问:
Set of first category in $\mathbb{R}$ such that the intersection with an open set has positive Lebesgue measure

There exists and $F_\sigma$-set $B\subset \mathbb{R}$ that is of first category in $\mathbb{R}$ which has the further property that $\lambda(U\cap B)>0$ and $\lambda(U\cap B')>0$ for every nonvoid open subset $U$ of $\mathbb{R}$. ($\lambda$ is the Lebesgue measure on $\mathbb{R}$.)

I have some questions on the proof of this theorem, specifically in the construction to find such a set. One starts by enumerating all the nonvoid open intervals $\lbrace I_n \rbrace_{n=1}^\infty$ having rational endpoints. First we choose $s_1=\lambda(I_1)$ and $B_0=\emptyset$. Inductively, when a compact nowhere dense set $B_{n-1}$ and positive numbers $s_1,\dots,s_n$ have been selected ($n$ is a natural number), we choose a compact nowhere dense set $A_n\subseteq I_n\cap B'_{n-1}$ with $0<\lambda(A_n)<2^{-n}\min \lbrace s_1,\dots,s_n\rbrace$. Write $B_n = B_{n-1}\cup A_n$ and $s_{n+1}=\lambda(I_{n+1})-\lambda(I_{n+1} \cap B_n )$. (proof continues)

So far I think I'm missing the big picture. Even if $\lambda(A_n)<2^{-n}({\rm etc})$, it seemes to me that the $A_n$ can have a rather wild measure. I guess they are chosen such that the measure goes to zero, but I cannot see this. Connected to this, I don't really see the motivation to choose $s_{n+1}$ they way it is, it might be $I_{n+1}\cap B_n=\emptyset$ but $\lambda(I_{n+1})$ can still be quite big.

答: There are cantor sets (nowere dense sets) that have positive measure (i can provide an example if you want), so take that set $C$ in $[0,1]$ such that $\lambda(C)=a>0$, also take $C/n$ (where $A/n=\lbrace a/n:a\in A\rbrace$) that has measure $a/n$, now define the set $C_n=\bigcup_{i=0}^{n-1}C/n+i/n$ this is a set contained in $[0,1]$, now take the set $K_0=\bigcup C_n\subset [0,1]$, since $K$ is a countable union of nowhere dense sets it is of first category, and $K_n=K_0+n$, $K=\bigcup_{n\in\mathbb Z} K_n$, since countable union of first category sets are of first category you have that $K$ is of first category. Now take an interval $(a,b)$, then there exists a rational $p/q\in [0,1)$, and $n\in\mathbb Z$ such that $\left[n+\frac{p}{q},n+\frac{p+1}{q}\right]\subset (a,b)$, then $C_q+p/q+n$ is contained in $(a,b)$, since $C_q+p/q+n\subset K$ and $\lambda(C_q+p/q+n)=a/q>0$ you have your proof.


问:
An equality involving a moving surface $\Sigma (t)$
Given a surface $\Sigma(t)$ that moves in space over time, a vector field $ \boldsymbol{B}=\boldsymbol{B}\left( x,y,z,t \right) $, and $\mathrm{div} \boldsymbol{B}=0$. $ \boldsymbol{V}$ is the velocity field on the surface.

Prove that
$$
\frac{\text{d}}{\text{d}t}\iint_{\Sigma \left( t \right)}{\boldsymbol{B}\cdot \text{d}\boldsymbol{S}}=\iint_{\Sigma \left( t \right)}{\left[ \frac{\partial \boldsymbol{B}}{\partial t}+\left( \nabla \cdot \boldsymbol{V} \right) \boldsymbol{B}+\left( \boldsymbol{V}\cdot \nabla \right) \boldsymbol{B}-\left( \boldsymbol{B}\cdot \nabla \right) \boldsymbol{V} \right] \cdot \text{d}\boldsymbol{S}}
$$

It seems difficult for me to deal with problems involving moving surfaces. Also the formula above is seemed to relate to Stokes' theorem. How to prove the formula? And can we generalize it to manifolds?

答:
I will give a sketch on how to find such a general formula for manifolds, but for those who don't know much about differential geometry and Lie derivatives, this answer will be pretty pointless, and definitely not enlightening at all. Not sure if this is what you are searching for, but here we go:

Let us work in $\mathbb{R}^3$ as ambient space (not that this is needed in the actual math going on, but just to fix the ideas) and let $\omega_t \in \Omega^2(\mathbb{R}^3)$ be a smoothly varying family of two forms. Moreover let $M_t = \phi_t(M)$, where $M$ is a two dimensional manifold and $\phi_t$ is the flux of some velocity vector field $X$. Finally, we denote by $\phi_t^{*} \omega$ the classical pull-back of the form $\omega$ by the diffeomorphism $\phi_t$. Then we have
\begin{align}
\frac{d}{dt}\int_{M_t} \omega_t &= \frac{d}{dt}\int_{M} \phi_{t}^*(\omega_t) = \lim_{h \to 0} \int_{M} \frac{\phi^*_{t+h}(\omega_{t+h}) - \phi^*_{t}(\omega_{t})}{h } \\ &= \lim_{h \to 0} \int_{M} \frac{\phi^*_{t+h}(\omega_{t+h}) - \phi^*_{t}(\omega_{t+h}) + \phi^*_{t}(\omega_{t+h})- \phi^*_{t}(\omega_{t})}{h } \\ &= \lim_{h \to 0} \int_{M} \phi_{t}^* \left( \frac{\phi_h^*(\omega_{t+h})-\omega_{t+h}}{h} \right) + \phi_t^*\left(\frac{\omega_{t+h}-\omega_t}{h} \right) \\ &= \lim_{h \to 0} \int_{M_t}  \left( \frac{\phi_h^*(\omega_{t+h})-\omega_{t+h}}{h} \right) + \left(\frac{\omega_{t+h}-\omega_t}{h} \right) \\ &= \int_{M_t} \mathcal{L}_{X}(\omega_t) + \dot{\omega}_t \, ,
\end{align}
where $\mathcal{L}_{X}(\omega_t)$ is the Lie derivative of $\omega_t$ with respect to the vector field $X$. By Cartan's magic formula then
\begin{align}
\frac{d}{dt}\int_{M_t} \omega_t &= \int_{M_t} \mathcal{L}_{X}(\omega_t) + \dot{\omega}_t = \int_{M_t} i_{X}(d\omega_t) + \int_{M_t} d(i_{X}\omega_t) + \int_{M_t} \dot{\omega}_t \, ,
\end{align}
and this is kind of the most general formula you can hope for. Now note that:

The $\omega_t$ corresponds to the term $\frac{\partial \textbf{B}}{\partial t}$ in your right hand side.

Since $\text{div}(\textbf{B})=0 $ then $d\omega_t=0$ and the first term cancels.

Finally, to be honest as a mathematician I have no clue on what the notation
$$ \left( \nabla \cdot \boldsymbol{V} \right) \boldsymbol{B}+\left( \boldsymbol{V}\cdot \nabla \right) \boldsymbol{B}-\left( \boldsymbol{B}\cdot \nabla \right) \boldsymbol{V} $$
actually is supposed to mean, but I strongly believe that this is what comes out by the term $ \int_{M_t} d(i_{X}\omega_t) \,$.

I hope that this gives at least the idea of where all these magic formulas about the integration of moving things come from. Long story short, once you have defined integration of forms, then integrating on moving surfaces is just the same as pulling back the integrand via the velocity flux, then one just computes the derivative inside the integral, since the domain of integration does not depend on time anymore.

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问:
Global sections of powers of a line bundle which is of the form $L' \otimes L$, where $L'$ is globally generated and $L$ is a torsion element.
Let $Y$ be an irreducible complex projective variety of $\mathrm{dim}(Y) \geq 1$ and $X$ an irreducible complex projective variety. Assume that $M \in \mathrm{Pic}(Y)$ is a torsion element, say of order $e\neq 1$, and let $M'$ be a very ample line bundle on $X$. Consider the line bundle $L' \otimes L$ on $X \times Y$, where $L':=\pi_1^{*}M'$ and $L=\pi_2^{*}M$ and $\pi_1, \pi_2$ are the projections.
What can we say about $H^0(X \times Y ,(L'\otimes L)^m)$)?

We know that $L'$ is globally generated and that $L$ is torsion and non-trivial, as it should be true that $X \times Y \to Y$ is an algebraic fibre space, and so $\mathrm{Pic}(Y)\subset \mathrm{Pic}(X \times Y)$. This means that if $m$ is a multiple of $e$, than $(L' \otimes L)^m \cong (L')^m$ has global sections. It seems to me that for $m$ that is not a multiple of $e$ we shouldn't have global sections. Is this true? If yes how to prove it? At least it should be true that we cannot have global sections of the type $s \otimes t$ (for $m$ that isn't a multiple of $e$), as $L$ is torsion.

Moreover, I think that we can just consider $M'$ to be globally generated.

答:
Use Künneth's Theorem.
$$H^0(X\times Y, M'\boxtimes M)\simeq H^0(X,M')\otimes H^0(Y,M).$$
The $\boxtimes$ notation is the tensor product of pullbacks from each projection.

It works more generally over separated schemes over fields with $M'$ and $M$ quasicoherent sheaves, and I'm sure it is even more general than that.


问:
Joint distribution - question on limits - find $$ so that $(,)=$ will be a joint density, where $0<,1 0<<1, x+y<1$
I know there are some similar questions but I am having trouble setting up the integral for this problem. I can't decide if it is correct or not to have the integral set up $$\int_0^1\int_0^1kxy\,dy\,dx$$ or
$$\int_0^1\int_0^{1-x}kxy\,dy\,dx$$

I am leaning towards the latter because we have that $$x+y<1   \implies   y<1-x$$  but I am not confident. Can somebody please explain which is correct and why. Thank you!

答:
The reason you are correct with your reasoning is that you are integrating over a triangle (plot your bounds). So if you think about the integral written above, we are letting $x$ range from 0 to $1-x$ and y range from 0 to 1, which results in the max at any point being $x+y < 1$. If it helps you visualize, this is equivalent to
$$ \int_0^1 \int_0^{1-y} kxy dx dy,$$
where now we are letting $y$ range from 0 to $1-y$ and x range from 0 to 1.


问:
For 52 cards in a line, what is the probability that the card immediately to the right of the first queen is a queen or a seven?
For a standard deck of 52 cards, lay down all 52 cards in a line. What is the probability that the card immediately to the right of the first
queen is a queen or a seven?

I tried 51!*28/52! but I dont think that I'm accounting for finding the first queen.

51! ways to order the items so that queen is next to another queen or seven (treat as 1 item).

28 ways to rearrange first queen with another queen or seven

52! possible outcomes.

答:
Break into cases:

The first queen is followed by another queen.

First, pick the order that the queens arrive in: $4!$ choices

Next, gluing the first two queens together, pick the three positions occupied by the queens in the $51$ positions for the deck: $\binom{51}{3}$ choices

Finally, from front to back, choose how all other positions are occupied by the remaining cards: $48!$ choices, giving then a total of $4!\cdot \binom{51}{3}\cdot 48!$ in this case

The first queen is followed by a $7$.

First, pick which queen appears first and which $7$ it was: $4^2$ choices.

Gluing the $7$ to that queen, choose the order in which the remaining queens occur: $3!$ choices.

Choose the positions occupied by the queens (remembering the $7$ is glued to the queen in the front): $\binom{51}{4}$ choices.

Finally, choose how the remaining cards are arranged in the remaining positions: $47!$ choices, for a total of $4^2\cdot 3!\cdot \binom{51}{4}\cdot 47!$ choices.

This is out of the $52!$ different options for how the deck could be arranged for a final probability of:

$$\frac{4!\cdot\binom{51}{3}\cdot 48!+4^2\cdot 3!\cdot \binom{51}{4}\cdot 47!}{52!}$$


问:
Discussion questions involving Liouville numbers
I'm working through Measure and Category by Oxtoby with some friends, but since the book doesn't have any exercises, we needed to come up with our own to discuss. My friend found the following problems on the Baire Category theorem (from Rudin and Sally), which were a good difficulty level for us.

Prove that $\mathbb{Q}$ is not a countable intersection of open sets.

Let $ \{ f_n\} $ be a sequence of continuous functions on $\mathbb{R}$ so that for every $x$, $\lim_{n \to \infty}f_n(x)$ exists and is finite. Prove that there is an interval $(a, b)$ of positive length so that the set $\{|f_n(x)| : n \in \mathbb{N}, x \in (a,b)\}$ is bounded above.

Does there exist a sequence of continuous positive functions $f_n$ on $\mathbb{R}$ so that the set $\{f_n(x)\}$ is unbounded if and only if $x$ is rational? What if rational is replaced with irrational?

Prove that if $ \{ f_n\} $ is a sequence of continuous functions from $ \mathbb{R}$ to $ \mathbb{R} $ so that for each $x$, $ f(x) = \lim_{n \to \infty}f_n(x)$ exists and is finite, then for each $\varepsilon > 0 $ there is a nonempty open set $U $ and a large $N$ so that $ |f(x) - f_n(x)| < \varepsilon$ for all $n \geq N$, $x \in U $.

Prove that if $f$ is a continuous function from positive reals to positive reals so that $f(x), f(2x), f(3x), \ldots $ tends to $0$ for all $x$, then $f(t) \to 0$ as $t \to \infty$.

It would be greatly appreciated if someone could provide a reference to some problems of similar difficulty involving Liouville numbers.

答:
Here are a couple of results about Liouville numbers that you can take for exercises, if you like. Both are from Burger and Tubbs, Making Transcendence Transparent.

Theorem 1.6. Every real number can be expressed as a sum of two Liouville numbers.

Corollary 6.24. Let $\alpha$ be a nonzero algebraic number, and $L$ a Liouville number. Then $e^{\alpha}+L$ is transcendental.

The second is considerably harder than the first (but I don't see why $\alpha$ has to be nonzero).

Erdos proved that every nonzero real number can be written as a product of two Liouville numbers. I don't know how hard this is.


问:
Subring $R \subset \Bbb{Q}[X]$. Evaluation hom. $R/(f) \to \Bbb{Z}/(f(a))$ gives me a weird result (not true), but everything looks correct.
Let $R = \text{Int}[X] \subset \Bbb{Q}[X]$ be the subring of integer-valued polynomials.

Fix an $a \in \Bbb{Z}$.

Let $f \in R$ be one such irreducible, monic polynomial.  Then $\psi: R/(f) \to \Bbb{Z}/(f(a)), \ \psi(g + (f)) = g(a) + (f(a))$ is a surjective ring homomorphism.  If $g(X) = h(X) \pmod f$ then $f(X) \mid g(X) - h(X) \implies f(a) \mid g(a) - h(a)$.

$\psi(h + g) = h(a) + g(a) + (f(a)) = \psi(h) + \psi(g)$.  Similarly $\psi(hg) = \psi(h)\psi(g)$.

$\psi$ is surjective since we can achieve any $z \in \Bbb{Z}$ by evaluating integer polynomials $X + d$ over all $d \in \Bbb{Z}$ which are in $R$.

But $f \in R \subset \Bbb{Q}[X]$ so if $f$ is irreducible in $\Bbb{Q}[X]$ then it is also irreducible in $R$, so that $(f)$ (in $R$) is also maximal in $R$ and $R/(f)$ is a field.

Then $\psi$ is actually an isomorphism since it's kernel has to be $(0)$.  Thus $R/(f) \simeq \Bbb{Z}/(f(a))$ for any $a \in \Bbb{Z}$, so you get weird isomorphisms for all the different moduli $f(a)$.

Therefore, something in my understanding is not correct.  Can you help me suss this out?

答:
There are a couple of problems: your assertion that $R/(f)$ is a field is unwarranted. Your assertion that $\psi$ is one-to-one is unwarranted.

You try to deduce that $R/(f)$ is a field from the conclusion that $f$ is irreducible in the case at hand ($f$ monic and irreducible in $\mathbb{Q}[x]$). But the implication
$$f\text{ is irreducible}\implies (f)\text{ is maximal}$$
holds in a PID, but not in an arbitrary domain. You can only conclude that $(f)$ is maximal among principal ideals. For instance, $2$ is irreducible in $\mathbb{Z}[x]$, but $(2)$ is not maximal: it is properly contained in $(2,x)$. Likewise, here $(f)$ is generated by an irreducible, but unless $R$ is a PID (and I’m pretty sure it is not), then you have no warrant to assert that $R/(f)$ is a field.

Your assertion that $\psi$ is an isomorphism of fields assumes that (1) the target is a field; and (2) $\psi$ is one to one. It’s possible that the target is trivial (if $f(a)=\pm 1$); in which case you don’t get an isomorphism of fields. And, more relevant, $\psi$ need not be one-to-one: in general, its kernel will contains $(x-a)+(f(x))$. If we are assuming that $f(x)$ is irreducible in $\mathbb{Q}[x]$ and monic, then $x-a\in(f(x))$ only if $x-a=f(x)$, which will not be the case in most instances. So $\psi$ usually has non-trivial kernel, which is okay since $(f)$ is only maximal among principal ideals, but not necessarily all ideals of $R$.


问:
Sum the series $\frac{3}{1⋅2⋅4}+\frac{4}{2⋅3⋅5}+ \frac{5}{3⋅4⋅6}+...\text{(upto n terms)}$
$\frac{3}{1⋅2⋅4}+\frac{4}{2⋅3⋅5}+ \frac{5}{3⋅4⋅6}+...\text{(upto n terms)}$

The general term seems to be
$$T_r= \frac{r+2}{r(r+1)(r+3)}.$$

I see no way to telescope this because the factors of the denominator of the general term are not in arithmetic progression.

Do I have to use something else? Or am I missing some easy manipulation?

答:
HINT:

Using Partial Fraction Decomposition,

$$\frac {r+2}{r(r+1)(r+3)}=\frac Ar+\frac B{r+1}+\frac C{r+3}$$ wherere $A,B,C$ are arbitrary constants

$$\implies r+2=A(r+1)(r+3)+Br(r+3)+Cr(r+1)$$

We can put $r=0,-1,-3$

or we can compare the coefficients of the different powers of $r$

to determine $A,B,C$

ContentLicense: CC BY-SA 3.0


问:
Combinatorics prove $\frac{(2n)!}{2^{n}}$
Use combinatorics consideration to prove that $\frac{(2n)!}{2^{n}}$ is an integer number

I have no idea what the question wants?
Does it want to prove it combinatoricaly or with binomial thoerm?

答:
Maybe you should convert it into a word problem. (Number of ways of arranging / permuting $2n$ objects). I believe you can figure that out. I will give an example below.

The above number $\frac{(2n)!}{2^n}$ is the number of distinct permutations of two copies of $n$ objects. Thus, it is an integer.


问:
Showing that $x (1-x) y (1-y) (1 - xy)^{-2}$ is less than or equal to 1.
Let $x,y \in [0,1]$ and consider the inequality $ f(x,y) \equiv x (1-x) y (1-y) (1 - xy)^{-2} \leq 1$. One can show this easily by finding $argmax_x f(x,y)$ and then plugging into $f(x,y)$ to show that $\max_y \max_x f(x,y) \leq 1$. But I wonder if this can be shown using a more powerful approach? Is this an implication of some well known inequality or general mathematical result? My ultimate goal is to extend this to an analogous inequality for matrices.

Body:
Alternative proof using AM-GM

$$x(1-x)y(1-y)-(1-xy)^2=(x-x^2)(y-y^2)-(1-2xy+x^2y^2)\\
=xy-xy^2-x^2y+x^2y^2-1+2xy-x^2y^2=xy(3-x-y)-1 \\
\le \left(\frac{x+y+3-x-y}{3}\right)^3-1=0.\blacksquare$$


问:
Non-numerical proof of an inequality

Let $ 0 < s < 1 $ be the unique solution to the equation $$ \frac{1}{2^s} + \frac{1}{6^s} + \frac{1}{12^s}=1, $$ and show that $$ \frac{1}{6^s}+\frac{1}{12^s} \geq \left(\frac{2}{7}\right)^s. $$

It is easy do this numerically; you can compute $ s \simeq 0.7584 $ with a calculator, and show that the desired inequality holds with this value plugged in.

I'm looking for a more general proof, maybe geometric, using some calculus, or using concavity of the function $ x \mapsto x^s $ for $ 0 < s < 1 $.

Title: Non-numerical proof of an inequality

Body:
Denote $a = \frac{1}{2^s}$ and $b = \frac{1}{3^s}$.

We have $a + a b + a^2 b = 1$ which results in $b = \frac{1 - a}{a + a^2}$.

Also, since $a^2 < b$, we have $\frac{1 - a}{a + a^2} > a^2$
or $\frac{(a^2 + 1)(1 - a - a^2)}{a + a^2} > 0$ which results in $1 - a - a^2 > 0$.

Since $x\mapsto x^s$ is concave on $x>0$, we have
$(\frac{6}{7})^s + (\frac{8}{7})^s \le 2(\frac{\frac{6}{7} + \frac{8}{7}}{2})^s = 2$
which results in $(\frac{2}{7})^s \le \frac{2}{4^s + 3^s}$.

It suffices to prove that $\frac{1}{6^s} + \frac{1}{12^s} > \frac{2}{4^s + 3^s}$ or
$ab + a^2 b > \frac{2}{\frac{1}{a^2} + \frac{1}{b}}$
or $\frac{(1-a)^2(a+1)(1 - a - a^2)}{a^4 + a^3 - a + 1} > 0$ which is true.

We are done.


问:
Can $X - Y A^\dagger Y^T\succ0$ be written as an LMI where $A^\dagger$ is a pseudoinverse?
I have the constraint

\begin{align}
X - Y A^\dagger Y^T\succ0,
\end{align}

where $A^\dagger$ is the pseudoinverse of $A\succeq0$. Can we still use the Schur complement to write the constraint as an LMI?

Explicitly, can we show something like:

\begin{align}
X - Y A^\dagger Y^T\succeq0 &\iff \begin{pmatrix}X&Y\\Y^T&A\end{pmatrix}\succeq0
\end{align}

Answer: It is true only when $A$ is invertible, please see answer below on the general case.

答:
As mentioned in the comments, in the general case where $A\succeq0$ may not be invertible there is an orthogonality condition that reads as follows:
\begin{align}
X-YA^\dagger Y^T\succeq0 \ \ \text{and}\ \  Y(I - A A^\dagger)=0 \iff \begin{pmatrix} X&Y\\ Y^T &A\end{pmatrix}\succeq0.
\end{align}


问:
Some Results in $\mathbb{Z} [\sqrt{10}]$
This is a question from an old Oxford undergrad paper on calculations in $\mathbb{Z} [\sqrt{10}]$. We equip this ring with the Eucliden function $d(a+b\sqrt{10})=|a^2-10b^2|$. I want to prove the following results:

If $d(x)=1$, then $\frac{1}{x} \in \mathbb{Z} [\sqrt{10}]$

Any non-zero element of $\mathbb{Z} [\sqrt{10}]$ which is not a unit can be expressed as a product of finitely many irreducibles in $\mathbb{Z} [\sqrt{10}]$

The ideal generated by $2$ and $\sqrt{10}$ is not principal in $\mathbb{Z} [\sqrt{10}]$

Thought so far

Suppose $x=a+b\sqrt{10}$. Clearly if $x$ is a unit then $d(x)=1$, though I'm not sure if this helps. Are we OK simply to note that $\frac{1}{x}=\frac{a-b\sqrt{10}}{a^2-10b^2}$ and since $d(x)=1$ then the deonminator is either $1$ or $-1$.

I know this is true in general in a principal ideal domain and every Euclidean ring is a principal ideal domain, but this proof is lengthy. Is there any calculation one can perform in $\mathbb{Z} [\sqrt{10}]$ to demonstrate this property more quickly.

Any help would be appreciated; I'm not actually too sure what this ideal looks set. Could someone put it in a set notation for me?

Many thanks.

答:
The set Z[10] makes use of half-primes, which must occur in even numbers.  This is pretty normal for many of these sorts of systems.  These half-primes live in some other part that have this same ratio, ie $a\sqrt{5}+b\sqrt{2}$.  In some systems, like z[3], there are even uits in there, eg $(\sqrt{6}+\sqrt{2})/2$.

In the case of Z[10] (and many other schemes), there are subspaces that hold primes, but there is no unit to move to that space.

An example is that of Z[210], which has three subspaces $a\sqrt{3}+b\sqrt{70}$ and $a\sqrt{2}+b\sqrt{105}$ and $a\sqrt{6}+b\sqrt{35}$, and a unit $\sqrt{15}+\sqrt{14}$, which transitions to another set of subspaces.  This group has no fewer than eight sub-spaces, linked in pairs, and primes can be individually found in any of the main group, or the three subgroups (3), (2), (6).  A number appears in the main group, if the product of the special primes of the subgroup, multiply up to a square.  For example the prime factors for 29 are $\sqrt{35}+\sqrt{6}$. and for 7, one can find it in $\sqrt{105}+7\sqrt{2}= \sqrt{7}*(\sqrt{15}+\sqrt{14})$, but since 6*2 gives 12, one needs a further '3' type prime to make it appear in Z[210].

That is pretty much the order of the day for composite numbers in this type.


问:
Is every ideal containing a non zero divisor a indecomposable module?
Let $R$ be a commutative Noetherian local ring. Let $I\subseteq R$ be an ideal containing a non zero divisor. Is $I$ an indecomposable $R$-module? Is it true at least if $\dim(R)=1$? I see this is true if $R$ is a domain: if $I\simeq J\oplus K$ for ideals $J,K$, then $JK\subseteq J\cap K=0$. Thanks.

答: In general the answer is no. Example: if $F$ is a field, then $\big[F[x,y]\big/(xy)\big]_{(\overline{x},\overline{y})}$ is Noetherian of dimension $1$ and local, with unique maximal ideal $M=\left(\overline{x}\big/\overline{1},\overline{y}\big/\overline{1}\right)$. Then $M$ contains the non-zero divisor $(\overline{x}+\overline{y})\big/\overline{1}$, but we can write $M=\left(\overline{x}\big/\overline{1}\right)\oplus\left(\overline{y}\big/\overline{1}\right)$, so $M$ is decomposable.


问:
Why isn't every finite locally free morphism etale?
I have been reading the notes here. There, a finite locally free morphism of schemes is defined as as a morphism of schemes $f: X \rightarrow Y$ which is finite and for which the sheaf $f_{*} \mathcal{O}_{X}$ is locally free as an $\mathcal{O}_{Y}$-module. A finite etale morphism is then defined as a morphism of schemes $f: X \rightarrow Y$ which is finite locally free, and for which the fiber over any point $q \in Y$ is an etale $\kappa(q)$-algebra.

I am struggling to understand what the part about the fibers being etale algebras adds at all. It seems like every finite locally free morphism would trivially satisfy this.

Take $f: X \rightarrow Y$ to be a finite locally free morphism of schemes. For any $q \in Y$, choose an affine $\operatorname{spec}A$ around $q$ small enough so that $f_{*} \mathcal{O}_{X}$ is free, say of degree $d$. Then since $f$ is finite, it is in particular affine, so the morphism looks locally like $\operatorname{spec}(A^{\oplus d}) \rightarrow \operatorname{spec}A$. Then the fiber over $q$ is just $\operatorname{spec}(\kappa(q)^{\oplus d})$. This seems to trivially be an etale $\kappa(q)$-algebra, since tensoring with some algebraic closure $\Omega$ would give $d$ copies of $\Omega$.

Where is the flaw in my reasoning? Or is it just the case that every finite locally free morphism is etale?

Further to this, the same notes say that for a finite etale morphism, the degree of the morphism at a point $q \in Y$ is the same as the cardinality of the fiber over $q$. But again the above reasoning seems to show this for any finite locally free morphism. Where does etale come into it at all?

答:
Here is an example of a finite locally free morphism which is not etale: take spec of the natural inclusion $\Bbb F_2(t^2)\subset \Bbb F_2(t)$. This fails to be etale because it's a non-separable field extension.

The flaw in your reasoning is that your identification of $f_*\mathcal{O}_X$ as $A^d$ locally is only as a module - it needs to be as a ring to say anything interesting. (You should also note that even in the case of $\Bbb R\subset\Bbb C$, your idea is wrong - this is a point mapping to a point, while your reasoning would have two points mapping to one point.)


问:
Trouble with finding variance of $(\hat{Y_{0}} - Y_{0})$ in terms of regression estimation
This question comes from Rice's Mathematical Statistics Ch. 14 - (14):

Values which may be of use:

$\hat{\beta_{0}} = \bar{y} - \hat{\beta_{1}}\bar{x}$

$Var(\hat{\beta_{1}}) = \frac{\sigma^{2}}{S_{xx}}$

So my problem is in finding the expression correctly. I think my issue is I'm not treating certain parameters and values in the correct way.

Attempt

$$\hat{Y} - Y_{0} = \bar{y} - \hat{\beta_{1}}\bar{x} - \beta_{0} + x_{0} \hat{\beta_{1}} - x_{0}\beta_{1} - e_{0} \\
= \bar{y} - \beta_{0} + \hat{\beta_{1}}(x_{0} - \bar{x}) - x_{0}\beta_{1} - e_{0}$$

Taking the variance:

$$Var(\hat{Y} - Y_{0}) = Var(\bar{y}) + Var(\beta_{0}) + (x_{0} - \bar{x})^{2} Var(\hat{\beta_{1}}) + Var(x_{0}\beta_{1}) + Var(e_{0}) \\
= \frac{\sigma^{2}}{n} +  Var(\beta_{0}) + (x_{0} - \bar{x})^{2}\frac{\sigma^{2}}{S_{xx}} + Var(x_{0}\beta_{1}) + \sigma^{2} $$

This is where I'm stuck. I'm not sure how to treat $ Var(\beta_{0})$ and $Var(x_{0}\beta_{1})$. They are not constants, at least I don't think they are. So I'm missing an assumption I should be applying. Fortunately I do have a solution of what I'm supposed to arrive at:

But I can't seem to work out how to get to this final step based on what I have. What is it I'm forgetting to do?

答:
Best idea is to use matrix algebra here both for the derivation and expressing the results.

For example, the topic of prediction intervals is covered on Pages 136 to 137 of this educational reference.

Succinctly, quoting from the source, the variance expression for the prediction interval is given by:

${Var[Y^*- \hat β'x^∗ |\space X \space ] = \sigma^2 [1 + V_{x^*}]}$

where ${V_{x^*}}$ was previously given on Page 134 as:

${V_{x^*} = {x^*}'[X'X]^{-1}x^*}$

and for a simple Least-Squares model with one variable and an intercept term, the corresponding prediction point vector, as represented by ${x^*}$, is given by: ${[ 1 \space x^*]}$.


问:
Vector space over $\mathbb{R}$ implies Vector space over $\mathbb{C}$
If $V$ is a finite dimensional space over $\mathbb{R}$ then is it also a vector space over $\mathbb{C}$.Also is the converse true ?

I was thinking of using Countability argument to disprove this ,but I can't seem to find any way. Please give some hint.

答:
Any vector space over $\mathbb C$ is automatically a vector space over $\mathbb R$, since multiplication by scalars is well defined.

Moreover, if $v_1,..., v_n$ is a basis of $V$ over $\mathbb C$ then $v_1,..,v_n, iv_1,.., iv_n$ is a basis of $V$ over $\mathbb R$.

This tells you not only that for a vector space over $\mathbb R$ the scalar multiplication to make it a vector space over $\mathbb C$ is not typically defined, but more, for a vector space of ODD dimension, we cannot even extend the scalar multiplication to make it a vector space over $\mathbb C$.


问:
Show that $A = \lambda I_n$ if $\lambda$ is an eigenvalue of $A$ with multiplicity $n$.
Let $A$ be a diagonalizable $n \times n$ matrix, and suppose that $\lambda$ is an eigenvalue of $A$ with multiplicity $n$. I want to show that $A= \lambda I_n$.

I started by saying that the determinant $\det (A- \lambda I_n)$ yields a characteristic polynomial with root $\lambda$. Thus, the eigenvalues of A are just $\lambda$ with multiplicity $n$. To find the eigenvectors now, we need to find the null space of $A- \lambda I_n$.

This is where I get confused. In order for $A = \lambda I_n$ to be true, $A - \lambda I_n = 0$ must also be true. Moreover, let $\vec v$ be an eigenvector of $A$ belonging to an eigenvalue $\lambda$. Then $v ∈ N(A − λI)$. But I'm not sure how to tie these concepts together or solve for the eigenvectors of A.

Any help is greatly appreciated!

答:
You have to use the diagonalizability somewhere, otherwise a matrix like $\begin{pmatrix} \lambda & 1 \\ 0 & \lambda\end{pmatrix}$ provides a counterexample, since it has $\lambda$ as all its eigenvalues but is not equal to $\lambda I_n$.

Suppose $A$ is diagonalizable, say $A = P^{-1}DP$. Then we know that $D$ has all the eigenvalues of $A$ on its diagonal, but every eigenvalue is $\lambda$ so $D = \lambda I_n$, and therefore $A = P^{-1}(\lambda I_n) P = \lambda I_n (P^{-1}P) = \lambda I_n$ since the identity matrix commutes with every matrix.


问:
$f^n(x) = \frac{5-10n}{2}*(7-5x)^{-1} * f^{n-1}(x)$. Prove valid for all $n \in \mathbb{N}$
$f^n$(x)$ = \frac{5-10n}{2}*(7-5x)^{-1} * f^{n-1}(x)$. Prove valid for all $n \in \mathbb{N}$
Where $f^n(x)$ is the $n^{th}$ derivative of $f(x)$. And $f(x) = (7-5x)^{\frac{1}{2}}$.
So far I have made my base case of $n=1$ and showed that it equals the derivative of $f(x)$ and then I assumed n=k. I then state that: $f^{k+1} = \frac{d}{dx}(f^k)$. I know that my end goal is to get: $f^{k+1} = \frac{5-10(k+1)}{2}*(7-5x)^{-1} * f^{(k+1)-1}(x)$ which equals $f^{k+1} = \frac{-5-10k}{2}*(7-5x)^{-1} * f^{k}(x)$.
I've tried plugging the equation for $f^{k+1}$ into $f^{k+1} = \frac{d}{dx}(f^k)$ but couldn't seem to get the result I wanted, as I ended up with:
$-25*(7-5x)^{-2}*f^k(x)$
Help would be much appreciated.

答:
The constant term in $f^n(x)$ should be $\frac{10n-15}{2}$, not $\frac{5-10n}{2}$.

It's best to rewrite the equation as
$$f^n(x) = a_n \frac{f^{n-1}(x)}{(f(x))^2}$$
This will greatly simplify our calculation.

Induction step from $n$ to $n+1$: First notice that

$$f^{n-1} (x) = \frac{1}{a_n} (f(x))^2 f^n(x)\tag1$$
and
$$f'(x) = a_1 \frac{f(x)}{(f(x))^2} \implies f(x)f'(x)= a_1\tag2$$
with $a_1=-\frac{5}{2}$.

Then
$$f^{n+1}(x) = \frac{d}{dx} f^n(x) = \frac{d}{dx} \left(a_n \cdot \frac{f^{n-1}(x)}{(f(x))^2}\right)\\
=a_n \cdot \frac{f^n(x) (f(x))^2 - f^{n-1}(x) \cdot 2f(x)f'(x)}{(f(x))^4}\\
=a_n \cdot \frac{f^n(x) (f(x))^2 - \frac{1}{a_n} (f(x))^2 f^n(x) \cdot (2a_1)}{(f(x))^4}\\
= a_n \left(1-\frac{2a_1}{a_n}\right) \frac{f^n(x)}{(f(x))^2}\\
= (a_n-2a_1) \frac{f^n(x)}{(f(x))^2}
$$

Then we have $a_{n+1}=a_n-2a_1$, an arithmetic sequence. Therefore $a_n=(n-1)(-2a_1)+a_1 = \frac{10n-15}{2}.\blacksquare$


问:
can a line integral of domain $C$ be negative when $C$ is the boundary of the region in the upper half plane? ($y>0$)
ok so here is the question:

evaluate $\oint \left(3y^2+e^{\cos x}\,dx\right) + \left(\sin y+5x^2\,dy\right)$, where $C$ is the boundary of the region in the upper half-plane ($y\ge 0$)between the circle $x^2+y^2=a^2$ and $x^2+y^2=b^2$ where $0\lt a\lt b$

note: the region can be expressed in polar coordinates $0\le\theta\le\pi$ and $a\le r \le b$  good time for Greens theorem?

I dont know if it is the right answer because it is negative.

first I took $\partial$$F_2$/$dx$ and got $10x$ then $\partial F_1/\partial y$ and got $6y$

I converted $\iint_D (10x-6y)\,dA$ to $\int_0^\pi\!\int_a^b(10r\cos\theta-6r\sin\theta)r\,drd\theta$

then when I integrated I got $-4b^3+4a^3$

答: You used correctly the Green's Theorem for that line integral and your last result is also correct. In fact, we have $$\iint_D(Q_x-P_y)dA=\iint_D (10x-6y)$$ in which $P=3y^2+e^{\cos x},~~Q=\sin y+5x^2$</p>


问:
If a self mapping linear functional is a projection, is it reducible?
$V$ is a vector space and $T\in L(V,V)$.

If $T$ is a projection then is $T$ reducible?
If $T$ is reducible then is $T$ a projection?

I am not sure if the first question is true since I am unsure of the case where $V$ is infinite dimensional and I have no way to show the second question is true.

答:
The first assertion is true unless $T$ is either trivial or the identity map.

Since $T^2 = T$, for every $v \in V, T(Tv) = Tv$, so $\operatorname{im} T = T(V)$ is invariant under $T$, and $T(v - Tv) = 0$, so $v - Tv \in \ker T$. $\ker T \cap \operatorname{im} T = 0$ so $V = \operatorname{im} T\oplus \ker T$. $\ker T$ is invariant under $T$, so as long as neither $\operatorname{im} T$ nor $\ker T$ is trivial, $T$ must be reducible. $\operatorname{im} T$ being trivial means $T \equiv 0$. And $\ker T$ being trivial means that $v - Tv = 0$ for all $v$, making $T$ the identity map.

Note that this applies without any regard to the dimension of $V$.

For your second assertion, consider $$T = \begin{bmatrix}1&0\\0&2\end{bmatrix}$$


问:
Show that the ideal $AB$ is prime iff $A = \{0_R\}$ or $B = \{0_R\}$ as follows
Let $A$ and $B$ be ideals in a principal ideal domain $R$, where $A \ne R$ and $B \ne R$.
Show that the ideal $AB$ is prime iff $A = \{0_R\}$ or $B = \{0_R\}$.

I know that since $R$ is PID then $A$ and $B$ are generated by one element. But, how to start the proof?

Any ideas ?

答:
If $A$ and $B$ are ideals in a PID $R$, then let $A = (a), B = (b)$. Then $AB = (ab)$ (this needs to be proven, but is not so hard in a PID). Clearly if $A = (0)$, then $a = 0$, so $AB = (0b) = (0)$ is prime and the symmetric argument holds for $B$.

On the other hand, suppose that $AB$ is prime. We know that that product $a\cdot b \in (ab)$, so by the primeness of $AB$ either $a \in AB$ or $b \in AB$. WLOG, $a \in AB$. This means there exists $r \in R$ such that $a = rab$. If $a = 0$ then we're done. If not, since $R$ is a PID, and hence a domain, this implies $rb = 1$, so $b$ is a unit, but this contradicts our assumption that $B \neq R$.


问:
Gravitational Potential Energy MDOF system
I'm trying to find the Equation of motion for a MDOF system using the energy method. Using Lagrange's equation, I can get almost everything except for the potential energy of the bar. Using newton's method, and assuming small angle, $\sin(\theta)$ = $\theta$, and $\cos(\theta)$ = 1. I tried that in Lagrange's equation and there is no $\theta$ or u term to take a derivative of. This is a textbook example so I Find that they are using $\sin(\theta)$=$\theta$ and $\cos(\theta)$ = (1-$\frac{1}{2}\theta^2$). Can someone please explain to me how in the world are they getting (1-$\frac{1}{2}\theta^2$).Worked Example

答:
The MacLaurin series for $\sin \theta$ and $\cos \theta$ are

$$\begin{align}
\sin \theta &= \theta -\frac{\theta^3}{3!} + \frac{\theta^5}{5!} - \cdots \\
\\
\cos \theta &= 1 - \frac{\theta^2}{2!} + \frac{\theta^4}{4!} - \cdots
\end{align}$$

so for small $\theta$ we get

$$ \sin \theta \approx \theta \quad \text{and} \quad \cos \theta \approx 1 - \frac{\theta^2}{2}$$


问:
Trigonometric graph analysis - point determination
I don't fully understand this question as an explanation and example weren't given to it in the book.

I'm fluent in the material itself just don't understand its application in this sense

could you help explain it to me?

Thank You,

答:
Hint, as you have not displayed much effort to solve.

Draw horizontal line segments (parallel to the x-axis) at say point 'a'.

Note, such a segment intersects the right part of the cosine curve just prior to the x-axis point of ${2\pi}$.

Per the provided information contained in the question, it is an x point apparently close to ${2\pi}$.

Which one based on symmetry?


问:
Does every compact (not necessarily convex) set have extreme points?
Let $A$ be a set in a normed vector space. Call a point $p$ in $A$ extreme if there do not exist $p_0, p_1$ in $A$, distinct from $p$, such that $\lambda p_0+(1-\lambda)p_1=p$, for $\lambda \in [0,1]$. Can you give an example of a compact $A$ which has no extreme points? (I'd prefer an example in finite-dimensional Euclidean space if possible!)

The context for my question is that I was wondering why we need convexity for the statement, "a linear objective is maximized at an extreme point of a compact, convex set." I figured maybe it is because non-convex sets need not have extreme points, but couldn't think of an example.

Thank you! And please let me know if my question is "missing the point" -- I've seen it suggested that "extreme points don't make sense for non-convex sets," but I don't understand why.

答:
I don't understand the infinite-dimensional situation so please permit me to assume that we are working in $\mathbb{R}^n$ (equivalently, since we don't need the norm except to define the topology and every norm topology is equivalent in the finite-dimensional case, any finite-dimensional normed vector space). In that case we can say the following:

Every (non-empty!) compact subset of $\mathbb{R}^n$ has an extreme point.

A linear objective is maximized at an extreme point of a compact subset of $\mathbb{R}^n$, as LinAlg says in the comments; this doesn't require convexity.

The following argument will prove both of these. We induct on the dimension $n$. When $n = 0$ the statement is clear ($\mathbb{R}^0$ is a point and hence $K$ is a point and hence an extreme point). Let $K \subset \mathbb{R}^n$ be compact (and non-empty). Consider any nonzero linear functional $f : \mathbb{R}^n \to \mathbb{R}$. By compactness, $f$ attains on $K$ a maximal value $m$. The subset $f^{-1}(m) \cap K$ of points of $K$ at which $p_1$ attains its maximum value is the intersection of $K$ with a hyperplane of dimension $n-1$, so by the inductive hypothesis has an extreme point $p$. (Picture $K$ as a polyhedron and $f^{-1}(m) \cap K$ as a face of it, which is the interesting case; generically $f^{-1}(m) \cap K$ is a point, e.g. if $K$ is an ellipsoid.)

This point is necessarily also an extreme point of $K$, since if $p = \lambda p_0 + (1 - \lambda) p_1$ where $p_0, p_1 \in K$ then $f(p) = \lambda f(p_0) + (1 - \lambda) f(p_1)$; by maximality $f(p_0), f(p_1) \le f(p)$ and hence $\lambda f(p_0) + (1 - \lambda) f(p_1) \le f(p)$ with equality iff $f(p_0) = f(p_1) = f(p) = m$, in which case $p_0, p_1 \in f^{-1}(m) \cap K$. So if $p$ weren't an extreme point of $K$ then it wouldn't be an extreme point of $f^{-1}(m) \cap K$.

So $K$ has an extreme point, and $f$ is maximized at it.


问:
Circle packing representation of a given graph
Based on the Circle packing theorem: "For every connected simple planar graph G there is a circle packing in the plane whose intersection graph is (isomorphic to) G."

I would like to draw the circle packing representation of this graph (the original graph in the picture).

I am trying using geogebra, but I am new at it and I am not able to get it done. Do you have any suggestion about other software or plug-ins of geogebra that can help.

UPDATE: 19/May/2011 - This one includes the circular packing for the graph (thanks to geogebra user forum - http://www.geogebra.org/forum/viewtopic.php?f=2&t=21841):

NOTE:
This question is related to: https://mathoverflow.net/questions/63861/representations-of-regular-maps-four-color-theorem

答:
You might want to post this question to the GeoGebra user forum, with a little more info about the obstacles you're running into.

Also, have you ever worked with the Geometer's Sketchpad?  It's available for a trial, and if you're happy with it, it costs ~$30.00?


问:
Doing modular division when denominator and modulus not coprime
So normally if you calculate $n/d \mod m$, you make sure $d$ and $m$ are coprime and then do $n[d]^{-1}\mod m$ , all $\mod m$. But what if $d$ and $m$ are not coprime? What do you do?</p>

答:
You are trying to solve the congruence $xd\equiv n \pmod{m}$. Let $e$ be he greatest common divisor of $d$ and $m$. Since $e$ divides $d$ and $m$, if the congruence has a solution, $e$ must divide $n$.  If $e$ does not divide $n$, division is not possible.

So let us assume that $e$ divides $n$. Then division is sort of possible, but as we shall see, not entirely satisfactory.

Let $d=d_1e$, $m=m_1e$, and let $n=n_1e$. Then
$$xd\equiv n\pmod{m}\quad\text{if and only if}\quad xd_1\equiv n_1\pmod{m_1}.$$
Since $d_1$ and $m_1$ are relatively prime, the congruence on the right has a unique solution modulo $m_1$, found in the usual way.

Call the solution $x_0$. Then the solutions modulo $m$ are $x_0+im_1$, where $i$ ranges from $0$ to $e-1$. Thus modulo $m$ division is possible, but it has several answers.


问:
Confused about when to treat probability calculation as small interval and use $\lambda$ in Poisson process
I am doing MITx 6.431x and I am really confused about Poisson process, in particular about when to treat the calculation as a small interval and when not.

Suppose we are merging two Poisson processes, say Red and Green. Then P(the kth arrival is red)=$\frac{\lambda}{\lambda_{Red}+\lambda_{Green}}$.

Why? This is as if we are looking at a small interval here. Precisely speaking, we are taught to use the following when $\tau$ as an interval is 'small'.

$P(k,\tau) \approx
\begin{cases}
\ 1-\lambda\tau,  & \text{if $k$ =0} \\[2ex]
\lambda\tau, & \text{if $k$ =1} \\[2ex]
\ 0, & \text{if $k$ >1}
\end{cases}$

$\frac{\lambda}{\lambda_{Red}+\lambda_{Green}}$ is different from $\lambda\tau$ obviously, but in the lecture the former is derived from the latter, which seems to suggest we are still assuming a small interval. Furthermore, why shouldn't we use, say, $p_{N_\tau}(k)=P(k,\tau)=\frac{(\lambda \tau)^ke^{-\lambda \tau}}{k!}$ to calculate, where $p_{N_\tau}(k)$ means the probability of having $k$ arrivals in interval $\tau$?

In the lecture, $\lambda$ is used to calculate probability when it is explicitly said that we are only concerning ourselves with small intervals. But it seems really vague when exactly do we use $\lambda$, consider the following:

All ships travel at the same speed through a wide canal. Each ship
takes $t$ days to traverse the length of the canal. Eastbound ships
(i.e., ships traveling east) arrive as a Poisson process with an
arrival rate of $λ_E$ ships per day. Westbound ships (i.e., ships
traveling west) arrive as an independent Poisson process with an
arrival rate of $λ_W$ ships per day. A pointer at some location in the
canal is always pointing in the direction of travel of the most recent
ship to pass it.

What is the probability that the next ship to pass will be westbound?

The solution here tells me that P(next ship is westbound)=$\frac{\lambda_W}{\lambda_{E}+\lambda_{W}}$. But nothing in the question suggests that we are considering a 'small' interval. Why shouldn't we use $p_{N_\tau}(k)=P(k,\tau)=\frac{(\lambda \tau)^ke^{-\lambda \tau}}{k!}$?

答:
This may help: let $T_{r},T_{g}$ be the first arrival times for the "red" and "green" processes, denoted $N_r(t)$ and $N_g(t)$.  Then, $P(T_{r}>t)=P(N_{r}(t)=0)=e^{-\lambda_{r}t}$, so $T_{r}$ has an exponential distribution with rate $\lambda_{r}$; similarly $T_{g}$ is exponential with rate $\lambda_{g}$.

So, to see the probability "red" arrives first, assuming the "red" and "green" processes are independent, we can compute $$P(T_{r}<T_{g})=\iint_{\{(s,t):0<s<t\}} \lambda_{g}e^{-t\lambda_g} \lambda_r e^{-s\lambda_r}dsdt
$$
$$
\int_0^\infty\int_0^t\lambda_{g}e^{-t\lambda_g} \lambda_r e^{-s\lambda_r}dsdt = \int_0^\infty \lambda_ge^{-\lambda_g t}(1-e^{-\lambda_rt})dt
$$
$$
=1-\frac{\lambda_g}{\lambda_g+\lambda_r} = \frac{\lambda_r}{\lambda_r+\lambda_g}.
$$


问:
Why is Newton's method faster than gradient descent?
Can you provide some intuition as to why Newton's method is faster than gradient descent?

Often we are in a scenario where we want to minimize a function f(x) where x is a vector of parameters. To do that the main algorithms are gradient descent and Newton's method. For gradient descent we need just the gradient, and for Newton's method we also need the hessian. Each iteration of Newton's method needs to  do a linear solve on the hessian:

$$x \leftarrow x - \textrm{hess}(f,x) \backslash \textrm{grad}(f,x)$$

Where \ indicates doing a linear solve (like in matlab). In many cases the hessian is sparse, and in that case we would use an iterative method like conjugate gradient descent to do the linear solve. So what we're actually doing is this:

Compute a quadratic approximation of f around x.

Minimize that quadratic approximation with conjugate gradient descent.

Make a step in the direction of that minimum.

Compare this to normal gradient descent, where we just run (conjugate) gradient descent directly on the unapproximated original function f.

My question is: why is it faster to repeatedly do a quadratic approximation, then minimize that approximation with gradient descent, and then do a step in that direction, than to simply run gradient descent on the original function?

答:
Suppose we're minimizing a smooth convex function $f: \mathbb R^n \to \mathbb R$. The gradient descent iteration (with step size $t > 0$) is
$$
x_{k+1} = \arg \min_x \,f(x_k) + \langle \nabla f(x_k), x - x_k \rangle + \frac{1}{2t} \| x - x_k \|_2^2.
$$
At each iteration, we minimize a linear approximation to $f$ (with an additional quadratic penalty term that prevents us from moving too far from $x_k$).

The Newton's method iteration is
$$
x_{k+1} = \arg \min_x f(x_k) + \langle \nabla f(x_k), x - x_k \rangle + \frac12 (x - x_k)^T \nabla^2 f(x_k)(x - x_k).
$$
At each iteration, we are minimizing a quadratic approximation to $f$. The quadratic approximation is more accurate than the linear approximation that gradient descent uses, so it's plausible that Newton's method converges faster to a minimizer of $f$.


问:
How to show that two ergodic non-mutually singular measures equal
Let $X$ be a compact metric space and let $T: X\to X$ be continuous.  Let $\mu$ and $\nu$ be two Borel probability measures on $X$ which are ergodic for $T$.  If $\mu$ and $\nu$ are not mutually singular, how to show that $\mu = \nu$?

答:
Welcome to MSE!

Hint:

The Ergodic Theorem tells you that for any $f : L^1(X)$ and for almost every $x$

$$\lim_{n \to \infty} \frac{1}{n} \sum_{i<n} f(T^ix)$$

equals something. Notice this limit doesn't depend on $\mu$ or $\nu$, whereas the something I'm alluding to does.

Can you leverage this (say by choosing $f = \chi_A$ to be a characteristic function) in order to prove $\mu = \nu$?

I hope this helps ^_^


问:
Why can we find a Cramér-von Mises test of level $\alpha$?
I'm reading through Shao's Mathematical Statitics section 6.5.2. on Cramér-von Mises Tests and there is something I don't understand.

Context :

Let $X_1,...,X_n$ be iid from a continuous cdf $F$. We wish to test if the sample comes from some continuous cdf $F_0$:
$$
\begin{cases}
H_0 : F = F_0 \\
H_1 : F \neq F_0.
\end{cases}
$$

Cramer-von Mises tests are tests which reject $H_0$ when $C_n(F_0) > c$ where

$$ C_n(F) = \int (F_n(x)-F(x))^2 \; dF(x).$$

Question :

On page $448$ we read the following " ... the distribution of $C_n(F)$ does not depend on $F$ (exercice). Hence a Cramer-von Mises test of size $\alpha$ can be found. "

I've managed to solve the exercice by showing that

$$ C_n(F) =  \frac{1}{12n^2} + \frac 1 n \sum_{i=1}^{n}\left(U_i-\frac{2i-1}{2n}\right)^2$$

where $U_i = F(X_{(i)})$ is uniformaly distributed on $[0,1].$ However I don't understand why this implies that a test of level $\alpha$ can be found. Why is this the case ?

If I'm not mistaken this means we can find $c$ such that

$$P_{H_0}(RH_0) = P(C_n(F_0) > c ) = \alpha.$$

How do we know such a $c$ exists ?

答:
If I'm not wrong, this goes back to the idea of a pivot (or a pivotal quantity).
When you want to construct a statistical test it is necessary to construct a confidence interval and pivots are the theoretical way to do this.
A pivot is a function of a sample that does not depend on the underlying distribution function.
For example let $X_1,\ldots,X_n$ be iid with common cdf $F_X$.
Then a pivot $R = R(X_1,\ldots,X_n;F_X)$ is such that
\begin{align}
F_R(x) = \mathbb{P}(R(X_1,\ldots,X_n;F_X) \leq x)
\end{align}
does not depend on $F_X$.
The best example for a pivot is
$$R = \sqrt{n}\frac{\bar{X}_n - \mu}{\sigma}$$
when $X_i \sim N(\mu,\sigma^2)$ and $\bar{X}_n = \frac{1}{n} \sum_{i=1}^n X_i$.
Here we clearly have a dependence on the sample, while $\mu$ and $\sigma$ are functions of $F_X$.
Nonetheless, the distribution of $R$ does not depend on $F_X$ in any way since we have $R \sim N(0,1)$ irrespective of the mean and variance of $F_X$.

In case of a pivot we can find (because $F_R$ is strictly increasing) a constant $c$ independent of $F_X$ such that $1-F_R(c) = \mathbb{P}(R(X_1,\ldots,X_n;F_X) > c) \geq \alpha$.
The important thing here to note is that $c$ is independent of $F_X$ so $c$ does not depend on the unknown $F_X$.

In your specific case $C_n(F)$ does not depend on $F$ (as long as $F$ is continuous I would guess), so the distribution function $G$ of $C_n(F)$ is also independent of $F$.
Moreover in this specific case $G$ is also continuous.
Therefore you can pick $c$ (irrespective of $F$) as a quantile of $G$ to get
$$
\mathbb{P}(C_n(F) > c) = \alpha.
$$


问:
Evaluate $\lim_{x\to+∞}\frac{(\sum_{n=0}^∞{(\frac{x^n}{n!})^2})^2}{(\sum_{n=0}^∞{(\frac{x^n}{n!})^1}) (\sum_{n=0}^∞{(\frac{x^n}{n!})^3})}$
Prove the following limit:
$$
\lim_{x\rightarrow +\infty} \frac{\left( \sum_{n=0}^{\infty}{\left( \frac{x^n}{n!} \right) ^2} \right) ^2}{\left( \sum_{n=0}^{\infty}{\left( \frac{x^n}{n!} \right) ^1} \right) \left( \sum_{n=0}^{\infty}{\left( \frac{x^n}{n!} \right) ^3} \right)}=\frac{\sqrt{3}}{2} \tag{1}
$$

Mathematica tells me that
$$
\sum _{n=0}^{\infty } \left(\frac{x^n}{n!}\right)^2=I_0(2 x)\\
\sum _{n=0}^{\infty } \left(\frac{x^n}{n!}\right)^3 = \, _0F_2\left(;1,1;x^3\right)
$$
but they don't make sense to me for calculating the limits, since I don't have any knowledge about Special Functions.

How can I prove $(1)$ in an elementary way(not involving special functions)?

答:
We investigate a heuristic idea that lead the correct answer. Consider a Poisson random variable $N$ with rate $x$, so that

$$ \mathbb{P}(N = n) = \frac{x^n}{n!} e^{-x} $$

for $n \geq 0$. Then by the (local) central limit theorem, we know that $Z = \frac{N-x}{\sqrt{x}}$ approximates the standard normal distribution, loosely in the sense that

\begin{align*}
\mathbb{P}(N = n)
&= \mathbb{P}\left( \left| N - n \right| < \frac{1}{2}\right) \\
&\approx \mathbb{P}\biggl( \left| Z - \frac{n-x}{\sqrt{x}} \right| < \frac{1}{2\sqrt{x}} \biggr) \\
&\approx \phi\left(\frac{n-x}{\sqrt{x}}\right) \frac{1}{\sqrt{x}}
\end{align*}

where $\phi(t) = \frac{1}{\sqrt{2\pi}} e^{-t^2/2}$ is the p.d.f. of the standard normal distribution. So we expect that, for each given $\alpha > 0$, as $x\to\infty$,

\begin{align*}
\sum_{n=0}^{\infty} \left(\frac{x^n}{n!}\right)^{\alpha}
&\sim e^{\alpha x} \sum_{n=0}^{\infty} \left[ \phi\left(\frac{n-x}{\sqrt{x}}\right) \frac{1}{\sqrt{x}} \right]^{\alpha} \\
&\sim \frac{e^{\alpha x}}{(2\pi x)^{(\alpha-1)/2}} \int_{-\infty}^{\infty} \frac{1}{\sqrt{2\pi}} e^{-\alpha t^2/2} \, \mathrm{d}t \\
&= \frac{e^{\alpha x}}{(2\pi x)^{(\alpha-1)/2}} \cdot \frac{1}{\sqrt{\alpha}}. \tag{*}
\end{align*}

Assuming that $\text{(*)}$ indeed holds, then the limit easily follows.

Indeed, some hard analysis using the idea of Laplace's method shows that, for each fixed $\alpha > 0$ and sufficiently small $\epsilon > 0$,

$$ \sum_{n=0}^{\infty} \left(\frac{x^n}{n!}\right)^{\alpha}
= \frac{e^{\alpha x}}{(2\pi x)^{(\alpha-1)/2}} \cdot \frac{1}{\sqrt{\alpha}} \bigl( 1 + \mathcal{O}(x^{-\frac{1}{2}+3\epsilon}) \bigr) \tag{$\diamond$} $$

This can be established by splitting the sum into two parts, one over the range $\left|n - x\right| \leq x^{\frac{1}{2}+\epsilon}$ that realizes the leading term of $(\diamond)$, and the other over the range $\left|n - x\right| \geq x^{\frac{1}{2}+\epsilon}$ that only contribute to the relative error $(\diamond)$.

As the computation is a bit lengthy, let me estimate the first sum only:

Fix a sufficiently small $\epsilon > 0$ and consider the case $n \in \mathcal{I} = \{ n \in \mathbb{N}_0 : \left| n - x \right| \leq x^{\frac{1}{2}+\epsilon} \}$. Also, introduce a new parameter $u$ which is related to $n$ by

$$n = x + u\sqrt{x}.$$

In particular, $\left| u \right| \leq x^{\epsilon}$ and $ \frac{n}{x} = 1 + \frac{u}{\sqrt{x}} $. Now by the Stirling's approximation,

\begin{align*}
\frac{x^n}{n!}
&= \exp\left( n \log x - \frac{1}{2}\log(2\pi n) + n - n \log n + \mathcal{O}\bigl(n^{-1}\bigr) \right) \\
&= \exp\left( - \frac{1}{2}\log(2\pi x) - \frac{1}{2}\log\frac{n}{x} + x \left( \frac{n}{x} -  \frac{n}{x} \log \frac{n}{x} \right) + \mathcal{O}\bigl(x^{-1}\bigr) \right) \\
&= \exp\left( - \frac{1}{2}\log(2\pi x) + \mathcal{O}\bigl(x^{-\frac{1}{2}+\epsilon}\bigr) + x \left( 1  - \frac{u^2}{2x} + \mathcal{O}\bigl(x^{-\frac{3}{2}+3\epsilon}\bigr) \right) + \mathcal{O}\bigl(x^{-1}\bigr) \right) \\
&= \exp\left( - \frac{1}{2}\log(2\pi x) + x  - \frac{u^2}{2} +  + \mathcal{O}\bigl(x^{-\frac{1}{2}+3\epsilon}\bigr) \right) \\
&= \bigl( 1 + \mathcal{O}\bigl(x^{-\frac{1}{2}+3\epsilon}\bigr) \bigr) e^{x} \phi(u) \frac{1}{\sqrt{x}}.
\end{align*}

So it follows that, for each given $\alpha > 0$,

\begin{align*}
\sum_{n \in \mathcal{I}} \left( \frac{x^n}{n!} \right)^{\alpha}
&= \bigl( 1 + \mathcal{O}(x^{-\frac{1}{2}+3\epsilon}) \bigr) \frac{e^{\alpha x}}{(2\pi x)^{(\alpha-1)/2}} \sum_{n \in \mathcal{I}} \frac{1}{\sqrt{2\pi x}} e^{-\alpha u^2/2}.
\end{align*}

Now the summation part in the right-hand side can be regarded as the Riemann sum. Indeed, using the fact that

$$ \left| \sum_{a \leq n \leq b} f(n) - \int_{a}^{b} f(x) \, \mathrm{d}x \right| \leq 3\sup_{a \leq x \leq b} \left| f(x) \right| $$

holds for any continuous unimodal function $f : [a, b] \to \mathbb{R}$,

$$ \sum_{n \in \mathcal{I}} \frac{1}{\sqrt{2\pi x}} e^{-\alpha u^2/2}
= \int_{-x^{\epsilon}}^{x^{\epsilon}} \frac{1}{\sqrt{2\pi}} e^{-\alpha u^2/2} \, \mathrm{d}u + \mathcal{O}(x^{-1/2})
= \frac{1}{\sqrt{\alpha}} + \mathcal{O}(x^{-1/2}). $$

Combining all the estimates, we obtain the desired leading term in $(\diamond)$.


问:
Existence of $x_0$ s.t. $\frac{f''( x_0 )}{2}=\frac{f( a )}{( a-b ) ( a-c )}+\frac{f( b )}{( b-a ) ( b-c )}+\frac{f( c )}{( c-a ) ( c-b )}$

Let $f$ be continuous on $[a,b]$, and $f''(x)$ exists on $(a,b)$. Prove that $\forall c\in(a,b), \exists x_0\in(a,b)$, s.t.
$$
\frac{f''\left( x_0 \right)}{2}=\frac{f\left( a \right)}{\left( a-b \right) \left( a-c \right)}+\frac{f\left( b \right)}{\left( b-a \right) \left( b-c \right)}+\frac{f\left( c \right)}{\left( c-a \right) \left( c-b \right)}
$$

My Approach: Assume $$\dfrac{K}{2} = \frac{f\left( a \right)}{\left( a-b \right) \left( a-c \right)}+\frac{f\left( b \right)}{\left( b-a \right) \left( b-c \right)}+\frac{f\left( c \right)}{\left( c-a \right) \left( c-b \right)}$$
Let
$$
g\left( x \right) =\frac{K}{2}\left( \left( a-x \right) \left( x-c \right) \left( c-a \right) \right) +f\left( a \right) \left( x-c \right) +f\left( x \right) \left( c-a \right) +f\left( c \right) \left( a-x \right)
$$
Then $g(a)=g(c)=g(b)=0$.
Since $g$ is continuous on $[a,b]$ and differentiable on $(a,b)$, there exist $x_1\in(a,c),x_2\in(c,b)$ s.t. $g'(x_1)=g'(x_2)=0$.
And since $g'$ is continuous on $[a,b]$ and differentiable on $(a,b)$, there exists $x\in(x_1,x_2)\subset (a,b)$, s.t. $g''(x_0)=0$.

From
$$
g''\left( x_0 \right) =\frac{K}{2}\cdot 2\left( a-c \right) +f''\left( x_0 \right) \left( c-a \right)
$$
We have $K=f''(x_0)$.

My Question is, is the proof right? And can we generalize the original claim(that is, change $f''(x)$ to $f^{(k)}(x)$)?

Other proofs of the original claim is also welcomed.

答:
Your formula can be written as

$$\frac{\left | \begin{matrix} 1& 1& 1\\ a  & b & c \\ f(a) & f(b) & f(c) \end{matrix} \right|}{
\left| \begin{matrix} 1& 1& 1\\ a  & b & c \\ g(a) & g(b) & g(c) \end{matrix} \right|}=\frac{f''(\xi)}{g''(\xi)}
$$
(with $g(x) = x^2$).

Let $K$ be the value of the quotient. Then the function
$$F(x) =\left | \begin{matrix} 1& 1& 1\\ x  & b & c \\ f(x) & f(b) & f(c) \end{matrix} \right| - K \cdot \left | \begin{matrix} 1& 1& 1\\ x  & b & c \\ g(x) & g(b) & g(c) \end{matrix} \right|$$
is $0$ at $x=a$. Clearly it is also $0$ at $x=b$, and $x=c$. So it has $3$ distinct zeroes. Applying Rolle twice we find $\xi$ such that $F^{(2)}(\xi)=0$.

It is easy to generalize this to $n+1$ distinct $a_i$'s. One can also pass to the limit in these formulas, grouping several $a_i$ together. In particular, if $n$ of them are "equal", we get the quotient of Taylor series approximations.


问:
4.8 Example of 0 chapter of Do Carmo's Riemannian Geometry
Picture below is from 23th page of Do Carmo's Riemannian geometry. I don't know why $\pi_1^{-1}\circ \pi_2$ is coincide with $\varphi_g$  on $x_2(W)$.  Since in my view, it  is needed that proving  $g$ is independent to $p_2$. But there is not proof about this.

答:
Assume that there are two different group elements, say $g$ and $g'$, which map $2$ distinct points of $\mathbf{x}_2(W)$ to $2$ points of $\mathbf{x}_1(W)$ respectively, then you get that

$$(g'^{-1}g).(\mathbf{x}_2(W)) \cap \mathbf{x}_2(W) \neq \phi,$$

which gives a contradiction with the definition of a properly discontinuous action (since $g'^{-1}g \neq 1$, by our assumption).


问:
Can't find $h_1(y,z)$, $h_2(x,z)$, and $h_3(x,y)$.

Given vector field $F(x,y,z)=(3x^2y, x^3+y^3,0)$. Find $f(x,y,z)$ such
that $\nabla f=F$.

I have tried as below.

\begin{alignat*}{2}
  &&\nabla f &= F\\
   \iff&&\left(\dfrac{\partial f}{\partial x}, \dfrac{\partial f}{\partial y}, \dfrac{\partial f}{\partial z}\right)&=(3x^2y, x^3+y^3,0)
 \end{alignat*}
We have PDEs $\dfrac{\partial f}{\partial x}=3x^2y$, $\dfrac{\partial f}{\partial y}=x^3+y^3$, and$\dfrac{\partial f}{\partial z}=0$.
\begin{align*}
 \dfrac{\partial f}{\partial x}=3x^2y\iff \partial f=3x^2y\partial x
 \iff \int\partial f=\int 3x^2y\partial x
 \iff f=x^3y+h_1(y,z).
\end{align*}
\begin{align*}
 \dfrac{\partial f}{\partial y}=x^3+y^3&\iff \partial f=\left(x^3+y^3 \right)\partial y
 \iff \int\partial f=\left(x^3+y^3 \right)\partial y\\
 &\iff f=x^3y+\dfrac{1}{4}y^4+h_2(x,z).
\end{align*}
\begin{align*}
 \dfrac{\partial f}{\partial z}=0\iff \partial f=0\partial z
 \iff \int\partial f=\int 0\partial z
 \iff f=h_3(x,y).
\end{align*}
Thus,
\begin{align*}
 x^3y+h_1(y,z)=x^3y+\dfrac{1}{4}y^4+h_2(x,z)\iff h_1(y,z)=\dfrac{1}{4}y^4+h_2(x,z),
\end{align*}
\begin{align*}
 x^3y+h_1(y,z)=h_3(x,y),
\end{align*}
\begin{align*}
 x^3y+\dfrac{1}{4}y^4+h_2(x,z)=h_3(x,y).
\end{align*}

Now I can't find $h_1(y,z)$, $h_2(x,z)$, and $h_3(x,y)$. How to find it?

答: From $$\frac{\partial f}{\partial x} = 3x^2y$$ we get that $f(x,y,z) = x^3y+g(y,z)$, and then $$x^3+y^3 = \frac{\partial f}{\partial y} = x^3 + \frac{\partial g}{\partial y} \quad \leadsto \quad \frac{\partial g}{\partial y} = y^3 \quad \leadsto \quad g(y,z) = \frac14 y^4+h(z).$$ Thus $$f(x,y,z) = x^3y+\frac14 y^4+h(z)$$ and finally $$0 = \frac{\partial f}{\partial z} = h'(z) \quad \leadsto \quad h(z) = c$$ where $c$ is a constant.


问:
Second order stochastic dominance + mean preserving spread
Let $X$ be a random variable and $F(X)$ associated distribution, and Let $Y = \alpha X +(1-\alpha),\alpha \in (0,1)$ and $G(Y)$ associated distribution. By linearity of expectation, $E[Y] = E[X]$, and by property of variance $Var(Y) = \alpha^2 Var(X)<Var(X) $ as $\alpha \in (0,1)$.

To me, it looks like $X$ is mean preserving spread of $Y$, and my question is that "Does $G(Y)$ second order stochastic dominates $F(X)$?

答:
So mean preserving spread $\implies$ SOSD. However, $E(X)=E(Y)$ AND $Var(X) > Var(Y) \nRightarrow Y$ SOSD $X$ (see here). However, an equivalent definition of SOSD is the following:

$Y$ SOSD $X$ iff for all $X$: $$\int_a^X F(x)dx \geq \int_a^X G(y)dy$$

where $F(a) = G(a) = 0$

Now consider your case (a little modified): $Y=aX+(1-a)\mu$, where $\mu=E(X)$ and $a\in (0,1)$. So we have that

\begin{align}
G(y) &= \mathbb P(Y \leq y) \\
&=\mathbb P(aX+(1-a)\mu \leq y) \\
&=\mathbb P\bigg(X \leq \frac{y-(1-a)\mu}{a}\bigg) \\
&=F\bigg(\frac{y-(1-a)\mu}{a}\bigg)
\end{align}

Now let $X$ (and so consequently $Y$) have finite support $[b,c]$ and define $S_f(x) = \int_b^xF(z)dz$ and $S_g(x) = \int_b^yG(z)dz$.

\begin{align}
S_g(y) &= \int_b^y G(z)dz \\
&=\int _b^y F\bigg(\frac{z-(1-a)\mu}{a}\bigg) dz \\
&=a\int _{b_1}^{y_1} F(\theta) d\theta \tag{$\theta =\frac{z-(1-a)\mu}{a}$ }
\end{align}

Now see that, $$\mu \geq b \implies  b \geq b_1\equiv \frac{b-(1-a)\mu}{a} , \forall a\in(0,1)$$

Therefore we have that:
\begin{align}
S_g(y) &=a\int _{b}^{y_1} F(\theta) d\theta \\
&=aS_f(y_1)
\end{align}

where $y_1 = (y-(1-a)\mu)/a$

Now consider the region where $c>y>\mu$. It's easy to see that $y_1>y$. Further, since $F(.)$ is a positive increasing function, $S_f(.)$ is also increasing. Also, let $1>a>S_f(y)/S_f(y_1)$

Therefore, for $y>\mu$,$$S_g(y) = aS_f(y_1) > S_f(y)$$

Now,  This proves, that $Y$ does not SOSD $X$, at least not for all values of $a$.


问:
In a Clifford algebra over an $n$-dimensional positive-semidefinite space, is a product of any number of vectors a product of at most $n$ vectors?
This is related to the Cartan-Dieudonne theorem.

In a Clifford algebra over an $n$-dimensional real vector space $V$ with quadratic form $a\cdot a\geq0$ for all $a\in V$, can any product of vectors $a_1,a_2,\cdots,a_m\in V$ be written as

$$a_1a_2\cdots a_m=a_1'a_2'\cdots a_k'$$

for some $k\leq n$ and $a_1',a_2',\cdots,a_k'\in V$?

Due to the grading on the algebra, necessarily $k\equiv m\bmod2$. It suffices to prove that a product of $n+1$ vectors can be reduced to a product of $n-1$ vectors.

For $n=1$ dimension, a product of $2$ vectors is a scalar: $ab=a\cdot b+a\wedge b=a\cdot b+0$, since $a$ and $b$ are linearly dependent. The empty product of $0$ vectors is $1$, not necessarily $a\cdot b$; but we can allow the product to have a scalar in front, which doesn't make a difference in higher dimensions because it can be absorbed into one of the vectors.

For $n=2$, a product of $3$ vectors is

$$abc=(a\cdot b+a\wedge b)c=(a\cdot b)c+(a\wedge b)\cdot c+(a\wedge b)\wedge c$$

$$=(a\cdot b)c+a(b\cdot c)-(a\cdot c)b+a\wedge b\wedge c$$

$$=(a\cdot b)c-(a\cdot c)b+(b\cdot c)a+0$$

which is a sum of vectors and thus a single vector $a'$.

Now let's assume that $V$ is positive-definite, and that we've proven the result for dimensions less than $n$ for induction. We have $n+1$ vectors $a_i$ in $n$ dimensions, which are thus linearly dependent:

$$c_1a_1+c_2a_2+\cdots+c_{n-1}a_{n-1}+c_na_n+c_{n+1}a_{n+1}=0$$

for some scalars $c_i$ not all $0$. Use this to define a vector

$$v=c_1a_1+\cdots+c_{n-1}a_{n-1}=-c_na_n-c_{n+1}a_{n+1}.$$

If $a_n,a_{n+1}$ are dependent, then their product is a scalar, and $a_1\cdots a_{n-1}a_na_{n+1}=a_1\cdots a_{n-1}(a_n\cdot a_{n+1})$, and the problem is solved. So assume $a_n,a_{n+1}$ are independent.

Similarly, if $a_1,\cdots,a_{n-1}$ are dependent, then they're contained in a subspace of dimension $\leq n-2$, and by induction we've solved the problem in that case: $a_1\cdots a_{n-1}=a_1'\cdots a_{n-3}'$, so  $a_1\cdots a_{n+1}=a_1'\cdots a_{n-3}'a_na_{n+1}$ is a product of $n-1$ vectors. So assume $a_1,\cdots,a_{n-1}$ are also independent.

If $v=0$ then these independences give $c_1=\cdots=c_{n-1}=0$ and $c_n=c_{n+1}=0$, a contradiction. Thus $v\neq0$, and by positive-definiteness $v$ has an inverse $(v\cdot v)^{-1}v$. Then we can write

$$a_1\cdots a_{n-1}a_na_{n+1}=(a_1\cdots a_{n-1}v^{-1})(va_na_{n+1});$$

the factor on the left is a product of $n$ vectors in $n-1$ dimensions, which by induction is a product of $n-2$ vectors; and the factor on the right is a product of $3$ vectors in $2$ dimensions, which is a single vector; so we get a product of $n-2+1=n-1$ vectors. This completes the proof.

Is the result still true when $V$ is only positive-semidefinite?

答:
No, it's not true. Take $V$ to have signature $+00$, that is, an orthogonal basis with $e_1\!^2=1,\,e_2\!^2=0,\,e_3\!^2=0$. Then the product of four unit vectors

$$e_1(e_1+e_2)(e_1+e_3)(e_1-e_2+e_3)=1+e_2e_3$$

cannot be written as a product of two vectors; their wedge product would have to be $e_2e_3$, so they'd be in the span of $e_2,e_3$, but then their dot product would be $0$, not $1$.

This phenomenon requires a 2D degenerate subspace. If $V$ has signature of the form ${++}\cdots{+0}$, with just a 1D degenerate subspace, the result is true.

Consider two vectors $a,b$. If $a$ is null ($a^2=0$) and $b$ is not, then we can move the null vector to the right in the product:

$$ab=b(b^{-1}ab)=ba'$$

where $a'^2=a^2=0$.

If $a$ and $b$ are both null, then they're in the same 1D degenerate subspace, and their product is $0$.

So in the product $a_1a_2\cdots a_{n+1}$ any null factors can be moved to the right side, and the product vanishes if there's more than one such factor.

Suppose there's exactly one null factor $a_{n+1}\propto e_n$. All of the other vectors can be written in the form $a_i=a_i^++a_i^\circ e_n$ where $a_i^+$ is in the span of the other basis vectors $e_1,\cdots,e_{n-1}$ and $a_i^\circ$ is a scalar. Then $a_ia_{n+1}=a_i^+a_{n+1}$ and, anticommuting $a_{n+1}$ past the near factors as necessary to reach the far factors, the product simplifies to

$$a_1a_2\cdots a_na_{n+1}=(a_1^+a_2^+\cdots a_n^+)a_{n+1}.$$

The expression in parentheses is a product of $n$ vectors in a positive-definite space of dimension $n-1$, which the OP showed is a product of $n-2$ vectors. Including $a_{n+1}$ makes a product of $n-1$ vectors.

So in what follows we can assume that there are no null factors; that each $a_i^+\neq0$.

Consider two vectors $a,b$, neither null. As before write $a=a^++a^\circ e_n$, where $a^+$ is in the span of $e_1,\cdots,e_{n-1}$ and $a^\circ$ is a scalar, and $b=b^++b^\circ e_n$. We desire to have $a=a^+$, that is, $a^\circ=0$; if $a^\circ\neq0$ then we'll try to find vectors $x,y$ such that $ab=xy$ and $x^\circ=0$. But since $a$ and $b$ are invertible, $x$ and $y$ are also invertible, so $ab=xy$ is equivalent to $x^{-1}ab=y$. And since $y$ is an arbitrary vector, this is equivalent to $x^{-1}ab$ having a vanishing trivector part: $x^{-1}\wedge a\wedge b=0$. This is further equivalent to $x$ being in the span of $a,b$ (assuming they're independent; but if they're dependent then trivially $ab=a\cdot b=\big((a\cdot b)e_1\big)e_1$). Thus, for some scalars $c_1,c_2$,

$$x=c_1a+c_2b=(c_1a^++c_2b^+)+(c_1a^\circ+c_2b^\circ)e_n$$

and the coefficient of $e_n$ must vanish. This has a solution

$$x=b^\circ a-a^\circ b=b^\circ a^+-a^\circ b^+$$

which is non-zero because $a^\circ\neq0$; hence $x$ is invertible as required.

So in the product $a_1a_2\cdots a_{n+1}$ any pairs $a_ia_{i+1}$ with $a_i^\circ\neq0$ can be rewritten with $a_i'^\circ=0$. This leaves only $a_{n+1}'$ possibly not in that form. Now we have, as before, a product of $n$ vectors $a_1'a_2'\cdots a_n'$ in a positive-definite space of dimension $n-1$, which reduces to a product of $n-2$ vectors, and the remaining $a_{n+1}'$ makes a product of $n-1$ vectors.


问:
Does the split-step operator method work for a PDE in cylindrical coordinates?
I am trying to numerically solve the 3D Schrodinger equation with an extra non-linear term (the Gross-Pitaevskii equation) using the split-step operator method. This is a follow up to a previous question which was answered nicely, but now I would like to do the same in cylindrical coordinates (to exploit a symmetry that is already in my problem).

The equation is of the form
$$
\frac{\partial\Psi(\rho,z,t)}{\partial t} = -\frac{i}{\hbar}\hat{H}\;\Psi(\rho,z,t) \tag{1}
$$
where
$$
\hat{H} = \hat{T} + \hat{V} = -\frac{\hbar^2}{2m}\nabla^2 + V(\rho,z), \tag{2}
$$
and $\rho$ and $z$ are the cylindrical coordinates (I am assuming no angular dependence in the equation), and $\nabla^2$ is the 3D laplacian.

If I split the operator $\hat{H}$ symetrically (Strang), then when solving for the $\hat{T}$ part (kinetic) over a full step $\Delta t$, I obtain the following
$$
\begin{align}
\frac{\partial \psi}{\partial t} &= -\frac{i}{\hbar} \hat{T} \psi \tag{3}\\
&= \frac{i\hbar}{2m}\nabla^2\psi . \tag{4}
\end{align}
$$
In cartesian coordinates, the equation is easy to solve by transforming the whole equation into Fourier space, as described here. However, if I try to follow that approach here using the expression for the Laplacian in cylindrical coordinates I obtain
$$
\begin{align}
\frac{\partial \psi}{\partial t} &= \frac{i\hbar}{2m}\nabla^2\psi \tag{5}\\
&= \frac{i\hbar}{2m} \Bigg( \frac{\partial^2\psi}{\partial \rho^2} + \frac{1}{\rho}\frac{\partial\psi}{\partial \rho}  + \frac{\partial^2\psi}{\partial z^2}  \Bigg) \tag{6}\\
\implies \frac{\partial \hat{\psi}}{\partial t} &= \mathcal{F}\Bigg\{ \frac{i\hbar}{2m} \Bigg( \frac{\partial^2\psi}{\partial \rho^2} + \frac{1}{\rho}\frac{\partial\psi}{\partial \rho}  + \frac{\partial^2\psi}{\partial z^2}  \Bigg) \Bigg\} \tag{7}\\
&= -\frac{i\hbar}{2m} \Bigg( k_\rho^2 \hat{\psi} + k_z^2\hat{\psi} + \mathcal{F}\Bigg\{ \frac{1}{\rho}\frac{\partial\psi}{\partial \rho} \Bigg\}  \Bigg) \tag{8}
\end{align}
$$
where $\hat{\psi} = \hat{\psi}(k_\rho,k_z,t)$ is the Fourier transform of $\psi(\rho,z,t)$, and $k_\rho$ and $k_z$ are the corresponding k-vectors. The last term here now poses a problem - the equation does not have the simple exponential solution as in the cartesian case.

Is there a known way to handle this situation? Or is the split-step method somehow only suited to cartesian coordinate systems? Any advice would be great!

(As mentioned, the aim is to numerically solve the equation eventually, where in the cartesian case one would usually use the FFT algorithm.)

答:
This is a partial answer that should get you going in the right direction. I have some uncertainty about satisfying the (unstated) boundary conditions and the sign conventions in this field, so there is significant room for clarifying this answer.

The result from taking the Fourier transform of both sides can also be arrived at by substituting the spectral representation of the wave function:
$$
\psi(\mathbf{x},t) = \int\hat\psi(\mathbf{k},t)\exp(-i \mathbf{k}\cdot\mathbf{x}) \;d\mathbf{k}
$$
I'm not 100% sure about the Fourier/time/phase convention in this field, but the sign convention above is one way to do things. In cartesian, we have
$$
\nabla^2 \psi(\mathbf{x},t) = \int\hat\psi(\mathbf{k},t)\nabla^2\Big(\exp(-i \mathbf{k}\cdot\mathbf{x})\Big) \;d\mathbf{k},
$$
and
$$
\nabla^2\Big(\exp(-i \mathbf{k}\cdot\mathbf{x})\Big) = -k^2\exp(-i \mathbf{k}\cdot\mathbf{x}),
$$
and so the Fourier transform turned all the spatial derivatives into constants, simplifying things enough to directly write a solution for $\hat\psi(\mathbf{k},t)$ as the solution to
$$
\frac{\partial \hat\psi}{\partial t}=-\frac{i\hbar}{2 m}k^2\hat\psi
$$
Now in cylindrical, I think we need a spectral representation that makes the cylindrical Laplacian into a constant. In other words, we're looking for eigenfunctions of the cylindrical Laplacian (with this extra symmetry that nothing depends on the azimuthal angle).
$$
\nabla^2\psi = -k^2\psi,
$$
with
$$
\nabla^2 = \frac{\partial^2}{\partial\rho^2} + \frac{1}{\rho}\frac{\partial}{\partial\rho} + \frac{\partial^2}{\partial z^2}
$$
That's the Helmholtz equation and simple separation of variables gives a general solution to the eigenfunction problem:
$$
\psi = \Big(A\exp(-ik_z z)+B\exp(ik_z z)\Big)\Big(C\mathrm{H}_0^{(1)}(k_\rho \rho)+D\mathrm{H}_0^{(2)}(k_\rho \rho)\Big).
$$
The H functions are the usual Hankel functions. I've also named my separation constants such that $k_\rho^2 + k_z^2 = k^2=|\mathbf{k}|^2$. Okay, so some superposition of waves of the above form with various values for the four constants will be the general solution for $\psi$ which will be our replacement for the Fourier transform. It's here that we usually start appealing to the physical nature of the problem and boundary conditions. There are probably sources of the wavefunction in this problem, and those sources make waves that propagate away from the sources and off to infinity. In classical EM (the field I'm more familiar with) we appeal to some kind of radiation condition, like a boundary condition at infinity, to say that waves can't enter from infinity, only go out. That effectively excludes one of the exponential functions from the solution. Which one is excluded depends on if $z<0$ or $z>0$. For the Hankel functions, I assume you want finiteness of $\psi$ at $\rho=0$ in general since there is nothing special there, so you have to have $C=D=1/2$ to cancel the Bessel functions of the second kind that have a singularity at $\rho=0$, so taken together we can write our superposition integral of the "physical" eigenfunctions (considering only $z>0$, the upper half-plane) as
$$
\psi(\mathbf{x},t) =
\int \hat\psi(\mathbf{k},t)\exp(-i k_z z)J_0(k_\rho\rho)\;d\mathbf{k}
$$
This is our new spectral transform that replaces the Fourier transform. Note I could be wrong about the sign in the exponential, because ultimately that's a matter of convention. I also could be wrong about the presence of the $J_0$ Bessel function; I think the exact form you want here depends exactly on your boundary conditions. In any case, you can modify accordingly with $J_0$, $Y_0$, $H_0^{(1)}$, and $H_0^{(2)}$. The same goes for the exponential, I guess it could be replaced with $\exp(i k_z z)$, $\cos(k_z z)$, or $\sin(k_z z)$ depending on exactly what you're doing and your radiation convention / boundary condition at infinity. For a source of waves sitting at the origin, you can even replace $z$ with $|z|$ for certain kinds of Green's function problems with an impulse at the origin. But lets plug my form into the equation of interest:
$$
\frac{\partial \psi}{\partial t}=\frac{i\hbar}{2 m}\nabla^2\psi
$$
Term by term, we have:
$$
\frac{\partial \psi}{\partial t} = \int \frac{\partial\hat\psi(\mathbf{k},t)}{\partial t}\exp(-i k_z z)J_0(k_\rho\rho)\;d\mathbf{k}
$$
$$
\nabla^2\psi = \int (-k^2)\hat\psi(\mathbf{k},t)\exp(-i k_z z)J_0(k_\rho\rho)\;d\mathbf{k}
$$
This last one is by construction and was the point of finding the eigenfuctions, so that this Laplacian turned into multiplication by a constant. Any of the alternate forms with Hankel functions or sines and cosines also of course have this property.
Putting those two together:
$$
\int \frac{\partial\hat\psi(\mathbf{k},t)}{\partial t}\exp(-i k_z z)J_0(k_\rho\rho)\;d\mathbf{k} = \int \left(-k^2\frac{i\hbar}{2 m}\hat\psi(\mathbf{k},t)\right)\hat\psi(\mathbf{k},t)\exp(-i k_z z)J_0(k_\rho\rho)\;d\mathbf{k}
$$
and collecting on one side
$$
\int \left(\frac{\partial\hat\psi(\mathbf{k},t)}{\partial t}+k^2\frac{i\hbar}{2 m}\hat\psi(\mathbf{k},t)\right)\exp(-i k_z z)J_0(k_\rho\rho)\;d\mathbf{k} = 0
$$
Its sufficient for the term in parentheses to be identically zero
$$
\frac{\partial\hat\psi(\mathbf{k},t)}{\partial t}+k^2\frac{i\hbar}{2 m}\hat\psi(\mathbf{k},t) = 0
$$
So we have transformed the equation into a spectral domain where the solutions are indeed simple exponentials in time; we just had to use Hankel AND Fourier transforms. Now when you go to implement this numerically, that's a 1D Fourier transform (can use the FFT) followed by a 1D Hankel transform (you will have to artificially divide the thing you're transforming by a factor of $k_\rho$ to get it in the right form). It turns out there are indeed "fast" Hankel transforms (FHT) buts good codes are much more niche than the FFT.

You can continue to generalize to all four terms instead of retaining the single term as I did. Then you'd have a transform that maps a single function of space and time to a vector function on four dimensions of wavenumber and time. I'm not really sure here but wanted to give you some ideas.


问:
Can a Measure be Literally Continuous?
Say that $(X,M)$ is a measurable space (i.e., $0,X \in M$ and $A \in M \implies X \setminus A \in M$ and, finally, $M$ is closed w.r.t. countable unions/intersections). Suppose that $\mu$ is either a signed or complex measure on $(X,M)$. In particular, $\mu$ is a function on $M$.

I know that if $\mu$ maps into $[0,\infty)$, then $d(A,B) = \mu(A \Delta B)$ is a metric modulo the equivalence relation $A \sim B \iff \mu(A\Delta B) = 0$. This was mentioned in baby Rudin, for instance.

More generally, however, if we are given a topology $T$ on $X$ does there consequently exist a canonical topology on $M$ induced by $T$ such that $\mu$ is continuous?

Of course, I would prefer if the topology on $M$ were not just the discrete topology.

Further, it would be nice if it were metrizable. If the answer is affirmative, a follow up question might be, for instance, "how coarse can the topology be while still making $\mu$ continuous?"
Has anyone thought about this or seen this before?

Thanks.

答:
Take the pseudometric $d$ on $M$ such that $d(A,B)=|\mu|(A\Delta B)$. Take $B(a,r)\subset \mathbb C$ and a set $A\in M$ such that $\mu(A)\in B(a,r)$, also take $\varepsilon>0$ such that $B(\mu(A),\varepsilon)\subset B(a,r)$. Take $B$ in the ball of radius $\varepsilon$ centered in $A$, so $|\mu|(A\Delta B)=|\mu|(A\setminus B)+|\mu|(B\setminus A)<\varepsilon$. Now $$|\mu(A)-\mu(B)|=|\mu(A\setminus B)-\mu(B\setminus A)|\leq |\mu(A\setminus B)|+|\mu(B\setminus A)|\leq|\mu|(A\setminus B)+|\mu|(B\setminus A)=|\mu|(A\Delta B)<\varepsilon$$
So $\mu(B(A,\varepsilon))\subset B(\mu(A),\varepsilon)\subset B(a,r)$. This implies that every point in the preimage of $B(a,r)$ is an interior point, so $\mu^{-1}(B(a,r))$ is open which implies that $\mu:(M,d)\to(\mathbb C,\tau_{\mathbb C})$ is continuous.

Title: Can a Measure be Literally Continuous?


问:
How to compute conditional expectation of a log function
I've been studying the Expectation Maximization algorithm. According to the formula shown here, what I have to do in the M step is to compute a new $\theta$ that maximizes the conditional expectation of the log function, which is $\ln P[X, z|\theta] $:

Formally we have,
$$
\begin{aligned}
\theta_{n+1} &=\arg \max_{\theta}\left\{l\left(\theta \mid \theta_{n}\right)\right\} \\
&=\arg \max _{\theta}\left\{L\left(\theta_{n}\right)+\sum_{\mathbf{z}} \mathcal{P}\left(\mathbf{z} \mid \mathbf{X}, \theta_{n}\right) \ln \frac{\mathcal{P}(\mathbf{X} \mid \mathbf{z}, \theta) \mathcal{P}(\mathbf{z} \mid \theta)}{\mathcal{P}\left(\mathbf{X} \mid \theta_{n}\right) \mathcal{P}\left(\mathbf{z} \mid \mathbf{X}, \theta_{n}\right)}\right\}
\end{aligned}
$$
Now drop terms which are constant w.r.t. $\theta$
$$
\begin{aligned}{l}
=\arg \max _{\theta}\left\{\sum_{\mathbf{z}} \mathcal{P}\left(\mathbf{z} \mid \mathbf{X}, \theta_{n}\right) \ln \mathcal{P}(\mathbf{X} \mid \mathbf{z}, \theta) \mathcal{P}(\mathbf{z} \mid \theta)\right\} \\
=\arg \max _{\theta}\left\{\sum_{\mathbf{z}} \mathcal{P}\left(\mathbf{z} \mid \mathbf{X}, \theta_{n}\right) \ln \frac{\mathcal{P}(\mathbf{X}, \mathbf{z}, \theta)}{\mathcal{P}(\mathbf{z}, \theta)} \frac{\mathcal{P}(\mathbf{z}, \theta)}{\mathcal{P}(\theta)}\right\} \\
=\underset{\theta}{\arg \max }\left\{\sum_{\mathbf{z}} \mathcal{P}\left(\mathbf{z} \mid \mathbf{X}, \theta_{n}\right) \ln \mathcal{P}(\mathbf{X}, \mathbf{z} \mid \theta)\right\} \\
=\arg \max _{\theta}\left\{\mathbf{E}_{\mathbf{Z} \mid \mathbf{X}, \theta_{n}}\{\ln \mathcal{P}(\mathbf{X}, \mathbf{z} \mid \theta)\}\right\}
\end{aligned}
$$
In Equation $(17)$ the expectation and maximization steps are apparent. The EM algorithm thus consists of iterating the:

E-step: Determine the conditional expectation $\mathrm{E}_{\mathbf{Z} \mid \mathbf{X}, \theta_{n}}\{\ln \mathcal{P}(\mathbf{X}, \mathbf{z} \mid \theta)\}$

M-step: Maximize this expression with respect to $\theta$.

( The excerpt above can be acquired in page 8 of this tutorial: http://www.seanborman.com/publications/EM_algorithm.pdf )

However, in the coin toss example below:
http://www.nature.com/nbt/journal/v26/n8/full/nbt1406.html?pagewanted=all

$\ln P[X, z|\theta] $ is nowhere to be found, and they don't prove how the new $\theta^{t+1}$ they got after each iteration is better than the $\theta^t$ previously acquired.

答: Note that $$\sum_z P[z|X,\theta_n] \ln P[X,z|\theta] \\=\sum_\color{red}z\bigg(\ln P[X,z|\theta]\bigg)\color{red}{P[z|X,\theta_n]} \\=E_{Z|X,\theta_n}\bigg(\ln P[X,z|\theta]\bigg)$$ Here probability distribution is $P[z|X,\theta_n]$ and summing over $z$. So, by the definition of expectation we get the desired result.</p>


问:
Classical number theoretic applications of the $p$-adic numbers
I am sure we can all agree that the $p$-adic numbers are highly fascinating objects in their own right - just as the closely related theory of valuations.

Having independently read up on the $p$-adic numbers for a few weeks now, I have so far only seen one application of them to what I would call classical number theory - namely the proof given in Serre's Cours d'arithmétique that a natural number is expressible as the sum of $\leq 3$ squares if and only if it is not of the form $4^a(8b-1)$ for some $a,b \in \mathbb{N}$.

Since I have a tendency to appreciate the value of the higher theories of mathematics in proportion to their applications to elementary number theory, I immediately found myself wondering if there are any other applications.

So my question to the community is: What are the most delightful applications of the $p$-adic numbers and the theory of valuations to elementary number theory?

Many thanks.

P.s.: I am aware that there are already several posts on the forum about the applications of the $p$-adic numbers, but none that refers to elementary number theory specifically.

Edit: I agree that I have been too vague in what I mean by "elementary number theory", so I will try to be a little more specific: By a classical "elementary" number theoretic proposition, I mean a number theoretic proposition that Fermat might have come up with. Thus, the above proposition about the sum of three squares is an elementary number theoretic proposition, as is e.g. Fermat's Last Theorem and the Twin Prime Conjecture, while e.g. the BSD Conjecture or the Class Number Problem are not.

Edit 2: Thank you for all the answers below - they are all excellent! In case anyone should come up with another one, I should like to say that bonus-points are given for results that have so far only been proven using the theory of $p$-adic numbers, or whose proof using $p$-adic numbers is far more conceptual and insightful than the original / more elementary one.

答:
Eisenstein criterion: if $f=x^n +p \sum_{m=0}^{n-1} a_m x^m\in \Bbb{Z}[x]$ with $p\nmid a_0$ then any root of $f$ in $\overline{\Bbb{Q}}_p$ must have valuation $1/n$

(if $v(\beta)>1/n$ then $v(f(\beta))= v(pa_0)$, if $v(\beta)<1/n$ then $v(f(\beta))= v(\beta^n)$)

If $h | f$ in $\Bbb{Q}_p[x]$ then $h(0)$ has valuation $\deg(h)/n$ so that  $\deg(h)=0$ or $n$ ie. $f$ is irreducible in $\Bbb{Q}_p[x]$ and hence in $\Bbb{Q}[x]$.

And Hensel lemma of course: given a polynomial $\in \Bbb{Z}[x]_{monic}$, there is some $k$ such that having a root in $\Bbb{Z}/p^k\Bbb{Z}$ is the necessary and sufficient condition to have a root in every $\Bbb{Z}/p^n\Bbb{Z}$.


问:
Trigonometric graph analysis - point determination
I don't fully understand this question as an explanation and example weren't given to it in the book.

I'm fluent in the material itself just don't understand its application in this sense

could you help explain it to me?

Thank You,

答:
Hint:

The query can be solved by exploring this.


问:
Factoring third degree polynomials using long division
I am sure there is a better strategy that someone smarter than me would use, but I am not that person.

I am trying to factor

$$x^2 - 3x^2 - 4x + 12$$

I don't know how so I attempt to guess with long division. I cheat and look at the answer so find out one of the factors to save myself time, so I try x - 2

I am not sure how to type out long division but I get

$$ x - 2  |  x^3 - 3x^2 - 4x + 12$$

so I know I can have an $x^2$ for how many times x goes into the leading term.

Subract it all and I am left with

$$ x - 2  |  -x^2 - 4x + 12$$

I know that the leading term goes into the inner leading term $-x$ times or however you say that.

$$ x - 2  |   - 2x + 12$$

Now -2

$$ x - 2  |   8$$

Now I don't know what to do, how did this go so wrong? I have $x^2 - x -2$ ontop and I have a remainder of 8. This can't be right, I cheated so I know that this should be a factor.

答:
As Andrew Salmon's comment points out, the easiest thing to try first to factor a cubic polynomial is to try to pair off the terms and see if they share a common factor.  If paired off in the order in which they are written, this becomes

$$x^2(x-3)-4(x-3)$$

or if the odd and even exponents are paired together

$$x(x^2-4)-3(x^2-4)$$

Then simply take out the common factor.

Of course, this won't always work, when it doesn't, the method of guessing roots is known as the rational root theorem.  First you'll want to try plugging in factors of the constant term to see if you get zero.  $\pm1$ is the easiest to check, then in this case you'd also have $\pm2,\pm3,\pm4,\pm6$ and $\pm12$.  If those all failed, you'd want to test any of those values divided by a factor of the leading term.  However, since your leading term is $1$, there isn't anything extra to test here.  If you find one root, you should be able to handle the remaining quadratic.  I doubt you'll be expected to factor any cubic polynomials with no rational roots, possibly excepting the form

$$x^3\pm a$$

at which point you could use the formula for the sum or difference of cubes even if $a$ is not a perfect cube.

Finally, although it's not foolproof, if you suspect a polynomial is a multiple of $x-2$, then plugging in $2$ should yield $0$.  Since you're dividing by $x-2$, your remainder at each step should also be a multiple of $x-2$.  Testing your polynomials at each step reveals

$$2^3-3(2)^2-4(2)+12=8-12-8+12=0$$
$$-(2)^2-4(2)+12=-4-8+12=0$$
$$-2(2)+12=-4+12=8$$

This implies there was an error in the remainder $-2x+12$


问:
If $f\colon \mathbb{R} \to \mathbb{R}$ is such that $f (x + y) = f (x) f (y)$ and continuous at $0$, then continuous everywhere

Prove that if $f\colon\mathbb{R}\to\mathbb{R}$ is such that $f(x+y)=f(x)f(y)$ for all $x,y$, and $f$ is continuous at $0$, then it is continuous everywhere.

If there exists $c \in \mathbb{R}$ such that $f(c) = 0$, then
$$f(x + c) = f(x)f(c) = 0.$$
As every real number $y$ can be written as $y = x + c$ for some real $x$, this function is either everywhere zero or nowhere zero. The latter case is the interesting one. So let's consider the case that $f$ is not the constant function $f = 0$.

To prove continuity in this case, note that for any $x \in \mathbb{R}$
$$f(x) = f(x + 0) = f(x)f(0) \implies f(0) = 1.$$

Continuity at $0$ tells us that given any $\varepsilon_0 > 0$, we can find $\delta_0 > 0$ such that $|x| < \delta_0$ implies
$$|f(x) - 1| < \varepsilon_0.$$

Okay, so let $c \in \mathbb{R}$ be fixed arbitrarily (recall that $f(c)$ is nonzero). Let $\varepsilon > 0$. By continuity of $f$ at $0$, we can choose $\delta > 0$ such that
$$|x - c| < \delta\implies |f(x - c) - 1| < \frac{\varepsilon}{|f(c)|}.$$

Now notice that for all $x$ such that $|x - c| < \delta$, we have
$$\begin{align*}
|f(x) - f(c)| &= |f(x - c + c) - f(c)|\\
 &= |f(x - c)f(c) - f(c)|\\
 &= |f(c)| |f(x - c) - 1|\\
 &\lt |f(c)| \frac{\varepsilon}{|f(c)|}\\
 &= \varepsilon.
\end{align*}$$
Hence $f$ is continuous at $c$. Since $c$ was arbitrary, $f$ is continuous on all of $\mathbb{R}$.

Is my procedure correct?

答: Use sequential criterion for continuity of a function at a point and the additive property of a function to show f(x) is continuous for an arbitrary x€R</p>


答: You could also take the pseudometric in $M$ such that $d(A,B)=|\mu(A)-\mu(B)|$, this is obviously continuous since $\mu^{-1}(B_{\mathbb C}(\mu(A),\varepsilon))=B_{M}(A,\varepsilon)$ and this is the weakest topology in $M$ such that $\mu$ is continuous, since $\tau_d=\{\mu^{-1}(A):A\in \tau_{\mathbb C}\}$, one property this topology has is that sets $A,B$ such that $\mu(A)=\mu(B)$ are not separable.

Title: Can a Measure be Literally Continuous?

Body:
Say that $(X,M)$ is a measurable space (i.e., $0,X \in M$ and $A \in M \implies X \setminus A \in M$ and, finally, $M$ is closed w.r.t. countable unions/intersections). Suppose that $\mu$ is either a signed or complex measure on $(X,M)$. In particular, $\mu$ is a function on $M$.

I know that if $\mu$ maps into $[0,\infty)$, then $d(A,B) = \mu(A \Delta B)$ is a metric modulo the equivalence relation $A \sim B \iff \mu(A\Delta B) = 0$. This was mentioned in baby Rudin, for instance.

More generally, however, if we are given a topology $T$ on $X$ does there consequently exist a canonical topology on $M$ induced by $T$ such that $\mu$ is continuous?

Of course, I would prefer if the topology on $M$ were not just the discrete topology.

Further, it would be nice if it were metrizable. If the answer is affirmative, a follow up question might be, for instance, "how coarse can the topology be while still making $\mu$ continuous?"
Has anyone thought about this or seen this before?

Thanks.


答:
You write that there are "no posts" on this forum that refer to using the $p$-adics in an elementary number theory setting. A universal claim can be refuted with a single counterexample, so look at the answers here for some elementary applications of $p$-adics, including one I mention there about determining the primes in the denominators of binomial coefficients $\binom{r}{n}$ for $r \in \mathbf Q$ by using $p$-adic continuity of polynomial functions on $\mathbf Q$.
This also came up in another math.stackexchange post here and is described in general terms here.

An application to linear recursions taking on specific values (very similar to what Alex gives in his answer) is here and an interpretation of the result in terms of solving the exponential Diophantine equation $3^m = 1 + 2x^2$ is in the appendix here. Another application along the same lines, for integral solutions of the Diophantine equation $x^3 - 2y^3 = 1$, is here.

A use of $p$-adics to explain the structure of $(\mathbf Z/p^k\mathbf Z)^\times$ for odd primes $p$ (that it is cyclic for all $k \geq 1$) is here.  The key point is to rewrite the group as a quotient of actual multiplicative groups $\mathbf Z_p^\times/(1 + p^k\mathbf Z_p)$ so that the multiplicative structure of $\mathbf Z_p^\times$ can be exploited.  It is intriguing that to explain the behavior of a finite abelian group we pass to a $p$-adic compact group like $\mathbf Z_p^\times$, study it, and then take its quotient by an open subgroup.  In the language of elementary number theory, this problem would be about showing odd prime-power moduli have a "primitive root" (old-fashioned terminology for a generator of the units for some modulus).

While not an actual use of $p$-adic completions, a cute use of an extended form of the $p$-adic absolute value is a proof of Gauss' lemma in $\mathbf Z[x]$: if a polynomial in $\mathbf Z[x]$ is reducible in $\mathbf Q[x]$ then it is reducible in $\mathbf Z[x]$ with factors of the same degrees as in $\mathbf Q[x]$. The idea of the $p$-adic proof is to extend the $p$-adic absolute value from $\mathbf Q$ to $\mathbf Q[x]$.  See here.

One of the standard proofs that the harmonic sums $H_n = 1 + 1/2 + \cdots + 1/n$ are not integers for $n \geq 2$ is by showing these rational numbers are not $2$-adically integral (there is a unique term of largest $2$-adic size greater than $1$).  See here.

In Koblitz's book on $p$-adic analysis and zeta-functions, he uses $p$-adic integration to explain $p$-power congruence properties of Bernoulli numbers that had been proved by Kummer, Clausen, and von Staudt in the 19th century by completely different methods.

Title: Classical number theoretic applications of the $p$-adic numbers

Body:
I am sure we can all agree that the $p$-adic numbers are highly fascinating objects in their own right - just as the closely related theory of valuations.

Having independently read up on the $p$-adic numbers for a few weeks now, I have so far only seen one application of them to what I would call classical number theory - namely the proof given in Serre's Cours d'arithmétique that a natural number is expressible as the sum of $\leq 3$ squares if and only if it is not of the form $4^a(8b-1)$ for some $a,b \in \mathbb{N}$.

Since I have a tendency to appreciate the value of the higher theories of mathematics in proportion to their applications to elementary number theory, I immediately found myself wondering if there are any other applications.

So my question to the community is: What are the most delightful applications of the $p$-adic numbers and the theory of valuations to elementary number theory?

Many thanks.

P.s.: I am aware that there are already several posts on the forum about the applications of the $p$-adic numbers, but none that refers to elementary number theory specifically.

Edit: I agree that I have been too vague in what I mean by "elementary number theory", so I will try to be a little more specific: By a classical "elementary" number theoretic proposition, I mean a number theoretic proposition that Fermat might have come up with. Thus, the above proposition about the sum of three squares is an elementary number theoretic proposition, as is e.g. Fermat's Last Theorem and the Twin Prime Conjecture, while e.g. the BSD Conjecture or the Class Number Problem are not.

Edit 2: Thank you for all the answers below - they are all excellent! In case anyone should come up with another one, I should like to say that bonus-points are given for results that have so far only been proven using the theory of $p$-adic numbers, or whose proof using $p$-adic numbers is far more conceptual and insightful than the original / more elementary one.


问:
Show that exists $m, n \in \mathbb{Z}$ such that $|m+n \sqrt{2}|<\frac{1}{10}$.
Show that there exists $m, n \in \mathbb{Z}$ such that $$|m+n \sqrt{2}|<\frac{1}{10}$$.

Try assuming that there are no $m$ and $n$ that satisfy the inequality

Then, $|m+n \sqrt{2}|\geq \frac{1}{10} \forall m,n \in \mathbb{Z}$

Then $\frac{1}{10}$ is a lower bound of  {$|m+n \sqrt{2}|$ for $m,n \in \mathbb{Z}$}

but here I'm stuck and I don't know what to do next!

答:
There are already some great answers given in the comments of the question that answer the question as stated. But just for fun, let's do a proper analysis proof. Also, $\frac1{10}$ is totally arbitrary; we're going to show that $|m+n\sqrt2|$ can always get below any arbitrary positive real $r$.

There is, of course, the trivial solution $m=n=0$, but let's suppose for the sake of argument that you want something else.

With this in mind, let's consider the set $S=\{|m+n\sqrt2|\colon m,n\in\mathbb{Z}, (m,n)\neq(0,0)\}$. Every element in this set is bounded from below (by $0$), and so there must be some real number $r$ which serves as a greatest lower bound for $S$. Let's assume that $r>0$. Then there are two cases: either $r\in S$ or $r\not\in S$.

In the first case, there are integers $m,n$ such that $m+n\sqrt2=r$. (Notice that we can drop the absolute value, because if $m+n\sqrt2=-r$ then $-m-n\sqrt2=r$.) Now, I claim that $S=\{kr\colon k\in\mathbb{Z}\}$, in other words, $S$ is just all of the multiples of $r$. The proof of this is as follows:

Clearly every integer multiple of $r$ is an element of $S$. Suppose that there was some $s=a+b\sqrt 2\in S$ such that $\frac sr\not\in\mathbb{Z}$. Then $\frac sr=k+x$, where $k\in \mathbb{Z}$ and $x\in(0,1)$. So $s=a+b\sqrt2=r(n+x)=kr+rx$. But that means that $s-kr=(a-m)+(b-n)\sqrt2=rx$. Therefore, we have $s-kr\in S$ and $0<s-kr<r$. But this contradicts the fact that $r$ is a greatest lower bound, so such an $s$ cannot exist, and $S$ is the set of integer multiples of $r$.

But we know that $1\in S$ and $\sqrt 2\in S$, and so therefore there are integers $k_1,k_2$ suchthat $1=k_1r$ and $\sqrt2=k_2r$, so $\frac{\sqrt2}1=\sqrt2=\frac{k_1}{k_2}$. So $\sqrt2$ is the ratio of two integers, and therefore $\sqrt2$ is a rational number. This is a contradiction, so we can move on to case $2$.

In our second case, we have $r\not\in S$. But because $r$ is a greatest lower bound of $S$, for every real $\epsilon>0$ there exists some $s\in S$ such that $s\in (r,r+\epsilon)$. (Again, notice that we are justified in dropping the absolute value.) So let $\epsilon=r$, and then we have some $s_1=m_1+n_1\sqrt2\in S$ such that $s_1\in (r,2r)$; in other words, $s_1=r+\epsilon_1$, where $0<\epsilon_1<r$.

But now, we do the same trick, this time letting $\epsilon=\epsilon_1$. What that means is that there is some $s_2=m_2+n_2\sqrt2\in S$ such that $s_2\in (r,r+\epsilon_1)$. So we have $s_2=r+\epsilon_2$, where $0<\epsilon_2<\epsilon_1$. But then, $s_1-s_2=(m_1-m_2)+(n_1-n_2)\sqrt2\in S$, and $s_1-s_2=\epsilon_1-\epsilon_2$. But we know that $0<\epsilon_2<\epsilon_1<r$, and so $0<\epsilon_1-\epsilon_2<r$. In other words, we have again found an element of $S$ which is less than $r$, contradicting the "lower-boundedness" of $r$!

So clearly, such a greatest lower bound cannot exist, and therefore, for any real $r$, we can find integers $m,n$, not both zero, such that $|m+n\sqrt2|<r$. As a special case, letting $r=\frac1{10}$ solves the problem given.


问:
If a function $F$ is defined as $F(x)=F(2x/3) + 1$, where $F(1) = 1$, how can I define the function non-recursively?

If a recursive function $F$ is defined as $F(x)=F(2x/3) + 1$, where $F(1) = 1$, how can one define the function non-recursively?

For example, $$F(1.5)=F(3/2)=F\left(\frac{2\cdot(3/2)}3\right)+1=F(1)+1=2$$How can I find the non-recursive version of this function?

答:
Let $a=2/3$. Thus $F(x)=F(ax)+1$. As was mentioned in the comments, if one defines $$G(x)=F\left(a^x\right),$$
then $$G(x)=F(a^{x+1})+1=G(x+1)+1.$$
Then, upon differentiation, we have that $$G'(x)=G'(x+1),\qquad G(0)=1.$$
This differential equation has a lot of solutions. Perhaps the simplest is of the form $G(x)=\alpha x+\beta$. Suppose this is the case. Then $$\alpha x+\beta=1+\beta+\alpha(x+1)\Rightarrow \alpha=-1\\ G(0)=1\Rightarrow \beta=1.$$
Thus we may write $G(x)=1-x$ as a solution. Indeed, this satisfies the differential equation. Anyway, if $y=\log_ax$, then $G(y)=F(a^y)=F(x)$. Therefore we can write
$$F(x)=1-\log_{2/3}x.$$