问:
Can a prime number be bigger than the sum of adding the previous twin primes (other than 13)?
A simple heuristic of the first million primes shows that no prime number can be bigger than the sum of adding the previous twin primes.

Massive update:
@mathlove made a comment that leaves me completely embarrassed. $13 > 7 + 5$ I don’t know how I missed it and I deeply apologize to everyone!

I ask anyone qualified to suggest any edits for the question.

$7 < 5 + 3$

$11 < 7 + 5$

$17 < 11 + 13$

$23 < 17 + 19$

At larger numbers:

$4886639 < 4886489 + 4886491$

$5389451 < 5388869 + 5388871$

$3155597 < 3154757 + 3154759$

I assume that if it could be proved, it would prove the twin prime conjecture of whether twin primes exist forever.

So I am not exactly seeking for a proof, but rather for possible explanations or references for why it is assumed true (or not)?

Also as the list grows, there seems to be a range for how small or big can a prime be in comparison to the sum of adding the previous twin primes.

As the list grows, a prime is usually never bigger or smaller than slightly above $50\%$ of the sum of the previous twin primes. Any references for such a range will be appreciated too.

*Update: When mentioning "the previous twin primes", I am implying to:  $(107, 109), 113, 127, 131, (137, 139)$.

$131 < 107 +109$

答:
Let's modify the OP's observation as follows:

Let $(p,p+2)$ and $(q,q+2)$ be consecutive pairs of twin primes,
e.g., $(107,109)$ and $(137,139)$. Then (conjecturally) $q\lt2p+2$.

This is, essentially, a Bertrand's Postulate for twin primes, and it's not hard to confirm that it holds for entries at the outset for the sequence $3,5,11,17,29,41,59,\ldots$.  The basic explanation can be found in the heuristic twin-prime "theorem" $\pi_2(x)\sim2C_2x/(\ln x)^2$, although arguing that it (conjecturally) holds for all twin primes, not just for ones that are sufficiently large -- i.e., giving a (heuristic) twin-prime analog of the proof of Bertrand's Postulate -- seems problematic.

Remark: Amusingly, the modification proposed above of the OP's observation is technically agnostic with regard to the twin prime conjecture. Indeed, it would be easiest, in principle, to prove (or disprove) if there were an identifiable last pair of twin primes.


问:
Results that came out of nowhere.
Most big results in mathematics are built on years and years of groundwork by the author and other mathematicians, such as Wiles' proof of FLT or the classification of finite simple groups.  Nevertheless, there is this big romantic idea in pop culture where a young genius comes up with a brilliant proof of a major conjecture, based on some creative new idea which flies in the face of the mathematical community.

What are some examples where this has actually happened?  That is, which results stem from independent work by a mathematician who came along "out of nowhere" and solved a huge problem by surprise through nonstandard techniques?

I read that the recent proposed proof of the ABC conjecture comes from years of autonomous theory by Prof Mochizuki, most of which ventures far outside of the current literature; this would seem to qualify as one example provided the proof turns out to be true.

答:
I remember when Apéry's proof that $\zeta(3)$  is irrational appeared. There had been essentially no progress on values of
$$1+\frac{1}{2^n}+\frac{1}{3^n}+\frac{1}{4^n}+\cdots$$
for odd $n$ since the time of Euler.  </p>

ContentLicense: CC BY-SA 3.0

Title: Results that came out of nowhere.


问:
Almost sure convergence for expected value of indicator function
Let $Z$ a random variable which is $\mathscr{G}$-measurable.

If $\mathcal{C}$ is a class of events, closed under intersection which generates the $\sigma$-field $\mathscr{G}$ (i.e $\mathscr{G}=\sigma(\mathcal{C})$) and $\mathbb{E}[{Z \mathbf{1}_B}]=\mathbb{E}[Z]=0$ for every $B \in \mathcal{C} $

and  $\mathbb{E}[{Z \mathbf{1}_B}]=0$ for every $B \in \mathcal{G}$ then how show that :
$Z = 0 $ almost sure.

Proposal

If $Z\geq 0$, then we may argue that
$$
\mathscr{G}\ni \mathcal{C}\mapsto \mathbb{E}[Z \mathbf{1}_{B}]
$$
are measures that agree on the $\pi$-system $\mathscr{G}$ and thus also on the sigma-algebra $\sigma(\mathcal{C})=\mathscr{G}$.

Now for a general $Z\in L^1(\Omega,\mathcal{C},P)$ we have
$$
\mathscr{G}\ni \mathcal{C} \mapsto \mathrm{E}[Z^{+}|B]+\mathrm{E}[Z^-| B]
$$

are measures that agree on the $\pi-\lambda$-system.

答: Since $Z$ is $\mathcal G$ measurable an $EZ1_B=0$ for all $B \in \mathcal G$ we have $EZ1_{Z>0}=0$ and $EZ1_{Z<0}=0$. But this implies that $P(Z>0)=0$ and $P(Z<0)=0$ so $Z=0$ almost surely.

ContentLicense: CC BY-SA 4.0


问:
How do calculate a fee from a given percentage
Scenario:

I want to buy a share in a company.

Cost per share = 50,000USD

Number of shares to buy = 0.5

Transaction fees = 0.2%

Calculations:

Convert fee percentage to a decimal : 0.2 / 100 = 0.002

Convert decimal fee to amount to subtract from shares : 1 - 0.002 = 0.998

Calculate shares bought after fees : 0.5 * 0.998 = 0.499

Two part question:

Is the calculation correct?

Is there a better, simpler way of doing this?

答:
I could be mistaken, but I don't think transaction fees are calculated in that way.  If you buy two shares, spending $\$100,000$ then you will have to pay an additional $\$200$ in transaction fees.

Therefore, the fraction of outlay that goes toward the transaction fee is

$$\frac{200}{100200} ~\text{rather than}~ \frac{200}{100000}.$$

The natural question is, since my proposed interpretation leads to less profit for the bank/brokerage house, why would they accept such an interpretation?  Answer : because the math is easier for their bookkeepers to understand.

Actually, I could be mistaken; I simply suspect that my interpretation is how the calculations are actually done.


问:
Global existence of ODE $f'(x)=-f(x)^3$, $f(0)=a$
I am studying a PDE and as simplification I begin with studying the ODE
\begin{equation*}
f(t)'=-f(t)^3,  f(0)=a.
\end{equation*} I want to develop global existence for this ODE, .i.e. for any $T>0$ $$\sup_{t\in[0,T]}|f(t)|<\infty.$$
Of course we can just write down the solution which is given by $f(t)=\pm \frac{1}{\sqrt{1/a^2+2t}}$ with + for $a>0$ and - for $a<0$. However such an explicit solution can't be hoped for in the PDE case and thus I want to know why global solutions exist. This clearly depends on the minus sign in the equation as $g(t)'=g(t)^3$ blows-up in finite time. However any Lipschitz argument or norm bound will treat both equations in the same manner and not account for the negativity.
Of course the idea would be to use some bounds to repetitively apply the local solution theory and glue solutions together, however one needs to show that these solution intervals don't decrease in size.

I can see how the equations behave differently from a plot but I am lacking a convincing argument. If we view the equation as $f(t)'=F(f(t))$ with $F(x)=-x^3$ and assume that $a>0$ then have have $F(a)<0$ and thus $f(0)'<0$ thus for some (small) $t_1>0$ $f(t_1)<f(0)$. If we assume we get a sequence $(t_n)_{n\in\mathbb{N}}$ for which we always have $0<f(t_n)$ we get a decreasing sequence and our solution won't blow-up in finite time. While this argument is intuitive it isn't rigours at all, but it shows that the local Lipschitz constant in a uniqueness proof wouldn't increase but even decrease.

Could someone develop a rigours argument, show how the solutions can be glued together and why the solution intervals don't decrease or point at a different approach to the global existence.

答:
Solutions to $f(t)'=-f(t)^3$ are locally unique (because $F(y) = -y^3$ is locally Lipschitz), so that solutions with different initial values do not intersect. In particular, a solution with $f(0) = a > 0$ on $[0, T]$ is strictly positive (and therefore decreasing) everywhere.

Similarly, solutions with $f(0) = a < 0$ are strictly negative and increasing everywhere on their interval of existence.

It follows that $\sup_{t\in[0,T]}|f(t)| = |a|$ for all $T > 0$.  There is no “blow up” and the maximal interval of existence is $[0, \infty)$.


答:
Suppose you want to spend $d$ dollars for shares and that amount should include the commission $c$ (expressed as a fraction, not a percent). Let $s$ be the amount that will go to shares. Then
$$
(1+c)s = d
$$
so
$$
s = \frac{d}{1+c}.
$$

That is not the same as what you calculated, which is $s(1-c)$. But when $c$ is very small (as it is in this case) they are very close. You can check the difference numerically.

Title: How do calculate a fee from a given percentage

Body:
Scenario:

I want to buy a share in a company.

Cost per share = 50,000USD

Number of shares to buy = 0.5

Transaction fees = 0.2%

Calculations:

Convert fee percentage to a decimal : 0.2 / 100 = 0.002

Convert decimal fee to amount to subtract from shares : 1 - 0.002 = 0.998

Calculate shares bought after fees : 0.5 * 0.998 = 0.499

Two part question:

Is the calculation correct?

Is there a better, simpler way of doing this?


答:
Consider $V(y)=y^2$, then
$$
\frac{d}{dt}V(f(t))=2f(t)f'(t)=-2f(t)^4<0.
$$
Thus this simple Lyapunov function is decreasing, which implies that it and with it $f$ is bounded.

Title: Global existence of ODE $f'(x)=-f(x)^3$, $f(0)=a$

Body:
I am studying a PDE and as simplification I begin with studying the ODE
\begin{equation*}
f(t)'=-f(t)^3,  f(0)=a.
\end{equation*} I want to develop global existence for this ODE, .i.e. for any $T>0$ $$\sup_{t\in[0,T]}|f(t)|<\infty.$$
Of course we can just write down the solution which is given by $f(t)=\pm \frac{1}{\sqrt{1/a^2+2t}}$ with + for $a>0$ and - for $a<0$. However such an explicit solution can't be hoped for in the PDE case and thus I want to know why global solutions exist. This clearly depends on the minus sign in the equation as $g(t)'=g(t)^3$ blows-up in finite time. However any Lipschitz argument or norm bound will treat both equations in the same manner and not account for the negativity.
Of course the idea would be to use some bounds to repetitively apply the local solution theory and glue solutions together, however one needs to show that these solution intervals don't decrease in size.

I can see how the equations behave differently from a plot but I am lacking a convincing argument. If we view the equation as $f(t)'=F(f(t))$ with $F(x)=-x^3$ and assume that $a>0$ then have have $F(a)<0$ and thus $f(0)'<0$ thus for some (small) $t_1>0$ $f(t_1)<f(0)$. If we assume we get a sequence $(t_n)_{n\in\mathbb{N}}$ for which we always have $0<f(t_n)$ we get a decreasing sequence and our solution won't blow-up in finite time. While this argument is intuitive it isn't rigours at all, but it shows that the local Lipschitz constant in a uniqueness proof wouldn't increase but even decrease.

Could someone develop a rigours argument, show how the solutions can be glued together and why the solution intervals don't decrease or point at a different approach to the global existence.


问:
Does nuclear norm decrease as some elements in a matrix are set zero?
Let $M$ be a generic nonzero real matrix and $M_{0}$ is constructed by replacing some elements in $M$ with zero. Let $||\cdot||_{*}$ denotes the nuclear norm operator. Is it true that $||M||_{*}\geq ||M_{0}||_{*}$ regardless of the replacement rule?

答:
No. Consider
$$
A=\pmatrix{1&1&1\\ 1&1&1\\ 1&1&1},\ B=\pmatrix{1&0&1\\ 0&1&1\\ 1&1&1}.
$$
Since $A$ is positive semidefinite, it nuclear norm is equal to its trace, which is $3$. The matrix $B$ is symmetric but indefinite. Hence its nuclear norm is the sum of absolute values of all its eigenvalues, i.e. $|1|+|1+\sqrt{2}|+|1-\sqrt{2}|=1+2\sqrt{2}>3$.

P.S. I asked a similar question before about the induced $2$-norm (and the answer is also negative).


问:
What is the sample space of the best of seven series?
Original question: In a soccer game, two teams are playing against each other in a best of seven series. The game ends when one team wins four games and each game corresponds to a win for the team. What is the sample space of the best of seven series?

Answer: $2(1+4+10+20)=70$

My question is whether my answer or the answer above is correct. Even though I think my answer makes more sense I also think it is wrong because my answer is very different from the given answer.

My method is to use the following formula and add the total number of ways to win then multiply the sum by $2$ to include the total number of ways to lose:

If the number of games $= 4$: $$\frac{n!}{n_1!n_2!...n_k!}$$

If the number of games $> 4$: $$\frac{n!}{n_1!n_2!...n_k!}-1$$

where $n$ is the total number of letters and $n_1,n_2,...,n_k$ are possible duplicate letters.

$W$ = {win}

$L$ = {lost}

Team wins $4$ games out of $7$: (1 way)

$$WWWW$$

Team wins $5$ games out of $7$: (4 ways)

$$LWWWW,WLWWW,WWLWW,WWWLW$$

$WWWWL$  (removed since this is the same as $WWWW$)

$$\frac{5!}{4!}-1=4$$

Team wins $6$ games out of $7$: (14 ways)

$$\frac{6!}{4!2!}-1=14$$

Team wins $7$ games out of $7$: (34 ways)

$$\frac{7!}{4!3!}-1=34$$

So the sample space (total number of ways of winning/losing) is

$$S=2(1+4+14+34)=106$$

答:
In a best of seven series, the winning team must win the last game.

Say team A wins.  We can double our result afterwards to account for the possibility that team B wins.

Team A wins in four games:  This can occur in one way: $W_AW_AW_AW_A$, where $W_A$ means that team A wins the game.

Team A wins in five games:  For this to occur, team A must win exactly three of the first four games and the fifth game, which can occur in
$$\binom{4}{3}\binom{1}{1} = 4$$
ways.  The elements in the sample space are $$W_AW_AW_AW_BW_A,\qquad W_AW_AW_BW_AW_A,\qquad W_AW_BW_AW_AW_A,\qquad W_BW_AW_AW_AW_A$$

Team A wins in six games:  For this to occur, team A must win exactly three of the first five games and the sixth game, which can occur in
$$\binom{5}{3}\binom{1}{1} = 10$$
ways.  I will leave listing the corresponding elements in the sample space to you.

Team A wins in seven games:  For this to occur, team A must win exactly three of the first six games and the seventh game, which can occur in
$$\binom{6}{3}\binom{1}{1} = 20$$
ways.  Again, I will leave listing the corresponding elements in the sample space to you.

Hence, there are $$1 + 5 + 10 + 20 = 35$$ ways for team A to win the series and
$$2(1 + 5 + 10 + 20) = 70$$ ways for one of the teams to win the best of seven series.


问:
Prove the covariance matrix can be obtained from the quadratic form of a pdf that has the same form as a Gaussian function
From chapter 2 in "Pattern Recognition and Machine Learning, Bishop", a multivariate Gaussian pdf is defined as,
$$
N(x | \mu, \Sigma) = \frac{1}{(2\pi)^{0.5D}} \frac{1}{|\Sigma|^{0.5}} \exp\big(-0.5(x-\mu)^T \Sigma^{-1} (x-\mu) \big) \space \space (1)
$$

In section 2.3.1, he proved a pdf has the form of a Gaussian function, with C and D are constant,
$$p(x) = C\exp(-0.5(x-b)^TS^{-1}(x-b) + D) \space \space (2)$$

and then concluded the pdf is Gaussian pdf with $S$ as the covariance matrix.

Why can the author be sure that without checking the constant C and D, (2) can be arranged to have the form as (1), and so it is a Gaussian pdf?

答:
The portion inside the brackets of the $\exp(\cdot)$ is just a quadratic. In one dimension, the portion from (2) is $-(x-b)^2/2S + D$ which can be factorised by difference of squares into $-(x-\mu)^2/2\sigma$ to match (1). In multiple dimensions it’s a little more involved, with diagonalisation of $S^{-1}$, but it’s the same principle.

Geometrically, the equivalence of (2) with (1) is a change of coordinates: the origin of the coordinate system is moved to the mean of the Gaussian, and the coordinates are then rotated so that the axes align with the principal components of the Gaussian.


问:
Balancing a dice game - probability question
I am currently developing a TTRPG game system, and I am trying to determine the probability of success in various situations to see how balanced it might be.

The system is fairly simple, you roll a number of 10 sided dice, and if any of them are below your score, you succeed. However, if any of the dice you roll land on $10$ (or $0$ as notated on most d10s), you remove the lowest die rolled, which I guess you could call an imploding dice rule (as opposed to exploding dice, which would allow you to roll extra on a critical success)

For example, my score is $5$, so to succeed I need to roll a $<5$. With one dice, and without the $10$ rule, that is easy to work out as $40$% chance of success, and as I add more dice, I can work it out as by using the chance of failure:

$1-(1-0.4)^n$ where $n$ is the number of dice.

But I am struggling to work out how to introduce the imploding dice to my probability calculations. The chance of rolling a $10$ is $10$% for each die I roll, but when I do, I am essentially reducing the amount of dice I originally rolled, right?

/edit: clarification, if you roll a $10$, you only remove one die, not one for every $10$ rolled

答:
Problem-statement: We have $n$ dice and a given $s$ score. We roll all the dice and we get results form $1$ to $9$ or a special result, called WILDCARD. If we have rolled a number of $k$ wildcards, the $k$ lowest dice are eliminated from the table. If any of the $n - k$ is less than $s$, we have WON. Otherwise, we have LOST.

Probability evaluation: The probability of having $k$ wildcards out of $n$ dice is $\frac{1}{n} \cdot \frac{1}{n - 1} \cdot \dots \cdot \frac{1}{n - k + 1} = \frac{(n - k)!}{n!} = \frac{1}{A_n^k}$. Next, from $n - k$ dice we have the probability to draw greater equal than $s$: $\frac{(10 - s)^{n - k}}{9^{n - k}}$, and the probability to have at least one dice less than $s$ will be $1 - \frac{(10 - s)^{n - k}}{9^{n - k}}$.

Result: By summating all probabilities (considering $k$ variating from $0$ to $n$):
$$p = \sum_{k = 0}^n \frac{1}{A_n^k} \cdot \bigg(1 - \frac{(10 - s)^{n - k}}{9^{n - k}}\bigg)$$

That's all!


答:
Let us fix a team and try to find the number ways for this team to win.

The answer is:$$\sum_{k=3}^6\binom{k}3=1+4+10+20=35$$where $\binom{k}3$ corresponds with the number of games that end after $k+1$ meetings.

We denote the results as e.g. $WLWWLW$ where the team won at meetings 1,3,4,6.

Note that in that situation where the game ends after $k+1$ meetings a $W$ must stand on the $k+1$-spot and exactly 3 letters $W$ must be among the $k$ former spots. So this results in $\binom{k}3$ possibilities.

More directly again looking at a fixed team we can think of 7 open spots of which 4 must be filled up with a letter $W$. The other spots stay empty if there is no letter $W$ on RHS and are filled with a letter $L$ otherwise. This gives (again):$$\binom74=35$$possibilities.

(actually a proof of the hockey-stick identity).

There are two teams that can win so the final answer is:$$2\times35=70$$

Body:
Original question: In a soccer game, two teams are playing against each other in a best of seven series. The game ends when one team wins four games and each game corresponds to a win for the team. What is the sample space of the best of seven series?

Answer: $2(1+4+10+20)=70$

My question is whether my answer or the answer above is correct. Even though I think my answer makes more sense I also think it is wrong because my answer is very different from the given answer.

My method is to use the following formula and add the total number of ways to win then multiply the sum by $2$ to include the total number of ways to lose:

If the number of games $= 4$: $$\frac{n!}{n_1!n_2!...n_k!}$$

If the number of games $> 4$: $$\frac{n!}{n_1!n_2!...n_k!}-1$$

where $n$ is the total number of letters and $n_1,n_2,...,n_k$ are possible duplicate letters.

$W$ = {win}

$L$ = {lost}

Team wins $4$ games out of $7$: (1 way)

$$WWWW$$

Team wins $5$ games out of $7$: (4 ways)

$$LWWWW,WLWWW,WWLWW,WWWLW$$

$WWWWL$  (removed since this is the same as $WWWW$)

$$\frac{5!}{4!}-1=4$$

Team wins $6$ games out of $7$: (14 ways)

$$\frac{6!}{4!2!}-1=14$$

Team wins $7$ games out of $7$: (34 ways)

$$\frac{7!}{4!3!}-1=34$$

So the sample space (total number of ways of winning/losing) is

$$S=2(1+4+14+34)=106$$


问:
What type of input does trigonometric functions take in
I see in my Book that 45 deg is equivalent of π/4 . Ι also do the conversion if I simply convert degrees into radians like this

45* π/180 = π/4 radians

and back again

π/4 * 180/π = 45 deg

.
So I think that I grasp the Idea finally that π/4 is another way of saying 45 degrees. Now if I use my calculator and I say sin(45) witch I assume that 45 is degrees I get the number 0.7071067812. If I do the calculation sqrt(2)/2 I will get the same number hence sin(45)=sqrt(2)/2 . But when I input sin(π/4) I get the number 0.0137073546. So I say to my self that probably that function does not get that type of input. But then again I see in my book that sin(π/4) = sqrt(2)/2 . So this forces me to ask the question, "What type of input do trigonometric functions take in?" Is the calculator not working properly ?

Excuse me if this looks simple to you but I am so bad in math and the calculator is the only validator I have at the moment and this confuses me allot.

答:
The problem is that there are two different sine functions, and each takes in a different kind of input!

Your calculator is set to be in degree mode, which means it is taking in degree inputs.  When you put in 45, it assumes you mean 45 degrees ($\pi/4$ radians).  And when you put in $\pi/4$, your calculator calculates $\pi/4 \approx .7854$, and then it computes the sine of $.78$ degrees!  Hopefully you can visualize that $.78$ degrees is just a very small angle, while $\pi/4$ radians is in fact $45$ degrees, which is a much larger angle.

You can't expect your calculator's sine function to know whether you are talking about degrees or radians, because $\pi / 4$ as a number can be both a number of degrees or a number of radians.  So you have to tell your calculator which sine function you want -- the degree input one or the radian input one -- and then it will know what you mean.

In fact, in actual mathematics there are no degrees -- just radians.  When you say, for example, $\sin 45^\circ$, what the $\circ$ really means is, convert this number to radians.  The reason for having to specify degrees with the $\circ$ symbol is to remove the ambiguity of the input which you are complaining about.  Also, $\sin x$ with $x$ in radians turns out to be a much nicer function in the end from the perspective of real and complex analysis.


问:
Difficult vector identity using Levi Civita
I have to prove the following:

$$[3(\vec{p}\cdot\hat{r})\hat{r}-\vec{p}]\times[3(\vec{m}\cdot\hat{r})\hat{r})-\vec{m}]=-2\vec{p}\times\vec{m}+3\hat{r}[\hat{r}\cdot(\vec{p}\times\vec{m})]$$

I am given the hint to use the BAC-CAB identity on the following triple product:
$$\hat{r}\times(\vec{p}\times\vec{m})=\vec{p}(\hat{r}\cdot\vec{m})-\vec{m}(\hat{r}\cdot\vec{p})$$

And then using that to evaluate the following in two ways, one directly and the other again using BAC-CAB:

$$\hat{r}\times[\hat{r}\times(\vec{p}\times\vec{m})]=\hat{r}\times\vec{p}(\hat{r}\cdot\vec{m})-\hat{r}\times\vec{m}(\hat{r}\cdot\vec{p})$$
$$=\hat{r}[\hat{r}\cdot(\vec{p}\times\vec{m})]-(\vec{p}\times\vec{m})$$

From here I decided to convert to levi civita and set the= two lines equal to each other:
$$\epsilon_{ijk}r_jp_kr_lm_m\delta_{lm}-\epsilon_{ijk}r_jm_kr_lp_m\delta_{lm}=r_mr_l\epsilon_{ijk}p_jm_k\delta_{li}-\epsilon_{ijk}p_jm_k$$

But from here I have no idea where to go. I can see pieces of what I want but I have no idea how to rearrange this at all to get the desired equality. Any help would be much appreciated.

答: The hint is useful. The problem's left-hand side expands to$$-3(r\cdot p)r\times m+3(r\cdot m)r\times p+p\times m,$$so you just need to multiply your BAC-CAB inference$$(r\cdot m)r\times p-(r\cdot p)r\times m=r[r\cdot p\times m]-p\times m$$by $3$, then rearrange, no Levi-Civita required.


答:
The integral of any probability density function $\ p\ $ over the whole of its domain must be $1$:
$$
\int_{\mathbb{R}^n}p(x)dx=1\ .
$$
So if $\ p(x)=kf(x)\ $, where $\ k\ $ is constant,then we must have
\begin{align}
1&=\int_{\mathbb{R}^n}p(x)dx\\
&=k\int_{\mathbb{R}^n}f(x)dx\ ,
\end{align}
and therefore $\ k=\left(\int_{\mathbb{R}^n}f(x)dx\right)^{-1}\ $.  In particular, if
\begin{align}
p(x)&=Ce^{-\frac{(x-b)^TS^{-1}(x-b)}{2}+D}\\
&=Ce^De^{-\frac{(x-b)^TS^{-1}(x-b)}{2}}\
\end{align}
where $\ x\in\mathbb{R}^n\ $, then
\begin{align}
Ce^D&=\left(\int_{\mathbb{R}^n}e^{-\frac{(x-b)^TS^{-1}(x-b)}{2}}dx\right)^{-1}\\
&=(2\pi)^{-\frac{n}{2}}\det(S)^{-\frac{1}{2}}\ ,
\end{align}

Title: Prove the covariance matrix can be obtained from the quadratic form of a pdf that has the same form as a Gaussian function

Body:
From chapter 2 in "Pattern Recognition and Machine Learning, Bishop", a multivariate Gaussian pdf is defined as,
$$
N(x | \mu, \Sigma) = \frac{1}{(2\pi)^{0.5D}} \frac{1}{|\Sigma|^{0.5}} \exp\big(-0.5(x-\mu)^T \Sigma^{-1} (x-\mu) \big) \space \space (1)
$$

In section 2.3.1, he proved a pdf has the form of a Gaussian function, with C and D are constant,
$$p(x) = C\exp(-0.5(x-b)^TS^{-1}(x-b) + D) \space \space (2)$$

and then concluded the pdf is Gaussian pdf with $S$ as the covariance matrix.

Why can the author be sure that without checking the constant C and D, (2) can be arranged to have the form as (1), and so it is a Gaussian pdf?


问:
Analysis Textbook like Zorich
I'm an undergraduate student currently studying mathematical analysis.

Our professor uses Zorich's Mathematical Analysis, but I found the text too difficult to understand.

After exploring some textbooks, I found that Abbott was easier to follow, so I studied Abbott until I realized that there's a significant amount of content in Zorich that Abbott doesn't cover.

So I was wondering if there's a book out there that covers as much content as Zorich but is more readable?

Thank you for any help.

答:
True that Zorich is difficult to understand.
BUT I still think if you go through with it, you will learn a lot in the end.
However, I think there’s also an other book you could consider about, which is Mathematical Analysis written by Apostol.Or, consider Courant’s book.
They are both good books.
Good luck!

Title: Analysis Textbook like Zorich


问:
When does a system of polynomial equations have infinitely many solutions?
It seems intuitively correct to say that a system of polynomial equations has finitely many solutions if there are as many equations as there are variables in the system. However, how would you prove that a system of polynomial equations with fewer equations than variables has infinitely many solutions (corresponding to the "under-determined" case in linear systems)? How would you characterize when a subset of equations "coincide" (you can combine several to get another, like linear dependence in linear systems)?

For context, I'm in the middle of an abstract algebra course, and we've just finished Chapter 7 of Dummit & Foote (though this question isn't yet directly related to the course, that's just my background knowledge).

答:
Question: "However, how would you prove that a system of polynomial equations with fewer equations than variables has infinitely many solutions?"

Answer: There is a result in commutative algebra (look at the Matusumura book "Commutative ring theory", theorem 14.1, page 105) saying that if $(A,\mathfrak{m})$ is a noetherian local ring and $x_1,..,x_r$ is a system of parameters then

$$M1.\text{   }dim(A/(x_1,..,x_i))=r-i.$$

Intuitively view each element $x_i$ as an "equation" in $r$ variables, and $dim(A/(x_1,..,x_i))$ as the dimension of the "space of solutions" to the system of equations

$$M2. \text{    }x_1=\cdots =x_i=0,$$

then M1 says that the system M2 has an "infinite set of solutions" iff $i<r$.

"Intuitively" if $I:=(f_1, \cdots , f_l)\subseteq B:=k[x_1,..,x_n]$ is an ideal of polynomials in the variables $x_i$ over a field $k$, we let $A:=k[x_1,..,x_n]/I$ be the coordinate ring of the algebraic variety/scheme
defined by the set of polynomials $f_i$. The ring $A$ is a commutative unital ring reflecting properties of the "set of solutions" of the system of equations

$$M3.\text{   }f_1=\cdots =f_l=0.$$

We define $X:=Spec(A)$ as the "set of prime ideals" in the ring $A$ and give this set a topology (the Zariski topology). If the ideal $I$ is a prime ideal
it follows $X$ is an irreducible topolpogical space. The dimension of $X$
as an algebraic variety may be defined using the ring $A$: we define $dim(X):=dim(A)$ where $dim(A)$ is defined as the supremum of the set of strictly decreasing sequences of prime ideals

$$ \mathfrak{p}_r \subsetneq \mathfrak{p}_1 \subsetneq \cdots \subsetneq \mathfrak{p}_0 \subseteq A.$$

Hence with this definition, your system of equations should have a "finite set of solutions" iff $dim(A)=0$ - hence you must calculate the dimension $dim(A)$ in your case.

Given any maximal ideal $\mathfrak{m}\subseteq A$, it follows there is a maximal ideal $\mathfrak{n}\subseteq B$ mapping to $\mathfrak{m}$, and you may consider the localization $R:=B_{\mathfrak{n}}$ which is a local ring of dimension $n$. The image $g_i$ of the generators $f_i$ generates an ideal $J \subseteq R$ and if the ideal $J$ can be generated by a regular sequence
$J=(a_1,..,a_i)$ with $i<n$ it follows $dim(R/J)=dim(A) >0$. Hence the set
of solutions to the system is not finite. I believe there are algorithms implemented on computers that solves this problem, but I have no precise reference. In general it is difficult to check if an explicit ideal is a prime ideal, and it is also difficult to construct a regular sequence for an explicit ideal $J$.

Another more elementary approach is the Noether normalization lemma: It says that there is a set of element $t_1,..,t_d \in A$ with the property that the subring

$$S:=k[t_1,..,t_d] \subseteq A$$

is a polynomial ring, $dim(A)=d$ and $S \subseteq A$ is an integral extension.

Hence $dim(A)=0$ iff $k \subseteq A$ is an integral ring extension. Hence $dim(A)=0$ iff there is a finite set of elements $a_1,..,a_k\in A$ generating $A$ as $k$-algebra where $a_i$ satisfies a monic polynomial with coefficients in $k$. So your problem is reduced to constructing a set of integral elements $a_1,..,a_k$ generating $A$.

Lemma. The system M3 has a finite set of solutions iff $k \subseteq A$ is an integral ring extension iff $dim_k(A)< \infty$.

Proof. The first equivalence is proved above. The last equivalence follows from the fact that $A$ is finitely generated as $k$-algebra. Hence $k \subseteq A$ is an integral ring extension iff $dim_k(A)< \infty$. QED

Note 1. If you try to calculate $dim_k(A)$ "by hand" in some examples, you will find that it is a difficult problem to construct a basis for $A$ as $k$-vector space. I believe there are computer programs solving this problem.
To prove that $k \subseteq A$ is not an integral extension you must find an element $t\in A$ that generates a polynomial sub ring $k[t]\subseteq A$.

Note 2. If the base field $k$ is not algebraically closed, this method detects solutions in arbitrary finite extensions $k \subseteq K$. This is the "Hilbert Nullstellensatz". For a maximal ideal $\mathfrak{m}\subseteq A$ it follows the extension $k \subseteq A/\mathfrak{m}$ is a finite extension, and a maximal ideal $\mathfrak{m}\subseteq A$ corresponds to a solution to the system M3 in the field $\kappa(\mathfrak{m}):=A/\mathfrak{m}$.

Example 1. Let $k$ be the field of real numbers and let $f:=x^2+y^2+1\in k[x,y]$ and consider the "system"

$$S1.\text{   } f(x,y)=x^2+y^2+1=0.$$

The system S1 has no real solutions but it has many (a set of dimension 1) complex solutions. Given an arbitrary real number $\theta$ and let $x:=isin(\theta), y:=icos(\theta)$ it follows

$$x^2+y^2=(isin(\theta))^2+(icos(\theta))^2=-sin^2(\theta)-cos^2(\theta)=-1$$

hence $(isin(\theta), icos(\theta))\in \mathbb{C}^2$ is a solution to S1 for any $\theta$.
The field extension $\mathbb{R} \subseteq \mathbb{C}$ is finite and the dimension of the ring $A:=k[x,y]/(f)$ is 1, hence the system S1 has an infinite set of solutions (in the real and complex number field). For every $(isin(\theta), icos(\theta))$ you get a well defined surjective map

$$ \phi_{\theta}: A \rightarrow \mathbb{C}$$

defined by $\phi_{\theta}(x):=isin(\theta), \phi_{\theta}(y):=icos(\theta)$,
and the kernel $\ker(\phi_{\theta})\subseteq A$ will be a maximal ideal. Hence solutions to S1 correspond to maximal ideals in $A$.

Note 3. If you work with system M3 on a computer you should know that you are studying solutions to M3 in a finite field of large characteristic. This is because a computer has a finite memory. If your system M3 is defined by polynomials with rational coefficients $\frac{a_i}{b_i}$, it follows there is a large prime $p$ not dividing any of the $b_i$'s. The computer converts the system M3 to a system over $\mathbb{F}_p:=\mathbb{Z}/p\mathbb{Z}$ (or $\mathbb{F}_{p^r}$). Hence if you ask the question

$$\text{"Does M3 have a finite set of solutions?"}$$

to a computer, the computer will answer:

$$\text{It always has a finite set of solutions since we are working over a finite field $\mathbb{F}_p$}.$$

If $k$ is a finite field and you seek solutions to M3 in $k^n$ it follows the set of solutions $S$ to M3 is a subset $S \subseteq k^n$ of $k^n$ which is a finite set, hence $S$ is trivially a finite set. In a sense:

$$\text{A computer cannot "understand infinity".}$$

Example. If the memory $M$ of your computer $HAL$ has $2^n$ elements for $n \geq 1$ an integer, it cannot "work" with the ring of integers $\mathbb{Z}$ which has a countably infinite set of elements: There is no way to "embed" $\mathbb{Z}$ as a subset of $M$ since $M$ is a finite set. There is a number $m\geq 0$ with the property that an integer $a\in \mathbb{Z}$ with more than $m$ digits cannot be stored on $HAL$. Hence if $p,q$ are two prime numbers with more than $m$ digits it follows $p,q$ and the product $pq$
cannot be calculated by $HAL$. Hence $HAL$ cannot work with very large primes. There is an infinite set of prime numbers and only a finite set of prime numbers has less than $m$ digits. Hence there is an infinite set of prime numbers that cannot be calculated by $HAL$.

Hence by "understand" I mean the following: You may always construct two prime numbers $p,q$ and ask your computer $HAL$ to calculate $pq$, and your computer will not be able to do this because it does not have enough memory.

Question: "How would you characterize when a subset of equations "coincide" (you can combine several to get another, like linear dependence in linear systems)?"

Answer: If $I:=(f_1,..,f_l)$ is an ideal generated by polynomials $f_i$ and $J:=(f_{i_1},..,f_{i_n})$ is a subset of $I$ it follows (this is a definition) $I$ and $J$ "coincide" iff they generated the same ideal in $k[x_1,..,x_n]$.


问:
Consequences of direct comparison test
Check whether the statement is true:

If $\left(n^{2} a_{n}\right)$ is convergent sequence, then there is a subsequence $\left(x_{n_k}\right)$ of $x_{n}=\frac{\sin (n)}{n}$ such that the series $\sum_{k=1}^{\infty} x_{n_k} a_{k}$ is convergent

I tried the direct comparison test with $b_k=\frac{1}{k^2}$, but confused with $n_k$ and $k$?

Any role for Bolzanno- Weierstrass theorem?

答:
Since $(n^2 a_n)$ converges, there is $C > 0$ such that
$$|a_n| \le \frac{C}{n^2}$$
for all sufficiently large $n$. Therefore,
$$\sum_n |a_n| < +\infty.$$
Moreover,
$$\frac{|\sin(n) a_n|}{n} \le |a_n|, $$
and so
$$\sum_n x_na_n <+\infty.$$
It seems to work without taking a subsequence of $x_n$..

Title: Consequences of direct comparison test


答:
I completely agree with the other responses, but would like to point out a shortcut.  Assume that Team-1 wins, perform the enumeration, and then multiply by 2, re Team-2 could also win.

There is a bijection between each sequence of games that end immediately after Team-1 achieves its 4th win and each 7 game sequence where Team-1 wins exactly 4 games.  Simply take each sequence where Team-1 has its 4th win prior to the 7th game and then extend the sequence with Team-1 losing all of the remaining games (if any).

Note that $2 \times \binom{7}{4} = 2 \times 35 = 70.$

Title: What is the sample space of the best of seven series?


问:
$V=\{A\in M_{3\times 3}(\mathbb{R}):\text{trace}(A)=0\}$ is isomorphic to $\text{span}\{AB-BA:A,B\in V\}$
Background: Let
$$V:=\{A\in M_{3\times 3}(\mathbb{R}):\text{trace}(A)=0\}$$
be the vector space of $3\times 3$ real matrices with vanishing trace, and let $[\cdot,\cdot]:V\times V\to V$ be defined by
$$[A,B]=AB-BA$$
Note that this is well defined since $\text{trace}(AB-BA)=\text{trace}(AB)-\text{trace}(AB)=0$ for any matrices $A,B$. Finally, let
$$\hat{V}:=\text{span}\{[A,B]:A,B\in V\}$$

Question: I have to give an isomorphism
$$\varphi:(V,[\cdot,\cdot])\to(\hat{V},[\cdot,\cdot]),$$
i.e. a bijective map such that $\varphi([A,B])=[\varphi(A),\varphi(B)]$, $\forall A,B\in V$.

I am aware that this is related to Lie algebra; namely $V=\mathfrak{sl}(3,\mathbb{R})$. However, I did not study Lie algebra yet. I am suppose to be able to prove it without referring to any well-known theorem of Lie algebra (unless I prove it first).

I used a very computational argument. I am wondering if

My argument is correct.

There is a more fundamental way to prove it, i.e. without having to tediously compute all possible $[X_i,X_j]$ from a given basis.

Atempt: Let

$$
X_1 = \left(
\begin{array}{ccc}
 1 & 0 & 0 \\
 0 & -1 & 0 \\
 0 & 0 & 0
\end{array}
\right),\quad
X_2 = \left(
\begin{array}{ccc}
 1 & 0 & 0 \\
 0 & 0 & 0 \\
 0 & 0 & -1
\end{array}
\right),\quad
X_3 = \left(
\begin{array}{ccc}
 0 & 1 & 0 \\
 0 & 0 & 0 \\
 0 & 0 & 0
\end{array}
\right),$$
$$\quad
X_4 = \left(
\begin{array}{ccc}
 0 & 0 & 1 \\
 0 & 0 & 0 \\
 0 & 0 & 0
\end{array}
\right),\quad
X_5 = \left(
\begin{array}{ccc}
 0 & 0 & 0 \\
 0 & 0 & 1 \\
 0 & 0 & 0
\end{array}
\right),\quad
X_6 = \left(
\begin{array}{ccc}
 0 & 0 & 0 \\
 1 & 0 & 0 \\
 0 & 0 & 0
\end{array}
\right),$$
$$\quad
X_7 = \left(
\begin{array}{ccc}
 0 & 0 & 0 \\
 0 & 0 & 0 \\
 1 & 0 & 0
\end{array}
\right),\quad
X_8 = \left(
\begin{array}{ccc}
 0 & 0 & 0 \\
 0 & 0 & 0 \\
 0 & 1 & 0
\end{array}
\right)
$$
Then, $\{X_1,\ldots,X_8\}$ forms a basis for $V$, and we find
$$[X_1,X_2]=0,\quad[X_1,X_3]=2X_3,\quad[X_1,X_4]=X_4,\quad[X_1,X_5]=-X_5$$
$$[X_1,X_6]=-2X_6,\quad[X_1,X_7]=-X_7,\quad[X_1,X_8]=X_8,\quad[X_2,X_3]=X_3$$
$$[X_2,X_4]=2X_4,\quad[X_2,X_5]=X_5,\quad[X_2,X_6]=-X_6,\quad[X_2,X_7]=-2X_7$$
$$[X_2,X_8]=-X_8,\quad[X_3,X_4]=0,\quad[X_3,X_5]=X_4,\quad[X_3,X_6]=X_1$$
$$[X_3,X_7]=-X_8,\quad[X_3,X_8]=0,\quad[X_4,X_5]=0,\quad[X_4,X_6]=-X_5$$
$$[X_4,X_7]=X_2,\quad[X_4,X_8]=X_3,\quad[X_5,X_6]=0,\quad[X_5,X_7]=X_6$$
$$[X_5,X_8]=X_2-X_1,\quad[X_6,X_7]=0,\quad[X_6,X_8]=-X_7,\quad[X_7,X_8]=0$$
Thus, we have
$$
\begin{array}{cccc}
X_1=[X_3,X_6],& X_2=[X_4,X_7],& X_3=[X_2,X_3],& X_4=[X_1,X_4] \\
X_5=[X_2,X_5],& X_6=[X_5,X_7],& X_7=[X_7,X_1],& X_8=[X_7,X_3]
\end{array}
$$
which shows that $V$ is a subspace of $\hat{V}$, and since $\hat{V}$ is clearly a subspace of $V$, then $V$ and $\hat{V}$ are in fact the same vector space. Thus, we can take $\varphi$ to be simply the inclusion map. Indeed, $\varphi$ is then clearly bijective and we have
$$\varphi([A,B])=[A,B]=[\varphi(A),\varphi(B)]$$

答: I didn't check all your computations, but the idea is certainly right.  You can reduce the amount of computation by noticing that $X_2$ is just $X_1$ with the coordinates relabeled (the second row and column are switched with the third), and that all of $X_4$ through $X_8$ are likewise obtainable from $X_3$ by suitable relabeling.  So, once you've obtained $X_1$ and $X_3$ as commutators $[A,B]$, you don't have to worry about the rest, as they could be obtained by applying the same relabeling to $A$ and $B$.  (To sound fancy, one can say "conjugation by a permutation matrix" instead of "relabeling", but I find it easier to think in terms of relabeling the coordinates.)</p>


问:
Can you prove that if $\mid D_n \mid = m$, where $m$ is odd, then $n$ is a perfect square.
$n,m \in \mathbb{N}$ and $D_n$ is the set that contains all divisors of $n$, including $1$ and $n$.

My Claim:

if $\mid D_n \mid=$ an odd number $\iff n$ is a perfect square. and indeed Using Ordinal Numbers.the number $$\frac{\mid D_n \mid+1}{2}.$$
in the set $D_n$: is the $\sqrt{n}$.

I didn't found a counterexample so far. if it exists just post it and I will delete the question.

Thank You.

答:
Note that if $d$ is a divisor of $n$, so is $\frac{n}{d}$. Also we have that $d=\frac{n}{d}$ iff $d^2=n$. Hence if $n$ is not a square then all divisors come in pairs of the form $(d,\frac{n}{d})$, so $|D_n|$ is even. On the other hand if $n$ is a square then we get another divisor that is not in such a pair, namely $d=\sqrt{n}$.

Another way to see it is to look at the prime factorization: If $n=p_1^{a_1}\dots p_k^{a_k}$, then $|D_n|=(a_1+1)\dots(a_k+1)$, so $D_n$ is odd iff all $a_i$ are even iff $n$ is a square.


问:
Necessary condition for local minima; non-negative Hessian matrix
The problem I have is the following. Any results on Taylor expansions etc. can be assumed:

Let F : R^n -> R  be a C2 function. Let x_0 be a local minimum of F. Prove that the Hessian matrix of F is non-negative.

Thanks for your help.

答:
In elementary calculus for a function to have a local minimum the following condition must hold around $x_0$
$$f(x_0+\epsilon )\gt f(x_0)\Rightarrow f(x_0+\epsilon )- f(x_0)\gt 0$$
By Taylor's approximation we have
$$f(x_0+\epsilon)\approx f(x_0)+\frac{df}{dx}|_{x_0}\epsilon+\frac 12\frac{d^2f}{dx^2}|_{x_0}\epsilon^2+O(\epsilon^3)$$
We can substitute the approximation of $f(x_0+\epsilon)$ to the definition of minimum
$$f(x_0+\epsilon )- f(x_0)=f(x_0)+\frac{df}{dx}|_{x_0}\epsilon+\frac 12\frac{d^2f}{dx^2}|_{x_0}\epsilon^2-f(x_0)\gt 0$$
$$\frac{df}{dx}|_{x_0}\epsilon+\frac 12\frac{d^2f}{dx^2}|_{x_0}\epsilon^2\gt 0$$
We can generalize it to n variables such as
$$\nabla f(x_0)\vec \epsilon+ \vec \epsilon^T \frac 12\nabla^2 f(x_0)\vec \epsilon\gt 0$$
Since it is given that $\nabla f(x_0)=0$ the condition reduces to
$$\vec \epsilon^T \nabla^2 f(x_0)\vec \epsilon\gt 0$$
Since the term $\vec \epsilon^T (I)\vec \epsilon\gt 0$ is always positive definite for every $\epsilon$ the condition for local minimum reduces to
$$\nabla^2 f(x_0)\gt 0$$</p>


问:
About Zariski closure
Let $A=\left\{\left(a^{2}+1, a^{3}+a\right) \mid a \in \mathbb{R}\right\} \subset \mathbb{R}^{2}$.

Find Zariski closure of $A$ .

I think Zariski closure of $A$ is $\mathbb{R}^{2}$ because let $ u=a^{2}+1$   and $ v= a^{3}+a  $ and let $ f(x,y) \in \mathbb{R}[x,y]$ then $f(u,v) =0$ now we must show $f$ is zero polynomial .

答:
Note that for any $a \in \Bbb R$:

$$(a^2 + 1)^3 - (a^2+1)^2 - (a^3+a)^2 = \\
a^6 + 3a^4 + 3a^2 + 1 - (a^4 + 2a^2 +1) - (a^6+ 2a^4 + a^2) = 0$$ so $A \subseteq \{(x,y)\mid x^3 - x^2 - y^2 = 0\}$ which is a Zariski closed subset by definition.

So the closure is at most $Z(x^3 - x^2- y^2)$, maybe even equal to it, thought I cannot show it yet. It probably is, though.

Title: About Zariski closure


问:
Conditional expectation of a bounded almost sure random variable
Let $Z$ a random variable which is $\mathscr{p}$-measurable  and bounded almost sure, ie, there is a positive number $M$ such that $|Z| \leq M $ a.s.Then show thatfor the conditional expectation $$Y = \mathbb{E}[X|\mathscr{p}]$$ we have :
$$\mathbb{E}[YZ] = \mathbb{E}[XZ]$$

and more general if $Z$ is $\mathscr{p}$-measurable and $$\mathbb{E}[|ZX|]< \infty$$ then the equality to be proven is still true.

答:
We know that for integrable r.v.'s $X$, the defining equation for its conditional expectation $Y=\mathbb{E}[X|\mathscr{G}]$ given some sub-$\sigma$-field $\mathscr{G}$ is
$$\mathbb{E}[X\cdot1_A]=\mathbb{E}[Y\cdot1_A],\ \forall A\in\mathscr{G}\quad(1)$$

Changing to:
$$\mathbb{E}[XZ]=\mathbb{E}[YZ],\ \forall\text{ nonnegative, bounded,} \mathscr{G}\text{-measurable r.v. }Z\quad(2)$$

Now $(2)\implies(1)$ is obvious. For the other direction. My proof is as follows:

Since $Z$ is $\mathscr{G}$-measurable, nonnegative and bounded, we can find an increasing sequence of simple functions $z_n\geq0$ such that $z_n\nearrow Z$ a.s. Moreover, $z_n$ has the form
$$z_n=\sum_{j=1}^{nk}a_j\cdot1_{A_{kj}},\text{ where}\ A_{kj}\in\mathscr{G}$$
Clearly, $\mathbb{E}[X z_k]=\mathbb{E}[Y z_k]$ by $(1)$. Let $k\to\infty$ and by dominated convergence we have $\mathbb{E}[XZ]=\mathbb{E}[YZ]$.


问:
rank$(R\otimes_R S)$ is the composition of $f^∗$ and rank$(P)$.
This is Weibel's K-book exercise 2.12

If $f : R →S $ is a homomorphism of commutative rings, there is a continuous map $f^*: Spec(S) → Spec(R)$ sending $p$ to $f^{-1}(p).$

If $P$ is a finitely
generated projective R-module, show that rank$(P\otimes_R S)$ is the composition
of $f^∗$ and rank$(P)$.

I don't know what's the mean of "composition of $f^∗$ and rank$(P)$."Since $P\otimes_R S$ can be viewed a $S(or\ R)$-module. I guess it means rank$(P\otimes_R S)$ calculated on some $q\in Spec S,$ then pulback to $R$. Right?

If you can tell me the correct meaning, I will thank you very much.

答:
Citation from the exercises (Exercise 2.11): "$M$ is finitely generated,$M_p$ is free for every prime ideal $p$ of $R$, and $rank(M)$ is a continuous function on $Spec(R)$."

Hence

$$rank(-): Spec(R) \rightarrow \mathbb{Z}$$

is a continuous map.

Question: "If you can tell me the correct meaning, I will thank you very much."

Given a map $f:R \rightarrow S$ of commutative rings and corresponding map $f^*: Spec(S) \rightarrow Spec(R)$ of affine schemes,  you get according to Ex.2.11 a composed map

$$ rank(-) \circ f^*: Spec(S) \rightarrow \mathbb{Z}.$$

I have not done the exercise, but it seems to me Weibel wants you to prove the formula

$$F1.\text{   }rank(\tilde{P\otimes_R S})=rank(f^*(\tilde{P}))$$

where $\tilde{P}$ is the sheafification of $P$. Hence it seems to me you must do Exercise 2.11 in the book and verify formula F1.


问:
Preservation of chain conditions
Suppose $\kappa<\lambda$ are regular cardinals, $P$ is a $\kappa$-c.c. poset, and $Q$ is a $\lambda$-c.c. poset.  Does forcing with $P$ preserve the $\lambda$-c.c. of $Q$?

答:
Not in general. In the paper Remarks on cellularity in products, Todorcevic shows that after adding a Cohen (or random) real, there is a ccc forcing $\mathbb P$ so that $\mathbb P\times\mathbb P$ is not $2^\omega$-cc (apparently this generalises earlier work by Fleissner and Roitman). In particular, if one blows up the contiuum to some large regular $\lambda$ by adding Cohen reals then there is a ccc $\mathbb P$ so that after forcing with $\mathbb P$, $\mathbb Q=\mathbb P$ is not $\lambda$-cc.

There are some restrictions though. Using the special case
$$(2^\kappa)^+\rightarrow (\kappa^+)^2_2$$
of the  Erdős-Rado theorem one can show that if $\mathbb P$ and $\mathbb Q$ are $\kappa^+$-cc then $\mathbb P\times\mathbb Q$ is $(2^\kappa)^+$-cc (similar to how one shows that weakly compact cardinals $\mu$ are productive using the partition relation $\mu\rightarrow(\mu)^2_2$). Thus under $\mathrm{GCH}$ there are no counterexample with $\mathbb P=\mathbb Q$ and $\kappa$ a successor cardinal (note that this is the case above): If $\mathbb P$ is $\eta^+=\kappa$-cc then $\mathbb P\times\mathbb P$ is $(2^
\eta)^+=\kappa^+\leq\lambda$-cc.

This also gives a bound on how badly $\lambda$-cc of $\mathbb Q$ can fail in this context: After any $\lambda$-cc forcing $\mathbb Q$ has at least the $(2^\lambda)^+$-cc (and even $(2^\eta)^+$-cc if $\lambda$=$\eta^+$).

It would be interesting to know whether the answer to your question is yes under $\mathrm{GCH}$.

Title: Preservation of chain conditions


问:
Gosper Formula for inv $\pi$, properties.
I need to understand very good how the properties of this formula

$\frac{4}{\pi} = \frac{5}{4} + \sum_{N \geq 1} \left[ 2^{-12N + 1} \times(42N + 5)\times {\binom {2N-1} {N}}^3 \right] $

Taken from the paper "Radian Reduction for Trigonometric Function" (Hanek Payne Algorithm)

Some remarkable properties are stated, specifically these four ones

The $k^{th}$ term of the formula is exactly representable in $6k$ bits;

The first $n$ terms of the sum can be represented exactly in $12n$ bits;

The most significant bit of the $k^{th}$ term has weight at most $2^{1-6k}$ and hence each successive term increases the number of valid bits in the sum by at least $6$;

If $12k < m + 1 \leq 12(k+1)$, then the $m^{th}$ bit of $\frac{4}{\pi}$ may be computed using only terms beyond the $k^{th}$.

My questions are:
1. How to prove the formula?
2. How to prove the properties stated above?

PS. I guess with terms the paper means the generic term $a_N$ of the sum...

答:
The formula in question can be rewritten as $$\frac {1}{\pi}=\sum_{j=0}^{\infty} \binom{2j}{j}^3\frac{42j+5}{2^{12j+4}}\tag{1}$$ It was first given by Ramanujan in his famous paper Modular equations and approximations to $\pi$ in 1914.

Luckily Ramanujan gave a proof of the general theory of such series and left the calculations for reader (except for the simplest of the series).

I have presented the details of Ramanujan's theory alongwith proofs in my blog posts (see part 1, part 2, part 3). Here I will present a brief picture and the details of calculation related to the series in question.

Let $k, l$ (called elliptic moduli) be numbers in interval $(0,1)$ and let $K, L, K', L'$ denote the complete elliptic integrals of first kind $$K(k) =\int_{0}^{\pi}\frac{dx}{\sqrt {1-k^2\sin^2x}},\\ K=K(k), L=K(l),k'=\sqrt {1-k^2},l'=\sqrt{1-l^2},K'=K(k'),L'=K(l')\tag{2}$$ It can proved that the as $K'/K$ is a strictly decreasing function of $k$ and maps $(0,1)$ to $(0,\infty) $. Thus if $n$ is a positive integer then for a given $k\in(0,1)$ there is a unique $l\in(0,1)$ such that $L'/L=nK'/K$. Jacobi also proved that in this case $l$ is an algebraic function of $k$ ie there is a polynomial $P(x, y) $ with integer coefficients such that $P(k, l) =0$ (called modular equation of degree $n$) and further the ratio $K/L$ can also be expressed as an algebraic function of $l, k$.

Let $q=\exp(-\pi K'/K) $ denote the nome corresponding to elliptic modulus $k$ and then $q^n=\exp(-\pi L'/L) $ corresponds to $l$. Ramanujan considered the function $$P(q) =1-24\sum_{j=1}^{\infty} \frac{jq^{2j}}{1-q^{2j}}\tag{2}$$ which is related to Dedekind eta function $$\eta(q) =q^{1/24}\prod_{j=1}^{\infty} (1-q^j)\tag{3}$$ via $$P(q) = 12q\frac{d}{dq}\{\log\eta(q^2)\}\tag{4}$$ and proved that $$nP(q^n) - P(q) =\frac{4KL}{\pi^2}F_n(k,l)\tag{5}$$ where $F_n$ is an algebraic function depending on $n$. Ramanujan gave explicit formulas for $F_n$ for many values of $n$.

Another key identity related to $P(q) $ is $$nP(e^{-\pi\sqrt{n}}) +P(e^{-\pi/\sqrt{n}})=\frac{6\sqrt{n}}{\pi}\tag{6}$$ Using $(5),(6)$ Ramanujan established that $$P(e^{-\pi\sqrt{n}}) - \frac{3}{\pi\sqrt{n}}=\left(\frac{2K}{\pi}\right)^2A_n\tag{7}$$ where $A_n$ is an algebraic number and elliptic integral $K$ corresponds to nome $q=e^{-\pi\sqrt{n}}$.

Next Ramanujan considered the well known formula $$\left(\frac{2K}{\pi}\right) ^2=1+\sum_{j=1}^{\infty} \left(\frac{1\cdot 3\cdot 5\dots(2j-1)}{2\cdot 4\cdot 6\dots(2j)}\right) ^3(2kk')^{2j}=\sum_{j=0}^{\infty} \binom{2j}{j}^32^{-6j}(2kk')^{2j}\tag{8}$$ It is also well known that $$\eta(q^2)=2^{-1/3}\sqrt{\frac {2K}{\pi}}(kk')^{1/6}\tag{9}$$ Using $(8),(9)$  we get $$\eta^4(q^2)=\left(\frac{kk'}{4}\right)^{2/3}\sum_{j=0}^{\infty} \binom{2j}{j}^32^{-6j}(2kk')^{2j}$$ Logarithmic differentiation of the above equation with respect to $k$ gives $$P(q) =(1-2k^2)\sum_{j=0}^{\infty} (3j+1)\binom{2j}{j}^32^{-6j}(2kk')^{2j}\tag{10}$$ Using $(7),(8),(10)$ one can get a series for $1/\pi$.

Let us now do the calculations for $n=7$ and obtain the series in question. We start with the modular equation of degree $7$ given by $$(kl) ^{1/4}+(k'l')^{1/4}=1\tag{11}$$ If $k, l$ correspond to nomes $e^{-\pi\sqrt{7}},e^{-\pi/\sqrt {7}}$ then we have $l=k', k'=l$ and then from $(11)$ we get $kk'=1/16$ or $$k^2(1-k^2)=\frac{1}{256}$$ or $$k^2=\frac{1-\sqrt {1-(1/64)} }{2}=\frac{8-3\sqrt{7}}{16}$$ (we need the smaller root here as the larger root is the value of $k'^2$). Thus from equation $(10)$ we get $$P(e^{-\pi\sqrt {7}})=\frac{3\sqrt{7}}{8}\sum_{j=0}^{\infty}(3j+1)\binom{2j}{j}^32^{-12j}\tag{12}$$ Next Ramanujan derived that $$7P(q^7)-P(q)=\frac{12KL}{\pi}(1+kl+k'l')$$ Putting $q=e^{-\pi/\sqrt{7}}$ we get $$7P(e^{-\pi\sqrt{7}})-P(e^{-\pi/\sqrt {7}})=\frac{27\sqrt {7}}{8}\left(\frac{2K}{\pi}\right)^2\tag{13}$$ where $K$ corresponds to nome $e^{-\pi\sqrt{7}}$. Adding the above to equation $(6)$ (with $n=7$) we get $$P(e^{-\pi\sqrt{7}})=\frac{3}{\pi\sqrt{7}}+\frac{27\sqrt {7}}{112}\left(\frac{2K}{\pi}\right)^2$$ Comparing this equation with $(12)$ and using $(8)$ (with $2kk'=1/8$) we can see that $$\frac{3}{\pi\sqrt{7}}=\frac{3\sqrt {7}}{8}\sum_{j=0}^{\infty} \left(3j+1-\frac{9}{14}\right)\binom{2j}{j}^32^{-12j}$$ or $$\frac{1}{\pi}=\sum_{j=0}^{\infty} \binom{2j}{j}^3\frac{42j+5}{2^{12j+4}}$$


问:
Index / Angle of the Real Root in a Cubic Equation
I am trying to find the roots of a general cubic equation, but with a twist:

In case there is only one real root (the discriminant of the polynomial is less than zero), I want to separate the real root from the pair of complex conjugate roots and treat them differently (replace them by other complex roots specific to the actual problem I am solving).

I also want to stick to complex numbers (as opposed to explicit trigonometry) because I have to factor out all divisions at the end to represent the roots as homogeneous coordinates.

I currently use the approach described on Wikipedia. But because I use floating point arithmetic (not real numbers) it suffers from a bit of numeric instability, which makes it hard to tell which of the three results was supposed to be the real root (in case the curve is really flat and they are all close to the real line).

My current approach selects the root with the smallest absolute imaginary component. However I think there should be a better way to tell them apart before calculating the result, maybe related to the angles of the root of unity, so that I can get the index ($k$) of the real root ($x_k$) sooner.

答:
For the general cubic equation
$\space ax^3+bx^2+cx+d\space $
the cubic formula provides a single real root allowing the other two roots, real or complex, to be solved by the quadratic equation.

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

There will sometimes be complex intermediate results requiring robust software like "D", IDL, Matlab, Octave, and Fortran,  Python, or WolframAlpha to resolve. (I know because Excel could sometimes not handle this formula as shown here). The advantage is that, with sufficient variable manipulation, computations can be kept in rational form for however long the programmer chooses.


问:
If $\{F_n\}$ is a decreasing sequence of compact subsets of $X$ Hausdorff and $f$ is continuous, then $f(\cap F_n) = \cap f(F_n)$
I have to prove the following statement:

Let $X$ and $Y$ be Hausdorff Spaces and $f:X\to Y$ a continuous function. If $\{F_n\}$ is a decreasing sequence of compact subsets of $X$, then $f(\cap F_n) = \cap f(F_n)$.

I already proved it, however, my proof doesn't use the hypothesis that $Y$ is Hausdorff, and that makes me think it might be wrong, but I don't find any mistakes. I would be really thankful if someone could verify it.

Proof. First we see that $f(\cap F_n) \subseteq \cap f(F_n)$. If $x\in f(\cap F_n)$, there must exists $a\in \cap F_n$ such that $f(a)=x$. It is clear that $a\in F_n$, for all $n\in\mathbb{N}$ and in consequence $x\in f(F_n)$, for all $n\in\mathbb{N}$, which implies $x\in\cap f(F_n)$.

Now we are going to prove that $\cap f( F_n) \subseteq f(\cap F_n)$. Let $x\in\cap f( F_n)$, this implies that $x\in f( F_n)$ for all $n\in\mathbb{N}$. From here we get for each $n\in\mathbb{N}$ there exists $y_n\in F_n$ such that $f(y_n)=x$. Since $\{F_n\}$ is a decreasing sequence, we have that $F_n \subseteq F_1$ for all $n\in\mathbb{N}$ and this implies that $y_n\in F_1$ for all $n\in\mathbb{N}$. This way, $\{y_n\}$ is a sequence in $F_1$. Because $F_1$ is compact, $\{y_n\}$ must have a convergent subsequence. Lets say this subsequence is $\{y_{n_k}\}$, and that it converges to $y$. Futhermore, since $X$ is Hausdorff, this limit is unique.

We see that $y\in\cap F_n$. Let $i\in\mathbb{N}$. There exists a $j\in\mathbb{N}$ such that $i<n_j$. Now lets consider the sequence $\{a_k^j\}$ given by $a_k^j=y_{n_{k+j}}$. It is clear that $a_k^j=y_{n_{k+j}}\in F_{n_{k+j}}\subseteq F_{n_{j}}\subseteq F_i$, which implies that the sequence $\{a_k^j\}$ is a sequence in $F_i$. Also, this sequence is just a shift of $\{y_{n_{k}}\}$, and because of this it must also converge to $y$. Because $F_i$ is compact, and $X$ is Hausdorff, $F_i$ is closed, and since $\{a_k^j\}$ is in $F_i$, we have that $y\in F_i$. As $i\in\mathbb{N}$ is arbitrary, we get that $y\in\cap F_n$.

Finally, using the continuity of $f$ and the uniqueness of the limit of $\{y_{n_k}\}$ we get that $f(y)=f(\text{lim}_{k\to\infty}y_{n_k})=\text{lim}_{k\to\infty}f(y_{n_k})=\text{lim}_{k\to\infty}x=x$, from where we get that $x=f(y)\in f(\cap F_n)$.

In conclusion, $f(\cap F_n) = \cap f(F_n)$

答:
For the non-trivial backward direction I'd do a sequence-free proof because not all compact Hausdorff spaces are sequentially compact.

So let $y \in \bigcap_n f[F_n]$. And define $C_n = f^{-1}[\{y\}] \cap F_n$ which is a non-empty closed subset of $X$ (for every $n$, some $x \in F_n$ maps onto $y$ as $y \in f[F_n]$, closed as singletons are closed in $Y$ and $f$ is continuous.)

As the $F_n$ are decreasing so are the $C_n$ ($x \in C_{n+1}$ then $f(x)=y$ and $x \in F_{n+1}$, so $x \in F_n$ and still $f(x)=y$ so $x \in C_n$).

So the $C_n$ form a family of closed subsets with the FIP inside the compact set $F_1$, so there is a point $x \in \bigcap_n C_n$.

Clearly $x \in \bigcap_n F_n$ too and for any $n$, $x \in C_n$ so that $f(x)=y$ and so $x$ witnesses that $y \in f[\bigcap_n F_n]$ as required.

So it seems we can get ayway with $Y$ being merely $T_1$ instead of Hausdorff.


问:
Terminologie of y1/y0
If I set a list of points: {2,3,5}

"Derivative" will be {3-2=1, 5-3=2}

How about {3/2=1.5, 5/3~=1.666}.  What is the name of this ?

答:
In math, we usually use curly braces to denote an unordered set, so that $\{2,3,5\}$ is not a list. But $(2,3,5)$ would be fine for a list/"finite sequence".

$(1,2)$ is not the "derivative", but "the (sequence of) (first) differences (between adjacent terms)". (That said, there are connections between finite differences and derivatives.)

I don't recall seeing a calculation like $(\frac32,\frac53)$ before. But I might call it "the list of quotients" if I could easily give an example to explain, or something like "the sequence of quotients of adjacent terms" if I wanted to be very precise in words alone.

Title: Terminologie of y1/y0


问:
Find what values of $n$ give $\varphi(n) = 10$

For what values of $n$ do we get $\varphi(n) = 10$?

Here, $\varphi$ is the Euler Totient Function. Now, just by looking at it, I can see that this happens when $n = 11$. Also, my friend told me that it happens when $n = 22$, but both of these are lucky guesses, or educated guesses. I haven't actually worked it out, and I don't know if there are any more.

How would I go about answering this question?

答:
If a number $n$ has prime factorization $p_1^{x_1} p_2^{x_2} p_3^{x_3} \cdots $, then the following formula holds:
$$
\varphi(n) = \left( p_1^{x_1} - p_1^{x_1 - 1}\right)\left( p_2^{x_2} - p_2^{x_2 - 1}\right) \left( p_3^{x_3} - p_3^{x_3 - 1}\right) \cdots
$$</p>


问:
Understanding L1 and L2 norms
I am not a mathematics student but somehow have to know about L1 and L2 norms. I am looking for some appropriate sources to learn these things and know they work and what are their differences. I am asking this question since there are lots of stuff on this matter in the internet and I am looking for a simple applicable one to understand this.

To add more, I want to understand why/how/when these norms are used in optimization problems? And maybe, how these cost functions are solved.

答:
Just the norm? The norm acts on a vector and any norm has to satisfy three properties. Given a vector space V over a field F, the norm must

$\|x\|\ge0$ for all x and $x=0\Leftrightarrow\|x\|=0$

$\|ax\|=|a|\|x\|$ for all $a\in\mathbb F, x\in V$

$\|x+y\|\le \|x\|+\|y\|$ (triangle inequality)

You ask about the L1 and L2 norms. The L1 norm is the sum of the absolute value of the entries in the vector. The L2 norm is the square root of the sum of the entries of the vector. In general, the Lp norm is the pth root of the sum of the entries of the vector raised to the pth power.

$$\begin{split}\|x\|_1&=\sum |x_i|\\
\|x\|_2&=\sqrt{\sum x_i^2}\\
\|x\|_p&=(\sum |x_i|^p)^{1/p}\end{split}$$

It is fairly common to square the l2 norm to get $\|x\|^2_2=\sum x_i^2$

For more use Wikipedia and go to the examples section.

Title: Understanding L1 and L2 norms


问:
Number of possible ways to go from point $A$ to $B$ using three types of train?
We have three trains to choose from.

Train $1$ - Stops at every station.

Train $2$ - Stops at every alternative station.

Train $3$ - stops at every third station.

Now we have to go from point $A$ to $B$ using any of the three trains but we cannot go backwards.

How many ways are there to go from point $A$ to $B$?

答:
Based on OP's responses to my comments:

Assume that each station is assigned an index number, and that the difference between the index numbers assigned to stations $B$ and $A$ is $n > 0.$

As suggested in one of my comments, let

$$c = \left\lfloor \frac{n}{3}\right\rfloor.$$

Then, clearly, the trip will involve $a$ Train-3 trips, where

$$a \in \{0,1,2,\cdots, c\}.$$

Here, I am construing (for example) the Train-3 trip from 1 to 4 to 7 as 2 separate Train-3 trips.

For each $a \in \{0,1,2\cdots,c\}$, let

$$B_a = \left\lfloor \frac{n-3a}{2}\right\rfloor.$$

Then, for each $a$, there can be $b$ Train-2 trips, where

$$b \in \{0,1,2,\cdots, B_a\}.$$

For a given combination of $a$ Train-3 trips and $b$ Train-2 trips, there will be $(n - 3a - 2b)$ Train-1 trips, so there will be a total of

$(n-3a-2b) + a + b  = (n-2a-b)$ trips.

These $(n-2a-b)$ trips can be arranged in $\binom{n-2a-b}{a} \times \binom{[n-2a-b]-a}{b}$ ways.

Let $f(a,b)$ denote $(n-2a-b).$

Then the number of possible arrangements of train trips is

$$\sum_{a=0}^c \sum_{b=0}^{B_a} \left[\binom{f(a,b)}{a} \times \binom{f(a,b)-a}{b}\right].$$


问:
Geometric Interpretation of Linear Regression
I was reading a tutorial written on Linear Regression by Avi Kak (https://engineering.purdue.edu/kak/Tutorials/RegressionTree.pdf). There is a part about geometric interpretation of linear regression on pg.19.

The optimum solution for β~ that minimizes the cost function C(β~) in
Eq. (14) possesses the following geometrical interpretation: Focusing
on the equation ~y = Xβ~, the measured vector ~y on the left resides
in a large N dimensional space. On the other hand, as we vary β~ in
our search for the best possible solution, the space spanned by the
product Xβ~ will be a (p+1)-dimensional subspace (a hyperplane,
really) in the N dimensional space in which ~y resides. The question
now is: which point in the hyperplane spanned by Xβ~ is the best
approximation to the point ~y which is outside the hyperplane. For any
selected value for β~, the “error” vector ~y − Xβ~ will go from the
tip of the vector Xβ~ to the tip of the ~y vector. Minimization of the
cost function C in Eq. (14) amounts to minimizing the norm of this
difference vector.

I could not understand how to relate N-dimensional space and (p+1)-dimensional subspace. B vector defines a (p+1) dimensional subspace but I could not understand why N dimensional space contains the (p+1) subspace. As I understand in (p+1) each dimension means features but in N dimensional space each dimension means a data point. I'm a lot confused about the idea. Are there any other resource that explains the idea in a much more detail? or Could anyone explains the idea how these spaces relate?

答:
The matrix $X$ is an $N\times (p+1)$ matrix. Its row space is at most of dimension $N$ and its column space is at most of dimension $p+1$. In your notes, it is assumed that $N>p+1$ and the columns of $X$ are linearly independent. So the rank of $X$ is $p+1$. A theorem in linear algebra says that the dimension of the row space and that of the column space are the same. So the row space is also of dimension $p+1$.

By matrix multiplication, the vector $y=X\beta$ is an $N\times 1$ matrix. This is why they say $y$ is in a large $N$ dimensional space (since there are $N$ rows).

On the other hand, $y=X\beta$ implies that the vector $y$ is a linear combination of the column vectors of $X$, and hence in the column space of $X$, which is of dimension $p+1$.

Here is an example.

Let
$$
X= \begin{pmatrix}
1 & 0 & 1\\
0 & 1 & 1\\
0 & 0 & 1\\
1 & 1 & 1
\end{pmatrix}
$$
$N=4$ and $p=2$.

On the one hand, for any $\beta=(b_1,\cdots,b_4)^T$, $y=X\beta$ is a vector in a large space $\mathbb{R}^4$, which is of dimension $N=4$, it is also a vector in a subspace (of $\mathbb{R}^4$) of dimentions $p+1=3$, namely, the space span by the column vectors of $X$.


答:
Not sure about my solution, but here's what I thought:

As others already mentioned, if we set $A(k) = \{(a^{2} + 1, a^{3} + a) | a \in k\}$ and $B(k) = \{(x, y)| y^{2} - x^{3} -x^{2} =0, (x, y) \in k^{2}\}$ for a field $k$, then we have $A(\mathbb{R}) \subset B(\mathbb{R})$. Unfortunately, two sets aren't the same since $(0, 0) \in B(\mathbb{R}) \backslash A(\mathbb{R})$. However, we can show $A(\mathbb{C}) = B(\mathbb{C})$: if $(x, y) \in B(\mathbb{C})$, then for $x\neq 0$, we have $(x, y) = (t^{2} + 1, t^{3}+ t)$ when $t = y/x$. For $x =0$, we can treat some cases by hands. What this shows is that the Zariski closure of $A(\mathbb{C})$ in $\mathbb{A}_{\mathbb{C}}^{2}$ is indeed $Z(y^{2} - x^{3} - x^{2})$. From this, we obtain that (which is the part that I'm not convinced enough) any polynomial $f(x, y) \in \mathbb{C}[x, y]$ with $f(t^{2} +1, t^{3} + t)=0$ for all $t\in \mathbb{C}$ can be divided by $y^{2} - x^{3} - x^{2}$, i.e. $f(x, y) = (y^{2}- x^{3} - x^{2})g(x, y)$ for some $g \in \mathbb{C}[x, y]$. *Now if $f(x, y) \in \mathbb{R}[x, y]$, then $f = \bar{f}$ (take conjugation of all coefficients) gives $g = \bar{g}$ and so $g \in \mathbb{R}[x, y]$, which concludes that the similar statement for real polynomial is also true.

Title: About Zariski closure

Body:
Let $A=\left\{\left(a^{2}+1, a^{3}+a\right) \mid a \in \mathbb{R}\right\} \subset \mathbb{R}^{2}$.

Find Zariski closure of $A$ .

I think Zariski closure of $A$ is $\mathbb{R}^{2}$ because let $ u=a^{2}+1$   and $ v= a^{3}+a  $ and let $ f(x,y) \in \mathbb{R}[x,y]$ then $f(u,v) =0$ now we must show $f$ is zero polynomial .


问:
In whiche case do we have $\left(\sum_{i=1}^n x_iy_{i}\right)^2 = \sum_{i=1}^n (x_i)^2 \sum_{i=1}^n (y_{i})^2$?
The exercice asked me first to prove that $( \sum_i x_i y_i )^2 \leq (\sum_i x_i^2)(\sum_i y_i^2)$ ( Cauchy inequality) and i managed to prove but then it asked me in which case we have the equality ($\left(\sum_{i=1}^n x_iy_{i}\right)^2 = \sum_{i=1}^n (x_i)^2 \sum_{i=1}^n (y_{i})^2$) but i don't Know how ????

答: Equality is attained in the Cauchy-Schwartz inequality $\vert \langle x, y \rangle \vert \leq \vert \vert x \vert \vert \cdot \vert \vert y \vert \vert$ if and only if $x$ and $y$ are linearly dependent and this holds in any space which admits an inner product.


问:
What is the remainder when $4^{96}$ is divided by 6
What is the remainder when $4^{96}$ is divided by 6

My approach :
When 4 is divided by 6 gives 4 remainder, when $4^2$ is divided by 6 gives remainder 4 ,... is it for the complete series.. that means when $4^{96}$ is divided by 6 gives 4 as remainder or is there any short cut or other answer of this question.

Please suggest if there is any sort of short cut to such problems.

Thanks.

答:
As was mentioned, since Euler is not directly applicable, you can use CRT.  Putting together that $4^{96}\equiv0\bmod2$ and $4^{96}\equiv 1\bmod3$, the latter by Fermat's little theorem, or the binomial theorem, we have using Bezout that $4^{96}\equiv -2\cdot1+0\cdot3=\equiv-2\equiv4\bmod6$.

As it turns out, CRT is not necessary here, as @BillDubuque points out.


问:
Figuring the volume of a partially filled cone without the radius of the material inside the cone
Say I have a cone. For simplicities sake, this cone is at the bottom of a storage silo. Is has a flat bottom, flat top and angled sides. I know the height of the material inside the cone but don't know the angle of the cone nor do I know exactly how to figure that out. Ultimately I need to figure out the formula used to calculate the occupied volume of this cone (I think it's referred to as a frustum?). I'm positive some trig is involved but my days of high school Pre-calc with Trig are gone away (and just when I needed them, shucks).

```<code>|<------ D ------>|
-------------------  =================
\                /              ^
 \              /               |
  \ |<-- r ->| /                |
   \==========/      ========   H
    \xxxxxxxx/         h ^      |
     \xxxxxx/            V      V
      ------         =================
     |< d >|
</code>```

I know H.
I know h.
I know D.
I know d.
I do NOT know r and I do not know the occupied volume (the x's above). My boss has tasked me with designing a spreadsheet that you punch in the values known and it outputs the volume. Can anyone assist?

Thanks!

答:
Do either of you now have a spreadsheet to do these calculations?  Seems a LOT of time to spend to reinvent the wheel.

Actually, this is a variation on the theme.  We're mixing 2 different liquids, at 2:1 ratio, and all we have is 5-gal buckets.  It's 13" tall, with radius at the bottom (Rb) = 5", and radius at the top (Rt) = 5.75".

The question becomes:  In order to keep the 2:1 as precise as possible, and because this gets used several times/day for significantly varying volumes, the shop wants a table  that says (effectively), "If I pour in ingredient A to the 3 1/4" line, to what ht. do I pour in ingredient B to have a precise 2:1 mix?

If is was a true cylinder, this is a no-brainer, but with that slope...

Normally, I'd just slog through and build my own, but I'm on a bit of a time crunch here.

Mark


答:
You could proceed this way: $6 = 2\cdot 3$.  When you divide $4^{96}$ by $2$ the remainder is $0$.  To divide by $3$, note that $4=3+1$ so that
$4^{96} = (3+1)^{96}$ which expands to a multiple of $3$ plus $1$, so the remainder is one.

Now you need a number which is $1$ more than a multiple of $3$ and divisible by $2$ and less than $6$.  That would be $4$.

Body:
What is the remainder when $4^{96}$ is divided by 6

My approach :
When 4 is divided by 6 gives 4 remainder, when $4^2$ is divided by 6 gives remainder 4 ,... is it for the complete series.. that means when $4^{96}$ is divided by 6 gives 4 as remainder or is there any short cut or other answer of this question.

Please suggest if there is any sort of short cut to such problems.

Thanks.


问:
Inequality : $ab^2+bc^2+cd^2+da^2 \leq 4$
I have a excercise like that: let $a;b;c \geq 0$ and $a+b+c=3$, prove: $ab^2+bc^2+ca^2 \leq 4$

We can assume $b=mid\{a;b;c\}$, then $(a-b)(b-c) \geq 0$, by that way, the problem is solved.

But I have a question: what should I do if the problem is for 4 variables, eg:

Let $a;b;c;d \geq 0$ and $a+b+c+d=3$, prove: $ab^2+bc^2+cd^2+da^2 \leq 4$

Thank you.

答:
We have

If $x,\,y,\,z$ are non-negative real numers satisfy $x+y+z=3,$ then
$$xy^2+yz^2+zx^2\leqslant 4.$$
Now, suppose $b = \max \{a,\,b,\,c,\,d\}.$ From this inequality we put $x=a,\,y=b,\,z=c+d,$ we get
$$a^2(c+d)+ab^2+b(c+d)^2 \leqslant 4.$$
But
$$a^2(c+d)+ab^2+b(c+d)^2$$
$$= ab^2+bc^2+cd^2+da^2 + c(a^2+2bd)+d^2(b-c)$$
$$ \geqslant ab^2+bc^2+cd^2+da^2.$$
Therefore
$$ab^2+bc^2+cd^2+da^2 \leqslant 4.$$
The proof is completed.


问:
Does $G/H\cong \mathbb{Z}^n$ with $H\cong \mathbb{Z}^m$ imply that $G\cong \mathbb{Z}^{m+n}$?
Let $G$ be a finitely generated group and $H\subseteq G$ a normal subgroup such that $H\cong \mathbb{Z}^m$ holds for some positive integer $m$. Suppose that we know that there exist a non-negative integer $n$ such that $G/H\cong \mathbb{Z}^n$ holds. Do we then necessarily have $G\cong \mathbb{Z}^{m+n}$? If it helps, we may assume that $G$ is Abelian (but I'm not certain if that will help).

I couldn't really find a counterexample to this, but I'm not particularly close to a proof either. I tried to come up with an isomorphism, but how do we even construct a homomorphism from G to $\mathbb{Z}^{m+n}$ that is promising? I was also thinking about whether or not the claim that $G$ is finitely generated is necessary, but this also didn't get me any further. Can anyone help me with this problem?

答:
Yes.  The case $G$ abelian follows immediately from structure theorem for finitely-generated abelian group.  Or if you want to be fancy, you can indeed drop the assumption $G$ or $H$ being finitely generated and appeal to the result

Fact: $\mathbb{Z}^n$ is a projective $\mathbb{Z}$-module

to get a splitting of $0\to H\to G\to\mathbb{Z}^n\to 0$.

If $G$ is not abelian, the discrete Heisenberg group
$$
H_3(\mathbb{Z}):=\left\{\begin{pmatrix}1&a&c\\0&1&b\\0&0&1\end{pmatrix}: a,b,c\in\mathbb{Z}\right\}
$$
provides a counterexample as @jMdA commented above.


问:
Generalized nontransitive dice
Let $X_1, \ldots, X_n$ be a collection of random variables. Consider the directed graph with vertex set $\{ 1, 2, \ldots, n \}$ where there is a directed edge $i \to j$ if $\mathbb{P}(X_i > X_j) > \frac{1}{2}$.

Question 1: What directed graphs can arise in this way? Certainly they must be simple and have no loops. Is that the only restriction? Alternatively, $\mathbb{P}(X_i > X_j) > \frac{1}{2}$ defines an irreflexive antisymmetric relation on $\{ 1, 2, ... n \}$. Which such relations arise in this way?

Question 2: Does the answer change if we require the $X_i$ to all be defined on a finite sample space?

Question 3: What if we require the $X_i$ to be independent?

It is known (see nontransitive dice) that this graph can have directed cycles even if the $X_i$ are independent; in particular, the corresponding relation need not be transitive.

答:
joriki has already presented a solution to Questions 1 and 2; but maybe the following alternate solution has some insight to provide. I'll describe some $n$-tuples of random variables, as sets of vectors in $\mathbb R^n$, with the vectors selected uniformly from that set. For example, with $n=3$, the three random variables $X_1,X_2,X_3$ could be given by the set of vectors $\{(0,1,2), (1,2,0), (2,0,1)\}$ (as in Michael Hardy's comment). Then each $X_i$ individually would be uniformly chosen from $\{0,1,2\}$, but the $X_i$ are far from independent. Notationally, let $e_j = \{0,\dots,0 ,1,0,\dots,0\}$ denote the $j$th standard basis vector in $\mathbb R^n$. Let $f=(1,2,\dots,n)$ and $b=(n,n-1,\dots,1)$.

To begin, consider the following set $U_n = \{u_{ij}\colon 1\le i,j \le n,\, i\ne j\}$ of $n(n-1)$ vectors in $\mathbb R^n$, indexed by ordered pairs of distinct integers from $1$ to $n$, where:
$$
u_{ij} = ne_i + ne_j + \begin{cases}
f, &\text{if } i<j, \\ b, &\text{if } i>j.
\end{cases}
$$
For example, with $n=7$, $u_{52} = (0,0,0,0,7,0,0) + (0,7,0,0,0,0,0) + (7,6,5,4,3,2,1) = (7,13,5,4,10,2,1)$, while $u_{25} = (1,8,3,4,11,6,7)$. Note that for each unordered pair ${i,j}$, there are exactly 2 elements of $U_n$, namely $u_{ij}$ and $u_{ji}$, for which the $i$th and $j$th coordinates are the two largest (each one gets to be largest once). Note also that if $i<j$, then the sign of the $i$th coordinate of $u_{kl}$ minus the $j$th coordinate of $u_{kl}$ always equals the sign of $k-l$; in other words, the random variables $X_1,\dots,X_n$ defined by the set $U_n$ have the property that $\mathbb P(X_i>X_j) = \frac12$ for all $i,j$.

Now we modify the set $U_n$ to account for a given directed graph $G$. Let $V_n = \{v_{ij}\colon 1\le i,j \le n,\, i\ne j\}$, where

$$
v_{ij} = \begin{cases}
u_{ij}+ne_i, &\text{if there's a directed edge from $i$ to $j$ in $G$,} \\
u_{ij}+ne_j, &\text{if there's a directed edge from $j$ to $i$ in $G$,} \\
u_{ij}, &\text{otherwise.}
\end{cases}
$$
For example, with $n=7$, suppose that there is a directed edge from 5 to 2 in $G$. Then $v_{52} = (7,13,5,4,17,2,1)$, while $v_{25} = (1,8,3,4,18,6,7)$. With this modification, it's easy to see (after grokking the notation) that
$$
\mathbb P(X_i>X_j) = \begin{cases}
\frac12+\frac1{n(n-1)}, &\text{if there's a directed edge from $i$ to $j$ in $G$,} \\
\frac12-\frac1{n(n-1)}, &\text{if there's a directed edge from $j$ to $i$ in $G$,} \\
\frac12, &\text{otherwise.}
\end{cases}
$$

Title: Generalized nontransitive dice


问:
Computing limit involving integration (application of FTC)
The question asks:

Let $f(x)$ be continuous with domain $\mathbb{R}$. Compute the following limit:
$$\lim_{x \to 0} \int_{0}^{x} \frac{(x-t)f(t)}{x^2} \,dt$$

I sense I have to use Fundamental Theorem of Calculus somehow but I don't know how. Could anyone help? Thank you so much.

答:
As $x^2$ doesn't depend on $t$ I take it out of the integral
$$\lim_{x\to 0}\frac{\int_0^x f(t) (x-t) \, dt}{x^2}$$
Apply twice  L'Hopital rule
$$\lim_{x\to 0}\frac{\int_0^x f(t) \, dt}{2x}=\lim_{x\to 0}\frac{f(x)}{2}=\frac{f(0}{2}$$


问:
Help understanding the proof: $\forall k \in \mathbb N [(a_1+...+a_k)+a_{k+1}=a_1+...+a_{k+1}]$
In Spivak's Calculus, one is asked to prove that $(a_1+...+a_k)+a_{k+1}=a_1+...+a_{k+1}$

I will write his proof (from the solution manual) verbatim, but then I have some pedantic questions at the end, which are the primary purpose of this post.

The only extra piece of information Spivak provides is:

$\text{P1}$. $a+(b+c) = (a+b)+c \quad$ Associative Law for Addition

and the following definition:

$a_1+a_2+a_3+...+a_{n-2}+a_{n-1}+a_n:= a_1+(a_2+(a_3+...+(a_{n-2}+(a_{n-1}+(a_n)))...)$

However, it seems to me that he should have also mentioned that his bracket symbols $[$ and $]$ are presumably meant to represent different layers of the parentheses symbols $($ and $)$.

Base Case:

$a_1+a_2 = a_1 + a_2 $

Assume:

$(a_1+...+a_k) + a_{k+1}=a_1+...+a_k+a_{k+1}$

Prove the $k+1$ case: $(a_1+...+a_k + a_{k+1})+a_{k+2}=a_1+...+a_k+a_{k+1}+a_{k+2}$

\begin{align}
\text{ If the equation holds for } k \text{ then } \\
(a_1+...+a_k + a_{k+1})+a_{k+2}&=[(a_1+...+a_k)+a_{k+1}]+a_{k+2} \\
& \quad \quad \text{ by P1} \\
&=(a_1+...+a_k)+(a_{k+1}+a_{k+2}) \\
&\quad \quad \text{since the equation holds for } k \\
&=a_1+...+a_k+(a_{k+1}+a_{k+2}) \\
&\quad \quad \text{by the definition of } a_1+...+a_{k+2}\\
&=a_1+...+a_{k+2} \\
\end{align}

First Question

I do not understand Spivak's base case. Shouldn't his base case start as $(a_1)+a_2$? It seems to me the only way of going beyond this sentence is to first make the definition that $(x):=x$. However, in making this definition, doesn't the base case already solve everything? Because any finite expression of $k$ terms can be lumped together, called $x$, and then you'd have $(x)+a_{k+1}=x+a_{k+1}$...which is the whole point of this proof.

Edit: A work around for this seems to be to dictate that this proof only operate on $k\geq 2$. Because if so:

We could say $(a_1+a_2) +a_3=a_1+a_2+a_3$ because by definition,  $a_1+a_2+a_3=a_1+(a_2+a_3)$ and then we can invoke the associative property to PROVE .$(a_1+a_2) +a_3=a_1+a_2+a_3$.

Second Question

The offending transition that has confused me is when Spivak writes:
\begin{align}
&=(a_1+...+a_k)+(a_{k+1}+a_{k+2}) \\
&\quad \quad \text{since the equation holds for } k \\
&=a_1+...+a_k+(a_{k+1}+a_{k+2}) \\
\end{align}

Now, I imagine what Spivak is trying to convey is that we can think of $a_{k+1}+a_{k+2}$ as being a single term...say $\alpha$...and that our assumption in our Assume step can be reframed as $(a_1+...+a_k) + \alpha=a_1+...+a_k+\alpha$.

From what I've learned in set theory, $a$ is actually a function with a domain of some subset in $\mathbb N$ and, in this context, a range that is a subset of $\mathbb R$. Given this, I do not see why at all our initial assumption case of $(a_1+...+a_k) + a_{k+1}=a_1+...+a_k+a_{k+1}$ can be applied to $(a_1+...+a_k)+(a_{k+1}+a_{k+2})$. In general, the quantity $(a_{k+1}+a_{k+2}) \neq a_{k+1}$.

So how exactly is this step actually being applied?

Edit: The only thing I can think to do here is invoke ordinals; I will explain. Originally, the formal definition of the proof I thought was: $\forall k \in \mathbb N [(a_1+...+a_k)+a_{k+1}=a_1+...+a_{k+1}]$. However, to get around the issue of $(a_{k+1}+a_{k+2}) \neq a_{k+1}$, it seems like I should instead write the proof as:

$\forall k \in \mathbb N \Big [ \forall a \Big( [\text{func}(a) \land \text{dom}(a) \in \omega \land \text{ran}(a) \subseteq \mathbb R] \rightarrow (a_1+...+a_k)+a_{k+1}=a_1+...+a_{k+1} \Big) \Big]$

In doing this, when I arrive at my assumption case, I assume for some $k$ that:

$\forall a \Big( [\text{func}(a) \land \text{dom}(a) \in \omega \land \text{ran}(a) \subseteq \mathbb R] \rightarrow (a_1+...+a_k)+a_{k+1}=a_1+...+a_{k+1} \Big)$

WITH THIS, when we arrive at the situation of:

\begin{align}
&=(a_1+...+a_k)+(a_{k+1}+a_{k+2}) \\
&\quad \quad \text{since the equation holds for } k \\
&=a_1+...+a_k+(a_{k+1}+a_{k+2}) \\
\end{align}

we can use our updated formula to tackle this. Specifically, because $\forall a \Big( [\text{func}(a) \land \text{dom}(a) \in \omega \land \text{ran}(a) \subseteq \mathbb R] \rightarrow (a_1+...+a_k)+a_{k+1}=a_1+...+a_{k+1} \Big)$, there definitely exists one function, call it $a'$, that has the first $k$ terms in common with $a$ but, as its $k+1$ term, has $a_{k+1} + a_{k+2}$. That is to say: $a'_{k+1}=a_{k+1} + a_{k+2}$. Then we can absolutely apply our assumption to produce: $a_1+...+a_k+(a_{k+1}+a_{k+2})$

答:
Your question fails to mention an important part of the statement of the problem: Spivak defines $a_1+a_2+\cdots+a_n$ as$$a_1+\bigl(a_2+\bigl(+a_3+\cdots+\bigl(a_{n-2}+(a_{n-1}+a_n)\bigr)\bigr)\cdots\bigr).$$
First question: If $k=2$, this simply means $a_1+a_2$. So, the case $k=2$ is simply to prove that $a_1+a_2=a_1+a_2$, which is trivial. And there is no need to write $(a_1)+a_2$ since the goal of the parentheses is to tell us the order by which the sum should be done and there is no sum inside $(a_1)$.
Second question: The equality$$(a_1+\cdots+a_k)+(a_{k+1}+a_{k+2})=a_1+\cdots+a_k+(a_{k+1}+a_{k+2})$$holds because of the reason that you mentioned: Spivak, at this point, is usin the induction hypothesis and he is dealing with $a_{k+1}+a_{k+2}$ as a single number. And he is doing this because it is a single number. There is nothing wrong with that.


问:
Solving 2 simultaneous equation
Find the value of x and y .The complete question can be found below.

```<code>Find the value of x and y which satisfy the simultaneous equation if:

8ˣ/128 = 1/4ʸ
And
49ˣ/7ʸ = 1
</code>```

so I tried to "eliminate" variable (1) by substituting that variable into (2) with multiplying (1) with 2 and (2) with 3

and I got these
(-4y X 14 = 6x) - (3y = 6x) =  -7y X (-14-3y) = 0
and y = 0

Can you tell me what is my mistake and how to solve this equation?

答: For the second equation you can write $49^x$ as $7^{2x}$, and then $\frac{7^{2x}}{7^y}=1$. Now rewrite this equality as $7^{2x-y}=1$. What should $2x-y$ be equal to so that the equation is solved? You can do the same for the first equation , write everything as a power of $2$ , then you get two equations for $x$ and $y$. I assume you can solve it yourself from here.

Title: Solving 2 simultaneous equation


问:
Prove that 2 inner products are related like this when they are equal for orthogonal vectors
I'm working through Axler's "Linear Algebra Done Right", but I'm stuck on Exercise 6.B Q11:

Suppose $<.,.>_1$ and $<.,.>_2$ are inner products on $V$ such that $_1 = 0 \iff _2 = 0$ Prove that $<v,w>_1 = c<v,w>_2$ for some fixed $c>0$ for any vectors v and w.

I've been able to simplify the problem as so:

Let $k$ be a scalar such that $u-kv$ is orthogonal to $v$, so we have

$$ \frac{_1}{_2} = \frac{<kv,v>_1 + _1}{<kv,v>_2 + _2}$$
$$ = \frac{<v,v>_1}{<v,v>_2}$$

So it would suffice to show that $||v||_1 = c||v||_2$. However I'm stuck on how to show this.

答:
Thanks to Gram-Schmidt you can take a basis $\{v_1,...,v_n\}$ which is orthonormal with respect to the first inner product. Now, thanks to orthonormality, you have:
$$<v_i+v_j,v_i-v_j>_1=<v_i,v_i>_1-<v_j,v_j>_1=1-1=0$$
Since $<.,.>_2$ has the same orthogonal vectors $v_i+v_j,v_i-v_j$ are also orthogonal w.r.t $<.,.>_2$. So:
$$0=<v_i+v_j,v_i-v_j>_2=<v_i,v_i>_2-<v_j,v_j>_2 \Rightarrow <v_i,v_i>_2=<v_j,v_j>_2=c$$
So, $\{\frac{v_1}{c},...,\frac{v_n}{c}\}$ is an orthonormal basis w.r.t the second inner product and, by linearity, for every vectors $v$ and $w$ you have $<v,w>_1=c<v,w>_2$


问:
Making tacit assumptions in solve for x problems explicit
When proving an identity such as
$$\frac{a}{c} + \frac{b}{c} = \frac{a + b}{c}\text{,}$$
the full statement being proven is actually the more pedantic
$$\frac{a}{c} + \frac{b}{c} = \frac{a + b}{c} \text{ if } c \neq 0\text{.}$$

Since I am permitted to assume the condition $c \neq 0$, I can multiply every term by $c$ and prove the identity.

I can also check this more pedantic version at $c = 0$. The condition becomes false, which makes the identity true without having to evaluate division by zero in the consequent.

I'm looking for an analog for solve for $x$ style problems. That is, the pedantic version of solve for $x$ style problems which makes explicit what to do with division by zero. Take the equation
$$\frac{6}{x} = 3$$
who's only solution is understood to be $x=2$.

I can't multiply by $x$ to solve it because I don't know $x \neq 0$. Using the idea,  "when solving for $x$, values which make a denominator zero are not solutions", I can write the problem as the more pedantic system of equations
$$\cases{\frac{6}{x} = 3 \\ x \neq 0\text{.}}$$

Now while working on $\frac{6}{x} = 3$ I can assume $x \neq 0$ and multiply both sides by $x$ to solve.

This version can also be checked at $x = 0$. One of the equations in the system of equations becomes false, which makes the entire statement false without having to evaluate division by zero.

Can I set up problems this way? I'm not suggesting $x \neq 0$ can be added based on some rule of algebra. I'm suggesting it can be added as a way to formally set up a solve for $x$ problem in a pedantic style which can then be solved using algebra.

The reason I am asking this is I've seen the concept of tacit conditions for identities in math books, but I've never seen the analogous idea of tacit requirements for solve for $x$ problems. Using it helps me understand what's going on, but I want to make sure it's right.

答:
Assuming you are talking about real numbers, the statement $\frac{a}{c}+\frac{b}{c}=\frac{a+b}{c}$ is meaningless for $c=0$. It is not that when you prove it you say "if $c\ne0$" but that the statement per se includes such condition. If you want to prove such identity from the first principle, you need to define first what fraction and addition mean.

Exercises in algebra-precalculus usually assume functions with natural domains. So equations such as
$$
\frac{6}{x}=3
$$
are implicitly assuming that both sides are meaningful and thus the equation is actually
$$
\frac{6}{x}=3,\quad (x\ne 0).
$$

To see one more example, it is implicitly assumed in the following equation that $x>2$:

Solve the equation $\frac{1}{\sqrt{x-2}}=4$


答:
Just to give a different approach, note that ${n\choose k}$ is even for $1\le k\le n-1$ when $n$ is a power of $2$ (i.e., Pascal's triangle mod $2$ looks like Sierpinski's triangle). Consequently

$$4^{32}=(3+1)^{32}\equiv3^{32}+1\equiv3+1=4\mod 6$$

and thus $4^{96}\equiv4^3=64\equiv4$ mod $6$.

Title: What is the remainder when $4^{96}$ is divided by 6


问:
Lipschitz constant estimation of continuous selection of upper hemicontinuous multivalued function
I'm reading a book, Set-Valued Analysis, written by Aubin.

I'd like to estimate the Lipschitz constant of the continuous selection of upper hemicontinuous multivalued function by Theorem 9.2.1 of the book.
It says that if $F:X \rightarrow Y$ is an upper hemicontinuous multivalued function from a compact metric space $X$ to a Banach space $Y$ and the valued of $F$ are nonempty and convex, then for every $\epsilon > 0$, there exists a locally Lipschitz single valued map $f_{\epsilon}: X \rightarrow Y$ such that $\text{Graph}(f_{\epsilon}) \subset B(\text{Graph}(F), \epsilon)$.

Now I consider the case when $Y$ is also compact, which implies $f_{\epsilon}$ is Lipschitz continuous on $X$.
I wonder if we can estimate the Lipschitz constant of such selection ($f_{\epsilon}$) w.r.t. $\epsilon$.

In the proof of the theorem in the paper, the authors construct the selection $f_{\epsilon}$ by introducing locally Lipschitz functions denoted by $a_i:X \rightarrow [0, 1]$. Unfortunately, the authors left a reference for the construction, theorem 1.1.2 in this book, but I cannot found it in the cited book.

Can anybody help me 1) how to construct such $a_i$'s and 2) a way to estimate the Lipschitz constant of $f_{\epsilon}$ if it's possible?

答:
Actually, my purpose is to show that given a sequence of u.h.c. multivalued functions $F_j:X \rightarrow Y$ we can choose an equi-Lipschitz sequence of functions $f_{\epsilon, j}$ such that $f_{\epsilon, j}$ is an $\epsilon$-selection of $F_j$.
Here is my idea if we can actually choose such $a_i$'s described in the proof of the theorem.

As in the proof of theorem 9.2.1. in Set-Valued Analysis (if it's correct), we can choose the finite open cover $ \{ \text{int}(B(x_i, \delta_{x_i}/4) \}_{i \in I}$ of $X$ and $a_i$'s described in the proof, independent of $j$.
The selection is given as $f_{\epsilon, j} = \sum_{i \in I} a_i(x) y_i$ with an associated point $y_j \in F_j(B(x_i, \delta_i/4))$.
Each $a_i$ is locally Lipschitz in a compact space $X$, then $a_i$ is Lipschitz continuous on $X$ (namely, with Lipschitz constant $L_i$; I don't understand why the authors did not conclude the theorem with Lipschitzness of $f_{\epsilon}$ because local Lipschitzness on a compact space $X$ implies Lipschitzness on the same space. Probably it's due to the generalised version of the theorem as commented above the theorem in the book). Then, for any $x, x' \in X$, $|f_{\epsilon, j}(x) - f_{\epsilon, j}(x)| \leq \max_{x \in X}\lVert x \rVert \cdot \sum_{i \in I}|a_i(x) - a_i(x')| \leq \tilde{L} \lVert x - x'\rVert$ where $\tilde{L} = \max_{x \in X} \lVert x \rVert \sum_{i \in I} L_i$. This implies that $\{ f_{\epsilon, j} \}_{j \in \mathbb{N}}$ is equi-Lipschitz.

My curiosity is quite resolved but I would be appreciated if someone let me know 1) we can actually choose such $a_i$'s and 2) explicit estimation of the Lipschitz constant of $f_{\epsilon}$.


问:
Solving periodic ODE with finite difference having strange results...
I was wondering if you could help me understand.

I am trying to solve this eigenvalue problem numerically via finite difference (I've solved it already using a spectral method).

\begin{equation}
i \frac{\partial u}{\partial x} = \lambda u
\end{equation} with boundary conditions $u(0) = u(1)$.

I've used the following discretisation of the operator on the left-hand-side
\begin{equation}
\frac{u_n - u_{n-1}}{\Delta x}
\end{equation} and with this, I simply use eig (in python) on the following matrix

\begin{equation}
\begin{pmatrix} -1 & 1 & 0 &... & 0 \\ 0 & -1 & 1 &... & 0 \\ \vdots  &  & \ddots  & \ddots  & 0 \\ \vdots  &  & \ddots  & \ddots  & 0 \\ 1 &  & 0  & 0  & -1 \\  \end{pmatrix}
\end{equation}, but this way I find that the xvalues are not prescribed and it gives me eigenvalues in a circle (they should be real). I don't even know what part of maths this is... Help please!

Kindest regards,

Catherine

答:
So you get $$u_n=(1-iλΔx)^{-n}u_0,$$ and if periodicity means $u_N=u_0$, then $(1-iλΔx)^N=1$. This has solutions $$1-iλΔx=e^{iα}=\cos(α)+i\sin(α)$$ with $αN=k2\pi\iff α=2k\pi\frac{Δx}{L}$. Or
$$
λ=\frac{1-e^{iα}}{iΔx}\approx \frac{2k\pi}{L}
$$
the last for small values of $k$.

Yes, the eigenvalues lie on a large circle (very large for larger $N$), but the smaller eigenvalues lie close to the real axis.

If you want real eigenvalues, use the more symmetric trapezoidal discretization
$$
i\frac{u_n-u_{n-1}}{Δx}=λ\frac{u_n+u_{n-1}}{2}.
$$


问:
Showing that ln$(xy)$ = ln $x$+ln $y$
Let ln $x\  =\ \int_1^x \frac{1}{t}\ dt.$

How do I show that ln$(xy)$ = ln $x$+ln $y$ where $x$ and $y$ are positive reals.
I read the following proof from Limaye book.

Fix $y\in (0,\infty).$ Consider $f(x)=$ ln $xy$ -ln $x$. Then
$f'(x)= \frac{1}{xy}.y-\frac{1}{x}=0 \ \forall\  x \in (0,\infty). $

$\therefore f$ is constant.
 $f(x)=f(1)=$ ln $y$ - ln $1 = $ ln $y.$

$ \therefore f(x) =  $ln$ (xy) - $ln$ x = $ln$  y\ \implies\   $ln$ (xy) = $ln$ \ x + $ ln$\   y$

I feels that this proof is so constructive.

Im wondering to know some alternative proofs.

Thanks in Advance...

答:
Let $p_1,p_2>0$.

$p_1p_2=e^{r_1}e^{r_2}=e^{r_1+r_2}$, $r_1,r_2\in\mathbb{R}$.

So, $\ln(p_1p_2)=r_1+r_2=\ln(p_1)+\ln(p_2)$.

I ignored uniqueness here. Constructive criticism is welcomed.

Title: Showing that ln$(xy)$ = ln $x$+ln $y$


问:
Why are randomly drawn vectors nearly perpendicular in high dimensions
I am struggling understanding this finding. Can somebody explain intuitively why randomly drawn high-dimensional vectors will tend to be mutually orthogonal? I realize that intuition in high dimensions might be too much to ask for, still, an explanation without having to integrate over several pages of symbols would be preferred.  </p>

答:
In statistics orthogonality is deeply linked to correlation. Random vectors are orthogonal when the values they contain are uncorrelated, meaning that in the $x_i$ values of $\vec{v_1}$ and $\vec{v_2}$ do not appear common patterns. The higher the dimension is the higher is $i$ and so the probability is higher that $x_{i}$ value of $\vec{v_1}$ and the $x_i$ values of $\vec{v_2}$ have no common pattern, because the higher the dimension the higher the possibilities (in an urn problem increase the number of balls ...). Equivalent to the statement that $\vec{v_1}$ and $\vec{v_2}$ have no common patterns is the statement that they are uncorrelated what is equivalent to the statement that they are orthogonal.

So the key to understand why randomly drawn vectors are perpendicular in high dimensions is the deep link between orthogonality and statistical correlation :

https://stats.stackexchange.com/questions/171324/what-is-the-relationship-between-orthogonal-correlation-and-independence

This is linked to a more philosophical question if real randomness actually exists. There are strong hints that this is not the case, i.e. nonlinear dynamical systems (chaotic systems) are able to generate white noise in a deterministic way :
https://ieeexplore.ieee.org/document/1236865 - Chaotic maps generating white noise


问:
Additivity of valuations
This feels like a way too easy question, but I've been sitting on this problem for a few days now, and can't make much progress.
We define a valuation ring by two equivalent properties: For an integral domain A and its field of fractions K we have

a) for every $x \in K^*$, one has $x\in A$ or $x^{-1}\in A$ and

b) the partially ordered group $(\Gamma, \geq) = (K^*/A^*, x \geq y$ iff $ xy^{-1} \in A)$ is totally ordered and abelian.

Then we call A a valuation ring. We define a valuation $v: K \rightarrow \Gamma \cup \infty$ , $ x \mapsto xA^*$ and $0 \mapsto \infty$.
I have already shown that $v(x+y) \geq min(v(x),v(y))$. Now I want to show that v is additive, ie that $v(xy) = v(x) + v(y)$. We can assume that either $v(xy) \geq v(x) + v(y)$, or the converse, since $\Gamma$ is totally ordered, and then try to prove the other direction. So I started with assuming $v(xy) \geq v(x) + v(y)$. Now, I have trouble formulating this relation in terms of the definition of the $\geq$ - relation. I tried writing this as $xy(x+y)^{-1} \in A$, but that didn't help me in proving that $(x+y)y^{-1}x^{-1}\in A$. So I figured I'd made a mistake: because if I assume $v(xy) \geq v(x+y)$, THAT should be translated to $xy(x+y)^{-1} \in A$, and $v(x+y)$ and $v(x)+v(y)$ are two different things.

So now how do I go about this? What would be the correct translation of my assumption that $v(xy) \geq v(x) + v(y)$?

答:
Turning my comment into an answer to get it off the unanswered list. The group structure on $\Gamma$ that we denote $+$ is in fact not defined as $xA^* + yA^* = (x + y)A^*$, but as $xA^* + yA^* = (xy)A^*$. With that definition, the fact that $v(xy) = v(x) + v(y)$ is fairly straightforward.

As to whether multiplicative notation would be more sensible, I guess that makes sense. One source of valuations (though they're typically Archimedean) is by taking an absolute value $|\cdot|$ on your ring, and writing $v(x) = e^{-|x|}$. (I suppose this is why valuations are also called exponential valuations.) The group $\Gamma$ is then $\mathbb{R}_{>0}$ with the multiplicative structure. Surely a $\cdot$ would be more sensible than a $+$ here.

But of course, it's all just convention, anyway. :)

Title: Additivity of valuations


问:
Find the length of C which has parametric equations $ x=9 \cos (t)-2, \quad y=9 \sin (t)+1, \quad-\frac{\pi}{6} \leqslant t \leqslant \frac{\pi}{2} $
I solved it by first finding the Cartesian equation of $C$ and its domain & range.
$$
(x+2)^{2}+(y-1)^{2}=81,\;\;-2 \leq x \leq 7,\;\;-\frac{7}{2} \leq y \leq 10
$$

After sketching the curve we can see that the angle subtended by an arc $C$ at the centre is equal to $
\frac{\pi}{2}+\theta
$.

I then found the distance between the two marked points on the circle, and used the cosine law to find the angle $\theta$. Of course, $$
\begin{aligned}
\text {arc length C } &=\theta r \\
&=\left(\frac{\pi}{2}+\frac{\pi}{6}\right) \times 9 \\
&=6 \pi
\end{aligned}
$$

The answer booklet seems to suggest there is a much faster way of finding $\theta.$

If this is the case, could someone enlighten me please?

答:
The parametric system defines a circle with radius $r=9.$ The parametrization is defined by the angle with vertex at the center of the circle, therefore the coordinates of the center do not influence the computation.

The interval $-\frac{\pi}{6}\leq t \leq \frac{\pi}{2}$ has a length $$l=\frac{\pi}{2}+\frac{\pi}{6}=\frac{2\pi}{3},$$ which corresponds to one third of a circle. Therefore, the length of $C$ is
$$l(C)=r\times l =6\pi.$$