问:
How to estimate the true rank of a matrix, given an estimation of the matrix?
Suppose I have a matrix $A$ and an estimation of the matrix $\hat{A} = A + \varepsilon$ where $\varepsilon$ is some noise.

I want to estimate $m = \text{rank}(A)$, but only have access to $\hat{A}$. How could I do it? I have tried to look at the singular values, but it is not clear where to put the threshold.

答:
It is impossible to predict the rank of $A$ exactly unless $\hat{A}$ has full rank because for every $n$ less than the full rank the set $\{A: \mathrm{rank}(A) > n\}$ is dense in the set of all matrices. However, you can estimate the rank from below if $\varepsilon$ is sufficiently small. This is because
$$\|s_k(A) - s_k(\hat{A})\| \leq \|A - \hat{A}\| \leq \varepsilon$$
for all (ordered) singular values $s_k$. Hence if the largest $n$ singular values of $\hat{A}$ are larger than $\varepsilon$, then $A$ has at least rank $n$.


问:
Variation in density of the Farey sequence in the interval $[0,1]$
I am curious about how the Farey sequence of order $M$ — which is to say, the sequence of rational numbers from $0$ to $1$, which may be expressed with denominator at most $M$ — is distributed on the real line; and in particular how the 'density' of the Farey sequence varies in this interval as $M \to \infty$.

Consider the following diagram. For varying $M$, we plot the members of the Farey sequence of order $M$. Of course, they are not evenly distributed, and as $M$ increases we see some very obvious clustering behaviour (e.g., among the lowest and highest elements of the sequence, albeit excluding $0$ and $1$ themselves).

Because the Farey sequences (for any given $M$) are finite, clearly we cannot ask meaningful questions literally concerning their 'density' at $x$. But we can consider related concepts — for example, it might be meaningful to consider how many rationals with denominator bounded by $M$, there are in the closed interval $[x\!-\!\tfrac{1}{2M},x\!+\!\tfrac{1}{2M}]$. As $M$ increases and the width of that interval decreases, we approach questions of density or measure. (And in any case, we can make statements about how the number of bounded-denominator rationals vary spatially in such neighborhoods of non-zero width.)

It is easy to imagine a range of similar questions, which in some cases might admit an analytic or combinatorial solution, or which might give rise to 'density' plots which in an asymptotic limit converge to some more-or-less cauliflowery fractal. Other big-picture quantitative perspectives on the variation of density of bounded-denominator rationals might also be available.

What can be said, or what is known, about the 'density' of Farey sequences of order $M$, as $M \to \infty$?

(N.B. This question has been edited to reflect the terminology suggested  in the comments, but is otherwise unchanged.)

答:
My original question is trying to get at some concepts of 'density', which may be contrasted with some coarse-grained results about the asymptotic distribution of the Farey sequence, as indicated in the comments to the question.

The following summarizes what can be said about the asymptotic distribution of the Farey sequence, which is fairly well-known (if slightly poorly documented on Wikipedia article on the Farey sequence). I post this in order to unpack the observations made in the comments, and also to contrast with some elementary observations about the idea which I would like to explore, which is about finer-grained notions of density which may be pertinent for finite $M$. For the time being I have no answers regarding that more precise notion. In the mean time, casual mathematicians such as I am  may find the better-known, coarse-grained notions of "the distribution of the Farey sequence"  of interest.

Coarse-grained distribution

As pointed out by @GerryMyerson in the comments, there is a meaningful sense in which, despite first appearances, the Farey sequences approach a near-uniform distribution over the unit interval.
(Indicative results are mentioned in the Wikipedia article on the Farey sequence, but a number of the results below are taken from Leodan [1] with minor changes in notation; results without citations are ones which Ref. [1] seems to represent as classic.)

I. First-order analysis

From the scaling of the length of the Farey sequence $F_M$ of order $M$,
$$ \lvert F_M \rvert = \frac{3 M^2}{\pi^2} + O(M \log M) $$
it immediately follows that $N_M := \lvert F_M \rvert$ is the average density of the Farey sequence over the entire interval $[0,1]$. Remarkably, to first order, this is also the density over any interval of constant size as well.
For an interval $I = [\alpha,\beta] \subseteq [0,1]$, if we let
$$ N_{M,I} = \lvert F_M \cap I \rvert $$
it then follows that
$$ \delta_{M,I} := \frac{N_{M,I}}{N_M} = \bigl(\beta - \alpha\bigr)\Bigl(1 + O\bigl(\tfrac{\log M}{M}\bigr)\Bigr) + O(M^{-1+\epsilon}) $$
for some (all?) $0 < \epsilon < 1$.
That is to say, up to minor corrections in $M$, it depends only on the measure of the interval $I$.
In this sense, on any interval $I \subseteq [0,1]$ of constant size, the Farey sequence $F_M$ converges relatively quickly to an approximately uniform distribution as $M \to \infty$.

II. On the discrepancy of the coarse-grained Farey sequence from first order distribution

One can ask just how close the Farey sequence is to being uniform over intervals of constant width. Generalising $\delta_{M,I}$ to arbitrary intervals $I \subseteq [0,1]$, let
$$ R_M(x) = \bigl\lvert \delta_{M,(0,x]} - x \bigr\rvert $$
be the 'local discrepancy' of the Farey sequence from uniform, on the interval $I = (0,x]$ for $0 \leqslant x \leqslant 1$.
Then, we may define
$$ \mathbf D_M := \sup_{0 \leqslant x \leqslant 1} R_M(x) $$
which measures the supremum of all 'local discrepancies' of the Farey sequence from uniform, over $(0,x]$ in the unit interval.
Remarkably, Dress [2] showed the tight result $ \mathbf D_M = 1/M $: then for these constant-width intervals $(0,x]$, the deviation of the density from uniform can be bounded quite precisely.

Beyond this — for instance, whether deviations of magnitude 'close' to $1/M$ are rare, or whether instead the distribution tends to be much closer to uniform — it is difficult to say more without running into important unproven conjectures.

Define the quantitites
$$\begin{align*}
S_M &:= \!\! \sum_{1 \leqslant j \leqslant N_M} R^2_M(f_{\!M,j}) \\[3ex]
T_M &:= \!\! \sum_{1 \leqslant j \leqslant N_M} R_M(f_{\!M,j})
\end{align*}$$
where $f_{\!M,j}$ is the $j^{\;\!\text{th}}$ element of $F_M$.
That is, $T_M$ is the cumulative discrepancy of the Farey sequence from uniform, over the intervals $(0,f_{\!M,j}]$, for each element $f_{\!M,j}$ of the sequence $F_M$; and $S_M$ is the corresponding cumulative discrepancy-squared.
One might wonder what can be said about these cumulative values, to help bound the deviation of the Farey sequences from the uniform distribution.

From Dress' result above and the scaling of $N_M$, we know that $S_M \leqslant 3/\pi^2 + O(M^{-1} \log(M))$ and $T_M \leqslant 3M/\pi^2 + O(\log M)$. So we do have some bounds on these quantities.

However, Franel [3] proved that the infimum of $\theta$ for which $S_M = O(M^{-2+2\theta})$, is the supremum of the real parts of the zeroes of the Riemann zeta-function. Thus, the Riemann Hypothesis is equivalent to $\forall \epsilon > 0: S_M = O(M^{-1+\epsilon})$.

Franel's result was subsequently strengthened by Landau [4] who showed that
the Riemann Hypothesis is equivalent to $\forall \epsilon > 0 : T_M =
   O(M^{1/2 + \epsilon})$.

That is to say: some questions of precisely how uniformly, the Farey sequence converges to the uniform distribution over intervals of constant size, run against well-known frontiers in mathematics.

Initial observations about fine-grained distribution

These observations regarding the uniformity of the distribution of the Farey sequence over intervals of constant width are all well and good. However, they largely correspond to integration of something like a density function over an interval, which may cover a multitude of sins.
Furthermore, this asymptotic behaviour fails to capture interesting features of the distribution of Farey sequences on a finer-grained level, which is relevant for representing real numbers in data structures.

For $0 \leqslant j < N_M$, consider the sizes of the gap $g_{M,j} = f_{\!M,{j+1}} - f_{\!M,j}$ between Farey neighbours $f_{\!M,j}$ and $f_{\!M,{j+1}}\,$. Then we have
$$\begin{align*}
 g_{M,0} &= 1/M \\
 g_{M,1} &= 1/M(M-1)
\end{align*}$$
which is a rather significant change in the spacing of Farey neighbours. Asymptotically, this jump would be invisible, occurring just about the origin as it does; this and other erratic behaviour at scales close to $1/M$ would simply average out.

In order to consider something like the Farey sequence in neighborhoods of arbitrary size about any real number, let $\mathcal F_M$ be the extended Farey set of all rational numbers $q$ (not just $0 \leqslant q \leqslant 1$) which may be expressed with a denominator at most $M$.
Define
$$ \rho_{M,\epsilon}(x) = \frac{1}{\epsilon N_M}\cdot \Bigl\lvert \mathcal F_M \cap \bigl[x\!-\!\tfrac{1}{2}\epsilon, x\!+\!\tfrac{1}{2}\epsilon\bigr] \Bigr\rvert$$
to be the "$\epsilon$-scale density" of the Farey numbers $\mathcal F_M$ about $x$. For any fixed $\epsilon > 0$, if we integrate $\rho_{M,\epsilon}$ over the unit interval, we should obtain precisely $1$; and the coarse-grained distribution results above imply that again for fixed $\epsilon > 0$ and $x \in \mathbb R$, $\rho_{M,\epsilon}(x) = 1 + O(M^{-1} \log(M))$. But if we let $w = 1/M$ to consider the "density" of Farey numbers in small neighborhoods about $x$, we may obtain precise (and quite distinct) values for the local density:
$$\begin{alignat}{2}
 \rho_{M,w}(0) &= \qquad\frac{M \cdot 1}{N_M} &&= \frac{\pi^2}{3 M} + O(M^{-2}) ; \\[3ex]
 \rho_{M,w}\Bigl(\tfrac{3}{2M}\Bigr) &= \frac{M \cdot \bigl(\lceil M/2 \rceil + 1\bigr)}{N_M} &&= \frac{\pi^2}{6} + O(M^{-1}).
\end{alignat}$$
That is to say: the density jumps from vanishing with $M$, to a value which differs slightly from $\pi^2/6 \approx 1.645$, and does so in a fairly small span. This structure will simply be invisible asymptotically for an interval of fixed size, and shows that inasmuch as we are interested in the behaviour of a density function as opposed to a mass function on intervals, the convergence of the distribution of the Farey sequence to a uniform distribution cannot, itself, be uniform.

In light of the fact that some of this structure will be invisible as $M \to \infty$ for intervals of fixed size, there is a question of how even to consider the asymptotic behaviour of the density in this fine-grained sense. It is possible that some of this structure could be fruitfully described only by making non-trivial choices of the parameterisation of the interval $[0,1]$. However, even without such devices, one might ask such questions as the measure of the set $\{ x \in [0,1] : \rho_{M,w}(x) > 1 + kw\}$ for fixed constants $k$, e.g., for $w=1/M$, as $M \to \infty$.

Summary

The asymptotic behaviour of the Farey sequence on any given non-empty interval in $[0,1]$ is "uniform", to first-order approximation, in the sense that 'integrating' it over any fixed non-empty interval (counting the elements and dividing by the interval width and $N_M = \lvert F_M \rvert$) converges to $1$ for any interval. Furthermore, the convergence is relatively quick, with some tight known bounds on convergence for intervals of the form $(0,x]$ for $0 \leqslant x \leqslant 1$. (Ask for more sophisticated bounds on convergence than this, and you may end up asking questions equivalent to the Riemann Hypothesis.)

However, this does not address more fine-grained notions of 'density' of the Farey sequence, for which elementary observations yield results which show that these well-known coarse-graining results arise from averaging out some quite significant and possibly interesting variations in small-scale behaviour (e.g., in neighbourhoods of size $\epsilon = 1/M$) for finite $M$.

References

[1] A. H. Ledoan. The discrepancy of Farey series. Acta Math. Hungar. 156 (465–480), 2018. https://doi.org/10.1007/s10474-018-0868-x

[2] F. Dress. Discrépance des suites de Farey. J. Théor. Nombres Bordeaux, 11 (345–367), 1999.

[3] J. Franel. Les suites de Farey et les problèmes des nombres premiers. Nachr. von der Ges. der Wiss. zu Göttingen, Math.-Phys. Kl., (198–201), 1924.

[4] E. Landau. Bemerkungen zu der vorstehenden Abhandlung von Herrn Franel. Nachr.
von der Ges. der Wiss. zu Göttingen, Math.-Phys. Kl., (202–206), 1924.

ContentLicense: CC BY-SA 4.0


问:
Showing that an open unit disk is homeomorphic to the surrounding space
I was wondering that can I show that an open unit disk in $\mathbb{R}^n$ is homeomorphic to the surrounding space $\mathbb{R}^n$ by arguing that all of the half-axes of the space are homeomorphic to the half-axes of the unit disk. By half-axes of the space I refer to the two parts, $(-\infty, 0], [0, \infty)$ of each coordinate axis, and by half-axes of the open unit disk I refer to the intervals $(-1, 0], [0, 1)$ along all of the coordinate axis of the surrounding space.

So essentially I would argue that by some univariate functions $f$ and $-f$ all $[0, 1) \approx [0, \infty)$ and $(-1, 0] \approx (-\infty, 0]$. Then the mapping $g(x_1, x_2,\dots,x_n)$ from the $n$-dimensional open disk to the $\mathbb{R}^{n}$ could be constructed by considering the sign of the arguments, so that if $x_i \in [0, 1)$ then the $i$th coordinate of the image of $g$ is determined by $f$, and if $x_i \in (-1, 0]$ then it is determined by $-f$.

答:
Map $x \in \Bbb R^n$ to $f(x)=\frac{1}{1+\|v\|}\cdot v$, where $\cdot$ is scalar multiplication and the norm is the standard Euclidean one. So every vector is scaled down (in the same direction) so that its norm becomes

$$\left\| \frac{v}{1+\|v\|} \cdot v \right\| = \left(\frac{1}{1+\|v\|}\right)\|v\| = \frac{\|v\|}{1+\|v\|} < 1$$

and so $f$ maps $\Bbb R^n$ into the open unit ball $B(0,1)$.

$f$ is clearly continuous as the norm as a function is continuous and division in $\Bbb R$ and scalar multiplication are too.

$f$ is invertible as we can solve $f(tv)=v$ for $v \in B(0,1)$ uniquely and it's easy to see we get $f^{-1}(v)=\frac{1}{1-\|v\|}\cdot v$ which is also continuous (and well-defined on $B(0,1)$).


问:
Angle bisected in a parallelogram

$ABCD$ is a parallelogram and $ID =JB$.

the point $P$ is the intersection of $DJ$ and $IB$.
Prove that the angle $X=Y$

$a,b,α,β$ are angles.

What I realized:

$a=b$.

and if we sum up these angles.
$$180=180-(x+α)+180-(y+β)+a $$
$$\iff 180=x+y+α+β-a \iff 360+2a=2(x+y)+2(α+β)$$

Please Help.

答:
It is straightforward by sine law,

$\angle PID = 180^0 - \beta, \angle PJB = 180^0 - \alpha, \angle DPI = \angle BPJ = \gamma$

So in $\triangle PID,  \frac{\sin \beta}{PD} = \frac{\sin \gamma}{ID}$

Similarly in $\triangle PBJ,  \frac{\sin \alpha}{PB} = \frac{\sin \gamma}{BJ}$

As $BJ = ID$, $ PD \sin \alpha = PB \sin \beta$

Now in $\triangle PCD,  \frac{\sin X}{PD} = \frac{\sin \alpha}{PC}$

In $\triangle PBC,  \frac{\sin Y}{PB} = \frac{\sin \beta}{PC}$

So, $\frac{\sin X}{\sin Y} = \frac{PD \sin \alpha}{PB \sin \beta} = 1$

$\therefore \angle X = \angle Y$

Title: Angle bisected in a parallelogram


问:
How can I find out which function is greater problem
I asked a similar question before but I got no answer so I will try my luck again with a different one.

We have the following function
$$f(x)=2\ln(x)-\frac{c}{x}$$ where $c$ is a positive number.
Which is greater?

$f(e)$

$f(\pi)$

they equal

we cannot determine because we don't know the value of $c$.

What I tried was:

I tried finding the derivative first , and I got $f'(x)=$$\frac{2x+c}{x^2}$ , then I tried to find the maximum or minimum thought it might give me a lead I compared $f'(x)=0$ and got that Maxima/minima is $\frac{c}{2}$ I don't know if this info will help but I kept going to put the value I found in the original function $f($$\frac{c}{2})$ ,  $f($$\frac{c}{2})=2\ln(c)-\ln(2)-2$ we know that this point can be a minimum or maximum but as I said I don't know if this info will help..

I also tried moving the graph a on desmos and found out the as long as c is positive $f(\pi)$ is greater
but I wanna know why .. I cant seem to figure it out.

I need help and advice on how to approach these type of questions, please I would appreciate it if anyone can give me tips for questions like that , the final answer is not that important.

thank you.

答:
This is for your original version of the problem, where $f(x)=2\ln(x)-\frac{c}{2}$.

You are making the problem more complicated than necessary.

Note that the logarithm is an increasing function and $e<\pi$. So you have $\ln(e)<\ln(\pi)$, which implies that $2\ln(e)+C<2\ln(\pi)+C$, where $C$ is any constant.

Now your problem is changed to
$$
f(x)=2\ln(x)-\frac{c}{x}\quad (c>0)
$$

Taking the derivative, for any $x>0$, you have (since $c>0$)
$$
f'(x)=\frac{2}{x}+\frac{c}{x^2}>0
$$
So $f$ is increasing for $x>0$ and thus $f(e)<f(\pi)$.


问:
Proving non-uniform-continuity for f,g when g(x) > f(x) for all x
Suppose $f, g$ are continuous functions on $\mathbb{R}$, s.t $f$ is not uniformly continuous on $\mathbb{R}$ and $g(x) > f(x)$ for all x. Is g not uniformly continuous?

This seems true, and I'm trying to prove this with an 'epsilon delta' proof, without much success.

I tried to negate the uniform continuity condition, which means I need to find x,y that will make $g(x)$ and $g(y)$ further apart than $f(x)$ and $f(y)$ are (so the distance will be bigger than/equal to epsilon).

Any help at all is appreciated :) Preferably a hint

答: This is not true. Consider $f(x) = \sin(x^2)$ and $g(x) = 2$.


问:
What's the area of the shaded blue cresent
This was a posted problem on a Facebook group. It was very challenging that how can you  identify the smaller radius and the larger circle's radius if the given values aren't given either on the two radiuses can anyone provide me an answer?? Both geometric  and trigonometry method?

答:
From the geometric mean theorem, see https://en.wikipedia.org/wiki/Geometric_mean_theorem, we get
$$(R-25)^2=(R-40)\cdot R,$$
that is, $R=62.5$.  From $2r=2R-40$ we have $r=42.5$.


问:
Why is the direct product of two groups $G_1 \times G_2$ too small to be their coproduct?
We’re looking at the following definition in my class.

Definition. The coproduct $X_{1} \amalg X_{2}$ of $X_{1}$ and $X_{2}$, together with the morphisms $i_{j}: X_{j} \rightarrow X_{1}\amalg X_{2}$, is characterized by the following universal property: Given any object $Y$ with morphisms $f_{j}: X_{j} \rightarrow Y,$ there exists a unique $f: X_{1} \amalg X_{2} \rightarrow Y$ such that $f_{j}=f \circ i_{j}$.

In $G_1 \times G_2$, there exist two subgroups isomorphic to $G_1$ and $G_2$, namely the sets of elements $\{(g_1,e_2): g_1 \in G_1\}$ and $\{(e_1,g_2): g_2 \in G_2\}$, respectively. These two subgroups have the property that their members commute with each other: $(g_1,e_2) \cdot (e_1,g_2) = (g_1,g_2) = (e_1,g_2) \cdot (g_1,e_2)$.

The reason given for the direct product not being the coproduct is that, in a group $Y$ with maps $f_1: G_1 \to Y$ and $f_2: G_2 \to Y$, elements of $Y$ don’t commute with each other. I’m not yet able to derive this.

My understanding of the argument is that, given the property of the two subgroups above, I ought to be able to derive $f_1(g_1) \cdot_Y f_2(g_2) = f_2(g_2) \cdot_Y f_1(g_1)$, which would not be true for a general group $Y$ (unless $Y$ is abelian).

Now if I also assume $G_1 \times G_2$ to be the coproduct, I’d have:
$$f_1(g_1) \cdot_Y f_2(g_2) = (f \circ i_1)(g_1) \cdot_Y (f \circ i_2)(g_2) = f(g_1,e_2) \cdot_Y f(e_1,g_2)$$

It’s here that I don’t know how to keep going to get the relation above.

答:
As you observed, the main issue is that elements of $G_1$ always commute with elements of $G_2$ in $G_1\times G_2$, so it is not a free construction.

Notably, the direct product does coincide with the coproduct (satisfies its universal property) in the category of Abelian groups.

Choose specifically $X_1=X_2=\Bbb Z$, and consider the cocone $X_1\to X_1\times X_2\leftarrow X_2$ with $n\mapsto (n,0)$ and $n\mapsto (0,n)$.

Then a homomorphism $f_i:X_i\to Y$ basically just selects an arbitrary element $y_i$ of $Y$ (taking $y_i:=f_i(1)$), so in order for the diagram to commute, as $(1,0)$ and $(0,1)$ commute in $G_1\times G_2$, we would need that $y_1y_2=y_2y_1$ in $Y$:

If an $f:X_1\times X_2\to Y$ makes the diagram commutative, we have
$$f(1,0)=f(i_1(1))=f_1(1)=y_1\\
f(0,1)=f(i_2(1))=f_2(1)=y_2\\
f(1,1)=f((0,1)+(1,0))=f(1,0)f(0,1)=y_1y_2\\
f(1,1)=f((1,0)+(0,1))=f(0, 1)f(1,0)=y_2y_1\,.$$

So, there is no arrow $X_1\times X_2\to Y$ making the diagram of the universal property commutative, whenever $y_1,y_2$ are noncommutative in $Y$.


问:
Proving path length, transitive closure
Set A is finite with n elements. Suppose a and b are elements of a set A with a != b. Let R be a relation on the set A so that there is a path from a to b of length at least 1. Show there is a path from a to b of length not exceeding n − 1.

I understand the proof for when a does equal b. And though I understand conceptually how this proof works, I am unable to put it in words.

答:
I will try to explain using less formal language.

Assume there is a path at least one in relation $R$ on Set $A$ from $a$ to $b$. Let $m$ be the shortest length of such path, so $x_0,x_1,x_2,...,x_{m-1},x_m$ is the path where $x_0 = a$ and $x_m = b$.

In the case when $a = b$, suppose $m > n$, we have $n$ vertices in total in the set $A$, so among the $m$ vertices $x_0,x_1,x_2,...,x_{m-1}$, there will be at least two vertices that are equal, according to the pigeonhole principle.

If two vertices are equal, then there will be a loop in the path and we can get rid of the loop to obtain the shortest path, so there will be at least $n$ path (i.e., $m = n$).

In the case when $a \not = b$, among the $m$ vertices $x_0,x_1,x_2,...,x_{m-1}$, no vertices can be equivalent to $b$, otherwise $m$ will not be the shortest path from $a$ to $b$. So there will be $n-1$ choices of vertices for the $m$ vertices. If $m > n - 1$, then by the pigeonhole principle, at least two vertices will be the same. Then if two vertices are equal, we can again get rid of the loop and obtain the shortest path, so there will be at least $n-1$ path.

ContentLicense: CC BY-SA 3.0

Title: Proving path length, transitive closure


问:
Where to find the proof for the isomorphism between the (2n-1)_th homotopy group of Aut(V) to Z?
I am learning Bott-Seeley's comment article on the Callias' index theorem. They quoted a theorem, that is, $\pi_{2n-1}(Aut(V_{C}^{n}))\simeq Z$ or $\pi_{2n-1}(GL_n(C))\simeq Z$. Can you tell me the key to the proof for this theorem or direct me to some reference books? Thanks a lot!</p>

答:
Since $U(n)$ is the maximal compact subgroup of $GL_n(\mathbb{C})$, they are homotopy equivalent, so $\pi_{2n-1}(GL_n(\mathbb{C}))=\pi_{2n-1}(U(n))$.

For any positive integer $k$, we have $U(n+k)/U(n+k-1)=S^{2n+2k-1}$, so we have the exact sequence of homotopy groups: $\pi_{2n}(S^{2n+2k-1})\rightarrow \pi_{2n-1}(U(n+k-1))\rightarrow \pi_{2n-1}(U(n+k))\rightarrow \pi_{2n-1}(S^{2n+2k-1})$.

Now $2n+2k-1>2n$, so $\pi_{2n}(S^{2n+2k-1})=\pi_{2n-1}(S^{2n+2k-1}) =0$, which means $\pi_{2n-1}(U(n+k))\cong\pi_{2n-1}(U(n+k-1))\cong\cdots\cong\pi_{2n-1}(U(n))$. So $\pi_{2n-1}(U)\cong \pi_{2n-1}(U(n))$.

Here comes the key point: By Bott periodicity, $\pi_{2n-1}(U)\cong \pi_1(U)\cong\mathbb{Z}$, so $\pi_{2n-1}(U(n))\cong \mathbb{Z}$.


答:
$2R-2r=40 \implies R-r = 20$ .. (1)

Use Pythagoras to find x

$x = \sqrt{2Rr-R^2}$

Now, $x + 25 = R$

$ \implies \sqrt{2Rr-R^2} = R - 25$ .. (2)

Finally, solve (1) and (2) to get $R = 62.5$ ; $r = 42.5$

Title: What's the area of the shaded blue cresent

Body: This was a posted problem on a Facebook group. It was very challenging that how can you  identify the smaller radius and the larger circle's radius if the given values aren't given either on the two radiuses can anyone provide me an answer?? Both geometric  and trigonometry method?


问:
differentiability $|f(x)|$
If a given function $f(x)$ is differentiable at $x=a$, then is $|f(x)|$ necessarily differentiable at $x=a$?

What about the the opposite direction of the statement (I'm quite sure that it isn't true, just want to make sure).

Thank you!

答:
No to both. $f(x) = x$ is differentiable at $x = 0$, but $|x|$ is not. The sign function
$$f(x) = \begin{cases} 1 & \text{if } x \geq 0 \\ -1 & \text{if } x < 0 \end{cases}$$
is not even continuous at $0$, but $|f(x)| = 1$ for all $x$, therefore $|f|$ is differentiable everywhere.

Title: differentiability $|f(x)|$


问:
Simplifying exponents..
Use the properties of rational exponents to simplify the expression

$(3^\frac{1}3 \cdot 4^\frac{1}{4})^3$

I got $3^1 \cdot 4^\frac{3}{4}$

I just wasn't sure if this was the most simplified expression. It says to do it without a calculator so can someone confirm if I'm right?

Title: Simplifying exponents..

Body:
$$(3^\frac{1}3 \cdot 4^\frac{1}{4})^3$$
$$=(3^\frac{1}3 \cdot 4^\frac{1}{3}\cdot 4^\frac{-1}{12})^3$$
$$=(12^\frac{1}3\cdot 4^\frac{-1}{12})^3$$
$$=12(4^\frac{-1}{12})^3$$
$$=\frac{12}{(4^\frac{1}{12})^3}$$
$$=\frac{12}{4^\frac{1}{4}}$$
$$=\frac{12}{\sqrt{2}}$$
$$=6\sqrt{2}$$

So after simplifying the expression reduces to $$6\sqrt{2}$$


Body:
No need to find $r$

$(R-25)^2+20^2=(R-20)^2$

$R=62.5$

$S=\pi R^2-\pi (R-20)^2$

$S=2100\pi$


问:
which function and integral are greater problem
My question is about kind of comparing values in a function, for example in this question

given 'c' a positive constant we have the following function $f(x)=cx^3+x^5-1$ , determine which is greater $f(\sin(x))$ or $f(\cos(x))$ for the values $\frac{\pi}{4}$$<x<$$\frac{\pi}{3}$

my try was that knowing how the graphs look, we know that in $\frac{\pi}{4}$$<x<$$\frac{\pi}{3}$
the slope of $\sin(x)$ is positive while the slope of $\cos(x)$ is negative and they both intersect at $\frac{\pi}{4}$ so my guess was that  $\sin(x)$ would be greater(would appreciate if anyone can confirm or have another way).
the second part of the question was similar we also need to determine if the integrals are equal or find which is smaller or greater.
$$\int_1^{\frac{\pi}{2}}\cos^8(x) dx $$ and $$\int_\frac{\pi}{2}^{\pi}\cos^8(x) dx $$
how can I find out which of these is greater/equal or less than the other? for the first part, I think it was easier because there was a way to compare since it is a $\sin(x)$ and $\cos(x)$ but I seem to be lost with the integrals.

So I am asking for advice and tips on how to approach and solve these questions, the answer is not important as I have a lot of these questions I just want to practice the topic and find a way to approach it.

答:
$f(x) = x^5 + cx^3 - 1$

$f'(x) = 5x^4 + 3cx^2$

We can see that $f'(x) \geq 0 \forall x \in \Bbb R$, therefore $f(x)$ is increasing for all x.

From the graphs of $\cos x$ and $\sin x$, one can clearly mention that they both intersect at $x=\pi/4$, after which for $x \in (\pi/4, \pi/3)$

$\sin x > \cos x$

$\implies f(\sin x) > f(\cos x)$

We know that the graph of $\cos x$ looks like this for $x \in (0,\pi)$:

So, we can say that graph of any even power of $\cos x$ would look something like this:

Now, because $\cos^8(\frac{\pi}{2} + x) = cos^8(\frac{\pi}{2} - x)$, We can say that $\cos^8x$ is symmetrical about $x=\pi/2$

Therefore, $\int_0^{\pi/2}\cos^8xdx = \int_{\pi/2}^{\pi}\cos^8xdx$

Now, As $\cos^8x$ is always non-negative, Area bounded by it in $x \in (0,\pi/2)$ would be greater than the area bounded by it in $x \in (1,\pi/2)$

$$\implies \int_0^{\frac{\pi}{2}}\cos^8xdx > \int_1^{\frac{\pi}{2}}\cos^8 xdx$$

$$\implies \int_{\frac{\pi}{2}}^{\pi}\cos^8xdx > \int_1^{\frac{\pi}{2}}\cos^8 xdx$$


问:
Using the property "onto" in a proof
Definition (Minimal Open Set): A nonempty open set $U$ of $X$ is said to be a minimal open set if and only if any open set which is contained in $U$ is $∅$ or $U$.

Definition (Minimal continuous map): Let $X$ and $Y$ be the topological spaces. A map $f: X → Y$ is called minimal continuous if $f^{-1}(M)$ is an open set in $X$ for every minimal open set $M$ in $Y$.

Definition (T$_{min}$ Space): A topological space $(X,\tau)$ is said to be T$_{min}$ space if every nonempty proper open subset of $X$ is minimal open set.

Theorem: Let : $f:X\rightarrow Y$  be a minimal continuous, onto map and Y be a T$_{min}$ space.
Then $f$ is continuous.

Proof: Let be $f$ a minimal continuous, onto map. Let $N$ be any nonempty proper open set in $Y$. By
hypothesis, $Y$ is T$_{min}$ space. It follows that $N$ is a minimal open set in $Y$. Since $f$ is minimal continuous, $f^{-1}(N)$ is an open in X. Therefore is a continuous.

Now, where did we use that $f$ is onto? I think if $f$ were not be onto, the proof would still work.

答:
In order to show that $f$ is continuous, we need to show that every open subset $O$ of $Y$ has an open inverse image.

But $f^{-1}[Y]=X$ for any function $f:X \to Y$, by definition and $X$ is always open in $X$. Likewise $f^{-1}[\emptyset] = \emptyset$ always and $\emptyset$ is always open in $X$, so the open sets $O=\emptyset,Y$ have been taken care of.

So we only need to consider $O$ with $\emptyset \neq O \neq Y$. But then indeed as $Y$ is $T_{\text{min}}$ by definition $O$ is minimal open in $Y$ and then as $f$ is "minimal continuous" we also have  that $f^{-1}[O]$ is open in $X$.

So always $f^{-1}[O]$ is open for all $O \subseteq Y$ open. So $f$ is continuous.

I don't see where ontoness would be needed, maybe the author thought it would be needed for $f^{-1}[Y]=X$ but this is not the case! $f^{-1}[Y] = \{x \in X\mid f(x)\in Y\} = X$ as $f$ maps $X$ into $Y$ by definition. Ontoness is saying $f[X]=Y$.

FWIW: I think these concepts of $T_{\text{min}}$ spaces and minimally continuous maps are a bit trivial and unnatural. There are hardly interesting natural examples. Nothing in "real life" fulfills it. The indiscrete space and some trivial variations thereof. Nothing worth bothering about. My 2 cts.

Title: Using the property "onto" in a proof


问:
Finding conditional probability distribution (X|Y) from (Y|X)
$(Y,X)$ has a joint distribution where the marginal of $X$ is a standard normal and $Y|X \sim U \left[|X|-\frac{1}{2},|X|+\frac{1}{2}\right] $ where $U[a,b] $ means uniform in the interval [a,b]. How can I find the distribution of $X$ given $Y$ and conditional expectation function $E(X|Y)$? I tried to multiply p.d.f. of $Y|X$ and $X$ to find joint distribution first, but support of $Y|X $ confused me. Thanks in advance :) </p>

答:
There are a number of issues here:

From A.S.'s comments $$f_{X\mid Y}(x\mid y)=\frac {f_X(x)f_{Y\mid X}(y\mid x)}{\int f_X(x)f_{Y\mid X}(y\mid x)\,dx}$$

where $f_X(x) =\phi(x)$ and $f_{Y\mid X}(y\mid x) = \mathbb 1_{\left[|x|-\frac12 \le y \le |x|+\frac12\right]}$

That last condition $|x|-\frac12 \le y \le |x|+\frac12$ can be rewritten as $y-\frac12 \le |x| \le y+\frac12$, and so when $x \ge 0$ as $\max\left(0,y-\frac12\right) \le x \le y+\frac12$ or when $x \lt 0$ as $-y-\frac12 \le x \le \min\left(0,-y+\frac12\right)$.

Here $y$ can take any value down to $-\frac12$, but when $-\frac12 \le y \lt \frac12$ the two possible intervals for $x$ meet giving $-y-\frac12 \le x \le y+\frac12$. Otherwise they are distinct and give two intervals for $x$ given $y$. So

When $-\frac12 \le y \lt \frac12$ and $-y-\frac12 \le x \le y+\frac12$ (one interval): $$f_{X\mid Y}(x\mid y)=\frac {\phi(x)}{\Phi\left(y+\frac 1 2\right)-\Phi\left(-y-\frac 1 2\right)}$$

When $y>\frac12$ and $y-\frac12 \le |x| \le y+\frac12$ (two intervals): $$f_{X\mid Y}(x\mid y)=\frac {\phi(x)}{2\left(\Phi\left(y+\frac 1 2\right)-\Phi\left(y-\frac 1 2\right) \right)}$$

and otherwise $f_{X\mid Y}(x\mid y)=0$. Note the different signs and the factor of $2$ in one of the denominators. Here is what the conditional densities look like for $y=0.6$ (red), $y=0.4$ (blue), and $y=-0.3$ (green):


问:
CDF graph of X random variable in game
We toss a coin. If we get head, we win 5 dollars. If we get tail, we roll a dice and have to pay $2 \cdot n$ where $n$ is the number we got on dice. Let $X$ be the value we win in this game.
I want to draw a CDF of variable $X$. I defined it as:
$$ X = \cases{{5 \ for \ (H, \omega) \ where \ \omega \in \{ 1,2,3,4,5,6\} } \\ {-2 \cdot \omega \ for \ (T, \omega) \ where \in \{1,2,3,4,5,6\} }}$$
The first case looks like this because I need the sample space to be the same for both of them - anyway we don't really care about the second coordinate in the first case. It's obvious that $$P(X=5) = \frac{1}{2} $$
and
$$P(X=-2 \cdot \omega) = \frac{1}{12} $$

My main problem here is that I have two coordinates and I'm a little lost. Should I have three axes? I've never seen such CDF graph.

EDIT:
I tried to do something:

答: I do not think you need three axes - you have defined the pmf already! So you know that $P(X = -2\omega) = \frac{1}{12}\ \forall \ \omega \in \{1,2,3,4,5,6\}$, and $X$ takes values $-12, -10, -8, -6, -4, -2, 5$. You can define the CDF piecewise, by adding the pmf as required.

Title: CDF graph of X random variable in game


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

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

I don't know if this question has been studied

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

答:
Update (2021/02/05)

We need to solve the optimization problem:
\begin{align}
&\max\quad A_m\\
&\mathrm{s.t.}\quad \sum_{k=1}^n A_k\cos (kx) \le 1, \ \forall x\in [0, 2\pi], \\
&\qquad\quad A_k \in \mathbb{R}, \forall k.
\end{align}
It can be viewed as a "Linear Programming" with uncountable many linear constraints.
However, as in the special cases ($n=2, 3, 4, 5, 6$ etc. see below), perhaps we can choose a few linear constraints such that $\max A_m$ is the same.
In other words, the following optimization problem has the same maximum as the previous one.
\begin{align}
&\max\quad A_m\\
&\mathrm{s.t.}\quad \sum_{k=1}^n A_k\cos (kx_j) \le 1, \ j = 1, 2, \cdots, M, \\
&\qquad\quad A_k \in \mathbb{R}, \forall k.
\end{align}
where $x_j$'s are chosen in advance (see the special cases below).

How to choose a set of $x_1, x_2, \cdots, x_M$?

Conjecture 1: $\max A_m = 1$ for $m = \lfloor \frac{n}{2}\rfloor + 1, \cdots, n$.

(Edit (2021/02/08): I have proved Conjecture 1 in my another answer.)

$\phantom{2}$

Some special cases $n=3, 4$

Remark: I will add more cases if possible.

Case $n=3$

$\max A_1 = \frac{\sqrt{5} + 1}{2} \approx 1.6180$

Letting $x = 0$ and $x = \frac{2\pi}{5}$ respectively, we have
\begin{align}
A_1 + A_2 + A_3 &\le 1, \tag{1}\\
A_1\frac{\sqrt{5}-1}{4} - (A_2 + A_3)\frac{\sqrt{5}+1}{4} &\le 1. \tag{2}
\end{align}
$(1)\times \frac{\sqrt{5} + 1}{4} + (2)$ gives $A_1 \le \frac{\sqrt{5}+1}{2}$.

Also, when $A_1 = \frac{\sqrt{5}+1}{2}, A_2 = - \frac{2}{\sqrt{5}},
A_3 = \frac{1}{2} - \frac{\sqrt{5}}{10}$, we have
\begin{align}
&A_1\cos x + A_2\cos 2x + A_3 \cos 3x - 1\\
=\ & \frac{5 - \sqrt{5}}{40}(1-\cos x)(\sqrt{5} - 1 - 4\cos x)^2 \le 0.
\end{align}
The desired result follows.

$\max A_2 = 1$ and $\max A_3 = 1$ (has been solved by @open problem)

Letting $x=0$, $x = \frac{2}{3}\pi$ and $x = \pi$ respectively, we have
\begin{align}
A_1 + A_2 + A_3 &\le 1, \tag{3}\\
-\frac{1}{2}A_1 - \frac{1}{2}A_2 + A_3 &\le 1, \tag{4}\\
-A_1 + A_2 - A_3 &\le 1. \tag{5}
\end{align}
$(3) + (5)$ gives $A_2 \le 1$, and $(3) + 2\times (4)$ gives $A_3 \le 1$.

The remaining is easy.

$\phantom{2}$

Case $n=4$

$\max A_1 = \sqrt{3}$

Letting $x = \frac{\pi}{6}$ and $x = \frac{\pi}{2}$ respectively, we have
\begin{align}
\frac{\sqrt{3}}{2} A_1 + \frac{1}{2}A_2 - \frac{1}{2}A_4 &\le 1, \tag{6}\\
-A_2 + A_4 &\le 1.\tag{7}
\end{align}
$(6)\times 2 + (7)$ gives $A_1 \le \sqrt{3}$.

Also, when $A_1 = \sqrt{3}, A_2 = - \frac{7}{6},
A_3 = \frac{1}{\sqrt{3}}, A_4 = - \frac{1}{6}$, we have
\begin{align}
&A_1\cos x + A_2\cos 2x + A_3 \cos 3x + A_4\cos 4x - 1\\
=\ & -\frac{\cos^2 x}{3} (\sqrt{3} - 2\cos x)^2\\
\le\ & 0.
\end{align}
The desired result follows.

$\max A_2 = \sqrt{2}$

Letting $x = \frac{\pi}{8}$ and $x = \frac{7\pi}{8}$ respectively, we have
\begin{align}
A_1 \cos \frac{\pi}{8} + \frac{A_2}{\sqrt{2}} + A_3\cos \frac{3\pi}{8} &\le 1, \tag{8} \\
-A_1 \cos \frac{\pi}{8} + \frac{A_2}{\sqrt{2}} - A_3\cos \frac{3\pi}{8} &\le 1. \tag{9}
\end{align}
$(8) + (9)$ gives $A_2\le \sqrt{2}$.

Also, when $A_1 = 0, A_2 = \sqrt{2}, A_3 = 0, A_4 = -\frac{1}{2}$, we have
\begin{align}
&A_1\cos x + A_2\cos 2x + A_3 \cos 3x + A_4\cos 4x - 1\\
=\ & -\frac{1}{4} (2 + \sqrt{2} - 4\cos^2 x)^2\\
\le\ & 0.
\end{align}
The desired result follows.

$\max A_3 = 1$ and $\max A_4 = 1$

Letting $x = 0$, $x = \frac{\pi}{2}$, $x = \frac{2\pi}{3}$ and $x=\pi$ respectively, we have
\begin{align}
A_1 + A_2 + A_3 + A_4 &\le 1, \tag{10}\\
-A_2 + A_4 &\le 1, \tag{11}\\
-\frac{1}{2}A_1 - \frac{1}{2}A_2 + A_3 - \frac{1}{2}A_4 &\le 1, \tag{12}\\
-A_1 + A_2 - A_3 + A_4 &\le 1. \tag{13}
\end{align}
$(10) + (12)\times 2$ gives $A_3 \le 1$, and $(10) + (11)\times 2 + (13)$ gives $A_4 \le 1$. The remaining is easy.


答:
Hints

For the first problem, note that for $\ x\ge0\ $, $\ f(x)\ $ is a strictly increasing function of $\ x\ $, so the larger of $\ f(\sin\theta)\ $ or $\ f(\cos\theta)\ $ is completely determined by which is the larger of $\ \sin\theta\ $ and $\ \cos\theta\ $.  Over the interval $\ \frac{\pi}{4}\le\theta\le\frac{\pi}{3}\ $ $\ \sin\theta\ $ strictly increases from $\ \frac{1}{\sqrt{2}}\ $ to $\ \frac{\sqrt{3}}{2}\ $ while $\ \cos\theta\ $ strictly decreases from $\ \frac{1}{\sqrt{2}}\ $ to $\ \frac{1}{2}\ $. Therefore
$$
\frac{1}{2}<\cos{\theta}<\frac{1}{\sqrt{2}}<\sin\theta<\frac{\sqrt{3}}{2}
 $$ for all $\ \theta\in\left(\frac{\pi}{4},\frac{\pi}{3}\right)\ $.

For the second problem, note that $\ \cos(\pi-\theta)=-\cos\theta\ $ and therefore $\ \cos^8(\pi-\theta)=\cos^8\theta\ $. Therefore, a change of variables in the integral $\ \int_\limits{\frac{\pi}{2}}^\pi\cos^8(x)dx\ $ from $\ x\ $ to $\ y=\pi-x\ $ gives
$$
  \int_\limits{\frac{\pi}{2}}^\pi\cos^8(x)dx=\int_0^\frac{\pi}{2}\cos^8(y)dy\ .
 $$

Title: which function and integral are greater problem

Body:
My question is about kind of comparing values in a function, for example in this question

given 'c' a positive constant we have the following function $f(x)=cx^3+x^5-1$ , determine which is greater $f(\sin(x))$ or $f(\cos(x))$ for the values $\frac{\pi}{4}$$<x<$$\frac{\pi}{3}$

my try was that knowing how the graphs look, we know that in $\frac{\pi}{4}$$<x<$$\frac{\pi}{3}$
the slope of $\sin(x)$ is positive while the slope of $\cos(x)$ is negative and they both intersect at $\frac{\pi}{4}$ so my guess was that  $\sin(x)$ would be greater(would appreciate if anyone can confirm or have another way).
the second part of the question was similar we also need to determine if the integrals are equal or find which is smaller or greater.
$$\int_1^{\frac{\pi}{2}}\cos^8(x) dx $$ and $$\int_\frac{\pi}{2}^{\pi}\cos^8(x) dx $$
how can I find out which of these is greater/equal or less than the other? for the first part, I think it was easier because there was a way to compare since it is a $\sin(x)$ and $\cos(x)$ but I seem to be lost with the integrals.

So I am asking for advice and tips on how to approach and solve these questions, the answer is not important as I have a lot of these questions I just want to practice the topic and find a way to approach it.


问:
Evaluate the following Limit (CALC I)
Evaluate the following limit

$\lim_{x \rightarrow  0} \frac{1}{x^2} * \int_{0}^{x} \frac{t+t^2}{1+sin(t)}$

When I apply the limit I get the indeterminate form $ \frac{0}{0}$ allowing me to use l'Hospital. My book says the derivative of this expression is

$\frac{\frac{x+x^2}{1+sin(x)}}{{2x}}$

The numerator comes from the FTC, but the denominator I don't understand. If I apply product rule get

$\frac{x+x^2}{1+sin(x)}*\frac{-2+x}{x^3}
\neq \frac{\frac{x+x^2}{1+sin(x)}}{{2x}} $

If I use quotient rule it still does not equal. Did I miss something?

答:
It is not that "the derivative of this expression", but that you are taking the quotient of the derivatives of the numerator and the denominator when applying L'Hopital.

In general, when L'Hopital is applicable,
$$
\lim_{x\to0}\frac{f(x)}{g(x)}=\lim_{x\to0}\frac{f'(x)}{g'(x)}
$$
NOT
$$
\lim_{x\to0}\frac{f(x)}{g(x)}=\lim_{x\to0}\left(\frac{f(x)}{g(x)}\right)'
$$

The derivative of $x^2$ is $2x$. That is where the denominator is from:

$$
\lim_{x \rightarrow  0} \frac{1}{x^2} \cdot \int_{0}^{x} \frac{t+t^2}{1+\sin(t)}dt
=\lim_{x \rightarrow  0}\frac{\frac{d}{dx}\int_{0}^{x} \frac{t+t^2}{1+\sin(t)}dt}{\frac{d}{dx}(x^2)}
$$

Title: Evaluate the following Limit (CALC I)


Title: Evaluate the following Limit (CALC I)

Body:
Evaluate the following limit

$\lim_{x \rightarrow  0} \frac{1}{x^2} * \int_{0}^{x} \frac{t+t^2}{1+sin(t)}$

When I apply the limit I get the indeterminate form $ \frac{0}{0}$ allowing me to use l'Hospital. My book says the derivative of this expression is

$\frac{\frac{x+x^2}{1+sin(x)}}{{2x}}$

The numerator comes from the FTC, but the denominator I don't understand. If I apply product rule get

$\frac{x+x^2}{1+sin(x)}*\frac{-2+x}{x^3}
\neq \frac{\frac{x+x^2}{1+sin(x)}}{{2x}} $

If I use quotient rule it still does not equal. Did I miss something?

Body: $$\frac{1}{x^2} \cdot \int_{0}^{x} \frac{t+t^2}{1+sin(t)}dt= \frac{\int_{0}^{x} \frac{t+t^2}{1+sin(t)}dt}{x^2}.$$


答:
$$(3^{1/3}\cdot 4^{1/4})^3$$

distribute the exponent using the rule that $(a\cdot b)^c = a^c\cdot b^c$ and the rule that $(a^b)^c = a^{(bc)}$ (which as alluded to are only guaranteed for when dealing with positive numbers).  This leads us to:

$$(3^{3/3}\cdot 4^{3/4})$$

Simplify the fraction $3/3$ and recognize that this would leave it with an exponent of $1$ and exponents of $1$ can be omitted for brevity.  This leads us to:

$$3\cdot 4^{3/4}$$

This is a perfectly fine place to stop if you like.  If you prefer, you can continue on however.  Recognizing $4$ as $2^2$, we could have continued as:

$$3\cdot (2^2)^{3/4}$$

$$3\cdot 2^{3/2}$$

and if you prefer writing things with surds

$$3\cdot 2\cdot \sqrt{2}$$

$$6\sqrt{2}$$

Title: Simplifying exponents..

Body:
Use the properties of rational exponents to simplify the expression

$(3^\frac{1}3 \cdot 4^\frac{1}{4})^3$

I got $3^1 \cdot 4^\frac{3}{4}$

I just wasn't sure if this was the most simplified expression. It says to do it without a calculator so can someone confirm if I'm right?


问:
Is the minimum distance of a linear code equal to the minimum distance to any single element?
I know that for a linear code we have that the minimum distance $d$ under the Hamming metric is equal to the minimal weight of a non-zero element but this minimum weight is just the minimum distance to a particular element in the code (the zero element).

The proof of this just notices that if there is a minimal distance then it is the weight of the difference of two codewords $u''=u-u'$ which is itself a codeword as the code is linear.

I feel as if we ought to be able to generalise this but I've not been able to figure out how to do so.

答: Alright I figured this out, simply take the vector you want, lets call it $x$, then  $d(x,x+u'')=w(x-(x+u''))=w(u'')=d(u,u')=d$.


问:
Trace of a differential operator
Given the differential operator:
$$A=\exp(-\beta H)$$
where $$H=\frac{1}{2}\left( -\frac{d^2}{dx^2}+x^2 \right)$$
and $\beta\gt 0$
How can I get the trace of this operator?
Thanks in advance.</p>

答:
The trace of this operator is easily obtained in the following way:
$$
   Z={\rm Tr}\exp(-\beta H).
$$
that is equivalent to
$$
 Z=\sum_n \langle n|\exp(-\beta H)|n\rangle.
$$
Assuming $H|n\rangle=E_n|n\rangle$, this is just
$$
  Z=\sum_n\exp(-\beta E_n).
$$
Your case is the harmonic oscillator $E_n=n+\frac{1}{2}$ and the sum is just a geometric series easy to perform.</p>

Title: Trace of a differential operator


问:
Taylor's series of $\sin \frac { 1 }{x }$
Is it possible to built the Taylor's series of the function $\sin \frac {1}{x}$ in $x=0$ also if $f (x )$ is not defined in this point?

答:
No, because $\sin\frac1x$ has an essential singularity at $\infty$. If you have $\frac{\sin x}{x}$ instead, then yes, since the singularity at $x=0$ is removable.

Notes.

I see you use the real-analysis tag. Note that any real analytic function can be extended as a complex analytic one.


问:
Isomorphism map between $\frac{\mathbb{Z}_3[x]}{(x^2+x+2)}$ and $ \frac{\mathbb{Z}_3[x]}{(x^2 + 1)} $
$\frac{\mathbb{F}_3[x]}{(x^2+x+2)} \cong \frac{\mathbb{F}_3[x]}{(x^2 + 1)} $

I know $\frac{\mathbb{F}_3[x]}{(x^2+x+2)} $ and $ \frac{\mathbb{F}_3[x]}{(x^2 + 1)} $ is isomorphic because they are both 2-degree extension of $ \mathbb{F}_3 $ .

But I cannot contract explicit isomorphism between them.

Could you show me isomorphism map between $\frac{\mathbb{F}_3[x]}{(x^2+x+2)}$  and $ \frac{\mathbb{F}_3[x]}{(x^2 + 1)} $?

Thank you for your help.

答:
Try $\phi:\mathbb{F}_3[X]\to\mathbb{F}_3[X]$ defined by $\phi:X\mapsto X+1$ whilst leaving everything in $\mathbb{F}_3$ alone.

This sends $X^2+X+2$ to $X^2+1$, so it sends the ideal $(X^2+X+2)$ to the ideal $(X^2+1)$, so when you quotient out $\mathbb{F}_3[X]$ by $(X^2+X+2)$ or by $(X^2+1)$, you get isomorphic fields.


问:
Homeomorphism from two open sets
Let $f:Y\rightarrow X$ be a continuous map. Let $V_1,V_2\subset Y$ open sets and $U\subset X$ open set such that $f:V_i\rightarrow U$ is a homeomorphism for $i=1,2$. Is it true that $V_1\cap V_2\neq\emptyset$ $\Rightarrow$ $V_1=V_2$?

What if f is a covering map?

答: It is not true, not even for covering maps. As an example, let $X = \{1,2\}$ with the discrete topology, $Y = X \times X$ and $f : X \times X \to X, f(x,x') = x$. This is a two-sheeted covering map. Let $V_1 = \{(1,1), (2,1)\}$ and $V_2 = \{(1,1),(2,2)\}$. Both are mapped by $f$ homeomorphically onto $U = X$, but $V_1 \ne V_2$ and $V_1 \cap V_2 \ne \emptyset$.

Title: Homeomorphism from two open sets


问:
Runge-Kutta method and step doubling
I am studying Runge-Kutta and step size control and came up with a few doubts. Because they are related with this integration method, I will divide it in two parts. First, allow me to introduce the problem.

$1^{st}$ part - questions about Runge-Kutta Method

Consider a $2^{nd}$ order Runge-Kutta with general form:

$k_{1}=hf(x_{n},y_{n})$

$k_{2}=hf(x_{n}+\frac{1}{2}h,y_{n}+\frac{1}{2}k_{1})$

$y_{n+1}=y_{n}+k_{2}+O(h^{3})$

where $f(x_{n},y_{n}) = y'(x_{n})$

Now, if we are to consider $4^{th}$ order Runge-Kutta we would get

$k_{1}=hf(x_{n},y_{n})$

$k_{2}=hf(x_{n}+\frac{1}{2}h,y_{n}+\frac{1}{2}k_{1})$

$k_{3}=hf(x_{n}+\frac{1}{2}h,y_{n}+\frac{1}{2}k_{2})$

$k_{4}=hf(x_{n}+h,y_{n}+k_{3})$

$y_{n+1}=y_{n}+\frac{1}{6}k_{1}+\frac{1}{3}k_{2}+\frac{1}{3}k_{3}+\frac{1}{6}k_{4}+O(h^{5})$

This leads me to the following questions:

I understand why the 2nd order method has the $y_{n+1}$ indicated above. Yet, shouldn't the $4^{th}$ order version of the method be expressed only as $y_{n+1}=y_{n}+k_{4}+O(h^{5})$ given that $k_{4}$ implicitly has the values of $k_{1}$ to $k_{3}$?

Why does the $4^{th}$ order version have those fractional coefficients?

$2^{nd}$ part - questions about step-size control

Consider the exact solution for an advance from $x$ to $x+2h$ by $y(x+2h)$ and the two approximate solutions by $y_{1}$ (one step 2h) and $y_{2}$ (two steps each of size $h$). Considering the $4^{th}$ order method we have:

$y(x+2h)=y_{1}+(2h)^{5}\phi+O(h^{6})+...$

$y(x+2h)=y_{1}+2(h)^{5}\phi+O(h^{6})+...$

The difference between these estimates permits estimating the truncation error:

$\Delta=y_{2}-y_{1}$

Which we then use to improve the numerical estimate of the true solution:

$y(x+2h)=y_{2}+\frac{\Delta}{15}+O(h^{6})$

This brings me to the following questions:

In the two expressions of y(x+2h) at the top, does $\phi$ include the terms $k_{1}...k_{4}$ in the first part of this post?

Why is the final expression $5^{th}$ order, if the original problem was $4^{th}$ order?

Where does the coefficient $\frac{1}{15}$ come from?

Thank you for all the insight!

答:
For the first part, the Wikipedia article http://en.m.wikipedia.org/wiki/Runge%E2%80%93Kutta_methods has a detailed discussion of the different members of the family along with a derivation of RK-4. The key idea is that you can do whatever you want in terms of writing down the form of your approximation; all that matters is that doing a Taylor expansion makes all the error terms up to the order of approximation vanish. This is how the coefficients are chosen.

Then in the next part, $\phi$ is the the expression for the error made in the step, without the $(2h)^5$ term. It should be thought of as an expression in terms of the derivatives of $y$ at $x$, but its exact form isn't important, just its independence from $h$. It's the same in both lines to first order in $h$ so the errors are hidden away in the correction term. Note that fourth order means the error term is $h^5$ per step which it is throughout. You then use two fourth order approximations to get a better one by calculating the error approximately. This is reasonable since you've used more data.

As to the $1/15$ factor, that simply arises because $(2h)^5-2(h)^5=(16-1)h^5=15h^5$ and if you make only a step of size $h$ you get a factor of $h^5$. Try carefully defining and writing down the different things you work out, then you can see why you take the difference in this way.

Title: Runge-Kutta method and step doubling


问:
What properties does one study for a subset $S$ inside $R = \mathbb{Z}[x]$ and has absorbing property like ideal
Let us assume that $S \subset\mathbb{Z}[x]$ is non-empty. This set $S$ has the property that $\mathbb{Z}[x] S \subset S$. However, $S$ is not an ideal because sum of two elements from $S$ is not guaranteed to be in $S$.

$S$ is actually a semigroup in $\mathbb{Z}[x]$, which amounts to saying that $S S \subset S$, which is weaker than $\mathbb{Z}[x] S \subset S$. I have not studied much about semi-groups; hence this question below:

From what I could find in literature and some searching, one could study elasticity of this semigroup, which is a constant related to the factorization of elements of $S$ into irreducibles of $S$. What other properties of semigroups could one study, given that this is a semigroup of polynomials? If one could direct me to a reference that studies semigroups of polynomials, I would be grateful. Thanks.

答:
As @yesterday pointed out, your set $S$ is an ideal of the multiplicative monoid ${\Bbb Z}[X]$.

It is not too difficult to describe the algebraic structure of this monoid. First observe that the polynomial $0$ is a zero of ${\Bbb Z}[X]$ and that $M = {\Bbb Z}[X] \setminus \{0\}$ is a submonoid of ${\Bbb Z}[X]$.

For each non-zero polynomial $p$, let $c(p)$ denote its  coefficient of higher degree. For instance,
$$
c(-1 -2X + 3X^3) = 3 \quad\text{and}\quad c(1 + X - X^2) = -1
$$
Let $P$ be the set of polynomials $p$ such that $c(p)$ is positive. Let also $I$ be the set of irreducible polynomials of $P$. Since ${\Bbb Z}[X]$ is factorial, $P$ is the free commutative monoid generated by $I$, and since $I$ is countable, it is isomorphic to the additive monoid $({\Bbb N}[X],+)$.

On the other hand, the set of units of ${\Bbb Z}[X]$ is the cyclic group $U = \{-1, + 1\}$. Finally observe that if $p$ is a non-zero polynomial, then either $p \in P$ or $-p \in P$, depending on the sign of $c(p)$.
I let you verify that in fact, the monoid $M$ is isomorphic to $U \times P$, the product of the monoids $U$ and $P$.

Altogether, ${\Bbb Z}[X]$ is isomorphic to $(U \times ({\Bbb N}[X],+))^0$, the monoid obtained by adjoining a zero to the monoid $U \times ({\Bbb N}[X],+)$.


问:
Convergence of sum $1+c_1(1+c_2..$
I was working on an exercise and I encountered this

$1 + c_1(1+c_2(1+c_3(1 + ...$

$0<c_i<1$

Now for some reason I feel like I recognize this from something I've watched or read but I looked all over and I can't find anything like it, does this converge to anything meaningful? Or does some version of this converge to something? (If each $c_i$ is constant for example)

答:
What you have is the infinite series
$$
1 + c_1 + c_1c_2 + c_1c_2c_3 + \cdots = 1 + \sum_{n=1}^\infty c_1 c_2 \cdots c_n \, .
$$
As pointed out in the comments, this is a geometric series if the sequence $(c_n)$ is constant.

There are other special cases where the sum can be computed explicity, see for example What is $\sum_{n=1}^{\infty}a_1 a_2...a_n$?.

Not much can be said in the general case. Of course the usual convergence criteria can be applied. For example, the ratio test tells us that the series is convergent if $ \limsup_{n\to \infty} c_n < 1$.

Setting $b_n = c_1 c_2 \cdots c_n$ we have $1 + \sum_{n=1}^\infty b_n$ with an arbitrary strictly decreasing positive sequence $(b_n)$, which demonstrates that the series may converge or diverge.

Title: Convergence of sum $1+c_1(1+c_2..$


答:
As it was pointed out in the comments, we are interested in deciding on the convergence of $1+\sum_{n=1}^{+\infty} (c_1 \cdots c_n)$. The ratio test guarantees convergence if $\limsup_{n\to \infty} c_n <1$, which is not necessarily true for any sequence satisfying $0 < c_i < 1$. Apart from that, if $0 < l_1 \leq c_i \leq l_2 < 1$, you can say that
$$
\dfrac{1}{1-l_1} = \sum_{n=1}^{+\infty} l_1^n \leq1+\sum_{n=1}^{+\infty} (c_1 \cdots c_n)\leq 1+\sum_{n=1}^{+\infty} l_2^n=\dfrac{1}{1-l_2}.
$$

Title: Convergence of sum $1+c_1(1+c_2..$

Body:
I was working on an exercise and I encountered this

$1 + c_1(1+c_2(1+c_3(1 + ...$

$0<c_i<1$

Now for some reason I feel like I recognize this from something I've watched or read but I looked all over and I can't find anything like it, does this converge to anything meaningful? Or does some version of this converge to something? (If each $c_i$ is constant for example)


问:
Method of Undetermined Coefficients when ODE does not have constant coefficients
Even though the Method of Undetermined Coefficients is usually taught as a method for solving nonhomogeneous linear ODEs with constant coefficients, it seems to also work if the coefficients are not constant. For example, let

$$
t^2y''-2y=t.
$$

The solution to the associated homogeneous equation, $t^2y''-2y=0$, is spanned by the functions $t^2$ and $t^{-1}$. Using the Method of Undetermined Coefficients, one can guess a particular solution of the form $y_p(t)=At+B$, where $A$ and $B$ are constants. Substituting $y_p(t)$ into the differential equation and solving gives $A=-\frac{1}{2}$ and $B=0$, so that the general solution to $t^2y''-2y=t$ is
$$y(t)=C_1t^2+C_2t^{-1}-\frac{1}{2}t.$$

However, if we consider the equation
$$t^2y''-2y=t^2$$
instead, then our initial guess for a particular solution, $y_p(t)=At^2+Bt+C$ ($A$, $B$, $C$ constants) "overlaps" with part of our solution to $t^2y''-2y=0$. This suggests that we should look for a particular solution of the form $y_p(t)=At^5+Bt^4+Ct^3$ instead. Substituting this into the ODE to try to solve for $A$, $B$, and $C$ doesn't give anything useful; in fact, through Variation of Parameters, we get a particular solution $y_p(t)=\frac{1}{3}\ln(t)t^2$, which the Method of Undetermined Coefficients has no hope of finding.

I'm not really sure what's going on here. Why does the Method of Undetermined Coefficients fail to give me the correct particular solution in the second case? Is there anything general that can be said here about applying this method in the nonconstant coefficient case?

答:
For Euler-Cauchy DE you have to exploit the relation to DE with constant coefficients given by the exponential function. This means what is $t^2$ here would be $e^{2t}$ there, and thus the ansatz for the resonance case changes from $Ate^{2t}$ there to $A\ln(t)t^2$ here.

But the method of undetermined coefficients is only valid for linear DE with constant coefficients, and Euler-Cauchy DE due to this relationship. It does not work for general linear DE.


答: The method of undetermined coefficients works only if you know what form a solution has in general case. So when coefficients are not constant, you don’t know the general answer. Look in Arnold’s Ordinary Differential Equations, for example. You can find there a formal proof for linear equations with constant coefficients that any solution is a quasi-polynomial (par. 26). You can also find some information about equations with non constant coefficients.</p>

Title: Method of Undetermined Coefficients when ODE does not have constant coefficients

Body:
Even though the Method of Undetermined Coefficients is usually taught as a method for solving nonhomogeneous linear ODEs with constant coefficients, it seems to also work if the coefficients are not constant. For example, let

$$
t^2y''-2y=t.
$$

The solution to the associated homogeneous equation, $t^2y''-2y=0$, is spanned by the functions $t^2$ and $t^{-1}$. Using the Method of Undetermined Coefficients, one can guess a particular solution of the form $y_p(t)=At+B$, where $A$ and $B$ are constants. Substituting $y_p(t)$ into the differential equation and solving gives $A=-\frac{1}{2}$ and $B=0$, so that the general solution to $t^2y''-2y=t$ is
$$y(t)=C_1t^2+C_2t^{-1}-\frac{1}{2}t.$$

However, if we consider the equation
$$t^2y''-2y=t^2$$
instead, then our initial guess for a particular solution, $y_p(t)=At^2+Bt+C$ ($A$, $B$, $C$ constants) "overlaps" with part of our solution to $t^2y''-2y=0$. This suggests that we should look for a particular solution of the form $y_p(t)=At^5+Bt^4+Ct^3$ instead. Substituting this into the ODE to try to solve for $A$, $B$, and $C$ doesn't give anything useful; in fact, through Variation of Parameters, we get a particular solution $y_p(t)=\frac{1}{3}\ln(t)t^2$, which the Method of Undetermined Coefficients has no hope of finding.

I'm not really sure what's going on here. Why does the Method of Undetermined Coefficients fail to give me the correct particular solution in the second case? Is there anything general that can be said here about applying this method in the nonconstant coefficient case?


问:
Find the radius of convergence for $\sum_{n=0}^\infty (n^2+a^n)z^n$
Find the radius of convergence for $\sum_{n=0}^\infty (n^2+a^n)z^n$, where $a$ is a complex number.

I'm not sure what the best way to solve this is, I know that $\sum(n^2+a^n)z^n=\sum a^nz^n+\sum n^2z^n$

Which both the sums on the right have radius of convergence $R=1$.

And I know there is a formula $1/R=\lim\sup\vert n^2+a^n\vert^{1/n}$ but I'm not sure how to compute that.

答:
You are not right: for $a \ne 0$ the power series  $\sum a^nz^n$ has radius of convergence $\frac{1}{|a|}.$
Now discuss two cases:

(1)  $|a| \le 1$ and (2) $|a| >1.$


问:
Need help understanding central limit theorem
I am very confused about CLT and have searched on the internet but found nothing that solved my confusion. How can I solve a problem like this with CLT?

Let $Y =\operatorname{Pois}(n)$.  Using Normal approximation, aka the CLT, give an
estimate of the probability

$$p\Big[|Y-n| \geq 2\sqrt{n}\Big].$$

答:
A Poisson distribution with parameter $n$ has mean $n$ and standard deviation $\sqrt{n}$

so the expression $\mathbb P\Big[|Y-n| \geq 2\sqrt{n}\Big]$ is the probability of being two standard deviations or more away from the mean.

For a normal distribution, this probability would be $\Phi(-2)+1-\Phi(2) \approx 0.0455$, suggesting this might be a reasonable estimate for a Poisson distribution.

How reasonable? It tends to get better as $n$ increases, but for $n=100$ the actual Poisson probability is about $0.0509$ while for $n=101$ the actual Poisson probability is about $0.04112$; this jump happens near values of $n$ which are near squares or halfway between squares.  As an illustration with R

```<code>n <- 1:101
plot(n, ppois(floor(n-2*sqrt(n)), n) + 1 - ppois(ceiling(n+2*sqrt(n)) -1, n))
abline(h=pnorm(-2) + 1 - pnorm(2), col="red")
</code>```

Body:
A Poisson distribution with parameter $n$ has mean $n$ and standard deviation $\sqrt{n}$

so the expression $\mathbb P\Big[|Y-n| \geq 2\sqrt{n}\Big]$ is the probability of being two standard deviations or more away from the mean.

For a normal distribution, this probability would be $\Phi(-2)+1-\Phi(2) \approx 0.0455$, suggesting this might be a reasonable estimate for a Poisson distribution.

How reasonable? It tends to get better as $n$ increases, but for $n=100$ the actual Poisson probability is about $0.0509$ while for $n=101$ the actual Poisson probability is about $0.04112$; this jump happens near values of $n$ which are near squares or halfway between squares.  As an illustration with R

n <- 1:101
plot(n, ppois(floor(n-2*sqrt(n)), n) + 1 - ppois(ceiling(n+2*sqrt(n)) -1, n))
abline(h=pnorm(-2) + 1 - pnorm(2), col="red")


问:
How to find number of ways one can move from $(0,0)$ to $(a,b)$ along the directed path?
In 2D space the point $(a,b)$ are said to be a lattice point if $a$, $b$ are integers. Two lattice points $(a,b)$ and $(c,d)$ are neighbours if they agree in one of the coordinates and differ in the other coordinate by $1$ . Every pair of neighbours is connected by a directed line with direction coinciding with the positive direction of the coordinate axis to which it is parallel. In how many ways one can move from $(0,0)$ to $(a,b)$ along the directed paths ?

答:
If we pay close attention to the situation, we have to make $a$ moves in the right direction, and $b$ moves in the upward direction.

We will find ourselves on $(a,b)$ after making those moves, no matter in what order we make those moves.

Now, this problem has switched to: In how many ways can we arrange $a$ R's and $b$ U's.

For instance, if we have to find the number of ways one can move from $(0,0)$ to $(4,5)$, We will have to arrange $\underbrace{RRRR}_\text{4 times} \overbrace{UUUUU}^\text{5 times}$ in all possible ways, i.e., $\frac{9!}{4!5!}$

Now, generalising it for $(a,b)$, the number of ways = $\frac{(a+b)!}{a! \times b!}$


问:
Why there exist a solution when Evans constructs a parabolic equation in proving strong maximum principle
Why there exist a solution when Evans constructs a parabolic equation in proving the strong maximum principle theorem 11 in chapter 7.

The theorem 11 is under below .

Theorem 11 (Strong maximum principle ) . Assume $u\in C_1^2(U_T)\cap (\overline{U}_T))$ and $c\equiv 0 \text{ in}  U_T$
Suppose also $U$ is connected.$\cdots$

Proof. 1. Assume $u_t+Lu\leq 0  \text{in}  U_T$ and $u$ attains its maximum at some point $(x_0,t_0)\in U_T$ . Select a smooth , open set $W\subset\subset U$, with $x_0\in W$ . Let $v$ solve
$$\left\{
 \begin{align}
  v_t+Lv&=&0 \ \text{in}   W_T \\
  v&=&u \  \text{on} \ \Delta _T
 \end{align}
\right.
 $$
where $\Delta _T$ denotes the parabolic boundary of $W_T$ $\cdots$

My doubts :why there is a solution $v$? Existence of weak solution can guarantee if the condiction $v=0$ on $\partial U×[0,1]$. Could someone give me some advice, thank you!

答:
The essential point is what Evans means by "parabolic boundary."  It is not understood as the set $\partial U \times [0,T]$ as you write.  Instead, if you check how this is defined in the book you'll see that
$$
W_T = W \times (0,T] \text{ and } \Delta_T = \overline{W_T} \backslash W_T = (W \times \{0\}) \cup (\partial W \times [0,T]).
$$
The solvability theory for the PDE with data specified on $\Delta_T$ is exactly what Evans establishes in the material prior to Theorem 11 in Section 7.1, though there he specializes to the case that the data on $\partial W \times [0,T]$ is zero.  To recover the general case, though, we can simply subtract $u$.  In other words we first solve
$$
\begin{cases}
w_t + L w = -u_t - L u & \text{in } W_T \\
w =0 & \text{on } \Delta_T
\end{cases}
$$
using the theory in Section 7.1.  We then define $v = w + u$, which then solves

$$
\begin{cases}
v_t + L v = 0 & \text{in } W_T \\
v =u & \text{on } \Delta_T.
\end{cases}
$$


问:
Is there a homogeneous concave function that is not a monomial?
Does there exist a concave function $f:(0,\infty)\to(0,\infty)$ with the following properties?

$f$ is $r$-homogeneous for some $r>0$, i.e., $f(\lambda x)=\lambda^r f(x)$ for all $x>0$

$\lim_{x\to 0}f(x)=0$

$\lim_{x\to\infty}f(x)/x=0$

$f$ is not a monomial, i.e., $f$ is not of the form $f(x)=x^m$ for some $m$

We also put $f(0)=0$.

答:
Condition 1 alone gives
$$\tag{1}
f(x)=f(x\cdot 1) = Cx^r, \qquad \forall x>0, $$
for $C=f(1)$. This is not exactly a monomial because $r$ needs not be integer, though.

Condition 2 implies that $r>0$ and condition 3 implies that $r<1$. So the only functions with the properties you seek are of the form (1) for some $C>0$ and some $r\in (0, 1)$.


问:
Measuring entangled quantum state
Just starting out with QM here. I believe I'm missing something simple but can't get my head around it.

Consider the state:

$$
\left | \psi \right> = \frac{1}{\sqrt 2}\left|0\right> + \frac{i}{\sqrt 2}\left|1\right>
$$

This should be normalised, since:

$$
\sqrt{|\alpha|^2 + |\beta|^2} = 1
$$

But when I try to calculate $|\left<\psi | \psi \right>|^2$

I get:
$$
\left<\psi | \psi \right> = \frac{1}{\sqrt 2}*\frac{1}{\sqrt 2}\left<0|0\right> + \frac{i}{\sqrt 2}*\frac{i}{\sqrt 2}\left<1|1\right>
$$
$$
\left<\psi | \psi \right> = \frac{1}{2}-\frac{1}{2}=0
$$
$$
|\left<\psi | \psi \right>|^2=0
$$

I would have expected $\left<\psi | \psi \right>$ to be 1 since it's normalised. By interpretation, I'm measuring the probability of the state in the SAME state, I would expect to get 1 as well.

Am I doing something wrong?

My reference article which I'm learning from is here by the way: https://qiskit.org/textbook/ch-states/representing-qubit-states.html#2.-The-Rules-of-Measurement-

答:
If

$$ \left| \psi \right>=\frac{1}{\sqrt{2}} \left| 0 \right> +\frac{i}{\sqrt{2}} \left| 1 \right> $$

Then, taking the adjoint

$$\left< \psi \right|=\frac{1}{\sqrt{2}} \left< 0 \right| +\frac{-i}{\sqrt{2}} \left< 1 \right| $$

So that, multiplying out and using $\left< 0 \middle| 1 \right>=0$, we have

$$ \left< \psi \middle| \psi \right>=\frac{1}{\sqrt{2}}\frac{1}{\sqrt{2}} \left< 0\middle| 0\right>+\frac{i}{\sqrt{2}}\frac{(-i)}{\sqrt{2}}\left< 1\middle| 1\right>=\frac{1}{2}+\frac{1}{2}=1$$

Title: Measuring entangled quantum state


问:
Sum over product of scalar products with indicator functions
Let $(h_n)_{n\in\mathbb{N}}$ be the Haar orthonormal basis. In our script we have the following equation as part of a proof for the existence of a Brownian Motion:

$\sum_{n=0}^{\infty} \langle h_n , \mathbb{1}_{[0,t]} \rangle \langle h_n, \mathbb{1}_{[0,s]} \rangle = \langle \mathbb{1}_{[0,t]}, \mathbb{1}_{[0,s]} \rangle $

Where the brackets represent the standard $L_2$ scalar product (i.e. integral over $\Omega$) and $s < t$ are in an intervall $[0, T]\subset \mathbb{R}$. I know of this equation for general orthonormal basis:

$x = \sum_{n=0}^{\infty} \langle x, h_n \rangle h_n$

But I still can't figure out what happened here. This might be an error in the script but I can't tell. Can someone clarify please?

答:
Choose $x = 1_{[0, t]}$ and scalarly mutiply your "known" equation by $x$:
$$
\langle x, x \rangle = \left \langle \sum_{n = 1}^\infty \langle x_n, h_n \rangle h_n, x\right \rangle
$$
We use continuity of the scalar product:
$$
\left \langle \sum_{n = 1}^\infty \langle x_n, h_n \rangle h_n, x\right \rangle = \left \langle \lim_{m \rightarrow \infty}\sum_{n = 1}^m \langle x_n, h_n \rangle h_n, x\right \rangle = \lim_{m \rightarrow \infty} \left \langle \sum_{n = 1}^m \langle x_n, h_n \rangle h_n, x \right \rangle = \lim_{m \rightarrow \infty} \sum_{n = 1}^m \langle \langle x_n, h_n \rangle h_n, x \rangle = \lim_{m \rightarrow \infty}\sum_{n = 1}^m \langle x_n, h_n \rangle \langle x_n, h_n \rangle = \sum_{n = 1}^\infty \langle x_n, h_n \rangle \langle x_n, h_n \rangle
$$
(The last sum exists in $\mathbb{R}$ according to Bessel's inequality.)


问:
Petersen's Book notational issue on $\nabla_X S$
The following notation is from Petersen book (p. 57 3rd Edition but not in previous edition):

it seems reasonable to define $\nabla S$ as $$\nabla S(X,Y)=(\nabla_X S)Y=\nabla_X(S(Y))-S\nabla_XY.$$
In other words $\nabla_XS=[\nabla_X,S]$.

But if $[\nabla_X,S]$ act on vector fields then $[\nabla_X,S]Y=(\nabla_X S-S\nabla_X)Y=(\nabla_X S)Y-S(\nabla_XY)$ that is not $(\nabla_X S)Y$. But if we evaluate $\nabla_X$ and $S$ at $Y$  first then compute its Lie bracket then that works! But this make no sense to me!! Why we should act on $Y$ first then apply the Lie bracket?

答:
Here $[\nabla_X, S]$ denotes the commutator $[\nabla_X, S] = \nabla_X\circ S - S\circ\nabla_X$ so

$$[\nabla_X, S](Y) = (\nabla_X\circ S)(Y) - (S\circ\nabla_X)(Y) = \nabla_X(S(Y)) - S(\nabla_XY).$$

What you've done is inconsistent. When you write $\nabla_XS$, you interpret that as $\nabla_X(S)$ (as opposed to $\nabla_X\circ S$), but when you write $S\nabla_X$ you interpret this as $S\circ\nabla_X$.

The Lie bracket of two vector fields $[X, Y]$ can also be viewed as a commutator when one regards vector fields as acting on smooth functions.


问:
Can we solve 4 unknows with 3 equations?

recently I am solving this 3 equations:
$$\begin{cases}
w+y+x+z=0\\
w-y+x-z=2\\
wy+xz=0\\
\end{cases}
$$

I can get
$$x=1-w$$
However, when I put $x = 1 - w$ into different equations, I can't still get the result.
There is no any restrictions of $w,x,y$ and $z$. How can I solve those unknowns? Thank you.

答:
The reason is that there are infinitely many solutions. If you are looking for integer solutions: $w = 0, y = -1, x = 1, z = 0$ and $w = 1, y = 0, x = 0, z = -1$ both seem to work. By substitution $x$, $y$, $z$ can be expressed in terms of $w$: $$x=1-w$$
$$y=\frac{1-w}{2w-1}$$
$$z=-\frac{w}{2w-1}$$


问:
Prove that $U_{n+1} + 1 \geqslant \tfrac{3}{4}(U_{n}+1)$
$U_{n}$ is a sequence where $n$ is natural.

$U_{n+1} = \tfrac{U_{n}^{2}+1}{U_{n}-1}$

$-3 \leqslant U_{n} < -1$

And the question is: Prove that $U_{n+1} + 1 \geqslant \tfrac{3}{4}(U_{n}+1)$

My attempt was:

$U_{n+1} + 1 = \tfrac{U_{n}}{U_{n}-1}(U_{n}+1)$

And $\tfrac{-1}{2} < \tfrac{1}{U_{n}-1} \leqslant \tfrac{-1}{4}$ ... (1)

And $-3 \leqslant U_{n} < -1$ ... (2)

And I wanted to multiply (1) by (2) but the signs are diffrent so I can't do that.

So how I solve it?

Body:
Let $-3 \le a <-1$

You have to show that $\frac{a^2+1}{a-1}+1 \ge \frac{3}{4}(a+1).$

We have $\frac{a^2+1}{a-1}+1=\frac{a(a+1)}{a-1}.$ thus

$$\frac{a^2+1}{a-1}+1 \ge \frac{3}{4}(a+1) \iff \frac{a(a+1)}{a-1} \ge \frac{3}{4}(a+1).$$

Since $a+1<0$ we derive

$$\frac{a}{a-1} \le \frac{3}{4}.$$

Since $a-1 <0$ we see that

$$a \ge \frac{3}{4}(a-1).$$

This last inequality is equivalent t0 $$a \ge -3.$$


问:
When is a Moishezon manifold rational?
We  talk about the question only on  complex manifold category.

A variety $Y$ is called rational if it is birationally equivalent to $\mathbf{P}^{n}$ for some $n$. Moishezon manifolds have many equivalent definitions. Here we use a well known one,  characterizing  Moishezon manifolds as manifolds which are bimeromorphic to projective algebraic varieties.

Question: When is a Moishezon manifold rational? If there is no a general answer, how about for a low-dimensional  Moishezon manifold? Any comment and answer is welcomed.

答:
Here is an answer for dimension $3$:

By Theorem 3 of Timmerscheidt's article, a three dimensional complex manifold $X$ is rational if and only if $X$ is Moishezon, ruled, $H^1(X,\mathbb Q) = 0$ and $H^2(X,\mathbb Q)$ is generated by the cohomology classes of ruled surfaces contained in $X$.

Title: When is a Moishezon manifold rational?


答:
First of all this is not a linear system. The third equation is a $2$nd degree equation.

Furthermore it has not a unique solution, because to have one and only one solution a necessary condition is that the number of equations was at least equal to the number of unknowns.

Anyway, if you add the first two equations you get
$$2w+2x=2\to w=1-x$$
then you substitute in the first and get
$$ x-x+y+z+1=0\to z=-1-y$$
the third equation becomes
$$(1 - x) y + x (-1 - y) =0$$
and you solve wrt $x$
$$x=\frac{y}{2 y+1};\;z=-y-1,w= \frac{y+1}{2 y+1}$$

Title: Can we solve 4 unknows with 3 equations?

Body:
recently I am solving this 3 equations:
$$\begin{cases}
w+y+x+z=0\\
w-y+x-z=2\\
wy+xz=0\\
\end{cases}
$$

I can get
$$x=1-w$$
However, when I put $x = 1 - w$ into different equations, I can't still get the result.
There is no any restrictions of $w,x,y$ and $z$. How can I solve those unknowns? Thank you.


问:
Quadratic factoring
What is a general rule of thumb that we can use to check whether a quadratic expression can be factored without the use of the quadratic formula since using the quadratic formula can be time consuming in exam situations?

Is it even possible to tell whether an expression will require application of the quadratic formula even before beginning to factor it?

答: The criterion is to check whether the discriminant $$b^2-4ac$$ of the equation $$ax^2+bx+c=0$$ is a perfect square (assuming the coefficients are integers). The equation can be factorised in $\mathrm Q$ if and only if this is the case. You can quickly check whether the square root of the discriminant is an integer if you're allowed a calculator.

Title: Quadratic factoring


问:
Fibonacci sequence proof.
Given the Fibonacci Sequence $F_{0}=0,F_{1}=1,F_{n+2}=F_{n+1}+F_{n+1};\;n\geq0$

Prove the following:

$\sum_{i=1}^{n}\frac{F_{i-1}}{2^{i}} = 1-\frac{F_{n+2}}{2^{n}}$

Base case:

$\frac{F_{0}}{2^{0}}= 1-\frac{F_{2}}{2^{0}}\\ \frac{0}{1} = 1-\frac{1}{1} \\ 0 =0$

Which is true.

Now, for the inductive step with $n=k$ we have

$\sum_{i=1}^{k}\frac{F_{i-1}}{2^{i}} = 1-\frac{F_{k+2}}{2^{k}}$

We want to prove then that for $n=k+1$

$\sum_{i=1}^{k+1}\frac{F_{i-1}}{2^{i}} = 1-\frac{F_{k+3}}{2^{k+1}}$

We rewrite the LHS such that

$\frac{F_{k}}{2^{k+1}}+\sum_{i=1}^{k}\frac{F_{i-1}}{2^{i}} = 1-\frac{F_{k+3}}{2^{k+1}}$

From the inductive hypothesis it follows that

$\frac{F_{k}}{2^{k+1}}+ 1-\frac{F_{k+2}}{2^{k}}= 1-\frac{F_{k+3}}{2^{k+1}}$

I get stuck here. Any hint will be appreciated.

Title: Fibonacci sequence proof.

Body: Since $\sum_{i=1}^k\frac{F_{i-1}}{2^i}=1-\frac{F_{k+2}}{2^k}$,$$\sum_{i=1}^{k+1}\frac{F_{i-1}}{2^i}=1-\frac{F_{k+2}}{2^k}+\frac{F_k}{2^{k+1}}=1-\frac{2F_{k+2}-F_k}{2^{k+1}},$$so you just need to prove $2F_{k+2}-F_k=F_{k+3}$. With $a:=F_k,\,b:=F_{k+1}$ we get$$F_{k+2}=a+b,\,F_{k+3}=a+2b\implies2F_{k+2}-F_k=2(a+b)-a=a+2b=F_{k+3}.$$


问:
Cumulative distribution function of quotient
Let's consider two independent variables $X \sim Exp(4)$ and $Y \sim Exp(12)$. I want to calculate $$P(\frac{X}{X-3Y} \le \frac19)$$

My work so far

$$P(\frac{X}{X-3Y} \le \frac19) = P(\frac{9X-X+3Y}{X-3Y} \le 0)= P(\frac{8X+3Y}{X-3Y} \le 0)= $$
$$ = P(\{8X+3Y \ge 0\} \cap \{X-3Y\le0\}) + P(\{8X+3Y \le 0\} \cap \{X-3Y\ge0\})$$

I'm not sure what I should do next... Intuitevly I would just decompose $P(\{8X+3Y \ge 0\} \cap \{X-3Y\le0\}) = P(\{8X+3Y \ge 0\}) \cdot  P(\{X-3Y\le0\})$ but thinking more it's not so obvious for me that these events are independent.

Am I going in the right direction ?

答:
Because the support of $X$ and $Y$ is $[0, \infty)$, $P(\{8X+3Y \le 0\}=0$

$$ P(\{8X+3Y \ge 0\} \cap \{X-3Y\le0\}) + P(\{8X+3Y \le 0\} \cap \{X-3Y\ge0\})= P(\{8X+3Y \ge 0\} \cap \{X-3Y\le0\}) = P(\{X-3Y\le0\}),$$

$$P(\{X-3Y\le0\}) = \int_0^{\infty}\int_0^{3y}f_y(y)f_x(x)dxdy= \\\int_0^{\infty}\int_0^{3y}12e^{-12y}4e^{-4x}dxdy = 0.5$$

Interpretation: Exp($\lambda$) represents the waiting times between Poisson($\lambda$)-distributed events. $E_Y$, the Poisson event associated with $Y$, is $\frac{12}{4} = 3$ times more frequent than $E_X$, the Poisson event associated with $X$. The probability that $E_X$ will occur once before $E_Y$ happens three times $\big(P(\{X-3Y\le0\})\big)$ is 50%, because the waiting times between different $E_Y$s are independent.


问:
find the space left in a triangle with an almost inscribed circle
is there a way to find the length of the gap that is left if a triangle is almost inscribed by a circle then the top bit of the circle is cut off? (sorry I'm bad at explaining stuff) Here's what I mean: is there a way to find p?</p>

答:
In the attached diagram below, AB and AC are tangent to the circle at B and C respectively.  Therefore, triangle ABC is isosceles, and since AD is perpendicular to BC, it must bisect the angle BAC, and its extension must pass through the centre of the circle at O.  Now assuming that $\ p\ $ is the length of AD (this was not completely clear in the diagram cited in the question), then we have
\begin{align}
p&=|\text{AD}|\\
&=|\text{AC}|\cos\frac{z}{2}\\
&=|\text{OC}|\frac{\cos\frac{z}{2}}{\tan\frac{z}{2}}\\
&=\frac{d\cos\frac{z}{2}}{2\tan\frac{z}{2}}\ ,
\end{align}
where $\ d=2|\text{OC}|\ $ is the diameter of the circle.


问:
A prime $p = 2a^2 – 2ab + 3b^2$ for integers $a$ and $b$, then $p$ cannot be congruent to $11 \pmod{20}$.

A prime $p = 2a^2 – 2ab + 3b^2$ for integers $a$ and $b$, then $p$ cannot be congruent to $11 \pmod{20}$.

I need to prove the conclusion and I tried by constructing a contradiction:

First, I suppose $$p \equiv 11 \pmod{20}$$ and factorized the prime as $$(a+3b)(a-b)$$

So I get two possible cases:

$(a+3b)\equiv 1 \pmod{20}$ and $(a-b) \equiv 11 \pmod{20}$

$(a+3b)\equiv 11\pmod{20}$ and $(a-b) \equiv 1 \pmod{20}$

However I do not know how to deal with it.

Is this a right direction or there are other good methods to prove the statement? Looking forward to any suggestions!

答:
Well, initially I didn't see that your factorization was wrong, so I have written the answer according to the cases you got in the end. This is how you would do If the factorization was correct.

Just subtract the two equations you got to get $$4b \equiv \pm 10 \pmod{20}$$ which means $4b\pm 10$ is divisible by $20$, but this is not possible since $20$ is of the form $4k$.


答:
We have clearly that $2|b-1$ so it is clear aswell that $4|2a(a-b)$ so we have
$$4|p-3$$
in order to prove the problem , we just need to show the following doesnt hold
$$5|p-1$$
now for the sake of contradiction , we assume it is true
which gives
$$5|2p-2=(2a-b)^2+5b^2-2$$
so we need $2$ to be a square in $Z_5$ which is wrong.
so it is impossible to have
$$20|p-11$$

Body:
A prime $p = 2a^2 – 2ab + 3b^2$ for integers $a$ and $b$, then $p$ cannot be congruent to $11 \pmod{20}$.

I need to prove the conclusion and I tried by constructing a contradiction:

First, I suppose $$p \equiv 11 \pmod{20}$$ and factorized the prime as $$(a+3b)(a-b)$$

So I get two possible cases:

$(a+3b)\equiv 1 \pmod{20}$ and $(a-b) \equiv 11 \pmod{20}$

$(a+3b)\equiv 11\pmod{20}$ and $(a-b) \equiv 1 \pmod{20}$

However I do not know how to deal with it.

Is this a right direction or there are other good methods to prove the statement? Looking forward to any suggestions!


问:
If $F$ is closed and $x \not\in F$, show that there are disjoint open sets $U,V$ such that $x \in U$ and $F \subset V$
If $F$ is closed and $x \not\in F$,

(a) show that there are disjoint open sets $U,V$ such that $x \in U$ and $F \subset V$.

(b) Can $U$ and $V$ be chosen such that their closures $\bar{U} , \bar{V}$are disjoint as well ?

Attempt:
Assume $d(x,F) = t$

Consider the function $d(y,F)$ which is continuous, so the set $A = \{y | d(x,F) > t/2\}$ is open and contains $x$, similarly the set $B = \{y | d(x,F) < t/2\}$ is also open and contains F, both A and B are disjoint hence the required sets.

Can I consider the closure of the A and B for the second part (b) of the question?

答:
Let r =d(x,F)/100

Now we define U=B(x,r) and V=UB(a,r)
a€F

Clearly, x€U and F is contained in V.

Then we claim that U and V are disjoint.
If not, then let there exists z€UnV.
=> d(x,z)<r and d(z,y) < r
⇒ d(x,y) < 2r= d(x,F)/50,which is a contradiction .
Hence our claim "U and V are disjoint " is true. Q.E.D


问:
$\begin{align}\color{#c00}a&\mid b\\\color{#c00}c&\mid d\end{align}\Rightarrow\ \color{#c00}{ac}\mid bd$ $\ \ \bf\small [Divisibility\ Product\ Rule]$
I just need to check the reasoning in my proof is correct, I think it is valid although I'm not totally convinced because I can't follow the logic; does proving that $x$ is an integer prove that $ac|bd$?

Theorem: Let $a$, $b$, $c$, $d$ be integers. If $a|b$ and $c|d$, then $ac|bd$.

Proof: $aj=b$ and $ck=d$ for integers $j$, $k$. Then, $ac|bd$ implies $acx=ajck$ and thus $x=jk$ for some $x$. Since the product $jk$ is an integer, $x$ is an integer and thus $acx=ajck$ and thus $ac|bd$.

答:
Your idea is right, but there are some difficulties of wording.  When you write "Then, $ac|bd$ implies" it begins to sound as if you are going in the wrong direction.  "If and only if" would have been OK.

You should have written instead something like this, which by the way is also shorter.

"Then $bd=(aj)(ck)=(ac)(jk)$. So if we let $x=jk$, then $(ac)x=bd$.  It follows that $ac$ divides $bd$."

Let me repeat, you knew exactly what was going on.  But there was some awkwardness in the exposition that somewhat masked that fact.


问:
How to obtain the minimum value of equation 6.7 in Pattern Recognition and Machine Learning, Bishop
Equation (6.5), with $a$ and $t$ are vectors, K is a symmetric matrix, and $\lambda$ is real value.
$$
J(a) = 0.5 a^T KK a - a^T K t + 0.5 t^T t + 0.5 \lambda a^T K a
$$
The book provide the minimum solution,
$$
a = (K + \lambda I)^{-1} t
$$

I derived the gradient of (6.5) with respect to a as below,
$$
J^{'}(a) = a^T KK - Kt + \lambda a^T K
$$
and set the gradient to zero to find the minimum solution, but it does not look like the provided solution,
$$
a = (KK + \lambda I)^{-1} (Kt)^T = K^-1 (K + \lambda I)^{-1} t^T K
$$

答:
In the gradient you mix row and column vectors
$$J^{'}(a) = KKa - Kt + \lambda Ka =K((K+\lambda I)a-t)=0$$
If $K$ is invertible, you get $(K+\lambda I)a-t=0$.


问:
What is the smallest polyomino that can't surround a $1\times 1$ hole?
Given a polyomino $P$, we can ask if it is possible for disjoint copies of $P$ to surround a single cell in the square grid - i.e., for the complement of their union to have a connected component of size $1$.

We can further refine this into polyominoes that weakly surround a cell (just covering the four edge-adjacent cells) and those that strongly surround a cell (covering all $8$ squares which share a vertex with the hole). Thanks to Julian Rosen for clarifying this distinction.

My original intuition was that this is always possible, but I was having trouble proving it; after enough struggling to show it was true, I started searching for counterexamples. Here is a polyomino with $48$ cells which does not even weakly surround a one-celled region, which is not too difficult to verify by hand:

After some modifications, I've reduced this down to a simply-connected solution with $26$ cells:

I have a size-$23$ example of a polyomino which does not strongly surround a hole:

What is the smallest polyomino that cannot surround a single-celled hole? I am interested in this question for both the weak and strong cases.

I've written some code to explore this, and have confirmed that all of the $1,227,708$ free polyominoes on at most $14$ cells can strongly surround a hole. How much can we tighten these bounds?

答:
This is by no means a complete answer, but as I mentioned in my last comment, we can make some headway by restricting a computer search to particular symmetry classes.  I've been able to check all the free polyominoes of size $N \le 23$ that have a horizontal or vertical symmetry axis (a class that contains OP's size-$23$ non-strongly-surrounding example).  Within this symmetry class, there are no non-weakly-surrounding examples of size $N \le 23$, and the smallest non-strongly-surrounding examples have size $N=23$; so OP's example is minimal.  Of the $464188$ size-$23$ free polyominoes with a horizontal or vertical axis of symmetry, there are just two that cannot strongly surround a single cell: the example given in the question, and the polyomino pictured below.

Heuristically, one might expect minimal examples to have some symmetry, because symmetric polyominoes have the fewest "different-looking local regions", and hence the fewest different possibilities for packing local regions around a hole.  Next steps, then, might be to check the other two large symmetry classes for free polyominoes: $180^\circ$ rotational symmetry around a point, and reflection symmetry across a diagonal.

Update:
Checking the polyominoes with rotational symmetry around a point turned up the following minimal (within that class) non-strongly-surrounding example, with size $20$:

Second update:
Checking the polyominoes with a diagonal reflection symmetry, there are exactly two minimal (within that class) non-strongly-surrounding examples with size $19$.  One is formed by removing an interior cell from the previous example (as already pointed out in a comment), and the other is new:

Third update:
In order to make progress on the non-weakly-surrounding case (which is a more stringent criterion, so the minimal examples are at least as large), I looked at polyominoes with both vertical and horizontal reflection symmetries.  The size-$25$ example below provides an improved upper bound for this case.


问:
Parallelizable spheres and whitney classes of bundles
In Milnor's paper, He asserts the following corollary numbered 2

If $\mathbb{R}^n$ has bilinear product with no zero divisors, then $n=1,2,4$ or $8$.

In his proof he constructs a vector bundle $E$ on $S^n$ by clutching map $$x \mapsto (\text{left multiplication by }x),$$ and then asserts that that stiefel-whitney class $w_n(E) \neq 0$.

My question: How to prove that $w_n(E) \neq 0$ ?

答:
For definiteness, the definition of $E$ is that $E = (D^n\times \mathbb{R}^n)\cup_f (D^n\times \mathbb{R}^n)$ where $f$ is a map from $\partial(D^n\times \mathbb{R}^n) = S^{n-1}\times \mathbb{R}^n$ to the other copy given by $f(x,y) = (x,xy)$, where the product $xy$ is the bilinear product with no zero divisors.  The projection $\pi:E\rightarrow S^n$ maps the first $D^n$ to the north pole and the second $D^n$ so the south pole of $S^n$.

Because $GL_n$ deformation retracts onto $O(n)$, we may assume that for each $x\in S^{n-1}$, the map $y\mapsto xy$ is an isometry.  In particular, the multiplication map restricts to map $\mu:S^{n-1}\times S^{n-1}\rightarrow S^{n-1}$.

Proposition 1:  Under the obvious identification $H_{n-1}(S^{n-1}\times S^{n-1})\cong \mathbb{Z}\oplus \mathbb{Z}$, the induced map $\mu_\ast:H_{n-1}(S^{n-1}\times S^{n-1})\rightarrow H_{n-1}(S^{n-1})$ is, up to changing orientations, addition:  $\mu_{\ast}(a,b) = a+b$.

Proof:  Fix $y\in S^{n-1}$.  The self map of $S^{n-1}$ given by $x\mapsto xy$ is linear (since the multiplication is bilinear) and has trivial kernel (since there are no zero divisors), hence is an isomorphism.  In particular, this self map is a diffeomorphism so has degree $\pm 1$.  By fixing orientations correctly, it has degree $+1$.  As a consequence, we see the map $\mu_\ast(a,0) = a$.

Likewise, the self map given by $y\mapsto xy$ has degree $+1$, so $\mu_\ast(0,b) = b$.  Since $\mu$ is a homomorphism, we are done.  $\square$

Let $E'\subseteq E$ be the unit sphere bundle of $E$.  In terms of the description above, this simply means that we replace the copies of $\mathbb{R}^n$ with $S^{n-1}$.

Proposition 2:  $H_\ast(E')\cong H_\ast(S^{2n-1})$.

Proof:  We will use Mayer-Vietoris with $U = D^n\times S^{n-1}$, $V$ the other copy of $D^n\times S^{n-1}$.  Then $U\cap V$ has the homotopy type of $S^{n-1}\times S^{n-1}$.

Since $H_\ast(U) = H_{\ast}(V) = 0$ for $\ast \neq 0,n-1$, the Mayer-Vietoris sequence easily shows thats $H_\ast(E')\cong H_\ast(S^{2n-1})$ when $\ast \neq n-1, n$, so let's focus on those cases.

We get the exact sequence $$...\rightarrow H_n(U)\oplus H_n(V)\rightarrow H_n(E')\rightarrow H_{n-1}(U\cap V) \rightarrow H_{n-1}(U)\oplus H_{n-1}(V)\rightarrow H_{n-1}(E')\rightarrow H_{n-2}(U\cap V)\rightarrow ...$$ which is just $$0\rightarrow H_n(E')\rightarrow \mathbb{Z}\oplus \mathbb{Z}\rightarrow \mathbb{Z}\oplus \mathbb{Z}\rightarrow H_{n-1}(E')\rightarrow 0.$$

Let's figure out the map from $\mathbb{Z}^2$ to itself.  To that end, let $u\in H_{n-1}(U)$ and $v\in H_{n-1}(V)$ be generators, and let $a,b\in H_{n-1}(U\cap V)$ be generators corresponding to the two factors of $U\cap V \cong S^{n-1}\times S^{n-1}$.

I am going to view $U\cap V$ as living in $U$, and then attached to $V$ via $f$.  What this means this that the inclusion $U\cap V\rightarrow U$ induces the map $b\mapsto u$ and $a\mapsto 0$.

What about the map $U\cap V\rightarrow V$?  Well, this is really a composition $U\cap V\xrightarrow{f} U\cap V\subseteq V$.  From Proposition $1$, it easily follows that $f_\ast(a,b) = (a,a+b)$.  Since the inclusion $U\cap V\subseteq V$ maps $a$ to $0$ and $b$ to $v$, the composition is the map $a\mapsto v$ and $b \mapsto v$.

Returning to the Mayer-Vietoris sequence, the map $\mathbb{Z}\times\mathbb{Z}\rightarrow \mathbb{Z}\times \mathbb{Z}$ is $a\mapsto (0,v)$ and $b\mapsto (u,v)$.  Now it is easy to verify that this map is a bijection, which finishes the proof of Proposition 2.  $\square$

Now, we'll compute the homology of $E'$ in a different way, and use this to compute $w_n(E)$.  To that end, recall the Grassmannian of $n$-planes in $\mathbb{R}^{\infty}$, $BO(n)$ has a tautological $n$-plane bundle over it, and that our bundle $E\rightarrow S^n$ is the pull back of this tautological bundle under some (homotopy class of) map $\phi:S^n\rightarrow BO(n)$.  Also, recall that $H^\ast(BO(n);\mathbb{Z}/2\mathbb{Z})\cong \mathbb{Z}/2\mathbb{Z}[\omega_1,...,\omega_n]$ and that $w_i(E) = \phi^\ast(\omega_i)$.
Note that if we restrict the tautological bundle to its unit sphere and pull back, we get $E'$.

Proposition 3:  The unit sphere bundle in the tautological bundle over $BO(n)$ has total space homotopy equivalent to $BO(n-1)$.

Proof:  One definition of this tautological bundle is that we start with the $O(n)$ principal bundle $EO(n)\rightarrow BO(n)$ and use the Borel construction $\mathbb{R}^n\times_{O(n)} EO(n)$, where $O(n)$ acts on $\mathbb{R}^n$ in the usual fashion.  If we restrict this to the unit sphere bundle, we just get $S^{n-1}\times_{O(n)} EO(n)$.  Since the $O(n)$ action on $S^{n-1}$ is transitive with isotropy group $O(n-1)$, this space is homeomorphic to $EO(n)/O(n-1) = BO(n-1)$.  $\square$

So, now we have a commutative diagram of bundles $\require{AMScd}$
\begin{CD}
S^{n-1} @>>> S^{n-1}\\
@VVV @VVV\\
E' @>>> BO(n-1)\\
@VVV @VVV\\
S^n @>>> BO(n)\\
\end{CD}

Consider the $\mathbb{Z}/2\mathbb{Z}$-cohomology spectral sequence of the bundle on the right.  As shown in these notes, beginning on page 85, the differential $d_{n-1} = d$ has the property that $dx = \omega_n\in H^\ast(BO(n);\mathbb{Z}/2\mathbb{Z})$.  By naturality, it follows that in the bundle on the left, we have $d(x) = \phi^\ast(\omega_2) = w_2(E)$.

On the other hand, from Proposition 2 (and universal coefficients), we know that $H^\ast(E')\cong H^\ast(S^{2n-1})$.  Thus, in the spectral sequence for the bundle on the left, we must have $d(x)\neq 0\in H^n(S^n;\mathbb{Z}/2\mathbb{Z}) = E_{n-1}^{n,0}$.

Thus, $w_n(E) = d(x)\neq 0$, establishing Milnor's claim.


问:
Angle bisected in a parallelogram

$ABCD$ is a parallelogram and $ID =JB$.

the point $P$ is the intersection of $DJ$ and $IB$.
Prove that the angle $X=Y$

$a,b,α,β$ are angles.

What I realized:

$a=b$.

and if we sum up these angles.
$$180=180-(x+α)+180-(y+β)+a $$
$$\iff 180=x+y+α+β-a \iff 360+2a=2(x+y)+2(α+β)$$

Please Help.

答:
Extend $CD$ to meet $BI$ in $Q$.

$$\triangle PJB \sim \triangle PDQ$$

Let $JB=x$ and $DQ=tx$, $t$ being the similarity ratio. So if $BP=y$, $PQ=ty$.

Let $CD=a=AB$. Now $\triangle IDQ \sim \triangle IAB$. So
$$\frac{ID}{DQ}=\frac{IA}{AB}\Rightarrow IA=a/t$$

$\therefore BC=AD=x+a/t$ and $CQ=tx+a$.

We see $$\frac{CQ}{BC}=t=\frac{PQ}{BP}\Rightarrow \angle PCD = \angle PCB$$

Title: Angle bisected in a parallelogram


问:
If $x \in [-1,1]$ and $y\in [0,1]$, given that $|x-y| < z$, what values can we constrain $z$ to?

We have $x \in [-1,1]$ and $y\in [0,1]$. Then consider the function

$$  f(x,y;z)=
    \begin{cases}
       \frac{1}{N}(x-y)^2 & \quad\text{if}\quad |x-y|\le z \\
       \frac{z }{N}|x-y| - \frac{z^2}{2}&\quad\text{else}
     \end{cases}
$$

So, for which values of $z$ does $f = \frac{1}{n}(x-y)^2$ and for which values of $z$ does $f=\frac{z }{N}|x-y| - \frac{z^2}{2}?$

We clearly see $(x-y) \in [-2,1],$ so $0<|x-y|<2$. What happens in this interval?

答:
That function puts no limitations on $z$. That is, the definition is sensible for every real number $z$. As I mentioned in the comment, $z$ can even be negative, in which case
$f(x,y; z) = \frac \alpha N|x - y| - \frac{\alpha^2}2$ for every $x, y$.

There is no basis in what you have given for restricting $z$ to any interval.


问:
$u_{tt} = 2(u_{xx}+ u_{yy}) $
I was given this equation $$u_{tt} = 2(u_{xx}+ u_{yy}) $$ and asked which one of the following will be the solution.

$1)$  $u(x , y , t) = t \sin (x+y^2)$

$2)$ $u(x , y , t) = t \cos (x+y^2) \sin (x+y^2)$

$3)$ $u(x , y , t) = \cos x \cos y \cos 2t$

I saw $u(x , y , t) = \cos x \cos y \cos 2t$ is satisfying the given partial differential equation and told him that this is the solution.

Have I gone wrong anywhere ?

答: No, you are correct. When you plug $u(x,y,t) = \cos x \cos y \cos (2t)$ into the equation, you see it is satisfied. So, this is a solution. It is not the solution because no initial or boundary conditions are specified (there are other solutions).

Title: $u_{tt} = 2(u_{xx}+ u_{yy}) $


问:
Let $(a_n)$ be a Fibonacci sequence , Prove that $a_1+a_3+a_5+\cdots+a_{2n-1} = a_{2n}$

Let $(a_n)$ be a Fibonacci sequence , Prove that $a_1+a_3+a_5+\cdots+a_{2n-1} = a_{2n}$

In this question $a_n$ is a Fibonacci sequence, I tried solving it with the PMI , problem arose when i was solving it for $P(n)=k+1$ , I could not make both of the sides equal.

答:
Assuming $a_0=0,a_1=1$ we can always add $a_0$ for free, so we get
$$
\begin{split}
(a_0 + a_1) + a_3 + a_5 + \ldots + a_{2n-1}
 &= (a_2 + a_3) + a_5 + \ldots + a_{2n-1} \\
 &= a_4 + a_5 + \ldots + a_{2n-1} \\
 &= a_{2n}
\end{split}
$$
To make this formal, you can use Mathematical Induction.


问:
Evolution of the projected angle with dihedral angle
I have a question about how angles behave under projection. Consider 2 lines making an angle of $\theta$ that are rotated out of the plane (say by an angle $\alpha$). Now, the projected angle $\theta_p$ (the angle between the red lines below) is larger than the initial $\theta$, and reaches a maximum value of $\pi$ for rotation $\alpha=\pi/2$.

Can someone please give me an equation giving the projected angle $\theta_p$ for a given $\alpha$? And also tell me how to derive it. I am guessing it involves basic trigonometry, but that only relates the sides under rotation. How do I then go on to relate the angles?

EXAMPLE:

This is what the projection looks like for $\theta=\pi/4$ and rotation angle $\alpha=\pi/4$. The blue lines are the initial lines (lying in the x-y plane say), and the red lines are the rotated lines:

Side view:

ADDENDUM TO ANSWER BELOW:

The accepted response below is perfectly correct. Let me just complement with some more details. Here is the full intended picture:

Error alert: I mislabeled $\alpha$ in the above picture. It should be $\angle AOC = \alpha$, not $\angle BOD$.

Now, by symmetry, we only need to consider half of the triangle, hence the apex angle is $\theta/2$. Also, AO is the perpendicular bisector to the base, so $\angle  AOB = \pi/2$.

Triangle COD is the projection of triangle AOB. Thus, CO = AO $\cos \alpha$.

The crucial fact we need to employ is that the base length AB is unchanged under the rotation $\alpha$. This is because it is parallel to the axis of rotation.

The rest follows is simple, see below.

Check limits: For $\alpha=0$, $\theta_p = \theta$. For $\alpha=\pi/2$, $\theta_p=\pi$. Perfect!

答:
Consider $AOB$ to be a half-angle $\theta/2$ projected onto plane $DCO$ (all angles at $C$ are right). $\angle AOC=\alpha$.

If $OA=1$, then $AB=\tan(\theta/2)$ and $CO=\cos\alpha$. Since $AB=CD$, then in triangle $COD$, $\tan\angle COD = CD/CO = \tan(\theta/2)/\cos\alpha$. Finally,
$$
\tan\frac{\theta_p}2 = \frac{1}{\cos\alpha}\tan\frac\theta2
$$


问:
Inequalities of Two i.i.d. Random Variables

The Problem

Let $X_1$ and $X_2$ be a continuous i.i.d. random variables.  By definition, we know that...

Independence: $\mathbb P(X_1 \in A \ \cap \ X_2 \in B)=\mathbb P(X_1 \in A) \ \mathbb P(X_2 \in B)$

Identically Distributed: $\mathbb P(X_1 \le x) = \mathbb P(X_2 \le x) \ \ \forall \ \ x \in \mathbb R^1$

My hypothesis is that:
$$\mathbb P(X_1 > X_2) = 0.5$$
I have attempted to prove this by using the above definitions to form the expression:
$$\mathbb P(X_1 \ge x \ \cap \ X_2 < x)=\mathbb P(X_2 \ge x \ \cap \ X_1 < x) \ \ \forall \ \ x \in \mathbb R^1$$

My Questions

Is this a valid proof, and am I missing anything?

This seems like a fundamental property of i.i.d. random variables.  Is this already a named theorem?

答:
Hint

(From Andrew Yung)

There is a standard trick involved here. Instead of giving you the full solution, let me point you in the right direction instead. Note that the probability you are interested in is equivalent to $\mathbb{P}(X_1 - X_2 > 0).$ Hence it would be enough to determine the cumulative distribution function of the variable $X_1 - X_2.$ For this it would be enough to first determine the CDF of $-X_2$ and then determine the CDF of the difference by explicitly computing the density of $X_1 - X_2$ in terms of the densities of $X_1$ and $-X_2$ (which you would have determined in the previous step). Is there any important Theorem that comes to mind when looking at it this way?
Please do let me know if you manage to fill in the blanks by yourself and if not, I shall do that.

Solution

Thank you for the hint, Andrew.  With your help I was able to find a solution.  Please let me know if you were thinking of an easier path to the answer.

Because $X_1$ and $X_2$ are Identically Distributed, the CDF and PDF of each random variable can be represented as:
$$F_X(x)=\mathbb P(X_1 \le x)=\mathbb P(X_2 \le x)$$
$$f_X(x)=\frac d {dx} F_X(x)$$

Note that:
$$F_{-X_2}(x)=\mathbb P(-X_2 \le x)=\mathbb P(X_2 > -x)=1-F_X(-x)$$

Also, the CDF of the sum of two Independent Continuous Random Variables ($X$ and $Y$) is:
$$F_{X+Y}(a) = \mathbb P(X+Y \le a) = \int_{-\infty}^{+\infty} f_X(x)F_Y(a-x)dx$$

Therefore:
$$\mathbb P(X_1 \le X_2)=\mathbb P(X_1 -X_2 \le 0)=$$
$$F_{X_1-X_2}(0) = F_{X_1+(-X_2)}(0) =$$
$$\int_{-\infty}^{+\infty} f_X(x)\Bigl(1-F_X\bigl(0-(-x)\bigr)\Bigr)dx =$$
$$\int_{-\infty}^{+\infty} f_X(x)dx+\int_{-\infty}^{+\infty} f_X(x) \Bigl(-F_X(x)\Bigr)dx=$$
$$1-\int_{-\infty}^{+\infty} f_X(x) \ F_X(x)dx = \ ...$$

Then you can use Integration by Substution where $u=F_X(x)$:
$$1-\int_{x=-\infty}^{x=+\infty} \biggl(\frac {du} {dx}\biggr)(u) dx=1-\int_{u=0}^{u=1} (u) du=1-\biggl[ \frac {u^2} {2} \biggr]_{0}^{1}=$$
$$1-(0.5-0)=0.5$$