问:
Find $z$ such that $|z-w| - |z-w^2|$ is maximal where $z$ lies on a given ellipse

Find $z$ such that $|z-w| - |z-w^2|$ is maximal where $z$ lies on the curve $|z-4| + |z+4| = 10$, where $w$ is a non real cube root of unity.

I found the locus of the $z$ from the given equation, which is an ellipse but am unable to proceed after it.

The locus that I found was $ \frac{x^2}{25} + \frac{y^2}{9} = 1$ where I assumed $z = x + iy$ .

I am a class 12 student who came across this problem in one of the online exams that I was supposed to give for my coaching.

答:
Denote $F_1, F_2$ the points with complex coordinates $F_1(-4),F_2(4)$ or, if you prefer, $F_1(-4+0i),F_2(4+0i).$ Then $$|z-4|+|z+4|=10$$ is equation of an ellipse with foci $F_1, F_2$ and $2a=10.$

Cubic roots of unity are $1,w,w^2,$ where $w=e^{i2\pi/3}=-\frac{1}{2}+i\frac{\sqrt3}{2}.$

For a fixed real $m$ satisfying $\;0<|2m|<|w-w^2|,$ $$|z-w| - |z-w^2|=2m \tag{H}$$ is equation of one branch of hyperbola with foci in $w$ and $w^2.$

We want to maximize $|z-w| - |z-w^2|,$ thus we assume that $m>0$ and we maximize $m.$ The corresponding branches of hyperbolas are lower.

The common center of hyperbolas lies inside the given ellipse, therefore each branch of hyperbola cuts the ellipse in two pints. Note that the distance of vertices of the hyperbola $(H)$ is $2m.$ The value $m$ increases when the vertices approach $w$ and $w^2$ becomes $m.$

Maximal value of $m$ is obtained when the hyperbola is degenerate (green line in the figure). In this case $2m=|w-w^2|.$ Only one point on the ellipse satisfies
$$|z-w| - |z-w^2|=|w-w^2|=\sqrt3$$

The point on the ellipse is the green point in the figure.

Can you find its complex coordinate?


问:
write an expression for the nth term of the sequence....
write an expression for the nth term of the sequence 0, 7, 16, 27, 40
It is neither geometric or arithmetic

My teacher gave us a key and I can't read his writing...please any help would be great!!

ContentLicense: CC BY-SA 3.0

Body:
This is a sequence defined by (n+6)n:

(0+6)0, (1+6)1, (2+6)2, (3+6)3, ..., (n+6)n


问:
Surjection Vs Surjective geometric morphism
Is it true that a map between ${\bf T1}$ topological spaces $f:X \to Y$ is surjective iff the induced geometric morphism $f:Sh(Y) \to Sh(X)$ is a surjection (i.e. its inverse image part $f^*$ is faithful)?

In "Sheaves in Geometry and Logic" a proof is given, but the the "if" part leaves me a bit unsatisfied (and, btw, the "only if" part holds for any topological space).
Thanks in advance!

答:
Here is some elucidation on Mac Lane and Moerdijk's proof.

"Suppose a point $y \in Y$ is not in the image of $f$."
This makes $f^{-1}(Y - \{y\}) = f^{-1}(Y) = X$, since all points of $X$ are mapped into $Y - \{y\}$.

"Since $Y$ is assumed to be $T_1$, $y$ is a closed point, so $Y - \{y\}$ is open."
That's because all points are closed in $T_1$ spaces.

"Therefore $Y - \{y\} = Y$, since $f^{-1}$ is faithful."
As @Zhen_Lin mentioned in the comments, we actually need the extra information that $f^{-1}$ is conservative to reach this conclusion, or at least that $f^{-1}$ is injective on subobject lattices, so you were right to be skeptical. While this is proved in Lemma VII.4.3 of the book using subobject classifiers, it is much easier to observe that faithful functors reflect monomorphisms and epimorphisms, and so the fact that $f^*$ is faithful and that a morphism in a topos is an isomorphism if and only if it is monic and epic implies that $f^*$ is conservative (and hence so is its restriction $f^{-1}$).

The strategy for an arbitrary space is to consider the stalks across all points, showing that for $\alpha,\beta:E \to F$ in $\mathrm{Sh}(Y)$ with $f^*(\alpha) = f^*(\beta)$ we have $\alpha_y = \beta_y$ for all points $y$ and hence $\alpha = \beta$, because everything is determined by the values on stalks. As you and they point out, holds for any space.

ContentLicense: CC BY-SA 4.0


问:
Notation Confusion (Specifically: "Such That", & Multiple $\exists$)
There are a couple of notation-related matters that I find myself clumsily working around quite often when writing proofs, and I was hoping for some help with dealing with them:

Let's say that $L$, $L_1$ and $L_2$ are regular languages.

I claim:

$w \in (L_1 \cup L_2) \cdot L \iff \exists w_1 \in (L_1 \cup L_2), w_2 \in L$ such that $w=w_1\cdot w_2$.

So I have two questions about this:

How should I handle that "there exist several $w_i$ such that..."? That is to say: is it okay to separate them with a comma as I did ($\exists w_1 \in ..., w_2 \in ...$), or would it be better to add a new $\exists$ per $w_i$ (e.g. $\exists w_1 \in ... \wedge \exists w_2 \in ...$)?

Is there any kind of standard notation to use as a substitute for "such that"? I gather that $|$ and $:$ are used in set-builder notation and not in cases such as this one; so it'd be off to write something like: "$w \in (L_1 \cup L_2) \cdot L \iff \exists w_1 \in (L_1 \cup L_2), w_2 \in L: w=w_1\cdot w_2$". Is that correct?

Looking at some questions & answers on SE and on Google, my understanding was that the common consensus seems to be that writing "such that" is the preferred method, usually, but my native language is written RTL, and that makes it a bit awkward sometimes when I want to add "such that" in the middle of a LTR mathematical statement (I usually find myself adding line breaks in unfortunate places to deal with it), so I was hoping for some clear, creative solution based on standard notation for this.

Does it look okay if I write something like: "$\exists w_1, w_2$ such that $w_1\in ... \wedge w_2 \in ... \wedge w = w_1 \cdot w_2$"? That is, to concentrate all the "exists" in one place, and all their "descriptions" in another?

I get a strong sense that I'm overthinking it all, but I dislike the way a lot of my notation usages feel so clumsy to me. If you can see some sort of root cause (maybe a mindset issue or something more general like that) for my confusion & are able to point it out, that'd be great, too.

答:
I’d write your example as

$$w \in (L_1\cup L_2)\cdot L \iff \exists w_1 \in (L_1\cup L_2): \exists w_2 \in L: w_1\cdot w_2 = w$$

The “such that “ is implicit in the final “:” (after the quqtifiers, before the term) for me, it need not be written out in natural language, and I prefer to give each variable it’s own quantifier, though other texts will do it differently. It depends also on how strictly “first order logic” you want to be, some will even not like the bounded quantifers and replace a statement like $\exists x \in A:P(x)$ by $\exists x: (x \in A) \land P(x)$ as well.

Like I said, it’s a matter of taste, largely. Just make sure it’s understandable.


问:
How to deduce trigonometric formulae like $2 \cos(\theta)^{2}=\cos(2\theta) +1$?
Very important in integrating things like $\int \cos^{2}(\theta) d\theta$ but it is hard for me to remember them. So how do you deduce this type of formulae? If I can remember right, there was some $e^{\theta i}=\cos(\theta)+i \sin(\theta)$ trick where you took $e^{2 i \theta}$ and $e^{-2 i \theta}$. While I am drafting, I want your ways to remember/deduce things (hopefully fast).

[About replies]

About TPV's suggestion, how do you attack it geometrically??

$\cos^{2}(x) - \sin^{2}(x)=\cos(2x)$

plus $2\sin^{2}(x)$, then

$\cos^{2}(x)+\sin^{2}(x)=\cos(2x)+2\sin^{2}(x)$

and now solve homogenous equations such that LHS=A and RHS=B, where $A\in\mathbb{R}$ and $B\in\mathbb{R}$. What can we deduce from their solutions?

答:
For Proving $\sin(\alpha+\beta)=\sin\alpha\cdot \cos\beta + \cos\alpha \cdot \sin\beta$ you can see this link:

http://www.math.wisc.edu/~leili/teaching/math222s11/problems/quizzes/trig.pdf

By the above by substituting $\beta=\alpha$ you have

$\sin{2\alpha} = \sin\alpha \cdot \cos\alpha + \cos\alpha \cdot \sin\alpha =2 \sin\alpha \cos\alpha$

$\sin{2\alpha} = \frac{2 \tan\alpha}{1+\tan^{2}\alpha} = \displaystyle 2\tan{\alpha}\cdot\cos^{2}\alpha=2\sin\alpha \cdot \cos\alpha$

We have $\cos(\alpha + \beta) = \cos\alpha\cdot \cos\beta - \sin\alpha \cdot \sin\beta$. Using this you can obtain the following:

$\cos{2 \alpha} = \cos\alpha\cdot \cos\alpha - \sin\alpha\cdot \sin \alpha =\cos^{2} \alpha - \sin^{2} \alpha$

$\cos{2 \alpha}= 1 - 2 \sin^{2}\alpha \qquad \Bigl[ \because \ \sin^{2}\theta = 1 - \cos^{2}\theta \Bigr]$

$ \cos{2 \alpha}=2 \cos^{2}\alpha - 1$

$\cos{2 \alpha}=\displaystyle \cos^{2}\alpha \cdot \Bigl[1-\frac{\sin^{2}\alpha}{\cos^{2}\alpha}\Bigr]=\displaystyle \frac{1-\tan^{2}\alpha}{1+\tan^{2}\alpha} \  \Bigl[\because \ \cos^{2}\alpha = \frac{1}{\sec^{2}\alpha}\Bigr]$.

Using the above formulas for $\sin(\alpha + \beta)$ and $\cos(\alpha + \beta)$ we have $$\tan(\alpha+\beta) = \frac{\sin(\alpha+\beta)}{\cos(\alpha+\beta)} = \frac{\sin\alpha\cdot \cos\beta + \cos\alpha \cdot \sin\beta}{\cos\alpha\cdot \cos\beta - \sin\alpha\cdot \sin\beta}$$

Dividing the last term by $\cos\alpha \cdot \cos\beta$ gives, $$\tan(\alpha + \beta) = \frac{\tan\alpha + \tan \beta}{1 - \tan\alpha \cdot \tan\beta}$$

From this if you put $\beta = \alpha$ you get $$\tan{2\alpha} = \frac{\tan\alpha + \tan\alpha}{1 - \tan\alpha \cdot \tan\alpha} = \frac{2 \tan \alpha}{1 - \tan^{2}\alpha}$$

Now moving on to formulas for $3\alpha$.

\begin{align*}
\sin{3\alpha} &= \sin(2\alpha + \alpha) = \sin(2\alpha)\cdot\cos{\alpha} + \cos(2\alpha)\cdot\sin\alpha \\ &= 2\sin\alpha\cdot \cos^{2}\alpha + (\cos^{2}\alpha - \sin^{2}\alpha) \cdot \sin\alpha \\ &= 2\sin\alpha \cdot (1-\sin^{2}\alpha) + (1-\sin^{2}\alpha)\sin\alpha - \sin^{3}\alpha \\ &= 3\sin{\alpha} - 4 \sin^{3}\alpha
\end{align*}

Again proceed similarly for $\cos{3\alpha}$ and convert all the $\sin^{2}\alpha$'s which appear in the expression to $(1-\sin^{2}\alpha)$. Then you get the value as:

$\cos(3\alpha) = 4\cos^{3}\alpha - 3 \cos\alpha$.

$\displaystyle \tan{3\alpha} = \frac{3\tan\alpha - \tan^{2}\alpha}{1-3\tan^{2}\alpha}$. Look here

ContentLicense: CC BY-SA 2.5


问:
Quadratic polynomials describe the diagonal lines in the Ulam-Spiral
I'm trying to understand why is it possible to describe every diagonal line in the Ulam-Spiral with an quadratic polynomial $$2n\cdot(2n+b)+a = 4n^2 + 2nb +a$$ for $a, b \in \mathbb{N}$ and $n \in 0,1,\ldots$.

It seems to be true but why?

Wikipedia says: "The pattern also seems to appear even if the number at the center is not 1 (and can, in fact, be much larger than 1). This implies [WHY?] that there are many integer constants b and c such that the function: $4n^2+bn+c$ as $n$ counts up $\{1, 2, 3, ...\}$, a number of primes that is large by comparison with the proportion of primes among numbers of similar magnitude."

I can't find a source with a detailed explanation.

I found these equations:

So here is the solution:

\begin{align*}
 y_t - y_{t+1} - (y_{t+1} - y_{t+2}) &= 8\\
 y_t - 2y_{t+1} + y_{t+2} &= 8\\
 y_{t+2} - 2y_{t+1} + y_t &= 8
\end{align*}

1) We solve $y_{t+2} - 2y_{t+1} + y_t = 0.$

Let $y_t = A\beta^t$ \begin{align*}
 A\beta^{t+2} - 2A\beta^{t+1} + A\beta^t &= 0\\
 A\beta^{t}\cdot (\beta^2 - 2\beta + 1) &= 0
\end{align*}
$\beta^2 - 2\beta + 1 = 0$ has two identical solutions $\beta_{1,2} = 1$.
So with $A_1$ and $A_2t$ we get $$y_t = A_1 + A_2t.$$

2) $1 + a_1 + a_2 = 0$ and $a_1 = -2$ so let $y_t = ct^2$
\begin{align*}
 c\cdot(t+2)^2 - 2c\cdot(t+1)^2 + ct^2 &= 8\\
 c\cdot\big(t^2+4t+4 - 2\cdot(t^2+2t+1) + t^2\big) &= 8\\
 c\cdot(t^2+4t+4 - 2t^2-4t-2 + t^2) &= 8\\
 2c &= 8\\
 c &= 4
\end{align*}
So $y_t = 4t^2$

3) The complete solution is $$y_t = 4t^2 + A_2t + A_1.$$

The "exclusion lines" seem to be interesting too:
$$4n^2+n$$ $$4n^2+3n$$ $$4n^2+3n-1$$ $$4n^2-n$$ seem not to have any primes at all.

Useful website I found a bit late: http://ulamspiral.com

答:
Reading georgmierau comment, I've been trying to reach this "arbitrary" 8, so here what I've done:

Let $A(n)$ be the cardinal number of the $n^{th}$ ring in Ulam's spiral.
$$A(1)=1\\A(2)=8=(2\cdot2-1)^2-1^2\\A(3)=16=(2\cdot3-1)^2-3^2\\ \vdots \\ A(n)=(2n-1)^2-(2n-3)^2\Rightarrow A(n)=8n-8, n\neq1$$

So, $A(n+1)-A(n)=8$

That means that each time we pass to the next ring, its cardinal number will be the same of the last one added by 8.


问:
The tangent mapping in algebra and algebraic geometry - when does it exist?
In differential geometry, for a smooth map $f:M\to N$ one gets the pushforward of the tangent sheaf $T_f:TM\to f^* TN$. Question is does the same work in algebra, that is, if I have a map $f:A\to B$ of commutative algebras, does it yield a map $Tf: Der (B)\to B\otimes_A Der(A)$ without additional requirements like the module of Kahler differentials being dualizable or something like that? I guess in the later case it should be a consequence of the usual map $B\otimes_A \Omega_A\to \Omega_B$.

Edit: I just found a very general solution to the problem: Dual Commutes with Base Change

答:
You may find this in Matsumuras book om commutative rings: If $k\rightarrow A \rightarrow^f B$ are maps of commutative unital rings, there is always a cotangent sequence

$$C1.\text{  } B\otimes_A \Omega^1_{A/k}\rightarrow \Omega^1_{B/k} \rightarrow \Omega^1_{B/A}\rightarrow 0,$$

and dualizing you get the following sequence:

$$ 0 \rightarrow Der_A(B) \rightarrow Der_k(B) \rightarrow Hom_B(B\otimes_A \Omega^1_{A/k},B).$$

If $A\rightarrow B$ is flat there is (Mat, 7.11) an isomorphism

$$Hom_B(B\otimes_A \Omega^1_{A/k},B) \cong B\otimes_A Der_k(A).$$

You get the "tangent sequence"

$$ Tf: Der_k(B) \rightarrow B\otimes_A Der_k(A).$$

The sequence C1 exists in complete generality and for this reason people usually use C1.

Example 1. Let $k$ be an algebraically closed field and let $X:=Spec(B), Y:=Spec(A)$ be non-singular varieties of finite type over $k$ in the sense of Hartshorne Chapter III Proposition 10.4. Assume the induced map $f: X\rightarrow Y$ is smooth of relative dimension $n:=dim(X)-dim(Y)$. Then $f$ is flat and you can define the tangent map. The tangent map $Tf$ is surjective at the level of tangent spaces: If $x\in X$ and $y:=f(x)$ it follows

$$Tf_x: T_x(X) \rightarrow f^*T_y(Y)$$

is surjective. Hence the algebraic geometric situation is similar to the differential geometric situation. In many cases you want to study maps that are non-flat and this is why you use C1 in general.

Note: The map $f$ is smooth of relative dimension $n$ iff $\Omega^1_{B/A}$ is locally free of rank $n$, and then you may dualize. There is a canonical isomorphism of $B$-modules

$$ \Omega^1_{B/A}\cong (\Omega^1_{B/A})^{**}$$

between the relative cotangent module and its double dual. Hence the cotangent module and tangent module determine each other.


问:
Inequalities on convexity
Let $f$ be second-order differentiable on $(-\infty,+\infty)$, and $|f(x)|^3$ be convex on $(-\infty,+\infty)$. Proved that for any $x$, $$f(x)f''(x)+2[f'(x)]^2\geqslant0.$$
I try to deform the inequality,and got it
$$f(x)f''(x)+(f'(x))^2\geqslant-(f'(x))^2\implies(f(x)f'(x))'\geqslant-(f'(x))^2$$
But then I don't know how to deal with it, and I don't know what "$|f(x)|^3$ be convex" can do...

答:
The idea is that either $f(x) = 0$ (in which case the inequality holds trivially) or $|f(x)| = f(x)$ or $|f(x)| = -f(x)$ in a neighbourhood of $x$ (in which case we can compute the second derivative).

Consider first the case that $f(x_0) > 0$. Then $|f(x)|^3 = f(x)^3 > 0$ for all $x$ in an open interval $I$ containing $x_0$. It follows that
$$
 0 \le \frac{d^2}{dx^2} (f(x))^3 = 3f(x) \bigl(2f'(x)^2 + f(x) f''(x) \bigr)
$$
for all $x \in I$, and in particular $2f'(x_0)^2 + f(x_0) f''(x_0) \ge 0$.

The case $f(x_0) < 0 $ works similarly.

Finally, if $f(x_0) = 0$ then $2f'(x_0)^2 + f(x_0) f''(x_0) = 2f'(x_0)^2 \ge 0$.

Title: Inequalities on convexity


问:
Since the Curvature tensor depends on a connection (not metric), is it the relevant quantity to characterize the curvature of Riemannian manifolds?
1) The definition of the Riemann curvature tensor does not include a metric. So, if we have a smooth manifold(not a Riemannian manifold), we can define the Riemannian curvature tensor for it by just giving it a connection(not the Levi-Civita connection). No metric is needed.

Now, if we also assign a metric to the smooth manifold, we can take traces of the Riemann curvature tensor and get the Ricci scalar. Does this imply that, for a Riemannian manifold(not just smooth), the relevant quantity that measures the curvature is the Ricci scalar and not the full Riemannian curvature tensor? Because to define the Riemann curvature tensor we need a connection(not a metric) and to define the Ricci scalar we also need a metric.

2) Lastly, since the Riemann curvature tensor depends on the connection and not the metric and the connection gives the way to parallel transport vectors, does it mean that parallel transporting the same vector along the same closed curve on two different Riemannian manifolds that correspond to the same smooth manifold (but we assign the same connection but different metric to each one), we will get the same angle of rotation for that vector at the end-point(which is also the starting-point) of the curve?

Note: When talking about a connection, I do not mean the Levi-Civita connection which comes from a metric. The question is about the difference in the role that the connection and metric play(independently) in the Riemann curvature tensor.

答:
This question is somewhat old and has already received answers - one thing I feel the need to point out is that while the curvature tensor (of any connection) characterizes the parallel transport only, it is fairly easy to see in a somewhat intuitive manner why the Levi-Civita (LC) connection also characterizes metric properties.

Consider a manifold $M$ with metric $g$, and a linear connection $\nabla$. Suppose that

$\nabla$ is flat, i.e. $R(X,Y)Z=0$;

$\nabla$ is metric compatible, i.e. $\nabla g=0$;

$\nabla$ is torsionless, i.e. $T(X,Y)=\nabla_XY-\nabla_YX-[X,Y]=0$.

Consider now an orthonormal cobasis $\theta^1_p,...,\theta^n_p\in T^\ast_p M$ at $p\in M$. Suppose that $\mathscr U\subseteq M$ is an open neighborhood of $p$ with coordinate functions $x^1,...,x^n$ and that in this coordinate system $\mathscr U$ is star-shaped with respect to $p$. Parallel transport the cobasis elements $\theta^a$ from $p$ to all points of $\mathscr U$ along radial lines (with respect to this coordinate system of course), thus we obtain a coframe $\theta^1,...,\theta^n$ defined in the entirety of $\mathscr U$.

The consequence of flatness is that the coframe is parallel, i.e. $\nabla\theta^a=0$ ($a=1,...,n$).

The consequence of metric compatibility is that the coframe is orthonormal at each point of $\mathscr U$, since the parallel transport induced by a metric compatible connection is a linear isometry of the tangent (thus cotangent) spaces.

The consequence of torsionlessness is that the coframe $\theta^a$ is holonomic. To see this let $\xi\in\Omega^1(\mathscr U)$ be a parallel 1-form on $\mathscr U$. Then its covariant derivative in components is $$ 0=\xi_{i;j}=\xi_{i,j}-\Gamma^k_{ji}\xi_k, $$ and due to torsionlessness the connection coefficients $\Gamma^k_{ij}$ are symmetric in the lower indices, thus $$ \xi_{i,j}-\xi_{j,i}=0. $$ Then by Poincaré's lemma there is a function $\phi\in C^\infty(\mathscr U)$ (the domains is star-shaped so Poincaré's lemma holds) such that $\xi=d\phi$. It follows that for the coframe there are functions $y^a$ on $\mathscr U$ such that $$ \theta^a=dy^a. $$

Putting these together gives that the metric is $$ g=\eta_{ab}\theta^a\otimes\theta^b=\eta_{ab}dy^a\otimes dy^b, $$ where $\eta_{ab}=\mathrm{diag}(-1,...,-1,1,...,1)$, therefore we obtain that the metric is flat.

For this result to hold, the connection $\nabla$ has to satisfy all three conditions, flatness, metric compatibility and torsionlessness. But we know from Riemannian geometry that a metric compatible and torsionless connection must uniquely be the LC connection of the metric.

We thus see that for a connection to characterize the flatness of the metric it must uniquely be the LC connection.


问:
Adjoint of the differential in Morse-Novikov cohomology
Let $M^n$ be a smooth manifold and we can define the Morse-Novikov cohomology group $H^k (M^n,\theta)$ with co-boundary differential operator $d_\theta (w) = dw + \theta \wedge w$  where $0\neq[\theta]∈ H_{dR}^1(M^n)$. This cohomology shares many properties with the ordinary de Rham cohomology and only depend on $[\theta]$.

As it has been shown that the Hodge theory works nicely on this cohomology by defining the adjoint operator $d_\theta^*$ ( I suppose it defines as usual adjoint $\delta$ of de Rham operator $d$ ) and defining the corresponding Laplacian of these operators.

I have the following question: It seems to me that the Stokes’ theorem does not work well with this operator $d_\theta$ because we will have an extra  term which is $\int_M[\theta \wedge w]$and this implies that the Green formula will not work (i.e. $d_\theta^*$ will no longer be adjoint of $d_\theta$). Am I right?

答:
In the paper mentioned in the comments, the operator $d_{\theta}$ is slightly different than the one you mention in your question: $d_{\theta}(w) = dw - \theta\wedge w$. In addition, the operator $\delta_{\theta}$ is defined as $-\ast d_{-\theta}\ast$. In that paper, Otiman is concerned with Inoue surfaces which have real dimension four. In even dimensions, the expression $-\ast d_{-\theta}\ast$ is indeed the adjoint of $d_{\theta}$. In odd dimensions, the adjoint is given by $-\ast d_{(-1)^k\theta}\ast$ when acting on $k$-forms.

To see this, let $\alpha \in \Omega^{k-1}(M)$ and $\beta \in \Omega^k(M)$, then

$$\langle d_{\theta}\alpha, \beta\rangle = \langle d\alpha - \theta\wedge\alpha, \beta\rangle = \langle d\alpha, \beta\rangle - \langle\theta\wedge\alpha, \beta\rangle = \langle\alpha, d^*\beta\rangle - \langle\theta\wedge\alpha, \beta\rangle.$$

Now note that

\begin{align*}
\langle\theta\wedge\alpha, \beta\rangle &= \int_M\theta\wedge\alpha\wedge\ast\beta\\
&= \int_M(-1)^{k-1}\alpha\wedge\theta\wedge\ast\beta\\
&= \int_M(-1)^{k-1}\alpha\wedge(-1)^{(n-k+1)(k-1)}\ast\ast(\theta\wedge\ast\beta)\\
&= (-1)^{(n-k)(k-1)}\int_M\alpha\wedge\ast\ast(\theta\wedge\ast\beta)\\
&= (-1)^{n(k-1)}\langle\alpha, \ast(\theta\wedge\ast\beta)\rangle.
\end{align*}

Therefore

\begin{align*}
\langle d_{\theta}\alpha, \beta\rangle &= \langle\alpha, d^*\beta\rangle - \langle\theta\wedge\alpha, \beta\rangle\\
&= \langle\alpha, -\ast d\ast\beta\rangle - (-1)^{n(k-1)}\langle\alpha, \ast(\theta\wedge\ast\beta)\rangle\\
&= \langle\alpha, -\ast(d\ast\beta + (-1)^{n(k-1)}\theta\wedge\ast\beta)\rangle\\
&= \langle\alpha, -\ast(d\ast\beta - (-1)^{n(k-1)+1}\theta\wedge\ast\beta)\rangle\\
&= \langle\alpha, -\ast d_{(-1)^{n(k-1)+1}\theta}\ast\beta\rangle
\end{align*}

so the adjoint of $d_{\theta} : \Omega^{k-1}(M) \to \Omega^k(M)$ is $-\ast d_{(-1)^{n(k-1)+1}\theta}\ast$. If $n$ is even, this reduces to $-\ast d_{-\theta}\ast$ (the operator Otiman calls $\delta_{\theta}$), and if $n$ is odd, this reduces to $-\ast d_{(-1)^k\theta}\ast$.

OwnerDisplayName: Mathematics


问:
Need help to understand the basic concept of this
Its may be a silly question but Can someone please explain how this works? ?
how to expand two terms where one is raised to some power and other is just 1?
the example is below.

This is what I want understand, for example (x^6 -1) = (x.x.x.x.x.x - 1) right? so how does it convert into (x^4 + x^2 + 1) (x^2 - 1)

答:
There is nothing crazy about it. Let $y = x^k$, and then
$$x^{kl}-1 = y^l-1 = (y-1)(y^{l-1}+y^{l-2} + \cdots + y + 1).$$
The second equality comes from just expanding and seeing the terms in between cancelling out, or from doing a partial sum of a geometric sequence. Now changing $y$ back to $x^k$ would give you your formula.


问:
Expected Value of a quite complex coin flipping game.
I have a rather complex game, whose expected value I need to find.
Given is a binary random value with probability for success p (lets say an unfair coin with probability of landing heads p and tails 1-p, p<<0,5)
The rules of the games are the following: you start by betting 1 euro, a multiplier = 1 and 15 tosses.
Each time you land heads (less likely option) you receive 15 more tosses (added to the remaining one you have) and your multiplier increases by 1. Landing tails doesn't give you anything, you've just lost one toss. Your multiplier can grow up to 5, and each time you land a head after achieving a multiplier of 5 adds new 15 tosses, but does not increase the multiplier further.
I am seeking for the expected multiplier of the game.

I started by building a Markov Chain table A[i,j] for the probability of spin j to have multiplier i:
A[1,1] = 1,
A[1,2] = 1-p,
A[2,2] = p,
A[1,3] = (1-p)*(1-p),
etc... We have 15 initial spins then each
A[i, 15+i] = 0.
But I am still stuck of deriving the Avg MP from the table.
Any help will be much appreciated!

Thanks!

EDIT:
I forgot to say, at each toss you win your bet x Multiplier, so actually, I am looking for the EV of the game, but I think it can be presented as finding the average MP of the game and the average number of tosses, so in the end the EV should be their product. I know how to find the Avg number of tosses, and my main issue is the AVG MP.

答:
Foreword.

Probably this answer is not what you expect. It aims actually at a more complicated task of computing the probability that the game ends at the $(n+1)$-th batch of trials. Knowing the probability one can of course compute the expected value of any quantity which depends only on the number of played batches. Yet  the computation of an expected value often does not require the knowledge of the individual probabilities. It may be the case also here. Nevertheless I have decided to post the answer since it reveals some beauty and can be possibly useful for other problems.

Preliminaries.

Let $\tilde n$ be a weak composition of a non-negative number $n$ into $n+1$ non-negative parts:
$$
\tilde n=(n_0,n_1,\ldots,n_{n}),\tag1
$$
and define the exceedance of the composition as:
$$
X(\tilde n)=\sum_{m=0}^{n}\mathbf{1}_{N_m>m}\tag2
$$
where $\mathbf1_A$ is the indicator function and $N_m=\sum_{i=0}^m n_i$ are the partial sums of $\tilde n$. Expressing simply the exceedance is the number of valid inequalities $N_m> m$ for $m$ ranging from $0$ to $n$. The exceedance can take on $n+1$ values from $0$ to $n$ (observe that $N_n=n$).

Solution.

Let $t$ be the length of a batch of trials (in your example $t=15$). Then the probability that the game ends at the $(n+1)$-th batch is:

$$W(n,t)\left(pq^{t-1}\right)^n q^t,\tag3
$$
where $q=1-p$ is the probability to toss a tail and $W(n,t)$ is the number of ways to distribute $n$ heads between $t(n+1)$ tosses in a way which ensures
the condition that there is sufficient number of previously tossed heads to proceed to every next batch.

Let $n_i$ in (1) be the number of heads tossed in the $i$-th batch. Then the above mentioned condition is simply:
$$
X(\tilde n)=n.\tag4
$$

It follows then that
$$
W(n,t)=\frac1{n+1}\binom{t(n+1)}n.\tag5
$$

The formula can be obtained in the following way. There are in total
$\binom{t(n+1)}n$ ways to distribute $n$ heads among $t(n+1)$ tosses. Consider now some particular distribution and cyclically shift it by $k$ batches ($0\le k \le n$) so that $n'_i=n_{(i+k)\pmod{n+1}}$. It appears that all $n+1$ cyclically shifted compositions have distinct exceedances so that only one of them has the required exceedance $n$ (an excellent proof can be found  here). This explains the prefactor $\frac1{n+1}$.

Appendix.

The overall probability that the game ends after a finite number of batches is:
$$
P_t(q)=q^t\sum_{n=0}^\infty\frac{1}{n+1}\binom{t(n+1)}n\left(pq^{t-1}\right)^n.\tag6
$$

The following figure demonstrates the behavior of the function $P_t(q)$ for $t=1,2,3,4$.

Observe that $P_t(q)=1$ only if $p\le\frac1t$ (the inequality is strict for $t=1$). This means that for larger $p$ there is a certain probability that the game never ends. This is however irrelevant for the expected value of a constant.

Hope this helps.


问:
Solve equations $\sqrt{t +9} - \sqrt{t} = 1$
Solve equation:  $\sqrt{t +9} - \sqrt{t} = 1$

I moved - √t to the left side of the equation $\sqrt{t +9} = 1 -\sqrt{t}$

I squared both sides $(\sqrt{t+9})^2  = (1)^2 (\sqrt{t})^2$

Then I got $t + 9 = 1+ t$

Can't figure it out after that point.

The answer is $16$

答:
The solution of such a problem such a problem lies in finding a number for which the equation holds.  Suppose you didn't know the answer was 16, how might you solve the problem?

Well, you have two square roots in the problem $\sqrt{t +9}$ and $\sqrt t$.  Their squares differ by 9.  Thus, if you can find two square numbers which differ by 9, then the larger square number will equal (t +9) and the smaller square number will equal t.  The set of square numbers, as you might know, consists of the sequence (1, 4, 9, 16, 25, ...).  25 and 16 differ by 9, and 16 is the smaller.  Thus, we have found 16 as the solution to this equation.


问:
Known bounds on $\int_{-\infty}^{\infty}|f(x)-f(x-a)|dx$?
Suppose that $f(x)$ is continuous on $\mathbb{R}$ such that $|f(x)|\leq M$. Are there any know result for estimating the quantity
$$
I(a)=\int_{-\infty}^\infty|f(x)-f(x-a)|dx.
$$

I have the following results: (i) If $f$ is strictly increasing (equivalently $f(-x)$ strictly decreasing),
$$
\begin{align}
I(a)&=\int_{f(\mathbb{R})} \left[f^{-1}(y)+a-f^{-1}(y)\right]dy \\
&\leq\int_{[-M,M]} a\,dy \\
&= 2Ma
\end{align}
$$

(ii) If $f\geq0$, $f$ strictly increasing on $(-\infty,0)$ and strictly decreasing on $(0,\infty)$. By the intermediate value theorem, there exist $c\in(0,a)$ such that $f(c)=f(c-a)$, then
$$
\begin{align}
I(a)&=\int_{-\infty}^c \left[f(x)-f(x-a)\right]dx+\int_c^\infty\left[f(x-a)-f(x)\right]dx \\
&=\int_{-\infty}^cf(x)dx-\int_{-\infty}^{c-a}f(x)dx+\int_{c-a}^\infty f(x)dx-\int_c^\infty f(x)dx \\
&=2\int_{c-a}^c f(x)dx \\
&\leq 2Ma
\end{align}
$$

Is there a generalization of such bound? My guess would be that for any bounded continuous function that has finitely many extrema,
$$
I(a)\leq 2aMn,
$$
where $n$ is the number of extrema. Any insight would be greatly appreciated.

答:
First note the integral does converge if $f$ has a finite number of local maxima. Indeed, in this case the function is eventually monotone and bounded, thus the integral telescopes, thus is finite.

Hint for a less sharp estimate:

Let $f$ be an increasing(or decreasing) function on $(-\infty, r]$(or $[r, \infty)$).  Prove that  $I_\tilde{f}(a)\le 3M a$ for $\tilde{f}= f \chi_{[-\infty,r]}.$

Let $f$ be an increasing function on $[r_1,r_2],$ Let $\tilde{f}= f \chi_{[r_1,r_2]},$ prove that $I_\tilde{f}(a)\le 4M a.$

Now for a function $f$ that has finite local maxima, let $\{-\infty =r_0 <r_1< ...<r_{n-1}< r_n= \infty\}$ be the set points at which $f$ attain its local maxima. Write $$f_i= f \chi_{[r_i,r_{i+1}]},$$ then by the triangle inequality $$I_f(a) \le \sum_{i=0}^n I_{f_i}(a).$$

Use the estimates in (1) and (2).

Note that if you manage to get better bounds for (1) and (2), you can get a better bound in (3). I hope this helps.


问:
Absolute error from root in false position method
Suppose I know that $f(x_1)$ and $f(x_2)$ have opposite signs, so $f(x)=0$ has a root $x\in[x_1,x_2]$. Then using the false position method, I have a guess for the root
$$x_3=\frac{f(x_2)x_1-f(x_1)x_2}{f(x_2)-f(x_1)},$$
and I can iterate on either $[x_1,x_3]$ or $[x_3,x_2]$ depending on the sign of $f(x_3)$.

Usually we terminate the process when $|f(x_n)|<\epsilon$ for some specified $\epsilon$.

How can I pick $\epsilon$ so that I am certain that my guess for the root $x_n$ is within $\delta$ of the true value of the root, i.e. $|x_n-x|<\delta$?

I am not sure how to pick such an $\epsilon$ when we don't even know the true value $x$ of the root. When $\delta$ is sufficiently small, something like $\epsilon=\delta f'(x)$ could work, but obviously this requires that you (a) know the true value of the root and (b) know the derivative of the function, two assumptions that are definitely not true in general.

答:
Let $f : \mathbb{R} \rightarrow \mathbb{R}$ and let us consider the problem of terminating an iterative method that is being used to solve the non-linear equation
$$ f(x) = 0$$
It is vital we consider the underlying application and what is actually needed in order to satisfy the user.

Suppose that the objective is to compute the square root of $\alpha>0$, then $f(x) = x^2- \alpha$ is a natural choice and a very natural objective is to compute $x$ such that $|x - \sqrt{\alpha}|<\delta$ where $\delta>0$ is chosen by the user. A bracketing method such as the bisection method or the false position method systematically shrinks a bracket which is certain to contain at least one root. Specifically, if $f(a)f(b)<0$ and $f$ is continuous in the interval $[a,b]$, then $f$ has a root $r \in (a,b)$. In the absence of any additional information our best estimate of $r$ is the midpoint, i.e., $$\mu = \frac{a+b}{2}$$ and the error is bounded by $$|r-\mu| < \frac{1}{2}|b-a|.$$ We should therefore terminate the iteration when $$|a-b| \leq 2 \delta.$$ It is entirely possible that the user is more interested in a small relative error. In this case we must of eliminate the possibility that $r=0$ is a root, but this is easy. If $ab>0$, then $r=0$ is not possible and the relative error can be bounded by $$\frac{|r-\mu|}{|r|} < \frac{\frac{1}{2}|a-b|}{\min\{|a|,|b|\}}.$$

Suppose the objective is to compute the elevation $\theta$ needed to hit a target with a howitzer. Then $$f(\theta) = r(\theta) - d$$ is the natural choice. Here $r(\theta)$ is the range of the howitzer when fired with elevation $\theta$ and $d$ is the distance to the target. Let $$\theta_1, \theta_2, \dotsc, \theta_j $$
denote the approximations of $\theta$ produced by our solver. When should we terminate the iteration. For this application, tiny errors $\theta-\theta_j$ are utterly irrelevant. The crew manning the howitzer cannot make minute adjustments to the elevation anyway, because howitzers are heavy objects. The only thing that matters is the residual, i.e.,
the value $y = f(\theta_j)$. In this situation, it is proper to terminate the iteration when $$|f(\theta_j)| < \rho,$$ where $\rho$ is kill radius of the exploding shell.

In general, it is not viable to terminate the iteration when it appears to be stagnating, i.e., when
$$|x_j - x_{j+1}| < \delta.$$ What is required to defeat this criteria in the context of the false position method? We need a continuous function $f$ and two points $a$ and $b$ such that $f(a)$ is large and negative and $f(b)$ is tiny and positive. The false position method will return an approximation $c$ which is very close to $b$. This is a major problem if there is only a single root $r \in (a,b)$ and $r$ is close to $a$. If we are using, say, Newton's method, then this criteria can be defeated by functions satisfying $$f(x) \approx e^{-\lambda x}, \quad f'(x) \approx -\lambda f(x)$$ where $\lambda>0$ because
$$x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} \approx x_n + \frac{1}{\lambda} \rightarrow \infty, \quad n \rightarrow \infty, \quad n \in \mathbb{N}.$$


问:
What does $w' := <M'>$ mean in the context of the Halting Problem?
I am studying the Halting Problem and I came across the following notation. I am not sure what it means. The context is as follows:

To prove that the Halting Problem is undecidable, we employ proof by contraction. Assume that the language $K = \{w \in \{0,1\}^* | M_w \ \text{stops for input w}\}$, is decidable. This means that the characteristic function $\chi_K$ is computable using a Turing Machine $M$. Now we can construct a Turing Machine $M'$ that simulates $M$. When the output is $0$, the $M$ stops and when the output is $1$, then $M$ goes in an endless loop. Let $w' := <M'>$ ...

What does $w' := <M'>$ mean in this context? Is it just that $M'$ is a Turing Machine that is encoded by $w'$?

答:
Notice that Turing Machines are powerful enough to simulate other Turing Machines. We encode Turing Machines $M$ as words $\langle M \rangle$ and we can formulate the Halting Problem as a word problem for the language $L=\{\langle M \rangle \epsilon \langle \omega \rangle\ |\ M \text{ halts on input } \omega\}$ whereby $\langle M \rangle$ and $\langle \omega \rangle$ are encodings for $M$ and $\omega$ and $\epsilon$ is a separator.

First denote $(M, \omega)$, whereby $M$ is a Turing Machine and $\omega$ some word of our input alphabet of $M$. Then the encoding of such a pair is often denoted by $\langle M, \omega \rangle$.

Since $\omega' \in \{0,1\}^*$ (which is our word encoding for which we could also write $\langle \omega' \rangle$), and $\langle M' \rangle$ denotes the encoding of the Turing Machine $M'$, it is indeed encoded by the binary word $\omega'$.

To summarize, we write $\langle M' \rangle$ to denote the encoding of the Turing Machine $M'$, and since $\omega'$ is an encoded word, $\omega' := \langle M' \rangle$ says that the encoding of $M'$ (which is $\langle M' \rangle)$ is given by $\omega'$.


问:
Showing ring homomorphism
How to show that the map $f: \Bbb{Z_{10}} \rightarrow \Bbb{Z_{20}}$ defined as

$$f(x) = 16x$$ is a ring homomorphism.

I am trying to use the fundamental theorem of ring homomorphism.

I define $g : \Bbb{Z} \rightarrow \Bbb{Z_{20}}$ as $g(x) = 16x$

How can I show that $g$ is a ring homomorphism?

Now I can see that $10\Bbb{Z}$ is contained in Ker$g$.

Now I consider the canonical projection $\Pi$ : ${\Bbb{Z}}$ to $\Bbb{Z_{10}}$

Then there will exist a unique ring homomorphism from $h$: $\Bbb{Z_{10}}$ to $\Bbb{Z_{20}}$ such that $g = h \pi$  .

How to show that $h = f$?

答:
To show homomorphism you need to show two properties:

$f(a+b)=f(a)+f(b)$

$f(a\cdot b)=f(a)\cdot f(b)$

So here we have:

$f(a+b)=16(a+b)=16a+16b=f(a)+f(b)$

$f(a\cdot b)=16(a\cdot b)=16a\cdot16b=f(a)\cdot f(b)$

Therfore, $f$ is ring homomorphism.

Title: Showing ring homomorphism


问:
Showing that a graph is planar
G is a graph with 5 vertices with degrees 3,4,5,6,7 (one each), n vertices with degree 2, and m vertices with degree 1.

Prove that G is planar.

I know that to prove G is planar, according to Kuratowski's and Wagner's theorems, I need to show it does not contain any minor of $K_5$ or $K_{3,3}$. But am unsure how to go about it.
For $K_5$, If I'm not wrong, I could use the fact there are only 4 vertices for which $deg_G(v)\geq4$, So it cannot contain a minor of $K_5$.
About $K_{3,3}$, maybe I could show that G contains an odd circle, therefore is not bipartite and does not contain any minor of a bipartite graph. But I do not know how to show it contains an odd circle.

I would appreciate any clarifications about this question! Thank you!

EDIT: Is it true that for G to include a minor of $K_{3,3}$ it needs to have at least 6 vertices with degree at least 3 each? If so, then it sorts that out for me. Thank you.

答:
A graph satisfying the given degree conditions may be planarly embedded as follows:

Unsubdivide to remove degree-$2$ vertices

Remove degree-$1$ vertices

Find a planar embedding of the remaining graph, which now contains only those vertices which originally had degrees $3,4,5,6,7$. This is always possible: because of the originally degree-$3$ vertex this "core" (minus loops and with multiple edges identified) will now always be a subgraph of $K_5-e$, which is planar

Reverse step 2 (add back degree-$1$ vertices; this preserves planarity)

Reverse step 1 (subdivide edges to restore degree-$2$ vertices; this preserves planarity)

Title: Showing that a graph is planar


问:
Number of functions from naturals to reals (${\aleph_1}^{\aleph_0}$)
The number of functions from $\mathbb{N}$ to itself is $|\mathbb{N}|^{|\mathbb{N}|}={\aleph_0}^{\aleph_0}=\aleph_1$. The number of functions from $\mathbb{R}$ to itself is $|\mathbb{R}|^{|\mathbb{R}|}={\aleph_1}^{\aleph_1}=\aleph_2$. Finally, the number of functions from $\mathbb{R}$ to $\mathbb{N}$ is $|\mathbb{N}|^{|\mathbb{R}|}={\aleph_0}^{\aleph_1}=\aleph_2$. My question is whether the number of functions from $\mathbb{N}$ to $\mathbb{R}$ ($|\mathbb{R}|^{|\mathbb{N}|}={\aleph_1}^{\aleph_0}$) is $\aleph_1$ or $\aleph_2$. Thanks!

EDIT:

Please note that I assume the generalised continuum hypothesis.

答:
I will write $\mathfrak c = 2^{\aleph_0}$.

Compute
$$\aleph_1^{\aleph_0} \ge 2^{\aleph_0} = \mathfrak c
\\
\aleph_1^{\aleph_0} \le \mathfrak c^{\aleph_0} = \big(2^{\aleph_0}\big)^{\aleph_0}
= 2^{\aleph_0\times\aleph_0}= 2^{\aleph_0} = \mathfrak c
$$
therefore
$$
\aleph_1^{\aleph_0} = \mathfrak c
$$


问:
Is connectedness relative property
I am bit confused after reading definition 2.45 from baby Rudin.

2.45   Definition   Two subsets A and B of a metric space X are said to be separated if both A∩B¯ and A¯∩B are empty, i.e., if no point of A lies in the closure of B and no point of B lies in the closure of A. A set E⊂X is said to be connected if E is not a union of two nonempty separated sets.

Suppose, $E\subset X$. Is statement '$E$ is connected in $X$' makes sense? I mean is connectedness is also relative property? I consider example $E=(0,1) \cup (1,2)$ as subset of $R$. Then $E$ is not connected in $E$, but connected in $R$. So in Rudin's definition it is not clear to me if author saying connected 'in E' or 'in X'.

答: If you see $E$ as a subset of $\Bbb R$, the $E$ is disconnected, since $\overline{(0,1)}\cap(1,2)=(0,1)\cap\overline{(1,2)}=\emptyset$. And, no, connectedness is not a relative property. That is, if $(X,\tau)$ is a topological space and if $Y\subset X$, then $Y$ is connected in $Y$ with respect to the subspace topology if and only if $Y$ is connected in $X$.

Title: Is connectedness relative property


问:
Integrating multivariate densities with variable bounds.
I am struggling how to deal with certain bounds when it comes to integrating multivariate densities.

How would you specify the bounds to integrate a function such as
$\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} f(x,y)  \,dx\,dy$ with bounds like

(i) $0<x<y<2$

(ii) $0<x<1$, $0<y<\sqrt(x)$

I don't know whether these examples make sense, I just made them up to go one step further from what we did in class. I know how to deal with for example $0<x$, $0<y$, $x+y<1$, but as soon as the bounds get more complex I fail to understand how to approach it. Is there a general method to go about more complex bounds as I have outlined above?

答:
Presumably you are trying to handle $\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} f(x,y)  \,dx\,dy$ where you know $f(x,y)=0$ outside the bounds.  A sensible way to approach this is with indicator functions, but sometimes the answer is sufficiently clear anyway.

In your two particular examples:

(i) $0<x<y<2$ can be $$\int_{y=0}^{2}\int_{x=0}^{y} f(x,y)  \,dx\,dx$$ or equivalently $$\int_{x=0}^{2}\int_{y=x}^{2} f(x,y)  \,dy\,dx$$

(ii) $0<x<1$, $0<y<\sqrt x$ can be $$\int_{x=0}^{1}\int_{y=0}^{\sqrt x} f(x,y)  \,dy\,dx$$


问:
Integrating $\int_0^\infty \frac{\sin (ax)}{x^3}dx$
I am trying to integrate this integral,

$$I_3 = \frac{ -4 }{ \pi} \int_0 ^{\infty} \frac{d\lambda}{\lambda^4} \cdot \sin(p_1 \lambda)\cdot \sin(p_2 \lambda) \left(-p_3 \lambda\cos\left(p_3 \lambda\right)\right)\cdot\sin\left(p_4 \lambda\right) \\
    = \frac{4p_3 }{ \pi} \int_0 ^{\infty} \frac{d\lambda}{\lambda^3} \cdot \sin(p_1 \lambda)\cdot \sin(p_2 \lambda) \cos\left(p_3 \lambda\right)\sin\left(p_4 \lambda\right)$$
I changed this trigonometric product to a sum of  $\sin\left(\left(dp_4+cp_3+bp_2+ap_1\right)\lambda\right)$ where $a,b,c,d= \pm1$ But then integral takes the form $\int_0 ^{\infty} \frac{d\lambda}{\lambda^4} \sin\left(k\lambda\right)$ Which seems divergent to me. Is there any other way to integrate this?

Background: I am trying to calculate $$ D_2(p_3,p_4) =\frac{-4}{\pi} \int \frac{d\lambda}{\lambda^4} \cdot \sin(p_1 \lambda)\cdot \sin(p_2 \lambda)
 \left(\sin\left(p_3 \lambda\right)-p_3 \lambda\cos\left((p_3 \lambda\right)\right)\cdot\left(\sin\left(p_4 \lambda\right)-p_4 \lambda\cos\left((p_4 \lambda\right)\right)$$ which has a finite value according to Appendix C of this paper and $I_3$ is one of the four terms after expanding it. If you can suggest any other simpler way to calculate $D_2$, that would also be great.

Thanks in advance :)

答:
$$\int\frac{\sin (ax)}{x^4}dx=a^3 \int\frac{\sin (t)}{t^4}dt$$ Now, you need a few integrations by parts to get
$$\int\frac{\sin (t)}{t^4}dt=-\frac{\cos (t)}{6 t^2}+\frac{\left(t^2-2\right) \sin (t)}{6 t^3}-\frac{\text{Ci}(t)}{6}$$
$$I(\epsilon)=\int_\epsilon^\infty\frac{\sin (t)}{t^4}dt=\frac{\cos (\epsilon)}{6 \epsilon^2}-\frac{\left(\epsilon^2-2\right) \sin (\epsilon)}{6 \epsilon^3}+\frac{\text{Ci}(\epsilon)}{6}$$
$$I(\epsilon)=\frac{1}{2 \epsilon ^2}+\frac{1}{36} (6 \log (\epsilon )+6 \gamma -11)-\frac{\epsilon
   ^2}{240}+\frac{\epsilon ^4}{20160}+O\left(\epsilon ^6\right)$$


问:
Manipulating a product term inside an integral
I have an expression of the form
$$P(x) = \int_0^\infty \prod_{i=0}^{n} e^{-G(a, x_i)}\,\mathrm{d}a $$
and I was wondering if there was any way that I could swap the order of the product and the integral? I suspect not but its been a while since I had to manipulate integrals for myself. Ideally P(x) will represent the total probability of making i measurements from a Possion distributed system with zero occurences each time.  </p>

答:
This is a CW answer intended to remove this question from the Unanswered queue.

As AlexE remarks, there is no general way to swap products and integrals.

This means that in general, evaluating integrals like the one stated can be quite hard, especially if the mapping $G$ is not in some way "nice".


问:
Geometric surjections are not stable under pullback
It is mentioned at the end of the introduction to Johnstone's Factorization systems for geometric morphisms, I that the pullback of a geometric surjection need not be a surjection, and hence that the surjection-inclusion factorisation system is not stable under pullback.

Is there a concrete counter-example out there for this statement?

答:
Let $L$ be any non-trivial locale with no points; we can take the 'locale of surjections $\mathbb{N} \to A$, for $A$ an uncountable set' found in Example C1.2.8 of Johnstone's Sketches of an Elephant, for example. Then the presheaf category $[\mathcal{O}(L)^{\mathrm{op}},\mathbf{Set}]$ has enough points, so we have a geometric surjection to it from $\mathbf{Set}/K$ for some set $K$. Since geometric inclusion are stable under pullback, and all subtoposes of $\mathbf{Set}/K$ are of the form $\mathbf{Set}/K'$ for some $K' \subseteq K$, we find that:
$$\require{AMScd}$$
\begin{CD}
0 @>>> \mathrm{Sh}(L);\\
@VVV @VVV \\
\mathbf{Set}/K @>>> [\mathcal{O}(L)^{\mathrm{op}},\mathbf{Set}],
\end{CD}
where $0$ is the degenerate topos, is a pullback diagram, and the upper morphism is certainly not a geometric surjection. So in a way, the fact that surjections are not stable under pullback arises from the same quirk that allows frames to not have enough points!


问:
Does bijectivity imply the same order type?
Question: Given an ordered set A which must have bijective correspondences with set B, there must be a correspondence that preserve order?

I know it's wrong. For example, 1.the dictionary order on $\mathbb{Z}_+\times\mathbb{Z}_+$, 2.$(x_0,y_0)<(x_1,y_1)$ if either $x_0+y_0<x_1+y_1$, or $x_0+y_0=x_1+y_1\,and\,y_0<y_1$. They must have bijective correspondence because they have same amount of elements, but are in different order type.

But I feel uncomfortable with that because we all have an intuition that for two groups with same amount of elements( For example, if I can match 1~11, 2~12, 3~13....9~19, 10~20 for sets {1, ... ,10} {11, ... ,20}, I can actually match 1 with 12 with other every elements in {1, ... ,10} matches with elements in {11, ... ,20} ), we can build random correspondence(some correspondence preserves order in this case) between them.

What's wrong with this intuition.

答:
Everything is wrong with that intuition. In fact, it doesn't work if your sets have more than two elements!

Look at the structure of divisibility, $m\mid n\iff\exists k(k\cdot m=n)$. So for example, $2\mid 4$ and $4\mid 8$, but $4\nmid 3$. Then $\{1,\dots,10\}$ with $|$ is very different from $\{11,\dots,20\}$ with $|$. The latter is an antichain, whereas the former has nontrivial comparisons (as we pointed out: $1\mid 2\mid 4\mid 8$).

But you might want to say, I didn't sign up for preserving divisibility. I only wanted to preserve a linear order! But that's the point. Bijections don't have to preserve any "natural structure".

For that matter, setting $f\colon\{0,1\}\to\{2,3\}$ by $f(0)=3$ and $f(1)=2$ also does not preserve the order. It flips it. Of course those two are isomorphic, and you could argue there that that's fine. But that only means that a function $f\colon\Bbb{Z\to N}$ needs to be an isomorphism with respect to some structure, not all structure. And that's true, given any bijection from $X$ to $Y$ we can use it to define a structure on $X$ (or on $Y$) such that $f$ is an isomorphism.

And I mean, what's the order on $\{x,y\}$ when I only tell you that $x\neq y$? Did I tell you what are these objects and how they relate to each other? Not all orders are linear.

Remember: almost any naive intuition that you're coming with is wrong, in almost any field of mathematics. That is why we have definitions. Work with the definitions, step by step, and very very carefully, and you'll develop the correct definitions.

Or to quote John von Neumann, "in mathematics you don't understand things, you just get used to them."


问:
Telegraph equation - energy method
i'm struggling to prove solution uniqueness with this one.
thanks for the assistance.

$$u_{tt}+u_{t}=u_{xx} $$
$$u_x(0,t)=u_x(\pi,t)=0 $$
$$u(x,0)=1$$
$$u_t(x,0)=\cos^2(x)$$

Prove the uniqueness of the solution

well, i defines two solutions, $$w(x,t)=u(x,t)-v(x,t) $$ in aim to show $$w(x,t)=0 $$ and therefore $$ u(x,t)=v(x,t)$$

i multiplied each side of the equation with $w$ and integrated it, to get:
$$\int_0^\pi  w(w_{tt}+w_{t}) dx=\int_0^\pi  ww_{xx} dx$$
$$\int_0^\pi  ww_{tt}+\int_0^\pi ww_{t} dx=\int_0^\pi  ww_{xx} dx$$
where,
$$ww_{t}=\frac{d}{dt}\big[\frac{1}{2}w^2\big] \rightarrow \frac{d}{dt}\frac{1}{2} \int_0^\pi w^2 dx $$
by integration by part i got the follows:
$$E'(t) = \frac{d}{dt}\frac{1}{2} \int_0^\pi w^2 dx = \int_0^\pi w_t^2-w_x^2 dx$$

am i in the right way?
asking your as guidance

答:
Maybe it is better to write down an explicit energy first and do the calculations with that one. Try this definition of an energy for your $w$:
$$E(w(\cdot,t)):=\int_0^\pi (w_x)^2 + (w_t)^2\, dx.$$
This $E$ is always nonnegative. Now show $\frac{d}{dt}E(w(\cdot,t))\leq 0$ and $E(w(\cdot,0))=0$ (these will need the equation, the initial and boundary data and partial integration). I leave these steps for you to complete yourself.

On another note: It seems to me, you used partial integration to infer the following:
$$\int_0^\pi w w_{tt}\, dx = - \int_0^\pi (w_t)^2\, dx.$$
Since the integration is with respect to $x$ and the derivative with respect to $t$, this step is not correct.

Title: Telegraph equation - energy method


答:
To be precise, what you mean is to find a double integral of the form
$$
\iint_D f(x,y)dxdy
$$
where $D$ is a region defined by inequalities or to find $$
\iint_{\mathbb{R}^2} f(x,y)dxdy
$$
where $f$ is "supported" on the region $D$.

For those integrals that can be calculated by hand, your goal is to write it as iterated integrals. To figure out the bounds, drawing a picture would be helpful. See below for your examples.

How to find the bounds should be a very standard materials you see in a textbook when you learn double integrals the first time.

In general, the "easy" integrals can be written as one of the following ways:
$$
\int_a^b\left(\int_{c(x)}^{d(x)}f(x,y)dy\right)dy,\quad
\int_a^b\left(\int_{c(y)}^{d(y)}f(x,y)dx\right)dy \tag{1}
$$

So if the region $ D$ has been written in one of the following forms:
$$
D=\{(x,y): c(x)<y<d(x), a<x<b\},\quad D=\{(x,y): c(y)<x<d(y), a<y<b\}
$$
you should be happy since it tells you directly the bounds in (1).

Your third example,
$$
x>0,\quad y>0,\quad x+y<1
$$
can be written as
$$
D=\{(x,y): 0<y<1-x, 0<x<1\}
$$
or
$$
D=\{(x,y): 0<y<1, 0<x<1-y\}
$$

Title: Integrating multivariate densities with variable bounds.

Body:
I am struggling how to deal with certain bounds when it comes to integrating multivariate densities.

How would you specify the bounds to integrate a function such as
$\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} f(x,y)  \,dx\,dy$ with bounds like

(i) $0<x<y<2$

(ii) $0<x<1$, $0<y<\sqrt(x)$

I don't know whether these examples make sense, I just made them up to go one step further from what we did in class. I know how to deal with for example $0<x$, $0<y$, $x+y<1$, but as soon as the bounds get more complex I fail to understand how to approach it. Is there a general method to go about more complex bounds as I have outlined above?


问:
Example on Correspondences
Giva an example of correspondence $F : \mathbb{R} \rightarrow  \mathbb{R}$ such that the closure of $F$ is $ \overline{F}: \mathbb{R} \rightarrow  \mathbb{R}$, upper semi continuous on $\mathbb{R}$, but $F$ is not, if any.

I got stuck with this question. Any help will be appreciated.

答:
$\def\mto{\multimap}\def\R{\mathbb R}$Define $F \colon \R \mto \R$ by
$$ F(x) = \begin{cases} [0, 1] & x > 0\\ [0,1) & x \le 0\end{cases} $$
Then $F$ isn't upper semi-continuous, as $(-1,1)$ is an open neighbourhood of $F(0)$, but for each open neighbourhood $U$ of $0$ we have $F(U) \not\subseteq (-1,1)$. The closure
$$ \bar F(x) = [0,1], \quad x \in \mathbb R $$
is upper semi-continuous, as it is constant.</p>

Title: Example on Correspondences


问:
Solving a sparse $n \times n$ binary matrix efficiently
My motivation is solving arbitrary Lights Out puzzles with $n$ cells with a method given in MathWorld: basically imagine the game as a square grid graph and construct an "adjacency matrix" $A$. Then for initial configuration $b$, solving $Ax = b$ gives you button presses to solve the game.

For a matrix of size $n \times n$, normally Gaussian elimination takes $O(n^3)$ time, but maybe having a sparse binary matrix (by binary I mean all operations are mod 2, over field $\mathbb Z / 2 \mathbb Z$) can make the operation faster. I know the matrix is very sparse because every cell only has at most 4 neighbors, so almost all the matrix entries are zero. Any references to algorithms or software implementations that can solve binary sparse matrices efficiently are appreciated.

答:
Lights Out solvers are already known, see, for instance,  here or here, so I assume you plan to deal with big puzzle boards.

But the dimension of the matrix to deal can be reduced to $O(\sqrt{n})$ by the following row-by-row solving algorithm. Let the board has $n_1$ rows and $n_2$ columns. Suppose that $n_2\le n_1\ge 2$. Then $n_2\le \sqrt{n}$.

Given a lit instance, going down from the second to the bottom rows and switching cells just below lit cells in the previous row, we switch off all lights but, possible, some of the bottom row. Let $b’$ be the binary vector corresponding to the lit instance of the bottom row.

Now we shall look for a binary vector $x_1$ with the following property. Assume that all lights at the board are switched off. Now switch the lights in the first row according to the vector $x_1$. Again go down from the second to the bottom rows and switch cells just below lit cells in the previous row. For each natural $i$ from $1$ to $n_1$ let $x_i$ be a vector corresponding to lights switched  in $i$-th row. Put $x_0=0_{n_1}$. It is easy to see that for each natural $i$ from $1$ to $n_1-1$, $x_{i+1}=Bx_i+x_{i-1}$, where $B=\|b_{ij}\|$ is an $n_2\times n_2$-matrix such that $b_{ij}$ equals $1$ if $|i-j|\le 1$, and equals $0$, otherwise. Then $x_i=B_ix_1$ for each $i$, where $B_0=0$, $B_1=I$, and $B_{i+1}=BB_{i}+B_{i-1}$ for each $1\le n_1-1$. It follows that $B_i=f_i(B)$,
where $f_i$ is the $i$-th Fibonacci polynomial defined over $\Bbb Z_2$ by $f_0=0$, $f_1=1$, and $f_k=xf_{k-1}+f_{k-2}$ for $k\ge 2$, see Theorem 2 in [GCT]. Our aim is to obtain in the bottom row a lit instance corresponding to $b$, that is to achieve $Bx_{n_1}+x_{n_1-1}=b$. This means that $x_1$ is a solution of an equation  $Ax=b$, where $A=f_{n_1+1}(B)=BB_{n_1}+B_{n_1-1}$.

Remark that since the matrix $A$ does not depend on $b$, it can be calculated once for a board, at a preprocessing step. Moreover, a calculation of the polynomial $f_{n_1+1}$ can be shortcut using the equalities from Lemma 4 of [GCT].

References

[GCT] J. Goldwasser, W. Klostermeyer, and G. Trapp, Characterizing Switch Setting Problems, Linear and Multilinear Algebra, 43:1–3 (1997) 121–136.


问:
The sides of a triangle are in A.P.
One of the angles of a triangle is $120^\circ$. The sides of the triangle are in A.P. Find the ratio of the sides.

 \begin{align} 3:5:7 \end{align}

So the solution in my book:

Let the sides are $a-d,a,a+d$ where $a>0,0<d<a$. Obviously the angle opposite to the largest side is $120^\circ,$ so we can write: $(a+d)^2=a^2+2a(a-d)\cos120^\circ$ and we get $d=\dfrac25a$. So the sides are $\dfrac35a,\dfrac55a, \dfrac75a$.

I would like to ask why we can write the arithmetic progression in this way WLOG. And why do we do so? What do we get from it? Why the restrictions $a>0$ and $0<d<a$? Thank you in advance!

答: As a general rule, it is usually a good idea to parametrize the problem in the simplest way possible. Naming the middle element of the arithmetic progression $a$ and the difference $d$, it is $a-d, a, a+d$. You may assume that the difference is nonzero (otherwise the triangle is equilateral having no angle of size $120^\circ$). Assuming the difference to be positive is then nothing special: if it were negative, you would just list the sides in the reverse order. The side $a$ has to be positive, as it is the side of a triangle. Finally, if $d\geq a$, then the side $a-d$ would not be positive.

Title: The sides of a triangle are in A.P.


问:
A bag contains toys marked 1,2,3,...n.Two tags are chosen at random.
A bag contains toys marked 1,2,3,...n.Two tags are chosen at random.

Find P such that the markers will be consecutive integers if the tags are chosen

1)without replacement

2)with replacement

I am having a solution that goes like this:

For case 1:(Without replacement)

There are $nC2$ ways of choosing an element.

Take n=5, you get (1,2)(2,3)(3,4)(4,5) Generalising you can get (n-1) cases of getting consecutive tags.

So the required probability stands $(n-1)/nC2$

For case 2:(With replacement)

There is a sample space of $nC1 * nC1$ or $n^2$.

Here my instructor says that the number of choices will be 2(n-1).

Therefore, the probability of choosing consecutive tags is $\frac{2(n-1)}{n^2}$.

But I don't understand why it's $2*(n-1)$.

Is it simply because we now take both the tags without order for example: 1,2 in two ways (1,2) and (2,1) and they still remain conseceutive.

答:
Here is another way to look at the second part of the question (which is with replacement) -

Let's say you pick the first tag and that is number $k$. If $k$ is $1$ or $n$, there is only possibility to get the consecutive number that we pick $2$ if the first number is $1$ and we pick $(n-1)$ if the first number is $n$. But if $k$ is any of the remaining $(n-2)$ numbers, you have two possibilities for the second number, it is either $(k-1)$ or it is $(k+1)$.

So the probability is $\frac{1}{n}(\frac{1}{n} + \frac{1}{n} + (n-2) \frac{2}{n}) = \frac{2(n-1)}{n^2}$


问:
coordinate form of an affine transformation
I am having trouble with this seemingly simple question: Write the standard coordinate form of an affine transformation in $\mathbb{A}^{2}(\mathbb{R})$ that maps
the point (1, −2) to the point (0, 10), and the lines $10x_{1} − 4x_{2} = 1$ and $3x_{1} − 3x_{2} = −7$ to the lines $x_{1} − 2x_{2} = −3$ and $x_{1} − x_{2} = 6$, respectively.

Now as I understand for the point transformation
$\begin{pmatrix}0\\10\\0\end{pmatrix}=\begin{pmatrix}a_{11} & a_{12} & b_{1}\\a_{21} & a_{22} & b_{2}\\0 & 0 & 1\end{pmatrix}\begin{pmatrix}1\\-2\\0\end{pmatrix}$ and I need to find this matrix of transformation, this matrix equation gives me two equations with 6 unknowns. So I guess 4 more equations will be coming from the transformations of the two lines. Now here is where I am having trouble, I don't know how to get those equations. Any kind of hints would be really helpful. Thanks in advance.

答:
Here is a different method. Let $A$ be the looked for matrix.

First of all, let us determine the intersection of the "initial" axes and of the "final" axes. It is rather easy to find that the first two axes intersect at $(14/9,35/9)$ and the second pair of axes at $(15,9)$.
Therefore, $(15,9,1)$ is the image of $(14/9,35/9,1)$ by $A$.

Besides, directing vectors of the initial axes are

$$\binom{2}{5} \ \ \text{and} \ \ \binom{1}{1}$$

and directing vectors of the final axis are:

$$s\binom{2}{1} \ \ \text{and} \ \ t\binom{1}{1}$$

(parameters $s$ and $t$ are there in order to take into account the change of scale)

With all these informations, we can say that

$$A : \begin{pmatrix}2\\5\\0\end{pmatrix} \to \begin{pmatrix}2s\\1s\\0\end{pmatrix}, \ \begin{pmatrix}1\\1\\0\end{pmatrix} \to \begin{pmatrix}1t\\1t\\0\end{pmatrix}, \ \begin{pmatrix}14/9\\35/9\\1\end{pmatrix}\to
\begin{pmatrix}15\\9\\1\end{pmatrix} $$

Otherwise said :

$$A \times \underbrace{\begin{pmatrix}2&1&14/9\\5&1&35/9\\0&0&1\end{pmatrix}}_{\begin{array}{c}\text{initial affine frame}\\B\end{array}} = \underbrace{\begin{pmatrix}2s&1t&15\\1s&1t&9\\0&0&1\end{pmatrix}}_{\begin{array}{c}\text{final affine frame}\\C\end{array}} $$

which is equivalent to :

$$A=C \times B^{-1} = \dfrac{1}{3}\begin{pmatrix}5t-2s&2s-2t&45-14s/3\\
5t-s&s-2t& 27-7s/3\\
                 0&                 0&            3\end{pmatrix}\tag{1}$$

Now, the last step is expressing that

$$A\begin{pmatrix}1\\-2\\1\end{pmatrix}=\begin{pmatrix}0\\10\\1\end{pmatrix}$$

which yields a system of two equations in two unknowns $s$ and $t$, finally giving :

$$s=9 \ \ \text{and} \ \  t=17/3$$

that have to be plugged into (1) to give the final answer:

$$A=\begin{pmatrix} 31/9&20/9&1\\ 58/9&-7/9& 2\\0&0&1\end{pmatrix}$$

Remark: the interest of this method is that all the computations can be done using a Computer Algebra System ; here is what I have done with Matlab:

```<code>syms x1 x2 s t
[X1,X2]=solve(10*x1-4*x2==0,...
               3*x1-3*x2==-7,...
               x1,x2)
B=[2, 1, X1;
   5, 1, X2;
   0, 0, 1];
[X1,X2]=solve(x1-2*x2==-3,...
              x1  -x2==6,...
              x1,x2)
C=[2*s, t, X1 ;
   s,   t, X2;
   0,   0, 1];
A=C*inv(B);
[S,T]=solve(A*[1,-2,1]'==[0,10,1]',s,t);
subs(A,{s,t},{S,T})
</code>```

Body:
Here is a different method. Let $A$ be the looked for matrix.

First of all, let us determine the intersection of the "initial" axes and of the "final" axes. It is rather easy to find that the first two axes intersect at $(14/9,35/9)$ and the second pair of axes at $(15,9)$.
Therefore, $(15,9,1)$ is the image of $(14/9,35/9,1)$ by $A$.

Besides, directing vectors of the initial axes are

$$\binom{2}{5} \ \ \text{and} \ \ \binom{1}{1}$$

and directing vectors of the final axis are:

$$s\binom{2}{1} \ \ \text{and} \ \ t\binom{1}{1}$$

(parameters $s$ and $t$ are there in order to take into account the change of scale)

With all these informations, we can say that

$$A : \begin{pmatrix}2\\5\\0\end{pmatrix} \to \begin{pmatrix}2s\\1s\\0\end{pmatrix}, \ \begin{pmatrix}1\\1\\0\end{pmatrix} \to \begin{pmatrix}1t\\1t\\0\end{pmatrix}, \ \begin{pmatrix}14/9\\35/9\\1\end{pmatrix}\to
\begin{pmatrix}15\\9\\1\end{pmatrix} $$

Otherwise said :

$$A \times \underbrace{\begin{pmatrix}2&1&14/9\\5&1&35/9\\0&0&1\end{pmatrix}}_{\begin{array}{c}\text{initial affine frame}\\B\end{array}} = \underbrace{\begin{pmatrix}2s&1t&15\\1s&1t&9\\0&0&1\end{pmatrix}}_{\begin{array}{c}\text{final affine frame}\\C\end{array}} $$

which is equivalent to :

$$A=C \times B^{-1} = \dfrac{1}{3}\begin{pmatrix}5t-2s&2s-2t&45-14s/3\\
5t-s&s-2t& 27-7s/3\\
                 0&                 0&            3\end{pmatrix}\tag{1}$$

Now, the last step is expressing that

$$A\begin{pmatrix}1\\-2\\1\end{pmatrix}=\begin{pmatrix}0\\10\\1\end{pmatrix}$$

which yields a system of two equations in two unknowns $s$ and $t$, finally giving :

$$s=9 \ \ \text{and} \ \  t=17/3$$

that have to be plugged into (1) to give the final answer:

$$A=\begin{pmatrix} 31/9&20/9&1\\ 58/9&-7/9& 2\\0&0&1\end{pmatrix}$$

Remark: the interest of this method is that all the computations can be done using a Computer Algebra System ; here is what I have done with Matlab:

syms x1 x2 s t
[X1,X2]=solve(10*x1-4*x2==0,...
               3*x1-3*x2==-7,...
               x1,x2)
B=[2, 1, X1;
   5, 1, X2;
   0, 0, 1];
[X1,X2]=solve(x1-2*x2==-3,...
              x1  -x2==6,...
              x1,x2)
C=[2*s, t, X1 ;
   s,   t, X2;
   0,   0, 1];
A=C*inv(B);
[S,T]=solve(A*[1,-2,1]'==[0,10,1]',s,t);
subs(A,{s,t},{S,T})


问:
I need help understanding factoring better for this problem
Here is the equation:
$$\frac{-3(x+h)^2-6(x+h)+4+3x^2+6x-4}{h}$$

I am getting the wrong answer every time.
\begin{align*}
f & = \frac{-3(x^2+3xh+h^2)-6x+6h+4+3x^2+6x-4}{h}\\
f & = \frac{-3x^2-6xh-3h^2-6x+6h+4+3x^2+6x-4}{h}\\
f & = \frac{-6xh-3h^2}{h}\\
f & = 3(-2x-h^2)
\end{align*}

Two questions, what am I doing wrong when factoring down this problem and where is the best place to really learn factoring super well?

答:
Your approach (expanding, collecting terms, and canceling) seems fine but you made four errors. Luckily, two of them cancel each other out: $$-6(x+h) \not = -6x \color{red}{+} 6h \tag{line 1 to 2}$$
$$(x+h)^2 \not = x^2 + \color{red}{3}xh + h^2\tag{line 1 to 2}$$
$$-3(x^2 + 3xh + h^2) \not = -3x^2 -\color{red}{6}xh -3 h^2\tag{line 2 to 3}$$ Finally, you managed to lose a term when you reduced ($6h$, which should be $-6h$).

To improve your factoring:

Watch someone else do it. Khan academy has excellent short videos. Your textbook may have examples of some particular method.

Practice. Preferably early and often. Preferably in a way that allows non-punitive, immediate feedback. A textbook will have factoring exercises with selected answers.

Use pen/pencil & paper.

Go slow.

Investigate your errors.

Try again.

Seek out positive reinforcement (from yourself if need be). By the way, nice post! You're already ahead of the curve.

Teach someone else. If this is impossible, consider a study group where you might find opportunities to trade teaching roles.

Avoid overly competitive or punitive educational environments. If this is impossible, do your learning before you arrive. If you show up at boot camp ready to start your weight loss, you're probably going to wash out of the military. If you do it repeatedly, you'll have zero probability of success.


问:
Factoring a multivariate polynomial.
Let $P(x_1,\ldots,x_n)\in \mathbb{C}[x_1,\ldots,x_n]$ a polynomial in complex $n$-variables. There is a method to find out if this polynomial can be written as
$$P(x_1,\ldots,x_n)=m(x_1)\ldots m(x_n)$$
with $m(x)$ a polynomial in one variable?

答:
The polynomial has the seprated form
$$P(x_1,\dots,x_n) = p_1(x_1)\cdots p_n(x_n), $$
if and only if the PDE
$$P^{n-1}\frac{\partial^n}{\partial x_1 \cdots \partial x_n} P =  \frac{\partial P}{\partial x_1} \cdots \frac{\partial P}{\partial x_n} $$
holds.
Conditional on that fact, you may also check whether the polynomial is invariant under all permutations in $S_n$.

Title: Factoring a multivariate polynomial.


答:
If $3$ terms are in AP (arithmetic progression), you may list them down as $$a, a+d, a+2d$$ where $a$ is the first term and $d$ is the common difference. This is what you usually do, i.e. based on the fact that the $n$th term of an AP with first term $a$ and difference $d$ is $a+(n-1)d$

Now, I'll show you why you can switch to the other notation WLOG. Substitute $a + d = b$. The three terms are now:
$$b-d,b,b+d$$ which look like what you want. This can be done WLOG because you can always find a $b$ such that $b = a+d$.

Why is this a good idea?

A lot of times when dealing with APs, one wants to talk about the sum of the terms. Choosing variables in a way that the sum looks simpler, makes your life easier.

If you have an odd number of terms, consider choosing variables such that the AP looks like:
$$\ldots,a-2d,a-d,a,a+d,a+2d,\ldots$$

If you have an even number of terms, consider choosing variables such that the AP looks like:
$$\ldots,a-3d,a-d,a+d,a+3d,\ldots$$
Note how in the even case I've chosen a difference of $2d$ instead of $d$. Convince yourself why you can always do this! (the argument is similar to the above one about choosing representation WLOG)

Title: The sides of a triangle are in A.P.

Body:
One of the angles of a triangle is $120^\circ$. The sides of the triangle are in A.P. Find the ratio of the sides.

 \begin{align} 3:5:7 \end{align}

So the solution in my book:

Let the sides are $a-d,a,a+d$ where $a>0,0<d<a$. Obviously the angle opposite to the largest side is $120^\circ,$ so we can write: $(a+d)^2=a^2+2a(a-d)\cos120^\circ$ and we get $d=\dfrac25a$. So the sides are $\dfrac35a,\dfrac55a, \dfrac75a$.

I would like to ask why we can write the arithmetic progression in this way WLOG. And why do we do so? What do we get from it? Why the restrictions $a>0$ and $0<d<a$? Thank you in advance!


问:
Understanding $F\phi \to FF\phi$ as density of time
I'm talking about the basic temporal language in modal logic.
Suppose we insist that the following is always true:
$$F\phi \to FF\phi$$
It is supposed to be interpreted as between any 2 time instants, there is always a third, i.e. density of time. I'm not able to understand how.

$F\phi$ means $\phi$ will be the case in the future, and $FF\phi$ means $F\phi$ will be the case in the future. How do I understand this as density?

答:
See Temporal Logic for the basic definition and the condition expressing "dense" in term of the precedence operator: $<$.

Suppose that the temporal model $\mathcal M$ is dense.

Consider a valuation $v$ such that   $\mathcal M, t \vDash \text F p[v]$. By the clause for $\text F$ operator we have that $\mathcal M, t' \vDash p[v]$ for some $t' > t$ [a valuation is a function assigning to each atomic proposition $p$ the set of time instants $v(p)$ at which $p$ is true.]

By density, there is some $t''$ such that: $t < t'' < t'$ and the fact above implies that $\mathcal M, t'' \vDash \text F p[v]$.

But this in turn implies: $\mathcal M, t \vDash \text F \text F p[v]$.

Thus, if the temporal model is dense, we have that: $\mathcal M, t \vDash \text F p \to \text F \text F p$.

Conversely, we show that the formula $\text F \text F p$ is not valid on a discrete ordering of time.

Consider two successive points $t_0$ and $t_1$ and consider the valuation $v^*$ that makes $p$ true only at $t_1$ [i.e. $v^*(p) = \{ t_1 \}$].

Obviously, $\text F p$ holds at $t_0$; but since there is no point in time between $t_0$ and $t_1$, the formula $\text F \text F p$ cannot be true at $t_0$.

Thus, in a discrete model of time, the formula $\text F p \to \text F \text F p$ is not satisfied by valuation $v^*$.


问:
We need vector space structure on $\mathbb{R^n}$ in order to define the standard topology?
I was watching a lecture on topology and the professor said that you don't need the vector space structure on $\mathbb{R^n}$ in order to define the standard topology for $\mathbb{R^n}$. But in the definition of the open ball (that he used to define the standard topology), he wrote something like this:

$$B_r(x):=\bigg\{y=(y_1, ...,y_n) \in \mathbb{R^n}:\sqrt{\sum_{i=1}^n(y_i-x_i)^2}<r\bigg\}$$

But how can we perform "$(y_i-x_i)^2$" in $\mathbb{R^n}$, only using its set and topological structure? And if I need only an aditional structure other than a vector space one, what is that?

P.S.: Sorry for the broken English.

答:
The values $y_i$ and $x_i$ in the expression are both real values, i.e., defined on $\mathbb R$. Therefore, the expression $y_i - x_i$ is not an operation on the vector space $\mathbb R^n$, but an operation in the field $\mathbb R$.

This means that even if $\mathbb R^n$ is viewed only as a set (and not as a vector space), the definition of the set $B_r(x)$ in your question is still perfectly valid.


问:
1D Wave PDE with Nonzero Initial and Boundary Conditions
I'm not sure how to start this PDE since the initial and boundary conditions are nonzero. May someone point me in the right direction?

This is the problem:

$$u_{tt} = u_{xx}$$
$$u(x,0) = \frac{1}{2+ \sin(x)}$$
$$u_t(x,0) = -\frac{\cos(x)}{(2+ \sin(x))^2}$$
$$u(0,t) = u(2\pi,t)= \frac{1}{2+ \sin(t)}$$

答:
Firstly, you should get homogenous boundary conditions. Let $v(x, t) = \frac{1}{2+ \sin(t)}$. We will seek the solution in the form $u(x, t) = v(x, t) + w(x, t)$, so $w(x, t) = u(x, t) - v(x, t)$. After substitution $u$ into equation, we get $\partial_t^2 w - \partial_x^2 w = -\frac{4 \sin (t)+\cos (2 t)+3}{2 (\sin (t)+2)^3}$. In this case, the initial conditions will take the form $w(x, 0) = u(x, 0) - v(x, 0) = \frac{1}{2+ \sin(x)} - \frac{1}{2}$ and $\partial_t w(x, 0) = \partial_t u(x, 0) - \partial_t v(x, 0) = -\frac{\cos(x)}{(2+ \sin(x))^2} + \frac{1}{4}$, and the boundary conditions $w(0, t) = w(2 \pi, t) =0$. We will find $w$ as $w = y + z$, where

$$\partial_t^2 y - \partial_x^2 y = 0, y(x, 0) = \frac{1}{2+ \sin(x)} - \frac{1}{2}, \partial_t y(x, 0) = -\frac{\cos(x)}{(2+ \sin(x))^2} + \frac{1}{4}, y(0, t) = y(2 \pi, t) =0$$
$$\partial_t^2 z - \partial_x^2 z= -\frac{4 \sin (t)+\cos (2 t)+3}{2 (\sin (t)+2)^3}, z(x, 0) = \partial_t z(x, 0) = z(0, t) = z(2 \pi, t) =0$$

These functions can be found using separation of variables method.


问:
Intuition for degree of maps ($S^n \rightarrow S^n$)
I understand the definition and some properties, but i was just wondering, is there some intuition behind the degree of a map $f:S^n \rightarrow S^n$?

The identity has degree $1$, the antipodal map has degree $-1$. What are maps with degree $0$? Are they constant?

答:
It helps to use the case $n = 1$ as a motivating example. The map $S^1 \to S^1$ in which the domain gets wrapped around the target $k$ times has degree $\pm k$, where the sign depends on whether or not you throw in a reflection. The same idea is more or less true for higher-dimensional spheres, though it gets slightly harder to visualise.

And yes, the special case of degree $0$ is the case in which you wrap around the target $0$ times. Up to a continuous deformation that's the same as a constant map.


问:
Matrices Multiple choice question
Let $P = \begin{bmatrix}
1&1&1\\
0&2&2\\
0&0&3
\end{bmatrix};Q = \begin{bmatrix}
2&x&x\\
0&4&0\\
x&x&6
\end{bmatrix};R = PQ{P^{ - 1}}$. Then which of the following is/are correct,

(A) There exists a real number x such that $PQ = QP$

(B) $\det R = \det \begin{bmatrix}
2&x&x\\
0&4&0\\
x&x&5
\end{bmatrix} + 8$, for all $x\in R$

(C) For x=0, if $R\begin{bmatrix}
1\\
a\\
b
\end{bmatrix}= 6\begin{bmatrix}
1\\
a\\
b
\end{bmatrix}$, then a+b=5

(D)  For x = 1, there exists a unit vector $\alpha \hat i + \beta \hat j + \gamma \hat k$ for which $R\begin{bmatrix}
\alpha \\
\beta \\
\gamma
\end{bmatrix} = \begin{bmatrix}
0\\
0\\
0
\end{bmatrix}$

This is a multiple choice question I have checked only B and it is correct how do I verify other option

My approach for option B is as follow

$\det R = \det P \times \det Q \times \det {P^{ - 1}} = \det Q = 2\left( {24 - 0} \right) - x\left( {0 - 0} \right) + x\left( {0 - 4x} \right) = 48 - 8{x^2}$

$J = \begin{bmatrix}
2&x&x\\
0&4&0\\
x&x&5
\end{bmatrix} \Rightarrow \det J + 8 = 2\left( {20 - 0} \right) - x\left( {0 - 0} \right) + x\left( {0 - 4x} \right) = 40 - 8{x^2} + 8 = 48 - 8{x^2}$ .

How do I verify other options.

Title: Matrices Multiple choice question

Body:
A. FALSE

$$PQ=\left(
\begin{array}{ccc}
 x+2 & 2 x+4 & x+6 \\
 2 x & 2 x+8 & 12 \\
 3 x & 3 x & 18 \\
\end{array}
\right);\;QP=\left(
\begin{array}{ccc}
 2 & 2 x+2 & 5 x+2 \\
 0 & 8 & 8 \\
 x & 3 x & 3 x+18 \\
\end{array}
\right)$$

C. TRUE

$$R(0)=\left(
\begin{array}{ccc}
 2 & 1 & \frac{2}{3} \\
 0 & 4 & \frac{4}{3} \\
 0 & 0 & 6 \\
\end{array}
\right)$$
$$R(0)(1,a,b)=6(1,a,b)\to \left(a+\frac{2 b}{3}+2,4 a+\frac{4 b}{3},6 b\right)=(6,6 a,6 b)
$$
$$
\begin{cases}
a+\frac{2 b}{3}+2=6\to a=4-\frac{2 b}{3}\\
4 a+\frac{4 b}{3}=6a \to b=3;\;a=2\\
6b=6b\\
\end{cases}
$$

D.FALSE

$$R(1)=\left(
\begin{array}{ccc}
 3 & \frac{3}{2} & \frac{1}{3} \\
 2 & 4 & \frac{2}{3} \\
 3 & 0 & 5 \\
\end{array}
\right)$$
$$R(1)(\alpha,\beta,\gamma)=(0,0,0)\to \alpha=\beta=\gamma=0$$
since $\det R(1)=44\ne 0$ the homogeneous system has only the trivial solution $(0,0,0)$.


问:
Determine function from Reimann sum

Determine which of the following is equal to
$$
\lim_{n\to \infty} \frac{34}{n} \sum_{i=1}^n \left(\frac{34^2i^2}{n^2} + 1 \right)
$$
(Answer: the area of the region above the $x$-axis below $f(x)=(x-1)^2+1$, from $x=1$ to $x=35$.)

I got a correct answer only because I knew the final $x$ and the initial $x$ have to have a difference of $34$. I have no idea how this function came about though and I would like some help

答:
The form of the sum suggests you are adding $n$ rectangles of width $34/n$ and different heights given by the sum. The common width, as you mention, suggests that you have $n$ rectangles over an interval of width $34$, which we can represent as $(a,a+34)$.

As for the heights, the height of rectangle $i$, over the interval $\left( a+\frac{34(i-1)}{n}, a+\frac{34i}{n}\right)$ is given by the $i$-th term of the sum, namely $\frac{34^2i^2}{n^2}+1$, which sounds a lot like one of those endpoints plugged into the function $f(x)=x^2+1$.

Can you now finish?

UPDATE

Imagine writing down a Riemann sum of $n$ rectangles for $\int_a^{a+L} \left(x^2+1\right) dx$. The width of one rectangle is $L/n$, and the $i$-th interval is $(a + (i-1)L/n, a+iL/n)$, so if you are using the RHS endpoint the height will be $(a+iL/n)^2+1$ and if you want the LHS endpoint, you get $(a + (i-1)L/n)+1$.

So the total sum for your integral becomes
$$
\int_a^{a+L} \left(x^2+1\right)dx
 = \lim_{n \to \infty}
   \frac{L}{n} \sum_{i=1}^n \left[\left(a+\frac{iL}{n}\right)^2+1\right]
$$

Title: Determine function from Reimann sum


问:
(R - an equivalence relation) if and only if R*R^-1 u I_A = R
Given relation $R:A\rightarrow A$

I need to prove that $R$ - equivalent relation $\iff (R\circ R^{-1}) \cup I_A = R$, where $I_A$ is an identity relation

My attempt:
$(R\circ R^{-1}) \cup I_A = R \iff {
\begin{cases}
R \subset (R\circ R^{-1}) \cup I_A \\
(R\circ R^{-1}) \cup I_A \subset R
 \end{cases}} \iff
{\begin{cases}
R \subset (R\circ R^{-1}) \cup I_A \\
(R\circ R^{-1}) \subset R \\
I_A \subset R
 \end{cases}}
$

Here I can already state that relation is reflexive. However, I have no clue how to show other two properties, namely symmetry and transitive.

答:
Facts: (i) $(R\circ S)^{-1}=S^{-1}\circ R^{-1}$, (ii) $(R^{-1})^{-1}=R$, (iii) $R\subseteq S\implies R^{-1}\subseteq S^{-1}$.

Then, by (i) and (ii), $(R\circ R^{-1})^{-1}=R\circ R^{-1}$ is symmetric; so is $I$, and the union of symmetric relations is symmetric.

It is given that $(R\circ R^{-1})\subseteq R$, hence by (iii), $(R\circ R^{-1})\subseteq R^{-1}$ as well. Thus
$$(R\circ R^{-1})\circ(R\circ R^{-1})\subseteq R\circ R^{-1}$$ which means transitivity of $R\circ R^{-1}$. It then follows easily that $(R\circ R^{-1})\cup I$ is also transitive.


问:
Evaluating Factorial Using Gamma Function
The gamma function is the function $\Gamma: \mathbb{R}_{>0} \rightarrow \mathbb{R}_{\geq 0}$ defined by
$$
\Gamma(t)=\int_{0}^{\infty} x^{t-1} \exp (-x) \mathrm{d} x
$$
8.1: If we use the fundamental theorem of calculus, we get $\Gamma(1)=1 .$

8.2: The gamma function has the following useful property:
$$
\forall t>0, \quad \Gamma(t+1)=t \Gamma(t)
$$
How could we show that $\Gamma(r)=(r-1) ! $  for every $r \in \mathbb{N}$ using  Exercise 8.1 and (8.2)

答: Nothe that  if $\Gamma(z+1)=z\Gamma(z)$, with $\Gamma(1)=1$ then if $z$ is a positive integer we have, $$\Gamma(n)=\Gamma((n-1)+1)=(n-1)\Gamma(n-1)=(n-1)\Gamma((n-2)+1)=\ldots =(n-1)!$$


问:
Find the joint distribution of $(X, Y)$
Let $X$ be a Bernoulli random variable with a probability of success $p$ and let $Y = 1 - X$.

Find the cumulative joint distribution of $(X, Y)$

I have tried to show the joint distribution is the product of the marginal distributions, but apparently, I can not use it since both random variables are not independent.

答:
the requested joint distribution can be represented by the following table

Y=0
Y=1
Total

X=0

1-p
1-p

X=1
p

p

Total
p
1-p
1

Title: Find the joint distribution of $(X, Y)$


答:
Going back to the definition of (discrete) joint probability distribution, you want to find the quantity:
$$
P(X=x,Y=y)
$$
Note that the event $(X=x,Y=y)$ means $(X=x\textrm{ and }Y=y)$. This event zero probability when $x+y\ne 1$ because $Y=1-X$. When $x+y=1$,
you have
$$
P(X=x,Y=y)=P(X=x,Y=1-x)=P(X=x),\quad x\in\{0,1\}
$$

Now you should know how to go on.

Notes.

In the calculation above, $P(X=x,Y=1-x)=P(X=x)$ because $(X=x,Y=1-x)$ and $(X=x)$ are the same event since $Y=1-X$.

Title: Find the joint distribution of $(X, Y)$

Body:
Let $X$ be a Bernoulli random variable with a probability of success $p$ and let $Y = 1 - X$.

Find the cumulative joint distribution of $(X, Y)$

I have tried to show the joint distribution is the product of the marginal distributions, but apparently, I can not use it since both random variables are not independent.


问:
Central elements in the Dihedral Group of order $ 2n.$
Continuing my independent journey through "Abstract Algebra" (see this previous question for context and notation), I am working on:

If $n = 2k$ is even and $n \geq 4$, show that (a) $z = r^k$ is an element of order 2 which (b) commutes with all elements of $D_{2n}$ (Dihedral group of a regular n-gon). Show also that (c) $z$ is the only non-identity element of $D_{2n}$ which commutes with all elements of $D_{2n}$.

(a)  $z^2 = (r^k)^2 = r^{2k} = r^n = 1$, so the order of $z$ divides 2.

$\quad\quad$  Does the hypothesis that $n \geq 4$ come into play here?

(b)  Elements in $D_{2n}$ are of the form $r^i$ or $sr^i$, $0 \leq i \leq n - 1 $

$\quad\quad$  For elements $r^i$, we have: $zr^i = r^kr^i = r^{k + i} = r^{i + k} = r^ir^k = r^iz$,

$\quad\quad$  so $z$ commutes with these elements.

$\quad\quad$ For elements $sr^i$, I am a little confused (yet still trying to remember what needs to be shown!):

$\quad\quad$  $zsr^i = r^ksr^i = (r^ks)r^i = (sr^{-k})r^i = sr^{-k + i}$

$\quad\quad$ Yet $sr^iz = sr^ir^k = sr^{k + i}$

$\quad\quad$ The two don't seem equal to me, so I must be missing something?

(c) The only thing that comes to mind is the argument that the identity in a group is unique,

$\quad\quad$  but I'm having a hard time generalizing/extending that concept to this group.

Thanks in advance for any hints!

答:
For $(a)$, if $n$ is even and less than $4$, then $n = 2 $ (and you have the non-cyclic group of order $4$, the symmetry group of the slightly bizarre "digon") or $n$ is not positive, and there is no dihedral group. Some people don't even like calling the dihedral group of order $ 4$ a dihedral group. The conclusion of $(a) $ and $(b)$ are true when $n=2$, but $(c)$ is not.

For $(b)$, I assume you are worried about claiming $r^k = r^{-k}$, as the rest of the expression is the same (and so could be canceled).  They certainly look different, but if consider the special relation k and n have, I think you'll see they are the same.

For $(c)$, I admit I would use the description of dihedral groups of order 2n as consisting of n rotations and n flips.  "Adjacent" flips (for $n\ge3$) do not commute, so no flip is in the center.  If a rotation is in the center, then it must be the same as its flip.  Only one rotation $r^K$ could possibly satisfy $r^K = r^{-K}$, and this is the key idea of $(b)$ again: $K=k $ has a very special relationship to n.

A conceptual thing for later: when you look at "normalizers" and "centralizers" one really bizarre and useful thing is that the normalizer of a group of order $2 $is always equal to its centralizer.  This is why $z$ is in the center: $z^*z = 1$, and $z$ is normalized by the flips.


问:
Showing that an absolute integrable monotone decreasing function $f: [1,\infty[ \rightarrow \mathbb{R}$ is in $L^p([1,\infty[)$
For an exercise in my analysis course, I have to show that: if $f: [1,\infty[ \rightarrow \mathbb{R}$ is monotone decreasing and $f \in L^1([1,\infty[)$, then $f \in L^p([1,\infty[)$ for every $p > 1$.

Define $f_+ = \max(f,0)$ and $f_{-} = \max(-f,0)$

I tried several things: My guess is that I have to use the integral of $|f|$ to provide an upper bound for the integral of $|f|^p$. I tried the following:

$|f| = f_+ + f_-$. Then $f_+$ is monotone decreasing and $f_-$ is monotone increasing. Moreover $|f|^p = f_+^p + f_{-}^p$ since $f^+$ and $f_{-}$ are never nonzero together. Hence
$$\int_{[1,\infty[} |f|^p d\lambda = \int_{[1,\infty[} f_+^p d\lambda + \int_{[1,\infty[} f_+^p d\lambda$$

Now I want to try to get the monotonicity into play. Let $g_k = f_+ \cdot \chi_{[k,k+1[}$, then $f_+ = \sum^\infty_{k=0} g_k$. We can find a similar sequence for $f_{-}$, let's call that one $h_k$.

\begin{align*}\int_{[1,\infty[} f_+^p d\lambda + \int_{[1,\infty[} f_+^p d\lambda &= \int_{[1,\infty[}(\sum^\infty_{k=0} g_k)^p d\lambda+ \int_{[1,\infty[}(\sum^\infty_{k=0} h_k)^p d\lambda \\&= \int_{[1,\infty[} \sum^\infty_{k=0} g_k^p d\lambda+ \int_{[1,\infty[}\sum^\infty_{k=0}h_k^p d\lambda
\\& = \sum^\infty_{k=0} \int_{[1,\infty[} g_k^p d\lambda+ \sum^\infty_{k=0}\int_{[1,\infty[}h_k^p d\lambda\end{align*}
We are allowed to bring in the exponent because for every $x$ only one of them is nonzero.

\begin{align*}
\sum^\infty_{k=0} \int_{[1,\infty[} g_k^p d\lambda+ \sum^\infty_{k=0}\int_{[1,\infty[}h_k^p d\lambda &\leq \sum^\infty_{k=0} \int_{[1,\infty[} f_{+}(k)^p\chi_{[k,k+1[} d\lambda+ \sum^\infty_{k=0}\int_{[1,\infty[}f_{-}(k+1)^p\chi_{[k,k+1[} d\lambda\\
&=  \sum^\infty_{k=0}(f_{+}(k)^p + f_{-}(k+1)^p)\end{align*}

Now I am not sure how to proceed. Am I looking into the wrong direction? Please only provide hints!

答:
First note that $f_- = 0$, as if $f_-(x^*) > 0$, then for each $x \ge x^*$, we have $f(x) \le f(x^*) < 0$, which implies that $f$ isn't integrable. Now choose $x^*$ such that for $x \ge x^*$ we have $f(x) \le 1$. Then
\begin{align*}
  \int_1^\infty f(x)^p\, dx &= \int_1^{x^*} f(x)^p\, dx + \int_{x^*}^\infty f(x)^p\, dx\\
       &\le \int_1^{x^*} f(1)^p\, dx + \int_{x^*}^\infty f(x)\, dx\\
       &\le (x^*-1) \cdot f(1)^p + \int_1^\infty f(x) \, dx\\
       &< \infty.
\end{align*}</p>


问:
Having problems with computing two-sided limtis
I am having problems calculating sided limits. I understand what a sided limit is (both graphically and analytically), but I'm having problems calculating them and I really need advice.

Here's an example:

$f(x) = \frac{x^2-2x}{(x^3-3x^2)^{\frac{2}{3}}}$

WolframAlpha returns that the left limit is $\infty$, and that the right limit is $-\infty$. But, when I try to calculate them, I get the opposite results.

For example, if I simply try to "plug in $0_{+}$, I get this:

$$\frac{ (0_{+})^2 - 2\cdot0_{+}}{(0_{+})^3 -3\cdot (0_{+})^3)^{\frac{2}{3}} }$$

I'm not quite sure how to calculate this. I'm not even sure if it can be done by simply plugging in the value. I'd really appreciate if anyone could shed some light on this example and give advice on how to calculate them generally speaking.

答:
You have to state in what intervals is positive
$$f(x)=\frac{x^2-2x}{(x^3-3x^2)^{2/3}}$$
Denominator is positive for any $x$ so the function is positive where the
numerator is positive, that is, in $(-\infty,0)\cup(2,+\infty)$

$$\underset{x\to 0^+}{\text{lim}}\;f(x)=-\infty$$
because for $x>0$ the function is negative

$$\underset{x\to 0^-}{\text{lim}}\;f(x)=+\infty$$

Therefore the limit at $x=0$ doesn't exist, while the limit at $x=3$ is $+\infty$ from both directions.


问:
Does existential import rejects subcontraries?
In Categorical logic (AEIO) I and O are subcontraries. They both can not be False at the same time.

What happens after integrating existential import?

My book says:

Two propositions are subcontraries if they cannot be false together.
Yet I- and O-propositions are both false when there is no S. If there
is no S, then there is no S that is P and there is no S that is not P.
It just is nothing at all. So I- and O-propositions are no more
subcontraries.

Lee, Siu-Fan. Logic: A Complete Introduction: Teach Yourself (Complete
Introductions) (p. 158). John Murray Press. Kindle Edition.

But I and O propositions always have existential import. And this post confirms that subcontraries retain.

答:
In Categorical logic (AEIO) I and O are subcontraries. They both can not be False at the same time.

What happens after integrating existential import?

Seems like you got this just the wrong way around.

With (i.e. after integrating existential import) Existential import (i.e. that the domain cannot be empty), any I and O pair of statements cannot both be false, and hence are subcontraries.

Without existential import (i.e. before integrating existential import), it is possible for a domain to be empty, and in such a world any existential is false, meaning that any I and O statements would be false, and thus without existential import, any I and O pair of statements are not subcontrary.

That website, by the way, that states that for universals you can do away with existential import, but that for existentials, existential imposrt is retained ... is really weird!  This would mean that if I gave you a number of statements involving both existentials as well as universals, then given the existential you do have existential import, but for the universal you don't?!?  That's nuts: any logic should have a consistent semantics; you can't just choose to have a different semantics for one statements and a different semantics for another. Indeed, I have never seen anything like what this website is claiming you can do.


答: The standard topology can also be defined as the product topology on the $n$th power and the topology on $\Bbb R$ just by the order on it. No group structure even on the factor is needed. But if you want a metric, you'll probably need some formula, and hence some algebraic structure, most likely. But purely from a topological viewpoint the order (plus set-theoretic Cartesian product structure) suffice.

Title: We need vector space structure on $\mathbb{R^n}$ in order to define the standard topology?

Body:
I was watching a lecture on topology and the professor said that you don't need the vector space structure on $\mathbb{R^n}$ in order to define the standard topology for $\mathbb{R^n}$. But in the definition of the open ball (that he used to define the standard topology), he wrote something like this:

$$B_r(x):=\bigg\{y=(y_1, ...,y_n) \in \mathbb{R^n}:\sqrt{\sum_{i=1}^n(y_i-x_i)^2}<r\bigg\}$$

But how can we perform "$(y_i-x_i)^2$" in $\mathbb{R^n}$, only using its set and topological structure? And if I need only an aditional structure other than a vector space one, what is that?

P.S.: Sorry for the broken English.


问:
$J(u+v)=J(u)$ for J convex and homogeneous
Suppose J is a convex and p-homogeneous (p>0) functional. That means $J(\lambda u +(1-\lambda)v) \leq \lambda J(u) + (1-\lambda) J(v)$ and $J(\lambda u)=|\lambda|^p J(u)$ for $u,v$ in the domain and $\lambda$ a real number.

I need to show $J(u+v)=J(u)$ for all $v \in \mathcal{N}(J)$ (the nullspace of $J$, i.e. $J(v)=0$). I have managed to prove this for $p\leq 1$ but am having trouble with the case $p>1$. Here is what I've done: ($p \leq 1$)

$J(u+v) = 2^{p} J(\frac{u}{2} +\frac{v}{2}) \leq 2^{p}(\frac{1}{2} J(u)+ \frac{1}{2} J(v))=2^{p-1} J(u)\leq J(u)$;

$J(u)=J(u+v-v)=2^p J(\frac{u+v}{2}-\frac{v}{2})\leq 2^{p} (\frac{1}{2} J(u+v)+ \frac{1}{2} J(-v))=2^{p-1} (J(u+v)+J(v))=2^{p-1} J(u+v) \leq J(u+v)$.

答:
I've actually just figured this out.

Take any $\lambda \in (0,1)$, then

$J(u+v) = J(\lambda \frac{u}{\lambda} +(1-\lambda) \frac{v}{1-\lambda}) \leq \lambda J(\frac{u}{\lambda})+ (1-\lambda) J(\frac{v}{\lambda})=\lambda^{1-p} J(u) + (1-\lambda)^{1-p} J(v)= \lambda^{1-p} J(u)$.

Take the limit as $\lambda \rightarrow 1$ and we get $J(u+v) \leq J(u)$. The other direction is similar.


问:
Pairs in a deck of cards
I had this question, and I was wondering how to do it. The question was what is the probability of not drawing a pair out of a deck of cards, and I wasn't sure how to do it could someone help me out</p>

答:
Since you are drawing two cards without replacement, you can determine the probability of them not being a pair through examining the second card drawn.

After drawing the first card there will now be 51 cards remaining in the deck (since you have drawn one), and there are only 3 cards remaining that can form a pair with your first card.

Now you can easily find the probability of not drawing one of those three cards, and consequently, the probability of not having a pair.

Title: Pairs in a deck of cards


问:
Does bijectivity imply the same order type?
Question: Given an ordered set A which must have bijective correspondences with set B, there must be a correspondence that preserve order?

I know it's wrong. For example, 1.the dictionary order on $\mathbb{Z}_+\times\mathbb{Z}_+$, 2.$(x_0,y_0)<(x_1,y_1)$ if either $x_0+y_0<x_1+y_1$, or $x_0+y_0=x_1+y_1\,and\,y_0<y_1$. They must have bijective correspondence because they have same amount of elements, but are in different order type.

But I feel uncomfortable with that because we all have an intuition that for two groups with same amount of elements( For example, if I can match 1~11, 2~12, 3~13....9~19, 10~20 for sets {1, ... ,10} {11, ... ,20}, I can actually match 1 with 12 with other every elements in {1, ... ,10} matches with elements in {11, ... ,20} ), we can build random correspondence(some correspondence preserves order in this case) between them.

What's wrong with this intuition.

答:
$\mathbb{Z}$ and $\mathbb{N}$ are bijective but there's no order-preserving map between them. Loosely speaking, this is to do with the fact that $\mathbb{N}$ has a starting number, but $\mathbb{Z}$ doesn't.

$\mathbb{Z}$ and $\mathbb{Q}$ are bijective but there's no order-preserving map between them. This has to do with $\mathbb{Q}$ being dense, whereas $\mathbb{Z}$ isn't.

Some sets of cardinality $|\mathbb{R}|$ also have same property: I don't think there is an order-preserving map between $f: [-1,0] \cup [1,2] \to [0,1]$ because $f(0) < f(1)$, which means that for  $f(0) < b < f(1)$ there is no $a \in [-1,0] \cup [1,2]$ such that $f(a) = b$.

I'm not sure how to characterise all these results, but I'm sure it can be done.

Anyway, hopefully I have shown that it is clear that your intuition that "Given an ordered set $A$ which is bijective to set $B$, there must be a correspondence that preserves order" is incorrect. I understand why you have this intuition, but if it's incorrect then it's incorrect.


问:
Finding density function of reciprocal of Rayleigh random variable
I have a Rayleigh distributed random variable $X$. How can I find pdf of another random variable $Y= \frac{1}{X}$.

Using some knowedge of transformations I found that pdf of $Y$,  $f_Y(y) = \frac{-1}{x^2}f_X(\frac{1}{x^2})$, but I am not sure whether I am correct.

答:
Hint.

The PDF is the derivative of CDF. By the definition of CDF
$$
P(Y\le y)=P(X\ge 1/y)=1-P(X<1/y)\quad y>0
$$

Now use the Rayleigh distribution assumption.

For your edited question, you make two mistake on the right: the variable should by $y$, not $x$; inside the density, you should use $\frac1y$.


答:
Order mattering, all events considered:  There are $52\times 51$ possible ways to draw two cards.  The number of ways to draw two cards such that a pair is not drawn is $52\times 48$, seen by picking the first card and then picking the second card such that it is not the same rank as the first.  The probability is then $\dfrac{52\times 48}{52\times 51}$

Equivalently, one could pick the rank of the first and then the suit of the first, then the rank of the second and the suit of the second giving $\dfrac{13\times 4\times 12\times 4}{52\times 51}$

Order not mattering, all events considered:  There are $\binom{52}{2}$ possible ways to draw two cards.  The number of ways to draw two cards such that a pair is not drawn is $\binom{13}{2}\times 4\times 4$, seen by picking the two ranks simultaneously then picking the suit of the lower selected rank and then the suit of the higher selected rank.  The probability is then $\dfrac{\binom{13}{2}\times 4\times 4}{\binom{52}{2}}$

Order mattering, events simplified:  Given whatever first card was drawn, there are $48$ cards in the $51$ remaining cards which do not match the same rank as the first.  The probability is then $\dfrac{48}{51}$

All of these of course give the same answer.  Which you choose with regards to order mattering or not, when you account for suits as opposed to ranks, etc... are largely irrelevant so long as you are consistent.  Ideally, you pick whichever explanation makes the most sense to you and/or simplifies arithmetic the most.

Title: Pairs in a deck of cards

Body: I had this question, and I was wondering how to do it. The question was what is the probability of not drawing a pair out of a deck of cards, and I wasn't sure how to do it could someone help me out</p>