问:
Example of a finitely generated flat module which is not free
I couldn't come up with an example of a finitely generated flat module which is not free. I know that over local rings, freeness and flatness are equivalent. So the ring cannot be a local ring. </p>

答:
Here is an example that one can come up with from algebraic number theory. Consider $K = \Bbb{Q}(\sqrt{-6})$ and the ideal $I = (2,\sqrt{-6}) \subseteq \mathcal{O}_K$. This ideal is clearly finitely generated as an $\mathcal{O}_K$ module and furthermore not only is it flat but it is a projective module: we have

$$I \oplus ( 3,\sqrt{-6}) \cong (\sqrt{-6}) \oplus \mathcal{O}_K \cong \mathcal{O}_K \oplus \mathcal{O}_K.$$

If you're interested in knowing how we have such isomorphisms you can look at my answer here. But now $I$ cannot possibly be free because the generators are linearly dependent over $\mathcal{O}_K$:

$$3 \cdot (2) + \sqrt{-6} \cdot (\sqrt{-6}) = 0.$$


问:
Show that a linear system of equations has a solution if and only one of the unknowns is set equal to a known value
Consider the following linear system
$$
\begin{cases}
Y_{1}+Y_2 & =&2\alpha_1+\gamma_1+\gamma_2\\
Y_{3}+Y_4&=&2\alpha_2+2\gamma_1\\
Y_{5}+Y_6&=&2\alpha_3+\gamma_2+\gamma_3\\
Y_{7}+Y_8&=&2\alpha_4+2\gamma_3\\
Y_1+Y_3+Y_4&=&\alpha_1+2\alpha_2+3\gamma_1\\
Y_2+Y_6&=&\alpha_1+\alpha_3+2\gamma_2\\
Y_5+Y_7+Y_8&=&\alpha_3+2\alpha_4+3\gamma_3\\
\end{cases}
$$
with unknowns $x\equiv (\alpha_1,\alpha_2,\alpha_3,\alpha_4, \gamma_1,\gamma_2, \gamma_3)$.

Question: Could you help me to show that this system has a unique solution with respect to $x$ if and only if one component among $\{\alpha_1,\alpha_2,\alpha_3,\alpha_4, \gamma_1,\gamma_2, \gamma_3\}$ is set equal to a known value?

My attempt: I have tried (without success, could you help if you think this is the right path?) to show that one equation among those listed above is a linear combination of the others. Once that is shown, we are left with a linear system in $6$ equations and $7$ unknowns. How do we go from there?

答:
The way I approached this problem was to write the equations in ${\alpha_i, \gamma_i}$ as a matrix. This is, I think, the best way to determine the linear dependence of your system.
The matrix I got was:
$$
\left[
\begin{array}{l}
2& 0& 0& 0& 1& 1& 0\\
0& 2& 0& 0& 2& 0& 0\\
0& 0& 2& 0& 0& 1& 1\\
0& 0& 0& 2& 0& 0& 2\\
1& 2& 0& 0& 0& 3& 0\\
1& 0& 1& 0& 0& 2& 0\\
0& 0& 1& 2& 0& 0& 3
\end{array}
\right]
$$

Then, I just took it's reduced row echelon form, which according to Julia this is: $$
\begin{array}{l}\alpha_1& \alpha_2& \alpha_3& \alpha_4& \gamma_1& \gamma_2& \gamma_3\end{array}
$$
$$
\left[\begin{array}{l}
 1&  0&  0&  0&  0&  0&   1\\
 0&  1&  0&  0&  0&  0&   1\\
 0&  0&  1&  0&  0&  0&   1\\
 0&  0&  0&  1&  0&  0&   1\\
 0&  0&  0&  0&  1&  0&  -1\\
 0&  0&  0&  0&  0&  1&  -1\\
 0&  0&  0&  0&  0&  0&   0
\end{array}
\right]
$$
As can be clearly seen, on the rightmost column of this vector that one parameter must in fact be a free parameter. That is, every other parameter is defined in terms of this free parameter (can actually be any element of the set, but as soon as it's set you're stuck with the choice). Therefore, this system necessarily needs a known value for the value of the free parameter to be uniquely soluble.

ContentLicense: CC BY-SA 4.0


问:
Proving something is an inner product
I am faced with the following question:

Prove that this is an inner product on V.

I understand to show something is an inner product I must verify the following conditions:

$\langle\,f,g\rangle = \overline{\langle\,g,f\rangle}$ (Conjugate symmetry)

$\langle\,\overline\alpha (f + g),h\rangle = \overline\alpha(\langle\,f,h\rangle + \langle\,g,h\rangle)$ (Linearity in the first argument)

$\langle\,f,f\rangle \ge 0$ with equality iff $f=0$

(Note: conditions (1) and (2) are slightly different to usual as we are dealing with complex numbers)

I have seen other examples of this being done, however, I am not sure how to do it in this particular case. Any help with this would be greatly appreciated.

答:
If you have $A$ and $B$, compute $C_{ij} := \sum_k A^*_{ki}B_{kj}$. That's what $A^\dagger B$ is all about, and with it you can compute its trace, giving
\begin{align}
\langle A, B\rangle &= \sum_{i} A^*_{ki}B_{ki}.
\end{align}

Straightforward this way to work through each of the conditions you list for an inner product.

Title: Proving something is an inner product


问:
Is $2\mathrm p_z$ orbital wave function three- or four-dimensional?
The wave function of the $2\mathrm{p}_z$ orbital is

$$Ψ = \frac{1}{4\sqrt{2π}}\left(\frac Z a\right)^{5/2} r \mathrm e^{-Zr/a}\cos θ.$$

I'm confused if this function will be a three-dimensional function or a four-dimensional function?

My professor told me that

$$z/r = \cos θ,$$

where $θ$ is the angle made by $r$ with the $z$ axis. So, I guess it will be four-dimensional since we would have to then plug in

$$r = \sqrt{x^2 + y^2 + z^2}$$

in the equation.

But when he plotted the function, he plotted a three-dimensional one! What is the correct answer and why?

答:
Hydrogen-like orbitals are always 3-dimensional. But, since we can't directly visualize functions $\mathbb R^3\mapsto \mathbb R$, to plot them you need some trick.

One of such tricks it to take a cross section of the 3D space, e.g. xOz plane, and plot the function inside this plane. Then, knowing that the function is symmetric with respect to rotation around the $z$ axis, you can imagine what it could look like in other cross sections (or just plot these too). Here's an example:

Another way is to draw the function in a density plot. This will look like a cloud of colored points, with their opacity depending on the function value. Example (here blue is negative, orange is positive):


问:
How can I prove that A is the associated Matrix to g(x)
What are nice ways to prove that

$$A= I_{n}+w v^{T}$$ is the associated matrix of the following function
$$g (x)=x+\left(v^{T} x\right) w$$ and $v,w \in \mathbb R^n$

My idea is to show that A is just the picture of the unit vectors but somehow I am stuck.
Thanks for your help.

答:
The expression $\left(v^{T} x\right) w$ doesn't use actual matrix multiplication all throughout (the dimensions don't fit, as $v^Tx$ is $1\times 1$ and $w$ is $n\times 1$). However, turning it around and writing $w\left(v^{T} x\right)$, now it does. So we don't need the parentheses any more, as matrix multiplication is associative, and we can instead just write $wv^Tx$.

This gives
$$
g(x) = x + wv^Tx = I_nx + wv^Tx = (I_n + wv^T)x
$$
which is what we wanted to show.


问:
Computing the convolution of random variables to determine $Z=X+Y$
I am trying to understand the calculations in the following solved exercise:

Let $X$ and $Y$ two independent random variables with pdf $f(x)=2x, x\in (0,1); f(x)=0, \text{otherwise}$. Determine the pdf of $Z=X+Y$.

For $z\in (0,1)$:
$$f_Z(z)=\int_{\mathbb{R}}2x 1_{(0,1)}(x) 2(z-x)1_{(z-1,z)}(x)dx=\int_{0}^{z}2x2(z-x)=\frac{2}{3}z^3.$$

For $z\in (1,2)$: $$f_Z(z)=\int_{\mathbb{R}}2x 1_{(0,1)}(x) 2(z-x)1_{(z-1,z)}(x)dx=\int_{z-1}^{1} 2x2(z-x)dx=\frac{2}{3}(6z-z^3-4).$$

I know the definition of convolution and I get why we are using it to compute the pdf of the sum of two random variables but I don't get why are the endpoints of the integral in the first case $0$ and $z$ and why are they in the second case $z-1$ and $1$?

答:
In short, it is simply because

for $z\in(0,1)$,
$$
1_{(0,1)}(x)1_{(z-1,z)}(x)=1_{\color{red}{(0,z)}}(x)
$$

for $z\in(1,2)$,
$$
1_{(0,1)}(x)1_{(z-1,z)}(x)=1_{\color{red}{(z-1,1)}}(x)
$$

Notes.

Since $X$ and $Y$ are i.i.d, the joint density is
$$
f_{X,Y}(x,y)=f(x)f(y)
$$
and the pdf of $Z$ is given by
$$
f_Z(z)=\int_{-\infty}^\infty f_{XY}(x,z-x)dx=
\int_{-\infty}^\infty f(x)f(z-x)dx
$$

For any given $z\in\mathbb{R}$, the integrand $f(x)f(z-x)$ is nonzero if and only if
$$
0<x<1\quad\textrm{and }\quad 0<z-x<1
$$
or equivalently,
$$
0<x<1\quad\textrm{and }\quad z-1<x<z\tag{1}
$$

If the constraint (1) is nonempty, then necessarily
$$
z-1<1\quad \textrm{and }\quad z>0
$$

That is why $f_Z(z)$ is only possibly nonzero for $0<z<2$.

If $z\in(0,1)$, then $z-1<1$ and thus the constraint (1) is the same as
$$
0<x<z
$$

If $z\in (1,2)$, then $z>1$ and thus the constraint (1) is the same as
$$
z-1<x<1
$$


问:
Question about null-homotopic function
How to prove that every continuous function $f:M\to X$ from the Möbius strip $M$ into a simply connected space $X$ is null-homotopic? Thanks in advance.</p>

答:
$$ \newcommand{\ra}[1]{\kern-1.5ex\xrightarrow{\ \ #1\ \ }\phantom{}\kern-1.5ex}
\newcommand{\da}[1]{\bigg\downarrow\raise.5ex\rlap{\scriptstyle#1}} $$

The Möbius strip $M$ is the quotient space of $I^2$ obtained by the identification $(0,y)\sim(1,1-y)$.
\begin{array}{ccc}
I^2\times I &\ra d &I^2\\
\da{q\times Id} & &\da q\\
M\times I &\ra{\tilde d} &M
\end{array}
The square $I^2$ deformation retracts onto $I\times\left\{\frac12\right\}$ via $d((x,y),t)=tr(x,y)+(1-t)(x,y)=\left(x,\ \frac t2+(1-t)y\right)$ where $r(x,y)=\left(x,\frac12\right)$ is the retraction. Since $d((0,y),t)\sim d((x,1-y),t)$ (i.e. it respects the identification) and $q\times Id$ is a quotient map, it induces a deformation retraction $\tilde d$ of $M$ onto $\left(I\times\left\{\frac12\right\}\right)/\sim\ \approx S^1$. Composing $\tilde d$ with $f$ we obtain a homotopy between $f$ and $f\circ r$, the latter being the image of a circle in $X$. This circle can be contracted in $X$ since it is simply connected. Concatenating these homotopies gives a homotopy between $f$ and a constant map.

ContentLicense: CC BY-SA 3.0

Title: Question about null-homotopic function


问:
Which number is greater? $2\sqrt{2}$ or $e$
I have to determine which number is greater, $2\sqrt{2}$ or $e$.

I had a similar question as well, it was $2^\sqrt{2}$ compared to $e$.

For that one I managed to prove the inequality by using the increasing sequence converging to $e$: $(1+\frac1n )^n $

So I just searched for a value to assign to n such that $(1+\frac1n )^n \gt 2^\sqrt2$

I tried to proceed in a similar way with $2\sqrt{2} \gt \lt e$ , but it seems I can't get nowhere (I used the sequence decreasing and converging to $e$:  $(1+\frac1n )^{n+1}$ )

Is there another way to prove the inequality without the use of the calculator and maybe using derivatives? The question was in a derivatives file, so I'm wondering is there's a way to get to the end using them.

Any hint would be much appreciated, thanks.

Body:
$$\ln (2×\sqrt 2)=\ln 2+\dfrac 12 ×\ln 2=\dfrac 32 \ln2$$

$$\ln 2 = \frac{2}{3} +\frac{3}{4}\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n(n+1)(n+2)(n+3)}>\dfrac 23$$

$$ \implies \ln (2×\sqrt 2)>1=\ln e \Longrightarrow 2×\sqrt 2 >e.$$


Title: Which number is greater? $2\sqrt{2}$ or $e$

Body:
I have to determine which number is greater, $2\sqrt{2}$ or $e$.

I had a similar question as well, it was $2^\sqrt{2}$ compared to $e$.

For that one I managed to prove the inequality by using the increasing sequence converging to $e$: $(1+\frac1n )^n $

So I just searched for a value to assign to n such that $(1+\frac1n )^n \gt 2^\sqrt2$

I tried to proceed in a similar way with $2\sqrt{2} \gt \lt e$ , but it seems I can't get nowhere (I used the sequence decreasing and converging to $e$:  $(1+\frac1n )^{n+1}$ )

Is there another way to prove the inequality without the use of the calculator and maybe using derivatives? The question was in a derivatives file, so I'm wondering is there's a way to get to the end using them.

Any hint would be much appreciated, thanks.

Body:
\begin{gather*}
3^5 = 243 < 256 = 2^8, \
\therefore\ \log_32 > \frac58; \\
\log_e3 = \log_e\left(1 + \frac12\right) - \log_e\left(1 - \frac12\right) > 2\left(\frac12 + \frac13\left(\frac12\right)^3\right) = \frac{13}{12}; \\
\therefore\ \log_e2 = \log_32 \cdot \log_e3 > \frac58\cdot\frac{13}{12} = \frac{65}{96} > \frac{64}{96} = \frac23, \
\therefore\ 2 > e^{2/3}, \
\therefore\ 2^{3/2} > e.
\end{gather*}


问:
Application with differential calculus
An airline will fill $100$ seats of its aircraft at a fare of $\$200.$ For every $ \$5$ increase in the fare, the plane loses two passengers. For every decrease of $\$5$, the company gains two passengers. What price maximizes revenue?

答:
$$\text{Revenue} = (200+5k)(100-2 k).$$

To maximize revenue, find its derivative with respect to $k$.  To know you are at a maximum, rather than at a minimum, look at the second derivative.  Alternatively, graph the function.

By the way, one subtlety is that the number of passengers is integer.  You might want to think about why this method works, nevertheless.

UPDATE

As @AndrewChin wrote in a comment, the most straightforward way to look at this is to understand that this is a quadratic function (a parabola if you expand the support to the reals).  Writing the revenue as:

$$\text{Revenue} = 10(2025-(5-k)^2),$$ it should be clear what $k$ maximizes this.

Title: Application with differential calculus


问:
$\lim_{n \rightarrow \infty}\int_0^1 |f_n(x)| dx = 0$ implies there is subsequence such that $\{f_{n_i}(x)\}$ converges to $0$ for a.e. $x$
Given that $\{f_n\}$ be a sequence of Lebesgue-measurable real-valued functions on $[0, 1]$ such that
\begin{equation}
    \lim_{n \rightarrow \infty}\int_0^1 |f_n(x)| dx = 0
\end{equation}

I want to show there is a subsequence of $\{f_n\}$ such that $\{f_{n_i}(x)\}$ converges to $0$ for a.e. $x$.

Im thinking of using the inequality $\int_0^1 |f_n(x)| dx > \mu(\{x: |f_n(x)| > \epsilon\})\epsilon$ but I didnt' get very far past that.

答:
Note that with $A_n = \{x : |f_n(x)| > \epsilon\}$,

$$\mu(A_n) \leqslant \frac{1}{\epsilon}\int_{A_n} |f_n| \leqslant  \frac{1}{\epsilon}\int_{[0,1]} |f_n|\underset{n \to \infty} \longrightarrow 0 ,$$

and $f_n \to 0$ in measure. Now you can use the well-known theorem that convergence in measure implies a.e. convergence of a subsequence proved here.


答:
It is easy to show (by induction) that

$${2^n\over n!}\lt{1\over2^{n-4}}$$

for all $n\ge0$. It follows that

$$\begin{align}
e^2&=1+2+{2^2\over2!}+{2^3\over3!}+{2^4\over4!}+{2^5\over5!}+\cdots\\
&=1+2+2+{4\over3}+{2\over3}+\left({2^5\over5!}+{2^6\over6!}+\cdots \right)\\
&=7+\left({2^5\over5!}+{2^6\over6!}+\cdots \right)\\
&\lt7+\left({1\over2}+{1\over4}+{1\over8}+\cdots\right)\\
&=7+1\\
&=8
\end{align}$$

so $e\lt\sqrt8=2\sqrt2$.

Remark: The induction proof for $2^n/n!\lt1/2^{n-4}$, best rewritten as $4^n\lt16n!$, requires verifying the first few "base" cases; the induction itself kicks in when $4\le n+1$:

$$4^n\lt16n!\implies4^{n+1}=4\cdot4^n\lt4\cdot16n!\le(n+1)16n!=16(n+1)!$$


问:
Relative Motion of Three Particle System
Here is the question.

Particle A moves North with a constant speed of $V$ and Particle B moves South with the same constant speed of $V$. Another particle C moves to the South-east with respect to particle A and and the same particle C moves making an angle of $30$ degrees with North, with respect to particle B. What is the velocity of particle C?

I have written two relative velocity equations for both A and B.

$\vec V_{C|A}=\vec V_{C}-\vec V_{A}$

$\vec V_{C|B}=\vec V_{C}-\vec V_{B}$

I can't seem to figure out what to do next. There seems to be three unknowns whereas I only have two equations. How should I solve this ?

答:
Vector $\vec V_C$ is the median of a triangle with base $2V$ and base angles of $30°$ and $45°$ (see figure). With a little trig you can then compute:
$$
V_C=\sqrt{11-6\sqrt3}\ V.
$$


问:
Better way to come up with a singular 3x3 matrix than trial and error
I am given a task to give an example of a singular matrix, for a 2x2 is very simple to try out some combinations of numbers to eventually get a determinant=0. But I find it a time consuming to go by trial and error for a 3x3 matrix.

Criteria:

No zero entries

No rows are multiple of other rows

What I know:

Formula to find determinant of 3x3 matrix.

Zero determinant means Singular matrix.

答:
How about the zero matrix?

How about the matrix $$\begin{pmatrix} 1&0&0\\0&1&0\\0&0&0\end{pmatrix}?$$

Or any matrix with a row of zeros.

More generally, a matrix will be singular if its rows or columns are linearly dependent. So choose any two rows you want, and then choose some multiples of each and add them together to get the third row. This will give a less obvious singular matrix.

For example, suppose I choose $(1, 2, 3)$ and $(1, -2, 2)$ as my first two rows, and then subtract them for the third. This gives

$$\begin{pmatrix}1&2&3 \\ 1&-2&3 \\ 0 & 4 & 1 \end{pmatrix},$$

which is perhaps less obviously singular.


问:
Is it possible to factor a polynomial over a ring which is not an integral domain?
Is it possible to factor a polynomial over a ring which is not an integral domain? I assumed not, but I'm wondering about the following scenario.

Let $f \in \mathbb{Z}[x]$ be monic and irreducible but factor completely over primes $p$ and $q$. Given the roots of $f$ mod $p$ and $q$ we can find roots in $\mathbb{Z}/pq\mathbb{Z}$ by CRT. (It may not be apparent which roots to combine via CRT to get a meaningful result, but it seems possible nonetheless).

Can we find these roots by factoring $f$ over $\mathbb{Z}/pq\mathbb{Z}$ directly? If so, is there an algorithm or reference?

答:
Example:

Take $p = 2$, $q = 3$ and $f = x^2 - x \in \Bbb Z / 6\Bbb Z[x]$.

We then have $f = x(x - 1) = (x - 3)(x - 4)$. Which one should be "the" factorization of $f$?

As you can see, the problem is that $f$ is a polynomial of degree $2$, yet has $4$ different roots in $\Bbb Z / 6 \Bbb Z$. Thus elements of the ring $\Bbb Z/6\Bbb Z[x]$ cannot be factored uniquely.

This also means that you cannot find all the roots by "factoring" the polynomial over $\Bbb Z/6\Bbb Z$.


问:
Birthday Problem: Expected number of people in a room
If an infinite amount of people enter a room one by one, what is the expected number of people in the room when you first find two that share the same birthday?
(Assuming no leap years and every birthday is equally likely).</p>

答:
Whenever $N$ is a random variable which takes values in the non-negative integers $\{0,1,2,3,\ldots\}$, there's a nice formula for the expected value $\mathbb{E}[N]$ of $N$:
$$\mathbb{E}[N] = \sum_{n=1}^\infty \mathbb{P}(N \geq n) = \sum_{n=0}^\infty \mathbb{P}(N > n)$$

In this case, the second expression is the nicest, because if $N$ is the number of the first person who enters with a birthday matching a birthday in the room, then
$$\begin{align*}\mathbb{P}(N > n) &= \text{ Probability that none of the first } n \text{ people has a birthday in common with another} \\ &= \left(\frac{365}{365}\right)\left(\frac{364}{365}\right)\cdots\left(\frac{366-n}{365}\right) \\ &= \frac{365!}{365^n(365-n)!} \qquad \text{ if } 0\leq n \leq 365 \text{ and zero otherwise}\end{align*}$$

Then
$$\mathbb{E}[N] = \sum_{n=0}^{365} \frac{365!}{365^n(365-n)!} \approx 24.616586$$

Note that this is close to, but not equal, the well-known median of $23$.  In fact, if you scroll down that page to here, you'll see the same formula we derived along with the same result we've found for the expected value.


答:
Here is a method that, in principle, can give you any rank-$2$ invertible matrix. Select any two matrices for which each row is a multiple of the first (i.e. rank-$1$ matrices), and add them together.

For example: consider the matrices
$$
\pmatrix{1 & 2 & 3\\2&4&6\\4&8&12}, \quad \pmatrix{1&1&1\\2&2&2\\1&1&1}.
$$
Adding these together yields
$$
\pmatrix{2&3&4\\4&6&8\\5&9&13}.
$$
This matrix has rank $2$.

The only thing that can go wrong with this method is that we "accidentally" end up with a matrix for which condition 2 fails to hold. In order to make sure that this does not happen, it suffices to make sure that the rows of the first matrix are not multiples of the rows of the second and the columns of the first matrix are not multiples of the columns of the second.

Title: Better way to come up with a singular 3x3 matrix than trial and error

Body:
I am given a task to give an example of a singular matrix, for a 2x2 is very simple to try out some combinations of numbers to eventually get a determinant=0. But I find it a time consuming to go by trial and error for a 3x3 matrix.

Criteria:

No zero entries

No rows are multiple of other rows

What I know:

Formula to find determinant of 3x3 matrix.

Zero determinant means Singular matrix.


问:
How to prove there is no bijection between two different natural numbers?
I am trying to draw a contradiction from "there exists a bijection between natural numbers $\boldsymbol{3}$ and $\boldsymbol{4}$" using the pigeon-hole theorem. This theorem is given by

For any $\boldsymbol{n} \in \mathbb{N}$, if a map $f: \boldsymbol{n} \to \boldsymbol{n}$ is one-one, then it is onto.

Here is my argument. As
\begin{equation*}
\boldsymbol{3} = \left\{\boldsymbol{0},\boldsymbol{1},\boldsymbol{2}\right\},
\end{equation*}
\begin{equation*}
\boldsymbol{4} = \left\{\boldsymbol{0},\boldsymbol{1},\boldsymbol{2},\boldsymbol{3}\right\},
\end{equation*}
and there exists a bijection between $\boldsymbol{3}$ and $\boldsymbol{4}$, then there has to be a bijection $f: \boldsymbol{4} \to \boldsymbol{3}$, which is one-one and onto. As
\begin{equation*}
\boldsymbol{3} \subsetneq \boldsymbol{4},
\end{equation*}
then there exists a map $g: \boldsymbol{3} \to \boldsymbol{3}$, which is one-one but not onto. The existence of $g$ contradicts the pigeon hold theorem. Thus, there cannot be a bijection between $\boldsymbol{3}$ and $\boldsymbol{4}$.

I am wondering how to prove a more general proposition as follows (although it seems quite obvious):

For any $\boldsymbol{m},\boldsymbol{n}\in \mathbb{N}$, if $\boldsymbol{m} \neq \boldsymbol{n}$, then there does not exist a bijection between $\boldsymbol{m}$ and $\boldsymbol{n}$.

In my understanding, this proposition has to be proved using mathematical induction. We should conduct mathematical induction on one variable while keep the other one fixed.

答: Yes so by induction works fine. Consider $m,n\in\mathbb{Z}^+$, $n\neq m$. Without loss of generality, assume that $n > m$. We can apply induction on $M:= max\{m,n\}$. We can prove the result for $M = 2$ by exhausting all of the possible maps easily. Assume the result holds for $M = N \geq 3$. Now suppose $M = N + 1$ and take a map $f:\textbf{m}\to \textbf{n}$ (with $n > m$; so $n = N+ 1$ here). Assume this map is bijective. Then the restriction will be a map on $\textbf{m - 1}$ which will be bijective onto $\textbf{n - 1}$ (just by inspecting where the element we "remove" from $\textbf{m}$ gets sent to; relabel $\textbf{n}$ if necessary), contradicting the induction hypothesis.


问:
Elementary proof of Power Rule For differentiation $f'(x)=rx^{r-1}$ for $f(x)=x^r$
I could  not find any elementary proof of the Power Rule for differentiation:

Given $x,r \in R$ ,$x>0$ and a function $f(x)=x^r$, then its derivative is $f'(x)=rx^{r-1}$

Defintion:Let a sequence of rational numbers {$r_n$} tend to $r$ then $x^r=\lim_{r_n \to r}x^{r_n}$

All the proofs I had seen utilized the derivative of $logx$ but I dont think this is  elementary because one has to show that for $e=\text{Euler's number}$ , $t \in R$and $n \in N$ $\lim_{n \to \infty}(1+\frac{1}{n})^n=\lim_{t \to 0}(1+t)^\frac{1}{t}=e$ which can be proved using the Power Rule

Ofcourse the proof for the Power Rule is straight forward if one is familiar with the result of Power Rule if $r$ is a rational number and theorem  regarding  derivative of a sequence of derivatives that are uniformly convergent.

答:
For $r\in\mathbb{R}$, to prove that $f(x)=x^{r}\implies f'(x)=rx^{r-1}$ We need two lemmas :

Lemma $(1)$ : Consider the function $E(x)=\displaystyle\sum_{n=0}^{+\infty}\frac{x^{n}}{n!}$. We claim that $E'(x)=E(x)$.

Proof :
Let $x\in\mathbb{R}$ with $h\neq0$, we define $\tau(h)=\displaystyle\frac{f(x+h)-f(x)}{h}=\frac{E(x+h)-E(x)}{h}$. Since $E(x+y)=E(x)E(y)$, then we have that :
$$
\tau(h)=\displaystyle\frac{f(x+h)-f(x)}{h}=\frac{E(x+h)-E(x)}{h}=E(x)\cdot\frac{E(h)-1}{h}
$$
Thus we need to prove that $\displaystyle\lim_{h\to 0}\frac{E(h)-1}{h}=1$. Note that :
\begin{align*}
\frac{E(h)-1}{h}&=\frac{1}{h}\left(\left(1+h+\frac{h^{2}}{2!}+\cdots\right)-1\right) \\
&=1+\frac{h}{2!}+\frac{h^{2}}{3!}+\frac{h^{3}}{4!}+\cdots \\
\implies\left|\frac{E(h)-1}{h}-1\right|&=\left|\frac{h}{2!}+\frac{h^{2}}{3!}+\frac{h^{3}}{4!}+\cdots\right| \\
&\leq\frac{|h|}{2!}+\frac{|h|^{2}}{3!}+\frac{|h|^{3}}{4!}+\cdots \\
&=\underbrace{|h|}_{\to0}\underbrace{\left(\frac{1}{2!}+\frac{|h|}{3!}+\frac{|h|^{2}}{4!}+\cdots\right)}_{\text{bounded}}
\end{align*}
Since $\displaystyle\lim_{h\to 0}|h|=0$, and if we can show that the other part is bounded, then the overall limit is $0$. Let $0<|h|<1$, we note that the chosen interval of $h$ is due to our interest of $h$ being close to $0$. We have that :
\begin{align*}
&\left|\frac{E(h)-1}{h}-1\right|=|h|\left(\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+\cdots\right)=|h|\underbrace{\sum_{n=2}^{+\infty}\frac{1}{n!}}_{\text{converges}} \\
\implies\lim_{h\to 0}&\left|\frac{E(h)-1}{h}-1\right|=\lim_{h\to 0}|h|\sum_{n=2}^{+\infty}\frac{1}{n!}=0
\end{align*}
Therefore, by sandwich theorem :
$$
\lim_{h\to 0}\left|\frac{E(h)-1}{h}-1\right|=0\implies\lim{h\to 0}\frac{E(h)-1}{h}=1
$$
Hence $\displaystyle\lim_{h\to 0}\tau(h)=\lim_{h\to 0}E(x)\cdot\frac{E(h)-1}{h}=E(x)$, $\forall x\in\mathbb{R}$ with $E'(x)=E(x)$.

Lemma $(2)$ : Let $g(x)=\ln(x)$, we claim that $\displaystyle g'(x)=\frac{1}{x}$, $\forall x>0$.

Proof : Let $x_{0}\in(0,+\infty)$ with $x_{0}\neq x$, let $\displaystyle\phi(x):=\frac{\ln(x)-\ln(x_{0})}{x-x_{0}}$. Moreover, there exists $t_{x}\in\mathbb{R}$ with $e^{t_{x}}=x$ and there exists $t_{0}\in\mathbb{R}$ with $e^{t_{0}}=x_{0}$. Thus, we have that :$$\displaystyle\phi(x)=\frac{t_{x}-t_{0}}{e^{t_{x}}-e^{t_{0}}}=\displaystyle\frac{1}{\displaystyle\frac{e^{t_{x}}-e^{t_{0}}}{t_{x}-t_{0}}}$$
Notice that $\ln(x)$ is continuous and that $\ln(x)=t_{x}$ and $\ln(x_{0})=t_{0}$. Thus, we have that :
\begin{align*}
&\lim_{x\to x_{0}}\ln(x)=\ln(x_{0})=t_{0} \\
\implies&\lim_{x\to x_{0}}\phi(x)=\lim_{x\to x_{0}}\frac{\ln(x)-\ln(x_{0})}{x-x_{0}}\\
&\;\qquad\qquad=\frac{1}{(e^{t})'|_{t=t_{0}}}=\frac{1}{e^{t_{0}}}=\frac{1}{x_{0}}
\end{align*}
Therefore, $g$ is differentiable at $x_{0}$ and $g'(x_{0})=\displaystyle\frac{1}{x_{0}}$, $\forall x_{0}\in(0,+\infty)$.

Conclusion : We have that $f(x)=x^{r}=e^{\ln(x^{r})}=e^{r\ln(x)}$. We note that $f$ is the composition of $e^{x}$ and $r\ln(x)$ that are both differentiable. Thus :
$$
f'(x)=(r\ln(x))'(e^{x})'|_{x=r\ln(x)} =\frac{r}{x}e^{r\ln(x)}=\frac{r}{x}x^{r}=rx^{r-1}
$$


问:
Books for Geometry processing
Please suggest some basic books on geometry processing.

I want to learn this subject for learning algorithms in 3d mesh generation and graphics.

Please suggest me subjects or areas of mathematics i have learn in order to be understanding 3d mesh generation. I am doing self study and i am very new to this topic

Please suggest me online links to any videos or resources, if available.

答:
See these books:

Polygon Mesh Processing by Botsch et al.

Geometry and Topology for Mesh Generation by Edelsbrunner

and these courses:

Mesh Generation and Geometry Processing in Graphics, Engineering, and Modeling

Geometry Processing Algorithms

Title: Books for Geometry processing


答:
The simplest way to devise matrices with certain properties is in general by $PDP^{-1}$ or $PJP^{-1}$ where $D$ is diagonal or $J$ the Jordan form, and $P$ any invertible matrix.

In your case you want $M$ to be singular, so take a $D$ or $J$ with a $0$ or more in the diagonal (i.e. a zero eigenvalue).

Then multiply by your prefered $P$ (hint, the simplest one is $P=I$).

It might be a bit uneasy to come with some $P,P^{-1}$ couples with integer values, that would give you an example $M$ a bit more "interesting" than just a diagonal one. but for $3\times3$ or $2\times2$ it is possible to devise some manually.

But when we just want a singular matrix, notice there is an easier way, by ensuring the matrix is not full rank. An obvious way to reduce the rank is to choose two columns at random and then calculate the third one to be a linear combination of the first two.

E.g. $\begin{pmatrix}2 & -5 & .\\3 & 1 & .\\7 & 4 & .\end{pmatrix}\overset{C1-4C_2}{\longrightarrow}\begin{pmatrix}2 & -5 & 22\\3 & 1 & -1\\7 & 4 & -9\end{pmatrix}$


问:
Spectral radius of a matrix in a Banach algebra
Let $A$ be a $n \times n $ matrix.

Let $S(A)= \{ T: T= S^{-1} A S\ , $ for some invertible  $ n\times n $ matrix $ S\}$.

Show that there exists $T \in \overline{S(A)}$ such that $\| T\|= \rho(T)= \rho(A) $, where $\rho(A)= \sup \{ |\lambda| : \lambda \in \sigma(A)\}$ is the spectral radius of $A$.

Is it possible to find $T \in S(A)$ such that  $\| T\|= \rho(T)= \rho(A) $?

Things I know:

We know that the set of $n \times n$ matrices, $\mathbb{M}_{n \times n}$, can be identified as a  Banach algebra.

Let $A \in \mathbb{M}_{n \times n}$.

By the spectral radius formula, $\rho(A) = \lim_{n \rightarrow \infty}\| A ^n\|^{1/n}$.

I'm not sure how to construct such $T \in \overline{S(A)}$, as I'm not sure what the closure of $S(A)$ is like.

Thank you in advance!

答:
Without loss of generality we may assume that $A$ is in canonical Jordan form,  namely
$$
  A =
  \pmatrix{
    A_1 & 0 & \ldots & 0 \cr
    0 & A_2 & \ldots & 0 \cr
    \vdots & \vdots & \ddots  & \vdots  \cr
    0 & 0 & \ldots &  A_k
  },
  $$
where each $A_i$ is a Jordan block, that is
$$
  A_i=\pmatrix{
    \lambda _i & 0 & 0 & \ldots & 0 & 0 \cr
    1 & \lambda _i & 0 & \ldots & 0 & 0 \cr
    0 & 1 & \lambda _i & \ldots & 0 & 0 \cr
    \vdots & \vdots & \ddots & \ddots  & \vdots  & \vdots  \cr
    0 & 0 & 0 & \ldots & \lambda _i &  0 \cr
    0 & 0 & 0 & \ldots & 1 &  \lambda _i
  }
  $$
The Jordan decomposition is not guaranteed to take place over an orthonormal basis but, by conjugating $A$ with a
matrix sending its Jordan basis to the canonical basis of $\mathbb C^n$, we may assume that $A$ is indeed in Jordan form
relative to the orthonormal canonical  basis.

Choose a parameter $\alpha >0$ and, for each Jordan block $A_i$, consider the invertible matrix
$$
  U^\alpha _i=
  \pmatrix{
    \alpha  & 0 & \ldots & 0 \cr
    0 & \alpha ^2 & \ldots & 0 \cr
    \vdots & \vdots & \ddots  & \vdots  \cr
    0 & 0 & \ldots &  \alpha ^p
  },
  $$
where $p$ is the size of $A_i$, and define $U^\alpha $ to be the block-diagonal matrix
$$
  U^\alpha  =
  \pmatrix{
    U^\alpha _1 & 0 & \ldots & 0 \cr
    0 & U^\alpha _2 & \ldots & 0 \cr
    \vdots & \vdots & \ddots  & \vdots  \cr
    0 & 0 & \ldots &  U^\alpha _k
  }.
  $$
We then have that $U^\alpha A(U^\alpha )^{-1}$ is the block diagonal matrix with blocks
$$
  U^\alpha _iA_i(U^\alpha _i)^{-1} =
  \pmatrix{
    \lambda _i & 0 & 0 & \ldots & 0 & 0 \cr
    \alpha  & \lambda _i & 0 & \ldots & 0 & 0 \cr
    0 & \alpha  & \lambda _i & \ldots & 0 & 0 \cr
    \vdots & \vdots & \ddots & \ddots  & \vdots  & \vdots  \cr
    0 & 0 & 0 & \ldots & \lambda _i &  0 \cr
    0 & 0 & 0 & \ldots & \alpha  &  \lambda _i
  },
  $$
where one should note that the 1's in $A_i$ have been changed to $\alpha $'s.

Taking the limit of $U^\alpha A(U^\alpha )^{-1}$ as $\alpha \to 0$, we have that the limit matrix lies in the closure of $S(A)$ and it has all
of the desired properties, including the fact that its spectrum is the same as that of $A$ (with different multiplicities).

The answer to question 2 is no,  a counter example being
$$
  A =   \pmatrix{
    1 & 0 \cr
    1 & 1
  }.
  $$
The reason is that
$$
  N:= A-I
  $$
is a nonzero nilpotent matrix,  and we claim that  for any invertible matrix $S$, one
has that the matrix $T:= SAS^{-1}$ has norm strictly bigger than 1.
To see this, notice that
$$
  T =   SAS^{-1} =   S(I+N)S^{-1} =  I+SNS^{-1} = I+S',
  $$
where $S'=SNS^{-1}$ is likewise a nonzero nilpotent matrix.

Arguing  by contradiction,  if $\|T\|\leq 1$,  then also $\|T^k\|\leq 1$, for all $k$, but
$$
  T^k = (I+S')^k = I+kS',
  $$
whose norm tends to infinite as $k\to \infty $.


问:
Prove that $\left\{x^{\lambda_i}\right\}_{i=0}^n$ is a Chebyshev system on $(0,\infty)$
The definition of Chebyshev system is as follows.

A linearly independent system of $n$ basis functions $\varphi_1,\ldots,\varphi_n$ on $[a,b]$ is a Chebyshev system on $[a,b]$ if every non-trivial linear combination $\displaystyle\sum_{k=1}^n a_k\;\varphi_k$ has at most $n-1$ zeros on $[a,b]$.

Problem

The problem is to show that for any real $0=\lambda_0 < \lambda_1 < \ldots < \lambda_n$, the basis functions $\left\{x^{\lambda_i}\right\}_{i=0}^n$ form a Chebyshev system on $(0,\infty)$.

Approach

The first thing that I should prove is that $\left\{x^{\lambda_i}\right\}_{i=0}^n$ form a basis of the space $C(0,\infty)$, right? I don't know how to see that.

Assuming that they form a basis. I think that the proof of being a Chebyshev system is equivalent to prove that

$$\det\left(\begin{matrix}1 & x_1^{\lambda_1} & \ldots & x_1^{\lambda_n}\\ \vdots & \vdots &  & \vdots\\ 1 & x_{n+1}^{\lambda_1} & \ldots & x_{n+1}^{\lambda_n}\end{matrix}\right)\neq 0$$

for all pairwise distinct $x_1, \ldots, x_{n+1}\in(0,\infty)$ because it means that there is no linear combination that evaluated in one of those points is zero.

I need a bit of guidance in this exercise. Thank you

答:
As I mentioned in the comment, you do not need to show that $x^{\lambda}$ forms a basis of $C(0,\infty).$ Obviously, it is not going to be true since you cannot span $C(0,\infty)$ using finite number of functions. There is a beautiful theorem due to Muntz which sates that $\sum\frac{1}{\lambda_i}=\infty$ is a equivalent to the system  $span\{x^{\lambda_n}\}^{\infty}_{n=0}$ being dense in $C[a,b],$ $a>0.$

As to the original problem, just use Rolle's theorem and induction on $n.$ Indeed, base case is trivial. Now to prove the step, we assume that $f(x)=\sum_{i=0}^ka_ix^{\lambda_i}=0$ has $k+1$ zeros in $(0,\infty).$ By Rolle's theorem, $f'(x)=\sum_{i=0}^k(\lambda_i)a_ix^{\lambda_i-1}=x^{\lambda_1-1}(\sum_{i=1}^k(\lambda_i)a_ix^{\lambda_i-\lambda_1})$ has at least $k$ zeros in $(0,\infty).$ This contradicts to the fact that $\{x^{\lambda_i-\lambda_1}\}_{i=1}^{k}$ forms a Chebyshev system.


问:
Let $X\subset \mathbb{R^2}$ be bounded and convex.Show that $\mathbb{R^2} \setminus X$ is path connected

Let $X\subset \mathbb{R^2}$ be bounded and convex.Show that $\mathbb{R^2} \setminus X$ is path connected.

I understand the proposition, I was trying proof by contradiction,but I don´t get inspired, My other idea is let $x,y\in \mathbb{R^2} \setminus X$ and notice that since $X$ is bounded exits $r \in \mathbb{R}$ such that
$ X \subset B_{r}(0)$  and from here begin to construct a path from $x$ to $y$ without pass for $X$.
but I don´t know how use that $X$ is convex.

Any sugestion, idea, intuition was very helpful.And in case of be possible, may possibly give advice for be more crative in the problems that requiere a construction. Thanks in advice

答: Let $P$ and $Q$ be two points in the complement of $X$ and let $O\in X$. Join $O$ to $P$ and $Q$. Observe that on the line $OP$, on the side of $P$ opposite to that where $O$ belongs, there are no other points of $X$ (otherwise convexity would be violated at $P$). For the same reason, on the side of $OQ$ opposite to that where $O$ belongs there are no other points of $X$. By boundedness you can now pick a circle centered at $O$ with radius much larger than $|OP|$ and $|OQ|$ such that $X$ is entirely contained in a disk of half the radius centered at $O$. Let this circle intersect $OP$ at $P_1$ and $OQ$ at $Q_1$. Then $PP_1Q_1Q$ is a path connecting $P$ and $Q$, entirely contained in the complement of $X$.


问:
Linear regression vector calculus confusion
For linear regression in the form

$$|y\rangle = X|\beta\rangle +|\epsilon\rangle $$

Where,

$$|y\rangle = \begin{bmatrix} y_1 \\\vdots \\ y_n\end{bmatrix}, \ X=\begin{bmatrix} 1 & x_1 \\ \vdots & \vdots  \\ 1 & x_n\end{bmatrix}, \ |\beta\rangle = \begin{bmatrix}\alpha \\ \delta \end{bmatrix}, \ |\epsilon\rangle = \begin{bmatrix} \epsilon_1 \\ \vdots \\ \epsilon_n \end{bmatrix} $$

We calculate the best fit using least squares,

$$\langle \epsilon|\epsilon\rangle = (\langle y | - \langle \beta|X^T)(|y \rangle-X|\beta \rangle$$

calculating the derivative we get

$$\frac{\partial \langle \epsilon|\epsilon\rangle }{\partial \langle  \beta|} = -2X^T| y \rangle + (X^{T} X + XX^{T})| \beta \rangle $$

Now this is the bit im confused on, in material I find online they simplify $ (X^{T}X +XX^{T})| \beta \rangle = 2 X^TX| \beta \rangle$ so that we end up with (after setting the derivative to zero) $|\beta \rangle = (X^TX)^{-1}X^T|y \rangle$ however this is only the case when X is symmetric in the case of X shown above this is not necessarily true correct? Furthermore, for $X$ above $XX^T$ is a $n\times n$ matrix so $XX^T |\beta \rangle$ doesn't make sense?

答:
So I quickly realised my mistake

$$ \frac{\partial \langle\beta|X^TX|\beta \rangle}{\partial\langle \beta| } = (X^TX+(X^TX)^T)|\beta\rangle = 2X^TX|\beta\rangle$$

I basically forgot when transposing $X^TX$ that they also switch positions at the same time i.e. $(X^TX)^T = X^TX$.


问:
There are 50 rooms in a line. If there are 26 rooms with girls, prove there are two girls exactly 5 rooms apart.
There are 50 rooms in a line. If there are 26 rooms with girls, prove there are two girls exactly 5 rooms apart.

My idea was place 25 girls in into pairs of rooms, and there is no scenario which there are 5 girls apart. Then, if you place one more, by pigeonhole principle, there's a girl exactly 5 away - but I want a more formal way of explaining this. Does anyone have a way? Cheers.

答: A hint: Color the rooms with five colors according to the remainder mod 5 of the room number.</p>


问:
Logic: 10 company directors
I need to solve following logical task:

10 people sit on the Board of Directors of a company that recieved 1 share as a profit. There is a linear hierarchy among directors: from the most senior (chief) to the most junior. The profits are shared as follows: the chief proposes who to give a share to, then everyone, including the chief, votes "for" or "against". If at least half of the directors vote "for", the decision of the chief is accepted, but if not, the chief is excluded from the Board of Directors, the next in the hierarchy becomes the chief and the process starts from the beginning according to the same principle. Everyone wants to stay on the Board of Directors, and also wants to get a share. All other things being equal, a director will vote for the option that leaves fewer colleagues on the Board. How many directors will remain on the Board after profit sharing?

My answer is 4, because by $\geqslant5$ all, except chief and new share holder would vote "against". And by 4 two of them would vote "for" to get the share and not to be excluded.

But I can't prove it more clearly. Any ideas?

答:
Assume n directors, n >= 1.

If n = 1 or n = 2, the chief wins the vote and gets the share.

If n = 3 the chief can’t propose himself because he will just be voted out. If he proposes the lowest ranking board member, he’ll get two votes. He’ll keep his job, and the lowest ranking director gets the share. If he proposes the middle board member, he’ll vote for this, the middle member will vote against, and since the lowest board member wants to reduce the numbers, he’ll vote against. The chief loses the job and #2 will first become chief and then get the share. The best the chief can do is vote for the lowest board member and keep his job, while the lowest board member gets the share.

If n = 4, the board consists of chief, x, y, and z. If the chief gets voted out, x becomes chief and z gets the share; y will get nothing. So if the chief proposes y, chief and y will vote for it, chief stays chief and y gets the share. If he proposes z then z knows he gets the share anyway, so z votes against. y votes against because he can’t get anything, and x votes against, so x becomes chief and z gets the share. If he proposes x, then y and z will vote against. x has the choice of voting “for” and getting the share, or voting against and becoming chief, he’ll choose what he values more. So the safe bet for the chief is proposing y and remain chief, while y gets the share.

If n >= 5 then a majority will vote the chief out. This repeats, so the end result is that the board will be reduced to the four lowest ranking board members chief, x, y and z. Chief will remain chief, and y gets the share.

So you were almost correct, except it will be the second lowest board member who gets the share.

Title: Logic: 10 company directors


问:
Finding a linear transformation such that $T((4,1)) = (0,21)$ and $T((1,5))=(-19,10)$
Finding a linear transformation such that $T((4,1)) = (0,21)$ and $T((1,5))=(-19,10)$

So, i'm trying to find multiple different ways to do this. I know that every linear transformation can be represented by a matrix, so i'm letting $A$ be a $2 \times 2$ matrix and defining $T(x) = Ax$ and trying to figure out what $A$ has to be. Is this a good way to go about this problem?

I'm ending up with two seperate systems of linear equations, each with four variables and two equations, t hus I have four variables and four equations, so that's good I guess, but I feel that I'm getting lost in the weeds and there may be an easier way to do this...

Would appreciate multiple perspectives on this one... Thank you!!

答:
Let's do this the way you initially tried (and yes, it's a good way to go about the problem). Let
$$A = \left[ \begin{array}{cc} a & b \\ c & d \end{array} \right].$$
We are given
$$A \left[\begin{array}{c}4\\1 \end{array} \right] = \left[\begin{array}{c} 0\\21 \end{array} \right], \text{ and}$$
$$A \left[\begin{array}{c} 1\\5 \end{array} \right] = \left[\begin{array}{c} -19\\10 \end{array} \right].$$
That is, we have
$$\left[ \begin{array}{cc} a & b \\ c & d \end{array} \right] \left[\begin{array}{c}4\\1 \end{array} \right] = \left[\begin{array}{c} 0\\21 \end{array} \right], \text{ and}$$
$$\left[ \begin{array}{cc} a & b \\ c & d \end{array} \right] \left[\begin{array}{c} 1\\5 \end{array} \right] = \left[\begin{array}{c} -19\\10 \end{array} \right],$$
giving us the system
$$4a+b=0$$
$$4c+d=21$$
$$a+5b=-19$$
$$c+5d=10.$$

Focussing on the system of two equations with the $a$ and $b$ variables, i.e. the system
$$4a+b=0$$
$$a+5b=-19,$$
we have an easy system of two simultaneous equations with two variables. Solving this, we have $a=1$ and $b=-4$.

Now do the same with the two equations involving the $c$ and $d$ variables, and you're done. Of course, the strategy outlined in the comments is equally valid, but you asked for multiple perspectives!


答:
There exist $n_1,n_2<\cdots$ such that $\int |f_{n_k}|d\mu <\frac  1 {2^{k}}$. This gives $\int \sum_k |f_{n_k}|d\mu=\sum_k \int |f_{n_k}|d\mu<\infty$. Hence, $\sum_k |f_{n_k}|<\infty$ almost everywhere. This implies that $|f_{n_k}| \to 0$ almost everyhwere.

The interchange of sum and integral above can be justified using  either Tonelli's Theorem or Montone Convergence Theorem.

Title: $\lim_{n \rightarrow \infty}\int_0^1 |f_n(x)| dx = 0$ implies there is subsequence such that $\{f_{n_i}(x)\}$ converges to $0$ for a.e. $x$

Body:
Given that $\{f_n\}$ be a sequence of Lebesgue-measurable real-valued functions on $[0, 1]$ such that
\begin{equation}
    \lim_{n \rightarrow \infty}\int_0^1 |f_n(x)| dx = 0
\end{equation}

I want to show there is a subsequence of $\{f_n\}$ such that $\{f_{n_i}(x)\}$ converges to $0$ for a.e. $x$.

Im thinking of using the inequality $\int_0^1 |f_n(x)| dx > \mu(\{x: |f_n(x)| > \epsilon\})\epsilon$ but I didnt' get very far past that.


问:
Compact formula for the $n$th derivative of $f\left(\sqrt{x+1}\right)$?
I'm interested in a general formula for

$$\frac{d^n}{dx^n}\Big[f\left(\sqrt{x+1}\right)\Big].$$

In particular, Fàa di Bruno's formula gives

$$\frac{d^n}{dx^n}\Big[f\left(\sqrt{x+1}\right)\Big]=\sum_{k=1}^n f^{(k)}\left(\sqrt{x+1}\right)B_{n,k}\big(g'(x),g''(x),\dots,g^{(n-k+1)}(x)\big),$$

where $g(x)=\sqrt{x+1}$ and $B_{n,k}$ denote Bell polynomials.

I suspect we can do better than this, using the identities

$$\frac{d^k}{dx^k}\Big[\sqrt{x+1}\Big]=\left(\frac{1}{2}\right)^\underline{k}x^{1/2-k}=-\frac{(2k-3)!!}{(-2)^k}x^{1/2-k}=\frac{\sqrt{\pi}}{2\Gamma(3/2-k)}x^{1/2-k}.$$

In particular, every term in $B_{n,k}$ for fixed $n$ and $k$ will be some numerical factor times $(x+1)^{k/2-n}$ (see notes below), so the result is always of the form

$$\frac{d^n}{dx^n}\Big[f\left(\sqrt{x+1}\right)\Big]
=\sum_{k=1}^n a_{n,k} \frac{f^{(k)}\left(\sqrt{x+1}\right)}{(x+1)^{n-k/2}}$$

for some constants $a_{n,k}.$ Is there an explicit expression for the $a_{n,k}$ above? For example, it seems likely that there is some expression using Stirling numbers, factorials, binomial coefficients, etc., given the combinatorial nature of the problem.

Current progress:

Using the first identity above for derivatives of $\sqrt{x+1}$ [where $(\cdot)^\underline{k}$ denotes the $k$th falling factorial], we have

$$B_{n,k}\big(g'(x),g''(x),\dots,g^{(n-k+1)}(x)\big)
=B_{n,k}\left[\left(\frac{1}{2}\right)^\underline{1}x^{1/2-1},\left(\frac{1}{2}\right)^\underline{2}x^{1/2-2},\dots,\left(\frac{1}{2}\right)^\underline{n-k+1}x^{1/2-(n-k+1)}\right].$$

The definition of the Bell polynomials guarantees that the power of $x$ in every term of $B_{n,k}(\cdots)$ comes out to $x^{k/2-n}$, with pre-factor given by

$$a_{n,k}=B_{n,k}\left[\left(\frac{1}{2}\right)^\underline{1},\left(\frac{1}{2}\right)^\underline{2},\dots,\left(\frac{1}{2}\right)^\underline{n-k+1}\right],$$

so the remaining task is to simplify the above expression, if possible.

答:
Thanks to G Cab's inference of the correct answer and Gary's comment on the recursive formula for the $a_{n,k}$'s, I have now found a proof by induction of the result. I'll post it here for posterity.

Define $a_{n,k}=0$ if $k=0$ or $k>n$, and otherwise let it be defined as in the original post. Then we can write

$$\frac{d^n}{dx^n}\Big[f\left(\sqrt{x+1}\right)\Big]
=\sum_{k=0}^\infty a_{n,k} \frac{f^{(k)}\left(\sqrt{x+1}\right)}{(x+1)^{n-k/2}},$$

and differentiating both sides gives the recurrence relation $a_{n+1,k}=\frac{1}{2}a_{n,k-1}+(k/2-n)a_{n,k}$ for all $n,k≥1.$ This is equivalent to

$$a_{n+1,k+1}=\frac{1}{2}a_{n,k}+\left(\frac{k+1}{2}-n\right)a_{n,k+1}\qquad\Bigg|\qquad n≥1,k≥0.$$

Now define

$$c_{n,k}:=\frac{(-1)^{n-k}}{2^n}\frac{(2n-1-k)!}{(2n-2k)!!(k-1)!}$$

for $n≥1,1≤k≤n$, and $c_{n,k}=0$ otherwise. (In fact, the first expression also works for these cases if we interpret negative factorials as $\infty$ and $1/\infty$ as zero.) It is straightforward to check that this formula for $c_{n,k}$ is equivalent to $(1)$ in G Cab's answer.

Induction Part 1

It is quick to check that $c_{n,k}=0=a_{n,k}$ for $k=0$ or $k>n$, and that $c_{1,1}=1/2=a_{1,1}.$

Induction Part 2

Let $n≥1,k≥0$ be fixed, and assume that $c_{n,k}=a_{n,k}$ and $c_{n,k+1}=a_{n,k+1}$ for these particular values. We will show that $c_{n+1,k+1}=a_{n+1,k+1},$ which proves the desired result by induction.

$\begin{align*}
a_{n+1,k+1}&=\frac{1}{2}a_{n,k}+\left(\frac{k+1}{2}-n\right)a_{n,k+1}
\\\\
&=\frac{1}{2}c_{n,k}+\left(\frac{k+1}{2}-n\right)c_{n,k+1}
\\\\
&=\frac{1}{2}\frac{(-1)^{n-k}}{2^n}\frac{(2n-1-k)!}{(2n-2k)!!(k-1)!}
+\left(\frac{k+1}{2}-n\right)\frac{(-1)^{n-k-1}}{2^n}\frac{(2n-2-k)!}{(2n-2k-2)!!k!}
\\\\
&=\left[
\frac{k}{2n-k}+(2n-k-1)\frac{2n-2k}{(2n-k)(2n-k-1)}
\right]
\frac{(-1)^{n-k}}{2^{n+1}}\frac{(2n-k)!}{(2n-2k)!!k!}
\\\\
&=\left[
\frac{k}{2n-k}+\frac{2n-2k}{2n-k}
\right]
c_{n+1,k+1}
\\\\
&=c_{n+1,k+1}
\end{align*}$


答:
Since the vectors $(4,1)$ and $(1,5)$ are linearly independent (not multiples of each other, in dimension two), you can write the matrix rel the basis that they form simply as $$\begin{pmatrix}a&b\\c&d\end{pmatrix}$$, where $\begin{cases}a(4,1)+c(1,5)=(0,21)\\b(4,1)+d(1,5)=(-19,10)\end{cases}$.

If you wish to get the matrix in the standard basis, conjugate by the change of basis matrix $$\begin{pmatrix}4&1\\1&5\end{pmatrix}$$.

Body:
Finding a linear transformation such that $T((4,1)) = (0,21)$ and $T((1,5))=(-19,10)$

So, i'm trying to find multiple different ways to do this. I know that every linear transformation can be represented by a matrix, so i'm letting $A$ be a $2 \times 2$ matrix and defining $T(x) = Ax$ and trying to figure out what $A$ has to be. Is this a good way to go about this problem?

I'm ending up with two seperate systems of linear equations, each with four variables and two equations, t hus I have four variables and four equations, so that's good I guess, but I feel that I'm getting lost in the weeds and there may be an easier way to do this...

Would appreciate multiple perspectives on this one... Thank you!!


问:
Algebraic of degree $n$ over $F$ implies algebraic of degree at most $n$ over $F(a)$?
Suppose $a, b$ in the field $K$ are algebraic over the subfield $F \subseteq K$. Suppose $a$ is algebraic of degree $m$ over $F$ and $b$ is algebraic of degree $n$ over $F$. Let $F(a)$ be the field obtained by adjoining $a$ to $F$.

My question is, is it true that $b$ is algebraic of degree at most $n$ over $F(a)$? If so, how come?

答: Yes. Call $f$ the minimal polynomial of $b$ in $F$. Then, $f(b)=0$ and $f$ is in $F(a)[X]$. So the minimal polynomial of $b$ in $F(a)$ divides $f$, so it must have degree $\leq n$.


答:
$A\begin{bmatrix} 4&1\\1&5\end{bmatrix} = \begin{bmatrix} 0&-19\\21&10\end{bmatrix}$

Multiply both on the right by $\begin{bmatrix} 4&1\\1&5\end{bmatrix}^{-1}$

Title: Finding a linear transformation such that $T((4,1)) = (0,21)$ and $T((1,5))=(-19,10)$


问:
Question about ambiguity of BNF
The BNF is defined as follows:

```<code><S> -> <S>a<S>a<S> | b
</code>```

This is my review question for a quiz, and according to my teacher, this grammar is ambiguous. However, I realized this grammar is not.

For example, given a sentence:

```<code>babababab
</code>```

There are three ways to generate this sentence depending on which <S> we use, but its parse-tree is unique.

Am I right in this case? Any feedback would be greatly appreciated.

Thanks,

Chan

答:
Starting with your formal grammar consisting of one production rule, $S \to SaSaS | b$, I'm going to modify it be adding parenthesis to demonstrate the the possible ways of generating your objective string $babababab$.

Let the demonstrative formal grammar be given by the production rule $S \to (S)a(S)a(S) | b$.

Starting with $(S)a(S)a(S)$ we can produce the following strings: $((b)a(b)a(b))a(b)a(b)$, $(b)a((b)a(b)a(b))a(b)$ and $(b)a(b)a((b)a(b)a(b))$.

If we remove the paratheses, (effectively performing the same set of operations using your formal grammar), then we arrive at your objective string $babababab$.

By definition, any formal grammar that yields the same string by applying the production rules in multiple ways, is ambiguous.

Therefore, your formal grammar is ambiguous. (the demonstrative formal grammar is not, since each string is unique.)

ContentLicense: CC BY-SA 2.5

Title: Question about ambiguity of BNF


问:
Find the limiting distribution
Find the limiting distribution for $n\rightarrow \infty   \text{ of} \prod\limits^n_{i=1}X_i$. Given is that $f(x)=\frac{1}{2x\sqrt{2\pi}}e^{-\frac{1}{8}(\ln x-\theta)^2}, x\geq 0$.</p>

答:
The density of the individual $X_i$ is nearly the density of a normal distribution, but it has $\log x$ where one would expect $x$ and that additional $\frac{1}{x}$-factor. That means that the $X_i$ are log-normally distributed, i.e. $\log X_i$ is normally distributed. See http://en.wikipedia.org/wiki/Log-normal_distribution.

By taking the logarithm of the product $S_n = \prod_{i=1}^n X_i$, you get $$
  \log S_n = \log \prod_{i=1}^n X_i = \sum_{i=1}^n \log X_i \text{.}
$$
$\log S_n$ is thus normally distributed too, and $S_n$ is thus log-normally distributed.

The mean and variable of $\log S_n$ are simply $n$ times the mean respectively variance of $\log X_i$, which can easily be found by looking at the density of $X_i$.

Using that $\log S_n \sim N(\theta n,4n)$, i.e. that $P(\log S_n \leq x) = \Phi\left(\frac{x-n\theta}{2\sqrt{n}}\right)$, where $\Phi$ is the CDF of the standard normal distribution, you get $$
  F_n(x) := P(S_n \leq x) = \Phi\left(\frac{\log x-n\theta}{2\sqrt{n}}\right) \text{.}
$$
for the CDF of $S_n$.

If $S_n$ is to converge in distribution to $S$ with CDF $F$, then $F_n \to F$ pointwise, except for discontinuity points of $F$.

You thus have the following cases.

If $\theta > 0$, $F_n(x) \to 0$ as $x \to \infty$ for all $x > 0$ and thus by monotonicity any possible limit distribution would have to be identically zero, which is impossible. The product thus doesn't converge in this case.

If $\theta < 0$, $F_n(x) \to 1$ as $x \to \infty$ for all $x > 0$. In other words, $F(x)=1$ if $x > 0$, and by right-continuity of CDFs also $F(0)=1$. On the other hand, $F_n(x) = 0$ if $x < 0$, and thus $$
  F(x) = \begin{cases} 0& \text{if $x < 0$} \\1 &\text{if $x \geq 0$}\end{cases}
$$

If $\theta = 0$, $F_n(x) \to \frac{1}{2}$ for all $x > 0$. This is also impossible, since you need $F(x) \to 1$ for $x \to \infty$ if $F$ is to be a CDF.

The product $S$ thus converges exactly if $\theta < 0$, and the limit distribution is a point distribution with $P(S=0)=1$, $P(S\neq0)=0$ in the case.

Intuitively, this makes sense. If the means of the logarithms are less than zero, on average multiplying will make the numbers smaller and smaller. Otherwise, the numbers get larger and larger and the product diverges.

Title: Find the limiting distribution


答: If $b$ is algebraic of degree $n$ over $F$, then there is a polynomial $P$ of degree $n$ with coefficients in $F$ such that $P(b) = 0$. This polynomial has coefficients in $F$, but it also has coefficients in $F(a)$, so we have that $b$ is algebraic of degree at most $n$ in $F(a)$.

Title: Algebraic of degree $n$ over $F$ implies algebraic of degree at most $n$ over $F(a)$?

Body:
Suppose $a, b$ in the field $K$ are algebraic over the subfield $F \subseteq K$. Suppose $a$ is algebraic of degree $m$ over $F$ and $b$ is algebraic of degree $n$ over $F$. Let $F(a)$ be the field obtained by adjoining $a$ to $F$.

My question is, is it true that $b$ is algebraic of degree at most $n$ over $F(a)$? If so, how come?


答:
Let have you change of base matrix $P=\begin{pmatrix}4&1\\1&5\end{pmatrix}$ and its inverse $P^{-1}=\frac 1{19}\begin{pmatrix}5&-1\\-1&4\end{pmatrix}$

Since $\det(P)=4\times 5-1\times 1=19$ and for $2\times 2$ matrices the inverse is obtained by permuting diagonal elements and negate the others.

Let's now have or result matrix $R=\begin{pmatrix}0&-19\\21&10\end{pmatrix}$

By hypothesis we have: $\quad TP=R$

Therefore $T=RP^{-1}=\frac 1{19}\begin{pmatrix}0&-19\\21&10\end{pmatrix}\begin{pmatrix}5&-1\\-1&4\end{pmatrix}=\frac 1{19}\begin{pmatrix}19&-76\\95&19\end{pmatrix}=\begin{pmatrix}1&-4\\5&1\end{pmatrix}$


答:
For positive integer $r$, we can use induction.

True for $r=1$:
$(x^1)' = x' = 1 = 1x^0$.

If true for $r$,
$(x^{r+1})' = (x\ x^r)'
= x'x^r+x(x^r)'
=x^r+x(rx^{r-1})
=x^r+rx^r = (r+1)x^r
$.

Title: Elementary proof of Power Rule For differentiation $f'(x)=rx^{r-1}$ for $f(x)=x^r$

Body:
I could  not find any elementary proof of the Power Rule for differentiation:

Given $x,r \in R$ ,$x>0$ and a function $f(x)=x^r$, then its derivative is $f'(x)=rx^{r-1}$

Defintion:Let a sequence of rational numbers {$r_n$} tend to $r$ then $x^r=\lim_{r_n \to r}x^{r_n}$

All the proofs I had seen utilized the derivative of $logx$ but I dont think this is  elementary because one has to show that for $e=\text{Euler's number}$ , $t \in R$and $n \in N$ $\lim_{n \to \infty}(1+\frac{1}{n})^n=\lim_{t \to 0}(1+t)^\frac{1}{t}=e$ which can be proved using the Power Rule

Ofcourse the proof for the Power Rule is straight forward if one is familiar with the result of Power Rule if $r$ is a rational number and theorem  regarding  derivative of a sequence of derivatives that are uniformly convergent.


问:
$S(t)=S_{0}e^{-R_{0}(R(t)-R_{0})/N}$

Show that from the system of equations generated by the SIR model we have $$S(t) = S(0) \exp\left(-R_{0}(R(t)-R(0))/N \right)$$ where the model SIR equations are as follows $$\begin{aligned} \frac{dS}{dt} &= -\beta \frac{SI}{N}\\ \frac{dI}{dt} &=  \beta \frac{SI}{N}-\gamma I\\ \frac{dR}{dt} &= \gamma I \end{aligned}$$ where $t \geq 0^{+}$.

答:
Sort of an odd problem, but here goes. If $S$, $I$, and $R$ satisfy this system, your first equation tells you that $S'/S = -(\beta/N)I$. So
\begin{align}
S'/S &= -\frac{\beta}{N}I \\
&= -\frac{\beta}{N\gamma} R',
\end{align}
the last line following from your third relation.

And you know that $S'/S = (\log S)'$, so
$(\log S)' = -(\beta/N\gamma)R'$ which tells you
$$ \log S = C -(\beta/N\gamma)R.$$

where $C$ is a constant of integration. Knowing the initial conditions for $S$ and $R$ let you figure out what $C$ is, and you get
$$
S = S_0 \exp(\frac{\beta}{\gamma N}(R_0 - R)),$$
which probably is what you meant to type in the title.

I imagine in practice you wouldn't know $R$ beforehand. Least squares estimation is one way to get a handle on the parameters if you happen to have measurements. Let's suppose you have measurements of $S$ and $I$:
$(s_1, i_1), (s_2, i_2), \dots$. You might set up some sort of error functional
\begin{align}
E(s(0), i(0), \beta/N, \gamma)
=& \sum_n (S(t_n; s(0), i(0), \beta/N, \gamma) - s_n)^2 \\
&+ \sum_n(I(t_n; s(0), i(0), \beta/N, \gamma) - i_n)^2
\end{align}
and adjust the parameters
$s(0)$, $i(0)$, $\beta/N$, and $\gamma$ until you get the accuracy you want, solving the system numerically to generate the associated $S$ and $I$ each time you change them, or better yet take the advice of @CarlChristian and use the exact solution!

Title: $S(t)=S_{0}e^{-R_{0}(R(t)-R_{0})/N}$


Body:
$5(4,1)−(1,5)=(19,0);(4,1)−4(1,5)=(0,−19)$,

so $(1,0)=\frac1{19}[5(4,1)-(1,5)]$ and $(0,1)=-\frac1{19}[(4,1)-4(1,5)]$,

so $T((1,0))=\frac1{19}[5T((4,1))-T((1,5))]$ and $T((0,1))=-\frac1{19}[T((4,1))-4T((1,5))]$,

so $T((1,0))=\frac1{19}[5(0,21)-(-19,10)]=(1,5)$

and $T((0,1))=-\frac1{19}[(0,21)-4(-19,10)]=(-4,1),$

so the matrix is $\pmatrix{1&-4\\5&1}$.


问:
Prove that matrix is non-negative
Problem:

Given $A_{1}, A_{2}, ..., A_{n}$ - finite sets and $a_{ij} = |A_{i}\cap A_{j}|$ - number of elements in intersection of sets. Prove, that matrix  $(a_{ij})_{i=1,2,..,n}^{j=1,2,.., n}$ is non-negative.

I've cleared out that this matrix is symmetric and the largest elements are on its diagonal. I need to prove that $x^{T}Ax \ge 0$ or

$$\sum_{i=1}^n\sum_{j=1}^na_{ij}x_{i}x_{j} \ge 0$$

or its eigenvalues are non-negative.

I don't know, what need I do next? Any hint will be useful!

答:
Hint:

Claim: A symmetric matrix $X=(x_{ij})_{VV}$ (here  $V=\{1,2,\ldots, k, \ldots, n\}$ ) is positive semidefinite if and only if all of its principal minors $det(x_{ij})_{UU}$ (here $\emptyset \neq U\subset V$) are nonnegative.

Claim: A symmetric matrix $\begin{pmatrix}A& C^T\\ C& D\end{pmatrix}$  is positive semidefinite if and only if $D$ and  $A - C^T D^{-1} C$ are positive semidefinite. In particular for $v\in\mathbb{R}^{n}$, $a \geq 0$ and $X_{n-1}\in\mathbb{R}^{(n-1)\times (n-1)}$
a symmetric matrix $\begin{pmatrix}a& v^T\\ v& X_{n-1}\end{pmatrix}$  is positive semidefinite if and only if $a$ and  $a - v^T X_{n-1}^{-1} v$ are positive semidefinite.

For proofs see  Meyer, Matrix Analysis and Applied Linear Algebra or Matrix Analysis by Roger A. Horn. Use induction on these facts to get the desired result.

Title: Prove that matrix is non-negative


问:
Solve differential in respect to a function rather than a variable
In an exercise I have to solve this differential. Now what confuses me is that we take the derivative in respect to a function rather then a variable. Now I am not sure how to proceed?

$$\frac{d}{d(\cos (\theta))} \cos(n\theta)$$

答:
Note: I'd rather call this a 'derivative' as opposed to 'differential'; in my experience, a differential refers to something like '$dy$' or '$dx$' considered on its own.

Hint: to solve this problem, use the chain rule:
$$
\frac{d\cos n\theta}{d\cos\theta}=\frac{d\cos n\theta}{d\theta} \cdot \frac{d\theta}{d\cos\theta} \, .
$$

As for your more general question about what
$$
\frac{d\cos n\theta}{d\cos\theta}
$$
actually means, this is a prime example of implicit differentiation. The equation
$$
x^2+y^2=1
$$
gives us a simpler example of this. There is no single way to write $y$ as a 'function of $x$'. Either
$$
y(x) = \sqrt{1-x^2}
$$
or
$$
y(x)=-\sqrt{1-x^2} \, .
$$
Implicit differentiation tells us that regardless of which of these two functions we are considering, the following relationship holds:
$$
\frac{dy(x)}{dx} = -\frac{x}{y(x)} \, .
$$
In your example, it might be difficult or impossible to find an explicit formula connecting $\cos n\theta$ and $\cos \theta$; that is, there is no single way to write $\cos n\theta$ as a 'function of $\cos\theta$'. However, we can still find an 'implicit' equation that tells us how $\cos n\theta$ changes for infinitesimal changes in $\cos\theta$, treating $\cos\theta$ as a variable just like any other.

In Lagrange notation, these manipulations would read differently. Let $f$ be an arbitrary function satisfying
$$
f(\cos\theta) = \cos n\theta
$$
for all $\cos\theta$. To make this more readable, let $g(\theta)=\cos \theta$. Then,
$$
\cos n \theta = f(g(\theta)) \, .
$$
By the chain rule,
\begin{align}
-n\sin n \theta &= f'(g(\theta))g'(\theta) \\
&= -f'(\cos\theta)\sin\theta
\end{align}
which implies
$$
f'(\cos\theta) = \frac{n\sin n \theta}{\sin\theta} \, .
$$

Finally, another way to solve this equation is to parameterise $\cos n\theta$ and $\cos \theta$ with respect to $\theta$. In general, parameterisation is where you express two variables $x$ and $y$ in terms of a third variable $t$. In general, if we find two functions $f$ and $g$ such that $x=f(t)$ and $y=g(t)$, then
$$
\frac{dx}{dt}=f'(t) \text{ and } \frac{dy}{dt}=g'(t)
$$
which implies that
$$
\frac{dy}{dx} = \frac{dy}{dt} \cdot \frac{dt}{dx} = \frac{g'(t)}{f'(t)} \, .
$$
In this case, if we let $y=\cos n\theta$ and $x=\cos \theta$, then
$$
\frac{dy}{d\theta}=-n\sin n\theta \text{ and } \frac{dx}{d\theta} = -\sin\theta \, .
$$
Hence,
$$
\frac{dy}{dx} = \frac{-n\sin(n\theta)}{-\sin\theta}=\frac{n\sin n\theta}{\sin\theta} \, .
$$
This method is in essence the same as the first one I showed you, but it's helpful to get another perspective.


问:
Find all primes such that $p$ and $p^2$ are of the form $2k^2-1$
Find all prime numbers such that $\dfrac{p+1}{2}$ and $\dfrac{p^2+1}{2}$ are both squares of an integer.

What makes it slightly annoying is that it is not just prooving that there is none, because 7 is such a prime.

答:
Let $p=2x^2-1$ and $p^2=2y^2-1$. Then $p(p-1)=2(y-x)(y+x)$.

Since $p>p-1$, $p$ must be a factor of $y+x$.

However, $y<\frac{p+1}{\sqrt 2}$ and $x<\frac{\sqrt {p}+1}{\sqrt 2}$ and so $x+y<\frac{p+\sqrt {p}+2}{\sqrt 2}$.

Therefore $p+\sqrt {p}+2>\sqrt 2p$ which means $p\le 13$ and these primes can be quickly checked. The only solution is $p=7$.


问:
Does it make sense to take the derivative of f(x) with respect to g(x)?
Let $f(x) = \sin(x)$ and $g(x) = \cos(x)$.

Then I want to find the derivative of sin(x) with respect to cos(x).

So I have:

$$\frac{d\sin(x)}{d\cos(x)}$$

$d/dx \sin(x) = \cos(x)$, so $d\sin(x) = \cos(x)dx$.

$d/dx \cos(x) = -\sin(x)$, so $d\cos(x) = -\sin(x)dx$.

Then,

$$\frac{d\sin(x)}{d\cos(x)} = \frac{\cos(x)dx}{-\sin(x)dx} = -\cot(x)$$

Is this legit or am I doing something wrong here?

答:
We have:$$\frac{d(\sqrt{1-\cos^2x})}{d(\cos x)}$$Substitute $u=\cos(x)$:

$$\frac{d}{du}(\sqrt{1-u^2})$$
Can you continue from here?


问:
Let p be a prime number and n an integer such that 1 ≤ n ≤ p. Show that $(p-n)!(n-1)! \equiv (-1)^{n} \mod p$

Let p be a prime number and n an integer such that 1 ≤ n ≤ p. Show that.

$$(p-n)!(n-1)! ≡ (-1)^n\mod p.$$

I know it's related to Wilson's theorem ( If p is a prime number, then (p - 1)! ≡ -1 (mod p)) but I don't know, how to prove it?

答:
By Wilson´s theorem, $(p-1)!≡-1 \mod p$. That is, $(p-1)(p-2)\dots n(n-1)!≡-1 \mod p$. Now using the $p-n$ congruences modulo $p$

$p-1≡-1$;

$p-2≡-2$;

$\dots$;

$n≡p-(p-n)≡-(p-n)$;

we can substitute in the previous equality and, we obtain$(-1)\cdot(-2)\dots(-(p-n))(n-1)!≡-1 \mod p$, or equivalently,

$(-1)^{(p-n)}(p-n)!(n-1)!≡-1 \mod p$.

This is the same equality as the one in the problem, using that $p$ is odd.


问:
How can I prove $\sqrt{3}+\sqrt{4}+...+\sqrt{n}<\frac{1}{4}n^2$ for all whole numbers $n \geq 3$?
$\sqrt{3}+\sqrt{4}+...+\sqrt{n}<\frac{1}{4}n^2 ; n \geq 3$

Induction first step:

$n=3$

$\sqrt{3}<\frac{9}{4}$ which is true

Induction second step:

$n=k$

$\sqrt{3}+\sqrt{4}+...+\sqrt{k}<\frac{1}{4}k^2$

Induction third step:

$n=k+1$

$\sqrt{3}+\sqrt{4}+...+\sqrt{k}+\sqrt{k+1}<\frac{1}{4}(k+1)^2$

From the assumption in the second step we know that: $\sqrt{3}+\sqrt{4}+...+\sqrt{k}<\frac{1}{4}k^2$

From this we can assume that: $\sqrt{3}+\sqrt{4}+...+\sqrt{k}+\sqrt{k+1}<\frac{1}{4}k^2+\sqrt{k+1}$

So we know that $\frac{1}{4}k^2+\sqrt{k+1}$ is bigger than $\sqrt{3}+\sqrt{4}+...+\sqrt{k}+\sqrt{k+1}$. We can then replace $\sqrt{3}+\sqrt{4}+...+\sqrt{k}+\sqrt{k+1}$ in the statement that we want to prove with $\frac{1}{4}k^2+\sqrt{k+1}$

So:

$$\frac{1}{4}k^2+\sqrt{k+1}<\frac{1}{4}(k+1)^2$$

$$16k+16<4k^2+4k+1$$

$$0<4k^2-12k-15$$

When I solve this I get $\left( \frac{3+2\sqrt{6}}{2};+\infty \right)$ for positive numbers

--Edit--

$3\not\in\left( \frac{3+2\sqrt{6}}{2};+\infty \right)$, but we manually proved above that the statement is true for $3$.

In this case, we proved by induction that this statement is true for $n \geq 4$. We don't need to rewrite our proof but only include a new first induction step.

*Induction first step:

$n=4$

$\sqrt{3}+\sqrt{4}<\frac{16}{4}$ which is true

And because we manually proved for $3$ we can say that this statement is true for all whole numbers $n \geq 3$

Is this correct?

--

Thanks

答:
You have a statement $S(n)$. You attempt to show $S(3)$, and that $S(n)$ implies $S(n+1)$ for all $n \geq 3$. Unfortunately this works only for $n \geq 4$.

You solve the conundrum by showing both $S(3)$ and $S(4)$, and since the induction statement is true for $n \geq 4$, $S(n)$ is also true for all $n \geq 5$.

It happens quite often that the induction statement is only true for $n \geq N$. In that case you prove $S(N)$ by hand and this proves $S(n)$ for all $n \geq N$. Then you try to prove $S(n-1)$, $S(N-2)$ and so on.

But $S(N)$ might also be false. In that case, if you find an $m \geq N$ where $S(m)$ is true, you know $S(n)$ is true for all $n \geq m$, and you try to prove $S(m-1)$ and so on as well. That would happen if you tried to prove $\text{LHS}<\text{RHS}-1000000$. You might not find any such $m$ because the statement is false for all $n$.


答:
Since $X$ is bounded, there exists $r > 0$ such that $\lVert x \rVert < r$ for all $x \in X$. Choose a point $\xi \in X$. Then $\lVert x - \xi \rVert < r + \lVert \xi \rVert = R$ for all $x \in X$.

The circle $C_R(\xi) = \{x \in \mathbb R^2 \mid \lVert x - \xi \rVert = R \}$ with center $\xi$ and radius $R$ is connected since it is the image of $\phi : [0,2\pi] \to \mathbb R^2, \phi(t) = \xi  + (R \cos(t), R \sin(t))$. Obviously $C_R(\xi) \cap X  = \emptyset$.

For $y \in \mathbb R^2 \setminus X$ and $t \in \mathbb R$ define $p(t) = \xi + t(y - \xi)$. Let $R_y = \{ p(t) \mid t \ge 0 \}$ denote the ray beginning at $\xi$ and going through $y$. It intersects $C_R(\xi)$ in the point $p(R/\lVert y - \xi \lVert)$. The set $R_y \cap X$ is a bounded convex subset of $R_y$, hence an interval containing $\xi$ as a boundary point. Therefore also the complement $A_y = R_y \setminus X \subset T_y$ is an (unbounded) interval, in particular it is connected. By construction it contains $y$. Hence
$$\mathbb R^2 \setminus X = C_R(\xi) \cup \bigcup_{y \in \mathbb R^2 \setminus X} A_y $$
is connected.

Body:
Let $X\subset \mathbb{R^2}$ be bounded and convex.Show that $\mathbb{R^2} \setminus X$ is path connected.

I understand the proposition, I was trying proof by contradiction,but I don´t get inspired, My other idea is let $x,y\in \mathbb{R^2} \setminus X$ and notice that since $X$ is bounded exits $r \in \mathbb{R}$ such that
$ X \subset B_{r}(0)$  and from here begin to construct a path from $x$ to $y$ without pass for $X$.
but I don´t know how use that $X$ is convex.

Any sugestion, idea, intuition was very helpful.And in case of be possible, may possibly give advice for be more crative in the problems that requiere a construction. Thanks in advice


问:
Suppose a random sample of size 25 is selected from a population with mean μ, the value of which is unknown.
The sample statistics are mean=  6.4, s = 10. In the test:

H0: μ = 10
H1: μ < 10 using α = .05.
Which of the following statements do you KNOW is correct?
A. A type 1 error has been committed.
B. H0 is rejected.
C. H0 is not rejected.
D. Statements (A) and (B) are correct.
E. None of the above.

Just super confused as to how to solve this

答:
What is the probability that a normally distributed random variable with mean = 10 (the hypothesized mean) and standard deviation 10 (the observed sample deviation) would produce in 25 observations a mean of 6.4?

The observed mean is $(6.4-10) = -3.6$ away from expectations.

The deviation of the average over 25 samples is: $\frac {s}{\sqrt n} = \frac {10}{5} = 2$

$z = \frac {-3.6}{2} = 1.8$

What is $P(z\le -1.8)$?

We look it up on our standard normal table, or run it through our statistical calculator and get $P(z\le -1.8)\approx 3.8\%$  But, are we running a 1 tailed test or a two tailed test?  Running a two tailed test, we would require $z < -2$ or $z > 2$ to reject the null hypothesis.


问:
How do i find the quotient and remainder of this problem?
$x^3 - a^3$ divided by $x - a$

Would really appreciate a step by step explanation of how to find the quotient and remainder because I am still learning this concept.

Thanks so much!

答:
You should get familiar with synthetic division and the rational root theorem.
Here, the possible rational roots are contained in $a^3$ and, if you worked the problem with synthetic division or found the roots(factors), you would see

$$\frac{x^3-a^3}{x-a}=\frac{(x-a)(x^2+ax+a^2)}{x-a}=x^2+ax+a^2\quad \text{with no remainder}$$


问:
How to take the limit $\lim_{n\to\infty}2\cos^2(nx)$
I wish to take this limit, but I don't know how to approach it. I am not very used to taking limit of a formula in which there are one index and one variable. A limit with just variable $x$ is more familiar to me.

Thank you!

$$\lim_{n\to\infty}2\cos^2(nx)$$

答:
Does not exist using the limit in terms of sequences (Heine)

$$ a_k =  \frac{2\pi \left [ k \right ]}{n} ~~~ \lim a_k = \infty \\ f(a_k) = 2 \cos(\frac{2 \pi \left [ k \right ]}{n} \cdot n)^2 = 2\cos(2 \pi \left [ k \right ])^2 \equiv 2$$

Choosing:
$$ a_k =  \frac{\pi}{2n} + \frac{nk}{n} ~~~ \lim a_k = \infty \\ f(a_k) = 2 \cos( (\frac{\pi}{2n} + \frac{nk}{n})n)^2 = 2 \cos( \frac{\pi}{2} + k \pi)^2 \equiv 0$$

We have two different limits thus the limit does not exist.


问:
How to find the zero of a rational function with a numerator that cannot be factored?
I'm a tutor working with an Algebra II student on rational functions, and this problem is stumping both of us:

$$\frac{x^3-2x^2-8}{x^2-3x}=0$$

She is supposed to find the zero of the function by hand, but we cannot figure out a good way to do so. Google tells me that we could use the Newton-Raphson method, but that seems quite advanced for an Algebra II class. I also read about the cubic equation, but that also seems very tedious and complex.

Does anyone have any thoughts? I'm thinking there might just be a typo in the problem given by her teacher, or maybe we are missing an obvious way to solve it! Any suggestions would be helpful :)

答:
The only real solution is

$$x = \frac13\left(2 + \sqrt[3]{116 - 12\sqrt{93}} + 2^{2/3}\sqrt[3]{29 + 3\sqrt{93}}\right)$$

My guess is there is a typo in the problem. It seems unlikely an Algebra II student would be expected to find this.


问:
How is the formula for calculating probability derived?
Why does the probability  = the number of favorable outcomes divided by the total number of outcomes? It is a really dumb question but it has been stuck in my head for days now.</p>

答:
The formula is incorrect.

The probability of an event is able to be said to be equal to the total number of favorable outcomes divided by the total number of outcomes only in the case that it is known that the outcomes are equally likely to occur.  If it is unknown that the outcomes are equally likely to occur, then we may not use this as a method of calculation.

Why does it work in such a case?  Because the outcomes are equally likely to occur!

If our sample space is $\Omega = \{\omega_1,\omega_2,\omega_3,\dots,\omega_n\}$ then we have $\Pr(\{\omega_i\})=\Pr(\{\omega_j\})$ for each $i,j$ by our hypothesis that each outcome is equally likely to occur.

We see then that since $\Pr(\Omega)=1$ by definition and that $\Pr(\Omega)=\Pr(\{\omega_1,\omega_2,\dots,\omega_n\})=\Pr(\{\omega_1\})+\Pr(\{\omega_2\})+\dots+\Pr(\{\omega_n\})=n\cdot \Pr(\{\omega_1\})$ that we have as a result $\Pr(\{\omega_i\})=\frac{1}{n}$ for each $i$.

Finally, given that an event $A = \{\omega_1,\omega_2,\dots,\omega_k\}$ we have $\Pr(A)=\Pr(\{\omega_1\})+\dots+\Pr(\{\omega_k\})=k\cdot\frac{1}{n} = \frac{k}{n}$, the aforementioned formula.  This all, of course, still being predicated on the hypothesis that the outcomes were equally likely to occur.  Without that, we could not have come to this conclusion.

As an example of a case where the formula does not work... there are only two outcomes when playing the lottery.  You either win or you lose.  However, you don't win the lottery with probability $\frac{1}{2}$ despite winning the lottery being one of the two possible outcomes.


答:
The root in the numerator in the answer from @MPW probably comes from Wolfram alpha. I don't think this can be what was expected of the student.

The graph there shows it's very near $3$. If it were $3$ (which is also a root of the denominator) you could proceed formally.

That suggests a typo. The $-8$ should probably be $-9$.

Even if that's the case there is a problem. The function's value everywhere other than $x=3$ and $x=0$ is
$$
\frac{x^2 + x + 3}{x}.
$$
The quadratic in the numerator has no real roots, so neither does the original.

Title: How to find the zero of a rational function with a numerator that cannot be factored?

Body:
I'm a tutor working with an Algebra II student on rational functions, and this problem is stumping both of us:

$$\frac{x^3-2x^2-8}{x^2-3x}=0$$

She is supposed to find the zero of the function by hand, but we cannot figure out a good way to do so. Google tells me that we could use the Newton-Raphson method, but that seems quite advanced for an Algebra II class. I also read about the cubic equation, but that also seems very tedious and complex.

Does anyone have any thoughts? I'm thinking there might just be a typo in the problem given by her teacher, or maybe we are missing an obvious way to solve it! Any suggestions would be helpful :)


问:
Is there any meaning or basis-free description for coevaluation in FDVect?
It is a fact that the category $\textbf{FDVect}_k$ of finite-dimensional vector space is a rigid category, i.e. it is symmetric monoidal and has the notion of dual object and (co)evaluation satisfying some condition. In particular, the coevaluation of $V\in\textbf{FDVect}_k$ is given by
$$\text{coev}_V:k\to V\otimes V^*,\quad 1\mapsto \sum_{i=1}^n v_i\otimes v_i^*$$
where $\{v_1,\cdots,v_n\}$ is a basis of $V$ and $\{v_i^*,\cdots,v_n^*\}$ is the corresponding basis of $V^*$ (given by $v_i^*(v_j)=\delta_{ij}$). The coevaluation is, in fact, independent from the choice of basis. So is there any "fundamental" relationship (or meaning) between this mysterious quantity $\text{coev}_V(1)=\sum_{i=1}^n v_i\otimes v_i^*$ and $V$? Or alternatively, is there any basis-free description for $\text{coev}_V(1)$?

答:
Worth adding to Quiachu Yuan's answer that the action on $V$ is described in the post is equivalent to an (injective) morphism $V\otimes V^* \to \mathrm{End}(V)$, even if $V$ is infinite dimensional.

If $X$ is a basis of $V$, we can identify $\mathrm{End}(V)$ with the subvector space of $\mathrm{Fun}(X\times X,k)$ consisting with functions $M$ such that for any $x\in X$, $M(x,-)$ has finite range. The isomorphism is given by
$$
[f \mapsto[(x,y)\mapsto y^*f(x)]]
$$
where $y^*(f(x))$ is the $y$-coordinate of the vector $f(x)$.
The inverse of this map is given by
$$
[M\mapsto [x\mapsto \sum_{y\in X}M(x,y)y]].
$$

Under this isomorphism the space $V\otimes V^*$ is the space of functions $M$ such that there is a finite subset $J\subset X$, such that $\forall (x,y) \in X\times (X-J)$, $M(x,y) = 0$.

If you think of matrices $x$ representing rows, $y$ representing columns, the endomorphisms are given by matrices such that each row has finite range, and elements of $V\otimes V^*$ are given by matrices that have a finite number of non zero columns. They both coincide if and only if $X$ is finite and if and only if the identity is in $V\otimes V^*$.

If the isomorphism holds, then since the space is of finite dimension, the sum $\sum_{x\in X} x\otimes x^*$ makes sense, and note that for any other base $Y$, $\sum_{x\in X} x\otimes x^* = \sum_{y\in Y} y\otimes y^*$. Let's then of course note that element $id_V$. One way to capture how many elements there are in the basis is then to compute $ev_{V}(id_V)$, this is the dimension of the vector space $V$, and it can be defined without ever referencing a basis, you just need the evaluation and coevaluation. Note that of course if you have a basis $X$, then it is necesseraly finite, and $ev_{V}(id_V) = card(X)$.

If $id_V$ is in $V\otimes V^*$, write $id_V$ as a finite sum of pure tensors, for example $id_V = \sum_{(x,y)\in X \times Y }x\otimes y$, where $X\subset V$ is a finite subset, and $Y\subset V^*$ is a finite subset. Then $X$ is a generating family since for every $v \in V$, $v = \sum_{(x,y)\in X\times Y} y(v)x$. From there, a finite base can be extracted.

So maybe to show that the dimension is an integer, when the field is of caracteristic 0, one has to show the theorem of extraction of basis from a generating family.


问:
Combinatorial proof of a Stirling number identity.
Consider the identity
$$\sum_{k=0}^n (-1)^kk!{n \brace k} = (-1)^n$$
where ${n\brace k}$ is a Stirling number of the second kind. This is slightly reminiscent of the binomial identity
$$\sum_{k=0}^n(-1)^k\binom{n}{k} = 0$$
which essentially states that the number of even subsets of a set is equal to the number of odd subsets.

Now there is an easy proof of the binomial identity using symmetric differences to biject between even and odd subsets. I am wondering if there is an analogous combinatorial interpretation for the Stirling numbers. The term $k!{n\brace k}$ counts the number of set partitions of an $n$ element set into $k$ ordered parts. Perhaps there is something relating odd ordered partitions with even ordered partitions?

As an added note, there is a similar identity
$$\sum_{k=1}^n(-1)^k(k-1)!{n\brace k}=0$$
for $n \geq 2$.
A combinatorial interpretation of this one would also be appreciated.

答:
Perhaps there is something relating odd ordered partitions with even ordered partitions?

There is indeed. Let's try to construct an involution $T_n$, mapping odd ordered partitions of $n$-element set to even and vice versa:
if partition has part $\{n\}$, move $n$ into previous part; otherwise move $n$ into new separate part.

Example: $(\{1,2\},\{\mathbf{5}\},\{3,4\})\leftrightarrow(\{1,2,\mathbf{5}\},\{3,4\})$.

This involution is not defined on partitions of the form $(\{n\},\ldots)$, but for these partitions one can use previous involution $T_{n-1}$ and so on.

Example: $(\{5\},\{4\},\{1,2\},\{\mathbf{3}\})\leftrightarrow(\{5\},\{4\},\{1,2,\mathbf{3}\})$.

In the end only partition without pair will be $(\{n\},\{n-1\},\ldots,\{1\})$. So our (recursively defined) involution gives a bijective proof of $\sum_{\text{k is even}}k!{n \brace k}=\sum_{\text{k is odd}}k!{n \brace k}\pm1$ (cf. 1, 2).

Upd. As for the second identity, the involution $T_n$ is already defined on all cyclically ordered partitions, so $\sum_{\text{k is even}}(k-1)!{n \brace k}=\sum_{\text{k is odd}}(k-1)!{n \brace k}$.

P.S. I can't resist adding that $k!{n \brace k}$ is the number of $(n-k)$-dimensional faces of an $n$-dimensional convex polytope, permutohedron (the convex hull of all vectors formed by permuting the coordinates of the vector $(0,1,2,\ldots,n)$). So $\sum(-1)^{n-k}k!{n \brace k}=1$ since it's the Euler characteristic of a convex polytope.


问:
Polynomial ring isomorphic to Quotient space
I'm relatively new to rings, and I also don't have a very good understanding of quotient spaces, but I'd like to understand the following problem.

Show that for $A \in M_n(\mathbb R ),$ there is some $p \in \mathbb R [X]$ such that $\mathbb R [A]$ is isomorphic to $\mathbb R [X]/ \langle p \rangle$. Show that if $p,q \in \mathbb R [X]$ are quadratics with two distinct real roots then $\mathbb R [X]/ \langle p \rangle \cong \mathbb R [X]/ \langle q \rangle$. Justify which, if any, of $\mathbb R [A]$, $\mathbb R [B]$, $\mathbb R [C]$ and $\mathbb R [D]$ are isomorphic as rings.
$$
A:=
    \left( \begin{matrix}
    2 & 0 & 0 \\
    0 & 1 & 0 \\
    0 & 0 & 1 \\
    \end{matrix} \right),
B:=
    \left( \begin{matrix}
    4 & 1 \\
    0 & 3 \\
    \end{matrix} \right),
C:=
    \left( \begin{matrix}
    2 & 1 & 0 \\
    0 & 2 & 0 \\
    0 & 0 & 2 \\
    \end{matrix} \right),
D:=
    \left( \begin{matrix}
    1 & 1 \\
    0 & 1 \\
    \end{matrix} \right).
$$

I already struggle to understand the question... I know that $\mathbb R [X]$ are the polynomials with real coefficients, and we can also plug matrices in and evaluate the polynomials as we are used to. So, I guess $\mathbb R [A]$ is the space of all matrices we can get by adding powers of $A,$ etc. I definitely lack an intuition what $\mathbb R [X]/ \langle p \rangle$ is. Could somebody offer some insight? And then maybe a hint about how to solve that?

As for the second part, I guess if we have isomorphisms, they are in the same dimension. Since $A$ is diagonal and $C$ is not, I don't think $\mathbb R [A]$ and $\mathbb R [C]$ are isomorphic. But how would I go about solving that part as well?,

答:
Unfortunately, I am not sure that the following explanation will mean very much to you, if you are not so familiar with ring theory; however, I will provide as much scaffolding as possible so that you can develop an understanding and eventually return to this post with full comprehension.

Let us begin with an $n \times n$ matrix $A$ with entries in $\mathbb R.$ Henceforth, let us use $I$ to denote the $n \times n$ identity matrix. Convince yourself that the collection of "polynomials" $$\mathbb R[A] = \{c_n A^n + \cdots + c_1 A + c_0 I \,|\, n \geq 0 \text{ is an integer and } c_0, c_1, \dots, c_n \in \mathbb R\}$$ in the matrix $A$ with real coefficients is a unital ring with respect to matrix addition and multiplication: the zero matrix is the additive identity; the identity matrix $I$ is the multiplicative identity. Consequently, the map $\varphi_A : \mathbb R[X] \to \mathbb R[A]$ defined by $$\varphi_A(c_n X^n + \cdots + c_1 X + c_0) = c_n A^n + \cdots + c_1 A + c_0 I$$ is a surjective unital ring homomorphism. By the First Isomorphism Theorem, therefore, we have that $\mathbb R[A] \cong \mathbb R[X] / \ker \varphi_A,$ where we denote by $$\ker \varphi_A = \{p(X) \in \mathbb R[X] \,|\, \varphi_A(p(X)) \text{ is the zero matrix} \}$$ the kernel of the ring homomorphism $\varphi_A.$ Until now, I hope that things have made sense.

But here is where you will have to cross-reference a few things. Considering that $\mathbb R$ is a field, it follows that $\mathbb R[X]$ is a Euclidean domain. Every Euclidean domain is a principal ideal domain (PID), so $\mathbb R[X]$ is a PID. By definition, therefore, for every ideal $J$ of $\mathbb R[X],$ there exists a polynomial $f(X)$ in $J$ such that $J = f(X) \mathbb R[X],$ i.e., for every polynomial $h(X)$ in $J,$ there exists a polynomial $g(X)$ such that $h(X) = f(X) g(X).$ By the First Isomorphism Theorem, we have that $\ker \varphi_A$ is an ideal of $\mathbb R[X],$ hence there exists a polynomial $p(X)$ in $\ker \varphi_A$ such that $\ker \varphi_A = p(X) \mathbb R[X] = \langle p(X) \rangle$ (in your notation). Consequently, the first claim is established. Given that $p(X)$ is not monic, its leading coefficient $C$ is a unit in $\mathbb R[X]$ (because $\mathbb R$ is a field), hence $\mu(X) = C^{-1} p(X)$ is a monic polynomial with $\langle \mu(X) \rangle = \langle p(X) \rangle.$ Often, in the literature, the monic polynomial $\mu(X)$ that generates $\ker \varphi_A$ is referred to as the minimal polynomial of $A$ because it is the unique monic polynomial of least degree such that $\mu(A)$ is the zero matrix.

Consequently, for any real matrices $M$ and $N$ with the same minimal polynomial $\mu(X)$ in $\mathbb R[X],$ we have that $\mathbb R[M] \cong \mathbb R[X] / \langle \mu(X) \rangle \cong \mathbb R[N].$ By the Chinese Remainder Theorem, if $\mu(X)$ is a product of $k$ distinct linear factors (i.e., $\mu(X)$ has distinct real roots), we have that $\mathbb R[X] / \langle \mu(X) \rangle \cong \mathbb R^k.$ We can therefore determine which rings $\mathbb R[A],$ $\mathbb R[B],$ $\mathbb R[C],$ and $\mathbb R[D]$ are isomorphic by computing the minimal polynomials of $A, B, C,$ and $D.$

Computing the minimal polynomial of a matrix can be difficult in certain situations; however, the matrix $A$ is diagonal, so its minimal polynomial is easily deduced. On the other hand, all three of $B, C,$ and $D$ are both upper-triangular, hence the determinants $\det(xI - B),$ $\det(xI - C),$ and $\det(xI - D)$ are easy to compute. We refer to $\chi_B(x) = \det(xI - B)$ as the characteristic polynomial of $B.$ Observe that if $B$ is an $n \times n$ matrix, then the degree of the characteristic polynomial of $B$ is $n.$ By the Cayley-Hamilton Theorem, the characteristic polynomial of a matrix annihilates the matrix (i.e., $\chi_B(B)$ is the zero matrix), hence by the above paragraph, we must have that $\mu_B(x) \,|\, \chi_B(x),$ where $\mu_B(x)$ is the minimal polynomial of $B.$ Consequently, if $B$ is an $n \times n$ matrix with $n$ small (e.g., $n \leq 3$), then one may compute the minimal polynomial of $B$ by finding the smallest product of the irreducible factors of $\chi_B(x)$ that annihilate $B.$


问:
Radius of Convergence with coefficient doesnt have a limit
I want to know when the series
$$ \sum _ { n = 1 } ^ \infty \frac { x ^ n } { \sin ( n \theta ) } $$
is convergent for $ \theta = t \pi $ such that $ t $ is irrational, e.g.  $ \theta = \sqrt 3 \pi $.

It is a really interesting question because it will tell something about how the $ \sin \left( \sqrt 3 \pi n \right)$ goes to zero for some sequence goes to infinity (since $ \sqrt 3 $ is irrational sin will have a dense image in $ [ 0 , 1 ] $).

答:
It converges when $x = 0$

$\lim_\limits{n\to\infty} \frac {x^n}{\sin n\theta}$ does not exist for $x\ne 0$


问:
Number of divisors
How can I find number of divisors of N which are not divisible by K.

($2 \leq N$, $k \leq 10^{15})$

One of the most easiest approach which I have thought is to first calculate total number of divisors of 'n' using prime factorization (by Sieve of Eratosthenes) of n and then subtract from it the Number of divisors of the number 'n' that are also divisible by 'k'.

In order to calculate total number of divisors of number n that are also divisible by 'k'
I will have to find the total number of divisors of (n/k) which can be also done by its prime factorization.

My problem is that since n and k can be very large, doing prime factorization twice is very time consuming.

Please suggest me some another approach which requires me to do prime factorization once.

答:
I use this method in recriprocal tables, like the sumerian 2:30  48   (which multiply to 60).

http://z13.invisionfree.com/DozensOnline/index.php?showtopic=782&view=findpost&p=22089607

This code looks for all of the numbers whose prime factors are a known set (listed in the code with longish logs), that fall below some $b^{n+1}$.  So, for example, given "prime 10 3", it would print all of the mantissa (once for 2, 20, 200, 2000, etc), of numbers of the form $2^a 5^b$, under 1000.

The output is sorted at the first or fifth column, to get a simple list (eg 1, 2, 4, 5, 8, 16, 25, 27, ...), or interspersed decimally (1, 1.6, 2, 2.5, 3.2, 4, ...).

It's designed to handle largish numbers, but the outputs can be quite large if there are lots of divisors, eg $510510 , 3$ produces a file over 1.6 million rows, and 150 MB.

The code can be modified to restrict it to divisors of $N$, rather than to be let run through the maximum allowable.  For example, the limits are set with the routine crd2log(), but one can select values from am array of divisor-powers of $N$.

Suitable output from this has been used elsewhere in the quoted thread, to find interesting numbers about bases.

Title: Number of divisors


答:
The simplest solution is to multiply both sides by the denominator and then solve using the cubic formula

$$\frac{x^3-2x^2-8}{x^2-3x}=0\implies x^3-2x^2-8=0\\
\implies \qquad a=1\qquad b=-2\qquad c=0 \qquad d=-8
$$

$$x=\sqrt[\Large{3}]{\bigg(\frac{-b^3}{27a^3 }+\frac{bc}{6a^2}-\frac{d}{2a}\bigg)+\sqrt{\bigg(\frac{-b^3}{27a^3}+\frac{bc}{6a^2}-\frac{d}{2a}\bigg)^2+\bigg(\frac{c}{3a}-\frac{b^2}{9a^2}\bigg)^3}}\\
+\sqrt[\Large{3}]{\bigg(\frac{-b^3}{27a^3 }+\frac{bc}{6a^2}-\frac{d}{2a}\bigg)-\sqrt{\bigg(\frac{-b^3}{27a^3}+\frac{bc}{6a^2}-\frac{d}{2a}\bigg)^2+\bigg(\frac{c}{3a}-\frac{b^2}{9a^2}\bigg)^3}}-\frac{b}{3a}$$

This will give you a real solution and, divided into the original, will give you a second degree equation that can be solved with the quadratic formula.

You can also go to WolframAlpha here to find the answer the easy way: $x≈2.9311....$


答:
Perhaps the teacher is pushing students to imagine a method based on iterations. Newton-Rapson is such a method, but it seems too hard for a novice.

Other, simple, iteration method is just trying values and see when the function changes its sign, a good signal that we are getting close to the solution.

The zeros are those of the numerator. But watch out the denominator: x=0 or x=3 make infinity, so can't use them.

Let's build a table:

x  |  f(x) numerator

_____________________

 -1| -11

 1 |  -9

 2 |  -8

 3 |   1 (but forbbiden by denominator)

 4 |  24

So, the solution goes between 2 and 3.

Then build a new table with 2.2, 2.4, 2.6, 2.8 and 2.9

The next table may use 2.92, 2.94, 2.96, 2.98

And continue this method until you get some decimals.


问:
How to prove $||X||^p,$ $p>1$ is strictly convex?
How to prove $||X||^p,$ $p>1$ is strictly convex? Here $X\in\mathbb{R}^d$, and $||x||$ is the $l_2$-norm.

I know how to prove $||X||^p$, $p\geq1$ is convex, for example, using Holder. But how to prove the strict convexity?

答:
One approach is to use the following:

Claim: Let $f:I\to {\mathbb R}$, for $I\subset {\mathbb R}$ an open interval. If $f'$ is strictly increasing on $I$, then $f$ is strictly convex on $I$.

If you need to prove this, you can use the mean value theorem.


问:
asymptotic expansion of $\ln(\cosh(x))$
I am currently looking into some C++ code which approximates the function in the title as follows

$$\ln\cosh(x))\approx x- \ln(2); \qquad x \geq 12$$

The approximation is plausible to me, since plotting the difference between the function results in an  $error < 10^{-3}$. However how does one arrive at such an approximation? The usual way of calculating the Taylor expansion at $x_0=12$ does not result in such a nice expression for me.

答:
Here is a useful approximation if we expand in both $x$ and $z=e^{-x}$.

\begin{align*}
\log(\cosh(x))&=\log(e^x+e^{-x})-\log(2)\\
&=x-\log(2)+\log(e^x+e^{-x})-x\\
&=x-\log(2)+\log(z+z^{-1})+\log(z)\\
&=x-\log(2)+\log(1+z^2)\\
&=x-\log(2)+\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}{z^{2n}}\\
&=x-\log(2)+\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}e^{-2xn}
\end{align*}

This came up in Cardy Scaling and Renormalization in Statistical Physics, problem 3.3, where $\log(2\cosh(K))$ is a free energy density and you care only about its non-analytic parts in the infinite sum.

Title: asymptotic expansion of $\ln(\cosh(x))$


问:
The Soup Problem: how to asymptotically fairly split a geometric series and a constant one using a single pattern?
Literally every time I'm serving some soup I'm thinking of this little mathematical problem I devised.

Imagine you have a very large (= infinite, for the purposes of the actual problem) bowl of soup and want to divide it into two halves using a (finite) ladle. The trouble is that some of the good stuff floated to the surface and every scoop takes a part $p$ of it, $p < \frac12$ (so its amount goes down geometrically with a quotient of $1-p$). The other good stuff is evenly dissolved in the volume. The protocol is that you always take a full scoop in this manner, so the only freedom is in what order do you empty the ladle into $A$ or $B$'s plate. What order would you take so that, asymptotically, both parties get fair shares in both respects?

Obviously the answer can't be $ABABAB\ldots$, because $A$ would always get $1/(1-p)$-times more than $B$. $\overline{ABBA}$ does not work either because $1+(1-p)^{-3}$ is always greater than $(1-p)+(1-p)^2$. In fact (except for a countable number of special values of $p$ (*)) it can't be any periodic infinite word of $A$ and $B$ for a slight variant of the same reason. One more thing I found is that the greedy algorithm (go $A$ until it has more than $B$, then swap) won't work either because while it would always satisfy splitting of the geometrical part, it would break the linear one (number of scoops). That's all I know at the moment.

Rephrased: for a given $1-p =: q \in (\frac12, 1)$, find a subset $M ∈ \mathbb{N}$ such that both

$\displaystyle\lim_{N→∞} \frac{\#\left( M \cap \{1,\ldots,N\} \right)}{N} = \frac12$,

$\displaystyle\sum_{n=1}^∞ q^n \chi_M(n) = \frac12 \sum_{n=1}^∞ q^n$.

I'm pretty sure the answer will lie somewhere in the theory of non-integer base systems or combinatorics on words or both, but I know little of those.

(*) For example, for $q = φ - 1$ the equality $q + q^2 + q^3 = 1 + q^4 + q^5$ holds, so for $p = 2 - φ \approx 0.381966$ the sequence $\overline{ABBBAA}$ would work.

答:
I wrote a short paper devoted to your problem.

Solving it, I introduced the following notion, considered in a separate question. A number $q\in (1/2, 1)$ is approximating, if there exist non-negative numbers $A$ and $N$ such that for each $x_0\in [0,A]$ there exist $n\le N$ and a polynomial $P(x)$ of the form $\sum_{i=1}^n \pm x^i$ such that $P(1)=0$ and $|x_0-P(q)|\le Aq^n$.

Proposition. For each approximating number $q$ there exists a subset $M$ of natural numbers such that $\lim_{N\to\infty} |M \cap \{1,\ldots, N\} |/N = 1/2$, and $\sum_{n\in M} q^n=\tfrac 12\sum_{n=1}^\infty q^n$.

Proof. Let the number $q$ is approximating with the constants $A$ and $N$. Put $k_0=0$. Then we can inductively build an increasing sequence $(k_n)$ of natural numbers such that $k_{n+1}-k_n\le N$ for each $n$, assigning signs to numbers $q^i$ for $k_{n-1}<i\le k_n$ (a half of the assigned signs are “$+$” and the other half are “$-$”) assuring $\sum_{i=1}^{k_n} \pm q^i\le Aq^{k_n}$. Let $M$ be the set of natural $n$ such that $q^n$ has “$+$” sign. $\square$

In the paper is shown that each $q\in (q_\infty,1)$ is approximating, where $q_\infty=0.5845751\dots$ is a unique positive root of a polynomial $Q_\infty(x)=x^4+x^3+2x^2-1$.


问:
If $f:\mathbb R \to \mathbb R$ then for any $A\subset \mathbb R$, $\mathbb R-f^{-1}(A)=f^{-1}(\mathbb R-A)$
Let $f:\mathbb R\to \mathbb R$ and $A\subset \mathbb R$. Show that $f^{-1}(\mathbb R-A)=\mathbb R-f^{-1}(A)$

I never got this question when I was reading Set theory & functions. But while in Continuity there is a Theorem in my book which used this.

答:
$(\subseteq)$ :

Let $x \in f^{-1}(\mathbb{R} \backslash A) .$ Then $f(x) \in \mathbb{R}$ and $f(x) \notin A .$ Notice that $x \in \mathbb{R}$ and so $x \notin f^{-1}(A)$ follows from $f(x) \notin A .$ Thus $x \in \mathbb{R} \backslash f^{-1}(A)$.

$(\supseteq)$ :

Let $x \in \mathbb{R} \backslash f^{-1}(A) .$ Then $x \in \mathbb{R}$ and $x \notin f^{-1}(A) .$ Clearly, $f(x) \in \mathbb{R}$ and note that $f(x) \notin A$ since $x \notin f^{-1}(A)$. Thus $f(x) \in \mathbb{R}\backslash A$ and therefore $x \in f^{-1}(\mathbb{R} \backslash A)$


问:
Can $f(\Theta)$ be open in $\mathbb{R}^k$, where $\Theta\subset\mathbb{R}^{k-1}$?
Can $f(\Theta)$, the image of $f$ defined as $\{f(\theta),\theta\in\Theta\}$ be open in $\mathbb{R}^k$, where $\Theta\subset\mathbb{R}^{k-1}$? $f$ is a function from $\Theta$ to $\mathbb{R}^k$.

I don't think so, otherwise $f(\Theta)$ would contain a open ball in $\mathbb{R}^k$, which is not possible. Is this correct? If not, can you please give me an example?

答: Consider the Peano Curve $\gamma:[0,1]\to Q:=[0,1]\times [0,1]$ where $Q$ is the unit square in the plane. The curve is surjective (the image of the curve is the whole square). Therefore, just pick an open subset $A$ included in the square and set $\Theta:=\gamma^{-1}(A)$.


问:
Can integral transforms be viewed as change of basis formulas?
Forgive any lack of rigor.

If you have a countable  orthonormal basis $B$ for a Hilbert space $H$ , then any function $f \in H$ can be expressed as
$$ f(t) = \sum\limits_{g \, \in \, B} \langle f, \, g \rangle \, g(t) $$
where $\langle f, g \rangle$ might be something like $$ \int_{t_1}^{t_2}f(t) \, g(t) \, \mathrm{d}t $$

So essentialy $ (a_1, a_2,  \dots, a_{N-1})$ is the coordinate vector of $f$ for basis $B$.

Now my question is: are integral transforms the continuous version of this? Can the kernel function of an integral transform be viewed as a sort of basis? For example suppose you do have a basis $B' = \{ K_u(t) : u \in (u_1, u_2) \} $, and a function $f'$. Can you use a similar expression for calculating the coordinate vector of $f'$ relative to $B'$ as with $f$? If so what conditions does $B'$ have to satisfy?

Here's an example of the finite case using sines and cosines for the basis -
http://en.wikipedia.org/wiki/Fourier_series#Hilbert_space_interpretation

Thanks for helping.

答: I would (belatedly) say yes. In particular, regarding Cameron's comment, I'd note that there's strictly speaking nothing preventing you from using a transformation whose Jacobian has a nontrivial kernel -- that's basically just a projection operator. It won't be invertible, but while that may or may not rigorously align with what mathematicians call a "basis" or "basis transformation" it certainly has a sensible and consistent meaning.