问:
Compactness implies limit point compact
In the proof of theorem 28.1 of Munkres that says compactness implies the limit point compact.

Proof: Let X be a compact. Given a set A of X, we wish to prove that if A is infinite, then A has a limit point. We prove the contraposition-if A has no limit point, then A must be finite.
So suppose A has no limit point. Then A contains all its limit points, so that A is closed. For each a in A we can choose a neighborhood U_{a} of a that intersect A in a alone. ...

My question is why we can choose such a neighborhood that does not intersect A-{a}?In other word, does such a neighborhood exists for each a in A?

答: If $x\in X\setminus A$, then $X\setminus A$ is an open nbhd of $x$ disjoint from $A$. If $a\in A$, then $a$ is not a limit point of $A$, so by the definition of limit point $a$ has an open nbhd $U_a$ that contains no other point of $A$. In other words, $U_a\cap A=\{a\}$.

Title: Compactness implies limit point compact


问:
Looking for a Banach space in which projection on a closed convex set is empty?
It is well-known that projection on closed convex sets is nonempty in Hilbert spaces. I am interested in seeing a Banach space where the projection is empty on a closed convex set.

Definition of Projection of $x $ on set $C$:

$$   P_{C} (x) := \{ y \in C ~| ~ d(x , C ) = \|x -y\|  \}  $$

答:
This MO answer solves your question. First, a small technicality: if $C$ is empty, it is a closed convex set such that all projections on $C$ are empty. This is technically a solution, but we want to exclude it.

Let $X$ be a non-reflexive Banach space (e.g. $\ell^1$). Let $f : X \to \mathbb{R}$ be a continuous linear functional of norm $1$ on $X$ which does not attain a maximum on the closed unit ball of $X$ (e.g. $(x_n) \mapsto \sum_{n \geq 1} \frac{n-1}{n} x_n : \ell^1 \to \mathbb{R}$). Let $C = \{x \in X : f(x) = 1, \lVert x \rVert \leq 2\}$.

$C$ is closed and convex, since it is the intersection of the closed convex sets $f^{-1}(1)$ and $\{x \in X : \lVert x \rVert \leq 2\}$.

Since $\lVert f \rVert = \sup_{\lVert x \rVert = 1} \lvert f(x) \rvert = 1$, there is some $a \in X$ such that $f(a) = 1/2$ and $\lVert a \rVert \leq 1$. Then $2a \in C$, so $C$ is nonempty.

Now suppose for contradiction that $P_C(0)$ is nonempty; let $y \in P_C(0)$ -- in other words, $y \in C$ has minimal norm.

Since $f(y) = 1$, we have $y \neq 0$. Thus, let $z = \frac{1}{\lVert y \rVert} y$. Since $f$ does not attain a maximum on the unit ball, there exists some $z'$ such that $\lVert z' \rVert = 1$ and $f(z') > f(z)$. Now let $y' = \frac{1}{f(z')}z'$, so that $f(y') = 1$. Moreover,
$$\lVert y \rVert \leq 2 \implies f(z) = \frac{1}{\lVert y \rVert} f(y) = \frac{1}{\lVert y \rVert} \geq \frac{1}{2},$$
so
$$\lVert y' \rVert = \frac{1}{\lvert f(z') \rvert} \lVert z' \rVert = \frac{1}{f(z')} < \frac{1}{f(z)} \leq 2.$$
Thus, $y' \in C$. Finally,
$$\lVert y' \rVert = \frac{1}{\lvert f(z') \rvert} \lVert z' \rVert = \frac{1}{f(z')} < \frac{1}{f(z)} = \frac{1}{\frac{1}{\lVert y \rVert} f(y)} = \lVert y \rVert.$$
This contradicts the prior conclusion that $y \in C$ has minimal norm.

ContentLicense: CC BY-SA 4.0


问:
Gradient of an optimization function - Frobenius norm and Hadamard product
I am trying to solve a problem for my Optimization class, in which it is asked to calculate the gradient of the following function:

$$g(P)=\frac{1}{2}||1_K\circ(R-Q^0P)||_F^2+\frac{\rho}{2}||Q^0||_F^2+\frac{\rho}{2}||P||_F^2$$

where $\rho, R, Q^0$ and $1_K$ are given constants in this case.

But I am totally stuck particularly on the first term (the one that icludes the Frobenius norm and the Hadamard prouct).

I tried to use the definition of the Frobenius norm $||A||_F=\sqrt{Tr(AA^H)}$ but I do not know how to handle it in this situation.

Thank you very much.

答:
Let's rename the variables using lowercase letters for vectors and uppercase letters for matrices, and omit all of the subscripts and superscripts.
$$\eqalign{
p=P,\quad Q=Q^0,\quad r=R,\quad u={\tt1}_K
}$$
Let's also define the auxiliary vectors
$$\eqalign{
s &= Qp-r \\
w &= u\circ s \\
}$$
We'll also need the trace/Frobenius product
$$\eqalign{
A:B &= {\rm Tr}(A^TB) = B:A \\
A:A &= \big\|A\big\|^2_F \\
}$$
Conveniently, the Frobenius product commutes with the Hadamard product, i.e.
$$A:(B\circ C) = (A\circ B):C$$
Use the above to rewrite the objective function, then calculate the gradient as follows.
$$\eqalign{
g &= \tfrac 12(w:w) + \tfrac \rho2(p:p) + \tfrac \rho2(Q:Q) \\
\\
dg &= (w:dw) + \rho(p:dp) + 0 \\
 &= w:(u\circ ds) + \rho(p:dp) \\
 &= (u\circ w):ds + \rho(p:dp) \\
 &= (u\circ w):Q\,dp + \rho(p:dp) \\
 &= Q^T(u\circ w):dp + \rho(p:dp) \\
 &= \Big(Q^T(u\circ w)+ \rho p\Big):dp \\
 &= \Big(\rho p + Q^T\big(u\circ u\circ(Qp-r)\big)\Big):dp \\
\\
\frac{\partial g}{\partial p}
 &= \rho p + Q^T\big(u\circ u\circ(Qp-r)\big) \\
}$$


问:
Inverse Laplace Transform for $ sK_0(\sqrt{as}) $?
Are there any ways to manipulate the following in order to make it invertible from Laplace space analytically?

$$ sK_0(\sqrt{as})$$

where $a$ is a constant.

答:
The following Laplace transform pair appears in Erdelyi's bible of integral transforms (eq. (30) on p.516) :

$$f(t)=\frac{1}{2t}\exp\left(-\frac{a}{4t}\right)\leftrightarrow F(s) = K_0(\sqrt{as})=\int_0^{\infty}e^{-s t}f(t)\,dt.$$

I write the explicit transform to recall that integration-by-parts implies

$$F(s)=-\left[\frac{1}{s}e^{-st}f(t)\right]_{0}^\infty+\frac{1}{s}\int_0^\infty e^{-st}f'(t)\,dt$$
The behavior as $t\to 0$ isn't as nice as one would like, but the exponential factor suppresses the $1/t$ divergence and so $f(0)=0$. Hence we have

$$s K_0(\sqrt{as})=s F(s)=\int_0^\infty e^{-s t}f'(t)\,dt$$

and therefore the requisite inverse Laplace transform is $$f'(t)=\left(-\frac{1}{2t^2}+\frac{a}{8t^3}\right)\exp\left(-\frac{a}{4t}\right)=\frac{a-4t}{8t^3}\exp\left(-\frac{a}{4t}\right).$$

Here is a partial proof of the identity. (I regard it as a partial proof because I can't justify some of the formulas used.)

I begin with the integral representation

$$F(s)=K_0(\sqrt{as})=\int_0^\infty \frac{\cos x}{\sqrt{x^2+as}}\,dx.$$

According to Mathematica, one has $$\mathcal{L}^{-1}\left[\frac{1}{\sqrt{x^2+as}}\right]=\frac{e^{-tx^2/a}}{\sqrt{a \pi t}}$$ and therefore the inverse transform of $F(s)$ may be represented as $$f(t)=\int_0^\infty \frac{e^{-t x^2/a}}{\sqrt{a \pi t}}\cos x\,dx.$$

Leaving this up to Mathematica, we obtain $f(t) = (2t)^{-1} e^{-a/4t}$ as claimed.

Notes:

(1) is not the standard form of this representation (compare equation 10.32.E6, but I've rescaled the integration variable to suit my purpose. For the derivation, the verification shown in the answer to this earlier question may provide a lead.

The functions in (2) are simple enough that one may perhaps verify the Laplace transform by hand and the inverse Laplace transform via the Bromwich integral.

If I was being rigorous, I'd need to justify writing $$\int_0^\infty f(x)\mathcal{L}[g(x,t)](s)\,dx=\mathcal{L}\left[\int_0^\infty f(x) g(x,t)\,dt\right](s)$$
...but you'd better believe I can't be bothered.

If we consider $\cos x$ as the real part of $e^{i x}$, then we may view (3) as the real part of a computable Gaussian integral. So this is tractable by hand. (We may also regard it as a Fourier cosine transform, and as such is almost certainly in Erdelyi's table as well.)

I couldn't resist one more try, and the result is worth it. We start from the integral representation

$$K_0(\sqrt{as})=\frac12 \int_0^\infty\exp\left(-x-\frac{as}{4x}\right)\frac{dx}{x}$$

(see eqn. 10.32.10 of the DLMF). We then make the substitution $t=a/(4x)$, obtaining

$$K_0(\sqrt{as})=\frac12 \int_0^\infty\exp\left(-\frac{a}{4t}-st \right)\frac{dt}{t}$$

which upon reflection is nothing other than the Laplace transform of $f(t)=\dfrac{1}{2t}\exp\left(-\dfrac{a}{4t}\right)$.


问:
Uniform convergence and uniform boundedness
I try to understand a demonstration from a book, but I have a problem with a line.
We have the series
$$u(x,t) = \sum_{k=0}^\infty \frac{g^{(k)}(t)}{(2k)!}x^{2k} \qquad (*)$$
where
$$g(t) = \left\{ \begin{array}{ll}
        \exp\left(-\frac{1}{t^2}\right) & \mbox{if } t > 0 \\
        0 & \mbox{if } t \leq 0
    \end{array}
\right.
$$
and we try to prove that it's well defined for $x$ and $t$ real, that we can take the derivatives, and so on. We have already proved that when $t > 0$,
$$\left|\frac{g^{(k)}(t)}{(2k)!}x^{2k}\right| \leq \frac{|x|^{2k}}{k!(\theta t)^k}\exp\left[-\frac{1}{2t^2}\right]$$
for some $\theta > 0$. We then have directly
$$\sum_{k=0}^\infty \left|\frac{g^{(k)}(t)}{(2k)!}x^{2k}\right| \leq \exp\left[\frac{|x|^2}{\theta t} - \frac{1}{2t^2}\right]$$
So they say that, by comparison, our series $(*)$ converges and that $\lim_{t\rightarrow 0}u(x,t) = 0$, I agree on that.

Here is the point I don't understand. It's said :

"The series $(*)$ as a power series in $x$ is majorised (they say that $f$ is majorised by $F$ if the derivatives of $f$ in $0$ are less than the corresponding derivatives of $F$ in $0$) by the power series for
$$U(x,t) = \left\{ \begin{array}{ll}
        \exp\left(-\frac{x^2}{\theta t} - \frac{1}{2t^2}\right) & \mbox{if } t > 0 \\
        0 & \mbox{if } t \leq 0
    \end{array}
\right.$$
Since $U(x,t)$ is bounded uniformly for bounded complex $x$ and all real $t$, the series $(*)$ converges uniformly in $(x,t)$ for bounded $x$ and real $t$, and the same holds for the series obtained by the term by term $x$-differentiations."

I actually don't understand what they mean when they say that $U(x,t)$ is bounded uniformly. Do they mean that the power series of $U$ is uniformly bounded? But which series? The one with respect to $x$ ? And even with that, I don't understand how we can conclude that $(*)$ converges uniformly...

Don't hesitate to ask if something is unclear in what I said!

I hope you could help me,

Thank you in advance :)

答:
I believe a uniform bound on $U$ simply means there exists $C > 0$ such that for all $x \in (-1,1),~t \in \mathbb{R}$ we have $|U(x,t)| < C$ (in vector space terminology, its supremum-norm is finite).

Can you see why this implies the convergence of the power series?

EDIT: Okay, so let $B\subset\mathbb{C}$ be bounded, consider the domain $D:=B\times\mathbb{R}$. You want to show that for any $\varepsilon>0$ there is $N\in\mathbb{N}$ such that for all $n>N$:
\begin{equation}\sup\limits_{(x,t)\in D}|u(x,t)-u_n(x,t)|<\varepsilon.\end{equation}
We know that for any $(x,t)\in D$,
\begin{align}
|u(x,t) - u_n(x,t)| &= \left|\sum\limits_{k=n}^\infty \frac{g^{(k)}(t)}{(2k)!}x^{2k}\right| \\
&\leq \sum\limits_{k=n}^\infty \left|\frac{g^{(k)}(t)}{(2k)!}x^{2k}\right|
\end{align}
Can you take it from there?

ContentLicense: CC BY-SA 3.0


问:
Properties of the Weyl vector $\rho = \frac{1}{2} \sum_{\alpha > 0} \alpha$
Let $G$ be a compact connected Lie group with Lie algebra $\mathfrak{g}$. The group has a maximal torus $T$ with Lie algebra $\mathfrak{t}$. Let $\rho = \frac{1}{2}\sum_{\alpha > 0} \alpha$ be the half-sum of positive roots, the so-called "Weyl vector."

When one considers the classical compact groups $U(n)$,$Sp(n)$,$SO(2n)$,$SO(2n+1)$ one gets the following Weyl vectors
$$\left(\frac{n-1}{2}, \frac{n-3}{2}, \dots, -\frac{n-1}{2} \right)$$
$$\left(n, n-1, \dots, 1 \right)$$
$$\left(n-1, n-2, \dots, 0 \right)$$
$$\left(n-\frac{1}{2}, n-\frac{3}{2}, \dots, \frac{1}{2} \right)$$
I have two questions: firstly I note that the difference between any two components is an integer, $\rho_i - \rho_j \in \mathbb{Z}$. Is there a general way of proving this?

Of course I am implicitly choosing a basis of $\mathfrak{t}$ and $\mathfrak{t}^\ast$ when I write the above vectors. And this is my second question. I am taking a basis $e_1, \dots, e_n \in \mathfrak{t}$, $e_1^\ast, \dots,  e_n^\ast \in \mathfrak{t}^\ast$, such that $\exp(2\pi e_j) = \mathrm{Id}$ and such that $\langle e_i , e_j \rangle = \delta_{ij}$, where $\langle \cdot, \cdot \rangle$ is an Ad-invariant inner product. In the groups considered I don't see any systematic reason why such a basis (with these properties) should exist, though in each case a computation shows that it does. Is there some property that all these groups share that makes this possible?

答:
The basis you have chosen to write $\rho$ in is the basis of fundamental weights, so for $U(n)$ for example you are saying that $$\rho = \frac{n - 1}{2} \varpi_1 + \frac{n - 3}{2} \varpi_2 + \cdots - \frac{n - 1}{2} \varpi_{n - 1}.$$
We can then rewrite your question: since the fundamental weights are dual to the simple coroots, why is it true for any two simple coroots $\alpha^\vee, \beta^\vee$ that $\langle \rho, \alpha^\vee - \beta^\vee \rangle$ is an integer? In fact, more is true: each of $\langle \rho, \alpha^\vee \rangle$ and $\langle \rho, \beta^\vee \rangle$ will be an integer.

Let $S \subseteq \Delta^+ \subseteq \Delta \subseteq \mathfrak{t}^*$ denote the simple roots, positive roots, and roots. For every simple root $\alpha \in S$, the simple reflection $s_\alpha$ permutes the other positive roots $s_\alpha(\Delta^+ - \{\alpha\}) = \Delta^+ - \{\alpha\}$ and of course acts on $\alpha$ as $s_\alpha(\alpha) = -\alpha$, hence we get that $s_\alpha(\rho) = \rho - \alpha$. By definition the reflection $s_\alpha(\rho) = \rho - \langle \rho, \alpha^\vee \rangle \alpha$ (where $\alpha^\vee \in \mathfrak{t}$ is the coroot corresponding to $\alpha$) and therefore we see that for $\alpha \in S$, $\langle \rho, \alpha^\vee \rangle = 1$.

If the Lie algebra $\mathfrak{g}$ is semisimple, then we can replace the fact "$\langle \rho, \alpha^\vee \rangle = 1$ for all simple coroots $\alpha^\vee$" with the equivalent (and easier to remember fact) $\rho = \varpi_1 + \cdots + \varpi_n$.


问:
Help understanding the result of a Cylindrical Algebraic Decomposition of $V\left( {x,y,z} \right) = {x^2} + {y^2} + {z^2} < 1$?
From https://mathworld.wolfram.com/CylindricalAlgebraicDecomposition.html it is said that:

Define a cell in ${\mathbb{R}^1}$ as an open interval or a point. A cell in ${\mathbb{R}^{k + 1}}$ then has one of two forms:

Form 1: $\left\{ {\left( {x,y} \right):x \in C,f\left( x \right) < y < g\left( x \right)} \right\}$

Form 2: $\left\{ {\left( {x,y} \right):x \in C,y = f\left( x \right)} \right\}$

Where $x = \left\{ {{x_1},{x_2},...{x_k}} \right\}$ and $C$ is a cell in ${\mathbb{R}^{k}}$.

$f$ and $g$ are either:

1/ Continuous function on $C$ such that for some polynomials $F$ and $G$,  $F\left( {x,f\left( x \right)} \right) = 0$ and $G\left( {x,g\left( x \right)} \right) = 0$

2/ $\pm \infty$ and $f\left( x \right) < g\left( x \right)$ for all $x \in C$

A cylindrical algebraic decomposition of $S \subset {\mathbb{R}^n}$ is a representation of $S$ as a finite union of disjoint cells. Let $F$ be finite set of polynomials in $n$ variables.

A cylindrical algebraic decomposition of $S \subset {\mathbb{R}^n}$ is said to be $F$-invariant if each of the polynomials from $F$ has a constant sign on each cell of the decomposition.

An example of the decomposition of the decomposition of ${x^2} + {y^2} + {z^2} < 1$ is:

$\left\{ {\begin{array}{*{20}{c}}
  {\begin{array}{*{20}{c}}
  { - 1 < x < 1} \\
  { - \sqrt {1 - {x^2}}  < y < \sqrt {1 - {x^2}} } \\
  { - \sqrt {1 - {x^2} - {y^2}}  < z < \sqrt {1 - {x^2} - {y^2}} }
\end{array}}&{\begin{array}{*{20}{c}}
  {\left( 1 \right)} \\
  {\left( 2 \right)} \\
  {\left( 3 \right)}
\end{array}}
\end{array}} \right.$

1/ From the result of the example does it mean that (1),(2) and (3) are the three cell of this decomposition ?

2/ In this example does it mean that $S$ is ${x^2} + {y^2} + {z^2} < 1$ , what is $F$ and what does it mean by $F$-invariant ?

3/ From what I know the Cylindrical Algebraic Decomposition could be thought as some form of quantifier elimination. Then does that mean the result in (1)(2)(3) is a quantifier free of this statement:

$\exists x,y,z \in {\mathbb{R}^3},{x^2} + {y^2} + {z^2} < 1$

4/ Is there anyway to rigorous proof that integrating over (3)(2)(1) yield the spherical volume of  ${x^2} + {y^2} + {z^2} < 1$ ? By inspection it seem that way.

Please help me !

Thank you for your enthusiasm !

答:
1/ Three cells are specified here.  The first cell, which we will call $C_1$, is in $\Bbb{R}^1$ and is given by (1), in Form 1, and is $(-1,1) \subseteq \Bbb{R}$.

The second cell, which we will call $C_2$, is in $\Bbb{R}^2$ (so take $k = 1$ for this cell), has $C = C_1$, and is given by (2) in Form 1,
$$  \{(x,y) : x \in C_1, -\sqrt{1-x^2} < y < \sqrt{1-x^2}\}  \text{.}  $$

The third cell is in $\Bbb{R}^3$ (so take $k = 2$) for this cell), has $C = C_2$, and is given by (3) in Form 1,
$$  \{((x,y),z) : (x,y) \in C_2, -\sqrt{1 - x^2 - y^2} < z < \sqrt{1-x^2-y^2}  \text{.}  $$

2/ Yes, $S$ is the set in $\Bbb{R}^3$ given by $x^2 + y^2 + z^2 < 1$.

Regarding $F$-invariance...  The use of $F$ in the specification of the cell forms is unrelated to the use of $F$ in $F$-invariance.

In the first cell, we need a polynomial $\hat{F}_1(x)$ that has constant sign on $C_1$.  An example if $\hat{F}_1(x) = x^2 - 1$.  Notice that this polynomial comes directly from $x^2 + y^2 + z^2 < 1$ by ignoring the $y$ and $z$ terms, $x^2 < 1$, then moving everything to the left: $x^2 - 1 < 0$.  This means $\hat{F}_1$ is negative exactly on $C_1$.  However, to match the usage in the example, we should negate this: $F_1(x) = 1 - x^2$.  Notice that this is exactly the form of $x$ that appears in the bounds of the inequality in the second cell.  In particular,
$$  -F_1(x) < y^2 < F_1(x)  $$  (The existence of these polynomial relations for the boundary of $C_2$ in terms of $F_1(x)$ and $y$ is the content of the text "$F(x,f(x)) = 0$ and $G(x,g(x)) = 0$".)

Continuing this pattern, $F_2(x,y) = 1 - x^2 - y^2$ is positive exactly on $C_2$ and $F_3(x,y,z) = 1 - x^2 - y^2 - z^2$ is positive exactly on $S$.  Then the given example CAD is $\{F_1, F_2, F_3\}$-invariant.

3/ Not necessarily.  If the quantifier order and the decomposition order do not match, a CAD may not be very helpful in eliminating quantifiers.

This particular example is easy, though since the variables are quantified in the same order as the variable order of the CAD.  It is trivial to determine that there is an $x$ satisfying (1).  Then, this choice of $x$ gives an interval of choices for $y$ in (2).  Then a choice of $y$ in that interval gives an interval of choices for $z$.  Notice that this is easy when the variable order $x$, $y$, $z$ is the same as the quantifier order.  If we quantify on $z$ first, this CAD is not particularly helpful.  If you construct a new CAD with variable order matching quantifier order, this is easy.

More discussion of quantification and elimination is present in http://www3.risc.jku.at/publications/download/risc_4231/s65kauers.pdf  (Also presented: a nontrivial example and an example of how much changes when you replace the present example of the open unit ball with the closed unit ball.)

4/  Yes.  This is immediate.  The volume $S$ is exactly given by the CAD, so the question you ask is equivalent to "is integration over $S$ given by this one inequality the same as integration over $S$ given by this CAD?"  Yes; they're both $S$.


问:
Handling non-convex inequality constraint in otherwise convex problem
I have an inequality constrained optimization problem, which I believe to be convex except for a single constraint I am having trouble handling. The objective function looks like this:

$\sum_{i=1}^{n} \frac{c_i}{x_i + y_i}$

Where $x,y \geq 0$ are the optimization variables and $c_i \geq 0$ are constants. For brevity, I'm going to omit some of the other constraints in the problem, which are mostly lower and upper bounds on the variables along with a single convex inequality constraint. However, there is one final set of constraints which I'm not sure how to handle, given by:

$\left(\frac{y_i - 1}{x_i + y_i - 1}\right) a_i - b_i \leq 0$

Here $a_i $ and $b_i$ are positive constants and my other constraints enforce that the fraction is also positive. From what I have read, this looks like a linear fractional function, which is quasilinear. I'm an optimization novice and have had difficulty finding references for handling these kinds of functions in constraints. What I have thought about so far:

I found some resources for these functions when they are the objective of the problem, but I don't see how those techniques are applicable here. Maybe there is a change of variable I could consider?

I could consider some kind of affine approximation of this function, which would make the overall problem convex. I would then need to quantify the worst case distance between the approximation and the actual function.

I could pretty easily relax the fraction to be $\frac{y_i}{x_i + y_i}$ but it's not immediately obvious how that might simplify things.

Looking for guidance or further references for handing this situation.

答:
You asked me to formalize my comment as an answer: Your constraints are (for some subset of indices $i$):
$$ \frac{(y_i-1)a_i}{x_i+y_i-1} \leq b_i$$
There is a potential singularity if the denominator of the fraction is zero.  However, if we assume that other constraints ensure that $x_i+y_i-1>0$ always, then we can multiply the inequality by the positive number $x_i+y_i-1$ to equivalently get
$$ (y_i-1) a_i \leq b_i(x_i+y_i-1)$$
which is a linear (hence convex) inequality constraint.


问:
How do I write the rule for this sequence that is neither arithmetic nor geometric as an equation?
I have the sequence $3, 5, 9, 17, 33, 65,\dots$

I know the rule is multiply by $2$ then subtract one. $3$ times $2$ is $6$ minus $1$ is $5$ and so on. Easy enough.

How do I write the rule for this sequence as an equation?  I know it is neither arithmetic or geometric.

答:
$$a_n=2a_{n-1}-1 \implies a_n - 1 = 2(a_{n-1}-1)$$
Hence $a_n-1$ is a geometric sequence, and
$$a_n-1 = 2^{n-1}(a_1-1) = 2^n \implies a_n = 2^n+1.\blacksquare$$


问:
Are there any examples of two-valued logics that are not classical?
What are some examples of nonclassical two-valued logics? How would such a logic work? What relationship do the non-classical two-valued logics have to ordinary two-valued logic?

This answer includes an interesting parenthetical comment.

Also, as a fairly subtle point, you can have a $2$-valued logic that isn't classical.

The question under consideration is specifically about propositional logic, so I'm assuming that this answer is also talking about different propositional logics.

The only things I can really think of that might qualify as non-classical two-valued logic are simple, somewhat bizarre systems like a fragment of classical logic where the only connective is $\land$, or one where the only connective is $\to$. Neither example system has truth value constants $\top$ and $\bot$. I'm not at all convinced that these examples are really two-valued, or are not better analyzed with different, non-two-valued, semantics.

答:
First of all, fragments of classical propositional logic gotten by reducing the language can actually be rather interesting. So those examples aren't unnatural. (They're discussed IIRC in Blok/Pigozzi's text which I'll cite more centrally below.)

That said, here's one you may find more compelling, since it involves $(i)$ an increase in strength and $(ii)$ a new "type" of logical operation. Let $\mathcal{L}$ be the usual language of propositional modal logic (so we have a set $P$ of propositional atoms, the usual classical connectives, and the new unary symbols $\Box$ and $\Diamond$). Equip $\mathcal{L}$ with the "pointed Kripke frames" semantics:

A $\mathcal{L}$-structure is a tuple $(W, R, w, \nu)$ where:

$W$ is a set of things called worlds and $R\subseteq W^2$ is any relation;

$w\in W$ is the distinguished world; and

$\nu$ is a valuation map from $P$ to $\mathcal{P}(W)$.

The "models" relation $\models_\mathcal{L}$ is based off the usual Kripke semantics: we say $(W,R,w,\nu)\models\varphi$ iff $\varphi$ is made true at world $w$ in the Kripke frame $(W,R)$ by the valuation $\nu$.

Now this is arguably two-valued: for every $\mathcal{L}$-structure $\mathfrak{A}$ and every $\mathcal{L}$-sentence $\varphi$, exactly one of $\mathfrak{A}\models_\mathcal{L}\varphi$ or $\mathfrak{A}\models_\mathcal{L}\neg\varphi$ holds. However, it's definitely not classical. Specifically, the operations $\Box$ and $\Diamond$  - thought of as functions on the set of sentences - are non-truth-functional, and that's not something which can occur in classical logic. Moreover, this isn't a silly example at all; Kripke frames with a distinguished world are definitely reasonable things to consider.

OK, but now we come to a follow-up:

I'm not at all convinced that these examples are really two-valued, or are better analyzed with different, non-two-valued, semantics.

In particular, one might object that the non-classicality of the "pointed frame logic" above, stemming from the non-truth-functionality of its syntax, is actually an indictment of its claimed two-valued nature. Presumably this objection would go something like this:

A "truth value" is supposed to be a total description of the relationship a given sentence has to truth. When we have a putative notion of truth value with respect to which our linguistic constructions are not truth functional, this is a sign that that putative notion is not capturing all the truth-related data of the sentences in our language.

(Weirdly I don't have a citation for this at the moment, although it's an obvious line of argument; I'll add one when I can find it. FWIW I believe this is not generally held amongst non-classical logicians, although I personally have a lot of sympathy for it.)

This whole line of thought takes us into rather odd territory: is there even a meaningful way to compare different semantics (semanticses? semantices? .... semantixoi!) for a given logic, or to talk about which semantics is "right?"

It turns out that there is actually a lot to be said about this! This is one of the topics treated in algebraic logic, and specifically in the context of (proto-)algebraizability of a logic. Blok/Pigozzi is a great introduction to this topic - it's well-written, full of examples, and extremely short. Nonclassical logic for the bus ride, indeed!

(And if by contrast what you want is a new dining room table, I can't not recommend Humberstone's book. Fun fact: you can use BP as a bookmark in H.)

If I recall correctly, if we interpret things strictly enough you can indeed prove that classical propositional logic is the strongest two-valued propositional logic in a certain sense. But care is always required with such results (e.g. we have to rule out things like infinitary conjunction), so I'm hesitant to outright claim it without a citation ... which I can't find at the moment. Regardless, the "algebraization" context provides a framework for understanding this type of question which I think will prove valuable to you.


问:
Given positive numbers $x_1,...,x_n$ such that $x_1\cdots x_n=1$ prove that $\frac{1}{n-1+x_1}+\cdots+\frac{1}{n-1+x_n}\le 1$

Given positive numbers $x_1,\dots,x_n$ such that $x_1\cdots x_n=1$ prove that $\frac{1}{n-1+x_1}+\cdots+\frac{1}{n-1+x_n}\le 1$.

I tried solving this question through substitution, e.g. $a_1=\sqrt[n]{x_1},...,a_n=\sqrt[n]{x_n}$, hence $a_1...a_n=1$. Also for $k=1,2,...,n$ we have that $x_k=a_k^n=a_ka_k^{n-1}=\frac{a_k^{n-1}}{a_1...a_n}$.

This is where I got stuck. My intuition tells me that I should be able to finish it off from here using AM-GM, however it isn't working out for me. Could you please explain to me how to finish the question off?

答:
According to page 8, Problem 25 in [1], the equivalent problem was published by Vasile Cirtoaje on Gazeta Matematica, Seria B, No. 10, 1991.

Five solutions are given in [1]. Here is one of them:

Let $r = \frac{n-1}{n}$. By AM-GM, we have, for $i=1, 2, \cdots, n$,
$$\sum_{j\ne i} x_j^r \ge (n-1) \left(\prod_{j\ne i} x_j\right)^{r/(n-1)} = (n-1) x_i^{r - 1}$$
which results in $x_1^r + x_2^r + \cdots + x_n^r \ge (n-1)x_i^{r - 1} + x_i^r$ and
$$\frac{x_i}{x_i + n-1} \ge \frac{x_i^r}{x_1^r + x_2^r + \cdots + x_n^r}. \tag{1}$$
Summing up (1) for $i = 1, 2, \cdots, n$, we have
$$\sum_{i=1}^n \frac{x_i}{x_i + n-1} \ge 1$$
which is written as
$$\sum_{i=1}^n \frac{1}{x_i + n-1} \le 1.$$
We are done.

Reference

[1] Vasile Cirtoaje, “Algebraic Inequalities: Old and New Methods”, 2006.


问:
Why convolution regularize functions?
There is a tool in mathematics that I have used a lot of times and I'm still not confortable with. In fact I can't figure out (by this I mean that I cannot understand it geometrically) why does convolution regularize things. It is know for example that if $u\in L_{loc}^1(\mathbb{R}^N)$ and $f\in C_0^k(\mathbb{R}^N)$, then $f\ast u\in C^k(\mathbb{R}^N)$ and $$D^\alpha (f\ast u)=(D^\alpha f)\ast g$$

Well, I have a lot of questions about convolution that might help me understand it better and I would appreciated one who points me out references to read and some insight about it.

I - For what reason was convolution implemented in mathematics? Was it for regularizing functions? Who used it the first time?

II - Is there any intuitive way to find out why convolution make functions more smooth?

I can understand the analytical proof of II, but in the end it is just algebraic manipulations and I dont get the essence of the thing. Thank you for your help.

答:
Concerning II:

When $f$ is a smooth function then all its translates $T_xf$, defined by $T_xf(t):=f(t-x)$, are equally smooth. Convolution of $f$ with an arbitrary $u$ can be viewed as a linear combination of such translates: $$f*u=\int_{-\infty}^\infty u(x)\>T_x f\ dx\ ,$$
therefore we expect that it inherits this smoothness. Looking now what this process has done to $u$ we get the feeling that $u$ has been smoothed out in this way.

Title: Why convolution regularize functions?


问:
Homeomorphism group of compact metric space $K$ is $G_{\delta}$ in $C(K,K)$
Let $(K,d)$ be a compact metric space and denote $\ C(K,K):=\{ f:K \to K \ ; \space f \space \text{is continuous}  \}$. Endow the set $C(K,K)$ with the supremum metric $\ \rho (f,g):=\mathrm{sup} \ \{ d(f(x),g(x)) \ ; \ x \in K \}$.

I'd like to show that the group of homeomorphisms of $(K,d)$ is a $G_{\delta}$ subset of the metric space $( C(K,K),\rho)$.

I managed to show that it is enough to prove that the set $\{ f \in C(K,K) \ ; \space f \space \text{is injective} \}$ is $G_{\delta}$, but then I got stuck.

I'll appreciate any help.

答: If you want to show that the set $\{ \,f \in C(K,K)\,\,:\,\,f{\text{ is injective }}\}$  is $Gδ$, note that for $f∈C(K,K)$, $f$ is injective iff $f \in \bigcap\limits_{x \ne y} {\{ \,f \in C(K,K)\,\,:\,\,f(x) \ne f(y)\,\} }$.


问:
How can I find a generalized expresion for $A^{n}$?
I have to find a generalized expresion for $A^{n}$, $\forall n\in \mathbb{N}$
\begin{align}
A=
\begin{pmatrix}
-5 & 3\\
-6 & 7
\end{pmatrix}
\end{align}

Do you have any idea of how can I do that?

答:
Write in form of jordan
$$\left|\begin{matrix}
-5-\lambda & 3 \\
-6 & 7-\lambda
\end{matrix}\right|=0\\\lambda_1=-3\sqrt2+1\\\lambda_2=3\sqrt2+1$$solve it for eigen vectors
$$(\left(\begin{matrix}
-5 & 3 \\
-6 & 7
\end{matrix}\right)-(-3\sqrt2+1)I)X=\bar{0}\to V_1=\left(\begin{matrix}
\frac{\sqrt{2}+2}{2} \\
1
\end{matrix}\right)$$,
$$(\left(\begin{matrix}
-5 & 3 \\
-6 & 7
\end{matrix}\right)-(3\sqrt2+1)I)X=\bar{0}\to V_2=\left(\begin{matrix}
\frac{-\sqrt{2}+2}{2} \\
1
\end{matrix}\right)$$,
$$M=\left(\begin{matrix}
\frac{\sqrt{2}+2}{2} & \frac{-\sqrt{2}+2}{2} \\
1 & 1
\end{matrix}\right)$$
and wite it as
$$\left(\begin{matrix}
-5 & 3 \\
-6 & 7
\end{matrix}\right)=\\
\left(\begin{matrix}
\frac{\sqrt{2}+2}{2} & \frac{-\sqrt{2}+2}{2} \\
1 & 1
\end{matrix}\right)\left(\begin{matrix}
-3\sqrt{2}+1 & 0 \\
0 & 3\sqrt{2}+1
\end{matrix}\right)\left(\begin{matrix}
\frac{\sqrt{2}}{2} & \frac{-\sqrt{2}+1}{2} \\
\frac{-\sqrt{2}}{2} & \frac{\sqrt{2}+1}{2}
\end{matrix}\right)\\A=MDM^{-1}$$so
$$A^2=MDM^{-1}MDM^{-1}=MD^2M^{-1}\\A^3=MD^2M^{-1}MDM^{-1}=MD^3M^{-1}\\...\\A^N=MD^NM^{-1}$$


问:
Solving a Complex Diophantine Equation with One Known
I am struggling to find a way to implement Euclid's Algorithm in order to solve this diophantine equation. The $N$ will be known and the set of solutions I wish to find will be a set of decreasing $X$ solutions that satisfy $$x^2 - 4  y^2 = n.$$

When simplifying I have concluded that the solution could be as follows...
$$x^2 - 4 y^2 = (x - 2y)  (x + 2y)$$ when $n = 12$,
$$12 = (x - 2y) (x + 2y)$$
would it be possible to use Euclid's Algorithm to solve for the GCD in decreasing order?
The poposed solutions are a set of $n=12$, $x = 4$ and $y = 1$.

答:
In response to your request for the detailed translated answer, you could would it out with algebra but I have more experience with these equations so it will be easier for me. I never heard of "discord" until you asked if I used it and the answer is "no". WolframAlpha is popular on this forum.

In the other answer we identified the "adjusted" Euclid's formula
$$x=m^2+k^2\qquad y=mk\qquad n=(m^2-k^2)^2$$

$$n=(m^2-k^2)^2
\quad\implies\quad k = \sqrt{m^2-\sqrt{n} }\\
 \text{for}\quad
\big\lceil\sqrt{\sqrt{n} - 1}\space\big\rceil
\le
 m\le
\frac{\sqrt{n}+1}{2}$$

The lower limit ensures that $k\in\mathbb{N}$ and the upper limit insures that $k<m$

For $n=225$ we know the solutions are $\quad(17,4,225)\land(113,56,225)$

$$n=225\implies \big\lceil\sqrt{\sqrt{225} - 1}\space\big\rceil =4
\le m \le
\frac{\sqrt{225}+1}{2}=8\\
\quad\land\quad m\in\{4,8\}\implies k\in\{1,7\}$$

$$
x=m^2+k^2\qquad y=mk\qquad n=(m^2-k^2)^2\implies$$
$$F(4,1)\rightarrow x=4^2+1^2=17\quad y=4(1)=4\quad n=(4^2-1^2)^2=225\\
F(8,7)\rightarrow x=8^2+7^2=113\quad y=8(7)=56\quad n=(8^2-7^2)^2=225$$


问:
Math blogs, pros and cons for writers?
I regularly read blogs by three mathematicians, and occasionally run into others. Definitely they help me a lot studying mathematics.

But now I am more interested in the writers' perspective, and I wonder whether writing a math blog help study mathematics. What are the pros and cons for the writer himself, especially for a first-year graduate student?

A lot of users have tons of experience writing blogs. I really want to hear your opinion on this.

Thanks!

Maybe I just start writing some of my own thoughts: cons: 1. writing a blog consumes lots of time. pros: 1. writing a blog forces you to think deeply and clearly; 2. it helps you learn how to communicate mathematics; 3. it might help you connect to others who share your passion.

答:
There are many pros of writing a blog or any other website to discuss math. I can see the following in my case:

Prepare a set of notes for a course you are taking, which in turn helps you with learning the material.

As a graduate student you shall be expected to write, so writing in a blog gives you some habit and makes your style clearer and also makes it possible for you to judge which articles people like and how you wrote them,

Communicating good, solid and rare ideas to the community is an extremely good habit and would give you a sense of accomplishment.

There will be cons too, I just mention one is that it will take you substantial time before you can do it in an efficient way and in a manner that is accessible to a person who is not much conversant with your field of expertise.

Title: Math blogs, pros and cons for writers?


问:
Is the cardinality of the topology of a separable infinite metric space uncountable
Given a metric space $X$ which is separable and infinite, can we say that $\tau$ (its topology) has cardinality greater than that of the natural numbers? And if so how to go about it.

答:
HINT: If $X$ is discrete, then $\tau=\wp(X)$, so $|\tau|\ge|\wp(\Bbb N)|=2^{\aleph_0}>\aleph_0=|\Bbb N|$.

If not, then there is an $x\in X$ such that $x\in\operatorname{cl}(X\setminus\{x\})$.

Explain why there is a sequence $\langle x_n:n\in\Bbb N\rangle$ of distinct points converging to $x$.

Let $D=\{x_n:n\in\Bbb N\}$.

Show that $D$ is discrete: for each $n\in\Bbb N$ there is an open nbhd $U_n$ of $x_n$ such that $U_n\cap D=\{x_n\}$.

Show that if $A,B\subseteq\Bbb N$, and $A\ne B$, then $\bigcup_{n\in A}U_n\ne\bigcup_{n\in B}U_n$.

Conclude that the map $$\varphi:\wp(\Bbb N)\to\tau:A\mapsto\bigcup_{n\in A}U_n$$ is an injection, and hence $|\tau|\ge|\wp(\Bbb N)|=2^{\aleph_0}>\aleph_0=|\Bbb N|$.


问:
Linear Algebra: Using Gaussian Elimination to obtain Row Echelon Form Matrix
I am learning Linear Algebra and I understand that we can use Gaussian Elimination to transform an augmented matrix into its Row Echelon Form using Elementary Row Operations. (ERO)

One thing that is not very clear to me is this: When using EROs, are we restricted to only using the rows in the current iteration of the augmented matrix (that is in the process of being transformed?)

For example, consider I have a matrix with 3 rows, Row A, Row B, and Row C. Now, I perform an ERO and my new augmented matrix (Still not in Row Echelon Form) is of the form Row A, Row A+B, Row C.

To further perform EROs on this new augmented matrix, am I limited to using Row A, Row A+B, Row C, or can I still use Row B (Which is no longer in my current matrix but was in a previous iteration)? And is there a reason why we can/cannot reuse it?

Many thanks in advance!

答:
In your example, you can continue from Row A, Row A+B, Row C. In this particular example, using Row B instead of Row A+B is a step backwards. There is no reason to do that.

I don't know how far you have studied into this subject, but there is an elementary matrix associated with each Elementary Row Operation. At each step of your operation, the interim matrix that you calculate is the result of multiplication of that step's elementary matrix by the previous step's (interim) matrix. In this way, as you perform a series of Elementary Row Operations, you are in fact multiplying a series of elementary matrices into your original matrix. I hope this gives you a sense why you can't cut back into the middle of the chain of operations and continue from one of the previous steps.


答: You cannot use row $B$. If you do, you could do $A\to A+B$. Now you have two equal rows, which in the next step gives you a zero row. And you have put your system in a state which is not equivalent to the original one.

Title: Linear Algebra: Using Gaussian Elimination to obtain Row Echelon Form Matrix

Body:
I am learning Linear Algebra and I understand that we can use Gaussian Elimination to transform an augmented matrix into its Row Echelon Form using Elementary Row Operations. (ERO)

One thing that is not very clear to me is this: When using EROs, are we restricted to only using the rows in the current iteration of the augmented matrix (that is in the process of being transformed?)

For example, consider I have a matrix with 3 rows, Row A, Row B, and Row C. Now, I perform an ERO and my new augmented matrix (Still not in Row Echelon Form) is of the form Row A, Row A+B, Row C.

To further perform EROs on this new augmented matrix, am I limited to using Row A, Row A+B, Row C, or can I still use Row B (Which is no longer in my current matrix but was in a previous iteration)? And is there a reason why we can/cannot reuse it?

Many thanks in advance!


问:
How to show $\int_{B_1} \frac{(p|\omega|)^m}{m!} \le 2^{-m} \Rightarrow \int_{B_1} e^{p_0 |\omega|} dx \le C$?
Assume $\omega \in H^1_{loc}(\mathbb R^n)$, if there is a constant $p>0$ depend on $n$ such that
$$
\int_{B_1} \frac{(p|\omega|)^m}{m!} \le 2^{-m},~~~~~~m=2,3,...
$$
where $B_1$ is unit ball. Then how to show there is constant $p_0 >0, C>0$  such that
$$
\int_{B_1} e^{p_0 |\omega|} dx \le C
$$

This problem is from the proof of Harnack inequality (of weak solution of Laplace equation) in a book written by Chinese, I can't understand this step, so ask here, thanks for any help.

答:
Your integral at the beginning does not have a differential, but the later one has $dx$.  I'll assume both have the same.

$e^{p|\omega|}=\sum_{m=0}^\infty \frac{(p|\omega|)^m}{m!}$

$\int_{B_1}e^{p|\omega|}=\sum_{m=0}^\infty \int_{B_1}\frac{(p|\omega|)^m}{m!}\le \frac{1}{2}+\int_{B_1}(1+p|\omega|)=C$.

Note $\frac{1}{2}=\sum_{m=2}^\infty 2^{-m}$  The last integral is dependent on $n$.


问:
Convergence of $\displaystyle\sum_{n=0}^{\infty} \left(\dfrac{n}{2n+1}\right)^{2n-1}x^n$
$\displaystyle\sum_{n=0}^{\infty} \left(\dfrac{n}{2n+1}\right)^{2n-1}x^n$ How can I solve this question? I tried Root and Ratio test, but I encounter with $\displaystyle\lim_{x\to\infty}$ $\left(\dfrac{n}{2n+1}\right)^{\dfrac{-1}{n}}$ so I can't solve this limit, are there anyway to evaluate this limit or sum?

答:
The ratio test works nice. Let
$$a_n=\left(\dfrac{n}{2n+1}\right)^{2n-1}\implies \log(a_n)=(2n-1)\log\left(\dfrac{n}{2n+1}\right)$$ By Taylor
$$\log\left(\dfrac{n}{2n+1}\right)=-\log (2)-\frac{1}{2 n}+\frac{1}{8 n^2}+O\left(\frac{1}{n^3}\right)$$
$$\log(a_n)=-2 n \log (2)+(\log (2)-1)+\frac{3}{4 n}+O\left(\frac{1}{n^2}\right)$$ Apply it twice and continue with long division
$$\log(a_{n+1})-\log(a_n)=-2 \log (2)-\frac{3}{4 n^2}+O\left(\frac{1}{n^3}\right)$$
$$\frac{a_{n+1} }{a_n}=e^{\log(a_{n+1})-\log(a_n)}=\frac{1}{4}-\frac{3}{16 n^2}+O\left(\frac{1}{n^3}\right)$$


问:
A question about the polar equation $r=ln(\theta)$
Take the polar equation r=ln(theta) for theta between $0$ and $2\pi$.
1
In this, we have a loop that connects at a point $(x,y)$ for $2$ unique points $(r_1,\theta_1)$ and ($r_2,\theta_2)$. There are $2$ questions about this I would like to know the answer to and how you would find it.

At what point $(x,y)$ does this loop have the intersection shown in the image, and

What is the area of the loop within this equation? I understand the equation is $1/2$ of the integral of $r^2 d\theta$ from $\theta_1$ to $\theta_2$, I just don't know how the get the $\theta_1$ and $\theta_2$ for this equation.

答:
Hint

The point of intersection represents two points with the same absolute value of $r$, but with different values of $\theta$ seperated by an integer multiple of $\pi$. Since we are restricted to $[0,2\pi]$, this would be $\pi$

$$r_1 = \ln\theta_1$$

$$r_2 = \ln\theta_2 = \ln(\theta_1 + \pi)$$

Since log is an increasing function, we have $r_2 = -r_1, r_1 < 0$

$$\ln(\theta_1) + \ln(\theta_1 + \pi) = 0$$

$$\implies \ln(\theta_1(\theta_1 + \pi)) = 0$$

$$\implies \theta_1 = \frac{1}{\theta_1 + \pi}$$


问:
Interpreting the logarithm as a sum of simple poles along the negative real axis
I've heard it remarked that you can basically consider $\log z$ to be a function which has simple poles everywhere on the negative real axis (with a constant "residue density" at each pole).  This would be something like
$$ \log z = \int_0^\infty \frac{dx}{z + x} $$
But of course the integral on the right-hand side actually diverges.  We actually get
$$ \int_0^\infty \frac{dx}{z + x} = \lim_{b \to \infty} \int_0^b \frac{dx}{z + x} = \lim_{b \to \infty} \left( \log(z + b) - \log(z) \right) = \infty $$

In physics, there are a variety of methods for subtracting out the divergent part of such a limit to get a finite answer (various flavors of regularization and renormalization).  I'm wondering whether there is a standard approach here so that something similar can be done to "rescue" the first equation above from the divergent part of the integral.

Another way of phrasing the problem above is that I showed that the Stieltjes transform of a constant on the interval $(-\infty, 0]$ does not exist.  But perhaps there is another density function $\rho$ so that $\log z$ is the Stieltjes transform of $\rho$.
$$ \log z = \int_0^\infty \frac{\rho(x)}{z + x} dx $$
What is $\rho$?  Well, the Stieltjes inversion formula says that it should be given by
$$ \begin{align} \rho(x) &= \lim_{\epsilon \to 0} \frac{\log(x+i\epsilon) - \log(x-i\epsilon)}{2\pi i} \\ &= \frac{(\log |x| + \pi i) - (\log |x| - \pi i)}{2\pi i} \\ &= 1 \end{align} $$

But this gets me exactly back to the integral that I started with, which is divergent!  Hopefully I am just missing something obvious.

Edit: Alternate statement of question

There has been a lot of confusion in the comments below about what I am looking for, so let me restate it in a very narrow way.  I would be satisfied with either of the following:

A sequence of meromorphic functions $f_n(z)$ with simple poles along the negative real axis with the following properties:

a. The poles become dense in the limit $n \to \infty$.

b. $\lim_{n \to \infty} f_n(z) = \log z$

A proof that there is no such sequence.

答:
By fiddling around with my original suggestion above, I have one solution that meets my criteria:
$$ f_n(z) = - \sum_{k=0}^{n} \frac{1}{n\frac{z}{1-z} + k} $$
This has poles at $z=-\frac{k}{n-k}$ for $0 \le k < n$ which becomes dense on the negative real axis as $n \to \infty$.

If we let $a=\frac{nz}{1-z}$, then
$$ f_n(z) = \psi(a) - \psi(a+n+1) $$
where $\psi$ is the digamma function.  As $n \to \infty$, $a \to \infty$, and the leading asymptotic behavior of $\psi(z)$ is $\log z$ as $z \to \infty$.  So we get
$$ f_n(z) \sim \log\left( \frac{a}{a+n+1} \right) \sim \log\left( \frac{1}{1 + \frac{1-z}{z} + 0} \right) = \log z $$

This works in a technical sense, but it seems more complex than it needs to be.  I would accept any answer which is more elegant, especially if it can be expressed in a Stieltjes-transform-like form.


问:
Modifying known probability density function using another pdf as an initial condition
My question is for a brownian motion type process with two physically different regions. Region 1 is between points A and B and region 2 is between points B and C. The time taken to travel between two points A and B is a stochastic quantity $T_1$. Say I know that the time taken to travel between A and B follows a gaussian distribution i.e.

$$P(T_1) = \frac{1}{\sigma \sqrt{2 \pi}} \exp\bigg[ -\frac{1}{2} \frac{(T_1 - \langle T_1\rangle)^2}{\sigma^2} \bigg]$$

where $\sigma$ is the standard deviation of the distribution as expected. And that I know once you enter region 2 the time to travel between the points B and C is another stochastic quantity $T_2$ with its own pdf $G(T_2)$ that is generically not gaussian. As an example

$$G(T_2) = \alpha \cdot T_2 \cdot e^{-\beta\cdot T_2^2}$$

but in practice it could be any kind of pdf.

My question is how can you combine the knowledge of the pdf for $T_1$ and seperately the pdf for $T_2$ to work out the pdf for the time taken to travel the total distance from A to B and then to C, $T_3$. Intuitively given that $P(T_1)$ is a simple gaussian distribution. I would expect the full pdf to just be a more spread out version of $G(T_2)$ but I am not sure how to show this explicitly or quantify it.

Any help to how I can combine pdfs in this way (even if it has to be numerically) would be greatly appreciated

答:
Welcome to math.stackexchange!

This looks like a convolution problem. You can look up what convolutions are and how they are used (here's one resource), but it's also useful to show how one can obtain the main convolution formula from simple arguments since such arguments are not often shown.

Let $P(t_1)$ be the probability density for moving from $A$ to $B$ in time $t_1$ and $G(t_2)$ be the probability density for moving from $B$ to $C$ in time $t_2$. Then, let's denote the probability density for moving from $A$ to $B$ to $C$ as $H(t_\text{tot})$. Noting that $t_{\text{tot}} = t_1 + t_2$ by constraint, we can find the probability density for $H$ by integrating our distributions of $P(t_1)$ and $G(t_2)$ over their domains subject to that constraint. We then have

\begin{align}
H(t_\text{tot}) &= \int^{\infty}_{-\infty} dt_1 \int^{\infty}_{-\infty} dt_2\, \delta(t_\text{tot}- t_1 - t_2) P(t_1) G(t_2)\\[.5em]
& = \int^{\infty}_{-\infty} dt_1\, P(t_1) G(t_{\text{tot}}-t_1)
\end{align}

which is the standard formula for convolutions (https://en.wikipedia.org/wiki/Convolution#Definition).


问:
3D software like GeoGebra
Does it exist a free interactive geometry software, like GeoGebra, which works for 3D geometry? I would be able to draw spheres, great circles, and so on.</p>

答:
I think you can, to a limited extent, use Google Sketchup, but that's not really the same.

Cabri3D is probably the most like what you want, but it's commercial.  Autograph, also commercial, may do some of that kind of stuff, but I'm not sure.

ContentLicense: CC BY-SA 2.5

Title: 3D software like GeoGebra


问:
Minimum of two exponentially distributed random variables
Let $X_1 ~ \sim \text{Exp}(\lambda_1)$ and $X_2 \sim \text{Exp}(\lambda_2)$ be two independent exponentially distributed random variables.

Show that $\mathbb{P}\{\min(X_1, X_2) > t\} = \exp(-(\lambda_1 + \lambda_2)t)$, and hence that $\min(X1, X2) \sim \text{Exp}(\lambda_1 + \lambda_2)$.

答:
It is instructive to make the following observation, giving intuition for the result.

Let $N_1$ and $N_2$ be independent Poisson processes with rates $\lambda_1$ and $\lambda_2$, respectively.
Let $X_i$, $i=1,2$, denote the waiting time until the first event to occur in the process $N_i$. Then $X_i \sim {\rm Exp}(\lambda_i)$, with $X_1$ and $X_2$ independent. Next, let $N = N_1 + N_2$. The process $N$ is a Poisson process with rate $\lambda_1 + \lambda_2$. Finally, let $Z$ be the waiting time until the first event to occur in the process $N$. Thus, on the one hand, $Z \sim {\rm Exp}(\lambda_1 + \lambda_2)$, and on the other hand, $Z = \min \{X_1,X_2\}$. Hence, $\min \{X_1,X_2\} \sim {\rm Exp}(\lambda_1 + \lambda_2)$.


问:
Rudin RCA Problem 6.4
I'm trying to solve the following Problem 6.4 from Rudin's RCA:

Suppose that $1\leq p\leq \infty$, and $q$ is the exponent conjugate to $p$. Suppose that $\mu$ is a $\sigma$-finite measure and $g$ is a measurable function such that $fg\in L^1(\mu)$ for every $f\in L^p(\mu)$. Prove that then $g\in L^q(\mu)$.

I can get to the point where $\int_{E}|g|d\mu<\infty$ for every finite measure set $E$, but I'm not sure where to proceed. I would appreciate any suggestions.

Thanks!

答:
Here is an argument using Uniform Boundenes Principle: Let $\mu(A_n) <\infty$ and $A_n$ increase to the whole space. Define $T_nf=fg\chi_{A_n \cap \{x: |g(x)| \leq n\}}$. Verify that $T_n$ is a bounded linear functional and $\|T_n\|^{q}=\int_{{A_n \cap \{x: |g(x)| \leq n\}}}|g|^{q} d\mu$. By hypothesis $\sup_n \|T_nf\|<\infty$ for each $f \in L^{p}$. It follows that $\sup_n \|T_n||<\infty$ But this is precisely the satatment that $g \in L^{q}$.

[For computation of $\|T_n\|$ just follow the argument in $(L^{p})^{*}=L^{q}$].

Title: Rudin RCA Problem 6.4


问:
Double integral over $f(x,y) = y^2x+x^3$

Compute the integral $$\int_D f \ dx \ dy$$ where $f : D\to \mathbb{R}, f(x,y) = y^2x+x^3$ and $D= \{(x,y) \mid \sqrt{x^2+y^2} \leqslant 5, y >0\}$

Converting to polar coordinates I got $f(r,\theta) = r^3\cos(\theta)$ and $D = \{(r,\theta) \mid 0 \leqslant r \leqslant 5, 0\leqslant \theta \leqslant 2\pi \} $ So the integral would become $$\int_{0}^{2\pi}\int_{0}^{5} r^4\cos(\theta) \ dr \ d\theta = \int _0^{2\pi }\cos \left(θ\right)\cdot \:625 \ d\theta = 0.$$

I guess I have a mistake here since isn't this just a cone whose projection on to the $xy$ plane is a circle of radius $5$ and the area shouldn't evaluate to zero?

答:
Please note that $f(x,y) = y^2x+x^3$ is an odd function with respect to $x$ which means

$f(-x,y) = y^2(-x) + (-x)^3 = - (y^2x+x^3) = - f(x, y)$

So if you integrate over a region which is symmetric to $y-$axis, integral over negative values of $x$ will cancel out integral over positive values of $x$ and that is correct answer. Here you are integrating it over a circle $(-5 \leq x \leq 5, x^2 + y^2 = 25)$ which is symmetric to $y-$ axis.

If you were given the condition $x \geq 0$, the integral would not be zero as the function is even wrt $y$.

I hope this clarifies.


问:
Basic probability problem!

"A boss plans a business meeting at Starbucks with the two engineers below him. However, he fails to set a time, and all three arrive at Starbucks at a random time between $2:00$ and $4:00$ p.m. When the boss shows up, if both engineers are not already there, he storms out and cancels the meeting. Each engineer is willing to stay at Starbucks alone for an hour, but if the other engineer has not arrived by that time, he will leave. What is the probability that the meeting takes place?"

What answer did you get?
Thanks!

答:
If the engineers arrive in the first hour and the boss arrives in the second hour (p=1/8), then the meeting will definitely be held (p=1).
If all three arrive in the same hour (first or second) (p=1/8+1/8=1/4), then the meeting will be held only if the boss (who may arrive first, second, or last) arrives last (p=1/3)
If one of the engineers arrives in the first hour while the other engineer and the boss arrive in the second hour (p=1/4), then the meeting will be held only if the second engineer arrives before the boss and before the first engineer departs (the order of the arrival of the boss and the departure of the first engineer does not matter) (p=1/3)
Adding these probabilities, the total probability of the meeting taking place = (1/8)1+(1/4)(1/3)+(1/4)*(1/3) = 7/24

Title: Basic probability problem!


问:
Calculation of Christoffel symbols of Schwarzschild metric using Lagrangian
I know that the Schwarzschild line-element is:

$$ds^2=c^2\left(1-\frac{2\mu}{r}\right)dt^2-\left(1-\frac{2\mu}{r}\right)^{-1}dr^2-r^2d\theta^2-r^2\sin^2\theta d\phi^2$$

Dividing by $d\lambda^2$, where $\lambda$ is an affine parameter, I get:

$$\left(\frac{ds}{d\lambda}\right)^2=c^2\left(1-\frac{2\mu}{r}\right)\dot{t}^2-\left(1-\frac{2\mu}{r}\right)^{-1}\dot{r}^2-r^2\dot{\theta}^2-r^2\sin^2\theta \dot{\phi}^2$$

where $\dot{X}$ is $\frac{dX}{d\lambda}$.

To my understanding, this is the Lagrangian, so Euler-Lagrange equations hold.

So I must have:

$$\frac{d}{d\lambda} \left( \frac{\partial L}{\partial \dot{\phi}} \right)=\frac{\partial L}{\partial \phi}$$

Evaluating LHS: $-2r^2\sin^2\theta\ddot{\phi}$.

RHS: $\frac{\partial L}{\partial \phi} = 0$.

Ie: $\ddot{\phi}=0$

Geodesic equation for $\phi$:

$$\frac{d\phi^2}{d\lambda^2}+\Gamma^{\phi}_{\mu\nu}\frac{dx^\mu}{d\lambda}\frac{dx^nu}{d\lambda}=0$$

So I conclude that $\Gamma^{\phi}_{\mu\nu}$ for any $\mu, \nu$.

This, however, is in conflict with what I find on the internet. Equation 5.4 from here, Wikipedia, disagrees with this result.

What am I doing wrong?

答:
Perhaps should be moved to physics stackexchange?

Commend you for trying to re-derive the equation of motion directly from the length geodesic for the Schwarzchild metric, but you applied the EL equations incorrectly. The "Lagrangian" in this case is a function of more than just $\phi$. It is also a function of $t$, $r$, and $\theta$, since each of these are functions of $\lambda$, so you would need to set up reciprocal equations of motion for these coordinates.


问:
Ballistic Motion in Space with One Gravitating Body.
I had an interesting math problem presented to me some time ago by a friend (he stated it in non-mathematical terms). At what angle would you launch a projectile from a spaceship/satellite such that it left that object and went on to hit another orbiting object? Then as a supplemental question he asked at what angle would you launch that projectile to hit the other orbiting object in the least amount of time?

I assumed that the objects were only acted upon by a single spherically symmetric mass distribution such that I could treat it as a one-body problem for each object. Further, I assumed it all took place in the plane with a polar coordinate system so that I ended up with this simple system of nonlinear autonomous ODE's,

$$ \begin{bmatrix} \frac{d \theta}{dt}  \\ \frac{dv}{dt}\\ \frac{dr}{dt} \end{bmatrix} =  \begin{bmatrix} \frac{h}{r^{2}} \\ \frac{h^{2}}{r^{3}} -\frac{\mu}{r^{2}} \\ v \end{bmatrix}.$$

Where the inital conditions for the projectile would be $\{ \theta_{i} , r_{i}, v_{\beta}\cos(\phi - \theta_{i}) + v_{i}\}$ with an $h_{\phi} = r_{i} (v_{\beta}\sin(\phi - \theta_{i}) + v_{\theta i})$ and the target objects' initial conditions are $\{ \theta_{i}^{'} , r_{i}^{'}, v_{i}^{'}\}$ with $h'= r^{'}_{i} v_{\theta i}^{'}$. Where $\phi$ is the launch angle from the polar axis while $v_{\beta}$ is the magnitude of the projectiles velocity (which I assume to not change, only its launch direction) and $\mu$ is a constant relating to the gravitational field strength of the attracting object. Below is a picture depicting the general initial and final conditions.

After all these preliminaries, i'm basically asking if there is a variational calculus or other simpler solution to solving this problem as perhaps a boundary problem of sorts mixed with an initial ODE problem. That is, aside from computationally pouring on through thousands of trajectories with minutely differing $\phi$'s then numerically guessing at the appropriate approximate launch angle or angles that solve my question(s).

Which. . . isn't what I exactly want to do and I desire to know if there is an equation, a single or system of ODE's, that I could solve themselves for this launch angle which gives a least time of travel or gives launch angles that would lead to a hit (irrespective of the time of travel). If you can help in anyway, this would be most appreciated. I'm a sophomore College student with little knowledge of solving robustly ODE's or even programming solvers for them.

答:
So, $v_\beta$ is fixed? There's just two parameters, $\phi$ and the launch time $t_0$?

For any given $\phi$, the set of $t_0$ which results in a collision will be a discrete set. The reverse is not quite true, but is true if you insist the collision occur within some time limit, or some finite number of orbits - for any given $t_0$, the set of $\phi$ that results in a collision within [time limit] is discrete.

Let's say you fix $t_0$ and solve for $\phi$. Once you have one solution, the travel time is determined, and you can (in principle) find the derivative of the travel time with respect to $t_0$. This allows you to use efficient optimisation methods (such as conjugate gradient), even if you can't solve it analytically.

Finding a neat representation of orbits (such as polar coordinates) really helps. However, the polar coordinate representation isn't the neatest. There's another way (and I can't find a link to it, alas) where (by modifying the time variable) all orbits become great circles on a 4-dimensional sphere $x^2+y^2+z^2+T^2=r^2$. If you can find that, it might make the problem even easier to solve, but alas, I can't help you find it.


问:
Understanding what $d(f(x))$ means
I know that $d(f(x))$ is small change in $f(x)$. But why is $\int d(f(x))=f(x)$? We know $\int dx$ means adding little bit of $x$. But what does adding little bit of $d(f(x))$ actually mean? Since it is not on the $x$ axis?

答: It may be easier to write $y$ instead of $f(x)$ and $dy$ instead of $d(f(x))$. You are right: $f(x)$ is not on the $x$ axis. On an $xy$ plane, $f(x)$ and $d(f(x))$ are on the $y$ axis. Now, if you are comfortable with integrating $dx$, then you can integrate $dy$ in the same way.

Title: Understanding what $d(f(x))$ means


问:
Given $\frac{dS}{dt}=-\frac{SI}{S + I + R}$ and $R = -(SI + S + I)/ \frac{dS}{dt}$ Determine $\frac{d^2S}{dt^2}$
I am given a system of equations:

\begin{align}
\frac{dS}{dt} &= -\frac{SI}{S + I + R}\\
\frac{dI}{dt} &= \frac{SI}{S + I + R} -I \\
\frac{dR}{dt} &= I - R\\
\end{align}

I am to subsitute $R = -\left(SI + S + I\right) / \frac{dS}{dt} $ to determine $\frac{d^2S}{dt^2}$

I chose to work only with the first equation.
When I substitute what is given for $R$ into the denominator I get:

\begin{align}
S + I + R &= S + I + R = -\left(SI + S + I\right) / \frac{dS}{dt}\\
&=\frac{\frac{dS}{dt}(S+I)-(SI+S+I)}{\frac{ds}{dt}}
\end{align}

I substitute this into the equation for $\frac{dS}{dt}$ and I get:

\begin{align}
\frac{dS}{dt} &= \frac{-SI \left(\frac{dS}{dt}\right)}{\frac{ds}{dt}\left(S+I\right)-(SI+S+I)}\\
\left(\frac{dS}{dt}\right)^2(S+I)-\frac{dS}{dt}\left(SI+S+I\right) &= -SI\left(\frac{dS}{dt}\right)\\
\left(\frac{dS}{dt}\right)^2(S+I) &= \left(\frac{dS}{dt}\right)\left(S+I\right)\\
\left(\frac{dS}{dt}\right)^2 &= \left(\frac{dS}{dt}\right)\\
\end{align}

How do I get the desired result of
$$
\frac{d^2S}{dt^2}= \frac{2}{S}\left(\frac{dS}{dt}\right)^2
$$

答:
Try differentiating $\frac{dS}{dt}$ directly.

$\begin{align}\frac{d^2S}{dt^2} &= -\frac{(S+I+R)\frac{d(SI)}{dt}- SI\frac{d(S+I+R)}{dt}}{(S+I+R)^2}\\
\\&=-\frac{(S+I+R)\left[I\frac{d(S)}{dt}+S\frac{d(I)}{dt}\right]- SI\frac{d(S+I+R)}{dt}}{(S+I+R)^2}\\
\\&=-\frac{(S+I+R)\left[-I\frac{SI}{S + I + R}+S\frac{SI}{S + I + R} -SI\right]+SIR}{(S+I+R)^2}\\
\\&=-\frac{(S+I+R)\left[\frac{-SI(R+2I)}{S + I + R}\right]+SIR}{(S+I+R)^2}\\
\\&=\frac{2SI^2}{(S+I+R)^2} \\
\\& = \frac{2}{S}\left(\frac{SI}{S+I+R}\right)^2\\
\\&=\frac{2}{S}\left(\frac{dS}{dt}\right)^2\end{align}$


问:
Difficulties solving this integral: $ \int_0^1 \frac{\ln(x+1)} {x^2 + 1} \, \mathrm{d}x $ by differentiation under the integral sign
So in the book Advanced Calculus Explored, by Hamza E. Asamraee. The next integral appears as an exercise to solve by differentiating under the integral sign:

$$ \int_0^1 \frac{\ln(x+1)} {x^2 + 1} \, \mathrm{d}x $$

I have solved this integral before by substitution and change in the limits of integration, but in this chapter the book asks to solve it by differentiation under the integral sign. I have tried several ways of solving this, but the only one that i thought it was leading me somewhere was:

$$f(a) = \int_0^1 \frac{\ln(x+a)} {x^2 + 1} \, \mathrm{d}x $$

So that:

$$f'(a) = \int_0^1 \frac{1} {(x+a)(x^2 + 1)} \, \mathrm{d}x $$

Then i tried to separate this last integral by partial fractions, my result on this was:

$$\frac {1} {(x+a)(x^2 + 1)} = \frac{1} {a^2 + 1} \left(\frac {1} {x+a} - \frac{x-a} {x^2+1}\right)$$

And the integral reduces to:

$$f'(a) = \int_0^1 \frac{1} {a^2 + 1} \left(\frac {1} {x+a} - \frac{x-a} {x^2+1} \right) \, \mathrm{d}x $$

Then this last expression evaluates to:

$$f'(a) = \frac{1} {a^2 + 1} (\ln(a+1) - \ln(a) - \ln(4)+ \frac{π}{4} a)$$

Then integrating from 0 to 1 with respect to $a$ we will get:

$$f(1) - f(0) = \int_0^1 \frac{\ln(a+1)} {a^2 + 1} \, \mathrm{d}a - \int_0^1 \frac{\ln(a)} {a^2 + 1} \, \mathrm{d}a $$

(The last two terms of $f'(a)$ cancel each other after the integration so i didn't wrote them)

But then the two integrals on the right hand side are equal to $f(1) - f(0)$ so the differentiation under the integral led nowhere.

Do i need some other approach? Or did i made any mistake?
Any help is appreciated.

Body:
Even if I think that @Tavish's solution is the most efficient, wht you did must work.

$$f(a) = \int_0^1 \frac{\log(x+a)} {x^2 + 1}  \,dx$$
$$f'(a) = \frac{1} {a^2 + 1} \left(\log(a+1) - \log(a) - \ln(4)+ \frac{\pi}{4} a\right)$$ Now, you must use $a^2+1=(a+i)(a-i)$ to make
$$ \int f'(a)\,da=\frac{1}{8} \pi  \log \left(a^2+1\right)+\frac{1}{2} i \text{Li}_2(-i
   a)-\frac{1}{2} i \text{Li}_2(i a)-$$ $$\frac{1}{2} i
   \text{Li}_2\left(\left(\frac{1}{2}-\frac{i}{2}\right) (a+1)\right)+\frac{1}{2} i
   \text{Li}_2\left(\left(\frac{1}{2}+\frac{i}{2}\right) (a+1)\right)-$$ $$\frac{1}{2} i
   \log ((1+i)-(1-i) a) \log (a+1)+\frac{1}{2} i \log (-(1+i) (a+i)) \log
   (a+1)-$$ $$\log (a) \tan ^{-1}(a)-\frac{1}{2} \log (2) \tan ^{-1}(a)$$

$$ \int_0^1 f'(a)\,da=\frac{1}{2} (2 C-i (\text{Li}_2(1-i)-\text{Li}_2(1+i))+\pi  \log (2))-\frac{1}{2} i
   \left(\text{Li}_2\left(\frac{1}{2}+\frac{i}{2}\right)-\text{Li}_2\left(\frac{1}{
   2}-\frac{i}{2}\right)\right)=C+\frac{1}{8} \pi  \log (2)$$ Since
$$\int_0^1 \frac{\log(x)} {x^2 + 1}  \,dx=-C$$ then the result.


问:
I can't solve this integral ( $\int 3\sec^4(3x)\;dx$ ). Is there something wrong with the problem?
$\displaystyle\int 3\sec^4(3x)\;dx$
I couldn't solve this integral

I've tried a lot

But it don't know how to solve it

答:
Hint: substitutions (u- and trig).

First, use the substitution $u=3x, \text{d}u=3\text{d}x$ to make the original integral a more general form, $\displaystyle\int\sec^4 u \text{d}u$.

Notice that $\text{d}\tan u = \sec^2 u \text{d}u$ and $\sec^2 u=1+\tan^2 u$, we may write the integral as $\displaystyle\int(1+\tan^2 u) \text{d}\tan u$, which is simply $\tan u+\frac{1}{3}\tan^3 u+ C$. Substitute $u=3x$ back, we get the final answer, $\tan 3x+\frac{1}{3}\tan^3 3x+ C$.


问:
If $E$ is compact and the graph of $E$ is compact then $f$ is continuous.
This is a question from Rudin and I know there are many answers to the question but I am trying to solve rudin on my own and needed someone to go through the proof which I think is probably not correct.Any help to rectify it will be appreciated.

Here's my attempt :

Let $E$ be a compact set and the graph of $E$ is defined as $(x, f(x))$ which is compact. We need to show that $f$ is continuous.

Now, we assume that $f(x)$ is discontinuous at the point $x_1$.Then there is a sequence $(x_n)$ which converges to $x_1$ such that $f(x_n)$ is not convergent at $f(x_1)$ . Now we consider the sequence $(x_n,f(x_n))$ which has a convergent subsequence say $(x_{n_k},f(x_{n_k})$ which has to converge to $(x_1,f(x_1))$ . Since the original sequence $f(x_n)$ is discontinuous at $x_1$ so $|f(x_n)-f(x_1)| \ge \epsilon$ for all $n \ge K_1$.
However $f(x_{n_k}) \to f(x_1)$ so $|f(x_{n_k})-f(x_1)| < \epsilon$ for all $n_k \ge k \ge K_2$.
This is what the contradiction that I have found out. I know this is wrong and I need some one to rectify it.

答: First of all you using $x_1$ with two meanings.  Suppose $f$ is not continuous at $x$. Then there exists $r>0$ and  $x_n \to x$ such that $|f(x_n)-f(x)| \geq r$ for all $n$. Now, Since the graph is compact, $(x_n,f(x_n))$ has a subsequence $(x_{n_k}, f(x_{n_k}))$ converging to some point $(u,f(u))$ on the graph. This implies $x_n\to u$ so we must have $u=x$ and $f(x_{n_k}) \to f(u)=f(x)$. But $|f(x_{n_k})-f(x)| \geq r$ for all $k$ and we have arrived at a contradiction.


问:
Is the structure of "solid angle" conformal maps more interesting in higher dimensions than just conformal maps?
In the complex plane there exists a very large set of smooth conformal maps, (The famous Riemann Mapping Theorem states that between any two simply connected open sets in $\mathbb{C}$ there exists a biholomorphic mapping which is also conformal).

The situation is much more bleak as soon you go up to 3 dimensions and higher where the only smooth conformal maps are the mobius transformations. This is the famous Liouville's Theorem

I was thinking about trying to find a generalization of the concept of an "angle" so that the "set of all generalized-angle preserving maps" in $\mathbb{R}^3$ would have an interesting structure larger than merely mobius transformations.

There is the obvious generalization, solid angles. In 3 dimensions these are defined for collections of 4 points at a time (we select 1 anchor point and consider the 3 vectors formed by the anchor to the 3 other non collinear points. And then take the area of the spherical triangle spanned by these 3 vectors). The solid angle can be calculated explicitly using a formula given here.

So that leads to our question.... is the set of all smooth solid-angle-conformal functions from $\mathbb{R}^3 \rightarrow \mathbb{R}^3$
a larger set than the set of mobius transformations? And does there exist a generalization of the Riemann Mapping Theorem for them?

答:
The basic question here is a linear algebra question: which orientation-preserving linear maps $T:\mathbb{R}^3\to\mathbb{R}^3$ preserve solid angles formed by triples of vectors? (You are then asking about smooth maps whose derivative at every point is such a linear map.)  Let $G\subset GL_3(\mathbb{R})$ be the group of such maps, and let $H=\mathbb{R}^+\times SO(3)\subset GL_3(\mathbb{R})$ be the group of conformal linear maps (i.e., compositions of rotations and scalings).  It is clear that $H\subseteq G$; I claim that in fact $G=H$, so your maps are just the same as conformal maps.

Suppose $T\in G$; we wish to show that $T\in H$.  Let $e_1,e_2,e_3$ be the standard basis vectors for $\mathbb{R}^3$.  Taking a singular value decomposition of $T$, we may compose $T$ with rotations (which does not change whether $T\in H$) to assume that $T$ is diagonal.  Since $T$ is orientation-preserving, we can compose it with a further rotation to assume the diagonal entries are positive, and we can compose with a scaling to assume that $T(e_1)=e_1$.  We have $T(e_2)=be_2$ and $T(e_3)=ce_3$ for positive scalars $b$ and $c$; I claim that $b=c=1$ so $T$ is just the identity.

To prove this consider the solid angle formed by the unit vectors $e_1,\frac{e_1+e_2}{\sqrt{2}},$ and $e_3$.  $T$ maps these vectors to $e_1,\frac{e_1+be_2}{\sqrt{2}},$ and $ce_3$, which form the same solid angle as the unit vectors $e_1,\frac{e_1+be_2}{\sqrt{1+b^2}},$ and $e_3$.  But now observe that as $b$ grows from $0$ to $\infty$, the spherical triangle formed by these vectors grows strictly (we are taking the spherical triangle formed by $e_1,e_2,$ and $e_3$ and replacing the $e_2$ vertex with a point on the edge between $e_1$ and $e_2$).  So, the only way the area of this triangle would be the same as the area of the triangle formed by $e_1,\frac{e_1+e_2}{\sqrt{2}},$ and $e_3$ is if $b=1$.  Swapping the roles of $e_2$ and $e_3$, we similarly conclude that $c=1$.

Similar arguments show that for any $n\geq m\geq 2$, the orientation-preserving linear maps $T:\mathbb{R}^n\to\mathbb{R}^n$ that preserve "$m$-dimensional solid angles" formed by $m$ vectors are the same as the conformal linear maps.


答:
I acknowledge this answer is very similar to that of our colleague Ak, with perhaps some differences in formatting and added words of explanation.

We start with the given equations

$\dot S = -\dfrac{SI}{S + I + R}, \tag 1$

$\dot I = \dfrac{SI}{S + I + R} - I, \tag 2$

$\dot R = I - R; \tag 3$

we differentiate (1) (here I have for the moment used "${}^\prime$" in lieu of "$\dot {}$" on the right-hand side):

$\ddot S = -\dfrac{(SI)'(S + I + R) - (SI)(S + I + R)'}{(S + I + R)^2}, \tag 4$

whence

$\ddot S = -\dfrac{(\dot S I + S\dot I)(S + I + R) - (SI)(\dot S + \dot I + \dot R)}{(S + I + R)^2}; \tag 5$

we add (1) and (2):

$\dot S + \dot I = -I, \tag 6$

and to this add (3):

$\dot S + \dot I + \dot R = -R, \tag 7$

and substitute this into (5):

$\ddot S = -\dfrac{(\dot S I + S\dot I)(S + I + R) + RSI}{(S + I + R)^2}; \tag 8$

again from (1) and (2):

$\dot SI + S\dot I = -\dfrac{SI^2}{S + I + R} + \dfrac{S^2I}{S + I + R} - IS$
$= \dfrac{S^2I - I^2S}{S + I + R} - IS, \tag 9$

whence

$(\dot SI + S\dot I)(S + I + R) = S^2I - I^2S - IS(S + I + R)$
$= S^2I - I^2S - IS^2 - I^2S - ISR = -2I^2S - ISR; \tag{10}$

thus,

$(\dot SI + S\dot I)(S + I + R) + RSI = -2I^2S; \tag{11}$

finally we substitute this into (8) and execute some very simple algebraic maneuvers:

$\ddot S = \dfrac{2I^2S}{(S + I + R)^2} = \dfrac{2I^2S^2}{S(S + I + R)^2}, \tag{12}$

$\ddot S = \dfrac{2}{S} \left (\dfrac{IS}{(S + I + R)} \right )^2 = \dfrac{2}{S} \dot S^2, \tag{13}$

$OE\Delta$.

Title: Given $\frac{dS}{dt}=-\frac{SI}{S + I + R}$ and $R = -(SI + S + I)/ \frac{dS}{dt}$ Determine $\frac{d^2S}{dt^2}$

Body:
I am given a system of equations:

\begin{align}
\frac{dS}{dt} &= -\frac{SI}{S + I + R}\\
\frac{dI}{dt} &= \frac{SI}{S + I + R} -I \\
\frac{dR}{dt} &= I - R\\
\end{align}

I am to subsitute $R = -\left(SI + S + I\right) / \frac{dS}{dt} $ to determine $\frac{d^2S}{dt^2}$

I chose to work only with the first equation.
When I substitute what is given for $R$ into the denominator I get:

\begin{align}
S + I + R &= S + I + R = -\left(SI + S + I\right) / \frac{dS}{dt}\\
&=\frac{\frac{dS}{dt}(S+I)-(SI+S+I)}{\frac{ds}{dt}}
\end{align}

I substitute this into the equation for $\frac{dS}{dt}$ and I get:

\begin{align}
\frac{dS}{dt} &= \frac{-SI \left(\frac{dS}{dt}\right)}{\frac{ds}{dt}\left(S+I\right)-(SI+S+I)}\\
\left(\frac{dS}{dt}\right)^2(S+I)-\frac{dS}{dt}\left(SI+S+I\right) &= -SI\left(\frac{dS}{dt}\right)\\
\left(\frac{dS}{dt}\right)^2(S+I) &= \left(\frac{dS}{dt}\right)\left(S+I\right)\\
\left(\frac{dS}{dt}\right)^2 &= \left(\frac{dS}{dt}\right)\\
\end{align}

How do I get the desired result of
$$
\frac{d^2S}{dt^2}= \frac{2}{S}\left(\frac{dS}{dt}\right)^2
$$


问:
Solving parametric in rationals
I want to find all values of a such that the equation

$$\sqrt{a - x} = a - x^2$$

has at least one real root and none of its roots are irrational. I made some decent progress (I think) but stuck at the latest stages.

Here's what I deduced myself so far:

First, we must have $a \ge 0$ because on the left hand side we have a square root which should be non-negative and if $a < 0$ then the right hand side will be strictly negative since $x^2$ is non-negative. Now, for the special case when $a=0$ we have $\sqrt{-x} = -x^2$ which has one real solution $x=0$ and it satisfies the condition. So, $a=0$ works.

Let's now have $a>0$. I solve the equation directly this way:

Square both sides. We get $a-x = (a-x^2)^2$, so $a-x = a^2-2ax^2+x^4$ and this is quadratic in a :

$a^2-(2x^2+1)a+(x^4+x)=0$ for which we have $D=(2x^2+1)^2-4(x^4+x)=4x^2-4x+1=(2x-1)^2$ so it's a perfect square and we get the solutions in a :

$a_{1,2} =\frac{2x^2+1\pm\sqrt{D}}{2} = \frac{2x^2+1\pm(2x-1)}{2}$ (we don't have to use absolute value here because we got $\pm$ anyways). Now, considering both values separately we arrive at two different equations:

$x^2-x+(1-a)=0$ and $x^2-x-a=0$ and we're off to the final stage here

The final stage is to ensure that these equations have solutions and if they do, all of those are not irrational. First, the easier case is - when only one of this equations has a solution. We have:

$$D_{1} = 1 - 4(1-a) = 4a - 3$$
$$D_{2} = 1 - 4(-a) = 4a + 1$$
($D_{1,2}$ stand for the respective discriminants of these equations). We have a strict inequality $4a-3 \lt4a+1$ therefore if it is the case of only one equation having a real root, it must be the second one. This means we need to solve the following system of inequalities:
$$4a-3\lt0$$
$$4a+1\ge0$$
which yields $a\in(0,\frac{3}{4})$ - (as established above, a must be non-negative and we exclude $a=0$ since we handled that case earlier) for these values of a, so long as 4a+1 is a square of a rational number, things will work. We can set $4a+1=\frac{p^2}{q^2}$ where p and q have no common divisors and get $a=\frac{p^2-q^2}{4q^2}$ ; Now, substituting our restriction for a we get: $0\lt a\lt\frac{3}{4}$, so $0\lt 4a\lt3$, so $0\lt\frac{p^2-q^2}{q^2}\lt3$ and finally $q^2\lt p^2 \lt 4q^2$

Finally, the most interesting is the case when both of the equations will have a solution. Starting with the value of $a=\frac{3}{4}$ (the second discriminant is equal to 0). We can check that in this case the first discriminant is equal to $4\times\frac{3}{4}+1 = \frac{12}{4}+1 = \frac{16}{4} = 4$ which is a perfect square, hence the solutions will be rational numbers. Thus $a=\frac{3}{4}$ works. Now, let's consider $a\gt\frac{3}{4}$ In this case our goal is to make sure that both discriminants are perfect squares.

It is sort of "obvious" that there can be no solutions to this when a is a natural number. This is because $4a-3$ and $4a+1$ are separated by 4, but no perfect squares of natural numbers have such gap: the smallest one is $4-1=3$ followed by $9-4=5$ and then it will only be increasing, thus the difference of 4 between perfect squares of two natural numbers is an impossibility. (We can also prove that more rigorously if we set $D_{1}=n^2$ and $D_{2}=m^2$, then subtracting one from another and factoring it) ; This leaves us with the only possibility - that a is a rational number. Then a is expressible as $a=\frac{p}{q}$ in lowest terms (so p and q are co-primes). Then we have $4a-3=4\times\frac{p}{q}-3=\frac{4p-3q}{q}$ but since p and q are co-primes, then so are 4p-3q and q and since we want a perfect square of a rational number, the denominator must be itself a perfect square. Thus for $q=k^2$ for some integer k. Rewriting these both using k we get
$$4a-3=\frac{4p-3k^2}{k^2}$$
$$4a+1=\frac{4p+k^2}{k^2}$$
Since our denominators are perfect squares, then the numerators must be too. And we can also notice that the first numerator differs from the second one by $4k^2$ which itself is $(2k)^2$ , hence - a perfect square. In the end we have a Pythagorean triple of numbers $4p-3k^2, (2k)^2, 4p+k^2$ as a solution here and... that's it. I don't really know how to arrive at some closed form for a in terms of p and q . I tried with Euclid Formula, but I didn't go far with it. I know for sure that such solutions exist. For instance $\frac{p}{q}=\frac{21}{16}$ works, but I want to find all of the values of a in some closed form which make this work in the case when both equations have real roots.

Any ideas on how to finish it off? You can, of course, offer a completely different way of solving it, but I would also like to know how to complete my take on the problem.

Title: Solving parametric in rationals

Body:
You're on the right track with the discriminant but use quadratic equation.

$$\sqrt{a - x} = a - x^2\\
\implies a-x =a^2 - 2 a x^2 + x^4\\
\implies a^2 - 2ax^2 - a + x^4+x=0\\
\implies a^2 -(2x^2+1)a+(x^4+x)=0\\
\implies a=\frac{(2x^2+1)\pm\sqrt{(2x^2+1)^2-4(1)(x^4+x)}}{2(1)}\\
=\frac{(2x^2+1)\pm\sqrt{4 x^2 - 4 x + 1}}{2}\\
=\frac{(2x^2+1)\pm (2x-1)}{2}$$
$$\implies a_1=\frac{2 x^2 + 2 x}{2}=x(x+1)\qquad\quad x\in\mathbb{Q}
\quad (+)\tag{1}$$
$$\implies a_2=\frac{2 x^2 - 2 x + 2}{2}=x^2-x+1\quad x\in\mathbb{Q}
\quad (-)\tag{2}$$

\begin{align*}
\text{For example}\quad    x &\in\{
2/2,\space
2/3,\space
2/4,\space
2/5,\space
2/6,\space
2/7,\space
2/8,\space
2/9,\space
2/10,\space  \}\\
\implies a_1 &\in\{
  2,\space
  1\space1/9,\space
  3/4,\space
  14/25,\space
   4/9,\space
  18/49,\space
   5/16,\space
  22/81,\space
   6/25\} \\
\implies a_2 &\in\{
1,\space
7/9,\space
   3/4,\space
  19/25,\space
   7/9,\space
  39/49,\space
  13/16,\space
  67/81,\space
  21/25\}
\end{align*}

We can solve for $x$ to find valid $a$-values in terms of $a=p/q$
$$\sqrt{a-x} = a-x^2\implies x = \frac{\pm1\pm\sqrt{4 a + 1}}{2}$$
To be rational $\quad 4a+1\quad$ must be a perfect square.

$$\text{Let}\qquad\qquad 4a+1=\frac{4p}{q}+1=\frac{4p+q}{q}
=\sqrt{ \frac{4\frac{j}{k}+\frac{l}{m}}{\frac{l}{m}}}
=\sqrt{\frac{4 j m + k l}{k l}}\\
\implies  4 j m + k l=r^2\quad\land\quad  k l=s^2\quad r,s\in\mathbb{N}$$

Some examples of $j,k,l,m$ are

$$(1,1,1,6)\quad (1,1,1,12)\quad (1,1,1,20)\quad (1,1,1,30)\quad (1,1,1,42),\space m\uparrow 2n+4$$
$$(1,1,4,3)\quad (1,1,4,8)\quad (1,1,4,15)\quad (1,1,4,24)\quad (1,1,4,35),\space m\uparrow 2n+3$$
$$(1,1,9,4)\quad (1,1,9,10)\quad (1,1,9,18)\quad (1,1,9,28)\quad (1,1,9,40),\space m\uparrow 2n+2$$
$$(1,4,1,3)\quad (1,4,1,8)\quad (1,4,1,15)\quad (1,4,1,24)\quad (1,4,1,35)\quad $$
$$(1,4,4,5)\quad (1,4,4,12)\quad (1,4,4,21)\quad (1,4,4,32)\quad (1,4,4,45)\quad $$
$$(1,4,9,7)\quad (1,4,9,16)\quad (1,4,9,27)\quad (1,4,9,40)\quad (1,4,9,55)\quad $$
$$(1,9,9,10)\quad (1,9,9,22)\quad (1,9,9,36)\quad (1,9,9,52)\quad (1,9,9,70)\quad  $$

$$(2,1,1,3)\quad (2,1,1,6)\quad (2,1,1,10)\quad (2,1,1,15)\quad (2,1,1,21)\quad  $$

$$(2,1,4,4)\quad (2,1,4,12)\quad (2,1,4,24)\quad (2,1,4,40)\quad (2,1,4,60)\quad  $$

$$(2,4,4,6)\quad (2,4,4,16)\quad (2,4,4,30)\quad (2,4,4,48)\quad (2,4,4,70)\quad $$

$$(3,4,4,4)\quad (3,4,4,7)\quad (3,4,4,15)\quad (3,4,4,20)\quad (3,4,4,32)\quad $$

In the first four example I noted what I think is the increment between $m$-values as $m\uparrow\cdots$ but this interval changes with the values of $j,k,l$. Still, you may be able to find patterns in this data to help you understand what makes an $a$-value valid.

Let's take the case for $\quad(4,4,4,3)\quad $ where
$$x=\frac{\pm1\pm\sqrt{4 a + 1}}{2}
=\frac{\pm1\pm\sqrt{\sqrt{\frac{4 j m + k l}{k l}}   }}{2}
=\frac{\pm1\pm\sqrt{\sqrt{\frac{4(4)( 3) + 4 (4)}{4(4)}}}}{2}
\tag{3}
$$
We now substitute the innermost part
$\quad\sqrt{4a+1}= \sqrt{\frac{4(4)( 3) + 4 (4)}{4(4)}}=2$

We can now see that
$$x=\frac{\pm1\pm\sqrt{4 a + 1}}{2} =\frac{\pm 1 \pm2}{2}$$
is rational.

On the surface, it appears to me that

$$j\in\mathbb{N}\quad k,l\in\mathbb{N^2}\quad m=f(n)\\
\quad\text{where}\quad f(n)$$ is a function of a that will generate the first and subsequent numbers as a series including an interval such as those intervals I suggested the $\uparrow$. The function would include a starting point as a function of $j,k,l$ but I leave it up to you to figure out what that function may be.

We can see that finding $a$ as a function of $x$ is easy. It seems that we might be able to extrapolate a pattern from the data provided by my dual solutions for $a$ above. I used a spreadsheet to find these modest amounts of data. A computer program using these solutions could generate unlimited samples for your analysis in finding valid $a$-values.


问:
Proving Open/Closed sets
Let A = {$x \in \mathbb{R} | 2 \leq f^2(x) \leq 4 $}. Let B = {$(x,y) \in \mathbb{R}^2| 1 \leq |x| + |y| < 2$}. Prove if they're open,closed, or whatever they are.

Firstly, I would like to do an intuition check. {$a\leq b\leq c$}, will always be closed? And the same for {$a<b<c$}, will it automatically be open? And if {$a < b \leq c$}, is it not open nor closed?

For the set A, is it acceptable if I use 4 different sequences, to show that the complement is open? Also, is not knowing the points specifically acceptable for it to be a sequence? For example, $x_n := \{x|{f(x) \xrightarrow -2}$}. I wanted to use these 'gimmicky' sequences to show that A is open.

For B, is fixing one variable a legitimate sequence? For example, $(x=0,y_n) := (0,\{2- \frac{1}{n} \})$

答: An example for the 1st Q. Let $A$ be $any$ subset of $\Bbb R$ such that $2\in A.$ Let $f(x)=2$ if $x\in A.$ Let $f(x)=0$ if $x\in \Bbb R \setminus A.$

Title: Proving Open/Closed sets


问:
Language usage: "for" vs. "for all"
Every student of math learns about the universal quantifier "for all/each/every". On the other hand, in mathematical writing, it is fairly common to read e.g. "for $i=1, \ldots n$, let $\ldots$". I had long assumed that the "for" here is identical to "for each" (less commonly, it might more reasonably stand for the existential quantifier "for some"), but where the author is perhaps getting sloppy with their language.

Now, based on things I've encountered over time, I'm less sure about my initial understanding. (Something closely related is the usage of "for" as one of the standard looping structures in programming languages.) So I want to clear this up once and for all. Does the word "for" in this context have any meaning or usage separate from standing for the universal quantifier? To ask this in a different way, is there ever a situation where you can write "for $i=1,\ldots, n$ but not "for all $i=1,\ldots,n$"?

答:
No.  The use of "for" rather than "for all" here is stylistic; the second way simply sounds strange.

It is analogous to the standard use of "if" instead of "if and only if" when making definitions in mathematics, which is also something that people ask about (see here).  When we write "a set $G$ with a binary operation on it is called a group if [put axioms here]" the intended meaning is "... called a group if and only if..." but nobody writes the second version; it sounds very strange in English to write like that.

You might want to read here for a discussion of "any" vs. "all". That is a situation with some ambiguity for people learning mathematics, especially if English is not their native language.

Title: Language usage: "for" vs. "for all"


答:
Nope, they're the same. The clearest rephrasing in my opinion would be

for each $i\in\{1,...,n\}$, [stuff]

that is, explicitly say the "each" (or whatever) and shift from equality to elementhood. Another rephrasing would be

for each $i$, if $i=1$ or $i=2$ or ... or $i=n$ then [stuff],

which reveals the "disjunctive" nature of the notation "$i=1,...,n$."

Title: Language usage: "for" vs. "for all"

Body:
Every student of math learns about the universal quantifier "for all/each/every". On the other hand, in mathematical writing, it is fairly common to read e.g. "for $i=1, \ldots n$, let $\ldots$". I had long assumed that the "for" here is identical to "for each" (less commonly, it might more reasonably stand for the existential quantifier "for some"), but where the author is perhaps getting sloppy with their language.

Now, based on things I've encountered over time, I'm less sure about my initial understanding. (Something closely related is the usage of "for" as one of the standard looping structures in programming languages.) So I want to clear this up once and for all. Does the word "for" in this context have any meaning or usage separate from standing for the universal quantifier? To ask this in a different way, is there ever a situation where you can write "for $i=1,\ldots, n$ but not "for all $i=1,\ldots,n$"?


问:
Find a circle $O_{\rho}(0)$ containing all the roots of polynomial $z^{4} - 4z + 1$
Given polynomial
$$z^{4} - 4z + 1$$

How can I find the radius of the circle in which all of its roots lie? I was trying to find the exact forms of the roots, but it's too complicated.

答:
Claim:$\;$If $z\in\mathbb{C}$ is a root of $z^4-4z+1$, then $|z| < 2$.

Proof:

Suppose instead that we have $z^4-4z+1=0$ and $|z| \ge 2$.
\begin{align*}
\text{Then}\;\;&
z^4-4z+1=0
\\[4pt]
\implies\;&
z^4-4z=-1
\\[4pt]
\implies\;&
(|z|)(|z^3-4|)=1
\\[4pt]
\implies\;&
2(|z^3-4|)\le 1
\\[4pt]
\implies\;&
|z^3-4|\le {\small{\frac{1}{2}}}
\\[4pt]
\implies\;&
|z^3|-4\le {\small{\frac{1}{2}}}
\\[4pt]
\implies\;&
8-4\le {\small{\frac{1}{2}}}
\end{align*}
contradiction, which validates the claim.


问:
Computing the total variation of a function $f:[0,1] \to \mathbb{R}$ using uniform partitions
Let $f:[0,1] \to \mathbb{R}$ be a function of bounded variation and let

$$
V_0^1(f) = \sup \{ \ \sum_{i=1}^{N-1} |f(x_i)-f(x_{i+1})| : 0<x_1 < \cdots < x_N < 1 \ \}
$$

be its total variation where the supremum runs over all partitions of $[0,1]$.

Can we caculate the $V_0^1(f)$ using uniform partitions? That means we set
$$
x_i = \frac{i-1}{N-1}, \ \ i=1, \dots,N
$$
and
$$
v_N = \sum_{i=1}^{N-1} |f(x_i)-f(x_{i+1})|.
$$
Does the sequnce $(v_N)_N$ converges to $V_0^1(f)$ (maybe under stronger conditions such as continuity of $f$) ?

答:
This is a complement to the excellent answer by user RRL. I present a counter-example to show that continuity is essential for the result (proved in other answer) to hold.

Let $f:[a, b] \to\mathbb {R} $ be a function which is continuous and strictly increasing on $[a, b] $ and let $c\in(a, b) $. And consider another function $g:[a, b] \to\mathbb {R} $ such that $f(x) =g(x) $ for all $x\in[a, b]\setminus \{c\} $ and $f(c) \neq g(c) $. Then $g$ has a removable discontinuity at $c$ and is continuous at all other points of $[a, b] $.

It should be easy to observe that both $f, g$ are of bounded variation and $$V_a^b(g) =V_a^b(f) +2|f(c)-g(c)|$$ The  relation above should be obvious if one takes a partition which includes point $c$. Points close to $c$ and left of $c$ will lead to a term $|f(c) - g(c) | $ in the total variation for $g$ and similar is the case for points near and to the right of $c$.

Next note that if we try to find the limit of sum $$V(g, P) =\sum_{k=1}^{n}|g(x_k)-g(x_{k-1})|$$ corresponding to partition $$P=\{x_0,x_1,x_2,\dots,x_n\}$$ as norm of partition $P$ tends to $0$ then depending on whether the partition contains $c$ (or not) the limit will (or will not) include the extra $2|f(c)-g(c)|$ and thus the limit does not exist (one can say it oscillates finitely with two limit points $V_a^b(f) $ and $V_a^b(f) +2|f(c)-g(c)|$).

Hence we must ensure continuity of a function before we can use the limit via norm approaching $0$ to find total variation of the function.

This is one of the cases where the limit using norm of partition and limit using refinement of partition differ.

The above counter-example is given as a hint in the lecture notes by Christopher Heil. See Exercise 19 (the exercise has a typo as it is supposed to find the counter-example to theorem 17 and not theorem 18 as mentioned). IMHO the lecture notes are good and would be worth going through.


问:
Combinatorial Games
There are n coins on a table and 2 players take turns to remove 1,3 or 4 coins. A player wins if he removes the last coin. Find all winning and losing positions.

b.t.w. I am a little confused about whether 7 is a winning or losing position. Also does there ever exist an undetermined position in such combinatorial games?

答:
If you only have doubts for $n=7$, then notice that either $n=6,4,3$ are forced victories for the person who has the turn, in which case the first to go loses by force, or one of them is a forced loss for the person who has the turn, in which case the first to go may choose any of those (I advise the smallest, because maybe he has to go eat).

Similarly, there are no "unclear positions", because if all the positions with $n<k$ have been determined, then so is the one with $k$ coins, and (strong?) induction concludes (notice that $k=0$ is a forced loss).

Title: Combinatorial Games


问:
Integrate $\int_0^1\frac{\ln(1-x)\ln(x+x^2)}{1+x^2}dx $
I have been trying to derive

$$\int_0^1\frac{\ln(1-x)\ln(x+x^2)}{1+x^2}dx = \frac{\pi^3}{64} +\frac\pi{16}\ln^22-G\ln2$$

with $G$ being the the Catalan constant.

I noticed that a similarly-looking integral is posted and solved here. Although the solution is applicable and meritorious in itself, it seems an overkill to resort to the special function $\operatorname{Li}_3(z)$ given the elementary result.

Body:
Write the numerator of the integrand as
$$\ln(1-x)\ln(x+x^2)=\ln(1-x)\ln x+\ln(1+x)\ln\frac{1-x}{1+x}+\ln^2(1+x)
$$
and, correspondingly

$$I=\int_0^1\frac{\ln(1-x)\ln(x+x^2)}{1+x^2}dx
= I_1+I_2+I_3
$$
where
\begin{align}
I_1 &= \int_0^1\frac{\ln(1-x)\ln x}{1+x^2}dx
= \frac12\int_0^1\frac{\ln^2(1-x)+\ln^2x-\ln^2\frac x{1-x}}{1+x^2}dx\\
I_2 &=\int_0^1\frac{\ln(1+x)\ln \frac{1-x}{1+x}}{1+x^2}dx
\overset{ \frac{1-x}{1+x} \to x} = -\int_0^1\frac{\ln x\ln(1+x)}{1+x^2}dx-G\ln2 \\
I_3 &= \int_0^1\frac{\ln^2(1+x)}{1+x^2}dx
= \int_0^\infty\frac{\ln^2(1+x)}{1+x^2}dx- \int_1^\infty \frac{\ln^2(1+x)}{1+x^2} \overset{x\to 1/x} {dx }\\
&= \int_0^\infty\frac{\ln^2(1+x)}{1+x^2}dx- \left(
I_3 + \int_0^1 \frac{\ln^2x}{1+x^2}dx-2 \int_0^1 \frac{\ln x\ln(1+x)}{1+x^2}dx\right)\\
&=\frac12 \int_0^\infty\frac{\ln^2(1+x)}{1+x^2}dx
 - \frac12 \int_0^1\frac{\ln^2 x}{1+x^2}dx
+\int_0^1 \frac{\ln x\ln(1+x)}{1+x^2}dx
\end{align}
Then, add up the three integrals above to obtain
\begin{align}
I=& \ I_1+I_2+I_3\\
=& \ \frac12 \int_0^1\overset{t=1-x}{\frac{\ln^2(1-x)}{1+x^2}}dx
 +\frac12\int_0^\infty\overset{t=1+x}{ \frac{\ln^2(1+x)}{1+x^2}}dx
 - \frac12\int_0^1\overset{t=x/(1-x)}{ \frac{\ln^2\frac x{1-x}}{1+x^2}}dx-G\ln2 \\
=& \ 2\int_0^\infty \underset{t^2\to 2t}  {\frac{t\ln^2t}{t^4+4}dt } -G\ln2
=\frac{\pi^3}{64} +\frac\pi{16}\ln^22-G\ln2
\end{align}


问:
Check whether a given graph is bipartite except for one edge
Given any graph $G=(V,E)$, how to decide if deleting just one of its edges can make it bipartite?

Trivial solution: check for each edge whether deleting it makes the graph bipartite. This is in $O(|E|\cdot(|E|+|V|))$.

Are there more efficient ways?

答:
First, find an odd cycle. Then, try sequentially deleting and re-inserting each of the edges of the cycle, keeping track of whether any of the deletions produces a bipartite graph, using the dynamic bipartiteness testing algorithms from

"Sparsification—A technique for speeding up dynamic graph algorithms".
D. Eppstein, Z. Galil, G.F. Italiano, and A. Nissenzweig.
J. ACM 44 (5): 669–696, 1997. https://dl.acm.org/doi/abs/10.1145/265910.265914

or

"Randomized fully dynamic graph algorithms with polylogarithmic time per operation". Henzinger and King. J. ACM, 1999. https://dl.acm.org/doi/abs/10.1145/320211.320215

The 1997 paper should give you $O(m + n^{3/2})$ and the 1999 one should give you randomized expected time $O(m\log^3 n)$. There's also a non-random dynamic algorithm from

"Poly-Logarithmic Deterministic Fully-Dynamic Algorithms for Connectivity, Minimum Spanning Tree, 2-Edge, and Biconnectivity". Jacob Holm, Kristian  De Lichtenberg, and Mikkel Thorup. J. ACM 2001. https://dl.acm.org/doi/abs/10.1145/502090.502095

that is probably something like $O(m + n\log^{O(1)} n)$ but they only mention bipartiteness in passing in a single line so I am not entirely sure of what their bounds are. Some of their polylogs have been subsequently improved by other papers but I didn't see bipartiteness listed among the improvements.


问:
Let $ a \in (0,1) $ and S = $ \{a^{n}\}_{n=1}^{\infty}$. Then $ \inf S = 0$
I was proving the statement, 'Let $ a \in (0,1) $ and S = $ \{a^{n}\}_{n=1}^{\infty}$. Then $ \inf S = 0$'

In that process, I need to prove other statement, but I couldn't. It is following one.

Given a number $\varepsilon \gt 0 $, there exists a natural number $n$ such that $a^{n} \lt \varepsilon $.

To prove it, I divided by the case,

(i) $\varepsilon \ge 1 $ (ii) $ a \lt \varepsilon \lt 1 $ (iii) $\varepsilon = a $ (iv) $ 0 \lt \varepsilon \lt a $

However, I can't prove the case (iv)

If someone knows how or other way to prove it, please help me.
Thank you.

答:
Here is an elementary argument: By the Bernoulli inequality we have $$\frac{1}{a^n}=(1+(\frac{1}{a}-1))^n\geq1+n(\frac{1}{a}-1)\geq n(\frac{1}{a}-1)$$
Therefore: $$a^n\leq K\frac{1}{n}$$
with $K=\frac{a}{1-a}>0$. By the Archimedian property there is some $n\in \Bbb N$ with $$\frac{K}{\varepsilon}<n$$
Can you finish from here?


问:
Any way to simplify this gcd totient function
I have the following expression

$$\frac{gcd(a,b)}{\varphi(gcd(a,b))}$$

$a,b$ are known positive integers. Is there any way to rephrase this or simplify it?

答:
In general, $$\frac{n}{\phi(n)} = \prod_{p|n} \frac{p}{p-1}$$

where the product is over all primes that divide $n$. So in your case, with $n=(a,b)$ this becomes:

$$\prod_{p|a \land p|b} \frac{p}{p-1}$$

In don't think there is any better formula. Since $\gcd(a,b)$ can be any natural number, there is no real trick.


问:
How to find $f(x)$ such that $f(y)-f(x)-(y-x)f'(\frac{x+y}{2})+\frac{(x-y)^2}{4}=0$

Given $f(x)$ to be differentiable such that $f(y)-f(x)-(y-x)f'(\frac{x+y}{2})+\frac{(x-y)^2}{4}=0$, then find $f(x)$

From the given equation, we can't plug in $x=y$ as it just gives $0=0$. For some reason I think $f(x)$ is a quadratic equation ( I dunno why tho ). So I set $f(x)=ax^2+bx+c$ and put it in the given equation.
\begin{equation}
ay^2+by+c-(ax^2+bx+c)-(y-x)[a(x+y)+b]+\frac{x^2-2xy+y^2}{4}=0\\
\Rightarrow x^2-2xy+y^2=0\\
\Rightarrow x=y
\end{equation}

So does that mean for all quadratic equation it satisfies this functional equation?.

答:
I'm assuming here the equation has to be satisfied for all $x,y$. Showing that $f$ being quadratic implies $x=y$ means that $f$ cannot be quadratic, as we could choose $x\neq y$.

As the problem is currently stated there is no such $f(x)$. First, let $x=a, y=b$.
The equation becomes
$$f(b) - f(a) - (b-a)f'\left(\frac{a+b}{2}\right)+\frac{(a-b)^2}{4} = 0$$

Then, letting $x=b, y=a$, we have
$$f(a) - f(b) - (a-b)f\left(\frac{b+a}{2}\right)+\frac{(b-a)^2}{4} = 0$$

Adding the two together, we get
\begin{align}
&\frac{(a-b)^2}{2} = 0\\
\implies& a=b
\end{align}

Which means we have to choose $x=y$ regardless of $f$. If we relax the problem so that the equation only has to be satisfied for $x=y$, it becomes trivial as you have shown.


问:
Compactness implies limit point compact
In the proof of theorem 28.1 of Munkres that says compactness implies the limit point compact.

Proof: Let X be a compact. Given a set A of X, we wish to prove that if A is infinite, then A has a limit point. We prove the contraposition-if A has no limit point, then A must be finite.
So suppose A has no limit point. Then A contains all its limit points, so that A is closed. For each a in A we can choose a neighborhood U_{a} of a that intersect A in a alone. ...

My question is why we can choose such a neighborhood that does not intersect A-{a}?In other word, does such a neighborhood exists for each a in A?

答:
$x$ is a limit point of $A$, by definition if

for all (open) neighbourhoods $U$ of $x$, $U \cap (A\setminus\{x\}) \neq \emptyset$ or every (open) neighbourhood $U$ of $x$ contains a point in $A$ that is not equal to $x$.

So negating this

There is an (open) neighbourhood $U_x$ of $x$ such that the only possible point of intersection with $A$ is $x$, which says that $A \cap U_x = \emptyset$ or $A \cap U_x = \{x\}$, depending on whether $x \in A$ or not. In short, we conclude $A \cap U_x \subseteq \{x\}$. No need to observe that $A$ would be closed, we can then just apply compactness directly to the cover $\{U_x: x \in X\}$ and get a similar contradiction with the infiniteness of $A$.

In fact this allows us to generalise a bit:

Call $x \in X$ a strong limit point of an infinite set $A$ if for all (open) neighbourhoods $U$ of $x$ we have $|U \cap A|=|A|$. ($|\cdot|$ denoting set cardinality..) Then if $X$ is compact we have that every infinite $A$ has a "strong limit point" in $X$. In this general formulation this property of $X$ is even equivalent to compactness which limit point compactness is not.

Title: Compactness implies limit point compact


问:
Taylor expansion of a function
I would like to approximate the function

$f(x)=\frac{2x}{1-e^{-2x}}$

analytically for both small and large $x$.
But when I use the formula for the Taylor expansion, I run into the problem that the function and its derivative are not defined for $x=0$. How can I get around this problem?

For large $x$, my idea was to substitute $y:=1/x$ and then expand the function $g(y)=\frac{2}{y(1-e^{-2x})}$ around $y=0$. However, here I run into the same problem as above.

How do you proceed in such a case?

答:
Manipulate the denominator to show that

$$\frac{1}{1-e^{-2x}}=\frac 12 (1+\coth(x))$$
$$f(x)=\frac{2x}{1-e^{-2x}}=x(1+\coth(x))=\sum_{m=0}^\infty \frac{B^+_n }{n!}x^n=\sum_{m=0}^\infty \frac{2^n B_n(1)}{n!}x^n$$

Title: Taylor expansion of a function