问:
Finitely generated pp-types in saturated modules
I'm working through Mike Prest's "Model Theory and Modules" and struggling a bit with one of the early exercises. Let $\mathcal{L}=\mathcal{L}_\text{ring}\cup\{r\}_{r\in R}$ be the language of right modules over a ring $R$, and fix a module $M$ in the language $\mathcal{L}$ and a subset $A\subseteq M$.

Recall that, to the stone space $S_n^M(A)$, there is the associated "positive primitive part" of $S_n^M(A)$, denoted $S_n^{+}(A)$, consisting of all pp-formulas in $S_n^M(A)$. Now, for any $p\in S_n^{+}(A)$, define $\mathcal{G}(p)=\{\varphi(v_1,...,v_n, \bar{0}):\varphi(\bar{v}, \bar{a})\in p\text{ for some }A\text{-tuple }\bar{a}\}$, and further define $G(p)=\bigcap_{\varphi\in\mathcal{G}(p)}\varphi(M^n,\bar{0})\leqslant M^n$. (A subgroup, though of course not necessarily a submodule if $R$ is not commutative.)

Finally, recall that a partial type $p(\bar{v})$ is "finitely generated" in $M$ if there is a formula $\varphi(\bar{v})$ (perhaps with parameters) such that $M\models \varphi(\bar{v})\rightarrow\psi(\bar{v})$ for any $\psi(\bar{v})\in p$. The exercise is then the following:

(i) Show that if $p\in S_n^{+}(A)$, $M$ is $|A|^+$-saturated, and $G(p)$ is pp-definable, then $p$ is finitely generated. Give an example to show that one needs some kind of saturation assumption on $M$.

(ii) Improve on (i) by showing that $|T|^+$-saturation is enough.

($T$ is not explicitly defined by Prest but presumably he is taking $T$ to be the complete $\mathcal{L}$-theory of $M$.) A counterexample without any saturation assumption is easy; for instance, consider $M=\mathbb{Z}$ as an abelian group, and let $p(v)$ be (the pp-part of a completion of the partial type consisting of $\{v\neq0\}$ unioned with) $\{\exists w (v=w n^k)\}_{k\in\mathbb{N}}$ for any $n>1$. Then $G(p)=\{0\}$, which is of course pp-definable, but $p(v)$ is clearly not finitely generated. I believe I've also shown the desired result in case that $M$ is $|A|^{+}$-saturated, but I'm unable to improve the bound. My solution is as follows (for simplicity I will restrict to the case where $p(v)$ is a 1-type):

Suppose $G(p)$ is cut out by a (pp-)formula $\theta(v)$. Now, define $C=\bigcap_{\varphi(v,\bar{a})\in p}\varphi(M,\bar{a})$. Each $\varphi(M,\bar{a})$ is a coset of $\varphi(M,\bar{0})$, so $C$ is either empty or a coset of $G(p)$. Since $M$ is $|A|^{+}$-saturated, and $p$ is a type with parameters from $A$, $p(v)$ is realized in $M$, so $C$ is non-empty, and thus it is a coset of $G(p)$; say $C=m+G(p)$ for some $m\in M$. Now clearly $C$ is (pp) definable over $\{m\}$, by the formula $\psi(v, m)=\exists w(\theta(w)\wedge v=w+m)$.

Now consider the set of formulas $\Sigma(v)=p(v)\cup\{\neg\psi(v, m)\}$. Certainly $\Sigma(v)$ is not realized in $M$, for by construction $M\models\psi(x, m)\iff x\in C\iff M\models p(x)$ for any $x\in M$. However, $\Sigma(v)$ is defined with parameters from $A\cup\{m\}$, a set with cardinality $<|A|^{+}$, and hence by the saturation assumption we must have that $\Sigma(v)$ is not (finitely) consistent with $M$.

Hence, taking conjunctions, there is a formula $\varphi(v, \bar{a})\in p(v)$ such that $M\models\forall v\neg(\varphi(v, \bar{a})\wedge\neg\psi(v, m))$, ie $M\models\forall v(\varphi(v, \bar{a})\rightarrow\psi(v, m))$, from which it is clear that $\varphi(v, \bar{a})$ generates $p(v)$, as desired. So this proves the problem when $M$ is $|A|^{+}$-saturated.

However, part (ii) seems very surprising to me, and I'm struggling to think of where to begin with it. Does anyone have any insight or hints?

Edit: Thanks to Alex Kruckman for pointing out a very silly mistake, now corrected.

答:
In fact maybe it's not so surprising... any two distinct cosets of a subgroup are of course disjoint, and so – because $p(v)$ is consistent – for any pp-formula $\varphi(v, \bar{0})$, there can be at most one coset of $\varphi(M, \bar{0})$ cut out by formulas of $p(v)$. Hence, removing duplicate formulas from $\Sigma(v)$ (ie formulas that cut out the same coset of a subgroup as another formula of $\Sigma(v)$), we obtain a new set of formulas $\Sigma'(v)$ over some set of parameters $A_0\cup\{m\}\subseteq A\cup\{m\}$ such that (i) $\Sigma'(v)$ and $\Sigma(v)$ are logically equivalent modulo $M$, and (ii) for any pp-formula $\varphi(v, \bar{0})$, $\varphi(v, \bar{a})$ appears in $\Sigma'(v)$ for at most one $A_0$-tuple $\bar{a}$. (We further remove from $A_0$ any elements not appearing in some formula of $\Sigma'(v)$.)

This condition (ii) guarantees that $|\Sigma'(v)|\leqslant |\{\text{pp-formulas over }\mathcal{L}\}|$. But now note, for any distinct pp-formulas $\varphi_1(v, \bar{0})$ and $\varphi_2(v, \bar{0})$ over $\mathcal{L}$, there are the distinct corresponding $\mathcal{L}$-sentences $\exists v \varphi_1(v, \bar{0})$ and $\exists v \varphi_2(v, \bar{0})$, and both are vacuously satisfied in $M$ (eg by $0$) and hence elements of $\text{Th}(M)$. So in fact we have $|\Sigma'(v)|\leqslant |\{\text{pp-formulas over }\mathcal{L}\}|\leqslant |\text{Th}(M)|$.

Finally, since only finitely elements of $A_0\cup\{m\}$ appear in any element of $\Sigma'(v)$, and every element of $A_0\cup\{m\}$ appears in at least one element of $\Sigma'(v)$, we have $|A_0\cup\{m\}|\leqslant \aleph_0 |\Sigma'(v)|\leqslant \aleph_0 |\text{Th}(M)|=|\text{Th}(M)|$, and so by considering $\Sigma'(v)$ instead $\Sigma(v)$ we need only $|\text{Th}(M)|^+$-saturation. Does this argument seem right?


问:
Probability of number formed from dice rolls being multiple of 8

A fair 6-sided die is tossed 8 times. The sequence of 8 results is recorded to form an 8-digit number. For example if the tosses give {3, 5, 4, 2, 1, 1, 6, 5}, the resultant number is $35421165$. What is the probability that the number formed is a multiple of 8.

I solved this by listing all possibilities for the last 3 digits that give multiples of 8, and found this to be $\frac{1}{8}$.

The solution key agrees with my answer, but also says that "There are quicker ways to solve the problem using a more advanced knowledge of number theory"

What would be a faster way to solve this using number theory?

答:
Let's use Zhuli's notation.

$c$ has a probability of $\frac{1}{2}$ to be even.

Now if $\frac{c}{2}$ is even (i.e., $4|c$), then $4|2b+c \iff b$ is even. If $\frac{c}{2}$ is odd, then $4|2b+c \iff b$ is odd. Both conditional probabilities are $\frac{1}{2}$ therefore the marginal probability $4|2b+c$ is $\frac{1}{4}$.

Next we apply the same reasoning on $2a + (b+\frac{c}{2})$ we get the marginal probability that $4 | 2a + (b+\frac{c}{2})$, or equivalently, $8 | 4a + 2b +c$, is $\frac{1}{8}$. Q.E.D.

============================

Alternatively we could avoid conditional probabilities and just use independence and marginal probabilities.

First note that $\Pr(2|a)=\Pr(2|b)=\Pr(2|c)=\frac 12$ and $a,b,c$ are independent.

Next,
$$
\Pr(4|2b+c)=\Pr(2|c,4|c,2|b) + \Pr(2|c, 4\nmid c, 2\nmid b)
$$
$$
=\Pr(2|c, 4|c) \Pr(2|b)+\Pr(2|c, 4\nmid c)\Pr(2\nmid b) \tag{independence}\\
= \frac 12 \left( \Pr(2|c, 4|c) + \Pr (2|c, 4\nmid c)\right) = \frac 12 \Pr(2|c)=\frac 14.
$$

Finally,
$$
\Pr(8|4a+2b+c)=\Pr(4|2b+c,8|2b+c,2|a) + \Pr(4|2b+c, 8\nmid 2b+c, 2\nmid a)\\
=\Pr(4|2b+c,8|2b+c)\Pr(2|a) + \Pr(4|2b+c, 8\nmid 2b+c)\Pr(2\nmid a)\\
=\frac 12 \left(\Pr(4|2b+c,8|2b+c) + \Pr(4|2b+c, 8\nmid 2b+c)\right)\\
=\frac 12 \Pr(4|2b+c)=\frac 12 \cdot \frac 14 = \frac 18.\blacksquare
$$

Remark: We can prove, via induction, the generalization proposed by Anant using the technique above.

ContentLicense: CC BY-SA 3.0

ContentLicense: CC BY-SA 4.0


问:
Cantor Set self homeomorphism
Suppose that $a, b \in C$ where $C$ denotes the cantor set. I want to show that there exists a homeomorphism from $C$ onto itself such that $f(a) = b$. I don't know how to start with this. Is there any hint or suggestion?

答:
Let $a$ and $b$ have the ternary representations $0.a_1a_2a_3\ldots$ and $0.b_1b_2b_3\ldots$, respectively, where $a_n,b_n\in\{0,2\}$ for $n\in\Bbb Z^+$, so that

$$a=\sum_{n\ge 1}\frac{a_n}{3^n}\quad\text{and}\quad b=\sum_{n\ge 1}\frac{b_n}{3^n}\,.$$

Let $D=\{n\in\Bbb N:a_n\ne b_n\}$. For $c\in\{0,2\}$ let $\bar c=2-c$, and define

$$h:C\to C:\sum_{n\ge 1}\frac{c_n}{3^n}\to\sum_{n\ge 1}\frac{c_n'}{3^n}\,,$$

where

$$c_n'=\begin{cases}
\bar c_n,&\text{if }n\in D\\
c_n,&\text{otherwise.}
\end{cases}$$

Clearly $h(a)=b$ (and $h(b)=a$), and it’s very easy to see that $h$ is a bijection that is its own inverse. Thus, all that remains is to show that $h$ is continuous, and for now, at least, I’ll leave it to you to try to do that. (Intuitively speaking, this map simply interchanges two of the closed intervals at each level $n$ of the construction of $C$ for which $n\in D$.)

Title: Cantor Set self homeomorphism


问:
Prove $ A \vdash \neg \neg A $ (Natural Deduction)
From this set of rules:

$ (\wedge E) $  $A \wedge B \vdash A $

$ (\wedge I) $ $A, B \vdash A \wedge B $

$ (\vee E) $ $ A \vee B, A \rightarrow C, B \rightarrow C \vdash C $

$ (\vee I) $ $ A \vdash A \vee B $

$ (\neg E) $ $ \neg A \rightarrow B, \neg A \rightarrow \neg B \vdash A $

$ (\neg I) $ $ A \rightarrow B, A \rightarrow \neg B \vdash \neg A $

$ (\rightarrow E) $ $ A, A \rightarrow B \vdash B $

$ (\rightarrow I) $ $ Premises \vdash A \rightarrow B $

With the last rule ($ \rightarrow I $), you can introduce any implication you like as long as you prove in a sub-proof the consequent of the implication under the assumption of the hypothesis of the implication e.g.

$ P \rightarrow Q \quad $ Premise

$ \neg Q \quad\quad\space\space $ Premise

$ P \rightarrow \neg Q \quad \rightarrow I,$ subproof below

3.1 $ P \quad\space\space\space $ Assumption

3.2 $ \neg Q \quad $ from (2)

$ \neg P \quad $ from (1)(3), $ \neg I $

Prove $ A \vdash \neg \neg A $

$ A \quad Premise $

$ \neg (\neg \neg A) \rightarrow A \quad \rightarrow I, subproof $

2.1 $ \neg (\neg \neg A) \quad Assumption $

2.2 $ A \quad\quad\quad (1) $

$ \neg (\neg \neg A) \rightarrow \neg (\neg \neg A) \quad \rightarrow I, subproof $

3.1 $ \neg (\neg \neg A) \quad Assumption $

3.2 $ \neg (\neg \neg A) \quad (3.1) $

$ \neg \neg A \quad (2)(3), \neg E $

This is the proof I have currently and I'm unsure about it because of line 3 (specificially, the consequent of the implication which would usually just be $\neg A$) and subsequently line 4 which uses line 3 as a premise.

Is my proof correct?

答:
(Posted after another answer was accepted)

Using a form of natural deduction...


问:
Does there exist a formula for $\int_0^{\infty} t^{k} {\tt sech}(t)dt$ that is correct whenever the real part of k is greater than negative 1?
The formula $\int_0^{\infty} t^{k} {\tt sech}(t)dt=\frac{(-1)^k}{2^{2k+1}}      \left(
     \psi^{(k) } \left(   \frac {3} {4}   \right)   -\psi^{(k)}\left(  \frac {1} {4}  \right)           \right)  $ is interesting; however, it is only true whenever k is a nonnegative integer. Does there exist a formula for $\int_0^{\infty} t^{k} {\tt sech}(t)dt$ that is correct whenever the real part of k is greater than negative 1?

$\space$

$\bf{\int_0^{\infty} t^{k-1} f(t)dt=(-1)^k \left[   F^{(k-1)}(s)  \right]_{s=0}^{s=\infty}}$.

Consider the Laplace transform of f(t) to be F(s), given by F(s)=$\int_0^{\infty} f(t) e^{-st} dt$.
We have that the Laplace transform of $t^n f(t)$ is $(-1)^n F^{(n)}(s)$, and we have that $\int_0^{\infty} \frac {f(t)} {t} dt = \int_0^{\infty} F(s)ds$. Hence, $\int_0^{\infty} \frac {t^k f(t)} {t} dt = \int_0^{\infty} (-1)^k F^{(k)}(s) ds$. This can be rewritten as $\int_0^{\infty} t^{k-1} f(t) dt = (-1)^k
  \int_0^{\infty} F^{(k)}(s) ds $; furthermore, $\int_0^{\infty} t^{k-1} f(t)dt=(-1)^k \left[   F^{(k-1)}(s)  \right]_{s=0}^{s=\infty}$.

$\space$

The Laplace transform of ${\tt sech}$(t) is $\bf{\frac {1} {2}\left(
     \psi^{(0) } \left(   \frac {s+3} {4}   \right)   -\psi^{(0)}\left(  \frac {s+1} {4}  \right)           \right) }$.

In the next section, we must know the Laplace transform of ${\tt sech}(t)$. Consider the fact that ${\tt sech}(t)=\frac {2} {e^t+e^{-t}}$. Then, $F(s)=\int_0^{\infty} \frac {2} {e^t+e^{-t}} e^{-st} dt$. Consider the digamma function, $\psi^{(0)}(z)=\int_0^{\infty} \frac {e^{-t}} {t}-\frac {e^{=zt}} {1-e^{-t}} dt$. $\int_0^{\infty} \frac{2} {e^t+e^-t}e^{-st}dt$ = $2 \int_0^\infty \frac {e^t} {e^{2t}+1} e^{-st}dt$ = $\frac {1} {2} \int_0^\infty \frac {e^{\frac {1} {4} t}} {e^{\frac {t} {2}}+1} e^{-\frac{1}{4}st}dt$ = $\frac {1} {2} \int_0^{\infty} \frac { e^{\frac{1} {4}(t-st)}} {e^{ \frac{t} {2}  }+1}dt$ = $\frac {1} {2} \int_0^{\infty} \frac { e^{t-\frac{1} {4}(s+3)t}} {e^{ \frac{t} {2}  }+1}dt$ = $\frac {1} {2} \int_0^{\infty} \frac {e^{-\frac{s+1}{4}t}-e^{-\frac{s+3}{4}t}} {1-e^{-t}}dt$ = $\frac{1}{2}\int_0^{\infty} \frac {e^{-\frac{s+1} {4}t}} {1-e^{-t}}-\frac{e^{-\frac{s+3}{4}t}}{1-e^{-t}}dt$ = $\frac {1} {2}\left(
     \int_0^{\infty} \frac {e^{-t}} {t} - \frac {e^{-\frac {s+3} {4}t}} {1-e^{-t}}dt  -  \int_0^{\infty} \frac {e^{-t}} {t} - \frac {e^{-\frac {s+1} {4}t}} {1-e^{-t}}dt
           \right) $ = $\frac {1} {2}\left(
     \psi^{(0) } \left(   \frac {s+3} {4}   \right)   -\psi^{(0)}\left(  \frac {s+1} {4}  \right)           \right) $

$\space$

$\bf{\int_0^{\infty} t^{k} {\tt sech}(t)dt=(-1)^k \frac{1}{2^{2k+1}}      \left(
     \psi^{(k) } \left(   \frac {3} {4}   \right)   -\psi^{(k)}\left(  \frac {1} {4}  \right)           \right)  }$

Consider $f(t)={\tt sech}(t)$. Then, we have $\int_0^{\infty} t^{k-1} {\tt sech}(t)dt=(-1)^k \left[   \frac{1}{2}\frac{d^{k-1}} {ds^{k-1} }  \left(
     \psi^{(0) } \left(   \frac {s+3} {4}   \right)   -\psi^{(0)}\left(  \frac {s+1} {4}  \right)           \right)  \right]_{s=0}^{s=\infty}$. Hence, $\int_0^{\infty} t^{k-1} {\tt sech}(t)dt=(-1)^k \frac{1}{2^{2k-2}} \left[  \frac{1}{2}   \left(
     \psi^{(k-1) } \left(   \frac {s+3} {4}   \right)   -\psi^{(k-1)}\left(  \frac {s+1} {4}  \right)           \right)  \right]_{s=0}^{s=\infty}$. First, we take the limit at infinity: $\lim_{s\rightarrow\infty} \frac{1}{2}\left(
     \psi^{(k-1) } \left(   \frac {s+3} {4}   \right)   -\psi^{(k-1)}\left(  \frac {s+1} {4}  \right)           \right) = \lim_{s\rightarrow\infty} \int_{0}^{\infty} \frac {2} {e^t+e^{-t}}e^{-st}dt$, which is zero. Thus, we have:  $\int_0^{\infty} t^{k-1} {\tt sech}(t)dt=-(-1)^k \frac{1}{2^{2k-1}}      \left(
     \psi^{(k-1) } \left(   \frac {3} {4}   \right)   -\psi^{(k-1)}\left(  \frac {1} {4}  \right)           \right)  $, which can be alternatively written as $\int_0^{\infty} t^{k} {\tt sech}(t)dt=(-1)^k \frac{1}{2^{2k+1}}      \left(
     \psi^{(k) } \left(   \frac {3} {4}   \right)   -\psi^{(k)}\left(  \frac {1} {4}  \right)           \right)  $.

答:
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
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 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
The answer you are looking for is related to the
Euler Number $\ds{E_{n}}$ because
$$
\on{sech}\pars{x} = \sum_{n = 0}^{\infty}{E_{2n} \over \pars{2n}!}\,x^{2n}
$$
I was trying to use the
Ramanujan's Master Theorem by rewriting the above expression as
$$
\on{sech}\pars{\root{x}} =
\sum_{n = 0}^{\infty}
\color{red}{{\Gamma\pars{1 + n}\cos\pars{n\pi} \over \Gamma\pars{1 + 2n}}\,E_{2n}}
\,{\pars{-x}^{n} \over n!}
$$
For this purpose, it was obvious that we need a $\ds{E_{\nu}}$ analytical continuation. Indeed, I found a paper where the author claims he found the coveted continuation -as related to the Riemman $\ds{\zeta}$ function-. However, I was unable to reproduce his results. My "checking" didn't agree with his claim. I hope this "big comment/no answer" will be helpful to you.


问:
What is the dimension of a vector space without a basis?
It's my understanding that without the axiom of choice, the vector space of $\mathbb{R}$ over $\mathbb{Q}$ lacks a set of basis vectors. Now it seems straight forward that with the axiom of choice that the following holds.
$$\text{dim }(\mathbb{R}) = \aleph_1$$
However, what happens if we assume the axiom of choice is false? Does the dimension simply become undefined? As we can't define any basis for the space, it seems like the question of dimension becomes meaningless; you can't have the cardinality of something that doesn't exist. Alternatively, do we have an alternate way of defining dimension for these cases?

答:
Careful: with the axiom of choice we get

$$\dim_{\mathbb{Q}}(\mathbb{R}) = 2^{\aleph_0} = \mathfrak{c}$$

(the cardinality of the continuum). To assert that $2^{\aleph_0} = \aleph_1$ requires the continuum hypothesis.

Without the axiom of choice, if we define the dimension as the size of a basis, then yes, without a basis this particular notion of dimension is undefined. So we might try to use others.

For example, we can talk about the rank, which is the minimal size of a set of generators of $\mathbb{R}$ as a $\mathbb{Q}$-vector space. Since $\mathbb{R}$ is uncountable no countable subset generates. And on the other hand $\mathbb{R}$ itself clearly generates. So assuming the continuum hypothesis the rank is $2^{\aleph_0} = \mathfrak{c}$ again. I don't know if it's consistent with ZF that the rank is somewhere strictly between $\aleph_0$ and $2^{\aleph_0}$.

There is also a somewhat different notion of rank given by the maximal size of a linearly independent subset. We can explicitly exhibit a linearly independent subset of $\mathbb{R}$ of cardinality $2^{\aleph_0}$ in ZF (see this MO answer). So it's provable in ZF that the rank in this sense is $2^{\aleph_0}$.

(In the absence of choice there's a subtle distinction to be made between "minimal size of a set of generators" vs. "size of a minimal set of generators" since without Zorn's lemma a minimal set of generators need not exist, and similarly for "maximal size of a linearly independent subset" vs. "size of a maximal linearly independent subset." Either a minimal set of generators or a maximal linearly independent subset must be a basis. I suppose in the absence of choice cardinals don't need to be totally ordered either though. So either way the rank might end up undefined in general. Welp.)


问:
Laurent series and apparent essential singularity
After reading "Mathematics for physicists" by Susan M. Lea I encountered a subtlety that I can't turn my head around (p. 128). Consider function

$$f(z)=\frac{1}{z^2-1} = \frac{1}{2}\left[\frac{1}{z-1}-\frac{1}{1+z}\right]$$

which clearly has 2 simple poles at $z=\pm 1$. If we want to compute its Laurent series around $z=1$, we would get:

$$\frac{1}{1+z}=\frac{1}{2+(z-1)} = \frac{1}{(z-1)\left(1+\frac{2}{z-1}\right)}$$
$$=\frac{1}{z-1}\sum_{n=0}^\infty\frac{(-2)^n}{(z-1)^{n}}$$
and thus:
$$\frac{1}{z^2-1} = \frac{1}{2}\left[\frac{1}{z-1}-\sum_{n=0}(-1)^n\frac{2^n}{(z-1)^{n+1}}\right] = \frac{1}{2}\sum_{n=1}^\infty\frac{(-2)^n}{(z-1)^{n+1}}$$

What I see here is an infinite number of negative powers of (z-1) and thus, I would be tempted to conclude that there is an essential singularity at this point. What prevents me from concluding this as it is clear that there is no essential singularity from the analytic function?

答:
As I mentioned in the comments, the series you derived does not exists in a punctured neighborhood of the point $z=1$ as it only converges in $|z-1|>2$.

You should have expanded $\frac{1}{1+z}$ via
$$\frac{1}{1+z}=\frac{1}{2-(1-z)}=\frac{1}{2}\frac{1}{1-\frac{1-z}{2}}=\frac12\sum_{n=0}^\infty\left( \frac{1-z}{2}\right)^n$$
and it converges in $|z-1|<2$. So about $z=1$ your full Laurent series is
$$f(z)=\frac{1}{2}\left[-\frac{1}{1-z}-\frac12\sum_{n=0}^\infty \left(\frac{1-z}{2}\right)^n \right]=-\frac{1}{4}\sum_{n=-1}^\infty \left(\frac{1-z}{2}\right)^n. $$


问:
Probability that a 5-card poker hand contains exactly four cards of the same suit?
I'm not that confident in my ability, so I've come here.

This is what I have:

$\frac{C(4,1) * C(13,4) * C(3,1) * C(13,1)}{C(52,5)}$

My thinking is that $C(4,1)$ selects one suit from the 4 suits and $C(13,4)$ selects 4 cards from that suit.
Then $C(3,1)$ selects another suit from the remaining 3 suits and $C(13,1)$ selects one card from that suit.

Then divide by $C(52,5)$.

Is this correct?

答:
Yes, that is correct.  As a sanity check, an alternative approach is to choose the two suits, choose which one contains the four cards, and then choose four cards from that one suit and one from the other suit:
$$\binom{4}{2}\binom{2}{1}\binom{13}{4}\binom{13}{1}=111,540$$


问:
How do I find an explicit formula, using backtracking, for $C_n =3C_{n-1} -2C_{n-2}$ with initial values $C_1=5$ and $C_2=3$?
I have the recurrence $C_n =3C_{n-1} -2C_{n-2}$ with initial values $C_1=5$ and $C_2=3$, and want to find an explicit formula for $C_n$.

I made a table with the terms up to the $6$th term and noticed:
\begin{align}
C_3 &= C_2-4\\
C_4 &= C_3-8\\
C_5 &= C_4-16\\
C_6 &= C_5-32
\end{align}
But I still can't find the correct formula.

答:
I present to you the overkill method.

Suppose we have the general recurrence $f_n$ given by $$f_{n+2}=af_{n+1}+bf_n,\tag1$$
where $f_0,f_1,a,b$ are known constants. Define the generating function
$$f(x)=\sum_{n\ge0} f_nx^n=f_0+f_1x+f_2x^2+\dots$$
Multiply $(1)$ by $x^{n+2}$ and sum over $n\ge0$:
$$
\begin{align}
\sum_{n\ge0}f_{n+2}x^{n+2}&=ax\sum_{n\ge0}f_{n+1}x^{n+1}+bx^2\sum_{n\ge0}f_nx^n\\
\sum_{n\ge2}f_nx^n &=ax\sum_{n\ge1}f_nx^n+bx^2f(x)\\
f(x)-f_1x-f_0&=ax(f(x)-f_0)+bx^2f(x)\\
f(x)&=\frac{f_0+(f_1-af_0)x}{1-ax-bx^2}.\tag2
\end{align}
$$
We then write
$$
\frac{f_0+(f_1-af_0)x}{1-ax-bx^2}=\frac{\xi_1}{1-\zeta_1x}+\frac{\xi_2}{1-\zeta_2x},\tag3
$$
for constants $\zeta_1,\zeta_2,\xi_1,\xi_2$. Adding together the rational functions on the right hand side of $(3)$ and comparing coefficients on the numerator and denominator of each side, we get
\begin{align*}
    f_0&=\xi_1+\xi_2, \qquad \qquad &a&=\zeta_1+\zeta_2,\\
    af_0-f_1&=\xi_1\zeta_2+\xi_2\zeta_1, \qquad \qquad &b&=-\zeta_1\zeta_2.
\end{align*}
The solution to this system is
\begin{equation}
\begin{bmatrix}
\zeta_1 \\
\zeta_2 \\
\xi_1 \\
\xi_2 \\
\end{bmatrix}
=
\begin{bmatrix}
\tfrac{a+\sqrt{D}}{2}\\
\tfrac{a-\sqrt{D}}{2}\\
\tfrac{1}{\sqrt{D}}\left(f_1+\tfrac{-a+\sqrt{D}}{2}f_0\right)\\
\tfrac{1}{\sqrt{D}}\left(-f_1+\tfrac{a+\sqrt{D}}{2}f_0\right)
\end{bmatrix},\qquad D=a^2+4b.
\end{equation}
Then from $(3)$ we can write $$f(x)=\xi_1\sum_{n\ge0}(\zeta_1x)^n+\xi_2\sum_{n\ge0}(\zeta_2x)^n=\sum_{n\ge0}(\xi_1\zeta_1^n+\xi_2\zeta_2^n)x^n, \qquad |x|\le 1/\min(|\zeta_1|,|\zeta_2|),$$
so that $$f_n=\xi_1\zeta_1^n+\xi_2\zeta_2^n,\qquad n\ge0.$$
Noticing that we have $C_n=f_{n-1}$, with $(f_0,f_1,a,b)=(5,3,3,-2)$ we get
\begin{equation}
\begin{bmatrix}
\zeta_1 \\
\zeta_2 \\
\xi_1 \\
\xi_2 \\
\end{bmatrix}
=
\begin{bmatrix}
2\\
1\\
-2\\
7
\end{bmatrix}.
\end{equation}
Hence $$C_n=-2\cdot2^{n-1}+7\cdot1^{n-1}=7-2^n.$$


问:
Prove A is symmetric matrix iff $A$ is square and $x^T Ay = (Ax)^T y$
Prove A is a symmetric matrix iff $A$ is square and $x^T Ay = (Ax)^T y$. (for all $x,y \in \mathbb{R}^n$)

Going from the assumption that it is symmetric to the two conditions is fairly straightforward.

However, going the other way, I am stuck at proving $A^T = A$ using the second and condition, being stuck at $X^T (A-A^T)y=0$.

Note T is for transpose!

答:
First, lets remember the rules of transposing the product: $(AB)^T = B^T A^T$

Using this, we start with the giver equation $ x^TAy=(Ax)^Ty$ and apply the rule above which yields
$x^TAy = x^TA^Ty$.

Note that the equation is true for all $x, y$, so the only way that is possible is if $A = A^T$.


问:
Proving a function is quasi-concave
This is from economics, but I think there's a lot of math involved and I want to make sure I didn't mess anything up.  There is a utility function U = $x_1$ + $\ln(x_2)$ such that $x_1$ and $x_2$ are nonnegative.

How do I prove this function is quasi-concave?

I was looking online for any notes about this, and it tells me I need to create this

https://mjo.osborne.economics.utoronto.ca/index.php/tutorial/index/1/qcc/t

"bordered Hessian matrix"

As in for this problem it seems to me I need to calculate

\begin{bmatrix}
0 & f'_{1}(x) & f'_{2}(x)\\
f'_{1}(x) & f''_{11}(x) & f''_{12}(x) \\
f'_{2}(x) & f''_{21}(x) & f''_{22}(x) \\
\end{bmatrix}

So I do that for the above function and I get

\begin{bmatrix}
0 & 1 & \frac{1}{x_2} \\
1 & 0 & 0 \\
\frac{1}{x_2} & 0 & -\frac{1}{x_{2}^{2}} \\
\end{bmatrix}

Then it says I need to find the determinant and if n = even, I need $D_{n} \geq 0$.  So here $D_{2}$ for the above matrix I got to be $\frac{1}{x_{2}^{2}}$.  Since $x_2$ is nonnegative, this means that the determinant is $\geq 0$.

Does this prove this function is quasi-concave?  Is my general approach right or am I totally missing the point of this question?

What about strictly quasi-concave?   The online notes I found never mentions this, but is that just when it is strictly greater than 0?  So in this case, this function is also strictly quasi-concave too?  If someone can verify if my guess is correct, that would be great.

Finally, on an unrelated note, how do I prove this function is continuous using math?  I know it is when I graph it, but is there a mathematical proof for this?

Thank you so much for the help!

答:
I actually am not familiar with the method you're using, so I'm using a different method.

The function is actually concave on the (natural) domain $\Bbb{R} \times (0, \infty)$. Note that the function $(x_1, x_2) \mapsto \ln(x_2)$ is concave, because the function $\ln$ is concave (check its second derivative). The function $(x_1, x_2) \mapsto x_1$ is an affine function, and hence is concave (and convex). Summing two concave functions produces a concave function, and every concave function is quasiconcave.

The only question remaining is strictness. Neither of the above functions are strictly (quasi)concave, so we need a separate argument. Strict quasiconcavity means that, for all $(x_1, x_2), (y_1, y_2)$ in the domain, and any $\lambda \in (0, 1)$, we have
$$f(\lambda x_1 + (1 - \lambda)y_1, \lambda x_2 + (1 - \lambda)y_2) > \min \{f(x_1, x_2), f(y_1, y_2)\}.$$
Quasiconvexity means the above with $\ge $ substituted for $>$. So, let's suppose that we have equality. Without loss of generality, assume $f(x_1, x_2) \le f(y_1, y_2)$. Then,
$$\lambda x_1 + (1 - \lambda)y_1 + \ln(\lambda x_2 + (1 - \lambda)y_2) = x_1 + \ln(x_2).$$
By the strict concavity of $\ln$ (again, examine the second derivative), we have
\begin{align*}
&x_1 + \ln(x_2) > \lambda x_1 + (1 - \lambda)y_1 + \lambda\ln(x_2) + (1 - \lambda)\ln(y_2) \\
\iff \, &(1 - \lambda)x_1  + (1 - \lambda)\ln(x_2) > (1 - \lambda)y_1 + (1 - \lambda)\ln(y_2) \\
\iff \, &x_1  + \ln(x_2) > y_1 + \ln(y_2) \iff f(x_1, x_2) > f(y_1, y_2),
\end{align*}
which contradicts $f(x_1, x_2) \le f(y_1, y_2)$. Therefore, $f$ is quasiconcave.

Title: Proving a function is quasi-concave


问:
Left and right product with a PD diagonal can be expressed as a PD quadratic form?
I conjecture the following: If $D$ is diagonal, $D \succeq I$, and $V$ has linearly independent columns,
then there exists $R \succeq I$ such that
$$
 V^T R V
 =
 D V^T V D.
$$
I can prove the theorem when $V$ is square. For the case where V is "tall", I have tested this numerically with randomly chosen values of $V$ and $D$.
The numerical solution performs a linear least-squares solve of
$$
\DeclareMathOperator{\Vec}{Vec}
\DeclareMathOperator{\rank}{rank}
 (V^T \otimes V^T) \Vec(R)
 =
 \Vec(D V^T V D)
$$
where $\otimes$ denotes the Kronecker product.
This is equivalent to solving for $R$ in the equality stated in the conjecture.
I check that the solution satisfies $\Vec^{-1}(R) \succeq I$ up to a numerical tolerance.

It is true that
$$
\rank(A \otimes B) = \rank(A) \rank(B),
$$
so we can assert that the solution $\Vec(R)$ exists,
but proving that $R \succeq I$ seems harder.

答:
This is false. Consider $V=\pmatrix{1&1\\ 0&1}$ and $D=\pmatrix{1\\ &2}$. Then
$$
R=(V^T)^{-1}DV^TVDV^{-1}=\pmatrix{1&1\\ 1&5}
$$
and clearly $R-I$ is indefinite.


问:
Is the sum of two coprime numbers also coprime to its summands?
I was comparing the statement of the ABC conjecture on Wikipedia to an article I found geared towards laymen. The propositions differ slightly, but they would be equivalent if the following were true:

If $a$ and $b$ are coprime, and $a + b = c$, is $c$ coprime to both $a$ and $b$?

By division algorithm, if (without loss of generality) $a > b$, then $a$ and $c$ must be coprime: the first iteration of the algorithm subtracts $c = 1*a + b$, and then you're left with $a$ and $b$, which are coprime by proposition and so generate 1 as the final result.

If $b$ is prime, I can imagine a proof by contradiction via the division algorithm.

For more complicated $b$ I am not sure it holds.

答:
$4$ is coprime to $7$, and their sum $11$ is coprime to both.

On the other hand, if $p\mid a$ and $p\mid a+b$, then $p\mid (a+b)-a=b$.


问:
Estimating number of infected people and getting bounds of its probability based on a few samples.
This question came on my exam a few week ago, and I've been stuck on it ever since.

Say a hospital has received 500 blood samples for COVID testing. We want to estimate how many of the the samples are of infected people before testing them all. Also, we can assume that are no false positives/negatives.

Let m be the total no. of infected samples out of these 500. Now, 20 samples are randomly chosen, and it is found that n of these are of infected people. These are the following questions we have to answer:

What is the estimate of m, given the value of n? That is, $E[M|N=n]$ where N denotes the random variable  of infected samples from the 20 randomly chosen ones, and M denotes the number of infected samples out of 500.

Say, 4 out of these 20 are of those infected. What is $P[M>E[M] + 1| N = 4]$? This is to find out how reliable the expected value is.

Any other assumptions required can be made.

My approach:

$P(M=m, N=n) = \frac{\binom{m}{n}\binom{500-m}{20-n}}{\binom{500}{20}}$, so we can find $E[M|N=n]$ as $\sum_{m=n}^{500} m\frac{P(M=m, N=n)}{Pr(N=n)}$, where $Pr(N=n) = \frac{1}{21}$(I don't think this part is correct, but since $n ={0,1,2....20}$, I wrote this down)

Thanks in advance, this question has really been bugging me!

答:
Your expression $\ {500\choose20}\ $ is the number of subsets of size $20$ that can be drawn from your set of $500$.  The expression $\ {m\choose n}{500-m\choose 20-n}\ $ is, for a given subset of size $\ m\ $, the number of subsets of size $\ 20\ $ which include exactly $\ n\ $ members of that given subset and exactly $\ 20-n\ $ members of its complement.  So, if each subset of size $\ 20\ $ is equally likely, the fraction $\ \frac{{m\choose n}{500-m\choose 20-n}}{{500\choose20}}\ $ is the probability that your randomly chosen subset of size $\ 20\ $ contains exactly $\ n\ $ members of the given subset of size $\ m\ $.

The problem here is that the description of the problem makes $\ m\ $ a non-random number whose value is already determinable, although unknown.  Its exact value could be obtained by testing all $500$ blood samples.  Non-Bayesians would consider it inappropriate to treat it as a random variable and if they had to estimate its value from a random sample of size $20$ they would probably use some sort of significance test.

That your exam question does treat it as a random variable implies, I presume, that you're required to adopt a Bayesian approach, which would entail your assigning a prior distribution to that random variable. For the moment, let's treat this prior, $\ \pi\ $, say, as arbitrary:
$$
\pi_m=P(M=m)\ .
$$
You can then obtain the posterior distribution of $\ M\ $, given $\ N=n\ $ , from Bayes's theorem:
\begin{align}
P(M=m\,|\ N=n\,)&=\frac{P(N=n\ |\,M=m\,)P(M=m)}{P(N=n)}\\
&=\cases{ \frac{{m\choose n}{500-m\choose 20-n}\pi_m}{\sum_{i=n}^{480+n} {i\choose n}{500-i\choose 20-n}\pi_i}& if $\ n\le m\le n+ 480$\\
0&otherwise .}
\end{align}
Your formula for $\ E[M|N=n]\ $ is correct, except that \begin{align}
P(M=m, N=n)&=\frac{{m\choose n}{500-m\choose 20-n}\pi_m}{{500\choose20}}\ \text{ and}\\
P(N=n)&= \displaystyle\sum_{i=n}^{480+n} \frac{{i\choose n}{500-i\choose 20-n}\pi_i}{{500\choose20}}\ .
\end{align}
The value of $\ P\big(M>E[M]+1\,|\,N=4\big)\ $ is given by
\begin{align}
P\big(M>E[M]+1\,|\,N=4\big)&=\sum_{m=
\max(\lfloor E[M] \rfloor +2, 4)}^{484}P(M=m\,|\,N=4\,)\\
&=\frac{\sum_{m=
\max(\lfloor \sum_{m=0}^{500}m\pi_m \rfloor +2, 4)}^{484} {m\choose 4}{500-m\choose 16}\pi_m}{\sum_{i=4}^{484} {i\choose 4}{500-i\choose 16}\pi_i}\ .
\end{align}
Coming to the vexed question of what to choose for $\ \pi\ $, the choice of priors in Bayesian statistics is nearly always going to be somewhat subjective.  Since I have no idea how this topic was treated in your course, I also have little idea how your examiners would have expected you to handle it in the exam question you've quoted.  Also, while I have used statistics professionally, I certainly wouldn't claim to have ever been a professional statistician, let alone one with a good working knowledge of Bayesian statistics.  Please bear that in mind while reading the following suggestion.

If your $\ 500\ $ blood samples were taken somewhat randomly from a large population, it seems to me that a reasonable choice for $\ \pi\ $ would be $\ \text{Binomial}(500,p)\ $:
$$
P(M=m)={500\choose m}p^m(1-p)^m
$$
for some value of $\ p\ $, which would be the proportion of the large population that are infected.  For COVID-$19$, $\ p\ $ will not be known exactly, but if you know the population from which the sample was drawn you may have a reasonable estimate that you could use for the value of $\ p\ $.  Otherwise, the best you're likely to be able to do is to get expert epidemiologists to suggest a range $\ [a,b]\ $ in which they think $\ p\ $ is $90\%$ (say) likely to lie, and choose a suitable prior distribution $\ \Pi\ $ for $\ p\ $ such that  $\ \Pi\big([a,b]\big)=0.9\ $. Your prior distribution for $\ M\ $ will then be
$$
P(M=m)={500\choose m}\int_0^1p^m(1-p)^md\Pi(p)\ ,
$$
and $\ E[M]=500E[p]=500\displaystyle\int_0^1pd\Pi(p)\ $.


问:
Find $\lim_{(x\rightarrow 0^{+})} \frac{\ln(x+\arccos^{3}x)-\ln x}{x^{2}}$.
Find  $\lim_{(x\rightarrow 0^{+})} \frac{\ln(x+\arccos^{3}x)-\ln x}{x^{2}}$.

After some basic steps, I reached to $\frac{\ln(x+\arccos^{3}x)-\ln x}{x^{2}}= \frac{\ln(1+\frac{\arccos^{3}x}{x})}{x^{2}}$. I intended to use Sandwich theorem because $0\leq  \arccos^{3}x\leq  \pi ^{3}$, but it did't seem to work. Can I have others ways to approach for the problem? Please help me, thank you so much!

Body:
If you compose Taylor series around $x=0$
$$\cos ^{-1}(x)=\frac{\pi }{2}-x+O\left(x^3\right)$$
$$\cos ^{-1}(x)^3=\frac{\pi ^3}{8}-\frac{3 \pi ^2 x}{4}+O\left(x^2\right)$$
$$1+\frac{\cos ^{-1}(x)^3}{x}=\frac{\pi ^3}{8 x}+1-\frac{3 \pi ^2}{4}+O\left(x^1\right)$$


问:
Use Euler's Formula to prove the following equation $ y = Ae^{\sqrt{\lambda}ix} + Be^{-\sqrt{\lambda}ix} $
I'm having a difficult time understanding how one would take $ y = Ae^{\sqrt{\lambda}ix} + Be^{-\sqrt{\lambda}ix} $ and end up with $ y = A \cos{\sqrt{\lambda}x} + B\sin{\sqrt{\lambda}x} $ knowing that $ \lambda > 0 $.

Would it be easiest to start out by just plugging in Euler's Formulas $e^{i\theta} = \cos(\theta) + i\sin(\theta)$ and $e^{-i\theta} = \cos(\theta) - i\sin(\theta)$? The only problem I have is I end up with this very long equation of $ y = A \cos{\sqrt{\lambda}x} + Ai\sin{\sqrt{\lambda}x} + B\cos{\sqrt{\lambda}x} - Bi\sin{\sqrt{\lambda}x}$. I'm really confused as to how this long expression can be reduced to the $ y = A \cos{\sqrt{\lambda}x} + B\sin{\sqrt{\lambda}x} $ it is asking for.

Any help is appreciated!

答: You can just add the complex coeffiecients together so that  $$y = A \cos(t) + Ai\sin(t) + B\cos(t) - Bi \sin(t)= (A + B)\cos(t) + (Ai - Bi)\sin(t).$$


答:
As pointed out by @Andrei , when you carefully observe, the expression is of the form

$$\frac{\infty}{0}\to\infty$$

Body:
Find  $\lim_{(x\rightarrow 0^{+})} \frac{\ln(x+\arccos^{3}x)-\ln x}{x^{2}}$.

After some basic steps, I reached to $\frac{\ln(x+\arccos^{3}x)-\ln x}{x^{2}}= \frac{\ln(1+\frac{\arccos^{3}x}{x})}{x^{2}}$. I intended to use Sandwich theorem because $0\leq  \arccos^{3}x\leq  \pi ^{3}$, but it did't seem to work. Can I have others ways to approach for the problem? Please help me, thank you so much!


答: Yes. You seek the probability for obtaining one from four suits with four from its thirteen cards, and one from the remaining three suits with one from its thirteen cards, when selecting any five from all fifty-two cards.$$\def\cbinom#1#2{\mathop{^{#1}\mathrm C_{#2}}}\dfrac{\cbinom 41\cbinom {13}4\cbinom 31\cbinom{13}1}{\cbinom{52}5}$$

Title: Probability that a 5-card poker hand contains exactly four cards of the same suit?

Body:
I'm not that confident in my ability, so I've come here.

This is what I have:

$\frac{C(4,1) * C(13,4) * C(3,1) * C(13,1)}{C(52,5)}$

My thinking is that $C(4,1)$ selects one suit from the 4 suits and $C(13,4)$ selects 4 cards from that suit.
Then $C(3,1)$ selects another suit from the remaining 3 suits and $C(13,1)$ selects one card from that suit.

Then divide by $C(52,5)$.

Is this correct?


问:
Proving the function g is continuous without using continuous function theorem
Suppose that $f$ is continuous at $a$, $f(a) = 0$, and $g(x) = 5f(x)^{2}$. Give an $\epsilon - \delta$ proof that g is continuous at a. *Do not use the continuous function theorem.

I am confused as to how I can choose a $\delta$.

答:
$g(a)=0$, hence you have to prove that $\forall\epsilon>0\;\;\exists\delta>0$ such that
$$
|g(x)|=5f(x)^2<\epsilon\;\;\;\forall|x-a|<\delta.
$$
Since $f$ is continuous at $a$ by hypotesis, for every $\epsilon>0$ there exists $\delta$ such that
$$
|f(x)|<\sqrt{\epsilon/5}\;\;\;\forall|x-a|<\delta
$$
which allows you to conclude.


问:
Evaluating exponential function without overflow
I have the following function:
$$
f(x) = \frac{1}{A} \cdot \frac{e^{\sqrt{A}x} + e^{\sqrt{A}-\sqrt{A}x}}{B\sqrt{A}e^{\sqrt{A}} - B\sqrt{A} + e^{\sqrt{A}} + 1}
$$

where $A = 10^{6}$ and $B = 10^{-7}$. I am attempting to do several calculations and visualizations of $f$ for $x \in [0,1]$. I obviously get overflow errors due to the presence of $e^{\sqrt{A}}$ and $e^{\sqrt{A}x}$, but I know for a fact that $f$ is bounded in the region I am evaluating it in. I tried to fix this by doing the following manipulation:
$$
f(x) = \frac{e^{\sqrt{A}x - \sqrt{A}}}{A} \cdot \frac{1 + e^{\sqrt{A} - 2\sqrt{A}x}}{B\sqrt{A} - B\sqrt{A}e^{-\sqrt{A}} + 1 + e^{-\sqrt{A}}}.
$$

But the issue is that for small $x$, $e^{\sqrt{A} - 2\sqrt{A}x}$ still overflows. So I'm wondering is there some clever manipulation I could do that allows me to fully evaluate $f$ in the desired region?

答:
When $x$ is small, use Taylor
$$e^{\sqrt{A} (x-1)}=e^{-\sqrt{A}}\left(1+\sqrt{A} x+\frac{A }{2}x^2+O\left(x^3\right) \right)$$


问:
Probability of an event with two or more parameters
So I'm trying to create a model that will predict car accident probability in the following year using driver's personal parameters, such as age, driving experience, sex etc.

So far I have calculated accident probabilities for each parameters, for example, I know that if a person is male, he'll get in an accident with probability $P_{\text{male}}$, if a driver is female, probability of an accident will be $P_{\text{female}}$ etc.

But I don't understand how to combine these parameters together, for example how to calculate probability of a car accident of a 20-year old male person having:

$P_{\text{male}}$ = 0.1

$P_{\text{20yr}}$ = 0.15

$P_{\text{20yr male}}$ = ?

答:
The problem you describe has been solved by actuaries many, many years ago in the form of predictive rating models.  These typically use a number of rating factors that are empirically adjusted based on historical data.

What you have not considered is that there are correlations between different rating factors.  Age and sex are (weakly) correlated, for instance.  The marginal risk probabilities do not obey assumptions of independence; e.g., the probability of a 50 year old male driver having an accident is not the product of the individual probabilities for a 50-year old having an accident, and a male having an accident.  Instead, if a book of business is large enough, the insurer can directly estimate the risk factors for this subset of policyholders by looking at their historical loss experience.  For those combinations that are not well-represented by a sufficiently large cohort, other methods are used, typically looking at a larger subset and extrapolating the risk through a (semi)parametric model; e.g., proportional hazards.

Insurers can specify a set of rules, or tables of rating factors, to instruct a regulator how a book of business is rated.  In the present day, these are programmed into a computer system.  Such tables may be hundreds of pages of output, covering every combination of every rating factor.

So, how are these tables created?  As I pointed out, they are based off historical loss data.  More recent data is weighted more heavily than older data.  A commonly employed rating mechanism in personal lines auto insurance is the "bonus-malus" system.  The scope of this topic is too broad to cover in a single answer, and is not particularly on-topic for this site; therefore you would be best advised to research the topic of "insurance ratemaking" online.


问:
Surface Integral Question
I have this question:

Evaluate $\int r . dS$ over the surface of a sphere, radius a, centred at the origin.

I'm not really sure what '$r$' is supposed to be? I would guess a position vector? If so, I would have $r . dS$ as $(asin\theta cos\phi, a sin\theta sin\phi, acos\theta) . (a^2sin\theta d\theta d\phi)$ which doesn't seem right. Any pointers appreciated, thanks!

答:
It's very easy, you don't even need to change coordinates or anything of this kind. Just note that at each point of the sphere the normal unit vector $n$ points outwards, so he is radially pointing away from the sphere. However, the position vector of this point also points radially outwards. This means that both $r$ and $n$ have the same direction. Also, you know that the magnitude of $r$ is $a$. Why you know that? Well, $r$ is the position of each point on the sphere, and by definition each point is at a distance $a$ from the origin.

Now, you have $r\cdot n = |r||n|\cos\theta$, however $\theta =0$ since they're paralel, $|n|=1$ because $n$ is unitary and $|r| = a$ by the preceding discussion. So $r\cdot n = a$. Now you'll get:

$$\int_S r \cdot n \ = a \int_S 1$$

And recall that the integral of the constant function $1$ over any surface gives you the area of the surface which you know is $4\pi a^2$. So you get:

$$\int_S r \cdot n = 4\pi a^3$$

Title: Surface Integral Question


问:
Proving Vector Identities
Let r=$(x,y,z)$ and $r=$||r||.

(A) Prove that $\nabla^2r^3=12r$.

(B) Is there a value of $p$ for which the vector field f(r) = r/$r^p$ is solenoidal?

What I have tried:

For part (a) I think that $r^3=\sqrt{27} $ but I am unsure of what to do next in terms of the del operator.

答:
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
$\ds{\large\pars{A}:}$
\begin{align}
\nabla^{2}r^{3} & = \nabla\cdot\pars{\nabla r^{3}} =
\nabla\cdot\pars{\totald{r^{3}}{r}\,{\vec{r} \over r}} =
3\nabla\cdot\pars{r\,\vec{r}}
\\[5mm] & =
3\bracks{\pars{\nabla r}\cdot\vec{r} + r\nabla\cdot\vec{r}}
\\[5mm] & =
3\bracks{\pars{\totald{r}{r}\,{\vec{r} \over r}}
\cdot\vec{r} + r\pars{3}}
\\[5mm] & = 3r + 9r = \bbx{12r} \\ &
\end{align}

$\ds{\large\pars{B}:}$
\begin{align}
\nabla\cdot\pars{\vec{r} \over r^{p}} & =
\pars{\nabla r^{-p}}\cdot\vec{r} +
r^{-p}\,\,\nabla \cdot\vec{r}
\\[5mm] & =
\pars{\totald{r^{-p}}{r}\,\,{\vec{r} \over r}}\cdot\vec{r} + r^{-p}\pars{3}
\\[5mm] & =
\underbrace{\pars{-pr^{-p - 1}\,\,\,{\vec{r} \over r}}\cdot\vec{r}}_{\ds{-pr^{-p}}}\
+\ 3r^{-p}
\\[5mm] & =
\pars{-p + 3}r^{-p}
\end{align}
Then,
$$
\nabla\cdot\pars{\vec{r} \over r^{p}} = 0 \implies
\bbx{p = 3} \\
$$

Title: Proving Vector Identities


问:
Is it possible to find general solution to 3 linear equations with 4 variables having only 2 particular solutions.
The question is simple - I would like to know if it's possible to describe general solution in the system of 3 linear equations with 4 variables ONLY with an information that (1, 2, 3, 4) and (2, 0, 0, 1) are particular solutions. I've tried to do it for so long that at this point I am not sure if it's even possible.

Any help would be much appreciated.

答: I don't think constraining yourself to only these two possible solutions is practical because the possible solutions depend on the rank of the equations and variables when you put them into a matrix. From you description, if you put them into an m by n matrix, the m will be 3 and n will be 4. If your matrix has rank lesser than m, it will have no solution or only one solution but if the rank is greater than m but less than n it would have practically an infinite number of solutions because of the null space. But since the n > m, then you are guaranteed to have in either case an infinite number of solutions. So, what you are looking at in the worst case is a matrix that has 0, or infinite solutions. (1,2,3,4) and (2,0,0, 1) might be among the infinite solutions.</p>


问:
An integer $n \geq 2$ is called square-positive- proof?
An integer $n \geq 2$ is called square-positive if there are $n$ consecutive positive integers whose sum is a square. Determine the first four square-positive integers.

So I have found the first four square-positive numbers, but I need to prove that why it $4$ is not a square-positive number and I also need to write a general formula for determining whether a number is square-positive or not. I have tried to write the sum of consecutive positive integers like this $a + a +1 + a + 2 + a+3 \dots a - 1$ and I wrote it like this for all numbers, and part of the proof for why $4$ isn't a square-positive number is that $4a + 6$ is not divisible with $4$. But I haven't got so far.

Here is my answer:

2 : 4 + 5 = 9  which is 3^2

3 : 2 + 3 + 4 = 9 which is 3^2

5: 18 + 19 + 20 + 21 + 22 = 100  which is 10^2

6: 35 + 36 + 37 + 38 + 39 + 40 = 225  which is 15^2

Interesting fact is that for all odd numbers and some even numbers like 6 and 10, you can find out which number is the first (the one you start with and then go forward here like 3, 2, 18 and 35) using this formula :

(I show it in an example because I still can't write it algebraically):

For example: the sum of 95 subsequent numbers is 5n + 10

(10^2 - 10) /5 = 18

So your first number is 18

And if you keep adding, 18 + 19 + 20 + 21 + 22 you get 100 which is 10^2, the same number you squared.

答:
We prove the following result:

If $n=2^b \cdot d, d \text{ is odd, then } n \text{ is square-positive if and only if } b=0  \text{ or } b  \text{ is odd}$.

$\underline{\text{Case 1}}$: $b$ is even and $b>0$. Let $b=2c$. We prove by contradiction $n$ cannot be square-positive. If it were, then there exist $n$ consecutive positive integers,
$$
a, a+1, \ldots, a+n-1
$$

such that the sum of these numbers is a square, i.e.,

$$
S=a+(a+1)+\cdots + (a+n-1) = n\frac{n+2a-1}{2} =T^2
$$

for some $T\in \mathbb{N}$.

Note that $n=2^{2c}\cdot d$ is even, and
$$
T^2 = S = 2^{2c-1}\cdot d \cdot (n+2a-1),
$$

which cannot be a perfect square because 1) $d(n+2a-1)$ is odd and, 2) $2c-1$, the highest power of its factor $2$ is odd, a contradiction.

$\underline{\text{Case 2}}: b=0$. Let $n=d=2m+1$, then the sum of the following $n$ consecutive numbers is a perfect square:

$$
n-m, n-m+1, \ldots, n+m.
$$

Indeed,

$$
S = (n-m)+(n-m+1)+\cdots +(n+m) = n \cdot \frac{(n-m)+(n+m)}{2}=n^2.
$$

$\underline{\text{Case 3}}: b=2c+1 $ for some $c\geqslant 0$, then $n=2^{2c+1}d$. We show that the sum of the following $n$ consecutive numbers is a perfect square:

$$
a, a+1, \ldots, a+(n-1),
$$

where

$$
a=\frac{1+d(K^2-2^{2c+1})}{2}
$$

and $K$ is some odd positive integer such that

$$
K^2 > 2^{2c+1}.
$$

For example, if $c\geqslant 1$, we can pick $K=2^{c+1}-1$. If $c=0$, we can pick $K=3$. Also note that $a$, the first number in the sequence, is a positive integer by construction.

Finally,
$$
S=n\frac{n+2a-1}{2} = \frac{2^{2c+1}d}{2} \left[ 2^{2c+1}d + d(K^2-2^{2c+1}) \right] = 2^{2c} d [ d K^2 ] = (2^c d K)^2. \blacksquare
$$

Example: $n=8=2^3 \cdot 1, c=1, d=1, K=3 \Rightarrow a=1$, and the sum of the sequence is $1+2+\cdots+8=36=6^2$.


问:
Prove there exists $c,d\in(a,b)$ such that $\frac{f'(c)}{f'(d)}=\frac{e^b-e^a}{b-a}e^{-d}$
Assume that $f$ is differentiable on $(a,b)$ and continuous on $[a,b]$, $f'(x)\neq0$.

Obviously it's related to the mean value theorem, so I was thinking
Let $g(x)=e^xf(x)$ and applying the mean value theorem, but that didn't work out well--does the $f'(x)\neq0$ mean that strict monotonicity will be needed to prove this? Or is it just because of the denominator in the question?
Any halp would be appreciated! Thanks！

答:
We recognize that the term $\frac{e^b - e^a}{b-a}$ resembles the form given by the Mean Value Theorem, namely $f'(c) = \frac{f(b) - f(a)}{b-a}$ for $a < c < b$.

Let $f(x) = e^x$ and note that $f$ is differentiable on $(a, b)$, continuous on $[a, b]$ and $f'(x) \ne 0$. Then by the Mean Value Theorem, there exists $c$ in $(a,b)$ such that

\begin{align*}
f'(c) &= \frac{f(b) - f(a)}{b-a} \\
&= \frac{e^b - e^a}{b-a}
\end{align*}

Dividing both sides by $f'(d) = e^d$, we obtain

\begin{align*}
\frac{f'(c)}{f'(d)} &= \frac{e^b - e^a}{b-a} \cdot \frac{1}{f'(d)} \\
&= \frac{e^b - e^a}{b-a} \cdot e^{-d}
\end{align*}


问:
Proof: Mean and Variance of the squared distance of a random walk in n-dimensional space
consider a $x$ step random walk starting from origin in $n$-dimensional space where each step is taken into a random direction and has a distance of 1, i.e., each step is a vector on the $n$-dimensional unit sphere.

What is the mean and the variance of the squared euclidean distance from the origin after $x$ steps?

Empirically and with some stochastic juggling, I was able to deduce the following properties:

$\mu = x$

$\sigma^2 = 2x^2/n$

However, I need these properties for a paper and thus have to prove them. My "juggeling" is not a good proof and it is too lengthy to be included in the paper. So is there a short proof for these properties or can you give me a reference where this is proven (then I would just cite it)?

答:
Assume that the $k$th step is $x_k$ in $\mathbb R^n$ with $E[x_k]=0$, $\|x_k\|=1$ almost surely, and that the steps $(x_k)$ are independent. Let $s_N=\|x_1+\cdots+x_N\|^2$.

Then $s_N=N+\sum\limits_{k\ne\ell} x_k\cdot x_\ell$. By independence, $E[x_k\cdot x_\ell]=E[x_k]\cdot E[x_\ell]=0$ for every $k\ne\ell$ hence
$$
E[s_N]=N.
$$
Likewise, $s_N^2=N^2+2N\sum\limits_{k\ne\ell} x_k\cdot x_\ell+t_N$ with
$
t_N=\left(\sum\limits_{k\ne\ell} x_k\cdot x_\ell\right)^2=\sum\limits_{k\ne\ell}\sum\limits_{i\ne j}(x_k\cdot x_\ell)(x_i\cdot x_j).
$
The only terms in the summation whose mean is not zero are such that $\{k,\ell\}=\{i,j\}$. There are $2N(N-1)$ such terms hence $E[t_N]=2N(N-1)\alpha$ with $\alpha=E[(x_1\cdot x_2)^2]$. Finally,
$$
\mathrm{var}(s_N)=E[t_N]=2N(N-1)\alpha.
$$
This is an exact formula, valid for every $N\geqslant1$. The value of the parameter $\alpha$ depends on the specifics of the distribution of the displacement $x_1$ since, considering $x_1=(x_1^{(i)})_{1\leqslant i\leqslant n}$,
$$
\alpha=\sum\limits_{i=1}^nE[(x_1^{(i)})^2]^2.
$$
If the coordinates of $x_1$ are identically distributed then $E[(x_1^{(i)})^2]=\frac1n$ for every $i$ by symmetry hence $\alpha=\frac1n$ and $\mathrm{var}(s_N)=\frac2nN(N-1)\sim\frac2nN^2$ when $N\to\infty$.


问:
Length of a circuit
I am reading an article about "Longest Paths in Digraphs". In the proof there is a step that they considered as a trivial step, it seems easy but I am not being able to write or concieve an exact proof for it.

We have a strong digraph $D$, that has minimum in degree $h$ and minimum out degree $k$. Let $C$ be the longest circuit in $D$ of length $c$; then $c\ge \max(h,k)+1$.

Why is that true?

I think it is solved by contradiction and we can find a larger circuit.

Suppose to contradiction that  $c< \max(h,k)+1$ and suppose without loss of generality that  $\max(h,k)=k$, then $c \le k$ which means that every vertex on the circuit has at least $1$ out neighbor outside $C$.

But how can we continue to find a larger circuit?

答:
I will use @Mike idea in his answer, but write a much shorter version of the answer.

Let $P=v_p\cdots v_1$ be the longest path in $D$, (starting from $v_p$ to $v_1$). All the out neighbors of $v_1$ are on $P$, since otherwise we get a longer path. $d^+(v_1)\ge k$, let $c$ be the maximum integer such that $(v_1,v_c)\in E(D)$ then we have a circuit $C=v_cv_{c-1}\cdots v_1$ of length greater than or equal to $k+1$.

Title: Length of a circuit


问:
Can anyone please tell me how to say that the following graph is $\frac{1}{x^2 - 4}$?
Can anyone please tell me how to tell that  the following graph is $\frac{1}{x^2 - 4}$ ?

My Attempt : From the graph , we can see $\lim_{x \to 2+} = \infty = \lim_{x \to -2-} $ and $\lim_{x \to 2-} = -\infty = \lim_{x \to -2+} $ .

But I can not conclude from these that the graph will be $ \frac{1}{x^2 - 4}$.

Can anyone please tell me how to show that the graph will be $ \frac{1}{x^2 - 4}$.

答: I think in addition to the limits to infinity that @PNDas gave above, you need to state the table of values for selected points on the graph. You can show the graph with its limits, limits to infinity and table of values at important points for x.</p>


问:
Why write computational complexity as $O\left( m n \log (n^2 / m) \right)$?
On slide 40 of this presentation, an algorithm is described as having a complexity of $O\left( m n \log (n^2 / m) \right)$. But since $$\log (n^2 / m) = 2 \log n - \log m$$
is itself $O(\log n)$, wouldn't it be equivalent to write the original complexity as $O\left( m n \log n \right)$?

Is this notation meant to emphasize that this algorithm is marginally worse than the $O\left( m n \log n \right)$ algorithm listed above it?

答:
The factor $\log(n^2/m)$ can be better than $O(\log n)$, because the number of edges $m$ can vary to some extent independently of the number of nodes $n$.

On the low end, we have $m \ge n-1$ for any reasonable network. If a node other than $t$ doesn't have any edges out of it, we can delete it without changing the problem, so we may assume every node other than $t$ has at least one edge leaving it - and then $m \ge n-1$.

On the high end, we have $m \le (n-1)^2 < n^2$ because each node other than $t$ can have at most $n-1$ edges out of it (and $t$ shouldn't have any, because they're pointless).

So for sparse networks where $m = O(n)$, $\log (n^2/m)$ is indeed as large as $\Omega(\log n)$, and the two algorithms with complexity $O(mn \log (n^2/m))$ and $O(mn \log n)$ are equivalent.

On the other hand, for dense networks where $m = \Omega(n^2)$, $\log(n^2/m)$ is just $O(1)$, and the algorithm with complexity $O(mn \log(n^2/m))$ is better: it's just $O(mn)$.

This is also the reason why we write the factor $\log (n^2/m)$ the way we do: the ratio $\frac{m}{n^2}$ measures the density of the network, so it's an independently interesting parameter.


问:
Prove that $B^{c} \subseteq A^{c} \Rightarrow A \subseteq B$
How do I prove that $B^{c} \subseteq A^{c} \Rightarrow A \subseteq B$?

I tried:

Take $x \in B^{c}$. We want to prove $\forall x \in A, x \in B$

By definition of the subset, $\forall x \in B^{c}, x \in A^{c}$

Then I got stuck not knowing where to go...

答:
I think the contrapositive can be used, modus tollens.

Modus tollens tells us that if $p \implies q$, then $\neg q \implies \neg p$, where $\neg$ indicates logical negation.

$x\in B^c \implies x \in A^c$ by hypothesis.

Here, $p=x \in B^c$, so $\neg p= x \in B$

We also have $q=x\in A^c$, so $\neg q=x \in A$.

Modus tollens gives us $\neg q \implies \neg p$, thus we conclude $x \in A \implies x\in B$. The conclusion follows by definition of a subset.

Alternative approach:
$p \implies q$ is logically equivalent to the disjuction $q \vee  \neg p$.

So we have as our hypothesis  $x \in A^c \vee x \in B$. The veracity of the disjunction implies at least one of the two statements is true. If we have that $x\in A$, then the first half of the disjunction is false, so the second half must be true, thus $x \in B$.


问:
Question about definition of Ramsey number
So I went through the definition of Ramsey number and I have a basic question.

Definition:
For any given number of colours, $c$, and any given integers $n_1, …, n_c$, there is a number, $R(n_1, …, n_c)$, such that if the edges of a complete graph of order $R(n_1, ..., n_c)$ are coloured with c different colours, then for some i between 1 and c, it must contain a complete subgraph of order ni whose edges are all colour i.

Question:
Is the multicolour Ramsey number $R(n_1,n_2,..n_c) $same as $R(n_2,n_1,..n_c) $or any other permutation of$ {n_1,n_2,..n_c}$? The definition seems to imply so, I just want to verify if I'm thinking right. I have read that $R(m,n)=R(n,m)$ but nothing about symmetricity of multicolour Ramsey numbers.

答: Yes. It is just a matter of using different colours. For example, In $R(3,4,5)$ we want either a red $K_3$, a blue $K_4$ or a green $K_5$ whereas in $R(5,3,4)$ we want either a red $K_5$, a blue $K_3$ or a green $K_4$, which doesn't make a difference at all.


问:
Find the equation of line $AP$ in triangle $ABC$ where $P$ will make $PA=PB=PC$
I was recently doing COMC (Canadian Opening Mathematic Challenge) past year exams. Here is another question that I want to ask!

Here is the question:
Triangle $ABC$ has its sides determinded in the following way: side $AB$ by line $3x-2y+3 = 0$, side $BC$ by line $x+y-14 = 0$, and side $AC$ by line $y=3$. If the point $P$ is chosen so that $PA=PB=PC$, determine the equation of the line containing $A$ and $P$

So I successfully figured out the three coordinates of the triangle $ABC$, which are (1,3), (5,9), (11,3). Now I can't find the point $P$ that makes $PA=PB=PC$. Hope someone would like to help me!

Thank you very much!

This question is from COMC 1998 Part B Question 1.

答:
In other words, the point $P$
is the center of the circumscribed circle,
and
the coordinates of the circumcenter
is known to be found as

\begin{align}
P&=
\frac{a^2(b^2+c^2-a^2)\cdot A+b^2(a^2+c^2-b^2)\cdot B+c^2(b^2+a^2-c^2)\cdot C}
{a^2(b^2+c^2-a^2)+b^2(a^2+c^2-b^2)+c^2(b^2+a^2-c^2)}
\tag{1}\label{1}
,
\end{align}

where $a=|BC|$, $b=|AC|$, $c=|AB|$.

For $A=(1,3),\ B=(5,9),\ C=(11,3)$
the coordinates of the circumcenter
is therefore $P=(6,4)$,
and the question is simplified to:
find the equation of the line through
the two distinct points with known coordinates.


问:
first order differential question
Mr. John want to buy a house and he must borrow $150,000 from the bank. He wants a 30-year mortgage and he has 2 choice.

Choice #1, he can borrow money at 7% per year with no point ( each point is 1% of the amount of the loan that Mr. John has to pay at the beginning of the loan).

Choice 2: he can borrow money at 6.5% per year with charge of 3 point ( this means he has to pay $4,500 at the beginning to get the loan).

Let the model of the amount owed is

$dM/dt=rM-p$

for $M(t)$ is the amount owed at time $t$,$r$ is the annual interest rate , and $p$ is annual payment

a) which deal is better over the entire time of the loan (assume Mr. John doesn't invest the money he would have paid in points)?

b) if Mr.John invest the $4,500 he would have paid in points for the second mortgage at 5% compound continuously, which is better deal?

答:
This is not a differential equation, because 30-year mortgages are simply not compounded continuously. Using a geometrical series, one may derive that the total amount paid on a loan of principal $P$ at annual interest $i$, compounded monthly for $n$ years, is

$$M = \frac{P\, i\, n}{1-\left(1+\frac{i}{12}\right)^{-12 n}}$$

for the numbers above and $i=0.07$, the total paid out is $M \approx \$359,263$.  For $i=0.065$, $m \approx \$341,317$, if the points are paid up front. If however, the points are rolled into the principal, the total amount paid is $\$351,556$.  So using the points here is better even when rolled into the principal.

For the second part, if the debtor invests the $\$4,500$ at $5\%$ compounded continuously, then the debtor would have

$$4,500 \, e^{(0.05) 30} \approx \$20,167.60$$

which trumps the savings on the points even in the best-case scenario of paying the points up front.

Title: first order differential question


问:
Finding the inverse distribution for a standard cauchy distribution: Is my method correct?
The standard cauchy distribution was given as follows $f_X(x) = \frac{1}{\pi} \frac{1}{1+x^2} \;\;\;\;\; x\in \mathbb{R}$

I need to find the distribution for $Y= \frac{1}{X}$ and show it is the same as $X$.

I managed to do the problem easily for $y \neq 0$ that $f_Y(y) = (F_X(\frac{1}{y}))'(\frac{1}{y})' = f_X(y)$.

For $y = 0$, can I use the right continuous property of CDFs?

I was hoping to do $F_Y(0) = \lim\limits_{t \to 0+} F_Y(t)$ and then go on to differentiate from there.

Is what I am doing correct? I am unsure because we haven't verified that $F_Y$ is in fact a CDF or something.

If the random variable $X$ is standard Cauchy then so is $1/X$ : In this question, they used a theorem called inverse-jacobian which we have not done yet.

Cumulative distribution function of Cauchy distribution : I also saw this and realise that there are correct methods I can compare to.

答:
$f_Y(y)=f_X(y)$ almost everywhere (w.r.t. Lebesgue measure) implies that $X$ and $Y$ have the same distribution. There is no need to worry about $F_Y(0)$.

Some basics: If the densities $f$ and $g$ of $X$ and $Y$ are equal almost everywhere then $P(X \leq x)=\int_{-\infty}^{x} f(t)dt=\int_{-\infty}^{x} g(t)dt=P(Y \leq x)$ for all $x$. Changing the value of an integrand at one point or on a set of measure zero does not change the value of the integral.


问:
How to determine the isolated singularity?

problem:
Calculate integral   :    $\int _0 ^\infty \frac{\log x}{x^a(x-1)}dx$

0＜a＜1

We should determine the isolated singularity,i think it only possibly has  0.since we know $\lim_{z\ to  1} \frac{\log z}{(z-1)} =0$.But hint indicate 1 is a simple pole,The following is the prompt words：

It extends analytically to$ (0,\infty)$ from above apparent singularity at $z = 1$ is removable. However, the extension to $(0, \infty) $
from below has a simple pole at z = 1.

答:
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\bbox[5px,#ffd]{\int_{0}^{\infty}{\ln\pars{x} \over x^{a}\pars{x - 1}}\,\dd x} =
\left.\partiald{}{\nu}\int_{0}^{\infty}{x^{\nu} - 1 \over x - 1}\,\dd x\,\right\vert_{\ \nu\ =\ -a}
\\[5mm] = &\
\left.\partiald{}{\nu}\int_{0}^{\infty}{x^{\color{red}{1^{-}}\ -\ 1}\ -\ x^{\pars{\color{red}{\nu + 1}} - 1} \over 1 - x}\,\dd x\,\right\vert_{\ \nu\ =\ -a}
\\[5mm] = &\
\left.-\partiald{}{\nu}
\Gamma\pars{\nu + 1}\Gamma\pars{-\nu}
\cos\pars{-\bracks{\nu + 1}\pi}
\right\vert_{\ \nu\ =\ -a}
\\[5mm] = &\
-\left.\partiald{}{\nu}
\pi\cot\pars{\nu\pi}\right\vert_{\ \nu\ =\ -a} =
\left.
\pi^{2}\csc^{2}\pars{\nu\pi}\right\vert_{\ \nu\ =\ -a}
\\[5mm] = &
\bbx{\pi^{2}\csc^{2}\pars{\pi a}} \\ &
\end{align}


答:
By the MVT, there exists some $c \in (a, b)$ such that
$$f'(c) = \frac{f(b) - f(a)}{b - a}.$$
Also, consider $g(x) = f(\ln(x))$, on the interval $[e^a, e^b]$. Then, also by the MVT, there exists some $d$ such that $e^d \in (e^a, e^b)$ and
$$e^{-d}f'(d) = \frac{1}{e^d} f'(\ln(e^d)) = \frac{f(\ln(e^b)) - f(\ln(e^a))}{e^b-e^a} = \frac{f(b) - f(a)}{e^b - e^a}.$$
Taking the quotient yields the desired result.

Title: Prove there exists $c,d\in(a,b)$ such that $\frac{f'(c)}{f'(d)}=\frac{e^b-e^a}{b-a}e^{-d}$

Body:
Assume that $f$ is differentiable on $(a,b)$ and continuous on $[a,b]$, $f'(x)\neq0$.

Obviously it's related to the mean value theorem, so I was thinking
Let $g(x)=e^xf(x)$ and applying the mean value theorem, but that didn't work out well--does the $f'(x)\neq0$ mean that strict monotonicity will be needed to prove this? Or is it just because of the denominator in the question?
Any halp would be appreciated! Thanks！


问:
Prove this geometric inequality
Given a triangle $\triangle ABC$, let $D$ and $E$ be on points $BC$ such that $BD=DE=EC$.The line $p$ intersects $AB,AD,AE,AC$ at $K,L,M,N$ respectively. Prove that $KN ≥ 3LM$

My attempt: I think cross ratio can be used to prove it though I am not sure how.
Let $a=KL,b=LM,c=MN$
We have to proove $a+b+c\geq 3z$, i.e. $a+b\geq 2c$
Since $R(K,N;L,M)=R(B,C;D,E)$ so $R(K,N;L,M)= \frac{1}{2}:\frac{1}{2}$ thus $4ac=(a+b)(b+c)$.

I am not able to move further from here.

答:
Let $K$ and $N$ be placed on the sides $AB$ and $AC$ respectively.

Also, let $\frac{AK}{AB}>\frac{AN}{AC}$ and $G\in NC$ such that $KG||BC$, $KG\cap AD=\{P\}$ and $KG\cap AE=\{Q\}$.

Thus, $KP=PQ=QG$, which says that it's enough to solve our problem, when $K\equiv B$.

Now, let $AN=kAC$, where $0<k<1$ and $F\in EC$ such that $NF||AE.$

Thus, $$\frac{BM}{BN}=\frac{BE}{BF}=\frac{\frac{2}{3}BC}{\frac{2}{3}BC+\frac{k}{3}BC}=\frac{2}{2+k}.$$
Now, let $G\in NC$ such that $EG||BN.$

Thus, $$\frac{AM}{ME}=\frac{AN}{NG}=\frac{kAC}{\frac{2}{3}(1-k)AC}=\frac{3k}{2(1-k)}.$$
Now, let $I\in DE$ such that $MI||AD$.

Thus, $$\frac{LM}{BM}=\frac{DI}{BI}=\frac{DI}{2DI+EI}=\frac{\frac{DI}{EI}}{\frac{2DI}{EI}+1}=$$
$$=\frac{\frac{AM}{ME}}{2\cdot\frac{AM}{ME}+1}=\frac{\frac{3k}{2(1-k)}}{2\cdot\frac{3k}{2(1-k)}+1}=\frac{3k}{2(2k+1)}.$$
Id est, $$\frac{LM}{BN}=\frac{LM}{BM}\cdot\frac{BM}{BN}=\frac{3k}{2(2k+1)}\cdot\frac{2}{2+k}=\frac{3k}{(2k+1)(k+2)}$$ and it's enough to prove that:
$$\frac{3k}{(2k+1)(k+2)}\leq\frac{1}{3}$$ or $$(2k+1)(k+2)\geq9k,$$ which is true by AM-GM:
$$(2k+1)(k+2)\geq3\sqrt[3]{k^2}\cdot3\sqrt[3]k=9k.$$

Title: Prove this geometric inequality


问:
If $f$ surjects from $A$ to $B$ then there is a bijection $\phi$ from a subset of $A$ to $B$
Let $A$ be a set and let $ f: A \to B $ be a surjective function. Prove that there exists a subset $ C \subseteq A $ and a function $ \phi: C \to B $ such that $ \phi $ is bijective.

I did it by giving particular examples, giving $ A = \{1,2,3 \} $ and $ B = \{1,2 \} $ and the function $f(x)=\begin{cases}
 1& \text{ if } x=1 \\
 1& \text{ if } x=2 \\
 2& \text{ if } x=3
\end{cases}$ it is clearly a surjective function. Now if $ C = A \setminus \{1 \} $ what is desired remains. Guiding me from particular examples, I thought to consider the set $ C $ of all the images that are repeated because the function is not necessarily injective, as in the previous example removing 1, since it repeated with 2. But I don't know if the idea is well and I don't know how to express that set either. I tried as follows, $ C = A\setminus \{f (a) = f (b) \wedge a \neq b \mid \forall a, b \in A \} $ but I don't know if it's ok. Help.

答:
You can’t actually prove it by giving examples; the best that you can do is illustrate the concept that you have in mind. You’ve already recognized that what you have to worry about are the elements $b\in B$ such that the set $\{a\in A:f(a)=b\}$ has more than one element. What if for each $b\in B$ you let $A_b=\{a\in A:f(a)=b\}$? You know that each of these sets is non-empty; why? Thus, from each of them you can single out an element $a_b$. (Of course when $A_b$ has only one element, that element will automatically be $a_b$. In the absence of any other information about $A$ there is no way to specify how the others are chosen.) Now use these elements $a_b$ to define a suitable $C$.

Added: I should probably mention that in general choosing the elements $a_b$ requires the axiom of choice, though in specific cases it may not be necessary.


问:
Connections between number theory and abstract algebra.
I haven't taken abstract algebra yet, but I am curious about connections between number theory and abstract algebra.  Do the proofs of things like Fermat's little theorem, the law of quadratic reciprocity, etc. rely upon techniques found in abstract algebra?

Thanks.

答: Major results like the ones you cite have many proofs, including ones using abstract algebra, but also ones using combinatorics.  See George Andrew's number theory book for an excellent example of the latter.</p>


问:
Evaluating exponential function without overflow
I have the following function:
$$
f(x) = \frac{1}{A} \cdot \frac{e^{\sqrt{A}x} + e^{\sqrt{A}-\sqrt{A}x}}{B\sqrt{A}e^{\sqrt{A}} - B\sqrt{A} + e^{\sqrt{A}} + 1}
$$

where $A = 10^{6}$ and $B = 10^{-7}$. I am attempting to do several calculations and visualizations of $f$ for $x \in [0,1]$. I obviously get overflow errors due to the presence of $e^{\sqrt{A}}$ and $e^{\sqrt{A}x}$, but I know for a fact that $f$ is bounded in the region I am evaluating it in. I tried to fix this by doing the following manipulation:
$$
f(x) = \frac{e^{\sqrt{A}x - \sqrt{A}}}{A} \cdot \frac{1 + e^{\sqrt{A} - 2\sqrt{A}x}}{B\sqrt{A} - B\sqrt{A}e^{-\sqrt{A}} + 1 + e^{-\sqrt{A}}}.
$$

But the issue is that for small $x$, $e^{\sqrt{A} - 2\sqrt{A}x}$ still overflows. So I'm wondering is there some clever manipulation I could do that allows me to fully evaluate $f$ in the desired region?

答:
The numerator may be written as $$e^{\sqrt{A}(x-1/2)}e^{\sqrt{A}/2} + e^{-\sqrt{A}(x-1/2)}e^{\sqrt{A}/2} = 2 e^{\sqrt{A}/2} \cosh ((x-1/2) \sqrt{A}).$$  The denominator may be written as
$$2 A e^{\sqrt{A}/2} ( B \sqrt{A} \sinh \sqrt{A}/2 + \cosh \sqrt{A}/2).$$  Cancellation yields $$f(x) = \frac{\cosh ((x-1/2)\sqrt{A})}{A (B \sqrt{A} \sinh \sqrt{A}/2 + \cosh \sqrt{A}/2)}.$$  We are further inspired to employ a hyperbolic trigonometric angle addition identity:  $$\cosh(w+z) = \sinh w \sinh z + \cosh w \cosh z$$ with the choice $w = \cosh^{-1} (1- B^2 A)^{-1/2}$ and $z = \sqrt{A}/2$ gives $$f(x) = \frac{\cosh((x - 1/2) \sqrt{A})}{A \sqrt{1 - B^2 A} \cosh (w + \sqrt{A}/2)}.$$  From here we take logarithms to obtain $$\log f(x) = \log \cosh ((x - 1/2) \sqrt{A}/2) - \log \cosh (w + \sqrt{A}/2) - \log A \sqrt{1 - B^2 A}.$$  The last term is a fixed constant, and for your choice of $A, B$, is well within machine precision to compute.

The second term is, to a high degree of precision, approximately equal to $w + \frac{\sqrt{A}}{2} - \log 2$, since for sufficiently large $z$, $$\log \cosh z = \log \frac{e^{z} + e^{-z}}{2} \approx z - \log 2.$$  Again, for your choice of $A, B$, this is tractable.

The only thing remaining is computing the first term.  If your overflow limit is on the order of $e^M$, where $M$ is your largest exponent your computation's floating point arithmetic can handle, then when $$|x - 1/2| \sqrt{A} \ge M,$$ you want to use the approximation above; i.e., $$\log \cosh ((x-1/2) \sqrt{A}) \approx \left|x - \frac{1}{2}\right|\sqrt{A} - \log 2.$$  If there is no overflow, then you simply evaluate the function as is.


问:
What is the number of weekly tables that generate the most profit?

A cabinetmaker designs and sells $q$ wooden tables per week. The relation $p=125\frac{\ln(q)}{q}$ is the price, in dollars, when $q$ tables are demanded. The average cost of producing $q$ units is:
$C(q)=\frac{q^2}{3}$ with $2\leq q\leq 9$

What is the number of weekly tables that generate the most profit?

In order to find the maximum value of $U$, the critical points are determined in the usual way and then its nature is investigated.

$I$ = income

$C$ = costs

$U$ = profit

$C'(q)=\frac{2q}{3}$

$I(q)=pq=125\frac{\ln(q)}{q}q$

$I'(q)=\frac{125}{q}$

$U'(q)=I'(q)-C'(q)$

$0=\frac{375-2q^2}{3q}$

$q=13.69$

答: $C(q)$ is the average cost of producing $q$ units, so the total cost is $qC(q)=q^3/3$. The profit function is $125\ln q-q^3/3$ which maximizes at $q=5$.


问:
Why 90% of central region of normal distribution is $\mu\pm qnorm(0.95) \cdot \sigma $
Question: The average height in a sample is $\mu=170.035$. Using simulation, it can be shown that the probability of the sample average of the height falling within 2 centi-meter of the population average is approximately equal to $\sigma=0.925$.

Using the normal approximation we get that the central region that contains 90% of the distribution of the sample average is of the form $170.035\pm z \cdot 1.122$. The value of $z$ is:

Its answer: $1.644854$

Its explanation: The structure of the central region that contains 90% of the Normal distribution is:

$\mu \pm qnorm(0.95)\cdot \sigma$

However, $\mu = 170.035$ and $\sigma=1.122$.

Therefore, $z=$ qnorm(0.95) $=1.644854$.

My thought:

Should it be qnorm(0.90) $=\mu\pm(z\cdot \sigma)$? Why not?

答:
The central region that contains 90% of the normal distribution is the one that excludes 5% in the left tail  and 5% in the right one.

Thus the two quantiles are

qnorm(0.05)=-1.644854

qnorm(0.95)=1.644854


问:
Details in mollifiers proof - Evans' PDE
I've been reading Evans' PDE book but I stumbled upon the proof of the following theorem (screenshot here):

THEOREM 7 (Properties of mollifiers).

(i) $f^{\epsilon} \in C^{\infty}\left(U_{\epsilon}\right)$.

(ii) $f^{\epsilon} \rightarrow f$ a.e. as $\epsilon \rightarrow 0$.

(iii) If $f \in C(U),$ then $f^{\epsilon} \rightarrow f$ uniformly on
compact subsets of $U$.

(iv) If $1 \leq p<\infty$ and $f \in L_{\text {loc }}^{p}(U),$ then
$f^{\epsilon} \rightarrow f$ in $L_{\text {loc }}^{p}(U)$

And this is part of the proof (screenshot here):

Assume now $f \in C(U)$. Given $V \subset \subset U$, we choose $V \subset \subset W \subset \subset U$ and note that $f$ is uniformly continuous on $W .$ Thus the limit (4) holds uniformly for $x \in V .$ Consequently the calculation above implies $f^{\epsilon} \rightarrow f$ uniformly on $V$.

The proof itself is quite clear but the bit that I'm not quite following is how does he choose $W$ satisfying that particular property. I've been trying to use compactness of $\overline{V}$ and openness of $U$ but I'm still not getting it. Some help would be appreciated.

答:
$A\subset\subset B$ is read "$A$ is compactly contained in $B$", sometimes also written with \Subset$\Subset$. According to Evans (2nd ed., pg 698), it means that, for open sets $A,B$, we have $A\subset \overline A\subset B$ and that $\overline A$ is compact:

(vi) $U, V,$ and $W$ usually denote open subsets of $\mathbb{R}^{n} .$ We write
$$
V \subset \subset U
$$
if $V \subset \bar{V} \subset U$ and $\bar{V}$ is compact, and say $V$ is compactly contained in $U$.

As $V\Subset U \subset \mathbb R^n$, in particular $\overline V\subset U$. I believe $U$ is (as in the above) implicitly an open set; thus $d(\overline V,U^c) \in (0,\infty]$. Set $l = \min(1,d(\overline V,U^c))\in(0,1]$.

For each $x\in \overline V$, let $B_x := \mathbb B(x,l/2) $ be the open ball of radius $l/2$ around $x$. Clearly $\{B_x\}_{x\in \overline V}$ is a cover of $\overline V$, so by compactness there is a finite number of them, $B_1,\dots,B_N$ that cover $\overline V$. Now we can take
$W:= B_1\cup \dots\cup B_N$. Since each $\overline{B_i}$ is a subset of $U$ (there is no intersection with $U^c$), $\overline W\subset U$. $\overline W$ is a finite union of closed and bounded balls; hence it is closed and bounded in $\mathbb R^n$, and hence compact. Summary: we created a second open subset $W$ of $U$ such that $V\Subset W \Subset U$. Written using normal subset notation, we have
$$ V\subset \overline V \subset W \subset \overline W \subset U.$$

The fact that $\overline W$ is compact immediately implies the uniform continuity of $f$ on $\overline W$, and hence on $W$.

You might ask why we needed $W$ in the first place since $V$ is already precompact; the answer is that we need some space in order to mollify and get a function defined on $V$, for all $\epsilon$ sufficiently small.


答: First is an even function. Next it is discontinnuous at $x=\pm 2$. Check $f(1.9)$ is large negative and $f(2.1)$ is large positive. $f'(x)=-(x^2-4)^{-2} -2x$, so $f'(0)=0$ and the first derivative changes sign when we pass from say $x= -0.1$ to $x=0.1$ from positive to negative, hence there is local max at $x=0$ and $f_{max}=f(0)=-1/4$. Lastly, $f(\pm \infty)=0$. No wonder it looks like as it does.

Title: Can anyone please tell me how to say that the following graph is $\frac{1}{x^2 - 4}$?

Body:
Can anyone please tell me how to tell that  the following graph is $\frac{1}{x^2 - 4}$ ?

My Attempt : From the graph , we can see $\lim_{x \to 2+} = \infty = \lim_{x \to -2-} $ and $\lim_{x \to 2-} = -\infty = \lim_{x \to -2+} $ .

But I can not conclude from these that the graph will be $ \frac{1}{x^2 - 4}$.

Can anyone please tell me how to show that the graph will be $ \frac{1}{x^2 - 4}$.


问:
Is the sum of two coprime numbers also coprime to its summands?
I was comparing the statement of the ABC conjecture on Wikipedia to an article I found geared towards laymen. The propositions differ slightly, but they would be equivalent if the following were true:

If $a$ and $b$ are coprime, and $a + b = c$, is $c$ coprime to both $a$ and $b$?

By division algorithm, if (without loss of generality) $a > b$, then $a$ and $c$ must be coprime: the first iteration of the algorithm subtracts $c = 1*a + b$, and then you're left with $a$ and $b$, which are coprime by proposition and so generate 1 as the final result.

If $b$ is prime, I can imagine a proof by contradiction via the division algorithm.

For more complicated $b$ I am not sure it holds.

答:
$$
(2\cdot3\cdot13) + (5\cdot7\cdot11)
$$
If the whole thing were divisible by $2$, then since the first term is divisible by $2$, so would be the second term.  And similarly for the other primes involved, and similarly if the roles of the two terms are reversed.  So the sum cannot be divisible by any of the primes involved.  Therefore other primes besides those must exist.  So this is another way to prove the infinitude of primes, PROVIDED you know about uniqueness of prime factorizations, which is tacitly relied on above.</p>


问:
complex integral giving different results
If I have an integral $\int_C \frac{1}{z} dz$ where $C$ is the left half of the circle from $i$ to $-i$, then I can choose the branch cut to be the positive axis, and have the antiderivative to be log(z). Then using the antiderivative, I can evaluate the integral:

$\int_C \frac{1}{z} dz = log(-i) - log(i)$

However, if I let $0 \leq arg(z) < 2\pi$, my integral will evaluate to $\frac{3\pi}{2} - \frac{\pi}{2} = \pi$. If I choose $-\pi \leq arg(z) < \pi$, then my integral will evaluate to $\frac{-\pi}{2} - \frac{\pi}{2}  = -\pi$. I feel like I should get the same value. What am I doing wrong here.

答:
Solving the integral by parametrization yields:

$$\int_{C} \frac{dz}{z} = \int_{\pi/2}^{3\pi/2} \frac{1}{e^{i \theta}} ie^{i \theta} \, d \theta=i \pi$$

You may arrive at the same result by choosing the branch of the logarithm given by

$$\log_0 (x) = \text{Log} |z| + i \arg_0(z),$$

where $\arg_0(z) \in [0, 2\pi)$, and using it as a primitive for $\frac{1}{z}$. As mentioned, we choose this branch since the contour passes through the negative real axis. This branch of the logarithm however, has its branch cut at the positive real axis. We get

$$\int_{C} \frac{dz}{z} = \log_0 (-i) - \log_0(i) =i (\frac{3\pi}{2}-\frac{\pi}{2})=i \pi$$

Using the principal branch of the logarithm instead will give a wrong result because it is not a primitive of the function we are integrating in an open set containing the contour over which we integrate.


问:
Finding the closed form of a recursive sequence
I am looking mostly for keywords, to search and study on examples and techniques on how one goes on finding the closed form expression on a recursive sequence.

I have a problem to solve, which looks like

$${ a_{n+1} = \frac{ma_n + x}{a_n + y} }$$

(I am not writing the exact values, because I am not looking for a solution, just guidance.)

I have seen a solution here, which is amazing, but I do not understand how one thinks about it.

答:
This is a first-order rational difference equation (have a look here).

$${ a_{n+1} = \frac{ma_n + x}{a_n + y} }=m+\frac{x-m y}{a_n+y}$$
Let $b_n=a_n-m$ to make
$$\frac{m y-x}{b_n+m+y}+b_{n+1}=0$$ Let $c_n=b_n+m+y$ to make
$$\frac{m y-x}{c_n}+c_{n+1}=m+y$$ Let $p=m y-x$ and $q=m+y$ to make
$$\frac{p}{c_n}+c_{n+1}=q$$

If you follow the steps given in the linked page, you should get coething like
$$c_n=\frac{A_n}{B_n}$$ with
$$A_n=K \left(q^2-2 p\right) \left(\frac{\sqrt{q^2-4 p}-q}{p}\right)^n-K q
   \sqrt{q^2-4 p} \left(\frac{\sqrt{q^2-4 p}-q}{p}\right)^n+\left(q^2-2 p\right)
   \left(-\frac{\sqrt{q^2-4 p}+q}{p}\right)^n+q \sqrt{q^2-4 p}
   \left(-\frac{\sqrt{q^2-4 p}+q}{p}\right)^n$$
and
$$B_n=\sqrt{q^2-4 p} \left(\left(-\frac{\sqrt{q^2-4 p}+q}{p}\right)^n-K
   \left(\frac{\sqrt{q^2-4 p}-q}{p}\right)^n\right)+q \left(K
   \left(\frac{\sqrt{q^2-4 p}-q}{p}\right)^n+\left(-\frac{\sqrt{q^2-4
   p}+q}{p}\right)^n\right)$$


问:
$tr(xy)$ is non-degenerate if and only if $A$ is absolutely semisimple
Suppose $A$ is any algebra over any groundfield $k$, and let $tr(u)$ denote the trace of the endomorphism of the underlying vector-space of $A$, $L_{u} : x \mapsto ux$. Then, I want to prove that :

$tr(xy)$ is a non-degenerate bilinear form if and only if $A$ is absolutely semisimple, i.e., it is semisimple and remains semisimple under any extension of the groundfield.

I remember proving something similar for field extensions (non-degenerate iff the extension is separable), but I can't see how it will hold for algebras. I encountered this stated as a fact while going through a paper on Algebraic groups, so any reference would also be welcome.

答:
You need $A$ to be finite-dimensional to define the trace.

Given that, it's a classic exercise to show that if the trace form is nondegenerate then $A$ is semisimple; this is done in this math.SE answer using the fact that the Jacobson radical must be contained in the radical of the trace form, so if the trace form is nondegenerate then the Jacobson radical vanishes. So $A$ is Artinian + semiprimitive which is equivalent to semisimple.

Next we argue that the nondegeneracy of the trace form passes to extensions of scalars $A \otimes_k L$. Maybe the most concrete way to see this is to note that nondegeneracy is equivalent to the determinant of a Gram matrix $\det \text{tr}(e_i e_j)$ being nonzero where $\{ e_i \}$ is a basis, and this condition is clearly preserved (and reflected) by extension of scalars. So now it follows that if the trace form is nondegenerate then $A \otimes_k L$ is semisimple for every field extension $k \to L$; this says that $A$ is geometrically semismple.

To prove the converse we note that the argument above showing that nondegeneracy of the trace form passes to extension of scalars also shows that nondegeneracy of the trace form can be checked by checking it on any extension of scalars; in particular it can be checked for the extension of scalars $A \otimes_k \bar{k}$ to the algebraic closure. By Artin-Wedderburn we have

$$A \otimes_k \bar{k} \cong \prod_i M_{n_i}(\bar{k})$$

and we can check by hand that the trace form is nondegenerate for any such algebra. Alternatively, $M_n(k)$ is simple so has no nontrivial two-sided ideals, but the radical of the trace form is a two-sided ideal and the trace form is nonzero, so the radical must be zero. And the direct product of two algebras whose trace forms are nondegenerate again has nondegenerate trace form.

It's also worth knowing that the geometrically semisimple $k$-algebras can be classified exactly:

Theorem: A finite-dimensional $k$-algebra $A$ is geometrically semisimple iff it is a finite product $\prod_i M_{n_i}(D_i)$ of matrix algebras over finite-dimensional division algebras $D_i$ whose centers are finite separable extensions of $k$.

For a proof see this blog post. In particular if $A$ is commutative then it's geometrically semisimple iff it's a finite product of finite separable extensions of $k$, which says exactly that $A$ is an étale algebra over $k$, so we recover as a special case that a finite field extension $L/k$ is separable iff it's geometrically semisimple iff $L \otimes_k \bar{k}$ is isomorphic to a finite product of copies of $\bar{k}$ iff $L \otimes_k \bar{k}$ has no nontrivial nilpotents iff the trace form on $L$ is nondegenerate.


问:
Proving a function is quasi-concave
This is from economics, but I think there's a lot of math involved and I want to make sure I didn't mess anything up.  There is a utility function U = $x_1$ + $\ln(x_2)$ such that $x_1$ and $x_2$ are nonnegative.

How do I prove this function is quasi-concave?

I was looking online for any notes about this, and it tells me I need to create this

https://mjo.osborne.economics.utoronto.ca/index.php/tutorial/index/1/qcc/t

"bordered Hessian matrix"

As in for this problem it seems to me I need to calculate

\begin{bmatrix}
0 & f'_{1}(x) & f'_{2}(x)\\
f'_{1}(x) & f''_{11}(x) & f''_{12}(x) \\
f'_{2}(x) & f''_{21}(x) & f''_{22}(x) \\
\end{bmatrix}

So I do that for the above function and I get

\begin{bmatrix}
0 & 1 & \frac{1}{x_2} \\
1 & 0 & 0 \\
\frac{1}{x_2} & 0 & -\frac{1}{x_{2}^{2}} \\
\end{bmatrix}

Then it says I need to find the determinant and if n = even, I need $D_{n} \geq 0$.  So here $D_{2}$ for the above matrix I got to be $\frac{1}{x_{2}^{2}}$.  Since $x_2$ is nonnegative, this means that the determinant is $\geq 0$.

Does this prove this function is quasi-concave?  Is my general approach right or am I totally missing the point of this question?

What about strictly quasi-concave?   The online notes I found never mentions this, but is that just when it is strictly greater than 0?  So in this case, this function is also strictly quasi-concave too?  If someone can verify if my guess is correct, that would be great.

Finally, on an unrelated note, how do I prove this function is continuous using math?  I know it is when I graph it, but is there a mathematical proof for this?

Thank you so much for the help!

答:
Your approach goes in the right direction.
But you should verify $D_2>0$ and not $D_2\geq0$.
In theory, one should also verify that $D_1<0$, but this is very easy to see.

As for strictly quasi-concave, you should not use the theorem about bordered Hessian matrix
because it does not discuss this case
(I would still conjecture that the same result can also be shown for strictly quasi-convex functions,
but I do not have a proof).
Instead, you could try to use the definition.

Title: Proving a function is quasi-concave


问:
Application of the r-combination with repetition formula seems to fail
I was taught that the formula for r-combination with repetitions being allowed is

$${r + n -1 \choose r}$$

The intuition for that was the "Stars & Bars" method.

Based on this method, I  took $r$ as the number of elements you choose and $n$ as the number of possible values an element can be.

This thought process seemed to work for 2 examples

Example #$1$:

There are $4$ stalls selling $1$ different food each(apple, banana, coke, nuggets). If
you are buying snacks for $20$ friends, how many different selections can you make?

Answer:
$r$(number of elements you choose) is $20$ here as we are choosing $20$ snacks for each of the friends. $n$(number of possible values an element can be) is $4$ as there as only $4$ stalls. Thus ${20 + 4 -1 \choose 20}$ works fine.

Example #$2$:

Choose $3$ numbers from ${1, 2, 3, 4}$.

Answer: $r$(number of elements you choose) is $3$ as we are choosing $3$ numbers.$n$(number of possible values an element can be) is $4$ as there as only $4$ numbers to choose from.Thus ${3 + 4 -1 \choose 3}$ works fine.

Now my thought process failed for this example, seen frequently on this SE:

How many solutions are there for $x_1 + x_2 + x_3 = 11$ where the $x_i$s are non-negative?

My wrong answer:$r$(number of elements you choose) is $3$ as we are choosing $3$ integers, $1$ for each $x$ from $11$ integers.$n$(number of possible values an element can be) is $11$ as there as $11$ possible integers to choose from. Obviously, ${3 + 11 -1 \choose 3}$ is wrong, $r$ is supposed to be $11$.

But why does my "thought" process fail for this example?

PS: How is $r$, $11$ in this case? Don't we only want to choose $3$ integers. All the solutions I see, treat $11$ as $r$ (ie the stars) which doesn't make sense as that would mean each star is of value $1$.

答:
You should think of your final example as the number of ways to choose $11$ objects, each of which is of one of $3$ types: $x_1$ is the number of Type $1$ objects that you choose, $x_2$ the number of Type $2$ objects, and $x_3$ the number of Type $3$ objects that you select. Thus, you have $r=11$ and $n=3$. This is actually very similar to your example with the fruit.

I prefer a slightly different model: $r$ is the number of indistinguishable objects that you’re distributing amongst $n$ numbered containers. The $r$ objects are the stars, and the $n$ containers are the spaces between $n-1$ bars. This fits your last example very well, and it’s not hard to make it fit your other examples as well. In your first example the types of fruit are the containers: imagine labelling a jar with each of the types and distributing $20$ pebbles amongst the $4$ jars to describe a given selection. In your second example the containers are the numbers $1,2,3$, and $4$, and you’re distributing $3$ pebbles amongst those containers.


问:
Constructive vs computable real numbers
I find it confusing that all of the following statements are true :

The computable real numbers are countable. $-\hspace{-3pt}-$ Alan Turing, "On Computable Numbers, with an Application to the Entscheidungsproblem"

In constructive analysis, the real numbers are uncountable. $-\hspace{-3pt}-$ Errett Bishop, Foundations of Constructive Analysis

"every mathematical statement [in constructive analysis] ultimately expresses the fact that if we perform certain computations within the set of positive integers, we shall get certain results" $-\hspace{-3pt}-$ Ibid.

Perhaps I am misunderstanding something.

I suppose I really have two questions. In constructive analysis :

Why isn't every real number computable?

How is it possible to construct an uncountable set?

答:
It is consistent with the constructive analysis that every real is computable. (In fact, it is consistent with intuitionistic ZF $\mathsf{IZF}$.) This follows from that the consistency of Church's thesis, which claims every total function from $\mathbb{N}$ to $\mathbb{N}$ is computable.

However, it does not mean constructive analysis can prove every real is computable. We know that classical analysis is a superset of constructive analysis (as a theory), that is, every statement that is provable from the constructive analysis is also a theorem of the classical analysis. And classical mathematics proves not every real is computable. Hence it would be accurate to say whether every real is computable is independent of constructive analysis.

To add some comment, I think it does not mean Bishop is simply wrong. Bishop's constructive analysis is minimal in the sense that it is contained in classical analysis, Brouwer's intuitionistic mathematics, and Recursive constructive analysis (also known as Russian constructivism.) The latter one reflects the behavior of recursive mathematics, so Bishop's claim would be true in that sense.

$\mathbb{R}$ is uncountable in the sense that there is no bijection between $\mathbb{N}$ and $\mathbb{R}$. The proof is available from Bishop's Constructive analysis. (Theorem 2.19 of Bishop and Bridges.) Here is a rough proof:

Theorem. There is no bijection from $\mathbb{N}$ to an interval $[x_0,y_0]=\{z\in\mathbb{R}\mid x_0\le z\le y_0\}$.
(Caution: $x\le y$ is not ($x< y$ or $x=y$.))

Proof. The proof uses diagonalization argument. Assume that $f:\mathbb{N}\to\mathbb{R}$ is a function. We will find a Cauchy real $x\in [x_0,y_0]$ such that $f(n)\neq x$ for all $n$.
We will construct sequences of natural numbers $(x_n)$ and $(y_n)$ recursively such that

$x_0\le x_n\le x_m<y_m\le y_n\le y_0$ if $1\le n\le m$,

$x_n>f(n)$ or $y_n<f(n)$, and

$y_n-x_n<1/n$.

Assume that $x_i$ and $y_i$ are given for all $i<n$.
Then we have either $f(n)>x_{n-1}$ or $f(n)<y_{n-1}$. (This follows from the following constructively valid theorem: if $x<y$ are reals, then either $z<y$ or $x<z$.)

Assume that we have $f(n)>x_{n-1}$. (The remaining case is analogous.)
Choose a rational number $x_n$ and $y_n$ such that $x_{n-1}<x_n<\min(a_n,y_{n-1})$ and $x_n<y_n<\min(a_n,y_{n-1},x_n+1/n)$. Then the mentioned inequalities are satisfied.

Hence $(x_n)$ forms a Cauchy sequence of rational numbers. Let $x$ be a limit of $(x_n)$, then $x$ satisfies the desired properties.

Note that the above proof makes use of the countable choice that Bishop accepted. As far as I know, it is open whether $\mathbb{R}$ is countable or not without the countable choice. Also, note that $\mathbb{R}$ can be subcountable, i.e. $\mathbb{R}$ can be an image of a subset of $\mathbb{N}$. (Unfortunately, I forgot whether the subcountability of $\mathbb{R}$ is compatible with $\mathsf{IZF}$, although I believe it is true.)

Title: Constructive vs computable real numbers


问:
$f$ is continuous and differentiable in $(-1,1)$, $f(1) = 0$, $f(x)>0 (x∈(-1,1))$. Show $∃c∈(-1,1); mf(c)f'(-c) = nf'(c)f(-c)$ in which $m∈N, n∈N$
My problem is the following: "Suppose that $f\colon[-1,1]\to\mathbb{R}$ is continuous and differentiable on $(-1,1)$, $f(1) = 0$ and $f(x)>0$ for $x\in(-1,1)$. Show that there exists $c\in(-1,1)$ such that $mf(c)f'(-c) = nf'(c)f(-c)$, in which $m\in N$ and $n\in N$."

I can prove if when $M=N$: there exists $c\in(-1,1)$ for which $f(c)f'(-c) = f'(c)f(-c)$ using $g(x) = f(x)f(-x)$, which satisfies $g(1) = g(-1) = 0$ and Rolle's theorem, but I can't prove it when $m ≠ n$. Can someone please tell me how to prove it?

答: Apply your argument to $g(x) =(f(x))^{n}(f(-x))^{m}$.


问:
order of stabilizer
$G$ is a group which acts transitively on a set $S$. $G_s$ is the stabilizer of $s$. How to see the map from $G/G_s$ to $S$ given by $g*G_s \to g(s)$ is one to one?</p>

答: $g(s)=h(s)\iff h^{-1}g\in G_s\iff gG_s=hG_s$.</p>

Title: order of stabilizer


问:
$h:B\to B'$ is a homomorphism, then $h(x\lor y) = h(x) \lor h(y)$
I was able to complete the proof, but I'm afraid I may have used a circular argument or made too many assumptions somewhere. A better/simpler proof would be nice.

Consider two boolean algebras $(B,\leq, \lor, \land,^c,0,1)$ and $(B',\leq, \lor, \land,^c,0,1)$. A map $h:B\to B'$ is a homomorphism, if for all $x,y\in B$ we have:

$h(x\land y) = h(x) \land h(y)$

$h(x^c) = (h(x))^c$

Then show that $h(x\lor y) = h(x) \lor h(y)$ for all $x,y \in B$, $h(0) = 0$ and $h(1) = 1$.

My work:

I started with $h(x\land y) = h(x) \land h(y)$ and got $(h(x\land y))^c = (h(x) \land h(y))^c$. Then, using De-morgan's laws (is it okay to use these?) and $h(x^c) = (h(x))^c$, I ended up with $h(x^c\lor y^c) = h(x^c) \lor h(y^c)$. Now, we know that a Boolean algebra is a complemented lattice, i.e. every element $x \in B$ has a complement $x^c \in B$. So $h(x^c\lor y^c) = h(x^c) \lor h(y^c)$ essentially means $h(p\lor q) = h(p) \lor h(q)$ for all $p,q \in B$, which proves what is desired. Is this last line of reasoning okay?

Lastly, putting $y = x^c$ in $h(x\land y) = h(x) \land h(y)$ and $h(x\lor y) = h(x) \lor h(y)$ gives $h(0) = 0$ and $h(1) = 1$.

Thanks!

答:
Everything seems fine. Double negation makes your life easier when combined with De Morgan: $$h(x\vee y) = h(x^{cc}\vee y^{cc}) = h((x^c \wedge y^c)^c) \stackrel{(\ast)}{=} (h(x)^c\wedge h(y)^c)^c = h(x)\vee h(y),$$where in $(\ast)$ we used the two homomorphism axioms in one shot.

Then for $0$, you have $$h(0) = h(x \wedge x^c) = h(x)\wedge h(x)^c = 0,$$and for $1$ we have $$h(1)= h(x \vee x^c) = h(x)\vee h(x)^c = 1.$$


问:
Root system of a semisimple Lie algebra spans the dual.
Let $L$ be a semisimple Lie algebra and $H$ a Cartan subalgebra. The roots of $L$ with respect to $H$ are the elements of
$$
\Phi = \{\alpha \in H^*\setminus 0 : L_\alpha \neq 0\}
$$
where $L_\alpha = \{x \in L : [h,x] = \alpha(h)x \text{ for all } h \in H\}$. According to Humphrey's lie algebras book, $\Phi$ spans $H^*$. But the proof somehow uses duality which I cannot seem to understand. This question has been asked before but I also did not understand the accepted answer.

The proof starts like this: if $\Phi$ does not span, then (by duality) there exists a nonzero $h\in H$ such that $\alpha(h) = 0$ for all $\alpha \in \Phi$. This is the part of the proof I don't get.

答:
Of course the other answer is optimal. Let me just add one way to see the result without all the duality machinery (which, however, is a good thing to learn anyway):

Let $V$ be an $n$-dimensional vector space over a field $k$ and $\alpha_1, ... ,\alpha_r$ be linearly independent elements of $V^*$ i.e. nonzero linear maps $V\rightarrow k$.

Then $\mathrm{dim} \left( \cap \mathrm{ker}(\alpha_i) \right) \ge n-r$.

(Proof: Induction over $r$.)

And now use that $V$ and $V^*$ have the same dimension (which is duality in a nutshell).