问:
Jacobians meaning geometrically
When tackling double and triple integrals, we always use the Jacobian matrix when changing variables, such as to polar coordinates.

What actually is the Jacobian and what does it represent or mean geometrically? Also what does it represent geometrically in terms of infinitesimal areas $$ \delta x \delta y $$.

答:
The Jacobian matrix of a differentiable function $f\colon \mathbb{R}^m \to \mathbb{R}^n$ at a point $p \in \mathbb{R}^m$ is the matrix representing the linear transformation which is the best linear approximation to $f(x) - f(p)$ for $x \in \mathbb{R}^m$ near $p$. More precisely, if $J_p$ is the Jacobian matrix of $f$ at $p$, then we have
$$f(x) = f(p) + J_p(x - p) + o(x - p),$$
where the last term is little-o notation and means a function that approaches the zero vector (note that both sides of the above equation take values in $\mathbb{R}^n$) faster than linearly as $x \to p$. (This is a higher-dimensional analogue of the constant and linear terms of the Taylor series.) Stated without asymptotic notation,
$$\lim_{x \to p} \frac{\lVert f(x) - f(p) - J_p(x - p) \rVert}{\lVert x - p \rVert} = 0,$$
where $\lVert v \rVert$ is the Euclidean distance between a vector $v$ and the zero vector.

Geometrically, the graph of $f$ is an $m$-dimensional subset of $\mathbb{R}^m \times \mathbb{R}^n$, and the graph of $J_p$ is the tangent space to the graph of $f$ at the point $(p, f(p)) \in \mathbb{R}^m \times \mathbb{R}^n$.

The reason why the determinant of the Jacobian shows up when you do a change of coordinates is that the absolute value of the determinant of a linear transformation is the factor by which the transformation multiplies volumes. (For example, if you scale by a factor of $2$ in every direction in $\mathbb{R}^3$, then this multiplies volumes by a factor of $8$. If you apply a rigid motion of Euclidean space, it doesn't change volumes at all, and indeed, rigid motions have determinant $\pm 1$.) So the determinant of the Jacobian $J_p$ tells you the factor by which volumes are multiplied by $f$ locally at $p$ (which may vary with $p$).

Title: Jacobians meaning geometrically


问:
Intuition for Brownian motion time-inversion formula
Background: I am studying a course in Stochastic Financial Models, and have been introduced to Brownian motions (BMs) and Gaussian processes.

The BM time-inversion formula states that if $W_t$ is a BM, then $$B_t := tW_{1/t}$$ is also a BM.

I know how to prove the result, through Gaussian processes, similar to the answer here.

However, the proof is unenlightening and does not offer any intuition. In fact, my course lecturer does not offer any intuition either.

Question: Please provide some intuition why the formula “should be” true.

Example of my intuitions for another formula:

“Scaling” formula: if $W_t$ is a BM, then $cW_{t/c^2}$ is also a BM.

Intuition: picture time on the $x$ axis and the motion on the $y$ axis. BMs “don’t go that far up or down”, i.e. their standard deviation is “only” sd$(W_t)=\sqrt t$, so to compensate for the factor of $c$ stretch in the $y$ axis, we have to stretch by a factor of $c^2$ in the $x$ axis.

答:
The idea is that we're essentially turning the $t$-axis inside out around the point $t=1$, i.e. we're taking $B_t$ for $1 \le t < \infty$ and compressing it into the interval $[0,1]$, and taking $B_t$ for $0 < t \le 1$ and stretching it out to the interval $[1,\infty)$.  This is sort of like the reason that $\sin(1/x)$ is discontinuous at $x=0$: we're taking all of the infinitely many oscillations between $-1$ and $1$ that $\sin(x)$ experiences on $[1,\infty)$ and compressing them down into $[0,1]$.

When we take all the variation in $B_t$ on $[1,\infty)$ and compress it into $[0,1]$, we need to rescale the size of the fluctuations to account for the fact that they're happening on a much smaller time scale.  The amount that we need to rescale by turns out to be a factor of $t$, which is probably easiest to find by making the variance correct, but makes sense: For small $t$, we need to compress $B_{1/t}$ by a lot because it came from farther along originally in the Brownian motion path and hence had more time to fluctuate.

Similarly, when we take the fluctuations in $B_t$ on $[0,1]$ and stretch them out to $[1,\infty)$ we need to rescale to make them larger.  The intuition for the factor of $t$ is essentially the same: For large $t$, we need to stretch $B_{1/t}$ by a lot because it originally came from close to $0$ and hence didn't have time to fluctuate much, so we need to amplify the fluctuations it did have by more.

ContentLicense: CC BY-SA 4.0


问:
regular expression
I would like to write the regular expression for the set of
              all binary strings where
              there are no three consecutive 0's.

The following strings are part of the language:

a) strings of the form $1,11,111,...$

b) strings $\{ 0, 01,011,0111,...\}$

d) strings $\{00, 001,0011,00111,..\}$

So,the regular expression is:
$$ (1+01+001)^*$$

Is this a correct regular expression for the given language?

答:
The correct expression would be
$$ (1+01+001)^*(\epsilon+0+00)$$
so that it also accepts words ending in up to two $0$s.</p>

ContentLicense: CC BY-SA 3.0

Title: regular expression


问:
Do you find this proof convincing?
I must show that the function $f:\mathbb R^n \to \mathbb R$ defined as $ f(\mathbf x)=\sum_{j=1}^{k} \Vert \mathbf x -\mathbf{a}_j\Vert^2 $ for fixed $\mathbf a_1,...,\mathbf a_k \in\mathbb R^n$ has a global minimum. Here's my reasoning:

The function $f$ is continuous.
Pick $\delta \gt 0$, by definition closed balls (neighborhoods) are compact; so define $\bar{B}_n(\mathbf 0,n\delta) =\{ \mathbf x\in\mathbb R^n :\Vert\mathbf x\Vert \le n\delta\}$ for $n \in \mathbb N$, and define $f_n:\bar{B}_n(\mathbf 0,n\delta) \to \mathbb R$ as $f_n=\sum_{j=1}^{k} \Vert \mathbf x -\mathbf{a}_j\Vert^2$.

By the maximum/minimum value Theorem there is a $\mathbf y \in \bar{B}_n(\mathbf 0,n\delta) $ such that $f_n(\mathbf y)\le f_n(\mathbf x)$ for all $\mathbf x \in \bar{B}_n(\mathbf 0,n\delta)$. Call $f_n(\mathbf y)= y_n$.

For all $k \in \mathbb N$ we have: $y_{n+1} \le y_n$ because $\bar{B}_k(\mathbf 0,k\delta) \subset \bar{B}_{k+1}(\mathbf 0,(k+1)\delta)$, and $y_k \ge 0$ because $f_k(\mathbf x) \ge 0$ for all $\mathbf x$s in the domain; so the sequence $\{y_n\}$ is bounded below and non increasing, hence it converges, as $n \to \infty$, to a number $y$.

$y$ is the global minimum of $f(\mathbf x)=\sum_{j=1}^{k} \Vert \mathbf x -\mathbf{a}_j\Vert^2$.

I'm not so sure if I can claim that $y$ is indeed the global minimum of the function $f$

EDIT:

I don't seem to be able to complete the proof with my method, also I'm thinking that if I prove there is a closed ball big enough (with all the $\mathbf a_i$s in it) such that every element $\mathbf q$ outside the ball has its $f( \mathbf q)$ greater than the minimum inside then all the work above is superfluous.

People in the comments and in the answers have provided a more efficient strategy:

Let $f(\mathbf a_1)=\sum_{j=2}^{k} \Vert \mathbf a_1 - \mathbf a_j\Vert^2=M$, and define the closed ball $\bar{B}(\mathbf a_1,\sqrt{M})= \{ \mathbf x \in \mathbb R^n: \Vert \mathbf x - \mathbf a_1 \Vert \le \sqrt{M} \}$.

By the minimum value theorem there is some $\mathbf y$ in $\bar{B}(\mathbf a_1,\sqrt{M})$ such that $f(\mathbf y) \le f(\mathbf x)$ for all $\mathbf x$ in $\bar{B}(\mathbf a_1,\sqrt{M})$.

Pick any element $\mathbf q \notin \bar{B}(\mathbf a_1,\sqrt{M})$, which means that $0 \le \sqrt{M} \lt \Vert \mathbf q - \mathbf a_1 \Vert$ and thus $M \lt \Vert \mathbf q - \mathbf a_1 \Vert ^2$.

By definition we have $f(\mathbf y) \le f(\mathbf a_1) = M$, but it's also true that $\Vert \mathbf q - \mathbf a_1 \Vert \le f(\mathbf q)$. Therefore $f(\mathbf y) \lt f(\mathbf q)$ which completes the proof.

答:
Okay, so, alternative answer: for $M >> 1$ and a compact set $K \supset B_M(\mathbf{a}_1)$, you can see that in $K^c$ we have $f \geq \|\mathbf x -\mathbf a_1\|^2 \geq M$.  Now, since $f$ is smooth and

$$
\partial_j f = 2 \cdot \sum_i x_j-a_{ij} = 2\cdot \left(nx_j-\sum_i a_{ij}\right),
$$

the equation $\nabla f = 0$ has one only solution,

$$
x_j = \frac{1}{n}\sum_{i}a_{ij}.
$$

That is, the solution is the baricenter of the points, as expected:

$$
\mathbf b = \frac{1}{n}(\mathbf{a}_1+\cdots +\mathbf{a}_n).
$$

Let $c = f(\mathbf b)$. Now consider some $M > c$ and $K =B_M(\mathbf a_1)$ with $\mathbf b \in K$. Hence in $K^c$ we have $f \geq f(\mathbf{b})$.

Inside $K$ there must exist a minimum, and we have shown it is unique. This has to be a global minima, as we have seen that it is below every other value in $K^c$ (Edit: I have changed the compact set to be exactly $B_M(\mathbf a_1)$. This avoids the problem of the minimum being attained at $\partial K$)

Title: Do you find this proof convincing?


问:
Green function derivation
My professor derived the free-space dyadic Green function from the general Green function:
$$\tag{1}
\stackrel{\leftrightarrow}{\boldsymbol{G}}=\left[\stackrel{\leftrightarrow}{\mathbb{I}}+\frac{1}{k^{2}} \nabla \nabla\right] G_{0}(R)$$
With the scalar Green function:
$$\tag{2}
G_{0}(R)=\frac{e^{i k R}}{4 \pi R}, \quad R \equiv\left|\boldsymbol{r}-\boldsymbol{r}^{\prime}\right|$$
In the derivation, he states the following:
$$\tag{3}
\nabla \hat{\mathbf{R}}=\nabla\left(\frac{\mathbf{R}}{R}\right)=\frac{\nabla \mathbf{R}}{R}+\mathbf{R} \nabla \frac{1}{R}$$
where $\boldsymbol{R}=\boldsymbol{r}-\boldsymbol{r}^{\prime}, \quad \widehat{\boldsymbol{R}}=\frac{\boldsymbol{R}}{R}$. He says we can see from eq. (3) that $\nabla \mathbf{R}=\mathbb{I}$, but I don't see that? Can someone explain how he gets to that conclusion?

答:
Well, what is $\nabla \mathbf{R}$? In 3 space, $\mathbf{R} = \{x-x', y-y',z-z'\}$, then $$\nabla \mathbf{R} =
\begin{bmatrix}
\partial_x \\
\partial_y \\
\partial_z \\
\end{bmatrix} \cdot
[x-x',y-y',z-z']
  $$

Which is a $3 \times 1$ matrix times a $1 \times 3$ matrix which results in a $3 \times 3$ matrix. If you compute this "outer product" you will get the identity matrix.

Title: Green function derivation


问:
Finitely generated flat modules that are not projective
Over left noetherian rings and over semiperfect rings, every finitely generated flat module is projective. What are some examples of finitely generated flat modules that are not projective?

Compare to our question f.g. flat not free where all the answers are f.g. projective not free.

答:
Following Jack's gentle nudging, I'm posting the part of my comments not covered by rschwieb's nice post more prominently as an answer:

In general, this type of question can be answered quickly by consulting Lam's Lectures on Modules and Rings, Springer Graduate Texts, Vol. 189, whose numerous exercises are solved and put into further context by Lam himself in the accompanying text Exercises in Modules and Rings. The section §4E, Finitely generated flat modules of the Lectures contains some sufficient conditions on when f.g. flat modules are projective.

Here's exercise 4.17 with solution, giving Vasconcelos's example of a principal ideal $(a)$ in a commutative ring such that $R/(a)$ is flat but not projective:

This appears as Example 3.2 of Wolmer V. Vasconcelos, On finitely generated flat modules, Trans. Amer. Math. Soc. 138 (1969), 505-512.


问:
Independence vs. Measurability
As someone who is relatively new to measure-theoretic probability, I'm really struggling to get to grips with the concepts of measurability and independence.

If we have a random variable $X$ defined on a probability space $(\Omega,\mathscr{F},\mathbb{P})$ and some sigma algebra $\mathscr{G}\subset\mathscr{F}$, and $X$ is independent of $\mathscr{G}$, then can we deduce that $X$ is not $\mathscr{G}$-measurable?

When I look at the definitions of measurability and independence they seem pretty distinct from one another, but when heuristically considering sigma algebras as "information", it seems like if our random variable is independent of our information (i.e. $X$ independent of $\mathscr{G}$), then we don't have enough information to determine the value of our random variable (i.e. $X$ is not $\mathscr{G}$-measurable).

If someone could please sort out my confusion it would be much appreciated!

答:
At least assuming that $X$ is non-trivial, you are correct that $X$ is independent of $\mathscr G$ implies $X$ is not $\mathscr G$-measurable.  For a proof, suppose to the contrary that $X$ is independent of $\mathscr G$ and $X$ is $\mathscr G$-measurable.  Then $\mathbb P(A \cap B) = \mathbb P(A) \mathbb P(B)$ for all $A \in \sigma(X)$ and $B \in \mathscr G$.  But since $X$ is $\mathscr G$-measurable, $\sigma(X) \subseteq \mathscr G$ so we can choose $B = A$ and hence have $$\mathbb P(A) = \mathbb P(A \cap B) = \mathbb P(A) \mathbb P(B) = \mathbb P(A)^2$$ for all $A \in \sigma(X)$.  But $\mathbb P(A) = \mathbb P(A)^2$ implies $\mathbb P(A) = 0$ or $\mathbb P(A)=1$, so $X$ would have to be trivial.

The other implication is not true, i.e. $X$ is not $\mathscr G$-measurable does not imply $X$ is independent of $\mathscr G$.  You could consider a simple example of tossing a fair coin twice, $X$ is the total number of heads, and $\mathscr G$ is the $\sigma$-algebra generated by the first coin toss.

Title: Independence vs. Measurability


问:
Closeness wrt Hausdorff distance implies closeness of the measure of the two sets
Given two measurable disjoint subsets $A,B\subset\Bbb R^n$, which are surfaces of some dimension $k<n$ (e.g. two curves in $\Bbb R^2$ or in $\Bbb R^3$) of finite $k$-dim Lebesgue measure such that
$$
d_H(A,B)<\epsilon
$$
where $d_H$ is the Hausdorff distance.

Can we deduce that
$$
|\mu_k(A)-\mu_k(B)|<\epsilon
$$
?

答:
Take $k = 1$ and $n=2$. Let $A = [0] \times \{i/N \mid i = 0,1,\ldots, N\}$ and $B = ([0] \times [0,1]) \backslash A$.
Then $d_H(A,B)  \to 0$ as $N \to \infty$, but $A$ has $1$-dimensional Lebesgue measure $0$ and $B$ has $1$.


问:
Derivate as a division of differentials
It's always said that $\frac{df(x)}{dx}$ is the derivate of a function with respect to x and it musn't be understood as a division of $df$ and $dx$.

I've read that it is true that $\frac{df(x)}{dx}$ is the quotient of $df$ and $dx$. But this cannot be generalised for $\frac{d^2f(x)}{dx^2}$.

Could someone explain me why does this happen?

答:
The derivative $\frac{df}{dx}$ is the incremental ratio:
$$\frac{df}{dx}=\frac{f(x+\Delta x)-f(x)}{\Delta x}$$ when $\Delta x\to 0$
The quotient $\frac{d^2 f}{dx^2}$ is the incremental ratio of $\frac{df}{dx}$, that means:
$$\frac{d^2 f}{dx^2}=\frac{d}{dx}\frac{df}{dx}$$
so:
$$\frac{d^2f}{dx^2}=\frac{f(x+2\Delta x)-2f(x+\Delta x)+f(x)}{\Delta x^2}$$ when $\Delta x\to 0$ which still is a quotient or the incremental ratio</p>

Title: Derivate as a division of differentials


问:
Show that either $(x_1,\dots,x_n) = (0,\dots,0)$ is the optimal solution or the LP is unbounded.
\begin{array}{llclccclcl}
\text{maximize}   & c_1x_1    & + & c_2x_2    & + & \dots & + & c_nx_n \\\text{subject to} & a_{11}x_1 & + & a_{12}x_2 & + & \dots & + & a_{1n}x_n & \le & 0\\                  & a_{21}x_1 & + & a_{22}x_2 & + & \dots & + & a_{2n}x_n & \le & 0\\                  &  & \vdots &  & \vdots &  & \vdots & & \vdots &  \\                  & a_{m1}x_1 & + & a_{m2}x_2 & + & \dots & + & a_{mn}x_n & \le & 0\\                  &  &  &  &  & x_1, & \dots, & x_n & \ge & 0 \\
\end{array}

Show that either $(x_1,\dots,x_n) = (0,\dots,0)$ is the optimal solution or the LP is unbounded.

答: If $x=0$ is not optimal, that means that there is an $x \geq 0$ for which $c^Tx > 0$ and $Ax \leq 0$. Then $\alpha x$ ($\alpha \geq 0$) is also feasible, and by $\alpha \to \infty$ the objective can be made arbitrarily large.


问:
Doubt in finding probability of an event from the conditional probability
I am studying about conditional probability and see a problem given as

In the solution$P(B_1)=P(B_2)=P(B_3)=1/3$ is taken.

From set point of view, event $B_1$ can be seen as ${\{R,R,R,R,G,G,G,B,B,B,B,B}\}$ and $R\cap B_1={\{R,R,R,R}\}$.

So $P(R\cap B_1)=|R\cap B_1|/|B_1|=P(R\cap B_1)/P(B_1)$.

Here our sample space is like ${\{R,...,R,G,...,G,B,...,B}\}$.

If we have to calculate P(R), so shouldn't we have to use $P(B_1)=|B_1|/|S|= 12/30=4/10$ instead of 1/3.

Because the formula used in the solution, holds only when we don't change our sample space from ${\{R,...,R,G,...,G,B,...,B}\}$ to ${\{B_1,B_2,B_3}\}$.

Also in the solution probability of choosing red ball is incorrect not equal to  number of red balls divided by this total number of balls. If we use $P(B)=|\text{Number of balls in box B}|/|\text{Total number of balls}|$, then we get $P(R)= 1/3$.

Is the solution incorrect, or my way of thinking is incorrect.
Please clarify the doubt.

答:
Your way of thinking is incorrect.  You are assuming that since the total number of balls in the 3 boxes is 30, each of the balls has a $(1/30)$ chance of being picked.

This is wrong.  In fact, because box 2 contains only 8 balls and box 1 contains 12 balls, each ball in box 2 is 1.5 times more likely to be chosen than a ball in box 1.

To take an extreme case, suppose box 2 had 1 ball (of any color), and boxes 1 and 3 each had 100 balls (of any color).  Then the ball in box 2 would be 100 times more likely to be chosen than any of the individual balls in either box 1 or box 3.

Addendum

Per OP's request.

Responding to his comment/question to Math Lover:

You questioned the formula

$$p(R) ~=~ \sum_{i=1}^3 p(B_i)p(R|B_i).\tag1 $$

Note that I have rearranged the factors in each term, with the $p(B_i)$ factor shown 1st to make the formula easier to intuit.

If I understand your question, you are questioning what would happen if the number of red balls in one of the boxes were to change.

Let's take an extreme case.

Suppose that Box 1 has 100 balls, all blue.

Suppose that Box 2 has 1 ball, which is red.

Suppose that Box 3 has 100 balls, all green.

According to the summation formula, the chance that a red ball would be selected is

$$\left(\frac{1}{3} \times \frac{0}{100}\right) ~+~
\left(\frac{1}{3} \times \frac{1}{1}\right) ~+~
\left(\frac{1}{3} \times \frac{0}{100}\right) = \frac{1}{3}.$$

In fact, in this extreme case, the probability that the ball chosen is red is in fact $\frac{1}{3},$ so the formula continues to be accurate.


问:
**trajectories** of the permutation of $S_6$ , $\sigma = (1 2 356) (345)$
I have been asked to find the trajectories of the permutation of $S_6$  , $\sigma = (1 2 356) (345)$ and I have to write $\sigma$ as the product of  discrete cycles.

Can anyone please tell me what are the meanings of the bold words ?

答:
In this question they call it the orbit, which is I think what's meant: what is the image of $i \in \{1,2,3,4,5,6\}$ under successive applications of $\sigma$? So the orbit of $1$ is $\{1,2,3,4,6\}$ (this is also the orbit of all $i \neq 5$ of course) and the orbit of $5$ is $\{5\}$.

For the second: it's a standard fact that any permutation can be canonically written as a product of cycles such that different cycles use different numbers; normally called disjoint cycles.

If you look at $\sigma$, $1 \to 2 \to 3 \to 4 \to 6 \to 1$ and $5$ is a fixpoint. So we can write $\sigma$ simply as $(1 2 3 4 6)(5)$ (we can omit the $(5)$) and its powers are then easy to write down..

Both tasks are closely related as you can see.


问:
change of basis conflicted definitions
the 2 Definitions I'm talking about are
$P_{B->B'}$ is the change of basis matrix from B to $B'$ such as $B=(x_1,x_2,x_3)$ and $ B'=(x_1',x_2',x_3')$
the two Definitions that I'm talking about are:
$$ \begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix}=P_{B->B'}\begin{pmatrix}x_1'\\x_2'\\x_3'\end{pmatrix}
$$
And
$$
P_{B->B'}=Mat_{B',B}(Id_E)
$$
Let's take an example:
$$
\begin{equation}
    \begin{cases}
     x_1'=x_1+3x_2+4x_3\\x_2'=x_2+2x_3\\x_3'=x_3
    \end{cases}\,.
\end{equation}
$$
$$\begin{pmatrix}x_1'\\x_2'\\x_3'\end{pmatrix}=\begin{bmatrix}1&3&4\\0&1&2\\0&0&1\end{bmatrix}\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix}$$
so
$$
 P_{B->B'}={\begin{bmatrix}1&3&4\\0&1&2\\0&0&1\end{bmatrix}}^{-1}={\begin{bmatrix}1&-3&2\\0&1&-2\\0&0&1\end{bmatrix}}
$$
using the  second definition :
$$ Id_{E}(x_1')=(1,3,4) \hspace{0.25cm} in \hspace{0.25cm} B\\
Id_{E}(x_2')=(0,1,2) \hspace{0.25cm} in \hspace{0.25cm} B\\
Id_{E}(x_3')=(0,0,1) \hspace{0.25cm}in \hspace{0.25cm}B\\
$$so$$ P_{B->B'}=Mat_{B',B}(Id_E)={\begin{bmatrix}1&0&0\\3&1&0\\4&2&1\end{bmatrix}}
$$
Equations source:Wikipedia(in frensh)

答:
Maybe this helps:
As written in the wikipage, let $E$ be a $K$-vector space of dimension $d$. Let us for  simplicity assume that $K$ is either $\mathbb{R}$ or $\mathbb{C}$. Now, the first thing to understand is, that $E$ must not be $\mathbb{R}^d$ nor $\mathbb{C}^d$, that is, the elements in $E$ could be anything as long as the axioms of a vector space are fulfilled. So, for instance $E$ could be the space of polynomials with degree smaller or equal to $d$ over the real numbers. However, one awesome result of linear algebra is that this vector space is still isomorphic to $\mathbb{R}^d$. In other words, you can uniquely identify every polynomial in that vector space with a $d$-tuple of real numbers. However, this identification depends on your choice of bases! But it this identification that makes it possible to still identify every linear map on a finite dimensional vector space with a matrix. Notice that this is not trivial.

Now, to the concrete problem. I think the confusion comes from the fact that you wrote down an equality that isn't defined. But you want the following:
$$
\begin{pmatrix}x_1' & x_2' & x_3'
\end{pmatrix}
=
P_{B'\to B}
\begin{pmatrix}
x_1 & x_2 & x_3
\end{pmatrix}.
$$
Notice, that like this the dimension fit. Suppose that $E=\mathbb{R}^3$ then on
the left there is written a $3\times 3 $ matrix. And, if and only if $P_{B'\to B}$
is a $3\times 3$ matrix then so it the right side a $3 \times 3$ matrix. In your case, that didn't
work, because $\begin{pmatrix} x_1 & x_2 & x_3 \end{pmatrix}^T$ is a $9$ tuple.

Now, the matrix $Mat_{B,B'}(Id_E)$ is constructed by taking that columns to be
the bases elements of $B'$ written in terms of the bases of $B$. So, that in
your case
$$
Mat_{B,B'}(Id_E) =
\begin{pmatrix}
1 & 0 & 0
\\
3 & 1 & 0
\\
4 & 2 & 1
\end{pmatrix}.
$$

You see that then indeed

$$
P_{B' \to B} = Mat_{B,B'}(Id_E)
\quad
\text{or equivalently}
\quad
P_{B\to B'} = Mat_{B',B}(Id_E).
$$

Title: change of basis conflicted definitions


答:
If we want to keep thing clean and formal, then $\frac{df(x)}{dx}$ is very simply not a fraction. Any similarity with the notation of fractions is simply a notational similarity and nothing more. You could just as well denote the same thing by $f'(x)$, and nothing would change mathematically. This is more fully discussed in the question "Is $dy/dx$ not a ratio" alluded to in the comment.

Of course, the story does not end there, because people chose this (slightly misleading) notation with good reason. If you allow yourself some degree of inexactness, or use non-standard analysis, you can make sense of heuristics such as $\frac{df(x)}{dx} = \frac{f(x+dx)-f(x)}{dx}$. If you just pretend that $dx$ is a variable, and pass to the limit $dx \to 0$, then you retrive the standard definition of derivative. In numerical methods, people seem to actually do this with finite $dx$, up to some accuracy.

There is nothing to stop you from doing the same with second derivative (except the same doubts as for the first derivative, of course):
$$ \frac{d^2f(x)}{dx^2} = \frac{d(f(x+dx) - f(x))}{dx^2} = \frac{f(x+2dx)-2f(x+dx)+f(x)}{dx^2}$$
For numerical purpses, this should work, at least in principle. If $f$ is twice differentiable, then in the limit $dx \to 0$ (again, treating the last term as a fraction) you will recover the usual second derivative. If you do non-standard analysis, and take $dx$ infinitesimal, then you recover the second derivative up to infinitesimal error. One difference is perhaps that you have to be careful with interpretation of $d^2 f$.

Later addition: A difference between first and second derivative is that for the first, you have an "if and only if" type of statement. Not only is it true that if $f$ is differentiable then $\frac{df(x)}{dx} = \lim_{dx \to 0} \frac{f(x+dx)-dx}{dx}$, but also $f$ is differentiablee at $x$ if the limit exists. This fails for the second derivative. For instance, if $f(x) = |x|$ then $\frac{f(x+2dx)-2f(x+dx)+f(x)}{dx^2} = 0$ if $dx$ is sufficiently small (depending on $x$), but $f$ is not even once differentiable at $x=0$. I believe it may also fail for once differentiable $f$, but I can't think of an example (translating the ordinary definition would lead to $\lim_{h_1 \to 0} \lim_{h_2 \to 0} \frac{f(x+h_1+h_2) -f(x+h_1)-f(x+h_2) + f(x)}{h_1h_2})$

Title: Derivate as a division of differentials

Body:
It's always said that $\frac{df(x)}{dx}$ is the derivate of a function with respect to x and it musn't be understood as a division of $df$ and $dx$.

I've read that it is true that $\frac{df(x)}{dx}$ is the quotient of $df$ and $dx$. But this cannot be generalised for $\frac{d^2f(x)}{dx^2}$.

Could someone explain me why does this happen?


问:
What is the probability for each number in a set of random numbers to be the highest number in the set?
I have viewed/researched many problems on this website but this is my first question.  I'm a little apprehensive by the length of the question but I have tried to explain it fully including a little of my thought process.  The problem is ...

I have n numbers in a set (N1, N2, N3 ... Nn). For each number I choose a random number (R1, R2, R3 ... Rn) between 1 and the number (e.g. 0 < R1 < N1, 0 < R2 < N2, 0 < R3 < N3 ... 0 < Rn < Nn ).  My quest is to seek the probability for each random number that it should be the largest of the random numbers in the set i.e. P(R1=max(R1 ... Rn)), P(R2=max(R1 ... Rn)) etc. up to P(Rn=max(R1 ... Rn)).

For example and by way of a Monte Carlo simulation (MCS) with a count of 10,000 I get the following ...

N1=300, N2=250, N3=450

R1 is Max(R1,R2,R3) 2576 times i.e. P(R1=Max(R1,R2,R3)) = 0.2576

R2 is Max(R1,R2,R3) 1526 times i.e. P(R2=Max(R1,R2,R3)) = 0.1526

R3 is Max(R1,R2,R3) 5898 times i.e. P(R3=Max(R1,R2,R3)) = 0.5898

Obviously the MCS will produce a slightly different result each time it is run. On 10 runs the MCS returned results of ...

```<code>  N1   N2  N3
</code>```

Min 2504 1499 5854

Ave 2546 1534 5921

Max 2591 1564 5972

Interestingly using some Mathematica code as provided by the Wolfram forum the probabilities were calculated as ...

P(R1=Max(R1,R2,R3))= 0.256173

P(R2=Max(R1,R2,R3))= 0.154321

P(R3=Max(R1,R2,R3))= 0.589506

So it seems there is probably a formula to return the expected result from an infinite number of MCS runs but it's proving difficult to find it!

A helpful Wolfram forumite from another time zone promised to help me after a good night's sleep but unfortunately they deleted my thread before he woke up because I was not seeking Mathematica support!  :(

Anyway if someone could help with a formula or point to where I may find one I would be very grateful.  I need to be able to resolve this using Excel, SQL or VBA because the result needs to link with an existing SQL database so the MCS option is not really suitable.  That was only for testing and is not a viable option because I have c.150,000 number sets with 75% of them having between 5 and 12 numbers to each set and the thought of performing that number of MCS's throws my computer into a cold sweat! :)

Well if you have got this far thank you for reading my question ... it is long but hopefully most of the content is for explanation and not question.

答:
Let ask $p(R_1>R_2)$.
It will be the sum of  all the $R_1$ values multiply by the chance its bigger then $R_2$.

$p(R_1>R_2) = \sum_{i \in [N_1]}\frac{1}{N_1}\cdot \frac{i-1}{N_2}=\frac{1}{N_1\cdot N_2}\cdot \sum_{i\in [N_1]}i-1=\frac{1}{N_1\cdot N_2}\cdot \frac{N_1^2-N_1}{2}=\frac{N_1-1}{2\cdot N_2}$

From here, $P(R_1 > R_2,R_3,...,R_n)=\frac{N_1-1}{2\cdot N_2}\cdot \frac{N_1-1}{2\cdot N_3}\cdot ...\cdot \frac{N_1-1}{2\cdot N_n}$


问:
explaining the derivative of $x^x$
You set the following exercise to your calculus class:

Q1. Differentiate $y(x) = x^x$.

A student submits the following solution:

Let $g(a)=a^x$ and $f(x)=x$.  Then $y(x) = g(f(x))$, so by the chain rule, $y'(x) = f'(x) g'(f(x)) = 1 \cdot x \cdot (x^{x-1}) = x^x$.

How would you explain to the student why their solution is incorrect?

To be clear, I know why this is wrong but am interested in good ways to explain it to undergraduate or high school students.

In this question someone has problems differentiating $x^x$, but they didn't take the approach of my hypothetical student.

答: I would try to explain that the notation $h'$ doesn't just mean the derivative of $h$ but the derivative of $h$ with respect to some particular variable, the variable being determined by the context. In the student's answer, $y'$ is $dy/dx$, but $g'$ is $dg/da$, when what's needed is $dg/dx$. </p>

Title: explaining the derivative of $x^x$


答:
I'll give my example.

The $0$-dimensional measure is just the usual counting measure.  In $\mathbb{R}$, consider $A=\{0\}$ and $B$ a finite set of points in $[-\epsilon,\epsilon]-\{0\}$.  Then $d_H(A,B)\leq\epsilon$ but $\#A=1$ and $\#B$ is arbitrary.

Now you can take product with $[0,1]^k\times\{0\}^{n-k-1}$ to get example for any $k<n$.

Title: Closeness wrt Hausdorff distance implies closeness of the measure of the two sets

Body:
Given two measurable disjoint subsets $A,B\subset\Bbb R^n$, which are surfaces of some dimension $k<n$ (e.g. two curves in $\Bbb R^2$ or in $\Bbb R^3$) of finite $k$-dim Lebesgue measure such that
$$
d_H(A,B)<\epsilon
$$
where $d_H$ is the Hausdorff distance.

Can we deduce that
$$
|\mu_k(A)-\mu_k(B)|<\epsilon
$$
?


问:
Calculate Rate of Change
One kilogram of Polonium radioisotope $Po_{84}^{218}$ reduces to 500g in 3 seconds.
a) Calculate the rate per second at which Polonium is decomposing. b) Calculate the mass to which 1kg of this radioisotope is reduced in 15 seconds.

答:
Assuming the decomposition is exponential(since it is radioactive decomposition), let $N_t = N_0 e^{-\lambda t}$ where $N_t$ is the amount of the radioisotope that is decomposed after $t$ seconds, $N_0$ is the initial amount of the radioisotope and $\lambda$ is a  constant.

It is given that the half-life of the radioisotope is $3$ seconds. Hence $N_3/N_0 = 0.5 = e^{-\lambda 3}$. After taking the natural logarithm of each side and rearranging terms you get:
$$ \lambda =  - \ln (0.5)/3 \approx 0.231$$

The rate of the decomposition can be found by taking the derivative with respect to $t$:

$$\dfrac{d}{dt} N_t = -0.231e^{-0.231t}$$

And, the amount(mass) left after $t = 15$ is $N_{15} = 1000e^{-0.231 \cdot 15} \approx 31.3\ g$ or $0.0313 kg$.

Title: Calculate Rate of Change


答:
My attempt would proceed along the following lines.

You defined $g(a) = a^x$; since the $x$ does not occur as an argument to the function, it is taken to be constant. This is not consistent with differentiating our expression with respect to $x$.

Therefore, the function $g$ should actually admit $x$ as a parameter as well: $g(a,x) = a^x$.

At which point we run into the difficulty of the chain rule for partial differentiation being unavailable (despite this approach giving a very clean, universal solution to differentiating expressions with multiple occurrences of the relevant variable).

I can imagine two reasonable courses of action from here:

Indicate the "standard" high school solution via $x^x = \exp(x\log x)$;

Introduce the more complicated chain rule, illustrating it e.g. by deriving the product rule.

Title: explaining the derivative of $x^x$

Body:
You set the following exercise to your calculus class:

Q1. Differentiate $y(x) = x^x$.

A student submits the following solution:

Let $g(a)=a^x$ and $f(x)=x$.  Then $y(x) = g(f(x))$, so by the chain rule, $y'(x) = f'(x) g'(f(x)) = 1 \cdot x \cdot (x^{x-1}) = x^x$.

How would you explain to the student why their solution is incorrect?

To be clear, I know why this is wrong but am interested in good ways to explain it to undergraduate or high school students.

In this question someone has problems differentiating $x^x$, but they didn't take the approach of my hypothetical student.


问:
Notation of topological space under group action
I met the following notation in a book I was reading which I've never seen before

$$A \times_{G}B$$

Here $G$ is a group, $A,B$ are topological spaces. How do we define this notation?

P.S. I have a guess according to the context but I'm not sure if it is right. In the context there is a group action of $G$ on both $A$ and $B$. Is this space defined to be the orbit space of the product space $A \times B$?

答:
Here $A,B$ are $G$-spaces and
$$A\times_GB=(A\times B)/G$$
is the quotient by the diagonal $G$-action on $A\times B$. To make sense of this you should assume that $G$ acts on both $A,B$ either from the left, or from the right.

On the other hand this notation is most frequently encountered when $A$ is a right $G$-space and $B$ is a left $G$-space. Then $A$ is turned into a left $G$-space by letting $g\cdot a=ag^{-1}$, and the construction is the same. Thus in this case $A\times_GB$ is the quotient of $A\times B$ by the $G$-action
$$g\cdot (a,b)=(ag^{-1},gb).$$


问:
If $c_{n}=b_{n}-a_{n}$ is a geometric sequence with ratio $q_{c}$ then $q_{a}=q_{b}=q_{c}$
Let $a_{n},b_{n}$ be two geometric sequences with ratios $q_{a},q_{b}$
such that $a_{1},b_{1}\neq0$ and also $a_{1}\neq b_{1}$. If $c_{n}=b_{n}-a_{n}$ is a geometric sequence with ratio $q_{c}$
then $q_{a}=q_{b}=q_{c}$

Attempt:

If $c_{n}$ is a geomtric sequence with ratio $q_{c}$ then:
\begin{align*}
 & q_{c}=\frac{c_{n+1}}{c_{n}}=\frac{a_{n+1}-b_{n+1}}{a_{n}-b_{n}}=\frac{a_{n}q_{a}-b_{n}q_{b}}{a_{n}-b_{n}}\\
\Rightarrow\quad & q_{c}\left(a_{n}-b_{n}\right)=a_{n}q_{a}-b_{n}q_{b}\\
\Rightarrow\quad & a_{n}\left(q_{c}-q_{a}\right)=b_{n}\left(q_{c}-q_{b}\right)
\end{align*}
I thought this last equation would help me, but still im stuck.

EDIT: In the question it was stated that $q_b = 2$ but I thought it doesn't change anything, so I just used $q_b$ instead.

答:
After your last step, you get that the ratio $\frac{a_n}{b_n}$ is constant for all $n$, and you know that both $a$ and $b$ are geometric, so you can get :
$$\forall n\in\mathbb{N}, \left(\frac{q_a}{q_b}\right)^n \text{ is constant }$$

And that is only possible if $q_a=q_b$, and from that you can deduce that $q_a=q_b=q_c$


问:
The time complexity of the binary search algorithm
Describe the time complexity of the binary search algorithm in terms of number of comparisons?

For simplicity, assume there are $n = 2^k$ elements in the list $a_1, a_2, . . . , a_n$, where $k$ is a
nonnegative integer.

Question 1: Without writing the whole proof, I would like to clarify the first step of this proof. At the first stage the search is restricted to a list with $2^{k−1}$ terms. So far, two comparisons
have been used. At the second stage, the search is restricted to a list with $2^{k−2}$ terms, etc. Can you please explain how each time the lest is restricted by subtracting from $k$ the number $1$?

Question 2: why the whole search requires the ceiling of $log$: $2\left\lceil \log(n)\right\rceil + 2$. One comparison to exist the loop and another to check if the value is $x$ or the key plus $2\left\lceil \log(n)\right\rceil$ whole comparisons before existing the loop.

答:
About Question 1. Lets say we play a game called: I guess any number between 0 and 1024 in just 10 questions (so our list starts with 1024 possible solutions).

My first question is: Is this number larger than 512?

Regardless of your answer, I effectively reduce the amount of possible solutions to 512 (either larger than 512, or smaller or equal to 512). As such, I have halved the list of numbers the solution is in. If I repeat this procedure (so I ask, is it larger than 256), I half the list again. If your list starts at size $2^k$, then the half of that list equals $0.5 \cdot 2^k = 2^{-1} \cdot 2^k = 2^{k-1}$.

Regarding question 2:
Suppose the requested solution is larger than the list (to use the example above, the solution would be 1024). This would mean that the answer to each question is simply "yes". At the final iteration, the guessed value will be $512+256+128+64+32+16+8+4+2+1 = 1023$. Hereby, it is obvious that it does not equal the solution, as such the binary search algorithm includes this additional question that checks if the solution is inside the search range. This check can be performed before the while loop to save iterations, but that is not done here.


问:
Average-case analysis of linear search
For linear search algorithm, if $x$ is the $ith$ term, two comparisons will be used at each of the $i$ steps of the loop, and one outside the loop, so a total of $2i+1$ comparisons are used, which is fine.

The average number of comparisons used equals:

$$\frac{3+5+7+\textit{...}+(2n+1)}{n} = \frac{2(1+2+3+4+\textit{...}+n)+n}{n}$$

Question: How we got that $\frac{3+5+7+\textit{...}+(2n+1)}{n}$ equals $\frac{2(1+2+3+4+\textit{...}+n)+n}{n}$?

答:
You have
$$
3 + 5+7 + \ldots (2n+1)
 = \sum_{k=1}^n (2k+1)
 = 2 \sum_{k=1}^n k + \sum_{k=1}^n 1
 = 2 \sum_{k=1}^n k + n
$$

In simpler words, there are exactly $n$ terms in your sum, so if you subtract one from each of them, you get $n$. Then,
$$
3+5+\ldots+(2n+1)
 = [2+4+\ldots +2n] + n
 = 2[1 + 2 + \ldots + n] + n
$$

Title: Average-case analysis of linear search


问:
From which set does the number $\sqrt[3]{-1}$ belong to?
I was trying to draw the function $f(x)=\sqrt[3]{x^2(6-x)}$ by hand (I'm in my first year of engineering; having Calculus I; this drawing is actually an exercise given for my class) and used WolframAlpha to see if I've got it right. Well, I was expecting to miss a few things on my first try, but I didn't even consider the possibility that I would start getting its domain wrong (I thought it was all the real numbers, however, the program said it was all real numbers equal to or below $6$). So, trying to understand what I'd missed, I've concluded that Wolfram must see $\sqrt[3]{-1}$ as only as a complex number when I swore it was at least real (since the equation $x^3 = -1$ has at least a real number that solves it, i.e. $-1$). I mean, if Wolfram is right (and I'm supposing it is), then $f(x=7)$ does not return the "real number" $\sqrt[3]{-49}$ (I don't even know if this number is real anymore, and I'm getting more and more confused whilst writing this text), and my professor has put a wrong graph as an answer for the exercise above. Look, although I wanted the ability to understand what I'm not understanding so that I could be, at least, didactic, I'll try to resume my confusion with this question (and with it, I'll try to understand on my own why Wolfram didn't consider numbers like $x=7$ for $f(x)$ domain): what in heavens is this $\sqrt[3]{-1}$?

$\\$

PS: sorry if I've said anything wrong... I don't speak English fluently.

答: You are completely right. $x^3=a$, with $a<0$ has a real root, so the problem must lay in the way you are writing it in Mathematica. Try using the function CubeRoot, and see if it works. If not,  I would advise you to share your code in the question.


问:
Tangents to two sets
Find all the affine tangents that are simultaneously tangent to the set $E$ and $H$: $$H=\{(x,y)\in \mathbb R^2:xy=-5\}, E=\{(x,y)\in \mathbb R^2:\frac{x^2}{9}+\frac{y^2}{4}=1\}$$

I know that when the line is tangent to the set A at point a: $T_aA=\{v:<grad(h(a)),v>=0\}$ but I don't know how to combine this condition for both sets $ E, H $ to find all tangents

答:
Any tangent to the given ellipse is $$y=mx \pm \sqrt{9m^2+4}$$, insert this in the Eq. of hyperbola:$xy=-5$, to get
$$mx^2+\pm\sqrt{9m^2+4}x+5=0$$
For tangency we impose $B^2=4AC$, Then
$$9m^2+4=20m \implies m=2, 2/9$$
The tangents are $$y=2x\pm 2\sqrt{10},~~ y=2x/9\pm2\sqrt{10}/3.$$

Title: Tangents to two sets


问:
On isomorphism of two quotient of polynomial rings
$\mathbf{The \ Problem \ is}:$ Show that  the ring $R_1=\mathbb{C}[x,y,z]/(xy-z^2)$ is not isomorphic to $R_2 = \mathbb{C}[x,y,z]/(xy-z)$ .

$\mathbf {My \ approach}:$ There was a hint stating to show that $R_2$ is a UFD and $x$ is irreducible in $R_1$.

Now, $x$ is not prime in $R_1$ as $x\mid xy=z^2$ but then primality of $x$ would mean $x\mid z$, then $R_1$ will not be a UFD.

I have only read up to free modules and Gauss' lemma, a proof involving those things only will be very helpful for me.

A small hint is warmly appreciated.

答:
Hint: Consider the surjective map $x \mapsto x, y \mapsto y, z \mapsto xy$ from $\mathbb C[x,y,z]$ to $\mathbb{C}[x,y]$. Can you show that its kernel is $(xy-z)$? This would give an iso $R_2 \simeq \mathbb C[x,y]$.

 This shows that $R_2$ is a UFD, since $A$ being a UFD implies $A[T]$ is an UFD.

Hint': Now we're left with showing that $R_1$ is not a UFD. By its very definition, the equality $xy = z^2$ holds in this ring (I'm abusing notation and dropping the classes). Do you see why this equation is troubling, factorization-wise?

 Indeed, we have $z^2 = z \cdot z = xy$. These are two factorizations for the same element, prove this by showing that $x,y,z$ remain irreducible in the quotient and are not associates


问:
Resolve $x+\frac{1}{x}$
While reading an old mathematics question paper of the Unified International Mathematics Olympiad, I hit a question with two very interesting patterns. Here I first mention them before presenting my question.

Pattern 1

When $0 < x < 1$, then the value of $x+\frac{1}{x}$ is greater than $2$.

Here, the value of $x$ and $x+\frac{1}{x}$ are in inverse proportion. I tried putting the values and here are the results:

$$0.000001 + \frac{1}{0.000001} = 1000001$$
$$0.999999 + \frac{1}{0.999999} = 2.000001000001$$

The value of $x+\frac{1}{x}$ got near about two, but not equal to it.

Pattern 2

When $1 < x < 2$, then the value of $x+\frac{1}{x}$ is smaller than $2.5$.

This pattern also works the same way as the first one did. Its just that here the value of $x+\frac{1}{x}$ will always evaluate to a number small than $2.5$.

$$1.000001 + \frac{1}{1.000001} = 1.999999000001$$
$$1.999999999999999 + \frac{1}{1.999999999999999} = 2.499999999999999$$

The value of $x+\frac{1}{x}$ got near about $2.5$, but not equal to it.

Question

How can we prove the statements, "When $0 < x < 1$, then the value of $x+\frac{1}{x}$ is greater than $2$" and "When $1 < x < 2$, then the value of $x+\frac{1}{x}$ is smaller than $2.5$." mathematically?

Any help is appreciated. Thank you!

答:
$x+\frac{1}{x}\geq2$ holds for all $x>0$, with equality at $x=1$ only. As a hint, expand
$$\left(\sqrt{x}-\frac{1}{\sqrt{x}}\right)^2\geq0.$$
For the second claim, observe that for $1<x<2$, we have
$$(x-2)(x-1)<0\implies x^2+2<3x\implies x+\frac{1}{x}<\frac{x}{2}+\frac{3}{2}<\frac{5}{2}.$$

Title: Resolve $x+\frac{1}{x}$


答:
If you're familiar with calculus, there's a very snappy way to analyze this sort of situation: taking derivatives (and thinking about what the derivative says about the function in question).

The derivative of $f(x)=x+{1\over x}$ is $1-{1\over x^2}$. For $x\in (0,1)$ this is negative, so $f$ is decreasing on $(0,1)$. In particular, this means that for each $a\in (0,1)$ we have $f(a)>f(1)$. Now just calculate $f(1)$.

This also works for analyzing the interval $(1,2)$. In this interval - and indeed for all $x>1$ - the derivative of $f$ is positive and so $f$ is increasing. So e.g. for every $x\in (1,2)$ we have $$f(1)<f(x)<f(2)$$ or more snappily $$1<f(x)<2.5.$$

Similarly, we get $$f(17)<f(x)<f(42),$$ or $$f(x)\in (17+{1\over 17}, 42+{1\over 42}),$$ for all $x\in (17,42)$. And so on.

Title: Resolve $x+\frac{1}{x}$

Body:
While reading an old mathematics question paper of the Unified International Mathematics Olympiad, I hit a question with two very interesting patterns. Here I first mention them before presenting my question.

Pattern 1

When $0 < x < 1$, then the value of $x+\frac{1}{x}$ is greater than $2$.

Here, the value of $x$ and $x+\frac{1}{x}$ are in inverse proportion. I tried putting the values and here are the results:

$$0.000001 + \frac{1}{0.000001} = 1000001$$
$$0.999999 + \frac{1}{0.999999} = 2.000001000001$$

The value of $x+\frac{1}{x}$ got near about two, but not equal to it.

Pattern 2

When $1 < x < 2$, then the value of $x+\frac{1}{x}$ is smaller than $2.5$.

This pattern also works the same way as the first one did. Its just that here the value of $x+\frac{1}{x}$ will always evaluate to a number small than $2.5$.

$$1.000001 + \frac{1}{1.000001} = 1.999999000001$$
$$1.999999999999999 + \frac{1}{1.999999999999999} = 2.499999999999999$$

The value of $x+\frac{1}{x}$ got near about $2.5$, but not equal to it.

Question

How can we prove the statements, "When $0 < x < 1$, then the value of $x+\frac{1}{x}$ is greater than $2$" and "When $1 < x < 2$, then the value of $x+\frac{1}{x}$ is smaller than $2.5$." mathematically?

Any help is appreciated. Thank you!


问:
Third statement in example 2.3.3 in Velleman's How to Prove It.
I am self-studying this text to try to finally learn good proof writing and to learn some more abstract math. It's been going fairly well, but I'm stuck on the statement mentioned in the title. Velleman asks the reader to analyze the logical form of the statement $$B \in \left\{\mathscr{P}(A)\mid A \in \mathcal{F}\right\}.$$ This means that $B \in \mathscr{P}(A)$ where $A \in \mathcal{F}$. If $B \in \mathscr{P}(A)$, then $B$ is a subset of $A$. Then, this means the same thing as $\exists A \in \mathcal{F} (B = \mathscr{P} (A))$. This comes from the opening of the section. He says that $$x \in \left\{n^2 \mid n \in \mathbb{N}\right\}$$ is equivalent to $$\exists n \in \mathbb{N} (x = n^2).$$

This sort of makes sense, but when I think about it harder it becomes confusing. Since $n$ is a natural number, the square of any $n$ will be a perfect square, and thus I don't understand why this statement isn't instead $\forall n \in \mathbb{N} (x = n^2).$ Does this have something to do with the fact that there are an infinitely many natural numbers?

Further, regarding 2.3.3, if $B \subseteq A$, then $B$ is made up of elements of $A$. But the power set of $A$ is made up of sets of elements in $A$. To me this makes this seem like $B$ is two different types of objects and the main source of my confusion. I couldn't find anything on this example beyond the textbook. I see that Velleman is applying the equivalence given in the opening of the section, but I haven't convinced myself that it's true yet. I'm thankful for a nudge in the right direction. I've been going in circles with this one.

答:
First let's establish that we know what $\{a \mid P(a) \}$ means for some property $P$

$\{a \mid P(a) \}$ is a set, containing everything which satisfies the property $P$

When we write $x \in \{a \mid P(a) \}$ we're effectively saying that $x$ satisfies property $P$, since $x$ is equal to some $a$ satisfying $P$

Building slightly on this notation, we can construct sets like the following one
$\{f(a)\mid P(a)\}$ for a function $f$ and a property $P$

$\{f(a)\mid P(a)\}$ is the set obtained by applying $f$ to every element of $\{a \mid P(a) \}$ and putting everything you get into a set

When we write $x \in \{f(a) \mid P(a) \}$ what this means is that there is some element $a$ satisfying $P$ such that when you apply $f$ to it, the result is $x$

i.e. that $\exists a$ such that $P(a)$ (is true) and $f(a) = x$

Note that both of the sets in your examples are of the above form

In the first one, $f$ is $$, and in the second one, $f$ is the function $n \mapsto n^2$ (i.e. the function mapping numbers to their squares)

With this in mind, looking at these two examples

When you write "This means that $B \in (A)$ where $A \in \mathcal{F}$" this is not correct

What we instead have is that $B = (A)$ where $A \in \mathcal{F}$

$B$ has to "match the pattern on the left-hand side of the vertical bar" in order for it to belong to the set, which means it must be equal to $(A)$ for some $A$, not be an element of it.

I think part of the confusion here comes from the fact that this set $\{(A) \mid A \in \mathcal{F} \}$ is a set, whose elements are themselves sets of sets. Just remember that this $\{f(a) \mid P(a) \}$ notation doesn't care whether $f(a)$ is a set, a set of sets, or anything else. It treats it as a "pattern to match with" in the same way irrespective of how much internal structure it has.

For the second example, $x$ again has to "match the pattern", so it has to be the square of a natural number

The "for all" statement that you wrote $\forall n \in \mathbb{N} \: (x = n^2)$ is asserting that $x$ is (simultaneously) the square of every single natural number

This is a much stronger condition than what we require (so strong that it's impossible to fulfill)

Instead we want that $x$ is the square of some natural number (that's enough to match the pattern and establish that it belongs to the set), so we only need $\exists n \in \mathbb{N} \: (x = n^2)$

If we wanted to write that "the square of every natural number is itself a square", then we'd instead write this as $\forall n \in \mathbb{N} \: \exists m \in \mathbb{N} \: (m = n^2)$

This is true, but not relevant to what's going on in this example. All that's being said is that $x$ belongs to the set of all square numbers, which is the same as saying that $x$ is a square number. Nothing more.


问:
isomorphic automorphism groups
As I am sure some of you may have noticed I'm doing a series of exercises by Rotman and I am finding difficulties. Now I unbeaten into this problem

Give an example of an abelian and a non-abelian group with isomorphic automorphism groups.

Can you help me?

Title: isomorphic automorphism groups

Body: Try $A=\mathbb{F}_2^2$, and $G=S_3$.


答:
First of all, the student introduces the function $g$ that is dependent on $a$ and $x$. But he overlooks the $x$ dependence in his solution. But if he had done it correctly it could have worked:

$$g:\mathbb{R}^2 \to \mathbb{R}:(a,x)\to a^x$$

Then $h(x)=g(f(x),x)$ and

$$h'(x)=\frac{\partial g}{\partial a}\!\! \left(f(x),x\right)\frac{df}{dx}+\frac{\partial g}{\partial x}\!\! \left(f(x),x\right)\frac{dx}{dx}\\
= x x^{x-1}+x^x \ln x = x^x(1+\ln x) \; .$$

Of course, this is an explanation he might only get completely if he is familiar with partial derivatives.

Title: explaining the derivative of $x^x$


答:
$
x+\dfrac{1}{x}
$ is increasing on $[1,2]$. So the maximum is attained when $x=2$.

Title: Resolve $x+\frac{1}{x}$


答:
I just have a problem with one thing you wrote. You have

$$0.000001 + \frac{1}{0.000001} = 1000001$$

But without doing any calculations, I know this can't be true. We have here
$$10^{-6} + \frac{1}{10^{-6}} = 1000001$$
$$10^{-6}+10^6=1000001$$
But that means
$$\text{(not an integer)}+\text{(integer)}=\text{(integer)}$$
Which is impossible.

Title: Resolve $x+\frac{1}{x}$


答:
Wolfram Alpha is slightly eccentric in this regard. The cube root function is a perfectly respectable function from $\Bbb R$ to $\Bbb R$, but the default behaviour of Wolfram Alpha is to evaluate cube roots of negative reals as complex numbers.

However, you can override this behaviour, by clicking on "Use the real-valued root instead". Then you can have Wolfram Alpha plot the function for any range you like.

Title: From which set does the number $\sqrt[3]{-1}$ belong to?

Body:
I was trying to draw the function $f(x)=\sqrt[3]{x^2(6-x)}$ by hand (I'm in my first year of engineering; having Calculus I; this drawing is actually an exercise given for my class) and used WolframAlpha to see if I've got it right. Well, I was expecting to miss a few things on my first try, but I didn't even consider the possibility that I would start getting its domain wrong (I thought it was all the real numbers, however, the program said it was all real numbers equal to or below $6$). So, trying to understand what I'd missed, I've concluded that Wolfram must see $\sqrt[3]{-1}$ as only as a complex number when I swore it was at least real (since the equation $x^3 = -1$ has at least a real number that solves it, i.e. $-1$). I mean, if Wolfram is right (and I'm supposing it is), then $f(x=7)$ does not return the "real number" $\sqrt[3]{-49}$ (I don't even know if this number is real anymore, and I'm getting more and more confused whilst writing this text), and my professor has put a wrong graph as an answer for the exercise above. Look, although I wanted the ability to understand what I'm not understanding so that I could be, at least, didactic, I'll try to resume my confusion with this question (and with it, I'll try to understand on my own why Wolfram didn't consider numbers like $x=7$ for $f(x)$ domain): what in heavens is this $\sqrt[3]{-1}$?

$\\$

PS: sorry if I've said anything wrong... I don't speak English fluently.


答:
When working in the real numbers, $\sqrt[3]{x}$ denotes the unique number $a$ such that $a^3=x$. Since $(-1)^3=-1$, we have $\sqrt[3]{-1}=-1$.

When working in the complex numbers, the notation $\sqrt[3]{z}$ is imprecise. Every complex number has three cube roots, and there is no natural way of defining the 'principal' cube root of a complex number. It is sensible, however, to define $\sqrt[3]{z}$ in the complex numbers if $z$ also happens to be a real number. Then, $\sqrt[3]{z}$ can be defined in the way above.

That being said, sometimes we define the 'principal' $n$-th root of a complex number in the following way. If $z=re^{i\theta}$, where $\theta$ is the principal argument of $z$*, and $r$ is the magnitude of $z$, then
$$
\sqrt[n]{z} = \sqrt[n]{r} \cdot e^{i\theta/n} \, ,
$$
where $\sqrt[n]{r}$ denotes the principal real root of $r$. This means that $z \mapsto \sqrt[n]{z}$ is not a continuous function, but we still might choose to adopt the notation $\sqrt[n]{z}$ as a matter of convenience. This is probably the root that Wolfram Alpha was referring to:
$$
\sqrt[3]{-1} = \sqrt[3]{1e^{i\pi}} = \sqrt[3]{1} \cdot e^{(i\pi)/3} = e^{(i\pi)/3} \approx 0.5 + 0.866i \, .
$$

*Again, the 'principal' argument of a complex number involves a branch cut, where we require that $\theta \in (-\pi,\pi]$. Again, this does not define a continuous function.


问:
2 trams are coming every 5 and 10 minutes. On average, how long do I have to wait for any tram to come?
I want to follow the approach from Christian Blatter here

3 trams are coming every 10, 15 and 15 minutes. On average, how long do I have to wait for any tram to come?

using "the assumption that the trams arrive on time with unknown but equidistributed phases. In the following I shall treat model". However, my theoretical result do not match my simulation result.

A random point $P=(X,Y)$ is chosen in the block $$B:=\{(x,y)\>|\>0\leq x\leq5,\ 0\leq y\leq 10,\}\ .$$ The waiting time $T$ is then given by $T=\min\{X,Y\}$.

The points $P$ with waiting time between $t$ and $t+dt$ is a panel of thickness $dt$ and having a distance $t$ from the planes $x=0$ and $y=0$. (Here the union of pairwise buses is one because there are only two buses.) The area of the panel is $(10-t)(5-t)$.

Therefore, the pdf $f_T$ of the waiting time is given by
$$f_T(t)={1 \over 5 \cdot 10}(5-t) \cdot (10-t) \qquad(0\leq t\leq5)\ .$$
$$f_T(t)={1 \over 50}(50-15t+t^2) \qquad(0\leq t\leq5)\ .$$

From this we obtain the expected waiting time as
$$E(T)=\int_0^{5} t \cdot f_T(t) dt=  \int_0^{5} \frac{1}{50}[50t-15t^2+t^3] = \frac{1}{50}[50/2t^2-15/3t^3+1/4t^4)]^{5}_0 = 3.125.$$

However, from my simulation I get around 2.09, I guess that is 25/12:

I did a R simulation with the assumptions that buses arrive promptly every 5 and 10 minutes, respectively. However, the interval time among buses is fixed with uniformly distributed starting times. I simulate buses between [0,10000] minutes and the guy arriving at the bus stop at time t in [50,9950] and check the minimum time to next bus (which arrives first out of 2).

```<code>   upperbound <- 10000
   waitVector <- vector()
   nbRuns <- 10000
   for(count1 in 1:nbRuns) {
     seq1 <- 10*runif(1) + seq(from=0,to=upperbound,by=10)
     seq2 <- 5*runif(1) + seq(from=0,to=upperbound,by=5) # uniformly distributed, but fixed distance among both buses
     seq3 <- sort(c(seq1,seq2))
     arrivalTime <- runif(n=1,min=50,max=(upperbound-50))
     minSeq <- seq3 - arrivalTime
     minSeq <- minSeq[minSeq > 0] # cant catch bus who left before I arrived
     minTime <- min(minSeq)
     waitVector <- c(waitVector,minTime)
}
minAvgWaitTime <- mean(waitVector)
print(range(waitVector))
print(paste("Minimum avg wait time base on ",nbRuns," simulations is: ",minAvgWaitTime,sep=""))
</code>```

答:
Personally, I find it much easier to think in CDF terms. In this example, the cumulative probability of waiting for a tram for up to $t$ corresponds to the area of the shaded region in the chart below.

In other words, $F_T(t) = \frac{1}{50} (5t + 10t - t^2)$, and $f_T(t) = \frac{1}{50} (15 - 2t)$. We thus avoid thinking in $dt$ terms.

This could have been a comment to @joriki answer, but I need to include the chart.


问:
Find the perimeter of Triangle ABC
This practice engineering board exam which I've tried seems very tricky (question no.2). The inscribed circle of $\triangle ABC$ is tangent to $AB$ at $P$ . If the radius is $21$, $AP=23$ and $PB=27$, find the perimeter of $\triangle ABC$ .I've attempted to use the Pythagorean Theorem and Heron's Formula but the problem is that I've missed the values of the two sides $BC$ and $CA$ .Can you please provide solutions by two methods(geometry and trigonometry)?

答:
Another approach:

$\tan(\frac{\alpha}2)=\frac{21}{23}$

$\tan(\frac{\beta}2)=\frac{21}{27}$

$\frac{\alpha}2+\frac{\beta}2+\frac{\gamma}2=90^o$

$\tan(\frac{\alpha}2+\frac{\beta}2)=\tan(90-\frac{\gamma}2)=cotan(\frac{\gamma}2)$

Puting values we get $\tan(\frac{\gamma}2)=\frac6{35}$

Now we use this formula:

$\tan(\frac{\gamma}2)=\frac r{p-c}$

Where r=21, p is half perimeter, and $c=AB=23+27=50$

Plugging these values we get:

$p=\frac r{tan(\frac{\gamma}2)}+c=\frac {21}{\frac 6{35}}+50=\frac {7\cdot 35}{2}+50$

So perimeter P is:

$P=2\times \left(\frac {7\cdot 35}{2}+50\right)=7\cdot 35+100=\boxed {345}$

Title: Find the perimeter of Triangle ABC


问:
Minimal Polynomial of $i\sqrt[4]{2}$ over $\mathbb{Q}(\sqrt[4]{2})$
I see that $$i\sqrt[4]{2} \notin \mathbb{Q}(\sqrt[4]{2})=\{a+b\sqrt[4]{2}:a,b\in \mathbb{Q}\}$$ And so the minimal polynomial has degree $\deg(m_{i\sqrt[4]{2}})\neq 1$.

Let $\alpha=i\sqrt[4]{2}$, from here I do $$ \alpha^2 = -\sqrt{2} = -(\sqrt[4]{2})^2$$

Here is where I'm not sure what I'm doing is correct:
\begin{equation}
    -(\sqrt[4]{2})^2 = (\underbrace{-\sqrt[4]{2}}_{\in \mathbb{Q}(\sqrt[4]{2})}) \cdot
    \underbrace{\sqrt[4]{2}}_{\in \mathbb{Q}(\sqrt[4]{2})} \hspace{2cm} (1)
\end{equation}
so I can say that $$-(\sqrt[4]{2})^2\in
    \mathbb{Q}(\sqrt[4]{2}) \hspace{3.73cm} (2)$$ and from here, that $$ m_{i\sqrt[4]{2}}=x^2+(\sqrt[4]{2})^2 \text{ over } \mathbb{Q}(\sqrt[4]{2})$$

Then I argue that because $x^2+(\sqrt[4]{2})^2$ is an Eisenstein polynomial over $\mathbb{Q}(\sqrt[4]{2})$ $\Rightarrow$ $x^2+(\sqrt[4]{2})^2$ is irreducible $\Rightarrow$ $x^2+(\sqrt[4]{2})^2$ is the minimal polynomial $m_{\alpha}$ over $\mathbb{Q}(\sqrt[4]{2})$.

My only actual question is if step $(1)$ and $(2)$ really are correct in this framework. Thanks in advance.

答:
Once you know that $\alpha \notin \mathbb{Q}(\sqrt[4]{2})$, but $\alpha^{2} \in \mathbb{Q}(\sqrt[4]{2})$, then it is clear that the minimal polynomial of $\alpha$ over $\mathbb{Q}(\sqrt[4]{2})$ has degree two, because there you have found a polynomial of degree two that vanishes on $\alpha$, but no polynomial of degree one does.

Also, write $\beta = \sqrt[4]{2}$. Then
$$\mathbb{Q}(\beta)=\{a_{0} + a_{1} \beta + a_{2} \beta^{2} + a_{3} \beta^{3} : a_{i} \in \mathbb{Q}\}$$
has degree four over $\mathbb{Q}$. So you see that $\alpha^{2} = - \beta^{2} \in \mathbb{Q}(\beta)$.


答:
Try going through the incorrect argument with $g(a) = \dfrac ax$.

Then we have $g(f(x)) = 1$ but $f'(x)\cdot g'(f(x)) = \dfrac 1x$.

At this point most students will try to explain why your argument is different from theirs, and why their argument works but yours doesn't. If not you can give them a little encouragement to find the hole in your version of the argument.

That way the student will do all the work for you, which is a good way of learning.

Title: explaining the derivative of $x^x$


问:
Determine the 100th term of the sequence
I want to find the 100th term of the following sequence:

$$\bbox[5px,border:2px solid #C0A000]{\small a_0 = 1}$$
$$\bbox[5px,border:2px solid #C0A000]{\small a_n = a_{n-1} + 2^{-n}}$$

The answer is supposed to be $\frac{2^{101} - 1}{2^{100}}$ but I'm not quite sure how to arrive to this conclusion. I've tried playing around with it but I don't really see it. I would really appreciate some help, thanks in advance!

Body: Hint:$$a_n-a_{n-1}=2^{-n}$$ $$\sum_{i=1}^{100} a_i-a_{i-1}=\sum_{i=1}^{100} 2^{-i}$$ $$\implies a_{100}-a_0=?$$


问:
Definition of a compact set correct?
I am suspicious about the following definition of a compact set:

Let X bei a normed vector space and $M\subset X$. M ist called compact, if every sequence $(x_n)_{n\in\mathbb{N}}\in M$ has a convergent subsequence (with limes in M).

Doesn't there need to be something like a bounded sequence?

Thank you!

答:
This is correct.

Actually, the boundedness is a consequence of the compactness of the set : a compact set in a metric space is always closed and bounded (and the converse is true iff the dimension is finite).

Title: Definition of a compact set correct?


问:
Addition of vector but with same direction
I was playing around with vectors and i got stuck at thing that i think should work.Please correct me if i m wrong as it was today that i realised that vectors are nothing but just graphical representation of co ordinates in a way .

I took two position vectors in first quadrant.

r1=2i+4j

r2=5i+5j

So adding r1 and r2 should give me a third vector let us say r3 which should have length 7i+9j which turns out to be $\sqrt130$ and this should be equal to length of r1+r2 which is $\sqrt20$ + $\sqrt50$ but it turns out that $\sqrt130$ is not exactly equal to $\sqrt20$ + $\sqrt50$
Why is it so it should be exactly same right ?

Blockquote

答: That's because the angle between those two vectors is small but not $0$. If it was $0$ (that is, if $r_2=\lambda r_1$ for some $\lambda>0$), then the length of the sum would indeed be the sum of the lengths. But in your case the sum is the diagonal of a parallelogram (see the picture below) and the length of such a diagonal is always strictly smaller than the sum of the lengths of two non-parallel sides.


问:
Finding the inverse of an integer a in Z_n
So I'm trying to find the inverse of $24$ in $\mathbb{Z}_{35}$.
I know this means that there must be some $ab \equiv 1 \pmod{n}$
But when I try this:

$24b \equiv 1 \pmod{35}$

$1 = 35*0 + 1$

$24b = 1$

$b = 1/24$

But this isn't an integer, so I'm not sure what I'm doing wrong here.

答: Since $2\cdot 18 \equiv 36 \equiv 1 \pmod{35}$, you have an inverse for $2$.   Since $3\cdot 12 \equiv 36 \equiv 1\pmod{35}$ you have an inverse for $3$.  Since $24 = 2^3 3$ you can easily construct its inverse.


Title: Finding the inverse of an integer a in Z_n

Body:
So I'm trying to find the inverse of $24$ in $\mathbb{Z}_{35}$.
I know this means that there must be some $ab \equiv 1 \pmod{n}$
But when I try this:

$24b \equiv 1 \pmod{35}$

$1 = 35*0 + 1$

$24b = 1$

$b = 1/24$

But this isn't an integer, so I'm not sure what I'm doing wrong here.

Body:
Start by computing $\text{gcd}(24, 35)$ using the Euclidean algorithm. We have that:
\begin{align*}
&35 = 24(1) + 11 \\
&24 = 11(2) + 2 \\
&11 = 2(5) + 1
\end{align*}

So $\text{gcd}(24, 35) = 1$. Our goal now is to work backwards:

\begin{align*}
&11 = 2(5) + 1 \implies 1 = 11 - 2(5) \\
&24 = 11(2) + 2 \implies 1 = 11 - 5[24 - 11(2)] = 11(11) - 5(24) \\
&35 = 24 + 11 \implies 1 = 11(35 - 24) - 5(24) = 11(35) - 16(24)
\end{align*}

So $11(35) - 16(24) = 1$. Thus, $24^{-1} \equiv -16 \pmod{35}$. Note that $-16 \equiv 19 \pmod{35}$.


Title: Determine the 100th term of the sequence

Body:
I want to find the 100th term of the following sequence:

$$\bbox[5px,border:2px solid #C0A000]{\small a_0 = 1}$$
$$\bbox[5px,border:2px solid #C0A000]{\small a_n = a_{n-1} + 2^{-n}}$$

The answer is supposed to be $\frac{2^{101} - 1}{2^{100}}$ but I'm not quite sure how to arrive to this conclusion. I've tried playing around with it but I don't really see it. I would really appreciate some help, thanks in advance!

Body:
$$a_n = \sum_{k=0}^n \left( \frac{1}{2} \right)^n$$

This is a geometric series, which sums to

$$a_n = \frac{1-(1/2)^{n+1}}{1/2}=2-(1/2)^n.$$

$$a_{100} = 2-(1/2)^{100}=\frac{2^{100}}{2^{100}} \cdot \left[ 2-(1/2)^{100} \right]=\frac{2^{101}-1}{2^{100}}.$$


问:
Is a basis that almost diagonalizes a matrix 'close' to its eigenbasis?
Let $A\in\mathbb{R}^{n\times n}$ be diagonalisable (over $\mathbb{C}$) with pairwise distinct eigenvalues $\lambda_1,\ldots,\lambda_n\in\mathbb{C}$, and suppose that

$$\tag{1}S^{-1}\cdot A\cdot S = \mathrm{diag}[\lambda_1,\cdots,\lambda_d]=:\Lambda \qquad \text{for some invertible } S\in\mathbb{C}^{n\times n}.$$

Assume that there is another matrix $T\in\mathbb{R}^{n\times n}$ which 'quasi-diagonalises' $A$ in the sense that

$$\tag{2}T^{-1}\cdot A\cdot T =\Lambda + \Delta \qquad \text{for some} \qquad \Delta\in\mathbb{R}^{n\times n} \ \text{ with } \ \|\Delta\|<\varepsilon$$
where $\varepsilon>0$ is 'small' and $\|\cdot\|$ is a matrix norm of your choice (on $\mathbb{C}^{n\times n}$).

Question: Can we infer that for $\varepsilon>0$ small enough, the matrices $S$ and $T$ are close to one another in the sense that

$$\tag{3}\mathrm{inf}\{\|T\cdot S^{-1} - D\cdot P\| \mid \text{$D\in\mathbb{C}^{n\times n}$ diagonal & invertible}, \  \text{$P$ permutation matrix}\} \ \lesssim \ \|\Delta\|$$

In other words, can we infer that for $\Lambda$ and $\tilde{\Lambda}:=\Lambda + \Delta$ almost identical, the columns of $S$ and $T$ (which define a basis for the almost identical representations $\Lambda$ and $\tilde{\Lambda}$ of the endomorphism $A$) almost coincide up to order and scale?

Any references or hints, or indeed counterexamples, are appreciated.

Edit: As demonstrated in Ruy's answer, the desired conclusion doesn't apply without additional conditions on $A$; are you aware of any such conditions that are as mild as possible?

答:
As stated, the answer is no,  at least for the operator norm.

For a counter example choose any $\varepsilon >0$,
and
take $A$ to be a
diagonal matrix whose diagonal entries $\lambda _1, \lambda _2, \ldots , \lambda _n$ are all distinct,  but satisfy
$$
  |\lambda _i-1|<\varepsilon/2 ,\quad \forall i=1,\ldots ,n.
  $$

Since $A$ is already diagonal, we can take $S=I$  and $\Lambda =A$, in order to satisfy the required relation:
$$
  S^{-1}AS=\Lambda .
  $$

Choose any unitary operator $T$,
whatsoever, and notice that
$$
  \|T^{-1}AT - \Lambda \| =
  \|T^{-1}AT - A\| \leq  $$$$\leq
  \|T^{-1}AT - I\| +  \|I - A\| =
  \|T^{-1}(A-I)T \|  +  \|I - A\| = $$$$ =
  2\|A-I\| < \varepsilon,
  $$
so all of the required conditions are verified.

Neverteless,
given the arbitrary choice of $T$, there is certainly no reason to expect  the existence of a diagonal matrix $D$,
and a permutation matrix $P$, such that
$$
  \|TS^{-1} - D P\| =   \|T - D P\|
  $$
is small.

I suspect that one needs to require the eigenvalues of $A$ to be spread far apart if one is to hope for an affirmative
answer.


答:
Hint:

First, $1/24$ is an integer in $\mathbb{Z}_{35}$. It just doesn't look like it yet because you haven't performed the division operation. This is similar to how $0.\overline{9}$ or $1/.5$ might not look like integers, but they are.

The easiest way is to simply go through every element of $\mathbb{Z}_{35}$ and see if it is the inverse of $24$.

$1\cdot 24=24$, so $1/24\neq 1$.

$2\cdot 24=48\equiv 13$, so $1/24\neq 2$.

$\vdots$

Again, we are looking for a value that when we multiply by 24, we get 1.


问:
Building a model using the power set of $Ord$
I have a question where working in ZF I need to build a transitive model of Z containing $Ord$ (the class of all ordinals) s.t it does not model Zermelo definition of an infinite set:

$$\exists x\left(\emptyset\in x\wedge \forall z\in x, \{z\}\in x\right)$$

I know (from a previous question) that defining

$$X_0 = \omega,\ X_{n+1}=\mathcal{P}(X_n),\ X=\bigcup_{n<\omega} X_n$$

then $X$ is a transitive model of Z without a Zermelo Infinite set (not containing $Ord$ of course). I thought about using the above but taking

$$X_0 = Ord,\ X_{n+1}=\mathcal{P}(X_n),\ X=\bigcup_{n<\omega} X_n$$

this gives me that $X$ the desired model. What I'm having a problem with is taking the power set of the class $Ord$, $\omega$ many times.

Can we use a "power class" as we would use a power set?

答:
Yes, we can define the idea of a "power class", of a class $A$, but then you need to clarify if you mean:

The second-order (or rather, third-order) object of all the subclasses of $A$, or

The collection of all sets which are subclasses of $A$.

The latter is a first-order object, if $A$ is defined by $\varphi$, then $\mathcal P(A)$ is given by $\{x\mid\forall y(y\in x\to\varphi(y))\}$. You want to iterate this power set, and then it's obviously a question whether or not we can even consider $\bigcup_{n<\omega}\mathcal P^n(A)$. The answer is again yes, since we are quantifying on an internal $\omega$, rather than writing the scheme of $\mathcal P^n(A)$ explicitly, we use the recursion theorem.

But there's a clearer way to see why we can do it: for $\alpha$, let $A_\alpha=A\cap V_\alpha$, then $A_\alpha$ is a set. So we can talk about $\mathcal P^\omega(A_\alpha)=\bigcup_{n<\omega}\mathcal P^n(A_\alpha)$, and we can prove that $\mathcal P^\omega(A)=\bigcup_{\alpha\in\mathrm{Ord}}\mathcal P^\omega(A_\alpha)$.


问:
Expected value of dice sum
We roll a dice. If we get 6, then we roll again. What's the expected value of sum of rolls?

Let $X$ be the sum. In this game we can only obtain sum in form $6k + x$ where $k \in \mathbb{Z}$ and $x \in \{1,2,3,4,5\}$. If $n$ is in this form then probability of obtaining it is equal $(\frac{1}{6})^{[\frac{n}{6}]}\cdot\frac{1}{6}$. To get the expected value I would to sum only these $n$ which are in this form but I don't know how to properly do it. Or maybe my idea is wrong?

答:
Here's a trick: let $E$ be the expected value. When you roll a $6$, you win $6+E$ points. Therefore
$$E = \frac16\left(1+2+3+4+5+(6+E)\right)$$
Now solve for $E$.

Title: Expected value of dice sum


问:
Proving $\sin^2A+\sin^2B+\sin^2C-2\cos A \cos B \cos C = 2$.
I have a question like this.

Let $A, B$ and $C$ be angles of a triangle where $A+B+C=180^\circ$. Prove that $\sin^2A+\sin^2B+\sin^2C-2\cos A \cos B \cos C = 2$.

So, I said to prove that equality it is the same as proving
$\sin^2A+\sin^2B+\sin^2C=2+2\cos A \cos B \cos C $.

After that, I manipulated the LHS by using some formulas and identities until I ended up with the RHS.

But, someone told me that I was wrong because I manipulated both sides to prove that equality. In my opinion, after I moved the part "$-2 \cos A \cos B \cos C$" to the RHS and then worked with the new LHS until I get the LHS is the same as RHS, it is okay. It seems like we work with new question.

What do you guys think? Please share your opinion. Thank you! (I'm sorry for my bad English)

Body:
Note $\cos C= \cos(\pi-A-B)=-\cos(A+B)$. Then
\begin{align}
&\sin^2A+\sin^2B+\sin^2C\\
=& \frac12(1-\cos 2A )+\frac12(1-\cos2B)+(1-\cos^2C )\\
=& 2-\cos(A+B)\cos(A-B) +\cos^2(A+B)\\
 =& 2-\cos(A+B)[\cos(A-B) +\cos(A+B)]\\
 =& 2-\cos(A+B)\cdot2\cos A\cos B\\
 =& 2+2\cos C\cos A\cos B\\
\end{align}


问:
How to find limit of $\lim_{x \to \infty} a_n$, where $a_{n+1}= \sin a_n$
The answer shows that

let $\lim a_n=A$

since $a_{n+1}= \sin a_n$, adding limit both side will get $A= \sin A$.

But how does it works?

I guess $A= \lim a_n =\lim \sin a_n \neq \sin \lim a_n = \sin A$

What did I miss?

答:
Since $\sin(x)$ is continuous and defined everywhere, we know that if $b_n$ is a sequence and $\lim_{n\to\infty}b_n = b$, then $\lim_{n\to\infty}\sin(b_n) = \sin\left(\lim_{n\to\infty}b_n\right) = \sin(b)$.

So, if $a_{n+1}=\sin(a_n)$, then we know that if $\lim_{n\to\infty}a_n$ exists and equals $A$, then
$$\begin{align*}
A &= \lim_{n\to\infty}a_n\\
&=\lim_{n\to\infty}a_{n+1}\\
&=\lim_{n\to\infty}\sin(a_n)\\
&= \sin\left(\lim_{n\to\infty}a_n\right)\\
&= \sin(A).
\end{align*}$$
Of course, this assumes that you already know that $\lim_{n\to\infty} a_n$ exists.

In fact, one can prove that, regardless of the value of $a_1$, the sequence defined by $a_1=k$, $a_{n+1}=\sin(a_n)$ for all $n\geq 1$, will converge. Say it converges to $A$; then, as above, we get that $\sin(A)=A$. Since this means $-1\leq A\leq 1$, and since the function $g(x) = x-\sin(x)$ has $g'(x) = 1-\cos(x)\geq 0$ for all $x$ in $[-1,1]$, equal to $0$ only at $x=0$, the function is strictly increasing on $[-1,1]$; thus, $g(x)=0$ has only one solution on $[-1,1]$, which is at $x=0$, and so we will conclude, once we prove that $\lim_{n\to\infty}a_n$ exists, that in fact $A=0$.


答:
Okay I found the hints:

Heine Theorem

if function continuous, $\lim f(x) = f(\lim x)$

where the latter one is easy to proove by definition.

Title: How to find limit of $\lim_{x \to \infty} a_n$, where $a_{n+1}= \sin a_n$

Body:
The answer shows that

let $\lim a_n=A$

since $a_{n+1}= \sin a_n$, adding limit both side will get $A= \sin A$.

But how does it works?

I guess $A= \lim a_n =\lim \sin a_n \neq \sin \lim a_n = \sin A$

What did I miss?


答:
If $\sin(a_1)=0$ then obvioulsy $a_n=0$ for all $n\geq 2$.

Otherwise, let $a_2=\sin(a_1)\in[-1,1]\setminus\{0\}$:

If $\sin(a_1)\in(0,1]$, then $a_{n+1}\in(0,1]$ and $a_{n+1}\leq a_n$ for all $n\in\mathbb{N}$. The proof for this is easy. Assume it's true for $n-1$. Then $a_{n}\in(0,1]\subset(0,\frac{\pi}{2}]$, so $a_{n+1}=\sin(a_n)\in(0,1]$. Also, because of the famous inequality $|\sin(x)|\leq |x|$ you have $a_{n+1}=\sin(a_n)\leq a_n$. Then $(a_n)_{n\in\mathbb{N}}$ is decreasing and bounded.

Analogously you can prove that if $\sin(a_1)\in[-1,0)$, then $(a_n)_{n\in\mathbb{N}}$ is a increasing and bounded sequence.

In any case $(a_n)_{n\in\mathbb{N}}$ is an monotone bounded sequence, so it must converge.  Let $l:=\lim_{n\to\infty} a_n$. Because of the convergence of $(a_n)_{n\in\mathbb{N}}$, $\lim_{n\to\infty}a_{n+1}-a_n=\sin{a_n}-a_n=0$.  Then $\sin(l)=l$, which implies $l=0$.


问:
How many ways to read the palindromic word ROTATOR in this design?
The problem is as follows:

The figure from below shows the word $\textrm{ROTATOR}$ arranged in a
peculiar way. How many ways can this word be read assuming the equal
least distance from one letter to another?.

The alternatives given in my book are as follows:

$\begin{array}{ll}
1.&\textrm{490 ways}\\
2.&\textrm{480 ways}\\
3.&\textrm{245 ways}\\
4.&\textrm{400 ways}\\
\end{array}$

I noticed that the word is palindromic, hence it can be read back and forth, thus this means that I should account for these possibilites.

In order to keep the right track for this purpose I used an auxiliary numbers atop the letters to account for these as shown in the diagram from below.

After doing all of that I reached the conclusion that:

$\textrm{ways}=(74+74+96)\times 2 =488$

But this doesn't appear in any of the alternatives. Did I mess up something or what? Can someone help me here? Please I require a step by step explanation as I feel lost if my method did worked out properly?

答:
Starting from the top A there are $1+2+1+3=7$ routes to an R. So there are $7^2$ ways through that A.

Starting from second A down there are $(1+3+3)\times2=14$ routes to an R. So there are $14^2$ ways through that A.

So the total number of ways is $7^2+7^2+14^2+14^2=490$.


问:
Which one is the correct series expansion?
Is

$$p^{n+1} = p^0+p^1+ \dots + p^n$$

or

$$p^{n+1} = p^0\times p^1\times \dots \times p^n\text{ ?}$$

I am confused.

please explain the correct one.

答: None of them. $p^{n+1} = p \times p^n = p × \dots \times p$ (n times).</p>


问:
kinematics question. this question is about using time in a kinematics question.
question : particles A and B are projected vertically upwards with velocities 20m/s and 25m/s. particle B is projected 0.4 seconds later than particle A.

find the velocities of A and B at the instant when the particles are at the same height.

here is my problem:
in my book it finds the displacements of both A and B using a suvat formula. and then it equates the two obtained equations (in terms of t ) as follows :

-particle A : s = 20t + 1/2 (-10) t^2 --> -10 for gravity
-particle B : s = 25 (t-0.4) + 1/2 (-10) (t-0.4)^2

and then it uses another formula to find velocity of each.

here is my question : as B is projected 0.4 seconds later than A , should it be t+0.4 and not t-0.4?

i really can't wrap my head around it. thanking you in advance for answering.

答: When $t=0.4$ you want the height that the original function gives at $t=0$, so you evaluate the original function at $t-0.4$ (i.e., indeed at $0.4-0.4=0$ when $t=0.4$)


问:
How to find top and bottom function for finding area between two curves?
When I am trying to find the area between two curves below the $\mathbf{x}$-axis, should I always have the function furthest away from the x-axis as the function you are subtracting from, or should it just be the function that is on top? And when finding the area between two curves, is the answer always positive even when it's below the x-axis?

答:
One way to think about it is to start by writing down the problem is an order-agnostic way: if your two curves are $y_1(x)$ and $y_2(x)$, you can calculate the area between them by
$$\int_{x_0}^{x_1} \left|y_1(x)-y_2(x)\right|\,dx.$$

This formula is true regardless of whether $y_1 > y_2$ or vice versa, since the absolute value will take the positive distance between the $y$ coordinates in both cases.

Now, you can think about improvements that make this (correct, but difficult) integral more easy to compute. For example, you can get rid of the absolute value if you know that $y_1(x) > y_2(x)$ for all values of $x$ (note that this means that $y_1$ is greater (more positive) than $y_2$, not farther from the $x$-axis than $y_1$.

You need to be extra careful if the two curves cross somewhere. For example, if $y_1(x) > y_2(x)$ for $x\in [x_0,x_1)$ and $y_2(x) > y_1(x)$ for $x\in (x_1,x_2]$, you will need to break up your integral into two pieces in order to correctly compute the total area under the curves:

$$\int_{x_0}^{x_2} \left|y_1(x)-y_2(x)\right|\,dx = \int_{x_0}^{x_1} \left(y_1(x)-y_2(x)\right)\,dx + \int_{x_1}^{x_2} \left(y_2(x)-y_1(x)\right)\,dx.$$


问:
Euclidean norm minimization using rotations, i.e., $\min\limits_\mathbf\Phi \lVert \mathbf{A \Phi b} \rVert_2^2$
Given the following minimization $$\min\limits_\mathbf\Phi \lVert \mathbf{A \Phi b} \rVert_2^2$$
where $\mathbf A$ and $\mathbf b$ are a complex matrix and vector, respectively, and
$$ \mathbf \Phi = \begin{bmatrix}
e^{j\phi_1}     & \cdots & 0\\
\vdots    & \ddots & \vdots\\
0       & \cdots & e^{j\phi_N}\,
\end{bmatrix}$$
is a diagonal matrix of rotations.

How would the rotations $\{\phi_1,\ldots,\phi_N\}$ that minimize the cost function be found? I tried minimizing each rotation independently, but sometimes it leads to a local minimum.

What is the intuition about why the rotations would produce a smaller norm? Is there a geometric explanation?

Is there a reference where this problem has been studied?

答:
$\def\p#1#2{\frac{\partial #1}{\partial #2}}\def\d{{\rm diag}}\def\D{{\rm Diag}}$For
typing convenience, let $P=\Phi$ and also define the associated vector $p$
$$\eqalign{
p &= \d(P) \quad&\iff\quad P &= \D(p) \quad\implies\quad Pv = p\odot v\\
p_k &= e^{i\phi_k} \quad&\implies\quad dp &= ip\odot d\phi \;=\; iP\,d\phi \\
}$$
where $\odot$ denotes the elementwise/Hadamard product. And let's denote the trace/Frobenius product using a colon, i.e.
$$\eqalign{
A:B &= {\rm Tr}(A^TB) \\
A^*:A &= \big\|A\big\|^2_F \\
}$$
Finally, it will also be convenient to define the matrices
$$\eqalign{
M &= (bb^H)^*\odot A^HA \;\;=\;\; b^*b^T\odot A^HA \\
Q &= A^HAPbb^H \quad\iff\quad \d(Q) = Mp \\
}$$

Write the cost function, calculate its gradient, set it zero, and solve for the optimal $p$ vector.

$$\eqalign{
{\cal C} &= (APb)^*:APb \\
d{\cal C} &= A^*P^*b^*:A\,dP\,b &+\quad APb:A^*\,dP^*\,b^* \\
 &= Q^*:dP \quad&+\quad conj \\
 &= Q^*:\D(dp) \quad&+\quad conj  \\
 &= \d(Q^*):dp \quad&+\quad conj  \\
 &= M^*p^*:ip\odot d\phi \quad&+\quad conj  \\
 &= ip\odot M^*p^*:d\phi \quad&+\quad i^*p^*\odot Mp:d\phi  \\
 &= 2\;{\rm Real}(ip\odot M^*p^*):d\phi \\
 &= 2\;{\rm Imag}(p^*\odot Mp):d\phi \\
\p{\cal C}{\phi}
 &= 2\;{\rm Imag}(p^*\odot Mp) \;\doteq\; 0 \\
}$$
Since $(p^*\odot p)$ is a real vector, this implies that if $Mp$ is a multiple of $p$ it will satisfy the zero gradient condition. Or in other words
$$Mp = \lambda p$$
and therefore $p$ is an eigenvector of $\;(bb^H)^*\odot A^HA\;$ and the required angles are the elementwise logarithms
$$\phi_k = -ip_k$$

To follow the above derivation, it's  helpful to know that the Hadamard and Frobenius products commute with one another
$$\eqalign{
B\,:\,C &= C\,:\,B \\
B\odot C &= C\odot B \\
A\odot B\,:\,C &= A\,:\,B\odot C \\
}$$
and that the cyclic property of the trace allows the terms in a Frobenius product to be rearranged in a number of different ways, e.g.
$$\eqalign{
A:B &= B:A = B^T:A^T \\
CA:B &= C:BA^T = A:C^TB \\
}$$


答:
Welcome to MSE!

You have asked some good questions, and I will try to answer them all. I won't address what happens when using calculus to find the area between the $x$ axis and a curve when the curve crosses the $x$ axis bewteen the two relevant endpoints, unless you specifically want me to.

Firstly, when using calculus to find the area between a curve and the $x$ axis between two $x$ values , if the curve is completely below the $x$ axis then using calculus will yield a negative area. So if you wanted the actual area, ie the absolute magnitude of the answer you got just ignore the $-$ sign.

Now suppose you had $2$ curves, both (completely) below the $x$ axis between our two $x$ values we are looking at, and they don't intersect, and we want to find the area bounded by them in this region. Then the negative area can be found by finding the integral of the lowest curve bewteen the two $x$ values and then subtracting from it the integral of the upper curve bewteen the two $x$ values; this will give you the negative of the area that you are looking for. Again, if you wanted the actual area, ie the absolute magnitude of the answer you got just ignore the $-$ sign.

I hope that was helpful. If you have any further queries please don't hesitate to ask!

Title: How to find top and bottom function for finding area between two curves?

Body: When I am trying to find the area between two curves below the $\mathbf{x}$-axis, should I always have the function furthest away from the x-axis as the function you are subtracting from, or should it just be the function that is on top? And when finding the area between two curves, is the answer always positive even when it's below the x-axis?


答:
HINT:

$$a^m\cdot a^n=a^{m+n}$$

and $$\sum_{0\le m\le n}ar^m=\frac{a(r^{m+1}-1)}{r-1}$$

Now, judge yourself

Title: Which one is the correct series expansion?

Body:
Is

$$p^{n+1} = p^0+p^1+ \dots + p^n$$

or

$$p^{n+1} = p^0\times p^1\times \dots \times p^n\text{ ?}$$

I am confused.

please explain the correct one.


问:
Find the extrema points of $ f\left(x\right)=\prod_{i=1}^{n}x_{i} $
Let $ f:\mathbb{R}^n \to \mathbb{R} $ given by

$ f\left(x\right)=\prod_{i=1}^{n}x_{i} $

I have to find the extrema points of $ f $ under the following constraint:

$ S=\left\{ \left(x_{1},...,x_{n}\right):\sum_{i=1}^{n}x_{i}^{p}=1\right\}  $

Where $ p $ is some positive even integer.

I tried to solve using Lagrange multipliers methood but it bacame too complicated so I doubt that thats the way to solve the question.

I'd appreaciate some help. Thanks in advance.

答:
I think Lagrange's multipliers work fine.
Namely, there is $\lambda$ such that $$L(x_{1},x_{2},...) = x_{1}*x_{2}*...*x_{n} - \lambda(x_{1}^{p} + x_{2}^{p} + ... x_{n}^{p}) $$
has all derivatives zero at the critical point.
Taking derivate with respect to $x_{1}$ one gets
$$x_{2}*x_{3}*...x_{n} -p*\lambda *x_{1}^{p-1} = 0$$
Multiplying both sides by $x_{1}$ gives
$$x_{1}*x_{2}*x_{3}*...x_{n} -p*\lambda *x_{1}^{p} = 0$$
Doing the same for each $k$ and then adding up leads to
$$nx_{1}*x_{2}*x_{3}*...x_{n} - p* \lambda(x_{1}^{p} + x_{2}^{p} + ... x_{n}^{p}) = 0$$
Which means, that at any extreme value one has
$$x_{1}*x_{2}*x_{3}*...x_{n} = p* \lambda *\frac{1}{n}$$
So, $$x_{2}*x_{3}*...x_{n} = p*\lambda  *\frac{1}{n * x_{1}}$$
Substituting this to the second equation from the top we finally get
$$ p*\lambda  *\frac{1}{n * x_{1}} = p*\lambda* x_{1}^{p-1}$$
and $$ x_{1}^{p} = \frac{1}{n}$$
Now, since $p$ is even $$x_{1} = +-\frac{1}{n^{\frac{1}{p}}}$$
A comment is in order, in the calculations above a few times we used the fact that
$x_{k}$ and $\lambda$ are not $0$ but this is obvious because at extreme points these values can not to be $0$ (otherwise max or min would be $0$ which is not the case).

To finish, we notice that to form a maximum even $x_{k}$ have to come with sing $-$ and for minimum an odd number has to have sign $-$.


问:
What is the probability for each number in a set of random numbers to be the highest number in the set?
I have viewed/researched many problems on this website but this is my first question.  I'm a little apprehensive by the length of the question but I have tried to explain it fully including a little of my thought process.  The problem is ...

I have n numbers in a set (N1, N2, N3 ... Nn). For each number I choose a random number (R1, R2, R3 ... Rn) between 1 and the number (e.g. 0 < R1 < N1, 0 < R2 < N2, 0 < R3 < N3 ... 0 < Rn < Nn ).  My quest is to seek the probability for each random number that it should be the largest of the random numbers in the set i.e. P(R1=max(R1 ... Rn)), P(R2=max(R1 ... Rn)) etc. up to P(Rn=max(R1 ... Rn)).

For example and by way of a Monte Carlo simulation (MCS) with a count of 10,000 I get the following ...

N1=300, N2=250, N3=450

R1 is Max(R1,R2,R3) 2576 times i.e. P(R1=Max(R1,R2,R3)) = 0.2576

R2 is Max(R1,R2,R3) 1526 times i.e. P(R2=Max(R1,R2,R3)) = 0.1526

R3 is Max(R1,R2,R3) 5898 times i.e. P(R3=Max(R1,R2,R3)) = 0.5898

Obviously the MCS will produce a slightly different result each time it is run. On 10 runs the MCS returned results of ...

```<code>  N1   N2  N3
</code>```

Min 2504 1499 5854

Ave 2546 1534 5921

Max 2591 1564 5972

Interestingly using some Mathematica code as provided by the Wolfram forum the probabilities were calculated as ...

P(R1=Max(R1,R2,R3))= 0.256173

P(R2=Max(R1,R2,R3))= 0.154321

P(R3=Max(R1,R2,R3))= 0.589506

So it seems there is probably a formula to return the expected result from an infinite number of MCS runs but it's proving difficult to find it!

A helpful Wolfram forumite from another time zone promised to help me after a good night's sleep but unfortunately they deleted my thread before he woke up because I was not seeking Mathematica support!  :(

Anyway if someone could help with a formula or point to where I may find one I would be very grateful.  I need to be able to resolve this using Excel, SQL or VBA because the result needs to link with an existing SQL database so the MCS option is not really suitable.  That was only for testing and is not a viable option because I have c.150,000 number sets with 75% of them having between 5 and 12 numbers to each set and the thought of performing that number of MCS's throws my computer into a cold sweat! :)

Well if you have got this far thank you for reading my question ... it is long but hopefully most of the content is for explanation and not question.

答:
Assume we are interested in the probability that the number $R_m$ is the largest one.

The probability of this is:
$$
P(R_m)=\frac1{N_m}\sum_{i=1}^{N_m}\prod_{k\ne m}\frac{\min(i,N_k)}{N_k}.\tag1$$

Particularly for your example one obtais:
$$P_1=0.257285,\ P_2=0.590619,\ P_3=0.155248.$$

This differs slightly from your result and besides does not sum to $1$. The reason for this is the fact that the result of a trial is ambiguous if there are two or more largerst numbers, so that the events "$R_1$ is the largest number" and "$R_2$ is the largest number" do not exclude each other if $R_1=R_2$.

Actually it is very strange that you did not report about such cases, because the probability of an event with multiple "winners" in your example is $0.0031$ and one would expect about 31 such events in 10000 trials. In your statistics however they are not present. I suspect you either discard such trials or somehow assign the largest property to one of the "winners".

UPDATE for real random numbers.

In the case of real random numbers the equation (1) turns into:
$$
P(R_m)=\frac1{N_m}\int_0^{N_m}\prod_{k\ne m}\frac{\min(x,N_k)}{N_k}dx\tag2
$$

This expression can be simplified as follows. Let
$N_1\le N_2\le \dots\le N_n$. Then
$$\begin{align}
P(R_m)&=\frac1{\prod\limits_{k}N_k}
\Big[\int_0^{N_1}x^{n-1}dx+N_1\int_{N_1}^{N_2} x^{n-2}dx\\
&+N_1N_2\int_{N_2}^{N_3} x^{n-3}dx+\cdots+
N_1N_2\dots N_{m-1}\int_{N_{m-1}}^{N_m} x^{n-m}dx
\Big]\\
&=\frac{\prod_{k>m}\frac{N_m}{N_k}}{(n-m+1)}
-\sum_{i=1}^{m-1}\frac{\prod_{k>i}\frac{N_i}{N_k}}{(n-i+1)(n-i)}.
\tag3
\end{align}$$

Particularly in your example one obtains:
$$
P(R_2)=\frac{50}{324}\approx0.154321,\quad P(R_1)=\frac{83}{324}\approx0.256173,\quad P(R_3)=\frac{191}{324}\approx0.589506.
$$
in agreement with reported Mathematica result.


问:
How to calculate the circumference of a circle, excluding the part intersecting another circle?

I have one circle of $11.63$ inch bricks that is $21.25$ ft in radius (green in the picture). I already purchased the bricks for this circle already.

I need to add another circle to the side (marked in red) of bricks. The size is scaled by the golden ratio of $1.618$, or $13.133$ ft in radius. The center of this red circle falls directly on the edge of the green circle.

I need to purchase bricks to form the second (red) circle, but I don't need bricks to go inside the green circle.

How do I find the circumference of circle B, minus the part intersected with circle A, so I can calculate how many bricks to buy?

答:
The $36^{\circ}-72^{\circ}-72^{\circ}$ $\triangle ABC$ is called the golden triangle as its sides are $a/\phi,a,a$.

Desired red circumference is
$$2\pi\cdot \frac{a}{\phi} \cdot\left(1-\frac{2\cdot72}{360}\right)$$
$$=2\pi\cdot \frac{a}{\phi}\cdot \frac{3}{5}$$
$$=\frac{6\pi}{5}\cdot \frac{a}{\phi} $$

where $a=21.25$


问:
Pigeon Hole: The row of numbered houses problem, why couldn't I include the mentioned ranges in my pigeon holes options?
The problem is as follow:

A row of houses are randomly assigned distinct numbers between 1 and 50 (inclusive). How many houses must there be to insure that there are 5 houses numbered consecutively?

Its solution:

Split the numbers into 10 pigeonholes: 1-5, 6-10, 11-15, 16-20… There must be at least           =41 “pigeons”=houses

my problem is that: why haven't we include the ranges (2-6, 3-7, 4-8, 5-9, ... 12-16 ..) within the ones mentioned in the solution? they're consecutive 5 numbers as well, so why not?

答:
So, taking "there are five houses numbered consecutively", which is slightly ambiguous, to mean "there are houses which use five consecutive numbers":

The solution given certainly demonstrates that there is no way to avoid five consecutive numbers given $41$ houses, based on the strong pigeonhole principle, and a small demonstration (e.g. remove the smallest of each group) shows that $40$ houses are not sufficient.

The use of the non-overlapping "pigeonholes" that use all the available numbers means there is less complexity about the proof presented, and since it achieves the result there is no need for further assessment.


Body:
Hints:

$$\begin{align*}\bullet&\;\;\;\;1+q+q^2+\ldots+q^n=\frac{q^{n+1}-1}{q-1}\\\bullet\bullet&\;\;\;\;1+2+\ldots+n=\frac{n(n+1)}2\\\bullet\bullet\bullet&\;\;\;a\cdot a^2\cdot\ldots\cdot a^n=a^{1+2+\ldots+n}\end{align*}$$


答:
The solution requires two parts:

With forty houses (namely all except $5,10,15,\ldots$), it is possible to not have five consecutives

With more than forty houses, we will always have five consecutive numbers

It does not matter how we prove the second point if only we do it in a way that wprks.
If we want to use the pigeon-hole principle, we better find a couple of disjoint sets as "holes" into which we place our houses/pigeons. It suggests itself to use sets of five consecutive numbers as holes. We could use sets such as $\{1,2,3,4,5\}$ and/or $\{2,3,4,5,6\}$ and/or $\{3,4,5,6,7\}$, but to achieve disjointness, we better take $\{1,2,3,4,5\}$, $\{6,7,8,9,10\}$, $\{11,12,13,14,15\}$ and so on up to $\{46,47,48,49,50\}$. Fortunately, this results in $10$ "holes" so that of $41$ "pigeons", there will be at least five entering the same hole and we are done.

Other five-sets are simply not helpful.

Title: Pigeon Hole: The row of numbered houses problem, why couldn't I include the mentioned ranges in my pigeon holes options?

Body:
The problem is as follow:

A row of houses are randomly assigned distinct numbers between 1 and 50 (inclusive). How many houses must there be to insure that there are 5 houses numbered consecutively?

Its solution:

Split the numbers into 10 pigeonholes: 1-5, 6-10, 11-15, 16-20… There must be at least           =41 “pigeons”=houses

my problem is that: why haven't we include the ranges (2-6, 3-7, 4-8, 5-9, ... 12-16 ..) within the ones mentioned in the solution? they're consecutive 5 numbers as well, so why not?


问:
Probability of throwing exactly 3 times 6 in 5 die rolls
Suppose you roll a die until you roll three six. Calculate the probability that you need exactly 5 throws for this.

My solution is the following, however it supposed to be wrong since the 'correct' answer is $0.0193$ (a colleague verified this with a simulation).

Solution:
Suppose
$$G= \text{exactly five throws is needed to throw three times 6}$$
It is clear that $G$ is determined by all events where the fifth throw is a $6$ and where within the first four throws there are exactly two $6$. You can partition $G$ in two sets, namely

a) $\omega=(6,6,x,x,6)$ where $x$ is a number from $\{1,2,3,4,5\}$, or

b) $\omega'=(6,6,x,y,6)$ where $x,y$ are different numbers from $\{1,2,3,4,5\}$.

The probobality of an a) like event is
$$P(\omega)=\frac{1}{6}\cdot\frac{1}{6}\cdot \frac{5}{6}\cdot \frac{1}{6} \cdot \frac{1}{6}=\left(\frac{1}{6}\right)^4\cdot \frac{5}{6} $$ since the 5 throws are statistically independent.

The probablity of an b) like event is
$$P(\omega')=\left(\frac{1}{6}\right)^3\cdot \frac{5}{6}\cdot \frac{4}{6}.$$

To compute $P(G)$ we need to take the arrangements into account. With a) you have  $\overline{P}_4^{2,2}=\frac{4!}{2!2!}$ different arrangements. And with b) there are $\overline{P}_4^{2,1,1}=V_4^2=\frac{4!}{2!}$ different arrangements.\

The probablity we are after is
$$P(G)=\frac{4!}{2!2!}\cdot \left(\frac{1}{6}\right)^4\cdot \frac{5}{6}+\frac{4!}{2!}\left(\frac{1}{6}\right)^3\cdot \frac{5}{6}\cdot \frac{4}{6}=0.0347  $$

答:
I have called out the error in your working in the end. Throw of die will follow the binomial distribution and here is a simple way to look at this -

Of the first four throws, two are $6$ with probability $\frac{1}{6}$ for a throw and other two are any of the remaining five ($1, 2, 3, 4, 5$) with a probability $\frac{5}{6}$ for a throw.

Any of the two out of four throws can be $6$ and remaining two throws are decided automatically by that. Also the last throw has to be six and probability of that is again $\frac{1}{6}$.

So the desired probability is $ = {4 \choose 2} \times (\frac{1}{6})^2 \times (\frac{5}{6})^2 \times \frac{1}{6} = \frac{25}{1296} \approx 0.0193$

EDIT: Coming to mistake in your working - in the second part of your working, please note that you are choosing two non-six numbers by $5 \cdot 4$ which assumes both positions are distinct. For example, if the first number is $5$, four choices of second number will give you $5, 1; 5, 2; 5, 3; 5, 4$. But when the first number is say $1$, the second number can be $5$. So you get both $5, 1$ and $1, 5$. Then when you are arranging positions of first $4$ numbers, you will again rearrange $5, 1$ as $1, 5$ and vice versa. That leads to duplicates. So there are two ways to fix your working. Keep the positions unnamed by dividing $5 \cdot 4$ by $2!$ or while making arrangements, go with $\frac{4!}{2! \ 2!}$. One of $2!$ is to account for the fact that two non-six numbers do not need to be arranged within and only need to be arranged with respect to six.


问:
compute $\int_\gamma\frac{-y^2dx+2xy\ dy}{x^2+y^4}$

The question is:
$$
\int_\gamma\frac{-y^2dx+2xy\ dy}{x^2+y^4}
\quad \gamma:r(t)=(t,2t^2-2), \quad -1\leq t\leq 1
$$

I have tried to solve it like this:
since $Q_x=P_y$ it's potential vector field but the singularity is at origin, first i thought that i could evaluate it from $(-1,0)$ to $(1,0)$ along $x$ axis but the singularity is the problem, how should i proceed in this case?

Any suggestion would be great Thanks

答:
$$
\int_\gamma\frac{-y^2dx+2xy\ dy}{x^2+y^4}=\int_\gamma\frac{-y^2}{x^2+y^4}\,dx+\int_\gamma\frac{2xy}{x^2+y^4}\,dy$$
$$\gamma(t)=(t,2 t^2-2);\;t\in[-1,1];\;\gamma'(t)=(1, 4 t)$$
Substitute
$$\int_{-1}^1 -\frac{\left(2 t^2-2\right)^2}{\left(2 t^2-2\right)^4+t^2}\cdot 1\,dt+\int_{-1}^1 \frac{2 t \left(2 t^2-2\right)}{\left(2 t^2-2\right)^4+t^2}\cdot(4t)\,dt=$$
$$=\int_{-1}^1 \frac{4 \left(3 t^4-2 t^2-1\right)}{16 t^8-64 t^6+96 t^4-63 t^2+16}\,dt$$
Integrating is quite hard

denominator $16 t^8-64 t^6+96 t^4-63 t^2+16=(16 t^8-64 t^6+96 t^4-64 t^2+16)+t^2=16 \left(t^2-1\right)^4+t^2$

numerator can be factored $4 \left(3 t^4-2 t^2-1\right)=4\left(t^2-1\right) \left(3 t^2+1\right)$

and integral can be written as
$$\int_{-1}^1\frac{4\left(t^2-1\right) \left(3 t^2+1\right)}{16 \left(t^2-1\right)^4+t^2}\,dt$$
Now divide numerator and denominator by $16 \left(t^2-1\right)^4$
$$\int_{-1}^1\frac{\frac{4\left(t^2-1\right) \left(3 t^2+1\right)}{16 \left(t^2-1\right)^4}}{1+\frac{t^2}{16 \left(t^2-1\right)^4}}\,dt=\int_{-1}^1\frac{\frac{ 3 t^2+1}{4 \left(t^2-1\right)^3}}{1+\frac{t^2}{16 \left(t^2-1\right)^4}}\,dt$$
Now substitute $\frac{t}{4\left(t^2-1\right)^2}=u$

extremes become $t=-1\to u=-\infty;\;t=1\to u=+\infty$

while differentiating we get $\frac{3 t^2+1}{4 \left(t^2-1\right)^3}\,dt=-du$

Therefore the integral becomes
$$-\int_{-\infty}^{\infty} \frac{du}{1+u^2}=-\left[\arctan u\right]_{-\infty}^{\infty}=-\pi$$