问:
Are all Givens rotations linear transformations?
I've read that Givens rotations are linear transformations, but as we know linear transformations preserve the length of the vector transformed. However when I tried this with a 3 dimensional Givens rotation matrix G(1,2,45°) it seems that the length of the vector transformed changes, and when putting in it into a 3d grapher it the transformation resambles a plan projection.

Is that normal?

答:
Welcome here Saad!

First point: linear transformations do not necessarily preserve the length. Take for example the map $A(x,y,z) = (3x,3y,3z)$: the length of the result is three times the original length!

Second point: luckily, rotations do preserve the length, so there should be something wrong with your matrix $G(1, 2, 45°) $ . How is it defined? If it is meant to be the rotation around the z-axis of 45°, it should be

$$ \begin{pmatrix}
\sqrt{2}/2 & \sqrt{2}/2 & 0 \\
- \sqrt{2}/2 & \sqrt{2}/2 & 0 \\
0 & 0 & 1
\end{pmatrix}
$$


问:
Comparing $\sqrt{1001}+\sqrt{999}\ , \ 2\sqrt{1000}$
Without the use of a calculator, how can we tell which of these are larger (higher in numerical value)?

$$\sqrt{1001}+\sqrt{999}\ , \ 2\sqrt{1000}$$

Using the calculator I can see that the first one is 63.2455453 and the second one is 63.2455532, but can we tell without touching our calculators?

答:
And one more... I'm reflecting at the concavity and monotonicty of the curve of the sqrt-function and conclude, that whatever greater or smaller the relation is, it should be the same if I decrease the arguments down to zero. Since we do not yet know whether the relation is $\gt$ or $\lt$ I introduce the "indeterminate comparision symbol" $\mathcal C$ which can assume the greater or smaller-relation and we rewrite the original question
$$ \sqrt{1001}  +  \sqrt{ 999} \qquad  \mathcal C \qquad  2\sqrt{1000} \tag 1 $$
as
$$ \sqrt{1001}  -  \sqrt{1000} \qquad \mathcal C \qquad  \sqrt{1000} - \sqrt{999} \tag 2$$
Then we use the monotonicity and concavity of the sqrt-function (from x=1001 down to x=0) and write the  consecutive differences

$ \displaystyle \qquad \begin{array} {lll}
&\sqrt{1001} & - & \sqrt{1000} & \mathcal C &  \sqrt{1000} &-& \sqrt{999} \\
&\sqrt{1000} & - & \sqrt{ 999} &  \mathcal C&  \sqrt{ 999} &-& \sqrt{998} \\
&\sqrt{ 999} & - & \sqrt{ 998} &  \mathcal C&  \sqrt{ 998} &-& \sqrt{997} \\
&\cdots \\
&\sqrt{ 2}   & - & \sqrt{ 1} &  \mathcal C &  \sqrt{   1} &-& \sqrt{  0}  \\
\end{array} $

and compare  the whole sums which are nicely telescoping

$ \displaystyle \qquad \begin{array} {lll}
\sum =&\sqrt{1001} &-&1 &  \mathcal C & \sqrt{1000}  &&& \end{array}  $    $ \tag 3$

which can then be rewritten by

$ \displaystyle \qquad \begin{array} {rrrrrrr}
&\sqrt{1001} &-&\sqrt{1000} &  \mathcal C & 1 && & // * (\sqrt{1001} + \sqrt{1000})\\
& 1001 & - & 1000  &  \mathcal C & \sqrt{1001} &+& \sqrt{1000} \\
&   &  & 1  &  \mathcal C & \sqrt{1001} &+& \sqrt{1000} &\sim 2\sqrt{1000} \\
\end{array} $

Here, in the last comparision, the geater/smaller-relation  is obvious and thus our operator $ \mathcal C = "\lt" $
and we have the result

$ \displaystyle \qquad \begin{array} {lll}
&\sqrt{1001} & + & \sqrt{ 999} & \lt &  2 \cdot \sqrt{1000}
\end{array} $ $ \tag 4$

ContentLicense: CC BY-SA 3.0


问:
Gradient of Gaussian entropy w.r.t. $W$
I have been struggling with deriving a gradient in matrix form for the following function. (w.r.t. $w$)

$$ H(w) = \sum_{i = 1}^{n} p(y_i - x_i^T*w) \log (p(y_i - x_i^T w)) $$

where $$p(x)$$ is the Gaussian,

$$ p(x) = \frac{1}{\sqrt{2 \pi \sigma^2}} \exp(-\frac{x}{2 \sigma^2}) $$

$X$ is $n \times m$, $y$ is $n \times 1$ and $w$ is $m \times 1$

So far I have used the chain rule to figure out the derivative but the dimensions do not add up if I want the gradient for $w$. How can I obtain this gradient w.r.t. $w$?

答:
Let $z = y - X w $ (where $y,z,w$ are column vectors)

Then $z_i = y_i - \sum_j w_j X_{i,j} = y_i-x_i^T w $ (where $x_i$ is the $i-$ row of $X$) and $$\frac{\partial z_i }{\partial w_j}=-X_{i,j}$$

Recall that $ (p \log p) ' = (\log p +1) p'$

Then $$\frac{\partial H(w)}{\partial w_j}= - \sum_i (\log(p(z_i)+1) \frac{\partial p(z_i)}{\partial z_i} X_{i,j} $$

$$\frac{\partial H(w)}{\partial w} = - X^T s$$

where $s_i = (\log(p(z_i)+1) \frac{\partial p(z_i)}{\partial z_i}$

Can you go on from here?

ContentLicense: CC BY-SA 4.0

Title: Gradient of Gaussian entropy w.r.t. $W$


问:
Peano axioms proof
Given that $K$ is an ordered field satisfying the least upper bound property and $1$ as the multiplicative identity, the set of natural numbers $\mathbb{N}_K$ in $K$ is defined as: $1 \in \mathbb{N}_K$ and $N + 1 \in \mathbb{N}_K$ if $N \in \mathbb{N}_K$. In this case, $N = 1 + ... + 1 \}$ $N$ times. Show that $\mathbb{N}_K$ satisfies the Peano axioms.

I am new to analysis and was not really taught about the Peano axioms and can't seem to find anything on the web about exactly what I need to prove in this problem. Can someone please briefly list and/or describe the axioms that need to be shown in order to complete the above proof? Any assistance is much appreciated.

答:
Suppose you are allowed to sum numbers, and if you are in the natural numbers, this is quite natural. And define the operation $S(x) = x+1$ to be it, then we have to believe:

$S(x) = S(y) $, meaning that our $x$ equals $y$, so $S$ is injective;

1 is the only element, that doesn't succed any other, then there is no $x$ in natural numbers such that $S(x) = 1$;

Suppose there is a subset of natural numbers with 1 inside it, then if in this subset there is another, then it's because $S(x) \in X$, and so, this subset is the natural numbers set.

Title: Peano axioms proof


问:
What is the mutual incoherence of a matrix?
While working on an exercise, I stumbled upon the term mutual incoherence of a matrix. Upon googling, I didn't find much information outside of some of its properties, but nowhere any definitions. The only thing I found is the definition of a mutual coherence of a matrix.

答:
I am also not sure about the terminology, but it seems to be the same as the mutual coherence. In the following article (eq. 2) the mutual incoherence is defined similarly as the maximal inner product in absolute value between different columns.
http://math.mit.edu/~liewang/OMP.pdf</p>


问:
Finding out whether a predicate logic formula is true or false.
I'm given the formula: ∃x∃y∃z(P(x, y) ∧ P(z, y) ∧ P(x, z) ∧ ¬P(z, x)).

The universe is all natural numbers. R is the relation corresponding to P. R = {<x, x+1> : x >= 0}

When putting values into the formula I've thought that if x = x then y = x + 1. By using this logic z is equal to x for P(z, y), but z has to be equal to y for P(x, z). I can't think of any other way to do this.

How do I show whether the formula is true or false? Could anyone point me in the right direction?

答:
When putting values into the formula I've thought that if x = x then y = x + 1. By using this logic z is equal to x for P(z, y), but z has to be equal to y for P(x, z). I can't think of any other way to do this.

Indeed.   That is the way to do this.

Let's tidy up your work a bit.

You have $P:=\{\langle x, x+1\rangle: x\in\Bbb N^+\}$  Which means $P(x,y)$ is substitutable with $y=x+1$.   Okay, so we shall do that.

$$\exists x~\exists y~\exists z~(P(x, y) \land P(z, y) \land P(x, z) \land\lnot P(z, x))\\\equiv\\\exists x~\exists y~\exists z~(y=x+1 \land y=z+1 \land z=x+1 \land x\neq z+1)$$

Now, assuming this was true, we could find naturals $x,y,z$ where:

$y=x+1$ and $z=x+1$, so $y=z$

Also $y=z+1$, so $\underline{y=y+1}$

So...


问:
How do I show that if $f$ is integrable on $[c,b]$ for all $c \in [a,b]$, then $f$ is integrable on $[a,b]$?
Let $a < b$. Suppose that the function $f: [a,b] \to \mathbb{R}$ is bounded and Riemann integrable on $[c,b]$ for every $a < c < b$. Prove that $f$ is Riemann integrable on $[a,b]$ and that

$$\int_{a}^{b} f(x) dx = \lim_{c \to a} \int_{c}^{b} f(x) dx$$

I can see that it is true, but I honestly have no idea how to prove that.

My thought would be to show that the upper and lower sums are equal:
$$U(f) = L(f) = \int_{a}^{b} f(x) dx$$
In terms of epsilon-delta, I would want to show that $U(f,P) - L(f,P) \leq \epsilon$ for any $\epsilon > 0$.

答:
Once you establish that $f$ is Riemann integrable on $[a,b]$ (for example as shown by Xander Henderson) there are several ways to proceed in showing that

$$\int_a^b f(x) \, dx = \lim_{c \to a+} \int_c^b f(x) \, dx$$

(1) Use the fact that if $f$ is Riemann integrable on $[a,b]$, then it is integrable on any subinterval and

$$\left|\int_a^b f(x) \, dx - \int_c^b f(x) \, dx \right| = \left|\int_a^c f(x) \, dx \right| \leqslant \int_a^c|f(x)| \, dx \\\leqslant \sup_{x \in [a,b]}|f(x)|\cdot (c-a) \underset{c \to a+}\longrightarrow 0$$

This argument obviously invokes several basic properties of the Riemann integral without proof.

For more of a "first principles" proof ...

(2)  There exists a partition $P: a = x_0< x_1=c< \ldots < x_n =b$ such that for any Riemann sum $S(P,f) = \sum_{j=1}^n f(\xi_j)(x_j - x_{j-1})$, we have

$$\left|\int_a^bf(x) \, dx-S(P,f)\right|= \left|\int_a^bf(x) \, dx-\sum_{j=1}^n f(\xi_j)(x_j - x_{j-1})\right|< \frac{\epsilon}{3},$$

and by refining the partition if necessary we can assume that with $M = \sup_{x \in [a,b]}|f(x)|$,

$$c-a = x_1-a < \frac{\epsilon}{3M}, \\ \left|\int_c^bf(x) \, dx-S(P',f)\right|=\left|\int_c^bf(x) \, dx-\sum_{j=2}^n f(\xi_j)(x_j - x_{j-1})\right|< \frac{\epsilon}{3}$$

Thus, for every $a< c < a+\frac{\epsilon}{3M}$,

$$ \left|\int_a^bf(x) \, dx-\int_c^bf(x) \, dx \right|\\ \leqslant \left|\int_a^bf(x) \, dx-S(P,f)\right|+ |S(P,f) - S(P',f)|+\left|\int_c^bf(x) \, dx-S(P',f)\right|\\ \leqslant \frac{2\epsilon}{3} + |f(\xi_1)|(x_1-a)\leqslant \frac{2\epsilon}{3} + M(x_1-a)< \epsilon,$$

and, therefore,

$$\lim_{c \to a+} \int_c^b f(x) \, dx = \int_a^b f(x) \, dx$$


答:
While there are certainly slicker proofs than what I am about to present, it does not seem unreasonable to work directly from the definitions.  My feeling is that an elementary proof gives one a better idea of what the definitions and theorems actually mean, and are therefore, in some respects, more enlightening.

Definitions:  Let $f : [a,b] \to \mathbb{R}$ be bounded.  Let
$$P = \{ a = x_0 < x_1 < \dotsb < x_n = b\} $$
be a partition of $[a,b]$, and for each $j = 1, 2, \dots, n$, define
$$ m_j = \inf_{x_{j-1} \le x \le x_j} f(x)
\qquad\text{and}\qquad
M_j = \sup_{x_{j-1}\le x \le x_j} f(x). $$
The lower and upper Riemann sum subordinate to $P$ are
$$
L(f,[a,b],P) = \sum_{j=1}^{n} m_j (x_{j} - x_{j-1})
$$
and
$$
U(f,[a,b],P) = \sum_{j=1}^{n} M_j (x_{j} - x_{j-1}),
$$
respectively.  The lower and upper Riemann integrals are
$$ L(f,[a,b]) = \sup \{ L(f,[a,b],P) : \text{$P$ is a partition of $[a,b]$} \} $$
and
$$
U(f,[a,b]) = \inf \{ U(f,[a,b],P) : \text{$P$ is a partition of $[a,b]$} \},
$$
respectively.  If $L(f) = U(f)$, then we say that $f$ is Riemann integrable on $[a,b]$, and define the Riemann integral to be the common value, i.e.
$$ \int_{a}^{b} f(x) \,\mathrm{d}x = L(f) = U(f). $$

Note that for any partition $P$,
$$ L(f,[a,b],P) \le L(f, [a,b]) \le U(f,[a,b]) \le U(f,[a,b],P). $$
In order to show that a function is Riemann integrable, it is therefore sufficient to show that
$$ U(f,[a,b]) - L(f,[a,b]) \le U(f,[a,b],P) - L(f,[a,b], P) $$
can be made arbitrarily small by choosing an appropriate partition of $[a,b]$.  It is also worth noting that if we refine a partition, the upper and lower sums "behave nicely".  This is encapsulated in the following lemma, which I will state without proof.

Lemma: Let $P$ be a partition of $[a,b]$, and let $P'$ be a refinement of $P$ (that is, $P'$ contains all of the points in $P$, and possibly more).  Then
$$ L(f,[a,b],P) \le L(f,[a,b],P') \le U(f,[a,b],P') \le U(f,[a,b],P). $$

The basic idea of the proof is that
$$ \bigl( \sup_{x_{j-1} \le x \le x_j} x \bigr) \bigl( x_j - x_{j-1} \bigr)
\le
\bigl( \sup_{x_{j-1} \le x \le x'} x \bigr) \bigl( x_j - x' \bigr) + \bigl( \sup_{x' \le x \le x_j} x \bigr) \bigl( x_j - x' \bigr) $$
for any $x' \in [x_{j-1}, x_j]$.  Thought about carefully, this lets us bound $L(f,[a,b],P)$ from above by $L(f,[a,b],P')$.  As similar argument will work for the upper Riemann sums.

In order to prove the desired result, we need to get control on the lower and upper Riemann sums, which will allow us to control the lower and upper Riemann integrals.  The essential idea is that we can control the sums on $[a,c]$ by choosing $c$ sufficiently close to $a$, and we can control the sums on $[c,b]$ via the Riemann integrability of $f$ on every such interval.

More precisely, choose $\varepsilon > 0$.  As $f$ is bounded, there exist $m$ and $M$ such that $m \le f(x) \le M$ for all $x \in [a,b]$.  Chose $x_1 > a$ so that
$$ (M-m) (x_1 - a) < \frac{\varepsilon}{2}. $$
Since $f$ is Riemann integrable on $[x_1,b]$, there exists a partition $P_1$ of $[x_1,b]$ such that
$$ \int_{x_1}^{b} f(x)\,\mathrm{d}x
= L(f,[x_1,b])
< L(f,[x_1,b],P_1) + \frac{\varepsilon}{4}. $$
Note that this follows from the definition of $L(f,[x_1,b])$ as a supremum.  Similarly, there is a partition $P_2$ of $[x_1,b]$ such that
$$ \int_{x_1}^{b} f(x)\,\mathrm{d}x
= U(f,[x_1,b])
> U(f,[x_1,b],P_1) - \frac{\varepsilon}{4}. $$
Take $P' = \{ x_1 < x_2 < \dotsb < x_n = b\}$ to be any common refinement of $P_1$ and $P_2$, e.g. $P' = P_1 \cup P_2$, and define
$$ P = \{ a = x_0 < x_1 < \dotsb < x_n = b\}. $$
Note that $P$ is a partition of $[a,b]$, and that
\begin{align*}
&U(f,[a,b]) - L(f,[a,b]) \\
&\qquad\le U(f,[a,b],P) - L(f,[a,b],P) \\
&\qquad= \sum_{j=1}^{n} M_j (x_{j} - x_{j-1}) - \sum_{j=1}^{n} m_j (x_{j} x_{j-1}) \\
&\qquad= (M-m) (x_1 - a) + \sum_{j=2}^{n} M_j (x_{j} - x_{j-1}) - \sum_{j=2}^{n} m_j (x_{j} x_{j-1}) \\
&\qquad< \frac{\varepsilon}{2} + U(f,[x_1,b],P') - L(f,[x_1,b], P') \\
&\qquad\le \frac{\varepsilon}{2} + U(f,[x_1,b],P_2) - L(f,[x_1,b], P_1) && \text{(by the lemma)} \\
&\qquad< \frac{\varepsilon}{2} + \left( U(f,[x_1,b]) + \frac{\varepsilon}{4} \right) - \left( L(f,[x_1,b]) - \frac{\varepsilon}{4} \right) \\
&\qquad= \varepsilon,
\end{align*}
with the last equality following from the fact that $L(f,[x_1,b]) = U(f,[x_1,b])$.  This completes the proof, as we have shown that the difference between the lower and upper Riemann integrals can be made arbitrarily small by choosing an appropriate partition.

Title: How do I show that if $f$ is integrable on $[c,b]$ for all $c \in [a,b]$, then $f$ is integrable on $[a,b]$?

Body:
Let $a < b$. Suppose that the function $f: [a,b] \to \mathbb{R}$ is bounded and Riemann integrable on $[c,b]$ for every $a < c < b$. Prove that $f$ is Riemann integrable on $[a,b]$ and that

$$\int_{a}^{b} f(x) dx = \lim_{c \to a} \int_{c}^{b} f(x) dx$$

I can see that it is true, but I honestly have no idea how to prove that.

My thought would be to show that the upper and lower sums are equal:
$$U(f) = L(f) = \int_{a}^{b} f(x) dx$$
In terms of epsilon-delta, I would want to show that $U(f,P) - L(f,P) \leq \epsilon$ for any $\epsilon > 0$.


问:
How do you check if a segment intersects a line?
As the title says, how can you check if a line segments intersects an infinite line. Note that this is not line-line intersection, or segment-segment intersection.

For example, how can you check whether a segment $A$ or $B $ crosses line $L $ as shown in Fig.1.

答:
Let the line segment endpoints be $\vec{p}_1 = (x_1 , y_1)$ and $\vec{p}_2 = (x_2 , y_2)$, and the line pass through points $\vec{p}_3 = (x_3 , y_3)$ and $\vec{p}_4 = (x_4 , y_4)$.  Then, if and only if
$$\Bigl( (\vec{p}_4 - \vec{p}_3) \times (\vec{p}_1 - \vec{p}_3) \Bigr) \Bigl( (\vec{p}_4 - \vec{p}_3) \times (\vec{p}_2 - \vec{p}_3) \Bigr) \le 0$$
does the line segment intersect the line.  In Cartesian coordinate form,
$$\Bigl( (x_4 - x_3)(y_1 - y_3) - (x_1 - x_3)(y_4 - y_3) \Bigr) \Bigl( (x_4 - x_3)(y_2 - y_3) - (x_2 - x_3)(y_4 - y_3) \Bigr) \le 0$$
The sign of the first multiplicand depends on which side $\vec{p}_1$ is to the line, and the sign of the second multiplicand depends on which side $\vec{p}_2$ is to the line. If the product is zero, $\vec{p}_1$ and/or $\vec{p}_2$ is on the line.  If the product is negative, then the two points must be on different sides of the line.

If the two points are on different sides of the (infinitely long) line, then the line segment must intersect the line.  If the two points are on the same side, the line segment cannot intersect the line.

Mathematically, this relies on the fact that the angle $\varphi$ between two vectors $\vec{a}$ and $\vec{b}$ fulfills
$$\sin\varphi = \frac{\vec{a}}{\left\lVert\vec{a}\right\rVert} \times \frac{\vec{b}}{\left\lVert\vec{b}\right\rVert}$$
i.e.
$$\left\lVert\vec{a}\right\rVert \left\lVert\vec{b}\right\rVert  \sin\varphi = \vec{a} \times \vec{b} \tag{1}\label{G1}$$
where $\times$ is the 2D analog of vector cross product,
$$(x_a, y_a) \times (x_b, y_b) = x_a y_b - y_a x_b$$
so that the sign of $\eqref{G1}$ corresponds to the sign of $\varphi$ when $-180° \lt \varphi \lt +180°$.


问:
What is an example that allows you to remember if the Hessian criterion for strict convexity is necessary or sufficient?
I know that given a strictly convex function $f$

There exists a one-way condition that relates the strict convexity of $f$ with the positive-definiteness of the Hessian $\nabla^2 f(x)$.

However I just cannot remember which way it was.

Is it,

Hessian PD $\implies$ strict convexity

or

Strict convexity $\implies$ Hessian PD

$f(x) = x^2$ is strictly convex, its Hessian is $1$, hence PD. In this case it seems both way works!

答: The Hessian being strictly positive definite is sufficient for strict convexity. It isn't necessary (even if you assume sufficient regularity that the Hessian makes sense) as you can see from, for example, the one dimensional $f(x)=x^4$.


问:
What is the probability of Adam being first in line or Beth being at the end of the line if each probability is 1 / n?
Given this type of problem, it's safe to assume there is a 1 in n chance of Adam being first because there are n students total, including Beth. To me, that means if we are doing a permutation, there are n! possible ways to arrange all the students including Adam and Beth, (n - 1)! ways to arrange the students if you pick Adam to be first, (n - 1)! ways to arrange the students if you pick Beth last, and (n - 2)! ways to arrange the students if you pick Adam first and Beth last, which you must subtract to remove duplicate permutations from one of the above ways where Adam is picked to be first or Beth is picked to be last.

In total, the probability is as follows: $$\frac{2(n - 1)! - (n - 2)!}{n!} = \frac{2}{n}-\frac{1}{n(n-1)}$$.

However, we can also think of it as $P(A)$ means the probability that Adam is first in line, which is $\Large{\frac{1}{n}}$, and that $P(B)$ means the probability that Beth is last, which is also $\Large{\frac{1}{n}}$. Knowing this, we can say the following: $$P(A\ or\ B) = P(A) + P(B) - P(A\ and\ B) = P(A) + P(B) - P(A)P(B)=\frac{1}{n}+\frac{1}{n}-\frac{1}{n^2}=\frac{2}{n}-\frac{1}{n^2}$$

This is where I get confused. My two probabilities don't equal: $$\frac{2}{n}-\frac{1}{n(n-1)}\neq\frac{2}{n}-\frac{1}{n^2}$$

Where am I going wrong? Is it with the first method of using permutations, the second method using laws of probability, or do I just need to use algebra to manipulate these two expressions into equivalent forms?

答:
Remember first that $P(A \cup B) = P(A) + P(B) - P(A \cap B)$

$P(A) = P(B) = \frac{1}{n}$

Now, your second answer was wrong because you assumed they were independent, aka $P(B | A) = P(B)$. In reality, the generalized solution is:

$P(A \cap B) = P(A) * P(B | A)$

Now let's say A is Adam being first in line and B is Beth being last. If Adam is first in line, there are only $n - 1$ spots Beth can be in, so:

$P(B | A) = \frac{1}{n-1}$

So final probability:

$P(A\cup B) = \frac{2}{n} - \frac{1}{n(n-1)}$


问:
If $A, B$ are self adjoint operators and $A \le B$, is $imA \subset imB$? is $ker(B) \subset ker(A)$?
If $A, B$ be self adjoint operators and $A \le B$, is $imA \subset  imB$? is $ker(A) \subset ker(B)$?

Let $H$ be a Hilbert space is $A,B$ self adjoint operators on $H$, such that $A \le B$.

Is $ImA \subset ImB$ ?

from the definition of operator order I get that for every $x \in H$:
$$<Ax, x> \le <Bx,x>$$

but I don't see how that helps me to find the images of $A$ and $B$.

Help would be appreciated.

Now when also $A \ge 0$, is $Ker(B) \subset Ker(A)$ ?

I got to $<Ax,x> = 0$ for every $x \in Ker(B)$, since $<Ax,x> \ge 0$ and $<Ax,x> \le <Bx,x> = <0,x> = 0$

But I'm not sure how to continue.

答:
Regarding ranges:

Simply assuming that $A\leq B$ does not imply that $\text{Ran}(A)\subseteq \text{Ran}(B)$, a counter-example being
$$-1\cdot I \leq 0\cdot I.$$

Perhaps assuming, in addition, that $A\geq 0$, might improve the chances of an affirmative answer.

Regarding kernels:

If $0\leq A\leq B$, and if $x\in\text{Ker}(B)$, then
$$
  \Vert A^{1/2}x\Vert^2 =
  \langle A^{1/2}x,A^{1/2}x \rangle =
  \langle Ax,x \rangle \leq
  \langle Bx,x \rangle = 0,
  $$
so $A^{1/2}x=0$, and hence also
$$
  Ax =
  A^{1/2}A^{1/2}x = 0,
  $$
so $x\in\text{Ker}(A)$.


问:
DFA Even a's and b's or Odd a's and b's
Is it possible to make a DFA that accepts an even number of a’s and an even number of b’s or an odd number of a’s and an odd number of b’s just using two states such as (EE/OO) which would be the accepted state and (EO/OE) would be the second state where 'E' means Even and 'O' means odd?

I know it can be done with 4 states but not sure if it's possible with just two combined states.

答: Yes, you can do it with two states just as you suggest, because a word is of the desired form if and only if its length is even. If $w$ is the input word, $|w|=|w|_a+|w|_b$ is even if and only if $|w|_a\equiv|w|_b\pmod2$.

Title: DFA Even a's and b's or Odd a's and b's


问:
Prove that $\sum_{n=1}^{\infty}\frac{q^n}{1+q^{2n}}=\sum_{n=1}^{\infty}(d_1(n)-d_3(n))q^n$
Prove that $$\sum_{n=1}^{\infty}\frac{q^n}{1+q^{2n}}=\sum_{n=1}^{\infty}(d_1(n)-d_3(n))q^n$$ where $d_1(n)$ is the number of divisors of $n$ congruent to $1$ modulo $4$ and $d_3(n)$  is the number of divisors of $n$ congruent to $3$ modulo $4$. I tried using the geometric series and interchanging the sum but that did not yield anything useful. I would prefer a algebraic solution rather than a one that depends on partitions but both are welcome.

Thanks!

答:
There is a small typo: $d_1(n) $ is the number of divisors congruent to 1 $\pmod{4}$, and similarly for $d_3(n) $.

Let's open the geometric series:

$$ \sum_n \sum_k q^n (-1) ^k q^{2nk} = \sum_{n, k} q^{n(2k+1) } (-1) ^k$$

Now we should group the terms depending on the exponent of $q$. For a given $m$ , which $n, k$ are such that $n(2k+1) =m$ ?

This happens exactly when we have an odd divisor $d=2k+1$ of $m$. The number $n$ will be then determined by $n= m/d$. Now we group the summands depending on the exponent of $q$ as we promised, then we use the substitution $d=2k+1$ :

$$ \sum_m \sum_{n(2k+1)= m } q^{m} (-1) ^k = \sum_m q^m \sum_{d \mid m \text{ odd}} (-1) ^{\frac{d-1}{2}}$$

Let's further separate the odd divisors in two classes. If $d\equiv 1 \pmod{4}$, then the quantity $(d-1) /2$ is even and the sign appearing is $+1$.   If $d\equiv 3 \pmod{4}$, then the sign appearing is $-1$. On balance we will have

$$\sum_{ d \mid m  \text{ odd}} (-1) ^{\frac{d-1}{2}} = \sum_{ d \mid m  \ \ d \equiv 1 } (+1) + \sum_{ d \mid m  \ \ d \equiv 3 } (-1) = d_1(m) -d_3(m) $$

Plugging this into the original equation we get

$$\sum_m q^m \sum_{d \mid m \text{ odd}} (-1) ^{\frac{d-1}{2}} = \sum_m (d_1(m) -d_3(m)) q^m$$

As desired.


问:
Proving the remainder is $1$ if the square of a prime is divided by $12$
Given, $p$ is a prime number and $p>3$. How do we prove that the remainder $r$ is always $1$ if $p^2$ is divided by $12$?</p>

答: Hint: consider $0,1,2,\ldots, 11$, which are all the possible remainders modulo 12.  Take out the ones that couldn't be congruent to a prime bigger than 3.  Square all the ones that are left modulo 12, and see what you get.</p>


答: A function $f$ bounded it is Riemann integrable if and only if it is continuous almost everywhere, i.e., the set of discontinuities has measure $0$. Since for all $c>a$, $f$ it is Riemann integrable on $[c,b]$ it follows that the set of discontinuities on $[c,b]$ has measure $0$, therefore on $[a,b]$.


答:
If $b+c<2a$ then $$LHS \geq 0 \ge RHS.$$
If $b+c \geq 2a.$ We write inequality as
$$4(a^3+b^3+c^3-3abc) \geq (b+c-2a)^3,$$
or
$$4(a+b+c)(a^2+b^2+c^2-ab-bc-ca) \geq  (b+c-2a)^3.$$
Because $a+b+c = 3a+(b+c-2a) \geq b+c-2a,$ so we will show that
$$4(a^2+b^2+c^2-ab-bc-ca)\geq (b+c-2a)^2,$$
equivalent to
$$3(b-c)^2 \geq 0.$$
Which is true. Equality holds when $ a=b=c$ or $ a=0,\,b=c.$

Note. We have
$$a^3+b^3+c^3-3abc-{2\left(\frac{b+c}{2}-a\right)^3}$$
$$=(b-c)^2+\frac{3\,a}{2} \Big[(a-b)^2+(b-c)^2+(c-a)^2\Big] \geq 0.$$

Title: Proof then $a^3+b^3+c^3-3abc\geq \left(\displaystyle\frac{b+c}{2}-a\right)^3$

Body:
If $a,  b,  c$ are non-negative numbers, prove  then

$a^3+b^3+c^3-3abc\geq \left(\displaystyle\frac{b+c}{2}-a\right)^3$

$8a^3+8b^3+8c^3-24abc\geq\displaystyle(b^3+3b^2c+3bc^2+c^3-6a(b^2+2bc+c^2)+12a^2(b+c)-8a^3)$

$16a^3+7b^3+7c^3\geq 3b^2c+3bc^2-6ab^2+12abc-6ac^2+12a^2b+12a^2c$

$16a^3+7b^3+7c^3+6ab^2+6ac^2\geq 3b^2c+3bc^2+12abc+12a^2b+12a^2c$

$AM-GM$

$\frac{6a^3+6ab^2}{2}\geq 6\sqrt{a^4b^2}$

$6a^3+6ab^2\geq 12a^2b...(1)$

$\frac{6a^3+6ac^2}{2}\geq 6\sqrt{a^4c^2}$

$6a^3+6ac^2\geq 12a^2c...(2)$

$\frac{4a^3+4b^3+4c^3}{3}\geq 4^3\sqrt{a^3b^3c^3}$

$4a^3+4b^3+4c^3\geq 12abc...(3)$

$\frac{b^3+b^3+c^3}{3}\geq ^3\sqrt{b^6c^3}$

$2b^3+c^3\geq 3b^2c...(4)$

$\frac{b^3+c^3+c^3}{3}\geq ^3\sqrt{b^3c^6}$

$b^3+2c^3\geq 3bc^2...(5)$

$(1)+(2)+(3)+(4)+(5)$, we get

$16a^3+7b^3+7c^3+6ab^2+6ac^2\geq 3b^2c+3bc^2+12abc+12a^2b+12a^2c$

Proved.


问:
Conceptual (homological) interpretation of a theorem about flat modules
I just learned the following beautiful result about flat modules from J. S. Milne's book Étale Cohomology:

Lemma 2.10(b'): Let $M$ be any flat $A$-module. If
$$\sum_i a_ix_i=0,$$
$a_i\in A$, $x_i\in M$, then there are equations
$$x_i = \sum_j a_{ij}x_j'$$
with $x_j'\in M$, $a_{ij}\in A$, such that
$$\sum_i a_ia_{ij} = 0$$
for all $j$.

It is stated under the hypothesis that $A$ is a commutative, noetherian ring, but as far as I can tell, the proof does not use noetherianity. (Actually I am not sure it even uses commutativity, but I will leave that matter alone for the purposes of this question.)

The proof given in Milne is essentially computational. However, the result struck me as almost a statement that something is exact (I'll make this precise in a moment), and of course flatness is all about exactness, so I wondered, is there a conceptual argument, or at least some reasonably satisfying conceptual handwaving, that gives us the claimed result in terms of an assertion that something is exact, or some (co)homology vanishes, in a way that is straightforwardly connected to $M$'s flatness?

(I'm using the soft-question tag because, while there could easily be a fully precise answer [which would be best possible], I will probably be happy with what I'm calling "satisfying conceptual handwaving," and what I mean by "straightforwardly" isn't precise either. Apologies in advance about this vagueness.)

Here is what I mean when I say the result struck me as "almost a statement that something is exact". The tuple $(a_i)_{i=1,\dots,r}$ defines a map $\varphi: M^r\rightarrow M$ by

$$(y_i)_{i=1,\dots,r} \mapsto \sum_i a_iy_i.$$

If $A$ is commutative, this is a map of $A$-modules. Then the equation

$$\sum_i a_ix_i=0$$

asserts that $(x_1,\dots,x_r)\in M^r$ lies in the kernel of this map. Similarly, the matrix $(a_{ij})$ defines a map $\psi: M^s\rightarrow M^r$ by

$$(y'_j)_{j=1,\dots,s}\mapsto \left(\sum_j a_{ij}y'_j\right)_{i=1,\dots,r},$$

whereupon the equations

$$\sum_i a_ia_{ij} = 0$$

assert that the composed map $\psi\circ\varphi = 0$, and the equations

$$x_i = \sum_j a_{ij}x_j'$$

assert that $(x_1,\dots,x_r)$ lies in the image of $\psi$. Thus the lemma "comes close" to asserting that we have a complex $M^s\xrightarrow{\psi} M^r \xrightarrow{\varphi} M$ which is exact in the middle.

It is not actually asserting this because the matrix $(a_{ij})$ is allowed to depend on the tuple $(x_i)$ (and the proof indeed requires the data of the $(x_i)$ for the construction of the $(a_{ij})$). So the real statement is: given an $A$-linear map $\varphi: M^r\rightarrow M$ defined by a tuple $(a_1,\dots,a_r)$, and given an element $(x_1,\dots,x_r)$ of the kernel of this map, there is an $A$-linear map $\varphi: M^s\rightarrow M^r$ defined by a matrix $(a_{ij})$, such that $M^s\rightarrow M^r\rightarrow M$ forms a complex and the particular given $(x_1,\dots,x_r)\in M^r$ lies in the image of $\psi$. There is no assertion that the entire kernel of $\varphi$ is exhausted by the image of $\psi$.

Nonetheless, these musings are sufficiently suggestive to me that I wanted to ask here: can you explain Milne's Lemma 2.10(b') in terms of the exactness of something, or the vanishing of some homology or cohomology, in a way that is evidently linked to $M$'s flatness over $A$?

答:
Maybe you already know this and are looking for something more conceptual, but here's a way to connect this statement to a more homologically flavoured one.

A module $M$ over some commutative ring $R$ is flat precisely if $\mathbf{Tor}_1^R(R/I,M)$ vanishes for all finitely generated ideals $I \subset R$.

A handwavy proof: $M$ is flat if $\mathbf{Tor}_1^R(N,M)$ vanishes for all $N$. By cocontinuity using that every module is the colimit of its finitely generated submodules, flatness is equivalent to the vanishing of $\mathbf{Tor}_1^R(N,M)$ for all $N$ finitely generated. Such a module sits in a filtration with cyclic quotients and allows us to assume $N$ to be cyclic, hence of the form $R/I$. "$\square$"

Moreover, in the same proposition it is shown that $M$ is flat if for any finitely generated ideal $I \subset R$ tensoring the inclusion gives an injective map $I \otimes M \to M$. This is a matter of using the aforementioned equivalence and the long exact sequence for $\mathbf{Tor}$ applied to the short exact sequence

$$
0 \to I \to R \to R/I \to 0.
$$

Indeed, since $\mathbf{Tor}_1^R(R,M) = 0$ we have

$$
0 \to \mathbf{Tor}_1^R(R/I,M) \to I \otimes M \to R \otimes M \simeq M
$$
so the vanishing of $\mathbf{Tor}$ is exactly characterized the injectivity of $I \otimes M \to M$.

Therefore, flatness is intrinsically related to all the kernels of inclusion-induced maps $\langle a_1, \ldots, a_n\rangle \otimes M \to M$. Concretely, for $M$ to be flat we need that whenever

$$
\sum_{i}b_i a_i m_i = 0
$$
in $M$,
$$
\sum_{i} b_i a_i \otimes m_i = 0
$$
in $\langle a_1, \ldots, a_n\rangle \otimes M$. This already looks connected to the equational criterion, and in fact the connection comes from the following lemma on zero elements of certain tensor products,

Lemma. Let $R$ be a commutative ring, $M, N$ two $R$-modules and $\{n_i\}_{i \in I}$ a finite generating set for $N$. Given $\{m_i\}_i \subset M$, we have $\sum_i m_i \otimes n_i = 0$ if and only if there exists a finite set $J$ and elements $r_{ij} \in R, \widetilde{m}_j \in M$ such that
$$
\sum_{j} r_{ij}\widetilde{m}_j = m_i, \quad \sum_{i \in I}r_{ij}n_i = 0 \quad (i \in I, j \in J).
$$

All of this can be found in section $6.3$ of Eisenbud's Commutative Algebra with a View Toward Algebraic Geometry.


问:
Solution for Vlasov equation PDE with delta function source term
I'm interested in finding a solution (or an approximate solution) to the following PDE:

$f_t + (a-bt^2)f_v = c\delta(v)$

This is the Vlasov equation for a distribution function $f$ with no spatial dependence, a delta function source term, and an acceleration of the form $(a-bt^2)$.

答:
Formally speaking, one can always solve an equation like this. Just apply the method of characteristics to the general equation

$$f_t+(a-bt^2)f_v=g(t,v)$$
which reads
$$dt=\frac{dv}{a-bt^2}=\frac{df}{g(t,v)}$$
The next step is to find two distinct integrals of motion. These are readily found since

$$dv-(a-bt^2)dt=0\iff v-at+\frac{bt^3}{3}=C_1$$

and with the aid of this integral of motion we can find a second one:

$$df=g(t,v)dt\iff f-\int^tdt'g(t',v-at+\frac{bt^3}{3}+at'-\frac{bt'^3}{3})=C_2$$

From these then one can find the most general solution to the equation above to be

$$f(t,v)=\int^tdt'g(t',v-at+\frac{bt^3}{3}+at'-\frac{bt'^3}{3})+ H\left(v-at+\frac{bt^3}{3}\right)$$

where $H$ is an arbitrary function. The first term is a solution to the inhomogeneous equation and the 2nd one is a solution to it's homogeneous part. To solve your particular problem, $g(t,v)=c\delta(v)$ you need to solve a cubic equation, which is not fun because of the casework and the lack of physical interpretation and boundary conditions. Regardless, the form of the solutions is going to look like

$$f(t,v)=\frac{c}{a}\sum\frac{\theta\left(t-2\sqrt{\frac{a}{b}}r_i(X)\right)}{|1-4r_i^2(X)|}$$
where $r_i(x)$ represents the set of real roots of the equation $4u^3-3u-x=0$ and $X$ being the following dimensionless quantity

$$X=\frac{3b^{1/2}[v-at+b\frac{t^3}{3}]}{2a^{3/2}}$$

EDIT: Now that the physical picture has been added it is much easier to provide a concrete solution based on the above. We posit that at time zero, there are no particles present, thus $f(0,v)=0$. Applying the boundary condition to the general solution we find after a short calculation that the solution is

$$f(t,v)=c\int_0^tdt'\delta(v-at+\frac{bt^3}{3}+at'-\frac{bt'^3}{3})$$

which yields a solution exactly as above but with the added restriction that the roots of the equation have to be taken into account are the ones that are positive!

This picture is supported further when one solves the problem by physical considerations only, monitoring the velocity of every particle generated at an arbitrary time by the source. The velocity of the particle generated at time $T$ as a function of time is given by

$$v(t,T)=a(t-T)-\frac{b}{3}(t^3-T^3)$$

We ask: At a certain time t, given that the stationary particle production rate is $\frac{dN}{dT}=c$, how many particles are contained in the velocity interval $(v,v+dv)$? It obviously is the ones that were generated at the correct moments $(T, T+dT)$ so that their velocities are the ones required. Formally

$$dN(v\leq v(t,T)\leq v+dv)=dN(v-at+\frac{bt^3}{3}\leq -aT+\frac{bT^3}{3}\leq v-at+\frac{bt^3}{3}+dv)$$

which yields that $T$ must lie in the (possibly multiple) intervals $r\leq T\leq r+\frac{dv}{br^2-a}$ where $r$ is a positive (since time of production is positive) root of the equation $-ar+\frac{br^3}{3}=v-at+\frac{bt^3}{3}$. Importantly, this root also has to be less than $t$ itself, since the time of production of the particle has to precede the time at which a particle is measured to have the purported velocity $v$.With these in mind, from the rate of particle production we get that
$$dN(v\leq v(t,T)\leq v+dv)=\sum_n dN(r_n\leq T\leq r_n+\frac{dv}{br_n^2-a})=cdv\sum_n\frac{1}{|br_n^2-a|} \\~,~r_n\in[0,t]$$

which finally implies

$$\frac{dN}{dv}=f(t,v)=c\sum_n\frac{1}{|br_n^2-a|}$$

This is precisely what we expect from the integration of the delta functions of each distinct positive root of the cubic presented before the edit (modulo rescaling of various prefactors).

Of course, to obtain the global properties of this distribution we need to do more work. In particular, we need to monitor how much time it actually takes for the first particle to arrive at a particular velocity. Before that time the distribution is actually identically zero at that point. Assuming for simplicity that $a,b>0$ and sparing you the details we have that for $v<0 ~, ~t\geq t_0(v)$ where $t_0$ is the only positive solution of the equation $at_0-\frac{bt_0^3}{3}=v$, for $v>\frac{2a}{3}\sqrt{\frac{a}{b}}$, the distribution is identically zero for all times, and for $0\leq v\leq\frac{2a}{3}\sqrt{\frac{a}{b}}$ there are two branches with different arrival times $t_1(v), t_2(v)$ both solving the same cubic. In this last case, only the smallest root contributes for $t_1\leq t \leq t_2$ and after $t_2$ both actually contribute.


问:
Showing a polynomial has no roots in $\mathbb{F}_p[x]$
So I'm given the polynomial $$x^5+x^2+1$$ and asked to prove that it is irreducible in $\mathbb{Z}[x]$ by first reducing it modulo 2 and then showing it has no roots in $\mathbb{F}_2$.

After reducing mod 2, the coefficients of the polynomial remain unchanged. How can I then show that $x^5+x^2+1$ has no roots in $\mathbb{F}_2$, meaning it is also irreducible in $\mathbb{Z}[x]$? I've thought about just writing out all elements of $\mathbb{F}_2[x]$ and showing that none of them divide the given polynomial, but I'm not really sure how I'd go about doing that.

答:
You have to prove it has no root in $\mathbf F_2$, so it can't have a linear factor. Therefore the only possibility for the polynomial to be reducible is that it decomposes as the product of a quadratic and a cubic factors (both irreducible).

The only irreducible quadratic polynomial in  $\mathbf F_2[X]$ is $X^2+X+1$, and it is not long to obtain the result of the Euclidean division:
$$X^5+X^2+1=(X^2+X+1)(X^3+X^2)+1.$$


答:
If we were to have $f = gh$ in $\Bbb Z[X]$ for $g,h$ polynomials of positive degree, reducing the coefficients modulo $p$ would give $\overline{f} = \overline{g}\overline{h}$. Also $g$ and $h$ have positive degree (they largest degree coefficient of $g$ and $h$ is $\pm 1$, since $f$ was monic).

Hence whenever $f$ is monic being irreducible modulo $p$ implies that $f$ is irreducible in $\Bbb Z[X]$. So far so good, but we have to show now that $x^5+x^2+1$ is irreducbile in $\Bbb F_2[X]$. Note that it has no roots in $\Bbb F_2$ (just plug in $1$ and $0$), so were $f$ to be reducible we should have $q \mid f$ with $\deg q = 2$.

By the previous remark; the polynomial $q$ has to be irreducible, and there's only one such polynomial, namely $q = x^2+x+1$.

Therefore your problem is reduced to showing that $q \not \mid f$ in $\Bbb F_2[X]$. Can you take it from here?

Title: Showing a polynomial has no roots in $\mathbb{F}_p[x]$

Body:
So I'm given the polynomial $$x^5+x^2+1$$ and asked to prove that it is irreducible in $\mathbb{Z}[x]$ by first reducing it modulo 2 and then showing it has no roots in $\mathbb{F}_2$.

After reducing mod 2, the coefficients of the polynomial remain unchanged. How can I then show that $x^5+x^2+1$ has no roots in $\mathbb{F}_2$, meaning it is also irreducible in $\mathbb{Z}[x]$? I've thought about just writing out all elements of $\mathbb{F}_2[x]$ and showing that none of them divide the given polynomial, but I'm not really sure how I'd go about doing that.


问:
Cusps on quadratic Bezier curve
A quadratic Bezier curve is given in parametric form by:
$$C(t) = (1-t)^2P_0 + 2(1-t)tP_1 + t^2P_2.$$

My points are: $(1,1)$, $(2,2)$ and $(3,3)$. How do I show that this curve has cusps?

Best regards,
Sergey

答:
I’m assuming that a “cusp” is defined to be a place where the curve’s tangent direction is discontinuous; i.e. there’s a sharp corner.

Your particular curve is a straight line, so it certainly doesn’t have a cusp.

In general, a quadratic Bézier curve is a parabola, and parabolas don’t have cusps, either.

It might seem that cusps occur at points where the first derivative vector has zero length. But this isn’t true. Take the quadratic Bézier curve with control points $(0,0)$, $(0,0)$, $(1,1)$, for example. It has zero first derivative at the origin, but it’s a straight line, so it doesn’t have a cusp.

Generally, you can’t learn about cusps by looking at derivatives and their magnitudes, because these depend on how the curve is parameterized, whereas the presence of a cusp is a purely geometric thing that does not depend on any particular parameterization.

Title: Cusps on quadratic Bezier curve


答:
Because this polynomial has no roots in $\mathbb{F}_{2}$ we know that it has no linear factors.

There are 4 polynomials in $\mathbb{F}_{2}[x]$ of degree 2 and 8 polynomials in $\mathbb{F}_{2}[x]$ of degree 4 - and some of them have roots in $F_{2}$, so they cannot divide the polynomial that you presented. So you only have to ckeck a small number of possible products of polynomials. Seems a very reasonable way to do this, and it follows your own idea of "writing out all elements of $\mathbb{F}_{2}[x]$ and showing that none of them divide the given polynomial".


问:
Integration by parts integrating with respect to CDF
I am told that the following equality follows from integration by parts:

$$\int_{R-k}^{1}(\theta-R)dG(\theta)-G(R-k)k=\int_{R-k}^{1}(1-G(\theta))d\theta-k$$ Where $(R-k)\in[0,1)$ and $G$ is the CDF of $\theta$ which is distributed on $[0,1]$. I cannot, however, see how this is the case. Might someone explain how integration by parts has been used? Thank you.

答:
Integration by parts takes the form of:

$\int_a^b v * du - (u*v) |_a^b = \int_a^b u * dv$

In this case, $u = -(1 - G(\theta))$ and $dv = d\theta$

$du = dG(\theta)$

$v = \theta$

$\int_a^b \theta dG(\theta) - (- ((1 - G(\theta)) * \theta))|_a^b - k = \int_a^b (1-G(\theta)) d\theta - k$

$\int_{R-k}^1 \theta dG(\theta) - (- ((1 - G(\theta)) * \theta))|_{R-k}^1 - k = \int_{R-k}^1 (1-G(\theta)) d\theta - k$

$\int_{R-k}^1 \theta dG(\theta) + (1 - G(1)) - ((1 - G(R - k)) * (R -k) - k = \int_{R-k}^1 (1-G(\theta)) d\theta - k$

$G(1) = 1$

$\int_{R-k}^1 \theta dG(\theta)  - R + k + G(R - k) * (R - k)- k = \int_{R-k}^1 (1-G(\theta)) d\theta - k$

$\int_{R-k}^1 \theta dG(\theta)  - R * (G(R-k) - 1)  - G(R - k)* k = \int_{R-k}^1 (1-G(\theta)) d\theta - k$

$\int_{R-k}^1 \theta dG(\theta)  - (R * G(\theta))|_{R-k}^1  - G(R - k)* k = \int_{R-k}^1 (1-G(\theta)) d\theta - k$

$\int_{R-k}^1 \theta dG(\theta)  - \int_{R-k}^1 R dG(\theta)  - G(R - k)* k = \int_{R-k}^1 (1-G(\theta)) d\theta - k$

$\int_{R-k}^1 (\theta - R) dG(\theta) - G(R - k)* k = \int_{R-k}^1 (1-G(\theta)) d\theta - k$


问:
Preparation of aqueous solution

In Chemistry class, Samantha finds that she can make a certain solution by mixing $.04\,\rm L$ of chemical $\mathrm{A}$ with $.02\,\rm L$ of water (this gives her $.06\,\rm L$ of the solution). She wants to make a total of $.48\,\rm L$ of this new solution. To do so, how many liters of water will she use?

How can I set this up? I'm having a hard time understanding this.

答: Firstly, we find that $\frac{.02}{.06}=\frac{1}{3}$ is the fraction of water in the solution. Thus, since Samantha wants $.48$ total liters of the solution, we have that she must use $.48 \times \frac{1}{3}=\boxed{0.16}$ liters of water.

Title: Preparation of aqueous solution


问:
Find the $n$th derivative of the function
I was trying to find the $n$th derivative of the function
$$y=e^x\tanh^{-1}x$$

I managed to find a formula for the $n$th derivative by inspection that relates the $n$th derivatives with its lower order derivatives :
$$(x^2-1)(y^{(n)}-y^{(n-1)})+2x(n-1)(y^{(n-1)}-y^{(n-2)})+n(n-1)(y^{(n-2)}-y^{(n-3)})+e^x=0$$

This applies only for values of $n \geq 3$

But is there a way to find the $n$th derivative explicitly in terms of $x$ and $n$ only?

答:
Hint.

The general Leibniz rule says that
$$
(fg)^{(n)}=\sum_{k=0}^n\binom{n}{k}f^{(n-k)}g^{(k)}
$$

Now let $f(x)=e^x$, $g(x)=\tanh^{-1}(x)$. So all you need is to find the $n$-th derivative for $g$.

Note that
$$
g'(x)=\frac{1}{1-x^2}=\frac12\left(\frac{1}{x+1}-\frac{1}{x-1}\right)
$$
By linearity, you only need to find the $(n-1)$-th derivative for the following two functions:
$$
\frac{1}{x+1},\quad \frac{1}{x-1},
$$
which should be straightforward by induction.


问:
Peano axioms proof attempt
Given that $K$ is an ordered field satisfying the least upper bound property and $1$ as the multiplicative identity, the set of natural numbers $\mathbb{N}_K$ in $K$ is defined as: $1 \in \mathbb{N}_K$ and $N + 1 \in \mathbb{N}_K$ if $N \in \mathbb{N}_K$. In this case, $N = 1 + ... + 1 \}$ $N$ times. Show that $\mathbb{N}_K$ satisfies the Peano axioms.

My attempt:

Let $S(N) = N + 1$. We now go ahead and prove each axiom:

For all natural numbers $N \in \mathbb{N}_K$, $S(N) \in \mathbb{N}_K$.

Assume that $N \in \mathbb{N}_K$. As $1, N \in \mathbb{N}_K$, we have that $N + 1 \in \mathbb{N}_K$. Therefore, $S(N) \in \mathbb{N}_K$.

For all $N \in \mathbb{N}_K$, $S(N) \neq 1$.

Assume that $1 < N \in \mathbb{N}_K$ and $S(N) = 1$. We get $S(N) = 1 \implies N + 1 = 1 \implies N < 1$. Contradiction.

$N = N' \iff S(N) = S(N')$.

Assume that $N, N', S(N)$, and $S(N')$ are natural numbers in $\mathbb{N}_K$. We get that

$$N = N' \iff N + 1 = N' + 1 \iff S(N) = S(N')$$

We conclude that $\mathbb{N}_K$ satisfies the Peano axioms.

QED.

Is this proof correct? I am new to analysis and Peano axioms. So, can somebody correct me if I am wrong? Any assistance much appreciated.

答:
It's basically correct.

You should, however, spell out some parts. Namely, you can use subtraction since $K$ is a field, so

$N+1=1\implies N=0$ and in an ordered field we indeed have $0<1$ and therefore $0<1+1+\dots+1=N$ for any $N\in\Bbb N_K$, i.e. $K$ must have characteristic $0$.

$N+1=N'+1\implies N=N'$ by subtracting $1$ from both sides.

Title: Peano axioms proof attempt


问:
Prove that a cut edge is in every spanning tree of a graph

Given a simple and connected graph $G = (V,E)$, and an edge $e \in E$. Prove:

$e$ is a cut edge if and only if  $e$ is in every spanning tree of $G$.

I have been thinking about this question for a long time and have made no progress.

答:
Hint ("only if"): Imagine you have a spanning tree in the graph which doesn't contain the cut-edge. What happens to the graph if you remove this cut edge? What happens to the spanning tree?

Hint ("if"): What happens if you remove this "indispensable" edge (the one which is in every spanning tree)? Can the resulting graph have any spanning tree? What kinds of graphs don't have any spanning tree?


问:
Prove that $\sigma(am) < \sigma(am + 1)$ for infinitely many positive integers $m$
Given a positive integer $a$, prove that $\sigma(am) < \sigma(am + 1)$ for infinitely many positive integers $m$.

($\sigma(n)$ is the sum of all positive divisors of the positive integer number $n$.)

I have two problem when trying to use the hint.

Let $p_1,p_2,...,$ be the primes which do not divide $a$. Let $k$ be the minimum integer such that $\dfrac{2\sigma(a)}{a} < \prod_{i=1}^k \left (1 + \dfrac{1}{p_i} \right )$. How can I prove $\prod_{i=1}^\infty \left (1 + \dfrac{1}{p_i}\right )$ diverges to show $k$ exist.

By Dirichlet's Theorem there exists infinitely many primes $q$ such that $q \equiv -a^{-1} \pmod{p_i}$ for $1 \le i \le k$. Choose $m=q$, then
$\sigma(aq+1) \ge (aq+1) \cdot \prod_{i=1}^k \left (1 + \dfrac{1}{p_i}\right ) > \dfrac{2(aq+1)\sigma(a)}{a} > 2q\sigma(a) > \sigma(aq)$.

The problem is done after that, but how can I prove $q$ exists without using Dirichlet's Theorem (which are too much for a High School Olympiad problem).

答:
First, it is known that $\prod\left(1+\frac{1}{p}\right)$ diverges, see e.g. Wikipedia. Alternatively, if you know that $\sum\frac{1}{p}$ diverges, then you can apply the result of this thread. In any case such a $k$ exists.

We claim that we can find solutions to the system $m\equiv-a^{-1}\pmod{p_i}$ for $1\leq i\leq k$ such that $\sigma(m)/m\to1$, which would clearly solve the problem.

Let $P=\prod_{i=1}^kp_i$, so by CRT we want solutions $m\equiv-a^{-1}\pmod P$. The key idea to avoiding Dirichlet is that we know there are residues $r_1,\dots,r_\ell$ mod $P$ that contain infinitely many primes, and we claim that we can choose $m$ so that its prime factors are all congruent to one of the $r_i\pmod P$. Indeed, by CRT we can find some $m\equiv-a^{-1}\pmod P$ that is coprime with all the primes that are not congruent to one of the $r_i\pmod P$.

Therefore, we can find a solution $m\equiv-a^{-1}\pmod P$ that is the product of $M$ distinct primes, each of which are congruent to some $r_i\pmod P$. Since there are infinitely many primes in each of these congruence classes, we can take the primes as large as we like, so that $\sigma(m)/m$ is arbitrarily close to $1$.


问:
Is a constructive proof known for Tutte's Theorem?
I have been reading the Chapter on Matchings in Graph Theory, R. Diestel, Fifth Edition, and have just encountered the proof of Tutte's theorem. The statement of the theorem is:

A graph $G$ has a $1$-factor if and only if $q(G-S) \leq |S|$ for all $S \subseteq V(G)$.

The forward direction for the proof is trivial. For the other direction, the first step says:

Let $G = (V, E)$ be a graph without a 1-factor. Our task is to find a bad set $S ⊆ V$, one that violates Tutte’s condition.

Thus, the contrapositive is being proven, which is not a constructive step. I tried searching for proofs of Tutte's theorem online that differed from the Diestel version but was not successful. Is a constructive proof known for Tutte's Theorem?

The motivation behind the question is that Hall's Theorem (the preceding section in the book) has several constructive proofs that offer insight into how to find maximal matchings but I haven't found the proof for Tutte's theorem to be particularly insightful (yet).

答:
The next theorem in the book (Graph Theory, Diestel, 5th Ed.) is a stronger result that implies Tutte's theorem. The theorem says:

Every graph $G = (V,E)$ contains a vertex set $S$ with the following two properties:
(i) $S$ is matchable to $C(G−S)$;
(ii) Every component of $G − S$ is factor-critical.
Given any such set $S$, the graph $G$ contains a 1-factor if and only if $|S| = |C(G−S)|$.

The proof of this theorem in the book is constructive: the largest set $T$ that satisfies the following property also satisfies the two properties in the theorem:

$$\forall U \subseteq V(G): q(G-U) - |U| \leq q(G-T) - |T|$$

If $|T| = |C(G-T)|$, then a 1-factor exists for $G$ that contains all the (contracted) edges between $T$ and $C(G-T)$; the remaining parts of the 1-factor may be constructed by recursively repeating the process for all the components in $C(G-T)$. Thus, an interesting algorithm for finding a 1-factor is present in this constructive proof.

I wish I had been more patient with the book before asking the question, but this answer may be useful for others reading the book.


问:
$Φ : V → V$ all such linear mappings have same determinant, what does it mean?
In the Book that I'm reading (Mathematics for Machine Learning), the following para is given, while listing the properties of a matrix determinant:

Similar matrices (Definition 2.22) possess the same determinant.
Therefore, for a linear mapping $Φ : V → V$ all transformation matrices
$A_Φ$ of $Φ$ have the same determinant. Thus, the determinant is invariant
to the choice of basis of a linear mapping.

I know that matrices $A$ and $B$ are similar if they satisfy $B=C^{-1}AC$.
I can prove that determinants of such $A$ and $B$ are equal using other properties of a determinant.

But beyond that I don't understand what this paragraph is saying. I can understand all matrices $Y$ such that $Y=X^{-1}AX$ have the same determinant as $A$, for varying $X$s.

But how do I connect this to linear mappings of the form $Φ : V → V$. What does $Φ : V → V$ mean here? Maybe someone can give me an example.

EDIT:
This video is pretty basic, but it helped me understand better
https://www.youtube.com/watch?v=s4c5LQ5a4ek

答:
Suppose that $V$ is the vector space of quadratic polynomials and $\Phi$ is multiplication by $(1 + x)$ mod $x^3$.

Now let us consider two different bases for $V$. One will be $1, x, x^2$ and the other will be $1, (1 + x), (1 + x)^2$.

In the first basis we have

\begin{align}
\phi(1) &= 1 + x \\
\phi(x) &= \phantom{1 + {}}x + x^2 \\
\phi(x^2) &= \phantom{1 + x +{}} x^2.
\end{align}

In the second basis,

\begin{align}
\phi(1) &= \phantom{1 - 3} (1 + x) \\
\phi(1 + x) &= \phantom{1 - 3(1 + x) + 3}{}(1 + x)^2 \\
\phi((1 + x)^2) &= (1 + x)^3 \equiv 1 +3x+3x^2 \pmod{x^3} \\
&= 1 - 3(1 + x) + 3(1 + x)^2
\end{align}

The two matrices we get out of this are
$$
\begin{pmatrix}
1 & 0 & 0 \\
1 & 1 & 0 \\
0 & 1 & 1
\end{pmatrix}
\text{ and }
\begin{pmatrix}
0 & 0 & 1 \\
1 & 0 & -3 \\
0 & 1 & 3
\end{pmatrix}
$$

And the statement/fact in question is that these two matrices are similar. If we were to write this as $A = PBP^{-1}$, the matrix $P$ would be the matrix corresponding to writing $\{1,(1+x),(1+x)^2\}$ in terms of the first basis. I.e.

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


问:
Pdf of sum of exponential random variables
I am trying to understand the following:

Let $X$ and $Y$ be independent and exponential random variables with parameter $1$, so $f_X(x) =e^{-x} $ and $f_Y(y) =e^{-y} $. Let $Z=X+Y, X<Y$ and $Z=X-Y, X\geq Y$. Determine the pdf of $Z$.

The solution to the exercise first calculates the cdf $F_Z(z)=P(Z\leq z, X<Y)+P(Z\leq z, X\geq Y)$ and then takes its derivative to obtain $f_Z(z)=\frac{e^{-z}}{2}(1+z)$.

This is fine but it is a bit long so I would like to do the exercise by computing the convolution to obtain $f_Z(z)$ directly, so I did: $f_Z(z)=\int_{0}^{z}e^{-x}e^{-(z-x)}dx=ze^{-z}$ for the $X<Y$ part but when doing the $X\geq Y$ part I get $f_Z(z)=\int_{0}^{x}e^{-z-y}e^{-y}dy=\frac{1}{2}(e^{-z}-e^{-z-2x})$ which is wrong.

Could someone please explain to me what I am doing wrong?

Thanks

答:
Note: $X$ and $Y$ are non-negative random variables.

We have $Z=X+Y$  (ie $Y=Z-X$), exactly when $Z>X$, which is only supported when also $X<Z-X$, ie $2X<Z$.

We have $Z=X-Y$ (ie $Y=X-Z$), exactly when $Z\leqslant X$, which isn't further restricted by the support.

Thus we have the support joint pdf for $X$ and $Z$, and this gives the bounds for your integration.

$$\begin{align}f_{X,Z}(x,z) &= f_{X,Y}(x,z-x)\cdot\mathbf 1_{0\leqslant 2x<z}+f_{X,Y}(x,x-z)\cdot\mathbf 1_{0\leqslant z\leqslant x}\\[1ex] &=\mathrm e^{-z}\cdot\mathbf 1_{0\leqslant 2x<z}+\mathrm e^{-2x+z}\cdot\mathbf 1_{0\leqslant z\leqslant x}\\[2ex] f_{Z}(z)&=\left( \int_0^{z/2}\mathrm e^{-z}\mathrm d x+\int_z^\infty \mathrm e^{-2x+z}\mathrm d x\right)\cdot \mathbf 1_{0\leqslant z}\\[1ex]&=\tfrac {z+1}2\mathrm e^{-z}\cdot\mathbf 1_{0\leqslant z}\end{align}$$


问:
Proof using truth tables and an equivalence
I am solving this problem:

Show that $p \to q \equiv \neg p ∨ q$ using a truth table.

I'm lost. Any help would be appreciated. I don't known how to start.

答:
Seems like you are starting with Logic for the first time, the idea is to prove that when one of the propositions is true then the other one as well, and when one is false the other one as well.

So we start by seeing that  as proposition $\neg p ∨ q$ is false only when both of components is false, in case one of them is true then it is automatically true. Similar $p \to q$ is false only when the premise or hypothesis is true but the conclusion is not, in any other case it is true.

We can see this reasoning in a table:

$$\begin{array}{r|r|r|r} p& q & \neg p & p\to q & \neg p \vee q \\
 T& T & F & T & T \\
  T& F & F & F & F\\
 F& T & T & T & T \\
 F& F & T & T & T
\end{array} $$

As you can see now, both propositions have the same values of truth, then they are equivalent.


答:
Given a linear transformation $T:V\to V$, if we choose a basis $\mathcal{B}$ for $V$ we get an induced matrix $\Phi_{\mathcal{B}}:\Bbb{R}^n\to \Bbb{R}^n$ representing the linear transformation with respect to this basis. Given another choice of basis $\mathcal{B}'$, there exists a change of basis matrix $P:\Bbb{R}^n\to \Bbb{R}^n$ such that $\Phi_{\cal{B}}=P^{-1}\Phi_{\cal{B}'}P$. Hence, in light of this fact about determinants of similar matrices, we get
$$
\det \Phi_{\cal{B}}=\det(P^{-1}\Phi_{\mathcal{B}'}P)=\det\Phi_{\mathcal{B}'}.
$$
Remark: You can think of choosing a basis as choosing coordinates on $V$ (like you might do for a surface in calculus). Then the point is that we can define the determinant by choosing any coordinate system and writing the matrix representation of $T$ in coordinates, then computing there. The fact above says that this is independent of coordinate system and hence well defined as an invariant of the transformation $T$. The same applies for the trace of a map which you will probably encounter later if you haven't yet.

Title: $Φ : V → V$ all such linear mappings have same determinant, what does it mean?

Body:
In the Book that I'm reading (Mathematics for Machine Learning), the following para is given, while listing the properties of a matrix determinant:

Similar matrices (Definition 2.22) possess the same determinant.
Therefore, for a linear mapping $Φ : V → V$ all transformation matrices
$A_Φ$ of $Φ$ have the same determinant. Thus, the determinant is invariant
to the choice of basis of a linear mapping.

I know that matrices $A$ and $B$ are similar if they satisfy $B=C^{-1}AC$.
I can prove that determinants of such $A$ and $B$ are equal using other properties of a determinant.

But beyond that I don't understand what this paragraph is saying. I can understand all matrices $Y$ such that $Y=X^{-1}AX$ have the same determinant as $A$, for varying $X$s.

But how do I connect this to linear mappings of the form $Φ : V → V$. What does $Φ : V → V$ mean here? Maybe someone can give me an example.

EDIT:
This video is pretty basic, but it helped me understand better
https://www.youtube.com/watch?v=s4c5LQ5a4ek


问:
How to show that every complex matrix with orthonormal columns can be supplemented into an unitary matrix?

Show that every matrix $A \in M_{n,k}(\mathbb{C})$ whose columns are orthonormal vectors in $M_{n1}(\mathbb{C})$ can be supplemented with additional n-k columns to an unitary matrix $U \in M_{n}(\mathbb{C})$

$A = \begin{bmatrix} a_{11} & a_{12} & ... & a_{1k}
\\ a_{21} & a_{22} & ... & a_{2k}
\\ ... & ... & ... & ...
\\ a_{n1} & _{n2} & ... & a_{nk} \end{bmatrix}$

Now, I can mark the columns with $c[i]$ for example where

$c[i] = \begin{bmatrix} a_{1i} \\ a_{2i} \\ ... \\ a_{ni}  \end{bmatrix} ; i \in {1, ..., k}$

We know that $|| c[i] || = \langle c[i], c[i] \rangle = 1$ and $\langle c[i], c[j] \rangle = 0 \space \forall i \ne j$

Since there is an bijection between the spaces $M_{n,1}(\mathbb{C})$ and $C^n$ we can view the set $\{ c[1], c[2], ... , c[k] \}$ as an orthonormal set in the space $C^n$ and by applying the Gram-Schmidt process we can get an orthonormal base for $C^n ... \{ c[1], c[2], ... , c[k], v[k+1], v[k+2], ... , v[n] \}$

And by supplementing the vectors $v[k+1], ..., v[n]$ as columns to the original matrix $A$ we get:

$A' = \begin{bmatrix} a_{11} & a_{12} & ... & a_{1k} & v_{1,k+1} & ... & v_{1n}
\\ a_{21} & a_{22} & ... & a_{2k}  & v_{2,k+1} & ... & v_{2n}
\\ a_{31} & a_{32} & ... & a_{3k}  & v_{3,k+1} & ... & v_{3n}
\\ a_{41} & a_{42} & ... & a_{4k}  & v_{4,k+1} & ... & v_{4n}
\\ ... & ... & ... & ... & ... & ... & ...
\\ a_{n1} & _{n2} & ... & a_{nk}  & v_{n,k+1} & ... & v_{nn} \end{bmatrix}$

$(A')^* = \begin{bmatrix} \overline{a_{11}} & \overline{a_{21}} & ... & \overline{a_{k1}} & \overline{v_{k+1,1}} & ... & \overline{v_{n1}}
\\ \overline{a_{12}} & \overline{a_{22}} & ... & \overline{a_{k2}}  & \overline{v_{k+1,2}} & ... & \overline{v_{n2}}
\\ \overline{a_{13}} & \overline{a_{23}} & ... & \overline{a_{k3}}  & \overline{v_{k+1,3}} & ... & \overline{v_{n3}}
\\ \overline{a_{14}} & \overline{a_{24}} & ... & \overline{a_{k4}}  & \overline{v_{k+1,4}} & ... & \overline{v_{n4}}
\\ ... & ... & ... & ... & ... & ... & ...
\\ \overline{a_{1n}} & \overline{a_{2n}} & ... & \overline{a_{kn}}  & \overline{v_{k+1,n}} & ... & \overline{v_{nn}} \end{bmatrix}$

And by multiplying the matrices, we notice that we can write the elements of the matrix as:

$(A')^*A' = \begin{bmatrix}
\sum_{i=1}^na_{i1}\overline{a_{i1}} & \sum_{i=1}^na_{i1}\overline{a_{i2}} & ... & \sum_{i=1}^na_{i1}\overline{v_{in}}
\\ \sum_{i=1}^na_{i2}\overline{a_{i1}} & \sum_{i=1}^na_{i2}\overline{a_{i2}} & ... & \sum_{i=1}^na_{i2}\overline{v_{in}}
\\ ... & ... & ... & ...
\\ \sum_{i=1}^na_{in}\overline{a_{i1}} & \sum_{i=2}^na_{in}\overline{a_{i2}} & ... & \sum_{i=1}^na_{in}\overline{v_{in}}
\end{bmatrix}$

The standard inner product for $\mathbb{C}^n$ is $\sum_{i=1}^n \alpha_i \overline{\alpha_i}$ and since the elements of the matrix are summations which consist of multiplications of elements of the corresponding columns which form vectors which are part of an orthonormal base of $C^n$, we can apply the standard inner product for the summations/elements of the matrix $\implies (A')^*A' = I $

But the following multiplication remains:

$A'(A')^* = \begin{bmatrix}
\sum_{i=1}^na_{1i}\overline{a_{1i}} & \sum_{i=1}^na_{1i}\overline{a_{2i}} & ... & \sum_{i=1}^na_{1i}\overline{v_{ni}}
\\ \sum_{i=1}^na_{2i}\overline{a_{1i}} & \sum_{i=2}^na_{2i}\overline{a_{2i}} & ... & \sum_{i=1}^na_{2i}\overline{v_{ni}}
\\ ... & ... & ... & ...
\\ \sum_{i=1}^na_{ni}\overline{a_{1i}} & \sum_{i=2}^na_{ni}\overline{a_{2i}} & ... & \sum_{i=1}^na_{ni}\overline{v_{ni}}
\end{bmatrix}$

The resulting matrix consists of summations of multiplications of the corresponding elements of the rows of the original matrix. But since we are dealing with a matrix, we know that the number of linear independent columns equals the number of linear independent rows and conclude that the rows of the matrix also form vectors which on the other hand form a basis for $C^n \implies A'(A')^* = I$

Since $A'(A')^* = (A')^*A' = I$ we can take $ U = A' $. The statement follows.

Is my reasoning correct ?
If yes, do I need to add some details/be more formal in some places?
If no, it would be great if you could give me some hints.

答: Seems about right to me. The point is that a matrix is unitary if and only if its columns form an orthonormal basis with respect to the complex Euclidean inner product. In fact, if $U = [v_1 | \cdots | v_n]$ then $(UU^\dagger)_{ij} = \langle v_i, v_j \rangle$ (or $U^\dagger U$ depending on your convention). Thus your question really boils down to: can any orthonormal set be extended to an orthonormal basis (in finite dimensions)? Yes, by Gram-Schmidt, as you indicated.</p>


答:
Given an abstract finite dimensional (real) vector space $V$, a linear transformation $\Phi:V\to V$ is a map such that for any $\lambda\in\mathbb{R}$, and any $v,w\in V$:
$$
\Phi(v+w)=\Phi(v)+\Phi(w),\quad \Phi(\lambda v)=\lambda\Phi(v)
$$

Note that in this definition, no matrix is mentioned and $V$ is not necessarily $\mathbb{R}^n$ (although $V$ is isomorphic to $\mathbb{R}^n$).

If you choose an ordered basis $\beta$ for $V$, then $\Phi$ can be represented as an $n\times n$ matrix $B=[\Phi]_{\beta}$. If you choose another ordered basis $\alpha$ for $V$, then $\Phi$ can be represented as another $n\times n$ matrix $A=[\Phi]_{\alpha}$. In linear algebra, it is known that the two matrices $A$ and $B$ are "similar", and thus have the same determinant.

Since the determinant is invariant to the choice of basis, we can talk about the determinant of a linear mapping.


问:
Intersection between two parametric cubic functions?
If I have two parametric cubic functions in 2D space defined by
$$\vec r(t) = \vec A_1t^3 + \vec B_1t^2 + \vec C_1t + \vec D_1$$
and
$$\vec q(t) = \vec A_2t^3 + \vec B_2t^2 + \vec C_2t + \vec D_2$$
how can I find all the points of intersection $$\vec P_0, \vec P_1, ..., \vec P_n$$ between them?

答: Look at these notes by Tom Sederberg. Chapter 7 is all about intersecting polynomial curves.</p>


问:
Checking if a given matrix is orthogonal
Given an orthogonal matrix

$$A=
    \begin{pmatrix}
    1 & 1 & 1 &1 \\
    1 & 1 & -1 &-1\\
    1 & -1 & 0 &0\\
    0 &0&1&-1
    \end{pmatrix},
$$

then values of $[AA^{T}]^{-1}$ is ?

My question is how come it is an orthogonal matrix $1^2+1^2+1^2+1^2\ne1$

Sum of squares of rows should be one isnt it?

答:
This is going to depend on your definitions.  While most people require the rows to be orthonormal, some people will say that a matrix with orthogonal rows is "orthogonal."  This matrix falls into that category.  If you divide all the rows by their lengths, you will get an actual orthogonal matrix.

It is worth noticing that the columns here are NOT orthogonal, but they would be if you divide the rows by their lengths.  Right now, you only have that $AA^T$ is diagonal.  After normalizing, you would have that $AA^T=I$.


问:
In how many ways can a necklace with $18$ different beads be cut into two necklaces with at least six beads on each?
I have a circular necklace with $18$ beads on it. All the beads are different. Making two cuts with a pair of scissors, I can divide the necklace into two strings of beads. If I want each string to have at least $6$ beads, how many different pairs of strings can I make?</p>

答: There are $18$ possible positions for the first cut. After we make the first cut, we have a string of $18$ beads in a row. There are $7$ places we can make the second cut (it can be after the $6^{\rm th}$, $7^{\rm th}$, $8^{\rm th}$, $9^{\rm th}$, $10^{\rm th}$, $11^{\rm th}$, or $12^{\rm th}$ bead). So, we have $18\cdot 7$ choices for the two cuts. However, it doesn't matter which cut we made first and which cut we made second, so we've counted every possible end result twice. Thus, the number of pairs of strings we can make is $(18\cdot 7)/2$, which is $\boxed{63}$.


问:
The deep reason why $\int \frac{1}{x}\operatorname{d}x$ is a transcendental function ($\log$)
In general, the indefinite integral of $x^n$  has power $n+1$.  This is the standard power rule.  Why does it "break" for $n=-1$?  In other words, the derivative rule $$\frac{d}{dx} x^{n} = nx^{n-1}$$ fails to hold for $n=0$.
Is there some deep reason for this discontinuity?</p>

答:
For me, the key observation has always been that $\int \frac{dx}{x}$ is "scale-free" in the following sense.  Suppose that you perform a scaling operation on $x$ by defining $u=ax$ for some constant $a$.  How does the integral change?  You get:
$$ \int \frac{dx}{x} = \int \frac{\frac{du}{a}}{\frac{u}{a}} = \int \frac{du}{u} $$
It doesn't change!  This would not be true for any other power of $x$, since the numerator and denominator would not cancel out perfectly.

This is fundamentally the same observation as the highest-voted answer by @Anonymous above, and there are hints of it in the other answers as well.  But I think that this way of explaining it is much more concise.

In physical terms, applying a scaling factor could represent measuring $x$ in feet rather than meters, for example.  It's unusual that the result does not depend at all on the units that you use, and this is ultimately the reason why the logarithm has all of its interesting properties, such as $\log ab = \log a + \log b$.

(For those who are interested, the term "scale-free" is a buzzword in statistical physics to describe some similar behavior in physical systems near a phase transition.  I have intentionally misused it here, because in physics you typically don't care about creating a constant factor when doing the scaling, so all power laws in physics are "scale-free," but here we are talking about strict equality.)


答:
Suppose $e$ is not a cut edge. Then $G\setminus e$ is connected. Now, considering any spanning tree $T$ of $G\setminus e$ we see that $T$ is a spanning tree of $G$ as well.

Now let $e$ be a cut edge and let $T$ be a spanning tree of $G$ with $e\notin T$. Then $T$ must be a spanning tree of a disconnected graph, a contradiction.

Title: Prove that a cut edge is in every spanning tree of a graph

Body:
Given a simple and connected graph $G = (V,E)$, and an edge $e \in E$. Prove:

$e$ is a cut edge if and only if  $e$ is in every spanning tree of $G$.

I have been thinking about this question for a long time and have made no progress.


问:
Maximum Degree of a Vertex in the Square of a Graph
I'm stuck on the following problem:

Given a simple graph with no loops or multiple edges, $G$, with $e$ edges, show that the vertex with the largest degree in $G^2$ (number of edges between $v_i$ and $v_j$ is the number of walks of length $2$ connecting $v_i$ and $v_j$), is at most $2e$.

How do I approach such a problem? I've been trying for a while and can't seem to figure out the issue.

答:
Hint: the degree of $v$ in $G^2$ is the amount of vertices $w$ which are connected to $v$ via a length $2$ path. Your task is to show that this quantity is bounded above by $2e$, regardless of $v$. The Handshake Lemma states that

$$
2e = \sum_{v \in G}d(v).
$$

 As in the comments, fix a length $2$ path $vxy$. Hence $x \in N(v), y \in N(x)$. There are $d(v)$ options for $x$ and at most $d(x)$ options for $z$ (they are $d(x)-1$ since we can't repeat $v$). Hence the amount of possible paths is bounded by $\sum_{x \in N(v)}d(x) \leq \sum_{x \in G}d(x) = 2e$. This in turn bounds the amount of possible $y$, as desired.


问:
Reducing fractions?
I want to reduce the two following fractions:

$$
\frac{2x + 2y}{x + y}
$$

$$
\frac{3ab^2}{12ab}
$$

I fully understand the concept of reduce fractions of this type:

$$
\frac{15}{20}
$$

but i do not know what steps to take for reducing fractions like the two above. Anyone that can explain the steps needed, or point me to a website explaining it?

答:
For the first example, note that $2x+2y = 2(x+y)$.

So, $\dfrac{2(x+y)}{x+y}= 2$.
Simple enough?

For the second, $\dfrac{3ab^2}{12ab}$, you can do a similar thing. You can cancel like terms by expanding the numerator as follows:

$3ab^2= 3abb= 3(ab)b$.

So, $3(ab)b / 12ab$ means you cancel "$ab$" and you get $3b / 12$.

Also notice that $3$ and $12$ are both divisible by 3. So divide both numerator and denominator by three to get $\dfrac{3}{3}b$ which is $b$, and $12/3$ which is $4$.

So the answer is $b/4$.

Hope this helped! :)

Title: Reducing fractions?


问:
Proof of the formula $\chi=2-2g$ for closed orientable surfaces
Is there any simple proof for the formula $\chi=2-2g$, where $\chi$ is the Euler characteristic and $g$ the genus?  This is likely to be answered before, but I cannot find it.
I use the definitions $\chi = \sum (-1)^k \text{rk} (H_k(X;\mathbb Z))$ and $g$ the maximum number of cuttings along non-intersecting closed simple curves without disconnecting the space $X$.

At the moment, the only method I can think of is the following steps:

Prove the classification theorem that every orientable closed surface is $S^2$ or a sum of tori.

Compute the homology groups $H_n(X;\mathbb Z)$ using 1.

Plug in the definition $\chi = \sum (-1)^k \text{rk} (H_k(X;\mathbb Z))$ to calculate the Euler characteristic.

This seems quite long and could take 20 pages.

Question: Is there a more direct way to prove $\chi=2-2g$ without going through this long process?

答:
I know this is a very old question so you may have found your answer already but will share this anyway since I think its quite a nice idea.

This does still require you to first prove that any surface is the sum of tori but does not require any homology at all.

I think you can see that its true using the poincare-hopf theorem though this is my own idea so it may have errors. Also this is very visual and non rigorous so apologies of its not well explained.

Take your surface (make it a torus for now) then take a plane with a vortex vector field on it spinning around the origin. Place the torus in $\mathbb{R}^3$ with the hole pointing along the z axis. Then sweep the plane perpendicular to the x axis with the origin on the x axis. As it sweeps over the torus we look at the intersection of the plane with the torus and imagine leaving the vector field behind then projecting each vector onto the tangent plane at its base thus defining a vector field on the torus.

Now we look at where the zeros of the vector field are. Obviously there are 2 index 1 zeros, one as the plane first hits the torus and one as the plane leaves. However while the plane is passing along the hole the intersection will split into 2 disjoint circles, and at the exact point of splitting i.e. when the intersection looks like a figure 8 we will get another 0. If you draw it, it should be clear that this is zero has index -1.

Clearly we get another similar zero when the intersection merges back into 1 circle on the other side.

So in total you have $\sum(index)= 1-1-1+1=0$ this is expected since $\chi(T^2)=0$. But try this now with an n hole torus S it should be clear that for every hole you get 2 zeros of index -1 and then 2 of index 1 (when the plane enters and exits the surface). Therefore since g is the number of holes $\chi(S)=\sum(index)=2-2g$.


问:
The Principle of inclusion and exclusion to find probability
A 5-card hard is dealt from a standard deck of 52 cards. Find the Probability of at least 1 heart and 1 spade is among the 5 cards, Using Inclusion and exclusion.
So far I have 52C5-2*47C5 (47C5 from the ways to have no spades and no hearts) but from here I am in desperate need of help.</p>

答:
I see that you are trying to count selections with at least 1 from 13 hearts and at least 1 from 13 spades, by counting the complement for obtaining no hearts or no spades. However, you should be using the Principle of Inclusion and Exclusion. Also your card counts were a little off.

Here's the correction.

$$\def\ch#1#2{{^{#1}\mathrm C_{#2}}}\begin{align}\nu(N_\heartsuit{\geqslant}1\cap N_\spadesuit{\geqslant}1)=& \nu(\Omega)-\nu(N_\heartsuit{=}0\cup N_\spadesuit{=}0)\\[2ex]=&~\nu(\Omega)-\nu(N_\heartsuit{=}0)-\nu(N_\spadesuit{=}0)+\nu(N_{\heartsuit}{=}0\cap N_\spadesuit{=}0)
\\[2ex]=&~\ch{52}{5}-2\cdot\ch{(52-13)}{5}+\ch{(52-13\cdot 2)}5\\[2ex]=&~\ch{52}5 -2 \cdot\ch{39}5+\ch{26}5\end{align}$$


问:
Prove $(SFG)$ is tangent to $(ABC)$

Let the circumcircle of triangle $ABC$ intersect the circle with diameter $AI$ at $S$ where $I$ is the incentre of triangle $ABC$.
Let $M$ be the midpoint of major arc $BAC$. Let the circle $BIC$ intersect circumcircle $SIM$ at $X$. Let $MX$ and $MI$ meet $BC$ at $G$ and $F$ respectively. Prove that circle $SFG$ is tangent to circle $ABC$.
The way I had in mind was to use power of a point wirh a point circle at $S$,  sadly I couldn't find the solution.  Please help..

答:
Here's a solution using some advanced tools of projective geometry. I apologise if this is not at all helpful -- perhaps tomorrow I'll see if there's an easier solution...

Let $N$ be the midpoint of minor arc $BC$. Let the incircle touch sides $BC$, $CA$, $AB$ at $P$, $Q$, $R$.

First, an inversion centre $N$ radius $NB=NI=NC$ shows that $N$, $P$ and $S$ are collinear. By the radical axis theorem, $MS$, $IX$ and $BC$ concur at $K$. Then
$$-1=S(B,C;M,N)=(B,C;K,P),$$
which implies that $QR$ goes through $K$ too.

Consider cyclic quadrilateral $BICX$. By Brokard's theorem, the polar of $K$ in circle $(BIC)$ passes through $Y=BX\cap CI$ and $Z=BI\cap CX$. But the polar of $M$ in circle $(BIC)$ is the line $BC$, so by La Hire's theorem the polar of $K$ goes through $M$ too. Now $(B,C;P,K)=-1$ implies that the polar of $K$ passes through $P$ as well. So $M,P,Y,Z$ are collinear.

Now by the dual of Desargues' involution theorem on $BICX$ and point $M$, we have that $(MB,MC)$, $(MI,MX)$, $(MY,MZ)$ are pairs of an involution. This induces the involution $(B,C)$, $(G,F)$, $(P,P)$ on line $BC$. But the unique involution that swaps $B$, $C$ and fixes $P$ is harmonic conjugation in the points $K$, $P$. Hence $(K,P;F,G)=-1$.

Together with $\angle KSP=90^{\circ}$, this implies that $SP$ is the angle bisector of $\angle FSG$. Since $SP$ goes through $N$, a homothety at $S$ implies the desired result.

Title: Prove $(SFG)$ is tangent to $(ABC)$


问:
Solving $\sqrt{7x-4}-\sqrt{7x-5}=\sqrt{4x-1}-\sqrt{4x-2}$
Where do I start to solve a equation for x like the one below?

$$\sqrt{7x-4}-\sqrt{7x-5}=\sqrt{4x-1}-\sqrt{4x-2}$$

After squaring it, it's too complicated; but there's nothing to factor or to expand?

Ideas?

Body:
Divide $$(7x-4)-(7x-5)=(4x-1)-(4x-2)\ \ \ \ \ (1)$$ by the original equation, which is $$\sqrt{7x-4}-\sqrt{7x-5}=\sqrt{4x-1}-\sqrt{4x-2}\ \ \ \ \ (2)$$  You should get:$$\sqrt{7x-4}+\sqrt{7x-5}=\sqrt{4x-1}+\sqrt{4x-2}\ \ \ \ \ (3)$$Now add (2) and (3), you get:
\begin{align}
2\sqrt{7x-4}&=2\sqrt{4x-1}
\\ \\
7x-4&=4x-1
\\ \\
x&=1
\end{align}


问:
Hard limit - integral with trigonometric function in the exponent yield $\pi$
The function:

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

Looks like it is built out of these bricks, which keep on forever:

The "length" of each one of them is $\sim2 \pi$ meaning that each block sits on $[ 2d \pi , 2 (d+1) \pi]$ for all $d \in \mathbb{N}$

But because I am highly confident these bricks are not the same - because they depend on $x$ and for each value of $x$ the block will be more and more "fixed".

I tried to calculate using a program and found that:

$$ \int_{2 \pi}^{4\pi} \frac{1}{x^{2 \cos (x)} + 1} ~ dx \approx \pi$$

But not quite, it only goes to pi as we increase the bounds (but keep them $2 \pi$ from each other), and so I thought to declare it as:

$$ \lim_{a \to \infty} \int_{2 a \pi}^{2(a+1)\pi} \frac{1}{x^{2 \cos (x)} + 1} ~ dx = \pi$$

One thing to note: even by removing the $2$ from the exponent, keeps the answer close to $\pi$. Meaning that also:

$$ \int_{2 \pi}^{4\pi} \frac{1}{x^{\cos (x)} + 1} ~ dx \approx \pi$$

Is it written correct (mathematically) because I need $a$ to be a whole number, but how can I write it in this limit form?

Is there any way to prove this? a rigorous proof or any simplifications to do here? trigonometric exponents are pretty tricky and hard to deal with, and I've tried all the ways I could think of - $u$-sub, trig-sub, Feynman's method etc..

Renaming $\cos(x) \to t = \frac{t}{1} \\ - \sin(x) dx = dt \\ dx = \frac{1}{\sqrt{1-t^2}} dt$

$$ \int \frac{1}{x^{2 \cos (x)} + 1} ~ dx  = -\int \frac{1}{(x^{2t} +1)} \cdot \frac{1}{\sqrt{1-t^2}} ~ dt $$

The problem is that we are stuck with this $x$ and not $t$.. Any help would be appreciated!

This is not for a specific class, no main "topic" other than integrals and calculus as far as I know.

I unfortunately don't have a reference / contex.

答:
More instructive would be to consider the generalization
$$f(x,y) = (y^{2 \cos x} + 1)^{-1}, \quad x \in [0,2\pi), \quad y \ge 0.$$  In essence, we decouple the periodicity of the exponent and investigate the pointwise limiting behavior of $f$ with respect to increasing values of $y$.

As $y$ gets "large," what does $g_y(x) = y^{2 \cos x}$ look like?  If $\cos x = 0$, then $g_y = 1$ irrespective of $y$.  If $\cos x = -1$, then $\lim_{y \to \infty} g_y = 0$.  And if $\cos x = 1$, then $\lim_{y \to \infty} g_y = \infty$.  In general, we note that $$\lim_{y \to \infty} y^m = \begin{cases} 0, & m < 0 \\ 1, & m = 0 \\ \infty & m > 0 \end{cases}$$  so we can see that the pointwise behavior is $$\lim_{y \to \infty} g_y(x) = \begin{cases} 0, & \pi/2 < x < 3\pi/2 \\ 1, & x \in \{\pi/2, 3\pi/2\} \\ \infty, & 0 \le x < \pi/2 \cup 3\pi/2 < x < 2\pi. \end{cases}$$  Therefore we have for $f(x,y)$ $$\lim_{y \to \infty} f(x,y) = \begin{cases} 1, & \pi/2 < x < 3\pi/2 \\ 1/2, & x \in \{\pi/2, 3\pi/2\} \\ 0, & 0 \le x < \pi/2 \cup 3\pi/2 < x < 2\pi. \end{cases}$$  This suggests that for integers $a$, $$\lim_{a \to \infty} \int_{x=2\pi a}^{2\pi(a+1)} f(x,x) \, dx = (3\pi/2 - \pi/2) = \pi.$$


问:
Symmetric Matrices of Multinormal Distributions
I'm working on some computational mathematics for development in three.js and have been working  with out the first parameters of multinormal distributions in Mathematica 12.2. (They're useful in procedural terrain generation because they have nice organic looking properties when mixed together.)

However, after looking pretty deeply at the properties of the covariance matrix such as its eigenvalues, variance, and standard deviation, I found a trait that reduces the randomness of the outputted distributions. They all seem to tilt bottom-left to top-right. (The variance and standard dev. are listed with each plot)

The properties of the covariance matrix used in a multinormal distribution is that it is positive and symmetric. However, I'm wondering what controls the rotation of the spread, and if a top-left, bottom-right orientation is possible? Or is this a fundamental property of positive and symmetric matrix? I thought I knew a lot of matrices but obviously that was wrong, and there is a gap in my knowledge. Investigating on my own it seems fundamental to its nature but I wanted to do a bit more indepth research because my plan of action for the project depends on the root cause of this orientation behaviour.

Updated 2021-01-29

What I was missing was the concept of correlation coefficients from statistics (described minimally in the accepted answer but it is the answer I used in the project). Understanding the relationship between "x", "y", and "p" does return the expected results. Below "p" is a negative exponent.

答:
$${\bf \Sigma} =
\begin{pmatrix}
\sigma_x^2, \rho \sigma_x \sigma_y  \\ \rho \sigma_x \sigma_y, \sigma_y^2
\end{pmatrix}$$

where $\rho \in [-1,1]$.


答:
Note that

$$ \sqrt{7x-4} - \sqrt{7x-5} = \frac{(7x-4)-(7x-5)}{\sqrt{7x-4} + \sqrt{7x-5}} = \frac{1}{\sqrt{7x-4} + \sqrt{7x-5}} $$

and likewise

$$ \sqrt{4x-1} - \sqrt{4x-2} = \frac{1}{\sqrt{4x-1} + \sqrt{4x-2}}. $$

Plugging these to our equation, we obtain

$$ \sqrt{7x-4} + \sqrt{7x-5} = \sqrt{4x-1} + \sqrt{4x-2}. $$

Thus we have

$$ \sqrt{7x-4} = \sqrt{4x-1} \quad \text{and} \quad \sqrt{7x-5} = \sqrt{4x-2}, $$

yielding $x = 1$.

Title: Solving $\sqrt{7x-4}-\sqrt{7x-5}=\sqrt{4x-1}-\sqrt{4x-2}$

Body:
Where do I start to solve a equation for x like the one below?

$$\sqrt{7x-4}-\sqrt{7x-5}=\sqrt{4x-1}-\sqrt{4x-2}$$

After squaring it, it's too complicated; but there's nothing to factor or to expand?

Ideas?


问:
Example of set that fails to be a field in one specific axiom
We know that a set $F$ together with two binary operations $+$ and $×$ is a field if and only if:

addition $+$ is associative;

addition $+$ is commutative;

addition $+$ has a neutral element $0$;

For every $a∈F$ there exists $b∈F$ such that $a+b=0$.

$×$ is associative;

$×$ is commutative;

$×$ distributes over $+$;

There is an element $1≠0$ such that $a×1=a$ for all $a∈F$;

For each $a∈F$, if $a≠0$ then there exists $x∈F$ such that $a×x=1$.

But I need to find an example of a set or structure that successfully fulfills all the requirements but the number $8$ (meaning there is not an element $1≠0$ such that $a×1=a$ for all $a∈F$). Any thoughts? Thank you in advance.

答: The trivial ring is the obvious example, i.e. $\left\{0\right\}$, with the obvious operations. In this ring, $0=1$, so there is no element not equal to $0$.


答: The set $2\mathbb Z$ of all even integers with the usual addition and multiplication fulfills $1$ up to $7$.

Title: Example of set that fails to be a field in one specific axiom

Body:
We know that a set $F$ together with two binary operations $+$ and $×$ is a field if and only if:

addition $+$ is associative;

addition $+$ is commutative;

addition $+$ has a neutral element $0$;

For every $a∈F$ there exists $b∈F$ such that $a+b=0$.

$×$ is associative;

$×$ is commutative;

$×$ distributes over $+$;

There is an element $1≠0$ such that $a×1=a$ for all $a∈F$;

For each $a∈F$, if $a≠0$ then there exists $x∈F$ such that $a×x=1$.

But I need to find an example of a set or structure that successfully fulfills all the requirements but the number $8$ (meaning there is not an element $1≠0$ such that $a×1=a$ for all $a∈F$). Any thoughts? Thank you in advance.


问:
Prove that $\frac{\sum_{i=1}^n (X_i - \bar{X})^2}{\sigma^2}\,\sim \,\chi_{(n-1)}^2$
If $X_1, \ldots, X_n$ are iid $\sim N(\mu,\sigma^2)$,

$\frac 1 {\sigma^2} \Big((X_1-\bar X)^2 + \cdots + (X_n - \bar X)^2 \Big) \sim \chi^2_{n-1}$.

This has been shown with linear algebra projections, and is the starting point to derive the density of the $t$-Student distribution.

I would really like to get a proof without the advanced math in linear algebra projections (if possible), or with a detailed explanation of the steps followed in the derivation (if the use of projections is essential). Thanks!

答:
Here's a writeup of this exact problem I made a few months ago. This is only the special case for i.i.d $\mathsf{N}(0,1)$, but it should only be off by a few multiplicative constants from the general case.

Derivation of the Chi-Squared distribution

The seek to find the CDF of the Chi-Squared distribution, that is, a function such that for $n$ i.i.d normally distributed CRVs $Z_1,...,Z_n\sim\mathsf{N}(0,1)$
$$\Xi(\chi^2)=\Pr\left(\sum_{i=1}^n {Z_i}^2< \chi^2\right)$$To start we consider the random vector
$$\underline{Z}=(Z_1,...,Z_n)^{\mathrm{T}}$$Which is distributed according to the PDF
$$\varphi(\underline{z})= \frac{1}{\left(\sqrt{2\pi }\right)^n}\exp\left(\frac{-\Vert\underline{z}\Vert^2}{2}\right)$$So then
$$\Pr\left(\Vert\underline{Z}\Vert^2<\chi^2\right)=\int\limits_{\mathcal{B}(0,\chi)}\varphi(\underline{z})\mathrm{d}V$$To do this we assume $n>3$ and convert to hyperspherical coordinates. The $n=1,2$ cases are easy and can be worked through individually.$$z_1=r\cos(\theta_1)~;~z_n=r\prod_{i=1}^{n-1}\sin(\theta_i)$$And for $k\notin \{1,n\}$
$$z_k=r\cos(\theta_k)\prod_{i=1}^{k-1}\sin(\theta_i)$$In this coordinate system the volume element is
$$\mathrm{d}V=r^{n-1}\mathrm{d}r\mathrm{d}\theta_{n-1}\prod_{k=1}^{n-2}\sin^{n-k-1}(\theta_k)\mathrm{d}\theta_k$$Then our integral is
$$\frac{1}{\left(\sqrt{2\pi}\right)^n}\underbrace{\int\limits_0^\pi \cdots\int\limits_0^\pi}_{n-2\text{ of these}}\int\limits_0^{2\pi} \int\limits_0^\chi \exp\left(\frac{-r^2}{2}\right)r^{n-1}\mathrm{d}r\mathrm{d}\theta_{n-1}\prod_{k=1}^{n-2}\sin^{n-k-1}(\theta_k)\mathrm{d}\theta_k$$We can express the $r$ integral in terms of the lower incomplete Gamma function
$$\int_0^\chi \exp\left(\frac{-r^2}{2}\right)r^{n-1}\mathrm{d}r=\left(\sqrt{2}\right)^{n-2}\gamma\left(\frac{n}{2},\frac{\chi^2}{2}\right)$$Hence after integrating with respect to $\theta_{n-1}$
$$\Pr(\Vert\underline{Z}\Vert^2<\chi^2)=\frac{\gamma\left(\frac{n}{2},\frac{\chi^2}{2}\right)}{\sqrt{\pi}^{n-2}} \int\limits_0^\pi \cdots\int\limits_0^\pi\prod_{k=1}^{n-2}\sin^{n-k-1}(\theta_k)\mathrm{d}\theta_k$$From here we make use of the integral identity$$\int_0^\pi \sin^m(x)\mathrm{d}x=\sqrt{\pi}\frac{\Gamma\left(\frac{m}{2}+\frac{1}{2}\right)}{\Gamma\left(\frac{m }{2}+1\right)}\tag1{}$$So
$$\Pr(\Vert\underline{Z}\Vert^2<\chi^2)=\gamma\left(\frac{n}{2},\frac{\chi^2}{2}\right)\prod_{k=1}^{n-2}\frac{\Gamma\left(\frac{n-k}{2}\right)}{\Gamma\left(\frac{n-k+1}{2}\right)}\tag{2}$$The above product (2) simplifies quite nicely, and we get the chi-squared CDF
$$\Xi(x;n):=\Pr(\Vert\underline{Z}\Vert^2<x)=\frac{\gamma\left(\frac{n}{2},\frac{x}{2}\right)}{\Gamma(n/2)}$$Which we can differentiate to get the chi squared PDF:
$$\xi(x;n)=\frac{x^{n/2-1}e^{-x/2}}{2^{n/2}\Gamma(n/2)}$$

PROOFS OF IDENTITIES (1),(2)
We'll start with the integral -
$$I_m=\int_0^\pi \sin^m(x)\mathrm{d}x$$Integrate by parts. Let $u=\sin^{m-1}(x)$, $\mathrm{d}v=\sin(x)\mathrm{d}x$ $$I_m=\int_0^\pi u~\mathrm{d}v=\underbrace{(uv)\big|^\pi_0}_{\sin0=\sin\pi=0}-\int_0^\pi v~\mathrm{d}u$$$$I_m=(m-1)\int_0^\pi\cos^2(x)\sin^{m-2}(x)\mathrm{d}x$$$$\frac{I_m}{m-1}=\int_0^\pi \sin^{m-2}(x)\mathrm{d}x-\int_0^\pi \sin^m(x)\mathrm{d}x$$So we have a recurrence relation$$\frac{I_m}{m-1}+I_m=I_{m-2}\text{ or, equivalently }I_{m+2}=\frac{m+1}{m+2}I_m$$With initial data $I_0=\pi=\Gamma(1/2)^2$, $I_1=2$. One can check that$$I_m=\sqrt{\pi}\frac{\Gamma\left(\frac{m+1}{2}\right)}{\Gamma\left(\frac{m}{2}+1\right)}$$Satisfies the recurrence. You also could get the solution constructively by using the recursive property of Gamma, namely, $\Gamma(z+1)=z\Gamma(z)$ on the denominator and the Legendre duplication formula on the numerator:$$\Gamma(2z)=\frac{2^{2z-1}\Gamma(z)\Gamma(z+1/2)}{\sqrt{\pi}}$$This gives that for integer $m$,
$$\Gamma\left(\frac{m}{2}+\frac{1}{2}\right)=\frac{\sqrt{\pi}(m-1)!}{2^{m-1}\Gamma(m/2)}$$So one could write the solution as $$I_m=\frac{\pi}{2^{m-2}m}\frac{(m-1)!}{\Gamma(m/2)^2}~~|~~m\neq0$$Perhaps this is more readily obtainable given our initial conditions.

Now the proof of the product identity,
$$P_n=\prod_{k=1}^{n-2}\frac{\Gamma\left(\frac{n-k}{2}\right)}{\Gamma\left(\frac{n-k+1}{2}\right)}=\frac{1}{\Gamma(n/2)}$$ We can do some index shifting:$$\prod_{k=1}^{n-2}\frac{\Gamma\left(\frac{n-k}{2}\right)}{\Gamma\left(\frac{n-k+1}{2}\right)}=\prod_{k=2}^{n-1}\frac{\Gamma(k/2)}{\Gamma\left(\frac{k+1}{2}\right)}$$We can index shift again to
$$P_n=\prod_{k=1}^{n-2}\frac{\Gamma\left(\frac{k+1}{2}\right)}{\Gamma\left(\frac{k+2}{2}\right)}=\frac{\prod_{i=2}^{n-1}\Gamma(i/2)}{\prod_{j=3}^{n}\Gamma(j/2)}$$Expanding,
$$P_n=\frac{\Gamma(2/2)\Gamma(3/2)...\Gamma((n-1)/2)}{\Gamma(3/2)...\Gamma((n-1)/2)\Gamma(n/2)}$$Hence
$$P_n=\frac{1}{\Gamma(n/2)}.$$


问:
Nonattacking "queens" on tiled triangles
I'll start with what my question is not. My question is not Nonattacking rooks on a triangular chessboard or The number of spacing $k$ non-attacking towers on the board $\left\{(i,j):1 \le i \le j \le n \right\}$. Those questions are about taking a square grid and making a "triangular chessboard" by removing the squares of a chessboard above the diagonal.

My question is about tiled triangles of the following form (not truncated square grids):

I will define a "queen" on this tiled triangle to attack in the three directions parallel to the triangle on which the queen is located. Below, I have placed a queen on the darkest blue triangle and the light blue triangles are where the queen attacks.

I am interested in the maximal number of non-attacking queens I can place on this tiled triangle. For example, I can imagine placing down queens until every triangle is under attack:

Or, as another example,

My questions are: For a tiled equilateral triangle of side length $n$, what is the maximal number of non-attacking queens I can add to the triangle? How many configurations have the maximal number of non-attacking queens? By side length, I mean for example that the above tiled triangles have a side-length of five.

I think my trouble has been that I'm unsure of a good coordinate system for the tiled triangle. It's two dimensional, but I take out three straight lines corresponding to the horizontal, diagonal, and antidiagonal directions when I add a queen. Further, knowing which horizontal row and diagonal I am in does not typically fix the antidiagonal I am in (there are typically two choice for the antidiagonal).

From inspection, it looks like the maximum number of non-attacking queens on a tiled triangle of sidelength $n$ is $\lceil n/2 \rceil$. Maybe induction on the length of the sides could work? It looks like adding row of triangles at the base of a maximal-number-of-queens odd $n$ triangle gives a maximal number-of-queens even $n+1$ triangle, but I am unsure of how to show this.

答:
[This isn't true.]

Here's a simple proof that the maximal number of non-attacking queens on a tiled triangle of sidelight $n$ is $ \lceil n/2 \rceil $.

Consider the bottom row of triangles that share a common base with the original triangle.

There are $n$ of these "bottom triangles".

Let there be $b$ queens in the bottom row. For each such queen, project it along the diagonal and anti-diagonal to 2 distinct bottom triangles.

Let there be $a$ queens in the bottom row. Clearly $a \leq 1$. If the queen is a bottom triangle, then it projects to 1 bottom triangle. Otherwise, it projects to 2.

Hence $ a + 2b \leq n$ and $ a\leq 1$, so $ 2a + 2b \leq n+1$.

Thus, $a+b \leq \lfloor (n+1) /2 \rfloor = \lceil n/2 \rceil $.

As for the number of such placements, I can bound it above by $n!$, by considering pairs of bottom triangles that define the queen (Each pair of bottom triangles could define 2 queens).

This overcounts though.

I don't have a clear idea how to proceed.


问:
Evaluating $I(z,s)=\int_0^1\int_0^1\left(1-\frac{(1-x)(1-y)}{(1-(1-z)x)(1-(1-z)y)}\right)^{s-2}\,\mathrm dx\mathrm dy$
I came across the following double integral in a statistics problem:

For $z>0$
$$
I(z,s)=\int_0^1\int_0^1\left(1-\frac{(1-x)(1-y)}{(1-(1-z)x)(1-(1-z)y)}\right)^{s-2}\,\mathrm dx\mathrm dy.
$$
All I was interested in was the values of $s$ for which $I$ converges $(s>0)$. That said, the integral looked interesting enough to try and evaluate.  I was successful in evaluating the special case $z=1$ (see below) but could not solve for the general solution. Any ideas on how to solve it?

Case for $z=1$:

We have
$$
I(1,s)=\int_0^1\int_0^1\left(1-(1-x)(1-y)\right)^{s-2}\,\mathrm dx\mathrm dy.
$$
Substituting $(u,v)=(1-(1-x)(1-y),x)$, the resulting integral in $v$ is easily evaluated in terms of the natural logarithm; hence,
$$
I(1,s)=-\int_0^1 u^{s-2}\log(1-u)\,\mathrm du.
$$
Observe that $-u^{s-2}\log(1-u)\sim u^{s-1}+\mathcal O(u^s)$ as $u\nearrow 0$, which tells us the integral should converge for $s>0$. Using the series expansion for the logarithm we then integrate termwise to find
$$
I(1,s)=\sum_{k=1}^\infty\frac{1}{k(k+s-1)}=\frac{H_{s-1}}{s-1},
$$
where $H_{s-1}=\psi(s)+\gamma$ is the generalized harmonic number, $\psi(z)=\partial_z\log\Gamma(z)$ is the digamma function and $\gamma=0.577\dots$ is the Euler–Mascheroni constant.

答:
Assuming that $s$ is an integer $\geq 2$, each term in the binomial expansion of the integrand is separable into a product of integrals over $x$ and $y$. Defining $N=s-2$ and $z'=1-z$ for notational simplicity,
$$
I(1-z',N+2) = \int_{0}^1 dx \int_{0}^1 dy \sum_{k=0}^N \binom{N}{k} \left( -\frac{(1-x)(1-y)}{(1-z'x)(1-z'y)}\right)^k \\
= \sum_{k=0}^N \binom{N}{k} (-1)^k \left\{
\left(
    \int_{0}^1 \left( \frac{1-x}{1-z'x}\right)^k \, dx
\right)^2
\right\}
$$
As a check, when $z=1$ and thus $z'=0$, the inner integral is $1/(1+k)$, and so
$$
I(1,N+2) = \sum_{k=0}^N \binom{N}{k} \frac{(-1)^k}{(1+k)^2}
$$
which sums (as verified by Mathematica) to the same result derived by you. For general $z'$, define
$$
\xi(z) =\int_0^1 \left( \frac{1-x}{1-(1-z)x} \right)^k \, dx
$$
Unfortunately I cannot evaluate this integral, but it can be expanded to linear order around $z=1$ to give
$$
\xi(z) = \frac{1}{1+k} + \frac{k}{(k+1)(k+2)} (1-z) + \mathcal{O}((1-z)^2)
$$
yielding (for integral $s \geq 2$ at least) (edit and with some corrections to converting the expansion of $\xi(z)$ into that of $\xi^2(z)$)
$$
I(z,s) = \frac{\psi(s)+\gamma}{s-1} + 2 \frac{s(\psi(s+1)+\gamma)-2s+1}{s(s-1)} (1-z) + \mathcal{O}((1-z)^2)
$$

ADDITION

As pointed out by the OP, from Mathematica or integral tables (c.f. https://dlmf.nist.gov/15.6#E1) we have
$$
\xi(z) = \int_0^1 \left(\frac{1-x}{1-(1-z)x}\right)^k \, dx = \frac{1}{k+1} F(1,k;k+2;1-z)
$$
(Note that the referenced table represents the identity using the Olver form of the hypergeometric function $\mathbf{F}(a,b;c;z)=F(a,b;c;z)/\Gamma(c)$, and that $F(a,b; c; z)$ is symmetric in $a$ and $b$). This gives the result, for integral $s \geq 2$,
$$
\left. I(z,s)\right|_{s \in \mathbb{N}, s\geq 2} = \sum_{k=0}^{s-2} \binom{s-2}{k} \frac{(-1)^k}{(1+k)^2} F(1,k;k+2,1-z)^2
$$

At least formally, without worrying about convergence, this could be generalized to non-integer $s$ (and $s=0,1$) by extending the series to infinity (with generalized binomial coefficients), but this is probably less useful. The Taylor series of the squared hypergeometric function is
$$
F(1,k;k+2;1-z)^2 = \left( \sum_{n=0}^\infty \frac{(1)_n (k)_n}{(k+2)_n} \frac{(1-z)^n}{n!} \right)^2 \\
= \left( \sum_{n=0}^\infty \frac{k(k+1)}{(k+n)(k+n+1)} (1-z)^n \right)^2 \\
= \sum_{n=0}^\infty \left\{ \sum_{m=0}^n \frac{k(k+1)}{(k+m)(k+m+1)} \frac{k(k+1)}{(k+n-m)(k+n-m+1)} \right\} (1-z)^n
$$
This inner sum is somewhat messy, but can be broken up with partial fractions to yield
$$
\sum_{n=0}^\infty \frac{2k^2(k+1)^2}{2k+n+1} \left(  \frac{H_{k+n}-H_{k-1}}{2k+n} - \frac{H_{k+n+1}-H_k}{2k+n+2}  \right) (1-z)^n
$$
Substituting into the expression for $I(s,z)$ and exchanging the order of summation, the Taylor series $I(z,s) = \sum_{n=0}^\infty C_n (1-z)^n$ has coefficients
$$
C_n = \sum_{k=0}^{\infty} \binom{s-2}{k} \frac{2k^2 (-1)^k}{2k+n+1} \left(  \frac{H_{k+n}-H_{k-1}}{2k+n} - \frac{H_{k+n+1}-H_k}{2k+n+2}  \right)
$$
This reduces to the special cases above for $n=0,1$, e.g.
$$
C_0 = \sum_{k=0}^{\infty} \binom{s-2}{k} \frac{2k^2 (-1)^k}{2k+1} \left(  \frac{1}{2k^2} - \frac{1}{2(k+1)^2}  \right) = \sum_{k=0}^{\infty} \binom{s-2}{k} \frac{k^2 (-1)^k}{(k+1)^2} \\
C_1 = \sum_{k=0}^{\infty} \binom{s-2}{k} \frac{2k^2 (-1)^k}{2k+2} \left(  \frac{1}{k(k+1)} - \frac{1}{(k+1)(k+2)}  \right) = \sum_{k=0}^{\infty} \binom{s-2}{k} \frac{2k (-1)^k}{(k+1)^2(k+2)}
$$
and so on.


答:
Maybe helpful, I don't really know, sometimes a different perspective sparks something. If you center the integration with the changes of variable
$x = (1 + u)/2$, $y = (1 + v)/2$ the integral becomes

\begin{equation}
z^{s - 2}\int_{-1}^{1}\int_{-1}^{1}
\left( \frac{u  v z - 2uv + uz + vz + z + 2}{(uz -u + z +1)(vz - v + z + 1)}\right)^{s-2}\,du\;dv.
\end{equation}
For a given $z$ the denominator vanishes when either
$u$ or $v$ equal $(1 + z)/(1 - z)$.

Interesting stuff!

Added fun fact

If computer algebra systems are to be trusted we have
\begin{equation}
I(z, 3) =
{\left(z^{2} + z \log z - 3 \, z + 2\right)} {\left(z - \log z - 1\right)} z
{\left(z - 1\right)}^{-4}.
\end{equation}
I haven't checked if that matches with what we've seen so far, but for $z$ near $1$ this looks like $\frac{3}{4} + \frac{1}{6}(z - 1) - \frac{1}{9}(z - 1)^2 + \cdots$.

As you work your way higher there seems to be a pattern of the form
\begin{equation}
I(z, n) = (1 - z)^{-2n + 2}z \left( A_n^{(2n-1)}(z) + B_n^{(2n-3)}(z)\log^2 z + C_n^{(2n-3)}(z)\log z \right)
\end{equation}
where $A_m^{(k)}(z)$, $B_m^{(k)}(z)$, and $C_m^{(k)}(z)$ are polynomials of degree $k$ in $z$. Would be fun to see if a three-term recurrence relation for them is hiding somewhere.

Title: Evaluating $I(z,s)=\int_0^1\int_0^1\left(1-\frac{(1-x)(1-y)}{(1-(1-z)x)(1-(1-z)y)}\right)^{s-2}\,\mathrm dx\mathrm dy$

Body:
I came across the following double integral in a statistics problem:

For $z>0$
$$
I(z,s)=\int_0^1\int_0^1\left(1-\frac{(1-x)(1-y)}{(1-(1-z)x)(1-(1-z)y)}\right)^{s-2}\,\mathrm dx\mathrm dy.
$$
All I was interested in was the values of $s$ for which $I$ converges $(s>0)$. That said, the integral looked interesting enough to try and evaluate.  I was successful in evaluating the special case $z=1$ (see below) but could not solve for the general solution. Any ideas on how to solve it?

Case for $z=1$:

We have
$$
I(1,s)=\int_0^1\int_0^1\left(1-(1-x)(1-y)\right)^{s-2}\,\mathrm dx\mathrm dy.
$$
Substituting $(u,v)=(1-(1-x)(1-y),x)$, the resulting integral in $v$ is easily evaluated in terms of the natural logarithm; hence,
$$
I(1,s)=-\int_0^1 u^{s-2}\log(1-u)\,\mathrm du.
$$
Observe that $-u^{s-2}\log(1-u)\sim u^{s-1}+\mathcal O(u^s)$ as $u\nearrow 0$, which tells us the integral should converge for $s>0$. Using the series expansion for the logarithm we then integrate termwise to find
$$
I(1,s)=\sum_{k=1}^\infty\frac{1}{k(k+s-1)}=\frac{H_{s-1}}{s-1},
$$
where $H_{s-1}=\psi(s)+\gamma$ is the generalized harmonic number, $\psi(z)=\partial_z\log\Gamma(z)$ is the digamma function and $\gamma=0.577\dots$ is the Euler–Mascheroni constant.