问:
Recurrence relation next step after finding root
I have the recurrence relation:

$a_n = 4 a_{n-1}  - 4 a_{n-2}  + 2n $. With the initial conditions $a_0 = 9$ and $a_1 = 10$

I have found the characteristic polynomium:

$r^2 - 4r + 4$ which is also $(r-2)^2 $, which have root 2.

Now is my solution to the recurrence relation is $\alpha_1 2^n + \alpha_2 2^n$?

My question simply is, am I on the right track in the first place? And what to do with the $2^n$?

答:
Your approach goes in the right general direction though, as @octave points out, the general solution looks different in the case of repeated roots.  (You can't simply repeat $2^n$.)

The $2n$ term makes the recurrence non-homogeneous and here a good strategy is to find a particular solution first, ignoring the initial condition for the time being.  Since $2n$ is a linear function of $n$, try $a_n = cn + d$: substituting in the recurrence shows that it works for $c = 2$ and $d=8$.  Then the general solution to the non-homogeneous recurrence is $a_n = 2n + 8 + b_n$, where $b_n$ is the general solution to the associated homogeneous recurrence $b_n = 4b_{n-1} - 4b_{n-2}$.  For double root 2, we have $b_n = \alpha_1 2^n + \alpha_2 n \cdot 2^n$, and the initial conditions $b_0 = a_0 - 2\cdot0 - 8 = 1$ and $b_1 = 0$ give $\alpha_1 = 1$ and $\alpha_2 = -1$, so $a_n = 2n+8+2^n-n \cdot2^n$.


问:
Hypothesis testing - When to subtract one during type I and type II testing?
I'm currently studying statistics and I'm reviewing notes I took during a class last week. However, there's something I'm confused about.
My professor subtracted 1 from the "successes" and I can't seem to figure out why.
Essentially, the scenario was like this:

Someone flipped a coin 16 times.

$$H_0:p=0.5\text{ v.s. }H_a:p=0.55$$

Test 1: Reject $p = 0.50$ if 10 or more heads are observed out of 16.

$\text{Pr}(X \ge  10 \text{ when } p= 0.5)$ where X is a binomial with $n = 16$ and $p = 0.50$.

This is the part I don't understand. When using R, he did the following:

```<code>1 - pbinom(9, 16, 0.5)
</code>```

Where did the 9 come from? Why did he subtract 1 from the initial 10 tosses? If I was in a scenario where I had:

$$\text{Pr}(X \le 15 \text{ when } p = 0.50)$$

Would I still subtract 1? Would the R solution be pbinom(15, 16, 0.50) or pbinom(14, 16, 0.50)? How do we decide when to subtract 1 like this?

( Sorry if my formatting is bad! I've never posted math equations like this to these forums and I can't seem to figure it out)

答:
$P(X \ge 10) = 1 - P(X \le 9) = 0.2272491.$ In R, pbinom is a binomial CDF and dbinom a binomial PDF.

For $X \sim \mathsf{Binom}(n=16,p=.5)$ here are three ways to find $P(X \ge 10)$ in R.
The last, and most direct method, subtracting the CDF, is the one shown in class.

```<code>sum(dbinom(10:16, 16, .5))
[1] 0.2272491
1 - sum(dbinom(0:9, 16, .5))
[1] 0.2272491
1 - pbinom(9, 16, .5)
[1] 0.2272491

x = 0:16;  PDF = dbinom(x, 16, .5)
plot(x, PDF, type="h", lwd=2, main="PDF of BINOM(16, .5)")
 abline(v=0, col="green2");  abline(h=0, col="green2")
 abline(v = 9.5, col="red", lwd=2, lty="dotted")
</code>```

The desired probability is the sum of the heights of the bars to the
right of the dotted red line.

ContentLicense: CC BY-SA 4.0

Body:
$P(X \ge 10) = 1 - P(X \le 9) = 0.2272491.$ In R, pbinom is a binomial CDF and dbinom a binomial PDF.

For $X \sim \mathsf{Binom}(n=16,p=.5)$ here are three ways to find $P(X \ge 10)$ in R.
The last, and most direct method, subtracting the CDF, is the one shown in class.

sum(dbinom(10:16, 16, .5))
[1] 0.2272491
1 - sum(dbinom(0:9, 16, .5))
[1] 0.2272491
1 - pbinom(9, 16, .5)
[1] 0.2272491

x = 0:16;  PDF = dbinom(x, 16, .5)
plot(x, PDF, type="h", lwd=2, main="PDF of BINOM(16, .5)")
 abline(v=0, col="green2");  abline(h=0, col="green2")
 abline(v = 9.5, col="red", lwd=2, lty="dotted")

The desired probability is the sum of the heights of the bars to the
right of the dotted red line.


问:
Limit of product of kth roots
For $n\geq 1$ , let $$A(n)=\prod_{k=1}^{n}(1+\frac{k}{n})^\frac{1}{k}.$$ I'd like to determine $$ \lim_{n->\infty} A(n).$$ Taking logarithms and using the following crude  bound gives $$\log(A(n))=\sum_{k=1}^{n}\frac{1}{k}\log(1+\frac{k}{n}) \geq \sum_{k=1}^{n}\frac{1}{n}\log(1+\frac{k}{n}).$$ By interpreting this latter sum as a Riemann sum for $\log(1+x)$ on $[0,1]$, we see that the desired limit is at least $4/e$. However, this lower bound is far from sharp and is clearly not strong enough to determine the actual limit. Appreciate if someone here could point me in the right direction.

答:
$\log A(n)$ is the sequence of Riemann sums of $$\int_0^1 x^{-1}\log(1+x)dx= \int_0^1 \sum_{m\ge 1} \frac{(-1)^{m+1} }mx^{m-1}dx$$
What do you get next?

Title: Limit of product of kth roots


问:
Calculating the geometric realization of a non-representable functor

Background

Let $\mathcal{F} : \textbf{CRing} \to \textbf{Set}$ be a functor and denote by $\textbf{P}_\mathcal{F}$ the category of points of $\mathcal{F}$ whose objects are pairs $(R , \rho)$ where $R$ is a ring and $\rho : h^R \to \mathcal{F}$ is a morphism (here $h^R := \operatorname{Hom}_{\textbf{CRing}}(R , -) \in [\textbf{CRing} , \textbf{Set}]$ is the Yoneda embedding $h^-$ applied to $R$), and whose morphisms $(R , \rho) \to (S , \sigma)$ are ring homomorphisms $\varphi: R \to S$ such that $\rho \circ h^\varphi = \sigma$. Define the diagram $D_\mathcal{F} : \textbf{P}_\mathcal{F}^\text{op} \to \textbf{LRS}$ to be the composition of the forgetful functor $\textbf{P}_\mathcal{F}^\text{op} \to \textbf{CRing}^\text{op}$ and the spec functor $\operatorname{Spec} : \textbf{CRing}^{\text{op}} \to \textbf{LRS}$. Explicitly it sends $(R , \rho) \mapsto \operatorname{Spec}(R)$.

In Demazure & Gabriel's Introduction to Algebraic Geometry and Algebraic Groups, the geometric realization of the functor $\mathcal{F}$ is defined by $| \mathcal{F} | := \operatorname{colim} D_\mathcal{F}$. In this book it is proved that the geometric realization functor $| - | : [\textbf{CRing} , \textbf{Set}] \to \textbf{LRS}$ is left adjoint to the functor $\textbf{LRS} \to [\textbf{CRing} , \textbf{Set}]$ which sends $X \mapsto \operatorname{Hom}_{\textbf{LRS}}(\operatorname{Spec}(-) , X)$.

In particular if $\mathcal{F}$ is representable by some scheme $Y$, i.e. $\mathcal{F} \cong \operatorname{Hom}_{\textbf{LRS}}(\operatorname{Spec}(-) , Y)$, then I think that $|\mathcal{F}| \cong Y$.

Question

My question is how does one go about computing the geometric realization of a functor which is not representable by schemes? For example what is the geometric realization of the functor $R \mapsto R^{\oplus \mathbb{N}}$? (This is basically the only non-representable functor I know.)

Small colimits in $\textbf{Set}$ are easy to construct (take the disjoint union and mod out by some equivalence relation), but the colimit in question here is certainly not small. Of course Demazure & Gabriel use Grothendieck universes throughout their book, so then $\textbf{P}_\mathcal{F}$ is a small category, however I'm not sure how to find a suitable Grothendieck universe to explicitly calculate this colimit for my example.

答:
If your goal is to read Demazure and Gabriel's book, then (as I explained in the comments) your question is based on false premises and the solution is to read the definitions carefully.
But let me address your question as written, since it will illuminate why Demazure and Gabriel use the definitions they do.

First, as you observe, the category of elements of an arbitrary functor $F : \textbf{CRing} \to \textbf{Set}$ is not always small.
Actually, it is almost never small, because $\textbf{CRing}$ itself is not small: as soon as $F (A)$ is non-empty for all rings $A$, then the category of elements of $F$ will be at least as big as $\textbf{CRing}$.
This is not actually fatal for the problem at hand (though it does introduce many complications).
It sometimes happen that the functor $F$ you are interested in is a colimit of a small diagram of representable functors, i.e. there is a small diagram $A : \mathcal{I}^\textrm{op} \to \textbf{CRing}$ such that $F (B) \cong \varinjlim_\mathcal{I} \textbf{CRing} (A, B)$ naturally in $B$.
In that case, you can compute the colimit $\left| F \right|$ you seek as $\varinjlim_\mathcal{I} \operatorname{Spec} A$.

The biggest complication is that $\left| F \right|$ is not well defined for arbitrary $F$, so you only get a partially defined functor $\left| - \right|$.
If you try to restrict to a full subcategory of functors $\textbf{CRing} \to \textbf{Set}$ on which $\left| - \right|$ is well defined everywhere, you then have the complication that the putative right adjoint may not have image contained in that subcategory.
I am not aware of any good way to resolve this dilemma; I think you have no choice but to settle for a partially defined adjoint.

Now for some good news: there is a clean necessary and sufficient condition for a functor $F : \textbf{CRing} \to \textbf{Set}$ to be a colimit of a small diagram of representable functors.

Definition.
Let $\kappa$ be an infinite regular cardinal.
A $\kappa$-accessible functor is a functor that preserves $\kappa$-filtered colimits.

Proposition.
Let $F : \textbf{CRing} \to \textbf{Set}$ be a functor.
The following are equivalent:

$F$ is $\kappa$-accessible.

$F$ is the left Kan extension of a functor $\textbf{CRing}_\kappa \to \textbf{Set}$ along the inclusion $\textbf{CRing}_\kappa \hookrightarrow \textbf{CRing}$, where $\textbf{CRing}_\kappa$ is the full subcategory of $\kappa$-presentable rings (i.e. rings presentable by $< \kappa$ generators and $< \kappa$ relations).

There is a small diagram $A : \mathcal{I}^\textrm{op} \to \textbf{CRing}$ such that $F \cong \varinjlim_\mathcal{I} \textbf{CRing} (A, -)$ and, for each $i$ in $\mathcal{I}$, $A (i)$ is a $\kappa$-presentable ring.

The functor $R \mapsto R^{\oplus \mathbb{N}}$ you mention is easily seen to preserve filtered colimits (i.e. be an $\aleph_0$-accessible functor).
It is just as easy to see that it is the colimit of a small (indeed, countable!) diagram of representable functors, namely,
$$\textbf{CRing} (\mathbb{Z}, -) \longrightarrow \textbf{CRing} (\mathbb{Z} [x_1], -) \longrightarrow \textbf{CRing} (\mathbb{Z} [x_1, x_2], -) \longrightarrow \cdots$$
where the maps are the ones induced by the homomorphisms $\mathbb{Z} [x_1, \ldots, x_n, x_{n+1}] \to \mathbb{Z} [x_1, \ldots, x_n]$ that send $x_i$ to $x_i$ for $1 \le i \le n$ and $x_{n+1}$ to $0$.
Thus, the geometric realisation of $R \mapsto R^{\oplus \mathbb{N}}$ is the colimit $\varinjlim_n \mathbb{A}^n$.

I suppose I owe you an example of a functor $\textbf{CRing} \to \textbf{Set}$ that is not accessible.
Choose an ordinal-indexed sequence of fields, $K_\alpha$, such that $K_\alpha$ is strictly smaller in cardinality than $K_\beta$ whenever $\alpha < \beta$.
Let $F (R) = \coprod_{\alpha} \textbf{CRing} (K_\alpha, R)$ for non-zero rings $R$ and let $F (\{ 0 \}) = 1$.
Since any ring homomorphism $K_\alpha \to R$ is injective when $R$ is non-zero, $\textbf{CRing} (K_\alpha, R)$ is empty for sufficiently large $\alpha$, so $F (R)$ is indeed a set.
On the other hand, it is clear that $F$ cannot be the left Kan extension of any functor $\textbf{CRing}_\kappa \to \textbf{Set}$: if it were, it would be impossible to distinguish between this $F$ and the one where we cut off the disjoint union at some ordinal $\beta$ such that $K_\beta$ is not $\kappa$-presentable.


问:
Proof for generator of vector space
Given

$A \in M_{mn}(\mathbb{F})$ is a static matrix

$W = \{w \in M_{m1}(\mathbb{F})\ | $there is one $x \in M_{n1}(\mathbb{F})$ such that $Ax = w\}$ is a subspace of $M_{m1}(\mathbb{F})$. Proof that the column vectors in $A$ are a generator for $W$.

Attempt

$w = Ax = \begin{pmatrix}\sum_{i=1}^n A_{1i} x_i \\\ ... \\\ \sum_{i=1}^n A_{mi} x_i\end{pmatrix}_{m1}$

We can see that each element $x_i$ in vector $x$ is a coefficient for the i-th column vector in $A$. The linear combination of these column vectors makes up any vector $w \in W$. ◻

Is my approach correct and if so, is this enough for a proof?
I hope that I get a better sense for math during the next days.

答:
In response to the last comment:

There is no standard notation for the columns, and you can use whichever you like as long as you say what it means. Let's say that the symbol $A_{:,j}$ denotes the $j$-th column of $A$ $($so $A_{:,j} \in M_{m1}(\mathbb F))$. Now, your solution is correct, if $x_1,\dots,x_n$ are the coordinates of $x$, then $Ax = \sum_{j=1}^n x_jA_{:,j}$ $($for example $$\begin{pmatrix} A_{11} & A_{12} & A_{13} \\ A_{21} & A_{22} & A_{23} \\ A_{31} & A_{32} & A_{33} \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \\ x_n \end{pmatrix} = x_1 \begin{pmatrix} A_{11} \\ A_{21} \\ A_{31} \end{pmatrix} + x_2 \begin{pmatrix} A_{12} \\ A_{22} \\ A_{32} \end{pmatrix} + x_3 \begin{pmatrix} A_{13} \\ A_{23} \\ A_{33} \end{pmatrix}).$$

Title: Proof for generator of vector space


问:
Looking for alternative ways to solve $3^{2^x}=2^{3^x}$

If $\log2=m$ and $ \log3=n$, What is the answer of
$\large3^{2^x}=\large2^{3^x}$? $1)\log
 mn\quad\quad\quad\quad\quad\quad2)\frac{m-n}{\log m-\log
 n}\quad\quad\quad\quad\quad\quad3)\frac{\log \frac
 mn}{m-n}\quad\quad\quad\quad\quad\quad4)\frac{m+n}{\log mn}$

Here is my approach:

To solve the equation I took logarithm in base $3$ of both sides of the equation:

$$2^x=\log_32^{3^x}=3^x\log_32=3^x\times\frac{m}{n}$$
So we have $(\frac23)^x=\frac mn$. by taking logarithm in base $\frac23$ we have:

$$x=\log_{\large\frac23}(\frac mn)=\frac{\log(\frac mn)}{\log(\frac23)}=\frac{\log \frac mn}{m-n}$$

My questin: Is there other approach to solve this problem?

答:
Semi-alternative approach : defer taking logarithms to the end.

$\log 2 = m, ~\log 3 = n \implies e^m = 2,
~e^n = 3 \implies $

$\displaystyle 2^x = (e^m)^x = e^{mx}, ~3^x = (e^n)^x = e^{nx}.$

Therefore,

$\displaystyle 3^\left({2^x}\right) = 3^{\left(e^{mx}\right)}
= \left(e^{nx}\right)^{\left(e^{mx}\right)}
= e^{(nx) \times \left(e^{mx}\right)}.$

Similarly,

$\displaystyle 2^\left({3^x}\right) = 2^{\left(e^{nx}\right)}
= \left(e^{mx}\right)^{\left(e^{nx}\right)}
= e^{(mx) \times \left(e^{nx}\right)}.$

Therefore,

$$ e^{(nx) \times \left(e^{mx}\right)}
= e^{(mx) \times \left(e^{nx}\right)}.$$

Now, you can take logarithms of each side:

$\displaystyle (nx) \times \left(e^{mx}\right)
= (mx) \times \left(e^{nx}\right) \implies $

$\displaystyle \frac{m}{n} = \frac{mx}{nx}
 = \frac{e^{mx}}{e^{nx}}.$

Taking logarithms again,

$\displaystyle \log{\frac{m}{n}} = mx - nx = (m-n)x \implies
\frac{\log{\frac{m}{n}}}{m - n} = x.$


问:
Let X be a random variable with pdf given by $\beta^2xe^{-\frac{1}{2}\beta^2x^2}$
Let X be a random variable with pdf given by $\beta^2xe^{-\frac{1}{2}\beta^2x^2}$ where $\beta>0, x >0$

Find $E[X]$.

I believe this is a function from the gamma distribution. Here is what I have thus far

$$\int_0^\infty\beta^2xe^{-\frac{1}{2}\beta^2x^2}dx$$
$$\beta^2\int_0^\infty xe^{-\frac{1}{2}\beta^2x^2}dx$$
$$u=-\frac{1}{2}\beta^2x^2$$ $$du=-\frac{1}{2}\beta^2x^2dx$$ $$dx=-\frac{du}{\beta^2x}$$

$$=\beta^2\int_0^\infty xe^{u}*-\frac{du}{\beta^2x}$$
$$=-\int_0^\infty e^{u}du$$

I just wanted to know if my steps are correct thus far. I haven't done integration in a while, so I am not sure if my steps are correct/incorrect.

答:
No, it is not a gamma but a reileigh with expectation

$$\mathbb{E}(X)=\frac{1}{\beta}\sqrt{\frac{\pi}{2}}$$

How to calculate it:

$$\mathbb{E}(X)=\int_0^{\infty}\beta^2x^2e^{-(\beta^2x^2)/2}dx$$

set $y=\beta x$ and the result is immediate using standard gaussian's second moment

In fact you get

$$\mathbb{E}(X)=\frac{1}{\beta}\int_0^{\infty}y^2e^{-y^2/2}dy=\frac{\sqrt{2\pi}}{\beta}\int_0^{\infty}\frac{1}{\sqrt{2\pi}}y^2e^{-y^2/2}dy=$$
$$=\frac{\sqrt{2\pi}}{2\beta}\underbrace{\int_{-\infty}^{\infty}\frac{1}{\sqrt{2\pi}}y^2e^{-y^2/2}dy}_{=1}=  \frac{1}{\beta}\sqrt{\frac{\pi}{2}}    $$

I can say that

$$\int_{-\infty}^{\infty}f(y)dy=2\int_{0}^{\infty}f(y)dy$$

because $f(y)$ is even.

thus the latest integral is $E(Y^2)=1$ because $Y\sim N(0;1)$


问:
Need clarification on differences/products of Modular Arithmetic
I am completing a discrete math assignment and noticed in my textbook a solution regarding modular arithmetic and the Pigeonhole Principle:

Using the Pigeonhole Principle, show that in every set of 100 integers, there
exist two whose difference is a multiple of 37.

Since there are 100 pigeons and only 37 pigeonholes,
two pigeons must go in the same pigeonhole. This means k1 rem 37 = k2 rem 37,
which implies that k1 − k2 is a multiple of 37.

Now for my questions. How does the mod and rem function differ?

Can someone clarify how two integers with the same remainder of another integer implies that the difference will be a multiple?

In other words, suppose we have $a,b,c \in Z$. How does $a  \mod c = b\mod c$ imply $a-b$ is a multiple of $c$?

I feel like this is some elementary principle/theorem that is not coming to me at the moment. Any help would be appreciated.

答: $a\equiv b\pmod c$ means that there is some integer $n$ such that $a=nc+b$.  If also $d\equiv b \pmod c$, then there is some integer $m$ such that $d=mc+b$.  Now, $$a-d = nc+b -(mc+b) = nc-mc =(n-m)c$$ and we see that $a-d$ is a multiple of $c$.


问:
Probability of going to school $A_1$ when there are 9 more other possibilities
I am  curious to an extension of the previous question Probability of going to school A
and  want to extend the schools from 3 to 10.

Now the story begins : I commute to teach at one of $A_i$ (i from 1 to 10) schools everyday by train. I live at the terminal station where every 5 x i mins (i.e.,5,10,15,20,... mins)there is a train leaving for school $Ai_i$ punctually, but the actual departure times are unknown to me.  I will jump on the earliest train available at the station and teach at that school(if they arrive the same time, I have equal probability for each one of them).

My question is: What is the probability I teach at school A$_1$? I can manually do it  for cases of  2,3,4 schools following the procedure provided from the answer. Is there a formula I can use to directly calculate when the school options grow to 10?

EDIT:

Actually, when there are only 2 schools $A_1$ and $A_2$ to choose
from, there is a 2D geometry solution (as shown below):

Points within the 5 min by 10 min rectangle represent all possible
next departure  times of $A_1$ and $A_2$.For example P(1.5,3) means
next $A_1$ departs at 1.5 min later, and $A_2$ departs at 3 min later.

Any points above  the 45$^\circ$ line represent cases when $A_1$
departs earlier than $A_2$ (when points are on this 45$\circ$ line,
the probabilities for me to take each is both 50%.

So the probability to go to school  $A_1$ is the (1- $\frac{S_\triangle}{S\square}$)= 75%.

But, I find it hard to extend to 3D,4D,...

答:
Let $X_1,\dots,X_{n}$ be independent random variables with $X_i$ distributed uniformly in $[0,i]$. You're asking for $P(\min_iX_i=X_1)$. It's fairly clear that this is
$$
\left.
\int_0^1\int_{x_1}^2\int_{x_1}^3\dots\int_{x_1}^n dx_n\dots dx_2dx_1
\middle/
\int_0^1\int_0^2\int_0^3\dots\int_0^n dx_n\dots dx_2dx_1
\right.
\\=\frac1{n!}\int_0^1\prod_{i=2}^n(i-x_1) dx_1.
$$

The integrand is a polynomial in $x_1$, so for a given value of $n$, you can expand it (using e.g. Vieta's formulas) in order to evaluate the integral. Maybe someone else will find a simpler closed-form answer.


问:
On whether or not $\sum na_n$ converges.
If $a_n$ is absolutely convergent, then does $\sum na_n$ also converge? I have an objection with $a_n = \frac{(-1)^n}{n^2}$, where it is absolutely convergent, but $\sum na_n=\frac{(-1)^n}{n}$, which is conditionally convergent. Can I have another example that also has that property?

Another question naturally arises is whether given a random conditionally convergent sum, $\sum a_n$, then is $\sum (-1)^n a_n$absolutely convergent? I feel that this is a yes. Different opinions?

答:
No. Take $a_n = \frac{1}{n^2}$ for all $n\geq 1$. $\sum_{n=1}^\infty a_n < \infty$ (absolute convergence, since all terms are positive), but $$\sum_{n=1}^\infty na_n = \sum_{n=1}^\infty\frac{1}{n} = \infty$$

For your second question, also no. Take $a_n = \frac{(-1)^n}{n}$. Then $\sum_{n=1}^\infty a_n$ is conditionally convergent; but $$\sum_{n=1}^\infty (-1)^n a_n = \sum_{n=1}^\infty\frac{1}{n} = \infty$$

For your second question, even more generally, what you are asking is exactly

If $\sum_n a_n$ is conditionally convergent, then is it also absolutely convergent?

which is clearly false. Why is it equivalent to your question? Well, because $|(-1)^n a_n|=|a_n|$, so $\sum_n (-1)^n a_n$ is absolutely convergent iff $\sum_n a_n$ is.


答:
$\mod{}$ is not actually a function.

The easiest way to learn and understand what $42 \equiv 79\pmod {37}$ means is that $42$ and $79$ have the same remainder when divided by $37$.  And $42 = 1\cdot 37 + 5$ and $79 = 2\cdot 37 + 5$ that is certainly true.

However the technical definition of $a\equiv b \pmod n$ is that $n$ divides into $a-b$ and so $42\equiv 79 \pmod {37}$ means $37$ divides into $42-79$.  And as $42-79 = -37$ that is also certainly true.

These two definitions are exactly this same.

.....

Every number can be written  $a = m\cdot q + r$ where $r$ is the remainder and is between $0$ and $q$ (i.e.  $0 \le r < q$).

So if $a = m\cdot q + r$ and $b = w\cdot q + r$ have the same remainder, then $a-b = (m\cdot q + r)-(w\cdot q +r) = q(m-w)$ so $q$ divides into $a-b$.

And if $q$ divides evenly into $a-b$,  then if we let $a = m \cdot q + r$ and $b = w \cdot q+ s$,  then $q$ divides evenly into $a-b = (m\cdot q + r) - (w\cdot q+s) = q(m-w) + (r-s)$.  So $q$ divides evenly into $r-s$.  But $0 \le r < q$ and $0 \le s < q$ so $-q < r-s < q$ and the only number between $-q$ and $q$ (NOT including $-q$ or $q$) that is divisible by $q$ is... $0$ [because $0 = 0\cdot q$]. So if $q|r-s$ and $r-s < q$ and $-q < r-s$ the only way that is possible is if $r-s=0$ or in other words if $r =s$.  So in other words.... $a$ and $b$ have the same remainder.

Title: Need clarification on differences/products of Modular Arithmetic

Body:
I am completing a discrete math assignment and noticed in my textbook a solution regarding modular arithmetic and the Pigeonhole Principle:

Using the Pigeonhole Principle, show that in every set of 100 integers, there
exist two whose difference is a multiple of 37.

Since there are 100 pigeons and only 37 pigeonholes,
two pigeons must go in the same pigeonhole. This means k1 rem 37 = k2 rem 37,
which implies that k1 − k2 is a multiple of 37.

Now for my questions. How does the mod and rem function differ?

Can someone clarify how two integers with the same remainder of another integer implies that the difference will be a multiple?

In other words, suppose we have $a,b,c \in Z$. How does $a  \mod c = b\mod c$ imply $a-b$ is a multiple of $c$?

I feel like this is some elementary principle/theorem that is not coming to me at the moment. Any help would be appreciated.


问:
Find all subgroups of $\mathbb{Z_7}\times\mathbb{Z_5}$
Find all subgroups of $\mathbb{Z_7}\times\mathbb{Z_5}$ without repeating the same subgroup. You should specify a generator of each subgroup.

I understand the question. However the only way I know how to find all the subgroups is to individually calculate the subgroup generated by each element of $\mathbb{Z_7}\times\mathbb{Z_5}$. That process is long, annoying and will definitely turn up some identical subgroups. Is there some method for doing this that is quicker/more concise? I suspect it has something to do with prime numbers but we've never actually covered a quicker method in class.

答:
The method of writing down the subgroup generated by each element is not as long a process as you anticipate, because it is fairly easy to recognize repeated subgroups. For that we use a consequence of Lagrange theorem wherein any non-identity element of $\mathbb{Z}_7$ generates the whole of $\mathbb{Z}_7$. Similarly for $\mathbb{Z}_5$.

Thus, we see that $(0, 0)$ generates the trivial subgroup. Next, $(0, k)$  for $k\in\{1, \dots, 4\}$ generates $\{0\}\times\mathbb{Z}_5$. Similarly, $(k, 0)$ for $k\in\{1, \dots, 6\}$ generates $\mathbb{Z}_7\times\{0\}$. Finally, $(i, j)$ for $i\in\{1, \dots, 6\}$ and $i\in\{1, \dots, 4\}$ generates $\mathbb{Z}_7\times \mathbb{Z}_5$, because $5$ and $7$ are coprime.

We conclude that $\mathbb{Z}_7\times \mathbb{Z}_5$ has four subgroups: $\{(0, 0)\}$, $\{0\}\times\mathbb{Z}_5$, $\mathbb{Z}_7\times\{0\}$ and $\mathbb{Z}_7\times \mathbb{Z}_5$.


问:
Show that H is a subset of the normalizer
My Abstract Algebra professer assigned us this homework problem which I'm assuming he created himself:

Suppose $H$ is a subgroup of $G$. Show that $H$ is a subset of the normalizer of $H$.

The normalizer of $H$ is defined as $N(H)=\{x \in G: x^{-1}Hx=H\}$. I've spent more time than I'd like trying just to understand the logic behind the statement and I end up confusing myself. Thanks in advance for any and all help.

答: Let $x \in H$. We need to show $x^{-1}Hx = H$. Suppose $h\in H$, and consider $x^{-1}hx \in x^{-1}Hx$. Since $x \in H$, we have $x^{-1} \in H$, and so $x^{-1}hx \in H$. Hence, $x^{-1}Hx\subseteq H$ for $x\in H$. Now suppose $h \in H$ and consider $k := xhx^{-1}\in H$. Then, we have $x^{-1}kx = h \in x^{-1}Hx$. Hence, $H\subseteq x^{-1}Hx$, and so $x^{-1}Hx = H$. Hence, $x \in N(H)$.</p>

ContentLicense: CC BY-SA 3.0


答:
$\mathbb Z_7×\mathbb Z_5\cong\mathbb Z_{35} $.

And for a cyclic group,  the subgroups are the cyclic groups of all orders dividing the order of the group.

That gives us the trivial group,  $\mathbb Z_5,\mathbb Z_7$, and the whole group.

For generators, use a generator, say $(1,1) $ for $\Bbb Z_{35} $.  Then $(0,0),(7,7),(5,5) $ and $(1,1) $ will do.

Body:
Find all subgroups of $\mathbb{Z_7}\times\mathbb{Z_5}$ without repeating the same subgroup. You should specify a generator of each subgroup.

I understand the question. However the only way I know how to find all the subgroups is to individually calculate the subgroup generated by each element of $\mathbb{Z_7}\times\mathbb{Z_5}$. That process is long, annoying and will definitely turn up some identical subgroups. Is there some method for doing this that is quicker/more concise? I suspect it has something to do with prime numbers but we've never actually covered a quicker method in class.


问:
normal distribution finding b
For part b, I wrote the answer like this:

Pr(− < ( − )/ < ) = 0.90

Pr(− < Z < ) = 0.90

b=（1.64＋1.65)/2 =1.645

Am I correct doing like this? I'm not sure.
Do I need to do the inversion of centralise by making it times  and plus ?

Question: If random variable  follows Normal distribution

(a)If we know () = 75 and () = 100 please find Pr(X < 60) and
Pr(70 < X < 100)

(b)  Find  so that Pr(− < ( − )/ < ) = 0.90

答:
If you use R (or other software) you typically don't
need to standardize in order to solve such problems.

Here are answers in R: The second parameter is the standard deviation $\sigma = 10.$

If $X \sim \mathsf{Norm}(\mu=75, \sigma=10),$ then $P(70 < X < 100).$

```<code>diff(pnorm(c(70,100), 75, 10))
[1] 0.6852528
pnorm(100, 75, 10) - pnorm(70, 75, 10)
[1] 0.6852528
</code>```

Find $b$ such that $P(− < Z < ) = 0.90.$ By symmetry, this amounts to
finding $P(Z < b) = 0.95.$ Then $b = 1.644854$ often rounded to $1.645.$ In R, qnorm is the the
quantile function of a normal distribution. If no
mean or variance is given, then $\mu=0,\sigma=1$ are assumed.

```<code>b = qnorm(.95);  b
[1] 1.644854
diff(pnorm(c(-b,b)))
[1] 0.9
</code>```

Notes: (1) Advantages of using software are that standardizing is often not necessary and rounding to use printed standard normal CDF tables can be avoided. Software sometimes gives more decimal places of accuracy than are needed in practice.

(2) When you look in the body of a printed table for
a desired probability and then get the z-score from
the margins of the table, you are doing an inverse CDF
lookup.

Title: normal distribution finding b

Body:
If you use R (or other software) you typically don't
need to standardize in order to solve such problems.

Here are answers in R: The second parameter is the standard deviation $\sigma = 10.$

If $X \sim \mathsf{Norm}(\mu=75, \sigma=10),$ then $P(70 < X < 100).$

diff(pnorm(c(70,100), 75, 10))
[1] 0.6852528
pnorm(100, 75, 10) - pnorm(70, 75, 10)
[1] 0.6852528

Find $b$ such that $P(− < Z < ) = 0.90.$ By symmetry, this amounts to
finding $P(Z < b) = 0.95.$ Then $b = 1.644854$ often rounded to $1.645.$ In R, qnorm is the the
quantile function of a normal distribution. If no
mean or variance is given, then $\mu=0,\sigma=1$ are assumed.

b = qnorm(.95);  b
[1] 1.644854
diff(pnorm(c(-b,b)))
[1] 0.9

Notes: (1) Advantages of using software are that standardizing is often not necessary and rounding to use printed standard normal CDF tables can be avoided. Software sometimes gives more decimal places of accuracy than are needed in practice.

(2) When you look in the body of a printed table for
a desired probability and then get the z-score from
the margins of the table, you are doing an inverse CDF
lookup.


答:
There is an $$e^{-\frac{\beta^2}{2}x^2}$$ in the pdf: looks like we may be able to link that to a Gaussian distribution with mean $0$ and variance $\sigma^2 = 1/\beta^2$. Let's do that.

The pdf of that Gaussian would be
$$
f(x) = \frac{\beta}{\sqrt{2\pi}} e^{-\frac{\beta^2}{2}x^2}, \qquad x\in\mathbb{R}
$$
and if $Y$ is distributed according to this, then $\mathrm{Var}[Y] = \mathbb{E}[Y^2]$ (the mean being zero), so
$$
\frac{1}{\beta^2} = \mathbb{E}[Y^2] = \int_{-\infty}^\infty y^2 f(y)dy
= \frac{\beta}{\sqrt{2\pi}} \int_{-\infty}^\infty y^2 e^{-\frac{\beta^2}{2}y^2}dy
= \frac{2\beta}{\sqrt{2\pi}} \int_{0}^\infty y^2 e^{-\frac{\beta^2}{2}y^2}dy \tag{1}
$$

Interesting. what we want to compute, with that other probability distribution we had for some r.v. $X$, was
$$
\mathbb{E}[X] = \int_{0}^\infty x\cdot \beta^2 x e^{-\frac{\beta^2}{2}x^2} dx
= \beta^2\int_{0}^\infty x^2 e^{-\frac{\beta^2}{2}x^2} dx \tag{2}
$$

Great! We are done. Combining (1) and (2), we have
$$
\mathbb{E}[X] = \beta^2 \cdot \left(\frac{2\beta}{\sqrt{2\pi}}\right)^{-1}\mathbb{E}[Y^2]
=  \beta^2 \cdot \frac{\sqrt{2\pi}}{2\beta} \cdot \beta^{-2}
=  \boxed{\frac{1}{\beta}\sqrt{\frac{\pi}{2}}}
$$
and we never needed to know what the distribution of $X$ was named: just basic properties of gaussians.

Body:
Let X be a random variable with pdf given by $\beta^2xe^{-\frac{1}{2}\beta^2x^2}$ where $\beta>0, x >0$

Find $E[X]$.

I believe this is a function from the gamma distribution. Here is what I have thus far

$$\int_0^\infty\beta^2xe^{-\frac{1}{2}\beta^2x^2}dx$$
$$\beta^2\int_0^\infty xe^{-\frac{1}{2}\beta^2x^2}dx$$
$$u=-\frac{1}{2}\beta^2x^2$$ $$du=-\frac{1}{2}\beta^2x^2dx$$ $$dx=-\frac{du}{\beta^2x}$$

$$=\beta^2\int_0^\infty xe^{u}*-\frac{du}{\beta^2x}$$
$$=-\int_0^\infty e^{u}du$$

I just wanted to know if my steps are correct thus far. I haven't done integration in a while, so I am not sure if my steps are correct/incorrect.


问:
calculate $[T^t]^{b^{\ast }}_{g^{\ast }}$ directly from $T^t(f)$
Let $V=P_1(\mathbb{R})$ and $W=\mathbb{R}^2$ with the corresponding basis $b= \left \{ 1,x \right \}$ and $g= \left \{ e_1,e_2 \right \}$ We define the linear transformation $T:V\rightarrow W$ $$T(p)=(p(0)-2p(1),p(0)+{p}'(0))$$

if $f \in  W^{\ast } $ with: $$f(a,b)=a-2b$$

Calculate $T^t(f)$ and $[T^t]^{b^{\ast }}_{g^{\ast }}$.

$T^t(f)=f(T)$ then $f(T)=p(0)-2p(1)-2(p(0)+{p(0)}')$ plugging in the $b$ basis we get $-3,-4$ so

$T^t(f)=-3a-4b$
but if i want to calculate $[T^t]^{b^{\ast }}_{g^{\ast }}$ it's not working. What I mean, if I put $T^t(f(e_1))=-3$ and $T^t(f(e_2))=-4$ which gives me this matrix $\begin{pmatrix}
-3 &-4 \\
 0 &0
\end{pmatrix}$ and I should get  $[T^t]^{b^{\ast }}_{g^{\ast }}=\begin{pmatrix}
-1 &1 \\
 -2& 1
\end{pmatrix}$

What am I missing?

答:
First of all, $T^t(f) = f \circ T \in V^*$, and then $T^t(f)$ is the function from $V$ to $\mathbb R$ that maps each $p \in V$ to the real number $$\begin{align} T^t(f)(p) &= f(T(p)) = f(p(0)-2p(1),p(0)+p'(0)) \\ &= \big( p(0)-2p(1) \big) -2 \big( p(0)+p'(0) \big) \\ &= -p(0)-2p(1)-2p'(0). \end{align}$$ Now, $g^* = \{\pi_1,\pi_2\}$ where $\pi_i \in W^*$ $(i=1,2)$ is the function from $W$ to $\mathbb R$ that maps each $(x_1,x_2) \in W$ to $x_i$, that is, $\pi_i$ is the projection onto the $i$-th coordinate (this is by merely definition of dual basis). Hence, you need to compute $T^t(\pi_1)$ and $T^t(\pi_2)$ (note that these are functions) and then write it as a linear combination of the functions $p_1$ and $p_2$ (what is a linear combination of functions?), where $b^* = \{p_1,p_2\}$ is the dual basis of $b$.

I hope this clarify the things a little!


问:
$\mu = m = \frac{q_1 +q_3}{2}$
Let $X$ be a random variable with p.m.f./p.d.f. $f_X (x)$ that is symmetric about
$\mu (\in R),$ i.e., $f_X (x + \mu) = f_X (\mu − x)$, $\forall x \in (−\infty, \infty).$

If $q_1 , m$ and $q_3$ are respectively the lower quartile, the median and the upper
quartile of the distribution of $X$ then show that $\mu = m = \frac{q_1 +q_3}{2}$

How to prove.....

any Hints.

答:
Just want to add another interesting tidbit - if $E[X] < \infty$, then $$E[X] = \mu = m = \frac{q_1+q_3}{2}$$
$$\begin{align}E[X] &= \int_{-\infty}^\infty xf_X(x) dx\\
&= \int_{-\infty}^\infty (2\mu - x)f_X(x)dx\\ &= 2\mu\int_{-\infty}^\infty f_X(x) dx - \int_{-\infty}^\infty xf_X(x)dx \\&= 2\mu - E[X]\\
\implies E[X] &= \mu
 \end{align}$$

Title: $\mu = m = \frac{q_1 +q_3}{2}$


问:
Given a metric function between a set of abstract points, what is the best way to plot them on a 2D space?
I have a list of several entities, all of each have a numerical relationship to each other that defines an abstract distance.

Is there a mathematical way to plot all of these on a 2D space, turning the abstract distance into a 2D euclidian distance, and preserving all of their distances?

答: As Jonas Meyer and others pointed out, you cannot expect to embed your points in the plane and preserve their distances.  If you want to at least come close, there is a notion of distortion for an embedding of metric spaces, and algorithms for computing a minimum distortion embedding.  I am no expert at this, so I won't attempt to suggest useful references, but searching those keywords may be helpful to you.

ContentLicense: CC BY-SA 2.5


问:
Will there be a proper subset B(x,d) of B(x,D) in a metric space?
I am working on this:

If $(X,\rho)$ is a non-discrete metric space, $x\in X$ is not an isolated point. Assume $U$ is an open set which contains $x$.
It's obvious that $\exists B(x,D) \subset U.$

I suppose that there must be a proper subset $B(x,d) \subsetneqq B(x,D)$

Here is my proof:

First of all, U can't be finite. If so, assume $U=\{x,x_1,x_2,...,x_n\}$, then take $0<\varepsilon<min \{\rho(x,x_1),...,\rho(x,x_n)\}$.

Then take $B(x,\varepsilon)$. By defination of $\varepsilon$, $B(x,\varepsilon)=\{x\}$. However, $x$ is not an isolated point. So U is not finite.

Then,take $y\ne x \in B(x,D)$, and $\frac{\rho(x,y)}{2}$ as d, then  $B(x,d) \subsetneqq B(x,D)$.

My tutor said there is something wrong with my proof. But I can't see where's the problem.

If there really something wrong? If so ,where is the mistake?

答: I think the issue is with your argument that $U$ is infinite; it's correct, but it's not really relevant.  All you really need is the last line ("take $y \ne x$..."), and to get there all you need is to show that $B(x,D)$ contains at least one point other than $x$.  But that's immediate because if it didn't then $x$ would be isolated.


问:
How to decompose a rational function into 2 functions?
Suppose that $h(x) = (f ◦ g)(x)$ and $ \frac{27x^3+3x}{3x-1}$. Solve for $f(x)$ and $g(x)$.

I'm having a hard time decomposing this rational function. I knew a bit about the composition functions. I tried my best to find the functions and I also used trial-error but I can't still find the result. I'm still in basic math specifically functions.

I tried this:
$
h(x) = \frac{3x(9x^2+1)}{3x-1}
$ I factored the numerator. I also intend to let $g(x) = 9x^2+1$, but there are still terms in $x$ which I cannot manipulate.

If I also let $g(x) = x$ in given $h(x)$, the function $f(x)$ will equal to $h(x)$ itself. Is this the correct answer? Is there anything else to get the function aside from $func(x) = x$?

What is the best technique to decompose this give rational function into 2 functions?

答:
You might find the question difficult because there's really no right answer; there are (infinitely) many ways to decompose $h$ into two other functions, and there's no such thing as the correct answer or the "best technique".

You mention trying
$$h(x) = \frac{3x(9x^2+1)}{3x-1}$$

From here, notice that $9x^2 = (3x)^2$, so let's try
$$f(g(x)) = \frac{3x((3x)^2+1)}{3x-1}$$

Now with this common element of $3x$ in several places, you could let $g(x)=3x$, so that

$$f(g(x)) = \frac{g(x)((g(x))^2+1)}{g(x)-1}$$

In other words, we have
$$f(x) = \frac{x(x^2+1)}{x-1}, \quad g(x)=3x$$

That's one possible answer, but keep in mind that there are a vast number of other answers. For example, consider $g(x) = 10\left(\frac{27x^3+3x}{3x-1}\right)$ and $f(x) = \frac1{10}x$, etc.


问:
Laplace transformation with $(t+3)^2$
The problem says that: $$F(s) = \frac{-2 \times {(s)}^{-1/2}}{1}$$

I need to find the laplace transformation of ${e}^{-8t} \times f(t) \times ({t+3})^{2} $

I know that if it was $({t})^{2} $ instead of $({t+3})^{2} $, I would have to: 1st) substitute s for s+8 (in my country we say translation, right?), than I would have to find the second derivate of it. I've already done the 1st step (s+8 thing) and got $$F(s) = \frac{-2 \times {(s+8)}^{-1/2}}{1}$$ What am I supposed to do now? Sorry for my english, I hope you understand my doubt.

答:
To solve this question, you need to use a combination of some properties of the Laplace Transform: linearity, frequency shift and frequency differentiation. Let's remember them first.

$$
\newcommand{\tl}[3]{#1 & #2 & #3 \\\hline}
\newcommand{\bs}[0]{\,\,\,\,\,}
\begin{array}{|c|c|c|} \hline
  \rule[0]{0pt}{2.5ex}
  \rule[0]{0pt}{-1.5ex}
  \tl {{\mathbf{\bs Property \bs}}}
      {{\mathbf{\bs f(t) \bs}}}
      {{\mathbf{\bs F(s) \bs}}}
  \rule[0]{0pt}{2ex}
  \rule[0]{0pt}{-1ex}
  \tl {\text{Linearity}} {a\!\cdot\! f(t) + b\!\cdot\! g(t)} {a \!\cdot\! F(s) +b\!\cdot\! G(s)}
  \rule[0]{0pt}{2ex}
  \rule[0]{0pt}{-1ex}
  \tl {\text{Frequency Shift}} {e^{-at}f(t)} {F(s+a)}
  \rule[0]{0pt}{2ex}
  \rule[0]{0pt}{-1ex}
  \tl {\text{Frequency Differentiation}} {t^nf(t)} {(-1)^n \,F^{(n)}(s)}
\end{array}
\\ \small\text{(Note: a,b}\in \mathbb{R} \, \text{and} \, F^{(n)} \,\text{denotes the} \, \text{n$^{\text{th}}$} \, \text{derivative of F)}
$$

Let $g(t) = e^{-8t} \cdot f(t) \cdot(t+3)^2$. Rewriting the expression, in order to use the above properties, we have:

$$
g(t) =\underbrace{t^2e^{-8t}f(t)}_{\text{(I)}} \,\,+  \,\,\underbrace{6te^{-8t}f(t)}_{\text{(II)}}  \,\,+ \,\, \underbrace{9e^{-8t}f(t)}_{\text{(III)}}
$$

For $\text{(I)}$, we use frequency shift combined with frequency differentiation. For $\text{(II)}$, we use the previous ones plus linearity. And for $\text{(III)}$ we use frequency shift and linearity. Then $G(s)$ is:

$$
G(s) = (-1)^2 F^{(2)}(s+8) \,\, + \,\, 6\cdot (-1)^1 F^{(1)}(s+8) \,\, + \,\, 9F(s+8)
$$

Since $F(s) = -2\cdot (s)^{-1/2}$, we have:

$$
\begin{alignat}{1}
G(s) &= -\frac{-3/2}{\sqrt{s+8}\cdot (s+8)^2}-\frac{6}{\sqrt{s+8}\cdot (s+8)} -\frac{18}{\sqrt{s+8}}
\\[6pt] &= -\frac{-3/2 \cdot (12s^2+196s+801)}{\sqrt{s+8}\cdot (s+8)^2}
\end{alignat}
$$

Title: Laplace transformation with $(t+3)^2$


问:
Elementary proofs about $f(t)=\lim_{n\to\infty}(1+\frac tn)^n$
Please read carefully what I mean by elementary before marking the question as duplicate.

I'm starting calculus lessons in High School; that is: no series, no derivatives, no limits of functions, no continuity... I have just stated the axiom of supreme and defined sequence and limit of sequence. I have proved that a monotone, bounded sequence has limit. Just that.

Under so 'elementary' conditions, I'd like to prove:

For each real $t$, the sequence $\left(1+\dfrac tn\right)^n$ converges. (I have problems for $t<0$). Then I'll define $e^t$ as the limit of this sequence.

For every pair $t,s$, $e^{t+s}=e^te^s$.

Any ideas or bibliography?

答:
For brevity let's write $$f_n(t) =\left(1+\frac{t}{n}\right)^n\tag{1}$$ Using binomial theorem we can observe (after some algebra) that $$f_n(t) =1+t+\dfrac{1-\dfrac{1}{n}}{2!}t^2+\dfrac{\left(1-\dfrac{1}{n} \right)\left(1-\dfrac{2}{n}\right)}{3!}t^3+\dots\tag{2}$$ The series on right is finite and consists of $(n+1)$ terms. If $t>0$ then as $n$ increases each term in above series increases and the number of terms also increases. Hence $f_n(t) $ is an increasing sequence if $t>0$. The case $t<0$ will be handled later.

Let's first start with the simplest case when $t=1$ (note $t=0$ gives the limit $1$ as the sequence is constant with each term being $1$). We prove that $f_n(t) $ is bounded above by $3$ if $t=1$. Using equation $(2)$ we have $$f_n(1)\leq 1+1+\frac {1}{2!}+\dots+\frac{1}{n!}$$ which does not exceed $$1+1+\frac{1}{2}+\dots+\frac{1}{2^{n-1}}=1+2(1-(1/2^n))<3$$ Hence $f_n(1)$ converges to some limit and since $f_n(1)\geq 2$ the limit $f(1)$ lies in interval $[2,3]$. All we need to know is that $f(1)>0$.

Next let $t$ be a positive integer. We can write $$f_n(t) =\prod_{i=1}^{t}\dfrac{\left(1+\dfrac{1}{n+i-1}\right)^{n+i-1}}{\left(1+\dfrac{1}{n+i-1}\right)^{i-1}}\tag{3}$$ Each numerator above is a subsequence of $f_n(1)$ and hence tends to $f(1)$. Each denominator on the other hand tends to $1$ and thus $f_n(t) $ tends to $(f(1))^t$. Since the sequence is increasing it also follows that $$f_n(t) \leq (f(1))^t,\forall t, n\in\mathbb {N}\tag{4}$$ Now let's consider the case when $t>0$ is a real number. And let $m$ be a positive integer with $t<m$. Then using $(4)$ we have $$f_n(t) <f_n(m) \leq (f(1))^m$$ so that the sequence is bounded above and hence converges. Thus we have shown that $f(t) $ exists for all $t\geq 0$ and one can check $f(t) \geq 1+t$ and $f(t) =(f(1))^t$ if $t$ is a positive integer.

Consider $t$ as a positive rational number of the form $t=p/q$ where $p, q$ are positive integers. Then we know that $f_n(t) $ converges to $f(t) $. Further $$(f_n(t)) ^q=\left(1+\frac {p} {qn} \right) ^{qn} $$ is a subsequence of $f_n(p) $ and hence converges to $f(p) = (f(1))^p$. It follows that $(f(t)) ^q=(f(1))^p$. We have thus proved that $f(t)=(f(1))^t$ if $t$ is a positive rational number.

Let us now deal with $t<0$ so that $-t>0$. We can use Bernoulli's inequality and get $$1-\frac{t^2}{n}\leq\left(1-\frac {t^2}{n^2}\right)^n\leq 1\tag{5}$$ whenever $n>|t|$. By squeeze theorem it follows that $(1-(t^2/n^2))^n\to 1$ (alternatively it can be proved using a lemma of Thomas Andrews). In other words we have $$\left(1+\frac{t}{n}\right)^n=\dfrac {\left(1-\dfrac {t^2} {n^2}\right)^n} {\left(1+\dfrac{(-t)} {n} \right)^n} \to \frac{1}{f(-t)}$$ Thus we have proved that $f(t) $ exists for all real $t$ and we have $f(t) f(-t) =1$. This also proves that $f(t) >0$ for all real $t$.

To prove the exponential property of $f(t) $ we can consider the sequence $$a_n=\dfrac{1+\dfrac{s+t} {n} } {\left(1+\dfrac{s} {n} \right)\left(1+\dfrac{
t} {n} \right)} \tag{6}$$ It can be checked that $n(a_n-1)\to 0$ and hence by lemma of  Thomas Andrews the sequence $a_n^n\to 1$. Thus $f_n(s+t) $ converges to $f(s) f(t) $ and we have $f(s+t) =f(s) f(t) $ for all real $s, t$.

The case when $t\in\mathbb {C} $ is handled in this answer.


问:
Find all functions $f: \mathbb{R} \to \mathbb{R}$ such that $f(f(x+y)) = f(x)+f(y)$
The problem is to find the set of all functions $f: \mathbb{R} \to \mathbb{R}$ such that for all $x,y \in \mathbb{R}$ we have $f(f(x+y)) = f(x)+f(y)$.
 We first notice that $f(x)=x+c$ is a solution. With some work it turns out that it is enough to find solutions with $f(0)=0$ and from that to find all solutions, and moreover, it is not that hard to show that in this case of $f(0)=0$ the functional equation is equivilent to the two conditions:

$$(1): f(x+y)=f(x)+f(y) \\ (2) :f(f(x))=f(x) $$
I was expecting that the only solutions will be $f(x) = x, f=0$ but with a bit more thought I think the set of solutions is bigger. I would like to know if I'm right. Here's what I thought: take a basis $\{e_\alpha\}_{\alpha \in I}$ of $\mathbb{R}$ as a vector space over $\mathbb{Q}$.
 Consider a division of this basis to disctint pairs $(e_\alpha, e_\beta)$ [each element of the basis appears in one and exactly one of those pairs). Define $f$ on the basis elements as $f(e_\alpha) = e_\beta, f(e_\beta)=e_\beta$. Thus we defined $f$ on the basis elements. Now, we simply continue $f$ linearly, to define it on all $\mathbb{R}$: $f(\sum a_i e_i) = \sum a_i f(e_i)$. This makes sure $(1)$ holds and $f$ is additive. By definition, it is trivial that $(2)$ also works if $x$ is a basis element,and therefore, because $f$ is additive, it also holds for any real number (just write it as a linear combination of the basis elements and do the algebra). Therefore I think this is a construction which shows that there are plenty of pathological solutions. Am I right? Is there a nice way to characterize all solutions?

答:
Your analysis is incomplete.

You misunderstood what is necessary for $ f(f(x)) = x$.

While having $f(e_\alpha) = e_\beta, f(e_\beta) = e_\alpha$ is a sufficient condition, it is not a necessary condition. For example, the solution $f(x) = 0 $ corresponds to mapping each basis element to 0.

The first condition tells us that with any basis $ \{e_\alpha\}$ of $ \mathbb{R}$ over $\mathbb{Q}$, if $ f( e_\alpha ) = f_\alpha$, then $ f(\sum a_\alpha e_\alpha) = \sum a_\alpha f_\alpha$.

The second condition tells us that the image of $f$ satisfies $ a \in $ Im $f \Rightarrow f(a) = a $.

So, the correct classification of solutions is:

Given any solution $f$,

Let $ \{ e_\alpha\}$ be a $\mathbb{Q}$ basis for the image of $f$, then for every $a \in Im$  f, $ f(a ) = a$

Let $\{e_\alpha\} \cup \{ f_\beta\}$ be a $\mathbb{Q}$ basis for $\mathbb{R}$. Then, $f( f_\beta) = \sum c_{\beta\alpha} e_\alpha$.

Finally, if $ r = (\sum a_\alpha e_\alpha ) + (\sum b_\beta f_\beta)$, then $f(r) = (\sum a_\alpha e_\alpha )+ (\sum b_\beta c_{\beta\alpha} e_\alpha)$.

It is clear that any such solution satisfies the 2 conditions.


答:
Suppose $h, x \in H$. Can you find $k\in H$ such that $x^{-1} k x =h$? Also, is it possible that for some $m\in H$ you have $x^{-1} k x \not \in H$?
Can you see why this answers the question?</p>

Title: Show that H is a subset of the normalizer

Body:
My Abstract Algebra professer assigned us this homework problem which I'm assuming he created himself:

Suppose $H$ is a subgroup of $G$. Show that $H$ is a subset of the normalizer of $H$.

The normalizer of $H$ is defined as $N(H)=\{x \in G: x^{-1}Hx=H\}$. I've spent more time than I'd like trying just to understand the logic behind the statement and I end up confusing myself. Thanks in advance for any and all help.


问:
$x,y\in \mathbb Z^+$ Show that $P(x)=P(y) \implies x=y$

Let $x,y\in \mathbb Z^+$, we define $P(a)$ as the product of all the divisors of $a$ including $1$ And $a$, Show that $$P(x)=P(y)\implies x=y.$$

Im gonna show my Attempt, which is based on cases.

Case 1: if  $x,y$ Are primes.
$$x,y\in \mathbb P\implies \cases{P(x)=x\\P(y)=y}\implies (P(x)=P(y) \iff x=y)$$
Where $\mathbb P$ Is the set of prime numbers.

Case 2: if $x,y$ are some powers of prime.
$$\cases{x=p^n\\ y=q^m } \implies \cases{P(x)=1\cdot p\cdot p^2...p^n=p^{\frac{n(n+1)}{2}}\\P(y)= 1\cdot q\cdot q^2...q^m= q^{\frac{m(m+1)}{2}}} $$ So to prove the statement We set $$p^{\frac{n(n+1)}{2}} =q^{\frac{m(m+1)}{2}}$$
The next step is to show that this statement implies $x=y$, I have no idea how to do that. (By the way the next case is to consider $x,y$ As a product of primes.)

答:
Recall that if $ N = \prod p_i ^ {a_i}$ is the prime factorization of $N$, then $N$ has $\tau(N) = \prod (a_i + 1)$ factors, and $P(N) = N^{\tau(N)/2 } $ (by pairing up the factor $d$ with $N/d$).

Let $M = \prod p_i ^{b_i}$ (with the same ordering of primes, and finitely many exponents are non-zero).

Suppose that $P(N) = P(M)$.

Then $N^{\prod(a_i + 1) } = M^{\prod (b_i + 1) }$.

Considering prime $p_k$, this gives us $ a_k \prod(a_i + 1 ) = b_k \prod(b_i + 1)$ for all $k$.

Hence $a_k / b_k$ is the constant $\prod (b_i+1)/(a_i+1)$.

If $a_k / b_k < 1$, then clearly $a_k \prod (a_i + 1) < b_k \prod (b_i + 1)$, which is a contradiction. Likewise for $a_k / b_k  >1$.

Hence, $a_k / b_k = 1$ and so $N = M$.


问:
Is this intuitive by graph ? Or its needs rigorous proof
A double differentiable function $f : \mathbb{R} \to \mathbb{R}$ satisfy $f(x)f^{\prime \prime}(x) \ne 0 \ \forall x \in \mathbb{R}$. Can we conclude $f(x)f^{\prime \prime}(x) >0$ for all $\mathbb{R}$ , even though continuity of $f''(x)$ is not given. I thought of making a contradiction by graph supposing $f(x)>0$ and $f''(x) <0$ , I thought graph would need to cut x axis hence after that $f(x)<0$ ( contradiction) but I don't have a good calculus based proof on this fact. As continuity of $f''(x)$ not given is also a problem for all $\mathbb{R}$

答:
As you observe that $f(x)$ and $f^{\prime \prime }(x)$ can't change sign as the derivative has the darboux property Now we can assume that $f(x) > 0 \ \forall x \in \mathbb{R} $ as if it was $<0$ we can apply it to $-f$.
For the sake of contradiction assume $f^{\prime \prime}(x) < 0 \ \forall x \in \mathbb{R} $

Consider $c = f^{\prime}(0)$ if $c =0$ then $\exists \alpha>0$ such that $f^{\prime}(\alpha) < 0$(WHy?)

Or $f(x) - f(\alpha) < f^{\prime}(\alpha) (x- \alpha) \ \forall x > \alpha$ And take $x \to \infty$ to contradict that $f(x) >0$

If $c >0$ then $f(0) - f(x)>-f^{\prime}(0)x \ \forall x <0$ again take $x \to - \infty$ and contradict that $f(x) >0$

Now I think you can proceed


问:
Proving a formula for $\sum_{j=1}^n\frac{x_j^k}{f'(x_j)}$ for $f$ an $n$-th degree polynomial with $n$ distinct real roots $x_j$
If polynomial $f(x)=a_0 x^n+a_1 x^{n-1}+\cdots+a_{n-1} x+a_n$ has $n$ different real roots
$x_1,x_2,\cdots,x_n$, prove the following fomular:
$$
\sum_{j=1}^n \frac{x_j^k}{f^{\prime}(x_j)}=\left\{\begin{array}{ll}
0, & 0 \leq k \leq n-2 \\
a_0^{-1}, & k=n-1
\end{array}\right.
$$

My answer for the first case is :

I use the following Identity
$$
\sum_{j=1}^{n}\left(x_{j}-x\right)^{k+1} l_{j}(x) \equiv 0,0 \leq k \leqslant n-2
$$
where $l_j(x)$ is Lagrangian Basis, more detail
$$
l_{j}(x)=\prod_{i=1 \atop i \neq j}^{n} \frac{\left(x-x_{i}\right)}{\left(x_{j}-x_{i}\right)}
$$
we know from the problem conditions that $f(x)$ can be writed as
$$f(x) = a_0(x-x_1)(x-x_2)\cdots(x-x_n)$$
then we can get the following identity
$$
\sum_{j=1}^{n}\left(x_{j}-x\right)^{k+1} \frac{f(x)}{\left(x-x_{j}\right) f^{\prime}\left(x_{j}\right)}\equiv 0
$$
Since $f(x) \not\equiv 0$, let $x = 0$ , then we get
$$\sum_{j=1}^n \frac{x_j^k}{f^{\prime}(x_j)} = 0 $$ for the $0\le k \le n-2$

Now I am stucking in the second case and don't know how to prove it, I think it
can be prooved from the following Identities
$$
\begin{array}{l}
\sum_{j=0}^{n} x_{j}^{k} l_{j}(x) \equiv x^{k}, \quad k=0,1, \cdots, n \\
\sum_{j=0}^{n}\left(x_{j}-x\right)^{k} l_{j}(x) \equiv 0, k=1,2, \cdots, n
\end{array}
$$
where $x_j$ are different points

答:
A very close approach is to use partial fraction decomposition. For any $k\geqslant 0$ we have $$\frac{x^{k+1}}{f(x)}=p_k(x)+\sum_{j=0}^n\frac{a_{k,j}}{x-x_j},$$ where $p_k(x)$ is a polynomial of degree $k+1-n$ if $k\geqslant n-1$, and $p_k(x)\equiv 0$ otherwise. And $$a_{k,j}=\lim_{x\to x_j}(x-x_j)\frac{x^{k+1}}{f(x)}=\frac{x_j^{k+1}}{f'(x_j)},$$ giving $$\frac{x^{k+1}}{f(x)}=p_k(x)+\sum_{j=1}^n\frac{x_j^{k+1}}{f'(x_j)}\frac{1}{x-x_j}.$$

Observe that $p_{n-1}(x)$ is a constant, easily seen to be $1/a_0$ after taking $x\to\infty$.

Taking $x\to 0$, we get what we need. One has to be careful if $k=0$ and $f(0)=0$ though. The LHS is then equal to $1/f'(0)$, and the RHS is missing exactly the term for which $x_j=0$.


问:
Existence of derivative at a point.
Assume that f : (a, b) → R is differentiable on (a, b) except possibly at c ∈ (a, b). Assume that lim$_{x→c}$ f '(x) exists. Prove that f '(c) exists and f ' is continuous at c.

I am getting a counter example for this statement.
Consider the function f(x) defined on (-1,1) by

f(x) = x , x ∈ (-1,0) U (0,1)

```<code>    5 , x = 0
</code>```

Here f '(x) is 1 for all x ∈ (-1,0) U (0,1).Thus lim$_{x→0}$ f '(x) exists. But we know f(x) is not continuous at 0. Thus , f'(0) doesn't exist.

I don't know where I have made a wrong assumption in the example. Please help.

答: You're right. For the statement to be true, you need to add the assumption that $f$ is continuous at $c$ (and then you can prove it using the mean value theorem, as has been done many times on this site, for example here).

Title: Existence of derivative at a point.


问:
Does the power rule come from the generalised binomial theorem or the other way around?
Remember how we say $lim \frac{\sin{x}}{x}=1$ doesn't come from the L Hospital's rule, because the differentiation of $sin x$ from first principles uses the fact that $lim \frac{\sin{x}}{x}=1$?

A similar situation is arising for the power rule and the generalised binomial theorem.

The generalised binomial theorem can be proven as the Taylor expansion of $(1+x)^r$. This derivation uses the power rule, as we need the power rule to calculate derivatives of $(1+x)^r$

The derivation of the power rule from first principles would need the generalised binomial theorem, as we'd need to expand the term $(x+h)^r$

So which of these two results is more fundamental? Which one comes from the other and why?

答:
I'm guessing:

$\lim_{h\rightarrow 0} \frac{(x+h)^r-x^r}{h}$

=$\lim_{h\rightarrow 0} \frac{e^{r\ln{x+h}}-x^r}{h}$

Now we use the L Hospital's rule

$=\lim_{h\rightarrow 0} e^{r\ln{x+h}} r \frac{1}{x+h}$

$= rx^{r-1}$

So the generalised power rule does not depend on the generalised binomial theorem.


问:
Introducing terms in the conclusion (natural deduction, propositional logic)
I am just wanting to wrap my head around the following question;

$$A\to B \vDash ((A \land B) \to C) \to (A\to C)$$

So far all I am getting from the conclusion is that $A\to C$, if I introduce the conclusion as an assumption, but I am not sure how to tie it all together to go from the premise to the conclusion.

to clarify - I understand it is a tautology and I can translate it into English, I am struggling to show this via a fitch-style proof.

答:
The statement
$$(A \land B) \Rightarrow ((A \land B \Rightarrow C) \Rightarrow (A \Rightarrow C))$$
is a tautology, as you can see from the following truth table:


问:
How to test the convergence of the series $\sum_{n=1}^\infty n^{-1-1/n}$?

How to test the convergence of the series $\displaystyle\sum_{n=1}^\infty\frac{1}{n^{1+1/n}}?$

Help me. I'm clueless.

答:
Here's another approach using a higher order ratio test. </p>

<p>Let $a_n = 1/n^{1+1/n}$.
Examining the ratio of successive terms for large $n$ we find
$$\frac{a_{n+1}}{a_{n}} = 1 - \frac{1}{n} + O\left(\frac{\log n}{n^2}\right).$$
Therefore, the series diverges by Gauss's test. </p>

Body:
Here's another approach using a higher order ratio test.

Let $a_n = 1/n^{1+1/n}$.
Examining the ratio of successive terms for large $n$ we find
$$\frac{a_{n+1}}{a_{n}} = 1 - \frac{1}{n} + O\left(\frac{\log n}{n^2}\right).$$
Therefore, the series diverges by Gauss's test.


Title: How to calculate the sum of a sequence of reciprocal of factorial

Body:
I wonder how to calculate the sum of the below sequence

\begin{align}
S & = \frac{1}{2!} - \frac{2}{3!} + \frac{3}{4!} - \frac{4}{5!} + \frac{5}{6!} - ... \\
& = \frac{2 - 1}{2!} - \frac{3 - 1}{3!} + \frac{4 - 1}{4!} - \frac{5 - 1}{5!} + ...\\
& = 1 - \frac{1}{2!} - \frac{1}{2!} + \frac{1}{3!} + \frac{1}{3!} - \frac{1}{4!} - \frac{1}{4!} + \frac{1}{5!}...\\
& = 1 - 2 * (\frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - \frac{1}{5!}...）\\
& = ？
\end{align}

There is a similar sequence, which is easy to solve

\begin{align}
S & = \frac{1}{2!} + \frac{2}{3!} + \frac{3}{4!} + \frac{4}{5!} + \frac{5}{6!} + ... \\
& = \frac{2 - 1}{2!} + \frac{3 - 1}{3!} + \frac{4 - 1}{4!} + \frac{5 - 1}{5!} + ...\\
& = 1 - \frac{1}{2!} + \frac{1}{2!} - \frac{1}{3!} + \frac{1}{3!} - \frac{1}{4!} + \frac{1}{4!} - \frac{1}{5!}...\\
& = 1 - \frac{1}{n!}
\end{align}

Body: $$\sum_{n!=2}^\infty (-1)^{n-1}\frac1{n!} = 1 - \frac1e.$$


问:
Prove that every plane triangulation has at least $n(n \ge 4)$ edges $e$ with the property that the only triangles containing $e$ are faces.
I am trying to prove the following theorem.

Theorem  Every maximal planar graph of order 4 or more is 3 connected.

I looked up similar questions(prove-that-every-maximal-planar-graph-of-order-4-or-more-is-3-connected) and found that Raoul provided a detailed answer details.pdf with the help of the following lemma.

Lemma Every plane triangulation with $n ≥ 4$ vertices has at least $n$ edges $e$ with the property that the only triangles containing $e$ are faces.

I did not understand one of the steps of the proof.

Thus, at most  $3$ edges with the desired property are lost in the
combining, leaving us with  $|V_1| + |V_2| − 3 = n$ such edges, as desired.

I feel that if all edges  on $t$   satisfy the property the only triangles containing them are faces
whether inside or outside.
Then it should become at least $|V_1| + |V_2| − 6$ in the combining not $|V_1| + |V_2| − 3$.

For example, in graph G below, we choose $C_3= uvw$, Any edge of the circle satisfies the condition in $C^{int}$ and $C^{out}$ but all lose the desired property in $G$.

If this lemma is incorrect, how should the theorem be proved?

In order to eliminate the confusion caused by the possible invalidation of the pdf file link, I re-copy the reference answer here.

Fix an embedding of the triangulation.

First, we prove that every plane triangulation with  $n ≥ 4$
vertices has at least $n$ edges $e$ with the property that the only
triangles containing $e$ are faces.  The proof is by induction on
$n$ . The base is case is $K_4$ and holds trivially because all  $6$
edges have this property.

For the induction, pick any triangle $t$ that is not a face. We split
the graph into two parts,  $G_1 = (V_1, E_1)$ and $G_2 = (V_2, E_2)$ ,
where  $G_1$ is the induced graph on  $t$ and all of the vertices
inside  $t$  in the embedding and $G_2$ is the induced graph on  $t$
and all of the vertices outside $t$ . Both graphs are plane
triangulations and both have fewer than n vertices. So, by induction,
there are $|V_1| + |V_2| = n + 3$ edges in the two graphs with the
desired property.  When we combine the graphs, the only edges that can
lose this property are the edges of $t$ because no edges pass between
any of the other vertices of  to any of the other vertices of $G_2$ .
Thus, at most  $3$ edges with the desired property are lost in the
combining, leaving us with  $|V_1| + |V_2| − 3 = n$ such edges, as
desired.

答:
I agree that the proof of the lemma is bad. The lemma might still be true, but I don't see how to prove it offhand.

I don't think we need the full power of the lemma. It's enough to show that:

Lemma. Every plane triangulation with $n \ge 4$ vertices has at least $6$ edges $e$ with the property that the only triangles containing $e$ are faces.

Proof. We do the same induction, except that $G_1$ and $G_2$ both contain $6$ of these edges. $3$ of those edges in $G_1$ might be edges of $t$ (so they don't count), and the same for $G_2$, but that still leaves $3$ edges in $G_1$ and $3$ in $G_2$ that do count, for a total of $6$. $\square$

We can carry out the proof that all plane triangulations with $n\ge 4$ vertices are $3$-connected, even if we only have the weaker version. All we need is that:

If $\{u,v\}$ is a cut of $G$, then we don't contract edge $uv$.

If we can partition $V(G)-\{u,v\}$ into $X \cup Y$ with no edges between $X$ and $Y$, then we shouldn't destroy the last vertex of $X$ or $Y$ (by contracting an edge incident to that vertex).

The first only excludes one edge at any moment. The second one doesn't constrain us at all unless $|X|=1$ or $|Y|=1$. But first of all, if $X = \{x\}$, that's at most two edges: $xu$ and $xv$ (and two more if $Y = \{y\}$). Second, I don't think that can actually happen, because at that point, the graph is pretty blatantly not a triangulation.

So having $6$ edges we can choose from to contract at any time is plenty to complete the proof.


问:
how should I calculate $\int_{0}^{\infty}\exp(-x-a/x)dx$
This question already has been solved here:
How to evaluate $\int_{0}^{\infty}\exp(-x^2-1/x^2)dx$?

And there is also an answer for How should I calculate $\int_0^\infty e^{-\frac{1}{2}(x^2+a^2/x^2)}\,dx$
But I think the answer is incorrect, if this problem solved, and I can try to replace x by u^2 in my question, even this would still be a little different because there would have another more 2u in the equation, but would it still be some help to solve my problem?

答:
The answer can be expressed in terms of Bessel functions, thus:
$$\fbox{$2 \sqrt{a} K_1\left(2
   \sqrt{a}\right)\text{ if }\Re(a)>0$},$$ since one of the standard integral formulas for Bessel K functions is

$$K_a(x) = \int_0^\infty \exp(-x \cosh t) \cosh a t \, dt$$ The substitution is fairly obvious.


答:
If you know that B follows from A, and C follows from A and B, then C follows merely from A. After all, A gets you B, and then both get you C.

An example, for context:

If it’s raining, then I carry an umbrella. Therefore, if it’s raining and I’m carrying an umbrella means that I’m dry, then if it’s raining, I’m dry.

You can deduce this  as follows:

$\def\fitch#1#2{~~~~\begin{array}{|l}#1\\\hline#2\end{array}}
{\fitch{1.\text{ Premise }A \to B}{\fitch{2.\text{ Assume }(A \wedge B) \to C)}{\fitch{3.\text{ Assume }A}{
4.\text{ Deduce }B\text{ from 1 and 3 }({\to}\mathsf E)\\
5.\text{ Deduce }A \wedge B\text{ from 3 and 4 }(\wedge\mathsf I)\\
6.\text{ Deduce }C\text{ from 2 and 5 }({\to}\mathsf E)}\\
7.\text{ Conclude }A \to C\text{ via 3 through 6 }({\to}\mathsf I)}\\
8.\text{ Conclude }((A \wedge B) \to C) \to (A \to C)\text{ via 2 through 7 }({\to}\mathsf I)}\\
\therefore\quad A\to B\vdash ((A\wedge B)\to C)\to(A\to C)}$

Title: Introducing terms in the conclusion (natural deduction, propositional logic)

Body:
I am just wanting to wrap my head around the following question;

$$A\to B \vDash ((A \land B) \to C) \to (A\to C)$$

So far all I am getting from the conclusion is that $A\to C$, if I introduce the conclusion as an assumption, but I am not sure how to tie it all together to go from the premise to the conclusion.

to clarify - I understand it is a tautology and I can translate it into English, I am struggling to show this via a fitch-style proof.


问:
Proving that free modules are flat (without appealing projective modules)
Suppose $R\neq 0$ is a commutative ring with $1$. Let $M$ be a free $R$-module. I would like to prove that $M$ is a flat $R$-module. Everywhere I have looked (mostly online) this is proved by first proving that every free module is projective, and then proving that every projective module is flat. Unfortunately, Atiyah & Macdonald's "Introduction to Commutative Algebra" (Chapter 2) does not discuss projective modules. But the result that every free module is flat comes very handy in the exercises.

So my question is,

Is it possible to prove that every free module is flat just by
  definitions and without appealing to projective modules?

Thanks!

答:
You can just show that the functor $M\otimes_R(-)$ is left-exact. Write $M=\bigoplus_{i\in I} R$ and let
$$0\to N'\xrightarrow{\quad\iota\quad} N\to N''\to 0$$
be an exact sequence of $R$-modules. Note that $M\otimes N=\bigoplus_{i\in I} R\otimes_R N = \bigoplus_{i\in I} N$, so the sequence

$$0 \to M\otimes N' \xrightarrow{\quad\mathrm{id}\otimes\iota\quad} M\otimes N \to  M\otimes N'' \to 0$$

is the same as

$$0 \to \bigoplus_{i\in I} N' \xrightarrow{\textstyle\quad\bigoplus_{i\in I} \iota\quad} \bigoplus_{i\in I}N \to  \bigoplus_{i\in I}N'' \to 0$$

and the morphism $\bigoplus_{i\in I} \iota$ is clearly injective if $\iota$ is injective.


问:
Make the denominator rational
I am doing the homework of a recorded video in a course, and I came across this question which I am not able to solve:

Simplify the denominator:
$\dfrac{12}{3+\sqrt5-2\sqrt2}$

And I am not able to understand the solution:

Thanks in advance.

答:
What we are doing is a process called "rationalizing" the denominator. The process relies on the property that

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

So when you have a fraction as follows

$$F = \frac{12}{3+\sqrt5 - 2\sqrt2}$$

Let $a = 3+\sqrt5$, $b = 2\sqrt2$

$$F = \frac{12}{a-b}$$

Now multiplying numerator and denominator by $a+b$, we have

$$F = \frac{12(a+b)}{(a-b)(a+b)} = \frac{12(3+\sqrt5 + 2\sqrt2)}{(3+\sqrt5)^2-(2\sqrt2)^2} = \frac{12(3+\sqrt5+2\sqrt2)}{6(1+\sqrt5) }$$

$$\implies F =\frac{2(3+\sqrt5+2\sqrt2)}{1+\sqrt5} $$

Now you can repeat the process to simplify it further

Title: Make the denominator rational


答:
Since $f(x)f''(x)\ne 0$, it follows that for all real $x, f''(x)\gt 0$ or $f''(x)\lt 0$ for all real $x$.

Because if there exist $x_1,x_2$ such that $f''(x_1)\gt 0$ and $f''(x_2)\lt 0$ then by Darboux theorem, there exists $c\in (x_1, x_2)$ such that $f''(c)=0$, which is a contradiction. (Why? Refer Note)

Similarly $f(x)\gt 0$ for all real $x$ or $f(x)\lt 0$ for all real $x$.

WLOG suppose that $f"(x)\gt 0$ for all $x$ and $f(x)\lt 0$ for all $x$, it follows that $f(x)f"(x)\lt 0$ for all $x$.  There exist $x'$ such that $f'(x')\ne 0$.

Case 1: $f'(x')\lt 0$

By LMVT,
For $x\lt x', f'(x)\lt f'(x')$

So for $x\lt x', f(x)-f(x')\gt xf'(x')\implies f(x)\gt f(x')+xf'(x')\implies f(x)$ will be positive when $x$ is negative enough. This will contradict our assumption that $f(x)\lt 0$

Case 2: $f'(x')\gt 0$

Can you take it from here?

Note: Define $g(x)=f'(x)$. Since $g$ is continuous on $[x_1,x_2]$, it must attain maximum on the interval.Let $c\in [x_1,x_2]$ be the point where maximum is attained. If $c=x_1$, then $g'(c)=f''(x_1)=\lim_{x\to x_1^+}\frac {f'(x)-f'(x_1)}{x-x_1}\le 0$, contradiction (Note that since $f'(x_1)$ is maximum, Numerator is not positive) and similarly $c\ne x_2$. So $c\in (x_1,x_2)$. Since $c$ is a point of extrema of $g$ on $[x_1,x_2]$, we must have $g'(c)=f''(c)=0$.

Title: Is this intuitive by graph ? Or its needs rigorous proof

Body: A double differentiable function $f : \mathbb{R} \to \mathbb{R}$ satisfy $f(x)f^{\prime \prime}(x) \ne 0 \ \forall x \in \mathbb{R}$. Can we conclude $f(x)f^{\prime \prime}(x) >0$ for all $\mathbb{R}$ , even though continuity of $f''(x)$ is not given. I thought of making a contradiction by graph supposing $f(x)>0$ and $f''(x) <0$ , I thought graph would need to cut x axis hence after that $f(x)<0$ ( contradiction) but I don't have a good calculus based proof on this fact. As continuity of $f''(x)$ not given is also a problem for all $\mathbb{R}$


问:
Evaluating $\lim_{n \to \infty}\prod_{r=1}^{n}[1+(r/n)^2]^{1/r}$ by expressing it as a definite integral
I have been asked to find the limit of the following series by expressing it as definite integral:

If $na=1$ always and $n$ tends to infinity, find the limiting value
$$\lim_{n \to \infty}\left(\prod_{r=1}^{n}[1+(ra)^2]^{1/r}\right)$$

The answer given is $\exp\left(\dfrac{\pi^2}{24}\right)$.

I took the log of both sides and set up my integral as $$\int_{0}^{1}\log(1+x^2)dx$$

But my answer came out wrong.

Thanks in advance

答:
Since $ na = 1$ we can write the product as $\displaystyle \mathcal{P} := \prod_{r=1}^\infty\left(1+\frac{r^2}{n^2}\right)^{1/r}.$

Then taking logs and using the continuity of logs

\begin{align} \log(\mathcal P) & =
\log\left(\lim_{n\to\infty}\prod_{r=1}^n\left(1+\frac{r^2}{n^2}\right)^{1/r}\right)
\\&=\lim_{n\to\infty}\log\left(\prod_{r=1}^n\left(1+\frac {r^2}{ n^2}\right)^{1/r}\right)\\
&=\lim_{n\to\infty}\sum_{r=1}^n\frac1r\log\left(1+ \frac {r^2}{ n^2}\right)\\
&=\lim_{n\to\infty}\frac{1}{n}\sum_{r=1}^n\frac{n}{r}  \log\left(1+ \frac {r^2}{ n^2}\right)\\
&=\int_0^1\frac1x\log(1+x^2)\,\mathrm{d}x\\
&=\int_0^1\sum_{k=0}^\infty(-1)^k\frac{x^{2k+1}}{k+1}\,\mathrm{d}x\\
&=\sum_{k=0}^\infty\frac{(-1)^k}{(k+1)(2k+2)} \\& = \frac{1}{2}\sum_{k=0}^\infty\frac{(-1)^k}{(k+1)^2}\\& = \frac{1}{4}\sum_{k=0}^\infty\frac{1}{(k+1)^2}
\\&=\frac{\pi^2}{24}.
\end{align}

Therefore $\displaystyle \mathcal P = \exp\left(\frac{\pi^2}{24}\right).$


问:
Sized Biased Picking Distribution
I am having trouble understanding the following proof on sized biased picking. We have the following situation:

Let $ X_1, \cdots , X_n $ be i.i.d. and positive, and $S_i = X_1 + \cdots + X_i$ for $ 1 \leq i \leq n $. The values of $S_i/S_n$ are used to partition the interval $[0,1]$, each sub-interval has size $Y_i = X_i/S_n$. Suppose $U$ is an independent uniform r.v. on $(0,1)$, and let $\hat{Y}$ denote the length of the sub-interval containing $U$. We aim to calculate the distribution of $\hat{Y}$.

The claimed result is $ \mathbb{P}(\hat{Y} \in dy) = n y \, \mathbb{P}(Y \in dy) $, where the notation means $\mathbb{P}(\hat{Y} \in A) = \int_A n y \mu(y) \, dy$ with $\mu$ the law of $Y$ (I think perhaps they mean the density function $f_Y(y)$).

The proof given is
\begin{align*}
\mathbb{P}(\hat{Y} \in dy) &  = \sum_{i=1}^n \mathbb{P}\left(\hat{Y} \in dy, \frac{S_{i-1}}{s_n} \leq U < \frac{S_i}{S_n}\right) \\
& = \sum_{i=1}^n \mathbb{P}\left(\frac{X_i}{S_n} \in dy, \frac{S_{i-1}}{S_n} \leq U < \frac{S_i}{S_n}\right) \\
&  =  \sum_{i=1}^n \mathbb{E}\left[\frac{X_i}{S_n} \, 1\left(\frac{X_i}{S_n} \in dy\right)\right] \tag{$\ast $} \\
& = \sum_{i=1}^n y \, \mathbb{P}\left(\frac{X_i}{X_n} \in dy \right) \tag{$\ast  \ast $} \\
& =ny \, \mathbb{P}(Y \in dy) \\
\end{align*}

I do not understand the equalities $(\ast)$ and $(\ast \ast)$, and do not fully understand the notation given in the proof.

答:
Eh, that notation looks like quite a bit of a scam; I assume they wanted to write densities, since that probability is certainly $0$ for most continuous distributions. If you see a question like this, you should $100\%$ do it the way @Misha has done in their answer. If you want to manipulate densities properly, you will want to invoke Radon-Nikodym theorem at some point, but since all you want to know what those stars are, here is their semi-formal "justification":

For $(*)$, the informal justification is that the probability that $U$ is in the $i$th interval is $\frac{X_i}{S_n}$ so it comes out of that expectation because of independence of $U$. Semi-formally, we first rewrite:

\begin{align*}
&\mathbb{P}\left(\frac{X_i}{S_n} \in dy, \frac{S_{i-1}}{S_n} \leq U < \frac{S_i}{S_n}\right) \\
&\stackrel?= \lim_{\epsilon \rightarrow 0} \frac{1}{2\epsilon}\mathbb{P}\left(\frac{X_i}{S_n} \in [y-\epsilon, y + \epsilon], \frac{S_{i-1}}{S_n} \leq U < \frac{S_i}{S_n}\right) \\
&=\lim_{\epsilon \rightarrow 0} \frac{1}{2\epsilon}\mathbb{E}\left[1\left(\frac{X_i}{S_n} \in [y-\epsilon, y + \epsilon], \frac{S_{i-1}}{S_n} \leq U < \frac{S_i}{S_n}\right)\right] \\
&=\lim_{\epsilon \rightarrow 0} \frac{1}{2\epsilon}\mathbb{E}\left[1\left(\frac{X_i}{S_n} \in [y-\epsilon, y + \epsilon]\right) 1\left(\frac{S_{i-1}}{S_n} \leq U < \frac{S_i}{S_n}\right)\right]\\
&=\lim_{\epsilon \rightarrow 0} \frac{1}{2\epsilon}\mathbb{E}\left[\mathbb{E}\left[1\left(\frac{X_i}{S_n} \in [y-\epsilon, y + \epsilon]\right) 1\left(\frac{S_{i-1}}{S_n} \leq U < \frac{S_i}{S_n}\right)\bigg|X, S\right]\right]\\
&=\lim_{\epsilon \rightarrow 0} \frac{1}{2\epsilon}\mathbb{E}\left[1\left(\frac{X_i}{S_n} \in [y-\epsilon, y + \epsilon]\right) \mathbb{E}\left[ 1\left(\frac{S_{i-1}}{S_n} \leq U < \frac{S_i}{S_n}\right)\bigg|X, S\right]\right]\\
&=\lim_{\epsilon \rightarrow 0} \frac{1}{2\epsilon}\mathbb{E}\left[1\left(\frac{X_i}{S_n} \in [y-\epsilon, y + \epsilon]\right) \frac{X_i}{S_n}\right]\\
&=\mathbb{E}\left[\frac{X_i}{S_n} \, 1\left(\frac{X_i}{S_n} \in dy\right)\right]
\end{align*}

Now, for $(**)$, informally, that indicator is only $1$ if that ratio is exactly $y$, so that random variable is either $y$ with probability equal to that ratio being equal to $y$, or $0$ otherwise. Semi-formally, we can write:

\begin{align*}
\mathbb{E}\left[\frac{X_i}{S_n} \, 1\left(\frac{X_i}{S_n} \in dy\right)\right] &= \lim_{\epsilon \rightarrow 0} \frac{1}{2\epsilon}\mathbb{E}\left[1\left(\frac{X_i}{S_n} \in [y-\epsilon, y + \epsilon]\right) \frac{X_i}{S_n}\right]
\end{align*}

Note that

$$
\begin{align*}
&(y-\epsilon)\mathbb{E}\left[1\left(\frac{X_i}{S_n} \in [y-\epsilon, y + \epsilon]\right)\right] \\
&\leq \mathbb{E}\left[1\left(\frac{X_i}{S_n} \in [y-\epsilon, y + \epsilon]\right) \frac{X_i}{S_n}\right] \\
&\leq (y + \epsilon)\mathbb{E}\left[1\left(\frac{X_i}{S_n} \in [y-\epsilon, y + \epsilon]\right)\right]
\end{align*}$$

and by plugging it into above and taking the limit as $\epsilon \rightarrow 0$, you can show that that equality holds.

Title: Sized Biased Picking Distribution


问:
Permutation and Combinations with conditions
Hallo :) This is a question about permutations but with conditions.

2 boys and 4 girls are to be arranged in a straight line.
In how many ways can this be done if the two boys must be separated? (The order matters)

Thank You.

答:
Total number of ways of arranging the people = 6!

Cases when boys are together:

(2B) G G G G

G (2B) G G G

G G (2B) G G

G G G (2B) G

G G G G (2B)

Each of the above combinations can be arranged in 2 * 4! ways.
(The factor of 2 is accommodated since the boys themselves could be interchanged, as they are on a straight line.)

Number of ways to separate the boys = (6! - (2 * 5 * 4!)) = 5! * (6 - 2) = 480


问:
Evaluating orbits of symmetric groups
$$G = S_4\quad\text{and}\quad a = (1234)(34)^3$$

The actual question is to find the order of $a$ and to then find [$G:\lt a \gt]$. However the part I am confused about is how my professor evaluated $(1234)(34)^3$.

His working:

$$(1234)(34)^3 = (123)^3 = 1$$

I understand the last step. What I don't understand is why the $4$ disappeared.

Here is my theory:

$$(1234)(34)^3 = (123)(4)^3$$

because the way $a$ is defined, $4$ ends up just going to $4$ repeatedly. We can simplify by removing orbits of $1$ element ie. we remove $(4)$ to get $(123)^3$ as seen. Is this correct?

Title: Evaluating orbits of symmetric groups

Body:
It's because $(1\ \ 2\ \ 3\ \ 4)(3\ \ 4)=(1\ \ 2\ \ 3)$. That's so because:

$(3\ \ 4)$ maps $1$ into $1$, and $(1\ \ 2\ \ 3\ \ 4)$ maps $1$ into $2$;

$(3\ \ 4)$ maps $2$ into $2$, and $(1\ \ 2\ \ 3\ \ 4)$ maps $2$ into $3$;

$(3\ \ 4)$ maps $3$ into $4$, and $(1\ \ 2\ \ 3\ \ 4)$ maps $4$ into $1$;

$(3\ \ 4)$ maps $4$ into $3$, and $(1\ \ 2\ \ 3\ \ 4)$ maps $3$ into $4$.


问:
Symmetric Group $S_n$ and Artin Braid Group $B_n$: Differences in definition of composition?
I have been starting to look more closely at braid groups recently and I am slightly confused on the notion of "composition" in $S_n$ versus $B_n$. Function composition in $S_n$ is the usual, where $f\circ g$ means to feed $g$ into $f$. Like for cycles in $S_5$:
$$(1 \; 2\; 3)\circ(3\; 5)$$

However, for $B_n$, it seems that "braid composition" means to feed the first braid into the second as the order would imply (say we're working in $B_4$):
$$\sigma_1\circ \sigma_2$$

Is this what is actually going on? Or do I understand this exactly backwards? I have read that if we forget braid twists, there is a surjective group homomorphism $B_n\to S_n$. How do we conserve the different notions of composition (if such a difference actually exists) in this homomorphism? Are we forced to reverse the order of products?

答:
In both $S_n$ and $B_n$, the group operation is essentially composition of functions, and there's no inherent reason $f \circ g$ has to mean "$g$, then $f$" as opposed to "$f$, then $g$". Either convention is fine if it is done consistently.

When reading, you have to pay attention to the author's conventions, but it is not part of the inherent definition of the groups.

When writing, if working with both $S_n$ and $B_n$ in the same text, it would be less confusing to the reader to choose conventions such that the order of composition does not need to be reversed to obtain the homomorphism you indicated in the original post.


问:
Cauchy in product space

Here is the theorem for the completely metrizibility of Product spaces. For that each $(X_{i},\rho_{i})$ to be completely metrizable, The author said that for any cauchy sequence taken in $(X_{j},\rho_{j})$, $\{p_{j}^{* -1}(x_{i})\}$ is Cauchy in $X_{j},\rho_{j}$, Can we say that? I think being Cauchy is not a topological property? Or here is there any relation between that $\{p_{j}^{* -1}(x_{i})\}$  is Cauchy and our metrics are bounded by 1? or any easy way for the proof of each $(X_{i},\rho_{i})$ is completely metrizable? (metric on the product is $\rho(x,y)=\rho((x_{i}),(y_{i}))=\displaystyle \sum_{i=1}^{\infty}\frac{1}{2^{i}}\rho_{i}(x_{i},y{i})$

答: The metrics on $X_j^\ast$ and $X_j$ are related (it's not merely the fact that they are homeomorphic): The former is the restricted $\rho$ (and all points in coordinates except $j$ are constant (the $A_i$) so $\rho(p_j^{-1}(x),p_j^{-1}(x')) = \frac{1}{2^j} \rho_j(x,x')$ for all $x,x' \in X_j$, is what the metric on $X_j^\ast$ is: just differing by a constant with $(X_j, \rho_j)$, so a Cauchy sequence in one is also one in the other. (It's also a quick extra argument why the restricted $p_j$ is a homeomorphism: it's an isometry-up-to-a-constant, a "scaling".)

Title: Cauchy in product space


问:
Can I find the equation of an ellipse with these points?
How can I solve an ellipse with its major axis on the $x$-axis, given one focus, and two points on the ellipse, one of which I know to be on the major axis?</p>

答:
Let $F$, $V$, and $P$ be the given focus, the given vertex (point on the major axis), and the other point. We'll assume that $V$ is the "closer" vertex to $F$ (that is, $|VF| = a-c$, where $a$ is the semi-major axis, and $c > 0$ is the focal radius; otherwise, $|VF|=a+c$, and the argument proceeds similarly).

The (fraction-free) standard form of the ellipse is
$$b^2 ( x - h )^2 + a^2 ( y - k )^2 = a^2 b^2$$
where $(h,k)$ is the center, $a>0$ is the semi-major axis, and $b>0$ is the semi-minor axis. With the major axis given to be on the $x$-axis, we have $k=0$. Let us assume that the focus $F$ is positioned at the origin, and that the ellipse's center is to the left of that focus, so that $h = -c < 0$. The equation becomes
 $$b^2 ( x + c )^2 + a^2 y^2 = a^2 b^2$$
We therefore need to determine the three quantities $a$, $b$, $c$.

Let us position focus $F$ at the origin, and assign coordinates to the other points thusly: $V(v,0)$ and $P(p,q)$ (with $v > 0$ and $v \ge p$). We have three equations:
$$\begin{align}
v &= a - c &(1) \\
a^2 &= b^2 + c^2 &(2) \\
b^2 ( p + c )^2 + a^2 q^2 &= a^2 b^2 &(3)
\end{align}$$
We use $(1)$ to eliminate $c$ from $(2)$ and $(3)$:
$$\begin{align}
b^2 = a^2 - c^2 = ( a - c )( a + c ) &= v ( 2a - v ) &(4) \\
b^2(p+a-v)^2+a^2q^2 &= a^2 b^2 &(5)
\end{align}$$
... and then use $(4)$ to eliminate $b^2$ from $(5)$:
$$v(2a-v)(p+a-v)^2+a^2q^2 = a^2 v(2a-v)$$
whence
$$a^2 \left(p^2 + q^2- (2v-p)^2\right) + 2 a v\left(v-p\right)\left(2v-p\right) - v^2 \left(v-p\right)^2 = 0$$
so that
$$a = \frac{v \left(v-p\right) \left(-\left( 2v - p \right) \pm \sqrt{p^2+q^2}\right)}{p^2 + q^2- (2v-p)^2} = \frac{v\left(v-p\right)}{2v-p \pm \sqrt{p^2+q^2}}$$
with the "$\pm$" chosen to make $a > v > 0$, and
$$b = \sqrt{\;v(2a-v)\;} \qquad\qquad c = a - v$$
(You should double-check my algebra.)


问:
convexity of function having set domain
Suppose we have a finite set $G \subset \mathbb{R}$. And it has $n$ elements. $|G| = n$

And we have a function $f: P(G) \to \mathbb{R}$, where $P(G)$ is the powerset of $G$.

I'm wondering how can I approach the proof of convexity or not to this function $f$?

For a real function $g: \mathbb{R} \to \mathbb{R}$, i can check the second derivative is positive.

Also if it helps, the explicit form of my function is $f(S):= u(S)-l(S)$.

Here $u$ is the sum of the upper half of $S$, and $l$ is the sum of the lower half of $S$.

If $S$ has an odd number of elements, its median is skipped.

To reuse the standard definition of convexity, I can use $\{0, 1\}^n$ vectors as input domain, and apply some sophisticated operation to achieve the same output $f': \{0, 1\}^n \to \mathbb{R}$. would this $f'$ be convex?

答:
I don't have enough reputation to add comments, but feel free to move this to a comment or delete it if it is not useful. Why don't you define convexity using
Minkowski sums?. Then you can say $f$ is convex if
$$f(\theta X\oplus (1-\theta)Y)\leq \theta f(X)+(1-\theta)f(Y)$$
holds for all $X,Y\in P(G)$ and all $\theta\in[0,1]$. Strict convexity could be defined for the above relation with strict inequality and $\theta\in(0,1)$ and all $X\neq Y$. Here, $\oplus$ is the Minkowski sum
$X\oplus Y = \{ x+y | x\in X, y\in Y\}$. Then you could use the usual convexity machinery just adapted for Minkowski sums.

Title: convexity of function having set domain


问:
Number of valuation ring of a given field
Let $K$ be a field. I'd like to know the number of valuation ring of $K$.

My conjecture;   The number of valuation ring of $K$ is $1$ or $2$ or
$∞$.

Let $K$ be $\mathbb{Q}$, then $\mathbb{Z}_{(p)}$ is a valuation ring of $\mathbb{Q}$, so there are infinitely many valuation ring according to the number of prime $p$.

Let $K$ be a local field, then valuation ring is just $K$ and its integer ring, so in this case the number is $2$.

Let $K$ be a finite field,I believe the number is $1$ because $K$ itself is the only valuation ring of finite field.

I couldn't find a field $K$ whose number of valuation ring is natural number more than $3$.

I know there is a $1$ to $1$ correspondence between the places and valuation rings of a field.
So I can confirm above claims from the view point of $places$, but I still cannot find field $K$ which contains more than $3$ valuation rings.

Thank you in advance.

答:
For a non-negative integer $a$ let

$$K=\Bbb{Q}(B), \qquad B=\{ (1+ma)^{1/n}, n\ge 1,\gcd(n,a)=1,m\ge 0\}$$

If $K=Frac(R)$ with $R$ a DVR then the valuation $v$ on $R$ extends the $p$-adic valuation for some $p$.

If $p\nmid a$ then $p$ divides some $1+ma$ and $v((1+ma)^{1/n})=v(1+ma)/n$ gives that $v(K)=\Bbb{Q}$ ie. $v$ is not discrete.

Whence $p\ |\ a$.

With $O_K$ the algebraic integers $\subset K$ then $O_K$ has a lot of prime ideals above $p$, and since $K$ is unramified at $p$ we get that any valuation above $p$ is discrete, ie. infinitely many choices for $v$.

To get only one choice for $v$ we need to add more elements to $B$:

For each $b\in B$, $f\in \Bbb{Z}[x]$,$n\ge 1,\gcd(n,a)=1$: add $(1+(b-1)f(b))^{1/n}$ to $B$, and then repeat iteratively, choosing $b$ also in the newly added elements.

This time only one valuation above $p$ is discrete: the one such that $v(b-1)>0$ for all $b\in B$, which gives a natural embedding $K\to \Bbb{Q}_p$.

The other valuations are not discrete because if $v(b-1)=0$ then for some $f\in \Bbb{Z}[x]$ we'll have $v(1+(b-1)f(b))>0$ and when adding the $n$-th roots of $1+(b-1)f(b)$ we'll get a non-discrete valuation.


问:
Will there be a proper subset B(x,d) of B(x,D) in a metric space?
I am working on this:

If $(X,\rho)$ is a non-discrete metric space, $x\in X$ is not an isolated point. Assume $U$ is an open set which contains $x$.
It's obvious that $\exists B(x,D) \subset U.$

I suppose that there must be a proper subset $B(x,d) \subsetneqq B(x,D)$

Here is my proof:

First of all, U can't be finite. If so, assume $U=\{x,x_1,x_2,...,x_n\}$, then take $0<\varepsilon<min \{\rho(x,x_1),...,\rho(x,x_n)\}$.

Then take $B(x,\varepsilon)$. By defination of $\varepsilon$, $B(x,\varepsilon)=\{x\}$. However, $x$ is not an isolated point. So U is not finite.

Then,take $y\ne x \in B(x,D)$, and $\frac{\rho(x,y)}{2}$ as d, then  $B(x,d) \subsetneqq B(x,D)$.

My tutor said there is something wrong with my proof. But I can't see where's the problem.

If there really something wrong? If so ,where is the mistake?

答:
We know that $B(x,D) \neq \{x\}$ because we assumed $x$ is not an isolated point. So there is some $y \neq x$ such that $y \in B(x,D)$.
But then, setting $d=\rho(x,y)>0$, we know $B(x,d) \subsetneqq B(x,D)$: the inclusion part follows from $z \in B(x, d) \to \rho(x,z) < d = \rho(x,y) < D$ and the $\neq$-part follows from $y \notin B(x,d)$ (as $\rho(x,y) = d \not< d$). QED. No infinite or not distinction is needed (though it is true that any ball around a non-isolated point must be infinite in a metric space; this essentially uses your first argument).


问:
Prove $\pi=\lim_{n\to\infty}2^{4n}\frac{\Gamma ^4(n+3/4)}{\Gamma ^2(2n+1)}$
How could it be proved that
$$\pi=\lim_{n\to\infty}2^{4n}\frac{\Gamma ^4(n+3/4)}{\Gamma ^2(2n+1)}?$$

What I tried

Let
$$L=\lim_{n\to\infty}2^{4n}\frac{\Gamma ^4(n+3/4)}{\Gamma ^2(2n+1)}.$$
Unwinding $\Gamma (n+3/4)$ into a product gives
$$\Gamma \left(n+\frac{3}{4}\right)=\Gamma\left(\frac{3}{4}\right)\prod_{k=0}^{n-1}\left(k+\frac{3}{4}\right).$$
Then
$$\lim_{n\to\infty}\frac{(2n)!}{4^n}\prod_{k=0}^{n-1}\frac{16}{(3+4k)^2}=\frac{\Gamma ^2(3/4)}{\sqrt{L}}.$$
Since
$$\frac{(2n)!}{4^n}\prod_{k=0}^{n-1}\frac{16}{(3+4k)^2}=\prod_{k=1}^n \frac{4k(4k-2)}{(4k-1)^2}$$
for all $n\in\mathbb{N}$, it follows that
$$\prod_{k=1}^\infty \frac{4k(4k-2)}{(4k-1)^2}=\frac{\Gamma ^2(3/4)}{\sqrt{L}}.$$
But note that this actually gives an interesting Wallis-like product:
$$\frac{2\cdot 4\cdot 6\cdot 8\cdot 10\cdot 12\cdots}{3\cdot 3\cdot 7\cdot 7\cdot 11\cdot 11\cdots}=\frac{\Gamma ^2(3/4)}{\sqrt{L}}.$$

I'm stuck at the Wallis-like product, though.

答:
$$a_n=2^{4 n}\frac{ \Gamma^4 \left(n+\frac{3}{4}\right)}{\Gamma^2 (2 n+1)
}$$
$$\log(a_n)=4 n \log (2)+4 \log \left(\Gamma \left(n+\frac{3}{4}\right)\right)-2 \log (\Gamma  (2 n+1))$$
Applying Stirling approximation twice and continuing with Taylor series
$$\log(a_n)=\log (\pi )-\frac{1}{8 n}+\frac{1}{32 n^2}+O\left(\frac{1}{n^3}\right)$$
$$a_n=e^{\log(a_n)}=\pi \left(1-\frac{1}{8 n}+\frac{5}{128 n^2} \right)+O\left(\frac{1}{n^3}\right)$$


问:
Rationalising term
What will be the rationalizing term for $30\sqrt 2 + 24 + 20 \sqrt 6$? It cannot be $30 \sqrt 2 - 20\sqrt 6$. Neither can it be $30 \sqrt 2+ 20 \sqrt 6 - 24$ because the product of these with the denominator will again contain a square root.
Please help. Thank you.

答:
Take $(30\sqrt2+24+20\sqrt6)\times(a+b\sqrt2+c\sqrt3+d\sqrt6)$
$=30a\sqrt2+60b+30c\sqrt6+60d\sqrt3+24a+24b\sqrt2+24c\sqrt3+24d\sqrt6+20a\sqrt6+40b\sqrt3+60c\sqrt2+120d$
$=(60b+24a+120d)+(30a+24b+60c)\sqrt2+(60d+24c+40b)\sqrt3+(30c+24d+20a)\sqrt6$

From this, we get
$$5a+4b+10c=0$$
$$15d+6c+10b=0$$
$$15c+12d+10a=0$$
Since we have $4$ variables and $3$ equations, we need to assume some value. Say $a=1$
$$b={{245}\over{604}}, c={{-100}\over{151}}, b={{-5}\over{906}}$$
Since multiplying won't affect rationalisation, let us multiply all 4 variables by  $1812$.
$$a=1812, b=735, c=-1200, d=-10$$
Therefore, your rationalising factor is $(1812+735\sqrt2-1200\sqrt3-10\sqrt6)$

Title: Rationalising term


答:
Assume our free module $M$ is of the form $\oplus_{i\in S}R_{i}$, then the statement just follows. The details may be best to leave out to you. </p>

<p>For your question on the projective module, you can consider the projective module as a direct summand of a free module and "piece" up the maps together. </p>

Title: Proving that free modules are flat (without appealing projective modules)

Body:
Suppose $R\neq 0$ is a commutative ring with $1$. Let $M$ be a free $R$-module. I would like to prove that $M$ is a flat $R$-module. Everywhere I have looked (mostly online) this is proved by first proving that every free module is projective, and then proving that every projective module is flat. Unfortunately, Atiyah & Macdonald's "Introduction to Commutative Algebra" (Chapter 2) does not discuss projective modules. But the result that every free module is flat comes very handy in the exercises.

So my question is,

Is it possible to prove that every free module is flat just by
  definitions and without appealing to projective modules?

Thanks!

Body:
Assume our free module $M$ is of the form $\oplus_{i\in S}R_{i}$, then the statement just follows. The details may be best to leave out to you.

For your question on the projective module, you can consider the projective module as a direct summand of a free module and "piece" up the maps together.


问:
show that $P(X\cap Y)\cup P(X\cap Z)=P(X\cap Y)+P(X\cap Z)$ if $X$ and $Y$ are independent,$X$ and $Z$ are independent, and $P(Y\cap Z) = 0$

Show that $P(X\cap Y)\cup P(X\cap Z)=P(X\cap Y)+P(X\cap Z)$   if $X$ and $Y$ are independent, $X$ and $Z$ are independent, and $P(Y\cap Z) = 0$

This makes sense intuitively if one draws a venn diagram. But how can one justify that $X\cap Y$ and $X\cap Z$ are mutually exclusive with probability rules ?

答:
Part 1. For the venn diagram, what you're doing is thinking of disjoint. Instead, think of almost disjoint.

$Y$ and $Z$ aren't disjoint but they're almost disjoint (defined as that probability of their intersection is zero). In this case, we're not sure that $X \cap Y$ and $X \cap Z$ are disjoint but we're sure that they're also almost disjoint. Then $P((X \cap Y) \cup (X \cap Z)) = P(X \cap Y) + P(X \cap Z)$.

Remark: Here, I do not use any independence assumption, I think.

Part 2. Oh you said in comments you wanted to show $X$ and $Y \cup Z$ are independent. Well you can use $\sigma$-algebras or...(wait I think you actually figured out the following but anyway)

$P(X \cap Y) = P(X)P(Y)$

$P(X \cap Z) = P(X)P(Z)$

Then

$P(X \cap (Y \cup Z)) = P(X)(P(Y)+P(Z))$

Again use almost disjoint to say

$= P(X) P(Y \cup Z)$


问:
How to solve for matrix $X$ in equation?
There are two matrices $$A = \begin{bmatrix}3&2\\1&4\end{bmatrix}$$ and $$B = \begin{bmatrix}2&1&-4\\0&-2&2\end{bmatrix}$$
I have to solve for matrix $X$ in this equation $$6X + B^T = XA$$
I tried getting rid of $A$ through multiplying both sides by the inverse of $A$, but I can't get any further than that. What is it that I'm missing?

答: Note that$$6X+B^T=XA\iff B^T=X(A-6\operatorname{Id}_2).$$So, take$$X=B^T(A-6\operatorname{Id}_2)^{-1}=\begin{bmatrix}-1 & -1 \\ 0 & 1 \\ \frac{3}{2} & \frac{1}{2}\end{bmatrix}.$$


问:
Existence of a descent direction
Let $d,k\in\mathbb N$ and $f:\mathbb R^d\to\mathbb R^k$. $x^\ast\in\mathbb R^d$ is called

Pareto critical (for $f$) if there is no $x\in\mathbb R^d$ with $f(x)<f(x^\ast)$ (component-wisely);

locally Pareto critical if there is a neighborhood $N$ of $x^\ast$ s.t. $x^\ast$ is Pareto critical for $\left.f\right|_N$.

Moreover, $v\in\mathbb R^d$ is called descent direction at $x\in\mathbb R^d$ if ${\rm D}f(x)v<0$ (component-wisely).

Question 1: How can we show that $x^\ast\in\mathbb R^d$ is locally Pareto critical if and only if there is no descent direction at $x^\ast$?

The "$\Rightarrow$" direction is easy to show, but I'm struggle to obtain "$\Leftarrow$", while the intuition is clear: Assuming the contrary, we could infer that $x^\ast$ is not Pareto critical for $\left.f\right|_{N_n}$, where $N_n:=B_{\frac1n}(x^\ast)$, for all $n\in\mathbb N$. Hence, there would be a $x_n\in N_n$ with $f(x_n)<f(x^\ast)$ for all $n\in\mathbb N$. We could now define $v_n:=x_n-x^\ast$ for $n\in\mathbb N$ and we should be able to find a $n_0\in\mathbb N$ such that ${\rm D}f(x^\ast)v_{n_0}<0$. But how do we conclude the existence of a descent direction rigorously?

Question 2: If $k=1$, then it's obvious that $x^\ast$ is (local) Pareto critical if and only if $x^\ast$ is a (local) minimum.

But now I've read that $x^\ast$ is locally Pareto critical if and only if $\nabla f(x^\ast)=0$. Why is that the case?

We clearly know that if $x^\ast$ is locally Pareto critical (hence a local minimum), then $\nabla f(x^\ast)=0$. But why should the other direction hold as well?

Please note that we can generally show that $x^\ast$ is Pareto critical if and only if there is a $\lambda\in[0,\infty)^k$ with $\langle\lambda,1\rangle=1$ and ${{\rm D}f(x^\ast)}^\ast\lambda=0$.

答:
Answer 1: (for $k\geq 2$)

It's obvious that $0\in\mathbb{R}^d$ can't be a descent direction. So we can substitute the hypersphere $S^{d-1}$ for $\mathbb{R}^d\setminus\{0\}$.

If there is no descent direction at $x^*$,
$$\forall v\in S^{d-1}, Df(x)v\geq 0\text{ for some component}\tag{1}$$
holds.
If there exists $v$ s.t. $Df(x)v=0$ (component-wisely), by your note at the bottom, $x^*$ is locally Pareto critical. (This note obviously doesn't hold when k=1, this is why we need $k\geq 2$.)

If such $v$ doesn't exist, we can constrcut the following map.
$$g:S^{d-1}\rightarrow\mathbb{R};v\mapsto \sup\{t\geq 0|f(x^*+\lambda v)\geq f(x^*) \text{ for some component},\forall \lambda \in [0,t]\}.$$
Because (1) holds for all $v$, and $f$ is continuous(given that it is derivable), we have $g(v)>0, \forall v$, and $g$ is a continuous function. Given that $S^{d-1}$ is a bounded closed set, i.e., a compact set in $\mathbb{R}^d$, $\delta:=\min\{g(v)|v\in S^{d-1}\}$ exists. Then we have $\delta>0$, and that $x^*$ is locally Pareto critical in $B_\delta(x^*)$.

Answer 2: I believe that the statement is wrong for $k=1$. $\nabla f(x^*)=0$ can't assure that $x^*$ is a local minimum, so it may not be locally Pareto critical.

Sorry for my poor English grammar...

Title: Existence of a descent direction


问:
Projected image of sequences.

Problem: Let $\mathbb{N}$ be the set of all natural numbers.

And $\forall n \in \mathbb{N},$ let $ X_n $ be a non-empty set and $f_n: X_{n+1} \rightarrow X_n.$
And let $\forall n \in \mathbb{N},\operatorname{Pr}_n : \prod_{i\in N}X_i \rightarrow X_n $ be the canonical projection. (i.e. $n$-th component of the given element.)

Let $X=\{x\in \prod_{i\in \mathbb{N}}X_i : \forall n\in \mathbb{N}, \operatorname{Pr}_n(x)=f_{n}(\operatorname{Pr}_{n+1}(x)) \}$.

Then, is it true that ($\forall A\subset X, \forall B\subset X,\space ((\forall a\in \mathbb{N}, \operatorname{Pr}_a(A)=\operatorname{Pr}_a(B) )\rightarrow A=B))$?

For any $x_1, x_2 \in X$, if $\space \exists n\in \mathbb{N}$ s.t. $\operatorname{Pr}_n(x_1)=\operatorname{Pr}_n(x_2),$ then $\forall m \le n, \operatorname{Pr}_m(x_1)=\operatorname{Pr}_m(x_2).$

i.e. all of the formal components of the sequence of $(X_i)_{i\in \mathbb{N}}$ is determined uniquely by the latter component.

For trivial example, let $X_n=\mathbb{N}, f_n : \mathbb{N} \rightarrow \mathbb{N}$ be the injective function like an identity map. Then, any components of $x \in \prod_{i\in \mathbb{N}}$ is deternmined by its 1st component, so it's true for this example.

The motivation of this problem was from the concept of an inverse system, and the above problem is the simpler one of the more general problem and I guess the original problem is false, so I tried to find a counterexample with more simpler problem with which more ones will feel familiar.

For the reference, the more general problem is like this.

Original Problem :
Let $I$ be a directed index set, (i.e. $\forall a, b\in I, \exists c \in I, a\le c\wedge b\le c.$)

$\forall i \in I, \space X_{i}$ be a non-empty set.

And $\forall a\le b\in I,$ let $f_{ab} : X_{b} \rightarrow X_{a}$ s.t. $\forall x\le y\le z\in I,\space f_{xy}\circ f_{yz}=f_{xz}$.

And let $\forall a \in I, \operatorname{Pr}_a : \prod_{i\in I}X_i \rightarrow X_a $ be the canonical projection. (i.e. a-th component of the given element and $\forall x, y\in \prod_{i\in I}X_i, (x=y \iff \forall a\in I, \operatorname{Pr}_a(x)=\operatorname{Pr}_a(y))$)

Let $X=\{x\in \prod_{i\in I}X_i : \forall a\le b\in I, \operatorname{Pr}_a(x)=f_{ab}(\operatorname{Pr}_b(x)) \}$.

Then, prove or disprove that
$\forall A\subset X, \forall B\subset X,\space (\forall a\in I, \operatorname{Pr}_a(A)=\operatorname{Pr}_a(B)\rightarrow A=B)$

where $\operatorname{Pr}_a(A)=\{ \operatorname{Pr}_a(x):x\in A\}, \operatorname{Pr}_a(B)=\{ \operatorname{Pr}_a(x):x\in B\}.$

答:
If I understand your problem correctly, then no.

Let $X_n$ be the set of all binary sequences of length $n$ (i.e., $X_n$ is the set of all functions from $\{0,1,\cdots, n-1\}$ to $\{0,1\}$.) You can see that $X$ is isomorphic with the set of all infinite binary sequences, and $\operatorname{Pr}_n$ corresponds to the function that cuts off the sequence to that of length $n$. Now consider

$A=\{s\in X \mid \text{some digits of the sequences of $s$ repeat eventually.} \}$

$B=\{s\in X \mid s\notin A\}.$

(If you are familiar with $2$-adic integers, you may identify $A$ with the set of rational $2$-adic integers, and $B$ with the irrational $2$-adic integers. The idea is that we cannot distinguish the set of rationals and irrationals by just looking at some finite digits of their elements.)

Then $A\neq B$. However, we can see that $\operatorname{Pr}_n A = \operatorname{Pr}_n B$ is the set of all binary sequences of length $n$.

Title: Projected image of sequences.


答:
$P(A\cup B)=P(A)+P(B)$ is true if $P(A\cap B)=0$. This is because we always have $P(A\cup B)+P(A\cap B)=P(A)+P(B)$.

Now just take $A=X\cap Y$ and $B=X \cap Z$.  Note that $A\cap B \subset Y\cap Z$ so $P(A\cap B)=0$.

Title: show that $P(X\cap Y)\cup P(X\cap Z)=P(X\cap Y)+P(X\cap Z)$ if $X$ and $Y$ are independent,$X$ and $Z$ are independent, and $P(Y\cap Z) = 0$

Body:
Show that $P(X\cap Y)\cup P(X\cap Z)=P(X\cap Y)+P(X\cap Z)$   if $X$ and $Y$ are independent, $X$ and $Z$ are independent, and $P(Y\cap Z) = 0$

This makes sense intuitively if one draws a venn diagram. But how can one justify that $X\cap Y$ and $X\cap Z$ are mutually exclusive with probability rules ?


问:
Calculating higher homotopy groups of (complements of) knots
There are techniques to calculate the group of a knot, i.e. the fundamental group of its complement in a manifold, but are there techniques to calculate its higher homotopy groups?

Can anyone suggest a reference on this topic?

答:
In

```<code>MR0090053 (19,761a) Reviewed
Papakyriakopoulos, C. D.
On Dehn's lemma and the asphericity of knots.
Ann. of Math. (2) 66 (1957), 1–26.
55.0XMore links
</code>```

It is proven that knot completements in $S^3$ are aspherical (all higher homotopy groups vanish)

Body:
In

MR0090053 (19,761a) Reviewed
Papakyriakopoulos, C. D.
On Dehn's lemma and the asphericity of knots.
Ann. of Math. (2) 66 (1957), 1–26.
55.0XMore links

It is proven that knot completements in $S^3$ are aspherical (all higher homotopy groups vanish)


答: The girls can be arranged in $4!$ ways. They determine $5$ "gaps," of which $3$  are real gaps between the girls, and the other $2$ are end "gaps." The oldest boy can be inserted into any of these gaps. For each choice, there are $4$ gaps for the other boy, for a total of $(5)(4)(4!)$. The idea obviously generalizes.</p>

Title: Permutation and Combinations with conditions

Body:
Hallo :) This is a question about permutations but with conditions.

2 boys and 4 girls are to be arranged in a straight line.
In how many ways can this be done if the two boys must be separated? (The order matters)

Thank You.