问:
Solution of $1+x<e^x$

Show that $1+x<e^x$ for all $x>0$

My attempt:

Consider the function $f(x)=e^x-x-1$ and fix $x>0$. By the MVT there exists a point $c\in(0,x)$ such that $\frac{f(x)-f(0)}{x-0}=f'(c)~~~(*)$.

Note that $f(0)=e^0-0-1=0$ and $f'(x)=e^x-1$. This implies that $f'(c)=e^c-1$.

The equality $(*)$ can be rewritten as $\frac{e^x-x-1-0}{x}=e^c-1$.

Since $c$ is a positive number, we have that $e^c>1$. So, the difference $e^c-1>0$.

This means that $\frac{e^x-x-1-0}{x}>0$. Then, $e^x>1+x$ for $x>0$.

Is this solution correct?

答:
Others have already commented on your solution.

Alternatively, if $x>0$, then $e^x=1+x+\dfrac{x^2}2+\dfrac{x^3}{3!}+\cdots=1+x+\text{ positive terms } > 1+x$.

Title: Solution of $1+x<e^x$


问:
Show that $(X \times Y)^{\ast}$ with the resulting quotient topology is homeomorphic to $Y$

Let $X \times Y$ be partitionde into subsets of the form $X\times \lbrace y \rbrace $ for all $y\in Y$.If we let $ (X \times Y)^{\ast}$ denote the collection of sets in the partition, Show that $(X \times Y)^{\ast}$ with the resulting quotient topology is homeomorphic to $Y$

My attempt

Consider the map $h: (X \times Y)^{\ast} \longrightarrow Y $ given by
$h(X \times \lbrace y \rbrace)=y$, notice that $g:Y \longrightarrow  (X \times Y)^{\ast}$ given by $g(y)=X \times \lbrace y \rbrace$ is the inverse of $h$ since

$(h \circ g)(y)=h(g(y))=h(X \times \lbrace y \rbrace)=y=id_Y$

$(g \circ h)(X \times \lbrace y \rbrace)=g(h(X \times \lbrace y \rbrace))=g(y)=X \times \lbrace y \rbrace=Id_{(X \times Y)^{\ast}}$
therefore the function is bijective.

$h$ is continuous function since is a projection
$g$ is continuous since $C=(X \times \lbrace y \rbrace)$ closed in $(X \times Y)^{\ast}$ implies $g^{-1}(C)=y$ closed.

答:
This is close, but there are a couple of problems. The first is that $h$ isn’t actually the projection map $\pi_Y:X\times Y\to Y$. The domain of $h$ is $(X\times Y)^*$, which is $\big\{X\times\{y\}:y\in Y\big\}$: the elements of this set are the sets $X\times\{y\}$ with $y\in Y$, while the elements of $X\times Y$, the domain of $\pi_Y$, are ordered pairs $\langle x,y\rangle$ with $x\in X$ and $y\in Y$. Thus, we can’t simply say that $h$ is continuous because $\pi_Y$ is continuous.

Let $U\subseteq Y$ be open;

$$h^{-1}[U]=\big\{X\times\{y\}:y\in U\big\}\,,$$

and we want to show that this is open in $(X\times Y)^*$. If

$$q:X\times Y\to(X\times Y)^*:\langle x,y\rangle\mapsto X\times\{y\}$$

is the quotient map, then $h^{-1}[U]$ is open in $(X\times Y)^*$ if and only if $q^{-1}\big[h^{-1}[U]\big]$ is open in $X\times Y$. And clearly

$$q^{-1}\big[h^{-1}[U]\big]=q^{-1}\left[\big\{X\times\{y\}:y\in U\big\}\right]=X\times U$$

is open in $X\times Y$, so $h$ is indeed continuous.

As a bit of an aside, it is certainly true that $h$ and $\pi_Y$ are closely related. Let $U$ be any open set in $Y$. Then

$$\pi_Y^{-1}[U]=X\times U=\bigcup\big\{X\times\{y\}:y\in U\big\}=\bigcup h^{-1}[U]\,,$$

and it’s easy to check that $h\circ q=\pi_Y$. Thus, $$q^{-1}\big[h^{-1}[U]\big]=(h\circ q)^{-1}[U]=\pi_Y^{-1}[U]\,,$$ which of course is open because $\pi_Y$ is continuous.

The proof that $g$ is continuous also has a problem. If you’re going to do this by showing that $g^{-1}[C]$ is closed in $Y$ for each closed $C\subseteq(X\times Y)^*$, you really do need to consider all closed subsets of $(X\times Y)^*$. In fact you’ve not considered even one: $X\times\{y\}$ is a single point of $(X\times Y)^*$.

I would show instead that $g^{-1}[U]$ is open in $Y$ for each open $U\subseteq(X\times Y)^*$, because you know exactly which subsets of $(X\times Y)^*$ are open: $U\subseteq(X\times Y)^*$ is open if and only if $q^{-1}[U]$ is open in $X\times Y$. Check that $q^{-1}[U]=X\times h[U]$; this implies that $q^{-1}[U]$ is open in $X\times Y$ if and only if $h[U]$ is open in $Y$. Finally, $g^{-1}=h$, so $U\subseteq(X\times Y)^*$ is open if and only if $q^{-1}[U]$ is open in $X\times Y$, which is the case if and only if $g^{-1}[U]$ is open in $Y$.

Notice that this is actually more than we needed: we were trying to show that if $U$ is open in $(X\times Y)^*$, then $g^{-1}[U]$ is open in $Y$, and we actually showed that the implication goes both ways and hence that both $g$ and $g^{-1}(=h)$ are continuous and therefore homeomorphisms. In other words, the work before the aside was unnecessary: our proof that $g$ is continuous also shows that $h$ is continuous. I kept the first part for two reasons. First, of course, I wanted to address your attempt. But there’s also a useful lesson here: once you’ve proved something, it’s worth going back over your argument to see whether it can be tightened up or presented more clearly. Here we could actually omit half of it.

ContentLicense: CC BY-SA 4.0


问:
Question about the chance of at least one packet of something being faulty
From a previous question, I calculated the probability of something being defective being 0.05175. Now I want to know the probability of at least one packet being faulty when someone buys 13 of them. So I was thinking that I could set is up using a binomial with $n=13$ and $p=0.05265$. Then using $P(Y\ge1)= 1-P(Y\lt1) = 1 - \binom{13}{0}*(0.05265)^{13}$ But this gives me an answer of 1 which obviously isn't right. So I'm not sure what to do.

答:
The error is that
$$
\mathbf{P}(Y < 1) = \mathbf{P}(\text{all 13 are not faulty}) = (1 - 0.05265)^{13}
$$
since $Y < 1$ means that all 13 of them will not be faulty.


问:
If $\dim_\varphi \mathfrak{M}=\dim_\varphi\mathfrak{N}$ then there is an elementary bijection $\varphi(\mathfrak{M})\rightarrow\varphi(\mathfrak{N})$
Suppose we have a complete theory $T$ and a model $\mathfrak{M}\models T$. Recall that, if $\varphi(v)$ is a strongly minimal formula with parameters in some set $A_0\subseteq\mathfrak{M}$, then we define a pregeometry $(\varphi(\mathfrak{M}),\text{cl})$ by $\text{cl}(X)=\text{acl}^\mathfrak{M}(X\cup A_0)\cap\varphi(\mathfrak{M})$ for any $X\subseteq\varphi(\mathfrak{M})$. We then denote the dimension of this pregeometry by $\dim_\varphi(\mathfrak{M})$. I am having a small hangup in the proof of the following result:

Let $\mathfrak{M}$ and $\mathfrak{N}$ models of $T$ containing some set $A_0$, and $\varphi(v)$ a strongly minimal formula with parameters in $A_0$. Then, if $\dim_\varphi(\mathfrak{M})=\dim_\varphi(\mathfrak{N})$, there exists an elementary bijection from $\varphi(\mathfrak{M})$ to $\varphi(\mathfrak{N})$ fixing $A_0$ pointwise.

I will run through my understanding of the proof of this. It follows the treatment given by Tent and Ziegler in section 5.7 of their textbook, but I have added in my understanding of a number of details they glazed over. The result in particular is Lemma 5.7.6, on page 84.

Lemma 1: If $\mathfrak{M}\models T$ and $A\subseteq\mathfrak{M}$, and $\varphi(v)$ is a strongly minimal formula with parameters in $A$, then for any elementary extension $\mathfrak{N}\succcurlyeq\mathfrak{M}$ and any $A\subseteq B\subseteq \mathfrak{N}$ there is a unique non-algebraic type in $S_1^\mathfrak{N}(B)$ containing $\varphi(v)$.

Proof: Consider the set $p(v)=\{\psi\in\text{Form}(\mathcal{L}_B):\varphi\wedge\neg\psi\text{ is algebraic}\}$. Since $\varphi$ is non-algebraic, $p(v)$ does not contain any algebraic formulas, and since $\varphi$ is strongly minimal $p(v)$ indeed contains one of $\psi$ and $\neg\psi$ for any $\mathcal{L}_B$ formula $\psi$.  If $p(v)$ were not consistent there would be $\psi_1,\ldots,\psi_n$ such that each $\varphi\wedge\neg\psi_i$ is algebraic and $\mathfrak{N}\models\forall v\neg(\varphi(v)\wedge\bigwedge_{i=1}^n\psi_i(v))$, ie $\mathfrak{N}\models\forall v(\varphi(v)\rightarrow\bigvee_{i=1}^n\neg\psi_i(v))$, contradicting that $\varphi$ is non-algebraic. So $p(v)$ is a non-algebraic type in $S_1^\mathfrak{N}(B)$ containing $\varphi$, and it is clearly the unique such type.

Lemma 2: Suppose $A\subseteq\mathfrak{M}$ and that $p(v)\in S_1^\mathfrak{M}(A)$ is a non-algebraic type containing a strongly minimal $\mathcal{L}_A$ formula $\varphi(v)$. Then we have the following for any $\mathfrak{N}\succcurlyeq\mathfrak{M}$: if $x_1,\ldots,x_n\in\mathfrak{N}$ are realizations of $p$ such that each $x_i\notin\text{acl}(\{x_1,\ldots,x_{i-1}\}\cup A)$, and $y_1,\ldots,y_n\in\mathfrak{N}$ are realizations of $p$ satisfying the same, we have $\text{tp}^\mathfrak{N}(x_1,\ldots,x_n/A)=\text{tp}^\mathfrak{N}(y_1,\ldots,y_n/A)$.

Proof: By induction on $n$. The case $n=1$ is clear, for $\text{tp}^\mathfrak{N}(x_1/A)=p(v)=\text{tp}^\mathfrak{N}(y_1/A)$. Now suppose the results holds for $n-1$. To show $\text{tp}^\mathfrak{N}(x_1,\ldots,x_n/A)=\text{tp}^\mathfrak{N}(y_1,\ldots,y_n/A)$ it suffices to show (i) that $\text{tp}^\mathfrak{N}(x_1,\ldots,x_{n-1}/A)=\text{tp}^\mathfrak{N}(y_1,\ldots,y_{n-1}/A)$ and (ii) letting $p(v)=\text{tp}^\mathfrak{N}(x_n/\{x_1,\ldots,x_{n-1}\}\cup A)$ and $q(v)=\text{tp}^\mathfrak{N}(y_n/\{y_1,\ldots,y_{n-1}\}\cup A)$, that the set $r(v):=\{\varphi(y_1,\ldots,y_{n-1},v):\varphi(x_1,\ldots,x_{n-1},v)\in p(v)\}$ equals $q(v)$, where the $\varphi(\bar{w},v)$ are $\mathcal{L}_A$ formulas.

(i) is just the induction hypothesis, and for (ii) we claim that $r(v)$ is a non-algebraic type in $S_1^\mathfrak{N}(A\cup\{y_1,\ldots,y_{n-1}\})$. To see that it is a type, suppose for contradiction that $r(v)$ were not consistent. Then there would be $\varphi_1(\bar{w},v),\ldots,\varphi_k(\bar{w},v)\in\mathcal{L}_A$ such that $\mathfrak{N}\models\forall v(\neg\bigwedge_{i=1}^k\varphi_i(y_1,\ldots,y_{n-1},v))$. But then $\forall v(\neg\bigwedge_{i=1}^k\varphi_i(w_1,\ldots,w_{n-1},v))\in\text{tp}(y_1,\ldots,y_{n-1}/A)$, so by the induction hypothesis $\forall v(\neg\bigwedge_{i=1}^k\varphi_i(w_1,\ldots,w_{n-1},v))\in\text{tp}(x_1,\ldots,x_{n-1}/A)$, a contradiction.

A similar argument shows that $r$ is non-algebraic; if we had $\varphi(y_1,\ldots,y_{n-1},v)\in r(v)$ algebraic – say with size $k\in\omega$ – then we would have $\neg\exists^{>k}v(\varphi(w_1,\ldots,w_{n-1},v))\in\text{tp}(x_1,\ldots,x_{n-1}/A)$, contradicting that $x_n\notin\text{acl}(\{x_1,\ldots,x_{n-1}\}\cup A)$. Thus $r(v)$ is a non-algebraic type, and so $r(v)$ and $q(v)$ are two non-algebraic types in $S_1^\mathfrak{N}(A\cup\{y_1,\ldots,y_{n-1}\})$ containing $\varphi(v)$, whence by lemma 1 we have $q(v)=r(v)$, as desired.

Now we may prove the main result. Let $U\subseteq\varphi(\mathfrak{M})$ and $V\subseteq\varphi(\mathfrak{N})$ be bases for $\varphi(\mathfrak{M})$ and $\varphi(\mathfrak{N})$, respectively. Since the dimensions are the same, we may find $f:U\rightarrow V$ a bijection. Now, this is the point where I get confused: Tent and Ziegler now claim that $f$ is an "$A_0$-elementary" map, which they have not defined but which I assume means that the induced map $f:A_0\sqcup U\rightarrow A_0\sqcup V$ (acting on $A_0$ as the identity) is partial elementary. If this map $f$ is partial elementary, then we have a elementary bijection between $A_0\sqcup U$ and $A_0\sqcup V$, which lifts to an elementary bijection between $\varphi(\mathfrak{M})=\text{acl}(A_0\sqcup U)$ and $\text{acl}(A_0\sqcup V)=\varphi(\mathfrak{N})$ fixing $A_0$ pointwise, where these last equalities hold since $U$ (resp. $V$) is a spanning set for $\varphi(\mathfrak{M})$ (resp. $\varphi(\mathfrak{N})$). So the result would be proved.

However, I'm not quite sure why $f$ is partial elementary. For this to be the case, it suffices to show that $\text{tp}^\mathfrak{M}(u_1,\ldots,u_n/A_0)=\text{tp}^\mathfrak{N}(v_1,\ldots,v_n/A_0)$ for any (pairwise distinct) $u_i\in U$, $v_i\in V$. This is almost but not quite clear to me. Indeed, since $U$ is a basis and thus independent, we have $u\notin\text{cl}(U\setminus\{u\})=\text{acl}(A_0\cup U\setminus\{u\})$ for any $u\in U$. In particular, $u_i\notin\text{acl}(\{u_1,\ldots,u_{i-1}\}\cup A_0)$ for any $i$. The same holds for the $v_i$, and so we are nearly in a position to apply lemma 2.

Lemma 2, however, does not deal with arbitrary substructures, and for it to apply I believe we would in particular need $\text{tp}^\mathfrak{M}(A_0)=\text{tp}^\mathfrak{N}(A_0)$. Then we would have $S_1^\mathfrak{M}(A_0)=S_1^\mathfrak{N}(A_0)$, and so could argue as follows: for any $u\in U$, $u$ is algebraic over $A_0$, and so the type $p(v)=\text{tp}^\mathfrak{M}(u)\in S_1^\mathfrak{M}(A_0)$ is non-algebraic. However, this type also contains the strongly minimal formula $\varphi(v)$, and so by lemma 1 is the unique non-algebraic type in $S_1^\mathfrak{M}(A_0)$ containing $\varphi(v)$. Likewise, for any $v\in V$, $q(v)=\text{tp}^\mathfrak{M}(v)\in S_1^\mathfrak{N}(A_0)$ is the unique non-algebraic type in $S_1^\mathfrak{N}(A_0)$ containing $\varphi(v)$. Since $S_1^\mathfrak{M}(A_0)=S_1^\mathfrak{N}(A_0)$, this means that $p(v)=q(v)$, so all the $u_i$ and $v_i$ are realizations of the same non-algebraic type $p$, and hence we could argue as in lemma 2.

So all we need to show is that $\text{tp}^\mathfrak{M}(A_0)=\text{tp}^\mathfrak{N}(A_0)$. However, I'm at a complete loss as to why this will be the case in general. Can anyone help with this last step of the proof? Sorry for the excessively long post, I wanted to give the context properly to make sure I'm not missing anything.

答:
When Tent and Ziegler write "Let $\mathfrak{M}$ and $\mathfrak{N}$ be models containing $A_0$", they implicitly mean that $\text{tp}^{\mathfrak{M}}(A_0) = \text{tp}^{\mathfrak{N}}(A_0)$. I agree that this should have been made explicit.

In Section 6.1, you will be introduced to the notion of a monster model $\mathfrak{C}$ and the common convention "all models are elementary submodels of the monster model". Under this convention, the condition on $A_0$ is automatic, since $\mathfrak{M}\prec \mathfrak{C}$ and $\mathfrak{N}\prec\mathfrak{C}$, so the truth value of a formula on a tuple from $A_0$ transfers up from $\mathfrak{M}$ to $\mathfrak{C}$ and then back down to $\mathfrak{N}$. This convention is so commonly used in model theory that I suspect Tent and Ziegler forgot that it was not in place yet when they wrote this section. (But note that we can't read the whole book as if this convention were in place: If it were, the characterization of quantifier elimination given in Theorem 3.2.5 would be trivially satisfied in every theory. For reasons like this, I prefer to avoid talking about two models $M$ and $N$ having a common substructure $A$ whenever possible, replacing the equality $A = A$ by an isomorphism or a partial elementary map between substructures $A\subseteq M$ and $A'\subseteq N$.)


答:
Note that:

$$P(\text{some packet faulty}) = 1 - P(\text{no packets faulty}) = 1 - P(\text{packet not faulty})^{13}$$

$$P(\text{some packet faulty}) = 1 - (1 - 0.05175)^{13}\approx\boxed{0.49882}$$

Title: Question about the chance of at least one packet of something being faulty

Body: From a previous question, I calculated the probability of something being defective being 0.05175. Now I want to know the probability of at least one packet being faulty when someone buys 13 of them. So I was thinking that I could set is up using a binomial with $n=13$ and $p=0.05265$. Then using $P(Y\ge1)= 1-P(Y\lt1) = 1 - \binom{13}{0}*(0.05265)^{13}$ But this gives me an answer of 1 which obviously isn't right. So I'm not sure what to do.


问:
Is there a way to prove the Cayley-Hamilton Theorem without the use of cofactors, adjoints, etc?

Is there a way to prove the Cayley-Hamilton Theorem without the use of cofactors, adjoints, etc?

Like is there another way to natural prove general matrix will satisfy its own characteristic polynomial?

答: If done in the right way, one can simply substitute $A$ for $x$ in $\rm{det}(xI-A)=0$.  Look at my question On the Cayley-Hamilton theorem


问:
Binomial Series of form $(1-x^2) ^{1/2}$
A question asks to use binomial series to find Maclaurin series for following:

$$f(x) = \frac{1}{{(4 + x^2)}^{1/2}}$$

Considering the x inside the radical is squared, the question is not of form: $${{(1 + x)}^{k}}$$

In general, how does one use binomial series when the x inside the expression is squared? I've attempted splitting denominator into
$$f(x) = \frac{1}{{(2-x)}^{1/2}}\frac{1}{{(2+x)}^{1/2}}$$

Getting power series of one and dividing end result by other denominator, but doesn't come out to the textbook's answer which is:

$$\frac{1}{2}\left(1 + \sum_{n=1}^{\infty}\frac{{(-1)^n(1)(3)(5)...(2n-1)x^{2n}}}{2^{3n}n!}\right)$$

答:
For $f(x) = \frac{1}{{(4 + x^2)}^{1/2}}=\left(4+ x^2 \right)^{-\frac{1}{2}}$ is possible to use Newton's_generalized_binomial_theorem for real power $-\frac{1}{2}$:

$$\left(4+ x^2 \right)^{-\frac{1}{2}} = \sum\limits_{k=0}^{\infty}\binom{-\frac{1}{2}}{k}4^{-\frac{1}{2}-k}x^{2k}$$
where $\binom{n}{k}=\frac{n(n-1)\cdots (n-k+1)}{k!}$


问:
What Is The Derivative Of f(x^3)
I want to find the derivative of $f'(x^3)$. I tried using chain rule and I got $3x^2f'(x^3)$. Is that correct?

答:
If your function is $g(x)=f'(x^3)$, then it would be by the chain rule $g'(x)=f''(x^3)3x^2$.

Otherwise, if you meant $f(x^3)$, it would be $f'(x)3x^2$.

Title: What Is The Derivative Of f(x^3)


问:
Integration over non-orientable Riemannian manifold
I am confused a bit by the following theorem that integrates over non-orientable manifold. How is that possible? (see this post).

Theorem. Let $M^n$ be a compact Riemannian non-orientable manifold, and $\omega$ a
differential 1-form. Let $\pi:(\widetilde{M},\tilde{g})\to (M,g)$ be the two-sheeted covering map, where $\tilde{g}=\pi^*g$ and let $\widetilde{A}=\pi^*A$. Then
$$\int_\widetilde{M}\tilde{\delta} \widetilde{\omega}\widetilde{dV}=2\int_M\delta\omega dV.$$

答:
It suffices to consider the case when $M$ is connected. Since all what you care about are integrals, just choose an open topological disk $U\subset M$ whose complement has measure zero, then pick an orientation on $U$ any way you want. For instance, you can take $U$ to be the complement to the cut-locus of a point in $M$. (If you do not like this, observe that $M$ is a connected sum of an oriented surface and the projective plane. Hence, $M$ contains a 1-sided smooth loop $c$ such that $M'=M-c$ is orientable. You can use $M'$ instead of a disk $U$.)

This will define a volume form $dV$, the Hodge star and $\delta$ on $U$. Then check that the integrand
$$
\delta\omega dV
$$
is independent of the chosen orientation on $U$ (you will be changing the sign twice) and the integral
$$
\int_U \delta\omega dV
$$
is independent of the choice of $U$ (since $M\setminus U$ has measure zero). This is how you make sense of    the integral
$$
\int_M \delta\omega dV.
$$

Now, consider the 2-fold covering $\pi$ and observe that $\pi^{-1}(U)$ has two components $U_1, U_2$ such that $\pi|_{U_i}: U_i\to U$ is a diffeomorphism, $i=1,2$. Both integrals
$$
\int_{U_i} \tilde\delta\tilde\omega \widetilde{dV}$$
will equal to
$$
\int_U \delta\omega dV.
$$
Thus, you get your formula.

The same trick works in higher dimensions as well.


问:
Proving $\,a\mid b^2\mid a^3\mid b^4\mid a^5\ldots\implies a=b$.
https://math.stackexchange.com/questions/704048/theory-number-problems
After I saw that post i wanted to solve the first one which is
$a\mid b^2,b^2\mid a^3,a^3\mid b^4,b^4\mid a^5\cdots$ Prove that $a=b$

Now i started by proving that $a$ and $b$ have same prime divisors,proving that is trivial,after doing so I checked do those factors have to have the same power.I tried with
$$a=p^2,b=p^3$$
I noticed it exactly goes for 3 terms,or that $a^3\mid b^4$
doing $$a=p^{n-1},b=p^n\\p^{n-1}\mid p^{2n}\\p^{2n}\mid p^{3n-3}\\p^{3n-3}\mid p^{4n}\\p^{4n}\mid p^{5n-5}\\\cdots\\p^{n(n-1)}\mid p^{n(n-1)}\\p^{n(n-1)}\mid p^{n(n+1)}\\p^{n(n+1)}\not\mid p^{(n-1)(n+1)}$$
Now yeah I guess that would be a proof,but wouldn't setting $n=n+1$ infinitely many times make every term dividable?

答:
Let $\nu_p(n)$ be the $p$-adic valuation of $n$ for a prime $p$.

Note for each $n \in \mathbb{N}$, we have
$(2n+1) \nu_p(a) \leq (2n+2) \nu_p(b)$ and
$(2n+2) \nu_p(b) \leq (2n+3) \nu_p(a)$ so that
$$1 - \frac{1}{2n+3} = \frac{2n+2}{2n+3} \leq \frac{\nu_p(a)}{\nu_p(b)} \leq \frac{2n+2}{2n+1} = 1+ \frac{1}{2n+1}.$$
As $n \to \infty$, $\frac{1}{2n+3} \to 0$ and $\frac{1}{2n+1} \to 0$, so $\frac{\nu_p(a)}{\nu_p(b)} \to 1$ and $\nu_p(a) = \nu_p(b)$ for each $p$. Thus, $a = b$ as desired.


问:
Find all the least squares solutions of Ax=b
Find all least squares solutions of  A x = b, where A =

\begin{bmatrix}
       1 & 3           \\[0.3em]
       -2 &-6          \\[0.3em]
       3 & 9
     \end{bmatrix}

and b = \begin{bmatrix}
       1           \\[0.3em]
       0          \\[0.3em]
       1
     \end{bmatrix}

and confirm that all
the solutions have the same error vector (and hence the same least squares error).
Compute the least-squares error.

The system that corresponds to the reduced row echelon form of the augmented matrix is $x_1 + 3x_2 =0, 0=1, 0=0 $ since the second equation cannot be solved the system is inconsistent

I'm not sure how to calculate the least squares solutions of A x = b because the inverse of $A^TA$ does not exist and so I can't solve the normal equations $A^TAx=A^Tb$ for x

答: Even though $A^TA$ is singular, you can solve the normal equations. Note that, from a theoretical perspective, $A^Tb$ is in the column space (image) of $A^T$, and the column space of $A^TA$ is the same subspace — again, for emphasis, always.

ContentLicense: CC BY-SA 3.0


问:
Is $x+y\geq2\sqrt{xy}$ for every $x,y\geq0$?
According to Wolframalpha, $\forall{x,y\geq0}:x+y\geq2\sqrt{xy}$.

How can I easily prove it?

I think that it has to do with the fact that for a given sum of $x+y$, the value of $xy$ reaches a maximum when $x=y$, thus leading to $x+y=2x=2\sqrt{x^2}=2\sqrt{xy}$.

But I'm not sure how to formulate this proof.

Thank you.

答:
$(\sqrt x - \sqrt y)^2\ge0$,

so $x-2\sqrt x \sqrt y +y\ge0$,

so $x+y\ge2\sqrt{xy}.$


问:
How many injective functions $f$ are there from set $\{1, 2, . . . , n\}$ to the set $\{1, 2, . . . , n\}$ so that $f(x) = x$ for some $1 ≤ x ≤ n$?
Using Inclusion/Exclusion and to solve this question, I am confused from the word "some" in the wording.

So far, I have determined the inclusion part to be the total number of injective functions with no constraints. This is $\boldsymbol {n!}$. Is the exclusion for the constraint just $\boldsymbol{-1}$ so that the total number of functions is $\boldsymbol{n! - 1}$?

Any guidance would be great.

答:
The some means there is at least one $x_i$ such that $f(x_i)=x_i.$

If you want to do inclusion/exclusion you could say that there are $(n-1)!$ functions where $f(1)=1$ because you can order the rest of the numbers any way you want.  You have $n$ choices of the fixed point, so this would give $n((n-1)!)=n!$ functions, but you have counted the ones with two fixed points twice so have to subtract them once.  Then the ones with three fixed points have been counted three times and removed three times, so have to be added once.

The approach in N. F. Taussig's comment is easier.


答:
Since $(x-y)^2$ is nonnegative, we have

$$0 \leq (x-y)^2 = x^2 - 2xy + y^2 = (x^2 + 2xy + y^2) -4xy = (x+y^2)-4xy$$

Hence,

$$ 4xy \leq (x+y)^2 \Longrightarrow xy \leq \frac{(x+y)^2}{4}.$$

Applying square-roots,

$$ \sqrt{xy} \leq \frac{x+y}{2} \leq x+y$$

Title: Is $x+y\geq2\sqrt{xy}$ for every $x,y\geq0$?

Body:
According to Wolframalpha, $\forall{x,y\geq0}:x+y\geq2\sqrt{xy}$.

How can I easily prove it?

I think that it has to do with the fact that for a given sum of $x+y$, the value of $xy$ reaches a maximum when $x=y$, thus leading to $x+y=2x=2\sqrt{x^2}=2\sqrt{xy}$.

But I'm not sure how to formulate this proof.

Thank you.


问:
Distance of sets proof
Problem: Let $K$ and $L$ be non-empty compact sets, and define $$d=\inf\{|x-y|:x \in K, y \in L\}$$

(a) If $K$ and $L$ are disjoint, show $d>0$ and that $d=|x_0-y_0|$ for some $x_0\in K$ and $y_0\in L$

My proof:

Let $D=\inf\{|x-y|:x \in K, y \in L\}$, we will fist show D is compact. Let $d \in D $ then $d=|x-y|$. Because $K$ and $L$ are bounded then for any $x\in K$ and $y\in L$ there exists $M_1$ and $M_2$ s.t:
$$|x|<M_1$$
$$|y|<M_2$$
therefore:
$$d=|x-y|<|x|+|y|=M_1+M_2$$
so D is bounded.

Now let $\{d_n\} \subset D$ be a convergent sequence of in D. To show D is closed we must show $d_n \to d$ for some $d\in D$. Notice $$d_n = |x_n - y_n|$$ where $(x_n)$ and $(y_n)$ are convergent sequences in $K$ and $L$ respectively. Since  $K$ and $L$ are closed, every convergent sequence converges somewhere in that set. Hence let $(x_n) \to x$ and $(y_n) \to y$ where $x \in K$ and $y \in L$.

By the reverse triangle inequality and triangle inequality :
$$||x_n-y_n|-|x-y||<|x_n-x+y-y_n|<|x_n-x|+|y_n-y|<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon $$
Hence $D$ is compact so $\inf(D)$ exists by the Axiom of completeness and $\inf(D) \in D$ therefore $d=|x_0-y_0|$ for some $x_0\in K$ and $y_0\in L$ as desired.

We will now show $d>0$ suppose for a contradiction that $d=0$ then there would exist some $x \in K$ and $y \in L$, such that $|x-y|=0$ therefore $x=y$ but this contradicts the fact that $K$ and $L$ are disjoint.

Why I am confused: The next part of the problem asks me to prove the following:

(b) Show that it is possible to have $d=0$ if we assume only that the disjoint sets $K$ and $L$ are closed

What am I confused about: I showed that the set $D$ was closed using the fact that $K$ and $L$ were closed so I think I might have done something wrong there but I am not sure what exactly.
Any help would be appreciated.

Thank you!!

答:
For a counterexample, try $C=\{(x,1/x):x>0\}$ and the $x$-axis. I do not understand your proof, but I can offer some hints on one way to do it: in fact, prove the more general result, in which $L$ is only assumed to be closed.

$1).\ $ Set $d=\inf\{|x-y|:x \in K, y \in L\}.$ Find sequences $(x_n)\subseteq K,(y_n)\subseteq L$ such that $|x_n-y_n|\to d.$

$2).\ $ Pick a point $x_0\in K$ and $y_0\in L$ and show that

$\inf\{|x-y|:x \in K, y \in L\}=\inf\{|x-y|:x \in K, y \in L\cap \overline B(x_0,2|x_0-y_0|)\}.$

$3).\ $ There is a subsequence $(x_{n_k})$ that converges to some $x\in K.$

$4).\ (y_{n_k})$ is a sequence in $L\cap \overline B(x_0,2|x_0-y_0|)$, which is compact, so there is a subsequence $(y_{n_{k_l}})$ that converges to $y\in L.$

$5).\ $ Conclude that $(x_{n_{k_l}}-y_{n_{k_l}})\to (x,y)$, that $|x-y|=d$ and that $d>0.$

Remark: the slick way to do it is to note that $d:X\times X\to \mathbb R:(x,y)\mapsto d(x,y)$ is continuous. Since $d(x,y)>0$ whenever $x\neq y$ and since $K\times L$ is compact, $d$ must have a strictly positive minimum on $K\times L$.

Title: Distance of sets proof


问:
Comparison test for series $\sum_{n=1}^{\infty}\frac{n}{n^3 - 2n + 1}$
I am trying to prove the convergence of the series $\sum_{n=1}^{\infty}\frac{n}{n^3 -2n +1}$ with the simple comparison test. I know it can be done with other tests but this question came up in my homework for the comparison test before the other tests were discussed so I'm being stubborn and want to use that test.

I know it converges but I haven't been able to find a rigorous justification. This is my reasoning so far.

$\sum_{n=1}^{\infty}\frac{n}{n^3 -2n +1}$ converges because $\sum_{n=1}^{\infty}\frac{n}{n^3}$ = $\sum_{n=1}^{\infty}\frac{1}{n^2}$ and I know the latter converges by the p-series criterion.

To use the comparison test as my justification I must show $\frac{n}{n^3−2n+1} \leq \frac{n}{n^3} \ \ \forall \  n$. This would be obvious if the denominator on the left were greater than the denominator on the right but that can't be true as on the left I am subtracting from $n^3$ and on the right $n^3$ is unchanged.

So I switch to proving $\frac{n}{n^3−2n+1} \leq \frac{1}{n^2}$ and then I'm stuck.

If the numerators were both 1 then $n^3−2n+1$ is eventually going to be larger than $n^2$ even if we're subtracting a little bit from $n^3$.

That is, at n=1 we'll get $1^3−2+1=0 \leq 1$ but thereafter $n^3− 2n +1  \geq n^2$ is always true.

But the statement of the comparison test says $a_n \leq b_n \forall n$ where $\sum b_n$ is known to converge.

So that's where I am so far. The problem comes down to showing the necessary inequality holds and is it enough the it holds "eventually".

Any ideas?

答:
How about the Limit Comparison Test? From Wikipedia:

Suppose that we have two series $\sum_n a_n$ and $\sum_n b_n$  with $a_n, b_n > 0$ for all $n$.

Then if $\lim_{n\to\infty}\frac{a_n}{b_n} = c$ with $0 < c < \infty$, then either both series converge or both series diverge.

For your problem, use:
$$a_n=\frac{n}{n^3 -2n +1}\\
b_n = \frac {1}{n^2}$$

Then:
$$\lim_{n\to\infty} \frac {a_n}{b_n} = \lim_{n\to \infty}\frac {\frac{n}{n^3 -2n +1}} {\frac 1 {n^2}} = \lim_{n\to \infty} {\frac{n^3}{n^3 -2n +1}} = 1\\$$

The last limit is a consequence of L'Hôpital's rule (see below).

Thus, because $0 < 1 < \infty$, and $\sum \frac 1 {n^2}$ converges (P-series), the given series must also converge.

The L'Hôpital's rule derivation is:

$\lim_{n\to\infty} \frac{n^3}{n^3 - 2n + 1} = \frac \infty \infty$ so we can take the derivative of the numerator and denominator.

$\lim_{n\to\infty} \frac{3n^2}{3n^2 - 2} = \frac \infty \infty$ so we can do it again.

$\lim_{n\to\infty} \frac{6n}{6n} = \frac \infty \infty$ so we do it once more.

$\lim_{n\to\infty} \frac{6}{6} = 1$ and we're done.


问:
Solving $A\vec x=\vec b$ for matrices
I have the following matrix
$$
A=
\begin{bmatrix}
  1 & 3 & 0 & 1 & 0\\
  1 & 2 & 1 &-2 & 1\\
 -1 & 1 &-1 & 3 & 2\\
\end{bmatrix}
$$
I have calculated the column space and I've got $\text{ Col}(A) = \big\{(1,1,-1)^T,(3,2,1)^T,(0,1,-1)^T\big\}$

Now the question is if the vector $\vec b = (1,1,1)^T$ is a part of the column space of $A$?

And what is the solution set for $A\vec x = \vec b$?

Tried to do Gauss-elimination with the values of the vector on the right side. Is that the correct first step? If not, can someone explain how it is done?

答:
$b\in C(A)$ if and only if you can write it as a linear combination of vectors that are in $C(A)$.

In that case, you can set up a system of equations $\alpha_1 v_1 + \alpha_2 v_2 + \alpha_3 v_3=(1,1,1)$ ($\alpha_i\in\mathbb{R}, v_i \in C(A), i\in[|1,3|]$).

Another approuch would be that $C(A)$ you got are linearly indpendent vectors (it should be if you got it right), and then the answer is yes, because it spans $\mathbb{R}^3$.

Title: Solving $A\vec x=\vec b$ for matrices


问:
How to you model a game when the rules you are given are wrong?
The title-question is the main focus of my confusion but I have several supporting questions at the end that try to clarify.

To illustrate my question, I will base it off of a game I played a while back during a team-building exercise at a university camp. I can't remember exactly how the rules were phrased. Basically all the students were blindfolded and led to a place where we were given a singular, long-taught rope to hold.

These were the constraints that were given to us or that could be inferred:

We were told that when the whistle blew we could begin moving.

We could move anywhere in space as long as we were touching the rope at any given moment (it was possible to pass other people with some talking and physical co-ordination).

If we thought we'd found the end of the rope we could raise our hand and some authority would tell us whether we had actually found it. There was no limit to how often we could do this.

If we found the end of the rope, we won.

We could ask to surrender; the game ended for us and we left with our blindfolds off.

It might not be the best formulation, but I have decided to model it as a one-player game with two actions, namely: surrender or continue searching for the end of the rope. Let $k$ be a quotient representing how much patience and/or resolve we have. And let $t$ be how much time has elapsed in minutes. Given the following decision matrix.

$$\begin{array}{c|c|}
\hline
Surrender & 0\\
Continue & 15-(t/k)\\
\hline
\end{array}$$

As time went on in the game, I heard as people asked if they had found the rope and being told that they hadn't. Soon, people with lower patience $k$ started surrendering. Everything fit the model of the rules explained so far.

The problem arose when I reached a knot in the rope. I recognised the knot from my starting point and realised that I had come full-circuit. The implication of this was that the theoretical win-condition stated at the beginning was un-achievable.

Much later, after the game, the facilitator told us that the solution was to surrender. The game was supposedly designed to show us how self-dependent we try to be. Before that, however, after I had realised that the actual rules were different to the way I had perceived them, I tried to formulate a new game with my new information. Since I wasn't satisfied to end with a solution that gave me the minimum utility.

Therefore, considering I was effectively playing against the patience of the facilitators, I adapted it as minimally as I could to a two-player game like the utility matrix bellow (where $j$ is the patience and/or resolve of the facilitator).

$$\begin{array}{c|c|c|}
& Surrender & Continue\\
\hline
Surrender & 0,0 & 0,5\\
Continue & 5,0 & 15-(t/k),15-(t/j)\\
\hline
\end{array}$$

The logic was that if I could motivate myself to have $k > j$, then I would have to win. Which, if this game was the measure, I eventually did, to the annoyance of the facilitator.

So what I want to know is, what was the actual game?

At what point did the rules change (if at all)?

If they did change, was my adaptation from the original one more reasonable or did the facilitator's "admission" about lying about the win condition not remove credence from their new win condition?

Is there a better way to model the two games (together maybe)?

This all borders on philosophy because it is unclear who determines the rules but I basically want to know what the "mathiest" way to talk about it is.

If there is any way this question can be clearer, feel free to edit it or suggest a change. I am just a hobbyist mathematician.

答:
When you're modeling a game, it's all about identifying strategies and outcomes. I think in your case, both of these are a bit underbaked; here's what I mean by that.

Strategies

When identifying strategies available to each player, you must be exhaustive. Every action that's available to each player at any point in the game must be represented somehow. In the simple game you've written above, there are two strategies: Surrender and Continue. But these don't represent the sum total of actions available to you, for instance you could continue until finding the knot and then surrender. Or you could surrender immediately. Or you could continue forever. Or you could spend up to 10 minutes searching, and then surrender. Likewise, the facilitator has many options.

I see two natural ways to capture this without changing your framework too much: continuous strategies and repeated play. For the first, one approach might be to define two new variables, $d_s$ and $d_f$, where $d_s$ is the total duration of time you're willing to search, and $d_f$ is the duration the facilitator is willing to let you search.

For repeated play, you could model the game as a series of decision points, at each one the players have the option to stop the game or let it continue. If the actions available at each decision point are the same, then the game can be represented by the single matrix you have, and some analysis can be done on the options available to the players. In the end, a "meta"strategy in the repeated game is going to look a lot like the continuous strategy above, since if at any point either player surrenders the game is over.

Outcomes

I think another source of difficulty is that you've tried to jump straight to the payoffs for each player. It's usually sounder to start by explicitly listing the outcomes, and then assigning payoffs to the outcomes afterward. For instance, some of the possible outcomes here are: The game ends quickly and both players are satisfied; the game drags on and both players are dissatisfied; the game drags on and the searcher feels vindicated but the facilitator is annoyed, etc., etc., and the actual length of the game may have some bearing on how satisfied / dissatisfied the players are.

The considerations about whether the rules changed, what it means that the facilitator lied about the nature of the game, etc., can (and should) be completely accounted for in the outcomes, and thus in the payoffs. For instance, in the continuous strategy example I gave above, in the case that $d_s > d_f$, I get the sense you would call that a victory for the searcher, so the payoffs may be $(10-d_f, -d_f)$ to indicate how annoyed you are that the facilitator made you play for so long, and how annoyed the facilitator is that they didn't get their preferred outcome and it took so long.

One thing that's very important to keep in mind is that in this kind of game theory, the concept of "winning" is basically meaningless. There is only the value (utility) that each player gets from whatever outcome occurs. This is one of the most difficult mental hurdles new game theorists must get over.

I hope this helps you think about the problem!


问:
We needed to find an example where $\sum x_{n}$ and $\sum y_{n}$ do not converge absolutely, but $\sum x_{n}y_{n}$ does.
1. Question

We needed to find an example where $\sum x_{n}$ and $\sum y_{n}$ do not converge absolutely, but $\sum x_{n}y_{n}$ does.

Answer:

I thought that is $\sum x_{n} = \sum_{n=0}^{n=\infty}\frac{1}{n^{3}}$ and $\sum y_{n} = \sum_{n=0}^{n=\infty}n$, then it works right? Since $x_{n}y_{n} = \frac{1}{n^{2}}$, and this converges by the p-series test, but individually $x_{n}$ does not converge.

2. Question

Find an example where $\sum x_{n}, \sum y_{n}, \sum x_{n}y_{n}$are absolutely convergent, are not just finite sums, and

($\sum x_{n}$) ($\sum y_{n} )$ $\ne$ ($\sum x_{n}y_{n}$).

Answer:

For this I used the following resource:

https://tutorial.math.lamar.edu/classes/calcii/absoluteconvergence.aspx

And my examples are $\sum x_{n} = \sum\frac{1}{n^{2}} , \sum y_{n} = \sum \frac{sin(n)}{n^{3}}, \sum x_{n}y_{n} = \sum \frac{sin(n)}{n^5}$.

But what is a finite sum? Does my example satisfy all properties ( $\sum x_{n}, \sum y_{n}, \sum x_{n}y_{n}$are absolutely convergent, are not just finite sums, and ($\sum x_{n}$)($\sum y_{n}) $ $\ne$ ($\sum x_{n}y_{n}$)?

答:
Your example for the first question doesn’t work, because $\sum_{n\ge 0}\frac1{n^3}$ does converge absolutely. Try letting $x_n=y_n=\frac1n$ instead. Or let $x_n=y_n=\frac{(-1)^n}n$, in which case $\sum_{n\ge 0}x_n$ and $\sum_{n\ge 0}y_n$ converge, but not abolutely.

A finite sum is a series with only finitely many non-zero terms. Examples for the second question are trivial if we allow finite series; for instance, $$(1+1+0+0+0+\ldots)^2\ne 1\cdot1+1\cdot1+0\cdot0+0\cdot0+\ldots\,.$$ By the way, you’ve a misplaced parenthesis in the question: it should be

$$\left(\sum_nx_n\right)\left(\sum_ny_n\right)\ne\sum_nx_ny_n\,.$$

For a straightforward example let $x_n=y_n=\frac1{2^n}$. Then

$$\sum_{n\ge 0}x_n=\sum_{n\ge 0}y_n=\sum_{n\ge 0}\left(\frac12\right)^n=2\,,$$

but

$$\sum_{n\ge 0}x_ny_n=\sum_{n\ge 0}\left(\frac12\cdot\frac12\right)^n=\sum_{n\ge 0}\left(\frac14\right)^n=\frac43\ne 2^2\,.$$


问:
Is it true that $\mathscr{O}_{X,x}=\mathscr{O}_{\operatorname{Spec} A,x}$ for any $\operatorname{Spec} A$ with $x\in\operatorname{Spec} A$?
Let $X$ be a scheme and $\operatorname{Spec} A\subset X$ be an affine with $x\in\operatorname{Spec} A$,

Is it true that $\mathscr{O}_{X,x}=\mathscr{O}_{\operatorname{Spec} A,x}$?

I ask this question because I feel that many "obvious" remarks made in Gortz-Wedhorn and Vakil FOAG seem to rely on this or a similar fact: such as in Vakil

"We say a ringed space is a locally ringed space if its stalks
are local rings. Thus Exercise 4.3.F shows that schemes are locally ringed spaces." (Where 4.3.F is Show that the stalk of $\mathscr{O}_{\operatorname{Spec} A}$ at the point $[\mathfrak{p}]$ is the local ring $A_{\mathfrak{p}}$).

Is this the case? If not, why is the above statement immediate from 4.3.F?

答: Yes that's true, the statement you want to prove is that if $U$ is any open subset of any ringed space $X$ then for $x\in U$ the natural map $\mathcal O_{U,x}\to\mathcal O_{X,x}$ is an isomorphism. It's pretty straightforward, for instance for surjectivity if you have an element $f_x\in\mathcal O_{X,x}$ then it is represented by a function $f\in\mathcal O_X(V)$ for some $V$ containing $x$, and then the stalk of $f|_{U\cap V}$ in $\mathcal O_{U,x}$ is the element you want mapping to $f_x$.


问:
Proving existence of unique polynomial satisfying some integral
I came across a problem that I was having some trouble with.

Fix a positive integer $n$. Let $f(x) \in C(R)$ be a (real valued) continuous function.
Show that there exists a unique polynomial $q(x) \in P_n(R)$, such that for every
polynomial $p(x) \in P_n(x)$ we have $\int^1_0 p(x)f(x)dx=\int^1_0 p(x)q(x)dx$.

I know that typically to solve uniqueness questions, you need to assume another $g(x)$ exists and then manipulate the resulting equation(s) to show that $g(x)=q(x)$. However, it's not so clear to me here how I should go about this here. I think that the $p(x)$ might be confusing me in this question.

If anyone can give me some pointers, it would be greatly appreciated. Thanks!

答:
Here's how you can claim uniqueness: Fix $f\in C(\Bbb{R})$, and suppose that both $g$ and $q$ satisfy the condition in question. This means we have
$$
\int_0^1\!p(x)f(x)\,dx = \int_0^1\!p(x)g(x)\,dx = \int_0^1\!p(x)q(x)\,dx
$$
for all $p\in P_n(\Bbb R)$. In particular, this property holds for $g(x) - q(x)$ (which is again in $P_n(\Bbb R)$). Then we have
$$
\int_0^1\!\left(g(x) - q(x)\right)g(x)\,dx = \int_0^1\!\left(g(x) - q(x)\right)q(x)\,dx
$$
so that
\begin{align*}
0 &= \int_0^1\!\left(g(x) - q(x)\right)g(x)\,dx - \int_0^1\!\left(g(x) - q(x)\right)q(x)\,dx\\
&= \int_0^1\!\left(g(x) - q(x)\right)g(x) - \left(g(x) - q(x)\right)q(x)\,dx\\
&= \int_0^1\!\left(g(x) - q(x)\right)\left(g(x) - q(x)\right)\,dx\\
&= \int_0^1\!\left(g(x) - q(x)\right)^2\,dx.
\end{align*}
However, $\left(g(x) - q(x)\right)^2$ is continuous and $\left(g(x) - q(x)\right)^2\geq 0$ for all $x\in [0,1]$, so that $\left(g(x) - q(x)\right)^2 = 0$ for all $x\in [0,1]$ (because if $h$ is continuous and nonnegative on $I = [a,b]$ and $\int_a^b\!h = 0$, then $h\equiv 0$ on $[a,b]$). This implies that $g(x) = q(x)$ on $[0,1]$, and since they are polynomials, they must then be the same polynomial.

Alternatively, we can view this same proof in terms of linear algebra language:
$\langle p,q\rangle = \int_0^1 pq$ defines an inner product on $P_n(\Bbb R)$. In the same manner as before, if $g,q$ both satisfy your desired condition, then
\begin{align*}
\langle q - g,q\rangle &= \langle q - g,g\rangle\\
\implies 0 &= \langle q - g,q\rangle - \langle q - g,g\rangle\\
&= \langle q - g,q - g\rangle\\
&= \left|\left| q - g\right|\right|^2,
\end{align*}
which implies that $q - g = 0$, or $q = g$. (Added because you tagged this "linear algebra.")


问:
Proving an inequality from a given condition
Consider the following exercise:

Suppose that $f$ satisfies $$f\left(\frac{x+y}{2}\right)\leq\frac{f(x)+f(y)}{2}.$$ Show that $$f(kx+(1-k)y)\leq kf(x)+(1-k)f(y)$$ whenever $k$ is a rational number between $0$ and $1$, and of the form $\frac{m}{2^n}$

I tried using induction. Suppose the result holds for any $k$ of the form $\frac{m_0}{2^n}$ with $m_0<2^n$. Now suppose that $k=\frac{m}{2^{n+1}}$ with $m<2^{n+1}$. However, I don't know how to proceed.

答:
Here is an example for $n=2$ and I hope this will give you some ideas about the induction:
\begin{align*}
f\left(\frac{x}{4}+\frac{3y}{4}\right)=f\left(\frac{x+y}{4}+\frac{y}{2}\right)&\leq\frac{1}{2}f\left(\frac{x+y}{2}\right)+\frac{f(y)}{2} \\
&\leq\frac{f(x)}{4}+\frac{f(y)}{4}+\frac{f(y)}{2} \\
&=\frac{f(x)}{4}+\frac{3f(y)}{4}.
\end{align*}


问:
Commutation of $c_0$-semigroups
Task is: given two $c_0$-semigroups $T(t)$ and $S(t)$ in Banach space X with property $\forall \;t \geq 0 : T(t)S(t) = S(t)T(t)$, show that
$\forall \; t, s \geq 0 : T(t)S(s) = S(s)T(t)$.

I've tried using some addition-subtraction tricks and differentiating
$\varphi(t,s) = T(t)S(s)x - S(s)T(t)x$ (defined for arbitrary $x \in X$), but none of these methods succeeded.

答:
Hint: Fix $t \in (0,\infty)$ and set
$$
  A := \{s \in [0,\infty): T(t)S(s) = S(s)T(t)\}.
$$
Show that if $s \in A$ and $k \in \mathbb{N} := \{1,2,\dots\}$, then $ks \in A$ and $s/k \in A$, too.

Title: Commutation of $c_0$-semigroups


问:
Spivak sum of three numbers
I'm trying to study calculus on my own, and in the very first page of Spivak's Calculus he states that "For the sum of three numbers, a, b, and c, this may be done in two different ways [...] a+(b+c) or  [...] (a+b)+c"

My question: is there a reason for him not to consider the case (a+c)+b? I would think that as another way to perform the sum of three numbers, so I'd say there are actually three ways to sum three numbers.

Thank you!

答:
There are actually $12$ ways, if one counts all $3!=6$ possible permutations of $a,b$, and $c$ and, for each of them, both possible ways of parenthesizing the expression. However, he is interested here specifically in the associative property of addition, the one that says that for all $a,b$, and $c$,

$$a+(b+c)=(a+b)+c\,.\tag{1}$$

In other words, it says that adding from right to left always gives the same result as adding from left to right. It’s this fact that allows us to dispense with parentheses altogether and write simply $a+b+c$, something that we cannot do with subtraction: $(a-b)-c$ is not in general equal to $a-(b-c)$.

It’s important to separate this from relations like

$$a+(b+c)=(a+c)+b\tag{2}$$

that also involve changing the order of the numbers themselves, because there are mathematical operations that satisfy $(1)$ but not $(2)$. One of them is matrix multiplication: if $A,B$, and $C$ are $n\times n$ matrices, it’s always true that $A(BC)=(AB)C$, but it is not always true that $A(BC)=(AC)B$, because it’s not always true that $BC=CB$.

Title: Spivak sum of three numbers


问:
What is the meaning of "independent events " and how can we logically conclude independence of two events in probability?

What is the meaning of  "independent events " in probability

For eg: Two events (say A and B)are independent , what I understand is the occurrence of A is not affected by occurrence of B .But I am not comfortable with this understanding , thinking this way every events I meet upon are independent!, does there exist a more mathematical definition( I don't want the formula associated with it)

Another thing I want to know in a real case or a problem how do we
  understand(logically) if two events are independent?

That is without verifying that $P(A\cap B) = P(A)P(B) \Large\color{blue}{\star}$  how do we conclude that two events $A$ and $B$ are independent or dependent?.

Take for eg an example by "Tim"

Suppose you throw a die. The probability you throw a six(Event $A$) is $\frac16$ and the
  probability you throw an even number(Event $B$) is $\frac12$.

And event $C$ such that $A$ and $B$ both happen would mean $A$ should happen(As here $A$ is a subset of $B$) hence it's too $\frac16$

My thought:

The above example suggest something like this "Suppose I define two events $A$ and $B$ and if one of the even is a subset of the other then the events are not independent". Of course such an argument is not a sufficient condition for dependency , for eg: consider an event $A$ throwing a dice and getting an odd number and event $B$ getting 6. They aren't independent ($\Large\color{blue}{\star}$ isn't satisfied) . So again I improve my suggestion "Suppose I define two events $A$ and $B$ and if one of them is a subset of the other or their intersection is a null-set then the events are not independent $\Large\color{red}{\star}$"

So at last is $\Large\color{red}{\star}$ a sufficient condition?. Or does there exist a suffcient condition (other than satisfying the formulas $\Large\color{blue}{\star}$? And what is the proof, I can't prove my statement as the idea for me is not that much mathematical .

答:
As pointed out, the question you pose is both mathematical (in which case the answer is clear and unambiguous) and philosophical (in which case, it isn't).

For a starter on the philosophy of probability - start here.

Your statement $\Large\color{red}{\star}$ is a sufficient condition is correct but it is highly restrictive. The statement $\Large\color{blue}{\star}$ is a necessarily and sufficient condition - indeed the necessarily and sufficient condition.

Consider the two events "Arsenal will win their next game" and "Manchester United will win their next game". Are these events independent? If they are playing each other - clearly not, if they are playing other teams then clearly yes.

So there exists a third event "Arsenal are playing Manchester United in their next game". The first two events are clearly dependent on this event.

The probability of this third event is interesting because it strikes at the heart of how one interprets probability.

This probability is either 0 or 1 - I could go online now and find out. But if I do not know -  can I determine a probability for it that is not 0 or 1. If I can, how can this be interpreted? Further, what if nobody knows (e.g. the draw has not yet been determined)?

For some interpretations of this state of indeterminacy see Schrodinger's Cat

I think the point is that mathematics is mathematics, physics is physics, music is music and dance is dance - what they mean is down to humanity.


答:
Spivak's introduction gives you a feel for how certain "obvious" properties may not be obvious at all. These different ways of evaluating the sum of three numbers relate to the property of associativity. Not all operations satisfy this very natural property, so it is important to be careful.

A different property which you bring up in your question is the idea of commutativity (whether $a+b = b+a$), and indeed it may not be true that the objects you are working with commute! For addition of real numbers, it is true that they satisfy both associativity and commutativity, but I believe Spivak provides these to you as axioms (fundamental "truths" that you can accept) in the following pages.

Title: Spivak sum of three numbers

Body:
I'm trying to study calculus on my own, and in the very first page of Spivak's Calculus he states that "For the sum of three numbers, a, b, and c, this may be done in two different ways [...] a+(b+c) or  [...] (a+b)+c"

My question: is there a reason for him not to consider the case (a+c)+b? I would think that as another way to perform the sum of three numbers, so I'd say there are actually three ways to sum three numbers.

Thank you!


问:
Contour integration and Green's function
I have a question which is related to changing an improper integral into a contour integral that 'looks' similar, but is not the same. The author (D.H. Griffel) says that it is 'right' to do it, but I don't understand why. It appears this is related to the standard thesis in the use of Green's functions. The author and I want to 'solve' the PDE
$$
u_{tt} - \nabla^2 u = f.
$$
The author rewrites the problem as
$$
(\partial_t ^2 - \nabla_x^2) w(\vec{x},t;\vec{y},s) = \delta(x-y) \delta(t-s).
$$
By using the Fourier transform he finds
$$
-(\omega^2 - |\vec{k}|^2) v = e^{i(\vec{k}\cdot \vec{y} - \omega s)}
$$
with $v$ being the Fourier transform of $w$. Now the question of contour integration arises. The author says that
$$
w(\vec{x},t;\vec{y},s) = \frac{1}{4 \pi^3 |\vec{y} - \vec{x}|} \int_{0}^\infty k \sin k |\vec{y} - \vec{x}| I(k)\,  \, dk
$$
with
$$
I(k) = \int_{- \infty}^\infty \frac{e^{i\omega(s-t)}}{k^2 - \omega^2} d\omega.
$$
The author can't evaluate this integral because it has poles at $\omega = \pm k$. Instead, the author evaluates different contour integral that avoids the poles and is closed in the upper- or lower half-plane.

But my 'gripe' is that these integrals are not the same integral as $I(k)$. The author says that each choice of integration contour corresponds to a different $w$. My question:

Why is it relevant to look at these alternative integral contours; after all, we're not interested in them. I can't see a reason why they are related to the integral we want to determine that would give us $w$. Is this anecdotal, or something rigorous?

How can the author prove that each choice of contour corresponds to different boundary conditions?

What do I mean by 'the integrals are not the same':

$I(k)$ is not a complex number for almost any $k$ (in the sense of lebesgue or riemann)

The integral of the integrand of $I$ over a contour avoiding the poles is a complex number (in the sense of riemann).

In the case so-called $i \varepsilon$-prescription is something mathematical, it would be great if you could provide a reference to the theorem, or just post the theorem and idea for the proof here.

答:
Finally, something I understand!

The basic idea is that all the information about a contour integral is encoded in its poles. This is why the solution to all sorts of integrals are determined by residues (which are just scaled line integrals around all poles within whatever contour you choose).

Now, for your question, the poles are located ON the contour, so the question we need to ask is whether or not the pole is inside or outside the contour of interest? It’s actually a trick question because a simple pole is “half-in, half out”. Try it! Manually compute the integral of a meromorphic function around the little deformation contour and as long as the pole is first order (simple), you will always get negative one half the expected residue.

So poles contain all the information, and your contour only contains half the pole, as the pole is “half-in, half-out”. This means you’re losing information, and if you solve whatever contour integral you want, this typically shows as a constraint on your solution. In order to avoid losing information, just take generate a second contour that covers the remaining half of the pole. You’ll find that the constraints on your integral then tell the full story.

In summary: Always use the full pole or you will lose information. If a pole is on the contour, generate as many instances of a contour as it takes to fully cover it. The restrictions on the solution are typically additive to the final solution.

I hope you found my answer useful


问:
Can a tree graph have only one vertex? And if so, that means that a tree graph has at minimum one leaf?
Also, if the tree is two vertices connected by an edge, does the root count as a leaf too? Since it's also a vertex of degree one?

I've had trouble clarifying this online and from my textbook. Thank you.

答:
Yeah
a graph only has one vertice is a tree
but only a tree with at least 2 vertices has a leaf</p>


问:
Proving a matrix has no real eigenvectors?
How does one go about proving that a matrix has no real eigenvectors or values? This would be for a general case, and all that is known is the nxn matrix A is invertible, and that A^2=-I

I tried using determinants and the characteristic polynomial, but as it is a general case with no specific matrix I can not figure out how to actually prove this. Thanks!

答:
You need to specify that $\ A\ $ be a real matrix for your question to have a satisfactory answer.  Stinking Bishop's comment shows that the eigenvalues of $\ A\ $ can only be $\ \lambda=\pm i\ $, so if $\ A\ $ is real, and $\ Ax=\lambda x\ $ with $\ x\ne0\ $, then $\ x\ $ cannot be real, because the expression on the left of the equation would have to be real, but the one on its right would have to have at least one non-zero imaginary entry.

If $\ A\ $ is not required to be real, however, it can have real eigenvectors. If
$$
A=\begin{pmatrix}0&i\\i&0\end{pmatrix}\ ,
$$
for example, then $\ A^2=-I\ $, and $\ A\ $ has eigenvectors $\ \begin{pmatrix}1\\1\end{pmatrix}\ $, corresponding to the eigevalue $\ \lambda=i\ $, and $\ \begin{pmatrix}1\\-1\end{pmatrix}\ $, corresponding to the eigevalue $\ \lambda=-i\ $.


问:
Proof about the kernel of two matrices
Prove or disprove that if $\ker A \neq \{ 0\}$ where $A \in \mathbb{K}^{n\times m}$, then $\ker (AB) \neq\{ 0\}$ for all Matrices $B \in \mathbb{K}^{m\times k}$

Thanks in advance

答:
If $\ker(A)$ is nontrivial, then $Ax=0$ for some nonzero $x \in \mathbb{R}^m$. Consider a simple case. For example, if $n=m=2$, then

$$Ax=
\begin{pmatrix}
a_{11} & a_{12}\\
a_{21} & a_{22}
\end{pmatrix}
\begin{pmatrix}
x_{1}\\
x_{2}
\end{pmatrix}
=
\begin{pmatrix}
x_1 a_{11} + x_2a_{12}\\
x_1a_{21} + x_2a_{22}
\end{pmatrix}.
$$

Since we're considering the case when $Ax=0$, we have

\begin{align*}
x_1 a_{11} + x_2a_{12} &= 0,\\
x_1a_{21} + x_2a_{22} &=0.
\end{align*}

So we can choose

$$A=
\begin{pmatrix}
1 & 1\\
1 & 1
\end{pmatrix}$$

and $x=(1,-1)^T$ to satisfy these equations. With this choice of $A$, we see that $\ker(A)$ is nontrivial. Now we want to try to find a matrix

$$B=\begin{pmatrix}
b_{11} & b_{12}\\
b_{21} & b_{22}
\end{pmatrix}
$$

such that $ABy=0$ has only the trivial solution $y=(0,0)^T$ (i.e. $\ker(AB) = 0$), for our choice of $A$ above. Can you find such a matrix $B$?

Title: Proof about the kernel of two matrices


问:
Find a diffeomorphism of R that is not a linear function.
this question from In V.I.Arnold's book about ordinary differential equations.

Examples from Boyce and di Prima:

Find a diffeomorphism of $\Bbb R$ that is not a linear function.

I know that a diffeomorphism is an isomorphism of smooth manifolds. It is an invertible function that maps one differentiable manifold to another such that both the function and its inverse are smooth.

Since any manifold can be locally parametrised, we can consider some explicit maps from $\Bbb R^2$ into $\Bbb R^2$.

But i couldn't solve it, so please help me. Thanks

答: $f(x)=x^3+x$ is such a diffeomorphism. $f'(x)=3x^2+1>0$ implies that $f$ is strictly increasing, $lim_{x\rightarrow -\infty}f(x)=-\infty$, $lim_{x\rightarrow +\infty}f(x)=+\infty$ implies that $f$ is surjective since its image is connected, thus an interval.


问:
Countinuous spectrum of multiplication operator on $L^{2}(\mathbb{R})$
Let $f:\mathbb{R} \rightarrow \mathbb{R}$ be a continuous function and define the multiplication operator $M_{f}:\text{dom}(M_{f})\subset L^{2}(\mathbb{R}) \rightarrow L^{2}(\mathbb{R})$ as $M_{f}g = fg$ for all $g \in \text{dom}(M_{f}) := \left\{g \in L^{2}(\mathbb{R}): fg \in L^{2}(\mathbb{R})\right\}$. Show that if $f$ is bounded and strictly monotonic, then $\text{sup}_{x\in \mathbb{R}}f(x) \in \sigma_{c}(M_{f})$. Here $\sigma_{c}(M_{f})$ denotes the continuous spectrum of $M_{f}$.

Here is what I´ve done: I know that $\sigma(M_{f}) = \overline{\left\{f(x) : x \in \mathbb{R}\right\}}$. Since $M_{f}$ is self-adjoint, I also know that $\sigma_{r}(M_{f}) = \emptyset$ and $\sigma_{p}(M_{f}) = \left\{\lambda \in \mathbb{R} : \mu\left(\left\{x \in \mathbb{R} : f(x) = \lambda\right\}\right)>0\right\}$. So I end up with $\sigma_{c}(M_{f})= \sigma(M_{f})\setminus\sigma_{p}(M_{f})$. How do I show that $\text{sup}f \in \sigma_{c}(\mathbb{R})$, any ideas?

答:
I'm assuming you are using Lebesgue measure. Since $f$ is strictly monotonic, $\sigma_p(M_f)=\varnothing$. Thus $$\sigma_c(M_f)=\sigma(M_f)=\overline{f(\mathbb R)}.$$

As $f$ is real valued, continuous, and strictly monotonic, $$\sup\{f(x):\ x\in\mathbb R\}=\lim_{x\to\infty}f(x)\in\overline{f(\mathbb R)}=\sigma(M_f)=\sigma_c(M_f).$$


问:
Extreme Value Theorem proof help
Extreme Value Theorem: If $f$ is a continuous function on an interval [a,b],

then $f$ attains its maximum and minimum values on [a,b].

Proof from my book:  Since $f$ is continuous, then $f$ has the least upper   bound, call it $M$. Assume there is no value $c \in [a,b]$ for which $f(c)=M$.

Therefore, $f(x)<M$ for all $x \in [a,b]$. Define a new function $g$ by

$g(x)=\frac{1}{M-f(x)}$

Observe $g(x)>0$ for every $x\in[a,b]$ and that $g$ is continuous and bounded on [a,b]. Therefore there exists $K>0$ such that $g(x)\le K$ for every $x\in [a,b]$. Since for each $x \in [a,b]$,

$g(x)= \frac{1}{M-f(x)} \le K$ is equivalent to $f(x)\le M-\frac{1}{K}$,

we have contradicted the fact that $M$ was assumed to be the least upper bound of $f$ on [a,b].

Hence, there must be a balue $c\in[a,b]$ such that $f(c)=M$.

Q: Where does the function $g$ come from? Is there a popular alternative proof?

答:
You asked for a "popular alternative proof". This is an alternative proof. I don't know how popular it is, but I like it. It uses the Bolzano-Weierstrass theorem (convergent subsequences) but hardly anything else, no least upper bounds; it skips the step of proving boundedness, going straight for the maximum. It could be shortened by using the fact that the set of rational numbers is countable, but that seems unnecessarily sophisticated.

Given a continuous real-valued function $f$ on $[a,b]$, we will show that
the set $Y = f([a,b])$ has a greatest element.

For each positive integer $n$, define a finite set $Q_n = \{\frac{p}{q}: p,q \text{ integers, } 0 < q \le n, |p| \le n\}$.

Choose $y_n\in Y$ so as to maximize the number of elements in the set $\{r\in Q_n: y_n > r\}$, and choose $x_n\in[a,b]$ with $f(x_n) = y_n$.

The sequence $\{x_n\}$ has a subsequence converging to a point $c\in[a,b]$. Since $f$ is continuous, the corresponding subsequence of $\{y_n\}$ converges to $f(c)$. We will show that $f(c)$ is the greatest element of $Y$.

Assume for a contradiction that $f(c)<y\in Y$. Choose a rational
number $r$ so that $f(c)<r<y$. Because of the way $y_n$ was chosen, we
have $y_n > r$ whenever $r\in Q_n$. Since $r\in Q_n$ for all
sufficiently large $n$, we have $y_n > r > f(c)$ for all sufficiently
large $n$. But this is absurd, since $\{y_n\}$ has a subsequence converging
to $f(c)$.

Title: Extreme Value Theorem proof help


问:
Let the joint pdf of 2 continuous rv's X,Y be f(x,y) = λμe^(-λx-μy) , for (x,y) > 0, and 0 otherwise.
You can see what I did for parts a and b of the question, I believe I did them correctly (if not please let me know). I don't have a solution for this exercise. I am struggling with part c) which asks to find Var(Y) using the total variance formula. I am struggling to find E[Y^2|X] as I somehow set up my integral wrong.

Clearly, my integral diverges so I messed up somewhere along the way. I would appreciate any help! Thanks

答:
Here $X$ and $Y$ are independent exponential random variables: $$f(x,y)=\lambda\mu e^{-\lambda x-\mu y}=\lambda e^{-\lambda x}\cdot\mu e^{-\mu y}=:f_X(x)\cdot f_Y(y). $$
The conditional distribution of $Y$ given $X$ (or $X$ given $Y$) is precisely the exponent distribution of $Y$ (of $X$, resp.)

Besides, the condition distribution $f_{Y\mid X}(y\mid x)$ you found at the above is right, but you plugged a wrong distribution into your last integral. This should be $\mu e^{-\mu y}$ instead of $1/\mu$.


问:
How to solve this cubic equation (where I must take the square root of a negative)?
I am trying to solve the following set of equations for $L'$ as described in this publication excerpt:

So basically, to re-write that, I have the following series of equations:

$a=X/2$

$b=-(75X^2 + 147Y^2)/80$

$c = -9X^3/16$

$P = 2a^3−9ab+ 27c$

$Q = P^2 -4(a^2-3b)^3$

$w_1=−0.5 +\frac{i\sqrt{3}}{2}$

$w_2=−0.5 -\frac{i\sqrt{3}}{2}$

Which lead to the following three possible solutions:

$L'_1 = -\frac{1}{3} * [a+\sqrt[3]{0.5*(P+\sqrt{Q})} +\sqrt[3]{0.5*(P-\sqrt{Q})}]$

$L'_1 = -\frac{1}{3} * [a+w_2\sqrt[3]{0.5*(P+\sqrt{Q})} +w_1\sqrt[3]{0.5*(P-\sqrt{Q})}]$

$L'_1 = -\frac{1}{3} * [a+w_1\sqrt[3]{0.5*(P+\sqrt{Q})} +w_2\sqrt[3]{0.5*(P-\sqrt{Q})}]$

I am now trying to solve $L'$, but I'm not actually sure how to do it.

My problem is for any input of $X$ or $Y$, $P$ I believe can end up positive or negative, but $Q$ is always a negative.

So let's say hypothetically $P = 2$ and $Q = -3$, then how do you solve:

$\sqrt[3]{0.5*(P-\sqrt{Q})} = \sqrt[3]{1-0.5*i\sqrt{3}}$

What would be the next step?

Basically, I don't know how to solve these equations if $Q$ is negative or if $0.5*(P-\sqrt{Q})$ is negative. I don't know much about complex math. I've used Euler's formula in other situations but I'm not sure here.

Any help? Thanks.

答:
In the next step, you want the cube root of the complex number $z = 1-i\sqrt{3}/2$. First express this number in the form $z = re^{i\theta}$ using Euler's identity. For a number $z = a + i b$ we have $ r = \sqrt{a^2 + b^2}$ and $\theta = \tan^{-1}(b/a)$:

$$ z = \sqrt{1^2 + (\sqrt{3}/2)^2} e^{i\tan^{-1}(-\sqrt{3}/2) + 2\pi i n} = \frac{1}{2}\sqrt{7}e^{i \tan^{-1}(\sqrt{3}/2) + i 2\pi n} .$$
Here $n$ is an arbitrary integer which accounts for the periodicity of $e^{i x}$.

Then you can take the cube root:

$$ z^{1/3} = \frac{7^{1/6}}{2^{1/3}} e^{i \tan^{-1}(-\sqrt{3}/2)/3 + i 2 \pi n/3} =  \frac{7^{1/6}}{2^{1/3}} \big[ \cos( \tan^{-1}(-\sqrt{3}/2)/3) + i \sin (\tan^{-1}(-\sqrt{3}/2)/3)\Big] e^{i 2 \pi n/3} $$

which you can simplify as desired. Notice it's multi-valued ($n = 0, \pm 1, \dots$). These different values are branches.


问:
How to find variance for given problem?
Suppose the expected earnings for a farm depends on the number of worms per apple. The number of worms per apple has a Poisson distribution with mean 0.32.

a) The expected earnings per apple depends on number of defects: 13-5X. What is the expected earning for 10 apples?
answer: I have done this part, where I do  10 * (13-5*0.32) which is the right answer.

b) Find the variance of earnings for 10 apples.

This is where I'm stuck. Thanks for any help

答:
Hints:

If $X$ has variance $\sigma^2$ and $Y=aX+b$ then $Y$ has variance $a^2 \sigma^2$

The variance of the sum of independent random variables
is equal to the sum of their variances

The variance of a Poisson distribution is equal to its mean

Title: How to find variance for given problem?


问:
where to find a proof of the Lebesgue Density Theorem
The other day I was browsing the site and found the question. I was trying to follow up with Topologieeeee, but clearly [s]he has not shown up for quite a while. So I wonder if anybody knows where to find the proof of the FACT referred in [s]he's question?

Thanks.

Edit

The FACT, from the old post, is the following:

Let $E$ be a Lebesgue measurable subset of $\mathbb{R}^n$. Almost every $x\in E$ satisfies $\lim\limits_{m(B)\to 0,~x\in B}\frac{m(B\cap E)}{m(B)}=1$ i.e. limit is taken over the ball $B$ containing $x$ with shrinking it.

答: This is called the Lebesgue Density Theorem.  With that knowledge in hand it should be easy to search for and find a proof.  I have made a nice proof, due to C.-A. Faure, from a recent Monthly article available here.


问:
How to compute $ P.V. \int_{-\infty}^{\infty} \frac{e^{iwx}}{|w|} \,dw $?
I came across the integral:
$$
\mbox{P.V.}\int_{-\infty}^{\infty} {\mathrm{e}^{\mathrm{i}\omega x} \over
\left\lvert\omega\right\rvert}
\,\mathrm{d}\omega
$$
in my research, and can’t figure it out after several hours of trial and error.

I didn’t know how to use Contour integration techniques on it because the pole is an absolute value, so I tried generating a power series for the exponential kernel, but ended up with a log function that blew up.

May you please explain how I can evaluate such an integral $?$. Wolfram has the answer, but I care more about the thought process than anything.

Many thanks in advance $!$.

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]{\mbox{P.V.}\int_{-\infty}^{\infty} {\expo{\ic\omega x} \over \verts{\omega}}\,\dd\omega}
\\[2mm] \stackrel{\mrm{DEF.}}{=}\,\,\,&
\lim_{\epsilon \to 0^{+}}\,\,\pars{%
\int_{-\infty}^{-\epsilon}{\expo{\ic\omega x} \over
-\omega}\,\dd\omega +
\int_{\epsilon}^{\infty}{\expo{\ic\omega x} \over
\omega}\,\dd\omega}
\\[5mm] = &\
\lim_{\epsilon \to 0^{+}}\,\,\pars{%
\int_{\epsilon}^{\infty}{\expo{-\ic\omega x} \over
\omega}\,\dd\omega +
\int_{\epsilon}^{\infty}{\expo{\ic\omega x} \over
\omega}\,\dd\omega}
\\[5mm] = &\
\lim_{\epsilon \to 0^{+}}\,\,\bracks{2%
\int_{\epsilon}^{\infty}{\cos\pars{\omega x} \over
\omega}\,\dd\omega}
\\[5mm] = &\
2\int_{0}^{\infty}{\cos\pars{\omega\verts{x}} \over
\omega}\,\dd\omega =
2\int_{0}^{\infty}{\cos\pars{\omega} \over
\omega}\,\dd\omega
\\[2mm] &\
\mbox{which}\ diverges\ \mbox{logarithmically.}
\end{align}


问:
Computing the Integral of a Complicated Matrix Exponential
For each $i = 1,2,\cdots,m$,
let $A_i \in \mathbb{R}^{m\times m}$ be full rank,
and $v_i \in \mathbb{R}^m$.
Let us use the notation $[v_i]_{i=1}^m$ for the $m\times m$ matrix
created by stacking $v_i$ as column vectors.

Is it possible to compute the following integral in closed form?

$$ \int_0^t
  \exp\left(
  \left[ [A_i]^{-1}
   ( \exp( -A_i s ) - I_m )
   v_i
  \right]_{i=1}^m
  \right)
  \, ds \,.
$$

If it helps, we know that the matrix $[v_i]_{i=1}^m$ is symmetric positive semidefinite,
and the tensor formed by $[A_i]_{i=1}^m = [a_{ijk}]_{i,j,k=1}^m$ is symmetric in the $i,j$ directions.

答:
This integral likely does not have a closed form solution. In the scalar case (i.e. $m=1$), denoting $A_i = a$ and $v_i = v$, the integral can be computed (via Wolfram Alpha) as

$$ \int_0^t \exp( a^{-1}(e^{-at}-1)v ) = \frac{e^{-v/a}}{a} ( \text{Ei}(v/a) - \text{Ei}(e^{-at}v/a) ) \,,
$$

where $\text{Ei}(x)=\int_{-\infty}^x t^{-1}e^{t} dt$ is the exponential integral, which is not an elementary function.

Consequently, it's unlikely the higher dimension cases ($m > 1$) will admit a closed form solution.


问:
Let R be the relation defined on the set of bijective functions from the set (1 to n) to the set (1 to n) by $fRg$ whenever $f = g^{-1}$
I am not too sure as to what the relation is but I think $R = \{(1, 2), (2, 3), (3, 4), ..., (n - 1, n)\} $.

Any guidance would be appreciated.

答:
To show that $R$ is reflexive you would need to show that $f R f$ for every $f.$ Is it the case that $f=f^{-1}$ for every $f?$

To show that $R$ is symmetric you would need to show that $f R g$ implies $g R f$ for every $f, g.$ Is it the case that $f=g^{-1}$ implies $g=f^{-1}$ for every $f, g?$

To show that $R$ is antisymmetric you would need to show that $f R g$ implies it is not the case that $g R f$ for every $f, g.$ Is it the case that $f=g^{-1}$ implies $g\neq f^{-1}$ for every $f, g?$

To show that $R$ is transitive you would need to show that $f R g$ and $g R h$ implies $f R h$ for every $f, g, h.$ Is it the case that $f=g^{-1}$ and $g=h^{-1}$ implies $f=h^{-1}$ for every $f, g, h?$


问:
Comparing variances of conditional expectation terms
In this article, equation 2.2 is defined by

$$
\text{Var}(\mathbb{E}[Y|X,Z]) - \text{Var}(\mathbb E[Y|Z])
$$
and is said to be positive, unless $\mathbb{E}[Y|X,Z] = \mathbb{E}[Y|Z]$ almost surely. The only assumption here is that the expectations exist, there are no assumptions about the random variables $X,Z,Y$.

Is it obvious why this quantity is non-negative? Could someone please explain why, and if possible provide a reference to some readings to help better understand what is going on here?

答:
Write $\theta=E(Y|X, Z)$, so $E(Y|Z)=E(\theta|Z)$. Since $E(\theta)=E(E(Y|Z))=E(Y)$, we see
$$
\text{Var}(\theta)=E(\theta^2)-E(Y)^2, \ \ \ \
\text{Var}(E(Y|Z))=E(E(Y|Z)^2)-E(Y)^2.
$$
By conditional Jensen's inequality,
$$
\Big(E(\theta|Z)\Big)^2\leq E(\theta^2|Z),
$$
Integrate this we see
$$
E(E(Y|Z)^2)\leq E(\theta^2).
$$
That proves the inequality.


问:
Comparison test for series $\sum_{n=1}^{\infty}\frac{n}{n^3 - 2n + 1}$
I am trying to prove the convergence of the series $\sum_{n=1}^{\infty}\frac{n}{n^3 -2n +1}$ with the simple comparison test. I know it can be done with other tests but this question came up in my homework for the comparison test before the other tests were discussed so I'm being stubborn and want to use that test.

I know it converges but I haven't been able to find a rigorous justification. This is my reasoning so far.

$\sum_{n=1}^{\infty}\frac{n}{n^3 -2n +1}$ converges because $\sum_{n=1}^{\infty}\frac{n}{n^3}$ = $\sum_{n=1}^{\infty}\frac{1}{n^2}$ and I know the latter converges by the p-series criterion.

To use the comparison test as my justification I must show $\frac{n}{n^3−2n+1} \leq \frac{n}{n^3} \ \ \forall \  n$. This would be obvious if the denominator on the left were greater than the denominator on the right but that can't be true as on the left I am subtracting from $n^3$ and on the right $n^3$ is unchanged.

So I switch to proving $\frac{n}{n^3−2n+1} \leq \frac{1}{n^2}$ and then I'm stuck.

If the numerators were both 1 then $n^3−2n+1$ is eventually going to be larger than $n^2$ even if we're subtracting a little bit from $n^3$.

That is, at n=1 we'll get $1^3−2+1=0 \leq 1$ but thereafter $n^3− 2n +1  \geq n^2$ is always true.

But the statement of the comparison test says $a_n \leq b_n \forall n$ where $\sum b_n$ is known to converge.

So that's where I am so far. The problem comes down to showing the necessary inequality holds and is it enough the it holds "eventually".

Any ideas?

答:
$\text{First, you should have your summation from $n=2$, since the denominator is $0$ for $n=1$.}$ $\text{Once you have this, note that}$
$$a_n = \dfrac{n}{n^3-2n+1} = \dfrac{n-1+1}{(n-1)(n^2+n-1)} = \dfrac1{n^2+n-1} + \dfrac1{(n-1)(n^2+n-1)}$$
$\text{Now note that for $n \geq 2$, we have}$
$n^2+n-1 > n^2$ and $(n-1)(n^2+n-1) > n^2$. $\text{Hence,}$
$$a_n < \dfrac2{n^2}$$
$\text{Now use the fact that }$$$\sum_{n=2}^{\infty} \dfrac1{n^p}$$ $\text{converges for $p>1$.}$</p>


问:
A question on conditional probability (Cambridge Admissions Exam)
I was solving the following problem

and I got to part (b). I thought of approaching this the following way:

There are two types of cartons, one with a 60 per cent probability of being selected and the other with 40 per cent. I split it intwo scenarios and I calculated each of the conditional probabilities and then added them together (alongside with the probability of each carton being selected). Here is what I have done:

First consider the case where he carton is of skimmed milk.

$$P(X>500 | X<505)= \frac{P(500 < X <505)}{P(X<505)}=\frac{b-\frac{1}{2}}{b}=\frac{2b-1}{2b}$$

Case for full-fat milk

$$P(X>500 | X<505)= \frac{P(500 < X <505)}{P(X<505)}=\frac{b-a}{a}$$

Thus the total probability must be

$$=\frac{6}{10} \frac{2b-1}{2b} + \frac{4}{10} \frac{b-a}{a} $$

However, this is wrong according to the markscheme given here:

My question: Why is it wrong to consider the different cases separately for part (i)?

答:
The error in your argument is that you are not considering the fact that "60% of the cartons it sells contain skimmed milk, and the rest contain full-fat milk".

With this fact we consider
$$X=\mbox{"Volume of skimmed milk in a carton",} \qquad \mbox{with }\mu_{X}=500,\:\:\sigma_{X}=10.$$
$$Y=\mbox{"Volume of full-fat milk in a carton",} \qquad \mbox{with }\mu_{Y}=495,\:\:\sigma_{Y}=10.$$
$$R\sim\mathrm{Bernoulli}(p=0.4), \quad \mbox{that is, }\mathbb{P}(R=-1)=0.6,\:\:\mathbb{P}(R=1)=0.4.$$
where both $X$ and $Y$ satisfy the relationships for $a$ and $b$ established at the beginning of the exercise and $R$ is independent of $X$ and $Y$.

Therefore we define
$$Z=\mathbf{1}_{\{R=-1\}}X+\mathbf{1}_{\{R=1\}}Y.$$
Note that $Z$ is the volume of milk in a carton. It is clear that $Z$ is a continuous random variable. So, it all comes down to calculating the following probability

\begin{align}
\mathbb{P}(Z>500|Z<505) &= \frac{\mathbb{P}(500<Z<505)}{\mathbb{P}(Z<500)}\\
&= \frac{\mathbb{P}(R=-1)\mathbb{P}(500<X<505)+\mathbb{P}(R=1)\mathbb{P}(500<Y<505)}{\mathbb{P}(R=-1)\mathbb{P}(X<505)+\mathbb{P}(R=1)\mathbb{P}(Y<505)} \\
&= \frac{0.6\mathbb{P}(\mu_{X}<X<\mu_{X}+\frac{1}{2}\sigma_{X})+0.4\mathbb{P}(\mu_{Y}+\frac{1}{2}\sigma_{Y}<Y<\mu_{Y}+\sigma_{Y})}{0.6\mathbb{P}(X<\mu_{X}+\frac{1}{2}\sigma_{X})+0.4\mathbb{P}(Y<\mu_{Y}+\sigma_{Y})} \\
&= \frac{0.6(\mathbb{P}(X<\mu_{X}+\frac{1}{2}\sigma_{X})-\mathbb{P}(\mu_{X}\leq X))+0.4(\mathbb{P}(Y<\mu_{Y}+\sigma_{Y})-\mathbb{P}(\mu_{Y}+\frac{1}{2}\sigma_{Y}\leq Y))}{0.6b+0.4b} \\
&=\frac{0.6(b-\frac{1}{2})+0.4(a-b)}{b}\\
&=\frac{4a+2b-3}{10b}.
\end{align}

For part (ii) of (b) what they are giving you are the following two equations.
\begin{align}
0.7&=\mathbb{P}(Z\geq 505)\\
\frac{1}{3}&=\mathbb{P}(R=1|Z\geq 495)
\end{align}
From these two equations, you must solve for $a$ and $b$, this is easy using the law of total probability in the same way as used in the previous calculation.

Hint: The solution of (ii) of (b) is $a=\frac{3}{4}$ and $b=\frac{2}{3}$.


问:
Is there a way to plot separate functions in 3D using Wolfram Alpha?
Is there a way to plot separate functions in 3D using Wolfram Alpha? Am I doing something wrong?

Similar problem with GeoGebra

答: You can plot multiple functions in GeoGebra. It looks like it doesn't handle inequalities well in 3D; try writing them as strict equalities.


问:
probability password problem (6 characters) 4 digits 2 letters
A password is made from 6 characters , 4 of them are digits (0-9) and 2 are letters a-z (26 letters only lower case) each digit and letter can appear once at most.

Someone choose a random password he remembers the letters and digits that make his password however he does not remember their order , what is the probability of getting it right in a single try?

He remembers that the letters were used as the first 2 characters but does not remember the letters and digits what is the probability of getting it right in a single try?

what is the probability of the password having the letter a and at least one of the digits 1,2?

I will start with "2" because I might have got it right ( not sure if it is right or just a coincidence ) if the first 2 characters are letters then we have $26*25*10*9*8*7$ and since we want him to get it right from a single try we will do $\frac 1 {26*25*10*9*8*7}$ $=$ $3.05*10^{-7}$ which is right according to the book but I am not sure if it is a coincidence.

for "1" since he knows the digits and letters he used we only need to sort them so my thought would be $6!$ and divide by $26*25*10*9*8*7$ but it is not right so I tried also sorting the digits first $4!$ and then the letters so we get $4!*2!$ and again divide by  $26*25*10*9*8*7$ but again it is not right , then tried picking the digits first ${6 \choose 4}$ and multiply by the 2 characters left ${2 \choose 2}$ and divide by $26*25*10*9*8*7$ which is also not right.. the answer should be $1.4*10^{-3}$

And for "3" I thought about using Complement so my try was $1-$ $\frac {{6 \choose 4}8*2} {26*25*10*9*8*7}$ but I am not even close as the answer should be $0.051$

Thank you for any help , tips and hints all is appreciated!

答:
Your answer and reasoning for problem 2 are both correct.

A password is composed of six characters.  Four of them are digits (0 - 9) and two are letters (a - z).  Each digit and letter can appear at most once.  He remembers the letters and digits that make his password.  However, he does not remember their order. What is the probability of getting the password right in a single try?

Since he knows the letters and digits used in the password, the only thing he needs to guess is the order in which they appear.  There are $6!$ possible orders in which six distinct characters could appear, of which only one is correct.  Hence, the probability that he guesses correctly the first time is
$$\frac{1}{6!} = \frac{1}{720}$$

A password is composed of six characters.  Four of them are digits (0 - 9) and two are letters (a - z).  Each digit and letter can appear at most once.What is the probability of the password having the letter a and at least one of the digits 1, 2?

There are $\binom{26}{2}$ ways to select two distinct letters, $\binom{10}{4}$ ways to select four distinct digits, and $6!$ ways to arrange six distinct characters, so there are
$$\binom{26}{2}\binom{10}{4}6!$$
possible passwords.

We will count the favorable cases using the Inclusion-Exclusion Principle.

The number of passwords which contain the letter a and the digit 1 is
$$\binom{1}{1}\binom{25}{1}\binom{1}{1}\binom{9}{3}6! = \binom{25}{1}\binom{9}{3}6!$$
since we must choose the letter a, one of the other $25$ letters, the digit 1, and three of the other nine digits, then arrange the six distinct characters.

By symmetry, the number of passwords which contain the letter a and the digit 2 is also
$$\binom{25}{1}\binom{9}{3}6!$$

However, if we add these two cases, we will have counted those cases in which the password contains the letter a and both the digits 1, 2 twice, once when we counted passwords containing the letter a and the digit 1 and once when we counted passwords containing the letter a and the digit 2.  We only want to count them once, so we must subtract them from the total.

The number of passwords which contain the letter a and both the digits 1 and 2 is
$$\binom{1}{1}\binom{25}{1}\binom{2}{2}\binom{8}{2}6! = \binom{25}{1}\binom{8}{2}6!$$
since we must select the letter a, one of the other $25$ letters, both the digits 1 and 2, and two of the other eight digits, then arrange the six distinct characters.

By the Inclusion-Exclusion Principle, the number of favorable passwords is
$$\binom{25}{1}\binom{9}{3}6! + \binom{25}{1}\binom{9}{3}6! - \binom{25}{1}\binom{8}{2}6!$$
so the probability that an admissible password will contain the letter a and at least one of the digits 1, 2 is
$$\frac{2\dbinom{25}{1}\dbinom{9}{3}6! - \dbinom{25}{1}\dbinom{8}{2}6!}{\dbinom{26}{2}\dbinom{10}{4}6!} = \frac{2\dbinom{25}{1}\dbinom{9}{3} - \dbinom{25}{1}\dbinom{8}{2}}{\dbinom{26}{2}\dbinom{10}{4}}$$
Note that the answer depends only on which characters appear in the password, not the order in which they appear.


问:
Find factor of $x^{11} + x + 1$ of degree 2 in $\mathbb{Z}[x]$

Find factor of $x^{11} + x + 1$ of degree 2 in $\mathbb{Z}[x]$

To be honest, it is the first time I see problem like this. I was trying to find divisor by just taking different polinomials of degree 2 and trying to divide this one by them. That was disgusting, I was mixing up in calculations and couldn't finally get answer.

So, Is there any better way to find such a polinomial (divisor of this one with degree 2)? It would be awesome if proposed way would not require super advanced knowledge. Thanks in advance!

答:
Now that I know the answer, thanks to @Will Jagy, here is my solution:

$$x^{11} + x^2 + 1 = (x^{11} - x^2) + (x^2 + x + 1) = \\
= x^2(x^9-1) + (x^2 + x + 1)$$

Note that the first term contains the factor $x^9-1$ which is divisible by $x^3-1= (x-1)(x^2+x+1)$.

Therefore, our polynomial is also divisible by $x^2 + x + 1$.

Generalization: If $a_1$,$\ldots$, $a_n$ are natural numbers that give different remainders when divided by $n$ then the polynomial
$$x^{a_1} + \cdots + x^{a_n}$$
is divisible by $x^{n-1} + \cdots + x + 1$.

Even more,  a polynomial $ \sum c_k x^k$ is divisible by $x^{n-1} + \cdots + x + 1$ if and only if we have

$$s_0 = s_1 = \cdots = s_{n-1}$$

where $$s_r = \sum_{k \equiv r \mod n} c_k$$


问:
Understanding the divergence of an asymptotic series
In class, we recently developed an asymptotic expansion for a perturbed Gaussian integral and I was wondering why it was divergent.

First, the $e^{-gx^4}$ term was Taylor expanded, then integral and summation swapped, and finally the formula of the $n$-th momentum of a Gaussian used to obtain:
$$
\int_{-\infty}^{+\infty} e^{-x^2-gx^4} dx \rightarrow \sum_{n=0}^\infty \frac{(-g)^n}{n!}
\int_{-\infty}^{+\infty} e^{-x^2} x^{4n}=\sqrt{\pi}\sum_{n=0}^{\infty}\frac{(-g)^n\cdot(4n-1)!!}{n!\cdot2^{2n}}
$$
By the Ratio Test this diverges for all $g\neq0$ because
$\frac{a_{n+1}}{a_n}=\frac{(-g)\cdot(4n+3)(4n-1)}{(n+1)\cdot2^2}\rightarrow\infty$.

So far I had only came across a bunch of asymptotic expansions and developed the (probably unjustified) intuition that the divergence of the asymptotics came from expanding a function beyond its radius of convergence:
$$
\int_0^\infty \frac{e^{-x}}{1+x}dx=\sum_{n=0}^\infty(-1)^n\int_0^\infty e^{-x}x^ndx=\textrm{divergent}
$$

because the expansion $\frac{1}{1+x}=\sum (-x)^n$ is not justified for the integration domain $\int_1^\infty$.

However, in the cause of the perturbed Gaussian, the Taylor expansion of $e^{-gx^4}$ is well justified for $\forall x$ and for $g\ge0$ the integral converges. I was expecting the series to converge for $g\ge 0$.

Even though I am aware that limits cannot arbitrarily be interchanged ($\int\sum^\infty \neq \sum^\infty\int$), I was wondering whether the divergence of the Gaussian series would admit some "intuitive" explanation appart from that.

答:
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
$\ds{\bbox[5px,#ffd]{}}$

\begin{align}
&\bbox[5px,#ffd]{\int_{-\infty}^{\infty}
\expo{-x^{2}\ -\ gx^{4}}\,\,\dd x}
\,\,\,\stackrel{\mrm{as}\ g\ \to \infty}{\sim}\,\,\,
2\int_{0}^{\infty}
\expo{-\ gx^{4}}\pars{1 - x^{2}}\,\,\dd x
\\[5mm] = &\
2\Gamma\pars{5 \over 4}{1 \over g^{1/4}} -
{1 \over 2}\,\Gamma\pars{3 \over 4}{1 \over g^{3/4}}
\end{align}


问:
Let $T$ be a model complete theory and let $\mathfrak{M}$ be a model of $T$ which embeds into every model of $T$. Show that $T$ is complete.
I need to prove the statement but am not sure about my proof. Here it is:

First we recall some definitions and basic results that we will use to prove the statement. At theory is model complete if for all models $\mathfrak{M}^{1}$ and $\mathfrak{M}^{2}$ of $T$, if $\mathfrak{M}^{1} \subset \mathfrak{M}^{2}$ then $\mathfrak{M}^{1} \prec \mathfrak{M}^{2}$. A theory $T$ is complete if and only if all models of $T$ are elementarily equivalent. Whenever $\mathfrak{M}^{1}$, $\mathfrak{M}^{2}$ are models of $T$ with a common $\mathcal{L}$-substructure $\mathfrak{A}$, then Th($\mathfrak{M}_{A}^{1}$)=Th($\mathfrak{M}_{A}^{2}$). We can now say that $\mathfrak{M}$ is a common elementary substructure since it embeds into every other model of $T$. Now since all those models have a common $\mathcal{L}$-substructure, the equivalence of each theory of each model follows. Hence $T$ is complete.

Could anyone validate it or give me a hint on how to tackle it?

答:
Whenever $\mathfrak{M}^{1}$, $\mathfrak{M}^{2}$ are models of $T$ with a common $\mathcal{L}$-substructure $\mathfrak{A}$, then $\text{Th}(\mathfrak{M}_{A}^{1})=\text{Th}(\mathfrak{M}_{A}^{2})$.

This is false in general! It's equivalent to saying that $T$ has quantifier elimination, and there are model complete theories which do not have quantifier elimination.

Now since all those models have a common $\mathcal{L}$-substructure, the equivalence of each theory of each model follows.

Instead of using the fact that $\mathfrak{M}^1$ and $\mathfrak{M}^2$ have a common $\mathcal{L}$-substructure, use the (much stronger) fact that they have a common elementary substructure.

Suppose a sentence $\varphi$ is true in $\mathfrak{M}^1$. Then it's true in $\mathfrak{M}$. Then it's true in $\mathfrak{M}^2$. Do you see why? And do you see why this implies $T$ is complete?


问:
Showing $\alpha(\beta+1)\leq\frac{5}{3}\alpha^2+\frac{1}{3}\beta^2$ for nonnegative integers $\alpha$ and $\beta$.
I found this lemma in a paper I was reading and it was not proved. There doesn't seem to be any obvious factorized form so how would one go about proving the inequality holds?

$$\alpha(\beta+1)\leq\frac{5}{3}\alpha^2+\frac{1}{3}\beta^2$$

$$\frac{5}{3}\alpha^2+\frac{1}{3}\beta^2-\alpha(\beta+1) \geq 0$$

$$5\alpha^2+\beta^2-3\alpha\beta-3\alpha \geq 0$$

答:
Your required inequality can be adjusted as follows:

$$\begin{equation}\begin{aligned}
0 & \le 5\alpha^2 + \beta^2 - 3\alpha\beta - 3\alpha \\
& = 4\alpha^2 - 4\alpha\beta + \beta^2 + \alpha^2 + \alpha\beta - 3\alpha \\
& = (2\alpha - \beta)^2 + \alpha(\alpha + \beta - 3)
\end{aligned}\end{equation}\tag{1}\label{eq1A}$$

Since $(2\alpha - \beta)^2 \ge 0$ and $\alpha \ge 0$, then \eqref{eq1A} can only possibly not be true if

$$\alpha + \beta - 3 \lt 0 \implies \alpha + \beta \lt 3 \tag{2}\label{eq2A}$$

If $\alpha = 0$, then \eqref{eq1A} is always true. If $\alpha = 1$, then \eqref{eq2A} means either $\beta = 0$, which gives the RHS of \eqref{eq1A} to be $2$, or $\beta = 1$, which gives the RHS of \eqref{eq1A} to be $0$. Next, for $\alpha = 2$, \eqref{eq2A} shows only $\beta = 0$ is possible, with this giving the RHS of \eqref{eq1A} to be $2$. Finally, for $\alpha \ge 3$, \eqref{eq2A} shows there's no non-negative value for $\beta$ possible.

This confirms \eqref{eq1A} is always true for all non-negative integer values of $\alpha$ and $\beta$.


问:
Integrating $\int^{\infty}_0 e^{-x^2}\,dx$ using Feynman's parametrization trick
I stumbled upon this short article on last weekend, it introduces an integral trick that exploits differentiation under the integral sign. On its last page, the author, Mr. Anonymous, left several exercises without any hints, one of them is to evaluate the Gaussian integral
$$
\int^\infty_0 e^{-x^2} \,dx= \frac{\sqrt{\pi}}{2}
$$
using this parametrization trick. I had been evaluating it through trial and error using different paramatrizations, but no luck so far.

Here are what I have tried so far:

A first instinct would be do something like:$$
I(b) = \int^\infty_0 e^{-f(b)x^2}\,dx
$$
for some permissible function $f(\cdot)$, differentiating it will lead to a simple solvable ode:
$$
\frac{I'(b)}{I(b)} = -\frac{f'(b)}{2f(b)}
$$
which gives:
$$
I(b) = \frac{C}{\sqrt{f(b)}}.
$$
However, finding this constant $C$ basically is equivalent to evaluating the original integral, we are stuck here without leaving this parametrization trick framework.

A second try involves an exercise on the same page:
$$
I(b) = \int^\infty_0 e^{-\frac{b^2}{x^2}-x^2}dx.
$$
Taking derivative and rescaling the integral using change of variable we have:
$$
I'(b) = -2I(b).
$$
This gives us another impossible to solve constant $C$ in:
$$
I(b) = C e^{-2b}
$$
without leaving this framework yet again.

The third try is trying modify Américo Tavares's answer in this MSE question:
$$
I(b) = \int^\infty_0 be^{-b^2x^2}\,dx.
$$
It is easy to show that:
$$
I'(b) = \int^\infty_0 e^{-b^2x^2}\,dx - \int^\infty_0 2b^2 x^2 e^{-b^2x^2}\,dx = 0
$$
by an integration by parts identity:
$$
\int^\infty_0 x^2 e^{- c x^2}\,dx = \frac{1}{2c}\int^\infty_0 e^{- c x^2}\,dx .
$$
Then $I(b) = C$, ouch, stuck again at this constant.

Notice in that Proving $\displaystyle\int_{0}^{\infty} e^{-x^2} dx = \frac{\sqrt \pi}{2}$ question, Bryan Yocks's answer is somewhat similar to the idea of parametrization, however he has to introduce another parametric integration to produce a definite integral leading to $\arctan$.

Is there such a one shot parametrization trick solution like the author Anonymous claimed to be "creative parameterizations and a dose of differentiation under the integral"?

答:
Just basically independently reinvented Bryan Yock's solution as a more 'pure' version of Feynman.

Let $$I(b) = \int_0^\infty \frac {e^{-x^2}}{1+(x/b)^2} \mathrm d x = \int_0^\infty \frac{e^{-b^2y^2}}{1+y^2} b\,\mathrm dy$$ so that $I(0)=0$, $I'(0)= \pi/2$ and $I(\infty)$ is the thing we want to evaluate.

Now note that rather than differentiating directly, it's convenient to multiply by some stuff first to save ourselves some trouble. Specifically, note

$$\left(\frac 1 b e^{-b^2}I\right)' = -2b \int_0^\infty e^{-b^2(1+y^2)} \mathrm d y = -2 e^{-b^2} I(\infty)$$

Then usually at this point we would solve the differential equation for all $b$, and use the known information at the origin to infer the information at infinity. Not so easy here because the indefinite integral of $e^{-x^2}$ isn't known. But we don't actually need the solution in between; we only need to relate information at the origin and infinity. Therefore, we can connect these points by simply integrating the equation definitely; applying $\int_0^\infty \mathrm d b$ we obtain

$$-I'(0)= -2 I(\infty)^2 \quad \implies \quad I(\infty) = \frac{\sqrt \pi} 2$$


问:
How to quickly select some vectors from a given set to make the sum of them has the smallest $\ell_2$ norm?
The question is how to design an effective algorithm to solve this problem:

$$
\underset{k_1,k_2,\cdots ,k_s}{\min}\lVert \sum_{i=1}^s{\mathbf{x}_{k_i}} \rVert _2,
$$

where $l (l >> s)$ is the set size, $n$ is the vector size, $ k_i $s are different numbers in $\left\{ 1,2, \cdots, l \right\}$. The set $\left\{ \mathbf{x}_i \right\} _{i=1}^{l}$ is known.

答: This is an $n$-dimensional version of the (NP-complete) subset sum problem, so you are unlikely to find an efficient algorithm.


问:
Is it possible to solve this recurrence equation?
I wish to solve $y_{n+2}-y_{n+1}-2y_{n}=n2^n$ by first reducing the problem to first order equations with the help of the characteristic equation, which yields the solutions: $r=-1,2$. I have learned to solve first orders with the following formula: $\forall n \geq 0: y_{n+1}=r y_{n}+q_{n} \implies y_{n+1}= r^{n+1}y_{0}+\sum_{k=0}^{n} r^{n-k} q_{k}$

So:

$y_{n+2}-y_{n+1}-2y_{n}=(y_{n+2}+y_{n+1})-2(y_{n+1}+y_{n})$, with help of substitution $z_{n}=y_{n+1}+y_{n}: z_{n+1}-2z_n=n2^n \implies z_{n+1}=2^{n+1}z_0+\sum_{k=0}^{n}2^{n-k}k2^k=2^{n+1}z_0+\frac{1}{2}2^n n(n+1) \implies z_{n}=2^nz_0+\frac{1}{2}2^{n-1}n(n-1)$

Subsequently solve in the same way $y_{n+1}+y_n=z_n \implies y_{n}=(-1)^n y_0 +\sum_{d=0}^{n} (-1)^{n-d} (2^dz_0+\frac{1}{2}2^{d-1}d(d-1))$

As you can see it gets quite messy at the end, making me think that I have made a few errors along the way or used this method the wrong way.

答:
As you can see, doing $y_{n+1}= r^{n+1}y_{0}+\sum_{k=0}^{n} r^{n-k} q_{k}$ once is probably OK but doing it one more time will usually lead to messiness.

In addition to Gerry Myerson's way, another method for this problem is to make it homogeneous:
$$
y_{n+2} - y_{n+1} -2 y_n = n2^n \tag1
$$

$$ y_{n+1} - y_{n} -2 y_{n-1} = (n-1) 2^{n-1} \tag2
$$

$$(1)-2 \times (2) \implies y_{n+2}-3y_{n+1} + 4y_{n-1}=2^n \tag3 $$
$$y_{n+1}-3y_n + 4y_{n-2}=2^{n-1} \tag4 $$

$$(3)-2\times(4) \implies y_{n+2} - 5y_{n+1} + 6y_n+4y_{n-1}-8y_{n-2} = 0 \tag 5$$

At first glance (5) may look scary, but you already know the homogeneous part of (1) has two roots -1 and 2; and by doing $(1)-2\times(2)$ and $(3)-2\times(4)$ you know there are 2 more duplicate roots of 2. Indeed it's easy to verify that
$$
y^4-5y^3+6y^2+4y-8=(y+1)(y-2)^3.
$$

If you use forward shift operator it's even more obvious:

$$
(\mathbb{E}+1)(\mathbb{E}-2)y_n=n2^n \\
\implies (\mathbb{E}+1)(\mathbb{E}-2)^2 y_{n-1} = n2^n - 2(n-1)2^{n-1}=2^n\\
\implies (\mathbb{E}+1)(\mathbb{E}-2)^3 y_{n-2} = 2^n-2\cdot 2^{n-1} = 0.
$$

For other examples on use of the forward shift operator, see Finding explicit formula for recursive relation


问:
How do I prove that the following sequence is not a Cauchy sequence?
$x(1)=1$ and $x(n)=x(n-1)+(\frac1n)$ for $n>1$

So far, I tried looking at $|x(n+1)-x(n)|$ to get $|x(n)+(\frac1n)-x(n)|$.

I also know that $\frac1n$ diverges.  I am stuck on what to do next.

答:
You have the two things. Just put them together. $|x_n - x_m| = |\sum_{i=n}^m 1/i |$ (why is this true?). Thus, it is not true that for all $n,m > N$, that quantity can be made small. Fix $n$, make $m$ as large as is necessary to exceed $\epsilon$.

I leave it to you to write down a detailed proof that would meet the standards of the question.


问:
Is it true that $\mathscr{O}_{X,x}=\mathscr{O}_{\operatorname{Spec} A,x}$ for any $\operatorname{Spec} A$ with $x\in\operatorname{Spec} A$?
Let $X$ be a scheme and $\operatorname{Spec} A\subset X$ be an affine with $x\in\operatorname{Spec} A$,

Is it true that $\mathscr{O}_{X,x}=\mathscr{O}_{\operatorname{Spec} A,x}$?

I ask this question because I feel that many "obvious" remarks made in Gortz-Wedhorn and Vakil FOAG seem to rely on this or a similar fact: such as in Vakil

"We say a ringed space is a locally ringed space if its stalks
are local rings. Thus Exercise 4.3.F shows that schemes are locally ringed spaces." (Where 4.3.F is Show that the stalk of $\mathscr{O}_{\operatorname{Spec} A}$ at the point $[\mathfrak{p}]$ is the local ring $A_{\mathfrak{p}}$).

Is this the case? If not, why is the above statement immediate from 4.3.F?

答:
Yes, as long as $\operatorname{Spec} A$ is an open affine set in $X$.

More generally if $U$ is an open subset of a ringed space $(X,\mathcal O_X)$, and $\mathcal O_U$ is the sheaf of rings on $U$ defined by $\mathcal O_U(V) = \mathcal O_X(V)$ for all open sets $V \subset U \subset X$, then $\mathcal O_{X,x} = \mathcal O_{U,x}$ for all $x \in U$.  This is because

$$\mathcal O_{X,x} = \varinjlim_V \mathcal O_X(V)$$
as $V \subset X$ runs through the open neighborhoods of $x$ ordered by reverse inclusion, and
$$\mathcal O_{U,x} = \varinjlim_V \mathcal O_U(V) = \varinjlim_V \mathcal O_X(V)$$

as $V \subset U$ runs through the open neighborhoods of $x$.  Although the direct system of rings defining $\mathcal O_{U,x}$ is a subsystem of the one defining $\mathcal O_{X,x}$, it is easy to see that they define the same direct limit.  It is the same principle by which the infinite polynomial ring $\mathbb C[x_1, x_2, x_3, ...]$ is the direct limit of the direct system

$$\mathbb C[x_1] \subset \mathbb C[x_1, x_2] \subset \mathbb C[x_1, x_2,x_3] \subset \cdots$$

as well as the subsystem

$$\mathbb C[x_1] \subset \mathbb C[x_1,x_2,x_3] \subset \mathbb C[x_1,x_2,x_3,x_4,x_5] \subset \cdots.$$