问:
The closure of Banach-valued simple functions
Let $(X, \mathcal{A})$ be a measurable space and $B$ a separable Banach space. Let us call a $B$-valued function $f$ on $X$ a simple function if it is of the form $\sum_{i=1}^n b_i \mathbb{1}_{A_i}$ for some $b_i \in B$ and $A_i \in \mathcal{A}$.

Without imposing any further conditions or endowing the measurable space with a measure, is there a nice characterization of the uniform closure of the class of simple functions? For example, we know that if we were dealing with real valued simple functions, then any bounded $\mathcal{A}$-measurable function could be uniformly approximated by simple functions. Obviously, one can imitate the argument and show that in the Banach-valued setting, any compact range function lies in the uniform closure of simple functions. Can we go further than that? What other functions do we have in the closure?

答:
The range a function in the closure needs not be compact, but it needs to be totally bounded. And it is not difficult to show that the Banach space of totally bounded measurable functions (with the uniform norm) is precisely the uniform closure of the class of simple functions, if you already know how to show that any measurable function with compact range can be uniformly approximated by simple functions. And the assumption of being separable is irrelevant.

Let me add some details. First, to show that the space $\mathcal{M}_{tb}(\mathcal{A};B)$ of $B$-valued (Borel) measurable functions with totally bounded range is closed in the uniform norm topology, let $(f_{n})_{n=1}^{\infty}$ be a sequence in $\mathcal{M}_{tb}(\mathcal{A};B)$ uniformly convergent to a measurable function $f$. Recall that any sequential pointwise limit of Borel measurable functions taking values in a perfectly normal space (in particular, a metric space) should be Borel measurable. To show that the range of $f$ is totally bounded, take any $\epsilon>0$ and choose $n$ such that $\|f_{n}-f\|_{\infty}\leq\epsilon/2$. Then since the range of $f_{n}$ is totally bounded, we can find finitely many $v_{1},\ \cdots\ ,v_{K}\in B$ such that for any $x\in X$, $f_{n}(x)\in B_{\epsilon/2}(v_{k})$ for some $k=1,\ \cdots\ ,K$. Then clearly, any $f(x)$ is in some $B_{\epsilon}(v_{k})$, so the range of $f$ is indeed totally bounded.

To show that any measurable function with totally bounded range can be uniformly approximated by simple functions, take such a function $f$ and let $\epsilon>0$ be given. Then we can find $v_{1},\ \cdots\ ,v_{K}\in B$ such that $B_{\epsilon}(v_{k})$'s cover the range of $f$. Take a measurable partition $\{A_{1},\ \cdots\ ,A_{K}\}$ of $X$ such that $f(x)\in B_{\epsilon}(v_{k})$ for all $x\in A_{k}$, then the simple function $g=\sum_{k=1}^{K}v_{k}\mathbb{1}_{A_{k}}$ is uniformly approximating $f$ within $\epsilon$.


问:
Show that all real number can be approximated by the difference between square root of two different integers.
Let $x \in \mathbb R$, show that for any $\epsilon > 0$, there exists $a,b \in \mathbb Z$ such that:
$$
\left|x - \left(\sqrt{a} - \sqrt{b}\right)\right| < \epsilon
$$

Not sure how to approach this, how do you show the difference between square roots of integers can be arbitrary.

any hints would be helpful!

答:
There are several ways to approach this. Here's one of them:

Hint: For any integers $n, k$, show that there exists integers $a, b$ such that

$$ \frac{k}{2^n } < \sqrt{a} - \sqrt{b} < \frac{k+1}{2^n}. $$

Corollary: Given any $x$, we can first approximate it by $ \frac{k}{2^n}$, then by $ \sqrt{a} - \sqrt{b}$.

Fill in the gaps as needed. If you're stuck, write out your working and thought process to demonstrate where you're at.

Here's another approach.

Hint: For $ x \geq 0$, let $ b_n = n^2$ and $ a_n = \lceil  (n+x)^2 \rceil $. Show that

$$ |x - ( \sqrt{a_n} - \sqrt{b_n} )| < \frac{1}{ n}. $$

(We could get a tighter bound, but this is all that we need.)

ContentLicense: CC BY-SA 4.0


问:
Show that a set is not a complex vector space
Let vector space $W$ be $\{p(z)\in P_{3}(\mathbb{C}):p(a)=0 \Rightarrow a\in \mathbb{R}\}$ under usual addition and usual scalar multiplication.

How to show that W is not a vector space over $\mathbb{C}$?

It should violate one or more of the 10 axioms of complex vector space, but I am stuck here, I tried and it all seems that 10 axioms holds, how can W not be an complex vector space?

答: You mention $10$ axioms of a complex vector space, but you only need one for the vast majority of cases. Assuming scalar multiplication and usual addition we have that $$W \text{ is VS } \iff p, q\in W \implies \alpha p+\beta q \in W$$ for every $\alpha, \beta \in \Bbb C$. You simply need to find two polynomials with real zeros whose combination has a non-real zero, and Jose's answer does exactly this.


问:
Proving that $\int_{0}^{\infty}\frac{\sin^{2n+1}(x)}{x} \ dx=\frac{\pi \binom{2n}{n}}{2^{2n+1}}$ by induction
I need to prove

$$\int_{0}^{\infty}\frac{\sin^{2n+1}(x)}{x} \ dx=\frac{\pi \binom{2n}{n}}{2^{2n+1}}.$$

I've seen other demonstrations of this, but they use some identities that I don't understand. So I asked my calculus teacher and he told me I can prove that by induction. I've already done the case $n=1$, but I can do the rest, could you help me please?

答:
I have a way of showing this directly, using Laplace transforms. In case you are new to this, a Laplace transform of a function $f(t)$ is

$$\hat{f}(s) = \int_0^{\infty} dt \, f(t) \, e^{-s t}$$

We begin with the following interesting formula.  If $\hat{f}$ and $\hat{g}$ are the Laplace transforms of $f$ and $g$, respectively, then

$$\int_0^{\infty} du \, f(u) \hat{g}(u) = \int_0^{\infty} du \, \hat{f}(u) g(u)$$

Here, let $f(t) = \sin^{2 n+1}(t)$ and $\hat{g}(s) = 1/s$.  Hopefully, it is clear that $g(t) = 1$.  The computation of $\hat{f}(s)$, however, is a bit messy, albeit possible:

$$\begin{align} \hat{f}(s) &= \int_0^{\infty} dt \, \sin^{2 n+1}{(t)} \, e^{-s t} \\ &= \frac{1}{(2 i)^{2 n+1}} \sum_{k=0}^{2 n+1} (-1)^k \binom{2 n+1}{k} \int_0^{\infty} dt \, e^{i(2 n+1-2 k)t} e^{-s t}\\&=\frac{1}{(2 i)^{2 n+1}} \sum_{k=0}^{2 n+1} (-1)^k \binom{2 n+1}{k} \frac{1}{s-i(2 n+1-2 k)}\\&= \frac{(-1)^n}{2^{2 n}} \sum_{k=0}^n (-1)^k \binom{2 n+1}{k} \frac{2 n+1-2 k}{s^2+(2 n+1-2 k)^2}\end{align}$$

For that last step, I combined the $k$th and $(n-k)$th terms in the previous sum.  The quantity we seek is

$$\begin{align}\int_0^{\infty} dx \frac{\sin^{2  n+1}{x}}{x} &= \frac{(-1)^n}{2^{2 n}} \sum_{k=0}^n (-1)^k \binom{2 n+1}{k} \int_0^{\infty} du \frac{2 n+1-2 k}{u^2+(2 n+1-2 k)^2}\\&= \frac{(-1)^n \pi}{2^{2 n+1}} \sum_{k=0}^n (-1)^k \binom{2 n+1}{k} \end{align}$$

It turns out that

$$\sum_{k=0}^n (-1)^k \binom{2 n+1}{k} = (-1)^n \binom{2 n}{n}$$

which fortuitously leads us to the stated result:

$$\int_0^{\infty} dx \frac{\sin^{2 n+1}{x}}{x} = \frac{\pi}{2^{2 n+1}} \binom{2 n}{n}$$

ADDENDUM

For a proof of the above sum, we use Knuth's negative index trick:

$$(-1)^k \binom{r}{k} = \binom{k-r-1}{k}$$

along with the well-known analog of a definite integral of a monomial:

$$\sum_{k=0}^n \binom{k+m}{k} = \binom{n+m+1}{n}$$

so that we have

$$\begin{align}\sum_{k=0}^n (-1)^k \binom{2 n+1}{k} &= \sum_{k=0}^n \binom{k-2 n-2}{k}\\&= \binom{n-2 n-1}{n} \\ &= \binom{-(n+1)}{n} \\ &=  (-1)^n\binom{n+n+1-1}{n} \\ &= (-1)^n \binom{2 n}{n}\end{align}$$

ContentLicense: CC BY-SA 3.0


问:
Integrating factor for the ODE $(3x+2y+y^2)dx + (x+4xy+5y^2)dy = 0$
I was asked to find an integrating factor $\mu = \mu(x+y^2)$ for the ODE
$$(3x+2y+y^2)dx + (x+4xy+5y^2)dy = 0.$$

So the natural approach was define $P = 3x+2y+y^2$ and $Q = x+4xy+5y^2.$ Then,
$$\partial_yP = 2+2y,~~~\partial_xQ = 1+4y$$
and hence $\partial_yP - \partial_xQ = 1-2y.$ Let us consider then
$$\frac{\partial_yP-\partial_xQ}{Q} = \frac{1-2y}{x+4xy+5y^2}$$
and define $z = x+y^2.$ Thus,
$$\frac{\partial_yP - \partial_xQ}{Q} = \frac{1-2y}{(z-y^2)(1+4y)+5y^2}.$$

But from here I really don't know how to proceed. Any hint?

Body:
$$(3x+2y+y^2)dx + (x+4xy+5y^2)dy = 0$$
That you can rewrite as:
$$(x+y^2)dx+2(x+y)dx + (x+y^2)dy+4y(x+y)dy = 0$$
$$(x+y^2)d(x+y)+2(x+y)(dx+2ydy) = 0$$
$$(x+y^2)d(x+y)+2(x+y)d(x+y^2) = 0$$
$$\dfrac {d(x+y)}{x+y}+2\dfrac {d(x+y^2)}{x+y^2} = 0$$
Integrate.
$$(x+y)(x+y^2)^2 = C$$


问:
Why are the extensions of valuations in a Galois extension conjugate under the action of the Galois group?
I'm reading the following paper by Hansen about Galois Coverings of Curves: http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.31.9776&rep=rep1&type=pdf (see page 37)
Unfortunately, I'm having problems with the following argument:

Let $K/L$ be a finite Galois extension and let $\nu$ be a valuation of $L$.
$G=\operatorname{Gal}(K/L)$ acts on the valuations $\nu^\prime$ of K by $\sigma(\nu^\prime)=\nu^\prime\circ \sigma$ for all $\sigma\in G$.
We see that $\sigma(\nu^\prime)|_L=\nu^\prime\circ\sigma_L=\nu^\prime|_L=\nu$ for a valuation $v^\prime$ of $K$ with $\nu^\prime|_L=\nu$

Hansen argues that, since $G$ consists of all automorphisms of $K$ fixing $L$, the valuations $\{\nu^\prime\text{ of }K: \nu^\prime|_L=\nu\}$ are conjugate under $G$'s action.
I don't understand why this follows from the above. I would be very thankful if you could give me a hint.

答:
For $K/L$ a (finite) Galois extension of any kind of fields, $w$ a valuation on $L$ and $v_j$ (the equivalence classes) of valuations on $K$ extending $w$ . There are finitely many such valuations.

$G=Gal(K/L)$ permutes the $v_j$ as $v_j \circ \sigma$ is another valuation above $w$.

The main claim is that there is $\pi_1\in K$ such that $v_1(\pi_1)>0,v_j(\pi_1)=0$.

Since $K/L$ is Galois then $N_{K/L}(\pi_1)=\prod_{\sigma\in G} \sigma(\pi_1)\in L$.

$$w(N_{K/L}(\pi_1))=v_1(N_{K/L}(\pi_1))=\sum_{\sigma\in G}v_1 \circ \sigma(\pi_1)\ge v_1(\pi_1) > 0$$

Whence for all $j$, $v_j(N_{K/L}(\pi_1))> 0$ so that $v_j \circ \sigma(\pi_1)>0$ for some $\sigma$ which implies that $v_1 = v_j\circ \sigma$.

If $w$ is a discrete valuation then once we know that there are finitely $v_j$ the main claim follows easily. The $v_j$ are discrete valuations because $[L:K]$ is finite. Look  at $S_1 = \{ a\in L, v_1(a)=0,\forall j, v_j(a)\ge 0\}$. For each $j$, if it exists take some $a_{j,1}\in S_1,v_j(a_{j,1})>0$ and let $A_1 = \prod_j a_{j,1}$. If some $a_{i,1}$ doesn't exist then take some element $b\in L,v_1(b)>v_i(b)$. Letting $c=bA_1^n$, for $n$ large enough then the minimum valuation of $c$ is at some $v_m$ such that $a_{m,1}$ doesn't exist. Taking $k\in K,w(k)\ne 0$ then for some $r> 0,s$, $r v_m(c) =s v_m(k)$,
$d=c^rk^{-s} \in S_m$ and $a_{m,1} = 1-d$ exists. Thus in fact all the $a_{m,1}$ exist and $\pi_1=1-A_1$ works in the main claim.


问:
Composite of continuous functions is continuous
I would like some feedback and corrections to the proof below.

Let $X$ and $Y$ be subsets of $\mathbb{R}$. Let $f\colon X\to \mathbb{R}$ a continuous functions at $x_0\in X$. Suppose $f(X)\subset Y$ and that $g\colon Y\to \mathbb{R}$ a continuous functions at $y_0=f(x_0)$. Show that $h:=g\circ f\colon X\to \mathbb{R}$ is a continuous function at $x_0 \in X$ using $\varepsilon$-$\delta$.

We have that $f\colon X\to\mathbb{R}$ is continuous at $x_0\in X$, that is, for all $\varepsilon_f > 0$ there is a $\delta_f > 0$ such that

$$x\in X,\ \left|x-x_0\right| < \delta_f \implies \left|f(x)-f(x_0)\right| < \varepsilon_f$$

We also have that $g\colon Y\to \mathbb{R}$ is continuous at $y_0=f(x_0)$, that is, for all$\varepsilon_g > 0$ there is a $\delta_g > 0$ such that

$$y\in Y, \left|y-y_0\right| < \delta_g \implies \left|g(y)-g(y_0)\right| < \varepsilon_g \iff  \left|g(y)-g(f(x_0))\right| < \varepsilon_g$$

As $f(X)\subset Y$ and we've defined a function $h := g\circ f\colon X\to \mathbb{R}$, we create a restriction $g|_{f(X)}\colon f(X) \to \mathbb{R}$.

Knowing that every restriction of a continuous function is continuous and that $x_0 \in f(X)$, then $g|_{f(X)}$ is continuous, that is, for all $\varepsilon > 0$ there is a $\delta > 0$ such that

$$f(x)\in f(X)\subset Y, \left|f(x)-f(x_0)\right| < \delta \implies \left|g(f(x)) - g(f(x_0))\right| < \varepsilon$$

Therefore, the function $h$ is continuous at $x_0$. $\blacksquare$

答: That is not correct. What you are supposed to prove is not that final expression. It is: for every $\varepsilon>0$, there is a $\delta>0$ such that$$(\forall x\in X):|x-x_0|<\delta\implies\bigl|g\bigl(f(x)\bigr)-g\bigl(f(x_0)\bigr)\bigr|<\varepsilon.\tag1$$Given $\varepsilon>0$, take $\delta'>0$ such that$$(\forall y\in Y):\bigl|y-f(x_0)\bigr|<\delta'\implies\bigl|g(y)-g\bigl(f(x_0)\bigr)\bigr|<\varepsilon$$and take $\delta>0$ such that$$(\forall x\in X):|x-x_0|<\delta\implies\bigl|f(x)-f(x_0)\bigr|<\delta'$$and then $(1)$ will hold.


问:
Inverses where $f(g(x))=x\neq g(f(x))$
If $f: \mathbb{Z\to Z}$ and $G: \mathbb{Z\to Z}$, find $f$, and $g$ such that $f(g(x))=x\neq g(f(x))$.

I can find lots of $f$ and $g$ that aren't equal when composed with each other, but I have no idea how to proceed. A hint would be helpful.

Thanks in advance.

答: Hint: If for all $x$ in the domain of $g$, $f(g(x))=x$, then $g$ is one-to-one, and $f$ is onto. Conversely, if $g\colon S\to T$ is one-to-one, then there is a function $f\colon T\to S$ with $f(g(x))=x$ for all $x\in S$. And also, if $f\colon T\to S$ is onto, then there is $g\colon S\to T$ with $f(g(x))=x$ for all $x\in S$. All of these statements are fairly easily proved except the last, which is equivalent to the Axiom of Choice. So, as @CutieKrait has said, you need a one-to-one function that’s not onto for the one of them, and its “semi-inverse” to be onto but not one-to-one.

Title: Inverses where $f(g(x))=x\neq g(f(x))$


问:
Integral equations
Are there functions $f(z)$ that satisfy the following:

$$\int_1^\infty \frac{\mathrm{d}z}{f(z)}  = \frac{1}{\int_{1}^{\infty} f(z) \, \mathrm{d}z} $$

答:
Partial Answer: There are no solutions for $f$ continuous.

First we can say WLOG that $f>0$ in $[1,\infty]$ (since $f$ has to be of constant sign) and so therefore for both integrals to converge we have, on the LHS and RHS respectively:
$$\lim_{x\to\infty}\frac{1}{f(x)}=0\quad \text{and}\quad \lim_{x\to\infty}f(x)=0$$

which is a contradiction.

Seems much harder for discontinuous functions...

Title: Integral equations


问:
Angles and ellipse (proof)
First of all, sorry for my poor English! Can you please help me? I'm trying to prove that, given a point P at an ellipse.

Please help me prove that the angles are equal.

Thanks!

答: It is not true for any point $P$. See the image:

Title: Angles and ellipse (proof)


问:
Find area of right triangle △ with circumcircle $O$ and tangent circle $K$.
The triangle $ABC$, with $\angle A = 90^\circ$,
is inscribed in the circle $\mathcal{O}$.
The circle $\mathcal{K}$ is tangent to the circle $\mathcal{O}$,
the sides $AB$ and $AC$.
Suppose that the radius of $\mathcal{O}$ is $3$
and the radius of $\mathcal{K}$ is $2$. Find the area of $\triangle ABC$.

I drew this problem out by making circle $\mathcal{K}$
internally tangent to circle $\mathcal{O}$.
I was able to make it tangent to $AB$ and $AC$.
However, I have gotten nowhere using trigonometry to solve this.

I also tried to name the sides and angles
to find the area, which did not work, either

答:
Start with arbitrary point $L$ on the circle $\mathcal{O}(O,R)$
and draw the circle $\mathcal{K}(K,r_k)$,
tangent to $\mathcal{O}(O,R)$
at $L$.

Let $L=(0,-R)$, $K=(0,-R+r_k)$.

Since $\mathcal{K}$ is inscribed in $\angle CAB$,
and $\angle CAB=90^\circ$, the distance $|KA|=r_k\sqrt{2}$,
so we can find the point $A$ as the intersection of
the $\mathcal{O}$
and the circle centered at $K$ with the radius $r_k\sqrt2$.

$AK$ is the bisector of $\angle CAB$, hence
the tangent point $D= AB\cap \mathcal{K}$
can be easily found, as $\triangle ADK$ is isosceles.

When $R=3$, $r_k=2$, $\triangle OKA$ is right angled,
since in this case $|OK|^2+|KA|^2=|OA|^2=R^2=9$,
which means that

\begin{align}
|AD|=|KD|&=r_k=2
,\\
|OE|&=\frac{r_k}{\sqrt2}=\sqrt2
,\\
|DE|&=\frac{r_k}{\sqrt2}-(R-r_k)=\sqrt2-1
,\\
|OD|&=
\sqrt{R^2-(2+\sqrt2)\,r_k\,(R-r_k)}
=\sqrt{5-2\sqrt2}
.
\end{align}

Then consider $\triangle ABO$,
for which we already know values of $AD$, $|OD|$
and also that $|OB|=|OA|=3$, so we can apply
the Stewart’s theorem

\begin{align}
\triangle ABO:\quad
&|OB|^2\cdot|AD|+|OA|^2
\cdot(|AB|-|AD|)
\\
&=
|AB|\cdot(|OD|^2+|AD|\cdot(|AB|-|AD|))
,
\end{align}

which gives the unique solution: $|AB|=4+\sqrt2$.
Then
\begin{align}
|AC|&=\sqrt{4R^2-|AB|^2}
=4-\sqrt2
\end{align}

and the sought area then is

\begin{align}
S_{ABC}&=\tfrac12\,|AB|\cdot|AC|=
\tfrac12(16-2)=7
.
\end{align}


问:
Single source shortest paths problem
Assume a directed graph with positive integer values for edge weights. Also, we know that the cost of the longest shortest path is $K$.

The objective is to construct an algorithm that finds the shortest paths from a source node $s$ to all the other nodes in $(K + V + E)$ time.

I cannot see how the cost of the longest shortest path can be used by the algorithm to produce the required shortest paths. Can someone provide me with a hint?

答:
Here are a few hints. I suggest you think about each hint before moving on to the next one, but it's really up to you.

Hint 1:

 The distance from $s$ to an arbitrary node can only take on $K + 1$ different values.

Hint 2:

 Make a modification of Dijkstra's algorithm that runs in $O(K + V + E)$ time and uses $O(K + V + E)$ memory.

Hint 3:

 Can you think of an algorithm to sort $n$ integers in the range $\{0,1,\ldots,k\}$ using $O(n + k)$ time and $O(n + k)$ memory?

Title: Single source shortest paths problem


问:
$\pm$ in solving differential equations by separation of variables
I know how to do separation of variables with differential equations, like $\frac{dy\left(x\right)}{dx}=4y\left(x\right)$ when $y\left(0\right)=4$ will get me $y=4e^{4x}$ in the end.

But when I'm solving for y, once I get to the stage where $\left|y\left(x\right)\right|=e^{4x} e^{C_1}$, do I replace as $C=\pm e^{C_1}$ or is that $\pm$ unnecessary?

Generally, it seems unnecessary since $e^x$ is always positive but couldn't it be the case that, if I had $y\left(0\right)=-4$, then I'd need to have that negative sign to get $y=-4e^{4x}$

答:
$$y'=4y$$
With integrating factor method:
$$y'-4y=0$$
$$(ye^{-4x})'=0$$
$$ye^{-4x}=C$$
$$y(x)=Ce^{4x}$$
With $C \in \mathbb{R}$. You don't really need the $\pm $.


问:
Finding all normal subgroups of a group
On my homework today, we had to find all the normal subgroups of $D_{n}$, the dihedral group of order 2n. I solved the problem by looking at how the conjugacy classes change based on whether n is even or odd and then constructed the normal subgroups as unions of the conjugacy classes.

I have 2 questions:

(i) Is there a better way to approach the problem than looking at the conjugacy classes?

(ii) Could someone explain why we want to find all the normal subgroups of a particular group?   How does this provide us additional insight into the structure of the group we are studying? (Right now, this exercise feels more like a 'computation' to me than a way of understanding $D_{n}$)

Thanks :)

答:
Finding normal subgroups of a group is a first step towards understanding how it "factors" into simpler groups, analogous to finding prime factors of a number.  To understand a group it then suffices to understand i) the pieces it's built out of (the simple groups in its composition series) and ii) how those pieces fit together.

In your case it is also just a good exercise in becoming familiar with groups and with basic constructions associated to groups.  Also, like Weltschmerz says, you should think of knowing the normal subgroups as the same thing as knowing all the quotient groups of a group, and the quotient groups of a group are like its "shadows" - they reveal some, but not all, of the structure of the group, and putting them together one gets a picture of the whole thing.

ContentLicense: CC BY-SA 2.5

Title: Finding all normal subgroups of a group


问:
Proof of inequality
I have problems with proving inequality :

$${a^{2}}+b^2+c^2+\frac{2}{5}abc<50$$

where $a,b,c$ are the lengths of triangle's sides, and the circumference of the triangle is $10$.

Thanks.

答:
Consider the polynomial $(x-a)(x-b)(x-c)$. Multiplied out, this is

$$
\begin{eqnarray}
(x-a)(x-b)(x-c)
&=&
x^3-(a+b+c)x^2+(ab+ac+bc)x-abc
\\
&=&
x^3-10x^2+(ab+ac+bc)x-abc\;.
\end{eqnarray}
$$

We also have

$$10^2=(a+b+c)^2=(a^2+b^2+c^2) + 2(ab+ac+bc)\;,$$
$$ab+ac+bc=\frac{100-(a^2+b^2+c^2)}{2}\;,$$

and thus

$$
(x-a)(x-b)(x-c)
=
x^3-10x^2+\frac{100-(a^2+b^2+c^2)}{2}x-abc\;.
$$

Now since $a$, $b$ and $c$ form a triangle with perimeter $10$, they must all be less than $5$. Thus the value of the polynomial for $x=5$ is positive, that is,

$$
5^3-10\cdot5^2+\frac{100-(a^2+b^2+c^2)}{2}\cdot5-abc>0\;,
$$

which upon rearrangement becomes your inequality.

Title: Proof of inequality


答:
You may want to introduce your integrating factor at an earlier stage of the process.  Put it into the partial derivative equation:

$\partial_y(P\mu)=\partial_x(Q\mu)$

From the product rule for differentiation:

$\partial_yP(\mu)+P\partial_y(\mu)=\partial_xQ(\mu)+Q\partial_x(\mu)$

To apply this equation given $\mu=\mu(z),z\equiv x+y^2$ we need to convert the partial derivatives $\partial_x\mu, \partial_y\mu$ into the total derivative $\mu'=d\mu/dz$.  This is done with the Chain Rule, thus

$\partial_y\mu=\partial_y(x+y^2)(\mu')=2y\mu'$

$\partial_x\mu=\text{reader fills that in}$

With the partial derivatives of $\mu$ thus resolved, we substitute in $P,Q$ and their derivatives and we should get

$2y\mu'(3x+2y+y^2)+\mu(2+2y)=\mu'(x+4xy+5y^2)+\mu(1+4y)$

We should be able to rearrange this to get

$\mu'\color{blue}{(2y^3-y^2+2xy-x)}-\mu(2y-1)=0$

Then after factoring the expression in blue ultimately get an ordinary differential equation containing only $\mu$ and $z$, which is easy to solve:

$\mu'z-\mu=0$

and plug $z=x+y^2$ into the identified solution.

I checked the result directly and the partial derivative equation is satisfied with the solution I chose for the ordinary differential equation.

Title: Integrating factor for the ODE $(3x+2y+y^2)dx + (x+4xy+5y^2)dy = 0$

Body:
I was asked to find an integrating factor $\mu = \mu(x+y^2)$ for the ODE
$$(3x+2y+y^2)dx + (x+4xy+5y^2)dy = 0.$$

So the natural approach was define $P = 3x+2y+y^2$ and $Q = x+4xy+5y^2.$ Then,
$$\partial_yP = 2+2y,~~~\partial_xQ = 1+4y$$
and hence $\partial_yP - \partial_xQ = 1-2y.$ Let us consider then
$$\frac{\partial_yP-\partial_xQ}{Q} = \frac{1-2y}{x+4xy+5y^2}$$
and define $z = x+y^2.$ Thus,
$$\frac{\partial_yP - \partial_xQ}{Q} = \frac{1-2y}{(z-y^2)(1+4y)+5y^2}.$$

But from here I really don't know how to proceed. Any hint?


问:
Path integral and vector fields
I have that $f(x,y) = (-y,x)$, and $\gamma(t)=(\cos(t),\sin(t)),$ where $t\in I=[0,2\pi]$. The function $f$ maps a vector to each point in the $(x,y)$ plane. The result is a vector field, which curls clockwise and is such that the magnitude of each vector increases as we move away from the origin.

The continuous mapping $\gamma$ is a parametrization for the unit circle ( a curve ), which lies in this same $(x,y)$ plane. I think this is correct. But what is the relationship/significance between $f$ and $f\circ \gamma$? ( They should form a vector field of sorts ) The relevant path integral is $$\int_\gamma f dt=\int_I\langle(-\sin(t),\cos(t)),(-\sin(t),\cos(t)\rangle dt.$$

A path integrals determines the area of the function under the curve, or something. What does this path integral tell us?

答:
The path is a unit circle $\vec{\gamma}(t) = (\cos t, \sin t)$ or $(x^2 + y^2 = 1)$

Vector field $\vec{F} = (-y, x)$ or $(-\sin t, \cos t)$.

Let's take a point on the circle, $(1, 0)$. Our vector field at that point is $(0,1)$ pointing upward and tangent to the circle. You can check similarly at other points that the vector field is tangent to the circle. So it is a rotational vector field (say some velocity field for example).

Now $\vec{\gamma} \,'(t) = (-\sin t, \cos t)$ which is the gradient of the curve and slope of the tangent in $2D$. You can see that our vector field is working in the same direction.

So if you see the line integral, it becomes

$\displaystyle \int_\gamma \vec{F}(\gamma(t))\cdot \vec{\gamma} \,'(t) dt =  \int_{0}^{2\pi} (-\sin t, \cos t) \cdot (-\sin t, \cos t) dt =  \int_{0}^{2\pi} 1 \, dt$

and which is the magnitude of our rotational vector field (unit) working on the unit circle.

Title: Path integral and vector fields


答:
I will assume that $f > 0$. Then by the Cauchy-Schwarz inequality, for $a > 1$,

$$ \left( \int_{1}^{a} 1 \, \mathrm{d}x \right)^2
= \left( \int_{1}^{a} f(x)^{1/2} f(x)^{-1/2} \, \mathrm{d}x \right)^2
\leq \left( \int_{1}^{a} f(x) \, \mathrm{d}x \right)\left( \int_{1}^{a} \frac{1}{f(x)} \, \mathrm{d}x \right). $$

So by letting $a \to \infty$,

$$ \left( \int_{1}^{\infty} f(x) \, \mathrm{d}x \right)\left( \int_{1}^{\infty} \frac{1}{f(x)} \, \mathrm{d}x \right) = \infty. $$

In other words, it is impossible to make both integrals simultaneously finite.

If we allow sign changes and interpret both integrals in improper integral sense, then we actually have an example.

In order to describe the example, for each bounded open interval $I = (a, b)$ we define

$$ \phi_{I}(x) = \frac{\sqrt{(x-a)(b-x)}}{b-a}. $$

Then it is not hard to check that

$$ \int_{I} \phi_I(x) \, \mathrm{d}x = \frac{\pi}{8}(b-a) \qquad\text{and}\qquad \int_{I} \frac{1}{\phi_I(x)} \, \mathrm{d}x = \pi(b-a). $$

Now let $H_n = 1 + \frac{1}{2} + \cdots + \frac{1}{n} = \sum_{k=1}^{n} \frac{1}{k} $ denote the $n$th harmonic number, and define

$$ f(x) = \begin{cases}
(-1)^{n-1} A \phi_{(H_n, H_{n+1})}(x) & \text{if $x \in (H_n, H_{n+1})$ for some $n$}, \\
0, & \text{otherwise},
\end{cases} $$

where $A$ is a non-zero constant to be determined later. Then it follows that

$$ \int_{1}^{\infty} f(x) \, \mathrm{d}x = \frac{\pi A}{8} \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n+1} = \frac{\pi}{8}(1-\log 2) A $$

and

$$ \int_{1}^{\infty} \frac{1}{f(x)} \, \mathrm{d}x = \pi A \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n+1} = \pi (1-\log 2) A. $$

So by solving the equation

$$ \left( \frac{\pi}{8}(1-\log 2) A \right) \left( \pi (1-\log 2) A \right) = 1, $$

we find that $f$ with $A = \frac{\sqrt{8}}{\pi(1-\log 2)}$ satisfies the given condition.

Remark. A similar construction should work if we begin with some function $\phi$ over an interval $I$ such that both $\int_{I} \phi$ and $\int_{I} \frac{1}{\phi}$ converge.

Title: Integral equations

Body:
Are there functions $f(z)$ that satisfy the following:

$$\int_1^\infty \frac{\mathrm{d}z}{f(z)}  = \frac{1}{\int_{1}^{\infty} f(z) \, \mathrm{d}z} $$


问:
Proof that a well-ordering is $\omega$-saturated only if it is finite
I wanna prove that an infinite well-ordering is not $\omega$-saturated. How can I do this? Does the proof involve Lowenheim-Skolem theorem?

答:
A useful fact is that an $\aleph_0$-saturated structure realizes any (partial) type in $\omega$-many variables with parameters from a finite set. Ie, if $\mathfrak{M}$ is an $\aleph_0$-saturated $\mathcal{L}$-structure, $A\subseteq M$ is a finite subset, and $\Sigma(v_i)_{i\in\omega}$ is a finitely satisfiable set of $\mathcal{L}_A$ formulas in $\omega$-many free variables, then $\Sigma(v_i)_{i\in\omega}$ is realized in $\mathfrak{M}$. See eg here for a proof. (The analogous result holds for all infinite cardinals.)

This fact allows us to show your result. Let $(M,\leqslant)$ be an $\aleph_0$-saturated well-order, and consider the collection of formulas $\Sigma(v_i)_{i\in\omega}=\{v_i>v_j\}_{i<j\in\omega}$; in other words, $\Sigma(v_i)_{i\in\omega}$ expresses that the $v_i$ form a strictly descending chain. Since well orders are total orders, if $M$ is infinite then $\Sigma(v_i)_{i\in\omega}$ is finitely satisfiable, and hence by $\aleph_0$-saturation is satisfied by elements $m_0>m_1>\dots$ of $M$. But now the set $\{m_i\}_{i\in\omega}$ is a subset of $M$ without a least element, contradicting that $<$ is a well-order.


问:
Is this operator bounded in $C^1[0,1]$?

Let $Ax(t)=x(\sqrt{t})$ in $C^1[0,1]$ with norm $\|x\|_1=\max\limits_{t\in[0,1]}|x(t)|+\max\limits_{t\in[0,1]}|x'(t)|$. It is required to check it for boundedness.

All I know is that the operator is not defined on the whole space, but is defined on the set of functions that satisfy the condition $\lim\limits_{t\to+0}\dfrac{x'(t)}{t}=A\in\mathbb{R}.$

I can't get an estimate like $\|Ax\|\leq c\|x\|$, but I also can't find the sequence $x_n(t)$ in unit ball, such that $\|Ax_n(t)\|\to\infty.$

I will be grateful for any help or hint on how to proceed.

答:
I believe $A$ is unbounded. We have

$$\|Af\| = \sup_{x\in[0,1]} |f(\sqrt{x})| + \sup_{x\in[0,1]} \left|f'(\sqrt{x})\cdot \frac1{2\sqrt{x}}\right| = \sup_{x\in[0,1]} |f(x)| + \frac12\sup_{x\in[0,1]} \left|\frac{f'(x)}x\right|$$

and in general we cannot dominate the last expression by $\|f\|_\infty$ and $\|f'\|_\infty$, even if $\lim_{x\to0^+} \frac{f'(x)}x$ is finite.

For $n \ge 2$ define $f_n(x) = -\cos(n\pi x)$. Then $f_n \in C^1[0,1]$ and $$f_n'(x)=n\pi \sin(n\pi x) \implies \lim_{x\to 0^+} \frac{f_n'(x)}{x} = \lim_{x\to 0^+} \frac{n\pi \sin(n\pi x)}{x} = n^2\pi^2$$
so $f_n$ is admissible. We have
$$\sup_{x\in[0,1]} |f_n(x)| = 1, \quad \sup_{x\in[0,1]} |f_n'(x)| = f_n'\left(\frac2{n\pi}\right) = n\pi, \quad \sup_{x\in[0,1]} \left|\frac{f_n'(x)}x\right| = n^2\pi^2$$

and hence $$\|f_n\| = 1+n\pi, \quad \|Af_n\| = 1+\frac12n^2\pi^2 \implies \frac{\|Af_n\|}{\|f_n\|} = \frac{1+\frac12n^2\pi^2}{1+n\pi} \xrightarrow{n\to\infty} +\infty.$$

Title: Is this operator bounded in $C^1[0,1]$?


问:
Is $\mathcal{U}(\mathfrak{g})$ semisimple (as a module over itself)? (and related examples)
Recall that an object in an abelian category is semisimple if it is a (finite) direct sum of simple objects. An abelian category is semisimple if every object is semisimple.

In studying representations of semisimple lie algebras, one often restricts to the subcategory called $\mathcal{O}_{int}$ inside the category left modules for the universal enveloping algebra, $\mathcal{U}(\mathfrak{g})$.  The main theorem about $\mathcal{O}_{int}$ says it is semisimple (or, at least, that all the objects in $\mathcal{O}_{int}$ are semisimple. Perhaps one needs to check several other things to see that $\mathcal{O}_{int}$ is abelian as a subcategory).

I am new to this area and would like to motivate this restriction.

A ring $R$ is semisimple in the category of left $R$ modules iff the category of left $R$ modules is semisimple. Thus,

Is $\mathcal{U}(\mathfrak{g})$ semisimple as a left $\mathcal{U}(\mathfrak{g})$-module (for $\mathfrak{g}$ a semisimple Lie algebra)?

As a bonus, are there other nice examples of $\mathcal{U}(\mathfrak{g})$-modules that aren't semisimple?

答:
Here is an alternate, somewhat indirect approach. The Artin-Wedderburn theorem completely classifies semisimple rings as finite products $\prod_i M_{n_i}(D_i)$ of matrix rings over division rings. The PBW theorem implies that $U(\mathfrak{g})$ has no nontrivial zero divisors (see, for example, this math.SE answer), but a semisimple ring $\prod_i M_{n_i}(D_i)$ has nontrivial zero divisors unless it's a division ring $D$.

$U(\mathfrak{g})$ is never isomorphic to a division ring unless $\mathfrak{g} = 0$; this is because it has a counit map $U(\mathfrak{g}) \to k$ to the underlying field, so can only be a division ring if this map is an isomorphism.


问:
Derivation of M step in EM model with GMM
I follow these notes, specifically page 9, about the EM algorithm. We assume observations $\{x^{1},\ldots,x^{n}\}$ come from a mixture of $k$ gaussians with parameters $\{\mu_j,\Sigma_j\}_{j=1}^{k}$, with a latent variable model, where for each observation $x^i$ there's a r.v. $z^i\sim\text{Mult}(\phi)$. We iteratre through two steps: in the first step, we compute $$w^{i}_j:=p(z^{i}=j|x^i;\phi,\mu,\Sigma)$$ In the second step we take derivative of the following function, called ELBO, w.r.t.  $\phi, \mu,\Sigma$
$$\sum_{i=1}^{n}w_j^{i}*\log(\frac{\mathcal{N(x^i;\mu_j,\Sigma_j)*\phi_j}}{w^i_j})$$

My problem is that when taking the derivative, we treat the $w^i_j$ as constant, and not as dependent on the parameters $\phi, \mu,\Sigma$. Here's a calculation. So for example, $\frac{\partial}{\partial\mu_j}w^i_j\cdot\mu_j=w^i_j$, where I understand that $w^i_j$ is a function of $\phi, \mu,\Sigma$, and hence we must use the chain rule and get a more complicated result. The statement that "we computed $w^i_j$ at a previous step and therefore it is constant" seems wrong, since we might as well have plugged in the definition of $w^i_j$ into the function we're taking derivative of, instead of precomputing it. How does breaking up the computation into two steps allows us to take simpler derivatives?

Note - my problem isn't specific at all to EM, but it arose in this context, and can be stated in simpler terms.

答:
The notes are correct - the posteriors/conditional distributions $w_j^i$ can be treated as constants in the M-step, and there is no need for the chain rule in what you have outlined above.

The problem you have outlined can be solved by focusing on how EM works algorithmically. To emphasise this, I'm going to amend what you've written to make the role of each iteration more explicit.

Start with current estimates of the the parameters $\theta^{(t)} = \{ \phi^{(t)}, \mu^{(t)}, \Sigma^{(t)} \}$, which in the case $t=0$ is an initial guess.

E-step: Compute

$$w_j^{(i)} = Q_i(z^{(i)} = j) = P(z^{(i)} = j | x^{(i)}; \phi^{(t)}, \mu^{(t)}, \Sigma^{(t)})$$

Notice that in the E-step you are the "computing the probability of $z^{(i)}$ taking the value $j$ under the distribution $Q_i$", conditional on observed values $x^{(i)}$ and fixing the values of the parameters at current parameter estimates $\phi^{(t)}, \mu^{(t)}, \Sigma^{(t)}$.

M-step: Maximise the following with respect to $\phi, \mu, \Sigma$ to yield new estimates $\theta^{(t+1)} = \{\phi^{(t+1)}, \mu^{(t+1)}, \Sigma^{(t+1)} \}$:

$$\begin{align}
\phi^{(t+1)}, \mu^{(t+1)}, \Sigma^{(t+1)} &= \underset{\phi, \mu, \Sigma}{\text{argmax}}
\sum^n_{i=1} \sum^k_{j=1} w_j^{(i)} \log \frac{\mathcal{N}(x^{(i)}; \mu_j, \Sigma_j) \cdot \phi_j}{w_j^{(i)}} \\
&= \underset{\phi, \mu, \Sigma}{\text{argmax}} \sum^n_{i=1} \sum^k_{j=1} P(z^{(i)} = j | x^{(i)}; \phi^{(t)}, \mu^{(t)}, \Sigma^{(t)}) \log \frac{\mathcal{N}(x^{(i)}; \mu_j, \Sigma_j) \cdot \phi_j}{P(z^{(i)} = j | x^{(i)}; \phi^{(t)}, \mu^{(t)}, \Sigma^{(t)})}
\end{align}$$

In this step, you now allow the parameters $\phi, \mu, \Sigma$ to vary, and you choose them to maximise the the above quantity. However, the parameters are only allowed to freely vary everywhere (i.e. in the numerator of the fraction) other than in the conditional distribution $w^i_j$, which is held fixed at the values of the parameters $\phi = \phi^{(t)}, \mu = \mu^{(t)}, \Sigma = \Sigma^{(t)}$ from the previous step (and also computed using your observations $x^{(i)}$). Therefore $w^i_j$ does not contain $\phi, \mu, \Sigma$ as arguments, and hence there is no need for the chain rule.

Only the numerator of the fraction, that is $\mathcal{N}(x^{(i)}; \mu_j, \Sigma_j) \cdot \phi_j$, which is evaluated at $x^{(i)}$ will contain $\phi, \mu, \Sigma$ as arguments, which are treated as unknown variables, and need to be estimated. In order to set $\phi, \mu, \Sigma$ for maximisation, you now compute the derivative with respect to $\phi, \mu, \Sigma$, set to 0 and solve to yield $\theta^{(t+1)} = \{\phi^{(t+1)}, \mu^{(t+1)}, \Sigma^{(t+1)} \}$.


问:
Triangle Inequation proof
I have to show for all x,y ∈ ℝ that ||x| - |y| ≤ |x-y|

I know that |x| = |x - y + y| ≤ |x - y|+|y|

Does it suffice to put in numbers now or how would I proof that?

答:
We use the fact that

$$a\le b \; and \; -a\le b \;\implies$$
$$-b\le a\le b \implies |a|\le |b|$$

So, as you wrote
$$|x|=|x-y+y|\le |x-y|+|y| \implies $$

$$|x|-|y|\le |x-y|$$
and by symetry,
$$|y|-|x|\le |x-y|$$

take $ a=|x|-|y|\; and \; b=|x-y|$.

Title: Triangle Inequation proof


答:
The class of well-orderings has a very nice property: it's exactly the class of structures which $(i)$ satisfy a particular first-order theory (the theory of linear orders) and $(ii)$ do not have a certain countable "bad configuration." Re: $(ii)$, the relevant "bad configuration" is:

 An infinite descending sequence.

The "saturation slogan" is, "all possible configrations exist." Of course it's more complicated than that - how much saturation you have determines the size of the conigurations the slogan applies to - but the basic idea is sound. This suggests the following line of attack: show that in an infinite $\omega$-saturated linear order, the countable "bad configuration" above must occur. This will be a corollary of a more general fact:

 Suppose $\mathcal{M}$ is $\omega$-saturated and $p$ is a finitely satisfiable $\omega$-type over $\mathcal{M}$. Then $p$ is realized in $M$.

Title: Proof that a well-ordering is $\omega$-saturated only if it is finite

Body: I wanna prove that an infinite well-ordering is not $\omega$-saturated. How can I do this? Does the proof involve Lowenheim-Skolem theorem?


问:
Specify a bijective function $f:[0,1] \longrightarrow [0,1)$
Specify a bijective function $f:[0,1] \longrightarrow [0,1)$

I know that such a function can't be continuous, so I've tried to find a function with one or more discontinuities, but with no progress so far.

答:
Notice that it's fairly east to find a bijection between two countable infinite set.
Now simply choose any countable infinite subset of $[0,1)$, put the extra end $1$ to $A$ to get a new set $B$, and you can have a bijective map from $A$ to $B$. Keep other elements unchanged and you just found a bijection form $[0,1)$ to $[0,1]$.


问:
How to change from a vector parametric description to a cartesian when describing a 2 dimensional flat in 5 dimensions?

I'm completely stuck on how to start this question; I'm not sure If I can get one cartesian to describe this flat or if I need multiple?

答:
We are in a $5$-dimensional vector space $V$ with two free parameters, so our vector subspace $U$ has dimension $2$.

If we now want to obtain its cartesian representation we need to find the scalar parameters $s$ and $t$ in function of the variables $v,w,x,y,z$ in order to obtain a system of three equations that describes a plane in the vector space $V$.
$$\begin{cases}
  v = 1+s \\
  w= 2+s \\
  x =3\\
  y=4+t\\
  z=5 +t
\end{cases}
\iff
\begin{cases}
  s = v-1 \\
  w= 2+s \\
  x =3\\
  t=y-4\\
  z=5 +t
\end{cases}
\iff
\begin{cases}
  s = v-1 \\
  t= y-4 \\
  x =3\\
  z=5+y-4\\
  w=2+v-1
\end{cases}
\Longrightarrow
\pi\equiv\begin{cases}
  x-3=0 \\
  y-z+1=0 \\
  v-w+1=0
  \end{cases}$$


问:
$\nabla_C f(Cu)$ where $C$ is a matrix
Let $f: \mathbb{R}^n \to \mathbb{R}$.  I am trying to find the
\begin{align}
\nabla_C f(Cu)
\end{align}
where $C \in \mathbb{R}^{n \times k}$ and $u\in \mathbb{R}^k$.  In other words, we are trying to find the derivative with respect to the matrix. This  question is about chain rule for matrix differentiation.

What I tried:

I found some links here but I couldn't really follow everything.   It appears to suggest that
\begin{align}
\frac{ \partial f(Cu)}{\partial C_{ij}}={\rm tr} \left( \nabla f(Cu) \left( \frac{\partial Cu}{\partial C_{ij}} \right)^T \right)
\end{align}

答:
$\def\p#1#2{\frac{\partial #1}{\partial #2}} \def\e{\eqalign}$
Define the $(x,g)$ vectors as
$$\e{
x &= Cu \quad\implies\quad &dx = dC\,u \\
g &= \p{f}{x} &\big({\rm gradient\,of\,}f{\rm\,wrt\,}x\big) \\
}$$
and introduce a convenient product notation for the trace using a colon, i.e.
$$A:B = {\rm Tr}(A^TB)
 \;\doteq\; \sum_{i=1}^m\sum_{j=1}^n A_{ij}B_{ij}
\;=\; B:A$$
This product is also defined for vectors; just treat them as rectangular matrices.

Write the differential in terms of $x$, then change the independent
variable from $x\to C$.
$$\e{
df &= g:dx \\&= g:dC\,u \\&= gu^T:dC \\
\p{f}{C} &= gu^T &\big({\rm gradient\,of\,}f{\rm\,wrt\,}C\big) \\\\
}$$
As you have discovered, the difficulty with using the chain rule in matrix calculus is that the intermediate quantities are often third and fourth order tensors, which are difficult to work with.

The nice thing about the differential approach is that the differential of a matrix acts like a matrix. In particular, it follows all of the rules of matrix algebra.

And using the trace (aka the colon product) eliminates a whole category of transposition errors, which arise with other methodologies.

Title: $\nabla_C f(Cu)$ where $C$ is a matrix


问:
Finding the x value after a matrix multiplication?
I have the following solution of a problem, and I was wondering about a hopefully quite simple thing in it:

I was wondering how do they get from [5,10,5] to 5x? I am pretty sure there is a simple explaination for that.

Thanks in advance..

答:
Because $5\left[\begin{smallmatrix}1\\2\\1\end{smallmatrix}\right]=\left[\begin{smallmatrix}5\\10\\5\end{smallmatrix}\right]$.

I wanted to put this in the comments, but it wouldn't compile properly.


问:
Limit with parameter $p$
Find $ p $ that will make limit convergent
$$ \lim _ { x \to \infty } \left( \left( n + \left( \frac { x ^ 2 + 2 } { x ^ 2 - x + 1 } \right) ^ { 2 x + 1 } \right) \cdot \sin \left( \frac { x \pi } 2 \right) \right) $$

I know how to solve most of the basic limits. In this problem I don't know if I'm allowed to separate it and make $ 3 $ or $ 4 $ small ones. I thought about making $ p \sin.. + \sin*(big bracket)$ and then separating again because $ p $ is parameter and can go in front of limit and I wanted to separate second one so I can find big limit (it's $e^2$ or something but that doesn't matter).

答:
$$ \lim _ { x \to \infty } \left( \left( n + \left( \frac { x ^ 2 + 2 } { x ^ 2 - x + 1 } \right) ^ { 2 x + 1 } \right) \cdot \sin \left( \frac { x \pi } 2 \right) \right) $$

since as mentioned the SIN is diverging because it keeps oscillating. Therefore, the first part has to converge to 0.

We have that the first part converges to $(n+e^2)$ Therefore your n should be $-e^2$.

Detailing the Calculation

$$\lim\limits_{x\to\infty}\left(\frac{x^2+2}{x^2-x+1}\right)^{2x+1}$$
$$=\lim\limits_{x\to\infty}\exp\left(\ln\left(\left(\frac{x^2+2}{x^2-x+1}\right)^{2x+1}\right)\right)$$
$$=\exp\left(\lim\limits_{x\to\infty}\ln\left(\left(\frac{x^2+2}{x^2-x+1}\right)^{2x+1}\right)\right)$$

$$\left(\exp\left(\left(1+2x\right)\ln\left(\frac{1+\frac{2}{x^{2}}}{1-\frac{1}{x}+\frac{1}{x^{2}}}\right)\right)\right)$$

Here we have to use the definition of $ln(1+x)=x$ if x goes to 0 and subtract the numerator from the denominator after applying the small transformation.

$$\exp\left(\left(2x+1\right)\left(\frac{1}{x^{2}}+\frac{1}{x}\right)\right)$$

$$\exp\left(\frac{2x}{x^{2}}+\frac{2x}{x}\right)$$

$$exp(2)$$

Title: Limit with parameter $p$


问:
Uniform convergence of a Fourier series and calculating and infinite series
I was given the function $f(x) = |x|$, for $-\pi < x < \pi$.

I was able to find the Fourier series corresponding to f and I obtained that
$$f\sim\frac{\pi}{2} + \frac{2}{\pi} \sum_{n=1}^{\infty} \frac{(-1)^n-1}{n^2}cos(nx)$$
I was also able to conculde that the Fourier series of this function uniformly converges to $f$ using the Weierstrass M test.

Now I am asked to show that $\frac{\pi^2}{8} = 1 + \frac{1}{3^2} + \frac{1}{5^2} +...$ using the uniform convergence of the Fourier series of $f$.

I recognize the right term of the last expression is $\sum_{n=1}^{\infty}\frac{1}{(2n+1)^2}$, and I'm also familiar with the Parseval's Identity but I haven't been able to put all of this information toghether in order to get to the desired conclusion.

Thanks in advace!

答:
Substituting $x=0$ for the function and it's fourier expansion we get:

$$0 = \frac{\pi}{2} + \frac{2}{\pi}\sum_{n=1}^{\infty}\frac{(-1)^n-1}{n^2}$$

Since $cos(0)=1$.

Now, notice that the term $(-1)^n-1$ will be zero whenever $n$ is odd, and will be $-2$ if $n$ is even. Therefore we can ignore the even terms and subtitute $n=2k+1$ summing over $k$, going from zero to infinity.

$$0 = \frac{\pi}{2} + \frac{2}{\pi}\sum_{k=0}^{\infty}\frac{-2}{(2k+1)^2}$$

Rearranging terms:

$$0 = \frac{\pi}{2} - \frac{4}{\pi}\sum_{k=0}^{\infty}\frac{1}{(2k+1)^2}$$

And finally:

$$\frac{\pi^2}{8} = \sum_{k=0}^{\infty}\frac{1}{(2k+1)^2}$$


问:
Does the dimension of a vector space tell us about vectors it contains?
In particular, suppose I have a vector space W and I know that $\dim (W)$ = $0$. Does that automatically imply that $W = \{0\}$? Or does one need to go about checking the actual basis of $W$ to determine whether $W = \{0\}$?

In other words, does $\dim(W) = 0 \implies W = \{0\}$?

Please let me know. Thanks!

答: A $0$-dimensional vector space has only one element: its zero vector. If there were any non-zero vector $v$, then this vector by itself would constitute a one-element linearly independent family, hence would witness that $\dim W$ is at least $1$.


问:
How can I prove that $\|\;\|_2$ is a norm on the vector space of $n \times m$ matrices with entries in $\mathbb{R}$?
Let $A \in \mathbb{R}^{n \times m}$ be a matrix and $\|A\|_2 =  \sqrt{\sum \limits_{i,j = 1}^{n,m} |a_{ij}|^2}$.

How can I prove that $\|\;\|_{2}$ is a norm on the vector space of $n \times m$ matrices with entries in $\mathbb{R}$?

I guess this is the Frobenius norm, but I don't know how to prove it.

答:
$ \displaystyle || A ||_2 = \sqrt{\sum_{i,j=1}^{n,m} | a_{ij}|^2 } \ge 0\; \forall A \in \mathbb{R}^{n \times m} $

Because $ |a_{ij}| $ and $ |a_{ij}|^2 $ are not negative, so their sum and root won't be negative.

Let $ \displaystyle ||A||_2 = \sqrt{\sum_{i,j=1}^{n,m} | a_{ij}|^2 } = 0 $

$ \displaystyle \Rightarrow \sum_{i,j=1}^{n,m} | a_{ij}|^2 = 0 $

Because $ |a_{ij}|^2 \ge 0\;\forall i,j\in \mathbb{N} $, it means $ |a_{ij}|^2 = 0 = a_{ij}\;\forall i,j \in \mathbb{N} $

$ \Rightarrow A = 0 $, because all entries are $0$

2.

$ \displaystyle ||\alpha \cdot A||_2 = \sqrt{\sum_{i,j=1}^{n,m}|\alpha \cdot a_{ij}|^2} = \sqrt{\sum_{i,j=1}^{n,m}|\alpha|^2 \cdot |a_{ij}|^2} = \sqrt{|\alpha|^2 \cdot \sum_{i,j=1}^{n,m} |a_{ij}|^2} = |\alpha| \cdot \sqrt{ \sum_{i,j=1}^{n,m} |a_{ij}|^2} = |\alpha| \cdot ||A||_2 $

3.

Let $ B \in \mathbb{R}^{n\times m} $

$ \displaystyle (||A+B||_2)^2 = \sum_{i,j=1}^{n,m} |a_{ij}+b_{ij}|^2 \underset{binomial}{=} \sum_{i,j=1}^{n,m} a_{ij}^2 + 2 \cdot a_{ij} \cdot b_{ij} + b_{ij}^2$

$ \displaystyle = \sum_{i,j=1}^{n,m}|a_{ij}|^2 + \sum_{i,j=1}^{n,m} 2 \cdot a_{ij} \cdot b_{ij} + \sum_{i,j=1}^{n,m} |b_{i,j}|^2$

$ \displaystyle \underset{*}{\le}
\sum_{i,j=1}^{n,m}|a_{ij}|^2 + 2 \cdot \sqrt{(\sum_{i,j=1}^{n,m}|a_{ij}|^2)\cdot(\sum_{i,j=1}^{n,m}|b_{ij}|^2)} + \sum_{i,j=1}^{n,m} |b_{i,j}|^2 \underset{binomial\:theorem}{=} (\sqrt{\sum_{i,j=1}^{n,m} |a_{ij}|^2} + \sqrt{\sum_{i,j=1}^{n,m}|b_{ij}|^2})^2 = (||A||_2 + ||B||_2)^2 $

$\Leftrightarrow ||A+B||_2 \le ||A||_2 + ||B||_2$

*: $ \displaystyle (\sum_{i,j=1}^{n,m} 2 \cdot a_{ij} \cdot b_{ij})^2  \underset{Cauchy-Schwarz}{\le} 4 \cdot (\sum_{i,j=1}^{n,m}|a_{ij}|^2)\cdot(\sum_{i,j=1}^{n,m}|b_{ij}|^2) = ( 2 \cdot \sqrt{(\sum_{i,j=1}^{n,m}|a_{ij}|^2)\cdot(\sum_{i,j=1}^{n,m}|b_{ij}|^2)})^2 $


问:
Surface integral of a vector field equal to volume V
Let V be a volume in  $\mathbb{R}^3$  bounded by a simple closed piecewise-smooth surface S with outward pointing normal vector n. For which one of the following vector fields is the surface integral $\iint\limits_{S}\mathbf{f}\cdot \mathbf{n}dS$ equal to the volume of V ?

A: $\mathbf{f}(\mathbf{r})=(1,1,1)$

B: $\mathbf{f}(\mathbf{r})=\frac{1}{2}(x,y,z)$

C: $\mathbf{f}(\mathbf{r})=(2x,-y^2, 2yz-z)$

D: $\mathbf{f}(\mathbf{r})=(z^2,y^2,1-2yz)$

What I have tried:

So for A and D I have worked out that the curl is $0$ so that rules those 2 out. I don't know how to tackle the other 2.

答:
hint

For that, you need
$$\iint_S\vec{f}\cdot \vec{n}ds=\iiint_Vdxdydz$$

or simply that $$\nabla \cdot \vec{f}=1$$

For example, in the case $ C,$
$$\nabla \cdot (2x,-y^2,2yz-z)=$$
$$2-2y+2y-1=1$$


问:
Matrix Algebra Over the Opposite Ring
I'm trying to show that $M_n(R^{opp}) \cong M_n(R)^{opp}$ as rings. My thought was that $\varphi : M_n(R)^{opp} \to M_n(R^{opp})$ defined by $\varphi (A) = A^{T}$ is the desired isomorphism. But I'm a little confused as to what there is to verify...

Clearly $\varphi^{-1} = \varphi$, so it is bijective, $\varphi (A+B) = (A+B)^{T} = A^{T} + B^{T}$, and $\varphi (A \star B) = \varphi (BA) = (BA)^{T} = A^{T} B^{T} = \varphi (A) \varphi (B)$, where $\star$ is the opposite product of $M_n(R)^{opp}$.

Is that it? Seems like a cheesy proof. I tried consulting some of the users in the chatroom. The only one to proffer a critique was Thorgott. He said that "I don't think $\varphi^{-1} = \varphi$, given that codomain and domain are different." After thinking about this for some time, I don't think this is a problem, because although $M_n(R^{opp})$ and $M_n(R)^{opp}$ denote different rings, the underlying sets are the same.

答:
You might consider that $M_n(R)$ is (isomorphic to) the ring of endomorphisms of $R_R^n$ (right $R$-module).

Thus $M_n(R^{\mathrm{op}})$ is the ring of endomorphisms of $(R^{\mathrm{op}})^n_{R^{\mathrm{op}}}$, which is the same as the ring of endomorphisms of $_RR^{n}$ (left $R$-module), which is $M_n(R)^{\mathrm{op}}$.

Title: Matrix Algebra Over the Opposite Ring


问:
Calculating matrix exponential

Given matrix $$M = \begin{pmatrix} 7i& -6-2i\\6-2i&-7i\end{pmatrix}$$ how do I calculate matrix exponential $e^M$?

I know I can use that $e^A=Pe^DP^{-1}$ where $D=P^{-1}AP$. I computed the characteristic polynomial of the above matrix as

$$P(\lambda)=\lambda^2+89$$

Is there an easier way to do this than trying to compute the diagonalized matrix?

答:
Via Cayley-Hamilton,

$${\rm M}^2 = - 89 \, {\rm I}_2 = \left( i \sqrt{89} \right)^2 {\rm I}_2$$

Hence, matrix ${\rm A} := \frac{{\rm M}}{i \sqrt{89}}$ is involutory, i.e., ${\rm A}^2 = {\rm I}_2$. Using Euler's formula,

$$\begin{aligned} \exp({\rm M}) = \exp \left( i \sqrt{89} {\rm A} \right) &= \cos \left( \sqrt{89} {\rm A} \right) + i \sin \left( \sqrt{89} {\rm A} \right)\\ &= \cos \left( \sqrt{89} \right) {\rm I}_2 + i \sin \left( \sqrt{89} \right) {\rm A}\\ &= \color{blue}{\cos \left( \sqrt{89} \right) {\rm I}_2 + \frac{\sin \left( \sqrt{89} \right)}{\sqrt{89}} {\rm M}}\end{aligned}$$

where $\cos \left( \sqrt{89} {\rm A} \right) = \cos \left( \sqrt{89} \right) {\rm I}_2$ and $\sin \left( \sqrt{89} {\rm A} \right)=  \sin \left( \sqrt{89} \right) {\rm A}$ because ${\rm A}$ is involutory.

Title: Calculating matrix exponential


问:
Sum of $5$ primes is $105$ times less than their product
I am trying to find a "legitimate" solution to this problem. For primes $p_n$:

$105(p_1+p_2+p_3+p_4+p_5) = p_1 p_2 p_3 p_4 p_5$

I've figured out that it's the best to start with breaking the $105$ down to prime factors: $105 = 3 \times 5 \times 7$, by which I know that $p_1 = 3, p_2 = 5, p_3 = 7$.

With that out of the way, I have

$105(15+p_4+p_5) = 105p_4 p_5$, and therefore $15 + p_4 +p_5=p_4 p_5$.

It's obvious that $p_4 p_5 > 15$, therefore $p_4, p_5 ≥ 5$.

And this where I am stuck. My only idea was to consider that the right hand side is increasing faster than the left hand side, and therefore $p_4,p_5$ cannot be big numbers, and start testing numbers starting from $p_4 = 5, p_5 = 5$, which coincidentally happened to be the solution.

Is there a more legitimate way to solve this problem, different from what I did here?

答:
Starting from
$$ 15+p_4+p_5=p_4p_5,$$
we transform to
$$ (p_4-1)(p_5-1)=p_4p_5-p_4-p_5+1=16.$$
From the known factorizations of $16$, we find that $(p_4,p_5)$ is one of $(2,17)$, $(3,9)$, $(5,5)$, $(9,3)$, $(17,2)$. After removing composites and up to symmetry, we are left with $(2,17)$ and $(5,5)$. So in total, we have (up to permutation) two solutions
$$(2,3,5,7,17)\qquad\text{and}\qquad  (3,5,5,5,7). $$


答:
To formalize your idea that $p_4$ and $p_5$ cannot be big numbers, one way to proceed would be as follows.

Without loss of generality, assume that $p_4 \le p_5$. Then, from
$$p_4p_5 = 15 + p_4 + p_5,$$
divide both sides by $p_5$ to get
$$
\begin{align}
p_4 &= \frac{15}{p_5} + \frac{p_4}{p_5} + 1\\
&\le \frac{15}2 + 1 + 1,\\
\end{align}$$
which implies that $p_4$ can be no larger than $7$. You can then see what happens when you set $p_4 = 2, 3, 5, 7$.

Title: Sum of $5$ primes is $105$ times less than their product

Body:
I am trying to find a "legitimate" solution to this problem. For primes $p_n$:

$105(p_1+p_2+p_3+p_4+p_5) = p_1 p_2 p_3 p_4 p_5$

I've figured out that it's the best to start with breaking the $105$ down to prime factors: $105 = 3 \times 5 \times 7$, by which I know that $p_1 = 3, p_2 = 5, p_3 = 7$.

With that out of the way, I have

$105(15+p_4+p_5) = 105p_4 p_5$, and therefore $15 + p_4 +p_5=p_4 p_5$.

It's obvious that $p_4 p_5 > 15$, therefore $p_4, p_5 ≥ 5$.

And this where I am stuck. My only idea was to consider that the right hand side is increasing faster than the left hand side, and therefore $p_4,p_5$ cannot be big numbers, and start testing numbers starting from $p_4 = 5, p_5 = 5$, which coincidentally happened to be the solution.

Is there a more legitimate way to solve this problem, different from what I did here?


问:
Random Poisson Sample, Probability in terms of $\vartheta$
If $X_1, X_2, \ldots, X_n$ are a random sample from a Poisson Distribution with mean $\vartheta>0$, how do you find $P(X\le 1)$ in terms of $\vartheta$?

I've proven that summing $X_i$ for $i=1,\ldots,n$ is sufficient for $\vartheta$, which I expected would be helpful, but I've got no clue where to take it from here.  Not even sure that was a productive step, but it's helped things fall into place before.

Thanks guys!

答:
If by $X$ you mean $X_i$ then the answer is just
$$
  P[X_i \leq 1]
= P[X_i = 0] + P[X_i = 1]
= e^{-\theta}(1 + \theta).
$$
If by $X$ you mean $X = \sum_{i=1}^n X_i$, then since $X_i$ $i=1,...,n$ is a random sample, $X$ is a Poisson($n\theta$) r.v. and so
$$
  P[X \leq 1]
= P[X = 0] + P[X = 1]
= e^{-n \theta}(1 + n\theta).
$$
No one can give a definite answer until you clarify what $X$ is though.</p>


答:
Your equation can be rewritten

$$p_4 p_5 - p_4 - p_5 - 15 = 0$$

or

$$p_4 p_5 - p_4 - p_5 +1 = 16$$

or

$$(p_4-1)(p_5-1) = 16.$$

Assume $p_4 \leq p_5$, so $p_4 = 1, 2,$ or $4$.  This gives you all the solutions.


问:
What is the expected value of the maximum of two independent exponentially distributed random variables?
the density for an exponentially distributed RV X with parameter λ>0

is given as:

f(x)=(1/λ )exp(−x/λ)

for non-negative x and 0 otherwise. The expected value of the maximum of two independent exponentially distributed random variables X  and Y  with parameters λ1=1 and λ2=2  is...?

答:
Whenever you need to find the Expected Value of the Maximum of a random variables it is always wise and easy to solve the problem in the following 3 steps:

Find the Cumulative distribution $F_M(x)$ of the maximum of the random variables

Differentiate this function to find the Probability Density Function $f_M(x)$

Find the expected value of the maximum using the formula: $\mathbb{E}[M] = \int_{-\infty}^{+\infty}xf_M(x)dx$

Lets do this together!

$\textbf{Step 1:}$

The cumulative density function  $F_M(x)$ is the probability that the random variable $M$ (defined to be the maximum of these 2 R.Vs) is less than $x$. I.E $F_M(x):= \mathbb{P}(M\leq x)$ In order for the maximum of the two random variables to be less than $x$ we need $\textbf{both}$ of them to be less than $x$. This should be obvious but let us just stop to think about it for a second . If we both have a dice and I roll a 4 and you roll a 6 is the maximum of our rolls less than or equal to  5? No. Why not? Because they are not $\textbf{both}$ less than 5. Hence  $ \mathbb{P}(M\leq x) = \mathbb{P}(X\leq x \cap Y\leq x) $  And as they are independent:  $ \mathbb{P}(X\leq x \cap Y\leq x)= \mathbb{P}(X \leq x) \cdot \mathbb{P}(Y \leq x)$ If you are familiar with the exponential distribution you should know that $\mathbb{P}(X \leq x) = 1 - e^{-\lambda_1 x}$ and equally  $\mathbb{P}(Y \leq x) = 1 - e^{-\lambda_2 x}$

Putting this all together completes Step $1$ and yields: $F_M(x):= \mathbb{P}(M\leq x) = \mathbb{P}(X\leq x \cap Y\leq x) = \mathbb{P}(X \leq x) \cdot \mathbb{P}(Y \leq x) = (1 - e^{-\lambda_1 x}) \cdot (1 - e^{-\lambda_2 x}) = 1 -e^{-\lambda_1x} - e^{-\lambda_2x} + e^{-(\lambda_1+\lambda_2)x}$ And using $\lambda_1 = 1$ and $\lambda_2 = 2$ yields the final CDF : $F_M(x) = (1-e^{-x}-e^{-2x}+e^{-3x})$

$\textbf{Step 2:}$

It is lightwork from here. We differentiate the CDF with to yield the PDF:

$f_M(x) = \frac{d}{dx}F_m(x) = \frac{d}{dx}(1-e^{-x}-e^{-2x}+e^{-3x}) = e^{-x} + 2e^{-2x}-3e^{-3x} $

$\textbf{Step 3:}$

We are after the expected value of $M$ (the maximum) so all we would need to do is integrate:

$\mathbb{E}[M] = \int_{0}^{+\infty}x\cdot f_M(x)dx = \int_{0}^{+\infty}x\cdot (e^{-x} + 2e^{-2x}-3e^{-3x})dx $

This is fairly easy to do via parts however I am very lazy so lets be naughty and recognise that this is the "sum" of exponential R.V's:
$ \int_{0}^{+\infty}x\cdot (e^{-x} + 2e^{-2x}-3e^{-3x})dx = \int_{0}^{\infty}x\cdot e^{-x} + \int_{0}^{\infty}x\cdot2e^{-2x} - \int_{0}^{\infty}x\cdot3e^{-3x}  $

Which we can see the simply be the calculations required to calculate the expected value of $\lambda = 1 ,  \lambda = 2, \lambda = 3$ exponential R.Vs respectively. And as the Expected value of an exponential R.V is $\frac{1}{\lambda}$ we have:
$  \int_{0}^{\infty}x\cdot e^{-x} + \int_{0}^{\infty}x\cdot2e^{-2x} - \int_{0}^{\infty}x\cdot3e^{-3x}  = \frac{1}{1} + \frac{1}{2} - \frac{1}{3} = \frac{6+3-2}{6} = \frac{7}{6}$

And hence the expected value is $\frac{7}{6} = 1 + \frac{1}{6} \approx 1.167 $


问:
Quintic Polynomials and finding the fifth root
I have a quintic Polynomial that I am having trouble finding the fifth root.

I have -8, 7, -3, and 1 as roots

The leading coefficient is -9

coefficient of $x^4$ is 9

I think it is 4 however, I'm having difficultly with the formula to find the fifth root.

Body: $$(x-7)(x-1) (x+3)(x+8) ( x - A) = x^5 + (3-A)x^4 - (3A + 57)x^3 + (57A - 115)x^2 + (115A + 168)x - 168A $$


答:
Here's one way to do it
$$\lim\limits_{x\to\infty}\left(\frac{x^2+2}{x^2-x+1}\right)^{2x+1}$$
$$=\lim\limits_{x\to\infty}\exp\left(\ln\left(\left(\frac{x^2+2}{x^2-x+1}\right)^{2x+1}\right)\right)$$
$$=\exp\left(\lim\limits_{x\to\infty}\ln\left(\left(\frac{x^2+2}{x^2-x+1}\right)^{2x+1}\right)\right)$$
By series expansion, we have
$$\exp\left(\lim\limits_{x\to\infty}\left[2+\frac4x+\frac1{6x^2}+O\left(\frac1{x^3}\right)\right]\right)$$
$$=\exp\left(2\right)$$
I think you can take it from here.

Title: Limit with parameter $p$

Body:
Find $ p $ that will make limit convergent
$$ \lim _ { x \to \infty } \left( \left( n + \left( \frac { x ^ 2 + 2 } { x ^ 2 - x + 1 } \right) ^ { 2 x + 1 } \right) \cdot \sin \left( \frac { x \pi } 2 \right) \right) $$

I know how to solve most of the basic limits. In this problem I don't know if I'm allowed to separate it and make $ 3 $ or $ 4 $ small ones. I thought about making $ p \sin.. + \sin*(big bracket)$ and then separating again because $ p $ is parameter and can go in front of limit and I wanted to separate second one so I can find big limit (it's $e^2$ or something but that doesn't matter).


答:
Hint: Use 1. Vieta formula:

$$x_1+x_2+x_3+x_4+x_5= -{b\over a}$$

where polynomial $$ax^5+bx^4+....$$

Title: Quintic Polynomials and finding the fifth root

Body:
I have a quintic Polynomial that I am having trouble finding the fifth root.

I have -8, 7, -3, and 1 as roots

The leading coefficient is -9

coefficient of $x^4$ is 9

I think it is 4 however, I'm having difficultly with the formula to find the fifth root.


问:
Help in factorization (Explain if I was wrong)

Task: Factorize $ax^2-a-x^2+x$.

And my answer is:

$$ a(x^2-1)-x(x-1)=a(x-1)(x+1)-x(x-1)$$

Am I right? If not, explain why and give the correct solution.

答:
Unfortunatly, not correct.
I could be downvoted

$$ax^2-a-x^2+x=$$
$$a(x^2-1)-(x^2-x)=$$
$$a(x+1)\color{red}{(x-1)}-x\color{red}{(x-1)}=$$

$$\color{red}{(x-1)}\Bigl(a(x+1)-x\Bigr)$$


问:
Prove or disprove the 2 norm for any column of an orthogonal matrix is smaller or equal to 1
Is this a true statement or under some condition this is a true statement? I have try a dozen of MATLAB randomly generated real value square matrix, than take the QR decomposition to them, and all of orthogonal matrix obtained from QR decomposition seems satisfied this property, but I don't understand why this happened. Could anyone help me on this?</p>

答: Hint. When $Q$ is an orthogonal matrix, how is the $j$-th diagonal element of $Q^TQ$ related to the entries of the $j$-th column of $Q$?


答:
"Factoring" usually means to write the whole expression as a product, not just factoring some terms.

Can see that $(x-1)$ can be factored out of your last answer now?

Title: Help in factorization (Explain if I was wrong)

Body:
Task: Factorize $ax^2-a-x^2+x$.

And my answer is:

$$ a(x^2-1)-x(x-1)=a(x-1)(x+1)-x(x-1)$$

Am I right? If not, explain why and give the correct solution.


问:
Find $f(x)$, given $f(x+y) - 2f(x-y) + f(x) - 2f(y) = y - 2$
For a differentiable function $f(x)$, $$f(x+y) - 2f(x-y) + f(x) - 2f(y) = y - 2.$$ Then find $f(x)$.

I am not able to think of any method which is working. I tried replacing $x$ and $y$, putting $x$ or $y = 0$, putting $y = x$, etc.
Can someone help me

Answer given at back is $f(x)=x+1$

答:
Solution based on the one provided by user FhoToN in this AoPS post. (no differentiability needed nor used)

Let $P(x,y)$ be the assertion that $$f(x+y)-2f(x-y)+f(x)-2f(y)=y-2.$$

From $P(x,0)$ we get $-2f(0)=-2$, and so $$f(0)=1.$$

From $P(0,y)$ and previous we get
$$f(y)+2f(-y)=3-y.\tag{1}$$

From $P(0,-y)$ and previous we get
$$f(-y)+2f(y)=3+y.\tag{2}$$

Solving $(1)$ and $(2)$ as a system of equations (with variables $f(y)$ and $f(-y)$), we get $f(y)=y+1$ (notice that $y$ was arbitrary chosen and so this holds for all $y$). We can verify this is a solution, and thus the only one.


问:
How to prove Chebyshev–Gauss quadrature integrate polynomial of degree less than $2n-1$ exactly
What I want to ask is mentioned in the title.

For example: how can we show that $\int_{-1}^1\frac{f(x)}{\sqrt{1-x^2}}dx=\frac{\pi}{5}\left(f(0)+f(\cos{\frac{3\pi}{10}})+f(\cos{\frac{5\pi}{10}})+f(\cos{\frac{7\pi}{10}})+f(\cos{\frac{9\pi}{10}})\right)$ is exact for all polynomials of degree 9.

If someone can explain to me how do we even get the weight and nodes formula as stated in Wikipedia, I would really appreciate that!

To check formula, please follow here http://en.wikipedia.org/wiki/Chebyshev%E2%80%93Gauss_quadrature

答:
One needs to solve a moment problem:

http://en.wikipedia.org/wiki/Moment_problem


问:
Triangle Inequation proof
I have to show for all x,y ∈ ℝ that ||x| - |y| ≤ |x-y|

I know that |x| = |x - y + y| ≤ |x - y|+|y|

Does it suffice to put in numbers now or how would I proof that?

答:
You have $|x| = |x - y + y| ≤ |x - y|+|y|$ so $|x|-|y| \leq |x-y|$

Also $|y| = |y-x+x| \leq |y-x|+|x|$
and by subtracting $|y-x|$ and $|y|$ from both sides one gets $-|x-y|=-|y-x| \leq |x|-|y|$

Since $-|x-y| \leq |x|-|y| \leq |x-y| $,$ ||x|-|y|| \leq |x-y|$.

Title: Triangle Inequation proof


问:
Under what restrictions do piecewise continuous functions in higher dimensions form vector spaces?
Definition 1. A function $ f : [a,b] \to \mathbb R $ is called piecewise continuous if $ [a,b] $ may be broken up into a finite number of subintervals $ [t_i,t_{i+1}] $, $ i = 1,2,\dots, n $, such that $ f $ is continuous on each open subinterval $ (t_i,t_{i+1}) $ and has finite limits at their endpoints.

A natural extension to higher dimensions could be formulated as:

Definition 2. A function $ f : \Omega \subseteq \mathbb R^d \to \mathbb R $ is called piecewise continuous if there is a finite set of open and connected sets $ \omega_1,\omega_2,\dots, \omega_n \subset \Omega $ satisfying $ \bigcup_{i=1}^n\overline{\omega_i} = \Omega $ and $ \omega_i\cap\omega_j=\emptyset $ for $ i\neq j $, and to each set $ \omega_i $ a function $ g_i \in C(\overline{\omega_i}) $ with $ g_i = f $ on $ \omega_i $.

Let us for simplicity assume $ d = 2 $.

Let $ PC(\Omega) $ be the set of all piecewise continuous functions $ f : \Omega \to \mathbb R $, as per the above Definition 2.

Proposition. The set $ V = PC(\mathbb R^2) $ is not a vector space.

Proof: Take for instance
$$ f(x_1,x_2) = \begin{cases} 1 & \text{if $ x_2 \geq \sin(x_1) $} \\ 0 & \text{otherwise},
\end{cases} $$
and
$$ g(x_1,x_2) = \begin{cases} 1 & \text{if $ x_2 \leq -\sin(x_1) $} \\ 0 & \text{otherwise}, \end{cases} $$
each in $ V $. Clearly, $ f+g $ will have the value 2 in an infinite number of disjoint sets, even if both $ f $ and $ g $ were defined by only two sets each. So $ f+g \not\in V $. $ \square $

What can we do to get a vector space? Restricting ourselves to compactly supported functions does not fix the problem.

Proposition. The set $ V = PC(\Omega) $ for a bounded set $ \Omega \subset \mathbb R^2 $ does not form a vector space in general.

Proof: Define
$$ \psi(s) = \begin{cases} 0 & \text{if $ s \leq 0 $} \\ -\sin\left(\frac{s}{1-s^2}\right) & \text{if $ s \in (0,1) $},\end{cases} $$
and consider
$$ f(x_1,x_2) = \begin{cases} 1 & \text{if $ \max(|x_1|,|x_2|) < 1 $ and $ x_2 \geq \psi(x_1) $} \\ 0 & \text{otherwise}, \end{cases} $$
and
$$ g(x_1,x_2) = \begin{cases} 1 & \text{if $ \max(|x_1|,|x_2|) < 1 $ and $ x_2 \leq -\psi(x_1) $} \\ 0 & \text{otherwise}. \end{cases} $$
Clearly, each are compactly supported on in the square domain $ \{(x_1,x_2) : \max(|x_1|,|x_2|) < 1+\epsilon \} $ for any $ \epsilon > 0 $, but again $ f+g $ takes the values 2 on an infinite number of disjoint open sets. So still we don't have a vector space. $ \square $

Question! Is there a simple requirement that immediately yields a vector space. It seems that to get these infinitely interweaving sets on a bounded set must require increasingly sharp turning. <s>Would restricting to compact support along with the boundary of each $ \omega_i $ being Lipschitz continuous be strong enough? I would assume yes, but is there a lesser requirement, which is sufficient?</s>

Edit #1: Clearly, from mihaild's excellent example, which is both Lipschits continuous and infinitely differentiable that assumption was wrong. mr_e_man suggests strictly convex sets $ \omega_i $, but that is incredibly restrictive as very few convex sets can tile together to fill an area. <s>However, it made me think, what if each set $ \omega_i $ was constructible by taking unions and differences of finitely many convex sets. Certainly, that should at least eliminate all these "easy" counterexamples with oscillating boundaries.</s>

Edit #2: Turns out even even my convex set idea is problematic. Consider the set $ \Omega = [-1,1]\times[0,2] $ and let
$$ \theta_i = \pi\left(1-\frac{1}{2^i}\right), i = 0,1,2,\dots,n-1\quad\text{and}\quad\theta_n = \pi, $$
and
$$ \varphi_{i+1} = \theta_i + \frac{\theta_{i+1}-\theta_i}{2}, i = 0,1,2,\dots,n-1,\quad \varphi_0 = 0, \quad\text{and}\quad \varphi_{n+1}=\pi. $$
Consider the point sets
$$ \mathcal P_n = \{(\cos(\theta_i),\sin(\theta_i)) : i = 0,1,\dots,n \} $$
and
$$ \mathcal Q_n = \{(\cos(\varphi_i),\sin(\varphi_i)) : i = 0,1,\dots,n+1 \}. $$
Take $ \mathcal P = \bigcup_{n=1}^\infty\mathcal P_n $ and $ \mathcal Q = \bigcup_{n=1}^\infty \mathcal Q_n $ and let $ \omega_1 $ be the convex hull of $ \mathcal P $ and $ \omega_2 =  \Omega\backslash\omega_1 $ and $ \eta_1 $ the convex hull of $ \mathcal Q $ and $ \eta_2 = \Omega\backslash\eta_1 $. Let
$$ f(\mathbf x) = \begin{cases} 1 & \text{if $ \mathbf x \in \omega_1 $} \\ 0 & \text{if $ \mathbf x \in \omega_2 $} \end{cases} \quad g(\mathbf x)
= \begin{cases} 0 & \text{if $ \mathbf x \in \eta_1 $} \\ 1 &\text{if $\mathbf x \in \eta_2$}. \end{cases}$$
Then the set of $ f $ can be constructed by difference of finitely many (just 2) convex sets and likewise for $ g $ and $ f+g $ will again have infinitely many disjoint sets with value 2.

Visual example here:

Edit #3: More restrictive idea: Each set $ \omega_i $ must be a polygon! That has to do it, right?

答: I think you just need to restrict the allowed subsets (on which the function is continuous) so that they're closed under intersection. These could be axis-aligned rectangles, for example. Or you could require the subsets to be open and convex, not just connected, since the intersection of convex sets is convex.


问:
Möbius function of subpermutation ordering of $[n]$, is equal to number of derangements of $[n]$.
Let $P_n$ be the poset of subpermutations of $[n]$, ordered by the relation $$x\preceq y \iff x \text{ is a subsequence of } y$$ (e.g. for $P_7$, we have that $(3,7,6)\preceq (2,3,7,1,6)$ ).

Let us assume now the poset $\hat{P_n}=P_n\cup \{ \hat{0},\hat{1}\} $, constructed by adding a minimum ellement $\hat{0}$ and a maximum ellement $\hat{1}$.
We want to show that $$|\mu(\hat{0},\hat{1})|=D_n,$$  where $D_n$ is the number of derangements of $\left[n\right]$ (i.e. the number of permutations of $\left[n\right]$ without fixed points) and $\mu$ is the Möbius function of $\hat{P_n}$.

I have managed to show that $P_n$ is a finite graded poset of rank $n-1$. I also observed that, for any $k\in\{0,1,\cdots,n\}$ $$\#\left\{ x\in\hat{P_{n}}\mid\hat{p}\left(x\right)=k\right\} =\binom{n}{k}k!$$and thus $$D_n=\sum_{k=0}^{n}\left(-1\right)^{k}\binom{n}{k}\left(n-k\right)!=\sum_{k=0}^{n}\left(-1\right)^{n-k}\#\left\{ x\in\hat{P_{n}}\mid\hat{p}\left(x\right)=k\right\}$$ where $\hat{p}$ is the rank function of $\hat{P_n}$. But I do not know how to procced.

Moreover, by Philip Hall’s theorem (Proposition 3.8.5, Stanley's "Enumerative Combinatorics Vol 1"(2011).) I know that $$\left|\mu\left(\hat{0},\hat{1}\right)\right|=\left|\sum_{i=1}^{n+1}\left(-1\right)^{i}c_{i}\right|,$$ where $c_i$ be the number of chains $\hat{0}<t_0<t_1<\cdots<t_i=\hat{1}$ of length $i$
between $\hat{0}$ and $\hat{1}$.

I have shown that $$c_{i}=\sum_{1\le k_{1}<\ldots<k_{i-1}\le n}\binom{k_{2}}{k_{1}}\cdots\binom{k_{i-1}}{k_{i-2}}\cdot k_{i-1}!$$ and it seems that the requested equality holds, but I cannot show it.

Can anyone help me?

答:
First show by induction on $k$ that $\mu(\hat 0,x)=(-1)^{\hat p(x)}$ for each $x\in\hat P_n$: if $\hat p(x)=k>0$, and $0\le\ell<k$, then $x$ has $\binom{k}\ell$ predecessors of rank $\ell$ in $\hat P_n$, so

$$\begin{align*}
\mu(\hat 0,x)&=-\sum_{\hat 0\preceq y\prec x}\mu(\hat 0,y)\\
&=-\sum_{\ell=0}^{k-1}(-1)^\ell\binom{k}\ell\\
&=-\left(\sum_{\ell=0}^k(-1)^\ell\binom{k}\ell-(-1)^k\binom{k}k\right)\\
&=(-1)^k\,.
\end{align*}$$

Thus,

$$\begin{align*}
\left|\mu(\hat 0,\hat 1)\right|&=\left|-\sum_{\hat 0\preceq x\prec\hat1}\mu(\hat 0,x)\right|\\
&=\left|\sum_{k=0}^n(-1)^k\binom{n}kk!\right|\\
&=\left|\sum_{k=0}^n(-1)^{n-k}\binom{n}{n-k}(n-k)!\right|\\
&=\left|\sum_{k=0}^n(-1)^{n-k}\binom{n}k(n-k)!\right|\\
&=\left|\sum_{k=0}^n(-1)^k\binom{n}k(n-k)!\right|\\
&=|D_n|\\
&=D_n\,.
\end{align*}$$


问:
Find $x_1$ and $x_2$ of the equation:
Without solving the equation

$$3x^2 - x-7=0$$

Find:

$$ \begin{array}{cl}
\mathrm{a)} & x_1^3 + x_2^3 \\
\mathrm{b)} & x_1^3 x_2^3 \\
\mathrm{c)} & 4 x_1^3 + 3 x_1^2 x_2 + 3 x_1 x_2^2 + 4 x_2^3
 \end{array} $$

I have the answers but I don't know the way. My solutions are:
a) $(\frac{4}{3})^3$, b) $-(\frac{7}{3})^3$, c) $\frac{193}{27}$.

I tried to use Vieta's Formulas...

Title: Find $x_1$ and $x_2$ of the equation:

Body:
hint

For $ a)$

Let $ x_1 $ and $ x_2 $ be the roots of the equation
$$ax^2+bx+c=3x^2-x-7=0$$

we know that
$$x_1+x_2=\frac{-b}{a}=\frac 13$$
and
$$x_1.x_2=\frac ca=\frac{-7}{3}$$

But
$$(x_1+x_2)^3=$$
$$x_1^3+x_2^3+3x_1x_2(x_1+x_2)$$

So
$$x_1^3+x_2^3=(\frac 13)^3-3.\frac{-7}{3}.\frac 13$$


问:
prove that any positive integer-valued random variable with memoryless property has the geometric distribution for some $p$
How to prove that any positive integer-valued random variable with memoryless property has the geometric distribution for some $p$.

By memoryless property,

$$P(X=i+s | X>i)=P(X=s)$$

How to get distribution for X from the above ?

答: Let $p=P[X\geqslant2]$, then $P[X\geqslant i+1\mid X\geqslant i]=p$ for every $i\geqslant1$ hence $P[X\geqslant i]=p^{i-1}$ for every $i\geqslant1$. Surely you can deduce the distribution of $X$ from this observation.</p>


问:
How can I show that the convergence of a sequence of functions is strictly increasing?
I am taking an advanced course in calculus and I am having hard times with this question.
I have that $\{ r_1, r_2,...\}$ is an enumeration of $Q$ and $f_n(x)= \frac{1}{n^2} $ if $x>r_n$ and $f_n(x)= 0 $ if $x\le r_n$ and f(x) = $\sum_{n=1}^{\infty}f_n(x)$ and the question is to show that $f$ is strictly increasing and continuous at every $x \in I$, irrationals. I really dont know where to start. I have looked at my literature (advanced calculus) and im not getting any smarter. Any help is welcome.

答:
Hints:

$1).\ $ for the first part, fix $x<y$ and an $f_n$. Consider three cases: $r_n\le x;\ x<r_n\le y;\ y\le r_n.$

$2).\ $For the second part, suppose $x$ is irrational and pick rationals $\ r_n<x<r_m.$ Since $f$ is increasing, $|f(x)-f(r_n)|<|f(r_m)-f(r_n)|=\left|\sum_{\{j|r_j>r_m\}}\frac{1}{j^2}-\sum_{\{j|r_j>r_n\}}\frac{1}{j^2}\right|=\left|\sum_{\{j|r_n<r_j\le r_m\}}\frac{1}{j^2}\right|.$ To finish note that there are arbitrary many rationals arbitrarly close to $x$, to the left and to the right of $x$.

Notice if $p=r_m$ is rational, then we can take a subsequence $r_{n_k}\to p^{-}$ and then

$|f(p)-f(r_{n_k})|=\left|\sum_{\{j|r_j>p\}}\frac{1}{j^2}-\sum_{\{j|r_j>r_{n_k}\}}\frac{1}{j^2}\right|=\left|\sum_{\{j|r_{n_k}<r_j\le p\}}\frac{1}{j^2}\right|>1/m^2$ so $f$ is not continuous at $p.$


答:
First, observe that
$$
\mathbb{P}\{X>t\}=\sum_{n > t}\mathbb{P}\{X=n\}
$$
so one can show that the memorylessness property implies ($s,t>0$)
$$
\mathbb{P}[X>s+t| X > t]=\mathbb{P}\{X> s\}
$$
Note also that since $s>0$
$$
\mathbb{P}[X>s+t| X > t\} = \frac{\mathbb{P}\{X>s+t\}}{\mathbb{P}\{X > t\}}
$$

Define $f\colon t\in\mathbb{R}_+\mapsto\mathbb{P}\{X>t\}\in[0,1]$. We have $f\searrow$, and furthermore $\forall t,s\geq 0$
$$
 \frac{f(t+s)}{f(t)}=f(s)
$$

i.e
$$
 f(s)f(t)=f(s+t)\qquad \forall s,t \geq0
$$
Now, remains to solve this functional equation (with additional condition that $f$ is non-increasing) to get the only possible form(s) for $\mathbb{P}\{X> t\}$ — which characterizes $X$'s law.

Body:
How to prove that any positive integer-valued random variable with memoryless property has the geometric distribution for some $p$.

By memoryless property,

$$P(X=i+s | X>i)=P(X=s)$$

How to get distribution for X from the above ?


问:
Fundamental group of the torus from hexagon with opposite sides identified
The hexagon with opposite sides identified is the topological torus, see here and here.

This would suggest to me that the fundamental group of the torus could be written as $\langle x,y,z|xyzx^{−1}y^{−1}z^{−1}\rangle$, but apparently this is not true (as mentioned here, for example). I don't understand why not.

Can the fundamental group of the torus be found starting from the hexagon with opposite sides identified?

答:
The method that you used to write down a presentation using the hexagon gluing diagram is valid only under a strong hypothesis: the gluing diagram has one vertex cycle.

But the hypothesis of one vertex cycle is not satisfied by the hexagon gluing diagram, which has two vertex cycles. Writing the vertices in order around the hexagon as A, B, C, D, E, F, then A, C, E forms one vertex cycle, and B, D, F forms the second vertex cycle.

The reason for this hypothesis is that when you form the quotient surface of the gluing diagram by gluing edge pairs as indicated, the image of the edges forms the 1-skeleton of the quotient surface, and you need a unique vertex in the 1-skeleton in order for each edge to close up into a generator of the group.

You can, on the other hand, obtain a presentation even when there are two or more vertex cycles, but to do that you must first choose a maximal tree in the 1-skeleton of the quotient surface, color those edges "red", then color the corresponding edge pairs of the polygon "red", and then you simply ignore the red edges when you write the presentation (what's happening topologically is that you are taking a further quotient by collapsing the maximal tree to a point).

So, for instance, with the hexagon gluing, one can choose a maximal tree consisting of a single red edge whose corresponding edge pair in the hexagon boundary is $z,z^{-1}$. So following the formula, you ignore $z$ in the generators and your ignore $z$ and $z^{-1}$ in the relators, and you get the presentation
$$\langle x, y \mid x y x^{-1} y^{-1} \rangle
$$


问:
Finding probability density function for $|2M+1|$, with $M \in [-1,4]$
Let $M$ be a number randomly selected from $[-1,4]$. Find the density function of $N=|2M+1|$.

My Attempt: I'll use the method of distribution functions.

First I note that

$(i)$ $N = -2X-1 : X \in[-1,\frac{-1}{2}) \Rightarrow Y \in (0,1]$

$(ii)$ $N= 2X+1 : X \in [\frac{-1}{2},4] \Rightarrow Y \in [0,9]$

So for $(i)$:

$F_Y(y) = P(Y \le y) = P(-2X-1 \le y) = P(x \le \frac{y+1}{-2}) = F_X(\frac{y+1}{-2})$

Since $X$ is a uniform continuous random variable we have that $F_X(x) = \frac{x-a}{b-a} for x\in[a,b]$. Thus,

$F_X(\frac{y+1}{-2}) = \frac{\frac{y+1}{-2} - (-1)}{\frac{-1}{2}-(-1)} = -y-1$

Thus, $F_Y(y) = -y-1 \Rightarrow f_y(y) = F_Y^{'}(y)=-1$

$\Rightarrow f_y(y) = -1, y\in (0,1]$

By similiar methods for $(ii)$:

$F_Y(y) = P(x \le \frac{y-1}{2}) = F_X(\frac{y-1}{2}) = \frac{y}{9} \Rightarrow f_y(y) = \frac{1}{9}, y \in[0,9]$

Combining these together I get that $f_y(y) = -1+ \frac{1}{9}, y \in (0,1], \frac{1}{9}, y \in (1,9] \cup \{0\}$

But this isn't a proper probability density function because the integral from $-\infty$ to $+ \infty$ is not equal to $1$.

答:
Lets have a nice go at this together. We shall do this in two steps. Firstly let $Y = |2M+1|$

Find the cumulative density $F_Y(x)$

Differentiate the cumulative density to find the probability density $f_Y(x)$

$\textbf{Step 1:}$

The cumaltive density of $Y:$$ F_Y(x) := \mathbb{P}(Y\leq x) = \mathbb{P}(|2M+1| \leq x) = \mathbb{P}( -x\leq2M+1\leq x)=\mathbb{P}(-x-1\leq 2M \leq x -1) =\mathbb{P}(\frac{-(x+1)}{2} \leq M \leq \frac{x-1}{2} ) = \mathbb{P}(M \leq \frac{x-1}{2}) - \mathbb{P}(M \leq \frac{-(x+1)}{2})$

This might get quite fiddly with all the different ranges. Firstly it's wise to note that $M$ ranges from $-1$ to $4$ and hence if $m\in[-1,4]$ then $\mathbb{P}(M \leq m ) = \frac{m+1}{5}$ if $m \geq 4 $ then $\mathbb{P}(M \leq m ) =1 $ and if $m \leq -1 $ then $\mathbb{P}(M \leq m ) = 0 $

Let us first deal with evaluating $\mathbb{P}(M \leq \frac{x-1}{2})$

Firstly  $\frac{x-1}{2} \geq 4 \iff x \geq 9 $ and $\frac{x-1}{2} \leq -1 \iff x \leq -1$ Hence for $x\in[-1,9]$ $ \mathbb{P}(M \leq \frac{x-1}{2}) = \frac{\frac{x-1}{2}+1}{5} = \frac{x+1}{10}$

And for $x$ less than $-1$ or greater than $9$ the probability for $M$ to be less than $\frac{x-1}{2}$ is respectively $0$ and $1$

Complete $\mathbb{P}(M \leq \frac{-(x+1)}{2})$ the exact same way and then differentiate for the density.


问:
Permutations of Number sequence with Minimum cost
There are numbers from 1 to N and our task is to create a permutation such that the cost of the permutation is minimum. For each number, there is a left and right cost. To put number p (i<=p<=N) at the  ith (1-based index) index, it costs Lp*(i-1) + Rp*(N-i) where L[] (left array) and R[] (right array) cost is given. (size of Left[] and Right are same = N)

We do not need to find that permutation but we need to find the total minimum possible cost of the permutation.

Example:- N = 3
L[] (left array) = {1,2,6}
R[] (right array)  = {4,7,2}

Answer = 13..because possible permutation is {3,1,2} and cost = (0 * 6 + 2 * 2) + (1 * 1 + 1 * 4) + (2 * 2 + 0 * 4) = 13 which is minimum possible cost.

My approach:-

```<code>As Lp*(i-1) + Rp*(N-i) is a Summation from i = 1 to N, tried to figure out a common formula for this as

Lp*(1-1) + Rp*(N-1) + Lp*(2-1) + Rp*(N-2) + Lp*(3-1) + Rp*(N-3) ...... Lp*(N-1) + Rp*(N-N)

=> Lp*(0) + Rp*(N-1) + Lp*(1) + Rp*(N-2) + Lp*(3) + Rp*(N-3) ..... Lp*(N-1) + Rp*(0)

=> Lp*(((N-1) * N)/2) + Rp*(((N-1) * N)/2) OR (((N-1) * N)/2)*(Lp + Rp)
</code>```

After that, I put Lp and Rp values from the Left array and right array respectively to get minimum cost, but this is not working.

Can any please help me to solve these questions? (Or please tell me if I am mistaking somewhere or breaking maths rules in my derivation).

答:
This is an instance of the linear assignment problem.  For $(p,i)\in\{1,\dots,N\}^2$, let $$c_{p,i} = L[p](i-1)+R[p](N-i)$$ be the cost of assigning number $p$ to position $i$.  Let binary decision variable $x_{p,i}$ indicate whether this assignment is made. The problem is to minimize $\sum_{p,i} c_{p,i} x_{p,i}$ subject to
\begin{align}
\sum_i x_{p,i} &= 1 &&\text{for all $p$}\\
\sum_p x_{p,i} &= 1 &&\text{for all $i$}\\
\end{align}
For your example, the cost matrix is
$$c=\begin{pmatrix}
8 &5 &2\\
14 &9 &4\\
4 &8 &12
\end{pmatrix}$$ and the problem is to minimize
$$8x_{1,1} + 5x_{1,2} + 2x_{1,3}
+ 14x_{2,1} + 9x_{2,2} + 4x_{2,3}
+ 4x_{3,1} + 8x_{3,2} + 12x_{3,3}
$$ subject to
\begin{align}
x_{1,1} + x_{1,2} + x_{1,3} &= 1 \\
x_{2,1} + x_{2,2} + x_{2,3} &= 1 \\
x_{3,1} + x_{3,2} + x_{3,3} &= 1 \\
x_{1,1} + x_{2,1} + x_{3,1} &= 1 \\
x_{1,2} + x_{2,2} + x_{3,2} &= 1 \\
x_{1,3} + x_{2,3} + x_{3,3} &= 1 \\
\end{align}
The optimal solution has $x_{1,2}=x_{2,3}=x_{3,1}=1$ and all other $x_{p,i}=0$, yielding optimal objective value $c_{1,2}+c_{2,3}+c_{3,1}=5+4+4=13$.


问:
Show that $F^4=I$ if $F$ is the Fourier Transform.
Let the Fourier Transfor $F:L_2(\mathbb{R})\to L_2(\mathbb{R})$ defined by $\displaystyle Fg(x)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}g(t)e^{ixt}dt\,\,$ for $\,\,g\in L^2(\mathbb{R})$.

I need to show that  $F^4=I$. Can you give me some suggestions to try this please

I know that $\displaystyle F^{-1}h(x)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}h(t)e^{-ixt}dt\,\,$ but
but I'm not sure if that could be used to prove that $F^4=I$.

答:
Given $g\in L_2(\mathbb R)$, you can prove that $(F \circ F)g(x)=g(-x)$ (as can be seen in this question: Fourier transform of the Fourier transform?. Here, they define Fourier Transform without $\frac{1}{\sqrt{2\pi}}$, but I'm sure you can deal with that).

Then, you just need to see that $F^4=F^2\circ F^2$.


问:
Extension of identity map on projective curve
I was reading the following proof from a lecture note:

Theorem: Let X and Y be two projective non-singular curves which are birationally equivalent. Then they are isomorphic.

Proof: $X$ and $Y$ are birationally equivalent $\implies$ $\exists$ open sets $U \subseteq X$ and $V \subseteq Y$ such that there is an isomorphism $\phi: U \to V$. Since $Y$ is projective and $X$ is non-singular, we can obtain a  morphism $\psi_1: X \to Y$ extending $\phi$. Similarly, there is morphism $\psi_2:Y \to X$ extending $\phi^{-1}$. Finally, X and Y are Hausdorff spaces and since $\psi_1\circ\psi_2:Y \to Y$ and $\psi_2\circ\psi_1: X \to X$ extend $\phi\circ\phi^{-1} = id_V$ and $\phi^{-1}\circ\phi = id_U$ respectively, we have $\psi_2\circ\psi_1 = id_Y$ and $\psi_1\circ\psi_2 = id_X$. Thus X and Y are isomorphic.

I'm completely fine with the Algebraic geometry part of the proof. What I'm struggling with is the Topology part. X is a Hausdorff space and $\psi_2\circ\psi_1: X \to X$ is an extension of $\phi^{-1}\circ\phi = id_U$ for open set $U \subseteq X$. Why does this imply $\psi_1\circ\psi_2 = id_X$? Can someone explain?

My definition of Projective variety and curve: A subset $Y$ of $\mathbb{P}^n$ is an algebraic set if it is zero set of a homogeneous polynomial in $k[x_0,x_1,...,x_n]$. A projective variety is an
irreducible algebraic set in $\mathbb{P}^n$ under Zariski topology. A projective curve is a projective variety of dimension 1.

Edit1: Is it true that open subset of a projective variety is always dense? Then I can see why the proof follows. But is this fact even true?

Edit2: I want to prove the above theorem using the following lemma:

Let $X$ be a curve, $Y$ a projective variety and $P \in X$ a non-singular point. Any rational map $\phi: X \to Y$ is defined at $P$.

As you can see, the above proof used this lemma when it said "we can obtain a  morphism $\psi_1: X \to Y$ extending $\phi$". So even if the above proof doesn't make sense, can you give another easy proof of this theorem that uses the above lemma (with the definition of projective curve I've mentioned)?

答:
For the question you ask in edit 1, any open subset of an irreducible topological space is dense. As your definition of variety includes irreducibility, the answer is yes.

For the method of solution of the problem using the fact you wrote in your second edit, the key statement is that the locus of agreement of two morphisms of varieties is closed(*). The strategy of the outlined proof is fine - if we let $U\subset X$ and $W\subset Y$ be the open sets where our rational morphisms $f_1:X\dashrightarrow Y$ and $f_2:Y\dashrightarrow X$ are defined, then $f_1\circ f_2: Y\to Y$ and $f_2\circ f_1:X\to X$ are rational maps which are equal to the identity on some open subset of $Y$ and $X$ respectively. By the lemma, $f_1\circ f_2$ and $f_2\circ f_1$ can be extended to honest morphisms $Y\to Y$ and $X\to X$. But this means that $f_1\circ f_2$ and $id_Y$ coincide on a dense open subset, so they're equal; similar reasoning for $f_2\circ f_1$ and $id_X$ gives that $X\cong Y$.

(*) This statement needs further conditions to be true once we upgrade to schemes - see here once you're at that level. If the quoted proof had said "separated" instead of Hausdorff, that would be a nod to this fact.


问:
For primes, $p_1 + p_2 +p_3+p_4 = p_1 p_2 p_3 p_4 - 15$
I am looking for primes that would satisfy this equation:

$p_1 + p_2 +p_3+p_4 = p_1 p_2 p_3 p_4 - 15$

I started by proving that an odd amount of these primes (two or four) cannot be equal to $2$ due to the parity of the left hand side and right hand side.

Then I assumed that exactly three of these primes would be equal to  $2$ and easily found a solution of $(2,2,2,3)$ and its permutations.

My problem now is, I cannot say that there isn't a fitting set of primes, out of which exactly one would be equal to $2$, or for all primes being odd. That leaves me with

$p_1+p_2+p_3=2p_1 p_2 p_3 - 17$

and

$p_1 + p_2 +p_3+p_4 = p_1 p_2 p_3 p_4 - 15$

respectively, and I am not sure how to go about those.

Is there a way to solve this assuming one of the primes is equal to $2$ and none of them is equal to  $2$?

答: Hint: Say $p_4\geq p_3\geq p_2\geq p_1\geq 2$, then we get $$4p_4\geq 8p_4-15$$


问:
Does $\sum_{k=1}^{\infty} \frac{1}{k}\cdot(\frac{1+i}{1-i})^k$ converge, absolutely converge?
$\sum_{k=1}^{\infty} \frac{1}{k}\cdot(\frac{1+i}{1-i})^k$, $a_k=\frac{1}{k}\cdot(\frac{1+i}{1-i})^k \in\mathbb{C}$

What I figured out so far:

$a_k=\left\{\begin{array}{0} \frac{1}{k}, & k\equiv0 \text{ (mod } 4) \\
         \frac{1}{k}i, & k\equiv1 \text{ (mod } 4)\\
-\frac{1}{k}, & k\equiv2 \text{ (mod } 4) \\
-\frac{1}{k}i, & k\equiv3 \text{ (mod } 4)\end{array}\right. $

I know that the series converges if the sequence of partial sums $s_n:=\sum_{k=1}^{n} a_k$ converges.

So $s_1 = -1, s_2=-\frac{3}{2}, s_3=\frac{5}{6}, s_2=\frac{13}{12}, s_2=\frac{53}{60},s_2=-\frac{7}{60}$.

Therefore I assume that $\lim_{n\to\infty} s_n$ exists.

I don't know how to handle the different cases for $a_k$ in my $s_n$.

I appreciate any hints.

答:
hint

Observe that
$$\frac{1+i}{1-i}=\frac{(1+i)(1+i)}{(1-i)(1+i)}$$
$$=\frac{2i}{2}=i$$

and
$$\left|\frac 1k \left(\frac{1+i}{1-i}\right)^k\right|=\frac 1k$$

with
$$\sum \frac 1k \;\; divergent$$


问:
Find the maximum and minimum values of of f(x,y,z) with the following constraints
Find the maximum and minimum values of $f(x,y,z) = x^3 -3y^2 + 4z$ subject to the constraint $g(x,y,z)=x + yz - 4 = 0$.

So far I've plugged this into Mathematica and it returned 6 complex conjugate coordinates and one real
$(4.00087, 0.0832872,-0.0104076)$.

I did a bit of manipulation of the constraint and found a second point $(4.00087, -0.0832872, 0.0104076)$ but when I evaluated both the values differed by less than 1.

Does this indicate that the function has no extreme values given the constraint?

答: The function is unbounded below and above. If you set $x=4$ and $y=0$, any $z$ satisfies the constraints while the objective function is $64+4z$.


答:
The left hand side grows linearly in the variables, while the right hand side grows as a degree four polynomial. We can use this to get a bound on the variables.
More precisely:

Without loss of generality assume that $p_1 \le p_2 \le p_3 \le p_4$ (all the other solutions can be obtained by permuting such solutions).

Since $p_4$ is a prime, $p_4 \ge 2$ and hence $8p_4 \ge 16 > 15$.
Now $p_1 p_2 p_3 p_4 = 15 + p_1 + p_2 + p_3 + p_4 < 8p_4 + p_4 + p_4 + p_4 + p_4 = 12 p_4$, and hence $p_1 p_2 p_3 < 12$.

If $p_3 \ge 3$, then we get $12 > p_1 p_2 p_3 \ge 2 \cdot 2 \cdot 3 = 12$, a contradiction. Hence $p_3 \le 2$. But $p_1 \le p_2 \le p_3$, so the only possible solutions is $p_3 = p_2 = p_1 = 2$.

Now the original equation becomes $6 + p_4 = 8p_4 - 15$, which yields $p_4 = 3$.
Clearly $(p_1, p_2, p_3, p_4) = (2, 2, 2, 3)$ is indeed a solution, so the solutions are

$(p_1, p_2, p_3, p_4) \in \{(2, 2, 2, 3), (2,2,3,2), (2,3,2,2), (3,2,2,2)\}$.

Title: For primes, $p_1 + p_2 +p_3+p_4 = p_1 p_2 p_3 p_4 - 15$

Body:
I am looking for primes that would satisfy this equation:

$p_1 + p_2 +p_3+p_4 = p_1 p_2 p_3 p_4 - 15$

I started by proving that an odd amount of these primes (two or four) cannot be equal to $2$ due to the parity of the left hand side and right hand side.

Then I assumed that exactly three of these primes would be equal to  $2$ and easily found a solution of $(2,2,2,3)$ and its permutations.

My problem now is, I cannot say that there isn't a fitting set of primes, out of which exactly one would be equal to $2$, or for all primes being odd. That leaves me with

$p_1+p_2+p_3=2p_1 p_2 p_3 - 17$

and

$p_1 + p_2 +p_3+p_4 = p_1 p_2 p_3 p_4 - 15$

respectively, and I am not sure how to go about those.

Is there a way to solve this assuming one of the primes is equal to $2$ and none of them is equal to  $2$?


答:
The question reduces to the following:

Find $p_1,p_2,p_3,p_4$ such that the equation

$$p_1p_2p_3p_4 - (p_1+p_2+p_3+p_4) = 15.$$

We claim however that there aren't too many possibilities to check here. Let us assume WLOG that $p_1 \le p_2 \le p_3 \le p_4$. Then

$$p_1p_2p_3p_4 -(p_1+p_2+p_3+p_4) = p_4(p_1p_2p_3 - \frac{p_1+p_2+p_3}{p_4}-1$$
$$ \ge p_4(8-4-1) = 3p_4.$$

[This because $p_1p_2p_3$ has to be at least 8 because each of $p_1,p_2,p_3$ is at least 2, and as $p_4 \ge p_1,p_2,p_3$, the fraction $\frac{p_1+p_2+p_3}{p_4}$ cannot be more than 3.]

So $3p_4$ cannot be more than 15 which implies that the largest prime $p_4$ cannot be more than 5.

Can you finish from here.


问:
Set of all functions $f:\mathbb R\to \mathbb R$ that are non-zero only finitely many times
I am looking for cardinality of set $S=\{f:\mathbb R\to \mathbb R\mid f(x)\not=0, \text{only in finitely many $x \in \mathbb R$}\}$.

My hunch tells me it is only uncountable. Obviously $F:\mathbb R\to S$ given by:
$$F(x)(t)=f_x(t)=\begin{cases}x,&t=1,\\0, & t \not = 1\end{cases}$$
is a well defined injection.

For other injection, I thought of mappings such as:
$$f \mapsto\left(\sum_{f(x)\not = 0}e^x,\sum_{f(x)\not = 0}e^{|x|},\sum_{f(x)\not =0}e^{-x},\sum_{f(x)\not = 0}f(x),\prod_{f(x)\not = 0}f(x)\right)$$
but I am not sure if $f$ is an injection or not. Any hint would be appreciated.

答:
Note that your set $S$ is nothing but the union
$$S=\bigcup_{F\in\mathcal{F}}\{f\colon\mathbb{R}\to\mathbb{R}:f|_{\mathbb{R}\setminus F}\equiv 0\}$$
where $\mathcal{F}$ is the family of all finite subsets of $\mathbb{R}$. For given $F$, the set $\{f\colon\mathbb{R}\to\mathbb{R}:f|_{\mathbb{R}\setminus F}\equiv 0\}$ has the cardinality $|\mathbb{R}^{F}|$, which is equal to the cardinality of $\mathbb{R}$ since $F$ is a finite set. Now, $\mathcal{F}$ is nothing but the union
$$\mathcal{F}=\bigcup_{n=1}^{\infty}\{S\subseteq\mathbb{R}:|S|=n\},$$
where for each $n$ the cardinality of the set $\{S\subseteq\mathbb{R}:|S|=n\}$ is $|\mathbb{R}^{n}|$, which is again equal to the cardinality of $\mathbb{R}$. Countable union of sets of caridnality $|\mathbb{R}|$ is of cardinality $|\mathbb{R}|$, so the set of all finite subsets of $\mathbb{R}$ has the cardinality equal to $|\mathbb{R}|$. Therefore, $S$ is a union, over a family $\mathcal{F}$ whose cardinality is $|\mathbb{R}|$, of sets of cardinality $|\mathbb{R}|$, thus we should have $|S|=|\mathbb{R}|$.


答:
Your injection $\Bbb{R} \to S$ implies $|S| \ge \mathfrak{c}$.

Define an injection $$S \to \bigcup_{A\subseteq \Bbb{R}\text{ finite}} (\Bbb{R}\setminus \{0\})^A$$

which maps $f \in S$ to a function $$\{x \in \Bbb{R} : f(x) \ne 0\} \to \Bbb{R}\setminus\{0\}, \quad x \mapsto f(x).$$

The latter set is
$$\bigcup_{A\subseteq \Bbb{R}\text{ finite}} (\Bbb{R}\setminus \{0\})^A = \bigcup_{n=0}^\infty \bigcup_{A\subseteq \Bbb{R},\,|A|=n} (\Bbb{R}\setminus \{0\})^n.$$
There are $\mathfrak{c}$ subsets of $\Bbb{R}$ with cardinality $n$, and $|(\Bbb{R}\setminus \{0\})^n| = \mathfrak{c}^n = \mathfrak{c}$
so
$$|S| \le \aleph_0 \cdot \mathfrak{c}\cdot \mathfrak{c} = \mathfrak{c}.$$

We conclude $|S|=\mathfrak{c}$.

Title: Set of all functions $f:\mathbb R\to \mathbb R$ that are non-zero only finitely many times

Body:
I am looking for cardinality of set $S=\{f:\mathbb R\to \mathbb R\mid f(x)\not=0, \text{only in finitely many $x \in \mathbb R$}\}$.

My hunch tells me it is only uncountable. Obviously $F:\mathbb R\to S$ given by:
$$F(x)(t)=f_x(t)=\begin{cases}x,&t=1,\\0, & t \not = 1\end{cases}$$
is a well defined injection.

For other injection, I thought of mappings such as:
$$f \mapsto\left(\sum_{f(x)\not = 0}e^x,\sum_{f(x)\not = 0}e^{|x|},\sum_{f(x)\not =0}e^{-x},\sum_{f(x)\not = 0}f(x),\prod_{f(x)\not = 0}f(x)\right)$$
but I am not sure if $f$ is an injection or not. Any hint would be appreciated.


问:
Finding the largest of $S_1,S_2,S_3,\cdots,S_{13}$
Let $S_n$ be the sum of the first $n$ terms of a arithmetic sequence $\{a_n\}.$ If $a_3 = 12,  S_{12} > 0$ and $S_{13} <0,$ then find the largest out of $S_1, S_2, S_3, \ldots, S_{13}.$

I first let $a$ be the first term and $d$ be the common difference. Then, I got that
\begin{align*}
S_1 &= a \\
S_2 &= 2a + d \\
S_3 &= 3a + 3d \\
S_4 &= 4a + 6d \\
S_5 &= 5a + 10d \\
S_6 &= 6a + 15d \\
S_7 &= 7a + 21d \\
S_8 &= 8a + 28d \\
S_9 &= 9a + 36d \\
S_{10} &= 10a + 45d \\
S_{11} &= 11a + 55d \\
S_{12} &= 12a + 66d \\
S_{13} &= 13a + 78d.
\end{align*}
However, I am unsure how to proceed from here. Can someone help please?

答: You have been given that $a_3=12$, that $S_{12}>0$ and that $S_{13}<0$. What does that tell you about $a$ and $d$? Can you express $S_n$ in terms of $a$ and $n$?


答: Such a function can be "encoded" by a finite sequence of reals, namely the places where $f\ne0$ and the values there. The set of such finite sequence has same cardinality as $\Bbb R$.


答:
You have $\left|\frac{1+i}{1-i}\right|=\frac{|1+i|}{|1-i|}=1$ and so, for each $k\in\Bbb N$,$$\left|\frac1k\left(\frac{1+i}{1-i}\right)^k\right|=\frac1k.$$Therefore your series doesn't converge absolutely.

But it converges, by Dirichlet's test: the sequence $\left(\frac1k\right)_{k\in\Bbb N}$ is monotonic and converges to $0$ and the partial sums of the series $\sum_{k=1}^\infty\left(\frac{1+i}{1-i}\right)^k$ are bounded, since, for each $K\in\Bbb N$,$$\left|\sum_{k=1}^K\left(\frac{1+i}{1-i}\right)^k\right|=\left|\frac{\frac{1+i}{1-i}-\left(\frac{1+i}{1-i}\right)^{K+1}}{1-\frac{1+i}{1-i}}\right|\leqslant\frac2{|1-i|}=\sqrt2.$$

Title: Does $\sum_{k=1}^{\infty} \frac{1}{k}\cdot(\frac{1+i}{1-i})^k$ converge, absolutely converge?

Body:
$\sum_{k=1}^{\infty} \frac{1}{k}\cdot(\frac{1+i}{1-i})^k$, $a_k=\frac{1}{k}\cdot(\frac{1+i}{1-i})^k \in\mathbb{C}$

What I figured out so far:

$a_k=\left\{\begin{array}{0} \frac{1}{k}, & k\equiv0 \text{ (mod } 4) \\
         \frac{1}{k}i, & k\equiv1 \text{ (mod } 4)\\
-\frac{1}{k}, & k\equiv2 \text{ (mod } 4) \\
-\frac{1}{k}i, & k\equiv3 \text{ (mod } 4)\end{array}\right. $

I know that the series converges if the sequence of partial sums $s_n:=\sum_{k=1}^{n} a_k$ converges.

So $s_1 = -1, s_2=-\frac{3}{2}, s_3=\frac{5}{6}, s_2=\frac{13}{12}, s_2=\frac{53}{60},s_2=-\frac{7}{60}$.

Therefore I assume that $\lim_{n\to\infty} s_n$ exists.

I don't know how to handle the different cases for $a_k$ in my $s_n$.

I appreciate any hints.


问:
Does a Space with the Finite Closed Topology have Fixed Point Property?
I wanted to know if a space $(X, \tau)$ with $\tau$ the finite-closed topology, have the Fixed Point Property, that is if every continuous mapping $f$ of $(X, \tau)$ into itself has a fixed point.

答: The natural numbers $\mathbb{N}$ with the cofinite( so that means finite sets are closed) topology does not have the fixed point property, since the $\sigma(n)=n+1$ map does not have fixed points, however takes open sets to open sets.


问:
arithmetic progression sequence , $\gcd(a,b)=1$
I have this question about arithmetic progression.

for a Natural Number $k>1$, the sequence :
$$1+L ,   1+2L   ,  1+3L ,\dots, 1+KL$$

its length is $K$

I need to pick $L$ > 0 Natural Number that makes every number in the sequence relatively prime.

and $a[i]-a[i-1]=d$ static

(no common divisor with any other number in the sequence $\gcd(a,b)=1$)

答:
Let $a_i = iL + 1$ for $i = 1,\ldots K$.

For any $i \ne j$,  let $d = \gcd(a_i,a_j)$.

Since $d$ divides both $a_i$ and $a_j$, $d$ divides $ia_j - ja_i = i-j$.

Since $1 \le i,j \le K$, we have $1 \le |i-j| \le K-1$. This implies
$$(i-j) | (K-1)!\quad\implies\quad d | (K-1)!$$

If we choose $L$ to be any multiple of $(K-1)!$, then $d|L$. As a result,

$$d | a_i\quad \iff\quad d |( iL + 1 ) \quad \implies \quad d | 1
\quad\implies\quad d = 1
$$

Since $i, j$ are arbitrary, this means whenever $L$ is a multiple of $(K-1)! {}^{\color{blue}{[1]}}$, all $a_i, a_j$ are pairwise relative prime to each other.

Note

$\color{blue}{[1]}$ - If you want a smaller $L$, you can replace $(K-1)!$ by ${\rm lcm}(1,2,\ldots,K-1)$ and that also work.


答:
The series can be written as
$$\sum _{k=1}^{\infty } \frac{i^k}{k}=\sum _{k=1}^{\infty } \frac{i^{2 k}}{2k}+\sum _{n=0}^{\infty } \frac{i^{2 k+1}}{2k+1}$$
The first one can be rewritten as
$$\sum _{k=1}^{\infty } \frac{i^{2 k}}{2k}=\sum _{k=1}^{\infty } \frac{(-1)^k}{2k}=-\frac{\log 2}{2}$$
which converges for the Leibniz test.

The second one can be rewritten as
$$\sum _{n=0}^{\infty } \frac{i^{2 k+1}}{2k+1}=i\sum _{n=0}^{\infty } \frac{(-1)^k}{2k+1}=\frac{i\pi}{4}$$
Thus the series converges to $$\sum _{k=1}^{\infty } \frac{i^k}{k}=-\frac{\log (2)}{2}+\frac{i \pi }{4}$$
As detailed in the answer above, the series does not converge uniformly because the absolute values series is the harmonic series which diverges.