问:
Simple solution to computing the eigenvalues/vectors of $N \times N$ matrix with diagonal entries $N$ and off-diagonal entries 1
Versions of this question have been asked before (here, here, here). However, I am asking again for two reasons:

My question is slightly different—it's about eigenvalues/vectors, not determinants—, and I don't know enough linear algebra to see the obvious connections, if they exist.

I was asked this during a job interview and therefore assume the interviewer did not want me to appeal to theorems, e.g. here.

The question: Find the eigenvalues/eigenvectors of the $N \times N$ matrix

$$
A = \begin{bmatrix}
N & 1 & 1 & \dots & 1
\\
1 & N & 1 & \dots & 1
\\
1 & 1 & N & \dots & 1
\\
\vdots & \vdots & \vdots & \ddots & \vdots
\\
1 & 1 & 1 & \dots & N
\end{bmatrix}
$$

Is there an simple (interview-level) way to reason about this question? If I'm missing some obvious connection to the answer here, I'm happy to learn and to then close this question.

答:
Call $E$ the all-ones $N\times N$ matrix. Notice that $A = (N-1)I + E$

Let now $v_1,v_2,\dots,v_n$ be a base of eigenvectors for $E$ with eigenvalues $\lambda_1,\lambda_2,\dots,\lambda_n$. Notice that
$$
Ev_i = \lambda_iv_i \implies Av_i = [(N-1)I + E]v_i = (N-1)v_i + \lambda_i v_i = (N-1+\lambda_i)v_i
$$
so the eigenvalues of $A$ are $N-1+\lambda_i$ with eigenvectors $v_i$.

You can now focus on finding eigevalues/vectors of $E$, that has been answered already


问:
Is there a missing definition for addition of vector fields?
In proving $\mathfrak X(U)$ is a $C^{\infty}U$-module, for an open subset $U$ of $\mathbb R^n$  my book defines scalar multiplication of smooth vector fields $U$ by smooth functions on $U$ as

$$[fX]_p := f(p)X_p$$

I could not find a definition for addition. Is it $[X+Y]_p  := X_p + Y_p$ ? Otherwise, with what definition do we prove the "Abelian group" part of showing $\mathfrak X(U)$ is a $C^{\infty}U$-module?

My book is An Introduction to Manifolds by Loring W. Tu.

答:
As Charlie pointed in the comments, it is just your regular pointwise addition:
$$(X+Y)_p=X_p+Y_p$$

ContentLicense: CC BY-SA 4.0


问:
Finding the eigenvectors of a matrix.
Find the normalized eigenvectors of this matrix:

$A=\begin{bmatrix}2&1&1\\1&2&1\\1&1&2\end{bmatrix}$

My work:

$$\det(A-\lambda I) =(\lambda-1)^2(\lambda-4)$$

$$\lambda_1=1, \lambda_2=1,\lambda_3=4$$

$$\lambda_1\iff x + y +z=0$$
$$\lambda_2\iff x + y +z=0$$
$$\lambda_3\iff -2x + y +z=0$$
$$\lambda_3\iff x + -2y +z=0$$
$$\lambda_3\iff x + y +-2z=0$$

When my lecturer did this the equations were easy to solve... How do I find the normalized eigevectors from these equations?

Please Help :)

答:
To find the eigenvectors of the matrix $A$ you find the null space of $(A-\lambda_1 I)$, $(A-\lambda_2 I)$ and $(A-\lambda_3 I)$. Since $\lambda_1=\lambda_2$, to find the eigenvectors corresponding to those eigenvalues it suffices to find the null space of $(A-\lambda_1 I)x=0$.

Thus
$$(A-\lambda_1 I)x=0$$
$$\longrightarrow\begin{bmatrix}1&1&1\\1&1&1\\1&1&1\\\end{bmatrix}x=0$$
$$\longrightarrow x_1=\begin{bmatrix}-1\\0\\1\\\end{bmatrix} \text{ and } x_2=\begin{bmatrix}-1\\1\\0\\\end{bmatrix}$$
Similarly
$$(A-\lambda_3 I)x=0$$
$$\longrightarrow\begin{bmatrix}-2&1&1\\1&-2&1\\1&1&-2\\\end{bmatrix}x=0$$
$$\longrightarrow x_3=\begin{bmatrix}1\\-1\\1\\\end{bmatrix}$$
Finally, to answer the question you just normalize them to get
$$x_1=\begin{bmatrix}-1/\sqrt{2}\\0\\1/\sqrt{2}\end{bmatrix} \text{, } x_2=\begin{bmatrix}-1/\sqrt{2}\\1/\sqrt{2}\\0\end{bmatrix}\text{, } x_3=\begin{bmatrix}1/\sqrt{3}\\-1/\sqrt{3}\\1/\sqrt{3}\end{bmatrix}$$

ContentLicense: CC BY-SA 3.0

Title: Finding the eigenvectors of a matrix.


问:
Proof by deduction using backward reasoning
I have a question:
Solve the following by deduction using backward reasoning to prove gt(5,2). I found from wikipedia that backward reasoning is same as backward chaining. Book says that backward chaining is same as goal dependent search. Suppose we are given the following facts:

Somebody please guide me. Sorry I don't have any idea. Should I try by doing the replacements provided at the end of each level of the tree? If somebody can help me I really appreciate.

I got one solution from my friend. Kindly check it and explain if possibe.

Linked concepts and Questions: By goal driven search it means that we have to start at the current state. How is it different from resolution proof?

答:
See Backward chaining.

Regarding the handwritten proof sketch, the first step is to apply the substitution $\{ 5/x, 2/z \}$ to the clause in fact 1) to get:

$gt(5,y) \land gt(y,2) \to gt(5,2)$.

Thus, in order to prove $gt(5,2)$ (by Modus Ponens) we have to derive the antecedent: $gt(5,y) \land gt(y,2)$.

This is the new goal, that we split into: $gt(5,y)$ and $gt(y,2)$.

Consider the first one: $gt(5,y)$. Using again fact 1) with substitution $\{ 5/x, y/z \}$ we get: $gt(5,y) \land gt(y,y) \to gt(5,y)$.

The new goal is:

$gt(5,y) \land gt(y,y)$.

But the third fact is: $\forall x \lnot gt(x,x)$.

Thus, there is no way to derive $gt(y,y)$.


问:
How this system transferred into this state variables
Take a look at this transfer function:
$$
\frac{Y(s)}{U(s)} = \frac{1}{s^2 + 2s + 1}
$$

The author of the book I'm reading says, the transfer function can be represented as

$$
\frac{Y(s)}{U(s)} = \frac{1}{s^2 + 2s + 1} = \frac{1}{(s+1)} \frac{1}{(s+1)}
$$

The block diagram is

Then the author states the following equations

$$
\begin{align}
\dot{x}_1 &= x_2 - x_1 \\
\dot{x}_2 &= -x_2 + u \\
y &= x_1
\end{align}
$$
Notice: the last equation is given. I'm aware of the fact that state variables representation is not unique but I failed to understand this method. How did he come up with this representation?

答:
Calling $X_1(x)$ a variable between the two transfer blocks and $X_2(s)$ between the last transfer block and $Y(s)$ we have

$$
\cases{
X_1(s) = \frac{1}{s+1}U(s)\\
X_2(s) = \frac{1}{s+1}X_1(s)\\
Y(s) = X_2(s)
}
$$

and equivalently

$$
\cases{
\dot x_1 + x_1 = u\\
\dot x_2 + x_2 = x_1\\
y= x_2
}
$$

etc.


问:
WANTED : Fast-growing 1st order statistics
Is there any distribution that satisfies these two conditions :

Bounded variance

$E[ $( Maximum of K i.i.d samples )$] = \Omega(\sqrt{K})$?

Or is there any proof that it is not possible?

.

I think if such distribution exists, it should be a very, very heavy-tailed distribution.

As long as I know, the maximum of K standard normal samples grows like $\log K$.

Please help!

答:
I doubt it is possible.

I suspect that $E[ $( Maximum of K i.i.d samples )$] = \Omega(\sqrt{K})$ leads to something broadly like $F(x) = \exp(-c/x^2)$ for $x>0$ (at least in the positive tail); this example has a density of $f(x)=\frac{2c}{x^3}\exp(-c/x^2)$ and a finite mean of $\mathbb E[X]=\sqrt{\pi c}$ but an infinite variance

This example does meet the stated requirement for the expectation of the maximum of $K$ i.i.d. samples, since $Y=\max(X_1,\ldots,X_k)$ has $F(Y) = F(X)^K=\exp(-cK/x^2)$ and mean $\mathbb E[Y]=\sqrt{\pi c K}$ which is $\Omega(\sqrt{K})$


答:
Another answer for this problem beside the accepted answer. Let
$$
\begin{align}
x_1 &= y \tag{1} \\
x_2 &= \dot{y} + y \tag{2}
\end{align}
$$
The actual dynamic equation is (i.e. this is given from the problem)
$$
\ddot{y} = u -2\dot{y} - y \tag{3}
$$

Taking the derivative of Eq(1) and Eq(2) with respect to time, we get
$$
\begin{align}
\dot{x}_1 &= \dot{y} \tag{4} \\
\dot{x}_2 &= \ddot{y} + \dot{y} \tag{5}
\end{align}
$$

From Eq(1) and Eq(4), we get
$$
\dot{x}_1 = x_2 - x_1
$$

From Eq(3) and Eq(5) and Eq(2), we get
$$
\begin{align}
\dot{x}_2 &= u -2\dot{y} - y + \dot{y} \\
&= u -\dot{y} - y  \\
&= u -(\dot{y} + y) \\
&= u - x_2
\end{align}
$$

Title: How this system transferred into this state variables

Body:
Take a look at this transfer function:
$$
\frac{Y(s)}{U(s)} = \frac{1}{s^2 + 2s + 1}
$$

The author of the book I'm reading says, the transfer function can be represented as

$$
\frac{Y(s)}{U(s)} = \frac{1}{s^2 + 2s + 1} = \frac{1}{(s+1)} \frac{1}{(s+1)}
$$

The block diagram is

Then the author states the following equations

$$
\begin{align}
\dot{x}_1 &= x_2 - x_1 \\
\dot{x}_2 &= -x_2 + u \\
y &= x_1
\end{align}
$$
Notice: the last equation is given. I'm aware of the fact that state variables representation is not unique but I failed to understand this method. How did he come up with this representation?


问:
Seems that I just proved $2=4$.
Solving $x^{x^{x^{.^{.^.}}}}=2\Rightarrow x^2=2\Rightarrow x=\sqrt 2$.

Solving $x^{x^{x^{.^{.^.}}}}=4\Rightarrow x^4=4\Rightarrow x=\sqrt 2$.

Therefore, $\sqrt 2^{\sqrt 2^{\sqrt 2^{.^{.^.}}}}=2$ and $\sqrt 2^{\sqrt 2^{\sqrt 2^{.^{.^.}}}}=4\Rightarrow\bf{2=4}$.

What's happening!?

答:
In general, you can solve the equation in terms of the Lambert W function as

$$ y=x^{x^{x^{.^{.^.}}}} \implies \ln(y)=y\ln(x) \implies y = -\frac{W(-\ln(x))}{\ln(x)}.  $$

Try to use this closed form to see what the problem is. Note this, if you ask maple to solve the equations

$$  -\frac{W(-\ln(x))}{\ln(x)}=2,\quad  -\frac{W(-\ln(x))}{\ln(x)}=4, $$

you will get the same answers

$$ x=1,\sqrt{2} $$

More generally, the solution of

$$ -\frac{W(-\ln(x))}{\ln(x)}=a $$

is given by

$$ \left\{ x=1,x={{\rm e}^{{\frac {\ln  \left( a \right) }{a}}}}\right\}. $$

Now, if you use the above solution with $a=2,4$, you will see why?

Title: Seems that I just proved $2=4$.


问:
How to tell if the product of two incidence (edges vs.vertices) matrices will be an incidence matrix
I'm trying to determine if there are properties of certain incidence matrices that mean that their multiple will create a valid incidence matrix, i.e., the total number of non zero entries in each row is either 2 or 0.

In the set of matrices $\mathbb{N}_{2}^{3\times3}$, there are three valid columns: $$\begin{bmatrix}1\\1\\0\end{bmatrix}, \begin{bmatrix}1\\0\\1\end{bmatrix}, \begin{bmatrix}0\\1\\1\end{bmatrix}$$
Which can make $3^{3}$ possible matrices (or $4^{4}$ if the zero vector / 'missing edges' is included) but only some will multiply together to make a third valid incidence matrix.

So in all, I'd like to know if there are any ways to determine if the product of two matrices will preserve the property of being a valid incidence matrix.

答:
It would be highly dependent on how the edges and vertices are ordered. I am struggling a bit to see why this would be something interesting (but that doesn't mean you don't have a reason to be interested in it).

You just need to phrase your question in terms of what the rows/columns of the incidence matrix means, and what it means to multiply matrices... this will give a very convoluted combinatorial condition like "for each vertex $v_{i}$ of the first graph, let $I_{i}$ be the set of edge indices for the edges adjacent to $v_{i}$; then for each edge $e_{j}$ of the second graph, there can be at most one vertex (of the second graph) incident with $e_{j}$ whose index is in $I_{i}$; and furthermore, there must be either $0$ or $2$ edges $e_{j}$ incident with a vertex whose index is in $I_{i}$."


问:
How to interpret $ \vec{v} \in [0,1]^{N+1}$
How to interpret $\vec{v} \in[0,1]^{N+1}$

Dose it mean a vector of size $N+1$ where the values inside are between $0$ and $1$? or is it a binary vector?

答:
First interpretation is correct.

In second case it should be writen like this $\{0,1\}^{N+
1}$ or  $ \mathbb{Z}_2^{N+1}$ or $\mathbb{F}_2^{N+1}$


问:
$\frac{\partial u}{\partial t} = \frac{\partial^2 u}{\partial x^2}$ with $u(x,0)=\frac{e^{2x}-1}{e^{2x}+1}$. Then $\lim_{t \to \infty}u(1,t)$ equals
Let $u(x,t)$ be a bounded solution of $\frac{\partial u}{\partial t} = \frac{\partial^2 u}{\partial x^2}$ with $u(x,0)=\frac{e^{2x}-1}{e^{2x}+1}$. Then $\lim_{t\to \infty}u(1,t)$ equals:

In the above problem, the boundary is not given but let's assume that the boundary conditions are as follows: $u(0,t)=u(\pi,t)=0$. Then how to solve for the bounded equation and how to find the above limit?

I will show my working below: with the given (assumed) initial conditions,
$$u(x,t)=\sum_{n=1}^{\infty}a_n e^{-\frac{n^2\pi^2t}{\pi^2}}\sin(\frac{n\pi x}{\pi})=\sum_{n=1}^{\infty}a_n e^{-{n^2t}}\sin(nx)$$
Now,  $u(x,0)=f(x)=\frac{e^{2x}-1}{e^{2x}+1}=\sum_{n=1}^{\infty}a_n \sin(nx)$, which is a fourier sine series, thus the constants $a_n$ are given by:
$$a_n=\frac{2}{\pi}\int_{0}^{\pi}f(x)\sin(nx)dx$$

But I am unable to determine $a_n$ or solve the problem. Can someone please help me solve this problem?
Thanks in advance.
The ans is :

$ -\frac{1}{2}$

答:
I don't know if this is what you want. It is easy to see
$$ |a_n|\le\frac{2}{\pi}\int_{0}^{\pi}\frac{e^{2x}-1}{e^{2x}+1}dx\le2.$$
So for $t\ge N>0$,
$$\sum_{n=1}^{\infty}\bigg|a_n e^{-{n^2t}}\sin(nx)\bigg|\le2\sum_{n=1}^{\infty} e^{-{n^2N}}$$
converges and hence $\sum_{n=1}^{\infty}a_n e^{-{n^2t}}\sin(nx)$ converges uniformly for any $t\ge N$ and $x\in[0,\pi]$. Thus
$$ \lim_{t\to\infty}\sum_{n=1}^{\infty}a_n e^{-{n^2t}}\sin(nx)=\sum_{n=1}^{\infty}\lim_{t\to\infty}a_n e^{-{n^2t}}\sin(nx)=0. $$
From this, you can see $\lim_{t\to\infty}u(1,t)=0$.


问:
Exercise 2.7.5 of the probabilistic method
Let $H$ be a graph, and let $n > |V(H)|$ be an integer.Suppose there is a graph on $n$ vertices and $t$ edges containing no copy of $H$, and suppose that $tk>n^2\log_en$. Show that there is a coloring of the edges of the complete graph on $n$ vertices by $k$ colors with no monochromatic copy of $H$.

答:
About $3$ years ago, I downloaded a solution manual from https://radimentary.files.wordpress.com/2017/03/solutions_compilation.pdf, but this link no longer works.  The author was Xiaoyu He.  Maybe you can find it.

Anyway, here's a sketch of his solution to this problem.

Let $G$ be the graph with $n$ vertices and $t$ edges containing no copy of $H$.  Consider $K_n$ as a labeled graph, make $k$ independent random relabelings of $G$ and color them with colors $1$ through $k$.  Viewing the labeled copies of $G$ as subgraphs, of the labeled $K_n$, color each edge of $K_n$ with the color of the first copy of $G$ that it shows up in.  Then $K_n$ contains no monochromatic $H$, because none of the $G$'s does.

To complete the proof, it must be shown that there is a non-zero probability that the $G$'s cover all the edges of $K_n$ so that the coloring described above can actually be carried out.  A straightforward calculation shows that the expected number of uncolored edges is $<1$, so there must be a complete coloring.


问:
Jump of an infinite step function
Example: Let $0<x$ be real and let $n$ be natural numbers. Consider the sequence $\left\{ x^n  \right\}_{n=1}^\infty$. Whenever $0<x<1$ one finds $x^n \to 0$ as $n\to\infty$. When $x=1$ one finds $x^n\to 1$, and whenever $1<x$ one finds $x^n\to\infty$. So, the sequence $\left\{ x^n  \right\}_{n=1}^\infty$ tends to $0$, $1$ or $\infty$ for positive $x$. We notice here that $x^n$ tends to $1$ at the jump, between values $0$ and $\infty$. This result can be summarized as follows.

$$
\lim\limits_{n\to\infty} x^n
=
\begin{cases}
0\text{, }0<x<1\\
1\text{, }x=1\\
\infty\text{, }1<x
\end{cases}
$$

Question: Let $a_n(x)$ be strictly positive smooth real function. As $n\to\infty$, let $a_n(x)\to 0$ for $0<x<1$, and $a_n(x)\to \infty$ for $1<x$. Define number $L$ by $L=\lim\limits_{n\to\infty} a_n(1)$. Then $0<L<\infty$.

So, in other words, is it true that for all such $a_n(x)$ the magnitude at the jump is strictly positive finite? Or are there sequences $\left\{ a_n(x)  \right\}_{n=1}^\infty$ such that $L=0$ or $L=\infty$?

答:
Instead of having a constant on $[0, 1/n]$, take a line from $(1/n, 1/n)$ to $(-1, 0)$ and then for $x<-1$, keep all $a_n=0$ (do the same thing for the other family of functions). This way the $a_n$ form a family of continuous functions.

Now, fix some integer $n\geq 2$. Since $a_n$ is continuous on $[-2n, 2n]$, there is a polynomial (by Weierstrass approximation) $p_n$ approximating $a_n$ uniformly within $\epsilon_n$ where $\epsilon_n$ is chosen so that $a_n-\epsilon_n>0$ for $x>0$ (exists because $a_n$ is non zero on the positive side).

This way, $p_n>0$ when $x>0$. Next, take a smooth (positive) bump function $\phi_n$ attaining $1$ on $[-n, n]$ with support in $[-2n, 2n]$ and consider $p_n\phi_n$. This is a smooth function and equal to $p_n$ on $[-n, n]$, however it is $0$ for large $|x|$.

It is clear that the sequence $p_n\phi_n$ is smooth, nonnegative on $x\geq 0$ and has the required limits (note that the $\epsilon_n$ converge to $0$). To obtain strictly positive functions, simply consider $p_n\phi_n+1/n$.

This seems to work. Could you elaborate your comment regarding non standard analysis, I am ignorant on the subject.

Title: Jump of an infinite step function


问:
Prove that $(n+1)^{n-1}<n^n$
How would one prove that  $$(n+1)^{n-1}<n^n \ \forall n>1$$

I have tried several methods such as induction.

答:
For completeness' sake, here is a proof coming from combinatorics.  The terminology and original proof can be attributed to Konheim and Weiss (1966) and the proof reproduced here by Pollak.  I first encountered the problem in Stanley's textbook Enumerative Combinatorics.  See also here for more information and references.

Imagine the scenario of a parking lot with $n$ labeled spaces all in a row where traffic may only flow in one direction through the parking lot.  You have $n$ cars, each of whom have their preferred space in the lot.  The sequence of preferences we call a preference sequence.  When a car enters the lot, they drive until they have reached their preferred space.  If it is unoccupied, they park in the space.  Otherwise, they continue driving and park in the next available unoccupied space.  If all spaces from that point on were occupied, then they leave and were unable to successfully park.

It is clear that some preference sequences lead to all cars successfully parking (for trivial example, when everyone prefers parking in the first space by the entrance) and is also clear that some preference sequences lead to few cars successfully parking (for trivial example, when everyone prefers parking in the last space by the exit).  We call a preference sequence a parking lot sequence or parking lot function if the preference sequence leads to all cars successfully parking.

We ask the question of how many parking lot sequences there are for $n$ cars and $n$ spaces.

To see this, imagine briefly that you were to wrap the row of parking spaces around to form a circle and that you were to add an additional $n+1$'st space.  The cars act the same as otherwise however now rather than exiting once passing the $n$'th space they continue and loop around the circle taking one of the otherwise earlier spaces.  It is clear that every preference sequence here leads to all cars parking in this circular lot and that exactly one space will be left unoccupied.  Further, it is clear that a preference sequence for this circular lot will be a parking lot sequence if and only if that unoccupied space is the newly added $n+1$'st space.  Finally, by symmetry, there will be exactly as many preference sequences which leave one space unoccupied at the end as any other.

It follows then that the number of parking lot sequences is $(n+1)^n / (n+1) = (n+1)^{n-1}$.

The probability of a preference sequence being a parking lot sequence for our original lot then, if the preference sequences are each equiprobable, will be:

$$\frac{(n+1)^{n-1}}{n^n}$$

Since this is a probability, we know it to be between $0$ and $1$.  Since for $n>1$ it is possible for a preference sequence to not be a parking lot sequence we know the probability can not be exactly $1$.

This proves then that $(n+1)^{n-1}<n^n$ for all $n>1$

Title: Prove that $(n+1)^{n-1}<n^n$


问:
Boundary Value Problem of Holomorphic Functions
Let $U \subset \mathbb{C}$ be a bounded open connected set. Let $f: \partial U \rightarrow \mathbb{R}$ be a continuous function. Is it always possible to find continuous function $u: \overline{U} \rightarrow \mathbb{C}$ such that $u: U \rightarrow \mathbb{C}$ is holomorphic and $u =f $ on $\partial U$?

We have Poisson formula and Cauchy integral formula, and we sometimes talk about the boundary value problem for Laplace equation. But I'm not sure what happens if we consider holomorphic functions, instead of harmonic functions.

Any help will be appreciated. Thanks.

答:
When $f$ is real-valued (as Martin R) pointed out, there is never a nonconstant solution $u$. Indeed, the imaginary part of $u$ will be continuous on $\overline{U}$, harmonic in $U$ and zero on the boundary, so it will vanish identically by the maximum principle. So $u$ will be real-valued and holomorphic hence constant.

When $f$ is complex-valued, the answer is still in general no (some nontrivial $f$ will still work, even though “most” won’t). Take $U$ to be the unit disk, and $f$ be the complex conjugation. Assume that there is some satisfactory $u$. Then, for any $0 < r < 1$, $\int_0^1{u(re^{2i\pi\theta})e^{2i\pi\theta}\,d\theta}=0$ because $u$ is holomorphic inside $U$.

Thus (by continuity) $0=\int_0^1{e^{2i\pi\theta}u(e^{2i\pi\theta}\,d\theta}=\int_0^1{d\theta}=1$, a contradiction.


问:
Are there coprime integers $x,y$ ( greater than 1 in absolute value) such that $3y(4x^3-y^3)$ is a square?
Are there any coprime integers $x,y$ ( greater than 1 in absolute value) such that $$3y(4x^3-y^3)$$ is a square?

I performed a search on wolfram, I could not find any. Please share if you can find any such pair using a different program or explain why there is none. I would really appreciate it. Thanks.

答: You are looking for integer solutions $(x,y,a)$ with $\lvert x\rvert, \lvert y\rvert>1$ and gcd$(x,y)=1$ such that $$3y\cdot (4x^3-y^3)=a^2$$ Notice that this implies \begin{align*}(3y^2+a)^3+(3y^2-a)^3&=54y^6+18a^2y^2\\&=54y^6+18y^2\cdot 3y\cdot (4x^3-y^3)\\&=18y^2\cdot 3y\cdot 4x^3\\&=216\cdot x^3y^3=(6xy)^3\end{align*} Hence, we have concluded that $$(3y^2+a)^3+(3y^2-a)^3=(6xy)^3$$ This is nice, since we can now use Fermat's Last Theorem to infer that one among $3y^2+a, 3y^2-a, 6xy$ is zero. Furthermore $\lvert x\rvert, \lvert y\rvert>1\implies 6xy\not=0$ and, thus, $a=\pm 3y^2$. But it is straightforward to check that this yields $x=y$, which violates gcd$(x,y)=1$. Hence, no solutions.


问:
Is it possible to divide n in d parts where each part can have a specific range of values?
Given a value n and number of parts d.
Each part has max and min values it can have.

Is it possible to divide n in d parts fullfilling the max and min value criteria
?
Example

d=2 n=5

Part 1 range-> 0 to 1
Part 2 range -> 3 to 5.
Answer
is YES .
Part1 can have value 1 and Part2 can have value 4

答:
This clearly depends on the given ranges. Let's fix some notation: We have a given number $n$, which is to be split into $d$ parts, where the $i$-th part has to have a value between $\ell_i$ and $u_i$ (inclusive).

Your example would be $n = 5$, $d = 2$, $\ell_1 = 0$, $u_1 = 1$, $\ell_2 = 3$, $u_2 = 5$.

Another example would be $n = 20$, $d = 3$, $\ell_1 = \ell_2 = \ell_3 = 0$, $u_1 = 5$, $u_2 = 6$, $u_3 = 7$. This does not have a solution: Assume that $p_1, p_2$ and $p_3$ are given such that $0 \le p_1 \le 5$,$0 \le p_2 \le 6$, $0 \le p_3 \le 7$. Then we clearly have $p_1 + p_2 + p_3 \le 5 + 6 + 7 = 18 < 20$, so $p_1 + p_2 + p_3 \neq n$.

More generally, we can divide $n$ in $d$ parts as above if $\sum_{i=1}^d \ell_i \le n \le \sum_{i=1}^d u_i$.


问:
Uniqueness of determinant function

Prove that there can exist one only function $f: F^n \to F$ that satisfies the properties of multilinearity, antisymmetry, and normalization.

I cannot figure out how to get started. I would appreciate any help.

答:
Here is a sketch. We basically want to show that given any matrix $A$, for any two functions $f, g$ satisfying those three axioms, $f(A) = g(A)$. We know that for any matrix $A$, there exist a sequence of invertible row operations $O_1, O_2, \cdots, O_k$ that when applied to $A$, bring it into reduced row-echelon form.

Now any reduced row-echelon form can be obtained via a sequence of operations $P_1, P_2, \cdots, P_l$ from the identity matrix. However, this time these operations may be elementary row operations or multiplying by zero. If $\text{rref}(A)$ is $n \times n$ and has $m \leq n$ pivot rows, then we can simply apply operations to $I_n$ until the first $m$ rows are equal to the first $n$ rows of $\text{rref}(A)$, and then multiply the remaining $n - m$ rows by zero.

It then follows that applying the sequence of operations
$$P_1, P_2, \cdots, P_l, O_k^{-1}, O_{k-1}^{-1}, \cdots, O_1^{-1}$$
in that order, from left to right, will result in a sequence of operations transforming $I$ to $A$. However, each of these operations do one of three things:

Multiply a row by a constant $c$ (which may be zero)

Replace a row with itself plus another row

Switch two rows

all of which have a predictable outcome on the determinant. The first operation we know from multilinearity will multiply the determinant by $c$. The second operation we know won't change the determinant. Finally, from antisymmetry we know the third operation will negate the determinant. As $f(I) = g(I) = 1$, it must follow that $f(A) = g(A)$. $\square$

Title: Uniqueness of determinant function


问:
Reference for extreme value theorem.
I look for a reference of the Extreme Value Theorem for semicontinuous functions defined on a topological space.
I know the proof, but I want to cite this result in my work.</p>

答:
The theorem for semi-continuous functions is stated in Nicolas Bourbaki, "General Topology", Ch. IV, § 6, no. 2, theorem 3.

In the 1966 edition Elements of Mathematics. General Topology. Part 1 (Ch. 1-IV), by Hermann / Addison-Wesley that's p. 361.

Title: Reference for extreme value theorem.


问:
Probability question - horse racing
The probability of a certain horse finishing in the top 5 of a race is 20%. The probability of him finishing in the top 10 is 45%.

A. What is the probability of him finishing anywhere from 6-10th?
B. What is the probability of the horse finishing anywhere from 6-10th given that he finished in the top 10?

Got some comments saying to post what I have tried so far so I will do that. For part B here is what I tried -
Let P(T5) and P(T10) be the probability of finishing top 5 and top 10.
We know P(T5 and T10) = .2
We also know P(T5 and T10)= P(10) * P(T5|T10)
So .2 = .45 * P(T5|T10)
Then P(T5|T10) = .2/.45
Therefore the probability of finishing anywhere from 6-10th given that finished in top 10 is 1-P(T5|T10)= 1-.2/.45

If there is not enough information to answer - what other info do we need?

Thanks,
Jack

答:
Let's suppose that the question is "probability to finish in the top $[6;10]$

Thus the probability is

A. $45\%-20\%=25\%$

B. $\frac{25}{45}=\frac{5}{9}\approx 55.56\%$

@Jack: the lack of information is that you do not know the probability to arrive exactly 10th. In any case the results will not be very far from the ones I showed you.

Edit: further explanation on B.

Set the events:

A: your horse arrives in the interval $[6;10]$

B: your horse arrives in the top 10, say in the interval $[1;10]$

The question is to calculate

$$\mathbb{P}[A|B]=\frac{\mathbb{P}[A \cap B]}{\mathbb{P}[B]}=\frac{\mathbb{P}[A]}{\mathbb{P}[B]}=\frac{0.25}{0.45}$$

It is self evident that $\mathbb{P}[A \cap B]=\mathbb{P}[A]$ as $A\subset B$

Title: Probability question - horse racing


问:
Probability density function of $W = X + Y$ where $X \sim \mathrm{Unif}[0,1]$ and $Y \sim \mathrm{Exp}(\lambda)$ are independent random variables?
This is the question that I was given:

And this was the provided solution:

I can't seem to make sense of it - firstly, what exactly is the question asking you to do? How did they know to divide the range of $w$ in that way? And why did they use those particular limits for the integrals?

答:
A tip: draw a plot of $X$ against $Y$ to find the domain of $W$. Then you will find that you cannot do the integration with only one integral; you need to divide it into two. Since $Y=W-X$, it looks like this:

As you can see, there is a triangle and then an infinite rectangle. The situation is different depending on which one you are considering. The rationale is this: you know that $y \in [0, \infty)$ and that $x\in[0, 1]$. Hence, $w\in[0, \infty)$. Thus, the limits are:

$$
0<x<w, \quad \text{for } w<1\\
0<x<1, \quad \text{for } w>1
$$

Say that $w=0.6$. Then, obviously, $x$ cannot be 0.7 for example. If $w=14$, then $x$ can be anything in its support, that is on the interval 0 to 1.


问:
Convergence of equation in late time
Suppose $$u_{t}=u_{xx}+u_{yy}$$ on the region $0\leq x,y \leq 1$ and that $u=0$ on the boundary.

Show that $$\int^1_0\int^1_0u^2dxdy\xrightarrow{}0, \text{ as } t\xrightarrow{}\infty $$

I'm really lost as to begin here. Any help would be appreciated.

答:
By Poincare's Inequality,
$$ \|\nabla u\|_{L^2}^2\ge\lambda_1\|u\|_{L^2}^2. $$
Here $\lambda_1>0$ is the first eigenvalue of $-\Delta$. Multiplying both sides of the equation by $u$ and integrating in $\Omega$, one has
$$ \frac12 (\|u(t)\|_{L^2}^2)'=-\|\nabla u(t)\|^2_{L^2}\le -\lambda_1\|u(t)\|_{L^2}^2$$
and hence
$$ (\|u(t)\|_{L^2}^2)'+2\lambda_1\|u(t)\|_{L^2}^2\le0. $$
So
$$ (e^{2\lambda_1t}\|u(t)\|_{L^2}^2)'\le0 $$
Integrating this from 0 to $t$, one has
$$ \|u(t)\|_{L^2}^2\le e^{-2\lambda_1t}\|u(0)\|_{L^2}^2. $$
Now it is easy to see
$$ \lim_{t\to\infty}\|u(t)\|_{L^2}^2=0. $$

Title: Convergence of equation in late time


问:
Volume with spherical polar coordinates
Determine the volume between the surface $z=\sqrt{4-x^2-y^2}$ and the area of the xy plane determined by $x^2+y^2\le 1,\ x+y>0,\ y\ge 0$.

I convert to spherical polar coordinates.

$$K=0\le r\le 1,\ 0\le \phi \le \frac{3\pi}{4},\ 0\le \theta \le 2\pi$$

$$\iiint_{K} (\sqrt {4-r^2\sin^2\phi \cos^2\theta-r^2\sin^2\phi \sin^2\theta)}r^2\sin\phi drd\phi d\theta$$

I can't figure out how to take $\int_{K} (\sqrt {4-r^2\sin^2\phi \cos^2\theta-r^2\sin^2\phi \sin^2\theta)}r^2\sin\phi dr$, which makes me think I made a mistake somewhere.

EDIT: Thanks for all the answers.

Now I understand how the limits of $\theta ,r,z$ works.

I don't fully understand where the function "disappear".

$\sqrt {4-x^2-y^2} =\sqrt {4-r^2}$

Why isn't it then:

$\int \int \int _{K} {\sqrt {4-r^2}rdzdrd\theta }$

答:
Using spherical coordinates, you would have to split up $K$ into two regions,

$$K_1=\left\{(r,\theta,\phi)\mid 0\le r\le2,0\le\theta\le\frac{3\pi}4,0\le\phi\le\frac\pi6\right\}$$

$$K_2=\left\{(r,\theta,\phi)\mid0\le r\le\sqrt{\csc\phi},0\le\theta\le\frac{3\pi}4,\frac\pi6\le\phi\le\frac\pi2\right\}$$

(where $x=r\cos\theta\sin\phi$, $y=r\sin\theta\sin\phi$, and $z=r\cos\phi$). The upper limit on $\phi$ for $K_1$ and lower limit for $K_2$ come from the intersection of the cylinder $x^2+y^2=1$ and the sphere $z=\sqrt{4-x^2-y^2}$. On the sphere, $r=2$, so we have

$$2\cos\phi=\sqrt3\implies\phi=\cos^{-1}\left(\frac{\sqrt3}2\right)=\frac\pi6$$

The upper limit for $r$ in $K_2$ is obtained by converting the equation of the cylinder $x^2+y^2=1$ into spherical coordinates:

$$(r\cos\theta\sin\phi)^2+(r\sin\theta\sin\phi)^2=r^2\sin^2\phi=1\implies r=|\csc\phi|=\csc\phi$$

Then the volume is

$$\int_0^{\frac\pi6}\int_0^{\frac{3\pi}4}\int_0^2r^2\sin\phi\,\mathrm dr\,\mathrm d\theta\,\mathrm d\phi+\int_{\frac\pi6}^{\frac\pi2}\int_0^{\frac{3\pi}4}\int_0^{\csc\phi}r^2\sin\phi\,\mathrm dr\,\mathrm d\theta\,\mathrm d\phi$$

The first integral is trivial. For the second, integrating with respect to $r$ yields

$$\int_{\frac\pi6}^{\frac\pi2}\int_0^{\frac{3\pi}4}\int_0^{\csc\phi}r^2\sin\phi\,\mathrm dr\,\mathrm d\theta\,\mathrm d\phi=\frac13\int_{\frac\pi6}^{\frac\pi2}\int_0^{\frac{3\pi}4}\csc^2\phi\,\mathrm d\theta\,\mathrm d\phi$$

and observing that $\csc^2\phi=\frac{\mathrm d}{\mathrm d\phi}(-\cot\phi)$, it turns out the second integral is, too.

Title: Volume with spherical polar coordinates


问:
Inverse of a symmetric positive diagonally dominant, doubly stochastic matrix
I am trying to prove that the inverse of a $\mathbf{positive}, \mathbf{symmetric}, \mathbf{diagonally-dominant}, \mathbf{doubly-stochastic}$ matrix and having the following $\mathbf{property (3)}$ -   has positive diagonal elements and negative non-diagonal elements.

I'll describe the below definitions.

1)Diagonally dominant matrix - $|a_{ii}| > \sum_{i \neq j}|a_{ij}| \quad \forall i$

2) Doubly stochastic matrix - sum of entries in any row or column is unity.  i.e., $\quad \sum_{i} a_{ij} = \sum_{j} a_{ij} = 1 \quad \forall i,j$

3) Let $b_1, b_2,..,b_{N-1}$ be a decreasing sequence of positive numbers. The non-diagonal elements of the matrix satisy:  $$\qquad a_{ij} = b_{|i-j|} \qquad \forall i \neq j$$

Need to show that the inverse of such a matrix $\mathbf{A}$ will have positive diagonal and negative non-diagonal elements.

I have come till proving that the row sum and column sum of $\mathbf{A}^{-1}$ is 1. But not able to prove further.

Body:
Here is a $3\times3$ counterexample:
$$
A=\frac{1}{17}\pmatrix{12&4&1\\ 4&9&4\\ 1&4&12},
\quad A^{-1}=\frac{1}{55}\pmatrix{92&-44&7\\ -44&143&-44\\ 7&-44&92}.
$$


问:
Show $x \leq y$ iff $y^c \leq x^c$ in a lattice which is a boolean algebra

Show $x \leq y$ iff $y^c \leq x^c$ where $x,y$ are elements of a lattice $(B,\leq,\lor, \land)$. $(B,\leq)$ is a partially ordered set, and $\lor$ and $\land$ refer to the l.u.b and g.l.b operations respectively.

I started off with the definition of the complement $^c$, which is:

$x \land x^c = 0$

$x \lor x^c = 1$

Then, assuming $x \leq y$, I am trying to show $y^c \leq x^c$ - or equivalently, $x^c \leq y^c$ will result in a contradiction. However, I'm unable to proceed. Could someone please help?

答:
If $x \leq y$, we know that $x \land y = x$ and $x \lor y = y$. Also, from De-morgan's laws we know that $(x \land y)^c = x^c \lor y^c$ and $(x \lor y)^c = x^c \land y^c$.

To show $y^c \leq x^c$, we see that:

$y^c \land x^c = (x\lor y)^c = y^c$

$y^c \lor x^c = (x \land y)^c = x^c$

Similarly, starting with $y^c \leq x^c$, we can show $x \leq y$, which completes the proof.


问:
How to prove $\lim_{n \to +\infty} \sqrt{n}\int_0^\pi{\cos(\frac{t}{2})^n}dt>0$
I want to prove
$$\lim_{n \to +\infty}\sqrt{n}\int_0^\pi{\cos\left(\frac{t}{2}\right)^n}dt>0.$$
First, I consider $$\lim_{n \to +\infty}\sqrt{n}\int_0^\pi{\cos\left(\frac{t}{2}\right)^n}\sin\left(\frac{t}{2}\right)dt,$$ which is smaller than what I want, but the second integral leads to $$\lim_{n \to +\infty}\frac{\sqrt{n}}{2(n+1)}.$$ So it does not work.</p>

答:
This integral can be analyzed using Laplace's Method.  Write

$$\cos{\left(\frac{t}{2}\right)}^n = e^{n \log{\cos{(t/2)}}}$$

Note that $\log{\cos{(t/2)}}$ has a maximum at $t=0$; as $n \to \infty$ there is an increasingly narrow and high peak about $t=0$.  Thus we may Taylor expand about $t=0$ and ignore contributions away from a small neighborhood of $t=0$:

$$\int_0^{\pi} dt \, \cos{\left(\frac{t}{2}\right)}^n \sim \int_0^{\epsilon} dt \, e^{n \log{(1-t^2/8)}} $$

where $\epsilon = O\left(n^{-1/2}\right)$.  Because contributions away from $[0,\epsilon]$ are exponentially small, we may extend the integral out to infinity within the error of the leading-order approximation.  Taylor expanding the log term, we find

$$\int_0^{\pi} dt \, \cos{\left(\frac{t}{2}\right)}^n \sim \int_0^{\infty} dt \, e^{-n t^2/8}  = \frac12 \sqrt{\frac{8 \pi}{n}}$$

Therefore

$$\lim_{n \to \infty} \sqrt{n} \int_0^{\pi} dt \, \cos{\left(\frac{t}{2}\right)}^n =\sqrt{2 \pi}$$


问:
Convert parametric to cartesian
could someone help me in this problem? i wanna convert $$x=t\sin t+\cos t,\;\;\;y=\sin t-t\cos t$$
to cartesian, but i'm not finding a way to get rid of t. I tried to square it all and add them up but i got $x^2+y^2=t^2+1$. t is still there. Any tips would be amazing. thanks in advance

答:
But... you have the solution !

$$t=\pm\sqrt{x^2+y^2-1}$$

and

$$x=\sqrt{x^2+y^2-1}\sin\sqrt{x^2+y^2-1}+\cos\sqrt{x^2+y^2-1}$$

is an implicit Cartesian equation.

Title: Convert parametric to cartesian


问:
What is a typical basis for the finite complement topology $\cal{T}_f$? (James R. Munkres "Topology 2nd Edition")
I am reading "Topology 2nd Edition" by James R. Munkres.

On p.77, Example 3:

Let $X$ be a set; let $\cal{T}_f$ be the collection of all subsets $U$ of $X$ such that $X-U$ either is finite or is all of $X$. Then $\cal{T}_f$ is a topology on $X$, called the finite complement topology.

I wonder what a typical basis for $\cal{T}_f$ is.

I only know $\cal{T}_f$ is a basis for $\cal{T}_f$.

Please tell me a typical basis for $\cal{T}_f$.

答:
$\cal{T}_f$ is a typical basis for $\cal{T}_f$.

If $X$ is finite then $\big\{\{x\}\big\}_{x\in X}$ will be a basis for $\cal{T}_f$, i.e. $(X,\cal{T}_f)$ is discrete when $X$ is finite.

For any $n\in\mathbb{N}$ the set $\big\{ A\subseteq X\ \big|\ n\leq |X-A|<\infty\big\}$ will be a basis for $\cal{T}_f$. And these are pairwise different if $X$ is infinite.


问:
Is there an adjoint functor to the hom bifunctor in the category of $R$-modules, $R$ is commutative unital ring.
I've seen that for $W\in R-\mathrm{\!Mod}$ ,both $\mathrm{Hom}(W,-)$ and $\mathrm{Hom}(-,W)$  have adjoint functors.

adjoint of Hom(-,W)

this leaves me with one question:

$C:= R-\mathrm{\!Mod}$

does $\mathrm{Hom}(-,-):C^{op}\times C \to C$ have an adjoint functor?

答:
I found a counterexample:

$C:= \mathrm{\mathbb{Z}-Mod}, D:=C^{op}\times C$

$R(-):=Hom(-,-) : D\to C$ , $L$ is the left adjoint of $R$.

Then $Hom_D(L(-),-)\cong Hom_C(-,R(-)):C^{op}\times D\to Set$

$\mathbb{Z}_i := \mathbb{Z}/i\mathbb{Z}$

$Hom_D(L(\mathbb{Z}_r),(\mathbb{Z}_n,\mathbb{Z}_m))\cong Hom_C(\mathbb{Z}_r,R(\mathbb{Z}_n,\mathbb{Z}_m))$

Suppose $L(\mathbb{Z}_r)=(M,N) \in D$

$Hom_D((M,N),(\mathbb{Z}_n,\mathbb{Z}_m))\cong Hom_{C^{op}}(M,\mathbb{Z}_n)\times Hom_D(N,\mathbb{Z}_m)\cong Hom_{C}(\mathbb{Z}_n,M)\times Hom_D(N,\mathbb{Z}_m)$

$|Hom_C(\mathbb{Z}_r,Hom(\mathbb{Z}_n,\mathbb{Z}_m)|=gcd(r,n,m)$

$|Hom_{C}(\mathbb{Z}_n,M)|\times |Hom_D(N,\mathbb{Z}_m)|=gcd(r,n,m)$

take n, we get $|Hom_D(N,\mathbb{Z}_m)|$ must be $1$.

by the same way $|Hom_{C}(\mathbb{Z}_n,M)|=1$

we have $gcd(r,n,m)=1$ for all n,m , that leads contradiction.

(I'm not sure whether I made a mistake or not)


答:
Area on XY plane is bound by $x^2 + y^2 \leq 1, y \geq 0, x + y \geq 0$

This is a sector of the circle $x^2 + y^2 \leq 1$ bound between positive $X$-axis and line $y = -x$ in the second quadrant. This comes from the fact that $y \geq 0$ so part of the circle in third and fourth quadrant of $XY$ plane is not included. $x + y \geq 0$ is true for quarter of the circle in the first quadrant as both $x$ and $y$ are positive. It is also true for part of the circle in the second quadrant above line $y = -x$ as $|y| \geq |x|$.

Now you are asked to find the volume between this area on XY plane and $z = \sqrt{4-x^2-y^2}$. So it is essentially a cylinder ($\frac{3}{8}$ cross section of a cylinder of radius $1$) cut out of the sphere of radius $2$ above $XY$ plane.

So here is how it will look in cylindrical coordinates -

$\displaystyle \int_{0}^{3\pi/4} \int_{0}^{1}  \int_{0}^{\sqrt{4-r^2}} r \, dz \, dr \, d\theta$

Title: Volume with spherical polar coordinates

Body:
Determine the volume between the surface $z=\sqrt{4-x^2-y^2}$ and the area of the xy plane determined by $x^2+y^2\le 1,\ x+y>0,\ y\ge 0$.

I convert to spherical polar coordinates.

$$K=0\le r\le 1,\ 0\le \phi \le \frac{3\pi}{4},\ 0\le \theta \le 2\pi$$

$$\iiint_{K} (\sqrt {4-r^2\sin^2\phi \cos^2\theta-r^2\sin^2\phi \sin^2\theta)}r^2\sin\phi drd\phi d\theta$$

I can't figure out how to take $\int_{K} (\sqrt {4-r^2\sin^2\phi \cos^2\theta-r^2\sin^2\phi \sin^2\theta)}r^2\sin\phi dr$, which makes me think I made a mistake somewhere.

EDIT: Thanks for all the answers.

Now I understand how the limits of $\theta ,r,z$ works.

I don't fully understand where the function "disappear".

$\sqrt {4-x^2-y^2} =\sqrt {4-r^2}$

Why isn't it then:

$\int \int \int _{K} {\sqrt {4-r^2}rdzdrd\theta }$


问:
Limit with variable: non-defined expression
I have a given limit that depends on a variable $a$:

$$\lim_{x \rightarrow \infty} \left (\frac{e^{ax}}{1 - ax} \right)$$

I understand cases for $a < 0 \implies \lim = 0$ and $a > 0 \implies \lim = -\infty$. However, for the case $a = 0$, the expression $ax$ which is basically $0\cdot \infty$ in undefined. I somehow know, that the result will be $\lim (\frac{e^0}{1}) = 1$ but I am not sure how to justify that $0\cdot\infty$ is $0$ in this case.

Thanks for any ideas or an explanation!

答:
There is no “indetermination” in $e^{ax}$ when $a=0$: it just means $e^0=1$, because $0x=0$.

You don't compute such a limit by plugging in $\infty$ in place of $x$, which wouldn't make sense. You can, however, use that

$\lim_{x\to\infty}e^{ax}=\infty$ (for $a>0$);

$\lim_{x\to\infty}e^{ax}=0$ (for $a<0$).

But this is different from simply plugging in $\infty$. For instance, in the case of $a<0$, you can conclude that
$$
\lim_{x\to\infty}\frac{e^{ax}}{1-ax}=0
$$
because the numerator has $0$ limit and the denominator has $\infty$ limit.

On the other hand, you cannot immediately draw a conclusion in case $a>0$, since the numerator and the denominator have limit $\infty$ and $-\infty$, respectively. For this you can do a simple application of L’Hôpital's theorem:

$$
\lim_{x\to\infty}\frac{e^{ax}}{1-ax}
\overset{(H)}{=}
\lim_{x\to\infty}\frac{ae^{ax}}{-a}
=
\lim_{x\to\infty}-e^{ax}=-\infty
$$

For the case $a=0$ you simply have
$$
\lim_{x\to\infty}\frac{e^{ax}}{1-ax}
=
\lim_{x\to\infty}\frac{1}{1}=1
$$


问:
Find positive definite $X$ such that $AXA^T=\alpha X$.
Given an invertible real matrix $A\in\mathbb{R}^n$, is it possible to obtain a (real) positive definite matrix $X$ (and scalar $\alpha>0$) such that
$$
AXA^T=\alpha X?
$$

Attempt: The previous equation can be written as $AXA^T-X +Q=0$ with $Q=(1-\alpha)X$. So, if $(1-\alpha)>0$ this looks like the Lyapunov equation here. However, the analytic solution to the Lypunov equation has the form
$$
X = \sum_{k=0}^\infty A^kQ(A^T)^k = (1-\alpha)\sum_{k=0}^\infty A^kX(A^T)^k
$$
So, even if such equation have a solution, I don't have a clue on how to obtain it from here.

Do you have other thoughts/suggestions?

答:
Using vectorization and Kronecker product properties this may be rewritten as
$$
(A \otimes A) \operatorname{vec} X = \alpha \operatorname{vec} X
$$
This is a regular eigenvalue problem for matrix $A \otimes A$. Moreover, eigenvalues of this matrix are $\{\lambda_i \lambda_j\}_{i,j=1,n}$ with corresponding eigenvectors $X$ being Kronecker products of eigenvectors $x_i \otimes x_j$. Here $\lambda_i, x_i$ are eigenvalues and eigenvectors of $A$.

Converting back from vectorized form to matrix form we obtain
$$
X_{(i,j)} = x_i x_j^\top\\
\alpha_{(i,j)} = \lambda_i \lambda_j
$$

There might be problems with this approach when $A$ has complex-valued eigenvalues or does not have full set of eigenvectors in case of eigenvalues with multiplicity greater than one.

As you see, matrix $X_{(i,j)}$ are rank-1 matrices, so they cannot be (strictly) positive definite. To obtain a positive definite matrix you need to linearly combine $n$ of them, all with the same value $\alpha$. This is rather rare case.


问:
last two digits of $14^{5532}$?
This is a exam question, something related to network security, I have no clue how to solve this!

Last two digits of $7^4$ and $3^{20}$ is $01$, what is the last two digits of $14^{5532}$?

答:
Consider the following class of exam questions:

If $n$ is even find the last two digits of $n^k$.

There is a quick way of cranking out the answers by giving the multiplicative semigroup,

contained in $\mathbb {Z} / \text{100} \mathbb {Z}$, a central/absorbing role.

We know that $76 \times 76 \equiv 76 \pmod{100}$ and more generally that for any $a$ in the semigroup

$\quad 76 \times a \equiv a \pmod{100}$

(we actually have a group with the identity being $76$)

We know that there is an easy calculation algorithm that can be used to multiply any two of these residue numbers.

Example: For $56 \times 96 \pmod{100}$ the tens digit is equal to the residue $5 + 9 + 3 \pmod{10}$:

$\quad 56 \times 96 \equiv 76 \pmod{100}$

We know that given an even integer $n$ such that $n \not\equiv 0 \pmod{10}$, at least one of the residue classes

$\quad [n], [n^2], [n^4]  \quad \pmod{100}$

belongs to our semigroup.

We know that given an even integer $n$ such that $n \not\equiv 0 \pmod{10}$,

$\quad n^{20} = 76 \pmod{100}$

Now with these preliminaries we can mentally turn the crank (recall that $14^2 = 196$) ,

$\; 14^{5532} \equiv 76 \times 14^{12} \equiv  76 \times (14^2)^6 \equiv 76 \times (96)^2  \times (96)^2 \times (96)^2 \equiv$

$\quad \quad  \quad \quad 76 \times 16  \times 16 \times 16 \equiv 96 \pmod{100}$

Title: last two digits of $14^{5532}$?


问:
Definition of Differentiability for $f:\Bbb{R}^{3}\to\Bbb{R}^{2}$
A function $f:\Bbb{R}^{3}\to\Bbb{R}^{2}$ is said to be $\mathbf{differentiable}$ at $a\in\Bbb{R}^{3}$ if there exists a $2\times3$ matrix $A$ such that

\begin{align}
\dfrac{\lvert f(a+h)-f(a)-Ah\rvert}{\lvert h\rvert}\to0
\end{align}

as $h\to0$. Show that the partial derivatives $\partial_{x_{j}}f^{i}(a)$ exist at $a$ where we write $f=(f^{1},f^{2})$ where each element $f^{j}$ is a real-valued function of three variables and that

\begin{align}
A=(\partial_{x_{j}}f^{i}(a))
\end{align}

is the $\mathbf{Jacobian}$ matrix of $f$.

I'm not too sure where to start, and yes, I am aware that a similar question has been asked before but I am still quite confused considering this is a map from $\Bbb{R}^{3}$ to $\Bbb{R}^{2}$. Appreciate if anyone could provide with some pointers, thanks.

答:
Denote $\{e_1, e_2, e_3\}$ $\mathbb R^3$  canonical basis. For $u\in \mathbb R$ and $1 \le j \le 3$ you have

$$\begin{aligned}\frac{\Vert f(a+u.e_j) -f(a) -A(u.e_j)\Vert}{\Vert u.e_j \Vert} &= \frac{\Vert f(a+u.e_j) -f(a) -A(ue_j)\Vert}{\vert u \vert}\\
&= \frac{\Vert f(a+u.e_j) -f(a) -u A_j\Vert}{\vert u \vert}\\
\end{aligned}$$

where $A_j$ is the $j$-th colum of $A$. And if you look at the $i$-th coordinate, you get

$$\lim\limits_{u \to 0} \frac{ f^i(a+u.e_j) -f^i(a) }{ u } = A_j^i$$ which means exactly

$$\begin{align}
A=(\partial_{x_{j}}f^{i}(a))
\end{align}$$


问:
Fundamental group of a bouquet of circles
Consider a bouquet of $n$ circles, centered at $(1,0), (2,0),...,(n,0)$ $$ W_n= \bigcup_1^nC_r$$ with $$C_r:(x-r)^2+y^2=r^2$$

I want to compute the fundamental group of this space.

$W_n$ is path connected. Moreover I know
$\pi_1(W_1)=\pi_1(C_1)=\mathbb{Z}.$ Seeing $W_n= W_{n-1}\cup C_n$ we
have both $W_{n-1}$ and $C_n$ are path connected and open. Moreover
$W_{n-1} \cap C_n $ is just the origin which has trivial fundamental
group. Hence
$$\pi(W_n)=\pi(W_{n-1})*\pi(C_1)=\pi(W_{n-1})*\mathbb{Z}=...=\pi_1(W_1)*\mathbb{Z}...*\mathbb{Z}=$$$$\pi(C_1)*\mathbb{Z}*...*\mathbb{Z}=\mathbb{Z}*...*\mathbb{Z}$$
so it is the $n$-times free product of $\mathbb{Z}$ with itself.

Does it work? First I want to know if my solution is correct and only secondly a different one thanks.

答:
Your proof is correct but a bit sloppy. It is true that the sets $W_{n-1}$ and $C_1$ are path-connected, but they are not open. Clearly no real harm has occured, but you should be careful to meet the requirements of the exact statement of the Seifert-van Kampen Theorem which you clearly applying.

The space $W_n$ is a wedge of $n$ circle $\vee^nS^1$. There are slight more sophisticated van Kampen Theorems which can compute the fundamental group of such spaces directly (i.e. by working with a family of many path-connected open sets and their intersections rather than just two).

Another way to compute its fundamental group is to construct its universal cover. There is a standard way to describe the universal covers of wedge sums. The case for $n=1,2$ is clear, but I don't even want to think about what the space looks like for higher $n$.


问:
$Q\le \prod \frac{5+2x}{1+x}\le P$ find $P,Q$

if $x,y,z,$ are positives and $x+y+z=1$ and $$Q\le \prod_{cyc} \frac{5+2x}{1+x}\le P$$ find  maximum value of $Q$ and minimum value of $P$

This is actually a question made up myself ,so i don,t know whether a nice solution exists!?.

Finding P seems easier:AM-GM results $$\prod \frac{5+2x}{1+x}\le \frac{{\left(\sum_{cyc} \frac{5+2x}{1+x} \right)}^3}{27}$$

we write $$\sum_{cyc} \frac{5+2x}{1+x}=6+\sum \frac{3}{1+x}$$

But neither Jensen nor the tangent line method help as sign of inequality is reversed.

In fact i am more interested in finding $Q$.

I am looking for  a solution that avoids using computational aid(SOS) or ,uvw.

答:
Finding Q via Jensen's:

Let $f(x) = \frac{ 5 + 2x } { 1 + x } $.

Let $ g(x) = \ln \frac{ 5+2x}{1+x} $, then $ g'(x) = - \frac{3}{ 2x^2 + 7x + 5 } $ and $ g'' (x) = \frac{ 3 ( 4x + 7 ) } { (2x^2 + 7x + 5 )^2 }$.

When $ x \geq 0$, $g''(x) \geq 0$, so we may apply Jensens to conclude that

$$ \sum g(x) \geq 3 g ( \frac{1}{3} ).$$

Taking $e$ to the power of both sides, we conclude that

$$ \prod f(x) \geq f(\frac{1}{3} ) ^ 3. $$

Note: The tangent line approach is similar on $g(x)$, and it works because of the condition  $ g''(x) \geq 0$.

Finding P via Step-wise smoothing

A good guess is that the maximum occurs at the extreme end (which isn't necessarily the case), which others have shown. As such, step-wise smoothing is often good approach.

Claim: When $a,b \geq 0$, $f(a) f(b) \leq f(0 ) f( a+b)$.

This follow by cross multiplying to get $ 6x^2 + 21 xy + 6y^2 \geq 0 $ which is obviously true.

Hence, $f(x)f(y) f(z) \leq f(0)f(x+y)f(z) \leq f(0) f(0) f(x+y+z) = \frac{175}{2}$.

It remains to show that this is indeed the supremum, which you can verify with $ x = y \rightarrow 0^+$.

Note: Karamata's inequality (which is a generalization of Jensen's) also gives this result directly. In fact, the above approach is a specific case of Karamata (but doesn't require as much power).


问:
How to use induction to prove this argument?

It is obvious that this grammar will always return an equal number of both a's and b's. But I was wondering how to prove it using induction? I understand induction, but I was finding it hard to apply to this situation.

The ε character denotes an empty.

答: Hint: use induction on number of steps made to get string of a,b,S,T</p>


问:
The area of a triangle, if the vertices are complex numbers
I need to check for this task, if my logic is correct ?

The task is:

Complex numbers are given, which represent the points of a triangle.

$z_1=2+i$

$z_2=i$

$z_3=\lambda-2i$

$z_1$ and $z_2$ represent fixed points, while $z_3$ depends on $\lambda$.

Calculate the area of the triangle

I tried this:

I took that $\lambda$ for example $1$.

$P_{\triangle z_1z_2 z_3}=\frac{1}{2}\left|\begin{vmatrix} 2 & 1 & 1\\ 0 & 1 & 1\\ 1 & -2 & 1 \end{vmatrix}\right|=\frac{1}{2}\cdot (2+1+0-(1-4+0))=\frac{1}{2}\cdot 6=3$

By taking any other number for the $\lambda$, the same result is obtained.

Can it be done in any other way?

Thanks in advance !

答:
We will assume that $\lambda$ varies over the reals.

Let segment $z_{1}z_{2}$ be the base of this triangle, and note that it has length $2$. It is parallel to the real axis and has imaginary coordinate $i$.

Now note that $z_{3}$ has fixed imaginary coordinate $-2i + \text{Im}(\lambda) = -2i$, so the triangle will have base $2$ and height $3$ without regard to the value of $\lambda$. Thus, $\boxed{\triangle z_{1}z_{2}z_{3}\text{ has area }3.}$

For a general triangle, I would either use determinants or the Shoelace Formula (which is equivalent).


答:
On Cartesian plane, $z_1=A(2,1)$, $z_2=B(0,1)$ and $z_3=C(\lambda, -2)$ i.e., $C$ is any point on the line $y=-2$ (for real $\lambda$).

$\triangle ABC$ is a triangle with base $AB$ of length $2$ and height (distance of $C$ from horizontal line $AB$) $3$.

$$\therefore [ABC] = \dfrac{1}{2}\cdot 2\cdot 3 = \boxed{3}$$

It becomes clear now why the area is independent of $\lambda$. The height of triangle doesn't depend on $\lambda$ (location of $C$ on line $y=-2$).

Title: The area of a triangle, if the vertices are complex numbers

Body:
I need to check for this task, if my logic is correct ?

The task is:

Complex numbers are given, which represent the points of a triangle.

$z_1=2+i$

$z_2=i$

$z_3=\lambda-2i$

$z_1$ and $z_2$ represent fixed points, while $z_3$ depends on $\lambda$.

Calculate the area of the triangle

I tried this:

I took that $\lambda$ for example $1$.

$P_{\triangle z_1z_2 z_3}=\frac{1}{2}\left|\begin{vmatrix} 2 & 1 & 1\\ 0 & 1 & 1\\ 1 & -2 & 1 \end{vmatrix}\right|=\frac{1}{2}\cdot (2+1+0-(1-4+0))=\frac{1}{2}\cdot 6=3$

By taking any other number for the $\lambda$, the same result is obtained.

Can it be done in any other way?

Thanks in advance !


问:
How to find $n \in \mathbb{N}$ such that, for some $m \in \mathbb{N}$, $n!$ is the largest factorial less than or equal to $m$
Say I have some natural number $m$.  I want to find the natural number $n$ with the property that $n!$ is the largest factorial less than $m$.  For instance, if $m = 17$, then $n = 3$ since $3! = 6$ but $4! = 24$.  Is there an elegant way to do that?

答:
$n!=\Gamma(n+1)$

As $n\to +\infty$ we have
$$\Gamma(n+1)\sim e^{-n} n^n \left(\sqrt{2 \pi  n}+\frac{1}{6} \sqrt{\frac{\pi }{2 n}}\right)$$

Thus if we want $n$ such that $n!<10^6$ we solve numerically
$$ e^{-n} n^n \left(\sqrt{2 \pi  n}+\frac{1}{6} \sqrt{\frac{\pi }{2 n}}\right)=10^{20}$$
$n\approx 21.2185$. Take $\lfloor n \rfloor=21$

Indeed $21!\approx 5.1\times 10^{19}<10^{20}$


问:
$f\in L^1(K)$ implies $f\in L^\infty (K_0)$ and $\int_{K\setminus K_0}f\leq\delta$, $\delta>0$ and some $K_0$.
Suppose that $K\subset\mathbb{R}^N$ is a measurable set and $f\in L^1(K)$ with $f\geq 0$. For given $\delta>0$, is it possible to find a measurable set $K_0\subset K$ such that $f\in L^\infty (K_0)$ and $$\tag{1}\int_{K\setminus K_0}f\leq\delta$$

Remark: I can see that the set of all measurable subsets $K_0$ of $K$, for which $(1)$ is valid is non empty. What I cannot prove, is that for some $K_0$ satisfying $(1)$, we have that $f\in L^\infty (K_0)$.

Thank you

答: Hint: take $f_k:=f\chi_{\{f\leqslant k\}}$. Then $f_k\to f$ almost everywhere and by monotone convergence, in $L^1(K)$. And $f_k$ is bounded on $\{f\leqslant k\}$.</p>


问:
How to find the position vector given $3$ colinear points?
$P$, $Q$ and $R$ are $3$ collinear points in space, with position vectors $p$, $q$ and $r$ respectively. Point $Q$ lies between $P$ and $R$ such that the distance from $P$ to $Q$ is twice the distance from $Q$ to $R$.

Which of the following is true?

How can I compute this? is $PQ$ = $2QR$?

答:
From the data given, $$Q=P+\frac 23 \vec{PR}$$
In other words $$Q=P+\frac 23(R-P)$$
so $$3Q = P + 2R$$


问:
Two sigma-algebras coincide
We have a polish space $\mathbb{R}^{+\infty}$ with a metric $\rho(x, y) = \sum_{k=1}^{+\infty} 2^{-k} {|x_k - y_k| \over 1 + |x_k - y_k|}$. Show that in this space Borel $\sigma$-algebra coincides with cylindrical one.

答: You may argue that the topology induced by $\rho$ is equivalent to the product topology, $\tau$, on $\mathbb{R}^{\infty}$. Then we know that $\sigma(\tau)$ equals to the cylindrical $\sigma$-algebra, $\mathcal{C}$. To see the latter assertion, note that the coordinate functions are continuous (w.r.t. $\tau$) and, therefore, $\sigma(\tau)$-measurable ($\therefore \mathcal{C}\subseteq\sigma(\tau)$). On the other hand, $\tau$ is second countable and the sets in its base belong to $\mathcal{C}$ ($\therefore \sigma(\tau)\subseteq\mathcal{C}$).

Title: Two sigma-algebras coincide


问:
Confusion with modelling shortest paths using linear programming
When looking around for formulating the shortest path problem as a LP I found this:

https://www.cs.princeton.edu/~rs/AlgsDS07/22LinearProgramming.pdf

At slide 39 they give an example

example

But in my eyes, this solution doesn't satisfy the inequalities.
For example,

```<code>x_7 + 44 <= x_t
with x_7 = 15 and x_t = 50 leads to
59 <= 50
</code>```

What is my misunderstanding?

答:
The constraint inequalities all seem to be backwards to me. Take the first one, $x_s+9\leq x_2$.  We know a way to get from $s$ to $2$ in $9$ so the shortest path to $2$ is at most the shortest distance to $s$ plus $9$.

The constraint should read $x_2\leq x_s +9$.


问:
Taylor series about $x=0$ or Maclaurin series

Find the Taylor series of $\frac{2+x}{1-x}$ at $x=0$.

I tried to find the Taylor expansion of $\frac{1}{1-x}$ and then multiplied by $2+x$, so my final answer is $$T(x) = \sum_{n=0}^{\infty}(2x^n + x^{n+1}).$$
Is that correct?

答:
We know that Maclauren series can be represented as $$M(x) = c_0 + \sum_{n=1}^{\infty}\frac{f^{(n)}(0)(x)^n}{n!}$$
where $c_0 = f(0)$.
So let's find the derivative of $f(x) = \frac{2+x}{1-x}$:
$$\dfrac{d}{dx}\frac{2+x}{1-x} = \frac{3}{(1-x)^2}$$
Then we find the second order derivative
$$\dfrac{d^2}{dx^2}\frac{2+x}{1-x} = \frac{3(2)}{(1-x)^3}$$
So for $n$-th derivative, we should get something like $$\dfrac{d^n}{dx^n}\frac{2+x}{1-x} = \frac{3(n!)}{(1-x)^{n+1}}$$

Substitute everything and we should get$$M(x) = 2 + \sum_{n=1}^{\infty} 3x^n$$

which is almost identical with what you've got.


问:
Finding the tangents common to two rotated ellipses?
Is there a way to find the four tangents that two rotated ellipses share?

I believe that if two ellipses do not intersect and do not contain one another, they will have four tangents in common and I wish to find these tangents.
I know how to complete the same process with circles and wish to do the same with ellipses.

答:
Assuming, for the sake of argument, the ellipse is centered and rotated about the origin (if it weren't, the same principle would apply, but the math would be more complex), the original, un-rotated ellipse can be modeled as a parametric equation:

$$x(t)=a\cos(t),y(t)=b\sin(t)\space 0\le t<2\pi$$

Now rotate it by multiplying it by the rotation matrix:

$$
        \begin{bmatrix}
        \cos(\theta) & -\sin(\theta) \\
        \sin(\theta) &  \cos(\theta) \\
        \end{bmatrix}\begin{bmatrix}
        a\cos(t) \\
        b\sin(t) \\
        \end{bmatrix}
$$

The resulting parametric equation for the rotated ellipse is

$$x(t)=a\cos(\theta)\cos(t)-b\sin(\theta)\sin(t)$$
$$y(t)=a\sin(\theta)\cos(t)+b\cos(\theta)\sin(t)$$

The tangent to any parametric curve is equal to $\frac{y'(t)}{x'(t)}$, so take the derivatives of each set of parametric equations, equate their ratios, and solve for t (remember that $\theta$ is a constant). That should give you the point where both their tangents are equal.


问:
Program won't accept $2(-5x+1)(-4x+3)$ as an answer to factoring $40x^2-38x+6$.
Here is what I have done so far:

$2(20x^2-19x+3)$ //factored the GCF

$60x² $//Multiplied $3$ into $20x²$

$-15x -4x$ //These two numbers add into $-19x$, and multiply into $60x²$

$2((20x^2-15x)(-4x+3))$ //Put it into form again.

$2(-5x(-4x+3)1(-4x+3))$ //Factored the GCF out of each grouping.

$2(-5x+1)(-4x+3)$ //Reformatted to try and get the final answer.

I am stuck on this step.

Any help would be greatly appreciated. I am a high school math student, so I am obviously not an expert in this stuff. I just want to understand what I did wrong or what the computer wants and learn from it.

答:
Your calculation seems fine, you could try to insert as answer the equivalent

$$40x^2-38x+6=2(5x-1)(4x-3)$$

by a simple change in sign for the two factors.


问:
Evaluate $\prod_{k=1}^{\infty}\left(2\cos\left(\frac{\pi}{3^{k+1}}\right)-1\right)$ from polynomial
I found the problem online (without solution) that to evaluate $$\prod_{k=1}^{\infty}\left(2\cos\left(\frac{\pi}{3^{k+1}}\right)-1\right)$$
given that we know how to solve polynomial
$$4x^3-3x+1 = 0$$

Right now I know that $$0<\prod_{k=1}^{\infty}\left(2\cos\left(\frac{\pi}{3^{k+1}}\right)-1\right)\leq 1$$ due to the fact that $\lim_{k\rightarrow\infty}\cos\left(\frac{\pi}{3^{k+1}}\right) = 1$ so we should have something like $0<(0.8...)...(\text{almost 1})...<1$ (my guess is that our answer is around $0.8$ to $0.9$) and I also know that the solution for the polynomial is $x = -1, \frac{1}{2}$.

One thing I found out interesting is that the polynomial can be express as $(x+1)(2x-1)^2$ in which is quite the same with our product if we let $x = \cos\left(\frac{\pi}{3^{k+1}}\right)$.

So how can I solve this problem using the polynomial? Could anyone provides me a hint or some guideline to solve this problem?

答:
Hint: $\cos 3 \theta = \cos \theta ( 2 \cos 2 \theta - 1)$.

Thus, we can use a telescoping series to evaluate the product (fill in the missing gap) to calculate a final value of $ \cos \frac{\pi}{6} = \frac{ \sqrt{3}}{2} \approx 0.866 $.

I'm not quite sure how it relates to solving the cubic (or even if that is relevant).


答:
\begin{align}
T(x)&=\sum_{n=0}^\infty 2x^n+x^{n+1} \\
&=\sum_{n=0}^\infty2x^n+\sum_{n=0}^\infty x^{n+1} \\
&=2+\sum_{n=1}^\infty 2x^n+\sum_{n=1}^\infty x^n \\
&=2+\sum_{n=1}^\infty 3x^n.
\end{align}

Hence, the Taylor series has the form $T(x)=\sum_{n=0}^\infty a_n x^n$, where
$$
a_n=\begin{cases}2,&n=0 \\ 3, & \text{else} \end{cases}
$$

Title: Taylor series about $x=0$ or Maclaurin series

Body:
Find the Taylor series of $\frac{2+x}{1-x}$ at $x=0$.

I tried to find the Taylor expansion of $\frac{1}{1-x}$ and then multiplied by $2+x$, so my final answer is $$T(x) = \sum_{n=0}^{\infty}(2x^n + x^{n+1}).$$
Is that correct?


问:
A logical implication in which both sides are unrelated
Is this proposition true or not:

$$ \forall x(x=x) \rightarrow \forall x( x^2 -a^2 = (x+a)(x-a))$$

The left side $x=x$ does not have to do anything with the right side $x^2 - a^2 = (x+a)(x-a),$ but they are both true in any case; does this mean that a tautology is implying a tautology, and thus this statement is true?

答:
Some definitions $\large($tautology $\subseteq$ validity $\subseteq$ truth$\large)$:

a tautology is a formula that is true in its
truth-functional form regardless of interpretation;

a validity is a formula that is true
regardless of interpretation;

a (synthetic or analytic) truth is a formula that is true in a particular
interpretation.

Thus in the given statement $$\forall x \left(x=x\right)\rightarrow \forall x\forall y \left(x^2 -y^2 = (x+y)(x-y)\right),$$

the antecedent (call it $L$) is valid (and therefore
true in the standard interpretation), but not tautological;

the consequent (call it $R$) is true in the standard interpretation, but neither valid nor tautological;

the statement as a whole is true in the standard interpretation
$\left(\text{so we can write }\,L\Rightarrow R\,\right),$ but neither
valid nor tautological.

P.S. I prefer to call $\,\large\rightarrow\,$ the material conditional and $\,\large\implies\,$ the implication symbol. (Why I don't call $\,\rightarrow\,$ “implication”.)


问:
Difference between biconditional and logical equivalence
A biconditional can also be stated as "P is equivalent to Q," whereas a logical equivalence can also be stated as "P is logically equivalent to Q." I never understood intuitively what the difference between them was. I get that biconditional is true when both P and Q are true or false, and logical equivalence means P and Q have all of the same possible truth values, but I cannot grasp intuitively what they really mean and what the difference between them is.

答:
Think of the material biconditional $\LARGE\leftrightarrow$ as a logical operation on two sentences rather than as equivalence. Unlike the latter, it is not making an assertion nor in a metalangauge.

The statement $\Large P\equiv Q$ asserts the logical equivalence of $P$ and $Q.$ It means that $\large P\leftrightarrow Q$ is a (logical) validity, i.e., is true regardless of interpretation.

What's the difference between $↔$ and $≡$ ?

The statement $\Large P {\iff} Q$ asserts the (not necessarily logical) equivalence of $P$ and $Q.$ This means that that $\large P \leftrightarrow Q$ is either analytically (in an axiom system) or synthetically true , i.e., that $P$ and $Q$ have the same truth value in the given interpretation.

$\Large P {\iff} Q$ can be read as ‘$P$ is true if and only if $Q$ is true’ or ‘$P$ being true is a necessary and sufficient condition for $Q$ to be true’.

For example, the sentences $\:0=0\:$ and $\:3^2=9\:$ are mathematically but not logically equivalent.

P.S. Symbolic logic is an area rife with conflicting notation, terminology and even notions; my understanding is eclectically evolving.


问:
Proving Vector Identities
Let r=$(x,y,z)$ and $r=$||r||.

(A) Prove that $\nabla^2r^3=12r$.

(B) Is there a value of $p$ for which the vector field f(r) = r/$r^p$ is solenoidal?

What I have tried:

For part (a) I think that $r^3=\sqrt{27} $ but I am unsure of what to do next in terms of the del operator.

Title: Proving Vector Identities

Body:
$$
\nabla^2r^3=(\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}+\frac{\partial^2}{\partial z^2})(x^2+y^2+z^2)^{3/2}\\
= \frac{\partial }{\partial x }(3x(x^2+y^2+z^2)^{1/2})+\frac{\partial }{\partial y }(3y(x^2+y^2+z^2)^{1/2})+\frac{\partial }{\partial z }(3z(x^2+y^2+z^2)^{1/2}) \\
=\frac{3(2x^2+y^2+z^2)}{(x^2+y^2+z^2)^{1/2}}+\frac{3(x^2+2y^2+z^2)}{(x^2+y^2+z^2)^{1/2}} +\frac{3( x^2+y^2+2z^2)}{(x^2+y^2+z^2)^{1/2}}\\=3\frac{ (4 x^2+4y^2+4z^2)} {(x^2+y^2+z^2)^{1/2}} =12r
$$
$$
\nabla\cdot\big[\frac{(x, y, z)}{(x^2 + y^2 + z^2)^{p/2}}\big] =\frac{3-p}{(x^2+y^2+z^2)^{p/2}}=\frac{3-p}{r^{p }}
$$
So $\nabla\cdot\frac{{\bf r}}{r^p}=0$ when $p=3$.


问:
Software for solving fractional differential equations numerically
I have been trying to find information on how to solve fractional differential equations numerically with the usual maths software (Mathematica, Maple, Matlab, etc). Or to find an alternative program to do that.

I had no success.

Thanks.

答:
Use the MATLAB fde12 function. You may download it here:

http://www.mathworks.com/matlabcentral/fileexchange/32918

Its the easiest way.

Another example: Software for Fractional Differential Equations and related problems:


问:
Vector Identities Proof
Let $\textbf{r}=(x,y,z)$ and $r=||\textbf{r}||$

Show that $\nabla \cdot (r\space\textbf{r})=4r.$

So, I know that we can use a Vector property to expand the LHS so we get:
$$(\nabla r)\cdot\textbf{r}+r(\nabla\cdot\textbf{r})$$
and that $r=\sqrt{3}$

But I'm not too sure where to go from here, and to how to calculate $\nabla \cdot \sqrt{3}$

Any help would be great thanks!

答:
We need two facts: $$\nabla r=(\frac{\partial r}{\partial x},\frac{\partial r}{\partial y},\frac{\partial r}{\partial z})=\mathbf{r}/r$$ since $\frac{\partial}{\partial x}\sqrt{x^2+y^2+z^2}=\frac{x}{\sqrt{x^2+y^2+z^2}}$, etc.
$$\nabla\cdot\mathbf{r}=\frac{\partial x}{x}+\frac{\partial y}{y}+\frac{\partial z}{z}=3$$

Then combining together $$\nabla\cdot(r\mathbf{r})=(\nabla r)\cdot\mathbf{r}+r(\nabla\cdot\mathbf{r})=\frac{\mathbf{r}\cdot\mathbf{r}}{r}+3r=4r$$

Title: Vector Identities Proof


问:
Prove that a parametrization is a injective function
I have the following problem:

Let $\textbf{x}$: $(-\pi,\pi)\times(-\pi,\pi)\rightarrow R^3$ such
that

$$\textbf{x}(\theta,t)=((2+\cos(t))\cos(\theta),(2+\cos(t))\sin(\theta),\sin(t))$$

I want to prove that $\textbf{x}$ is a injective function. I tried by the definition of injective function, but I didn't get it. That problem are in the frame of differential geometry

My attemp:

Let  $(\theta_{1},t_{1}),(\theta_{2},t_{2})\in (-\pi,\pi)$ such that $\textbf{x}(\theta_{1},t_{1})=\textbf{x}(\theta_{2},t_{2})$.
Then

$$(2+\cos(t_{1}))\cos(\theta_{1})=(2+\cos(t_{2}))\cos(\theta_{2})$$

$$(2+\cos(t_{1}))\sin(\theta_{1})=(2+\cos(t_{2}))\sin(\theta_{2})$$

$$\sin(t_{1})=\sin(t_{2})$$

We can say, from the last equality:

$t_{1} = \pi - t_{2}$ , if $t_{1},t_{2}\geq 0$

$t_{1} = -\pi - t_{2}$ , if $t_{1},t_{2}\leq 0$

$t_{1}=t_{2}$

We want to prove that if any of the first two happens, then (1) or (2) is not true.

答:
As you said, you have your system of equations. You can create a new one by using $(1)^2+(2)^2 $, which yields :

$$(2+cos(t_1))^2=(2+cos(t_2))^2 $$
And since everything is positive, you have :
$$cos(t_1)=cos(t_2), sin(t1)=sin(t_2), (3) $$
So you have $t_1=t_2$, then, you can plug it in the first 2 equations and simplify by $(2+cos(t_{1,2}))$ (since they are the same), and you also get :
$$sin(\theta_1)=sin(\theta_2), cos(\theta_1)=cos(\theta_2) $$
Then you can conclude that :

$$x(t_1,\theta_1)=x(t_2,\theta_2) \Rightarrow (t_1,\theta_1)=(t_2,\theta_2)$$

So your function is injective


答:
If you want to express the Taylor series in the form $\sum_{n=0}^\infty a_nx^n$, you can use the formula

$$a_n=2+\mathrm{sgn}(n)$$

where $\mathrm{sgn}(n)=0$ if $n=0$ and $1$ if $n\ge1$. But this is really just another way of expressing zugzug's answer, giving a name to a function that has a piecewise definition.


问:
Using "implies" to refer to material conditional
Is it acceptable to translate the binary connective "$\let\ f\rightarrow$" into English with "implies"? I'm unsure because "implies" for me immediately brings to mind logical implication, but I've seen some places use it for the material conditional (including wikipedia, in the opening sentence of this article).

For example, does mathematical convention, in principle, permit the following formulation of the standard definition for functional continuity?

$f$ is continuous at $c$ if for any $\epsilon > 0 :$ there exists $\delta > 0:$ for any $x$ in Domain[$f$]$:$ $|x-c|<\delta$ implies $|f(x)-f(c)|<\epsilon$.

答:
$\huge\rightarrow$ (the material conditional) is a logical connective that operates on its antecedent $P$ and its consequent $Q$ to form the truth function $$P\rightarrow Q,$$ which is false precisely when $P$ is true but $Q$ false.

$\huge\Rightarrow$ (implication) similarly operates on its antecedent $P$ and its consequent $Q;$ here, the result $$P\Rightarrow Q$$ lives in a specific (usually implicitly understood) interpretation/context and signifies that $P\rightarrow Q$ is true in it. Some ways to read $P\Rightarrow Q:$

If $P$ is true, then $Q$ is true.

$P$ being true is a sufficient condition for $Q$ to be true.

$P$ being true implies that $Q$ is true.

$P$ is true only if $Q$ is true.

$Q$ being true is a necessary condition for $P$ to be true.

When $P\not\Rightarrow Q,$ then it must be that $P$ is true yet $Q$ false.

In the given formulation $$|x-c|<\delta\;\; \textbf{implies} \;\;|f(x)-f(c)|<\varepsilon,$$ “implies” is not the material conditional $\large\rightarrow\normalsize$ per se, but rather mathematical implication $\large\Rightarrow\normalsize;$ it analytically (from mathematical axioms and a given context) asserts that its right side can be derived from its left.

P.S. I distinguish between implication $\,\large\Rightarrow\,$ and logical implication $\,\large\vDash\,,$ which is often used to mean first-order implication, i.e., that $P\rightarrow Q$ is true regardless of interpretation.

P.P.S. Symbolic logic is an area rife with conflicting notation, terminology and even notions; my understanding is eclectically evolving.


问:
Every continuous map $f:\mathbb{C}P(2) \to \mathbb{C}P(2)$ has a fixed point, without Lefschetz theorem.
I would like to know if there is a nice proof of the fact that every continuous map $f:\mathbb{C}P(2) \to \mathbb{C}P(2)$ has a fixed point, without use of the Lefschetz fixed point theorem.

答:
Through any two distinct points in $\mathbb{C}P^n$ there is a unique (complex) geodesic. Therefore  from any fix point free self map $f$ , we have a 1 dimensional complex subbundle of the tangent bundle by taking the subspace above a point $p$ to be the tangent space of the geodesic from $p$ to $f(p)$ at $p$.

This implies that the total Chern class of $\mathbb{C}P^n$ has a linear factor. If $n=2$ this implies that the total Chern class $1+3x+3x^2$ has two real (integer) roots. However, this is easily checked to be false since the discriminant is negative.

As a sanity check, this should be different if $n=3$. In that case the total Chern class is $1+4x +6x^2 +4x^3$ and this equals $(2 x + 1) (2 x^2 + 2 x + 1)$, as expected.

I imagine if you are better with polynomials than I, you can get this to work for any even $n$.


问:
Solving the equation $\displaystyle \frac{e^x}{x}=\int_n^{n+1}f(t)\,dt$
Suppose the equation $\displaystyle \frac{e^x}{x}=\int_n^{n+1}f(t)\,dt$ as $f(t)=\frac{e^t}{t}$ and $n\in \mathbb{N} \setminus{0}$.

How to prove that:

The equation above has a unique solution $U_n$ on $[n,n+1]$.

The $\displaystyle \lim_{n \to +\infty} \frac{U_n}{n}=1$.

答:
Take $F(x)=\int_0^x f(t)\, dt$

Then $\int_n^{n+1}f(t)\,dt = F(n+1)-F(n)$ and $f=F'$

So by the mean value theorem, you have $U_n \in [n,n+1]$ so that $F'(U_n)=\cfrac{F(n+1)-F(n)}{(n+1)-n}$

$n \le U_n \le n+1$

$1 \le \cfrac{U_n}{n} \le 1+\cfrac{1}{n} \underset{n\to +\infty}{\longrightarrow} 1$

So $\cfrac{U_n}{n}\underset{n\to +\infty}{\longrightarrow} 1$


问:
Minimum number of points in soccer game.
Eight soccer teams participate in a soccer tournament, so that each team plays with every other team exactly once. If one teams wins against another, then the winning team gets two points, and the losing team gets zero points. If the two teams tie, then both teams get one point each.

After the tournament, the teams are ranked by the total number points each team won. What is the minimum number of points a team must win, to ensure that it is in the top four teams?

I think that the four teams in the top have to have the same number of points to maximize the number of points the other team needs to get in, but I'm not sure.

答:
With an eight team round robin, $28$ games will be played in total (The total edges in $K_8$).

Since each game will produce two points regardless of outcome, $56$ total points will be shared among the eight teams. In the absence of a well-defined tie breaking system, we assume that ties do not qualify as a guarantee. We define a team's point total as "ensuring" fourth place when the fifth place team could not possibly have that many points.

\begin{align}
Team && Points && && Team && Points\\
A && 10 && && A && 11\\
B && 10 && && B && 11\\
C && 10 && && C && 11\\
D && 10 && && D && 11\\
--- && --- && && --- && ---\\
E && 10 && && E && 11\\
F && 2 && && F && 1\\
G && 2 && && G && 0\\
H && 2 && && H && 0\\
\end{align}

Ten points are not enough and eleven points must be enough. In one extreme case, teams $A,B,C,D$ and $E$ all draw each other and win against $F,G,$ and $H$ (who also draw each other), giving a fifth place team with ten points. If we have a fifth place team with eleven points then teams $A,B,C,D,$ and $E$ share at least $55$ points, meaning teams $F, G,$ and $H$ share at most one point. However, $F, G,$ and $H$ play each other and must share at least six points, a contradiction. Therefore, $11$ points guarantees fourth place at a minimum. $\blacksquare$


问:
Conditional expectation with countably many disjoint events
In his probability book Bauer gives the following prelude to the definition of conditional expectation:

Let $X$ be an integrable real random variable on a probability space $(\Omega,\mathcal{A},P)$, and let $\Omega=\bigcup_{i\in \mathbb{N}} B_i$ be a partition of $\Omega$ into countably many pairwise disjoint events $B_i\in \mathcal{A}$ having positive probability $P(B_i)>0$. Define the conditional expectation of $X$ given $B_i$ by

$$E_{B_i}(X):=\frac{1}{P(B_i)}\int_{B_i}XdP.$$

This leads to a new random variable

$$X_0:=\sum_{i\in \mathbb{N}}E_{B_i}(X)1_{B_i}.$$

This random variable is measurable with respect to the sub-$\sigma$-algebra $\mathcal{C}:=\sigma(B_i:i\in\mathbb{N})$, which consists of all sets of the form $\bigcup_{i\in J} B_i$ with $J\subset\mathbb{N}$. Now

$$\int_{B_i}X_0 dP=E_{B_i}(X) P(B_i)=\int_{B_i}X dP \hspace{1cm} i\in \mathbb{N},$$

and because of the special form of $\mathcal{C}$ it follows that

$$\int_{C}X_0 dP=\int_{C}X dP \hspace{1cm} \text{for all }  C\in \mathcal{C}.$$

Question: Why does this last equality hold?

My reasoning is the following:

We see that $E_{B_i}(X)=E_{B_i}(X^+)-E_{B_i}(X^-)$, and so $X_0:=\sum_{i\in \mathbb{N}}E_{B_i}(X^+)1_{B_i}-\sum_{i\in \mathbb{N}}E_{B_i}(X^-)1_{B_i}.$ Now since

$$\sum_{i=1}^n E_{B_i}(X^+)1_{B_i} \uparrow \sum_{i\in \mathbb{N}}E_{B_i}(X^+)1_{B_i} \hspace{0.5cm} \text{pointwise as } n\to \infty$$

the MCT implies that

$$\int \sum_{i\in \mathbb{N}}E_{B_i}(X^+)1_{B_i} dP = \lim_{n\to \infty} \sum_{i=1}^n E_{B_i}(X^+)P(B_i)$$

But $\sum_{i=1}^n E_{B_i}(X^+)P(B_i)=\sum_{i=1}^n \int_{B_i}X^+ dP$ and $\sum_{i=1}^n X^+ 1_{B_i} \uparrow X^+$ pointwise as $n\to \infty$. Therefore another application of the MCT delivers the relation

$$\int \sum_{i\in \mathbb{N}}E_{B_i}(X^+)1_{B_i} dP=\int X^+ dP < \infty$$

The same argument with $X^-$ also gives

$$\int \sum_{i\in \mathbb{N}}E_{B_i}(X^-)1_{B_i} dP=\int X^- dP < \infty$$

Putting everything together we get that $X_0$ is integrable and $\int X_0 dP=\int X dP$.

This covers the case $C=\bigcup_{i\in J} B_i=\Omega$. If $J\neq \mathbb{N}$ is infinite then $X_0 1_C=\sum_{i\in J}E_{B_i}(X)1_{B_i}$ and we see that the same argument gives $\int_C X_0 dP=\int_C X dP$. Finally if  $J$ is finite then the result is immediate.

Is this correct?

答:
If $C=\bigcup_{i\in J}B_i$ for some $J\subseteq\mathbb N$ then for nonnegative $Y$ we find:$$\int_CYdP=\sum_{i\in J}\int_{B_i}YdP=\sum_{i\in J}\int_{B_i}Y_0dP=\int_CY_0dP$$

This can be used to find that: $$\int_CX^+dP=\int_C{X_0}^+dP\text{ and }\int_CX^-dP=\int_C{X_0}^-dP$$

From this we conclude that:$$\int_CXdP=\int_CX_0dP$$


问:
similarity matrix tridiagonal
I am solving a problem, in the middle of which I reached the following block-tridiagonal matrix.

\begin{bmatrix}
0 & B_1 & &0  \\
B_1 & 0 & \ddots  \\
&\ddots& \ddots&B_n \\
 0 & &  B_n & 0
\end{bmatrix}
Where $B_i$ is an invertible square matrix and size $0$ is equal to size $B_i$. Can I make this matrix similar to the following  tridiagonal block matrix as shown below?

\begin{bmatrix}
0 & C & &0  \\
C & 0 & \ddots  \\
&\ddots& \ddots&C \\
 0 & &  C & 0
\end{bmatrix}

in which size $C$ and $0$ equals $B_i$. What about if size $B_i$ is 1?

答:
This is not always true, even if the $B_i$ have size $1 \times 1$. For example, if $n = 3$ we have the following matrix and characteristic polynomial:
$$ A = \begin{pmatrix} 0 & b_1 & 0 & 0 \\ b_1 & 0 & b_2 & 0 \\ 0 & b_2 & 0 & b_3 \\ 0 & 0 & b_3 & 0 \end{pmatrix}, \qquad p_A(\lambda) = \lambda^4 - (b_1^2 + b_2^2 + b_3^2)\lambda^2 + b_1^2 b_3^2. $$

Therefore:

If $(b_1,b_2,b_3) = (1,2,1)$, the characteristic polynomial is $\lambda^4 - 6\lambda^2 + 1$.

If $(b_1,b_2,b_3) = (c,c,c)$, the characteristic polynomial is $\lambda^4 - 3c^2\lambda^2 + c^4$.

Thus, for the matrices
$$ \begin{pmatrix} 0 & 1 & 0 & 0 \\ 1 & 0 & 2 & 0 \\ 0 & 2 & 0 & 1\\ 0 & 0 & 1 & 0 \end{pmatrix} \qquad \text{and} \qquad \begin{pmatrix} 0 & c & 0 & 0 \\ c & 0 & c & 0 \\ 0 & c & 0 & c\\ 0 & 0 & c & 0 \end{pmatrix} $$
to be similar, it is necessary that $c^2 = 2$ but $c^4 = 1$. This is impossible.

Title: similarity matrix tridiagonal


答: A more concise approach to (A) uses$$\nabla^2r^3=\frac{1}{r^2}\frac{d}{dr}\left(r^2\frac{d}{dr}r^3\right)=\frac{1}{r^2}\frac{d}{dr}\left(3r^4\right)=12r.$$Similarly, for (B) use$$\nabla(r^{-p})=\frac{d}{dr}(r^{-p})\hat{\mathrm{r}}=-pr^{-p-2}\mathrm{r}$$to deduce$$\nabla\cdot(r^{-p}\mathrm{r})=r^{-p}\nabla\cdot\mathrm{r}+\nabla(r^{-p})\cdot\mathrm{r}=(3-p)r^{-p},$$i.e. the field is solenoidal iff $p=3$ (which would surprise no physicists familiar with the differential form of Gauss's law).

Title: Proving Vector Identities

Body:
Let r=$(x,y,z)$ and $r=$||r||.

(A) Prove that $\nabla^2r^3=12r$.

(B) Is there a value of $p$ for which the vector field f(r) = r/$r^p$ is solenoidal?

What I have tried:

For part (a) I think that $r^3=\sqrt{27} $ but I am unsure of what to do next in terms of the del operator.


问:
Is there any name for this family of functions?
Let $\Phi$ be the class of all fonctions  $\phi$ (let's say from $\mathbb{R}^+$ to $\mathbb{R}^+$) such that:

For each sequence $\{t_n\}\subset \mathbb{R}^+$ that converges, the sequence $\{\phi(t_n)\}$ converges also.

Is there any specific name for this class of functions?

It will be interesting to see some interesting examples/counterexamples.

答: There is a tiny gap between the stated condition and the standard definition of a continuous function, but that gap can be filled as follows: if $\{ t_n\}$ converges to $a$, consider the sequence $\{s_n\}$ with terms $t_1, a, t_2, a, t_3, a, \ldots $. Then $\{ s_n\}$ converges (to $a$) and the convergence of $\{ f(s_n) \}$ implies that $\{f(t_n)\}$ converges to $f(a)$.


问:
Find infimum, supremum and max, min of a set if they exist
I need to find infimum, supremum and max, min of a set if they exist:$$E=\left\{1-\sum_{m=1}^{n}\frac{1}{4^m}\ : n\in \mathbb{N}\right\}$$
It seems that $\max E=\sup E=\frac{3}{4}$ and $\inf E=\frac{2}{3}$.

However, I am not sure if I can claim that $\min E=\frac{2}{3}$.

Intuitively, I'm certain that this set does not have a minimum value.

答:
Consider the sequence $$a_n=1-\sum_{m=1}^n\frac{1}{4^m}$$
Note that $E=\{a_n\}_n$, where that sequence is strictly decreasing and bounded. Therefore you have a maximum (because its strictly decreasing) and an infimum with no minimum because if it ever reached the minimum (suppose it's reached by $a_{n_0}$), it contradicts the fact it is the minimum because the sequence is strictly decreasing.


问:
Compactly supported test functions
I am considering the test function:
$$f(x)=\frac{1}{\cos(\pi x)},$$
My question is whether the test function is compactly supported? I am aware of the definition of compact support, namely:
"A function has compact support if it is zero outside of a compact set".
I believe the domain of this function is: $$x\in \mathbb{R}\setminus \left\{\pm\frac{n}{2},n\in\mathbb{N}\right\},$$
This is a closed and bounded set, so also compact. But how do I show that this function is zero outside of this set? as it surely takes no value outside of this set? I'm fairly sure it is compactly supported, but I don't know how to show it.

答:
As suggested by @Fred , @OlivierMoschetta and @Chrystomath this test function is not compactly supported. However, this function can be used in the context of the composition integral, as per @md2perpe 's comment as it is smooth on a neighbourhood of the support of $\delta \circ g$
(I posted this as answer just so the thread can be closed)

Title: Compactly supported test functions


问:
Tangent plane through $(1,1,1)$
Determine the equation of the tangent plane to the surface $x^2y+z^2y-2z-y=-1$ that passing through the point $(1,1,1)$. Is there any point on the surface where the tangent plane is parallel to the $xy$ plane? Which?

$$f'_1=2xy,\ f'_2=x^2+z^2-1\\$$
Equation of tangent plane: $Z=f(a,b)+f'_1(a,b)(x-a)+f'_2(a,b)(y-b)$

$$Z=1+2(x-1)+z^2(y-1)=2x+z^2y-z^2-1$$

Or shall I also put $z=1$ in $f'_2(a,b)$?

$$Z=1+2(x-1)+1(y-1)=2x+y-2$$

答:
Your formula does not account for the z-dimension, computing the directional derivatives

$f_1'(x,y,z) = 2xy$

$f_2'(x,y,z) = x^2 + z^2 - 1$

$f_3'(x,y,z) = 2zy - 2$

And evaluating these at $(1, 1, 1)$

$f_1'(1, 1, 1) = 2$

$f_2'(1, 1, 1) = 1$

$f_3'(1, 1, 1) = 0$

The equation of the tangent plane at $(a, b, c)$ is given by

$ 0 = (x-a)f_1'(a, b, c) + (y-b)f_2'(a, b, c) + (z-c)f_3'(a, b, c) $

$ 0 = (x-1)f_1'(1, 1, 1) + (y-1)f_2'(1, 1, 1) + (z-1)f_3'(1, 1, 1) $

$ 0 = 2(x-1) + y-1 $

$ 3 = 2x + y $

note that in general the plane is not parallel to the $xy$-plane and you do have to compute the directional derivatives for every input. In this example since $f_3'(1, 1, 1) = 0$ we have that it is in fact parallel to the $xy$-plane.

Title: Tangent plane through $(1,1,1)$


问:
sequence and series for telescoping terms
$x=\frac{1}{16 \sqrt{17}+17 \sqrt{16}}+\frac{1}{17 \sqrt{18}+18 \sqrt{17}}+\ldots \ldots+\frac{1}{144 \sqrt{145}+145 \sqrt{144}}$ the value of $\mathrm{X}$ is approximately.

(A) 0.09

(B) 0.17

(C) 0.29

(D) 0.41

Please help me in the above.
I tried but got to sqrt(a)-sqrt(b)/ab  where a is the bigger of each term in the fraction
but how to get the series to have a common denominator to cancel the terms.

答:
Hint (re telescoping terms):

Try multiplying the first term by
$$\frac{17\sqrt{16} - 16\sqrt{17}}{17\sqrt{16} - 16\sqrt{17}}$$ and see what happens.


问:
Is this alternating series divergent?
I have this series and I need to determine whether it is divergent or convergent, are my calculations correct and it is divergent or am I not seeing something

$$
\sum_{n=2}^\infty (-1)^n \left( \frac{n}{2n+1}\right)^2
$$
since it is an alternating series I find the limit of  a_n
$$
\lim_{n\to \infty}\left( \frac{n}{2n+1}\right)^2 = \frac{n^2}{(2n+1)^2} = \frac{n^2}{4n^2+4n+1} = \frac{1}{4}
$$
and since it does not equal to 0 by the alternating divergence test it s divergent ??

答:
Since $a_n \not \to 0$ we can conclude that the series doesn't converges but we don't need to invoque any alternating divergence test since $a_n \to 0$ is a necessary condition for any series to converge, that is

$$\sum_n a_n \quad \text{converges} \implies a_n \to 0$$

therefore

$$a_n \not \to 0 \implies \sum_n a_n \quad \text{doesn't converge}$$

Refer also to the related

Limit of infinite series

Title: Is this alternating series divergent?


问:
Find the nature of $\sum_{n = 1}^\infty \frac{(2n - 1)!!}{(2n)!! (2n + 1)}$
I need to find whether the following series converges or diverges:

$$\sum_{n = 1}^\infty \frac{(2n - 1)!!}{(2n)!! (2n + 1)}$$

It seems to converge but I don't really have any good idea on how to prove it. I tried to use D'Alembert Criterion, but the limit is $1$, so it doesn't help. Can you give me a hint, please?

答:
You can also just note that
\begin{align*}
\frac{(2n-1)!!}{(2n)!! (2n+1)}& = \frac{1\times 2 \times \cdots \times (2n-1)!}{(2n+1) \times 1 \times 2 \times \cdots \times (2n-1)! \times ((2n-1)!+1)\times \cdots \times (2n)!}\\
= & \frac{1}{(2n+1)\times [(2n-1)!+1] \times [(2n-1)!+2] \times \cdots (2n)!}  \\
< & \frac{1}{2n^2}
\end{align*}

and that $\sum \frac{1}{2n^2}$ is convergent.


答:
Okay.

So if $S_n = 1-\sum_{m=1}^n \frac 1{4^n}$ its easy to see that  $S_{k+1} = S_k -\frac 1{4^{k-1}} < S_k$ so $S_1 >S_2 > S_3.....$.

So every $S_k \le S_1$ and so $\sup S_n = \max S_n = S_1$.

And because, for every $S_k$ there is always $S_{k+1} < S_k$ we can't have have a smallest $S_k$ so $\min S_n$ can not exists.

So what remains to answer are:  Are the $S_k$ bounded below and if so what is $\inf S_n$?

Are you familiar with geometric series?

$1 + r + r^2 + r^3 +........ + r^n = M_n$

Then $M_n(1-r) = (1 + r + r^2 + r^3 +........ + r^n) - (r+r^2 + ...... + r^n + r^{n+1}) = 1-r^{n+1}$ so

$M_n = \frac {1-r^{n+1}}{1-r}$.

So $\sum_{m=1}^n \frac 14 = (\sum_{m=0}^n \frac 14) -1=\frac {1-(\frac 14)^{n+1}}{1-\frac 14} -1=$

$\frac {4(1-(\frac 14)^{n+1}}3 -1= \frac {4-(\frac 14)^n}3 - 1=\frac {1-\frac 1{4^n}}3$

So $S_n = 1-\frac {1-\frac 1{4^n}}3= \frac {2+\frac 1{4^n}}3$.

And as $\frac 1{4^n} >0$ we can see than indeed $S_n > \frac 23$ and the sequence is bounded below by $\frac 23$.

If we take any number larger than $\frac 23$, say $\frac 23 + \iota$ then we can find a $\frac 1{4^n}$ so that $\frac 23 < \frac {2+\frac 1{4^n}}3 < \frac 23 + \iota$.  (Why?)

So any number larger than $\frac 23$ is not a lower bound, so $\frac 23$ is the greatest lower bound.

And $\inf S_n = \frac 23$.

Title: Find infimum, supremum and max, min of a set if they exist

Body:
I need to find infimum, supremum and max, min of a set if they exist:$$E=\left\{1-\sum_{m=1}^{n}\frac{1}{4^m}\ : n\in \mathbb{N}\right\}$$
It seems that $\max E=\sup E=\frac{3}{4}$ and $\inf E=\frac{2}{3}$.

However, I am not sure if I can claim that $\min E=\frac{2}{3}$.

Intuitively, I'm certain that this set does not have a minimum value.


问:
Property of a divergent sequence $\{a_n\}$ of positive real numbers satisfying $\lim_{n\to\infty}(a_{n+1}-a_n)=0$
Let $\{a_n\}$ be an unbounded sequence of positive real numbers s. t. $\lim\limits_{n\to\infty}(a_{n+1}-a_n)=0$.

Question:

Is $a_n$ necessarily equal to a partial sum $\sum_{k=1}^n b_k$ of some sequence $\{b_k\}$ s. t. $\lim\limits_{k\to\infty}b_k=0$, but $\sum_{k=1}^\infty b_k=\infty$?

Thoughts:

Since $\{a_n\}$ is an unbounded sequence of positive real numbers, I thought it must diverge.

On the other hand, $\lim\limits_{n\to\infty}(a_{n+1}-a_n)=0$ reminded me of the limits of the form $\lim\limits_{n\to\infty}\frac{x_n}{y_n}$, where also $x_n\to\infty$, which could be evaluated via Stolz-Cesaro, but the only examples of divergent sequences $\{a_n\}$ satisfying $\lim\limits_{n\to\infty}(a_{n+1}-a_n)=0$ I could think of were the partial sums $\sum_{k=1}^nb_k$ of a sequence $b_k$ converging to $0$.

However, my statement seems fragile and I'm pretty sceptical.

Is there any counter-example?

I also took into account sequences that contain infinite products, but, if the logarithms are involved, we could transform the product into a sum.

Thank you in advance!

答:
As already stated in the comment, if $a_n=\sum_{k=1}^n{b_k}$ then $b_n=a_n-a_{n-1}$ is the only possibility. Therefore, you have $\lim\limits_{k\to\infty}b_k=\lim\limits_{k\to\infty}a_k-a_{k-1}=0$.

The second part is more problematic since $\sum_{k=1}^\infty{b_k} = \infty \iff \lim\limits_{k\to\infty}a_k=\infty$ which is not guaranteed.
Check the following counter example:
$$a_n=\ln(n)(1+\sin(\sqrt{n})), n\ge1, a_0=0$$


问:
Limit of $\min \{a\in \mathbb{N} : \sum_{i=1}^{a}\frac1{i}\geq n\}$ equals $e$
Given $a_n = \min \{a\in \mathbb{N} : \sum_{i=1}^{a}\frac1{i}\geq n\}$ $\forall n \in \mathbb{N}$, I want to prove that:

$$\lim_{n\to\infty}\frac{a_{n+1}}{a_n}=e$$

For that purpose, I am trying to prove that $\log a_{n+1}-\log a_n \to 1$.

From the definition of $a_n$, we have that $1+\frac12+\ldots+\frac1{a_n}\geq n$, and we have:

$$n\leq 1+\frac12+\ldots+\frac1{a_n}\leq \int_1^{a_n+1}\frac1{x}dx = \log(a_n+1)$$

In the same way, for $a_{n+1}$ we have $n+1\leq \log(a_{n+1}+1)$.

Combining both inequalities we have:

$$\log (a_{n+1}+1)-\log (a_n+1) \geq 1$$

However, I don't know how to get the other inequality. Can somebody help me? Many thanks in advance!

Body:
Let's note as (harmonic series)
$$H_{a_n}=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{a_n}$$
From the definition of $a_n$ we have
$$H_{a_n} -\frac{1}{a_n}< n\leq H_{a_n} \Rightarrow
0< H_{a_n} - n< \frac{1}{a_n} \tag{1}$$

Given this we also have:
$$\ln{a_n}<H_{a_n} -\frac{1}{a_n} < n\leq H_{a_n}<\ln{a_n}+1\Rightarrow
\\a_n<e^n<a_n\cdot e \Rightarrow\\
\frac{1}{e^{n}}<\frac{1}{a_n}<\frac{1}{e^{n-1}} \tag{2}$$

From $(1)$ and $(2)$
$$0< H_{a_n} - n< \frac{1}{e^{n-1}} \tag{3}$$

A s a result, combining $0< H_{a_n} - n< \frac{1}{e^{n-1}}$ and $0< H_{a_{n+1}} - (n+1)< \frac{1}{e^{n}}<\frac{1}{e^{n-1}}$
$$|(H_{a_{n+1}}-(n+1))-(H_{a_{n}}-n)|=|H_{a_{n+1}}-H_{a_{n}}-1|<\frac{1}{e^{n-1}}$$
or
$$\lim\limits_{n\to\infty} (H_{a_{n+1}}-H_{a_{n}})=1 \tag{4}$$

And finally
$$(H_{a_{n+1}}-H_{a_{n}})-(\ln{a_{n+1}}-\ln{a_{n}})=\\
(H_{a_{n+1}}-\ln{a_{n+1}})-(H_{a_{n}}-\ln{a_{n}})\to \gamma -\gamma =0, n\to\infty$$
which, from $(4)$ and $|x-y|\geq {\bigg |}|x|-|y|{\bigg |}$, means
$$\ln{a_{n+1}}-\ln{a_{n}}\to 1, n\to\infty$$


问:
Proving (via Hilbert's axioms) that, when two parallel lines are cut by a third line, they make congruent angles.

Prove that, when two parallel lines are cut by a third line, they make congruent angles.

I'm not using Euclid's axioms, but instead I'm using Hilbert's. This is Theorem 19 of Hilbert's "The Foundations of Geometry" (PDF link via berkeley.edu).

Theorem 19. If two parallel lines are cut by a third straight line, the alternate-interior angles and also the exterior-interior angles are congruent. Conversely, if the alternate-interior or the exterior-interior angles are congruent, the given lines are parallel.

The definition of parallel lines is simply two lines who don't meet.

The definition of angle is a bit long it is on page 9. I think the important bit is that there is a bijection between angle and rays from a certain point.

And we have the (Euclid's) Axiom of Parallelism (page 7): Given a line $r$ and a point $A \notin r$ we can always draw one, and only one, line through $A$ parallel to $r$.

We can use that angles opposite on a vertex are congruent.

My attempt was this:

Given two concurrent lines $r$ and $s$, s.t. $r \cap s = A$ let's take a point on $s$ different from $A$ and draw the one parallel line to $r$ from it, call it $h$.

supose $\angle (h,s) < \angle (r,s)$ and let $h'$ be the ray (line) such that $\angle (h',s) = \angle (h,s)$ can we prove that $h'$ is another parallel line or that it is line $s$?

I think if we assume the angles formed are different than we would have two parallel lines through $B$ but I'm out of ideas.

答:
Proving this in full detail from Hilbert's axioms takes a lot of work, but here is a sketch.  Suppose $\ell$ and $m$ are parallel lines and $n$ is a line that intersects both of them.  Say $n$ intersects $m$ at $P$.  Now let $m'$ be the line through $P$ which forms angles with $n$ that are congruent with the the angles that $n$ forms with $\ell$ (using axiom IV,4).  If we can prove that $m'$ is parallel to $\ell$, then we must have $m=m'$ by axiom III.

So, suppose $m'$ was not parallel to $\ell$.  Then the lines $\ell, m',$ and $n$ would form a triangle.  Since $\ell$ and $m'$ form the same angles with $n$, this triangle will have two angles (the two angles on $n$) which add up to a straight angle.  Now you can prove this is impossible by essentially the same argument as in Euclid; see https://mathcs.clarku.edu/~djoyce/java/elements/bookI/propI17.html, for instance.  (Note that Euclid's argument relies on the existence of midpoints, which Euclid proves by constructing equilateral triangles by intersecting circles.  Doing this with Hilbert's axioms requires the use of the completeness axiom and is pretty complicated.  Alternatively, without the completeness axiom, it is still possible to construct an isosceles triangle with a given base, which is enough to obtain the midpoint of the base.)


问:
Determine the screw axis of a $3D$ rigid motion given by $F(x)=Ax+c$

Determine the screw axis of a $3D$ rigid motion given by $F(x)=\begin{pmatrix}0&1&0\\1&0&0\\0&0&-1\\\end{pmatrix}x+\begin{pmatrix}1\\2\\3\\\end{pmatrix}$

Characteristic polynomial is $P(x)=-(\lambda-1)(\lambda+1)^2$

So I found the eigenvalues were $\lambda_1=1, \lambda_2=-1$ which had multiplicity $2$ for the matrix

Then the eigenvectors are $v_1=\begin{pmatrix}1\\1\\0\\\end{pmatrix}$

And $v_2=\begin{pmatrix}-1\\1\\0\\\end{pmatrix}$ and $v_3=\begin{pmatrix}0\\0\\1\\\end{pmatrix}$

I'm not really sure what to do from here, I thought the axis should correspond to one of these $3$ lines given by the eigenvectors, and that $\begin{pmatrix}1\\2\\3\\\end{pmatrix}$ should then lie on one of the lines. But that doesn't happen.

答:
The determinant is $1$ so the matrix is a pure rotation, with no reflection.

The eigenvectors determine the reference system in which the action of the matrix is  a stretching along the axes: in that reference the matrix becomes diagonal.

The product of the eigenvalues shall be $1$ as well. In this case they are $1, -1, -1$.

Clearly a vector along the rotation axis shall remain the same, and that is $(1,1,0)$ which is at $\pi /4$ in the $x,y$ plane.

The rotation is reversing the direction of the $z$ axis: so it is a rotation of $\pi$.

This will sweep the position of the $x$ and $y$ axes, which in fact is what the matrix does.

--- note in reply to your comment  ---

A rigid rotation matrix must have at least one eigenvalue $=1$.

If it has two at $1$, then it must have all three ad it is a zero rotation matrix, otherwise the determinant is not one,
and it is not a "pure" ("rigid body") rotation, but contains some stretching and/or reflection.


问:
Gröbner basis and generating set
I have come across the following past exam question...

Define an ideal $J:=(z^2x+y^2-2y,x^3+y^3+z^3,x^2+2z^2) \subseteq \mathbb{Q}[x,y,z].
$

Compute a generating set for $J \cap \mathbb{Q}[y]$.

Compute a generating set for $J \cap(y)$.

Compute a Gröbner Basis for $(J:y)$ using lexicographic order with $x>y>z$.

Any help in computing the generating sets and then the Gröbner basis would be greatly appreciated as I'm a bit lost.

答:
This might help you : with the help of my computer I found out that

$$
J \cap {\mathbb Q}[y]=(y^6 + 4y^5 - \frac{7}{2}y^4 - 18y^3 + 18y^2){\mathbb Q}[y] \tag{1}
$$

at that the Groebner basis for $(J:y)$ (with respect to the graded lex order with
$x \gt y \gt z$) consists of six elements, namely

$$
\begin{eqnarray}
h_1 & = & x^2 + 2z^2, \\
h_2 & = & \left( \frac{y}{2} - 1\right)x + (y^2 + 2y - 4)z, \\
h_3 & = & z^3 + (y^3 + 2y^2 - 4y), \\
h_4 & = & z^2x + (y^2 - 2y), \\
h_5 & = & (y^2 + 2y - 4)x + (-y + 2)z, \\
h_6 & = & \left( -\frac{y^2}{2} + y\right)x + z^4 \\
\end{eqnarray}
$$

Perhaps there is a reasonable hand-made computation that leads to (2), but it seems
unlikely that (1) can be reached without a computer. So there might be a typo in your
original question.

Regarding the second question, number your original generators :
$$
g_1=z^2x+y^2-2y, \ \ g_2=x^3+y^3+z^3,\ \ g_3=x^2+2z^2\tag{3}
$$

Let $f\in J\cap {\bf\big (}y {\bf\big )}$. Then $f$ can be written
$f=p_1g_1+p_2g_2+p_3g_3$, where each $p_i$ is a polynomial in $x,y,z$. For each $i$
we can write $p_i=q_i+r_iy$ where $q_i,r_i$ are polynomials and $q_i$ does not contain
a $y$ (in fact, $q_i(x,z)=p_i(x,0,z)$). Then the hypothesis $f \in {\bf\big (}y {\bf\big )}$ yields $p_1{g^{*}}_1+p_2{g^{*}}_2+p_3{g^{*}}_3=0$ where for each $i$ we put
${g^{*}}_i(x,z)=g_i(x,0,z)$.

So we are naturally lead to consider the following ideal $I_1$ of ${\mathbb Q}[x,z]^2$ :

$$
I_1=\lbrace (q_1,q_2,q_3) \in {\mathbb Q}[x,z]^2 \ | \
q_1{g^*}_1+q_2{g^*}_2+q_3{g^*}_3=0
 \rbrace \tag{4}
$$

To study $I_1$, we first look at the projection

$$
I_2=\lbrace q_2 | (q_1,q_2,q_3) \in I_1\rbrace \tag{5}
$$

First, we have $x\in I_2$ because

$$
(2x-z)(z^2x)+x(x^3+z^3)+(-x^2)(x^2+2z^2)=0 \tag{6}
$$

and we also have $z\in I_2$ because

$$
(\frac{x}{2}+2z)(z^2x)+z(x^3+z^3)+(-\frac{z}{2})(x^2+2z^2)=0 \tag{7}
$$

and as it is easy to see that $1\not\in I_2$, we deduce that $I_2={\bf\big (}x,z{\bf\big )}$ and hence that $I_1$ is generated as an ideal by
$(2x-z,x,-x^2),(\frac{x}{2}+2z,z,-\frac{z}{2})$ and $(-{g^*}_3,0,{g^*}_1)$.

Plugging this result back into the decomposition of $f$, we deduce the following set of six generators for $J\cap {\bf\big (}y {\bf\big )}$ :

$$
yg_1,yg_2,yg_3, (2x-z)g_1+xg_2+(-x^2)g_3,
(\frac{x}{2}+2z)g_1+zg_2+(-\frac{z}{2})g_3,
(-{g^*}_3)g_1+{g^*}_1g_3.
$$

Title: Gröbner basis and generating set


问:
Proof of $\log(n!) < (n-1)\log(n)$

Prove that $\log(n!) \leq (n-1)\log(n)$ directly and by induction.

I'm having a hard time with this one. I tried:

$$\log(n!) + \log(n+1) \leq (n-1)\log(n+1) + \log(n+1)$$
$$\log((n+1)!) \leq n\log(n) - \log(n) + \log(n+1)$$
$$\log((n+1)!) \leq n\log(n) - \log(n) + \log(n+1) \leq n\log(n+1) - \log(n) + \log(n+1)$$
$$\log((n+1)!) \leq n\log(n+1) - \log(n) + \log(n+1) \leq n\log(n+1) + \log(n+1)$$
$$\log((n+1)!) \leq (n+1)\log(n+1)$$

But that's not what is asked, and I can't find a way of getting there.

答: Hint: You can use $n!\leqslant n^{n-1}$

Title: Proof of $\log(n!) < (n-1)\log(n)$

Body: Hint: You can use $n!\leqslant n^{n-1}$