问:
Generalising positive semi-definiteneness of 2×2 to N×N matrix
If I have some function $G(x), \quad G:X\to\mathbb{R}$, which is used to construct a matrix as follows
$$
A_{ij} = G\Bigl(\frac{x_i + x_j}{2}\Bigr)
$$

and I can show that the $2\times2$ matrix has all non-negative eigen values, is there a way to generalise this to an $N\times N$ matrix, perhaps via some inductive argument, to show that the matrix $A_{ij}^N$ (n indicating the size of the matrix, not an exponent) is positive semi-definite?

I understand another way to show PSD is by showing that the determinants of all the principle minors are +ve, and therefore all the pivots are non-negative, and there is a recursive argument on determinants of an $N\times N$ matrix, but this particular route seems very unwieldy.  However, if this is the way to go, please could you provide details?  Similarly, the are plenty of other ways of showing PSD, but I have abandoned them as unweidly (for example, the Cholesky decomposition).  Thank you

答:
It seems the best way is to go via the Cholesky decomposition, as given by this post here

How to prove the existence and uniqueness of Cholesky decomposition?


问:
Can anyone help me finding recurrence relation in combinatoric?
Guys, I am having trouble finding recurrent relation.

A codeword is made up of the digits $0,1,2,3$ (order is important!). A codeword is deﬁned as legitimate if and only if it has an even number of $0$’s. Let $a_n$ be the number of legitimate codewords of lenth $n$. Write a recurrence for $a_n$.

The answer is $a_n = 3a_{n-1} + 4^{n-1} - a_{n-1}$

I know where $a_n=3a_{n-1}$ came from, but the problem is, I can't figure out where the
rests ($4^{n-1} - a_{n-1}$) came from!

Please can anyone tell me how to approach the question?

Thank you!

答:
So lets look at a word $w = w_1 \ldots w_n$ of length $n$ with $w_i \in \{0,1,2,3\}$ and denote by $w' = w_2 \ldots w_n$ the word consisting of all but the first letter. $w$ is legimate in the following cases:

$w_1 \ne 0$ and $w'$ is legitimate (as then the number of $0$s in $w$ equals that in $w'$. There are 3 possiblities for $w_1$ and $a_{n-1}$ for $w'$, hence $3a_{n-1}$ for $w$ in this form.

$w_1 = 0$, and $w'$ is illegitimate (as then the number of $0$s in $w'$ has to be odd) there are $4^{n-1}$ wors of 4 letters with length $n-1$, of which $a_{n-1}$ are legitimate, hence $4^{n-1}-a_{n-1}$ illegitimate. So $4^{n-1}-a_{n-1}$ possibilities for $w$ in this case.

ContentLicense: CC BY-SA 3.0


问:
Does the orientation bundle not depend on the choice of cover?
In Bott and Tu's book on page 84 they give a cover $(U_{\alpha},\phi_{\alpha})$ of some smooth manifold $M$ and define the "orientation bundle" to be the line bundle over $M$ given by transition functions sgn $J(g_{\alpha \beta})$ where $J(g_{\alpha \beta})$ denotes the Jacobian determinant of $g_{\alpha \beta} = \phi_{\alpha} \circ \phi_{\beta}^{-1}$.

Does this bundle not depend on the choice of open cover? I assume it must not as they go on to say "$M$ is orientable if and only if its orientation bundle is trivial". I also do not see how this claim follows either (they write that it follows "directly from the definition").

答:
I will sketch a way to prove this but you will have to fill in the details. They define in page 48

The transition functions $g_{\alpha \beta}: U_{\alpha} \cap U_{\beta} \rightarrow G$ satisfy the cocycle condition:
$$
g_{\alpha \beta} \cdot g_{\beta y}=g_{\alpha \gamma}
$$
Given a cocycle $\left\{g_{\alpha \beta}\right\}$ with values in $G$ we can construct a fiber bundle $E$ having $\left\{g_{\alpha \beta}\right\}$ as its transition functions by setting
$$E=\left(\coprod U_{\alpha} \times F\right) /(x, y) \sim\left(x, g_{\alpha \beta}(x) y\right)$$
for $(x, y)$ in $U_{\beta} \times F$ and $\left(x, g_{\alpha \beta}(x) y\right)$ in $\cdot U_{\alpha} \times F$

In page 54 they define equivelnce of cocycles $g_{\alpha\beta}$ and $g'_{\alpha \beta}$ if there exist a maps $\lambda_\alpha:U_\alpha \to GL(n,\mathbb{R})$ such that
$$
g_{\alpha \beta}=\lambda_{\alpha} g_{\alpha \beta}^{\prime} \lambda_{\beta}^{-1}\qquad \text { on } U_{\alpha} \cap U_{\beta}
$$
Then they give an exercise 6.2

Show that two vector bundles on $M$ are isomorphic if and only if
their cocycles relative to some open cover are equivalent.

Recall that a refinement of an open cover $\mathcal{U}=\left(U_{i}\right)_{i \in I}$ is an open cover $\mathcal{U}^{\prime}=$ $\left(U_{\alpha}^{\prime}\right)_{\alpha \in A}$ such that there exists a map $\varphi: A \rightarrow I$ with the property
$$
U_{\alpha} \subset U_{\varphi(\alpha)}, \quad \forall \alpha \in A
$$
We write this $\mathcal{U}^{\prime} \prec_{\varphi}$ U. Given a gluing cocycle $g_{i j}$ subordinated to $\mathcal{U}$ then its restriction to $\mathcal{U}^{\prime}$ is the gluing cocycle $g \mid \bullet$ defined by
$$
\left.g\right|_{\alpha \beta}=\left.g_{\varphi(\alpha) \varphi(\beta)}\right|_{U_{\alpha \beta}}
$$
You can prove using exercise 6.2 that the bundles $\left(g_{\bullet \bullet}, u, W\right)$ and $\left(g_{\bullet \bullet}^{\prime}, \mathcal{U}^{\prime}, W\right)$ are isomorphic if and only if there exist an open cover $\mathcal{V} \prec \mathcal{U}, \mathcal{U}^{\prime}$ such that the restrictions of $g$ and $g'$ to $\mathcal{V}$ are equivalent.

Let $(U_\alpha,\psi_\alpha)_{\alpha\in A}$ and $(U_{\alpha'},\psi_{\alpha'})_{\alpha'\in A'}$ be to different covers of $M$. We can take the refinement of both cover to be given by taking the intersections of the open covers and the maps of the refinement to be $\varphi(\alpha\beta')=\alpha \in A$ and $\varphi'(\alpha\beta')=\beta' \in A'$. Now we can construct the equivalence to show that both bundles are isomorphic. To construct $\lambda_{\beta \beta'}$ use the fact that the determinant is multiplicative and

$$\psi_{\alpha}\circ \psi_{\beta}^{-1}=\psi_{\alpha}\circ \psi^{-1}_{\alpha'}\circ \psi_{\alpha'}\circ\psi^{-1}_{\beta'}\circ \psi_{\beta'}\circ \psi_{\beta}^{-1}$$
so we will get $\lambda_{\beta \beta'}=sgn J(\psi_{\beta'}\circ \psi_{\beta}^{-1})$.

for the statement "M is orientable if and only if its orientation bundle is trivial". Assume you have an orientation you can choose the transition maps to be constant $1$ and this will be equivalent to the trivial line bundle by choosing the $\lambda$'s to be constant. In the other direction you can multiply the first coordinate of the maps $\psi_{\alpha}^{-1 }$ with the $\lambda_{\alpha}$ to get that the sign will be constant as
$$1=\lambda_{\alpha}\cdot \lambda_{\beta}\cdot g_{\alpha \beta}$$
by the equivalence condition.

ContentLicense: CC BY-SA 4.0


问:
Laplace transformations and Convolutions
This question is question number 37 of Apostol's Mathematical Analysis.

Show that the convolution $h=f\ast g$ assumes the form $h(t)=\int_{0}^{t} f(x) g(t-x) \;\mathrm dx$ where both $f$ and $g$ vanish on the negative real axis. Use convolution theorem for Fourier Transforms to prove that $\mathcal L(f*g)= \mathcal L(f)\cdot \mathcal L(g)$.

I proved that $h=f\ast g$ assumes the form $h(t)=\int_{0}^{t} f(x) g(t-x) \;\mathrm dx$. but I am unable to get ideas on how to prove that $\mathcal L(f\ast g)= \mathcal L(f)\cdot \mathcal L(g)$.

Here LHS will be given by $\int_{0}^{\infty} e^{-zt} \int_{0}^{t} f(x) g(t-x) \;\mathrm dx \;\mathrm dt $ but how to prove it as given in the RHS? Any hints.

答: IIRC, change of variables are very useful in this proof, i.e. $y=t-x$ and change the order of integration (consider why you're allowed to do this).


问:
Best strategy to optimize probability of winning in card drawing game with opponent
I am curious about the solution to a probability question that was asked in a trading interview:

You and your opponent choose a suit (of a standard 52 cards deck). Then one card after another is drawn (without replacement) until either your suit or your opponents suit have appeared 5 times (not necessarily in a row). You win if your suit is the one which was drawn 5 times first. Now, your opponent lets you decide if you want to choose your suit before the game starts, after 1, or after 2 cards are revealed and he will choose his suit afterwards. Which strategy gives you the best chances to win?

So as an example, I decide to choose after the first card, it comes hearts, I choose hearts, he chooses any of the other suits and from that point I will only need 4 more hearts whereas he still needs 5 of his suit.

This makes it obvious that choosing before the game starts is not optimal, as you get an advantage by choosing the suit of the first card.

But I am not sure how to determine whether choosing after the first or after the second card is better. In the scenario where I choose after the second there are two possibilities: Either the first 2 cards have the same suit, then I will only need 3 more or they have a different suit. In that case I choose one of the two suits and my opponent chooses the other and our chances will be equal again.

To summarize, my chances of winning if I choose after the second card are
$$ P(\textrm{first 2 cards have same suit}) P(\textrm{3 out of 11 before 5 out of 13}) \\
+ \frac{1}{2}P(\textrm{first 2 cards have different suit}) \\
= \frac{12}{51}P(\textrm{3 out of 11 before 5 out of 13}) + \frac{1}{2}\frac{39}{51}
$$
And if I choose after the first card my chances are simply
$$
P(\textrm{4 out of 12 before 5 out of 13})
$$

But in both cases I have no idea how to come up with solutions for the missing probabilities, especially because the game is played without replacement and I can not use a binomial distribution approach as it were possible if the game was for example played with coin tosses and you can choose head or tails.

答:
Update:
Notice that if we had only two suits to begin with, the two strategies would yield the same winning probability. So in effect, having two more suits diluted the weight of $p$ in $P_2$.

There's little calculation needed.

Denote $P_i$: probability you win if you choose suit after seeing $i$ card(s).

Denote $p$: probability you win if you have already seen two cards of your suit, and no cards from opponent's suit. Then $p>\frac 12$.

And

$$P_1 = \frac{12}{25}p+\frac{13}{25}\cdot \frac 12\\
P_2 = \frac {12}{51} p + \frac{39}{51} \cdot \frac 12\\
\implies P_1 - P_2 = \left(\frac{12}{25}- \frac{12}{51} \right) p + \frac 12 \left( \frac{13}{25}-\frac{39}{51}\right)\\
> \left(\frac{12}{25}- \frac{12}{51} \right) \frac 12 + \frac 12 \left( \frac{13}{25}-\frac{39}{51}\right) = 0. \blacksquare
$$


问:
Proving that for $f\geq0$ on $X$, $\int_X f d\mu = 0$ iff $f = 0$ a.e.
Okay, so the question is the following:
Suppose $f \geq 0$ is a measurable function on the measure space $(X,\Sigma,\mu)$. Prove that

\begin{align} \int_X f d\mu = 0 \text{ if and only if } f = 0 \text{ almost everywhere.} \end{align}

I've sort of finished the proof, but my version is not very elegant or simple. I was wondering if there is a simple proof of the statement. It's meant to be using only the basics of Lebesgue integration, i.e., simple functions, etc. Thanks!

答:
The (almost) only fact needed is

Fact:  If $0\le f\le g$, then $\int f\le \int g$.

One direction of your assertion is easy:

Suppose $f$ is nonnegative and that the set
$E=\{ x : f(x)>0\}$ has measure zero. Then we have, using the Fact: $$0\le\int f\le\int \infty\cdot\chi_E=\infty\cdot\mu(E)=0.$$

For the other direction, prove the contrapositive:

Assume the set $E=\{ x : f(x)>0\}$ has positive measure. We proceed as suggested by my comment above.  For $n$ a positive integer, define the set $E_n=\{x: f(x)>1/n \}$. Note that

$\ \ \ 1)$ $E=\bigcup\limits_{n=1}^\infty E_n$

and

$\ \ \ 2)$  $E_1\subset E_2\subset E_3\subset\cdots\,$.

From $1)$ and $2)$ (and a result referenced by the "almost" of the first paragraph) it follows that $\mu(E)=\lim\limits_{n\rightarrow\infty} E_n$. Consequently, it follows from our assumption that $\mu(E)>0$  that there is some $n$ with $\mu(E_n)>0$.

So, with this in hand, using the Fact again, we have:
$$
\int f \ge \int_{E_n} f\ge \textstyle{1\over n}\cdot\mu(E_n)>0.
$$

 Whether or not this is "elegant" is debatable ...


问:
Exercise 6, section 2.7 of Hammack's Book of Proof
I have two questions about exercise 6 of section 2.7 of  Book of Proof from Hammack.

The exercise asks the reader to write the following statement in English and then to say if the statement is true or false.

$$ \exists \ n \in \mathbb{N}, \forall X \in \mathcal{P}(\mathbb{N}), \vert X \vert < n $$

My first question is whether or not I am reading this correctly.
I interpret this statement as saying "there exists a positive integer greater than the cardinality of all subsets of $\mathbb{N}$". Is that correct?

My second question, assuming my understanding of the statement is correct, is whether the statement is true or not.
My fist line of thought was that the statement is false because for any positive integer $n$ there will always be a subset of $\mathbb{N}$ with cardinality equal or larger than $n$ since the set of natural numbers is infinitely large. However, the converse reasoning seems as valid: for any subset of $\mathbb{N}$, however large it is, there will always exist a positive integer larger than this subset's cardinality. It seems to me now that we cannot really say if the statement is true or false: the statement itself is sort of paradoxical. Would this assessment be the correct answer? Am I missing something here?

答:
My fist line of thought was that the statement is false because for any positive integer $n$ there will always be a subset of $\mathbb{N}$ with cardinality equal or larger than $n$ since the set of natural numbers is infinitely large.

Yes, pretty much ... but you want to be a little more explicit/concrete about exactly what subset you can always pick such that its cardinality is not smaller than $n$ for whatever number $n$ you may have picked initially. But it is easy to see what works: you can pick any infinite-sized subset, such as all even numbers ... or even just simply $\mathbb{N}$ itself ... which is also an element of $\mathcal{P}(\mathbb{N})$

However, the converse reasoning seems as valid: for any subset of $\mathbb{N}$, however large it is, there will always exist a positive integer larger than this subset's cardinality.

First of all, you are now considering the truth of the statement:

$$ \forall X \in \mathcal{P}(\mathbb{N}), \exists \ n \in \mathbb{N}, \vert X \vert < n $$

which is a different statement. But, like the original one, this one is also false. Again, you can use any infinite-sized subset as a counterexample.


问:
Continuity of $x^2$ at $x=1$
In proving that $x^2$ is continuous at $x=1$ (or any non-zero real number), is the following sufficient as a proof?

$f(x)$ is continuous at $x=1$ if $\forall \varepsilon > 0 \exists \delta > 0 : |x-1|<\delta \implies |x^2-1| < \varepsilon$.

Pick $\delta = \sqrt{\varepsilon + 1}-1$.

Then if $|x-1| < \delta$,  $|x^2-1| = |x-1||x+1| = |x-1||(x-1)+2| <|x-1|(|x-1|+2) < \delta(\delta+2) = \varepsilon$

If this is correct, why do you most commonly see proofs of this involving taking $\delta$ to be the minimum of 1 and something else? Is that considered simpler?

答:
One common misconception in the $\epsilon$-$\delta$ proofs for beginners is that they try to find $\delta$ by solving inequalities of the form $g(\delta;x,y,\cdots)<\epsilon$ directly.

This works for some simple cases but it usually makes things unnecessarily complicated and sometimes undoable. (Imagine how you would give an $\epsilon$-$\delta$ proof for continuity of $f(x)=x^3+x^2+3x+5$ at $x=1$. (:)

For your particular case, just look at the goal: $|x^2-1|<\epsilon$. This is equivalent to
$$
|x-1||x+1|<\epsilon\tag{1}
$$

The estimate (1) tells you that if $x$ is close to $1$, then you can make the whole quantity on the left small since $|x-1|$ would be small and $|x+1|$ is bounded.

Say you first restrict that $\delta\le1$, then if $|x-1|<\delta$, you have $$|x+1|<|x-1+2|\le |x-1|+2<3.$$

Now if you make further $\delta$ smaller so that
$$
\delta\cdot 3\le\epsilon\tag{2}
$$
you get the desired estimate in (1).

To wrap things up, you can set $\delta=\min(1,\epsilon/3)$.

Title: Continuity of $x^2$ at $x=1$


问:
A discrete random variable question

The probability distribution of the discrete random variable $X$ is
$$
\mathbb{P}[X=x] = \begin{cases}
k/x,    & x \in \{1,2,3\}\\
m/(2x), & x \in \{6,9\}\\
0,      & \text{otherwise}
\end{cases}
$$
where $k,m$ are positive constants. Given that $\mathbb{E}[X] = 3.8$, find $\mathbb{Var}(X)$.

How does one do this question? I tried but am stuck, could somebody please help?

答:
Since there are 2 unknowns, we need 2 equations to solve. First we might use the fact that $\mathbb{P}(X=x)$ is a PMF. We can use the fact that $$\sum_{i=1}^{n}\mathbb{P}(X=x_i) = 1$$
From your question it is obvious that $x_i's$ are $\{1,2,3,6,9\}\ $ for rest of them values are $0$. We get:
$$k+\frac{k}{2}+\frac{k}{3}+\frac{m}{12}+\frac{m}{18}=1$$
$$\frac{11k}{6}+\frac{5m}{36}=1$$
or
$$66k+5m=36$$
For the next part, we know that $$\mathbb{E}(X)= \sum_{i=1}^{n}x_i\mathbb{P}(X=x_i)=3.8$$
$$1\times k+2\times \frac{k}{2}+3\times \frac{k}{3} + 6\times \frac{m}{2\times6}+ 9\times \frac{m}{2\times 9}=3.8$$
$$3k+m=3.8$$
Solving both equations you get, $k=\frac{1}{3},m=\frac{14}{5}$
To calculate the variance, we calculate:
$$\mathbb{E}(X^2)=\sum_{i=1}^{n}x_i^2\mathbb{P}(X=x_i)=1\times k+4\times \frac{k}{2}+9\times \frac{k}{3}+ 36\times \frac{m}{12}+81\times \frac{m}{18}$$
$$\mathbb{E}(X^2) = 6k+7.5m=6\times \frac{1}{3}+ 7.5\times \frac{14}{5}=23$$
$$\mathbb{V}ar(X)=\mathbb{E}(X^2)-[\mathbb{E}(X)]^2=23-(3.8)^2=8.56$$
Hope this helps...

Title: A discrete random variable question


问:
How to divide two inequalities in the form $a - \epsilon < x < a + \epsilon$?
I have the following inequalities
$$
a - \epsilon_x < x < a + \epsilon_x \\
b - \epsilon_y < y < b + \epsilon_y
$$
with $x,y,a,b,\epsilon_x,\epsilon_y\in {\rm I\!R}$, $|\epsilon_x|>0$ and $|\epsilon_y|>0$.

How to bound $\frac{x}{y}$?

I first tried to inverse the second inequality
$$
\frac{1}{b + \epsilon_y} < \frac{1}{y} < \frac{1}{b - \epsilon_y}
$$
I am not sure if it is correct, but I then used this answer to obtain
$$
\min\left( \frac{a-\epsilon_x}{b+\epsilon_y}, \frac{a-\epsilon_x}{b-\epsilon_y}, \frac{a+\epsilon_x}{b+\epsilon_y} \right) < \frac{x}{y} < \max\left( \frac{a+\epsilon_x}{b-\epsilon_y}, \frac{a+\epsilon_x}{b+\epsilon_y}, \frac{a-\epsilon_x}{b-\epsilon_y} \right)
$$
Can I conclude that
$$
\frac{a-\epsilon_x}{b+\epsilon_y} < \frac{x}{y} < \frac{a+\epsilon_x}{b-\epsilon_y}
$$
or should I detail every special cases depending on the sign of $a$, $b$... ?

I assume that $0\notin(b-\epsilon_y, b+\epsilon_y)$.
I tried to detail the four cases

if $a > 0$ and $b > 0$ then $\frac{a-\epsilon_x}{b+\epsilon_y} < \frac{x}{y} < \frac{a+\epsilon_x}{b-\epsilon_y}$

if $a > 0$ and $b < 0$ then $\frac{a+\epsilon_x}{b+\epsilon_y} < \frac{x}{y} < \frac{a-\epsilon_x}{b-\epsilon_y}$

if $a < 0$ and $b > 0$ then $\frac{a-\epsilon_x}{b-\epsilon_y} < \frac{x}{y} < \frac{a+\epsilon_x}{b+\epsilon_y}$

if $a < 0$ and $b < 0$ then $\frac{a+\epsilon_x}{b-\epsilon_y} < \frac{x}{y} < \frac{a-\epsilon_x}{b+\epsilon_y}$

But it seems different from @mathcounterexamples.net 's answer (see below).

答:
If $0 \in (b- \epsilon_y, b + \epsilon_y)$, there is no way to bound $\frac{x}{y}$.

Otherwise you have $$\vert y \vert \gt m = \begin{cases}
b-\epsilon_y & \text{ if } b - \epsilon_y \gt 0\\
-(b + \epsilon_y) & \text{ if } b + \epsilon_y\lt 0
\end{cases}$$As $\vert x \vert \lt \vert a \vert + \epsilon_x$, you get

$$\left\vert \frac{x}{y} \right\vert \lt \frac{\vert a \vert + \epsilon_x}{m}.$$


问:
Trees are contractible.
I was studying some covering spaces and the notion of graph appeared in  a way that I've never seen before:

To define the idea of a graph, let's start with a discrete set of points $G^0$, those will be the vertices of the graph. Then let's create a set $G^1$ of spaces $J_i$ homeomorphic to $[0,1]$. This spaces will be called edges of the graph. For convention let's define that $u_i$ ($v_i$) is the point of $J_i$ that is identified with $0$ ($1$) in this homeomorphism. A graph $G$ is the set $G = G^0 \sqcup G^1 / \sim$, where $\sim$ is the relation that takes points in $\partial J_i$ for all $i$ and identifies them with elements of $G^0$. The topology on $G$ will be the one that satisfies $$A \subset G \textrm{ is open in } G \iff A \cup J_i \textrm{ is open in } J_i \textrm{ for all } i.$$

Almost immediately the text introduces the notion of connected graphs and circuits.

A graph is connected if for any two vertices $g,h \in G^0$ there are a finite number of distinct edges $J_1, \dots, J_n$ such that $g \sim u_1$ and $h \sim v_n$. A circuit in a graph $G$ is a finite set of distinct edges $J_1, \dots, J_n$ such that $v_i \sim v_{i+1}$ for all $i < n$ and $v_n \sim u_1$.

Fianlly, the definition of a tree

A tree is a graph $G$ that is connected and contains no circuits.

The main question is: how to prove that every tree is contractible? If the tree is finite, the proof follows easily from the fact that every tree has a free vertex (that is, a vertex that is connected to only one edge), I'm having a lot of trouble in coming with a proof for the infinite case.

答:
I suppose you can just take any point, "connect" it to all other points and then contract all paths simultaneously.

That is, take a point $a$ and denote by $ab$ the only path (a sequence of edges plus the part of the last edge up to $b$) connecting $a$ and some other point $b$. You have a well-defined length for this path and you can shorten this path by the factor $t, t \in [0, 1]$. Set $f_t:G \to G$ by $b \mapsto end(t \cdot ab)$. This is a well-defined homotopy connecting $id$ and the map to $a$.

Title: Trees are contractible.


答: Normally you would assume that the epsilons are small enough not to make the sums and differences cross zero.  You need to specify that in your answer.  Then, yes, you do need to worry about the signs of $a,b$.  Your result is correct if they are both positive but not otherwise.  For most purposes you are interested in bounding the absolute value of $\frac xy$.  If that is true, you can just take absolute values of $a,b$ and be done.

Title: How to divide two inequalities in the form $a - \epsilon < x < a + \epsilon$?

Body:
I have the following inequalities
$$
a - \epsilon_x < x < a + \epsilon_x \\
b - \epsilon_y < y < b + \epsilon_y
$$
with $x,y,a,b,\epsilon_x,\epsilon_y\in {\rm I\!R}$, $|\epsilon_x|>0$ and $|\epsilon_y|>0$.

How to bound $\frac{x}{y}$?

I first tried to inverse the second inequality
$$
\frac{1}{b + \epsilon_y} < \frac{1}{y} < \frac{1}{b - \epsilon_y}
$$
I am not sure if it is correct, but I then used this answer to obtain
$$
\min\left( \frac{a-\epsilon_x}{b+\epsilon_y}, \frac{a-\epsilon_x}{b-\epsilon_y}, \frac{a+\epsilon_x}{b+\epsilon_y} \right) < \frac{x}{y} < \max\left( \frac{a+\epsilon_x}{b-\epsilon_y}, \frac{a+\epsilon_x}{b+\epsilon_y}, \frac{a-\epsilon_x}{b-\epsilon_y} \right)
$$
Can I conclude that
$$
\frac{a-\epsilon_x}{b+\epsilon_y} < \frac{x}{y} < \frac{a+\epsilon_x}{b-\epsilon_y}
$$
or should I detail every special cases depending on the sign of $a$, $b$... ?

I assume that $0\notin(b-\epsilon_y, b+\epsilon_y)$.
I tried to detail the four cases

if $a > 0$ and $b > 0$ then $\frac{a-\epsilon_x}{b+\epsilon_y} < \frac{x}{y} < \frac{a+\epsilon_x}{b-\epsilon_y}$

if $a > 0$ and $b < 0$ then $\frac{a+\epsilon_x}{b+\epsilon_y} < \frac{x}{y} < \frac{a-\epsilon_x}{b-\epsilon_y}$

if $a < 0$ and $b > 0$ then $\frac{a-\epsilon_x}{b-\epsilon_y} < \frac{x}{y} < \frac{a+\epsilon_x}{b+\epsilon_y}$

if $a < 0$ and $b < 0$ then $\frac{a+\epsilon_x}{b-\epsilon_y} < \frac{x}{y} < \frac{a-\epsilon_x}{b+\epsilon_y}$

But it seems different from @mathcounterexamples.net 's answer (see below).


问:
Permutation with Grouped Objects and Repetitions
I have letters AAABCD. The letters A and B must be together. How many ways can I arrange these 6 letters?

At first, I thought the answer would be to group one of the A's and B together, and treat it as one object, so that's 5! ways of arranging {A,B} A A C D. Then divided by 2! to account for the repeated A's outside of the group then multiplied by 2! to account for the number of ways one can arrange the A and B inside the group.

However, I later realized that certain permutations are in fact duplicates, e.g. {A,B}AACD and A{B,A}ACD and the number of duplicates seem to be n! where n is the number of repeated objects, however I'm not sure how to prove this.

答:
The duplicates can be excluded by counting the permutations of four symbols : $ABA$, $A$, $C$, $D$ which is simply $4!$

Desired answer is $5!-4!=96$


问:
Find x for which for every "a" the equation has solution
$$a^{31x} \equiv a \mod 271$$

I need to find x variable, for which the equation has solution with any a. How can I do this?

Generaly, modular equations have solutions when $GCD(a^{31x}, 271) = 1$, or $GCD(a^{31x}, 271) = d > 1; d|a$

It also looks like I could use Little Fermat's Theorem...

But I can't come up with anything...

Body:
If $271|a,271|(a^{31x}-a)$ for integer $x>0$

Else $(271,a)=1\implies 271|(a^{31x}-a)\iff 271(a^{31x-1}-1)$

Now as $271$ is prime, it has at least one primitive root,

the highest $ord_{271}(a)$ will be $271-1=270$ and the order of other numbers will be divisor of $270$

$\implies 270$ must divide $31x-1\iff 31x\equiv1\pmod {270}$

Now, $$\frac{270}{31}=8+\frac{22}{31}=8+\frac1{\frac{31}{22}}=8+\frac1{1+\frac9{22}}$$
$$=8+\frac1{\frac{31}{22}}=8+\frac1{1+\frac1{\frac{22}9}}=8+\frac1{1+\frac1{2+\frac49}}=8+\frac1{1+\frac1{2+\frac1{\frac94}}} =8+\frac1{1+\frac1{2+\frac1{2+\frac14}}}$$

So, the previous convergent of $\frac{270}{31}$ is $$8+\frac1{1+\frac1{2+\frac1{2}}}=\frac{61}7$$

Using Theorem $3$ of this,
$$ 31\cdot 61 -270\cdot7=1\implies 31\cdot61\equiv1\pmod{270} $$

$$\implies x\equiv61\pmod{270}$$


问:
Trying to find a generating function for the PDF $f(x) = 6(1 - x)x$
I need to find the generating function for this PDF:

$$f(x) = 6(1 - x)x$$

I know that the generating function for $f(x)$ is $\frac{12}{t^3} - \frac{12e^t}{t^3} + \frac{6}{t^2} + \frac{6e^t}{t^2}$

I also know that I need to integrate $e^{tx} \cdot f(x)$ on $[0, 1]$ in order to find this function.

I have been given a tip that I can calculate $\int_0^1 x \cdot e^{tx}\, dx$ and $\int_0^1 x^2 \cdot e^{tx}\, dx$ by calculating $\int_0^1 e^{tx}\, dx$ and taking the derivative of the result twice.

When I do this, I get $$\frac{e^t}{t} - \frac{e^t}{t^2} + \frac{2e^t}{t^3} - \frac{2}{t^3}.$$ When I multiply this by $6$, the result is close to what it's supposed to be, but not quite.

Any advice for how to fix this? Thanks in advance.

Body:
Integration by parts $$I(x)=\int x e^{t x} \, dx=x\left(\frac{e^{t x}}{t}\right)-\int \frac{e^{t x}}{t}\,dx= \frac{x e^{t x}}{t}-\frac{e^{t x}}{t^2}+C$$
and
$$\int x^2 e^{t x} \, dx=x^2\left(\frac{e^{t x}}{t}\right)-\int (2x)\left(\frac{e^{t x}}{t}\right)\,dx=\frac{x^2e^{t x}}{t}-\frac{2}{t}\int x e^{t x} \, dx=\frac{x^2e^{t x}}{t}-\frac{2}{t}I(x)=$$
$$=\frac{x^2e^{t x}}{t}+\frac{2x e^{t x}}{t^2}+\frac{2e^{t x}}{t^3}+C$$
And finally
$$6\int_0^1 (1-x) x e^{t x} \, dx=6\left[-\frac{2 e^{t x}}{t^3}+\frac{2 x e^{t x}}{t^2}-\frac{e^{t x}}{t^2}-\frac{x^2 e^{t x}}{t}+\frac{x e^{t x}}{t}\right]_0^1=$$
$$=\frac{12}{t^3}-\frac{12 e^t}{t^3}+\frac{6 e^t}{t^2}+\frac{6}{t^2}$$

Hope this helps


问:
Prove that $x \leq \frac{a x+b y}{a +b} \leq y$ whenever $x \leq y$.
Prove that $$\forall x,y\in \mathbb{R}, \forall a, b \in \mathbb{R}_{> 0}, \, x \leq y \Longleftrightarrow x \leq \frac{a x+b y}{a+b} \leq y \, \text{.}$$

答: Hint: simplify $\frac{ax+by}{a+b}-x,\,y-\frac{ax+by}{a+b}$, then use $a+b>0$.


答:
\begin{align*}
x = \frac{ax + bx}{a+b} \leq \frac{ax + by}{a + b} \leq \frac{ay + by}{a + b} = y
\end{align*}

Title: Prove that $x \leq \frac{a x+b y}{a +b} \leq y$ whenever $x \leq y$.

Body: Prove that $$\forall x,y\in \mathbb{R}, \forall a, b \in \mathbb{R}_{> 0}, \, x \leq y \Longleftrightarrow x \leq \frac{a x+b y}{a+b} \leq y \, \text{.}$$


问:
Finding Pythagorean triple where the squares of sides are weighted
Suppose $a, b, c \in \mathbb{Z}_{>0}$. The following is a variation of a Pythagorean triple, but with weighted squares:

$$a^2 + 3 b^2 = 4 c^2$$

Can I find something like Euclid's formula that can generate all solutions for this formula? I'm not sure how to approach this problem.

答:
After discovering Find all integers satisfying $m^2=n_1^2+n_1n_2+n_2^2$, the paper Abdelalim, S., & Dyani, H. (2014). The Solution of the Diophantine Equation x2+ 3y2= z2. International Journal of Algebra, 8(15), 729–732. gives an easy answer:

Theorem 2.1 Let $E : x^2 + 3 y^2 = z^2$ diophantine equation and $(x, y, z) \in \mathbb{Z}^3$ with $x \wedge y = 1$, $y$ is even and $x z \wedge 3 = 1$. Then the following properties are equivalent:

$(x, y, z)$ is the solution of $E$.

$| z | = 3 y_1^2 + y_2^2, | x | = 3 y_1^2 - y_2^2, | y | = 2 y_1 y_2$ with $y_1 \wedge y_2 = 1$.

Theorem 2.2 Let $E : x^2 + 3 y^2 = z^2$ diophantine equation and $(x, y, z) \in \mathbb{Z}^3$ with $x \wedge y = 1$, $y$ is odd and $x z \wedge 3 = 1$. Then the following properties are equivalent:

$(x, y, z)$ is the solution of $E$.

$| z | = \frac{3 y_1^2 + y_2^2}{2}, | x | = \frac{3 y_1^2 - y_2^2}{2},
  | y | = y_1 y_2$ with $y_1 \wedge y_2 = 1$.

Scaling the above primitive solutions $(x, y, z)$, and in our case restricting to $2 \mid z$, gives all solutions.

The following is a method that is less explicit, as it requires dividing by a $\gcd$.

Thanks to Daniel Hast's comment, I arrived at the following:

Finding such $(a, b)$ is equivalent to finding a single rational point on the ellipse $x^2 + 3 y^2 = 4$, then finding all rational lines through it. A rational point is $(1, 1)$, and a rational line through that is $t y = s x - s + t$ where $\gcd (s, t) = 1$. The other intersection with the ellipse is at:

$$ (x, y) = \left( \frac{3 s^2 - 6 s t - t^2}{3 s^2 + t^2}, - \frac{3 s^2 + 2 s t - t^2}{3 s^2 + t^2} \right) $$

So:

$$ \begin{array}{rll}
 \left( \frac{3 s^2 - 6 s t - t^2}{3 s^2 + t^2} \right)^2 + 3 \left(
 \frac{3 s^2 + 2 s t - t^2}{3 s^2 + t^2} \right)^2 & = & 4\\
 (3 s^2 - 6 s t - t^2)^2 + 3 (3 s^2 + 2 s t - t^2)^2 & = & 4 (3 s^2 +
 t^2)^2\\
 4 (3 s^2 + t^2)^2 - 3 (3 s^2 + 2 s t - t^2)^2 & = & (3 s^2 - 6 s t -
 t^2)^2
\end{array} $$

Note that the numerator and denominators of $x, y$ can be scaled while the
rational point stays the same. Let $d
= \gcd (3 s^2 + t^2, 3 s^2 + 2 s t - t^2)$. For $k \in \mathbb{Z}$, our
solutions are then:

$$ a = \frac{k}{d} \cdot (3 s^2 + 2 s t - t^2) \quad b = \frac{k}{d} \cdot (3 s^2 + t^2) $$


问:
Integer solutions to $r\cdot a^2+t=s\cdot b^2$ forms sequences with $a_{n+1}\cdot b_n-a_n\cdot b_{n+1}$ is invariant
Conjecture:

Given natural numbers $r,s,t$. Suppose there are an infinite number of solutions $a_n,b_n \in \mathbb N$ to $ra_n^2+t=sb_n^2$, where $a_n, b_n$ correspond to increasing sequences of solutions, then $a_n\cdot b_{n+1} - a_{n+1}\cdot b_n$ is invariant equal for each $n$ greater than some $N>0$.

I've done a lot of tests without exceptions. The number $N$ is typically $\le 1$. I would like to see ideas how to proceed.

Some $(r,s,t)$ values plus sets of $(a,b)$ values:

```<code>(1,2,1)
{(1,1),(7,5),(41,29),(239,169),(1393,985),(8119,5741)}

(2,1,1)
{(0,1),(2,3),(12,17),(70,99),(408,577),(2378,3363)}

(1,2,2)
{(0,1),(4,3),(24,17),(140,99),(816,577),(4756,3363)}

(3,2,2)
{(0,1),(4,5),(40,49),(396,485),(3920,4801)}

(2,3,1)
{(1,1),(11,9),(109,89),(1079,881),(10681,8721)}
</code>```

There seems to be a matrix

$
\begin{pmatrix}
\alpha & ms \\
mr & \alpha \\
\end{pmatrix}
$

that transform $(a_{n},b_{n})$ to $(a_{n+1},b_{n+1})$, but I haven't proved that and I'm insecure about $\alpha$ and $m$. It could be that $\alpha=\sqrt{m^2rs+1}$ and that the determinant is unital.

The conjecture is false with counter-example

```<code>(1,2,7)
{(1,2),(5,4),(11,8),(31,22),(65,46),(181,128),(379,268)}
</code>```

What's left is

Given $r,s,t\in\mathbb R_+$ and $S=\{(a,b)\in\mathbb R_+^2|rs^2+t=sb^2\}$, then it exist a sequence $(a_n,b_n)\in S$ such that:

$$a_{n+1}b_n-a_nb_{n+1}=2t \tag 1$$
$$
\begin{pmatrix}
a_{n+1} \\
b_{n+1} \\
\end{pmatrix}
=
\begin{pmatrix}
\sqrt{4rs+1}& 2s \\
2r & \sqrt{4rs+1} \\
\end{pmatrix}
\begin{pmatrix}
a_n\\
b_n \\
\end{pmatrix}
\tag 2$$
which I consider as virtually proved by the accepted answer.

答:
Let $a,b$ solve the equations and let $\begin{pmatrix}
\alpha & s \\
r & \alpha \\
\end{pmatrix}\begin{pmatrix}a \\b\\\end{pmatrix}=\begin{pmatrix}A \\B\\\end{pmatrix}$ with $\alpha$ as in your post and $m=1$.

Then straightforward algebra shows that $A,B$ also solve the equation.

Then $Ab-aB=(\alpha a+sb)b-a(ra+\alpha b)=sb^2-ra^2=t$, as you require.

Note. The above proof does not assume anything about the integer nature of the parameters. One can multiply each of $r,s,t$ by any number $m$ say and obtain the corresponding result. In that case the invariant value will of course be $mt$.

I assume that is what you you are interested in and that you will wish to choose $m$ so that $(m,\alpha)$ is an integer solution of the Pell equation $$(rs)x^2+1=y^2.$$

Furthermore, so as not to 'miss' solutions, $(m,\alpha)$ should be the fundamental solution.


问:
The normalizer of a proper sub-algebra properly contains the sub-algebra in a nilpotent Lie algebra.
So I am just making my way into the theory of Lie Algebras. The question at hand comes from page 14 of Humphreys, Introduction to Lie Algebra and Representation Theory.

Given a finite dimensional nilpotent Lie algebra $L$, and a proper sub-algebra $K$ of $L$, prove that the normalizer of $K$ properly contains $K$.

I keep trying to use Engel's theorem, but I have only been able to show there is a vector such that $[v\ K]$ is in the center, which much be in $K$. Though the condition for $v$ was only that it is in $L$.

A proof or anything else would be helpful. Thank you.

答:
Here is my comment as an answer:

We do this by induction on the length of the ascending central series (which is finite since $L$ is nilpotent).

If $K$ does not contain $Z(L)$ then we are done, as clearly $Z(L)$ normalizes $K$. So we assume that $Z(L)\subseteq K$ and look at the quotient $L/Z(L)$ which is nilpotent and has a shorter ascending central series than $L$.

$K$ corresponds to a subalgebra $\overline{K}$ of $L/Z(L)$ since $Z(L)\subseteq K$, and by induction, the normalizer of $\overline{K}$ strictly contains $\overline{K}$. But anything in the preimage of this normalizer in $L$ will normalize $K$, and we are done.


问:
If $\sum a_n$ is convergent , then $\sum {a_n}^2$ convergent.
I found proofs telling that if $\sum a_n$ with $a_n>0$ is convergent (hence absolutely), then  $\sum {a_n}^2$ always convergent.

But what happens if we do not suppose that $a_n>0$?

My guess is that as $a_n \to 0$, $\exists n_0$ such that $\forall n \geq n_0, -\frac{1}{2} \leq a_n \leq \frac{1}{2} \implies 0 \leq a_{n}^2 \leq \frac{1}{2}a_n < a_n$ and conclude with comparison test. Is it right ?

答: counterexample: $a_n=(-1)^n \cdot \frac{1}{\sqrt{n}}$


问:
If H, H' are subgroups of G and $H \bigcup H'$ is also a subgroup of G, then it holds that $H \subset H' \lor H' \subset H$.
I've come up with this proof but I'm not sure that the last step is correct, where I state that the last case is impossible.

Statement:

If H, H' are subgroups of G and $H \bigcup H'$ is also a subgroup of G, then it holds that $H \subset H' \lor H' \subset H$.

Proof:

Assume that $H \not \subset H'$ (proof for the case $H' \not \subset H$ is identical to the one below).

Let's take $a \in H \setminus H'$, implying by the definition of a subgroup that in this case also $a^{-1} \in H \setminus H'$.

Since H, H' are subgroups of G then it also holds that $e \in H \bigcap H' \subset H \bigcup H' \subset G$.

Then for any $b \in H'$ it'd hold that $b = be = b(aa^{-1}) = (ba)a^{-1}$. Since $a^{-1} \in H$, it is only left to show that also $(ba) \in H$.

Since $b,a \in H \bigcup H'$, then $ba \in H \bigcup H'$. So there are these cases to check:

$$
1) \ ba \in H \land ba \not \in H'\\
2) \ ba \in H \land ba \in H' \\
3) \ ba \not \in H \land ba \in H' \\
$$

The first two cases correspond to $ba \in H$. Hence it is only required to check the last case.
That case is, however, not possible, since $a \not \in H'$. Therefore, it holds that since $b = (ba)a^{-1}$, then $b \in H$, so it follows that $H' \subset H$.

$ \blacksquare $

答:
The proof is correct. You have $b\in H'$ and $ba\in H'$, therefore $a=b^{-1}ba\in H'$ which is a contradiction.

But your proof can be modified shorter.

Suppose $a\in H\backslash H'$. Take arbitary $b\in H'$. Since $a\notin H'$ and $a=abb^{-1}$, $ab$ can't be in $H'$. Therefore $ab\in H$ and $b=a^{-1}ab\in H$.


问:
Transfer function of a state space with correlated noises
Assume a standard MIMO state-space
\begin{align}
x_{i+1}&= Fx_i + Gw_i\\
y_i&= Hx_i + v_i,
\end{align}
where the sequence $(w_i,v_i)$ is Gaussian i.i.d. with $\mathbb{E}[v_iw_i^T] = L$.

What is the corresponding transfer function?

This takes me to my main question whether we can "approximate" any continuous transfer function with a state-space with uncorrelated noise, i.e., $L=0$.

答:
If we disregard the fact that the sequence $(w_i, v_i)$ is Gaussian or correlated.
Then by considering $(w_i, v_i)$ as the input of the State-Space, one could rewrite it to the following:
$$x_{i+1} = Fx_i + \begin{bmatrix}G & 0\end{bmatrix}\begin{bmatrix}w_i \\ v_i\end{bmatrix}$$
$$y_{i} = Hx_i + \begin{bmatrix}0 & I\end{bmatrix}\begin{bmatrix}w_i \\ v_i\end{bmatrix}$$

Which looks like a regular discrete-time state-space model. Therefore, the corresponding Transfer function would be:
$$P(z) = \frac{Y(z)}{U(z)} = H(zI-F)^{-1}\begin{bmatrix}G & 0\end{bmatrix} +  \begin{bmatrix}0 & I\end{bmatrix}$$
$$Y(z) = H(zI-F)^{-1}G W(z) + V(z)$$

Since I am not using any probabilistic properties, the fact that they are correlated is not important. This can however become troublesome if you try to control either $w$ or $v$, as the other one will change as well due to this correlation. With that, I should also note that this is also problematic if you try to identify a system with this structure, then we have to use alternative techniques (depending on what you can measure of course).


问:
Show $\dfrac{2nm}{4n^2+m^2}$ is bounded
Show $\dfrac{2nm}{4n^2+m^2}$ is bounded and determine its supremum and infimum, where $n$ and $m$ are natural numbers.  I'm not to sure where to start.

答:
Hint

For upper bound, check two cases $m\le n$ and $m>n$. The infimum is straightforward.

Title: Show $\dfrac{2nm}{4n^2+m^2}$ is bounded


问:
Line $L$ is tangent to $f(x)=x^4+2x^3−11x^2−13x+35$ at $x=p$ and $x=q$. Find the equation of $L$.

Question: Line $L$ is tangent to $f(x)=x^4+2x^3−11x^2−13x+35$ at $x=p$ and $x=q$. Find the equation of $L$.

I'm not allowed to graph or use a calculator. I believe I am allowed to assume for this question that both tangents are the same line, and that $x=p$ and $x=q$ are the turning points. (Checking Desmos confirms this, and that $L= -x-1$)

I got that $f(x) - L = (x-p)^2(x-q)^2$, but I'm not sure how I would find $p$ or $q$ without graphing, and I'm pretty sure I need that to solve for $L$. Can someone point me in the right direction? I'm pretty sure this has an easy solution, but I just can't see it.

答:
$$\begin{array}{rcl}(x-p)^2(x-q)^2&=&(x^2-2px+p^2)(x^2-2qx+q^2)\\&=&x^4-2(p+q)x^3+(p^2+4pq+q^2)x^2+(\ldots)x+(\ldots)\end{array}$$

So, if we can fit $p$, $q$ so that the coefficients of $2$nd, $3$rd and $4$th degree fit, then $L$ will do the rest of the job, as it is linear - it can be tailored to produce the correct coefficient of degree $0$ and $1$.

Thus, you only need to solve:

$$p+q=-1$$
$$p^2+4pq+q^2=-11$$

From the second equation you get that $(p+q)^2+2pq=-11$, i.e. $pq=-6$, and then Vieta formulas tell you that $p,q$ are the solutions of the equation $x^2+x-6=0$, i.e. $p,q=2,-3$ in some order.

Now, knowing $p$ and $q$, you can easily determine the equation for the line $L$.


问:
A sequence of Operators converging to the Range projection

The book in question is "C* algebras and their automorphism groups" from Petersen et al. The question is simple and I imagine it should be easy, as they do not say much about that, but I am having difficulties proving that the sequence, for $x\geq 0$, $(1/n+x)^{-1}x$ converges (I suppose they mean strongly) to $[x]$, where $[x]$ is the range projection (it is defined in the picture).
I can see that the sequence is monotonically increasing and bounded, so it should converge strongly to an operator $y\in B(H)$, but why $[x]$?

答:
The functions $f_n:[0, \infty )\to {\mathbb R}$ defined by
$$
  f_n(t)=(1/n+t)^{-1}t
  $$
are bounded by 1 and converge pointwise to the characteristic function $1_{(0,\infty )}$.  Therefore, for every positive
operator $x$,
one has  by [1, Lemma (3.5.5)] that
$f_n(x)$ converges strongly to  $1_{(0,\infty )}(x)$, which turns out to be  the range projection of $x$.

[1] Sunder, V. S., Functional analysis: spectral theory, Birkhäuser Advanced Texts. Basel: Birkhäuser. ix, 241 p. (1997). ZBL0919.46002.


答: In fact, the only functions $f: \mathbb R \to \mathbb R$ such that $\sum_n f(a_n)$ is convergent whenever $\sum_n a_n$ is convergent are those that are linear in a neighbourhood of $0$.  See here

Title: If $\sum a_n$ is convergent , then $\sum {a_n}^2$ convergent.

Body:
I found proofs telling that if $\sum a_n$ with $a_n>0$ is convergent (hence absolutely), then  $\sum {a_n}^2$ always convergent.

But what happens if we do not suppose that $a_n>0$?

My guess is that as $a_n \to 0$, $\exists n_0$ such that $\forall n \geq n_0, -\frac{1}{2} \leq a_n \leq \frac{1}{2} \implies 0 \leq a_{n}^2 \leq \frac{1}{2}a_n < a_n$ and conclude with comparison test. Is it right ?


答: It may be interesting to note that $\frac{a}{a+b}$ and $\frac{b}{a+b}$ are both ratios between $0$ and $1$ , and their sum is $1$ . Using these ratios to mix $x$ and $y$ gives us a weighted average of $x$ and $y$ , which is somewhere between $x$ and $y$.


问:
Let $A \in M(n, \mathbb{R})$. Prove that for every matrix $A$ exists a polynomial $f \in \mathbb{R}[X]$, such that $f(A) = 0$.
Let $A \in M(n, \mathbb{R})$ ($n$x$n$ matrix under $\mathbb{R}$). Prove that for every matrix $A$ exists a polynomial $f \in \mathbb{R}[X]$, such that $f(A) = 0$.

Where, $f(A) = a_nA^n + ... + a_1A + a_0E_n$.

I tried to use Polynomial remainder theorem, but no more idea.

答:
We can avoid Cayley-Hamilton with the structure theorem for finitely-generated modules over a PID.

By the structure theorem, we have
$$\mathbb R^n=\bigoplus_{i=1}^m\mathbb R[x]/(f_i).$$
Now take $f=\prod_{i=1}^mf_i$, then $f$ acts by $0$ on $\mathbb R^n$, i.e. $f(A)=0$. Comparing dimensions tells us that
$$n=\dim\mathbb R^n=\sum_{i=1}^m\dim\mathbb R[x]/(f_i)=\sum_{i=1}^m\deg f_i=\deg f.$$

Here's a neat method to show that some polynomial must annihilate $A$.

Suppose not. Then $I,A,\dots,A^{n^2-1}$ are linearly independent, so form a basis for $M_n(\mathbb R)$.

Hence every matrix commutes with $A$, so $A=\lambda I$. Then take $f(x)=x-\lambda$ for a contradiction.


问:
$\mathbb{C}$ and $\mathbb{Q}(i)$ are isomorphic as vector spaces but not as fields
This question is from Thomas Hungerford algebra  Pg 241 and I was unable to solve this problem.

$\mathbb{C}$ and $\mathbb{Q}(i)$ are isomorphic as vector spaces but not as fields.

If I could prove that $\mathbb{C}$ and $\mathbb{Q}(i)$ have same degree as fields over $\mathbb{Q}$ then they would be isomorphic but the degree of $\mathbb{C}$ over $\mathbb{Q}$ is not defined and degree of $\mathbb{Q}(i)$ over $\mathbb{Q}$ is 2. So, I think I should use some other method.

Also, To disproving that they are not isomorphic as fields I thought of defining f : $\mathbb{C} \to \mathbb{Q(i)}$ and obtain some contradiction using f and some elements but failed to do so.

So, Can you help me with both these questions . Just give hints, not complete solutions.

答:
The degrees over $\Bbb Q$ are different. We have $[\Bbb Q(i):\Bbb Q]=2$ with vector space basis $\{1,i\}$ whereas the other degree is infinite: The field extension $\mathbb{C}/\mathbb{Q}$ is infinite.

Another fact is that $\Bbb C$ is algebraically closed, but $\Bbb Q(i)$ is not.

And as Gae.S. says, $\Bbb C$ and $\Bbb Q(i)$ cannot be isomorphic as vector spaces over a field because of cardinality reasons. Is there a typo? The "usual" homework is:

"$\Bbb Q(i)$ and $\Bbb Q(\sqrt{2})$ are isomorphic vector spaces, but not isomorphic fields."


答:
The expression is equivalent to

$$\frac{2q}{q^2+4}$$ where $q$ is a positive rational. We have

$$\frac{2q}{q^2+4}\le\frac q2,$$ and $$\frac{2q}{q^2+4}\le\frac2q$$ so that the expression is bounded.

As the expression is positive and zero for $q=0$, this must be the infimum.

Now the inverse of the expression can be written

$$\frac q2+\frac 2q.$$ This is the sum of a growing and a decreasing term, which achieves an extremum when the terms are equal, i.e. at $q=2$.

Title: Show $\dfrac{2nm}{4n^2+m^2}$ is bounded

Body: Show $\dfrac{2nm}{4n^2+m^2}$ is bounded and determine its supremum and infimum, where $n$ and $m$ are natural numbers.  I'm not to sure where to start.


问:
Find a rational number that is greater than $t$ where $t^2<2$
I have trouble in proving this, that is if $t\in Q\land t^2<2$, then there exists another rational number $q$ such that $q>t\land q^2<2$. May I ask how to find this rational number?

答: Let $q=t+h$ where WLOG $0<h<1$ and $q>0$. Then we want $q^2+2qh+h^2 <2$ but $h^2\leq h$ so it is enough to have $q^2 +2qh+h<2$  or $$h<\frac{2-q^2}{1+2q}$$


问:
Manipulating a symmetric sum indices
$$ \sum_{}^{} \binom{2n+1}{i}( 6 \sqrt{6})^i 14^{2n+1-i}$$

Over the sequence,

$$ i=  \{1,3,5,..,2n+1 \}$$

Suppose I do the substitution, $ 2n+1 - i = j$,

$$ \sum_{j}^{} \binom{2n+1}{j} ( 6 \sqrt{6})^{2n+1 -j} 14^{j}$$

Note that the value of $j$ we are summing over is same here but it turns out in this new sum, the power of $ (6 \sqrt{6} )$ is even..? (Note: that $2n+1-j$ is even because $2n+1$ and $j$ are odd.)

Title: Manipulating a symmetric sum indices

Body:
The answer is: $2n + 1 - j$ is odd. Because $j$ is even.

You can check writing for $n = 1$.

The first sum:
$$
\binom{3}{1}(6\sqrt{6})^1(14)^2 +
\binom{3}{1}(6\sqrt{6})^3(14)^0
$$
Then, the second sum ...
$$
\binom{3}{2}(6\sqrt{6})^{3 - 2}(14)^2 +
\binom{3}{0}(6\sqrt{6})^{3 - 0}(14)^0
$$


问:
Find $\lim_{x\to +\infty}\frac{\sin(1-x)}{2-2x}$
I have an assignment about limits:

$$\lim_{x\to +\infty}\frac{\sin(1-x)}{2-2x}$$

It tells me to calculate the value of the limit.
I observed that it comes up $\frac{\sin (-\infty)}{-\infty}$, so the sin of infinite can't exist. I was just about to solve it, when I decided to try to use Microsoft Math Solver to verify that I was right. But, surprisingly, it told me that the result is 1. Can someone explain why?

Thanks a lot.

答:
The sine function only takes real numbers as input and it does not make sense to write $\sin(+\infty)$.

Instead, you should get that
$$
\lim_{x\to+\infty}\frac{1}{2-2x}=0
$$
and $\sin(1-x)$ is bounded.

Now look for the property in your notes/book that
$$
\lim_{x\to+\infty}f(x)g(x)=0
$$
if $\lim_{x\to+\infty}f(x)=0
$
and $g(x)$ is a bounded function.


问:
Profinite groups are Galois groups
I tried to understand the proof on page 195 (in PDF page 17) here.

At the beginning $T$ is denoted as disjoint union of sets $G/U$.
Is it important for the proof that they are disjoint ?

After the definition of $G_{k}$ are some $t_{i}\in G/U_{i}$ taken, that appear in the rational expression of $k$. What does that mean?
Take they some $U_{i}$ so that $k$ mod $U_{i}$ equal $k$ or what is meant by that?
And why are there finally many $U_{i}$.

答:
The simplest relevant example is $$\Bbb{Z_p} =\varprojlim \Bbb{Z/p^kZ}$$

For each $k$ and $a\bmod p^k$ create a variable $T_{a\bmod p^k}$ and let $$K = \Bbb{Q}(\{ T_{a\bmod p^k},k\ge 1, a\in 0,\ldots p^k-1\})$$ which is a field of rational functions in infinitely many variables. Each $b\in \Bbb{Z}_p$ acts naturally as an automorphism $g_b\in Aut(K)$, switching the variables as $$g_b(T_{a\bmod p^k})=T_{a+b\bmod p^k}$$
It remains to look at $L$ the subfield fixed by $G=\{ g_b,b\in \Bbb{Z}_p\}\subset Aut(K)$ and find that $Gal(K/L)=G$ which is done in your linked text and cannot be made simpler in my opinion, the key argument being that for any $y\in K$ the Galois orbit  $G.y = \{g_b(y),b\in\Bbb{Z}_p\}$ is finite so that $L$ is generated over $\Bbb{Q}$ by the coefficients of the polynomials $$f_y(X)=\prod_{z\in G.y} (X-z) \in L[X]$$

Title: Profinite groups are Galois groups


问:
How do I find the volume of a solid rotated around y = 3?
I had a college textbook problem which states:

The region bounded by the curves y = x and y = x^2 is rotated about the line y = 3. Find the volume of the solid.

I know how to find the volume if it is not rotated by y = 3. But when it states rotated about the line y = 3. I have no idea how to do it.

Thanks for reading!

答:
HINT

Very similar idea. Each vertical cross-section of the region in question, lying on some constant value $x$ has height of $h(x) = x - x^2$ and rotates in a circle of radius $3-x$. What is the circumference of the rotational circle? What is the surface area of the cylindrical slice generated at this $x$? Can you now compute the volume?


问:
When do I decide a logistic model with no roots equals 0?
I have a logistic model function:

$$ y=\frac{424.92}{1+0.37027e^{0.000715x}}$$

This exponential model has an asymptote at $y=0$, and hence doesn't ever meet the $x$-axis. However, I need to discern when the quantity being measured ceases to exist (i.e. when it reaches zero). Is there some way I can calculate this anyway? Is there some cut-off point where I decide that the $y$-value becomes negligible?

答:
The quantity is never $0$, but you can find the value of $x$ after which it becomes arbitrarily small. Say we want to know when we have $y<\varepsilon$ for some small positive number $\varepsilon$. Then we have $$\frac{424.92}{1+0.37027 e^{0.000715x}}<\varepsilon.$$ Taking reciprocals: $$\frac{1+0.37027 e^{0.000715x}}{424.92}>\varepsilon$$ and solving for $x$:$$\frac{424.92\varepsilon-1}{0.37027}<e^{0.000715x}$$ $$\log{\left(\frac{424.92\varepsilon-1}{0.37027}\right)}/0.000715<x.$$ So for any $x$ larger than this quantity, you will have $y<\varepsilon$.
So, you can decide what value of $\varepsilon$ can be considered as "negligible" in your model and then find what value of $x$ makes your $y$ this small.


问:
Show that the extension field $\mathbb{Q}(\sqrt{2},\sqrt{3})$ is a field
We just started field extensions and I need to show that $\mathbb{Q}(\sqrt{2},\sqrt{3})=\{a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6} : a,b,c,d\in\mathbb{Q}\}$ is a field.

It's easy for $\mathbb{Q}(\sqrt{2})$, by just showing that the multiplicative inverse has the form $a+b\sqrt{2}$, but how can I do it for $\mathbb{Q}(\sqrt{2},\sqrt{3})$. How can I show that the inverse has the form $a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6}$ or is there an easier approach.

答:
As @lhf given a link, you can see that  there$\Bbb{Q}(\sqrt{2},\sqrt{3})=\Bbb{Q}(\sqrt{2}+\sqrt{3})$.

One can verify that $\sqrt{2}+\sqrt{3}$ is a root of the polynomial $p(x)=x^4-10x^2+1$. In fact $p(x)$ is an irreducible (what if not!) polynomial over $\Bbb{Q}$ and  ideal $\left \langle p(x) \right \rangle $ generated by $p(x)$ is a maximal ideal.

You know that possibly,

$R/M$ is a field $\Leftrightarrow$ $M$ is a maximal ideal.
Hence
$\Bbb{Q}[x]/\left \langle p(x) \right \rangle$ is a field. and $\Bbb{Q}[x]/\left \langle p(x) \right \rangle\cong \Bbb{Q}(\sqrt{2},\sqrt{3})$.


问:
Using a definite integral, to create a specific recurrence relation.
Hello i have the integral:

$$y_n=\int_0^1\frac{x^n}{x+5}dx$$ where $ n=1,2,3,4,....,\infty$

I need to show that the integral can be represented by the recurrence relation below;

$$y_n= \frac1n-5y_{n-1}$$

I started by doing integration by parts but still cannot find the relation. I am assuming it has to do with the integration by parts method where i am going wrong.

Body:
HINT:

If $$y_n=\int_0^1\frac{x^n}{x+5}dx$$

$$\int_0^1\frac{x^n}{x+5}=\int_0^1\frac{x^{n-1}(x+5-5)}{x+5}dx=\int_0^1x^{n-1}dx-5\int_0^1\frac{x^{n-1}}{x+5}dx$$  as $x+5\ne0$

Alternatively,

$$\text{So,}y_n+5y_{n-1}$$
$$=\int_0^1\frac{x^n}{x+5}dx+5\int_0^1\frac{x^{n-1}}{x+5}dx$$
$$=\int_0^1\frac{x^{n-1}(x+5)}{x+5}dx=\int_0^1x^{n-1}dx$$  as $x+5\ne0$


问:
Shapley Value for Voters in Electoral College
Consider (a simplified version of) the USA electoral college. Each state has millions of voters and tens of electors. The candidate with the most electors wins and each state allocates electors according the popular vote in that state (some are winner take all and some are proportional). There is lots of work on algorithms for computing the Shapley-Shubik value (voting power) for each state, but how do you compute the Shapley-Shubik value by voters? ie how much power does a voter in one state have compared to another? Note that this is not simply a weighted voting game.</p>

答:
I think the Shapley-Shubik power of any given voter is the Shapley-Shubik power of the district the voter is in, divided by the amount of voters in the district.

Definition

First, let's define what we're talking about. We will define the Shapley-Shubik power of a voter $i$ as the proportion of permutations of voters such that a coalition of voters up to and including $i$ is sufficient to win the election. In other words, if every voter up to and including $i$ votes for candidate A and all others vote for B, A should win.

Note that this is a slight deviation of the "real" Shapley-Shubik power, or alternatively of the United States electoral system. The problem is that there is no set point at which one wins an election, as one's candidate might still win without a majority of the electoral votes, either because there is a tie, because a third party has won some states, or because some electors don't properly do their job. A faithful representation of the American system would have to account for this by giving a value to any given coalition based on the probability of winning with that coalition, after which we could calculate the Shapley value for each voter, instead of the Shapley-Shubik power. Clearly however this is not feasible, so we instead assume that there are only two candidates, ties don't occur, and everything works "nicely".

Which voter is the pivot voter?

Now then, we need to find the pivot voter for a given permutation. One way to do so is to find the pivot voter for each separate district, throw out all the other ones, and then consider the remaining voters as being districts. The pivot "district" in this new scenario will be the pivot voter of the old scenario.

For example, consider a system with three districts ($a$, $b$, $c$), all with three voters and with the same amount of votes/power. One possible permutation would be:

abbcaccab

First, we only consider the middle voter of each district:

abbcaccab $\rightarrow$ bac

Now, because each district has the same amount of power, district $a$ is the pivot district, and the second voter from this district is the pivot voter in the original permutation.

What is the probability of a district being pivotal?

This is the most important part of this answer, as I argue that the probability of a district being pivotal is the same, regardless of whether we look at districts or at voters as the entities to which we ascribe Shapley-Shubik power.

This is equivalent to saying that for a permutation of voters from a number of districts, the permutations of districts derived by only considering the pivotal voters for each district will yield all possible permutations of districts equally often. I believe the truth of this statement is fairly clear for 2 districts, and I'm pretty sure you can prove it through induction for more districts, but I have not done so. However, if you disagree with this, or if you cannot find a proof, you might want to ask this subquestion on this very site again.

As for this answer, we will state without proof that the probability of a voter from district $a$ being pivotal is equal to district $a$ being pivotal if we consider the district as voters themselves.

Conclusion

Because the probability of a voter from a district being pivotal is equal to the district itself being pivotal, the sum of the Shapley-Shubik powers of the voters must equal the power of the district. Furthermore, as the voters are all identical, they must through symmetry all have the same power. Therefore, the power of a voter is the power of a district divided by the number of voters.


问:
Calculate $\int_{|z|=3} \frac{z^{17}}{(z^2+2)^3(z^3+3)^4}dz$
Calculate $\int_{|z|=3} \frac{z^{17}}{(z^2+2)^3(z^3+3)^4}dz$

Firstly we see that all the singular points lie in the circle $|z|<3$. However I don't have idea how to calculate it.

答:
A variant on @Conrad's approach:

With $z=3e^{i\theta}$ the integral becomes $i\int_0^{2\pi}\frac{(3z)^{18}e^{i\theta}d\theta}{(9e^{2i\theta}+2)^3)(27e^{3i\theta}+3)^4}$. Discounting the factor of $i$ I've moved outside, the numerator and denominator are both degree-$18$ polynomials in $e^{i\theta}$, with leading coefficient $3^{18}$. The integrand is therefore a rational function that has constant term $1$, making the final result $i\int_0^{2\pi}d\theta=2\pi i$, since other powers of $e^{i\theta}$ integrate to $0$.


问:
Is changing limits in this fashion okay? What is the justification for it?
I am currently working through chapter $5$ on limits of Spivaks calculus book. I have come to exercise $15$ix. I feel like I can answer this question correctly, but there is one small step I know to be true but can't seem to show it concisely.

I would like to note, I am aware similar questions have been asked, but I am querying a specific part of this question.

The question:

Assume that: $$\lim_{x\to{0}}{}\frac{\sin(x)}{x}=\alpha.$$

Find, in terms of $\alpha$, the following limit:

$$\lim_{x\to1}\frac{\sin(x^2-1)}{x-1}.$$

My attempt:

Firstly, multiplying bottom and top by $x+1$ yields:

$$\lim_{x\to1}\frac{\sin(x^2-1)(x+1)}{(x-1)(x+1)}=\lim_{x\to1}\frac{\sin(x^2-1)(x+1)}{x^2-1}$$

Now, IF $(*):=\lim_{x\to1}\frac{\sin(x^2-1)}{x^2-1}$ exists, we can split the above into:

$$\lim_{x\to1}\frac{\sin(x^2-1)}{x^2-1}\times{}\lim_{x\to1}{x+1},$$

using algebra of limits. Which gives us:

$$\lim_{x\to1}\frac{\sin(x^2-1)}{x^2-1}\times{}2.$$

So I'm essentially left to show $(*)$ exists and is equal to $\alpha$. Which means the original limit is equivalent to $2\alpha$.

Now clearly as $x\to{1}$ we have that $(x^2-1)\to{0},$ so can we write this limit as:

$$\lim_{x\to0}\frac{\sin(x)}{x}\text{ or }\lim_{x^2-1\to0}\frac{\sin(x^2-1)}{x^2-1}?$$

If so, why exactly? It just feels a bit imprecise.

I guess what I'm asking exactly is:

$$\lim_{x\to0}\frac{\sin(x)}{x}\overset{?}{=}\lim_{x\to1}\frac{\sin(x^2-1)}{x^2-1}$$

答:
Yes, it is as long as we keep into the definition domain of the function. One way to see this is making substitutions: Substitute $\;t:=x^2-1\;$ , and observe that $\;x\to1\implies t\to0\;$ , so

$$\lim_{x\to1}\frac{\sin(x^2-1)}{x^2-1}=\lim_{t\to0}\frac{\sin t}t=1$$


答:
Since all the singularities are inside $|z|=3$, one can move the integration circle to $|z|=R>3$ and let $R \to \infty$ which gives a nice result by the "coincidence" that the numerator has degree one less than the denominator, so only the ratio of leading terms matters.

Concretely, using $dz=izd\theta$ one gets that:

$I=\int_{|z|=3} \frac{z^{17}}{(z^2+2)^3(z^3+3)^4}dz=\int_{|z|=R} \frac{z^{17}}{(z^2+2)^3(z^3+3)^4}dz=\int_0^{2\pi}\frac{iz^{18}}{(z^2+2)^3(z^3+3)^4}d\theta, z=Re^{i\theta}$, so

$I=\int_0^{2\pi}\frac{i}{(1+2/z^2)^3(1+3/z^3)^4}d\theta, z=Re^{i\theta}$ hence $I=I_R \to 2\pi i$ as $|z|=R \to \infty$

so the integral is $2\pi i$

Body:
Calculate $\int_{|z|=3} \frac{z^{17}}{(z^2+2)^3(z^3+3)^4}dz$

Firstly we see that all the singular points lie in the circle $|z|<3$. However I don't have idea how to calculate it.


问:
Series expansion and Laurent series
I need to find the series expansion at $$z=0$$ and the Laurent series  at $$  \sqrt{ 2/3} <|z| < \infty $$
for the function $$ f(z)= \frac{1}{(2-3z^{2})^2} $$
I guess I first need to  first find the partial fraction decomposition

Title: Series expansion and Laurent series

Body: Hint:  $\;f(z)=\dfrac{\left(\dfrac12\right)^2}{\left(1-\dfrac32z^2\right)^2}=\dfrac {\left(\dfrac1{3z^2}\right)^2}{\left(1-\dfrac2{3z^2}\right)^2}\;$ and $\;\dfrac 1{(1-x)^2}=1+2x+3x^2+\cdots$


问:
Formula to increase/decrease a relative number based on a fixed number
Being a fairly good math student in high school, this is humbling. But my knowledge about graphs and formulas has greatly diminished since then.

I'm trying to write a formula that calculates a percentage, depending on the total amount. My base is $15\%$ of $100$. Basically the percentage needs to go up as the total goes down, and vice versa. So some examples:

```<code>10000 =>  5% (500)
 1000 =>  7% (70)
  300 => 10% (30)
  100 => 15% (15)
   80 => 20% (16)
   40 => 30% (12)
   20 => 40% (8)
    5 => 60% (3)
</code>```

There would also be a limit I think, in that $0\%$ must not be possible. The upper boundary is infinite in theory, but in practice more likely to be around $100000$.

Without the deviation, the formula would be simple:

$ y = 100/x*15 $

But as far as the deviation goes, I could use some help. The numbers are just an example, I'll probably tinker with the variables a bit to get the exact formula.

答:
You seem to be looking for a function $f\colon[1,\infty)\to(0,1]$ such that $f(1)=100\%$, $f(100)=15\%$ and $\lim_{x\to\infty}f(x) = 0\%$.

A nice family of functions to try would be $f(x) = \frac{a}{bx+c}$. Your conditions then fix $f(1) = \frac{a}{b+c}=100\%$ and $f(100)=\frac{a}{100b+c}=15\%$. Some numbers that would work are $b=17$, $c=280$ and $a=297\cdot 100\%$.

Hence the formula would be $y = \frac{297}{17x+280}\cdot 100\%$.

Since you remark that this formula decreases too fast for large $x$ you might alternatively try $f(x) = \frac{a}{b\sqrt[n]{x}+c}$ for some $n\in\{1,2,\ldots\}$.

Now your conditions on $x=1$ and $x=100$ fix $f(1) = \frac{a}{b+c}=100\%$ and $f(100) = \frac{a}{\sqrt[n]{100}b+c}=15\%$. Some numbers satisfying these requirements are $a=(3\sqrt[n]{100}-3)\cdot100\%$, $b=17$ and $c=3\sqrt[n]{100}-20$.

Hence we now have
$$ y = \frac{3\sqrt[n]{100}-3}{17\sqrt[n]{x}+3\sqrt[n]{100}-20}\cdot 100\%.$$
For $n=1$ this nicely reduces to the original formula, but for $n=2$ say, we get
$$ y = \frac{27}{17\sqrt{x}+10}\cdot 100\%.$$


问:
Is the following function a bijection?
During my work on hash functions, I try to prove/disprove that the a linear map function $f:\mathtt{Z}_2^n\to \mathtt{Z}_2^n$, where $n=64$,  is bijection.

The function is defined as followed
$$f(x_0,x_1,...,x_{n-1})=(y_0,y_1,...,y_{n-1}),$$ where
$$y_i=x_i+x_{i+49}+x_{i+24} \mod 2,$$
for $0\leq i\leq n-1$  (the indices $i+49$ and $i+24$ are taken modulo $n$).

I tried to find the determinant of the transformation,  but did not succeed.

答:
I solved my problem :)

I used simply the following python computer program

```<code>import numpy as np
from scipy.linalg import det

n=64

mat=np.zeros((n,n))

for i in range(n):
    j1=i
    j2=(i+24)%n
    j3=(i+49)%n

    mat[i][j1]=1
    mat[i][j2]=1
    mat[i][j3]=1

print(det(mat)%2)
</code>```

The answer is 1. thus, the transformation must be a bijection.

Title: Is the following function a bijection?

Body:
I solved my problem :)

I used simply the following python computer program

import numpy as np
from scipy.linalg import det

n=64

mat=np.zeros((n,n))

for i in range(n):
    j1=i
    j2=(i+24)%n
    j3=(i+49)%n

    mat[i][j1]=1
    mat[i][j2]=1
    mat[i][j3]=1

print(det(mat)%2)

The answer is 1. thus, the transformation must be a bijection.


答:
Using the fact that $|\sin(x)| \le 1$ you have the following:

$$\lim_{x\to \infty} \left| \frac{\sin(1-x)}{2-2x} \right| \le \lim_{x\to \infty} \frac{1}{|2-2x|} = 0$$

This concludes that $$\lim_{x\to \infty} \frac{\sin(1-x)}{2-2x} = 0$$

Title: Find $\lim_{x\to +\infty}\frac{\sin(1-x)}{2-2x}$

Body:
I have an assignment about limits:

$$\lim_{x\to +\infty}\frac{\sin(1-x)}{2-2x}$$

It tells me to calculate the value of the limit.
I observed that it comes up $\frac{\sin (-\infty)}{-\infty}$, so the sin of infinite can't exist. I was just about to solve it, when I decided to try to use Microsoft Math Solver to verify that I was right. But, surprisingly, it told me that the result is 1. Can someone explain why?

Thanks a lot.


问:
A balancing pigeon puzzle
While designing a new logo for our company I had to solve a version of the geometric problem below. I have repackaged it here in the form of a puzzle that I hope others might enjoy!

The pigeon drawn above in the Cartesian plane:

consists of a straight line and 5 circle arcs

is perfectly smooth along its perimeter except for the points labeled B,C,D and E on the graph

tiles the plane

Assuming that the weight of the pigeon is distributed uniformly across its area, what are the co-ordinates of E such that it balances exactly on its foot?

答: Not really an answer: I've just prepared the figure below, with the relevant geometric information. But finding the center of mass is too tedious for me...

Title: A balancing pigeon puzzle


问:
Scale Space - Scales and Octaves
So I'm desperately trying to understand scale space for signals, specifically for 2D images...

I'm having trouble with algorithms that discuss creating a pyramid. Specifically, I don't understand how we go between resolutions (octaves). Within in a pyramid, I realize that an octave is a difference in resolution (or subsampled) but I don't understand how it relates to the scale parameter. (I understand that smoothing enough times is equivalent to subsampling).

For example, if we have five signals/images in an octave which have iteratively been smoothed with a Gaussian, how do the images at the edges relate to each other? (Image 5 and Image 6). Is Image 6 subsampled from Image 1 or from Image 5?

What about the scale parameter, discussed often throughout many articles on Scale Space, how does it relate to subsampling?

I've been reading papers all day, my head is swimming and I'm still confused.

From Wikipedia, I'm having intense trouble understanding this, but answers what I'm trying to understand:

An image pyramid is a discrete representation in which a scale space is sampled in both space and scale. For scale invariance, the scale factors should be sampled exponentially, for example as integer powers of 2 or root 2. When properly constructed, the ratio of the sample rates in space and scale are held constant so that the impulse response is identical in all levels of the pyramid. Fast, O(N), algorithms exist for computing a scale invariant image pyramid in which the image or signal is repeatedly smoothed then subsampled. Values for scale space between pyramid samples can easily be estimated using interpolation within and between scales.

答: Here: http://www.vlfeat.org/api/sift.html you can find all the details you need.</p>

Title: Scale Space - Scales and Octaves


问:
Commutativity of a prime ring
Question :

Show that $R$ is a prime ring containing two commuting non-zero left ideals $I$ and $J$ $\implies$ $R$ is commutative, where "commuting ideals” means $ij=ji$ for all $i \in I$, $j \in J$.

My attempt : Let $x, y \in R, \quad xIyJ=yJxI$ and I need a hint to accomplish the proof.

答:
Since $R$ is prime, and $I,J$ are nonzero left ideals, we know that $xI=0$ or $xJ=0$ implies $x=0$.

Take any $x\in R$, $i\in I$ and $j\in J$. We have
$$ i(xj)=(xj)i=x(ji)=x(ij) $$
so that $(ix-xi)j=0$. Thus $(ix-xi)J=0$, and hence $ix=xi$.

Now take any $x,y\in R$ and $i\in I$. We have
$$ y(xi)=(xi)y=x(iy)=x(yi) $$
so that $(xy-yx)i=0$. Thus $(xy-yx)I=0$, and hence $xy=yx$.

We conclude that $R$ is commutative.

Title: Commutativity of a prime ring


答: Does the above diagram help? See the shaded area that you are rotating around $y = 3$. Do you see that the radius you are interested in is between curves $y = x$ and $y = x^2$ and is function of $x$? So all you need to do is to find the limits of $r$ and set up the integral as $r \ dr \ dx \ d\theta$ and you know the limits of $x$.

Title: How do I find the volume of a solid rotated around y = 3?

Body:
I had a college textbook problem which states:

The region bounded by the curves y = x and y = x^2 is rotated about the line y = 3. Find the volume of the solid.

I know how to find the volume if it is not rotated by y = 3. But when it states rotated about the line y = 3. I have no idea how to do it.

Thanks for reading!


问:
Does it make sense for a constant function to have infinitely many critical points?
So I've been pondering over constant functions, that is, functions of the form

$$ f(x)=c$$

where $c$ is a constant. $f'(x) = 0$, so every value you put into $f$ is a critical point. That means there are infinitely many critical points.

But does it make sense for it to have infinitely many critical points (and that every point, everywhere is a critical point?)

答:
One way to think about it is that constant functions are flat, so every point is a "maxima/minima".

And, local and global maxima and minima occur at critical points. So, it does make sense.

$\sin(x)$ is a more meaningful example of a function with infinitely many critical points


问:
Confirming a easy proof: the product of two consecutive numbers is always even.
Can someone confirm if my prove is right?

Theorem. The product of two consecutive integers is always even.

Proof. Define a number $n$ such that $n:=2k$ where $k\in\mathbb{Z}$, this ensures that $n$ is an even number. Define a second number $p$ such that $p:=n+1=2k+1$ this ensures that $p$ is an odd number and the numbers $n$ and $p$ are consecutive because they differ by $1$. There product is given by:

$$np=2k(2k+1)=2(2k^2+k)=4k^2+2k\tag1$$

Which is clearly a multiple of $2$.

Define a number $h$ such that $h:=2m+1$ where $m\in\mathbb{Z}$, this ensures that $h$ is an odd number. Define a second number $z$ such that $z:=n-1=2m$ this ensures that $z$ is an even number and the numbers $h$ and $z$ are consecutive because they differ by $1$. There product is given by:

$$hz=2m(2m+1)=2(2m^2+m)=4m^2+2m\tag2$$

Which is clearly a multiple of $2$, this proves that $hz$ is an even number.

This proves this theorem.

If I am wrong how can I make my proof valid?

答:
You proved that the product of an even number and the next number is even.

Since any number is even or odd, you could complete the proof of the theorem

by proving that the product of an odd number and the next number is also even.


问:
Integration-finding an upper limit of integration.
I'm given the value of an integral: $$\int_1^a (x-\frac{w^3}{x^2})\,dx=1.5$$ and told that the total area between the graph of $$y=x-\frac{w^3}{x^2}$$ and the x-axis is 6.5 units squared. w and a are both integers greater than 1. How can I find the values of w and a?

答:
$$\int_1^a \left(x-\frac{w^3}{x^2}\right) \, dx=\left[\frac{w^3}{x}+\frac{x^2}{2}\right]_1^a=\frac{a^2}{2}+\frac{w^3}{a}-w^3-\frac{1}{2}=\frac{3}{2}$$
The second condition is unclear. The graph of the family of functions $$y=x-\frac{w^3}{x^2}$$
is like the one in the image below. There is no finite area.

When OP will tell what is the area we are talking about I'll try to help further.

$$...$$


问:
What rule prevents us from inspecting the limit of terms individually?
I am just curious why we are not allowed to conclude that whenever we have a term like $\frac{1}{n}$ or any other term that converges to $0$, that then the whole expression converges to $0$, since $0\cdot x = 0$.

Here is some context:
When evaluating the limit of $\lim_{n\to \infty}\left(n^2\cos(2/n)\frac{2}{n^2}\right)$ one should see that $n^2$ cancels out and thus $\lim_{n\to \infty}\left(2\cos(2/n)\right)=2$. However, if we just inspect the limit of the term $\frac{2}{n^2}$ then this limit is $0$ and then one could (wrongly) conclude that $\lim_{n\to \infty}\left(n^2\cos(2/n)\frac{2}{n^2}\right)=0\neq 2$. So I know that there is some rule which does not allow us to do that, but I don't know what rule that is.

Edit: Is it because $0\cdot \infty$ is undefined?

答:
It is actually lack of rule that does not allow you to do this.

If you have $A(n) \cdot B(n)$ and you know that $A(n) \to 0$ as $n$ goes to infinity, it is not necessarily true that the product $A(n) \cdot B(n)$ goes to zero.

You need to know something about the other term.

E.g. if $B(n)$ is limited/bounded then the conclusion is true. Even if $A(n) \to \infty$, the conclusion may still be true, given that $A(n)$ goes faster to zero than $B(n)$ goes to infinity. So... you just need to know the behavior of the other term or at least some of its properties (boundedness e.g.).

There's just no rule/theorem which based just on the behavior of one term concludes something about the behavior of the product.


问:
Inequality for definite integral.
In a proof in my class we used the following inequality: $$\left| \int_0^{\varepsilon}te^{ut}\,du\right|\le |\varepsilon| |t|(e^{\varepsilon_0t}+e^{-\varepsilon_0t})\quad \forall |\varepsilon| <\varepsilon_0.$$However, I'm not quite sure how we got that inequality so I need some help.

答:
$$\left| \int_0^{\varepsilon}te^{ut}\,du\right| \le |t|  \int_0^{\varepsilon}|e^{ut}|\,du  = |t| \int_0^{\varepsilon}e^{ut}\,du $$

Consider $$f(\varepsilon) := \int_0^\varepsilon e^{tu} \ du.$$ This is a convex function and for the fundamental theorem of calculus  $f'(\varepsilon) = e^{\varepsilon t}$.

Fix $\varepsilon_0 > 0$, then for all $0\le \varepsilon \le \varepsilon_0$ you have $f(0) \ge f(\varepsilon) + f'(\varepsilon)(0-\varepsilon)$ so

$$\int_0^\varepsilon e^{tu}\ du \le \varepsilon \ e^{\varepsilon t} \le \varepsilon \ e^{\varepsilon_0 t}$$

Concluding

$$\left| \int_0^{\varepsilon}te^{ut}\,du\right| \le |t| \varepsilon \ e^{\varepsilon_0t} \le |t| |\varepsilon| \left( \ e^{\varepsilon_0t} + e^{-\varepsilon_0t}\right)$$

Since in the last passage I just added a strictly positive term.

Title: Inequality for definite integral.


问:
Find a volume of a figure defined with: $y=zx, y\geq 0, x=z, x=z^3$
Find a volume of a figure defined with: $y=zx, y\geq 0, x=z, x=z^3$.

This is my approach:

First I drew a picture in 3D, and I changed the orientation of coordinates so that $y$-axis is in $z$'s place (up), $z$ is in $x$'s place(left) and $x$ is in $y$'s place(right).

$V=\begin{gather*}
    \iint_D y(x,z) \,dx\,dz=-\frac{1}{16}
\end{gather*}$

Where I saw from the picture that $D$ is $ \{  z \leq x \leq z^3, 0 \leq z \leq 1$

Is this correct since the result is negative? Is there any other way of solving?

答: The region $\mathcal{D}$ in the $xz-$plane bounded by $x=z$ and $x=z^3$ consists of two disjoint regions: $$\mathcal{D}_1=\{x^{1/3}<z<x,-1<x<0\}$$ $$\mathcal{D}_2=\{x<z<x^{1/3},0<x<1\}$$ Notice that $y=xz$ is positive on both $\mathcal{D}_1$ and $\mathcal{D}_2$. Therefore the volume of your solid equals $$\int_{-1}^0 \int _{x^{1/3}}^xxzdzdx+\int_0^1 \int _{x}^{x^{1/3}}xzdzdx=\frac{1}{8}$$


答:
Let $f(x)=x-\frac{w^3}{x^2}$.

We know that the function is increasing on $(0,\infty)$ and that it has $x$-intercept $(w,0)$. So we have

$$ \int_1^w f(x)\,dx+\int_w^a f(x)\,dx=1.5\tag{1} $$

$$ -\int_1^w f(x)\,dx+\int_w^a f(x)\,dx=6.5\tag{2} $$

Would you like me to work it further or do you see how to proceed?

Here is the next step:

Subtract equation (2) from equation (1) and divide by 2 to get

$$ \int_1^w f(x)\,dx =-2.5\tag{3} $$

but you can also find the value of the integral directly

$$ \int_1^w f(x)\,dx =-w^3+\frac{3}{2}w^2-\frac{1}{2}\tag{4} $$

giving

$$ 2w^3-3w^2-4=0 $$

which has a real solution $w=2$.

Now if you add equations (1) and (2) and divide by $2$ you will get

$$ \int_2^a f(x)\,dx=4 $$

If you evaluate this integral directly you will get an equation which you can solve for $a$, which is also an integer.

Title: Integration-finding an upper limit of integration.

Body: I'm given the value of an integral: $$\int_1^a (x-\frac{w^3}{x^2})\,dx=1.5$$ and told that the total area between the graph of $$y=x-\frac{w^3}{x^2}$$ and the x-axis is 6.5 units squared. w and a are both integers greater than 1. How can I find the values of w and a?


问:
Help solving this Diophantine equation
For a problem that I'm working on, I need to solve this Diophantine equation:-

$ -2a^3 + b^3 + c^3 = 36650$, where $a, b, c > 0$ are all DISTINCT positive integers, and $a, b, c \notin$ { 2, 9, 15, 16, 33, 34}

How does one go about solving this? Is brute-force the only possible way? Or could there be a case that no integer solutions exist for this equation?

Also, are there any online computing engines, that allow me to set constraints, and solve Diophantine equations of this sort?

Any and all help is appreciated! Thanks!

答:
For your modified question $a^3 + b^3 + c^3 - 3d = -83449$, there are in fact infinite solutions.

Considered modulo 3,

$$a^3 + b^3 + c^3 \equiv 2 \mod 3$$

Now, we know that modulo 3, any cube is congruent to itself, or $x^3 \equiv x \mod 3$.

Therefore, the above equation reduces to

$$a+b+c \equiv 2 \mod 3$$

One triplet of $(a,b,c)$ that satisfies the above equation is $(1,3,4)$. We can then calculate the value of $d$, which is simply $27847$ in this case.

Rearranging the above equation, since $3d=a^3+b^3+c^3+83449$, $d$ is positive for any positive values of $a,b,c$, allowing us to conclude that there are infinite solutions.

Title: Help solving this Diophantine equation


问:
If L and M are algebraic over K then so is LM
Edit: My question was linked with the following question and closed as duplicate:$M /K \land L /K$ algebraic $\implies ML/K$ algebraic?

But if one just sees and compares the statements in the two questions, it is easy to see that the both are not same. So, kindly reopen it.

This question was asked in my abstract algebra mid term exam ( in November) and I was uunable to solve this particular question.

Let L and M be intermediate fields in the extension $K\subset F$. Then if L and M are algebraic over K then so is LM.

I was unable to use the fact that L and M are given algebraic over K. So, I revised through my notebook today also but still couldn't  proceed. I need help in this so can you please outline main details of solution?

For Theory I have been Using Textbook by Hernstein.

答:
If $M$ is algebraic over $K$, it is for sure also algebraic over $L$. So $LM=L(M)$ is algebraic over $L$ (since it is generated by algebraic elements). Since $L/K$ is algebraic and the property 'algebraic' is transitive we see that $LM/K$ is algebraic.

Another way to see the claim is by noting that the elements of $LM$ are quotients of sums of products of elements in $L$ and $M$. And the product/sum/quotient of algebraic elements is algebraic.


问:
Prove that $(x^2+2x+1)$ is continuous by using a $\epsilon$ - $\delta$ argument
I want to prove that the following function is continuous with an Epsilon Delta Argument.
$$ f: \mathbb R \rightarrow \mathbb R \;  (x^2+2x+1)$$

So I started with
$$\left\lvert f(x)-f(y) \right\rvert= \left\lvert (x^2+2x+1) - (y^2+2y+1) \right\rvert = \left\lvert (x^2+2x)-(y^2+2y) \right\rvert$$

$$\le 2\left\lvert \left\lvert \frac{x^2-y^2}{2} \right\rvert + \left\lvert x- y\right\rvert \right\rvert $$
$$\le 2\left\lvert \left\lvert \frac{(x+y)(x-y)}{2} \right\rvert + \delta \right\rvert$$
$$= 2\left\lvert \frac{\left\lvert(x+y)(x-y)\right\rvert}  {2} + \delta \right\rvert $$
$$ \le 2\left\lvert \frac{\left\lvert(x+y)\delta\right\rvert}  {2} + \delta \right\rvert $$
$$= 2\left\lvert \frac{[\left\lvert(x-y)\right\rvert+\left\lvert2y\right\rvert]\delta}  {2} + \delta \right\rvert $$
Now I define $\delta \le1.$
$$= 2\left\lvert \frac{(1+\left\lvert2y\right\rvert)\delta}  {2} + \delta \right\rvert $$
$$= \left\lvert(1+\left\lvert2y\right\rvert)\delta + 2\delta \right\rvert $$
$$= \delta \left\lvert(1+\left\lvert2y\right\rvert) + 2 \right\rvert $$
$$= \delta \left\lvert\left\lvert2y\right\rvert + 3 \right\rvert = \epsilon $$
And so my $\delta$ is $\frac{\epsilon}{\left\lvert\left\lvert2y\right\rvert + 3 \right\rvert}$ and $\delta \le 1$

Can anyone tell my if this is correct or where I went wrong I am studying for exam. :)
Thanks in advance.

答:
I don't know why you want to factor the number $2$ out of your estimate. But the following may be a more natural way to think about how to find a needed $\delta$.

Notice that $$
|x^2+2x-y^2-2y|\le|x+y||x-y|+2|x-y|=(|x+y|+2)|x-y|
\tag{1}
$$

Suppose you want to show that your function is continuous at $y$. If $|x-y|<\delta$, you want
$$
(|x+y|+2)|x-y|\le \epsilon. \tag{2}
$$

On the other hand,

$$
(|x+y|+2)|x-y|\le (|x-y|+2|y|+2)|x-y|\le (\delta+2|y|+2)\delta\tag{3}
$$

So if you let $\delta<1$ and $(2|y|+3)\delta<\epsilon$, you get what you want.

To wrap things up, set $\delta=\min(1,\frac{\epsilon}{2|y|+3})$.

Your solution is ok except for several places. See below.

So I started with (*assuming $|x-y|<\delta$ *)
$$\left\lvert f(x)-f(y) \right\rvert= \left\lvert (x^2+2x+1) - (y^2+2y+1) \right\rvert = \left\lvert (x^2+2x)-(y^2+2y) \right\rvert$$

$$\le 2\left\lvert \left\lvert \frac{x^2-y^2}{2} \right\rvert + \left\lvert x- y\right\rvert \right\rvert $$
$$< 2\left\lvert \left\lvert \frac{(x+y)(x-y)}{2} \right\rvert + \delta \right\rvert$$
$$= 2\left\lvert \frac{\left\lvert(x+y)(x-y)\right\rvert}  {2} + \delta \right\rvert $$
$$\color{red}{\le} 2\left\lvert \frac{\left\lvert(x+y)\delta\right\rvert}  {2} + \delta \right\rvert $$
$$= 2\left\lvert \frac{[\left\lvert(x-y)\right\rvert+\left\lvert2y\right\rvert]\delta}  {2} + \delta \right\rvert $$
Now if I set $\delta \le1$ then
$$= 2\left\lvert \frac{(1+\left\lvert2y\right\rvert)\delta}  {2} + \delta \right\rvert $$
$$= \left\lvert(1+\left\lvert2y\right\rvert)\delta + 2\delta \right\rvert $$
$$= \delta \left\lvert(1+\left\lvert2y\right\rvert) + 2 \right\rvert $$
$$= \delta \left\lvert\left\lvert2y\right\rvert + 3 \right\rvert = \epsilon $$
And so my $\delta$ is $\min(1,\frac{\epsilon}{\left\lvert2y\right\rvert + 3 })$.


问:
What is a characteristic subspace (w.r.t a linear operator)?
I am working through a book on linear operators and, in the context of a subspace being invariant under an operator, they mention "characteristic subspaces".

```<code>"All characteristic subspaces of an operator are invariant"
</code>```

The book never defined such a term, and a quick google search didn't really provide me with a formal definition. I assume it is related to characteristic functions in some way?

答:
In the French speaking community, the notion of "Characteristic subspace" (=sous-espace caractéristique) is taught in every advanced textbook on linear algebra. Linguistically, it corresponds to the notion of "generalised eigenspace" in the US textbooks:

https://fr.wikipedia.org/wiki/Sous-espace_caractéristique (in French)

https://en.wikipedia.org/wiki/Generalized_eigenvector

For the book you are referring too, I suspect either a poor translation, or a foreign author not fully aware of the subtleties of the mathematical English language ;-).


问:
Questions about Trigonometry
I just learned the very basics of Trigonometry yesterday. I can answer all the questions on the homework but feel uneasy because I still have some question about it. Here are the questions:

What does it mean to take the sin, cos, or tan of an angle? I know it has do do with the ratios but how does that apply to triangles? I just don't fully understand that concept.

Also how does the calculator know for example what the sin(53) is. When finding missing side of non special triangles, at some point I type a random number into the calculator and press either sin/tan/cos. How does it know that though?

Thanks! I'm very new to this subject and so excited to learn more.

答:
Firstly, good questions! Many people (in my experience) just learning trig don't actually understand what the trig functions are, so well done for asking!

Firstly, you need to understand similarity of shapes. More specifically, when talking about right triangles, if I keep all angles in the triangle the same but I change the length of one the sides by some factor, then all the sides of the triangle will be changed by that factor.

For example, if I had a right angled triangle with side lengths $a,b$ and $c$, then if I doubled the length of the side that was $a$ to $2a$ then all the other lengths of the triangle will also be doubled. This is where trigonometric functions come in.

Consider a right angles triangle containing an angle A, as shown in the image below.

We define the (basic) trigonometric functions as follows:
$$\sin A=\frac{a}{c}=\frac{\text{Opposite}}{\text{Hypotenuse}}$$
$$\cos A=\frac{b}{c}=\frac{\text{Adjacent}}{\text{Hypotenuse}}$$
$$\tan A=\frac{a}{b}=\frac{\text{Opposite}}{\text{Adjacent}}$$

So in fact the trigonometric functions for a given angle give us the ratio of any two relevant sides of this right angled triangle. Note that because of similarity as discussed before this ratio is constant however large this right angled triangle is, as long as the angles are kept constant.

This definition cannot be extended to angles larger than $90$ degrees (or negative ones for that matter); if you want I could do this for you but it is more complicated.

As for your second question about how the calculator 'knows' what values these trigonometric functions have for a given angle, I'm afraid you'll need to wait until you learn some calculus in order to be able to understand how we can approximate values of trig functions. (We can use Maclaurin series if you know what that is; if not don't worry.)

I hope I have helped you. Please feel free to ask anything that may still be bothering you.

Title: Questions about Trigonometry


问:
Calculate $\lim _{x\to 0^+}\frac{\left(e^x-e^{-x}\right)}{\sin x}$
***How to find the this limit without using L'Hopital's Rule.

$\lim _{x\to 0^+}\frac{\left(e^x-e^{-x}\right)}{\sin x}$

答:
Hint:

Use maclaurin series for $e^x$ and $e^{-x}$

then apply $$\lim_{x\to 0} \dfrac{\sin x}{x}=1$$


问:
Show that $e(\omega ^i)=0$ or 1.
Let $\omega$ be a primitive nth root of unity. Let $C=\langle g(x)\rangle=[e(x)]$ be a cyclic code in $R_n$. Show that $e(\omega ^i)=0$ or 1.

Proof:(i) Let $C=<g(x)>=<e(x)>$ be a cyclic group in $R_n$. Since $\omega$ is a primitive nth root of unity and since $e(x) \equiv a(x)g(x)$, we deduce that $g(\omega^i)=0$ which implies that $e(\omega^i)=0$. Now, suppose $e(\omega^i) \neq 0$. By a theorem and since $\omega$ is a primitive nth root of unity, we have $e(\omega^i)$ is a unity. Hence, $e(\omega^i)=1$.

I am no sure of this proof. May I know if this is correct?
And from (i), how would I say that the dimension of C is just equal to the number of $\omega^i$ such that $e(\omega^i)=1$?

答:
I assume the $e(x)$ you are talking about here is the generating idempotent of the cyclic code. What happens when $e(\omega^i)\neq0$? Can you recall the definition of an idempotent?

Also, if $e(\omega^i)=0$, then that just means $\omega^i$ is a zero of $e(x)$. How do these $n$th roots of unity relate to the zeros of $C$, considering that $e(x)$ generates $C$?

Title: Show that $e(\omega ^i)=0$ or 1.


问:
Find x for which for every "a" the equation has solution
$$a^{31x} \equiv a \mod 271$$

I need to find x variable, for which the equation has solution with any a. How can I do this?

Generaly, modular equations have solutions when $GCD(a^{31x}, 271) = 1$, or $GCD(a^{31x}, 271) = d > 1; d|a$

It also looks like I could use Little Fermat's Theorem...

But I can't come up with anything...

Body:
Since $271$ is prime, $\phi(271)=270$. Therefore,
$$
a^{31x}\equiv a\pmod{271}\tag{$\ast$}
$$
when
$$
31x\equiv1\pmod{270}
$$
We can solve $270y+31x=1$ using the Euclid-Wallis Algorithm:
$$
\begin{array}{r}
&&8&1&2&2&4\\\hline
1&0&1&-1&3&-7&31\\
0&1&-8&9&-26&61&-270\\
270&31&22&9&4&1&0
\end{array}
$$
Therefore, $61\times31\equiv1\pmod{270}$. Thus, if $x=61\pmod{270}$, then $(\ast)$ holds.


问:
A monotone sequence and finite series sum - characterization
Is there a characterization of the set of all sequences $\{x_{n}\}_{n=1}^{\infty}$ of real numbers such that
$$1\leq x_{n}\leq x_{n+1}$$
for all $n\geq 1$, and
$$\sum_{n=1}^{\infty}\frac{1}{x_{n}^{2}}\leq 1.$$

( A trivial example for such sequence could be $\{\frac{\pi n}{\sqrt{6}}\}$)

Edit:  It would be good to have such sequences which increase "slowly"?

答:
As you noted, using the identity
$$\sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6}$$
we get
$$\sum_{n=1}^\infty \frac{1}{\left(\frac{\pi}{\sqrt 6}n\right)^2} = 1$$
Now by comparison, we see that the statement holds for all $x_n \ge \frac{\pi}{\sqrt 6}n$. Because
$$\begin{align}
x_n \ge \frac{\pi}{\sqrt 6}n &\implies \frac{1}{x_n^2} \le \frac{1}{\left(\frac{\pi}{\sqrt 6}n\right)^2} \\
&\implies \sum_{n=1}^\infty\frac{1}{x_n^2} \le \sum_{n=1}^\infty \frac{1}{\left(\frac{\pi}{\sqrt 6}n\right)^2} = 1
\end{align}$$

Since you want a class of such sequences which grow slowly, you can take any polynomial of the form
$$\boxed{x_n = \frac{\pi}{\sqrt 6}\sum_{i=0}^k a_in^i}  \tag{$\ge\frac{\pi}{\sqrt 6}n$}$$ where $a_i \ge 0, ~ (a_k \ne 0)$ and $k\ge 1$.

Of course, the larger you choose $k$, the faster the sequence increases as demonstrated in the picture below:


问:
Plotting inverse laplace function in matlab
I'm currently working on an assignment where we want to plot an output function.

Since our input signal( v(t) )is in the Time domain and our transfer function ( H(s) ) is in the S domain.
we want to bring our input signal to the S domain so we can simply use multiplication and then bring it back to the time domain to plot.

The error I'm getting is:
Error using plot
Data must be numeric, datetime, duration or an array convertible to double.

```<code>syms s C R L T n

C = (36)*10^(-6); % Our given values for capacitance, inductance and resistance
L = 0.18;
R = 100;
T = 1/50; % Period
H = 1/(C*R*s + C*L*s^2 + 1);        %Transfer function
y = 3/2;                           %c_0

for n = -5:1:5
    c_k = (((-1)^n)/(pi*(2*n+1)))*exp(1i*(2*n+1)*100*pi*t);
    y = y + c_k;
end
V = laplace(y);
Vc = V * H;
vc = ilaplace(Vc);

vpa(vc);
plot(t,vc);<span class="math-container">```
```

答:
I see different issues in this RLC filter circuit.

Here are the modifications you should operate at the end of your program:

```<code>vc = real(ilaplace(Vc));%"real" is compulsory even if... it is already real (very small imaginary parts way cree on)
vvc = string(vc);
close all;hold on;axis([-1,1,-3,3]);
f=str2func(['@(t)(',vvc,')']);%in order to get a function
for t=-1:0.1:1
    plot(t,f(t),'.')
end;
</code>```

You have to change types : "syms" type has to be converted into "string" type

You have to convert to real.

"t" is symbolic: Matlab cannot use it as it is; you know that you must give to $t$ a range of values, for example $t=-1:0.1:1$.

We use here the so-called "immediate definition" of a function (https://uk.mathworks.com/help/matlab/matlab_prog/anonymous-functions.html)

What is this function $y$ ? It looks like a truncated Fourier series expansion of a ramp function ?

vpa(...) isn't useful. Instead of for example $2/3$, it gives you $0.66666666666666666666666666...$ with thirty decimals... It's better to keep the fractional form.

The value of T isn't used.

Body:
I see different issues in this RLC filter circuit.

Here are the modifications you should operate at the end of your program:

vc = real(ilaplace(Vc));%"real" is compulsory even if... it is already real (very small imaginary parts way cree on)
vvc = string(vc);
close all;hold on;axis([-1,1,-3,3]);
f=str2func(['@(t)(',vvc,')']);%in order to get a function
for t=-1:0.1:1
    plot(t,f(t),'.')
end;

You have to change types : "syms" type has to be converted into "string" type

You have to convert to real.

"t" is symbolic: Matlab cannot use it as it is; you know that you must give to $t$ a range of values, for example $t=-1:0.1:1$.

We use here the so-called "immediate definition" of a function (https://uk.mathworks.com/help/matlab/matlab_prog/anonymous-functions.html)

What is this function $y$ ? It looks like a truncated Fourier series expansion of a ramp function ?

vpa(...) isn't useful. Instead of for example $2/3$, it gives you $0.66666666666666666666666666...$ with thirty decimals... It's better to keep the fractional form.

The value of T isn't used.


问:
Regularity of generic initial ideals: The only associated prime of $I$ containing $x_r$ is the maximal ideal.
I was going through Eisenbud's proof of a result of Bayer and Stillman concerning the regularity of an ideal and its generic initial ideal, and I'm having trouble understanding his proof. The theorem I present below is a simplified version (Eisenbud proves the theorem for graded free modules).
$\newcommand{\reg}{\operatorname{reg}}
\newcommand{\gin}{\operatorname{Gin}}
\newcommand{\In}{\operatorname{In}}
\newcommand{\m}{\mathfrak{m}}$

Theorem. Let $S=k[x_1,\ldots,x_r]$ be a polynomial ring over an infinite field $k$. Let $I$
be a homogeneous ideal of $S$ and $>$ be the reverse lexicographic order. Then $$\reg S/I=\reg S/\gin(I).$$

Proof. Rechoosing the variables $x_1,\ldots,x_r$ generically, we may assume in
particular that $\In(I) = \gin(I)$. Since the only associated prime of $I$
containing $x_r$ is the maximal ideal, we see that $(I:x_r)$ is of finite length....

It looks like Eisenbud is assuming $x_r$ is a zerodivisor on $S/I$. I think this is done via the generic change of coordinates, but I'm not sure. Anyway, even by assuming $x_r$ is a zerodivisor on $S/I$, I don't see the argument.

We know that the elements $x_r,\ldots,x_s$ form a regular sequence on $S/I$ iff $x_r,\ldots,x_s$ form a regular sequence on $S/\In(I)$. In particular, $x_r$ is a zerodivisor on $S/I$ iff $x_r$ is a zerodivisor on $S/\In(I)$. Moreover, since $\In(I)$ is Borel fixed, the associated primes of $\In(I)$ are all of the form $(x_1,\ldots,x_j)$ for $1\leq j\leq r$. Using these observations, we see that if $x_r$ is a zerodivisor on $S/I$ then $\m$ is an associated prime of $\In(I)$. Does this imply the above claim?

答:
I still don't see Eisenbud's argument, but the ultimate purpose of the argument is to prove that $(I:x_r)/I$ is of finite length. I present a different proof of this claim below.

From the above discussion, we see that $\newcommand{\reg}{\operatorname{reg}}
\newcommand{\gin}{\operatorname{Gin}}\newcommand{\In}{\operatorname{In}}\newcommand{\m}{\mathfrak{m}}\newcommand{\p}{\mathfrak{p}}\m$ is the only associated prime of $\In(I)=J$ containing $x_r$. Let $\p\neq \m$ be a prime ideal. Since the associated primes of $J_\p$ correspond to the associated primes of $J$ contained in $\p$ (reason: the set of associated primes commutes with localisation), we see that $x_r$ is a nonzerodivisor on $M_\p=(S/J)_\p$ (the set of all zerodivisors is equal to the union of all associated primes). In other words, $$\ker(M_\p\xrightarrow{x_r}M_\p)=((J:x_r)/J)_\p=0$$ for all $\p\neq \m$. This means that the annihilator of $(J:x_r)/J$ is not contained in any prime $\p\neq \m$. Hence the radical of the annhilator of $(J:x_r)/J$ is precisely $\m$ (radical of $\mathfrak a=\cap_{\mathfrak a\subset \p}\p$). Now it is clear that $(J:x_r)/J$ is a module over $S/\m^n$ for some $n$ and hence has finite length.

Since $(J:x_r)/J$ is a finite length module, we have $(J:x_r)_d=J_d$ for $d\gg0$. Since we are using the reverse lexicographic order, we have $(J:x_r)_d=J_d$ if and only if $(I:x_r)_d=I_d$ (see Lemma 2.2 here). Hence we see that $(I:x_r)/I$ has finite length.


问:
Theorem on quotient rings modulo an ideal
I've to prove the following theorem:

Let R a ring, I one of its two-sided ideal and π: R ⟶ R/I, the
canonical projection. Then

S subring (ideal) of R ⇒ π(S) = (S + I)/I is a subring
(ideal) of R/I;

S subring (ideal) of R/I ⇒
π−1(S') is a subring (ideal) of R containing I.

The map $\pi$ is $r ⟼ r + I$ with $r \in R$, so considering $s ⟼ s + I$ with $s \in S$, it should follow S + I ⊂ R + I, since S ⊂ R.

But I'm confused on why $\pi(S) = (S + I)/I$ and not $\pi(S) = S/I$.

答: Yes $S+ I\subset R+I$ (since the cosets of S in I will certainly all be cosets of R in I). I would just consider that ring homomorphism $\pi:S\to \pi(S)$. If we show that the kernel of this homomorphism is $S\cap I$, then we're done by the first isomorphism theorem (since $S/(S\cap I) \cong (S+I)/I$).


问:
Extensions and Limits of Inner Amenable Groups
A group $G$ is said to be inner amenable if there exists a positive linear functional $m$ on $\ell^{\infty}(G)$ with $m(1)=1$ that is invariant under conjugation; i.e., $m(g^{-1}fg) = m(f)$ for every $f \in \ell^{\infty}(G)$, where $(g^{-1}fg)(x) = f(g^{-1}xg)$.

I have two related questions. First, if $(G_{n}, f_{ij})$ is a injective directed system of groups, does it follow that $\displaystyle \lim_{\rightarrow} G_n$ is inner amenable? Second, if $N$ is an inner amenable normal subgroup of $G$ such that $G/N$ is inner amenable, does it follow that $G$ is inner amenable?

I know that these questions can be answered affirmatively if all the groups are actually amenable. But I am not sure about the inner amenable case.

答:
The first question has a positive answer.

Indeed, let $m_n$ be conjugacy invariant on $G_n$ with $m_n(\{e\})=0$, and push it forward to $\mu_n$ on the direct limit. Then $\mu_n(\{e\})=0$ using injectivity, and $\mu_n$ is $G_n$-invariant. Hence any cluster point of $(\mu_n)$ is a conjugacy-invariant mean as required.


问:
How many 10 letter 'words' can be made with no repeated letters and at least 8 consonants?
Need a little bit of help with this one.

How many 10 letter 'words' can be constructed which have no repeated letters and at least 8 consonants?

where 'word' is defined as any combination of the 26 letters of the english alphabet, and by consonant, (obviously) a letter of the english alphabet that are not either a,i,u,e, or o.

This is where I am at:

We need to choose the positions for
the 8 consonants which is $c(10,8)$

Then we need to choose the 8
consonants which I think is
$p(21,8)$? (not sure about this one)

Choose the remaining 2 letters: We have already used 8 letters so we only have 18 letters to choose from so $p(18, 2)$

What else needs to be calculated? Do we need to choose the position of the remaining 2 letters?

答:
The number of 10 letter words for any given 10 unique letters is simply 10!. Once you notice that the problem becomes how many unique sets of 10 letters are there for the problem at hand.

Breaking it down:

Calculate the number of unique sets of ten letters with 8-10 consonants and 0-2 vowels.

Multiply by the number of position permutations for a list with 10 items (10!).

The first step has three cases that need to be added together: words with 0, 1 or 2 vowels. So 21*20*19*18*17*16*15*14*13*12 + 5*21*20*19*18*17*16*15*14*13 + 5*4*21*20*19*18*17*16*15*14 = 1977336748800. Multiply that by 10! and you get: 7,175,359,594,045,440,000.

Or in your notation: (p(21,10) + p(5,1)*p(21,9) + p(5,2)*p(21,8)) * p(10)