答:
If $S$ is a left inverse of $T$, then $ST = \operatorname{id}_V$, which means that $\ker T = \{0\}$ (since $Tv = 0$ implies that $v = S(Tv) = S(0) = 0$). Thus, by the rank-nullity theorem we have $$\dim W = \dim V = \underbrace{\dim (\ker T)}_0 + \dim(\operatorname{im} T)$$ and then $T$ is surjective, that is, $T$ is invertible. Hence $TS = \operatorname{id}_W$, since the inverse is unique.

Now suppose that $S$ is a right inverse of $T$. Reversing the roles of $S$ and $T$ in the previous dot shows that $S$ is also a left inverse of $T$.

Title: Linear transformations: left and right inverses

Body:
I want to prove the following statement.

Let $V$ and $W$ be $n$-dimensional vector space and $T \in \mathcal{L}(V,W)$. Prove that $S \in \mathcal{L}(W,V)$ is a left inverse if and only if it is a right inverse.

My first thought was that I can use the following facts:

(1) A function $f: A \to B$ is injective if and only if it possesses a left inverse and is surjective if and only if it possesses a right inverse.

(2) A linear map $T \in \mathcal{L}(V)$ is injective if and only if it surjective if and only if it is invertible.

The first statement seems valid, but the 'link' between surjectivity and injectivity breaks down, I believe, because $V$ and $W$ only have the same dimension, but are not necessarily the same vector space. Finite-dimensional doesn't imply finite, so I cannot use the usual fact for functions that if $f: A \to B$ where $|A| = |B| = n$ is injective, then it is surjective (and vice-versa). I also can't use the fact that a function is bijective if and only if it is invertible because having a left or a right inverse says nothing about invertibility (with the key word being 'or').

Any help would be appreciated. I'd like to think I'm on the right track.


问:
Find all positive integers x such that x(x + 180) is a square.
$x^2 + 180x = k^2$

$k^2 - x^2 = 180x$

$(k+x)(k-x) = 180x$

Now i wrote $180x$ as $1(180x) , (2)(90x) , \dots , (180)(x)$ and i compared these values with $(k+x)(k-x)$ and found $12,60$ to be the only 2 integers , are there more ?
Also if you have a better method please share.

答: Hint: Rewrite it as $(x + 90 + k)(x + 90 - k) = 8100$. Now you can factorize $8100$.

ContentLicense: CC BY-SA 4.0


问:
Calculate Z Values from a plane normal
We need to fit a set of points (x,y,z) to plane and get fitted value of Z - Zi at a point (Xi,Yi).

We have used Eigen Library for plane fitting. https://gist.github.com/ialhashim/0a2554076a6cf32831ca

Now we have the normal to the plane and centroid of all points.

How can I can calculate the value of Z axis at a point (xi, yi) with the following information?

Centroid of all points

Normal to the fitted plane

答:
The equation of the plane with normal vector $\hat n$ passing though the point $p$ is $$\hat n \cdot v = \hat n \cdot p$$. This is a linear equation of the form $$ax + by + cz = d$$where $\hat n =(a, b, c), d = \hat n \cdot p$, and $v = (x,y,z)$. as long as $c \ne 0$, $$z = \frac dc - \frac acx -\frac bcy$$

If $c = 0$, then for all values of $z$, the point $(x_i, y_i, z)$ will be on the plane.

Title: Calculate Z Values from a plane normal


问:
Does the Probability holds the distributivity?
I am doing a basic probability problem from a textbook exercise. The question is as follows:

Consider two events A and B such that P(A) = 1/3 and P(B) = 1/2 and P(AB) = 1/8. Determine the value of P(BA^c). The way I did the problem is as,
       P(BA^c) =  P(B(1-A))  --> step 1
               =  P(B - BA)  --> step 2
               = P(B) - P(BA)  --> step 3
               = (1/2) - (1/8) = 3/8.

Answer from the book is same as mine.Is my approach correct? Does the Probability holds the distributivity as I did it from step 2 to step 3?

Please correct me if I were wrong with my terminology.

Thanks

答: You can consider the notation you're using as shorthand for operations on sets, where juxtaposition/multiplication represents intersection, subtraction represents the set-theoretic difference, and the $1$ represents the entire event space. Then the distributivity that you used (from step 1 to 2, not 2 to 3 as you wrote) is the distributivity of set intersection over set differences: $R\cap(S\setminus T)=(R\cap S)\setminus(R\cap T)$. </p>

ContentLicense: CC BY-SA 3.0


问:
The set of complex numbers of modulus $1$ is a group under multiplication
Show that $C=\{z\in \mathbb{C} \mid  |z|=1\}$ is a group under complex multiplication.

I'm a little confused because isn't the identity the only element with order $1$?  What is this set?

答: Regarding your second question: in $\mathbb{C}$, you have a lot of elements with modulus $1$ — all $e^{i\theta}$ for $\theta\in[0,2\pi)$.</p>


问:
How to mathematically solve for a binary vector whose dot product with a matrix gives a non-binary vector?
This is a bit of a long post, so two quick notes.

1.) I am a biologist by trade so I apologize if I misuse terminology or notation. I encourage clarifying edits to both the Title and Body!

2.) Thanks for taking the time to read through and consider my question!

My current project involves assigning simulated data to regions of a scatterplot based on the that regions local density.

I have a binary matrix, $A$, that represents whether a simulated point (column) falls in a region (row). Regions can overlap which is why columns can have multiple 1s in them.

$$\begin{equation*}
A_{m,n} =
\begin{pmatrix}
1 & 0 & \cdots & a_{1,n} \\
0 & 0 & \cdots & a_{2,n} \\
\vdots  & \vdots  & \ddots & \vdots  \\
1 & 1 & \cdots & a_{m,n}
\end{pmatrix}
\end{equation*}$$

I also have a non-binary vector, $Y$, of length $m$ (the number of regions) that represents the number of simulated points that I expect to see in each region:
$$\begin{equation*}
Y_{m,1} =
\begin{pmatrix}
7 \\
22  \\
\vdots  \\
y_{m}
\end{pmatrix}
\end{equation*}$$

I am trying to determine a binary vector, $X$, of length $n$ that when multiplied with $A$, gives Y:

$$\begin{equation*}
X_{n,1} =
\begin{pmatrix}
0 \\
1  \\
\vdots  \\
x_{n}
\end{pmatrix}
\end{equation*}$$

I think this is the correct formulation:
$$
A \cdot \textbf{X} = Y
$$

Here is an example:

$$\begin{equation*}
\begin{pmatrix}
1 & 0 & 0 & 1 \\
0 & 0 & 1 & 1 \\
0 & 1  & 0 & 0
\end{pmatrix}\begin{pmatrix}
1 \\
1\\
0 \\
1
\end{pmatrix} = \begin{pmatrix}
2 \\
1\\
1
\end{pmatrix}
\end{equation*}$$

In other words, I need (to sum the rows of) columns 1, 3, and 4 (AKA simulated points 1,3,4) to attain my "expected points in each region" vector.

My attempt: I understand that there are some scenarios where there are multiple solutions and other scenarios where there are no solutions. I wrote a metropolis algorithm that would randomly sample, $Y$, until it finds the best $Y$ that minimizes the cost function. But usually it takes a while to run...

My question: Given $A$ and $Y$, is there a way to mathematically solve for $\textbf{X}$?

答:
Another way of stating your problem: you want to express the vector $\bf Y$ as the sum of a subset of columns of the matrix $\bf A$.

The problem is NP-complete.  For example, the Exact cover problem is the case of your problem where all elements of $Y$ are $1$.  So we don't expect an efficient (polynomial-time) algorithm to exist.


问:
Exercise 12.9 book of Manetti (Topology)
I am trying to prove the following exercise:
Let $p:E\rightarrow X$ be a covering space. Prove that for every $Y$ connected and every continuos map $f:Y\rightarrow X$,
$$f^*p:\{(e,y)\in E\times Y|p(e)=f(y)\}\rightarrow Y,\, f^*p(e,y)=y,$$
is a covering space.

I think, I already proved that $f^*p$ is continuos and surjective. But I can't still find admisible open sets.

答:
I will write $E\times_X Y$ for $\{(e, y) \in E\times Y \mid p(e)=f(y)\}$. Let $f^*p : E\times_X Y \to Y$ be the projection on the $Y$ coordinate.

Given any $y \in Y$. Let $U$ be the evenly covered neighborhood of $f(y)$ in $X$. Then $f^{-1}(U)$ is a neighborhood of $y$. I claim that this $f^{-1}(U)$ is the evenly covered neighborhood of $y$. Note,

$(p^*f)^{-1}(f^{-1}(U)) = \{ (e, y) \in E\times_X Y \mid y \in f^{-1}(U) \} = \{(e, y) \in E \times Y \mid p(e) = f(y) \in U\}$. But since $U$ is evenly covered by $p$, $p^{-1}(U) = \bigsqcup_\alpha V_\alpha$, such that, $p|_{V_\alpha}$ is a homeomorphism. So,

$$(p^*f)^{-1}(f^{-1}(U)) = \bigsqcup_\alpha \{ (e, y) \in E\times Y \mid p(e) = f(y), e \in V_\alpha \} =: \bigsqcup_\alpha W_\alpha$$
Note that, $W_\alpha$ is open in $E\times_X Y$ since $W_\alpha = (V_\alpha\times Y) \cap (E\times_X Y)$. Now, you just need to check that $p|_{W_\alpha}$ is a homeomorphism, which should be easy.


问:
How do I show that $\int _0^\infty \! \frac{x\cos^3(x)}{t^2+x^2} \, \mathrm{d}x$ converges uniformly for all t > 0?
Until now, I've tried to solve by Dirichlet Criteria and Weierstrass M-test, but couldn't find an answer. Also tried to do inequalities, but didn't get anywhere.
Can somebody help me on this please?</p>

答:
According to the Dirichlet test for uniform convergence, the improper integral $\int_0^\infty f(x,t) g(x,t) \, dx$ is uniformly convergent for $t \in D$ if $\int_0^x g(y,t) \, dy$ is uniformly bounded for all $x >0$ and $t \in D$ and $f(x,t) \searrow 0$ as $x \to \infty$ monotonically with respect to $x$ and uniformly for $t \in D$.

In this case, the first hypothesis is true for all $x > 0$ and $t \in (0,\infty)$, since

$$\left|\int_0^x \cos^3y  \, dy\right| = \left|\frac{9}{12}\sin x + \frac{1}{12}\sin 3x  \right| \leqslant \frac{10}{12}$$

Considering the behavior of $f(x,t) = \frac{x}{x^2 +t^2}$ we first note that $|f(x,t)| \leqslant \frac{1}{x}$ which gives us uniform convergence to $0$ as $x \to \infty$ for all $t \in (0,\infty)$. However, $f(x,t)$ is monotone nonincreasing only when $x \geqslant t$ and we don't fulfill the uniformity of the monotonicity condition for all $t \in (0,\infty)$.

If we restrict the domain to $t \in (0,\alpha]$ for any finite $\alpha > 0$, then we have uniform convergence because

$$\int_0^\infty \frac{x}{x^2 + t^2} \cos ^3 x \, dx = \int_0^\alpha \frac{x}{x^2 + t^2} \cos ^3 x \, dx+ \int_\alpha^\infty \frac{x}{x^2 + t^2} \cos ^3 x \, dx,$$

and the first integral on the RHS is proper and for the second we have all the conditions of the Dirichlet test fulfilled  as $x \geqslant \alpha \geqslant t$ for all $t \in (0,\alpha]$.

There remains the question of uniform convergence for all $t > 0$.


问:
Example of compact upper semicontinuous map
can any one give me an exemple of compact multivalued upper semicontinuous maps in $L^{2}$ ?
thanks !

答: $S^1 \to \Bbb C : z \mapsto \sqrt{z}$ is perhaps the simplest.


问:
How to find critical F-Value?
So for a problem in the book I'm using I need to find $F_{0.025,10,11}$ and $F_{0.025,11,10}$

So in my book I used the F-distribution table to help me find

$F_{0.025,10,11} = 3.53$

But how do I find

$F_{0.025,11,10} = \text{???}$

My book only has degrees of freedom in the numerator go from 1-10, then 12, 15, $20,$ etc. to infinity completely skipping 11. Other tables I find on google have the same issue. So how do I calculate $F_{0.025,11,10} = \text{???}$

Is there another way to do so without a table that doesn't involve using a programming language like R?

(I can't find the F-distribution tag, so I'm just tagging this under statistics)

答:
Suppose $F\sim F_{10,11}$ and $G\sim F_{11,10}.$

Then $\Pr(F>3.525672) = 1 - 0.025.$

The distribution of $1/G$ is the same as the distribution of $F.$ Suppose $\Pr(G> a) = 1 -0.025$ and you want to find $a.$ So $\Pr(1/G < 1/a) = 1-0.025.$ You can find $1/a$ in the table for $F_{10,11}$ and then takes its reciprocal.

Title: How to find critical F-Value?


问:
How to prove this inequality? $ab+ac+ad+bc+bd+cd\le a+b+c+d+2abcd$
let $a,b,c,d\ge 0$,and $a^2+b^2+c^2+d^2=3$,prove that

$ab+ac+ad+bc+bd+cd\le a+b+c+d+2abcd$

I find this inequality are same as Crux 3059 Problem.

答:
Your constraints are $a,b,c,d > 0$ and $a^2+b^2+c^2+d^2 = 3$, by doing a change of variables, $x^2 = a^2/3, y^2 = b^2/3, z^2 = c^2/3, w^2 = d^2/3$. Then your constraints change to

$$x,y,z,w > 0, x^2+y^2+z^2+w^2 =1$$

and the problem is equivalent to proving $xy +xz +xw +yz+yw+zw \leq \sqrt{3}(x+y+z+w) + \frac{2}{3}xyzw$. Since $x^2+y^2+z^2+w^2 =1$ that first implies that $x,y,z,w \leq 1$ and in particular $x^2+y^2+z^2+w^2 \leq x+y+z+w$. Doing your standard expansion,

$$ (x+y+z+w)^2 = x^2+y^2+z^2+w^2 + 2(xy+xz+xw+yz+yw+zw) $$

The right hand side is bounded above by $x^2+y^2+z^2+w^2 + (x^2+z^2)+(x^2+w^2)+(y^2+z^2) +(y^2 + w^2) + 2(xy + wz) = 3(x^2+y^2+z^2+w^2) + 2(xy+wz)$

Balancing this with the RHS of the display we get

$$2(xy+xz+xw+yz+yw+zw) \leq 2(x^2+y^2+z^2+w^2) + 2(xy+wz)$$

Then we have that $xy \leq (x^2+y^2)/2$ and $wz \leq (w^2+z^2)/2$ and placing this in we have

$$xy+xz+xw+yz+yw+zw \leq 1.5(x^2+y^2+z^2+w^2) \leq \sqrt{3}(x+y+z+w) + \frac{2}{3}xyzw$$

since $\frac{2}{3}xyzw \geq 0$, $1.5 <\sqrt{3}$ and $x^2+y^2+z^2+w^2 \leq x+y+z+w$.


问:
If $2$ altitudes of a triangle are $9$ and $40$ then find the minimum perimeter

If $2$ altitudes of a triangle with integer side lengths are $9$ and $40$ units in length, then find the minimum possible perimeter of the triangle

Since the altitude is the shortest distance from a vertex to the opposite side, I got $a>9$ and $b>40$, where $a$ and $b$ are two of the sides of the triangle. For minimum perimeter I took $a_{min}=10,b=41_{min}$ which gave me $31<c<51 \implies c_{min}=32$. Hence I concluded that the minimum perimeter was $83$. However, the answer given is $90$. I suspect this is the right angled triangle of $(9,40,41)$ but I couldn't see why that would be the one with the smallest perimeter.

答:
I want to add something to Calvin Lin's comment to make it more rigorous.

By equating area we have $$40a=9b$$ Now since $a,b$ are integers and $\gcd(9,40)=1$ we must have $a=9k,b=40k$ for some natural $k$. If $k=1$ we see that the condition is satisfied and $a+b+c=90$.If $k\ge 2$ then we have $a+b+c>a+b=49k\ge 98$ hance minimum of $a+b+c$ is indeed $90$.

Note:Just because $a_{min}=9$ we cant say that the minimum value of $a+b+c$ occurs when this occurs(since a,b,c are not  completely independent variables)


问:
How to determine the most probable value of the perimeter of a quadrilateral?
The problem is as follows:

The diagonals of a quadrilateral measure $8\,cm$ and $10\,cm$. Using
this information find the most probable perimeter of this
quadrilateral indicated as a range.

The alternatives given in my book are:

$\begin{array}{ll}
1.&\textrm{Between 14 cm and 16 cm}\\
2.&\textrm{Between 18 cm and 36 cm}\\
3.&\textrm{Between 38 cm and 40 cm}\\
4.&\textrm{Between 42 cm and 50 cm}\\
5.&\textrm{Between 12 cm and 14 cm}\\
\end{array}$

In this problem I don't know what sort of strategy can be used?. Does it exist a relationship between the diagonals of a quadrilateral?.

So far what it come to my mind was to use the triangle inequality which states that the sum of the two sides of a triangle is greater than the third side.

Given a quadrilateral ABCD then this means the diagonals are from $AC$ and from $BD$.

$AC<AB+BC$

$BD<BC+CD$

$AC<CD+AD$

$BD<AB+AD$

Given that the diagonals are AC and BD then its sum must be less than:
By summing the expressions given:

$2AC+2BD<2AB+2BC+2CD+2AD$

$AC+BD<AB+BC+CD+AD$

This must be the upper bound in the perimeter of the quadrilateral. But how about the lower bound?. Well, for this part I'm using the reverse triangle inequality which states that any side is greater than the difference of the other two sides provided one is greater than the other.

In this case I'm assuming $AB>BC$ and $BC>CD$ and $CD>AD$ and $AB>AD$

Then this would mean:

$AC>AB-BC$

BD>BC-CD

$AC>CD-AD$

$BD>AB-AD$

The thing here is that you may not get the perimeter directly.

Suming these:

$2AC+2BD>2AB-2AD$

$AC+BD>AB-AD$

But we know that we're assuming $BD>AC$.

Then summing both gets:

$2BD>AB-AD$

The perimeter is $AB+BC+CD+AD$

But that's where I'm stuck. Can someone help me here to finish this proof?. I think I'm near but I don't know how to use those inequalities to get to the answer. Can someone help me?.

By the way, the official answer from my book states to be the second option. Which it kinds of confirms my initial finding for the upper bound. But it doesn't make sense for the lower bound. Shouldn't it be less?. Therefore, can someone help me?

答:
You found a lower bound, but not the lower bound.

Instead of adding your four inequalities all together, just add the ones that have $AC$ on the left. Make a separate sum of inequalities with $BD$ on the left.
The results are
\begin{align}
2AC &< AB + BC + CD + AD, \\
2BD &< AB + BC + CD + AD.
\end{align}
That is, the perimeter must be greater than twice the diagonal, for each diagonal.
In short,
$$ AB + BC + CD + AD > 2 \max \{ AC, BD \}. $$

Since in your example, $\max \{ AC, BD \} = 10,$
the minimum perimeter is $20.$

To put this intuitively, to get around the quadrilateral once starting at $A$ you must first travel from $A$ to $C$, which is a path at least $10$ units long since $A$ and $C$ are $10$ units apart, and then you must get back from $C$ to $A$, which is at least another $10$ units.

There is no upper bound if we allow quadrilaterals that are not convex. But if the quadrilateral is convex then the diagonals intersect.
We maximize the perimeter by intersecting the diagonals very near an end of each diagonal at a very shallow angle; in the limit we have the two diagonals collinear and joined at one endpoint, so that two vertices (say $B$ and $C$) are coincident, so that you go from $A$ to $B$ (distance $8$, since $B=C,$ assuming WLOG that $AC$ is the short diagonal),
from $B$ to $C$ (distance $0$), from $C$ to $D$ (distance $10$), and then $D$ to $A$
(distance $18$), for a total perimeter $36$.

In conclusion,

$$ 20 < \mathrm{perimeter} < 36. $$

This is not exactly the same as answer 2, which has a slightly larger range, but answer 2 is the only one that covers the actual minimum and maximum; all the other answers give perimeters that are much too short or much too long. In fact every other answer is giving a range of numbers for which we know the quadrilateral cannot have any of those values as its perimeter.


问:
Probability problem from CMU contest prep course
Your sports team is playing a best-of-15 series against a single opponent. Against this opponent, your
team wins with probability 40%, unless it is behind in the series (with strictly fewer wins than losses
so far in these 15-game finals), at which point your team wins with probability 60%. What is the
probability that your team wins?

I'm not great at math so I wrote a dynamic programming algorithm to solve it. If my algorithm is correct, the answer would be 40%. I found that if the number of matches is odd, the probability that you win is always 40%, for any number of matches.

Can anyone verify this and provide a probabilistic explanation of the solution?

Thanks!

答:
Note: What's crucial here is that $0.4 + 0.6 = 1$. Otherwise, this will not work.

Let the series be denoted by an element of $ S =  \{ L, W \} ^{15}$.

Hint: Create a bijection $f: S \rightarrow S$ between matches that you won, and matches that you lost

 - If you won the series, consider all of the matches (you won more games)

 - Take that series, and 1) Swap each of the W into a L and vice versa, and 2) Reverse the game order

 - This then becomes a series that you lost (since you lost more games)

Claim: P(You win) / 40% = P(You lost ) / 60%.

 Proof: Say you won with $s$.

 For the last match that you won, you won with probability 40%.

 Now consider $f(s)$.

 For the first match that you lost, you lost with probability 60%.

 Ignoring that match in both series (which correspond to the same match), verify that the probability of the sequence happening is the same. (There's a small step here that you should verify the probabilities match up, which has to do with how the games are defined).

Corollary: P(You win) = 40%.

As an explicit example, consider a round of 5 games where $ s = WLLWW$ where you won.

Then $f(s) = LLWWL$ where you lost.

$P(s) = 0.4 \times 0.6 \times 0.6 \times 0.6 \times 0.4$.

Ignoring the last win, we get $P(s) / 0.4 = 0.4 \times 0.6 \times 0.6 \times 0.6 $.

$P(f(s) ) = 0.6 \times 0.4 \times 0.6 \times 0.6 \times 0.6 $

Ignoring the first loss, we get $ P (f(s) ) / 0.6 = 0.4 \times 0.6 \times 0.6 \times 0.6$.

Hence, the probabilities match up.


问:
Find the gradient of $f$ at a point $x_{0}$
This question was part of my analysis assignment and I was unable to solve it. So, I am asking for help here.

Let $f: \mathbb{R}^{4} \rightarrow \mathbb{R}$ be defined by $f(x)=x^{t} A x$
where $A$ is a $4 \times 4$ matrix with real entries and $x^{t}$ denotes the transpose of $x$. The gradient of $f$ at a point $x_{0}$ necessarily is

$2 A x_{0}$

$A x_{0}+A^{t} x_{0}$

$2 A^{t} x_{0}$

$A x_{0}$

I used the definition of gradient and assume $x =[x, y, z, w]$ but couldn't match any of the options. So, it's my humble request to you to help me. Thanks!!

答:
Let $x=(x_1,x_2,x_3,x_4)$. By definition $(Ax)_i = \sum_{j=1}^4 A_{ij} x_j$, and therefore
$$f(x) = x^t Ax = x\cdot Ax = \sum_{i=1}^4 x_i (Ax)_i = \sum_{i,j}  A_{ij} x_ix_j,$$
where $\sum_{i,j}$ means $\sum_{i=1}^4\sum_{j=1}^4$. Differentiating with respect to $x_k$, we have by ordinary product rule in 1D,
\begin{align} \partial_{x_k} f(x) &= \partial_{x_k} \sum_{i,j}  A_{ij} x_ix_j
\\
&=  \sum_{i,j}  \partial_{x_k} (A_{ij} x_ix_j )
\\
&= \sum_{i,j}  A_{ij} [(\partial_{x_k} x_i) x_j + x_i (\partial_{x_k} x_j)]
\\
&= \sum_{i,j}  A_{ij} (\partial_{x_k} x_i) x_j +  \sum_{i,j}A_{ij}x_i (\partial_{x_k} x_j).
\end{align}

Since $\partial_{x_k} x_i=1$ if $k=i$ and $0$ if $k\neq i$, i.e. $\partial_{x_k} x_i=\delta_{ik}$,
$$ \partial_{x_k} f(x)=\sum_j A_{kj} x_j + \sum_i A_{ik} x_i.  $$
The first term gives $Ax$. The second term gives $A^tx$. So the answer is 2.


问:
Transform roll, pitch, yaw from one coordinate system to another
I'm trying to figure out how to transform a pose given with Euler angles roll (righthanded around X axis), pitch (righthanded around Y axis), and yaw (lefthanded around Z axis) from the Unreal Engine into a pose in a purely right-handed North, West, Up coordinate frame.

I know the full rotation matrix is $R = R_z R_y R_x$, as Euler angles are applied in that order.

But, irritatingly, the Y axis switches direction, and the whole coordinate system is rotated.

How can I elegantly determine pose in my target reference frame? I'm open to quaternions.

答:
The relationship between the two frames as you’ve drawn them is simple:  rotate by an angle of $\pi/2$ radians counterclockwise around the z axis, then reflect through the plane perpendicular to y.

There isn’t going to be any elegance about it with Euler angles. Someone has already done the repulsive conversion from unreal to 3x3 matrix or to quaternions for you, so find that.  Just pick one and stick with it.

By “handed” it sounds like you mean the choice of sign when expressing a rotation around an axis.  If that bothering you then obviously you can swap direction by swapping signs. But like I said, if it’s coming out of unreal someone probably already has written a conversion to a normal right handed coordinate system.  (That’s what I think of as handedness: an orientation assigned to a coordinate system, not a type of rotation.)

Performing the rotation and reflection is trivial using matrices or quaternions. Rotations by quaternions are of course famous. Reflections might be lesser known but they are pretty simple to describe: take the unit normal $n$ to your plane expressed a s a quaternion: then $x\mapsto nxn $ is the reflection.


问:
Confused about combinatorials
How do I solve 4$\cdot$6 = 8$\cdot$3 by a combinatorial proof? How can I start this proof? I know that I can show a two pictures that represent 24 but I'm not entirely sure how to go about this. Please help!</p>

答: Here is a combinatorial/pictorial proof. The image was made using the awesome TikZ package in LaTeX.

Title: Confused about combinatorials


问:
How to show that Hardy-Littlewood maximal function is lower semicontinous
I'm trying to show this by the absurd. If it's not, then, for every neighborhood of a given point $x$, exists a point $y$ such that $M(f)(y) > M(f)(x) + \epsilon$ so if I take a cube $Q_x$ such that $\frac{1}{|Q_x|}\int_{Q_x}|f(t)|dt$ is close to $M(f)(x)$ so the cube $Q_y$ that I take to approximate $M(f)(y)$ can't have the point $x$, but I don't know what to do if $Q_y$ is small enough to not have $x$. Also, I think the approach may not be good since analyzing points when I am talking about things with measure seems like a bad idea

Thanks

答:
Fix $\lambda>0.$ Let $E(\lambda) = \{x\in \mathbb R^n: Mf(x) > \lambda\}.$  You want to show that $E(\lambda)$ is open. So fix $x\in E(\lambda)$ and find an open cube $Q\ni x$ such that $\frac{1}{m(Q)}\int_{Q}|f(t)dt>\lambda.$ Now just observe that if $y\in Q,$ then by definition of the maximal function, $Mf(y)>\frac{1}{m(Q)}\int_{Q}|f(t)dt>\lambda$ so $Q\subseteq E(\lambda).$

Remark: I am using the definition of $Mf$ and follows:

for $f\in L_{\text{loc}}(\mathbb R^n),\ Mf(x)=\sup\{\frac{1}{m(Q)}\int_Q|f(t)|dt:\ Q\ \text{is a cube and}\ x\in Q\}.$ There is another common definition, namely,

$\tilde Mf(x)=\sup\{\frac{1}{m(Q)}\int_Q|f(t)|dt:\ Q\ \text{is a cube centered at}\ x\}.$

These two definitions are equivalent because it easy to show that

$\tilde Mf(x) ≤ Mf(x) ≤ 2^n\tilde Mf(x).$


问:
Exhausting integral $ \int_{0}^{\infty}\frac{\ln^3(x)}{(x-1)^3} \, \mathrm{d}x $
I would like to evaluate following integral:

$$ \int\limits_{0}^{\infty}\frac{\ln^3(x)}{(x-1)^3} \, \mathrm{d}x $$

by complex means. I integrate

$$f(z)=\frac{\ln^4(z)}{(z-1)^3}$$

(where ln is analytic in $\mathbb{C} \setminus \mathbb{R_+}$ with $\ln(-1)=i\pi$) over following contour: $[0;R]$, $C_0(R)$ - circle of radius $R$, $[R, 1+\varepsilon]$, semicircle in lower half-plane $C_1(\varepsilon)$ and $[1-\varepsilon,0]$. After evaluation and taking $\lim\limits_{\varepsilon \to 0}$ I came to correct answer, but my question is:

Is it possible to avoid that exhausting evaluation?

Thanks in advance.

答:
The computation can be made simpler if we use two versions of the complex logarithms:

$\log_{-}(z)$ denotes the complex logarithm with $\arg \in (-2\pi, 0)$.

$\log_{+}(z)$ denotes the complex logarithm with $\arg \in (0, 2\pi)$.

In particular, for each $x > 0$, we have

\begin{align*}
\log_-(x+i0^+) &= \log x - 2\pi i, &
\log_-(x-i0^+) &= \log x, \\
\log_+(x+i0^+) &= \log x, &
\log_+(x-i0^+) &= \log x + 2\pi i.
\end{align*}

Now define

$$ f(z) = \frac{\log^2_+(z) \log^2_-(z)}{(z-1)^3}. $$

This choice simplifies the computation in two ways:

The symmetry allows us to better relate the contour integral of $f$ to the real integral in the question.

Regardless of whether $z$ is above or below the $x$-axis, either $\log_+(z)^2$ or $\log_-(z)^2$ cancels out $(z-1)^2$ in the denominator. This is particularly helpful when computing the limit of the integral along the semicircle around $z=1$.

Now consider the modified keyhole contour $\mathcal{C}(\varepsilon,R)$ consisting of:

$\mathsf{L}_{+,1} = [i0^+, 1-\varepsilon + i0^+]$,

clockwise upper semicircle $\mathsf{C}_{+}(1,\varepsilon)$,

$\mathsf{L}_{+,2} = [1+\varepsilon + i0^+, R + i0^+]$,

counter-clockwise circle $\mathsf{C}(0,R)$,

$\mathsf{L}_{-,2} = [R - i0^-, 1+\varepsilon - i0^+]$,

clockwise lower semicircle $\mathsf{C}_{-}(1,\varepsilon)$,

$\mathsf{L}_{-,1} = [1-\varepsilon + i0^-, i0^-]$.

Then it is easy to check that

\begin{align*}
&\int_{\mathsf{L}_{+,1} \cup \mathsf{L}_{-,1}} f(z) \, \mathrm{d}z + \int_{\mathsf{L}_{+,2} \cup \mathsf{L}_{-,2}} f(z) \, \mathrm{d}z \\
&= \int_{0}^{1-\varepsilon} \frac{(-8\pi i)\log^3 x}{(x-1)^3} \, \mathrm{d}x
+ \int_{1+\varepsilon}^{R} \frac{(-8\pi i)\log^3 x}{(x-1)^3} \, \mathrm{d}x.
\end{align*}

Moreover, since $(z-1)f(z)$ is bounded near $z = 1$,

$$ \lim_{\varepsilon \to 0^+} \int_{\mathsf{C}_{+}(1,\varepsilon)} f(z) \, \mathrm{d}z
= (-\pi i) \lim_{\substack{z \to 1 \\ \operatorname{Im}(z) > 0}} (z-1)f(z)
= (-\pi i) (-2\pi i)^2
= 4\pi^3 i $$

and similarly

$$ \lim_{\varepsilon \to 0^+} \int_{\mathsf{C}_{-}(1,\varepsilon)} f(z) \, \mathrm{d}z
= (-\pi i) \lim_{\substack{z \to 1 \\ \operatorname{Im}(z) < 0}} (z-1)f(z)
= (-\pi i) (2\pi i)^2
= 4\pi^3 i. $$

Finally, it is clear that

$$ \lim_{R \to \infty} \int_{\mathsf{C}(0, R)} f(z) \, \mathrm{d}z = 0. $$

Now, since $f$ is analytic on and inside of $\mathcal{C}(\varepsilon,R)$, it follows that

$$ 0 = \lim_{\substack{R \to \infty \\ \varepsilon \to 0^+}} \int_{\mathcal{C}(\varepsilon,R)} f(z) \, \mathrm{d}z
= (-8\pi i)\int_{0}^{\infty} \frac{\log^3 x}{(x-1)^3} \, \mathrm{d}x + 8\pi^3 i $$

and therefore

$$ \int_{0}^{\infty} \frac{\log^3 x}{(x-1)^3} \, \mathrm{d}x = \pi^2. $$


问:
How is the AKS primality test Rosetta code so simple?
Skip to the end to see alternative question.

The following is a Python implementation of the AKS Primality Test.

```<code>from sympy import *

def expand_x_1(n):
    # This version uses a generator and thus less computations
    c = 1
    for i in range(n//2 + 1):       # // means flooring after divide
        c = c*(n - i)//(i + 1)
        yield c

def aks(p):
    if p==2:
        return True

    for i in expand_x_1(p):
        if i % p:
            # we stop without computing all possible solutions
            return False
    return True

for n in range(2, 10000):
    primality = aks(n)
    primality1 = isprime(n)
    if primality != primality1:
        print("Fails @", n)  # Never prints
        assert (0)
    else:
        print(primality)
</code>```

How is it possible that they took that much more in-depth pseudocode of the algorithm (that involves polynomial operations), and converted it into this 10-line version?

Is the above really the AKS primality test?  I got it from:

https://rosettacode.org/wiki/AKS_test_for_primes#Python

Let the input be called $n$, not $p$.

The code in expand_x_1(n) must be computing:

$$c_0 = 1, c_i = \lfloor \dfrac{c_{i-1}(n-i)}{i + 1}\rfloor$$

Where $c_i = $ the $i$th yielded value.  The other code using this value simply tests whether $c_i \neq 0 \pmod n$, in which case (if true) it returns False for composite.  Else if for all $c_i$ values at $i = 0..\lfloor \dfrac{n}{2} \rfloor + 1$ we have $c_i = 0 \pmod n$, then True is returned.

The recursion plus this test don't seem at all like what makes up the AKS algorithm.  So I was hoping an analytic number theorist could explain the formula.

Alternatively, if you can't answer the above, then:

How can we study the formula for $c_i$; can you think of any rearrangments it has? Such as maybe the denominators combining across recursive subcalls that have floor etc.

This is so I don't have to open up another question regarding this formula.

For example, I modified the code to:

```<code>def expand_x_1(n):
   c = 1
   d = 1
   for i in range(n//2 + 1):
       d *= (i + 1)
       c = c*(n - i)
       yield c//d
</code>```

Therefore, since it's getting no failures when I run it, I can somewhat safely assume that "denominators can be combined" algebraically, i.e. there is some
identity made use of that derives from the basic properties of floor.

What else can we say and how does this formula relate to polynomial arithmetic?

答: The numbers you’ve labelled as $c_i$ are the binomial coefficients $\binom ni$; the code checks whether $\binom ni \equiv 0 \pmod n$ for all $0 < i \le \frac n2$.  This is not the AKS algorithm.  It’s the exponential-time brute force algorithm listed in the Wikipedia article to motivate the AKS algorithm.


问:
What is the terminology for a product of a ring with a group (like the quaternions) or (more generally) with a monoid (like a polynomial ring)?
I don't think there is much for me to elaborate beyond the title question: "What is the terminology for a product of a ring with a group (like the quaternions) or (more generally) with a monoid (like a polynomial ring)?"

I know the quaternions require a field, and fields are rings, so the example holds.  Also, it is popular example and was the first to come to mind.  Though, if you think I should put something else instead or in addition to, please let me know.

So, if you know of terminology for any of the four products of a field or (preferably) ring with a group or (preferably) monoid, please let know each one you know. (In case of confusion, the four products I'm referring to are field with group, field with monoid, ring with a group, and ring with monoid.)

If you can think of anything useful for me to elaborate, please let me know.

Edit: Right before I was about to submit this, I remembered some of the answer.  A product of a ring with a group is called a group ring, right?  Ring with a monoid is a monoid ring, right?  Is there terminology for the other two?

答:
I do not think anyone calls the group ring $R[G]$ “the product of a group and a ring”, but evidentially that is what you are talking about.  Yes, the same construction using a monoid instead of a group Is called a “monoid ring $R[M]$.”

When the ring is a field, you could also call it “the group algebra $k[G]$” or “the monoid algebra $k[M]$, respectively.

There are also semigroup rings and semigroup algebras which do not take much imagination to define starting from this point.

What is the terminology for a product of a ring with a group (like the quaternions)

One more comment to address this: the quaternions themselves are not a group ring over a nontrivial group.  The problem is that it is a simple ring, and $R[G]$ is not simple unless $|G|=1$ and $R$ is simple.

I imagine you thought $\mathbb R[Q]$ is related to $\mathbb H$, where $Q$ is the quaternion group. It is, but not directly: $\mathbb H$ is a quotient of $\mathbb R[Q]$ and not a group ring itself.


问:
Prove that any set of 27 six-digit positive integers must have three subsets with the same sum.
So I know this is a pigeon hole problem, however, I am struggling with what the holes and pigeons should be. I originally was thinking that there should be 13 holes because there's 27 numbers and we needed to fit at least two into each hole (26) plus another number to get the third subset. But this doesn't make sense to me any more because that would mean we are acting as if there's only 27 possible subsets.

Now I believe there are 55 possible sums for each group of 27 six-digit numbers. I got this by doing 9+9+9+9+9+9 + 1 for 0 this would make sense if we now use the numbers as holes and the possible sums as pigeons because there are 27 numbers and 27*2=54 and then the +1 means that there is a third number in the subset. However this doesn't make sense to me as much now because then we are acting as if the last 5 digits can't be zero. So, does anyone have some ideas of what I should use for pigeons/holes?

答:
Note: I suspect you're not stating the problem correctly / as intended. If you've rephrased the problem, let me know.

(Fill in the gaps as needed. If you're stuck, write out your working and thought process to demonstrate where you're at.)

There are $ 2^{27}$ subsets of these 27 six-digit positive integers. This is our pigeons.

For our holes, they are the possible values of these subsets, which range from $0$ to $27 \times 999999$.

Hence, by PP, one of these holes must contain at least $X$ pigeons. Hence we are done.


问:
Find all positive integers x such that x(x + 180) is a square.
$x^2 + 180x = k^2$

$k^2 - x^2 = 180x$

$(k+x)(k-x) = 180x$

Now i wrote $180x$ as $1(180x) , (2)(90x) , \dots , (180)(x)$ and i compared these values with $(k+x)(k-x)$ and found $12,60$ to be the only 2 integers , are there more ?
Also if you have a better method please share.

Body:
WLOG $k\ge0\implies x+90+k>x+90-k$

$\implies(x+90+k)^2>(x+90-k)(x+90+k)=8100$

$\implies(x+90+k)>\sqrt{8100}=90\ \ \ \ (1)$

As $x+90+k+x+90-k=2(x+90)$ is even, $x+90+k,x+90-k$ have the same parity

As $(x+90-k)(x+90+k)=8100$  is even

$$\dfrac{x+90-k}2\cdot\dfrac{x+90+k}2=2025$$

By $(1),\dfrac{x+90+k}2>45$ and must divide $2025$  so, must belong to $\in[75, 81, 135, 225, 405, 675, 2025]$


问:
Order of Element in Factor Group
Give the order of the element in the factor group:  $$26 + \langle 12 \rangle \text{ in }\mathbb Z_{60} /\langle 12 \rangle.\;    (26 + \langle 12 \rangle) = (2 + \langle 12 \rangle) = \{2, 14, 26, 38, 52\} $$ yeah? This has order $5$ so I think that $(26 + \langle 12 \rangle)$ has order 5 in $\mathbb Z_{60}/ \langle 12\rangle$
but that is wrong apparently. The answer is that it has order $6$. Why?</p>

答:
<12>= {12, 24, 36, 48, 0}
Keep generating 26 until we find an element of <12> in <26>.
<26>={26, 52, 18, 44, 10, 36}
36 is in <12>, so we count the number of elements in the set which is 6.

So, the order is 6.

Title: Order of Element in Factor Group


问:
Transitivity of union of two transitive relations
I have a question concerning proving properties of Relations. The question is this: How would I go about proving that, if R and S (R and S both being different Relations) are transitive, then R union S is transitive?

The answer is actually FALSE, and then a counter example is given as a solution in the book.

I understand how the counterexample works as explained in the book, but what I don't understand is, how exactly they arrive to the conclusion that the statement is actually false.

Basically I can see myself giving a proof that if that for all values (x,y,z) in R and S, if (x,y) is in R and (y,z) is in R, (x, z) is in R since R is transitive. And if (x,y) is in S and (y,z) is in R, (x,z) is in S since S is transitive. Since (x,z) is in both R and S, the intersection is true. But why wouldn't the union of R and S be true as well?

Is it because that the proof cannot be ended with "since (x,z) is in both R and S, (x,z) can be in R or S"?

Basically, that a proof can't be ended with an OR statement at the end?

Hope that all makes sense.

答:
"How exactly they arrive to the conclusion that the statement is actually false" is by giving a counterexample. A counterexample is all it takes to prove that the statement is false. I think the question you meant to ask is, what goes wrong if you try to prove that the union of two transitive relations is transitive? Let's see.

Suppose $R$ and $S$ are transitive relations, and let's try to prove that $R\cup S$ is transitive. For that we have to show that, if $(x,y)\in R\cup S$ and $(y,z)\in R\cup S$, then $(x,z)\in R\cup S$.

Since $(x,y)\in R\cup S$, we know that either $(x,y)\in R$ or $(x,y)\in S$.

Since $(y,z)\in R\cup S$, we know that either $(y,z)\in R$ or $(y,z)\in S$.

We have four cases to consider.

Case 1. $(x,y)\in R$ and $(y,z)\in R$. Since $R$ is transitive, it follows that $(x,z)\in R$, and so $(x,z)\in R\cup S$.

Case 2. $(x,y)\in S$ and $(y,z)\in S$. Since $S$ is transitive, it follows that $(x,z)\in S$, and so $(x,z)\in R\cup S$.

Case 3. $(x,y)\in R$ and $(y,z)\in S$. Oops! Now what?

Case 4. $(x,y)\in S$ and $(y,z)\in R$. Oops! Now what?


Title: Terms in GP series finding trigonometry summation

Body:
Let ${a_r}\left( {{a_r} > 0\forall r \in N} \right)$ be the $r^{th}$ term of a GP such that $\sqrt {{a_1}.{a_3}.{a_5}.{a_7}}  = {\left( {18{a_1} + {a_1}{r^2}} \right)^2}$ and $\sum\limits_{r = 1}^\infty  {\frac{1}{{{a_r}}} = \frac{3}{4}} $. The value of $\sum\limits_{r = 1}^{10} {\tan \left( {{{\sin }^{ - 1}}\left( {\frac{{{a_{r + 1}}}}{{\sqrt {{a_r}^2 + {a_{r + 1}}^2} }}} \right)} \right)} $=_____

My approach is as follow

Let $r\in N$

${a_1} = {a_1};{a_3} = {a_1}{r^2};{a_5} = {a_1}{r^4};{a_7} = {a_1}{r^6}$

$\sqrt {{a_1}^4.{r^{12}}}  = {\left( {18{a_1} + {a_1}{r^2}} \right)^2}$

${a_1}^2{r^6} = {a_1}^2{\left( {18 + {r^2}} \right)^2} \Rightarrow {r^6} = {\left( {18 + {r^2}} \right)^2}$

${r^6} = 324 + 36{r^2} + {r^4} \Rightarrow {r^6} - {r^4} - 36{r^2} - 324 = 0$

How do I proceed from here

Body:
Hint:$$r^6={(18+r^2)}^2$$ $$r^3=18+r^2,r^3=-18-r^2$$
$$r^3=18+r^2\Rightarrow r^2(r-1)=(3)3(2)\Rightarrow \boxed{r=3}$$ Also $r^3=-18-r^2\Rightarrow r^2(r+1)=(-3)(-3)(-3+1)\Rightarrow r=-3$  hence rejected

Using $r=3$ can you do the rest?


问:
The contraction of $B_p\cdot q$ is $p\cdot A_p$
I am trying to prove the following:

Let $A\subset B$ with $B$ integral over $A$. Let $q\subset q' \subset B$ be prime ideals. If $q^c=q'^c\subset A$ then $q=q'$.

Here is my attempt with the points I am confused about inside brackets:
Let $p=q^c$ and take the localisation at $p$. Then $p\cdot A_p$ is a maximal ideal in $A_p$.
Then it should be that $p\cdot A_p$ is the contraction of $q \cdot B_p$ (I'm not sure how to see this).
Since $p\cdot A_p$ is maximal it gives that $q \cdot B_p$ will be maximal.
This gives that $q'\cdot B_p$ is either $q \cdot B_p$  or all of $B_p$. (I can't see why this cannot be all of $B_p$).
Now if we know that $q\cdot B_p=q\cdot B_q'$ (how can we lift this to get that $q=q'$).

答:
I change your notation slightly for convenience; in particular, I let $P$, $Q$, and $Q'$ denote what you refer to as $p$, $q$, and $q'$. Also, let $S=A\setminus P$.

For your first question, we wish to show that $P(S^{-1}A)=Q(S^{-1}B)\cap(S^{-1}A)$. The inclusion $\subseteq$ is clear, and for the other direction consider an arbitrary element $\tilde{q}$ in the right hand side. There must be some $q\in Q$, $a\in A$, $s_1,s_2\in S$ such that $\tilde{q}=q\big/s_1=a\big/s_2$. Thus, there is $t\in S$ such that $(as_1-qs_2)t=0$, so we have $as_1t=qs_2t\in Q$. Clearly also $as_1t\in A$, so in fact $as_1t\in Q\cap A=P$. Since $P$ is prime and $s_1,t\notin P$, this means that $a\in P$, whence $\tilde{q}=a\big/s_2\in P(S^{-1}A)$.

For your second question, suppose that $Q(S^{-1}B)=S^{-1}B$. Then there are $q\in Q$ and $s\in S$ such that $q\big/s=1\big/1$, and hence there is $t\in S$ such that $(s-q)t=0$. But this means $st=qt\in Q$, and again clearly $st\in A$, so – as above – this means $st\in Q\cap A=P$, a contradiction.

For your final question, let $q'\in Q'$ be arbitrary. Now, we have $$q'\big/1\in Q'(S^{-1}B)=Q(S^{-1}B),$$ and so there are $q\in Q$, $s\in S$ such that $q'\big/1=q\big/s$. This means there is some $t\in S$ such that $(q's-q)t=0$, whence $q'st=qt\in Q$. Since $st\in A\setminus P$, we have $st\notin Q$, and so since $Q$ is prime this means $q'\in Q$, as desired.


问:
Does this proof of the well-ordering principle subtly use induction?
My impression is that the well-ordering principle and induction are equivalent, so you must have to use induction to prove WOP, right? Does this proof use induction somewhere? Or just subtly fail?

Let $S$ be an arbitrary subset of the natural numbers, and take an arbitrary $k \in S$. Consider the subset $T = \{s \in S \mid s \leq k\}$. Since $T \subset \{1,2,\dotsc,k\}$ it's a finite set, and so it must have some least element $m$. We'll show $m$ is the least element of $S$. Take an arbitrary $s \in S$. If $s\leq k$ then $s \in T$ and so $m\leq s$. Otherwise if $s > k$ then $m \leq k < s$. In either case $m$ is less than your arbitrary element of $S$, so $m$ is the least element of $S$.

答:
How do you know a finite set has a least element?  That requires proof, and the usual proof would use induction (or something equivalent) somewhere, since "finite" means "in bijection with a natural number".  Moreover, induction would similarly be needed to prove that any subset of a finite set is finite.

(Alternatively, you could bypass talking about finite sets and just directly prove that for any $k$, every nonempty subset of $\{1,2,\dots,k\}$ has a least element.  But again, you would prove this by induction on $k$!)


问:
Prove that $|\sin 1| + |\sin 2| + |\sin 3| +\cdots+ |\sin 3n| > 8n/5$
So, the question is as follows:

Prove that $\left|\sin 1\right| + \left|\sin 2\right| + \left|\sin 3\right| +\cdots+ \left|\sin 3n\right| > 8n/5.$

I have tried quite a few approaches, including using the Taylor expansion or the A.M-G.M inequality, but to no avail. It would be of great help if this problem could be proven.

Edit: Angles are in radians, $n$ is a natural number.

答:
Lemma: The function

$$f(x)=|\sin(x)|+|\sin(x+1)|+\sin(x+2)|>\frac{8}{5}$$

for all $x\in\mathbb{R}$.

Proof: It suffices to show that the above equation holds for $x\in [0,2\pi]$. The function is piecewise differentiable except for

$$x\in \{0,\pi,\pi-1,2\pi-1,\pi-2,2\pi-2,2\pi\}$$

Then $f(x)$ can be rewritten

$$f(x)=\begin{cases}
      f_1(x)=\sin(x)+\sin(x+1)+\sin(x+2) & 0\leq x\leq \pi-2 \\
      f_2(x)=\sin(x)+\sin(x+1)-\sin(x+2) & \pi-2\leq x\leq \pi-1 \\
\vdots \\
      f_6(x)=-\sin(x)+\sin(x+1)+\sin(x+2) & 2\pi-1\leq x\leq 2\pi
   \end{cases}$$

We can then take each of these intervals and prove $f_i(x)>\frac{8}{5}$. For $i=1$, we have

$$f_1(x)=\sin(x)+\sin(x+1)+\sin(x+2)$$

$$=-\sin ^2(1) \sin (x)+\sin (x)+\cos ^2(1) \sin (x)+2 \sin (1) \cos (1)
   \cos (x)+\sin (1) \cos (x)+\cos (1) \sin (x)$$

Note that

$$f_1(0)=\sin (1)+2 \sin (1) \cos (1)>\left(1-\frac{1}{3!}\right)+2\left(1-\frac{1}{3!}\right)\left(1-\frac{1}{2!}\right)=\frac{5}{3}>\frac{8}{5}$$

$$f_1(\pi-2)=\sin (1)+\sin (2)>\left(1-\frac{1}{3!}\right)+\left(2-\frac{2^3}{3!}+\frac{2^5}{5!}-\frac{2^7}{7!}\right)=\frac{1097}{630}>\frac{8}{5}$$

(we have used the Taylor series expansions to get bounds for $\sin(1),\sin(2)$, and $\cos(1)$). Thus, at the endpoints of $[0,\pi-2]$ we know $f_1(x)>\frac{8}{5}$. Now, taking the derivative we get

$$f_1^{'}(x)=\cos (x)+\cos (x+1)+\cos (x+2)=(1+2 \cos (1)) \cos (x+1)$$

This is easily solved and we see that the only zero on the interval $[0,\pi-2]$ is $x=\frac{\pi }{2}-1$. The final step is to take the derivative one more time:

$$f_1^{''}(x)=-(1+2 \cos (1)) \sin (x+1)$$

Since

$$\cos(1)>1-\frac{1}{2!}=\frac{1}{2}>0$$

we know

$$f_1^{''}(x)=-(1+2 \cos (1)) \sin (x+1)<0$$

for $x\in [0,\pi-2]$. Putting it all together, we have

$$f_1(0)>\frac{8}{5}$$

$$f_1(\pi-2)>\frac{8}{5}$$

$$f_1^{'}(x)\text{ has a single zero on the interval}$$

$$f_1^{''}(x)<0\text{ on the interval}$$

These conditions imply that $f_1(x)>\frac{8}{5}$ for all $x\in[0,\pi-2]$. The remaining cases can be proven in much the same manner as the $i=1$ case. With this, the lemma is proved.

Theorem: The finite sum

$$\sum_{i=0}^{n-1} \bigg(|\sin(3i+1)|+|\sin(3i+2)|+|\sin(3i+3)|\bigg)>\frac{8}{5}n$$

Proof: By the lemma (with $x=3i-1$), we know that each portion of the sum is greater than $\frac{8}{5}$. Then

$$\sum_{i=0}^{n-1} \bigg(|\sin(3i+1)|+|\sin(3i+2)|+|\sin(3i+3)|\bigg)>\sum_{i=0}^{n-1}\frac{8}{5}=\frac{8}{5}n$$

and the theorem is proved.

EDIT: I included this after doing some numerical examples. It seems that

$$\lim_{n\to\infty} \frac{1}{n}\sum_{i=0}^{n-1} \bigg(|\sin(3i+1)|+|\sin(3i+2)|+|\sin(3i+3)|\bigg)=1.9098...$$

One point, this kinda looks like a Riemann sum (at least the fraction out in front of a finite sum). Second point, if the limit really does exist then the conjecture is true for all but a finite number of $n$ for all $x<1.9098...$. That is, if $x<1.9098...$ then for all but a finite number of $n$

$$\sum_{i=0}^{n-1} \bigg(|\sin(3i+1)|+|\sin(3i+2)|+|\sin(3i+3)|\bigg)>xn$$

It just so happens that $\frac{8}{5}$ is not a tight bound. In fact, a tighter bound which should work for all $n$ is $\frac{42}{25}$. That is

$$\sum_{i=0}^{n-1} \bigg(|\sin(3i+1)|+|\sin(3i+2)|+|\sin(3i+3)|\bigg)>\frac{42}{25}n$$

is true for all $n$. Proving this would just require a lot more terms of the Taylor series expansions of $\sin(1),\cos(1),$ and $\sin(2)$ (or some other expansion).

EDIT 2: Last edit, I realized that the limit up there (in EDIT 1) is similar to a Riemann sum. Specifically

$$\frac{1}{2\pi}\int_0^{2\pi} \bigg(|\sin(x)|+|\sin(x+1)|+|\sin(x+2)|\bigg)dx=\frac{12}{2\pi}=\frac{6}{\pi}=1.90986...$$

which is what the limit seemed to converge to. It would take some finess (you would probably have to use the fact that the natural numbers are equidistributed modulo $2\pi$), but I am now confident that the limit above really does exist and that it is equal to $\frac{6}{\pi}$.


问:
generalized eigenvector for 3x3 matrix with 1 eigenvalue, 2 eigenvectors

I am trying to find a generalized eigenvector in this problem. (I understand the general theory goes much deeper, but we are only responsible for a limited number of cases.)

I have found eigenvectors $\vec {u_1}$ and $\vec {u_2}.$

When I try $u_1$ and $u_2$ as $u_3$ into this equation:
$$ (A - I)u_4 = u_3$$
I get systems which are inconsistent.

How can I find the $u_3$? I've been told it has something to do with $(A - I)^3 = 0$, but that's about it.

答:
We are given the matrix:

$$\begin{bmatrix}2 & 1 & 1\\1 & 2 & 1\\-2 & -2 & -1\\\end{bmatrix}$$

We want to find the characteristic polynomial and eigenvalues by solving

$$|A -\lambda I| = 0 \rightarrow -\lambda^3+3 \lambda^2-3 \lambda+1 = -(\lambda-1)^3 = 0$$

This yields a single eigenvalue, $\lambda = 1$, with an algebraic multiplicity of $3$.

If we try and find eigenvectors, we setup and solve:

$$[A - \lambda I]v_i = 0$$

In this case, after row-reduced-echelon-form, we have:

$$\begin{bmatrix}1 & 1 & 1\\0 & 0 & 0\\0 & 0 & 0\\\end{bmatrix}v_i = 0$$

This leads to the two eigenvectors as he shows, but the problem is that we cannot use that to find the third as we get degenerate results, like you showed.

Instead, let's use the top-down chaining method to find three linearly independent generalized eigenvectors.

Since the RREF of

$$[A - 1 I] = \begin{bmatrix}1 & 1 & 1\\0 & 0 & 0\\0 & 0 & 0\\\end{bmatrix}$$

We have $E_3 = kernel(A - 1I)$ with dimension $= 2$, so there will be two chains.

Next, since

$$[A - 1 I]^2 = \begin{bmatrix}0 & 0 & 0\\0 & 0 & 0\\0 & 0 & 0\\\end{bmatrix}$$

the space Kernel $(A-1I)^2$ has dimension $=3$, which matches the algebraic multiplicity of $\lambda=1$.

Thus, one of the chains will have length $2$, so the other must have length $1$.

We now form a chain of $2$ generalized eigenvectors by choosing $v_2$ in kernel $(A-1I)^2$ such that $v_2$ is not in the kernel $(A-1I)$.

Since every vector is in kernel $(A-1I)^2$, and the third column of $(A-1I)$ is non-zero, we may choose:

$$v_2 = (1, 0, 0) \implies v_1 = (A-1I)v_2 = (1,1,-2)$$

To form a basis for $\mathbb R^3$, we need one additional chain of one generalized eigenvector. This vector must be an eigenvector that is independent from $v_1$. Since

$$E_3 = ~\text{span}~ \left(\begin{bmatrix}0\\1\\-1\\\end{bmatrix}, \begin{bmatrix}-1\\0\\1\\\end{bmatrix}\right).$$

and neither of these spanning vectors is itself a scalar multiple of $v1$, we may choose either one of them. So let

$$w_1 = (0, 1, -1).$$

Now we have two chains:

$$v_2 \rightarrow v_1 \rightarrow 0$$

$$w_1 \rightarrow 0$$

So, to write the solution, we have:

$\displaystyle 1^{st}$ Chain

$$x_1(t) = e^t \begin{bmatrix}1\\1\\-2\\\end{bmatrix}$$

$$x_2(t) = e^t\left(t \begin{bmatrix}1\\1\\-2\\\end{bmatrix} +  \begin{bmatrix}1\\0\\0\\\end{bmatrix}\right)$$

$\displaystyle  2^{nd}$ Chain

$$x_3(t) = e^t \begin{bmatrix}0\\1\\-1\\\end{bmatrix}$$

Thus:

$$x(t) = x_1(t) + x_2(t) + x_3(t)$$

Note, you can use this linear combination of $x(t)$ and verify that indeed it is a solution to $x' = Ax$.


问:
Measuring a subset of a null set
Which of these following statements is true?

a. There exists a non Lebesgue-measeruable set in $\mathbb{R}$ which gets measurable  with a (not necessarly disjoint) union with a null set.

b. Every subset of a null set is measurable.

c. Every subset of a measurable set is measurable.

I'd say that b is true and that c is wrong, I'm not quite sure about a, can someone help me?

答:
Your answers for b) and c) are both wrong.  b) is true and c) is wrong.

a): $A=(A\cup N \setminus N) \cup (A \cap N)$. If $A \cup N$ is measurable then  $A\cup N \setminus N$ and $A \cap N$ are both measurable, So $A$ cannot be non-measurable.

Title: Measuring a subset of a null set


问:
Integrating $\int \left( \frac{\cos\theta}{1+\sin^2\theta} \right)^n \, d\theta$, for $n=2$ and $n=3$
How do you integrate the following functions:
$$\int \left( \frac{\cos\theta}{1+\sin^2\theta} \right)^2 \, d\theta$$ and $$\int \left( \frac{\cos\theta}{1+\sin^2\theta} \right)^3 \, d\theta
$$
respectively?

Note: Initially, I tried integrating the function without the power and obtained the result below.
$$
\int \frac{\cos\theta}{1+\sin^2\theta} \, d\theta = \arctan(\sin\theta)+ C
$$
However, from here it is difficult to proceed. Integration by substitution doesn't seem to work.

How should I go on from here? Any pointers would be greatly appreciated.

答: $$ \begin{align*} \int \frac{\cos^2 x}{\left(1 + \sin^2 x\right)^2} \ dx &= \int \frac{\sec^2 x}{\left(\sec^2 x + \tan^2 x\right)^2} \ dx \\ &= \int \frac{1}{\left(2\tan^2x + 1\right)^2}\ d(\tan x) \\ \left(\tan x \mapsto \frac{\tan u}{\sqrt2} \right) &= \frac{1}{\sqrt{2}}\int \frac{\sec^2 u}{\sec^4 u} \ du \\&= \frac{1}{2\sqrt2}\int\cos(2u) + 1 \ du \\ &= \frac{1}{2\sqrt2}\left(\frac{\sin(2u)}{2} + u \right) + C \\ &=\frac{1}{2\sqrt2}\left(\frac{\sin(2\arctan(\sqrt2\tan x))}{2} + \arctan\left(\sqrt{2}\tan x\right)\right)+C\\&= \frac{\tan x}{2\left(2\tan^2x + 1\right)} + \frac{\arctan(\sqrt2 \tan x)}{2\sqrt{2}} + C\end{align*}$$ </p>


答:
Here's a way to do it only using the fact that $\int_{0}^{\infty}\frac{x^{s-1}}{e^{ax}-1}dx=\frac{\Gamma(s)\zeta(s)}{a^s}$.

First break the integral in two intervals and perform a substitution $x\to 1/x$ and then $x\to e^x$

$$\int_{0}^{\infty}\frac{\ln^3(x)}{(x-1)^3}dx=\left(\int_{0}^{1}+\int_{1}^{\infty}\right)\frac{\ln^3(x)}{(x-1)^3}dx=\int_{1}^{\infty}\frac{(1+x)\ln^3(x)}{(x-1)^3}=2\int_0^{\infty}dt\frac{t^3 e^t}{(e^t-1)^3}+\int_0^{\infty}dt\frac{t^3 e^t}{(e^t-1)^2}\equiv I_1+I_2$$

We can compute the integrals $I_1,I_2$ as follows. We see that

$$\frac{\partial }{\partial a}\int_{0}^{\infty}\frac{t}{e^{at}-1}dt=-\int_{0}^{\infty}\frac{t^2}{e^{at}-1}dt-\int_{0}^{\infty}\frac{t^2}{(e^{at}-1)^2}dt$$
Taking another derivative
$$\frac{\partial^2 }{\partial a^2}\int_{0}^{\infty}\frac{t}{e^{at}-1}dt+\frac{\partial}{\partial a}\int_{0}^{\infty}\frac{t^2}{e^{at}-1}dt=2\int_{0}^{\infty}dt\frac{t^3 e^{at}}{(e^{at}-1)^3}=I(a)$$
Note that $I(a=1)=I_1$. We find that it is equal to

$$I(a)=\frac{\partial^2}{\partial a^2}\frac{\zeta(2)}{a^2}+2\frac{\partial }{\partial a}\frac{\zeta(3)}{a^3}=6\frac{\zeta(2)-\zeta(3)}{a^4}$$

Similarly we note that

$$\frac{\partial }{\partial a}\int_{0}^{\infty}\frac{t^2}{e^{at}-1}dt=-\int_{0}^{\infty}\frac{t^3e^{at}}{(e^{at}-1)^2}dt=-J(a)$$

where $J(1)=I_2$. We conclude that

$$J(a)=\frac{6\zeta(3)}{a^4}$$.

Adding it all together we obtain that

$$\int_{0}^{\infty}\frac{\ln^3(x)}{(x-1)^3}=6\zeta(2)=\pi^2$$

PS instead of differentiation under the integral sign one can also use smart integration by parts. I don't know if this is a simpler way to do it but it definitely requires way less complex analysis and only a few basic identities.

Title: Exhausting integral $ \int_{0}^{\infty}\frac{\ln^3(x)}{(x-1)^3} \, \mathrm{d}x $

Body:
I would like to evaluate following integral:

$$ \int\limits_{0}^{\infty}\frac{\ln^3(x)}{(x-1)^3} \, \mathrm{d}x $$

by complex means. I integrate

$$f(z)=\frac{\ln^4(z)}{(z-1)^3}$$

(where ln is analytic in $\mathbb{C} \setminus \mathbb{R_+}$ with $\ln(-1)=i\pi$) over following contour: $[0;R]$, $C_0(R)$ - circle of radius $R$, $[R, 1+\varepsilon]$, semicircle in lower half-plane $C_1(\varepsilon)$ and $[1-\varepsilon,0]$. After evaluation and taking $\lim\limits_{\varepsilon \to 0}$ I came to correct answer, but my question is:

Is it possible to avoid that exhausting evaluation?

Thanks in advance.


问:
I need to solve $\phi (x,y) = \frac{2V}{\pi} \int_{0}^{\infty} \frac{\sin(kx)\cosh(ky) dk}{k\cosh(ka)}$
I start with a integral in complex plane $$\oint_c \frac{e^{izx} e^{zy} dz}{z\cosh(za)}$$ where $c$ is a countour starting in $z = -R$ along the real axis and jumping the pole at origin and continuing to $z = R$ and closing above the real axis with a semicircle circulating all the poles. I got $$\int_{0}^{\infty} \frac{e^{ky}\sin(kx)}{k\cosh(ky)}dk = \frac{\pi}{2} + 2\pi\sum_{n=1}^{\infty} (-1)^{n+1} \exp\left\{\frac{-xn\pi}{2a}\right\} \frac{\cos{\frac{yn\pi}{2a}}}{n}$$ and now i have no idea.</p>

答:
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
$\ds{\bbox[5px,#ffd]{\phi\pars{x,y} = {2V \over \pi} \int_{0}^{\infty}{\sin\pars{kx}\cosh\pars{ky}  \over k\cosh\pars{ka}}}\,\dd k}$

With $\ds{\alpha \equiv x/\verts{a}}$ and $\ds{\beta \equiv y/\verts{a}}$:
\begin{align}
&{\pi \over 2V\alpha}\,\phi\pars{x,y} \equiv
\bbox[5px,#ffd]{\int_{0}^{\infty}{\sin(k\alpha) \over k\alpha}{\cosh\pars{k\beta} \over \cosh\pars{k}}\dd k}
\\[5mm] = &\
\int_{0}^{\infty}{\sin(k\alpha) \over k\alpha}{2\sinh\pars{k}\cosh\pars{k\beta} \over \sinh\pars{2k}}\dd k
\\[5mm] & =
{\cal I}\pars{\alpha,1 + \beta} +
{\cal I}\pars{\alpha,1 - \beta}\label{1}\tag{1}
\end{align}
where
\begin{align}
{\cal I}\pars{\alpha,z} & \equiv
\int_{0}^{\infty}{\sin(k\alpha) \over k\alpha}{\sinh\pars{kz} \over \sinh\pars{2k}}\dd k
\\[5mm] = &\
\int_{0}^{\infty}\pars{{1 \over 2}
\int_{-1}^{1}\expo{\ic k\alpha q}\,\dd q}
{\expo{-k\pars{2 - z}} - \expo{-k\pars{z + 2}} \over
1 - \expo{-4k}}\dd k
\\[5mm] = &\
{1 \over 2}\int_{-1}^{1}\int_{0}^{\infty}
{\expo{-k\pars{2 - z - \ic\alpha q}}\,\, -
\expo{-k\pars{z + 2 - \ic\alpha q}} \over
1 - \expo{-4k}}\dd k\,\dd q
\\[5mm] = &\
{1 \over 8}\int_{-1}^{1}\int_{0}^{\infty}
{\expo{-k\pars{1/2 - z/4 - \ic\alpha q\,/4}}\,\,\,\, -
\expo{-k\pars{z/4 + 1/2 - \ic\alpha q\,/4}}\quad \over
1 - \expo{-k}}\,\dd k\,\dd q
\\[5mm] = &\
{1 \over 8}\int_{-1}^{1}
\bracks{%
\Psi\pars{{z \over 4} + {1 \over 2} - {\alpha\ic \over 4}\,q} -
\Psi\pars{{1 \over 2} - {z \over 4} - {\alpha\ic \over 4}\,q}}
\dd q
\end{align}

where I used the A & S Table ${\bf\color{black}{6.3.22}}$ identity.

Then,
\begin{align}
{\cal I}\pars{\alpha,z} & \equiv
\int_{0}^{\infty}{\sin(k\alpha) \over k\alpha}{\sinh\pars{kz} \over \sinh\pars{2k}}\dd k
\\[5mm] = &\
\left.{1 \over 8}\pars{{4 \over \alpha}\,\ic}
\ln\pars{\Gamma\pars{z/4 + 1/2 - \ic\alpha q/4} \over
\Gamma\pars{1/2 - z/4 - \ic\alpha q/4}}
\right\vert_{\,q\ =\ -1}^{\,q\ =\ 1}
\\[5mm] = &\
{\ic \over 2\alpha}\,
\ln\pars{{\Gamma\pars{1/2 + z/4 - \ic\alpha/4} \over
\Gamma\pars{1/2 - z/4 - \ic\alpha/4}}\,
{\Gamma\pars{1/2 - z/4 + \ic\alpha/4} \over
\Gamma\pars{1/2 + z/4 + \ic\alpha/4}}}
\\[5mm] = &\
{\ic \over 2\alpha}\,
\ln\pars{\sin\pars{\pi/2 - \pi z/4 - \ic\pi\alpha/4} \over
\sin\pars{\pi/2 + \pi z/4 - \ic\pi\alpha/4}}
\\[5mm] = &\
{\ic \over 2\alpha}\,
\ln\pars{\cos\pars{\pi z/4 + \ic\pi\alpha/4} \over
\cos\pars{\pi z/4 - \ic\pi\alpha/4}}
\\[5mm] = &\
{\ic \over 2\alpha}\,
\ln\pars{\cos\pars{\pi z/4}\cosh\pars{\pi\alpha/4}
-\ic\sin\pars{\pi z/4}\sinh\pars{\pi\alpha/4} \over
\cos\pars{\pi z/4}\cosh\pars{\pi\alpha/4} +
\ic\sin\pars{\pi z/4}\sinh\pars{\pi\alpha/4}}
\\[5mm] = &\
-\,{1 \over \alpha}\,\Im
\ln\pars{\cos\pars{\pi z/4}\cosh\pars{\pi\alpha/4}
-\ic\sin\pars{\pi z/4}\sinh\pars{\pi\alpha/4}}
\\[5mm] = &\
{1 \over \alpha}\,
\arctan\pars{\tan\pars{\pi z \over 4}\tanh\pars{\pi\alpha \over 4}}
\end{align}
Then ( see (\ref{1}) ),
\begin{align}
{\pi \over 2V\alpha}\,\phi\pars{x,y} & \equiv
\bbox[5px,#ffd]{\int_{0}^{\infty}{\sin(k\alpha) \over k\alpha}{\cosh\pars{k\beta} \over \cosh\pars{k}}\dd k}
\\[5mm] & =
{1 \over \alpha}\left\{%
\arctan\pars{\tan\pars{\pi\bracks{1 + \beta} \over 4}
\tanh\pars{\pi\alpha \over 4}}\right.
\\[2mm] &
\left.\phantom{{1 \over \alpha}\left\{\,\,\,\,\right.}
+ \arctan\pars{\tan\pars{\pi\bracks{1 - \beta} \over 4}
\tanh\pars{\pi\alpha \over 4}}\right\}
\\[5mm] & =
{1 \over \alpha}
\arctan\pars{\sinh\pars{\pi\alpha \over 2} \sec\pars{\pi\beta \over 2}}
\\[5mm] & =
{1 \over \alpha}
\arctan\pars{\sinh\pars{\pi x \over 2\verts{a}} \sec\pars{\pi y \over 2\verts{a}}}
\end{align}
Finaly,
\begin{align}
\phi\pars{x,y} & =
\bbx{{2V \over \pi}
\arctan\pars{\sinh\pars{\pi x \over 2\verts{a}}
\sec\pars{\pi y \over 2a}}} \\ &
\end{align}


问:
Confusion about how eigenvectors span the space
For the following matrix
$$ T =
\begin{pmatrix}
6 & 3 & 4 \\
0 & 6 & 2 \\
0 & 0 & 7
\end{pmatrix}
,$$
the eigenvalues are $\lambda = 6, 6, 7$. When $\lambda = 6$, the textbook $\textit{Linear Algebra Done Right}$ says the eigenvectors span the space $((1, 0, 0), (0, 1, 0))$. But when I plug in $\lambda = 6$ and solve the $(T - \lambda I) = 0$, I get $y = z = 0$ and $x =$ an arbitrary number for the eigenvector. So, shouldn't when $\lambda = 6$, it spans only $(1, 0, 0)$?

答:
First of all, I believe you copied down the matrix incorrectly. The video you linked has the 4 and 3 swapped. You are correct that the 6-eigenspace of this matrix is just spanned by $(1, 0, 0)$. However, at 2:38 in the video they write that the generalized eigenspace $G(6, T) = span\{(1, 0, 0), (0, 1, 0)\}$. Recall that an eigenvector is a nontrivial element of the kernel of $T - \lambda I$. A rank $m$ generalized eigenvector is defined to be an element of the kernel of $(T - \lambda I)^m$ that is not in the kernel of $(T - \lambda I)^{m-1}$. Axler is correct in his computation of the generalized eigenspaces.

Indeed, we can see clearly that $(0, 1, 0)$ is not a eigenvector with eigenvalue 6, so it is not in $ker(T-6I)$. However, it is in $ker(T-6I)^2$. Indeed, we can compute $(T-6I)^2 = \begin{pmatrix}
0 & 0 & 10\\
0 & 0 & 2\\
0 & 0 & 1
\end{pmatrix}$.

Thus $(0, 1, 0) \in ker(T-6I)^2 - ker(T-6I)$, so it is a generalized eigenvector of rank 2 for the eigenvalue 6. The generalized eigenspace for the eigenvalue 6 is spanned by all generalized eigenvectors for 6, so it is indeed $span\{(1, 0, 0), (0, 1, 0)\}$.


问:
The paraboloid $z = x^2 + y^2$ has no conjugate points at $p = (0,0,0)$
Can someone check if I am correct? This is exercise 2 in Do Carmo (Surface and curves) in chapter 5.5.

Let $S = \{ z = x^2 + y^2 \}$ with geodesic $\gamma(0) = p = (0,0,0)$ then clearly $T_pS = \{ (x,y,0) \}$.

Now since the paraboloid has positive Gaussian curvature, we cannot invoke the theorem involving empty conjugate locus. So I thought we have to go to Proposition5 to show that the exponential map $\exp_p:T_pS \to S$ is regular. That is it has no critical points (and hence no conjugate points relative to $p$). But I can't think of how to do it without knowing what the map should be.

So for the surface $\sigma (x,y) = (x,y,x^2 + y^2)$, the velocity of the curve on the surface is $\gamma' = (x',y',0)$. Now it is also equivalent to show that the Jacobi field along this geodesic is in fact $0$.

By Corollory 4 (of proposition 4), Jacobi fields along $\gamma$ with $J(0) = J(\ell) = 0$ must be orthogonal to the velocity, that is $(J(s),\gamma'(s)) = 0$.

Hence by inspection, one could take $J(s) = (-y',x',-)$. Here is where I am not sure if I can do this. Since $\gamma$ is a geodesic, $\gamma'' = (x'',y'',z'') = (0,0,0)$. Thus $J'(s) = (-y'',x'',-) = (0,0,0)$. Now I want to say something like since $J = 0$ on the end points and it has vanishing derivative that $J = 0$. But there is a lot of shaky arguments I have written.

EDIT: It just occured to me that in fact the last component of $J(s)$ is $0$. But whether $J(s)=(-y',x',0)$ is debatable since we can insert a smooth function $h(s)$ such that $J(s) = -h(s)(-y',x',0)$.

答:
Let $S$ be the paraboloid $S = \{(x,y,z) : z = x^2 + y^2\}$, and let $\gamma : [0, a] \to S$ be a geodesic starting at $p = (0,0,0)$. Let $J$ be a Jacobi field along $\gamma$ with $J(0) = 0$. By the definition of a Jacobi field given in do Carmo's Curves and Surfaces on page 363, we can find a smooth variation $f : (-\varepsilon, \varepsilon) \times [0, a] \to S$ of $\gamma$ such that

$J$ is the variational field of $f$, i.e. $J(t) = \frac{\partial f}{\partial s}(0, t)$, and

for each $s \in (-\varepsilon, \varepsilon)$, the variational curve $f_s(t) = f(s, t)$ is a geodesic starting from $p$.

Any geodesic of the paraboloid $S$ starting at $p = (0,0,0)$ must be a meridian, which is proven in Example 5 on page 258 of do Carmo. Therefore
$$
f_s(t) = (t \cos \theta(s), t \sin \theta(s), t^2)
$$
(up to reparametrization in $t$) for some angle function $\theta(s)$. We show that $\theta : (-\varepsilon, \varepsilon) \to \mathbb{R}$, defined by the above equation, is differentiable at $0$. If $\phi(u,v) = (v \cos u, v \sin u, v^2)$ is the parametrization of $S$ described in the aforementioned example, then for sufficiently small $s$, we can write the function $\theta$ as
$$
\theta(s) = (\pi_2 \circ \phi^{-1})(f(s, a)).
$$
As a composition of differentiable functions (for sufficiently small $s$, so that $f(s,a)$ is in a neighbourhood of $\gamma(a)$ in which $\phi$ is a diffeomorphism), this must be differentiable at $s = 0$. (In fact, it is differentiable for all sufficiently small $s$.)

Now, since $J$ is the variational field, we see that if $J(a) = 0$, then
$$
(0,0,0) = J(a) = \frac{\partial f}{\partial s}(0, a) = (-a \theta'(0) \sin \theta(0), a \theta'(0) \cos \theta(0), 0).
$$
Since $a \neq 0$, and since $\sin \theta(0)$ and $\cos \theta(0)$ are not simultaneously zero, we have that $\theta'(0) = 0$. Then, for any $t$,
$$
J(t) = \frac{\partial f}{\partial s}(0, t) = (-t \theta'(0) \sin \theta(0), t \theta'(0) \cos \theta(0), 0) = (0, 0, 0).
$$
But then $J \equiv 0$. Therefore no non-trivial Jacobi field along $\gamma$ can vanish at more than one point, which proves that there are no conjugate points to $p$ along $\gamma$.


问:
How can I rate convergence speed at infinity?
For example, I want to know how $1-e^{-x}$ converges to zero as $x \rightarrow \infty$. Can I rate the convergence speed at infinite (or approximation error) as in big-O notation in infinitesimal, such like $O(\frac{1}{x})$ or something? The taylor series $e^{-\frac{1}{x}}$ is not differentiable at zero, in above case.

答:
If you are asking for rate of convergence of $1-e^{-1/x}$ here is the answer:

$e^{x}=\sum \frac {x^{n}} {n!}$ for every real number $x$. Hence $e^{-1/x} =1-\frac 1  x+\frac 1 {2x^{2}}...$ for all $x \neq 0$ and $1-e^{-1/x}=O(\frac 1 x)$.


问:
Find the value of $\frac{\sum_{k<\frac n2} (n-2k){n \choose k}}{2^n-1}$
$$
{{\sum_{k<\frac{n}{2}}(n-2k){n \choose k}} \over {2^{n}-1}}
$$

This problem should be solved using the Vandermonde Identity. But I am not able to solve it. I don't understand how to apply the identity to solve this problem. Are there any other ways to solve this problem ?

答:
I am ignoring $\frac{1}{2^n-1}$ for now. We can add that back in the final result.

$ \displaystyle \sum \limits_{k=0}^{\lfloor \frac{n}{2} \rfloor} (n-2k) {n\choose k} = n\sum \limits_{k=0}^{\lfloor \frac{n}{2} \rfloor} {n\choose k} - 2\sum \limits_{k=0}^{\lfloor \frac{n}{2} \rfloor} k{n\choose k}$

$= n\sum \limits_{k=0}^{\lfloor \frac{n}{2} \rfloor} {n\choose k} - 2n\sum \limits_{k=1}^{\lfloor \frac{n}{2} \rfloor} {n-1 \choose k-1}$

If $n$ is odd,

$= \frac{n}{2} \sum \limits_{k=0}^{n} {n\choose k} - n\sum \limits_{k=1}^{n} {n-1 \choose k-1} + n {n-1 \choose {\lfloor \frac{n}{2} \rfloor}}$

$= \frac{n}{2} \sum \limits_{k=0}^{n} {n\choose k} - n\sum \limits_{k=0}^{n-1} {n-1 \choose k} + (n - {\lfloor \frac{n}{2} \rfloor}) {n \choose {\lfloor \frac{n}{2} \rfloor}}$

$ = ({\lfloor \frac{n}{2} \rfloor} + 1) {n \choose {\lfloor \frac{n}{2} \rfloor} + 1}$

$ = m {n \choose m} \, \,$ where $m = {\lfloor \frac{n}{2} \rfloor} + 1$

If $n$ is even,

$= \frac{n}{2} \sum \limits_{k=0}^{n} {n\choose k} + \frac{n}{2} {n \choose {\lfloor \frac{n}{2} \rfloor}} - n\sum \limits_{k=0}^{n-1} {n-1 \choose k}$

$= \frac{n}{2} {n \choose {\lfloor \frac{n}{2} \rfloor}} = (\lfloor \frac{n}{2} \rfloor + 1) {n \choose {\lfloor \frac{n}{2} \rfloor}+1}$

$ = m {n \choose m} \, \,$ where $m = {\lfloor \frac{n}{2} \rfloor} + 1$

So your answer is $\displaystyle \frac {m {n \choose m}}{2^n - 1} \,$ where $m = {\lfloor \frac{n}{2} \rfloor} + 1$


问:
Prove $W \cap W^\perp =\{\vec{0}\}$

If $W$ is a subspace of $\mathbb{R}^n$, then $W^\perp = \overline{W} =
 \{v \cdot w = 0, \forall w \in W\}$

Prove $W \cap W^\perp = \{\vec{0}\}$.

How do I fully prove this intersection is $\vec{0}$? I showed $\vec{0}$ is in $W$, $\overline{W}$, since they are both subspaces of $\mathbb{R}^n$, thus $\vec{0}$ is a subset of $W \cap \overline{W}$  which is also a subset of $\vec{0}$.

Then I must show $u = \vec{0}$ to complete.

I have let $u$ be an element of  $W \cap \overline{W}$  which implies $u$ is an element of the zero vector (because they are subsets of each other). But does this show $u$ must be equal to $\vec{0}$? If not, how do I show that part?

答:
Prove that $W ∩ W^{⊥} = {0}$

If we say u represents a vector that is present in both $W$ and $W^{⊥}$,

then $u \cdot u = 0$, or in other words $u^2 = 0$.

If we take the square root of both sides, we see that $u = 0$.

Title: Prove $W \cap W^\perp =\{\vec{0}\}$


问:
If $f :\mathbb{R}\to\mathbb{R}$ is measurable, then $E = \{x: f(x) \geq 3\}$ is measurable

Prove: Suppose $f : \mathbb{R}\to\mathbb{R}$ where $f$ is measurable and $E = \{x: f(x) \geq 3\}$.   Show $E$ is measurable.

I saw this statement while reading in a paper and thought this might be a good proof to know. It was about Lebesgue measures which I am sort of familiar with. I did not learn this sort of proof though.

Can someone please show me how this proof would look like?

I came up with this:

$$E=f^{-1}([3,\infty))$$
Is this correct? Is it possible to break this down further? I am not sure since I did not learn anything about this. I am just trying this for fun. I would still like to see a proof of it.

答: Note that $$E=\{x\mid f(x)\ge 3\}=f^{-1}([3,\infty))$$ where $f^{-1}$ denotes the pre-image. What is the definition of a measurable function?</p>


问:
$(X,Y) \sim U[0,1]^2, (X - Y), (X+Y) \sim $ what distribution?
$(X,Y) \sim U [0,1]^2$. Therefore $X,Y \sim U[0,1]$ each and $X$ and $Y$ are independent. What are the distributions of $X-Y$ and $X+Y$?

My approach: It is my intuitive understanding that $X-Y$ should follow a symmetric triangular distribution over $[-1,1]$, i.e. the triangular distribution with $a=-1,b=1,c=0$. I have tested this using discrete uniform distributions, e.g. the discrete uniform distribution described by: $\mathbb{P}(x,y)= \frac{1}{16}$ for $x,y \in \{\frac{1}{8},\frac{3}{8},\frac{5}{8},\frac{7}{8}\}$, $\mathbb{P}(x,y) = 0$, otherwise. We get $X-Y$ follows the distribution given by:

\begin{align*}
\mathbb{P}(x) &=\frac{4}{16} \;\text{if}\; x=0,\\
&=\frac{3}{16} \;\text{if}\; |x|=\frac{1}{4},\\
&=\frac{2}{16} \;\text{if}\; |x|=\frac{2}{4},\\
&=\frac{1}{16} \;\text{if}\; |x|=\frac{3}{4},\\
&=0, \;\text{otherwise.}\\
\end{align*}

Similarly, $X+Y$ should follow the symmetric triangular distribution with $a=0,b=2,c=1$.

However I'm not able to prove these.
Thank you in advance for your help.

Edit: The above discrete uniform example is just an example I played with to intuitively understand what the solution should be for the given problem. This example is not meant as a solution to the problem. I thought that was clear but apparently it was not - as somebody pointed out in a (now deleted) comment.

答:
Let's consider first $U=X+Y$

Let's set

$$
\begin{cases}
u=x+y, \\
v=x,
\end{cases}\rightarrow \begin{cases}
x=v, \\
y=u-v,
\end{cases}$$

The jacobian is 1 thus

$$f_{UV}(u,v)=\mathbb{1}_{[0;1]}(v) \cdot \mathbb{1}_{[v;v+1]}(u)= \mathbb{1}_{[0;1)}(u)\mathbb{1}_{[0;u]}(v) +\mathbb{1}_{[1;2]}(u)\mathbb{1}_{[u-1;1]}(v) $$

That is defined overthe following parallelogram.

Integrate

$$f_U(u)=\int_{\mathcal{V}} f_{UV}(u,v)dv$$

And obtain your triangular density.

Same reasoning for $X-Y$ (here the parallelogram is shifted)


问:
Do I need to analyse sequence given by $ x_{1+n} = \frac{1}{2 + x_{n}}$ without an equation with $0$?
I have a problem with exercises with sequences given by recursion when I need to "prove convergence and find limit if it exists" and I am given a recursion of that kind:

$$ x_{1+n} = \frac{1}{2 + x_{n}}, x_1 \in (0 ; \infty)$$

It is fairly easy to find the limit - I just assume that the limit exists in $ \mathbb{R}$ and then use arithmetic properties of limits:
$$\lim_{n \to \infty} x_{n+1} = \lim_{n \to \infty} x_{n}$$
$$\lim_{n \to \infty} x_{n} = l, l \in \mathbb{R}>0$$

Taking my recursion:
$$l = \frac{1}{2 + l}$$
$$l^2 +2l - 1 = 0$$
$$l_1 = \sqrt{2} - 1 \in D$$
$$l_2 = -1 - \sqrt{2} \notin D$$

So my only possible limit in $ \mathbb{R}$ is $l = \sqrt{2} - 1$. That is if I can actualy prove that the limit exists - that is: the sequence is monotonous and bounded. And here is my problem - it is just impossible to analyse without computer the difference of:

$$x_{1+n} - x_{n} = \frac{1}{2 + x_{n}} - x_{n}$$

In search of limits I just multiply both sides of equation by $ \lim_{n \to \infty} x_{n} = l$ and it is impossible to do so here, so I get:
$$x_{1+n} - x_{n} = \frac{-x_{n}^2-2x_n+1}{2 + x_{n}}$$

Then I can't tell when it is bigger than $0$ to analyse monotonicity and I can't the see for which values o $n$ which values of $n+1$ i get (to get the boundary) because min value gets crazy.

So I just waneted to ask - am I missing something? Is it possible to make here $x_{1+n} - x_{n} = \frac{-x_{n}^2-2x_n+1}{2 + x_{n}}$ an equality with $0$ and analyse simpler function (red one on the picture)?

答:
Although $x_1$ can be any positive number, all the terms starting from $x_2$ are less than $\frac 12$, so cannot be far from your limit.  An approach that can be useful is to write one term as the limit plus an error term, so here let $x_i=\sqrt 2-1+\epsilon$  Then
$$x_{i+1}=\frac 1{2+x_i}=\frac 1{1+\sqrt 2 + \epsilon}\\
x_{i+1}=\frac{\sqrt 2-1}{1+(\sqrt 2-1)\epsilon}\\
x_{i+1}\approx (\sqrt 2-1)-(\sqrt 2-1)^2\epsilon$$
where I have used the first order approximation to $\frac 1{1+\epsilon}$.  We see from this that the error is decreased by a factor about $6$ every step, so the sequence will converge.  To be more formal, you can bound the error from above using the fact that $x_i \in (0,\frac 12)$.  You won't get this fast a decrease, but any factor less than $1$ is good enough.


问:
n married couples, are randomly seated at a round table. Find probability for k couples to sit together.
A total of 2n people, consisting of n married couples, are randomly seated at a round table. Find the probability for k of the couples (k<n) to be seated together.

Let us first find the probability for 1 of the couples to be seated together.
Once the 2 persons are seated somewhere, the remaining $2n-2$ can be seated in $(2n-2)!$ ways, plus that the 2 members of the couple can be also seated in $2!$ ways. On the other hand, 2n people in a circular arrangement can be seated in $(2n-1)!$ ways, therefore the probability is
$\frac {2!(2n-2)!}{(2n-1)!} = \frac{2}{(2n-1)}$.

Now let's consider a second couple sitting together: Once the 1st couple has seated, we have $2n-2$ free seats, the 1st couple (which we consider as one unit) and a 2nd couple (also as one unit), so $2n-4+1$ can be arranged in $(2n-3)!$ ways, and the 2 couples can also be arranged in $2!.2!$ ways. So the probability for 2 couples to sit together (somewhere in the table) is $\frac{2!.2!.(2n-3)!}{(2n-1)!}$.

For the $k_{th}$ couple, the probability is $\frac{2!.2!.2!...2!.k!.(2n-2k-1)!}{(2n-1)!}$.

Is it correct?
I know this is a very common problem but haven't found this variation (with k couples) anywhere.
Can you help me? Thank you!

答:
Firstly, consider the $k$ couples as occupying one "double" seat each in $2$ ways, so there are $2n-2k +k = 2n-k\;$ seats available, and number of ways of seating them with $k$ couples together $= 2^k\cdot(2n-k-1)! $

And w/o restrictions, there are $= (2n-1)!\;$ways

hence $Pr = \frac{2^k\,\cdot\,(2n-k-1)!}{(2n-1)!}$


问:
If $\varrho$ has characteristic function $\hat\varrho(x)=e^{-\frac{\langle Ax,\:x\rangle}2}$, then $\int\varrho({\rm d}y)\langle y,x\rangle y=Ax$
Let $E$ be a $\mathbb R$-Banach space, $A:E'\to E$ be nonnegative ($\langle A\varphi,\varphi\rangle\ge0$ for all $\varphi\in E'$) and symmetric ($\langle A\varphi_1,\varphi_2\rangle=\langle A\varphi_2,\varphi_1\rangle$ for all $\varphi_1,\varphi_2\in E'$) and $\varrho$ be a probability measure on $\mathcal B(E)$ with characteristic function $$\hat\varrho(\varphi)=e^{-\frac{\langle A\varphi,\:\varphi\rangle}2}\;\;\;\text{for all }\varphi\in E'\tag1.$$

How can we show that $$\int\varrho({\rm d}x)\langle x,\varphi\rangle x=A\varphi\tag2$$ for all $\varphi\in E'$?

答:
It is given that $a\phi+b\psi \sim N(0, \langle A(a\phi+b\psi), a\phi+b\psi) \rangle$ for all $\phi$ and $\psi$, for all $a,b \in \mathbb R$.

Hence $E(a\phi+b\psi )^{2}=\langle A(a\phi+b\psi), a\phi+b\psi) \rangle$. Exapand the square on LHS. Consider the special cases $a=0, b=1$ and $a=1, b=0$ and conclude that $E \phi \psi =\langle A\phi, \psi \rangle$. Since $\psi$ is arbitrary this gives $E\phi =A\phi$, as required.


问:
How to find the angle formed by a cevian in a right triangle given a condition?
The problem is as follows:

In a right triangle $\triangle ABC$, right at $B$, it is traced the
height $BH$, in the triangle $BHC$, then is traced the interior cevian
$HM$ in such a way that $MC = AB$. Find the measure of the angle
$\angle MHC$ if it holds that $HC = BH + 2AH$.

The alternatives given in my book are as follows:

$\begin{array}{ll}
1.&\frac{53^{\circ}}{2}\\
2.&\frac{37^{\circ}}{2}\\
3.&53^{\circ}\\
4.&37^{\circ}\\
4.&30^{\circ}\\
\end{array}$

In this particular problem. I'm stuck at the very beginning. What's exactly the right drawing or interpretation for this?. Can someone help me here?. Can this be solved using congruency?. Which among the identities of congruency of triangles would apply?.

Please include a drawing in the answer because I don't know which sort of appropiate drawing would be for this particular problem. Can it be solved using only euclidean geometry postulates?.

Typically I would put or place an effort in the solution of this problem, but in this particular subject I'm stuck at the very beginning because I don't know how to translate what it is written in words into the problem with accuracy so that the problem can be solved, thus I really need help.

If it helps, I found this problem in a puzzles book which goes by the name Reason and logic which is from 2000s and includes topics probably borrowed from Martin Gardner's 50's book and Moise Geometry's textbook from the 70's.

I really wish this can suffice to give information. So far, as I indicated in the problem my best guess is that the problem uses triangle congruency, but I don't know how to make the right sketch for making this to happen.

Upon on my progress I could spot that there are going to be formed two congruent triangles and if this condition is used then I believe the problem can be solved, hence what might be needed is to establish a relationship between the altitude from $M$ to $HC$ would allow the answer. Such approach is mentioned in the answer given below.

Therefore, since I'm not good in this topic, it would help me a lot if the answer would include a step by step solution so I can understand.

答:
The interpretation of the problem is as follows:

Please see the diagram from below:

This first diagram is a representation of what its intended to say in the problem.

The key part here is that $MC=AB$ this will be used later in the problem.

The following diagram is the solution and the way to solve it is by tracing a line as the altitude from $MP$. In order to take track of the congruency. Angles $\alpha$, $\beta$ and $\omega$ as auxiliary are used.

Therefore the angle can be obtained from the triangle $\triangle HPM$.

I noted that there's congruence of triangles: This is the part where its used the information given in the problem regarding $AB=MC$.

$\triangle PMC \cong \triangle AHB$ because of $A-S-A$.
For additional information and reference this link can explain the case.

This means $PC=BH$ and $AH=PM$

Here's where we use the second clue given in the problem where:

$HC=BH+2AH$

Since:

$HP = HC-PC$ and $PC=BH$

$HP = BH+2AH-BH = 2AH$

and since $PM=AH$ we're left with a right triangle whose sides are in a proportion of $1:2$.

This means that:

$\tan \omega = \frac{AH}{2AH}=\frac{1}{2}$

$\angle MHC = \frac{53^{\circ}}{2}$

Without resorting to that trigonometric function it can easily be inferred that the intended triangle corresponds to:

The ratio of:

$1-2-\sqrt{5}$

is none other than $\frac{53^{\circ}}{2}$.


问:
Proof of Inequality Of Expectation
I am looking at a proof for subgaussian random variables but can't make sense for one of the steps:

$$\mathbb{E}[\exp(aX^2)]\le 1 + \int_0^{\infty}2at\cdot\exp (at^2)\mathbb{P}(|X|>t) \, dt$$

Can someone help clarify how this inequality is derived.

Edit: $a>0$

答:
The inequality is a special case of the following general formula:

Let $(\Omega,\mathscr{F},\mathbb{P})$ a probability space and $X:\Omega\to \mathbb{R}$ a random variable with $X\geq 0$. Then, for every continuously differentiable function $\phi:\mathbb{R}\to [0,\infty)$ with $\phi'\geq 0$, we have the following:
$$\tag{1}\mathbb{E}[\phi(X)]=\phi(0)+\int_{0}^{\infty}\phi'(t)\mathbb{P}\bigl(X>t\bigr)\,dt$$

The proof $(1)$ is based on Tonelli's theorem and the fundemental theorem of calculus: Denote by $F_X(x)$ the c.d.f function of $X$, then ($X$ is nonnegative)
\begin{align}\tag{2}
\mathbb{E}[\phi(X)]&=\int_{0}^{\infty}\phi(x)dF_X(x)
\end{align}
Fix some $x$ and write $\phi(x)=\int_{0}^{x}\phi'(t)dt+\phi(0)$. Substituting in $(2)$ we obtain
\begin{align}
\mathbb{E}[\phi(X)]&=\int_{0}^{\infty}\phi(x)dF_X(x)\\
&=\int_{0}^{\infty}\biggl(\int_{0}^{x}\phi'(t)dt\biggr)dF_x(x)+\phi(0)\\
&=\int_{0}^{\infty}\biggl(\int_{0}^{\infty}\phi'(t)\cdot \mathbb{1}_{\{t<x\}}dt\biggr)dF_X(x)+\phi(0)\tag{3}
\end{align}
Now, the function $(t,x)\mapsto \phi'(t)\cdot \mathbb{1}_{\{t<x\}}$ is nonnegative. Using Tonelli's theorem in $(3)$ and interchange the order of integration, we obtain
\begin{align}
\mathbb{E}[\phi(X)]&=\int_0^\infty\biggl(\int_0^\infty\phi'(t)\cdot \mathbb{1}_{\{t<x\}}\,dF_X(x)\biggr)dt+\phi(0)\\&=\int_0^\infty\phi'(t)\cdot\mathbb{P}(X>t)\,dt+\phi(0)
\end{align}
and $(1)$ is proven. Now, using $(1)$ with $\phi(t)=e^{\alpha t^2}$ and as nonnegative random variable the absolute value $|X|$ of $X$ you get the inequality you want and in fact its an equality.

Title: Proof of Inequality Of Expectation


问:
$9$-card-long (or longer) straight flush in a $13$-card hand
The other day at an online bridge site I got dealt a $13$-card hand containing $3,4,5,6,7,8,9,10,J$ of spades, i.e. a $9$-card-long straight-flush (s-f).  This is so rare that I decided to count how many $13$-card hands contain a s-f of length $9$ or longer.

For the purpose of this question:

A s-f is a consecutive sequence in the same suit, and aces always count high (since I was playing bridge!) so $A,2,3,4,5$ of spades is not a $5$-card s-f for this purpose.

Let us just count s-f in spades.  If we want to count all suits we can simply multiply by $4$ since a hand cannot contain two s-f's of different suits each of length $\ge 9$.

I was trying to use inclusion-exclusion but got very confused.  Finally I decided to count explicitly as follows.  My questions:

(1) Can someone verify my answer below?

(2) Can someone show how this can be counted via inclusion-exclusion, or explain why that's not a good idea?

My attempt:

$13$-long s-f: No. of hands $=1$.

exactly $12$-long s-f: There are two such ($2-K$ or $3-A$), and each must be completed with an extra card (to make a $13$-card hand) but the extra card must not complete a $13$-long s-f.  No. of hands $= 2 \times {39 \choose 1}$.

exactly $11$-long s-f: Three such; the $2-Q$ hand must not include the $K$ (leaving $40$ valid cards to choose $2$ to fill out the hand), the $3-K$ hand must not include $2$ or $A$ (leaving $39$ valid cards), and the $4-A$ hand must not include the $3$.  No. of hands $= {40 \choose 2} + {39 \choose 2} + {40 \choose 2}$.

exactly $10$-long s-f: Similar to above, no. of hands $= {41 \choose 3} + {40 \choose 3} + {40 \choose 3} + {41 \choose 3}$.

exactly $9$-long s-f: Similar to above, no. of hands $= 2 \times {42 \choose 4} + 3 \times {41 \choose 4}$.

According to wolfram alpha total $= 571130$.

答:
Inspired by the PIE-based answer of @RobPratt I found the following much simpler (computationally speaking) solution!

As usual all s-f's considered here are in spades.

Consider the value $5 \times {52-9 \choose 13-9}$.  This is the number of pairs of sets $(X,Y)$, where $|X| = 9$ and the contents form a $9$-long s-f and $|Y| = 4$ and the contents are arbitrary, as long as $X, Y$ are disjoint.  The first factor in the product is $5$ because there are $5$ possible $9$-long s-f's.

Now, every hand with an exactly $9$-long s-f can be represented as $(X,Y)$ in exactly one way.  Meanwhile, every hand with an exactly $10$-long s-f can be represented as $(X,Y)$ in exactly two ways, and every hand with an exactly $11$-long s-f can be represented as $(X,Y)$ in exactly three ways, etc.  Therefore:

$$5 \times {52 - 9 \choose 13 - 9} = N_9 + 2 N_{10} + 3 N_{11} + 4 N_{12} + 5 N_{13}$$

where $N_j$ is the no. of hands with an exactly $j$-long s-f.

By a completely analogous argument but this time considering $(W,Z)$ where $|W| = 10$ and the contents form a $10$-long s-f, and $|Z| = 3, W \cap Z = \emptyset$, we have

$$ 4 \times {52 - 10 \choose 13 - 10} = N_{10} + 2 N_{11} + 3 N_{12} + 4 N_{13}$$

The difference between the two gives the desired count, i.e. no. of hands with $9$-long (or longer) s-f:

$$N_9 + N_{10} + N_{11} + N_{12} + N_{13} = 5 \times {52 - 9 \choose 13 - 9} - 4 \times {52 - 10 \choose 13 - 10} = 571130 \,\,\,\,\,\,\square$$


问:
Sigma Value Ratio
What is the possible minimum sigma value (given $1\sigma$, $2\sigma$, $3\sigma$, $4\sigma$, $5\sigma$, $6\sigma$) one can use when we want a less than 0.3% defect ratio?

My approach was to implement $68-95-99.7$ rule, and get $3\sigma$ as a correct answer. Would it be right method, or there are missing parts?

答:
My answer would be to choose $3\sigma$ from the Z-table, but unsure whether it gives correct values for this problem

@StubbornAtom Regarding previous comment, does that question have an answer to this problem? I suppose there was not an exact answer to that.

Title: Sigma Value Ratio


问:
A question about  binomial-coefficients
How to compute the following

$\sum_{t=0}^{k}{2m-1-t \choose t}{2m-1-(k-t) \choose k-t}$, where $1\leq k\leq 2m-1$?

Please point me to some references if this has already been studied. Thanks for your time!

答:
Hint: For two power series $\sum_{n=0}^{\infty}a_nx^n$ and $\sum_{n=0}^{\infty}b_nx^n$ the formal multiplication $\left(\sum_{n=0}^{\infty}a_nx^n\right) \cdot \left(\sum_{n=0}^{\infty}b_nx^n \right)=\sum_{n=0}^{\infty}c_nx^n$ gives us $c_n=\sum_{t=0}^{n}a_tb_{n-t}$.

Denote $a_k=b_k=\binom{2m-1-k}{k}$. Then the required sum is exactly $c_k$, the coefficient of $x^k$ in $(f(x))^2$, where $f(x)=\sum_{k=0}^{\infty}a_kx^k=\sum_{k=0}^{2m-1}\binom{2m-1-k}{k}x^k$

Title: A question about  binomial-coefficients


问:
Complexity class of comparison of power towers
Consider the following decision problem: given two lists of positive integers $a_1, a_2, \dots, a_n$ and $b_1, b_2, \dots, b_m$ the task is to decide if $a_1^{a_2^{\cdot^{\cdot^{\cdot^{a_n}}}}} < b_1^{b_2^{\cdot^{\cdot^{\cdot^{b_m}}}}}$.

Is this problem in the class $P$?

If yes, then what is the algorithm solving it in polynomial time?

Otherwise, what is the fastest algorithm that can solve it?

Update:

I mean polynomial type with respect to the size of the input, i.e. total number of digits in all $a_i, b_i$.

$p^{q^{r^s}}=p^{(q^{(r^s)})}$, not $((p^q)^r)^s$.

答:
Recently I asked a very similar question at Mathematica.SE.
I assume you know it, because you participated in the discussion.

Leonid Shifrin suggested an algorithm that solves this problem for the majority of cases, although there were cases when it gave an incorrect answer. But his approach seems correct and it looks like it is possible to fix those defects. Although it was not rigorously proved, his algorithm seems to work in polynomial time. It looks like it would be fair if he got the bounty for this question, but for some reason he didn't want to.

So, this question is not yet settled completely and I am going to look for the complete and correct solution, and will start a new bounty for this question once the current one expires. I do not expect to get a bounty for this answer, but should you choose to award it, I will add it up to the amount of the new bounty so that it passes to whomever eventually solves this question.


问:
Number of ways to legally insert n pairs of parentheses into a string of k elements
Perhaps I'm not very good a googling questions like this, but I can't find anything to help me with this.

Suppose I have a string of $k$ elements, and now I want to 'legally' insert $n$ pairs of parenthesis. By 'legally' I mean that the parentheses must contain an element of the string, but they can be `double paired' and nested. For example, $()ab(c)$ is illegal, while $(a)(b)c, ((ab))c, (a(bc)d)$ are legal.

I have found formulae for inserting one or two pairs of parentheses into a string, and similarly, the Catalan numbers enumerate the ways pairs of parentheses can be arranged, but they don't answer this question in particular.

I feel this is a common question that has probably already made it to StackExchange, so if anyone could at least point me in the right direction that would be great, thankyou!

(This is not homework, but I'm also happy for you to make me work for my solution).

答:
Let the number be $F(k,n)$.

Either a sequence starts with a letter; there are $F(k-1,n)$ of those, or it starts with a bracket which eventally closes, so

$$F(k,n)=F(k-1,n)+\sum_{p=1}^{k}\sum_{q=0}^{n-1}F(p,q)F(k-p,n-1-q)$$
The generating function $$G(x,y)=\sum_k\sum_nF(k,n)x^ky^n$$  obeys the equation
$$G(x,y)=1+xG(x,y)+yG(x,y)(G(x,y)-1)$$
Solve this quadratic, and find G's Taylor series.


问:
In an integral, why does logarithmic function of an exponential completely drop out?
How to show that

$$-\int_{-\infty}^{\infty}  e^{-(x-\mu)^2 / 2\sigma^2} \ln\left[  e^{-(x-\mu)^2 / 2\sigma^2} \right] \mathrm{d} x$$

equals this:

$$-\int_{-\infty}^{\infty}  e^{-(x-\mu)^2 / 2\sigma^2}  \mathrm{d} x$$

If the exponential function appears twice in the first formula as its non-logged self multiplied by its logged self, why does the first formula reduce to an integral of only the non-logged exponential, while the logarithm completely drops out?

答: $\frac 1  {\sqrt  {2\pi}}\int_{-\infty}^{\infty} t^{2}e^{-t^{2} /2 } dt=\frac 1  {\sqrt  {2\pi}}\int_{-\infty}^{\infty} e^{-t^{2} /2 } dt$ since RHS is $1$ and LHS is the variance of the standard normal distribution which is also $1$.  Just make the substitution $t=\frac {x-\mu} {\sigma}$ or $x=\mu +\sigma t$ and you will get your identity.


问:
$\bar{x}$ irreducible in $F[x,y]/(y^2-x^3+x)$
I want to show that $\bar{x}$ is irreducible in $F[x,y]/(y^2-x^3+x)$ with $F$ a field of characteristic not equal to 2. I was thinking of construct homomorphism to $F[t]$ sending $x$ and $y$ to polynomials of $t$, but no choice I found was optimal. Any help is much appreciated.

答:
The trick here is to use a norm function to reduce factorization questions about your ring (which is not a UFD) to one which is a UFD. (You may have already seen examples of this with, say, $\mathbb{Z}[\sqrt{-5}]$ and $\mathbb{Z}$.)

Let $R$ be a commutative ring, and let $d \in R$. Consider the ring $R[\sqrt{d}] := R[Y]/\langle Y^{2} - d\rangle$; we will use $\sqrt{d}$ to denote the class $\overline{Y}$. Note that $R[\sqrt{d}]$ is a free $R$-module of rank $2$ with basis $1, \sqrt{d}$. One may accordingly define a norm map $N \colon R[\sqrt{d}] \to R$ which sends $a+b\sqrt{d}$ to $a^{2}-b^{2}d$ for $a, b \in R$. It is easy, if tedious, to check that $N$ is multiplicative, i.e. $N(uv) = N(u)N(v)$ for any $u, v \in R[\sqrt{d}]$.

Let $R = F[X], d = X(X-1)(X+1)$, so that your ring of interest is $A := R[\sqrt{d}]$. (To avoid possible confusion, $\sqrt{d}$ is playing the role of the residue class $\overline{y}$ in your notation.) Suppose $X = uv$ for some $u, v \in A$. Then $N(X) = X^{2} = N(u)N(v)$. Up to scaling by units of $F$ (which you may check does not affect the ensuing analysis), we have $N(u)$ must be one of $1, X, X^{2}$. If $N(u) = 1$, then $u$ is a unit; this is an easy exercise. Likewise, if $N(u) = X^{2}$, then $N(v) = 1$, so $v$ is a unit. It thus remains to show that we cannot have $N(u) = N(v) = X$.

Suppose there exist $f(X), g(X) \in R$ such that for $u = f(X)+g(X)\sqrt{d}$,

$$N(u) = f(X)^{2}-g(X)^{2}(X^{3}-X) = X.$$

Since $X$ divides the RHS and divides $g(X)^{2}(X^{3}-X)$, it must also divide $f(X)^{2}$, hence $f(X)$. If $f(X) = Xh(X)$, we obtain

$$Xh(X)^{2}-g(X)^{2}(X^{2}-1) = 1$$

after cancelling all redundant factors of $X$. If this were the case, then the leading terms of $Xh(X)^{2}$ and $g(X)^{2}(X^{2}-1)$ must cancel. But we can see that if $n := \deg(h), m := \deg(g)$, then $Xh(X)^{2}$ has degree $2n+1$, and $g(X)^{2}(X^{2}-1)$ has degree $2m + 2$; hence, the leading terms cannot cancel. This is our desired contradiction.


问:
Stochastic difference equation
I am a newbie in studying time series. Could anyone help solve the following problem:

Consider the second-order stochastic difference equation: $y_t=1.5y_{t-1}-0.5y_{t-2}+\varepsilon_t$. Given initial conditions for $y_0$ and $y_1$, find $y_t$ in terms of $\{\varepsilon_t\}$ sequence.

Find the forecast function for $y_{T+s}$ given the value of $y_T$ and $y_{T-1}$.

Title: Stochastic difference equation

Body:
\begin{matrix}
y_t-y_{t-1}&=&0.5(y_{t-1}-y_{t-2})+\varepsilon_t\\
0.5(y_{t-1}-y_{t-2})&=&0.5^2(y_{t-2}-y_{t-3})+0.5\varepsilon_{t-1}\\
&\vdots&\\
0.5^{t-2}(y_2-y_1)&=&0.5^{t-1}(y_1-y_0)+0.5^{t-2}\varepsilon_2
\end{matrix}
Summing up the LHS and RHS gives $$y_t-y_{t-1}+\sum_{i=1}^{t-2}0.5^i(y_{t-i}-y_{t-i-1})=\sum_{i=1}^{t-1}0.5^i(y_{t-i}-y_{t-i-1})+\sum_{i=0}^{t-2}0.5^i\varepsilon_{t-i}$$
so $$y_t-y_{t-1}=0.5^{t-1}(y_1-y_0)+\sum_{i=0}^{t-2}0.5^i\varepsilon_{t-i}$$
Then
\begin{align*}
y_t=\sum_{i=1}^t(y_{t-i+1}-y_{t-i})+y_0=&\sum_{i=1}^t\left(0.5^{t-i}(y_1-y_0)+\sum_{j=0}^{t-2}0.5^j\varepsilon_{t-j}\right)+y_0\\
=&(y_1-y_0)\sum_{i=1}^t0.5^{t-i}+t\sum_{i=0}^{t-2}0.5^i\varepsilon_{t-i}+y_0\\
=&\frac{1-0.5^t}{0.5}(y_1-y_0)+t\sum_{i=0}^{t-2}0.5^i\varepsilon_{t-i}+y_0\\
=&(2^{1-t}-1)y_0+(2-2^{1-t})y_1+t\sum_{i=0}^{t-2}0.5^i\varepsilon_{t-i}
\end{align*}


问:
How to solve $\int_{0}^{x} \frac{y^{m}}{(1+y/p)^{2m+2}} \, \mathrm{d}y$?
$$\int_{0}^{x} \frac{y^{m}}{(1+y/p)^{2m+2}} \, \mathrm{d}y$$

where $m$ and $p$ are constant. I tried to solve it using LIATE rule but getting stuck. Any help will be highly appreciated.

Body:
Hint

$$\frac 1 {\left(1+\frac{y}{p}\right)^{2 (m+1)}}=\sum_{n=1} \binom{-2 (m+1)}{n}\left(\frac{y}{p}\right)^n$$


问:
Function satisfying the relation $f(x+y)=f(x)+f(y)-(e^{-x}-1)(e^{-y}-1)+1$
Let f be the differentiable function satisfying the relation  $f\left( {x + y} \right) = f\left( x \right) + f\left( y \right) - \left( {{e^{ - x}} - 1} \right)\left( {{e^{ - y}} - 1} \right) + 1$; $\forall x,y \in R$ and $\mathop {\lim }\limits_{h \to 0} \frac{{f'\left( {1 + h} \right) + f\left( h \right) - {e^{ - 1}}}}{h}$ exist. The value of $\int\limits_0^1 {f\left( x \right)dx}  = \_\_\_\_\_\_\_$.

My approach is as follow

$f\left( {x + y} \right) = f\left( x \right) + f\left( y \right) - \left( {{e^{ - x}} - 1} \right)\left( {{e^{ - y}} - 1} \right) + 1$

$\Rightarrow f\left( {x + y} \right) - f\left( x \right) = f\left( y \right) - \left( {{e^{ - x}} - 1} \right)\left( {{e^{ - y}} - 1} \right) + 1$

$ \Rightarrow \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {x + y} \right) - f\left( x \right)}}{h} = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( y \right) - \left( {{e^{ - x}} - 1} \right)\left( {{e^{ - y}} - 1} \right) + 1}}{h}$

$ \Rightarrow \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {x + y} \right) - f\left( x \right)}}{h} = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( y \right) - \left( {{e^{ - x}} - 1} \right)\left( {{e^{ - y}} - 1} \right) + 1}}{h}$

y=h

$ \Rightarrow f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {x + h} \right) - f\left( x \right)}}{h} = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( h \right) - \left( {{e^{ - x}} - 1} \right)\left( {{e^{ - h}} - 1} \right) + 1}}{h}$

How will I proceed from here

答:
Let $g(x)=f(x)+e^{-x}$. Then given equation reduces to  Cauchy's equation $g(x+y)=g(x)+g(y)$. If $f$ is assumed to be continuous (or at least measurable) then $g(x)=cx$ for some constant $c$ and $f(x)=-e^{-x}+cx$.

The given limit exists if and only if $c=1$ and this gives $\int_0^{1} f(x)dx=\frac 1 e-\frac 1  2$.


Title: Do I need to analyse sequence given by $ x_{1+n} = \frac{1}{2 + x_{n}}$ without an equation with $0$?

Body:
I have a problem with exercises with sequences given by recursion when I need to "prove convergence and find limit if it exists" and I am given a recursion of that kind:

$$ x_{1+n} = \frac{1}{2 + x_{n}}, x_1 \in (0 ; \infty)$$

It is fairly easy to find the limit - I just assume that the limit exists in $ \mathbb{R}$ and then use arithmetic properties of limits:
$$\lim_{n \to \infty} x_{n+1} = \lim_{n \to \infty} x_{n}$$
$$\lim_{n \to \infty} x_{n} = l, l \in \mathbb{R}>0$$

Taking my recursion:
$$l = \frac{1}{2 + l}$$
$$l^2 +2l - 1 = 0$$
$$l_1 = \sqrt{2} - 1 \in D$$
$$l_2 = -1 - \sqrt{2} \notin D$$

So my only possible limit in $ \mathbb{R}$ is $l = \sqrt{2} - 1$. That is if I can actualy prove that the limit exists - that is: the sequence is monotonous and bounded. And here is my problem - it is just impossible to analyse without computer the difference of:

$$x_{1+n} - x_{n} = \frac{1}{2 + x_{n}} - x_{n}$$

In search of limits I just multiply both sides of equation by $ \lim_{n \to \infty} x_{n} = l$ and it is impossible to do so here, so I get:
$$x_{1+n} - x_{n} = \frac{-x_{n}^2-2x_n+1}{2 + x_{n}}$$

Then I can't tell when it is bigger than $0$ to analyse monotonicity and I can't the see for which values o $n$ which values of $n+1$ i get (to get the boundary) because min value gets crazy.

So I just waneted to ask - am I missing something? Is it possible to make here $x_{1+n} - x_{n} = \frac{-x_{n}^2-2x_n+1}{2 + x_{n}}$ an equality with $0$ and analyse simpler function (red one on the picture)?

Body:
$$X_{n+1}=\frac{1}{2+X_n} \implies 2 X_{n+1}+X_{n+1}X_n=1$$
Let $X_n=\frac{Y_{n-1}}{Y_n}$, then
$$2 \frac{Y_{n}}{Y_{n+1}}+\frac{Y_n}{Y_{n+1}}\frac{Y_{n-1}}{Y_n}=1 \implies 2Y_n+Y_{n-1}=Y_{n+1}.$$
Let $Y_n=t \implies t^2-2t-1=0 \implies t=1\pm \sqrt{2}.$
So $$Y_n=p(1+\sqrt{2})^n+q (1-\sqrt{2})^{n} $$
$$\implies X_n=\frac{(1+\sqrt{2})^{n-1}+r(1-\sqrt{2})^{n-1}}{(1+\sqrt{2})^{n}+r(1-\sqrt{2})^{n}}, r=q/p.$$
$$\lim_{n \to \infty}X_{\infty}=\frac{1}{1+\sqrt{2}}=\sqrt{2}-1$$