答:
Regardless of whether $b = 0$ or not, we can always write $$f(x) = \cos(x)  - (1 + ax^2)(1 - bx^2 + b^2 x^4 - b^3 x^6 + \dots)$$ which leads to $$f(x) = (-a + b - 1/2)x^2 + (ab - b^2 + 1/24)x^4 + (-ab^2 + b^3 - 1/720)x^6 + \dots$$

Therefore for $a = -5/12$ and $b = 1/12$, we have $f(x) = 1/480 x^6 + \dots$ which has order of vanishing $6$.

On the other hand, if $f$ has order $\geq 6$, then we must have $−a+b−1/2 = 0$ and $ab−b^2+1/24 = 0$. It is easy to solve from these two equations that $a = -5/12$ and $b = 1/12$ is the only possibility.

Title: Problem from Zorich's book

Body:
Choose real numbers $a$ and $b$ so that the function $f(x)=\cos x - \dfrac{1+ax^2}{1+bx^2}$ is an infinitesimal of highest possible order as $x\to 0$.

My solution: Actually this problem is not so difficult and I guess that I've almost solved it. We'll split the solution into some cases:

If $b=0$ and using Taylor's series one can write $f(x)=x^2(-1/2-a)+o(x^3)$ as $x\to 0$.

So we see that if $a=-1/2$ then $f(x)=o(x^3)$. If $a\neq -1/2$ then $f(x)=o(x)$ as $x\to 0$.

Suppose that $b\neq 0$ then again using Taylor's series one can derive that $$f(x)=\cos x-(1+ax^2)(1+bx^2)^{-1}=x^2\left(-\frac{1}{2}-a+b\right)+x^4\left(\frac{1}{24}-b^2+ab\right)+x^6\left(-\frac{1}{720}-ab^2+b^3\right)+o(x^7).$$

One can show that if $a=-\dfrac{5}{12}$ and $b=\dfrac{1}{12}$ then $f(x)=o(x^5)$ as $x\to 0$.

Definition: If $f=o(g)$ and g is itself infinitesimal as $x\to 0$, we say that $f$ is an infinitesimal of higher order than $g$ as $x\to 0$.

Question: Can anyone explain in details which one has the highest possible order? I guess it should be in the case when $a=-\dfrac{5}{12}$ and $b=\dfrac{1}{12}$ but cannot explain it in a rigorous way. If someone can explain it in a great detail that would be awesome!


问:
Problem from Zorich's book
Choose real numbers $a$ and $b$ so that the function $f(x)=\cos x - \dfrac{1+ax^2}{1+bx^2}$ is an infinitesimal of highest possible order as $x\to 0$.

My solution: Actually this problem is not so difficult and I guess that I've almost solved it. We'll split the solution into some cases:

If $b=0$ and using Taylor's series one can write $f(x)=x^2(-1/2-a)+o(x^3)$ as $x\to 0$.

So we see that if $a=-1/2$ then $f(x)=o(x^3)$. If $a\neq -1/2$ then $f(x)=o(x)$ as $x\to 0$.

Suppose that $b\neq 0$ then again using Taylor's series one can derive that $$f(x)=\cos x-(1+ax^2)(1+bx^2)^{-1}=x^2\left(-\frac{1}{2}-a+b\right)+x^4\left(\frac{1}{24}-b^2+ab\right)+x^6\left(-\frac{1}{720}-ab^2+b^3\right)+o(x^7).$$

One can show that if $a=-\dfrac{5}{12}$ and $b=\dfrac{1}{12}$ then $f(x)=o(x^5)$ as $x\to 0$.

Definition: If $f=o(g)$ and g is itself infinitesimal as $x\to 0$, we say that $f$ is an infinitesimal of higher order than $g$ as $x\to 0$.

Question: Can anyone explain in details which one has the highest possible order? I guess it should be in the case when $a=-\dfrac{5}{12}$ and $b=\dfrac{1}{12}$ but cannot explain it in a rigorous way. If someone can explain it in a great detail that would be awesome!

答: If $f\in o(x^5)$, $x \to 0$, then $f(x)=\varepsilon(x) x^5 $, where $\lim\limits_{x\to x_0} \varepsilon(x) = 0$. Hence we can write representation $f(x)=\varepsilon(x) x^2 \cdot x^3= \varepsilon_1(x)\cdot x^3$, where  $\lim\limits_{x\to x_0} \varepsilon_1(x) = 0$, so we can say, that $f\in o(x^3)$ i.e. $o(x^5) \subset o(x^3)$. So, being in $o(x^5)$ means to be more higher order, then being in $o(x^3)$.

ContentLicense: CC BY-SA 4.0

Title: Problem from Zorich's book


问:
Show the following satisfies bounded convergence theorem
Assume $f_n: [0,1] \to [0,\infty)$ is integrable for each $n$ and $(f_n)_{n = 1}^{\infty}$ converges pointwise a.e. to $f$. Prove that

$$\lim_{n \to \infty} \int_{[0,1]} f_n(x) e^{-f_n(x)}dx = \int_{[0,1]} f(x)e^{-f(x)}dx$$

I think I am very close. To satisfy the BND convergence theorem, we need the input to be of finite measure, which we have: $m([0,1]) = 1 - 0 = 1 < \infty$. We also need the sequence of functions to converge pointwise, which we have by assumption.

The only thing I believe I am missing is that $\exists c \in (0,\infty)$ s.t. $|f_n(x)| \leq c \ \forall n \in \mathbb{N}$. If I had this, then BND convergence theorem would apply.

Maybe I'm not seeing it, but what in this problem implies we have this bound? Can anyone point me in the right direction?

答:
Note that $f_n(x) e^{-f_n(x)}$ is uniformly bounded (EDIT: Since the functions $f_n$ are non-negative).

(For example, a plot of $x e^{-x}$ will show it immediately.) So the bounded convergence theorem applies as you are on a space of finite measure, and the sequence of functions you are integrating are uniformly bounded.

Note that the bounded convergence theorem is a special case of the Dominated Convergence Theorem.


问:
Why am I not getting total mortgage equal to $P \frac{r (1+r)^{n}}{(1 + r)^{n} - 1}$?
For a computer science exercise I had to make a simple mortgage calculator. Given the principial, annual interest rate and period in years the program should return the monthly payment. I was not familiar with mortgage calculation, but after doing some research most of the resources pointed to this popular formula:

$$M = P \frac{r (1+r)^{n}}{(1 + r)^{n} - 1}$$

I personally didn't find this a very intuitive formula. Out of curiosity I tried to use a more intuitive approach based on the basic principles of a how a fixed rate loan should work.

So basically every month you pay percentage interest and that gets substracted from the principal, leaving you with a smaller principal and therefore a little bit less interest to pay.

If P is your principal and R is your annual interest rate (in decimal form) and r = (R/12) your monthly interest rate, then it would make sense to me to say that
$$(1 - r) * P$$
is the principal left after the first payment.

If we know the total amount of payments (12 months for every year), let's call this n, we can write something like this: $$(1 - r)^{n}* P$$
and this should equal to the remaining principal after n payments. For a 30 year loan, there would be 360 monthly payments and this formula should give us the amount of principal left after consistently paying r interest monthly.

If we substract this remainder from the total principal, we get the total amount of interest paid over the period. The sum of the interest paid and the original principal would be the total cost of the loan.

For the sake of having one uniform monthly payment, you could divide the total cost by the amount of payments.

This made sense to me, but this is always several dollars off from the calculation done with standard formula I've given in the beginning.

Take this example.

Principal = $100000, Annual Interest Rate = 3.92%, Period of loan = 30 years.

Using the standard formula we get a monthly payment of 472.81 dollars, but with my approach we get $470.02386.

I'm assuming that my way of thinking is not totally wrong, but there is clearly something off. I am not disputing the correctness of the standard formula as I have seen the proof for it. I just would like for someone to point out why my approach wouldn't work or how to correct to it.

Many thanks in advance.

答: You have a basic misunderstanding.  You say that after one payment, the principal remaining should be $P(1-r)$, that is, that the interest is applied to reduce the principal.  This is the reverse of how things actually work.  Each month, you incur an interest charge, and that's the first thing you must pay.  The interest is subtracted from your payment, and any remainder is applied to reduce the principal.  If you didn't make a payment, or made only a partial payment that didn't cover the full interest charge, then the loan payment would, in theory, increase.  (I say "in theory" because I'm not sure how it's treated in law, but as far as the computation goes, and the amount you will owe in the future, it's exactly how this will work.  Legally, the classification as interest or principal may be different, but the amount of money will be the same.)


问:
Motivation for the definition of quotient space
I'm having some trouble understanding the concept of quotient space of a vector space. The definition my teacher gave us is the following:

Let $V$ be a finite-dimensional vector space over a field $\mathbb K$ and $V' \leq V$.

We can define the following equivalence relation: for any $u,v \in V, u \simeq v$ iff $v -u\in V'$. Let, for any $u\in V$, $[u]=\{v \in V: u \simeq v\}$.

Then we can define: $$V/V':=\{[u],u\in V\}$$

In the first question, my teacher asked us to determine the vector space $\mathbb R^2 / span(1,2)$ and I have no clue how to proceed because I'm still confused about the definition. Is there any intuition on why we define it this way or the meaning of the quotient space? And, how can I determine the space $\mathbb R^2 / span(1,2)$?

答:
Maybe a concrete example will help your intuition: imagine a $3$-dimensional scenario and a (scale $1{-}1$) $2$-dimensional map of that scenario. The map shows two points $(x, y, z)$ and $(x, y, z')$ as the same point, i.e., it identifies $(x, y, z)$ and $(x', y', z')$ if $(x, y, z) - (x', y', z') = (0, 0, z'')$ for some $z''$. So with $V= \Bbb{R}^3$ and $V' = \{ (0, 0, z'') \mid z'' \in \Bbb{R}\}$, you can think of the points on the map as the equivalence classes under the equivalence relation your teacher defined.

For $\Bbb{R}^2/\mathrm{span}\{(1, 2)\}$, you can describe the quotient space as the set of all lines equal to or parallel to the line $L$ comprising all points of the form $(t, 2t)$. You can identify any such line by its point of intersection with some chosen line $L'$ through $(0, 0)$ that is not parallel to $L$. $L'$ is a subspace of $\Bbb{R}^2$ and the vector space operations on $\Bbb{R}^2/\mathrm{span}\{(1, 2)\}$ are the same as those on $L'$ under this identification.


问:
Questions about SVD, Singular Value Decomposition
I am not a mathematician, so I need to understand what SVD does and WHY more than how it works exactly from the math perspective. (I understand at least what is the decomposition though).

This guy on youtube gave the only human explanation of SVD saying, that the U matrix maps "user to concept correlation" Sigma matrix defines the strength of each concept, and V maps "movie to concept correlation" given that initial matrix M has users in the rows, and movie (ratings) in the columns.

He also mentioned two concept specifically "sci fi" and "romance" movies. See the picture below.

My questions are:

How SVD knows the number of concepts. He as human mentioned two - sci fi, and romance, but in reality in resulting matrices are 3 concepts. (for example matrix U - that one with blue titles - has 3 columns not 2).

How SVD knows what is the concept after all. I mean, what If i shuffle the columns randomly how SVD then knows what is sci fi, what is romance. I mean, I suppose there is no rule, group the concepts together in the column order. What if scifi movie is the first and last one? and not first 3 columns in the initial matrix M?

What is the practical usage of either U, Sigma or V matrices? (Except that you can multiply them to get the initial matrix M)

Is there also any other possible human explanation of SVD than the guy up provided, or it is the only one possible function? Matrices of correlations.

答:
Here is a way to understand from a different point of view what the SVD means, using an algorithm based on a balanced weighting between rows and columns.

I will use two slides of the Linear Algebra lectures I have been giving for many years (adapted from its french version):

First slide.

It deals with the following SVD:

$$\underbrace{\begin{pmatrix} 0&1\\1&2\\3&3\end{pmatrix}}_A = \underbrace{\begin{pmatrix}  \color{red}{0.1595} &  \ \ 0.7077  & \ \ 0.6882\\
   \color{red}{0.4520} &  \ \ 0.5675  &  -0.6882\\
   \color{red}{0.8776}  &  -0.4208 &  \ \ 0.2294\end{pmatrix}}_U \underbrace{\begin{pmatrix}
    \ \ \color{red}{4.8146}    &     0\\
         0   & 0.9054\\
         0    &     0\end{pmatrix}}_{\Sigma}
\underbrace{\begin{pmatrix}\color{red}{0.6407}  & -0.7678\\
   \color{red}{0.7678} &  \ \ 0.6407\end{pmatrix}^T}_{V^T}$$

This algorithm (not the most efficient) produces, when stopped after a certain number of steps, a numerical approximation of the first singular element $(U_1,V_1,\sigma_1)$ where $U_1$ and $V_1$ are the first columns of $U$ and $V$ resp. and $\sigma_1$ the dominant singular value.

Why does this "alternate balancing" works ? Here is an explanation.

Second slide.

In fact, it is not surprizing that matrices $AA^T$ and $A^TA$ have "poped up". We find back a property that has been used for a long time to introduce the SVD:

If $A$ is $m \times n$ with $m>n$, the $m \times m$ matrix $AA^T$ and the $n \times n$ matrix $A^TA$, being squared symmetric semi-definite positive matrices have the resp. eigendecompositions :

$$AA^T=U \Sigma' U^T \ \ \ \& \ \ \ A^TA=V \Sigma'' V^T\tag{1}$$

with $U$ and $V$ orthogonal matrices and the same eigenvalues (set apart the zero eigenvalues). More exactly, $\Sigma''$ is the upper diagonal block of $\Sigma'$, the remaining (diagonal) entries of  $\Sigma'$ being filled by zeroes.

Remarks:

Of course, (1) is an immediate consequence of the multiplication of

$$A=U \Sigma V^T \ \ \ \text{by} \ \ \ A^T=(U \Sigma V^T)^T=V \Sigma^T V^T$$

due to the fact that $U^TU=I$ and $V^TV=I$ (orthogonality property).

Once $(U_2,V_2,\sigma_2)$ has been obtained, a so-called "deflation" $A'=A-\sigma_1U_1V_1^T$ is operated on matrix $A$, then the same algorithm is applied to $A'$ in order to get the second singular element $(U_2,V_2,\sigma_2)$. Here, from a numerical point of view, there is a drawback for big matrices : deflation accumulates errors. But efficiency and precision are not here our main concerns, our objective being a better insight about what SVD really is...

Matrix $\Sigma$ is a kind of "tradeoff" between $\Sigma'$ and $\Sigma''$.


问:
Need help understanding Erdős' proof about divergence of $\sum\frac1p$
I'm looking at proofs from Proofs from the Book (Martin Aigner, Günter M. Ziegler). The proof I'm having trouble is the sixth proof of the infinitude of the primes they give (on page 5; although I'll be reconstructing it here). I'll be highlighting the areas I'm having trouble with. They set out to prove two things; namely, and letting $P$ be the set of all primes:

$P$ is infinite.

$\sum_{p \in P} {1 \over p}$ diverges.

They begin by assuming the sum in 2. is convergent and letting $P = \{p_1, p_2, ...\}$ The first problem comes up here. By assuming the given form for $P$, haven't they already assumed that it is infinite? Then, by assumption,

$$\sum_{i > k} {N \over p_i} < {N \over 2}, \ \  (N \in \mathbb{Z}^+). \tag{1}$$

They then define $p_1, p_2, ..., p_k$ to be the small primes, and $p_{k + 1}, p_{k + 2}, ...$ to be the large primes, and define $N_b$ to be the number of positive integers $n \leq N$ which are divisible by at least one of the big primes; and, similarly, $N_s$ the number of positive integers $n \leq N$ which only have small prime divisors. By definition, $N_b + N_s = N$, and the intended contradiction will hinge on showing $N_b + N_s < N$.

Note, since $\left \lfloor {N \over p_i} \right \rfloor$ gives the number of positive integers $n \leq N$ which are multiples of $p_i$,

$$N_b \leq \sum_{i > k} \left \lfloor {N \over p_i} \right \rfloor < {N \over 2}. \tag{2}$$

Then, they let $n = a_n b_n^2$, where $n \leq N$ such that it has only small prime divisors. Here $a_n$ is the square free part. Then, every $a_n$ is a product of different small primes. Assuming that by this it is simply meant that every factor of $a_n$ is distinct, they assert there are exactly $2^k$ different square free parts. Notice, also, $b_n \leq \sqrt{n} \leq \sqrt{N} \implies b_n \leq \sqrt{N}$, and as such,

$$N_s \leq 2^k \sqrt{N}. \tag{3}$$

Here, is the second problem for me. I can see intuitively how $(3)$ holds; but, I'm not really sure, and I'd like to see it through a more rigorous approach, and also, why it's relevant. I'm not really sure, and I'd like a more rigorous approach.

They conclude the proof by showing that, since $(2)$ holds for any $N$, $2^k \sqrt{N} \leq {N \over 2}$ holds for $N = 2^{2k + 2}$. This is the third area I'm having trouble. I'm not really sure why this is relevant and how the contradiction $N_s + N_b < N$ follows from it. Maybe I missed it, but I also didn't see a proof of the infinitude of primes here.

Finally, I have a question as well. Looking back, how could one have guessed that this approach will lead us to a proof of the two claims above? In other words, what's the insight behind taking this particular approach?

Thanks to all in advance!

Note: I wasn't really sure what tags relating to "Proofs" would be appropriate here, so I've only used tags concerning number theory and primes. Please feel free to tag away.

答:
By writing $P = \{p_1, p_2, ...\}$, have they assumed the infinitude of primes

It is not specified whether this is an infinite or finite sequence. We often (and usually, for that matter) assume that $...$ means there are infinitely many items. But we could just as well assume that there are only finitely many terms, and they just don't know what number to give the last one. In particular, it does not impact the proof in any way to suppose instead that $P = \{p_1, p_2, \ldots, p_k\}$ for some unknown $k$, except to transform the proof into a proof by contradiction.

As a brief aside, I'd like to point out that Euclid's proof of the infinitude of primes was not by contradiction. Rather, given any finite set of primes, he produced another, showing that no finite set contains all the primes. This is slightly different than assuming there are finitely many primes and arriving at a contradiction. This same subtlety almost seems to be at work here - we don't assume there are finitely many or infinitely many - we just enumerate the sequence of primes as much as we can.

(Daniel McLaury just posted his answer, and he addresses this point just fine so I leave it to him)

The final contradiction they arrive at is by showing that $N_b + N_s < \frac{N}2 + \frac N2 = N$, which is a contradiction as $N_b + N_s = N$. Their estimates of $N_b$ and $N_s$ are just to get to this contradiction.

ContentLicense: CC BY-SA 3.0


问:
Show that first order Peano's axioms capture the natural numbers regarding satisfiablity
Denote be $\mathcal P_{MO}$ the set of the monadic second order axioms of Peano. Then as shown by Dedekind any two model are isomorphic to $\mathbb N$. Hence for a monadic second order sentence we have
$$
 \mathbb N \models \varphi \mbox{ iff } \mathcal P_{MO} \models \varphi
$$
i.e., the natural numbers fullfil $\varphi$ iff every model that fulfills $\mathcal P_{MO}$ also fulfills $\varphi$.

Let $\mathcal P_{FO}$ be the first order Peano arithmetic where the monadic induction axiom is replaced by a first order induction scheme over arbitrary predicates.

Let $\psi$ be some first order sentence, then how to show that $\mathcal P_{FO} \models \psi$ implies $\mathbb N \models \psi$? (the other direction is clear, so I am asking about the relation between satisfiability of the axioms vs satisfiability in the natural numbers).

Similar for other first order axiomatization like Robinson arithmetic? The problem is that as mentioned here, the first order theories are not categorical.

答:
To move this off the unanswered queue:

The fact that $\mathcal{P}_{FO}\models\psi\implies\mathbb{N}\models\psi$ is immediate from the definition of $\models$ once you know that $\mathbb{N}\models\mathcal{P}_{FO}$: by definition we have $\mathcal{P}_{FO}\models\psi$ iff every model of $\mathcal{P}_{FO}$ satisfies $\psi$, so since $\mathbb{N}\models\mathcal{P}_{FO}$ we must have $\mathbb{N}\models\psi$.

The other direction meanwhile is false, since $Th(\mathbb{N})$ is complete (every structure's theory is complete) but $\mathcal{P}_{FO}$ is not complete (per Godel - both incompleteness and completeness, the latter letting us freely switch between $\vdash$ and $\models$ when applying the former).

In general, whenever we have a theory and a model of that theory in mind, think of the theory as a probably-limited-approximation to the model: everything the theory knows is true (about that model), but there are probably lots of things the theory doesn't know.


问:
What is the structure of the probability space that makes the probabilty of coprimity equal to $\frac{6}{\pi^2}$
Let $X=\mathbb{N}\times \mathbb{N}$ be the product of positive integers. I wonder how to establish the sigma-algebra and the probability measure on $X$ in a formal way so that the probability of the set

$$\{(m,n):\gcd(m,n)=1\}$$

is equal to $\frac{6}{\pi^2}$.

If we start with the probability $\mathbb P$ on $\mathbb N$ and take the product, then this structure should make the probability of $\{m:p|m\}$ equal to $\frac{1}{p}$. Then what are the probabilities of the singletons $\{n\}$? (If the sigma algebra is discrete, then they shouldn't be all zero otherwise there would be a contradiction... So maybe we can use nondiscrete sigma algebra )

Reference: Probability that two random numbers are coprime is $\frac{6}{\pi^2}$

答: This is not a rigorous statement about probability in a fixed probability space. You could make it rigorous by saying that if $X = \{1,2,\dots,N\} \times \{1,2,\dots,N\}$ and we sample an ordered pair $(m,n)$ uniformly from $X$, then $$\lim_{N \to \infty} \Pr[\gcd(m,n) = 1] = \frac{6}{\pi^2}.$$


问:
What is the mean of the random variable $\min\{X_1,X_2\}$ and $\max\{X_1,X_2\}$, where each $X_i$ is distributed uniformly over $[0,1]$?
So as stated on the title, what is the mean of the random variable $\min\{X_1,X_2\}$ and $\max\{X_1,X_2\}$, where each $X_i$ is distributed uniformly over $[0,1]$?

I know it's going to be $1/3$ and $2/3$ but I cannot derive it analytically.

Thank you in advance

Paul

答:
The function $\min(x,y)$ can be detected by considering the function $x-y$. When $x-y\ge 0$, we have $\min(x,y)=y$, and when $x-y<0$ we have $\min(x,y)=x$.

So it suffice to consider the value of $x-y$ on the unit square $[0,1]\times [0,1]$. It is obvious that above the diagonal line, $x-y<0$, and below the diagonal line, $x-y\ge 0$. So to calculate the "mean" of $\min(x,y)$ it suffice to integrate $x$ on the upper half triangle and $y$ on the lower half triangle.

Now write integral along the upper half triangle as $$\int^{y}_{0}xdx\int^{1}_{0}dy=\int^{1}_{0}\frac{1}{2}y^{2}dy=\frac{1}{6}$$
The integral along the lower half triangle as
$$\int^{x}_{0}ydy\int^{1}_{0}dx=\int^{1}_{0}\frac{1}{2}x^{2}dx=\frac{1}{6}$$

So the whole "mean" is the sum $\frac{1}{3}$. I am leaving the second result for you to prove.


答:
Your formula amounts to:
$$M = P \frac{2-(1-r)^n}{n}. \tag{1}$$
For the sake of clarity, $r$ is the effective monthly rate of interest, and $n$ is the number of months in the loan.  Note that we immediately run into trouble if $r > 1$, which is certainly possible for especially usurious loans (e.g., a very short term payday loan could easily have effective annual rates of interest exceeding $300\%$).  Practicality aside, the mathematics must remain valid even when the rate of interest is especially high.  But in your case, if we set $r = 3$, then $(1-r)^n = (-2)^n$, which would be positive if $n$ is even and negative if $n$ is odd.  This would suggest your periodic payment $M$ would be negative if the interest rate is high enough and there were an even number of payment periods!

Here is a more realistic example, in which we might let $r = 0.05$ and $n = 100$.  Then on a principal of $P = 100000$, the correct formula gives $M = 5038.31$, whereas yours is much smaller, only $M = 1994.08$.  As you can see, the deviation is substantial.

So now that we know that this formula cannot be mathematically correct, why does this model you propose fail?  The problem is that you are confusing different time values of money, and not taking into account the compounding of interest.

In particular, when you say that after $n$ months, the amount of principal remaining is $(1-r)^n P$, this makes no sense, because it means that the loan is never extinguished; the quantity $(1-r)^n$ for any $0 < r < 1$ and $n > 0$ is always positive.  For example, if $r = 0.0001$ and $n = 1000$, $(1-r)^n \approx 0.904833$, meaning that your formula would say that over $90\%$ of the principal is still remaining after $1000$ payments.

Instead, what you should write is an equation of value that matches the present value of the periodic payments made in the future, against the present value of the principal of the loan.  What I mean by this is that if $r$ is the monthly rate, then a payment of $M$ made $t$ months into the future would have the present value $$M(1+r)^{-t}.$$  Then the sum of these "discounted" payments must equal the total principal of the loan.

Alternatively, we can say that if $M$ is the monthly payment, then after the first month has passed, the loan has accrued interest of $rP$, hence the total outstanding balance is $P + rP = (1+r)P$, and you pay off $M$ of this, so the remaining balance is $(1+r)P - M$.  Then another month passes, at which point the outstanding balance has increased by a factor of $1+r$, and you pay off another month:
$$(1+r)((1+r)P - M) - M = (1+r)^2P - (1+r)M - M.$$
And then after the third month, the same process happens:  accumulate interest on the outstanding balance, then subtract the payment $M$:
$$(1+r)((1+r)^2 P - (1+r)M - M) - M = (1+r)^3 P - (1+r)^2 M - (1+r) M - M.$$
So after $n$ payments, we get
$$(1+r)^n P - (1+r)^{n-1} M - (1+r)^{n-2} M - \cdots - (1+r) M - M.$$
And if, after the $n^{\rm th}$ payment, we have no balance left over, this gives us the equation of value that lets us compute what $M$ needs to be in order for the loan to be paid off.

Title: Why am I not getting total mortgage equal to $P \frac{r (1+r)^{n}}{(1 + r)^{n} - 1}$?

Body:
For a computer science exercise I had to make a simple mortgage calculator. Given the principial, annual interest rate and period in years the program should return the monthly payment. I was not familiar with mortgage calculation, but after doing some research most of the resources pointed to this popular formula:

$$M = P \frac{r (1+r)^{n}}{(1 + r)^{n} - 1}$$

I personally didn't find this a very intuitive formula. Out of curiosity I tried to use a more intuitive approach based on the basic principles of a how a fixed rate loan should work.

So basically every month you pay percentage interest and that gets substracted from the principal, leaving you with a smaller principal and therefore a little bit less interest to pay.

If P is your principal and R is your annual interest rate (in decimal form) and r = (R/12) your monthly interest rate, then it would make sense to me to say that
$$(1 - r) * P$$
is the principal left after the first payment.

If we know the total amount of payments (12 months for every year), let's call this n, we can write something like this: $$(1 - r)^{n}* P$$
and this should equal to the remaining principal after n payments. For a 30 year loan, there would be 360 monthly payments and this formula should give us the amount of principal left after consistently paying r interest monthly.

If we substract this remainder from the total principal, we get the total amount of interest paid over the period. The sum of the interest paid and the original principal would be the total cost of the loan.

For the sake of having one uniform monthly payment, you could divide the total cost by the amount of payments.

This made sense to me, but this is always several dollars off from the calculation done with standard formula I've given in the beginning.

Take this example.

Principal = $100000, Annual Interest Rate = 3.92%, Period of loan = 30 years.

Using the standard formula we get a monthly payment of 472.81 dollars, but with my approach we get $470.02386.

I'm assuming that my way of thinking is not totally wrong, but there is clearly something off. I am not disputing the correctness of the standard formula as I have seen the proof for it. I just would like for someone to point out why my approach wouldn't work or how to correct to it.

Many thanks in advance.


答:
The $\frac6{\pi^2}$ you reference is not really a probability but the limit of a sequence of probabilities, where the pairs are chosen uniformly from a finite set for each experiment in the sequence. So you're asking whether there's a single distribution over $\Bbb N\times\Bbb N$ that "does the same thing".

There is no uniform discrete distribution on an infinite set, since the probabilities must sum to $1$. If you allow non-uniform distributions, then many "silly" solutions are possible that give the desired probability, e.g.

$$P(X=x)=\begin{cases}\frac6{\pi^2}&\text{if }x=(1,1)\\1-\frac6{\pi^2}&\text{if }x=(2,2)\\0&\text{otherwise.}\end{cases}$$

Title: What is the structure of the probability space that makes the probabilty of coprimity equal to $\frac{6}{\pi^2}$

Body:
Let $X=\mathbb{N}\times \mathbb{N}$ be the product of positive integers. I wonder how to establish the sigma-algebra and the probability measure on $X$ in a formal way so that the probability of the set

$$\{(m,n):\gcd(m,n)=1\}$$

is equal to $\frac{6}{\pi^2}$.

If we start with the probability $\mathbb P$ on $\mathbb N$ and take the product, then this structure should make the probability of $\{m:p|m\}$ equal to $\frac{1}{p}$. Then what are the probabilities of the singletons $\{n\}$? (If the sigma algebra is discrete, then they shouldn't be all zero otherwise there would be a contradiction... So maybe we can use nondiscrete sigma algebra )

Reference: Probability that two random numbers are coprime is $\frac{6}{\pi^2}$


问:
Showing ${\rm Aut}(V) \cong S_3$
I am trying to prove that the automorphisms of the Klein four-group is isomorphic to $S_3$.

The first insight is the Klein group is an abelian group of four elements, three of which are order $2$, and must be mapped to other elements of order $2$, and the identity must of course be mapped to itself.

So any automorphism "permutes" the non-identity elements.

So there are exactly $3! = 6$ of them, matching the number of elements of $S_3$. Then if I call those non-identity maps $f_2, f_2, \ldots, f_6$, I can naturally map them to $S_3$.

For example, if the non-identity elements of the Klein group are $a$, $b$, $c$, I can regard these as $1$, $2$, and $3$ (I believe the labelling doesn't matter) and if I swapped $a$ and $b$ and fixed $c$, I can regard this as the transposition $(12)$.

By repeating this, I get a natural bijection of sets.

I don't know how to "prove" it's a bijection other than to observe the mapping and that it hits everything in $S_3$ exactly once.

I don't know, in particular, how to prove that this is a homomorphism, other than by checking every possible pair of isomorphisms, which is a bit laborious.

I assume there is a better, cleaner way to do it.

UPDATE: After working on this problem more, I have an updated attempt. It's incomplete, but I think it makes some progress.

We realize the Klein 4-group as $V = \{e, a, b, c\}$, where $a,b,c$ are the non-identity elements, each of which has order $2$. Any automorphism $f$ of $V$ must send elements of order $2$ to elements of order $2$, so necessarily it must permute the elements $a,b,c$. I claim that such a permutation is necessarily an automorphism. We observe that the Klein four-group has the property that if we multiply two non-identity elements, we get the third. Since $f$ maps from $V$ to $V$ bijectively, it is also the case that when $f(g)f(g') = f(g'')$, where $g$ and $g'$ are distinct elements and $g''$ is the third. That is, for any $g \neq g' \neq e$, we have
\begin{align*}
f(gg') = f(g'') = f(g)f(g'').
\end{align*}
If we take an arbitrary $g \in V$ and then consider multiplying with the identity $e$, we have
\begin{align*}
f(ge) = f(g) = e f(g) = f(e) f(g),
\end{align*}
since $f(e) = e$. This proves that $h$ is a homomorphism, and since it is bijective by definition, it is an automorphism. Since every automorphism must send elements of order $2$ to elements of order $2$ and there are exactly $3! = 6$ permutations of $\{a,b,c\}$, there are exactly six elements of Aut(G).

An automorphism is, therefore, uniquely determined on how it permutes the elements ${a,b,c}$, giving a natural bijective map to $S_3$, relabelling these non-identity elements as $1,2,3$. I call $e$ the identity permutation and $f_{abc}$ as the automorphism that maps ${a,b,c} \to {a,b,c}$ (i.e., the identity map) and so forth. The map is:
\begin{align*}
& f_{a,b,c} \longmapsto e & & \text{both have order $1$} \\
& f_{a,c,b} \longmapsto (23) & & \text{both have order $2$} \\
& f_{b,a,c} \longmapsto (12) & & \text{both have order $2$} \\
& f_{c,b,a} \longmapsto (13) & & \text{both have order $2$} \\
& f_{b,c,a} \longmapsto (123) & & \text{both have order $3$} \\
& f_{c,a,b} \longmapsto (132) & & \text{both have order $3$}
\end{align*}
This is clearly a bijection of sets: every permutation in $S_3$ was hit exactly once. I still do not know how to show that this is a homomorphism. I constructed the map so that it preserved the order of elements, but is that enough? I could check that $\varphi(xy) = \varphi(x) \varphi(y)$ for each pair, but there must be a way to do it that I'm missing.

答:
Good thing about $V$ is that $V \cong \mathbb{Z}_{2}\times\mathbb{Z}_{2}$. And there is a nice approach, because once you determined where $(1,0)$ and $(0,1)$ go the whole automorphism is defined. So let's represent automorphism as a matrix $$\left(\begin{array}{cc}
a & b\\
c& d
\end{array}\right),$$
where $(1,0)\mapsto (a,b)$ and $(0,1)\mapsto (c,d)$. It easy to show that such matrix is invertible. More than that, composition of two automorphism is given by multiplication of their matrices. So let $G$ be a group of invertible $2\times 2$ matrices with $\{0,1\}$ coefficients. $|G|=6$ and we can easily list its elements: $$\left(\begin{array}{cc}
1 & 0\\
0& 1
\end{array}\right),\left(\begin{array}{cc}
1 & 1\\
0& 1
\end{array}\right),\left(\begin{array}{cc}
1 & 0\\
1& 1
\end{array}\right),\left(\begin{array}{cc}
1 & 1\\
1& 0
\end{array}\right),
\left(\begin{array}{cc}
0 & 1\\
1& 1
\end{array}\right),\left(\begin{array}{cc}
0 & 1\\
1& 0
\end{array}\right)
$$
From here we can easily determine "transpositions", because they are of order 2. We can match them with matrices of order 2 and then extend to isomorphism (because transpositions generate symmetric group and such matrices satisfy relations). This is a bit laborious too, but gives some kind of systematic approach.

Title: Showing ${\rm Aut}(V) \cong S_3$


问:
Gradient of least-squares cost — how to compute it?
I have a system $A x ~ b$ where vector $b$ is not actually in the span of matrix $A$. I want to use a least squares approach to minimize the distance between the two vectors.

$$\begin{aligned} \min_x \Vert \mathbf Ax - b\Vert ^2 &= (\mathbf Ax-b)^T (\mathbf Ax-b) \\ &= x^T \mathbf A^T \mathbf A x - (\mathbf Ax)^Tb - b^T \mathbf Ax + b^Tb \\ &= x^T \mathbf A^T \mathbf A x - 2b^T \mathbf Ax + b^Tb \end{aligned}$$

I'm struggling to understand how to arrive at

$$0 = 2 \mathbf A^T \mathbf Ax -2 \mathbf A^T b$$

I understand that some sort of derivative with respect to $x$ has been taken in order to minimize the distance, but what does it mean to take a derivative of a matrix product with respect to a vector? I have never come across such a calculation?

Thanks in advance.

答:
Yes, you can indeed perform matrix calculus, which is exactly what it sounds like. In this case, we're differentiating a scalar value (since it's a norm, which is just a number) by the vector $x$, and the way you do that is surprisingly simple - you differentiate by each of the vector's components, and make a vector out of that. In other words, if $x = \left( x_1, x_2, \ldots, x_n \right)$ then $\frac{\partial}{\partial x} = \left(\frac{\partial}{\partial x_1}, \frac{\partial}{\partial x_2}, \ldots, \frac{\partial}{\partial x_n} \right)$.

It might not be entirely clear how that's going to work when you've got matrix operations happening under the hood there, but because matrices and derivatives are both linear operators (i.e. they behave nicely when you add things together or multiply them by constants) stuff mostly follows sensible rules, and in particular if you go to the section in that Wikipedia article on "Identities" you'll find the bit you're looking for under "Scalar-by-vector identities", in particular:

$\mathbf{A}$ is not a function of $\mathbf{x}$, $\mathbf{A}$ is symmetric: $\frac{\partial \mathbf{x}^\top\mathbf{A}\mathbf{x}}{\partial \mathbf{x}} = 2\mathbf{x}^\top\mathbf{A}$


答:
As saulspatz points out, you have a very fundamental misunderstanding about the finance of amortized mortgage payments. (Your principal isn't reduced by $r$ percent each month.) But precisely for that reason, it’s surprising that your answer is so close to the correct answer. Let’s see why mathematically.

Let $M$ be the monthly payment, $R$ the annual interest rate, $r:=R/12$, $P$ the principal, and $n$ the number of monthly payments.

Your formula is

$$M_{you}=\frac{P+\big(1-(1-r)^n\big)P}{n}=P\cdot\color{blue}{\frac{2-(1-r)^n}{n}}$$

whereas the correct formula is

$$M_{correct}=P\cdot\color{red}{\frac{r(1+r)^n}{(1+r)^n-1}}$$

The question is thus why

$$\color{blue}{\frac{2-(1-r)^n}{n}}\approx\color{red}{\frac{r(1+r)^n}{(1+r)^n-1}}$$

When $n$ is big and $r$ is small, $(1+r)^n\approx 1+rn$ (to first order, using the binomial theorem), so

$$\color{red}{\frac{r(1+r)^n}{(1+r)^n-1}}\approx\frac{r(1+rn)}{rn}=\frac{1+rn}{n}$$

and

$$\color{blue}{\frac{2-(1-r)^n}{n}}\approx\frac{2-(1-rn)}{n}=\frac{1+rn}{n}$$

That explains why you're close, despite a fundamental error.


问:
Bounded Analytic Functions on Parabolic Riemann surfaces are constant
I'm reading the book on Riemann Surfaces by Farkas and Kra (third edition). When proving uniformization theorem for nonhyperbolic simply connected Riemann surfaces, they say that bounded analytic functions on parabolic Riemann surfaces are constant. There is no result saying it explictly. Why is it true?</p>

答: The OP didn't define parabolic: a Riemann surface with no non-constant negative subharmonic functions. If $f$ is analytic bounded non-constant then $\Re(f)-C$ (with $C>\sup |f|$) is non-constant negative harmonic.


问:
Weierstrass substitution on an algebraic expression
I start with the trigonometric equation $3\tan{2x} + 16\cot^2{x} = 0$, although there is a particular solution pathway that I can follow by taking the double-angle identity to reduce the LHS to a cubic expression in terms of $\tan{x}$, I'm curious on a particular line of reasoning that follows by substituting $t = \tan(\frac{x}{2})$, and substituting the typical expressions after applying $$\tan{2x} = \frac{2\tan{x}}{1 - \tan^2{x}},~~\cot^2{x} = \frac{1}{\tan^2{x}},$$ where $\tan{x} = \frac{2t}{1-t^2}$.

My two questions posed are:

(a) Is the Weierstrass substitution here valid in this context of manipulating trigonometric equations algebraically?

(b) If so, is this particular case worthwhile trying to solve using such a substitution?

Post-note: A better question would be whether such substitutions are even well-formed in algebra, there is the constraint that $0 \leq x \leq \pi$, so I would've presumed there'd be no conflicts between domains/ranges if that is even a consideration.

答:
The Weierstrass substitution is always valid in whatever you're doing, be it integrating tricky trigonometric integrals or solving trig equations (eg the one you're solving). This is because the Weierstrass substitution $t=\tan\frac{x}{2}$ relies on identities, and identities are by definition always true.

However, as noted in the comments, the Weierstrass substitution can introduce extra solutions into the equation we are solving given the constraints you are dealing with, so if you use it always doublecheck if the solutions you obtained make sense.

In this particular case, without using the Weierstrass substitution we have the cubic equation
$$\frac{6\tan x}{1-\tan^2 x}+\frac{16}{\tan^2x}=0\iff3\tan^3x-8\tan^2x+8=0$$
By using the factor theorem and trying small values for $\tan x$, I quite quickly saw that $\tan x=2$ is a solution and therefore $\tan x-2$ is a factor of our expression. Hence, our expression is equivalent to
$$(\tan x-2)(3\tan^2x-2\tan x-4)=0$$
And the second factor is just a quadratic in $\tan x$ which we can solve easily.

So, all in all, in this case my personal preference would be not to use the Weierstrass substitution; it would've complicated an expression which turned out to be rather nice.

On the other hand, eg if we had the equation $3\sin x+2\cos x-9=0$ then the Weierstrass substitution is very useful if we don't want to resort to the $R\cos(x+\alpha)$ and $R\sin(x+\alpha)$ formulae (indeed I very rarely want to resort to them). In general, if we have the equation $a\sin x+b\cos x+c=0$ then I always recommend using the Weierstrass substitution, unless it's something really simple.

I hope that was helpful. If you have any questions please don't hesitate to ask :)


答:
$e^{t} \geq t$ for all $t \geq 0$. Hence, $te^{-t} \leq 1$. Taking $t=f_n(x)$ it follows that $f_n(x)e^{-f_n(x)} \leq 1$ for all $n$ and all $x$.

Two proofs of the inequality $e^{t}\geq t$ for $t \geq 0$:

Use series expansion of $e^{t}$.

Let $f(t)=e^{t}-t$. Then $f'(t)=e^{t}-1 \geq 0$ so $f$ is non-decreasing. But $f(0)=1$ so $f(t) \geq 1 >0$ for all $t \geq 1$.

Title: Show the following satisfies bounded convergence theorem

Body:
Assume $f_n: [0,1] \to [0,\infty)$ is integrable for each $n$ and $(f_n)_{n = 1}^{\infty}$ converges pointwise a.e. to $f$. Prove that

$$\lim_{n \to \infty} \int_{[0,1]} f_n(x) e^{-f_n(x)}dx = \int_{[0,1]} f(x)e^{-f(x)}dx$$

I think I am very close. To satisfy the BND convergence theorem, we need the input to be of finite measure, which we have: $m([0,1]) = 1 - 0 = 1 < \infty$. We also need the sequence of functions to converge pointwise, which we have by assumption.

The only thing I believe I am missing is that $\exists c \in (0,\infty)$ s.t. $|f_n(x)| \leq c \ \forall n \in \mathbb{N}$. If I had this, then BND convergence theorem would apply.

Maybe I'm not seeing it, but what in this problem implies we have this bound? Can anyone point me in the right direction?


问:
One-to-one correspondence between even spanning subgraphs and spanning subgraphs given parity of each vertex's degree
Given any connected graph $G = (V, E)$, the number of even spanning subgraphs is $2^{|E|-|V|+1}$. One nice proof of this fact is given here.

Recently, I came to know that the number of spanning subgraphs where we are given the parity of the degrees of each vertex of the subgraph is also $2^{|E|-|V|+1}$, provided that the number of odd vertices is even. Thus as a method of proving the fact, I wanted to show that there exists a one-to-one correspondence between them and the even spanning subgraphs of $G$.

However, I was only able to prove this when $G$ is a complete graph. If so, let us say that the odd vertices are $a_1, a_2, ... a_{2k}$ in some order. Consider the spanning subgraph $H = (V, E')$, where $E' = \{a_{2i-1}a_{2i}|1 \leq i \leq k\}$. Then, we can establish a one-to-one correspondence between any even spanning subgraph $H'$ to $H+H'$. (Here $H+H'$ refers to the graph with edge set the symmetric difference of edge sets of $H$ and $H'$, as in the link).

This proof may work for more general graphs, but it entirely breaks down if say the set of odd vertices form an indpendent set, as then no edge of $H'$ actually belongs to $G$. So how can I show this correspondence for any connected graph in general?

答:
The same idea still works. If $H$ is a fixed solution to the spanning-subgraph-given-parity problem, and $H'$ is any even spanning subgraph, then $H + H'$ will have the same parity, and so it will be another solution. Moreover, this will give you all the solutions, because the symmetric difference of two solutions is an even graph.

The only question is, how can we find an initial solution? To do this:

Pair up the $2k$ odd vertices into $k$ pairs, so that each pair is in the same connected component of $G$. (If this is not possible, then no solution $H$ exists, because each component of $H$ must have an even number of odd vertices.)

For $1 \le i \le k$, let $P_i$ be a path in $G$ joining the $i^{\text{th}}$ pair of vertices.

Let $H$ be the symmetric difference $P_1 + P_2 + \dots + P_k$.

This subgraph $H$ will have the degree parities you want, because $P_i$ has an even degree on all vertices except the $i^{\text{th}}$ pair of odd vertices.


问:
How to calculate $(\sum_{n=0}^{\infty} a_n x^n )^3$?
I know that for $$\left(\sum_{n=0}^{\infty} a_n x^n\right)^2=\sum_{n=0}^{\infty} \left(\sum_{j=0}^{n} a_j a_{n-j}\right)x^n,$$ however I'm not sure precisely how to extend this to a cube rather than a square. Certainly it'll be applying the above identity again, but I think the double sum is tripping me up!

I would really appreciate some help showing me how to index my sums at the end here!

答:
If we rewrite the formula for squaring in a slightly more symmetric way, it becomes slightly easier to see the pattern:

$$\left(\sum_{n=0}^{\infty} a_n x^n\right)^2 = \sum_{n=0}^{\infty} \left(\sum_{\substack{i+j=n \\ i,j \geq 0}} a_i a_j \right)x^n.$$

You can work out a similar formula for cubing, or taking 4th powers, etc.

$$\left(\sum_{n=0}^{\infty} a_n x^n\right)^3 = \sum_{n=0}^{\infty} \left(\sum_{\substack{i+j+k=n \\ i,j,k \geq 0}} a_i a_j a_k \right)x^n.$$

If you don't like the idea of summing over a subset of ordered triples, one can rewrite things in the following way, which is perhaps easier for a computer to implement but which I personally find more difficult to understand at a glance.

$$ \sum_{\substack{i+j+k=n \\ i,j,k \geq 0}} a_i a_j a_k = \sum_{i=0}^n \sum_{j=0}^{n-i} a_i a_j a_{n-i-j}. $$

The idea behind the conversion is that a priori we only know that i is between 0 and n, but for a given value of i, that puts more restrictions on j, and once we know both i and j, we know what k will be.  This is completely analogous to how one expresses an integral over a region as a double integral or triple integral.  With both sums and integrals, it is often conceptually convenient to express things over the region or set until actual computations need to be done, and only then to expand into multiple sums or multiple integrals.


答:
How this can be proved depends a bit on what knowledge you are allowed to use.

Here is one possible approach, assuming that you have some knowledge on linear algebra over general fields.

The klein group is, in fact, a two dimensional vector space over $\Bbb F_2$. If you compare the definitions of group homomorphisms and $\Bbb F_2$-linear maps, then it should be clear that they are the same thing: the extra requirement of being $\Bbb F_2$-linear is redundant.

Thus the automorphism group of the klein group is simply the group $\operatorname{GL}_2(\Bbb F_2)$ (canonical up to a choice of $\Bbb F_2$-basis). It is a standard exercise in linear algebra that the number of elements of $\operatorname{GL}_n(\Bbb F_p)$ is equal to $\prod_{i = 0}^{n - 1}(p^n - p^i)$, which in our case is $6$.

As the map sending any automorphism of the klein group to the permutation of the three non-identity elements is clearly an injective group homomorphism, by comparing the number of elements we conclude that it is an isomorphism of groups.

Title: Showing ${\rm Aut}(V) \cong S_3$

Body:
I am trying to prove that the automorphisms of the Klein four-group is isomorphic to $S_3$.

The first insight is the Klein group is an abelian group of four elements, three of which are order $2$, and must be mapped to other elements of order $2$, and the identity must of course be mapped to itself.

So any automorphism "permutes" the non-identity elements.

So there are exactly $3! = 6$ of them, matching the number of elements of $S_3$. Then if I call those non-identity maps $f_2, f_2, \ldots, f_6$, I can naturally map them to $S_3$.

For example, if the non-identity elements of the Klein group are $a$, $b$, $c$, I can regard these as $1$, $2$, and $3$ (I believe the labelling doesn't matter) and if I swapped $a$ and $b$ and fixed $c$, I can regard this as the transposition $(12)$.

By repeating this, I get a natural bijection of sets.

I don't know how to "prove" it's a bijection other than to observe the mapping and that it hits everything in $S_3$ exactly once.

I don't know, in particular, how to prove that this is a homomorphism, other than by checking every possible pair of isomorphisms, which is a bit laborious.

I assume there is a better, cleaner way to do it.

UPDATE: After working on this problem more, I have an updated attempt. It's incomplete, but I think it makes some progress.

We realize the Klein 4-group as $V = \{e, a, b, c\}$, where $a,b,c$ are the non-identity elements, each of which has order $2$. Any automorphism $f$ of $V$ must send elements of order $2$ to elements of order $2$, so necessarily it must permute the elements $a,b,c$. I claim that such a permutation is necessarily an automorphism. We observe that the Klein four-group has the property that if we multiply two non-identity elements, we get the third. Since $f$ maps from $V$ to $V$ bijectively, it is also the case that when $f(g)f(g') = f(g'')$, where $g$ and $g'$ are distinct elements and $g''$ is the third. That is, for any $g \neq g' \neq e$, we have
\begin{align*}
f(gg') = f(g'') = f(g)f(g'').
\end{align*}
If we take an arbitrary $g \in V$ and then consider multiplying with the identity $e$, we have
\begin{align*}
f(ge) = f(g) = e f(g) = f(e) f(g),
\end{align*}
since $f(e) = e$. This proves that $h$ is a homomorphism, and since it is bijective by definition, it is an automorphism. Since every automorphism must send elements of order $2$ to elements of order $2$ and there are exactly $3! = 6$ permutations of $\{a,b,c\}$, there are exactly six elements of Aut(G).

An automorphism is, therefore, uniquely determined on how it permutes the elements ${a,b,c}$, giving a natural bijective map to $S_3$, relabelling these non-identity elements as $1,2,3$. I call $e$ the identity permutation and $f_{abc}$ as the automorphism that maps ${a,b,c} \to {a,b,c}$ (i.e., the identity map) and so forth. The map is:
\begin{align*}
& f_{a,b,c} \longmapsto e & & \text{both have order $1$} \\
& f_{a,c,b} \longmapsto (23) & & \text{both have order $2$} \\
& f_{b,a,c} \longmapsto (12) & & \text{both have order $2$} \\
& f_{c,b,a} \longmapsto (13) & & \text{both have order $2$} \\
& f_{b,c,a} \longmapsto (123) & & \text{both have order $3$} \\
& f_{c,a,b} \longmapsto (132) & & \text{both have order $3$}
\end{align*}
This is clearly a bijection of sets: every permutation in $S_3$ was hit exactly once. I still do not know how to show that this is a homomorphism. I constructed the map so that it preserved the order of elements, but is that enough? I could check that $\varphi(xy) = \varphi(x) \varphi(y)$ for each pair, but there must be a way to do it that I'm missing.


问:
Simplifying $a(a-2) = b(b+2)$
I reduced a number theory problem to finding all ordered pairs $(a,b)$ that satisfy the equation $a(a-2) = b(b+2)$ in a certain range. After thinking about this for a while, I figured that either $a = b + 2$, $a = -b$, $a = b = 0$ or $a = 2 \text{ and } b = -2$. It is easy to prove that these values for $a$ and $b$ will all satisfy the equation, but how would I solve this equation had I not come up with these solutions? And how do I know I did not miss one? In short, can this equation be solved (more) rigorously?

Edit:
Silly of me I did not recognize that my last two solutions were already covered by my first two.

Title: Simplifying $a(a-2) = b(b+2)$

Body:
$a(a - 2) = b(b + 2)$

$a^2 - 2a = b^2 + 2b$

$a^2 - b^2 = 2(b + a)$

$(a + b)(a - b) = 2(a + b)$

$(a + b)(a - b - 2) = 0$


答:
If $$f(x) = \sum_{n=0}^\infty a_n x^n,$$ then $$(f(x))^3 = \sum_{n=0}^\infty b_n x^n,$$ where $$b_n = \sum_{i = 0}^n \sum_{j=0}^{n-i} a_i a_j a_{n-i-j}.$$

And in general, for $(f(x))^m$ with $m \ge 2$,
$$b_n = \sum_{c_1 = 0}^n \sum_{c_2 = 0}^{n-c_1} \cdots \sum_{c_{m-1}}^{n-c_1-\cdots-c_{m-2}} a_{c_1} a_{c_2} \cdots a_{c_{m-1}} a_{n-c_1-c_2-\cdots-c_{m-1}}.$$  This looks awfully ugly, but it makes a lot more sense if you write it like this:
$$b_n = \sum_{c_1 + \cdots + c_m = n} \prod_{i=1}^m a_{c_i},$$ where $c_1, c_2, \ldots, c_m \in \{0, 1, \ldots, n\}$ and the sum is taken over all possible sums that equal $n$.  So for instance, back to the case $m = 3$, we have
$$b_n = \sum_{c_1 + c_2 + c_3 = n} \prod_{i=1}^3 a_{c_i} = \sum_{c_1 + c_2 + c_3 = n} a_{c_1} a_{c_2} a_{c_3}.$$  Then we note that one way to run through all of the terms in this sum is to write it as a double sum as shown above.

Title: How to calculate $(\sum_{n=0}^{\infty} a_n x^n )^3$?

Body:
I know that for $$\left(\sum_{n=0}^{\infty} a_n x^n\right)^2=\sum_{n=0}^{\infty} \left(\sum_{j=0}^{n} a_j a_{n-j}\right)x^n,$$ however I'm not sure precisely how to extend this to a cube rather than a square. Certainly it'll be applying the above identity again, but I think the double sum is tripping me up!

I would really appreciate some help showing me how to index my sums at the end here!


问:
Find the GCD of a polynomial using the extended Euclidean algorithm and express it in the form $a(x)f(x)+b(x)g(x)$
I am working on a problem that I can not seem to finish.

Find the gcd of $f(x)=x^7+1$ and $g(x)=x^6+x^5+x^3+1$, and express it in the form $a(x)f(x)+b(x)g(x)$ using the extended euclidean algorithm.

I have performed the extended euclidean algorithm to find the following
$$f(x)=(x-1)g(x)+(x^5-x^4+x^3-x+2)$$
$$g(x)=(x+2)(x^5-x^4+x^3-x+2)+(x^4-x^3+x^2-3)$$
$$x^5-x^4+x^3-x+2=(x)(x^4-x^3+x^2-3)+(2x+2)$$
$$x^4-x^3+x^2-3=(\frac{1}{2}x^3-x^2+\frac{3}{2}x-\frac{3}{2})(2x+2)+0$$
This should give us the GCD as following
$$\gcd(f,g)=\gcd(x^5-x^4+x^3-x+2, 2x+2)=2x+2$$
And the monic GCD then becomes
$$\gcd(f,g)=x+1$$

This is where I get stuck. Now that I have this I am not sure how to algebraically get back to the form $a(x)f(x)+b(x)g(x)$

The answer as provided in the solutions is
$$\gcd(f(x),g(x))=x^4+x^3+x^2+1=xf(x)+(x^2+x+1)g(x)$$

I do not understand how to proceed and get to the answer. I have tried going backward but I am not getting anywhere. How do I go from $x+1$ to the correct solution?

答:
With some guessing, I figured out that the original question asks the $\gcd$ of these two polynomials over $\Bbb F_2$.

This means that $f, g$ are viewed as polynomials in $\Bbb F_2[x]$. Therefore the algorithm stops at the step $x^5-x^4+x^3-x+2=(x)(x^4-x^3+x^2-3)+(2x+2)$, as the remainder is now $0$.


问:
Proving $\sum \limits _{n=1}^{\infty}\frac{\sin (n)}{n}$ convergent
I need to prove that  $\sum \limits _{n=1}^{\infty}\frac{\sin (n)}{n}$ convergent only by Cauchy's test.

Cauchy's test for sequences convergence :

The $\sum \limits _{n=1}^{\infty}a_n$ convergent $\iff$ $\forall \varepsilon >0$ $ \exists$ $ n_0\in \mathbb{N}$ such that $\forall $ $n_0 \le n$ , $\forall$ $p\in \mathbb{N}$ the following holds $\mid S_{n+p}-S_n\mid = \mid a_{n+1}+\dots+a_{n+p}\mid<\varepsilon$

My Attempt:

Let $\varepsilon>0$,$\ p\in\mathbb{N}$. We define $n_0=\frac{p}{\varepsilon}$, and since $-1\le \sin(x)\le 1$ therefore, for each $n\le n_0$: $$\mid \frac{\sin (n+1)}{n+1}+\frac{\sin (n+2)}{n+2}+...+\frac{\sin (n+p)}{n+p} \mid \ \le \ \mid \frac{\sin (n+1)}{n+1}\mid + \mid \frac{\sin (n+2)}{n+2}\mid+...+\mid \frac{\sin (n+p)}{n+p} \mid \le \frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{n+p}\le \frac{1}{n}+\frac{1}{n}+...+\frac{1}{n}=\frac{p}{n}<\varepsilon$$

Is my proof correct? if not so how can I prove it only by Cauchy's test as I said earlier, without any further test - such as Dirichlet, Abel...

答:
There have been a couple of requests in the comments to show a way forward after I suggested applying summation by parts.  To that end, we proceed.

Since we asked to prove that the series $\sum_{n=1}^\infty \frac{\sin(n)}{n}$ converges, we begin with the partial sum $S_N$ defined by

$$S_N=\sum_{n=1}^N \frac{\sin(n)}{n}$$

Letting $T_n=\sum_{m=1}^n \sin(m)$ and summing by parts, we have for $N>M+1$

$$\begin{align}
\left|S_N-S_M\right|&=\left|\frac{T_N}{N+1}-\frac{T_M}{M+1}+\sum_{n=M+1}^N \left(\frac1n-\frac1{n+1}\right)T_n\right|\\\\
&\le \max_n (|T_n|)\left(\frac1{N+1}+\frac1{M+1}+\frac1{M+1}-\frac1{N+1}\right)
\end{align}$$

I showed in THIS ANSWER that $\max_n |T_n|\le \frac12(\csc(1/2)+\cot(1/2))\le 2$.

Using this bound for $|T_n|$, we can assert that for any given $\varepsilon >0$, $|S_N-S_M|<\varepsilon$ whenever $N>M+1>\frac{4}{\varepsilon}$.

From Cauchy's criterion, we conclude that $S_N$ converges.  And we are done!


问:
Is there a sequence of continuous functions which converges nowhere
find $\{f_n \}$ a sequence of continuous functions on $\mathbb{R}$ such that $0 \leq f_n(x) \leq 1$ and $\lim_{n \rightarrow \infty} \int_0^1 f_n(x)dx =0$, but $f_n(x)$ does not converge anywhere on $[0,1]$.

Everytime I am thinking with a sequence, I come with one satisfies some of the conditions but violate one condition. For example, $f_n(x)=x^n$ is continuous sequence, and $f_n \rightarrow 0$ so limit the integral is zero, but it is  pointwise convergent. , I tried to create a function similar to Dirchlet function, as they not converge pointwise, but they converge on part of the domain and they are not continuous.

答:
Rather than give formulas, I will describe the graphs of the functions, which are piecewise linear. Think of a “discretely moving wave”, where after the wave is done moving, we switch to a “thinner” wave.

Let $f_1$ go from $(0,1)$ to $(\frac{1}{2},0)$ to $(1,0)$. Let $f_2$ go from $(0,0)$ to $\frac{1}{2},1)$ to $(1,0)$. Let $f_3$ go from $(0,0)$ to $\frac{1}{2},0)$ to $(1,1)$.

Then let $f_4$ go from $(0,1)$ to $(\frac{1}{4},0)$ to $(1,0)$. Let $f_5$ go from $(0,0)$ to $(\frac{1}{4},1)$ to $(\frac{1}{2},0)$ to $(1,0)$. Then $f_6$ from $(0,0)$ to $(\frac{1}{4},0)$ to $(\frac{1}{2},1)$ to $(\frac{3}{4},0)$ to $(1,0)$. Then $f_7$ go from $(0,0)$ to $(\frac{1}{2},0)$ to $(\frac{3}{4},1)$ to $(1,0)$. Then $f_8$ from $(0,0)$ to $(\frac{3}{4},0)$ to $(1,1)$.

Next do the same thing with the partition $0\lt \frac{1}{8}\lt\frac{1}{4}\lt\frac{3}{8}\lt \frac{1}{2}\lt \frac{5}{8}\lt\frac{3}{4}\lt\frac{7}{8}\lt 1$. Then partitioning $[0,1]$ into $16$ equal subintervals, etc.

Each new batch of functions has a smaller integral, all positive, but converging to $0$. The functions are all continuous. And $f_n(x)$ does not converge for any $x$ because you can always find arbitrarily large values of $n$ where $f_n(x)=0$ and values where $f_n(x)$ is very close to $1$ (by approximating $x$ with a rational with denominator a power of $2$).


问:
Define the mass function
Consider the following experiment: A coin is tossed with probability $\alpha \in (0,1)$ of head and a dice is rolled at the same time. Define the following random variable:

\begin{equation*}
        X = \begin{cases}
        \text{number in the face of the dice} \ , \ \text{if heads} \\
        \text{number in the face of the dice + 2} \ , \text{if tails}
        \end{cases}
\end{equation*}

Define the probability mass function of $X$.

Well, i think that the elements of the random variable are something like $(1,H),(2,H),\dots,(6,H),(3,T),\dots, (9,T)$. The probability mass function would be something like $$P(X=x)=P(X=(n,H))=\alpha\cdot \frac{1}{6}$$ for $n \in \{1,2,\dots,6\}$. This makes sense? I'm really lost i'm this question.

答:
Let $D$ be the result of the die throw, and $T$ be the count of tails on the coin toss, and so: $X{=}D{+}2T$.

Now when the coin shows heads, $X$ may be $1,2,3,4,5,$ or $6$ with equal probability, else when the coin shows tails, $X$ may be $3,4,5,6,7,$ or $8$ with equal probability (ie $1/6$ each case).

So all up, $X$ may be $1,2,3,4,5,6,7,8$ with various probabilities; and so, because the coin is heads with probability $\alpha$ and tails with probability $(1-\alpha)$, we get:

$$\begin{align}\mathsf P(X{=}x)&=\mathsf P(T{=}0,D{=}x)+\mathsf P(T{=}1,D{=}x{-}2)\\[1ex]&=\mathsf P(T{=}0)\,\mathsf P(D{=}x)+\mathsf P(T{=}1)\,\mathsf P(D{=}x{-}2)\\[1ex]&=\alpha\,\mathsf P(D{=}x)+(1-\alpha)\,\mathsf P(D{=}x{-}2)\\[2ex]&=\begin{cases}?\qquad\qquad&:&x\in\{1,2\}\\?&:&x\in\{3,4,5,6\}\\?&:&x\in\{7,8\}\end{cases}\end{align}$$

Title: Define the mass function


问:
Blow-up, strict transform and tangent cone (Gathmann Notes, Exercise 9.22)
I'm studying Gatmann's Notes (version of 2014) https://www.mathematik.uni-kl.de/~gathmann/de/alggeom.php

I'm currently reading the Chapter 9. Birational Maps and Blowing Up. I'm trying to do exercise 9.22 which appears to be important.

Exercise 9.22 (Computation of tangent cones). Let $I\trianglelefteq K[x_1,\dots,x_n]$ be an ideal, and assume that the corresponding affine variety $X=V(I)\subseteq \mathbb{A}^n$ contains the origin. Consider the blow-up $\tilde{X}\subseteq \widetilde{\mathbb{A}^n}\subseteq \mathbb{A}^n\times \mathbb{P}^{n-1}$ at $x_1,\dots,x_n$, and denote the homogeneous coordinates of $\mathbb{P}^{n-1}$ by $y_1,\dots,y_n$.

(a) By example 9.15 we know that $\widetilde{\mathbb{A}^n}$ can be covered by affine spaces, with one coordinate patch being
\begin{align}
\mathbb{A}^n&\to \widetilde{\mathbb{A}^n}\subseteq \mathbb{A}^n\times \mathbb{P}^{n-1}\\
(x_1,y_2,\dots,y_n)&\mapsto((x_1,x_1y_2,\dots,x_1y_n),(1:y_2:\dots:y_n)).
\end{align}
Prove that on this coordinate patch the blow-up $\tilde X$ is given as the zero locus of the polynomials
\begin{equation}
\frac{f(x_1,x_1y_2,\dots,x_1y_n)}{x_1^{\min\deg f}}
\end{equation}
for all non-zero $f\in I$, where $\min\deg f$ denotes the smallest degree of a monomial in $f$.

(b) Prove that the exceptional hypersurface of $\tilde X$ is
\begin{equation}
V_p(f^{in}:f\in I)\subseteq \{0\}\times \mathbb{P}^{n-1}
\end{equation}
where $f^{in}$ is the initial term of $f$, i.e. the sum of all monomials in $f$ of smallest degree. Consequently, the tangent cone of $X$ at the origin is
\begin{equation}
C_0X=V_a(f^{in}:f\in I)\subseteq \mathbb{A}^n.
\end{equation}

(c) If $I=(f)$ is a principal ideal prove that $C_0X=V_a(f^{in})$. However, for a general ideal $I$, show that $C_0X$ is in general not the zero locus of the initial terms of a set of generators for $I$.

I'm stuck at (a), which I think it's related to (b) and (c). I have done the following but it might be wrong:

First, I'm gonna state two Lemmas that I think are right

Lemma 1: Let $X=X_1\cup\dots\cup X_r$ be the decomposition of a Noetherian space into irreducible subspaces. If $A$ is a closed subset of $X$ such that for each $i=1,\dots,n$, $X_i\not\subseteq A$, then $X\setminus A$ is dense in $X$.

Lemma 2: If $f\neq 0$, then
\begin{equation}
\frac{f(x_1,x_1y_2,\dots,x_1y_n)}{x_1^{\min\deg f}}\notin (x_1).
\end{equation}

Let's call $\phi:\mathbb{A}^n\to \widetilde{\mathbb{A}^n}$ the morphism defined in (a). If $\pi:\widetilde{\mathbb{A}^n}\to \mathbb{A}^n$ is the map associated to the blow-up, I believe I can prove the following equality

\begin{equation}
\phi(V(\frac{f(x_1,x_1y_2,\dots,x_1y_n)}{x_1^{\min\deg f}}:f\in I\setminus\{0\})\setminus V(x_1))=\pi^{-1}(X\setminus\{0\})\cap \phi(\mathbb{A}^n).
\end{equation}

From there, If I could prove that
\begin{equation}
\overline{V(\frac{f(x_1,x_1y_2,\dots,x_1y_n)}{x_1^{\min\deg f}}:f\in I\setminus\{0\})\setminus V(x_1)}=V(\frac{f(x_1,x_1y_2,\dots,x_1y_n)}{x_1^{\min\deg f}}:f\in I\setminus\{0\}),
\end{equation}
then the exercise would be done just by taking closures. I think Lemma 1 and 2 come into play here. The problem is that $V(\frac{f(x_1,x_1y_2,\dots,x_1y_n)}{x_1^{\min\deg f}}:f\in I\setminus\{0\})\setminus V(x_1)$ might not be dense in $V(\frac{f(x_1,x_1y_2,\dots,x_1y_n)}{x_1^{\min\deg f}}:f\in I\setminus\{0\})$ because it may happen that $X_i\subseteq V(x_1)$ for some irreducible component $X_i$ of $V(\frac{f(x_1,x_1y_2,\dots,x_1y_n)}{x_1^{\min\deg f}}:f\in I\setminus\{0\})$.

Again, I might have made a mistake, so please read critically. I am mainly interested in (a), but a complete answer is also welcome.

答:
A preliminary lemma:

Lemma: Let $I\subset k[x_1,\cdots,x_n]$ be an ideal and let $f\in k[x_1,\cdots,x_n]$ be  a nonzero element. The closure of $V(I)\cap D(f)$ is given by the vanishing locus of the ideal $$J=\left(\frac{e}{f^{\deg_f e}}\mid e\in I\right),$$ where $\deg_f e$ is the number of times $f$ divides $e$.

Proof: The morphism of varieties $V(I)\cap D(f) \to \Bbb A^n$ corresponds to the ring map $k[x_1,\cdots,x_n]\to k[x_1,\cdots,x_n]_f/I_f$, and the closure of $V(I)\cap D(f)$ is the variety cut out by the kernel of this ring morphism. As $I_f=\left\{ \frac{e}{f^m}\mid e\in I, m\in\Bbb Z_{\geq0} \right\}$, if $e\in I$ then if $f^m\mid e$, we have $\frac{e}{f^m}$ is an element in $k[x_1,\cdots,x_n]$ which maps to zero. As every such element is $f^l\cdot \frac{e}{f^{\deg_f e}}$ for some $l\geq 0$, we have the result. $\blacksquare$

On to the exercise.

By remark 9.11, we know that $\widetilde{X}$ is the closure of $\pi^{-1}(X\setminus\{0\})$ in $\widetilde{\Bbb A^n}$. On $U_1$, we have that $\pi^{-1}(X)$ is the variety cut out by the ideal $(f(x_1,x_1y_2,\cdots,x_1y_n)\mid f\in I)$, while $\pi^{-1}(0)=V(x_1)$. As $\pi^{-1}(X\setminus \{0\})$ is $\pi^{-1}(X)$ without $\pi^{-1}(0)$, we see that on $U_1$, the closure of $\pi^{-1}(X\setminus \{0\})$ is exactly the desired ideal by our lemma. This proves (a).

For (b), we use the computation from (a). The exceptional divisor is the intersection of $\widetilde{X}$ with the $\Bbb P^{n-1}$ living over the origin in $\widetilde{A^n}$, so we can find this on $U_1$ by setting $x_1=0$. As evaluating $\frac{f(x_1,x_1y_2,\cdots,x_1y_n)}{x_1^{\deg_{x_1} f}}$ at $x_1=0$ gives the initial term of $f(y_1,\cdots,y_n)$ and then sets $y_1=1$, we see that the exceptional set is cut out by the ideal generated by $f^{in}(y_1,\cdots,y_n)$ as $f$ runs over the polynomials in $I$.

In order to attack (c), expand $f=f_r+f_{r+1}+\cdots$ and $g=g_s+g_{s+1}+\cdots$ in to homogeneous parts: then $fg=f_rg_s+(f_rg_{s+1}+f_{r+1}g_s)+\cdots$, and thus $(fg)^{in}=f^{in}\cdot g^{in}$. If $I=(f)$, this shows that the ideal we constructed in (b) is just $(f^{in})$. The counterexample is a standard sort of trick: consider the ideal $(x,x-y^2)\subset k[x,y,z]$. Then the initial ideal of this is $(x,y^2)$, but using the set of generators $\{x,x-y^2\}$, the ideal which is generated by the initial terms is just $(x)$.


问:
Compute the p-value in hypothesis testing
The data generating process is
$$
f(x \mid \theta)=\frac{1}{\theta}, \quad 0 \leq x \leq \theta
$$
Let $H_1: \theta=5$ and $H_2: \theta \neq 5$. Prior probabilities of $\frac{1}{2}$ are assigned to the two hypotheses, and a sample of size $5$ yields a maximum value of $4$. Find the $p$-value here.

According to the definition of $p$-value,
$$
p=\mathbb{P}(LR \leq LR^* \mid H_1),
$$
and $LR^*=\left(\frac{4}{5}\right)^5$. I am not sure what to do next. Furthermore, intuitively, if the maxima is equal to $5$ exactly, shall we get an extremely big $p$ or a small one?

答:
The p-value is the probability of getting a value as extreme as the one observed under the null hypothesis, therefore should be $Pr(\max\{X_1,...,X_5\}\le 4|\theta=5)$. Since $H_0:\theta=5$ means that X has the continuous uniform distribution $X\sim \mathscr {Unif}\left(0, 5\right)$, we expect the pvalue to be $Pr(X_1\le4)*...*Pr(X_5\le4)=\left(\frac 4 5\right)^5=.32768$.

We should get a p-value close to 1 if the maximum is close to 5 and less than 5, and a p-value close to 0 if the maximum is close to 0 and positive. We should get p-values of 0 if the maximum is greater than 5, and it's impossible to get a maximum less than 0 because it's not in the support of the data-generating process.


问:
Find the rational points on $1 + 18 x + 81 x^2 + 44 x^3 = y^2$ with Sage
I'm trying to use Sage on-line,but I meet some trouble with the code of it.
I want to find the rational points on an ellipse curve,such as $$1 + 18 x + 81 x^2 + 44 x^3 = y^2,\tag1$$
I know that $(x,y)=(0,1)(1,\pm12)(-\frac{1}{11},0)$ are on the curve,but I don't know how to find more,so I try to solve it with the math-software Sage.

What should I input?Thanks in advance!

Edit: Multiply by $44^2$,we can difine the ellipse curve by this way,

E = EllipticCurve([0,81,0,44*18,44^2])

Then how to find the rational points on it?

答:
In case a curve has positive rank, you'd like to know its generators by E.gens().

In the case

```<code>E = EllipticCurve([0,121,0,22*52,52^2]);
</code>```

the rank is 1

```<code>sage: E.rank()
1
</code>```

and its generator is

```<code>sage: E.gens()
[(-104 : 260 : 1)]
</code>```

In this case you can construct infinitely many points on E, by adding a torsion point and some multiple of the generator, for instance

```<code>sage: T = E(0,52)
sage: P = E(-104,260)
sage: T + 2*P
(480/49 : 55796/343 : 1)
sage: T + 3*P
(-12662312/485809 : -65467251044/338608873 : 1)
sage: T - 5*P
(-70265016926482573/11385999920450401 : 5045808098009295785013222/1214945410097644049235601 : 1)
</code>```

and so on.

Body:
In case a curve has positive rank, you'd like to know its generators by E.gens().

In the case

E = EllipticCurve([0,121,0,22*52,52^2]);

the rank is 1

sage: E.rank()
1

and its generator is

sage: E.gens()
[(-104 : 260 : 1)]

In this case you can construct infinitely many points on E, by adding a torsion point and some multiple of the generator, for instance

sage: T = E(0,52)
sage: P = E(-104,260)
sage: T + 2*P
(480/49 : 55796/343 : 1)
sage: T + 3*P
(-12662312/485809 : -65467251044/338608873 : 1)
sage: T - 5*P
(-70265016926482573/11385999920450401 : 5045808098009295785013222/1214945410097644049235601 : 1)

and so on.


问:
Why is Normality an axiom in Modal Logic?
Why are we having $\square(A → B) → (\square A → \square B)$ as an axiom when we can prove* that if we have $\square(A → B)$ then we will have $(\square A → \square B)$?

*$B$ is true in every world by assumption which makes $\square B$ true and hence $\mathfrak{M}, w
 \nVdash \square A $ or $\mathfrak{M}, w \Vdash \square B$ for every $w$ in $\mathfrak{M}$ that is to say $\mathfrak{M}, w \Vdash \square A \rightarrow \square B$.

I am almost certain that I am completely wrong but I cannot find my mistake in my thinking.

Help, please.

答:
You are conflating the proof theory of modal logic with the semantics of modal logic. We can approach verifying statements in two ways: 1) formal proofs from the axioms. 2) informal arguments for their validity in terms of Kripke frames. (Which is not to say these are the only two approaches, just the two relevant to this question.)

The theorems of system K (the smallest normal modal logic) can correspondingly be thought of in two ways. The first way is that a statement holds in K iff it can be formally proved by applying the rules/axioms of propositional logic, combined with the necessitation rule and the normality axiom. The second way is that a statement holds in K iff it is valid in all Kripke frames. The hard work is showing that these two sets of statements are actually the same. This is an example of a soundness/completeness result: the statements provable in some formal system are exactly those valid with respect to some semantics.

So given that fact, it is completely expected that axiom K is valid in all Kripke frames, cause obviously it can be proved in a formal system that takes it as an axiom. But it is in no way superfluous, because without it you cannot prove (using the proof theory approach) every statement that is valid in all Kripke frames.

The next normal modal logic people usually learn is system T, which is an extension of K with more valid statements. This extension can be thought of in two ways: 1) adding the axiom $\square \phi\to \phi$ to the formal system. 2) No longer requiring a valid statement to hold in all Kripke frames, rather just the reflexive frames. So in the first case you get more valid statements since you have another axiom to prove things with, and in the second you get more because you require the statement to hold in a smaller class of frames. And it turns out that either way results in the same set of valid statements: another soundness/completeness result.

One other thing to note: they are less well-known, but there are non-normal modal logics that don't admit the normality axiom. Of course, these won't have semantics simply in terms of Kripke frames like normal modal logics.


答:
This looks to me like the distinction between Semantics and Syntax, as applied to modal logic.

What we want to do with formal logic is allow the manipulation of strings of symbols according to mechanical rules - the Syntax - to allow us to learn about logical facts about certain mathematical structures - the Semantics. From the point of view of the Semantics, the symbols we use have particular meanings that we all agree on, so that strings of symbols say things about mathematical structures. From the point of view of the Syntax, the symbols just start out as empty labels: they only gain their meanings because they obey certain mechanical rules that match what we expect those meanings to do.

For instance...

Semantically, $\rightarrow$ means "Implies". Syntactically, we only start out with $\rightarrow$ being some connective of a pair of terms, and have to explicitly define Modus Ponens as a rule in order to make $\rightarrow$ act like "Implies".

Semantically $\forall$ means "For All". In order to make the symbol $\forall$ act like "For All" syntactically, we have to define a rule that says that whenever we have $\forall x: P(x)$ we can substitute in any term $k$ to conclude $P(k)$.

In both these cases, we have to add those rules to our logical system in order to make the symbols act like we want them to.

You've shown that any symbol that means "Necessary" has to obey that rule, semantically. But syntactically, $\square$ just starts out as an empty symbol with no meaning behind it. If we didn't state our rules for using it, we wouldn't be able to do anything with it. That's why we have to state rules for using $\square$, one of which is $\square(A → B) → (\square A → \square B)$.

Title: Why is Normality an axiom in Modal Logic?

Body:
Why are we having $\square(A → B) → (\square A → \square B)$ as an axiom when we can prove* that if we have $\square(A → B)$ then we will have $(\square A → \square B)$?

*$B$ is true in every world by assumption which makes $\square B$ true and hence $\mathfrak{M}, w
 \nVdash \square A $ or $\mathfrak{M}, w \Vdash \square B$ for every $w$ in $\mathfrak{M}$ that is to say $\mathfrak{M}, w \Vdash \square A \rightarrow \square B$.

I am almost certain that I am completely wrong but I cannot find my mistake in my thinking.

Help, please.


问:
What's the condition required in terms of roots for getting exact three integral negative values for this quadratic?
The number of integral values of a for which $ax^2 - (4-2a)x - 8<0$ for exactly three integral value of x .what i did was find the roots of this equation lets say $x_1$ ,$x_2$ ,will this condition satisfy the criteria: $2<|x_1 -x_2|<4$ and a>0? Or is there some more elegant way to get the required condition in terms of roots?

答: You're kind of on the right track.  Let's suppose without loss of generality that $x_1 < x_2$ are the roots of $$f(x;a) = ax^2 - (4-2a)x - 8.$$  Then it is clear that we must have $a > 0$; otherwise, there will be infinitely many $x$ such that $f(x;a) < 0$.  Moreover, since we require exactly three integer solutions, there must be an integer $k$ such that $$k-2 \le x_1 < k-1 < k < k+1 < x_2 \le k+2.$$  Now write $$f(x;a) = a(x-x_1)(x-x_2) = ax^2 - a(x_1 + x_2)x + ax_1 x_2,$$ hence $$a(x_1 + x_2) = 4-2a, \\ a x_1 x_2 = -8.$$  Solving this system in terms of $a$, we obtain $$\{x_1, x_2\} \in \{-2, 4/a\}.$$  Since $a > 0$, we must have $x_1 = -2$, $x_2 = 4/a$, and we require $3 < x_2 - x_1 \le 4$.  This is because one root is already an integer.  It follows that $a \in \{2, 3\}$, and $k = 0$.


问:
Commutant built from the Unilateral Shift
Let $T: \ell^2 \rightarrow \ell^2$ be the unilateral shift, and let $\mathcal{T}= \{ I, T, T^* \}$.

Show that $\mathcal{T}' = \{\lambda I_{\ell^2}  : \lambda \in \mathbb{C} \}$ and $\mathcal{T}'' = B(\ell^2)$.

Attempt:

Since $T$ is the unilateral shift, we have that $Te_n = e_{n+1}$.

Also, by definition, we have $\mathcal{T}' = \{ S \in B(\ell^2) : ST =TS, \forall T \in \mathcal{T}\}$.

Let's look at the each element in $T$:

We know that the identity $I$ always has the property $IS=SI$ for $S \in B(\ell^2) $.

Also, let $S \in B(\ell^2)$. Then we have  $STe_n= Se_{n+1} =e_{n+1 }S= Te_n S = TSe_{n}$.

For $T^*$, I know that $T^*e_n= e_{n-1}$, for $n>1$. So don't we get a similar result as above, i.e $ST^*e_n =T^* S e_n$?

So I'm a bit confused on how to get the results we want. I feel like I got the definitions of $\mathcal{T}'$ and $\mathcal{T}''$ mixed up somewhere...

Thank you for your help!

答:
Hint: If an operator commutes with the the unilateral shift and its adjoint, then it commutes with the whole Toeplitz algebra. Furthermore, the Toeplitz algebra contains all the compact operators.

If you don't want to appeal to such facts, I would suggest showing by hand that the algebra generated by the unilateral shift and its adjoint contains every operator whose matrix with respect to the standard basis of $\ell^2(\mathbb{N})$ has only finitely many nonzero entries. Basically, figure out how to construct the matrix $E_{ij}$ which has a $1$ in the $(i,j)$ position and zeros elsewhere.


问:
$\mathbf K1.1$ is sound and strongly complete w.r.t. frames for which $R$ is a partial function

(Blackburn et al.'s Modal Logic, Ex 4.3.1): Let $1.1$ be the axiom $\diamond p\to\square p$. Show that $\mathbf K1.1$ is sound and strongly complete with respect to the class of all frames $(W,R)$ such that $R$ is a partial function.

Firstly, I am confused about the meaning of a partial function. After a quick Google search, I found out that a partial function $f$ from $X$ to $Y$ is essentially a function $f: S\to Y$ where $S\subset X$ (possibly improper). In other words, we are basically allowing the function to forget about some values in the domain. That being said, we know $R\subset W\times W$. If $R$ is a partial function, then the idea is that we can have isolated worlds, right? In addition, one world can be mapped to only one other (because $f: S\to Y$ is a function, so it is well-defined).

Just want to quickly verify my solution.

Suppose the class of all frames $(W,R)$ in which $R$ is a partial function is denoted by $P$. Then, for every frame $\mathfrak F \in P$, we show $\mathfrak F\Vdash \diamond p\to\square p$. $\mathfrak F\Vdash \diamond p\to\square p$ if and only if $\mathfrak F,w\Vdash \diamond p\to\square p$ for all $w\in W$. Consider arbitrary $w\in W$. Suppose $\mathfrak F,w\Vdash \diamond p$. Then, there exists $v\in W$ such that $wRv$ and $\mathfrak F,v\Vdash p$. Moreover, for all $v' \in W, v'\ne v$, we have $w\lnot Rv'$. This is because $R$ is a partial function. Hence, $\mathfrak F,w\Vdash \square p$. One direction is done!

Next, I want to show that if $\mathfrak F\Vdash \diamond p\to\square p$ (for all frames in $P$), i.e. $\mathfrak F,w\Vdash \diamond p\to\square p$ for all $w\in W$, then $R$ is a partial function. This is easily done too, and I am skipping the proof.

Another question I have is, where are soundness and completeness coming into play, exactly? When we say soundness, we want $\vdash \implies \Vdash$. For completeness, it's the other way round. Since $1.1$ is put into the axioms of $\mathbf K1.1$, perhaps the first part (1) of my solution corresponds with proving soundness, i.e. seeing if the axiom is valid in all frames of a particular type? If yes, then the other part (2) is probably completeness, but I do not really see how.

Could someone help me really understand if what I'm doing is right, and why I am doing it? I have difficulty understanding the bigger picture of completeness and soundness in modal logic. Thanks!

答:
Your proof sketch aims to show that the 1.1-formula defines the class of frames whose accessibility relations are partial functions. But that is a frame correspondence result and not a frame completeness result, which you are asked to show in the exercise.

But there is a connection here between frame correspondence and frame completeness. To show the completeness result you may use what Blackburn et al. call a completeness-via-canonicity argument: Prove that the 1.1-formula is canonical for partial functionhood. That is prove (i) that the canonical frame of every normal modal logic containing the 1.1-formula is partially functional and (ii) that the 1.1-formula is valid in every partially functional frame.

Now your frame correspondence result has already covered (ii). What is still left is (i). To show (i), let $ \diamond p \rightarrow \square p \in \Lambda$, where $\Lambda$ is any normal logic, and Let $\mathcal{F}^{\Lambda} = (W^{\Lambda}, R^{\Lambda})$ be the canonical frame for $\Lambda$. Suppose that $R^{\Lambda}st, R^{\Lambda}sv$, for arbitrary $s,t,v \in W^{\Lambda}$ and that $\psi \in t$. So, $\diamond \psi \in s$. As $\diamond \psi \rightarrow \square \psi \in s$, we then have that $\square \psi \in s$. As $R^{\Lambda}sv$ it follows that $\psi \in v$. Hence $t \subseteq v$. Analogously, it can be shown that $v \subseteq t$.


问:
Superposition calculus and equality factoring
Superposition with equality resolution and equality factoring is said to be a complete calculus for first-order logic. The purpose of equality factoring is basically to get rid of paraduplicate literals like $p(X)$ | $p(Y)$, which is fine.

But all the papers I have on the topic agree that factoring applies only to positive equality. In other words, it doesn't handle paraduplicate negative literals like $\lnot p(X)$ | $\lnot p(Y)$. How do you deal with those?

答: The answer (thanks to Stephan Schulz, author of the E theorem prover, personal communication) is that factoring on positive literals suffices; together with the other inference rules, this will generate clauses that will resolve with the negative literals to solve the problem, if it can be solved at all.</p>


问:
Proof that mirror coupling is maximal coupling
In this link p3, I read that the mirror coupling of Brownian motion is actually a maximal coupling. However, they do not prove this and I was wondering if anyone might have a reference to a proof of this fact.

To be more specific, I am asking for a proof that:
$$P\{T(X1, X2) ≥ t\} = \phi_t(|x1 − x2|)$$
where $T(X1, X2)$ is the earliest
time at which the two Brownian motions from the mirror coupling coincide and where $$\phi_t(r)=\frac{2}{\sqrt{2\pi t}} \int_0^{r/2}e^{-\rho^2/2}d\rho$$
Thank you in advance!

答:
$\phi_t(r)$ gives the tail probability of the first passage time of a 1D Brownian motion as they say on the first page (and whose proof is a direct application of reflection principle), i.e. $P(\tau_{r/2} \geq t) = \phi_t (r)$

You can reduce the hitting time of an n-dimensional Brownian motion to an n-1 dimensional hyperplane to the hitting time of a one-dimensional Brownian motion by spherical symmetry and essentially discarding all the directions orthogonal to the normal vector of the hyperplane. In particular, if an n-dimensional Brownian motion starts $r = |x_1 - x_2|/2$ away from the hyperplane, the distribution of its hitting time $\tau$ will be given by $P(\tau \geq t) = \phi_t (|x_1 - x_2|)$

$T(X_1, X_2) = \tau$, since as soon as $X_1$ and $X_2$ coincide on the hyperplane, they will stick together, so we just need to look at the first time $X_1$ hits the hyperplane, which is given by $\tau$, and thus they share the same CDF.


问:
Relationship between subsemigroups of GL2 and SL2
Let $A$ be a finite subset of $\mathbb{N}$.

Let $X$ be the semigroup generated by elements of form

$$\begin{pmatrix}
    0&1\\1&a
\end{pmatrix}$$
where $a\in A$

Let $Y$ be the semigroup generated by all products of the form

$$\begin{pmatrix}
    0&1\\1&a
\end{pmatrix} \cdot  \begin{pmatrix}
    0&1\\1&b
\end{pmatrix} $$
where $a,b \in A$.

Clearly $X\subseteq \mbox{GL}(2,\mathbb{Z})$ and $Y \subseteq \mbox{SL}(2,\mathbb{Z}) $.

How can one show also that $Y = X \cap  \mbox{SL}(2,\mathbb{Z})  $ ?

It is clear to me that $Y \subseteq X \cap  \mbox{SL}(2,\mathbb{Z}) $ but I do not know how to go about showing the reverse direction.

答:
An element in $X$ is a product of matrices  $A=\prod_{i=0}^nA_i=\prod_{i=0}^n\begin{pmatrix}
0 & 1 \\
1 & a_i
\end{pmatrix}$. In order for this to have determinant $1$, you need $n$ to be even. Therefore you can write $A=(A_1A_2)(A_3A_4)...(A_{n-1}A_n)$.


问:
Solve Non-lineaer Differential Equation (Step-by-Step)
Being able to solve Linear and Non-liner differentional equations is an important part of mathematics. I am trying to work out the exact solution for this non-linear differential equation and require some help from the community (Please could you provide the full step-by-step-solution), I want to undertstand the mathematics and be able to apply this do other questions.

$$\dfrac {du}{dx}+ u^2 = 0  \tag 1 $$

Boundary conditions $0<x<1; u(0)=1$

What is the exact solution?

答:
This is an answer to your question in the comments. Please, keep in mind for the future, provide any attempts you have made, any progress, everything you have tried to solve the problem. Where did you stuck, what was the difficulty? MSE is not a place to simply solve your homework for you ...

Regarding your question about the solution of $u^\prime + u = 0$. Apply the same logic as @Rem explained and also here, think about the problem first:

You want to find a function such that adding its own derivative would results in zero, i.e., they would cancel each other out. So, we need two identical terms with an opposite sign. How about something like -1 + 1 = 0? It seems like we are looking for a function which is very similar to itself when differentiating. Do you have a clue? If not, you can solve this ODE by separation of variables:

First, write your ODE in Leibnitz notation
$$
\frac{\mathrm{d}u}{\mathrm{dx}} + u = 0 \\
\frac{\mathrm{d}u}{\mathrm{dx}} = -u.
$$
We can already see that we are looking for a function which after differentiation stays the same, only changes its sign. Still no clue yet? Change the variables such that the terms containing the variable $u$ are on one side, and the terms containing the dependent variable $x$ on the other side:
$$
\frac{1}{\mathrm{d}u} = -1 \mathrm{d}x.
$$
Now you want to integrate both sides over $x$, so we change the dependent variable from $x$ to $s$ to avoid confusion and write
$$
\int_0^x\frac{1}{u}\mathrm{d}u = \int_0^x-1\mathrm{d}s.
$$
This is easy, think about what function differentiated results in the fraction you see on the l.h.s? Exactly, it's the logarithmic function. If $u = \ln(u)$, then $u^\prime = \frac{1}{u}$, hence we have
$$
\ln(u) + c_1 = -x + c_2.
$$
Bring the coefficient $c_1$ to the right side, apply the exponential function on both sides, et voilà, there's your solution:
$$
%\mathrm{e}^{\ln(u)} = \mathrm{e}^{-x} \cdot \underbrace{\mathrm{e}^{c_1 - c_2}}_{= c_3}\\[1em]
u(x) = \mathrm{e}^{-x} + c_3.
$$

If you want you can check your solution, just take the derivative of this function and add it to the function $u(x)$: $\mathrm{-e}^{-x}+\mathrm{e}^{-x} = 0$.

I forgot, you are given the boundary condition $u(0) = 1$. Apply this to your function and solve for the constant $c_3$:
$$
1 = \mathrm{e}^{-0} + c_3 \quad \Longleftrightarrow \quad c_3 = 0.
$$


问:
Unitary Operators are Connected in a $C^*$-algebra
Let $A$ be a unital $C^*$- algebra. Let $U = \{ u \in A : u^*u=uu^*=1\}$ be the unitary group of $A$.

Let $U'= \{ e^{ia_1}e^{ia_2} \cdots e^{ia_n} : a_k = a^*_k \in A, \text{for } 1\leq k \leq n \}$.

Show that $U'$ is the connected component of the identity in $U$.

If $A$ is commutative, show that $U' = \{ e^{ia} : a = a ^* \in A\}$.

I thought of using the theorems below from Banach Algebra Techniques in Operator Theory by Douglas.

Following the proof of 2.14, if $f = e^{ia} \in A$. Then $f\in U$. Consider $\phi: [0,1 ] \rightarrow e^(A)$ defined by $\phi(\lambda) = e^{i \lambda a}$, I don't see why $f \in U'$, and $e^A$ is contained in $U'$...

I also showed that if $\| u-1\| <2 $ , then we have $-1 \ne spec(u) $ and we can write $u = e^{ia}$ for some self-adjoint $a \in A$

Any help or suggestions will be appreciated.

Thank you!

答:
Any unitary sufficiently close to $1$ is the exponential of a self adjoint element. More precisely, if the distance from a unitary $u$ to $1$ is smaller than $2$, then $-1 \notin \mathrm{spec}(u)$, which allows you to define a logarithm for $u$ using continuous functional calculus.

Now if $u$ is connected to $1$ by a path $u_t$, $t \in [0,1]$, you can easily find
$$0 = t_0 < t_1 < \ldots < t_n = 1$$
such that the distance between $u_{t_i}$ and $u_{t_{i+1}}$ is always smaller than $2$.

Can you take it from there?


问:
Integrating the product $x|x|$
I want to integrate the product $x|x|$ with respect to $x$. That is, compute $\int x|x| dx$. I want to make sure I am doing this correctly.

So, $|x| = -x$ when $x < 0$ and $|x| = x$ when $x > 0$ and vanishes for $x = 0$. That means I should write $\int_{-\infty}^{0}-x\cdot x dx + \int_0^{\infty} x\cdot x dx = -\frac{x^3}{3} \mid_{-\infty}^{0} + \frac{x^3}{3}\mid_{0}^{\infty}$ but this isn't really making sense as the left term is zero and the right term goes to $\infty$.

答:
$$I:=\int x\lvert x\rvert\,\mathrm{d}x=\int x^2 \frac{x}{\lvert x\rvert}\,\mathrm{d}x$$
$u=\lvert x\rvert,\mathrm{d}x=\frac{\lvert x\rvert}{x}\,\mathrm{d}u$
$$I=\int u^2 \,\mathrm{d}u=\frac{u^3}{3}+C=\frac{\lvert x\rvert x^2}{3}+C$$

Title: Integrating the product $x|x|$


问:
Why this equation is able to be said as the condition for composing a constant function?
$a,b,c,d:=$non zero real number constants.

$t\geq0$

$f(t):=\frac{a+bt}{c+dt}$

The thing which I want to do is determine the values of the above $4$ constants to make $f(t)$ as a constant function.

The textbook gave the condition for it as below.

$\frac{a}{c}=\frac{b}{d}$

Why this equation is able to be said as the condition for composing a constant function?

答: Taking $t=0$ and $t=1$, we obtain that $\frac{a}{c} = \frac{b}{d}$ is a necessary condition. Conversely, if $a = \frac{bc}{d}$ then $\frac{bc/d + bt}{c+dt} = \frac{b}{d}$, a constant.


问:
How to tackle a statistics problem given a random variable X?
I'm having trouble getting an overall grasp of certain aspects of statistics, namely knowing when to use which formula/approach to solve a problem.

Let's consider a random variable X, after you determine whether the random variable is continuous or discrete, how would you know whether to know whether to use a binomial distribution, exponential distribution, Poisson distribution, moment generating function, etc.?

For example, I know that if I'm asked to model the time between random events I know I have an exponential distribution problem.

It's just that there are so many different mean, variance, and other equations, I got lost along the way on which equation to apply where. Some clarification would be great.

答:
I am not really qualified to answer this question since I am a topologist, but I think it really depends on the analytical nature of your problem instead of judging from its behavior mathematically alone. There are surprising applications of various distributions to problems in real life, but that usually comes with an understanding of the phenomenon first. </p>

<p>A classic in this regard is Feller's book, in which almost no statement is proposed without an applied "example" at mind. You probably can find the book in your local library. </p>

Body:
I am not really qualified to answer this question since I am a topologist, but I think it really depends on the analytical nature of your problem instead of judging from its behavior mathematically alone. There are surprising applications of various distributions to problems in real life, but that usually comes with an understanding of the phenomenon first.

A classic in this regard is Feller's book, in which almost no statement is proposed without an applied "example" at mind. You probably can find the book in your local library.


问:
Let $I,J$ be ideals of $R$. Is $I\star J= \{ij\mid i\in I, j\in J\}$ an ideal of $R$?
Attempt:
We have that $0\in I\star J$ because $0\in I$. If we consider $ij\in I\star J$ and $c\in R$, then  $ijc\in I\star J$ as $J$ is a right ideal of $R$; $jc\in J$. In addition, $cij\in I\star J$ as $I$ is a left ideal of $R$; $ci\in I$. Suppose we have another element $hg\in I\star J$. We now need to show that
$$ij-hg\in I\star J$$
to declare $I\star J$ an ideal of $R$ by the ideal test. I have thought about employing some factorization technique, but have failed to do so. Might there be some special trick for this last part? Or am I mistaken in that there is exists a contradiction?

Any input appreciated.

Edit: As per my comment below, my attempted proof fails because the statement is not true. For a contradiction, consider $R=\mathbb{R}[x,y]$ and the ideal(s) $I=J=(x,y)$. Since $x,y\in (x,y)$, $I\star J$ is not closed under addition: $x^2+y^2$ is not real factorizable.

答:
Usually the product of two ideals $I, J$ is defined as the ideal generated by all elements of the form $ij$ with $i \in I$ and $j \in J$.

This set in your question is in general not an ideal. Here is a counterexample.

Take $R = \Bbb Z[\sqrt{-5}]$ and $I = J = (2, 1 + \sqrt{-5})$.

We have $(1 + \sqrt{-5})\times(1 - \sqrt{-5}) = 6$ and $2 \times 2 = 4$ which are both in $I \star J$.

However, I claim that their difference $2$ is not an element of $I\star J$.

Suppose that $2 = (2a + (1 + \sqrt{-5})b)(2c + (1 + \sqrt{-5})d)$ with $a, b, c, d\in\Bbb Z$. We take norms of both sides and get $4 = ((2a + b)^2 + 5b^2)((2c + d)^2 + 5d^2)$. Clearly there are no integers $a, b, c, d$ satisfying this equation.


问:
Find the GCD of a polynomial using the extended Euclidean algorithm and express it in the form $a(x)f(x)+b(x)g(x)$
I am working on a problem that I can not seem to finish.

Find the gcd of $f(x)=x^7+1$ and $g(x)=x^6+x^5+x^3+1$, and express it in the form $a(x)f(x)+b(x)g(x)$ using the extended euclidean algorithm.

I have performed the extended euclidean algorithm to find the following
$$f(x)=(x-1)g(x)+(x^5-x^4+x^3-x+2)$$
$$g(x)=(x+2)(x^5-x^4+x^3-x+2)+(x^4-x^3+x^2-3)$$
$$x^5-x^4+x^3-x+2=(x)(x^4-x^3+x^2-3)+(2x+2)$$
$$x^4-x^3+x^2-3=(\frac{1}{2}x^3-x^2+\frac{3}{2}x-\frac{3}{2})(2x+2)+0$$
This should give us the GCD as following
$$\gcd(f,g)=\gcd(x^5-x^4+x^3-x+2, 2x+2)=2x+2$$
And the monic GCD then becomes
$$\gcd(f,g)=x+1$$

This is where I get stuck. Now that I have this I am not sure how to algebraically get back to the form $a(x)f(x)+b(x)g(x)$

The answer as provided in the solutions is
$$\gcd(f(x),g(x))=x^4+x^3+x^2+1=xf(x)+(x^2+x+1)g(x)$$

I do not understand how to proceed and get to the answer. I have tried going backward but I am not getting anywhere. How do I go from $x+1$ to the correct solution?

答:
Here is how the extended Euclidean algorithm works in this case (over $\mathbf F_2$). Keep  in mind that, in characteristic $2$, adding or subtracting is the same.

This extended algorithm is based on the remark that all remainders are linear combination of the two given polynomials, and the coefficients thereof are computed recursively.
\begin{array}{rccl}
r_n &u_n&v_n & q_n\\\hline
f(x) & 0 & 1 \\
g(x) & 1 & 0 & x+1 \\ \hline
x^5+x^4+x^3 + x & x+1 & 1 & x \\
x^4+x^3 + x^2 + 1 & 1+x(x+1) & x & x \\[-0.5ex]
&=\color{red}{x^2+x+1} & \color{red}x \\ \hline
0
\end{array}


问:
Calculate the Limit of Double Sum
Compute
\begin{equation}
L=\lim _{n \rightarrow \infty}\frac{1}{n} \sum_{a=1}^n \sum_{b=1}^n \frac{a}{a^2+b^2 }.
\end{equation}
My attempt:
Define \begin{equation} f(n,m)= \frac{1}{n} \sum_{a=1}^n \frac{1}{m}\sum_{b=1}^m \frac{\frac{a}{n}}{(\frac{a}{n})^2+(\frac{b}{m})^2 } = \frac{1}{m}\sum_{b=1}^m \frac{1}{n} \sum_{a=1}^n \frac{\frac{a}{n}}{(\frac{a}{n})^2+(\frac{b}{m})^2 }
\end{equation}
Now for any $\epsilon >0$ there exists some $B>0$ such that for any $n,m \in \mathbb{N}$ and $n,m\geq B$:
\begin{equation}
|f(n,m)-f(n,n)|<\epsilon
\end{equation}
Thus, we have \begin{equation}
L=\lim _{n \rightarrow \infty} f(n,n)=\lim _{m \rightarrow \infty} \lim _{n \rightarrow \infty} f(n,m)= \lim _{m \rightarrow \infty} \frac{1}{m}\sum_{b=1}^m \int_{0}^{1}\frac{x}{x^2+(\frac{b}{m})^2}dx=\\ \frac{1}{2}\lim _{m \rightarrow \infty} \frac{1}{m}\sum_{b=1}^m \ln\frac{1+(\frac{b}{m})^2}{(\frac{b}{m})^2}=\frac{1}{2}\int_{0}^{1}\ln\frac{1+x^2}{x^2}dx=\frac{2\ln2 +\pi}{4}
\end{equation}

答:
While this approach arrives at the correct value for the limit it is not easily justified.
At first glance,

$$f(n,n) = \frac{1}{n}\sum_{a=1}^n\sum_{b=1}^n\frac{a}{a^2 + b^2} = \frac{1}{n^2}\sum_{a=1}^n\sum_{b=1}^n\frac{a/n}{(a/n)^2 + (b/n)^2} $$

is a Riemann sum for the integral of $g:(x,y) \mapsto \frac{x}{x^2+ y^2}$ over $[0,1]\times[0,1]$.  However, the integrand is unbounded around $(0,0)$ and is not Riemann integrable.  Instead it must be considered as an improper integral which can be evaluated as an iterated integral
$$I = \int_0^1 \left(\int_0^1 \frac{x}{x^2 + y^2} \, dy\right )dx = \frac{2 \log 2 + \pi}{4}$$

Note that the inner integral is properly Riemann and we can write

$$I = \int_0^1 \lim_{m \to \infty} \left(\frac{1}{m}\sum_{b=1}^m \frac{x}{x^2 + (b/m)^2} \right)\, dx$$

At this point it would be nice to switch the integral and the limit and apply a second Riemann sum to obtain

$$I =  \lim_{m \to \infty} \frac{1}{m}\sum_{b=1}^m \int_0^1\frac{x}{x^2 + (b/m)^2} \, dx =  \lim_{m \to \infty} \left(\frac{1}{m}\lim_{n \to \infty} \frac{1}{n}\sum_{a=1}^n\sum_{b=1}^m \frac{a/n}{(a/n)^2 + (b/m)^2}\right)$$

However justifying that switch is not obvious.  For example, we see immediately that

$$\left|\frac{1}{m}\sum_{b=1}^m \frac{x}{x^2 + (b/m)^2} \right| \leqslant \frac{1}{m}\sum_{b=1}^m \frac{x}{x^2 } = \frac{1}{x},$$

but the dominating function $x \mapsto 1/x$ is not integrable on $[0,1]$.

Furthermore, there remains finding the justification for  $\lim_{n\to \infty} f(n,n) = \lim_{m\to \infty} \lim_{n\to \infty} f(n,m)$.

A easily justified approach is to apply the Stolz-Cesaro theorem to get

$$\lim_{n \to \infty} f(n,n) = \lim_{n \to \infty} \frac{\sum_{a=1}^{n+1}\sum_{b=1}^{n+1}\frac{a}{a^2 + b^2} - \sum_{a=1}^n\sum_{b=1}^n\frac{a}{a^2 + b^2} }{n+1 - n} \\ = \lim_{n \to \infty} \left(\frac{n+1}{(n+1)^2 + (n+1)^2}+ \sum_{a=1}^n \frac{a}{a^2 + (n+1)^2} + \sum_{b=1}^n \frac{n+1}{(n+1)^2 + b^2} \right) \\ = \lim_{n\to \infty} \sum_{a=1}^{n} \frac{a + n+1}{a^2 + (n+1)^2}  = \lim_{n\to \infty} \frac{1}{n+1}\sum_{a=1}^{n} \frac{1 + \frac{a}{n+1}}{1 + \left(\frac{a}{n+1}\right)^2}\\ = \int_0^1 \frac{1+x}{1+x^2} \, dx = \frac{2 \log 2 + \pi}{4}$$

Title: Calculate the Limit of Double Sum