问:
Domain and range of composite functions
Would I be correct in assuming that the domain of the composite function $(f∘g)(x)$ is the intersection of the range of $g$ and the domain of $x$? If not, how do I find the domain of a composite function?

Would I find the range of $(f∘g)$ as I would find the range of a non-composite function, or do I need to use the components $g$ and $f$?

P.S.: Are $D_f$ and $R_f$ accepted notations for the domain and range of $f$ respectively?

答:
The domain of a composite function $f(g(x))$ is the set of those inputs $x$ in the domain of $g$ for which $ g(x)$ is in the domain of $f$.

Think of it this way, In order to find domain of any function the value inside it must be permissible i.e for a $y=g(x)$ to be a function; for every $x$, $g$ should yield 1 value of $y$.

for a $y=f(g(x))$ to be a function; for every $g(x)$ , $f$ should yield a $y$. And for a $g(x)$ to be a function; for every $x$, $g$ should yield a $y'$(a value).
so, $x$, $g(x)$ will restrict domains.

for a $ y=h(f(g(x)))$ to be a function; for every $f(g(x))$, $h$ should yield a $y$. And if $f(g(x))$ is to be a function; for every $g(x)$, $f$ should yield a $y'$(a value). And for a $g(x)$ to be a function; for every $x$, $g$ should yield a $y''$(a value).
so, $x$, $g(x)$, $f(x)$ will all restrict domains.

This is how I understand it.

Title: Domain and range of composite functions


问:
Determine the largest domain to which this function can be analytically continued
Consider the following question which was asked in my Complex analysis Quiz( which is now over) :

Consider the following analytic function $f(z)= \sum_{n=1}^{\infty}\frac{1+c}{n+c} z^n $ Determine the largest domain to which f can be anaytically continued?

Determine an analytic continuation of f  from the unit disk to a larger domain?

I was not able to do any of the 2 questions and thought about it again today and not getting any idead except that the Radius of convergence of the series in |z|<1 but I think that this will not be a boundation in decieding analytic continuation as I have seen examples of series with ROC 1 but which can be continued to regions with |z|>1.

So, I request your help as I will not be able to do it myself. Thanks!
I have been following Serge's Lang Complex Analysis.

答:
There is no single answer.  Consider e.g. $c=0$, where $\sum_{n=1}^\infty z^n/n = -\log(1-z)$ for $|z|<1$.  This can be analytically continued to the complement of a branch cut, which can be any curve extending from $1$ out to $\infty$ and not intersecting the unit disk.  But none of these is "the largest".

EDIT: In general, $f(z)$ can be written in terms of the Lerch Phi function.
It satisfies the differential equation $$f'(z) + \frac{c}{z} f(z) = \frac{1+c}{1-z}$$
in the unit disk, i.e. $$ \dfrac{d}{dz} z^c f(z) = z^c \frac{1+c}{1-z}$$
Take any simply-connected open region $U$ of $\mathbb C$ not including $0$ or $1$, and an analytic antiderivative $F(z)$ of $z^c (1+c)/(1-z)$ on $U$, and
$f(z) = z^{-c} (F(z) + constant)$ (for a suitable value of the constant) will be an analytic continuation of $f$ on that region.

ContentLicense: CC BY-SA 4.0


问:
Give an example of the $a,b,c$ which satisfies conditions in the generating set
How to derive the specific case of the generating element of a group given its generating set. For example, when
$$G=\langle a,b,c|a^3=b^3=c^2=1,ab=ba,ca=a^2c,cb=b^2c\rangle$$
we can let $G\subset S_3\times S_3$, and let
$$a=((123),1),b=(1,(123)),c=((12),(12))$$
to get the desired result. However, when $3$ is replaced by general $p$, where $p$ is an odd prime,say,
$$G=\langle a,b,c|a^p=b^p=c^2=1,ab=ba,aca=c,bcb=c\rangle$$
 I cannot specifically give the example of $a,b$ and $c$, which satisfies "$a^p=b^p=c^2=1,ab=ba,aca=c,bcb=c$"  Is there any quick way to do so?</p>

答:
I think the correct generalization is
$$G=\langle a,b,c \mid a^p=b^p=c^2=1,ab=ba,cac=a^{-1},cbc=b^{-1}\rangle$$
where $p$ is any prime.

Take a vector space $V$ of dimension $2$ over $\Bbb{Z}_{p}$, written multiplicatively, with basis $a, b$. (In other words, $V$ is an elementary abelian group of order $p^{2}$ generated by $a, b$.) Let $c$ be the automorphism of $V$ given by inversion.

Now build $G$ as the semidirect product $V \langle c \rangle$.

ContentLicense: CC BY-SA 3.0


问:
Are there any rigid number fields?
A field is rigid if it has no nontrivial automorphisms.  A number field is a finite extension of $\mathbb{Q}$.  My question is, are there any rigid number fields?

I’m pretty sure that if $K$ is a finite Galois extension of $\mathbb{Q}$, then $K$ cannot be rigid.  But what about non-Galois extensions of $\mathbb{Q}$?

答: Every quadratic number field is Galois so the smallest possible counterexample is cubic. A cubic field is rigid iff it's not Galois: the simplest example is $\mathbb{Q}(\sqrt[3]{2})$.

Title: Are there any rigid number fields?


问:
Is $X^6$ prime element in $Z[X^2,X^3]$?
Is $X^6$ prime element in $Z[X^2,X^3]$?

$X^6=(X^2)^3$ so $X^6$ is not irreducible element and then $X^6$ is not prime element.

However I am not sure if my reasoning is correct because in the definition of irreductible element is: "such an irreversible element of a ring which cannot be represented as the product of two irreversible elements". But I do not understand what elements are irreversible in my ring and wish someone would explain it to me and tell me if my reasoning is correct.

答:
A unit in a commutative ring is an element $a$ such that there exists an element $b$ with $ab=1$. What you call "irreversible" elements, is usually referred to as a non-unit, that is, an element that is not a unit.

In $\mathbb Z[X]$ we have a notion of degree that satisfies $\deg(fg)=\deg(f)+\deg(g)$. In particular, since $\deg(1)=0$, in order for $fg=1$ to be true we must have $\deg(f)=\deg(g)=0$. Hence, the only units in $\mathbb Z[X]$ are the units in $\mathbb Z$, which are exactly $+1$ and $-1$.

The same is then of course true for the subring $\mathbb Z[X^2,X^3]$. Every element different from $\pm 1$ is a non-unit (irreversible).

Hence, $X^6 = (X^2)(X^2)(X^2)$ is indeed a decomposition of $X^6$ into irreversibles, so that $X^6$ is not irreducible and hence not prime.

Title: Is $X^6$ prime element in $Z[X^2,X^3]$?


问:
The negation of $\exists x \forall y \neg \forall z(P(x,y) \iff Q(x,y) \land R(x,y,z))$
I am trying to understand the negation of
$\exists x \forall y \neg \forall z(P(x,y) \iff Q(x,y) \land R(x,y,z))$

As a side example if I have a statement
$\neg \forall x P(x)$, then this is equivalent to $\exists x \neg P(x)$ where the negation is $\forall x P(x)$, motivating what I think is the answer to my overall question.

Then I believe the answer should be:
$\forall x \exists y \forall z(P(x,y) \iff Q(x,y) \land R(x,y,z))$ (i.e simply remove the $\neg$ from the $\forall z$)

but I am a bit paranoid that the answer could be:
$\forall x \exists y \forall z(P(x,y) \iff \neg Q(x,y) \lor \neg R(x,y,z))$, passing the negation through to the inner bracket.

Any insights which is correct appreciated.

答: Your initial feeling is right:$$\neg\exists x\forall y\neg\forall z\phi(x,\,y,\,z)\iff\forall x\neg\forall y\neg\forall z\phi(x,\,y,\,z)\iff\forall x\exists y\forall z\phi(x,\,y,\,z).$$Alternatively,$$\neg\exists x\forall y\neg\forall z\phi(x,\,y,\,z)\iff\neg\exists x\neg\exists y\exists z\phi(x,\,y,\,z)\iff\forall x\exists y\forall z\phi(x,\,y,\,z).$$


答:
Your  initial  answer  is right

The tool in the link is the most helpful thing I have found for such problems.

Title: The negation of $\exists x \forall y \neg \forall z(P(x,y) \iff Q(x,y) \land R(x,y,z))$

Body:
I am trying to understand the negation of
$\exists x \forall y \neg \forall z(P(x,y) \iff Q(x,y) \land R(x,y,z))$

As a side example if I have a statement
$\neg \forall x P(x)$, then this is equivalent to $\exists x \neg P(x)$ where the negation is $\forall x P(x)$, motivating what I think is the answer to my overall question.

Then I believe the answer should be:
$\forall x \exists y \forall z(P(x,y) \iff Q(x,y) \land R(x,y,z))$ (i.e simply remove the $\neg$ from the $\forall z$)

but I am a bit paranoid that the answer could be:
$\forall x \exists y \forall z(P(x,y) \iff \neg Q(x,y) \lor \neg R(x,y,z))$, passing the negation through to the inner bracket.

Any insights which is correct appreciated.


问:
Rank of coherent sheaf on complex manifolds
Let $X$ be a complex smooth connected manifold of dimension $n$ and let $\mathcal{O}_{X,x}$ be the ring of germs of holomorphic functions at $x$. Since the stalk of $\mathcal{O}_X$ at $x$ does not depend in which open subset is contained the point, then it can be identified with the ring of convergent power series in $n$ variables. Let $K(x)$ be the field of fractions of $\mathcal{O}_{X,x}$, i.e. it is the field of germs of meromorphic functions at $x$.

Definition: Let $\mathcal{F}$ be a sheaf over $X$, then we say that $\mathcal{F}$ is coherent if for each $x\in X$ there exist a neighborhood $U$ of $x$ and an exact sequence \begin{equation*}\mathcal{O}_U^{\oplus q}\rightarrow\mathcal{O}_U^{\oplus p}\rightarrow\mathcal{F}_{|U}\rightarrow 0\end{equation*}for some $p,q>0$ where $\mathcal{O}_U$ is the restriction to $U$ of the sheaf of holomorphic functions on $X$. In particular the sheaf $\mathcal{F}$ is locally finitely presented.

Definition: Let $\mathcal{F}$ be a coherent sheaf on $X$, since the stalk $\mathcal{F}_x$ is a finitely-generated $\mathcal{O}_{X,x}$-module then we can define the rank of $\mathcal{F}$ to be \begin{equation*}\text{Rank}(\mathcal{F}):=\text{dim}_{K(x)}\bigg(\mathcal{F}_x\bigotimes_{\mathcal{O}_{X,x}}K(x)\bigg)
\end{equation*}

Question 1

Since $X$ is assumed to be smooth and connected, is the rank of a
coherent sheaf $\mathcal{F}$ constant and equal to $r$ say?

Now suppose that $\mathcal{F}$ is coherent and torsion-free (i.e. $\mathcal{F}_x$ is a torsion-free $\mathcal{O}_{X,x}$-module for each $x\in X$), then it can be proved that there exist a closed complex manifold $Y\subset X$ of codimension at least $2$ sucht that $\mathcal{F}|_{X\setminus Y}$ is locally-free and in particular has finite constant rank (since over this open subset is a holomorphic vector bundle, am I right?).

Question 2

Adding the hypothesis of being torsion-free, what can we say about the rank over $Y$? Could it jump?

答:
To your first question: No, you can not in general assume that the rank of a coherent sheaf on a connected manifold is constant. In fact, vector bundles on connected manifolds are precisely the coherent sheaves with constant rank.

A little more involved example would be the cokernel sheaf of the following morphism. Let $B\subset \mathbb{C}^n$ be the unit ball and let $\phi\colon \mathcal{O}_B \to \mathcal{O}_B^n$ be given by $f\mapsto \left(z_1f,...,z_nf\right)$. The cokernel is coherent but not a vector bundle, since the cokernel is not free at $0$.

Explicitly, if it were a vector bundle then $\text{coker}\left(\phi\right)\cong \mathcal{O}_B^{n-1}$, since every holomorphic bundle over $B$ is trivial. And then the sequence

$0 \to \mathcal{O}_B \to \mathcal{O}_B^{n} \to \mathcal{O}_B^{n-1} \to 0$

would be spilt exact, meaning there would exist $g \colon \mathcal{O}_B^{n} \to \mathcal{O}_B$ such that $g\circ f = \text{id}$, which is clearly not possible.

To your second question: Even if your sheaf is torsion-free, the rank can still jump. One difference is that the singular set of a coherent sheaf has codimension at least $1$ and the singular set of a torsion-free coherent sheaf has codimension at least $2$. So for example, a torsion-free coherent sheaf on a connected complex $1$-dimensional manifold is locally free and thus has constant rank.

In general the rank of a coherent sheaf $F$ is upper-semicontinuous, i.e. for every $p\in M$ there exists an open subset $U$ such that $\forall q \in U \colon \text{rank}_q\left(F\right)\leq \text{rank}_p\left(F\right)$.

Edit: I realized your second question was not addressed by my answer, since you asked about the rank of torsion-free sheaf over its singular set. For this note that pullback preserves coherence and fiber rank is invariant under pullback. Hence again one can at most expect there to be a codimension $1$ subset in $Y$ such that $i^*_YF$ has constant rank away from this set. Here $i_Y$ denotes the inclusion of $Y$ in $M$.


问:
Has a dyadic logarithm been studied?
Define a dyadic logarithm of an odd dyadic integer $a$ as the limit of $\frac{a^z-1}{z}$ for $z\to0$ in the dyadic sense. I can prove, I'm assuming it's also well known, that dyadic exponentiation makes sense. (This is not true in general for p-adic integers.) A definition that works is to define $a^z$ as the limit of $a_n^{z_m}$ where $a_n$ is a sequence of positive integers that converges to $a$ in the dyadic sense and similarly $z_m \to z$.

I also have a proof that what I'm calling the dyadic logarithm exists and has some algebraic properties, such as the dyadic logarithm of a product is the sum of the dyadic logarithms. The dyadic logarithm of $-1$ and $1$ are both $0$.

I wonder if there are any known properties or uses of this function. Maybe cryptographic, or fast binary multiplication? Or perhaps it's used to prove some number theoretic result?

答:
Yes, this is rather well known, especially to folks in the formal-group racket.

You’re dealing with the formal group $\mathscr M(x,y)=x+y+xy=(1+x)(1+y)-1$, the formal group associated to multiplication, or in fancier (but perhaps rather sloppy) language, the formal group of the multiplicative group scheme $\mathbf G_{\mathrm m}$.

I hope you see that my formula for $\mathscr M$ is just a recoordinatization of multiplication, with the neutral element moved from $1$ to $0$. Your construction of the logarithm, which is a homomorphism from $\mathscr M$ to the additive formal group $\mathscr A(x,y)=x+y$, is perfect. In fact, it can be expressed by the $p$-adically convergent power series, $-\sum_1^\infty(-x)^n/n$, which you know from Calculus, but with the $1+x$ replaced by $x$. A wonderful thing about it is that it remains convergent whenever you substitute for $x$ something that’s closer to $0$ than $1$ is: in technical terms, $\log(a)$ is convergent for all $a$ with $|a|_p<1$, equivalently $v_p(a)>0$. So, for instance, you can plug $\sqrt2$ in for $a$ in your $2$-adic logarithm, and get a good value.

Only one other thing to say for now, and that’s that you don’t need to consider all $z\to0$ in your formula, you can just take powers of $2$ (powers of $p$ in the general case).

Title: Has a dyadic logarithm been studied?


问:
Number of possible Scrabble draws
I would like to compute the number of Scrabble draws, when starting a game. Translated to mathematics, this asks for the number of sub-multisets of a multiset.

Lets say we have a multiset $M$ over a set $N=\{1,...,n\}$. This means there are numbers $m_1,..m_n$, which indicate the multiplicity of the numbers $1$ to $n$. For $k \leq m := m_1+...+m_n$ we now want to draw a $k-$multiset from $M$, in other words numbers $k_1,...,k_n$ with the property $k_1+...+k_n=k$ and $0 \leq k_i \leq m_i$.

My Ansatz was drawing a number of $1$'s, this is $k_1$. In order to do so, we have $\min(m_1, 1+k)$ many possibilities.

Then we draw $k_2$, the number of $2$'s. For this, we have $\min(m_2, 1+k - k_1)$ many possibilities.

...

Finally, for $k_n$, there are $\min(m_n, 1+k- (k_1 + ... +k_{n-1}))$ many possibilities.

The problem is the obviously the self referentiality, i.e the number of possibilities for $k_i$ depending on $k_j$ for $j < i$. I'm sure there is a very clean way to do it via binomial coefficients or so but I'm not sure how to do it that way. Any help is appreciated. :-)

EDIT: bump

答:
For a general case, without specific values for $m_1, \ldots m_n$, see here and/or here, whereas, if we have $m_1=m_2=\ldots=m_n$ there are simpler formulas here or here ($k_i=0$ not allowed there).

If we have specific numeric values for $m_1, \ldots m_n$, then we use generating functions more easily to get the number of $k$-multisets as the coefficient of $x^k$ of the generating function:

$$f(x) = \prod_{i=1}^n (1+x+\ldots+x^{m_i}) = \prod_{i=1}^n \frac{x^{m_i+1}-1}{x-1}$$ which in our case is (using Wikipedia data, the same as @John L's answer):

$$f(x)=\frac{(x^2-1)^5(x^3-1)^{10}(x^4-1)(x^5-1)^4(x^7-1)^3(x^9-1)(x^{10}-1)^2(x^{13}-1)}{(x-1)^{27}}$$

And using Wolfram Alpha get some coefficients, e.g.:

$$[x^7]f(x)=3199724$$

like in @John L answer.

Note also that the total number of multisets of any size is:

$$\prod_{i=1}^n (1+m_i)$$

Title: Number of possible Scrabble draws


问:
Is this necessary that $k$ be algebraic close field in this theorem?
let $k$ be algebraic close field and
$A=k\left[x_{1}, \ldots, x_{n}\right]$

for any subset $Y \subseteq \mathbf{A}^{n}_k$,define the ideal of $Y$ in $A$ by

$
I(Y)=\left\{f \in A \mid f(P)=0 \text { for all } P \in Y\right\}$

and $Z(T)=\left\{P \in \mathbf{A}^{n}_k \mid f(P)=0\right.$ for all $\left.f \in T\right\}$

Theorem : For any ideal $a \subseteq A, I(Z(a))=\sqrt{a},$ the radical of $\mathfrak{a} .$

Is this necessary that $k$  be algebraic close field in the above theorem ?

答: Yes. Take $k=\mathbb R$ and $\mathfrak a=(x^2+1)\subset k[x]$ for example. We have have $Z(\mathfrak a)=\varnothing$ and $I(Z(\mathfrak a)) = k[x]$, which is not the radical of $\mathfrak a$.


问:
Does the integral converge $\int_{-\infty}^{+\infty}\frac{\exp(ibx)}{1+e^x}\,dx$ for $b>0$?
I try to calculate integral $$\int_{-\infty}^{+\infty}\frac{\exp(ibx)}{1+e^x}\,dx.$$ Does it converge for $b>0$? Wolfram Mathematica shows contradictory results: for $b=2$ it converges, for $b=1$ and $b=3$ it does not. I need to calculate it.

答:
This will be not a purely mathematical but rather a physical illustration via the integrand regularization.

Lets consider $f(x,a)=\frac{\exp(ibx+ax)}{1+\exp(x)}$, where $a$ is a small parameter which we will finally set to zero. Let's consider the integral over the contour C - a rectangle from $-R$ to $R$ and from $0$ to $2\pi{i}$; counter clockwise.

The integrand is a single-valued function in the area, so $I(a,b)=\oint_C\frac{\exp(ibx+ax)}{1+\exp(x)}dx$. Taking integral along every line of the contour and due exponent periodicity $\exp(x+2\pi{i})=\exp(x)$ we get:

$I(a,b)=\int_{-R}^{R}\frac{\exp(ibx+ax)}{1+\exp(x)}(1-\exp(-2\pi{b}+2{\pi}ia))dx+I_1+I_2=2{\pi}iRes_{x=\pi{i}}\frac{\exp(ibx+ax)}{1+\exp(x)}=$$ =-2\pi{i}\exp(-\pi{b}+\pi{ia})$

$|I_1|<\int_0^{2\pi}|\frac{\exp(ibR+aR-bt+iat)}{1+\exp(R+it)}|dt<const\exp(-(1-a)R)\to0$ as $R\to{\infty}$ ($a<<1$)

$I_2=i\int_{2\pi}^0\frac{\exp(-ibR-aR-bt+iat)}{1+\exp(-R+it)}dt\to-i\exp(-aR-ibR)\int_0^{2\pi}\exp(-bt+iat)dt=$$=-i\exp(-aR-ibR)\frac{1-\exp(2\pi(ia-b))}{b-ia}$ as $R\to{\infty}$

$$\int_{-R}^{R}\frac{\exp(ibx+ax)}{1+\exp(x)}dx\to-2\pi{i}\frac{1}{\exp(\pi{b}-\pi{ia})-exp(-\pi{b}+\pi{ia})}+i\frac{\exp(-aR-ibR)}{b-ia}$$

As soon as we set $a\to0$ we get $$\int_{-R}^{R}\to-\pi{i}\frac{1}{\sinh(\pi{b})}+i\frac{\exp(-ibR)}{b}$$

Second term is not strongly defined - the limit at $R\to{\infty}$ does not exist.
Physically speaking it is strongly oscillating but limited by value. Here we have the same situation, for instance, when we identify delta-function  $\delta(x)$ as $\frac{1}{2\pi}\int_{-\infty}^{\infty}exp(ikx)dk=\lim_{R\to{\infty}}\frac{1}{2\pi}\int_{-R}^{R}\exp(ikx)dk=\lim_{R\to{\infty}}\frac{\sin(xR)}{\pi{x}}$

Limit does not exist, but any further manipulations (for instance, integrating with a smooth function) shows that this is actually delta-function which is equal to zero at any point except for zero.

All this is not of course a math proof - just an illustration.


问:
Continuity of the real and imaginary parts of a continuus complex-valued function
If a complex-valued function is continuous, are the component real and imaginary parts $u(x,y)$ and $u(x,y)$ necessarily continuous? If so, why?</p>

答:
The functions $\operatorname{Re}, \operatorname{Im}: \mathbb{C} \to \mathbb{R}$ are continuous since $\operatorname{Re} (z_1+z_2) = \operatorname{Re} z_1 + \operatorname{Re} z_2$ and $|\operatorname{Re} z | \le |z|$, and similarly for $\operatorname{Im}$.

Hence $\operatorname{Re} \circ f$ and $\operatorname{Im} \circ f$ are continuous.


问:
Upper bounds for Lyapunov equation
Consider the Lyapunov equation
$$
 AX+XA^T+I=0
$$
for the given stable matrix $A$.

Is it possible to obtain the upper estimation for $\|X\|$ (in spectral or Frobenius norm) in terms of $A$?

答:
For the spectral norm: it is well known that the unique solution to this equation can be expressed as
$$
X = \int_0^\infty e^{At}e^{A^Tt}\,dt.
$$
We note that
$$
\|X\| = \left\|\int_0^\infty e^{At}e^{A^Tt}\,dt \right\| \leq
\int_0^\infty \left\|e^{At}e^{A^Tt}\right\|\,dt
= \int_0^\infty \|e^{At}\|^2\,dt
$$
Let $\alpha$ denote the maximum value of $\operatorname{Re}(\lambda)$ among the eigenvalues $\lambda$ of $A$. There exists a constant $C > 0$ such that $\|e^{At}\| \leq C \cdot e^{\alpha t}$ (in particular, if $A$ is diagonalizable and $SAS^{-1}$ is diagonal, then $C$ can be taken to be the condition number of $S$). By the stability of $A$, we must have $\alpha < 0$. With that, we have
$$
\| X\| \leq \int_0^\infty \|e^{At}\|^2\,dt \leq C\int_0^\infty e^{2 \alpha t}\,dt = \frac C{2|\alpha|}.
$$

For the Frobenius norm: we can write
$$
\operatorname{vec}(X) = -(I_n \otimes A + A \otimes I_n)^{-1} \operatorname{vec}(I).
$$
It follows that
$$
\|X\|_F = \|\operatorname{vec}(X)\| = \|(I_n \otimes A + A \otimes I_n)^{-1} \operatorname{vec}(I_n)\|
\\ \leq \|(I_n \otimes A + A \otimes I_n)^{-1}\| \cdot \|\operatorname{vec}(I_n)\|
\\ \leq \|(I_n \otimes A +  A \otimes I_n)^{-1}\| \cdot \sqrt{n}.
$$
Let $S$ be such that $D = S^{-1}AS$ is diagonal. We note that
$$
\|(I_n \otimes A + A \otimes I)^{-1}\| = \\
(S \otimes S)^{-1}(I \otimes D + D \otimes I)^{-1}(S \otimes S) \leq \\
\|(S \otimes S)^{-1}\|\cdot\|(I \otimes D + D \otimes I)^{-1}\| \cdot \|S \otimes S\| =\\
\|S^{-1}\|^2 \cdot \|(I \otimes D + D \otimes I)^{-1}\| \cdot \|S\|^2.
$$
Let $\alpha$ denote the maximum value of $\operatorname{Re}(\lambda)$ among the eigenvalues $\lambda$ of $A$. Let $C$ denote the condition number of $S$, i.e. $C = \|S\|\cdot\|S\|^{-1}$. We have
$$
\|(I \otimes D + D \otimes I)^{-1}\| \leq \frac 1{2|\alpha|},
$$
so that $\|X\|_F \leq \frac{C^2}{2|\alpha|}$.

Title: Upper bounds for Lyapunov equation


问:
Solving non-quadratic equations with mutliple solutions

Given that there is more than one solution value of $x$ to the equation $4^{ax} = b \times8^x$, find all possible values of $a$ and $b$.

While I understand majority of the approach to solve this question, outlined in the steps below, the reasoning for calculating all the possible values of $a$ and $b$ isn't apparent to me - which is the last step.

$4^{ax} = b \times8^x$ can be simplified by:

$(2^2)^{ax} = b \times (2^3)^x\\ 2^{2ax} = b \times 2^{3x}\\ \log_2(2^{2ax}) = \log_2(b\times2^{3x})\\ 2ax = \log_2(b) + 3x \\ x(2a-3) = \log_2(b)$

The question states that there are more than one solution values of $x$, to find the value of $a$ and $b$ the solution indicates to equate the simplified equation to $0$.

$2a-3=\log_2(b)=0$

This yields $a$ as $3/2$ and $b$ as $1$. I understand how placing these values in the above equation will equate it to $0$. What confuses me is why $x$ was removed in order to find the solution. Also, is there any general approach to solve such questions?

答:
From $x(2a-3)=\log_2(b)$ (let call this equation (*))

If $2a-3\ne0$ then you can solve this equation $x=\frac{\log_2b}{2a-3}$ and the solution is UNIQUE. So $2a-3$ HAVE TO be zero if you want more than one solution.

BUT in this case, (*) can be written as $0=\log_2b$.

So, the ONLY possibility for your initial equation to have more than one solution is that b$2a-3=0=\log_2 b$.


问:
Cluster points in products
Trying to prove the following.

If $\left(x_{\lambda}\right)_{\lambda\in\Lambda}$ is a net in $\prod X_{\alpha}$ having $x$ as a cluster point (limit point) then for each $\alpha$, $\left(\pi_{\alpha}\left(x_{\lambda}\right)\right)_{\lambda\in\Lambda}$ has $\pi_{\alpha}(x)$ for a cluster point.

In other words, I am trying to show that (I think?) for each $\alpha$,

$$\forall U\text{ open}:\pi_{\alpha}(x)\in U, \forall\mu\in\Lambda, \exists\lambda\in\Lambda:\pi_{\alpha}\left(x_{\lambda}\right)\in U.$$

Intuitivly, this makes sense as an extension of sequence, however, when actually trying to formulate a proof, I get immediately stuck. A proof of it or any remarks would be helpful.

答: Not quite: you need to show that if $U$ is an open nbhd of $\pi_\alpha(x)$, then for each $\mu\in\Lambda$ there is a $\lambda\in\Lambda$ such that $\color{red}{\mu\le\lambda}$ and $\pi_\alpha(x_\lambda)\in U$. This follows directly from the continuity of $\pi_\alpha$: $\pi_\alpha^{-1}[U]$ is an open nbhd of $x$ in the product, so for each $\mu\in\Lambda$ there is a $\lambda\in\Lambda$ such that $\mu\le\lambda$ and $x_\lambda\in\pi^{-1}[U]$, and hence $\pi_\alpha(x_\lambda)\in U$.

Title: Cluster points in products


问:
Modelling problem (ODE)
I have the below problem:

The water level, y, in a tank is modelled by the differential equation
$$\dot y(t) + y(t) = z(t)$$
where z denotes the inflow. The inflow is a function of a valve position, which in turn is controlled by the electric control signal u. The relation between control signal and flow is given by the differential equation
$$\ddot z(t) + \dot z(t) + z(t) = u(t)$$
What differential equation relates the water level y to the control signal u?

I am self-taught and did not do this kind of modelling problem before. I tried to google an example but I don't know which kind of problem is that exactly (modelling? state space form?). Can someone kindly give me some pointers on what to read to solve it?

答:
$z(t)=y(t)+\dot y(t)$.

$\dot z(t)=\dot y(t)+\ddot y(t)$.

$\ddot z(t)=\ddot y(t)+\dddot y(t)$.

By summing all this equations,

$$u(t)=z(t)+\dot z(t)+\ddot z(t)= y(t)+2\dot y(t)+2\ddot y(t)+\dddot y(t).$$
So tyhe EDO is
$$\dddot y(t)+2\ddot y(t)+2\dot y(t)+y(t)=u(t).$$

Title: Modelling problem (ODE)


问:
How to check statistical significance?
I have a machine which produces belts with normalized length of 30 cm. Now there is the assumption that the machine works incorrectly and produces belts with a slightly lesser length.

From a test with 100 measurements a mean value of 25 and a standard deviation of 3 cm have been acquired. How could one check these data against a significance level of 0.05? I am somehow stuck..

答:
From Minitab, which accepts summary data, here is output for an appropriate test. The null hypothesis that the population mean remains $\mu = 30,$ is strongly rejected. (Because only summary data are provided, there is no way to check whether data are normal, as required for this one-sample t test.)

```<code>One-Sample T

Test of μ = 30 vs < 30

  N    Mean  StDev  SE Mean  95% Upper Bound       T      P
100  25.000  3.000    0.300           25.498  -16.67  0.000
</code>```

Title: How to check statistical significance?

Body:
From Minitab, which accepts summary data, here is output for an appropriate test. The null hypothesis that the population mean remains $\mu = 30,$ is strongly rejected. (Because only summary data are provided, there is no way to check whether data are normal, as required for this one-sample t test.)

One-Sample T

Test of μ = 30 vs < 30

  N    Mean  StDev  SE Mean  95% Upper Bound       T      P
100  25.000  3.000    0.300           25.498  -16.67  0.000


问:
Food taste vs. health. Is this an optimization problem?
Consider restaurant menu choices:
\begin{array}{c|c|c}
\mathbf{Name} & \mathbf{Taste} & \mathbf{Health} \\
\hline
\mathrm{Soup}    &  7 & 6\\
\mathrm{Bread}   &  8 & 3\\
\mathrm{Eggs}    &  5 & 5\\
\mathrm{Fish}    &  3 & 10\\
\mathrm{Chicken} &  5 & 8 \\
\mathrm{Steak}   &  9 & 5\\
\mathrm{Candy}   & 10 & 1\\
\mathrm{Cake}    &  8 & 3\\
\hline
\end{array}

Is there a way to extract the ideal food to eat?
What kind of math can I do with this?

Is this related to basic optimization taught in Linear Algebra or Operations Research ?

答:
Main question would be how do you understand what is the ideal food? With 2 given data categories, you can, e.g., say ideal is

best tasting (in which case candy is best) or

best health (in which case fish is best), or

you design some sort of metric by which you can rank the foods, for example health per unit taste, in which case you divide the second number by the first and pick the one with the biggest ratio (and fish comes out best).

You can design more complex metrics, but everything comes down to what does ideal mean.

UPDATE

For ideal = taste + health, just compute that for every line and pick the best one. The standard optimization methods from operations research would be a better fit if you were trying to construct some diet of these foods, maximizing the ideal diet parameters over some constraints.


答:
It could be an optimisation problem, depending on how you define ideal.
Following is a simple linear model:
$$ \text{Benifit}(f) = \lambda T(f) + \beta H(f)$$
Where $f$ is a given food and $T(f)$ is the taste of food and $H(f)$ is the Health.
$\lambda$ and $\beta$ depend on the trade-off between health and taste your definition of ideal allows. And the optimisation problem is the following:
$$\arg \max_{f} \mathrm{Benefit(f)}$$

Title: Food taste vs. health. Is this an optimization problem?

Body:
Consider restaurant menu choices:
\begin{array}{c|c|c}
\mathbf{Name} & \mathbf{Taste} & \mathbf{Health} \\
\hline
\mathrm{Soup}    &  7 & 6\\
\mathrm{Bread}   &  8 & 3\\
\mathrm{Eggs}    &  5 & 5\\
\mathrm{Fish}    &  3 & 10\\
\mathrm{Chicken} &  5 & 8 \\
\mathrm{Steak}   &  9 & 5\\
\mathrm{Candy}   & 10 & 1\\
\mathrm{Cake}    &  8 & 3\\
\hline
\end{array}

Is there a way to extract the ideal food to eat?
What kind of math can I do with this?

Is this related to basic optimization taught in Linear Algebra or Operations Research ?


问:
Combinatorial problem on sum of integers
Let $\mathcal{A}_n$ be the collection of all sequences of positive integers $A_n:=\langle n_1, n_2, \ldots, n_k\rangle$, where $n:=n_1$, $~2\le k\le n-1$ and $n_1>n_2>\ldots>n_k$. We denote by $S(A_n)$ the sum $\sum_{i=1}^{k-1} \frac{n_i}{n_{i+1}}$, and by $A^*_n$ the sequence of $\mathcal{A}_n$ maximizing this sum.

Question: What is $S^*_n:=S(A^*_n)$?

Note: I think that the maximum for $S(A_n)$ is attained by the sequence $B_n:=\langle n, n-1, n-2, \ldots, 1\rangle$. If this is true, how can we prove that $A^*_n=B_n$?

答:
Suppose $A^*_n$ has successive terms $a,b$ with $a>b+1$. Consider the effect of inserting the integer $a-1$ into the sequence.

It is clear that $$\frac{a}{a-1}+\frac{a-1}{b}>\frac{a}{b}$$ and so $S$ has been increased. Similarly, we can increase $S$ by extending the sequence to $1$, if necessary.

Hence $A^*_n=B_n$, as you suspected.


问:
Student $t$ distribution table
Why is it impossible to find a table of the Student $t$ distribution without the confidence area ?

The normal distribution $\mathcal N(\mu,\sigma^2)$ for example, has these two tables :

$\mathbb P(Z\leq z)$ table of distribution.

$\mathbb P(|Z|\geq z)=\alpha$ the quantile table

The Student distribution doesn't have a table of the first kind.

答:
All Normal distributions can be linearly transformed to $\mathcal{N}(0,\,1)$, so the CDF depends only on the resulting standard score, typically denoted $z$. The family of Student $t$ distributions has one too many parameters for the same treatment.

You can, however, find differently formatted tables to circumvent this. Instead of using rows and columns to specify different digits in $z$ so as to report the CDF to very high precision, as in the Normal case, we have to use (say) rows for $\nu$ and columns for a $z$-like variable. The usual approach is to linearly transform to a vairiable we call $X$ instead of $Z$, of mean $0$ and variance $\nu/(\nu-2)\ne1$ (hence our avoiding $Z$), so the PDF is $\frac{\Gamma(\tfrac{\nu+1}{2})}{\sqrt{\nu\pi}\Gamma(\tfrac{\nu}{2})}(1+x^2/\nu)^{-(\nu+1)/2}$.

Edit: the examples I've found use columns for CDFs, thus computing in the opposite direction to the one I described above. See also my comment replying to the OP.

Title: Student $t$ distribution table


问:
Give an example of the $a,b,c$ which satisfies conditions in the generating set
How to derive the specific case of the generating element of a group given its generating set. For example, when
$$G=\langle a,b,c|a^3=b^3=c^2=1,ab=ba,ca=a^2c,cb=b^2c\rangle$$
we can let $G\subset S_3\times S_3$, and let
$$a=((123),1),b=(1,(123)),c=((12),(12))$$
to get the desired result. However, when $3$ is replaced by general $p$, where $p$ is an odd prime,say,
$$G=\langle a,b,c|a^p=b^p=c^2=1,ab=ba,aca=c,bcb=c\rangle$$
 I cannot specifically give the example of $a,b$ and $c$, which satisfies "$a^p=b^p=c^2=1,ab=ba,aca=c,bcb=c$"  Is there any quick way to do so?</p>

答:
I suspect that you are trying to find an embedding of your group $G \hookrightarrow S_p \times S_p$, the product of symmetric groups.  Let
$$
a = ((1\;2\; \cdots \; p), 1),
$$
$$
b = (1, (1\;2\; \cdots \; p)),
$$
$$
\text{and} \quad c = (r, r),
$$
where $r = (1, p-1)(2, p-2)\cdots$ is the reflection that fixes $p$.  (This actually embeds $G$ in $D_p \times D_p$, the product of dihedral groups.) </p>


问:
Seating 5 distinct men, women, and a dog around a circular table such that no two people of the same gender are consecutive
I would like to offer a known erroneous solution to the titular question. It would be fantastic if someone could point out what went wrong. Of course, feel free to offer your own creative solutions.

It can be proved that no two people of the same gender can be seated next to the dog. We will proceed initially without the dog, using a 10-seat table. We seat 5 men around the table, leaving gaps between one man and the next. There are $5!$ ways to do this. Next, we seat the women around the table in the preexisting gaps, there are also $5!$ ways to do so. We must also account for rotation, thus there is $$\frac{(5!)^{2}}{10}$$ ways to seat the men and women around the table. The dog can be inserted in any of the 10 gaps between each person, thus we have a total of $$\frac{(5!)^{2}}{10} \cdot 10 = (5!)^2$$ ways to seat all of the entities.

答: The error is that your original placing of the men has $10\times 4!$ possibilities.


问:
$d$ - dimensional point lies outside the unit sphere with high probability
I'm reading the section on high dimensional geometry in this book. The author presents the law of large numbers like this, around page 14.
Given a random variable $x$  with expected value $E(x)$,
$$Prob\Big(\Big|\frac{x_1 + \ldots + x_n}{n} - E(x)\Big| \geq \epsilon\Big) \leq \frac{Var(x)}{n \epsilon^2}$$

As an application, they write
"Let z be a $d$ dimensional random point whose coordinates are chosen from a zero mean, $\frac{1}{2\pi}$ variance Gaussian... By the law of large numbers, the square of the distance of z to the origin will be $\theta(d)$ with high probability. In particular, this means there is a vanishingly small probability that such a random point z will lie in the unit ball..."

I cannot understand this statement about "the square of the distance of z to the origin will be $\theta(d)$ with high probability". I've tried fiddling with the expression of the law of large numbers and making substitutions using algebraic rules for variance and expectation with no luck. Can you explain this calculation with gory detail?

答:
So the idea of the central limit theorem is this : samples concentrate around the mean.

What does that mean? Well, it means that if I had a distribution and wished to find its mean (for example, I had a coin and I wanted to find the probability with which it falls heads) , then I should collect many samples from the distribution (by flipping that coin many times and noting if I got a heads or tails) and take their average (number of coin flips which were heads, divided by total number of flips). According to the Central Limit Theorem, this number is , with high probability close to the mean.

Now, Kannan asks you to consider the following random variable : let $Z_1,Z_2,...$ be independent normal random variables with mean zero, variance $\frac 1{2 \pi}$. Then the random variable under consideration is $Z = Z_1^2 + Z_2^2 + ... + Z_d^2$. What is this? Well, if you put together the numbers $(Z_1,...,Z_d)$ as a tuple, you can think of it as a point in $\mathbb R^d$, or $d$-dimensional space. The quantity $Z$ is the squared distance of the random point $(Z_1,...,Z_d)$ from the origin $\mathbf{0} = (0,0,...,0)$.

Now the central limit theorem applies for $Z$. That is, if I sample a random point by picking coordinates with that normal distribution and consider the squared distance of this point from the origin, it should be close to the mean of $Z$ with high probability.

Note : The mean, and the expected value of a random variable are the same thing , denoted by $E[Z]$.

The mean of $Z$ can be calculated using linearity of expectation : $E[Z] = E[Z_1^2] + E[Z_2^2] + ... + E[Z_d^2]$. So, we need to know what $E[Z_i^2]$ is.

The point is, each coordinate is a zero mean , $\frac 1{2 \pi}$ variance Gaussian. Now, for any random variable $X$, the quantity $E[X]$ is called the mean, and the quantity $E[X^2] - E[X]^2$ is called the variance, denoted by $Var[X]$.

In particular, $E[X^2] = E[X]^2 + Var[X]$, so we have $E[Z_i^2] = E[Z_i]^2 + Var[Z_i]$. However, both these quantities are known : $E[Z_i] = 0$ and $Var[Z_i] = \frac 1{2 \pi}$ because the coordinates are given to have that mean and that variance.

Therefore, $E[Z_i^2] = \frac {1}{2 \pi}$ for each $i$ and thus $E[Z^2] = \frac{d}{2 \pi}$.

Note : I can understand that the main point of confusion may be in the above and below paragraph : if that is so, kindly inform me so I can elaborate.

Adding it all up, $E[Z] =\frac{d}{\sqrt{2 \pi}}$. This is a constant multiple of $d$, so the $\Theta$-notation explained here(check in and around the table for examples)  tells you that $E[Z] = \Theta(d)$. Basically, it means that as $d$ grows, $E[Z]$ grows as a constant multiple of $d$, which is obviously true.

Now, by the central limit theorem, $Z$ will be $\Theta(d)$ with high probability, which explains all but the last line.

Now, the author says that the probability that such a point will lie in the unit ball is vanishingly small. But now we see why : a point in the unit ball has squared distance at most $1$. Now for example, in dimensions $1000$, the squared distance to the origin of a random point , by the central limit theorem, would lie close to the mean which is $E[Z] = \frac{1000}{\sqrt{2 \pi}} \approx 399$.  So, with very low probability, the point can be less than $1$ squared distance from the origin. As $d$ goes up, the mean goes up with $d$, so the chances of this point being contained in the unit ball are vanishingly small.


答:
Given it is a circular table, when you are seating women first or men first, you need to fix one of the positions and make all arrangements relative to that (assuming rotational symmetry).

If so, there are only $4!$ ways to seat $5$ women leaving one seat in between each of them for men. Then there are $5!$ ways to seat men in those $5$ empty seats. Now you have $10$ places where you can seat the dog.

So total number of arrangements are $4! \times 5! \times 10 = 2 \times (5!)^2$.

Another way to look at this is to let the dog take the first seat as it wishes. Now we go clockwise seating women and men in alternate seats. There are $(5!)^2$ ways to do that starting with a woman as women can be arranged in $5!$ ways and so can be men. But we can also start with a man. So we have another $(5!)^2$ arrangements - in total $2 \times (5!)^2$ arrangements.

Title: Seating 5 distinct men, women, and a dog around a circular table such that no two people of the same gender are consecutive

Body:
I would like to offer a known erroneous solution to the titular question. It would be fantastic if someone could point out what went wrong. Of course, feel free to offer your own creative solutions.

It can be proved that no two people of the same gender can be seated next to the dog. We will proceed initially without the dog, using a 10-seat table. We seat 5 men around the table, leaving gaps between one man and the next. There are $5!$ ways to do this. Next, we seat the women around the table in the preexisting gaps, there are also $5!$ ways to do so. We must also account for rotation, thus there is $$\frac{(5!)^{2}}{10}$$ ways to seat the men and women around the table. The dog can be inserted in any of the 10 gaps between each person, thus we have a total of $$\frac{(5!)^{2}}{10} \cdot 10 = (5!)^2$$ ways to seat all of the entities.


问:
Unifying the connections between the trigonometric and hyperbolic functions
There are many, many connections between the trigonometric and hyperbolic functions, some of which are listed here. It is probably too optimistic to expect that a single insight could explain all of these connections, but is there a holistic way of seeing the parallels between $\sin$ and $\sinh$, $\cos$ and $\cosh$? Can all of these seemingly disparate connections be shown to be essentially the same, or at least very similar?

Geometric connections

Sine and cosine parameterise the unit circle $x^2+y^2=1$, just as hyperbolic sine and cosine parameterise the 'unit hyperbola' $x^2-y^2=1$. Both circles and hyperbolas are conic sections.

The sector of the circle connecting the points $(0,0)$, $(1,0)$, and $(\cos t,\sin t)$ has an area of $t/2$. The region of the hyperbola connecting the points $(0,0)$, $(1,0)$, and $(\cosh t,\sinh t)$ has an area of $t/2$. This can even be used to define the hyperbolic functions geometrically, and many authors do the same with the trigonometric functions.

Sine and hyperbolic sine are odd, whereas cosine and hyperbolic cosine are even. But sine and cosine are periodic functions, unlike the hyperbolic counterparts.

The analogue of the identity $\cos^2x+\sin^2x \equiv 1$ is $\cosh^2x-\sinh^2x \equiv 1$. The compound angle formulae are almost identical to their hyperbolic counterparts, save for a pesky minus sign:
\begin{align}
\sin(x+y) &= \sin(x)\cos(y)+\cos(x)\sin(y) \\
\sinh(x+y) &= \sinh(x)\cosh(y) + \cosh(x)\sinh(y) \\[4pt]
\cos(x+y) &= \cos(x)\cos(y) \color{red}{-} \sin(x)\sin(y) \\
\cosh(x+y) &= \cosh(x)\cosh(y) \color{blue}{+} \sinh(x)\sinh(y) \, .
\end{align}

In general, given a trigonometric function, it is possible to write down the corresponding hyperbolic identity using Osborn's rule: replace every occurrence of $\cos$ with $\cosh$; replace every occurrence of $\sin$ by $\sinh$; but negate the product of two $\sinh$ terms.

Analytic connections

$\sin$ is the unique solution to the initial value problem
\begin{align}
f''(x) &= \color{red}{-}f(x) \\
f'(0) &= 1 \\
f(0) &= 0 \, ,
\end{align}
and the corresponding initial value problem for $\sinh$ is the same, except $f''(x) = \color{blue}{+}f(x)$.

Likewise, the initial value problem for $\cos$ is
\begin{align}
f''(x) &= \color{red}{-}f(x) \\
f'(0) &= 0 \\
f(0) &= 1 \, ,
\end{align}
and again we see a mysterious sign change for $\cosh$: $f''(x) = \color{blue}{+}f(x)$.

It follows that the higher-order derivatives of $\sin$ and $\sinh$ form periodic sequences.

If we solve the initial value problems shown above, we obtain the exponential forms of all $4$ functions:
\begin{align}
\sin x &= \frac{e^{\color{green}{i}x}-e^{-\color{green}{i}x}}{2\color{green}{i}} \quad{} \cos x = \frac{e^{\color{\green}{i}x}+ e^{-\color{green}{i}x}}{2} \\[3pt]
\sinh x &= \frac{e^{x}-e^{-x}}{2} \quad{} \cosh x = \frac{e^x + e^{-x}}{2} \, .
\end{align}

All $4$ functions are analytic, and their Taylor series bear a striking resemblance to each other:
\begin{align}
\sin x &= x \color{red}{-} \frac{x^3}{3!} + \frac{x^5}{5!} \color{red}{-} \frac{x^7}{7!} + \ldots \\[4pt]
\sinh x &= x \color{blue}{+} \frac{x^3}{3!} + \frac{x^5}{5!} \color{blue}{+} \frac{x^7}{7!} + \ldots \\[4pt]
\cos x &= 1 \color{red}{-} \frac{x^2}{2!} + \frac{x^4}{4!} \color{red}{-} \frac{x^6}{6!} + \ldots \\[4pt]
\cosh x &= 1 \color{blue}{+} \frac{x^2}{2!} + \frac{x^4}{4!} \color{blue}{+} \frac{x^6}{6!} + \ldots
\end{align}

And Euler's formula
$$
e^{ix} = \cos x + i \sin x
$$
is replaced by the underwhelming
$$
e^x = \cosh x + \sinh x \, .
$$

答:
I my opinion I think that when you know

$$\sin(x) = \frac{e^{ix}-e^{-ix}}{2i} ~~~~~~~~~~~~~ \cos(x) = \frac{e^{ix} + e^{-ix}}{2}$$

you can derive all the circular trigonometric identities.

If you add the "Wick" transformation $ x \to ix$ then you will step into the hyperbolic world, with all the consequent identities.

$$\sin(ix) = \frac{e^{-x} - e^{x}}{2i} = i\frac{e^x - e^{-x}}{2} = i\sinh(x) ~~~~~ \longrightarrow ~~~~~ \sinh(x) = \frac{e^x + e^{-x}}{2}$$

And similarly for $\cosh(x)$.


问:
Number of labeled trees which have one vertex of degree 4
How many labeled trees which have one vertex of degree 4 can we have?

A tree with $10$ vertices has $9$ edges, so the sum of the degrees of the vertices is $2\cdot 9=18$. If there is one vertex of degree $4$, the sum of the degrees of the other $9$ vertices must be $14$. I don't know what to do next. Should I count all possible vertex degrees?

答:
This is due to Clarke (1958). Let $C(n,k)$ denote the number of (labelled) trees $T_n$ in which a given vertex, say the $n$th, has degree $d_n=k$.

Theorem (Clarke): If $1 \leq k \leq n-1$, then $C(n,k)= {n-2 \choose {k-1}} (n-1)^{n-k-1}$

You can find the proof here due to Moon. The idea is nicely described in the provided proof and I don't think that I could add much more value to it other than reformulating it.

For $k=4$ and $n=10$, we get
$$C(10,4)= {8 \choose 3} (9)^{5}=3306744$$


问:
Elements of Algebra (Euler) in comparison to similar texts
I have no problem with Elements of Algebra simply because of its age; in fact, I strongly agree with Abel's famous assertion regarding the proper place to look for knowledge.

What do you guys think about Elements of Algebra, in particular, and texts covering similar material in general?

答:
short answer: read euler.   E.g. would you recommend reading Shakespeare's Hamlet, or the Classics comics version?

I think Euler is probably the best introductory algebra book in existence.  I would recommend it for all gifted high school students.


问:
What are morphisms of functors
I am not been able to understand, what is a morphism between two functors. I have gone through the formal definition involving a commutative diagram. Can someone explain that to me in a bit more details or in pictures?</p>

答:
A natural transformation is essentially an "isomorphism" between the images of two functors, our examples being $F,G:{\cal C}\to{\cal D}$. This "isomorphism" is realized by connecting those objects in $\cal D$ in the image of $F$ to those in the image of $G$ via morphisms that already exist in $\cal D$, with the attendant condition that combining any of the morphisms or objects in $\cal D$ at play (meaning, in addition to the objects and morphisms in the image of $F$ and $G$, the "connecting" morphisms as well) one obtains a commutative diagram.

Perhaps a better way of thinking about it, as Marco says, is as "translating" the image of one functor to the image of another functor via morphisms within the codomain.

You can see the image of $\cal C$ under $F$ and $G$ in red and blue respectively over in $\cal D$ (our picture has "zoomed in" onto a particular "region" in the category $\cal D$). The "connecting" morphisms (denoted by $\eta_{\rm obj}$s in the Wikipedia article on natural transformations) are light purple above. The fact that $F$ and $G$ are functors is seen by the fact that all of the objects and arrows in their image are put together in (red and blue) commutative diagrams. The fact that the system of light purple arrows are a natural transformation $F\cong G$ is seen by the fact that they come together with the red and blue data and also still form commutative diagrams.

Title: What are morphisms of functors


答:
Yes; in fact they can all be traced to the same point, and it is a point that you brought up in your question. One important thing to note is that in Math there can be many equivalent definitions of concepts that can lead to each other, for instance one can define $\pi$ as the ratio of circumference to diameter; or area to radius squared. Both are equivalent.

Similarly we can define these functions geometrically,

And from that obtain differential equations that describe the functions, and from that obtain their exponential definitions; this can be used to unify all of the points you brought up, the Pythagorean Identity, sum of angles, Taylor Series, and exponential definition.

Alternatively we could have defined these functions in terms of their exponential definitions (As @Turing suggests), and from that obtain the Pythagorean Identity, sum of angles, Taylor Series, and geometric definition.

The latter is much easier, but the former is how we came to understanding this family of functions historically.

Title: Unifying the connections between the trigonometric and hyperbolic functions

Body:
There are many, many connections between the trigonometric and hyperbolic functions, some of which are listed here. It is probably too optimistic to expect that a single insight could explain all of these connections, but is there a holistic way of seeing the parallels between $\sin$ and $\sinh$, $\cos$ and $\cosh$? Can all of these seemingly disparate connections be shown to be essentially the same, or at least very similar?

Geometric connections

Sine and cosine parameterise the unit circle $x^2+y^2=1$, just as hyperbolic sine and cosine parameterise the 'unit hyperbola' $x^2-y^2=1$. Both circles and hyperbolas are conic sections.

The sector of the circle connecting the points $(0,0)$, $(1,0)$, and $(\cos t,\sin t)$ has an area of $t/2$. The region of the hyperbola connecting the points $(0,0)$, $(1,0)$, and $(\cosh t,\sinh t)$ has an area of $t/2$. This can even be used to define the hyperbolic functions geometrically, and many authors do the same with the trigonometric functions.

Sine and hyperbolic sine are odd, whereas cosine and hyperbolic cosine are even. But sine and cosine are periodic functions, unlike the hyperbolic counterparts.

The analogue of the identity $\cos^2x+\sin^2x \equiv 1$ is $\cosh^2x-\sinh^2x \equiv 1$. The compound angle formulae are almost identical to their hyperbolic counterparts, save for a pesky minus sign:
\begin{align}
\sin(x+y) &= \sin(x)\cos(y)+\cos(x)\sin(y) \\
\sinh(x+y) &= \sinh(x)\cosh(y) + \cosh(x)\sinh(y) \\[4pt]
\cos(x+y) &= \cos(x)\cos(y) \color{red}{-} \sin(x)\sin(y) \\
\cosh(x+y) &= \cosh(x)\cosh(y) \color{blue}{+} \sinh(x)\sinh(y) \, .
\end{align}

In general, given a trigonometric function, it is possible to write down the corresponding hyperbolic identity using Osborn's rule: replace every occurrence of $\cos$ with $\cosh$; replace every occurrence of $\sin$ by $\sinh$; but negate the product of two $\sinh$ terms.

Analytic connections

$\sin$ is the unique solution to the initial value problem
\begin{align}
f''(x) &= \color{red}{-}f(x) \\
f'(0) &= 1 \\
f(0) &= 0 \, ,
\end{align}
and the corresponding initial value problem for $\sinh$ is the same, except $f''(x) = \color{blue}{+}f(x)$.

Likewise, the initial value problem for $\cos$ is
\begin{align}
f''(x) &= \color{red}{-}f(x) \\
f'(0) &= 0 \\
f(0) &= 1 \, ,
\end{align}
and again we see a mysterious sign change for $\cosh$: $f''(x) = \color{blue}{+}f(x)$.

It follows that the higher-order derivatives of $\sin$ and $\sinh$ form periodic sequences.

If we solve the initial value problems shown above, we obtain the exponential forms of all $4$ functions:
\begin{align}
\sin x &= \frac{e^{\color{green}{i}x}-e^{-\color{green}{i}x}}{2\color{green}{i}} \quad{} \cos x = \frac{e^{\color{\green}{i}x}+ e^{-\color{green}{i}x}}{2} \\[3pt]
\sinh x &= \frac{e^{x}-e^{-x}}{2} \quad{} \cosh x = \frac{e^x + e^{-x}}{2} \, .
\end{align}

All $4$ functions are analytic, and their Taylor series bear a striking resemblance to each other:
\begin{align}
\sin x &= x \color{red}{-} \frac{x^3}{3!} + \frac{x^5}{5!} \color{red}{-} \frac{x^7}{7!} + \ldots \\[4pt]
\sinh x &= x \color{blue}{+} \frac{x^3}{3!} + \frac{x^5}{5!} \color{blue}{+} \frac{x^7}{7!} + \ldots \\[4pt]
\cos x &= 1 \color{red}{-} \frac{x^2}{2!} + \frac{x^4}{4!} \color{red}{-} \frac{x^6}{6!} + \ldots \\[4pt]
\cosh x &= 1 \color{blue}{+} \frac{x^2}{2!} + \frac{x^4}{4!} \color{blue}{+} \frac{x^6}{6!} + \ldots
\end{align}

And Euler's formula
$$
e^{ix} = \cos x + i \sin x
$$
is replaced by the underwhelming
$$
e^x = \cosh x + \sinh x \, .
$$


答:
We have two linear differential operators $\mathcal{D}_1$ and $\mathcal{D}_2$ such that

$$
\cases{
\mathcal{D}_1[y] = z\\
\mathcal{D}_2[z] = u
}
$$

so the answer is

$$
\mathcal{D}_2[\mathcal{D}_1[y]] = u
$$

Title: Modelling problem (ODE)

Body:
I have the below problem:

The water level, y, in a tank is modelled by the differential equation
$$\dot y(t) + y(t) = z(t)$$
where z denotes the inflow. The inflow is a function of a valve position, which in turn is controlled by the electric control signal u. The relation between control signal and flow is given by the differential equation
$$\ddot z(t) + \dot z(t) + z(t) = u(t)$$
What differential equation relates the water level y to the control signal u?

I am self-taught and did not do this kind of modelling problem before. I tried to google an example but I don't know which kind of problem is that exactly (modelling? state space form?). Can someone kindly give me some pointers on what to read to solve it?


问:
$ \lim_{n\to\infty} {2n\choose n}^{(1/n)}$
$$ \lim_{n\to\infty}{2n\choose n}^{(1/n)}$$
   I tried it with Riemann sum and I got that an interval that is divergent.
Can somebody provide me a solution using Riemann sum as I think that I mistaken something.

答:
We have the bounds $\displaystyle \frac{4^{n}}{2n+1}\leq \binom{2n}{n}\leq 4^{n}$ from the Binomial Theorem on $(1 + 1)^{2n}$. Then by the Squeeze Theorem we have:

$$\lim_{n\to\infty}\bigg(\frac{4^{n}}{2n+1}\bigg)^{\frac{1}{n}}\leq \lim_{n\to\infty}\binom{2n}{n}^{\frac{1}{n}}\leq\lim_{n\to\infty}\big(4^{n}\big)^{\frac{1}{n}}$$

$$\frac{4}{\displaystyle\lim_{n\to\infty}\sqrt[n]{2n+1}}\leq\lim_{n\to\infty}\binom{2n}{n}^{\frac{1}{n}}\leq 4$$

It is well known that the limit in the denominator is equal to $1$, so:

$$4\leq\lim_{n\to\infty}\binom{2n}{n}^{\frac{1}{n}}\leq 4$$

$$\boxed{\lim_{n\to\infty}\binom{2n}{n}^{\frac{1}{n}} = 4}$$


问:
Square bracket $[X]$ with finite fields and polynomial rings
I understand that parentheses are used for functional notation. I do not have any confusion about this one.

However, in some literature, I find square brackets ($[X]$)  after some finite field notation.

Source: Encyclopedia of Cryptography and Security, p356

Since, $Q$ has been mentioned as a polynomial, I understand that $F_q[X]$ must be some polynomial. Indeed, the previously cited article in Wikipedia mentions about polynomial ring. This is further elaborated here. This is where things go beyond me.

I was wondering, whether it would be possible to explain a polynomial ring in simple terms. Yes, something like Polynomial Ring for Dummies.

I hope that I am not asking for too much.

答: If $R$ is a ring, $R[X]$ denotes the ring of polynomials in the indeterminate $X$ whose coefficients are drawn from $R$.</p>


问:
Let $V$ be the vector space polynomials of degree less than $n$ over $\mathbb{R}$
Let $V$ be the vector space polynomials of degree less than $n$ over $\mathbb{R}$. In
other words

$$V = \{p\in R[x] \mid \deg p < n\}.$$

Let $T : V \rightarrow V$ be the map $T =\frac{d}{dx}$. From elementary calculus, $T$ is linear

and $T^n=0$ for some positive integer $n$. For $t\in \mathbb{R}$,
let $H_t : V \rightarrow V$ be the map $$H_t(p(x)) = p(x + t),$$ $p(x) \in V$.

We need to Show that
$$e^{tT}=H_t.$$
So, should I start as follows?
As, for any $p(x)\in \mathscr{C}^{\infty}$ and I should expand using Taylor expansion, $p(x+t)$ in the small neighborhood of t and which follows the desired result.

Any help will be highly appreciated.

Body:
I will do it when $n=4$; the general case is similar.

So, suppose that $p(x)=a_0+a_1x+a_2x^2+a_3x^3$. Then\begin{align}e^{tT}p(x)&=p(x)+tp'(x)+\frac{t^2}{2!}p''(x)+\frac{t^3}{3!}p'''(x)\\&=a_0+a_1x+a_2x^2+a_3x^2+t(a_1+2a_2x+3a_3x^2)+t^2(a_2+3a_3x)+t^3a_3\end{align}and, on the other hand,\begin{align}p(x+t)&=a_0+a_1(x+t)+a_2(x+t)^2+a_3(x+t)^3\\&=a_0+a_1x+a_2x^2+a_3x^2+t(a_1+2a_2x+3a_3x^2)+t^2(a_2+3a_3x)+t^3a_3.\end{align}


答:
No, $F_q[X]$ is not a polynomial, but rather the collection of all polynomial over the field $F_q$.

The $[X]$ notation denotes the formal variable, $3X^2+1$, for example, is a member of $F_5[X]$.

The set of all polynomials have naturally defined addition and multiplication operations, these make it into a structure called a ring, which is why this is also called the polynomial ring $F_q[X]$.

Title: Square bracket $[X]$ with finite fields and polynomial rings

Body:
I understand that parentheses are used for functional notation. I do not have any confusion about this one.

However, in some literature, I find square brackets ($[X]$)  after some finite field notation.

Source: Encyclopedia of Cryptography and Security, p356

Since, $Q$ has been mentioned as a polynomial, I understand that $F_q[X]$ must be some polynomial. Indeed, the previously cited article in Wikipedia mentions about polynomial ring. This is further elaborated here. This is where things go beyond me.

I was wondering, whether it would be possible to explain a polynomial ring in simple terms. Yes, something like Polynomial Ring for Dummies.

I hope that I am not asking for too much.


问:
How to get a rectangular pulse function over the entire domain of $\mathbb{R}$
I am reading this article:

https://lpsa.swarthmore.edu/Fourier/Series/ExFS.html#EvenPulse

The following plot is depicted there.

This is called a rectangular pulse function.

The article states then

$$x_T(t) = \left\{ {\matrix{
 {A,\quad |t| \le {{{T_p}} \over 2}}  \cr
 {0,\quad |t| \gt {{{T_p}} \over 2}}  \cr
 } } \right.,\quad \quad  - {T \over 2} < t \le {T \over 2}$$

However, I want this function to extend over the entire domain of $\mathbb{R}$, just as $\sin(t)$ and $\cos(t)$ do.

How can I extend this function so it covers the entire domain of $\mathbb{R}$?

答:
I have a proposed answer, although I'm sure someone can probably come up with a better one.

Let $h(t) = \left(\left(t-\frac{T}{2}\right)\text{ mod } T\right)-\frac{T}{2}$

$$x_T(t) = \left\{ {\matrix{
 {A,\quad |h(t)| \le {{{T_p}} \over 2}}  \cr
 {0,\quad |h(t)| \gt {{{T_p}} \over 2}}  \cr
 } } \right.$$

I think this should create a pulse function whose domain is $\mathbb{R}$.

Assume $A=5$, $T=10$, $T_p = 5$. Here's a graph:


问:
Why can’t I convert to polar for $\int_0^1 e^{-x^2}dx$
I tried to apply the common technique for the Gaussian integral but I changed a bound:
\begin{align}
I &= \int_0^1 e^{-x^2}dx \\[2mm]
I^2 &= \left( \int_0^1 e^{-x^2}dx\right) \cdot \left( \int_0^1 e^{-y^2}dy \right) \\[2mm]
&= \int_0^1\int_0^1 e^{-(x^2+y^2}dxdy \\[2mm]
\textrm{Changing to polar:}& \, x^2 + y^2 = r^2 \, ,\, \,\textrm{and} \,\, \, , dydx=dA = rdrd\theta \\[2mm]
 e^{-r^2} \,\textrm{ has circular symmetry}& \textrm{meaning that I can integrate it from} \, 0 \to 2\pi \\[2mm]
&= \int_0^{2\pi}d\theta\int_0^1 re^{-r^2}dr \\[2mm]
&= \pi\left(1-\frac{1}{e} \right) \\[2mm]
I &= \sqrt{\pi-\frac{\pi}{e}}
\end{align}
Which turns out to be wrong. Where did I go wrong. thanks

答:
You cannot use polar coordinates, for your integration range is a rectangle, as @Joshua Wang told you in a comment.

In this situation, a good way to attack the problem is to consider that, since you're bounded between $0$ and $1$, you can use Taylor series:

$$\int_0^1 \sum_{k = 0}^{+\infty} \frac{(-x^2)^k}{k!}\ \text{d}x = \sum_{k = 0}^{+\infty} \frac{(-1)^k}{k!} \int_0^1 x^{2k}\ \text{d}x = \sum_{k = 0}^{+\infty} \frac{(-1)^k}{k!} \frac{1}{2 k+1}$$

Now this series is well known in literature, and you get:

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

Where $\text{erf}(\cdot)$ denotes the Error Function (special function): https://en.wikipedia.org/wiki/Error_function

($\text{erf}(1) \approx 0.842701(...)$)


问:
The greatest number of points of intersection of n circles and m straight lines is-
The question is about combinatorics. I have no idea on how to start solving the problem. Please guide me.

$(a) 2mn+ {m \choose 2}$

$(b) \frac{1}{2}m(m-1)+n(2m+n-1)$

$(c) {m \choose 2}+2({n \choose 2})$

(d) none of these

I tried solving this question, but I couldn't even find a starting point??

答:
Solution 1:
Make a diagram. Two geometric figures intersect if they have one or more points in common. Draw two circles which intersect in $2$ points. Draw a line which intersects the two circles in $4$ points. Draw another line which intersects the two circles in $4$ points and also intersects the first line. There are $\boxed{11}$ points of intersection.

enter image description here

Solution 2:
Make a table of the maximum number of points of intersection.$$
\begin{array}{|c|c|}
\hline
\text{Geometric Figures} & \text{Number of Common Points} \\
\hline
\text{2 Circles} & 2 \\
\hline
\text{Line (1) and 2 circles} & 4 \\
\hline
\text{Line (2) and 2 circles} & 4 \\
\hline
\text{2 lines} & 1 \\
\hline
& \text{Total = \boxed{11}} \\
\hline
\end{array}
$$


问:
Variation of Parameters to solve
Told to solve $y''-2y'+y=e^xcos(x)$ using variation of parameters method. Solution then uses Wronskian, $W$, as $e^x \begin{vmatrix}1 & x\\
1 & 1+x
\end{vmatrix}$ which I understand. However, the next step is that $W\begin{bmatrix}c'_{1}\\c'_{2}\end{bmatrix}=\begin{bmatrix}0\\cos(x)\end{bmatrix}$ which I don't understand at all. Obviously comes out as the right answer for $c_{1}$ and $c_{2}$ as the coefficients of the two solutions of the homogeneous equation but I am not sure how they knew it was $0$ and $cos(x)$. If anyone could explain this step that would be great thanks!

答:
I'm answering your side question about how to find solutions for linear ODE, you can read first my answer here https://math.stackexchange.com/a/3431378/399263.

Here the roots of the characteristic equation $r^2-2r+1=0$ are $r=1$ with multiplicity $m=2$.

The general solution of the homogeneous ODE is then $y_c=(ax+b)e^x$

The RHS is $e^x\cos(x)$ which is associated to roots $(1\pm i)$, since these roots do not collide with the one of the homogeneous ODE, we can search for a particular solution of the form $y_p(x)=c\,e^x\cos(x)+d\,e^x\sin(x)$, i.e. same degree than RHS (=cst).

After reporting in the ODE, you find $c=-1,d=0$ therefore $y_p(x)=-e^x\cos(x)$.

Title: Variation of Parameters to solve


答:
The error is in the line:

$e^{-r^2}$ has circular symmetry, meaning that I can integrate it from $0$ to $2\pi$

Using the same (incorrect) argument, you would get
$$
1=\int_0^1\int_0^1 1 dxdy=
\int_0^{2\pi}d\theta \int_0^1rdr=\pi
$$

In general, you have
$$
\iint_{\color{red}{x^2+y^2\le 1}}f(x^2+y^2)dxdy=\int_0^{2\pi}d\theta\left(\int_0^1f(r)rdr\right)
$$
but
$$
\iint_{\color{red}{[0,1]\times[0,1]}}f(x^2+y^2)dxdy\ne\int_0^{2\pi}d\theta\left(\int_0^1f(r)rdr\right)
$$

Title: Why can’t I convert to polar for $\int_0^1 e^{-x^2}dx$

Body:
I tried to apply the common technique for the Gaussian integral but I changed a bound:
\begin{align}
I &= \int_0^1 e^{-x^2}dx \\[2mm]
I^2 &= \left( \int_0^1 e^{-x^2}dx\right) \cdot \left( \int_0^1 e^{-y^2}dy \right) \\[2mm]
&= \int_0^1\int_0^1 e^{-(x^2+y^2}dxdy \\[2mm]
\textrm{Changing to polar:}& \, x^2 + y^2 = r^2 \, ,\, \,\textrm{and} \,\, \, , dydx=dA = rdrd\theta \\[2mm]
 e^{-r^2} \,\textrm{ has circular symmetry}& \textrm{meaning that I can integrate it from} \, 0 \to 2\pi \\[2mm]
&= \int_0^{2\pi}d\theta\int_0^1 re^{-r^2}dr \\[2mm]
&= \pi\left(1-\frac{1}{e} \right) \\[2mm]
I &= \sqrt{\pi-\frac{\pi}{e}}
\end{align}
Which turns out to be wrong. Where did I go wrong. thanks


答:
Accounting for the shape of the domain, and using symmetry

$$\int_0^1\int_0^1e^{-(x^2+y^2)}dx\,dy=2\int_0^{\pi/4}\int_0^{1/\cos\theta}re^{-r^2}dr\,d\theta=\int_0^{\pi/4}\left(1-e^{-1/\cos^2\theta}\right)d\theta$$ and you are stuck.


答: Just to provide another straightforward solution to the integral in question, taking advantage of the fact that $$\frac{d}{dx}\operatorname{erfi}(x) = \frac{2}{\sqrt{\pi}}e^{-x^{2}}.$$  We have then $$\int_0^1 e^{-x^2}\,dx = \frac{\sqrt{\pi}}{2}\left(\operatorname{erfi}(1) - \operatorname{erfi}(0)\right) = \frac{\sqrt{\pi}}{2}\operatorname{erfi}(1).$$


问:
fast string fixing
Hi I'm trying to prove something and i would appreciate some help.

Given an uniformly 2n long randomly  selected string  of Parenthesis( "(",")" ) prove that
it is possible to generate  a valid sequence of parenthesis by changing  csqrt(nlogn) chars
at most where c is constant in probability of 1-o(1) .

答:
Given a string of parentheses, and $i\in \{1,\dots,2n\}$, let $o_i$ be the number of open parentheses in the first $i$ symbols of the string, and let $c_i$ be the number of closed parentheses in this prefix. There are two quantities of interest:

Let $X=o_{2n}-c_{2n}$ be the excess of open parentheses over closed parentheses.

Let $Y=\max_{0\le i\le 2n}(c_i-o_i)$ be the worst case excess of closed parentheses over open parentheses in any prefix.

I claim that you can make a permutation valid in $O(X+Y)$ moves. This can be done as follows:

As long as there is an unmatched closed parenthesis, flip the leftmost unmatched parenthesis. This decreases $Y$ by at least one, and increases $X$ by two.

Then, until the number of parentheses of each type are equal, flip the rightmost open parenthesis. This decreases $X$ by two without changing $Y$.

Therefore, all that remains is to show that $X+Y$ is at most $c\sqrt{n\log n}$ with probability at least $1-o(1)$. It is easy to see this is true of $X$, since $X$ is approximately normal with a mean of $0$ and a variance of $O(\sqrt{n})$. The variable $Y$ is much trickier. However, using the reflection principle, you can show that $Y$ shares the same probabilistic bound.

To give some details, identify parenthesis string with lattice walks, where an open parenthesis is the step $(1,1)$ and a closed parenthesis is the step $(1,-1)$. Then $Y$ is the absolute value of the lowest height of the path, while $X$ is the final height of the path. You can then show that, for any $k\ge 0$,
$$
P(Y> k)\le 2P(X> k)\tag{*}
$$
This is because paths whose lowest height is at least $k$ come in two flavors:

Those paths whose final height is less than $-k$; the probability of this is exactly $P(X>k)$.

Those whose final height is more than $-k$, but which attain $-k$ at some intermediate point. By taking the first point this path hits $-k$ and reflecting everything (that is, switching $(1,-1)$ with $(1,1)$), you get a path whose final height is less than $-k$. Since this process is reversible, the probability of these paths is at most $P(X>k)$.

As long as you believe $(*)$, you know now that $Y$ enjoys the same probability bound that $X$ does, which means the same is true of $X+Y$.

Title: fast string fixing


问:
Are order statistics implicit in the calculation of the expected value of a bottleneck random variable?
Two points are realized on a $l\times w$ rectangle. A point has higher chance of being realized in the top-left quadrant $Q_{t,l}$. Of interest is the maximum x-coordinate among the two points. Given that one point was realized in $Q_{t,l}$ and the other not, does it suffice to find the expected value of the point that is not in the quadrant, i.e.

$$\frac{1}{3} \times \frac{l}{4} + \frac{2}{3} \times \frac{3l}{4} $$

(*) or do I explicitly need to take care of the case where the point in $Q_{t,l}$ happens to be realized at a higher x-coordinate than the point that is not in $Q_{t,l}$?

How do you solve for order statistics when there is no closed-form survival function or c.d.f. (in the variation where it is known that both points are NOT realized in $Q_{t,l}$). Opposed to the following case where there is no discontinuity: Expected value of maximum of two random variables from uniform distribution.

(*) Three quadrants remain that are not in $Q_{t,l}$. The probability that it is in the buttom-left is 1/3, 2/3 that it is one the right. If it is on the right, its expected value is halfway between l/2 and l. If it is on the bottom left, it is halfway in between 0 and l/2. Ignore the point in $Q_{t,l}$ because the expected value of the point not in $Q_{t,l}$ is father away than the expected value of the point in $Q_{t,l}$.

答:
NOT AN ANSWER
Difficult to write in the comments, perhaps this can get you started...

Here the y-dimension is irrelevant.  Reducing it to one dimension and in the unit range

$$
E\{X\}=\int_0^1 P(X⩾x)dx = \int_0^1 (1-P(X \le x))dx  \\
E\{\text{max}(X_1,X_2)\}=\int_0^1 (1-P(X_1 \le x)P(X_2 \le x))dx = \int_0^1 (1-P(X \le x)^2)dx
$$

Assuming uniform distribution will give you $\int_0^1 (1-x^2)dx = \frac23$

scale with the actual dimension $l$.


答:
Given categories $\mathcal{C}$ and $\mathcal{D}$, a functor $F:\mathcal{C}\to\mathcal{D}$ consists of two pieces of information:

A rule assigning, to each object $x\in \mathrm{ob}(\mathcal{C})$, a corresponding $Fx\in\mathrm{ob}(\mathcal{D})$.

A rule assigning, to each morphism $f\in\mathrm{Hom}_{\mathcal{C}}(x,y)$, a corresponding $Ff\in\mathrm{Hom}_{\mathcal{D}}(Fx,Fy)$.

"Pictorally":
$${\mathrm{ob}(\mathcal{C})\xrightarrow{F\text{ (on objects)}}\mathrm{ob}(\mathcal{D})\atop x\;\;\;\;\;\;\;\;\to\;\;\;\;\;\;\;\; Fx}\qquad\qquad {\mathrm{Hom}_{\mathcal{C}}(x,y)\xrightarrow{F\text{ (on morphisms)}}\mathrm{Hom}_{\mathcal{D}}(Fx,Fy)\atop \begin{pmatrix} x\\ \downarrow\!\scriptsize f\\ y\end{pmatrix} \;\;\;\;\;\;\;\;\to\;\;\;\;\;\;\;\;\begin{pmatrix} Fx\\ \downarrow\!\scriptsize Ff\\ Fy\end{pmatrix} }$$
Of course, there are also properties that these rules must have, to ensure that we have some consistency: we must have $F(\mathrm{id}_x)=\mathrm{id}_{Fx}$ and $F(g\circ f)=Fg\circ Ff$. We can't just haphazardly move objects and morphisms around without respecting composition.

Now, what would a "morphism of functors" be? We need a sort of "meta-rule"; a rule for turning the rules defining $F$ into the rules defining $G$.

$$\begin{array}{ccc}
&  & Fx\\
&  \nearrow & \color{red}{\downarrow}\\
x & & \color{red}{\downarrow}\\
& \searrow & \color{red}{\downarrow}\\
& & Gx
\end{array}\qquad\qquad \begin{array}{ccc}
&  & (Fx\xrightarrow{Ff}Fy)\\
&  \nearrow & \color{red}{\downarrow}\\[-0.4in]
(x\xrightarrow{f} y) & & \color{red}{\downarrow}\\
& \searrow & \color{red}{\downarrow}\\
& & (Gx\xrightarrow{Gf}Gy)
\end{array}$$

So, to specify a morphism of functors (let's call it $\alpha$), we can figure that we'll need to do two things:

for each $x\in\mathrm{ob}(\mathcal{C})$, we want to choose a morphism $\alpha_x:Fx\to Gx$

for each $f\in\mathrm{Hom}_\mathcal{C}(x,y)$, we want to choose a "map of arrows" from $Ff$ to $Gf$. What is a map of arrows? Why, it's a choice of arrows on each end that make it a commutative square. And hey, we've already chosen maps $\alpha_x:Fx\to Gx$ and $\alpha_y:Fy\to Gy$, so we actually don't need to be choosing anything more; we just add the requirement that the $\alpha_x$'s we chose make a commutative diagram
$$\begin{array}{ccc}
Fx & \xrightarrow{Ff} & Fy\\
\scriptsize\alpha_x\!\normalsize\downarrow & & \downarrow \!\scriptsize\alpha_y\\
Gx& \xrightarrow[Gf]{} & Gy
\end{array}$$
for any morphism $f$ in the category $\mathcal{C}$.

This choice of arrows $\alpha_x$ for each $x\in\mathrm{ob}(\mathcal{C})$ defines a natural transformation $\alpha$ from $F$ to $G$.

There is actually a very deep connection between the notion of natural transformation, and the notion of homotopy (from the field of topology). In the gif animation below (taken from Wikipedia), what we start with are two different plane curves, $f:X\to \mathbb{R}^2$ and $g:X\to\mathbb{R}^2$, where $X$ stands for some closed interval. A homotopy is a "morphing" of one curve to the other.

But what does this really mean? For each $s\in X$, we have chosen a path $h_s:I\to \mathbb{R}^2$ that starts at $f(s)$ and ends at $g(s)$ ($I$ stands for the interval $[0,1]$). Using $t$ to denote the variable for these $h_s$'s, you can think of $t$ as time: at time $t=0$, we have $h_s(0)=f(s)$, and at time $t=1$, we have $h_s(1)=g(s)$, and the curve $h_s(t)$ is the path traced out as $f(s)$ moves to $g(s)$. Of course, there must be some consistency; nearby points on one curve can't have too wildly different paths, the maps $h_s$ must "continuously vary" with $s$.

Sounds familiar! In fact, there is a category analogous $I$, let's call it $\mathcal{I}$: it has two objects, named $0$ and $1$, and it has one non-identity morphism $T:0\to 1$. To be poetic, we might call it the "arrow of time". Just as a homotopy from a map $f:X\to Y$ to a map $G:X\to Y$ is a map $h:X\times I\to Y$ such that $h(x,0)$ is $f(x)$, $h(x,1)$ is $g(x)$, and for each $x\in X$, $h(x,t)$ is traces out a  curve from $f(x)$ to $g(x)$, a natural transformation $\alpha$ from a functor $F:\mathcal{C}\to\mathcal{D}$ to a a functor $G:\mathcal{C}\to\mathcal{D}$ can be identified with (or even alternatively defined as) a functor $H:\mathcal{F}\times \mathcal{I}\to\mathcal{G}$, where $H(x,0)=Fx$, and $H(x,1)=Gx$, and $H(x,T)=\alpha_x$.

See this MathOverflow thread Natural transformations as categorical homotopies and this nLab article geometric realization of categories.

Title: What are morphisms of functors

Body: I am not been able to understand, what is a morphism between two functors. I have gone through the formal definition involving a commutative diagram. Can someone explain that to me in a bit more details or in pictures?</p>


答:
I'll focus on three issues: why your reasoning was wrong, what a corrected version of your idea obtains, and why you'll still need a numerical method at some point.

The region $0\le x\le1,\,0\le y\le1$ is not equivalent to $0\le\theta\le2\pi,\,0\le r\le1$. It's actually equivalent to $0\le r\le\sqrt{2},\,\theta_\min(r)\le\theta\le\tfrac12\pi-\theta_\min(r)$ with $\theta_\min(r):=\arccos\min\{1,\,\tfrac1r\}$. So$$\begin{align}I^2&=\int_0^1re^{-r^2}dr\int_0^{\pi/2}d\theta+\int_1^2re^{-r^2}dr\int_{\arccos\tfrac1r}^{\pi/2-\arccos\tfrac1r}d\\&=\tfrac12\pi\int_1^2re^{-r^2}+\int_1^2re^{-r^2}(\tfrac12\pi-2\arccos\tfrac1r)dr\\&=\tfrac12\pi\int_0^2re^{-r^2}+2\int_1^2re^{-r^2}\arccos\tfrac1rdr\\&=\tfrac14\pi(1-e^{-2})+2\int_1^2re^{-r^2}\arccos\tfrac1rdr.\end{align}$$The last integral is the challenge! One way or another, we need numerical methods, as per e.g. @DMcMor or @Turing's answer.


问:
Average value of $g'$ on $[a, b]$
If $g$ is a continuous function on $[a, b]$ and if $g'$ is also continuous on $[a, b]$. How do I find the average value of $g'$ on $[a, b]$?

I know that if a function is continuous on $[a, b]$, then the average value is
$$g_{avg}(x) = \frac{1}{b-a}\int_{a}^{b}g(x)\ dx$$

I'm a little stuck at this point. Some tips and/or help is appreciated!

答:
hint

$ g'$ is continuous at $ [a,b] $, and is then integrable at $ [a,b]$.

By FTC, we have

$$\int_a^bg'(x)dx=\Bigl[g(x)\Bigr]_a^b=g(b)-g(a)$$

Title: Average value of $g'$ on $[a, b]$


问:
Euclid Division Lemma Proof for J < 0
I am reading George E. Andrews' Number Theory and am at the portion where he proves Euclid's Division Lemma.

The lemma states that

For any integers $k (k>0)$ and $j$, there exist unique integers $q$ and $r$ such that $0≤r<k$ and $j=qk+r.$

I am having difficulty understanding the final part of the proof, where the uniqueness of $q$ and $r$ when $j < 0$ is proven.

By this point it has been established that the lemma is true for $j>1$ because by the basis representation theorem, $j$ has a unique representation to the base $k$

$$j=a_sk^s+a_{s-1}k^{s-1}+...+a_1k+a_0$$
$$=k(a_sk^{s-1}+a_{s-1}k^{s-2}+...+a_1)+a_0$$
$$=kq+r$$

Where $0≤r=a_0<k$.

Supposing a second pair $q'$ and $r'$ exists, there is a representation for $q'$ to the base $k$:

$$q'=b_tk^t+b_{t-1}k^{t-1}+...+b_1k+b_0.$$

Thus,

$$j=kq'+r'$$
$$=b_tk^{t+1}+b_{t-1}k^t+...+b_1k^2+b_0k+r'.$$

But,

$$j=a_sk^s+a_{s-1}k^{s-1}+...+a_1k+a_0.$$

By the uniqueness of the representation of $j$ to base $k$, the following may be gleaned: $r'=r=a_0, \ a_{i+1}=b_i, \ \text{and} \ t=s-1$. Therefore,

$$q'=b_tk^t+...+b_1k+b_0$$
$$=a_sk^{a-s}+...+a_2k+a_1$$
$$=q.$$

Therefore the lemma is true when $j>0$. And if $j=0, \ q=r=0$ is the only possible solution where $0≤r<k$. It is left to prove that the lemma is true when $j<0$.

Here we run into my issue. He writes,

If $j<0$, then $-j>0$, and there exist unique integers $q'' \ \text{and} \ r''$ such that

$$-j=kq''+r''.$$

If $r''=0$, then $j=k(-q'')$; thus we may take $q=-q'' \ \text{and} \ r=0$.

If $r''\neq 0$, then

$$j=-kq''-r''$$
$$=k(-q''-1)+(k-r''),$$

and we may take $q=-q''-1, \ \text{and} \ r=k-r''$.

In either case, $q$ and $r$ satisfy equation $j=kq+r$. Uniqueness for negative $j$ follows from uniqueness for $-j$, which is then positive.

I understand the reason for using the equation $-j=kq''+r''$, since it has already been established that when the equation is greater than $0$, there exists unique integers $q''$ and $r''$. This is useful because it is equivalent to $j=-kq''-r''$, so if there exists unique integers $q''$ and $r''$ for $-j$, they must also exist for $j$, and so we may conclude that the lemma is true when $j<0$, correct? But this raises the question, why doesn't the proof end there? I am missing something.

There is a reason why we treat cases $r'' \neq 0$ and $r''=0$ separately. Why is that?

Furthermore, how is the fact that in the former case $q=-q''$ and $r=r''=0$ relevant to the proof? And in the latter case, how do $q=-q''-1$ and $r=k-r''$ satisfy the equation $j=kq+r$?

Finally, the phrasing of the last sentence is confusing:

Uniqueness for negative $j$ follows from uniqueness for $-j$, which is then positive.

It seems as if "negative $j$" and "$-j$" are different things, otherwise the statement would be begging the question. But what then is "negative $j$?" The "$j$" from the equation $j=kq+r$ multiplied by $-1$? I assume "$-j$" is from the equation $-j=kq''+r''.$

I realize I have asked a great many questions and probably have not been clear enough in stating my confusions. I am very new to number theory and am learning it independently, so I am very prone to misunderstanding the content. This is the only place I can reliably go for clarification. My apologies for the length of this question, I have tried to be as specific as possible to avoid miscommunication.

答:
E.g. in radix $\,k = 10\,$ we convert a $\rm\color{#c00}{negative}$ integer to sought form $\,q\cdot 10 +r\,$ with $\,\color{#0a0}{0\le r < 10}\,$ by simply $ $ borrowing $\,1\,$ from $\,q,\,$ which adds $10$ to the units digit $\,r\,$ to get it $\ge 0,\,$ e.g.

$$\begin{align} -2(10)\overbrace{\color{#c00}{-1}}^{\textstyle\color{#c00}{< 0}} &\,= -3(10)\!\!\!\!\!\!\!\!\!\overbrace{+\color{#0a0}9}^{\textstyle\color{#0a0}{0\le 9 < 10}}\\[.2em]
      {\rm i.e.} \ -\!(21) &\,= -\!3,9\end{align}\qquad\qquad$$

The same idea works in any radix to force the units digits (remainder) to be $\ge 0$, which is exactly what is done in the proof to reduce the case of negative dividends to positive dividends.

Remark $ $ Allowing negative digits (as $\,-\!3,9 := -3(10)+9\,$ above) often proves handy, e.g. see here for their use in simplifying divisibility test calculations.

Title: Euclid Division Lemma Proof for J < 0


问:
Prove with integration the inequality $e(\frac{n}{e})^n < n! < n \times e(\frac{n}{e})^n$
Prove with integration  the inequality, I need some advice about how to start prove it.

I know that if function is Monotonically increasing function so :

$$ f(1)+\int^n_1f(x)dx\leq f(1)+f(2)+....+f(n)\leq f(n)+\int^n_1f(x)dx$$

$$e(\frac{n}{e})^n < n! < n \times e(\frac{n}{e})^n$$

Thanks!

答:
Hint 1: Logarithm is your friend; it turns products into sums.

Hint 2: $\ln(n!)=\sum\limits_{i=1}^n \ln i$. Thus, $f(x)=\ln(x)$ might be a good candidate for trying to apply the inequalities you have listed.


问:
Find the Maximum Trigonometric polynomial coefficient $A_{k}$

Let $n,k$ be given positive integers and $n\ge k$. Let $A_i, i=1, 2, \cdots, n$ be given real numbers. If for all real numbers $x$ we have $$A_{1}\cos{x}+A_{2}\cos{(2x)}+\cdots+A_{n}\cos{(nx)}\le 1$$
Find the maximum value of $A_{k}$.

I don't know if this question has been studied

If  $n=2$ it is easy to solve it.

答:
COMMENT

May be this idea can help:

We use following identity:

$$\sin^2(x)+\sin^2(2x)+\sin^2 (3x)+ . . .+\sin^2(nx)=\frac{n\sin(x)-\sin(nx)\cos(n+1)x}{2\sin x}$$

Taking derivative of both sides we get:

$$2\sin(x)\cos(x)+4\sin(2x)\cos(2x)+6\sin(3x)\cos(3x)+ . . . 2n\sin(nx)\cos(nx)=\big(\frac{n\sin(x)-\sin(nx)\cos(n+1)x}{2\sin x}\big)'\leq 1$$

We solve derivative $\leq 1$, we get a relation between n and x which may be used to find maximum value of $A_k$.
In fact the proble can be reduced to:
If:

$\big(\frac{n\sin(x)-\sin(nx)\cos(n+1)x}{2\sin x}\big)'\leq 1$

Find maximum of  $A_n=f(x)= 2n\sin(nx)$


问:
Integral of a floor function $\iint_D [x + y] \, dx\, dy$
Given a domain
$$D=[0,2]×[0,2]$$
How can I calculate the integral $$\bbox[5px,border:2px solid #C0A000]{\iint_D [x + y]  \, dx\, dy}\quad$$

答:
Hint.

Observe that $0\le x+y\le 4$ on $D$. Consider the following cases
$$
m\le x+y<m+1,\quad m=0,1,2,3
$$
Then split your integral into four pieces.


Title: Integral of a floor function $\iint_D [x + y] \, dx\, dy$

Body:
Given a domain
$$D=[0,2]×[0,2]$$
How can I calculate the integral $$\bbox[5px,border:2px solid #C0A000]{\iint_D [x + y]  \, dx\, dy}\quad$$

Body:
How about breaking the domain into
$$
D=D_0\cup D_1\cup D_2
$$
where $$
{
D_0=\{(x,y): 0<x+y< 0.5\}
\\
D_1=\{(x,y): 0.5<x+y< 1.5\}
\\
D_2=\{(x,y): 1.5<x+y< 2\}
}
$$?


问:
Matrix inverse of block matrix $\begin{pmatrix} T & l \\l^*& S \end{pmatrix}$
Consider the special block matrix

$$A = \begin{pmatrix} T & l \\l^*& S \end{pmatrix},$$ where we assume that each entry is invertible and $l$ and $l^*$ denote a scalar (and its complex conjugate) times the identity, whereas $T$ and $S$ are full matrices of equal size.

Is there a simple formula for the matrix inverse $A^{-1}?$ I am aware there are general formulas for block matrix inverses, but I thought in this case, things should simplify?

Please let me know if you have any questions.

Body:
Applying the block-inversion formula linked in the comments, we have
$$
\pmatrix{\mathbf{A}^{-1}+\mathbf{A}^{-1}\mathbf{B}(\mathbf{D}-\mathbf{CA}^{-1}\mathbf{B})^{-1}\mathbf{CA}^{-1} & -\mathbf{A}^{-1}\mathbf{B}(\mathbf{D}-\mathbf{CA}^{-1}\mathbf{B})^{-1} \\ -(\mathbf{D}-\mathbf{CA}^{-1}\mathbf{B})^{-1}\mathbf{CA}^{-1} & (\mathbf{D}-\mathbf{CA}^{-1}\mathbf{B})^{-1}} = \\
\pmatrix{T^{-1}+|l|^2T^{-1}(S-|l|^2T^{-1})^{-1}T^{-1} &
-l\,T^{-1}(S-|l|^2 T^{-1})^{-1} \\
-l^*(S-|l|^2 T^{-1})^{-1}T^{-1} & (S-|l|^2T^{-1})^{-1}} = \\
\pmatrix{T^{-1}[1+|l|^2(TS-|l|^2)^{-1}] &
-l\,(ST-|l|^2 )^{-1} \\
-l^*(TS-|l|^2 )^{-1} & T(ST-|l|^2)^{-1}}
$$


问:
Continuity of the real and imaginary parts of a continuus complex-valued function
If a complex-valued function is continuous, are the component real and imaginary parts $u(x,y)$ and $u(x,y)$ necessarily continuous? If so, why?</p>

答:
Projections from any topological product space are continuous by the definition of a product space. Since $\mathbb C$ is topologically the same as $\mathbb R^2$, $\operatorname{Re}$ and $\operatorname{Im}$ are continuous. But the situation is entirely general—it applies also to such crazy things as $\mathbb R^\mathbb R$.

Applying composition as in copper.hat's answer answers your question.


答:
We can divide in four areas $A,B,C,D$ inside I indicated the value of $\lfloor x+y\rfloor$

Thus $I=0A+1B+2C+3D$ but notice that $A=D$ so $$I=(A+B+C+D)+(C+D)=Q+\frac 12Q$$

where $Q=4$ is the area of the square.


问:
Proving that $\tan{x}-x$ has an inverse and than determining it
Consider this problem;

Show that the function $\tan{x}-x$ can be inverted and find the corresponding inverse for which applies

$$ f^{-1} (y) = (f^{-1}(y) + y)^{-2}$$

So to prove that a function can be inverted it has to be bijective, which means it has to be surjective and injective

Now to prove for injective I derived the function $\tan{x}$ and got that it is constantly growing so it is injective. To prove that it is surjective we are already given a solution. Now what I did is derived the function $\tan{x}-x$ and got that its simply $\tan^2{x}$ since the $1$ and $-1$ cancel out. Then out of the starting equation I found a way to explicitly write $\tan{x}$;

$$\tan{x}=f(x)+x$$ and with that we can also determine this;

$$\tan^2{x}=(f(x)+x)^2$$

Now I've used the formula $$ (f^{-1})'(y) = \frac {1}{f'(x)} $$ and this is where I am stuck, the only progress I could make is if instead of $f'$ I insert $\tan^2{x}$ but that is it.

Looking forward to your answers !

EDIT: The domain of the function was also given (my bad)

$$\left(\mbox{-}\frac{\pi}{2}, \frac{\pi}{2}\right)$$

答:
COMMENT.-There is no difficulty in verifying that the function $f (x) = \tan (x) -x$ is a bijection of $(-\dfrac{\pi}{2},\dfrac{\pi}{2})$ on $\mathbb R$. Furthermore, as many points of the inverse function can be constructed in the Cartesian plane as desired. Since if $(x, f (x))$ is a point on the initial curve then the symmetric with respect to the first diagonal, $(f (x), x)$ is a point of $f^{-1}$. In the attached figure, the red points of the inverse function have been drawn as symmetrical to the green points on the curve $f (x) = \tan (x) -x$ and the curve corresponding to $f^{-1}$ has been drawn with imperfect drawing in blue color.

I don't think there is a simple expression for the inverse function. But you can find, as I repeat, as many points of the graph of $f^{-1}$ as you want.


问:
How do we find finite difference schemes to numerically solve this B.V.P?

\begin{equation}
\frac{\partial P}{\partial t}=\frac{\partial}{\partial x}\left[D(x)\frac{\partial P}{\partial x}\right]-v\frac{\partial P}{\partial x}-\tau P+S
\end{equation}
and
\begin{equation}
\frac{\partial Q}{\partial t}=\frac{\partial}{\partial x}\left[D(x)\frac{\partial Q}{\partial x}\right]-v\frac{\partial Q}{\partial x}-\tau P,
\end{equation}
where $t$ is time, $x$ is the spatial variable along the column, $D(x)$ is the diffusion rate, $v$ is the advection rate, $\tau$ is the decay rate, and $S(x)$ is the source rate. We will be looking for the late-time, steady state solution to these equations with $\partial P/\partial t=\partial Q/\partial t = 0$. The diffusion rate and source rate may be assumed to be smooth and non-negative. We will also assume $v$ is non-negative.

The $x$ domain is $x\in[0,L]$. Boundary conditions for $P$ and $Q$ are mixed at $x=0$, representing zero influx
\begin{equation}
\left.\left(D\frac{\partial P}{\partial x}-vP\right)\right|_{x=0}=\left.\left(D\frac{\partial Q}{\partial x}-vQ\right)\right|_{x=0}=0.
\end{equation}
At $x=L$ the boundary condition will be $P=Q=c$, where $c$ is a constant.

The question is we wish to find a finite difference scheme (break up the domain into $N$ gridpoints ranging from $0$ to $N-1$ and form index equations for each point) in order to create matrices such as $A$ where $AP = S$ then numerically solve this equation. I'm wondering which schemes would be most useful. I have thought of using centred/forward/backward finite difference scheme but I'm very confused about how to input the boundary conditions and which to use?

We require a tridiagonal matrix A. e.g as produced in this example. values of D(x) and S(x) are provided at all points on the index grid.

https://web.mit.edu/10.001/Web/Course_Notes/Differential_Equations_Notes/node9.html

答:
You can translate that almost literally. For the second derivative apply twice central difference quotients with step size $\Delta x$, that is, half-steps from the center. This will lead to
$$
\left.\frac{\partial}{\partial x}\left(D(x)\frac{\partial P(x,t)}{\partial x}\right)\right|_{(x,t)=(x_i,t_j)}
\approx \frac{D_{i+\frac12}(P^j_{i+1}-P^j_{i})-D_{i-\frac12}(P^j_{i}-P^j_{i-1})}{Δx^2}
$$
For the first derivative take the central difference quotient with step size $2Δx$,
$$
\left.\frac{\partial P(x,t)}{\partial x}\right|_{(x,t)=(x_i,t_j)}
\approx \frac{P^j_{i+1}-P^j_{i-1}}{2Δx}
$$
For the right boundary conditions you set that component as constant and remove it from the active computation. For the left condition it is probably easiest to introduce ghost cells at index $i=-1$ so that the PDE can be applied there and also the boundary condition can be constructed with central difference quotients.


问:
Points A & B have coordinates A (-2, 5) and B (3, 9). Find a unit vector parallel to the vector AB
This is what I got but it was wrong

I have attempted this question and attached an image of my attempt, and I don't see where I went wrong or what I could've done.

答:
HINT

Your vector $AB=(5,4)$ is correct.

Now you must find the length of this vector and then divide by that length.


问:
Scalar obtained from division of magnitudes of two vectors.
Consider we are having a scalar vector multiplication. Is it correct in concept (numerically it makes sense) to compute the scalar as division of the magnitude of the obtained vector from multiplication and the magnitude of the vector that was originally a parameter? For example, trying to compute the mass as a division of magnitudes between F(force) and a(acceleration) vectors. Numerically I know this fits, as well as scalar component division of the two vectors, but it would seem redundant to compute both divisions, since the same scalar is applied to both components in the first place. I'm trying to understand what would be the correct approach from a mathematical approach. Thank you in advance.</p>

答:
Suppose you know that, for example, $F=ma$.

Then you can find $m$ by dividing one specific component or the magnitude of the vectors as you suggest.

Both methods are mathematically correct.


问:
$(n+1)!>2^{(n+1)}$ for $n$, where $n \geq 4$. (Mathematical Induction)
By using Mathematical Induction prove that $(n+1)!>2^{(n+1)}$ for $n$, where $n \geq 4$.

Body: If true for some $k\geq{4}$, that is $(k+1)!>2^{k+1}$, then $(k+2)!=(k+2).(k+1)!>(k+2).2^{k+1}>2.2^{k+1}=2^{k+2}$


答: $n!=\overbrace{1\times 2\times 3\times 4}^{=24\, >\, 2^4}\times \overbrace{5}^{>2}\times \overbrace{6}^{>2}\cdots\times \overbrace{n}^{>2}>2^4\times 2^{n-4}$

Title: $(n+1)!>2^{(n+1)}$ for $n$, where $n \geq 4$. (Mathematical Induction)

Body: By using Mathematical Induction prove that $(n+1)!>2^{(n+1)}$ for $n$, where $n \geq 4$.


问:
Determining $\sin(15)$, $\sin(32)$, $\cos(49)$, etc.
How do you in general find the trigonometric function values? I know how to find them for 30 45, and 60 using the 60-60-60 and 45-45-90 triangle but don't know for, say $\sin(15)$ or $\tan(75)$ or $\csc(50)$, etc.. I tried looking for how to do it but neither my textbook or any other place has a tutorial for it. I want to know how to find the exact values for all the trigonometric functions like $\sin x$, $\csc x$, ... opposed to looking it up or using calculator. According to my textbook, $\sin(15)=0.26$, $\tan(75)=3.73$, and $\csc(50)=1.31$ but doesn't show where those numbers came from, as if it was dropped from the Math heaven!</p>

答:
Value of $\sin{x}$ with prescribed accuracy can be calculated from Taylor's representation
$$\sin{x}=\sum\limits_{n=0}^{\infty}{\dfrac{(-1)^n x^{2n+1}}{(2n+1)!}}$$ or infinite product
$$\sin{x}=x\prod\limits_{n=1}^{\infty}{\left(1-\dfrac{x^2}{\pi^2 n^2} \right)}.$$
For some partial cases numerous trigonometric identities can be used.</p>


问:
Does the integral converge $\int_{-\infty}^{+\infty}\frac{\exp(ibx)}{1+e^x}\,dx$ for $b>0$?
I try to calculate integral $$\int_{-\infty}^{+\infty}\frac{\exp(ibx)}{1+e^x}\,dx.$$ Does it converge for $b>0$? Wolfram Mathematica shows contradictory results: for $b=2$ it converges, for $b=1$ and $b=3$ it does not. I need to calculate it.

答:
This note explains the comment (to the OP) by @Maxim regarding $\mathscr{F}\left[\frac{1}{e^x+1}\right](\xi)$ of a tempered distribution, provided that we define it as $\mathscr{F}\big[f(x)\big](\xi)=\int_{-\infty}^\infty f(x)e^{i\xi x}\,dx$ for regular $f$.

Using the sign function, $$\frac{1}{e^x+1}=\frac{\operatorname{sgn}x}{e^{|x|}+1}+\frac{1-\operatorname{sgn}x}{2},$$ and the first term on the RHS has a regular Fourier transform $\color{LightGray}{\texttt{(todo: link)}}$: $$\mathscr{F}\left[\frac{\operatorname{sgn}x}{e^{|x|}+1}\right](\xi)=2i\int_0^\infty\frac{\sin\xi x}{e^x+1}\,dx=\frac{i}{\xi}-\frac{\pi i}{\sinh\pi\xi}.$$ The FT of the remainder is found using $\mathscr{F}[1](\xi)=2\pi\delta(\xi)$ and $\mathscr{F}[\operatorname{sgn}x](\xi)=2i$$\mathscr{P}\frac{1}{\xi}$.


问:
Does algebraic structure on a set has any influence on its topological structure?
In the beginning chapters of real analysis we learn the algebraic properties of the real no.s (i.e., operations defined on $\mathbb{R}$, order properties and completeness properties of $\mathbb{R}$) as an algebraic structure imposed over $\mathbb{R}$. And then suddenly we are introduced to the topology of real no.s where we learn about neighbourhood, open set, limit points etc. But doesn't the algebraic properties of real no.s like the total order property, play any role in the topology of real no.s? I feel like there's an abrupt gap between the bridging of algebra and topology of $\mathbb{R}$ and I want to know if there's any source which can explain to me how topology of real no.s (or in that sense general topology) flows out naturally from it's algebraic structure.

答:
In general, a mathematical object $\mathfrak{O}$ can have both algebraic and topological structure which are unrelated. However, in practice there are usually some connections. One of the standard ones is that the algebraic aspects of $\mathfrak{O}$ should be "topologically simple," e.g., all the relevant functions on $\mathfrak{O}$ should be continuous with respect to the relevant topology. See e.g. the notion of topological groups.

Actually, a structure can carry multiple different "natural" topologies at once, but I'm ignoring that for now. This happens frequently in functional analysis and algebraic geometry.

This happens with for example the fields of real and complex numbers: in each case there is a natural topology, and the field operations are continuous with respect to that topology (and the appropriate product topologies it generates).

In the case of $\mathbb{R}$ in fact there's an even tighter connection: addition and multiplication of real numbers alone determine the usual topology on $\mathbb{R}$. Specifically, the topology on $\mathbb{R}$ is the order topology coming from the usual ordering on $\mathbb{R}$, and that ordering is definable from addition and multiplication alone as $$x\le y\quad\iff\quad \mbox{ for some $z$ we have }x+(z\cdot z)=y.$$ But this is rather special to $\mathbb{R}$ - the usual topology on $\mathbb{C}$, for example, cannot be defined from addition and multiplication of complex numbers alone. (HINT: $\mathbb{C}$ has lots of discontinuous field automorphisms.)

Interestingly, the topology on $\mathbb{Q}$ is definable from the algebraic structure alone. Like $\mathbb{R}$ this goes by defining the order on $\mathbb{Q}$ from the algebraic operations, but things are much harder now since we can't take square roots. One way to do it (very briefly) is to first define $\mathbb{Z}$ in $\mathbb{Q}$, and building off that use the four squares theorem to define $\mathbb{Z}_{>0}$ in $\mathbb{Q}$. A rational is positive iff it is a ratio of two positive rationals, so this gets us $\mathbb{Q}_{>0}$, and this in turn gets us the ordering on $\mathbb{Q}$ since $x\le y$ iff $y-x$ is either $0$ or in $\mathbb{Q}_{>0}$. But there's a lot of work there (and there is probably a much simpler approach I'm missing here).

For a deeper connection between algebraic and topological structure, you may be interested in how topology can be applied to prove the fundamental theorem of algebra, which is of course a topology-free result. More generally, there is a whole subject of algebraic topology.


问:
Vectors and directed line segments
A vector is a quantity that has magnitude and direction.

A directed line segment is a line segment that has both a starting and an end point ,so it has a direction

From these definitions, a directed line segment and a vector are different, while some textbooks may treat them interchangeably, meaning that they are synonymous, but they are not the same. I can explain this from my perspective as follows:

A directed line segment is a geometrical object, a set of points that not only has length but also direction.

While vectors may be considered as a physical concept, or abstract mathematical object.

Accordingly, some notations really confuse me:
Considering $\overrightarrow{AB}$ and $\overrightarrow{BC}$ two directed line segments
Then

$\overrightarrow{AB} + \overrightarrow{BC} = \overrightarrow{AC}$

What really confuses me is how are we adding directed line segments? We can add vectors, but not segments.

We don't add normal line segments, we add their lengths.

Can someone explain this to me if I am getting something wrong?

答:
Actually it is possible to define a partial addition operation on the set of directed line segments by restricting the arguments to compatible segments such that the terminal point of the first matches the initial of the second. For these restricted pairs the sum can be defined by the triangle law $\overrightarrow{PR}=\overrightarrow{PQ}+\overrightarrow{QR}$.

It is then not too difficult to show that the partial operation extends to a well-defined binary operation (vector addition) on the set of equivalences classes.

Title: Vectors and directed line segments


问:
Why is a vector bundle called E?
Vector bundles are often denoted as $p:E \to B$, where $p$ is a projection map, $B$ is the base space and $E$ is the total space. Here the choice of the letters $p$ and $B$ is clear, but is there also a reason to denote the total space with $E$?
Is it possibly derived from the French 'espace totale'?

答:
Perhaps this should not be an answer, but I give it a try.

I am not sure whether the origin is French, but I believe it is sensible to think it could be for the following reason. In an old work by Daniel Bernard "Sur la géométrie différentielle des $G$-structures" (perhaps one of the oldest works I found on the subject) I see the notation $E(F,G,B)$ for what is called "espace fibré" (so not really "espace totale"). All other letters match the initials of the corresponding objects, "fibre", "group", "base". I hope someone else can comment on this.

EDIT: to be fair, I should also mention that the canonical frame bundle $L_m$ is called "espace de repères". So in this case there is no correspondence at all...

Title: Why is a vector bundle called E?