问:
Number of Dyck paths with bounded height and fixed number of peaks
The number of Dyck paths of length $2n$ and exactly $k$ peaks is given by the Narayana numbers:
\begin{equation}
 \frac{1}{n} {n\choose k}{n \choose k-1}
\end{equation}

In a OEIS A080934, a recurrence relation is given for the number of Dyck paths with maximum height less than or equal to $h$, as well as a closed form solution. Looking at the same logic as in a previous question, it seems that the number of Dyck paths with halflength $n$, $k$ peaks and bounded in height by $h$ (call it $H(n,k,h)$) should obey the recursion relation
\begin{equation}
H(n+1,k,h) = \sum_{i=0}^n\sum_{j=0}^i H(i,j,h)H(n-i,k-j,h-1)
\end{equation}
With a few initial conditions such as $H(0,k,h) = \delta_{k0}$, $H(n,0,h) = \delta_{n0}$, $H(n,k,0) = \delta_{n0}\delta_{k0}$ .
From this recursion relation, is it possible to derive a trivariate generating function $G(x,y,z) = \sum_{x,y,z}H(n,k,h)x^ny^kz^h$, or even a closed form solution for H (though that might be optimistic)?

Edit: I previously had the wrong bounds on the sum, we must include the possibility of the first return being the empty path

答:
Consider the bivariate generating function, where we keep $h$ fixed:
$$
F_h(x,y)=\sum_{n,m\ge 0}H(n,k,h)x^ny^k
$$
By decomposing a Dyck path with height at most $h$ into the times when it returns to the $x$ axis, we see that a path with height $h$ is an ordered sequence of paths with height $h-1$. If you sum (length + 1) over all sub-paths, you get the length of the whole path. If you sum the peaks of the sub-paths, you get the total number of peaks, with one exception; an empty path has zero peaks, but it contributes on peak to the total.

Translating all this to generating function language,
$$
F_h(x,y)=\frac1{1-x(F_{h-1}(x,y)-1+y)}
$$
Unfortunately, I do not see how to derive the trivariate generating function you mentioned. However, you can prove by induction that $F_h$ can be written in the form
$$
F_h(x,y)=\frac{P_h(x,y)}{Q_h(x,y)}
$$
for some polynomials $P_h,Q_h$.


问:
Find eigenvalues of $T$ and find a basis $\beta$ for $P_2(\mathbb{R})$ such that$[T]_\beta$ is a diagonal matrix.
$T(f(x)=3f(x)+xf'(x)+xf''(x)$ what I did was, $T(1)=3+0+0=3,T(x)=3x+x+0=4x,T(x^2)=3x^2+2x^2+2=5x^2+2$, then $[T]_\beta=\begin{pmatrix}3&0&0\\0&4&0\\0&2&5\end{pmatrix}$ how should I continue after this?

答:
Were you given a problem like this without knowing what eigenvalues and eigenvectors are?  That's very strange!

The eigenvalues satisfy the equation $\left|\begin{array}{ccc}3-\lambda & 0 & 0 \\ 0 & 4-\lambda & 0 \\ 0 & 2 & 5- \lambda \end{array}\right|= (3- \lambda)(4- \lambda)(5- \lambda)= 0$ so are 3, 4, and 5.

An eigenvector corresponding to eigenvalue 3 satisfies the equation $\begin{bmatrix} 3 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 2 & 5\end{bmatrix}\begin{bmatrix} x \\ y \\ z\end{bmatrix}= \begin{bmatrix}3x \\ 4y \\ 2y+ 5z\end{bmatrix}= \begin{bmatrix} 3x \\ 3y \\ 3z \end{bmatrix}$ so 3x= 3x, 4y= 3y, and 2y+ 5z= 3z.  The first equation, 3x= 3x, is t ue for all x.  4y= 3y is only true for y= 0, and 2y+ 5z= 5z- 4z is only  true for z= 0. So the eigenvectors corresponding to eigevalue 3 are of the form $\begin{bmatrix}x \\ 0 \\ 0 \end{bmatrix}= x\begin{bmatrix}1 \\ 0 \\ 0 \end{bmatrix}$.

Do the same to find eigenvectors corresponding to eigenvalues 4 and 5.

ContentLicense: CC BY-SA 4.0


问:
How to Factorise these expressions?
I'm dong some Factorisation revision and I'm a bit stuck. I've never done this before and I'm stuck a bit. First of all I know how to do small factorisations and I know you have to find the Highest Common Factor and that it's the opposite of expanding but these expressions are scaring me a bit.

e.g. how would you do this $x^2 + 8x + 16$

or this $9m^2 - 24mn + 16n^2$

答:
To solve these kinds of exercises all you need to know is
$$
(a+b)^2=a^2+b^2+2ab,\quad (a-b)^2=a^2+b^2-2ab
$$
and then be a little smart.

Suppose we are looking at $9m^2-24mn+16n^2$ and want to write it in one of the above forms. Since there is a minus-sign involved, we should bring it on the form $(a-b)^2$ for some $a$ and $b$ to be determined. Now $9m^2$ should correspond to the $a^2$ term, $16n^2$ should correspond to the $b^2$ term and $-24mn$ corresponds to $-2ab$.

Edit upon OP's now deleted comment: Aligning your expression with the expression we obtain from expanding $(a+b)^2$ we have:
$$
\begin{align}
&a^2\;\;\;+b^2\;\;\;\,-2ab\\
=&9m^2+16n^2-24mn
\end{align}
$$
meaning that you should pick $a$ and $b$ such that $a^2=9m^2$, $b^2=16n^2$ and $-2ab=-24mn$. Which $a$ and $b$ does this correspond to?

ContentLicense: CC BY-SA 3.0

Title: How to Factorise these expressions?


问:
Connections between Clifford algebra and polynomials?
The multi-vector idea in Clifford algebra really reminds me of polynomial multiplication.

Can we treat every basic vector as a degree one polynomial, and multi-vectors as simply high-degree polynomials?

Is there any research trying to bridge the gaps between those two branches?

答:
Both are constructed as important quotients of the tensor algebra over a vector space $V$. The polynomials are heavily linked to the symmetric algebra (via a duality), which is basically "the tensor algebra where you've injected commutativity". The exterior algebra is "the tensor algebra into which you've injected anticommutativity"; a Clifford algebra is a quantization of the exterior algebra. There are interesting morphisms between all algebras of this bunch.

The property (analogy) I think that you're noticing is that of a graded (or more generally filtered) algebra. As for their multiplication, it is what we call a discrete convolution (bilinear/distributive multiplication): that's because they're all quotients of the tensor algebra which is the "most general" way of defining a bilinear multiplication over vectors.

You may also want to look at Weyl algebras, which are quantizations of the symmetric algebra.


问:
If for the real numbers $a,b(a\ne b)$ it is true that $a^2=2b+15$ and $b^2=2a+15$, then what is the value of the product $ab$?
If for the real numbers $a,b(a\ne b)$ it is true that $a^2=2b+15$ and $b^2=2a+15$, then what is the value of the product $ab$?

I tried to solve it as follows:

I state that $p=ab$

$p^2=(2b+15)(2a+15)$

$p^2=4ab+30(a+b)+225$

$p^2=4p+30(a+b)+225$

and this is where I got stuck. I don't know how to get over this hurdle. could you please explain to me how to solve the question?

Body:
$a^2=2b+15$ and $b^2=2a+15$

Subtracting, $a^2-b^2 = -2(a-b)$. As $a \ne b$,

$a+b = - 2$

Also adding both equations, $a^2+b^2 = 2(a+b)+30 = 26$

$(a+b)^2 = a^2+b^2+2ab \implies 4 = 26 + 2ab$

$ab = -11$


问:
Rational solutions
Find $ x \in \mathbb{Q} $ such that $ 1 + \frac{2}{[x]}  + \frac{3}{{x - [x] }} = 4 {x}$. Initially I was thinking that is easy. If we separate the therms with integer part and fractionary part and consider  the functions with fractionary part she is continous functions and we have a solution all the time but we don't know if it is rational. I was trying to consider and an equation of degree two but doesn't  work.

答:
If we write $k = [x]$, then the equation translates to $1 + \frac 2 k + \frac 3 {x - k} = 4x$, which after simplification gives $$4k x^2 - (4k^2 + k + 2)x + k^2 - k = 0.$$ Since this quadratic equation in $x$ has a rational solution, its discriminant $\Delta$ must be the square of some rational number $r$ (note that $k \neq 0$).

Thus we write $$r^2 = \Delta = (4k^2 + k + 2)^2 - 4 \cdot 4k \cdot (k^2 - k) = 16k^4 - 8k^3 + 33k^2 + 4k + 4.$$ Since $\Delta$ is an integer, we know that $r$ is also an integer, which we shall assumed to be non-negative.

Completing the square, we get $r^2 = (4k^2 - k + 4)^2 + 12k - 12$.

There are three cases: $k = 1$, $k > 1$, $k < 1$.

For $k = 1$, we solve the equation and get $x = 0$ or $\frac 7 4$. Obviously $x = \frac 7 4$ is a possible solution to the original equation.

If $k > 1$, then we have $r^2 > (4k^2 - k + 4)^2$ which gives $r \geq 4k^2 - k + 5$ (note that $4k^2 - k + 4$ is strictly positive for any $k$). Thus we have $$(4k^2 - k + 5)^2 \leq r^2 = (4k^2 - k + 4)^2 + 12k - 12$$ which leads to $8k^2 - 14k + 21 \leq 0$, which is impossible.

If $k < 1$, then a similar argument gives $r \leq 4k^2 - k + 3$ and hence $$(4k^2 - k + 3)^2 \geq r^2 = (4k^2 - k + 4)^2 + 12k - 12$$ which leads to $8k^2 + 10k - 5 \leq 0$. It is easy to see that the only integers $k$ satisfying this inequality are $k = 0, -1$, and solving the original equation confirms that no rational solution $x$ exists in this case.

Therefore the only rational solution is $x = \frac 7 4$.

Title: Rational solutions


问:
How to calculate the limit of $(tu^{\alpha} + (1-t) v^{\alpha})^{1/\alpha}$ when $\alpha\to0$?
as written in the subject, I'm interested in a proof of

$$\lim_{\alpha \to 0} \left( tu^{\alpha} + (1-t)v^{\alpha} \right)^{1/\alpha} = u^t v^{1-t},$$

where $t \in [0,1]$ and $u,v\geq0$. Can anybody help me out or give a hint?

Thanks in advance!

Background:

I came across the definition of $\alpha$-concave measures, where the quantity

$$M_{\alpha,t}(u,v) = \left( (1-t) u^{\alpha} + t v^{\alpha} \right)^{1/\alpha}$$

plays a role. Here, $t \in [0,1]$, $u,v > 0$ and $\alpha \in [-\infty,\infty]$, where for the cases $\alpha \in \left\{ 0, \pm \infty \right\}$ one defines $M_{-\infty,t}(u,v) = \min \left\{ u,v \right\}$, $M_{0,t}(u,v) = u^{1-t}v^t$ and $M_{\infty,t}(u,v) = \max \left\{ u,v \right\}$. It is clear to me that

$$M_{\pm \infty,t}(u,v) = \lim_{\alpha \to \pm \infty } M_{\alpha,t}(u,v)$$

and I assume the same relation holds at zero but don't see how to prove it.

答:
Assume $0<t<1$ and $u,v>0.$ Consider $f(x)=(tu^{x} + (1-t) v^{x})^{1/x}.$ We have that $$\ln f(x)=\frac{\ln(tu^{x} + (1-t) v^{x})}{x}.$$ We have, using L'Hôpital's rule, that

$$\lim_{x\to 0} \ln f(x)=\lim_{x\to 0}\frac{tu^x\ln u+(1-t)v^x\ln v}{tu^{x} + (1-t) v^{x}}=t\ln u+(1-t)\ln v=\ln (u^tv^{1-t}).$$ Thus we are done.


问:
Find integer in the form: $\frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b}$

Let $a,b,c \in \mathbb N$ find integer in the form: $$I=\frac{a}{b+c}
 + \frac{b}{c+a} + \frac{c} {a+b}$$

Using Nesbitt's inequality: $I \ge \frac 32$

I am trying to prove $I \le 2$ to implies there $\nexists \ a,b,c$ such that $I\in \mathbb Z$: $$I\le 2 \\ \iff c{a}^{2}+3\,acb+{a}^{3}+{a}^{2}b+a{b}^{2}+{b}^{3}+c{b}^{2}+{c}^{2}b+{c
}^{3}+{c}^{2}a-2\, \left( b+c \right)  \left( c+a \right)  \left( a+b
 \right) \le 0 \\ \iff {a}^{3}+{b}^{3}+{c}^{3}\leq c{a}^{2}+{a}^{2}b+a{b}^{2}+c{b}^{2}+{c}^{2
}a+{c}^{2}b+abc
$$
and stuck.

EDIT: Look like prove $I \le 2$ not a good thinking :P

Body:
I find a solution to $n=4,$when

a=2332797891204725453580403814216955612718693675609518139813675622446336
  8530351921955206357565424226029748329737767516130520072674084336131550
  2597616224970927979227396663481447506019173462295157784788781420305046
  5201815993661680059006448575315523206103260762210944137954571975497854
  9786027663601160534574317253280344812956727894696796553762212813889660
  9065956718516224446015577143267128739011935697434909021669583635379832
  35022557869209259

b=1161640217306132458900911441651415023972393417197892812143262449233898
  8034221463466278254018560734492913221738943224762433374574861704275058
  0062902808034990817009121975186967451351814311101112040391014295321972
  8784138582766210837461563508481437266175417187186208008663435889653439
  7066554486263784443013141020886435995672339322997499528376940620045001
  1919735272479457688230567501843839892799164246003766614214017398378635
  0444307965016411

c=5054729227475450427274369484803239479825091305751388135572448603576037
  6549781961422098862259430557133842304461180359698183208339647924784255
  6816542651386138853492649101592171641096957016404851774814750638840260
  3496289958758089911825477669004739864966841494437579004665357462952425
  4130327474390635537868978719887059697148297723373566417781389238382736
  3204638301684342182024187145267526992579708085994452308601529371953916
  7125415529515145

This may be not the smallest solution to $n=4$,but I think the smallest one is also very huge.

I'd like to tell something about how I get this answer.

I estimate that there will be a solution to $n=4,$(I also tried $n=3$,but failed.)

Let $a,b,c$ be rational numbers,and $a+b+c=1,a,b,c>0,$we get
$$\frac{a}{1-a}+\frac{b}{1-b}+\frac{1-a-b}{a+b}=4.\tag{eq.1}$$
denote $\dfrac{a}{1-a}=s,\dfrac{b}{1-b}=t,$then $a=\dfrac{s}{1+s},b=\dfrac{t}{1+t},$
$eq.1$ is equivalent to $$s+t+\dfrac{1-st}{s+t+2st}=4,\tag{eq.2}$$
denote $s=x+y,t=x-y,$then$$\frac{4 x^3+3 x^2-4 x y^2+y^2+1}{2 \left(x^2+x-y^2\right)}=4$$
$$y^2=\frac{4 x^3-5 x^2-8 x+1}{4 x-9}$$
denote $x=\dfrac{9k+1}{4k},$then $$y^2=\frac{(13 k+1) \left(4 k^2+9 k+1\right)}{16 k^2}$$
denote $r=4ky,$then $$r^2=(13 k+1) \left(4 k^2+9 k+1\right)=1 + 22 k + 121 k^2 + 52 k^3$$
denote $X=52k,Y=52r,$then $$Y^2 = X^3 + 121 X^2 + 1144 X + 2704.\tag{eq.3}$$
So we just need to find a rational solution of $eq.3$ which makes $a,b,c>0.$

I find two basic solutions of $eq.3,(X,Y)=(-104,260)(52,728),$we denote $p=(-104,260),q=(52,728),$

At first,I find that $6q=O,$when I calculate $p,2p,3p,4p,\cdots$ ,I find that $21p$ makes $-52<X<-33.8564,$in fact,$X\approx -36.8062,$this is what we need,because when $-52<X<-33.8564,a,b,c$ will all be positive(it need some calculation).

Finally,we can get $a,b,c>0$ from $X,Y,$ and then multiply by the LCM of the denominators of $a,b,c,$ we are done.

You can see that I only use $p,$ but no $q$ or other points on $eq.3,$so I have to plus $21$ times by $p$,which makes the digits so huge.


Body:
If for the real numbers $a,b(a\ne b)$ it is true that $a^2=2b+15$ and $b^2=2a+15$, then what is the value of the product $ab$?

I tried to solve it as follows:

I state that $p=ab$

$p^2=(2b+15)(2a+15)$

$p^2=4ab+30(a+b)+225$

$p^2=4p+30(a+b)+225$

and this is where I got stuck. I don't know how to get over this hurdle. could you please explain to me how to solve the question?

Body:
Alternatively,

If $a≠b$, then

$a^2-b^2=2(b-a)$

$$ a+b =-2$$

$(-2-b)^2-2b-15=0$

$$b^2+2b-11=0$$

$a^2-2(-2-a)-15=0$

$$a^2+2a-11=0$$

Then, by Vieta

$$ ab=-11.$$


问:
ODE - Floquet theorem recommendations
Can someone please refer me to a good textbook or some papers about the Floquet Theorem (English)?

Thanks

答: Linear System Theory by Wilson j. Rugh</p>

Title: ODE - Floquet theorem recommendations

Body: Linear System Theory by Wilson j. Rugh


问:
Show that if $T\in\mathcal{L}(V,W)$ injective then there exists $S \in \mathcal{L}(W,V)$ such that $ST$ is the identity map on $V$

Let $W$ be a finite-dimensional and $T\in \mathcal{L}(V,W)$. Show that if $T\in\mathcal{L}(V,W)$ injective then there exists $S \in \mathcal{L}(W,V)$ such that $ST$ is the identity map on $V$. I saw some posts on this statement but I didn't find the same approach as mine. If someone could give a feedback on my proof, I would really appreciate it. Thank you in advance.

As $T\in \mathcal{L}(V,W)$ is injective, then by fundamental theorem of linear maps we have that $\dim V=\dim \text{null}\ T+\dim \text{range}\ T=0+\dim \text{range}\ T\le \dim W$. So, $\dim V\le \dim W$.

Let $(v_1,...,v_n)$ be a basis of $V$ and $(w_1,...,w_m)$ be a basis of $W$ (so $n\le m$). Define $T$ and $S$ as the following:

$T(v_i)=w_i, S(w_i)=v_i$ with $i=1,...,n$.

Both $T$ and $S$ are well defined as $n\le m$. Thus, $ST(v_i)=S(w_i)=v_i$which is exactly an identity map as each $v\in V$ can be expressed uniquely.

答:
The point is that every subspace $W'$ of $W$ has at least one complement, i.e. a subspace $W''$ of $W$ such that $W$ is the direct sum of $W'$ and $W''$.  This means that every $w \in W$ can be expressed uniquely as $w' + w''$ for $w' \in W'$ and $w'' \in W''$.

So if $W'$ is the image of $T$, the injectivity of $T$ implies that for each $w' \in W'$, there is a unique $v_{w'} \in V$ such that $T(v_{w'}) = w'$.  Now for $w = w'+w'' \in W$, define

$$S(w) = v_{w'}.$$
Then $S$ is a linear map with the required properties.


问:
Domain of characteristic function
Hey if $X$ is random variable, then what is the domain of characteristic function of $X$? In many books there is written that the domain is $\mathbb{R}$, but sometimes authors takes value of characteristic function at point $-i$ (which is complex number) when they caculate $\mathbb{E}(e^X)$. So what is the domain and if complex, what are the differences between these two domains?

答:
Continuing my comment, the "approach" you mentioned in your post is invalid for some very common characteristic function.

It is well known that the c.f. of the standard exponential distribution $f(x) = e^{-x}, x > 0$ is

\begin{align*}
\phi(t) = \frac{1}{1 - it}, t \in \mathbb{R}.
\end{align*}

When you literally plugging $t = -i$, the denominator will be $0$, hence inconclusive. In fact, the expectation of $e^X$ is not a finite number for this case.

Title: Domain of characteristic function


问:
Rolling Dice Probability Distribution

In this question do I have to find the probability of 1,2,3,4,5, and 6 individually or do I have to find their probability together? Also isn't this binomial distribution?

答:
As per my understanding they are requesting to get exactly twice 1,2,3,4,5,6 in 12 rolls thus it is a multinomial, not a binomial

You have to calculate the probability of all the events like this one

$$\{1,3,5,1,2,2,4,6,5,3,4,6\}$$

Thus the requested probability is

$$\binom{12}{2,2,2,2,2,2}\left(\frac{1}{6}\right)^{12}=\frac{12!}{2^6}\left(\frac{1}{6}\right)^{12}\approx 0.34\%$$

Title: Rolling Dice Probability Distribution


问:
Existence of an Element with a particular order
I've solved the following problem in what seems to me to be a very inelegant way:

Let $G$ be an abelian group containing elements $a$ and $b$ of orders $m$ and $n$, respectively, where $m$ and $n$ are not necessarily relatively prime.  Show that $G$ must contain an element $c$ whose order $\mathcal{O}(c)$ equals the least common multiple of $m$ and $n$.

My solution is the following:

Let $d=\gcd(m,n)$ and note that $m=dk$, $n=dl$ where $k$ and $l$ are relatively prime. Now, define
      $$
      r=\prod_{p\mid d, p\mid k}p^i,
    $$
      where $p^i\mid d$ but $p^{i+1}\nmid d$. Now, by construction, since $k$ and $l$ are relatively prime, $r$ and $l$ are relatively prime. Likewise, since $r$ takes all of the primes in $d$ that are in common with $k$, we know $r$ and $\frac{d}{r}$ are relatively prime. Now we have
      \begin{align}
        &m=dk=r\left(\frac{d}{r}\right)k\\
        &n=dl=r\left(\frac{d}{r}\right)l
    \end{align}
      Clearly the element $a^{d/r}$ has order $rk$ and the element $b^{r}$ has order $\frac{dl}{r}$. Now the orders are relatively prime, so from a previous problem (where we showed that elements with rel. prime orders has the property that their product has the order of the product of their individual orders), we know the order of $a^{d/r}b^{r}$ is $dkl$, which by definition is the least common multiple of $m$ and $n$.

This doesn't strike me as being an elegant proof, so I'm wondering if there are any other ways to prove the statement?

答:
An alternative way is to do the prime fiddling inductively as follows.

LEMMA $\ $ A finite abelian group $\rm\:G\:$ has an  lcm-closed  order set, i.e.

$$\rm X,Y \in G\ \ \Rightarrow\ \ \exists\ Z \in G:\  o(Z) = lcm(o(X),o(Y)),\,\ \ o(X) =\: order\ of\ X$$

Proof $\ \ $  By induction on  $\rm\, o(X)\ o(Y).\ $ If it is $\,1\,$ then $\rm\:Z = 1\, $ works. $\ $ Otherwise

write  $\rm\ o(X)\ =\ AP,\: \ \ o(Y) = BP',\ \ \ P'\mid P = p^m > 1,\ \ $  prime $\rm\: p\:$ coprime to $\rm\: A,B.$

Then  $\rm\: o(X^P) = A,\ \ o(Y^{P'}) = B.\ $  By induction there is a $\rm\: Z\:$ with $\rm \: o(Z) = lcm(A,B)$

so  $\rm\ \ o(X^A\: Z)\: =\: P\ lcm(A,B)\: =\: lcm(AP,BP')\: =\: lcm(o(X),o(Y)).\ \ $ QED


问:
Little $o$ and big $O$

The sum of a little $o$ and a big $O$ is given by the little $o$?

The sum of two big $O$ is a little $o$?

In studying I have encountered an implication that seems to imply these facts but I am not sure. If it is, someone can explain me why this holds? Thanks in advance.

答: Strictly speaking, $O(f)$ and $o(f)$ are sets of functions, and we can add them in the sense$$S+T:=\{g+h|g\in S,\,h\in T\},$$where $g+h$ is a pointwise addition of functions $g,\,h$. With that out of the way, $o(g)+O(h)$ is $O(h)$ if $g\in O(h)$ or $o(g)$ if $h\in o(g)$. In some cases, neither of these conditions hold. For a small-$x$ convention (so $x^2\in o(x)\subset O(x),\,o(x^2)\subset O(x^2)\subset o(x)$) example, $g(x):=x,\,h(x):=\sin\tfrac1x$ would be such a tricky case. Meanwhile, as @vitamin.d noted, $O(g)+O(h)=O(\max\{g,\,h\})$ (in the sense of a pointwise maximum).

Title: Little $o$ and big $O$


问:
two numbers chosen from 0 to 1, expected val of larger number
If I pick two numbers in the interval (0,1), what is the expected value of the larger number?

Please help. I know that the expected value of one number is 0.5 but I'm not sure how to find the expected vlaue of the larger number here.

答:
Imagine the scenario with two die:

```
6 6 6 6 6 6
5 5 5 5 5 6
4 4 4 4 5 6
3 3 3 4 5 6
2 2 3 4 5 6
1 2 3 4 5 6
```

The expected value here is:

$$\frac{1+6+15+28+45+66}{36}=\frac{161}{36}=4.47\dot2$$

$161$ is the sixth term in the sequence of hexagonal pyramid numbers ($A002412$).

These are given by:

$$\frac{n(n+1)(4n-1)}{6}$$

On the unit grid, we can do a similar trick. However if say we use a resolution of $k$, then to calculate the expectation we use $kP(X=k)$, and over $k^2$ squares, to give the result as:

$$\lim_{k\to\infty}\frac{k(k+1)(4k-1)}{6k^3}=\frac23$$

Body:
Imagine the scenario with two die:

6 6 6 6 6 6
5 5 5 5 5 6
4 4 4 4 5 6
3 3 3 4 5 6
2 2 3 4 5 6
1 2 3 4 5 6

The expected value here is:

$$\frac{1+6+15+28+45+66}{36}=\frac{161}{36}=4.47\dot2$$

$161$ is the sixth term in the sequence of hexagonal pyramid numbers ($A002412$).

These are given by:

$$\frac{n(n+1)(4n-1)}{6}$$

On the unit grid, we can do a similar trick. However if say we use a resolution of $k$, then to calculate the expectation we use $kP(X=k)$, and over $k^2$ squares, to give the result as:

$$\lim_{k\to\infty}\frac{k(k+1)(4k-1)}{6k^3}=\frac23$$


问:
Calculate the measure in degrees of the acute angle formed by $ NMA $?
Consider the $PAT$ triangle such that the angle $ \angle P = 36 $ degrees, $ \angle A = 56 $ degrees and $ PA = 10 $. Knowing that the points $ U $ and $ G $ belong, respectively, to the sides $ TP $ and $ TA $, so that $ PU = AG = 1 $. If $ M $ and $ N $ are the midpoints of the $ PA $ and $ UG $ segments, respectively. What is the measure in degree of the acute angle formed by $ NMA $?

Can anyone give me any tips?

答:
Draw $PG$ and let its midpoint be $D$. Draw $MN$, $MD$ and $ND$.

Observe that, $MD\parallel AG$ and $DN\parallel PU$. Also, $\triangle MDN$ is isosceles because $MD=\frac{1}{2} AG=\frac{1}{2} PU=DN$.

Now, $\angle MDN=\angle MDG+\angle GDN=\angle PGT+\angle GPT=92^{\circ}$

Thus, $\angle NMA=180-\angle NMD-\angle DMP=180-44-56=\boxed{80}$


问:
For what values of p is the limit test invalid?
Let $a_n=1-2^{\frac{-1}{n}}$, $b_n=\frac{1}{n^p}$. Does using the limit comparison test work when $p>1$? I'm pretty sure that the limit of $\frac{a_n}{b_n}$ diverges to $\infty$, and that $\sum b_n$ converges. So the limit comparison test doesn't tell me anything new, correct?

答:
Only if
$$0<\lim_{n\to\infty}\frac{a_n}{b_n}<\infty$$
you can conclude that  either both series converge or both series diverge. In your example the limit is equal to $0$, so the LCT is inconclusive. To show the convergence of $\sum b_n$ one could use the integral test. Note that
$$\sum_{n=1}^{\infty} {\frac{1}{n^p}}\leq \int\limits_{1}^{\infty} {\frac{1}{n^p}}\,\mathrm{d}n.$$
This integral can be easily evaluated and it can be shown that the integral, and therefore the sum, converges for $p>1$.

Addendum:

The one-sided version with the limit superior, where the zero also is included, is unfortunately inconclusive too since only from the convergence of $\sum b_n$ it follows that $\sum a_n$ converges (you want to have it the other way around).


问:
Uniform continuity on $(0, \infty)$ and limits of the function at $\lim$ at $0$ and $\infty$
Are the following statements True or False?

If $ f : (0, \infty) \rightarrow \mathbb{R}$ is uniformly continuous then $\lim_{x \rightarrow 0} f(x)$ exists.

If $ f : (0, \infty) \rightarrow \mathbb{R}$ is uniformly continuous then $\lim_{x \rightarrow \infty} f(x)$ exists.

For part 1, this is what I tried : consider $\{x_n\}$, $\{y_n\}$ be two Cauchy sequences converging to 0. Then by uniform continuity $\{f(x_n)\}$ and $\{f(y_n)\}$ are Cauchy and convergent to the same point $l = \lim_{n \rightarrow \infty} f(x_n)$. To show $l = f(\lim_{n \rightarrow \infty} x_n) = f(0)$, which is satisfied as f is continuous on $(0, \infty)$ hence the limits can be exchanged. Hence $\lim_{x \rightarrow 0} f(x)$ exists.

For the second part, by uniform continuity on $[b, \infty)$, $b$ sufficiently large, given $\epsilon > 0$, there exists $\delta > 0$  such that whenever $|x-y| < \delta$, $|f(x)- f(y)| < \epsilon$. Consider a sequence $x_n \rightarrow \infty$. I am stuck here.

Could you please check my ideas and help me finish the problems? Thank you.

答: The first assertion is indeed true. The second one is false. Take the sine function, for instance.</p>


答:
Sure it works. The limit comparison test says that if $a_n, b_n \ge 0$ and

$$ \lim \frac{a_n}{b_n} = L$$

and $0 < L < \infty$ then $\sum a_n$ converges if and only if $\sum b_n$ converges.

So for instance if $a_n = \frac{1}{n(n + 1)}$ and $b_n = \frac{1}{n^2}$ then $b_n$ converges ($p > 1$) and hence so does $a_n$.

There is one special case where instead of an "if and only if" you have one-sided implication: if $L = 0$ then $\sum b_n$ converges implies $\sum a_n$ converges. I don't believe one can say anything if $L = \infty$.

For your series, $\lim a_n = 1$ (which needs to be the first thing you should notice). So if $b_n$ is any $p$-series then $\lim b_n = 0$ and hence $L = \lim a_n / b_n = \infty$. So you'll never be able to say anything using the limit comparison test for that series because $L$ will always be $\infty$. This is why you should always check if $\lim a_n = 0$ before you do anything else.

Title: For what values of p is the limit test invalid?

Body: Let $a_n=1-2^{\frac{-1}{n}}$, $b_n=\frac{1}{n^p}$. Does using the limit comparison test work when $p>1$? I'm pretty sure that the limit of $\frac{a_n}{b_n}$ diverges to $\infty$, and that $\sum b_n$ converges. So the limit comparison test doesn't tell me anything new, correct?


问:
solving a fraction without calculator
Can anyone explain how to get to the following solution without using a calculator:

$$\arctan(\frac{-10}{-10√3}) = \frac{π}{6}$$

?

答: First the obvious: 10/10sqrt(3)= 1/sqrt(3). "Tangent" in a right triangle is "opposite side over near side" so we have a right triangle with legs of length 1 and sqrt(3). By the Pythagorean theorem, the hypotenuse has length c, given by $c^2= 1+  3= 4$ so c= 2.  If we put two such triangles together, along their sqrt(3) sides, we have a triangle with sides 2, 2, and, 1+ 1= 2-  an equilateral triangle.  Each angle is 180/3= 60 degrees or pi/3 radians.  The original triangle is half of that- the angle is (pi/3)/2= pi/6 radians.

Title: solving a fraction without calculator


问:
Improving Zhang's prime gap
I am referring to Zhang's paper.

Since the set $\cal{H}$ is a subset of $[3.5\times 10^6, 7\times 10^7]$, shouldn't the prime gap he obtained be less than
$ 7\times 10^7 - 3.5\times 10^6$ rather than $7\times 10^7$, as stated in his paper? Or am I missing something here?

答:
Indeed, the gap size can be brought down to below 60 million (in particular, to 59,874,594) by more careful choice of parameters, according to a short preprint by Timothy Trudgian.

Edit: Further work has reduced the bound to 13,008,612.

Edit: Following Maynard's work and the subsequent improvements of the Polymath project, the bound has been lowered to 246: Variants of the Selberg sieve, and bounded intervals containing many primes

Title: Improving Zhang's prime gap


问:
simple derivative question at a point
Use the limit process to find the slope of the graph of the function at the specified point.
$f(x) = 3x − 2x^2,    (3, −9)$

I found $$Msec = \frac{(3(3+h)-2(3+h)^2-9)}{h}
=\frac{-2h^2+9h+18}{h}.$$ Where do I go from here?

答:
You have a few little sign mistakes in your calculation. The equation should be
$$\frac{3(3+h)-2(3+h)^2+9}{h}\frac{-2h^2-9h}{h}$$
which you can write as
$$\frac{-2h-9}{1} = -2h-9 \to -9 \ (h \to 0)$$
So the answer is $-9$.

Title: simple derivative question at a point


问:
Solving an exponential recurrence relation
Let $f(t)$ be a sequence s.t. $e^{f(t)}\geq e^{\frac{2}{n}\log n}+e^{\frac{2}{n}f(t-1)}$ where $f(0)=e^{\frac{2}{n}\log n}, t\in \mathbb{N}, t\geq 1$. I am trying to find a lower bound on $f(t)$ as a function of $n,t$. So I was trying to think of this as a recurrence inequality but am not sure how to solve an exponential recurrence relation because I'm used to linear recurrence relations. Any ideas?

If we look at a few terms, it looks as follows:

\begin{align*}
e^{f(t)} &\geq e^{\frac{2}{n}\log n} + e^{\frac{2}{n}f(t-1)} \\
&= e^{\frac{2}{n}\log n} + \left( e^{f(t-1)} \right)^\frac{2}{n} \\
&\geq e^{\frac{2}{n}\log n} + \left( e^{\frac{2}{n}\log n} + e^{\frac{2}{n}f(t-2)} \right)^{\frac{2}{n}}\\
&\geq \cdots
\end{align*}

答:
First, there is some confusion from using $n$ and $t\in \mathbb{N}.$ So, let $n^{2/n} = c.$ Now, let $g(t) = \exp(f(t)),$ so we have:

$$g(t) \geq c + g(t-1)^{2/n}\geq c + (c+ g(t-2)^{2/n})^{2/n}\sim c + c^{2/n} (1+ \frac2n \frac{g(t-2)^{2/n}}{c}) $$ for $n$ large. This shows that the trivial bound given by blamethelag is close to sharp.


问:
The conjugate function $\overline{f(z)}$ has derivative $0$ w.r.t $z$.
In ahlfors text it states that(formally)

If the rules of calculus were applicable, we would obtain $$\frac{\partial{f}}{\partial{z}}=\frac12\left(\frac{\partial{f}}{\partial{x}}-i\frac{\partial{f}}{\partial{y}}\right), \frac{\partial{f}}{\partial{\overline{z}}}=\frac12\left(\frac{\partial{f}}{\partial{x}}+i\frac{\partial{f}}{\partial{y}}\right)$$

And we see that $f$ being analytic,

$$\frac{\partial{f}}{\partial{x}} = -i\frac{\partial{f}}{\partial{y}}$$ holds, which gives that,
$$\frac{\partial{f}}{\partial{\overline{z}}}=0$$

Now , in text following he states that,

the conjugate function $\overline{f(z)}$ has derivative $0$ w.r.t $z$.

That is ,
$$\frac{\partial{\overline{f}}}{\partial{z}}=0$$

I didn't get this.

Won't this mean , $\frac{\partial{\overline{f}}}{\partial{z}}=\frac12\left(\frac{\partial{\overline{f}}}{\partial{x}}-i\frac{\partial{\overline{f}}}{\partial{y}}\right)$ , and from $\frac{\partial{\overline{f}}}{\partial{x}} = -i\frac{\partial{\overline{f}}}{\partial{y}}$

I didn't see how it's coming, i suppose their a mistake of mine considering $\frac{\partial{\overline{f}}}{\partial{z}} = \overline {\frac{\partial{f}}{\partial{z}}}$

Please explain,  where's the mistake , and what  does actually is the difference between differentiating a analytic function and its conjugate.

答: The conjugate function is analytic as a function of $\overline{z}.$ So, letting $u =\overline{z},$ the computation you give implies that $\frac{\partial \overline{f}}{\partial \overline{u}} = 0.$ But $\overline{u} = z,$ so what your professor says is true.


问:
Proof for The Limit $\lim_{n\to\infty}\frac{3n^2-n}{4n^2+1}=\frac{3}{4},\space n\in\Bbb{N}$
I am revising my knowledge on the topic of real analysis by attempting some simple proofs. The question requires for the proof of the Limit for $\lim_{n\to\infty}\frac{3n^2-n}{4n^2+1}=\frac{3}{4}$ using the definition of convergence: $$\forall\epsilon>0\space\space\exists N\in\Bbb{R}\space s.t\space\space \forall n>N,|s_n-L|<\epsilon$$ Below is my working: $$|\frac{3n^2-n}{4n^2+1}-\frac{3}{4}|<\epsilon$$ $$|\frac{3n^2-n}{4n^2+1}-\frac{3n^2+\frac{3}{4}}{4n^2+1}|=|\frac{-n-\frac{3}{4}}{4n^2+1}|=\frac{|-n-\frac{3}{4}|}{4n^2+1}<\frac{4n}{4n^2}=\frac{1}{n}<\epsilon $$ $$\frac{1}{n}<\epsilon\rightarrow\frac{1}{\epsilon}<n$$ $$\forall n>N=\frac{1}{\epsilon},\space|\frac{3n^2-n}{4n^2+1}-\frac{3}{4}|<\frac{1}{n}<\epsilon$$ By the definition of convergence,$\space\frac{3n^2-n}{4n^2+1}\rightarrow\frac{3}{4}$ as $n\rightarrow\infty.$

Is this correct? Any feedback would be greatly appeciated.


问:
Given a circle $(O, R)$ and a chord $AB=2a$. If $CA,CB$ tangents to it, then what is the area of $\triangle ABC$ equal to, with respect to a and R?
Given a circle $(O, R)$ and a chord $AB=2a$. If $CA,CB$ tangents to it, then what is the area of $\triangle ABC$ equal to, with respect to a and R?

I tried solving this question as follows:

$AO=OB=R$

$\triangle ACO=\triangle BOC$

$CA=CB$

Since $OC\perp AB$ and $\triangle ACB$ iscosceles, we just need to find the hight of $\triangle ABO$, from point $O$. This is what gave me difficulties. Could you please explain to me how to solve this question?

答:
Let $M$ be the midpoint of $AB$. Then $OCA$ and $ OAM$ are similar, which leads to
$$[ABC]= \frac{AM^2}{OM^2} [OAB] =\frac{AM^2}{OM^2} \cdot \frac12  AB\cdot OM=\frac{AM^3}{OM}= \frac{a^3}{\sqrt{R^2 -{a^2}}}
$$


问:
A problem of permutations using the probabilistic method.
If found this problem in the book "The probabilist method" of Alon and Spencer:

Prove that there is an absolut constant $c>0$ with the following property: let $A$ be a $n\times n$  matrix with pairwise distinct entries. Then there is a permutation of the rows of $A$ so that no column in the permuted matrix contains an increasing subsequence of lenght at least $c\sqrt{n}$.

I have no idea of how to begin.

答:
I spent some time going through all the exercises of this book two years ago and here is the solution I found for this one.

We let $c$ be undetermined for the moment. Given a sequence of $n$ distinct numbers, we say that it is good if it has no increasing subsequence of length at least $l = \lceil{c\sqrt{n}}\rceil$, otherwise we call it bad. Suppose we have a random permutation $\pi$ of $[n]$. Now, if we fix some indices $1\leq i_1 < i_2< \ldots < i_{l}\leq n$, the chance that $\pi(i_1), \pi(i_2),\ldots, \pi(i_{l})$ is increasing is
\begin{align}
    \mathbb{P} \left[ \pi(i_1),\ldots, \pi(i_{l}) \text{ is bad}\right]  =  \frac{1}{l!}.
\end{align}
Therefore, the chance that $\pi(1),\ldots,\pi(n)$ is bad is bounded by
\begin{align}
    \mathbb{P}(\pi(1),\ldots, \pi(n) \text{ is bad}) &\leq   \bigvee\limits_{1\leq i_1 < \ldots < i_{l} \leq n}{\mathbb{P}  [ \pi(i_1),\ldots, \pi(i_{l}) \text{ is bad}} ]\\
    &= \binom{n}{l}\frac{1}{l!},
\end{align}
Now, given an $n\times n$ matrix of distinct numbers, the chance that a random permutation of the rows creates at least one bad column satisfies
\begin{equation}
    \mathbb{P} (\exists \text{ bad column}) \leq n \binom{n}{l} \frac{1}{l!}.
\end{equation}
Now, we have some freedom in choosing $c$, so let $c>e$, implying that
\begin{equation}
    e < \frac{\lceil{ c\sqrt{n}} \rceil}{\sqrt{n}} = \frac{l}{\sqrt{n}},\qquad \implies\qquad \frac{e}{l} < \frac{1}{\sqrt{n}}.
\end{equation}
Then we can use the bound $\binom{n}{l} \leq ((ne)/l )^l$ to obtain
\begin{align}
   \mathbb{P} (\exists \text{ bad column})\leq n \binom{n}{l} \frac{1}{l!}< n \left( \frac{n\cdot  e}{l} \right)^{l}\frac{1}{l!}< n  \frac{n^{\frac{l}{2}}}{l!}
\end{align}
We use Stirling's approximation in the form $l! \geq \sqrt{2\pi l }\cdot l^{l}e^{-l} $ to get the bound
\begin{equation}
    \mathbb{P} (\exists\text{ bad column})\leq n  \frac{n^{\frac{l}{2}}e^l}{\sqrt{2\pi l}l^{l} } <\frac{n\cdot n^{-\frac{l}{2}}}{\sqrt{2\pi l}} = \frac{n^{1 - \lceil{c\sqrt{n}\rceil}/2}}{\sqrt{2\pi \lceil{c\sqrt{n}\rceil}}}.
\end{equation}
Our choice of $c>e$ implies that $l = l(n) \geq 3$ for all $n\geq 1$, and the above bound becomes
\begin{equation}
    \mathbb{P} (\exists\text{ bad column})< \frac{ n^{-\frac{1}{2}}}{\sqrt{6\pi}} < \frac{1}{\sqrt{6 \pi}}.
\end{equation}
So, with positive probability, there are no bad columns, and so a permutation having no bad columns exists. Moreover, the above probability that a random permutation produces a bad column decreases at least as fast as $n^{1- e\sqrt{n}/2}$, so if we want to obtain an instance of such a permutation, a random one will very likely do the trick!


答:
I think you need more words, and that as it stands the second line isn't quite right.

I would write

Given $\epsilon>0$ we want to find an $N$ such that for $n>N$ we have
$$|\frac{3n^2-n}{4n^2+1}-\frac{3}{4}|<\epsilon.\tag 1 $$

Let us see how big the left hand side can be:
$$|\frac{3n^2-n}{4n^2+1}-\frac{3n^2+\frac{3}{4}}{4n^2+1}|=|\frac{-n-\frac{3}{4}}{4n^2+1}|=\frac{|-n-\frac{3}{4}|}{4n^2+1}<\frac{4n}{4n^2}=\frac{1}{n},$$
so that (1) will be true for all $n>N:=\frac{1}{\epsilon}$.

Title: Proof for The Limit $\lim_{n\to\infty}\frac{3n^2-n}{4n^2+1}=\frac{3}{4},\space n\in\Bbb{N}$

Body:
I am revising my knowledge on the topic of real analysis by attempting some simple proofs. The question requires for the proof of the Limit for $\lim_{n\to\infty}\frac{3n^2-n}{4n^2+1}=\frac{3}{4}$ using the definition of convergence: $$\forall\epsilon>0\space\space\exists N\in\Bbb{R}\space s.t\space\space \forall n>N,|s_n-L|<\epsilon$$ Below is my working: $$|\frac{3n^2-n}{4n^2+1}-\frac{3}{4}|<\epsilon$$ $$|\frac{3n^2-n}{4n^2+1}-\frac{3n^2+\frac{3}{4}}{4n^2+1}|=|\frac{-n-\frac{3}{4}}{4n^2+1}|=\frac{|-n-\frac{3}{4}|}{4n^2+1}<\frac{4n}{4n^2}=\frac{1}{n}<\epsilon $$ $$\frac{1}{n}<\epsilon\rightarrow\frac{1}{\epsilon}<n$$ $$\forall n>N=\frac{1}{\epsilon},\space|\frac{3n^2-n}{4n^2+1}-\frac{3}{4}|<\frac{1}{n}<\epsilon$$ By the definition of convergence,$\space\frac{3n^2-n}{4n^2+1}\rightarrow\frac{3}{4}$ as $n\rightarrow\infty.$

Is this correct? Any feedback would be greatly appeciated.


问:
Increasing net of projections in von Neumann algebra
In $B(H)$, there exists an increasing sequence of finite rank projections convergent to the identity in SOT.

Is there an analogous statement for a  von Neumann algebra $M\subseteq B(H)$?

答: No, $M$ need not contain any nonzero finite rank projections.  For a really simple example, $M$ could consist of only the scalar multiples of the identity.  Or, if $H=L^2(\mathbb{R})$, then $M$ could be $L^\infty(\mathbb{R})$, which contains no nonzero finite rank operators since Lebesgue measure is atomless.


答:
Given $\epsilon>0$ be any number.

Now
\begin{align}
&\left|\frac{3n^2-n}{4n^2+1}-\frac{3}{4}\right|\\
&=\left|\frac{-4n-3}{4(4n^2+1)}\right|\\
&=\frac{4n+3}{4(4n^2+1)}\\
&<\frac{n+1}{(4n^2+1)}\\
&<\frac{n+1}{(n^2+1)}\\
&<\frac{n+1}{n^2}<\epsilon\quad &\text{whenever}\quad n^2\epsilon-n-1>0\\
&&\text{i.e whenever}\quad n>\dfrac{1+\sqrt {1+4\epsilon}}{2\epsilon}\\
\end{align}

Now you can take $N=\left[{\dfrac{1+\sqrt {1+4\epsilon}}{2\epsilon}}\right]+1$ and this also allows $N$ to be a positive integer rather than just a real number which, in cases, can be difficult to find out.

Therefore

$\left|\frac{3n^2-n}{4n^2+1}-\frac{3}{4}\right|<\epsilon$ whenever $n>N$


问:
Understanding the definition of Lebesgue integral in "Real and complex analysis"-book
I'm reading the "Real and complex analysis"-book by Rudin. I'm currently in the section where the Lebesgue integral is being defined using simple functions. The definition of the Lebesgue integral is stated in the following way:

Definition: If $s:X\to[0, \infty)$ is a measurable simple function, of the form:

$$s(x)=\sum_{i=1}^n \alpha_i \textbf{1}_{A_i}(x),$$

where $\alpha_1, ...\alpha_n$ are the distinct values of $s$,
$\textbf{1}_{A_i}(x)$ is an indicator function, and if
$E\in\mathbb{R}$, we define

$$\int_E s\,d\mu=\sum_{i=1}^n \alpha_i\,\mu(A_i\cap E).$$

If $f:X\to [0, \infty]$ is measurable, and $E\in\mathbb{R}$, we define
the Lebesgue integral of $f$ over $E$ with respect to measure $\mu$
as:

$$\int_E f\,d\mu=\sup_{0\leq s\leq f} \int_E s\,d\mu$$

Now if I look at the definition of Lebesgue integral in wikipedia I get that:

$$\int f\,d\mu = \int_0^\infty \mu(\{x\,|\,f(x)>t\})\,dt,$$

where $dt$ is a differential change in the value of $f$, that is $dt = \lim_{\Delta x\to0}f(x+\Delta x)-f(x).$

Now the wiki definition makes sense to me ("summing the slices upwards"), but the definition in Rudin's book confuses me. How are these the same:

$$\sup_{0\leq s\leq f}\sum_{i=1}^n \alpha_i\,\mu(A_i\cap E)=\int_0^\infty \mu(\{x\,|\,f(x)>t\})\,dt\;\;\;?$$

I could understand this if the $\alpha_i$ would be the differential changes corresponding to $dt$.?

答:
They are in fact equal and you can set up the theory of the Lebesgue integral starting from either one and arrive at the other. We just had this last semester as part of calc III and the proof is not trivial (depending on how far you are into the topic), but very easy to follow once you have the basic tools to prove it assembled. We did it with measure theoretic induction, monotone convergence etc. In fact, we started from the definition you quoted from the book.

I have not read that particular piece, but I would not be surprised to see the author prove that equality throughout his book.

This explanation maybe does not directly give intuition for how to prove that both expressions are equal. But maybe it is convincing enough to suggest that both arrive at the same value, which is "computing the integral for functions and sets not as well-behaved as in the Riemann case". The integral of simple functions basically means you go through all values of that function's domain and sum up the corresponding area, i.e. per $\alpha_{i} $ you compute the area as $\alpha_{i}\cdot \mu(s=\alpha_{i})$ and take the sum to arrive at $\int s$. Then $\sup_{0\leq s\leq f} \int s$ basically equates to going through all approximations of f through simple functions s.t. $s \leq f$ (similar to the Riemann upper sum), computing their integrals, and taking the closest match.

Edit:

If you restrict this to the Lebesgue integral this follows quickly from Fubini's theorem and another result which states $\int_{\mathbb{R}^{n}}f d\mathcal{L}^{n}=\mathcal{L}^{n+1}(\{(x,y)\in \mathbb{R} \mid 0\leq y < f(x)\})$.

For general measure spaces the proof goes like this. If $\exists k\in\mathbb{N}\setminus{\{0\}}:\mu(f^{-1}((\frac{1}{k},\infty])=\infty \implies$ both sides are infinite.

So we can assume $\forall k \in\mathbb{N}\setminus{\{0\}}:\mu(f^{-1}((\frac{1}{k},\infty])<\infty$.
Thus, $f^{-1}((0,\infty])$ is $\sigma$-finite (will be relevant in Step 2). For functions where this is the case, the integral (defined as the sup of simple $s$ s.t. $s\leq f$) can be written as follows:
$\int_{E} f d \mu = sup\big\{\int_{F} \varphi d \mu \mid F \subseteq E, F \in
 \mathcal{S},\mu(F)<\infty, \varphi : F\rightarrow [0,\infty) measurable, \#\varphi(F)<\infty, 0\leq\varphi\leq f\big\}$.

The proof for the above is lengthier, so I am skipping that, but it's a useful result for proving your equality.

This proof follows measure theory induction (starting with simple functions, then working your way up to arbitrary measurable functions).

Step 1: let f be linear combination of indicator functions. Then $f=\sum_{j=1}^{J}c_{j}\chi_{F_j}$ s.t. $F_j \cap F_k=\emptyset$ for $j \neq k$. Then $f^{-1}((t,\infty])$ is the union of the $F_j$ s.t. $c_j > t$, therefore $\mu(f^{-1}(t,\infty])=\sum_{j} \mu(F_{j})\chi_{[0,c_{j})}$. Linearity of the integral implies $\int_{(0,\infty)}\mu(f^{-1}((t,\infty]))d\mathcal{L}^{1}(t)=\int_{(0,\infty)}\sum_{j}\mu(F_{j})\chi_{[0,c_{j})}(t)d\mathcal{L}^{1}=\sum_{j}\mu(F_{j})c{j}=\int_{E}f d\mu $

Step 2: let f be positive. Let $\{\varphi_{k}\}_{k\in\mathbb{N}}$ be a monotone series of integrable functions s.t. $\#\varphi_{k}(E)<\infty$ and $\varphi_{k} \rightarrow f$ pointwise. This series exists because $f^{-1}((0,\infty])$ is $\sigma$-finite (a result that is proven as part of the theory). Monotone convergence (another important result) now implies
$\int_{E}fd\mu=lim_{k\rightarrow \infty}\int_{E} \varphi_{k}d\mu$ [1]

Let $t\in(0,\infty)$, then $\varphi_{k}\leq\varphi_{k+1}$ implies $\{x\mid \varphi_{k}(x)>t\} \subseteq \{x\mid \varphi_{k+1}(x)>t\}$ and pointwise convergence leads to $\{x\mid \varphi_{k}(x)>t\}=\bigcup_{k\in\mathbb{N}}\{x\mid \varphi_{k}(x)>t\}$. Hence $\mu(\{x\mid \varphi_{k}(x)>t\})\rightarrow \mu(\{x\mid f(x)>t\})$ monotonically. Therefore, $\lim_{k\rightarrow \infty}\int_{0,\infty)}\mu(\{x\mid \varphi_{k}(x)>t\})d\mathcal{L}^{1}(t)=\int_{(0,\infty)}\mu(\{x\mid f(x)>t\})d\mathcal{L}^{1}(t)$

The equality immediately follows by applying Step 1 to every $\varphi_{k}$ and equation [1]


问:
Finding what the sum of $(2^n)(3^{(-n+1)})$ from $n=1$ to infinity converges to
Sorry for the lack of math symbols but I'm trying to find what the sum of
$f(x)=(2^n)(3^{-n+1})$ from $n=1$ to infinity converges to. I have tried using the sequence of partial sums but I'm too stupid to find what Sn is equal to in order to take the limit. Any help would be greatly appreciated.

答:
Define $$S=\sum_{n=1}^{\infty}2^n\cdot3^{-n+1}.$$

Observe that $$S=\sum_{n=1}^{\infty}2^n\cdot3^{-n+1}=-2^0\cdot3^{-0+1}+\sum_{n=0}^{\infty}2^n\cdot3^{-n+1}=-3+\underbrace{\sum_{n=0}^{\infty}2^n\cdot3^{-n+1}}_{:=S_0}.$$
Rewrite the summand:
$$S_0=3\sum_{n=0}^{\infty} \frac{2^n}{3^{n}}=3\sum_{n=0}^{\infty} \left(\frac{2}{3}\right)^n.$$
By the geometric series this equal to $$\frac{3}{1-\frac{2}{3}}=9.$$ Since
$S=S_0-3$, the final answer is $S=6$.


问:
Motivation for complex numbers
I understand that the complex numbers have a geometric representation as ordered pairs of real numbers. From what I understand complex numbers are very useful for working in two dimensions mathematically. The fact that $i^2 = -1$ follows from the definition of multiplication of complex numbers:

$(a,b)(c,d) = ac-bd+(bc+ad)i$

My question is this:
If one was simply working with ordered pairs of real numbers with no knowledge of $i=\sqrt{-1}$, would this definition of multiplication ever follow logically from a practically arising situation?

答:
While the trigonometric addition formulae do indeed imply complex multiplication, it is overkill, and in fact it is backwards. There is a perfectly natural motivation for complex multiplication, from which all of trig follows.

It is as follows. Given complex numbers $z$ and $z'$ in the first quadrant, let $z''$ be obtained by stacking $z'$ atop $z$, as in the figure linked here. (In the figure, $z'=x'+iy'$,  and the circle has center $w=x'z$ and radius $y'$.)

This is angle stacking. Use the figure to show the intersection of the circle with the arc is $zz'$ and $z/z'$.

Once you know this, define the argument $\theta(z)$ by the Archimedes bisection method: the limit of the increasing sequence of chord-length sums obtained by repeated bisection of the subtended arc on the unit circle. Then one shows directly

$$\theta(zz') = \theta(z) + \theta(z')$$

and $\theta(z)$ is a bijection of the punctured unit circle $z\not=-1$ with the interval $(-\pi,\pi)$, where $\pi=2\theta(i)$. Let $\sin\theta$ and $\cos\theta$ be the imaginary and real parts of the inverse $z(\theta)$. Then the trig identities follow.

Moral: Complex multiplication is a direct consequence of angle stacking.

Title: Motivation for complex numbers


问:
Quantitative trading
What math subjects are used in quantitative trading firms?</p>

答:
I stumbled upon Quantstart.com, lately. There is a series of articles, called "How to learn advanced mathematics without heading to university", which gives an answer to your question in great detail:

Part 1

Part 2

Part 3

Part 4

ContentLicense: CC BY-SA 2.5

Title: Quantitative trading


问:
Number of words with two "$A$" using two letters from "$RATA$" and three letters from "$TIERRA$"

Find the number of words with two "$A$" using two letters from "$RATA$"
  and three letters from "$TIERRA$".

What I did:

There are two cases, one where I choose both $A$ from $RATA$ and the rest of the letters from $TIERRA$, and the other one where I choose an $A$ from $RATA$ and the other $A$ from $TIERRA$:

1. Choosing both $A$ from $RATA$:

There are ${5}\choose{2}$ ways to put the $A$s on the word, and then I choose the other three letters from $TIERRA$: $\frac{5\cdot4\cdot3}{2!}$

So there are $5\choose2$$\frac{5\cdot4\cdot3}{2!}=300$ words using both $A$ from $RATA$.

2. Choosing an $A$ from $RATA$ and the other one from $TIERRA$:

Again, there are ${5}\choose{2}$ ways to put the $A$s on the word, there are only two options to choose a letter from $RATA$ (either $R$ or $A$), and then there are $\frac{5\cdot4}{2!}$ ways to choose the remaining two letters from $TIERRA$.

So there are ${5}\choose{2}$$\frac{5\cdot4}{2!}=100$ words in this case.

By the rule of sum, there are $300+100=400$ words for this problem.

But the solution to this problem is $430$. What am I doing wrong?

答:
The apparent complication of having two sources for letters is nullified by the restrictions, assuming that we require exactly two $A$s: the letters then are "AA" + three from "RRRTTEI".

If we allow three $A$s (and arguably a word with three $A$s has two $A$s), we only need to add two from "RRTIE" (all from TIERRA) once the three As are assumed chosen. Note the different remaining choices.

For a given unordered multiplicity pattern in the additional chosen letters, say $AA+xxy$, we can fill this from the choice of letters based on which letters occur at a suitable multiplicity. For the example $AA+xxy$, $x$ can be chosen form $R$ or $T$, then $y$ is a free choice from the remaining $3$ letters. Then the arrangements of this pattern are the multinomial $\binom {5}{2,2,1} = 30$.

So once we identify all patterns we can assess each in turn to get a total answer:
$$\begin{array}{|c|c|} \hline
\text{For this pattern:} & \text{options to fill} & \text{arrangements} &  \text{total options} \\[1ex] \hline
 AA+xxx & \binom 11  = 1  & \binom{5}{3,2} = 10  &  10 \\[1ex]
 AA+xxy & \binom 21 \binom 31  = 6  & \binom{5}{2,2,1} = 30  &  180 \\[1ex]
 AA+xyz & \binom 43 = 4  & \binom{5}{2,1,1,1} = 60  &  240 \\[1ex]\hline
 &   & \text{total for }AA &  430 \\[1ex]\hline
 AAA+xx & \binom 11 = 1  & \binom{5}{3,2} = 10  &  10 \\[1ex]
 AAA+xy & \binom 42 = 6  & \binom{5}{3,1,1} = 20  &  120 \\[1ex] \hline
 &   & \text{total, }AA\text{ or }AAA &  560 \\[1ex]\hline
\hline
\end{array}$$

give a base answer of $\fbox{430}$ options for exactly two $A$s and an extended possibility for the case when three $A$s are also allowed of $\fbox{560}$ options


问:
On proving that $\sum\limits_{n=1}^\infty \frac{n^{13}}{e^{2\pi n}-1}=\frac 1{24}$
Ramanujan found the following formula:

$$\large \sum_{n=1}^\infty \frac{n^{13}}{e^{2\pi n}-1}=\frac 1{24}$$

I let $e^{2\pi n}-1=\left(e^{\pi n}+1\right)\left(e^{\pi n}-1\right)$ to try partial fraction decomposition and turn the sum into telescoping, but methinks it doesn't lead anywhere and only makes things hairy.

How does one go about proving this? Thanks.

答:
This is not a strict solution but a heuristic proof using CAS. It also shows that there are other "magic" numbers instead of 13 which result in similar simple fractions. Playing around I also found the interesting conincidence that in some cases the sum and the related integral give the same results.

Defining

$$S(m) = \sum_{n=1}^{\infty} \frac{n^m}{e^{2 \pi n}-1}\tag{1}$$

we have to show that $S(13) = \frac{1}{24}$.

My first idea was to identify the denominator. I knew this expression from the Planck radioation formula and from the generating function of the Benoulli numbers.

But why so complicated? It is just the sum of a power series. Indeed we can write

$$\frac{1}{e^{2 \pi  n}-1} = \sum _{j=1}^{\infty } \exp (- 2 \pi j n)\tag{2}$$

Next, for the numerator we replace the power of $n$ by $z^n$ and consider the intermediate generating function

$$g_{0}(z,j) = \sum _{n=1}^{\infty } z^n \exp (- 2 \pi j n)=\frac{z}{e^{2 \pi  j}-z}\tag{3}$$

Summing over $j$ gives the generating function

$$\begin{align}g(z) = & \sum _{j=1}^{\infty } \frac{z}{e^{2 \pi  j}-z}\\=& -\frac{1}{2 \pi}\psi _{e^{-2 \pi }}^{(0)}\left(1-\frac{\log (z)}{2 \pi }\right)+\frac{1}{2 \pi} \log \left(1-e^{-2 \pi }\right)
\end{align}\tag{4}$$

Here $\psi _{q }^{(0)}(x)$ is the q-digamma function.

The the sums in question here can be found as derivatives of $g(z)$

$$S(m,z) = (z\frac{\partial}{\partial z})^m g(z) |_{z \to 1}\tag{5}$$

Using Mathematica for some values of $m$ I saw the pattern and found the general formula

$$S(m,z)=\frac{(-1)^{m+1} \psi _{e^{-2 \pi }}^{(m)}\left(1-\frac{\log (z)}{2 \pi }\right)}{2^{m+1} \pi ^{m+1}}\tag{6}$$

which gives at $z=1$ where the $\log$ vanishes the result

$$S(m)=S(m,z\to 1) = \frac{1}{(2 \pi )^{n+1}} \psi _{e^{-2 \pi }}^{(n)}(1)\tag{7}$$

Finally, the numerical results in Mathematica for $\frac{1}{S(m)}$ were particularly simple for 3 values of $m$. In the format {m,1/S(m)} we have

$$\{\{5, 504\}, \{9, 264\}, \{13,24\}\}\tag{8}$$

The case $m=13$ leads to the desired result. The "magic" values of $m$ have the form $1+4k$.

Discussion

It is always tempting with sums to look at the corresponding integral.

In our case we consider

$$i(m) = \int_0^{\infty } \frac{n^m}{\exp (2 \pi  n)-1} \, dn\tag{9}$$

where in contrast to the sum the integration starts at $n=0$.

The surprising observation is that we have (checked numerically)

$$i(m) = S(m),m=1+4k, k=1,2,...\tag{10}$$

which includes the "magic" values.

It would be nice to find a proof of this observation


问:
How to show that the polynomial $ab^3 + cd^3$ is irreducible?
I have to show that the polynomial:

$$ab^3 + cd^3 \in \mathbb{C}[a,b,c,d]$$

cannot be factorised into polynomials of lower degrees, i.e. it is not reducible. However, I'm quite unsure on how to proceed here. I thought I could try to factorise this into a linear times a cubic term and reach a contradiction but involves dealing with dozens of terms and I don't think it's the best strategy.

答: If $ab^3 + cd^3=f(a,b,c,d)g(a,b,c,d)$, then $\deg_af+\deg_ag=1$. Suppose $\deg_af=0$, and $\deg_ag=1$, so $f\in\mathbb C[b,c,d]$ and $g=h(b,c,d)a+k(b,c,d)$. We get $$ab^3 + cd^3=f(b,c,d)[h(b,c,d)a+k(b,c,d)].$$ Then $f(b,c,d)h(b,c,d)=b^3$ and $f(b,c,d)k(b,c,d)=cd^3$. What do we get from here?


问:
The subring test
This is how the Wikipedia article on subring defines the subring test

The subring test states that for any ring $R$, a nonempty subset of $R$ is a subring if it is closed under addition and multiplication, and contains the multiplicative identity of $R$.

When you follow the link for the subring test, it is stated as follows

In abstract algebra, the subring test is a theorem that states that for any ring, a nonempty subset of that ring is a subring if it is closed under multiplication and subtraction. Note that here that the terms ring and subring are used without requiring a multiplicative identity element.

My question is, is the first statement of subring test correct? This is also how a subring is "defined" in Atiyah-Macdonald. It seems incorrect to me as $\mathbb{R}_+$ satisfies those conditions and is not a subring unless I am missing something.

Looking at the responses I feel I should further clarify my question. Closure under subtraction and multiplication (with the added provision that the given subset contain the identity depending on how you define your rings), guarantees a subring, as in the second statement. I am comfortable with this statement as I know that closure under subtraction for a subset of a group (written additively) gives a subgroup. My question is whether the first statement is correct - is closure under addition and multiplication enough?

答:
As your counterexample shows, the first statement is incorrect. But there is a one character fix: require that the subset contains $\,{-}1\,$ vs. $\,1.\,$ Alternatively, require it to also be closed under negation (additive inverses). Perhaps they meant  $\rm\,S\,$ is closed under subtraction (vs. addition), since then  the subgroup test implies $\rm\,S\,$ is a subgoup of the additive group of $\rm\,R.\,$ Or,  explicitly

$$\rm r,s\in S\ \Rightarrow\ r-r\ =\ 0\in S\ \Rightarrow\ 0-r\ =\: -r\in S\ \Rightarrow\ s-(-r)\: =\ s+r\in R $$

Remark $ $ A similar error appears in Greuel and Pfister: A singular introduction to commutative algebra, p. $1$, where they define a subring as a subset containing $1$ that is closed under the induced ring operations. But negation is not explicitly listed as a ring operation in their definition of a ring. Rather, similar to Atiyah and MacDonald, they say that "R, together with addition, is an abelian group". Perhaps "addition" is meant to denote the full additive group structure, so that the ring is supposed to inherit the negation operation (or equivalent axioms for inverses) from the abelian group structure, when it inherits the addition operation.

Title: The subring test


答:
$CAOB$ is a kite and a cyclic quadrilateral as well. Let the point at which $CO$ intersects $AB$ be $M$, the midpoint of $AB$.

Now,  $CM\cdot MO=AM\cdot BM$ because $CAOB$ is cyclic.  $AM=BM=a$ and $MO$ can be gotten from $\triangle OMA$ by applying the Pythagorean theorem. Find $CM$ and then proceed to use the formula for area.

Body:
Given a circle $(O, R)$ and a chord $AB=2a$. If $CA,CB$ tangents to it, then what is the area of $\triangle ABC$ equal to, with respect to a and R?

I tried solving this question as follows:

$AO=OB=R$

$\triangle ACO=\triangle BOC$

$CA=CB$

Since $OC\perp AB$ and $\triangle ACB$ iscosceles, we just need to find the hight of $\triangle ABO$, from point $O$. This is what gave me difficulties. Could you please explain to me how to solve this question?


问:
Classifying the stationary point of $f(x,y) = 4y^2 + 6yx^2 + 17$
Given the function $f(x,y) = 4y^2 +6yx^2 + 17$, I found the single stationary point $(0,0)$.

I then tried to use the Hessian matrix to classify it as a local minimum/maximum/saddle point, but the determinant is zero. I'm not sure how else I can classify this point. Any help would be much appreciated!

答:
In this kind of situation the Hessian test is inconclusive so you can try to prove/disprove the existence of a local maxima/minima by definition.

For simplicity's sake I'll work instead with the function $f(x,y) = 4y^2 +6yx^2$ since the $17$ doesn't change the nature of the stationnary point.

With this new function we have $f(0,0) = 0$.

Notice that for $y \geq 0$ we have $f(x,y) \geq 0$. Indeed if $f(x,y)$ was negative then we would have
\begin{align*}4y^2+6yx^2 < 0 &\iff 4y^2 < -6yx^2 \\
& \iff y < -\frac{3}{2}x^2 < 0
\end{align*}
which is a contradiction.

Therefore $f$ cant have a local maximum at $(0,0)$ since the function takes larger value whenever $y>0$.

To show that $(0,0)$ is a local minimum we must show that $f$ is positive on a small open neighbourhood around $0$. That is

$$ \exists \epsilon >0 : \vert (x,y)\vert < \epsilon \implies f(x,y) \geq 0. $$

We've already shown that $f$ is positive when $y$ is positive. If $y\leq  0$  then we have

\begin{align*}
4y^2+6yx^2 \geq 0 &\iff 4y^2 \geq -6yx^2 \\
&\iff  y \leq -\frac{3}{2}x^2
\end{align*}

Let $1 > \epsilon >0$ then consider

$$\tilde x  = \epsilon/2$$
$$\tilde y = \frac{1}{2} \frac{-3}{2}\tilde x^2  > \frac{-3}{4}\tilde x^2$$

By our choice for $\tilde x$ and $\tilde y$ we have $f(\tilde x,\tilde y) < 0$ but also $\tilde x^2 + \tilde y^2 < \epsilon^2$ which means that $f$ cant be a local minimum.

Since $(0,0)$ isn't a local maxima or local minima it is a saddle point.


问:
Proving a triangle isosceles using Thales Theorem

Given that:

[BM) is the bisector of the angle ABC.

(BM) and (AN) are parallel straight lines.

I am trying to prove that the triangle ANB is an isosceles triangle with a main vertex B using Thales Theorem.

Here's what I have done:

Since BM and AN are parallel then by using Thales Theorem:
$MA/MC = NB/BC = BM/AN$

I don't know where to go from here.. Any help is appreciated. Thank you.

答:
Look for the alternate interior angles and corresponding angles in the figure: $\measuredangle MBA=\measuredangle BAN$ and $\measuredangle CBM=\measuredangle BNA$, so the triangle is isosceles.

If you insist to use Thales':

$$AM/CM = BN/BC$$

From interior angle bisector theorem:

$$AM/CM=AB/BC$$

Combining the two:

$$BN/BC=AB/BC \implies AB=BN$$


Title: Finding what the sum of $(2^n)(3^{(-n+1)})$ from $n=1$ to infinity converges to

Body:
Sorry for the lack of math symbols but I'm trying to find what the sum of
$f(x)=(2^n)(3^{-n+1})$ from $n=1$ to infinity converges to. I have tried using the sequence of partial sums but I'm too stupid to find what Sn is equal to in order to take the limit. Any help would be greatly appreciated.

Body:
$2^n3^{-n+1} = 2^n3^{-(n-1)} =2^n\cdot \frac 1{3^{n-1}} = 2^n\cdot \frac {3}{3^{n}} = 3\cdot \frac {2^n}{3^n} = 3(\frac 23)^n$.

So we need $\sum_{n=1}^{\infty} 3\cdot (\frac 23)^n$.

Can you take from there?

 Well known formula: If $|r| < r$ then $\sum_{n=\color{green}0}^{\infty} r^n = \frac 1{1-r}$. (Note the base step is $n = \large{\color{green}0}$ and not $n=\large{\color{red}1}$).

Derivation if you didn't know the above result.

 $(1 + r + r^2 + r^3 + ........+r^m)(1-r) = $

 $(1 + r + r^2 + r^3 + ........+r^m) - (r+ r^2 + r^3 + r^4 + ...... + r^{m+1})=$

 $1 - r^{m+1}$. So

 $ (1 + r + r^2 + r^3 + ........+r^m) =\frac {1-r^{m+1}}{1-r}$

 if $|r| < 1$ then $\lim_{m\to \infty} r^{m+1} = 0$ so

 $\sum_{n=0}^\infty r^k =\lim_{m\to\infty} \sum_{n=0}^m r^k = \lim_{m\to \infty}\frac {1-r^m}{1-r} = \frac 1{1-r}$.

Actual Answer (but only read if you really, really, really need it!)

  $\sum_{n=1}^{\infty} 3\cdot (\frac 23)^n =$

  $3 \sum_{n=1}^{\infty} (\frac 23)^n = $

  $3([\sum_{n=\color{red}{\large 0}}^{\infty}(\frac 23)^n]- (\frac 23)^{\color{red}{\large 0}}) =$

 $3(\frac 1{1-\frac 23} - 1)=3(\frac 1{\frac 13} - 1)=3(3-1)=3\cdot 2 = 6$

====

Might be useful to try to calculate

$2^1\cdot 3^{-1+1} + 2^2 \cdot 3^{-2+1} + ..........  + 2^m \cdot 3^{-m+1}$ for a finite $m$.

 $2^1\cdot 3^{-1+1} + 2^2 \cdot 3^{-2+1} + ..........  + 2^m \cdot 3^{-m+1}=$

 $(2^1\cdot 3^{-1+1} + 2^2 \cdot 3^{-2+1} + ..........  + 2^m \cdot 3^{-m+1} )\frac {1-\frac 23}{1-\frac 23} =$

 $\frac {(2^1\cdot 3^{-1+1} + 2^2 \cdot 3^{-2+1} + ..........  + 2^m \cdot 3^{-m+1} )(1- \frac 23)}{\frac 13} =$

 $3\cdot   (2^1\cdot 3^{-1+1} + 2^2 \cdot 3^{-2+1} + ..........  + 2^m \cdot 3^{-m+1} )(1- \frac 23)=$

 $3[(2^1\cdot 3^{-1+1} + 2^2 \cdot 3^{-2+1} + ..........  + 2^m \cdot 3^{-m+1} )- (2^2 \cdot 3^{-2+1} + 2^3 \cdot 3^{-3+1} + ........ + 2^{m+1}\cdot 3^{-m})]=$

 $3[2^1 \cdot 3^{-1+1} - 2^{m+1}\cdot 3^{-m})]=$

 $3[2 - 2\cdot \frac {2^m}{3^m}]=$

 $6 - 6(\frac 23)^m$.

 As $6(\frac 23)^m \to 0$

 we get the infinite sum is $6$.


问:
What conditions are sufficient for "Basically disconnectedness implies Extremally disconnectedness"?
Recall the definition of basically disconnected:

A space is basically disconnected if every cozero-set has an open closure.

There exists a Basically disconnected space which is not extremally disconnected; the one-point Lindelöfization of an uncountable discrete space is such a space.

But with what conditions basically disconnectedness does imply extremally disconnectedness?

Thanks.

答:
Completely regular plus basically disconnected gives extremely disconnectedness, since then all open sets are cozero sets. From the counterexamples given, it sounds like second countability and some separtion axiom may be a requirement (which would imply being normal), so I'm not sure that you can get anything sharper than that while being as general.

Edit: I realized I was wrong in the comments. I wanted to find conditions for all open sets to be cozero sets; however, if there are functions $f,g$ that are zero precisely on $A$ and $B$, then $f/(f+g)$ gives a function that is $0$ precisely on $A$ and 1 precisely on $B$, so we have to have normality. Also, such sets must be $G_{\delta}$'s, since they are the intersection of preimages of open sets in the interval. So a space must in fact be $T_6$ for all open sets to be cozero sets, so my approach won't work.


问:
Prove that {(-1)^n} is not a Cauchy Sequence by using the definition.
I need to use the definition of a Cauchy Sequence to prove that {$(-1)^n$} is not a Cauchy sequence.
I believe I have the main idea down by am lacking a part in my reasoning.

Proof:

For sake of contradiction assume that {$(-1)^n$} is a cauchy sequence. Take $\epsilon =1$. Then there is a $N \in \mathbb{N}$ such that for all m,n $\geq N$ $\Rightarrow |(-1)^m -(-1)^n|$. Let m= N and n = N+1.
Then, $|(-1)^m -(-1)^n| = |(-1)^N-(-1)^{N+1}| = |1-(-1)^{N+1}|$.

This is where I am stuck. I am not sure how to properly show that $(-1)^{N+1}$ reduces down to (-1) which give us |1-(-1)| = 2 >1 thus it is not a cauchy sequence.

答:
It isn’t necessarily true that $(-1)^N=1$: that is the case if and only if $N$ is even. You could divide the argument into two cases, one for even $N$ and one for odd $N$, but it’s neater simply to factor out $(-1)^N$:

$$\left|(-1)^N-(-1)^{N+1}\right|=\left|(-1)^N\right|\cdot|1-(-1)|=1\cdot 2=2\,.$$

If you choose to use cases, when $N=2m$ is even you have $$(-1)^N=(-1)^{2m}=\big((-1)^2\big)^m=1^m=1$$ and $$(-1)^{N+1}=(-1)^N(-1)=(1)(-1)=-1\,,$$

and when $N=2m+1$ is odd you have

$$(-1)^N=(-1)^{2m+1}=(-1)^{2m}(-1)=(1)(-1)=-1$$

and

$$(-1)^{N+1}=(-1)^N(-1)=(-1)(-1)=1\,.$$


问:
Are algebraic structures first order theories?
When we talk about algebraic structures, we can say things like: ($\mathbb{Z}$,+,$\cdot$), which I think is a ring; while in first-order theories, it is typical to see a theory defined like {0, 1, +, ·, =}, which I think is Peano arithmetic).

I see a close similarity between them, could anyone explain if they are the same or why not?

答:
There are three distinct concepts here:

A language $L$ (or "signature" or "similarity type" or "vocabulary" - these are all synonyms) is a collection of symbols naming distinguished elements and operations and relations. The language does not put any restriction on how these elements and operations and relation behave, it just specifies the symbols. Example: $L = \{0,1,+,\cdot\}$ is a language.

A theory $T$ (in the language $L$) is a collection of axioms, i.e., sentences of first-order logic, using the symbols in $L$. Example: The ring axioms are a theory in the language $L$. Peano Arithmetic is another theory in the language $L$.

A model (of a theory $T$) is a set equipped with interpretations of the symbols in the language $L$, which satisfies the sentences in $T$. Example: $\mathbb{Z}$, with the symbols $0$, $1$, $+$, and $\cdot$ interpreted in the usual ways, is a model of the ring axioms, but it is not a model of Peano Arithmetic. $\mathbb{N}$, with the symbols $0$, $1$, $+$, and $\cdot$ interpreted in the usual ways, is a model of Peano Arithmetic, but it is not a model of the ring axioms. We could also take $\mathbb{Z}$, interpret $+$ as multiplcation, interpret $\cdot$ as subtraction, interpret $0$ as $7$, and interpret $1$ as $42$. The resulting structure is neither a model of the ring axioms nor a model of Peano Arithmetic. But my point is that the symbols in the language $L$ can be interpreted completely arbitrarily, and it is up to the axioms in the theory to determine whether the resulting structure is a model.

I've just given you a very quick summary. For the precise definitions, pick up literally any introductory logic textbook that covers first-order logic.


问:
Distribution of continous random variable
Given the following

$Y_1 \sim \mathcal{N}(μ, σ^2 )$

and

$Y_2=α+βY_1+U \;where \; Y_1 \;and \;U\;is\;independent\;and\;U∼\mathcal{N}(0,v^2)$

Let $μ=350$ and $σ^2 =12365$

How do i find the distribution of $Y_2$?

答:
Using properties of Gaussians, Expectation and variance you get that

$$Y_2\sim N(\alpha+\beta\mu;\beta^2\sigma^2+v^2)$$


答:
If you like, define $Y_3 = \alpha + \beta Y_1$.  Use properties of expectation.  If $E[X] = \mu,$ then because expectation is a linear sum, $E[bX] = b\mu$, and $E[X+c] = \mu + c$   For example, if you double every number, the average will double, and if you add $5$ to every single number, the average goes up by $5$ as well.  Using those, you can find that $Y_3$ is distributed  $N(\alpha + \beta \mu, \beta^2 \sigma^2$)

Now you have a sum $Y_2 = Y_3 + U$ of independent variables.  The mean of a sum is the sum of the means.  The variance of the sum is the sum of the variances for independent variables.  In formulas, $$E[X+Y] = E[X] + E[Y]$$ and $$V[X+Y] = V[X] + V[Y] + Cov[X,Y]$$. and for independent variables, the covariance is zero.  Remember that standard deviation is the square root of the variance.

Title: Distribution of continous random variable

Body:
Given the following

$Y_1 \sim \mathcal{N}(μ, σ^2 )$

and

$Y_2=α+βY_1+U \;where \; Y_1 \;and \;U\;is\;independent\;and\;U∼\mathcal{N}(0,v^2)$

Let $μ=350$ and $σ^2 =12365$

How do i find the distribution of $Y_2$?


问:
Optimisation of options under a budget constraint
I don't really know where to begin with solving this problem. I think I'll probably have to use some code but I'm not even sure how to approach the code.

I have two categories and from category 1, I have to pick 5 options and from category 2, I have to pick 1 option. My total budget is 100.0 and each option has a price and an expected value shown below:

Category 1 (Expected Value/Price/EV per unit price):

50.5/33.5/1.51

37.5/24.8/1.51

28.5/23.6/1.21

24.5/18.4/1.33

26.5/17.3/1.53

17.5/13.1/1.28

19/16.8/1.13

16/16.2/0.99

14.5/15.6/0.93

6/14.4/0.42

3/13.9/0.22

3/10.1/0.30

8/11.7/0.68

3/8.8/0.34

8/9.6/0.83

7/6.2/1.13

2/7.9/0.25

2/6.5/0.31

7/5.5/1.27

2/5.8/0.34

Category 2 (Expected Value/Price):

71/38.0/1.87

59/25.9/2.28

37/18.9/1.98

20/18.1/1.10

14/17.6/0.80

11/15.4/0.71

6/12.7/0.47

5/8.9/0.56

4/6.3/0.63

4/6.1/0.66

Is there any strategy I should follow to optimise my expected value while sticking to the budget. There is no reason not to use my entire budget.

Would appreciate it if anyone had any ideas!

答:
This is a variant of the 0-1 knapsack problem.

You can solve it via integer linear programming as follows.  Let binary decision variable $x_i$ indicate whether option $i$ is selected.  You want to maximize $\sum_i v_i x_i$ subject to linear constraints
\begin{align}
\sum_{i \in \text{category 1}} x_i &= 2 \\
\sum_{i \in \text{category 2}} x_i &= 1 \\
\sum_i p_i x_i &\le 100 \\
\end{align}

You can also solve it via dynamic programming.


问:
Does $\lim_{k \to \infty}[(\sum_{n=1}^{k}\frac{1}{n})-\ln k]$ exists?
Let $k>2$, is a natural number, does $\lim_{k \to \infty}[(\sum_{n=1}^{k}\frac{1}{n})-\ln k]$ exists? If $k \to \infty$, then I think no, because $\ln k$ will go to infinite. And $\sum \frac{1}{n}$ diverge too, and so it's $\infty - \infty$. But someone say the limit exists? and it's positive. How?

答: Yes. It's called the Euler-Mascheroni constant and it is positive. This is because the harmonic numbers are approximately logarithmic so the difference between them is small.</p>


问:
Show that every rank-deficient matrix has a full rank matrix arbitrarily close to it
I came across the following exercise in preparation for a test.

Let $A \in \mathbb{R}^{n \times m}$ with rank $r < \min\{n,m\}$. Use SVD of $A$ to show that for every $\epsilon >0$ (no matter how small), there existsa full-rank matrix $A_{\epsilon} \in \mathbb{R}^{n\times m}$ such that $|| A - A_\epsilon ||_{2} < \epsilon$.

I am really puzzled on how to approach this exercise. From my understanding
$$ A - A_\epsilon = \underset{n\times n}{U} \ \underset{n \times m}{\Sigma} \ \underset{m\times m}{V^{T}} - \underset{n\times n}{U} \ \underset{n \times m}{\Sigma_\epsilon} \ \underset{m\times m}{V^{T}} = U(\Sigma - \Sigma_\epsilon)V^{T}.$$
Then $\Sigma$ is diagonal matrix with singular values ranging from $\sigma_1$ to $\sigma_{\min\{n,m\}}$ and $\Sigma_\epsilon$ having singular values from $\sigma_1$ to $\sigma_m$ since $A_\epsilon$ is of rank $m$.

Then since $\min\{n,m\} \le m$
$$ \Sigma - \Sigma_\epsilon = \begin{pmatrix} \sigma_1 &  &  & \\
& \ddots &  \\
 & & \sigma_n &  \\
& & & 0\end{pmatrix} - \begin{pmatrix} \sigma_1 &  &  & \\
& \ddots &  \\
 & & \sigma_{\min\{n,m\}} &  \\
& & & 0\end{pmatrix} =  \begin{pmatrix} 0&  &  & \\
& \ddots &  \\
 & & \sigma_? &  \\
& & & 0\end{pmatrix}$$
So since a singular value exists for the matrix difference above, $|| A - A_\epsilon ||_2 = \sigma_?$ which has to be small since $\sigma_1 > \sigma_2 > \dots > \sigma_r$, and here this $\sigma_?$ is among among the last in the sequence.

I am most likely writing non-sense, I am aware of that, I don't know how to proceed, can anyone give me some hint or idea ?

答:
You are very close. The proof needs a couple of ingredients:

What is the norm of A (say the 2-norm, since they are all equivalent) in terms of $\Sigma$? And what's the norm of $\Sigma$ in terms of the $\sigma_i$'s?

The simples way to perturb $\Sigma$ is to let $\Sigma_\varepsilon$ have the first $r$ singluar values matching those of $\Sigma$.

If you write $\Sigma_\varepsilon$ as described above, what would $\|\Sigma_\varepsilon-\Sigma\|$ be? And therefore, what aboud $\|A-A_\varepsilon\|$?

Edit: read further for more details.

You are free to build $A_\varepsilon$ in whatever way you want. The goal is to produce one matrix that is close to A, however peculiar its construction may be. So pick same $A_\varepsilon=U\Sigma_\varepsilon V^T$, with $\Sigma_\varepsilon$ with the same first $r$ singular values as $\Sigma$, and $\sigma_i<\varepsilon$ for $i=r+1,...,p$. (Note: you could pick any values for $\sigma_{r+1},...,\sigma_p$, so long as they are all positive, and all smaller than $\varepsilon$. Then $\|\Sigma-\Sigma_\varepsilon\|=\sigma_{r+1}<\varepsilon$. This proves that $A_\varepsilon$ is arbitrarily close to A. The fact that all its singular values are non zero (small as they may be), implies that $A_\varepsilon$ is full rank. QED.


答:
This question might also be a good excuse to mention Gauss' lemma and Eisenstein's criterion for irreducibility. Most often these seem to be mentioned just for $\mathbb Z[x]$ (and $\mathbb Q[x]$), but they do apply to $\mathbb C[x,y,z,w]$ and other unique factorization domains, as well.

This puts a slightly different emphasis on the direct argument give by @user26857. Yes, based on degree considerations, a linear polynomial in $x$, with coefficients in the field of rational functions $\mathbb C(y,z,w)$, is irreducible in the PID $\mathbb C(y,z,w)[x]$. With a little more detail about those coefficients, Gauss' lemma gives the irreducibility in $\mathbb C[x,y,z,w]$.

Gauss' lemma and Eisenstein's criterion do also apply to algebraic examples wherein explicit formulaic discussion becomes much messier. (Indeed, those two results nicely package a useful family of computational ideas.)

Title: How to show that the polynomial $ab^3 + cd^3$ is irreducible?

Body:
I have to show that the polynomial:

$$ab^3 + cd^3 \in \mathbb{C}[a,b,c,d]$$

cannot be factorised into polynomials of lower degrees, i.e. it is not reducible. However, I'm quite unsure on how to proceed here. I thought I could try to factorise this into a linear times a cubic term and reach a contradiction but involves dealing with dozens of terms and I don't think it's the best strategy.


问:
Solve $(xp-y)^{2}(x^2+y^2)^{-3/2}=a(1+p^2)$
Question is to solve this differential equation:-
$$(xp-y)^{2}(x^2+y^2)^{-3/2}=a(1+p^2)$$where $$p = \frac{dy}{dx}$$.

Question is given in the clairaut's form in my notebook.
But seems some proper substitution would do the trick .
Hints/leads are appreciated.

答:
$$\left(x\frac{dy}{dx}-y\right)^{2}=a+a\left(\frac{dy}{dx}\right)^2(x^2+y^2)^{3/2}\tag 1$$
In polar coordinates
$\begin{cases}
x=r\cos(\varphi)\\
y=r\sin(\varphi)
\end{cases} \quad\implies\quad
\begin{cases}
dx=\cos(\varphi)dr-r\sin(\varphi)d\varphi\\
dy=\sin(\varphi)dr+r\cos(\varphi)d\varphi
\end{cases}$

Putting them into Eq.$(1)$ and after simplification leads to :
$$\left(\frac{dr}{d\varphi}\right)^2=\frac{r}{a}-r^2 \tag 2$$
Eq.$(2)$ agrees with the equation already found by Lutz Lehmann.
$$d\varphi=\pm\sqrt{\frac{1}{\frac{r}{a}-r^2}}dr$$
$$\varphi=\pm\int\frac{1}{\sqrt{\frac{r}{a}-r^2}}dr=\pm\sin^{-1}(2ar-1)+\text{constant}$$
The inverse function is :
$$\boxed{r(\varphi)=\frac{1}{2a}\left(\pm\sin(\varphi+c)+1 \right)}$$
This is the equation of a set of cardioids.

Back to Cartesian coordinates :
$$2a\sqrt{x^2+y^2}=\pm\frac{y\cos(c)+x\sin(c)}{\sqrt{x^2+y^2}}+1$$


答:
Not using Thales's intercept theorem, but:

$AN \parallel BM$, so from $AB$ crossing $\color{#092}{\angle NAB} = \color{#02E}{\angle ABM}$ and from $CN$ crossing $\color{#092}{\angle ANB} = \color{#02E}{\angle MBC}$.

Since $BM$ bisects $\angle ABC$, $\color{#02E}{\angle ABM =  \angle MBC}$ so $\color{#092}{\angle NAB =\angle ANB }$, giving $\triangle BAN$ isosceles.

Title: Proving a triangle isosceles using Thales Theorem

Body:
Given that:

[BM) is the bisector of the angle ABC.

(BM) and (AN) are parallel straight lines.

I am trying to prove that the triangle ANB is an isosceles triangle with a main vertex B using Thales Theorem.

Here's what I have done:

Since BM and AN are parallel then by using Thales Theorem:
$MA/MC = NB/BC = BM/AN$

I don't know where to go from here.. Any help is appreciated. Thank you.


问:
Evaluate : $\int\limits_0^{\pi/2} \cot x \ln(\sec x) dx$
I want to solve this integral :
$$\int_0^{\pi/2} \cot x \ln(\sec x) dx$$
I tried the following substitution : $\ln(\sec x)=t$ which means $dt=\tan x dx$
$$I=\int_0^\infty \frac{\cot x}{\tan x}tdt=\int_0^\infty \frac{t}{\tan^2 x}dt$$
I'm really disturbed by the $\tan^2 x$, I tried also to substitute $\sec x =t$ but it's not helpful either. Any helpful approach to solve this problem ?

答:
General case

After $u=\sec x$ the integral becomes
$$\int\dfrac{\ln\left(u\right)}{\left(u-1\right)u\left(u+1\right)}\,\mathrm{d}u.$$
With partial fraction decomposition:
$$=\class{steps-node}{\cssId{steps-node-3}{\dfrac{1}{2}}}{\displaystyle\int}\dfrac{\ln\left(u\right)}{u+1}\,\mathrm{d}u-{\displaystyle\int}\dfrac{\ln\left(u\right)}{u}\,\mathrm{d}u+\class{steps-node}{\cssId{steps-node-4}{\dfrac{1}{2}}}{\displaystyle\int}\dfrac{\ln\left(u\right)}{u-1}\,\mathrm{d}u.$$
After some trivial substitutions and two integration by parts we can see that this is equal to
$$\dfrac{\ln\left(u\right)\ln\left(u+1\right)}{2}-\dfrac{\ln^2\left(u\right)}{2}+\dfrac{\operatorname{Li}_2\left(-u\right)}{2}-\dfrac{\operatorname{Li}_2\left(1-u\right)}{2}.$$

Boundaries $0$ and $\pi/2$

To finish the general case you only have to substitute back (and add the constant). I will not do this now because the expression will get unnecessarily messy.

Remember we set $u=\sec x$, which means that $0\to 1$ and $\pi/2^- \to \infty$. Final solution is (please forgive me for the use of "$\infty$", will add the limits later on)

$$\left[\dfrac{\ln\left(\infty\right)\ln\left(\infty+1\right)}{2}-\dfrac{\ln^2\left(\infty\right)}{2}+\dfrac{\operatorname{Li}_2\left(-\infty\right)}{2}-\dfrac{\operatorname{Li}_2\left(1-\infty\right)}{2}\right]-\left[\dfrac{\ln\left(1\right)\ln\left(1+1\right)}{2}-\dfrac{\ln^2\left(1\right)}{2}+\dfrac{\operatorname{Li}_2\left(-1\right)}{2}-\dfrac{\operatorname{Li}_2\left(1-1\right)}{2}\right]$$
$$=\left[\underbrace{\dfrac{\operatorname{Li}_2\left(-\infty\right)}{2}-\dfrac{\operatorname{Li}_2\left(1-\infty\right)}{2}}_{=0}\right]-\left[\underbrace{\dfrac{\operatorname{Li}_2\left(-1\right)}{2}}_{-\pi^2/24}-\underbrace{\dfrac{\operatorname{Li}_2\left(0\right)}{2}}_{0}\right]$$
Finally we see that our integral is equal to $\frac{\pi^2}{24}.$


问:
Basic word problem
Friends Matt and Gary are planning to study together. Matt buys some pencils and erasers. Gary spends the same amount of money to buy thrice as many pencils and five less erasers. Now they are set for the group study. If cost of one eraser is two more than the cost of one pencil, what is the minimum number of pencils bought by Matt and Gary together?

I took the variables as -

$p$ (number of pencils bought by Matt)

$e$   (number of erasers bought by Matt)

then,

$3p$  (number of pencils bought by Gary)

$e-5$ (number of erasers bought by Gary)

also,

$C_p$ (Cost of a pencil)

$C_e$  (cost of an eraser)

Since they both spent the same amount of money,

$$ pC_p + eC_e = (3p)C_p + (e-5)C_e $$

and $$C_e = 2 + C_p$$

On solving we get,

$$\frac{5(2+C_p)}{2C_p} = p$$

The value that we need is $p+3p$. Therefore, we need the minimum value of $4p$ (Number of pencils bought by Matt and Gary)

$$\frac{10(2+C_p)}{C_p} = p$$

I then differentiated it with respect to $C_p$ and equated it to $0$, to find the minimum value. But after differentiation I got

$$\frac{-20}{C_p^2}$$

Which I can’t equate to $0$

Can you point out what’s wrong in this method, and suggest me a better solution, explaining the concept used in detail?

The answer given is $12$

答:
From your equation:$$2pC_p=5(2+C_p)$$
calculate $C_p$:
$$(2p-5)C_p=10$$
You know $C_p>0$, so $(2p-5)>0$. The minimum integer that satisfies that is $p=3$. Then the total number of pencils is $4p=12$.

Title: Basic word problem


问:
$63^{63^{63}} \mod 100$
I need to find $63^{63^{63}} \bmod 100$.

This is what I've got so far:

Since $\gcd(63,100)=1$ we can use Euler's theorem. We have $\phi (100)=40$ so $63^{40} \equiv 1 \mod 100$

Again $\gcd(16,100)=1$ and $\phi (40)=16$, that is $63^{16} \equiv 1 \mod 40$

Using this I got that $63^{63} \equiv 7 \mod 40 $ which led me to $63^{63^{63}} \equiv 63^7 \mod 100$

I'm stuck here and don't know what to do next, what could I do now?

答:
You have done the hard work which can be reached by Carmichael Function as well

Now, $63^2=(60+3)(60+3)\equiv2\cdot60\cdot3+3^2\equiv70-1$

$\displaystyle 63^7=63(70-1)^3\equiv63(-1+\binom31\cdot70^1)\pmod{100}\equiv(60+3)\cdot9\equiv40+27$

Title: $63^{63^{63}} \mod 100$


答:
Note that
$$
 - \log k =  - \log \left( {\frac{2}{1} \cdots \frac{{k - 1}}{{k - 2}}\frac{k}{{k - 1}}} \right) =  - \sum\limits_{n = 2}^k {\log \left( {\frac{n}{{n - 1}}} \right)}  = \sum\limits_{n = 2}^k {\log \left( {1 - \frac{1}{n}} \right)} .
$$
Thus,
$$
\sum\limits_{n = 1}^k {\frac{1}{n}}  - \log k = 1+\sum\limits_{n = 2}^k {\left( {\frac{1}{n} + \log \left( {1 - \frac{1}{n}} \right)} \right)} .
$$
Now if $0<x\leq \frac{1}{2}$, then
$$
\left| {x + \log (1 - x)} \right| = \left| { - \int_0^x {\frac{t}{{1 - t}}dt} } \right| \le 2\int_0^x {tdt}  = x^2 .
$$
Therefore,
$$
\left| {\sum\limits_{n = 2}^k {\left( {\frac{1}{n} - \log \left( {1 - \frac{1}{n}} \right)} \right)} } \right| \le \sum\limits_{n = 2}^k {\left| {\frac{1}{n} - \log \left( {1 - \frac{1}{n}} \right)} \right|}  \le \sum\limits_{n = 2}^k {\frac{1}{{n^2 }}}  < \sum\limits_{n = 2}^\infty  {\frac{1}{{n^2 }}}  < \infty .
$$
Consequently, the sequence
$$
1 + \sum\limits_{n = 2}^k {\left( {\frac{1}{n} - \log \left( {1 - \frac{1}{n}} \right)} \right)}  = \sum\limits_{n = 1}^k {\frac{1}{n}}  - \log k
$$
is convergent.

Title: Does $\lim_{k \to \infty}[(\sum_{n=1}^{k}\frac{1}{n})-\ln k]$ exists?

Body: Let $k>2$, is a natural number, does $\lim_{k \to \infty}[(\sum_{n=1}^{k}\frac{1}{n})-\ln k]$ exists? If $k \to \infty$, then I think no, because $\ln k$ will go to infinite. And $\sum \frac{1}{n}$ diverge too, and so it's $\infty - \infty$. But someone say the limit exists? and it's positive. How?


问:
What's the differential of a couple
Suppose we have a function $f : \Bbb R^3\to\Bbb R^2$ such that $f:(x,y,z)\to(x+y^2,xy^2z) $

What's the differential of $f$?

The result of my attempt :
$D_f(x,y,z)(h,k,l)=(h+2yk,y^2zh+2xyzk+xy^2l)$

What I did was simply calculating the partial derivative of each function ( $f_1=x+y^2$ and $f_2=xy^2z$ ) but I'm not sure if that's the right way to go ( as in writing the result in a couple like I did).
Is the reasoning correct?

答:
Yes, that is correct. In general the default format is column vectors, so that
$$
f([x,y,z]^T)=\pmatrix{x+y^2\\xy^2z}
$$
and
$$
J_f([x,y,z]^T)\pmatrix{h\\k\\l}=\pmatrix{h+2yk\\y^2zh+2xyzk+xy^2l}
=\pmatrix{1&2y&0\\y^2z&2xyz&xy^2}\pmatrix{h\\k\\l}
$$

Title: What's the differential of a couple


问:
Does SSA congruence criterion work if the non-included angle is obtuse?
I was trying to make some triangles when $2$ sides and the non-included angle was given. When the angle given was below $90$ degrees, there were two possible triangles which could be made. But, when it went above $90$ degrees, only one possibility came. I was using a compass to determine the third side's possibilities. I checked the internet if there were questions like this, but there were none. Also, no site said specifically that it works for $90+$ angles.

  Does this mean that the SSA congruence can work for angles greater than $90$ degrees? If so, I was taught very wrong in school.

答: If another triangle has parts congruent to angle A and sides AB and BC of our triangle, and angle A is not acute, then the two triangles are congruent. If angle A is acute, and BC > or = to AB, then they are congruent. Also congruent if either angle B or C is a right angle. Otherwise (i.e., angle A is acute, ABC is not a right triangle, and BC < AB) there are two triangles, the smaller of which has an obtuse angle at C which is supplementary to its cousin of the larger triangle.  For this latter case, there is only congruence if we also know both original triangles (ours and the other one) are obtuse or both acute.</p>


Title: Evaluate : $\int\limits_0^{\pi/2} \cot x \ln(\sec x) dx$

Body:
I want to solve this integral :
$$\int_0^{\pi/2} \cot x \ln(\sec x) dx$$
I tried the following substitution : $\ln(\sec x)=t$ which means $dt=\tan x dx$
$$I=\int_0^\infty \frac{\cot x}{\tan x}tdt=\int_0^\infty \frac{t}{\tan^2 x}dt$$
I'm really disturbed by the $\tan^2 x$, I tried also to substitute $\sec x =t$ but it's not helpful either. Any helpful approach to solve this problem ?

Body:
Substitute $t= \tan^2 x$

\begin{align}
\int_0^{\pi/2} \cot x \ln(\sec x) dx= & \>\frac14\int_0^1\frac{\ln(1+t)}{t(1+t)}dt
 + \frac14\int_1^\infty \underset{t\to1/t}{\frac{\ln(1+t)}{t(1+t)}dt}\\
=&\>\frac14 \int_0^1 \frac{\ln(1+t)}{t}dt-\frac14 \int_0^1  \frac{\ln t}{1+t}dt \\
=& \>\frac12 \int_0^1 \frac{\ln (1+t)}{t}dt
=\frac12\cdot \frac{\pi^2}{12}=\frac{\pi^2}{24}
\end{align}

where $\int_0^1 \frac{\ln (1+t)}{t}dt =\frac{\pi^2}{12}$


问:
f(n) tends to infinity and $f(n)=O(g(n))$ then $\log{(f(n))}=O(\log{(g(n))})$
I need to prove that if $f(n)$ tends to infinity and $f(n)=O(g(n))$ then $\log{(f(n))}=O(\log{(g(n))})$. I tried to use the fact that $g(n)/f(n)  \geq 1/c$, and then it's limit is bigger than $0$, but I don't know how to continue.

答:
Since $f(n)\le Cg(n)$ and $f(n)\to+\infty$ then $g(n)\ge \frac 1Cf(n)\to+\infty$ (since $C>0$).

Now use that $\ln\nearrow$ so $\ln(f(n))\le\ln(Cg(n))=\ln(C)+\ln(g(n))$

But since $\ln(g(n))\to+\infty$ for $n$ large enough it is greater than $\ln(C)$.

Therefore $\ln(f(n))\le 2\ln(g(n))$.

Your path was not wrong; you can continue like this:

$\ln(\frac{g(n)}{f(n)})\ge \ln(\frac 1C)\iff \ln(g(n))-\ln(f(n))\ge -\ln(C)\ge -\frac 12\ln(f(n))$

Similarly invoking that $\ln(C)$ is smaller than $\frac 12\ln(f(n))$ for $n$ large enough.


问:
How to find if a number has any odd divisor (greater than 1)?
I want to build an algorithm which tells if a number has any odd divisor or not.

for example, 6 has an odd divisor 3 (greater than 1),
4 doesn't have any odd divisors, and so on.

I know that I can find all the pairs of divisors by only traversing till square root of the number.

for example, if n%i == 0 then i and n/i both are divisors.

for example, if the number given is 100 using this method I can
find all the divisors and check if any divisor is odd or not i.e. 1, 100, 2, 50, 4, 25, 5, 20, 10.

Therefore this algorithms solves the problem in O(sqrt(n)) complexity.
But I'm looking to optimize this.

Please comment if any more clarity is required from my side as I'm new to this forum.

Thanks !

答:
Just keep dividing by $2$ until you get an odd number; either it’s $1$, in which case the number is a power of $2$ and has no other positive odd divisor, or it’s an odd divisor greater than $1$. This is faster, $\Theta(\log_2n)$.

Added: And see the comment by JMoravitz for what may be an even more efficient solution, depending on just how you’re implementing the test.


问:
What is the purpose of the first test in an inductive proof?
Learning about proof by induction, I take it that the first step is always something like "test if the proposition holds for $n = \textrm{[the minimum value]}$"

Like this:

Prove that $1+2+3+...+n = \frac{n(n + 1)}{2}$ for all $n \ge 1$.

Test it for $n = 1$:

$$n = \frac{n(n + 1)}{2} \implies 1 = \frac{2}{2}\textrm{, which holds.}$$

* The rest of the proof goes here *

So, I do it all the time (like a standard). But I never really thought about why. Why do I do such test? I can see that if the test does not hold, it cannot be proven by induction, I guess. But is there another reason we do this?

答:
There's another way of viewing a proof by induction.  It's a demonstration that the set of counterexamples of your statement is empty because it has no least element.  Since any non-empty set of natural numbers must have a least element, if the set of counterexamples has no least element, it must be empty.

Viewed from this perspective, the inductive step shows us that candidate $n+1$ can't possibly be the smallest counterexample of the proposition $P(k)$ because the contrapositive of $P(n) \Rightarrow P(n+1)$ is $\lnot P(n+1) \Rightarrow \lnot P(n)$.  So the inductive step proves that if $n+1$ is indeed a counterexample to our proposition, $n$ also must be a counterexample, so $n+1$ can't possibly be the smallest counterexample.

Why doesn't this complete the proof of $P(k)$ for all $k$?  Because there's a number out there, $1$ (or $0$) that can't be expressed in the form $n+1$ (where $n \in \Bbb N^+$ or $n \in \Bbb N$).  So to be sure your proposition really doesn't have any counterexamples, you need to show that this special number, the base case, also isn't a counterexample fo $P(k)$.


问:
Existence of an Element with a particular order
I've solved the following problem in what seems to me to be a very inelegant way:

Let $G$ be an abelian group containing elements $a$ and $b$ of orders $m$ and $n$, respectively, where $m$ and $n$ are not necessarily relatively prime.  Show that $G$ must contain an element $c$ whose order $\mathcal{O}(c)$ equals the least common multiple of $m$ and $n$.

My solution is the following:

Let $d=\gcd(m,n)$ and note that $m=dk$, $n=dl$ where $k$ and $l$ are relatively prime. Now, define
      $$
      r=\prod_{p\mid d, p\mid k}p^i,
    $$
      where $p^i\mid d$ but $p^{i+1}\nmid d$. Now, by construction, since $k$ and $l$ are relatively prime, $r$ and $l$ are relatively prime. Likewise, since $r$ takes all of the primes in $d$ that are in common with $k$, we know $r$ and $\frac{d}{r}$ are relatively prime. Now we have
      \begin{align}
        &m=dk=r\left(\frac{d}{r}\right)k\\
        &n=dl=r\left(\frac{d}{r}\right)l
    \end{align}
      Clearly the element $a^{d/r}$ has order $rk$ and the element $b^{r}$ has order $\frac{dl}{r}$. Now the orders are relatively prime, so from a previous problem (where we showed that elements with rel. prime orders has the property that their product has the order of the product of their individual orders), we know the order of $a^{d/r}b^{r}$ is $dkl$, which by definition is the least common multiple of $m$ and $n$.

This doesn't strike me as being an elegant proof, so I'm wondering if there are any other ways to prove the statement?

答:
Let $m=dm'$ and $n=dn'$ with $gcd(m',n')=1$.

Define

$$f: Z \times Z \to G \,;\, f(k,l)=a^{kd}b^{ld} \,.$$

We claim $\ker(f)=m'Z \times n'Z$. Since $\frac{Z \times Z}{m'Z \times n'Z} \sim Z_{m' n'}$ pick a generator $a^{id}b^{jd}$ for $Im(f)$ and then look to $a^ib^j$.

We prove now the claim

$$f(k,l)=0 \Rightarrow a^{kd}b^{ld}=e \,.$$

Thus, since $m',n'$ are relatively prime

$$b^{n'dl}=e \Rightarrow dm'|dln' \Rightarrow m'|ln' \Rightarrow m'|l \,,$$
$$a^{m'dk}=e \Rightarrow dn'|dkm' \Rightarrow n'|km' \Rightarrow n'|k \,,$$


问:
Prove the following expression in linear algebra
Let $S=\{(x,y,0)\in\mathbb{K}^3,x,y\in\mathbb{K}\}$. Cosider a vector subspace $P\subset\mathbb{K}^3$ such that $K^3=S\bigoplus P$.

Prove that their exists $x_0,y_0\in\mathbb{K}$ such that
$$(x_0,y_0,1)\in P$$

Prove that $P=span\{(x_0,y_0,1)\}$

Conversely, prove that if $x_0,y_0\in\mathbb{K}$, then
$$S\bigoplus span\{(x_0,y_0,1)\}=\mathbb{K}^3$$
Note $\mathbb{K}=\mathbb{R}\cup\mathbb{C}$

For my idea that I used to think this problem. I start from
$$S\bigoplus P=\mathbb{K}^3\Longleftrightarrow \begin{cases} S\cap P=\{0\}\\S+P=\mathbb{K}^3
\end{cases}$$
Please kindly help me to solve this or give a recommendation idea for me to prove this. Thank in advance.

答:
$P$ has an element with $3$rd coordinate nonzero, otherwise $P \subseteq S$ and so $\mathbb{K}^3 \neq S \oplus P$. So, just multiply the element by the multiplicative inverse of the $3$rd coordinate.

$S$ is $2$-dimensional, so $P$ is $1$-dimensional: $3 = \dim \mathbb{K}^3 = \dim (S \oplus P) = \dim S + \dim P = 2 + \dim P \implies \dim P = 1$. From (1) we have produced a non-zero vector in $P$ so it spans $P$.

Since $S \cap \text{span}\{(x_0, y_0, 1)\} = \{0\}$, the sum is direct. The dimension of the sum is thus $3$ and hence the sum is $\mathbb{K}^3$.


问:
Contact form in Polar Coordinates
While going through Etnyres Lectures on Contact Topology (which can be found here) Example 2.8 two questions came up

He uses cylindrcal coordinates $(r,\theta, z)$ to define a 1-form $\alpha_2 = dz +r^2d\theta$ on $\mathbb{R}^3$ and it is not clear to me how this is well defined. I would interpret this as follows: $\theta$ should be a function $\mathbb{R}^3\to\mathbb{R}$ sending a point $p$ to the angle of the projection onto the xy-plane of p with respect to the ray along the x-axis going out to + $\infty$, $d\theta$ would then be the exterior derivative. However, $\theta$ does not  define a smooth map on all of $\mathbb{R}^3$ because of problems at the origin. My question is then, how should one interpret this definition of $\alpha_2$.

He claims that $\alpha_2$ defines a contact form, that is $\alpha_2\wedge d\alpha_2\neq 0$,
but his computation shows that
$$\alpha_2\wedge d\alpha_2 = 2rdr\wedge d\theta\wedge dz$$
which gives the trivial form at the origin since $r(0,0,0) = 0$. I thought the contact condition implies that the form is non-trivial at every point $p\in\mathbb{R}^3$, or is it enough that the form is non-trivial at a single point?

答:
Your point is well-taken. However, note that
$$r^2\,d\theta = (x^2+y^2)\left(\frac{-y\,dx+x\,dy}{x^2+y^2}\right) = -y\,dx + x\,dy$$
is in fact globally defined, even though the polar coordinate system breaks down at the origin.

Then $\alpha_2 = dz-y\,dx+x\,dy$ and $$\alpha_2\wedge d\alpha_2 = 2\,dz\wedge dx\wedge dy = 2(dx\wedge dy)\wedge dz = 2(r\,dr\wedge d\theta)\wedge dz,$$ as promised.

Title: Contact form in Polar Coordinates


问:
Does Morita equivalence commute with tensor products or passing to polynomial rings?
As in the title, I am wondering whether Morita equivalence commutes with tensor products or passing to polynomial rings. More specifically, are any of the following true?

If $A$ is Morita equivalent to $B$, are the rings $A\otimes C$ and $B\otimes C$ Morita equivalent? Here if $A,B$ are rings, the tensor product is over $\mathbb{Z}$, and if $A,B$ are $k$-algebras, it is over $k$.

If $A$ is Morita equivalent to $B$, is $A[x]$ Morita equivalent to $B[x]$?

1 implies 2, and I think the answer is positive for 2. In that case, $A\cong e\mathbb{M}^n(B)e$ for some full idempotent $e$ of the matrix ring $\mathbb{M}^n(B)$, and $e$ stays a full idempotent in the ring $\mathbb{M}^n(B)[x]=\mathbb{M}^n(B[x])$, so we have $A[x]\cong e\mathbb{M}^n(B[x])e$. But is 1 true at all? I don't think it is, but can't come up with a counterexample?

答:
In fact, $1$ is true.

Here's why:

A left $A \otimes_{k} C$-module is the same thing as a left $A$-module $M$ together with a $k$-algebra homomorphism $C \to \operatorname{End}_A(M)$.

Likewise, a left $B \otimes_{k} C$-module is the same thing as a left $B$-module $M$ together with a $k$-algebra homomorphism $C \to \operatorname{End}_B(M)$.

So, if $F:_A\mathrm{Mod} \to _B\mathrm{Mod}$ is a $k$-linear equivalence, then any $k$-algebra homomorphism $C \to \operatorname{End}_A(M)$ includes a $k$-algebra homomorphism $C \to \operatorname{End}_B(F(M))$, which by the above remarks, gives a functor from left $A \otimes_{k} C$-modules to left $B \otimes_{k} C$-modules. This functor is then easily verified to be a $k$-linear equivalence, and so the $k$-algebras $A \otimes_{k} C$ and $B \otimes_{k} C$ are Morita equivalent.

The above applies for any commutative ring $k$, including the special case $k=\mathbb{Z}$.


问:
Exercise on $\boldsymbol{\Delta}_2^0$ sets in the Baire space $\omega^\omega$
I want to prove the following statement:

Given a set $A\subseteq \omega^\omega$, if there exists a continuous function $g: 2^\omega \rightarrow \omega^\omega$ s.t. $f(\{z \in 2^\omega \mid \exists n \forall m\ge n \ (z_m = 0)\}) \subseteq A$ and $f(\{z \in 2^\omega \mid \exists n \forall m\ge n \ (z_m = 1)\}) \cap A = \emptyset$ then $A$ is not $\boldsymbol{\Delta}_2^0$

I tried to prove that there exists a continuous reduction of $\{z \in 2^\omega \mid \exists n \forall m\ge n \ (z_m = 0)\}$, which is $\boldsymbol{\Sigma}_2^0$-complete, onto $A$, but without much success. Any hint?

Thanks

答: Consider the set $B=f^{-1}(A)\subseteq 2^\omega$, separating the set $\{z\mid \text{cofinitely }0\}$ from the set $\{z\mid \text{cofinitely }1\}$. It suffices to show that $B$ cannot be $\mathbf{\Delta^0_2}$. Assume that $B$ is, and we find that $B$ is both dense $G_\delta$ and codense $F_\sigma$, meaning that $B$ is both comeager and meager. This violates the Baire category theorem.


问:
Magnitude of projection proof
I don't get how the magnitude of projection formula follows from the projection vector.

$proj_v(u) = \frac{u \cdot v}{|v|} \frac{v}{|v|}$

$|proj_v(u)| = \frac{|u \cdot v|}{|v|}$

答: Hint: The norm $\lVert w \rVert$ of a vector $w$ has the property that $\lVert aw \rVert = |a|\lVert w \rVert$ for all scalars $a$.

Title: Magnitude of projection proof