i} $ where $P_k = \sum_{1 \leq i_1<\cdots \le i_k \leq n} P(A_{i_1} \cap\cdots \cap A_{i_k})$

Attempt:  I was trying to understand this term on the RHS : $\binom{m+i}{m} P_{m+i}$. And then the inclusion-exclusion principle shall follow.

This is what I thought to be able to use inclusion-exclusion but I am not very sure if this is the right approach.

$m$ events can occur either from the first $m$ events, or from the first $m+1$ events, or from the first $m+2$ events, $\cdots$, or from the n events.

$$E_k = \text{the $m$ events occur from the first $m+k$ events}$$

Then $P(E_k) = \binom{m+k}{m} \sum_{1\leq i_1<\cdots i_m\leq m+k} P(A_{i_1} \cap \cdots  \cap A_{i_m})$

But, then I could not move forward from here. I tried understanding the solution given here, but I would really appreciate something more intuitive.

Could someone please explain how do I move forward from here?

答: I discovered a marvelous proof. However, since it's long, I haven't been able to type it. Posting clear images for the sake of anyone who might be searching for the same answer.

ContentLicense: CC BY-SA 4.0


问:
Show that $n^2\log\left(1+\frac{1}{n}\right)$ does not converge to $1$

How to show that $n^2\log\left(1+\dfrac{1}{n}\right)\to 1$ is false?

I have to show that $\left(1+\dfrac{1}{n}\right)^{n^2}$ doesn't tend to $e.$

ContentLicense: CC BY-SA 3.0

Body: $$n^{2}\log \left( 1+\frac{1}{n}\right) \sim n.$$


问:
My attempt at $S = \overline{S}$ iff $S$ is closed
$E$ is a metric space and $S$ a set in it.

$\overline{S}$ denotes the closure of $S$. The definition for closure that I am using is:

The closure of a set $A$ is the intersection of all closed sets containing $A$. More explicitly:

$\displaystyle \overline{A} = \bigcap_{i \in I} C_i $ , where for each $i \in I$ the set $C_i$ is closed in $E$ and $S \subset C_i$.

$I$ is an index set.

Here is my write-up of a proof for the statement in the title. It seems a bit too facile and terse to me, so I wanted to ask for some feedback on its correctness.

Proof:

Suppose $S = \overline{S}$. The set $\overline{S}$ is an intersection of closed sets, and such intersections are always themselves closed. Therefore $S$ is closed.

Suppose now $S$ is closed. Since $S$ contains itself, it qualifies as $S = C_j$ for some $j \in I$.

$S$ is a subset of each $C_i$ so $S$ is necessarily in the intersection of all such sets, that is $S \subset \overline{S}$.

Next, for each $x \in \overline{S}$:

$x \in \overline{S} = \displaystyle \bigcap_{i \in I} C_i$ means $x \in C_i$ for each $i \in I$. In particular, $x \in C_j = S$. Therefore $\overline{S} \subset S$.

Hence $ \overline{S} = S$. We are done. $\blacksquare$.

Any thoughts or comments appreciated! Peace!

EDIT:

After José's feedback, I suppose the definition below is incorrect since it does not capture every possible closed set containing $S$:

More explicitly:

$\displaystyle \overline{A} = \bigcap_{i \in I} C_i $ , where for each $i \in I$ the set $C_i$ is closed in $E$ and $S \subset C_i$.

$I$ is an index set.

The index set is not specified, so if there are infinitely many closed sets containing $S$ but $I$ is finite, we're missing some sets.

I think this ought to be the correct way to state it explicitly:

$\overline{S} = \displaystyle \bigcap_{C \in A} C$

where $ C = \{  X \subset E: X \text{ is closed and } S \subset X \} $

答:
Actually, the second part of that definition of $\overline A$ doesn't make sense (and it is not equivalent to the first part). If the definition was as in the second part, a set could have several distinct closures (a closure for each family of closed subsets of $E$ which contain $A$). The correct definition is the one from the first part: $\overline A$ is the intersection of all closed sets which contain $A$.

Other than that, your proof is fine.


问:
Find whether $\sum_{n=1}^\infty\frac{(2n+1)!}{4^n+\frac{1}{π^n}}$ is convergent or divergent
In this exercise I have to find whether this series is convergent or divergent, and I do not really know what to do with denominator, that is why I would be grateful for any tips or solutions.

$$\sum_{n=1}^\infty\frac{(2n+1)!}{4^n+\frac{1}{π^n}}$$

答:
Big hint: Consider the general term:

its numerator is clearly $> \ n!$

its denominator is less than $ \ < 5^n$. (you will have to provide a rigorous proof)

Therefore the general term is $>\dfrac{n!}{5^n}$ which tends to $\infty$...

As a consequence, the given series is divergent.


问:
Infinite product of $\prod_{n=0}^{\infty }(1+\frac{1}{4^{n}})$
Is there a closed formula for the infinite product of $$\prod_{n=0}^{\infty }\left(1+\frac{1}{4^{n}}\right)\:?$$

答:
In terms of the $q$-Pochhammer symbol $(a,\,q)_n$, if $|r|<1$ then$$\prod_{n\ge0}(1-r^n)=(r;\,r)_\infty.$$Since $1+\tfrac{1}{4^n}=\tfrac{1-\tfrac{1}{16^n}}{1-\tfrac{1}{4^n}}$, your product is $\frac{(\tfrac{1}{16};\,\tfrac{1}{16})_\infty}{(\tfrac14;\,\tfrac14)_\infty}$.

As @Startwearingpurple notes, we can instead write$$\prod_{n\ge0}(1-r^n)=r^{-1/24}\eta(\tfrac{-i\ln r}{2\pi})$$in terms of the Dedekind eta function, so your product is $2^{1/12}\frac{\eta\left(\tfrac{2i\ln2}{\pi}\right)}{\eta\left(\tfrac{i\ln2}{\pi}\right)}$.


问:
Binomial coefficients identity : $\sum_{k=1}^{n-m+1} k\binom{n-k+1}{m}=\binom{n+2}{m+2}$
For any positive integer m&n.$n\ge m$ , let $\left( {\begin{array}{*{20}{c}}
n\\
m
\end{array}} \right) = {}^n{C_m}$. Prove that $\left( {\begin{array}{*{20}{c}}
n\\
m
\end{array}} \right) + 2\left( {\begin{array}{*{20}{c}}
{n - 1}\\
m
\end{array}} \right) + 3\left( {\begin{array}{*{20}{c}}
{n - 2}\\
m
\end{array}} \right) + .. + \left( {n - m + 1} \right)\left( {\begin{array}{*{20}{c}}
m\\
m
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
{n + 2}\\
{m + 2}
\end{array}} \right)$

My approach is as follow
$n=4, m=2$

$\left( {\begin{array}{*{20}{c}}
4\\
2
\end{array}} \right) + 2\left( {\begin{array}{*{20}{c}}
3\\
2
\end{array}} \right) + 3\left( {\begin{array}{*{20}{c}}
2\\
2
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
6\\
4
\end{array}} \right) = 15$

$6 + 6 + 3 = 15 \Rightarrow LHS = RHS$

But not able to solve it via property

Body:
Starting from

$$\sum_{k=1}^{n-m+1} k {n-k+1\choose m}$$

we re-write as

$$\sum_{k=1}^{n-m+1} k {n-k+1\choose n-m+1-k}
= [z^{n-m+1}] (1+z)^{n+1}
\sum_{k=1}^{n-m+1} k \frac{z^k}{(1+z)^k}$$

Here the coefficient extractor enforces the upper limit of the sum and
we obtain

$$[z^{n-m+1}] (1+z)^{n+1}
\sum_{k\ge 1} k \frac{z^k}{(1+z)^k}
\\ = [z^{n-m+1}] (1+z)^{n+1}
\frac{z/(1+z)}{(1-z/(1+z))^2}
= [z^{n-m+1}] (1+z)^{n+1}
z (1+z)
\\ = [z^{n-m}] (1+z)^{n+2} = {n+2\choose n-m}
= {n+2\choose m+2}$$

as claimed.


问:
Maximum Sum of Character Stats with Multiple Categories
So I've got a type of generic question about a topic I'm not sure how to word (so I struggled to find it online) but I'll give a specific example first to help describe it.

Consider you have $30$ characters, each with a score in $6$ different categories? Is there some mathematics method to help work out which characters in which category yields the highest overall sum? I'm proposing to code an algorithm to do this so I'll refer to the first character as character $0$ and the last character as character $29$. Also, I wasn't sure which Stack Exchange site to post this question on, I'm sure there's some mathematical theory to benefit me, I just don't know what it would be called.

Eventually, I want to get up to $4$ characters in each of the $6$ categories. So far I've managed to do it by brute force (in reasonable time) for only the first $4$ categories or so, that is trying all of $(0,0,0,0)$ through to $(29,29,29,29)$ and ignoring any coordinate with some repeating integer. Clearly using brute force for the $16$ characters would result in going through $\frac{30!}{14!}\approx 3.04 \times 10^{21}$ from $M_{1 \times 16}(0)$ through to $M_{1 \times 16}(29)$, again ignoring any coordinate with a repeating integer. Not shockingly, way too many numbers for my computer to find a maximum in.

And then by working out a running maximum and the maximum score in each category, I was able to significantly make my code more efficient by dismissing any (say) first $i$ coordinates such that adding the absolute maximum score in each of the remaining categories would still make it less than the largest actual (non-repeating) combination sum found yet. So say that $(0,5,7)$ had a total so far of $1000$ and the sum of the absolute largest scores in each of the remaining $3$ categories is $1000$, but I've already found an actual (non-repeating) sum of $2100$, then I'd dismiss any coordinates $(0,5,7,x,y,z)$ with $x,y,z\in\{1,2,3,4,6,8,9,...,29\}$ and $x\neq y\neq z \neq x$.

This worked in reasonable time for $6$ characters ($1$ in each category) and then further for $8$ characters ($3$ characters for the first category and $1$ character in each of the remaining categories). Also, note that as I get higher with the number of characters it becomes impossible for me to verify that my code is yielding the largest possible actual (non-repeating) sum.

This is similar to having $>4$ swimmers each with a personal best time in each of the $4$ swimming strokes and try to work out the medley relay (one swimmer for each stroke) with the fastest overall time. I've recently started learning graph theory including proper vertex colouring where you can arrange (say) exams in a way with the least possible number of time slots so that no student has two exams at the same time. This question seems somewhat similar.

So does anyone know if this is a common type of question? In which case I'd appreciate a simple point in the right direction. Or rather could someone give me some tips on answering this specific question more efficiently.

答:
This can be solved using integer linear programming, for which there are a large number of exact and approximation algorithms. Number the people $0$ to $29$ and the categories $0$ to $5$. Let $x_{i,j}$ be a variable which is $1$ if you put person $i$ in category $j$, and zero otherwise. You want to maximize
$$
\sum_{i,j}s_{i,j}x_{i,j}
$$
where $s_{i,j}$ is the score in the $j^{th}$ category of person $i$ (thanks to RobPratt for catching a mistake I had here). This is constrained by
$$
\sum_{i=0}^{29}x_{ij}=4\qquad \text{for each }j\in \{0,\dots,5\}\\
\sum_{j=0}^{5}x_{ij}\le 1\qquad \text{for each }i\in \{0,\dots,29\}\\
x_{i,j}\ge 0
$$
The first batch of constraints ensure that there are $4$ people chosen for each category, and the second batch ensures each person is put in at most one category.


问:
About actions of ${\rm Aut}(G)$, conjugacy classes and inner automorphisms of said group
Let $G$ be a group, show that $\text{Aut}(G)$ acts on the set $\mathscr C$ of the conjugacy classes of $G$ with the action given by $\varphi(\mathscr C_x)=\mathscr C_{\varphi(x)}$ with $\varphi\in\text{Aut}(G)$ and $\mathscr C_x$ the class of $x$.

Also we have to show that $\mathscr F(G)=\{\varphi\in\text{Aut}(G)\mid \varphi(\mathscr C_x)=\mathscr C_x\}$ is normal in $\text{Aut}(G)$ and that $\mathscr F(G)$ contains the inner automorphisms given by conjugation.

__

This was an assignment for my group theory course. I've done this course, rep theory and rings and modules.
The problem I'm having is that I don't know how to tackle this exercise with the things we've been taught on the group theory course, because we skimmed over the automorphisms (and their actions) part, and focused more on cyclic and Sylow groups and that kind of stuff.

I've tried seeing that the difference of two elements in the same coclass is sent into that class itself and to expand the product of the action, but can't go further than that

答:
As the action is on a set of classes (necessarily dealt with by means of representatives), we have first to prove that the map is well defined, namely that $x'\in \mathscr C_x \Rightarrow \varphi\cdot\mathscr C_{x'}=\varphi\cdot\mathscr C_{x}$. This in indeed the case, because:

\begin{alignat}{1}
\mathscr C_{\varphi(x')} &= \{g\varphi(x')g^{-1}, g\in G\} \\
&= \{g\varphi(g'xg'^{-1})g^{-1}, g\in G\} \\
&= \{g\varphi(g')\varphi(x)\varphi(g')^{-1}g^{-1}, g\in G\} \\
&= \{(g\varphi(g'))\varphi(x)(g\varphi(g'))^{-1}, g\in G\} \\
&= \{g''\varphi(x)g''^{-1}, g''\in G\} \\
&= \mathscr C_{\varphi(x)} \\
\end{alignat}

where the last but one equality follows from $g\mapsto g\varphi(g')$ being onto $G$, for every $g'\in G$. Next, $\varphi\cdot\mathscr C_x\in \mathscr C$ by definition. Furthermore, $Id_G\cdot \mathscr C_x=\mathscr C_{Id_G(x)}=\mathscr C_x$. Finally, $(\varphi\psi)\cdot\mathscr C_x=\mathscr C_{(\varphi\psi)(x)}=\mathscr C_{\varphi(\psi(x))}=\varphi\cdot(\psi\cdot\mathscr C_x)$. So, this is indeed a group action and $\mathscr F(G)=\operatorname{Stab}(\mathscr C_x)$ is a subgroup of $\operatorname{Aut}(G)$. Then:

\begin{alignat}{1}
\psi\operatorname{Stab}(\mathscr C_x)\psi^{-1} &= \{\psi\varphi\psi^{-1}\mid \varphi\cdot\mathscr C_x=\mathscr C_x\} \\
&= \{\rho\mid (\psi^{-1}\rho\psi)\cdot\mathscr C_x=\mathscr C_x\} \\
&= \{\rho\mid \rho\cdot (\psi\cdot\mathscr C_x)=\psi\cdot\mathscr C_x\} \\
&= \operatorname{Stab}(\psi\cdot\mathscr C_x)
\end{alignat}

so $\mathscr F(G)$ is not normal in $\operatorname{Aut}(G)$. Rather, here we get that the stabilizers of the points of one same orbit are all conjugate to each other. Last, if $\varphi\in\operatorname{Inn}(G)$, then $\exists g\in G$ such that $\varphi\cdot\mathscr C_x=\mathscr C_{\varphi(x)}=\mathscr C_{gxg^{-1}}=\mathscr C_x$, whence $\varphi\in\mathscr F(G)$. Therefore,  $\operatorname{Inn}(G)\le\mathscr F(G)$. (Note that this holds for every conjugacy class considered in $\mathscr F(G)$, and hence $\operatorname{Inn}(G)\le\operatorname{ker}\phi$, where $\phi$ is the homomorphism $\operatorname{Aut}(G)\to S_\mathscr C$ equivalent to your action.)

Comment. If the action is defined by $\varphi\cdot\mathscr C_x:=\mathscr C_{\varphi(x)}$, then the proof of the good definition focuses on the invariance w.r.t. representative's choice, being the membership of $\varphi\cdot\mathscr C_x$ to $\mathscr C$ trivial (by definition); this is the above approach. Conversely, if the action is defined by $\varphi\cdot\mathscr C_x:=\varphi(\mathscr C_x)$, then the proof of the good definition focuses on the membership of $\varphi\cdot\mathscr C_x$ to $\mathscr C$, being the invariance w.r.t. to representative's choice trivial (as $x'\in\mathscr C_x\Rightarrow \mathscr  C_{x'}=\mathscr C_x$). The effort in both approach is the same, and equal to proving that $\varphi(\mathscr C_x)=\mathscr C_{\varphi(x)}$.


问:
Be $I \subset \mathbb{R}$ a interval and $f: I \to \mathbb{R} $ a convex function.
Be $I \subset \mathbb{R}$ a interval and $f: I \to \mathbb{R} $ a convex function. Prove that if f has maximum value in a point $a \in int I$, then f is a constant function.

My answer:

Suppose that f is convex but isn't constant. Taking $a$ like the only point of maximum in I. Notice that how f is convex , we have the followed inequality that says:

\begin{align}
m < a < n \text{  in I} \Rightarrow \frac{f(a) - f(m)}{a - m} \leq \frac{f(n) - f(m)}{n- m} \leq \frac{f(a) - f(n)}{a - n}
\end{align}

But We had suppose that a was the only point of maximum, which implies in a contradition.

Correct?

答:
Let us prove it by contradiction.

In order to do so, we shall assume that $f$ is not constant.

If $a\in I$ is a local maximum, there exists a neighborhood $(a-\delta,a+\delta)$ such that $f$ is increasing in $(a-\delta,a)$ and $f$ is decreasing in $(a,a+\delta)$.

Take $x_{1}\in(a-\delta,a)$ and $x_{2}\in(a,a+\delta)$. Hence we can say that $f(x_{1}) < f(a)$ and $f(x_{2}) < f(a)$.

On the other hand, given that $f$ is convex, we should have
\begin{align*}
f(\lambda x_{1} + (1-\lambda)x_{2})\leq \lambda f(x_{1}) + (1-\lambda)f(x_{2})
\end{align*}

Given that $a\in[x_{1},x_{2}]$, there is some $\lambda_{0}$ such that $a = \lambda_{0}x_{1} + (1-\lambda_{0})x_{2}$. Thus we have that
\begin{align*}
f(a) \leq \lambda_{0}f(x_{1}) + (1-\lambda_{0})f(x_{2})
\end{align*}

On the other hand, we also have that
\begin{align*}
\lambda_{0}f(x_{1}) + (1-\lambda_{0})f(x_{2}) < \lambda_{0}f(a) + (1 - \lambda_{0})f(a) = f(a)
\end{align*}
which contradicts our previous claim.

Therefore $f$ must be constant, and we are done.


问:
Probability of a median being $x_i$
I am trying to derive a probability function and here are the assumptions.

Let $S = \{x_{(1)},x_{(2)},...,x_{(7)}\}$ be a set of distinct values that are ordered.

Let $S*=\{x_1^*,x_2^*,...,x_7^*\}$ be a replication of $S$, as in, each elements $x_i^*$ are randomly chosen from $S$ with equal probability.

ex), $S^*$ could be $\{x_1,...,x_1\}$, or it could be an exact replica of $S$, etc.

The sampling process are mutually independent.

My goal is to find the probability that the median of $S*$, say, $M(S*)$, is equal to $x_{(i)}$.

I would like to use the notation

$$B\left[j;n,p\right] ={n \choose j}p^j(1-p)^{n-j}$$

associated to the binomial distribution.

Here is my thought process.

To find $p(i)=Pr[M(S*)=x_{(1)}]$ we need at least 4 elements from $S*$ to be $x_{(1)}$ so

$$p(1) = \sum_{j=4}^7 B\left[j;7,\frac{1}{7}\right]$$

in other words,

$$\therefore = \sum_{j=0}^3 B\left[j;7,\frac{6}{7}\right]$$.

To find $p(2)$ we need at least the least 4 elements to be less than or equal to $x_{(2)}$, but take away the probability $p(1)$, so with similar argument I can get

$$p(2) = \sum_{j=0}^3 B\left[j;7,\frac{5}{7}\right]-p(1)$$

I am hoping that the general case would look something like

$$p(i) = \sum_{j=0}^3 \left(B\left[j;7,\frac{i-1}{7}\right]-B\left[j;7,\frac{i}{7}\right]\right)$$

but I am not too confident...

May I have some assistance please?

A different result is also more than welcome as long as it is accurate and makes sense.

Title: Probability of a median being $x_i$

Body:
There are $7^7=823543$ equally likely possibilities for the resampling with replacement.

This is a tractable number and it is possible just to count how many give which order statistic as the median:

 x_(1)  x_(2)  x_(3)  x_(4)  x_(5)  x_(6)  x_(7)
  8359  80809 196519 252169 196519  80809   8359

and so probabilities

     x_(1)      x_(2)      x_(3)      x_(4)      x_(5)      x_(6)      x_(7)
0.01015005 0.09812360 0.23862628 0.30620016 0.23862628 0.09812360 0.01015005

If you wanted this analytically, you want the probability that at least half of the sample is less than or equal to $x_{(n)}$ but are not less than or equal to $x_{(n-1)}$, i.e.  $$\sum\limits_{j=4}^7 {7 \choose j}\frac{n^j (7-n)^{7-j}}{7^7}- \sum\limits_{j=4}^7 {7 \choose j}\frac{(n-1)^j (8-n)^{7-j}}{7^7} \\= \frac1{7^7}\sum\limits_{j=4}^7 {7 \choose j}\left({n^j (7-n)^{7-j}}- {(n-1)^j (8-n)^{7-j}}\right)$$

or using R

n <- 1:7
(1-pbinom(3,7,n/7)) - (1-pbinom(3,7,(n-1)/7))
# 0.01015005 0.09812360 0.23862628 0.30620016 0.23862628 0.09812360 0.01015005

as already found


问:
Weighing scale error and the minimum number of same weights
Problem: Philip’s weighing scale has an error of ±10 grams. What is the smallest number of identical copies of a book that Philip should weigh together so that the error of one book will be ±0.5 grams?

Attempted method:
It's a contest math problem and there is no solution available on the internet. I started off with assuming that it's a manual weighing scale. I add a weight on one side and a book on the other side, the error is 10gm per book. If I add one more book to each side of the scale, the error is 20gm per 3 books. I tried this method but can't get to the right answer.

答:
Suppose the true weight is $W$, and you weigh $n$ copies of the same book. The weight you will see will be in the range
$$
nW\pm 10
$$
Dividing by $n$, the weight of a single book is now known to be
$$
W\pm \frac{10}n
$$
How should you choose $n$ so this is $W\pm 0.5$?


答:
Write the Euclidean distance (squared) as $((x_1,f(x_1)) - (x_2, g(x_2)))^2$.  Now you have a straightforward optimization in two variables, where you take derivatives, set to zero, ...

Here's a graph of your two functions:

where I've added the minimizing points (and joining line)... read on...

Pick an arbitrary point $x_1$ on $f(x)$ and another point $x_2$ on $g(x)$.  Draw a line between them.  That's the distance between those two points.  Right?

This Euclidean distance is a function of the two variables.  Since you tagged your question with "calculus" you surely must know how to find the (local) minimum of a function of two variables.

Apply that to your distance function.  (Actually, do it for the square of the distance function... simpler.)

You'll get many local minima and maxima, but the graphs will guide you to finding the global minimum.  (Also, you can use second-order derivatives.)

Here's a "closeup" of the solution:

Title: Shortest distance between two general curves

Body:
Have to find shortest distance between
$$f(x)=-0.000024729x^7+0.0011461x^6-0.019857x^5+0.15402x^4-0.46314x^3-0,068261x^2+2.2506x-8.3436$$
and
$$g(x)=0.000016282x^8-0.00094728x^7+0.022526x^6-0.28122x^5+1.9699x^4-7.6833x^3+15.691x^2-14.557x+0.22000$$

I need all the help I can get! Thanks.


问:
Roots of a certain sixth order polynomial
I am looking for the roots (or basically any information regarding them) of the sixth order polynomial
$$p(x):=ax^6+(a+1)x^4+2bx^3-b^2$$
for positive, real constants $a,b$.
Since $p(0)=-b^2<0$ and $\lim_{x\to\pm\infty}p(x)=+\infty$ we have at least two real roots $x_0,x_1$. Moreover, plotting the function for several parameters suggests, that the remaining four roots are non-real (i.e. two pairs of complex conjugate solutions).

Unfortunately this is as far as I got. I am especially interested in the complex solutions and their behavior for large parameters $a,b$.
Any idea on how to tackle this problem is very welcome!

答:
$$ p(x) = ax^6+(a+1)x^4+2bx^3-b^2$$
I earlier thought the polynomial was solvable, until I checked the Galois group of $6x^6+7x^4+10x^3-25 = 0$
$$ G \in [720,-1,1,"S6"]$$
If the Galois group of any polynomial above quartic is the Symmetric group, then the polynomial is unsolvable, and you don't need to ask for its root, but instead their approximation

Your polynomial is in two variable, so lucky enough some would be solvable, I wrote a program on pari/gp to check

for( a=-20, 20, for(b= -20, 20 , f= ax^6+(a+1)x^4+2bx^3-b^2; if(a!=0&&polisirreducible(f)&&polgalois(f)!=[720,-1,1,"S6"], print(a, " ", b, " " , polgalois(f)))))

The result is just

$$ 9 , -1 , [72, -1, 1, "F_36(6):2 = [S(3)^2]2 = S(3) wr 2"]$$
$$9 , 1 , [72, -1, 1, "F_36(6):2 = [S(3)^2]2 = S(3) wr 2"]$$

So if $a= 9$, $b= 1$ , then $9x^6+10x^4+4x^3-1 = $ is solvable and irreducible among range $ -20 \le a,b \le 20$

Now in case you still want to see a formula to $ ax^6+(a+1)x^4+2bx^3-b^2$ I can get it for you but would require a resolvent of degree $15$

EDIT:

R < x > := PolynomialRing (RationalField ()); f := 9x^6+10x^4+2*x^3-1; G := GaloisGroup (f); G; IsSolvable (G);

Working on magma, the result is

Permutation group $G$ acting on a set of cardinality $6$
Order = $72$ = $2^3 \cdot 3^2$

$(1, 2)(3, 4)(5, 6)$

$(1, 3, 6)$

$ (1, 3)$

true


问:
Is it true that for every prime number $p >3$, the next one will be less than $p+p/2$?
I know that for every prime $p$, the next prime is less than $2p$. Can we improve this statement? Can it be less than $(3/2)p$? What is the best function of $p$ for which this is true (for every prime, not just infinitively many)? (assume $p>3$, or $p>$ than a fixed $p$ if necessary).</p>

答:
The wikipedia article on "prime gap" has a good deal of useful information.

In particular, it give the history of the result
that $p_{n+1} < p_n+p_n^c$
for various values of $c$
(all less than 1)
for sufficiently large $n$.

The best so far is
$c < 3/4+\epsilon$
for any $\epsilon > 0$.

Another result is that
there is always a prime between
$n^3$
and
$(n+1)^3$
for sufficiently large $n$.


答: Your proof works ! but you have to say more. Starting with the double inequality you stated in your post, we have: $f(a) \ge f(m)$, and $f(a) \ge f(n)$ since $f(a)$ is the max value. Furthermore, $a - m > 0$ and $a - n < 0$. Thus $\dfrac{f(a) - f(m)}{a-m} \ge 0$, and $\dfrac{f(a) - f(n)}{a-n} \le 0$. So putting together: $0 \le \dfrac{f(m) - f(n)}{m-n} \le 0\implies \dfrac{f(m) - f(n)}{m-n} = 0\implies f(m) = f(n)=f(a)$. This means for any $m, n \in I$ and $m \neq n$, $f(m) = f(n)=f(a)$. Thus $f$ must be constant and this constant is $f(a)$.

Body:
Be $I \subset \mathbb{R}$ a interval and $f: I \to \mathbb{R} $ a convex function. Prove that if f has maximum value in a point $a \in int I$, then f is a constant function.

My answer:

Suppose that f is convex but isn't constant. Taking $a$ like the only point of maximum in I. Notice that how f is convex , we have the followed inequality that says:

\begin{align}
m < a < n \text{  in I} \Rightarrow \frac{f(a) - f(m)}{a - m} \leq \frac{f(n) - f(m)}{n- m} \leq \frac{f(a) - f(n)}{a - n}
\end{align}

But We had suppose that a was the only point of maximum, which implies in a contradition.

Correct?


问:
Proving that a set is a vector subspace of a complex vector space
I would like to solve the following question

W  = { $\left[\begin{array}{cc} a&b\\ c&d \\ \end{array}\right] $ | $ a = \overline{d}$ }

I need to show that

a. W is NOT a complex vector subspace of M$_2(C)$

b. W is a vector subspace of real vector space  M$_2(C)$

I do not understand the difference between the notations of M$_2(C)$ in both the parts because if we are talking about a real vector space then shouldn't it be  M$_2(R)$ in the second part ?

答: Every complex vector space is also a real vector space. In particular, ${\rm M}_2(\Bbb C)$. You have to check that $W$ is not closed under multiplication by complex scalar, but it is closed under multiplication by real scalar. So, once you have shown that $W$ is closed under multiplication by real scalar, it suffices to show that if $A \in W$ has $ad\neq 0$, then ${\rm i}A \not\in W$.


问:
Is every commutative ring having the invariant basis number property equivalent to AC?
The proof I know that every commutative ring has the invariant basis number property involves quotienting by a maximum ideal to get a field, and so reducing to the case where the commutative ring is a field, for which the result is already proven. I wondered if there might be a "direct" proof, and tried to formalize it like this.

The fact that every commutative ring has a maximal ideal is proven with Zorn's lemma, and I suspect is in fact equivalent to Zorn's lemma. If we work in a set theory without Zorn's lemma/AC, is it still true that every commutative ring has the invariant basis number?

答: We do not need axiom of choice to show that commutative rings have invariant basis number property. We can prove it with the help of exterior algebra $\wedge(V)$. Let $A$ be a commutative ring. First of all, it is obvious that if $m>n$, the $A$-module $\wedge^m(A^n)=0$ by considering a basis $p_1,p_2,...p_n$ and writing each element of $A^n$ in terms of the basis. Now we will show that $\wedge^n(A^n)$ is not $0$, by providing a surjective homomorphism onto $A$. Take a labelled basis $p_1,p_2,...p_n$ of $A^n$. Now for any permutation $\sigma$ in $S_n$ and elements $a_1,a_2,...a_n$ in $A$, send $a_1*p_{\sigma(1)} \wedge a_2*p_{\sigma(2)} .... \wedge a_n*p_{\sigma(n)}$ to $s(\sigma)*a_1a_2....a_n$ where $s(\sigma)$ is the sign of the permutation. Check that this uniquely extends to a surjective homomorphism $\wedge^n(A^n) \rightarrow A$ by bilinearity and the $v \wedge v=0$ property of the exterior algebra. Thus if $A$ is commutative and not trivial, $\wedge^n(A^n)$ isn't trivial. Now if $A^m$ and $A^n$ are isomorphic for some $m \neq n$, if we take $m>n$ WLOG, we see that there is a surjective homomorphism from $0=\wedge^m(A^n)=\wedge^m(A^m)$ to $A$, contradiction as $A$ is non-trivial. Thus non-trivial commutative rings satisfy IBN property.


问:
Solving $x\left(y^2-z^2\right)u_x-y\left(x^2+z^2\right)u_y+z\left(x^2+y^2\right)u_z=0$
Or $x\left(y^2-z^2\right)\frac{\partial u}{\partial x}-y\left(x^2+z^2\right)\frac{\partial \:u}{\partial \:y}+z\left(x^2+y^2\right)\frac{\partial \:u}{\partial \:z}=0$ if you are not familiar with $u_x$ and such

This is what I have:

$$\frac{dx}{x\left(y^2-\:z^2\right)}=\frac{dy}{y\left(x^2+z^2\right)}=\frac{dz}{z\left(x^2+y^2\right)}$$

$$\frac{dx}{x\left(y^2-\:z^2\right)}=\frac{-dy}{-y\left(x^2+z^2\right)}=\frac{dz}{z\left(x^2+y^2\right)}$$

$$\frac{dx-dy+dz}{x\left(y^2-z^2\right)-y\left(x^2+z^2\right)+z\left(x^2+y^2\right)}=ds$$

$$\frac{dx-dy+dz}{0}=ds$$

$$dx-dy+dz=0$$

The potential I find is $v\left(x,y,z\right)=x-y+z+C_1$ thus:

$$C_1=x-y+z$$

But now I tried to calculate $C_2$ but can't find a proper way to do that

Also, my $C_1$ is apparently wrong, I don't know why, because our teacher said that we can calculate it like that, that we just try to find a potential v(x,y,z) here (I'd be interested why we can do this though because I've applied this technique before and it worked without any problems)

答:
$$\frac{dx}{x\left(y^2-\:z^2\right)}=\frac{dy}{y\left(x^2+z^2\right)}=\frac{dz}{z\left(x^2+y^2\right)}$$
Mulitply all the fractions by $2$. And the numerator and denominator by $y$ for  the second fraction and by $z$ for the third fraction:
$$\frac{2dx}{x\left(y^2-\:z^2\right)}=\frac{2ydy}{y^2\left(x^2+z^2\right)}=\frac{2zdz}{z^2\left(x^2+y^2\right)}$$
$$\frac{2dx}{x\left(y^2-\:z^2\right)}=\frac{dy^2}{y^2\left(x^2+z^2\right)}=\frac{dz^2}{z^2\left(x^2+y^2\right)}$$
$$\frac{2dx}{x\left(y^2-\:z^2\right)}=\frac{dy^2-z^2}{x^2\left(y^2-z^2\right)}$$
$$2x{dx}={d(y^2-z^2)}$$
$${dx^2}={d(y^2-z^2)}$$
$$x^2-y^2+z^2=C_1$$

For the second constant of integration you can use the constant $C_1$
$$\frac{dx}{x\left(y^2-\:z^2\right)}=\frac{dy}{y\left(x^2+z^2\right)}$$
Use the fact that:
$$x^2-y^2+z^2=C_1$$
$$z^2=C_1-(x^2-y^2)$$
Then solve the differential equation.
$$\frac{dx}{x\left(x^2-C_1\right)}=\frac{dy}{y\left(C_1+y^2\right)}$$


问:
How to find a basis of the subspace of$ M_2×_2$ spanned by the matrix $A = \begin{bmatrix} -b & b \\ -d & d \end{bmatrix}$
The matrices  are of the form  $A = \begin{bmatrix} -b & b \\ -d & d  \end{bmatrix}$

Where b and d are real numbers.
By my intuition, the basis of this matrix is $$\{\begin{bmatrix} -1 & 1 \\ 0 & 0  \end{bmatrix},\begin{bmatrix} 0 & 0 \\ -1 & 1  \end{bmatrix}\} $$

But I do not know how to prove it, or if it is enough, since I think it is trivial.

Besides that, another question is what is the dimension of the vector subspace of A is.
I would again guess that because we have only two matrices for basis, the dimension is $1\times2$ or ?

(Note: we have not taken eigenvectors yet)

I was more comfortable doing the same thing with vectors, and I have more trouble with matrices.

答:
I suppose that your question is:

Find a basis for the subspace $$H=\left\{\begin{pmatrix}-b & b \\ -d & d \end{pmatrix}; b,d\in \mathbb{R} \right\}$$

In that case, you can find a basis for $H$ as:

$$\begin{pmatrix}-b & b \\ -d & d \end{pmatrix}=b\begin{pmatrix} -1 & 1 \\ 0 & 0\end{pmatrix}+d\begin{pmatrix}0 & 0 \\ -1 & 1 \end{pmatrix}.$$
Therefore, a basis for $H$ is given by $$\beta_{H}=\left\{ \begin{pmatrix} -1 & 1 \\ 0 & 0\end{pmatrix},\begin{pmatrix}0 & 0 \\ -1 & 1 \end{pmatrix}\right\}$$
and since that $$\dim(H)=\#\beta_{H}=\operatorname{cardinality}(\beta_{H}).$$
So, you have that $$\dim(H)=2.$$

Remark:

Matrices have no bases, that question makes no sense. Due to the context of your questions, I assume that the correct question is the one I asked at the beginning.

[You can prove it] On the other hand, the method you use to calculate a base of the vector subspace is sufficient to guarantee:

2.1 Linear independence.

2.2 $\operatorname{span}(\beta_{H})=H$.


问:
Integral of $m$ products of identical normal probability density functions
I am looking for a closed form solution of the integral from $m$ products of identical normal distribution probability density functions (PDF's). The mean value $\bar{x}=0$ and $\sigma$ identical for all distributions. Giving:

$$p(m)=\int_{-\infty}^{\infty} \left[ {\frac {1}{\sigma {\sqrt {2\pi }}}}e^{-{\frac {1}{2}}\left({\frac {x}{\sigma }}\right)^{2}} \right]^m dx$$

Empirical with Wolfram Alpha [WA] after puzzling I found the following closed function:

$$p(m)= \frac{1}{\sqrt{m} \cdot (\sigma \sqrt{2 \pi})^{(m-1)} }$$

I am looking for proof of this expression. The question is related to: SE, this function fit's perfect my Monte Carlo simulations.

答:
I was thinking to complicated initially. With help of the derivation normal distribution I was able to find the answer see: Wiki.

$$\int _{-\infty }^{\infty }e^{-x^{2}}\,dx=\sqrt{\pi}$$

My question basically is:

$$\frac{1}{(\sqrt{\pi})^m} \int _{-\infty }^{\infty }  e^{-mx^{2}}\,dx$$

Define:

$$u=\sqrt{m}x$$
$$dx=\frac{1}{\sqrt{m}} du$$

$$\frac{1}{\sqrt{m}(\sqrt{\pi})^m} \int _{-\infty }^{\infty }  e^{-u^{2}}\,du$$

$$\frac{\sqrt{\pi}}{\sqrt{m}(\sqrt{\pi})^m}$$

$$\frac{1}{\sqrt{m}(\sqrt{\pi})^{(m-1)}}$$


问:
Converting word problem into algebra
A microprocessor plant that manufactures computer chips costs $\$400,000$ per day to operate. It costs $\$30$ to produce one chip which sell for $\$190$ dollars. How many chips does the plant have to make to break even on one day operation?

Let $X$ be the number of chips I need to make.

$400,000 = 190X-30X$ (but I'm not sure this is even right). I'll continue anyway, I'll check it later...

$400000 = 160X$

$X=2500\quad$ ?

$2500 * 190 = 475000$ (its wrong)

Can someone help me? I think that formula I have is very wrong and I can't figure out another one...

答: You seem to have the correct answer (definitely the correct method to solve this). Why do you think "2500 * 190 = 475000" means you are wrong? Remember that the number of chips also takes into account their manufacture cost.</p>

Title: Converting word problem into algebra


问:
Given a group homomorphism $f:G\to H$, if $m$ is relatively prime to $|H|$ and $x^m\in\ker f$, then $x\in \ker f$

Let $f:G\to H$ be a homomorphism, and let  $m$ be an integer such that $m$ and $|H|$ are relatively prime. For any $x \in G$, if $x^m \in \ker f$, then $x \in \ker f$.

My proof step:  if $x^m \in \ker f$, $e=f (x^m)=( f(x))^m$, $f(x)\in H$, $\operatorname{ord} f (x)\leq m$

Any suggestions? Thanks.

答:
Ok so $e = f(x^m) = f(x)^m$ by definition.

This tells you that $f(x)$ has order dividing $m$ in $H$. But Lagrange's theorem tells us that the order of $f(x)$ divides $|H|$. Since they are coprime this order must be $1$, hence $f(x) = e$, telling us that $x$ is in the kernel.


问:
Calculate $e^A$ where $A \in M_{2\times 2}$
Let  $A=\begin{pmatrix}
0 & 1 \\
-1& 0
\end{pmatrix}$ calculate $e^A$.

i find the jordan form of $A$ and i said that $A \sim \begin{pmatrix}
i & 0 \\
0& -i
\end{pmatrix}=B$

so $e^A\sim e^B$

then from the algebra of that diagonal matrix $B$ we can say that $e^B= \begin{pmatrix}
 \sum_{n=0}^{\infty }\frac{i^n}{n!}& 0 \\
0&\sum_{n=0}^{\infty }\frac{(-i)^n}{n!}
\end{pmatrix}$

Can I find where $ \sum_{n=0}^{\infty }\frac{i^n}{n!}$ converges? Or do I leave it such as ?

答:
By Cayley-Hamilton, $A^2=-I.$  Let
$$e^{At}=\alpha I+\beta A$$ for functions
$\alpha$ and $\beta$ of $t.$
$${d\over dt}e^{At}=\alpha' I+\beta' A =Ae^{At}=$$  $$\alpha A + \beta (-I)=-\beta I + \alpha A.$$
Therefore, $\alpha'=-\beta$ and $\beta'=\alpha$ so that $\alpha$ is the solution of
$$(D^2-1)y=0;\ y(0)=1,\ y'(0)=0.$$
$$\text{Therefore, }\ \alpha={1\over 2}(e^t+e^{-t}) \ \ \text{and}$$ $$\beta = -\alpha' =-{1\over 2}(e^t-e^{-t}).$$
Replace $t$ by $1$ to get $e^A.$


问:
I came across this book in Numerical Methods. I need the title. Only found scanned pages of the book. But I don't know its title
I came across this book in Numerical Methods. I need the title. Only found scanned pages of the book. But I don't know its title. Attached are the scanned images.

For more images look here: $1$,$2$,$3$,$4$.

答: The title appears in the images: Numerical Methods in Engineering and Science.  The author is B.S. Grewal.


问:
Prove that $\lim\limits_{n\to\infty} \prod\limits_{k=1}^n \sin(k)$ converges to $0$
$$\lim_{n\to\infty} \prod_{k=1}^n \sin(k) =0$$
How can I show that this in fact is true? I know that when I take the absolute value of this product than it is strictly decreasing and bounded between 0 and 1, but I want to show that this limit is actually 0.

答:
Group $\sin(2k-1)\sin(2k)=\frac1{2}(\cos(1)-\cos(1-4k))$

You have then

$$\prod\limits_{k=1}^{\infty}\sin(k)=\prod\limits_{k=1}^{\infty}\frac1{2}(\cos(1)-\cos(1-4k))$$

Since $\cos(1)-1\leq\cos(1)-\cos(1-4k)\leq \cos(1)+1$ and

$$\prod\limits_{k=1}^{\infty}\frac{\cos(1)-1}{2}=0$$
$$\prod\limits_{k=1}^{\infty}\frac{\cos(1)+1}{2}=0$$

the limit of $\prod\limits_{k=1}^{\infty}\sin(k)$ by sandwich theorem must be $0$ as well.


问:
Relationship between union of cartesian products and cartesian product of unions two sets
It's given

For sets $A, B, C$ and $D$, let

$$L::= (A \cup B) \times(C \cup D)$$ $$R::= (A \times C) \cup (B \times D).$$ Prove that $R \subseteq L$.

Attempted solution: $$\forall(x, y): (x, y) \in (A \times C) \cup (B \times D) \Rightarrow (x,y)\in (A \times C) \vee(x,y) \in (B \times D) \Rightarrow$$ $$[x\in A \wedge y \in C ] \vee [x\in B \wedge y \in D ] \Rightarrow$$  $$x \in (A \cup B) \wedge y \in (C \cup D) \wedge (x\in B \vee y \in C) \wedge (x\in A \vee y \in D) \Rightarrow$$ $$\bullet (x,y) \in ((A\cup B) \times (C \cup D)) \wedge (x\in B \vee y \in C) \wedge (x\in A \vee y \in D) \Rightarrow $$ $$(x,y) \in (A\cup B) \times (C \cup D) \iff(x,y) \in L$$ $$\blacksquare$$
Is there a problem with the last line of the proof, if yes, where? I have doubt on the last implication. Thanks

答:
$\begin{align}
&(x,y)\in(A \times C) \cup (B \times D)\\
&\implies (x,y)\in(A \times C) \vee (x,y)\in   (B \times D)\\
&\implies (x\in A \wedge y \in C) \vee (x\in B \wedge y\in D)\\
&\implies (x\in A \vee x\in B) \wedge (y\in C \vee y \in D)\\
&\implies (x\in(A \cup B) \wedge (y\in C\cup D))\\
&\implies (x,y)\in (A \cup B) \times (C \cup D)
\end{align}$


问:
Using the Epsilon-N Definition of limits of sequences
I think I have a proper understanding of the general procedure, but I'm having difficulty manipulating my inequality so that I can isolate $n$ by itself. Sadly I wasn't given many examples to model my answer on.

Prove that $$\lim_{n\to\infty} \frac{1}{n(n-1)} = 0.$$

Body: Hint: $\dfrac{1}{n(n-1)} < \dfrac{1}{n}$.


问:
How would I graph this implicit function in geogebra?
I am trying to graph the below equation in geogebra.

\begin{equation}
 x\cos { (xy) } = 4 -y
\end{equation}

I was able to get wolframalpha to graph it but I am unsure on how to in geogebra. I am assuming I should be using an implict type function. Anyone one know how I can do this?

http://www.wolframalpha.com/input/?i=xcos%28xy%29+%3D+4+-+y

答: Write the equation in standard form ($f(x,y)=0$) and use the ImplicitCurve command:


问:
Logic behind base to number conversion
I am trying to understand and visualize the logic behind the base conversion method.

By "base" I mean how many numbers in a number system:

The decimal number system we use every day has 10 digits
{0, 1, 2, 3, 4, 5, 6, 7, 8, 9} and so it is Base 10

A binary digit can only be 0 or 1, so is Base 2

A hexadecimal digit can be {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C,
D,    E, F},  so is Base 16

And we can use letters {A, B, C, ..., X, Y, Z}, and we get Base 26

Like in base 26 for AAA its  1* 26*26 + 1*26 + 1

Why formula is:-  digit * ( base )^position index

答:
Because it works.

We need a way to express every possible whole numbers.

By having a base $b$ (assuming $b$ is a whole number larger than one) we can simply use the $b$ different digits to list the first $b$ whole numbers from $0$ to $(b-1)$.

(You might ask why $0$ to $b-1$ rather than $1$ to $b$.  Technically it is arbitrary.  But the indexing works out better if we start at "nothing")

That's the first group of $b$ numbers.  We can do a second batch of $b$ numbers by putting a $1$ in front and relisting the $b$ digits after the $1$ for $1\color{blue}0$ to $1\color{blue}{(b-1)}$.  This next group of number will go from $1\color{blue}0= b$ to $1\color{blue}{(b-1)} = b + (b-1) =2*b - 1$.

So those are the first $2b$ numbers from $0$ to $2b -1$.

We can do a third, fourth and fifth group of $b$ numbers (assuming our $b$ is that large by placing a $2,3,4$ befor and repeating the $b$ digits.  We can do this for $b$ groups of $b$ from:

$\underbrace{\underbrace{0, 1,2 ,...,(b-1)}_{b} + \underbrace{10, 11,12 ,...,1(b-1)}_{b}+ \underbrace{20, 21,22 ,...,2(b-1)}_{b}+...... \underbrace{(b-1)0, (b-1)1,(b-1)2 ,...,(b-1)(b-1)}_{b}}_{b\text{ groups of }b\text{ numbers each}}$

This $b$ groups of $b$ numbers so this is $b^2$ numbers from $0$ to $b^2 -1$.  The the number $ka$ would the int $k+1$th group which has the numbers from $k*b$ to $k*b + (b-1)$ and this is the $a+1$ number in the group and is the number $k*b + a$.  (Okay, I admit it.  The indexing from $0$ is confusing sometimes... but trust me... on the whole it is easier.

Now we have run out of two digit combos after $b^2$ numbers.  (Which makes sense.  We can have $b$ options for the first digit and $b$ for the second of $b^2$ total.)

But we can do another set of $b^2$ numbers but putting a $1$ before the two digits.

And if we put the $b$ digits before the two digits we can get $b$  groups of $b^2$ number making a total of $b^3$ numbers ranging from $0$ to $b^3-1$ but having $0...(b-1)$ being the numbers $0$ to $b-1$. $10$ to $(b-1)(b-1)$ being so that the number $an = a*b + n$ being the numbers from $b$ to $(b-1)*b + (b-1) = b^2-1$.  And then numbers $100$ to $(b-1)(b-1)(b-1)$ where $amn$ would be the $a+1$th group of $b^2$ numbers which go from $ab^2$ to $a*b^2 + (b^2-1)$ and within that group it is the $m+1$th group of $b$ numbers that go from $a*b^2 + m*b$ to $a*b^2 + m*b + (b-1)$ and it is the $n+th$ number in the group so is $n$ more than $a*b^2 + m*b$ or is $a*b^2 + m*b + n$.

And we bootstrap up.  We can get $(b-1)$ more groups of $b^3$ numbers but putting $1$ to $(b-1)$ before them to represent the first $b^4$ numbers with $4$ digits and so on.

It works.

And that's really all there is that can be said.

We do it because it will work.

Title: Logic behind base to number conversion


问:
Gagliardo-Nirenberg interpolation inequality when $p=N\geq 2$ and the order of highest derivative is $1$
I am going through the proof of Gagliardo-Nirenberg interpolation inequality, in A First Course in Sobolev Spaces, by Giovanni Leoni, Chapter 12. Here I am stuck with a Exercise problem comes in the middle of the proof (page 376):

Let $u\in \dot{W}^{1,N}(\mathbb{R}^N)$, $N\geq 2$, with $u\in L^q(\mathbb{R}^N)$ for some $1\leq q<\infty$. Let $v:=|u|^t$, where $t>1$ is such that $v\in W^{1,1}(\mathbb{R}^N)$. Prove that
$$\|u\|_{L^{tN/(N-1)}}\leq c\|u\|_{L^{(t-1)N/(N-1)}}^{(t-1)/t}\|\nabla u\|_{L^N}^{1/t}.$$

Here $\dot{W}^{1,N}$ denots homogeneous Sobolev space.

My best try was that, by using $L^p$ type interpolation inequality to $\|v\|_{L^{N/(N-1)}}$ as $\frac{1}{\frac{N}{N-1}}=\frac{1-\frac{1}{t}}{\left(1-\frac{1}{t}\right)\frac{N}{N-1}}+\frac{\frac{1}{t}}{\infty}$,
$$\|u\|_{L^{tN/(N-1)}}=\|v\|_{L^{N/(N-1)}}^{1/t}\leq\|v\|_{L^{(1-1/t)N/(N-1)}}^{(1-1/t)1/t}\| v\|_{L^\infty}^{1/t^2}\\=\|u\|_{L^{(t-1)N/(N-1)}}^{(t-1)/t}\| v\|_{L^\infty}^{1/t^2}.$$
I believe such $v\in L^\infty$, however I cannot proceed further than this. I will deeply appreciate if somebody can provide some help.

答:
Assume $u\in C_0^\infty({\mathbb R}^N)$, apply the classical Sobolev inequality
$$
\Big(\int_{{\mathbb R}^N} (u^t)^{\frac{N}{N-1}}\Big)^{\frac{N-1}{N}}
\leq C\int_{{\mathbb R}^N} |\nabla(u^t)|,
$$
then apply Holder on the right hand side:
$$
\int_{{\mathbb R}^N} |\nabla(u^t)|=t\int_{{\mathbb R}^N} |u^{t-1}\nabla u|
\leq t\Big(\int_{{\mathbb R}^N} |u^{t-1}|^{\frac{N}{N-1}}\Big)^{\frac{N-1}{N}}\Big(
\int_{{\mathbb R}^N} |\nabla u|^N\Big)^{\frac{1}{N}},
$$
thus
$$
\|u\|_{\frac{tN}{N-1}}^t\leq Ct\|u\|_{\frac{(t-1)N}{N-1}}^{t-1}\cdot \|\nabla u\|_N.
$$
Take $\frac 1t$ power.


问:
Probability of a median being $x_i$
I am trying to derive a probability function and here are the assumptions.

Let $S = \{x_{(1)},x_{(2)},...,x_{(7)}\}$ be a set of distinct values that are ordered.

Let $S*=\{x_1^*,x_2^*,...,x_7^*\}$ be a replication of $S$, as in, each elements $x_i^*$ are randomly chosen from $S$ with equal probability.

ex), $S^*$ could be $\{x_1,...,x_1\}$, or it could be an exact replica of $S$, etc.

The sampling process are mutually independent.

My goal is to find the probability that the median of $S*$, say, $M(S*)$, is equal to $x_{(i)}$.

I would like to use the notation

$$B\left[j;n,p\right] ={n \choose j}p^j(1-p)^{n-j}$$

associated to the binomial distribution.

Here is my thought process.

To find $p(i)=Pr[M(S*)=x_{(1)}]$ we need at least 4 elements from $S*$ to be $x_{(1)}$ so

$$p(1) = \sum_{j=4}^7 B\left[j;7,\frac{1}{7}\right]$$

in other words,

$$\therefore = \sum_{j=0}^3 B\left[j;7,\frac{6}{7}\right]$$.

To find $p(2)$ we need at least the least 4 elements to be less than or equal to $x_{(2)}$, but take away the probability $p(1)$, so with similar argument I can get

$$p(2) = \sum_{j=0}^3 B\left[j;7,\frac{5}{7}\right]-p(1)$$

I am hoping that the general case would look something like

$$p(i) = \sum_{j=0}^3 \left(B\left[j;7,\frac{i-1}{7}\right]-B\left[j;7,\frac{i}{7}\right]\right)$$

but I am not too confident...

May I have some assistance please?

A different result is also more than welcome as long as it is accurate and makes sense.

答:
Let $F(k,n,p)$ denote the binomial CDF.

To keep things general, let's use $n$ as the (odd) number of elements instead of $7$.
Then,
$$
p(i)=F((n - 1) / 2, n, 1 - i / n) - \sum_{j < i} p(j).
$$
In plain English, the probability of the median being $i$ is the probability of resampling a number greater than $x_{(i)}$ at most $(n - 1) / 2$ times minus the probability of the median being any one the previous elements.

The script (below) also gives the same answer as that of @Henry.
Note that $p(i) = p(n - i + 1)$ whenever $i < (n - 1) / 2$ and so you only have to compute the first $(n + 1) / 2$ elements of $p$.

```<code>from scipy.stats import binom
import numpy as np

def probs(n):
    assert n % 2 == 1  # Make sure n is odd
    p = np.empty([n + 1])
    p[0] = 0.
    mid = (n + 1) // 2
    cumsum = 0.
    for i in range(1, mid + 1):
        p[i] = binom.cdf(mid - 1, n, 1. - float(i) / n) - cumsum
        cumsum += p[i]
    p[mid + 1:] = p[mid - 1:0:-1]
    return p

n = 7
p = probs(n)
for i in range(1, n + 1):
    print('p({}) = {}'.format(i, p[i]))

# p(1) = 0.010150046809941905
# p(2) = 0.0981235952463927
# p(3) = 0.23862627695214456
# p(4) = 0.3062001619830417
# p(5) = 0.23862627695214456
# p(6) = 0.09812359524639269
# p(7) = 0.010150046809941915
</code>```

Title: Probability of a median being $x_i$

Body:
Let $F(k,n,p)$ denote the binomial CDF.

To keep things general, let's use $n$ as the (odd) number of elements instead of $7$.
Then,
$$
p(i)=F((n - 1) / 2, n, 1 - i / n) - \sum_{j < i} p(j).
$$
In plain English, the probability of the median being $i$ is the probability of resampling a number greater than $x_{(i)}$ at most $(n - 1) / 2$ times minus the probability of the median being any one the previous elements.

The script (below) also gives the same answer as that of @Henry.
Note that $p(i) = p(n - i + 1)$ whenever $i < (n - 1) / 2$ and so you only have to compute the first $(n + 1) / 2$ elements of $p$.

from scipy.stats import binom
import numpy as np

def probs(n):
    assert n % 2 == 1  # Make sure n is odd
    p = np.empty([n + 1])
    p[0] = 0.
    mid = (n + 1) // 2
    cumsum = 0.
    for i in range(1, mid + 1):
        p[i] = binom.cdf(mid - 1, n, 1. - float(i) / n) - cumsum
        cumsum += p[i]
    p[mid + 1:] = p[mid - 1:0:-1]
    return p

n = 7
p = probs(n)
for i in range(1, n + 1):
    print('p({}) = {}'.format(i, p[i]))

# p(1) = 0.010150046809941905
# p(2) = 0.0981235952463927
# p(3) = 0.23862627695214456
# p(4) = 0.3062001619830417
# p(5) = 0.23862627695214456
# p(6) = 0.09812359524639269
# p(7) = 0.010150046809941915


问:
Binomial theorem from first principles?
I suppose I'll give a little context to this...

I was at first really excited by the prospect that both Bernoulli and Taylor's versions of $e^x$ actually ammount to the same thing (where, when you expand Bernoulli's definition $\lim_{n\to\infty}(1+\frac{x}{n})^n$ it simplifies to the Taylor series $\sum_{n=0}^{\infty}\frac{x^n}{n!}$ - unlike, say, trigonometric functions like $\sin(x)$ which look a bit more unsatisfying). However, upon further reflection, to say that one identity 'simplifies' to the other seems almost circular given it presupposes binomial theorem.

So, I decided to do a little scouting online, and found that binomial theorem could be proven using proof by induction. But once again, I felt unsatisfied as it feels as though you've just divined up some formula which works. How did you come up with it in the first place?

Well, that's when I thought of probability theory. We use a binomial expression $^nC_{r}$ when we are looking for the number of combinations we have for arranging things. We can very intuitively derive this formula, however, as I don't want this question to be unecessarily large, I'm not going to go into detail about this here. What counts, is that we can figure out this formula from first principles - we're not getting it from nowhere.

As I'm sure you already know, $^nC_{r}$ is key to figuring out binomial distributions. Once again I'm not going to explaim this, but an example probability calculation becomes $^{10}C_{4}(\frac{1}{6})^4(\frac{1}{6})^{10-4}$ for rolling 4 sixes in 10 total rolls.

According to (Kolmogorov's) axioms of probability, the probability of all total outcomes $P(\Omega)$ is 1. Let's consider the context of total binomial outcomes of 3 trials (I pick this number for simple and illustrative purposes), where we designate the probablity of success, $\frac{1}{5}$, and that of failure, $\frac{4}{5}$, (where the sum of all different outcomes, here success and failure = 1). Well, we know that becuase we have 3 trials, logically, we'll have 4 possibilities: 0 successes, 1 success, 2 successess, or 3 success; we can write out our probability as: $^{3}C_{0}(\frac{1}{5})^3(\frac{4}{5})^0$ + $^{3}C_{1}(\frac{1}{5})^2(\frac{4}{5})^1$ + $^{3}C_{2}(\frac{1}{5})^1(\frac{4}{5})^2$ + $^{3}C_{3}(\frac{1}{5})^0(\frac{4}{5})^3$ = $(\frac{1}{5})^3$ + $3(\frac{1}{5})^2(\frac{4}{5})$ + $3(\frac{1}{5})(\frac{4}{5})^2$ + $(\frac{4}{5})^3$ $= 1$.

This can of course be generalised to $\sum_{k=0}^{n}$$^{n}C_{k}{p}^k{q}^{n-k}$ where $n$ is the total number of trials $k$ the number of successes, $p$ the probability of success, $q$ that of failure.

Now let's consider the binomial expansion of $(a + b)^3$. Well, supposing we know nothing about binomial theory and do it by hand, our algebra nonetheless simplifies to $a^3 + 3a^2b + 3b^2a + b^3$. Hmmm... this looks suspicously like our total probability forumla expanded above. It almost looks as though $a$ could be representing our successes, and $b$ our failures.

So my reasoning is such: it just so happens that in the context of probability, the total probability of your outcomes add up to 1. If $\Omega = \left\{ a, b \right\}$, then, by Kolmogorov's axioms, $a + b = 1$. What we're effectively saying is that for any binomial expansion, we can treat them as 'probability outcomes' where $a + b \ne 1$.

Is this a reasonable proof?

答:
Here is one way to understand the binomial theorem that doesn't use induction. I'll explain the links between this and probability at the end of this post. First, because multiplication is both distributive over addition and commutative, we know that
$$
a(x+y)=(x+y)a=ax+ay \, .
$$
When expanding a double bracket, the same principle applies. To expand $(a+b)(c+d)$, we can let $k=c+d$, and so
$$
(a+b)k=ak+bk=a(c+d)+b(c+d)=ac+ad+bc+cd \, .
$$
This justifies the 'FOIL' method for expanding a double bracket:
$$
(a+b)(c+d)=\underbrace{ac}_{\text{First}}+\underbrace{ad}_{\text{Outside}}+\underbrace{bc}_{\text{Inside}}+\underbrace{bd}_{\text{Last}} \, .
$$
At this point, we spot something interesting. The expansion of $(a+b)(c+d)$ is simply the sum of all the different ways of selecting a single element from each bracket. This pattern repeats for three brackets:
\begin{align}
(a+b)(c+d)(e+f) &= (ac+ad+bc+bd)(e+f) \\[4pt]
&= e(ac+ad+bc+bd)+f(ac+ad+bc+bd) \\[4pt]
&=ace+ade+bce+bde+acf+adf+bcf+bdf \, .
\end{align}
Consider how every time we add on a new bracket, the number of terms in the expansion doubles. When we multiplied $(a+b)(c+d)$ with $(e+f)$, the resulting expansion is simply the previous one with $e$ tacked on at the end of each term, followed by the previous one with $f$ tacked on at the end of each term. This means that it is still the case that the expansion represents the sum of all of the different ways of selecting a single element from each bracket. If you're still not convinced, then notice how the expansion of $(a+b)(c+d)$ is $ac+ad+bc+bd$. These four terms represent all of the different ways you can select a single element from the brackets $(a+b)$ and $(c+d)$. There are eight ways of picking a single element from the brackets $(a+b)$, $(c+d)$, and $(e+f)$. And we can find these eight ways by drawing a tree diagram that 'branches out' three times. When we consider the first bracket, there are two combinations: $a$ and $b$. When we consider the second bracket, there are four combinations: $ab$, $ac$, $bc$, and $bd$. When we consider the third bracket, there are eight combinations: $abe$, $abf$, $ace$, $acf$, $bce$, $bcf$, $bde$, and $bdf$. The eight combinations come from making a choice about whether to tack on $e$ at the end or an $f$ at the end, which is exactly happens when we multiply $(ac+ad+bc+bd)$ with $(e+f)$. Once you accept this, it is not too difficult to understand, at least intuitively, that the expansion of several brackets always results in a sum of all of the different possible combinations.

The expansion of $(a+b)^n$ is then the result of adding up all of the different ways you can select a single element from each bracket. The difference here is that many of the terms are repeated.
$$
(a+b)^n = \underbrace{(a+b)(a+b)(a+b)\dots(a+b)}_{\text{$n$ times}} \, .
$$
Let's go through the expansion step-by-step. If we pick $a$ zero times and $b$ $n$ times, then the term we end up with is $b^n$. If we pick $a$ $1$ time and $b$ $(n-1)$ times, then the term we end up is $ab^{n-1}$. However, there are $n$ different ways of picking $a$ $1$ time and $b$ $(n-1)$ times. Hence, the term $ab^{n-1}$ is repeated $n$ times over, meaning that we end up with $nab^{n-1}$. If we pick $a$ $2$ times and $b$ $(n-2)$ times, then the term we get is $a^2b^{n-2}$, and the coefficient is $^nC_2$. (By definition, $^nC_r$ is the number of ways you can select $r$ elements out of group of $n$.) The pattern continues. In general, if we pick $a$ $r$ times, and $b$ $(n-r)$ times, then the term that we get is $a^rb^{n-r}$, and the coefficient is $^nC_r$. This gives us the binomial theorem:
$$
(a+b)^n = \sum_{r=0}^{n}{n \choose r}a^rb^{n-r} \, .
$$
How does this link to this probability? Imagine you're flipping a fair coin ten times. Each individual sequence is as likely as any other. It is just as likely for you to get $\mathrm{HHHHHHHHHH}$ as it is $\mathrm{HHTHHHHTTH}$. However, it is much more likely for you to get $5$ heads as opposed to $0$ heads. Why? Because there are many ways in which a group of $10$ tosses can result in $5$ heads, compared with only one way to get $10$ heads. To be precise, there are $^{10}C_5=252$ ways of getting $5$ heads, but only $1$ way of getting $10$ heads.

Title: Binomial theorem from first principles?


问:
Root of expectation
I have the following expression: $\mathbb{E}[(\hat{\eta}(x)-\eta(x))^2] \leq \frac{1}{4k}$, from which I have to show that: $\mathbb{E}[|\hat{\eta}(x)-\eta(x))|] \leq \frac{1}{2\sqrt{k}}$. But the question is: how? I was first going to take a square root on both sides, however the square root of the expectation does not equal the expectation of the square root (Jensens inequality does not hold since square root is not convex).

Does anybody have a good idea? Thanks in advance :)

答:
Applying the Jensen's inequality to the convex function $f(x)= x^2$ (not $\sqrt{x}$ !)

$$\mathbb{E}^2[|\hat{\eta}(x)-\eta(x)|] \le \mathbb{E}[(\hat{\eta}(x)-\eta(x))^2]$$

Hence,

$$\mathbb{E}^2[|\hat{\eta}(x)-\eta(x)|] \le  \frac{1}{4k}$$
or
$$\mathbb{E}[|\hat{\eta}(x)-\eta(x))|] \leq \frac{1}{2\sqrt{k}}$$

Title: Root of expectation


问:
Find the thousandth number in the sequence of numbers relatively prime to $105$.

Suppose that all positive integers which are relatively prime to $105$ are arranged into a increasing sequence: $a_1 ,  a_2 ,  a_3, . . . .$ Evaluate $a_{1000}$

By inclusion exclusion principle I found the following equation:

$n - \left( \left\lfloor \dfrac {n} {3} \right\rfloor + \left\lfloor \dfrac {n} {5} \right\rfloor + \left\lfloor \dfrac {n} {7} \right\rfloor \right) + \left( \left\lfloor \dfrac {n} {15} \right\rfloor + \left\lfloor \dfrac {n} {21} \right\rfloor + \left\lfloor \dfrac {n} {35} \right\rfloor \right) - \left\lfloor \dfrac {n} {105} \right\rfloor=1000$

I tried to solve the equation by setting $n=105k+r ; 0 \leq 0<105$.  But this method is so tiring and lengthy to solve. I thought to invoke inequalities like $a-1 <\lfloor a \rfloor \leq a$, but I couldn't do as the equation contains positive as well as negative terms .Please provide me any other method to solve such equations involving floor function. Any help would be appreciated.

答:
The number of integers $n \in \{1,2,3,\dots, 105\}$ that are relatively prime to $105$ is

$$\phi(105) = \phi(3 \times 5 \times 7) = 2 \times 4 \times 6 = 48$$

We know that $\gcd(105A + n, 105) = \gcd(n, 105)$.

Since $1000 = 20\times 48 + 40$, we know that there are $20 \times 48 = 960$ numbers relatively prime to $105$ in the interval $[1, 2100]$ (where $2100 = 20 \times 105$).

$86$ is the $40^{\text{th}}$ number in that range that is relatively prime to $105$.

So the thousandth number relatively prime to $105$ is $2100 + 86 = 2186$.


问:
Does the stationary distribution of this Markov Chain exist?
To find the stationary distribution of a Markov Chain, I believe I must solve for $\vec{s} = \langle s_0, s_1 \rangle$ in $\vec{s} = \vec{s}Q$, where $Q$ is the transition matrix.

$Q$, in my case, is

$$
\left( \begin{array}{cc}
p & 1-p \\
1-q & q \end{array} \right)
$$

where $Q_{ij}$ is the probability of moving from state $i$ to state $j$ (row $i$, column $j$). When I solve for $s_0$ and $s_1$, however, I get

$$
s_0 = s_0 p + s_1 (1-q) \\
s_1 = s_0 (1 - p) + s_1 q
$$

Subsequently,

$$
s_0 (1 - p) = s_1 (1 - q) \\
s_1 (1 - q) = s_0 (1 - p)
$$

These two equations look identical. Does that mean there are an infinite number of stationary distributions for this Markov chain?

Thanks for helping a Markov Chain newb :)

答:
As in the comments, the condition you are missing is that the probabilities must sum to 1 ($s_0+s_1=1$).

However, this shouldn't surprise you. There is always "one more equation than you need and one missing" when you write them out like this. That missing condition is the normalisation, that the probabilities must sum to 1. Why is there an extra equation?

It's a property of the $Q$ matrix. The $Q$ matrix has rank 1 less than its dimension (here rank 1) because the final column is redundant. If it were left blank you could 'fill in the gaps':
$$\begin{pmatrix} p & *\\ 1-q & ** \end{pmatrix}.$$
You know that each row is a probability vector because it describes what can potentially happen. When in state $0$ the chain can remain in state $0$ or move to state $1$. As these are the only things that can happen the probabilities $p+*=1$ so $*=1-p$. Similarly $**=q$.


问:
Lebesgue outer measure - Trouble understanding proof that outer Lebesgue measure of an interval is its length
I am stuck understanding Royden's version of the proof that an interval's outer Lebesgue measure is equal to its length, specifically just the first part where we consider a closed, bounded interval $[a,b]$, and want to prove that $m^*([a,b]) = b-a$. (full proof is here)

I understand the first step, which we just consider an arbitrary open interval $(a-\varepsilon, b+\varepsilon)$ and arrive at $m^*([a,b]) \leq b-a$. From here, I agree that if we can show $m^*([a,b]) \geq b-a$, then we have $m^*([a,b]) = b-a$, which is what we want for closed bounded intervals.

For the converse, I understand the application of Heine-Borel on an arbitrary open cover of $[a,b]$ and obtaining the subsequent sum of lengths of the intervals in a finite subcover. We get from these calculations that for any arbitrary finite subcover $\{I_k\}_{k=1}^N$, we have $\sum_{k=1}^N l(I_k) > b-a$.

The connection I'm not making here is why $\sum_{k=1}^N l(I_k) > b-a$ implies $m^*([a,b]) \geq b-a = l([a,b])$. The former is a statement about the sum of lengths over a finite subcover of $[a,b]$, so it seems like if anything, I want to say that $m^*([a,b]) \leq b-a$ since $m^*(A) = \inf\{\sum_k l(I_k)|A\subseteq I_k\}$.

EDIT: for reference, Royden defines Lebesgue outer measure of a set $A$ of real numbers as $m^*(A) = \inf\{\sum_{k=1}^\infty l(I_k) | A \subseteq \bigcup_{k=1}^\infty I_k\}$, where $l$ of an interval is defined to be the difference of its endpoints if the interval is bounded, and $\infty$ otherwise.

答:
$m^*([a,b])$ is defined as the infimum over all covers of $[a,b]$. If for any cover finite cover of $A$ $\sum_{k=1}^N l(I_k)>b-a$. Then:
$$m^*([a,b])=\inf \{\sum_{k=1}^\infty l(I_k) | [a,b] \subset  \bigcup_k I_k \} \geq \inf \{\sum_{k=1}^N l(I_k)| [a,b] \subset  \bigcup_k I_k, N \in \mathbb{N} \} \geq b-a$$
because the infimum is the biggest lower bound.


问:
Solve the recurrence relation $f(n) = f(n - 1) + f(n - 2) + f(n - 3)$
This is a problem I was playing with that troubled me greatly.

$f(n) = f(n - 1) + f(n - 2) + f(n - 3)$

$f(1) = f(2) = 1$

$f(3) = 2$

So, the goal is to try and find a solution for f(n).

I tried using the regular method that is given in Johnsonbaugh , which involves finding a solution to an alternative recurrence relation, given by:

$S_{n} = t^{n}$

$S_{n} = S_{n - 1} + S_{n - 2} + S_{n-3}$

$t^{n} = t^{n - 1} + t^{n - 2} + t^{n - 3}$

$t^{n} - t^{n - 1} - t^{n - 2} - t^{n - 3} = 0$

$t^{3} - t^{2} - t - 1 = 0$

Now, you're supposed to find the roots of the above auxiliary equation, but you need all of the roots, and two of them are complex

$t = 1.8393$,

and also:

$t = -0.41964 + 0.60629i$

and its conjugate

$t = -0.41964 - 0.60629i$

Therefore,

$S_{n} = a(1.8393)^{n} + b(-0.41964 + 0.60629i)^{n} + c(-0.41964 - 0.60629i)^{n}$

Now, it is a matter of solving the set of linear equations for $a$, $b$, $c$ given by the initial conditions.

$S_{1} = 1 = a(1.8393) + b(-0.41964 + 0.60629i) + c(-0.41964 - 0.60629i)$
$S_{2} = 1 = a(1.8393)^{2} + b(-0.41964 + 0.60629i)^{2} + c(-0.41964 - 0.60629i)^{2}$
$S_{3} = 2 = a(1.8393)^{3} + b(-0.41964 + 0.60629i)^{3} + c(-0.41964 - 0.60629i)^{3}$

Now, this is supposed to yield a solution for f(n), if I didn't mess anything up, where

$f(n) = a(1.8393)^{n} + b(-0.41964 + 0.60629i)^{n} + c(-0.41964 - 0.60629i)^{n}$

Now, how do I solve this, and is it right?

Body:
In matrix form, we can construct a matrix $A=\left( \begin{array}{ccc}
1 & 1 & 1 \\
1 & 0 & 0 \\
0 & 1 & 0 \end{array} \right)$ such that

$\left( \begin{array}{c}
f(n+1) \\
f(n) \\
f(n-1)
\end{array} \right)
= \left( \begin{array}{ccc}
1 & 1 & 1 \\
1 & 0 & 0 \\
0 & 1 & 0
\end{array} \right)
%
\left( \begin{array}{c}
f(n) \\
f(n-1) \\
f(n-2)
\end{array} \right)
= A
\left( \begin{array}{c}
f(n) \\
f(n-1) \\
f(n-2)
\end{array} \right)$

Generalizing, we shall have,

$\left( \begin{array}{c}
f(n+k) \\
f(n+k-1) \\
f(n+k-2)
\end{array} \right)
= A^k
\left( \begin{array}{c}
f(n) \\
f(n-1) \\
f(n-2)
\end{array} \right)$

Let's put $k=n−3$ and $n=3$, then we shall have

$\left( \begin{array}{c}
f(n) \\
f(n-1) \\
f(n-2)
\end{array} \right)
= A^{n-3}
\left( \begin{array}{c}
f(3) \\
f(2) \\
f(1)
\end{array} \right)$

With spectral decomposition of $A=V\Lambda V^{-1}$, where we have

$\Lambda=\left( \begin{array}{ccc}
1.8392868 & 0 & 0 \\
0 & -0.4196434+0.6062907i & 0  \\
0 & 0 & -0.4196434-0.6062907i
\end{array} \right)=\left( \begin{array}{ccc}
1.8392868 & 0 & 0 \\
0 & (0.7373527)e^{(2.176234)i} & 0  \\
0 & 0 & (0.7373527)e^{-(2.176234)i}
\end{array} \right)$

and

$V = \left( \begin{array}{ccc}
-0.8503402 & -0.1411941-0.3752032i & -0.1411941+0.3752032i \\
-0.4623206 & -0.3094252+0.4470501i & -0.3094252-0.4470501i \\
-0.2513586 & 0.7373527 & 0.7373527
\end{array} \right)$

so that

$\Lambda^n=\left( \begin{array}{ccc}
1.8392868^n & 0 & 0 \\
0 & (0.7373527)e^{(2.176234)ni} & 0  \\
0 & 0 & (0.7373527)e^{-(2.176234)ni}
\end{array} \right)$

Hence, we have

$\left( \begin{array}{c}
f(n) \\
f(n-1) \\
f(n-2)
\end{array} \right)
= V\Lambda^{n-3}V^{-1}
\left( \begin{array}{c}
f(3) \\
f(2) \\
f(1)
\end{array} \right)
= V\Lambda^{n-3}V^{-1}
\left( \begin{array}{c}
2 \\
1 \\
1
\end{array} \right)$

For example, for $n=20$, we have,

$\left( \begin{array}{c}
f(20) \\
f(19) \\
f(18)
\end{array} \right)
= V\Lambda^{17}V^{-1}
\left( \begin{array}{c}
2 \\
1 \\
1
\end{array} \right)
=\left( \begin{array}{c}
66012 \\
35890 \\
19513
\end{array} \right)$


问:
Finding µ and σ using N (µ, σ^2)
Suppose that 10% of the probability for a certain distribution that is N (µ, σ^2 ) is below 60 and that 5% is above 90. What are the values of µ and σ?

This is what I have so far:

```<code>P(x < 60) = 0.1 // since 10% of the probability is below 60

P(x > 90) = 0.05 // since  5% of the probability is above 90
</code>```

$${P({(x-µ)\overσ} < {(60-µ)\overσ} = 0.1)} $$

$${P({(x-µ)\overσ} < {(90-µ)\overσ} = 0.95)} $$

Step 2:

$$(1) {P(Z < {(60-µ)\overσ} = 0.1)} $$
$$(2) {P(Z < {(90-µ)\overσ} = 0.95)}$$
// Why is (2) = 0.95 and not 0.9?

Step 3:

$${{(60-µ)\overσ} = -1.282)} $$
$${{(90-µ)\overσ} = 1.645)} $$

Step 4:

$${{(60-µ)} = -1.282σ)} $$
$${{(90-µ)} = 1.645σ)} $$

The final answer is provided but I am unsure on how to get to it. Any help on this would be appreciated

Solution µ = 73.14 and σ = 10.25

答: Hint: find values $z_1,\,z_2$ with$$\mu+z_1\sigma=60,\,\mu+z_2\sigma=90$$by solving$$\Phi(z_1)=0.1,\,\Phi(z_2)=0.95,$$with $\Phi$ the $N(0,\,1)$ CDF. Now solve simultaneous equations for $\mu,\,\sigma$.

Title: Finding µ and σ using N (µ, σ^2)


问:
What is the Euler characteristic of the edge word $ abc^{-1}b^{-1}da^{-1}d^{-1} c $?

Identify the Euler characteristic of the edge word
  $ abc^{-1}b^{-1}da^{-1}d^{-1} c $.

The Euler characteristic is
$$ X=V-E+F$$
where $V$, $E$ and $F$ are the vertices, edges and faces respectively.

The solution to the problem is given below.

I do not understand how to find the number of vertices from the polygon model or edge wordl.

答:
You ask how to find the number of vertices for this edge word.

We begin with $8$ distinct vertices. Identifying the two edges labeled $a$ eliminates $2$ vertices (since two vertex pairs are glued together). Then the identification of the two edges labeled $d$ identifies these two vertices (look at the $a$ edge in the lower right), so we subtract $1$ again.

The two vertices at the $c$ edge in the upper right have already been identified, so gluing the two $c$ edges eliminates $2$ more vertices. Finally, we see in the same way that gluing the $b$ edges eliminates $2$ more.

In this way, we see that all $7$ vertices not equal to the one in the lower right, where $a$ and $d$ meet, are identified to this vertex. So all vertices, after identification, become one vertex.


问:
Understanding $(12)(23)=(132)$ in the symmetric group $S_3$.
I am studying symmetric groups, as an example for the symmetric group $S_3$ I have the equality:

$$(12)(23)=(132)$$

It may be very trivial but I cannot derive it: how can this be done?

答:
Given the answer $(1\ 3\ 2)$, I assume that you are meant to compose these from left to right (contrary to the more common convention).

You might find the following figure to be helpful.

If anyone is curious, here is the code I used to generate this in Tikz

```<code>\begin{tikzpicture}[mystyle/.style={circle,draw,fill=gray!40}]
%"mystyle" is a variable that specifies the node's style when used as optional arg
  \foreach \i in {1,...,3}
    \foreach \j in {1,2,3}
      {\pgfmathtruncatemacro{\label}{(\i)}
      \node[mystyle]  (\i\j) at (6*\j,-1*\i) {\label};}

  \foreach \i [count = \ct] in {2,1,3}              %first permutation
    \draw[->,ultra thick,draw=red!70] (\ct 1)--(\i 2);
  \foreach \i [count = \ct] in {1,3,2}              %second permutation
    \draw[->,ultra thick,draw=olive] (\ct 2)--(\i 3);
  \node[draw] at (9,0) {<span class="math-container">$(1\ \ 2)$</span>};
  \node[draw] at (15,0) {<span class="math-container">$(2\ \ 3)$</span>};

\end{tikzpicture}

\begin{tikzpicture}[mystyle/.style={circle,draw,fill=gray!40}]
  \foreach \i in {1,...,3}
    \foreach \j in {1,2}
      {\pgfmathtruncatemacro{\label}{(\i)}
      \node[mystyle]  (\i\j) at (12*\j,-1*\i) {\label};}

  \foreach \i [count = \ct] in {3,1,2}
    \draw[->,ultra thick,draw=blue!70] (\ct 1)--(\i 2);
  \node[draw] at (18,-1) {<span class="math-container">$(1\ \ 2)(2\ \ 3)$</span>};

\end{tikzpicture}
</code>```

Body:
Given the answer $(1\ 3\ 2)$, I assume that you are meant to compose these from left to right (contrary to the more common convention).

You might find the following figure to be helpful.

If anyone is curious, here is the code I used to generate this in Tikz

\begin{tikzpicture}[mystyle/.style={circle,draw,fill=gray!40}]
%"mystyle" is a variable that specifies the node's style when used as optional arg
  \foreach \i in {1,...,3}
    \foreach \j in {1,2,3}
      {\pgfmathtruncatemacro{\label}{(\i)}
      \node[mystyle]  (\i\j) at (6*\j,-1*\i) {\label};}

  \foreach \i [count = \ct] in {2,1,3}              %first permutation
    \draw[->,ultra thick,draw=red!70] (\ct 1)--(\i 2);
  \foreach \i [count = \ct] in {1,3,2}              %second permutation
    \draw[->,ultra thick,draw=olive] (\ct 2)--(\i 3);
  \node[draw] at (9,0) {$(1\ \ 2)$};
  \node[draw] at (15,0) {$(2\ \ 3)$};

\end{tikzpicture}

\begin{tikzpicture}[mystyle/.style={circle,draw,fill=gray!40}]
  \foreach \i in {1,...,3}
    \foreach \j in {1,2}
      {\pgfmathtruncatemacro{\label}{(\i)}
      \node[mystyle]  (\i\j) at (12*\j,-1*\i) {\label};}

  \foreach \i [count = \ct] in {3,1,2}
    \draw[->,ultra thick,draw=blue!70] (\ct 1)--(\i 2);
  \node[draw] at (18,-1) {$(1\ \ 2)(2\ \ 3)$};

\end{tikzpicture}


问:
Gradient of vector field notation
Working in 3D. I know that the gradient is a vector operator defined as $\nabla = [\frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial}{\partial z}]$.

The gradient of a scalar scalar-valued function $f(\vec{x})\in\mathbb{R}$ is $\nabla f(\vec{x}) = [\frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial}{\partial z}]f(\vec{x}) = [\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z}]$. This makes sense to me.

But if we take the gradient of a vector field, say $\vec{f} = [f_1,f_2,f_3]$, I know that this is

$\displaystyle
\nabla \vec{f} =
\begin{bmatrix}
\frac{\partial f_1}{\partial x} &
\frac{\partial f_1}{\partial y} &
\frac{\partial f_1}{\partial z} \\
\frac{\partial f_2}{\partial x} &
\frac{\partial f_2}{\partial y} &
\frac{\partial f_2}{\partial z} \\
\frac{\partial f_3}{\partial x} &
\frac{\partial f_3}{\partial y} &
\frac{\partial f_3}{\partial z} \\
\end{bmatrix}
$.

But how did we get to this? Since both $\nabla$ and $\vec{f}$ are vectors, this seems a bit like an outer product, but writing $\nabla \otimes\vec{f}$ turns out to be the transpose of what I want i.e.

$\displaystyle
\nabla \otimes\vec{f}=\nabla\vec{f}^T
=
\begin{bmatrix}
\frac{\partial}{\partial x}\\
\frac{\partial}{\partial y}\\
\frac{\partial}{\partial z}
\end{bmatrix}
\begin{bmatrix}
f_1 & f_2 & f_3
\end{bmatrix}
=
\begin{bmatrix}
\frac{\partial f_1}{\partial x} &
\frac{\partial f_2}{\partial x} &
\frac{\partial f_3}{\partial x} \\
\frac{\partial f_1}{\partial y} &
\frac{\partial f_2}{\partial y} &
\frac{\partial f_3}{\partial y} \\
\frac{\partial f_1}{\partial z} &
\frac{\partial f_2}{\partial z} &
\frac{\partial f_3}{\partial z} \\
\end{bmatrix}
$
Am I not understanding something correctly? What am I doing wrong?

答:
Your first idea is incomplete, the result of the gradient operator is a column vector, or at least the same type of vector as used in the point space. The relation between them is that the directional derivative $df(v)=\frac{\partial f}{\partial v}$ for a scalar function $f$ can also be written as the scalar product $∇f·v$.

With $\vec f$ likewise a column vector, $∇\vec f$ does not make much sense, the usual constructs are the divergence $∇·\vec f$ and the curl $∇\times\vec f$.

One can of course also write the derivative of the vector field as $D\vec f=∇^T\vec f$. While $∇\otimes \vec f$ has the same components, it has a slightly different geometrical interpretation as a bi-vector.

There could also be some mix-up in the assignment of row and column vectors to space points and translations on one side and linear functionals on the other. In the first half you use rows for the points as is it quite common in computer graphics. Then also transformations act as vector-matrix products. In the second half discussing the Jacobian you switch to the convention more usual in analytical and differential geometry, where (coordinate) vectors are columns and linear forms are represented by row vectors.

Title: Gradient of vector field notation


问:
Show the diagonal is Lebesgue measurable
Let $A\subset \Bbb R$ and we set $D\subset \Bbb R^2$ by $D = \{ {(x, x) : x\in A}\}.$  Show that $D$ is Lebesgue measurable calculate it measure $m(D)$.

答:
Let us show that $m(\bar{D}) = 0$ when $\bar{D} =  \{{(x, x) : x\in \mathbb{R}\}}$, then as $D \subseteq \bar{D}$ and $m(\bar{D})=0$ we have that $m(D) = 0$ and so $D$ is measurable.

Notice that:
$$m(\bar{D}) = \int \limits_{\mathbb{R}^2} \chi_{\bar{D}} \, dm = \int \limits_{\mathbb{R}} \underbrace{ \int \limits_{\mathbb{R}} \chi_{\bar{D}}(x,y) \, dx} _{\star} \, dy
$$

But $\star = 0$ because when we hold $\tilde{y}$ as a constant, we have that:
$$\chi_{\bar{D}} (x,\tilde{y}) = \begin{cases}
1 & x =\tilde{y} \\
0 & x \neq \tilde{y} \\
\end{cases} \implies \chi_{\bar{D}} (x,\tilde{y}) = 0 \; a.e. \implies \int \limits_{\mathbb{R}} \chi_{\bar{D}}(x,\tilde{y}) \, dx = 0 \implies m(\bar{D}) = 0$$
q.e.d.


问:
Problem 6.16 Ordinary Differential Equations and Dynamical Systems Teschl
While reading the book Ordinary Differential Equations and Dynamical Systems by Gerald Teschl, I got stuck at Problem 6.16. It states:

Consider the system
\begin{eqnarray*}
\dot{x} = x-y-x(x^2+y^2)+\dfrac{xy}{\sqrt{x^2+y^2}} \\
\dot{y} = x+y-y(x^2+y^2)-\dfrac{x^2}{\sqrt{x^2+y^2}}.
\end{eqnarray*}
Show that $(1,0)$ is not stable even though $\lim\limits_{t \to +\infty} |\phi(t,x) - x_0| = 0$.

I was able to show the second part, but i could not show that $(1,0)$ is not stable. By plotting the system on mathlab I could see that it isn't stable, but how do I show it using the definition of stability?

Body:
Hint.

From

\begin{eqnarray*}
\dot{x} = x-y-x(x^2+y^2)+\dfrac{xy}{\sqrt{x^2+y^2}} \\
\dot{y} = x+y-y(x^2+y^2)-\dfrac{x^2}{\sqrt{x^2+y^2}}.
\end{eqnarray*}

we have

\begin{eqnarray*}
x\dot{x} = x^2-xy-x^2(x^2+y^2)+\dfrac{x^2y}{\sqrt{x^2+y^2}} \\
y\dot{y} = xy+y^2-y^2(x^2+y^2)-\dfrac{x^2y}{\sqrt{x^2+y^2}}.
\end{eqnarray*}

and after addition

$$
(x^2+y^2)'=(x^2+y^2)(1-(x^2+y^2))
$$

or making $r=\sqrt{x^2+y^2}$

$$
\dot r = r(1-r^2)
$$


问:
Matrix notation and multiplication
I have a question regarding the index set notation for matrices.
Let $Q$ be a matrix indexed by $\{0,1\}^N\times\{0,1\}^N$ that is for $n,m\in \{0,1\}^N$ we can define $Q(n,m)\in\mathbb{R}$. How can I now write down matrix vector multiplication?
In particular let $G:\{0,1\}^N\rightarrow\mathbb{R}$ be a function how do I interpret $QG(n)$?
For example in the case $G(n)=n_i$, i.e. coordinate projection to the $i$-the coordinate the formula is supposed to be
$$QG(n)=\sum_mQ(n,m)m_i$$
I don't understand this formula at all and can't think of this "multidimensional" matrix notation in any way.

答:
If $A$ is indexed by sets $I,J$ and $B$ is indexed by $J,K$, then the product $AB$ of these matrices is indexed by $I,K$ with entries defined by
$$
[AB](i,k) = \sum_{j \in J} A(i,j)B(j,k).
$$
With that said, we will have
$$
[QG](n) = \sum_{m \in \{0,1\}^N}Q(n,m)G(m).
$$

Title: Matrix notation and multiplication


问:
What is a proof?
I am just a high school student, and I haven't seen much in mathematics (calculus and abstract algebra).

Mathematics is a system of axioms which you choose yourself for a set of undefined entities, such that those entities satisfy certain basic rules you laid down in the first place on your own.

Now using these laid-down rules and a set of other rules for a subject called logic which was established similarly, you define certain quantities and name them using the undefined entities and then go on to prove certain statements called theorems.

Now what is a proof exactly? Suppose in an exam, I am asked to prove Pythagoras' theorem. Then I prove it using only one certain system of axioms and logic. It isn't proved in all the axiom-systems in which it could possibly hold true, and what stops me from making another set of axioms that have Pythagoras' theorem as an axiom, and then just state in my system/exam "this is an axiom, hence can't be proven".

EDIT : How is the term "wrong" defined in mathematics then? You can say that proving Fermat's Last Theorem using the number theory axioms was a difficult task but then it can be taken as an axiom in another set of axioms.

Is mathematics as rigorous and as thought-through as it is believed and expected to be? It seems to me that there many loopholes in problems as well as the subject in-itself, but there is a false backbone of rigour that seems true until you start questioning the very fundamentals.

答: Excellent Question.  The concept of a proof carries authority; the conclusion of a proof is demonstrated from the premises.  Proofs of theorems still assume a proof theory, which, as you've noted, depend on accepting a logic and some axioms. One can think of a proof as a convincing demonstration (informal sense) or as a mathematical object (formal sense).  I take it that you are more interested in the idea of a convincing demonstration.  All demonstrations must make certain assumptions.  An opponent can't be put in checkmate unless they accept the rules of chess.  Similarly, one can't offer a convincing demonstration unless one accepts the premises from which the demonstration proceeds.  The more general question of whether anything is ever convincingly demonstrated is a philosophical question, but first-order logic and the Peano axioms provide many natural cases of convincing demonstrations.  The Stanford Encylopedia of Philosophy has a nice article on the development of proof theory here.

Title: What is a proof?


问:
Complex differentiation of a function
Assume $ f: \mathbb{C} \to \mathbb{C} $ is $ \mathbb{C} $ differentiable. Denote its derivative as $ f'(z) $.

Next, if we denote $ f(z)=u(z)+iv(z) $ where $u,v $ are functions to $ \mathbb{R} $, then the partial derivatives of $$ f\left(x,y\right)=\begin{pmatrix}u\left(x,y\right)\\
v\left(x,y\right)
\end{pmatrix} $$

Satisfies:

$$ \frac{\partial f}{\partial x}\left(z\right)=-i\frac{\partial f}{\partial y}\left(z\right)=f'\left(z\right) $$

Why is this true?

I couldnt find formal proof for this equation, just intuitive explanations (that werent very convincing).

I'll be glad to see a formal proof for this equation.

Thanks in advance.

答:
Since $f'(z)=\lim_{h\to0}\frac{f(z+h)-f(z)}h$, then, in particular,$$f'(z)=\lim_{h\to0,\,h\in\Bbb R}\frac{f(z+h)-f(z)}h=\frac{\partial f}{\partial x}(z).$$On the other hand, if $z=a+bi$, then$$\frac{\partial f}{\partial x}(z)=\frac{\partial u}{\partial x}(a,b)+\frac{\partial v}{\partial x}(a,b)i.\tag1$$But you can use here the Cauchy-Riemann equations $\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}$ and $\frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x}$. So, $(1)$ tells you that\begin{align}\frac{\partial f}{\partial x}(z)&=\frac{\partial v}{\partial y}(a,b)-\frac{\partial u}{\partial y}(a,b)i\\&=-\left(\frac{\partial u}{\partial y}(a,b)+\frac{\partial v}{\partial y}(a,b)i\right)i\\&=-i\frac{\partial f}{\partial y}(z).\end{align}
If you want to use this to deduce the Cauchy-Riemann equations, then note that you also have\begin{align}f'(z)&=\lim_{h\to0,\ h\in\Bbb R}\frac{f(z+hi)-f(z)}{hi}\\&=-i\lim_{h\to0,\ h\in\Bbb R}\frac{f(z+hi)-f(z)}h\\&=-i\frac{\partial f}{\partial y}(z).\end{align}

Title: Complex differentiation of a function


问:
Huffman's Algortihm Probability Distributions
I am currently studying a bit of coding and information theory. I just learned Huffman's algorithm. I was solving some problems and came up with this one:

Given a code $C=\{0,10,110,111\}$ as a result of Huffman's algorithm construction, what are its probability distributions $P=\{p_1,p_2,p_3,p_4\}$?

My approach is to think of $C$ as a binary tree and analyze it from the bottom up using Huffman's algorithm. $110$ and $111$ have the same ascendant vertex. So,
$$p_4\leq p_1,p_2\text{ and }p_3\leq p_1,p_2.$$
Also,
$$p_2\leq p_1\text{ and }p_3+p_4\leq p_1.$$
My question(s) is(are): Is $p_3,p_4\leq p_2\leq p_1$ and $p_3+p_4\leq p_1$ enough of an answer? Is this correct? Is there more to explore? In advance, many thanks!

答:
I believe you're approach and result are both correct.

Edit: In general, if two symbols have Huffman codeword lengths $\ell_i$ and $\ell_j$, then

$$ \ell_i > \ell_j \implies p_i \le p_j$$
$$ \ell_i > \ell_j +1 \implies p_i < p_j$$

while $ \ell_i  = \ell_j $ does not imply anything.

This also applies to grouped symbols. That is, if two codewors are equal except for the last bit, so that $\ell_j =  \ell_k$, then you can combine them into a new pseudo-symbol $m$ with $\ell_m =\ell_j-1$ and $p_m=p_j+p_k$, and the same relationships above apply.


问:
Trace $x=y$ when plotting a paraboloid
When plotting regions in $\mathbb{R}^{3}$, it is a common strategy to draw the region for some constant parameter (like $y=0$ or $x=0$ or $z=0$).
When plotting $z=y^{2}+x^{2}$, I tried to plot for when $x=y$ to see what happened on this plane (or at least that was my intention). In this case, $z= 2x^{2}$, but this is relatively to the x and z-axis, because I know that the figure drawn in the $x=y$ plane is a normal parabola ($z="x"^{2}$). So is there a way that allows us to see the figures as they happen and not distorted by the axis representation? (I don't know if I'm being explicit enough?)

答:
In this case you could think about transforming $(x,y)$ in to radial coordinates, $(r,\theta)$.  For $x = y$, you have $(r,\theta)  = (\sqrt{2}x,\frac{\pi}{4})$.

$z = r^2$

$z$ is independent of the angle.

Title: Trace $x=y$ when plotting a paraboloid


Title: Finding µ and σ using N (µ, σ^2)

Body:
Suppose that 10% of the probability for a certain distribution that is N (µ, σ^2 ) is below 60 and that 5% is above 90. What are the values of µ and σ?

This is what I have so far:

```<code>P(x < 60) = 0.1 // since 10% of the probability is below 60

P(x > 90) = 0.05 // since  5% of the probability is above 90
</code>```

$${P({(x-µ)\overσ} < {(60-µ)\overσ} = 0.1)} $$

$${P({(x-µ)\overσ} < {(90-µ)\overσ} = 0.95)} $$

Step 2:

$$(1) {P(Z < {(60-µ)\overσ} = 0.1)} $$
$$(2) {P(Z < {(90-µ)\overσ} = 0.95)}$$
// Why is (2) = 0.95 and not 0.9?

Step 3:

$${{(60-µ)\overσ} = -1.282)} $$
$${{(90-µ)\overσ} = 1.645)} $$

Step 4:

$${{(60-µ)} = -1.282σ)} $$
$${{(90-µ)} = 1.645σ)} $$

The final answer is provided but I am unsure on how to get to it. Any help on this would be appreciated

Solution µ = 73.14 and σ = 10.25

Body:
$P(X<60) $

$= P(Z < \frac{60-\mu}{\sigma})=0.1=\Phi(-1.281552)$

$=P(Z < -1.281552)$

$\implies 60-\mu = -1.281552\sigma$

Similarly,

$P(X\leq 90)=1-P(X>90)=1-0.05=0.95$

$=P(Z \leq \frac{90-\mu}{\sigma}) =\Phi(1.644854)$

$=P(Z \leq 1.644854)$

$\implies 90-\mu = 1.644854\sigma$

Now, solving the $2 \times 2$ system of linear equations

$\mu -1.281552\sigma = 60$

$\mu + 1.644854\sigma = 90$

you have $\sigma = \frac{30}{1.281552 + 1.644854}=10.25148$ and

$\mu=60+1.281552*10.25148=73.1378$


答:
Call $w$ the axis in the direction $x=y$. One unit along $w$ projects to $\dfrac1{\sqrt 2}$ unit along $x$. Hence the parabola is

$$z=w^2.$$

Title: Trace $x=y$ when plotting a paraboloid

Body:
When plotting regions in $\mathbb{R}^{3}$, it is a common strategy to draw the region for some constant parameter (like $y=0$ or $x=0$ or $z=0$).
When plotting $z=y^{2}+x^{2}$, I tried to plot for when $x=y$ to see what happened on this plane (or at least that was my intention). In this case, $z= 2x^{2}$, but this is relatively to the x and z-axis, because I know that the figure drawn in the $x=y$ plane is a normal parabola ($z="x"^{2}$). So is there a way that allows us to see the figures as they happen and not distorted by the axis representation? (I don't know if I'm being explicit enough?)


问:
Locally noetherian schemes
I'm reading Algebraic Geometry written by R. Hartshorne. There is a proposition in section 3, I have few problems with a part of its proof

A scheme $X$ is locally noetherian iff for every open subset $U=\operatorname{Spec} A$, $A$ is noetherian.

Let $X=\operatorname{Spec} A$ be an affine scheme, we want to prove $A$ is noetherian. In this regard let $U=\operatorname{Spec} B$, where $B$ is noetherian, be an open subset. There exists $f\in A$ so that $D(f)\subset \operatorname{Spec} B$; Moreover, suppose $\bar{f}$ is the image of $f$ in $B$ then we have $A_f\cong B_{\bar{f}}$.

My problem is that I'm not sure how $f$ is chosen and why $A_f\cong B_{\bar{f}}$ is correct.
My attempt: Let $f_1\in A-B$, for each $p\in \operatorname{Spec} A$ for which we have $p\in D(f_1)$, let $g$ be a member in $p-B$ and define $f_2=f_1g$. Continue these steps until there is no such $p\in \operatorname{Spec} A$.

答:
The existence of $f$ follows from the fact that the open sets $D(f)$ form a base for the topology on $\operatorname{Spec} A$: every open set in $\operatorname{Spec} A$ is the union of some sets of the form $D(f)$, so if $\operatorname{Spec} B\subset\operatorname{Spec} A$ is an open subset, there's an $f$ so that $D(f)\subset \operatorname{Spec} B$.

Next, to see that $D(f)\subset \operatorname{Spec} A$ is the same set as $D(\overline{f})\subset \operatorname{Spec} B$, the first thing to note is that $\overline{f}$ is the restriction of $f\in\mathcal{O}_{\operatorname{Spec} A}(\operatorname{Spec} A)$ to $\mathcal{O}_{\operatorname{Spec} A}(\operatorname{Spec} B)$. So if $p$ is a point in $\operatorname{Spec} B$, $f$ and $\overline{f}$ have the same image in $\mathcal{O}_{\operatorname{Spec} A,p}=\mathcal{O}_{\operatorname{Spec} B,p}$.
As a point $p$ is in $D(f)$ iff $f$ is not in the maximal ideal of the local ring at $p$, we see that $D(\overline{f})=\operatorname{Spec} B \cap D(f)$, and as $D(f)\subset \operatorname{Spec} B$, we get that $D(\overline{f})=D(f)$.

Finally, as $\mathcal{O}_{\operatorname{Spec} A}(D(f))\cong A_f$ and $\mathcal{O}_{\operatorname{Spec} B}(D(\overline{f}))\cong B_{\overline{f}}$, we see that $A_f\cong B_{\overline{f}}$.

Title: Locally noetherian schemes


问:
What does $\prod_{p}$ mean without the upper value?
I was reading this paper and on page 2, I saw this: $$\prod_{p}\Bigl(\frac{1}{1-p^{-z}}\Bigr)$$ I know how to use a Pi when there is a value on the top, but what does it mean when there is no value on the top and $p$ is not assigned a value?
Happy pi day!

答:
Turning comments into a formal answer, the notation $\prod_p$ in the linked-to paper means the product over all primes. To be hyperprecise, it refers to the limit of the partial products when the primes are ordered by size, i.e., $\lim_{N\to\infty}\prod_{p\le N}$. This slightly cryptic notation is quite common in number-theoretic treatments of the Riemann zeta function. It's done partly for convenience -- it reduces the amount of work the typesetter has to do -- and partly for visual appeal, eliminating "clutter" from the page. (That said, some authors prefer notation such as $\prod_{p\in\mathscr{P}}$, with $\mathscr{P}$ denoting the set of primes.)

In general, any time you run into notation you don't understand, it's a good idea to look back at preceding pages to see if the notation has been used earlier; sometimes you'll find text that explains the notation. In this case, $\prod_p$ first appeared on page 2, in Lemma 2.1, in the formula

$$\zeta(z)=\prod_p{1\over1-p^{-z}}$$

There's no text that stipulates the meaning of the notation, but that is presumably because the author thought the meaning was clear from context. (You might note, they also never explicitly said the variable $p$ refers to primes.)


问:
Hint for exercise on differential geometry
I am again having problems solving some exercise for my classical differential geometry class:

Let $\alpha : I \longrightarrow \mathbb{R}^2$ be a curve (parameterised by arc length) and let $\dot{\alpha}(s) = \big( f(s), g(s) \big)$. Show that there exists a smooth function $\varphi : I \longrightarrow \mathbb{R}$ such that $\dot{\alpha}(s) = \big( \cos \varphi(s), \sin \varphi(s) \big)$. Approach this problem as follows:

a) $\quad$ Set
$$\varphi(s) = \int_{s_0}^s \big( f \dot{g} - \dot{f} g \big)$$
and show that
$$F := f \cos \varphi + g \sin \varphi$$
as well as
$$G := f \sin \varphi - g \cos \varphi$$ are constants.

b) $\quad$ Calculate the values of $F$ and $G$ and show that this implies the assumption.

What I have so far is, that because of $\; || \, \dot{\alpha}(s) \,|| = 1 \;$, the equations

$$ f^2 + g^2 = 1 \quad , \qquad f \dot{f} + g \dot{g} = 0 \qquad \qquad (\ast)$$

both hold. Now, when calculating the derivatives of $F$ and $G$ with respect to $s$, one finds that by using equations $(\ast)$ the derivatives vanish and therefore $F$ and $G$ are constant.

Moreover, I assume that $F = 1$ has to hold. Why is this? Because if $\dot{\alpha}$ can be expressed as $\big( \cos \varphi(s), \sin \varphi(s) \big)$, then $F$ is nothing but $|| \, \dot{\alpha}(s) \, ||^2$.

Furthermore, I assume that $G = 0$ has to hold. Why is this? Because if $\dot{\alpha}$ can be expressed as $\big( \cos \varphi(s), \sin \varphi(s) \big)$, then $G$ is nothing but a scalar product of $\dot{\alpha}$ with a perpendicular vector of same length.

However, I have difficulties showing that $F = 1$ and $G = 0$. I have already done some lengthy calculations -- all of them lead nowhere. I have the feeling that I am missing some obvious and easy point/approach here. Does anyone have a hint for me?

答: Assuming $s_0$ is a real number such that $(\cos s_0, \sin s_0) = (1,0)$, calculate $F(s)$ and $G(s)$ at $s=s_0$. They are the constants you are looking for.


问:
Let $\zeta=\{E \in \mathcal{P}(X): card(E)=2 \}$ proof that $\sigma(\zeta)=\mu_{c}$
Consider $X$ an uncountable set and $\mu_{c}:=\{A\in \mathcal{P}(X): A\quad\text{is countable or} \quad X-{A}\quad \text{is countable} \}$

Let $\zeta=\{E \in \mathcal{P}(X): card(E)=2 \}$

proof that $\sigma(\zeta)=\mu_{c}$

it is clear that $\mu_{c}$ is a sigma-algebra that contains $\zeta$, indeed, given $E\in \zeta$ then $Card(E)$=2 so, $E$ is countable and therefore $E \in \mu_{c}$  and we know that $\sigma (\zeta)$ is the smallest sigma algebra that contains $\zeta$,

therefore, $\sigma (\zeta) \subset \mu_{c} $

for the other contention I expressed $\sigma(\zeta)$ as follows:

$\sigma(\zeta)= \bigcap_{g \in G} g$, $G$ is the set of all sigma algebras that contain $\zeta$

then, to prove that $\mu_{c} \subset \sigma (\zeta)$ take a $s\in \mu_{c}$ and $g\in G$ But I have not been able to find a good reasoning that leads me to show that $s \in g$

答:
If $A\in\mu_c$ is countable, let $\{P_n:n\in\Bbb N\}$ be a partition of $A$ into $2$-element sets; then $P_n\in\zeta$ for each $n\in\Bbb N$, so $A=\bigcup_{n\in\Bbb N}P_n\in\sigma(\zeta)$.

For that matter, we could simply let $\mathscr{P}$ be the family of all $2$-element subsets of $A$: $A$ is countable, so $\mathscr{P}$ is also countable, and $A=\bigcup\mathscr{P}\in\sigma(\zeta)$.

Thus, every countable member of $\mu_c$ belongs to $\sigma(\zeta)$, and $\sigma(\zeta)$ is a $\sigma$-algebra, so it must contain the complements of these sets as well. It follows that $\mu_c\subseteq\sigma(\zeta)$ and hence that $\mu_c=\sigma(\zeta)$.


问:
Finding a unitary matrix that diagonalizes a given matrix
Let $$T=\begin{pmatrix}5 & 0 & 0 \\ 0 & 2 & i\\ 0 & -i & 2 \end{pmatrix}$$ be a Hermitian matrix.

I found the eigenvalues and eigenvectors already and they are $1,3,5$ and $\begin{pmatrix}0\\-i\\1\end{pmatrix}$,$\begin{pmatrix}0\\i\\1\end{pmatrix}$, and $\begin{pmatrix}1\\0\\0\end{pmatrix}$.

Normalizing each vector, I get $<1,0,0>$, $<0,\frac{-i}{\sqrt{2}},\frac{1}{\sqrt{2}}>$, and $<0,\frac{i}{\sqrt{2}},\frac{1}{\sqrt{2}}>$.

I need to find a matrix $P$ such that $P^*AP$ is diagonal.  My first idea was $P=\begin{pmatrix}1 & 0 & 0\\ 0 & -i & i\\ 0 & 1 & 1 \end{pmatrix}$.  $P^*=P^{-1}$, but $P^*P\neq I$.

I'm having a some problems trying to find a $P$ that will allow me to diagonalize $T$.

答:
Let's check your idea:

$$P=\begin{pmatrix}1 & 0 & 0\\ 0 & \!\!-i & i\\ 0 & 1 & 1 \end{pmatrix}\;,\;\;P^*=\begin{pmatrix}1 & 0 & 0\\ 0 & i & 1\\ 0 & \!\!-i & 1 \end{pmatrix}$$

and since

$$PP^*\neq I\;\;\text{then}\;\;P^*\neq P^{-1}$$

The problem is this matrix's columns (rows) aren't orthonormal though they're orthogonal. We must apply Gram-Schmidt (I assume the usual euclidean inner product and let's write all the vector as row ones, for simplicity)):

$$u_1=(1,0,0)$$

$$w_2=(0,-i,1)-\langle\;(1,0,0),(0,-i,1)\;\rangle\,(1,0,0)=(0,-i,1)-0\cdot(1,0,0)=(0,-i,1)$$

and since

$$||w_2||=\sqrt 2\implies u_2=\frac1{\sqrt 2}(0,-i,1)$$

$$w_3:=(0,i,1)-\langle\;(0,i,1),(1,0,0)\;\rangle\,(1,0,0)-\left\langle\;(0,i,1),\frac1{\sqrt 2}(0,-i,1)\;\right\rangle\,\frac1{\sqrt 2}(0,-i,1)=$$

$$=(0,i,1)-\frac13\cdot 0=(0,i,1)$$

and since

$$||w_2||=\sqrt 2\implies u_3=\frac1{\sqrt2}(0,i,1)$$

Thus, our $\,P\,$ is (well, let's call it $\,Q\,$ to avoid confussion):

$$Q=\begin{pmatrix}1&0&0\\
0&-\frac{i}{\sqrt 2}&\frac1{\sqrt 2}\\
0&\frac i{\sqrt 2}&\frac1{\sqrt 2}\end{pmatrix}\;\implies\;Q^*=\begin{pmatrix}1&0&0\\
0&\frac{i}{\sqrt 2}&-\frac i{\sqrt 2}\\
0&\frac 1{\sqrt 2}&\frac1{\sqrt 2}\end{pmatrix}$$

It is now a simple exercise to verify that indeed $\,QQ^*=UI\,$ and, thus, $\,Q^{-1}=Q^*\,$ ...

Note: The above G-S process was almost trivial because the columns of $\,P\,$ were already orthogonal. The above stuff shows how to carry on the process in general, though in this case it was sufficient to divide each vector by its norm.


问:
Is it possible to find a parabola knowing the average value.
My question: Is it possible to find a specific parabola when you know the average value, and the x intercepts? The parabola I am trying to find has x intercepts at 0 and 18 with an average value of 2.34

Can you reverse engineer the parabola using an average value equation?

答:
In a slightly different notation from @K.defaoite's comment:

If $A>0$ and $a<b$, the parabola $y=A(x-a)(b-x)$ has area from $x=a$ to $x=b$ given by$$A\int_a^b(-ab+(a+b)x-x^2)=A(b-a)^3/6,$$as can be computed easily with $y=\frac{x-a}{b-a}$. The average on this interval is $\mu=A(b-a)^2/6$, so $A=\frac{6\mu}{(b-a)^2}$.


问:
Is there a matrix $A \in \mathbb{Q}^{n \times n}$ such that $P(A)=\mathbf{0}$, where $P$ is a monic polynomial with rational coefficients?
Given a monic polynomial of degree $k \leq n$ with rational coefficents, can I always find a matrix $A \in \mathbb{Q}^{n \times n}$ that is a root of $P$?

What I have already tried:

Using the Cayley-Hamilton theorem, without success.

I know that the minimal polynomial of $A$ (if $A$ exists) must divide $P$.

I also tried to relate rational roots (if any exists) of $P$ with the existence of $A$ (by using the rational root theorem), but I don't know how to make any progress from there.

答: The companion matrix of a polynomial $p$ has characteristic polynomial $p$.